Geotechnical Engineering, Murthy

Published on May 2016 | Categories: Documents | Downloads: 92 | Comments: 0 | Views: 943
of x
Download PDF   Embed   Report

Comments

Content


Dedicated to the Cause of Students
FOREWORD
Geotechnical Engineering: Principles and Practices of Soil Mechanics and Foundation
Engineering i s a lon g titl e befittin g a majo r work. I a m please d t o introduc e thi s super b volume
destined fo r a readershi p o f students , professors, an d consultants . What make s thi s text differen t
from othe r books o n these subject s that appear eac h yea r an d why am I recommending i t to you? I
have been workin g an d teachin g in the are a o f geotechnica l engineering fo r 2 5 years . I have rea d
and use d score s o f textbook s i n m y classe s an d practice . Dr . Murthy' s tex t i s b y fa r th e mos t
comprehensive tex t I have found. You will fin d tha t his organizatio n of th e subjec t matter follows
a logical progression. Hi s example problems ar e numerous and, like the text, start from fundamenta l
principles an d progressivel y develo p int o mor e challengin g material . The y ar e th e bes t se t o f
example problems I have see n i n a textbook. Dr . Murthy has include d ampl e homewor k problem s
with a rang e o f difficult y mean t t o hel p th e studen t ne w t o th e subjec t t o develo p his/he r
confidence an d t o assis t th e experience d enginee r i n his/he r revie w o f th e subjec t an d i n
professional development .
As the technical editor I have read the entire manuscript three times. I have been impressed by
the coverage , th e clarit y o f th e presentation , and th e insight s int o the hows an d why s o f soi l an d
foundation behavior . Ofte n I have been astonishe d at Dr. Murthy's near-conversational approac h t o
sharing helpfu l insights . Yo u ge t th e impressio n he' s righ t ther e wit h yo u guidin g yo u along ,
anticipating your questions, and providing instruction and necessar y informatio n as the next step s
in th e learning process. I believe you wil l enjoy thi s book an d that it wil l receive a warm welcom e
wherever it is used.
I thank Dr. Murthy for his commitment to write this textbook an d for sharing his professional
experience wit h us. I thank him for his patience i n making corrections and considering suggestions .
I thank Mr. B. J. Clark, Senior Acquisitions Editor at Marcel Dekke r Inc. , for the opportunity to be
associated wit h such a good book. I likewise express my appreciation to Professor Pierr e Foray of
1'Ecole National e Superieur e d' Hydrauliqu e e t d e Mecaniqu e d e Grenoble , Institu t Nationa l
Polytechnique d e Grenoble , Franc e for hi s enthusiasti c and unflagging support whil e I edited th e
manuscript.
MarkT. Bowers, Ph.D., P. E.
Associate Professo r o f Civi l Engineerin g
University o f Cincinnat i
FOREWORD
It give s m e grea t pleasur e t o writ e a forewor d fo r Geotechnical Engineering: Principles an d
Practices o f Soil Mechanics an d Foundation Engineering. Thi s comprehensive, pertinen t an d up-
to-date volum e i s wel l suite d fo r us e a s a textboo k fo r undergraduat e student s a s wel l a s a
reference boo k fo r consultin g geotechnica l engineer s an d contractors . Thi s book i s wel l writte n
with numerous example s o n applications o f basic principles to solve practical problems .
The earl y histor y o f geotechnica l engineerin g and the pioneerin g wor k of Kar l Terzaghi i n
the beginning of the las t century are described i n Chapter 1 . Chapter s 2 and 3 discuss methods of
classification o f soil and rock, th e chemical and the mechanical weathering of rock, an d soil phas e
relationships an d consistenc y limit s fo r clay s an d silts . Numerou s example s illustrat e th e
relationship betwee n th e differen t parameters . Soi l permeabilit y an d seepag e ar e investigate d i n
Chapter 4 . Th e constructio n o f flo w net s an d method s t o determin e th e permeabilit y i n th e
laboratory an d i n th e fiel d ar e als o explained .
The concep t o f effective stres s an d the effec t o f por e wate r pressur e o n effective stres s ar e
discussed i n Chapter 5 . Chapter 6 is concerned wit h stress increas e i n soi l cause d b y surfac e loa d
and methods t o calculate stress increase cause d by spread footings, rafts, an d pil e groups. Several
examples ar e give n i n Chapter 6 . Consolidation o f soil s an d th e evaluatio n of compressibilit y i n
the laborator y b y oedomete r test s ar e investigate d i n Chapte r 7 . Determinatio n o f draine d an d
undrained shea r strengt h b y unconfine d compression, direc t shea r o r triaxia l test s i s treate d i n
Chapter 8.
The importan t subjec t o f soi l exploratio n i s discusse d i n Chapter 9 , includin g th e us e o f
penetration tests such as SPT and CPT in different countries . The stabilit y of slopes i s investigated
in Chapte r 10 . Methods usin g plai n an d circula r sli p surface s t o evaluat e stabilit y ar e describe d
such a s th e method s propose d b y Bishop , Fellenius , Morgenstern , an d Spencer . Chapte r 1 1
discusses method s t o determin e activ e an d passive eart h pressure s actin g o n retainin g an d shee t
pile walls .
Bearing capacit y an d settlemen t o f foundation an d th e evaluatio n of compressibilit y i n th e
laboratory b y oedomete r test s ar e discusse d i n Chapter s 12 , 13 , and 14 . The effec t o f inclination
and eccentricit y o f th e loa d o n bearing capacit y is als o examined . Chapte r 1 5 describes differen t
pile types , th e concep t o f critica l depth , method s t o evaluat e th e bearin g capacit y o f pile s i n
cohesive an d cohesionless soils , an d pile-driving formulae. The behavior of laterally loade d pile s
is investigated in Chapter 1 6 for piles i n sand and i n clay. The behavior of drilled pier foundation s
V I I
viii For ewor d
and th e effec t of th e i nst al l at i o n method on bearing capacity an d upl i f t ar e analyze d i n Chapter 17 .
Foundations on swel l i n g and col l apsi bl e soils ar e treate d i n Chapter 1 8 as ar e method s tha t can b e
used t o reduc e heave . Thi s i s a n importan t subject , seldo m treate d i n textbooks . Th e desig n o f
retaining wall s i s covere d i n Chapte r 19 , a s wel l a s th e differen t factor s tha t affec t activ e an d
passive eart h pressures . Differen t applications of geotextiles ar e covered i n thi s chapte r a s wel l a s
the topi c o f reinforce d earth . Cantilever , anchored, an d strutte d sheet pil e wall s are investigate d in
Chapter 20 , a s ar e method s t o eval uat e st abi l i t y an d th e momen t di st ri but i on . Differen t soi l
improvement methods , suc h a s compactio n o f granula r soils , san d compactio n piles ,
vibroflotation, preloading , an d ston e columns , ar e describe d i n Chapte r 21 . Th e chapte r als o
discusses lim e an d cemen t st abi l i zat i on . Appendi x A provide s a lis t o f S I uni t s , an d Appendi x B
compares method s tha t have been proposed .
This textboo k b y Prof . V . N. S . Murt h y i s hi ghl y recommended fo r student s specializin g i n
geotechnical engineerin g an d fo r practicin g civil engineer s i n th e Unite d State s an d Europe . Th e
book include s recen t development s suc h a s soi l improvemen t an d stabilizatio n method s an d
applications o f geotextile s t o control settlement s an d latera l eart h pressure . Numerou s graph s an d
examples illustrat e th e mos t importan t concept s i n geotechnica l engineering . Thi s textboo k
should serv e a s a val uabl e reference boo k fo r man y years t o come.
Bengt B. Br oms , Ph. D.
Nanyang Technical University , Singapore (retired) .
PREFACE
This boo k ha s th e followin g objectives :
1. T o explain the fundamental s of th e subjec t from theor y t o practice i n a logical wa y
2. T o be comprehensiv e an d mee t th e requirement s o f undergraduat e student s
3. T o serve a s a foundation course fo r graduat e student s pursuing advanced knowledg e i n the
subject
There ar e 21 chapter s i n thi s book. The firs t chapte r trace s the historical backgroun d o f the
subject an d the second deal s wit h the formation an d mineralogical compositio n o f soils. Chapte r 3
covers th e inde x propertie s an d classificatio n of soil . Chapter s 4 an d 5 explai n soi l permeability ,
seepage, an d th e effec t o f wate r o n stres s condition s i n soil . Stresse s develope d i n soi l du e t o
imposed surfac e loads , compressibilit y an d consolidatio n characteristics , an d shea r strengt h
characteristics o f soi l ar e deal t wit h i n Chapter s 6,7 , an d 8 respectively. Th e firs t eigh t chapter s
develop th e necessar y tool s fo r computin g compressibilit y an d strengt h characteristics o f soils .
Chapter 9 deals with methods fo r obtainig soil samples in the field for laboratory tests and for
determining soi l parameter s directl y b y us e o f fiel d tests . Chapter s 1 0 t o 2 0 dea l wit h stabilit y
problems pertainin g t o eart h embankments , retaining walls, and foundations. Chapter 2 1 explains
the various methods b y which soi l i n sit u can be improved. Many geotechnical engineer s hav e not
appreciated th e importanc e o f thi s subject . N o amoun t o f sophisticatio n i n th e developmen t o f
theories wil l hel p the designers i f the soi l parameters use d i n the theor y ar e not properly evaluate d
to simulat e fiel d conditions . Professors wh o teac h thi s subject shoul d stres s thi s topic .
The chapter s i n thi s book ar e arrange d i n a logical wa y fo r th e developmen t o f th e subjec t
matter. There is a smooth transitio n from on e chapter t o the next and the continuit y of the material
is maintained . Each chapte r start s wit h an introduction to the subjec t matter , develop s th e theory,
and explain s it s applicatio n t o practica l problems . Sufficien t example s ar e wo
r1
:ed ou t t o hel p
students understan d th e significanc e o f th e theories . Man y homewor k problem s ar e give n a t th e
end o f each chapter .
The subjec t matte r deal t wit h i n eac h chapte r i s restricte d t o th e requirement s o f
undergraduate students. Half-baked theorie s an d unconfirmed test results ar e not developed i n this
book. Chapter s ar e up-to-dat e a s pe r engineerin g standards . Th e informatio n provide d i n
Chapter 1 7 on drilled pier foundations is the latest availabl e at the time of thi s writing. The desig n
Preface
of mechanicall y stabilized eart h retaining walls i s als o current . A ne w method fo r predictin g th e
nonlinear behavior o f laterall y loade d vertica l and batter pile s i s described i n Chapter 16.
The boo k i s comprehensive , rational , an d pertinen t t o th e requirement s o f undergraduat e
students. It serves as a foundation cours e for graduate students, and is useful a s a reference book for
designers an d contractor s i n the fiel d o f geotechnica l engineering .
ACKNOWLEDGEMENTS
It i s my pleasur e t o than k Marce l Dekker , Inc. , fo r acceptin g me a s a singl e author fo r th e
publication o f my book. The ma n who was responsibl e fo r thi s wa s Mr. B.J . Clark, the Executive
Acquisition Editor . I t wa s m y pleasur e t o wor k unde r hi s guidance . Mr . Clar k i s a refine d
gentleman personified , polished , an d clea r sighted . I than k hi m cordiall y fo r th e courtesie s an d
help extended t o me during the course of writing the manuscript. I remain ever grateful t o him.
Writing a book fo r American Universities by a nonresident o f America i s not a n easy task.
I neede d a n America n professo r t o edi t m y manuscrip t an d guid e m e wit h regard s t o th e
requirements of undergraduate student s i n America. Dr. Mar k T. Bowers , Associate Professor o f
Civil Engineering , Universit y o f Cincinnati , accepte d t o become m y consultan t an d chie f editor .
Dr. Bower s i s a man o f honest y an d integrity . He i s dedicated t o the caus e o f his profession. He
worked har d fo r over a year in editing my book an d helped me t o streamline t o make i t acceptable
to the undergraduate student s o f American Universities . I thank Dr . Bower s fo r the help extende d
to me.
There ar e man y i n Indi a wh o helpe d m e durin g th e cours e o f writin g thi s book . Som e
provided m e usefu l suggestion s an d other s wit h references . I acknowledg e thei r service s wit h
thanks. Th e members are :
Mr. S. Pranesh Managin g Directo r
Prism Books Pv t Ltd
Bangalore
Dr. K.S.Subba Ra o Professo r o f Civil Engineerin g
Indian Institut e of Science Bangalor e
Dr. T.S. Nagara j Professo r of Civil Engineering
(Emeritus), Indian Institute o f Science,
Bangalore
Professor o f Civil Engineering
Dr. C. Subba Rao India n Institut e of Technology
Kharagpur
Chaitanya Graphics , Bangalore , provide d th e artwor k fo r th e book . I than k M r S.K .
V ijayasimha, th e designer, fo r the excellent job don e by him.
My so n Prakas h wa s associate d wit h th e boo k sinc e it s inception . H e carrie d o n
correspondence wit h th e publisher s o n m y behal f an d sen t referenc e book s a s needed . My wif e
Sharadamani wa s mainl y responsibl e fo r keepin g m y spiri t hig h durin g th e year s I spen t i n
writing the book. I remain grateful t o my son and my wif e fo r al l they did.
I sincerel y than k Mr . Brian Black fo r his continuous effort s i n the production o f this book. I
immensely than k Mr . Janardha n an d Mr . Rajeshwar , compute r engineer s o f Aicra Inf o Mate s Pvt
Ltd., Hyderabad, for their excellent typesetting wor k on this book.
V .N.S. Murthy
CONTENTS
Foreword Mar k T . B ower s v
Foreword Beng t B . Brom s vi i
Preface i x
CH APTER 1 I NTRODU CTI O N 1
1.1 Genera l Remark s 1
1.2 A Brie f Historica l Developmen t 2
1.3 Soi l Mechanic s an d Foundatio n Engineerin g 3
CH APTER 2 SOI L FORM ATI ON AND CH ARACTERI Z ATI ON 5
2.1 Introductio n 5
2.2 Roc k Classificatio n 5
2.3 Formatio n o f Soil s 7
2.4 Genera l Type s o f Soil s 7
2.5 Soi l Particl e Siz e an d Shap e 9
2.6 Compositio n o f Cla y Mineral s 1 1
2.7 Structur e of Cla y Mineral s 1 1
2.8 Cla y Particle-Wate r Relation s 1 4
2.9 Soi l Mas s Structur e 1 7
XI
x ii Con t en t s
CH APTER 3 SOI L PH ASE REL ATI ONSH I PS, INDEX PROPERTI ES
AND CL ASSI FI CATI ON 1 9
3.1 Soi l Phas e Relationship s 1 9
3.2 Mass-V olum e Relationships 2 0
3.3 Weight-V olum e Relationship s 2 4
3.4 Comment s o n Soi l Phas e Relationship s 2 5
3.5 Inde x Propertie s o f Soil s 3 1
3.6 Th e Shap e an d Siz e o f Particle s 3 2
3.7 Siev e Analysi s 3 3
3.8 Th e Hydromete r Metho d o f Analysi s 3 5
3.9 Grai n Siz e Distributio n Curves 4 3
3.10 Relativ e Densit y o f Cohesionles s Soil s 4 4
3.11 Consistenc y o f Cla y Soi l 4 5
3.12 Determinatio n o f Atterber g Limit s 4 7
3.13 Discussio n o n Limit s an d Indice s 5 2
3.14 Plasticit y Char t 5 9
3.15 Genera l Consideration s fo r Classificatio n o f Soil s 6 7
3.16 Fiel d Identificatio n o f Soil s 6 8
3.17 Classificatio n o f Soil s 6 9
3.18 Textura l Soi l Classificatio n 6 9
3.19 AASHT O Soi l Classificatio n Syste m 7 0
3.20 Unifie d Soi l Classificatio n Syste m (USCS ) 7 3
3.21 Comment s o n th e System s o f Soi l Classificatio n 7 6
3.22 Problem s 8 0
CH APTER 4 SOI L PERM EAB I L I TY AND SEEPAGE 8 7
4.1 Soi l Permeabilit y 8 7
4.2 Darcy' s La w 8 9
4.3 Discharg e an d Seepag e V elocitie s 9 0
4.4 Method s o f Determinatio n o f Hydrauli c Conductivit y o f Soil s 9 1
4.5 Constan t Hea d Permeabilit y Tes t 92
4.6 Fallin g Hea d Permeabilit y Tes t 9 3
4.7 Direc t Determinatio n o f k o f Soil s i n Plac e b y Pumpin g Tes t 9 7
4.8 Borehol e Permeabilit y Test s 10 1
4.9 Approximat e V alue s o f th e Hydrauli c Conductivit y of Soil s 10 2
4.10 Hydrauli c Conductivit y i n Stratifie d Layer s o f Soil s 10 2
4.11 Empirica l Correlation s fo r Hydrauli c Conductivit y 10 3
4.12 Hydrauli c Conductivit y of Rock s b y Packe r Metho d 11 2
4.13 Seepag e 11 4
4.14 Laplac e Equatio n 11 4
Con t en t s x ii i
4.15 Flo w Ne t Constructio n 11 6
4.16 Determinatio n o f Quantit y o f Seepag e 12 0
4.17 Determinatio n o f Seepag e Pressur e 12 2
4.18 Determinatio n o f Uplif t Pressure s 12 3
4.19 Seepag e Flo w Throug h Homogeneou s Eart h Dam s 12 6
4.20 Flo w Ne t Consistin g o f Conjugat e Confoca l Parabola s 12 7
4.21 Pipin g Failur e 13 1
4.22 Problem s 13 8
CH APTER 5 EFFECTI V E STRESS AND PORE WATER PRESSU RE 1 4 3
5.1 Introductio n 14 3
5.2 Stresse s whe n N o Flo w Take s Plac e Throug h th e
Saturated Soi l Mas s 14 5
5.3 Stresse s Whe n Flo w Take s Plac e Throug h th e Soi l
from To p t o Botto m 14 6
5.4 Stresse s Whe n Flo w Take s Plac e Throug h th e Soi l
from Botto m t o To p 14 7
5.5 Effectiv e Pressur e Du e t o Capillar y Wate r Ris e i n Soi l 14 9
5.6 Problem s 17 0
CH APTER 6 STRES S DI STRI B U TI ON I N SOI L S
DUE TO SU RFACE L OADS 1 7 3
6.1 Introductio n 17 3
6.2 Boussinesq' s Formul a fo r Poin t Load s 17 4
6.3 Westergaard' s Formul a fo r Poin t Load s 17 5
6.4 Lin e Load s 17 8
6.5 Stri p Load s 17 9
6.6 Stresse s Beneat h th e Corne r o f a Rectangula r Foundatio n 18 1
6.7 Stresse s Unde r Uniforml y Loade d Circula r Footin g 18 6
6.8 V ertica l Stres s Beneat h Loade d Area s o f Irregula r Shap e 18 8
6.9 Embankmen t Loading s 19 1
6.10 Approximat e Method s fo r Computin g c r 19 7
6.11 Pressur e Isobar s 19 8
6.12 Problem s 20 3
CH APTER 7 COM PRESSI B I L I T Y AN D CONSOL I DATI ON 20 7
7.1 Introductio n 20 7
7.2 Consolidatio n 20 8
7.3 Consolidomete r 21 2
x iv Con t en t s
7.4 Th e Standar d One-Dimensiona l Consolidatio n Tes t 21 3
7.5 Pressure-V oi d Rati o Curve s 21 4
7.6 Determinatio n o f Preconsolidatio n Pressur e 21 8
7.7 e-logp Fiel d Curves for Normally Consolidate d and
Overconsolidated Clay s of Low to Medium Sensitivit y 21 9
7.8 Computatio n of Consolidatio n Settlemen t 21 9
7.9 Settlemen t Du e t o Secondar y Compressio n 22 4
7.10 Rat e o f One-dimensiona l Consolidation Theor y o f Terzagh i 23 3
7.11 Determinatio n of th e Coefficien t o f Consolidatio n 24 0
7.12 Rat e o f Settlemen t Du e t o Consolidatio n 24 2
7.13 Two - and Three-dimensional Consolidatio n Problem s 24 3
7.14 Problem s 24 7
CH APTERS SH EA R STRENG TH O F SOI L 2 5 3
8.1 Introductio n 25 3
8.2 Basi c Concep t o f Shearin g Resistanc e an d Shearin g Strengt h 25 3
8.3 Th e Coulom b Equatio n 25 4
8.4 Method s o f Determinin g Shea r Strengt h Parameter s 25 5
8.5 Shea r Tes t Apparatu s 25 6
8.6 Stres s Conditio n a t a Poin t i n a Soi l Mas s 26 0
8.7 Stres s Condition s i n Soi l Durin g Triaxia l Compressio n Tes t 26 2
8.8 Relationshi p Betwee n th e Principa l Stresse s an d Cohesio n c 26 3
8.9 Moh r Circl e o f Stres s 26 4
8.10 Moh r Circl e o f Stres s Whe n a Prismati c Elemen t i s Subjecte d t o
Normal an d Shea r Stresse s 26 5
8.11 Moh r Circl e o f Stres s fo r a Cylindrica l Specime n
Compression Tes t 26 6
8.12 Mohr-Coulom b Failur e Theor y 26 8
8.13 Moh r Diagra m fo r Triaxia l Compressio n Tes t a t Failur e 26 9
8.14 Moh r Diagra m fo r a Direc t Shea r Tes t a t Failur e 27 0
8.15 Effectiv e Stresse s 27 4
8.16 Shea r Strengt h Equatio n i n Terms o f Effectiv e Principa l Stresse s 27 5
8.17 Stress-Controlle d an d Strain-Controlled Test s 27 6
8.18 Type s o f Laborator y Test s 27 6
8.19 Shearin g Strengt h Test s o n San d 27 8
8.20 Unconsolidated-Undraine d Tes t 28 4
8.21 Unconfine d Compressio n Test s 28 6
8.22 Consolidated-Undraine d Tes t o n Saturate d Cla y 29 4
8.23 Consolidated-Draine d Shea r Strengt h Tes t 29 6
8.24 Por e Pressur e Parameter s Unde r Undraine d Loadin g 29 8
8.25 V an e Shea r Test s 30 0
Contents x v
8.26 Othe r Method s fo r Determinin g Undraine d Shea r Strengt h
of Cohesiv e Soil s 30 2
8.27 Th e Relationshi p Betwee n Undraine d Shea r Strengt h an d
Effective Overburde n Pressur e 30 4
8.28 Genera l Comment s 31 0
8.29 Question s an d Problem s 31 1
C H A P T E R S S O I L E X P L O R A T I ON 3 1 7
9.1 Introductio n 31 7
9.2 Borin g o f Hole s 31 8
9.3 Samplin g i n Soi l 32 2
9.4 Roc k Cor e Samplin g 32 5
9.5 Standar d Penetratio n Tes t 32 7
9.6 SP T V alue s Relate d t o Relativ e Densit y o f Cohesionles s Soil s 33 0
9.7 SP T V alue s Related t o Consistenc y o f Cla y Soi l 33 0
9.8 Stati c Con e Penetratio n Tes t (CPT ) 33 2
9.9 Pressuremete r 34 3
9.10 Th e Fla t Dilatomete r Tes t 34 9
9.11 Fiel d V an e Shear Tes t (V ST ) 35 1
9.12 Fiel d Plat e Loa d Tes t (PUT ) 3 51
9.13 Geophysica l Exploratio n 35 2
9.14 Plannin g o f Soi l Exploratio n 35 8
9.15 Executio n o f Soi l Exploratio n Progra m 35 9
9.16 Repor t 36 1
9.17 Problem s 36 2
CH APTER 1 0 STAB I L I T Y O F SL OPES 36 5
10.1 Introductio n 36 5
10.2 Genera l Consideration s an d Assumptions in th e Analysi s 36 7
10.3 Facto r o f Safet y 36 8
10.4 Stabilit y Analysi s o f Infinit e Slope s i n San d 37 1
10.5 Stabilit y Analysis o f Infinit e Slope s i n Cla y 37 2
10.6 Method s o f Stabilit y Analysis of Slope s o f Finit e Heigh t 37 6
10.7 Plan e Surfac e o f Failur e 37 6
10.8 Circula r Surface s o f Failur e 37 8
10.9 Failur e Unde r Undraine d Condition s ( ( f>
u
= 0 ) 38 0
10.10 Friction-Circl e Metho d 38 2
10.11 Taylor' s Stabilit y Numbe r 38 9
10.12 Tensio n Crack s 39 3
10.13 Stabilit y Analysis b y Metho d o f Slice s fo r Stead y Seepag e 39 3
x vi Con t en t s
10.14 Bishop' s Simplifie d Metho d o f Slice s 40 0
10.15 Bisho p an d Morgenster n Metho d fo r Slop e Analysi s 40 3
10.16 Morgenster n Metho d o f Analysi s fo r Rapi d Drawdow n Conditio n 40 5
10.17 Spence r Metho d o f Analysi s 40 8
10.18 Problem s 41 1
CH APTER 1 1 L ATERA L EARTH PRESSU RE 4 1 9
11.1 Int roduct i o n 41 9
11.2 Latera l Eart h Pressur e Theor y 42 0
11.3 Latera l Eart h Pressur e fo r a t Res t Conditio n 42 1
11.4 Rankine' s State s o f Plasti c Equilibriu m for Cohesionles s Soil s 42 5
11.5 Rankine' s Eart h Pressur e Agains t Smoot h V ertica l Wal l wit h
Cohesionless Backfil l 42 8
11.6 Rankine' s Activ e Eart h Pressur e wit h Cohesiv e Backfil l 44 0
11.7 Rankine' s Passiv e Eart h Pressur e wit h Cohesiv e Backfil l 44 9
11.8 Coulomb' s Eart h Pressur e Theor y fo r San d fo r Activ e Stat e 45 2
11.9 Coulomb' s Eart h Pressur e Theor y fo r San d fo r Passiv e Stat e 45 5
11.10 Activ e Pressur e b y Culmann' s Metho d fo r Cohesionles s Soil s 45 6
11.11 Latera l Pressure s b y Theor y o f Elasticit y fo r Surcharg e Load s
on th e Surfac e o f Backfil l 45 8
11.12 Curve d Surface s o f Failur e fo r Computin g Passiv e Eart h Pressur e 46 2
11.13 Coefficient s o f Passiv e Eart h Pressur e Table s an d Graph s 46 4
11.14 Latera l Eart h Pressur e o n Retainin g Wall s Durin g Earthquake s 46 7
11.15 Problem s 47 6
CH APTER 1 2 SH AL L O W FOU NDATI ON I:
U L TI M ATE B EARI NG CAPACI TY 4 8 1
12.1 Int roduct i o n 48 1
12.2 Th e Ultimat e Bearin g Capacit y o f Soi l 48 3
12.3 Som e o f th e Term s Define d 48 3
12.4 Type s o f Failur e i n Soi l 48 5
12.5 A n Overvie w o f Bearin g Capacit y Theorie s 48 7
12.6 Terzaghi' s Bearin g Capacit y Theor y 48 8
12.7 Skempton' s Bearin g Capacit y Facto r N
C
49 3
12.8 Effec t o f Wate r Tabl e o n Bearin g Capacit y 49 4
12.9 Th e Genera l Bearin g Capacit y Equatio n 50 3
12.10 Effec t o f Soi l Compressibilit y o n Bearin g Capacit y o f Soi l 50 9
12.11 Bearin g Capacit y o f Foundation s Subjecte d t o Eccentri c Load s 51 5
12.12 Ultimat e Bearin g Capacit y o f Footing s Base d o n SP T V alue s ( N ) 51 8
12.13 Th e CP T Metho d o f Determinin g Ultimat e Bearin g Capacit y 51 8
Con t en t s x vi i
12.14 Ultimat e Bearin g Capacit y o f Footing s Restin g o n Stratifie d
Deposits o f Soi l 52 1
12.15 Bearin g Capacit y o f Foundation s o n To p o f a Slop e 52 9
12.16 Foundation s o n Roc k 53 2
12.17 Cas e Histor y o f Failur e o f th e Transcon a Grai n Elevato r 53 3
12.18 Problem s 53 6
CH APTER 1 3 SH AL L O W FOU NDATI ON II :
SAFE B EARI NG PRESSU RE AND SETTLEMENT CAL CU L ATI ON 54 5
13.1 Introductio n 54 5
13.2 Fiel d Plat e Loa d Test s 54 8
13.3 Effec t o f Siz e o f Footing s o n Settlemen t 55 4
13.4 Desig n Chart s fro m SP T V alue s for Footing s o n San d 55 5
13.5 Empirica l Equation s Base d o n SP T V alue s for Footing s o n
Cohesionless Soil s 55 8
13.6 Saf e Bearin g Pressur e fro m Empirica l Equation s Base d o n
CPT V alue s fo r Footing s o n Cohesionles s Soi l 55 9
13.7 Foundatio n Settlemen t 56 1
13.8 Evaluatio n o f Modulu s o f Elasticit y 56 2
13.9 Method s o f Computin g Settlement s 56 4
13.10 Elasti c Settlemen t Beneat h th e Corne r o f a Uniforml y Loade d
Flexible Are a Base d o n th e Theor y o f Elasticit y 56 5
13.11 Janbu , Bjerru m an d Kjaernsli' s Metho d o f Determinin g
Elastic Settlemen t Unde r Undraine d Condition s 56 8
13.12 Schmertmann' s Metho d o f Calculatin g Settlemen t i n Granula r
Soils b y Usin g CPT V alue s 56 9
13.13 Estimatio n o f Consolidatio n Settlemen t b y Usin g Oedometer
Test Dat a 57 5
13.14 Skempton-Bjerru m Metho d o f Calculatin g Consolidatio n
Settlement (1957 ) 57 6
13.15 Problem s 58 0
CH APTER 1 4 SH AL L O W FOU NDATI ON III :
COMBINED FOOTI NG S AND M AT FOU NDATI ONS 58 5
14.1 Introductio n 58 5
14.2 Saf e Bearin g Pressure s fo r Ma t Foundation s o n San d an d Clay 58 7
14.3 Eccentri c Loadin g 58 8
14.4 Th e Coefficien t o f Subgrad e Reactio n 58 8
14.5 Proportionin g o f Cantileve r Footin g 59 1
x viii Con t en t s
14.6 Desig n o f Combine d Footing s b y Rigi d Metho d (Conventiona l
Method) 59 2
14.7 Desig n o f Ma t Foundatio n b y Rigi d Metho d 59 3
14.8 Desig n o f Combine d Footing s b y Elasti c Lin e Metho d 59 4
14.9 Desig n o f Ma t Foundation s b y Elasti c Plat e Metho d 59 5
14.10 Floatin g Foundatio n 59 5
14.11 Problem s 60 3
CH APTER 1 5 DEE P FOU NDATI ON I :
PILE FOU NDATI ON 60 5
15.1 Introductio n 60 5
15.2 Classificatio n of Pile s 605
15.3 Type s o f Pile s Accordin g t o th e Metho d o f Installatio n 60 6
15.4 Use s o f Pile s 60 8
15.5 Selectio n o f Pil e 60 9
15.6 Installatio n o f Pile s 61 0
PART A- V ERTI CAL L OAD B EARI NG CAPACI TY OF ASI NG LE V ERTI CAL PIL E 61 3
15.7 Genera l Consideration s 61 3
15.8 Method s o f Determinin g Ultimat e Loa d Bearin g Capacit y o f a
Single V ertica l Pil e 61 7
15.9 Genera l Theor y fo r Ultimat e Bearin g Capacit y 61 8
15.10 Ultimat e Bearin g Capacit y i n Cohesionles s Soil s 62 0
15.11 Critica l Dept h 62 1
15.12 Tomlinson' s Solutio n fo r Q
b
in San d 62 2
15.13 Meyerhof' s Metho d o f Determinin g Q
b
for Pile s i n San d 62 4
15.14 V esic' s Metho d o f Determinin g Q
b
62 5
15.15 Janbu' s Metho d o f Determinin g Q
b
62 8
15.16 Coyl e an d Castello' s Metho d o f Estimatin g Q
b
in San d 62 8
15.17 Th e Ultimat e Skin Resistanc e of a Singl e Pil e i n Cohesionles s Soi l 62 9
15.18 Ski n Resistanc e Q
f
by Coyl e an d Castell o Metho d (1981 ) 63 1
15.19 Stati c Bearin g Capacit y o f Pile s i n Cla y Soi l 63 1
15.20 Bearin g Capacit y o f Pile s i n Granula r Soil s Base d o n SP T V alu e 63 5
15.21 Bearin g Capacit y o f Pile s Base d o n Stati c Con e Penetratio n
Tests (CPT ) 65 2
15.22 Bearin g Capacit y o f a Singl e Pil e b y Loa d Tes t 66 3
15.23 Pil e Bearing Capacit y fro m Dynami c Pil e Driving Formulas 66 6
15.24 Bearin g Capacit y o f Pile s Founde d o n a Rock y Be d 67 0
15.25 Uplif t Resistanc e o f Pile s 67 1
Contents x i x
PART B-PILE G ROU P 67 4
15.26 Numbe r an d Spacin g o f Pile s i n a Grou p 67 4
15.27 Pil e Grou p Efficienc y 67 6
15.28 V ertica l Bearin g Capacit y o f Pil e Group s Embedde d i n
Sands an d Gravel s 67 8
15.29 Settlemen t o f Pile s an d Pil e Group s i n Sand s an d Gravel s 68 1
15.30 Settlemen t o f Pil e Group s i n Cohesiv e Soil s 68 9
15.31 Allowabl e Load s o n Group s o f Pile s 69 0
15.32 Negativ e Frictio n 69 2
15.33 Uplif t Capacit y o f a Pil e Grou p 69 4
15.34 Problem s 69 6
CH APTER 1 6 DEE P FOU NDATI ON II :
B EH AV I OR OF L ATERAL LY L OADED V ERTI CAL AND
B ATTER PILES 69 9
16.1 Introductio n 69 9
16.2 Winkler' s Hypothesi s 70 0
16.3 Th e Differentia l Equatio n 70 1
16.4 Non-dimensiona l Solution s fo r V ertical Pile s Subjecte d t o
Lateral Load s 70 4
16.5 p- y Curve s fo r th e Solutio n o f Laterall y Loade d Pile s 70 6
16.6 Broms ' Solution s fo r Laterall y Loade d Pile s 70 9
16.7 A Direc t Metho d fo r Solvin g th e Non-linea r Behavio r o f
Laterally Loade d Flexibl e Pil e Problem s 71 6
16.8 Cas e Studie s fo r Laterall y Loade d V ertica l Pile s i n San d 72 2
16.9 Cas e Studie s fo r Laterall y Loade d V ertica l Pile s i n Cla y 72 5
16.10 Behavio r o f Laterall y Loade d Batte r Pile s i n San d 73 1
16.11 Problem s 73 9
CH APTER 1 7 DEE P FOU NDATI ON III :
DRILLED PIER FOU NDATI ONS 74 1
17.1 Introductio n 74 1
17.2 Type s o f Drille d Pier s 7 41
17.3 Advantage s an d Disadvantage s o f Drille d Pie r Foundation s 74 3
17.4 Method s o f Constructio n 74 3
17.5 Desig n Consideration s 75 1
17.6 Loa d Transfe r Mechanis m 75 2
17.7 V ertica l Bearin g Capacit y o f Drille d Pier s 75 4
17.8 Th e Genera l Bearin g Capacit y Equatio n fo r th e Bas e Resistanc e
=
75 5
"
x x Con t en t s
17.9 Bearin g Capacit y Equation s fo r th e Bas e i n Cohesiv e Soi l 75 6
17.10 Bearin g Capacit y Equatio n fo r th e Bas e i n Granula r Soi l 75 6
17.11 Bearin g Capacit y Equation s fo r th e Bas e i n Cohesiv e IG M o r Roc k 75 9
17.12 Th e Ultimat e Ski n Resistanc e o f Cohesiv e an d
Intermediate Material s 76 0
17.13 Ultimat e Ski n Resistanc e i n Cohesionles s Soi l an d Gravell y Sand s 76 3
17.14 Ultimat e Sid e an d Tota l Resistanc e i n Roc k 76 4
17.15 Estimatio n o f Settlement s o f Drille d Pier s a t Workin g Load s 76 5
17.16 Uplif t Capacit y o f Drille d Pier s 77 7
17.17 Latera l Bearin g Capacit y o f Drille d Pier s 77 9
17.18 Cas e Stud y o f a Drille d Pie r Subjecte d t o Latera l Load s 78 7
17.19 Problem s 78 7
CH APTER 1 8 FOU NDATI ON S ON COL L APSI B LE AND
EXPANSI V E SOI L S 7 9 1
18.1 Genera l Consideration s 79 1
PART A- COL L APSI B L E SOI L S 79 3
18.2 Genera l Observation s 79 3
18.3 Collaps e Potentia l an d Settlemen t 79 5
18.4 Computatio n o f Collaps e Settlemen t 79 6
18.5 Foundatio n Desig n 79 9
18.6 Treatmen t Method s fo r Collapsibl e Soil s 80 0
PART B - EXPANSI V E SOI L S 80 0
18.7 Distributio n of Expansiv e Soil s 80 0
18.8 Genera l Characteristic s o f Swellin g Soil s 80 1
18.9 Cla y Mineralog y an d Mechanis m o f Swellin g 80 3
18.10 Definitio n o f Som e Parameter s 80 4
18.11 Evaluatio n o f th e Swellin g Potentia l o f Expansiv e Soil s b y Singl e
Index Metho d 80 4
18.12 Classificatio n o f Swellin g Soil s b y Indirec t Measuremen t 80 6
18.13 Swellin g Pressur e b y Direc t Measuremen t 81 2
18.14 Effec t o f Initia l Moistur e Conten t an d Initia l Dr y Densit y o n
Swelling Pressur e 81 3
18.15 Estimatin g the Magnitud e o f Swellin g 81 4
18.16 Desig n o f Foundation s i n Swellin g Soil s 81 7
18.17 Drille d Pie r Foundation s 81 7
18.18 Eliminatio n o f Swellin g 82 7
18.19 Problem s 82 8
Con t en t s x x i
CH APTER 1 9 CONCRET E AND M ECH ANI CAL L Y STAB I L I Z ED
EARTH RETAI NI NG WAL L S 8 3 3
PART A- CONCRETE RETAI NI NG WAL L S 83 3
19.1 Introductio n 83 3
19.2 Condition s Unde r Whic h Rankin e an d Coulom b Formula s Ar e
Applicable t o Retainin g Wall s Unde r th e Activ e Stat e 83 3
19.3 Proportionin g o f Retainin g Wall s 83 5
19.4 Eart h Pressur e Chart s fo r Retainin g Wall s 83 6
19.5 Stabilit y o f Retainin g Wall s 83 9
PART B -M ECH ANI CAL L Y STAB I L I Z ED EARTH RETAI NI NG WAL L S 84 9
19.6 Genera l Consideration s 84 9
19.7 Backfil l an d Reinforcin g Material s 85 1
19.8 Constructio n Detail s 85 5
19.9 Desig n Consideration s fo r a Mechanicall y Stabilize d Eart h Wal l 85 7
19.10 Desig n Metho d 85 9
19.11 Externa l Stabilit y 86 3
19.12 Example s o f Measure d Latera l Eart h Pressure s 87 5
19.13 Problem s 87 7
CH APTER 2 0 SH EE T PILE WAL LS AND B RACED CU TS 88 1
20.1 Introductio n 88 1
20.2 Shee t Pil e Structure s 88 3
20.3 Fre e Cantileve r Shee t Pil e Wall s 88 3
20.4 Dept h o f Embedmen t o f Cantileve r Wall s i n Sand y Soil s 88 5
20.5 Dept h o f Embedmen t o f Cantileve r Wall s i n Cohesiv e Soil s 89 6
20.6 Anchore d Bulkhead : Free-Eart h Suppor t Method—Dept h o f
Embedment o f Anchore d Shee t Pile s i n Granula r Soil s 90 8
20.7 Desig n Chart s fo r Anchore d Bulkhead s i n San d 91 3
20.8 Momen t Reductio n fo r Anchored Shee t Pil e Wall s 91 6
20.9 Anchorag e o f Bulkhead s 92 5
20.10 Brace d Cut s 93 1
20.11 Latera l Eart h Pressur e Distributio n o n Braced-Cut s 93 5
20.12 Stabilit y o f Brace d Cut s i n Saturate d Cla y 93 8
20.13 Bjerru m an d Eid e Metho d o f Analysi s 94 0
20.14 Pipin g Failure s i n San d Cut s 94 5
20.15 Problem s 94 5
XXI I Con t en t s
CH APTER 2 1 SOI L I M PROV EM ENT
21. 1 Int roduct i o n
21.2 Mechanica l Compactio n
21.3 Laborator y Test s o n Compactio n
21.4 Effec t o f Compactio n o n Engineerin g Behavio r
21.5 Fiel d Compactio n an d Contro l
21.6 Compactio n fo r Deepe r Layer s o f Soi l
21.7 Preloadin g
21.8 San d Compactio n Pile s an d Ston e Column s
21.9 Soi l Stabilizatio n b y th e Us e o f Admixtures
21.10 Soi l Stabilizatio n by Injectio n o f Suitabl e Grout s
21.11 Problem s
9 5 1
951
952
953
959
962
973
974
980
981
983
983
APPENDI X A S I UNITS I N G EOTECH NI CAL ENG I NEERI NG 9 8 7
APPENDI X B SL OP E STAB I L I TY CH ARTS AND TAB L ES 9 9 3
REFERENCES 1 0 0 7
I NDEX 1 0 2 5
CHAPTER 1
INTRODUCTION
1 .1 G ENERA L REM ARK S
Karl Terzaghi writing in 1951 (Bjerrum, et. al., 1960), on 'The Influence of Modern Soi l Studies on
the Design an d Construction o f Foundations' commente d o n foundations as follows:
Foundations can appropriately be described as a necessary evil. If a building is to be
constructed on an outcrop of sound rock, no foundationis required. Hence, in contrast to the
building itself which satisfies specific needs, appeals to the aesthetic sense, and fills its
matters with pride, the foundations merely serve as a remedy for the deficiencies of whatever
whimsical nature has provided for the support of the structure at the site which has been
selected. On account of the fact that there is no glory attached to the foundations, and that
the sources of success or failures are hidden deep in the ground, building foundations have
always been treated as step children; and their acts of revenge for the lack of attention can be
very embarrassing.
The comment s mad e b y Terzagh i ar e ver y significan t an d shoul d b e take n not e o f b y al l
practicing Architect s an d Engineers. Architect s or Engineers wh o do not wis h t o make us e of the
growing knowledge of foundation design ar e not rendering true service t o thei r profession. Sinc e
substructures ar e a s important as superstructures, persons wh o ar e wel l qualified i n the design of
substructures shoul d alway s b e consulte d an d th e ol d prover b tha t a 'stitc h i n tim e save s nine '
should always be kept i n mind.
The design of foundations i s a branch of Civil Engineering. Experience ha s shown that most
of thes e branche s hav e passed i n succession throug h two stages , th e empirical an d the scientific ,
before the y reached th e present one which may be called th e stage of maturity.
The stag e of scientifi c reasoning i n the design of foundations started wit h the publication of
the book Erdbaumechanik (mean s Soil Mechanics) by Karl Terzaghi i n 1925. This book represent s
the first attemp t to treat Soi l Mechanics o n the basis of the physical properties o f soils. Terzaghi' s
Chapt er 1
contribution for the development o f Soi l Mechanic s an d Foundation Engineering i s so vast that he
may trul y be calle d th e Father o f Soil Mechanics, Hi s activit y extended ove r a period o f about 5 0
years startin g from th e year 1913 . He was bor n on October 2 , 188 3 i n Prague an d die d o n Octobe r
25, 196 3 i n Winchester, Massachusetts, USA. Hi s amazing caree r i s wel l documented i n the boo k
'From Theory t o Practice in Soil Mechanics' (Bjerrum , L., et . al. , 1960) .
Many investigator s i n th e fiel d o f Soi l Mechanic s wer e inspire d b y Terzaghi . Som e o f th e
notable personalitie s wh o followe d hi s footstep s ar e Ralp h B . Peck , Arthu r Casagrande ,
A. W. Skempton, etc. Because of the unceasing efforts o f these and other innumerable investigators,
Soil Mechanics and Foundation Engineerin g ha s come to stay as a very important par t o f the Civi l
Engineering profession .
The transitio n o f foundatio n engineering fro m th e empirica l stag e t o tha t o f th e scientifi c
stage starte d almos t a t the commencemen t o f th e 20t h century . The desig n o f foundation s durin g
the empirica l stag e wa s base d mostl y on intuition and experience. Ther e use d t o be man y failures
since th e procedur e o f design was onl y by trial and error .
However, i n the present scientifi c age, th e design o f foundations based on scientifi c analysis
has receive d a muc h impetus . Theories hav e been develope d base d o n fundamenta l properties o f
soils. Stil l one ca n witnes s unsatisfactory performance o f structure s constructed even on scientific
principles. The reasons fo r such poor performance ar e many. The soil mass on which a structure is to
be buil t i s heterogeneou s i n characte r an d n o theor y ca n simulat e fiel d conditions . Th e
fundamental propertie s o f soi l whic h w e determin e i n laboratorie s ma y no t reflec t trul y th e
properties o f th e soi l in-situ. A judicia l combinatio n o f theor y an d experienc e i s essentia l fo r
successful performanc e o f any structure built on earth. Another method tha t is gaining popularity i s
the observational approach. Thi s procedur e consist s i n makin g appropriat e observation s soo n
enough during construction to detect sign s of departur e of the real conditions from thos e assume d
by th e designe r an d i n modifyin g either th e desig n o r th e metho d o f constructio n i n accordanc e
with th e findings.
1 .2 A B RI E F H I STORI CAL DEV EL OPM ENT
Many structure s tha t wer e bui l t centurie s ago ar e monument s o f curiosit y eve n today . Egyptia n
temples buil t three o r four thousand year s ag o still exist thoug h the design o f the foundations were
not based o n any presentl y known principles. Romans buil t notable engineering structure s such as
harbors, breakwaters , aqueducts , bridges, large publi c buildings and a vast network of durable and
excellent roads . Th e leaning tower o f Pisa i n Ital y complete d durin g th e 14t h centur y i s stil l a
center o f tourist attraction. Many bridges wer e als o built during the 15t h t o 17t h centuries. Timber
piles wer e use d fo r man y of the foundations.
Another marve l o f engineering achievemen t i s the construction o f the fame d mausoleu m Ta j
Mahal outsid e th e cit y of Agra. This was constructed in the 17t h century by the Mogul Empero r o f
Delhi, Shahjahan, to commemorate hi s favorite wife Mumtaz Mahal. The mausoleum i s built on the
bank of the river Jamuna. The proximit y of the river required special attention in the building of the
foundations. I t i s reporte d tha t masonr y cylindrica l well s hav e bee n use d fo r th e foundations . It
goes t o the credi t of the engineers who designe d and constructed this grand structur e which is stil l
quite soun d even afte r a lapse o f abou t three centuries.
The firs t rationa l approac h fo r th e computatio n o f eart h pressure s o n retainin g wall s wa s
formulated by Coulomb (1776) , a famous French scientist . He proposed a theory i n 177 6 called the
"Classical Eart h Pressur e Theory" . Poncele t (1840 ) extende d Coulomb' s theor y b y givin g a n
elegant graphica l metho d fo r fi ndi n g th e magnitud e of eart h pressur e o n walls . Later , Culman n
(1875) gav e th e Coulomb-Poncele t theor y a geometrica l formulation , thus supplyin g the metho d
with a broad scientifi c basis. Rankin e (1857) a Professor o f Civi l Engineering i n the Universit y of
In t roduct ion
Glasgow, propose d a new eart h pressur e theory , whic h is als o calle d a Classical Earth Pressure
Theory.
Darcy (1856) , on the basis o f his experiment s o n filter sands, proposed a law for the flow of
water in permeable material s and in the same year Stokes (1856) gave an equation for determining
the terminal velocit y of soli d particle s fallin g i n liquids. The rupture theory of Mohr (1900) Stres s
Circles ar e extensivel y use d i n th e stud y o f shea r strengt h o f soils . On e o f th e mos t importan t
contributions t o engineering scienc e wa s mad e by Boussines q (1885 ) wh o proposed a theor y for
determining stres s distributio n under loade d area s i n a semi-infinite , elastic, homogeneous , an d
isotropic medium .
Atterberg (1911) , a Swedish scientist , proposed simpl e tests for determining the consistency
limits o f cohesiv e soils . Felleniu s (1927 ) heade d a Swedis h Geotechnica l Commissio n fo r
determining the causes of failure of many railway and canal embankments . The so-calle d Swedish
Circle method or otherwise terme d as the Slip Circle method was the outcome of his investigation
which wa s publishe d i n 1927 .
The developmen t o f th e scienc e o f Soi l Mechanic s an d Foundatio n Engineerin g fro m th e
year 192 5 onwards was phenomenal. Terzaghi lai d down definite procedures i n his book published
in 192 5 for determinin g propertie s and the strengt h characteristic s of soils . The moder n soi l
mechanics wa s bor n i n 1925 . Th e presen t stag e o f knowledge i n Soi l Mechanic s an d th e desig n
procedures o f foundation s ar e mostl y du e t o th e work s o f Terzagh i an d hi s ban d o f devote d
collaborators.
1.3 SOI L M ECH ANI CS AND FOU NDATI ON ENG I NEERI N G
Terzaghi define d Soi l Mechanics a s follows:
Soil Mechanics is the application of the laws of mechanics and hydraulics to engineering
problems dealing with sediments and other unconsolidated accumulations of solid particles
produced by the mechanical and chemical disintegration of rocks regardless of whether or
not they contain an admixture of organic constituents.
The ter m Soil Mechanics i s no w accepte d quit e generall y t o designat e tha t disciplin e of
engineering scienc e whic h deals wit h the properties an d behavior o f soi l a s a structural material .
All structure s have t o be buil t on soils . Ou r mai n objective i n the study of soi l mechanic s i s
to lay down certain principles, theories an d procedures fo r the design of a safe and sound structure.
The subjec t o f Foundation Engineering deal s wit h th e desig n o f variou s type s o f substructure s
under differen t soi l an d environmenta l conditions .
During th e design , th e designe r ha s t o mak e us e o f th e propertie s o f soils , th e theorie s
pertaining t o th e desig n an d hi s ow n practica l experienc e t o adjus t th e desig n t o sui t fiel d
conditions. H e ha s t o dea l wit h natura l soi l deposit s whic h perfor m th e engineerin g functio n o f
supporting th e foundatio n an d th e superstructur e abov e it . Soi l deposit s i n natur e exis t i n a n
extremely errati c manne r producin g thereb y a n infinit e variet y o f possibl e combination s whic h
would affec t th e choice and design o f foundations. The foundation engineer mus t have the ability
to interpret the principles of soil mechanics t o suit the field conditions. The success or failure of his
design depend s upo n how muc h i n tune he i s wit h Nature.
CHAPTER 2
SOIL FORMATION AND CHARACTERIZATION
2.1 INTRODUCTIO N
The wor d 'soil ' has different meaning s for different professions . To the agriculturist, soi l is the top
thin laye r o f eart h withi n whic h organi c force s ar e predominant an d whic h is responsibl e fo r th e
support of plant life . To the geologist , soi l i s the materia l i n the top thi n zon e withi n which root s
occur. Fro m th e poin t o f vie w o f a n engineer , soi l include s al l eart h materials , organi c an d
inorganic, occurring i n the zone overlying the rock crust .
The behavio r o f a structur e depend s upo n the propertie s o f the soi l material s o n whic h the
structure rests . Th e propertie s o f the soi l material s depen d upo n the propertie s o f the rock s fro m
which they are derived. A brief discussion of the parent rocks is, therefore, quite essential i n order
to understand the properties o f soil materials .
2. 2 ROC K CLASSIFICATION
Rock can be defined as a compact, semi-har d t o hard mass of natural material composed of one or
more minerals. The rocks that are encountered at the surface of the earth or beneath, are commonly
classified int o three group s according t o their modes o f origin. They ar e igneous, sedimentar y and
metamorphic rocks.
Igneous rock s ar e considere d t o b e th e primar y rock s forme d b y th e coolin g o f molte n
magmas, o r by the recrystallization of older rock s under heat and pressure grea t enough to render
them fluid. The y have been forme d on or at various depths below the earth surface . There are two
main classes o f igneous rocks. They are :
1. Extrusiv e (poured out a t the surface) , and
2. Intrusiv e (large rock masse s whic h have not been formed in contact wit h the atmosphere) .
6 Chapt e r 2
Initially bot h classes o f rocks wer e i n a molten state. Their presen t stat e result s directl y fro m
the wa y i n which they solidified. Due t o violent volcanic eruptions in the past , some o f the molte n
materials wer e emitte d int o th e atmospher e wit h gaseou s extrusions . Thes e coole d quickl y and
eventually fel l o n the earth' s surfac e as volcanic ash and dust. Extrusive rocks ar e distinguished, in
general, by thei r glass-lik e structure .
Intrusive rocks , coolin g an d solidifyin g a t grea t depth s an d unde r pressur e containin g
entrapped gases , are wholl y crystalline in texture. Such rocks occu r i n masses o f great extent , often
going to unknown depths. Some o f the important rocks tha t belong t o the igneous group are granite
and basalt. Granit e i s primaril y composed o f feldspar , quart z an d mic a an d possesse s a massiv e
structure. Basal t i s a dark-colore d fine-graine d rock. I t i s characterize d b y th e predominanc e o f
plagioclase, th e presence o f considerable amount s of pyroxene and some olivin e and the absence of
quartz. Th e colo r varie s fro m dark-gre y t o black . Bot h granit e an d basal t ar e use d a s buildin g
stones.
When th e product s o f th e disintegratio n an d decompositio n o f an y roc k typ e ar e
transported, redeposited , an d partl y or full y consolidate d o r cemented int o a new roc k type , th e
resulting materia l i s classifie d a s a sedimentary rock. Th e sedimentar y rock s generall y ar e
formed i n quit e definitel y arrange d beds , o r strata, whic h can be see n t o have bee n horizonta l at
one time althoug h sometime s displace d throug h angle s u p t o 90 degrees. Sedimentar y rock s ar e
generally classifie d o n the basis o f grain size , texture and structure. From a n engineering point of
view, the most important rock s tha t belong t o the group ar e sandstones, limestones, and shales.
Rocks forme d b y th e complet e o r incomplet e recrystallizatio n o f igneou s o r sedimentar y
rocks b y hig h temperatures , hig h pressures , and/o r hig h shearin g stresse s ar e metamorphic rocks.
The rock s s o produce d ma y displa y feature s varyin g fro m complet e an d distinc t foliatio n o f a
crystalline structur e t o a fine fragmentary partiall y crystallin e stat e cause d b y direc t compressiv e
stress, including also the cementation of sediment particles by siliceous matter . Metamorphic rock s
formed withou t intense shea r actio n hav e a massiv e structure . Some o f th e importan t rock s tha t
belong t o thi s group ar e gneiss, schist, slate an d marble. Th e characteristi c featur e of gneis s i s its
structure, th e minera l grain s ar e elongated , o r platy , an d bandin g prevails . Generall y gneis s i s a
good engineerin g material . Schis t i s a finel y foliate d roc k containin g a high percentag e o f mica .
Depending upo n th e amoun t of pressure applie d b y the metamorphi c forces , schis t ma y be a very
good buildin g material . Slat e i s a dar k colored , plat y roc k wit h extremel y fin e textur e an d eas y
cleavage. Becaus e o f thi s easy cleavage , slat e i s spli t int o ver y thi n sheet s an d use d a s a roofing
material. Marble i s the end product of the metamorphism o f limestone and other sedimentar y rock s
composed o f calciu m o r magnesiu m carbonate . I t i s ver y dens e an d exhibit s a wid e variet y o f
colors. I n construction , marbl e i s use d fo r facin g concret e o r masonr y exterio r an d interio r wall s
and floors .
Rock M ineral s
It i s essentia l t o examin e th e propertie s o f th e roc k formin g mineral s sinc e al l soil s ar e derive d
through the disintegration or decomposition o f some parent rock. A 'mineral' i s a natural inorganic
substance o f a definit e structur e and chemica l composition . Som e o f th e ver y importan t physica l
properties o f mineral s ar e crysta l form , color , hardness , cleavage , luster , fracture , an d specifi c
gravity. Out of these onl y two, specific gravit y and hardness, ar e of foundation engineering interest.
The specifi c gravit y o f th e mineral s affect s th e specifi c gravit y o f soil s derive d fro m them . Th e
specific gravit y of most rock and soi l formin g minerals varies fro m 2.5 0 (som e feldspars ) and 2.6 5
(quartz) to 3.5 (augite or olivine). Gypsum has a smaller value of 2.3 and salt (NaCl) has 2.1. Some
iron mineral s ma y hav e higher values, for instance, magnetite ha s 5.2 .
It i s reported tha t about 9 5 percen t o f th e known part o f th e lithospher e consist s o f igneous
rocks and only 5 percent of sedimentary rocks. Soi l formation is mostly due to the disintegration of
igneous rock which ma y be terme d a s a parent rock.
Soil Form at io n an d Char act er izat io n 7
Table 2. 1 M in era l com pos it io n o f ig n eou s rock s
M in eral Percen t
Quartz 12-2 0
Feldspar 50-6 0
Ca, Fe and Mg, Silicates 14-1 7
Micas 4- 8
Others 7- 8
The average mineral composition of igneous rocks is given in Table 2.1. Feldspars ar e the most
common roc k minerals , whic h account for the abundance of clays derived from th e feldspar s on the
earth's surface . Quartz comes nex t in order of frequency. Mos t sand s are composed o f quartz.
2.3 FORM ATI O N O F SOI L S
Soil i s define d a s a natural aggregat e o f mineral grains , wit h or without organic constituents , that
can b e separate d b y gentl e mechanica l mean s suc h a s agitatio n i n water . B y contras t roc k i s
considered t o be a natural aggregate o f mineral grains connected by strong and permanent cohesiv e
forces. Th e proces s o f weatherin g o f the roc k decrease s th e cohesiv e force s bindin g th e minera l
grains an d lead s t o the disintegratio n o f bigger masse s t o smalle r an d smalle r particles . Soil s ar e
formed b y th e process of weathering o f the parent rock. The weathering of the rock s migh t be by
mechanical disintegration, and/o r chemica l decomposition .
M echanical Weathering
Mechanical weatherin g o f rock s t o smalle r particle s i s du e t o th e actio n o f suc h agent s a s th e
expansive force s o f freezing water i n fissures, due t o sudden changes o f temperature o r due t o the
abrasion o f roc k b y movin g wate r o r glaciers . Temperatur e change s o f sufficien t amplitud e an d
frequency brin g abou t change s i n th e volume o f th e rock s i n th e superficia l layer s o f th e earth' s
crust in terms of expansion and contraction. Such a volume change sets up tensile and shear stresse s
in the rock ultimatel y leading t o the fracture of even large rocks. This type of rock weatherin g takes
place in a very significant manner in arid climates where free, extreme atmospheri c radiation brings
about considerabl e variatio n i n temperature a t sunrise and sunset.
Erosion b y wind and rai n is a very important factor and a continuing event. Cracking force s
by growin g plants an d root s i n voids and crevasses o f rock ca n forc e fragment s apart .
Chemical Weathering
Chemical weatherin g (decomposition ) ca n transform hard rock mineral s int o soft , easil y erodable
matter. The principa l type s o f decompositio n ar e hydmtion, oxidation, carbonation, desilication
and leaching. Oxyge n and carbon dioxid e which are always present in the air readily combine with
the element s o f rock i n the presenc e o f water.
2.4 G ENERA L TY PES OF SOI L S
It ha s bee n discusse d earlie r tha t soi l i s forme d b y th e proces s o f physica l an d chemica l
weathering. The individua l size o f th e constituen t part s o f eve n th e weathere d roc k migh t range
from th e smalles t stat e (colloidal ) t o th e larges t possibl e (boulders) . Thi s implie s tha t al l th e
weathered constituent s of a parent roc k canno t be terme d soil . According t o thei r grai n size , soi l
8 Chapt e r 2
particles ar e classified a s cobbles, gravel , sand, sil t and clay. Grains havin g diameters i n the rang e
of 4.7 5 t o 76. 2 mm ar e calle d gravel . I f th e grain s ar e visibl e t o th e nake d eye , bu t ar e les s than
about 4.7 5 m m i n siz e th e soi l i s describe d a s sand . The lowe r limi t of visibilit y of grain s fo r th e
naked eye s i s about 0.075 mm. Soi l grains ranging from 0.07 5 t o 0.002 mm ar e termed a s silt and
those tha t ar e fine r tha n 0.002 mm a s clay. Thi s classificatio n i s purely based o n siz e whic h doe s
not indicat e th e propertie s o f fin e graine d materials .
Residual and Transporte d Soil s
On th e basi s o f origi n o f thei r constituents, soil s ca n b e divide d int o two larg e groups :
1. Residua l soils , an d
2. Transporte d soils .
Residual soils ar e thos e tha t remai n a t th e plac e o f thei r formatio n a s a resul t o f th e
weathering o f parent rocks . Th e dept h o f residua l soil s depend s primaril y o n climati c condition s
and th e tim e o f exposure . I n som e areas , thi s dept h migh t b e considerable . I n temperat e zone s
residual soil s ar e commonl y stif f an d stable. An important characteristic o f residual soi l i s that the
sizes of grains ar e indefinite . Fo r example , whe n a residual sampl e i s sieved , th e amoun t passin g
any give n siev e siz e depend s greatl y o n the time an d energy expende d i n shaking , because o f the
partially disintegrate d condition .
Transported soils ar e soil s tha t ar e foun d a t location s fa r remove d fro m thei r plac e o f
formation. The transporting agencies of such soils are glaciers, wind and water. The soils are named
according t o th e mod e o f transportation . Alluvial soil s ar e thos e tha t hav e bee n transporte d b y
running water . The soil s tha t have been deposite d i n quiet lakes, ar e lacustrine soils . Marine soils
are thos e deposite d i n se a water . Th e soil s transporte d an d deposite d b y win d ar e aeolian soils .
Those deposite d primaril y through the action o f gravitational force , a s in land slides , ar e colluvial
soils. Glacial soil s ar e thos e deposite d b y glaciers . Man y o f thes e transporte d soil s ar e loose and
soft t o a dept h o f severa l hundre d feet. Therefore, difficultie s wit h foundations and othe r type s of
construction ar e generall y associate d wit h transporte d soils .
Organic an d I norgani c Soil s
Soils i n genera l ar e furthe r classifie d a s organic o r inorganic. Soil s o f organi c origi n ar e chiefl y
formed eithe r b y growt h an d subsequen t decay o f plant s such a s peat , o r b y th e accumulatio n of
fragments o f th e inorgani c skeletons o r shell s of organisms . Henc e a soi l o f organi c origi n ca n b e
either organi c or inorganic. The term organic soil ordinarily refers to a transported soi l consisting of
the product s o f roc k weatherin g wit h a more o r les s conspicuous admixtur e of decayed vegetabl e
matter.
Names o f Some Soil s that ar e G enerall y U se d i n Practic e
Bentonite i s a cla y forme d b y th e decompositio n o f volcani c as h wit h a hig h conten t o f
montmorillonite. It exhibits the properties o f clay t o an extreme degree .
Varved Clay s consis t of thi n alternating layers of sil t and fat clays of glacial origin. They posses s
the undesirabl e propertie s o f bot h sil t and clay. The constituent s of varved clay s wer e transporte d
into fres h wate r lake s b y th e melte d ic e a t the clos e o f th e ic e age .
Kaol in, Chin a Cla y ar e ver y pur e forms o f whit e clay use d i n th e cerami c industry .
Boulder Cla y i s a mixtur e o f a n unstratifie d sedimente d deposi t o f glacia l clay , containin g
unsorted roc k fragment s of all sizes ranging from boulders , cobbles, an d gravel to finely pulverize d
clay material .
Soil Form at io n an d Charact erizat ion 9
Calcareous Soi l i s a soi l containin g calcium carbonate. Suc h soi l effervesce s whe n teste d wit h
weak hydrochloric acid.
Marl consist s of a mixture of calcareous sands , clays, or loam.
Hardpan i s a relativel y hard , densel y cemente d soi l layer , like rock whic h does not softe n whe n
wet. Boulder clays or glacial till is also sometime s named as hardpan.
Caliche i s an admixtur e of clay, sand, and gravel cemented by calcium carbonate deposite d fro m
ground water.
Peat i s a fibrou s aggregat e o f fine r fragment s o f decaye d vegetabl e matter . Pea t i s ver y
compressible an d one shoul d be cautious when using it for supporting foundations of structures.
Loam i s a mixture of sand, silt and clay.
Loess is a fine-grained, air-borne deposit characterized by a very uniform grain size, and high void
ratio. The size of particles ranges between about 0.01 t o 0.05 mm. The soi l can stand deep vertical
cuts because of slight cementation between particles. It is formed in dry continental regions and its
color i s yellowish light brown.
Shale is a material i n the state of transition from cla y to slate. Shale itself is sometimes considere d
a rock but, when it is exposed to the air or has a chance to take in water it may rapidly decompose .
2.5 SOI L PARTI CL E SI Z E AND SH AP E
The siz e o f particles a s explained earlier , ma y range fro m grave l t o the fines t siz e possible. Thei r
characteristics var y with the size. Soil particles coarser tha n 0.075 mm are visible to the naked eye
or ma y b e examine d b y mean s o f a hand lens. They constitut e the coarse r fraction s of th e soils .
Grains fine r tha n 0.075 mm constitut e the fine r fraction s o f soils . I t i s possibl e t o distinguis h the
grains lying between 0.075 mm an d 2 \ J L ( 1 [ i = 1 micron = 0.001 mm ) unde r a microscope. Grain s
having a size between 2 ji and 0.1 JL A can be observed under a microscope bu t their shapes cannot be
made out . Th e shap e o f grain s smalle r tha n 1 ja ca n b e determine d b y mean s o f a n electro n
microscope. The molecular structur e of particles can be investigated by means of X-ray analysis.
The coarse r fraction s of soil s consis t of grave l and sand. The individua l particles o f gravel ,
which ar e nothin g but fragment s of rock , ar e compose d o f on e o r mor e minerals , wherea s san d
grains contain mostly one mineral which is quartz. The individual grains of gravel and sand may be
angular, subangular , sub-rounded , rounde d o r well-rounde d a s show n i n Fig . 2.1. Grave l ma y
contain grains which may be flat . Som e sand s contain a fairly hig h percentage o f mica flakes that
give them the property of elasticity.
Silt and clay constitute the finer fractions of the soil . Any one grain of this fraction generally
consists of only one mineral. The particles may be angular, flake-shaped or sometimes needle-like .
Table 2. 2 give s th e particl e siz e classificatio n system s a s adopte d b y som e o f th e
organizations i n th e USA . Th e Unifie d Soi l Classificatio n Syste m i s no w almos t universall y
accepted an d has been adopted by the American Societ y for Testing and Materials (ASTM) .
Specific Surfac e
Soil is essentially a paniculate system, that is, a system in which the particles ar e i n a fine stat e of
subdivision or dispersion. In soils, the dispersed or the solid phase predominates an d th e dispersio n
medium, soil water, onl y helps to fil l th e pores between the soli d particles . The significanc e of the
concept of dispersion become s mor e apparen t when the relationshi p o f surfac e t o particl e siz e i s
considered. I n the cas e o f silt , sand and larger siz e particles th e rati o of the are a o f surfac e of the
particles t o the volume of the sampl e i s relatively small. This rati o becomes increasingl y large as
Chapt er 2
Angular Subangular Subrounded
Rounded Wel l rounded
Figure 2. 1 Shape s o f coars e r fract ion s o f s oil s
size decreases from 2 \ J L which i s the uppe r limit for clay-sized particles . A useful index o f relativ e
importance o f surfac e effect s i s the specific surface o f grain. The specifi c surfac e is defined a s th e
total are a o f th e surfac e o f th e grain s expresse d i n squar e centimeter s pe r gra m o r pe r cubi c
centimeter o f the disperse d phase .
The shap e o f th e cla y particle s i s a n important property fro m a physical poin t of view. The
amount o f surfac e pe r uni t mas s o r volume varie s wit h the shap e o f th e particles . Moreover , th e
amount of contact area per uni t surface changes with shape. It is a fact that a sphere has the smallest
surface are a pe r uni t volum e wherea s a plat e exhibit s th e maximum . Ostwal d (1919 ) ha s
emphasized th e importance of shape in determining the specific surface of colloidal systems . Sinc e
disc-shaped particle s ca n b e brough t mor e i n intimat e contact wit h eac h other , thi s shap e ha s a
pronounced effec t upo n the mechanical properties o f the system. The interparticl e force s between
the surfaces of particles have a significant effec t o n the properties o f the soi l mass i f the particles i n
the medi a belong t o the clay fraction. The surfac e activity depends no t only on the specifi c surface
but als o o n the chemical and mineralogical composition of the soli d particles. Sinc e clay particles
Table 2. 2 Part icl e s iz e cl as s ificat io n b y various s ys t em s
Na me o f t h e or gani z at i o n
Massachusetts Institut e of
Technology (MIT )
US Departmen t o f Agriculture (USDA)
American Associatio n of Stat e
Highway an d Transportatio n
Officials (AASHTO )
Unified Soi l Classificatio n System ,
US Burea u o f Reclamation, US Army
Corps of Engineer s an d American
Society fo r Testing an d Material s
Par t i cl e siz e ( mm)
Gr avel San d Sil t Cl a y
>2 2 to 0.06 0.0 6 t o 0.002 <
> 2 2 t o 0.05 0.0 5 t o 0.002 <
76.2 t o 2 2 to 0.075 0.07 5 t o 0.002 <
76.2 t o 4.75 4.7 5 t o 0.075 Fine s (silt s and clays)
< 0.075
0.002
0.002
0.002
Soil Form at io n an d Charact erizat io n 1 1
are th e activ e portion s o f a soi l becaus e o f thei r hig h specifi c surfac e an d thei r chemica l
constitution, a discussion on the chemical composition and structure o f minerals i s essential.
2.6 COM POSI TI O N OF CL AY M I NERAL S
The word 'clay' is generally understood to refer to a material composed o f a mass of small mineral
particles which, in association wit h certai n quantities of water , exhibit s the propert y o f plasticity.
According to the clay mineral concept, clay materials are essentially compose d o f extremely smal l
crystalline particle s o f on e o r mor e member s o f a smal l grou p o f mineral s tha t ar e commonl y
known a s cla y minerals . Thes e mineral s ar e essentiall y hydrou s aluminu m silicates , wit h
magnesium o r iro n replacin g wholl y or i n par t fo r th e aluminum , in some minerals . Man y clay
materials may contain organic materia l and water-soluble salts. Organi c material s occur either as
discrete particle s o f wood , lea f matter , spores , etc. , o r they may be present a s organi c molecule s
adsorbed on the surface of the clay mineral particles. The water-soluble salts that are present in clay
materials must have been entrapped i n the clay at the time of accumulation or may have developed
subsequently as a consequence of ground water movement and weathering or alteration processes.
Clays can be divided int o three genera l groups on the basis of their crystalline arrangement
and i t i s observe d tha t roughl y simila r engineerin g propertie s ar e connecte d wit h al l th e cla y
minerals belonging to the same group. An initial study of the crystal structure of clay minerals leads
to a better understanding of the behavior of clays under different condition s of loading. Table 2. 3
gives the groups of minerals and some of the important minerals under each group .
2.7 STRU CTU R E O F CL AY M I NERAL S
Clay mineral s ar e essentiall y crystallin e i n natur e thoug h som e cla y mineral s d o contai n
material whic h is non-crystalline (for example allophane) . Two fundamental building block s
are involved i n th e formatio n o f clay minera l structures . They are :
1. Tetrahedra l unit.
2. Octahedra l unit.
The tetrahedra l uni t consist s o f fou r oxyge n atoms (or hydroxyls , if neede d t o balanc e
the structure ) placed a t the apice s o f a tetrahedron enclosin g a silicon ato m whic h combine s
together t o for m a shell-lik e structur e wit h al l th e tip s pointin g i n th e sam e direction . Th e
oxygen a t the bases of al l th e unit s lie i n a common plane .
Each o f the oxyge n ions a t the bas e i s common t o two units . The arrangemen t i s shown in
Fig. 2.2. The oxygen atoms ar e negatively charged wit h two negative charges eac h and the silicon
with fou r positiv e charges . Eac h o f th e thre e oxyge n ions a t th e bas e share s it s charge s wit h the
Table 2. 3 Cla y m in eral s
Nam e o f m in era l St ruct ura l form ul a
I. Kaoli n group
1. Kaolinite Al
4
Si
4
O
10
(OH)
g
2. Halloysit e Al
4
Si
4
O
6
(OH)
16
II. Montmorillonit e grou p
Montmorillonite Al
4
Si
8
O
20
(OH)
4
nH
2
O
III. Illit e grou p
Illite K
y
(Al
4
Fe
2
.Mg
4
.Mg
6
)Si
8
_
y
Al
y
(OH)
4
0
20
1 2 Chapt er 2
adjacent tetrahedra l unit . The sharin g o f charge s leave s thre e negativ e charge s a t th e bas e pe r
tetrahedral uni t an d thi s alon g wit h tw o negativ e charges a t the ape x make s a tota l o f 5 negative
charges t o balance the 4 positive charges of the silicon ion. The process of sharing the oxygen ions
at the base wit h neighborin g units leaves a net charge of -1 per unit.
The second building block is an octahedral unit with six hydroxyl ions at apices of an octahedra l
enclosing an aluminum ion at the center. Iron or magnesium ions may replace aluminum ions in some
units. These octahedral units are bound together in a sheet structure with each hydroxyl ion common t o
three octahedral units. This sheet is sometimes called as gibbsite sheet. The Al ion has 3 positive charges
and each hydroxyl ion divides its -1 charg e wit h two other neighboring units. This sharing of negative
charge wit h other units leaves a total of 2 negative charges per unit [(1/3) x 6]. The net charge of a unit
with a n aluminu m ion a t th e cente r i s +1 . Fig . 2. 3 give s th e structura l arrangement s o f th e units .
Sometimes, magnesiu m replace s th e aluminu m atoms i n th e octahedra l unit s i n thi s case , th e
octahedral shee t i s called a brucite sheet.
Formation o f M ineral s
The combination of two sheet s of silica and gibbsite i n different arrangement s an d conditions lea d
to the formation of different cla y minerals as given in Table 2.3. I n the actual formation of the sheet
silicate minerals , th e phenomeno n o f isomorphous substitution frequentl y occurs . Isomorphou s
(meaning same form) substitution consists of the substitution of one kind of atom for another .
Kaoiinite Minera l
This i s th e mos t commo n minera l o f th e kaoli n group . Th e buildin g block s o f gibbsit e an d
silica sheet s ar e arrange d a s shown i n Fig. 2. 4 t o giv e th e structur e of the kaolinit e layer . Th e
structure i s compose d o f a singl e tetrahedra l shee t an d a singl e alumin a octahedra l shee t
combined i n unit s s o tha t th e tip s o f th e silic a tetrahedron s an d on e o f th e layer s o f th e
octahedral shee t for m a common layer . All the tips o f the silica tetrahedrons poin t i n the same
direction an d towards the center of the uni t made of the silica an d octahedral sheets . This gives
rise t o stron g ioni c bonds betwee n th e silica an d gibbsit e sheets . Th e thicknes s o f th e laye r i s
about 7 A (on e angstro m = 10~
8
cm) thick . Th e kaolinit e minera l i s forme d b y stackin g th e
layers on e above the other wit h the base of the silica sheet bondin g to hydroxyls of the gibbsit e
sheet b y hydroge n bonding . Sinc e hydroge n bond s ar e comparativel y strong , th e kaolinit e
(a) Tetrahedral uni t (b) Silica sheet
Silicons
Oxygen
] _ Symboli c representation
of a silica sheet
Figure 2. 2 B as i c s t ruct ural un it s i n t he s ilico n s hee t (Grim , 1959 )
Soil Form at io n an d Charact erizat io n 13
(a) Octahedral uni t (b) Octahedral shee t
0 Hydroxyl s
I Symboli c representatio n
of a octahedral shee t
Aluminums,
magnesium o r iro n
Figure 2. 3 Bas i c s t ruct ural unit s i n oct ahedral sheet (Grim , 1959 )
crystals consis t o f many sheet stacking s tha t are difficul t t o dislodge. Th e mineral is therefore,
stable, an d wate r canno t ente r betwee n th e sheet s t o expan d th e uni t cells . Th e latera l
dimensions o f kaolinit e particle s rang e fro m 100 0 t o 20,000 A and th e thicknes s varie s fro m
100 to 100 0 A . In the kaolinite mineral there is a very small amount of isomorphous substitution.
H alloysite Mineral
Halloysite minerals are made up of successive layers with the same structural composition a s those
composing kaolinite. I n this case, however , the successive unit s are randomly packed an d may be
separated by a single molecular layer of water. The dehydration of the interlayers by the removal of
the wate r molecule s lead s t o change s i n th e propertie s o f th e mineral . An importan t structural
feature of halloysite is that the particles appear to take tubular forms a s opposed t o the platy shape
of kaolinite.
Ionic bon d
Hydrogen bon d
Gibbsite shee t
Silica shee t
T
7 A
7 A
7 A
Figure 2. 4 St ruct ur e o f kaolinit e laye r
14 Chapt er 2
M ontmorillonite M inera l
Montmorillonite i s th e mos t commo n minera l o f th e montmorillonit e group . Th e structura l
arrangement o f thi s minera l i s compose d o f tw o silic a tetrahedra l sheet s wit h a centra l alumin a
octahedral sheet . All the tips of the tetrahedra poin t in the same direction and toward the center of the
unit. The silica and gibbsite sheets are combined in such a way that the tips of the tetrahedrons of each
silica shee t an d one o f th e hydroxyl layer s of the octahedral shee t for m a common layer . The atom s
common t o bot h th e silic a and gibbsit e layer become oxyge n instea d of hydroxyls. The thicknes s of
the silica-gibbsite-silica uni t is about 10 A (Fig. 2.5). In stackin g these combined unit s one above th e
other, oxyge n layer s o f eac h uni t ar e adjacen t t o oxyge n o f th e neighborin g unit s wit h a
consequence tha t ther e i s a ver y wea k bon d an d a n excellen t cleavag e betwee n them . Wate r ca n
enter between th e sheets, causing them to expand significantl y and thus the structure can break int o
10 A thic k structura l units. Soil s containin g a considerabl e amoun t o f montmorillonit e mineral s
wil l exhibi t high swelling and shrinkage characteristics. The lateral dimensions o f montmorillonite
particles rang e fro m 100 0 t o 500 0 A wit h thicknes s varyin g fro m 1 0 t o 5 0 A. Bentonit e cla y
belongs t o th e montmorillonit e group . I n montmorillonite , there i s isomorphou s substitutio n of
magnesium an d iro n for al umi num.
Illite
The basi c structura l uni t o f illit e i s simila r t o tha t o f montmorillonit e excep t tha t som e o f th e
silicons ar e always replaced b y al uminum atoms and the resultant charge deficienc y i s balanced by
potassium ions . Th e potassiu m ion s occu r betwee n uni t layers . Th e bond s wit h th e
nonexchangeable K
+
ions ar e weaker tha n the hydrogen bonds, bu t stronger tha n the water bond of
montmorillonite. Illite , therefore , doe s no t swel l a s muc h i n th e presenc e o f wate r a s doe s
montmorillonite. Th e latera l dimension s o f illit e cla y particle s ar e abou t th e sam e a s thos e o f
montmorillonite, 100 0 t o 500 0 A , bu t th e thicknes s o f illit e particle s i s greate r tha n tha t o f
montmorillonite particles, 50 t o 500 A. The arrangement of silica and gibbsit e sheet s ar e as shown
in Fig . 2.6 .
2 .8 CL A Y PARTI CL E- WATER REL ATI ONS
The behavior of a soil mass depends upo n the behavior of the discrete particle s composin g th e mas s
and th e patter n o f particl e arrangement . I n al l thes e case s wate r play s a n importan t part . Th e
Figure 2. 5 St ruct ur e o f m on t m orillon it e l aye r
Soil Form at io n an d Char act er izat io n 15
10A
T I I I I
T\
l oA
I I I I I I I I I I I I I I I I H T M — Potassium molecules
I I I M il I I I I I I I II I I I
\
1 1 1 1 1 1 1 1 1 1 1 1 I I 1 1 1 1 1 V «— Fairly stron g bond
Silica shee t
Gibbsite shee t
Figure 2. 6 St ruct ur e o f illit e laye r
behavior o f th e soi l mas s i s profoundl y influence d b y th e inter-particle-wate r relationships , th e
ability o f the soi l particle s t o adsor b exchangeabl e cation s an d the amount of water present .
Adsorbed Water j
The cla y particle s carr y a ne t negativ e charg e ^ n thei r surface . Thi s i s th e resul t o f bot h
isomorphous substitutio n and of a break i n the continuity of the structure at its edges. The intensity
of th e charg e depend s t o a considerabl e exten t o n tlji e mineralogica l characte r o f th e particle. Th e
physical an d chemica l manifestation s o f th e surfac e charge constitut e th e surfac e activit y of th e
mineral. Mineral s ar e sai d t o have high o r lo w surfac e activity , depending o n th e intensit y o f th e
surface charge . As pointed ou t earlier, th e surface activit y depends no t onl y on the specifi c surfac e
but also on the chemical and mineralogical compositio n of the solid particle. Th e surface activit y of
sand, therefore, wil l not acquire al l the properties of ^ true clay, even if it is ground to a fine powder .
The presenc e o f wate r doe s no t alte r it s properti e
changing it s uni t weight . However , th e behavio r o l
of coarse r fraction s considerabl y exceptin g
' a saturate d soi l mas s consistin g o f fin e san d
might chang e unde r dynami c loadings . Thi s aspec t o f th e proble m i s no t considere d here . Thi s
article deal s onl y wit h clay particle-wate r relations.
In natur e every soi l particl e i s surrounded b y \^ater . Since th e center s o f positive an d negative
charges o f water molecule s d o not coincide, th e molecules behave lik e dipoles . Th e negative charg e
on th e surfac e o f th e soi l particle , therefore , attract s th e positiv e (hydrogen ) en d o f th e wate r
molecules. Th e wate r molecule s ar e arrange d i n a definit e patter n i n th e immediat e vicinit y of th e
boundary between soli d and water. More than one layer of water molecules sticks on the surface with
considerable forc e an d thi s attractive force decrease s wit h the increas e i n the distanc e o f the wate r
molecule from the surface. The electrically attracte d water that surrounds the clay particle i s known as
the diffused double-layer of water. The wate r locate d withi n the zon e of influenc e is known as the
adsorbed layer as shown in Fig. 2.7. Within the zone of influence the physical properties o f the water
are very different fro m thos e of free o r normal water at the same temperature. Near the surface of the
particle the wate r has th e propert y o f a solid. At the middl e o f the laye r it resembles a very viscous
liquid an d beyon d th e zon e o f influence, the propenles o f the wate r become normal . Th e adsorbe d
water affects th e behavior of clay particles whe n subjected t o external stresses, sinc e it comes between
the particl e surfaces . To drive of f the adsorbe d water , the clay particl e mus t be heate d t o mor e tha n
200 °C, whic h woul d indicat e tha t th e bon d betwee n th e wate r molecule s an d th e surfac e i s
considerably greate r tha n that between norma l water molecules .
16
Particle surfac e
Chapt er 2
Adsorbed
water
Distance
Figure 2. 7 Ads orbe d wat e r laye r s urroun din g a s oi l part icl e
The adsorbe d fil m o f water on coarse particle s i s thin in comparison wit h the diameter o f the
particles. I n fine graine d soils , however , thi s layer of adsorbed wate r i s relatively much thicker and
might eve n excee d th e siz e o f the grain . The force s associate d wit h the adsorbe d layer s therefor e
play an important part in determining the physical properties of the very fine-grained soils, but have
little effect o n the coarser soils .
Soils in which the adsorbed film is thick compared t o the grain size have properties quit e differen t
from othe r soil s havin g th e sam e grai n size s bu t smalle r adsorbe d films . Th e mos t pronounce d
characteristic o f th e forme r i s thei r abilit y t o defor m plasticall y withou t cracking whe n mixe d wit h
varying amounts of water. This is due to the grains moving across one another supported by the viscous
interlayers of the films. Such soils are called cohesive soils, for they do not disintegrate with pressure but
can be rolled int o threads with ease. Here the cohesion is not due to direct molecular interaction between
soil particles at the points of contact but to the shearing strength of the adsorbed layer s that separate the
grains at these points.
Base Exchang e
Electrolytes dissociat e whe n dissolve d i n wate r int o positivel y charge d cation s an d negativel y
charged anions . Acids brea k u p int o cations o f hydrogen an d anion s suc h a s Cl o r SO
4
. Salt s an d
bases spli t into metallic cations such as Na, K or Mg, and nonmetallic anions. Even water itself is an
electrolyte, because a very small fraction o f its molecules always dissociates int o hydrogen ions H
+
and hydroxy l ions OH" . Thes e positivel y charged H
+
ions migrat e t o the surfac e of the negatively
charged particle s an d for m what is known as the adsorbed layer . These H
+
ions can be replaced b y
other cation s suc h as Na, K or Mg. These cation s enter th e adsorbe d layer s an d constitut e wha t is
termed a s an adsorption complex. Th e proces s o f replacing cations of one kind by those of another
in a n adsorptio n comple x i s known as base exchange. B y bas e exchang e i s meant th e capacit y of
Soil Form at io n an d Charact erizat io n 1 7
Table 2. 4 Ex chan g e capacit y o f s om e cla y m in eral s
M in eral g roup Ex chan g e capacit y (m eq per 100 g )
Kaolinites 3. 8
Illites 4 0
Montmorillonites 8 0
Table 2. 5 Cation s arran g e d in the orde r of decreas in g shear s t ren g t h o f cla y
NH/ > H
+
> K
+
> Fe
+++
>A1
+++
> Mg
+
> Ba+
+
> Ca
++
> Na
+
> Li
+
colloidal particle s t o change th e cations adsorbe d o n thei r surface . Thus a hydrogen cla y (colloi d
with adsorbe d H cations) can be changed t o sodium cla y (colloi d wit h adsorbed N a cations ) by a
constant percolation o f water containing dissolved Na salts. Suc h changes ca n be used t o decrease
the permeability of a soil. Not all adsorbed cations are exchangeable. Th e quantity of exchangeabl e
cations i n a soil i s termed exchange capacity.
The base exchange capacit y i s generall y define d in terms of the mass of a cation whic h may
be held on the surface of 10 0 gm dry mas s of mineral. It is generally more convenient to employ a
definition o f bas e exchang e capacit y i n milli-equivalent s (meq ) per 10 0 gm dr y soil . On e me q i s
one milligra m o f hydroge n o r th e portio n o f an y io n whic h wil l combin e wit h o r displac e
1 milligram of hydrogen.
The relative exchange capacit y of some o f the clay mineral s is given in Table 2.4 .
If on e element , suc h a s H, Ca, or Na prevail s ove r the othe r in the adsorptio n comple x o f a
clay, th e cla y i s sometime s give n th e name o f thi s element , fo r exampl e H-cla y o r Ca-clay . Th e
thickness and the physical properties o f the adsorbed fil m surroundin g a given particle, depen d to a
large extent on the character of the adsorption complex. These films ar e relatively thick in the case
of strongl y water-adsorbent cations suc h as Li
+
and Na
+
cations but very thi n for H
+
. The films of
other cations have intermediate values . Soils wit h adsorbed Li
+
and Na
+
cations are relatively more
plastic at low water contents and possess smalle r shea r strengt h because th e particles ar e separate d
by a thicker viscou s film. Th e cation s i n Tabl e 2. 5 ar e arrange d i n th e orde r o f decreasin g shea r
strength of clay.
Sodium clays i n nature are a product either o f the deposition o f clays i n sea water or of their
saturation by saltwate r flooding or capillary action . Calcium clay s are formed essentiall y b y fres h
water sediments . Hydroge n clay s ar e a resul t o f prolonge d leachin g o f a cla y b y pur e o r acidi c
water, with the resulting removal o f all other exchangeable bases .
2.9 SOI L M ASS STRU CTU RE
The orientatio n of particles i n a mass depends o n the siz e and shape o f the grains as wel l a s upon
the mineral s o f whic h th e grain s ar e formed . Th e structur e o f soil s tha t i s forme d b y natura l
deposition ca n b e altere d b y externa l forces . Figur e 2. 8 give s th e variou s type s o f structure s of
soil. Fig. 2.8(a ) i s a single grained structure whic h is formed by the settlemen t o f coarse graine d
soils i n suspensio n i n water . Fig. 2.8(b ) i s a flocculent structure forme d b y th e depositio n o f th e
fine soi l fractio n i n water . Fig . 2.8(c ) i s a honeycomb structure whic h i s forme d b y th e
disintegration o f a flocculen t structur e unde r a superimpose d load . Th e particle s oriente d i n a
flocculent structur e wil l hav e edge-to-fac e contac t a s show n i n Fig . 2.8(d ) wherea s i n a
Chapt er 2
(a) Singl e grai n st r uct ur e (b) Flocculent st ruct ure (c) Honeycomb structure
(d) Flocculated typ e structure
(edge t o face contact)
(e) Dispersed structur e
(face t o face contact)
(f) Undist urbe d salt wate r deposit (g ) Undisturbe d fresh wat e r deposit
Figure 2. 8 Schemat i c d i a g r a m s o f var i ou s type s o f st r uct ur e s ( La mbe , 1 958a)
honeycomb structure , the particle s wil l have face-to-fac e contac t a s shown i n Fig. 2.8(e). Natural
clay sediment s wi l l hav e mor e o r les s flocculate d particl e orientations . Marine clay s generall y
have a mor e ope n structur e t ha n fresh wate r clays. Figs. 2.8(f ) an d (g ) sho w th e schemati c views
of sal t wate r an d fres h wate r deposits .
CH APTER 3
SOIL PH ASE RELATIONSH IPS, INDEX
PROPERTIES AND CLASSIFICATION
3 .1 SOI L PH AS E REL ATI ONSH I P S
Soil mas s i s generall y a thre e phas e system . I t consist s o f soli d particles , liqui d and gas . Fo r al l
practical purposes, th e liquid may be considered t o be water (although in some cases , the water may
contain som e dissolve d salts ) an d th e ga s a s air . The phas e syste m ma y b e expresse d i n S I units
either i n term s o f mass-volum e o r weight-volum e relationships . Th e inte r relationship s o f th e
different phase s ar e important since they help to define the condition or the physical make-up of the
soil.
M ass- V olume Relationshi p
In S I units , the mas s M, i s normall y expresse d i n kg an d th e densit y p i n kg/m
3
. Sometimes , th e
mass and densities ar e also expressed i n g and g/cm
3
or Mg and Mg/m
3
respectively. The density of
water p
o
a t 4 °C is exactly 1.0 0 g/cm
3
(= 100 0 kg/m
3
= 1 Mg/m
3
). Since the variation i n density is
relatively smal l ove r th e range o f temperature s encountere d i n ordinar y engineering practice , th e
density o f wate r p
w
a t othe r temperature s ma y b e take n th e sam e a s tha t a t 4 °C . The volum e i s
expressed eithe r i n cm
3
or m
3
.
Weight- V olume Relationshi p
Unit weigh t o r weigh t pe r uni t volum e i s stil l th e commo n measuremen t i n geotechnica l
engineering practice . The density p, may be converted t o unit weight , 7by using the relationship
Y=pg (3.la )
The 'standard ' value of g i s 9.807 m/s
2
(= 9.81 m/s
2
for al l practical purposes) .
19
20
Chapt er 3
Conversion o f Densit y o f Water p
w
t o Uni t Weight
From Eq . (3.la)
•W~^V V °
Substituting p
w
= 1000 kg/m
3
and g = 9.81 m/s
2
, we have
r
= 1000^9.81^1=9810^
m
3
\ s
2
/ m
3
s
2
1 k g -m
Since I N (newton) = — , w e have,
rrr nr
cm -
I S 1
or 7 , = l x —s — x 9.81 = 9.81 kN/m
3
In general , th e uni t weight of a soil mass ma y be obtained fro m th e equation
y=9. 81pkN/ m
3
(3. If )
where i n Eq. (3. If), p i s in g/cm
3
. For example, if a soil mass has a dry density, p
d
= 1.7 g/cm
3
, the
dry uni t weight of the soi l i s
7^=9. 81 x 1. 7 = 16.6 8 kN/m
3
(3.1g )
3 .2 M ASS- V OL U M E REL ATI ONSH I P S
The phase-relationships in terms of mass-volume and weight-volume for a soil mass are shown by a block
diagram in Fig. 3.1. A block of unit sectional area is considered. The volumes of the different constituents
are shown on the right side and the corresponding mass/weights on the right and left sides of the block. The
mass/weight of air may be assumed as zero.
V olumetric Ratio s
There are three volumetri c ratios that are very useful i n geotechnical engineerin g an d these ca n be
determined directl y from th e phase diagram, Fig. 3.1.
Weight V olume
W
Air
Water
Solids
Mass
M
u
M
Figure 3. 1 B loc k diag r am —t hr e e phas e s o f a s oi l elemen t
Soil Phas e Relat ion s hips , Inde x Propert ie s an d Soi l Cl as s ificat io n 2 1
(3.2)
1. Th e void ratio, e, is defined as
«=^
S
where, V
v
= volume of voids, an d V
s
= volume of the solids .
The void ratio e is always expressed a s a decimal.
2. Th e porosity n is defined as
V
v
v _ _ - I (\C\G1 / O O \
n — x luu /o {->•->)
where, V -tota l volume of the soil sample .
The porosity n is always expressed a s a percentage.
3. Th e degree of saturation S i s defined a s
5 =
^L
X
100%
(34 )
v
where, V
w
= volume of water
It is always expresse d a s a percentage. When S = 0%, the soi l i s completely dry , and when
S = 100%, th e soi l i s full y saturated .
M ass- V olume Relationships
The othe r aspect s o f th e phas e diagra m connecte d wit h mas s o r weigh t ca n b e explaine d wit h
reference t o Fig. 3.1 .
Water Content , w
The wate r content, w, of a soil mas s i s defined a s the ratio of the mass of water, M
w
, i n the voids t o
the mass o f solids, M
s
, a s
M
The water content, which is usually expressed a s a percentage, ca n range from zer o (dr y soil) to
several hundred percent. The natural water content for most soils is well under 100%, but for the soils of
volcanic origin (for example bentonite) it can range up to 500%o r more.
Density
Another ver y usefu l concep t i n geotechnical engineerin g i s density (or , uni t weight ) whic h i s
expressed a s mas s pe r uni t volume . Ther e ar e severa l commonl y use d densities . Thes e ma y b e
defined a s the total (o r bulk), or moist density, p
r
; the dry density, p
d
; th e saturated density , p
sat
; the
density of the particles, soli d density, p
s
; an d density of water p
w
. Each of these densities i s defined
as follows with respect t o Fig. 3.1.
M
Total density, p
t
= — (3.6 )
22 Chapt e r 3
s
Dry density , p
d
= -y- (3.7 )
M
Saturated density , /?
sat
= — (3.8 )
f o r S= 100 %
M
?
Density of solids, p
s
= — -(3.9 )
M
w
Density of water , P
w
=
~77
L
(3.10 )
w
Specific G ravit y
The specifi c gravit y of a substance i s defined a s the rati o of it s mas s i n ai r t o the mas s o f an equal
volume of water at reference temperature , 4 °C. The specific gravity of a mass of soil (includin g air,
water and solids ) i s termed a s bul k specific gravit y G
m
. It is expressed a s
r -?< -
M

The specifi c gravit y of solids, G
s
, (excluding air and water ) i s expressed b y
_ P, _
M
,
(3J 2)
I nterrelationships o f Differen t Parameter s
We can establish relationships between the differen t parameter s define d by equations from (3.2) through
(3.12). In order to develop the relationships, the block diagram Fig. 3.2 is made use of. Since the sectiona l
area perpendicular t o the plane of the paper i s assumed as unity, the heights of the blocks wil l represent
the volumes . The volum e of solids ma y be represente d a s V
s
= 1 . When th e soi l i s full y saturated , the
voids are completely filled with water.
Relationship B etwee n e and n (Fig . 3 .2 )
1-n
l + e
(3.13)
Relationship B etwee n e, G
s
and S
Case 1 : When partiall y saturated ( S < 100%)
p
Soil Phas e Relat ion s hips , In de x Propert ie s an d Soi l Cl as s ificat io n 23
wG wG
Therefore, 5 = - o r e = -
(3.14a)
Case 2: When saturate d ( S = 100%)
From Eq. (3.14a), we have (for 5 =1)
e = wG. (3.14b)
Relationships B etwee n Densit y p and Other Parameter s
The densit y of soi l ca n be expressed i n terms of other parameters for case s o f soi l (1) partially
saturated ( 5 < 100%) ; (2 ) ful l y saturate d ( S = 100%); (3) Fully dry ( S = 0); an d (4 ) submerged.
Case 1 : For S < 100%
= Pt
~
V l + el + e
From Eq. (3. 1 4a) w = eS/G
s
; substitutin g for w in Eq. (3.15) , we have
p
'
=
1 «
Case 2: For S= 100 %
From Eq. (3.16)
Case 3 : For S = 0%
FromEq. (3.16)
l + e
(3.15)
(3.16)
(3.17)
(3.18)
Weight
W
Air
Water
V olume
Solids
I
V
V=e
-V= l+e
Mass
M
Figure 3. 2 B loc k diag r am —t hr e e phas e s of a s oi l elem en t
24 Chapt e r 3
Case 4: When the soi l is submerge d
If th e soi l i s submerged , th e densit y of th e submerge d soi l p
b
, i s equa l t o th e densit y o f th e
saturated soi l reduce d b y the densit y of water, that is
p ( G + e) p
EsJ
--
£s
Relative Densit y
The loosenes s o r densenes s o f sand y soil s ca n b e expresse d numericall y b y relative densit y D
r
,
defined b y th e equation
D
r=
e
e
maX
~l *
10Q
(3.20 )
max mi n
in whic h
e
max
= voi d rati o of sand i n its looses t stat e having a dry densit y of p
dm
e
mm
=V O
^
ra
ti°
m
i
ts
densest stat e having a dry density of p
dM
e = voi d rati o unde r in-situ condition having a dry densit y of p
d
From Eq. (3.18) , a general equation for e may be written as
Pd
Now substitutin g the corresponding dr y densities for e
max
, e
m
-
m
and e in Eq. (3.20) an d simplifying,
we have
n
_ PdM
v
Pd ~Pdmi
m
U— A A 1 \J\J/" 5 O 1 \
r
o o - o U- ^
1
)
rd VdM ^dm
The loosest state for a granular material can usually be created b y allowing the dry material t o
fall int o a container from a funnel hel d in such a way that the free fal l i s about one centimeter. Th e
densest stat e can be established b y a combination o f stati c pressure an d vibration of soi l packed i n
a container .
ASTM Tes t Designatio n D-2049 (1991) provides a procedure fo r determining th e minimum
and maximu m dr y uni t weight s (o r densities ) o f granula r soils . Thi s procedur e ca n b e use d fo r
determining D
r
i n Eq. (3.21) .
3 .3 WEI G H T- V OL U M E REL ATI ONSH I P S
The weight-volume relationships can be established fro m th e earlier equation s by substituting yfor
p and W for M. The variou s equations are tabulated below.
W
1. Wate r conten t w = - j ^
L
xl OO (3.5
a
)
s
W
2. Tota l uni t weight ^
=
17 (3.6a )
W
s
3. Dr y uni t weight y
d
=—j- (3.7a )
Soil Phas e Relat ion s hips , Inde x Propert ie s an d Soi l Cl as s ificat io n 2 5
W
4. Saturate d uni t weight y
sal
= — (3.8a )
W
s
5. Uni t weight of solids y
f
s
=
~y~ (3.9a )
w
6. Uni t weight of water YW
=
~V~~ (3.10a )
w
W
1. Mas s specifi c gravity G = -(3. 1 la)
'w
W
si_ S
8. Specifi c gravit y of solids
s
~Vv (3.12a )
s'w
G Y ( 1 + w)
9. Tota l uni t weight for 5 < 100 y =-£^z -(3.15a )
or
1 + e
Y ( G +e)
10. Saturate d uni t weight Y^=— — - -(3.17a )
1 + e
Y G
11. Dr y uni t weight y
d
= -
!K
— *- (3 . 1 8a)
1 + e
Y ( G -1 )
12. Submerge d uni t weight Yh=— — - -(3.19a )
l + e
n
_ Y dM „Yd ~ Y dm
13. Relativ e density
r
~ T~ v T ^ (3.21a )
'd idM f dm
3.4 COM M ENT S ON SOI L PH AS E REL ATI ONSH I P S
The voi d ratio s o f natura l sand deposit s depen d upo n the shap e o f th e grains , th e uniformit y of
grain size, and the conditions of sedimentation. The void ratios of clay soil s range from less than
unit y t o 5 or more. Th e soil s wit h higher void ratios have a loose structur e and generall y belon g
to the montmorillonit e group. The specifi c gravity of solid particles of most soil s varies from 2.5
to 2.9. For most of the calculations, G ca n be assumed as 2.65 for cohesionless soil s and 2.70 fo r
5
clay soils . The dr y uni t weight s ( y
d
) o f granula r soil s rang e fro m 1 4 to 1 8 kN/m
3
, whereas , th e
saturated uni t weights o f fin e graine d soil s ca n range fro m 12. 5 t o 22. 7 kN/m
3
. Tabl e 3. 1 give s
typical values of porosity, voi d ratio, water content (when saturated) and unit weights of various
types of soils.
26 Chapt er 3
Table 3 . 1 Poros it y , voi d rat io , wat er con t en t , an d unit weig ht s o f t ypica l s oil s i n
n at ural s t at e
Soil
no.
1
1
2
3
4
5
6
7
8
9
10
Descr i pt i on o f soi l
2
Uni form sand , loos e
Uniform sand , loos e
Mixed-grained sand , loos e
Mixed-grained sand , dens e
Glacial t i l l , mixe d grained
Soft glacia l cl ay
Soft gl acia l clay
Soft slightl y organic cla y
Soft hi ghl y organic clay
Soft bentonit e
Porosi t y
n
%
3
46
34
40
30
20
55
37
66
75
84
V oi d
r at i o
e
4
0.85
0.51
0.67
0.43
0.25
1.20
0.60
1.90
3.00
5.20
Wat er
content
w%
5
32
19
25
16
9
45
22
70
110
194
Un i t wei ght
k N/ m
3
r
d
6
14.0
17.0
15.6
18.2
20.8
11.9
16.7
9.1
6.8
4.2
' sat
7
18.5
20.5
19.5
21.2
22.7
17.3
20.3
15.5
14.0
12.4
Example 3 . 1
A sampl e o f we t silt y cla y soi l ha s a mas s o f 12 6 kg . Th e followin g dat a wer e obtaine d fro m
laboratory test s o n the sample: Wet density, p
t
= 2.1 g/cm
3
, G = 2.7, wate r content , w - 15%.
Determine (i ) dr y density , p
d
, (ii ) porosity, (iii ) void ratio, an d (iv ) degree of saturation .
Solution
Mass of sampl e M = 126 kg .
V olume V =
126
= 0.06 m
2.1 x l O
3
Now, M
s
+ M
w
= M, o r M
y
+ wM
y
= M
?
(l + w) = M
Therefore, M. = -^—= -— = 109.57 kg ; M ,= M
^ l + w 1.1 5
H
s
= 16.43 k g
V olume Mass
Air
Water
Sol i ds-
M,.
M,
M
Figure Ex . 3. 1
Soil Phas e Relat ion s hips , Inde x Propert ie s an d Soi l Clas s ificat io n 2 7
Now, V = = = 0.01643 m
3
;
w
p
w
100 0
=
0.04058
G
s
p
w
2.7x100 0
= V-V= 0.06000 -0.04058 = 0.01942 m
3
.
(i) Dr y density, / ? . = —* - =
= 182
6.2 kg/m
3
j ^ ^ y 0.0 6
(ii) Porosity , n^xl OO~
f t 01942 x l 0
° = 32.37 %
V 0.0 6
(iii) V oi d ratio, e = ^-=
Q
-
01942
=0.4786
V ; 0.0405 8
(i v) Degre e o f saturation, S = -*- x 1 00 = °'
01 43
x 1 00 = 84.6%
V. 0.0194 2
Example 3 . 2
Earth is required t o be excavated from borro w pit s for building an embankment. The wet unit weight
of undisturbed soil is 1 8 kN/m
3
and its water content is 8%. In order t o build a 4 m high embankment
with top width 2 m and side slope s 1: 1 , estimate th e quantity of earth required t o be excavated pe r
meter lengt h of embankment . Th e dr y uni t weight required i n the embankment i s 1 5 kN/m
3
wit h a
moisture conten t o f 10%. Assume th e specifi c gravit y of solid s a s 2.67 . Als o determin e th e voi d
ratios an d the degree o f saturation of the soi l i n both the undisturbed and remolded states .
Solution
The dr y uni t weight of soi l i n the borrow pi t is
7 , = -£- = — = 16.7 kN/m
3
d
l + w 1.0 8
V olume of embankment per meter lengt h V
e
2
The dr y uni t weight of soi l i n the embankment i s 1 5 kN/m
3
2m — |
Figure Ex . 3. 2
28 Chapt e r 3
V olume of eart h required t o be excavated V pe r mete r
V = 24 x— = 21.55m
3
16.7
Undisturbed state
V =-^- = — x— = 0.64 m
3
; V =1-0.6 4 = 0.36 m
3
5
Gj
w
2.6 7 9.8 1
n ^^ >
e = —-= 0.56, W =18.0-16. 7 = 1.3 kN
0.64
V
Degree of saturation , S = — —x 100 , wher e
V
= 0.133 m
3
9.8i
Now, £ = -9^x100 = 36.9%
0.36
Remolded state
V =-^ - = -- -= 0.57 m
3
s
Gy
w
2.67x9.8 1
y
v
= 1-0.57 = 0.43 m
3
0.43
e = = 0.75; 7 , = y
d
( 1 + w) = 15 x 1.1 = 16. 5 kN/m
3
0.57
Therefore, W =16.5-15. 0 = 1. 5 k N
V
w
= — = 0.153 m
3
w
9.8 1
0.43
Example 3 . 3
The moistur e conten t o f a n undisturbe d sampl e o f cla y belongin g t o a volcani c regio n i s 265%
under 100 %saturation. The specifi c gravit y of th e solid s i s 2.5. The dr y uni t weight i s 2 1 Ib/ft
3
.
Determine (i ) the saturate d uni t weight , (ii) the submerge d uni t weight, and (iii ) void ratio .
Solution
(i) Saturate d uni t weight, y
sat
= y
W=W + W = w W + W = W ( l + w)
w s a s s ^ '
W W
From Fig . Ex. 3.3, Y
t
= — = —=W. Henc e
Soil Phas e Relat ion s hips , Inde x Propert ie s an d Soi l Clas s ificat io n 29
V = l
Water
W=Y,
Figure Ex. 3. 3
Y
t
= 21(1 + 2.65) = 21 x 3.65 = 76.65 lb/ft
3
(ii) Submerged uni t weight, y
b
Y
b
= ^sat - Y
w
= 76.65 - 62. 4 = 14.25 lb/ft
3
(iii) V oid ratio, e
V = -^- = = 0.135 f t
3
5
G
t
r
w
2.5x62. 4
Since 5 = 100 %
v =v
=
W X s
=
2.65x— = 0.89 ft
3
Y
• V
K 0.8 9
62.4
V 0.13 5
= 6.59
Example 3. 4
A sample of saturated clay from a consolidometer tes t has a total weight of 3.36 I b and a dry weight
of 2.3 2 Ib : th e specifi c gravit y of th e soli d particle s i s 2.7. Fo r thi s sample , determin e th e wate r
content, void ratio, porosity and total unit weight.
Solution
W 336-23 2
w
= —a-
x
100%= = 44.9%= 45%
W. 2.3 2
0.45 x 2.7
e —
n =
1
1.215
= 1.21 5
= 0.548 o r 54.8 %
l + e 1 + 1.215
Y
w
( G
s
+e) 62.4(2. 7 + 1.215)
l + e 1 + 1.215
= 110. 3 lb/ft
3
30 Chapt e r 3
Example 3 . 5
A sampl e of silt y clay has a volume of 14.88cm
3
, a total mass o f 28.81 g, a dry mas s of 24.83 g, and
a specifi c gravit y of solids 2.7. Determine th e voi d rati o and the degre e o f saturation.
Solution
V oid rati o
M
s
24.8 3 „ „ ,
y _ -5__ -
= 92 cm
3
J
G
s
p
w
2.7(1 )
V = V- V = 14.88-9.2 = 5.68 cm
3
V
5
9. 2
Degree of saturation
M
w
28.81-24.8 3
w = — — = -= 0.16
M 24.8 3
0.618
= 0- 70o r 70%
Example 3 . 6
A soi l sampl e i n its natural state has a weight of 5.05 Ib and a volume of 0.041 ft
3
. I n an oven-drie d
state, th e dr y weigh t of the sampl e i s 4.49 Ib. The specifi c gravit y of the solid s i s 2.68. Determin e
the tota l uni t weight , wate r content, void ratio, porosity, an d degre e o f saturation.
Solution
V 0.04 1
5 05 - 4 49 3
-
U3
**
y
W 4.4 9
or 12.5 %
V W 44 9
= ^,V= -"—==0.026 8 f t
3
V
s
G 2.6 8 x 62.4
V =V-V= 0.041-0.0268 = 0.0142 f t
3
0.0268
r\£^'~)
n=— = —
:
--0.3464 o r 34.64%
\ +e 1 + 0.53
0125X168
0.53
Soil Phas e Rel at ion s hips , In de x Propert ie s an d Soi l Cl as s ificat io n 3 1
Example 3 . 7
A soi l sampl e has a total unit weight of 16.97 kN/m
3
and a void rati o of 0.84. The specifi c gravity
of solid s i s 2.70. Determine the moistur e content, dr y uni t weigh t and degre e o f saturatio n of th e
sample.
Solution
Degree o f saturation [from Eq . (3.16a) ]
= o r 1= =
' l + e1 + 0.84
Dry uni t weight (Eq. 3.18a)
d
l + e1 + 0.84
Water content (Eq. 3.14a1
Se 0.58x0.8 4
n i o
w
- — = -= 0.18 or 18 %
G 2. 7
Example 3 . 8
A soil sample in its natural state has, when fully saturated , a water content of 32.5%. Determine th e
void ratio, dry and total unit weights. Calculate the total weight of water required t o saturat e a soi l
mass of volume 10 m
3
. Assume G^ = 2.69.
Solution
V oid rati o (Eq. 3.14a)
=
^
=
32. 5 x 2.69
S ( l ) xl O O
Total uni t weight (Eq. 3.15a)
=
. )
=
2*9 (9-81)0 + 0323)
=
,
' l + e1 + 0.874
Dry uni t weight (Eq. 3.18a)
L &_ _ _ 2.69x9.8 1
= 14Q8kN/ m3
d
l + e1 + 0.874
FromEq. (3.6a), W=y
t
V= 18.6 6 x 10 = 186. 6 kN
From Eq . (3.7a) , W
s
= y
d
V= 14.0 8 x 1 0 = 140. 8 kN
Weight of water =W-W
S
= 186. 6 - 140. 8 = 45.8 kN
3.5 I NDE X PROPERTI E S OF SOI L S
The variou s properties o f soil s which would be considered a s index properties are:
1 . Th e siz e and shap e o f particles .
2. Th e relative density or consistency of soil.
32 Chapt e r 3
The inde x properties o f soil s ca n be studie d in a general wa y under two classes. The y are :
1. Soi l grai n properties .
2. Soi l aggregat e properties .
The principa l soi l grai n propertie s ar e th e siz e an d shap e o f grain s an d th e mineralogica l
character o f th e fine r fraction s (applied t o clay soils) . Th e mos t significan t aggregate propert y o f
cohesionless soil s i s the relativ e density , wherea s tha t o f cohesiv e soil s i s th e consistency . Wate r
content can als o be studied as an aggregat e propert y a s applied t o cohesive soils . The strengt h and
compressibility characteristic s o f cohesiv e soil s ar e function s o f wate r content . A s suc h wate r
content i s a n importan t facto r i n understandin g th e aggregat e behavio r o f cohesiv e soils . B y
contrast, water content does not alter the properties of a cohesionless soi l significantl y except when
the mas s i s submerged, i n which case onl y its uni t weigh t is reduced.
3 .6 TH E SH APE AND SI Z E OF PARTI CL ES
The shapes o f particles a s conceived b y visual inspection give only a qualitative idea of the behavior
of a soi l mas s compose d o f suc h particles. Sinc e particles fine r tha n 0.075 mm diamete r canno t b e
seen b y th e nake d eye , on e ca n visualiz e the natur e of th e coars e graine d particle s only . Coarse r
fractions compose d o f angula r grain s ar e capabl e o f supportin g heavie r stati c load s an d ca n b e
compacted t o a dens e mas s b y vibration . Th e influenc e o f th e shap e o f th e particle s o n th e
compressibility characteristic s of soils are:
1. Reductio n i n the volume of mass upo n the applicatio n o f pressure .
2. A smal l mixtur e of mica t o sand wil l resul t in a large increas e i n it s compressibility .
The classificatio n accordin g t o siz e divide s th e soil s broadl y int o two distinctiv e groups , namely ,
coarse graine d an d fin e grained . Since th e properties o f coarse graine d soil s are , t o a considerabl e
extent, based on grain size distribution, classification of coarse graine d soil s accordin g t o size would
therefore be helpful . Fin e graine d soil s are so much affected by structure , shape o f grain, geologica l
origin, an d othe r factor s tha t thei r grai n siz e distributio n alon e tell s littl e abou t thei r physica l
properties. However , on e can asses s th e nature of a mixed soi l o n the basis o f the percentage o f fin e
grained soi l present i n it. It is, therefore, essential t o classify the soi l according t o grain size .
The classificatio n o f soil s a s gravel , sand , sil t an d cla y a s pe r th e differen t system s o f
classification i s give n i n Table 2.2 . Soi l particle s whic h ar e coarse r tha n 0.075 mm ar e generall y
termed a s coarse grained an d the finer ones a s silt, clay and peat (organi c soil ) are considered fine
grained. Fro m a n engineerin g poin t o f view , thes e tw o type s o f soil s hav e distinctiv e
characteristics. I n coars e graine d soils , gravitationa l force s determin e th e engineerin g
characteristics. Interparticl e forces ar e predominan t i n fin e graine d soils . Th e dependenc e o f th e
behavior o f a soi l mas s o n th e siz e o f particle s has le d investigator s t o classif y soil s accordin g t o
their size .
The physica l separation o f a sampl e of soi l b y an y method int o two o r mor e fractions , eac h
containing only particles of certain sizes, is termed fractionation. Th e determination of the mass of
material i n fraction s containin g onl y particle s o f certai n size s i s terme d Mechanica l Analysis .
Mechanical analysi s is one o f the oldes t an d mos t common form s o f soi l analysis . It provide s th e
basic informatio n for revealing the uniformit y o r gradation o f the material s withi n established siz e
ranges an d for textural classifications. The results of a mechanical analysi s are not equally valuable
in differen t branche s o f engineering . The siz e o f th e soi l grain s i s o f importanc e i n suc h case s a s
construction of earth dams o r railroad and highway embankments, where earth i s used as a material
that shoul d satisf y definit e specifications . I n foundation s o f structures , dat a fro m mechanica l
analyses are generally illustrative ; other properties suc h as compressibility an d shearing resistanc e
are o f mor e importance . The norma l metho d adopte d fo r separatio n o f particle s i n a fin e graine d
Soil Phas e Rel at ion s hips , In dex Propert ie s an d Soi l Cl as s ificat io n 3 3
soil mas s i s the hydrometer analysi s and for the coarse graine d soil s the sieve analysis . These two
methods ar e described i n the following sections .
3 .7 SI EV E ANAL Y SI S
Sieve analysi s is carried ou t by using a set of standar d sieves . Sieve s ar e made by weaving two
sets o f wire s a t righ t angle s t o on e another . Th e squar e hole s thu s forme d betwee n th e wire s
provide th e limi t which determine s th e siz e o f the particles retaine d o n a particular sieve . Th e
sieve size s ar e give n i n terms o f the numbe r of openings per inch. The number of openings pe r
inch varie s accordin g t o different standards . Thus, a n ASTM 6 0 sieve has 60 openings pe r inch
widt h wit h each openin g of 0.250 mm. Table 3. 2 gives a set of ASTM Standar d Sieve s (same as
US standard sieves) .
The usua l procedure i s t o use a set of sieves whic h will yield equa l grai n siz e interval s on a
logarithmic scale. A good spacing of soil particle diameters on the grain size distribution curve will
be obtained if a nest of sieves i s used in which each sieve has an opening approximately one-hal f of
the coarser sieve above i t in the nest. If the soil contains gravel , the coarsest siev e tha t can be used
to separate out gravel from san d is the No. 4 Sieve (4.75 mm opening). To separate ou t the silt-clay
fractions fro m th e sand fractions, No. 200 sieve may be used. The intermediate sieves between the
coarsest an d th e fines t ma y be selecte d o n the basi s o f the principl e explained earlier . Th e nes t of
sieves consist s o f No s 4 (4.7 5 mm) , 8 (2.36 mm) , 1 6 (1.18 mm) 3 0 (600 jun) , 5 0 (300 pun) , 10 0
(150 jim), and 200 (75 |im).
The siev e analysi s i s carrie d ou t b y sievin g a known dry mas s o f sampl e throug h the nes t of
sieves placed one below the other so that the openings decrease i n size from th e top sieve downwards,
with a pan at the bottom of the stack as shown in Fig. 3.3. The whole nest of sieves is given a horizontal
shaking for about 10 minutes (if required, more) til l the mass of soil remaining on each sieve reache s
a constan t valu e (the shaking ca n be don e by han d or using a mechanical shaker , i f available) . Th e
amount of shaking required depends o n the shape and number of particles. I f a sizable portion o f soi l
is retained on the No. 200 sieve, it should be washed. This is done by placing the sieve with a pan at the
bottom and pouring clean water on the screen. A spoon may be used to stir the slurry. The soil which
is washed through is recovered, drie d and weighed. The mass of soil recovered i s subtracted fro m th e
mass retained on the No. 200 sieve before washin g and added t o the soil tha t has passed through the
No. 200 sieve by dry sieving. The mass of soil required for sieve analysis is of oven-dried soi l with all
Table 3. 2 U S St an dar d s ieves
Desi gnat i on
2 i n
l
l
/2 i n
%i n
3/8 i n
4
8
10
14
16
18
30
Openi ng
mm
50.80
38.10
19.00
9.51
4.75
2.36
2.00
1.40
1.18
1.00
0.60
Desi gnat i on
35
40
50
60
70
80
100
120
170
200
270
Openi ng
mm
0.50
0.425
0.355
0.250
0.212
0.180
0.150
0.125
0.090
0.075
0.053
34 Chapt e r 3
Table 3 . 3 Sam pl e s iz e fo r s iev e an al ys i s
M ax par t icl e s iz e Min . s am pl e s iz e i n g
3 i n 600 0
2 i n 400 0
1 i n 200 0
1/2 i n 100 0
No. 4 20 0
No. 1 0 10 0
the particles separate d ou t by some means . The minimum size of sample t o be used depends upo n the
maximum particl e siz e a s give n i n Tabl e 3. 3 (U S Army Corp s o f Engineers) . B y determinin g th e
mass o f soi l sampl e lef t o n each sieve , the following calculations can be made .
mass o f soi l retaine d
1. Percentag e retaine d o n an y siev e = ; x l O O
total soi l mas s
Figure 3 . 3 (a ) Sieve s hak e r an d (b ) a s e t o f s ieve s fo r a t es t i n t h e l aborat or y
(Cour t es y: Soil t es t , USA)
Soil Phas e Relat ion s hips , In de x Propert ie s an d Soi l Cl as s ificat io n 35
Gravel
100
90
80
70
60
c 5 0
40
30
20
10
Sand
Coarse to medium Fine
Silt
10 8 6 4 2 1 . 8 . 6 . 4 . 2 0.1.0 8 .0 6 .0 4
Particle size , mm (log scale)
Figure 3. 4 Part icle-s iz e dis t ribut ion curve
.02 0.01
2. Cumulativ e percentage
retained o n any sieve
3. Percentag e fine r tha n
any siev e size, P
Sum of percentages retaine d o n
all coarser sieves .
100 per cen t minus cumulative
percentage retained .
The results may be plotted i n the form of a graph on semi-log paper wit h the percentage fine r
on the arithmeti c scale and the particle diameter on the log scal e as shown in Fig. 3.4 .
3.8 TH E H Y DROM ETER METH O D OF ANAL Y SI S
The hydromete r metho d wa s originall y propose d i n 192 6 b y Prof . Bouyouco s o f Michiga n
Agricultural College , an d late r modifie d b y Casagrand e (1931) . Thi s metho d depend s upo n
variations in the densit y of a soi l suspensio n containe d in a 100 0 mL graduate d cylinder . The
density o f th e suspensio n i s measure d wit h a hydromete r a t determine d tim e intervals ; then th e
coarsest diamete r o f particle s i n suspensio n a t a give n time and th e percentag e of particle s fine r
than tha t coarsest (suspended ) diameter ar e computed. Thes e computations ar e based o n Stokes '
formula whic h i s describe d below .
36 Chapt e r 3
Stokes' L a w
Stokes (1856), an English physicist , proposed an equation fo r determining th e terminal velocity o f
a falling spher e i n a liquid. If a single sphere i s allowed t o fal l throug h a liquid of indefinit e extent ,
the termina l velocity, v can b e expressed as ,
v=
r
s
-r
w D2
18// ^
>ZZ;
in which,
distance L
v - termina l velocit y of fall o f a sphere through a liquid = = — J
F 5 M
tlm e f
Y
s
= unit weight of soli d spher e
Y
w
= unit weight of liquid
H = absolute viscosit y of liquid
D = diameter o f sphere .
From Eq . (3.22) , after substituting for v , we have
_ i -"/ - I ^
l t a- i )r
w
V 7
(3
-
23)
in whic h y
s
= G
s
y
w
If L i s i n cm, t i s i n min , y i n g/cm
3
, \ J i i n (g-sec)/cm
2
and D i n mm, the n Eq. (3.23 ) ma y b e
written as
D(mm)
or D=
' ^_i
) 7 w
V 7
= A
V 7
(3
-
24)
where, K = I
30/ /
(3.25 )
by assumin g Y
W
~ lg/cm
3
It ma y b e note d her e tha t the facto r K i s a function o f temperatur e T , specifi c gravit y G
s
of
particles an d viscosit y o f water . Tabl e 3.4 a give s th e value s of K fo r th e variou s value s o f G
s
at
different temperature s T . If i t i s necessary t o calculat e D without the us e o f Tabl e 3.4a we can us e
Eq. (3.24) directly. The variation of n with temperature i s required whic h is given in Table 3.4b .
Assumptions o f Stoke s L a w an d it s V alidity
Stokes' la w assume s spherica l particle s fallin g i n a liqui d o f infinit e extent , an d al l th e particle s
have th e same uni t weight y
s
- Th e particle s reac h constan t termina l velocit y withi n a few second s
after the y are allowe d t o fall .
Since particles ar e not spherical , the concept o f an equivalent diameter ha s been introduced .
A particl e is said to have an equivalent diameter D
g
, if a sphere of diameter D having the same uni t
weight as the particle, has the same velocity of fall a s the particle. For bulky grains D
e
~ D, wherea s
for flak y particle s DI D = 4 or more .
Soil Phas e Relat ion s hips , In dex Propert ie s an d Soi l Clas s ificat io n 37
Table 3.4 a Value s o f /(fo r us e in Eq. (3. 24) fo r s evera l s pecifi c g ravit y o f s olid s
and t em perat ure combination s
G
s
o f Soi l
Temp ° C
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
2.50
0.0151
0.0149
0.0148
0.0145
0.0143
0.0141
0.0140
0.0138
0.0137
0.0135
0.0133
0.0132
0.0130
0.0129
0.0128
Table 3.4b
Temp
4
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
2.55
0.0148
0.0146
0.0144
0.0143
0.0141
0.0139
0.0137
0.0136
0.0134
0.0133
0.0131
0.0130
0.0128
0.0127
0.0126
2.60
0.0146
0.0144
0.0142
0.0140
0.0139
0.0137
0.0135
0.0134
0.0132
0.0131
0.0129
0.0128
0.0126
0.0125
0.0124
2.65
0.0144
0.0142
0.0140
0.0138
0.0137
0.0135
0.0133
0.0132
0.0130
0.0129
0.0127
0.0126
0.0124
0.0123
0.0122
Solids
2.70
0.0141
0.0140
0.0138
0.0136
0.0134
0.0133
0.0131
0.0130
0.0128
0.0127
0.0125
0.0124
0.0123
0.0121
0.0120
Properties o f di st i l l e d wat e r (/ / =
°C Uni t wei gh t o f wat er ,
1.00000
0.99897
0.99880
0.99862
0.99844
0.99823
0.99802
0.99780
0.99757
0.99733
0.99708
0.99682
0.99655
0.99627
0.99598
0.99568
g/cm
3
2.75
0.0139
0.0138
0.0136
0.0134
0.0133
0.0131
0.0129
0.0128
0.0126
0.0125
0.0124
0.0122
0.0121
0.0120
0.0118
2.80
0.0139
0.0136
0.0134
0.0132
0.0131
0.0129
0.0128
0.0126
0.0125
0.0123
0.0122
0.0120
0.0119
0.0118
0.0117
2.85
0.0136
0.0134
0.0132
0.0131
0.0129
0.0127
0.0126
0.0124
0.0123
0.0122
0.0120
0.0119
0.0117
0.0116
0.0115
absolute viscosity )
V iscosity o f water , pois e
0.01567
0.01111
0.0108
0.0105
0.01030
0.01005
0.00981
0.00958
0.00936
0.00914
0.00894
0.00874
0.00855
0.00836
0.00818
0.00801
The effect of influence of one particle over the other is minimized by limiting the mass of soil
for sedimentatio n analysi s to 60 g in a sedimentation jar o f 10
3
cm
3
capacity.
38 Chapt e r 3
H ydrometer Analysi s
Figure 3. 5 show s a streamline d hydromete r o f th e typ e AST M 15 2 H use d fo r hydromete r
analysis. Th e hydromete r possesse s a lon g ste m an d a bulb . Th e hydromete r i s use d fo r th e
determination of unit weight of suspensions at different depth s and particular intervals of time. A
uni t volum e o f soi l suspensio n a t a dept h L an d a t an y tim e / contain s particle s fine r tha n a
particular diameter D. The valu e of thi s diameter i s determined b y applying Stokes' law whereas
the percentage fine r t han this diameter i s determined by the use of the hydrometer. The principl e
of the method i s that the reading of the hydrometer gives the uni t weigh t of the suspension a t the
center o f volume of the hydrometer. The firs t ste p in the presentation of this method i s to calibrat e
the hydrometer .
Let the sedimentation jar contai n a suspension of volume V with total mass o f solids M
s
. Le t
the jar b e kept verticall y on a table after the solids are thoroughly mixed. The initial density p
;
. of the
suspension a t an y depth z from th e surfac e a t time t = 0 may be expresse d a s
M M M M _ _
Pi=
~V
+ l
~~G^
P
»=~V
+ l
~^ (
where p
o
= density of water at 4°C an d p
w
densit y of water at test temperatur e T , and G
s
= specific
gravity of the solids . Fo r al l practica l purpose s p
o
= p
w
= 1 g/cm
3
.
After a lapse o f time t, a unit volume of suspension a t a depth z contains onl y particles fine r
than a particular diameter D, since particles coarser than this diameter hav e fallen a distance greate r
than z as per Stokes'law. Th e coarsest diamete r o f the particle i n a unit volume of the suspension at
depth z and time t is given by Eq. (3.24) wher e z = L . Let M
d
b e the mas s o f al l particles fine r tha n
D in the sample taken for analysis. Th e density of the suspension p, after a n elapse d time t may be
expressed a s
M
D
where -= Mas s o f particles o f diameter smalle r tha n diamete r D in the unit volum e of
suspension a t dept h z at a n elapse d time t.
From Eq . (3.26b ) we may write
"
=
- T )
P
f - (3.26 0
The ASTM 15 2 H type hydrometer, normally used for the analysis, is calibrated to read from
0 t o 60 g of soi l i n a 100 0 m L soil - water mixtur e wit h the limitation tha t the soi l particle s hav e a
specific gravit y G
s
= 2.65. Th e reading i s directly related t o the specifi c gravit y of the suspension .
In Eq. (3.26c ) the mas s o f the solids M
D
i n the suspension varie s fro m 0 to 60 grams. Th e readin g
R on the ste m o f the hydrometer (correcte d fo r meniscus) ma y be expressed a s
(3
.
26d)
where,
G
s
= 2.65, an d V= 100 0 m L
p,= densit y of suspension per uni t volume = specifi c gravity of the suspension .
Soil Phas e Relat ion s hips , Inde x Propert ie s an d Soi l Cl as s ificat io n 39
From Eq. (3.26d), it is clear tha t the ASTM 152 H hydrometer is calibrated i n such a way that
the reading on the ste m wil l be
R = 0 when p
f
= 1 , and R = 60 when p
f
= 1.037 4
The ASTM 15 2 H hydrometer give s the distance of any reading R on the stem t o the center of
volume and is designated a s L as shown in Fig. 3.5 . Th e distance L varies linearl y wit h the reading
R. A n expressio n fo r L ma y b e writte n a s follow s fo r an y readin g R fo r th e AST M 15 2 H
hydrometer (Fig . 3.5) .
£= A+Y (
3
-
27
)
where L
{
= distance fro m readin g R t o the to p o f the bul b
L
2
= lengt h of hydromete r bul b = 1 4 cm fo r ASTM 15 2 H hydromete r
When th e hydromete r i s inserted int o the suspension , the surfac e of th e suspensio n rise s a s
shown i n Fig . 3.6 . Th e distanc e L in Fig . 3. 6 i s th e actua l distanc e throug h whic h a particl e o f
diameter D ha s fallen . Th e point a t leve l Aj a t dept h L occupies th e position A
2
(whic h coincide s
with th e cente r o f volume of th e hydrometer ) i n the figur e afte r th e immersio n o f th e hydrometer
and correspondingl y th e surface of suspension rises fro m B
l
t o B
2
. The dept h L ' is therefore greate r
than L through which the particle of diameter D has fallen. The effective valu e of L can be obtaine d
from th e equation
T
Ra
L Meniscu s
60
X
V
Center of bul b
V
h
/Aj
V
h
/2Aj
Meniscus
L'
Figure 3. 5 AST M 152 H t ype
hydrom et er
Before th e immersion Afte r th e immersio n
of hydrometer o f hydromete r
Figure 3. 6 Im m ers io n correct io n
40
Table 3 . 5 Val ue s of L (effect ive dept h) fo r
part icl es fo r AST M s oi l
Chapt er 3
us e i n St ok es ' form ul a fo r diam et er s o f
hydrom et er 152 H
Or i g i n a l Or i g i n a l
hydr omet er hydr omet e r
r e a di ng Ef f e c t i v e r e a d i n g
(correct ed f o r dept h L (correct e d f o r
meni s cus o n l y ) c m meni s cu s o n l y )
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
L - L '
16.3
16.1
16.0
15.8
15.6
15.5
15.3
15.2
15.0
14.8
14.7
14.5
14.3
14.2
14.0
13.8
13.7
13.5
13.3
13.2
13.0
V, 1
h J \ J
— L i, \ L i~.
2A
j
2
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
y
A
;
Or i g i n a l
hydr omet er
Ef f e c t i ve r e a di n g Ef f e c t i v e
dept h L (correct e d fo r dept h L
cm meni s cu s o n l y) c m
12.9
12.7
12.5
12.4
12.2
12.0
11.9
11.7
11.5
11.4
11.2
11.1
10.9
10.7
10.5
10.4
10.2
10.1
9.9
9.7
9.6
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
9.4
9.2
9.1
8.9
8.8
8.6
8.4
8.3
8.1
7.9
7.8
7.6
7.4
7.3
7.1
7.0
6.8
6.6
6.5
(3.28)
where V
h
= volume of hydrometer (152 H) = 67 cm
3
; A. = cross-sectional are a o f the sedimentation
cylinder = 27. 8 cm
2
for 100 0 mL graduated cylinder .
For a n ASTM 15 2 H hydrometer, the valu e of L for an y reading R (correcte d fo r meniscus)
may b e obtaine d fro m
L = 16.3 -0.1641 R (3.29)
Table 3. 5 give s th e value s o f L fo r variou s hydromete r reading s o f R fo r th e 15 2 H
hydrometer.
Determination o f Percen t Fine r
The ASTM 15 2 H hydrometer is calibrated t o read from 0 to 60 g of soi l i n a 100 0 m L suspensio n
with th e limitation that the soi l has a specific gravit y G = 2.65. Th e reading is, of course, directl y
Soil Phas e Relat ion s hips , In de x Propert ie s an d Soi l Cl as s ificat io n 4 1
related t o the specifi c gravit y o f the suspension . The hydromete r give s readings pertainin g t o the
specific gravit y of the soil-water suspensio n at the center o f the bulb. Any soil particle s large r than
those stil l i n suspensio n i n the zone shown as L (Fig 3.5 ) hav e fallen below th e cente r o f volume,
and thi s constantly decreases th e specifi c gravit y of the suspensio n a t the center o f volume of the
hydrometer. Lesser th e specific gravit y of the suspension, th e deeper th e hydromete r will sink into
the suspension. It must also be remembered here , that the specific gravity of water decreases as the
temperature rises from 4° C. This will also cause the hydrometer to sink deeper into the suspension.
The reading s o f the hydrometer ar e affecte d b y the ris e i n temperatur e durin g the test . Th e
temperature correctio n i s a constant . Th e us e o f a dispersin g agen t als o affect s th e hydromete r
reading. Corrections fo r thi s can be obtained b y using a sedimentation cylinde r of water from th e
same sourc e an d wit h th e sam e quantit y o f dispersin g agen t a s tha t use d i n th e soil-wate r
suspension to obtain a zero correction. This jar of water should be at the same temperature as that of
the soi l wate r suspension.
A reading of less than zero in the standard jar of water is recorded a s a (-) correctio n value ; a
reading betwee n 0 and 6 0 i s recorded a s a (+) value . Al l the reading s ar e lake n t o th e to p of the
meniscus i n both the standar d jar (clea r water ) and soil suspension.
If th e temperatur e durin g the test i s quite high, the densit y of water wil l be equall y les s an d
hydrometer wil l sin k to o deep . On e ca n us e a temperatur e correctio n fo r th e soil-wate r
suspension. Table 3. 6 gives the values of temperature correlation C
r
The zero correction C
o
can be
(±) and the temperature correction als o has (±) sign .
The actual hydrometer readin g R
a
has to be corrected a s follows
1. correctio n fo r meniscus C
m
only for us e i n Eq. (3.24 )
2. zer o correction C
o
and temperature correctio n C
r
for obtaining percent finer .
Reading for use i n Eq. (3.24 )
R = R
a
+C
m
(3.30a )
Reading for obtaining percent fine r
R
c=
R
a-
C
o
+C
T(3.30b )
Percent Fine r
The 15 2 H hydrometer i s calibrated fo r a suspension wit h a specific gravity of solids G
s
= 2.65. I f
the specifi c gravit y of solids used i n the suspension is different fro m 2.65 , th e percent fine r ha s to
be corrected by the factor C expresse d a s
Table 3. 6 T em perat ur e correct io n fact or s C
T
Temp ° C
15
16
17
18
19
20
21
22
C
T
-1.10
-0.90
-0.70
-0.50
-0.30
0.00
+0.20
+0.40
Temp ° C
23
24
25
26
27
28
29
30
CT
+0.70
+ 1.00
+1.30
+ 1.65
+2.00
+2.50
+3.05
+3.80
42 Chapt er 3
1.65G
C = i —
58
2.65(G
?
-1)
(3. 31)
Typical value s of C
?
ar e given in Table 3.7 .
Now th e percent fine r wit h the correction facto r C
s
ma y b e expressed a s
Percent finer , P ' =
M
x l OO
(3.32)
where R
c
= gram s o f soi l i n suspensio n a t som e elapse d tim e t [correcte d hydromete r
reading fro m Eq . (3.30b) ]
M
s
= mas s o f soi l use d i n th e suspensio n i n gms (no t mor e tha n 60 g m fo r 15 2 H
hydrometer)
Eq. (3.32 ) give s th e percentage of particle s fine r tha n a particle diameter D i n the mas s of
soil M
s
use d i n the suspension . If M i s the mas s of soi l particle s passing throug h 75 micron siev e
(greater tha n M) an d M th e tota l mass take n fo r the combined siev e an d hydrometer analysis , the
percent fine r fo r th e entir e sampl e ma y b e expresse d a s
Percent finer(combined) , P = P'% x
M
(3.33)
Now Eq . (3.33 ) wit h Eq. (3.24 ) give s point s fo r plottin g a grai n siz e distributio n curve .
Test procedure
The suggeste d procedur e fo r conductin g the hydromete r tes t i s a s follows :
1. Tak e 6 0 g or les s dr y sampl e fro m th e soi l passin g throug h th e No. 20 0 siev e
2. Mi x thi s sampl e wit h 12 5 mL o f a 4%o f NaPO
3
solutio n i n a smal l evaporatin g dis h
3. Allo w the soi l mixtur e to stand for abou t 1 hour. At the end o f the soakin g perio d transfe r
the mixtur e t o a dispersio n cu p an d ad d distille d wate r unti l th e cu p i s abou t two-third s
ful l . Mi x fo r abou t 2 min .
4. Afte r mixing, transfer all the content s of the dispersion cu p t o the sedimentation cylinder ,
being carefu l no t t o los e an y materia l No w ad d temperature-stabilize d wate r t o fil l th e
cylinder t o th e 100 0 m L mark .
5. Mi x th e suspensio n wel l by placin g th e pal m o f th e han d ove r th e ope n en d an d turning
the cylinder upside down and back fo r a period o f 1 min. Set the cylinder down on a table.
6. Star t th e time r immediatel y afte r settin g th e cylinder . Inser t th e hydromete r int o th e
suspension j us t abou t 2 0 second s befor e th e elapse d tim e o f 2 min . an d tak e th e firs t
reading a t 2 min . Tak e th e temperatur e reading . Remov e th e hydromete r an d th e
thermometer an d plac e bot h o f the m i n th e contro l jar .
7. Th e contro l ja r contain s 100 0 m L o f temperature-stabilize d distille d wate r mixe d wit h
125 mL of the same 4%solutio n of NaPO
3
.
Table 3 . 7 Correct io n fact or s C fo r uni t weig h t o f s olid s
G
s
o f s oi l s olid s
2. 85
2.80
2. 75
2.70
Cor r ect ion f act o r C
0.96
0.97
0.98
0.99
G
s
o f s oi l s olid s
2.65
2.60
2. 55
2. 50
Correct ion fact o r C
1.00
1.01
1.02
1.04
Soil Phas e Relat ion s hips , In de x Propert ie s an d Soi l Cl as s ificat io n 4 3
8. Th e hydromete r reading s ar e take n a t th e to p leve l o f th e meniscu s i n bot h th e
sedimentation and control jars.
9. Step s 6 through 8 are repeated b y taking hydrometer and temperature readings at elapse d
times of 4, 8, 16 , 30, 60 min. and 2, 4, 8 , 16 , 32, 64 and 96 hr .
10. Necessar y computation s ca n b e mad e wit h th e dat a collecte d t o obtai n th e grain -
distribution curve.
3 .9 G RAI N SI Z E DI STRI B U TI ON CU RV E S
A typical set of grain size distribution curves is given in Fig. 3. 7 with the grain size D as the absciss a
on the logarithmic scale and the percent finer P as the ordinate on the arithmetic scale. On the curve C
{
the section AB represent s th e portion obtained by sieve analysis and the section B'C' by hydrometer
analysis. Since the hydrometer analysis gives equivalent diameters whic h are generally less than the
actual sizes, the section B'C' wil l not be a continuation of AB and would occupy a position shown by
the dotted curve. If we assume that the curve BC is the actual curve obtained by sketching it parallel to
B'C', then at any percentage finer , sa y 20 per cent, the diameters D
a
and D
e
represent th e actual and
equivalent diameters respectively. The ratio of D
a
t o D
g
ca n be quite high for flaky grains .
The shape s of the curves indicat e the natur e of the soi l tested. On the basis of the shape s we
can classif y soils as:
1 . Uniforml y graded or poorly graded.
2. Wel l graded .
3. Ga p graded .
Uniformly grade d soil s ar e represente d b y nearl y vertica l line s a s show n b y curv e C
2
i n
Fig. 3.7. Suc h soil s posses s particle s o f almost the same diameter . A well graded soil , represente d
by curv e C
p
possesses a wide range of particle sizes ranging from grave l t o clay size particles. A
gap graded soil , as shown by curve C
3
has some of the sizes of particles missing. On this curve the
soil particles fallin g withi n the range of XY ar e missing.
The grain distribution curves as shown in Fig. 3. 7 can be used to understand certain grain size
characteristics o f soils. Hazen (1893) has shown that the permeability of clean filter sands in a loose
state can be correlated wit h numerical values designated D
10
, the effective grain size. The effective
grain siz e correspond s to 10 per cen t fine r particles . Haze n foun d tha t the size s smalle r tha n the
effective siz e affected the functioning of filters more than did the remaining 90 per cent of the sizes.
To determine whether a material is uniformly graded or well graded, Hazen proposed th e following
equation:
_
D
60
where D
60
is the diameter of the particle at 60 per cent finer on the grain size distribution curve. The
uniformity coefficient, C
u
, is about one if the grain size distribution curve is almost vertical , and the
value increases wit h gradation. For all practical purpose s we can consider the following values for
granular soils .
C
u
> 4 fo r wel l graded gravel
C
u
> 6 fo r wel l graded san d
C < 4 fo r uniforml y graded soi l containing particles o f the same siz e
44 Chapt er 3
Particle diameter, mm
Figure 3 . 7 Grai n s iz e dis t ribut io n curve s
There is another step in the procedure to determine the gradation of particles. Thi s i s based on
the ter m calle d th e coefficient o f curvature whic h i s expressed a s
C =
X D
60
(3.35)
wherein D
30
is the siz e of particl e at 30 percent fine r o n the gradation curve . The soi l i s sai d t o be
well grade d i f C
c
lies between 1 and 3 for gravel s and sands .
Two samples o f soil s ar e sai d t o be similarl y graded i f their grain siz e distributio n curves ar e
almost parallel to each other on a semilogarithmic plot. Whe n the curves are almost parallel t o each
other th e ratio s o f thei r diameter s a t an y percentag e fine r approximatel y remai n constant . Suc h
curves ar e usefu l i n the design o f filter material s around drainag e pipes.
3 .1 0 REL ATI V E DENSI TY OF COH ESI ONL ESS SOI LS
The densit y o f granula r soil s varie s wit h th e shap e an d siz e o f grains , th e gradatio n an d th e
manner i n whic h th e mas s i s compacted . I f al l th e grain s ar e assume d t o be sphere s o f unifor m
size an d packe d a s show n i n Fig . 3.8(a) , th e voi d rati o o f suc h a mas s amount s t o abou t 0.90 .
However, i f the grains are packed a s shown in Fig. 3.8(b) , the voi d ratio of the mass i s about 0.35 .
The soi l corresponding t o the higher voi d ratio i s called loos e and that corresponding t o the lower
void rati o i s calle d dense . I f the soi l grain s ar e no t uniform, the n smalle r grain s fil l i n the spac e
between th e bigge r one s an d th e voi d ratio s o f suc h soil s ar e reduce d t o a s lo w a s 0.2 5 i n th e
densest state . I f th e grain s ar e angular , the y ten d t o for m loose r structure s than rounde d grain s
Soil Phas e Relat ion s hips , In de x Propert ie s an d Soi l Cl as s ificat io n 4 5
(a) Looses t stat e (b ) Denses t state
Figure 3. 8 Pack in g of g rain s o f un ifor m s ize
Table 3. 8 Clas s ificat io n o f s an d y s oil s
Rel at i ve densi t y , D
f
, %
0-15
15-50
50-70
70-85
85-100
Type o f soi l
V ery loos e
Loose
Medium dens e
Dense
V ery dens e
because thei r shar p edges and points hold the grains further apart . If the mass with angular grains
is compacted b y vibration , it forms a dense structure . Static load alon e wil l not alter the density
of grains significantly but i f it is accompanied b y vibration, there wil l be considerabl e chang e in
the density. The water present i n voids may act as a lubricant to a certain exten t for an increase in
the densit y unde r vibration. The chang e i n voi d rati o woul d change th e densit y an d thi s i n turn
changes the strengt h characteristics o f granular soils. V oid ratio or the uni t weight of soi l can be
used t o compare th e strengt h characteristics o f samples o f granular soils o f the same origin. The
term use d t o indicat e th e strengt h characteristic s i n a qualitativ e manne r i s terme d a s relativ e
density which is already expresse d b y Eq. (3.20). On the basis of relative density, we can classify
sandy soil s a s loose, medium or dense a s i n Table 3.8 .
3 .1 1 CONSI STENC Y OF CL AY SOI L
Consistency i s a term used t o indicate the degree o f firmness of cohesive soils . The consistency of
natural cohesive soi l deposit s i s expressed qualitativel y by suc h terms as very soft , soft , st i ff , ver y
stiff and hard . The physica l propertie s of clay s greatl y diffe r at differen t wate r contents . A soi l
which i s ver y sof t a t a higher percentag e o f wate r content becomes ver y har d wit h a decreas e i n
water content . However, i t has bee n foun d tha t at the same wate r content , two sample s o f clay of
different origin s ma y possess different consistency . One clay may be relatively soft whil e the other
may be hard. Further , a decrease i n water content may have littl e effect o n one sampl e o f clay but
may transform the other sampl e from almos t a liquid to a very firm condition. Water content alone,
therefore, i s no t a n adequat e inde x o f consistenc y fo r engineerin g an d man y othe r purposes .
Consistency of a soil can be expressed i n terms of :
1. Atterber g limit s of soil s
2. Unconfme d compressiv e strength s of soils .
46 Chapt e r 3
Atterberg L imit s
Atterberg, a Swedish scientist , considered th e consistency of soil s i n 1911 , and proposed a serie s
of test s fo r definin g the propertie s o f cohesive soils . These test s indicat e the rang e o f the plasti c
state (plasticit y is define d a s the propert y o f cohesive soil s whic h possess th e abi l i t y t o underg o
changes o f shape wit hout rupt ure) and other states. He showe d that if the water content of a thick
suspension of clay i s gradual ly reduced, th e clay water mixtur e undergoes change s fro m a liquid
state throug h a plasti c stat e and f i nal l y int o a soli d state . The differen t states throug h which the
soil sampl e passe s wit h the decreas e i n the moistur e conten t ar e depicted i n Fig. 3.9 . Th e wate r
contents corresponding t o the t ransit ion from one stat e to another ar e termed a s Atterberg L imits
and the tests required to determine the limits are the Atterberg L imit Tests. The testing procedure s
of Atterber g wer e subsequent l y improved b y A. Casagrande (1932) .
The transition stat e from the liqui d state t o a plastic state is called the liquid limit, w
r
A t this
stage al l soil s posses s a certai n smal l shea r strength . Thi s arbitraril y chose n shea r strengt h i s
probably th e smalles t value that i s feasible to measur e i n a standardized procedure . Th e transition
from th e plastic state to the semisolid state is termed the plastic limit, w . At this state the soil rolle d
into thread s o f abou t 3 mm diamete r just crumbles . Further decreas e o f th e wate r content s o f the
same wil l lead finall y t o the point where the sample ca n decrease in volume no further. At this point
the sampl e begin s t o dr y a t the surface , saturatio n i s n o longe r complete , an d furthe r decrease i n
water i n the voids occurs without change i n the void volume. The color of the soi l begins t o change
from dar k t o light. This wate r content is called th e shrinkage limit, w
s
. The limit s expressed abov e
are al l expresse d b y thei r percentages o f wate r contents . Th e rang e o f wate r content betwee n th e
liquid an d plasti c limits , which is an important measure o f plastic behavior , i s called the plasticity
index, I
}
, i.e. ,
I
P
= w
r
w
p (
3
-
36
)
Figure 3.1 0 depict s th e change s i n volum e fro m th e liqui d limi t t o th e shrinkag e limi t
graphically. Th e soi l remain s saturate d down t o th e shrinkag e limi t an d whe n onc e thi s limi t i s
crossed, th e soi l become s partiall y saturated. The air takes the place o f the moisture that is lost due
to evaporation. At about 105 ° t o 110°C , there will not be any normal wate r lef t i n the pores an d soi l
at thi s temperatur e i s sai d t o b e oven-dry. A soi l sampl e o f volum e V
o
an d wate r conten t w
o
i s
represented b y poin t A i n the figure .
As th e soi l lose s moistur e conten t ther e i s a correspondin g chang e i n th e volum e o f soils .
The volum e chang e o f soi l i s equa l t o th e volum e o f moistur e lost . Th e straigh t line , AE ,
therefore, give s th e volum e of th e soi l a t differen t wate r contents . Point s C an d D represen t th e
transition stage s o f soi l sampl e at liqui d and plasti c limit s respectively. As th e moistur e content is
reduced furthe r beyon d th e poin t D, th e decreas e i n volume of th e soi l sampl e wil l not b e linea r
States
Liquid
w,
Plastic
w
n
p
Semi soli d
. .. . w
Solid
Li mi t
Liquid l imi t
Plastic limi t
Shrinkage limi t . .
Consi st ency
V ery sof t
Soft
Stiff
.. V er y stif f
Extremely stif f
Hard
V ol ume change
!
Decrease in volume
i
Constant volum e
Figure 3 . 9 Differen t s t at e s an d con s is t en cy o f s oil s wit h At t erber g limit s
Soil Phas e Rel at ion s hips , In de x Propert ie s an d Soi l Cl as s ificat io n 47
( V
0
-V
S
) 2
Solid I Semi-soli d ^
state
stat e Plastic state
Liquid
~ stat e
A
V
s
= V olum e of solid s
V
a
= V olum e of ai r
V
d
= V olum e of dr y soi l
V
w
= V olum e of wate r
Water conten t
Figure 3 .1 0 Curv e s howin g t ran s it ion s t ag e s fro m t he liqui d t o soli d s t at e
with the decrease in moisture beyond a point E due t o many causes. On e possibl e caus e i s that air
might star t enterin g int o th e void s o f th e soil . Thi s ca n happe n onl y whe n th e norma l wate r
between th e particles i s removed. I f the normal water between some particle s i s removed, th e soi l
particles surrounde d by absorbe d wate r wil l come i n contact wit h eac h other . Greate r pressur e i s
required i f these particle s ar e t o be brought stil l closer . As such the change i n volume i s less than
the chang e i n moistur e content . Therefore , th e curv e DEBT depict s th e transitio n fro m plasti c
limit t o th e dr y conditio n o f soi l represente d b y poin t F. However , fo r al l practica l purposes , th e
abscissa o f th e poin t o f intersectio n B o f th e tangent s FB an d EB ma y b e take n a s th e shrinkag e
limit, w
s
. The straigh t lin e AB whe n extende d meet s th e ordinat e a t point M. Th e ordinat e o f M
gives th e volum e o f th e soli d particle s V , . Sinc e th e ordinat e o f F i s th e dr y volume , V
d
, o f th e
sample, th e volume o f ai r V
fl
, i s give n by ( V
d
- V
s
}.
3 .1 2 DETERM I NATI O N O F ATTERB ERG L I M I TS
Liquid Limi t
The apparatu s shown i n Fig. 3.11 i s the Casagrande Liqui d Limi t Devic e use d fo r determinin g th e
liquid limit s of soils. Figur e 3.12 shows a hand-operated liqui d limi t device. Th e device contain s a
brass cup which can be raised and allowed to fall on a hard rubber base by turning the handle. The cup
is raised by one cm. The limit s are determined on that portion of soil finer than a No. 40 sieve (ASTM
Test Designation D-4318). About 100 g of soil is mixed thoroughly with distilled water into a uniform
paste. A portio n o f th e past e i s place d i n th e cu p an d levele d t o a maximu m dept h o f 1 0 mm. A
channel o f th e dimension s o f 1 1 m m widt h an d 8 mm dept h i s cu t throug h th e sampl e alon g th e
48 Chapt er 3
Brass cu p
Sample ^ ,
Liquid limi t devic e
Hard stee l
Casagrandes groovin g t ool AST M groovin g t ool
Figure 3 .1 1 Cas ag r an de' s liqui d limi t apparat u s
symmetrical axi s of the cup. The grooving tool shoul d always be held normal t o the cup at the point of
contact. The handl e is turned a t a rate of about two revolutions per secon d an d th e numbe r of blows
necessary t o clos e th e groov e alon g th e botto m fo r a distanc e o f 12. 5 mm i s counted . Th e groov e
should b e close d b y a flow o f th e soi l an d no t by slippag e betwee n th e soi l an d th e cup. The wate r
content o f the soi l i n the cup is altered an d the tests repeated. At least fou r tests shoul d b e carried ou t
by adjustin g th e wate r contents i n such a way tha t th e numbe r of blows required t o close the groov e
may fal l withi n the range of 5 to 40. A plot of water content against the log of blows is made a s shown
in Fig. 3.13. Withi n the rang e o f 5 t o 40 blows , th e plotte d point s li e almos t o n a straigh t line . Th e
curve so obtained i s known as a 'flow curve'. The wate r content corresponding t o 25 blows i s terme d
the liquid limit. Th e equatio n of the flo w curv e can b e writte n as
= -I
f
\ ogN +C (3.37)
where, w = wate r conten t
/, = slop e o f the flo w curve , termed a s flow inde x
N = numbe r o f blows
C = a constant.
L iquid Limi t b y One- Point M etho d
The determinatio n o f l iqui d l imi t a s explained earlie r require s a considerabl e amoun t o f time an d
labor. W e ca n us e wha t i s terme d th e ' one-poin t method ' i f a n approximat e valu e o f th e limi t i s
required. Th e formul a used for thi s purpose i s
( N
(3.38)
Soil Phas e Relat ion s hips , In de x Propert ie s an d Soi l Cl as s ificat io n 4 9
Figure 3 .1 2 Han d-operat e d liqui d limi t devic e (Court es y : Soilt es t , USA)
where w is the water content corresponding t o the number of blows N, and n, an index whose value has
been foun d t o var y fro m 0.06 8 t o 0.121 . An averag e valu e o f 0.10 4 ma y b e usefu l fo r al l practica l
purposes. I t is, however, a good practice t o check this method with the conventional method as and when
possible.
Liquid Limi t b y the Us e o f Fal l Con e Penetromete r
Figure 3.1 4 show s th e arrangemen t o f th e apparatus . Th e soi l whos e liqui d limi t i s t o b e
determined i s mixe d wel l int o a sof t consistenc y an d presse d int o th e cylindrica l mol d o f 5 c m
diameter an d 5 cm high. The con e whic h has a central angl e of 31° and a total mas s of 14 8 g will
be kep t fre e o n th e surfac e of th e soil . The dept h o f penetratio n 3 ; of th e con e i s measure d i n mm
on th e graduate d scal e afte r 3 0 se c o f penetration. The liqui d limi t w
l
ma y b e compute d b y using
the formula ,
Wf = w
y
+ 0.01(25 - y)( w
y
+ 15) (3.39 )
where w i s the wate r conten t correspondin g t o the penetration y .
The procedur e i s based o n the assumptio n that the penetration lie s betwee n 2 0 and 30 mm.
Even thi s method ha s t o be used wit h caution.
Plastic Limi t
About 1 5 g o f soil , passin g throug h a No. 4 0 sieve , i s mixe d thoroughly . Th e soi l i s rolle d o n a
glass plat e wit h the hand, unti l it is about 3 mm i n diameter. This procedur e o f mixing and rolling
is repeated til l the soi l shows signs of crumbling. The water content of the crumbled portion of the
thread i s determined. Thi s is called the plastic limit .
50 Chapt er 3
O
N
o
^
c
o
4
^
o
o
s
N
J
o
Liquid limit
C
3 4 6 8 1 0 2 0 2 5
Log numbe r of blows N
40 6 0 100
Figure 3 .1 3 Det erm in at io n o f liqui d limi t
Shrinkage Limi t
The shrinkag e limi t of a soi l ca n be determine d b y eithe r of the following methods :
1. Determinatio n o f vv
s
, whe n the specifi c gravit y of th e solid s G
s
i s unknown.
2. Determinatio n o f vv
v
, whe n the specifi c gravit y of th e solids , G
s
is known.
Figure 3 .1 4 Liqui d limi t b y t h e us e o f t h e fal l con e pen et rom et er : (a ) a s chem at i c
diag ram , an d (b ) a phot og raph (Court es y : Soilt es t , USA)
Soil Phas e Relat ion s hips , Inde x Propert ie s an d Soi l Clas s ificat io n 51
M ethod I When G
f
i s U nknown
5
Three bloc k diagrams of a sample of soil having the same mass of solids M
s
, ar e given in Fig. 3.15.
Block diagram (a ) represents a specimen i n the plasti c state , whic h j ust fills a container of known
volume, V
o
. The mass of the specimen i s M
o
. The specimen i s then dried gradually, and as it reaches
the shrinkag e limit , th e specime n i s represente d b y bloc k diagra m (b) . Th e specime n remain s
saturated up to this limit but reaches a constant volume V
d
. When the specimen is completely dried,
its mas s wil l be M
s
wherea s its volume remains as V
d
.
These different state s are represented i n Fig. 3.10 . The shrinkage limit can be written as
M
w =
M,
where, M = M
(3.40)
M
s
- ( V
o
- V
d
) p
w
Therefore w =
M
x 100 % (3.41)
The volum e of th e dr y specime n ca n b e determine d eithe r b y th e displacemen t o f mercury
method or wax method. Many prefer the wax method because wax is non-toxic. The wax method is
particularly recommended i n an academic environment.
Determination o f Dr y V olume V
d
o f Sampl e b y Displacement i n M ercur y
Place a small dish fille d wit h mercury up to the top in a big dish. Cove r th e dish wit h a glass plate
containing three metal prongs in such a way that the plate i s entrapped. Remove the mercury spilt
over int o the bi g dis h an d tak e ou t the cover plat e fro m th e smal l dish . Plac e the soi l sampl e on
the mercury . Submerg e th e sampl e wit h the pronged glas s plat e an d mak e th e glas s plat e flus h
with the top of the dish. Weigh the mercury that is spilt over due t o displacement. Th e volume of
the sample is obtained by dividing the weight of the mercury by its specific gravity which may be
taken a s 13.6 . Figur e 3.16 show s the apparatus used fo r the determination o f dr y volume.
M ethod I I When G
0
i s Known
o
M
100 where, M
w
=( V
d
-V
s
)p
w
=
M
T
M
(a) (b ) (c )
Figure 3 .1 5 Det erm in at io n o f s hrin k ag e limi t
52 Chapt er 3
Glass plate
Figure 3 .1 6 Det erm in at io n o f dr y volum e by m ercur y dis placem en t m et ho d
Therefore, v v = •
M
- x l OO = x l OO
(3.42)
or
w = -^ --- -xl O O
(3.43)
where, p = 1 for all practical purposes .
3 .1 3 DI SCU SSI O N ON L I M I TS AND I NDI CES
Plasticity inde x an d liqui d limit are th e importan t factors tha t hel p a n enginee r t o understan d th e
consistency o r plasticit y of a clay . Shearin g strength , thoug h constan t a t liqui d limits , varie s a t
plastic limit s for al l clays . A highl y plastic clay (sometime s calle d a fat clay ) ha s highe r shearin g
strength at the plastic limit and the threads at this limit are rather hard to roll whereas a lean clay can
be rolled easil y at the plasti c limi t and thereby possesse s lo w shearing strength.
There ar e some fin e graine d soil s tha t appear simila r t o clays bu t they canno t b e rolle d int o
threads s o easily. Such material s are not really plastic. They ma y be j ust at the border lin e between
plastic an d non-plasti c soils . I n suc h soils , on e find s th e l iqui d limi t practically identical wit h the
plastic limi t and 1=0.
Two soil s ma y diffe r i n thei r physical properties eve n thoug h the y have the same liqui d and
plastic limits or the same plasticit y index. Such soils possess different flow indices. Fo r example i n
Fig. 3.1 7 ar e shown two flo w curve s C, and C
2
of two sample s o f soils. C
}
is flatter than C
2
. It may
be assume d fo r th e sak e o f explanatio n tha t bot h th e curve s ar e straigh t line s eve n whe n th e
moisture conten t in the soil i s nearer the plastic limit and that the same liquid limi t device is used to
determine th e numbe r o f blow s require d t o clos e th e groov e a t lowe r moistur e contents . Th e
plasticity inde x / i s taken t o be the same fo r bot h th e soils. I t can be see n fro m the figure that the
sample o f flo w curv e C , ha s liqui d and plasti c limit s o f 10 0 an d 8 0 percen t respectively , givin g
thereby a plasticity index / o f 20 per cent. The sample of flow curve C
2
has liquid and plastic limits
of 5 4 an d 3 4 percen t givin g thereb y th e sam e plasticit y inde x valu e o f 2 0 percent . Thoug h th e
plasticity indice s o f th e two sample s remai n the same , th e resistanc e offere d b y th e two sample s
for slippag e a t thei r plastic limit s is different . Sampl e on e takes 90 blows fo r slippage wherea s th e
second on e take s onl y 40 blows . This indicate s tha t at th e plasti c limits , the cohesiv e strengt h of
sample 1 with a lower flo w inde x is larger than that of sampl e 2 with a higher flo w index .
Soil Phas e Relat ion s hips , In de x Propert ie s an d Soi l Cl as s ificat io n 53
4 6 1 0 2 0 2 5 4 0 6 0 10 0
Log numbe r of blows N
Figure 3 .1 7 T w o s am ple s o f s oil s wit h differen t fl o w in dice s
Plasticity Inde x l
p
Plasticity inde x / indicate s th e degre e o f plasticit y o f a soil . Th e greate r th e differenc e betwee n
liquid an d plasti c limits , th e greate r i s th e plasticit y o f th e soil . A cohesionles s soi l ha s zer o
plasticity index . Suc h soil s ar e termed non-plastic . Fat clays ar e highly plastic an d possess a high
plasticity index. Soil s possessing larg e values of w, and / ar e said to be highly plastic or fat. Those
with low values ar e described a s slightl y plasti c or lean. Atterberg classifies th e soil s accordin g t o
their plasticity indice s a s in Table 3.9 .
A liqui d limi t greate r tha n 10 0 i s uncommo n fo r inorgani c clay s o f non-volcani c origin .
However, fo r clay s containin g considerabl e quantitie s o f organi c matte r an d clay s o f volcani c
origin, the liquid limit may considerably excee d 100 . Bentonite, a material consisting of chemically
disintegrated volcani c ash , has a liquid limit ranging from 400 to 600. I t contains approximately 70
percent o f scale-lik e particle s o f colloida l siz e a s compare d wit h abou t 3 0 pe r cen t fo r ordinar y
highly plasti c clays. Kaoli n an d mica powde r consis t partiall y or entirel y of scal e lik e particles of
relatively coars e siz e i n compariso n wit h highl y colloida l particle s i n plasti c clays . The y
therefore posses s les s plasticit y tha n ordinar y clays . Organi c clay s posses s liqui d limit s greate r
than 50 . Th e plasti c limit s of suc h soil s ar e equall y higher . Therefor e soil s wit h organi c conten t
have lo w plasticit y indice s correspondin g t o comparativel y hig h liqui d limits .
Table 3. 9 Soi l cl as s ificat ion s accordin g t o Plas t icit y Inde x
Plas t icit y in de x Plas t icit y
0
<7
7-17
Non-plastic
Low plasti c
Medium plasti c
Highly plasti c
54 Chapt e r 3
Toughness I ndex , l
t
The shearin g strengt h of a clay a t the plasti c limi t is a measure o f it s toughness. Tw o clays havin g
the same plasticit y index possess toughness which is inversely proportional t o the flow indices . An
approximate numerica l valu e for the toughness ca n be derived a s follows .
Let s
l
= shearin g strengt h corresponding t o th e liqui d limit, w
f
, whic h i s assumed t o b e
constant for al l plastic clays .
s = shearin g strengt h a t th e plasti c limit , whic h ca n b e use d a s a measur e o f
toughness of a clay.
Now Wj = -l
f
l ogAf , + C, w
p
= -I
f
logN
p
+ C
where N
(
an d N ar e the number of blows at the liqui d and plastic limit s respectively. The flo w curv e
is assumed t o be a straight line extending into the plastic range as shown in Fig. 3.17.
Let, N
{
= ms
r
N
}
= ms , wher e m is a constant.
We can write
w
l
= -I,\ ogms [ + C,w - -I,\ ogms + C
Therefore
l
p
=
w
i~
w
p
= I
f
( logms
p
-\ ogm
Sl
)= I
f
\ og-?-
s i
or
t=
T
= g
~ (
3
-
44
>
Since w e ar e intereste d onl y i n a relativ e measur e o f toughness , l
t
ca n b e obtaine d fro m
Eq. (3.44 ) a s th e rati o o f plasticit y inde x an d flo w index . Th e valu e o f I
(
generall y fall s betwee n
0 an d 3 for most cla y soils. When I
t
i s less than one, th e soi l i s friable at the plastic limit . I
t
i s quite
a usefu l inde x t o distinguish soils of different physica l properties .
L iquidity I nde x /,
The Atterberg limit s are found for remolded soi l samples. These limits as such do not indicat e
the consistenc y o f undisturbe d soils . Th e inde x tha t i s use d t o indicat e th e consistenc y o f
undisturbed soil s i s called th e liquidity index. Th e liquidit y inde x i s expressed a s
7
/=^—~ (3.45 )
where, w
n
is the natural moisture content of the soi l i n the undisturbed state . The liquidit y index of
undisturbed soi l ca n var y fro m les s tha n zer o t o greater tha n 1 . The valu e of I
{
varie s according t o
the consistenc y o f the soi l a s i n Table 3.10 .
The liquidit y index indicates the state of the soil in the field. If the natural moisture content of
the soil is closer t o the liquid limit the soil can be considered a s soft, and the soil is stiff i f the natural
moisture content is closer t o the plastic limit. There ar e some soil s whos e natura l moisture content s
are highe r tha n th e liqui d limits . Suc h soil s generall y belon g t o th e montmorillonit e grou p an d
possess a brittle structure . A soi l o f thi s type when disturbed b y vibratio n flows like a liquid. The
liquidity inde x value s o f suc h soil s ar e greate r tha n unity . On e ha s t o b e cautiou s i n usin g suc h
soils fo r foundation s of structures.
Soil Phas e Relat ion s hips , Inde x Propert ie s an d Soi l Clas s ificat io n 5 5
Table 3.1 0 Value s o f / / an d l
c
accordin g t o con s is t en c y o f soi l
Con s is t en cy / / l
c
Semisolid or soli d stat e Negativ e > 1
V ery stif f stat e ( w
n
= w
p
) 0 1
V ery sof t stat e ( w
n
= w
l
) 1 0
Liquid stat e (whe n disturbed ) > 1 Negativ e
Consistency I ndex , /
C
The consistenc y inde x may be defined as
/ (3.46 )
p
The index l
c
reflects the state of the clay soi l condition in the field i n an undisturbed state just in the
same wa y a s I
t
describe d earlier . Th e value s o f / fo r differen t state s o f consistenc y ar e give n i n
Table 3.10 along with the values I
r
I t may be seen that values of 7, and I
c
ar e opposite t o each othe r
for th e same consistency of soil .
From Eqs (3.45 ) an d (3.46) we have
w
l
— w
I
i
+I
c= j
P
=
l
(3.47 )
p
Effect o f Dryin g o n Plasticit y
Drying produces a n invariable change in the colloidal characteristic s o f the organic matte r in a soil.
The distinctio n between organi c an d inorgani c soil s ca n be mad e b y performing tw o liqui d limit
tests o n the same material . On e test i s made o n an air-dried sample an d the other o n an oven-drie d
one. I f the liqui d limit of the oven-dried sampl e i s less tha n about 0.75 time s tha t for the air-drie d
sample, the soils ma y be classed a s organic. Oven-drying also lower s the plasti c limit s of organi c
soils, but the drop i n plastic limi t is less tha n that for the liquid limit.
Shrinking and Swelling o f Soil s
If a mois t cohesiv e soi l i s subjecte d t o drying , i t lose s moistur e an d shrinks . Th e degree o f
shrinkage, S , is expressed a s
. , = - x (3.48a )
o
where,
V
o
= original volume of a soil sampl e at saturated stat e
V
d
= final volume of the sampl e a t shrinkage limit
On the basi s of the degree o f shrinkage , Scheidig (1934) classified soil s a s in Table 3.11.
Shrinkage Rati o SR
Shrinkage rati o i s define d a s th e rati o o f a volum e chang e expresse d a s a percentag e o f dr y
volume t o th e correspondin g chang e i n wate r content abov e th e shrinkag e limit .
56 Chapt e r 3
Table 3 .1 1 Soi l clas s ificat io n accordin g t o deg re e o f s hrin k ag e S
r
S
r
% Qualit y o f s oi l
< 5 Goo d
5-10 Mediu m goo d
10-15 Poo r
> 1 5 V er y poo r
( V -V,)/V,
SR=' °
d)l d
x l O O
(3-48b )
W
0~
W
S
where
V
o
= initial volume of a saturated soi l sampl e a t water conten t w
o
V
d
= the fina l volum e of the soi l sampl e at shrinkage limi t w
s
( w
o
-w
s
) = change i n the water content
M
d
= mass o f dry volume , V
d
, of the sampl e
Substituting fo r ( w
o
-w
s
) i n Eq (3.48b) and simplifying, we have
• ; - • - •
Thus the shrinkag e rati o of a soi l mas s i s equal t o the mas s specifi c gravit y of the soi l i n it s
dry state .
V olumetric Shrinkag e S
v
The volumetri c shrinkage or volumetric change is defined as the decrease i n volume of a soil mass ,
expressed as . a percentage of the dr y volume o f th e soi l mas s whe n th e wate r conten t i s reduce d
from th e initia l w
o
to the fina l w
s
at the shrinkag e limit.
d
(3
.
49)
L inear shrinkage ca n be computed fro m the volumetri c chang e b y the following equatio n
1/3
5.. +1.0
L S= l
~ c 1 m
Xl
°°
percen t
(3-50 )
The volumetri c shrinkag e S
v
i s use d a s a decima l quantit y i n Eq . (3.50) . Thi s equatio n
assumes tha t the reduction in volume is both linear and unifor m in al l directions.
Linear shrinkage can be directly determined by a test [thi s test has not yet been standardize d
in the United States (Bowles, 1992)] . Th e British Standard BS 137 7 used a half-cylinder of mold of
diameter 12. 5 mm an d lengt h L
o
= 140 mm. Th e we t sampl e fille d int o the mol d i s drie d an d th e
final lengt h L,i s obtained. Fro m this , the linea r shrinkag e L S i s computed a s
Soil Phas e Relat ion s hips , Inde x Propert ie s an d Soi l Clas s ificat io n 57
L S =
L -L .
(3.51)
Activity
Skempton (1953) considers that the significant change in the volume of a clay soil during shrinking
or swellin g is a function o f plasticit y inde x and the quantit y of colloidal cla y particle s presen t i n
soil. The clay soil can be classified inactive, normal or active (afte r Skempton, 1953) . The activity
of cla y is expressed a s
Activity A =
Plasticity index, /
Percent fine r tha n 2 micro n
(3.52)
Table 3.12 give s th e typ e o f soi l accordin g t o th e valu e o f A. The cla y soi l whic h ha s a n
activity value greater than 1. 4 can be considered as belonging to the swelling type. The relationship
between plasticity index and clay fraction i s shown in Fig. 3.18(a).
Figure 3.18(b ) show s result s o f som e test s obtaine d o n prepare d mixture s o f variou s
percentage o f particle s les s tha n an d greate r tha n 2 /^ . Severa l natura l soil s wer e separate d int o
fractions greate r and less than 2 /z and then the two fractions were combined a s desired. Fi g 3.18(c)
shows the results obtained on clay minerals mixed wit h quart z sand.
Table 3 .1 2 Soi l clas s ificat io n accordin g t o act ivit y
A Soi l typ e
<0.75 Inactiv e
0.75-1.40 Norma l
>1.40 Activ e
P
l
a
s
t
i
c
i
t
y

i
n
d
e
x
,

I
p



K
>

U
>
-
£
>
.
L
f
>

O
\
D

O

O

O

O

O

OActi
/
//
VQ soi l
/
//
/ 1
{
s
/
A
Nformal s o
/*
Inactiv
/
11
jr
^/l = (
e soi l
• 0
.75
10 2 0 3 0 4 0 5 0
Percent finer than 2 micron
60
Figure 3 .1 8 (a) Clas s ificat io n o f s oi l accordin g t o act ivit y
58 Chapt er 3
1UU
80
X
1
6 0
'o
1 40
a,
20
°
(
/
»&
y
/ /
^o>
/
y
<*
Shell
y
( 1
Lone
((
haven
33)
Ion cla y
).95)
Weald cla y
^ (0-95 )
Horten
(0.95)
) 2 0 4 0 6 0 8 0 1 G
500
400
300
200
100
mon
f
,
/
Sodium
tmorillo
(1.33)
/
/
^ „— — —""
aite
/
/
.---"KaoHnite
T
(A=0.9_)
Clay fractio n ( < 2/ /) (%)
(b)
20 4 0 6 0 8 0
Clay fractio n ( < 2 / / ) ( %)
(c)
100
Figure 3 .1 8 (b, c ) Rel at io n bet wee n plas t icit y in de x an d cl a y fract ion . Fig ure s i n
paren t hes es ar e t he activities o f t h e cl ay s (aft e r Sk em pt on , 1953 )
Consistency o f Soil s a s per the U nconfine d Compressiv e Strengt h
The consistenc y o f a natural soil is different fro m tha t of a remolded soi l at the same wate r content.
Remolding destroy s th e structur e of the soi l and the particl e orientation. The liquidit y index value
which i s an indirect measure of consistency is only qualitative. The consistenc y of undisturbed soil
varies quantitativel y o n th e basi s o f it s unconfine d compressiv e strength . Th e unconfme d
compressive strength , q
u
, i s define d a s th e ultimat e loa d pe r uni t cros s sectiona l are a tha t a
cylindrical specime n o f soi l (wit h height to diameter rati o of 2 to 2.5) can take unde r compressio n
without an y latera l pressure . Wate r conten t of th e soi l i s assume d t o remai n constan t durin g the
duration of the tes t which generally takes only a few minutes. Table 3.1 3 indicates the relationship
between consistenc y an d q
u
.
As explaine d earlier , remoldin g o f a n undisturbe d sampl e o f cla y a t th e sam e wate r
content alter s it s consistency, because o f the destruction of it s original structure . The degre e o f
disturbance o f undi st ur be d clay sampl e du e t o remolding ca n b e expresse d a s
Table 3 .1 3 Relat ion s hi p bet wee n con s is t en c y o f cl ay s an d q
u
Consi st ency
V ery sof t
Soft
Medium
q
u
, k N/ m
2
<25
25-50
50-100
Consi st ency
Stiff
V ery stif f
Hard
q
u
, k N/ m
2
100-200
200-400
>400
Table 3 .1 4 Soi l clas s ificat io n o n t he bas i s of s en s it ivit y (aft e r Sk em pt on an d
Northey, 1954 )
s
t
1
1-2
2-4
Na t u r e o f cl a y
Insensit ive cl ays
Low-sensitive clay s
Medium sensitiv e clay s
S
t
4-8
8-16
Na t u r e o f cl a y
Sensit ive clay s
Extra-sensitive clay s
Quick clay s
Soil Phas e Relat ion s hips , Inde x Propert ie s an d Soi l Clas s ificat io n 59
Sensitivity, S
r
=
q
u
, undisturbed
q'
u
, remolde d
(3. 53)
where q'
u
i s the unconfmed compressiv e strengt h of remolded cla y a t the same wate r content as that
of the undisturbe d clay.
When q'
u
is very low as compared t o q
u
the clay is highly sensitive. When q
u
= q'
u
the clay is said to
be insensitive to remolding. On the basis of the values of S
t
clays can be classified as in Table 3.14.
The clay s tha t have sensitivit y greater tha n 8 shoul d b e treate d wit h car e durin g constructio n
operations becaus e disturbanc e tends to transform them, at least temporarily , into viscous fluids. Suc h
clays belong t o the montmorillonite group and possess flocculent structure .
Thixotropy
If a remolde d cla y sampl e wit h sensitivit y greate r tha n on e i s allowe d t o stan d withou t furthe r
disturbance an d chang e i n wate r content , i t ma y regai n a t leas t par t o f it s origina l strengt h an d
stiffness. Thi s increas e i n strengt h i s due to the gradual reorientatio n o f the absorbed molecule s o f
water, and i s known as thixotropy (fro m th e Greek thix, meanin g 'touch' an d tropein, meanin g 'to
change'). Th e regainin g o f a par t o f th e strengt h afte r remoldin g ha s importan t application s i n
connection wit h pile-driving operations , an d othe r type s o f constructio n i n whic h disturbanc e o f
natural clay formations i s inevitable.
3 .1 4 PL ASTI CI T Y CH AR T
Many properties o f clays and silt s suc h as their dry strength, compressibilit y an d their consistenc y
near the plastic limit can be related wit h the Atterberg limits by means of a plasticity chart as shown
is Fig . 3.19 . I n thi s char t th e ordinate s represen t th e plasticit y inde x 7 an d th e abscissa s th e
40 6 0
Liquid limit , w,
Figure 3 .1 9 Plas t icit y char t
80 100
60 Chapt e r 3
corresponding liqui d limit w
r
The chart is divided into six regions, thre e above and three below lin e
A, The equatio n t o the l in e A i s
7
p
= 0.73 (
W/
- 20) (3.51 )
If a soi l i s known t o be inorgani c it s group affiliatio n ca n b e ascertaine d o n th e basi s o f th e
values o f/ an d w
l
alone. However , point s representing organi c clay s ar e usually located withi n the
same regio n a s thos e representin g inorgani c silt s of high compressibility , an d point s representin g
organic silt s i n the region assigne d t o inorgani c silt s of medium compressibility .
Casagrande (1932 ) studie d the consistency limits of a number of soil and proposed the plasticity
chart shown i n Fig. 3.19. The distribution of soils accordin g t o the regions ar e given below .
Regi on
Above / 4- l i n e
1
2
3
Below / 4- l i n e
4
5
6
Li q u i d l i mi t w
t
Less t ha n 3 0
30 < W j < 5 0
w, >5 0
w
l
<30
30<w,< 50
w, >5 0
Type o f soi l
Inorganic clays of lo w plasticit y an d cohesionles s soil s
Inorganic clay s of medium plasticit y
Inorganic clay s o f hig h plasticity
Inorganic silt s o f lo w compressibilit y
Inorganic silt s of mediu m compressibilit y and organi c silt s
Inorganic silt s of high compressibilit y an d organi c cla y
The uppe r limi t o f th e relationshi p betwee n plasticit y inde x an d liqui d limi t i s provide d b y
another lin e called th e [/-lin e whos e equatio n is
I = 0.9( w-&) (3.52 )
Example 3 . 9
Determine th e time s (? ) require d fo r particle s o f diameter s 0.2 , 0.02 , 0.0 1 an d 0.005 mm t o fal l a
depth o f 1 0 cm fro m th e surfac e i n water .
Given: \ J L = 8.15 x 10~
3
poises, G = 2.65. (Note : 1 poise = 10~
3
gm-sec/cm
2
)
Solution
H = 8.15 x 10~
3
x 10~
3
= 8.15 x lO^
6
gm-sec/cm
2
.
Use Eq. (3.24)
30// L 30X8. 15X10"
6
1 0 1.48 2 x!0~
3
.
-
x
— -= -- -m m
( G
s
- l ) D
2
(2.65-1 ) D
2
D
2
The time s require d fo r th e variou s value s of D ar e a s given below.
D (mm) t
0.2
0.02
0.01
0.005
2.22 se c
3.71 mi n
14.82 mi n
59.28 mi n
Soil Phas e Relat ion s hips , In de x Propert ie s an d Soi l Cl as s ificat io n 6 1
Example 3 .1 0
A sedimentatio n analysi s by th e hydromete r metho d (15 2 H) wa s conducte d wit h 5 0 g ( = A/
s
)of
oven dried soil . The volume of soil suspensio n is V = 10
3
cm
3
. The hydrometer readin g R
a
= 19.50
after a lapse o f 60 minutes afte r th e commencement o f the test .
Given: C
m
(meniscus) = 0.52, L (effective) = 14. 0 cm, C
o
(zero correction) = +2.50, G
s
= 2.70
and [ i = 0.01 poise .
Calculate th e smalles t particl e size , whic h woul d have settle d a dept h o f 14. 0 c m an d th e
percentage fine r tha n this size. Temperatur e o f test = 25° C.
Solution
From Eq . (3.24 )
D(mm) =
where ^ = 0.01 x 10~
3
(gm-sec)/cm
2
.
Substituting
SOxO. Ol xl O-
3
1 4
£>- -
X
J — = 0.0064 mm .
V (2.7-1 ) V 6 0
From Eq . (3.31 )
From Tabl e 3. 6 for T= 25 °C, C
r
= +1.3. Therefore ,
R
c
=19.5-2. 5 + 1.3=18.3
From Eq s (3.32) an d (3.31), w e have
CR 1.65G .
, C =
M
s
sg
2.65(G-1 )
=
1.65X2. 7
p
.
%
sg
2.65(2.7-1 ) 5 0
Example 3 .1 1
A 500 g sampl e o f dr y soi l wa s use d fo r a combined siev e an d hydromete r analysi s (15 2 H type
hydrometer). Th e soi l mas s passin g throug h th e 7 5 fi siev e = 12 0 g . Hydromete r analysi s wa s
carried ou t on a mass of 40 g that passed through the 75 ( J i sieve . The averag e temperatur e recorde d
during the test was 30°C.
Given: G
s
= 2.55, C
m
(meniscus) = 0.50, C
o
= +2.5, n = 8.15 x 10~
3
poises.
The actua l hydromete r readin g R
a
= 15.00 after a lapse o f 12 0 min afte r th e star t of the test .
Determine th e particl e siz e D and percent fine r P'% and P%.
Solution
From Eq . (3.29 )
L =16.3-0.16417?
62
where, R = R
a
+ C
m
= 15.0 + 0.5 = 15. 5
L = 16.3 -0.1641 x15.5 = 13.757
From Eq . (3.24 )
Chapt er 3
30x8. 15xl O-
6
13.75 7 ^
0
x . |=0.004 3 mm
2.55-1
From Eq . (3.32 )
C R
Percent finer , P'% =
S8 c
x 100
M
From Tabl e 3.7 , C = 1.02 for G
s
=2.5 5
From Tabl e 3.6, C
r
= +3.8 for T= 30 °C
Now, R
c
= R
a
- C
o
+ C
T
= 15 - 2. 5 + 3.8 = 16. 3
Now, / "=
L0 2 x l 6 3
x 100 = 41.57%
P% = 41.57 x
40

500
Example 3 .1 2
500 g of dry soil was used for a sieve analysis. The masses of soil retained on each sieve are given below:
US s t an dar d s iev e
2.00 m m
1 .40 m m
1.00mm
M as s i n g
10
18
60
US s t an dar d s ieve
500 f j .
250 j U
125/1
75 f j i
M as s i n
135
145
56
45
g
Plot a grain siz e distribution curve and comput e the following:
(a) Percentage s o f gravel , coars e sand , mediu m sand , fine san d an d silt , as pe r th e Unifie d
Soil Classificatio n System, (b) uniformit y coefficien t (c) coefficient of curvature.
Comment o n the type of soil .
Solution
Com put at ion o f percen t fin e r
US s t an d - Diam et er , D
ard s iev e o f g rain s i n m m
2.00 m m
1 .40 m m
1. 00m m
500/1
250 f j ,
125/1
75 p.
2.00
1.40
1.00
0.500
0.25
0. 125
0.075
M as s
ret ain ed i n g
10
18
60
135
145
56
45
%
ret ain ed
2.0
3.6
12.0
27.0
29.0
11.2
9.0
Cum ulat ive
%ret ain e d
2.0
5. 6
17.6
44.6
73. 6
84. 8
93. 8
%
fin er P
98.0
94.4
82. 4
55. 4
26.4
15.2
6.2
Soil Phas e Relat ion s hips , In de x Propert ie s an d Soi l Cl as s ificat io n 63
100
90
80
70
g 6 0
<5
<|5 0
<u
£4 0
OH
30
20
10
n
Gravel
Sand
Coarse to medium
«
\
\
60%
^
\
V
\
A*r
1
\
r\
309
£>3o
Fine
>,
N,
A
10%
\
D ~**
^
(
e
Silt + clay
108 6 4 2 1 . 8 . 6 . 4 0. 2 0.1.08.06.0 4 .0 2
Grain diameter, D in mm
Figure Ex. 3 .12
(a) Percentage coars e t o medium sand = 98 - 4 8 = 50 percent
Percentage fin e san d = 48 - 6. 2 = 41.8 percen t
Percentage sil t and clay = 6.2 percent .
ZXn
(b) Uniformit y coefficient C =
D
IQ
0.09 8
(c) Coefficient of curvatur e C =
= 5.92
(0.28)
2
i y
xD
6 0
0.098x0.5 8
The soi l i s just o n the border line o f well grade d sand .
= 1.38
Example 3 .1 3
Liquid limit tests on a given sample of clay were carried out . The dat a obtained ar e as given below.
T es t No . 1
Water content , %7 0
Number of blows , N 5
64 47
30
44
45
Draw the flow curv e on semi-log pape r and determine th e liquid limit and flow inde x of the soil.
Solution
Figure Ex . 3.1 3 give s the flo w curv e for th e give n sampl e o f clay soil . As pe r th e curve ,
Liquid limit , \ v
{
= 50%
Flow index , /, = 29
64 Chapt er 3
70
60
S3 5 0
I
40
30
\
\
No
\
2 4 6 81 0 202 5 4 0 608010 0
Number o f blow s
Figure Ex . 3 .13
Example 3 .14
The laborator y test s o n a sampl e o f soi l gav e th e followin g results:
w
n
- 24%, w, = 62%, w
p
= 28%, percentag e of particles les s than 2 J J L - 23 %
Determine: (a ) The liquidit y index , (b) activit y (c ) consistenc y an d natur e of soil .
Solution
(a) Plasticit y index, I
p
= w
l
- w
p
= 62 - 2 8 = 34%
w
n
-w 24-2 8
Liquidity index , 7 , = —
p
- = — = -0.12.
(b) Activity , A -
34
p
*P
=
3 4
of particles < 2/u 2 3
= 1.48.
(c) Comments :
( i) Since I
:
is negative, the consistency of the soil is very stiff to extremely stiff
( semisolid state).
( ii) Since I is greater than 17% the soil is highly plastic.
( Hi) Since A is greater than 1.40, thesoil is active and is subject to significant volume
change ( shrinkage and swelling).
Example 3 .1 5
Two soi l sample s teste d i n a soi l mechanic s laborator y gav e th e followin g results:
Sam ple n o . 1 Sam pl e n o. 2
Liquid limi t 50 %
Plastic l i mi t 30 %
Flow indices , /, 2 7
40%
20%
17
Soil Phas e Relat ion s hips , Inde x Propert ie s an d Soi l Clas s ificat io n 6 5
(a) Determine th e toughness indices an d
(b) comment on the types of soils .
Solution
w, - w
S~\ 7 _ ' P
Sample
,, , ,
=
Z .
==
0.74 ; Sampl e 2, /, = . = = 1.18
F
' 2 7 2 7 1 7 1 7
(b)
(i) Bot h the soils are clay soil s a s their toughness indices li e between 0 and 3.
(ii) Soi l on e i s friable at the plastic limi t since it s I
t
valu e is less tha n one .
(iii) Soi l two is stiffer tha n soil one at the plastic limit since the I
t
value of the latter is higher.
Example 3 .1 6
The natural moisture content of an excavated soi l is 32%. Its liquid limit is 60%and plastic limi t is
27%. Determine th e plasticity index of the soil and comment abou t the nature of the soil .
Solution
Plasticity index , I = \ v
t
- w
p
= 60 - 2 7 = 33%
The nature of the soil can be judged by determining it s liquidity index, /
;
from Eq . (3.45)
W
»-
W32
"
27
IP 3 3
since th e value of I
t
i s ver y close to 0, the natur e of the soi l a s per Table 3.10 is ver y stiff .
Example 3 .1 7
A soil wit h a liquidity index of-0.20 has a liquid limit of 56%and a plasticity index of 20%. What
is its natural water content? What i s the natur e of this soil ?
Solution
As per Eq. (3.45 )
Liquidity index ,
' p
Wp
= w
{
-1 = 56 - 2 0 = 36,
w
n
= l
t
l
p
+ w
p
=-0.20 x 20 + 36 = 32.
Since /, is negative , th e soi l i s i n a semisoli d o r soli d stat e a s per Table 3.10.
Example 3 .1 8
Four differen t type s o f soil s wer e encountere d i n a larg e project . Th e liqui d limit s (w
z
), plasti c
limits ( w ) , an d th e natura l moistur e content s ( w
n
) o f th e soil s ar e give n belo w
I
66 Chapt e r 3
Soil t yp e
1
2
3
4
w,%
120
80
60
65
w
p
%
40
35
30
32
w
n
%
150
70
30
25
Determine: (a) the liquidity indices l
t
of the soils, (b) the consistency o f the natural soil s an d (c) the
possible behavio r o f the soil s unde r vibrating loads.
Solution
(a) /, =
/
By substitutin g the appropriat e value s in this equation, we have
Type
1
2
3
4
I,
1.375
0.778
0
-0.21
(b) Fro m Tabl e 3.10 , Type 1 is i n a liquid state, Type 2 in a very sof t state , Typ e 3 in very
stiff state , an d Type 4 i n a semisoli d state .
(c) Soi l types 3 and 4 are not much affected b y vibrating loads. Type 1 is very sensitive even for
small disturbanc e and a s such is no t suitabl e for an y foundation. Type 2 i s also ver y soft ,
with greater settlement of the foundatio n or failure of the foundatio n due to development of
pore pressure under saturated condition taking place due to any sudden application of loads.
Example 3 .1 9
A shrinkag e limi t tes t o n a cla y soi l gav e th e followin g data . Comput e th e shrinkag e limit .
Assuming tha t the total volume o f dry soil cake is equal t o it s total volume a t the shrinkag e limit ,
what i s the degre e o f shrinkage ? Comment o n the natur e of soi l
Mass o f shrinkag e dish and saturate d soi l M , = 38.7 8 g
Mass o f shrinkage dis h and ove n dr y soi l M
2
= 30.4 6 g
Mass o f shrinkage dis h M
3
= 10.6 5 g
V olume of shrinkage dis h V
o
- 16.2 9 cm
3
Total volum e o f oven dr y soil cak e V
d
- 10.0 0 cm
3
Solution
Refer t o Fig. 3.1 5
M
The equatio n fo r shrinkag e limi t w
s
= ——
where M
w
= mass o f wate r i n the void s at the shrinkage limit .
M
o
= mass o f sampl e a t the plastic stat e = M
l
-M
3
= 38.78- 10.6 5 = 28.13 g
Soil Phas e Relat ion s hips , In de x Propert ie s an d Soi l Cl as s ificat io n 6 7
V olume of water lost from th e plastic state to the shrinkage limi t AV = ( V
o
- V
d
)
or AV = 16.29 - 10.0 0 = 6.29 cm
3
Mass of dry soil = M
s
= M
2
-M
2
= 30.46 - 10.6 5 = 19.81 g
Now, M
w
= M
o
- M
s
-( V
o
-V
d
)p
w
= 28.13 -19.81- (6.29) (1) = 2.03 g
( M - M ) - ( V -V,)p M
From Eq. (3.41) , v v = — -2 -
s
-— ^ -^^ = — ^= -- = 0.102 = 10.2%
4
' M ^ M
s
19.8 1
As per Eq. (3.48a) , the degree of shrinkage, S
r
is
Sf =
V ^L
x
,„„
=
(16.29- 10.0) x 100
=
V
0
16.2 9
From Tabl e 3.11 the soi l i s of ver y poor quality.
3 .1 5 G ENERA L CONSI DERATI ONS FOR CL ASSI FI CATI ON OF SOI L S
It has been state d earlie r tha t soil can be described a s gravel, sand, sil t and clay according t o grain
size. Mos t o f th e natura l soil s consis t o f a mixtur e o f organi c materia l i n th e partl y o r full y
decomposed state . The proportion s o f the constituent s in a mixture vary considerabl y an d there i s
no generall y recognize d definitio n concerning the percentage of , for instance, cla y particles tha t a
soil mus t have to be classified as clay, etc.
When a soil consist s o f the various constituents in differen t proportions , th e mixture is then
given the name of the constituents that appear to have significant influence on its behavior, and then
other constituents are indicated by adjectives. Thus a sandy clay has most of the properties of a clay
but contains a significant amoun t of sand.
The individual constituents of a soil mixture can be separated and identified as gravel, sand, silt
and cla y o n th e basi s o f mechanica l analysis . Th e cla y minera l tha t i s presen t i n a cla y soi l i s
sometimes a matter of engineering importance. According t o the mineral present, the clay soi l can be
classified a s kaolinite, montmorillonit e or illite. The mineral s presen t i n a clay ca n be identified by
either X-ray diffraction o r differential therma l analysis. A description of these methods i s beyond the
scope of this book.
Buildings, bridges, dam s etc . ar e buil t on natural soil s (undisturbe d soils), wherea s earthe n
dams for reservoirs, embankment s for roads an d railway lines, foundation bases for pavement s of
roads an d airport s ar e mad e ou t o f remolde d soils . Site s fo r structure s o n natura l soil s fo r
embankments, etc, wil l have to be chosen firs t o n the basis of preliminary examinations of the soi l
that can be carried ou t i n the field. An engineer shoul d therefor e be conversant wit h the fiel d test s
that woul d identif y th e variou s constituent s of a soi l mixture.
The behavio r of a soil mas s under load depend s upon many factors such as the properties o f
the various constituents present i n the mass, the density, the degree of saturation, the environmental
conditions etc. If soils are grouped on the basis of certain definit e principles and rated accordin g to
their performance, the properties o f a given soil can be understood t o a certain extent , on the basi s
of some simpl e tests . Th e objective s o f th e followin g section s o f thi s chapte r ar e t o discus s th e
following:
1 .Fiel d identificatio n o f soils.
2. Classificatio n of soils .
6 Chapt e r 3
3 .1 6 FI EL D I DENTI FI CATI O N OF SOI L S
The method s o f fiel d identificatio n of soil s ca n convenientl y be discusse d unde r th e heading s o f
coarse-grained an d fine-grained soil materials .
Coarse- G rained Soi l M aterial s
The coarse-graine d soi l material s ar e minera l fragment s that ma y b e identifie d primaril y o n th e
basis o f grai n size . Th e differen t constituent s of coarse-grained material s ar e san d an d gravel . As
described i n th e earlie r sections , th e siz e o f san d varie s fro m 0.07 5 mm t o 4.75 mm an d tha t o f
gravel fro m 4.75 mm t o 8 0 mm. San d ca n furthe r b e classifie d a s coarse, mediu m an d fine . Th e
engineer shoul d hav e a n ide a o f th e relativ e size s o f th e grain s i n orde r t o identif y th e variou s
fractions. Th e description of sand and gravel should include an estimate of the quantit y of material
in the different size ranges as well as a statement o f the shape an d mineralogical composition of the
grains. Th e minera l grain s ca n b e rounded , subrounded , angula r o r subangular . Th e presenc e o f
mica o r a wea k materia l suc h a s shal e affect s th e durabilit y o r compressibilit y o f th e deposit . A
small magnifying glass can be used t o identify th e small fragments of shale or mica. The propertie s
of a coars e graine d materia l mas s depen d als o o n th e uniformit y o f th e size s o f th e grains . A
well-graded san d i s mor e stabl e for a foundation base a s compared t o a uniform or poorl y grade d
material.
Fine-G rained Soi l M aterial s
Inorganic Soils: Th e constituent parts of fine-grained materials ar e the silt and clay fractions. Since
both these material s ar e microscopi c i n size, physical properties othe r than grain siz e mus t be used
as criteri a fo r fiel d identification . Th e classificatio n test s use d i n th e fiel d fo r preliminar y
identification ar e
1. Dr y strengt h test
2. Shakin g tes t
3. Plasticit y test
4. Dispersio n tes t
Dry strength: The strength of a soil in a dry state is an indication of its cohesion and hence of its nature.
It ca n b e estimate d b y crushin g a 3 mm siz e drie d fragmen t betwee n thum b an d forefinger . A cla y
fragment ca n be broken onl y with great effort , wherea s a silt fragment crushe s easily.
Shaking test : The shaking test is also calle d a s dilatancy test . It helps t o distinguish sil t from clay
since sil t i s mor e permeabl e tha n clay . I n thi s tes t a par t o f soi l mixe d wit h wate r t o a ver y sof t
consistency i s placed i n the pal m o f the hand. The surfac e of the soi l i s smoothed ou t wit h a knife
and the soi l pa t i s shaken by tappin g the back o f the hand. If the soi l i s silt , wate r wil l rise quickl y
to the surfac e an d give it a shiny glistenin g appearance . I f the pat i s deformed eithe r b y squeezin g
or b y stretching , th e wate r wil l flo w bac k int o th e soi l an d leav e th e surfac e wit h a dul l
appearance. Sinc e cla y soil s contai n muc h smalle r void s tha n silt s an d ar e muc h les s permeable ,
the appearanc e o f th e surfac e o f th e pa t doe s no t chang e durin g th e shakin g test . An estimat e o f
the relativ e proportion s o f sil t an d cla y i n a n unknow n soi l mixtur e ca n b e mad e b y notin g
whether th e reactio n i s rapid, slo w o r nonexistent .
Plasticity test : I f a sampl e o f mois t soi l ca n be manipulate d betwee n th e palms o f th e hand s an d
fingers an d rolle d int o a lon g threa d o f abou t 3 mm diameter , th e soi l the n contain s a significan t
amount of clay. Sil t cannot be rolled int o a thread o f 3 mm diamete r withou t severe cracking .
Dispersion test : This tes t i s useful fo r making a rough estimat e o f sand, sil t an d cla y presen t i n
a material . Th e procedur e consist s i n dispersin g a smal l quantit y of th e soi l i n wate r take n i n a
Soil Phas e Relat ion s hips , Inde x Propert ie s an d Soi l Clas s ificat io n 6 9
glass cylinder and allowing the particles t o settle. The coarser particle s settl e first followed by finer
ones. Ordinaril y san d particle s settl e withi n 30 seconds i f the dept h o f wate r i s abou t 1 0 cm. Sil t
particles settl e in about 1/ 2 to 240 minutes, whereas particles o f clay size remain i n suspension for
at least severa l hour s and sometimes severa l days.
Organic soils : Surface soils and many underlying formations may contain significant amounts of
solid matte r derived fro m organisms . While shel l fragment s and simila r soli d matte r ar e found a t
some locations, organic material i n soil is usually derived from plan t or root growt h and consists of
almost completel y disintegrate d matter , such as muck or mor e fibrou s material, suc h as peat. The
soils wit h organi c matte r ar e weaker an d mor e compressibl e tha n soil s havin g the sam e minera l
composition bu t lackin g i n organi c matter . Th e presenc e o f a n appreciabl e quantit y o f organi c
material ca n usuall y b e recognize d b y th e dark-gre y t o blac k colo r an d th e odo r o f decayin g
vegetation which it lends to the soil .
Organic silt : I t is a fine grained mor e o r less plasti c soi l containin g mineral particle s o f sil t siz e
and finel y divide d particle s o f organi c matter . Shell s an d visibl e fragment s o f partl y decaye d
vegetative matte r ma y als o b e present .
Organic clay : I t i s a cla y soi l whic h owe s som e o f it s significan t physica l propertie s t o th e
presence of finely divide d organic matter . Highly organic soil deposits suc h as peat or muck may be
distinguished by a dark-brown to black color , an d by the presence o f fibrous particles o f vegetabl e
matter in varying states of decay. The organic odor i s a distinguishing characteristic o f the soil. The
organic odor ca n sometimes b e distinguished by a slight amount of heat .
3 .1 7 CL ASSI FI CATI O N OF SOI L S
Soils i n natur e rarel y exis t separatel y a s gravel , sand , silt , clay o r organi c matter , bu t ar e usually
found a s mixtures with varying proportions of these components. Grouping of soils on the basis of
certain definit e principle s woul d help the engineer to rate the performance o f a given soi l either as
a sub-bas e materia l fo r road s an d airfiel d pavements , foundation s o f structures , etc . Th e
classification o r grouping of soils i s mainly based o n one or two index properties o f soi l which are
described i n detai l i n earlier sections . The methods that ar e used for classifying soil s ar e based o n
one or the other of the following two broad systems :
1. A textural system which is based onl y on grai n siz e distribution.
2. Th e systems that are based o n grain size distribution and limits of soil .
Many system s ar e i n us e tha t ar e base d o n grai n siz e distributio n an d limit s o f soil . Th e
systems that are quite popular amongs t engineer s ar e the AASHTO Soi l Classification Syste m and
the Unified Soil Classification System.
3 .1 8 TEXTU RA L SOI L CL ASSI FI CATI O N
U .S. Departmen t o f Agricultur e System (U SDA)
The boundarie s betwee n th e variou s soi l fraction s of thi s systems ar e give n i n Tabl e 3.15 .
By makin g us e o f th e grai n siz e limit s mentione d i n th e tabl e fo r sand , sil t an d clay , a
triangular classification char t has been developed a s shown in Fig. 3.2 0 for classifying mixed soils.
The firs t ste p i n the classification of soi l i s to determine the percentages o f sand, sil t and clay-siz e
materials in a given sample by mechanical analysis . With the given relative percentages o f the sand,
silt and clay, a point is located o n the triangular chart as shown in Fig. 3.20. Th e designation given
on th e char t fo r th e are a i n whic h the poin t fall s i s used a s the classificatio n o f th e sample . Thi s
method of classification does not reveal any properties o f the soil other than grain-size distribution .
Because o f it s simplicity , it i s widel y used by worker s i n the fiel d o f agriculture . On e significan t
70 Chapt er 3
Table 3.15 Soi l Fract ion s a s per U.S. Depart m en t o f Ag ricult ur e
Soil fr act io n Diam et er i n m m
Gravel
Sand
Silt
Clay
>2.00
2-0.05
0.05-0.002
<0.002
100
10
\V
100 9 0 8 0 7 0 6 0 5 0 4 0 3 0 2 0 1 0 0
Percentage o f sand
Figure 3.20 U. S . Depart m en t o f Ag ricul t ur e t ex t ura l cl as s ificat io n
100
disadvantage o f thi s metho d i s tha t th e textura l name a s derive d fro m th e char t doe s no t alway s
correctly expres s the physical characteristics of the soil. For example, sinc e some clay size particle s
are muc h les s activ e tha n others , a soi l describe d a s cla y o n th e basi s o f thi s syste m ma y hav e
physical propertie s mor e typica l of silt.
3 .1 9 AASH T O SOI L CL ASSI FI CATI ON SY STE M
This syste m was originally proposed i n 192 8 by the U.S. Bureau of Public Roads for use by highway
engineers. A Committe e o f highwa y engineers fo r th e Highwa y Research Board , me t i n 194 5 an d
made a n extensive revisio n o f th e PRA System. Thi s syste m i s known as the AASHTO (America n
Association o f Stat e Highwa y an d Transportatio n Officials ) Syste m (AST M D-3242 , AASHT O
Soil Phas e Rel at ion s hips , In de x Propert ie s an d Soi l Cl as s ificat io n 7 1
Method M 145) . Th e revise d syste m comprise s seve n group s of inorganic soils , A- l t o A-7 wit h 12
subgroups i n all . The syste m i s based o n the following three soi l properties :
1. Particle-siz e distributio n
2. Liqui d Limi t
3. Plasticit y Index .
A Grou p Inde x i s introduce d t o furthe r differentiat e soil s containin g appreciabl e
fine-grained materials . The characteristic s o f various groups ar e defined i n Table 3.16. The Grou p
Index ma y b e determine d fro m th e equation.
Group. Index ( GI) = 0.2a + O.OOSa c + 0.01 bd (3.56a )
in which ,
a = tha t portio n o f percentag e o f soi l particle s passin g No . 20 0 (ASTM ) siev e greate r tha n
35 = (F-35).
b = tha t portion of percentage of soil particles passing No. 200 sieve, greate r tha n 15 = ( F -15).
c = tha t portion of the liquid limi t greate r tha n 40 = ( w
l
-40).
d = tha t portion o f the plasticit y index greater tha n 1 0 = (7 -10) .
F = percen t passin g No . 200 sieve. I f F < 35, use ( F -35) = 0
It ma y b e note d her e tha t i f G I < 0, us e G I = 0. Ther e i s n o uppe r limi t fo r GI . Whe n
calculating the GI for soils that belong t o groups A-2-6 and A-2-7, use the partial grou p index ( PGI)
only, that is (From Eq . 3.56a)
PGI = O.Olb d = 0.01(F - 15)(7
p
- 10 ) (3.56b )
Figure 3.2 1 provide s a rapi d mean s o f usin g th e liqui d an d plasti c limit s (an d plasticit y
index 7 ) t o mak e determinatio n o f th e A- 2 subgroup s an d th e A- 4 throug h A- 7 classifications .
Figure 3.2 1 i s base d o n th e percen t passin g th e No . 20 0 siev e (whethe r greate r o r les s tha n 3 5
percent)
The group index i s a means of rating the value of a soil as a subgrade material withi n its own
group. I t i s no t use d i n orde r t o plac e a soi l i n a particula r group , tha t i s don e directl y fro m th e
results o f siev e analysis , th e liqui d limi t an d plasticit y index. The highe r th e valu e o f th e grou p
index, th e poore r i s th e qualit y of th e material . Th e grou p inde x i s a functio n o f th e amoun t of
material passin g the No. 200 sieve, the liquid limit and the plasticity index .
If the pertinent index value for a soil fall s below the minimum limit associated wit h a, b, c or d,
the value of the corresponding ter m i s zero, and the term drops ou t of the group index equation. The
group index value should be shown in parenthesis afte r a group symbol suc h as A-6(12) where 1 2 is
the group index.
Classification procedure : With the required dat a i n mind, procee d fro m lef t t o righ t i n the
chart. Th e correc t grou p wil l b e foun d b y a process of elimination . The firs t grou p fro m th e lef t
consistent wit h the test data i s the correct classification. The A-7 group is subdivided int o A-7-5 or
A-l-6 dependin g o n the plasticity index , 7 .
For A-7-5, l
p
<
w /
- 30
ForA-7-6, 7
p
>w
/
- 3 0
Table 3.16 AASHT O soil classification
General
classification
Group
classification
Sieve analysis percent
passing
No. 1 0
No. 40
No. 200
Characteristics of fractio n
passing No. 40
Liquid limit
Plasticity Inde x
Usual types of significant
constituent materials
General rating as
subgrade
Granular Material s
(35 percent or less of total sampl e passing No. 200 )
A-l
A-l-a A-l-b
50 max
30 max 5 0 max
15 max 2 5 ma x
6 max
Stone fragment s —
gravel and san d
A-3
51 mi n
10 ma x
N.P.
Fine
sand
A-2
A-2-4 A-2- 5
35 max 3 5 max
40 max 4 1 min
10 max 1 0 max
A-2-6 A-2- 7
35 max 3 5 max
40 max 4 1 min
1 1 min 1 1 ma x
Silty or clayey gravel an d sand
Excellent to good
Silt-clay Material s (More tha n 35 percent
of total sampl e passing No. 200 )
A-4 A- 5
36 min 3 6 min
40 max 4 1 min
10 max 1 0 max
Silty soil s
A-6 A-l
A-l -5
A-7-6
36 mi n 3 6 mi n
40 max 4 1 mi n
1 1min 1 1 min
Clayey soil s
Fair t o poo r
Soil Phas e Relat ion s hips , Inde x Propert ie s an d Soi l Clas s ificat io n 73
70
60
50
30
20
10
0
Note: A
3;
-2 s o
5 %f i
t
il sco
ner t
ntain
lan >
less
Jo. 2(
than
)0sk
A-6 and A-2-6
^-4 a id A-2-4
/
A-
/
1
7-6
/
\ -5i
/
•>
/
/ j
/
A-7-5 and A-2-7
ndA-2-5
/
/
/
/
/
0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 10 0
Liquid limi t w,
Figure 3.2 1 Char t fo r us e in AASHT O soil clas s ificat io n s ys t e m
3.20 UNIFIE D SOI L CL ASSI FI CATI ON SY STEM (U SCS)
The Unified Soil Classification System is based on the recognition of the type and predominance of
the constituent s considering grain-size, gradation, plasticity and compressibility. It divides soil into
three major divisions : coarse-grained soils , fine grained soils, and highly organic (peaty) soils. In the
field, identificatio n i s accomplishe d b y visua l examination for th e coarse-graine d soil s an d a few
simple hand tests for the fine-grained soils. In the laboratory, the grain-size curve and the Atterberg
limits ca n b e used . Th e peat y soil s ar e readil y identifie d by color , odor , spong y fee l an d fibrous
texture.
The Unifie d Soi l Classificatio n Syste m i s a modifie d versio n o f A. Casagrande' s Airfield
Classification (AC) System developed i n 194 2 for the Corps of Engineers. Since 194 2 the original
classification ha s bee n expande d an d revised i n cooperation wit h the Burea u o f Reclamation, s o
that i t applie s no t onl y t o airfields but als o t o embankments , foundations , and othe r engineerin g
features. This system was adopted in 1952. In 1969 the American Societ y for Testing and Materials
(ASTM) adopte d th e Unifie d Syste m a s a standar d metho d fo r classificatio n fo r engineerin g
purposes (ASTM Test Designation D-2487) .
Table 3.1 7 present s th e primar y factor s t o consider i n classifyin g a soi l accordin g t o th e
Unified Soi l Classificatio n system.
The followin g subdivisions ar e considere d i n the classification :
1. Gravels and sand s are GW, GP, SW, or S P
if les s tha n 5 percen t o f th e materia l passe s th e No . 20 0 sieve ; G = gravel; S = sand;
W = well-graded; P = poorly-graded. The well - or poorly-graded designation s depend o n
C. and C a s defined in section 3. 9 and numerical values shown in Table 3.16
74 Chapt e r 3
Table 3 .1 7 T h e Un ifie d Soi l Cl as s ificat io n Sys t e m (Source : B owl es , 1992 )
Major
divisions
C
o
a
r
s
e
-
g
r
a
i
n
e
d

s
o
i
l
s
(
m
o
r
e

t
h
a
n

h
a
l
f

o
f

m
a
t
e
r
i
a
l

i
s

l
a
r
g
e
r

t
h
a
n

N
o
.

2
0
0
)
F
i
n
e
-
g
r
a
i
n
e
d

s
o
i
l
s
f

o
f

m
a
t
e
r
i
a
l

i
s

s
m
a
l
l
e
r

t
h
a
n

N
o
.

2
0
0
)
(
m
o
r
e

t
h
a
n

h
a
s
;
o
a
r
s
e

f
r
a
c
t
i
o
n
4

s
i
e
v
e

s
i
z
e
)
G
r
a
v
e
(
m
o
r
e

t
h
a
n

h
a
l
f

o
f
i
s

l
a
r
g
e
r

t
h
a
n

N
o
.
S
a
n
d
s
a
n

h
a
l
f

o
f

c
o
a
r
s
e

f
r
a
c
t
i
o
n
e
r

t
h
a
n

N
o
.

4

s
i
e
v
e

s
i
z
e
)
(
m
o
r
e

t
h
i
s

s
m
a
l
G
r
a
v
e
l
s

w
i
t
h

f
i
n
e
s

C
l
e
a
n

g
r
a
v
e
l
s
(
a
p
p
r
e
c
i
a
b
l
e

(
l
i
t
t
l
e

o
r

n
o
a
m
o
u
n
t

o
f

f
i
n
e
s
)

f
i
n
e
s
)
S
a
n
d
s

w
i
t
h

f
i
n
e
s

C
l
e
a
n

s
a
n
d
s
(
a
p
p
r
e
c
i
a
b
l
e

(
l
i
t
t
l
e

o
r

n
o
a
m
o
u
n
t

o
f

f
i
n
e
s
)

f
i
n
e
s
)
S
i
l
t
s

a
n
d

c
l
a
y
s
(
l
i
q
u
i
d

l
i
m
i
t

<

5
0
)
i
l
l
s

a
n
d

c
l
a
y
s
c/o
i/i
A
6
T3
3
2"
£o
Group
symbol
GW
GP
GM
GC
d
u
SW
SP
SM
SC
u
ML
CL
OL
MH
CH
OH
Pt
Typical name s
Well-graded gravels, gravel-san d
mixtures, l i t t l e or no fines
Poorly grade d gravels , gravel-
sand mixtures , little or no fines
Silty gravels, gravel-sand-sil t
mixture
Clayey gravels , gravel-sand-clay
mixture
Well-graded sands , gravell y
sands, littl e or no fines
Poorly grade d sands , gravell y
sands, l i t t l e or no fine s
Silty sands , sand-sil t mixtur e
Clayey sands , sand-sil t mixture
Inorganic silt s an d very fine
sands, roc k flour , silty or
clayey fine sands , o r clayey
silts wi t h slight plasticit y
Inorganic clays of ver y low
to medium plasticity, gravelly
clays, sandy clays , silt y clays,
lean clay s
Organic silt s and organic silt y
clays o f lo w plasticity
Inorganic silts , micaceous or di-
atomaceous fin e sand y or silty
soils, elastic silt s
Inorganic clay s or high plasticity ,
fat clay s
Organic clay s of medium to high
plasticity, organic silt s
Peat an d other highl y organic soil s
Classification criteri a for
coarse-grained soil s
C
u
>4
1 < C
c
< 3
Not meetin g al l gradation requirement s
for GW ( C
u
< 4 or 1 >C, > 3)
Atterberg limit s
Abov e A line with
below A lin e or 4 < / < 7 are
IP
<
^borderlin e cases
Atterberg l i mi t s symbol s
above A line with
/ „>?
C
u
>6
1 < C
c
< 3
Not meeting al l gradation requirement s
for S W ( C
u
< 6 or 1 >C
c
> 3)
Atterberg limits
Abov e A
i
ine with
below A line or 4 < / < 7 are
'p
<
borderlin e cases
Atterberg limit s symbol s
above A line with
1 . Determin e percentages o f sand an d
gravel fro m grain-siz e curve .
2. Dependin g on percentages o f fine s
(fraction smalle r tha n 200 sieve size) ,
coarse-grained soil s ar e classified a s
follows:
Less than 5%-GW, GP, SW, SP
More tha n 12%-GM, GC, SM, SC
5 t o 12%-Borderlin e cases requiring
dual symbol s
c
-=ft
Gravels an d sands ar e GM, GC, SM, or SC
if mor e tha n 1 2 percent passe s th e No . 20 0 sieve ; M = silt ; C = clay. Th e sil t o r cla y
designation i s determined by performing the liquid and plastic limit tests on the (-) No . 40
fraction an d usin g th e plasticit y char t o f Fig . 3.22 . Thi s char t i s als o a Casagrand e
contribution t o th e US C system , and th e A lin e show n o n thi s char t i s sometime s calle d
Casagrande's A line .
Soil Phas e Relat ion s hips , Inde x Propert ie s an d Soi l Cl as s ificat io n
60
75
3.
4.
50
- 30
20 30 4 0 5 0
Liquid limi t w, percent
60 70 80
Figure 3.2 2 Plas t icit y char t fo r fin e-g rain e d s oil s
The char t a s presented her e ha s bee n slightl y modifie d based o n th e Corp s o f Engineer s
findings tha t no soil has so far been found wit h coordinates that lie above the "upper limit "
or U line shown. This char t an d lines ar e part of the ASTM D 2487 standard .
Gravels and sands are (not e using dual symbols)
GW-GC SW-S C GP-G C SP-SC , or GW-GM SW-S M GP-G M SP-S M
if between 5 and 1 2 percent of the material passes the No. 200 sieve . It may be note d that
the M o r C designatio n i s derive d fro m performin g plasti c limi t test s an d usin g
Casagrande's plasticit y chart .
Fine-grained soil s (mor e tha n 50 percent passes th e No. 200 sieve) are :
ML, OL, or CL
if th e liqui d limit s ar e < 50 percent ; M = silt ; O = organi c soils ; C = clay. L = Less tha n
50 percent fo r \ v
t
Fine grained soil s are
MH, OH, or CH
if th e liqui d limits are > 50 percent; H = Higher than 50 percent . Whether a soil i s a Clay
(C), Sil t (M), or Organic (O) depends on whether the soi l coordinates plo t abov e or below
the A line on Fig. 3.22 .
The organi c (O ) designatio n als o depend s o n visua l appearanc e an d odo r i n th e US C
method. I n th e ASTM metho d th e O designation is mor e specificall y define d b y usin g a
comparison o f the air-dr y liqui d limit vv
/
and the oven-dried w'
r
I f the ove n dried valu e is
0.75w
and the appearance and odor indicate s "organic" then classify th e soi l a s O.
76 Chapt er 3
Table 3 .1 8 Un ifie d Soi l Clas s ificat io n Sys t e m —fin e-g r ain e d s oil s (m or e t ha n hal f
of m at eria l i s l arg e r t han No . 200 s iev e s ize )
Soil
Silt
and
clays
Highly
organic
soils
Major
divisions
Liquid
limit less
than 5 0
Liquid
limit mor e
than 5 0
Group
symbols
ML
CL
OL
MH
CH
OH
Identification procedure s on
fraction smalle r tha n No. 4 0
sieve siz e
Dry
strength
None to
slight
Medium
to hig h
Slight to
medium
Slight t o
medium
High to
very hig h
Medium
to high
Dilatancy
Quick to
slow
None t o
very slo w
Slow
Slow t o
none
None
None to
very slo w
Toughness
None
Medium
Slight
Slight t o
medium
High
Slight t o
medium
Pt Readil y identifie d b y color , odor,
spongy fee l an d frequentl y
by fibrou s textur e
The liqui d and plastic limits are performed o n the (-) No . 40 sieve fractio n of all of the soils ,
including gravels , sands , and the fine-graine d soils. Plasticit y limi t test s ar e not require d fo r soil s
where th e percen t passin g th e No . 20 0 siev e < 5 percent . Th e identificatio n procedur e o f fin e
grained soil s ar e give n i n Tabl e 3.18 .
A visua l descriptio n o f th e soi l shoul d accompan y th e lette r classification . Th e AST M
standard include s some descriptio n i n terms o f sandy or gravelly, but color i s also ver y important .
Certain area s ar e underlai n wit h soi l deposit s havin g a distinctiv e colo r (e.g. , Bosto n blu e clay ,
Chicago blu e clay) which may be red, green , blue , grey, black, an d s o on. Geotechnical engineer s
should becom e familia r wit h th e characteristic s o f thi s materia l s o th e colo r identificatio n i s o f
considerable ai d i n augmenting the dat a bas e o n the soil .
3 .2 1 COM M ENT S ON TH E SY STEM S OF SOI L CL ASSI FI CATI ON
The various classificatio n system s described earlie r ar e based on :
1. Th e propertie s o f soi l grains .
2. Th e propertie s applicabl e t o remolded soils .
The system s d o no t tak e int o accoun t th e propertie s o f intac t material s a s foun d i n nature .
Since th e foundatio n material s o f mos t engineerin g structure s ar e undisturbed , th e propertie s o f
intact material s onl y determin e th e soi l behavio r durin g and afte r construction . The classificatio n
of a soi l accordin g t o any of the accepte d system s doe s no t i n itsel f enabl e detaile d studie s of soil s
to b e dispense d wit h altogether . Solvin g flow, compressio n an d stabilit y problem s merel y o n th e
basis o f soi l classificatio n ca n lea d t o disastrou s results . However , soi l classificatio n ha s bee n
found t o b e a valuabl e too l t o th e engineer . I t help s th e enginee r b y givin g genera l guidanc e
through makin g availabl e i n a n empirica l manne r th e result s o f fiel d experience .
Soil Phas e Relat ion s hips , In de x Propert ie s an d Soi l Clas s ificat io n 7 7
Example 3 .2 0
A s am pl e o f in org an i c soi l has t he followin g
g rain s iz e charact eris t ic s
Size (m m ) Percen t pas s in g
2.0 (No . 10 ) 9 5
0.075 (No . 200 ) 7 5
The liquid limi t is 56 percent, an d the plasticity inde x 25 percent. Classify the soil according t o the
AASHTO classification system.
Solution
Percent o f fine graine d soil = 75
Computation o f Group Inde x [Eq . (3.56a)] :
a = 75 - 3 5 = 40
b = 75 - 1 5 = 60
c = 56-40 = 16, d=25-W=15
Group Index, GI = 0.2 x 40 + 0.005 x 40 x 1 6 + 0.01 x 60 x 1 5 = 20. 2
On the basis of percent of fine-grained soils, liquid limit and plasticity index values, the soi l
is either A-7-5 or A-7-6. Since ( w
l
- 30) = 56 - 30 = 26 >/(25) , the soil classification isA-7-5( 20).
Example 3 .2 1
Mechanical analysis on four different sample s designate d a s A, B, C and D were carried out in a soil
laboratory. The results of tests are given below. Hydrometer analysis was carried ou t on sample D.
The soi l i s non-plastic.
Sample D: liquid limit = 42, plastic limit = 24, plasticity inde x =18
Classify th e soils per the Unified Soi l Classification System.
Samples
ASTM Siev e
Desi gnat i on
63.0 mm
20.0 mm
6.3
2. 0mm
600 JL I
212 j i
63 j i
20 n
6(1
2 |i
A
Percentage
100
64
39
24
12
5
1
B
f i n e r t ha n
100
98
90
9
2
C
93
76
65
59
54
47
34
23
7
4
D
100
95
69
46
31
78 Chapt er 3
0.001 0.01 0.075 0. 1 1
Particle size (mm)
Figure Ex . 3 .2 1
Cobbles
(> 76.2 mm)
100
Solution
Grain siz e distribution curve s of samples A, B, C and D are given i n Fig. Ex . 3.21. The values of C
u
and C
c
are obtained fro m th e curves as give n below.
Sampl e
A
B
C
D
10
0.47
0.23
0.004
^30
3.5
0.30
0.036
D
60
16.00
0.41
2.40
c
u
34.0
1.8
600.0
c
c
1.60
0.95
0.135
Sample A: Grave l siz e particle s mor e tha n 50%, fin e graine d soi l les s tha n 5%. C
u
, greater
than 4, and C
c
lies between 1 and 3. Well graded sand y gravel classifie d a s GW.
Sample/?: 96 %o f particle s ar e san d size . Fine r fractio n les s tha n 5%. C
u
= 1.8, C, is not
between 1 and 3. Poorly-graded sand , classified a s SP .
Sample C: Coars e graine d fractio n greate r tha n 66 %an d fin e graine d fractio n les s tha n
34%. Th e soi l i s non-plastic. C
u
is very high but C
c
is onl y 0.135. Gravel-sand -
silt mixture , classified a s CM.
Sample/): Fine r fractio n 95%wit h clay siz e particle s 31%. The point plot s j ust abov e th e
A-l i ne i n th e C L zon e o n th e plasticit y chart . Silty-cla y o f lo w plasticity ,
classified a s CL.
Example 3 .2 2
The followin g dat a refer s t o a silt y cla y tha t wa s assume d t o b e saturate d i n th e undisturbe d
condition. On the basis of these data determine th e liquidit y index, sensitivity , and voi d ratio o f the
saturated soil . Classif y th e soi l accordin g t o the Unified an d AASHTO systems. Assume G = 2.7.
Soil Phas e Relat ion s hips , Inde x Propert ie s and Soi l Clas s ificat io n 7 9
I nde x pr oper t y Undi s t ur be d
Unconfmed compressive
strength, q
u
kN/m
2
24 4 kN/m
2
Water content , %2 2
Liquid l i mi t , %
Plastic l i mi t , %
Shrinkage limit , %
%passing no. 200 sieve
Remol ded
144 kN/m
2
22
45
20
12
90
Solution
w
n
-w 22-2 0
Liquidity Index , / , = — == 0.08
' Wf -w 45-2 0
q undisturbe d 24 4
Sensitivity, 5 = — = = 1.7
q'
u
disturbe d 14 4
V
V oid ratio , e = —
V,
ForS=l,e = wG
s
= 0.22 x 2. 7 = 0.594.
Unified Soi l Classificatio n
Use the plasticity char t Fig. 3.22. w, = 45, / = 25. The point fall s above the A-line in the CL-zone ,
that is the soi l i s inorganic clay of low t o medium plasticity.
AASHTO Syste m
Group Inde x GI = 0.2a + 0.005ac + 0.01 bd
a = 90 - 3 5 = 55
£= 90-15 = 75
c
= 45 ~ 40 = 5
d = 25 - 1 0 = 15 (here I
p
= w
t
- w
p
= 45 - 2 0 = 25)
Group inde x GI = 0.2 x 55 + 0.005 x 55 x 5 + 0.01 x 75 x 15
= 1 1 + 1.315 + 11.2 5 =23.63 or sa y 2 4
Enter Table 3.1 5 wit h the following dat a
%passing 200 sieve = 90 %
Liquid limi t = 45 %
Plasticity inde x = 25 %
With this , the soil i s either A-7-5 o r A-7-6. Sinc e ( w
l
- 30 ) = (45 - 30 ) = 15 <25 (/ ) the soil
is classified a s A-7-6. Accordin g t o this system the soi l is clay, A-7-6 (24).
80 Chapt e r 3
3 .2 2 PROB L EM S
3.1 A soi l mas s i n its natural state i s partially saturated having a water conten t o f 17.5 %an d a
void rati o o f 0.87 . Determin e th e degre e o f saturation , tota l uni t weight , an d dr y uni t
weight. What i s the weight of water required t o saturat e a mass of 1 0 m
3
volume? Assume
G^ = 2.69 .
3.2 Th e voi d rati o o f a cla y sampl e i s 0. 5 an d th e degre e o f saturatio n i s 70%. Comput e th e
water content , dr y and wet uni t weight s of the soil . Assume G
s
= 2.7.
3.3 A sampl e o f soi l compacte d accordin g t o a standar d Procto r tes t ha s a uni t weigh t o f
130.9 lb/ft
3
a t 100 %compactio n an d a t optimu m wate r conten t o f 14%. Wha t i s th e dr y
unit weight? If the voids become fille d wit h water what would be the saturated uni t weight?
Assume G
s
= 2.67.
3.4 A sampl e o f san d abov e th e wate r tabl e wa s foun d t o hav e a natura l moistur e conten t o f
15%an d a uni t weigh t o f 18.8 4 kN/m
3
. Laborator y test s o n a drie d sampl e indicate d
values o f e
min
= 0.50 an d e
max
- 0.8 5 fo r th e denses t an d looses t state s respectively .
Compute th e degree o f saturation and the relative density. Assume G
s
= 2.65.
3.5 Ho w many cubic meters of fill can be constructed at a void ratio of 0.7 from 119,00 0 m
3
of
borrow materia l that has a voi d rati o of 1.2 ?
3.6 Th e natura l water content of a sampl e take n fro m a soi l deposi t wa s found t o be 11.5%. It
has been calculated tha t the maximum densit y for the soil wil l be obtained when the wate r
content reaches 21.5%. Comput e how much water must be added to 22,500 Ib of soil (i n its
natural state ) i n orde r t o increas e th e wate r conten t t o 21.5%. Assume tha t th e degre e of
saturation i n it s natura l stat e wa s 40%an d G = 2.7 .
3.7 I n an oi l wel l drilling project, drillin g mud wa s used t o retain th e side s o f the borewell . I n
one lite r of suspension i n water , the drilling mud flui d consist s o f the following material :
Material
Clay
Sand
Iron filing s
Mass
(g)
410
75
320
Sp. gr
2.81
2.69
7.13
Find th e densit y of the drilling fluid o f unifor m suspension .
3.8 I n a fiel d exploration , a soi l sampl e wa s collected i n a sampling tub e o f internal diamete r
5.0 cm below th e ground water table . The lengt h of the extracted sampl e wa s 10. 2 c m and
its mass wa s 387 g. If G
y
= 2.7, and the mass of the dried sampl e i s 313 g, find th e porosity,
void ratio, degree of saturation, and the dr y densit y of the sample .
3.9 A saturated sample of undisturbed clay has a volume of 19. 2 cm
3
and weighs 32.5 g . Afte r
oven drying , the weight reduces t o 20. 2 g . Determine th e following :
(a) wate r content , (b ) specifi c gravity , (c) voi d ratio, an d (d ) saturate d densit y o f th e cla y
sample.
3.10 Th e natura l total uni t weigh t o f a sandy stratu m i s 117. 7 lb/ft
3
an d ha s a water conten t of
8%. For determining of relative density, dried san d from the stratum was filled loosely int o
a 1.0 6 ft
3
mol d and vibrated t o give a maximum density. The loose weight of the sampl e i n
the mol d wa s 105. 8 Ib , an d th e dens e weigh t wa s 136. 7 Ib . I f G
9
= 2.66, fin d th e relativ e
density o f the san d i n it s natural state.
3.11 A n eart h embankmen t is to be compacted t o a density of 120. 9 lb/ft
3
a t a moisture conten t
of 1 4 percent . Th e in-sit u tota l uni t weigh t an d wate r conten t o f th e borro w pi t ar e
Soil Phas e Relat ion s hips , Inde x Propert ie s an d Soi l Clas s ificat io n 8 1
114.5 lb/ft
3
an d 8 %respectively . Ho w muc h excavatio n shoul d b e carrie d ou t fro m th e
borrow pi t fo r eac h ft
3
o f the embankment ? Assume G^ , = 2.68 .
3.12 A n undisturbed sampl e of soi l has a volume of 29 cm
3
and weighs 48 g. The dr y weight of
the sampl e i s 32 g. The value of G
s
= 2.66. Determin e th e (a) natural water content , (b) in-
situ voi d ratio, (c ) degree o f saturation, and (d) saturated uni t weight of the soil .
3.13 A mas s o f soi l coate d wit h a thi n layer o f paraffi n weigh s 0.98 2 Ib . Whe n immerse d i n
water i t displace s 0.01130 2 ft
3
o f water . Th e paraffi n i s peele d of f an d foun d t o weig h
0.0398 Ib . Th e specifi c gravit y o f th e soi l particle s i s 2. 7 an d tha t o f paraffi n i s 0.9 .
Determine th e voi d rati o o f the soi l i f its water content i s 10%.
3.14 22 5 g of oven dried soi l wa s placed i n a specific gravity bottle and then filled with water to
a constant volume mar k mad e o n the bottle. The mas s o f the bottl e wit h water and soi l i s
1650 g. The specifi c gravit y bottle wa s filled wit h water alone to the constant volume mark
and weighed. It s mas s wa s found to be 151 0 g . Determine th e specifi c gravit y of the soil .
3.15 I t i s required t o determine th e water content of a wet sampl e o f silty sand weighing 400 g.
This mas s of soil was placed i n a pycnometer an d water filled t o the top of the conical cu p
and weighe d (M
3
). It s mass wa s found t o be 2350 g. The pycnomete r wa s next filled with
clean wate r and weighe d an d it s mass wa s foun d t o be 2200 g (A/
4
). Assuming G^ . = 2.67 ,
determine th e wate r content of the soi l sample .
3.16 A clay sampl e i s found to have a mass o f 423.53 g in its natural state. I t is then dried i n an
oven at 10 5 °C. The dried mas s is found t o be 337.65 g . The specifi c gravity of the solids is
2.70 an d the density of the soi l mass in its natural state is 170 0 kg/m
3
. Determin e the water
content, degre e o f saturation and the dr y densit y of the mass i n its natural state .
3.17 A sampl e o f san d i n it s natura l stat e ha s a relativ e densit y o f 6 5 percent . Th e dr y uni t
weights of the sample at its densest and loosest state s ar e respectively 114. 5 and 89.1 lb/ft
3
.
Assuming th e specifi c gravit y o f th e solid s a s 2.64 , determin e (i ) it s dr y uni t weight ,
(ii) wet unit weight when full y saturated , an d (iii ) submerged uni t weight.
3.18 Th e mas s o f wet sampl e o f soi l i n a drying dish i s 462 g . The sampl e an d the dis h hav e a
mass o f 36 4 g afte r dryin g i n an ove n a t 11 0 °C overnight . Th e mas s o f th e dis h alon e i s
39 g. Determine th e wate r content of the soil .
3.19 A sampl e o f san d abov e th e wate r tabl e wa s found t o have a natura l moistur e conten t of
10%and a uni t weight o f 12 0 lb/ft
3
. Laborator y test s o n a drie d sampl e indicate d value s
e
mm ~ 0-45 , an d e
max
= 0.90 fo r th e denses t an d looses t state s respectively . Comput e th e
degree o f saturation , S, and the relative density, D
f
. Assum e G ^ = 2.65 .
3.20 A 50 cm
3
sample o f moist cla y wa s obtained b y pushing a sharpened hollo w cylinde r into
the wall of a test pit. The extruded sampl e had a mass of 85 g, and after oven drying a mass
of 60 g. Compute w, e, S, and p
d
. G
s
= 2.7.
3.21 A pi t sampl e o f mois t quart z sand wa s obtained fro m a pit by th e san d con e method . Th e
volume of the sample obtained wa s 15 0 cm
3
and its total mass was found t o be 250 g. In the
laboratory th e dr y mas s o f th e san d alon e wa s foun d t o b e 24 0 g . Test s o n th e dr y san d
indicated e
max
= 0.80 an d e
min
= 0.48. Estimat e p
s
, w, e, S, p
d
an d D
r
of the sand in the field.
Given G
s
= 2.67.
3.22 A n earthe n embankmen t unde r constructio n ha s a tota l uni t weigh t o f 99. 9 lb/ft
3
an d a
moisture content o f 1 0 percent. Comput e th e quantit y of wate r require d t o b e adde d pe r
100 ft
3
o f earth t o raise it s moistur e content t o 1 4 percent a t the same void ratio .
3.23 Th e wet uni t weight of a glacial outwash soi l is 12 2 lb/ft
3
, th e specifi c gravity of the solid s
is G
S
= 2.67, an d the moistur e conten t o f the soi l i s w = 12%by dr y weight . Calculat e (a )
dry uni t weight, (b) porosity, (c ) voi d ratio, and (d ) degree of saturation.
82 Chapt er 3
3.24 Deriv e th e equation e = wG
s
whic h expresses th e relationship between th e voi d ratio e, the
specific gravit y G
s
and th e moistur e content w for ful l saturatio n of voids .
3.25 I n a siev e analysi s of a give n sampl e o f san d th e followin g data wer e obtained . Effectiv e
grain siz e = 0.25 mm , uni formi t y coefficient 6.0, coefficien t of curvatur e = 1.0 .
Sketch th e curve on semilo g paper .
3.26 A siev e analysi s of a given sampl e o f san d wa s carrie d ou t b y makin g us e o f US standar d
sieves. Th e tota l weight of sand used fo r th e analysi s was 52 2 g . The followin g data wer e
obtained.
Sieve siz e i n mm 4.75 0 2.00 0 1.00 0 0.50 0 0.35 5 0.18 0 0.12 5 0.07 5
Weight retaine d
ing 25.7 5 61.7 5 67.00126. 0 57.7 5 78.7 5 36.7 5 36.7 5
Pan 31. 5
Plot th e grai n siz e distributio n curve on semi-lo g pape r an d comput e th e following:
(i) Percen t grave l
(ii) Percen t o f coarse , mediu m and fin e san d
(iii) Percen t o f sil t and clay
(iv) Uniformit y coefficient
(v) Coefficien t of curvatur e
3.27 Combine d mechanica l analysi s of a given sampl e of soi l wa s carried out . The tota l weigh t
of soi l use d i n th e analysi s wa s 35 0 g . Th e sampl e wa s divide d int o coarse r an d fine r
fractions b y washin g i t throug h a 7 5 micron s siev e Th e fine r tractio n wa s 12 5 g . Th e
coarser fraction was use d fo r th e sieve analysi s and 5 0 g of th e fine r fractio n wa s use d fo r
the hydrometer analysis . The tes t result s wer e a s given below:
Sieve analysis:
Part icl e s iz e
4. 75 m m
2.00 m m
1.40 m m
1.00mm
500 fi
M as s ret ain e d g
9.0
15. 5
10.5
10.5
35. 0
Part icl e s iz e
355 u
180 n
125 u
75 n
M as s
24.5
49.0
28. 0
43.0
ret ain ed g
A hydrometer (15 2 H type) was inserted int o the suspension j ust a few seconds before th e
readings wer e taken. It was next removed an d introduced j ust before eac h o f the subsequent
readings. Temperatur e of suspensio n = 25°C.
Hydrom et er an al ys is : Readin g s i n s us pen s io n
T im e, m i n
1/4
1/2
1
2
4
8
15
Readin g , R
g
28. 00
24.00
20. 50
17.20
12.00
8. 50
6. 21
T im e, mi n
30
60
120
240
480
1440
Readin g , R
g
5. 10
4. 25
3. 10
2. 30
1.30
0.70
Soil Phas e Relat ion s hips , In de x Propert ie s an d Soi l Cl as s ificat io n 8 3
Meniscus correction C
m
= +0.4, zero correction C
o
= +l.5,G
s
= 2.75
(i) Sho w (step b y step) al l the computations required fo r the combined analysis ,
(ii) Plo t the grai n size distribution curve on semi-log pape r
(iii) Determin e the percentages o f gravel, sand, and fine fraction s present i n the sampl e
(iv) Comput e the uniformit y coefficien t an d th e coefficien t o f curvature
(v) Commen t on the basi s of the test result s whether the soi l i s well graded o r not
3.28 Liqui d limit tests wer e carried ou t on two given samples of clay. The test data are as given
below.
Test No s
Sample no . 1
Water content %
Number o f blows , N
Sample no . 2
Water conten t %
Number o f blows , N
1
120
7
96
9
2
114
10
74
15
3
98
30
45
32
4
92
40
30
46
The plasti c limi t of Sampl e No. 1 is 40 percent and that of Sampl e No . 2 is 32 percent .
Required:
(i) The flow indice s of the two sample s
(ii) The toughnes s indices of the sample s
(iii) Comment on the type of soils o n the basi s o f the toughness index values
3.29 Fou r differen t type s o f soil s wer e encountere d i n a large project . Thei r liqui d limit s (w
;
),
plastic limit s ( w ) and thei r natural moisture contents (w
n
) wer e as given below:
S o i l t y pe
1
2
3
4
w,%
120
80
60
65
w
p
%
40
35
30
32
w
n
%
150
70
30
25
Required:
(i) The liquidit y indices of the soils, (ii) the consistency of the natural soils (i.e., whether soft ,
stiff, etc.)
(ii) and the possible behavior of the soils under vibrating loads
3.30 Th e soi l type s a s give n i n Proble m 3.2 9 contained soi l particle s fine r tha n 2 micron s a s
given below:
Soil typ e
Percent fine r
than 2 micron
1
50
2
55
3
45
4
50
Classify th e soil s according t o their activity values.
84 Chapt er 3
3.31 A sampl e o f clay ha s a water content of 40 percent a t ful l saturation . It s shrinkag e limi t is
15 percent . Assumin g G
s
= 2.70 , determin e it s degre e o f shrinkage . Commen t o n th e
quality of the soil .
3.32 A sampl e o f clay soi l has a liquid limi t of 62%an d it s plasticit y index i s 32 percent .
(i) Wha t i s th e stat e o f consistenc y o f th e soi l i f th e soi l i n it s natura l stat e ha s a wate r
content o f 34 percent ?
(ii) Calculat e th e shrinkag e limit i f the voi d rati o o f th e sampl e a t th e shrinkag e limi t i s
0.70
Assume G ^ = 2.70 .
3.33 A soi l wit h a l i qui di t y inde x of-0.20 has a liquid limit of 56 percent an d a plasticity index
of 20 percent . What i s it s natural water content?
3.34 A sampl e o f soi l weighin g 5 0 g i s disperse d i n 100 0 m L o f water . Ho w lon g afte r th e
commencement o f sedimentatio n shoul d th e hydromete r readin g b e take n i n orde r t o
estimate th e percentage o f particles les s tha n 0.002 mm effective diameter, i f the center of
the hydrometer i s 15 0 mm belo w th e surface of the water ?
Assume: G
s
= 2.1; ^ = 8.15 x 10"
6
g-sec/cm
2
.
3.35 Th e result s of a sieve analysi s of a soil wer e a s follows :
Sieve
s ize (m m )
20
12
10
6. 3
4. 75
2. 8
M as s
ret ain ed (g )
0
1.7
2.3
8.4
5.7
12.9
Sieve
s ize (m m )
2
1.4
0.5
0. 355
0.180
0.075
M as s
ret ain ed (g )
3. 5
1.1
30.5
45. 3
25. 4
7.4
The total mas s of the sampl e wa s 147. 2 g.
(a) Plo t th e particle-size distribution curve and describe th e soil . Commen t o n the flat par t
of the curv e
(b) Stat e th e effectiv e grai n siz e
3.36 A liquid limit test carried ou t on a sample of inorganic soil taken from belo w the water tabl e
gave th e following results:
Fall cone penetration y (mm) 15. 5 18. 2 21. 4 23. 6
Moisture content , w% 34. 6 40. 8 48. 2 53. 4
A plasti c limi t tes t gav e a value of 33%. Determin e th e average liqui d limit and plasticity
index o f thi s soil an d give its classification.
3.37 Th e ove n dr y mas s o f a sampl e o f cla y wa s 11.2 6 g . The volum e o f th e dr y sampl e wa s
determined b y immersin g i t in mercur y an d th e mas s o f the displace d liqui d was 80.2 9 g.
Determine th e shrinkag e limit , vv
y
, of th e clay assumin g G
s
= 2.70 .
Soil Phas e Relat ion s hips , Inde x Propert ie s an d Soi l Cl as s ificat io n 8 5
3.38 Particle s o f five differen t size s are mixed in the proportion shown below an d enough water
is added t o make 100 0 cm
3
of the suspension.
Part icle s iz e (m m )
0.050
0.020
0.010
0.005
0.001
Mass (g )
6
20
15
5
4 Total 50 g
It is ensured that the suspension is thoroughly mixed so as to have a uniform distribution of
particles. All particles have specific gravity of 2.7.
(a) Wha t is the largest particle size present at a depth of 6 cm after 5 mins from th e start of
sedimentation?
(b) Wha t i s the densit y of the suspension a t a depth o f 6 cm afte r 5 mins from th e star t of
sedimentation?
(c) Ho w long shoul d sedimentation be allowed s o that all the particles hav e settled below
6 cm? Assume ,u= 0.9 x 1Q-
6
kN-s/m
2
3.39 A sample of clayey sil t i s mixed at its liquid limit of 40%. I t i s placed carefull y in a small
porcelain dis h wit h a volume of 19. 3 cm
3
and weighs 34.67 g. After oven drying, the soi l
pat displaced 216. 8 g of mercury.
(a) Determine the shrinkage limit , w
s
, of the soil sampl e
(b) Estimate the dry unit weight of the soi l
3.40 Durin g the determinatio n o f the shrinkag e limi t of a sandy clay, th e following laboratory
data was obtained:
Wet wt. of soi l + dish = 87.8 5 g
Dry wt . of soi l + dis h = 76.9 1 g
Wt of dis h = 52.7 0 g
The volumetri c determination o f the soi l pat :
Wt. o f dish + mercur y = 430. 8 g
Wt. o f dis h = 244.62 g
Calculate th e shrinkage limit , assuming G
s
= 2.6 5
3.41 A sedimentation analysis by a hydrometer (152 H type) was conducted wit h 50 g of oven
dried soi l sample . The hydrometer reading in a 1000 cm
3
soil suspension 60 mins after th e
commencement o f sedimentatio n i s 19.5 . Th e meniscu s correctio n i s 0.5 . Assumin g
G
s
= 2.70 an d \ L - 1 x 10"
6
kN-s/m
2
fo r water , calculat e the smalles t particl e siz e which
would have settled durin g the time o f 60 mins and percentage o f particles fine r tha n this
size. Assume: C
0
= +2.0, an d C
T
= 1.2
3.42 Classif y the soi l give n below using the Unified Soi l Classification System.
Percentage passin g No. 4 sieve 7 2
Percentage passin g No. 200 sieve 3 3
Liquid limit 3 5
Plastic limit 1 4
86 Chapt e r 3
3.43 Soi l sample s collecte d fro m th e fiel d gave the following laboratory tes t results :
Percentage passin g No. 4 sieve 10 0
Percentage passin g No. 200 sieve 1 6
Liquid limi t 6 5
Plastic limi t 3 0
Classify th e soi l usin g the Unified Soil Classificatio n System .
3.44 Fo r a large project , a soi l investigatio n was carried out . Grai n siz e analysi s carried out on
the samples gav e the following average tes t results .
Sieve No . Percen t fin e r
4 9 6
10 6 0
20 1 8
40 1 2
60 7
100 4
200 2
Classify th e soi l b y using the Unifie d Soi l Classification System assumin g the soi l i s non-
plastic.
3.45 Th e sieve analysis of a given sample of soil gave 57 percent o f the particles passin g through
75 micro n sieve . Th e liqui d an d plasti c limit s o f th e soi l wer e 6 2 an d 2 8 percen t
respectively. Classif y th e soi l pe r th e AASHT O an d th e Unifie d Soi l Classificatio n
Systems.
CHAPTER 4
SOIL PERMEABILITY AND SEEPAG E
4 .1 SOI L PERM EAB I L I T Y
A material i s permeable i f it contains continuous voids. All materials suc h as rocks, concrete , soil s
etc. ar e permeable . Th e flo w o f wate r throug h al l o f the m obey s approximatel y th e sam e laws .
Hence, th e differenc e between th e flo w o f wate r throug h roc k o r concret e i s on e o f degree . Th e
permeability o f soil s ha s a decisiv e effec t o n th e stabilit y o f foundations , seepag e los s throug h
embankments of reservoirs, drainag e of subgrades, excavation of open cut s i n water bearing sand ,
rate of flow o f water int o wells an d many others .
H ydraulic G radien t
When wate r flows through a saturated soi l mas s ther e i s certain resistanc e fo r the flow becaus e o f
the presence o f solid matter. However, the laws of fluid mechanics whic h are applicable fo r the flo w
of fluid s throug h pipe s ar e als o applicabl e t o flo w o f wate r throug h soils . A s pe r Bernoulli's
equation, th e total head a t any point i n water under steady flo w conditio n ma y b e expressed as
Total hea d = pressure hea d + velocity head + elevation hea d
This principl e can be understood wit h regards t o the flow o f water through a sample o f soi l
of lengt h L and cross-sectional are a A as shown in Fig. 4. 1 (a). The heads o f water at points A and
B a s the wate r flows fro m A to B ar e given as follows (wit h respect t o a datum)
Total hea d a t A, H. = Z
A
+ —^ + -^-
Y 2g
p V
2
Total head at B, H
K
=Z
K
-\ —— + ——
87
88 Chapt er 4
Figure 4 .1 (a) Flo w o f wat e r t hroug h a s am pl e o f s oi l
As the water flows from A t o B, there is an energy loss which is represented b y the difference
in the total head s H, an d H
D
or
H
A
-H
B
=\ Z
A
PA PR »
c
o u i _ ,
where, p
A
an d p
B
= pressure heads , V
A
and V
B
= velocity, g - acceleratio n du e to gravity, y
w
= unit
weight of water , h = loss o f head.
For all practical purposes the velocity head is a small quantity and may be neglected. The loss
of head of h units is effected a s the water flows from A t o B. The loss of head per unit length of flow
may b e expressed a s
h
i =
(4.1)
where / i s called th e hydraulic gradient.
L aminar and Turbulent Flo w
Problems relatin g to the flo w o f fluids i n general ma y be divided into two mai n classes :
1. Thos e in whic h the flo w i s laminar.
2. Thos e i n which the flo w i s turbulent.
There i s a certain velocity, v
c
, below whic h for a given diameter o f a straight tube and for a
given flui d a t a particula r temperature, th e flo w wil l alway s remai n laminar . Likewise ther e i s a
higher velocity, v
r
above which the flow wil l always be turbulent. The lower bound velocity, v
p
of
turbulent flo w i s abou t 6. 5 time s th e uppe r boun d velocit y v o f lamina r flo w a s show n i n
Fig. 4.1(b). Th e uppe r boun d velocit y o f lamina r flow i s calle d th e lower critical velocity. Th e
fundamental law s tha t determin e th e stat e existin g fo r an y give n cas e wer e determine d b y
Reynolds (1883). H e foun d th e lowe r critical velocity is inversel y proportional t o the diamete r of
Soil Perm eabilit y an d Seepag e 89
l og/
Flow always
Flow alway s laminar -
laminar turbulen t
Flow alway s
turbulent
V
T
l ogv
Figure 4 .Kb) Relat ion s hi p bet wee n velocit y o f fl o w an d hydraulic g radien t fo r fl o w
of liquid s i n a pip e
the pipe an d gave the followin g general expressio n applicabl e fo r an y flui d an d for an y syste m of
units.
= 2000
where, A ^ = Reynold s Numbe r take n a s 200 0 a s th e maximu m valu e fo r th e flo w t o remai n
always laminar, D = diameter of pipe, v
c
= critical velocit y below whic h the flow always remains
laminar, y
0
= uni t weight of flui d a t 4 °C , fJ L = viscosity of fluid, g = acceleration du e t o gravity.
The principa l difference between lamina r flow an d turbulent flow i s that i n the forme r cas e
the velocity i s proportional t o the first power of the hydraulic gradient, / , whereas in the latter case
it is 4/7 the power of / . According to Hagen-Poiseuille's' Law the flow through a capillary tube may
be expressed a s
R
2
ai
(4.2a)
or (4.2b)
where, R = radius of a capillary tube of sectional area a, q = discharge through the tube, v = average
velocity through the tube, ^ = coefficient of viscosity.
4 .2 DARCY ' S L AW
Darcy i n 185 6 derive d a n empirica l formul a for th e behavior o f flo w throug h saturated soils . H e
found tha t th e quantit y o f wate r q pe r se c flowin g through a cross-sectiona l are a o f soi l unde r
hydraulic gradient / can be expressed b y the formula
q = kiA
or the velocit y of flo w ca n be written as
(4.3)
90 Chapt er 4
1 .0
1.6
1.4
1.0
0.8
n £
\
\
\
\
\
\
^^
k^
10 2 0
Temperature ° C
30
Figure 4 . 2 Rel at io n bet wee n t em perat ur e an d vis cos it y o f wat e r
v = -j = & (4.4 )
where k i s termed th e hydraulic conductivity (o r coefficient o f permeability)with unit s of velocity. A
in Eq. (4.4) is the cross-sectional area of soil normal to the direction of flow which includes the area of
the solid s an d th e voids , whereas th e area a i n Eq. (4.2 ) i s the are a o f a capillary tube. The essentia l
point i n Eq . (4.3 ) i s tha t th e flo w throug h th e soil s i s als o proportiona l t o th e firs t powe r o f th e
hydraulic gradient i as propounded by Poiseuille' s Law. From this , we are justified i n concluding that
the flo w o f wate r throug h th e pore s o f a soi l i s laminar . It i s foun d that , o n th e basi s o f extensive
investigations made sinc e Darcy introduced his law in 1856 , thi s law i s valid strictl y for fine grained
types of soils .
The hydrauli c conductivit y i s a measur e o f th e eas e wit h whic h wate r flow s throug h
permeable materials . I t i s inversel y proportional t o th e viscosit y o f wate r whic h decrease s wit h
increasing temperatur e a s shown i n Fig. 4.2 . Therefore, permeabilit y measurement s a t laborator y
temperatures shoul d b e correcte d wit h th e ai d of Fig. 4. 2 befor e applicatio n t o fiel d temperatur e
conditions by mean s o f the equation
k ~
~
(4.5)
where k
f
an d k
T
ar e th e hydrauli c conductivit y value s correspondin g t o th e fiel d an d tes t
temperatures respectivel y and /^,and ^
r
are the corresponding viscosities . It is customary to repor t
the value s of k
T
a t a standar d temperatur e of 20°C. The equatio n i s
^20
(4.6)
4 .3 DI SCH ARG E AND SEEPAG E V EL OCI TI ES
Figure 4. 3 show s a soi l sampl e o f lengt h L and cross-sectiona l are a A. Th e sampl e i s place d i n a
cylindrical horizonta l tube betwee n screens . Th e tub e i s connected t o two reservoir s R^ an d R
2
i n
which the water levels ar e maintained constant. The difference in head between R
{
an d R
2
i s h. This
difference i n head i s responsible fo r the flow o f water. Since Darcy' s la w assumes n o change i n the
Soil Perm eabilit y an d Seepag e 91
Screen
— Sampl e
B
\
Screen
Figure 4.3 Flo w o f wat e r t hroug h a s am ple o f soi l
volume of voids and the soi l i s saturated, the quantity of flow pas t sections AA, BB and CC should
remain the same fo r steady flow conditions . We may expres s th e equation of continuity as follows
Qaa = <lbb = 3cc
If the soi l be represented a s divided int o solid matte r and void space , the n the area availabl e
for th e passag e o f wate r i s onl y A
v
. I f v
s
i s th e velocit y of flo w i n th e voids , an d v , the averag e
velocity acros s th e section then , we have
A v = Av o r v = —v
s
A
A 1 l + e
Since, ~7 ~ = ~ =
A.. n e
+ e
(4.7)
Since (1 + e)le i s always greater tha n unity, v
s
is always greater tha n v. Here, v
s
is called th e
seepage velocity an d v the discharge velocity.
4.4 M ETH OD S OF DETERM I NATI ON O F H Y DRAU L I C
CONDU CTI V I TY OF SOI L S
Methods that are in common use for determining the coefficient o f permeability k can be classified
under laboratory an d fiel d methods .
Laboratory methods :
Field methods :
Indirect Method:
1. Constant head permeabilit y method
2. Falling hea d permeabilit y metho d
1. Pumping tests
2. Bore hol e test s
Empirical correlation s
The various types of apparatus which are used in soil laboratories for determining the permeability of
soils are called permeameters. The apparatus used for the constant head permeabilit y tes t is called a
constant head permeameter and the one used for the falling head test is a fallingheadpermeameter.
The soi l sample s use d i n laborator y method s ar e eithe r undisturbe d o r disturbed . Sinc e i t i s no t
92 Chapt er 4
possible t o obtai n undisturbe d sample s o f cohesionles s soils , laborator y test s o n cohesionles s
materials are always conducted on samples whic h are reconstructed t o the same densit y as they exist
in nature. The result s of tests on such reconstructed soils are often misleadin g since it is impracticable
to obtain representative samples and place them in the test apparatus to give exactly the same density
and structural arrangement of particles. Direct testing of soil s in place i s generall y preferred i n case s
where i t i s no t possibl e t o procur e undisturbe d samples . Sinc e thi s metho d i s quit e costly , i t i s
generally carried ou t in connection with major projects such as foundation investigation for dams and
large bridge s o r building foundation jobs wher e lowering of th e wate r tabl e i s involved. In plac e of
pumping tests, bore hole tests as proposed b y the U.S. Bureau of Reclamation are quite inexpensive as
these tests eliminate th e use of observation wells .
Empirical correlation s hav e bee n develope d relatin g grai n siz e an d voi d rati o t o hydraulic
conductivity an d wil l be discusse d late r on .
4 .5 CONSTAN T H EAD PERM EAB I L I TY TEST
Figure 4.4(a ) shows a constant head permeameter whic h consists o f a vertical tub e of lucite (or any
other material ) containin g a soil sampl e whic h is reconstructed o r undisturbed as the case may be.
The diamete r an d heigh t of the tube can be of any convenient dimensions. Th e head an d tai l wate r
levels ar e kep t constan t b y overflows . Th e sampl e o f lengt h L an d cross-sectiona l are a A i s
subjected t o a head h which is constant during the progress o f a test. A test is performed b y allowing
water t o flow through the sampl e an d measuring the quantit y of discharge Q in time t .
The valu e of k can b e computed directl y from Darcy' s la w expresse d a s follows
Supply
, c -
Screen
.Soil
sample
Filter skin
Graduated ja r
~\
T
h
1
(a) (b )
Figure 4.4 Con s t an t hea d perm eabilit y t es t
Soil Perm eabilit y an d Seepag e 9 3
Q = k—At (4.8 )
01 =
<
4
-
9
>
The constant head permeameter test is more suited for coarse grained soils such as gravelly sand
and coarse an d medium sand. Permeability test s i n the laboratory ar e generally subjecte d t o various
types of experimental errors. One of the most important of these arise s fro m th e formation of a filter
skin of fine material s on the surface of the sample. The constant head permeameter of the type shown
in Fig. 4.4(b) can eliminate the effect o f the surface skin. In this apparatus the loss of head is measured
through a distance in the interior of the sample, and the drop in head across the filter skin has no effec t
on the results.
4.6 FAL L I N G H EA D PERM EAB I L I TY TES T
A falling head permeameter i s shown in Fig. 4.5(a). The soil sampl e is kept in a vertical cylinder of
cross-sectional are a A. A transparen t stand pipe of cros s sectiona l area , a, is attache d to the tes t
cylinder. The test cylinder is kept in a container filled wit h water, the level of which is kept constant
by overflows . Befor e th e commencemen t o f th e tes t th e soi l sampl e i s saturate d b y allowin g the
water to flow continuousl y through the sample from th e stand pipe. After saturation is complete, th e
stand pipe is filled wit h water up to a height of h
Q
and a stop watch is started. Let the initial time be
t
Q
. The time t
l
when the water level drops from h
Q
to h
}
i s noted. The hydraulic conductivity k can be
determined on the basis of the drop in head ( h
Q
- hj an d the elapsed time ( t
l
- ?
0
) required for the drop
as explained below.
Let h be the head o f water at any time t. Let the head dro p by an amount dh in time dt. The
quantity of water flowing through the sample i n time dt from Darcy' s law i s
h
dQ = kiAdt = k—Adt(4.10 )
L
v
'
where, i = h/L th e hydrauli c gradient .
The quantit y of discharge dQ can be expressed a s
dQ = -adh(4.11 )
Since the head decreases as time increases, dh is a negative quantity in Eq. (4.11). Eq. (4.10)
can be equated t o Eq. (4.11)
h
-adh = k— Adt (4.12 )
The discharg e Q i n tim e ( t^ - f
Q
) ca n b e obtaine d b y integratin g Eq . (4.10 ) o r (4.11) .
Therefore, Eq . (4.12) can be rearranged an d integrated as follows
*i
Cdh kA C h
n
kA
-a \ — = —\ d t o r ' - °
The genera l expression for k is
94 Chapt er 4
dh
T
Stand pipe
1
L
I
E!
#Vc?**£
"X'^;<
V
J ?
k'v's&S
/>Scre e
J
Sample
(a) (b )
Figure 4.5 Fallin g hea d perm eabilit y t es t
k =
aL
A( t,
or k =
23aL
(4.13)
The setu p shown in Fig. 4.5(a) is generally used for comparatively fine material s suc h a s fin e
sand an d sil t where th e time require d for the drop i n head fro m h
Q
t o h
l
i s neither undul y to o lon g
nor too short for accurat e recordings. If the time is too long evaporation of water from the surface of
the water migh t take plac e and als o temperatur e variations might affec t th e volume of the sample .
These woul d introduc e seriou s error s i n th e results . Th e se t u p i s suitabl e fo r soil s havin g
permeabilities rangin g fro m 10~
3
t o 10~
6
c m pe r sec . Sometimes , fallin g hea d permeameter s ar e
used fo r coarse graine d soil s also . Fo r such soils, the cross sectiona l are a of the stand pipe i s mad e
the sam e a s th e tes t cylinde r s o tha t th e dro p i n hea d ca n convenientl y be measured . Fig . 4.5(b )
shows th e tes t se t up for coarse grained soils . When a = A, Eq . (4.13) i s reduced t o
2.3L , h
Q
log,
n

(4.14)
Example 4. 1
A constant head permeabilit y tes t was carried ou t on a cylindrical sampl e o f sand 4 in. i n diamete r
and 6 in. i n height . 1 0 in
3
of wate r wa s collecte d i n 1.7 5 min , under a head o f 1 2 in. Comput e th e
hydraulic conductivit y in ft/year and th e velocit y of flo w i n ft/sec .
Solution
The formul a for determinin g k i s
Ait
Soil Perm eabilit y an d Seepag e 9 5
4
2
Q = 10 in
3
, A = 3.14 x — = 12.56 in.
2
7 1 O
i = — = — = 2 , t = 105 sec .
L 6
Therefore f c = = 3.79 x 10~
3
in./se c = 31.58 x 10~
5
ft/se c = 9960 ft/yea r
12.56x2x105
V elocity of flow = Id = 31.58 x 10~
5
x2 = 6.316 x 10~
4
ft/se c
Example 4 . 2
A san d sampl e o f 3 5 cm
2
cros s sectiona l are a an d 2 0 c m lon g wa s teste d i n a constan t hea d
permeameter. Unde r a head o f 60 cm, the discharge wa s 12 0 ml i n 6 min. The dr y weight of sand
used for the test was 1 120 g, and G
s
= 2.68. Determin e (a ) the hydraulic conductivity in cm/sec, (b)
the discharge velocity , and (c) the seepage velocity .
Solution
Use Eq. (4.9), k = —
hAt
where Q = 120 ml, t = 6 min, A = 35 cm
2
, L = 20 cm, and h = 60 cm. Substituting , we have
k = -= 3.174 x 10~
3
cm/se c
60x35x6x60
Discharge velocity, v = ki = 3.174 x 10~
3
x — = 9.52 x 10~
3
cm/se c
Seepage velocity v
s
W 112 0
Y G G
From Eq . (3.1 8a), Y
f
t ~
w s o r e
~ — ~~^sinc e y = 1 g/cm
3
l + e y
d
Substituting, e = — - - 1 = 0.675
1.6
0.675
= 0.403
l + e1 + 0.675
v 95 2 x l O~
3
Now, v = — =— '• = 2.36 x 10"
2
cm/se c
J
n 0.40 3
Example 4. 3
Calculate the value of A : of a sample of 2.36 in . height and 7.75 in
2
cross-sectional area , if a quantity
of wate r of 26.33 in
3
flows down i n 1 0 min unde r an effective constan t hea d o f 15.7 5 in . On oven
96 Chapt e r 4
drying, th e tes t specime n weighe d 1. 1 Ib . Assuming G
s
= 2.65, calculat e th e seepag e velocit y of
water during the test .
Solution
From Eq. (4.9), k = ^-= -
2633x236
-
= 0
.8x 10~
3
in./se c
hAt 15.75x7.75x10x6 0
Discharge velocity, v = ki = k— = 0. 8xl O~
3
x —
:
—= 5. 34xl O~
3
in./se c
L 2.3 6
W 1 1
Y
= —s- = -: -= 0.0601 lb/in
3
= 103. 9 lb/ft
3
d
V 7.75x2.3 6
Y G
FromEq. (3.18a) , e = ^-^—\
Yd
62.4x2.65
or e
- --- 1 = 0.59 1 5
103.9
°-
5915
=0.37 2
l + e1 + 0.5915
v 5.3 4 x l O~
3
Seepage velocity , v = — =— -= 14.35 x 10~
3
in./se c
6 s
n 0.37 2
Example 4 . 4
The hydrauli c conductivity of a soi l sampl e wa s determine d i n a soi l mechanic s laborator y b y
making us e o f a fallin g hea d permeameter . Th e dat a use d an d th e tes t result s obtaine d wer e a s
follows: diamete r of sample = 2.36 in, height of sample = 5.91 in , diameter of stand pipe = 0.79 in ,
initial hea d h
Q
= 17.72 in . fina l hea d h
l
= 11.81 in . Time elapse d = 1 min 4 5 sec . Determin e th e
hydraulic conductivit y i n ft/day .
Solution
The formul a for determinin g k i s [Eq . (4.13)]
, . , , .
k = -log,
0
— -where tis the elapsed time .
A • 3.14x0.79x0.7 9
0
.
1 A 4
,
2
Area of stand pipe, a = -= 34 x 10
4
f t ^
4x 12 x 12
Area of sample, A =
3
-
14x2
-
36x236
= 304
x 10~
4
f t
2
4x 12 x 12
Height of sample, L = (
17
-
72
~
1L81
)
= 0
4925 ft
1 7 72 118 1
Head, /z
n
= -^— = 1.477 ft, h, = —= 0.984 ft
0
1 2
]
1 2
Soil Perm eabilit y an d Seepag e 9 7
Elapsed time , t = 105 sec = = 12.15 x 10~
4
day s
60x60x24
2.3x34x10-4x0.4925 , 1.47 7
t f t f l j
k = x log = 18 ft/day
304 x 10-
4
x 12.15 x ID"
4
0.98 4
4.7 DI REC T DETERM I NATI ON OF k OF SOI L S I N PLACE B Y
PUMPING TEST
The most reliabl e information concerning th e permeability of a deposit o f coarse graine d material
below the wate r tabl e ca n usuall y be obtaine d by conductin g pumping test s i n the field. Although
such tests have their most extensive application i n connection wit h dam foundations, they may also
prove advisable on large bridge or building foundation jobs where the water table must be lowered.
The arrangement consist s of a test well and a series of observation wells . The test wel l is sunk
through the permeable stratum up to the impermeable layer . A well sunk into a water bearing stratum,
termed a n aquifer, an d tappin g fre e flowin g groun d wate r having a fre e groun d wate r tabl e unde r
atmospheric pressure , i s termed a gravity or unconfined well. A well sunk into an aquifer where the
ground wate r flow i s confined between tw o impermeabl e soi l layers , an d i s under pressur e greate r
than atmospheric , i s terme d a s artesian o r confined well. Observatio n well s ar e drille d a t variou s
distances fro m th e test o r pumping well along two straight lines, one oriented approximatel y i n the
direction of ground water flow and the other at right angles to it. A minimum of two observation wells
and their distances from the test well are needed. These well s are to be provided on one side of the test
well i n the direction of the ground water flow.
The test consists of pumping out water continuously at a uniform rate from th e test well until the
water level s i n the tes t an d observation well s remain stationary . When thi s condition i s achieve d th e
water pumped out of the well is equal to the inflow int o the well from th e surrounding strata. The water
levels in the observation wells and the rate of water pumped out of the well would provide the necessary
additional data for the determination of k.
As the water from th e test wel l is pumped out, a steady stat e wil l be attained when the water
pumped out will be equal to the inflow int o the well. At this stage the depth of water in the well will
remain constant . Th e drawdow n resulting du e t o pumpin g i s calle d th e cone o f depression. Th e
maximum drawdown D
Q
i s i n the test well . It decreases wit h the increas e i n the distanc e fro m th e
test well . The depressio n die s ou t graduall y and forms theoretically, a circl e aroun d th e tes t wel l
called th e circle of influence. Th e radiu s o f thi s circle, /?. , i s calle d th e radius of influence o f th e
depression cone .
Equation fo r k fo r a n U nconfined Aquifer
Figure 4.6 gives the arrangement of test and observation wells for an unconfined aquifer. Only two
observation well s at radial distance s o f r
{
an d r
2
from th e test wel l are shown. When th e inflow o f
water int o th e tes t wel l i s steady , th e depth s o f wate r i n thes e observatio n well s ar e h
{
an d h
2
respectively.
Let h be the depth of water at radial distance r. The area of the vertical cylindrical surfac e of
radius r and depth h through which water flows i s
A = Inrh
The hydraulic gradient is i = —
dr
98
Ground level Test wel l
Chapt er 4
Observation well s
Figure 4.6 Pum pin g t es t i n an un con fin ed aquife r
As per Darcy' s la w the rat e o f inflow into the wel l whe n the water level s i n the well s remai n
stationary i s
q = kiA
Substituting for A an d / the rat e of inflo w acros s th e cylindrica l surface i s
, dh^ ,
q - k — 2nrh
dr
Rearranging th e terms , w e have
dr Inkhdh
r q
The integra l of the equation withi n the boundary limit s is
dr Ink
r q .
hdh
(4.15)
The equatio n fo r k afte r integratio n and rearrangin g i s
k = -
(4.16)
Proceeding i n th e sam e wa y a s befor e anothe r equatio n fo r k i n term s o f r
Q
, h
Q
and R
{
ca n b e
established a s (referring t o Fig. 4.6)
Soil Perm eabilit y an d Seepag e 99
2.3 <? /?, -
* log-
1
-
(4.17)
If we write h
Q
= ( H- D
0
) in Eq. (4.17), where D
Q
is the dept h of maximum drawdown in the
test well , we have
2.3 q
-log^-
\ Y
0
Now fro m Eq . (4.18) , the maximum yiel d from th e wel l may be written as
_ 7rD
0
k( 2H-D
Q
) I
q
~ 2 3
(4.18)
(4.19)
Radius of Influence R^ Base d on experience , Sichard t (1930 ) gav e an equatio n for
estimating the radius of influence for the stabilized flo w condition as
/?. = 3000D
0
V & meter s (4.20)
where D
Q
= maximum drawdown i n meter s
k = hydraulic conductivity in m/sec
Equation fo r k i n a Confined Aquifer
Figure 4. 7 show s a confine d aquife r wit h th e tes t an d observatio n wells . Th e wate r i n th e
observation well s rise s abov e th e top of the aquifer due t o artesian pressure . Whe n pumpin g fro m
such a n artesian wel l two cases migh t arise. They are:
Case 1 . The wate r leve l i n th e tes t wel l migh t remai n abov e th e roo f leve l o f the aquife r at
steady flo w condition .
Observation well s
\
Piezometnc
level durin g pumpin g
Case 1 h
0
> H
0
Case 2h
0
<H
0
Impermeable
Impermeable stratu m
Figure 4.7 Pumpin g t es t i n con fin ed aquife r
100 Chapt e r 4
Case 2. The water level in the test wel l might fall belo w th e roof leve l of the aquifer at steady
flow condition .
If H
Q
i s the thickness of the confined aquifer and h
Q
is the dept h of water i n the test wel l at the
steady flo w condition Cas e 1 and Case 2 may be state d as —
Casel. Whe n/ z
0
>/ /
0
. Cas e 2. When/ i
Q
</ /
0
.
Case 1 . Whe n h
0
> H
0
In this case, the area of a vertical cylindrical surface of any radius r does not change, sinc e the dept h
of the water bearing strat a i s limited to the thicknes s H
Q
. Therefore, th e discharge surfac e are a i s
(4.21)
A • • • . dh . ~ . _ , ,
Again writin g i - — , the now equation a s per Darcy s law is
dr
dh_
dr °
The integratio n o f the equation afte r rearranging th e terms yield s
dr_ a r
— o r ( A
2
- / i
1
) = Tr7^1og,— (4-22 )
The equatio n fo r k i s
, . ,
k = -log
2 - A, ) r ,
Alternate Equation s
As befor e w e ca n writ e the following equation fo r determining k
23q r ,
k =
i u ti. —TT
log
~~ (4.24a )
- ^ '
, .
or k- -- -log—
1
-
, .
t
OF
2xH D
g
~r~ 27rH
Q
L >
0
r
0
Case 2 . Whe n h
0
< H
0
Under the condition whe n h
Q
is less than H
Q
, the flow pattern close t o the well is similar to that of an
unconfmed aquife r whereas a t distances farthe r from th e wel l th e flo w i s artesian. Muska t (1946 )
developed a n equation to determine the hydrauli c conductivity. The equatio n i s
*,-
— log —
L
Soil Perm eabilit y an d Seepag e 101
4.8 B OREH OL E PERM EAB I L I TY TEST S
Two types of test s may be carried out in auger hole s for determining k. They ar e
(a) Fallin g water level metho d
(b) Risin g water level method
Falling Water L eve l M etho d (case d hol e and soi l flush wit h bottom)
In thi s tes t auge r hole s ar e mad e i n th e fiel d tha t exten d belo w th e wate r tabl e level . Casin g i s
provided down to the bottom of the hole (Fig. 4.8(a)). The casing is filled with water which is then
allowed t o see p int o th e soil . Th e rat e o f dro p o f th e wate r leve l i n th e casin g i s observe d b y
measuring the dept h of the water surface below the top of the casing at 1 , 2 and 5 minutes after th e
start of the tes t and a t 5 minutes intervals thereafter. These observation s ar e made unti l the rate of
drop become s negligibl e o r unti l sufficien t reading s hav e bee n obtained . Th e coefficien t o f
permeability i s computed a s [Fig. 4.8(a) ]
2-3 nr
Q
H
{
k = —log — -
-f , ) f f ,
(4.26)
where, H
{
= piezometric hea d ait = t
l
,H
2
= piezometric head at t - t
2
-
Rising Wate r L eve l M etho d (case d hol e an d soi l flus h wit h bottom)
This method , most commonl y referred t o as the time-lag method, consists o f bailing the water out
of the casing and observing the rate of rise of the water level in the casing at intervals until the rise
in water level becomes negligible. The rate is observed by measuring the elapsed time and the depth
of the wate r surfac e below th e top of the casing. Th e interval s a t which the readings ar e require d
will var y somewhat wit h the permeabilit y o f the soil . Eq . (4.26) i s applicabl e fo r thi s case , [Fig.
4.8(b)]. A rising water level test shoul d always be followed by sounding the bottom o f the holes t o
determine whethe r the test created a quick condition.
HI a t t =
H a t t = t
(a) Fallin g water head method (b ) Rising water head method
Figure 4.8 Fallin g and risin g wat e r m et ho d o f det erm in in g k
102 Chapt e r 4
4 .9 APPROXI M AT E V AL U ES OF TH E H Y DRAU L I C CONDU CTI V I TY
OF SOI L S
The coefficient s o f permeabilit y o f soil s var y accordin g t o thei r type , textura l composition ,
structure, voi d rati o and other factors. Therefore, n o single value can be assigned t o a soil purely on
the basis of soi l type. The possibl e coefficients of permeability of some soil s are given in Table 4. 1
4.1 0 H Y DRAU L I C CONDUCTI V I TY I N STRATI FI ED L AY ERS OF SOI L S
Hydraulic conductivity of a disturbed sampl e may be different fro m tha t of the undisturbed sample
even thoug h th e voi d rati o i s the same . Thi s ma y b e due t o a change i n the structur e or due t o the
stratification of the undisturbed soil or a combination of both of these factors. I n nature we may fin d
fine graine d soil s havin g either flocculate d o r disperse d structures . Tw o fine-graine d soil s a t th e
same voi d ratio, on e dispersed an d the other flocculated, wil l exhibit different permeabilities .
Soils ma y b e stratifie d b y th e depositio n o f differen t material s i n layer s whic h posses s
different permeabilit y characteristics . I n suc h stratifie d soils engineer s desir e t o have the averag e
permeability eithe r i n th e horizonta l o r vertica l directions . Th e averag e permeabilit y ca n b e
computed i f the permeabilitie s o f each laye r are determine d i n the laboratory. Th e procedur e i s as
follows:
k
{
, k
2
, ..., k
n
= hydraulic conductivities of individua l strata o f soi l eithe r i n th e vertica l o r
horizontal direction .
z
r
Z
2 • • •
z
n
=
thickness of th e correspondin g strata .
k
h
= average hydrauli c conductivity parallel t o the bedding plane s (usuall y horizontal).
k
v
- averag e hydrauli c conductivity perpendicular t o the bedding plane s (usuall y vertical).
Flow i n the H orizonta l Directio n (Fig . 4 .9 )
When th e flo w i s i n th e horizonta l directio n th e hydrauli c gradien t / remains th e sam e fo r al l th e
layers. Let V j , v
2
, ... , v
n
be the discharge velocitie s in the corresponding strata . The n
Q = kiZ = ( v^j + v
2
z
2
+ - - - +v
n
z
n
) = ( k
[
iz
l
+k
2
iz
2
+ ••• +k
n
iz
n
)
Therefore,
'"+k
n
z
n
) (4.27 )
Table 4.1 Hydrauli c con duct ivit y o f s om e s oil s
(aft er Cas ag ran d e an d Fadum , 1939 )
k (cm/ sec )
10
1
t o 10
2
10
1
10-' t o IO-
4
io-
5
io-
6
IO-
7
t o IO-
9
Soils t yp e
Clean gravel s
Clean san d
Clean san d an d grave l mixture s
V ery fin e san d
Silt
Clay soil s
Dr a i na ge condi t i on s
Good
Good
Good
Poor
Poor
Practically imperviou s
Soil Perm eabilit y an d Seepag e 103
//?$><z><$xv<x><??t^^
Zi / i t V " ~ V , , I , *,
Z2 * 2 V 2, 1 ,
V 3,
Z
4 kT
v
4
, i ,
Figure 4.9 Flo w t hroug h s t rat ifie d l ayer s of s oi l
Flow i n the V ertica l Directio n
When flo w i s i n the vertical direction, the hydraulic gradients fo r each o f the layer s ar e different . Le t
these be denoted by i
r
z'
2
, . .., i
n
. Let h be the total loss of head as the water flows from the top layer to the
bottom through a distance ofZ. Th e average hydraulic gradient is h/Z. The principle of continuity of flow
requires that the downward velocity be the same i n each layer . Therefore ,
h
v = k
v
- = kj
l
=k
2
i
2
=--- = k
n
i
n
If / Zj , hj, ..., h
n
, are the head losse s i n each o f the layers , we have
or = z
ll
+z
22
+ -+z
nn
Solving the above equation s we have
Z
k =•
(4.28)
It shoul d be noted tha t in all stratified layer s of soil s th e horizontal permeabilit y i s generall y
greater tha n th e vertica l permeability . V arve d cla y soil s exhibi t th e characteristic s o f a layere d
system. However , loes s deposit s posses s vertica l permeabilit y greate r tha n th e horizonta l
permeability.
4 .1 1 EMPIRICA L CORREL ATI ONS FOR H Y DRAU L I C CONDU CTI V I T Y
G ranular Soil s
Some o f the factor s tha t affec t th e permeabilit y ar e interrelated suc h a s grai n size , voi d ratio , etc.
The smalle r th e grai n size , th e smalle r th e voids whic h lead s t o the reduced siz e o f flow channel s
and lower permeability .
The averag e velocit y o f flow i n a pore channe l fro m Eq . (4.2b) is
8// 32/ /
where d is the averag e diamete r o f a pore channe l equa l t o 2R.
(4.29)
104 Chapt er 4
Eq. (4.29 ) expresse s fo r a given hydrauli c gradient / , the velocit y of wate r i n a circular por e
channel i s proportional t o the squar e of the diameter of the por e channel . The averag e diamete r of
the voids i n a soi l a t a given porosity increase s practicall y i n proportion t o the grai n size, D
Extensive investigations of filte r sand s by Hazen (1892 ) led t o the equation
k( m/s) = CD
2
(4.30)
where D
e
is a characteristic effectiv e grain size which was determined t o be equal t o D
10
(10%size).
Fig. 4.10 gives a relationship between k and effective grai n size D
10
of granular soil whic h validates
Eq. (4.30) . Th e permeabilit y dat a approximate s a straigh t lin e wit h a slop e equa l t o 2 consisten t
with Eq. (4.30). These dat a indicat e an average valu e of C - 10~
2
wher e k is expressed i n m/s and
D
10
i n mm. According t o the dat a i n Fig. 4.10 , Eq . (4.30) ma y underestimat e o r overestimat e th e
permeability o f granular soils by a factor of about 2.
Further investigation s on filte r sand s wer e carrie d ou t by Kenne y et al. , (1984) . The y foun d
the effective grain siz e D
5
woul d be a better choice compare d t o D
}Q
. Fig. 4.11 gives relationship s
between D
5
and k. The sand they used in the investigation had a uniformity coefficient ranging from
1.04 to 12 .
H ydraulic Conductivit y a s a Functio n o f V oi d Rati o fo r G ranula r Soil s
Further analysi s of hydraulic conductivity in granular soil s based o n Hagen-Poiseuille' s Eq . (4.2b)
leads t o interestin g relationship s between k and voi d rati o e. Three type s o f relationships ma y be
expressed a s follows.
It can be shown that the hydrauli c conductivity k can be expressed a s
k = kF( e) (4.31)
Silt
Silty
Sand
Sand
Fine Medium | Coarse
Gravel
10-
10
~
o 10 "
-Q
o
CJ
o
10"
10"
10"
0.002
Hazen equation
J t = 1/10 0 £)?
0
m/sec
0.01 0.1 1
D
10
(mm )
10
Figure 4 .1 0 Haze n equat ion an d dat a relat in g hydrauli c con duct ivit y an d D
10
of
g ran ular s oil s (aft e r Louden , 1952)
Soil Perm eabilit y an d Seepag e 105
Sand
Fine Medium] Coars e
Sand
10"
T3
8 io - -
3
2
•o
X
10-
10'
C, ,= 1 - 3
10- 10- 10° IO
1
D
5
(mm)
Figure 4 .1 1 Influenc e o f g radat io n o n permeabilit y o n granular soil s
(aft er Kenne y e t al. , 1984)
where k = a soil constant depending on temperature and void ratio e .
F( e) ma y be expressed a s
F( e) =
o
2e
l + e
(4.32)
When e = 1, F( e) ~ 1 . Therefor e k represent s th e hydrauli c conductivit y corresponding t o voi d
ratio e - 1 . Since k i s assumed a s a constant, k is a function o f e only.
By substitutin g in F( e), th e limiting values, ;c = 0, x = 0.25, an d x = 0.5, we get
For J c = 0 ,
x = 0.25,
(4.33)
(4.34)
x = 0.50
F,(e) represents the geometric mea n of F.( e) an d F.(
The arithmeti c mean of the functions F^e) an d F
3
(e) is
(4.35)
= e
2
(4.36)
106 Chapt er 4
1000
u
o o o
V oid rati o functio n
Figure 4 .1 2 Relat ion s hi p bet wee n voi d rat i o an d perm eabilit y fo r coars e g rain ed
s oils
B est V alu e fo r x fo r Coars e G raine d Soil s
From laborator y tests determine k for various void ratios e of the sample. Then plot curves k versus
2e
2(1+x)
/(l + e) fo r value s of x = 0, 0.25, 0. 5 an d k versus e
2
. The plo t that fit s wel l gives th e best
value of x. I t has been foun d fro m experimenta l results that the functio n
2e
3
l + e
(4.37)
gives better agreement than the other functions. However, the function F
4
( e) = e
2
is sometimes preferred
because o f its simplicity and it s fair degre e o f agreement wit h the experimental data . Fig. 4.12 present
experimental data in the for m o f k versus functions o f e.
Figure 4.1 3 I n situ perm eabilit y o f s of t cl ay s i n relat ion t o in it ia l voi d rat io , e
o
; clay
fract ion ; CF; an d act ivit y A (Aft e r M es r i e t al. , 1994)
Soil Perm eabilit y an d Seepag e 107
3.5
3.0
2.5
•|2.0
1.0
0.5
0
Clay
O Batisco n
A Berthiervill e
D St . Hilair e
V V osb y
• Bosto n blu e
10- 10" 10
,-8
Figure 4.1 4 Res ult s o f fallin g -hea d an d con s t an t -head perm eabilit y t es t s on
undisturbed s am ples o f s of t clay s (T erzag hi , Pec k an d Mesri, 1996 )
Fine G raine d Soil s
Laboratory experiment s hav e shown that hydrauli c conductivit y of ver y fin e graine d soil s ar e no t
strictly a function o f voi d rati o sinc e ther e i s a rapid decrease i n the value of k for clays belo w th e
plastic limit . This i s mostly due to the much higher viscosity of water i n the normal channel s which
results from th e fact that a considerable portio n of water is exposed t o large molecular attractions by
the closely adjacen t soli d matter . I t also depends upon the fabric of clays especiall y thos e o f marine
origin whic h ar e ofte n flocculated . Fig . 4.13 show s tha t th e hydrauli c conductivit y i n th e vertica l
direction, a t in situ void ratio e
Q
, is correlated wit h clay fraction (CF) finer than 0.002 mm an d with
the activity A ( = I
p
/CF).
Consolidation o f sof t clay s may involv e a significant decrease i n voi d rati o an d therefor e o f
permeability. The relationships between e and k (log-scale) fo r a number of soft clays ar e shown in
Fig. 4.14 (Terzaghi, Peck, an d Mesri 1996) .
Example 4. 5
A pumping test wa s carried ou t for determining th e hydraulic conductivity of soi l i n place. A well
of diamete r 4 0 c m wa s drille d dow n t o a n impermeabl e stratum . The dept h o f wate r abov e th e
bearing stratu m wa s 8 m. The yiel d fro m th e wel l wa s 4 mV mi n a t a steady drawdow n of 4. 5 m.
Determine th e hydrauli c conductivity of the soi l i n m/day i f the observed radiu s of influence was
150m.
Solution
The formul a for determining k is [Eq . (4.18) ]
k =
2.3 q
xD
0
( 2H-D
0
) r
0
q = 4 m
3
/min = 4 x 60 x 24 m
3
/day
D
0
= 4.5 m, H = 8 m, R . = 150 m, r
Q
= 0.2 m
108 Chapt er 4
k =
2 . 3x 4x 60x 2 4
3. 14x4. 5(2x8-4. 5) 0. 2
log = 234.4 m/da y
Example 4 . 6
A pumping test was made i n pervious gravels and sands extending to a depth o f 50 ft, where a bed
of cla y wa s encountered . The norma l groun d wate r leve l wa s a t th e groun d surface . Observatio n
wells were locate d a t distances of 1 0 and 25 ft from th e pumping well. At a discharge o f 761 ft
3
pe r
minut e fro m th e pumpin g well , a stead y stat e wa s attaine d i n abou t 2 4 hr . Th e draw-dow n a t a
distance o f 1 0 ft wa s 5. 5 f t an d a t 2 5 f t wa s 1.2 1 ft. Comput e th e hydrauli c conductivity in ft/sec.
Solution
Use Eq. (4.16) wher e
where = —
60
= 12.68 3 ft
3
/sec
= 10 ft, r = 25 ft , h = 50 -1.21 = 48.79 ft , h = 50 - 5. 5 = 44.5 ft
k =
2.3x12.683 25 ,
n o 1 A
_
r /
log — = 9.2 x10
J
ft/sec .
3.14(48.79
2
-44.5
2
) 1 0
Example 4 . 7
A fiel d pumpin g tes t wa s conducte d fro m a n aquife r o f sand y soi l o f 4 m thicknes s confine d
between two impervious strata. When equilibrium was established, 9 0 liter s o f water wa s pumpe d
Test wel l
Observation well s
1 2
Impermeable stratum
} i
y
.T
-^ — — r , = 3 m
^^
•^-
,..1 —" •
.
T
/i, =2. 1 m
1 / / /
^
— T
T
/i
?
= 2.7 m
1
Confined aquife r \
Impermeable
Fi gure Ex . 4 .7
Soil Perm eabilit y an d Seepag e 10 9
out per hour . The wate r elevation i n an observation wel l 3. 0 m away from th e tes t wel l was 2. 1 m
and another 6.0 m away was 2.7 m from the roof level of the impervious stratum of the aquifer. Find
the value of k of the soi l i n m/sec. (Fig. Ex. 4.7)
Solution
Use Eq. (4.24a )
,
23
<7 ,
r
2
k = -- -log—
q = 90 x 10
3
cm
3
/hr = 25 x KT
6
m
3
/sec
2. 3x25xl O~
6
, 6
11/ 1 0 i n
_
6
.
k = -log— = 1.148 x 10
6
m/se c
2x3.14x4(2.7-2.1) 3
Example 4. 8
Calculate th e yiel d pe r hou r fro m a wel l drive n int o a confine d aquifer . Th e followin g dat a ar e
available:
height of original piezometri c leve l from th e bed o f the aquifer , H = 29.53 ft,
thickness of aquifer, H
a
- 16.4 1 ft,
the dept h o f water i n the wel l at steady state , h
Q
= 18.05 ft ,
hydraulic conductivity of soi l = 0.079 ft/min ,
radius of well , r
Q
= 3.94 in . (0.3283 ft), radius of influence, R. = 574.2 ft.
Solution
Since h
Q
is greater tha n H
Q
th e equation for q (refer t o Fig 4.7) i s Eq. (4.24b )
where k = 0.079 ft/min = 4.74 ft/h r
2x3.14x4.74x16.41(29.53-18.05) ^
1 Mr
,
n
^^
n
Now q = -- --=75 1.87 ft
3
/hour « 752 ft
3
/hour
2.3 log(574.2 70.3283)
Example 4. 9
A sand deposit contain s three distinct horizontal layers of equal thickness (Fig. 4.9) . The hydraulic
conductivity o f th e uppe r an d lowe r layer s i s 10~
3
cm/sec an d tha t o f th e middl e i s 10~
2
cm/sec .
What ar e the equivalent values of the horizontal an d vertical hydrauli c conductivities o f the thre e
layers, and what is their ratio ?
Solution
Horizontal flow
~( ki+k
2
+k
3
) sinc e z \ = Z
2
= £3
110 Chapt e r 4
k
h
= -(10-
3
+10-2 + 10-
3
) = -(2x 10~
3
+10-2) = 4x 10-
3
cm/se c
Vertical flow
Z 3 3
i Li l i l _ L _ L _ L . 1 _ L
3kik, 3 x l O~
3
x l O~
2
1 / l < a i n
_
3
.
_ _ = 1.43 x 10 cm/se c
2k
2
+ ki
k
h
_ 4 x l O~
3
k
v
1.43x10 "
= 2.8
Example 4 .1 0
The followin g detail s refe r t o a test t o determine th e valu e of A ; of a soi l sample : sampl e thicknes s
= 2. 5 cm, diameter of soil sample = 7. 5cm, diameter of stand pipe = 10mm, initial head of water in
the stand pipe =100 cm, water level in the stand pipe afte r 3 h 20 min = 80 cm. Determine th e value
of k i f e = 0.75. What i s the valu e of k of the same soi l a t a void rati o e = 0.90?
Solution
Use Eq . (4 . 1 3) where, k = ' lo g
2
'
3aL
4
(I)
2
-0.785 cm
2
3 14
A = — (7.5)
2
= 44. 1 6cm
2
t= 1200 0 sec
By substitutin g the valu e of k for e
{
= 0.75
, , 2.3x0.785x2. 5 , 10 0
rtcv
^
i n
, ,
k = k,= -x log -= 0.826 x 10~
6
cm/se c
1
44.16x1200 0 8 0
For determining k at any other voi d ratio, us e Eq . (4.35 )
i
e
l
Now, k
2
= --x — x k
{
e
For e
2
= 0.90
1.75 ( 0. 9 V
=
190
X
"
X
°'
826 X 10 = l3146 X
Soil Perm eabilit y an d Seepag e 111
Example 4 .1 1
In a fal l i n g hea d permeameter , th e sampl e used i s 2 0 c m lon g havin g a cross-sectiona l are a
of 2 4 cm
2
. Calculat e th e tim e require d fo r a dro p o f hea d fro m 2 5 t o 1 2 c m i f th e cross -
sectional are a o f th e stan d pip e i s 2 cm
2
. Th e sampl e o f soi l i s mad e o f thre e layers . Th e
thickness o f th e firs t laye r fro m th e to p i s 8 cm an d ha s a value of k
{
= 2 x 10"
4
cm/sec, th e
second laye r o f thicknes s 8 cm ha s k
2
= 5 x 10~
4
cm/se c an d th e botto m laye r o f thickness
4 c m ha s &
3
= 7 x 10~
4
cm/sec . Assume tha t th e flo w i s takin g plac e perpendicula r t o th e
layers (Fig. Ex. 4.11) .
Solution
Use Eq. (4.28)
k =
20
- + —+ •
O O < 4
_l _ ______^^^^^ _ I _
2 x l O~
4
5xl O-
4
7xl O-
4
= 3. 24xl O~
4
cm/se c
Now fro m Eq . (4.13) ,
2.3aL , h
n
log—
or
2.3aL, h
Q
2. 3x2x2 0 , 2 5
log— = -log —
Ak /i , 24x3. 24xl O~
4
1 2
= 3771 sec = 62.9 minute s
8cm Laye r 1 I ^
=
2 x 10"
4
cm/sec
8cm Laye r 2 1 ^
2
= 5 x 10"
1
cm/sec
£
3
= 7xl 0^cm/ s ec
Figure Ex . 4 .11
Example 4 .1 2
The dat a given below relate t o two falling hea d permeameter test s performed o n two different soi l
samples:
(a) stan d pip e are a = 4 cm
2
, (b ) sampl e are a = 2 8 cm
2
, (c ) sampl e heigh t = 5 cm ,
(d) initial head i n the stand pipe =100 cm, (e ) final hea d = 20 cm, (f ) time required for the fal l o f
water level in test 1 , t = 500 sec, (g) for test 2, t = 15 sec.
Determine th e value s of k for eac h o f the samples . If thes e two type s of soil s form adjacent
layers i n a natura l stat e wit h flo w (a ) i n th e horizonta l direction , an d (b ) flo w i n th e vertica l
112 Chapt e r 4
direction, determine th e equivalent permeability for both the cases b y assumin g that the thickness
of eac h laye r i s equa l t o 15 0 cm.
Solution
Use Eq. (4.13)
. 23aL h
k, =— -lo g = 2.3x10
3
cm/se c
1
28x50 0 2 0
For test 2
2 . 3x 4x 5 , 10 0
28x15 2 0
F/ovv i n the horizontal direction
Use Eq. (4.27 )
3
= 76. 7xl O~
3
cm/se c
= — -(2. 3 x 150 + 76.7 x 150) x ID"
3
=39. 5xl Q-
3
cm/se c
,j L /w
F/ow in the vertical direction
Use Eq. (4.28 )
Z
2
15 0 15 0
fcT 2.3x10- 3
+
76.7x10-3
300
= 4.46 x 10"
3
cm/se c
4 .1 2 H Y DRAU L I C CONDU CTI V I TY OF ROCKS B Y PACKER METH OD
Packers ar e primaril y use d i n bor e hole s fo r testin g th e permeabilit y o f rock s unde r applie d
pressures. Th e apparatu s used fo r th e pressur e tes t i s comprised o f a wate r pump , a manually
adjusted automati c pressure relie f valve , pressure gage , a wate r mete r an d a packer assembly .
The packe r assembl y consist s o f a syste m o f pipin g t o whic h tw o expandabl e cylindrica l
rubber sleeves , calle d packers , ar e attached . Th e packer s whic h provide a mean s o f sealin g a
limited sectio n o f bor e hol e fo r testing , shoul d hav e a lengt h fiv e time s th e diamete r o f th e
hole. The y ma y b e o f th e pneumat i cal l y o r mechanicall y expandabl e type . Th e forme r ar e
preferred sinc e the y adap t t o a n oversize d hol e wherea s th e latte r ma y not . However , whe n
pneumatic packer s ar e used , th e tes t apparatu s mus t als o includ e a n ai r o r wate r suppl y
connected, throug h a pressur e gage , t o th e packer s b y mean s o f a highe r pressur e hose . Th e
piping o f a packe r assembl y i s designe d t o permi t testin g o f eithe r th e portio n o f th e hol e
between th e packer s o r the portion below th e lower packer . The packer s ar e usuall y set 50, 150
or 30 0 c m apart . Th e wide r spacing s ar e use d fo r roc k whic h i s mor e uni form . Th e shor t
spacing i s used t o test i ndi vi dua l joints whic h may be the cause o f high wate r loss i n otherwise
tight strata .
Soil Perm eabilit y an d Seepag e 11 3
Two types of packer method s ar e used for testing of permeability. They are:
1. Singl e packer method.
2. Doubl e packer method .
The singl e packe r metho d i s usefu l wher e th e ful l lengt h o f th e bor e hol e canno t stan d
uncased/ungrouted i n sof t rocks , such as sof t san d stone, clay shal e or due t o the highl y fractured
and sheared natur e of the rocks, or where it is considered necessary to have permeability values side
by sid e wit h drilling. Where th e rocks ar e sound and the ful l lengt h of the hol e ca n stan d without
casing/grouting, the double packer method may be adopted. The disadvantage of the double packer
method i s tha t leakag e throug h the lowe r packe r ca n g o unnoticed and lea d t o overestimation of
water loss.
Single Packe r M etho d
The method used for performing water percolation test s i n a section of a drilled hole using a single
packer i s show n i n Fig . 4.15a. I n thi s metho d th e hol e shoul d b e drille d t o a particula r dept h
desirable for the test. The core barrel shoul d then be removed and the hole cleaned wit h water. The
packer shoul d be fixe d a t th e desire d leve l abov e th e botto m o f th e hol e an d th e tes t performed .
Water should be pumped into the section unde r pressure. Each pressure shoul d be maintained until
the readings of water intake at intervals of 5 min show a nearly constant reading of water intake for
one particular pressure. The constant rate of water intake should be noted. After performing the test
the entire assembly shoul d be removed. The drilling should then proceed fo r the next test section.
Double Packe r M etho d
In thi s method the hole i s first drille d t o the final dept h and cleaned. The packer assembl y may be
fixed a t any desired tes t section a s shown in Fig. 4.15b.
Both packers ar e then expanded and water under pressure is introduced into the hole between
the packers. The test s ar e conducted as before.
Regardless o f which procedure is used, a minimum of three pressures shoul d be used for each
section tested. The magnitude of these pressures ar e commonly 100, 200 and 300 kPa. (1,2 and 3
kg/cm
2
) abov e th e natura l piezometri c level . Howeve r i n n o case shoul d th e exces s pressur e b e
greater tha n abou t 2 0 kP a pe r mete r o f soi l an d roc k overburde n abov e th e uppe r packer . Th e
limitation i s imposed t o insure against possible heav y damage t o the foundation.
c
E
=" / Groun d surface \
&$^&\ s///$\
, —Casing
/ ~T~
/-
Packer
Tes t
n *sectio n
Test
section
.J
c
§
§
^
1
V
1
^
/S7XXN
V
N
Packe r
^ Perforated pipe
^ Packe r
-Bottom of the hole—'
(a) (b )
Figure 4 .1 5 T es t s ect ion s fo r s ingl e an d doubl e pack e r m et hod s
114 Chapt e r 4
The formulae use d t o compute the permeability from pressure tes t dat a ar e (from US Burea u
of Reclamation , 1968 )
g
' °
r
-
r
» <438a )
k- -sinrr
1
-for 1 0 r
n
>L>r
n f A
TC M
27J L H 2r
Q
° ° (4.3Sb )
where
k = hydraulic conductivity
q = constant rat e o f flow int o the hol e
L = lengt h o f the tes t sectio n
H = differential hea d o n th e tes t sectio n
r
Q
= radius o f the bor e hol e
4 .1 3 SEEPAG E
The interactio n between soil s an d percolating wate r ha s a n important influence on:
1 . Th e desig n o f foundations and eart h slopes ,
2. Th e quantit y of water that wil l b e los t b y percolation throug h a dam o r it s subsoil .
Foundation failure s due t o 'piping' ar e quite common. Pipin g is a phenomenon b y whic h the
soil o n th e downstrea m side s o f some hydrauli c structures get lifte d u p du e t o exces s pressur e o f
water. The pressur e tha t is exerted on the soil due to the seepage of water is called th e seepage force
or pressure. In the stabilit y of slopes, the seepage forc e is a very important factor. Shear strength s of
soils ar e reduce d du e t o th e developmen t o f neutra l stres s o r por e pressures . A detaile d
understanding o f th e hydrauli c condition s i s therefor e essentia l fo r a satisfactor y desig n o f
structures.
The computatio n o f seepage loss under or through a dam, th e uplif t pressure s cause d b y the
water on the base of a concrete da m and the effect o f seepage on the stabilit y of earth slope s ca n be
studied by constructin g flo w nets .
Flow Ne t
A flow net for an isometric mediu m is a network of flow lines and equipotential lines intersecting at
right angles t o each other .
The pat h whic h a particl e of wate r follows i n it s course o f seepage throug h a saturate d soi l
mass is called a flow line.
Equipotential lines are lines that intersect the flow line s at right angles. At all points along an
equipotential line , th e wate r woul d ris e i n piezometri c tube s t o th e sam e elevatio n know n a s th e
piezometric head . Fig. 4.16 give s a typica l example o f a flow ne t for th e flo w belo w a shee t pil e
wall. The head o f water on the upstream sid e o f the sheet pil e i s h
t
and on the downstream sid e h
d
.
The hea d los t a s the wate r flows from th e upstream t o the downstream sid e i s h.
4 .1 4 L APL AC E EQU ATI ON
Figure 4. 16(a) illustrate s th e flo w o f wate r alon g curve d line s whic h ar e paralle l t o th e sectio n
shown. The figur e represent s a section throug h an impermeable diaphrag m extendin g t o a depth D
below th e horizonta l surfac e of a homogeneous stratu m of soi l o f dept h H .
Soil Perm eabilit y an d Seepag e 1 1 5
It i s assume d tha t th e differenc e h betwee n th e wate r level s o n th e tw o side s o f th e
diaphragm i s constant . The wate r enter s th e soi l o n the upstream sid e o f the diaphragm, flows
in a downward directio n an d rise s o n the downstrea m sid e towards th e surface .
Consider a prismatic element P shown shaded in Fig. 4.16(a) which is shown on a larger scale
in (b). The element i s a parallelepiped with sides dx, dy and dz. The x and z directions are as shown
in the figur e an d the y direction is normal to the section. The velocit y v of water which is tangential
to the strea m lin e can be resolved int o components v
x
and v
z
in the x an d z directions respectively.
Let,
dh
i
x
= ——, th e hydraulic gradient i n the horizontal direction .
dh
i
z
= —— , th e hydraulic gradient i n the vertical direction .
oz
k
x
= hydraulic conductivity in the horizontal direction.
k
z
= hydraulic conductivity in the vertical direction.
If we assume that the water and soil are perfectly incompressible, an d the flow i s steady, then
the quantit y of water that enters the element mus t be equal t o the quantity that leaves it .
The quantit y of water that enters th e side ab = v
x
dzdy
dv
The quantit y of water that leaves th e side cd =
v
+— —dx dzdy
x
dx.
The quantit y of water that enters th e side be = v
z
dxdy
The quantit y of water that leaves the side ad = v
z
+ —- dz dxdy
dz
Therefore, w e have the equation,
v
x
dzdy + v dxdy = v
x
+ —— dx dzdy + v + —- dz dxdy
dx dz
After simplifying , w e obtain,
dv
r
dv
-^ + -^ = 0 (4.39 )
ox oz
Equation (4.39 ) expresse s th e necessar y conditio n fo r continuit y o f flow . Accordin g t o
Darcy's La w we may write,
dh dh
v
v
= -k
r
—^, v = -k, T-
x x
ox
z l
oz
Substituting for v
x
and v
z
we obtain,
d dh d dh
dx * dx dz
z
dz
d
2
h d
2
h
ork
r
-^- + k
7
-^- = Q(4.40 )
x
ox
2
'
z
oz
116 Chapt e r 4
When k = k
x
, i.e., when the permeability is the same i n all directions, Eq . (4.40) reduce s t o
d
2
h d
2
h
ax.
2
- oz
= 0 (4.41 )
Eq. (4.41) i s the L aplace Equation fo r homogeneous soil . It says tha t the change o f gradient
in the jc-direction plus the change of gradient in the z-direction is zero. The solution of this equation
gives a family o f curves meeting at right angles t o each other. One family of these curves represent s
flow line s an d the other equipotentia l lines. For the simpl e case shown i n Fig. 4.16, the flo w line s
represent a famil y o f semi-ellipses an d the equipotential lines semi-hyperbolas .
Anisotropic Soi l
Soils i n natur e d o posses s permeabilitie s whic h ar e differen t i n th e horizonta l an d vertica l
directions. Th e permeabilit y i n the horizontal direction i s greater tha n in the vertical direction in
sedimentary deposit s an d in most eart h embankments . In loess deposit s th e vertical permeabilit y
is greater tha n the horizonta l permeability . The stud y of flow net s woul d be o f littl e value i f thi s
variation i n th e permeabilit y i s no t take n int o account . Eq . (4.40) applie s fo r a soi l mas s wher e
anisotropy exists . Thi s equatio n may be written in the for m
d
2
h d
2
h
^ &
2
^ (4-42 )
If we consider a new coordinate variabl e x
c
measured i n the same direction as x multiplied by
a constant , expressed b y
(4.43)
Eq. (4.42 ) ma y be written as
d
2
h d
2
h
~d^
+
~d^= °(4-44 )
c
Now Eq. (4.44) i s a L aplace equation in the coordinates x
c
and z. This equatio n indicates tha t
a cross-sectio n throug h a n anisotropi c soi l ca n b e transforme d t o a n imaginar y sectio n whic h
possesses th e same permeabilit y in all directions. The transformation of the section ca n be effected
as per Eq. (4.43) b y multiplying the ^-coordinates by J k
z
/k^ an d keeping th e z-coordinates a t the
natural scale. The flow net can be sketched on this transformed section . The permeability to be used
with the transformed sectio n i s
(4.45)
4 .1 5 FL O W NE T CONSTRU CTI O N
Properties o f a Flo w Ne t
The propertie s o f a flow net can be expressed a s given below:
1. Flo w an d equipotential lines ar e smoot h curves .
2. Flo w line s and equipotentia l lines meet at right angles t o each other .
Soil Perm eabilit y an d Seepag e 11 7
3. N o two flow line s cross eac h other .
4. N o two flow o r equipotential lines start fro m th e same point.
B oundary Condition s
Flow o f wate r throug h eart h masse s i s i n genera l thre e dimensional . Sinc e th e analysi s o f
three-dimensional flo w i s too complicated, th e flow problems ar e solved on the assumption that the
flow i s two-dimensional. All flow line s i n such a case ar e parallel t o the plane of the figure , an d the
condition i s therefore known as two-dimensional flow. Al l flow studie s dealt wit h herei n ar e for the
steady stat e case . Th e expressio n fo r boundar y condition s consist s o f statement s o f hea d o r flo w
conditions at all boundary points. The boundary conditions are generally four i n number though there
are only three in some cases. The boundary conditions for the case shown in Fig. 4. 16 are as follows :
1 .Lin e ab is a boundary equipotential line along which the head i s h
(
2. Th e line along the sheet pil e wal l is a flow boundary
3. Th e line xy i s a boundary equipotentia l line along which the head i s equal t o h
d
4. Th e line m n is a flow boundar y (at depth H below bed level).
If we consider an y flow line , say, p
1
p
2
p
3
i n Fig. 4.16, the potential head a t p
{
i s h
(
an d at
p
3
i s h
d
. The total head lost as the water flows along the line is h which is the difference between the
upstream an d downstrea m head s o f water . Th e hea d los t a s th e wate r flow s fro m p
l
t o equi -
potential line k is Ah which is the difference between the heads shown by the piezometers. Thi s loss
of hea d Ah i s a fraction o f the total head lost .
Flow Ne t Constructio n
Flow net s are constructed i n such a way as to keep the ratio of the side s o f each bloc k bounded by
two flow lines and two equipotential lines a constant. If all the sides of one such block are equal, then
the flow ne t must consist o f squares. The squar e block referred t o here does no t constitute a square
according to the strict meaning of the word, it only means that the average width of the square blocks
are equal. For example, i n Fig. 4.16, the width a
l
o f block 1 is equal t o its length b
}
.
The are a bounded by any two neighboring flow line s is called a/low channel. If the flow ne t
is constructed i n such a way that the ratio al b remains the same for all blocks, the n it can be shown
that there i s the same quantit y of seepage in each flo w channel . In order t o show this consider two
blocks 1 and 2 i n on e flo w channe l an d anothe r bloc k 3 i n anothe r flo w channe l a s show n i n
Fig. 4.16. Block 3 is chosen i n such a way that it lies within the same equipotential lines that bound
the block 2. Darcy's la w for the discharge through any block such as 1 per unit length of the section
may b e written as
Ah a
Aq = kia = — a = kAh —
b b
where Ah represents the head loss in crossing the block. The expressions in this form for each of the
three blocks unde r consideration ar e
Aq
{
= kAh—, Aq
2
= kAh
2

b
\
b
2
In th e abov e equatio n th e valu e o f hydrauli c conductivit y k remain s th e sam e fo r al l th e
blocks. I f the blocks ar e al l squares then
b
2
118 Chapt er 4
Piezometer tube s
(a) Flow ne t
Piezometer
Flow line
Equipotential
line
(b) Flow through a prismatic elemen t
Figure 4 .1 6 Flo w t hroug h a hom og en eous s t rat u m o f s oi l
Since blocks 1 and 2 are i n the same flow channel , we have &q
l
= Ag
2
. Since blocks 2 and 3
are withi n the same equipotential lines we have A/z
2
= A/?
3
. If these equations are inserted we obtain
the following relationship:
A# j = Ag
2
an d A/ZJ = A/z
2
This proves tha t the same quantity flows throug h each block and there is the same head drop
in crossin g eac h bloc k i f al l th e block s ar e square s or posses s th e sam e rati o alb. Flo w net s ar e
constructed by keeping the ratio alb the same in al l figures. Squar e flow net s are generall y used in
practice as this is easier t o construct.
Soil Perm eabilit y an d Seepag e 11 9
There ar e man y method s tha t ar e i n us e fo r th e constructio n o f flo w nets . Som e o f th e
important methods ar e
1. Analytica l method,
2. Electrica l analo g method ,
3. Scale d mode l method,
4. Graphica l method .
The analytica l method , base d o n th e Laplac e equatio n althoug h rigorousl y precise , i s no t
universally applicabl e i n al l case s becaus e o f th e complexit y o f th e proble m involved . Th e
mathematics involved even in some elementar y cases is beyond the comprehension of many design
engineers. Althoug h thi s approac h i s sometime s usefu l i n th e checkin g o f othe r methods , i t i s
largely of academi c interest .
The electrica l analog y metho d ha s bee n extensivel y made us e of i n man y important desig n
problems. However , i n mos t o f th e case s i n th e fiel d o f soi l mechanic s wher e th e estimatio n o f
seepage flow s an d pressures ar e generall y required , a mor e simpl e metho d suc h a s the graphica l
method i s preferred .
Scaled model s ar e ver y usefu l t o solv e seepag e flo w problems . Soi l model s ca n b e
constructed t o depict flo w o f water below concret e dam s or through eart h dams . These models ar e
very useful t o demonstrate th e fundamental s of flui d flow, bu t their use i n other respect s i s limited
because o f the large amoun t of time and effor t require d t o construct suc h models .
The graphica l metho d developed b y Forchheimer (1930 ) has been found to be very useful i n
solving complicated flo w problems . A. Casagrande (1937 ) improve d thi s method by incorporating
many suggestions . Th e mai n drawback o f thi s method i s that a good dea l o f practice an d aptitude
are essential to produce a satisfactory flow net . In spite of these drawbacks, the graphical metho d is
quite popular amon g engineers .
G raphical M etho d
The usual procedure fo r obtaining flow net s is a graphical, trial sketching method, sometimes calle d
the Forchheimer Solution . This method of obtaining flow net s is the quickest an d the most practica l
of al l th e availabl e methods . A . Casagrande (1937 ) ha s offere d man y suggestion s t o th e beginne r
who i s interested i n flow ne t construction. Some o f his suggestions are summarized below :
1. Stud y carefully the flo w ne t patter n of well-constructed flo w nets .
2. Tr y to reproduce th e same flo w net s without seeing them.
3. A s a first trial, use not more than four to five flow channels. Too many flow channel s would
confuse th e issue.
4. Follo w th e principle o f 'whol e to part', i.e. , one has to watch the appearance o f the entir e
flow ne t an d whe n once th e whol e ne t i s foun d approximatel y correct , finishin g touche s
can be given t o the details .
5. Al l flo w an d equipotentia l line s shoul d b e smoot h an d ther e shoul d no t b e an y shar p
transitions between straigh t and curved lines .
The abov e suggestions , thoug h quit e usefu l fo r drawin g flo w nets , ar e no t sufficien t fo r a
beginner. I n orde r t o overcome thi s problem, Taylo r (1948 ) propose d a procedur e know n a s the
procedure by explici t trials. Some o f the salient features of this procedure ar e given below:
1. A s a first step i n the explici t tria l method, on e trial flow lin e or one trial equipotential line
is sketched adjacen t to a boundary flow lin e or boundary equipotential .
2. Afte r choosin g th e firs t tria l lin e (say i t i s a flow line) , the flow pat h between th e line and
the boundary flow lin e i s divided int o a number of squares by drawing equipotential lines.
120 Chapt e r 4
These equipotential line s are extended t o meet the bottom flo w lin e at right angles keeping
in vie w that the lines drawn shoul d be smoot h withou t any abrupt transitions.
3. Th e remainin g flo w line s ar e next drawn, adhering rigorously t o squar e figures.
4. I f th e firs t tria l i s chose n property , th e ne t draw n satisfie s al l th e necessar y conditions .
Otherwise, th e las t drawn flo w lin e wil l cros s th e botto m boundar y flo w line , indicating
that the trial line chosen i s incorrect and needs modification.
5. I n such a case, a second tria l line shoul d be chosen an d the procedure repeated .
A typica l exampl e o f a flo w ne t unde r a shee t pil e wal l i s give n i n Fig . 4.16. I t shoul d b e
understood tha t the numbe r of flo w channel s wil l b e a n intege r onl y by chance . Tha t means , th e
bottom flo w lin e sketched migh t no t produc e ful l square s wit h the botto m boundar y flo w line . In
such a case the bottom flow channel will be a fraction of a full flo w channel . It should also be noted
that the figur e formed b y the first sketche d flow line with the last equipotential lin e in the region is
of irregular form. This figur e i s called a singular square. The basic requirement for such squares, as
for al l th e othe r squares , i s tha t continuou s sub-division o f th e figure s giv e a n approac h t o tru e
squares. Suc h singula r square s ar e forme d a t th e tip s o f shee t pil e wall s also . Square s mus t b e
thought of as vali d only where the Laplace equation applies . The Laplace equatio n applies t o soil s
which are homogeneous an d isotropic. When the soil is anisotropic, the flow ne t should be sketched
as befor e o n th e transforme d section . Th e transforme d sectio n ca n b e obtaine d fro m th e natural
section explaine d earlier .
4 .1 6 DETERM I NATI O N OF QU ANTI TY OF SEEPAG E
Flow net s ar e usefu l fo r determinin g th e quantit y of seepag e throug h a section . Th e quantit y of
seepage q is calculated per unit length of the section. The flow throug h any square can be written as
&q = kkh (4.46 )
Let th e numbe r o f flo w channe l an d equipotentia l drop s i n a sectio n b e N , an d N
d
,
respectively. Sinc e al l drops ar e equal, we can writ e
h
Since the discharge i n each flow channel is the same we can write ,
q = N
f
kq
Substituting for Ag and A/I , we have
N
f
1 = M—-(4.47 )
" d
Eq. (4.47) ca n also be used to compute the seepage through anisotropic section s by writing k
e
in plac e o f k . A s pe r Eq . (4.45), k
e
i s equa l t o J k
x
k
z
, wher e k
x
an d k
z
ar e th e hydrauli c
conductivities i n the x and z directions, respectively . The validit y of this relationship can be prove d
as follows. Conside r a figure bounde d by flow and equipotential line s i n which the flow is parallel
to the x direction. I n Fig. 4.1 7 the figure in question i s drawn t o a transformed scal e in (b) and the
same to the natural scale in (a).
In Fig. 4. 17(b) the permeability has the effective valu e k
e
in both the x and z directions and the
flow through the squar e according t o Eq. (4.46), i s
kkh (4.48 )
Soil Perm eabilit y an d Seepag e 121
T
a
\
\
k "
Flow »
kx
lines |
(a) Natural sectio n (b ) Transformed sectio n
Figure 4.17 Flo w t hr oug h ani sot ropi c soi l
In Fig. 4.17(a) the hydraulic conductivity k
x
i n the horizontal sectio n mus t appl y becaus e th e
flow i s horizontal and the sketch i s to the natural scale. The flow equatio n is, therefore ,
= k iA = k
(4.49)
Ref. Number s
3 2
Sheet
pile wal l
« »
o/2
(a) Natural sectio n
Multiplying factor = V1/4=1/2
(b) Transformed sectio n
Figure 4.18 Flowne t i n anisotropi c soi l
122 Chapt e r 4
Equating Eq . (4.48) an d (4.49), we obtain
k
, = A/*A~ (4-50 )
Flow Ne t i n Anisotropi c Soil s
To obtain a flow ne t fo r anisotropi c soi l conditions , the natura l cross-section ha s t o be redrawn t o
satisfy th e condition o f Laplace Eq . (4.41) .
The transforme d sectio n ma y b e obtaine d b y multiplyin g eithe r th e natura l horizonta l
distances by ^k
z
I k
x
o r the vertical distances b y ^k
x
I k
z
keepin g th e other dimensio n unaltered .
Normally th e vertical dimension s ar e kept a s they are but the horizontal dimensions ar e multiplied
by ^jk, I k
x
. The natura l section get s shortened o r lengthened i n the x- direction i n accordance wit h
the condition tha t k i s greater o r less than k .
Fig. 4.18(a ) i s a natura l sectio n wit h flo w takin g plac e aroun d a shee t pil e wall . Th e
horizontal permeabilit y i s assume d t o be 4 time s tha t of th e vertica l permeability . Fig . 4.18(b ) i s
transformed sectio n wit h th e horizonta l dimension s multiplie d b y a facto r equa l t o
TJ k
z
/k
x
= v l / 4 = l / 2. Thi s sectio n i s no w assume d t o posses s th e sam e permeabilit y o f
k
g
= J 4k
2
- 2 k i n al l directions. The flow net s ar e constructed o n thi s sectio n i n the usual way.
The sam e flo w ne t i s transferre d t o th e natura l sectio n i n (a ) o f Fig . 4.18 , b y multiplyin g th e
jt-coordinates o f point s o n th e flo w an d equipotentia l line s b y th e facto r 2 . O n th e natura l
cross-section th e flo w ne t wil l no t b e compose d o f square s bu t o f rectangle s elongate d i n th e
direction o f greate r permeability .
4 .1 7 DETERM I NATI O N OF SEEPAG E PRESSU RE
Flow net s ar e usefu l i n the determination o f the seepage pressure a t any point along th e flow path.
Consider the cubical element 1 in Fig. 4.16(a ) with all the sides equal to a. Let h
l
be the piezometri c
head actin g on the face k t and h
2
on face jo .
The tota l forc e o n face k t = P
[
=a
2
y
w
h
l
The tota l forc e on facey' o = P
2
= a
2
Y
w
h
2
The differentia l forc e actin g on the element i s
P
l
-P
2
= P
3
= a\ ( h
l
-h
2
)
Since ( h
l
- h
2
) i s the head dro p A/z , we can write
w w
where a
3
is the volume of the element . The forc e pe r uni t volume of the element is , therefore ,
This forc e exert s a drag o n the element known as the seepage pressure. It has the dimensio n
of unit weight, and at any point it s line of action is tangent to the flow line. The seepage pressure i s
a ver y important facto r i n the stabilit y analysis o f eart h slopes . I f the lin e of actio n o f the seepage
force act s i n the vertica l directio n upwar d a s on a n element adjacen t t o poin t ; c in Fig. 4.16(a) , th e
force tha t is acting downward t o keep th e element stabl e i s the buoyant unit weight o f the element .
When thes e two forces balance , the soil wil l j ust be at the point of being lifte d up , and ther e wil l be
Soil Perm eabilit y an d Seepag e 123
effectively n o grain-to-grain pressures. The gradient at which this occurs can be computed from th e
balance o f forces give n by Eqs. (3.19a ) and (4.51). Therefore we can writ e
or * „ = • (4.52)
The soi l wil l be i n quick condition a t thi s gradient, which is therefor e calle d i
c
, the critical
hydraulic gradient.
4 .1 8 DETERM I NATI O N O F UPLIFT PRESSU RES
Water that seeps belo w masonry dams or weirs founded on permeable soil s exerts pressures on the
bases of structures. These pressures are called uplift pressures. Uplif t pressures reduce the effectiv e
weight of the structur e and thereby cause instability. It is therefore ver y essential t o determine the
uplift pressure s o n the base of dams or weirs accurately. Accurate flow net s should be constructed
in cases where uplif t pressure s are required t o be determined. The method of determining the uplif t
pressures ca n be explained a s follows.
Consider a concrete dam Fig. 4.19a founded on a permeable foundatio n at a depth D below
the ground surface. The thickness of the permeable strat a is H. The depth of water on the upstream
side i s h
{
an d on the downstrea m sid e i s zero. Water flows fro m th e upstrea m t o th e downstream
Impervious
(a) Concrete dam
a b e d
u
b
(b) Uplift-pressure distribution
Figure 4.1 9 Uplif t pres s ur e o n t he bas e o f a con cret e da m
124 Chapt e r 4
side. It is necessary t o determine th e uplift pressure on the base of the dam by means o f flow nets as
shown i n th e figure .
The differenc e i n hea d betwee n th e upstrea m an d downstrea m wate r level s i s h
f
. Le t th e
number o f equipotentia l drops b e A^. The hea d los t pe r dro p b e Ah ( = h/N J . A s th e wate r flows
along the side and base o f the dam, there will be equal drops of head between the equipotential lines
that meet th e dam a s shown in the figure. A piezometer tub e at point a (coinciding with the corne r
of th e da m i n th e figure ) give s a pressur e hea d h . No w th e uplif t pressur e a t poin t a ma y b e
expressed a s
u
a
=h
a
r
w
=( h
t
+D-^h)r
w
(4.53a )
Similarly, the uplif t pressur e a t any other point , say e (see the figure), may be estimated from
the expressio n
u
e
=( h
t
+D-n
d
Mi)y
w
(4.53b )
where n
d
= the number of equipotential drops t o the point e.
Fig. 4.19b shows th e distribution of uplif t pressur e o n the base o f the dam.
Example 4 .13
In orde r t o comput e th e seepag e los s throug h th e foundatio n o f a cofferdam , flownet s wer e
constructed. Th e resul t o f th e flowne t stud y gave N ,= 6 , N
d
= 16. The hea d o f wate r los t during
seepage wa s 19.6 8 ft. I f the hydrauli c conductivity of th e soi l i s k = 13.12 x10~
5
ft/min, comput e
the seepag e los s pe r foot lengt h of dam pe r day.
Solution
The equation fo r seepag e los s i s
Substituting the given values,
q = 13.12 x l (T
5
x 19.68 x — = 9.683 xl (T
4
ft
3
/min = 1.39 ftV day pe r f t length of dam.
16
Example 4 .1 4
Two lines of sheet pile s wer e driven in a river bed as shown in Fig. Ex. 4. 14. The dept h of water over
the river bed i s 8.20 ft. The trench level withi n the sheet piles i s 6.6 ft below th e river bed. The water
level withi n th e shee t pile s i s kept a t trenc h leve l b y resorting t o pumping. I f a quantit y o f wate r
flowing int o the trench from outside is 3.23 ft
3
/hour per foot length of sheet pile, what is the hydraulic
conductivity of the sand? What i s the hydrauli c gradient immediatel y below th e trench bed?
Solution
Fig. Ex . 4.14 give s th e flo w ne t an d othe r details . Th e differentia l head betwee n th e botto m o f
trench an d th e wate r level i n the rive r i s 14. 8 ft .
Number o f channel s = 6
Number of equipotential drops =10
N
f
6
q = kh~+- o r 3.23= 14. 8 x — x f c
N
d
1 0
Soil Permeabilit y an d Seepag e 125
Figure Ex. 4.1 4
. 3.23x1 0
or & = x
1
= l xl O-
4
f t / s e c
14.8x6 60x6 0
The distanc e betwee n th e las t two equipotential s give n i s 2.95 ft . The calculate d hydrauli c
gradient is
A/z
i =
14.8
= 0.5 0
As 10x2.9 5
- ~
£r
r~
=
"TT = 2 < 5 t o 6 whic h is normally required for sand.
Example 4 .1 5
A concrete da m (Fig 4. 19) is constructed acros s a river over a permeable stratu m of soil o f limite d
thickness. The water heads ar e upstream sid e 16 m and 2 m on the downstream side. The flow ne t
constructed under the dam gives A^.= 4 and N
d
=l2. Calculat e the seepage los s through the subsoil
if the average valu e of the hydraulic conductivity is 6 x 10~
3
cm/sec horizontally and 3 x 10"
4
cm/
sec vertically. Calculate the exit gradient if the average lengt h of the last fiel d i s 0.9 m. Assuming
e = 0.56, an d G
s
= 2.65, determin e th e critical gradient . Comment on the stabilit y of the river bed
on the downstream side .
Solution
Upstream sid e h
{
= 16 m and downstream side h
2
= 2 m, therefore h= 16-2, = 14 m
k = 6 x 10~
3
cm/sec, k = 3 x 10"
4
cm/sec
= 1. 34xl O-
3
cm/sec
N
f
4
= kh-+- = (1.34 X 10-3) x (14 X 100) x — = 0.626 cm
3
/ sec
126 Chapt e r 4
h 1 4
The head los s pe r potential drop = — = — = 1.17 m
v
N
d
1 2
The exi t gradient i = — = —— = 1.30
5
/ 0. 9
As per Eq. (4.52) , the critical gradient i
e
is
G
s
-l 2.65- 1
i = = = i.(J 6
1 + e 1 + 0.56
Since th e exi t gradien t i s greater tha n the critical gradient , th e river be d o n the down strea m
side wil l be subjected to a quick condition. One solution would be to provide a sheet pile wall on the
upstream sid e belo w th e dam t o prevent this condition.
4.1 9 SEEPAG E FLOW TH ROU G H H OMOGENEOU S EART H DAM S
In almost al l problems concernin g seepage beneath a sheet pil e wal l or through the foundation of a
concrete da m al l boundary conditions ar e known. However, in the case of seepage through an earth
dam th e uppe r boundar y o r th e uppermos t flo w lin e i s not known . This uppe r boundar y i s a fre e
water surfac e an d wil l be referred t o as the line o f seepage o r phreatic line. The seepag e line may
therefore be defined as the line above which there is no hydrostatic pressure and below whic h there
is hydrostati c pressure . I n the design o f al l eart h dams , th e following factors ar e ver y important.
1. Th e seepag e line shoul d no t cut the downstream slope .
2. Th e seepag e los s through the dam shoul d be the mi ni mum possible.
The two important problems tha t are required t o be studie d in the design o f eart h dams are :
1. Th e predictio n o f th e position of the lin e of seepage in the cross-section .
2. Th e computatio n of the seepage loss .
If the line of seepage i s allowed t o intersect the downstream fac e much above the toe, mor e or
less serious sloughing may take place and ultimate failure ma y result. This mishap can be prevented
by providing suitable drainage arrangement s on the downstream sid e o f the dam.
The sectio n o f a n eart h da m ma y b e homogeneou s o r non-homogeneous . A homogeneou s
dam contain s th e sam e materia l ove r th e whol e sectio n an d onl y on e coefficien t o f permeabilit y
may b e assume d t o hol d fo r the entir e section. I n the non homogeneous o r the composit e section ,
two o r mor e permeabilit y coefficient s may hav e t o be use d accordin g t o the material s use d i n the
section. When a number of soil s of different permeabilitie s occur i n a cross-section, th e prediction
Phreatic line (seepage line )
Basic parabola
r
Figure 4.2 0 Basi c parabol a and the phreati c lin e for a homogeneous eart h dam
Soil Perm eabilit y an d Seepag e 12 7
of th e positio n o f th e lin e o f seepag e an d th e computatio n o f th e seepag e los s becom e quit e
complicated.
It ha s bee n notice d fro m experiment s o n homogeneou s eart h da m model s tha t th e lin e of
seepage assumes mor e or less the shape of a parabola a s illustrated in Fig. 4.20. I n some sections a
little divergence from a regular parabola is required at the surfaces of entry and discharge of the line
of seepage . I n some idea l section s wher e conditions ar e favorabl e the entir e seepage lin e may be
considered a s a parabola. When the entire seepage lin e is a parabola, al l the other flow line s will be
confocal parabolas . The equipotential lines for this ideal case will be conjugate confocal parabola s
as shown in Fig. 4.21. As a first ste p it is necessary to study the ideal case where the entire flow net
consists o f conjugate confocal parabolas .
4.20 FLO W NET CONSI STI NG OF CONJUGATE CONFOCAL PARABOLAS
As a prelude t o th e stud y of a n ideal flo w ne t comprisin g o f parabola s a s flo w an d equipotentia l
lines, i t is necessary t o understand the properties of a single parabola. The parabola ACV illustrated
in Fig. 4.21, is defined as the curve whose every point is equidistant from a point F called the focus
and a line DG called the directrix. I f we consider any point, say, A, o n the curve, we can write FA =
AG, wher e th e lin e A G i s norma l t o th e directrix . I f F i s th e origi n o f coordinates , an d th e
coordinates o f point A are (jc , y), we ca n writ e
AF =
-^- (4-54 )
where, y
Q
= F D
Eq. (4.54) i s the equation of the basi c parabola. I f the parabola intersect s the y-axis a t C, we
can writ e
FC=CE = y
0
Similarly for the vertex point V, the focal distance a
Q
is
FV = VD = a
0
= y
0
/2 (4.55 )
Figure 4.2 1 illustrate s th e idea l flo w ne t consistin g o f conjugat e confoca l parabolas . Al l th e
parabolas hav e a common focu s F.
The boundar y lines of such an ideal flow net are:
1 . Th e upstream face AB, an equipotential line, is a parabola.
2. Th e downstream discharge face FV, an equipotential line, is horizontal .
3. ACV, th e phreatic line, is a parabola .
4. BF, the bottom flow line, is horizontal.
The know n boundary conditions ar e onl y three i n number . They are , th e two equipotentia l
lines AB and FV, and the bottom flow line BF. The top flow line ACV is the one that is unknown. The
theoretical investigatio n of Kozen y (1931) reveale d tha t the flo w ne t fo r suc h a n idea l conditio n
mentioned above with a horizontal discharge face FV consists of two families of confocal parabola s
with a common focus F. Since the conjugate confocal parabolas shoul d intersect a t right angles t o
each other, all the parabolas crossin g the vertical line FC should have their intersection points lie on
this line.
128
Chapt er 4
Since th e seepag e lin e i s a line of atmospheri c pressur e only , the onl y typ e of head tha t can
exist along it is the elevation head . Therefore, ther e must be constant drop s in elevation betwee n th e
points at which successive equipotential s meet the top flow line , as shown in Fig. 4.21.
In al l seepag e problem s connecte d wit h flo w throug h eart h dams , th e focu s F o f th e basi c
parabola i s assumed t o li e at the intersection of the downstream discharge face FV and th e bottom
flow lin e BF a s show n i n Fig . 4.21. Th e poin t F i s therefor e known . The poin t A, whic h i s th e
intersection poin t o f th e to p flo w lin e of th e basi c parabol a an d th e upstrea m wate r level , i s als o
supposed t o be known. When the point A i s known, its coordinates ( d, K) wit h respect t o the origi n
F ca n b e determined . Wit h thes e tw o know n points , th e basi c parabol a ca n b e constructe d a s
explained below. We may writ e
(4.56)
Seepage L os s Through the Da m
The seepag e flo w q across an y section can be expressed accordin g t o Darcy' s la w as
q = kiA (4.57)
Considering th e sectio n F C i n Fig . 4.21, wher e th e sectiona l are a A i s equa l t o y
Q
, th e hydraulic
gradient / can b e determine d analyticall y as follows:
From Eq . (4.54) , th e equation of the parabola ca n be expressed a s
' o +^o
2
(4.58)
Directrix
Figure 4.2 1 Idea l flown e t con s is t in g of con j ug at e con focal parabola s
Soil Per meabi l i t y an d Seepag e 1 29
The hydraulic gradient i at any point on the seepage lin e in Fig. 4.21 can be expressed a s
dy yo
For the point C which has coordinates (0 , y
Q
), th e hydraulic gradient from Eq . (4.59) i s
Therefore, th e seepage quantit y across sectio n F C is
dy
(4.60)
Seepage Throug h H omogeneou s and I sotropi c Eart h Dams
Types o f Entr y an d Exi t o f Seepage lines
The flow ne t consisting of conjugate confocal parabolas i s an ideal case whic h is not generally met
in practice. Though the top flow lin e resembles a parabola for most of its length, the departure fro m
the basi c parabol a take s plac e a t the face s of entr y and discharge o f the flo w line . The departur e
from th e basi c parabol a depend s upo n th e condition s prevailin g a t th e point s o f entranc e an d
discharge o f the flow lin e a s illustrated i n Fig. 4.22 from (a ) to (e).
The seepage lin e shoul d be normal t o the equipotential lin e at the point of entry as shown in
Fig. 4.22(a). However , thi s conditio n i s violate d i n Fig . 4.22(b), wher e th e angl e mad e b y th e
upstream fac e AB wit h the horizonta l i s les s tha n 90°. It ca n be assume d i n thi s cas e th e coars e
material used to support the face AB i s highly permeable an d does not offer an y resistance fo r flow.
In suc h cases AB take n a s the upstream equipotentia l line . The top flo w lin e canno t therefor e b e
Seepage
line
Coarse /••'•''/*?
material St.*'-^''.' -
Seepage
line
(a) (b)
Discharge face
XN |
P
/ 3<90°
(c) (d ) (e )
Figure 4.2 2 Type s o f en t r y an d ex it o f s eepag e line s
130 Chapt er 4
normal t o the equipotential line. However, this line possesses zero gradient and velocity at the point
of entry . Thi s zer o conditio n relieve s th e apparen t inconsistenc y o f deviatio n fro m a norma l
intersection.
The condition s prevailing at the downstream toe of the dam affec t th e type of exit of the flo w
line at the discharge face . I n Fig. 4.22(c ) the material a t the to e i s the same a s i n the othe r part s of
the dam wherea s i n (d) and (e ) rock to e drains are provided. Thi s variatio n i n the soi l conditio n at
the to e affect s th e exi t patter n o f th e flo w line . The flo w lin e wil l mee t th e discharg e fac e F E
tangentially i n 4.22(c) . Thi s ha s t o b e s o becaus e th e particle s o f wate r a s the y emerg e fro m th e
pores a t the discharge fac e have to conform a s nearly a s possible t o the direction of gravity. But i n
cases wher e roc k to e drains ar e provided , the top flow lin e becomes tangential t o the vertica l lin e
drawn a t the poin t of exi t on th e discharge face as shown i n (d) and (e ) of Fig. 4.22 .
M ethod o f L ocatin g Seepage Lin e
The genera l metho d o f locatin g th e seepag e lin e i n an y homogeneou s da m restin g o n a n
impervious foundation may b e explained wit h reference t o Fig. 4.23(a) . As explained earlier , th e
focus F of the basi c parabol a i s taken as the intersection poin t of the bottom flow line BF an d the
discharge face EF . I n thi s cas e th e focu s coincide s wit h th e toe o f th e dam. On e mor e poin t i s
required t o construc t th e basi c parabola . Analysi s o f th e locatio n o f seepag e line s b y
A. Casagrande ha s reveale d tha t th e basi c parabol a wit h focu s F intersect s th e upstrea m wate r
surface a t A such that AA'= 0.3 m, where m is the projected lengt h of the upstream equipotentia l
line A'B o n the wate r surface . Point A is called th e corrected entranc e point . The parabol a APSV
may no w b e constructe d a s pe r Eq . (4.54) . Th e divergenc e o f th e seepag e lin e fro m th e basi c
parabola i s shown a s AT
1
and SD i n Fig. 4.23(a) . For dams wit h flat slopes , th e divergence s ma y
be sketche d b y ey e keeping i n view the boundary requirements . The erro r involve d in sketchin g
by eye , th e divergenc e o n the downstrea m side , migh t be considerabl e i f the slope s ar e steeper .
B' T
(a)
Basic parabol a
u.t
1
0.3
a
< 0.2
+
a
0.1
n
--.
----,
i
— ^
^
'^
-^_
^^
^
^
^\
(b)
30° 60 ° 90 ° 120 ° 150 ° 180 °
/5-Slope of discharge face
Figure 4.2 3 Con s t ruct io n of s eepag e lin e
Soil Perm eabilit y an d Seepag e 13 1
Procedures hav e therefor e bee n develope d t o sketc h th e downstrea m divergenc e a s explaine d
below. A s show n i n Fig . 4.23(a), E i s th e poin t a t whic h th e basi c parabol a intersect s th e
discharge face . Le t the distance ED be designated a s Aa and the distanc e DF as a. The values of
Aa an d a + Aa vary with the angle , j3, made b y the discharge fac e wit h the horizonta l measure d
clockwise. The angl e ma y var y fro m 30 ° t o 180° . Th e discharg e fac e i s horizonta l a s shown in
Fig. 4.22(e). Casagrande (1937 ) determined th e ratios of Aa / (a + Aa) for a number of discharg e
slopes varyin g from 30° t o 180 ° an d the relationshi p i s shown in a graphical for m Fig . 4.23(b) .
The distanc e ( a + Aa ) can b e determine d b y constructin g the basi c parabol a wit h F a s th e
focus. Wit h th e know n ( a + Aa ) an d th e discharg e fac e angl e j3 , A a ca n b e determine d fro m
Fig. 4.23(b) . The point D may therefore b e marked ou t at a distance of Aa from E. With the point D
known, the divergence DS ma y be sketche d by eye.
It should be noted that the discharge length a, is neither an equipotential nor a flow line , since
it i s a t atmospheri c pressure . I t i s a boundar y alon g whic h th e hea d a t an y poin t i s equa l t o th e
elevation.
Analytical Solutions fo r Determining a and q
Casagrande (1937) proposed th e following equation for determining a fo r j 8 < 30°
(4.61)
cos/? ^j cos
2
/ ? sin
2
/?
L. Casagrande (1932 ) gave the following equation for a when {$ lies between 30 ° and 90° .
(4.62)
The discharg e q pe r uni t lengt h through any cross-section o f th e da m ma y b e expresse d a s
follows:
For/?<30°, a = fcasin/?tan/ ? (4.63 )
For 30°</ ?<90°, a = fca si n
2
/? (4.64) .
4 .2 1 PIPIN G FAI L U R E
Piping failure s caused b y heav e ca n be expecte d t o occu r o n th e downstrea m sid e o f a hydraulic
structure whe n th e uplif t force s o f seepag e excee d th e downwar d force s du e t o th e submerge d
weight of the soil .
The mechanics of failure due to seepage was first presented b y Terzaghi. The principle of this
method ma y be explaine d wit h respect t o seepag e flo w belo w a shee t pil e wall . Fig . 4.24(a ) i s a
sheet pile wall with the flow ne t drawn. The uplif t pressure s actin g on a horizontal plan e ox can be
determined a s explained i n Sect. 4.18. The ordinates o f curve C in Fig. 4.24(b ) represent the uplif t
pressure at any point on the line ox. It is seen that the uplift pressur e is greatest clos e to the wall and
gradually becomes les s wit h an increase i n the distance from th e wall . When the upwar d forces of
seepage o n a portio n o f o x nea r th e wal l become equa l t o th e downwar d forces exerte d b y th e
submerged soil , th e surfac e o f th e soi l rise s a s show n i n Fig . 4.24(a). Thi s heav e occur s
simultaneously wit h a n expansio n o f th e volum e o f th e soil , whic h cause s it s permeabilit y t o
increase. Additional seepage cause s the sand to boil, which accelerates th e flow o f water and leads
to complet e failure . Terzaghi determine d fro m mode l test s tha t heave occur s withi n a distance of
about DI 2 (wher e D i s th e dept h o f penetratio n o f th e pile ) fro m th e shee t pil e an d th e critica l
section ox passes throug h the lower edge o f the sheet pile.
132
Chapt er 4
(a)
D
1
(b)
Sheet pile wall
D/2
d'
\ c
Figure 4.2 4 Pipin g failur e
Factor o f Safet y Against H eave
The prism aocd in Fig. 4.24(b) subjected to the possible uplif t has a depth of D and width D/2.
The average uplift pressur e on the base of prism is equal to Y
w
h
a
- The total uplift force per unit
length o f wal l is
_i
The submerge d weigh t of the prism aocd i s
1
( 4.65)
where y
b
i s the submerge d uni t weight o f the material . Th e facto r of safet y wit h respect t o piping
can therefor e b e expressed a s
F =
(4.66)
Soil Perm eabilit y an d Seepag e 133
If i t i s not economical t o drive the shee t piles deepl y enough t o prevent heave, the factor of
safety ca n be increase d b y placin g a weighted filte r ove r th e pris m aocd a s shown b y the pris m
aa'd'd. I f the weight of such a filter i s W
(
, the new factor of safet y can be written as
F =
u
(4.67)
Filter Requirement s t o Contro l Pipin g
Filter drain s ar e require d o n th e downstrea m side s o f hydrauli c structures an d aroun d drainage
pipes. A properly grade d filte r prevents the erosion o f soil i n contact wit h it due to seepage forces.
To prevent the movement of erodible soil s int o or through filters, the pore spaces betwee n the filte r
particles shoul d be smal l enough to hol d some o f the protecte d material s i n place. Taylor (1948 )
shows tha t if thre e perfec t sphere s hav e diameters greate r tha n 6.5 time s the diamete r o f a smal l
sphere, th e smal l sphere s ca n mov e throug h th e large r a s show n i n Fig . 4.25(a). Soil s an d
aggregates ar e always composed o f ranges of particle sizes, and i f pore space s i n filters ar e smal l
enough t o hold the 85 per cent size (D
85
) of the protected soi l i n place, the fine r particle s wil l also
be held i n place a s exhibited schematicall y i n Fig. 4.25(b).
The requirement s o f a filte r t o kee p th e protecte d soi l particle s fro m invadin g th e filte r
significantly ar e based o n particle size . These requirement s wer e developed fro m test s by Terzaghi
which were later extended by the U.S. Army Corps of Engineers (1953). The resulting filter specifica-
tions relate the grading of the protective filter t o that of the soil being protected b y the following;
D
5 filter
<4
. s filter ^50 filter
85 soil
D
15 soil
D
<25
50 soil
(4.68)
(a) Size of smallest spherical particl e which just fit s the space between large r spheres
Soil which has migrated into
filter an d is held by D
85
size
soil particles
(b) Condition of the boundary betwee n protecte d soi l and the filter materia l
Figure 4.25 Requi rement s o f a f i l t e r
134 Chapt er 4
10 1.0
Grain size D mm
0.1
= 0.015 mm
0.01
Figure 4.2 6 Grai n size distribution curves for grade d filter an d protected material s
The criteri a ma y be explained a s follows:
1. Th e 1 5 per cen t size ( D
15
) o f filte r materia l mus t be les s tha n 4 time s th e 8 5 per cen t siz e
(D
85
) of a protected soil . The ratio of D
15
of a filter to D
85
of a soil is called the piping ratio.
2. Th e 1 5 per cent size ( D
15
) o f a filter materia l shoul d be at least 4 times the 1 5 per cent size
(D
]5
) of a protected soi l but not mor e tha n 20 times of the latter .
3. Th e 50 per cent size ( D
5Q
) o f filter material shoul d be less than 25 times the 50 per cent size
(D
50
) of protected soil .
Experience indicate s that if the basic filter criteria mentioned abov e are satisfied in every part
of a filter, piping canno t occu r unde r even extremel y sever e conditions .
A typica l grai n siz e distributio n curv e o f a protecte d soi l an d th e limitin g size s o f filte r
materials fo r constructing a graded filter i s given in Fig. 4.26. Th e siz e o f filter material s mus t fal l
within the two curves C
2
and C
3
to satisf y th e requirements.
Example 4 .1 6
Fig. Ex . 4.1 6 give s th e sectio n o f a homogeneou s da m wit h a hydrauli c conductivit y
k = 7.87 4 x 10"
5
in/sec. Draw the phreatic line and compute th e seepage loss pe r foot lengt h of the
dam.
Soil Perm eabilit y an d Seepag e 135
13.12ft
d = 68.9 ft
Figure Ex . 4.1 6
Solution
The dept h h of water on upstream side = 32.81 ft .
The projecte d length of slope A 'B o n the wate r surface = 32.81 ft.
The point A o n the water level is a point on the basic parabola. Therefor e
AA' = 0. 3x32. 81=9. 84 ft.
F i s the focus of the parabola. The distanc e of the directri x from th e focus F is
v
0
= 4d
2
+h
2
- d
where d = 68. 9 ft , h = 32.81 ft. Therefore
y
0
= V (68.9)
2
+(32.81)
2
-68.9 = 7.413 ft
The distanc e of the vertex of the parabol a from F is
FV = a
-.
=
0
2 2
=
3J06 ft
The (jc , y) coordinates of the basi c parabola may be obtained from Eq . (4.58) a s
2y
Q
2x7.413 14.83
v
Given below ar e values of y for various value s of x
j t ( f t ) 0 15 30 45 68.9
y ( f t ) 7.41 6 16.6 5 22.3 6 26.8 8 32.8 1
The parabola has been constructed with the above coordinates as shown in Fig. Ex. 4.16.
From Fig. Ex. 4.16 A a + a = 24.6 f t
From Fig. 4.23, for a slope angl e ) 3 = 45°
136 Chapt e r 4
-^--035
a + Aa
or Aa = 0.35 ( a + Aa) = 0.35 x 24. 6 = 8.61 f t
From Eq . (4.60)
q = ky
Q
where k = 7.874 x 10~
5
in/se c or 6.56 x 10"
6
ft/sec an d y
Q
= 7.413 f t
q = 6.56 x 10-
6
x 7.413 = 48.63 x 10"
6
ft
3
/sec pe r f t lengt h o f dam.
Example 4 .1 7
An eart h da m whic h i s anisotropic i s given i n Fig. Ex . 4.17(a). The hydrauli c conductivitie s k
x
and
k
z
i n th e horizonta l an d vertica l direction s ar e respectivel y 4. 5 x 10~
8
m/ s an d 1. 6 x 10~
8
m/s .
Construct th e flo w ne t an d determin e th e quantit y o f seepag e throug h th e dam. Wha t i s the por e
pressure a t point PI
Solution
The transforme d sectio n i s obtaine d b y multiplyin g the horizonta l distance s b y ^J k
z
I k
x
an d b y
keeping th e vertical dimension s unaltered. Fig. Ex . 4.17(a) is a natural section o f the dam. The scale
factor fo r transformation i n the horizontal directio n i s
Scale factor = P - = J
L6xl
°"
8
B
= 0.6
] k
x
V 4. 5 X10-
8
The transforme d sectio n o f th e da m i s give n i n Fig . Ex . 4.17(b) . Th e isotropi c equivalen t
coefficient o f permeabilit y i s
k =
e
Confocal parabola s ca n be constructed with the focus of the parabola a t A. The basic parabol a
passes throug h point G such that
GC=0. 3 H C = 0 . 3 x 2 7 = 8. 10m
The coordinate s of G are:
x = +40.80 m, z = +18.0 m
7
2
- 4f l
2
As per Eq. (4.58)
x
= 9 . (a )
Substituting for x an d z, we get , 40.8 0 =
Simplifying we have, 4a
2
+ 163.2a
Q
- 324 = 0
Solving, a
Q
= 1. 9 m
Substituting for a
Q
i n Eq. (a ) above, we can writ e
Soil Perm eabilit y an d Seepag e 137
15.0m,
h = 18.0 m
Blanket drain
B S
(b) Transformed sectio n
Figure Ex. 4.17
Z
2
-14. 4
7.6
(b)
By using Eq. (b), the coordinates o f a number of points on the basic parabol a ma y be calculated .
7( m) -1. 9 0. 0 I o 10. 0 20. 0 30. 0
z( m) 0. 0 3. 8 7.2 4 9.5 1 12. 9 15.5 7
The basi c parabol a i s shown in Fig. Ex. 4.17(b).
The flowne t i s complete d b y makin g the entr y correction s b y ensurin g tha t th e potentia l
drops ar e equa l between the successiv e equipotentia l lines a t the to p seepag e lin e level .
As per Fig. Ex. 4.17(b), there are 3.8 flow channel s and 1 8 equipotential drops. The seepag e
per unit length of dam i s
N
f
3 8
-^- = ( 2. 7xl O-
8
) xl 8x—= l xl O-
7
m
3
/s
N 18
The quantity of seepage acros s section Az can also be calculated without the flownet by using
Eq. (4.60 )
q = k^Q = 2k
e
a
Q
= 2x2.7x 1Q-
8
x 1. 9 « 1 x 10~
7
m
3
/sec per meter
Pore pressur e at P
Let RS be the equipotential line passing through P. The number of equipotential drop s u p to point
P equals 2. 4
Total head loss = h = 18m, number of drops =18
138 Chapt e r 4
18
Head los s per drop = — = 1 m.
18
Therefore th e head at point P = 18 - 2.4(A/z ) = 18 - 2.4(1 ) = 15. 6 m
Assuming the base o f the dam a s the datum, the elevation head o f point P = 5.50 m.
Therefore th e pressure head at P = 15.6 - 5. 5 = 10. 1 m.
The por e pressur e at P is , therefore, u
w
= 10.1 x 9.81 = 99 kN/m
2
Example 4 .1 8
A sheet pil e wall was driven across a river to a depth of 6 m below the river bed. It retains a head of
water of 12. 0 m. The soi l below the river bed is silty sand and extends up to a depth of 12. 0 m where
it meet s a n impermeable stratum of clay. Flow net analysis gave A/,= 6 and N
d
- 12 . The hydraulic
conductivity of the sub-soi l is k = 8 x 10~
5
m /min. The average uplift pressur e head h
a
at the bottom
of th e pil e i s 3. 5 m. The saturate d unit weight of the soi l y
sat
= 19. 5 kN/m
3
. Determine :
(a) The seepag e less per meter length of pile per day.
(b) The facto r of safet y against heave on the downstream side o f the pile .
Solution
(a) Seepage loss ,
The los s o f head h = 12 m
N
f 6
q = kh—^- = ( 8x 10~
5
)x 12x — = 48x 10~
5
m
3
/min = 69.12x 10~
2
m
3
/day
N
d
1 2
(b) The F
s
a s per Eq. (4.67 ) i s (Ref . Fig 4.24 )
F
W
»
+W
>
D7
»
V
h
J
w
h
a
= 3.5 m
Yb = Xsat -Y
w
= 19-5-9.8 1 = 9.69 k N/ m
3
6 x 9.69
4 .22 PROB L EM S
4.1 A constan t head permeabilit y test was carried ou t on a cylindrical sample of sand of 1 0 cm
diameter and 1 5 cm height. 200 cm
3
of water was collected i n 2.25 min under a head of 30 cm.
Compute the hydraulic conductivity in m/sec.
4.2 Calculat e th e hydrauli c conductivity of a soi l sampl e 6 c m i n height an d 5 0 cm
2
cross -
sectional are a i f 43 0 m L o f wate r wa s collecte d i n 1 0 mi n unde r a constan t hea d o f
40cm.
On oven-drying the test specimen had a mass of 498 g. Assumin g G
s
= 2.65, calculat e the
seepage velocity.
4.3 A constan t head permeabilit y tes t wa s carried ou t on a sample o f sand. The diamete r an d
the lengt h o f th e sampl e wer e 1 0 an d 2 0 cm . respectively . Th e hea d o f wate r wa s
Soil Perm eabilit y an d Seepag e 139
maintained at 35 cm. If 1 10 cm
3
of water is collected i n 80 seconds, comput e the hydraulic
conductivity of the sand.
4.4 A falling head permeability tes t was performed on a sample of silly sand. The time required
for th e hea d t o fall i n the stand pipe fro m 6 0 cm to the 30 cm mark was 70 min. The cros s
sectional area of the stand pipe was 1 .25 cm
2
. If the height and diameter of the sample were
respectively 1 0 and 9 cm, determine th e value k in cm/min.
4.5 I n a falling head permeability test, the time taken for the head to fall fro m h
{
t o h
2
is t. If the
test is repeated wit h the same initial head h
r
what would be the final head in a time interval
oft/21
4.6 I n a falling head permeameter test the initial head at t = 0 is 40 cm. The head drops by 5 cm
in 1 0 minutes. Determine the time required t o run the test for the final hea d t o be at 20 cm.
Given: Heigh t o f sampl e = 6 cm; cros s sectiona l area s o f sampl e = 5 0 cm
2
an d stan d
pipe = 0. 5 cm
2
Determine th e hydrauli c conductivity in cm/sec.
4.7 Th e hydraulic conductivity of a soil sampl e at a temperature of 30°C wa s 8 x 10~
5
cm/sec.
4.8
4.9
Determine it s permeability at 20°C.
Given: V iscosit y o f wate r a t (a ) 3 0 °C
it,n = 10.0 9 x 10~
7
kN-sec/m
2
.
= 8. 0 x 10~
7
kN-sec/m
2
, an d (b ) 20°C ,
Fig. Prob. 4.8 gives a test well with observation well s for conducting a pumping tests. The
following dat a ar e available .
Maximum D
O
= 0.5 m, r
o
= 20 cm, H = 8m, & = 8 x lO^m/sec .
Determine th e maximum yield i n m
3
/hour.
Refer t o Fig. Prob . 4.8 .
Given: H = 52 ft, h
{
- 4 7 ft, h
2
= 50. 75 ft, discharge q under steady condition = 80 ft
3
/min,
r
{
= 10 ft, an d r
2
= 20 ft .
Required: The hydrauli c conductivity in ft/year .
Observation well s
Test wel l
Impermeable
Figure Prob . 4. 8
140 Chapt er 4
Observation wells
Test well
Figure Prob . 4 .12
4.10 Refe r t o Fig. Prob . .8. Determine th e hydraulic conductivity of the aquifer i n m/hr under a
steady stat e discharg e o f 240 m
3
/hr wit h th e following data:
H = 30.5m, h
l
= 26.5 m, h
2
= 29.8 m, r
}
= 10m, r
2
= 50 m. Diameter o f the test well = 20 cm.
4.11 Refe r to Prob. 4.10. For a maximum D
Q
= 4.9 m, and radius of influence R
t
= 30m, calculat e
the value of k.
4.12 Fig . Prob . 4.12 gives th e sectiona l profil e of a confined aquifer .
Given: H
Q
= 5m, D
Q
(max) = 4.5m, /?
(
. = 100m, radiu s of test wel l r
o
= 10 cm. an d H = 10m.
Determine th e hydrauli c conductivit y in cm/se c assumin g q = 1. 5 m
3
/min unde r stead y
state conditions.
4.13 Fo r the Prob 4.12 , i f D
o
(max) = 5.5m, determin e k. All the other dat a remai n th e same .
4.14 Calculat e th e yiel d i n ft
3
pe r hou r fro m a wel l drive n int o a confine d aquife r
(Fig. Prob. 4.12) .
Given: H = 3 5 ft , H
Q
= 15ft , h
Q
= 18 ft , k = 0.09 ft/min , r
Q
= 4in., R
{
= 600 ft .
14.15 Th e soi l investigatio n at a sit e reveale d thre e distinc t layer s o f sand y soi l (Fig . 4.9) . Th e
data availabl e are :
Layer No T hick n es s (m ) k (cm /s ec)
8 x 10~
3
6 x 10-
2
5 x 10-
3
Determine th e equivalent values o f k both i n the horizontal an d vertica l directions .
4.16 Laborator y test s o n a sample o f undisturbed silty sand gav e the following data :
void rati o = 0.62; k - 4 x 10~
2
cm/sec.
Estimate the valu e of k of anothe r similar sampl e whos e voi d ratio i s 1.05 .
Soil Perm eabilit y an d Seepag e 141
4.f
//A\\,
T
m
//A\\
Sheet pil e
<S
^ |f0. 5 m
i //A\\|//A\\
6 m
m
1
Sand
/Z\\V /A\\ //\\V /\\\
Figure Prob . 4.1 7
//\\V /A\\ /A\V /\\\
Figure Prob . 4.1 8
//\\V /\\\//\W/A\ \
4.17 Figur e Prob. 4.17 shows sheet piles driven into a permeable stratum. Construct the flow net
and determin e th e quantit y o f seepag e i n m
3
/hour pe r mete r lengt h o f piling . Assum e
& = 8 x 10-
4
cm/sec.
4.18 Fig . Prob . 4.18 gives a cross sectio n o f a concrete dam . The subsoil i s anisotropic an d has
permeabilities k
h
= 0.8 x 10"
6
in./se c an d k
v
- 2. 0 x 10~
7
in./sec . Fin d th e rat e o f flo w
beneath the dam pe r foot lengt h of the dam. Assume N ,= 4, and N
d
= 8.
-HsftH-
= 30 f t
T
= 40 ft
Permeable sand
k = 40 x 10"
3
in./sec
Impermeable sand
Figure Prob. 4.19
142 Chapt er 4
4.19 Construc t a flow net in Fig. Prob. 4.19 and estimate th e seepage los s i n ft
3
per hour per foot
length of weir .
4.20 A homogeneou s eart h da m i s show n i n Fig . Prob . 4.20 . Sketc h th e phreati c lin e an d
estimate th e quant i t y o f seepage .
20f t
Impervious bas e
Figure Prob. 4.2 0
fc . t •/.• ' [ • .\ _ -j X' N. .
32f t
CHAPTER 5
EFFECTIV E STRESS AND PORE WATER
PRESSURE
5 .1 I NTRODU CTI O N
The pressur e transmitte d throug h grai n t o grai n a t th e contac t point s throug h a soi l mas s i s
termed as intergranular or effective pressure. It is known as effective pressur e since thi s pressure
is responsibl e fo r th e decreas e i n th e voi d rati o o r increas e i n th e frictiona l resistanc e o f a soi l
mass.
If the pores of a soil mass ar e filled wit h water and if a pressure induced into the pore water,
tries t o separat e th e grains , thi s pressur e i s terme d a s pore water pressure o r neutral stress. The
effect o f this pressure is to increase the volume or decrease the frictional resistance of the soil mass.
The effect s o f the intergranular and pore water pressures on a soil mass can be illustrated by
means of simpl e practical examples .
Consider a rigid cylindrical mold, Fig. 5.1(a), in which dry sand is placed. Assume that there
is no side friction. Load Q is applied at the surface of the soil through a piston. The load applied a t
the surface is transferred to the soil grains in the mold through their points of contact. If the load is
quite considerable, i t would result in the compression of the soil mass in the mold. The compression
might be partly due to the elastic compression of the grains at their points of contact and partly due
to relative sliding between particles . If the sectional are a of the cylinder is A, the average stres s at
any leve l XY ma y be writte n as
-«=f (5.1 )
The stres s a
a
i s th e averag e stres s an d no t the actua l stres s prevailin g a t th e grai n t o grai n
contacts whic h is generall y ver y high. Any plane suc h as XY wil l not pass through all the point s of
contact an d man y o f th e grain s ar e cu t b y th e plan e a s shown i n Fig . 5.1(b) . Th e actua l point s of
143
144 Chapt er 5
contact exhibi t a wavy form. However , fo r al l practical purposes th e averag e stres s i s considered .
Since this stress is responsible for the deformation of the soil mass, it is termed the intergranular o r
effective stress. We may therefor e write,
a = (5.2)
where cr'i s the effective stress.
Consider no w anothe r experiment . Le t th e soi l i n th e mol d b e full y saturate d an d mad e
completely watertight. If the same load Q is placed on the piston, this load wil l not be transmitted to
the soi l grains as i n the earlier case. I f we assume that water is incompressible, th e external load Q
will be transmitted to the water in the pores. This pressure that is developed i n the water is called the
pore water or neutral stress u
w
as shown schematically in Fig. 5.1(c). This pore wate r pressure u
w
prevents the compression of the soi l mass. The valu e of thi s pressure i s
G
A
(5.3)
If th e valve V provided in the piston is opened, immediatel y there wil l be expulsion of water
through the hol e i n the piston. The flow of water continues for some time and the n stops .
The expulsion of water from th e pores decreases th e pore water pressure and correspondingly
increases th e intergranula r pressure. At an y stage th e tota l pressur e Q/A i s divide d between water
and the point s of contact of grains. A new equation may therefor e be written as
Total pressur e cr
[
- — = Intergranular pressure + pore water pressure
A
Piston
Rigid cylindrical
mold
(a) Soi l unde r load i n a rigid containe r
(b) Intergranular pressur e (c ) Porewater pressure,
Figure 5. 1 Effect iv e an d por e wat e r pres s ure s
Effect ive St res s an d Por e Wat e r Pres s ur e 14 5
or a
t
=<?'+u
w
(5.4 )
Final equilibriu m will be reached whe n there i s no expulsion of water. At this stage the por e
water pressure u
w
= 0. All the pressure wil l be carried b y the soi l grains . Therefore, w e can write,
a
t
= <r' (5.5)
The pore water pressure u
w
can be induced in the pores of a soil mass by a head of water over
it. When there is no flow o f water through the pores of the mass, the intergranular pressure remains
constant a t an y level . Bu t i f ther e i s flow , th e intergranula r pressur e increase s o r decrease s
according to the direction of flow. I n partially saturate d soil s part of the voi d space is occupied by
water and part by air. The pore water pressure u
w
must always be less than the pore air pressure ( u
a
).
Bishop (1955 ) propose d a n equatio n fo r computin g th e effectiv e pressur e i n partiall y saturated
soils. This equation contains a parameter which cannot be determined easily. Sinc e this equation is
only o f academi c interest , no furthe r discussio n is necessary here.
5.2 STRESSE S WH EN NO FL OW TAKE S PL ACE TH ROU G H TH E
SATU RATED SOI L M ASS
In Fig. 5. 2 the container A is filled wit h sand to a depth z
l
an d water to a dept h z
2
abov e th e sand
surface. A flexibl e tube connects the bottom of the container A to another containe r B. The water
levels are kept constant in these two containers.
The water surfaces i n both the containers i n Fig. 5.2(a) are kept at the same level . Under this
condition, no flow take s plac e fro m on e container to another.
Consider two points M and N as shown in the figure on a horizontal plane. The water pressure
at M shoul d be equal t o the pressur e a t N according t o the laws of hydraulics. Therefore,
the wate r pressure a t N =U
Z
= ( Z +z
2
)Y
w
(5-6 )
The pressur e u
z
is termed a s the por e water pressure acting on the grains at dept h z from th e
surface of the sample. However, the total pressure at point N is due to the water head plus the weight
of the submerged soi l above N. If y
b
is the submerged unit weight of the soil , the total pressure at N is
a
z
= zy
b
+( z + z
2
)y
w
(5.7 )
The intergranula r or effective pressure a t the point N i s the difference between th e total and
the por e wate r pressures. Therefore , th e effectiv e pressur e CF, ' is
<r
z
=<r
z
-u
z
=zr
b
+( z + z
2
)r
w
-( z + z
2
')r
w
=zr
b
(5.8a )
Equation (5.8a ) clearl y demonstrate s tha t th e effectiv e pressur e cr, ' i s independen t o f th e
depth of water z
2
above the submerged soi l surface. The total por e water and effective pressures at
the bottom of the soi l sampl e ar e as follows
Total pressure cr
t
= a
c
= (z, + Z
2
)Y
W
+ z\ Y
b
(5.8b )
Pore water pressure u
c
= ( zi +Z
2
)y
w
(5.8c )
Effective pressur e a'
c
- ( o
c
~u
c
} = z\ Yb(5.8d )
The stres s diagrams ar e shown in Fig. 5.2(b) .
146 Chapt er 5
Total
M
f
z
L
N
<•!
Stress diagrams
Pore wate r
o
Effective
o o'
(a) (b)
Figure 5. 2 St res s e s whe n n o fl ow t ak e s pl ac e
5 .3 STRESSE S WH EN FL OW TAKE S PL ACE TH ROU G H TH E SOI L
FROM TOP TO B OTTO M
In Fig . 5.3(a ) th e wate r surfac e i n containe r B i s kep t a t h unit s belo w th e surfac e i n A. Thi s
difference i n head permit s wate r t o flow fro m containe r A t o B.
Since containe r B wit h th e flexibl e tub e ca n b e considere d a s a piezomete r tub e freel y
communicating wit h th e botto m o f containe r A, the piezometri c hea d o r th e por e wate r pressur e
head at the bottom o f container A is (z, + z
2
- h) . Therefore, the pore water pressure u
c
at the bottom
level i s
u = (5.9)
As per Fig . 5.3(a) , th e por e wate r pressur e a t the bottom o f container A whe n n o flo w take s
place throug h the soi l sampl e i s
« =
(5.10)
It is clear from Eq. (5.9) and (5.10) that there i s a decrease in pore water pressure t o the extent
of hy whe n water flows through the soil sampl e from top to bottom. I t may be understood that this
decrease in pore water pressure i s not due to velocity of the flowing water. The value of the velocity
head V
2
/2g i s a negligible quantity even i f we take the highest velocit y of flow that is encountered
in natural soil deposits . As in Fig. 5.2(a), the total pressure o
c
at the bottom of the container i n this
case also remain s th e same . Therefore ,
The effectiv e pressure <J
C
' a t th e botto m o f th e containe r i s
-
u
(5.11)
(5.12)
Equation (5.12 ) indicate s tha t i n thi s cas e ther e i s a n increas e i n th e effectiv e pressur e b y
hy a t the botto m o f the containe r A a s compared t o the earlier case. The effectiv e pressur e a t the
top surfac e of th e sampl e i s zer o a s before. Therefore , th e effectiv e pressure cr. / at any dept h z ca n
be writte n as
Effect ive St res s an d Por e Wat e r Pres s ur e 147
r\
\
•I
Total
Stress diagrams
Pore water Effective
a'
(a) (b )
Figure 5. 3 St res s e s when flow t ak e s place from t o p t o bot t o m
Z\
(5.13)
Equation (5.13 ) indicate s that hzYjz
{
i s the increas e i n the effectiv e pressur e a s the water
flows fro m th e surfac e t o a depth z. This increas e i n effectiv e pressur e du e t o th e flo w o f water
through the pores o f the soi l i s known as seepage pressure. It may be noted that h is the total loss
of head a s the water flows from th e top surfac e of the sample t o a depth z
r
The correspondin g loss of head a t depth z is ( z/z^h. Sinc e (/z/Zj ) = / , the hydraulic gradient,
the loss of head at depth z can be expressed a s iz. Therefore th e seepage pressur e at any depth may
be expressed a s izy
w
- Th e effective pressur e a t depth z can be written as
<r'
z
=zr
b
+izr
w
(5.14) '
The distribution of pore water and effective pressure s are shown in Fig. 5.3(b). In normal soil
deposits whe n flow take s place i n the direction of gravity there wil l be an increase i n the effectiv e
pressure.
5.4 STRESSE S WH EN FL OW TAKES PL ACE TH ROU G H TH E SOI L
FROM B OTTOM TO TOP
In Fig. 5.4(a), th e water surface in container B i s kept above that of A by h units. This arrangement
permits water to flow upward s through the sample in container A. The total piezometric o r the pore
water head at the bottom of the sampl e i s given by
( z
1
+ z
2
+ / z )
Therefore, th e por e wate r pressure u
c
at the bottom of the sampl e is
(5.15)
As befor e th e tota l pressur e hea d o~
c
at the bottom o f the sampl e i s
(5.16)
148 Chapt er 5
Stress diagram s
\
I
Total Effective
z/b ~ *zy»
(a) (b )
Figure 5. 4 St res s e s whe n fl ow t ak e s plac e fro m bot t o m t o t o p
The effectiv e pressur e o~
c
' a t th e botto m o f sampl e is , therefore ,
= °c ~
u
c =
hy
As i n Eq . (5.14 ) th e effectiv e pressur e a t any dept h z ca n b e writte n a s
(5.17)
(5.18)
Equation (5.18 ) indicate s tha t ther e i s a decreas e i n th e effectiv e pressur e du e t o upwar d
flow o f water . At an y dept h z, zy
b
i s the pressur e o f the submerge d soi l actin g downwar d an d izy
b
is the seepag e pressur e actin g upward . The effectiv e pressure o~ ' reduces t o zer o whe n thes e tw o
pressures balance . This happens when
' = zy
b
- izy = 0 o r / = / = •
(5.19)
Equation (5.19 ) indicate s tha t th e effectiv e pressur e reduce s t o zer o whe n th e hydrauli c
gradient attain s a maximum value which is equal t o the rati o o f the submerge d uni t weight of soi l
and th e uni t weigh t of water . This gradien t i s known as the critical hydraulic gradient i
c
. In suc h
cases, cohesionless soil s lose al l of their shear strengt h and bearing capacit y and a visible agitation
of soi l grain s i s observed . Thi s phenomeno n i s known a s boiling o r a quick sand condition. By
substituting i n Eq. (5.19 ) for y
b
y
I
-i)
'
l + e
we have
G - 1
(5.20)
The critical gradient of natural granular soil deposits ca n be calculated i f the void ratios of the
deposits ar e known . Fo r al l practica l purpose s th e specifi c gravit y o f granula r material s ca n b e
assumed a s equa l t o 2.65 . Tabl e 5. 1 give s th e critica l gradient s o f granula r soil s a t different voi d
ratios ranging from 0. 5 t o 1.0 .
Effect ive St res s an d Por e Wat e r Pres s ur e 14 9
Table 5. 1 Crit ica l hydrauli c g radien t s o f g ran ula r soil s
Soil No . Voi d rat i o i
c
1 0. 5 1.1 0
2 0. 6 1.0 3
3 0. 7 0.9 7
4 0. 8 0.9 2
5 1. 0 0.8 3
It ca n b e see n fro m Tabl e 5. 1 tha t th e critica l gradien t decrease s fro m 1.1 0 b y abou t 2 5
percent onl y as the voi d ratio increases by 10 0 percent from a n initial value of 0.5 t o 1.0 . The void
ratio of granular deposits generall y lies withi n the range of 0.6 to 0.7 and as such a critical gradient
of unity can justifiably b e assumed for all practical purposes. It should be remembered tha t a quick
condition does not occur i n clay deposit s sinc e the cohesive force s between the grains prevent the
soil fro m boiling .
Quick condition s ar e commo n i n excavation s belo w th e groun d wate r table . Thi s ca n b e
prevented by lowerin g the ground water elevation b y pumping before excavation. Quick conditions
occur most often i n fine sands or silts and cannot occur in coarse soils. The larger the particle size, the
greater i s th e porosity. To maintain a critical gradien t of unity , th e velocit y a t which water must b e
supplied at the point of inflow varie s as the permeability. Therefore a quick condition cannot occur in
a coarse soi l unless a large quantity of water can be supplied.
5.5 EFFECTIV E PRESSURE DU E TO CAPILLARY WATER RISE I N SOI L
The ter m wate r level , wate r tabl e an d phreati c surfac e designat e th e locu s o f th e level s t o which
water rises in observation well s in free communication wit h the voids of the soi l at a site. The water
table can also be defined as the surface at which the neutral stress u
w
in the soi l is equal to zero.
If the water contained in the soil wer e subjected t o no force othe r than gravity, the soi l above
the wate r tabl e woul d b e perfectl y dry . I n reality , ever y soi l i n th e fiel d i s completel y saturate d
above thi s level up t o a certain height . The water that occupies th e voids of the soi l locate d abov e
the water table constitutes soil moisture.
If th e lowe r par t o f the mass o f dr y soi l comes int o contact wit h water , th e wate r rises in
the voids t o a certain heigh t above the free water surface. The upward flow into the voids of the
soil i s attributed t o the surface tension of the water . The heigh t to whic h water rise s above th e
water tabl e agains t th e forc e o f gravit y i s calle d capillary rise. The heigh t o f capillar y ris e i s
greatest fo r ver y fine graine d soi l materials . Th e wate r tha t rises abov e th e wate r tabl e attain s
the maximu m heigh t h
c
onl y i n th e smalle r voids . A fe w larg e void s ma y effectivel y sto p
capillary rise in certain part s of the soi l mass . As a consequence, onl y a portion o f the capillary
zone abov e th e fre e wate r surfac e remain s ful l y saturate d an d th e remainde r i s partiall y
saturated.
The sea t of the surfac e tension i s located a t the boundary between ai r and water . Within the
boundary zone the water is in a state of tension comparable t o that in a stretched rubbe r membrane
attached t o th e wall s o f th e void s o f a soil . However , i n contras t t o th e tensio n i n a stretche d
membrane, th e surfac e tensio n i n the boundar y fil m o f wate r i s entirel y unaffecte d b y eithe r th e
contraction or stretching of the film. The water held in the pores of soil above the free wate r surface
is retained i n a state of reduced pressure . Thi s reduce d pressur e is called capillary pressure o r soil
moisture suction pressure.
The existenc e o f surface tension can be demonstrated a s follows:
150 Chapt er 5
2 T
e
L cos a
Figure 5. 5 Needl e s m eare d wit h g reas e float in g o n wat e r
A grease d sewin g needle , Fig . 5.5 , ca n b e mad e t o floa t o n wate r becaus e wate r ha s n o
affinity t o grease, and , therefore , th e wate r surface curve s down unde r the needl e unti l the upward
component o f the surfac e tensio n i s large enoug h t o suppor t th e weigh t of the needle . I n Fig. 5.5 ,
7^ i s the surfac e tensio n pe r uni t lengt h of the needl e an d W
n
the weight of the needle . Th e upwar d
vertical forc e du e t o surface tension i s 2TL co s a , wher e L is the length of the needle. Th e needl e
floats whe n thi s vertical forc e i s greater tha n the weight of the needl e W
n
acting downwards .
Rise o f Wate r i n Capillar y Tube s
The phenomenon o f capillary rise can be demonstrated b y immersing th e lower end of a very smal l
diameter glas s tub e into water. Suc h a tube is known as capillary tube . As soon as the lower en d of
the tub e come s int o contact wit h water , the attraction betwee n th e glass an d th e wate r molecule s
combined wit h the surfac e tension of the water pulls the water up into the tube to a height h
c
above the
water leve l a s shown i n Fig. 5.6(a) . The height h
c
i s known a s the height o f capillar y rise . The uppe r
surface of water assumes the shape o f a cup, called the 'meniscus' tha t joins the wall s of the tube at an
angle a known as the contact angle.
On the othe r hand , if the tube is dipped int o mercury a depression o f the surfac e develop s i n
the tub e belo w th e surfac e o f th e mercury , wit h the formatio n of a conve x meniscu s a s shown i n
Fig. 5.6(b) . The reason fo r the difference between th e behavior o f water an d mercury resides in the
different affinit y betwee n th e molecule s o f th e soli d an d wate r o r mercury . I f ther e i s a stron g
affinity betwee n th e molecules o f the soli d and the liquid, the surface of the liqui d will climb up on
the wal l o f th e soli d unti l a definit e contac t angl e a i s established . Th e contac t angl e betwee n a
clean mois t glas s surfac e an d wate r i s zero , tha t is , th e wate r surfac e touche s th e glas s surfac e
tangentially. For the case of a dry glass surfac e and water, a is not a constant. I t may be as high as
45° a t first the n graduall y reducing t o much smalle r values . Probabl y th e inevitabl e contaminatio n
of surface s cleane d b y ordinar y methods , an d th e humidit y o f ai r ar e responsibl e fo r suc h
variations. Fig . 5.6(c ) show s th e contac t angle s betwee n wate r an d th e surface s unde r differen t
conditions.
Surface Tensio n
Surface tensio n i s a force that exists at the surface of the meniscus. Along the line of contact betwee n
the meniscus i n a tube and the walls of the tube itself, the surface tension, T
s
, is expressed a s the force
per uni t length acting i n the direction of the tangent as shown i n Fig. 5.7(a) . The component s o f this
force along the wal l and perpendicular to the wall are
Along the wal l = T co s a pe r uni t lengt h of wal l
Effect ive St res s an d Por e Wat e r Pres s ur e 151
Water
(a)
Meniscus
T,
Glass tub e
Meniscus
Glass tube
Convex meniscu s
\
Mercury
(b)
a = 00 < a < 45° a > 90°
Moist glas s Dr y glass Greas y glas s
surface surfac e surfac e
(c)
Figure 5. 6 Capillar y ris e an d m en is cus
Normal t o th e wal l = T
s
si n a pe r uni t lengt h o f wall .
The forc e norma l t o the wal l tries t o pull the walls of the tube together and the one along th e wall
produces a compressive forc e i n the tube below th e line of contact .
The meniscu s ca n b e visualize d a s a suspensio n bridg e i n thre e dimension s whic h i s
supported o n th e wall s o f th e tube . Th e colum n o f wate r o f heigh t h
c
belo w th e meniscu s i s
suspended fro m thi s bridg e b y mean s o f th e molecula r attractio n o f th e wate r molecules . I f th e
meniscus ha s stoppe d movin g upwar d i n th e tube , the n ther e mus t b e equilibriu m betwee n th e
weight of the column of water suspended from th e meniscus and the force wit h which the meniscus
is clinging to the wal l of the tube . We can writ e the following equation of equilibrium
TidT cos « = or h =
4T co s a
(5.21)
The surfac e tensio n T
s
for wate r at 20 °C can be taken as equal t o 75 x 10~
8
kN per cm. The
surface tension s o f some of the common liquid s ar e given in Table 5.2.
Equation (5.21 ) can be simplified by assuming a = 0 for moist glass and by substituting for T
s
.
Therefore, fo r the case of water, the capillary height h
c
can be writte n as
h =
47" 4x 75 x l O-
8
x l 0
6
03
d dy
w
dx9.Sl
In Eq. (5.22 ) h an d d are expressed i n cm, and, v= 9.81 kN/m
3
.
(5.22)
152 Chapt er 5
Table 5. 2 Surfac e t en s io n o f s om e liquid s a t 20 ° C
Liquids 7"k N/c m x 1CT
8
Et hyl Alcoho l
Benzene
Carbon Tetr a Chlorid e
Mercury
Petroleum
Water
22.03
28.90
26.80
573.00
26.00
75.00
Stress Distributio n i n Water B elo w the M eniscu s
Figure 5.7(b ) show s a capillar y tub e wit h it s botto m en d immerse d i n water . Th e pressur e i s
atmospheric a t point s A an d B. Sinc e point C is a t the same leve l a s A, according t o the laws of
hydraulics, the pressure at C is also atmospheric. Since the point D which is just below the meniscus
is higher than point C by the head h
c
, the pressure at D must be less than atmospheric by the amount
h
c
y
w
. Therefore, th e pressure a t any point in water between C and D i s less tha n atmospheric. That
means, the wate r above poin t C is i n tension i f we refer t o atmospheri c pressur e a s zer o pressure .
The tension in water at any height h above C is given by hy
w
. By contrast , the pressure i n the water
below th e fre e surfac e A i s abov e atmospheri c an d therefor e i s i n compression . Th e stres s
distribution i n water is given in Fig. 5.7(b) .
T
r
si n a -*-
a
Water
-*- T
s
si n a
a
- Glas s tub e
(a) Forces due t o surface tension
Tension
Stress
distribution
\
(b)
u
c
= 4 T
s
ld
Water
\Capillar y tub e wal l
under compressio n
(c)
Figure 5. 7 Capillar y pres s ure
Effect ive St res s an d Por e Wat e r Pres s ur e
Thus th e tension u
w
i n wate r immediatel y below th e meniscus is give n by
47 cos a
153
(5.23)
If r
m
i s the radiu s of the meniscus , Fig. 5.7(a) , we ca n write ,
d
r = or d = 2r co s a
m
2cos a
Substituting for d i n Eq. (5.23) , w e have
u = —
4T
s
co s a
2r cos a
2T
s
r
(5.24)
It may be noted here that at the level of the meniscus the magnitude o f the capillary pressur e
u tha t compresses th e wall of the tube i s also equal t o the capillary tensio n in the water just below
the meniscus. The magnitude of the capillary pressur e u
c
remains constant wit h dept h a s shown in
Fig. 5.7(c ) wherea s th e capillar y tension , u
w
, i n wate r varie s fro m a maximu m o f h
c
Y
w
a t th e
meniscus level to zero at the free wate r surface level as shown in Fig. 5.7(b) .
Capillary Ris e o f Wate r i n Soil s
In contrast t o capillary tubes the continuous voids in soils have a variable width. They communicat e
with each othe r i n all directions an d constitute an intricate networ k of voids. When water rises into
the network from below, the lower par t of the network becomes completel y saturated . I n the upper
part, however , th e water occupies onl y the narrowest void s and the wider area s remai n fille d wit h
air.
Fig. 5.8(a) shows a glass tube filled with fine sand. Sand would remain full y saturate d only up
to a heigh t h' whic h i s considerabl y smalle r tha n h
c
. A fe w larg e void s ma y effectivel y sto p
capillary rise i n certain parts . The water woul d rise, therefore, t o a height of h
c
only in the smalle r
voids. The zone between the depths ( h
c
- h'J wil l remain partiall y saturated .
Soil
sample
s
V
~
**,"/:•• :-v
ft
• V °-j :!••'£
/
1
r
h
cm
)ry zone f f 15 0
T
CM
T
e
w
Partially - g 10 0
h
c
- h'
c
saturate d &
zone = 3
c , '§ ,
?? s n
<+-
, , Saturate d 2
"c £
i zon e O B
i 3
y
/)
A
/
^
"10 1. 0 0. 1 0.0 1 0.00 1
Grain siz e (mm) log scale
(a) Height o f capillary ris e (b) Rate o f capillary ris e i n soi l consistin g o f
uniform quart z powde r
Figure 5. 8 Capillar y ris e i n soil s
154 Chapt e r s
Capillary ris e of wate r
/~z£r\
_ T ,
////////SS/////S/S/////////////////////////S//7///////S/7////S//S/7/S//S
Figure 5. 9 Capillar y s iphon in g
The heigh t o f the capillar y ris e i s greates t fo r ver y fin e graine d soil s materials , bu t th e rat e
of ris e i n suc h material s i s slo w becaus e o f thei r lo w permeability . Fig . 5.8(b ) show s th e
relationship between th e height of capillary ris e i n 24 hour s and the grai n siz e o f a unifor m quartz
powder. This clearl y show s tha t the rise i s a maximum for material s falling i n the categor y o f silt s
and fin e sands .
As th e effectiv e grai n siz e decreases, the siz e o f the void s als o decreases , an d th e heigh t of
capillary ris e increases. A rough estimation of the height of capillary ris e can be determine d fro m
the equation ,
C
h
^^D~
(5
'
25)
eL J
\ Q
in whic h e is the voi d ratio, D
IQ
i s Hazen's effective diameter i n centimeters, an d C is an empirical
constant whic h ca n have a value between 0. 1 and 0.5 sq. cm.
Capillary Siphonin g
Capillary force s ar e abl e t o raise wate r agains t the force o f gravit y not onl y int o capillary tube s or
the void s i n column s o f dr y soil , bu t als o int o narrow ope n channel s o r V -shaped grooves . I f th e
highest point of the groove i s located belo w the level t o which the surface tension can lif t th e water ,
the capillary force s wil l pul l the water into the descending par t of the groove an d wil l slowl y empty
the vessel . Thi s proces s i s known as capillary siphoning. Th e same proces s ma y als o occu r i n the
voids of soil . For example, water may flow ove r the crest o f an impermeabl e cor e in a dam i n spit e
of th e fac t tha t th e elevatio n o f th e fre e wate r surfac e i s belo w th e cres t o f th e cor e a s show n i n
Fig. 5.9.
Capillary Pressur e i n Soil s
The tensio n u
w
in water j ust below th e meniscus i s given by Eq. (5.23) as
4T cost f
Since thi s pressur e i s below atmospheri c pressure , i t draws the grains o f soil s close r t o eac h
other a t al l point s wher e th e menisc i touc h th e soi l grains . Intergranula r pressur e o f thi s typ e i s
called capillary pressure. The effectiv e o r intergranular pressure a t any point in a soi l mas s ca n be
expressed b y
o
f
= a-u,(5.26 )
Effect ive St res s an d Por e Wat er Pres s ur e
Capillary fring e
155
(b) (c ) (d ) (e )
Figure 5.1 0 Effec t o f capillar y pres s ur e u
c
o n soi l vert ical s t res s diag ra m
where o
t
i s th e tota l pressure, t f i s th e effectiv e o r the intergranula r pressure an d u
w
i s th e por e
water pressure . Whe n th e wate r i s i n compressio n u
w
i s positive , an d whe n i t i s i n tensio n u
w
is
negative. Sinc e u
w
is negative i n the capillar y zone , the intergranular pressure i s increased b y u
w
.
The equation , therefore, ca n b e writte n as
o
f
= a
t
-( -u
w
) = a
t
+ u
w
(5.27 )
The increas e i n th e intergranula r pressur e du e t o capillar y pressur e actin g o n th e grain s
leads t o greate r strengt h o f the soi l mass .
Stress Conditio n i n Soi l due to Surfac e Tensio n Force s
It i s to be assume d her e tha t the soi l abov e the ground water table remain s dr y prior t o the rise of
capillary water . The stress condition in the dry soil mass changes due to the rise of capillary water .
Now conside r th e soi l profil e given in Fig. 5.10(a) . When a dr y soi l mas s abov e th e GWT
comes i n contact with water, water rises by capillary action. Let the height of rise be h
c
and assume
that th e soi l withi n thi s zon e become s saturate d due t o capillar y water . Assume tha t th e menisc i
formed a t heigh t h
c
coincid e wit h th e groun d surface . Th e plan e o f th e menisc i i s calle d th e
capillary fringe.
The vertical stres s distribution of the dry soil mass is shown in Fig 5.10(b). The vertical stres s
distribution o f th e saturate d mas s o f soi l i s give n i n Fi g 5.10(d) . Th e tensio n i n th e wate r i s
maximum at the menisci level , say equal to u
w
and zero at the GWT level as shown in Fig. 5.10(e).
Prior t o capillary rise the maximum pressure of the dry mass, rf
d
, a t the GWT level is
where, y
d
= dry uni t weight of soil .
After th e capillar y rise , th e maximu m pressure o f the saturate d weigh t o f soi l a t the GWT
level is
Since th e por e wate r pressur e a t th e GWT leve l i s zero , i t i s obviou s tha t th e differenc e
between th e two pressure s o
/
sat
an d tf
d
represent s th e increas e i n pressur e du e t o capillar y ris e
which is actually the capillary pressure , whic h may be expressed a s
Mr ~~i-Wsa t ~ ~ I A) (.3. )
156 Chapt er 5
By substitutin g for
^, an d
l + e
in Eq . (a) , w e have , afte r simplifyin g
l + e
ct
(5.28)
where, e = void ratio ,
n = porosity
It i s clea r fro m Eq . (5.28 ) tha t the capillar y pressur e fo r soi l i s directl y proportional t o th e
porosity o f the soi l and this pressur e i s very much less than h./ whic h is used onl y for a fine bore
and unifor m diameter capillar y tube.
The distributio n o f capillar y pressur e u
c
(constan t wit h depth ) i s give n i n Fig . 5.10(c) . Th e
following equatio n for the pressure at any depth z may be written as per Fig. 5.10
(5.29)
Example 5 . 1
The dept h of water in a well is 3 m. Below the bottom of the well lies a layer of sand 5 meters thick
overlying a clay deposit . Th e specifi c gravit y of the solid s o f san d an d cla y ar e respectivel y 2.6 4
and 2.70. Thei r water contents are respectively 25 and 20 percent. Comput e the total, intergranular
and pore water pressures a t points A an d B shown in Fig. Ex . 5.1.
Solution
The formul a for th e submerged uni t weight is
l + e
Since the soi l i s saturated,
3m
5 m
2 m
2 m
Sand
Clay
Figure Ex . 5. 1
Effect ive St res s an d Por e Wat e r Pressur e 15 7
_ . 9.81(2.64-1 )
3
For sand, y, = = 9.7 kN/m
J
* 1 + 0.25x2.64
For clay, y ,
=
9
'
81(2JO
"^ = 10.83 kN/m
3
* 1 + 0.20x2.70
Pressure at point A
(i) Tota l pressur e = 3 x 9. 7 (sand ) + 6 x 9.81 = 29. 1 + 58. 9 = 88 kN/m
2
(ii) Effectiv e pressur e = 3 x 9. 7 = 29. 1 kN/m
2
(iii) Por e wate r pressur e = 6 x 9.81 = 58.9 kN/m
2
Pressure at point B
(i) Tota l pressur e = 5 x 9. 7 + 2 x 10.8 3 + 1 0 x 9.81 = 168. 3 kN/m
2
(ii) Intergranula r pressur e = 5 x 9. 7 + 2 x 10.8 3 = 70.2 kN/m
2
(iii) Por e wate r pressur e = 10x 9. 81= 98. 1 kN/m
2
Example 5. 2
If water in the wel l in example 5. 1 is pumped out up to the bottom o f the well, estimat e th e change
in the pressures a t points A an d B given in Fig. Ex . 5.1.
Solution
Change i n pressur e a t points A and B
(i) Chang e i n total pressur e = decreas e in water pressur e du e to pumping
= 3x9. 81=29. 4 3 kN/m
2
(ii) Chang e i n effective pressur e = 0
(iii) Chang e i n pore wate r pressure = decreas e i n water pressure du e to pumping
= 3x9. 81=29. 4 3 kN/m
2
Example 5. 3
A trench i s excavated i n fine san d for a building foundation, u p t o a depth o f 1 3 ft. The excavatio n
was carried ou t by providing the necessary sid e supports for pumping water. The water levels a t the
sides an d the botto m o f the trenc h ar e a s given Fig . Ex . 5.3 . Examin e whethe r th e botto m o f the
trench i s subjected t o a quick condition i f G
s
= 2.64 an d e = 0.7. I f so, what is the remedy ?
Solution
As per Fig. Ex. 5.3 the depth of the water table above the bottom of the trench = 10 ft. The sheeting
is taken 6.5 f t below th e bottom of the trench to increase th e seepage path.
G - 1
The equatio n for the critical gradient is / =
l + e
If the trench i s to be stable, th e hydraulic gradient, / , prevailing at the bottom shoul d be les s
than i . The hydraulic gradien t i is
158 Chapt er 5
//A\V /A\\
3 f t
//A\V /A\\
10f t
6. 5f t
_L
Figure Ex . 5. 3
There wil l be n o quic k conditio n if ,
L l + e
From th e give n dat a
2.64-1
=
1.6 4
1 + 0.7 1. 7
*= - *= , .54
L 6. 5
It i s obvious tha t h/L > i
c
. There wil l b e quic k condition .
Remedy:
(i) Increas e L t o a t leas t a 1 3 ft dept h belo w th e botto m o f trenc h s o tha t h/L = 0.77 whic h
gives a margi n o f facto r o f safety .
or (ii ) Kee p th e wate r tabl e outside the trench at a low level by pumping out water . This reduce s
the hea d h .
or (iii ) D o no t pump wate r u p t o the bottom leve l o f th e trench. Arrange th e wor k i n suc h a way
that th e wor k ma y be carrie d ou t wit h some wate r in the trench.
Any suggestio n give n above shoul d be considered b y keepin g i n vie w th e sit e condition s
and other practical considerations .
Example 5. 4
A cla y laye r 3.6 6 m thick rest s beneat h a deposit o f submerge d san d 7.9 2 m thick. The to p of the
sand i s locate d 3.0 5 m below th e surfac e of a lake. Th e saturate d uni t weight o f the san d i s 19.6 2
kN/m
3
an d of the clay i s 18.3 6 kN/m
3
.
Compute (a ) th e tota l vertica l pressure , (b ) th e por e wate r pressure , an d (c ) th e effectiv e
vertical pressur e a t mi d height of the clay layer (Refe r t o Fig. Ex . 5.4) .
Solution
( a) Total pressure
The tota l pressur e cr , over the midpoint of the clay i s due t o the saturate d weight s of clay an d
sand layer s plus the weigh t of water over th e bed o f sand, that is
Effect ive St res s an d Por e Wat e r Pres s ur e 159
T '
3.05m Lake
/psxXK^v&tfSS^^
7 . 9 2 m. ' • ' • • ' • ' • . ' '••':••':''•. Submerged sand :
:
' . . . ' • • ' [ • ' • ' . ' / •
3. 66m'
I
!
'6--''v'
f
' '/• '• •
?
. J '''--/*'.'"'•}.•••'•'.••'•£•'•: .v
:
'-;
:
'*'.'''*V j'l'
:
C *•••
_I_ ' • '•'•
:
-'-'- ' *••"''••' '-."• ..*•• ' ';'• ":• . " v • ."-'• .• '• ."• .. ' •• ' ':'• ":• . • *• ; '• ..' •
cr, =
3.66
Figure Ex . 5. 4
x 18.36 +7.92 x 19.62 + 3.05 x 9.81 = 33.6 + 155.4 + 29.9 = 218.9 kN/m
2
( b) Pore water pressure is due to the total water column above the midpoint.
That i s
3.66
u.. = x 9.81 + 7.92 x 9.81 + 3.05 x 9.81 = 125. 6 kN/m
2
( c) Effective vertical pressure
a-u = &' = 218.9-125.6 = 93.3 kN/m
2
Example 5. 5
The surfac e of a saturated cla y deposi t i s located permanentl y below a body o f water a s shown in
Fig. Ex. 5.5. Laborator y test s hav e indicate d tha t th e averag e natura l wate r conten t o f the cla y i s
41% an d that the specifi c gravity of the soli d matte r is 2.74. What is the vertical effective pressur e
at a depth o f 37 ft below the top o f the clay.
Solution
To find th e effective pressure, w e have t o find firs t th e submerge d uni t weight of soi l expresse d a s
Y
h
=
l + e
wG,
Now fro m Eq. (3.14a) , e = *- = wG
s
sinc e 5 = 1
or e = 0.47 x 2.74 = 1.29
Therefore,
(2.74-
1
.00)x62.4
=47411b/ft3
* 1 + 1.29
160 Chapt er 5
Lake
/ A\V A\V A\V A\\/ / > , \V A\V A\V A\\
Clay deposi t
37f t
D
/A\V A\\
A\V A\V A\V W\\
Figure Ex . 5. 5
Effective pressure , a' = 37 x 47.41 = 1754 lb/ft
:
Example 5 . 6
If the water level i n Ex. 5. 5 remains unchanged and an excavation i s made by dredging, wha t depth
of clay mus t be removed to reduce the effective pressure at point A at a depth of 37 ft by 1000 lb/ft
2
?
(Fig. Ex. 5.5)
Solution
As in Ex. 5.5, y
b
= 47.41 lb/ft
3
, le t the depth of excavation be D. The effective depth over the point
A i s (3 7 - D ) ft . Th e dept h o f D mus t b e suc h whic h give s a n effectiv e pressur e o f
(1754 - 1000 ) lb/ft
3
= 754 lb/ft
2
or (3 7 - D)x 47.41 =75 4
^ 37x47.41-75 4 _
1 i r
or D = = 21.1 ft
47.41
Example 5 . 7
The wate r tabl e i s lowered fro m a depth of 1 0 ft to a depth of 20 ft in a deposit o f silt . All the sil t is
saturated eve n afte r the water table is lowered. It s water content is 26%. Estimate th e increase i n the
effective pressur e a t a depth of 34 f t on account of lowering the wate r table. Assume G
s
= 2.7.
Solution
Effective pressur e befor e lowerin g the water table .
The wate r tabl e i s a t a depth of 1 0 ft and th e soi l abov e thi s dept h remains saturate d but no t
submerged. Th e soi l fro m 1 0 ft t o 2 0 f t remain s submerged . Therefore , th e effectiv e pressur e a t
34 f t dept h i s
(34-10)^
Effect ive St res s an d Por e Wat e r Pres s ur e 161
,
v
"^
34
-
20
ft
1
1
10ft
ft
i .
<
10ft
[ I
Silt deposi t
Oric
>
Figure Ex . 5. 7
Now Y = ——
'
Ysat
l + e ' '° l + e
Therefore, e = 0.26 x 2.7 = 0.70
62.4(2.7.0.7)
sat
1 + 0.7
62.4(2.7-1)
- , y
w
= 62.4 lb/ft
3
, e = wG
s
fo r S = 1
1 + 0.7
( j{ = 10 x 124.8 + 24 x 62.4 = 2745.6 lb/ft
2
Effective pressure after lowering of water table
After lowerin g the water table t o a depth of 20 ft, the soi l abov e thi s level remains saturate d
but effectiv e an d below thi s submerged. Therefore, th e altered effectiv e pressure i s
a
'i =
20
^sat
+
(
34
~
2
°)^fc
= 20 x124
-
8 +14 x62
-
4 = 3369
-
6 lb/ft2
The increas e i n the effective pressure i s
cr'
2
- CT{ = ACT' = 3369.6 - 2745. 6 = 624.0 lb/ft
2
Example 5. 8
Compute th e critica l hydrauli c gradient s fo r th e followin g materials : (a ) Coars e gravel ,
k = 10 cm/sec, G
s
= 2.67, e= 0.65 (b) sandy silt, k = IQr*cm/sec, G
5
= 2.67, e = 0.80
Solution
As per Eq. (5.20), th e critical gradien t i
c
may be expressed a s
G -1
l + e
162
( a) Coarse gravel
Chapt er 5
c
1 + 0.65
( b) Sandy silt
1 + 0.80
Example 5. 9
A larg e excavatio n i s mad e i n a stif f cla y whos e saturate d uni t weigh t i s 109. 8 lb/ft
3
. Whe n th e
depth o f excavation reache s 24.6 ft , cracks appear and water begins to flow upwar d to bring sand t o
the surface . Subsequen t boring s indicat e tha t th e cla y i s underlai n by san d a t a dept h o f 36. 1 ft
below th e original groun d surface .
What i s the dept h o f the water tabl e outside the excavation belo w th e original groun d level ?
Solution
Making an excavation i n the clay creates a hydraulic gradient between the top of the sand layer and
the bottom o f the excavation. As a consequence, wate r start s seepin g i n an upwar d directio n fro m
the san d laye r toward s th e excavate d floor . Becaus e th e cla y ha s a ver y lo w permeability , flo w
equilibrium can onl y be reached afte r a long period o f time. The solutio n mus t be considered ove r
a shor t time interval.
The floo r o f the excavatio n a t depth d i s stabl e onl y i f the wate r pressur e <J
w
a t the to p o f th e
sand laye r a t a dept h o f 36. 1 f t i s counterbalance d b y th e saturate d weigh t <7 , per uni t are a o f th e
clay abov e i t disregarding th e shear strengt h of the clay.
Let H = total thicknes s o f cla y laye r = 36. 1 ft , d = depth o f excavatio n i n cla y = 24. 6 ft ,
h = depth o f wate r tabl e from groun d surface , y = saturated uni t weigh t of the clay .
Stiff cla y stratum
H
H-d
i , . . } , < ! , . I
.'• ' '. . Sandy stratum
Figure Ex . 5. 9
Effect ive St res s an d Por e Wat e r Pres s ur e 163
( H -d) = 36.1 - 24. 6 = 11. 5 ft, the thickness of clay strat a below the bottom of the trench.
°
c
= r
sa
t (#-<*) = 109.8x11.5 = 1263 lb/ft
2
( T
w
= y
w
( H-h) = 62.4 x(36.1 -h) lb/ft
2
cracks ma y develo p whe n a = <J
or 126 3 = 62.4(36.1 -h\ o r A = 36.1-
1263
= 15.86 f t
Example 5 .1 0
The water table is located at a depth o f 3.0 m below the ground surface in a deposit o f sand 11. 0 m
thick (Fig. Ex. 5.10). The sand i s saturated above the water table. The total unit weight of the sand
is 20 kN/m
3
. Calculat e th e (a ) the total pressure, (b ) the por e wate r pressur e an d (c ) the effectiv e
pressure a t depth s 0 , 3.0 , 7.0 , an d 11. 0 m fro m th e groun d surface , an d dra w th e pressur e
distribution diagram.
Solution
y
sat
= 20 kN/m
3
, y
b
= 20 - 9.8 1 = 10.19 kN/m
3
Depth (m) Tota l pressur e Por e water pressure Effectiv e pressur e
cr
f
(kN/m
2
) w^kN/m
2
) <7'(kN/m
2
)
0
3
7
11
0
3 x 20 =60
7 x 20 =140.00
11 x 20 = 220.00
0
0
4x9.81 =39.24
8x9.81=78.48
0
60
100.76
141.52
60 kN/m
2
100.76 kN/m
2
o
t
= 220 kN/nr „„
= 78.48 kN/m
2
Figure Ex. 5.1 0
a' = 141.52 kN/m
2
164 Ch a p t e r s
The pressure distributio n diagrams of a
t
, u
w
and cr'are given in Fig. Ex. 5.10.
Example 5 .1 1
A cla y stratu m 8. 0 m thic k i s locate d a t a dept h o f 6 m fro m th e groun d surface . Th e natura l
moisture conten t of the clay is 56%and G^ = 2.75. The soil stratum between the ground surface and
the clay consists of fine sand. The water table is located at a depth of 2 m below the ground surface.
The submerge d uni t weight of fine sand is 10. 5 kN/m
3
, and it s moist uni t weight abov e th e water
table i s 18.6 8 kN/m
3
. Calculat e the effective stres s a t the center o f the clay layer.
Solution
Fine sand:
Above wate r table: y
t
= 18.68 kN/m
3
Below WT: y
b
= 10.5 kN/m
3
y
sat
= 10.5 + 9.81 = 20.31 kN/m
3
Clay stratum:
For 5= 1.0,
e = wG = 0.56 x 2.75 = 1.54
y
w
( G
s
+e) 9.81(2.7 5 + 1.54)
= 16.57 kN/m
3
l + e1 + 1.54
y
b
= 16.57-9.81 = 6.76 kN/m
3
At a depth 10. 0 m from GL, that is, at the center of the clay layer,
a
t
=2x18. 6 8 + 4x20. 31 + 4x16.57
= 37.36 + 81.24 + 66.28 = 184.88 kN/m
2
,
6
_c
//K&Q?^ . / . . ; . x^Sx^ V . . . //>^y/
1 ' i ' • ' L'
:
C^ir- _• ' • ' "J ' i '- ' L'
:
C^1
:
- -• ' • ' • ' •
m . • ; _ . , • ' • ' . - " . ' • ' ' . ' . • ; . , • ' • ' . - . . . \
:
!
- • • ' . ; _- . . ' • • ' ' . Sand-' ' - V -.' . ' • • ' V! :
; ;
: 4
/vVV^vXvvS /vVVOvVVv s /vVv^OVVvs , /VVNV \
* j
l<\\
2m
m
m
1..
Figure Ex . 5 .1 1
Effect ive St res s an d Por e Wat e r Pres s ur e 165
u
w
= 4 x 9.81 + 4 x 9.81 = 39.24 + 39.24 = 78.48 kN/m
2
Effective stress , cr' = a
t
-u
w
= 184.88 - 78.48 = 106.40 kN/m
2
Example 5 .1 2
A 39. 4 f t thick laye r o f relativel y imperviou s saturated cla y lies ove r a gravel aquifer . Piezomete r
tubes introduce d t o th e grave l laye r sho w a n artesia n pressur e conditio n wit h th e wate r leve l
standing in the tubes 9. 8 ft above the top surface of the clay stratum. The propertie s o f the clay ar e
e=l.2,G = 2.7 and v = 110.62 lb/ft
3
.
5 ' Sal
Determine (a ) the effectiv e stres s at the top of the gravel stratu m layer , an d (b) the dept h of
excavation tha t can be made i n the clay stratu m without bottom heave .
Solution
(a) At the top of the gravel stratu m
cr
c
= 39.4 x 110.62 = 4358.43 lb/ft
2
The por e wate r pressure a t the top of the gravel i s
u
w
= 62.4 x 49.2 = 3070 lb/ft
2
The effective stress a t the top of the gravel i s
a' = <j
c
- u
w
= 4358.43 -3070 = 1288.43 lb/ft
2
(b) If an excavation i s made int o the clay stratum as shown in Fig. Ex. 5.12, the depth mus t be such
that
Clay
39.4 ft
49.2 ft
Gravel
Figure Ex . 5 .1 2
1 6 6 Chapter s
a <u
c w
Let th e botto m o f th e excavatio n b e h f t abov e th e to p o f grave l layer . No w th e downwar d
pressure actin g at the top of the gravel laye r is
u
w
= 3070 lb/ft
2
3070
Now, 110.62/ z = 3070 o r / z = ——— = 27.75 f t
Depth of excavation, d = 39.4 - 27.7 5 = 1 1.65 ft
This i s just the dept h of excavation with a factor of safet y F
S
= 1 .0. If we assume a minimum
F
s
= 1.1 0
A=
3070XU
110.62
Depth of excavation = 39.4 - 30.5 2 = 8.88 ft
Example 5 .1 3
The diameter of a clean capillary tub e is 0.08 mm . Determine the expected rise of water in the tube.
Solution
Per Eq. (5.22), th e expected rise , h
c
, in the capillary tub e is
7-7 <
= 37. 5 cm „
c
d 0.00 8
where, d i s i n centimeters
Example 5 .1 4
The wate r table i s at a depth o f 1 0 m in a silt y soi l mass . The siev e analysi s indicates the effectiv e
diameter D
10
of th e soi l mas s i s 0.05 mm . Determin e th e capillar y ris e o f wate r abov e th e wate r
table an d th e maximu m capillar y pressur e (a ) b y usin g Eq . (5.23) an d (b ) b y usin g Eq . (5.28) .
Assume the voi d rati o e = 0.51.
Solution
Using Eq. (5.25) an d assuming C = 0.5, th e capillary rise of water is
C 0. 5
= 196cm
< D 0.51x0.00 5
(a) Per Eq. (5.23)
the capillar y pressur e i s u
w
= -h
c
Y
w
= -1.96 x 9.81 = -19.2 kN/m
2
(b) Per Eq. (5.28 )
Effect ive St res s an d Por e Wat er Pres s ur e 167
Porosity, n =
0.51
= 0.338
l + e1 + 0.51
u
w
= u
c
= -n hj
w
= -0.338 x 19. 2 = 6.49 kN/m
2
Example 5 .1 5
A layer of silty soi l of thickness 5 m lies below th e ground surface a t a particular sit e and below the
silt layer lies a clay stratum. The groun d water tabl e i s at a depth o f 4 m below th e ground surface .
The following data ar e availabl e fo r both th e sil t and clay layer s of soil .
Silt layer : £>
10
= 0.018 mm, e = 0.7, and G
s
= 2.7
Clay layer : e = 0.8 and G
s
= 2.7 5
Required: (a ) Height o f capillary rise , (b ) capillary pressure , (c ) the effective pressure a t the
ground surface, a t GWT level , at the bottom o f the silt layer and at a depth of H = 6 m below groun d
level, an d (d) at a depth 2 m below groun d level .
Solution
For the silt y soil :
m
l + e 1.7
/ s a t
= / m
l + e1. 7
-Y =19.62-9.8 1 = 9.81 k N/ m
3
GL Capillary fring e GL h-H
«
c
= 16.1 6 kN/m?
wOLUUL WAJUUV AAJV
/ i
c
-f m
1
,
h^lm
H-
Y
:
1m
'
;
Capillary
saturated zone
5
6 m Silt y layer
GWT
1 1
i '
m
A
\
\
A- o' = 47.36 kN/m
2
> s
\
\
\
a; = 88.37 kN/m
2
\
Clay
j- a\ =98 kN/m
2
*\
Effective pressur e distribution diagram
Figure Ex . 5 .1 5
168 Chapt e r s
In the clay stratum:
(2.75 + 0.8)9.81
r
s
*= -[i -=
19
-
35 k N/ m 3
Y
b
=19.35-9.8 1 = 9.54 k N/ m
3
(a) Height of capillary rise
/ z
c
= — -perEq. (9.5)
eD
w
Assume C = 0.5 sq. cm.
We have h
c
= -' • -= 397 cm or say 4.0 m
It i s clear fro m h
c
that the plane of menisci formed by the capillary wate r coincides wit h the
ground surfac e as the wate r table is also a t a depth of 4 m from groun d level .
(b) Capillary pressure u
c
0.7
or u
c
=— - x4x9. 81 = 16. 16kN/ m
2
(c) Th e effective pressure a t G L
Since th e plane of menisci coincides wit h the ground surface, the effective pressure a t GL is
equal t o the capillary pressure u
c
Total effective pressure at GWT level, o
f
sat
Per Fig. Ex . 5.1 5
0' =15. 6x 4 + 16.16 = 78.56 k N/ m
2
Sal
Total effective pressure at the bottom of the silt layer
The bottom of the silt layer is at a depth of 1 m below GWT level . The effective pressure due
to this dept h i s
cf = y
b
h
w
= 9.81 x 1 = 9.81 kN/m
2
Total effective pressure, o
f
l
= c/
sat
+ rf = 78.56 + 9.81 = 88.37 kN/m
2
Total effective pressure at a depth of 6m below GL
This poin t lies i n the clay stratum at a depth of 1 m below the bottom o f the silt y layer.
The increas e i n effective pressur e a t thi s depth i s
of = y
b
h
w
= 9.54 x 1 = 9.54 kN/m
2
The total effective pressure </ , = 88.37 + 9.54 = 97.91 kN/m
2
« 98 kN/m
2
(d) <J \ at 2 m below G L
0'
z
= u
c
+zY
d
= 16.16 + 2 x 15. 6 = 47.36 kN/m
2
Effect ive St res s an d Por e Wat er Pres s ur e 169
The pressure distribution diagram i s given in Fig. Ex. 5.15.
Example 5 .16
At a particular sit e lies a layer of fine sand 8 m thick below the ground surface and having a void ratio
of 0.7. The GWT is at a depth of 4 m below the ground surface. The average degree of saturation of the
sand above the capillary fringe is 50%. The soil is saturated due to capillary action to a height of 2.0 m
above the GWT level. Assuming G
5
= 2.65, calculat e the total effective pressures a t depths of 6 m and
3 m below the ground surface.
Solution
Y
d
=-r
2.65 x 9.8!
= m9 k N/ m3
l + e1. 7
( e + G
s
)y (0. 7 +2.65) x 9.81
= 19.33 k N/ m
3
l + e1. 7
Y
b
=
Psat ~Y
W
= 19.33-9.81 = 9.52 kN/ m
3
The mois t uni t weight of soi l abov e the capillary fringe is
l + e
Capillary pressure ,
u
c =
nh
cY
w
=
1.7
0.7
=77 x 2x9.81 = 8.08 k N/ m
2
Effective stresse s at different level s
8m
GL
= 2 m Mois t soi l
Capillary fring e
(a) Soi l profil e
Submerged
Fine sand
3 m
a'
0
= 34.62 kN/m
2
w
c
= 8.0 8 kN/m
2
o'
d
= 30.58 kN/m
2
o'
w
= 19.04 kN/m
2
a; = 92.32 kN/m
2
(b) Effective vertica l stres s diagram
Figure Ex. 5.16
170 Chapt er 5
(a) At ground leve l cf = 0
(b) Overburde n pressur e at fringe leve l = o
f
o
= h
c
y
m
= 2 x 17.3 1 = 34.62 kN/m
2
(c) Effective pressur e at fringe leve l = o
f
c
= o
f
o
+ u
c
= 34.62 + 8.08 = 42.70 kN/m
2
(d) Effective pressure at GWT level = o^ =rf
c
+o'
d
= 42.70 + 2 x 15.2 9
= 42.70 + 30.58 = 73.28 kN/m
2
(e) Effective pressure at 6 m below GL
< j = o f + h y . = 73.28 + 2 x 9.52 = 73.28+ 19.04 = 92.32 kN/m
2
I Sd t W' D
Effective stres s a t a depth 3 m below GL
Refer Fig . Ex . 5.16.
cf = cf
0
+ u
c
+ ( z - h
c
)y
d
= 34.62 + 8.08 + (3 - 2 ) x 15.2 9 « 58 kN/m
2
5.6 PROB L EM S
5.1 Th e dept h o f wate r i n a lake i s 3 m. The soi l propertie s a s obtained fro m soi l exploration
below th e bed o f the lake ar e as given below.
Depth fro m be d
of lake (m)
0-4
4-9
9-15
Type of
soil
Clay
Sand
Clay
V oid rati o
e
0.9
0.75
0.60
Sp. gr .
G,
2.70
2.64
2.70
Calculate the following pressures a t a depth o f 1 2 m below th e bed leve l o f the lake ,
(i) The total pressure, (ii ) the pore pressure an d (iii ) the intergranular pressure.
5.2 Th e wate r tabl e i n a certain deposi t o f soi l i s at a depth of 6. 5 f t below th e ground surface.
The soi l consist s of clay up t o a depth of 1 3 ft from th e ground and below whic h lies sand.
The clay stratu m i s saturated abov e the water table.
Given: Clay stratum: w = 30 percent, G
s
= 2.72; Sandy stratum: w = 26 percent, G
s
= 2.64.
Required:
(i) Th e tota l pressure , por e pressur e an d effectiv e pressur e a t a dept h o f 2 6 f t below th e
ground surface .
(ii) Th e change in the effective pressur e if the water table is brought down to a level of 1 3 ft
below th e ground surfac e by pumping.
5.3 Wate r flows from containe r B to A as shown in Fig. 5.4. The piezometric hea d at the bottom
of container A i s 2.5 m and the dept h of water above the sand deposit i s 0.25 m . Assuming
the dept h of the sand deposi t is 1.4 0 m, compute the effective pressur e at the middl e of the
sand deposit . Assume e = 0.65 an d G
s
= 2.64 fo r the sand.
5.4 I n orde r t o excavat e a trenc h fo r th e foundatio n of a structure , the wate r tabl e leve l wa s
lowered fro m a depth of 4 ft to a depth of 1 5 ft in a silty sand deposit. Assuming that the soi l
above th e wate r tabl e remaine d saturate d a t a moistur e conten t of 2 8 percent , estimat e th e
increase i n effectiv e stres s a t a depth of 1 6 ft. Give n G
s
= 2.68
Effect ive St res s an d Por e Wat e r Pres s ur e 171
El.B
5 m
«—Soil
El.A
(a) Saturated (b ) Submerged
Figure Prob. 5. 5
( //\\ \
5 f t
'
'
5 f t
! f t
1
• . - • ' • ' ' • • • • • ' . : • • '
• '• • ' • • ' • ' . *- • • ' .
' V
-
• '• '•' • >-' • • '
^
• . * • • .
• ' • -• '• -' . ' .':•-*-
•' .' '» " • *-"°
k
*°.
;
'
:
-
>

%
'"fl
f
*
V . '"• '{•; '• .'*.*«» "• ' •''•'!'
^ti^fa*
'i'-"£-
:
?:?'y^'^--
• '.-. V .'"'.*, . ' '•.;;'.' , "*'.'
'.'• '• ' V. .*. ' :<-'" .' . • '.'•' V • '
//A\\
y
sa
t
=
— San
— Cl a
Xsat =
at
=1201b/ft
3
//\\\^
Figure Prob. 5. 6
5.5 Soi l i s placed i n the containers shown in Fig. Prob. 5.5. The saturated unit weight of soil is
20 kN/m
3
. Calculate the pore pressure, and the effective stres s at elevation A, when (a) the
water table is at elevation A, and (b) when the water table rises to El.B .
5.6 Figur e Prob. 5. 6 give s a soi l profile . Calculate th e total and effective stresses a t point A.
Assume that the soi l above the water table remains saturated.
5.7 Fo r the soil profil e given in Fig. Prob. 5.6, determin e the effective stres s at point A for the
following conditions : (a ) water table at ground level, (b) water table at El.A. (assume the
soil above this level remains saturated), and (c) water table 6. 5 ft above ground level.
5.8 A glass tube, opened a t both ends, has an internal diameter of 0.002 mm. The tube is held
vertically an d wate r i s adde d fro m th e to p end . Wha t i s th e maximu m heigh t h o f th e
column of water that will be supported?
5.9 Calculat e (a) the theoretical capillary height and pressure h
c
, and (b) the capillary pressure,
«
c
, in a silt y soi l wit h D
10
= 0.04 mm. Assume the void ratio is equal to 0.50 .
5.10 Calculat e the height to which water will rise in a soil deposit consisting of uniform fine silt.
The dept h o f wate r belo w th e groun d surfac e i s 2 0 m. Assum e th e surfac e tensio n i s
75 x 10~
8
kN/cm and the contact angle is zero. The average size of the pores i s 0.004 mm.
CHAPTER 6
STRESS DISTRIBUTION I N SOILS DU E TO
SURFACE LOADS
6 .1 I NTRODU CTI O N
Estimation o f vertica l stresse s at any point i n a soil-mass du e t o external vertica l loading s ar e of
great significanc e in th e predictio n o f settlement s o f buildings , bridges, embankment s an d many
other structures . Equations have been developed t o compute stresses at any point in a soil mass on
the basi s o f th e theor y o f elasticity . Accordin g t o elasti c theory , constan t ratio s exis t betwee n
stresses an d strains . For th e theor y t o be applicable , th e real requiremen t i s not tha t the material
necessarily b e elastic , bu t ther e mus t be constan t ratios betwee n stresse s an d th e correspondin g
strains. Therefore, i n non-elastic soi l masses, the elastic theory may be assumed t o hold so long as
the stresse s induce d i n th e soi l mas s ar e relativel y small . Sinc e th e stresse s i n th e subsoi l o f a
structure having adequate factor of safet y agains t shear failur e ar e relativel y small i n comparison
with the ultimate strength of the material, the soil may be assumed to behave elastically under such
stresses.
When a loa d i s applie d t o th e soi l surface , i t increase s th e vertica l stresse s withi n the soi l
mass. The increase d stresse s are greatest directl y under the loaded area , but extend indefinitel y in
all directions. Man y formulas based o n the theory of elasticity have been used t o compute stresse s
in soils . The y ar e al l simila r an d diffe r onl y i n th e assumption s mad e t o represen t th e elasti c
conditions o f th e soi l mass . Th e formula s tha t ar e mos t widel y use d ar e th e Boussines q an d
Westergaard formulas . These formula s wer e firs t develope d fo r poin t loads actin g a t the surface.
These formula s have bee n integrate d t o giv e stresse s belo w unifor m stri p load s an d rectangular
loads.
The extent of the elastic layer below the surface loading s may be any one of the following:
1. Infinit e i n the vertical an d horizontal directions .
2. Limite d thickness i n the vertical direction underlai n with a rough rigid base such as a rocky
bed.
173
174 Chapt er 6
The load s a t th e surfac e ma y ac t o n flexibl e or rigi d footings . The stres s condition s i n th e
elastic laye r belo w var y accordin g t o th e rigidit y o f th e footing s an d th e thicknes s o f th e elasti c
layer. All the external load s considere d i n this book ar e vertical loads onl y as the vertical load s ar e
of practical importanc e fo r computing settlements of foundations .
6.2 B OU SSI NESCT S FORM U L A FOR POI NT L OADS
Figure 6.1 shows a load Q acting at a point 0 o n the surface of a semi-infinite solid. A semi-infinite
solid i s th e on e bounde d o n on e sid e b y a horizonta l surface , her e th e surfac e o f th e earth , an d
infinite i n all the other directions. The proble m of determining stresses at any point P at a depth z as
a resul t of a surface point laod wa s solve d by Boussines q (1885 ) on the following assumptions.
1. Th e soi l mas s i s elastic, isotropic , homogeneou s an d semi-infinite.
2. Th e soi l i s weightless.
3. Th e loa d i s a point load actin g on the surface .
The soi l is said t o be isotropi c i f there are identical elastic properties throughout the mass and
in ever y directio n throug h any point of it . The soi l i s sai d t o be homogeneous i f there ar e identica l
elastic propertie s a t every poin t of the mass i n identical directions .
The expressio n obtaine d by Boussines q for computin g vertical stres s <7 , at point P (Fig. 6.1 )
due t o a point loa d Q is
3(2
1
Q
(6.1)
where, r = the horizonta l distanc e between a n arbitrar y point P below th e surfac e an d th e vertica l
axis through the point loa d Q.
z = the vertica l dept h of the point P fro m th e surface .
1
I
R
- Boussines q stres s coefficien t = —
The value s of the Boussines q coefficient I
B
ca n b e determined fo r a number of value s of r/z.
The variatio n of /„ wit h r/z i n a graphical form i s given in Fig. 6.2 . I t can b e see n fro m thi s figur e
O
Q
\x \
\>WJ \
P
°Z
Figure 6. 1 Vert ica l pres s ur e wit hi n an eart h m as s
St res s Dis t ribut io n i n Soil s du e t o Surfac e Load s 175
that I
B
ha s a maximu m valu e o f 0.4 8 a t r/ z = 0 , i.e. , indicatin g thereb y tha t th e stres s i s a
maximum belo w th e poin t load .
6.3 WESTERG AARD' S FORM U L A FO R POI NT L OAD S
Boussinesq assume d tha t the soi l i s elastic, isotropi c an d homogeneou s fo r th e developmen t o f a
point loa d formula . However , th e soi l i s neithe r isotropi c no r homogeneous . Th e mos t commo n
type of soils that are met in nature are the water deposited sedimentar y soils . When the soil particle s
are deposited i n water, typical clay strata usually have their lenses o f coarser materials withi n them.
The soil s o f thi s typ e ca n b e assume d a s laterall y reinforce d b y numerous , closel y spaced ,
horizontal sheet s o f negligible thicknes s but of infinite rigidity, which prevent the mass a s a whole
from undergoin g lateral movement of soil grains . Westergaard, a British Scientist, proposed (1938 )
a formula for the computation of vertical stres s o
z
b y a point load, Q, at the surface a s
cr, -'
Q
,3/2 2
M
(6.2)
in whic h fj, i s Poisson' s ratio. I f fj, i s taken as zero fo r al l practical purposes , Eq . (6.2 ) simplifie s t o
Q
1
Q
[l
+
2(r/z)
2
]
3
'
2 (6.3)
where /,, , =
( II a)
[l + 2(r / z)
2
]
3 / 2
is th e Westergaar d stres s coefficient . Th e variatio n o f / wit h th e
ratios o f ( r/z) i s shown graphically i n Fig. 6. 2 along wit h the Boussinesq's coefficien t I
B
. The value
of I
w
a t r/ z = 0 i s 0.32 whic h is less tha n that of I
B
b y 3 3 per cent .
h
or 7
w
0 0. 1 0. 2 0. 3 0. 4 0. 5
r/z 1. 5
2.5
Figure 6. 2 Val ue s o f I
B
or /^f or us e in t he B ous s in es q o r Wes t erg aar d form ul a
176 Chapt e r s
Geotechnical engineer s prefe r t o us e Boussinesq' s solutio n a s thi s give s conservativ e
results. Furthe r discussion s ar e therefor e limite d to Boussinesq' s metho d i n thi s chapter .
Example 6 . 1
A concentrated loa d o f 100 0 k N i s applied at the ground surface . Comput e th e vertica l pressur e (i )
at a dept h o f 4 m belo w th e load , (ii ) at a distanc e o f 3 m a t th e sam e depth . Us e Boussinesq' s
equation.
Solution
The equatio n is
Q 3/2; r
_ _ —/ wher e /„ = rrj^-
Z 7 i i f ' t i f 9 p/Z
z
[ l + ( r / z )
2
\
Q 100 0
(i) When r/ z = 0, /„ = 3/2 n = 0.48, a = 0.48^- = 0.48 x —— = 30 kN/m
2
B z
z
2
4x 4
(ii) Whe n r/ z = 3/4 = 0.7 5
3/27T 0.156x100 0
I
R=~T ^ T = 0.156, a = — = 9.8 k N/ m
2
B
l + (0.75)
2
f
2 z 4 x 4
Example 6 . 2
A concentrated loa d o f 45000 I b acts at foundation level a t a depth of 6.56 f t below groun d surface .
Find th e vertica l stres s alon g th e axi s o f th e loa d a t a dept h o f 32. 8 f t an d a t a radial distanc e o f
16.4 f t a t the same dept h b y (a ) Boussinesq, and (b ) Westergaard formula e for n = 0. Neglec t th e
depth o f th e foundation .
Solution
(a) Boussinesq Eq . (6. la)
"
2
z
z
2 B
'
B
271 l + ( r / z )
2
Substituting the known values, and simplifyin g
I
B
= 0.2733 for r/ z = 0.5
=
_45000
x02733
^
n431b/ f t 2
z
(32.8)
2
(b) Westergaard (Eq . 6.3 )
13/2
Q
1
l + 2 ( r / z )
2
Substituting the known values and simplifying , w e have,
/ =0. 1733f or r / 7 = 0.5
St res s Dis t ribut io n i n Soil s due t o Surfac e Load s 177
therefore,
a =
(32.8)
x 0.1733 = 7.25 lb/ft
2
Example 6. 3
A rectangula r raf t o f siz e 3 0 x 1 2 m founde d a t a dept h o f 2. 5 m belo w th e groun d surfac e i s
subjected t o a unifor m pressur e o f 15 0 kPa . Assum e th e cente r o f th e are a i s th e origi n o f
coordinates (0 , 0). an d the corner s hav e coordinates (6 , 15) . Calculate stresse s at a depth o f 20 m
below the foundation leve l by the methods of (a) Boussinesq, and (b) Westergaard a t coordinates of
(0, 0), (0, 15) , (6, 0) (6, 15) and (10, 25). Also determine the ratios of the stresses a s obtained by the
two methods . Neglec t the effec t o f foundation dept h on the stresses (Fig . Ex . 6.3) .
Solution
Equations (a ) Boussinesq:
(b) Westergaard:
= — I
B
,
z
I
B
= '
l + <r / ^f
2
Q
0.32
The ratio s of r/ z at the give n locations fo r z = 20 m ar e as follows:
Location
(0, 0)
(6, 0)
(0, 15 )
r/z
0
6/20 = 0. 3
15/20 = 0.75
Locat i on
(6, 15 )
(10, 25)
(^
(V io
2
"
r/z
f 15
2
)/ 20 = 0.81
+ 25
2
) / 20 = 1.35
The stresse s a t th e variou s location s a t z = 20 m may b e calculate d b y usin g the equations given
above. The results are tabulated below for the given total load Q = qBL = 150 x 12 x 30 = 54000 kN
acting at (0, 0) coordinate. Q/z
2
=135.
(6,15) (6,0) (6,15)
.(0,0)
(0,15)
(6,15) (6,0)
Figure Ex . 6. 3
(6,15)
(10,25)
178 Chapt er 6
Locat i on r/ z B ous s in es q
I
0
crJ k Pa )
Wes t er g aar d
w
a/a,
w
(0, 0)
( 6, 0)
(0, 15 )
(6,15)
(10, 25 )
0
0.3
0.75
0.81
1.35
0.48
0.39
0.16
0.14
0.036
65
53
22
19
5
0.32
0.25
0.10
0.09
0.03
43
34
14
12
4
1.51
1.56
1.57
1.58
1.25
6.4 LIN E L OAD S
The basi c equatio n use d fo r computin g a, a t an y poin t P i n a n elasti c semi-infinit e mas s i s
Eq. (6.1) o f Boussinesq . B y applyin g the principl e o f hi s theory , th e stresse s a t an y poin t i n th e
mass du e t o a line loa d o f i nfi ni t e exten t acting at the surfac e ma y b e obtained. The stat e of stres s
encountered i n thi s cas e i s tha t o f a plan e strai n condition . Th e strai n a t an y poin t P i n th e
F-direction parallel t o the lin e load i s assumed equal t o zero. The stres s c r norma l t o the XZ-plane
(Fig. 6.3 ) i s the same at all sections an d the shear stresse s o n thes e section s ar e zero. By applying
the theor y o f elasticity , stresse s a t an y poin t P (Fig . 6.3 ) ma y b e obtaine d eithe r i n pola r
coordinates o r i n rectangula r coordinates . Th e vertica l stres s a a t poin t P ma y b e writte n i n
rectangular coordinate s a s
a =
z [ 1 + U / z )
2
]
2
z
z
where, / i s the influence factor equa l t o 0.637 a t x/z - 0 .
(6.4)
r — \ i x •" • + z
cos fc) =
Figure 6. 3 St res s e s due t o ver t ica l lin e loa d i n rect an g ul a r coordin at e s
St res s Dis t ribut io n i n Soil s du e t o Surfac e Load s 179
6.5 STRI P L OADS
The stat e of stres s encountere d i n thi s case also i s that of a plane strai n condition. Suc h condition s
are foun d fo r structure s extende d ver y muc h i n on e direction , suc h a s stri p an d wal l foundations,
foundations o f retaining walls, embankments, dams and the like. For such structures the distribution
of stresses i n any section (except for the end portions of 2 to 3 times the widths of the structures from
its end ) wil l be th e same a s i n the neighborin g sections , provide d tha t the loa d doe s no t change i n
directions perpendicular to the plane considered .
Fig. 6.4(a ) shows a load q pe r uni t area actin g o n a stri p o f infinit e lengt h and o f constant
width B. The vertical stress at any arbitrary point P due t o a line load of qdx actin g at j c = x ca n be
written fro m Eq . (6.4) as
~
2q
n [ ( x - x )
2
+z
2
]
(6.5)
Applying th e principl e o f superposition , th e tota l stres s o~
z
a t poin t P du e t o a stri p loa d
distributed ove r a width B( = 2b) may be written as
+b
[ ( x-x)
2
+z
2
}
2
dx
or
-b
q , z
a =— tan"
1
1
n x-b
tan"
2bz( x
2
-b
2
-z
2
)
x + b
(6.6)
The non-dimensiona l value s o f cjjq ar e give n graphicall y i n Fig. 6.5 . Eq. (6.6 ) can b e
expressed i n a more convenien t form as
=— [/?+sin/?cos(/?+2£)]
n
(6.7)
x O
(a) (b )
Figure 6. 4 Stri p loa d
180 Chapt er 6
( ajq) x 1 0
4 5 6 7 10
Figure 6. 5 Non -dim en s ion a l val ue s o f <j /q fo r s t ri p loa d
where / 8 an d S ar e th e angle s a s show n i n Fig . 6.4(b) . Equatio n (6.7 ) i s ver y convenien t fo r
computing o~ , sinc e the angle s ft and S can be obtaine d graphicall y fo r an y point P. The principa l
stresses o
{
an d o"
3
at any point P may b e obtained from the equations .
cr, = —(/ ?+si n/ ?)
n
(6.8)
0", = — ( p-sm,
TC
(6.9)
Example 6 . 4
Three parallel strip footings 3 m wide each and 5 m apart center to center transmit contact pressure s
of 200 , 15 0 an d 10 0 kN/m
2
respectively . Calculat e th e vertica l stres s du e t o th e combine d load s
beneath the centers of each footing at a depth of 3 m below the base. Assume the footings are placed
at a depth of 2 m below the groun d surface. Use Boussinesq' s method for lin e loads .
Solution
From Eq. (6.4), we have
2/;r _ q
_\2
St res s Dis t ribut io n i n Soil s due t o Surfac e Load s 181
XXX\\x\\
1
2
C „- .
30
1
1
c
50
1
l oot
xxx\\xc\\
J^/m
2
3 m
,
3 m
t y\
3 m
t
3 m
C
Figure Ex . 6.4 T hre e paral l e l foot in g s
The stres s a t A (Fig . Ex. 6.4 ) i s
(4 =
2 x 2 00F 1
3.14x3
2x100
2x150
3.14x3
1
_l + (5/ 3)
2
3. 14x3_l + (10/3)
2
= 45 k N/ m
2
The stres s at B
( "•}
\ z)
B
2x200
3x
1
_l + (5/ 3)
2
2x150
2x100
(0/3)
= 36. 3 kN / m
2
The stres s a t C
kt =
2x200
l + (10/3)
2
2x150 1
3^r l + (5/3)
2
2x100
= 23.74 k N/ m
2
6.6 STRESSE S B ENEATH TH E CORNER OF A RECTANG U L AR
FOU NDATI ON
Consider an infinitel y smal l unit of area of size db x dl , shown i n Fig. 6.6. The pressur e acting on
the smal l area ma y be replaced by a concentrated load dQ applied to the center of the area.
Hence
= qdb.dl (6.10)
The increas e of the vertical stress a du e to the load dQ can be expressed pe r Eq. (6.11) as
182 Chapt er 6
* ^r\
i:M
1
1
1
s
'
\
' \ \
<1
'
\
'
N
\'
'
Figure 6. 6 Vert ica l s t r es s under t h e corn e r o f a rect an g ular foun dat io n
dcr =
dQ 3z
3
(6.11)
The stres s produce d b y th e pressure q over the entir e rectangle b x I ca n then be obtained by
expressing dl, db an d r in terms of the angles a an d /3, and integrating
a=a
}
/?=/? ,
(6.12)
There ar e severa l form s o f solutio n for Eq . (6.12) . Th e on e tha t i s normall y use d i s o f th e
following for m
cr=q
or
2mn( m
2
+n
2
+1)
1/2
m
2
+n
2
+2
m
2
+n
2
+m
2
n
2
+l m
2
+n
2
+l
tan
_, 2mn( m
2
+n
2
+l)
l/2
m
2
+n
2
-m
2
n
2
+1
(6.13)
a
z
= ql (6.14 )
wherein, m = b/z, n = l/z, ar e pure numbers. / i s a dimensionless factor and represents th e influence
of a surcharge coverin g a rectangular are a on the vertical stres s at a point located at a depth z below
one of it s corners .
Eq. (6.14 ) i s presented i n graphical for m i n Fig. 6.7. Thi s char t helps t o comput e pressure s
beneath loade d rectangula r areas . Th e chart als o show s that the vertical pressur e i s not materiall y
altered i f the length of the rectangle is greater than ten times its width. Fig. 6.8 may also be used for
computing the influence value / based o n the values of m and n and may als o be used t o determine
stresses belo w point s that lie either inside or outside the loaded area s a s follows.
St res s Dis t ribut io n i n Soil s du e t o Surfac e Load s 183
z/b =
0.05
V alues of / = ojq
0.10 0.1 5 0.20 0.25
Figure 6. 7 Char t fo r com put in g G
Z
below t h e corn e r o f a rect an g ular foun dat io n
(aft er St ein bren n er , 1934 )
When the Poin t i s I nside
Let O be a n interior poin t of a rectangular loaded are a ABCD show n i n Fig. 6.9(a) . I t i s required
to comput e the vertica l stres s <J
z
belo w this point O at a depth z from th e surface . For thi s purpose,
divide the rectangl e ABCD int o fou r rectangle s marke d 1 to 4 i n the Fig. 6.9(a ) b y drawin g lines
through O . Fo r eac h o f thes e rectangles , comput e th e ratio s zfb. Th e influenc e valu e 7 may b e
obtained from Fig . 6. 7 or 6. 8 for each of these ratio s and the total stres s a t P is therefor e
_. / T . T . J. T \/ S IC\
&
7
= q Ui + h + M +
y
J (6.15 )
When the Poin t i s Outside
Let O be an exterior poin t of loaded rectangula r area ABCD show n in Fig. 6.9(b) . I t i s required to
compute the vertica l stres s <T
Z
below point 0 a t a depth z from th e surface.
Construct rectangles a s shown i n the figure. The point O is the corner point of the rectangl e
OB
l
CD
r
Fro m th e figure i t can be seen that
Area ABCD = OB
1
CD
1
- OB
{
BD
2
- OD
}
DA
{
+ OA
1
AD
2
(6.16 )
184
Chapt er 6
0.00
0.01 2 4 6 80. 1 2 4 6 81. 0
V alues of n = l/z
4 6 8 10
Figure 6. 8 Grap h fo r det erm in in g in fluen c e valu e fo r vert ica l n orm a l s t res s cr
z
at
point P l ocat ed ben eat h on e corn e r o f a un iform l y loade d rect an g ul a r area . (Aft e r
Fadum, 1948 )
O
f
b
I
1 2
^ 6
3 4
D C
(a) When th e point 'O' i s within the rectangl e (b ) When th e point 'O' is outside the rectangl e
Figure 6. 9 Comput at i o n o f ver t i ca l stres s belo w a poi n t
St res s Dis t ribut io n i n Soil s due t o Surfac e Load s 18 5
The vertica l stres s a t point P locate d a t a dept h z below poin t 0 du e t o a surcharge q per
unit are a o f ABCD i s equa l t o th e algebrai c su m o f th e vertica l stresse s produce d b y loadin g
each on e o f the area s liste d o n the right hand side o f the Eq. (6.16) wit h q per uni t of area. I f / j
to /
4
ar e th e influenc e factors o f eac h o f thes e areas , th e tota l vertica l stres s i s
(6.17)
Example 6. 5
ABCD i s a raf t foundatio n o f a multi-stor y buildin g [Fig . 6 . 9(b) ] wherei n AB = 65. 6 ft , an d
BC = 39.6 ft . The uniforml y distributed load q over the raft i s 73 10 lb/ft
2
. Determine cr
z
at a depth of
19.7 f t below point O [Fig. 6.9(b) ] wherei n AA, = 13.1 2 ft and A,0 = 19.6 8 ft . Use Fig. 6.8 .
Solution
Rectangles ar e constructed a s shown in [Fig . 6.9(b)] .
Area ABCD = OB
}
CD
l
- OB
}
BD
2
- OD
1
DA
1
+ OA
1
AD
2
Rect angl e
OB
1
CD
1
OB
1
BD
2
OD
1
DA
1
OA
{
AD
2
I
( ft)
85.28
85.28
52.72
19.68
b
( ft)
52.72
13.12
19.68
13.12
m
2.67
0.67
1.00
0.67
n
4.33
4.33
2.67
1.00
7
0.245
0.168
0.194
0.145
Per Eq. (6.17)
o
z
= q (/! - /
2
- /
3
+ /
4
) = 7310 (0.24 5 - 0.168 - 0.19 4 + 0.145) = 204.67 lb/ft
2
The same valu e can be obtained using Fig. 6.7 .
Example 6. 6
A rectangula r raf t o f siz e 3 0 x 1 2 m founde d on th e groun d surfac e i s subjecte d t o a uniform
pressure of 150 kN/m
2
. Assume the center of the area as the origin of coordinates (0,0) , and corners
with coordinate s (6 , 15) . Calculat e th e induce d stres s a t a dept h of 2 0 m b y th e exac t metho d a t
location (0, 0).
Solution
Divide the rectangle 1 2 x 30 m into four equa l part s of size 6 x 15m .
The stres s belo w th e corne r o f eac h footin g ma y b e calculate d b y usin g chart s give n i n
Fig. 6. 7 or Fig. 6.8 . Her e Fig . 6. 7 is used.
For a rectangle 6 x 1 5 m, z Ib = 20/6 = 3.34, l/b = 15/6 = 2.5.
For
z
/b = 3.34, l/b = 2.5, < r Iq = 0.07
Therefore, o ; = 4cr = 4 x 0.01 q= 4 x 0.07 x 15 0 = 42 kN/m
2
.
186
Chapt er 6
6 .7 STRESSE S UNDE R U NI FORM L Y L OADE D CI RCU L A R FOOTI N G
Stresses Along the V ertica l Axis o f Symmetry
Figure 6.1 0 show s a pla n an d sectio n o f th e loade d circula r footing . Th e stres s require d t o b e
determined a t any point P along the axi s is the vertical stres s cr, .
Let dA be an elementary area considered a s shown in Fig. 6.10. dQ ma y b e considered a s the
point loa d actin g on thi s area whic h is equal t o q dA. We may write
(6.18)
The vertica l stress d( J a t point P due t o point load dQ ma y be expressed [Eq . (6 . la)] as
3q
(6.19)
The integra l for m of the equation for the entire circular area ma y be writte n as
0=0 r= 0
3qz
3
( f rdOdr
~^~ J J
( r
2
+ z
2
) 5
,
0=0 r= 0
,3
On integration we have, (6.20)
o
R z
P
Figure 6 .1 0 Ver t ica l s t res s un de r un iform l y loade d circul a r foot in g
St res s Dis t ribut io n i n Soil s due t o Surfac e Load s 187
Influence valu e 7
Z
(xlOO)
1.0 1 0
Note: Number s o n curves
indicate value of r/R
Q
Figure 6 .1 1 In fluen c e diag ra m fo r vert ica l norma l s t res s a t various point s withi n
an elas t i c half-s pac e unde r a un iform l y loade d circula r area . (Aft e r Fos t e r an d
Ahlvin , 1954)
or
3/2
(6.21)
where, /. , i s th e Influence coefficient. Th e stres s a t an y poin t P o n th e axi s o f symmetr y o f a
circular loade d are a ma y b e calculate d b y th e us e o f Eq . (6.21 ) V ertica l stresse s o ~ ma y b e
calculated b y usin g the influenc e coefficient diagram give n i n Fig. 6.11.
Example 6. 7
A wate r tank i s require d t o b e constructe d wit h a circular foundation having a diameter o f 1 6 m
founded a t a dept h o f 2 m belo w th e groun d surface . Th e estimate d distribute d loa d o n th e
foundation i s 325 kN/m
2
. Assuming that the subsoi l extends t o a great dept h and i s isotropi c and
homogeneous, determine the stresses o
t
at points (i) z = 8 m, r = 0, (ii) z = 8 m, r = 8 m, (iii) z= 1 6
m, r = 0 and (iv) z = 1 6 m, r = 8m, where r is the radial distance from the central axis . Neglect the
effect o f the dept h of the foundation on the stresses . (Use Fig. 6.11)
Solution
q — 325 kN/m
2
, R
Q
= 8 m. The result s ar e given i n a tabular form a s follows :
(i)
(ii)
(iii)
(iv)
Point
(8, 0)
(8, 8)
(16, 0)
(16, 8)
z//?
0
1
1
2
2
r/H
Q
0
1.0
0
1.0
/
0.7
0.33
0.3
0.2
c r
z
k N/ m
2
227.5
107.25
97.5
65
188 Chapt e r 6
Example 6. 8
For a raft o f siz e 98. 4 x 39.36 ft, compute the stres s a t 65. 6 f t depth below the cente r of the raf t b y
assuming that the rectangle can be represented by an equivalent circle. The load intensit y on the raf t
is31331b/ft
2
.
Solution
The radiu s o f a fictitiou s circula r footin g o f are a equa l t o th e rectangula r footin g o f siz e
98.4 x 39.36 f t i s
= 98.4 x 39.36 = 3873 sq. ft o r R
Q
= p = 35.12 ft
V
Use Eq. (6.21) fo r computing a a t 35. 6 f t depth
65.6
35.12
Now, z/R
Q
= -^^= 1.9 , and r/R
Q
= 0. From Fig. 6.11 , 7
Z
= 0.3
Therefore, c r = 0.3 q = 0.3 x 3133 = 940 lb/ft
2
.
6.8 V ERTI CA L STRESS B ENEATH L OADE D AREAS OF I RREG U L AR
SH APE
Newmark's I nfluenc e Chart
When the foundation consists of a large number of footings or when the loaded mat s or rafts ar e not
regular i n shape , a char t develope d b y Newmar k (1942 ) i s mor e practica l tha n th e method s
explained before . I t i s based o n the following procedure. The vertica l stres s cr , below th e cente r of
a circular area o f radius R which carries uniforml y distribute d load q i s determined per Eq. (6.21) .
It may be seen fro m Eq . (6.21) that when Rlz = °°,a
z
/q=l, tha t is cr , = q. This indicate s that
if the loaded are a extends to infinit y, th e vertical stress i n the semi-infinite solid at any depth z is the
same as unit load q at the surface. If the loaded area i s limited to any given radius R\ i t is possible t o
determine fro m Eq . (6.21 ) the ratios Rlz fo r which the rati o of Gjq ma y hav e any specifie d value,
say 0.8 or 0.6. Tabl e 6.1 give s the ratios of Rlz for different value s of <j/q.
Table 6.1 may be used for the computation o f vertical stress <J
7
at any depth z below the center
of a circular loaded are a of radius R. For example, at any depth z, the vertical stress o^ = 0.8 q if the
radius o f th e loade d are a a t th e surfac e i s R = 1.38 7 z. At th e sam e depth , th e vertica l stres s i s
cr = 0.7 q if R = 1.110 z. If instead of loading the whole area, if only the annular space between th e
circles o f radi i 1.38 7 z and 1.11 0 z ar e loaded , th e vertica l stres s a t z a t th e cente r of th e circl e i s
ACT =0. 8 q-0.7 q = 0.lq. Similarly if the annular space between circle s of radii l . l l O z and 0.91 7
z are loaded , th e vertica l stres s a t the same dept h z is ACT , = 0.7 q-0.6 q = 0.1 q. We may therefor e
draw a series of concentric circle s o n the surface of the ground in such a way that when the annular
space betwee n an y two consecutive circles i s loaded wit h a load q per uni t area, th e vertica l stress
ACT produce d a t an y dept h z belo w th e cente r remain s a constan t fractio n o f q . We ma y write ,
therefore,
Aa
z
= Cq (6.22 )
where C is constant. I f an annular space between an y two consecutive concentri c circles is divided
into n equal blocks and i f any one such block i s loaded wit h a distributed load q, the vertical stress
produced a t the center is, therefore,
St res s Dis t ribut io n in Soil s due t o Surfac e Load s 18 9
Table 6. 1 Value s o f Rlz fo r differen t value s o f a' Iq
Aa
L
n
ajq
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
C
n '
Rlz
0.000
0.270
0.401
0.518
0.637
0.766
0.917
1.110
<V <7
0.80
0.90
0.92
0.94
0.96
0.98
1.00
-
Rlz
1.387
1.908
2.094
2.351
2.748
3.546
oo
-
(6.23)
z
- = C when< 7 = l .
n
l
A load q = 1 covering one of the blocks wil l produce a vertical stress C- . In other words , the
'influence value' o f each loaded block i s C
(
. If the number of loaded blocks i s N, and if the intensity
of load i s q per uni t area, the total vertical stres s at depth z below the center of the circle i s
o
t
= CN q (6.24 )
The graphical procedure for computing the vertical stress G
Z
du e t o any surface loading is as
follows.
Select som e definit e scale t o represent dept h z . For instanc e a suitabl e lengt h AB i n c m as
shown i n Fig . 6.1 2 t o represen t dept h z i n meters . I n suc h a case , th e scal e i s
1 cm = zlAB meters . The length o f the radius R
Qg
which corresponds to ajq = 0.8 is then equal t o
1.387 x AB cm, and a circle of that radius may be drawn. This procedure may be repeated for other
ratios of ajq, fo r instance, for ojq = 0.7, 0. 5 etc. shown in Fig. 6.12 .
The annular space between the circles may be divided into n equal blocks, and in this case n
= 20. The influence value C. is therefore equal to 0.1/20 = 0.005. A plan of the foundation is drawn
on a tracing pape r t o a scal e suc h that the distance AB o n the char t correspond s t o the dept h z at
which the stress c?
z
is to be computed. For example, if the vertical stress at a depth of 9 m is required,
and i f the length AB chosen is 3 cm, the foundation plan is drawn to a scale of 1 cm = 9/3 = 3 m. In
case the vertical stress at a depth 12 m is required, a new foundation plan on a separate tracing paper
is required. The scal e for thi s plan is 1 cm = 12/AB = 12/ 3 = 4 m.
This means that a different tracin g has to be made for each differen t dept h whereas the chart
remains th e same fo r all . Fig . 6.12(b ) give s a foundation plan, which i s loade d wit h a uniforml y
distributed loa d q pe r uni t area . I t i s no w require d t o determin e th e vertica l stres s &
z
a t dept h
vertically below point O shown in the figure. I n order t o determine cr
z
, th e foundatio n pla n i s laid
over the char t i n suc h a wa y tha t the surfac e point O coincides wit h th e cente r O' of th e char t a s
shown in Fig. 6.12. The number of small blocks covered by the foundation plan is then counted. Let
this number be N . Then th e valu e of G
Z
at depth z below O is
a
z
= C
i
N q, which is the same as Eq. (6.24).
190 Chapt er 6
Influence valu e = C. = 0.005
(a) (b )
Figure 6 .1 2 Newm ark ' s in fluen c e char t
Example 6. 9
A ring footing o f external diamete r 8 m and internal diamete r 4 m rests at a depth 2 m below the
ground surface. I t carries a load intensit y of 15 0 kN/m
2
. Find the vertical stress at depths of 2,4 and
8 m along the axi s of the footing below the footing base. Neglec t th e effect of the excavation on the
stress.
Solution
From Eq . (6.21 ) we have,
1
3/2
where q = contact pressur e 15 0 kN/m
2
, /. , = Influence coefficient.
The stres s o_ a t any dept h z on th e axi s of the rin g i s expressed a s
o; = cr. -U, = q( I, - / , )
Z ^ i <- 2 - i <-2
where cr , = stres s du e t o the circula r footing of diameter 8 m, an d / , = I
7
an d R
Q
/z =
cr = stress du e to the footing of diameter 4 m, / , = / an d RJ z = ( RJ z).
St res s Dis t ribut io n i n Soil s due t o Surfac e Load s 191
The value s of /. , may b e obtaine d fro m Tabl e 6. 1 for variou s value s of /?
0
/z. The stres s c r a t
depths 2, 4 and 8 m ar e given below:
Dept h (m ) R^lz
2 2
4 1. 0
8 0. 5
'*,
0.911
0.647
0.285
R
2
/z l
z
1.0 0.69 7
0.5 0.28 5
0.25 0.08 7
(/ - I
2
) q = a
z
k N/ m
2
39.6
54.3
29.7
Example 6 .1 0
A raf t foundatio n o f th e siz e give n i n Fig . Ex . 6.1 0 carrie s a uniforml y distribute d loa d o f
300 kN/m
2
. Estimate th e vertical pressure a t a depth 9 m below th e point O marked i n the figure.
Solution
The dept h a t which &
z
required i s 9 m.
Using Fig . 6.12 , th e scal e o f the foundation plan i s AB = 3 cm = 9 m or 1 cm = 3 m. The
foundation pla n i s require d t o b e mad e t o a scal e o f 1 cm = 3 m o n tracin g paper . Thi s pla n i s
superimposed o n Fig . 6.1 2 wit h O coinciding wit h the cente r o f th e chart . Th e pla n i s shown i n
dotted line s in Fig. 6.12 .
Number of loaded block s occupie d b y the plan, N = 62
Influence value , C
f
= 0.005, q = 300 kN/m
2
The vertical stress , cr
z
= C
{
Nq - 0.00 5 x 62 x 300 = 93 kN/m
2
.
1 8m-
16.5 m
3 m
= x
6 m
1
3 m =
O
[ •— 9 m —-|
Figure Ex . 6.1 0
6.9 EM B ANKM EN T L OADI NG S
Long earth embankments with sloping sides represent trapezoidal loads . When the top width of the
embankment reduce s t o zero , th e loa d become s a triangula r stri p load . Th e basi c proble m i s t o
determine stresses du e t o a linearly increasing vertical loading on the surface .
192 Chapt er s
L inearly I ncreasin g V ertical L oading
Fig. 6.13(a ) shows a linearl y increasing vertica l loadin g startin g fro m zer o at A t o a finit e value q
per uni t lengt h a t B. Conside r a n elementary stri p of widt h db at a distance b from A. The loa d pe r
uni t lengt h ma y b e writte n as
dq - ( q/d) b db
Ifdq i s considered a s a line load on the sur f ace, th e vert i ca l stres s dcr, at P [Fig . 6 . 1 3(a)]
due t o d q ma y b e wr i t t e n fro m Eq . (6.4 ) a s
dcr, =\—\ —
'
Therefore,
er
b=a
2q
[(x-,
/ 9
on integration, o - = 77" ~ ~ a-s m20\ = 07 (6.25 )
z
2/T \a y
z
where 7 i s non-dimensional coefficien t whose value s for variou s value s of xla and zla ar e given i n
Table 6.2 .
If th e point P lies in the plane BC [Fig . 6.13(a)] , then j 8 = 0 at j c = a. Eq. (6.25) reduces t o
v
z
=-( a) (6.26 )
<• n
Figs. 6.13(b ) an d (c ) sho w th e distributio n o f stres s e r o n vertica l an d horizonta l section s
under th e actio n o f a triangula r loadin g a s a functio n o f q. Th e maximu m vertica l stres s occur s
below th e cente r o f gravit y of the triangula r load a s shown i n Fig. 6.13(c) .
V ertical Stress Du e t o Embankmen t Loadin g
Many time s i t ma y b e necessar y t o determin e th e vertica l stres s e r beneat h roa d an d railwa y
embankments, an d als o beneat h eart h dams . Th e vertica l stres s beneat h embankment s ma y b e
Table 6. 2 / fo r t rian g ula r loa d (Eq . 6. 25 )
x/a
-1.500
-1.00
0.00
0.50
0.75
1.00
1.50
2.00
2.50
0.00
0.00
0.00
0.00
0.50
0.75
0.50
0.00
0.00
0.00
0.5
0.002
0.003
0.127
0.410
0.477
0.353
0.056
0.017
0.003
1.0
0.014
0.025
0.159
0.275
0.279
0.241
0.129
0.045
0.013
2/fl
1.5
0.020
0.048
0.145
0.200
0.202
0.185
0.124
0.062
0.041
2
0.033
0.061
0.127
0.155
0.163
0.153
0.108
0.069
0.050
4
0.051
0.060
0.075
0.085
0.082
0.075
0.073
0.060
0.049
6
0.041
0.041
0.051
0.053
0.053
0.053
0.050
0.050
0.045
St res s Dis t ribut io n i n Soil s due t o Surfac e Load s 193
0 0. 2 0. 4 0. 6 O. i
3a
(a) Triangular loading (b) V ertical stres s o n vertical section s
At z = l.Of l
(c) V ertical stress on horizontal sections
Figure 6.1 3 St res s e s in a s em i-in fin it e m as s due t o t rian g ula r loadin g o n t h e
s urface
determined eithe r by the method of superposition by making use of Eq. (6.26) or by making use of
a singl e formula which can be developed fro m firs t principles .
cr
z
by Metho d o f Superposition
Consider an embankment given in Fig. 6.14 . a a t P may be calculated as follows:
The trapezoida l sectio n o f embankment ABCD, ma y be divide d int o triangular sections b y
drawing a vertical line through point P as shown in Fig. 6.14. We may writ e
ABCD = AGE + FGB -EDJ - FJ C (6.27)
If < r
r
<T
z2
, G
zy
and <7
z4
are the vertical stresses a t point P due to the loadings of figures AGE,
FGB, EDJ an d FJ C respectively , th e vertica l stres s o"
z
due t o the loadin g o f figur e ABCD ma y b e
written as
o=o -o - o
Z2
Z
3
Z
(6.28)
By applyin g th e principl e o f superpositio n fo r eac h o f th e triangle s b y makin g us e o f
Eq. (6.26), w e obtain
194 Chapt er 6
//VCVC<\XX\V
GG
D X
0, .
Figure 6 .1 4 Vert ica l s t res s due t o em ban k m en t
K\
(6.29)
a=ql=-f( a/z,b/z)
(6.30)
where / is the influence facto r for a trapezoidal load which is a function of a/z and biz.
The value s o f /, fo r variou s value s of a/ z an d biz ar e give n i n Fig. 6.15. (After Osterberg ,
1957)
a
^
fro m a Singl e Formul a fo r Asymmetrica l Trapezoidal L oadin g
A single formul a can be developed fo r trapezoidal loadin g for computing CT
Z
at a point P (Fig. 6.16 )
by applying Eq. (6.26) . The origi n of coordinates i s as shown in the figure. The fina l equatio n ma y
be expressed as
( a, ( a, +
X
— («!
a
i
(6.31)
where a
r
a
2
, and «
3
are th e angle s subtende d a t th e poin t P i n th e supportin g mediu m b y the
loading and R = a,/a^. When R = 1, the stresses ar e due to that of a symmetrical trapezoidal loading .
St res s Dis t ribut io n i n Soil s due t o Surfac e Load s 195
0.50
0.40
0.30
0.20
0.10
0.01 2 4 6 8 0.1 2 4 6 8 1. 0 2 4 6 8 1 0
Figure 6.1 5 A g rap h t o det erm in e com pres s iv e s t res s e s from a load varyin g by
s t raig ht lin e la w (Aft e r Os t erberg , 1957)
b b
a
2
—^
Figure 6.1 6 T rapezoida l loads
196
Chapt er 6
When th e top widt h i s zero, i.e , when b = 0, a
2
= 0, the vertical stress < r wil l be due t o a triangular
loading. The expressio n fo r triangula r loading i s
(6.32)
Eq. (6.31 ) and Eq . (6.32 ) can b e use d t o comput e c r a t any point i n the supportin g medium.
The angle s a
{
, cc
2
, an d a
3
ma y convenientl y be obtaine d b y a graphica l procedur e wher e thes e
angles ar e expressed a s radians i n the equations.
Example 6 .1 1
A 3 m high embankment i s t o be constructed a s shown i n Fig. Ex . 6. 11 . If the uni t weight of soi l
used i n the embankment i s 19. 0 kN/m
3
, calculat e the vertical stres s due to the embankment loadin g
at points P
I;
P
2
, and P
y
M
3.0
F
\ y= 1 9 kN/m
f
'3. 0
i
Note: Al l dimensions ar e i n metre s
P2 P^
Figure Ex . 6 .11 Vert ica l s t res s e s a t P
v
P
2
&
Solution
q = yH = 19 x 3 = 57 kN/m
2
, z = 3 m
The embankment i s divided into blocks a s shown i n Fig. Ex. 6.11 for making us e of the graph
given i n Fig. 6 . 15 . The calculation s ar e arranged a s follows :
Point
p
{
P2
PI
Bl ock
ACEF
EDBF
AGH
GKDB
HKC
ML DB
MACL
b
( m)
1.5
4.5
0
7.5
0
10.5
1.5
a
( m)
3
3
1.5
3
1.5
3.0
3.0
biz
0.5
1.5
0
2.5
0
3.5
0.5
alz
1
1
0.5
1.0
0.5
1.0
1.0
'
0.39
0.477
0.15
0.493
0.15
0.498
0.39
St res s Dis t ribut io n i n Soil s due t o Sur fac e Loads 197
V ertical stress <J
z
At point P, , cr , =
At poin t P
2
, CF . =
At poin t P
y
&
z
=
(0.39 + 0.477) x 57 = 49.4 kN/m
2
0. 1 5 x (57/2) + 0.493 x 57 - 0.1 5 x (57/2) = 28.
(0.498 - 0.39 ) 57 = 6.2 kN/m
2
1 kN/m
2
6 .1 0 APPROXI M AT E M ETH ODS FOR COM PU TI NG o
2
Two approximate methods are generall y used for computing stresses i n a soi l mas s belo w loade d
areas. They ar e
1. Us e o f the point load formulas such as Boussinesq's equation.
2. 2 : 1 method which gives an average vertical stress <r a t any depth z. This method assumes
that th e stresse s distribut e from th e loade d edg e point s a t a n angl e o f 2 (vertical ) t o 1
(horizontal)
The firs t method if properly applied gives the point stress at any depth which compares fairl y
well wit h exac t methods , wherea s th e secon d doe s no t giv e an y poin t stres s bu t onl y give s a n
average stres s cr at any depth. The averag e stress computed by the second metho d has been foun d
to be i n error depending upon the depth at which the stress i s required.
Point L oa d M etho d
Eq. (6.1) may be used for the computation of stresses i n a soil mass due to point loads acting at the
surface. Sinc e loads occupy finite areas , the point load formula may stil l be used if the footings are
divided int o smalle r rectangle s o r square s an d a serie s o f concentrate d load s o f valu e q dA ar e
assumed t o act a t the center of each squar e or rectangle. Here dA is the are a of the smalle r block s
and q the pressur e pe r uni t area. The onl y principl e t o be followed i n dividing a bigger are a int o
smaller blocks i s that the width of the smaller block should be less than one-third the depth z of the
point at which the stress i s required to be computed. The loads acting at the centers o f each smalle r
area ma y b e considere d a s poin t load s an d Boussinesq' s formul a ma y the n b e applied . Th e
difference betwee n th e poin t loa d metho d an d th e exac t metho d explaine d earlie r i s clea r fro m
z/B
Figure 6 .1 7 c r b y poin t loa d metho d
198 Chapt er 6
Figure 6 .1 8 c r 2 : 1 m et hod
Fig. 6.17. In thi s figur e th e absciss a o f th e curv e C
l
represent s th e vertica l stres s (7. , at differen t
depths z below th e center of a square area B x B which carries a surcharge g per uni t area or a total
surcharge loa d o f B
2
q. This curve is obtained by the exact metho d explaine d under Sect. 6.6. The
abscissa o f the curve C
2
represents the corresponding stresse s du e t o a concentrated loa d Q = B
2
q
acting at the center of the squar e area. The figure shows that the difference between th e two curves
becomes ver y smal l fo r value s of z/B i n exces s o f three . Henc e i n a computatio n o f th e vertica l
stress cr , at a depth z below an area, the area should be divided into convenient squares or rectangles
such that the leas t widt h of any block i s not greate r tha n z/3.
2 : 1 M etho d
In thi s method , th e stres s i s assume d t o b e distribute d uniforml y ove r area s lyin g belo w th e
foundation. The siz e of the area a t any depth i s obtained by assuming that the stresses sprea d ou t at
an angle of 2 (vertical) to 1 (horizontal) from th e edges o f the loaded area s shown in Fig. 6.18. The
average stres s a t any dept h z i s
Q
( B+z)( L
(6.33)
The maximu m stres s o
m
b y a n exac t metho d belo w th e loade d are a i s differen t fro m th e
average stres s a a t the same depth . The value of cr/tr reache s a maximum of about 1. 6 at zlb =
0-5, wher e b = half width.
6 .1 1 PRESSU R E I SOB AR S
Definition
An isobar i s a lin e whic h connect s al l point s o f equa l stres s belo w th e groun d surface . I n othe r
words, a n isobar is a stress contour. We may draw an y number of isobars as shown in Fig. 6.19 for
any give n loa d system . Each isoba r represent s a fractio n o f the loa d applie d a t th e surface . Sinc e
these isobar s for m close d figure s an d resembl e th e for m o f a bulb , the y ar e als o terme d bulb of
pressure or simply the pressure bulb. Normall y isobars ar e drawn for vertical, horizontal and shear
stresses. The one that i s most importan t in the calculation of settlement s o f footings i s the vertica l
pressure isobar .
St res s Dis t ribut io n i n Soil s due t o Surfac e Load s 199
Lines of
equal vertical
pressure or
isobars
Figure 6.1 9 Bul b o f pres s ur e
Significant Dept h
In hi s openin g discussio n o n
settlement o f structure s a t th e
First Internationa l Conferenc e
on Soi l Mechanic s an d
Foundation Engineering (held in
1936 a t Harvar d Universit y i n
Cambridge, Mass , USA) ,
Terzaghi stresse d th e
importance o f th e bul b o f
pressure an d it s relationshi p
with th e sea t o f settlement . A s
stated earlie r w e ma y dra w an y
number o f isobar s fo r an y given
load system , bu t th e on e tha t i s
of practica l significanc e i s th e
one which encloses a soil mass which is responsible fo r the settlement of the structure. The depth of
this stressed zon e may be termed a s the significant depth D
S
which is responsible fo r the settlement
of the structure. Terzaghi recommended tha t for all practical purpose s one can take a stress contour
which represent s 2 0 per cent of the foundation contact pressur e q, i.e, equal t o Q.2q. The dept h of
such a n isobar ca n be taken a s the significant depth D
s
which represents th e sea t of settlement fo r
the foundation . Terzaghi's recommendatio n wa s based o n hi s observatio n tha t direc t stresse s ar e
considered o f negligibl e magnitud e when they ar e smalle r tha n 20 per cent o f the intensit y of the
applied stress from structura l loading, and that most of the settlement, approximately 80 per cent of
the total , takes plac e a t a depth less than D
s
. The depth D
s
i s approximately equa l t o 1. 5 times the
width o f squar e o r circula r footing s [Fig . 6.20(a)].
If severa l loade d footing s ar e space d closel y enough , the individual isobars o f eac h footing
in questio n woul d combin e an d merg e int o on e larg e isoba r o f the_intensit y a s show n i n
[Fig. 6.20(b)] . The combine d significan t dept h D i s equa l t o abou t 1. 5 B .
a
z
= Q.2q
D<=\ .5B\ Stresse d zone
Isobar
(a) Significan t dept h of stressed zone
for singl e footing
Isobar
Combined stressed zone
(b) Effect o f closely placed footing s
Figure 6.2 0 Si gni f i can t dept h o f stresse d zon e
200 Chapt er 6
Pressure I sobar s fo r Footings
Pressure isobar s o f square , rectangula r an d circula r footing s ma y convenientl y b e use d fo r
determining vertica l pressure, ( J
z
, a t any depth, z, below th e base of the footings. Th e depth s z from
the groun d surface , an d th e distanc e r (o r jc ) fro m th e cente r o f th e footin g ar e expresse d a s a
function o f the widt h of the footing B. In the case o f circular footing B represent s th e diameter .
The followin g pressur e isobar s ar e give n base d o n eithe r Boussines q o r Westergaard' s
equations
1. Boussines q isobar s for squar e an d continuous footings, Fig. 6.21 .
2. Boussines q isoba r fo r circular footings, Fig. 6.22 .
3. Westergaar d isobar s for squar e and continuous footings, Fig. 6.23 .
B/2=b BI2=b
Continuous
25
Figure 6 .2 1 Pres s ur e is obar s bas e d o n B ous s in es q equat io n fo r s quar e an d
con t in uous foot in g s
St res s Dis t ribut io n i n Soil s du e t o Surfac e Load s 201
Figure 6.2 2 Pres s ur e is obar s bas e d o n B ous s in es q equat io n fo r un iform l y loade d
circular foot in g s
B/2=b B/2=b
5b
6b
Continuous
IB 2 B 3 5
Figure 6.2 3 Pres s ur e is obar s bas e d on Wes t erg aar d equat io n for s quar e an d
con t in uous foot in g
202 Chapt er 6
Example 6 .1 2
A singl e concentrate d loa d o f 100 0 k N act s a t th e groun d surface . Construc t a n isoba r fo r
<7 = 40 kN/m
2
b y makin g use o f th e Boussines q equation.
Solution
From Eq . (6.la) we have
3(2
1
We may no w writ e by rearranging an equation for the radial distanc e r as
-1
Now fo r Q = 1000 kN, cr , = 40 kN/m
2
, we obtai n th e value s o f r
p
r
2
, r
y
etc . for differen t
depths z, , z
2
, z
v
etc . The values so obtained are
z ( m)
0.25
0.50
1.0
2.0
3.0
3.455
r ( m)
1.34
1.36
1.30
1.04
0.60
0.00
g=1000kN
a, = 40 kN/m
J
Isobar
3. 455
Figure Ex . 6 .1 2
St res s Dis t ribut io n i n Soil s due t o Surfac e Load s 203
The isoba r fo r cr
z
= 4 0 kN/m
2
ma y b e obtaine d b y plottin g z agains t r a s show n i n
Fig. Ex. 6.12 .
6 .1 2 PROB L EM S
6.1 A column of a building transfers a concentrated load of 225 kips to the soil in contact with
the footing . Estimat e th e vertica l pressur e a t th e followin g points b y makin g us e o f th e
Boussinesq and Westergaard equations.
(i) V erticall y below th e column load a t depths o f 5, 10 , and 1 5 ft.
(ii) A t radia l distances of 5 , 1 0 and 2 0 f t and a t a depth of 1 0 ft.
6.2 Thre e footings are placed at locations forming an equilateral triangle of 13 ft sides. Each of
the footings carries a vertical loa d of 112. 4 kips. Estimate th e vertical pressures b y means
of the Boussinesq equation at a depth of 9 ft at the following locations :
(i) V erticall y below th e centers of the footings ,
(ii) Belo w th e center o f the triangle.
6.3 A reinforced concret e wate r tank of siz e 2 5 f t x 2 5 f t an d restin g o n th e groun d surfac e
carries a uniforml y distribute d loa d o f 5.2 5 kips/ft
2
. Estimat e th e maximu m vertica l
pressures a t depths o f 37. 5 an d 60 ft by point loa d approximatio n below th e center o f th e
tank.
6.4 Tw o footings of sizes 1 3 x 1 3 ft and 1 0 x 1 0 ft are placed 30 ft center t o center apart at th e
same leve l an d carr y concentrate d load s o f 33 7 an d 281 kip s respectively . Comput e th e
vertical pressur e at depth 1 3 ft below point C midway between th e centers o f the footings.
6.5 A an d B ar e tw o footing s o f siz e 1. 5 x 1. 5 m eac h place d i n positio n a s show n i n
Fig. Prob . 6.5. Eac h o f th e footing s carrie s a column loa d o f 400 kN. Determin e b y th e
2. 5m
S x?Xs\
1
«
A
Q
1
1 '
[-*- 1. 5 m~H '
1
' f t ^ m
m
1
B
//X\N
Q - 400 kN
(2
1
(*- 1.5
Figure Prob . 6. 5
Boussinesq method , th e exces s loa d footin g B carrie s du e t o th e effec t o f th e loa d o n A.
Assume the loads a t the centers of footings act as point loads.
6.6 I f both footings A and B in Fig. Prob. 6. 5 are at the same level at a depth of 0.5 m below the
ground surface, compute the stress d, midway between the footings at a depth of 3 m fro m
the ground surface. Neglect the effect o f the size for point load method .
6.7 Thre e concentrated loads Q
l
= 255 kips, Q
2
= 450 kips and <2
3
= 675 kips act in one vertical
plane and they are placed i n the order Q
l
-Q
2
~Qy Thei r spacings are 1 3 ft-10 ft . Determine
204 Chapt e r 6
the vertica l pressur e a t a depth of 5 f t along th e cente r lin e of footing s using Boussinesq' s
point loa d formula .
6.8 A squar e footin g o f 1 3 x 1 3 f t i s founde d a t a dept h o f 5 f t belo w th e groun d level . Th e
imposed pressur e a t th e bas e i s 873 2 lb/ft
2
. Determin e th e vertica l pressur e a t a dept h o f
24 f t below th e ground surfac e on th e cente r line of the footing.
6.9 A lon g masonr y wal l footin g carrie s a uniforml y distributed loa d o f 20 0 kN/ m
2
. I f th e
width of the footing is 4 m, determine the vertical pressures a t a depth o f 3 m below th e (i)
center, an d (ii ) edge o f th e footing.
6.10 A long foundation 0.6 m wide carries a line load o f 100 kN/m. Calculat e th e vertical stres s
cr, a t a poin t P , the coordinate s o f whic h ar e x = 2.75 m , an d z = 1. 5 m , wher e th e x -
coordinate i s normal t o the lin e load fro m th e central line of the footing.
6.11 A stri p footin g 1 0 f t wid e i s loade d o n th e groun d surfac e wit h a pressur e equa l t o
4177 lb/ft
2
. Calculat e vertica l stresses a t depth s o f 3 , 6 , an d 1 2 ft unde r the cente r o f th e
footing.
6.12 A rectangular footing of size 2 5 x 40 f t carries a uniformly distributed loa d o f 5200 lb/ft
2
.
Determine th e vertical pressur e 2 0 ft below a point O which is located a t a distance of 35 ft
from th e cente r o f th e footin g o n it s longitudina l axi s b y makin g us e o f th e curve s i n
Fig. 6.8.
6.13 Th e center of a rectangular area a t the ground surface has cartesian coordinat e (0,0 ) an d the
corners hav e coordinate s (6,15) . Al l dimension s ar e i n foo t units . Th e are a carrie s a
uniform pressur e o f 300 0 lb/ft
2
. Estimat e th e stresse s a t a dept h o f 3 0 f t belo w groun d
surface a t each o f the following locations: (0,0), (0,15) , (6,0) .
6.14 Calculat e th e vertical stress a t a depth of 50 ft below a point 10 ft oubide th e corner (alon g
the longe r side ) o f a rectangula r loade d are a 3 0 x 8 0 f t carryin g a unifor m loa d o f
2500 lb/ft
2
.
6.15 A rectangular footing 6 x 3 m carries a uniform pressure o f 300 kN/m
2
on the surface o f a
soil mass . Determin e th e vertical stress a t a depth o f 4.5 m below th e surfac e on the cente r
line 1. 0 m inside the lon g edge o f the foundation.
6.16 A circular ring foundation for an overhead tan k transmits a contact pressur e o f 300 kN/m
2
.
Its internal diameter i s 6 m and external diamete r 10m . Comput e th e vertical stres s o n the
center lin e o f th e footin g du e t o th e impose d loa d a t a dept h o f 6. 5 m belo w th e groun d
level. The footin g i s founded a t a depth o f 2. 5 m.
6.17 I n Prob. 6.16, if the foundation fo r the tank is a raft o f diameter 1 0 m, determine th e vertical
stress a t 6. 5 m dept h on the center lin e of the footing. All the other dat a remai n th e same .
6.18 Ho w fa r apar t mus t tw o 2 0 m diamete r tank s b e place d suc h tha t thei r combine d stres s
overlap i s no t greate r tha n 10 %of th e surfac e contac t stres s a t a dept h o f 1 0 m?
6.19 A wate r towe r i s founded o n a circular ring type foundation. The widt h of the ring i s 4 m
and it s interna l radiu s i s 8 m. Assuming th e distribute d loa d pe r uni t are a a s 300 kN/m
2
,
determine th e vertica l pressure a t a depth o f 6 m below th e center o f the foundation.
6.20 A n embankmen t fo r roa d traffi c i s require d t o b e constructe d wit h th e followin g
dimensions :
Top width = 8 m, height = 4 m, side slopes = I V : 1. 5 Hor
The uni t weight of soi l unde r the wors t condition i s 21 kN/m
3
. The surcharg e loa d o n th e
road surfac e ma y b e take n a s 5 0 kN/m
2
. Comput e th e vertica l pressur e a t a dept h o f 6 m
below th e ground surfac e at the followin g locations :
(i) O n th e centra l l ongi t udi nal plan e of th e embankment ,
(ii) Belo w th e toe s of th e embankment.
St res s Dis t ribut io n i n Soil s due t o Surfac e Loads 205
6.21 I f the top width of the road given in Prob. 6.20 i s reduced t o zero, wha t would be the change
in the vertical pressur e a t the same points?
6.22 A squar e footin g o f siz e 1 3 x 1 3 f t founde d on th e surfac e carrie s a distribute d loa d o f
2089 lb/ft
2
. Determin e the increas e i n pressure at a depth of 1 0 ft by th e 2: 1 method
6.23 A loa d o f 33 7 kip s i s impose d o n a foundation 1 0 ft squar e a t a shallo w dept h i n a soi l
mass. Determin e th e vertica l stres s a t a poin t 1 6 f t belo w th e cente r o f th e foundation
(a) assuming th e loa d i s uniforml y distributed ove r th e foundation , and (b ) assumin g the
load act s a s a point load a t the center o f the foundation.
6.24 A total loa d o f 900 kN i s uniformly distributed over a rectangular footing of size 3 x 2 m.
Determine th e vertica l stres s a t a dept h o f 2. 5 m belo w th e footin g a t poin t C
(Fig. Prob . 6.24) , unde r on e corne r an d D unde r th e center . I f anothe r footin g o f siz e
3 x 1 m wit h a total load o f 450 kN i s constructed adjoinin g the previous footing, what is
the additional stres s a t the point C at the same dept h du e to the construction of the second
footing?
2 m
D
3m
1m
i
3 m
h— i m- H
Figure Prob. 6.2 4
6.25 Refe r t o Prob . 6.24 . Determin e th e vertica l stres s a t a dept h o f 2. 5 m belo w poin t E i n
Fig. Prob . 6.24. Al l the other dat a given in Prob. 6.2 4 remai n the same .
CHAPTER 7
COMPRESSIBILITY AND CONSOLIDATION
7 .1 I NTRODU CTI O N
Structures are built on soils. They transfe r loads t o the subsoi l through the foundations. The effec t
of th e loads i s fel t b y th e soi l normall y up t o a depth of about two t o three time s th e widt h of the
foundation. Th e soi l withi n thi s dept h get s compresse d du e t o th e impose d stresses . Th e
compression o f the soi l mas s lead s t o the decrease i n the volume of the mass whic h results i n the
settlement of the structure.
The displacement s that develop at any given boundary of the soil mass can be determined o n
a rational basi s by summing up the displacements of small elements of the mass resulting from th e
strains produce d b y a change i n th e stres s system . The compressio n o f th e soi l mas s du e t o th e
imposed stresse s ma y b e almos t immediat e o r tim e dependen t accordin g t o th e permeabilit y
characteristics o f th e soil . Cohesionles s soil s whic h ar e highl y permeabl e ar e compresse d i n a
relatively shor t perio d o f tim e a s compare d t o cohesiv e soil s whic h ar e les s permeable . Th e
compressibility characteristics o f a soil mass might be due to any or a combination of the following
factors:
1. Compressio n of the soli d matter.
2. Compressio n of water and ai r within the voids .
3. Escap e o f water and ai r from th e voids.
It i s quit e reasonabl e an d rationa l t o assum e tha t th e soli d matte r an d th e por e wate r ar e
relatively incompressible under the loads usually encountered in soil masses. The change in volume
of a mass under imposed stresses must be due to the escape of water if the soil is saturated. But if the
soil i s partiall y saturated , the chang e i n volume of the mas s i s partl y due t o the compressio n an d
escape of air from the voids and partl y due to the dissolution of air in the pore water .
The compressibilit y o f a soi l mas s i s mostl y dependent o n the rigidit y o f the soi l skeleton .
The rigidity , in turn , i s dependen t o n th e structura l arrangement o f particle s and , i n fin e graine d
207
208 Chapt e r 7
soils, o n th e degre e t o whic h adjacen t particle s ar e bonde d together . Soil s whic h posses s a
honeycombed structur e possess hig h porosity and as such are more compressible . A soil composed
predominantly o f fla t grain s i s mor e compressibl e tha n one containin g mostly spherica l grains . A
soil i n an undisturbed state i s less compressibl e tha n the same soi l i n a remolded state .
Soils ar e neither truly elasti c nor plastic. When a soil mass i s under compression, th e volume
change i s predominantl y due t o th e slippin g of grain s one relativ e t o anothe r . The grain s do no t
spring back t o their original positions upon removal of the stress. However , a small elastic rebound
under lo w pressure s coul d b e attribute d t o th e elasti c compressio n o f th e adsorbe d wate r
surrounding th e grains.
Soil engineering problems ar e of two types. The first type includes all cases wherei n there is
no possibilit y o f th e stres s bein g sufficientl y larg e t o excee d th e shea r strengt h o f th e soil , bu t
wherein th e strain s lead t o wha t may b e a serious magnitud e of displacement o f individua l grains
leading t o settlement s withi n the soi l mass . Chapte r 7 deals wit h thi s type of problem. The secon d
type includes cases in which there is danger of shearing stresses exceeding th e shear strengt h of the
soil. Problems o f thi s type ar e called Stability Problems whic h are dealt wit h under the chapter s of
earth pressure , stabilit y of slopes , an d foundations.
Soil i n natur e may b e foun d i n any of the following state s
1. Dr y state .
2. Partiall y saturated state.
3. Saturate d state .
Settlements o f structure s buil t o n granula r soil s ar e generall y considere d onl y unde r tw o
states, tha t is, either dry or saturated. The stress-strain characteristics o f dry sand, depend primaril y
on th e relative densit y of the sand , an d t o a much smaller degre e o n the shap e an d siz e o f grains .
Saturation does not alte r the relationshi p significantly provided th e wate r conten t o f th e san d ca n
change freely . However , i n ver y fine-graine d o r silt y sand s th e wate r content ma y remai n almos t
unchanged durin g a rapi d chang e i n stress . Unde r thi s condition , th e compressio n i s time -
dependent. Suitabl e hypotheses relating displacement and stress change s i n granular soils have not
yet bee n formulated . However , th e settlement s ma y b e determine d b y semi-empirica l method s
(Terzaghi, Peck an d Mesri , 1996) .
In the case of cohesive soils , the dry stat e of the soil s i s not considered a s this state i s only of
a temporar y nature . Whe n th e soi l become s saturate d durin g th e rain y season , th e soi l become s
more compressibl e unde r the same impose d load . Settlemen t characteristic s o f cohesive soil s are ,
therefore, considere d onl y under completely saturated conditions. It is quite possible tha t there ar e
situations wher e th e cohesiv e soil s ma y remai n partiall y saturated du e t o th e confinemen t of ai r
bubbles, gase s etc . Curren t knowledge o n th e behavior o f partiall y saturate d cohesiv e soil s unde r
external loads i s not sufficient t o evolve a workable theory to estimate settlement s of structures built
on suc h soils .
7 .2 CONSOL I DATI O N
When a saturate d clay-wate r syste m i s subjecte d t o a n externa l pressure , th e pressur e applie d i s
initially take n b y th e wate r i n th e pore s resultin g thereb y i n a n exces s por e wate r pressure . I f
drainage i s permitted, the resulting hydraulic gradients initiat e a flow o f water out of the clay mas s
and the mas s begin s to compress. A portion of the applied stres s i s transferred t o the soi l skeleton ,
which i n tur n cause s a reductio n i n th e exces s por e pressure . Thi s process , involvin g a gradua l
compression occurrin g simultaneousl y wit h a flo w o f wate r ou t o f th e mas s an d wit h a gradua l
transfer o f the applied pressur e from th e pore water to the mineral skeleton i s called consolidation.
The proces s opposit e t o consolidation i s called swelling, which involve s an increas e i n th e wate r
content du e t o a n increas e i n the volume of the voids .
Com pres s ibil it y an d Con s olidat io n 20 9
Consolidation ma y be due t o one or more o f the following factors:
1. Externa l stati c load s fro m structures .
2. Self-weigh t of the soi l such as recently placed fills .
3. Lowerin g o f the ground wate r table .
4. Desiccation .
The tota l compressio n o f a saturate d cla y strat a unde r exces s effectiv e pressur e ma y b e
considered a s the su m of
1. Immediat e compression ,
2. Primar y consolidation, and
3. Secondar y compression .
The portion of the settlement of a structure which occurs more or less simultaneousl y with the
applied load s i s referre d t o a s th e initial o r immediate settlement. Thi s settlemen t i s du e t o th e
immediate compressio n o f the soi l laye r under undrained condition and is calculated by assuming
the soi l mas s t o behave as an elastic soil .
If the rate of compression o f the soi l layer is controlled solel y by the resistance of the flow of
water unde r the induce d hydrauli c gradients, th e proces s i s referred t o a s primary consolidation.
The portio n o f th e settlemen t tha t i s du e t o th e primar y consolidatio n i s calle d primary
consolidation settlement or compression. A t the present time the onl y theory of practical valu e for
estimating time-dependent settlement du e t o volume changes, that is under primary consolidation
is the one-dimensional theory.
The thir d part o f the settlement is due t o secondary consolidatio n or compression o f the clay
layer. This compressio n i s supposed t o start after th e primary consolidation ceases , that is after th e
excess por e wate r pressur e approache s zero . I t i s ofte n assume d tha t secondar y compressio n
proceeds linearly with the logarithm of time. However, a satisfactory treatment of this phenomenon
has not been formulate d for computing settlement under thi s category .
The Proces s o f Consolidatio n
The proces s o f consolidatio n o f a clay-soil-wate r syste m ma y b e explaine d wit h th e hel p o f a
mechanical mode l a s described b y Terzaghi an d Frohlich (1936) .
The mode l consist s of a cylinder with a frictionless piston as shown in Fig. 7.1. The piston is
supported o n one or mor e helica l metalli c springs . The spac e underneat h the piston i s completel y
filled wit h water . The spring s represen t th e mineral skeleton i n the actua l soi l mas s an d th e water
below th e piston i s the por e wate r under saturated conditions in the soi l mass . When a load o f p i s
placed o n th e piston , thi s stres s i s full y transferre d t o th e wate r (a s wate r i s assume d t o b e
incompressible) an d the water pressure increases . The pressure in the water is
u = p
This i s analogou s t o por e wate r pressure, u , that woul d be develope d i n a clay-water syste m
under external pressures. I f the whole model i s leakproof without any holes i n the piston, ther e is no
chance for the water to escape. Such a condition represents a highly impermeable clay-water system
in which there i s a very high resistance for the flow o f water. It has been found i n the case of compact
plastic clays that the minimum initial gradient required t o cause flow ma y be as high as 20 to 30.
If a few holes ar e mad e i n the piston, th e water wil l immediatel y escape through the holes .
With the escape of water through the holes a part of the load carrie d b y the wate r i s transferred t o
the springs . This process of transference of load fro m wate r t o spring goes on unti l the flo w stop s
210
Chapt er 7
Piston
Spring
Pore water
Figure 7 . 1 M echan ica l m ode l t o ex pl ai n t h e proces s o f con s olidat io n
when all the load wil l be carried b y the spring and none by the water. The time required t o attain this
condition depend s upo n th e numbe r an d siz e o f th e hole s mad e i n th e piston . A fe w smal l hole s
represents a clay soi l wit h poo r drainag e characteristics .
When th e spring-wate r syste m attain s equilibriu m conditio n unde r th e impose d load , th e
settlement o f th e pisto n i s analogou s t o th e compressio n o f th e clay-wate r syste m unde r externa l
pressures.
One- Dimensional Consolidatio n
In man y instance s th e settlemen t of a structure is due t o the presence of one o r mor e layer s o f sof t
clay locate d betwee n layer s of sand or stiffer cla y as shown i n Fig. 7.2A . The adhesion betwee n th e
soft an d stif f layer s almos t completel y prevent s the latera l movement o f the sof t layers . The theor y
that wa s develope d b y Terzagh i (1925 ) o n th e basi s o f thi s assumptio n i s calle d th e
one-dimensional consolidation theory. I n the laboratory thi s condition is simulated most closel y by
the confined compression o r consolidation test.
The proces s o f consolidatio n a s explaine d wit h referenc e t o a mechanica l mode l ma y no w be
applied t o a saturated cla y laye r i n the field. I f the clay strat a shown i n Fig 7. 2 B(a ) i s subjecte d t o an
excess pressur e Ap due to a uniformly distributed load/? on the surface, the clay layer is compressed ove r
Sand
Sand
Drainage
faces
Figure 7 .2 A Cl a y l aye r s an dwiche d bet wee n s an d l ayer s
Com pres s ibil it y an d Con s olidat io n 211
Drainage
boundary
Ap = 55 kPa
Impermeable
boundary
10 2 0 3 0 4 0 5 0
Excess porewate r pressur e (kPa ) (a)
Properties o f clay:
w
n
= 56-61%, w, = 46%
w =24%,p
c
/p
0
=l3l
Clay fro m Berthier-V ille , Canada
3 4 5 6 7
Axial compression (mm)
(b)
Figure 7.2 B (a ) Obs erved dis t ribut ion o f ex ces s pore wat e r pres s ur e durin g
con s olidat ion o f a s oft cla y layer ; (b) obs erved dis t ribut io n o f vert ica l com pres s io n
durin g con s olidat io n o f a s oft cla y laye r (aft e r M es r i an d Choi , 1985, M es r i an d
Feng, 1986 )
time an d exces s por e water drain s ou t o f i t t o th e sand y layer . Thi s constitute s th e proces s o f
consolidation. At the instant of application of the excess load Ap, the load is carried entirely by water in
the voids of the soil. As time goes on the excess pore water pressure decreases, an d the effective vertical
212 Chapt er 7
pressure i n the layer correspondingly increases . At any point within the consolidating layer , the value u
of the excess por e wate r pressure at a given time may be determined fro m
u = M. -
where, u = excess por e wate r pressur e a t depth z at any time t
u
{
= initia l tota l pore wate r pressur e a t time t = 0
Ap, = effectiv e pressur e transferre d to the soil grains at depth i and time t
At the end of primary consolidation, the excess por e water pressure u becomes equal t o zero.
This happens when u = 0 at all depths.
The time taken for ful l consolidatio n depends upon the drainage conditions , the thickness of the
clay strata , the excess loa d at the top of the clay strata etc. Fig. 7.2B (a ) gives a typical example o f an
observed distributio n of excess pore water pressure during the consolidation of a soft cla y layer 50 cm
thick restin g o n a n impermeabl e stratu m wit h drainag e a t th e top . Figur e 7.2B(b ) show s th e
compression o f the strata with the dissipation of pore water pressure. I t is clear from the figure that the
time taken for the dissipation of pore wate r pressure may be quit e long, say a year or more.
7 .3 CONSOL I DOM ETE R
The compressibilit y o f a saturated , clay-wate r syste m i s determine d b y mean s o f th e apparatu s
shown diagrammatically in Fig. 7.3(a) . This apparatus is also known as an oedometer. Figure 7.3(b)
shows a table to p consolidation apparatus.
The consolidation test is usuall y performed a t room temperature , i n floating or fixed ring s of
diameter from 5 to 1 1cm and from 2 to 4 cm in height. Fig. 7.3(a ) is a fixed ring type. In a floating
ring type, the ring i s free t o move i n the vertical direction.
Extensometer
Water reservoi r
(a)
Figure 7. 3 (a ) A s chem at i c diag ra m o f a con s ol idom et e r
Com pres s ibilit y an d Con s ol idat io n 21 3
Figure 7. 3 (b ) T abl e t o p con s olidat io n apparat u s (Court es y : Soilt es t , USA)
The soil sample is contained in the brass ring between two porous stones about 1.25 cm thick. By
means o f th e porou s stone s wate r ha s free acces s t o an d fro m bot h surface s o f th e specimen . Th e
compressive loa d i s applie d t o the specime n throug h a piston, either by mean s of a hanger and dead
weights or by a system of levers. The compression i s measured on a dial gauge.
At the bottom of the soil sampl e the water expelled from th e soil flows through the filter ston e
into the wate r container . At th e top , a well-jacke t fille d wit h wate r i s place d aroun d th e ston e i n
order to prevent excessive evaporatio n fro m th e sample during the test. Water from the sample als o
flows int o the jacket through the upper filter stone . The soi l sample is kept submerged in a saturated
condition during the test .
7.4 TH E STANDARD ONE- DI M ENSI ONA L CONSOL I DATI O N TES T
The main purpose of the consolidation test on soil samples is to obtain the necessary informatio n about
the compressibilit y propertie s o f a saturate d soi l fo r us e i n determinin g th e magnitud e an d rat e o f
settlement o f structures . The followin g test procedur e i s applie d t o an y typ e o f soi l i n th e standar d
consolidation test .
Loads ar e applie d i n step s i n suc h a wa y tha t th e successiv e loa d intensity , p, i s twic e th e
preceding one. The load intensities commonl y used being 1/4 , 1/2,1 , 2, 4, 8, and 16 tons/ft
2
(25 , 50,
100,200,400, 800 and 1600 kN/m
2
). Each load is allowed to stand until compression ha s practically
ceased (no longer than 24 hours). The dial readings are taken at elapsed time s of 1/4 , 1/2 , 1,2,4, 8,
15, 30, 60, 120 , 240 , 48 0 an d 144 0 minute s from th e time the ne w incremen t of load i s put on the
sample (or at elpased times as per requirements). Sandy samples are compressed in a relatively short
time as compared t o clay samples an d the use of one day duration is common for the latter .
After th e greates t loa d require d fo r th e tes t ha s bee n applie d t o th e soi l sample , th e loa d i s
removed i n decrements t o provide data for plotting the expansion curve of the soi l in order t o learn
214 Chapt e r 7
its elasti c propertie s an d magnitude s o f plasti c o r permanen t deformations . Th e followin g dat a
should als o be obtained :
1. Moistur e conten t and weight of the soi l sampl e before th e commencement o f the test .
2. Moistur e conten t and weight of the sampl e after completio n of the test .
3. Th e specifi c gravit y of the solids .
4. Th e temperatur e of the room wher e the test i s conducted.
7.5 PRESSU RE- V OI D RATIO CU RV ES
The pressure-voi d rati o curv e can b e obtaine d i f th e voi d rati o o f th e sampl e a t th e en d o f eac h
increment o f loa d i s determined . Accurat e determination s o f voi d rati o ar e essentia l an d ma y b e
computed fro m th e followin g data :
1. Th e cross-sectional are a o f the sampl e A, whic h is the same a s that of the brass ring.
2. Th e specifi c gravity , G^ , o f the solids .
3. Th e dr y weight , W
s
, of the soi l sample.
4. Th e sampl e thickness , h, at any stage o f the test.
Let V
s
=• volume of th e solid s i n th e sampl e
where
w
where y
w
- uni t weight of water
We can als o writ e
V
s
=h
s
A o r h
s
=^
where, h
s
= thickness o f soli d matter.
If e i s the voi d ratio of the sample , the n
Ah -Ah, h- h.
e =
Ah.. h..
(7.1)
In Eq . (7.1) h
s
is a constant and only h is a variabl e which decreases wit h increment load. If
the thickness h of the sampl e is known at any stage of the test , the void ratio a t all the stages o f the
test ma y be determined .
The equilibrium voi d ratio at the end of any load increment may be determined by the change
of voi d rati o method as follows:
Change o f V oid- Rati o M ethod
In one-dimensiona l compressio n th e chang e i n heigh t A/ i pe r uni t o f origina l heigh t h equal s th e
change i n volume A Vper uni t of original volume V.
(7.2)
h V
V ma y no w be expressed i n terms of voi d rati o e.
Com pres s ibilit y an d Con s olidat io n 215
; I
V"\ >
(a) Initial condition (b ) Compressed conditio n
Figure 7. 4 Chang e o f voi d rat i o
We may writ e (Fig. 7.4) ,
V
Therefore,
A/i _
~h~
or
V-V
V
e-e
V l+e l + e
l + e
h
(7.3)
wherein, t± e = change in void ratio under a load, h = initial height of sample, e = initial void ratio of
sample, e' - voi d ratio after compressio n unde r a load, A/ i = compression o f sample unde r the load
which may be obtained fro m dia l gaug e readings .
Typical pressure-voi d rati o curve s fo r a n undisturbe d cla y sampl e ar e show n i n Fig . 7.5 ,
plotted bot h o n arithmetic an d on semilog scales . Th e curve on the log scal e indicate s clearl y two
branches, a fairly horizonta l initia l portion an d a nearly straigh t incline d portion . The coordinate s
of poin t A i n th e figur e represen t th e voi d rati o e
Q
an d effectiv e overburde n pressur e p
Q
corresponding t o a state of the clay in the field a s shown in the inset of the figure. When a sample i s
extracted by mean s of the bes t of techniques , the wate r conten t of the cla y doe s not chang e
significantly. Hence , th e voi d rati o e
Q
at the star t of the test i s practically identica l wit h that of the
clay i n the ground. When the pressure o n the sampl e i n the consolidometer reache s p
0
, the e-log p
curve should pass through the point A unless the test conditions differ i n some manner from thos e in
the field. In reality the curve always passes below point A, because eve n the best sampl e i s at leas t
slightly disturbed .
The curve that passes throug h point A is generall y terme d a s afield curve o r virgin curve. In
settlement calculations, the field curve is to be used.
216 Chapt er 7
V irgin
curve
A)
Figure 7. 5 Pres s ure-voi d rat i o curve s
Pressure- V oid Rati o Curve s fo r San d
Normally, n o consolidatio n test s ar e conducte d o n sample s o f san d a s th e compressio n o f san d
under external load i s almost instantaneous as can be seen i n Fig. 7.6(a ) which gives a typical curve
showing the time versus th e compression cause d b y an increment of load.
In thi s sampl e mor e tha n 90 pe r cen t o f the compression ha s take n plac e withi n a period o f
less than 2 minutes. The time lag is largely of a frictional nature. The compression i s about the same
whether the sand i s dry or saturated. The shape of typical e-p curves for loos e an d dense sand s are
shown i n Fig. 7.6(b) . The amount of compression even under a high load intensit y is not significant
as can be see n fro m the curves.
Pressure- V oid Rati o Curve s fo r Clay s
The compressibilit y characteristic s o f clays depend o n many factors .
The mos t importan t factors are
1. Whethe r th e clay i s normall y consolidated o r overconsolidate d
2. Whethe r th e clay i s sensitive or insensitive.
%

c
o
n
s
o
l
i
d
a
t
i
o
n
O
O
O

O
N

-
f
c
'
K
J
O

O

O

O

O

C
\
\
V
V 1
) 1 2 3 4 5
Time i n mi n
l . U
0.9
• 2 0. 8
£
*o
'|0. 7
0.6
0.5
£
\
^
Rebou
^
S~
\
\ r\
nd cu n
Comp
<^

/e
Dense
/
ression
— ^

sand
curve
j sand
^=^
) 2 4 6 8 1 (
(a)
Pressure i n kg/cm^
(b)
Figure 7. 6 Pres s ure-voi d rat io curve s fo r s an d
Com pres s ibilit y an d Con s olidat io n 217
Normally Consolidate d and Overconsolidated Clay s
A cla y i s sai d t o b e normall y consolidate d i f th e presen t effectiv e overburde n pressur e p
Q
i s th e
maximum pressure t o which the layer has ever been subjecte d at any time i n its history, whereas a
clay laye r i s sai d t o be overconsolidate d i f the laye r was subjecte d a t one time i n it s histor y t o a
greater effective overburde n pressure, /?
c
, than the present pressure, p
Q
. The rati o p
c
I p
Q
i s called th e
overconsolidation ratio (OCR).
Overconsolidation o f a cla y stratu m ma y hav e bee n cause d du e t o som e o f th e following
factors
1. Weigh t of an overburden o f soi l which has erode d
2. Weigh t of a continental ic e sheet that melted
3. Desiccatio n of layers close t o the surface.
Experience indicate s tha t th e natural moisture content , w
n
, is commonl y close t o the liquid
limit, vv
;
, for normall y consolidated cla y soi l wherea s for the overconsolidate d clay , w
n
is clos e t o
plastic limit w .
Fig. 7. 7 illustrates schematicall y th e difference betwee n a normally consolidated cla y strata
such as B on the left sid e of Section CC and the overconsolidated portio n of the same layer B on the
right side of section CC . Layer A is overconsolidated du e t o desiccation.
All of the strata located above bed rock were deposited i n a lake at a time when the water level
was located abov e the leve l of the present hig h groun d when parts of the strat a wer e removed by
erosion, th e wate r conten t i n th e cla y stratu m B o n th e righ t han d sid e o f sectio n C C increased
slightly, whereas that of the lef t sid e of section CC decreased considerabl y because of the lowering
of th e wate r tabl e leve l fro m positio n D
Q
D
Q
t o DD. Nevertheless , wit h respec t t o th e presen t
overburden, th e cla y stratu m B on the right hand side o f section C C is overconsolidated clay , and
that on the lef t han d side is normall y consolidated clay.
While the wate r table descended fro m it s original t o it s final positio n below th e floor of the
eroded valley , the sand strat a above and below the clay layer A became drained. As a consequence,
layer A graduall y drie d ou t du e t o exposur e t o outsid e heat . Laye r A i s therefor e sai d t o b e
overconsolidated by desiccation .
Overconsolidated by
desiccation
DO
C
Original wate r tabl e
Original groun d surfac e
Structure
Present groun d
surface
J
Normally consolidated clay
• . ; ^— Overconsolidate d clay/ _ • •
:
' .
Figure 7. 7 Diag ra m illustratin g t he g eolog ica l proces s leadin g t o overcon s olidat io n
of cl ay s (Aft e r T erzag h i an d Peck , 1967 )
218 Chapt er 7
7 .6 DETERM I NATI O N O F PRECONSOL I DATI ON PRESSU RE
Several method s hav e bee n propose d fo r determinin g th e valu e o f th e maximu m consolidatio n
pressure. The y fal l unde r the followin g categories . They ar e
1. Fiel d method ,
2. Graphica l procedur e based on consolidation tes t results .
Field M etho d
The fiel d metho d i s based o n geological evidence . The geolog y an d physiography o f the sit e may
help t o locat e th e original groun d level. The overburden pressur e i n the clay structur e wit h respect
to th e origina l groun d leve l ma y b e take n a s th e preconsolidatio n pressur e p
c
. Usuall y th e
geological estimat e of the maximum consolidation pressure is very uncertain. In such instances, the
only remaining procedure fo r obtaining an approximate value of p
c
i s to make a n estimate based o n
the result s o f laborator y test s o r o n som e relationship s establishe d betwee n p
c
an d othe r soi l
parameters.
G raphical Procedur e
There ar e a fe w graphica l method s fo r determinin g th e preconsolidatio n pressur e base d o n
laboratory tes t data . N o suitabl e criteri a exist s fo r appraisin g th e relativ e merit s o f th e variou s
methods.
The earlies t an d the mos t widel y use d method wa s the one proposed by Casagrande (1936) .
The metho d involve s locatin g th e point of maximum curvature , 5, o n the laborator y e-lo g p curv e
of a n undisturbe d sampl e a s show n i n Fig . 7.8 . Fro m B , a tangen t i s draw n t o th e curv e an d a
horizontal lin e is also constructed. The angle between thes e two lines is then bisected. The absciss a
of the poin t o f intersectio n o f thi s bisector wit h the upwar d extension o f th e incline d straight par t
corresponds t o the preconsolidation pressure/^,.
Tangent a t B
e-log p curv e
log p P c
Figure 7. 8 M et ho d o f det erm in in g pb y Cas ag ran d e m et hod
Com pres s ibilit y an d Con s olidat ion 21 9
7.7 e-log p FIEL D CU RV ES FOR NORM AL L Y CONSOL I DATED AND
OV ERCONSOL I DATED CL AY S OF L OW TO MEDIU M SENSI TI V I TY
It ha s bee n explaine d earlie r wit h referenc e t o Fig . 7.5 , tha t th e laborator y e-lo g p curv e o f a n
undisturbed sampl e does not pass throug h point A an d always passes below th e point . I t has bee n
found fro m investigatio n tha t th e incline d straigh t portio n o f e-log p curve s o f undisturbe d o r
remolded sample s o f cla y soi l intersec t a t one poin t a t a low voi d rati o an d correspond s t o 0.4e
Q
shown as point C in Fig. 7. 9 (Schmertmann, 1955). It is logical to assume the field curv e labelled as
K
f
shoul d als o pas s throug h thi s point . Th e fiel d curv e ca n b e draw n fro m poin t A, havin g
coordinates ( e
Q
, /?
0
), which corresponds t o the in-situ condition of the soil . Th e straigh t line AC i n
Fig. 7.9(a ) give s the fiel d curv e AT,fo r normall y consolidated cla y soi l of low sensitivity.
The fiel d curv e for overconsolidate d cla y soi l consist s of two straight lines, represented b y
AB an d BC i n Fig. 7.9(b) . Schmertmann (1955) has shown that the initial section AB o f the fiel d
curve is parallel t o the mean slope MN of th e rebound laboratory curve. Point B is the intersection
point of the vertical line passing through the preconsolidation pressur e p
c
o n the abscissa an d the
sloping lin e AB. Sinc e poin t C i s th e intersectio n o f th e laborator y compressio n curv e an d th e
horizontal line at void ratio 0.4e
Q
, line BC can be drawn. The slope of line MN whic h i s the slop e
of the rebound curv e i s called th e swell index C
s
.
Clay o f Hig h Sensitivit y
If th e sensitivit y S
t
i s greate r tha n abou t 8 [sensitivit y i s define d a s th e rati o o f unconfme d
compressive strength s of undisturbed and remolded soi l sample s refer to Eq. (3.50)], then th e clay
is sai d t o be highl y sensitive. Th e natura l water content s o f such cla y ar e mor e tha n the liqui d
limits. Th e e-log p curv e K
u
fo r a n undisturbe d sampl e o f suc h a cla y wil l hav e th e initia l
branch almost fla t a s shown in Fig. 7.9(c) , and after thi s it drops abruptl y int o a steep segmen t
indicating ther e b y a structura l breakdown o f th e cla y suc h tha t a sligh t increas e o f pressur e
leads t o a large decreas e i n voi d ratio. The curv e then passes through a point of inflectio n a t d
and it s slop e decreases . I f a tangent i s drawn a t th e poin t o f inflectio n d, i t intersect s th e lin e
e
Q
A a t b. Th e pressur e correspondin g t o b ( p
b
) i s approximatel y equa l t o tha t a t whic h th e
structural breakdown take s place . I n areas underlai n by sof t highl y sensitiv e clays , th e exces s
pressure Ap ove r the layer shoul d be limited t o a fraction of the difference o f pressure (p
t
-p
0
).
Soil o f thi s typ e belong s mostl y t o volcani c regions .
7.8 COM PU TATI O N OF CONSOL I DATI ON SETTL EM ENT
Settlement Equation s for Normall y Consolidate d Clays
For computin g th e ultimat e settlemen t o f a structur e founde d o n cla y th e followin g dat a ar e
required
1. Th e thicknes s o f the cla y stratum, H
2. Th e initia l void ratio, e
Q
3. Th e consolidatio n pressur e p
Q
o r p
c
4. Th e field consolidatio n curv e K,
The slope of the field curve K.on a semilogarithmic diagram is designated a s the compression
index C
c
(Fig. 7.9 )
The equation for C
c
may be written as
C
e
°~
e e
°~
e A g
Iogp-logp
0
logp/
Po
logp/p
Q
(7
'
4)
220 Chapt er 7
0.46>
0
Remolded
compression curv e
Laboratory
compression
curve of an
undisturbed
sample k
u
Field curve
K,
Po P
l Og/7
(a) Normally consolidated cla y soil
A b
Laboratory compressio n curv e
Ae
Field curve
or virgin
compression
curve
Po PC Po +
logp
(b) Preconsolidated cla y soi l
0.4 e
PoPb
e-log p curve
(c) Typical e-log p curve for an undisturbed sample o f clay of high sensitivit y (Peck e t al. , 1974)
Figure 7. 9 Fiel d e-lo g p curve s
In one-dimensiona l compression , a s pe r Eq . (7.2) , th e chang e i n heigh t A/ / pe r uni t o f
original H ma y b e writte n a s equa l t o th e chang e i n volum e AV pe r uni t o f origina l volum e V
(Fig. 7.10) .
Art _ A V
H ~ V
Considering a uni t sectiona l area o f the clay stratum, we may write
(7.5)
V
l
=H
l
= H
s
( e
Q
-
Com pres s ibilit y an d Con s olidat io n 221
H
f
n,
I J
A//
Figure 7.1 0 Chan g e of height due to on e-dim en s ion a l com pres s io n
Therefore,
Substituting for AWV in Eq. (7.5)
Ae
(7.6)
(7.7)
If we designate the compression A// of the clay layer as the total settlement S
t
of the structure
built on it , we have
A// = S =
l + e
r
(7.8)
Settlement Calculatio n fro m e-lo g p Curve s
Substituting for Ae in Eq. (7.8 ) we have
Po
or
• /flog-
Po
(7.9)
(7.10)
The ne t change i n pressure Ap produced b y the structur e a t the middl e o f a clay stratu m i s
calculated fro m th e Boussinesq o r Westergaard theorie s a s explained i n Chapter 6.
If th e thicknes s o f th e cla y stratu m i s to o large , th e stratu m may b e divide d int o layer s o f
smaller thickness not exceeding 3 m. The net change in pressure A/ ? at the middle of each layer will
have t o b e calculated . Consolidatio n test s wil l have t o b e complete d o n sample s take n fro m th e
middle of each of the strata and the corresponding compression indices wil l have to be determined.
The equation for the total consolidation settlement may be written as
(7.11)
222 Chapt er 7
where th e subscrip t ; ' refer s t o eac h laye r i n th e subdivision . If ther e i s a serie s o f cla y strat a o f
thickness H
r
//
2
, etc. , separate d b y granula r materials , th e sam e Eq . (7.10 ) ma y b e use d fo r
calculating the total settlement.
Settlement Calculatio n fro m e- p Curve s
We ca n plo t th e fiel d e- p curve s fro m th e laborator y tes t dat a an d th e fiel d e-\ og p curves . Th e
weight o f a structur e or o f a fil l increase s th e pressur e o n th e cla y stratu m from th e overburde n
pressure p
Q
t o th e valu e p
( )
+ A/? (Fig . 7.11). The correspondin g voi d rati o decreases fro m e
Q
t o e .
Hence, fo r th e rang e i n pressur e from p
Q
t o ( p
Q
+ A/?), we ma y writ e
-e -
or a
v
( c m
2
/ g m) = (7.12)
/?(cm
2
/gin)
where a
v
is called th e coefficient o f compressibility.
For a given difference in pressure, the value of the coefficient of compressibility decrease s as
the pressur e increases . No w substitutin g fo r Ae i n Eq. (7.8) fro m Eq . (7.12) , we have the equation
for settlemen t
a H
S
;
= —-—Ap = m
v
H A/ ?
(7.13)
where m
v
= a
v
/( 1 + e
Q
) i s known as the coefficient o f volume compressibility.
It represents th e compression o f the clay per uni t of original thickness due t o a uni t increas e
of the pressure .
Clay stratum
Po P
Consolidation pressure, p
Figure 7 .1 1 Set t lem en t cal cul at io n fro m e - pcurve
Com pres s ibilit y an d Con s olidat ion 22 3
Settlement Calculatio n fro m e-lo g p Curv e fo r Overconsolidate d Clay Soi l
Fig. 7.9(b) gives the field curve K
f
for preconsolidate d cla y soil. The settlemen t calculation depends
upon the excess foundatio n pressure Ap over and above the existing overburden pressur e p
Q
.
Settlement Computation , i f p
Q
+ A/0 < p
c
(Fig. 7 .9 (b))
In suc h a case, us e th e sloping lin e AB. I f C
s
= slope o f thi s line (also called th e swel l index), we
have
\ a
c =
l og
( p
o
+ Ap )
(7.14a )
Po
or A * = C, log^ (7.14b )
By substitutin g for A< ? in Eq . (7.8) , we have
(7.15a)
Settlement Computation , i f p
0
< p
c
< p
0
+ Ap
We may writ e from Fig. 7.9(b)
Pc
(715b)
In this case the slope of both the lines AB and EC in Fig. 7.9(b) are required t o be considered.
Now th e equation for S
t
ma y be written as [fro m Eq . (7.8) and Eq. (7.15b) ]
C
S
H p
c
C
C
H
log— + — -—log
*
Pc
The swel l index C
s
« 1/ 5 to 1/10 C
c
can be used as a check.
Nagaraj an d Murthy (1985) hav e proposed th e following equation for C
s
as
C =0.046 3 -^ - G
100
s
where w
l
= liquid limit, G
s
= specific gravity of solids .
Com pres s ion Inde x C
c
— Empirical Relat ion s hip s
Research worker s i n different part s of the worl d have established empirica l relationships betwee n
the compression index C and other soi l parameters. A few of the important relationships are given
below.
Skempton's Formul a
Skempton (1944) established a relationship between C, and liquid limits for remolded clay s as
C
c
= 0.007 ( w
l
- 10 ) (7.16 )
where w
l
i s in percent.
224 Chapt e r 7
Terzaghi and Pec k Formul a
Based o n th e wor k o f Skempto n an d others , Terzagh i an d Pec k (1948 ) modifie d Eq . (7.16 )
applicable t o normall y consolidated clays of low t o moderat e sensitivit y as
C
c
= 0.009 (w, -10) (7.17 )
Azzouz e t al. , Formul a
Azzouz e t al. , (1976 ) propose d a numbe r o f correlation s base d o n th e statistica l analysi s o f a
number of soils . The on e o f the many which is reported t o have 86 percent reliabilit y is
C
c
= 0.37 ( e
Q
+ 0.003 w
{
+ 0.0004 w
n
- 0.34 ) (7.18 )
where e
Q
= in-situ voi d ratio, w
f
an d w
n
ar e i n per cent . Fo r organi c soi l the y proposed
C
c
= 0. 115w
n
(7.19 )
H ough's Formul a
Hough (1957) , o n th e basi s o f experiment s o n precompresse d soils , ha s give n th e followin g
equation
C
c
= 0.3 ( e
0
- 0.27) (7. 20 )
Nagaraj an d Srinivas a M urthy Formul a
Nagaraj an d Srinivas a Murth y (1985 ) hav e develope d equation s base d o n thei r investigatio n a s
follows
C
c
= 0.2343 e, (7.21 )
C
c
= 0.39*
0
(7.22 )
where e
l
i s the voi d ratio at the liqui d limit , and e
Q
is the in-situ voi d ratio.
In th e absenc e o f consolidatio n test data, on e o f th e formula e given abov e ma y b e use d fo r
computing C
c
according t o the judgment of the engineer .
7.9 SETTL EM EN T DU E TO SECONDARY COM PRESSI ON
In certai n types of clays the secondar y time effect s ar e ver y pronounced t o the extent that in som e
cases th e entir e time-compressio n curv e ha s th e shap e o f a n almos t straigh t slopin g lin e whe n
plotted o n a semilogarithmi c scale , instea d o f th e typica l inverte d S-shap e wit h pronounce d
primary consolidation effects. Thes e so called secondar y time effects ar e a phenomenon somewha t
analogous t o th e cree p o f othe r overstresse d materia l i n a plasti c state . A delaye d progressiv e
slippage of grain upon grain as the particles adjust themselve s to a more dense condition, appears t o
be responsible for the secondary effects. Whe n the rate of plastic deformations of the individual soil
particles o r o f thei r slippag e on eac h othe r i s slowe r tha n th e rat e o f decreasin g volum e of void s
between th e particles , the n secondar y effects predominat e and thi s is reflected b y the shap e o f the
time compression curve. The factor s which affect th e rate of the secondary compression o f soils are
not ye t ful l y understood , and n o satisfactor y method ha s ye t bee n develope d fo r a rigorou s an d
reliable analysi s and forecas t o f the magnitude of thes e effects . Highl y organic soil s ar e normally
subjected t o considerable secondar y consolidation.
The rat e o f secondar y consolidatio n ma y b e expresse d b y th e coefficient o f secondary
compression, C
a
a s
Com pres s ibilit y an d Con s olidat ion 225
c =
c
n
or A e = C
a
log —
*\
(7.23)
where C
a
, the slop e o f th e straight-lin e portio n o f th e e-log t curve , i s known a s th e secondary
compression index. Numerically C
a
is equal to the value of Ae for a single cycle of time on the curve
(Fig. 7.12(a)) . Compressio n i s expresse d i n term s o f decreas e i n voi d rati o an d tim e ha s bee n
normalized wit h respect t o the duration t o f the primary consolidation stage . A general expressio n
for settlemen t due t o secondar y compressio n unde r the final stag e of pressure p
f
ma y be expresse d
as
5 = •H
(7.24)
The valu e of Ae fro m tit = 1 to any time / may b e determine d fro m th e e versus tit curv e
corresponding t o the final pressur e p
f
.
Eq. (7.23) may now be expressed a s
A<? = C
a
log —
For a constant value C
a
between t an d t, Equation (7.24) ma y be expressed a s
(7.25)
(7.26)
where, e
Q
- initia l voi d rati o
H = thickness o f the clay stratum.
The value of C
a
fo r normally loade d compressibl e soil s increases i n a general wa y with the
compressibility an d hence , wit h th e natura l wate r content , i n th e manne r show n i n Fig . 7.12(b )
(Mesri, 1973) . Althoug h th e rang e i n value s fo r a give n wate r conten t i s extremel y large , th e
relation give s a conceptio n o f th e uppe r limi t o f th e rat e o f secondar y settlemen t tha t ma y b e
anticipated if the deposit i s normally loaded or if the stress added by the proposed constructio n will
appreciably excee d th e preconsolidatio n stress . Th e rat e i s likel y t o b e muc h les s i f th e cla y i s
strongly preloade d o r i f the stres s afte r th e additio n o f the loa d i s smal l compare d t o the existing
overburden pressure. Th e rate i s also influenced by the length of time the preload ma y have acted,
Slope = C
a
r
2
= 1 0 f ,
Time (log scale)
Figure 7. 12(a ) e-lo g p tim e curve repres en t in g s econ dar y com pres s io n
226 Chapt er 7
100
10
1.0-
0.1
10
1. Sensitiv e marine clay, Ne w
Zealand
2. Mexic o city cla y
3. Calcareou s organic clay
4. Led a clay
5. Norwegia n plastic clay
6. Amorphou s and fibrous pea t
7. Canadia n muskeg
8. Organi c marine deposits
9. Bosto n blue clay
10. Chicag o blue clay
11. Organi c silty cla y
O Organi c silt , etc .
100
Natural water content, percent
1000 3000
Figure 7 .1 2 (b) Relat ion s hi p bet wee n coefficien t o f s econ dar y con s olidat io n an d
n at ural wat e r con t en t o f n orm all y loade d depos it s o f cl ay s an d various com pres s ibl e
org an ic s oil s (aft e r M es ri , 1973 )
by the existence of shearing stresses an d by the degree o f disturbance of the samples. The effect s of
these variou s factor s hav e not ye t been evaluated . Secondar y compressio n i s high i n plasti c clay s
and organi c soils . Table 7. 1 provides a classification of soi l base d o n secondary compressibility . If
' young, normally loaded clay' , having an effective overburden pressure o f p
0
i s left undisturbe d for
thousands o f years, ther e wil l be creep o r secondar y consolidation . Thi s wil l reduce th e voi d rati o
and consequentl y increas e th e preconsolidatio n pressur e whic h wil l b e muc h greate r tha n th e
existing effectiv e overburde n pressur e p
Q
. Suc h a cla y ma y b e calle d a n aged, normally
consolidated clay.
Mesri an d Godlewski (1977) report that for any soil the ratio C
a
/C
c
i s a constant (where C
c
is
the compression index) . This i s illustrated in Fig. 7.13 for undisturbed specimens o f brown Mexic o
City cla y wit h natural water content w
n
= 313 to 340%, vv
;
= 361%, w
p
= 9\ % andp
c
/p
o
= 1. 4
Table 7. 2 gives values of C
a
/C
c
for some geotechnica l material s (Terzaghi , et al. , 1996) .
It i s reported (Terzagh i e t al. , 1996 ) tha t fo r al l geotechnica l material s C
a
/C
c
range s fro m
0.01 t o 0.07. Th e value 0.04 i s the most common valu e for inorgani c clays and silts.
Com pres s ibilit y an d Con s olidat io n 227
0.3
0.2
•o
§
0.1
i i
Mexico City clay
c
a
lc
c
= 0.046
0 1 2 3 4 5 6
Compression index C
c
Figure 7 .1 3 A n ex am pl e o f t h e relat io n bet wee n C
a
and C
c
(aft e r M es r i an d
Godlews k i, 1977 )
Table 7. 1 Cl as s ificat io n o f soi l bas e d on s econ dar y com pres s ibilit y (T erzag hi ,
et al. , 1996)
C Secon dar y com pres s ibilit y
< 0.002
0.004
0.008
0.016
0.032
0.064
V ery lo w
Low
Medium
High
V ery high
Extremely high
Table 7. 2 Value s o f C
a
IC
c
for g eot echn ica l m at erial s (T erzag hi , et al. , 1996 )
M at erial
Granular soils includin g rockfill 0.0 2 ± 0.01
Shale an d mudstone 0.0 3 ± 0.01
Inorganic clay an d silts 0.0 4 ± 0.01
Organic clays and silt s 0.0 5 ± 0.01
Peat an d muskeg 0.0 6 ± 0.01
Example 7 . 1
During a consolidation test , a sample of fully saturate d clay 3 cm thick (= h
Q
) i s consolidated unde r
a pressure increment of 200 kN/m
2
. When equilibrium is reached, th e sampl e thickness i s reduced
to 2.60 cm. The pressur e i s then removed an d the sampl e i s allowe d t o expand an d absor b water .
The final thickness i s observed a s 2.8 cm (ft, ) and the final moistur e content is determined a s 24.9%.
228 Chapt er 7
K, = 0.672 cm
3
Figure Ex . 7. 1
If th e specifi c gravit y of th e soi l solid s i s 2.70, fin d th e voi d rati o o f th e sampl e befor e an d afte r
consolidation.
Solution
Use equation (7.3 )
-
h
A/z
1. Determination o f 'e*
Weight of solid s = W
s
= V
s
G
s
J
m
= 1 x 2.70 x 1 = 2.70 g .
W
W
= 0.249 o r W
w
= 0.249 x 2.70 = 0.672 gm, e
f
= V
w
= 0.672.
2. Changes i n thickness from fina l stage t o equilibrium stage with load o n
(1 + 0.672)0.20
A/i = 2.80 -2.60 = 0.20 cm ,
2.80
• = 0.119.
V oid rati o afte r consolidatio n = e,-&e = 0.672 - 0. 1 19 = 0.553.
3. Change i n void ratio from th e commencement t o the end o f consolidation
1 +0
-
553
(3.00 - 2.60) = x 0.40 = 0.239 .
2.6 2. 6
V oid rati o at the star t of consolidation = 0.553 + 0.239 = 0.79 2
Example 7 . 2
A recentl y complete d fil l wa s 32. 8 f t thic k and it s initia l average voi d rati o wa s 1.0 . Th e fil l wa s
loaded o n th e surfac e b y constructin g a n embankmen t coverin g a larg e are a o f th e fill . Som e
months after the embankment was constructed, measurements of the fill indicated a n average voi d
ratio of 0.8. Estimat e th e compressio n o f the fill .
Com pres s ibilit y an d Con s olidat ion 22 9
Solution
Per Eq. (7.7) , th e compression o f the fill may be calculated as
where AH = the compression, Ae = change in void ratio, e
Q
= initial void ratio, H
Q
= thickness of fill .
Substituting, A/ f =
L0
~
0
-
8
x 32.8 = 3.28 f t .
Example 7. 3
A stratum of normall y consolidate d cla y 7 m thick i s located a t a depth 12 m below groun d level .
The natura l moistur e conten t of th e cla y i s 40. 5 pe r cen t an d it s liqui d limi t i s 48 pe r cent . Th e
specific gravit y of the solid particles is 2.76. The water table is located a t a depth 5 m below ground
surface. The soil i s sand above the clay stratum. The submerged uni t weight of the sand is 1 1 kN/m
3
and the same weighs 1 8 kN/m
3
above the water table. The average increase i n pressure at the center
of the clay stratum is 12 0 kN/m
2
due to the weight of a building that will be constructed on the sand
above the clay stratum. Estimate the expected settlemen t o f the structure.
Solution
1 . Determinatio n o f e and y
b
for the clay [Fig . Ex. 7.3 ]
W
=1x2. 76x1 = 2.76 g
405
W = — x2.7 6 = 1.118 g
w
10 0

r
v
s
i
= UI& + 2.76 = 3.878 g
W
' 1 Q - J / 3
Y, = -= -= 1-8 3 g/c m
' 2.11 8
Y
b
=(1.83-1 ) = 0.83 g/cm
3
.
2. Determinatio n of overburde n pressur e p
Q
PO = y\
h
i + Y2
h
i + yA °
r
P
0
= 0.83x9.81x3.5 + 11x7 + 18x5 = 195.5 kN/m
2
3. Compressio n inde x [Eq . 11.17 ]
C
c
= 0.009(w, -10) = 0.009 x (48 -10) = 0.34
230 Chapt er 7
w
w.
5 m
7m J
7 m
I
(b)
Fig. Ex . 7. 3
4. Exces s pressur e
A;? = 12 0 kN/m
2
5. Tota l Settlemen t
C
s
t
=
0.34 _ _
n i
195. 5 + 120
00 0
x 700 log = 23.3 c m
2.118 " 195. 5
Estimated settlemen t = 23. 3 cm.
Example 7 . 4
A column of a building carries a load of 1000 kips. The load is transferred to sub soil through a square
footing o f siz e 1 6 x 1 6 ft founded at a depth of 6.5 f t below ground level . The soi l below the footing
is fine san d up t o a depth of 16. 5 ft and below thi s is a sof t compressibl e cla y of thickness 1 6 ft. The
water table is found at a depth of 6.5 ft below the base of the footing. The specifi c gravities of the solid
particles o f sand and clay are 2.64 and 2.72 and their natural moisture contents are 25 and 40 percent
respectively. The sand above the water table may be assumed t o remain saturated. If the plastic limi t
and the plasticity index of the clay are 30 and 40 percent respectively, estimate the probable settlement
of the footing (see Fig . Ex. 7.4)
Solution
1. Require d A/ ? at the middl e of the cla y layer using the Boussines q equatio n
24.5
16
= 1.53 <3.0
Divide the footing into 4 equal parts s o that Z/B > 3
The concentrate d loa d at the center o f each par t = 250 kips
Radial distance , r = 5.66 f t
By th e Boussines q equation the exces s pressur e A/ ? at dept h 24. 5 f t i s ( I
B
= 0.41 )
0.41 = 0. 683k/ f t
2
24.5'
Com pres s ibilit y an d Con s olidat io n 231
. - . *' . . •
.• 6.5 f t W.
CX
3
^^ferief,.
24.5 ft = Z
16f t
16ft -
r = 5.66 ft
Figure Ex. 7. 4
2. V oi d ratio an d uni t weights
Per the procedure explaine d i n Ex. 7. 3
For san d y , = 12 4 lb/ft
3
y
fc
= 61. 6 lb/ft
3
For cla y y
b
= 51.4 lb/ft
3
<?
0
= 1.0 9
3. Overburde n pressur e p
Q
p
Q
= 8 x 51. 4 + 1 0 x 62 + 1 3 x 12 4 = 2639 lb/ft
2
4. Compressio n inde x
w/
= I
p
+ w
p
= 40 + 30 = 70%, C
c
= 0.009 (70 - 10 ) = 0.54
0.54 ... . 263 9 + 683
ft/lia A A 0
, .
Settlement S , = . . _x! 6xl o g = 0.413 ft = 4. 96m.
1 + 1.09 2639
Example 7. 5
Soil investigation at a site gave the following information. Fine sand exists t o a depth of 10. 6 m and
below this lies a soft cla y layer 7.60 m thick. The water table is at 4.60 m below the ground surface.
The submerged uni t weight of sand y
b
i s 10. 4 kN/m
3
, and the wet unit weight above the water table
is 17. 6 kN/m
3
. Th e wate r conten t o f th e normall y consolidate d cla y w
n
= 40%, it s liqui d limi t
w
t
= 45%, an d th e specifi c gravit y o f th e soli d particle s i s 2.78. Th e propose d constructio n wil l
transmit a net stress of 12 0 kN/m
2
at the center of the clay layer. Find the average settlemen t of the
clay layer .
232 Chapt e r 7
Solution
For calculating settlement [Eq . (7.15a) ]
C p
n
+ A/?
S = — —H log^ --wher e &p = 120 kN /m
2
l + e
Q
p
Q
From Eq . (7.17), C
r
= 0.009 (w, - 10 ) =0.009(45 - 10 ) =0.32
wG
From Eq. (3. 14a), e
Q
= -= wG = 0.40 x 2.78 = 1.1 1 sinc e S = 1
tJ
Y
b
, th e submerge d uni t weigh t of clay, is foun d a s follows
MG.+«.)
=
9*1(2.78 + Ul)
3
' ^"t 1 , 1 . 1 1 1
l + e
Q
l + l . ll
Yb=Y^-Y
w
=18.1-9.8 1 = 8.28 k N/ m
3
The effectiv e vertical stres s p
Q
a t the mi d heigh t of the cla y layer i s
p
Q
= 4.60 x 17.6 + 6 x 10.4 + — x 8.28 = 174.8 kN / m
2
_ 0.32x7.60 , 174. 8 + 120
Now S
t
= -log -= 0. 26m = 26 cm
1
1+1.1 1 174. 8
Average settlemen t = 2 6 cm.
Example 7. 6
A soi l sampl e ha s a compressio n inde x o f 0.3. I f the voi d rati o e a t a stres s o f 2940 Ib/ft
2
i s 0.5,
compute (i ) the voi d rati o i f the stres s i s increase d t o 4200 Ib/ft
2
, an d (ii ) th e settlemen t o f a soi l
stratum 1 3 ft thick.
Solution
Given: C
c
= 0.3, e
l
= 0.50, /? , = 2940 Ib/ft
2
, p
2
= 4200 Ib/ft
2
.
(i) No w fro m Eq . (7.4) ,
p— p
C i%. "-)
C =
l
-
2

or e
2
= e
]
-
c
substituting the known values, we have,
e- = 0.5 - 0.31og - 0.454
2
294 0
(ii) Th e settlemen t pe r Eq. (7.10) is
c
c
c „ , Pi 0. 3x13x12 , 420 0
S = — —//l og— = -log -= 4.83 m.
p
l
1. 5 294 0
Com pres s ibilit y an d Con s olidat io n 23 3
Example 7. 7
Two points on a curve for a normally consolidated cla y have the following coordinates .
Point 1 : *, = 0.7,
Pl
= 2089 lb/ft
2
Point 2: e
2
= 0.6, p
2
= 6266 lb/ft
2
If th e averag e overburde n pressur e o n a 2 0 f t thic k cla y laye r i s 313 3 lb/ft
2
, ho w muc h
settlement wil l the clay layer experience du e to an induced stress of 3340 lb/ft
2
a t its middepth .
Solution
From Eq . (7.4) w e have
C -
e
^
e
> = °-
7
"
a6
-02 1
c
\ ogp
2
/p
l
lo g (6266/2089)
We need th e initia l void ratio e
Q
at an overburden pressur e o f 3133 lb/ft
2
.
e
n
-e~
C =—2 —T
2
— = 0.2 1
or ( e
Q
- 0.6 ) = 0.21 log (6266/3 133) = 0.063
or e
Q
= 0.6 + 0.063 = 0.663 .
Settlement, s =
Po
Substituting the known values, with Ap = 3340 lb/ft
2
„ 0. 21x20x12 , 313 3 + 3340
nee
.
$ = -log -= 9.55 i n
1.663
&
313 3
7 .1 0 RAT E OF ONE-DI M ENSI ONAL CONSOL I DATI ON TH EORY OF
TERZ AG H I
One dimensional consolidation theor y as proposed by Terzaghi i s generally applicable i n all case s
that arise in practice wher e
1. Secondar y compressio n i s not ver y significant,
2. Th e clay stratum is drained o n one or both the surfaces ,
3. Th e clay stratum is deepl y buried, and
4. Th e clay stratu m is thi n compared wit h the size of the loaded areas .
The following assumptions are made i n the development of the theory:
1. Th e voids of the soi l ar e completel y fille d wit h water,
2. Bot h water and soli d constituent s are incompressible ,
3. Darcy' s la w is strictly valid,
4. Th e coefficient of permeability i s a constant,
5. Th e time lag of consolidation i s due entirely t o the low permeability o f the soil , and
6. Th e clay i s laterally confined.
234 Chapt er 7
Differential Equatio n fo r One- Dimensiona l Flo w
Consider a stratu m o f soi l infinit e i n exten t i n th e horizonta l directio n (Fig . 7.14) bu t o f suc h
thickness // , tha t th e pressure s create d b y th e weigh t o f th e soi l itsel f ma y b e neglecte d i n
comparison t o the applied pressure .
Assume that drainage takes place onl y at the top and further assume that the stratum has been
subjected t o a uniform pressure o f p
Q
fo r suc h a long time tha t it i s completely consolidate d unde r
that pressur e an d tha t ther e i s a hydrauli c equilibriu m prevailing , i.e. , th e wate r leve l i n th e
piezometric tub e a t an y sectio n XY i n th e cla y stratu m stand s a t th e leve l o f th e wate r tabl e
(piezometer tub e in Fig. 7.14) .
Let a n incremen t of pressur e A/ ? b e applied . Th e tota l pressur e t o whic h th e stratu m i s
subjected i s
Pl
=p
Q
+ Ap (7.27 )
Immediately afte r the increment of load i s applied the water in the pore spac e throughout the
entire height , H , wil l carr y th e additiona l loa d an d ther e wil l b e se t u p a n exces s hydrostati c
pressure u
i
throughout the pore water equal t o Ap as indicated i n Fig. 7.14 .
After a n elapsed time t = t
v
some o f the pore wate r wil l have escaped at the top surface and as a
consequence, the excess hydrostati c pressure will have been decreased an d a part of the load transferred to
the soil structure. The distribution of the pressure between the soil and the pore water, p and u respectively
at any time t, may be represented by the curve as shown in the figure. It is evident that
Pi=p + u (7.28 )
at any elapsed time t and at any depth z, and u is equal to zero at the top. The por e pressure u , at any
depth, i s therefor e a function o f z an d / and ma y b e writte n as
u =f( z, t) (7.29)
Piezometers
Impermeable
(a) (b )
Figure 7 .1 4 On e-dim en s ion a l con s ol idat io n
Com pres s ibilit y an d Con s olidat ion 23 5
Consider a n element o f volume of the stratum a t a depth z, and thickness dz (Fig. 7.14) . Let
the bottom and top surfaces o f this element have unit area.
The consolidation phenomeno n i s essentially a problem o f non-steady flo w o f water through a
porous mass. The difference between th e quantity of water that enters the lower surface at level X'Y'
and the quantity of water which escapes the upper surface at level XY in time element dt must equal the
volume change of the material whic h has taken place in this element of time. The quantity of water is
dependent on the hydraulic gradient which is proportional t o the slope of the curve t .
The hydrauli c gradients a t level s XY an d X'Y' of the element ar e
, 1 d du 1 du ^ 1 d
2
u ,
iss
^-*
u+
**
=
T
w
Tz
+
T
w
^
dz (7
-
30)
If k is the hydraulic conductivity the outflow fro m th e element a t level XY i n time dt i s
k du
dq
l
=ikdt = ——dt(7.31 )
' W **
The inflow a t level X'Y' is
k du d
2
u
dq
2
= ikdt = ~^dt+ -^dzdt(7.32 )
The differenc e in flow i s therefor e
k
dq = dq^ -dq
2
= -— -r-^dzd t (7.33 )
• w
From th e consolidatio n tes t performe d i n th e laboratory , i t i s possibl e t o obtai n th e
relationship between th e void ratios corresponding t o various pressures t o which a soil is subjected.
This relationshi p i s expresse d i n th e for m o f a pressure-voi d rati o curv e whic h give s th e
relationship a s expressed i n Eq. (7.12)
de = a
v
dp (7.34 )
The chang e i n volum e Ad v o f th e elemen t give n i n Fig . 7.14 may b e writte n a s pe r
Eq. (7.7) .
de
Mv = Mz = - -dz (7.35 )
i + e
Substituting for de, we have
(7.36)
Here dp is the change in effective pressure at depth z during the time element dt . The increas e
in effective pressure dp i s equal t o the decrease in the pore pressure, du.
du
Therefore, d p = -du = -^~dt (7.37 )
at
236 Chapt e r 7
a
v
du du
Hence, Mv = —— dtdz = -tn
v
—dtdz (7.38 )
l + eat
v
dt
v
'
Since the soi l i s completely saturated , the volume change AJ v of the element o f thicknes s
dz i n time dt i s equa l t o th e chang e i n volume of wate r d q i n th e same elemen t i n time dt .
Therefore, d q = Mv (7.39 )
or
or
k d
2
u
-, - , di L
r
w
di
2
k( \ + e}d
2
u
Y
w
a
v
dz
2
a du
-it -i -if
u — , ^ az at
l + e dt
dt
v
dz
2
k
y a y m
' W V ' W V
is define d a s th e coefficient o f consolidation.
(7.40)
(7.41)
Eq. (7.40) is the differential equatio n for one-dimensional flow . Th e differential equation for
three-dimensional flo w ma y be developed i n the same way . The equatio n ma y b e written as
du l + ed
2
u d
2
u d
2
u
(7
'
42)
where k
x
, k
y
an d k
z
ar e th e coefficient s of permeabilit y (hydrauli c conductivity) i n th e coordinat e
directions o f jc , y an d z respectively .
As consolidatio n proceeds , th e value s o f k , e an d a
v
al l decreas e wit h tim e bu t th e rati o
expressed b y Eq . (7.41) ma y remai n approximatel y constant.
M athematical Solutio n fo r the One- Dimensiona l Consolidatio n Equatio n
To solve th e consolidation Eq . (7.40) i t is necessary t o set up the prope r boundar y conditions . Fo r
this purpose , conside r a layer o f soi l havin g a total thicknes s 2 H an d having drainage facilitie s a t
both th e to p an d bottom face s a s shown in Fig. 7.15. Under thi s condition n o flo w wil l take plac e
across th e cente r lin e a t dept h H. Th e cente r lin e ca n therefor e b e considere d a s a n imperviou s
barrier. Th e boundar y conditions for solving Eq. (7.40) ma y be writte n as
1 . u = 0 when z = 0
2. u = 0 when z = 2H
3. u = <\ p fo r al l depth s at time t = 0
On th e basi s o f th e abov e conditions , th e solutio n o f th e differentia l Eq . (7.40 ) ca n b e
accomplished b y means o f Fourier Series .
The solutio n i s
mz _
m
2
r
u
= - sin — e
ml
(7.43 )
m H
1)*c
v
t .
where m - -,/ = — — = a non-dimensional time factor .
2 H
2
Eq. (7.43 ) ca n b e expressed i n a general for m a s
Com pres s ibilit y an d Con s olidat io n 237
H
H
Clay
'• San d y..'•:.': I
P\
Figure 7 .1 5 B oun dar y con dit ion s
~Kp
=f
~H'
T (7
'
44)
Equation (7.44 ) ca n b e solve d b y assumin g T constan t fo r variou s value s o f z/H. Curve s
corresponding t o different value s of the time factor T may b e obtained a s given i n Fig. 7.16. It i s of
interest to determine how far the consolidation process under the increment of load Ap has progressed
at a time t corresponding t o the time factor T at a given dept h z. The term £/ , is used t o express thi s
relationship. It is defined as the ratio of the amount of consolidation whic h has already taken place t o
the total amount which is to take place under the load increment .
The curves in Fig. 7.16 shows the distribution of the pressure Ap between solid and liquid phases
at various depths. At a particular depth, say z/H = 0.5, the stress in the soil skeleton is represented by AC
and the stress in water by CB. AB represents the original excess hydrostatic pressure u
i
= Ap. The degree
of consolidation U
z
percent a t this particular dept h is then
AC Ap- « u
u % = ioox—- = -
t
-— = 100 i -—
z
A B A p A p
(7.45)
r
1.0
A/7- U
0.5
z/H 1. 0
r = o
1.5
2.0
T= o o
Figure 7.1 6 Con s olidat io n o f cla y laye r as a fun ct ion T
238
Chapt er 7
Following a similar reasoning, the average degre e o f consolidation U% fo r the entir e layer at
a time factor Tis equal t o the ratio of the shaded portion (Fig. 7.16) of the diagram t o the entire area
which i s equal t o 2H A/? .
Therefore
2H
U% =
u
_. . x l O O
2H
or £/ %= 2H —— udz
2H ^p
o
Hence, Eq. (7.46 ) afte r integration reduces t o
(7.46)
£/ %=100 1 - — -£
-m
2
T
(7.47)
It can be seen fro m Eq . (7.47) tha t the degree o f consolidation i s a function o f the time factor T
only which is a dimensionless ratio. The relationship between Tan d U% may therefore be established
once an d fo r al l b y solvin g Eq. (7.47 ) fo r variou s value s o f T . V alues thu s obtaine d ar e give n i n
Table 7. 3 and als o plotte d o n a semilog plo t a s shown in Fig. 7.17.
For value s o f U% betwee n 0 an d 60%, th e curv e i n Fig . 7.1 7 ca n b e represente d almos t
exactly b y the equation
T =
4 10 0
which i s the equation of a parabola. Substitutin g for T , Eq. (7.48 ) ma y b e writte n a s
U%
oo
(7.48)
(7.49)
u%
u
20
40
60
80
100
O.C
— • —„
--\
\,
k
\
\
^^
)03 0.0 1 0.0 3 0. 1 0. 3 1. 0 3. 0 1 0
Time factor T(log scale )
Figure 7 .1 7 U vers us T
Com pres s ibilit y an d Con s olidat io n 239
Table 7. 3 Relat ion s hi p bet wee n U an d T
u%
0
10
15
20
25
30
35
T
0
0.008
0.018
0.031
0.049
0.071
0.096
U%
40
45
50
55
60
65
70
T
0.126
0.159
0.197
0.238
0.287
0.342
0.405
U%
75
80
85
90
95
100
T
0.477
0.565
0.684
0.848
1.127
oo
In Eq . (7.49) , th e value s o f c
v
and H ar e constants. On e ca n determin e th e time require d t o
attain a given degre e of consolidation b y using this equation . I t shoul d b e note d tha t H represent s
half th e thicknes s o f th e cla y stratu m whe n th e laye r i s draine d o n bot h sides , an d i t i s th e ful l
thickness whe n drained o n one sid e only .
TAB L E 7. 4 Relat io n bet wee n U %an d T (Specia l Cas es)
Permeable Permeable
U%
Impermeable
Case 1
Impermeable
Case 2
T im e Fact ors , T
Con s olidat ion pres s ur e
in creas e wit h depth
Con s olidat ion pres s ur e
decreas es wit h dept h
00
10
20
30
40
50
60
70
80
90
95
100
0
0.047
0.100
0.158
0.221
0.294
0.383
0.500
0.665
0.94
1
o o
0
0.003
0.009
0.024
0.048
0.092
0.160
0.271
0.44
0.72
0.8
o o
240 Chapt e r 7
For values of U% greater than 60%, th e curve in Fig. 7.1 7 may be represented b y the equation
T= 1.78 1 - 0.93 3 log (10 0 - U%) (7.50 )
Effect o f B oundar y Condition s o n Consolidatio n
A layer of clay whic h permits drainage through both surfaces is called an open layer. Th e thickness
of suc h a laye r i s alway s represented b y th e symbo l 2H, i n contrast t o th e symbo l H use d fo r th e
thickness of half-close d layer s which can discharg e thei r excess wate r onl y throug h one surface .
The relationship expressed betwee n rand (/give n in Table 7. 3 applies to the fol l owing cases:
1. Wher e th e cla y stratu m i s draine d o n bot h side s an d th e initia l consolidatio n pressur e
distribution i s uni for m o r linearl y increasing or decreasing wit h depth.
2. Wher e th e cla y stratu m i s draine d on on e sid e bu t th e consolidatio n pressure i s uniform
with depth .
Separate relationship s between T and U ar e require d fo r hal f closed layer s wit h thickness H
where the consolidation pressure s increas e o r decrease wit h depth. Suc h cases ar e exceptional an d
as suc h not deal t wi t h i n detai l here. However , the relations between U% an d 7 " for thes e two cases
are given i n Table 7.4 .
7 .1 1 DETERM I NATI O N OF TH E COEFFI CI ENT OF CONSOL I DATI ON
The coefficien t o f consolidatio n c ca n b e evaluate d b y mean s o f laborator y test s b y fittin g th e
experimental curv e wi t h th e theoretical.
There ar e two laborator y methods t hat ar e in common us e for the determination of c
v
. They
are
1. Casagrand e Logarith m o f Time Fittin g Method.
2. Taylo r Squar e Roo t o f Time Fittin g Method.
L ogarithm o f Time Fittin g M etho d
This metho d wa s proposed b y Casagrande an d Fadum (1940) .
Figure 7.1 8 i s a plo t showin g th e relationshi p betwee n compressio n dia l readin g an d th e
logarithm of time of a consolidation test. The theoretical consolidation curve using the log scale for
the tim e facto r i s als o shown . Ther e i s a similarit y o f shap e betwee n th e tw o curves . O n th e
laboratory curve , th e intersectio n forme d b y th e fina l straigh t lin e produce d backwar d an d th e
tangent t o the curve at the point of inflection is accepted a s the 10 0 per cent primary consolidation
point an d th e dia l readin g i s designate d a s /?
100
. Th e time-compressio n relationshi p i n th e earl y
stages i s also paraboli c j ust as the theoretical curve. The dial reading a t zero primar y consolidatio n
R
Q
can be obtained b y selecting any two points on the parabolic portion of the curve where time s are
in th e rati o o f 1 : 4. The differenc e i n dial reading s betwee n thes e tw o point s i s the n equa l t o th e
difference betwee n th e fi rst point and the dial reading corresponding t o zero primar y consolidation.
For example , tw o point s A an d B whos e time s 1 0 and 2. 5 minute s respectively, ar e marke d o n th e
curve. Let z
{
b e the ordinat e difference between the two points. A point C is marked verticall y over
B such tha t BC = z
r
The n th e point C corresponds t o zero primary consolidation. The procedure i s
repeated wit h several points . An average horizonta l l ine i s drawn through these point s t o represen t
the theoretical zer o percen t consolidation line.
The interva l betwee n 0 an d 100 %consolidatio n i s divide d int o equa l interval s o f percen t
consolidation. Sinc e i t ha s bee n foun d tha t th e laborator y an d th e theoretica l curve s hav e bette r
Com pres s ibilit y an d Con s ol idat io n 241
Asymptote
R
f
\-
f i/
4
t
}
lo g (time )
(a) Experimental curve (b ) Theoretical curve
Figure 7 .1 8 Lo g of t im e fit t in g m et ho d
correspondence a t th e centra l portion , th e valu e of c
y
i s compute d b y takin g th e time t an d tim e
factor T at 50 percent consolidation . The equation to be used is
T
-~
H
lr (7.51 )
T - 1
5Q
c t
v 5 0
or
l
dr
where H
dr
= drainage path
From Table 7.3 , we have at U = 50%, T= 0.197. From th e initial height //. of specimen an d
compression dia l reading a t 50%consolidation, H
dr
fo r doubl e drainage i s
H: ~
(7.52)
where hH= Compression o f sampl e u p to 50%consolidation.
Now the equation for c ma y be written as
c = 0.19 7
H
(7.53)
Square Roo t o f Time Fittin g M etho d
This method wa s devised by Taylor (1948). In this method, the dial readings are plotted agains t the
square root of time a s given in Fig. 7.19(a) . The theoretica l curve U versus ^J T i s also plotte d and
shown i n Fig. 7.19(b) . On the theoretical curve a straight line exists up to 60 percent consolidatio n
while at 90 percent consolidation the abscissa of the curve is 1.15 times the abscissa o f the straight
line produced .
The fitting method consists of first drawing the straight line which best fits the early portion of the
laboratory curve. Next a straight line is drawn which at all points has abscissa 1.1 5 times as great as those
of th e firs t line . The intersectio n of thi s line and the laborator y curve is taken as the 90 percent ( R
QQ
)
consolidation point. Its value may be read and is designated as t
gQ
.
242 Chapt er 7
(a) Experimental curve (b ) Theoretical curve
Figure 7 .1 9 Squar e roo t o f t im e fit t in g m et ho d
Usually th e straight line through the early portion o f the laborator y curv e intersect s the zer o
time lin e a t a poin t ( R
o
) differin g somewha t fro m th e initia l point (/?
f
.). Thi s intersectio n poin t i s
called th e corrected zero point. I f one-nint h of th e vertica l distanc e betwee n th e correcte d zer o
point and the 90 per cent point is set off below the 90 percent point , the point obtained i s called the
"100 percent primary compression point" ( R
loo
). Th e compression betwee n zer o an d 10 0 per cent
point i s called "primary compression".
At the point of 90 percent consolidation, the value of T = 0.848. The equatio n of c
v
may now
be writte n as
H
2
c =0.848- ^
(7.54)
'90
where H, - drainag e pat h (average)
7 .1 2 RAT E OF SETTL EM ENT DU E TO CONSOL I DATI ON
It ha s bee n explaine d tha t th e ultimat e settlement S
t
o f a cla y laye r due t o consolidatio n ma y b e
computed b y usin g eithe r Eq . (7.10 ) o r Eq . (7.13) . I f S i s th e settlemen t a t an y tim e t afte r th e
imposition o f loa d o n th e cla y layer , th e degre e o f consolidatio n o f th e laye r i n tim e t ma y b e
expressed a s
U% = — x 100 percent
Since U is a funct ion o f the time factor T, we ma y writ e
(7. 55)
= —x l OO
O
(7.56)
The rat e o f settlemen t curv e o f a structur e buil t o n a cla y laye r ma y b e obtaine d b y th e
following procedure :
Com pres s ibil it y an d Con s olidat io n 243
Time t
Figure 7.2 0 T im e-s et t lem en t curv e
1. Fro m consolidation tes t data , comput e m
v
and c
v
.
2. Comput e th e total settlement S
t
that the clay stratum would experience wit h the increment
of loa d Ap.
3. Fro m the theoretical curve giving the relation between U and T, find T for different degree s
of consolidation , sa y 5, 10 , 20, 30 percent etc .
TH
2
,
4. Comput e from equatio n t = —— th e values of t for different value s of T. It may be noted
C
v
here that for drainage on both side s H
dr
i s equal t o half the thicknes s of the clay layer .
5. No w a curve ca n b e plotte d givin g the relatio n betwee n t and U% o r t and S a s shown i n
Fig. 7.20.
7.13 TWO - AND TH REE-DIMENSIONAL CONSOLIDATION
PROBLEMS
When th e thicknes s o f a cla y stratu m i s grea t compare d wit h th e widt h o f th e loade d area , th e
consolidation of the stratu m is three-dimensional . I n a three-dimensional proces s of consolidatio n
the flow occur s either i n radial planes or else the water particles travel along flow lines which do not
lie in planes. The problem of this type is complicated thoug h a general theor y of three-dimensiona l
consolidation exists (Biot, et al., 1941). A simple example of three-dimensional consolidation i s the
consolidation o f a stratu m o f sof t cla y o r sil t b y providin g san d drain s an d surcharg e fo r
accelerating consolidation .
The mos t important example of two dimensional consolidation i n engineering practic e i s the
consolidation of the case of a hydraulic fill dam. In two-dimensional flow, th e excess water drains
out of the clay in parallel planes . Gilboy (1934) has analyzed the two dimensional consolidation of
a hydrauli c fill dam.
Example 7. 8
A 2. 5 cm thic k sampl e o f clay was take n from th e fiel d for predicting th e time o f settlement fo r a
proposed building which exerts a uniform pressur e of 100 kN/m
2
over the clay stratum. The sampl e
was loaded t o 100 kN/m
2
and proper drainage was allowed from to p and bottom. It was seen tha t 50
percent o f the total settlemen t occurre d i n 3 minutes. Find th e time required for 5 0 percent o f the
244 Chapt e r 7
total settlemen t of the buil ding, if it is to be constructed on a 6 m thick layer of clay which extends
from th e groun d surfac e an d i s underlai n by sand .
Solution
Tfor 50 %consolidation = 0.197.
The la b sampl e i s drained on bot h sides . Th e coefficient o f consolidation c i s found from
TH
2
( 2 5)
2
1
c = — = 0.197 x —— x - = 10.25 x 10~
2
cm
2
/ min.
t 4 3
The tim e t for 50 %consolidatio n i n the fiel d wil l be foun d a s follows.
0. 197x300x300x100 , „„_ ,
t =
:
= 12 0 days .
10. 25x60x24
Example 7 . 9
The voi d rati o o f a clay sampl e A decreased fro m 0.572 t o 0.505 under a change i n pressure fro m
122 to 18 0 kN/m
2
. The voi d ratio of another sampl e B decreased from 0.61 to 0.557 under the same
increment o f pressure . Th e thicknes s o f sampl e A wa s 1. 5 times tha t o f B. Nevertheles s th e tim e
taken fo r 50 %consolidatio n wa s 3 time s large r fo r sampl e B tha n fo r A . Wha t i s th e rati o o f
coefficient o f permeabilit y of sampl e A t o that of Bl
Solution
Let H
a
= thicknes s o f sampl e A , H
b
= thicknes s o f sampl e B, m
va
= coefficien t o f volum e
compressibility o f sampl e A , m
vb
= coefficien t o f volum e compressibilit y o f sampl e B , c
va
=
coefficient o f consolidatio n fo r sampl e A, c
vb
= coefficient of consolidatio n fo r sampl e B, A/?
a
=
increment o f loa d fo r sampl e A , A/?
fe
= incremen t o f loa d fo r sampl e B, k
a
= coefficien t o f
permeability fo r sampl e A, an d k
b
= coefficient of permeability o f sampl e B.
We may writ e the following relationship
Ae 1 A<? , l
m = -
a
- --—,m
h
= -
b
- --—
va
\ + eA w
vb
l + e,Ap ,
a * a b rb
where e i s the void ratio of sample A at the commencement o f the test and Ae
a
is the change i n void
ratio. Similarl y e
b
and ke
b
apply t o sample B.
-, an d T =^K T , = ^fy-
wherein T
a
, t
a
, T
b
and t
b
correspond t o samples A and B respectively. We may writ e
c T H
2
t,

=
"F^p * «
= c
va
m
v
a
r
w
>
k
b =
c
vb
m
vb
y
w
vb b b a
k c m
™ r . a v a v a
Therefore,
k c m
b vb vb
Given e
a
= 0.572, an d e
b
= 0.61
A<? = 0.572-0.50 5 = 0.067, Ae . =0.610-0.55 7 = 0.053
a ' o
Compr essi bi l i t y an d Cons ol i dat i o n 24 5
A D = Ap , = 180-12 2 = 58kN/m
2
, H=l.5H,
But t
b
= 3t
a
Q.Q67 1 + 0.61 a
We have,
m
= 0.05 3 *1 + 0.572
= L29
vb
k
Therefore, T
2
-= 6.75 x 1.29 = 8.7
K
b
The rati o i s 8. 7 : 1.
Example 7 .1 0
A strat a o f normall y consolidate d cla y o f thicknes s 1 0 f t i s draine d o n on e sid e only . I t ha s a
hydraulic conductivit y of h = 1.86 3 x 10~
8
in/sec an d a coefficien t o f volum e compressibilit y
ra
v
= 8.6 x 10"
4
in
2
/lb. Determine the ultimate value of the compression o f the stratum by assuming
a uniforml y distribute d load o f 5250 lb/ft
2
an d also determine the time required for 20 percent an d
80 percent consolidation .
Solution
Total compression ,
S = m //A/ ? = 8. 6 x 10~
4
x 10 x 12 x 5250 x — = 3.763 in .
< v f
14 4
For determining the relationshi p between U% an d T for 20%consolidation us e the equation
n U%
2
3.1 4 2 0
2
T
= ^m
orT
= ~
x
m =
a
°
314
For 80%consolidation us e the equation
T = 1.781 - 0.93 3 log (100 - £/%)
Therefore T= 1.78 1 - 0.93 3 Iog
10
(100 - 80 ) = 0.567.
The coefficien t o f consolidation is
k 1.863xlO~
8
, m 4 • ? /
c = = - 6 x 10~
4
in
2
/ sec
y
w
m
v
3. 61xl O~
2
x8.
The times required fo r 20%and 80%consolidation ar e
H
2
dr
T (10xl2)
2
x0.0314
f
2
o = —
££L
— = ~
A
= 8.72 days
c
v
6x l O"
4
x 60x 60x 2 4
H
2
dr
T (1 0 x! 2)
2
x 0.567
?
so
= =
A
=
157. 5 days
c
v
6x l O"
4
x 60x 60x 2 4
246 Chapt e r 7
Example 7 .1 1
The loadin g perio d fo r a ne w building extended fro m Ma y 199 5 t o May 1997 . I n Ma y 2000, th e
average measure d settlemen t was found t o be 11.43 cm. It is known that the ultimate settlement wil l
be about 35.5 6 cm. Estimat e the settlement in May 2005. Assume doubl e drainag e t o occur .
Solution
For the majority of practical cases in which loading i s applied ove r a period, acceptabl e accurac y is
obtained whe n calculating time-settlement relationships b y assuming the time datum to be midway
through th e loading or construction period .
S
t
= 11.43 cm whe n t = 4 years an d 5 = 35.56 cm.
The settlement i s required for t = 9years, that is, up to May2005. Assuming as a starting point
that at t = 9 years, the degree o f consolidation wil l be = 0.60. Unde r these conditions per Eq. (7.48) ,
U= 1.1 3 V l
If S
t
= settlement at time t,, S, = settlement at time t,
'
l
= — sinc e
H
2
dr
IT
°
r
~
c m
_
where ~~ i s a constant. Therefore ~ T ~ A o" ' 2
V
L
^ , 1 -*
H
dr h
17.5
Therefore a t t = 9years, U = 7777 = 0.48
35.56
Since th e valu e o f U i s les s tha n 0.6 0 th e assumptio n i s valid . Therefor e th e estimate d
settlement i s 17.1 5 cm. I n the event of the degree of consolidation exceeding 0.60 , equatio n (7.50 )
has t o be used t o obtai n the relationshi p between T and U.
Example 7 .1 2
An oedometer test is performed o n a 2 cm thick clay sample. After 5 minutes, 50%consolidation i s
reached. Afte r ho w lon g a time woul d th e sam e degre e o f consolidatio n b e achieve d i n th e fiel d
where th e clay layer is 3.70 m thick? Assume the sample an d the clay layer have the same drainag e
boundary condition s (doubl e drainage) .
Solution
c, . t
The time factor T is defined as T -
where H
dr
- hal f the thickness of the clay for double drainage .
Here, the time factor T and coefficient of consolidation ar e the same fo r bot h th e sampl e an d
the fiel d cla y layer . The parameter tha t changes i s the time / . Let t
l
an d t
2
be the times require d t o
reach 50 %consolidatio n bot h i n the oedomete r an d fiel d respectively. t
{
= 5 min
=
Therefore t/ 2 t/ 2
n
dr( \ )
n
dr( 2)
2 j
H,
n
, 3 7 0 1 1
Now
f
i = t,= x 5 x — x —d a ys ~ 119 days.
WOW 2
H
d
' 2 6 0 2 4
y
Com pres s ibilit y an d Con s olidat io n 24 7
Example 7 .1 3
A laborator y sampl e o f cla y 2 cm thic k too k 1 5 mi n t o attai n 6 0 percen t consolidatio n unde r a
double drainag e condition . What time wil l be require d t o attai n the same degre e o f consolidation
for a cla y laye r 3 m thic k unde r th e foundatio n o f a buildin g fo r a simila r loadin g an d drainag e
condition?
Solution
Use Eq. (7.50) for U > 60%fo r determining T
T= 1.781-0.933 l og(l 00- £/ %)
= 1.781-0.93 3 log (100-60) = 0.286.
From Eq . (7.51 ) th e coefficient o f consolidation, c
v
is
TH
2
0.28 6 x (I)
2
c = • = 1.91xlO~
2
cm
2
/min.
15
The valu e o f c
v
remains constan t fo r bot h th e laborator y an d fiel d conditions . As such , w e
may write ,
lab \ J field
where H
dr
- hal f th e thicknes s = 1 cm fo r th e la b sampl e an d 150c m fo r fiel d stratum , an d
15 min.
Therefore,
t
lab
= 15 min.
or t
f
= (150)
2
x 0.25 = 5625 hr or 234 days (approx) .
for th e fiel d stratu m t o attai n the same degre e of consolidation .
7 .1 4 PROB L EM S
7.1 A bed of sand 10 m thick is underlain by a compressible o f clay 3 m thick under which lies
sand. The water tabl e i s at a depth of 4 m below the ground surface. The total unit weights
of sand below and above the water table are 20.5 an d 17. 7 kN/m
3
respectively. The clay has
a natura l water conten t o f 42%, liqui d limi t 46%an d specifi c gravit y 2.76 . Assumin g th e
clay t o be normall y consolidated, estimat e th e probabl e fina l settlemen t unde r a n averag e
excess pressure o f 10 0 kN/m
2
.
7.2 Th e effectiv e overburde n pressur e a t th e middl e o f a saturate d cla y laye r 1 2 f t thic k i s
2100 lb/ft
2
an d i s drained o n both sides . Th e overburden pressure at the middl e of the clay
stratum i s expecte d t o b e increase d b y 315 0 lb/ft
2
du e t o th e loa d fro m a structur e a t the
ground surface . An undisturbed sample o f clay 2 0 mm thic k is tested i n a consolidometer .
The tota l chang e i n thicknes s o f th e specime n i s 0.80 m m whe n th e applie d pressur e i s
2100 lb/ft
2
. Th e fina l wate r content of the sampl e i s 24 percent and the specifi c gravit y of
the solids i s 2.72. Estimat e th e probable fina l settlemen t o f the proposed structure .
248 Chapt e r 7
7.3 Th e followin g observation s refe r t o a standar d laborator y consolidatio n tes t o n a n
undisturbed sampl e o f clay.
Pres s ure
k N/m
2
0
50
100
200
Fin al Dia l Gaug e
Readin g x 10~
2
m m
0
180
250
360
Pres s ure
k N/m
2
400
100
0
Fin al Dia l Gaug e
Readin g x 10~
2
m m
520
470
355
The sampl e wa s 7 5 mm i n diamete r an d ha d a n initia l thicknes s o f 1 8 mm. Th e moistur e
content a t the end of the tes t wa s 45.5%; th e specifi c gravit y of solid s wa s 2.53 .
Compute th e voi d rati o a t the end o f each loadin g increment an d als o determin e whethe r
the soi l wa s overconsolidate d o r not . I f i t wa s overconsolidated , wha t wa s th e
overconsolidation rati o i f th e effectiv e overburde n pressur e a t th e tim e o f samplin g wa s
60 kN/m
2
?
7.4 Th e following point s are coordinates on a pressure-void rati o curve for an undisturbed clay.
p 0. 5 1 2 4 8 1 6 k i p s / f t
2
e 1.20 2 1.1 6 1.0 6 0.9 4 0.7 8 0.5 8
Determine (i ) C
c
, and (ii ) the magnitude of compression i n a 10 ft thick layer of this clay for
a load incremen t of 4 kips/ft
2
. Assume e
Q
= 1.320, andp
0
=1. 5 kips/ft
2
7.5 Th e thicknes s of a compressible layer , prior t o placing of a fil l coverin g a large area , i s 30
ft. It s origina l voi d rati o wa s 1.0 . Sometim e afte r th e fil l wa s constructe d measurement s
indicated tha t the average voi d rati o was 0.8. Determin e th e compression o f the soi l layer .
7.6 Th e wate r conten t o f a sof t cla y i s 54.2 %an d th e liqui d limi t i s 57.3%. Estimat e th e
compression index , by equations (7.17) and (7.18) . Given e
Q
= 0.85
7.7 A laye r o f normall y consolidate d cla y i s 2 0 f t thic k an d lie s unde r a recentl y constructe d
building. The pressure o f sand overlying the clay layer is 6300 lb/ft
2
, an d the new construction
increases th e overburde n pressur e a t th e middl e o f th e cla y laye r b y 210 0 lb/ft
2
. I f th e
compression inde x is 0.5, comput e the final settlement assuming vv
n
= 45%, G
s
= 2.70, an d the
clay is submerged wit h the wate r table at the top of the cla y stratum.
7.8 A consolidatio n tes t wa s mad e o n a sampl e o f saturate d marin e clay . Th e diamete r an d
thickness o f the sampl e wer e 5. 5 cm and 3.75 c m respectively. The sampl e weighe d 65 0 g
at the star t of the tes t and 480 g in the dry stat e afte r th e test . The specifi c gravit y of solid s
was 2.72 . Th e dia l reading s correspondin g t o th e fina l equilibriu m condition unde r eac h
load ar e give n below.
Pres s ure, k N/m
2
0
6.7
11.3
26.6
53. 3
DR c m x 10~
4
0
175
275
540
965
Pres s ure, k N/m
2
106
213
426
852
£>/?cm x 10-
4
1880
3340
5000
6600
(a) Comput e the void ratios and plot the e-\ og p curve .
(b) Estimat e th e maximum preconsolidation pressur e b y the Casagrande method.
(c) Dra w th e fiel d curv e and determine the compression index .
Com pres s ibilit y an d Con s olidat ion 24 9
7.9 Th e result s o f a consolidatio n tes t o n a soi l sampl e fo r a loa d increase d fro m 20 0 t o
400 kN/m
2
ar e given below:
Ti me i n Mi n.
0
0.10
0.25
0.50
1.00
2.25
4.00
9.00
Di al r eadi n g di vi s i o n
1255
1337
1345
1355
1384
1423
1480
1557
Ti me i n Mi n.
16
25
36
49
64
81
100
121
Di a l r e adi n g di vi si on
1603
1632
1651
1661
1670
1677
1682
1687
The thickness of the sample corresponding to the dial reading 125 5 is 1.561 cm. Determine
the value of the coefficient o f consolidation using the squar e root of time fitting metho d in
cm
2
/min. One division of dial gauge corresponds t o 2. 5 x lO^
4
cm. The sampl e i s drained
on bot h faces.
7.10 A 2. 5 cm thic k sampl e wa s teste d i n a consolidomete r unde r saturate d condition s wit h
drainage o n bot h sides . 3 0 percen t consolidatio n wa s reache d unde r a loa d i n
15 minutes. For the same conditions of stress but with only one way drainage, estimat e th e
time i n days it would take for a 2 m thick layer of the same soi l to consolidate i n the field to
attain the same degree of consolidation.
7.11 Th e dial readings recorded durin g a consolidation test at a certain load increment are given
below.
Tim e
min
0
0.10
0.25
0.50
1.00
2.00
4.00
8.00
Dial Readin g
cm x 10~
4
240
318
340
360
385
415
464
530
Tim e
min
15
30
60
120
240
1200
-
Dial Readin g
cm x 10~
4
622
738
842
930
975
1070
-
Determine c
v
by both the square root of time and log of time fitting methods . The thickness
of the sampl e a t DR 240 = 2 cm and the sampl e i s drained bot h sides .
7.12 I n a laborator y consolidatio n test a sampl e of clay wit h a thickness o f 1 in. reache d 50 %
consolidation i n 8 minutes. The sampl e wa s drained to p an d bottom. The cla y layer fro m
which th e sampl e wa s take n i s 25 f t thick. It i s covere d b y a layer of sand throug h which
water can escape and is underlain by a practically impervious bed of intact shale. How long
will the clay layer require t o reach 50 per cent consolidation?
7.13 Th e followin g data wer e obtaine d fro m a consolidation tes t performed o n a n undisturbed
clay sampl e 3 cm in thickness:
(i) p
l
= 3.5 kips/ft
2
, e
{
= 0.895
(ii) p
2
= 6.5 kips/ft
2
, e
2
= 0.782
250 Chapt e r 7
By utilizin g the know n theoretica l relationshi p between percen t consolidatio n an d tim e
factor, comput e an d plo t the decreas e i n thicknes s wit h time fo r a 1 0 f t thick layer of thi s
clay, whic h i s draine d o n th e uppe r surfac e only . Give n : e
Q
= 0.9 2 /?
0
= 4. 5 kips/ft
2
,
Ap = 1. 5 kips/ft
2
, c , = 4. 2 x 10~
5
ft
2
/min.
7.14 A structure built on a layer of clay settle d 5 cm in 60 days afte r it was built. If this settlemen t
corresponds t o 2 0 percen t averag e consolidatio n o f th e cla y layer , plo t th e time settlemen t
curve of the structur e for a period o f 3 years from th e time it was built . Give n : Thickness of
clay laye r = 3 m an d draine d on one sid e
7.15 A 3 0 f t thic k cla y laye r wit h singl e drainag e settle s 3. 5 in . i n 3. 5 yr . Th e coefficien t
consolidation fo r thi s cla y wa s foun d t o b e 8.4 3 x 10"
4
in.
2
/sec. Comput e th e ultimat e
consolidation settlemen t an d determin e ho w lon g i t wil l tak e t o settl e t o 90 %o f thi s
amount.
7.16 Th e tim e facto r T fo r a cla y laye r undergoin g consolidatio n i s 0.2 . Wha t i s th e averag e
degree of consolidation (consolidatio n ratio ) for the layer ?
7.17 I f the final consolidatio n settlement for the clay layer in Prob. 7.16 is expected to be 1. 0 m,
how muc h settlemen t ha s occurred whe n the time factor i s (a) 0.2 and (b ) 0.7 ?
7.18 A certai n compressibl e laye r ha s a thicknes s o f 1 2 ft . Afte r 1 yr whe n th e cla y i s 50 %
consolidated, 3 in. of settlement has occurred. For similar clay and loading conditions, how
much settlemen t woul d occur a t the end o f 1 yr and 4 yr , if the thicknes s of thi s new laye r
were 2 0 ft ?
7.19 A laye r o f normall y consolidate d cla y 1 4 f t thic k ha s a n averag e voi d rati o o f 1.3 . It s
compression inde x is 0.6. Whe n th e induced vertica l pressure o n the clay laye r is doubled,
what chang e i n thicknes s o f th e cla y laye r wil l result ? Assume : p
Q
= 120 0 lb/ft
2
an d
A/? = 60 0 lb/ft
2
.
7.20 Settlemen t analysis for a proposed structur e indicates that 2.4 in. of settlement wi ll occur in
4 y r an d tha t th e ultimat e total settlemen t wil l b e 9. 8 in . Th e analysi s i s base d o n th e
assumption tha t th e compressibl e cla y laye r i s draine d o n bot h sides . However , i t i s
suspected tha t ther e ma y no t b e drainag e a t th e botto m surface . Fo r th e cas e o f singl e
drainage, estimat e th e time required for 2. 4 in. of settlement .
7.21 Th e tim e t o reac h 60 %consolidatio n i s 32. 5 se c fo r a sampl e 1.2 7 cm thic k teste d i n a
laboratory unde r conditions o f double drainage . Ho w lon g wil l the correspondin g laye r i n
nature require t o reach th e same degree of consolidation i f it is 4.57 m thick and drained on
one side only?
7.22 A certai n cla y laye r 3 0 f t thic k i s expecte d t o hav e a n ultimat e settlement o f 1 6 in. I f th e
settlement was 4 in. after fou r years , how much longer wil l i t take to obtai n a settlement of
6 in?
7.23 I f the coefficient of consolidation o f a 3 m thick layer of clay i s 0.0003 cm
2
/sec, what is the
average consolidatio n o f that layer of clay (a ) i n one yea r wit h two-way drainage , an d (b )
the same a s above fo r one-way drainage .
7.24 Th e averag e natura l moisture conten t of a deposit i s 40%; th e specifi c gravit y of the soli d
matter i s 2.8 , an d th e compressio n inde x C
c
i s 0.36 . I f th e cla y deposi t i s 6. 1 m thic k
drained o n both sides, calculat e the final consolidatio n settlemen t S
t
. Given: p
Q
= 60 kN/m
2
and A/ ? = 3 0 kN/m
2
7.25 A rigi d foundatio n block , circula r in plan an d 6 m i n diamete r rest s o n a be d o f compac t
sand 6 m deep. Belo w th e san d i s a 1. 6 m thick layer of clay overlyin g on imperviou s bed
rock. Groun d wate r leve l i s 1. 5 m belo w th e surfac e o f th e sand . The uni t weigh t o f san d
above wate r tabl e i s 19. 2 kN/m
3
, the saturated uni t weight of sand i s 20.80 kN/m
3
, and the
saturated uni t weigh t of th e clay i s 19.9 0 kN/m
3
.
Com pres s ibilit y an d Con s olidat io n 25 1
A laborator y consolidatio n tes t o n a n undisturbe d sampl e o f th e clay , 2 0 mm thic k an d
drained to p and bottom, gave the following results:
Pressure (kN/m
2
) 5 0 10 0 20 0 40 0 80 0
V oid rati o 0.7 3 0.6 8 0.62 5 0.5 4 0.4 1
If th e contac t pressur e a t th e bas e o f th e foundatio n i s 20 0 kN/m
2
, an d e
Q
= 0.80 ,
calculate th e fina l averag e settlemen t o f th e foundatio n assumin g 2: 1 metho d fo r th e
spread o f th e load .
7.26 A stratu m of cla y i s 2 m thick and ha s a n initial overburden pressur e o f 5 0 kN/m
2
a t the
middle of the clay layer. The cla y i s overconsolidated wit h a preconsolidation pressur e of
75 kN/m
2
. Th e value s o f th e coefficient s o f recompressio n an d compressio n indice s ar e
0.05 an d 0.2 5 respectively . Assum e th e initia l voi d rati o e
Q
= 1.40 . Determin e th e fina l
settlement du e t o a n increas e o f pressure o f 40 kN/m
2
a t the middl e o f the cla y layer .
7.27 A cla y stratu m 5 m thic k ha s th e initia l voi d ratio n o f 1.5 0 an d a n effectiv e overburde n
pressure o f 12 0 kN/m
2
. Whe n th e sampl e i s subjecte d t o a n increas e o f pressur e o f
120 kN/m
2
, th e voi d rati o reduce s t o 1.44 . Determin e th e coefficien t o f volum e
compressibility and the final settlemen t of the stratum.
7.28 A 3 m thic k cla y laye r beneat h a buildin g i s overlai n b y a permeabl e stratu m an d i s
underlain by an impervious rock. The coefficient o f consolidation o f the clay was found t o
be 0.025 cm
2
/min. The fina l expecte d settlemen t for the layer i s 8 cm. Determine (a ) how
much tim e wil l i t tak e fo r 8 0 percen t o f th e tota l settlement , (b ) th e require d tim e fo r a
settlement of 2. 5 cm t o occur, and (c) the settlement tha t would occur i n one year .
7.29 A n area i s underlai n by a stratum of clay layer 6 m thick. The laye r i s doubl y drained and
has a coefficien t o f consolidatio n o f 0. 3 m
2
/month. Determin e th e tim e require d fo r a
surcharge loa d t o cause a settlement o f 40 cm i f the same loa d caus e a final settlemen t of
60cm.
7.30 I n an oedometer test , a clay specime n initiall y 25 mm thick attain s 90%consolidation i n
10 minutes. I n th e field , th e cla y stratu m fro m whic h th e specime n wa s obtaine d ha s a
thickness o f 6 m and i s sandwiche d betwee n two san d layers . A structur e constructe d o n
this clay experienced an ultimate settlement of 200 mm. Estimate the settlement at the end
of 10 0 days after construction .
CHAPTER 8
SH EAR STRENGTH OF SOI L
8 .1 I NTRODU CTI O N
One o f th e mos t importan t and th e mos t controversia l engineering propertie s o f soi l i s it s shea r
strength o r abilit y t o resis t slidin g along internal surface s within a mass. The stabilit y of a cut , the
slope of an earth dam, the foundations of structures, the natural slopes of hillsides and other structures
built on soi l depend upo n the shearin g resistance offere d b y the soi l alon g th e probabl e surface s of
slippage. Ther e i s hardl y a proble m i n th e fiel d o f engineerin g whic h doe s no t involv e the shea r
properties o f the soil i n some manner or the other .
8.2 B ASI C CONCEPT OF SH EARI NG RESI STANCE AND
SH EARI NG STRENG TH
The basi c concept of shearing resistance and shearing strengt h can be made clear by studying first
the basi c principle s o f frictio n betwee n soli d bodies . Conside r a prismati c bloc k B restin g o n a
plane surfac e MN a s show n i n Fig . 8.1 . Bloc k B i s subjecte d t o th e force P
n
whic h act s a t right
angles to the surfac e MN , an d the force F
a
tha t acts tangentiall y to the plane. The norma l force P
n
remains constant whereas F
a
graduall y increases from zer o to a value which will produce sliding. If
the tangentia l force F
a
i s relativel y small, block B wil l remai n a t rest , an d th e applie d horizonta l
force wil l be balanced by an equal and opposite forc e F
r
on the plane of contact. This resisting force
is developed as a result of roughness characteristics of the bottom of block B and plane surface MN .
The angl e 8 formed by the resultant R of the two forces F
r
and P
n
with the normal to the plane MN
is known a s the angle o f obliquity.
If th e applie d horizonta l forc e F
a
i s graduall y increased, th e resisting forc e F
r
wil l likewise
increase, alway s being equal i n magnitude and opposit e i n direction t o the applie d force. Bloc k B
will star t sliding along the plane when the force F
a
reaches a value which will increase th e angl e of
obliquity t o a certain maximu m value 8 . If block B an d plane surfac e MN ar e made o f the same
253
254 Chapt er 8
M N
Figure 8 . 1 B as i c con cep t o f s hearin g res is t an c e an d s t ren g t h.
material, th e angl e 8
m
i s equa l t o ( ft whic h i s terme d th e angle o f friction, an d th e valu e ta n 0 i s
termed th e coefficient of friction. I f block B and plane surface MN ar e made o f dissimilar materials ,
the angl e 8 i s terme d th e angle o f wall friction. Th e applie d horizonta l forc e F
a
o n bloc k B i s a
shearing forc e an d th e develope d forc e i s friction o r shearing resistance. The maximu m shearing
resistance whic h the material s ar e capabl e o f developing i s called th e shearing strength.
If anothe r experimen t i s conducte d o n th e sam e bloc k wi t h a highe r norma l loa d P
n
th e
shearing force F
a
wil l correspondingly be greater. A series of such experiments would show that the
shearing forc e F
a
i s proportional t o the normal load P
n
, that is
F =P ta n (8.1)
If A i s th e overal l contac t are a o f bloc k B o n plan e surfac e M/V , th e relationshi p ma y b e
written a s
F P
shear strength, s = —- = —- t an,
A A
or s = a tan (8.2)
8.3 TH E COU L OM B EQU ATI O N
The basi c concep t o f frictio n a s explained i n Sect . 8. 2 applies t o soil s whic h are purel y granular in
character. Soil s whic h ar e no t purel y granula r exhibi t a n additiona l strengt h whic h i s du e t o th e
cohesion betwee n th e particles. It is, therefore, stil l customary to separate th e shearing strengt h s of
such soils int o two components, on e due to the cohesion betwee n the soil particles and the other due
to th e frictio n betwee n them . Th e fundamenta l shea r strengt h equatio n propose d b y th e Frenc h
engineer Coulom b (1776 ) is
s = c + ( J ta n (8. 3)
This equatio n expresse s th e assumptio n tha t th e cohesio n c i s independen t o f th e norma l
pressure c r acting o n the plan e o f failure. At zer o normal pressure , th e shea r strengt h o f the soi l i s
expressed a s
s = c (8.4)
Shear St ren g t h o f Soi l 255
c
1
Normal pressure, a
Figure 8. 2 Coul omb' s la w
According t o Eq . (8.4) , th e cohesio n o f a soi l i s define d a s th e shearin g strengt h a t zer o
normal pressur e o n the plan e of rupture.
In Coulomb' s equatio n c an d 0 ar e empirica l parameters , th e value s o f whic h fo r an y soi l
depend upo n severa l factors ; the mos t importan t of these ar e :
1. Th e pas t histor y o f the soil .
2. Th e initia l state o f the soil , i.e. , whethe r it is saturated or unsaturated.
3. Th e permeabilit y characteristic s o f the soil .
4. Th e conditions of drainage allowe d t o take place during the test .
Since c an d 0 i n Coulomb' s Eq . (8.3 ) depen d upo n man y factors , c i s terme d a s apparent
cohesion an d 0 th e angl e o f shearin g resistance . Fo r cohesionles s soi l c = 0, the n Coulomb' s
equation become s
s = a tan (8.5)
The relationshi p betwee n th e variou s parameter s o f Coulomb' s equatio n i s show n
diagrammatically in Fig. 8.2 .
8.4 M ETH OD S OF DETERM I NI NG SH EAR STRENG T H
PARAM ETERS
M ethods
The shea r strengt h parameter s c and 0 of soil s eithe r i n the undisturbed or remolded state s ma y be
determined b y any of th e following methods:
1. L aboratory methods
(a) Direc t o r box shea r tes t
(b) Triaxia l compressio n tes t
2. Field method: V an e shear tes t or by any other indirect methods
Shear Parameter s o f Soil s in-sit u
The laborator y or the fiel d metho d tha t has to be chosen i n a particular case depend s upo n the type
of soi l an d th e accurac y required . Whereve r th e strengt h characteristic s o f th e soi l in-sit u ar e
required, laborator y test s ma y b e use d provide d undisturbe d sample s ca n b e extracte d fro m th e
256 Chapt er 8
stratum. However , soil s ar e subjec t t o disturbanc e eithe r durin g samplin g o r extractio n fro m th e
sampl i ng tube s i n th e laborator y eve n thoug h soi l particle s posses s cohesion . I t i s practicall y
impossible t o obtai n undi st urbe d samples o f cohesionles s soil s an d highl y pre-consolidate d cla y
soils. Sof t sensitiv e clays are nearl y always remolded durin g sampling. Laboratory method s may ,
therefore, b e use d onl y i n suc h case s wher e fairl y goo d undisturbe d sample s ca n b e obtained .
Where i t is not possibl e to extract undisturbed samples fro m the natural soil stratum, any one of the
fol l owi ng method s ma y hav e t o be use d accordin g t o convenience an d j udgment :
1. Laborator y test s on remolded sample s which coul d a t best simulat e fiel d condition s o f th e
soil.
2. An y suitabl e fiel d test .
The presen t tren d i s t o rel y mor e o n fiel d test s a s thes e test s hav e bee n foun d t o b e mor e
reliable tha n even th e mor e sophisticate d laborator y methods .
Shear Strengt h Parameter s o f Compacte d Fill s
The strengt h characteristic s o f f i l l s whic h are t o b e constructed , suc h a s eart h embankments , ar e
generally foun d i n a laboratory . Remolde d sample s simulatin g th e propose d densit y an d wate r
content o f th e fil l material s ar e mad e i n th e laborator y an d tested . However , th e strengt h
characteristics o f existin g fi l l s ma y hav e t o b e determine d eithe r b y laborator y o r fiel d method s
keeping i n view th e l i mi t at i on s of eac h method .
8 .5 SH EA R TES T APPARATU S
Direct Shea r Test
The origina l for m o f apparatu s for th e direc t applicatio n o f shea r forc e i s the shea r box . Th e bo x
shear test , t houg h simpl e i n principle , ha s certai n shortcoming s whic h wi l l b e discusse d late r on .
The apparatu s consist s of a squar e bras s bo x spli t horizontall y at the leve l o f th e cente r o f the soi l
sample, whic h is held between meta l grilles and porous stone s a s shown in Fig. 8.3(a) . V ertical load
is applie d t o th e sampl e a s show n i n th e figur e an d i s hel d constan t durin g a test . A graduall y
increasing horizontal load i s applied to the lower par t of the box unt i l the sampl e fail s in shear. Th e
shear loa d at failure is divided by the cross-sectional are a of the sample t o give the ultimat e shearing
strength. The vertica l load divided by the are a o f the sampl e give s the applie d vertica l stres s <7 . Th e
test may be repeated wi t h a few more sample s having the same initia l conditions as the first sample.
Each sampl e i s tested wi t h a di fferent vertica l load.
— Norma l loa d
Porous ston e
Proving ring
^^^^^^^^
<x><xxx><xxxp>^
Shearing
force
Rollers
Figure 8 .3 (a) Con s t an t rat e o f s t rai n s hea r bo x
Shear St ren g t h o f Soi l 25 7
Figure 8.3 (b) St rai n con t rolle d direc t s hea r apparat u s (Court es y : Soilt es t )
The horizonta l loa d i s applie d a t a constan t rat e o f strain . Th e lowe r hal f o f th e bo x i s
mounted o n rollers and i s pushed forwar d at a uniform rat e by a motorized gearin g arrangement .
The uppe r half of the box bears against a steel proving ring, the deformation o f which is shown on
the dial gauge indicatin g the shearin g force. To measure the volume change during consolidation
and durin g the shearing proces s anothe r dial gaug e i s mounted to sho w th e vertica l movement of
the to p platen . Th e horizonta l displacement o f th e botto m o f th e bo x ma y als o b e measure d b y
another dia l gaug e whic h is no t show n i n th e figure . Figur e 8.3(b ) shows a photograph o f strai n
controlled direc t shea r test apparatus.
Procedure fo r Determinin g Shearin g Strengt h o f Soi l
In the direct shea r test , a sampl e o f soi l i s placed int o the shea r box. The siz e of the box normally
used for clays and sands i s 6 x 6 cm and the sample i s 2 cm thick. A large box o f size 30 x 3 0 cm
with sampl e thicknes s o f 1 5 cm i s sometimes use d for gravell y soils .
The soil s use d fo r th e tes t ar e eithe r undisturbe d samples o r remolded . I f undisturbed , the
specimen has to be carefully trimme d and fitted int o the box. If remolded samples are required, the soil
is placed into the box in layers at the required initial water content and tamped to the required dry density.
After th e specime n i s place d i n the box, an d al l the othe r necessar y adjustment s are made , a
known norma l loa d i s applied . The n a shearin g forc e i s applied . The norma l loa d i s hel d constant
258 Chapt e r 8
throughout th e tes t bu t th e shearin g forc e i s applie d a t a constan t rat e o f strai n (whic h wil l b e
explained late r on) . The shearin g displacement i s recorded b y a dial gauge .
Dividing the normal load and the maximum applied shearing force by the cross-sectional are a of
the specimen a t the shear plane gives respectively the unit normal pressure cran d the shearing strengt h
s at failure o f the sample. These results may be plotted o n a shearing diagra m wher e cri s the absciss a
and s th e ordinate . Th e resul t o f a singl e tes t establishe s on e poin t o n th e grap h representin g th e
Coulomb formul a for shearing strength. In order t o obtain sufficient point s to draw the Coulomb graph ,
additional test s mus t b e performe d o n othe r specimen s whic h ar e exac t duplicate s o f th e first . Th e
procedure i n thes e additiona l test s i s th e sam e a s i n th e first , excep t tha t a differen t norma l stres s i s
applied each time . Normally, the plotted points of normal and shearing stresse s a t failure of the various
specimens wil l approximate a straight line. But in the case of saturated, highly cohesive cla y soils in the
undrained test , the graph of the relationship between th e normal stres s and shearing strengt h is usually
a curved line, especially at low values of normal stress. However , i t is the usual practice t o draw the best
straight line through the test points to establish the Coulomb Law. The slope of the line gives the angl e
of shearing resistance an d the intercept on the ordinate gives the apparent cohesion (See . Fig . 8.2) .
Triaxial Compressio n Tes t
A diagrammatic layout of a triaxial test apparatus is shown in Fig. 8.4(a) . I n the triaxial compressio n
test, thre e o r mor e identica l sample s o f soi l ar e subjecte d t o uniforml y distribute d flui d pressur e
around th e cylindrica l surface. Th e sampl e i s seale d i n a watertigh t rubber membrane . The n axia l
load i s applied t o the soi l sampl e unti l it fails. Although only compressive loa d i s applied t o the soi l
sample, i t fail s b y shea r o n interna l faces. I t i s possibl e t o determin e th e shea r strengt h o f th e soi l
from th e applie d load s a t failure . Figur e 8.4(b ) gives a photograph o f a triaxial tes t apparatus .
Advantages and Disadvantage s o f Direc t an d Triaxial Shea r Test s
Direct shea r test s ar e generall y suitable for cohesionless soil s except fin e san d an d sil t whereas th e
triaxial test i s suitable for all types of soils and tests. Undrained and consolidated undraine d tests on
clay sample s ca n b e mad e wit h th e box-shea r apparatus . Th e advantage s o f th e triaxia l ove r th e
direct shea r tes t are :
1. Th e stres s distributio n across th e soi l sampl e i s more unifor m in a triaxial test as compare d
to a direct shea r test .
2. Th e measuremen t o f vol ume changes i s more accurat e i n the triaxia l test .
3. Th e complet e stat e o f stress i s known at all stages durin g the triaxial test , whereas onl y the
stresses at failur e ar e known in the direc t shea r test .
4. I n th e cas e o f triaxia l shear, th e sampl e fail s alon g a plan e o n whic h th e combinatio n o f
normal stres s an d th e shea r stres s give s th e maximu m angl e o f obl iquit y of th e resultan t
with th e normal , wherea s i n th e cas e o f direc t shear , th e sampl e i s sheare d onl y o n on e
plane whic h i s the horizonta l plane whic h need no t be th e plane o f actua l failure .
5. Por e wate r pressure s ca n b e measure d i n th e cas e o f triaxia l shea r test s wherea s i t i s no t
possible i n direct shea r tests .
6. Th e triaxia l machine i s mor e adaptable .
Advantages o f Direc t Shea r Test s
1. Th e direc t shea r machin e i s simpl e an d fas t t o operate .
2. A thinne r soi l sampl e i s used i n th e direc t shea r tes t thu s facilitating drainage o f th e por e
water quickl y from a saturated specimen .
3. Direc t shea r requiremen t i s much les s expensiv e a s compared t o triaxial equipment .
Shear St ren g t h o f Soi l 259
Proving ring
Ram
Cell
Rubber membran e
Sample
(a) Diag ram m at i c layou t
Inlet Outlet
(b) Multiple x 50- E load
fram e t riax ia l t es t
apparat us (Court es y :
Soilt es t USA)
Figure 8. 4 T riax ia l t es t
apparat us
260 Chapt er 8
Original sampl e Failure wi t h
uniform strains
(a) Direct shea r test
/— Dea d zon e
Actual failur e
condition
_ Stresse d
zone
Zone wit h
large strains
Dead zon e
(b) Triaxial shear tes t
Figure 8.5 Con dit io n o f s am pl e durin g s hearin g i n direct an d t riax ial s hea r t es t s
The stres s condition s across th e soil sampl e i n the direct shea r test are very complex becaus e
of th e chang e i n th e shea r are a wit h th e increas e i n shea r displacemen t a s th e tes t progresses ,
causing unequa l distributio n of shea r stresse s an d norma l stresse s ove r th e potentia l surfac e o f
sliding. Fig. 8.5(a ) shows the sampl e condition before and after shearin g i n a direct shea r box. The
final sheare d are a A,i s less tha n the original area A.
Fig. 8.5(b) shows the stressed condition in a triaxial specimen. Because of the end restraints, dead
zones (non-stresse d zones ) triangular in section ar e formed a t the ends wherea s th e stres s distributio n
across th e sample midway between the dead zones may be taken as approximately uniform.
8.6 STRES S CONDI TI ON A T A POI NT I N A SOI L M ASS
Through ever y poin t i n a stressed bod y ther e ar e three planes a t righ t angle s t o each othe r whic h are
unique as compared t o all the other planes passing through the point, because they are subjected only to
normal stresse s wit h n o accompanying shearing stresse s actin g on the planes. These thre e plane s ar e
called principal planes, and the normal stresses actin g on these planes are principal stresses. Ordinarily
the three principal stresses a t a point differ i n magnitude. They may be designated a s the major principal
stress <TJ , th e intermediat e principa l stress o~
2
, an d th e mino r principal stres s <J y Principa l stresse s a t a
point in a stressed bod y are important because, once they are evaluated, the stresses o n any other plane
through the point can be determined. Many problems in foundation engineering can be approximated by
considering onl y two-dimensional stress conditions . The influenc e o f the intermediat e principal stress
( J
2
o n failur e ma y b e considered as not ver y significant .
A Two- Dimensiona l Demonstratio n o f the Existenc e o f Principa l Plane s
Consider th e body (Fig . 8.6(a) ) i s subjected t o a system of forces suc h as F
r
F
2
F
3
an d F
4
whos e
magnitudes an d line s of action ar e known.
Shear St ren g t h o f Soi l 261
D
dx
(c)
Figure 8.6 St res s at a poin t i n a body i n two dim en s ion a l s pac e
Consider a smal l prismati c elemen t P. The stresse s actin g on thi s elemen t i n the direction s
parallel t o the arbitraril y chose n axe s x and y are shown in Fig. 8.6(b) .
Consider a plane AA through the element, making an angle a with the jc-axis. The equilibrium
condition o f th e elemen t ma y be analyze d b y considerin g th e stresse s actin g o n th e face s o f th e
triangle ECD (shaded) which is shown to an enlarged scal e in Fig. 8.6(c) . The normal and shearing
stresses o n the faces o f the triangl e ar e also shown.
The unit stress in compression and in shear on the face ED are designated as crand T respectively.
Expressions fo r c r an d T ma y b e obtaine d b y applyin g th e principle s o f static s fo r th e
equilibrium condition o f the body. The su m of al l the forces i n the jc-direction i s
<J
x
dx ta n a + T dx+ rdx se c a cos a - crdx se c a sin a = 0
The sum of al l the forces i n the y-direction is
cr dx +T
X
d x tan a - T dx sec a sin a - crdx se c a cos a = 0
Solving Eqs. (8.6 ) an d (8.7 ) for cran d T , we have
(8.6)
(8.7)
a
V
+G
X
a - G J
— H — cos2 a + T ™ sm2 a
o i •* ?
T = —|CT
V
- c r
r
) si n2a- i
r v
cos2 a
fj \ y • * / - v
(8.8)
(8.9)
By definition, a principal plane is one on which the shearing stress is equal t o zero. Therefore,
when i is made equa l t o zero i n Eq. (8.9), the orientation of the principal plane s i s defined by the
relationship
tan2a =
2i,
(8.10)
262 Chapt er 8
Equation (8.10) indicates that there ar e two principal planes throug h the point P in Fig. 8.6(a )
and tha t the y ar e a t right angle s t o eac h other . By differentiating Eq. (8.8 ) wit h respec t t o a , an d
equating t o zero, w e have
— = - a. . sin 2a + a
r
sin 2a + 2t _. cos 2a = 0
da
y y
or
tan 2a =
a -G
X
(8.11)
Equation (8.11 ) indicate s th e orientatio n o f th e plane s o n whic h th e norma l stresse s e r ar e
maximum an d minimum. This orientatio n coincides wit h Eq. (8.10). Therefore, i t follows tha t the
principal plane s ar e also plane s o n which the normal stresse s ar e maximum and minimum.
8.7 STRES S CONDI TI ONS I N SOI L DU RI NG TRI AXI A L
COM PRESSI ON TES T
In triaxia l compressio n tes t a cylindrica l specime n i s subjecte d t o a constan t all-roun d flui d
pressure whic h is the minor principal stres s O"
3
since the shear stres s on the surface i s zero. The two
ends ar e subjected t o axial stres s whic h is the major principal stres s o
r
The stress conditio n i n the
specimen goe s o n changin g wit h the increas e o f th e majo r principa l stres s cr
r
I t i s o f interes t t o
analyze the stat e o f stres s alon g inclined sections passin g through the sampl e a t any stres s leve l ( J
l
since failur e occur s alon g incline d surfaces .
Consider th e cylindrica l specime n o f soi l i n Fig . 8.7(a ) whic h i s subjecte d t o principa l
stresses <7
{
an d <7
3
(<7
2
= <T
3
).
Now CD, a horizontal plane, is called a principal plane since i t is normal t o the principal stres s
<TJ an d th e shea r stres s i s zer o o n thi s plane . EF i s the othe r principa l plan e o n whic h th e principal
stress <7
3
acts. AA i s the inclined section o n which the stat e of stress i s required t o be analyzed.
Consider a s befor e a smal l pris m o f soi l show n shade d i n Fig . 8.7(a ) an d th e sam e t o a n
enlarged scal e i n Fig. 8.7(b) . All the stresses actin g on the prism ar e shown. The equilibrium of the
prism require s
Horizontal force s = cr
3
sin a dl - a sin a dl + T cos adl = (8.12)
A/
- D
E
(a) (b )
Figure 8. 7 St res s con dit io n i n a t riax ial com pres s io n t es t s pecim e n
Shear St ren g t h o f Soi l 26 3
£V ertical forces = o
{
cos a dl - a cos adl - i sin adl - 0 (8.13 )
Solving Eqs. (8.12 ) and (8.13) we have
<7, + <7 , <7 , — ( 7-,
cr = — - +— -cos2 « (8.14 )
2 2
1
r = -(cr
1
-<J
3
)sin2« (8.15 )
Let the resultant of <rand Tmake an angle 8 with the normal to the inclined plane. One should
remember that when ens less than 90°, the shear stres s Ti s positive, an d the angle S is also positive.
Eqs. (8.14 ) an d (8.15 ) ma y b e obtaine d directl y fro m th e genera l Eqs . (8.8) and (8.9)
respectively by substituting the following:
cr = <7 . , < T = ( T , a n d T = 0
8.8 REL ATI ONSH I P B ETWEE N TH E PRI NCI PAL STRESSE S AN D
COH ESI ON c
If the shearing resistanc e s of a soil depends o n both friction an d cohesion, slidin g failure occurs i n
accordance wit h the Coulomb Eq . (8.3), that is, when
T = s = c + c r t a n0 (8.16 )
Substituting for the values of erand rfrom Eqs . (8.14) and (8.15) into Eqs. (8.16) and solving
for <7 j w e obtai n
c + <7
3
tan </>
= <r , + ~ 5 (8.17 )
j -"'v-'^'v-cos ^ tftan^
The plane with the least resistance t o shearing along i t will correspond t o the minimum value
of <7 j whic h can produc e failur e in accordance wit h Eq. (8.17) . o
l
wil l be a t a minimum when the
denominator i n the second membe r o f the equation i s at a maximum, that is, when
d
— —(sin a cos a- cos
z
a tan <z>) = 0
da
Differentiating, an d simplifying, w e obtain (writin g a - a
c
)
«, = 45° + 0/2 (8.18 )
Substituting for a i n Eq. (8.17) an d simplifying, we have
CTj = CT
3
tan
2
(45 ° + 0/2) + 2c ta n (45 ° + 0/2) (8.19 )
or ( T
l
=v
3
N 0 + 2cN (8.20 )
where A^ = tan
2
(45° + 0/2) i s called th e flow value.
If th e cohesion c = 0, we have
°i = °I
N
*(8.21 )
If 0 = 0, we have
<T = < T + 2c (8.22 )
264 Chapt e r 8
If the side s o f the cylindrical specimen ar e no t acted o n by the horizontal pressur e <7
3
, the loa d
required t o caus e failur e i s calle d th e unconfme d compressiv e strengt h q
u
. I t i s obviou s tha t a n
unconfmed compressio n tes t can be performed onl y o n a cohesive soil . According t o Eq. (8.20) , th e
unconfmed compressiv e strengt h q i s equal to
<T = a — 2r N f 8 71\
u
i y « -\ ] </> ( o.Zj)
If 0 = 0, then q
u
= 2c (8.24a )
or the shea r strength
s = c = —(8.24b )
Eq. (8.24b ) show s on e o f th e simples t way s o f determinin g the shea r strengt h o f cohesiv e
soils.
8.9 M OH R CI RCL E OF STRESS
Squaring Eqs . (8.8 ) an d (8.9) an d adding, we have
i2 / _ ^ x 2
+ ^ = I "
2
j + *ly (8.25 )
Now, Eq. (8.25 ) i s the equation of a circle whose cente r ha s coordinate s
and whos e radiu s is — i /(c7 - c r ) -
2 v v
y
'
The coordinates o f points on the circle represent the normal and shearing stresses on inclined
planes a t a given point . The circl e i s called th e Mohr circle o f stress, after Mohr ( 1 900), who firs t
recognized thi s usefu l relationship . Mohr's metho d provide s a convenien t graphica l metho d fo r
determining
I . Th e norma l and shearin g stress on any plane through a point i n a stressed body .
2. Th e orientation of the principal planes i f the normal and shear stresses on the surface of the
prismatic elemen t (Fig . 8.6 ) ar e known . Th e relationship s ar e vali d regardles s o f th e
mechanical propertie s o f th e material s sinc e onl y th e consideration s o f equilibriu m ar e
involved.
If th e surface s o f th e elemen t ar e themselve s principa l planes , th e equatio n fo r th e Moh r
circle o f stres s ma y be writte n as
T + o
y
- - -= - y -- (
8
.26)
The center of the circle has coordinates T - 0 , and o= ( a
{
+ (T
3
)/2, and its radius is ( <J
l
- (T
3
)/2.
Again from Mohr' s diagram, the normal and shearing stresses o n any plane passing through a point
in a stressed bod y (Fig. 8.7 ) may be determined i f the principal stresses cr
l
an d (J
3
are known. Since
<7j an d O"
3
are alway s known in a cylindrical compression test , Mohr's diagram i s a very useful too l
to analyze stresses o n failur e planes .
Shear St ren g t h o f Soi l 265
8.1 0 MOH R CI RCL E OF STRESS WH EN A PRI SM ATI C ELEMENT
I S SU B JECTED TO NORM AL AND SH EA R STRESSE S
Consider first th e case of a prismatic element subjected to normal and shear stresse s a s in Fig. 8.8(a).
Sign Conventio n
1. Compressiv e stresse s ar e positive and tensile stresse s ar e negative.
2. Shea r stresse s ar e considere d a s positiv e if the y give a clockwis e momen t abou t a point
above the stresse d plane as shown in Fig. 8.8(b) , otherwise negative.
The norma l stresse s ar e take n a s absciss a an d th e shea r stresse s a s ordinates . I t i s
assumed th e norma l stresse s c r , c r an d th e shea r stres s r ( T = T ) acting o n th e surfac e o f
x y xy xy yx
the elemen t ar e known . Tw o point s P
l
an d P
2
ma y no w b e plotte d i n Fig . 8.8(b) , whos e
coordinates ar e
If the points P
}
an d P
2
are joined, the line intersects the abscissa a t point C whose coordinates
are [(0, +op/2, 0].
Minor principal
>
a
i plane
(a) A prismatic element subjected t o normal and shear stresses
( a
x
+ a
y
)/2
+ ve
(b) Mohr circle of stres s
Figure 8.8 Moh r s t res s circl e fo r a general cas e
266 Chapt e r 8
Point O is the origi n of coordinates fo r the center of the Mohr circl e o f stress . With center C
a circl e ma y no w be constructed with radius
This circl e whic h passe s throug h point s P
l
an d P
2
i s calle d th e Mohr circle o f stress. The
Mohr circl e intersect s the absciss a a t two point s E and F . The major and minor principa l stresse s
are o
l
( = OF) an d cr
3
( = OE) respectively .
Determination o f Norma l and Shea r Stresse s o n Plane AA [Fig . 8 .8 (a) ]
Point P
{
o n the circl e of stress i n Fig. 8 . 8(b) represents th e stat e of stress on the vertica l plane of the
prismatic element ; similarl y poin t P
2
represent s th e stat e o f stres s o n th e horizonta l plan e o f th e
element. If from poin t P
{
a line is drawn parallel to the vertical plane , it intersects the circle at point P
Q
and i f from th e poin t P
2
o n th e circle , a lin e is drawn parallel t o th e horizonta l plane, thi s lin e als o
intersects th e circl e a t point P
Q
. The poin t P
Q
s o obtained i s called th e origin o f planes or the pole. If
from th e pole P
Q
a line is drawn parallel t o the plane AA i n Fig. 8.8(a ) to intersect the circle at point P
3
(Fig. 8.8(b)) then the coordinates o f the point give the normal stress cran d the shear stres s To n plan e
AA as expressed b y equations 8.8 and 8.9 respectively. This indicates that a line drawn from the pole P
Q
at an y angl e a t o th e cr-axi s intersect s the circl e a t coordinates tha t represent th e norma l an d shea r
stresses o n the plane inclined at the same angl e t o the abscissa .
M aj or an d Mino r Principa l Planes
The orientation s o f the principal planes may be obtained b y joining point P
Q
t o the point s E and F
in Fi g 8.8(b) . P
Q
F i s the direction o f the major principal plane on which the major principal stres s
dj acts ; similarl y P
Q
E i s th e directio n o f th e mino r principa l plane o n whic h th e mino r principa l
stress <7
3
acts. It is clear from the Mohr diagram that the two planes P
Q
E and P
Q
F intersect at a right
angle, i.e. , angl e EP
Q
F = 90° .
8 .1 1 M OH R CI RCL E OF STRESS FO R A CY L I NDRI CAL SPECI M EN
COM PRESSI ON TEST
Consider th e case of a cylindrical specimen o f soi l subjected to normal stresses <7 j and <J
3
whic h are
the major and minor principal stresses respectivel y (Fig . 8.9 )
From Eqs . (8.14 ) and (8.15) , we may writ e
2 2
Again Eq. (8.27 ) i s the equation of a circle whos e cente r ha s coordinate s
<7, + CT , (7 , — (J-.
<J = — - --and T =0 and whose radiu s is
/ O /-*^T \
(8.27)
2 2
A circl e wit h radius ( o
{
- cr
3
)/2 wit h its center C on the abscissa a t a distance o f ( a
l
+ cr
3
)/2
may b e constructe d a s shown i n Fig . 8.9 . Thi s i s the Moh r circl e o f stress . Th e majo r an d mino r
principal stresse s ar e shown i n th e figur e wherei n cr , = OF an d <7
3
= OE .
From Fig. 8.8 , w e ca n writ e equations for cf j an d <7
3
and T
max
a s follow s
±
Shear St ren g t h o f Soi l 267
.A
Figure 8. 9 Moh r s t res s circl e fo r a cylin drical s pecim e n
(8.29)
where T
max
is the maximum shear stres s equa l to the radius of the Mohr circle.
The origin of planes or the pole P
Q
(Fig. 8.9) may be obtained as before by drawing lines fro m
points E and F parallel t o planes on which the minor and major principa l stresses act . In this case ,
the pole P
O
lies on the abscissa and coincides wit h the point E.
The norma l stress < J and shear stress T on any arbitrary plane AA makin g an angle a wit h the
major principa l plane may be determined a s follows.
From th e pole P
0
draw a line P
Q
P
l
paralle l t o the plane AA (Fig . 8.9) . The coordinates o f the
point P
l
giv e the stresses c r and i. From the stress circl e we may write
= 2a
cr, + cr, cr, - cr .
- (8.30)
Normal stress a
0° 15 ° 30 ° 45 ° 60 ° 75 ° 90 °
Angle of inclination of plane, a ^
Figure 8.1 0 Variat io n o f cran d r wit h a
268 Chapt e r 8
( j, -cr ,
r =
3
sin2 # (8.31 )
Equations (8.30) an d (8.31) ar e the same a s Eqs. (8.14 ) an d (8.15) respectively .
It i s of interest t o study the variation of the magnitude s of normal and shea r stresse s wit h the
inclination o f the plane.
Eqs. (8.30 ) an d (8.31 ) ar e plotted wit h a a s the absciss a show n i n Fig. 8.10 . The followin g
facts ar e clear fro m thes e curves:
1. Th e greates t an d leas t principa l stresse s ar e respectivel y th e maximu m an d minimu m
normal stresse s on any plane throug h th e point i n question .
2. Th e maxi mu m shear stress occur s o n planes a t 45° t o the principal planes .
8 .1 2 M OH R- COU L OM B FAI L U RE TH EORY
V arious theories relatin g t o the stress condition i n engineering material s at the time of failure are
available i n th e engineerin g literature . Eac h o f thes e theorie s ma y explai n satisfactoril y th e
actions o f certai n kind s o f material s a t th e time the y fail , bu t n o one o f the m i s applicabl e t o al l
materials. Th e failur e o f a soi l mas s i s mor e nearl y i n accordanc e wit h th e tenet s o f th e Moh r
theory o f failur e tha n wit h thos e o f an y othe r theor y an d th e interpretatio n o f th e triaxia l
compression tes t depend s t o a large extent on this fact. The Mohr theor y i s based on the postulat e
that a material wil l fail whe n the shearing stres s o n the plane alon g whic h the failur e is presume d
to occur i s a unique funct i on o f the normal stres s acting on that plane. The material fail s along th e
plane onl y whe n the angl e betwee n th e resultant of the shearin g stres s an d the norma l stres s i s a
maximum, tha t is , wher e th e combinatio n o f norma l an d shearin g stresse s produce s th e
maxi mum obliquit y angle 8.
According t o Coulomb' s Law , the condition of failur e i s that the shea r stres s
T ^ c + at an^ (8.32 )
In Fi g 8. 1 l(b) M
Q
N an d M
Q
N
l
ar e the line s that satisfy Coulomb' s conditio n o f failure. If the
stress a t a given poin t wi t hi n a cylindrical specimen unde r triaxia l compression i s represented b y
Mohr circl e 1 , it ma y b e note d tha t ever y plan e throug h thi s point ha s a shearin g stres s whic h i s
smaller than the shearing strength.
For example, i f the plane AA i n Fig. 8. 1 l(a) is the assumed failur e plane, the normal and shea r
stresses on this plane at any intermediate stage of loading ar e represented b y point b on Mohr circl e
1 where the line P
Q
b is parallel t o the plane AA. The shearing stres s o n this plane i s ab which is less
than th e shearing strengt h ac at the same norma l stres s Oa . Under thi s stres s conditio n there i s no
possibility o f failure . O n th e othe r han d i t woul d no t b e possibl e t o appl y th e stres s conditio n
represented b y Mohr stres s circl e 2 to this sampl e becaus e i t is not possible fo r shearing stresse s t o
be greate r tha n th e shearin g strength . At th e norma l stres s Of, th e shearin g stres s o n plan e AA is
shown t o b e fh whic h i s greate r tha n the shea r strengt h o f th e material s fg whic h i s no t possible .
Mohr circl e 3 i n th e figur e i s tangen t to th e shea r strengt h lin e M
Q
N an d M
Q
N j a t point s e an d e
{
respectively. O n th e sam e plan e AA at normal stres s Od , th e shearin g stres s d e i s th e sam e a s th e
shearing strengt h de . Failur e i s therefor e imminen t o n plan e A A a t th e norma l stres s O d an d
shearing stres s de. The equation for the shearing stres s de i s
s = de - de'+ e' e = c + crtan 0 (8.33 )
where 0 is the slope o f the line M
Q
N whic h is the maximum angl e o f obliquity on the failure plane.
The valu e of the obliquit y angle can neve r exceed <5
m
= 0, the angl e of shearing resistance , withou t
the occurrence of failure. The shear strengt h line M
Q
N whic h is tangent to Mohr circl e 3 is called th e
Shear St ren g t h o f Soi l 269
' i /
Rupture
plane Moh r
envelope N
Mohr circle of
rupture
(b)
Figure 8 .1 1 Diag ra m pres en t in g M ohr' s t heor y o f rupt ur e
Mohr envelope o r line of rupture. The Mohr envelope may be assumed a s a straight line although it
is curved under certain conditions . The Mohr circle which is tangential to the shear strengt h line is
called th e Mohr circle o f rupture. Thu s th e Moh r envelop e constitute s a shea r diagra m an d i s a
graph o f th e Coulom b equatio n fo r shearin g stress . Thi s i s calle d th e Mohr-Coulomb Failure
Theory. Th e principal objective of a triaxial compression tes t is to establish th e Mohr envelope fo r
the soil being tested. The cohesion and the angle of shearing resistance can be determined from thi s
envelope. Whe n th e cohesio n o f th e soi l i s zero, tha t is , whe n the soi l i s cohesionless , th e Moh r
envelope passes throug h the origin.
8 .1 3 M OH R DI AG RA M FOR TRI AXI AL COM PRESSI ON TEST AT
FAI L U RE
Consider a cylindrica l specime n o f soi l possessin g bot h cohesio n an d frictio n i s subjecte d t o a
conventional triaxia l compressio n test . I n th e conventiona l tes t th e latera l pressur e cr
3
i s hel d
constant an d th e vertica l pressur e <T J i s increase d a t a constan t rat e o f stres s o r strai n unti l th e
sample fails . I f cr
l
i s th e pea k valu e o f th e vertica l pressur e a t whic h th e sampl e fails , th e tw o
principal stresse s tha t ar e t o b e use d fo r plottin g th e Moh r circl e o f ruptur e ar e cr
3
an d o
r
I n
Fig. 8.12 the value s of cr
{
an d <7
3
ar e plotted on the er-axi s and a circle i s drawn with (o ^ - cr
3
) as
diameter. The center of the circle lies at a distance of ( <j
{
+ cr
3
)/2 from th e origin. As per Eq. (8.18) ,
the soi l fail s alon g a plane which makes a n angl e a, = 45° + 0/ 2 with the major principal plane. In
Fig. 8.1 2 the two line s P
Q
P
l
an d P
Q
P
2
(wher e P
Q
i s the origi n of planes) ar e the conjugat e rupture
planes. The two line s M
Q
N an d M
Q
N ^ draw n tangential to the rupture circle a t points P
l
an d P
2
ar e
called Mohr envelopes. I f the Mohr envelope can be drawn by some other means, the orientation of
the failure plane s ma y be determined .
The result s o f analysi s o f triaxia l compressio n test s a s explaine d i n Sect . 8. 8 ar e no w
presented i n a graphical for m i n Fig. 8.12 . Th e various information that can be obtaine d fro m th e
figure include s
1. Th e angl e of shearing resistance 0 = the slope of the Mohr envelope .
270 Chapt er 8
A a
Mohr envelope
(a, - a
3
)/2
Figure 8 .1 2 Moh r diag ra m fo r t riax ia l t es t a t fail ur e fo r c- 0 s oi l
Rupture
plane
T
c
I
Rupture
plane
0 = 0
0
(a) c = 0
C
(b) 0 = 0
Figure 8.1 3 Moh r diag ra m fo r s oil s wit h c = 0 an d = 0
2. Th e apparen t cohesio n c = the intercept of the Mohr envelop e o n the T-axis .
3. Th e inclinatio n of the rupture plane = a .
4. Th e angl e betwee n th e conjugate planes = 2a.
If th e soi l i s cohesionles s wit h c = 0 the Moh r envelope s pas s throug h the origin , an d i f the
soil is purely cohesive wit h 0 = 0 the Mohr envelope is parallel t o the abscissa. The Mohr envelope s
for thes e two type s o f soil s ar e shown i n Fig. 8.13.
8 .1 4 M OH R DI AG RAM FOR A DI RECT SH EA R TES T A T FAI L U R E
In a direct shea r tes t th e sampl e i s sheared alon g a horizontal plane . Thi s indicate s tha t the failur e
plane is horizontal. The normal stress do n thi s plane is the external vertical load divided by the area
of th e sample . Th e shea r stres s a t failur e i s th e externa l latera l loa d divide d b y th e are a o f th e
sample.
Point Pj on the stress diagram in Fig. 8.14 represents th e stress condition on the failure plane .
The coordinate s o f the point are
normal stres s = <7 , shea r stres s i- s.
Shear St ren g t h o f Soi l 271
Minor
Plane o f rupture
0
t
Majo r principal
plane
Figure 8.1 4 Moh r diag ra m fo r a direct s hea r t es t a t failur e
If i t i s assume d tha t th e Moh r envelop e i s a straigh t lin e passin g t hroug h th e ori gi n
(for cohesionles s soi l o r normal l y consolidate d cl ays) , i t follow s tha t th e maxi mu m
obl i qui t y 8
m
occurs o n the fai l ur e plane an d 8
m
= 0. Therefore th e l i n e OP
{
mus t be t angen t
to th e Moh r circle , an d th e circl e ma y b e const ruct e d a s follows :
Draw Pj C norma l t o OP
r
Poin t C whic h i s th e intersectio n poin t o f th e norma l wit h th e
abscissa i s th e cente r o f th e circle . CP
{
i s th e radiu s of th e circle . Th e Moh r circl e ma y no w b e
constructed whic h gives th e majo r and mino r principal stresses cr
{
an d <7
3
respectively .
Since th e failur e i s o n th e horizonta l plane , th e origi n o f plane s P
Q
ma y b e obtaine d b y
drawing a horizontal lin e through P
{
givin g P
Q
. P
Q
F an d P
Q
E giv e th e direction s o f th e majo r and
minor principal planes respectively .
Example 8. 1
What is the shearing strengt h of soil along a horizontal plane at a depth of 4 m in a deposit o f sand
having the following properties :
Angle of internal friction, 0 = 35°
Dry uni t weight, y
d
- 1 7 kN/m
3
Specific gravity , G
s
= 2.7.
Assume the groun d water tabl e i s at a depth of 2. 5 m from th e ground surface. Also fin d th e
change i n shear strengt h whe n the water tabl e rises t o the ground surface .
Solution
The effectiv e vertica l stres s a t the plane of interest is
<r ' =2. 50x y
d
+ l . SOx y
b
Given y
d
= 17 kN/m
3
and G
s
= 2. 7
We haver, = 17- = — X9.81
9A9
or lie = 26.5 - 1 7 = 9.49 o r e = —— = 0.56
Therefore, Y b =
l + e 1 + 0.56
*9.81 = 10.7 kN/m
3
272 Chapt er 8
Hence c / = 2. 5 x 1 7 + 1. 5 x 10. 7 = 58.55 kN/m
2
Hence, the shearin g strength of th e sand i s
5 = (/ tan 0 = 58.55 x tan 35° = 41 kN/m
2
If th e wate r tabl e rise s t o th e groun d surfac e i.e. , b y a heigh t o f 2. 5 m , th e chang e i n th e
effective stres s wil l be ,
Ao" = y
d
x 2. 5 -Y
b
*2. 5 = 1 7 x 2. 5 - 10. 7 x 2. 5 = 15.7 5 kN/m
2
(negative)
Hence th e decrease i n shear strengt h wil l be,
= Ac/ ta n 35 ° = 15.7 5 x 0.7 0 = 1 1 kN/m
2
Example 8 . 2
Direct shea r test s wer e conducte d o n a dr y sand . The siz e o f th e sample s use d fo r th e test s wa s
2 in. x 2 in. x 0.75 in . The tes t result s obtained ar e given below:
T es t No . Norm a l l oa d Norm a l s t res s a Shea r f or c e Shea r s t res s
(Ib) (Ib/ft
2
) a t fail ur e (Ib ) (Ib/ft
2
)
1 1 5 54 0 1 2 43 2
2 2 0 72 0 1 8 64 8
3 3 0 108 0 2 3 82 8
4 6 0 216 0 4 7 169 2
5 12 0 432 0 9 3 334 8
Determine th e shear
4000-
3000-
c/f
C/3
£ 200 0 -
C/3
j3
1000-
strength param et ers can d 0.
^L
S^ A ^"7 8°
/
/
y
/
/
1000 2000 3000 4000
Normal stress , a Ib/ft
2
Figure Ex. 8. 2
5000
Shear St ren g t h o f Soi l
Solution
273
The failur e shea r stresse s r^ a s obtained fro m th e tests ar e plotted agains t the normal stresses a , i n
Figure Ex 8.2. The shea r parameter s fro m th e graph are: c= 0, 0 = 37.8° .
Example 8. 3
A direct shear test , when conducted on a remolded sampl e of sand, gave the following observations
at th e time o f failure : Norma l loa d = 288 N; shea r loa d = 17 3 N. The cros s sectiona l are a o f the
sample = 36 cm
2
.
Determine: (i ) the angl e o f internal friction, (ii ) the magnitude and direction o f the principal
stresses i n the zone o f failure .
Solution
Such problem s ca n b e solve d i n tw o ways , namel y graphicall y an d analytically . The analytica l
solution has been lef t a s an exercise fo r the students.
Graphical Solutio n
173
(i) Shea r stres s T = = 4. 8 N/ c m
2
= 48 k N/ m
2
36
288
Normal stres s a = — = 8.0 N / cm
2
= 80 k N / m
2
36
We know one point on the Mohr envelope. Plot point A (Fig. Ex. 8.3) wit h coordinates 1 -
48 kN/m
2
, and o= 80 kN/m
2
. Since cohesion c = 0 for sand, the Mohr envelope OM passes
through the origin . The slop e o f OM give s the angl e o f internal frictio n ( j) =31° .
(ii) I n Fig. Ex. 8.3, draw lin e AC norma l t o the envelop e O M cutting the absciss a a t point C .
With C as center, and AC as radius, draw Mohr circle C
l
which cuts the abscissa a t points B
and D, whic h gives
120
80
40
Mohr circle C\
Major principa l plane
C
2
40 F 8 0 C 12 0
a, kN/m
2
160 200
Figure Ex . 8. 3
274 Chapt e r 8
major principa l stress = OB = ( J
l
= 163. 5 kN/m
2
minor principal stress = OD = <J
3
= 53.5 kN/m
2
Now, ZACB = 2cc = twic e th e angl e betwee n th e failur e plan e an d th e majo r principal
plane. Measurement gives
2a= 121 ° or a- 60.5 °
Since i n a direc t shea r tes t th e failur e plane i s horizontal , th e angl e mad e b y th e majo r
principal plan e wi t h th e horizonta l wil l b e 60.5° . Th e mino r principa l plan e shoul d b e
drawn a t a right angl e to the major principal plane.
The direction s o f th e principa l planes ma y als o b e foun d b y locatin g th e pol e P
o
. P
o
i s
obtained b y drawing a horizontal line from poin t A which is parallel t o the failur e plane in
the direct shear test . Now PE an d P
(
D giv e the directions of the major and minor principal
planes respectively.
8 .1 5 EFFECTI V E STRESSES
So far , th e discussio n ha s bee n base d o n consideratio n o f tota l stresses . I t i s t o b e note d tha t th e
strength an d deformatio n characteristic s o f a soi l ca n b e understoo d bette r b y visualizin g it a s a
compressible skeleto n o f soli d particles enclosing voids. The void s ma y completel y b e fille d wit h
water o r partl y wit h wate r and air . Shea r stresse s ar e t o b e carrie d onl y b y th e skeleto n o f soli d
particles. However , th e tota l norma l stresse s o n an y plan e are , i n general , th e su m o f tw o
components.
Total norma l stres s = component of stres s carrie d b y soli d particle s
+ pressur e i n the fl ui d i n the voi d space .
This visualizatio n of th e distributio n o f stresse s betwee n soli d an d flui d ha s tw o important
consequences:
1. Whe n a specimen of soil is subjected to external pressure, the volume change of the specime n
is not due t o the total normal stress but due t o the difference between the total normal stres s
and the pressure of the fluid i n the void space. The pressure in the fluid is the pore pressure u.
The difference whi ch i s called the effective stres s d ma y now be expressed a s
tf = cr-u (8.34 )
2. Th e shea r strengt h o f soils , a s o f al l granula r materials , i s largel y determine d b y th e
frictional force s arisin g during slip at th e contact s betwee n th e soi l particles . Thes e ar e
clearly a functio n o f th e componen t o f norma l stres s carrie d b y th e soli d skeleto n rathe r
than o f th e tota l norma l stress . Fo r practica l purpose s th e shea r strengt h equatio n o f
Coulomb i s given by th e expression
s
=
c
' + ( o-
U
) tan </)' = c' + a' tan </)' (8.35 )
where c' = apparent cohesio n i n terms of effective stresse s
0' = angl e o f shearing resistance i n terms of effective stresse s
<7 = total norma l pressure t o th e plan e considere d
u = pore pressure.
The effectiv e stres s parameter s c' an d 0 ' of a give n sampl e o f soi l ma y b e determine d
provided th e por e pressur e u develope d durin g the shea r tes t i s measured . Th e por e pressur e u i s
developed whe n th e testin g o f th e soi l i s don e unde r undraine d conditions . However , i f fre e
Shear St ren g t h o f Soi l 27 5
drainage take s plac e durin g testing, ther e wil l not b e an y developmen t o f por e pressure . I n suc h
cases, th e total stresse s themselve s ar e effective stresses .
8 .1 6 SH EA R STRENG T H EQU ATI O N I N TERM S O F EFFECTI V E
PRI NCI PAL STRESSE S
The principa l stresses ma y be expressed eithe r as total stresse s o r as effective stresse s i f the values
of por e pressur e ar e known.
If u is the por e pressur e develope d durin g a triaxial test, we may writ e as befor e
o = o, -u
where a j and <5'
3
ar e the effective principa l stresses. The equation for shear strengt h i n terms
of effective stresse s i s
<7,' — <7o G<— ( J -, <J , — ( T-.
s = —si n 2a = —si n 2a= —
;
co s 0(8.37 )
2 2 2
where 2a= 90° + 0'
Coulomb's equatio n i n terms o f effective stresse s i s
s = c''+ (<7- u) tan 0'
(7, — ( J ~.
Therefore, — cos<z> ' = c' + ( er-u) tan0'
Since, cr =
2 2
we hav e — co s ( /)' = c' H—
l
- ta n <f)'
+ — cos(9 0 + 0') ta n 0' - u tan 0'
Simplifying
<7, - cr , , . . . cr , + or, . , O" , - 1 . — -.
2 2 2
1 c ' cos$)' + (<7
3
— «) sin^'
- w s n
or
where (ci j - cr
3
) indicate s th e maximu m deviato r stres s a t failure . E q (8.38 ) ma y also b e
expressed i n a different for m a s follows by considering effectiv e principa l stresse s
1 , , c' cos^' + a. sin^'
— (<j, -cr, ) , = -- -
2
l 3 f
1-si n
or

276 Chapt e r 8
Simplifying, w e have
( o[ -o'
3
)
f
= ( o{ + o'
3
) sin ( /)' + 2c' cos 0' (8.39 )
8 .1 7 STRESS- CONTROL L E D AN D STRAI N- CONTROL L ED TESTS
Direct shear tests or triaxial compression test s may be carried ou t by applying stresses o r strains at
a particularly known rate. When the stress i s applied at a constant rate it is called a stress-controlled
test an d whe n th e strai n i s applie d a t a constan t rat e i t i s calle d a strain-controlled test. Th e
difference betwee n the two types of test s may be explained with respect t o box shea r tests .
In the stress-controlled test [Fig . 8.15(a) ] the lateral load F
a
whic h induces shear i s gradually
increased unti l complet e failur e occurs. Thi s ca n b e don e b y placin g weight s o n a hange r o r by
filling a counterweighte d bucke t o f origina l weigh t W a t a constan t rate . Th e shearin g
displacements ar e measured b y means of a dial gauge G as a function o f the increasing loa d F . The
shearing stres s a t any shearing displacement , is
where A i s th e cros s sectiona l are a o f th e sample . A typica l shap e o f a stress-strai n curv e o f th e
stress-controlled tes t i s shown in Fig. 8.15(a) .
A typical arrangement of a box-shear test apparatus for the strain-controlle d tes t i s shown in
Fig. 8.15(b) . Th e shearin g displacement s ar e induce d an d controlle d i n suc h a manne r tha t the y
occur at a constant fixed rate. This can be achieved by turning the wheel either by hand or by means
of an y electricall y operate d moto r s o that horizontal motio n i s induce d throug h th e wor m gea r B.
The dia l gauge G gives the desired constan t rate of displacement. The bottom o f box C is mounted
on frictionles s rollers D. The shearin g resistance offere d t o this displacement b y the soi l sampl e i s
measured b y the proving ring E. The stress-strai n curve s for thi s type of test have the shape shown
in Fig . 8.15(b) .
Both stress-controlle d an d strain-controlled types of test ar e used i n connection wit h all the
direct triaxia l an d unconfi ne d soi l shea r tests . Strain-controlle d test s ar e easie r t o perfor m an d
have the advantag e of readil y giving not onl y the peak resistance a s i n Fig. 8.1 5 (b ) but als o the
ultimate resistance whi c h is lower tha n the peak suc h a s point b i n the same figure, whereas th e
stress controlle d give s onl y the peak value s but not the smalle r value s after th e peak i s achieved .
The stress-controlle d tes t i s preferred onl y in some special problem s connecte d wit h research.
8 .1 8 TY PE S OF L AB ORATORY TEST S
The laborator y test s on soil s may be on
1. Undisturbe d samples, o r
2. Remolde d samples .
Further, the test s ma y be conducted on soil s that are :
1 . Full y saturated , or
2. Partiall y saturated.
The typ e of tes t t o be adopte d depend s upo n how best we ca n simulat e the fiel d conditions .
Generally speaking , th e various shear test s for soil s ma y be classified as follows :
Shear St ren g t h o f Soi l 277
Dial
gauge
Displacement
\_-
(b) Strain controlled
Figure 8 .1 5 St res s an d s t rai n con t rolle d bo x s hea r t es t s
1. U nconsolidated-U ndraine d Test s (U U )
The sample s ar e subjected t o an applied pressur e unde r conditions i n which drainage i s prevented,
and then sheared under conditions o f no drainage .
2. Consolidated-U ndraine d or Quick Test s (CD)
The sample s ar e allowe d t o consolidat e unde r a n applie d pressur e an d the n sheare d unde r
conditions of no drainage .
3. Consolidated-Draine d o r Slow Test s (CD)
The sample s ar e consolidate d a s i n the previous test , bu t the shearin g i s carried ou t slowl y under
conditions of no excess pressure i n the por e space .
The drainage conditio n o f a sampl e i s generall y th e deciding factor i n choosing a particular
type o f tes t i n th e laboratory . Th e purpos e o f carryin g ou t a particula r tes t i s t o simulat e fiel d
conditions a s fa r a s possible . Becaus e o f th e hig h permeabilit y o f sand , consolidatio n occur s
relatively rapidl y an d i s usuall y completed durin g the applicatio n o f th e load . Test s o n san d ar e
therefore generall y carried out under draine d conditions (drained or slow test) .
For soil s othe r tha n sand s the choice o f test conditions depend s upo n the purpose fo r which
the shea r strengt h i s required. The guidin g principl e i s that drainage conditions o f the test shoul d
conform a s closely as possible t o the conditions unde r which the soil s wil l be stresse d i n the field .
Undrained o r quic k test s ar e generall y use d fo r foundation s on cla y soils , sinc e durin g the
period of construction onl y a small amount of consolidation wil l have taken place and consequently
the moistur e content wil l have undergone littl e change. For clay slope s o r cuts undrained tests ar e
used bot h for design an d for th e investigation of failures.
Consolidated-undrained test s ar e used wher e changes i n moisture content ar e expected t o take
place due to consolidation befor e the soil is fully loaded. An important example i s the condition known
as "sudden drawdown" suc h as that occurs in an earth dam behind which the water level is lowered at
278 Chapt e r 8
a faster rate than at which the material of the dam can consolidate. I n the consolidated-undrained test s
used i n this type of problem, the consolidation pressure s ar e chosen t o represent the initial conditions
of the soil , and the shearing loads correspond t o the stresses calle d int o play by the action of sudden
drawdown.
As alread y stated , draine d test s ar e alway s used i n problems relatin g t o sand y soils . I n clay
soils drained tests are sometimes used in investigating the stability of an earth dam, an embankment
or a retaining wall after a considerable interva l of time has passed .
V ery fine sand , silt s and silt y sands also have poor drainage qualities. Saturated soil s of these
categories ar e likel y t o fai l i n the fiel d unde r conditions similar to those unde r which consolidated
quick test s are made .
Shearing Tes t Apparatu s fo r the V ariou s Types o f Test s
The various types of shear test s mentioned earlier may be carried ou t either by the box shear test or
the triaxial compression tes t apparatus. Tests tha t may be made by the two types of apparatus are :
B ox Shea r Test Apparatus
1. Undraine d and consolidated- undraine d tests on clay sample s only .
2. Draine d o r Slow test s on any soil.
The bo x shea r tes t apparatu s i s no t suite d fo r undrained o r consolidated-undrained test s o n
samples other than clay samples, because the other soil s are so permeable tha t even a rapid increase
of the stresse s i n the sampl e ma y caus e at least a noticeable chang e of the wate r content.
Triaxial Compressio n Test Apparatus
All types o f test s ca n convenientl y be carried ou t i n this apparatus.
8 .1 9 SH EARI N G STRENG TH TESTS ON SAND
Shear test s on sand may be made when the sand is either in a dry state or in a saturated state. No test
shall b e mad e whe n th e soi l i s i n a mois t stat e a s thi s stat e exist s onl y due t o apparen t cohesio n
between particle s whic h woul d b e destroye d whe n i t i s saturated . Th e result s o f shea r test s o n
saturated samples ar e almost identical with those on the same sand at equal relative densit y in a dry
state except tha t the angl e 0 is likely to be 1 or 2 degrees smalle r fo r the saturated sand .
The usua l typ e o f tes t use d fo r coars e t o mediu m san d i s th e slo w shea r test . However ,
consolidated undraine d tests ma y be conducte d on fin e sands , sand y silt s etc. whic h do no t allow
free drainag e unde r changed stres s conditions . If the equilibriu m of a large bod y o f saturate d fin e
sand i n a n embankmen t i s disturbe d by rapi d drawdown o f th e surfac e o f a n adjoinin g body o f
water, the change i n water content of the fil l lag s behind the change i n stress .
In all the shearing tests on sand, onl y the remolded samples are used as it is not practicable to
obtain undisturbed samples. The soi l samples ar e to be made approximatel y to the same dry density
as it exists in-situ and tested eithe r by direct shear or triaxial compression tests .
Tests on soils are generally carried out by the strain-controlled type apparatus. The principal
advantage o f thi s type of tes t on dens e san d i s tha t it s peak-shea r resistance , a s wel l a s th e shea r
resistances smalle r tha n the peak, can be observed an d plotted.
Direct Shea r Tes t
Only th e draine d o r th e slo w shea r test s o n san d ma y b e carrie d ou t b y usin g th e bo x shea r tes t
apparatus. The bo x i s filled wit h san d t o the required density . The sampl e i s sheared a t a constan t
Shear St ren g t h o f Soi l 279
vertical pressure a. The shear stresses ar e calculated at various displacements of the shear box. The
test i s repeated wit h different pressure s <7 .
If th e sampl e consist s o f loos e sand , th e shearin g stres s increase s wit h increasin g
displacement until failure occurs. If the sand is dense, the shear failure of the sample is preceded by
a decreas e o f th e shearin g stres s fro m a peak valu e t o a n ultimat e value (also known a s residual
value) lower than the peak value.
Typical stress-strain curves for loose and dense sands are shown in Fig. 8.16(a).
The shear stress of a dense sand increases from 0 to a peak value represented b y point a, and
then graduall y decreases and reaches an ultimate value represented b y point b. The sampl e of sand
in a dense stat e i s closely packed an d the number of contact points between the particles ar e more
than in the loose state. The soil grains are in an interlocked state. As the sample is subjected to shear
stress, th e stres s ha s t o overcom e th e resistanc e offere d b y th e interlocke d arrangemen t o f th e
particles. Experimenta l evidence indicate s that a significant percen t o f the pea k strengt h i s due t o
the interlocking of the grains. In the process o f shearing one grai n tries t o slide over the othe r and
the voi d rati o o f th e sampl e whic h i s th e lowes t a t th e commencemen t o f th e tes t reache s th e
maximum value at point a, i n the Fig 8.16(a) . The shear stres s als o reaches th e maximum value at
this level . An y furthe r increas e o f strai n beyon d thi s poin t i s associate d wit h a progressiv e
disintegration o f th e structur e of th e san d resultin g in a decreas e i n th e shea r stress . Experienc e
shows that the change i n voi d ratio due t o shear depends on both the vertical loa d an d the relative
density of the sand. At ver y low vertical pressure, the void ratio at failure is larger and at very high
pressure it is smaller than the initial void ratio, whatever the relative density of the sand may be. At
Peak value
Dense san d
b ultimat e value
Displacement
(a) Shear stres s v s displacement
0
Dense san d
Loose sand
(b) V olume change
Normal stress, a
(c) Shear strengt h vs normal stres s
Figure 8.1 6 Direc t s hea r t es t o n s an d
280 Chapt e r 8
Table 8. 1 T ypica l val ue s o f 0 and (j )
u
fo r g ran ula r s oil s
Types o f soi l
Sand: rounded grains
Loose
Medium
Dense
Sand: angula r grains
Loose
Medium
Dense
Sandy gravel
0 de g
28 t o 30
30 to 35
35 t o 38
30 t o 35
35 t o 40
40 t o 45
34 to 48
0
u
d e g
26 t o 30
30 t o 35
33 to 36
intermediate value s of pressure, the shearing force causes a decrease in the void ratio of loose sand
and a n increas e i n the voi d rati o o f dense sand . Fi g 8.16(b ) shows ho w th e volume of dens e san d
decreases u p t o a certai n valu e o f horizonta l displacemen t an d wit h furthe r displacemen t th e
volume increases , wherea s i n th e cas e o f loose san d th e volum e continue s t o decreas e wit h a n
increase i n th e displacement . I n saturate d san d a decrease o f th e voi d rati o i s associate d wit h an
expulsion of pore water, and an increase wit h an absorption o f water. The expansion of a soil due to
shear a t a constan t valu e o f vertica l pressure i s calle d dilatancy. At som e intermediat e stat e o r
degree of density in the process o f shear, the shear displacement does not bring about any change in
volume, that is, density. The densit y of sand a t which no change i n volume is brought about upon
the applicatio n o f shea r strain s i s calle d th e critical density. Th e porosit y an d voi d rati o
corresponding t o th e critica l densit y ar e calle d th e critical porosity an d th e critical void ratio
respectively.
By plottin g the shea r strength s corresponding t o th e stat e o f failur e i n th e differen t shea r test s
against the normal pressur e a straight line is obtained for loose sand and a slightly curved line for dense
sand [Fig . 8.16(c)] . However , fo r al l practica l purposes , th e curvatur e fo r th e dens e san d ca n b e
disregarded an d an average line may be drawn. The slopes of the lines give the corresponding angles of
friction 0 of the sand. The general equation for the lines may be written as
s = <J ta n ( f)
For a given sand , the angl e 0 increases wit h increasin g relativ e density . For loos e san d i t is
roughly equal t o the angle of repose, defined as the angl e between the horizontal and the slope of a
heap produced b y pouring clean dry sand from a small height. The angl e of friction varie s with the
shape of the grains. Sand sample s containing well graded angula r grains give higher values of 0 as
compared t o uniforml y graded san d wit h rounded grains . The angl e of frictio n </ > for dens e san d at
peak shea r stres s i s higher than that at ultimate shear stress . Table 8. 1 gives some typical values of
0 (at peak) an d 0
M
(at ultimate) .
Triaxial Compressio n Test
Reconstructed san d samples at the required density are used for the tests. The procedure of making
samples shoul d b e studie d separatel y (refe r t o an y book o n Soi l Testing) . Test s o n san d ma y b e
conducted eithe r i n a saturated stat e or i n a dry state. Slow or consolidated undraine d tests may be
carried ou t as required.
Drained o r Slo w Test s
At leas t thre e identica l sample s havin g the sam e initia l conditions ar e t o b e used . Fo r slo w test s
under saturate d conditions the drainage valve should always be kept open. Eac h sampl e shoul d be
Shear St ren g t h o f Soi l 281
': v•••:..-y<; A
jv:,:^-V^
-• ' -i . * • '' - '' •"'

v
••*». • '- x ' ' ' « ' "
••> ""•.. , \."' • ••> '
(a) Dens e san d (b ) Loos e san d
Figure 8.1 7 T ypica l s hape s o f den s e an d loos e s an d s a t fail ur e
Strain
(a) Stress-strai n curves for three samples at dense stat e
Mohr
envelope
(b) Mohr envelop e
Figure 8.1 8 Moh r en velop e fo r den s e s an d
2 8 2 Chapte r 8
tested unde r differen t constan t all-round pressures fo r example, 1 , 2 and 3 kg/cm
2
. Each sampl e i s
sheared t o failur e by increasing the vertica l loa d a t a sufficientl y slo w rat e t o prevent an y build up
of exces s por e pressures .
At an y stag e o f loadin g th e majo r principa l stres s i s th e all-roun d pressur e <7
3
plu s th e
intensity of deviator stress ( o
{
- cr
3
). The actually applied stresse s ar e the effective stresses i n a slow
test, tha t i s <7
}
= a\ and O"
3
= <r'
3
, Dense sample s fai l alon g a clearly define d ruptur e plane wherea s
loose sand sample s fai l alon g many planes which result in a symmetrical bulging of the sample. The
compressive strengt h o f a sampl e i s define d a s th e differenc e betwee n th e majo r an d mino r
principal stresse s a t failure ( G
I
- <T
3
),,. Typical shapes of dense and loose sand samples a t failure are
shown i n Fig. 8.17 .
Typical stress-strai n curve s for thre e sample s i n a dense stat e an d th e Mohr circle s fo r thes e
samples a t peak strengt h ar e shown in Fig. 8.18.
If the experimen t i s properl y carrie d ou t ther e wil l be one commo n tangen t t o al l these thre e
circles an d thi s will pass throug h the origin. This indicates that the Mohr envelop e i s a straight line
for san d and the sand has no cohesion. The angl e made by the envelope wit h the a-axi s is called th e
angle o f interna l friction . The failur e planes fo r eac h o f thes e sample s ar e show n i n Fig . 8.18(b) .
Each o f them mak e a n angl e a wit h the horizontal which i s approximately equa l t o
a = 45° + 0/2
From Fig . 8.18(b ) a n expression for th e angl e of internal friction ma y b e writte n as
- (J 3 (T j / <73 - 1
(840)
{Q
-™
}
Example 8. 4
Determine th e magnitude of the deviator stress i f a sample o f the same san d with the same voi d rati o
as given i n Ex. 8. 3 was teste d i n a triaxial apparatus wit h a confining pressur e o f 60 kN/m
2
.
Solution
In th e case of a triaxial tes t on an identical sampl e of san d a s given i n Ex. 8.3 , us e the same Moh r
envelope O M (Fig . Ex . 8.3) . No w th e poin t F o n th e absciss a give s th e confinin g pressur e
<7
3
= 60 kN/m
2
. A Mohr circl e C
2
may no w be drawn passing throug h point F and tangentia l t o the
Mohr envelop e OM . Th e poin t E gives th e majo r principa l stress <J
}
fo r th e triaxia l test.
Now cr j = O E = 188 kN/m
2
, <7
3
= 60 kN/m
2
Therefore a
l
- <7
3
= 188 - 6 0 = 128 kN/m
2
= deviator stres s
Example 8 . 5
A consolidated draine d triaxia l test was conducted on a granular soil. At failure cr'/o ^ = 4.0. Th e
effective mino r principa l stres s a t failur e wa s 10 0 kN/m
2
. Comput e 0 ' an d th e principa l stres s
difference a t failure.
Solution
-j -1 4- 1
3
+1 4 + 1
The principa l stres s differenc e at failure is
s i n<z$' = —; ~ = ~= 0.6 o r 6' - 37°
cr,7cr
3
+ 14 + 1
Shear St ren g t h o f Soi l 28 3
,
= <^ —- 1 =100(4-l ) = 300kN/m
2
^
Example 8. 6
A draine d triaxia l tes t o n san d wit h cr'
3
= 315 0 lb/ft
2
gav e ( a\ la
f
^)
f
= 3.7 . Comput e (a )
(b) (o- j - 0-3)^ , and (c) $'.
Solution
Therefore, o{ = 3.1 ( T
f
3
= 3.7 x 3 150 = 1 1,655 lb/ft
2
(b) (<T ! - o-
3
)
/
= (0-; -crp
/
= 1 1,655 -3150 = 8505 l b/ ft
2
l 3.7- 1
OT
Example 8. 7
Assume th e tes t specime n i n Ex . 8. 6 wa s sheare d undraine d a t th e sam e tota l cel l pressur e o f
3150 lb/ft
2
. The induced excess pore wate r pressure at failure u, was equal to 1470 lb/ft
2
. Compute :
(a) o\
f
(b) (cr , - 0
3
)
f
(c) 0 i n terms of total stress ,
(d) th e angle of the failure plane a,
Solution
(a) and (b) : Sinc e the voi d rati o afte r consolidatio n woul d be the same for thi s test as for Ex. 8.6,
assume §' is the same.
a'
cr, - a-, ) , = cr( , —
L
-1
7 J /
-
As before (
°3/
=
^3/ ~ «/ = 3150 -1470 = 1680 lb/ft
2
So ( ff
l
- cr
3
)
f
= 1680 (3.7 -1) = 4536 lb/ft
2
a(
f
= ( ff
l
-0
3
)
f
+ &'
3f
= 4536 + 1680 = 6216 lb/ft
2
(c) s i n <z >
f
, , = -
1
2 .
= =
0.59 o r 0
tn1
. = 36.17°
<"totai 621 6 + 1470
tota l
(d) From Eq. (8.18 )
284 Chapt e r 8
a
f
= 45° + — = 45° + — = 62.5°
7
2 2
where 0'i s taken fro m Ex . 8.6.
Example 8. 8
A saturate d specime n o f cohesionles s san d wa s teste d unde r draine d condition s i n a triaxia l
compression tes t apparatus and the sample failed at a deviator stress o f 482 kN/m
2
and the plane of
failure mad e a n angl e o f 60 ° wit h the horizontal . Fin d th e magnitude s o f th e principa l stresses .
What woul d be th e magnitude s o f the deviato r stres s an d th e majo r principa l stres s a t failur e fo r
another identical specimen o f sand i f i t is tested unde r a cell pressur e of 200 kN/m
2
?
Solution
Per Eq. (8.18), the angle of the failure plan e a i s expressed a s equal t o
Since a = 60°, w e have 0 = 30°.
From Eq . (8.40) , si n ^ = —
1
with 0 = 30°, an d (7, - cr
3
= 482 kN/m
2
. Substituting we have
o-j - <J
3
48 2
°"i
+
^3
=
~7 ~ ~ •
on o
= 964 kN/m
2
(a ) 1 J
si n^z ) si n 30
cr, - cr
3
- 482 kN/m
2
(b )
solving (a ) and (b ) we have
o
l
= 723 kN/m
2
, an d <J
3
= 241 kN/m
2
For the identical sampl e
0 = 30°, <T
3
= 200 kN/m
2
From Eq. (8.40) , we have
cr, - 200
Sin30
°=^7^
Solving fo r <T J we have a
l
= 600 kN/m
2
and (cr , - cr
3
) = 40 0 kN/m
2
8.20 U NCONSOL I DATED- U NDRAI NE D TES T
Saturated Cla y
Tests o n saturated cla y ma y be carried out either on undisturbed or on remolded soi l samples . The
procedure o f the tes t i s the same i n bot h cases . A series o f sample s (a t leas t a minimum of three )
having th e same initia l condition s ar e teste d unde r undraine d conditions . Wit h a
y
th e all-roun d
pressure, actin g o n a sampl e unde r conditions of no drainage, th e axial pressur e i s increased unti l
failure occur s a t a deviator stress (<7 , - (7
3
). From the deviator stress, the major principal stress cr , is
determined. I f th e othe r sample s ar e teste d i n th e sam e wa y bu t wit h differen t value s o f cr
3
, i t i s
Shear St ren g t h o f Soi l 285
found tha t fo r al l type s o f saturate d clay , th e deviato r stres s a t failur e (compressiv e strength ) i s
entirely independen t o f the magnitud e of cr
3
a s shown in Fig. 8.19 . The diameter s o f al l th e Moh r
circles are equal and the Mohr envelope is parallel to the cr-axi s indicating that the angle of shearing
resistance 0
U
= 0 . Th e symbo l 0
U
represent s th e angl e o f shearin g resistanc e unde r undraine d
conditions. Thus saturated clays behave as purely cohesive material s wit h the following properties :
(8.41)
where c
u
is the symbol used for cohesion unde r undrained conditions. Eq. (8.41) hold s true for the
particular case of a n unconfine d compressio n tes t i n whic h <7
3
= 0. Sinc e thi s tes t requires a very
simple apparatus , i t i s ofte n used , especiall y fo r fiel d work , a s a read y mean s o f measurin g th e
shearing strengt h of saturated clay, in this cas e
q
= —
!L
, wher e
(8.42)
Effective Stresses
If during the test, pore-pressures ar e measured, the effective principa l stresse s ma y be written as
<j( = CT j - U
(8.43)
where u is the pore water pressure measure d during the test. The effective deviato r stres s a t failur e
may be written as
Eq. (8.44) shows that the deviator stress i s not affected b y the pore water pressure. As such the
effective stres s circl e i s only shifted from th e position of the total stress circl e as shown in Fig. 8.19.
Partially Saturate d Cla y
Tests o n partiall y saturate d cla y ma y b e carrie d ou t eithe r o n undisturbe d o r o n remolde d soi l
samples. All the samples shall have the same initial conditions before the test, i.e., they should possess
the sam e wate r conten t an d dr y density . The test s ar e conducte d i n th e same wa y a s fo r saturate d
samples. Eac h sampl e i s teste d unde r undraine d condition s wit h differen t all-roun d pressures o~
3
.
T
C
u
A.
Effective stres s circl e
Total stres s circl e
Figure 8.1 9 Moh r circl e for un drain e d s hea r t est o n s at urat e d cla y
286 Chapt er 8
Figure 8.2 0 M oh r circl e fo r un drain e d s hea r t es t s o n part ial l y s at urat e d cl a y s oil s
Total stres s
circle
Figure 8 .2 1 Effect iv e s t res s circl e s fo r un drain e d s hea r t es t s o n part iall y s at urat e d
cl ay s oil s
Mohr circle s fo r thre e soi l sample s an d th e Moh r envelop e ar e shown i n Fig . 8.20 . Thoug h al l th e
samples ha d the same initia l conditions, the deviator stress increases with the increase i n the all-round
pressure o~
3
as shown in the figure. This indicate s that the strengt h of the soil increases wit h increasing
values of o~
3
. The degre e o f saturation also increases wit h the increase i n o~
3
. The Mohr envelope which
is curved at lower values of o~
3
becomes almos t parallel to the o*-axi s as full saturatio n is reached. Thus
it i s no t strictl y possibl e t o quot e singl e value s for th e parameter s c
u
an d §
u
fo r partiall y saturate d
clays, but over any range o f normal pressure cr
;
encountered in a practical example, the envelope can
be approximated b y a straight line and the approximate values of c
u
and 0
H
can be used in the analysis.
Effective Stresses
If th e por e pressures ar e measure d durin g the test , the effectiv e circle s ca n b e plotte d a s shown i n
Fig. 8.2 1 an d the parameter s c ' and 0' obtained. The envelope to the Mohr circles , whe n plotted i n
terms o f effective stresses , i s linear.
Typical undraine d shea r strengt h parameter s fo r partiall y saturate d compacte d sample s ar e
shown i n Table 8.2 .
8 .2 1 UNCONFI NE D COM PRESSI O N TEST S
The unconfme d compressio n tes t i s a specia l cas e o f a triaxia l compressio n tes t i n whic h th e all -
round pressur e o"
3
= 0 (Fig . 8.22) . Th e test s ar e carrie d ou t onl y o n saturate d sample s whic h ca n
stand without any latera l support . The test , is , therefore, applicabl e t o cohesive soil s only. The tes t
Shear St ren g t h o f Soi l 28 7
Table 8.2 Probabl e undraine d s hear s t ren g t h param et er s fo r part iall y s at urat e d s oil s
Types o f soi l
Sand wit h clay binder
Lean silt y clay
Clay, moderate plasticity
Clay, ver y plastic
c
u
( t s f)
0.80
0.87
0.93
0.87
4> u
23°
13°


c ' ( t s f )
0.70
0.45
0.60
0.67
0'
40°
31°
28°
22°
is an undrained test and is based on the assumption that there is no moisture loss during the test. The
unconfmed compressio n tes t is one of the simplest and quickest tests used for the determination of
the shear strengt h of cohesive soils . These test s can also be performed i n the field by making use of
simple loading equipment .
Figure 8.2 2 Un con fin e d com pres s io n t es t equipm en t (Court es y : Soilt es t )
288 Chapt e r 8
Any compressio n testin g apparatu s wit h arrangemen t fo r strai n contro l ma y b e use d fo r
testing th e sample s . The axia l load u\ ma y b e applied mechanically or pneumatically.
Specimens o f heigh t to diamete r ratio of 2 ar e normall y used fo r th e tests . The sampl e fail s
either by shearing on an inclined plane (if the soil is of brittle type) or by bulging. The vertica l stress
at any stage of loading is obtained by dividing the total vertical load by the cross-sectional area . The
cross-sectional are a of the sampl e increases wit h the increase i n compression. Th e cross-sectiona l
area A at any stage of loading of the sampl e may b e computed on the basi c assumption t hat the total
vol ume of the sampl e remai ns the same. That i s
AO/IQ = A h
where A
Q
, h
Q
= i ni t i al cross-sectiona l area an d heigh t of sampl e respectively.
A,h = cross-sectional are a an d height respectively at any stage o f loading
If Ah i s the compression o f the sample , th e strai n is
A/z
£
~ ~j~~ sinc e A/ z = h
0
- h, we may write
AO/ZQ = A(/Z
O
- A/z )
Therefore, A = -j^-= ^^ = ^(8.45 )
The averag e vertica l stress at any stage of loadin g may be writte n as
P P( l-e]
A A
( )
(8.46)
where P i s the vertica l load a t the strai n e.
Using the rel at i onship given by Eq. (8.46) stress-strai n curves may be plotted. The peak value
is taken as the unconfi ne d compressiv e strength q
ti
, that is
(
f f
i )
f
=
V
u
(8-47 )
The unconfine d compression tes t (UC ) i s a specia l cas e o f th e unconsolidated-undrained
(UU) triaxia l compression tes t (TX-AC). The onl y difference between th e U C tes t an d U U tes t i s
that a total confining pressure under which no drainage was permitted wa s applie d i n the latter test .
Because o f the absence of any confi ni ng pressur e in the UC test, a premature failure through a weak
zone may terminat e an unconf i ne d compressio n test. For typical soft clays, prematur e failure i s not
likely t o decreas e th e undrained shear strengt h by mor e tha n 5%. Fi g 8.23 show s a comparison of
undrained shea r strengt h values from unconfine d compressio n test s and fro m triaxia l compression
tests o n soft-Natsushima cla y fro m Toky o Bay. The propertie s o f the soi l are :
Nat ural moistur e content w = 80 t o 90 %
Liquid l imi t w, = 10 0 to 110 %
Pl ast icit y i nde x /
;
= 60%
There i s a uni que relationship between remolded undraine d shear strengt h an d th e liquidit y
index, / , as shown in Fig. 8.24 (afte r Terzaghi et al., 1996) . This plot includes sof t clay soi l and silt
deposits obtaine d fro m differen t part s of the world.
Shear St ren g t h o f Soi l 289
50
40
§30
: ?2 0
10
Natsushima Cla y
I
p
= 60%
c
u
= undrained strengt h
c
u
( UQ
° = 0.80
0 1 0 2 0 3 0 4 0 5 0
c
u
( TQ, kPa
Figure 8.2 3 Relat io n bet ween un drain e d s hear s t ren g t h s fro m un con fin e d
com pres s ion an d t riax ial com pres s ion t es ts o n Nat s us him a cl ay (dat a fro m Han zawa
an d Kis hida, 1982)
10
2
10'
M
r v
•0 10 °
2
"o
10-
10"
2 3 4
Liquidity inde x
Figure 8.2 4 Relat io n bet ween un drain e d s hear s t ren g t h an d liquidit y inde x o f cl ays
from aroun d t he worl d (aft e r T erzag h i e t al. , 1996 )
290 Chapt e r 8
Example S. 9
Boreholes revea l tha t a thi n laye r of alluvia l sil t exist s a t a dept h of 5 0 f t below th e surfac e of th e
ground. The soi l abov e thi s level has a n average dr y uni t weight of 96 lb/ft
3
an d a n average wate r
content of 30%. Th e water table is approximately at the surface. Tests o n undisturbed samples give
the followin g data : c
u
= 100 8 lb/ft
2
, 0
M
= 13° , c
d
= 861 lb/ft
2
, ( j)
d
= 23°. Estimat e th e shearin g
resistance of the sil t on a horizontal plane (a) when the shear stress builds up rapidly, and (b) when
the shea r stres s build s up ver y slowly.
Solution
Bulk uni t weight y
t
= y
d
( 1 + w) = 96 x 1. 3 = 124. 8 lb/ft
3
Submerged uint weight y
b
= 124.8- 62.4 = 62. 4 lb/ft
3
Total norma l pressure a t 50 f t dept h = 50 x 124. 8 = 6240 lb/ft
2
Effective pressur e a t 50 f t dept h = 50 x 62. 4 = 3120 lb/ft
2
(a) For rapi d build-up, use the properties o f the undrained state an d tota l pressure .
At a total pressur e of 6240 lb/ft
2
shear strength, s = c + crtan </ > = 100 8 + 6240 tan 13 ° = 2449 lb/ft
2
(b) For slo w build-up , use effective stress propertie s
At a n effectiv e stres s o f 312 0 lb/ft
2
,
shear strengt h = 861 + 3120 tan 23° = 2185 lb/ft
2
Example 8 .1 0
When a n undraine d triaxia l compressio n tes t wa s conducte d o n specimen s o f claye y silt , th e
following result s wer e obtained:
Specim en No . 1
cr
3
(kN/m
2
) 1 7 4 4 5 6
<T! (kN/m
2
) 15 7 20 4 22 5
M (kN/m
2
) 1 2 2 0 2 2
Determine th e value s o f shea r parameter s considerin g (a ) tota l stresse s an d (b ) effectiv e
stresses.
Solution
(a) Total stresse s
For a solution with tota l stresses , dra w Mohr circle s C
r
C
2
and C
3
for each o f the specimen s
using the corresponding principa l stresses a
{
an d cr
3
.
Draw a Moh r envelop e tangen t to thes e circle s a s show n i n Fig . Ex . 8.10 . No w fro m th e
figure
c- 48 kN/m
2
, 0= 15°
Shear St ren g t h o f Soi l 291
120
80
40
c = 48 kN/m
2
c' = 46kN/m
2
40 80 12 0
a, kN/m
2
»
Figure Ex . 8.1 0
200 240
(b) With effective stresse s
The effectiv e principa l stresse s ma y b e foun d b y subtractin g the pore pressure s u from th e
total principal stresses as given below.
Speci men No .
cr'
3
= (CT
3
- u) kN/m
2
o\ = (CT J - w ) kN/m
2
1
5
145
2
24
184
3
34
203
As befor e dra w Moh r circle s C', , C"
2
an d C"
3
fo r eac h o f th e specimen s a s show n i n
Fig. Ex. 8.10. No w fro m th e figur e
c' = 46 kN/m
2
, $'= 20°
Example 8 .1 1
A soi l ha s a n unconfine d compressiv e strengt h o f 12 0 kN/m
2
. I n a triaxia l compressio n tes t a
specimen of the same soi l when subjected to a chamber pressure of 40 kN/m
2
failed at an additional
stress of 16 0 kN/m
2
. Determine :
(i) The shea r strengt h parameters o f the soil, (ii) the angle made by the failure plane wit h the
axial stres s i n the triaxial test.
Solution
There i s one unconfined compressio n tes t result and one triaxial compression tes t result. Hence two
Mohr circles, C
p
and C
2
may be drawn as shown in Fig. Ex. 8.11. For Mohr circl e C
r
cr
3
= 0 and
CTj = 12 0 kN/m
2
, an d fo r Moh r circl e C
2
, O
3
= 40 kN/m
2
an d a
{
= (4 0 + 160 ) = 20 0 kN/m
2
. A
common tangen t to these two circles i s the Mohr envelope whic h give s
(i) c = 43 kN/m
2
and 0 = 19°
(ii) For the triaxial test specimen, A i s the point of tangency for Mohr circl e C
2
an d C is the
center of circle C
2
. The angle made by AC with the abscissa is equal to twice the angle between the
failure plan e and th e axi s o f the sampl e = 26. From Fig . Ex . 8.11 , 26 = 71 °and 6 = 35.5°. The
angle made by the failure plane wit h the e r -axis is a = 90°-35.5° = 54.5°.
292 Chapt er 8
80 120 160 20 0
o, kN/m
2

Figure Ex . 8 .1 1
Example 8 .1 2
A cylindrica l sampl e o f saturate d cla y 4 c m i n diamete r an d 8 cm hig h wa s tested i n an unconfined
compression apparatus . Fin d th e unconfine d compression strength , i f the specime n faile d a t a n axia l
load o f 360 N, when the axial deformation wa s 8 mm. Find th e shear strengt h parameter s i f the angl e
made by the failure plane wit h the horizontal plane was recorded a s 50°.
Solution
Per Eq. (8.46) , th e unconfine d compression strengt h of the soi l i s given by
where P =
A = = 12.56 cm
2
, = — = 0.1
8
50 100 15 0
a, kN/m
2
0= 10 °
200 250
Figure Ex . 8 .12
Shear St ren g t h o f Soi l 29 3
Therefore a , =
360(1
~°-
1)
= 25.8 N/ cm
2
= 2 5 8k N/ m
2
1
12.5 6
Now 0 = 2a - 90 ° (Refer to Fig. 8.12) where a = 50°.Therefore 0 = 2 x 50 - 90 ° = 10°.
Draw th e Mohr circl e a s shown in Fig. Ex. 8.12 (a
3
= 0 and o~ j = 258 kN/m
2
) and fro m th e
center C of the circle, draw CA at 2a = 100°. At point A, draw a tangent to the circle. The tangent is
the Mohr envelope whic h gives
c = 106 kN/m
2
, an d 0=10°
Example 8.1 3
An unconfme d cylindrica l specime n o f clay fail s under a n axial stres s o f 5040 lb/ft
2
. Th e failure
plane was inclined at an angle of 55° to the horizontal. Determine the shear strengt h parameters of
the soil.
Solution
From Eq. (8.20) ,
<r
l
=( T
3
N
</>
+2cjN ^, wher e ^ = tan
2
45 ° +|
since < T = 0, we have
= 2c tan ^45° + -, wher e ^ = 5040 lb/ft
2
(a )
From Eq . (8.18), the failure angle a i s
o
a = 45 + — , sinc e a = 55°, w e have
2
From Eq . (a),
c =
2t an45°- 4
2tan55
°
2
Example 8.1 4
A cylindrical sample of soil having a cohesion o f 80 kN/m
2
and an angle of internal friction of 20°
is subjecte d t o a cell pressure o f 10 0 kN/m
2
.
Determine: (i ) the maximum deviator stress ((jj- <7
3
) at which the sample will fail, and (ii) the
angle made by the failure plane wit h the axi s of the sample .
G raphical solution
<7
3
= 10 0 kN/m
2
, 0 = 20°, an d c = 80 kN/m
2
.
A Mohr circle an d the Mohr envelope can be drawn as shown in Fig. Ex. 8.14(a). The circl e
cuts the cr-axi s a t B ( = <7
3
), an d a t E ( = o^). Now <7 j = 433 kN/m
2
, and <7
3
= 10 0 kN/m
2
.
294 Chapt er 8
200
100
100 20 0 30 0
a, kN/m
2
» •
(a)
400 45 0
(b)
Figure Ex . 8.1 4
(<7, - cr
3
) = 433 - 10 0 = 333 kN/m
2
.
Analytical solution
Per Eq. (8.20 )
or, = a, tan
2
45 ° +— +2ct a n 45 ° + —
1 3
I 2 ) I 2 .
Substituting the known values, we have
tan(45° + 0/2) = tan (45° + 10 ) = tan 55° = 1.42 8
tan
2
(45° + 0/2 ) = 2.04 .
Therefore,
<7, = 10 0 x 2.04 + 2 x 8 0 x 1.42 8 « 433 kN/m
2
(CTj - <J
3
) = (433 - 100 ) = 333 kN/m
2
If 6 = angle made by failure plane s with the axi s of the sample, (Fig. Ex. 8.14(b) )
29 = 90 - 0 = 90 - 2 0 = 70° o r 6 = 35°.
Therefore, th e angl e made by the failur e plane with the cr-axi s i s
a- 9 0 -35 = 55°
8 .2 2 CONSOL I DATED- U NDRAI NE D TEST ON SATU RATED CL AY
Normally Consolidate d Saturated Cla y
If tw o cla y sample s 1 and 2 are.consolidate d unde r ambien t pressure s o f p
l
an d p
2
an d ar e then
subjected t o undraine d triaxial tests withou t further chang e i n cel l pressure , th e result s ma y b e
expressed b y th e tw o Moh r circle s C
L
an d C
2
respectivel y a s show n i n Fig . 8.25(b) . The failur e
envelope tangentia l to thes e circle s passe s throug h the origi n an d it s slop e i s define d b y 0
CM
, th e
angle of shearing resistance i n consolidated undraine d tests. I f the pore pressures ar e measured th e
effective stres s Moh r circle s C\ an d C'
2
can also be plotted an d the slop e o f thi s envelope i s 0'
cu<
The effectiv e principa l stresses are :
Shear St ren g t h o f Soi l 295
Axial strai n Axia l strain
(a) V ariation of ( a\ - a
3
) and u with axial strain
[- " 2 H
(b) Mohr envelope
Figure 8.2 5 Norm all y con s olidat e d cl a y unde r un drain e d t riax ial t es t
p\ P2 P^=Pa
Total stres s
circle
Effective
stress circl e
(^3)1 (^3) 2 (^3) 3
Figure 8.26 Con s olidat ed-un drain e d t es ts o n s at urat ed overcon s ol idat e d clay
296 Chapt er 8
°
=
°~
where u
l
an d w
2
ar e the por e wate r pressures fo r the sample s 1 and 2 respectively.
It i s a n experimenta l fac t tha t th e envelope s t o th e tota l an d effectiv e stres s circle s ar e
linear. Fig. 8.25(a ) shows the nature of the variation on the deviator stres s (<7 j - <7
3
) and the pore
water pressur e u i n th e specime n duri n g the tes t wit h the axia l strain . The por e wate r pressur e
bui l ds u p duri n g shearing wit h a corresponding decrease i n the volume of the sample .
Overconsolidated Cla y
Let a saturate d sampl e 1 be consolidate d unde r an ambien t pressur e p
a
an d the n allowe d t o swel l
under th e pressur e p
r
A n undraine d triaxial test i s carried ou t on thi s sampl e unde r the all-round
pressure p\ ( = <T
31
). Another sample 2 i s als o consolidate d unde r the same ambien t pressur e p
a
an d
allowed to swell under the pressure p
2
( = <7
32
). An undrained triaxial test is carried out on this sample
under th e sam e all-roun d pressur e p
2
. Th e tw o Moh r circle s ar e plotte d an d th e Moh r envelop e
tangential t o the circles is drawn as shown in Fig. 8.26. The shear strengt h parameters ar e c
u
and 0
CU
.
If por e wate r pressur e i s measured , effectiv e stress Moh r circle s ma y b e plotte d a s shown i n th e
figure. Th e strengt h parameters for effective stresse s ar e represented b y c' and §'.
8 .2 3 CONSOL I DATED- DRAI NE D SH EAR STRENG T H TEST
In drained triaxial tests the soil is first consolidated under an ambient pressure p
a
an d then subjected to
an increasing deviator stress unti l failur e occurs , the rate of strain being controlled i n such a way that
at no time is there any appreciable pore-pressure i n the soil . Thus at all times the applied stresse s ar e
effective, an d whe n th e stresse s a t failur e ar e plotte d i n th e usua l manner , th e failur e envelope i s
directly expresse d i n terms o f effectiv e stresses . Fo r normall y consolidated clay s an d fo r sand s th e
envelope i s linear for norma l workin g stresses an d passes throug h the origi n a s shown i n Fig. 8.27 .
The failur e criterio n fo r suc h soil s i s therefor e th e angl e o f shearin g resistanc e i n th e draine d
condition 0
d
.
The draine d strength is
- ( o -
1
- f f
3
) / =-
. - si n
(8. 48)
Eq. (8.48 ) i s obtained from Eq. (8.38 )
Figure 8.2 7 Drain e d t es t s o n n orm all y con s olidat e d cl a y s am ple s
Shear St ren g t h o f Soi l 297
Peak /O.C . clay
Ultimate
N.C. cla y
Axial strai n
3
I
D,
X
U
CX
o
u
O.C. cla y
Axial strai n
N.C. cla y
(a) V ariation of ( a\ - a
3
) with axial strai n
O.C. cla y
Q
N.C. cla y
Normal stress , o
(b) Mohr envelope
Figure 8.2 8 Draine d t es t s o n overcon s olidat e d clays
For overconsolidated clays , the envelope intersects the axis of zero pressure at a value c
d
. The
apparent cohesion i n the drained test an d the strengt h are given by the expression .
(8.49)
1- si ni
The Mohr envelope for overconsolidated clay s is not linear as may be seen i n Fig. 8.28(b). An
average line is to be drawn within the range of normal pressure cr
n
. The shear strengt h parameters c
d
and ( j)
d
ar e referre d t o thi s line.
Since the stresses i n a drained test are effective, i t might be expected tha t a given ( f)
d
woul d be
equal t o 0 ' a s obtaine d fro m undraine d test s wit h pore-pressur e measurement . I n normall y
consolidated clays and in loose sand s the two angles of shearing resistance ar e in fact closel y equa l
since the rate of volume change in such materials at failure in the drained test is approximately zer o
and there i s no volume change throughout an undrained test on saturated soils . Bu t i n dense sand s
and heavily overconsolidated clays ther e is typically a considerable rate of positive volume chang e
at failure i n drained tests , and work has t o be done not only i n overcoming th e shearing resistanc e
of the soils, bu t als o i n increasing th e volume of the specimen agains t th e ambient pressure . Ye t in
298 Chapt er 8
undrained test s o n th e sam e soils , th e volume change i s zer o an d consequentl y ( j)
d
fo r dens e sand s
and heavil y overconsolidated clay s i s greater tha n 0'. Fig. 8.28(a ) shows the natur e of variation of
the deviator stres s with axial strain . During the application o f the deviator stress, the volume of the
specimen graduall y reduces fo r normall y consolidated clays . However, overconsolidate d clay s g o
through some reductio n of volume initially bu t then expand.
8 .2 4 POR E PRESSU RE PARAM ETERS UNDER U NDRAI NE D
L OADI NG
Soils i n nature are at equilibrium under their overburden pressure. I f the same soi l i s subjected t o an
instantaneous additional loading, ther e will be development of pore pressur e i f drainage i s delaye d
under th e loading . The magnitud e of the por e pressur e depend s upo n the permeabilit y o f the soil ,
the manne r of application of load, the stres s histor y of the soil , and possibl y man y other factors . If
a load i s applied slowl y and drainage takes place wit h the application of load, ther e wil l practically
be no increase o f pore pressure . However , if the hydrauli c conductivity of the soi l i s quite low, and
if the loading i s relatively rapid, there will not be sufficien t tim e for drainage t o take place . I n such
cases, there wil l be an increase in the pore pressure in excess o f the existing hydrostatic pressure. I t
is therefore necessar y man y times to determine or estimate the excess por e pressur e fo r the various
types o f loadin g conditions . Por e pressur e parameter s ar e use d t o expres s th e respons e o f por e
pressure t o change s i n tota l stres s under undrained conditions. V alues of th e parameter s ma y b e
determined i n th e laborator y an d ca n b e use d t o predic t por e pressure s i n th e fiel d unde r simila r
stress conditions .
Pore Pressur e Parameter s U nder Triaxia l Test Condition s
A typical stress applicatio n on a cylindrical element of soil under triaxial test conditions i s shown in
Fig. 8.2 9 (Ad j > A<7
3
). AM is the increas e i n the pore pressur e withou t drainage. From Fig . 8.29 , w e
may writ e
AM
3
= 5A<7
3
, Aw j = Afi(Acr
1
- Acr
3
), therefore,
AM = AM j + AM
3
= #[A<7
3
+ /4(A(Tj - Acr
3
)]
or A M =BAcr
3
+ A(Aer, - A<r
3
)
where, A = AB
for saturate d soil s B = 1, s o
(8.50)
(8.51)
Aw = A< 7 - A<7
3
)
I ACT ,
(8.52)
ACT,
A<73 A<7 3
(ACT, - ACT
3
)
ACT, AM,
ACT, ACT,
(ACT, - ACT 3)
Figure 8.2 9 Ex ces s wat er pres s ur e unde r t riax ia l t es t con dit ion s
Shear St ren g t h o f Soi l 299
P
o
r
e

p
r
e
s
s
u
r
e

c
o
e
f
f
i
c
i
e
n
t

A
3

O

O

O

O
»
-
/
>

K
>

N
J

0
»

^
-
J

C
D

L
r
t

O
<
-
f
i

O

<
-
t

C
v
\
\
\
\,
*-^
0 2 4 1 0 2 0 4 0
Overconsolidation ratio (log scale)
Figure 8.3 0 Relat ion s hi p bet ween Overcon s olidat io n rat io an d por e pres s ure
coefficien t A
OQ
£
p
r
e
s
s
u
r
e
£
1.0
ob
o
60 70 80
S, percen t
90 100
Figure 8.3 1 T ypica l relat ion s hi p bet wee n B and deg ree of s at urat io n S.
where A an d B ar e called por e pressur e parameters. The variation of A under a failure condition
(A,) wit h th e Overconsolidatio n ratio, O
CR
, is give n i n Fig . 8.30 . Som e typica l value s of A, are
given i n Table 8.3. The valu e of B varies wit h the degre e of saturation as shown in Fig. 8.31 .
Table 8.3 T ypica l value s of A
f
Type o f Soi l
Highly sensitiv e cla y
Normally consolidate d cla y
Compacted sand y clay
Lightly overconsolidate d cla y
Compacted cla y grave l
Heavily overconsolidated cla y
V ol ume chang e
large contractio n
contraction
slight contractio n
none
expansion
expansion
A
t
+ 0.75 to + 1. 5
+ 0. 5 t o + 1. 0
+ 0.25 to + 0.75
+ 0.00 to + 0.5
- 0.2 5 to + 0.25
- 0. 5 to 0
300 Chapt e r 8
8 .2 5 V AN E SH EA R TEST S
From experienc e i t ha s bee n f oun d t ha t the van e tes t ca n be use d a s a reliabl e i n- si t u test fo r
det er mi ni ng th e shea r st r engt h of soft -sensi t i ve clays. I t i s i n deep beds o f suc h mat eri a l tha t
the van e tes t i s mos t val uabl e , for th e simpl e reaso n t ha t ther e i s a t presen t n o othe r metho d
known b y whi c h th e shea r st rengt h o f thes e clay s ca n b e measured . Repeate d attempts ,
par t i cul ar l y i n Sweden , hav e fai l ed t o obtai n undi st ur be d samples fro m depth s o f mor e t ha n
about 1 0 meter s i n nor mal l y consol idat e d clay s o f hi g h sensi t i vi t y eve n usi n g th e mos t
modern for m o f t hi n- wal l e d pist on samplers . I n thes e soil s th e van e i s indispensabl e . Th e
vane shoul d b e regarde d a s a metho d t o b e use d unde r th e f ol l owi n g conditions :
1. Th e cla y i s normall y consolidated and sensitive .
2. Onl y th e undrained shear strengt h is required.
It ha s been determine d tha t the vane gives result s similar t o thos e obtaine d fro m unconfme d
compression test s o n undisturbe d samples.
The soi l mas s shoul d be in a saturated condition i f the vane test i s to be applied. The van e test
cannot be applie d t o partially saturated soil s t o which the angl e of shearing resistanc e i s not zero .
Description o f the V an e
The van e consists o f a steel ro d having at one end fou r smal l projecting blade s or vanes parallel t o
its axis , an d situate d at 90° interval s around the rod. A post hol e bore r i s first employed t o bore a
hole up to a point j ust above the required depth. The rod is pushed or driven carefully until the vanes
are embedde d a t the require d depth . At the othe r en d o f the ro d abov e th e surfac e o f the groun d a
torsion head i s used t o apply a horizontal torque and this is applied at a uniform speed of about 0.1°
per se c unt i l th e soi l fails , thu s generating a cylinde r o f soil . Th e are a consist s o f th e periphera l
surface o f th e cylinde r an d th e tw o roun d ends . Th e firs t momen t o f thes e area s divide d b y th e
applied moment give s the unit shear value of the soil . Fig. 8.32(a) gives a diagrammatic sketc h o f a
field vane .
Determination o f Cohesio n o r Shear Strengt h o f Soi l
Consider th e cylinde r o f soi l generate d b y th e blade s o f th e van e whe n the y ar e inserte d int o th e
undisturbed soi l in-sit u and gradually turned or rotated abou t the axis of the shaf t or vane axis. The
turning moment applie d a t the torsion head above the ground is equal t o the force multiplied by the
eccentricity.
Let the forc e applie d = P eccentricit y (lever arm) = x units
Turning momen t = Px
The surfac e resisting the turning is the cylindrical surface of the soi l and the two end faces of
the cylinder.
Therefore,
resisting momen t = ( 2nr x L x c
u
x r + Inr
2
x c
u
x 0.67r) = 2nr
2
c
u
( L + 0.67r)
where r = radius of the cylinder and c
u
the undrained shear strength .
At failure the resisting moment of the cylinder of soil i s equal t o the turning moment applie d
at the torsion head .
Therefore, P x = 2/rr
2
c
u
( L + 0.67r)
Px
(8>53)
Shear St ren g t h o f Soi l 301
1. Straingauge fo r
reading torqu e
2. Rotation indicato r
3. 8-i n casing wit h side fins for
anchoring torqu e assembl y
4. Torque rod
5. A-rod fo r applying torque t o
vane. Made up i n 5-ft lengths
(a)
Torque ring
5° graduations
1.0
0.8
0.6
0.4
\
20 4 0 6 0 8 0
Plasticity index , l
p
100 120
6. BX casing for housing
torque rod and A rod
7. V ane rod
8. BX-casing-point containing
bearing an d water seal s for vane rod
9. V ane varying sizes
2 in dia by 4 in
3 in dia by 6 in
4 i n dia by 8 in
(b)
Figure 8.3 2 Van e s hear t es t (a ) diag ram m at ic s k et c h o f a fiel d van e , (b ) correct io n
fact or \i (B j errum , 1973)
The standar d dimension s o f fiel d vane s a s recommende d b y AST M (1994) ar e give n i n
Table 8.4.
Some investigator s believe that vane shear test s i n cohesive soi l gives a values of the shea r
strength abou t 1 5 per cen t greate r tha n i n unconfme d compressio n tests . Ther e ar e other s wh o
believe that vane tests give lower values.
Table 8.4 Recom m en de d dimensions o f fiel d van e s (AST M , 1994)
Cas in g s iz e
AX
BX
NX
Height,
m m (L )
76.2
101.6
127.0
Diam et er,
mm (d )
38. 1
50.8
63.5
Blade t hick n es s
mm
1.6
1.6
3.2
Diam et er o f ro d
mm
12.7
12.7
12.7
302 Chapt er 8
0.6
0.4
0.2
Based on
Bjerrum curves
00
20 40 60 80 100
Figure 8.3 3 Un drain e d s hea r s t ren g t hs fr o m fiel d van e t es t s o n in org an i c s of t
cl ays an d s ilt s (aft e r T aven as an d Leroueil , 1987 )
Bjerrum (1973) back computed a number of embankment failures on sof t cla y and concluded
that the vane shear strengt h tended t o be too high. Fig. 8.32(b ) gives correction factor s fo r the fiel d
vane tes t a s a function o f plasticit y index, / (Lad d e t al. , 1977) . We may writ e
c
u
(field ) = IJ L C
U
(vane ) (8.54 )
where \ i is the correction facto r (Fig. 8.32b) .
Fig. 8.3 3 giv e relationship s between plasticit y index / an d cjp' wher e c
u
i s the undrained
shear strengt h obtained by field vane and//t he effective overburde n pressure. This plot i s based on
comprehensive tes t dat a compile d o f Tavena s an d Lerouei l (1987) . Necessar y correctio n factor s
have been applie d t o the dat a as per Fig. 8.3 2 (b ) before plotting.
8.26 OTH E R M ETH ODS FO R DETERM I NI NG U NDRAI NE D SH EAR
STRENG TH OF COH ESI V E SOI LS
We have discussed i n earlier sections three methods for determining the undrained shear strengt h of
cohesive soils . The y ar e
1. Unconfme d compressio n tes t
2. U U triaxial test
3. V an e shear tes t
In thi s section two mor e method s ar e discussed. The instrument s used fo r thi s purpose ar e
1. Torvan e (TV )
2. Pocke t penetrometer (PP)
Torvane
Torvane, a modification of the vane, is convenient for investigating the strength of clays in the walls
of tes t pit s i n th e fiel d o r fo r rapi d scannin g o f th e strengt h o f tub e o r spli t spoo n samples .
Fig 8.34(a ) gives a diagrammatic sketch of the instrument. Figure 8.34(b ) gives a photograph of the
same. Th e vane s ar e presse d t o thei r ful l dept h int o th e cla y belo w a fla t surface , whereupo n a
torque i s applie d throug h a calibrate d sprin g unti l th e cla y fail s alon g th e cylindrica l surfac e
Shear St ren g t h o f Soi l 303
LJ UULJ U
(a)
Figure 8.3 4 T orvan e s hea r device (a ) a diag ram m at ic s k et ch , an d (b ) a phot og rap h
(Court es y: Soilt es t )
circumscribing the vanes and simultaneously along the circular surface constituting the base of the
cylinder. The value of the shear strength is read directl y from th e indicator on the calibrated spring .
Specification fo r three size s o f vanes ar e given below (Holt z et al. , 1981 )
Diam et er (m m )
19
25
48
Heig ht o f van e (m m )
3
5
5
M ax im um s hea r s t ren g t h (k Pa)
250
100 (standard)
20
Pocket Penetrometer
Figure 8.3 5 show s a pocke t penetromete r (Holt z e t al. , 1981 ) whic h ca n b e use d t o determin e
undrained shea r strengt h o f cla y soil s bot h i n th e laborator y an d i n th e field . Th e procedur e
consists i n pushing the penetromete r directl y int o the soi l an d notin g the strengt h marke d o n th e
calibrated spring .
304 Chapt e r 8
Figure 8 .3 5 Pock e t pen et rom et e r (PP) , a han d-hel d devic e whic h in dicat e s
un con fin ed com pres s iv e s t ren g t h (Cour t es y : Soilt es t , USA )
8 .2 7 TH E REL ATI ONSH I P B ETWEEN U NDRAI NE D SH EA R
STRENG TH AN D EFFECTI V E OV ERB U RDEN PRESSU RE
It ha s bee n discusse d i n previou s section s tha t th e shea r strengt h i s a functio n o f effectiv e
consolidation pressure . I f a relationshi p betwee n undraine d shea r strength , c
u
, an d effectiv e
consolidation pressure/?' can be established, we can determine c
u
i f //is known and vice versa . I f a
soil stratum i n nature is normally consolidated th e existing effective overburden pressurep
Q
' can b e
determined fro m th e know n relationship . Bu t i n overconsolidate d natura l cla y deposits , th e
preconsolidation pressur e /?/i s unknow n whic h ha s t o b e estimate d b y an y on e o f th e availabl e
methods. I f there i s a relationship between p
c
'and c
u
, c
u
can be determined fro m the known value of
p
c
". Alternatively, if c
u
i s known, p/can be determined. Som e o f the relationships between c
u
and
p' ar e presente d below . A typica l variatio n of undraine d shea r strengt h wit h dept h i s show n i n
Fig. 8.3 6 fo r bot h normall y consolidate d an d heavil y overconsolidate d clays . Th e highe r shea r
strength as shown i n Fig. 8.36(a ) for normally consolidated clay s close t o the ground surfac e i s due
to desiccation o f the to p laye r o f soil .
Skempton (1957 ) establishe d a relationship which may be expressed a s
^- = 0.10 + 0.004 /„ (8.55 )
Shear St ren g t h o f Soi l 305
Undrained shea r strengt h c
u
kN/m
2
50 0 5 0 10 0
N.C. Cla y Heavily O.C. Cla y
Figure 8.3 6 T ypica l variat ion s o f un drain e d s hea r s t ren g t h wit h dept h (Aft e r
B is hop an d Hen k el , 1962 )
He foun d a clos e correlatio n betwee n cjp' an d I a s illustrate d i n Fig . 8.37 . Thoug h th e
Eq. (8.55) wa s originall y mean t fo r normall y consolidate d clays , i t ha s bee n use d fo r
overconsolidated clay s also, //may be replaced b y p^as the existing effective overburde n pressur e
for normall y consolidate d clays , an d b y / ?
c
' for overconsolidate d clays . Pec k e t al. , (1974 ) ha s
extensively use d thi s relationshi p fo r determinin g preconsolidatio n pressur e p
c
'. Eq . (8.55 ) ma y
also be used for determining^'indirectly. If p^can be determined independently , the value of the
undrained shea r strengt h c
u
for overconsolidated clay s ca n be obtained fro m Eq. (8.55). The value s
of c s o obtaine d ma y b e checke d wit h th e value s determine d i n th e laborator y o n undisturbe d
samples o f clay .
Bjerrum and Simons (1960 ) proposed a relationship between cjp'and plasticit y inde x / a s
^• = 0.45(7,)* fo r I
p
> 5%
The scatte r i s expecte d t o be o f the orde r o f ± 25 percen t o f the compute d value .
(8.56)
u.o
0.4
0.2
A
I
^
i
1 _ ^
f
t .
^
^•
^
_ ^
'••
"*9
J
^
^
^
c
u
Ip' = 0.10 + 0.004 I
p
20 4 0 6 0 8 0
Plasticity Index , I
p
(%)
100 12 0
Figure 8.37 Rel at io n bet wee n cj p' an d plas t icit y in de x
306 Chapt er 8
Another relationshi p expresse d by them i s
- = 0.1 8 f o r / >0 . 5
(8
.57)
where I
{
i s the liquidit y index . The scatte r i s found to be of the orde r o f ± 3 0 percent .
Karlsson an d V iberg (1967 ) propose d a relationship as
— = 0.005w, fo r w, > 20 percen t
P'
(8. 58)
where \ v
l
i s the liqui d limi t in percent . The scatte r i s of the orde r o f ± 30 percent .
The enginee r ha s t o us e j udgmen t whil e selectin g an y on e o f th e form s o f relationship s
mentioned above .
cjp' Rati o Relate d t o Overconsolidatio n Rati o P
c
'lp
0
'
Ladd an d Foott (1974) presented a non-dimensional plot (Fig. 8.38) giving a relationship betwee n a non-
dimensional factor jV ,and Overconsolidation ratio OCR. Figure 8.38 is based o n direct simpl e shea r test s
carried out on five clays from differen t origins . The plot gives out a trend but requires further investigation.
The non-dimensiona l factor N
f
i s define d a s
(8.59)
oc
where p
Q
' = existing overburde n pressur e
OC = overconsolidate d
N C = normally consolidate d
From th e plo t i n Fig. 8.38 the shea r strengt h c o f overconsolidate d cla y ca n b e determine d
if p
Q
'and ( cJ p
0
')
N C
ar e known.
. 2 3 -
Upper l i mi t
Average l in e
Lower limi t
4 6 8
Overconsolidation rati o
10 12
Figure 8.3 8 Relat ion s hi p bet wee n N
f
an d Overcon s ol idat io n rat i o OCR (Ladd an d
Foot t , 1974)
Shear St ren g t h o f Soi l 30 7
Example 8 .15
A normall y consolidate d cla y wa s consolidate d unde r a stres s o f 315 0 lb/ft
2
, the n sheare d
undrained i n axia l compression. The principa l stres s differenc e at failure was 2100 lb/ft
2
, an d the
induced por e pressur e a t failur e wa s 184 8 lb/ft
2
. Determin e (a ) th e Mohr-Coulom b strengt h
parameters, i n term s o f bot h tota l an d effectiv e stresse s analytically , (b ) comput e ((T,/cr
3
), and
(<7
/
1
/cr
/
3
),, and (c ) determine the theoretical angl e of the failur e plane i n the specimen .
Solution
The parameters require d are: effective parameter s c' and 0', and total parameters c and 0.
(a) Given <T
3/
= 3150 lb/ft
2
, and (<TJ - a
3
)
f
= 2100 lb/ft
2
. The total principal stress at failure a
lf
is obtained fro m
ff
lf
= (CT j - aj
f
+ <7
3/
= 2100 + 3150 =525 0 lb/ft
2
Effective o
/
1/
= a
lf
- u
f
= 525 0 - 184 8 = 3402 lb/ft
2
°V
=
cr
3/
- "/= 3150 - 184 8 = 1302 lb/ft
2
Now cr j = <7
3
tan
2
(45° + 0/2) + 2c tan (45 ° + 0/2)
Since th e soi l i s normally consolidated, c = 0. As such
- - - t a n
2
( 4 5 ° -t an (4 5
or
T * I * " 1
210
°' I
21 0 0
1, 1 Co
Total 0 = si n
]
-= sin"
1
-= 14.5
5250 + 3150 840 0
^ . - i 210 0 . _ ! 2100 _ , _ „
Effective 0 = si n -= si n -= 26. 5
3402+1302 470 4
(b) The stres s ratio s at failure are
^L
=
5250 ^[
=
3402
= Z61
cr
3
315 0 <j'
3
130 2
(c) From Eq . (8.18 )
a = 45° + — = 45° + — = 58.25°
f
2 2
The above proble m can be solved graphicall y by constructing a Mohr-Coulomb envelope .
Example 8 .16
The following results were obtained at failure in a series of consolidated-undrained tests, wit h pore
pressure measurement, on specimens o f saturated clay. Determine th e values of the effective stres s
parameters c'and 0
x
by drawing Mohr circles .
a
3
k N/m
2
a , - o
3
k N/m
2
u
w
k N/m
2
150 19 2 8 0
300 34 1 15 4
450 50 4 22 2
308 Chapt er 8
300
200
100
800
o, kN/m
2
» -
Figure Ex . 8 .1 6
Solution
The value s of th e effectiv e principa l stresses <J \ and cr'
3
at failur e are tabulate d belo w
CT
S
k N/ m
2
150
300
450
a, k N/ m
2
342
641
954
cr'
3
k N/ m
2
70
146
228
a\ k N/ m
2
262
487
732
The Moh r circle s i n term s o f effectiv e stresse s an d th e failur e envelop e ar e draw n i n
Fig. Ex . 8.16. The shea r strengt h parameters a s measured ar e :
c' =16kN/ m
2
; 0' = 29°
Example 8 .1 7
The following results were obtained at failure in a series of triaxial tests on specimens of a saturated
clay initiall y 3 8 m m i n diamete r an d 7 6 m m long . Determin e th e value s o f th e shea r strengt h
parameters wit h respect t o (a) total stress, and (b) effective stress .
Type o f tes t <
(a) Undrained
(b) Drained
cr
3
k N/ m
2
200
400
600
200
400
600
Axi a l l oa d ( N )
222
215
226
467
848
1265
Axi a l compressi o n ( mm )
9.83
10.06
10.28
10.81
12.26
14.17
V ol ume chang e (cm
3
)
-
-
-
6.6
8.2
9.5
Solution
The principa l stres s differenc e at failur e in each tes t i s obtaine d b y dividin g the axia l load b y th e
cross-sectional are a of the specimen at failure. The corrected cross-sectiona l are a i s calculated from
Eq. (8.45). There is, of course, no volume change during an undrained test on a saturated clay. The
initial value s o f length , are a an d volum e for eac h specime n ar e h
Q
= 76 mm , A
0
= 11.3 5 cm
2
;
V
0
= 86. 0 cm
3
respectively.
Shear St ren g t h o f Soi l 309
'0 20 0 40 0 60 0 80 0 100 0 120 0
cr, kN/m
2
-
Figure Ex . 8.17
The Mohr circles a t failure and the corresponding failure envelopes for both series of tests are
shown in Fig. Ex . 8.17. In both cases the failure envelope is the line nearest t o the common tangent
to the Mohr circles . The total stres s parameter s representing the undrained strength of the clay are:
c
u
= 85 kN/m
2
; 0
u
= 0
The effectiv e stres s parameters , representin g th e drained strengt h of the clay, are :
c' = 20 kN/m
2
; 0 = 26°
a~ A/7//?
n
AWV
n
Are a (correct ed ) a, - a- a,
3 0 0 1 3 1
a
b
kN/ m
2
200
400
600
200
400
600
0.129
0.132
0.135
0.142 0.07 7
0.161 0.09 5
0.186 0.11 0
cm
2
13.04
13.09
13.12
12.22
12.25
12.40
kN/ m
2
170
160
172
382
691
1020
k N/ m
2
370
564
772
582
1091
1620
Example 8.1 8
An embankmen t i s being constructe d o f soi l whos e propertie s ar e c'-107 1 lb/ft
2
, 0' = 21° (all
effective stresses) , an d y= 99.85 lb/ft
3
. Th e pore pressur e parameter s a s determined fro m triaxia l
tests ar e A = 0.5, an d B = 0.9. Find the shear strengt h of the soi l at the base of the embankment just
after th e heigh t o f fil l ha s bee n raise d fro m 1 0 f t t o 2 0 ft . Assume tha t th e dissipatio n o f por e
pressure durin g this stage of construction i s negligible, and that the lateral pressur e a t any point is
one-half o f the vertical pressure .
Solution
The equation fo r pore pressur e i s [Eq. (8.51)]
A« = 5JAcr
3
+A( Ac Tj -Acr
3
)|
AcTj = V ertical pressure due t o 1 0 ft of fil l = 1 0 x 99.85 = 998.5 lb/ft
2
310 Chapt er s
9985
ACT, = ^^ = 499.25 lb/ft
2
3
2
Therefore, A w = 0.9[499.25 + 0.5 x 499.25] = 674 lb/ft
2
Original pressure, ^ = 10x99.85 = 998.5 lb/ft
2
Therefore, a' = <j
}
+ A<J
I
- A w
= 998. 5 + 998.5 -674 = 1323 lb/ft
2
Shear strength, s = c' + a'tan0' = 1071 + 1323tan21° = 1579 lb/ft
2
Example 8 .19
At a dept h o f 6 m belo w th e groun d surfac e a t a site , a van e shea r tes t gav e a torqu e valu e o f
6040 N-cm. The vane was 1 0 cm high and 7 cm across the blades. Estimat e the shear strength of the
soil.
Solution
Per Eq. (8.53 )
Torque (r )
c =
where T = 6040 N-cm, L = 1 0 cm, r = 3.5 cm.
substituting,
6040 , „
XT
, 7
c, = = 6.4 N / cm
2
- 6 4 kN/m
2
" 2 x 3.14 x 3.5
2
(10 + 0.67 x 3.5) ~ °
4 K1N/m
Example 8.20
A van e 11.2 5 c m long , an d 7. 5 c m i n diamete r wa s presse d int o sof t cla y a t th e botto m o f a
borehole. Torqu e wa s applie d t o caus e failur e of soil . Th e shea r strengt h o f cla y wa s foun d t o be
37 kN/m
2
. Determine the torque that was applied.
Solution
From Eq . (8.53) ,
Torque, T = c
u
[ 2nr
2
( L + 0.67r)] wher e c
u
= 37 kN/m
2
= 3.7 N/cm
2
- 3.7 [2 x 3.14 x (3.75)
2
(11.25 + 0.67 x 3.75)] = 4500 N -cm
8.28 G ENERA L COM M ENT S
One o f th e mos t importan t an d th e mos t controversia l engineerin g propertie s o f soi l i s it s shea r
strength. The behavior of soi l under external load depends on many factors suc h as arrangement of
particles i n th e soi l mass , it s mineralogica l composition , wate r content , stres s histor y an d man y
others. The types of laboratory test s to be performed o n a soil sampl e depend s upo n the type of soi l
Shear St ren g t h o f Soi l 31 1
and its drainage conditio n during the application of external loads i n the field. I t is practically very
difficult (i f no t impossible ) t o obtai n undisturbe d sample s o f granula r soil s fro m th e fiel d fo r
laboratory tests . I n suc h case s laborator y test s o n remolde d sample s ar e mostl y o f academi c
interest. Th e angl e o f shearin g resistanc e o f granula r soil s i s normall y determine d b y th e
relationships establishe d betwee n <j) an d penetratio n resistanc e i n th e field . Th e accurac y o f thi s
method i s sufficient fo r all practical purposes. The penetrometer use d may be standar d penetration
equipment or static cone penetrometer. Shear strengt h tests on cohesive soil s depend mostl y on the
accuracy wit h which undisturbed samples ca n be obtained from th e field .
Undisturbed samples are extracted i n sampling tubes of diameter 75 or 10 0 mm. Samples for
triaxial test s are extracted in the laboratory fro m the samples in the sampling tube s by using sampl e
extractors. Sample s ma y b e disturbe d a t bot h th e stage s o f extraction . I f w e ar e dealin g wit h a
highly overconsolidated cla y the disturbance is greater a t both the stages. Beside s ther e i s another
major disturbance which affects th e test results very significantly. A highly overconsolidated clay is
at equilibrium in its in-situ conditions. The overconsolidation pressures of such soils could be of the
order 100 0 kPa (10 tsf) o r more. The standar d penetration value N in such deposits coul d be 10 0 or
more. The shear strengt h of such a soil under the in-situ condition coul d be in the order of 600 kPa
or more . Bu t i f a n undisturbe d sampl e o f suc h a soi l i s teste d i n standar d triaxia l equipment , the
shear strengt h under undrained conditions coul d be ver y low. This i s mostly due t o the cracks that
develop o n th e surfac e o f th e sample s du e t o th e relie f o f th e in-situ overburde n pressur e o n th e
samples. Possibl y th e onl y way of obtaining the in-situ strengt h i n the laborator y i s t o bring back
the stat e o f th e sampl e t o it s origina l fiel d conditio n b y applyin g all-aroun d pressure s o n th e
samples equa l t o the estimated overconsolidatio n pressures . Thi s ma y not be possibl e i n standard
triaxial equipment due to its limitations. The present practice i s therefore t o relate the in-situ shea r
strength t o some of the field tests such as standard penetration tests, static cone penetration test s or
pressuremeter tests .
8.29 QU ESTI ON S AND PROB L EM S
8.1 Explai n Coulomb' s equation for shear strengt h of a soil. Discuss the factors that affect th e
shear strengt h parameters of soil .
8.2 Explai n th e metho d o f drawin g a Moh r circl e fo r a cylindrica l sampl e i n a triaxia l test .
Establish th e geometrica l relationship s betwee n th e stresse s o n th e failur e plan e an d
externally applie d principal stresses.
8.3 Classif y th e shea r test s base d o n drainag e conditions . Explai n ho w th e por e pressur e
variation an d volume change tak e place during these tests . Enumerate the field conditions
which necessitat e each o f these tests .
8.4 Wha t ar e th e advantage s and disadvantage s of a triaxial compression tes t i n compariso n
with a direct shea r test ?
8.5 Fo r what types of soils, will the unconfmed compressio n test give reliable results? Draw a
Mohr circle for this test. How do you consider the change in the area of the specimen whic h
takes place durin g the test i n the final results?
8.6 Wha t type s o f fiel d test s ar e necessar y fo r determinin g the shea r strengt h parameter s o f
sensitive clays ? Derive th e relationship s that ar e usefu l fo r analyzin g the observation s o f
this test.
8.7 Fo r loose and dense sands , draw the following typical diagrams :
(i) deviator stress vs . linear strain, and
(ii) volumetri c strain vs. linear strain. Discuss them.
8.8 Discus s the effect s of drainage conditions on the shear strengt h parameters of clay soil .
8.9 A direct shear tes t on specimens o f fine san d gave the following results :
312 Chapt e r s
Normal stres s (lb/ft
2
) 210 0 370 0 450 0
Shearing stres s (lb/ft
2
) 97 0 170 0 208 0
Determine :
(i) the angl e o f interna l frictio n o f the soil , an d
(ii) shea r strengt h of the soi l a t a depth of 1 5 ft fro m th e groun d surface .
The specifi c gravit y of solids i s 2.65, voi d ratio 0.7 and the ground water table is at a depth
of 5 f t from the groun d surface. Assume th e soi l above groun d wata r tabl e i s saturated .
A specime n o f clea n san d whe n subjecte d t o a direc t shea r tes t faile d a t a stres s o f
2520 lb/ft
2
whe n the norma l stres s wa s 3360 lb/ft
2
.
Determine:
(i) the angl e of interna l friction, an d
(ii) th e deviato r stres s a t whic h th e failur e wil l tak e place , i f a cylindrica l sampl e o f th e
same san d i s subjecte d t o a triaxial test wit h a cel l pressur e o f 2000 lb/ft
2
. Fin d th e angl e
made b y the failur e plane wit h the horizontal.
A specimen o f fine sand, when subjecte d t o a drained triaxial compressio n test , failed at a
deviator stres s o f 8400 lb/ft
2
. I t was observe d tha t the failur e plane mad e a n angl e o f 30°
with the axi s of the sample. Estimat e the valu e of the cel l pressur e t o which this specimen
would have been subjected .
8.12 A specimen o f sandy silt, when subjected to a drained triaxial test failed at major and minor
principal stresses o f 12 0 kN/m
2
and 50 kN/m
2
respectively. At what value of deviator stres s
would anothe r sampl e o f th e sam e soi l fail,i f i t wer e subjecte d t o a confinin g pressur e o f
75 kN/m
2
?
8.13 A san d i s hydrostaticall y consolidated i n a triaxia l tes t apparatu s t o 882 0 lb/ft
2
an d the n
sheared wit h the drainage valve s open. At failure, (c^ - <7
3
) i s 22 kips/ft
2
. Determin e th e
major an d mino r principal stresses a t failure and the angl e o f shearing resistance .
8.14 Th e same san d as in Prob. 8.13 is tested in a direct shea r apparatus unde r a normal pressur e
of 8820 lb/ft
2
. The sampl e fail s whe n a shear stres s of 5880 lb/ft
2
i s reached. Determin e th e
major an d mino r principal stresses a t failur e an d the angl e o f shearin g resistance . Plo t the
Mohr diagram .
8.15 A sample o f dense san d tested i n a triaxial CD test failed along a well defined failure plane
at an angle of 66° with the horizontal. Find the effective confining pressure of the test i f the
principal stres s differenc e at failure was 10 0 kPa.
8.16 A draine d triaxia l tes t i s performe d o n a san d wit h o^ , = 10. 5 kips/ft
2
. A t failur e
CTj'/cr^ = 4 . Find o^, , (<7 j - <7
3
)
f
and 0'.
8.17 I f the test o f Prob. 8.1 6 had been conducte d undrained , determine ( <J
l
- er
3
)
f
, 0' , 0
tota[
and
the angl e of the failure plane in the specimen. The por e wate r pressur e u = 2100 lb/ft
2
.
8.18 I f the tes t o f Prob. 8.1 6 i s conducte d a t an initial confining pressure o f 21 kips/ft
2
, estimat e
the principal stress difference and the induced pore pressur e a t failure.
8.19 A sample of silty sand was tested consolidated draine d in a triaxial cell wher e cr
3
= 475 kPa.
If th e tota l axia l stres s a t failure was 160 0 kP a whil e <7
3
= 475 kPa , comput e th e angl e of
shearing resistanc e an d th e theoretica l orientatio n of th e failur e plan e wit h respec t t o th e
horizontal.
8.20 A drained triaxia l test is to be performed o n a uniform dense sand wit h rounded grains. The
confining pressur e i s 4200 lb/ft
2
. At what vertical pressur e wil l the sampl e fail ?
8.21 Comput e the shearing resistance along a horizontal plane a t a depth of 6.1 mi n a deposit of
sand. The water table is at a depth of 2.13 m. The uni t weight of moist sand above the water
Shear St ren g t h o f Soi l 31 3
table is 18.5 4 kN/m
3
and the saturated weigh t of submerged san d i s 20.11 kN/m
3
. Assume
that the san d i s drained freel y and ( f)
d
fo r th e we t san d i s 32° .
A sampl e o f dr y san d wa s teste d i n a direc t shea r devic e unde r a vertica l pressur e o f
137.9 kN/m
2
. Comput e th e angl e o f interna l frictio n o f th e sand . Assum e shearin g
resistance = 96.56 kN/m
2
.
The san d i n a deep natural deposit ha s a n angl e o f interna l friction o f 40° i n the dr y stat e
and dr y uni t weight of 17.2 8 kN/m
3
. I f th e wate r tabl e i s a t a dept h o f 6. 1 m, wha t is th e
shearing resistanc e o f the material along a horizontal plane at a depth of 3.05 m? Assume:
G^ = 2.68 .
8.24 Compute the shearing resistance under th e conditions specifie d i n Prob. 8.23, i f the wate r
table i s at the groun d surface.
8.25 A draine d triaxia l tes t wa s conducte d o n dens e san d wit h a confinin g pressur e o f
3000 lb/ft
2
. Th e sampl e faile d a t an added vertica l pressure o f 11,00 0 lb/ft
2
. Comput e th e
angle o f interna l frictio n 0 an d th e angl e o f inclinatio n a o f th e failur e plane s o n th e
assumption that Coulomb' s la w is valid.
A saturated sampl e of dense sand was consolidated i n a triaxial test apparatus at a confining
pressure o f 143. 6 kN/m
2
. Furthe r drainage wa s prevented. Durin g the additio n o f vertical
load, the pore pressure in the sample was measured. At the instant of failure, it amounted to
115 kN/m
2
. The added vertical pressur e at this time was 138.85 kN/m
2
. What was the value
of 0 for the sand ?
An undraine d triaxial tes t wa s carried ou t o n a sampl e o f saturate d cla y wit h a confining
pressure o f 200 0 lb/ft
2
. Th e unconfme d compressive strengt h obtaine d wa s 730 0 lb/ft
2
.
Determine th e exces s vertica l pressur e i n additio n t o th e all-roun d pressure require d t o
make th e sampl e fail .
A series of undrained triaxial tests o n samples o f saturated soi l gav e the following results
cr
3
,kN/m
2
10 0 20 0 30 0
u, kN/m
2
2 0 7 0 13 6
(<TJ -cr
3
), kN/m
2
29 0 40 0 53 4
Find the values of the parameters c and 0
(a) with respect t o total stress , an d (b) wit h respect t o effective stress .
When an unconfmed compression tes t was conducted on a specimen of silty clay, it showed
a strengt h of 315 0 l b/ f t
2
. Determin e the shea r strengt h parameters o f the soi l i f the angl e
made b y the failure plane wit h the axi s of the specimen wa s 35°.
A normall y consolidate d cla y wa s consolidate d unde r a stres s o f 15 0 kPa, the n sheare d
undrained i n axial compression. Th e principa l stress differenc e at failur e wa s 10 0 kPa and
the induced por e pressure at failure was 88 kPa. Determine analytically (a) the slopes of the
total and effective Mohr stress envelopes, and (b) the theoretical angl e of the failure plane.
A normall y consolidate d cla y sampl e wa s consolidate d i n a triaxial shea r apparatu s a t a
confining pressur e of 21 kips/ft
2
an d then sheared under undrained condition. The ( <J
l
- <7
3
)
at failure wa s 2 1 ki ps/ ft
2
. Determin e 0
CM
and a .
8.32 A CD axial compression triaxia l test on a normally consolidated cla y failed along a clearly
defined failur e plane of 57°. The cel l pressur e during the test wa s 4200 lb/ft
2
. Estimat e 0' ,
the maximum o
/
/o
/
3
, and the principal stress difference at failure.
Two identical sample s o f sof t saturate d normall y consolidate d cla y wer e consolidate d t o
150 kPa i n a triaxial apparatus . One specime n wa s sheare d unde r drained conditions , an d
the principa l stres s differenc e a t failur e wa s 30 0 kPa . Th e othe r specime n wa s sheare d
undrained, and the principal stress difference at failure was 200 kPa. Determine 0, an d 0 .
314 Chapt e r s
8.34 Whe n a triaxia l compressio n tes t wa s conducte d o n a soi l specimen , i t faile d a t a n axia l
pressure o f 735 0 lb/ft
2
. I f th e soi l ha s a cohesion o f 105 0 lb/ft
2
an d a n angl e o f internal
friction o f 24° , wha t wa s th e cel l pressur e o f th e test ? Als o fin d th e angl e mad e b y th e
failure plan e wi t h th e directio n of <7
3
.
8.35 Give n the following triaxial test data, plot the result s in a Mohr diagra m an d determine 0 .
cr
3
k N/m
2
Pea k <
69
138
207
7, k N/m
2
190
376
580
<T
3
k N/m
2
276
345
414
Peak a , k N/m
2
759
959
1143
8.36 Tw o set s o f triaxia l test s wer e carrie d ou t o n two sample s o f glacia l silt . Th e result s ar e
(a) <7
n
= 400 kN/m
2
, <T
31
= 10 0 kN/m
2
(b ) cr
12
= 680 kN/m
2
, cr
32
= 200 kN/m
2
.
The angl e o f th e failur e plan e i n bot h test s wa s measure d t o b e 59° . Determin e th e
magnitudes of 0 and c.
8.37 A triaxial compressio n tes t o n a cylindrical cohesive sampl e gav e the followin g effective
stresses:
(a) Major principal stress, a
/
l
= 46,000 lb/ft
2
(b) Mino r principal stress, tr'
3
= 14,50 0 lb/ft
2
(c) The angl e of inclination of the rupture plane = 60° wit h the horizontal.
Determine analyticall y the (i) normal stress, (ii) the shear stress, (iii ) the resultant stress on
the rupture plane through a point, and (iv) the angle of obliquity of the resultant stress with
the shear plane .
8.38 Give n the result s of two set s of triaxial shear tests :
<7
n
= 180 0 kN/m
2
; cr
31
= 100 0 kN/m
2
<7
12
= 2800 kN/m
2
; d
32
= 2000 kN/m
2
Compute 0 an d c.
8.39 Wha t is the shear strength in terms of effective stres s on a plane within a saturated soi l mas s
at a point where th e norma l stres s i s 295 kN/m
2
an d the pore water pressur e 12 0 kN/rn
2
?
The effectiv e stress parameters for th e soi l ar e c' = 12 kN/m
2
, an d 0' = 30° .
8.40 Th e effective stress parameters for a fully saturate d clay are known to be c' = 315 lb/ft
2
an d
0' = 29° . I n a n unconsolidated-undraine d triaxial tes t o n a sampl e o f th e sam e cla y th e
confining pressur e wa s 525 0 lb/ft
2
an d th e principa l stres s differenc e a t failur e wa s
2841 lb/ft
2
. Wha t was th e value of the por e wate r pressure i n the sampl e a t failure?
8.41 I t i s believe d tha t the shea r strengt h o f a soi l unde r certain condition s i n th e fiel d wil l be
governed b y Coulomb' s law, wherei n c - 40 2 lb/ft
2
, an d 0 = 22°. Wha t minimu m lateral
pressure woul d b e require d t o preven t failur e o f th e soi l a t a give n poin t i f th e vertica l
pressure wer e 9000 lb/ft
2
?
8.42 Th e followin g data refe r t o thre e triaxia l test s performe d o n representativ e undisturbe d
samples:
T es t Cel l pres s ur e k N/m
2
Ax ia l dia l readin g
(divis ion ) a t fail ur e
1 5 0 6 6
2 15 0 10 6
3 25 0 14 7
Shear St ren g t h o f Soi l 31 5
The loa d dia l calibratio n facto r i s 1. 4 N pe r division . Eac h sampl e i s 7 5 m m lon g an d
37.5 mm diameter . Fin d b y graphica l means , th e valu e of th e apparen t cohesio n an d th e
angle of internal friction for thi s soil .
8.43 I n a triaxial test a soil specimen wa s consolidated unde r an allround pressure of 1 6 kips/ft
2
and a bac k pressur e o f 8 kips/ft
2
. Thereafter, unde r undraine d conditions , th e allroun d
pressure wa s raise d t o 1 9 kips/ft
2
, resultin g i n a pore wate r pressure of 10. 4 kips/ft
2
, the n
(with th e confinin g pressur e remainin g a t 1 9 kips/ft
2
) axia l loa d wa s applie d t o giv e a
principal stres s differenc e o f 12. 3 kips/ft
2
an d a por e wate r pressur e o f 13. 8 kips/ft
2
.
Calculate the values of the pore pressur e coefficient s A an d B.
8.44 I n an in-situ van e test on a saturated cla y a torque o f 35 N-m is required t o shear the soil .
The van e i s 5 0 mm i n diamete r an d 10 0 mm long . What i s the undraine d strengt h o f th e
clay?
8.45 I n a vane test a torque of 46 N-m is required t o cause failur e of the vane in a clay soil . The
vane is 150 mm long and has a diameter of 60 mm. Calculate the apparent shear strengt h of
the soi l fro m thi s tes t whe n a vane of 200 mm long and 90 mm i n diameter i s used i n th e
same soil and the torque at failure was 138 N-m. Calculate the ratio of the shear strengt h of
the clay i n a vertical directio n t o that in the horizontal direction .
8.46 A vane of 80 mm diameter an d 16 0 mm height has been pushe d into a soft cla y stratu m at
the bottom o f a bone hole . The torque required t o rotate th e vane was 76 N-m. Determin e
the undraine d shear strengt h of th e clay. After the tes t the van e wa s rotated severa l time s
and the ultimat e torque wa s found t o be 50 N-m. Estimate the sensitivit y of the clay.
8.47 A normall y consolidate d deposi t o f undisturbed clay extend s t o a dept h of 1 5 m from th e
ground surfac e wit h the ground water level a t 5 m depth from groun d surface . Laborator y
test o n th e cla y give s a plasticit y inde x o f 68%, saturate d an d dr y uni t weight s o f
19.2kN/m
3
an d 14. 5 kN/m
3
respectively . A n undisturbe d specime n fo r unconfine d
compressive strengt h i s take n a t 1 0 m depth . Determin e th e unconfine d compressiv e
strength of the clay.
8.48 A triaxial sample was subjected to an ambient pressure of 200 kN/m
2
, and the pore pressur e
recorded wa s 5 0 kN/m
2
a t a full y saturate d state . The n th e cel l pressur e wa s raise d t o
300 kN/m
2
. Wha t woul d be th e valu e of por e pressure ? At thi s stage a deviator stres s o f
150 kN/m
2
was applied t o the sample. Determine the pore pressure assumin g pore pressure
parameter A = 0.50 .
In a triaxial test o n a saturated clay , the sampl e wa s consolidated unde r a cell pressur e of
160 kN/m
2
. Afte r consolidation , th e cel l pressur e wa s increase d t o 35 0 kN/m
2
, an d th e
sample wa s then failed unde r undrained condition. I f the shea r strengt h parameter s o f the
soil ar e c' = 15.2 kN/m
2
, 0 " = 26°, B = 1, and A
f
= 0.27 , determin e th e effectiv e majo r and
minor principa l stresse s a t th e time o f failur e of th e sample .
A thi n laye r o f sil t exist s a t a dept h o f 1 8 m belo w th e surfac e o f th e ground . The sol i
above this level has an average dry unit weight of 15. 1 kN/m
3
and an average water content
of 36%. The water table is almost at the ground surface level. Tests on undisturbed samples
of the sil t indicate the following values:
c
u
= 45 kN/m
2
, 0
u
= 18° , c' = 36 kN/m
2
and 0' = 27°.
Estimate th e shearin g resistance o f the sil t on a horizontal plane whe n (a ) the shea r stres s
builds up rapidly, and (b) the shear stres s builds up slowly.
CHAPTER 9
SOIL EXPLORATION
9 .1 I NTRODU CTI O N
The stabilit y of the foundation of a building, a bridge, a n embankment or any other structur e built
on soi l depend s o n th e strengt h an d compressibilit y characteristic s o f th e subsoil . The fiel d an d
laboratory investigation s required t o obtai n the essential informatio n on the subsoi l i s called Soil
Exploration o r Soil Investigation. Soi l exploration happens to be one of the most important parts of
Foundation Engineerin g an d a t th e sam e tim e th e mos t neglecte d par t o f it . Terzagh i i n 195 1
(Bjerrum, et al., 1960 ) ha d rightl y remarked, tha t "Building foundations have always been treated
as step children". Hi s remark s ar e relevan t eve n today . Th e succes s o r failur e o f a foundation
depends essentiall y o n th e reliabilit y o f th e variou s soi l parameter s obtaine d fro m th e fiel d
investigation an d laborator y testing , an d use d a s a n inpu t int o th e desig n o f foundations .
Sophisticated theories alone will not give a safe and sound design.
Soil exploration i s a must in the present age for the design of foundations of any project. The
extent of the exploration depends upon the magnitude and importance of the project. Project s suc h
as buildings , power plants , fertilize r plants , bridge s etc. , ar e localize d i n area l extent . Th e are a
occupied b y suc h project s ma y var y fro m a fe w squar e meter s t o man y squar e kilometers .
Transmission lines , railwa y lines , roads an d other suc h project s extend alon g a narrow path. The
length o f suc h project s ma y b e severa l kilometers . Eac h projec t ha s t o b e treate d a s pe r it s
requirements. Th e principl e o f soi l exploratio n remain s th e sam e fo r al l th e project s bu t th e
program an d methodology ma y var y from projec t t o project.
The element s o f soi l exploratio n depen d mostl y o n th e importanc e an d magnitud e of th e
project, bu t generally shoul d provide the following:
1. Informatio n to determine the type of foundation required such as a shallow or deep foundation.
2. Necessar y informatio n wit h regards t o the strengt h and compressibilit y characteristic s o f
the subsoi l t o allow the Design Consultan t to make recommendations on the saf e bearin g
pressure or pile load capacity.
317
318 Chapt er 9
Soil exploratio n involve s broadly the following :
1. Plannin g of a program fo r soi l exploration .
2. Collectio n o f disturbe d and undisturbe d soi l o r rock sample s fro m th e hole s drille d i n the
field. Th e numbe r and depth s of holes depen d upo n th e project .
3. Conductin g al l th e necessar y in-situ test s fo r obtainin g the strengt h an d compressibilit y
characteristics o f the soi l or rock directl y or indirectly.
4. Stud y of ground-wate r conditions and collection o f wate r samples fo r chemica l analysis .
5. Geophysica l exploration , if required.
6. Conductin g al l the necessar y test s o n the sample s o f soi l /rock an d water collected .
7. Preparatio n o f drawings, charts, etc .
8. Analysi s of the dat a collected .
9. Preparatio n o f report .
9.2 B ORI N G O F H OL E S
Auger M etho d
H and Operate d Augers
Auger borin g i s the simples t o f the methods. Han d operate d or power drive n auger s ma y be used .
Two types o f hand operate d auger s ar e i n use as shown i n Fig. 9. 1
The depths of the holes are normally limited to a maximum of 10 m by this method. These augers
are generally suitabl e for all types of soil above the wate r table but suitable only in clayey soil below the
water tabl e (excep t fo r th e limitation s given below) . A strin g o f dril l rod s i s use d fo r advancin g th e
boring. Th e diameter s o f th e hole s normall y var y fro m 1 0 t o 2 0 cm. Han d operate d auger s ar e no t
suitable i n ver y stif f t o har d cla y no r i n granula r soil s belo w th e water table . Han d augerin g i s no t
practicable i n dense san d nor i n sand mixed with gravel even if the strata lies above the water table .
Power Drive n Augers
In man y countries th e us e o f power drive n continuous flight auger s i s the mos t popula r metho d o f
soil exploration fo r boring holes . The flights ac t as a screw conveyo r t o bring the soi l t o the surface .
Helical auger
Post hol e auger
Extension
rod
Figure 9. 1 Han d aug er s
Soil Ex pl orat ion 319
Sampler ro d
Sampler
(a) (b)
Figure 9. 2 Hollow-s t e m aug e r
(a) Plug g e d whil e advan cin g t h e aug er , an d (b ) plug rem ove d an d s am ple r in s ert e d
t o s am pl e s oi l belo w aug e r
This method ma y be used i n al l types of soi l including sandy soil s belo w th e water tabl e but is not
suitable i f the soi l i s mixe d wit h gravel , cobble s etc . The centra l ste m o f the auge r fligh t ma y b e
hollow or solid. A hollow ste m is sometimes preferre d sinc e standard penetration test s or sampling
may be done throug h the ste m withou t liftin g th e auge r fro m it s position i n the hole . Besides , th e
flight o f auger s serve s th e purpos e o f casin g th e hole . Th e hollo w ste m ca n b e plugge d whil e
advancing th e bor e an d th e plu g ca n b e remove d whil e takin g sample s o r conductin g standar d
penetration test s (t o be described) a s shown in Fig. 9.2 . The drilling rig can be mounted on a truck
or a tractor. Holes ma y be drilled by thi s method rapidl y t o depths o f 60 m or more .
Wash B orin g
Wash boring i s commonl y use d fo r boring holes . Soi l exploratio n belo w th e ground water tabl e i s
usually ver y difficul t t o perfor m b y mean s o f pit s or auger-holes . Was h boring i n suc h cases i s a
very convenient method provided the soil is either sand, silt or clay. The method i s not suitable if the
soil i s mixed wit h gravel o r boulders .
Figure 9. 3 shows the assembl y fo r a wash boring. To start with , the hol e i s advanced a shor t
depth by auger and then a casing pipe is pushed t o prevent the sides from cavin g in. The hole is then
continued by the us e of a chopping bi t fixed at the end of a string of hollow dril l rods . A stream of
water under pressure i s forced through the rod and the bit into the hole, which loosens the soil as the
water flows u p aroun d the pipe. The loosene d soi l i n suspension i n water i s discharged int o a tub.
The soi l in suspension settle s down in the tub and the clean water flows int o a sump which is reused
for circulation . The motiv e powe r fo r a was h borin g i s eithe r mechanica l o r ma n power . Th e bi t
which i s hollow i s screwed t o a string of hollow dril l rods supporte d o n a tripod b y a rope o r stee l
cable passin g ove r a pulley and operated b y a winch fixed o n one of the legs o f the tripod .
320 Chapt er 9
Swivel j oi nt
Rope
Pulley
Tripod
Winch
Suction pipe
Chopping bi t
Figure 9. 3 Was h borin g
The purpos e of wash boring is to drill holes only and not to make use of the disturbed washed
materials for analysis. Whenever an undisturbed sample i s required at a particular depth, the boring
is stopped, an d the chopping bit i s replaced b y a sampler. The sampler i s pushed into the soi l at the
bottom o f the hol e and th e sampl e is wi t hdrawn .
Rotary Drillin g
In th e rotar y drilling method a cutter bit o r a core barrel wit h a coring bi t attache d t o the end o f a
string o f dril l rod s i s rotate d b y a powe r rig . Th e rotatio n o f th e cuttin g bi t shear s o r chip s th e
material penetrated an d the material is washed ou t of the hole by a stream of water j ust as in the case
of a was h boring . Rotar y drillin g i s use d primaril y fo r penetratin g th e overburde n betwee n th e
levels of which samples ar e required. Coring bits, on the other hand, cut an annular hole aroun d an
intact core which enters th e barrel an d i s retrieved. Thus th e core barrel i s used primaril y in rocky
strata t o get rock samples .
As the rods wi t h th e attached bi t or barrel ar e rotated, a downward pressure i s applied t o the
drill string to obtai n penetration, and drilling fluid unde r pressure i s introduced int o the bottom of
the hol e through the hollow dril l rods and the passages i n the bit or barrel. Th e drilling fluid serves
the dual funct i on o f cooling the bi t as i t enters the hol e an d removing the cuttings from the bottom
of the hole as it returns to the surface in the annular space between the drill rods an d the walls of the
hole. I n an uncased hole, the drilling flui d als o serves t o support the wall s of the hole. When borin g
Soil Ex plorat io n 321
Water swive l
Tower mas t
Swivel hol e
Stand pip e
Yoke and Kell y driv e
Rotary driv e
Hoisting dr u
Overflow ditc h
Bit, replaced by sampling spoo n
during sampling operation s
Figure 9. 4 Rot ar y drillin g ri g (Aft e r Hvors lev , 1949 )
in soil, the drilling bit is removed and replaced by a sampler when sampling is required, but in rocky
strata th e coring bi t i s use d t o obtai n continuous rock samples . The rotar y drillin g rig o f th e type
given i n Fig. 9. 4 can als o be used for was h boring and auger boring.
Coring Bit s
Three basi c categories o f bits are in use. They are diamond, carbide insert , and saw tooth. Diamond
coring bit s ma y b e o f th e surfac e se t or diamon d impregnate d type . Diamond corin g bit s ar e the
most versatil e of all the coring bits since they produce high quality cores i n rock material s ranging
from sof t t o extremely hard. Carbide insert bits use tungsten carbide in lieu of diamonds. Bit s of this
type are used t o core sof t t o medium har d rock. They ar e less expensive than diamond bit s but the
rate o f drillin g i s slowe r tha n wit h diamon d bits . I n saw-toot h bits , th e cuttin g edge comprise s a
series o f teeth. The teet h ar e face d an d tipped wit h a hard metal allo y suc h a s tungsten carbide t o
provide wea r resistanc e an d thereb y increas e th e lif e o f th e bit . These bit s ar e les s expensiv e but
normally use d t o core overburden soi l an d ver y sof t rock s only.
Percussion Drillin g
Percussion drillin g is another method of drilling holes. Possibl y thi s is the only method fo r drilling
in river deposits mixe d with hard boulders of the quartzitic type. In this method a heavy drilling bit
322
Chapt er 9
is alternatively raised an d dropped i n such a manner that it powders th e underlying materials which
form a slurry with water an d ar e removed a s the boring advances .
9.3 SAM PL I N G I N SOI L
Soils me t i n natur e are heterogeneous i n character wit h a mixture of sand, sil t and clay i n differen t
proportions. I n water deposits, there are distinct layers of sand, silt and clay of varying thicknesses
and alternatin g with depth. We can brin g al l th e deposit s o f soi l unde r two distinc t groups fo r th e
purpose o f study, namely, coarse graine d an d fine graine d soils . Soil s wit h particles o f size coarse r
than 0.07 5 mm ar e brough t unde r the categor y o f coars e graine d an d thos e fine r tha n 0.075 mm
under fine graine d soils . Sandy soil falls i n the group of coarse grained , and sil t and clay soil s in the
fine grained group . A satisfactory design of a foundation depends upon the accuracy wit h which the
various soi l parameter s require d fo r th e desig n ar e obtained. Th e accurac y o f the soi l parameter s
depends upo n the accuracy wit h which representative soi l sample s ar e obtaine d from the field.
Disturbed Sample s
Auger samples ma y be used t o identify soi l strat a and for field classification tests, but are not usefu l
for laborator y tests . Th e cutting s o r choppin g fro m was h boring s ar e o f littl e valu e excep t fo r
indicating change s i n stratificatio n t o th e borin g supervisor . Th e materia l brough t u p wit h th e
drilling mu d i s contaminated an d usuall y unsuitable even fo r identification.
For prope r identificatio n and classificatio n of a soil , representativ e sample s ar e require d a t
frequent interval s along th e bore hole . Representativ e sample s ca n usuall y be obtaine d b y driving
or pushing into the strat a in a bore hol e an open-ended samplin g spoon called a split spoon sample r
(Fig. 9.5) whic h i s used fo r conducting standar d penetratio n test s (refe r Sect . 9.5). I t is made up of
a driving shoe an d a barrel. The barrel i s split longitudinally into two halves wit h a coupling at the
upper end for connection to the drill rods. The dimensions of the split spoon ar e given in Fig. 9.5. In
a tes t th e sample r i s drive n int o the soi l a measure d distance . Afte r a sampl e i s taken, th e cutting
shoe an d th e couplin g ar e unscrewe d an d th e tw o halve s o f th e barre l separate d t o expos e th e
material. Experienc e indicate s tha t sample s recovere d b y thi s devic e ar e likel y t o b e highl y
disturbed an d as such can onl y be used a s disturbed samples . The samples s o obtained ar e stored i n
glass o r plastic jars o r bags, reference d an d sent t o the laborator y fo r testing. If spoon sample s ar e
to be transporte d t o the laboratory withou t examination in the field, the barrel i s often cored out to
hold a cylindrical thin-walled tube known as a liner. After a sample has been obtained, the liner and
the sampl e i t contains are removed fro m the spoon an d the ends ar e sealed wit h caps o r wit h metal
discs an d wax . Sample s o f cohesionles s soil s belo w th e wate r tabl e canno t b e retaine d i n
conventional samplin g spoons wit hou t the addition of a spring core catcher.
3/4"
(19 mm)
3"
(76.2 mm ) 24" (60.96 cm)
8 Acme thread s per inch
Water por t 1/16 " di a
Make fro m 2 seamless
tubes t o give ful l diamete r
(38.1 mm )
3/8"(34.93 mm)
Tool stee l driv e shoe chise l
point tempered a t edg e
3/4" di a steel bal l
( 19mm)
Thread fo r
wash pip e
or A rod s
Figure 9. 5 Spli t barre l s am ple r fo r s t an dar d pen et rat io n t es t
Soil Ex plorat io n 323
Sampler head
Ball check valve
Rubber seat
Thin wall
sampling tube
Sampler
Many type s o f sampler s ar e i n us e fo r
extracting th e s o calle d undisturbe d samples .
Only two types of samplers ar e described here .
They are ,
1. Ope n driv e sampler,
2. Pisto n sampler .
Open Driv e Sample r
The wal l thicknes s o f th e ope n driv e sample r
used fo r samplin g ma y b e thi n o r thic k
according t o the soil conditions met in the field.
The samplers are made of seamless stee l pipes .
A thin-walle d tube sample r i s called a s shelby
tube sample r (Fig . 9.6) , consist s o f a thi n wal l
metal tub e connecte d t o a sample r head . Th e
sampler hea d contain s a bal l chec k valv e and
ports whic h permi t th e escap e o f wate r o r ai r
from th e sampl e tub e a s th e sampl e enter s it .
The thi n wal l tube , whic h i s normall y forme d
from 1/1 6 t o 1/ 8 inch metal , i s drawn i n a t the
lower en d an d i s reame d s o tha t th e insid e
diameter o f th e cuttin g edg e i s 0. 5 t o 1. 5
percent les s tha n that of the inside diamete r o f the tube. The exac t percentage i s governed b y th e
size an d wal l thickness o f the tube. The wal l thickness i s governed by the area ratio, A
r
, which is
defined a s
Figure 9. 6 Thi n wal l Shelby tube
s am pler
d
2
-d
2
A
r
= ° ' x 100 percent ,
(9.1)
where, d
i
- insid e diameter ,
d
o
- outsid e diameter .
A
r
is a measure of the volume of the soil displacement t o the volume of the collected sample . Well -
designed samplin g tubes have an area ratio of about 10 percent. However , the area ratio may have to
be muc h mor e tha n 1 0 percent whe n samples ar e t o be taken in very stif f t o hard clay soil s mixed
with stone s t o prevent the edges o f the sampling tubes from distortio n during sampling.
Sample Extraction
The thin-wal l tub e sample r i s primarily use d fo r samplin g i n sof t t o mediu m stif f cohesiv e soils.
The wal l thickness has t o be increased i f sampling is to be done in very stiff t o hard strata. For best
results i t is better t o push the sampler staticall y int o the strata . Sampler s ar e driven int o the strat a
where pushing is not possible or practicable. The procedure of sampling involves attaching a string
of drill rods t o the sampler tube adapter and lowering the sampler t o rest on the bottom of the bor e
hole which was cleaned of loose materials i n advance. The sampler is then pushed or driven into the
soil. Over driving or pushing shoul d be avoided. After th e sampler i s pushed t o the required depth,
the soi l a t the bottom o f the sample r i s sheared of f by giving a twist to the dril l rod a t the top. The
sampling tube is taken out of the bore hole and the tube is separated fro m th e sampler head. The top
and botto m o f th e sampl e ar e either seale d wit h molten wa x o r cappe d t o preven t evaporation o f
moisture. The samplin g tube s ar e suitably referenced for later identification.
324
Chapt er 9
Bore hol e
Drill
Sampler hea d
Piston
Pressure cylinder
Casing pipe
A Hollo w
piston rod
Fixed pisto n
Sampler hea d
Air vent
Water under
pressure
Water return_
circulation
Hole i n _
piston rod
_Thin- walled _
sampling tube
(a) (b) (c)
Figure 9.7 Os t erber g Pisto n Sample r (a ) Sampler i s set i n drille d hole , (b ) Sampl e tube
is pushe d hydraulicall y int o the soil , (c ) Pres s ur e is releas ed through hol e i n pisto n rod .
Piston Sample r (Afte r Osterber g 1 9 5 2 )
To improve th e qualit y of sample s and t o increase th e recover y o f sof t o r slightl y cohesive soils , a
piston sampler i s normally used. Suc h a sampler consist s of a thin walle d tub e fitted with a piston
that closes the end of the sampling tube until the apparatus is lowered t o the bottom of the bore hol e
(Fig. 9.7(a)). Th e samplin g tub e i s pushe d int o th e soi l hydraulicall y b y keepin g th e pisto n
stationary (Fig . 9.7(b)) . The presenc e o f th e piston prevent s th e sof t soil s fro m squeezin g rapidl y
into th e tub e an d thu s eliminate s mos t o f th e distortio n o f th e sample . Th e pisto n als o help s t o
increase th e lengt h of sample tha t can be recovered b y creating a slight vacuum that tends t o retain
the sample if the top of the column of soil begins to separate fro m th e piston. During the withdrawal
of the sampler, the piston also prevents water pressure from actin g on the top of the sample and thus
increases the chances o f recovery. The design of piston samplers ha s been refine d t o the extent that
it i s sometimes possibl e t o take undisturbed samples o f sand from belo w th e water table. However ,
piston samplin g is relatively a costly procedure and may be adopted onl y where its use is justified.
Example 9 . 1
The followin g dimensions ar e given for a shelby tube sampler :
External diameter = 51 mm
Internal diameter = 48 mm
Determine th e area rati o
Solution
Per Eq (9.1) the area rati o A
r
is
A. =
48
2
= 0.12 9 = 12.9%
Soil Ex plorat io n 325
Example 9. 2
75 mm is the external diameter of a sampling tube. If the area ratio required i s 20%, determine the
thickness of the sampling tube. In what type of clay would such a high area rati o be required?
Solution
15
2
-d
2
Solving
The wal l thickness
d
i
= 68.465 mm.
75.0-68.465
= 3.267 mm
When sample s ar e t o be take n i n ver y stif f t o har d cla y soil s mixe d wit h stones , samplin g
tubes wit h high area ratios ar e required.
9.4 ROC K COR E SAM PL I NG
Rock coring is the process in which a sampler consisting of a tube (core barrel ) wit h a cutting bit at
its lower end cuts an annular hole i n a rock mass, thereby creating a cylinder or core of rock which
is recovered i n the core barrel . Roc k cores are normally obtained by rotary drilling.
The primary purpose of core drilling is to obtain intact samples. The behavior of a rock mass is
affected b y th e presenc e o f fracture s i n th e rock . Th e siz e an d spacing of fractures , the degre e o f
weathering o f fractures , an d th e presenc e o f soi l withi n the fracture s ar e critica l items . Figur e 9.8
gives a schemati c diagra m o f cor e barrel s wit h corin g bit s a t the bottom. As discusse d earlier , th e
cutting element may consist of diamonds, tungsten carbide inserts or chilled shot. The core barrel may
consist of a single tube or a double tube. Samples taken in a single tube barrel ar e likely to experience
Drill rod Drill rod
Fluid passag e
Bearing
Outer barrel
Inner barrel
Core
lifter
a.
Corin
g
^—bit
(a) (b)
Figure 9. 8 Schem at i c diag ra m o f cor e barrel s (a) Singl e t ube , (b ) Doubl e t ube .
326 Chapt er 9
considerable disturbanc e due t o torsion, swellin g and contamination by th e drillin g fluid, bu t thes e
disadvantages ar e no t ther e i f th e corin g i s conducte d i n hard , intact , rock y strata . However , i f a
double tub e barre l i s used , th e cor e i s protecte d fro m th e circulatin g fluid . Mos t cor e barrel s ar e
capable of retaining cores u p to a length of 2 m. Single barrel i s used i n Calyx drilling. Standard roc k
cores range from abou t 1
1
A inches to nearly 6 inches in diameter. The more common size s are given in
Table 9.1.
The recovery ratio R
r
, define d a s th e percentag e rati o betwee n th e lengt h o f th e cor e
recovered an d th e lengt h o f th e cor e drille d o n a give n run , i s relate d t o th e qualit y o f roc k
encountered i n boring, bu t i t i s als o influence d by th e drillin g technique an d th e typ e an d siz e o f
core barrel used . Generall y the use of a double tube barrel result s in higher recovery ratio s tha n can
be obtaine d wit h singl e tub e barrels . A bette r estimat e o f in-sit u roc k qualit y i s obtaine d b y a
modified cor e recover y rati o known as the Rock Quality Designation (RQD) whic h i s expressed a s
RQD =
(9.2)
where, L
a
= tota l lengt h o f intac t har d an d soun d piece s o f cor e o f lengt h greate r tha n 4 in .
arranged i n it s proper position ,
L
t
= total lengt h of drilling.
Breaks obviousl y caused by drilling are ignored. The diameter of the core shoul d preferably be not less
than 2V
8
inches. Table 9. 2 gives the rock qualit y description a s related t o RQD.
Table 9. 1 St an dar d s ize s of cor e barrel s , dril l rods , an d com pat ibl e cas in g
( Pe ck e t al. , 1974 )
Core Ba r r e l
Symbol
EWX, EW M
AWX, AWM
BWX, BW M
NWX, NW M
2
3
/
4
x 3
7
/
g
Hol e
di a
( i n )
l>/2
I
15
/ 1
'1 6
2
3
/
8
3
3
7
/ j /
g
Core
di a
( i n )
1 6
1%
2' /
8
2
U
/
16
Dr i l l Ro d
Symbol
E
A
B
N
-
Out si de
di a
( i n )
115/
16
1 "V
I
7
/
2
3
/ L
' 8
-
Casi ng
Symbol
_
EX
AX
BX
NX
Out si de
di a
( i n )
_
113/
16
2V
4
2
?
/8
3V
2
I ns i de
di a
( i n )
_
l'/2
I
29
/ 1
'3 2
2
3
/ z /
8
3
Note: Symbo l X indicates singl e barrel , M indicate s doubl e barrel .
Table 9. 2 Relat io n o f RQD and in-situ Roc k Qualit y (Pec k et al. , 1974)
RQD % Rock Qualit y
90-100
75-90
50-75
25-50
0-25
Excellent
Good
Fair
Poor
V ery Poor
Soil Ex pl orat io n 32 7
9 .5 STANDAR D PENETRATI ON TES T
The SP T i s th e mos t commonl y use d i n situ tes t i n a bor e hol e i n th e USA . Th e tes t i s mad e by
making use of a split spoon sampler shown in Fig. 9.7. The method has been standardize d as ASTM
D-1586 (1997) wit h periodic revision since 1958 . Th e method of carrying out this test is as follows:
1. Th e spli t spoo n sample r i s connecte d t o a strin g o f dril l rod s an d i s lowere d int o th e
bottom o f th e bor e hol e whic h was drille d an d cleane d i n advance .
2. Th e sample r i s driven into the soi l strat a to a maximum depth of 1 8 in by making use of a
140 Ib weight falling freel y from a height of 30 in on to an anvil fixed on the top of drill rod.
The weigh t is guided to fal l alon g a guide rod. The weight is raised an d allowed t o fal l b y
means of a manila rope, one end tied to the weight and the other end passing over a pulley
on t o a hand operated winc h or a motor driven cathead.
3. Th e number of blows required t o penetrate each of the successive 6 in depths i s counted t o
produce a total penetration of 1 8 in.
4. T o avoid seating errors, the blows required for the first 6 in of penetration ar e not taken into
account; those required t o increase the penetration from 6 in to 18 in constitute the N -value.
As per some code s of practice i f the N -value exceed s 100 , i t is termed a s refusal, and the test
is stoppe d eve n i f th e tota l penetratio n fall s shor t o f th e las t 30 0 m m dept h o f penetration .
Standardization of refusal a t 10 0 blows allows all the drilling organizations to standardize costs s o
that highe r blow s i f require d ma y b e eliminate d t o preven t th e excessiv e wea r an d tea r o f th e
equipment. The SPT i s conducted normall y at 2. 5 t o 5 ft intervals. The interval s may be increase d
at greater depth s i f necessary .
Standardization o f SP T
The validit y of th e SP T ha s bee n th e subjec t o f stud y and researc h b y man y author s fo r th e las t
many years . Th e basi c conclusio n i s tha t th e bes t result s ar e difficul t t o reproduce . Som e o f th e
important factors tha t affec t reproducibilit y are
1. V ariatio n in the heigh t of fal l o f the drop weigh t (hammer) during the tes t
2. The numbe r of turns of rope around the cathead, and the condition of the manil a rope
3. Lengt h and diameter of drill ro d
5. Diamete r of bore hol e
6. Overburde n pressur e
There ar e man y mor e factor s tha t hampe r reproducibilit y of results . Normall y correction s
used t o be applie d fo r a quick condition i n the hol e botto m du e t o rapi d withdrawa l of the auger .
ASTM 158 6 ha s stipulate d standard s to avoi d suc h a quick condition . Discrepancie s i n the input
driving energ y an d it s dissipatio n aroun d th e sample r int o the surroundin g soil ar e th e principa l
factors fo r the wide range i n N values. The theoretical inpu t energy may be expressed as
E
in
= Wh (9.3 )
where W = weight or mass of the hamme r
h = height of fal l
Investigation ha s reveale d (Kovac s and Salomone , 1982 ) tha t the actua l energy transferre d t o th e
driving head and then to the sampler ranged from abou t 30 to 80 percent. I t has been suggeste d that
the SPT be standardize d to some energ y rati o R
g
keepin g i n mind the dat a collected s o far fro m th e
existing SPT. Bowle s (1996) suggest s that the observed SP T value N be reduced t o a standard blow
count corresponding t o 70 percent of standard energy. Terzaghi, et al., (1996) suggest 60 percent. The
standard energy ratio may be expressed a s
328 Chapt e r 9
Actual hammer energy t o sampler, E
a
ft, =
Input energy, E
in
^ '
Corrections t o the Observe d SPT V alue
Three type s o f correction s ar e normall y applie d t o th e observe d N values . The y are :
1. Hamme r efficienc y correctio n
2. Drillrod , sample r an d borehol e correction s
3. Correctio n du e t o overburde n pressure
1 . H amme r Efficienc y Correction, E
h
Different type s o f hammer s ar e i n us e fo r drivin g the dril l rods. Tw o type s ar e normall y use d i n
USA. They ar e (Bowles , 1996)
1. Donu t wit h two turns of manil a rope o n the cathead wit h a hammer efficienc y E
h
= 0.45.
2. Safet y wit h two turns of mani l a rope on the cathead wit h a hammer efficienc y a s follows:
Rope-pulley o r cathead = 0.7 t o 0.8 ;
Trip o r automati c hammer = 0. 8 t o 1.0 .
2. Dril l Rod , Sampler and B orehol e Correction s
Correction factor s ar e use d fo r correctin g th e effect s o f lengt h o f dril l rods , us e o f spli t spoo n
sampler wit h or withou t liner, and siz e of bor e holes . Th e variou s correction factor s ar e (Bowles ,
1996).
a) Dril l rod lengt h correction facto r C,
Length (m )
>10m
4-10 m
<4. 0m
Correct ion fact or (C
d
)
1.0
0.85-0.95
0.75
b) Sample r correction factor , C
s
Without line r C
x
= 1.0 0
With liner,
Dense sand , clay = 0.8 0
Loose san d = 0.9 0
c) Bor e hol e diamete r correction factor , C
b
B ore hol e diam et e r Cor r ect io n f act or , C,
60-120 mm 1. 0
150mm 1.0 5
200mm 1.1 5
3 . Correctio n Factor for Overburde n Pressure in G ranular Soils, C
N
The C
N
a s per Liao and Whitman (1986 ) is
Soil Ex plorat io n 32 9
"95.761
/2
.irj
(9
-
5)
where, p'
0
^effective overburde n pressure i n kN/m
2
There ar e a numbe r of empirica l relation s propose d fo r C
N
. However , th e mos t commonl y
used relationship i s the one given by Eq. (9.5).
N
cor
ma y be expressed as
"cor =
C
N
N E
h
C
d
C
s
C
b
(9-6 )
N
cor
i s related t o the standard energy ratio used by the designer. N
cor
ma y be expressed a s A^
70
or N ^Q accordin g t o the designer' s choice .
In Eq (9.6) C
N
N i s the corrected valu e for overburden pressure only . The value of C
N
as per
Eq. (9.5) is applicable fo r granular soils only , whereas C^ = 1 for cohesive soil s for al l depths .
Example 9. 3
The observed standar d penetration test value in a deposit of fully submerge d sand was 45 at a depth
of 6. 5 m. The averag e effectiv e uni t weight of the soi l i s 9.69 kN/m
3
. The othe r dat a give n are (a)
hammer efficienc y = 0.8, (b) drill rod lengt h correction facto r = 0.9, and (c ) borehole correctio n
factor = 1.05. Determine the corrected SP T value fo r standard energy (a) R - 6 0 percent, an d (b)
Solution
Per Eq (9.6), the equation for N
60
may be written as
( \ } N - C N F C C C
W '
V
60 ^N " ^h ^d S ^b
where N = observed SP T value
C
N
- overburde n correction
Per Eq (9.5) we have
1/2
=
95.7 6
N
Po
where p'
Q
= effectiv e overburden pressur e
= 6. 5 x 9.69 = 63 kN/m
2
Substituting fo r p'
Q
,
C
N
= ^^ =1.23 3
N
6 0
Substituting the known values, the corrected N
60
is
N
60
= 1.233 x 45 x 0. 8 x 0.9 x 1.0 5 = 42
For 70 percent standar d energy
W
70
=42 x^ = 36
70
0. 7
330 Chapt e r 9
9.6 SP T V AL U ES REL ATED TO REL ATI V E DENSI TY OF
COH ESI ONL ESS SOI L S
Although th e SP T i s not considere d a s a refined and completel y reliabl e metho d o f investigation,
the N
cor
value s giv e usefu l informatio n with regar d t o consistenc y o f cohesiv e soil s an d relativ e
density of cohesionles s soils . The correlation betwee n N ,
r
values and relative densit y of granular
soils suggeste d b y Peck, e t al. , (1974 ) i s given in Table 9.3 .
Before usin g Tabl e 9. 3 th e observe d N valu e ha s t o b e correcte d fo r standar d energ y an d
overburden pressure . The correlation s given in Table 9. 3 ar e just a guide and ma y var y according t o
the finenes s of the sand .
Meyerhof (1956) suggested th e following approximate equations for computing the angl e of
friction 0 from the known value of D
f
.
For granular soil wit h fine sand an d mor e tha n 5 percent silt ,
<p° = 25 + Q.15D
r
(9.7 )
For granular soils wit h fine sand an d less tha n 5 percent silt ,
0° = 30 + 0.15D
r
(9.8 )
where D
r
i s expressed i n percent .
9.7 SP T V AL U ES REL ATED TO CONSI STENCY OF CL AY SOI L
Peck e t al. , (1974 ) hav e give n fo r saturate d cohesiv e soils , correlation s betwee n N
cor
valu e an d
consistency. Thi s correlatio n i s quite useful bu t has t o be used accordin g t o the soi l condition s met
in th e field. Tabl e 9. 4 gives the correlations .
The N
cor
valu e t o be use d i n Tabl e 9. 4 i s the blow coun t corrected fo r standar d energy rati o
R
es
. The present practic e i s to relate q
u
wit h N
cor
a s follows,
q
u
= kN
cor
kP a (9.9 )
Table 9. 3 N an d 0 Relat ed t o Rel at iv e Den s it y
N
cor
0-4
4-10
10-30
30-50
>50
Compact nes s
V ery loos e
Loose
Medium
Dense
V ery Dens e
Table 9. 4
Consi st ency
V ery sof t
Soft
Medi um
Stiff
V ery Stif f
Hard
Re l a t i ve densi t y , D
r
( %)
Rel at i on
"cor
0-2
2-4
4-8
8-15
15-30
>30
0-15
15-35
35-65
65-85
>85
Between N
cor
an d q
u
q , kP a
^ u'
<25
25-50
50-100
100-200
200-400
>400

<28
28-30
30-36
36-41
>4 1
where q
u
i s th e unconfi ne d compressiv e strength .
Soil Ex plorat io n 33 1
or K
=
7T-
Kr a
(9.10 )
cor
where, k i s th e proportionalit y factor . A valu e o f k = 1 2 ha s bee n recommende d b y Bowle s
(1996).
Example 9. 4
For the corrected N values in Ex 9.3, determine the (a) relative density, and (b) the angle of friction .
Assume the percent of fines i n the deposit is less than 5%.
Solution
Per Table 9.3 the relative density and 0 are
For N
60
= 42, D
r
= 11%, 0 = 39°
For N
70
= 36, D
f
= 71%, 0 = 37.5°
Per Eq (9.8 )
For D
r
= 77%, 0 = 30 + 0. 15x77 = 41.5°
For D
r
= 71%, 0=30 + 0. 15x71=40. 7°
Example 9. 5
For the corrected value s of N given in Ex 9.4, determine the unconfined compressive strengt h q
u
in
a clay deposit .
Solution
(a) Fro m Tabl e 9. 4
For N ^ = 42\
For N = 361 Q
u
>
^00 kPa - The soil i s of a hard consistency.
(b) Pe r Eq_(9.9;
q
u
= kN
cor
, wher e k = 1 2 (Bowles, 1996 )
For N
M
= 42 , q =12x42 = 504 kPa
Du • * M
For yV
70
= 36, q
u
= 12 x 36 = 432 kP a
Example 9. 6
Refer t o Exampl e 9.3 . Determin e th e correcte d SP T valu e fo r R
es
=10 0 percent , an d th e
corresponding value s of D
r
and 0. Assume the percent of fine san d in the deposi t i s less than 5%.
Solution
From Example 9.3, N
60
= 42
„ °-
6
^
Hence Af,
m
= 2 x — ~ 25
1.0
From Tabl e 9.3, 0 = 34.5° and D
f
= 57.5%
From Eq . (9.8) for D
r
= 57.5%, 0 = 30 + 0.15 x 57. 5 = 38.6°.
332 Chapt e r 9
9 .8 STATI C CONE PENETRATI ON TEST (CRT)
The stati c con e penetratio n tes t normall y calle d th e Dutc h con e penetratio n tes t (CPT) . I t ha s
gained acceptanc e rapidl y i n man y countries . Th e metho d wa s introduce d nearl y 5 0 year s ago .
One o f the greates t value s of the CPT consist s o f it s function a s a scale model pil e test . Empirica l
correlations establishe d ove r man y year s permi t th e calculatio n o f pil e bearin g capacit y directl y
from th e CP T result s without the us e o f conventiona l soil parameters .
The CP T ha s prove d valuabl e for soi l profilin g a s th e soi l typ e ca n b e identifie d fro m th e
combined measuremen t of end resistance of cone and side friction o n a jacket. The tes t lends itself
to th e derivatio n of norma l soi l propertie s suc h a s density , frictio n angl e an d cohesion . V ariou s
theories hav e been develope d fo r foundation design .
The popularit y of the CPT can be attribute d to the following three important factors:
1. Genera l introductio n of th e electri c penetromete r providin g mor e precis e measurements ,
and improvement s in the equipment allowing deeper penetration.
2. Th e nee d fo r th e penetromete r testin g in-sit u techniqu e i n offshor e foundatio n
investigations i n vie w of th e difficultie s i n achievin g adequat e sampl e qualit y i n marin e
environment.
3. Th e additio n o f othe r simultaneou s measurements t o th e standar d frictio n penetromete r
such a s pore pressure and soi l temperature.
The Penetrometer
There ar e a variet y o f shape s an d size s o f penetrometer s bein g used . Th e on e tha t i s standar d i n
most countries i s the cone wit h an apex angl e of 60° and a base area of 1 0 cm
2
. The sleev e (jacket)
has becom e a standar d ite m o n th e penetromete r fo r mos t applications . O n th e 1 0 cm
2
con e
penetrometer th e friction sleev e shoul d have an area of 15 0 cm
2
as per standar d practice. The rati o
of sid e frictio n an d bearin g resistance , th e friction ratio, enable s identificatio n of th e soi l typ e
(Schmertmann 1975 ) an d provide s usefu l informatio n i n particula r whe n n o bor e hol e dat a ar e
available. Eve n whe n borings are made , th e frictio n rati o supplie s a check o n th e accurac y o f th e
boring logs .
Two types of penetrometers ar e used which are based o n the method use d for measuring con e
resistance and friction. They are,
1. Th e Mechanica l Type,
2. Th e Electrical Type .
M echanical Type
The Begemann Friction Cone Mechanical type penetrometer i s shown in Fig. 9.9. I t consists of a 60°
cone wit h a base diameter of 35.6 mm (sectional area 1 0 cm
2
). A sounding rod is screwed t o the base.
Additional rod s o f one mete r lengt h each ar e used . These rods ar e screwe d o r attache d togethe r t o
bear agains t eac h other . The soundin g rod s mov e insid e mantl e tubes . Th e insid e diamete r o f th e
mantle tub e i s just sufficien t fo r th e soundin g rods t o mov e freel y wherea s th e outsid e diamete r i s
equal to or less than the base diamete r of the cone. All the dimensions i n Fig. 9. 9 ar e in mm.
Jacking System
The rigs used for pushing down the penetrometer consist basically of a hydraulic system. The thrust
capacity for cone testing on land varies from 2 0 to 30 kN for hand operated rig s an d 10 0 to 200 kN
for mechanicall y operated rig s a s shown i n Fig. 9.10. Bourden gauges ar e provided i n the driving
mechanism fo r measurin g th e pressure s exerte d b y th e con e an d th e frictio n jacke t eithe r
individually or collectively during the operation. The rigs may be operated eithe r on the ground or
Soil Ex plorat io n 333
mounted on heavy duty trucks. In either case, the rig should take the necessary upthrust . For ground
based rig s scre w anchor s ar e provided t o take up the reaction thrust .
Operation o f Penetromete r
The sequenc e o f operation o f the penetrometer show n in Fig. 9.11 is explained a s follows:
Position 1 Th e cone an d friction jacket assembl y i n a collapsed position.
Position 2 Th e cone is pushed down by the inner sounding rods to a depth a until a collar engage s
the cone. The pressur e gauge records th e total force Q
c
to the cone. Normall y a = 40 mm.
Position 3 Th e sounding rod is pushed further t o a depth b. This pushes the friction jacket an d the
cone assembl y together ; the force i s Q
t
. Normally b = 40 mm.
Position 4 Th e outsid e mantl e tub e i s pushe d dow n a distanc e a + b whic h bring s th e con e
assembly an d the friction jacket t o position 1 . The total movement = a + b= 80 mm.
The proces s of operation illustrate d above i s continued unti l the proposed dept h i s reached .
The con e i s pushed a t a standard rat e o f 20 mm per second . Th e mechanical penetromete r ha s it s
advantage a s i t i s simpl e t o operat e an d th e cos t o f maintenanc e i s low. The qualit y of th e wor k
depends o n the skil l of the operator . Th e dept h of CPT i s measured by recording th e lengt h of the
sounding rods tha t have been pushe d int o the ground.
35.7
266
35.6
i 3 0 3 5
Note: All dimensions ar e in mm.
Figure 9.9 B eg em an n frict ion -con e m echan ica l t ype pen et rom et e r
334
Chapt er 9
iff
Hydraulically
operated cylinde r
High pressur e
manometer
Guide bow
Measuring
equipment
Road whee f
in raise d position
Base fram e
3. 5m
Upper suppor t
beam
Guide column
Low pressur e
manometer
LH maneuve r ng handl e
RH maneuverin g handl e
Control valv ;
Wooden sleepe r
1. 80m- 2. 00m
PLAN
Fig. 9 .1 0 St at i c con e pen et rat io n t es t in g equipm en t
The Electri c Penetromete r
The electri c penetromete r i s an improvement ove r th e mechanical one . Mechanica l penetrometer s
operate incrementall y whereas th e electri c penetromete r i s advanced continuously.
Figure 9.1 2 shows a n electric-static penetromete r wit h the friction sleev e j ust above th e bas e
of th e cone . Th e sectiona l are a o f th e con e an d th e surfac e are a o f th e frictio n jacke t remai n th e
same a s those o f a mechanical type . The penetrometer ha s buil t in load cell s tha t record separatel y
the cone bearing an d sid e friction . Strai n gauge s ar e mostl y use d fo r the loa d cells . Th e loa d cell s
have a norma l capacit y o f 5 0 t o 10 0 k N fo r en d bearin g an d 7. 5 t o 1 5 k N fo r sid e friction ,
depending o n th e soil s t o b e penetrated . An electri c cabl e inserte d throug h the pus h rods (mantl e
tube) connect th e penetrometer wit h the recording equipmen t a t the surface which produces graph s
of resistance versu s depth.
Soil Ex plorat ion 335
Position 1
\
\
\
\
\
,
\
%
>1
A
'. a+ b
1
r.
a
I "
V
n
r
1
; /
\
V
1
\
1
^
\
!
/
?
^
Sounding rod
Bottom mantl e tube
Friction jacket
r M
Cone assembl y
Position 2
Position 3 Positio n 4
Figure 9 .1 1 Fou r pos it ion s o f t h e s oun din g apparat u s wit h frict io n j ack e t
The electri c penetromete r ha s man y advantages . The repeatabilit y o f th e con e tes t i s ver y
good. A continuou s recor d o f th e penetratio n result s reflect s bette r th e natur e o f th e soi l layer s
penetrated. However, electronic con e testing requires skilled operators and better maintenance. The
electric penetromete r i s indispensabl e for offshore soil investigation.
Operation o f Penetrometer
The electri c penetromete r i s pushe d continuousl y a t a standar d rat e o f 2 0 m m pe r second . A
continuous record o f the bearing resistance q
c
and frictional resistance/ ^ against depth is produced
in th e for m o f a graph a t the surfac e in the recording unit .
Piezocone
A piezometer elemen t included in the cone penetrometer i s called a piezocone (Fig . 9.13). There i s
now a growing use of the piezocone for measuring pore pressures at the tips of the cone. The porous
element i s mounted normall y midwa y along the cone ti p allowing pore wate r t o ente r the tip . An
electric pressur e transduce r measures the pore pressur e during the operation o f the CPT. Th e por e
pressure record provide s a much more sensitive means to detect thin soil layers. This coul d be very
important in determining consolidation rate s i n a clay withi n the sand seams .
Chapt er 9
.
Piezocone.
1. Loa d cell
2. Friction sleeve
3. Water proof bushi n g
4. Cable
5. Strai n gase s
6. Connection with rods
7. Inclinometer
8. Porous stone (piezometer)
Figure 9 .1 2 An -el ect r ic-s t at i c con e pen et rom et e r
Temperature Cone
The temperatur e o f a soi l i s require d a t certai n localitie s t o provid e informatio n abou t
environmental changes. The temperature gradient with depth may offer possibilities to calculate the
heat conductivit y o f th e soil . Measuremen t o f th e temperatur e durin g CP T i s possibl e b y
incorporating a temperatur e sensor i n the electri c penetrometer . Temperatur e measurement s hav e
been mad e i n permafrost, under blast furnaces, beneath undercooled tanks , along marine pipe lines,
etc.
Effect o f Rat e o f Penetratio n
Several studie s hav e bee n mad e t o determine the effec t o f the rat e o f penetration o n con e bearin g
and sid e friction. Although the value s tend t o decrease fo r slowe r rates , th e genera l conclusio n i s
that the influence is i nsi gni fi cant for speeds betwee n 1 0 and 30 mm per second. The standar d rate of
penetration ha s bee n generall y accepted a s 20 mm per second .
Cone Resistanc e c r an d L oca l Side Frictio n f
c c »
Cone penetration resistanc e q
c
i s obtained b y dividing the total forc e Q
c
acting on the cone b y the
base area A o f the cone .
(9.11)
Probe mai n frame
Pressure transducer
Retainer
Housing
Tip (Upper portion)
Porous element
Tip (lower portion)
Apex angle
Figure 9 .1 3 Det ail s o f 60°/10 cm
2
piezocon e
Soil Ex pl orat io n 337
In th e same way , the loca l sid e frictio n f
c
i s
Q
f
f c=^^
A
f
(9.12)
where, Q
f
= Q
t
- Q
c
= force required t o push the friction jacket ,
Q
t
= the total forc e required t o push the cone and friction jacke t togethe r i n the cas e o f a
mechanical penetrometer ,
A
f
= surfac e area of the friction jacket .
Friction Ratio , R
f
Friction ratio, RAs expressed a s
K
fc
*/ - —' (9.13 )
where f
c
an d q
c
are measured at the same depth. RAs expressed a s a percentage. Frictio n rati o i s an
important parameter fo r classifying soil (Fig. 9.16) .
Relationship B etwee n q
o
, Relative Densit y D
r
and Friction Angl e 0 fo r Sand
Research carrie d ou t by many indicates that a unique relationship between cone resistance, relative
density an d frictio n angl e vali d fo r al l sand s doe s no t exist . Robertso n an d Campanell a (1983a )
have provided a set of curves (Fig. 9.14) which may be used to estimate D
r
based o n q
c
and effective
Cone bearing, q
c
MN/m
2
0 1 0 2 0 3 0 4 0 5 0
•a 30 0
350
400
D
r
expressed i n percent
Figure 9.1 4 Relat ion s hi p bet wee n rel at iv e den s it y D
r
an d pen et rat ion res is t an c e q
c
for un cem en t e d quart z s an d s (Robert s o n and Cam pan ella, 1983a )
338 Chapt er 9
Cone bearing , q
c
MN/m
2
10 2 0 3 0 4 0
400
50
Figure 9 .1 5 Relat ion s hi p bet wee n con e poin t res is t an c e q
c
an d an g l e o f in t ern a l
fr ict ion 0 for un cem en t e d quart z s an d s (Robert s o n an d Cam pan el l a, 1 983b)
overburden pressure . Thes e curves ar e supposed t o be applicabl e fo r normall y consolidate d clea n
sand. Fig . 9.1 5 gives th e relationshi p between q
c
and 0 (Robertson an d Campanella, 1983b) .
Relationship B etwee n q
c
and U ndrained Shea r Strength, c
u
o f Clay
The con e penetratio n resistance q
c
and c
u
may be related as
or
<lc - PC
(9.14)
where, N
k
= cone factor ,
p
o
- y? = overburden pressure.
Lunne an d Kelve n (1981 ) investigate d th e valu e o f th e con e facto r N
k
fo r bot h normall y
consolidated and overconsolidated clays . The value s of A^ as obtained are given below:
T ype o f cl a y Con e fact o r
Normally consolidated
Overconsolidated
At shallo w depths
At dee p depth s
11 t o 1 9
15 t o 2 0
12 t o 1 8
Soil Ex plorat io n 339
10
Heavily
over consolidate d
or cemented soil s
Sandy
gravel t o
gravelly
sand t o
sand
Sand t o
silty san d
Silty sand
to sandy sil t
Sandy sil t to
clayey silt
Clayey sil t to
silty cla y
to clay /
Clay t o
organic
clay
Highly
sensitive
soils
10'
1 2 3 4 5 6
Friction rati o (%)
Figure 9.1 6 A s im plifie d cl as s ificat io n char t (Doug las , 1984 )
Possibly a value of 2 0 fo r A^ for bot h type s o f clay s ma y b e satisfactory . Sanglera t (1972 )
recommends th e same valu e for al l cases wher e an overburden correction i s of negligibl e value.
Soil Classificatio n
One of the basic uses of CPT i s to identify an d classify soils. A CPT-Soil Behavior Type Prediction
System ha s bee n develope d b y Dougla s an d Olse n (1981 ) usin g a n electric-frictio n con e
penetrometer. Th e classificatio n i s base d o n th e friction ratio f/q
c
. Th e rati o f
(
/q
c
varie s greatl y
depending on whether it applies t o clays or sands. Their findings have been confirmed by hundreds
of tests .
For cla y soils , i t ha s bee n foun d tha t th e frictio n rati o decrease s wit h increasin g liquidity
index /, . Therefore , th e frictio n rati o i s a n indicato r o f th e soi l typ e penetrated . I t permit s
approximate identification of soi l type though no samples ar e recovered .
340 Chapt er 9
f
1
J c
0 5 0 10 0 15 0 20 0 25 0
/ an d qexpresse d in kg/cm
Friction ratio, R
f
i n %
0
8
cfl
£ 1 6
<u
c
J3
8- 2 4
32
/in
1 2 3 4 5 Soi l profil e
r
C
r
\
i
\
v_
^
^- —•
"•' ^
„ '
^
-^-*
^
-—
c~
^
^ —
)
"X
^— • --
>
v
^
Sandy sil t
Silts & silt y cla y
Silty san d
Silty clay
and
Clay
Sand
Silts & Clayey silt s
Sands
Silty san d & Sandy sil t
Figure 9 .1 7 A t ypica l s oun din g lo g
Douglas (1984 ) presented a simplified classification char t shown i n Fig. 9.16. Hi s chart use s
cone resistanc e normalize d ( q ) for overburde n pressur e usin g the equation
q - q (l-1.251ogp' )
"en "c^ o *o'
where, p' = effective overburden pressur e i n tsf, and q = cone resistanc e i n tsf,
(9.15)
In conclusion, CPT data provide a repeatable inde x of the aggregate behavior ofin-situ soil .
The CP T classificatio n method provide s a better pictur e of overal l subsurfac e conditions tha n i s
available wit h most othe r methods o f exploration.
A typica l sounding log i s given in Fig. 9.17 .
Soil Ex pl orat io n 34 1
Table 9. 5 Soi l clas s ificat io n bas e d o n frict ion rat i o R
f
(San g lerat , 1972)
R
f
% T ype o f s oi l
0-0.5
0.5-2.0
2-5
>5
Loose gravel fil l
Sands or gravel s
Clay san d mixtures and silt s
Clays, peats etc.
The friction ratio R, varies greatl y wit h the type of soil . The variation of R, for the various
types of soils i s generally o f the order give n in Table 9. 5
Correlation B etwee n SPT an d CPT
Meyerhof (1965) presented comparative data between SPT and CPT. For fine or silty medium loose
to medium dense sands , he presents the correlation a s
q
c
=OAN MN /m
2
His finding s ar e as given in Table 9.6 .
(9.16)
Table 9. 6 Approx im at e relat ion s hi p bet wee n relat iv e den s it y o f fin e s an d , t h e SPT ,
t he s t at i c con e res is t an c e an d t he an g l e of in t ern a l fract io n (M eyerhof , 1965 )
State o f san d D
r
N
cof
q
c
( MPa ) </> °
V ery loos e <0. 2 < 4 <2. 0 <3 0
Loose 0.2-0. 4 4-1 0 2- 4 30-3 5
Medium dense 0.4-0. 6 10-3 0 4-1 2 35-4 0
Dense 0.6-0. 8 30-5 0 12-2 0 40-4 5
V ery dens e 0.8-1. 0 >5 0 >2 0 4 5
10
9
8
c 7
l l
1
6
^
o"
5
1 4
3
2
1
0
o.c
q
c
in kg/cm
2
; N, blows/foo t
. A
^^
«*"^
k
^*
***^*
A
{*
A
A 4
**^n
A
k
A
A
/
/
A
A
)01 0.0 1 0. 1 1. 0
Mean grai n siz e D
50
, mm
Figure 9.1 8 Relat ion s hi p bet wee n qJN an d mean g rai n s iz e D
50
(m m ) (Robert s on
an d Cam pan ella, 1983a )
342 Chapt e r 9
The lowes t value s o f th e angl e o f interna l frictio n give n i n Tabl e 9. 6 ar e conservativ e
estimates for uniform, clean sand and they should be reduced b y at least 5° for clayey sand . These
values, as well as the upper values of the angles of internal friction which apply to well graded sand,
may b e increase d b y 5 ° for gravell y sand.
Figure 9.18 show s a correlation presente d b y Robertson an d Campanell a (1983 ) betwee n th e
ratio o f qJ N an d mea n grai n size, D
5Q
. I t can be see n fro m the figur e tha t the rati o varie s fro m 1 at
D
5Q
= 0.001 m m t o a maximum value of 8 at D
50
=1.0 mm. The soi l typ e also varie s fro m cla y t o
sand.
It i s clea r fro m th e abov e discussion s tha t the valu e o f n( = qJ N ) i s no t a constan t fo r an y
particular soil. Designers mus t use their own judgment while selecting a value for n for a particular
type of soil .
Example 9. 7
If a deposit a t a site happens to be a saturated overconsolidated cla y with a value of q
c
= 8.8 MN/m
2
,
determine th e unconfme d compressiv e strengt h of cla y give n p
Q
= 127 kN/m
2
Solution
Per E q (9 . 14)
_
c. , —
N " N
k k
Assume N
k
= 20. Substitutin g the known values and simplifying
2(8800-127) __, . . . ,
q
"
=
-20 -
=
If w e neglec t th e overburde n pressure p
Q
q
,
20
It i s clea r that , th e valu e of q
u
i s littl e affecte d b y neglectin g th e overburde n pressur e i n
Eq. (9.14 )
Example 9. 8
Static con e penetratio n test s wer e carrie d ou t a t a sit e b y usin g a n electric-frictio n con e
penetrometer. Th e followin g data were obtained a t a depth of 12. 5 m.
cone resistance q
c
=19.15 2 MN/m
2
(200 tsf )
f
D _
J
C _ 1 -J
friction rati o
A
/ ~ ~
l
-
J
"c
Classify th e soi l a s per Fig. 9.16. Assume ^effective) = 16. 5 kN/m
3
.
Solution
The value s of q
c
= 19.152 x 10
3
kN/m
2
and R
f
= 1.3 . Fro m Eq . (9.15 )
q =200 x 1-1.2 5 log
16
'
5x12
'
5
= 121 tsf
*cn & 1Q O
The soi l i s sand t o silt y san d (Fig . 9.16 ) fo r q
m
= 121 tsf and /?,= 1.3 .
Soil Ex plorat io n 34 3
Example 9. 9
Static cone penetration tes t at a site at depth of 30 ft revealed the following
Cone resistance q
c
= 12 5 tsf
Friction rati o R
f
= 1.3 %
The averag e effective uni t weight of the soi l i s 11 5 psf. Classify the soi l pe r Fig. 9.16.
Solution
The effectiv e overburde n pressure p'
0
= 3 0 x 11 5 = 3450 lb/ft
2
= 1.72 5 tsf
From Eq (9.15 )
q
m
= 125 (1-1.25 log 1.725 ) = 88 tsf
R
f
= 1.3%
From Fig. 9.16, the soi l i s classified a s sand to silt y sand for q
cn
= 88 tsf and R
f
= 1.3%
Example 9 .10
The stati c con e penetratio n resistanc e a t a sit e a t 1 0 m dept h i s 2. 5 MN/m
2
. Th e frictio n rati o
obtained from th e test is 4.25%. If the unit weight of the soi l is 18. 5 kN/m
3
, what type of soil exists
at the site.
Solution
q
c
= 2.5 x 100 0 kN/m
2
= 2500 kN/m
2
= 26.1 tsf
p'
Q
= 10 x 18. 5 = 18 5 kN/m
2
=1. 93 tsf
q
cn
=26.1 (1-1.25 log 1.93) =16.8 tsf
R
f
= 4.25 %
From Fi g 9.16, the soi l i s classified as clayey silt to silt y clay to clay
9.9 PRESSU REM ETE R
A pressuremete r tes t i s a n in-situ stress-strai n tes t performe d o n the wall s of a bore hol e usin g a
cylindrical prob e tha t ca n b e inflate d radially . The pressuremeter , whic h wa s firs t conceived ,
designed, constructe d an d used by Menar d (1957) o f France, ha s been i n use sinc e 1957 . The test
results are used either directly or indirectly for the design of foundations. The Menard test has been
adopted as ASTM Test Designation 4719. The instrument as conceived by Menard consists of three
independent chambers stacked one above the other (Fig. 9.19) with inflatable user membranes hel d
together a t top and bottom by steel discs wit h a rigid hollow tube at the center. The top and bottom
chambers protec t th e middl e chambe r fro m th e en d effect s cause d b y th e finit e lengt h o f th e
apparatus, and thes e ar e known as guard cells. Th e middl e chamber wit h the end cell s togethe r i s
called the Probe. The pressuremeter consists of three parts, namely, the probe, the control uni t and
the tubing.
The Pressuremete r Tes t
The pressuremeter test involves the following:
1. Drillin g of a hole
2. Lowerin g the probe int o the hole an d clamping it at the desired elevation
3. Conductin g the test
344
Chapt er 9
V olumeter
control uni t
Pressure gaug e
Gas
Guard cel l
Measuring cel l
Guard cel l
Guard cel l
Measuring cel l
Water
Central tube
Guard cel l
Gas
(a) (b)
Figure 9 .1 9 Com pon en t s o f M en ar d pres s urem et e r
Drilling and Positionin g o f Prob e
A Menar d pressur e tes t i s carrie d ou t i n a hol e drille d i n advance . Th e drillin g o f th e hol e i s
completed usin g a suitabl e drilling rig whic h disturbs the soi l th e least . Th e diamete r o f th e bor e
hole, D
h
, in which the test is t o be conducted shall satisf y th e condition
D
h
< l.20D
p
(9.17 )
where D i s the diameter o f the probe under the deflated condition.
Typical size s o f the probe an d bore hol e are given in Table 9.7.
The prob e i s lowere d dow n the hol e soo n afte r borin g t o th e desire d elevatio n an d hel d i n
position by a clamping device. Pressuremeter test s are usually carried ou t at 1 m intervals in all the
bore holes .
Conducting the Test
With th e probe i n position i n the bor e hole, the test i s started b y opening th e valve s i n the control
unit for admittin g water and gas (o r water ) to the measuring cel l an d the guar d cell s respectively .
The pressur e i n the guar d cell s i s normall y kept equal t o th e pressur e i n th e measurin g cell . Th e
pressures t o the soi l through the measuring cell ar e applied by any one of the following methods:
1 . Equa l pressur e incremen t method,
2. Equa l volume increment method.
Soil Ex plorat io n 34 5
Table 9. 7 T ypica l s ize s o f prob e an d bor e hol e fo r pres s urem et e r t es t
Hole dia.
des ig n at ion
AX
BX
NX
Probe dia.
(m m )
44
58
70
/o
(cm )
36
21
25
/
(cm )
66
42
50
B ore hol e dia.
Nom in al
(m m )
46
60
72
M ax .
(m m )
52
66
84
Note: /
0
= lengt h of measuring cell ; / = lengt h o f probe.
If pressur e i s applie d by th e firs t method , eac h equa l incremen t o f pressur e i s hel d constan t
for a fixed lengt h of time, usuall y one minute. V olume readings ar e made afte r on e minut e elapse d
time. Normally ten equal increments of pressure ar e applied for a soil t o reach the limit pressure, p
t
.
If pressur e i s applie d b y th e secon d method , th e volume of th e prob e shal l b e increase d i n
increments equa l t o 5 percent o f the nomina l volume of the probe (i n the deflated condition ) an d
held constant for 30 seconds. Pressur e reading s ar e taken afte r 3 0 seconds o f elapsed time .
Steps i n both th e methods ar e continued until the maximum probe volume to be used i n the
test i s reached. Th e tes t ma y continue at each positio n from 1 0 to 1 5 minutes. This mean s tha t the
test i s essentially an undrained test i n clay soil s and a drained tes t i n a freely drainin g material.
Typical Test Resul t
First a typical curve based on the observed readings i n the field ma y be plotted. The plot is made of
the volume of the wate r read a t the volumete r in the control unit , v, as abscissa fo r eac h increment
of pressure , /? , a s ordinate . Th e curv e i s a resul t o f th e tes t conducte d o n th e basi s o f equa l
increments of pressure an d each pressur e hel d constant for a period o f one minute. This curve is a
raw curv e whic h require s som e corrections . Th e pressuremete r has , therefore , t o b e calibrate d
before i t is used i n design. A pressuremeter ha s t o be calibrated fo r
1. Pressur e loss , p
c
,
2. V olum e loss v
c
,
3. Differenc e i n hydrostatic pressur e head H
w
.
Corrected Plo t o f Pressure- V olume Curv e
A typical corrected plot of the pressure-volume curve is given in Fig. 9.20. The characteristi c
parts of this curve are :
1. Th e initial part of the curve OA. This curve is a result of pushing the yielded wall of the hole
back t o th e origina l position . At poin t A, th e at-res t conditio n i s suppose d t o hav e bee n
restored. The expansio n o f the cavit y is considered onl y from poin t A. V
Q
is the volume of
water required t o be injected ove r and above the volume V
c
of the probe unde r the deflated
condition. I f V
Q
is the total volume of the cavity at point A, we can writ e
V
0
=V
c +
v
Q
(9.18 )
where v
o
is the abscissa of point A. The horizontal pressure at point A is represented asp
owj
.
2. Th e secon d par t o f th e curv e i s AB. This i s suppose d t o b e a straigh t line portio n o f th e
curve and may represent the elastic range. Since AB gives an impression of an elastic range,
it i s called th e pseudo-elastic phas e of the test . Point A i s considered t o be the star t of the
pressuremeter tes t in most theories. Point B marks the end of the straight line portion of the
curve. The coordinate s o f point B ar e pyand v « wher e py is known as the creep pressure.
346 Chapt e r 9
3. Curv e BC marks the final phase . The plastic phase i s supposed t o start from point fi, and the
curve become s asymptoti c a t poin t C a t a larg e deformatio n o f th e cavity . Th e limit
pressure, p
r
i s usual l y define d as the pressure tha t is required t o doubl e the initia l volume
of th e cavity . It occur s a t a volume such that
v / - v
0
= V
0
= V
c +
v
0
(9.19 )
or v
t
=V
c
+ 2v
0
(9.20 )
v
Q
i s normall y limite d t o abou t 30 0 cm
3
fo r probe s use d i n AX an d B X holes . Th e initia l
volume o f thes e probe s i s o n th e orde r o f 53 5 cm
3
. Thi s mean s tha t (V , + 2v
Q
) i s o n th e orde r o f
1135 cm
3
. These value s may var y according t o the design o f the pressuremeter .
The reservoi r capacit y i n th e contro l uni t shoul d b e o f th e orde r o f 113 5 cm
3
. I n cas e th e
reservoir capacit y i s limited and p
l
i s not reached withi n its limit , the test , ha s t o be stoppe d a t that
level. I n suc h a case, th e limi t value, p
r
ha s t o be extrapolated .
At- Rest H orizonta l Pressur e
The at-res t tota l horizonta l pressure, p
oh
, at any depth, z, under the in-situ condition before drillin g
a hol e ma y be expressed a s
P
oh
=( rz-u)K
Q
+u, (9.21 )
where u = pore pressur e a t depth z,
7= gross uni t weight of the soil ,
K
Q
= coefficient of eart h pressur e for th e at-res t condition .
The value s o f 7 and K
Q
ar e generall y assume d takin g int o account th e typ e an d conditio n of
the soil . The por e pressur e unde r the hydrostati c condition is
u = r
w
( z-h
w
), (9.22 )
where y
w
= unit weight of water ,
h
w
= depth o f wate r tabl e fro m the ground surface .
As pe r Fig . 9.20 , p
om
i s th e pressur e whic h corresponds t o th e volum e V
Q
a t th e star t o f th e
straight lin e portio n o f th e curve . Sinc e i t ha s bee n foun d tha t i t i s ver y difficul t t o determin e
accurately p
om
, p
oh
ma y no t be equa l t o p
om
. As such , p
om
bear s n o relation t o wha t th e tru e eart h
pressure at-res t is . I n Eq . (9.21 ) K
Q
ha s t o b e assume d an d it s accurac y i s doubtful . I n suc h
circumstances i t i s no t possibl e t o calculat e p
oh
. However , p
om
ca n b e use d fo r calculatin g th e
pressuremeter modulu s E . Th e experienc e o f man y investigator s i s tha t a self-borin g
pressuremeter give s reliabl e values for/?
o/ j
.
The Pressuremete r M odulu s £
m
Since th e curve between point s A and B i n Fig. 9.2 0 i s approximately a straight line, the soi l i n this
region ma y b e assume d t o behav e a s a mor e o r les s elasti c material . Th e equatio n fo r th e
pressuremeter modulu s may b e expressed a s
E
n
=2G ( l+u) = 2(1 + u)V — (9.23 )
p m
Av
where G
s
is the shear modulus .
If V i s the volume at mid poin t (Fig 9.20) , we ma y write ,
347
Injected volume
» ~ Cavit y volume
Figure 9.2 0 A t ypical correct e d pres s urem et e r curv e
v
o
+ v
/
(9.24)
where V
c
is the volume of the deflated portio n of the measuring cell at zero volume reading on the
volumeter i n the control unit.
Suitable values/or ^may be assumed in the above equation depending on the type of soil. For
saturated clay soils // is taken as equal to 0.5 and for freely drainin g soils, the value is less. Since G
s
(shear modulus ) is no t ver y muc h affected b y a smal l variatio n i n ^u , Menar d propose d a constant
value of 0.33 for /L As such the resulting deformation modulus is called Menard' s Modulu s E
m
. The
equation for E
m
reduces t o
E
m
= 2.66V
m
^- (9.25 )
The followin g empirica l relationshi p ha s bee n establishe d fro m th e result s obtaine d fro m
pressuremeter tests . Undraine d shea r strengt h c
u
a s a functio n o f th e limi t pressur e ~p
l
ma y b e
expressed a s
c - c
~
(9.26)
where p
t
= p
t
- p
oh
an d p
oh
= total horizontal eart h pressur e for the at rest condition .
Amar and Jezequel (1972) have suggested anothe r equation of the for m
Pi
c
«
=
T o
+ 2 5 k p a
where both p an d c ar e i n kPa.
(9.27)
348 Chapt er 9
Example 9 .1 1
A pressuremeter tes t was carried out at a site at a depth of 7 m below th e ground surface. The water
table leve l wa s a t a dept h o f 1. 5 m. The averag e uni t weigh t of saturate d soi l i s 17. 3 kN/m
3
. Th e
corrected pressuremete r curv e is given in Fig. Ex. 9.11 and the depleted volume of the probe i s V
c
- 53 5 cm
3
. Determine the following.
(a) Th e coefficien t o f eart h pressur e for th e at-rest condition
(b) Th e Menar d pressuremeter modulus E
m
(c) Th e undrained shear strengt h c
u
. Assume that p
oh
= p
om
i n this case
Solution
From Fig . Ex 9.11, p
oh
= p
om
= 105 kPa
The effectiv e overburden pressur e i s
P'
Q
= 17. 3x7- 5. 5x9. 81 =67.2 kPa
The effectiv e horizonta l pressure i s
p'
0h
= 105- 5. 5x9. 81 = 51.0kPa
(a) From Eq (9.21)
51.0
u
P'
0
67. 2
(b) From E q (9.25 )
= 0.76
£*=2. 66V
m
200 40 0 600 800 100 0
Volume cm
Figure Ex . 9 .1 1
Soil Ex pl orat io n 34 9
From Fig. Ex 9.11
v
f
= 200 cm
3
p
f
= 530 kP a
v
0
= 160 cm
3
p
om
= 10 5 kPa
From Eq (9.24) V
m
= 5
35 += 715cm
3
A 530-10 5
Av 200-16 0
Now E
m
= 2.66 x 715 x 10.62 5 = 20,208 kPa
(c) From Eq (9.26 )
9
From Fi g Ex 9.11
845
Therefore c
u
= — = 94 kPa
From Eq (9.27 )
c =^ - + 25 =— + 25 = 109.5 kPa
" 1 0 1 0
9.1 0 TH E FL AT DI L ATOM ETE R TES T
The/Zaf dilatometer i s a n in-situ testin g devic e develope d i n Ital y b y Marchett i (1980) . I t i s a
penetration devic e tha t include s a latera l expansio n arrangemen t afte r penetration . Th e test ,
therefore, combine s man y o f th e feature s containe d i n th e con e penetratio n tes t an d th e
pressuremeter test . Thi s tes t ha s bee n extensivel y used fo r reliable, economica l an d rapi d in-situ
determination o f geotechnica l parameters . Th e fla t plat e dilatomete r (Fig . 9.21) consists o f a
stainless steel blade wit h a flat circular expandable membrane of 60 mm diameter on one side of the
stainless steel plate, a short distance above the sharpened tip. The size of the plate is 220 mm long,
95 mm wide and 1 4 mm thick. When at rest the external surface of the circular membrane i s flus h
with the surroundin g flat surfac e of the blade.
The prob e i s pushe d t o th e require d dept h b y makin g us e o f a ri g use d fo r a stati c con e
penetrometer (Fig. 9.10). The probe is connected to a control box at ground level through a string of
drill rods, electri c wire s for power supply and nylon tubing for the supply of nitrogen gas. Beneat h
the membran e i s a measuring devic e whic h turns a buzzer of f i n th e contro l box. The metho d o f
conducting the DMT i s as follows:
1. Th e probe is positioned at the required level. Nitrogen gas is pumped into the probe. When
the membrane i s just flush wit h the side of the surface, a pressure reading is taken which is
called the lift-off pressure . Approximate zero corrections are made. This pressur e i s called
Pi-
2. Th e prob e pressur e i s increased unti l the membran e expands by a n amount A/ = 1. 1 mm.
The corrected pressur e i s p
r
350 Chapt er 9
3. Th e nex t step i s to decrease th e pressure unti l the membrane returns t o the lif t of f position.
This correcte d readin g i s p
y
Thi s pressur e i s relate d t o exces s por e wate r pressur e
(Schmertmann, 1986) .
The detail s of the calculation lead t o the following equations.
p, ,
1 . Materia l index , I = —
p
2
-u
2. Th e latera l stres s index , K
n
=
Pi
-u
3. Th e dilatomete r modulus , E
D
= 34.7 ( p
2
- p
{
) kN/m
2
(9.28)
(9.29)
(9.30)
where, p' = effective overburden pressure = y' z
u = pore wate r pressur e equa l t o stati c water leve l pressur e
Y - effectiv e unit weight of soil
z = dept h o f probe leve l from ground surface
The latera l stres s inde x K
D
i s relate d t o K
Q
(th e coefficien t of eart h pressur e fo r th e at-res t
condition) an d t o OCR (overconsolidatio n ratio).
Marchetti (1980 ) has correlate d severa l soi l propertie s a s follows
K,
0.47
1.5
0.6
(9.31)
(9.32)
Wire
Pneumatic
t ubi ng
14 m m
membrane
-\ -
[
L
c
J
4- Pi
Flexible
membrane
-V n

\/
1.1 m m
Figure 9 .2 1 Il l us t rat io n o f a fl at plat e dil at om et e r (aft e r M archet t i 1980 )
Soil Ex plorat io n 351
200
05
o.. 0.2 0. 3 0.60. 8 1. 2 1. 8 3. 3
Material index I
D
Figure 9.2 2 Soi l profil e bas e d o n dilat om et er t es t (aft e r Schm ert m an n , 1986 )
= ( 0.5K
D
)
1.6
(9.33)
(9.34)
where E
s
i s the modulus of elasticity
The soil classification as developed by Schmertmann (1986) is given in Fig. 9.22. /
D
is related
with E
D
i n the development of the profile.
9 .1 1 FIEL D V ANE SH EAR TEST (V ST)
The vane shear test is one of the in-sit u test s used for obtaining the undrained shear strengt h of sof t
sensitive clays. It is in deep beds o f such material that the vane test is most valuabl e for the simpl e
reason that there is at present no other method known by which the shear strength of these clays can
be measured. The detail s of the V ST have already been explained in Chapter 8.
9 .1 2 FIEL D PL ATE L OAD TEST (PL T)
The fiel d plat e tes t i s th e oldes t o f th e method s fo r determinin g eithe r th e bearin g capacit y o r
settlement of footings. The detail s of PLT are discussed unde r Shallow Foundations in Chapter 13 .
352 Chapt er 9
9 .1 3 G EOPH Y SI CA L EXPL ORATI ON
The stratificatio n o f soil s an d rock s ca n b e determine d b y geophysica l method s o f exploratio n
which measur e change s i n certai n physica l characteristic s o f thes e materials , fo r exampl e th e
magnetism, density , electrical resistivity , elasticity or a combination of these properties . However ,
the ut i l i t y of these method s i n the fiel d o f foundation engineering is very limited since the method s
do no t quant i f y th e characteristic s o f th e variou s substrata . V ita l informatio n o n groun d wate r
conditions i s usual l y lacking . Geophysica l method s a t bes t provid e som e missin g informatio n
between widel y spaced bore hole s bu t they can not replace bor e holes . Two methods o f exploration
which ar e some time s usefu l ar e discussed briefl y in this section. They ar e
D, D , D , D
A
D
V elocity V
}
V elocity V
2
Rocky strat a
V elocity V
3
(a) Schemati c representation o f refraction method
Layer 1
Layer 2
R
R,
Electrode spacin g Electrod e spacin g
(b) Schemati c representation of electrical resistivit y metho d
Figure 9 .2 3 Geophys ica l m et hod s o f ex pl orat io n
Soil Ex pl orat io n 35 3
1. Seismi c Refractio n Method,
2. Electrica l Resistivit y Method.
Seismic Refractio n M etho d
The seismi c refractio n metho d i s base d o n th e fac t tha t seismi c wave s hav e differen t velocitie s i n
different type s of soils (or rock). The waves refract whe n they cross boundaries between differen t type s
of soils. If artificial impulses are produced either by detonation of explosives or mechanical blows with
a heavy hammer a t the groun d surfac e or at shallow dept h withi n a hole, thes e shock s generat e thre e
types of waves. In general, only compression wave s (longitudinal waves) are observed. These waves are
classified as either direct, reflected or refracted. Direct waves travel in approximately straight lines from
the sourc e o f th e impuls e to the surface . Reflecte d o r refracted wave s underg o a change i n directio n
when they encounter a boundary separating media of different seismi c velocities. The seismic refraction
method is more suited to shallow exploration fo r civil engineering purposes .
The metho d start s b y inducin g impact o r shoc k wave s int o the soi l a t a particular location .
The shoc k wave s ar e picke d u p b y geophones . I n Fig . 9.23(a) , poin t A i s th e sourc e o f seismi c
impulse. The point s D^ throug h D
g
represent th e locations of the geophones o r detectors whic h are
installed i n a straigh t line . The spacing s o f the geophone s ar e dependen t o n the amoun t o f detai l
required and the depth of the strata being investigated. In general, th e spacing mus t be such that the
distance fro m D j t o D
8
i s thre e t o fou r time s th e dept h t o b e investigated . Th e geophone s ar e
connected b y cabl e to a central recording device . A series of detonations o r impact s ar e produce d
and th e arriva l tim e o f th e firs t wav e a t eac h geophon e positio n i s recorde d i n turn . Whe n th e
distance betwee n sourc e and geophone i s short, the arrival time wil l be that of a direct wave . When
the distance exceeds a certain value (depending on the thickness of the stratum), the refracted wave
will be the first t o be detected b y the geophone. Thi s i s because the refracted wave, although longer
than that of the direct wave, passes through a stratum of higher seismi c velocity .
A typical plo t of test result s for a three laye r system i s given i n Fig. 9.23(a ) wit h the arriva l
time plotte d agains t th e distanc e sourc e an d geophone . A s i n th e figure , i f th e source-geophon e
spacing i s mor e tha n the distanc e d
r
whic h i s the distanc e fro m th e sourc e t o point B, the direc t
wave reaches the geophon e i n advance o f the refracted wav e and the time-distanc e relationshi p i s
represented b y a straigh t lin e AB throug h th e origi n represente d b y A. I f o n th e othe r hand , th e
source geophon e distanc e i s greate r tha n d
{
, th e refracte d wave s arriv e i n advanc e o f th e direc t
waves and the time-distance relationship is represented b y another straight line BC which will have
a slop e differen t fro m tha t of AB. Th e slope s o f the lines AB an d BC ar e represented b y \ IV
r
an d
1/V
2
respectively , wher e V
{
an d V
2
ar e th e velocitie s o f th e uppe r an d lowe r strat a respectively .
Similarly, the slope of the thir d line CD i s represented b y 1/V
3
i n the thir d strata.
The genera l type s o f soi l o r rocks ca n be determined fro m a knowledge o f thes e velocities .
The dept h H
{
o f the top strat a (provide d th e thickness of the stratu m i s constant) can be estimated
from th e formul a
(9.35a)
2 ' " 1
The thickness of the second laye r (//
2
) i s obtained fro m
The procedure i s continued i f there ar e more tha n three layers .
If the thicknes s o f any stratum is not constant , average thicknes s i s taken.
354 Chapt er 9
Table 9. 8 Ran g e of s eis m i c vel ocit ie s i n s oil s n ea r t h e s ur fac e o r a t s hal l o w
dept hs (aft e r Pec k e t al. , 1974)
Ma t e r i a l
1. Dry silt , sand , loos e gravel , loam, loos e roc k talus , and
moist fine-graine d to p soi l
2. Compact t il l , indurate d clays, compact claye y gravel,
cemented san d an d san d clay
3. Rock, weathered , fracture d or partl y decompose d
4. Shale , soun d
5. Sandstone , soun d
6. Limestone, chalk, sound
7. Igneou s rock , soun d
8. Metamorphi c rock , soun d
V el oci t y
f t / sec m/se c
600-2500
2500-7500
2000-10,000
2500-11,000
5000-14,000
6000-20,000
12,000-20,000
10,000-16,000
180-760
760-2300
600-3000
760-3350
1500-4300
1800-6000
3650-6000
3000-4900
The followin g equations may be use d for determinin g the depths H, an d H
2
i n a three laye r
strata:
t,V,
2 cos a
(9.36)
2 cos/?
(9.37)
where t
{
= AB
r
(Fig . 9.23a) ; th e poin t B
l
i s obtaine d o n th e vertica l passin g throug h A b y
extending the straigh t line CB,
t
2
= (AC j - A5j) ; ACj i s the intercept on the vertical through A obtained by extending the
straight line DC,
a = sin~
l
(V /V
2
),
j8 = sin-
1
(V
2
/V
3
). (9.38 )
a an d ( 3 are the angle s of refraction of the firs t an d secon d stratu m interfaces respectively .
The formulae used to estimate the depths from seismi c refraction survey data are based on the
following assumptions:
1. Eac h stratu m is homogeneous an d isotropic .
2. Th e boundaries between strat a are either horizontal or inclined planes.
3. Eac h stratu m is of sufficient thicknes s to reflect a change i n velocity on a time-distance plot .
4. Th e velocit y o f wave propagation for each succeedin g stratum increases wit h depth.
Table 9. 8 give s typica l seismi c velocitie s i n variou s materials . Detaile d investigatio n
procedures fo r refraction studies are presented by Jakosky (1950).
Electrical Resistivit y M etho d
The method depends o n differences in the electrical resistanc e o f different soi l (and rock) types . The
flow o f curren t throug h a soi l i s mainl y du e t o electrolyti c actio n an d therefor e depend s o n th e
Soil Ex pl orat io n 35 5
concentration o f dissolve d salt s i n th e pores . Th e minera l particle s o f soi l ar e poo r conductor s o f
current. The resistivit y of soil , therefore , decreases as bot h water content and concentration o f salts
increase. A dense clea n san d above the water table, for example, woul d exhibit a high resistivity due
to it s lo w degree o f saturatio n and virtua l absence o f dissolve d salts . A saturate d cla y o f high voi d
ratio, on the other hand, would exhibit a low resistivity due to the relative abundance of pore water and
the free ion s i n that water.
There ar e several method s by whic h the fiel d resistivit y measurement s ar e made . Th e mos t
popular of the methods i s the Wenner Method.
Wenner M etho d
The Wenne r arrangemen t consist s o f fou r equall y space d electrode s drive n approximatel y 2 0 cm
into the ground as shown in Fig. 9.23(b) . In this method a dc current of known magnitude is passe d
between th e tw o oute r (current ) electrodes , thereb y producin g withi n th e soi l a n electri c field ,
whose patter n i s determine d b y th e resistivitie s o f th e soil s presen t withi n th e fiel d an d th e
boundary conditions. The potentia l drop E for the surface current flow line s i s measured b y means
of the inner electrodes. Th e apparent resistivity, R, i s given by the equatio n
R = — —(9.39 )
It i s customar y t o expres s A i n centimeters, E i n volts, / i n amperes, an d R ohm-cm. Th e apparen t
resistivity represent s a weighted averag e of true resistivity to a depth A in a large volume of soil , the
soil close to the surface being mor e heavil y weighted than the soil at greater depths . The presence of
a stratum of low resistivity forces the current to flow close r t o the surface resulting in a higher voltage
drop and hence a higher value of apparent resistivity. The opposite i s true if a stratum of low resistivity
lies below a stratum of high resistivity.
The metho d know n a s sounding i s use d whe n th e variatio n o f resistivit y wit h dept h i s
required. Thi s enable s roug h estimate s t o be mad e o f th e type s an d depth s o f strata . A serie s of
readings ar e taken , th e (equal ) spacin g o f th e electrode s bein g increase d fo r eac h successiv e
reading. However , th e cente r o f th e fou r electrode s remain s a t a fixe d point . A s th e spacin g i s
increased, the apparent resistivity is influenced by a greater dept h of soil. If the resistivity increase s
with th e increasin g electrod e spacings , i t can b e conclude d tha t a n underlyin g stratum of highe r
resistivity i s beginnin g t o influenc e th e readings . I f increase d separatio n produce s decreasin g
resistivity, on the other hand, a lower resistivit y is beginning to influence the readings .
Apparent resistivity is plotted agains t spacing, preferably, on log paper. Characteristic curve s
for a two layer structur e are shown i n Fig. 9.23(b) . For curve C
p
the resistivity of layer 1 is lower
than that of 2; for curve C
2
, layer 1 has a higher resistivity than that of layer 2. The curves becom e
asymptotic to lines representing the true resistance R
r
and R
2
of the respective layers . Approximate
layer thicknes s ca n b e obtaine d b y comparin g th e observe d curve s o f resistivit y versu s electrod e
spacing wit h a set of standard curves .
The procedure known as profilingi s used in the investigation of lateral variatio n of soil types.
A series of readings i s taken, the four electrodes bein g moved laterall y as a unit for each successiv e
reading; the electrode spacing remains constant for each reading o f the series. Apparent resistivity
is plotted agains t the center position of the four electrodes , t o natural scale; such a plot can be used
to locate th e position o f a soil o f high or low resistivity. Contours o f resistivity can be plotted ove r
a given area .
The electrica l metho d o f exploratio n ha s bee n foun d t o b e no t a s reliabl e a s th e seismi c
method a s the apparent resistivit y of a particular soi l or rock ca n vary over a wide range of values.
Representative value s of resistivity are given i n Table 9.9 .
356 Chapt er 9
Table 9. 9 Repres en t at iv e value s o f res is t ivit y . T h e value s ar e ex pres s e d i n unit s o f
10
3
ohm -c m (aft e r Pec k e t al. , 1974 )
M at erial Res is t ivit y ohm -c m
Clay an d saturate d sil t
Sandy cla y an d we t silt y san d
Clayey san d an d saturate d san d
Sand
Gravel
Weathered roc k
Sound roc k
0-10
10-25
25-50
50-150
150-500
100-200
150-4,000
Example 9 .1 2
A seismi c surve y wa s carried out for a large projec t t o determine th e natur e o f the substrata . Th e
results o f the surve y ar e give n i n Fig. Ex 9.12 i n the for m o f a graph. Determin e th e depth s o f the
strata.
Solution
Two methods may be use d
1. Us e of Eq (9.35)
2. Us e o f Eqs (9.36 ) an d (9.37 )
First we have t o determine th e velocities i n each stratu m (Fig . Ex . 9.12) .
I
V , ."/••/.':.'•>•'••" - Surface soi l • .• '. " ;:.i.'.-"
:
:
)
: H\
V
2
•. •''.';: . Sand an d loos e gravel
Rock
Afl, = 8.75 x 10'
3
sec
AC, = 33.75 x 10'
3
sec
AC
2
= 38.75 x 10~
3
sec
J, =2. 188m
d
2
= 22.5 m
A5=12. 75x 10'
3
sec
H,
20 d
^
3 0
Distance m
40 50
Figure Ex . 9 .1 2
Soil Expl or at i o n 35 7
distance 2.18 8
= — -
AB 12.75X1CT
3
= 172 m / sec
V = -2 -= --_= 750 m/ se c
7
2
AC-AB
l
(7.75 - 1.75)5
In th e same way , the velocit y i n the thir d stata can be determined. The velocit y obtained i s
V
3
= 2250 m/sec
Method 1
From Eq (9.35 a), the thickness H
{
o f the top layer is
2.188 /750-17 2
= 0.83 m
2 V 100 0
From Eq (9.35b) the thickness H
2
i s
i/ n o < no a
22
'
5
2250-75 0
H> > =0.85x0.8 3 + —
3000
= 0.71 + 7.955 = 8.67m
Method 2
From Eq (9.36 )
1
2 cos a
t
{
= AB
l
= 1.75 x 5 x 1Q-
3
sec(Fig.Ex.9.12)
i V , i 17 2
a = sin
!

L
= sin
l
= 13.26°
V
2
75 0
cosa = 0.973 3
_ _ 12. 75xl Q-
3
xl 72
//, = = 1.13 m
1
2x0.973 7
From Eq (9.37)
t
2
V
2
2
2cos/ 7
t
2
= 5 x 5 x10~
3
sec
, 75 0
/?= shr
1
—— = 19.47°; cos J 3= 0.942 8
2250
_5 x 5 x l O"
3
x 7 5 0
2
~ 2x0.942 8
n f t
.
= 9.94 m
358 Chapt e r 9
9 .1 4 PL ANNI N G O F SOI L EXPL ORATI ON
The planne r has t o consider th e following points before makin g a program:
1. Type , siz e an d importanc e of the project .
2. Whethe r th e sit e investigation is preliminary or detailed .
In the case of large projects , a preliminary investigation is normally required for the purpose of
1. Selectin g a sit e and makin g a feasibility study of the project ,
2. Makin g tentative designs an d estimates of the cos t o f the project .
Preliminary sit e investigation needs onl y a few bore hole s distribute d suitabl y over the are a fo r
taking samples. The data obtained from the field and laboratory tests must be adequate to provide a fairly
good ide a of the strengt h characteristics o f the subsoi l for making preliminary drawing s and design. In
case a particular site is found unsuitable on the basis of the study, an alternate site may have to be chosen.
Once a sit e i s chosen , a detaile d soi l investigatio n i s undertaken . Th e plannin g o f a soi l
investigation include s the following steps:
1. A detailed stud y of the geographica l conditio n of the are a whic h include
(a) Collectio n o f al l th e availabl e information about th e site , including the collectio n o f
existing topographical an d geological maps ,
(b) Genera l topographica l feature s of the site ,
(c) Collectio n o f th e availabl e hydrauli c conditions , suc h a s wate r tabl e fluctuations ,
flooding o f the sit e etc,
(d) Acces s t o the site .
2. Preparatio n o f a layout plan of the project .
3. Preparatio n o f a borehol e layou t pla n whic h include s the depth s an d th e numbe r o f bor e
holes suitabl y distributed over th e area .
4. Markin g on the layout plan any additional types of soi l investigation.
5. Preparatio n o f specifications and guidelines for the fiel d executio n of the various element s
of soi l investigation.
6. Preparatio n o f specification s an d guid e line s fo r laborator y testin g o f th e sample s
collected, presentatio n of fiel d an d laborator y tes t results, writing of report , etc .
The planner can make an intelligent, practical and pragmatic plan if he is conversant with the
various element s o f soi l investigation.
Depths and Numbe r o f B or e H ole s
Depths o f B or e H oles
The dept h u p to which bore hole s shoul d be driven is governed b y the dept h of soi l affecte d by the
foundation bearin g pressures . Th e standar d practic e i s t o tak e th e boring s t o a dept h (calle d th e
significant depth ) a t whic h the exces s vertica l stres s cause d b y a full y loade d foundatio n i s of the
order of 20 per cent or less of the net imposed vertica l stress a t the foundation base level. The dept h
the borehole a s per this practice works out to about 1. 5 times the least widt h of the foundation from
the bas e leve l o f th e foundation as shown i n Fig. 9.24(a). Where stri p o r pa d footing s ar e closel y
spaced whic h result s i n the overlappin g of th e stresse d zones , th e whol e loade d are a become s i n
effect a raft foundatio n with correspondingly deep borings as shown i n Fig. 9.24(b ) and (c). In the
case o f pil e o r pier foundation s the subsoi l shoul d be explored t o the depths require d t o cover th e
soil lyin g eve n belo w th e tip s o f pile s (o r pil e groups ) an d piers whic h ar e affecte d b y th e load s
transmitted t o th e deepe r layers , Fig . 9.24(d) . I n cas e roc k i s encountere d a t shallo w depths ,
foundations ma y hav e t o res t o n rock y strata . Th e borin g shoul d als o explor e th e strengt h
characteristics o f rocky strat a i n such cases.
Soil Ex pl orat io n 359
(a) Footings place d fa r apart (b ) Footings place d a t closed interval s
//x*//x\
1
U- B
_J
-T
i.;
1
B'
^^^ "'
•f
1
J-
i> »
*—
-
•>/
-
<
V ^^^<5\
1
(2/3) D 1
A D

L
(c) Raft foundation (d ) Pile foundatio n
Figure 9.24 Dept h o f bor e hole s
Number o f B or e H ole s
An adequat e number of bore holes i s needed t o
1. Provid e a reasonabl y accurat e determinatio n o f th e contour s o f th e propose d bearin g
stratum,
2. Locat e an y sof t pocket s i n the supporting soil which would adversely affec t th e safet y and
performance o f the proposed design .
The number of bore holes which need to be driven on any particular site is a difficult proble m
which i s closel y linke d wit h th e relativ e cos t o f the investigatio n an d th e projec t fo r whic h i t i s
undertaken. When the soi l is homogeneous over the whole area, the number of bore holes could be
limited, but i f the soi l condition is erratic, limiting the number would be counter productive.
9 .1 5 EXECU TI O N O F SOI L EXPL ORATI ON PROG RA M
The three limbs o f a soil exploration are
1. Planning ,
2. Execution ,
3. Repor t writing.
360 Chapt er 9
All thre e limb s ar e equall y importan t fo r a satisfactor y solutio n o f th e problem. However , th e
execution of the soi l exploration progra m act s a s a bridge betwee n plannin g and report writing , and as
such occupie s a n importan t place . No amoun t of plannin g would hel p repor t writing , i f the fiel d an d
laboratory work s ar e not executed wit h diligence an d care. I t is essential tha t the execution par t shoul d
always be entrusted to well qualified, reliabl e and resourceful geotechnical consultant s who wil l also be
responsible fo r report writing .
Deployment o f Personne l and Equipmen t
The geotechnical consultan t should have wel l qualified and experienced engineer s an d supervisor s
who complet e th e wor k pe r th e requirements . Th e fir m shoul d hav e th e capacit y t o deplo y a n
adequate numbe r of rigs an d personne l fo r satisfactor y completion o f the job o n time .
BOREHOLE LOG
Job No.
Project: Farakk a STP P
Date: 6-4-84
BHNo. : 1
GL: 64. 3 m
Location: W B
Boring Method : Shel l & Auge r
WTL: 63. 0 m
Supervisor: X
Dia. o f BH 15c m
Soil Typ e
Yellowish
stiff cla y
Greyish
sandy sil t
med. dens e
Greyish
silty san d
dense
Blackish
very stif f
clay
62.3
56.3
53. 3
- 1.0
3.3
-5. 0
- 7.5
-9.0
SPT
15
cm
14
15
15
cm
10
16
18
10
15
16
21
23
14
N
14
26
37
41
24
D
D
W
D
D
Remarks
D = disturbed sample ; U = undisturbed sample ;
W = water sample ; N = SPT valu e
Figure 9 .2 5 A t ypica l bore-hol e lo g
Soil Ex plorat io n 36 1
B oring L og s
A detailed recor d o f boring operations an d other test s carried ou t i n the field i s an essential par t of
the field work. The bore hole log is made during the boring operation. The soi l is classified base d on
the visual examination of the disturbe d sample s collected . A typical exampl e o f a bore hol e log i s
given i n Fig. 9.25. The log shoul d include the difficulties face d during boring operation s including
the occurrenc e o f sand boils, an d the presence o f artesian wate r conditions i f any, etc.
In-situ Test s
The field work may also involve one or more of the in-situ tests discussed earlier. The record should
give the detail s of the tests conducted wit h exceptional clarity.
L aboratory Testin g
A preliminary examination of the nature and type of soil brought to the laboratory i s very essentia l
before decidin g upon the type and number of laboratory tests . Normall y the SPT samples ar e used
for thi s purpose. Firs t th e SPT samples shoul d be arranged bor e wis e and dept h wise . Each o f the
samples shoul d be examined visually . A chart should be made giving the bore hole numbers and the
types of test s t o be conducte d o n each sampl e dept h wise . An experienced geotechnica l enginee r
can do this job wit h diligence and care .
Once the types of tests are decided, th e laboratory assistant shoul d carry out the tests wit h all
the care require d for each of the tests. The test results should next be tabulated on a suitable format
bore wis e an d th e soi l i s classifie d accordin g t o standar d practice . Th e geotechnica l consultan t
should examine each of the tests before being tabulated. Unreliable test results should be discarded .
G raphs an d Chart s
All the necessary graphs and charts are to be made based on the field and laboratory tes t results. The
charts and graphs should present a clear insight into the subsoil conditions over the whole area. The
charts mad e shoul d hel p th e geotechnica l consultan t t o mak e a decisio n abou t th e typ e o f
foundation, th e strengt h and compressibility characteristic s o f the subsoi l etc .
9 .1 6 REPOR T
A repor t i s th e fina l documen t o f th e whol e exercis e o f soi l exploration . A repor t shoul d b e
comprehensive, clea r and t o the point. Man y can writ e reports, but only a very few can produce a
good report . A report writer should be knowledgable, practical and pragmatic. No theory, books or
codes o f practice provide all the materials required to produce a good report . I t is the experience of
a numbe r of year s o f dedicate d servic e i n th e fiel d whic h help s a geotechnica l consultan t mak e
report writin g an art . A good repor t shoul d normally comprise th e following:
1. A general descriptio n o f the natur e of the project and it s importance .
2. A general description o f the topographical features and hydrauli c conditions o f the site .
3. A brief descriptio n o f the various field and laboratory test s carried out .
4. Analysi s and discussion o f the test results
5. Recommendation s
6. Calculation s for determining saf e bearing pressures, pil e loads, etc .
7. Table s containin g borelogs, an d other fiel d an d laboratory tes t result s
8. Drawing s which include an index plan, a site-plan, test result s plotted i n the form of chart s
and graphs , soil profiles , etc .
362
9 .1 7 PROB L EM S
Chapt er 9
9.1 Comput e the area rati o of a sampling tube given the outside diameter = 10 0 mm and inside
diameter = 94 mm. I n what types o f soi l ca n thi s tube be use d fo r sampling?
9.2 A standar d penetratio n tes t wa s carrie d ou t a t a site . Th e soi l profil e i s give n i n
Fig. Prob . 9. 2 wit h th e penetration values . The averag e soi l dat a ar e give n for each layer .
Compute th e correcte d value s of N an d plo t showing the
(a) variatio n of observe d value s with depth
(b) variatio n of corrected value s with depth for standar d energy 60 %
Assume: E
h
= 0.7, C
d
= 0.9, C
s
= 0.85 an d C
b
= 1.05
Depth (m)
0
2 m
2-
Sand
• y
s d t
= 1 85 k N/ m
3

4 m
Figure Prob . 9. 2
N-values
20
A
8-
10-
1 9
14
'V ' . '. ' ' ' '
''-; '.- • ' ' • ;
B
; • ''•:
'"' •"•' ' V - ?• '"• • ..' • ''".' i .'"• '• '-.v'
: V
-''
:
\'..'' •'•'.' i .'"•''•'•.'J''
J
v • "• •
:
' \j.
(
- /•- . '.*• > *. ,v - V
:
'i.^; .V
:
*y ... v - V ;'*; •
. 7 ^ y
a t
= is 5 kN/m
3
J'V 'c , • *'.;--• '•/ -. - .'/"j;":^
Sand • - . ;• • • • ' . . : . ; ' . ' .-.• ' • ' . ; • . '
. : y
s a l
=1981kN/ m
3
:/ * . . • . , . '- . - : - . - ' ? . ,
. . C . . - . . . - ; • • - • • • '
,"
m
t-
m
m
[. ' • '
. 3 0
' f " ^ *
/ /^
]
CV
1 5
^ /''A/- ' 19

9.3 Fo r the soi l profil e given in Fig. Pro b 9.2 , comput e th e corrected value s of W for standar d
energy 70%.
9.4 Fo r the soil profile given in Fig. Prob 9.2, estimate the average angl e of friction fo r the sand
layers base d o n the following:
(a) Tabl e 9. 3
(b) E q (9.8 ) b y assumin g the profil e contain s les s tha n 5 %fine s ( D
r
ma y b e take n fro m
Table 9.3 )
Estimate th e value s of 0 and D
r
for 60 percent standar d energy.
Assume: N
cor
= N
6Q
.
9.5 Fo r the corrected value s of W
60
given in Prob 9.2, determine th e unconfine d compressiv e
strengths o f cla y a t point s C an d D i n Fi g Pro b 9. 2 b y makin g us e o f Tabl e 9. 4 an d
Eq. (9.9) . Wha t i s the consistenc y o f the clay?
9.6 A stati c con e penetratio n tes t wa s carrie d ou t a t a sit e usin g a n electric-frictio n con e
penetrometer. Fig . Pro b 9. 6 give s th e soi l profil e an d value s o f q
c
obtaine d a t variou s
depths.
(a) Plo t the variation of q wit h depth
Soil Ex plorat io n 363
Depth (ft )
0
2-
Figure Prob. 9. 6
(b) Determin e th e relative densit y of the sand at the points marked i n the figur e by using
Fig. 9.14 .
(c) Determin e th e angl e o f interna l frictio n o f th e san d a t th e point s marke d b y using
Fig. 9.15 .
9.7 Fo r the soi l profil e given in Fig. Pro b 9.6, determin e the unconfmed compressiv e strengt h
of the clay at the points marked i n th e figure using Eq (9.14) .
9.8 A stati c con e penetratio n tes t carrie d ou t a t a sit e a t a dept h o f 5 0 f t gav e th e following
results:
(a) con e resistance q
c
= 250 t /ft
2
(b) averag e effectiv e unit weight of the soi l = 11 5 lb/ft
3
Classify th e soi l for friction ratio s of 0.9 and 2.5 percent .
9.9 A stati c con e penetratio n tes t wa s carrie d ou t a t a sit e usin g a n electric-frictio n con e
penetrometer. Classif y the soi l for the following data obtained from the sit e
q (MN/m
2
) Friction rati o R
f
%
25
6.5
12.0
1.0
5
0.50
0.25
5.25
Assume i n al l the above cases that the effective overburden pressur e is 50 kN/m
2
.
9.10 Determin e the relative density and the friction angle if the corrected SP T value 7V
60
at a site
is 30 from E q (9.16) and Tabl e 9.6. What are the values o/D
r
and 0 for N
J Q
1
9.11 Fi g Pro b 9.11 gives a corrected pressuremete r curve . The value s of p
om
, p
f
an d p
l
an d the
corresponding volumes ar e marked on the curve. The test was conducted at a depth of 5 m
below the ground surface. The average unit weight of the soil is 18. 5 kN/m
3
. Determine the
following:
364 Chapt er 9
1400
1200
1000
800
600
400
200
p
om
= 200 kPa, v
0
= 180 cm
3
; p
f
= 66 0 kPa; v
f
= 22 0 cm
3
;
p
t
= 1100kPa;v
/
= 700cm
3
Pan,
Pi
100 20 0 600 700
Figure Prob . 9 .1 1
(a) Th e coefficien t of eart h pressur e fo r the at-res t condition
(b) Th e Menar d pressuremete r modulus
(c) Th e undraine d shear strengt h c
u
9.12 A seismi c refractio n survey of an are a gav e the followin g data:
(i) Distanc e from impac t point to
geophone i n m 1 5 3 0
(ii) Tim e o f firs t wav e
arrival i n sec 0.02 5 0.0 5
60 80 100
0.10 0.1 1 0.1 2
(a) Plo t the time travel versus distance an d determine velocities of the top and underlyin g
layer o f soil
(b) Determin e th e thicknes s of the top layer
(c) Usin g the seismi c velocitie s evaluate the probabl e eart h material s i n the two layer s
CH APTER 1 0
STABILITY OF SLOPES
10.1 I NTRODU CTI O N
Slopes o f earth ar e o f two types
1. Natura l slopes
2. Ma n made slope s
Natural slope s ar e thos e tha t exist i n natur e and ar e formed b y natura l causes. Suc h slope s
exist in hilly areas. The sides of cuttings, the slopes of embankments constructed for roads, railway
lines, canal s etc . an d th e slope s o f eart h dams constructed fo r storin g water ar e example s o f man
made slopes . The slope s whethe r natural or artificial may be
1. Infinit e slope s
2. Finit e slopes
The term infinit e slop e i s used to designate a constant slope of infinite extent . The long slope
of the face of a mountain is an example of this type, whereas finite slope s ar e limited in extent. The
slopes of embankments and earth dams ar e examples of finit e slopes . Th e slope length depends on
the height of the dam or embankment.
Slope Stability: Slop e stabilit y i s a n extremel y importan t consideratio n i n th e desig n an d
construction of earth dams. The stability of a natural slope i s also important . The result s of a slope
failure ca n often be catastrophic, involvin g the loss of considerable propert y an d many lives.
Causes of Failure of Slopes: The important factors that cause instability in a slope and lead t o
failure ar e
1. Gravitationa l force
2. Forc e due t o seepage water
3. Erosio n o f the surfac e of slopes du e t o flowing wate r
365
366 Chapt er 1 0
4. Th e sudde n lowering of water adjacent to a slop e
5. Force s due t o earthquakes
The effec t o f al l th e force s liste d abov e i s t o caus e movemen t o f soi l fro m hi g h point s
to lo w poi nt s . The mos t i mpor t an t of suc h force s i s the componen t o f gravi t y t hat act s i n th e
direction o f probabl e mot i on . Th e var i ou s effects o f fl owi n g or seepin g wat e r ar e general l y
recognized a s ver y i mpor t an t i n st abi l i t y probl ems , bu t ofte n thes e effect s hav e no t bee n
properly i dent i fi ed . I t i s a fac t t ha t the seepag e occurrin g wi t hi n a soi l mas s cause s seepag e
forces, whi c h hav e muc h greate r effec t t ha n i s commonl y realized .
Erosion o n th e surfac e o f a slop e ma y b e th e caus e o f th e remova l o f a certai n weigh t of
soil, an d ma y thu s lea d t o an increase d stabilit y as far a s mass movemen t i s concerned. O n the
other hand , erosio n i n the for m o f undercut t i ng at the toe ma y increas e the heigh t o f th e slope ,
or decreas e th e lengt h o f th e i nci pi ent f ai l ure surface , t hu s decreasing th e stability .
When ther e i s a lowering of the ground water or of a freewater surface adjacen t t o the slope ,
for exampl e i n a sudde n drawdow n o f th e wate r surfac e i n a reservoi r ther e i s a decreas e i n th e
buoyancy o f th e soi l whic h i s i n effec t a n increas e i n th e weight . Thi s increas e i n weigh t cause s
increase i n th e shearin g stresse s tha t ma y o r ma y no t b e i n par t counteracte d b y th e increas e i n
Component of weight
C
Failure
surface
(a) Infinit e slope (b) An earth dam
Ground
water table
Seepage
parallel t o slope
(c) Seepage below a natural slope
Lowering of water
from leve l A t o B
Earthquake
force
(d) Sudden drawdown conditio n (e ) Failure due t o earthquake
Figure 1 0. 1 Force s t hat ac t o n eart h s lope s
St abilit y o f Slope s 36 7
shearing strength, depending upon whether or not the soil is able to undergo compression whic h the
load increas e tend s t o cause . I f a larg e mas s o f soi l i s saturate d an d i s o f lo w permeability ,
practically no volume changes will be able to occur except at a slow rate, and in spite of the increase
of load the strengt h increase ma y be inappreciable .
Shear a t constant volume ma y b e accompanie d b y a decreas e i n the intergranula r pressur e
and a n increas e i n the neutra l pressure. A failur e ma y b e caused b y suc h a condition in which the
entire soi l mas s passes int o a state of liquefaction and flows lik e a liquid. A condition of this type
may be developed i f the mass of soi l is subject to vibration, for example, due t o earthquake forces.
The various forces tha t act on slopes ar e illustrated in Fig. 10.1 .
1 0.2 G ENERA L CONSI DERATI ONS AND ASSU M PTI ONS I N TH E
ANAL Y SI S
There ar e three distinc t parts t o an analysi s of the stabilit y of a slope. The y are :
1. Testing o f sample s to determine the cohesio n and angl e o f internal frictio n
If the analysis is for a natural slope, it is essential tha t the sample be undisturbed. In such important
respects a s rate of shear application and stat e of initial consolidation, the condition of testing must
represent a s closel y a s possibl e th e mos t unfavorabl e conditions ever likel y t o occur i n the actual
slope.
2. The study o f items whic h ar e known to enter but whic h canno t b e accounted
for i n the computations
The mos t importan t o f suc h items i s progressive crackin g whic h wil l start a t the to p o f the slope
where th e soi l i s i n tension, and aide d b y wate r pressure, ma y progress t o considerabl e depth . I n
addition, ther e ar e th e effect s o f th e non-homogeneou s natur e o f th e typica l soi l an d othe r
variations from th e ideal conditions which must be assumed.
3. Computatio n
If a slope i s t o fai l alon g a surface, al l the shearin g strengt h must be overcome alon g tha t surface
which then becomes a surface of rupture. Any one such as ABC i n Fig. 10. 1 (b) represents one of an
infinite numbe r of possible trace s o n which failure migh t occur.
It is assumed tha t the problem i s two dimensional, which theoretically requires a long length
of slope normal to the section. However, i f the cross section investigate d holds for a running length
of roughly two or more times th e trace of the rupture, i t i s probable tha t the two dimensional cas e
holds withi n the required accuracy .
The shea r strengt h of soi l i s assumed t o follow Coulomb' s la w
s = c' + d ta n 0 "
where,
c' - effectiv e uni t cohesio n
d = effective norma l stress o n the surfac e of rupture = (cr - u)
o -tota l normal stres s on the surface of rupture
u - por e water pressur e on the surface of rupture
0' = effective angle of internal friction.
The ite m o f grea t importanc e i s the loss o f shearin g strengt h whic h many clays sho w when
subjected t o a large shearin g strain . The stress-strai n curve s for suc h clay s sho w th e stres s risin g
with increasin g strai n t o a maximu m value, afte r whic h i t decrease s an d approache s a n ultimat e
368 Chapt e r 1 0
value whic h ma y b e muc h les s tha n th e maximum . Sinc e a ruptur e surfac e tend s t o develo p
progressively rathe r tha n wit h all the point s a t the same stat e o f strain , i t i s generall y th e ultimat e
value that shoul d be use d fo r the shearing strengt h rather tha n the maximum value.
1 0.3 FACTO R OF SAFETY
In stabilit y analysis, two type s of factor s o f safet y ar e normall y used. They ar e
1. Facto r o f safet y wit h respect t o shearing strength.
2. Facto r o f safet y wi th respect t o cohesion. This i s termed th e factor of safet y with respect t o
height.
Let,
F
S
= facto r o f safet y wit h respect t o strengt h
F, = facto r o f safet y wit h respect t o cohesio n
F
H
= facto r of safet y wit h respect t o height
F, = facto r o f safet y wi t h respec t t o friction
c' = mobilize d cohesio n
m
0' = mobilize d angl e of friction
T = averag e valu e of mobilize d shearin g strengt h
s = maximu m shearin g strength.
The facto r o f safet y wit h respect t o shearing strength, F
5
, may be writte n as
s c' + <j' ta n <j)'
F
>=7 = ; -
The shearin g strengt h mobilize d a t each poin t on a failure surfac e may be writte n as
c' .
T — _ _ L /T
i - \ LJ
S
F
,
or r =c ; +<7' t a n 0 ; (10.2 )
c' ., tanfi
where c
m
-—, ta n fi
m
= -
m
p
Tm
p
Actually the shearing resistanc e (mobilize d value of shearing strength ) does not develop t o a
like degree at al l point s o n an incipient failure surface. The shearin g strain s var y considerabl y an d
the shearing stres s ma y be far from constant . However th e above expressio n i s correct o n the basi s
of averag e conditions .
If th e factor s o f safet y wit h respect t o cohesion an d frictio n ar e different , we ma y writ e th e
equation o f the mobilize d shearin g resistanc e a s
It wil l b e show n late r o n tha t F , depends o n th e heigh t o f th e slope . Fro m thi s i t ma y b e
concluded tha t the factor of safety wit h respect t o cohesion ma y be designated a s the factoro f safety
with respect t o height. This factor i s denoted by F
H
an d it is the ratio between th e critical heigh t an d
St abilit y o f Slope s 36 9
the actual height, the critical height being the maximum height at which it is possible fo r a slope t o
be stable . We may writ e from Eq. (10.3 )
(1Q4)
H
where F^ i s arbitraril y take n equa l t o unity.
Example 10. 1
The shearin g strengt h parameter s o f a soil ar e
c' = 26.1 kN/m
2
0' = 15°
c' = 17. 8 kN/m
2
Calculate th e factor of safety (a) with respect t o strength, (b ) with respect to cohesion an d (c)
with respect t o friction. The average intergranula r pressure t f o n the failure surface i s 102. 5 kN/m
2
.
Solution
On the basi s o f the give n data , the average shearin g strengt h o n the failur e surfac e i s
s = 26.7 + 102. 5 ta n 15 °
= 26.7 + 102. 5 x 0.268 = 54. 2 kN/m
2
and the average valu e of mobilized shearin g resistanc e i s
T= 17.8 + 102. 5 ta n 12°
= 17. 8 + 102. 5 x 0.212 = 39. 6 kN/m
2
F - - .
L 26
39.6 17.8 0 ta n 0.21 2
The abov e exampl e show s th e facto r o f safet y wit h respec t t o shea r strength , F
s
i s 1.37 ,
whereas th e factor s o f safet y wit h respec t t o cohesio n an d frictio n ar e different . Conside r tw o
extreme cases :
1 . Whe n th e factor o f safet y wit h respect t o cohesion i s unity.
2. Whe n th e factor o f safet y wit h respect t o friction i s unity.
Casel
=
26
.
70+
102.50x0.268
9
12.9 0
Case 2
= 2.1 3
T
= 39.60 = —— +102.50 tan 15
C
F
370 Chapt e r 1 0
26.70
F
• + 27.50
c
12.1 0
We can have any combination of F
c
an d F, betwee n these two extremes cite d abov e t o give
the same mobilize d shearing resistance of 39. 6 kN/m
2
. Some o f the combinations of F
c
and F
0
are
given below.
Com bin at ion o f F
c
an d F^
F
c
1.0 0 1.2 6 1.3 7 1.5 0 2.2 0
F
0
2.1 2 1.5 0 1.3 7 1.2 6 1.0 0
Under Case 2, the value of F
c
= 2.20 when F
0
- 1.0 . The factor of safet y F
C
= 2.20 i s defined
as the, factoro f safety with respect t o cohesion.
Example 1 0. 2
What wil l be th e factor s of safet y with respect t o average shearing strength , cohesion an d internal
friction o f a soil , fo r whic h th e shea r strengt h parameters obtaine d fro m th e laborator y test s ar e
c' = 32 kN/m
2
an d 0 ' = 18° ; th e expecte d parameter s o f mobilize d shearin g resistanc e ar e
c'
m
= 21 kN/m
2
and 0' = 13 ° and the average effective pressure on the failure plane is 1 10 kN/m
2
.
For the same valu e of mobilized shearin g resistance determine th e following:
1 . Facto r o f safet y with respect t o height;
2. Facto r o f safet y with respect t o friction whe n that with respect t o cohesion i s unity; and
3. Facto r o f safet y with respect t o strength.
Solution
The availabl e shear strengt h of the soi l is
s = 32 + 1 10 tan 18 ° = 32 + 35. 8 = 67. 8 kN/m
2
The mobilize d shearin g resistance of the soi l is
T = 2 1 + 11 0 tan 13 ° = 21 + 25. 4 = 46.4 kN/m
2
_ 67. 8 . .,
Factor of safety wit h respec t t o average strength, r
s
= ——- 1-4 6
46.4
32
Factor o f safet y wit h respec t t o cohesion, F
C
= — -= 1.52
_ _ ta n 18° _ 0.324 9 _
Factor o f safet y wit h respec t t o friction, F<t> - ~ T T ~ ~ ~ T T ~ n 2309
Factor of safet y wit h respect t o height, F
H
( = F
c
) will be at F
0
= 1 .0
. , . 3 2 110t anl 8 ° , . 3 2
i = 46. 4 = — + -,therefore , F = -= 3.0
F
c
1. 0 46.4-35. 8
Factor o f safet y wit h respect t o friction a t F = 1 .0 is
St abilit y o f Slope s 371
. , . 3 2 110tanl8 ° , . ^ 35. 8
r = 46.4 =— + -,therefore, F , = -= 2.49
1.0 F
0
* 46.4-3 2
Factor o f safet y wit h respect t o strengt h F
s
i s obtained whe n F
C
= F+. We may writ e
32 11 0 tan 18°
or F = 1.46
10.4 STAB I L I T Y ANAL Y SI S OF INFINITE SLOPES I N SAND
As an introduction to slope analysis , the problem of a slope of infinite extent i s of interest. Imagin e
an infinit e slope , a s show n i n Fig . 10.2, making a n angl e j 8 wit h th e horizontal . Th e soi l i s
cohesionless an d completel y homogeneou s throughout . Then th e stresse s actin g o n an y vertica l
plane in the soil are the same as those on any other vertical plane. The stress a t any point on a plane
EF paralle l to the surfac e at dept h z wil l be the same as at every point on thi s plane.
Now consider a vertical slic e o f material ABCD havin g a unit dimension norma l t o the page .
The forces acting on this slice are its weight W, a vertical reaction R on the base of the slice, and two
lateral force s P
{
actin g o n the sides . Sinc e th e slic e i s i n equilibrium, the weigh t an d reaction ar e
equal i n magnitud e an d opposit e i n direction . The y hav e a commo n lin e o f actio n whic h passe s
through th e center o f the base AB. The latera l force s mus t be equal an d opposit e an d thei r lin e of
action mus t be parallel t o the sloped surface .
The normal and shear stresse s o n plane AB ar e
a' = yzcos
2
fi
where cr'
n
= effective norma l stress ,
y = effective unit weight of the sand .
If full resistanc e i s mobilized on plane AB, the shear strength, s, of the soil per Coulomb's la w
is
s = a
f
n
ta n 0'
when T = s, substituting for s and tf
n
, we have
Figure 10. 2 St abilit y an alys i s o f infinit e s lop e i n s an d
372
or ta n / 3 = tan 0 '
Chapt er 1 0
(10.5a)
Equation (10.5a ) indicate s tha t the maximu m value of ( 3 i s limited t o 0' i f the slop e i s t o be
stable. Thi s conditio n hold s tru e fo r cohesionles s soil s whethe r th e slop e i s completel y dr y o r
completely submerge d unde r water.
The facto r of safet y o f infinit e slope s i n sand may b e writte n as
p =
tanfi
(10.5b)
10.5 STAB I L I T Y ANAL Y SI S O F I NFI NI TE SL OPES I N CL AY
The vertica l stres s <J
v
actin g on plane AB (Fig . 10.3 ) wher e
a
v
= yzcosfi
is represented b y OC i n Fig. 10. 3 i n the stres s diagram. The norma l stres s o n thi s plane i s OE and
the shearin g stres s i s EC. Th e lin e OC make s a n angl e ( 3 with the cr-axis .
The Moh r strengt h envelope is represented b y lin e FA whose equation i s
s = c' + cr' t an^'
According t o the envelope, the shearing strengt h i s ED wher e the normal stres s i s OE.
When / 3 i s greater tha n 0' the lines OC and ED meet. In this case th e two lines meet a t A. As
long a s th e shearin g stres s o n a plane i s les s tha n th e shearin g strengt h o n th e plane , ther e i s n o
danger o f failure. Figur e 10. 3 indicate s that at al l depths a t which the direct stres s i s less tha n OB,
there i s no possibility of failure. Howeve r at a particular depth a t which the direct stres s i s OB, the
O E B
Figure 10. 3 St abilit y an al ys i s o f in fin it e s lope s i n clay s oil s
St abilit y o f Slope s 37 3
shearing strengt h an d shearin g stres s value s ar e equal a s represented b y AB, failur e i s imminent.
This dept h a t which the shearing stres s and shearing strengt h are equal i s called th e critical depth.
At depths greate r tha n thi s critical value, Fig. 10.3 indicates that the shearing stres s i s greater than
the shearing strengt h but thi s is not possible. Therefor e i t may b e concluded tha t the slop e ma y be
steeper tha n 0' as long a s the dept h of the slope i s less tha n the critical depth .
Expression fo r the Stabilit y o f an I nfinit e Slop e o f Cla y o f Dept h H
Equation (10.2 ) gives th e developed shearin g stres s a s
T = c'
m
+( T'tan</> '
m
(10.6 )
Under condition s o f n o seepag e an d n o por e pressure , th e stres s component s o n a plane a t
depth H an d paralle l t o the surfac e of the slop e ar e
r=
<j' = yHcos
2
j3
Substituting these stres s expression s i n the equation above an d simplifying, we have
c'
m
= YHcos
2
0 (tan 0 - ta n 0'J
c'
or N = ^- = cos
2
/ ?(t anyt f - t an^) (10.7 )
yti
where H i s th e allowabl e heigh t an d th e ter m c'J y H i s a dimensionles s expressio n calle d th e
stability number an d is designated a s A^. This dimensionless number is proportional t o the required
cohesion an d is inversely proportional t o the allowable height. The solution is for the case when no
seepage is occurring. I f in Eq. (10.7) the factor of safety with respect t o friction i s unity, the stability
number wit h respect t o cohesion ma y be written as
8)
, c
where c
m
= —
The stabilit y number in Eq. (10.8) ma y be written as
where H
c
= critical height . From Eq . (10.9) , we have
Eq. (10.10) indicates tha t the factor of safet y with respect t o cohesion, F
c
, is the same a s the
factor o f safet y wit h respect t o height F
H
.
If ther e i s seepag e paralle l t o th e groun d surfac e throughou t the entir e mass , wit h th e fre e
water surfac e coinciding wit h the groun d surface , th e component s o f effective stresse s o n plane s
parallel to the surfac e of slopes at depth H are given as [Fig. 10.4(a)].
Normal stres s
( l O. l l a )
374 Chapt er 1 0
(a)
(b)
Figure 10. 4 An al ys i s o f in fin it e s lop e (a ) with s eepag e fl ow t hroug h t h e en t ir e
m as s , an d (b) wit h com plet el y s ubm erg e d s lope .
the shearing stress
T = y
sat
H si n /3 cos /3
Now substitutin g Eqs (10. 11 a) and (10. l i b) int o equation
( l O. l l b )
and simplifying , th e stabilit y expression obtained i s
-^2— = cos
2
0 ta n 0- -- tan </> '„
Y H Y 1
sat ' sat
(10.12)
As before, i f the factor of safet y with respect t o friction i s unity , the stabilit y number which
represents th e cohesion may be writte n as
N = •
FY H Y H
c' sat 'sat ,
C/
= cos
2
,tf tan^--^ -
' sat
(10.13)
If th e slop e i s completel y submerged , an d i f ther e i s n o seepag e a s i n Fig . 10.4(b) , the n
Eq. (10.13) become s
N = = cos
2
/?(tan f t ~ tan <}> ')
(10.14)
where y , = submerge d uni t weigh t o f th e soil .
St abilit y o f Slope s 37 5
Example 10. 3
Find the factor of safet y of a slope of infinite extent having a slope angl e = 25°. The slop e i s made
of cohesionless soi l wit h 0 = 30° .
Solution
Factor o f safet y
tan 30° 0.577 4
tan/? ta n 25° 0.466 3
Example 10. 4
Analyze the slope of Example 10. 3 if it is made of clay having c' - 3 0 kN/m
2
, 0' = 20°, e = 0.65 and
G
s
= 2.7 and under the following conditions: (i) when the soi l i s dry, (ii) when water seeps paralle l
to the surfac e of the slope , an d (iii ) when the slope i s submerged .
Solution
For e = 0.65 an d G = 2.7
=
27x^1
= =
(2. 7 + 0.65)x9.81
=
ld
1 + 0.65
/ sa t
1 + 0.65
y
b
= 10.09 kN/m
3
(i) For dr y soi l th e stabilit y number N
s
i s
c
N = ——— = cos
2
/?(tan/?- t a n <j> ' ) whe n F,=l
' d c
= (cos 25° )
2
(tan 25° - ta n 20°) = 0.084.
c'3 0
Therefore, th e critical height H = -= -= 22.25 m
16.05x0.084
(ii) Fo r seepage parallel t o the surface of the slope [Eq . (10.13) ]
c' 100 Q
N = — -—= cos
2
25° ta n 25°-^--- tan 20° =0.231 5
s
y
t
H
c
19. 9
H
c
=^=
3
° =6.5 1 m
c
y
t
N
s
19.9x0.231 5
(iii) Fo r the submerge d slop e [Eq . (10.14) ]
N = cos
2
25° (tan 25° - ta n 20°) = 0.084
c
y
b
N
s
10.09x0.08 4
376 Chapt e r 1 0
10.6 M ETH OD S OF STAB I L I TY ANAL Y SIS OF SL OPES OF FI NI TE
H EI G H T
The stabilit y of slope s o f i nfi ni t e exten t has bee n discusse d i n previous sections . A mor e commo n
problem i s the one i n which the failure occur s on curved surfaces . The mos t widel y used method of
analysis o f homogeneous , isotropic , finit e slope s i s th e Swedish method base d o n circula r failure
surfaces. Petterso n (1955 ) firs t applie d th e circl e metho d t o th e analysi s o f a soi l failur e i n
connection wit h th e failur e o f a quarr y wal l i n Goeteberg , Sweden . A Swedis h Nationa l
Commission, afte r studying a large number of failures, published a report i n 192 2 showing that the
lines o f failur e of mos t suc h slide s roughl y approached th e circumferenc e o f a circle . The failur e
circle migh t pass above the toe, throug h the toe or below it . By investigating the strengt h along the
arc o f a larg e numbe r o f suc h circles , i t wa s possibl e t o locat e th e circl e whic h gav e th e lowes t
resistance t o shear . Thi s genera l metho d ha s bee n quit e widel y accepte d a s offerin g a n
approximately correc t solutio n fo r th e determinatio n o f th e facto r o f safet y o f a slop e o f a n
embankment an d o f it s foundation . Developments i n th e metho d o f analysi s hav e bee n mad e b y
Fellenius (1947) , Terzagh i (1943) , Gilbo y (1934) , Taylo r (1937) , Bisho p (1955) , an d others , wit h
the resul t that a satisfactor y analysi s of th e stabilit y of slopes , embankment s an d foundation s by
means o f the circl e metho d i s no longer a n undul y tediou s procedure .
There ar e othe r method s o f historic interes t suc h as the Culmann method (1875 ) and the
logarithmic spiral method. Th e Cul man n metho d assume s tha t ruptur e wil l occu r alon g a
plane. I t i s o f interes t onl y a s a classica l sol ut ion , since actua l fai l ur e surface s ar e invariably
curved. Thi s metho d i s approximatel y correc t fo r steep slopes . Th e logarithmi c spira l metho d
was recommende d b y Rendul i c (1935 ) wit h th e rupt ur e surfac e assumin g th e shap e o f
logarithmic spiral . Thoug h thi s metho d make s th e proble m staticall y determinat e an d give s
more accurat e resul t s , th e greate r lengt h o f tim e require d fo r computatio n overbalance s thi s
accuracy.
There are several method s o f stabilit y analysis based o n the circular arc surfac e of failure. A
few o f the methods ar e described belo w
M ethods o f Analysis
The majorit y o f th e method s o f analysi s ma y b e categorize d a s limi t equilibriu m methods . Th e
basic assumptio n o f the limi t equilibrium approach i s that Coulomb' s failur e criterion i s satisfie d
along th e assume d failur e surface. A free body i s taken from the slope an d startin g from known or
assumed value s of the forces acting upon the free body, the shear resistance of the soil necessar y fo r
equilibrium i s calculated . Thi s calculate d shea r resistanc e i s the n compare d t o th e estimate d o r
available shea r strengt h o f the soi l t o give an indication of the factor of safety.
Methods tha t conside r onl y th e whol e fre e bod y ar e th e (a ) slop e failur e unde r undraine d
conditions, (b ) friction-circl e metho d (Taylor , 1937 , 1948 ) an d (c ) Taylor' s stabilit y numbe r
(1948).
Methods tha t divide the fre e bod y int o many vertical slice s an d conside r th e equilibriu m of
each slic e ar e th e Swedis h circl e metho d (Fellenius , 1927) , Bisho p metho d (1955) , Bisho p an d
Morgenstern metho d (1960 ) an d Spence r metho d (1967) . Th e majorit y o f thes e method s ar e i n
chart for m an d cove r a wide variet y of conditions.
10.7 PL AN E SU RFACE OF FAI L U RE
Culmann (1875 ) assume d a plan e surfac e of failur e for th e analysi s o f slope s whic h i s mainl y of
interest becaus e i t serves a s a test of the validit y of the assumptio n o f plane failure . In some case s
this assumptio n i s reasonable an d i n others i t is questionable .
St abilit y o f Slope s 377
Force triangle
Figure 10. 5 St abilit y o f s lope s b y Culm an n m et ho d
The metho d a s indicated abov e assume s tha t the critical surfac e of failur e i s a plane surfac e
passing through the toe of the dam as shown in Fig. 10.5 .
The forces that act on the mass above trial failure plane AC inclined at angle 6 with the horizontal are
shown in the figure. The expression for the weight, W, and the total cohesion C are respectively,
W = -yL H cose c /? sin(jtf- 0)
The us e of the la w of sine s i n the force triangle , shown i n the figure , give s
C _ sm(6>-f )
W ~ cos^ '
Substituting herei n fo r C and W, and rearranging we have
1
in which the subscript Q indicates tha t the stabilit y number is for the trial plane a t inclination 6.
The mos t dangerou s plane i s obtaine d by settin g th e firs t derivativ e o f the abov e equatio n
with respec t t o Q equal t o zero. Thi s operatio n give s
where & '
c
i s th e critica l angl e fo r limitin g equilibriu m an d th e stabilit y numbe r fo r limitin g
equilibrium may be written as
yH
c
4 sin/? cos 0'
where H i s the critical heigh t o f the slope.
(10.15)
378 Chapt e r 10
If we writ e
F -— F
t an
^'
c
~V ' <>~ t a n ^
where F
c
an d F^ ar e safet y factor s wit h respect t o cohesion an d frictio n respectively , Eq . (10.15 )
may b e modifie d fo r chose n value s of c an d 0' a s
^
=
4 sin/3 cos ( /)'
m
(10.16 )
The critica l angl e for an y assumed values of c'
m
an d 0'
m
i s
1
From Eq . (10.16) , th e allowabl e height of a slope i s
Example 1 0 .5
Determine b y Culmann' s method the critical height of an embankment having a slope angl e of 40°
and the constructed soi l having c' = 630 psf, 0' = 20° and effective uni t weight = 1 14 lb/ft
3
. Fin d the
allowable heigh t of the embankment if F, = F, = 1 .25.
Solution
4c'sin/?cos0' 4 x 630 x sin 40° cos 20°
H, = -- -—= -= 221 ft
y[ l-cos( 0-4> ')] 114(l-cos20° )
For F
c
= F. = 1.25,
c
'= —= —= 504 lb/ft
2
<( > m
' ta n 20°
and ta n #, = —- = — —= 0.291, f a = 16.23 °
,, , • , 4x50 4 sin 40° cos 16.23° ^
0 r
Allowable height , H = -= 128. 7 ft.
_ 114[l-cos(40 - 16.23°)]
10.8 CI RCU L A R SU RFACES OF FAI L U RE
The investigation s carried ou t i n Swede n a t th e beginnin g of thi s centur y have clearl y confirme d
that the surfaces of failure of earth slopes resembl e th e shape of a circular arc. When soi l slips along
a circula r surface , suc h a slid e ma y b e terme d a s a rotationa l slide . I t involve s downwar d an d
outward movemen t o f a slice o f eart h a s shown i n Fig. 10.6(a) an d slidin g occur s alon g th e entir e
surface o f contact betwee n th e slic e an d it s base. Th e type s of failur e tha t normally occu r ma y b e
classified a s
1. Slop e failure
St abilit y o f Slope s 37 9
2. To e failure
3. Bas e failure
In slope failure, the arc of the rupture surface meets the slope above the toe. This can happen
when the slope angl e / 3 is quite high and the soil close to the toe possesses hig h strength. Toe failure
occurs when the soil mass of the dam above the base and below the base is homogeneous. The base
failure occur s particularl y when the bas e angl e j 3 is low an d th e soi l below th e bas e i s softe r and
more plastic than the soi l above the base. The various modes of failure are shown in Fig. 10.6 .
Rotational
slide
(a) Rotational slide
(b) Slope failur e
(c) Toe failur e
(d) Base failur e
Figure 10. 6 T ype s o f failur e o f eart h dam s
380 Chapt er 1 0
10.9 FAI L U R E UNDER U NDRAI NE D CONDI TI ON S (0
M
= 0 )
A f ul l y saturate d cla y slop e ma y fai l unde r undraine d condition s (0
u
= 0 ) immediatel y afte r
construction. Th e stabilit y analysis is based o n the assumption that the soi l i s homogeneous an d the
potential failur e surface i s a circular arc. Two types of failures considered ar e
1. Slop e failur e
2. Bas e failure
The undraine d shea r strengt h c
u
of soi l i s assumed t o be constan t wit h depth. A tria l failure
circular surfac e AB with center a t 0 an d radius R is shown i n Fig. 10.7(a ) for a toe failure. The slope
AC and the chord AB mak e angles / 3 and a wit h the horizontal respectively. W is the weight per unit
Firm bas e
(a) Toe fail ur e (b) Base failur e
Figure 10. 7 Crit ica l circl e pos it ion s fo r (a ) s lope fail ur e (aft e r Fellen ius , 1927) , (b )
bas e fail ur e
1>
50
C
40
C
20°
10
90
C
70°
V alues of
(a)
60
C
50°
50 40° 30 ° 20 °
V alues o f ?
10° 0°
Figure 10. 8 (a ) Rel at io n bet wee n s lop e an g l e / 3 an d param et er s a an d Q fo r
locat ion o f crit ica l t o e circl e whe n /3 i s g reat e r t han 53° ; (b ) relat io n bet wee n s lop e
an g le /3 an d dept h fact o r n
d
fo r variou s val ue s o f param et e r n
x
(aft er Fellen ius , 1927 )
St abilit y o f Slope s 38 1
length of the soi l lying above the trial surface acting through the center of gravity of the mass. l
o
is
the leve r arm , L
a
i s th e lengt h o f th e arc , L
c
th e lengt h o f th e chor d AB an d c
m
th e mobilize d
cohesion fo r an y assumed surfac e of failure.
We may expres s th e facto r of safet y F^ as
(10.19)
For equilibrium of the soi l mass lying above the assumed failure surface, we may write
resisting moment M
r
= actuating moment M
a
The resisting moment M
f
= L
a
c
m
R
Actuating moment , M
a
= Wl
o
Equation for the mobilized c i s
W1
0
(10.20)
Now th e facto r o f safet y F fo r th e assume d tria l ar c o f failur e ma y b e determine d fro m
Eq. (10.19). This is for one trial arc. The procedure has to be repeated fo r several trial arcs and the
one that gives the least valu e is the critical circle .
If failure occurs along a toe circle, the center of the critical circle can be located by laying off
the angles a an d 26 as shown in Fig. 10.7(a) . V alues of a an d 6 for different slop e angles / 3 can be
obtained fro m Fig. 10.8(a) .
If there i s a base failur e a s shown in Fig. 10.7(b) , the trial circle wil l be tangential to the firm
base and as such the center of the critical circle lies on the vertical line passing through midpoint M
on slope AC. The following equations may be written with reference t o Fig. 10.7(b) .
D x
Depth factor , n
d
= —, Distanc e factor , n
x
=— (10.21 )
H H
V alues o f n
x
ca n b e estimate d fo r differen t value s o f n
d
an d j 8 b y mean s o f th e char t
Fig. 10.8(b) .
Example 10. 6
Calculate th e facto r o f safet y agains t shea r failur e along th e sli p circl e show n i n Fig . Ex . 10. 6
Assume cohesio n = 4 0 kN/m
2
, angl e o f interna l friction = zer o an d th e tota l uni t weigh t o f th e
soil = 20.0 kN/m
3
.
Solution
Draw the given slope ABCD a s shown in Fig. Ex. 10.6 . To locate the center of rotation, exten d th e
bisector o f line BC t o cu t th e vertica l line drawn from C at point O. With O as cente r and OC as
radius, draw the desired sli p circle.
2
Radius OC =R = 36.5 m, Area BECFB = - xEFxBC
2
= - x 4 x 32.5 = 86.7 m
2
Therefore W = 86.7 x 1 x 20 = 173 4 kN
W act s through point G which may be taken as the middle of FE.
382 Chapt er 1 0
s
s R = 36.5m
Figure. Ex . 10. 6
From the figur e we have, x = 15. 2 m, and 9= 53 °
3.14
Length o f arc EEC =R0= 36.5 x 53° x —— = 33.8 m
180
length of arc x cohesion x radius 33.8x40x36. 5
Wx 1734x15.2
1 0 .1 0 FRI CTI ON- CI RCL E M ETH OD
Physical Concept o f the M etho d
The principl e o f th e metho d i s explaine d wit h referenc e t o th e sectio n throug h a da m show n i n
Fig. 10.9 . A tria l circl e wit h center o f rotation O i s shown i n the figure . With cente r O and radiu s
Friction circl e
Trial circula r
failure surfac e
Figure 10. 9 Principl e o f frict io n circl e metho d
Stability o f Slope s 383
sin 0" , where R is the radius of the trial circle, a circle i s drawn. Any line tangent to the inner circl e
must intersec t th e tria l circl e a t a n angl e t f wit h R . Therefore , an y vecto r representin g a n
intergranular pressure a t obliquity 0' to an element o f the rupture ar c must be tangent to the inner
circle. Thi s inne r circle i s called the friction circle o r ^-circle. The frictio n circl e metho d o f slop e
analysis i s a convenien t approac h fo r bot h graphica l an d mathematica l solutions . I t i s give n thi s
name because th e characteristic assumptio n of the method refers t o the 0-circle.
The forces considere d i n the analysi s are
1. Th e total weight W of the mass above the trial circle acting through the center of mass. The
center o f mass ma y be determined b y any one of the known methods .
2. Th e resultan t boundar y neutra l force U. The vecto r U may be determine d b y a graphica l
method fro m flowne t construction .
3. Th e resultant intergranular force, P, acting on the boundary.
4. Th e resultant cohesive forc e C.
Actuating Force s
The actuating forces may be considered t o be the total weight W and the resultant boundary force U
as shown in Fig. 10.10 .
The boundary neutral force always passes through the center of rotation O. The resultant of W
and U, designated a s Q, is shown i n the figure.
Resultant Cohesiv e Forc e
Let the lengt h of arc AB b e designated a s L
a
, the length of chord AB b y L
c
. Let the arc length L
a
be
divided int o a number of small element s an d let the mobilized cohesiv e forc e on these element s b e
designated a s C
r
C
2
, C
3
, etc. as shown in Fig. 10.11 . The resultant of al l these forces i s shown by
the force polygon in the figure. The resultant is A'B' whic h is parallel an d equal to the chord length
AB. The resultant of all the mobilized cohesiona l force s along the ar c is therefor e
C = c'L
Figure 10.1 0 Act uat in g force s
384
Chapt er 1 0
(a) Cohesiv e forces o n a trial ar c (b ) Polygo n o f force s
Figure 10. 1 1 Resi st an t cohesiv e force s
We may write c'
m
- —
c
wherein c'= unit cohesion, F
C
= factor o f safet y wit h respect t o cohesion .
The line of action of C may be determined by moment consideration. Th e moment of the total
cohesion i s expressed a s
c' L R = c' L I
m a m c a
where l = moment arm. Therefore ,
(10.22)
It i s see n tha t the lin e of action of vector C is independent o f the magnitude of c'
m
.
Resultant o f B oundar y I ntergranula r Force s
The tria l ar c o f th e circl e i s divide d int o a numbe r o f smal l elements . Le t P
v
P
2
, P
y
etc . b e the
intergranular force s actin g o n these element s a s shown i n Fig. 10.12 . Th e frictio n circl e i s drawn
with a radius o f R si n ( j/
m
where
The lines of action of the i nt ergranul ar forces P
r
P
2
, P
y
etc . ar e tangential t o the friction
circle an d mak e a n angl e o f 0'
m
a t th e boundary . However , th e vecto r su m o f an y tw o smal l
forces ha s a lin e o f actio n t hroug h poin t D , missin g tangenc y t o th e 0'
m
-circle b y a smal l
amount . The resul t an t o f al l granul a r force s mus t therefor e mis s tangenc y t o th e 0'
m
-circle by
an amoun t whic h i s no t considerable . Le t th e distanc e o f th e resul t an t of th e granul ar forc e P
from th e cente r o f th e circl e b e designate d a s K R si n 0 ' (a s show n i n Fig . 10.12) . Th e
St abilit y o f Slope s 385
KRsin<p'
n
Figure 10.1 2 Res ult an t o f in t erg ran ula r force s
magnitude of K depends upon the type of intergranular pressure distributio n alon g the arc. Th e
most probabl e for m o f distributio n i s the sinusoida l distribution .
The variation of K with respect to the central angle a'is shown in Fig. 10.13 . The figure als o
gives relationship s betwee n of an d K fo r a unifor m stres s distributio n of effectiv e norma l stres s
along th e ar c of failure .
The graphical solutio n based o n the concepts explaine d abov e i s simpl e i n principle. Fo r the
three force s Q, C and P of Fig. 10.1 4 to be i n equilibrium, P must pass throug h the intersection of
1.20
1.16
1.12
1.08
1.04
1.00
o
x
£
^
7
1
str
^
Cent
For i
essc
/
ral angl e
anifo
istrih
/
^
rm
>utior
j
/
s<
i — S
/
/
/
s
/
y
/
For
tress
/
'
/
sinus
distr
/
/
/
oida
ibuti(
J
/
^n
20 4 0 6 0 8 0
Central angl e a ' in degree s
100 120
Figure 10.1 3 Relat ion s hi p bet wee n K an d cen t ral an g l e a'
386 Chapt er 1 0
Figure 1 0.1 4 Forc e t rian g l e fo r t h e frict ion -circl e m et ho d
the known lines of action of vectors Q and C. The line of action of vector P must also be tangent to
the circl e o f radiu s KR si n 0' . The valu e o f K ma y b e estimate d b y th e us e o f curve s give n i n
Fig. 10.13 , and the line of action offeree P may be drawn as shown in Fig. 10.14 . Since the lines of
action of all three forces and the magnitude of force Q are known, the magnitude of P and C may-be
obtained b y the force parallelogram construction that is indicated i n the figure. The circl e o f radius
of KR si n 0' i s called th e modified friction circle. T
rn j j
Determination o f Facto r o f Safet y Wit h Respec t t o Strengt h
Figure 10.15(a ) i s a section o f a dam. AB i s the tria l failur e arc. Th e forc e Q, the resultant o f W
and U is drawn as explained earlier . The lin e of action of C is also drawn. Let th e forces Q and C
D
(a) Friction circl e (b) Facto r of safet y
Figure 1 0.1 5 Graphica l m et ho d o f det erm in in g fact o r o f s afet y wit h res pec t t o
s t ren g t h
St abilit y o f Slope s 38 7
meet a t point D. An arbitrary first tria l using any reasonable $
m
value, which will be designated
by 0'
ml
i s give n b y th e us e o f circl e 1 or radiu s KR si n <j)'
ml
. Subscrip t 1 is use d fo r al l othe r
quantities of the first trial . The force P
l
i s then drawn through D tangent to circle 1 . C
l
i s parallel
to chor d an d poin t 1 is th e intersectio n of force s C
{
an d P
r
Th e mobilize d cohesio n i s equal
c'
m]
L
c
. Fro m thi s the mobilized cohesion c'
ml
i s evaluated. The factor s of safet y with respect t o
cohesion an d frictio n are determined from the expression s
c' tanfl'
F' = ——, an d F*,
These factor s ar e th e value s used t o plot poin t 1 in the grap h i n Fig. 10.15(b) . Similarly
other frictio n circles wit h radii KR si n <j/
m2
, K R si n 0'
m3
. etc. ma y b e drawn an d th e procedur e
repeated. Point s 2 , 3 etc. ar e obtaine d a s shown i n Fig. 10.15(b). The 45 ° line , representin g
F
c
= F., intersects th e curve t o give th e factor of safet y F
s
fo r thi s trial circle .
Several tria l circles mus t be investigated i n order t o locate the critical circle, which is the one
having the minimum value of F
5
.
Example 1 0.7
An embankment has a slope o f 2 (horizontal) to 1 (vertical) with a height of 1 0 m. I t i s made of a
soil havin g a cohesio n o f 3 0 kN/m
2
, a n angl e o f interna l frictio n o f 5 ° an d a uni t weigh t o f
20 kN/m
3
. Conside r an y sli p circl e passing through the toe. Us e the friction circle method t o fin d
the factor of safet y wit h respect t o cohesion .
Solution
Refer t o Fig. Ex . 10.7 . Le t EFB be the slop e an d AKB b e the sli p circl e drawn wit h center O and
radius R = 20 m.
Length of chord AB = L
c
= 27 m
Take J as the midpoint of AB, then
Area AKBFEA = area AKBJ A + area ABEA
= -ABxJ K + -ABxEL
3 2
= -x27 x 5.3 + -x 27 x 2.0 = 122.4 m
2
3 2
Therefore th e weight of the soi l mas s = 122. 4 x 1 x 20 = 2448 kN
It wil l act throug h point G, the centroi d o f the mass whic h can b e taken a s the mi d poin t of
FK.
Now, 0=85° ,
314
Length of arc AKB = L = RO = 20 x 85 x — = 29.7 m
6
18 0
L 29. 7
Moment ar m of cohesion, / = R— = 20 x —— = 22 m
L
c
21
From cente r O , a t a distanc e /
fl
, dra w th e cohesiv e forc e vecto r C , whic h i s paralle l t o th e
chord AB. Now fro m th e point of intersection of C and W, draw a line tangent to the friction circle
388 Chapt er 1 0
1.74m
/ / = 1 0 m
Figure Ex . 10. 7
drawn a t 0 wit h a radius of R sin 0' = 20 sin 5° = 1 .74 m. This line is the line o f action o f the thir d
force F .
Draw a triangle o f forces i n which the magnitude and th e direction fo r W is known and only
the directions o f the othe r two force s C and F are known.
Length ad give s th e cohesive forc e C = 520 kN
Mobilized cohesion ,
c
' = -= — = 17.51 kN/m
2
m
L 29. 7
Therefore th e factor of safet y wit h respect t o cohesion, F
c
, i s
F =11 = ^=1.713
F
C
wil l be 1 .7 13 if the factor o f safet y wit h respect to friction, F^ - 1 .0
t an5
c
If, F = 1.5, the n 0' =
F.
= 0.058 rad; o r 0' = 3.34°
St abilit y o f Slope s 38 9
The ne w radius of the friction circl e i s
r
{
= R si n 0'
m
= 20 x si n 3.3° = 1.1 6 m.
The directio n o f F changes an d the modified triangl e of force abd' gives ,
cohesive forc e = C = length ad' = 600 kN
C 60 0
Mobilised cohesino, c'
m
= ~— - - 20. 2 kN/mr
L J Z*yI /
c'3 0
Therefore, F = — = = 1. 5
c
c' 20. 2
10.1 1 TAY L OR' S STAB I L I TY NUMBER
If th e slop e angl e j8 , heigh t o f embankmen t H , th e effectiv e uni t weigh t o f materia l y , angl e o f
internal friction </>' , and unit cohesion c' ar e known, the factor of safety may be determined. I n orde r
to make unnecessary the more or less tedious stabilit y determinations, Taylo r (1937) conceive d th e
idea of analyzing the stabilit y of a large number of slopes throug h a wide range of slope angle s and
angles of internal friction, and then representing the results by an abstract numbe r which he calle d
the "stability number". This numbe r is designated a s A^. The expression use d i s
From thi s the factor of safet y wit h respect t o cohesion ma y be expressed a s
F
- = 7 <
10
-
24
>
Taylor published his results in the form of curves which give the relationship betwee n N
s
an d
the slop e angle s / ? for variou s value s o f 0 ' a s show n i n Fig . 10.16 . These curve s ar e fo r circle s
passing throug h th e toe , althoug h fo r value s o f 13 less tha n 53° , i t ha s bee n foun d tha t th e mos t
dangerous circle passes below the toe. However , these curves may be used without serious erro r for
slopes dow n t o f i = 14° . Th e stabilit y number s ar e obtaine d fo r factor s o f safet y wit h respec t t o
cohesion b y keeping th e factor of safety wit h respect t o friction (FJ equa l to unity.
In slopes encountere d i n practical problems , the depth to which the rupture circle ma y extend
is usually limited by ledge or other underlying strong material a s shown in Fig. 10.17 . The stability
number N
s
fo r the cas e whe n 0 " = 0 is greatly dependent o n the position o f the ledge. The dept h at
which the ledge o r strong materia l occur s ma y be expressed i n terms o f a depth factor n
d
which i s
defined a s
»
r f
=;|(10-25 )
where D - dept h of ledge belo w the top of the embankment, H = height of slope abov e th e toe.
For variou s value s o f n
d
an d fo r th e 0 = 0 cas e th e char t i n Fig . 10.1 7 give s th e stabilit y
number N
S
fo r various value s of slope angl e ft. In this case th e rupture circle may pass throug h the
toe o r belo w th e toe . Th e distanc e j c o f th e ruptur e circl e fro m th e to e a t th e to e leve l ma y b e
expressed b y a distance facto r n whic h is defined as
• a
S t a b i l i t y n u m b e r , N
s
C D
H °
| _ c u
o a
c r
- * C D
< ° o
^ ^
C
C D
_ j
O
O )
0 )
o "
V ) *
Q )
C T
( Q
C
S t a b i l i t y n u m b e r , N , .
Q )
- . < .
Q ) O
— f> - i
r - t - >
C D c n
~ " w
— J r - +
Q ) 0 )
< . E
o ^ ~
C D C
c o 3
- > J c r
' * * C D
- ^
c n
o e e
St abilit y o f Slope s 39 1
The char t i n Fig. 10.1 7 shows the relationship between n
d
and n
x
. If there i s a ledge o r other
stronger material a t the elevation of the toe, the depth factor n
d
for this case is unity.
Factor o f Safet y wit h Respec t to Strengt h
The developmen t of the stabilit y number is based o n the assumptio n tha t the factor of safet y with
respect t o frictio n F, , i s unity . The curve s giv e directl y th e facto r o f safet y F
c
wit h respec t t o
cohesion only . I f a true facto r of safet y F
s
wit h respec t t o strengt h i s required, thi s factor should
apply equall y t o bot h cohesio n an d friction . Th e mobilize d shea r strengt h ma y therefor e b e
expressed a s
s c ' a' tan ( /)'
In the above expression, we may write
— = c'
m
, ta n ( f> '
m
= —=—, o r # , = — (approx. )
(10
.27)
S 5 S
c'
m
and tf
m
may be described a s average values of mobilized cohesio n an d friction respectively .
Example 10. 8
The following particulars are given for an earth dam of height 39 ft. The slope i s submerged and the
slope angl e j 3 = 45°.
Y
b
= 69 lb/ft
3
c' = 550 lb/ft
2
0' = 20°
Determine th e factor of safet y F
S
.
Solution
Assume as a first tria l F
s
= 2. 0
20
<t> '
m
=
Y
=10
° (approx. )
For ( j)'
m
= 10°, an d ( 3 = 45° th e valu e of N
s
fro m Fig . 10.1 6 is 0.1 1, we ma y writ e
c
'
From Eq. (10.23 ) N = -,substitutin g
55Q
2 x 69x #
or H = — =36.2 3 ft
2x69x0. 11
20
If F = 1.9, $= —= 10.53° an d N = 0.105
5
1 9 *
392 Chapt e r 1 0
. 40f t
1.9x69x0.105
The compute d heigh t 40 f t i s almos t equal to the give n heigh t 3 9 ft . The compute d facto r of
safety i s therefor e 1 .9.
Example 1 0. 9
An excavatio n is t o be made i n a soil deposit wit h a slope o f 25° to the horizontal and t o a depth of
25 meters. Th e soi l ha s the following properties:
c'= 35kN/m
2
, 0' = 15 ° and 7= 2 0 kN/m
3
1 . Determin e th e facto r of safet y of th e slop e assuming ful l frictio n i s mobilized .
2. I f th e facto r o f safet y wit h respec t t o cohesio n i s 1.5 , wha t woul d be th e facto r o f safet y
with respec t t o friction?
Solution
1 . Fo r 0' = 15 ° and ( 3 = 25°, Taylor' s stabilit y number chart gives stabilit y number N
s
= 0.03.
- 233
0. 03x20x25
2. Fo r F = 1.5 , N = -- -- -— -= 0.047
J
F
c
xyxH 1. 5x20x2 5
For A ^ = 0.047 and ( 3 = 25°, w e have from Fig. 10.16 , 0'
m
= 1 3
tan0' ta n 15° 0.26 8
Therefore, F, = -— = -= -= 1.16
0
t an 0 ta n 13° 0.23 1
Example 1 0 .1 0
An embankmen t i s t o be mad e fro m a soi l havin g c' = 420 lb/ft
2
, 0 ' = 18 ° and y = 12 1 lb/ft
3
. Th e
desired facto r o f safet y wi t h respec t t o cohesio n a s wel l a s tha t wit h respec t t o frictio n i s 1.5 .
Determine
1 . Th e saf e height i f the desire d slop e i s 2 horizontal t o 1 vertical.
2. Th e saf e slop e angl e i f the desire d heigh t i s 50 ft .
Solution
, 0.32 5
tan 0' = tan 18° = 0.325, 0'
m
- ta n ' — -= 12.23°
1. Fo r 0' = 12.23 ° an d ( 3 =26.6° (i.e. , 2 horizontal and 1 vertical) the char t give s N
s
= 0.055
c' 42 0
Therefore, 0.05 5 =
F
c
yH 1. 5 x 121 x H
St abilit y o f Slope s 393
Therefore, # . =
420
2. Now , N
S
= •
safe
1.5x121x0.05 5
420
= 42 ft
= 0.04 6
F
c
yH 1.5x121x5 0
For N = 0.046 and 0' = 12.23° , slop e angl e P = 23.5
C
1 0 .1 2 TENSI O N CRACKS
If a dam i s built of cohesive soil , tensio n cracks ar e usually present at the crest. The dept h of such
cracks ma y be computed fro m th e equation
r
(10.28)
where z
0
= dept h of crack, c ' = unit cohesion, y = unit weight of soil .
The effective length of any trial arc of failure is the difference between the total lengt h of arc
minus the dept h of crack a s shown in Fig. 10.18.
1 0 .1 3 STAB I L I T Y ANAL Y SI S B Y M ETH OD O F SL I CES FOR
STEADY SEEPAG E
The stabilit y analysi s wit h stead y seepag e involve s th e developmen t o f th e por e pressur e hea d
diagram along the chosen tria l circl e o f failure. The simples t of the methods fo r knowing the por e
pressure hea d a t any point on the trial circl e i s by the use of flownets whic h is described below .
Determination o f Por e Pressur e wit h Seepage
Figure 10.1 9 shows the section o f a homogeneous da m with an arbitrarily chosen tria l arc. There i s
steady seepag e flo w throug h th e da m a s represente d b y flo w an d equipotentia l lines . Fro m th e
equipotential line s the por e pressur e ma y be obtaine d a t any point on the section . Fo r exampl e at
point a i n Fig. 10.19 the pressure hea d i s h. Point c is determined by settin g the radial distanc e ac
Tension crac k
Effective lengt h of
trial arc of failur e
Figure 10.1 8 T en s io n cr ac k i n dam s buil t o f cohes iv e s oil s
394 Chapt er 1 0
Trial circl e
- ' ' R =radius /
of trial circle/'
d/s sid e /
Phreatic lin e
Piezometer
Pressure hea d
at point a - h
Discharge fac e
\- Equipotential lin e
x
r ---- -'
Pore pressure head diagram -/
Figure 1 0.1 9 Det erm in at io n o f por e pres s ur e wit h s t ead y s eepag e
equal t o h. A numbe r of point s obtaine d i n th e same manne r as c giv e th e curve d lin e through c
which i s a pore pressur e hea d diagram.
M ethod o f Analysi s (graphica l method)
Figure 10.20(a ) show s th e sectio n o f a da m wit h a n arbitraril y chose n tria l arc . The cente r o f
rotation of the ar c is 0. The pore pressur e acting on the base of the ar c as obtained from flo w net s is
shown i n Fig. 10.20(b).
When th e soi l formin g the slop e ha s t o be analyze d under a condition wher e ful l o r partia l
drainage takes plac e the analysis must take into account both cohesive an d frictional soil properties
based o n effective stresses. Sinc e th e effectiv e stres s actin g across eac h elementa l lengt h o f th e
assumed circula r ar c failur e surfac e must be computed i n thi s case, th e metho d o f slice s i s one of
the convenient methods for thi s purpose. The metho d o f analysi s is as follows.
The soi l mass above the assumed sli p circle is divided into a number of vertical slices of equal
width. The numbe r of slices may be limited to a maximum of eight to ten to facilitate computation.
The force s use d i n the analysi s acting on the slices ar e shown in Figs. 10.20(a ) and (c). The forces
are:
1 . Th e weigh t W of th e slice .
2. Th e norma l an d tangentia l components o f th e weigh t W acting o n th e bas e o f th e slice .
They ar e designated respectively as N an d T .
3. Th e por e wate r pressure U acting on the base of the slice .
4. Th e effectiv e frictiona l an d cohesiv e resistance s actin g o n th e bas e o f th e slic e whic h i s
designated as S.
The force s actin g on the sides of the slices ar e statically indeterminate as they depend o n the
stress deformation properties of the material, and we can make only gross assumption s about their
relative magnitudes.
In the conventional slice method of analysis the lateral forces are assumed equal on both sides
of the slice. This assumption is not strictly correct. The error due t o this assumption on the mass as
a whol e i s about 1 5 percent (Bishop, 1955) .
St abilit y o f Slope s 395
(a) Total normal and tangential components
B ~ - - ^ C
(b) Pore-pressure diagram
Trial failur e
surface
f\ l
/ 7 " U } = «,/,
Pore-pressure
diagram
U2 = M2/2
U3 = M3/3
(c) Resisting force s o n the base o f slic e (d) Graphical representatio n o f al l the force s
Figure 10.2 0 St abilit y an alys i s o f s lop e by the metho d o f s lice s
396 Chapt e r 1 0
The forces tha t are actually considered i n the analysis are shown in Fig. 10.20(c) . The various
components ma y b e determined a s follows:
1 . Th e weight , W, of a slice per uni t lengt h of dam ma y b e compute d fro m
W=yhb
where, y = total uni t weigh t of soil, h = average heigh t of slice, b - widt h of slice .
If the width s of al l slice s ar e equal , an d i f the whol e mas s i s homogeneous , th e weigh t W
can b e plotted a s a vector AB passin g throug h the center o f a slice a s i n Fig. 10.20(a) . AB
may b e mad e equa l to the height of the slice .
2. B y constructing triangl e ABC, th e weight can be resolved int o a normal componen t N and
a tangential component T . Similar triangles can be constructed fo r all slices. The tangential
components o f the weight s cause th e mas s t o slide downward . The su m o f al l the weight s
cause th e mass _ to slid e downward . Th e su m o f al l th e tangentia l component s ma y b e
expressed a s T= I.T. If the trial surface is curved upward near its lower end, the tangential
component of the weight of the slice wil l act in the opposite directio n alon g the curve. The
algebraic su m o f T shoul d b e considered .
3. Th e averag e por e pressur e u acting on the bas e o f an y slic e o f lengt h / may b e foun d fro m
the pore pressure diagram shown in Fig. 10.20(b) . The total pore pressure, U, on the base of
any slic e i s
U=ul
4. Th e effective normal pressur e N ' acting on the base of any slice i s
N '=N - t/[Fig . 10.20(c) ]
5. Th e frictiona l force F
f
actin g on th e bas e o f any slic e resistin g th e tendenc y o f the slic e t o
move downwar d i s
F = ( N - U ) tan 0'
where 0' is th e effectiv e angl e o f friction . Similarl y th e cohesiv e forc e C " opposing th e
movement o f the slic e and acting at the base of the slic e is
where c i s the effective uni t cohesion. Th e tota l resisting forc e S acting o n the base of the
slice i s
S = C + F' = c'l + ( N - U ) tan 0'
Figure 10.20(c ) shows th e resisting forces actin g on the base of a slice .
The su m of al l the resisting forces actin g on the base o f each slic e ma y be expresse d a s
S
s
= c'I,l + tan 0' I(W- £/ ) = c'L + tan 0' X( N - U)
where £/ = L = lengt h of the curved surface .
The moment s o f th e actuatin g an d resistin g force s abou t th e poin t o f rotatio n ma y b e
written a s follows:
Actuating momen t = R~L T
Resisting momen t = R[ c'L + tan 0' £(j V - U)]
The facto r of safet y F
?
may no w be writte n as
(10.29)
St abilit y o f Slope s 397
The variou s component s show n in Eq. (10.29 ) can easil y be represente d graphicall y as
shown i n Fig. 10.20(d) . The line AB represent s t o a suitabl e scale Z,( N - U) . BC i s drawn
normal to AB at B and equal to c'L + tan 0' Z( N - U) . The line AD drawn at an angle 0'to
AB gives the intercept BD on BC equal to tan 0'Z(N- U) . The length BE on BC is equal to
IT. Now
F =
BC
BE
(10.30)
Centers fo r Trial Circles Through To e
The factor of safet y F
s
a s computed and represented by Eq. (10.29) applie s t o one trial circle. Thi s
procedure is followed for a number of trial circles until one finds the one for which the factor of safet y
is the lowest . This circl e that gives the least F
s
i s the one most likely to fail . The procedur e i s quite
laborious. The number of trial circles ma y be minimized if one follows th e following method.
For an y given slope angl e / 3 (Fig. 10.21) , th e cente r of the firs t tria l circl e cente r O may b e
determined a s proposed b y Fellenius (1927) . Th e direction angle s a
A
and a
B
may b e taken fro m
Table 10.1 . For the centers of additional trial circles, the procedure i s as follows:
Mark point C whose position i s as shown in Fig. 10.21 . Joi n CO. The center s o f additional
circles lie on the line CO extended. This method is applicable for a homogeneous ( c -</> ) soil. When
the soi l i s purely cohesive and homogeneous the direction angles given in Table 10. 1 directly give
the center for the critical circle.
Centers fo r Trial Circles Belo w To e
Theoretically i f the materials of the dam and foundation ar e entirely homogeneous, any practicable
earth dam slope may have its critical failure surface below the toe of the slope. Fellenius found that
the angle intersected at 0 in Fig. 10.22 for this case is about 133.5°. To find the center for the critical
circle below the toe, th e following procedur e i s suggested.
Locus of centers
of critical circle s
Curve of factor
of safet y
Figure 10.2 1 Locatio n o f center s o f critica l circl e passin g throug h toe o f dam
398 Chapt er 1 0
Figure 10.2 2 Cen t er s o f t ria l circle s fo r bas e failur e
Table 1 0. 1 Direct io n an g le s a°
A
an d a°o f o r cen t er s o f crit ica l circl e s
Slope Sl ope an g l e Direct ion an g l e s
0.6: 1
1 : 1
1.5: 1
2: 1
3: 1
5 : 1
60
45
33.8
26.6
18.3
11.3
29
28
26
25
25
25
40
37
35
35
35
37
Erect a vertical at the midpoint M of the slope. On this vertical wil l be the center O of the firs t
trial circle. In locating the trial circle use an angle (133.5°) betwee n the two radii at which the circle
intersects th e surfac e o f th e embankmen t an d th e foundation . After th e firs t tria l circl e ha s bee n
analyzed the center is some what moved to the left, the radius shortened an d a new trial circle drawn
and analyzed. Additional centers for the circles ar e spotted an d analyzed.
Example 1 0 .1 1
An embankmen t i s t o b e mad e o f a sand y clay having a cohesion o f 3 0 kN/m
2
, angl e o f internal
friction o f 20° an d a uni t weight of 1 8 kN/m
3
. The slop e an d height of the embankment are 1. 6 : 1
and 10m respectively. Determine the factor of safety by using the trial circle given in Fig. Ex. 10.1 1
by the metho d o f slices .
Solution
Consider th e embankment as shown in Fig. Ex.10.11. The center of the trial circle O is selected b y
taking a
A
= 26° and a
B
= 35° from Table 10.1 . The soi l mass abov e the sli p circle i s divided into 13
slices o f 2 m widt h each. The weigh t of each slic e pe r uni t lengt h of embankment i s given by W =
h
a
by
;
, where h
a
= average heigh t of the slice , b = width of the slice , y
t
= unit weight of the soil .
The weigh t o f eac h slic e ma y be represente d b y a vector o f height h
a
i f b and y, remain th e
same for the whole embankment. The vectors values were obtained graphically . The height vectors
St abilit y o f Slope s 399
Figure Ex . 1 0.1 1
may be resolved int o normal components h
n
and tangential components h
{
. The values of h
a
, h
n
and
h
t
for the various slices ar e given below i n a tabular form.
Values o f h
al
h
n
an d /?,
o / v r
Slice No .
1
2
3
4
5
6
7
h
a
( m)
1.8
5.5
7.8
9.5
10.6
11.0
10.2
h
n
( m)
0.80
3.21
5.75
7.82
9.62
10.43
10.20
h
t
( m]
1.72
4.50
5.30
5.50
4.82
3.72
2.31
Slice No .
8
9
10
11
12
13
h
a
( m)
9.3
8.2
6.8
5.2
3.3
1.1
h
n
( m)
9.25
8.20
6.82
5.26
3.21
1.0
h
t
( m)
1.00
-0.20
-0.80
-1.30
-1.20
-0.50
The sum of these component s h
n
and h
t
may be converted int o forces ZN and Irrespectively
by multiplying them as given below
Sfc
n
= 81.57m, Ui
t
= 24.87m
Therefore, ZN = 81.57 x 2 x 1 8 = 2937 k N
Zr = 24.87 x2x 1 8 = 895kN
Length o f arc = L = 31.8 m
'L + tonfiZ N 30x31. 8 + 0.364x2937
Factor of safet y =
895
= 2.26
400
1 0 .1 4 B I SH OP' S SI M PL I FI E D M ETH O D OF SL I CES
Chapt er 1 0
Bishop's metho d o f slice s (1955 ) is useful i f a slope consist s o f severa l types o f soi l wit h differen t
values of c and 0 and if the pore pressures u in the slope ar e known or can be estimated. The method
of analysi s i s a s follows:
Figure 10.2 3 gives a section of an earth dam having a sloping surface AB. ADC i s an assume d
trial circula r failur e surfac e with its center a t O. The soi l mas s abov e th e failur e surfac e i s divide d
into a number of slices . The forces actin g on each slic e are evaluated from limi t equilibrium of the
slices. The equilibriu m of the entire mass i s determined b y summatio n of the forces o n each o f the
slices.
Consider fo r analysi s a singl e slic e abed (Fig . 10.23a ) whic h i s drawn t o a large r scal e i n
Fig. 10.23(b) . The force s actin g on thi s slice ar e
W = weight of the slic e
N = total normal forc e o n the failur e surface cd
U = pore water pressur e = ul o n the failur e surface cd
F
R
= shear resistanc e actin g on the base of the slic e
E
r
E
2
- norma l force s on the vertical faces be and ad
T
r
T
2
= shear force s o n the vertical faces be and ad
6 = the inclinatio n of the failur e surface cd t o the horizontal
The syste m i s staticall y indeterminate . A n approximat e solutio n ma y b e obtaine d b y
assuming tha t th e resultan t o f £, an d T^ i s equa l t o tha t o f E
2
an d T
2
, an d thei r line s o f actio n
coincide. Fo r equilibrium of the system, the following equation s hold true .
O
(a) (b )
Figure 10.2 3 B is hop' s s im plifie d m et ho d o f an al ys i s
St abilit y o f Slope s 40 1
N =Wcos6
(10.31)
where F
(
= tangentia l component of W
The uni t stresses o n the failur e surface of length, /, may be expressed a s
Wcos6
normal stress , <r
n
= -
Wsin0 (10.32 )
shear stress , r
n
= -
The equation for shear strength , s, is
s = c' + cr' t an^' = c' + ( cr-u)tan0'
where r f = effective normal stres s
c' - effectiv e cohesio n
( ft = effective angle of friction
u = unit pore pressur e
The shearin g resistanc e t o sliding on the base o f the slice i s
si = c'l + ( Wcos 9 -ul) ta n $
where ul = U, the total por e pressur e on the base of the slice (Fi g 10.23b )
d = F At - r
R
The total resisting force and the actuating force on the failure surface ADC ma y be expressed
as
Total resisting force F
R
i s
F
R
= [c 7 + ( Wc os 0- M/ ) t a n0'] (10.33 )
Total actuatin g force F
t
i s
F
t
= Wsm0 (10.34 )
The facto r of safet y F
s
i s then given as
F
Eq. (10.35) is the same as Eq. (10.29) obtained b y the conventional metho d of analysis .
Bishop (1955 ) suggest s tha t the accurac y o f th e analysi s ca n b e improve d b y takin g int o
account the forces E and Ton the vertical faces of each slice. For the element i n Fig. 10.23(b) , we
may writ e a n expressio n fo r al l th e force s actin g i n th e vertica l directio n fo r th e equilibrium
condition a s
N ' co&0 = W + ( T^ -T
2
)-ulcos0- F
R
si n # (10.36 )
If th e slop e i s no t o n th e verg e o f failur e ( F
s
> 1) , the tangentia l forc e F
t
i s equa l t o th e
shearing resistance F
R
o n cd divided by F
g
.
402 Chapt er 1 0
c'l
(10.37)
where, N '=N -U,andU= ul.
Substituting Eq. (10.37) int o Eq. (10.36) an d solving for N \ we obtain
c'l
— si n< 9
F
cos <9 +
tan 0' sin 6
F..
(10.38)
where, AT= T
{
- T
r
For equilibrium of the mass above the failur e surface , we have by taking moments abou t O
Wsin0R = F
R
R (10.39 )
By substituting Eqs. (10.37) an d (10.38) into Eq. (10.39) an d solving we obtain an expressio n
f or F a s
F
where,
tan ( /> ' si n 9
F
(10.40)
(10.41)
The factor of safet y F
S
i s present i n Eq. (10.40) on both sides . The quantit y AT=T
{
- T
2
has
to b e evaluate d b y mean s o f successiv e approximatio n . Trial value s o f E^ an d T
l
tha t satisf y th e
equilibrium o f each slice , an d th e conditions
1.6
1.4
1.2
1.0
m
f)
= cos 6 + (sin 6tan d) )/F
i i i
Note: 0 i s + when slope o f failur e
arc i s i n the same quadrant
as ground slope .
0.6
-40 -3 0 -2 0 -1 0 0 1 0 2 0
V alues of 6 degrees
30 40
Figure 10.2 4 Val ue s o f m
f i
(aft e r J an bu e t al. , 1956 )
St abilit y o f Sl ope s 403
( E
l
-E
2
) = Q an d (r
l
- T
2
) = 0
are used . Th e valu e o f F
s
ma y the n b e compute d b y firs t assumin g an arbitrar y valu e for F
s
. Th e
value of F
s
may then be calculated by making use of Eq. (10.40). I f the calculated value of F
v
differ s
appreciably fro m th e assume d value , a secon d tria l i s mad e an d th e computatio n i s repeated .
Figure 10.2 4 developed b y Janbu et al . (1956) helps t o simplif y th e computation procedure .
It is reported that an error of about 1 percent will occur if we assume Z(Tj - T"
2
) tan0'= 0. But
if w e us e th e conventiona l method o f analysi s usin g Eq . (10.35 ) th e erro r introduce d i s abou t
15 percent (Bishop , 1955) .
10.15 BISHO P AND MORGENSTERN METHOD FOR SLOPE ANAL Y SI S
Equation (10.40) develope d base d o n Bishop' s analysi s of slopes, contain s the term por e pressur e
u. Th e Bishop and Morgenstern metho d (1960) proposes th e following equation for the evaluation
of u
yh
(10.42)
where, u = pore wate r pressur e a t any point on the assumed failur e surface
Y= unit weight of the soi l
h = the dept h o f the point in the soi l mas s below th e ground surface
The por e pressur e rati o r
u
i s assumed t o be constant throughout the cross-section, whic h is
called a homogeneous pore pressure distribution. Figur e 10.2 5 shows the various parameters use d
in th e analysis.
The facto r of safet y F i s defined as
F_ = m - nr,. (10.43)
where, m, n = stability coefficients.
The m and n values may be obtained eithe r from chart s in Figs. B. 1 to B.6 or Tables B1 to B6
in Appendix B. The dept h factor given in the charts or tables i s as per Eq. (10.25), that is n
d
= DIH,
where H = height o f slope , an d D = depth o f fir m stratu m from th e to p o f th e slope . Bisho p an d
Morgenstern (1960 ) limite d thei r chart s (o r tables ) t o value s of c'ly H equa l t o 0.000, 0.025, and
0.050.
Center of failure surfac e
Failure surfac e
y = unit weight of soi l
/^^^^^^^^//^f ^^^
Figure 1 0.2 5 Specificat ion s o f param et er s for B is hop-M org en s t er n m et ho d o f
an alys is
404 Chapt er 1 0
Extension o f the Bisho p an d M orgenster n Slop e Stabilit y Chart s
As state d earlier , Bisho p an d Morgenster n (1960 ) chart s o r table s cove r value s o f c'lyH equa l t o
0.000, 0.025, an d 0.050 only. These chart s do not cover the values that are normally encountered in
natural slopes . O' Conno r an d Mitchel l (1977 ) extende d th e wor k o f Bishop an d Morgenster n t o
cover value s of c'lyH equa l to 0.075 an d 0.100 for various values of depth factors n
d
. The metho d
employed i s essentiall y th e same a s that adopte d b y th e earlie r authors . The extende d value s ar e
given i n the for m o f chart s and table s from Figs . B. 7 t o B.14 and Tables B7 t o B14 respectively in
Appendix B.
M ethod o f Determinin g F
s
1. Obtai n th e value s of r
u
and clyH
2. Fro m th e tables in Appendix B, obtain the values of m and n for the known values ofc/yH,
0 and /3, and for n
d
- 0 , 1 , 1.2 5 and 1.5.
3. Usin g Eq. (10.43) , determine F
s
fo r each valu e of n
d
.
4. Th e require d value of F
s
i s the lowes t of the values obtained i n step 3.
Example 1 0 .1 2
Figure Ex . 10.1 2 gives a typica l sectio n o f a homogeneou s eart h dam . The soi l parameter s are:
0' = 30°, c' = 59 0 lb/ft
2
, an d y = 12 0 lb/ft
3
. Th e da m ha s a slop e 4: 1 an d a por e pressur e rati o
r
u
= 0.5. Estimate th e facto r of safet y F
s
b y Bisho p and Morgenster n metho d fo r a height o f da m
# =140 f t.
Solution
Height o f da m H= 140f t
c' 59 0
120x140
= 0.035
Given: 0' = 30°, slope 4: 1 and r
u
= 0.5.
Since c'lyH = 0.035, an d n
d
= 1.43 for H = 140 ft , th e F
s
fo r th e da m lie s betwee n c'lyH
0.025 an d 0.05 and n
d
between 1. 0 and 1.5 . The equatio n for F
s
is
= m-nr
Using th e Tables i n Appendix B, the following table can be prepared fo r th e give n values of
c'lyH, 0 , an d /3.
0' =30°
c' = 590psf
y - 12 0 pcf
/• „ =0.5 0
D = 200 f t
Alluvium (same propertie s a s above )
Figure Ex . 1 0.1 2
St abi l i t y o f Slope s 40 5
From Table s B 2 an d B3 for c'/yH =0.02 5
n
d
1.0
1.25
From Tabl e B4, B5
n
d
1.0
1.25
1.50
m
2.873
2.953
and B6 for c'ljH - 0.0 5
m
3.261
3.221
3.443
n
2.622
2.806
n
2.693
2.819
3.120
F,
1.562
1.55
F,
1.915
1.812
1.883
Lowest
Lowest
Hence n
d
= 1.2 5 i s th e mor e critica l dept h factor . Th e valu e o f F
s
fo r c'lyH = 0.035 lie s
between 1.5 5 (fo r c'lyH = 0.025) and 1.81 2 (fo r c'lyH = 0.05). By proportion F = 1.655 .
1 0 .1 6 M ORG ENSTER N M ETH OD O F ANAL Y SI S FOR RAPI D
DRAWDOWN CONDI TI O N
Rapid drawdow n of reservoi r wate r leve l i s one o f the critica l state s i n th e desig n o f eart h dams .
Morgenstern (1963 ) developed th e method of analysis for rapid drawdown conditions based o n the
Bishop and Morgenster n metho d o f slices . The purpos e o f thi s method i s to compute the facto r of
safety durin g rapid drawdown , which is reduced unde r no dissipation of pore wate r pressure. The
assumptions made i n the analysi s are
1. Simpl e slop e o f homogeneous materia l
2. Th e dam rest s o n an impermeabl e bas e
3. Th e slope i s completely submerge d initially
4. Th e pore pressur e doe s no t dissipat e during drawdown
Morgenstern use d the pore pressur e paramete r 5 a s developed b y Skempton (1954 ) whic h
states
5 = — (10.45 )
where cr , = y h
j- total uni t weight of soil or equal to twice the unit weight of water
h = height of soi l abov e th e lowe r leve l of water after drawdown
The chart s develope d tak e int o account the drawdown rati o whic h is defined as
(10.46)
where R
d
= drawdown rati o
// = height of drawdown
H = height of dam (Fig . 10.26 )
All the potential slidin g circles mus t be tangent to the base o f the section .
406 Chapt er 1 0
Full reservoi r level
" Drawdow n
/ l evel
H
Figure 10.2 6 Da m s ect ion fo r drawdow n con dit ion s
The stabilit y chart s ar e give n in Figs. 10.2 7 t o 10.2 9 coverin g a range of stabilit y numbers
c'/yH fro m 0.012 5 t o 0.050. Th e curves developed are for the values of 0'of 20°, 30°, and 40° for
different value s of B.
PL,
0.2 0. 4 0. 6 _0. 8
Drawdown rati o H/ H
(a) 0 = 2:1
1.0 0 0. 2 0. 4 0. 6 _0. 8
Drawdown rati o H/ H
\
<P
40°
30°
20°
0.2 0. 4 0. 6 _0. 8
Drawdown rati o H/ H
1.0
Figure 10.2 7
0.2 0. 4 0. 6 _0. 8 1. 0
Drawdown rati o H/ H
( d) ft = 5:1
Drawdown s t abilit y char t fo r c'/yH = 0. 012 5 (aft e r M org en s t ern ,
1963)
St abilit y o f Slope s 407
40°
30°
20°
0.2 0. 4 0. 6 _0. 8 1. 0
Drawdown rati o H/ H
( a) ft = 2:1
40°
30°
20°
0 0. 2 0. 4 0. 6 _0. 8 1. 0
Drawdown rati o H/ H
(b) ft = 3:1
0.2 0. 4 0. 6 _0. 8
Drawdown rati o H/ H
( c) ft = 4:1
<P
40°
30°
20°
1.0
i*,
>*
UH
(d) 0 = 5:
<P
40°
30°
20°
0 0. 2 0. 4 0. 6 _0. 8 1. 0
Drawdown rati o H/ H
Figure 10.2 8 Drawdow n s t abilit y char t fo r c'lyH = 0. 02 5 (aft e r M org en s t ern ,
1963)
Example 1 0 .1 3
It i s required t o estimat e th e minimum factor of safet y for th e complet e drawdow n of th e sectio n
shown i n Fig. Ex . 10.1 3 (Morgenstern, 1963 )
.*._./:
Water level befor e
drawdown
Water level afte r
drawdown
Figure Ex . 1 0.1 3
408 Chapt er 1 0
Solution
From the dat a give n i n the Fig. Ex . 10.1 3
N = — =
312
= 0.025
yH. 124.8x10 0
From Fig . 10.28 , fo r W = 0.025, 0= 3:1, </> ' = 30°, an d H/ H = 1,
F
s
= 1.20
It is evident tha n the critical circl e i s tangent t o the base o f the dam and no other leve l nee d b e
investigated sinc e thi s woul d onl y rais e th e effectiv e valu e o f N
S
resultin g i n a highe r facto r o f
safety.
1 0 .1 7 SPENCE R M ETH OD OF ANAL Y SI S
Spencer (1967 ) develope d hi s analysi s base d o n th e metho d o f slice s o f Felleniu s (1927 ) an d
Bishop (1955) . Th e analysi s i s i n term s o f effectiv e stres s an d satisfie s tw o equation s o f
X
0.2 0. 4 0. 6 _0. 8
Drawdown rati o H/ H
0 0.2 0. 4 0. 6 _0. 8
Drawdown rati o H/ H
40°
30°
20°
1.0 0.2 0. 4 0. 6 _0. 8
Drawdown rati o H/ H
n- 2
k
\
.
\l
\
\
X
X
\
X
X
X
\
x
.
^"•^
X
x^
» ^_

— —E^M
40°
30°
20°
0 0.2 0. 4 0. 6 _0. 8
Drawdown rati o H/ H
1.0
(c) ft = 4:1 (d) ft = 5:1
Figure 10.2 9 Drawdow n s t abilit y char t fo r c'lyH = 0. 0 5 (aft e r M org en s t ern , 1963 )
St abilit y o f Slope s 40 9
equilibrium, the first wit h respect t o forces and the second wit h respect t o moments. The interslic e
forces ar e assume d t o be parallel a s i n Fig. 10.23 . The facto r of safet y F
f
i s expressed a s
Shear strengt h availabl e
F = CI O 47 ) 5
Shea r strengt h mobilize d ' '
The mobilize d angl e of shear resistance and other factor s are expressed as
(10.48)
u
pore pressure ratio , r = — n n 49)
yh ^ ' '
c'
Stability factor, N
S
=—— (10.50 )
The charts developed b y Spencer for different values of N
s
, §'
m
and r
u
are given in Fig. 10.30 .
The us e of these chart s wil l be explained with worked ou t examples .
Example 1 0.1 4
Find th e slop e corresponding t o a factor of safet y of 1. 5 for a n embankment 10 0 ft high i n a soi l
whose properties ar e as follows:
c' = 87 0 Ib/s q ft , y= 12 0 Ib/ft3, </> ' = 26°, r
u
= 0. 5
Solution (b y Spencer' s M ethod)
N =
^L
=
87 0
5
F
s
ytl 1.5x120x10 0
t
., tan f 0.48 8 _ „ _
tan 0 = -— = -= 0.325
F 1. 5
Referring t o Fig. 10.30c , for which r =0.5 , the slope correspondin g t o a stabilit y number
of 0.048 is 3:1.
Example 1 0.1 5
What woul d be the change i n strengt h on sudden drawdown for a soil elemen t a t point P which is
shown i n Fig. Ex . 10.15 ? Th e equipotentia l line passin g throug h thi s elemen t represent s los s o f
water head o f 1. 2 m. The saturate d uni t weight of the fil l i s 21 kN/m
3
.
Solution
The dat a given are shown in Fig. Ex . 10.15 . Befor e drawdown,
The stresses a t point P are:
%= /A
+
nA = 9.81 x 3 + 21 x 4 = 11 3 kN/m
2
"o = Y
w
( h
w
+ h
c
- h'} = 9.81(3 + 4 - 1.2 ) = 57 kN/m
2
410 Chapt er 1 0
4:1 3: 1 2:1
0.12
0.10
0.08
?L
^0.06
\j
0.04
0.02
4:1 3: 1 2:1 1.5:
2 4 6 8 1 0 1 2 1 4 1 6 1 8 2 0 2 2 2 4 2 6 2 8 3 0 3 2 3 4
Slope angle/?, degree s
Figure 10.3 0 St abilit y chart s (aft e r Spen cer , 1967 )
Stability o f Slopes 411
Figure Ex . 10.1 5
Therefore tf
0
= ( J
Q
- U
Q
= 113 - 5 7 = 56 kN/m
2
After drawdown,
o= y
sat
h
c
= 21 x 4 = 84 kN/m
2
u = y
w
( h
c
- h'} = 9.81(4 - 1.2 ) = 27.5 kN/m
2
o
f
= a-u = S4-27.5 = 56.5 kN/m
2
The change i n strength is zero sinc e the effective vertica l stres s does no t change.
Note: There i s no change i n strengt h due to sudden drawdown but the direction o f forces of
the seepage wate r changes from a n inward direction befor e drawdown to an outward direction afte r
drawdown an d this is the mai n cause for the reduction i n stability .
1 0.1 8 PROB L EM S
10.1 Fin d the critical height of an infinite slope having a slope angl e of 30°. The slope is made of
stiff cla y havin g a cohesio n 2 0 kN/m
2
, angl e o f interna l frictio n 20° , voi d rati o 0. 7 an d
specific gravit y 2.7. Conside r th e following cases for the analysis.
(a) the soi l i s dry.
(b) the water seeps parallel t o the surface of the slope .
(c) the slope i s submerged.
10.2 A n infinit e slope ha s a n inclinatio n of 26 ° wit h the horizontal . I t i s underlai n by a fir m
cohesive soi l havin g G
s
= 2.72 an d e = 0.52. Ther e i s a thi n weak laye r 2 0 f t below an d
parallel to the slope ( c' - 52 5 lb/ft
2
, 0 ' = 16°) . Comput e the factors of safety when (a) the
slope i s dry, and (b) ground water flows paralle l t o the slope a t the slope level .
10.3 A n infinite slop e is underlain with an overconsolidated cla y having c' - 21 0 lb/ft
2
, 0 ' = 8°
and y
sat
= 12 0 lb/ft
3
. Th e slop e i s inclined at an angl e of 10 ° to the horizontal . Seepag e i s
parallel t o th e surfac e and the groun d water coincide s wit h the surface . I f the slop e fail s
parallel to the surface along a plane at a depth of 12 ft below the slope, determine the factor
of safety.
10.4 A deep cu t o f 1 0 m dept h i s made i n sandy clay for a road. The side s o f the cu t make an
angle o f 60 ° wit h th e horizontal . Th e shea r strengt h parameter s o f th e soi l ar e
c' - 2 0 kN/m
2
, f i = 25°, an d 7= 18. 5 kN/m
3
. I f AC is the failure plane (Fi g Prob . 10.4) ,
estimate the factor of safet y of the slope .
412
Chapt er 1 0
y = 18.5kN/m
3
Figure Prob . 10. 4
W = 1050 kN
Figure Prob . 10. 5
10.5 A 40° slop e is excavated t o a depth of 8 m in a deep laye r of saturated clay having strength
parameters c = 60 kN/m
2
, 0 = 0, and y= 1 9 kN/m
3
. Determin e th e factor of safet y for the
trial failur e surfac e shown in Fig. Prob . 10.5 .
10.6 A n excavation t o a depth of 8 m wit h a slope o f 1: 1 was made i n a deep layer o f saturate d
clay havin g c
u
= 65 kN/m
2
an d 0
M
= 0. Determine th e factor o f safet y for a trial sli p circl e
passing throug h the toe of the cut and having a center a s shown in Fig. Prob . 10.6 . The unit
weight of the saturate d clay i s 1 9 kN/m
3
. No tension crack correctio n i s required .
10.7 A 45 ° cu t wa s mad e i n a claye y sil t wit h c = 15 kN/m
2
, 0 = 0 an d y = 19. 5 kN/m
3
. Sit e
exploration reveale d th e presence o f a sof t cla y stratu m of 2 m thick having c = 25 kN/m
2
and 0 = 0 a s shown i n Fig . Prob . 10.7 . Estimat e th e facto r o f safet y o f th e slop e fo r th e
assumed failur e surface .
10.8 A cut was made in a homogeneous clay soil to a depth of 8 m as shown in Fig. Prob . 10.8 . The
total uni t weigh t o f th e soi l i s 1 8 kN/m
3
, an d it s cohesiv e strengt h i s 2 5 kN/m
2
.
St abilit y o f Slope s 413
<§ )
Figure Prob . 10. 6
Figure Prob . 10. 7
Assuming a 0 = 0 condition, determine the factor of safety with respect to a slip circle passing
through the toe. Consider a tension crack at the end of the slip circle on the top of the cut.
10.9 A dee p cu t o f 1 0 m dept h i s made i n natura l soil fo r th e constructio n o f a road. Th e soi l
parameters are : c' = 35 kN/m
2
, 0' = 15 ° and 7= 2 0 kN/m
3
.
Figure Prob . 10. 8
414
Chapt er 1 0
Figure Prob . 10. 9
The side s o f th e cut mak e angle s o f 45° wit h the horizontal . Compute th e facto r o f safet y
using friction circle method for the failur e surface AC shown i n Fig. Prob . 10.9 .
10.10 A n embankment is to be buil t to a height of 50 ft at an angle of 20° wit h the horizontal. The
soil parameter s are : c' - 63 0 lb/ft
2
, 0 ' = 18° and 7= 11 5 lb/ft
3
.
Estimate th e following;
1. Facto r o f safet y of the slop e assumin g full frictio n i s mobilized.
2. Facto r of safety with respect t o friction i f the factor of safety with respect to cohesion i s
1.5.
Use Taylor' s stabilit y chart.
10.11 A cut wa s mad e i n natura l soil fo r th e constructio n of a railway line. The soi l parameter s
are: c' = 700 lb/ft
2
, 0 ' = 20° and 7 = 11 0 lb/ft
3
.
Determine th e critical height of the cut for a slope of 30° wit h the horizontal by making use
of Taylor's stabilit y chart.
10.12 A n embankmen t i s t o b e constructe d b y makin g us e o f sand y cla y havin g th e followin g
properties: c' = 35 kN/m
2
, 0' = 25° and y= 19. 5 kN/m
3
.
The height of the embankment is 20 m with a slope of 30° wit h the horizontal a s shown in
Fig. Prob . 10.12 . Estimat e th e facto r o f safet y b y th e metho d o f slice s fo r th e tria l circl e
shown i n the figure .
10.13 I f an embankment of 1 0 m height is to be made fro m a soil having c' = 25 kN/m
2
, 0' = 15° ,
and 7=18 kN/m
3
, wha t will be the saf e angl e of slop e fo r a factor of safet y of 1.5 ?
10.14 A n embarkmen t i s constructe d fo r a n eart h da m o f 8 0 f t hig h a t a slop e o f 3:1 . Th e
properties o f th e soi l use d fo r th e constructio n are : c - 77 0 lb/ft
2
, 0 ' = 30° , an d
7=110 lb/ft
3
. Th e estimated pore pressue r ratio r =0.5 . Determine the factor of safet y by
Bishop an d Morgenstern method .
10.15 Fo r the Prob. 10.14 , estimate th e factor of safety for 0' = 20°. All the other dat a remai n the
same.
10.16 Fo r the Prob. 10.14 , estimat e the factor of safet y for a slope o f 2: 1 wit h all the oother dat a
remain the same .
St abilit y o f Slope s 415
Figure Prob. 10.1 2
10.17 A cu t o f 2 5 m dopt h i s mad e i n a compacte d fil l havin g shea r strengt h parameter s o f
c = 25 kN/m
2
, an d 0' = 20°. The tota l uni t weight of the materia l i s 1 9 kN/m
3
. The por e
pressuer ratio has an average value of 0.3. The slope of the sides i s 3:1. Estimate th e factor
of safet y usin g the Bishop and Morgenstern method .
10.18 Fo r the Prob. 10.17 , estimate the factor of safety for 0'= 30°, wit h all the other data remain
the same .
10.19 Fo r the Prob. 10.17 , esatimate the factor of safet y for a slope of 2: 1 with all the other dat a
remaining the same .
10.20 Estimat e th e minimum factor of safet y for a complete drawdown condition for the sectio n
of da m i n Fig. Prob . 10.20 . Th e ful l reservoi r leve l of 1 5 m dept h i s reduced t o zer o afte r
drawdown.
10.21 Wha t is the safet y facto r if the reservoi r leve l i s brought down from 1 5 m t o 5 m depth i n
the Prob . 10.20 ?
10.22 A n earth da m to be constructed at a site has the following soil parameters : c' = 60 0 lb/ft
2
,
y = 110 lb/ft
3
, an d 0' = 20°. The height of of dam H = 50 ft .
The pore pressure ratio r
u
= 0.5. Determine the slope of the dam for a factor of safety of 1. 5
using Spencer' s method (1967) .
c' = 15 kN/m
2
<f> ' = 30 °
y = 20 kN/m
3
Figure Prob . 10.2 0
416
Chapt er 1 0
O
R = 45 f t
15ft
Figure Prob . 10.2 4
10.23 I f the give n pore pressur e rati o i s 0.25 i n Prob. 10.22 , wha t wil l be th e slop e o f the dam?
10.24 A n embankment has a slope of 1. 5 horizontal to 1 vertical wit h a height of 25 feet. The soi l
parameters are :
c - 60 0 lb/ft
2
, 0 ' = 20°, an d 7= 11 0 lb/ft
3
.
Determine th e factor of safety using friction circl e method for the failure surface AC shown
in Fig . Prob . 10.24 .
10.25 I t i s required t o construct an embankment for a reservoir t o a height o f 2 0 m a t a slope of
2 horizontal t o 1 vertical. The soi l parameter s are :
c = 40 kN/m
2
, f = 18° , and 7= 17. 5 kN/m
3
.
Estimate th e following:
1. Factor o f safet y of th e slop e assumin g full frictio n i s mobilized .
2. Factor o f safet y wit h respect t o friction i f the factor of safet y with respect t o cohesion i s
1.5.
Use Taylor' s stabilit y chart.
10.26 A cuttin g o f 4 0 f t dept h i s t o b e mad e fo r a road a s show n i n Fig . Prob . 10.26 . The soi l
properties are :
c' = 500 lb/ft
2
, 0 ' = 15° , and 7= 11 5 lb/ft
3
.
Estimate the factor of safet y b y the method o f slices for the tria l circle shown i n the figure.
10.27 A n eart h da m i s t o b e constructe d fo r a reservior . Th e heigh t o f th e da m i s 6 0 ft . Th e
properties o f the soi l use d i n the construction are:
c = 400 lb/ft
2
, 0 ° = 20°, an d 7= 11 5 lb/ft
3
, an d ft = 2:1.
Estimate the mi ni mum factor of safety for the complete draw n from the ful l reservio r leve l
as shown i n Fig. Prob . 10.2 7 b y Morgenster n method .
10.28 Wha t i s the facto r of safet y i f the wate r level i s brought down from 60 f t t o 20 f t above th e
bed leve l o f reservoi r i n Prob. 10.27 ?
St abilit y o f Slope s 417
c' = 5001b/ft
2
0' =15°
y=1151b/ft
3
Figure Prob . 10.2 6
Full reservoir level
1
Figure Prob . 10.2 7
10.29 Fo r the dam given in Prob. 10.27, determine the factor o f safety fo r r «= 0.5 by Spencer' s
method.
CHAPTER 1 1
LATERAL EARTH PRESSURE
1 1 .1 I NTRODU CTI O N
Structures tha t ar e buil t t o retai n vertica l o r nearl y vertica l eart h bank s o r an y othe r materia l ar e
called retaining walls. Retainin g walls may be constructed of masonry or sheet piles . Some o f the
purposes fo r which retaining wall s are used ar e shown in Fig. 11.1 .
Retaining wall s ma y retai n wate r also . Th e eart h retaine d ma y b e natura l soi l o r fill . Th e
principal type s of retaining walls are give n in Figs. 11. 1 and 11.2 .
Whatever ma y b e th e typ e o f wall , al l th e wall s liste d abov e hav e t o withstan d latera l
pressures either from eart h or any other material on their faces. The pressures acting on the walls try
to mov e the wall s fro m thei r position. Th e wall s shoul d be s o designed a s to keep them stabl e i n
their position . Gravit y wall s resis t movemen t becaus e o f thei r heav y sections . The y ar e buil t of
mass concrete o r stone or brick masonry. No reinforcement is required i n these walls . Semi-gravity
walls ar e not a s heavy a s gravity walls. A smal l amoun t of reinforcement i s used fo r reducin g th e
mass of concrete. The stems of cantilever walls are thinner in section. The base slab is the cantilever
portion. Thes e wall s ar e made o f reinforced concrete . Counterfor t wall s ar e simila r t o cantileve r
walls excep t tha t th e ste m o f th e wall s spa n horizontall y betwee n vertica l bracket s know n a s
counterforts. Th e counterfort s ar e provide d o n th e backfil l side . Buttresse d wall s ar e simila r t o
counterfort wall s excep t th e bracket s o r buttres s wall s ar e provide d o n th e opposit e sid e o f th e
backfill.
In al l these cases , the backfill tries t o move the wall from it s position. The movement o f the
wall is partly resisted b y the wal l itself an d partly by soi l i n front o f the wall.
Sheet pil e wall s ar e more flexibl e tha n the other types . The eart h pressur e o n these wall s is
dealt wit h i n Chapte r 20 . Ther e i s anothe r typ e o f wal l tha t i s gainin g popularity . Thi s i s
mechanically stabilize d reinforce d eart h retainin g wall s (MSE) whic h wil l be deal t wit h later on .
This chapte r deal s wit h lateral eart h pressure s only.
419
420 Chapt er 1 1
(c) A bridge abutment (d) Water storage
.\\V\\\\\\I
(e) Flood wall s (f) Sheet pile wall
Figure 1 1 . 1 Us e of ret ain in g wall s
1 1 .2 L A T E R A L E A R TH P R E S S URE T H E O RY
There are two classical eart h pressur e theories. They ar e
1. Coulomb' s eart h pressure theory.
2. Rankine' s eart h pressur e theory.
The f i r s t ri gorou s anal ys i s o f th e proble m o f latera l eart h pressur e wa s publ i she d
by Coulom b i n (1776) . Ranki n e (1857 ) proposed a differen t approac h t o th e problem .
These theorie s propos e t o estimat e the magnitude s of two pressure s calle d active earth pressure
and passive earth pressure a s explained below.
Consider a rigi d retainin g wal l wit h a plan e vertica l face , a s show n i n Fig . 11.3(a), i s
backfilled wit h cohesionless soil . I f th e wal l doe s no t mov e eve n afte r bac k filling , th e pressur e
exerted o n the wal l i s terme d a s pressure fo r the at rest condition o f the wall . If suppos e th e wal l
gradually rotates abou t point A an d moves away from th e backfill , the uni t pressur e o n the wal l is
gradually reduced and after a particular displacement of the wal l at the top, the pressure reaches a
constant value . The pressur e i s the minimum possible. Thi s pressure i s termed th e active pressure
since th e weight of the backfil l is responsible fo r the movement o f the wall . If the wal l is smooth,
Lat eral Eart h Pres s ur e 421
Base sla b
Heel
(a) Gravity walls (b) Semi-gravit y walls (c) Cantilever wall s
Backfill
Counterfort Fac e o f wall —i
— Buttres s
Face
of wall
(d) Counterfort wall s (e) Buttressed wall s
Figure 11. 2 Principa l t ype s of rigi d ret ain in g wall s
the resultan t pressure act s normal t o the fac e of the wall . If the wal l i s rough, i t makes a n angl e < 5
with the normal on the wall. The angle 8 is called th e angle of wall friction. A s the wall moves away
from the backfill, the soil tends to move forward. When the wall movement is sufficient, a soil mass
of weight W ruptures along surfac e ADC show n in Fig. 11.3(a) . This surfac e i s slightl y curved. If
the surfac e i s assume d t o be a plane surfac e AC, analysi s woul d indicat e tha t thi s surfac e woul d
make an angl e of 45° + 0/2 with the horizontal.
If the wal l is now rotated abou t A towards the backfill, th e actual failure plane ADC i s also a
curved surfac e [Fig . 11.3(b)] . However , if the failur e surface i s approximate d a s a plane AC, thi s
makes an angle 45° - 0/ 2 with the horizontal and the pressure on the wall increases fro m th e value
of the at rest condition to the maximum value possible. The maximum pressure P tha t is develope d
is terme d th e passive earth pressure. Th e pressur e i s calle d passiv e becaus e th e weigh t o f th e
backfill opposes th e movement of the wall. It makes an angle 8 with the normal if the wall is rough.
The gradual decrease o r increase o f pressure on the wall with the movement of the wal l from
the at rest condition ma y b e depicted a s shown in Fig. 11.4 .
The movement A require d t o develop the passive state is considerably large r than A
Q
required
for th e active state .
1 1 .3 L ATERA L EARTH PRESSU RE FOR AT RES T CONDI TI ON
If the wall is rigid and does no t move with the pressure exerted on the wall, the soil behind the wall
will b e i n a stat e o f elastic equilibrium. Conside r a prismatic elemen t E i n th e backfil l at dept h z
shown i n Fig. 11.5 .
Element E i s subjected t o the following pressures.
V ertical pressur e = cr
v
= yz; latera l pressur e = <J
h
422
Chapter 1 1
(a) Active earth pressur e
(b) Passive eart h pressur e
Figure 1 1 . 3 Wal l m ovem en t fo r t h e developm en t o f act iv e an d pas s ive eart h
pres s ures
where yi s the effective unit weight of the soil. If we consider the backfill i s homogeneous the n both
cr
y
and o
h
increase linearly wit h depth z. In such a case, th e rati o of a
h
to <J
V
remains constant with
respect t o depth, that is
—- = —- = constant = AT ,
cr y z
(11-1)
where K
Q
i s called th e coefficient o f earth pressure for th e at rest condition o r at rest earth pressure
coefficient.
The latera l eart h pressur e o
h
acting on the wal l at any depth z may be expressed a s
cr, - (11.la)
Lat eral Eart h Pres s ur e 423
Passive pressure
Away fro m backfil l Into backfil l
Figure 11. 4 Developm en t o f act iv e an d pas s ive eart h pres s ure s
H
z z
H/3
L
A
Ph =
K
tiYZ
(a) (b )
Figure 1 1 . 5 Lat era l eart h pres s ur e fo r a t res t con dit io n
The expression fo r o
h
at depth H, the height of the wall , is
The distributio n of o
h
on the wal l i s given in Fig. 11.5(b) .
The total pressur e P
Q
for the soi l fo r the a t rest conditio n i s
(11.Ib)
(11.lc)
424 Chapt e r 1 1
Table 1 1 . 1 Coefficien t s o f eart h pres s ur e fo r a t res t con dit io n
Type o f soi l / K
Q
Loose sand , saturated
Dense sand , saturated
Dense sand , dry ( e = 0.6) -
Loose sand , dry ( e = 0.8) -
Compacted cla y 9
Compacted cla y 3 1
Organic silt y clay , undisturbe d ( w
{
= 74%) 4 5
0.46
0.36
0.49
0.64
0.42
0.60
0.57
The valu e of K
Q
depend s upo n the relative densit y of th e san d an d th e proces s by whic h th e
deposit wa s formed. I f thi s process does not involv e artificial tamping the valu e of K
Q
range s fro m
about 0.40 for loose sand t o 0.6 for dens e sand . Tamping th e layer s ma y increas e i t to 0.8.
The value of K
Q
may also be obtained on the basis of elastic theory. If a cylindrical sample o f soil
is acted upon by vertical stress CT, and horizontal stress a
h
, the lateral strai n e
{
ma y be expressed as
(11.2)
where E = Young's modulus, n = Poisson's ratio.
The latera l strai n e
{
= 0 when the eart h i s in the at rest condition. For thi s condition, we may
write
a
h V
or — = —(11.3 )
where ~T^~ = K
Q
, cr
v
=yz (11.4 )
According t o Jaky (1944) , a good approximatio n fo r K
0
i s given b y Eq. (11.5) .
K
Q
=l-sin0 (11.5 )
which fit s mos t o f the experimental data.
Numerical value s of K
Q
fo r some soil s ar e given i n Table 11.1 .
Example 1 1 . 1
If a retaining wal l 5 m hig h i s restrained fro m yielding , what wil l be th e at-res t eart h pressur e pe r
meter lengt h of the wall ? Given: the backfill is cohesionless soi l having 0 = 30° and y = 1 8 kN/m
3
.
Also determine th e resultant force fo r the at-rest condition.
Solution
From Eq. (11.5 )
K
Q
= l - s i n^= l-sin30° =0. 5
From Eq. (1 Li b), a
h
= KjH - 0. 5 x 1 8 x 5 = 45 kN/m
2
Lat eral Eart h Pres s ur e 425
From Eq. ( l l . l c)
P
Q
= -K
Q
y H
2
= ~x 0.5 x 18 x 5
2
= 112.5 kN/m length of wall
1 1 .4 RANKI NE' S STATES OF PL ASTI C EQU I L I B RI U M FOR
COH ESI ONL ESS SOI L S
Let AT in Fig. 11.6(a) represent th e horizontal surfac e of a semi-infinite mas s of cohesionless soil
with a unit weight y. The soi l is in an initial state of elastic equilibrium. Consider a prismatic block
ABCD. Th e dept h o f th e bloc k i s z an d th e cross-sectiona l are a o f th e bloc k i s unity . Sinc e th e
element is symmetrical wit h respect t o a vertical plane, the normal stress on the base AD i s
°
V
=YZ (11.6 )
o~
v
i s a principal stress . Th e norma l stres s o
h
on th e vertica l plane s AB o r DC a t dept h z ma y b e
expressed a s a function o f vertical stress .
<r
h
=
f( °v)
= K
orz (H.7 )
where K
Q
i s the coefficient o f earth pressure for the at rest condition which is assumed as a constant
for a particula r soil . Th e horizonta l stres s o
h
varie s fro m zer o a t th e groun d surfac e t o K
Q
yz a t
depth z.
Expansion
45° +
D A
T T T T T
Direction o f
major principa l
stress Stres s line s
(a) Active stat e
Compression
X
K
0
yz
— K
p
yz
Direction o f
minor principa l
stress
(b) Passive stat e
Direction of
minor principa l
stress
Direction o f
major principa l
stress
Figure 1 1 .6 (a , b) Ran k in e' s con dit io n fo r act iv e an d pas s ive failure s i n a semi-
infinite m as s o f cohes ion les s s oil
426 Chapt er 1 1
B' B C B B' C
Failure
plane
H
45° + 0/ 2
Failure
plane
45°-0/2
(c) Local activ e failur e (d ) Local passiv e failur e
A45°+0/2
Cn
(e) Mohr stres s diagra m
Figure 1 1 .6 (c , d, e) Ran k in e' s con dit io n fo r act iv e an d pas s iv e fail ure s i n a s em i -
in fin it e m as s o f cohes ion l es s s oi l
If we imagine that the entire mass i s subjected to horizontal deformation, such deformation i s
a plane deformation. Ever y vertical section through the mass represents a plane of symmetry fo r the
entire mass . Therefore, th e shea r stresse s o n vertical and horizontal sides o f the prism ar e equal t o
zero.
Due t o the stretching , the pressur e o n vertical sides AB an d C D of the pris m decrease s unti l
the condition s o f plastic equilibrium ar e satisfied , whil e th e pressur e o n th e bas e A D remain s
unchanged. Any further stretchin g merely causes a plastic flow withou t changing the state of stress .
The transition fro m th e stat e of plastic equilibrium t o the stat e o f plastic flow represent s th e failure
of th e mass . Sinc e th e weigh t o f th e mas s assist s i n producin g a n expansio n i n a horizonta l
direction, the subsequent failure i s called active failure.
If, o n the other hand, the mass of soil i s compressed, a s shown in Fig. 11.6(b) , in a horizontal
direction, the pressure on vertical sides AB an d CD of the prism increases whil e the pressure o n its
base remains unchange d a t yz. Since th e lateral compressio n o f the soil i s resisted by the weight of
the soil , the subsequen t failure b y plasti c flow i s called a passive failure.
Lat eral Eart h Pres s ur e 42 7
The proble m no w consist s o f determinin g th e stresse s associate d wit h th e state s o f plasti c
equilibrium in the semi-infinite mass and the orientation of the surface of sliding. The problem was
solved by Rankine (1857).
The plastic states which are produced by stretching or by compressing a semi-infinite mass of
soil parallel t o its surface are called active and passive Rankine states respectively. The orientation
of the planes may be found b y Mohr' s diagram .
Horizontal stretchin g o r compressin g o f a semi-infinit e mass t o develo p a stat e o f plasti c
equilibrium i s onl y a concept . However , loca l state s o f plasti c equilibriu m in a soi l mas s ca n b e
created by rotating a retaining wal l about its base either away from the backfill for an active stat e or
into the backfil l for a passive stat e i n the wa y shown in Figs. 1 1.3(c) and (d ) respectively. I n both
cases, th e soil withi n wedge ABC wil l be in a state of plastic equilibrium and line AC represents the
rupture plane.
Mohr Circl e fo r Active and Passiv e State s o f Equilibriu m i n Granular Soil s
Point P
{
o n the d-axi s in Fig. 1 1.6(e) represents th e state of stress on base AD of prismatic element
ABCD i n Fig. 1 1.6(a). Since th e shear stress o n AD i s zero, the vertical stress o n the bas e
is a principal stress. OA and OB are the two Mohr envelopes which satisfy th e Coulomb equation of
shear strengt h
j = crtan^ (11.9 )
Two circles C
a
and C ca n be drawn passing through P
l
an d at the same time tangential to the Mohr
envelopes OA and OB. When th e semi-infinit e mass is stretched horizontally , the horizontal stres s
on vertica l face s AB and CD (Fig. 1 1.6 a) at dept h z is reduced to the minimu m possible and thi s
stress is less than vertical stres s o
v
. Mohr circle C
a
gives the state of stress o n the prismatic element
at dept h z whe n the mas s i s in active failure. Th e intercept s OP
l
an d OP
2
ar e the major and minor
principal stresse s respectively.
When the semi-infinit e mass is compressed (Fig . 1 1.6 b), the horizontal stress on the vertical
face o f th e prismati c elemen t reache s th e maximu m value OP
3
an d circl e C i s th e Moh r circl e
which gives tha t state of stress .
Active Stat e o f Stress
From Moh r circl e C
a
Major principa l stres s = OP
{
= cr
l
= yz
Minor principal stres s = OP
2
= <7
3
(7, + <J~, <J, — (To
nn —
\J\J, —
1
2
cr. — <TT <j, + CT-,
From triangl e 00, C,, — = — si n i
1 J
2 2
( 1 + sin 0|
"
\
Therefore, p
a
= cr
3
=— -= yzK
A
( U.ll)
'V
where a, = yz, K
A
= coefficient o f earth pressure for th e active state = tan
2
(45° - 0/2) .
428 Chapt er 1 1
From poin t P
r
dra w a line parallel t o the base AD on which ( 7
{
acts . Since thi s line coincide s
with the cr-axis , point P
9
is the origin o f planes. Lines P
2
C
{
an d P^C \ gi
ye tne
orientations of the
failure planes. They make an angle of 45° + 0/2 with the cr-axis. The lines drawn parallel t o the lines
P
2
Cj an d P
2
C'
{
i n Fig. 11.6(a ) give the shear lines along which the soil slips in the plastic state. The
angle between a pair of conjugate shear lines is (90° - 0) .
Passive Stat e o f Stress
C i s the Moh r circl e in Fig. (11.6e ) for th e passive stat e and P
3
i s the origi n o f planes .
Major principa l stres s = ( j
}
= p = OP^
Minor principal stress = (7
3
= OP
l
= yz.
From triangl e OO^C
2
, o
{
= yzN ^
Since <J
l
- p an d <J
3
= yz, we have
n -yzN :-r7K ( ] ] ]?}
i n * Q) i f t \ L L * \. £ j
where K = coefficient o f earth pressure for th e passive state = tan
2
(45° + 0/2) .
The shea r failur e line s ar e P
3
C
2
an d P
3
C^ an d the y mak e a n angl e o f 45° - 0/ 2 wit h the
horizontal. The shea r failur e line s are drawn parallel t o P
3
C
2
an d P
3
C'
2
i n Fig. 11.6(b) . The angl e
between an y pair of conjugat e shear lines i s (90° + 0) .
1 1 .5 RANKI NE' S EARTH PRESSU R E AG AI NST SM OOTH
V ERTI CAL WAL L WI TH COH ESI ONL ES S B ACKFI L L
B ackfill H orizontal- Activ e Eart h Pressur e
Section AB i n Fig. 11.6(a ) i n a semi-infinite mass i s replaced b y a smooth wal l AB i n Fig. 11.7(a) .
The latera l pressur e actin g agains t smoot h wal l AB i s du e t o th e mas s o f soi l ABC abov e
failure lin e A C whic h make s a n angl e o f 45 ° + 0/ 2 wit h th e horizontal . Th e latera l pressur e
distribution o n wal l AB o f heigh t H increase s i n simpl e proportio n t o depth . Th e pressur e act s
normal t o the wal l AB [Fig . 11.7(b)] .
The latera l active pressure a t A is
(11.13)
B'B
W
45° +
(a) (b )
Figure 1 1 . 7 Ran k in e' s act iv e eart h pres s ur e i n cohes ion les s soi l
Lat eral Eart h Pres s ur e 429
The total pressure o n AB i s therefore
H H
zd Z=K
(11.14)
o o
where, K
A
= tan
2
(45° -
1 + sin^
V i
P
a
act s at a height H/ 3 abov e th e base of the wall .
B ackfill H orizontal- Passiv e Eart h Pressur e
If wal l AB i s pushed int o the mass to such an extent as to impart unifor m compressio n throughout
the mass, soi l wedg e ABC i n Fig. 11.8(a ) wil l be in Rankine's passiv e stat e of plastic equilibrium.
The inner rupture plane AC makes an angle 45° + 0/2 with the vertical AB. The pressure distribution
on wal l AB i s linear as shown in Fig. 11.8(b) .
The passiv e pressur e p a t A is
P
P
=YHK
p
the total pressure agains t the wall is
P
P =
(11.15)
where, K
p
= tan
2
(45° +
1 + sin ^
1 -si n 6
Relationship between K
p
and K
A
The rati o of K
p
an d K
A
ma y be written as
K
p
tan
2
(45
c
K
A
tan
2
(45
c
(11.16)
B B'
Inner rupture plane
' W
(a) (b)
Figure 11. 8 Ran k in e' s pas s iv e eart h pres s ur e i n cohes ionles s soi l
430 Chapt er 1 1
H
P + P
1
a
Tr
w
45° + 0/2
(a) Retaining wal l
\' -*
H- v
b
nK
A
(b) Pressure distribution
H
Figure 1 1 .9 Ran k in e' s act ive pres s ur e under s ubm erg e d con dit io n i n cohes ion les s
s oil
For example, i f 0 = 30°, we have,
K
P
-7T
1
= t an
4
60
0
=9 , o r K
p
=9K
A
KA
This simpl e demonstration indicate s tha t the value of K
p
i s quite large compare d t o K
A
.
Active Eart h Pressure- B ackfil l Soi l Submerged with the Surfac e H orizontal
When th e backfil l is ful l y submerged , two types of pressures ac t on wal l AB. (Fig . 1 1.9) They ar e
1. Th e active eart h pressur e du e t o the submerged weigh t of soi l
2. Th e lateral pressure due to water
At an y dept h z the total uni t pressur e on the wal l is
At dept h z = H, w e have
~p~ = y,HK. + y H
r a ID A ' w
where y
b
i s th e submerge d uni t weight of soi l an d y
w
th e uni t weight o f water . The tota l pressur e
acting on the wal l at a height H/3 abov e the base is
(11.17)
Active Eart h Pressure- B ackfil l Partly Submerged with a U niform Surcharge L oad
The groun d wate r tabl e i s a t a dept h o f H
l
belo w th e surfac e and th e soi l abov e thi s level ha s a n
effective mois t uni t weight of y. The soi l below th e water table is submerged wit h a submerged unit
weight y
b
. I n this case, th e total unit pressure ma y be expressed a s given below.
At dept h H
l
a t the level of the water tabl e
Lat eral Eart h Pres s ur e 431
At dept h H we have
or (11.18)
The pressur e distribution is given in Fig. 1 1.10(b). It i s assumed tha t the value of 0 remains
the same throughou t the depth H.
From Fig . 1 1.10(b), we may sa y that the total pressure P
a
actin g per uni t length of the wall
may b e written as equal t o
(11.19)
The point of application of P
a
abov e the base of the wall can be found by taking moments of
all the forces actin g on the wall about A.
Sloping Surface- Activ e Eart h Pressur e
Figure 1 1.1 1 (a) shows a smooth vertica l wall with a sloping backfill o f cohesionless soil . As in the
case of a horizontal backfill, the active stat e of plastic equilibrium can be developed i n the backfil l
by rotating the wal l about A away from th e backfill. Let AC be the rupture line and the soi l within
the wedge ABC b e i n an active stat e of plastic equilibrium.
Consider a rhombic elemen t E withi n the plasti c zone ABC whic h is shown to a larger scal e
outside. The base of the element i s parallel to the backfill surface which is inclined at an angle / 3 to
the horizontal. The horizonta l widt h of the element i s taken as unity.
Let o~
v
= the vertica l stres s actin g on a n elemental lengt h ab =
( 7
l
= the lateral pressur e actin g on vertical surface be of the elemen t
The vertica l stres s o~
v
can be resolved int o components <3
n
the normal stres s an d t the shea r
stress o n surface ab of element E. We may now write
H
g/unit area
I I I 1 I I I
P
a
(total ) =
(a) Retaining wal l (b) Pressure distribution
Figure 11.1 0 Ran k in e' s act ive pres s ur e in cohes ion les s back fil l unde r part l y
s ubm erg ed con dit io n wit h s urcharg e load
432
Chapt er 1 1
H
(a) Retaining wall (b) Pressur e distribution
O CT 3 0 , O n O]
(c) Mohr diagram
Figure 1 1 .1 1 Ran k in e' s act ive pres s ur e for a s lopin g cohes ion l es s back fil l
n
- <J
v
co s fi = yz cos /?cos fl=y z cos
2
j3
T = a sin/ ? =
(11.20)
(11.21)
A Mohr diagra m ca n be drawn as shown in Fig. 11. 1 l(c). Here , lengt h OA = yzcos/3 make s
an angl e ( 3 with the (T-axis. OD = o
n
- yzcos
2
/3 and AD = T= yzcosf} sin/3 . OM is the Mohr envelope
making an angl e 0 with the <7-axis . Now Mohr circle C
}
can be drawn passing throug h point A and
at the same time tangential to envelope OM. This circle cut s line OA at point B an d the CT-axis at E
andF.
Now OB = the lateral pressure o
l
=p
a
i n the active state .
The principal stresse s ar e
OF = CT j an d O E = a
3
The followin g relationships can be expressed wit h reference t o the Mohr diagram .
BC = CA =— -l -sm
2
j3
Lat eral Eart h Pres s ure 433
= OC-BC =
2
cr, +CT , cr, + cr,
i
2 2
Now we have (after simplification)
cos 0 -T] cos
2
ft -cos
cr
v
yzcosfi cos 0+J cos
2
fi -cos
2
0
or
cos B- A/cos
2
/?- cos
2
( b
-
v
c os / ? +c os
2
/ ? - c os
where, K. = cos fi x
(11.22)
(11.23)
(11.24)
is calle d a s th e coefficient o f earth pressure fo r th e activ e stat e o r th e activ e eart h pressur e
coefficient.
The pressure distribution on the wall is shown in Fig. 1 1 . 1l(b). The active pressure at depth H
is
which act s parallel t o the surface . The total pressure P
Q
pe r uni t length of the wal l is
(11.25)
which act s a t a heigh t H/3 fro m th e bas e o f th e wal l an d paralle l t o th e slopin g surfac e o f th e
backfill.
(a) Retaining wal l (b ) Pressure distribution
Figure 1 1 . 1 2 Ranki ne' s passiv e pressur e i n sl opi ng cohesionless b a c k f i l l
434
Chapt er 1 1
Sloping Surface- Passiv e Eart h Pressur e (Fig . 1 1 .1 2 )
An equation fo r P fo r a sloping backfil l surface can be developed i n the same wa y as for an active
case. Th e equatio n for P ma y b e expressed a s
(11.26)
n
cos fi +J cos
2
fl- cos
2
0
where, K
p
=cos] 3x /
cos /3 - ^cos
2
j3- cos
2
0
P act s a t a height H/3 abov e poin t A an d paralle l t o the slopin g surface .
(11.27)
Example 1 1 . 2
A cantilever retainin g wall of 7 meter height (Fig. Ex. 11.2 ) retains sand. The properties o f the sand
are: e - 0.5 , 0 = 30° and G^ = 2.7. Usin g Rankine' s theory determine th e active earth pressur e a t the
base when the backfill is (i) dry, (ii) saturated an d (iii) submerged, an d also the resultant activ e forc e
in eac h case . I n additio n determine the total wate r pressure unde r the submerge d condition.
Solution
e = 0.5 and G = 2.7, y, = -^= —— x 9.81 = 17.66 kN/m
3
d
l + e1 + 0.5
Saturated uni t weight
Backfill submerge d
Backfill saturate d
Water pressur e
p
a
= 48.81 kN/m"
= 68.67 kN/m
2
=
Pw
Figure Ex . 1 1 . 2
Lat eral Eart h Pres s ur e 43 5
=
sat
l + e1 + 0.5
Submerged uni t weight
r
b
= r
sal
-r
w
= 20.92-9.81 = 11.1 kN/m
3
l -si n^ 1 - sin 30° 1
For* =30, * A
Active earth pressur e a t the base i s
(i) fo r dr y backfill
Pa =
P = -K
A
r,H
2
= -x 41.2x7 = 144.2 k N/ mo f wa ll
a r\ A ' a rj
(ii) fo r saturated backfil l
P
a
=
K
A Ysat H = -x20.92 x 7 = 48.8 1 kN/m
2
p = -x 48.8 1x7 = 170.85 k N/ m of wall
a
2
( in) fo r submerge d backfil l
Submerged soi l pressur e
P
a
= K
/J b
H
= - x 1 1.1 x 7 = 25.9 kN/m
2
P = - x 25.9 x 7 =90.65 kN/ m of wall
a
2
Water pressur e
p
w
= y
w
H = 9.8 1 x 7 = 68.67 kN/m
2
P
w
=-
Yw
H
2
= -x 9.81 x7
2
=240.3 5 kN/ mof wal l
Example 1 1 . 3
For the earth retaining structure shown in Fig. Ex. 11.3 , construct the earth pressure diagram for the
active state and determine th e total thrust per uni t lengt h of the wall.
Solution
1-sin 30° 1
For <z ) =30°, K
A
: - = -
G Y 265 i
Dry unit weight Y
H
= —^^ = —
: x
62.4 = 100.22 lb/ fr
y d
l + e1 + 0.65
4 3 6 Chapter 1 1
q = 292 lb/ft
2
J I U J J H J
E--
//A\\
|
= i
32. 8ft > >
1
1
\J
1
9. 8f t
Sand G
s
= 2.65
e = 0.65
0 = 30°
(a) Give n syste m
Pl Pi P3
(b) Pressure diagram
Figure Ex. 11. 3
( G
s
-l)y
w
2.65- 1
7b
=-^—-
= T
-^
X
62.4 = 62.4 ,b/f, 3
Assuming the soi l abov e th e water tabl e i s dry, [Refer t o Fig. Ex. 11.3(b)] .
P! = K
A
y
d
H
l
=- x 100.22x9.8 = 327.39 lb/ft
2
p
2
= K
A
y
b
H
2
= -x 62.4 x 23 = 478.4 l b/ft
2
p
3
= K
A
xq = -x292= 97.33 lb/ft
2
P4 = (
K
A^
w
r
w
H
2 = 1x62.4x23 = 1435. 2 lb/ft
2
Total thrus t = summation o f the area s o f the differen t part s o f the pressur e diagra m
1 1 1
= ^PiH
l
+p
l
H
2
+-p
2
H
2
+p
3
( H
l
+H
2
) + -p
4
H
2
= -x 327.39 x 9.8 + 327.39 x 23 + -x 478.4 x 23 + 97.33(32.8) + -x 1435.2x23
2 2 2
= 34,333 lb/f t = 34.3 kips/ft o f wall
Example 1 1 . 4
A retainin g wal l wit h a vertical bac k o f heigh t 7.32 m support s a cohesionless soi l o f uni t weigh t
17.3 kN/m
3
an d a n angl e o f shearin g resistanc e 0 = 30°. The surfac e o f th e soi l i s horizontal .
Determine th e magnitud e an d directio n o f th e activ e thrus t pe r mete r o f wal l usin g Rankin e
theory.
Lat eral Eart h Pressure 437
Solution
For the condition given here, Rankine' s theor y disregards th e friction betwee n th e soil and the back
of the wall .
The coefficien t of active eart h pressur e K
A
i s
1-sind l-sin30 ° 1
Tf T_
A
1 + sin^ 1 +sin 30° 3
The lateral active thrust P
a
i s
P
a
= -K
A
yH
2
= -x-x 17.3(7.32)
2
= 154.5 kN/ m
Example 1 1 . 5
A rigid retaining wall 5 m high support s a backfill of cohesionless soi l with 0= 30°. The water table
is below th e bas e o f th e wall . Th e backfil l i s dr y an d ha s a uni t weigh t o f 1 8 kN/m
3
. Determin e
Rankine's passiv e eart h pressur e pe r meter lengt h of the wal l (Fig. Ex. 11.5) .
Solution
FromEq. (11.15a )
Kp =
1 + sin^ 1 + sin 30° 1 + 0.5
i n^ l-sin30 ° 1-0. 5
At the base level , the passive eart h pressur e i s
p
p
=K
p
yH = 3 x 1 8x5 = 270 kN/m
2
FromEq. (11.15 )
P
p
=- K
P
y H = -x 3 x1 8 x 5 = 675 kN/m length of wall
The pressure distributio n i s given i n Fig. Ex. 1 1.5.
Pressure distributio n
Figure Ex. 1 1 . 5
438 Chapt e r 1 1
Example 1 1 . 6
A counterfor t wal l o f 1 0 m heigh t retain s a non-cohesiv e backfill . Th e voi d rati o an d angl e o f
internal friction o f the backfill respectivel y are 0.70 and 30° in the loose state an d they are 0.40 and
40° i n the dense state . Calculate and compare activ e and passive eart h pressure s fo r bot h the cases.
Take th e specifi c gravit y o f solid s as 2.7.
Solution
(i) In the loose state, e - 0.7 0 which give s
/""*- .r\ i—j
=
_
I
^
L=
_ _
x 9
gj
= 15 6kN/m
3
d
l + e1 + 0.7
c j. ™ ° v l - si n 0 1-si n 30° 1 1
For 0 = 3 0, K , - = = — , and^
0
= = 3
' A - i * ' 1 * O /"\o O i TS
1 +sin 30 3 K,
Max. p
a
= K
A
y
d
H = - x 15.6 x 10 = 52 kN/m
2
Max. p = K
p
y
d
H = 3 x 15.6 x 10 = 468 kN/m
2
(ii) I n the dens e state , e = 0.40, which gives,
Y
= -22—
x
9.81 = 18.92 kN/m
3
d
1 + 0.4
1-sin 40° 1
For 0 = 40°, K=- — — = 0.217, K
p
=- — = 4.6
y A
1 +s in 40°
p
K.
f\
Max. p
f l
=K
A
y
d
H = 0.217x18.92x10 = 41.1 kN/m
2
and Max . p = 4.6 x 18.92 x 10 = 870. 3 kN/m
2
Comment: Th e compariso n o f th e result s indicate s tha t densificatio n o f soi l decrease s th e
active earth pressure an d increases the passive eart h pressure. Thi s is advantageous i n the sense that
active eart h pressur e i s a disturbing force an d passive eart h pressur e i s a resisting force .
Example 1 1 . 7
A wall of 8 m height retain s san d havin g a density o f 1.936 Mg/m
3
and an angle of internal frictio n
of 34°. I f the surface of the backfill slopes upwards at 15° to the horizontal, fin d the active thrust per
uni t lengt h of the wall . Use Rankine' s conditions.
Solution
There ca n be two solutions : analytical and graphical. The analytica l solution can be obtained fro m
Eqs. (11.25 ) and (11.24 ) vi z.,
Lat eral Eart h Pres s ure 439
Figure Ex . 1 1 .7 a
where K. = cos ft x
cos/?- ycos
2
ft - cos
2
COS/?+ yCOS
2
ft - COS
2
( f)
where f t = 15°, cos/? = 0.9659 an d cos
2
f t = 0.933
and ^ = 34° give s cos
2
( /) = 0.688
0.966 -V O.933- 0.688
= 0.3 1 1 Hence K
A
= 0.966 x .
A
0.96 6 + V O.933 -0.688
y = 1.936x9.81 = 19.0 kN/m
3
Hence P
a
= -x0.311x!9(8)
2
= 189 kN/ m wall
G raphical Solutio n
V ertical stres s a t a depth z = 8 m i s
7/ f c o s / ? = 19x 8x c o s l 5 ° = 147 kN/m
2
Now draw the Mohr envelope a t an angl e of 34° and th e ground line at an angl e of 15 ° with
the horizontal axi s as shown in Fig. Ex. 1 1.7b.
Using a suitable scale plot OP
l
= 147 kN/m
2
.
(i) th e center of circle C lies on the horizontal axis,
(ii) th e circle passes throug h point P
r
and
(iii) th e circl e i s tangent to the Mohr envelope
440 Chapt er 1 1
Ground line
16 18 x 1 0
Pressure kN/m
Figure Ex . 1 1 .7 b
The point P
2
at which the circle cuts the ground line represents the lateral earth pressure. The length
OP
2
measure s 47. 5 kN/m
2
.
Hence the active thrust per unit length, P
a
= -x 47.5 x 8 = 190 kN/m
1 1 .6 RANKI NE' S ACTI V E EARTH PRESSU RE WI TH COH ESI VE
B ACKFI L L
In Fig. 1 1.1 3(a) is shown a prismatic element i n a semi-infinite mas s wit h a horizontal surface. The
vertical pressur e o n the base AD o f the element at depth z is
The horizontal pressure on the element when the mass is in a state of plastic equilibrium may
be determined b y makin g use of Mohr's stres s diagra m [Fig . 1 1.13(b)].
Mohr envelope s O' A and O' E fo r cohesive soil s ar e expressed b y Coulomb' s equation
s - c +tan 0 (11.28)
Point P j o n th e cr-axi s represent s th e stat e o f stres s o n th e bas e o f th e prismati c element .
When the mass is in the active state cr , is the major principal stress Cf j . Th e horizontal stress o
h
is the
minor principa l stres s <7
3
. Th e Moh r circl e o f stres s C
a
passin g throug h P
{
an d tangentia l t o th e
Mohr envelope s O'A an d O'B represent s th e stres s condition s i n th e activ e state . Th e relatio n
between th e two principal stresses may be expressed b y the expression
(11.29)
(11.30)
<7, = <7, A
1 J V v y
Substituting O" , = 72, <7
3
=p
a
an d transposin g we hav e
rz 2 c
Lat eral Eart h Pres s ur e 441
45° + 0/2
Stretching
45° + 0/2
D B
C A
'"\, - ; \,-" \,-" \,-• ; t -
^A'-^ i " j Z:J A'
Tensile
zone
Failure shear lines
(a) Semi-infinit e mass
Shear line s
(b) Mohr diagra m
Figure 1 1 .1 3 Act iv e eart h pres s ur e of cohes iv e s oi l wit h horizon t al back fil l o n a
vert ical wal l
The activ e pressur e p
a
= 0 when
yz 2c
rt
(1 1 . 3 1 )
that is, p
a
i s zero a t depth z, such that
At dept h z = 0, the pressur e p
a
i s
2c
Pa - JTf ^
(11.32)
(11.33)
442 Chapt e r 1 1
Equations (1 1 .32) an d ( 1 1.33) indicat e tha t the activ e pressur e p
a
i s tensil e betwee n dept h 0
and Z
Q
. The Eqs. ( 1 1.32) and (1 1.33) can also be obtained fro m Mohr circles C
Q
and C
t
respectively .
Shear L ine s Patter n
The shea r line s are shown i n Fig. 1 1 . 13(a). Up t o depth Z
Q
they are shown dotted t o indicate that this
zone i s i n tension.
Total Active Eart h Pressur e o n a V ertical Sectio n
If AB i s the vertical sectio n [ 1 1.14(a)], the active pressure distributio n against thi s section o f height
H i s shown i n Fig. 1 1.1 4(b) as per Eq . ( 1 1.30). The tota l pressure agains t the sectio n i s
H H H
yz 2c
Pa =PZdz= ~dz- -r==dz
o
0
'
0
V
A
0
H
The shade d are a i n Fig. 1 1.14(b) gives the total pressur e P
a
. If the wal l has a height
the total eart h pressur e i s equal t o zero. This indicate s that a vertical bank of height smaller tha n H
can stand withou t lateral support . //, is called th e critical depth. However , th e pressure agains t the
wall increase s fro m - 2c/J N ^ a t the top to + 2c/jN ^ a t depth //,, whereas o n the vertical fac e of
an unsupported ban k the normal stres s i s zero at every point. Because o f this difference, the greates t
depth of which a cut can be excavated without lateral suppor t fo r its vertical side s i s slightly smalle r
than H
c
.
For sof t clay , 0 = 0, and N ^= 1
therefore, P
a
=± yH
2
-2cH (11.36 )
4c
and
HC=
~^
(1L37 )
Soil does not resist an y tension and as such it is quite unlikely that the soil would adhere t o the
wall withi n the tension zon e o f depth z
0
producing crack s i n the soil . I t is commonl y assume d tha t
the activ e eart h pressur e i s represented b y the shade d are a i n Fig. 1 1.14(c).
The tota l pressur e o n wal l AB i s equa l t o th e are a o f th e triangl e i n Fig. 11.14(c) whic h i s
equal t o
1 yH 2c
D
1 yH 2c „ 2c
or F =
"
H
"
Lat eral Eart h Pres s ure 443
Surcharge load q/unit are a
B \ \ l \ l \ \ C
2c
8
jH 2c
q
q
\
%v %
N
*
(a) (b ) (c ) (d )
Figure 1 1 .1 4 Act iv e eart h pres s ur e o n vert ical s ect ion s i n cohes ive s oil s
Simplifying, w e have
1 2c
2
2 N
*
For sof t clay , 0 = 0
P
a
= -yH
l
It may b e noted tha t K
A
= \ IN ^
(11.38c)
(11.39)
Effect o f Surcharg e an d Water Tabl e
Effect o f Surcharge
When a surcharge loa d q per uni t are a act s o n the surface , the latera l pressur e o n the wal l due t o
surcharge remain s constan t wit h dept h a s show n i n Fig . 11.14(d ) fo r th e activ e condition . Th e
lateral pressur e du e to a surcharge unde r the active stat e may be written as
The total active pressure du e to a surcharge load is ,
n _ &
(11.40)
Effect o f Water Tabl e
If th e soi l i s partl y submerged , th e submerge d uni t weigh t below th e wate r tabl e wil l have t o b e
taken int o account i n both the active and passive states .
444 Chapt er 1 1
Figure 11.15(a ) shows the case of a wall in the active state wit h cohesive materia l as backfill.
The wate r tabl e i s at a depth of H
l
belo w th e top of the wall . The dept h o f water i s //
2
.
The latera l pressur e o n the wall due t o partial submergence i s due t o soi l and water as shown
in Fig. 11.15(b) . The pressur e du e t o soi l = area o f the figur e ocebo.
The tota l pressur e du e t o soi l
P
a
= oab + acdb+ bde
1 2c
N
A
J N .
2c
N
(11.41)
2Cr—
After substitutin g for z
n
= — N
and simplifyin g we hav e
1
p (
v
jr2 , ,
A — »• » ,\/ - - * - * 1 l t
2c
The tota l pressur e on the wal l due t o water is
p
v
JJ2
~
n
2c
2
(11.42)
(11.43)
The poin t o f applicatio n o f P
a
ca n b e determine d withou t an y difficulty . Th e poin t o f
application P
W
i s at a height of H
2
/3 fro m the bas e o f the wall .
Cohesive soi l
7b
T
H
2
/3
_L
Pressure du e
to water
(a) Retaining wall (b) Pressure distribution
Figure 1 1 .1 5 Effec t o f wat e r t abl e o n l at era l eart h pres s ur e
Lat eral Eart h Pres s ur e 44 5
If the backfill materia l i s cohesionless, th e terms containing cohesion c in Eq. (11.42) reduce
to zero .
Example 1 1 . 8
A retaining wal l has a vertical back and i s 7.32 m high. The soi l i s sandy loam o f unit weight 17.3
kN/m
3
. It has a cohesion o f 1 2 kN/m
2
an d 0 = 20°. Neglectin g wal l friction, determine th e activ e
thrust on the wall . The upper surfac e of the fil l i s horizontal .
Solution
(Refer t o Fig. 11.14 )
When th e materia l exhibit s cohesion , th e pressur e o n th e wal l a t a dept h z i s give n b y
(Eq. 11.30 )
where K J_^iT = — = 0.49, I
K 0
.7
v A
1-sin 20°
1 +sin 20°
When th e dept h i s smal l the expressio n fo r z i s negative becaus e o f the effec t o f cohesion u p t o a
theoretical dept h z
0
. The soi l i s in tension and the soi l draws away from th e wall .
-— I —-—I
y v Y
1 + si n ( f) i
where K
p
= 7- 7 = 2.04, an d J K
P
= 1.43
p
- *
p
2x12
Therefore Z
Q
= "TTT"
x
1-43 = 1-98 m
The latera l pressure at the surface ( z = 0) i s
D = -2cJ xT = -2 x 12 x 0.7 = -16.8 kN/m
2
* u V • * »
The negative sign indicates tension .
The latera l pressur e a t the base o f the wal l ( z = 7.32 m) is
p
a
= 17.3 x 7.32 x 0.49 -16.8 = 45.25 kN/m
2
Theoretically th e are a o f th e uppe r triangl e i n Fig. 11.14(b ) t o th e lef t o f th e pressur e axi s
represents a tensile forc e whic h should be subtracted fro m th e compressive forc e on the lower par t
of the wal l below th e dept h Z
Q
. Since tension cannot be applied physicall y between th e soi l and the
wall, this tensile force i s neglected. I t is therefore commonly assumed that the active earth pressur e
is represented b y the shaded are a i n Fig. 1 1 . 14(c). The total pressure on the wall is equal t o the area
of the triangle i n Fig. 1 1.14(c).
= -(17.3 x 7.32 x 0.49 - 2 x 12 x0.7) (7.32- 1.98) = 120.8 kN/ m
446 Chapt er 1 1
Example 1 1 . 9
Find the resultant thrust on the wal l i n Ex. 11. 8 if the drains are blocked and water builds up behind
the wall unti l the water tabl e reaches a height of 2.75 m above th e bottom o f the wall .
Solution
For detail s refer t o Fig. 11.15 .
Per thi s figure ,
H
l
= 7.32 - 2.7 5 = 4.57 m, H
2
= 2.75 m, H, - Z
0
= 4.57 -1.98 = 2.59 m
The bas e pressur e i s detailed i n Fig. 11.15(b )
(1) Y
S
at
H
\
K
A -
2c
J K~A = !7.3x4.57x0.49-2x12x0.7 = 21.94 kN/m
2
(2) 7
b
H
2
K
A
- (17.3 -9.8l)x 2.75x0.49 = 10.1 kN/m
2
(3) y
w
H
2
= 9.81 x 2.75 = 27 kN/m
2
The tota l pressur e = P
a
= pressure due t o soil + water
From Eqs . (11.41) , (11.43) , and Fig. 11.15(b )
P
a
= oab + acdb+ bde+ bef
1 1 1
= - x 2.59 x 21.94 + 2.75 x 21.94 + -x 2.75 x 10.1 + - x 2.75 x 27
= 28.41 + 60.34 + 13.8 9 +37.13 = 139. 7 kN/ m or say 14 0 kN/m
The point of application of P
a
may be found by taking moments of each area and P
a
about the
base. Let h be the height of P
a
abov e the base. Now
1 97 5 97 5 3713x97 5
140x^ = 28.41 -X2.5 9 + 2.75 + 60.34 x —+ 13.89 x —+
3 2 3 3
16.8 kN/m
2
y
sat
= 17. 3 kN/m
0 = 20°
c= 1 2 kN/m
2
P,, = 14 0 kN/m
Figure Ex . 1 1 . 9
Lat eral Eart h Pres s ur e 447
or
= 102.65 + 83.0 +12.7 + 34.0 = 232. 4
232.4
140
= 1.66m
Example 1 1 .1 0
A rigid retaining wall 19.69 f t high has a saturated backfill of sof t cla y soil. The propertie s of th e
clay soil are y
sat
= 111.76 lb/ft
3
, and unit cohesion c
u
= 376 lb/ft
2
. Determin e (a) the expected dept h
of the tensile crack in the soil (b) the active earth pressure before the occurrence of the tensile crack,
and (c ) the active pressure after th e occurrence of the tensile crack. Neglect the effect o f water that
may collect i n the crack.
Solution
At z = 0, p
a
= -2c = -2 x 376 = -752 lb/ft
2
sinc e 0 = 0
Atz = H, p
a
= yH-2c=l\ \ .16x 19.6 9 - 2 x 376 = 1449 lb/ft
2
(a) From Eq. (11.32), the depth of the tensile crack z
0
is (for 0= 0)
_ 2c _ 2x37 6
Z
° ~ y~ 111.7 6
= 6.73 f t
(b) The active earth pressur e before the crack occurs.
Use Eq. (11.36) for computing P
a
1
19.69 f t
y=111. 76 lb/ft
3
c
u
= 376 lb/ft
2
752 lb/ft
2
6.73 f t
1449 lb/ft
2
(a) (b)
Figure Ex . 1 1 .1 0
448 Chapt e r 1 1
since K
A
= 1 for 0 = 0. Substituting , we have
P
a
= -x 1 1 1.76x(19.69)
2
- 2 x 376x19.69 = 21,664 -14,807 = 6857 lb/ ft
(c) P
a
afte r the occurrence o f a tensile crack.
UseEq. (11.38a) ,
Substituting
p
a
= 1(1 11.76 x 19.69- 2 x 376) (19.69- 6.73) = 9387 Ib/f t
Example 1 1 .1 1
A rigi d retaining wal l of 6 m height (Fig. Ex. 11.11 ) has two layer s of backfill . The to p laye r t o a
depth of 1. 5 m is sandy clay having 0= 20°, c= 12.15 kN/m
2
and y- 16. 4 kN/m
3
. The bottom laye r
is sand havin g 0 = 30°, c = 0, and y- 17.25 kN/m
3
.
Determine th e total active earth pressure acting on the wall and draw the pressure distribution
diagram.
Solution
For the top layer ,
70 1
K
A
= tan
2
45 ° - — = 0.49, K
p
= —5— = 2.04
A
2
p
0.4 9
The dept h o f th e tensil e zone, Z
Q
i s
2c r— 2X12.15V I0 4
16
.
4
=112m
Since th e dept h o f th e sand y cla y laye r i s 1. 5 m, whic h i s les s tha n Z
Q
, th e tensil e crac k
develops onl y t o a dept h of 1. 5 m.
K
A
fo r th e sand y layer is
At a depth z= 1.5 , the vertica l pressure G
V
is
cr
v
= yz= 16.4 x 1.5 = 24.6 kN/m
2
The activ e pressure i s
p = K
A
vz = -x24.6 = 8.2 kN/m
2
a A
3
At a depth o f 6 m, the effective vertical pressure i s
Lat eral Eart h Pres s ure 449
GL
\V/\\V/A\V/\\V/\
1.5m
4. 5m
Figure Ex . 1 1 .1 1
<j
v
= 1.5 x 16.4 + 4.5 x 17.25 = 24.6 + 77.63 = 102.23 kN/m
2
The active pressure p
a
i s
p
a
= K
A
a
v
= -x 102.23 = 34.1 kN/m
2
The pressur e distributio n diagram i s given i n Fig. Ex. 11.11 .
8.2 kN/m
2
• 34.1 kN/m
2
1 1 .7 RANKI NE' S PASSI V E EART H PRESSU RE WI TH COH ESI V E
B ACKFI L L
If the wall AB i n Fig. 1 1 . 16(a) is pushed towards the backfill, the horizontal pressure p
h
o n the wall
increases an d become s greate r tha n the vertica l pressur e cr
y
. When the wal l i s pushed sufficientl y
inside, th e backfil l attains Rankine' s stat e o f plasti c equilibrium. The pressure distributio n on the
wall ma y be expressed b y the equation
In th e passive state , th e horizonta l stres s G
h
is the majo r principa l stres s G
I
an d th e vertica l
stress o
v
is the minor principal stress a
3
. Since a
3
= yz, the passive pressur e a t any depth z may be
written as
(11.44a)
At depth z = O, p= 2c
At dept h z = H, p=rHN :+ 2cjN , =
7
HK
p
+ 2cJ K
f (11.44b)
450 Chapt er 1 1
q/unii are a
UM J i l I I I
H/2
(a) Wall (b ) Pressur e distributio n
Figure 1 1 .1 6 Pas s iv e earth pres s ur e on vert ical s ect ion s i n cohes ive s oil s
The distributio n of pressur e wit h respec t t o dept h i s show n i n Fig. 11.16(b). Th e pressur e
increases hydrostatically. The total pressure on the wall may be written as a sum of two pressures P'
o
This act s a t a height H/3 fro m the base .
H
p
;=
0
This act s at a height of H/ 2 fro m th e base .
The passiv e pressure du e t o a surcharge load o f q per uni t area i s
Ppq =
The tota l passive pressure due to a surcharge load is
which act s at mid-height of the wall.
It may b e noted her e that N . = K
p
.
(11.45a)
(11.45b)
(11.45c)
(11.46)
Example 1 1 .1 2
A smoot h rigi d retainin g wall 19.6 9 ft hig h carrie s a unifor m surcharg e loa d o f 25 1 lb/ft
2
. Th e
backfill i s clayey sand with the following properties:
Y = 102 lb/ft
3
, 0 = 25°, and c = 136 lb/ft
2
.
Determine th e passive eart h pressure and draw the pressure diagram.
Lat eral Eart h Pres s ur e
451
251 lb/ft
2
1047.5 lb/ft
2
19.69 f t
,
\V /A\v\V /A\V /\\V /\
0 = 25°
c = 13 6 lb/ft
2
y = 10 2 lb/ft
3
Clayey san d
7.54 f t
Figure Ex . 1 1 .1 2
Solution
For 0 = 25°, the value of K
p
i s
1 +sin^ 1 + 0.423 1.42 3
TS
~
1-0.423" 0.577
From Eq. ( 1 1.44a), p a t any dept h z is
p
p
= yzK
p
At depth z = 0, a
v
= 25 l i b/ f t
2
p
p
= 25 1 x 2.47 + 2 x 136V l47 = 1047.5 Ib / ft
2
At z = 19.69 ft , a-
v
= 25 1 + 19.69 x 102 = 2259 Ib/ f t
2
p
p
= 2259 x 2.47 + 2 x 136^247 = 6007 Ib / f t
2
The pressur e distribution is shown in Fig. Ex. 11.12 .
The total passive pressure P
p
actin g on the wal l is
P
p
= 1047.5 x 19.69 + -x 19.69(6007 -1047.5) = 69,451 Ib/ ft of wall * 69.5 kips/f t of wall.
Location o f resultan t
Taking moments about the base
P
x
h = -x (19.69)
2
x 1047.5 + - x (19.69)
2
x 4959.5
p
2 6
= 523,51 8 Ib.ft .
452 Chapt er 1 1
or h =
523,518 _ 523,51 8
~~P
n
~ 69.45 1
= 7.54f t
1 1 .8 COU L OM B ' S EARTH PRESSU RE TH EORY FOR SAND
FOR ACTI V E STATE
Coulomb mad e th e following assumptions in the development o f hi s theory :
1. Th e soi l i s isotropi c an d homogeneou s
2. Th e ruptur e surface i s a plane surfac e
3. Th e failur e wedge i s a rigid bod y
4. Th e pressure surfac e i s a plane surfac e
5. Ther e i s wal l friction o n the pressure surfac e
6. Failur e i s two-dimensional an d
7. The soi l is cohesionles s
Consider Fig . 11.17 .
1. A B i s the pressur e fac e
2. Th e backfil l surfac e BE i s a plane incline d at an angl e /3 with the horizonta l
3. a i s the angle mad e by the pressure fac e AB wit h the horizontal
4. H i s the height of the wal l
5. A C i s the assume d ruptur e plane surface, an d
6. 6 i s the angl e mad e b y the surface AC wit h the horizonta l
If A C i n Fig . 17(a ) i s th e probabl e ruptur e plane , th e weigh t o f th e wedg e W
length o f the wal l ma y b e written as
W = yA, where A = area o f wedge ABC
per uni t
(180°-d7-(y)
a -d = a>
W
(a) Retaining wal l (b) Polygon of force s
Figure 1 1 .1 7 Con dit ion s fo r fail ur e unde r act iv e con dit ion s
Lat eral Eart h Pres s ur e 45 3
Area of wedge ABC = A= 1/2 AC x BD
where BD is drawn perpendicular t o AC.
From th e law of sines, we have
H
AC = AB~—~~ , B D = A5sin(a + 9\ A B =
sm(# — p)
Making the substitution and simplifying we have,
yH
W=vA = .. ~—sin( a + >)-7—-— — (1147 ) /
2sm
2
a sm(#-/? ) ^
ii
^')
The variou s forces tha t are acting o n the wedge ar e shown i n Fig. 11.17(a) . As the pressur e
face AB move s awa y from th e backfill, there will be sliding of the soi l mas s along the wal l from B
towards A. The sliding of the soil mass i s resisted b y the friction of the surface. The direction of the
shear stress is in the direction from A towards B. lfP
n
i s the total normal reaction o f the soil pressure
acting on fac e AB, th e resultant of P
n
an d the shearin g stres s i s the activ e pressur e P
a
makin g an
angle 8 wit h the normal . Sinc e th e shearin g stres s act s upwards , th e resulting P
a
dip s belo w th e
normal. The angle 5 for this condition i s considered positive.
As the wedge ABC ruptures along plane AC, it slides alon g this plane. This is resisted by the
frictional forc e actin g betwee n th e soi l a t res t belo w AC , an d th e slidin g wedge . Th e resistin g
shearing stres s i s acting i n the directio n fro m A towards C . If W
n
is the norma l componen t o f th e
weight o f wedg e W o n plan e AC, th e resultan t o f th e norma l W
n
an d th e shearin g stres s i s th e
reaction R. This makes an angle 0 with the normal since the rupture takes place within the soil itself.
Statical equilibrium requires tha t the three forces P
a
, W, and R meet at a point. Since AC is not the
actual rupture plane, the three forces do not meet at a point. But if the actual surface of failure AC'C
is considered, al l three forces mee t at a point. However, th e error due to the nonconcurrence of the
forces is very insignificant and as such may be neglected .
The polygo n of forces i s shown in Fig. 11.17(b) . From th e polygon of forces, w e may writ e
°
r P
*
=
°- - <
1L48
>
In Eq . (11.48) , the onl y variabl e i s 6 and al l the other terms fo r a given cas e ar e constants .
Substituting for W, we have
yH
2
sin( 0 . ,
P = -*—; - - - -— -sm( a +
a
2si n
2
a sin(180 ° -a-
The maximu m value for P
a
i s obtained b y differentiatin g Eq. (11.49) wit h respect t o 6 and
equating the derivative to zero, i.e.
The maximum value of P
a
s o obtained ma y be written as
(11.50)
454 Chapt er 1 1
Table 11 . 2a Act i v e ear t h pressur e coef f i ci ent s K
A
fo r

8 =
8 =
8 =
8 =
15
0 0.5 9
+0/2 0.5 5
+/2/30 0.5 4
+0 0.5 3
20 2 5
0.49 0.4 1
0.45 0.3 8
0.44 0.3 7
0.44 0.3 7
30
0.33
0.32
0.31
0.31
Table 1 1.2b Act iv e eart h pressur e coef f i ci ent s K
A
fo r 8 =
+ 30° an d a f r o m 70 ° t o 110 °
0=
<t> =
0 =
0 =
-30° - 12 °
20° a =70 °
80°
90°
100
110
30° 70 °
80°
90°
100
110
40° 7 0
80
0.54
0.49
0.44
0.37
0.30
0.32 0.4 0
0.30 0.3 5
0.26 . 0.3 0
0.22 0.2 5
0.17 0.1 9
0.25 0.3 1
0.22 0.2 6
90 0.1 8 0.2 0
100 0.1 3 0.1 5
110 0.1 0 0.1 0

0.61
0.54
0.49
0.41
0.33
0.47
0.40
0.33
0.27
0.20
0.36
0.28
0.22
0.16
0.11
(3 = 0 an d
35
0.27
0.26
0.26
0.26
0, 13 var i e s
+ 12 °
0.76
0.67
0.60
0.49
0.38
0.55
0.47
0.38
0.31
0.23
0.40
0.32
0.24
0.17
0.12
a = 90 °
40
0.22
0.22
0.22
0.22
f r om -30 ° t o
+ 30 °
-
-
-
-
-
1.10
0.91
0.75
0.60
0.47
0.55
0.42
0.32
0.24
0.15
where K
A
i s the active earth pressur e coefficient .
sin
2
asin( a-S)
— —
J
t
sin(a - 8) sin(a + /?)
2
The tota l norma l component P
n
of the earth pressure on the back of the wall
1
2
p
n
= P
a
cos --yH 1 f
,COS*
(11.51)
is
(11.52)
If the wal l is vertical and smooth, and i f the backfill is horizontal, we have
J 3=S = 0and a = 90°
Substituting these value s in Eq. (11.51), we have
1-sin^ _ f <f\ 1
K. = — 7 = tan
2
45°-- J =
A
I 2 ) N ,
(11.53)
Lat eral Eart h Pres s ure 455
where = tan
2
1 45 ° + —
2
(11.54)
The coefficient K
A
i n Eq. (11.53) is the same as Rankine's. The effect o f wall friction is frequently
neglected wher e active pressures ar e concerned. Tabl e 11. 2 makes thi s clear. I t is clear from thi s table
that K
A
decrease s with an increase o f 8 and the maximum decrease is not more than 10 percent.
1 1 .9 C O UL O MB' S E A R TH P R E S S URE T H E O RY FOR S A ND FOR
P A S S I VE S T A TE
In Fig. 11.18 , the notations used ar e the same a s in Fig. 11.17 . As the wal l moves int o the backfill,
the soi l trie s t o mov e u p o n th e pressur e surfac e AB whic h i s resisted b y frictio n of th e surface .
Shearing stres s o n thi s surfac e therefor e act s downward . Th e passiv e eart h pressur e P i s th e
resultant of the norma l pressur e P an d the shearin g stress . Th e shearin g forc e i s rotated upwar d
with an angl e 8 which is again th e angle of wall friction. In this case S is positive.
As the rupture takes plac e alon g assume d plane surfac e AC, the soi l trie s t o move up the plane
which is resisted by the frictional force acting on that line. The shearing stress therefore, acts downward.
The reaction R makes an angle 0 with the normal and is rotated upwards as shown in the figure.
The polygo n o f force s i s shown i n (b ) of the Fig . 11.18 . Proceeding i n the same wa y a s for
active earth pressure , w e may writ e the following equations:
(11.55)
(11.56)
Differentiating Eq . (11.56 ) wit h respec t t o 0 an d settin g th e derivativ e t o zero , give s th e
minimum value of P a s
2
2 sin
2
a
.
sm(#-/?)
6 + a = a)
(a) Forces on the sliding wedge (b ) Polygon of forces
Figure 1 1 .1 8 Con dit ion s for failur e unde r pas s iv e s t at e
456 Chapt e r 1 1
(11.57)
where K i s called th e passive earth pressure coefficient.
K
p
=
sin
2
asin( a
(11.58)
Eq. (11.58) i s vali d for bot h positive and negative values of ft and 8.
The tota l normal component o f the passive eart h pressur e P o n the back o f the wal l i s
(11.59)
<•- - / ,
For a smoot h vertica l wall with a horizontal backfill, we have
N
t (11.60 )
Eq. (11.60) is Rankine's passive earth pressure coefficient. We can se e from Eqs. (11.53 ) and
(11.60) that
1
K
p
=
"l< y ^j - . ^i y
Coulomb slidin g wedg e theor y o f plan e surface s o f failur e i s vali d wit h respec t t o passiv e
pressure, i.e. , t o the resistance of non-cohesive soil s only. If wall friction is zero fo r a vertical wal l
and horizonta l backfill , the valu e of K
p
ma y b e calculate d usin g Eq . (11.59) . I f wal l frictio n i s
considered i n conjunctio n wit h plan e surface s o f failure , muc h to o high , .an d therefor e unsaf e
values o f eart h resistanc e wi l l b e obtained , especiall y i n th e cas e o f hig h frictio n angle s 0 . Fo r
example for 0= 8 = 40°, an d for plane surfaces of failure, K
p
= 92.3, wherea s for curved surfaces of
failure K
p
= 17.5 . However , i f S i s smalle r tha n 0/2 , th e differenc e betwee n th e rea l surfac e o f
sliding and Coulomb' s plan e surfac e is ver y smal l and we can comput e th e corresponding passiv e
earth pressur e coefficien t by means of Eq. (11.57) . If S is greater tha n 0/2, th e value s of K
p
shoul d
be obtaine d b y analyzin g curved surfaces of failure .
1 1 .1 0 ACTI V E PRESSU R E BY CU L M ANN' S M ETH OD FOR
COH ESI ONL ESS SOI L S
Without Surcharg e Lin e L oa d
Culmann' s (1875 ) metho d i s the same a s the trial wedge method . I n Culmann' s method , th e forc e
polygons ar e constructed directl y on the 0-line AE takin g AE a s the load line . The procedur e i s as
follows:
In Fig . 11.19(a ) AB i s the retainin g wal l drawn t o a suitabl e scale . The variou s steps i n th e
construction of the pressur e locus are :
1. Dra w 0 -lin e AE at an angl e 0 to the horizontal .
2. La y off on AE distances , AV, A1, A2, A3, etc. to a suitable scale to represent th e weight s of
wedges ABV, A51, AS2, AS3 , etc . respectively .
Lat eral Eart h Pres s ur e 457
Rupt
V ertical
(a) (b )
Figure 1 1 .1 9 Act iv e pres s ur e by Culm an n ' s metho d for cohes ion les s s oil s
3. Dra w lines parallel t o AD fro m point s V, 1, 2, 3 to intersect assume d ruptur e lines AV, Al,
A2, A3 at points V", I',2', 3', etc. respectively.
4. Joi n point s V, 1' , 2' 3' etc . b y a smoot h curv e which i s the pressure locus .
5. Selec t poin t C' o n th e pressur e locu s suc h tha t th e tangen t t o th e curv e a t thi s poin t i s
parallel t o the 0-line AE.
6. Dra w C'C paralle l t o th e pressur e lin e AD. Th e magnitud e of C'C i n it s natura l unit s
gives the active pressur e P
a
.
7. Joi n AC" and produce t o meet th e surface of the backfill at C. AC i s the rupture line.
For the plane backfill surface, the point of application o f P
a
is at a height ofH/3 fro m the base
of the wall .
Example 1 1 .1 3
For a retainin g wal l system , th e followin g dat a wer e available : (i ) Heigh t o f wal l = 7 m ,
(ii) Properties o f backfill: y
d
= 1 6 kN/m
3
, 0 = 35°, (iii ) angle of wal l friction, 8 = 20°, (iv ) back of
wall i s inclined at 20° t o the vertical (positiv e batter), and (v) backfill surface i s sloping a t 1 : 10.
Determine th e magnitude of the active eart h pressur e b y Culmann' s method.
Solution
(a) Fig . Ex. 11.1 3 shows the 0 line and pressure line s drawn to a suitable scale.
(b) Th e trial rupture lines Bc
r
Bc
2
, Bc
y
etc . ar e drawn by making Ac
l
= Cj C
2
= c
2
c
3
, etc.
(c) Th e lengt h of a vertical lin e from B t o the backfill surface is measured .
(d) Th e areas of wedges BAc
r
BAc
2
, BAc
y
etc . ar e respectively equal t o l/2(bas e lengths Ac
}
,
Ac
2
, Ac
y
etc. ) x perpendicular length.
458 Chapt er 1 1
Rupture plane
= 90 - ( 0 +<5) = 50°
Pressure lin e
Figure Ex . 1 1 .1 3
(e) Th e weight s o f th e wedge s i n ( d) abov e pe r mete r lengt h o f wal l ma y b e determine d b y
multiplying the areas b y the unit weight of the soil . The results ar e tabulated below :
Wedg e
BAc^
BAc
2
BAc
3
Weig ht , k N
115
230
345
Wedg e
BAc
4
BAc
5
Weig ht , k N
460
575
(f) Th e weight s of th e wedge s BAc
}
, BAc
2
, etc. ar e respectivel y plotte d ar e Bd
v
Bd
2
, etc . on
the 0-line .
(g) Line s ar e draw n paralle l t o th e pressur e lin e fro m point s d
{
, d
2
, d
3
etc . t o mee t
respectively th e trial ruptur e lines Bc
r
Bc
2
, Bc^ etc . a t point s e
}
, e
2
, e
y
etc .
(h) Th e pressure locu s i s drawn passing throug h points e\ , e
2
, e
y
etc .
(i) Lin e zz i s drawn tangential to the pressur e locu s a t a point a t whic h zz i s parallel t o the 0
line. This poin t coincides wit h the point e
y
(j) e
3
d^ give s th e activ e eart h pressur e whe n converted t o forc e units.
P
a
= 180 kN pe r mete r lengt h of wall ,
(k) Bc
3
i s th e critica l ruptur e plane.
1 1 .1 1 L ATERA L PRESSU RES BY TH EORY OF EL ASTI CI TY FOR
SU RCH ARG E L OADS ON TH E SU RFACE OF B ACKFI LL
The surcharge s o n th e surfac e o f a backfil l paralle l t o a retainin g wal l ma y b e an y on e o f th e
following
1. A concentrated loa d
2. A line loa d
3. A stri p loa d
Lat eral Eart h Pres s ure 459
_ x = mH_ |Q
Pressure distribution
(a) Vertical section (b ) Horizontal section
Figure 1 1 .2 0 Lat era l pres s ur e ag ain s t a rigid wal l due t o a poin t loa d
L ateral Pressur e at a Point i n a Semi-I nfinite Mass due to a Concentrated
L oad o n the Surfac e
Tests b y Spangle r (1938) , an d other s indicat e tha t latera l pressure s o n th e surfac e o f rigi d
walls ca n b e compute d fo r variou s type s o f surcharge s b y usin g modifie d form s o f th e
theory o f elasticit y equations . Latera l pressur e o n a n elemen t i n a semi -i nfi ni t e mas s a t
depth z from the surfac e may be calculate d by Boussines q theor y for a concentrated loa d Q
act i ng at a point o n th e surface . The equat i o n may b e expresse d a s (refe r t o Section 6. 2 fo r
not at i on)
Q
cos
2
/? 1 T I I—^Ll \ ^(J 5>
3 sin
2
ft cos
2
ft -± ^
1 +cos ft
(11.62)
as
If we writ e r = x in Fig. 6. 1 and redefine the terms as
jc = mH and , z = nH
where H - heigh t of the rigid wall and take Poisson's rati o \ J L = 0.5, we may write Eq. (11.62)
3<2
m n
2xH
2
( m
2
+
n
2
f
2
(11.63)
Eq. (11.63 ) i s strictl y applicabl e fo r computin g latera l pressure s a t a poin t i n a semi -
infinit e mass . However , thi s equation has t o be modifie d i f a rigi d wal l intervene s and break s
the continuit y of th e soi l mass . Th e modifie d form s ar e give n belo w fo r variou s type s o f
surcharge loads .
Lateral Pressur e o n a Rigi d Wall Due to a Concentrated Load o n the Surfac e
Let Q be a point load acting on the surface as shown in Fig. 11.20 . The various equations are
(a) For m > 0.4
Ph =
1.77(2
H
2
(11.64)
(b) For m < 0.4
460 Chapt er 1 1
0.28Q n
2
H
2
(0.1 6 + n
2
)
3
(11.65)
(c) Latera l pressur e a t point s alon g th e wal l o n eac h sid e o f a perpendicula r fro m th e
concentrated loa d Q to the wal l (Fig. 11.20b )
Ph = Ph cos
2
(l.la) (11.66)
L ateral Pressur e o n a Rigi d Wal l du e t o Lin e L oa d
A concrete block wall conduit laid on the surface, or wide strip loads may be considered a s a series of
parallel line loads as shown in Fig. 11.21 . The modified equations for computing p
h
ar e as follows:
(a) Fo r m > 0.4
Ph =
n H
(a) Fo r m < 0. 4
2 x 2 (11.67)
Ph =
0.203n
(0.16+ n
2
)
2 (11.68)
L ateral Pressur e o n a Rigi d Wal l du e t o Stri p L oa d
A stri p loa d i s a loa d intensit y wit h a finit e width , suc h a s a highway , railwa y lin e o r eart h
embankment whic h is parallel to the retaining structure. The applicatio n o f load i s as given i n Fig.
11.22.
The equatio n for computing p
h
i s
p
h
= — (/?-sin/?cos2«r)
(11.69a)
The tota l latera l pressur e pe r uni t lengt h of wal l du e t o stri p loadin g ma y b e expresse d a s
(Jarquio, 1981 )
x = mH
*"] q/unit lengt h
x
H
q/unit are a
Figure 1 1 .2 1 Lat era l pres s ur e ag ain s t a Figur e 1 1 .2 2 Lat era l pres s ur e ag ain s t a
rig id wal l due t o a lin e loa d rig i d wal l due t o a s t rip loa d
Lat eral Eart h Pres s ur e 461
(11.69b)
where a, = tan
l
— an d cc~ = tan'
i
H
2
A + B
Example 1 1 .1 4
A railway lin e i s laid parallel to a rigid retainin g wal l as shown in Fig. Ex. 11.14. The widt h of the
railway trac k and it s distance from th e wal l is shown in the figure. The height of the wal l is 10m.
Determine
(a) The uni t pressure at a depth of 4m from the top of the wal l due to the surcharge loa d
(b) Th e total pressur e actin g on the wal l due to the surcharge loa d
Solution
(a)FromEq(11. 69a)
The lateral eart h pressur e p
h
a t depth 4 m is
2q
p
h
=—(/?-sin/?cos2a)
2x60 18.44
3.14 18 0
x 3.14 - si n 18.44° cos 2x36.9 = 8.92 kN/m
2
(b)FromEq. (11.69b)
where, q = 60 kN/m
2
, H = 1 0 m
2m . 2 m
»T*
= A =B
Figure Ex . 1 1 .1 4 "
462 Chapt e r 1 1
,
A
,
2
a, = tan"
1
— = tan"
1
— = 11.31°
H 1 0
T
1
— ^t an"
1
— =21.80
C
H 1 0
=—[10(21.80-11.31)] « 70 k N / m
1 1 .1 2 CU RV E D SU RFACES O F FAI L U RE FOR COM PU TI NG
PASSI V E EART H PRESSU R E
It i s customary practic e t o use curved surfaces of failure for determining the passive eart h pressur e
P o n a retaining wall with granular backfill if § is greater tha n 0/3. If tables or graphs are available
for determinin g K fo r curved surfaces of failure the passive earth pressure P ca n b e calculated. If
tables o r graphs ar e no t availabl e for thi s purpose, P ca n b e calculate d graphicall y b y an y one of
the following methods .
1 . Logarithmi c spira l metho d
2. Frictio n circl e method
In bot h thes e methods , th e failur e surfac e clos e t o th e wal l i s assume d a s th e par t o f a
logarithmic spira l o r a part o f a circular ar c wit h the to p portion o f the failur e surfac e assume d a s
planar. Thi s statemen t i s vali d fo r bot h cohesiv e an d cohesionles s materials . Th e method s ar e
applicable fo r bot h horizonta l an d incline d backfil l surfaces . However , i n th e followin g
investigations i t wil l be assume d tha t the surface of the backfil l i s horizontal.
L ogarithmic Spira l M etho d o f Determinin g Passiv e Eart h Pressur e o f I dea l
Sand
Property o f a L ogarithmi c Spira l
The equatio n o f a logarithmic spira l may be expresse d a s
(11.70)
where
r
Q
= arbitrarily selecte d radiu s vector for referenc e
r = radius vector o f any chosen poin t on the spira l makin g an angl e 0 wit h r
Q
.
<j) = angl e of interna l friction o f th e material .
In Fig. 11.23a O i s th e origi n o f th e spiral . Th e propert y o f th e spira l i s tha t ever y radiu s
vector suc h a s Oa makes a n angl e of 90°-0 t o the tangen t of the spira l a t a or i n other words , th e
vector Oa makes a n angl e 0 wit h the normal to the tangent of the spira l a t a.
Analysis o f Force s fo r the Determinatio n o f Passiv e Pressur e P
p
Fig. 1 1 .23b gives a section through the plane contact face AB of a rigid retaining wall which rotates
about poin t A int o th e backfil l o f cohesionles s soi l wit h a horizonta l surface . B D i s draw n a t a n
angle 45°- 0/2 to the surface. Let O
l
be an arbitrary point selected o n the line BD as the center of
a logarithmi c spiral , an d le t O
}
A b e th e referenc e vecto r r
Q
. Assume a tria l slidin g surfac e Ae
l
c
l
which consists of two parts. The firs t par t is the curved part Ae
l
whic h is the part of the logarithmi c
Lat eral Eart h Pres s ur e 463
0
Tangent
V ,
(a) Properties of logarithmic spira l
Curve C
(c) Polygon o f forces
-0/2 /
B
(b) Methods o f analysis
Figure 11.2 3 Log arit hm i c s pira l m et ho d o f obt ain in g pas s ive eart h pres s ur e of s an d
(Aft er T erzag hi , 1943)
spiral wit h center a t O
l
an d the secon d a straight portion e
l
c
l
whic h i s tangentia l t o the spiral a t
point e
{
o n the line BD.
e^c\ meet s th e horizontal surfac e at C j a t an angl e 45°- 0/2. O
l
e
l
i s the end vector r
t
o f the
spiral which makes a n angle 6
l
wit h the reference vecto r r
Q
. Line BD makes an angle 90°- 0 with
line ^Cj whic h satisfies the propert y of the spiral .
It i s no w necessar y t o analyz e the force s actin g o n th e soi l mas s lyin g abov e th e assume d
sliding surface A^j Cj .
Within the mass of soil represented b y triangle Be
l
c
l
th e stat e of stress i s the same a s that in
a semi-infinit e mass i n a passive Rankin e state . The shearin g stresse s alon g vertica l section s ar e
zero i n thi s triangular zone. Therefore, w e can replace th e soi l mas s lyin g in the zone e
l
d
l
c
l
b y a
passive eart h pressur e P
d
actin g on vertical sectio n e
l
d
l
a t a height h
gl
/3 wher e h
g]
i s the height of
the vertica l sectio n e
{
d
{
. This pressur e i s equal to
p =
e\
(11.71)
464 Chapt e r 1 1
where W
0
= tan
2
(45 ° + 0/2 )
The bod y o f soi l mas s BAe
]
d
l
(Fig . 1 1.23b) i s acted o n by th e following forces :
1. Th e weigh t Wj o f th e soi l mas s actin g throug h the cente r o f gravit y of th e mas s havin g a
lever ar m /
2
wit h respec t t o O
r
the center of the spiral.
2. Th e passive eart h pressur e /^acting on the vertical sectio n e
l
d
}
havin g a lever ar m /
3
.
3. Th e passive eart h pressur e P j actin g on the surface AB a t an angle S to the normal and at a
height H/3 abov e A havin g a lever ar m l
{
.
4. Th e resultan t reaction forc e F
l
o n th e curved surfac e Ae
{
an d passin g throug h th e cente r
Det erm in at ion o f t h e Forc e /
>
1
Graphically
The directions o f all the forces mentioned above except tha t of F
l
ar e known. In order t o determine
the directio n o f F, combin e th e weigh t W
{
an d th e forc e P
el
whic h give s th e resultan t /? , (Fig .
1 1.23c). This resultant passes throug h the point of intersection n
l
o f W
{
an d P
el
i n Fig. 1 1.23b and
intersects forc e P
{
a t poin t n
2
. Equilibriu m requires tha t forc e F
{
pas s throug h th e sam e point .
According t o th e propert y o f th e spiral , i t mus t pas s throug h th e sam e point . Accordin g t o th e
property o f the spiral , i t must pass throug h the center O
l
o f the spiral also . Hence , th e direction of
Fj i s known and the polygon of forces show n i n Fig. 1 1.23c can be completed. Thus we obtain the
intensity o f the force P
}
require d to produce a sli p along surfac e Ae
l
c
l
.
Det erm in at ion o f /* , b y M om en t s
Force P
l
ca n b e calculate d b y takin g moment s o f al l th e force s abou t th e cente r O
{
o f th e spiral .
Equilibrium of the syste m require s that the sum of the moment s o f al l the force s mus t be equal t o
zero. Since th e direction of F
l
i s now known and since i t passes through O
l
, it has no moment. The
sum o f the moment s o f al l the othe r forces ma y be written as
P
1
/
1
+ W
1
/
2
+
J
P
1
/
3
= 0 (11.72 )
Therefore,
P
\
=
-7(^2
+ P
^) (11.73 )
l
i
P
l
i s thus obtained fo r an assumed failur e surface Ae^c^. The nex t ste p consist s i n repeatin g
the investigation for mor e tria l surfaces passin g through A whic h intersect lin e BD a t point s e
2
, e
3
etc. The values of P
r
P
2
P
3
et c so obtained ma y be plotted a s ordinates d
l
d{ , d
2
d'
2
etc. , as shown
in Fig. 1 1 .23b and a smooth curve C is obtained by joining points d{ , d'
2
etc . Sli p occurs alon g the
surface correspondin g t o th e mi ni mu m value P whic h i s represente d b y th e ordinat e dd'. Th e
corresponding failur e surface is shown as Aec i n Fig. 1 1.23b.
1 1 .1 3 COEFFI CI ENT S OF PASSI V E EARTH PRESSU RE TAB L ES
AND G RAPH S
Concept o f Coulomb' s Formul a
Coulomb (1776 ) computed th e passive eart h pressur e o f ideal san d o n the simplifyin g assumptio n
that the entire surfac e of sliding consists of a plane through the lower edge A o f contact fac e AB a s
shown i n Fig. 1 1.24a. Lin e AC represent s a n arbitrar y plan e sectio n throug h thi s lower edge. The
forces actin g on thi s wedge an d the polygon o f forces ar e shown i n the figure. The basi c equatio n
for computin g the passive eart h pressur e coefficien t ma y be developed a s follows :
Lat eral Eart h Pres s ur e 465
Consider a point on pressure surfac e AB a t a depth z from poin t B (Fi g 11.24a) . The normal
component of the eart h pressur e per uni t area of surface AB ma y be expressed b y the equation,
P
pn
= yzK
p
(11.74 )
where K
p
i s the coefficient of passive earth pressure. The total passive eart h pressure norma l
to surface AB, P
n
, is obtained from Eq . (11.74) as follows,
zdz
sin a si n a
o o
pn
sm«
(11.75)
where a i s the angle made by pressure surface AB wit h the horizontal.
Since the resultant passive earth pressur e P act s a t an angle 8 to the normal ,
p =
pn
- —
p
cos<5 2
K
s
sin cc cos o
(11.76)
H/3
(a) Principles o f Coulomb' s Theor y o f passive eart h pressure o f sand
30
C
^ 20°
"=3
<4-l
o
o i
35
C
10 1 5
V alues of K
P
40
C
20 25
(b) Coefficient of passive earth pressur e K
P
Figure 11.24 Diag ra m illus t rat in g pas s ive earth pres s ur e t heory o f s an d an d
relat ion bet wee n (j) , 8 an d K
p
(Aft e r T erzag hi , 1 943)
466 Chapt e r 1 1
Table 1 1 . 3 Pas s iv e eart h pres s ur e coefficien t K'
p
fo r curve d s urface s o f fail ur e
(Aft er Caquo t an d Keris e l 1948) .
0 =
3=0
(5=0/ 2
(5=0
8 = -0/ 2
10°
1.42
1.56
1.65
0.73
15°
1.70
1.98
2.19
0.64
20°
2.04
2.59
3.01
0.58
25°
2.56
3.46
4.29
0.55
30°
3.0
4.78
6.42
0.53
35°
3.70
6.88
10.20
0.53
40°
4.6
10.38
17.50
0.53
Eq. (11.76) may als o be expresse d a s
(11-77)
Kr,
where K'
p
= £— - (11.78 )
sin # cost)
Passive Eart h Pressur e Coefficien t
Coulomb develope d a n analytica l solution for determinin g K
p
base d o n a plane surfac e o f failure
and this is given i n Eq. (11.57). Figure 11.24(b ) gives curves for obtaining Coulomb' s value s of K
p
for variou s value s of 8 and 0 for plane surface s o f failure with a horizontal backfill . They indicat e
that fo r a give n valu e o f 0 th e valu e o f K
p
increase s rapidl y wit h increasin g value s o f 8 . Th e
limitations o f plan e surface s o f failur e ar e give n i n Sectio n 11.9 . Curve d surface s o f failur e ar e
normally use d fo r computin g P o r K
p
whe n the angl e o f wal l frictio n 8 exceeds 0/3 . Experienc e
indicates tha t the curved surfac e of failure may be taken either a s a part o f a logarithmic spira l or a
circular arc . Caquo t an d Kerise l (1948 ) compute d K'
p
by making us e of curved surface s o f failure
for variou s values of 0, 8, 0 and /3 . Caquot and Kerisel' s calculations for determining K'
p
for curved
surfaces o f failur e ar e availabl e in th e for m of graphs .
Table 11. 3 give s th e value s o f K'
p
for variou s value s o f 0 an d 8 fo r a vertica l wal l wit h a
horizontal backfil l (after Caquot and Kerisel , 1948) .
In the vast majority of practical cases the angle of wall friction ha s a positive sign , that is, the
wall transmit s t o a soi l a downwar d shearin g force . Th e negativ e angl e o f wal l frictio n migh t
develop i n th e cas e o f positiv e batte r pile s subjecte d t o latera l loads , an d als o i n th e cas e o f pier
foundations fo r bridges subjecte d t o lateral loads .
Example 1 1 .1 5
A gravity retaining wal l i s 1 0 ft high with sand backfill. Th e backface o f the wal l is vertical. Given
8= 20°, an d 0 = 40°, determin e the total passive thrust using Eq. (11.76) and Fig. 11.2 4 for a plane
failure. Wha t i s the passive thrust for a curved surface of failure? Assume y = 18. 5 kN/m
3
.
Solution
From Eq. (11.76 )
1
K
P' = -Y H
2 p
- wher e a = 90°
' 2 si n a cos S
From Fig. 11.2 4 (b) for 8 = 20°, an d 0 = 40°, w e have K
p
= 1 1
Lat eral Eart h Pres s ur e 46 7
P = -xl 8. 5x!0
2
= 10,828 k N/m
p
2 si n 90 cos 20°
From Tabl e 11. 3 K'
p
fo r a curved surfac e of failure (Caquot and Kerisel . 1948 ) fo r 0 = 40°
and 8 =20° is 10.38.
From Eq. (11.77)
p = -yH
2
K' = -x 18.5 x 10
2
x 10.38
p
2
p
2
= 9602kN/m
Comments
For S = $2, the reduction i n the passive eart h pressur e du e t o a curved surfac e o f failure is
10,828-9602
Reduction = — — x 100 = 11.32%
Example 1 1 .1 6
For th e dat a give n i n Exampl e 11.15 , determin e th e reductio n i n passiv e eart h pressur e fo r a
curved surfac e o f failur e if 8 = 30°.
Solution
For a plane surfac e of failur e P fro m Eq. (11.76 ) i s
P = - xl 8. 5x! 0
2
x — = 22,431 kN/m
p
2 sin90°cos30 °
where, K = 21 from Fig. 11.24 for § = 30° an d </ > = 40 °
From Tabl e 11. 3 for 8 = 30° an d </ » = 40 °
K
.
f =
10.38 + 17.50
From Eq ( l 1.77)
P = -x 18.5x!0
2
x 13.94 =12,895 kN/ m
p
2
o A .• • • 22,431-12,89 5 „„_ _ ,
Reduction i n passive pressure = = 42.5 %
22,431
It i s clea r fro m th e abov e calculations , tha t th e soi l resistanc e unde r a passiv e stat e give s
highly erroneous value s for plane surface s of failur e with an increase i n the valu e of S. This erro r
could lea d t o a n unsaf e condition becaus e th e compute d value s o f P woul d become highe r tha n
the actua l soi l resistance .
1 1 .1 4 L ATERA L EART H PRESSU RE ON RETAI NI NG WAL L S
DU RI NG EARTH QU AKE S
Ground motion s durin g an earthquak e ten d t o increas e th e eart h pressur e abov e th e stati c eart h
pressure. Retainin g walls with horizontal backfills designed wit h a factor of safet y of 1. 5 for static
468 Chapt er 1 1
loading ar e expected t o withstand horizontal accelerations u p to 0.2g. For larger accelerations , an d
for wall s wit h slopin g backfill , additiona l allowance s shoul d b e mad e fo r th e earthquak e forces .
Murphy (1960) shows that when subjected to a horizontal acceleratio n a t the base, failure occur s i n
the soi l mass alon g a plane inclined at 35° from th e horizontal . The analysi s of Mononobe (1929 )
considers a soi l wedg e subjecte d t o vertical and horizontal accelerations t o behave a s a rigid body
sliding over a plane sli p surface.
The curren t practice fo r earthquake design o f retaining walls is generally based on design rules
suggested b y See d an d Whitma n (1970) . Richard s e t al . (1979 ) discus s th e desig n an d behavio r of
gravity retainin g wall s wit h unsaturate d cohesionless backfill . Mos t o f th e paper s mak e us e o f th e
popular Mononobe-Okabe equation s as a starting point for their own analysis. They follow generally the
pseudoplastic approac h fo r solving the problem. Solution s are available for both the active and passive
cases with as granular backfil l materials . Though solutions for (c-0) soils have been presented by some
investigators (Prakas h an d Saran , 1966 , Saran an d Prakash , 1968) , thei r finding s have no t ye t bee n
confirmed, an d as such the solutions for (c-0) soils have not been taken up in this chapter .
Earthquake Effec t o n Active Pressur e wit h G ranula r B ackfil l
The Mononobe-Okab e metho d (1929 , 1926 ) for dynami c latera l pressur e o n retainin g wall s i s a
straight forwar d extensio n o f th e Coulomb slidin g wedg e theory . The force s tha t ac t o n a wedg e
under the active stat e ar e shown in Fig. 11.25
In Fig . 11.2 5 AC i n th e slidin g surfac e o f failur e o f wedg e ABC havin g a weigh t W with
inertial components k
v
W and k
h
W. The equation for the total active thrust P
ae
acting on the wall AB
under dynami c force condition s as per th e analysi s of Mononobe-Okabe i s
(11.79)
in which
K. =•
Ae
cos //cos
2
<9cos(# + 0+77) 1+
cos( 8+ 9+ /7)cos(/?- 9]
(11.80)
Figure 1 1 .2 5 Act iv e for c e o n a ret ain in g wal l wit h eart hquak e for ce s
Lat eral Eart h Pres s ur e 469
where P
ae
=dynami c component o f the tota l eart h pressur e P
ae
o r P
ae
= P
a
+ P
ae
K
Ae
= the dynami c earth pressur e coefficient
77 = tan "
(11.81)
P
a
= active eart h pressur e [Eq . (11.50)]
k
h
= (horizontal acceleration)/g
k
v
^(vertical acceleration)/ g
g = acceleration du e t o gravity
y= unit weight of soi l
0 = angle of friction of soi l
8 = angle of wall friction
/3 = slope o f backfill
6 = slop e o f pressur e surfac e o f retainin g wal l wit h respec t t o vertica l a t poin t B
(Fig. 11.25 )
H = height of wal l
The total resultant active earth pressure P
ae
due to an earthquake is expressed a s
P - P +P L
ae
L
a ^
l
ae
(11.82)
The dynami c component P
ae
i s expected t o act at a height 0.6H abov e the bas e wherea s the
static earth pressure act s at a height H/3. For all practical purposes i t would be sufficient t o assume
that the resultant force P
ae
acts at a height H/2 above the base with a uniformly distributed pressure.
0.7
0.6
0.5
50. 4
c
0.2
0.1
= 0,0 = 0,0 =
= 1/2 0
'0 = 30
C
'0 = 35 <
0 0. 1 0. 2 0. 3 0. 4 0. 5
kh
(a) Influence of soi l frictio n
on soi l dynami c pressure
u. /
n f\
O
c
"O n A
T3
C
v? A a
O 9
0
1
A
°(
//,
^x
) 0
0
/
/
//
X
1 0
= 20°
/ ft
A
//
/ s
2 0
/
= 10 °
ft =
/
=
k
v
-0
d =
l
/2
3 0
/
°v
-10°
:o
6> - 0
0
4 0 .
(b) Influence of backfil l slop e on
dynamic lateral pressure
Figure 11.2 6 Dyn am i c l at era l act ive pres s ur e (aft er Richard s et al. , 1979 )
470 Chapt er 1 1
It ha s bee n show n tha t the activ e pressur e i s highl y sensitiv e t o bot h th e backfil l slop e ( 3, an d th e
friction angl e 0 of the soi l (Fig . 11.26) .
It i s necessary t o recognize the significanc e of the expressio n
(11.83)
given unde r the roo t sig n i n Eq. (11.80) .
a. Whe n Eq . ( 1 1.83) i s negative no real solutio n i s possible. Henc e fo r stability , the limiting
slope o f th e backfil l mus t f ul f i ll th e condition
P<( tp-ri)
b. Fo r no earthquake condition, r|= 0. Therefore fo r stabilit y we have
p<q>
c. Whe n th e backfil l i s horizontal (3 = 0. For stabilit y we hav e
ri<( p
d. B y combinin g Eqs. ( 1 1.81) and ( 1 1.86), we have
(11.84a)
(11.85)
(11.86)
(11.87a)
From Eq . ( 1 1.87a), we ca n defin e a critical value for horizonta l acceleratio n k*
h
a s
^ = ( l - f c
v
) t a n ^ (11.87b )
V alues o f critica l acceleration s ar e give n i n Fi g 11.2 7 whic h demonstrate s th e sensitivit y of
the variou s quantities involved.
0.7
0.6
0.5
0.2
0.1
10 2 0 3 0 4 0
0 degree s
Figure 1 1 .2 7 Crit ica l val ue s o f horizon t a l accel erat ion s
Lat eral Eart h Pres s ur e 47 1
Effect o f Wal l L atera l Displacemen t o n the Desig n o f Retainin g Wal l
It is the usual practice of some designer s to ignore the inertia forces of the mass of the gravity
retaining wal l i n seismi c design . Richard s an d Elm s (1979 ) hav e show n tha t thi s approac h i s
unconservative sinc e i t i s th e weigh t o f th e wal l which provides mos t o f th e resistanc e t o latera l
movement. Taking into account al l the seismi c force s actin g on the wal l and a t the base the y have
developed a n expressio n fo r th e weigh t o f th e wal l W
w
unde r th e equilibriu m conditio n a s (for
failing b y sliding)
W
w
=± yH
2
( l-k
v
)K
Ae
C
IE
(11.88 )
in which,
cos(S + 6>) - sin( £+6>) tan S
1E
(l - &
v
)(t an£- t an77) (11.89 )
where W
w
= weight of retaining wall (Fig. 11.25 )
8 = angle of friction betwee n the wall and soi l
Eq. (11.89) is considerably affecte d b y 8. If the wall inertia factor is neglected, a designer will
have t o go t o an exorbitant expense t o design gravit y walls.
It is clear tha t tolerable displacemen t of gravity walls has to be considered i n the design. The
weight of the retaining wal l is therefore require d t o be determined t o limit the displacement t o the
tolerable limit . The procedure i s as follows
1. Se t the tolerabl e displacemen t Ad
2. Determin e the design value of k
h
by making use of the following equation (Richards et al., 1979)
0.2 A,
2
^
where A
a
, A
V
= acceleration coefficient s used i n the Applied Technology Counci l (ATC) Building
Code (1978) for various regions o f the United States. M i s in inches.
3. Usin g the values of k
h
calculated above, and assuming k
v
- 0 , calculate K
Ae
fro m Eq (11.80)
4. Usin g the valu e of K
Ae
, calculat e th e weight , W
w
, of the retainin g wal l by makin g us e of
Eqs. (11.88 ) and (11.89)
5. Appl y a suitable factor of safety, say, 1. 5 to W
w
.
Passive Pressur e Durin g Earthquake s
Eq. (11.79 ) give s a n expressio n fo r computin g seismi c activ e thrus t whic h i s base d o n th e wel l
known Mononobe-Okabe analysi s for a plane surface of failure. The corresponding expressio n fo r
passive resistance i s
P
p
e=
2
^-
k
^
K
P
e
(11. 91 )
K
Pe
= —
cosrjcos
2
0cos( S-0+Tj) 1- .
472 Chapt er 1 1
Figure 1 1 .2 8 Pas s iv e pres s ure o n a ret ain in g wal l durin g eart hquak e
Fig. 11.2 8 gives the various forces actin g on the wal l under seismic conditions . Al l the othe r
notations in Fig. 11.2 8 are the same a s those i n Fig. 11.25 . The effec t o f increasing the slope angl e
P is to increase th e passive resistance (Fig. 11.29) . The influenc e of the friction angle of the soi l (0)
on the passive resistance i s illustrated the Fig. 11.30 .
Figure 1 1 .2 9 In fluen c e o f back fil l s lop e an g l e o n pas s ive pres s ur e
Lat eral Eart h Pres s ur e 473
0 0. 2 0. 4 0. 6
Figure 11.30 In fluen c e o f s oi l frict io n an g l e o n pas s ive pres s ure
It ha s bee n explaine d i n earlie r section s o f thi s chapte r tha t th e passiv e eart h pressure s
calculated on the basis of a plane surface of failure give unsafe results if the magnitude of 6 exceeds
0/2. Th e erro r occur s becaus e th e actua l failur e plan e i s curved , wit h th e degre e o f curvatur e
increasing wit h a n increas e i n th e wal l frictio n angle . Th e dynami c Mononobe-Okab e solutio n
assumes a linear failure surface , as does th e stati c Coulomb formulation.
In order t o set right this anomaly Morrison and Ebelling (1995) assumed the failure surface as
an ar c o f a logarithmi c spira l (Fig . 11.31 ) an d calculate d th e magnitud e of th e passiv e pressur e
under seismic conditions .
It i s assume d her e tha t th e pressur e surfac e i s vertica l ( 9=0) an d th e backfil l surfac e
horizontal (j 3 = 0). The following charts have been presented by Morrison and Ebelling on the basis
of thei r analysis.
Logarithmic spira l
Figure 11.31 Pas s iv e pres s ur e from lo g s piral fail ur e s urface durin g eart hquak e s
474 Chapt er 1 1
LEGEND
Mononobe-Okabe
Log spiral
0 0.1 0 0.2 0 0.3 0 0.4 0 0.5 0 0.6 0
Figure 1 1 .3 2 K
pe
vers u s k
h
, effect o f 8
LEGEND
Mononobe-Okabe
Log spiral
k
v
= 0,6= (2/3)0
0.60
Figure 1 1 .33 K
pe
vers us k
h
, effect o f
1 . Fig . 1 1 .32 gives the effect o f 5 on the plot K
pe
versu s k
h
wit h k
v
= 0, for 0 =30°. The values
of § assume d ar e 0 , 1/ 2 (()) ) and(2/3<j)) . Th e plo t show s clearl y th e differenc e between th e
Mononobe-Okabe an d lo g spira l values . The differenc e betwee n th e tw o approache s i s
greatest a t k
h
= 0
Lat eral Eart h Pres s ur e 47 5
2. Fig . 11.33 show s th e effec t o f 0 o n K
pg
. Th e figur e show s th e differenc e betwee n
Mononobe-Okabe and log spiral values of K versu s k
h
with 8=( 2/30 ) and k
v
= 0. It is also
clear fro m th e figure th e difference betwee n the two approaches i s greatest for k
h
- 0 and
decreases wit h an increase in the value of k
h
.
Example 1 1 .1 7
A gravit y retaining wall is required t o be designed for seismi c condition s for the active state. The
following dat a ar e given:
Height of wall = 8 m 0=0° , 0=0, 0=30°, &= 15°, £, = 0, k
h
= 0.25 and y= 19kN/m
3
. Determine
P
ae
an d th e approximat e point o f application . What i s th e additiona l activ e pressur e cause d b y th e
earthquake?
Solution
From Eq. (11.79 )
P
ae
=\ rH
2
( l-k
v
)K
Ae
=^yH^K
Ae
, sinc e *
y
= 0
For 0 = 30°, 5= 15 ° and k
h
= 0.25, w e have from Fig . 1 1.26 a
K
Ae
= 0.5. Therefore
p
ag
= - ?- 19x8
2
x 0.5 = 304 k N/m
1 9
From Eq. (11.14) P
a
=-y H
2
K
A
2
° -
2
where K
A
= tan
2
(45° - ^ 2 ) = tan
2
30° = 0.33
Therefore P
a
= - x 19 x 8
2
x 0.33 = 202.7 kN/ m
&P
ae
= the additional pressure du e to the earthquake = 304 - 202. 7 = 101. 3 kN/m
For all practical purposes, the point of application of P
ae
may be taken as equal to H/2 above
the base of the wal l or 4 m above the base in this case.
Example 1 1 .1 8
For th e wal l give n i n Exampl e 11.17 , determin e th e tota l passiv e pressur e P
e
unde r seismi c
conditions. What is the additional pressure du e to the earthquake?
Solution
From Eq. (11.91) ,
P
ae
= rH*( l-k
v
)K
pe
=7H*K
pe
, sinc e *
v
= 0
From Fi g 1 1.32, (from M- O curves), K
pe
= 4.25 fo r 0 = 30°, an d 8= 15 °
Now/
3
=-/H
2
K = - x ! 9 x 8
2
x 4.25 = 2584 k N/ m
pe 2 P
e
9
476 Chapt er 1 1
FromEq. (11.15 )
p = -
7
H
2
K = - x ! 9 x 8
2
x 3 =
p
2
p
2
30
where # = t a n
2
4 5 ° + — |= tan
2
60° = 3
= (
P
pe -
P
Pe) =
2584
~
1824 = 76
°
kN
/
1 1 .1 5 PROB L EM S
11.1 Fig . Prob . 11. 1 shows a rigid retaining wal l prevented fro m latera l movements . Determin e
for thi s wal l th e latera l thrus t for th e at-res t conditio n an d th e poin t o f applicatio n o f th e
resultant force .
11.2 Fo r Prob 11.1 , determine th e active eart h pressur e distributio n fo r the following cases :
(a) when the wate r tabl e i s below th e base and 7= 1 7 kN/rn
3
.
(b) whe n th e wate r tabl e i s at 3m below groun d leve l
(c) when the wate r tabl e i s at ground leve l
11.3 Fig . Prob . 11. 3 gives a cantilever retaining wall wit h a sand backfill . The propertie s o f the
sand are :
e = 0.56, 0 = 38°, and G ^ = 2.65.
Using Rankin e theory , determin e th e pressur e distributio n wit h respec t t o depth , th e
magnitude an d th e point of applicatio n o f the resultan t active pressur e wit h the surcharg e
load bein g considered .
Ground surfac e
.:''. .'• : ..' San d • : ' . ' • . *• / .
3 m . ' . ' - ' ' ' " " ' ,
: • • • . • • : y= 1 7 kN/m
3
- . . ' • . - . • . . ' • : ..' San d •''.•':.:'
4. 5m V
;
' . '
;
' :• ".' ' '- .
. / y
s a t
=19. 8kN/ m
3
:•. ' • • • ' • • = 34 °
Surcharge, q = 500 lb/ft
2
1 1 1
Saturated san d
= 0.5 6
=
38° G
s
= 2.65
Figure Prob . 1 1 . 1 Figure Prob . 1 1 . 3
Lat eral Eart h Pres s ur e
477
11.4 A smooth vertica l wal l 3.5m high retains a mass of dry loose sand. The dr y unit weight of
the sand is 15. 6 kN/m
3
and an angle of internal friction 0is 32°. Estimate the total thrust per
meter acting against the wall (a) if the wall is prevented from yielding , and (b) if the wall is
allowed t o yield.
11.5 A wal l of 6 m height retains a non-cohesive backfil l of dr y uni t weight 1 8 kN/m
3
an d a n
angle o f interna l friction o f 30°. Us e Rankine' s theor y an d fin d th e tota l activ e thrus t per
meter lengt h o f th e wall . Estimat e th e chang e i n th e tota l pressur e i n th e followin g
circumstances:
(i) Th e top of the backfill carrying a uniformly distributed load of 6 kN/m
2
(ii) Th e backfil l under a submerged conditio n wit h the water table at an elevation o f 2 m
below the top of the wall. Assume G
s
- 2.65 , an d the soil above the water table being
saturated.
11.6 Fo r the cantileve r retaining wal l given i n Fig. Pro b 11. 3 wit h a sand backfill , determin e
pressure distribution with respect to depth and the resultant thrust. Given:
H
l
= 3m, H
2
= 6m, y
sat
= 19. 5 kN/m
3
q =25 kN/m
2
, and 0=36 °
Assume the soil above the GWT is saturated
11.7 A retaining wall of 6 m height having a smooth back retains a backfill made up of two strata
shown in Fig . Prob . 11.7 . Construc t the activ e eart h pressur e diagra m and fin d the
magnitude and point of application o f the resultant thrust. Assume the backfil l above WT
remains dry.
11.8 (a ) Calculate the total active thrust on a vertical wall 5 m high retaining sand of unit weight
17 kN/m
3
fo r whic h 0 = 35° . Th e surfac e i s horizonta l an d th e wate r tabl e i s below th e
bottom of the wall, (b) Determine the thrust on the wall if the water table rises to a level 2 m
below the surface of the sand. The saturated unit weight of the sand i s 20 kN/m
3
.
11.9 Figur e Proble m 11. 9 shows a retaining wall wit h a sloping backfill . Determin e th e active
earth pressur e distribution , the magnitude and the point of application o f the resultant by
the analytical method.
Cinder
H, - 2 m £. = «•
XV V V vXX/'WV S
Figure Prob. 1 1 . 7 Figure Prob . 1 1 . 9
478 Chapt er 1 1
j |j g = 50kN/m
:
\A ~ ~
Soil A 6 m H
l
• SO U t i I i
Figure Prob . 1 1 .1 0
11.10 Th e soi l condition s adjacen t t o a rigi d retainin g wal l ar e show n i n Fig . Prob . 11.10 , A
surcharge pressure o f 50 kN/m
2
is carried o n the surface behind the wall. For soil (A) above
the wate r table, c'= 0, 0' = 38°, y' = 18 kN/m
3
. For soil (B) below th e WT, c'= 1 0 kN/m
2
,
0'= 28°, and y
sat
= 20 kN/m
3
. Calculate the maximum unit active pressure behind the wall,
and the resultant thrust per uni t length of the wall.
11.11 Fo r the retaining wall given in Fig. Prob . 11.10 , assume the following data :
(a) surcharg e loa d = 100 0 lb/ft
2
, an d (b ) H
l
= 10 ft, H
2
= 20 ft ,
(c) Soi l A: c'= 50 0 lb/ft
2
, 0' = 30°, y = 110 lb/ft
3
(d) Soi l B: c'= 0 , 0'= 35°, y
sat
= 12 0 lb/ft
3
Required:
(a) Th e maximu m active pressure a t the base o f the wall .
(b) Th e resultan t thrust per uni t length of wall.
11.12 Th e depth s o f soi l behin d an d i n fron t o f a rigi d retainin g wal l ar e 2 5 f t an d 1 0 f t
respectively, bot h th e soi l surface s bein g horizonta l (Fig . Pro b 11.12) . Th e appropriat e
'A\\//A\ \ //\\\
0 = 22°
c = 600 lb/ft
2
y =110 lb/ft
3
Figure Prob. 1 1 .1 2
Lat eral Eart h Pres s ur e 479
shear strengt h parameters fo r the soi l ar e c = 600 lb/ft
2
, an d 0 = 22°, an d the uni t weight
is 110 lb/ft
3
. Using Rankine theory, determine the total active thrus t behind the wall and the
total passive resistance i n front o f the wall . Assume the water tabl e i s at a great depth .
11.13 Fo r the retaining wall given in Fig. Prob. 11.12 , assume the water table i s at a depth of 1 0 ft
below th e backfill surface . The saturated uni t weight of the soil i s 12 0 lb/ft
3
. The soi l above
the GWT i s also saturated . Comput e the resultant active and passive thrust s per unit length
of the wall.
11.14 A retaining wal l has a vertical bac k fac e an d i s 8 m high. The backfil l ha s th e following
properties:
cohesion c = 1 5 kN/m
2
, 0 = 25°, y = 18. 5 kN/m
3
The wate r tabl e i s a t grea t depth . Th e backfil l surfac e i s horizontal . Dra w th e pressur e
distribution diagra m an d determin e th e magnitud e an d th e poin t o f applicatio n o f th e
resultant active thrust .
11.15 Fo r the retaining wal l give n i n Prob. 11.14 , the wate r tabl e i s at a dept h o f 3 m below th e
backfill surface . Determine th e magnitude of the resultant active thrust .
11.16 Fo r the retaining wal l given i n Prob. 11.15 , comput e th e magnitude o f the resultant activ e
thrust, i f the backfil l surface carries a surcharge loa d o f 30 kN/m
2
.
11.17 A smooth retainin g wall is 4 m high and support s a cohesive backfil l with a unit weight of
17 kN/m
3
. Th e shea r strengt h parameter s o f the soi l ar e cohesio n =1 0 kPa an d 0 = 10° .
Calculate th e tota l activ e thrus t acting agains t the wal l an d the dept h t o th e point o f zer o
lateral pressure .
11.18 A rigi d retainin g wal l i s subjecte d t o passive eart h pressure . Determin e th e passive eart h
pressure distributio n and the magnitude and point of application o f the resultant thrust by
Rankine theory.
Given: Heigh t o f wal l = 1 0 m ; dept h o f wate r tabl e fro m groun d surfac e = 3 m ;
c - 2 0 kN/m
2
, 0 = 20° and y
sat
= 19. 5 kN/m
3
. The backfill carrie s a uniform surcharg e of
20 kN/m
2
.
Assume th e soi l abov e th e water table i s saturated.
11.19 Fig . Prob . 11.19 gives a retaining wal l wit h a vertical bac k fac e and a sloping backfill . All
the other data are given i n the figure. Determine the magnitude an d point of application of
resultant active thrust by the Culmann method.
y =115 lb/ft
3
0 = 38°
d = 25°
Figure Prob . 1 1 .1 9
480 Chapt er 1 1
6 ft **- 8f t — |
q= 12001b/ft
2
5 f t
2 5 f t
Figure Prob . 1 1 .2 0
11.20 Fig . Prob . 11.2 0 gives a rigid retaining wall wit h a horizontal backfill . The backfil l carries
a stri p load o f 120 0 lb/ft
2
a s shown i n the figure . Determin e th e following:
(a) Th e uni t pressur e o n th e wal l a t poin t A a t a dept h o f 5 f t belo w th e surfac e du e t o th e
surcharge load .
(b) Th e tota l thrus t on the wal l due t o surcharge load .
11.21 A gravit y retaining wall wit h a vertica l back fac e i s 1 0 m high . The followin g dat a ar e
given:
0=25°, S= 15° , and y = 1 9 k N/ m
3
Determine th e tota l passiv e thrus t using Eq (11.76) . What i s th e tota l passiv e thrus t for a
curved surfac e of failure?
11.22 A gravit y retaining wal l i s require d t o b e designe d fo r seismi c condition s fo r th e activ e
state. The bac k fac e i s vertical. The following data ar e given:
Height o f wal l = 30 ft, backfill surface i s horizontal; 0 = 40°, 8 = 20°, k
v
= 0, k
h
= 0.3, y =
120 lb/ft
3
.
Determine th e total active thrust on the wall . What i s the additional lateral pressur e du e t o
the earthquake?
11.23 Fo r the wal l give n in Pro b 11.22 , determine th e tota l passive thrus t durin g the earthquak e
What i s the change i n passive t hr us t due t o the earthquake? Assume $ = 30° and 8 = 15°.
CHAPTER 1 2
SH ALLOW FOUNDATION I :
ULTIMATE BEARING CAPACITY
1 2 .1 I NTRODU CTI O N
It i s the customar y practice t o regard a foundation as shallow i f the dept h o f the foundation is les s
than or equal t o the widt h of the foundation. The different type s of footings that we normally come
across ar e give n i n Fig . 12.1 . A foundatio n i s a n integra l par t o f a structure . Th e stabilit y o f a
structure depend s upo n th e stabilit y o f th e supportin g soil . Two important factor s tha t ar e t o b e
considered ar e
1. Th e foundation must be stabl e against shear failur e of the supporting soil .
2. Th e foundation must not settl e beyond a tolerable limi t to avoid damage t o the structure.
The othe r factor s that require consideratio n ar e the location an d dept h o f the foundation. In
deciding the location an d depth, one has to consider the erosions due to flowing water, underground
defects suc h as root holes, cavities , unconsolidate d fills, groun d water level, presence o f expansive
soils etc.
In selectin g a type o f foundation, one has t o consider th e functions of the structur e and the
load i t has t o carry, the subsurface condition of the soil , an d the cost of the superstructure .
Design load s als o pla y a n importan t par t i n th e selectio n o f th e typ e o f foundation . Th e
various load s tha t ar e likel y t o b e considere d ar e (i ) dea d loads , (ii ) live loads , (iii ) wind an d
earthquake forces, (iv) lateral pressures exerte d by the foundation earth on the embedded structura l
elements, and (v) the effect s o f dynamic loads .
In additio n t o th e abov e loads , th e load s tha t ar e du e t o th e subsoi l condition s ar e als o
required t o be considered . The y ar e (i ) latera l o r uplif t force s o n the foundatio n elements du e t o
high water table, (ii ) swelling pressures o n the foundations in expansive soils, (iii ) heave pressure s
481
482 Chapt er 1 2
(c) (d)
Figure 1 2 . 1 T ype s o f s hal l o w foun dat ion s : (a ) plain con cret e foun dat ion ,
(b) s t eppe d rein force d con cret e foun dat ion , (c ) rein forced con cret e rect an g ula r
foun dat ion , an d (d ) rein forced con cret e wal l foun dat ion .
on foundations in areas subjecte d to frost heav e and (iv) negative frictional drag on piles where pil e
foundations ar e used i n highly compressible soils .
Steps fo r the Selectio n o f the Typ e o f Foundatio n
In choosing th e type of foundation, the design engineer mus t perform five successiv e steps .
1. Obtai n th e required information concerning the natur e of the superstructur e and the load s
to be transmitted to the foundation.
2. Obtai n th e subsurfac e soil conditions.
3. Explor e th e possibilit y o f constructin g any on e o f th e type s o f foundatio n unde r th e
existing condition s by taking into account (i ) the bearing capacit y o f the soi l t o carr y th e
required load , an d (ii ) the advers e effect s o n th e structur e due t o differentia l settlements.
Eliminate i n thi s way, the unsuitabl e types.
4. Onc e one or two types of foundation ar e selected o n the basis of preliminary studies, make
more detaile d studies . Thes e studie s ma y requir e mor e accurat e determinatio n o f loads ,
subsurface condition s and footin g sizes . I t ma y als o b e necessar y t o mak e mor e refine d
estimates of settlement in order t o predict the behavior of the structure.
Shallow Foun dat io n I : Ult im at e B earin g Capacit y 483
5. Estimat e th e cos t o f eac h o f th e promising types o f foundation, and choos e th e typ e that
represents the most acceptabl e compromis e betwee n performance an d cost .
1 2 .2 TH E U L TI M ATE B EARI N G CAPACI TY OF SOI L
Consider the simplest cas e of a shallow foundation subjected to a central vertical load. The footing
is founded at a depth D
f
belo w the ground surface [Fig. 12.2(a)] . If the settlement , 5, of the footing
is recorded agains t the applie d load , Q, load-settlement curves , similar i n shap e t o a stress-strai n
curve, may be obtained as shown in Fig. 12.2(b) .
The shap e o f th e curv e depend s generall y o n th e siz e an d shap e o f th e footing , th e
composition o f the supportin g soil, an d the character , rate , an d frequenc y o f loading. Normall y a
curve will indicate the ultimat e load Q
u
that the foundation can support . If the foundation soil i s a
dense sand or a very stif f clay , the curve passes fairl y abruptl y to a peak value and then drops down
as shown by curve C
l
i n Fig. 10.2(b) . The pea k loa d Q
u
is quit e pronounced i n thi s case. O n the
other hand, if the soil is loose sand or soft clay, the settlement curve continues to descend on a slope
as show n b y curv e C
2
whic h show s tha t th e compressio n o f soi l i s continuousl y takin g plac e
without giving a definite value for Q
u
. On such a curve, Q
u
may be taken a t a point beyond which
there is a constant rate of penetration.
12.3 SOM E OF TH E TERM S DEFINED
It will be useful t o define, at this stage, some of the terms relating to bearing capacity of foundations
(refer t o Fig. 12.3) .
(a) Tota l Overburden Pressur e q
0
q
o
is the intensity of total overburden pressur e due to the weight of both soi l an d water a t the base
level of the foundation.
a — vD+ v D { "191 1
^n I
U
MU\ ~ I int^w \
i
^-
i
).
Q
L
Quit
Load
(a) Footing (b ) Load-settlement curve s
Figure 1 2. 2 T ypica l load-s et t lem en t curve s
484 Chapt er 1 2
D
f
GL
/ZXSs/^X^
GWT
V
v^-XvX-X-
I I I
,
'
Qu,,
^
1
5.1
GL
7
/sat
<?'o
i n
( D
f
-D
wl
) = D
w
, y - uni t weight of soil above GWT
y
sat
= saturated uni t weight of soi l belo w GWT
Vb
=
(/sat ~ 7w> - submerge d uni t weight of soil
y
w
= unit weight o f wate r
Figure 1 2. 3 T ot a l an d effect ive overburde n pres s ure s
(b) Effectiv e Overburden Pressure qr'
0
<?« is the effectiv e overburde n pressur e a t the base level of the foundation .
when O,
q
'
Q
= yD
w{
= yD
f
.
(12.2)
(c) Th e Ultimat e B earing Capacity o f Soil , q
u
q
u
is the maximu m bearing capacit y of soil at which the soi l fail s by shear .
(d) The Net Ultimat e B earing Capacity, q
nu
q
nu
is the bearing capacit y in excess o f the effective overburden pressure q'
Q
, expressed a s
(e) G ross Allowable B earin g Pressure, q
a
q
a
i s expressed a s
3a-jr
where F
s
= factor o f safety.
(f) Ne t Allowabl e B earin g Pressure, q
na
q i s expressed a s
(12.3)
(12.4)
_ q
nu
(12.5)
Shallow Foundation I : Ultimat e B earin g Capacity 485
(g) Saf e B earing Pressure, q
s
q
s
is defined as the net saf e bearin g pressur e which produces a settlement of the foundation which
does not exceed a permissible limit.
Note: I n the design of foundations, one has to use the least of the two values of q
na
and q
s
.
12.4 TY PE S OF FAI L U RE IN SOI L
Experimental investigation s hav e indicate d tha t foundation s o n dens e san d wit h relativ e
density greate r tha n 7 0 percen t fai l suddenl y wit h pronounce d pea k resistanc e whe n th e
settlement reache s abou t 7 percen t o f th e foundatio n width. The failur e i s accompanie d b y
the appearanc e o f failur e surfaces and b y considerabl e bul gin g of a sheared mas s o f san d a s
shown i n Fig. 12.4(a) . This typ e of failur e is designated a s general shea r fail ur e by Terzagh i
(1943). Foundations on sand of relative density lying between 35 and 70 percent d o not show
a sudden failure. As the settlement exceeds abou t 8 percent o f the foundation width, bul ging
of sand start s a t the surface . At settlement s of about 1 5 percent of foundat i on width, a visible
boundary o f sheare d zone s a t the surfac e appears. However , the peak o f base resistance ma y
never b e reached . Thi s typ e o f failur e i s terme d loca l shea r failure , Fig . 12.4(b) , b y
Terzaghi (1943) .
(a) General shear failur e
(b) Local shear failure
Quit
Load
Load
Load
(c) Punching shear failure
Figure 12. 4 M ode s o f bearin g capacit y failur e (Ves ic , 1963)
486
Chapt er 1 2
Foundations o n relatively loos e san d wit h relative densit y les s tha n 35 percent penetrat e int o
the soi l withou t an y bulgin g o f th e san d surface . Th e bas e resistanc e graduall y increase s a s
settlement progresses . Th e rat e o f settlement , however , increase s an d reache s a maximu m a t a
settlement o f abou t 1 5 t o 2 0 percen t o f th e foundatio n width . Sudde n jerk s o r shear s ca n b e
observed a s soon a s the settlement reaches abou t 6 to 8 percent of the foundation width. The failure
surface, whic h i s vertical o r slightl y inclined and follows the perimete r o f the base, neve r reache s
the san d surface . Thi s typ e o f failur e i s designate d a s punchin g shear failur e b y V esi c (1963 ) a s
shown i n Fig. 12.4(c) .
The thre e type s o f failur e describe d abov e wer e observe d b y V esi c (1963 ) durin g test s o n
model footings . I t ma y b e note d her e tha t as the relative depth/widt h rati o increases , th e limiting
relative densities a t which failure type s change increase. The approximat e limit s of types of failure
to be affecte d a s relative dept h DJ B, an d relative densit y of sand , D
r
, vary ar e shown i n Fig. 12. 5
(V esic, 1963) . Th e sam e figur e show s tha t ther e i s a critica l relativ e dept h belo w whic h onl y
punching shea r failur e occurs. For circular foundations , this critical relative depth , DJ B, i s around
4 an d for long rectangula r foundations around 8.
The surface s o f failure s a s observe d b y V esi c ar e fo r concentri c vertica l loads . An y smal l
amount of eccentricity i n the load application changes the modes o f failure and the foundation tilts
in the direction o f eccentricity. Tilting nearly always occurs i n cases of foundation failures becaus e
of th e inevitabl e variatio n in th e shea r strengt h and compressibilit y o f th e soi l fro m on e poin t t o
another and causes greate r yieldin g on one side or another of the foundation. This throws the cente r
of gravity of the load towards the side wher e yielding has occurred, thu s increasing th e intensity of
pressure o n thi s side followe d by further tilting.
A footing founded on precompressed clay s or saturated normall y consolidate d clay s wil l fail
in general shea r i f it is loaded s o that no volume change ca n take place and fail s by punching shear
if th e footin g i s founded on sof t clays .
o
.c
D- 7
<U *J
T3
O
Relative densit y o f sand, D
r
0.2 0. 4 0. 6 0. 8 1.0
Punching
shear
Local
shear
General
shear
Figure 12. 5 M ode s o f fail ur e o f m ode l foot in g s i n s an d (aft e r Ves ic , 1963 )
Shallow Foun dat io n I : Ultimat e B earin g Capacit y 48 7
1 2 .5 A N OV ERV I E W OF B EARI NG CAPACI TY TH EORI E S
The determinatio n o f bearin g capacit y o f soi l base d o n th e classica l eart h pressur e theor y o f
Rankine (1857) bega n wit h Pauker , a Russian militar y engineer (1889) , and was modified by Bell
(1915). Pauker's theory was applicable only for sandy soils but the theory of Bell took into account
cohesion also . Neithe r theor y too k int o accoun t th e widt h o f th e foundation . Subsequen t
developments le d t o the modification of Bell's theory to include width of footing also .
The methods of calculating the ultimate bearing capacit y of shallow stri p footings by plastic
theory develope d considerabl y ove r th e years sinc e Terzagh i (1943 ) firs t propose d a metho d b y
taking int o accoun t th e weigh t o f soi l b y th e principl e o f superposition . Terzagh i extende d th e
theory of Prandt l (1921). Prandt l developed a n equation based o n his stud y of the penetration of a
long har d meta l punc h int o softe r material s fo r computin g th e ultimat e bearin g capacity . H e
assumed the material was weightless possessing only cohesion and friction. Taylor (1948) extended
the equatio n o f Prandt l b y takin g int o accoun t th e surcharg e effec t o f th e overburde n soi l a t th e
foundation level .
No exact analytical solution for computing bearing capacity of footings is available at present
because th e basi c syste m o f equation s describin g th e yiel d problems i s nonlinear . On accoun t of
these reasons, Terzaghi (1943) first proposed a semi-empirical equatio n for computing the ultimate
bearing capacit y o f stri p footings by taking into account cohesion, frictio n and weight of soil , and
replacing th e overburde n pressur e wit h a n equivalen t surcharg e loa d a t th e bas e leve l o f th e
foundation. Thi s metho d wa s fo r th e genera l shea r failur e conditio n an d th e principl e o f
superposition wa s adopted . Hi s wor k wa s a n extensio n o f th e wor k o f Prandt l (1921) . Th e fina l
form o f the equation proposed b y Terzaghi i s the same a s the one given by Prandtl .
Subsequent t o the wor k by Terzaghi , man y investigators became interested i n thi s proble m
and presente d thei r ow n solutions . Howeve r th e for m o f th e equatio n presente d b y al l thes e
investigators remained th e same a s that of Terzaghi, but thei r methods o f determining the bearing
capacity factor s were different .
Of importanc e i n determinin g th e bearin g capacit y o f stri p footing s i s th e assumptio n o f
plane strai n inheren t i n the solutions o f strip footings. The angl e of internal frictio n a s determined
under an axially symmetric triaxia l compression stres s state, 0
f
, is known to be several degrees less
than that determined under plane strain conditions under low confining pressures . Thus the bearing
capacity o f a stri p footin g calculate d b y th e generall y accepte d formulas , using 0
r
, i s usuall y less
than th e actua l bearin g capacit y a s determine d b y th e plane strai n footin g test s whic h lead s t o a
conclusion that the bearing capacity formulas are conservative
The ultimat e bearing capacity , or the allowabl e soi l pressure , ca n b e calculate d eithe r fro m
bearing capacit y theorie s o r fro m som e o f the i n situ tests . Eac h theor y ha s it s own good an d bad
points. Some of the theories are of academic interes t only. However, it is the purpose o f the author
to presen t her e onl y suc h theorie s whic h ar e o f basi c interes t t o student s i n particula r an d
professional engineer s i n general . Th e applicatio n o f fiel d test s fo r determinin g bearin g capacit y
are als o presente d whic h ar e o f particula r importanc e t o professiona l engineer s sinc e presen t
practice i s t o rel y mor e o n fiel d test s fo r determinin g th e bearin g capacit y o r allowabl e bearin g
pressure o f soil .
Some o f the methods tha t are discussed i n this chapter ar e
1. Terzaghi' s bearing capacit y theor y
2. Th e general bearin g capacit y equatio n
3. Fiel d test s
488
12.6 TERZ AG H I ' S B EARI NG CAPACI TY TH EORY
Chapt er 1 2
Terzaghi (1943 ) use d th e same for m o f equation as propose d b y Prandt l (1921 ) an d extende d hi s
theory t o take int o account the weight of soi l and the effect o f soil abov e the base of the foundation
on th e bearin g capacit y o f soil . Terzagh i mad e th e followin g assumption s fo r developin g a n
equation for determining q
u
for a c-0 soil.
(1) The soi l is semi-infinite, homogeneous and isotropic, (2) the problem is two-dimensional,
(3) the base of the footing i s rough, (4 ) the failur e is by genera l shear , (5) the loa d i s vertical and
symmetrical, (6) the ground surface is horizontal, (7) the overburden pressure at foundation level is
equivalent to a surcharge load q'
0
= yD^ wher e y is the effective unit weight of soil, and D,, the depth
of foundation les s than the width B of the foundation, (8) the principle of superposition i s valid, and
(9) Coulomb' s la w i s strictl y valid, that is, <J - c + crtan</>.
M echanism o f Failure
The shapes of the failur e surface s under ultimate loading conditions ar e given in Fig. 12.6. The zones
of plastic equilibrium represented i n this figure b y the area gedcfmay b e subdivided int o
1 . Zon e I of elasti c equilibrium
2. Zone s I I of radial shear stat e
3. Zone s II I of Rankine passive stat e
When loa d q
u
per uni t area actin g on the base of the footing o f widt h B wit h a rough bas e i s
transmitted int o th e soil , th e tendenc y o f th e soi l locate d withi n zon e I i s t o sprea d bu t thi s i s
counteracted b y frictio n an d adhesio n betwee n th e soi l an d th e bas e o f th e footing . Du e t o th e
existence of this resistance against lateral spreading, th e soil locate d immediatel y beneat h the bas e
remains permanently i n a state of elastic equilibrium, and the soil located withi n this central Zone I
behaves a s i f it wer e a part of the footing and sink s with the footing under the superimpose d load .
The dept h o f thi s wedg e shape d bod y o f soi l abc remain s practicall y unchanged , ye t th e footin g
sinks. Thi s proces s i s onl y conceivabl e i f th e soi l locate d jus t belo w poin t c move s verticall y
downwards. This typ e of movement requires that the surface of sliding cd (Fig. 12.6 ) through point
c shoul d star t from a vertical tangent . The boundar y be of the zone o f radial shea r bed (Zon e II ) is
also the surface of sliding. As per the theory of plasticity, the potential surfaces of sliding in an ideal
plastic materia l intersec t eac h othe r i n ever y poin t o f th e zon e o f plasti c equilibriu m a t a n angl e
(90° - 0) . Therefore th e boundary be must rise a t an angle 0 to the horizontal provide d the friction
and adhesion between th e soi l and the base of the footing suffice t o prevent a sliding motion a t the
base.
45°-0/2 45°-0/2
Logarithmic spira l
Figure 1 2. 6 Gen era l s hear fail ur e s urfac e a s as s um e d b y T erzag h i fo r a s t rip
foot in g
Shallow Foun dat io n I : Ult im at e Bearin g Capacit y 48 9
The sinking of Zone I creates two zones of plastic equilibrium, II and III, on either side of the
footing. Zon e I I i s the radia l shea r zon e whose remot e boundarie s bd an d a/meet th e horizontal
surface at angles (45° - 0/2) , whereas Zone III is a passive Rankine zone. The boundaries de and/g
of these zones are straight lines and they meet the surface at angles of (45° - 0/2) . The curved parts
cd an d c f i n Zon e I I ar e part s o f logarithmi c spiral s whos e center s ar e locate d a t b an d a
respectively.
Ultimate B earin g Capacit y o f Soi l
Strip Footings
Terzaghi develope d hi s bearing capacit y equation for strip footings by analyzing the forces acting
on the wedge abc i n Fig. 12.6 . The equation for the ultimate bearing capacit y q
u
i s
where Q
ult
= ultimate load per unit length of footing, c = unit cohesion, /th e effective uni t weight of
soil, B = width of footing, D,= depth of foundation, N
c
, N
q
an d N
y
ar e the bearing capacit y factors .
They ar e functions o f the angl e of friction, 0 .
The bearing capacit y factor s are expressed b y the following equations
N =
2 cos
2
(45°+0/ 2)
:
a
~ = <?
/
7
t a n
«*. n = ( Q.75n:-d/2)
(12.7)
q
2 cos
2
(45°+0/ 2)
where a = e
r
> **
n
*,
N
v
=-
r
2
where K
p
= passive eart h pressure coefficient
Table 12. 1 give s th(
same i n a graphical form.
Table 12. 1 give s the value s of N
r
, N an d N fo r variou s values of 0 and Fig. 12. 7 gives the
Table 12. 1 B earin g capacit y fact or s o f T erzag h i
0° N
C
N N Y
0
5
10
15
20
25
30
35
40
45
50
5.7
7.3
9.6
12.9
17.7
25.1
37.2
57.8
95.7
172.3
347.5
1.0
1.6
2.7
4.4
7.4
12.7
22.5
41.4
81.3
173.3
415.1
0.0
0.14
1.2
1.8
5.0
9.7
19.7
42.4
100.4
360.0
1072.8
490 Chapt er 1 2
o
SO C O —
V alues of N
c
o o o
(N r o r f
o
o o o
vO O O —
HO
c/j 4 0
<u
2
S
1
1 5
3
-© -
a? J "
o
03
•" O C
W5
2
tiD i n
03
J= I S
•5 °
<*-,
O
01 1 n
00
c
< s
V i
c
/
3
3
*•
o
C
'
O
5
/
/
^
i
/
/
q r
N
c
/
^
\
/
^ C ^
y
/
X
/
o T
/
^*^
t
/
,
/
V J
/
-
/
0
''
s
oo
/
y^
/
0 C
» — r
X
"
3 C
^ r
^
2 C
1 r
3 C
f I /
^
3
1
X
v,
x
t
\ /
s
s
|
^
y
2 C
5 C
H r
^^
3 C
3 C
-J o
3 C
3 C
1 T
3 C
3 C
t i /
§ S
1 C
V alues of N an d
Figure 1 2. 7 T er zag hi' s bearin g capacit y fact or s fo r g en era l s hea r fail ur e
Equations fo r Square , Circular , an d Rectangula r Foundation s
Terzaghi's bearin g capacit y Eq . (12.6 ) ha s bee n modifie d fo r othe r type s o f foundation s b y
introducing the shape factors . The equation s ar e
Square Foundations
q
u
= l.3cN
c
+ yD
f
Circular Foundation s
q
u
= \ 3cN
c
+ YD
f
N
q
+ 0.3yBN
y
Rectangular Foundations
(12.8)
(12.9)
D
: Yj +/ D
/
A^+- r BA^l - 0. 2x- J (12.10 )
where B — width or diameter , L = lengt h of footing.
U ltimate B earin g Capacit y fo r L oca l Shea r Failur e
The reason s a s to why a soil fail s under local shea r hav e been explaine d unde r Sectio n 12.4 . When
a soi l fail s b y loca l shear , th e actua l shea r parameter s c an d 0 ar e t o b e reduce d a s pe r Terzagh i
(1943). Th e lower limitin g values of c and 0 are
Shallow Foun dat io n I : Ult im at e B earin g Capacit y 49 1
c = 0.67 c
and ta n 0 = 0.67 tan </> o r 0 = tan'
1
(0.67 tan <fi) (12.11 )
The equations for the lower bound values for the various types of footings ar e as given below.
Strip Foundatio n
q
u
= 0.61cN
c
+yD
f
N
c/
+ yBN
y
(12.12 )
Square Foundatio n
q
+ OAyBN
y
(12.13 )
Circular Foundatio n
q
u
= O.S61cN
c
+yD
f
N
( ]
+ 03yBN
y
(12.14 )
Rectangular Foundatio n
q
u
=Q.61c l + 0.3x|N
C
+yD
f
N
q
+± yBN
y
l-02x |
(12
.1
5
)
where N , N an d N ar e the reduced bearing capacity factors for local shear failure. These factor s
may b e obtaine d either from Tabl e 12. 1 or Fig. 12. 7 by makin g use o f the frictio n angl e ( f> .
Ultimate B earin g Capacit y q
u
i n Purel y Cohesionles s an d Cohesiv e Soil s
Under G enera l Shea r Failur e
Equations for the various types of footings for ( c - 0 ) soi l unde r general shea r failur e have been
given earlier. The same equations can be modified to give equations for cohesionless soil (for c = 0)
and cohesiv e soil s (for </ > = 0) a s follows.
It may be noted here that for c = 0, the value of N
c
= 0, and for 0=0, the value of N
C
= 5. 7 for
a stri p footing and N = 1 .
a) Stri p Footing
F o r c = 0, q
u
=yD
f
N
q
+-yBN
y
(12.16 )
For 0 = 0, q
u
=5.7c + yD
f
b) Squar e Footing
For c = 0, q
u
= yD
f
N
q
+OAyBN
r
(12.17 )
For 0 = 0, q
u
= 7.4c + yD
f
c) Circula r Footin g
For c = 0, q
u
= yD
f
N
q
+Q3yBN
y
(12.18 )
492 Chapt er 1 2
For 0 = 0, q
u
= 7.4c + yD
f
d) Rectangul a r Footin g
For c = 0, q
ii
=yD
f
N
q
+-yBN
Y
l -0. 2 x —
(12.19)
For 0- 0 , o =5.7 c l + 0. 3x— + YD,
" L
f
Similar type s o f equation s as presented fo r genera l shea r failur e can b e develope d fo r loca l
shear failur e also .
Transition fro m L ocal to G enera l Shear Failure in Sand
As already explained, loca l shear failur e normally occurs i n loose an d general shea r failur e occur s
in dens e sand . There i s a transition from loca l t o general shea r failur e as the stat e of san d change s
140
120
100
\
1 ' 1 '
^
/ery Loose
- Loos e
Medium
\
N
m
-
•^
\
>
y
/
"\ ^
^v
/
/
Dense V er y Dense
\
/
I/
V
\ ^
N
/
//
/
\
\
^
/
/
j
A
s
\
\
o
o
^
j
a
-
\
L
f
t
-
f
^
(
j
J
K
>
^
o
o
o
o
o
o
o
o
o
S
t
a
n
d
a
r
d

p
e
n
e
t
r
a
t
i
o
n

t
e
s
t

v
a
l
u
e
,

N
c
o
r
60
40
20
°28 3 0 3 2 3 4 3 6 3 8 4 0 4 2 4 4 4 6
Angle of internal friction, 0
Figure 12. 8 T er zag hi' s bearin g capacit y fact or s whic h t ak e car e of m ix e d s t at e o f
l ocal an d g en eral s hear fail ures i n s an d (Pec k e t al. , 1974)
Shallow Foun dat io n I : Ult im at e B earin g Capacit y 49 3
from loos e t o dense condition . There i s no bearing capacit y equation to account for thi s transition
from loos e to dense state. Peck et al. , (1974) hav e given curves for N an d N whic h automaticall y
incorporate allowance for the mixed stat e of local and general shea r failures as shown in Fig. 12.8 .
The curve s for N an d N ar e developed o n the following assumptions.
1. Purel y local shear failure occurs when 0 < 28°.
2. Purel y general shear failur e occurs whe n 0 > 38°.
3. Smoot h transition curves for values of 0 between 28° and 38° represent th e mixed stat e of
local an d general shea r failures.
N an d N
y
fo r value s of 0 > 38° are as given in Table 12 . 1 . V alues of N
q
an d N
y
fo r 0 < 28°
may be obtaine d from Table 12 . 1 by making use o f the relationshi p ( f> - tan"
1
(2/3) tan tf> .
In th e cas e o f purel y cohesive soi l loca l shea r failur e may b e assume d t o occu r i n sof t t o
medium stif f cla y wit h an unconfined compressive strengt h q
u
< 10 0 kPa.
Figure 12. 8 als o give s th e relationshi p between SP T valu e N
cor
an d th e angl e o f internal
friction 0 by means of a curve. This curve is useful t o obtain the value of 0 when the SPT value is
known.
Net Ultimat e B earin g Capacit y an d Safet y Facto r
The net ultimate bearing capacity q
nu
is defined as the pressure at the base level of the foundation in
excess of the effective overburden pressure q'
Q
= yD,as defined in Eq. (12.3). The net q
nu
for a strip
footing i s
Similar expressions ca n be written for square, circular, and rectangular foundations and als o
for loca l shea r failur e conditions.
Allowable Bearing Pressure
Per Eq. (12.4), the gross allowabl e bearing pressure is
1a
=
^~ (12.21a )
In the same way the net allowable bearing pressur e q
na
is
a
Vu-yDf q
nu
Vna - - p - - -p- (12.21b )
i S
where F
s
= factor of safet y which is normally assumed as equal to 3.
12.7 SKEM PTON' S B EARI NG CAPACI T Y FACTOR N
C
For saturated cla y soils, Skempto n (1951) proposed th e following equation fo r a strip foundation
q
u
=cN
c
+yD
f
(12.22a )
or
4
m
=4
U
- y°=
cN
c (12.22b )
494 Chapt er 1 2
Figure 12. 9 Sk em pt on ' s bearin g capacit y fact o r N
c
fo r cl a y s oil s
<lna
=
-p-
=
-J -(12.22C )
s s
The N
c
value s for stri p and square (or circular) foundations as a function of the DJ B rati o are
given i n Fig. 12.9 . Th e equatio n for rectangular foundation may be writte n as follows
0.84 +0.16 x
(12.22d)
where ( N
C
)
R
= N
C
fo r rectangula r foundation, ( N
c
)
s
= N
c
fo r squar e foundation .
The lowe r an d upper limiting values of N
c
fo r strip and squar e foundations may b e written as
follows:
Type o f foundation
Strip
Square
Ratio D
f
IB
0
>4
0
>4
V alue ofN
c
5.14
7.5
6.2
9.0
1 2.8 EFFEC T O F WATER TAB L E ON B EARI NG CAPACI T Y
The theoretica l equation s develope d fo r computin g th e ultimat e bearin g capacit y q
u
o f soi l ar e
based on the assumption tha t the water tabl e lies at a depth belo w th e base of the foundation equa l
to o r greate r tha n th e widt h B o f th e foundatio n o r otherwis e th e dept h o f th e wate r tabl e fro m
ground surfac e i s equal t o or greater tha n (D,+ B) . I n case th e wate r tabl e lie s a t any intermediat e
depth less than the depth ( D,+ B) , the bearing capacity equations are affected due to the presence of
the water table .
Shallow Foun dat io n I : Ult im at e B earin g Capacity 49 5
Two cases may be considered here .
Case 1. When the water tabl e lies abov e the base of the foundation.
Case 2. When the water tabl e lie s withi n depth B below th e base o f the foundation.
We wil l conside r th e two method s fo r determinin g th e effec t o f th e wate r tabl e o n bearin g
capacity a s given below.
M ethod 1
For any position of the water tabl e within the depth (ZX + B), we may writ e Eq. (12.6) as
q
u
=cN
c
+rD
f
N
q
R
wl+
± yBN
y
R
w2
(12.23 )
where R
wl
= reduction factor for water table above the base leve l of the foundation,
R
w2
= reduction factor for water table below the base leve l of the foundation,
7 = 7
sat
for al l practical purpose s i n both the second an d third terms o f Eq. (12.23).
Case 1 : Whe n th e wate r tabl e lie s abov e th e bas e leve l o f th e foundatio n or whe n D
wl
/D
f
< 1
(Fig. 12.10a ) the equation for R
wl
ma y be written as
For D
wl
/D
f
= 0 , we have R
wl
= 0.5, an d for D
wl
/D
f
= 1.0 , we have R
wl
= 1.0.
Case 2: When the water table lies below the base level or when D
w2
/B < 1 (12.1 Ob) the equation for
(12
'
24b)
For D
w2
/B = 0, we have R
w2
= 0.5, an d for D
w2
/B = 1.0, we have R
w2
= 1.0.
Figure 12.1 0 shows in a graphical for m the relations D
wl
/D,vs. R
wl
an d D
w2
/B v s R
w2
.
Equations (12.23a) and (12.23b) ar e based o n the assumption that the submerged uni t weight
of soi l i s equa l t o hal f o f th e saturate d uni t weigh t an d th e soi l abov e th e wate r tabl e remain s
saturated.
M ethod 2 : Equivalent effective unit weight metho d
Eq. (12.6) for the stri p footing may be expressed a s
V
u
=
cN
c
+
Yei
D
f
N
q
+\ r
e2
BN
Y
(12.25 )
where y
el
= weighted effectiv e uni t weight of soi l lying above the base level of the foundation
X?2
=
weighte d effectiv e uni t weigh t o f soi l lyin g withi n th e dept h B belo w th e bas e
level of the foundation
7 = moist o r saturated uni t weight of soil lyin g above WT (cas e 1 or case 2)
496 Chapt er 1 2
GWT
T
i
D
D
f
l . U
0.9
0.8
0.7
0.6
0.
5(
/
/
/
/
/
) 0. 2 0. 4 0. 6 0. 8 1
(a) Water tabl e above bas e leve l of foundation
D
f
GWT
i . U
0.9
0.8
0.7
0.6
0.5
(
^/
/
/
/
/
) 0. 2 0. 4 0. 6 0. 8 1
Submerged
(b) Water tabl e below base level of foundation
Figure 1 2 .1 0 Effec t o f WT o n bearin g capacit y
7
sat
= saturated uni t weight of soi l below th e WT (cas e 1 or case 2)
Y
b
= submerged uni t weight of soil = y
sat
- Y
w
Case 1
An equatio n fo r y
el
ma y b e writte n as
+
D^
' e\ ' b £)
f
(12.26a)
Case 2
Y
e
\ ~ y m
Shallow Foun dat io n I : Ult im at e B earin g Capacit y 497
Av2 /
' e2 > b f t \ 'm
v \
-Yb)
(12.26b)
Example 1 2. 1
A stri p footing of width 3 m is founded at a depth of 2 m below the ground surface i n a ( c - 0 ) soil
having a cohesion c = 30 kN/m
2
and angl e of shearing resistanc e 0 = 35°. The wate r tabl e is at a
depth of 5 m below groun d level . Th e mois t weigh t of soi l above th e wate r tabl e i s 17.2 5 kN/m
3
.
Determine (a ) the ultimate bearing capacity o f the soil , (b) the net bearing capacity, and (c) the net
allowable bearing pressure an d the load/ m for a factor of safet y of 3. Use the genera l shea r failur e
theory o f Terzaghi.
Solution
2 m
5 m
0
1 CO
= 35
Y= 17.2 5 kN/m
3
c = 30 kN/m
2
Figure Ex. 12. 1
Fo r 0 =35°, N
c
= 57.8, N =41.4, and N
y
= 42.4
Fro m Eq. (12.6),
= 30 x 57.8 + 17.25 x 2 x 41.4 + - x 17.25 x 3 x 42.4 = 4259 kN/m
2
^ =q
u
-yD
f
= 425 9 -17.25 x 2 -4225 kN/m
2
=
= 1408 x 3=4225 kN/m
498 Chapt e r 12
Example 1 2 . 2
If th e soi l i n Ex . 12. 1 fail s b y local shea r failure , determin e th e ne t saf e bearing pressure . Al l th e
other dat a given i n Ex. 12. 1 remain the same .
Solution
For local shea r failure :
0 = tan "'0.67 tan 35° =25°
c = 0.67'c = 0.67 x 30 = 20 kN/ m
2
From Tabl e 12.1 , fo r 0 = 25°, N
c
= 25.1, N = 12.7, N
y
= 9.7
Now fro m Eq . (12.12)
q
u
= 20 x 25.1 +17.25 x 2x 12.7 + -x 17.25x3x9.7 = 1191 kN/ m
2
q
m
=1191-17.25x 2 =1156.5 kN/m
2
1156.50
*« = — = 385. 5 kN/m
2
Q
a
=385. 5x3 = 1156. 5 kN/m
Example 1 2 . 3
If the water table i n Ex. 12. 1 rises to the ground level, determine the net safe bearing pressur e of the
footing. Al l the other dat a given in Ex. 12. 1 remai n the same. Assume the saturated uni t weight of
the soi l y
s a t
= 18. 5 kN/m
3
.
Solution
When th e WT i s at ground level we have t o use the submerged uni t weight of the soil .
Therefore y , = y - y = 18.5 -9.81 = 8.69 kN/m
3
' u ' Sa l * W
The ne t ultimat e bearing capacity i s
q
nu
= 30 x 57.8 + 8.69 x 2(41.4 - 1 ) + - x 48.69 x 3 x42.4 «2992 kN/m
2
2992
q
m
=—^- = 997.33 kN/ m
2
Q =997. 33x 3 = 2992 kN/m
Example 1 2. 4
If th e wate r tabl e i n Ex . 12. 1 occupies an y o f th e position s (a ) 1.2 5 m belo w groun d leve l o r
(b) 1.2 5 m below th e base leve l of the foundation, what will be the ne t saf e bearing pressure ?
Assume y
sat
= 18. 5 kN/m
3
, /(above WT) = 17. 5 kN/m
3
. All the other data remai n the same as
given i n Ex. 12.1 .
Shallow Foun dat io n I : Ult im at e B earing Capacit y 49 9
Solution
Method 1 —By making use of reduction factor s R
wl
an d R
w2
an d using Eqs. (12.20 ) an d (12.23) ,
we may writ e
1
nu c f q w 2 '
Given: N = 41.4, N
y
= 42.4 an d N
c
= 57.8
Case I—When the WTis 1 .25 m below the GL
From Eq. (12.24), we get R
wl
= 0.813 fo r D
w
/D
f
= 0.625 , R
w2
= 0.5 for D
w2
/B = 0.
By substitutin g the known values in the equation for q
nu
, we have
q
m
= 30x57.8 + 18.5x2x40.4x0.813 + -xl 8. 5x3x42. 4x0.5 = 3538 kN/m
2
= 1179 kN/m
2
3
Case 2—When the WTis 1 .25 m below the base of the foundation
R , = 1. 0 for D ,/ZX = 1 , R , = 0.71 fo r D J B = 0.42.
wl w l / ' w 2 w2
Now the net bearing capacit y is
q
m
= 30x57.8 + 18. 5x2x40. 4xl + -xl8.5x3x42.4x0.71 = 4064 kN/m
2
= 135 5
Method 2 —Using the equivalent effective uni t weight method .
Submerged uni t weight y
b
= 18.5 - 9.8 1 = 8.69 kN/m
3
.
Per Eq. (12.25 )
The net ultimate bearin g capacit y i s
q =c N + y , D
f
( N -l) + -y
7
BN
V
"nu c ' el f ^ q ' n ' e2 y
Case I—When D
wl
= 7.25 m (Fig. Ex. 1 2.4)
From Eq . (12.26a )
f
where y - y . = 18.5 kN/m
3
' 77 1 * Sal
125
y
el
= 8.69 +— (18.5-8.69) = 14.82 kN/m
3
r
e2
=r
b
=8.6 9 kN/m
3
o = 30 x57.8 + 14.82 x 2 x40.4 + - x 8.69 x 3 x42.4 = 3484 kN/m
2
500 Chapt er 1 2
GL
"1
2
1
m
J
'
s S/K
L
-* —
- 3 m-
_
H
C,., = 1.25
m
1 . T f
t
D
w2
= 1. 25 m
T
1
K\
Case 1
Case 2
Figure Ex . 12. 4 Effec t o f WT o n bearin g capacit y
3484
Case 2—When D
w2
=1 .25 m (Fig. Ex. 1 2.4)
FromEq. (12.26b )
y=y =18. 5kN/ m
3
*el ' m
1.25
y , = 8.69 + —-(18.5- 8.69) = 12.78 kN/m
3
€ L 4.
= 30x57.8 +18. 5x2x40. 4+ -x 12.78x3x42.4 = 4042 kN/m
2
4042
= 1347 kN/m
2
Example 1 2 . 5
A squar e footin g fail s b y genera l shea r i n a cohesionles s soi l unde r a n ultimat e loa d o f
Q
uh
- 1687. 5 kips . The footing i s placed at a depth of 6.5 ft below ground level. Given 0 = 35°, and
7=110 Ib/ft
3
, determin e the size of the footing i f the water table is at a great dept h (Fig . Ex. 12.5) .
Solution
For a square footing [Eq. (12.17) ] for c = 0, we have
For 0= 35° , N
q
= 41.4, an d W
y
= 42.4 fro m Tabl e 12.1 .
Shallow Foundatio n I : Ult im at e B earin g Capacit y 501
6. 5f t
Q
ul!
= 1687 . Ski p s
0 = 35°, c = 0
y = 11 0 lb/ft
3
BxB - |
Figure Ex . 12. 5
_ Q
u
_ 1687.5xlO
3
qu
~~B^~ 5 2
By substitutin g known values , we have
1 87
'
5xl
— = 110 x 6.5 x 41.4 + 0.4 x110 x 42.45
/?
= (29.601 +1.8665)10
3
Simplifying an d transposing, we have
5
3
+ 15.8635
2
-904.34 = 0
Solving thi s equation yields , 5 = 6.4 ft.
Example 1 2. 6
A rectangular footin g of size 1 0 x 2 0 f t i s founded at a depth o f 6 ft below th e groun d surfac e i n a
homogeneous cohesionles s soi l having an angle of shearing resistance 0 = 35°. The water table is at
a great depth. The unit weight of soil 7= 11 4 lb/ft
3
. Determine: (1) the net ultimate bearing capacity ,
y = 11 4 lb/ft
3
- 10x2 0 ft
Figure Ex . 12. 6
502 Chapt e r 1 2
(2) the net allowable bearing pressure for F
V
= 3, and (3) the allowable load Q
a
the footing can carry.
Use Terzaghi' s theory . (Refer to Fig. Ex. 12.6)
Solution
Using Eq . (12.19 ) and Eq. (12.20 ) fo r c - 0 , th e net ultimat e bearing capacit y fo r a rectangula r
footing i s expressed a s
From Tabl e 12.1 , N = 41.4, N
y
= 42.4 fo r 0 = 35°
By subst i t ut i n g the known values,
q =114x6(41. 4- l ) + - x l l 4 x l Ox 4 2 .4 l - 0. 2x — =49,38 5
2 2 0
49 385
q
na
= — = 16,462 lb/ft
2
Q
a
= ( B x L)q
na
= 10 x 20 x 16,462- 3,292 x l O
3
I b = 3292 kip s
Example 1 2 . 7
A rectangula r footin g of siz e 1 0 x 2 0 f t i s founde d at a dept h o f 6 f t belo w th e groun d leve l i n a
cohesive soi l ( 0 = 0) whi ch fails b y general shear . Given: y
sal
=114 lb/ft
3
, c = 945 lb/ft
2
. Th e wate r
table i s clos e t o th e groun d surface . Determin e q , q an d q
na
b y (a ) Terzaghi's method , an d (b )
Skempton's method . Us e F
v
= 3.
Solution
(a) Terzaghi' s metho d
Use Eq. (12.19)
For 0=0° , N
c
= 5.7, N = I
q
u
=cN
c
l+0. 3x |+y
h
D
f
Substituting th e known values ,
q
u
=945x5. 7 l + 0.3x— +(114- 62. 4)x 6 ='6,504 lb/ft
2
q
m
= ( q
u
~ y
b
D
f
) = 6504 - (11 4 - 62.4 ) x 6 = 6195 lb/ft
2
lb/ft
2
™ F
y
3
(b) Skempton' s method
From Eqs . (12.22a ) an d (12.22d ) we ma y write
Shal l ow Foundat i o n I : Ul t i ma t e Bear i n g Capaci t y 50 3
where N
cr
= bearing capacit y factor for rectangular foundation.
N
cr
= 0.8 4 + 0.16 x — xN
where N
cs
= bearing capacit y factor for a square foundation.
From Fig. 12.9, N
cs
= 7.2 for D
f
/B = 0.60.
Therefore N
c
= 0.8 4 + 0.16 x— x 7.2 = 6.62
20
Now q
u
= 945 x 6.62 + 1 14 x6 = 6940 lb/ft
2
q
nu
=( q
u
-YD") = 6940 -114x6 = 6,256 lb/ft
2
= 2
Note: Terzaghi's an d Skempton' s values are i n close agreement fo r cohesive soils .
Example 1 2.8
If th e soi l i n Ex . 12. 6 is cohesionles s ( c = 0) , an d fail s i n loca l shear , determin e (i ) th e ultimat e
bearing capacity , (ii ) the ne t bearing capacity, and (iii ) the net allowabl e bearing pressure . Al l the
other dat a remain the same .
Solution
From Eq . (12.15) an d Eq. (12.20), the net bearing capacit y for local shea r failur e for c - 0 is
q^bu-YDf^YDfWq-D + ^YBN y l~0.2x |
where JZ > = tan"
1
0.67 tan 35° - 25°, J V = 12.7, an d N = 9.7 fo r 0 = 25° fro m Tabl e 12.1.
By substitutin g known values, we have
tf
n
=114x6(12.7-l) + - x l l 4x l Ox 9.7 l-0.2 x — = 12,979 lb/ft
2
tnu
2
2 0
12979
V = - = 4326 lb/ft
2
1 2.9 TH E G ENERAL B EARI N G CAPACI T Y EQU ATI ON
The bearin g capacit y Eq . (12.6 ) develope d b y Terzaghi i s for a stri p footing unde r genera l shea r
failure. Eq . (12.6 ) ha s bee n modifie d fo r othe r type s o f foundation s such a s square , circula r an d
rectangular b y introducin g shap e factors . Meyerho f (1963 ) presente d a genera l bearin g capacit y
equation whic h take s int o accoun t th e shap e an d th e inclinatio n of load . Th e genera l for m o f
equation suggeste d by Meyerhof fo r bearing capacit y is
504 Chapt e r 1 2
where c - uni t cohesion
g'
o
= effectiv e overburde n pressure a t the bas e leve l of the foundatio n = Y®f
y = effectiv e uni t weigh t above the base leve l of foundatio n
7 = effectiv e uni t weight of soi l below th e foundation base
D, = dept h of foundation
s
c
, s , s = shap e factors
d
c
, d , d = dept h factor
/
c
, / , i = loa d inclination factors
B = widt h of foundatio n
N
c
, N , N = bearin g capacit y factors
Hansen (1970 ) extende d th e wor k o f Meyerho f b y includin g in Eq . (12.27 ) two additiona l
factors t o take care of base tilt and foundations on slopes. V esic (1973, 1974 ) use d the same form of
equation suggested by Hansen. All three investigators use the equations proposed b y Prandtl (1921)
for computin g th e values of N
c
an d N wherei n the foundation base i s assumed a s smoot h wit h the
angle a = 45° + 0/2 (Fig. 12.6) . However , the equations used by them for computing the values of
N ar e different . Th e equation s for N
c
, N an d N ar e
Table 1 2. 2 T h e val ue s o f A/ , /V . and M eyerho f (M ) , Han s e n (H) an d Ves i c (V ) N
c q 7
Fact ors
0
0
5
10
15
20
25
26
28
30
32
34
36
38
40
45
50
N
c
5.14
6.49
8.34
10.97
14.83
20.71
22.25
25.79
30.13
35.47
42.14
50.55
61.31
72.25
133.73
266.50
N
q
1.0
1.6
2.5
3.9
6.4
10.7
11.8
14.7
18.4
23.2
29.4
37.7
48.9
64.1
134.7
318.50
/V
X
(H)
0.0
0.1
0.4
1.2
2.9
6.8
7.9
10.9
15.1
20.8
28.7
40.0
56.1
79.4
200.5
567.4
/V
r
(M)
0.0
0.1
0.4
1.1
2.9
6.8
8.0
11.2
15.7
22.0
31.1
44.4
64.0
93.6
262.3
871.7
A/
y
(V)
0.0
0.4
1.2
2.6
5.4
10.9
12.5
16.7
22.4
30.2
41.0
56.2
77.9
109.4
271.3
762.84
Note: N
C
an d N ar e th e sam e fo r al l th e thre e methods . Subscript s identif y th e autho r fo r N .
Shallow Foundation I: Ult im at e B earin g Capacit y 505
T able 12. 3 Shape , dept h an d load in clin at io n fact or s o f M eyerhof , Han s e n an d
V esic
Fact ors M eyerhof Han s en Ves ic
1 + 0.2N , —
<* L
1 + 0.1A^ — fo r <f»\ Q°
L
s = s fo r 0> IQ°
s — s = 1 fo r 0 = 0
i— D
f
/ —
D
f
1 + O.UA^ —*- fo r
d
y
= d
q
fo r </> > 10°
d=d = l fo r ^ =
1
a
*
1 fo r any
90
i
q
= i
c
fo r any 0
1- — fo r
i
r
=0 fo r ^ =
5
1 + — tan^
1-0.4-
L
1 + 0.4
B
1 fo r al l <f>
Note; V esic' s s and Jfactors
= Hansen' s s and d factors
f o r 0>0
g
N - 1
i T
±
0.5 1--^-
2
for 0 = 0
The shap e an d dept h factor s
of V esic are the same as thos e
of Hansen .
Same a s Hansen fo r > 0
1--
N
y
= ( N
q
- 1) tan(1.40) (Meyerhof )
N
y
= l.5( N
q
- 1) tan 0(Hansen )
N
y
=2( N
q
+l)ten</> (V esic )
Table 12.2 give s th e value s o f th e bearin g capacit y factors . Equation s fo r shape , dept h an d
inclination factors ar e given in Table 12.3 . The tilt of the base and the foundations on slopes ar e not
considered here .
In Table 12. 3 The following terms ar e defined wit h regard t o the inclination factor s
Q
h
= horizonta l componen t o f the inclined loa d
Q
u
= vertica l componen t of the inclined load
506 Chapt e r 1 2
c
a
= uni t adhesion o n the bas e o f the footing
A, = effectiv e contac t area o f the footin g
_
m
~
m
~
_
m
~
m
~
_
B ~ i
+B/L
wit h Q
h
parallel t o B
2 + B/L
l +B/L
wit h Q
h
parallel to L
The genera l bearin g capacit y Eq . (12.27 ) ha s no t take n int o account the effec t o f th e wate r
table positio n o n th e bearin g capacity . Hence , Eq . (12.27 ) ha s t o b e modifie d accordin g t o th e
position o f water level i n the same wa y as explained in Section 12.7 .
V alidity o f the B earin g Capacit y Equation s
There is currently no method of obtaining the ultimate bearing capacit y of a foundation other than
as an estimate (Bowles, 1996) . There has been littl e experimental verification of any of the methods
except b y usin g mode l footings . Up t o a dept h o f D
f
~ B th e Meyerho f q
u
i s no t greatl y differen t
from th e Terzagh i valu e (Bowles , 1996) . Th e Terzagh i equations , being th e firs t proposed , hav e
been quit e popular wit h designers. Bot h the Meyerhof and Hansen method s ar e widel y used. Th e
V esic method has not been much used. It is a good practice t o use at least two methods and compare
the computed value s of q
u
. If the two values do not compare well , use a third method.
Example 1 2. 9
Refer t o Example 12.1 . Compute using the Meyerhof equation (a) the ultimate bearing capacit y of
the soil , (b) the ne t bearing capacity, and (c ) the net allowabl e bearing pressure. Al l the other dat a
remain th e same .
Solution
Use Eq . (12.27) . For i = 1 the equation for net bearing capacit y i s
- \ }s
q
d
q
From Table 12. 3
D
s
c
- 1 + 0.2A — = 1 fo r strip footing
n
5 = 1 + Q.\ N — = 1 fo r strip footing
JLv
d = l + 0. 2 / v7 — -= l + 0.2tan 45°+ — - =1.25 7
B
^ c r\
= 1 + 0.1 tan 45 °+ — - =1.12 9
2 3
d
y
=d
q
=U29
Shallow Foun dat io n I : Ult im at e B earin g Capacity 50 7
From Ex . 12.1 , c= 30 kN/m
2
, y = 17.25 kN/m
3
, D
f
= 2 m , B = 3 m.
From Table 12. 2 for 0= 35 ° we have A/
c
= 46.35, A T = 33.55, N
y
= 37.75 . Now substituting the
known values , we have
q
m
= 30 x 46.35 x 1 x 1.257 + 17.25 x 2 x (33.55 -1) x 1 x 1.129
+ - x 17.25 x 3 x 37.75 x 1 x 1.129
2
= 1,748 + 1,268 + 1,103 = 4,1 19 kN/m
2
1na=—^ = 1373 kN/m
2
There i s very close agreemen t between Terzaghi' s an d Meyerhof 's methods .
Example 1 2 .1 0
Refer t o Example 12.6 . Compute by Meyerhof 's method the net ultimate bearing capacit y and the
net allowabl e bearing pressur e for F
s
= 3. All the other dat a remain the same .
Solution
From Ex. 12. 6 we have B = 10 ft, L = 20 ft, D
f
= 6 ft, and y= 1 14 lb/ft
3
. Fro m Eq. (12.27) for c = 0
and / = 1 , we have
From Table 12.3
D O C 1 f\
s
a
= 1 +0.1AL - = 1 + 0.1 tan
2
45°+ — — =1.18 5
q
* L 2 2 0
d = l + O.L/A/7 -^ = l + 0.1tan 45°+ — — =1.11 5
v V 0 B 2 1 0
^=^=1. 115
From Tabl e 12. 2 for </ > = 35°, we hav e N = 33.55 , N = 37.75. B y substitutin g the known
values, we have
q
m
=114x6(33. 55-l )xl . l 85xl . l l 5 + - xl l 4xl Ox37. 75xl . l 85xl . l l5
= 29,417 + 28,431 = 57,848 lb/ft
2
lb/ft
2
By Terzaghi' s metho d q
m
= 16,462 lb/ft
2
.
Meyerhof's metho d give s a higher value for q
na
by about 17%.
508 Chapt e r 1 2
Example 1 2 .1 1
Refer t o Ex. 12.1 . Comput e by Hansen' s metho d (a ) net ultimat e bearing capacity, an d (b) the net
safe bearin g pressure . All the other dat a remain the same .
Given for a stri p footing
B = 3 m, D
f
= 2 m, c = 30 kN/m
2
and y = 17.25 kN/m
3
, F
s
- 3 .
From Eq . (12.27 ) fo r / = 1 , we have
From Tabl e 12. 2 for Hansen' s method, we have for 0 = 35
c
N
c
= 46.35, N
q
= 33.55, an d N
y
= 34.35 .
From Tabl e 12. 3 we have
o
s = 1 H — — = 1 fo r a stri p footing
N L
D
s = 1H— tan ( /) = 1 fo r a strip footing
D
s - 1 - 0.4 — = 1 fo r a strip footing
D
f 2
d = 1 + 0.4-^- = l + 0.4x- = 1.267
0
B 3
D
f
-^
= l + 2t a n35°( l - s i n35°)
2
x- =l + 2x0.7(1- 0. 574)
2
x-= 1.17
Substituting the known values, we have
q
nu
= 30 x 46.35 x 1 x 1.267 + 1 7.25 x 2 x (33.55 -1) x 1 x 1.1 7
+ - Xl 7. 25x3x34. 35xl xl
2
= 1,762 + 1,3 14 + 889 = 3,965 kN/m
2
= U22 kN/m
2
The value s of q
na
by Terzaghi, Meyerhof an d Hansen method s ar e
Example Autho r q
na
kN/m
2
12J Terzagh i 1,40 8
12.9 Meyerho f 1,37 3
12.11 Hanse n 1,32 2
Shallow Foun dat io n I : Ult im at e B earin g Capacit y 50 9
Terzaghi's method is higher than Meyerhof 's by 2.5%and Meyerhof 's higher than Hansen's by
3.9%. The difference betwee n the methods is not significant . Any of the three methods can be used.
Example 1 2 .1 2
Refer t o Example 12.6 . Comput e the net saf e bearin g pressur e by Hansen' s method . All the othe r
data remain the same .
Solution
Given: Size 1 0 x 20 ft , D
f
= 6 ft, c = 0, 0 = 35°, y= 1 14 lb/ft
3
, F
s
= 3.
For 0 = 35° we have from Tabl e 12.2 , N
q
= 33.55 and N
y
= 34.35
From Tabl e 12. 3 we have
/? 1 0
5 = 1 +— t a n 0=l + — xt an35 ° = 1.35
q
L 2 0
D i n
s
v
=1-0. 4 — =l - 0. 4x — = 0.80
7
L 2 0
d
q
= 1 + 2 tan 35° (1 - si n 35° )
2
x — = 1.153
Substituting the known values, we have
-l)
Sq
d
g
+
= 1 14 x 6(33.55 -1) x 1.35 x 1.153 + - x 1 14 x 10 x 34.35 x 0.8 x 1
2
= 34,655 + 15,664 = 50,319 lb/ft
2
50,319
The values of q
na
by other methods ar e
Example Autho r q kN/m
2
12.6 Terzagh i 16,46 2
12.10 Meyerho f 19,28 3
12.12 Hanse n 16,77 3
It can be seen from the above, the values of Terzaghi and Hansen are very close to each other ,
whereas the Meyerhof valu e is higher than that of Terzaghi by 1 7 percent.
1 2 .1 0 EFFEC T OF SOI L COM PRESSI B I L I T Y O N B EARI N G
CAPACI TY OF SOI L
Terzaghi (1943 ) develope d Eq . (12.6) based o n the assumptio n that the soi l i s incompressible. I n
order t o take into account the compressibility o f soil, he proposed reduce d strengt h characteristic s
c and 0 defined by Eq. (12.11). As per V esic (1973) a flat reduction of 0 in the case o f local and
510
Chapt er 1 2
punching shea r failure s i s too conservativ e an d ignore s th e existenc e o f scal e effects . I t has bee n
conclusively establishe d tha t the ultimate bearing capacit y q
u
of soil does not increase i n proportio n
to the increas e i n th e siz e o f th e footin g a s shown i n Fig. 12.1 1 or otherwis e th e bearin g capacit y
factor A f decrease s wit h th e increas e i n the siz e o f the footin g as shown i n Fig. 12.12 .
In order to take int o account the influence of soil compressibility and the related scal e effects ,
V esic (1973 ) propose d a modificatio n o f Eq . (12.27 ) b y introducin g compressibilit y factor s a s
follows.
q
u
= cN
c
d
c
s
c
C
c+
q'
o
N
q
d
q
s
q
C
q +
(12.28)
where, C
c
, C an d C ar e th e soi l compressibilit y factors . Th e othe r symbol s remai n th e same a s
before.
For th e evaluatio n o f th e relativ e compressibilit y o f a soi l mas s unde r loade d conditions ,
V esic introduce d a term calle d rigidity index I
r
, which is defined a s
/ =
c + qtar\ 0
(12.29)
where, G = shea r modulu s of soi l = ^( \ + u]
E
s
= modulu s of elasticit y
q = effectiv e overburde n pressur e a t a depth equa l t o (ZX + 5/2 )
600
400
200
100
80
60
3 4 0
20
10
8
6
4
2
1
I I T
Circular footing s
Chattahoochee san d (vibrated )
Dry uni t weight , 96. 4 lb/ft
3
Relative density , D
r
= 0.79
Standard triaxial , 0 = 39°
Deep penetration resistanc e
(Measured) (Postulated ) V
;
Test plat e size s
Dutch con e siz e - * *
I I I I I I I
Usual footin g size s
I I I I I I
0.2 0. 40. 60. 81. 0 2 4 6 8 10
Footing size , f t
20 40 608010 0 20 0
Figure 1 2 .1 1 Variat io n o f ult im at e res is t an c e o f foot in g s wit h s iz e
(aft er Ves ic , 1969 )
Shallow Foun dat io n I : Ult im at e B earin g Capacit y 511
800
600
400
200
Gent, y
k
= 1.674 ton/m
3
V esic,
y
k
= 1.538 ton/m
3
Meyerhof,
\y
k
= 1.70ton/m
3
v
v
s Golder ,
Circular plates
Square plates
Rectangular plates
-> Gent , y
k
= 1.619 ton/m
3
Gent , yf c= 1.50 9 ton/m
3
® Meyerhof , y
k
= 1.62 ton/m
3
Meyerhof, y
k
= 1.485 ton/m
3
0
V esic , y
k
= 1.440 ton/m
3
y
k
= 1.76 ton/m
C* /—vai n • V andeperre, y
k
=• 1.647 ton/m
Meischeider, y
k
=1/788 ton/m
3
'
A
Muhs
0.01 0.02 0.03 0.0 4
yB, kg/cm
2
0.05 0.06 0.07
•M,
100
Figure 1 2.1 2 Effec t o f s iz e on bearin g capacit y o f s urfac e foot in g s i n s an d (Aft e r
De B eer , 1965)
^ = Poisson' s rati o
c, 0 = shea r strengt h parameter s
Eq. (12.29 ) wa s develope d o n th e basi s of the theor y o f expansion o f cavities i n a n infinit e
solid wit h th e assume d idea l elasti c propertie s behavio r o f soil . I n orde r t o tak e car e o f th e
volumetric strai n A i n th e plasti c zone , V esi c (1965 ) suggeste d tha t th e valu e o f I
r
, give n b y
Eq. (12.29), b e reduced b y the following equation.
I=F
r
I (12.30 )
where F. = reduction facto r =
/ A
It is known that I
r
varie s with the stress level and the character o f loading. A high value of I
rr
,
for exampl e over 250, implie s a relatively incompressible soi l mass, whereas a low value of say 10
implies a relatively compressible soi l mass .
Based o n the theor y of expansion of cavities, V esic developed th e following equatio n for the
compressibility factors .
B 3.0 7 sinolog 21
C =ex p -4. 4 + 0.6— t an0 + ^
5 r
q
L l + sin<z >
For 0 >0, one can determine fro m th e theorem of correspondenc e
C = C -
c
For 0 = 0, we have
C =0.3 2 + 0.12— + 0.6 log/
L
(12.31)
(12.32)
(12.33)
512 Chapt er 1 2
/ =
G
r
c+qtan( j>
0
10 20 20 30 4 0 5 0 0 1 0
Angle o f shearin g resistance , (J >
Figure 1 2 .1 3 T heoret ica l com pres s ibilit y fact or s (aft e r Ves ic , 1970 )
For al l practical purposes , V esi c suggest s
C = C.. (12.34)
Equations (12.30 ) throug h Eq . (12.34) ar e vali d a s long a s the value s o f the compressibilit y
factors ar e les s tha n unity . Fig. 12.1 3 shows graphicall y th e relationshi p betwee n C ( = C ) and 0
for two extreme case s ofL /B > 5 (strip footing) and B/L = 1 (square) for different values of I
r
(V esi c
1970). V esi c give s anothe r expression calle d the critical rigidity index ( I
r
)
cr
expresse d a s
Table 12. 4 Val ue s o f crit ica l rig idit y in de x
An g le o f s hearin g res is t an c e Crit ica l rig idit y in de x fo r
St rip foun dat io n Squar e foun dat ion
(t) B/L = 0 B/L = 1
0
5
10
15
20
25
30
35
40
45
50
13
18
25
37
55
89
152
283
592
1442
4330
8
11
15
20
30
44
70
120
225
486
1258
Shallow Foun dat io n I : Ult im at e B earin g Capacit y 51 3
(Ur=f*p 3.3-0.45- ? cot(45-0/2 )
(1 2 35)
The magnitud e of ( I)
cr
fo r an y angl e o f 0 an d an y foundatio n shape reduce s th e bearin g
capacity because of compressibility effects. Numerica l values of ( l)
cr
fo r two extreme cases ofB/L
= 0 and BIL = 1 are given in Table 12. 4 for various values of 0.
Application o f /, (o r /„) and (/
r
)
crit
1 . I f l
r
(or I
rr
) > (/
r
)
crit
, assume the soi l i s incompressible and C, = C = C = I i n Eq. (12.28) .
2. I f I
r
(o r I
rr
) < (/
r
)
crit
, assume th e soi l i s compressible . I n suc h a case th e compressibilit y
factors C
c
, C and C are to be determined and used in Eq. (12.28) .
The concept an d analysis developed abov e by V esic (1973) are based o n a limited number of
small scal e model test s an d need verificatio n in fiel d conditions .
Example 1 2 .13
A squar e footing of siz e 1 2 x 1 2 ft i s placed a t a depth of 6 ft i n a deep stratum of medium dens e
sand. The following soil parameter s ar e available:
Y = 100 lb/ft
3
, c = 0, 0 = 35°, E
s
= 100 t/ft
2
, Poissons ' rati o n = 0.25.
Estimate the ultimat e bearing capacit y by taking into account the compressibility o f the soi l
(Fig. Ex. 12.13) .
Solution
E
Rigidity / = -
s
- -fo r c = 0 fro m Eq. (12.29)
r
q = y( D
f
+ 5/2) = 100 6 + — = 1,200 lb/ft
2
= 0.6 ton/ft
2
Neglecting the volume change i n the plastic zone
/ , = _!° ° _= 95
r
2( 1 + 0.25)0.6 tan 35°
From Tabl e 12.4 , (7
r
)
crit
= 12 0 for 0 = 35°
Since /. < (/
r
)
crit
, the soil i s compressible .
From Fig. 12.13, C
q
( = C
y
) = 0.90 (approx) for square footing for 0 = 35° and /. = 95.
From Tabl e 12.2 , N
q
= 33.55 and W
y
= 48.6 (V esic's value)
Eq. (12.28) ma y no w be written as
I
2
From Tabl e 12.3
u
= q'
0
N
q
d
q
s
q
C
q
+-yBN
y
d
r
s
r
C
r
D
s = 1 + —t a n ^ =l + tan35° =1. 7 fo r B = L
j = 1-0.4 = 0.6 forf i = L
514
Chapt er 1 2
d
n
=
q
2 t a n 35 °( l - s i n 35 °)
2
x — = 1.127
12
<7o = 100x6 = 600 l b/ft
2
Substituting
q
u
= 600 x 33.55 x 1.1 27 x 1.7 x 0.90 + - x 100 x1 2 x 48.6 x 1.0 x 0.6 x0.90
= 34,710 + 15,746 = 50,456 lb/ft
2
If th e compressibilit y factor s ar e no t take n int o accoun t (Tha t is , C = C = 1) the ultimat e
bearing capacit y q
u
is
q
u
= 38,567 + 17,496 = 56,063 lb/ft
2
,
6
1
ft
«=.
— *
c = 0 , y= 10 0 lb/ft
3
,
0 = 35°, £,= 10 0 ton/ft
2
ju = 0.25
— 1 2 x 12f t '•
Figure Ex . 1 2 .1 3
Example 1 2 .1 4
Estimate the ultimat e bearing capacity of a square footing of size 1 2 x 1 2 ft founded at a depth of 6
ft i n a deep stratum of saturated clay of soft t o medium consistency. The undrained shear strengt h of
the cla y i s 40 0 lb/ft
2
( = 0. 2 t/ft
2
) Th e modulu s o f elasticit y E
s
= 1 5 ton/ft
2
unde r undraine d
conditions. Assume ^ = 0.5 and y = 10 0 lb/ft
3
.
Solution
Rigidity I
r
=
15
= 25
2(1 + 0.5)0.2
From Tabl e 12.4 , (/.)
crit
= 8 for 0 = 0
Since /. >(/.)
crit
, the soil i s supposed t o be incompressible. Us e Eq. (12.28) for computing q
u
by puttin g C
c
= C
q
= 1 for 0 = 0
Shallow Foun dat io n I : Ult im at e B earin g Capacit y 51 5
From Tabl e 12. 2 for 0 = 0, N
c
= 5. 14, and N
q
- 1
N
B 1
5 4
From Tabl e 12. 3
s
c
=
d = 1 + 0.4-^- = 1 + 0. 4— =1. 2
B 1 2
Substituting and simplifying, we have
q
u
= 400 x 5.14 x 1.2 x 1.2 +100 x 6 x (1)(1)(1)
= 2,960 + 600 = 3,560 lb/ft
2
= 1.78 to n / f t
2
1 2 .1 1 B EARI N G CAPACI TY OF FOU NDATI ONS SU B JECTED TO
ECCENTRI C L OADS
Foundations Subj ecte d t o Eccentri c V ertica l L oad s
If a foundation is subjected to lateral loads and moments in addition to vertical loads, eccentricity in
loading results . Th e poin t o f applicatio n o f th e resultan t o f al l th e load s woul d li e outsid e th e
geometric cente r o f th e foundation , resulting i n eccentricit y i n loading . Th e eccentricit y e i s
measured fro m th e cente r o f th e foundatio n to th e poin t o f applicatio n norma l t o th e axi s o f th e
foundation. Th e maximu m eccentricit y normall y allowe d i s B/ 6 wher e B i s th e widt h o f th e
foundation. The basic problem i s to determine the effect o f the eccentricity o n the ultimate bearing
capacity of the foundation. When a foundation i s subjected to an eccentric vertica l load, as shown in
Fig. 12.14(a) , i t tilt s towards the side o f the eccentricity an d the contact pressur e increase s o n the
side of tilt and decreases on the opposite side. When the vertical load Q
ult
reaches the ultimate load,
there will be a failure of the supporting soil on the side of eccentricity. As a consequence, settlemen t
of th e footin g wil l b e associate d wit h tilting o f th e bas e toward s th e sid e o f eccentricity . I f th e
eccentricity is very small, the load required to produce this type of failure is almost equal to the load
required fo r producin g a symmetrica l genera l shea r failure . Failure occur s du e t o intens e radia l
shear on one side of the plane of symmetry, while the deformations in the zone of radial shear on the
other side are still insignificant. For this reason the failure is always associated wit h a heave on that
side toward s which the footing tilts.
Research an d observation s o f Meyerho f (1953 , 1963 ) indicate tha t effectiv e footin g
dimensions obtained (Fig . 12.14 ) as
L ' = L -2e
x
, B' = B-2e
y
(12.36a )
should be used i n bearing capacit y analysi s to obtain an effective footing area defined as
A' = B'L ' (12.36b )
The ultimate load bearing capacit y of a footing subjected to eccentric loads may be expresse d
as
Q'
ult
=^' (12-36 0
where q
u
= ultimate bearing capacity of the footing with the load acting at the center of the footing.
516
Chapt er 1 2
-- X
(a) (b)
- L '
A' = Shaded are a
(c) (d ) (e )
Figure 1 2 .1 4 Eccen t ricall y loade d foot in g (M eyerhof , 1953 )
Determination o f M aximu m and Minimum B as e Pressure s U nde r Eccentri c
L oadings
The method s o f determinin g th e effectiv e are a o f a footing subjecte d t o eccentri c loading s have
been discusse d earlier . I t i s no w necessar y t o kno w th e maximu m an d minimu m base pressure s
under the same loadings. Consider the plan of a rectangular footing given in Fig. 12.1 5 subjected t o
eccentric loadings .
Let the coordinat e axe s XX an d YY pass through the center O of the footing. If a vertical loa d
passes throug h O, the footing is symmetrically loaded. I f the vertical load passes throug h O
x
on the
X-axis, the footing i s eccentricall y loade d wit h one wa y eccentricity. The distanc e o f O
x
from O ,
designated a s e
x
, is called th e eccentricit y i n the X-direction. I f the loa d passe s throug h O o n the
7-axis, the eccentricity i s e i n the F-direction. If on the other hand the load passes throug h 0 th e
eccentricity i s called two-way eccentricity o r double eccentricity.
When a footin g i s eccentricall y loaded , th e soi l experience s a maximu m o r a minimum
pressure at one of the corners or edges o f the footing. For the load passing through O (Fig . 12.15) ,
the point s C and D at the corners of the footing experience th e maximum and minimum pressures
respectively.
Shallow Foun dat io n I : Ult im at e B earin g Capacity 517
M
r
Section
Q Q
Plan
X —
D
T
y
*~" "
T
-UJSSu.
i e.
-X B ' B
I
Q
Section
Figure 1 2 .1 5 Footin g s ubj ect e d t o eccen t ri c loadin g s
The general equatio n for pressure ma y be written as
nr
°
r
Q , M
X
M
= ± — -X±—
A I
A l
X± — -V
(12.37a)
(12.37b)
where q = contac t pressure at a given point (x, y)
Q = tota l vertical load
A = are a of footing
Qe
x
= M
x
= moment about axis YY
Qe = M = moment about axi s XX
/ , / = momen t of inertia o f the footing abou t XX an d YY axes respectivel y
for
we have
and <7
min
a t points C and D respectivel y ma y be obtained b y substitutin g in Eq. (12.37 )
12 12
L B
= —, y = —
2 2
(12.39a)
(12.39b)
Equation (12.39) may also be used for one way eccentricity by putting either e
x
= 0, or e = 0.
518 Chapt e r 1 2
When e
x
or e excee d a certain limit , Eq. (12.39) gives a negative value of q which indicates
tension betwee n th e soi l an d the bottom o f the footing. Eqs (12.39 ) ar e applicabl e onl y whe n the
load i s applied withi n a limited area whic h is known as the Kern as is shown shaded i n Fig 12.1 5 so
that the load ma y fall withi n the shaded are a t o avoid tension. The procedur e fo r the determination
of soi l pressur e whe n th e loa d i s applie d outsid e th e ker n i s laboriou s an d a s suc h not deal t with
here. However , chart s ar e availabl e for ready calculation s in reference s suc h a s Teng (1969 ) an d
Highter and Anders (1985) .
1 2 .1 2 U L TI M AT E B EARI NG CAPACI T Y OF FOOTI NG S B ASE D ON
SPT V AL U ES (N]
Standard Energy Rati o R
es
Applicabl e to N V alue
The effect s of field procedures an d equipment on the field values of N wer e discussed i n Chapter 9 .
The empirical correlation s establishe d i n the USA between N and soil properties indicat e the value
of N conforms t o certain standard energy ratios. Some sugges t 70%(Bowles, 1996 ) an d others 60%
(Terzaghi e t al. , 1996) . I n orde r t o avoi d thi s confusion, the autho r use s N
cor
i n thi s book a s th e
corrected value for standar d energy.
Cohesionless Soil s
Relationship B etwee n N
cor
and <| >
The relation betwee n A^ and 0 established b y Peck e t al. , (1974) i s give n in a graphical for m in
Fig. 12.8 . The value ofN
cor
t o be used for getting 0 is the corrected valu e for standard energy. The
angle 0 obtained b y thi s method can be used for obtaining the bearing capacit y factors , and henc e
the ultimat e bearing capacit y o f soil .
Cohesive Soil s
Relationship B etwee n N
cor
and q
u
(U nconfined Compressive Strength)
Relationships hav e bee n develope d betwee n N
cor
an d q
u
(the undraine d compressive strength ) for
the 0 = 0 condition . Thi s relationshi p give s th e valu e o f c
u
fo r an y know n valu e o f N
cor
. Th e
relationship may b e expressed a s [Eq. (9.12) ]
tf^^jA^CkPa) (12-40 )
where the valu e of the coefficien t & ma y var y from a minimum of 1 2 to a maximum of 25. A low
value of 1 3 yields q
u
given i n Table 9.4 .
Once q
u
i s determined , th e ne t ultimat e bearin g capacit y an d th e ne t allowabl e bearin g
pressure can be foun d followin g Skempton's approach .
1 2 .1 3 TH E CRT M ETH OD OF DETERM I NI NG U L TI M ATE B EARI NG
CAPACI TY
Cohesionless Soil s
Relationship B etwee n q
c
, D
r
and 0
Relationships betwee n th e stati c con e penetratio n resistanc e q
c
an d 0 hav e bee n develope d b y
Robertson an d Campanella (1983b) , Fig . 9.15. Th e value of $ can therefore b e determined wit h the
known valu e o f q . With th e know n valu e of 0 , bearin g capacit y factor s ca n b e determine d an d
Shallow Foun dat io n I : Ult im at e B earin g Capacit y 519
hence the ultimate bearing capacity. Experience indicates that the use of q
c
for obtaining 0 is more
reliable tha n the use of N .
B earing Capacit y o f Soi l
As per Schmertmann (1978), the bearing capacity factors N an d N fo r use in the Terzaghi bearin g
capacity equatio n can be determined b y the use of the equation
N =
(12.41)
where q
c
= cone point resistance i n kg/cm
2
(or tsf) average d over a depth equal to the width below
the foundation.
U ndrained Shea r Strengt h
The undraine d shea r strengt h c
u
unde r 0 = 0 conditio n ma y b e relate d t o th e stati c con e poin t
resistance q
c
a s [Eq . (9.16) ]
q
c
=
N
k
c
u
+
Po
or c,, =
(12.42)
where N
k
= con e factor , ma y b e take n a s equa l t o 2 0 (Sanglerat , 1972 ) bot h fo r normall y
consolidated an d preconsolidate d clays .
p
o
= tota l overburde n pressur e
When onc e c
u
i s known , th e value s o f q
m
an d q
na
ca n b e evaluate d a s pe r th e method s
explained i n earlier sections .
Example 1 2 .1 5
A water tank foundation has a footing of size 6 x 6 m founded at a depth of 3 m below groun d level
in a medium dense sand stratum of great depth. The corrected averag e SPT value obtained fro m the
site investigatio n i s 20 . Th e foundatio n is subjecte d t o a vertica l loa d a t a n eccentricit y o f fi/1 0
along one of the axes. Figur e Ex. 12.1 5 gives the soil profile with the remaining data. Estimat e the
ultimate load, Q
ult
, by Meyerhof's method .
s/*\ //^\
m
—1
QuU
^
€B
-
SPT
c = 0, y=18. 5kN/ m
3
,
0 = 33°, N
cor
= 20
Medium dense san d
B
10
fix 5 = 6 x 6m -H
Figure Ex . 1 2.1 5
520 Chapt e r 1 2
Solution
From Fig . 12.8 , 0=33° for N =20.
e '
r
cor
B' = B-2e = 6- 2(0.6 ) - 4.8 m
L ' = L = B = 6m
For c = 0 and i = 1, Eq. (12.28) reduces t o
From Tabl e 12. 2 for 0 = 33° we have
N
q
= 26.3, N
Y
= 26.55 (Meyerhof )
From Tabl e 12. 3 (Meyerhof )
s = l + 0.1Af — = l + 0.1tan
2
45°+ — (1 ) =1.34
q
* L 2
^=^=1. 34 fo r 0> 10°
/ —
D
f 3
d=l + Q.L N . -
L
= 1 + 0.1x1.84 — =1.11 5
q
V * B ' 4. 8
Substituting d
y
=d
q
= 1.1 15 fo r 0 > 10°
q'
u
= 18.5 x 3 x 26.3 x 1.34 x 1. 1 15 + - x 18.5 x 4.8 x 26.55 x 1.34 x 1.1 15
= 2, 1 81+ 1,76 1 = 3,942 kN/m
2
Q'
ult
=BxB'xq'
u
=6x4. 8x3942=113, 53 0 kN- 11 4 M N
Example 1 2 .16
Figure Ex . 12.1 6 gives the plan of a footing subjected t o eccentric loa d wit h two wa y eccentricity .
The footing is founded at a depth 3 m below the ground surface. Given e
x
= 0.60 m and e = 0.75 m ,
determine Q
u[ [
. The soi l propertie s are : c = 0, N
cgr
= 20, y = 18.5 kN/m
3
. The soi l i s medium dens e
sand. Use N (Meyerhof) fro m Tabl e 12. 2 and Hansen' s shape an d dept h factor s from Table 12.3 .
Solution
Figure Ex . 12.1 6 show s th e two-wa y eccentricity . Th e effectiv e length s an d breadth s o f th e
foundation fro m Eq . (12.36a ) i s
B' = B- 2e = 6 - 2 x 0.75 = 4.5 m.
L ' = L - 2e
x
= 6 - 2 x 0.6 = 4.8 m.
Effective area , A' = L ' x B' = 4.5 x 4. 8 = 21. 6 m
2
As i n Exampl e 12.1 5
Shallow Foun dat io n I : Ultimat e B earin g Capacit y 521
6 m
e
y
= 0.75 m
e = 0.6 m
y
6 m -
Figure Ex . 12.1 6
For 0 = 33°, N
q
= 26.3 an d N
y
= 26.55 (Meyerhof )
From Tabl e 12. 3 (Hansen )
B' 4 5
5 = 1 + —tan33° = l +—x 0.65 = 1.61
q
L ' 4. 8
R' 4 5
s
v
=1-0.4—=1-0. 4 x— = 0.63
Y
L ' 4. 8
d
a
= l + 2t an33°(l - si n33°)
2
x —
*4. 5
= 1 + 1.3x0.21x0.67 = 1.183
Substituting
1
q'
u
=18.5x3x26.3xl.61xl.l83+-xl8.5x4.5x26.55x0.63x(l )
= 2,780 + 696 = 3,476 kN/m
2
Q
uh
= A'q'
u
= 21.6 x 3,476 = 75,082 k N
1 2 .1 4 U L TI M AT E B EARI NG CAPACI TY O F FOOTI NG S RESTI N G
ON STRATI FI ED DEPOSI T S OF SOI L
All the theoretical analysis dealt with so far is based on the assumption that the subsoi l i s isotropic
and homogeneou s t o a considerabl e depth . I n nature , soi l i s generall y non-homogeneou s wit h
mixtures of sand, sil t and clay i n differen t proportions . I n the analysis , an average profil e of such
soils is normally considered. However, if soils are found i n distinct layers of different composition s
and strengt h characteristics, the assumption of homogeneity to such soils is not strictl y valid if the
failure surfac e cuts across boundaries of such layers.
522 Chapt e r 1 2
The present analysi s is limited to a system of two distinct soil layers . For a footing located i n
the upper layer at a depth D, below the ground level, the failure surfaces at ultimate load ma y either
lie completel y i n th e uppe r laye r o r ma y cros s th e boundar y o f th e tw o layers . Further , w e ma y
come acros s th e upper layer strong and the lower layer weak or vice versa . In either case, a general
analysis for (c - 0 ) will be presented an d will show the same analysi s holds true if the soil layers are
any on e of the categories belongin g t o sand or clay.
The bearin g capacit y o f a layere d syste m wa s firs t analyze d b y Butto n (1953 ) wh o
considered onl y saturate d cla y ( 0 = 0) . Late r o n Brow n an d Meyerho f (1969 ) showe d tha t th e
analysis o f Butto n lead s t o unsaf e results . V esi c (1975 ) analyze d th e tes t result s o f Brow n an d
Meyerhof an d other s an d gav e hi s own solutio n t o the problem .
V esic considere d bot h the types of soi l i n each layer , that i s clay an d (c - 0 ) soils. However ,
confirmations o f the validit y of the analysi s of V esic and others ar e not available. Meyerhof (1974 )
analyzed the two layer system consisting of dense san d on soft cla y and loose sand on stif f cla y and
supported hi s analysi s wit h som e mode l tests . Agai n Meyerho f an d Hann a (1978 ) advance d th e
earlier analysi s o f Meyerhof (1974 ) t o encompas s ( c - 0 ) soi l an d supporte d thei r analysi s wit h
model tests . The presen t sectio n deal s briefl y wit h the analyses of Meyerhof (1974 ) an d Meyerho f
and Hanna (1978) .
Case 1 : A Stronge r L aye r Overlyin g a Weaker Deposi t
Figure 12.16(a ) show s a stri p footin g o f widt h B restin g a t a dept h D, below groun d surfac e i n a
strong soi l layer (Layer 1) . The dept h to the boundary of the weak laye r (Layer 2) below th e base of
the footing is H. I f this depth H is insufficient t o form a full failur e plasti c zone i n Layer 1 under the
ultimate loa d conditions , a par t o f thi s ultimat e load wil l be transferre d t o the boundar y leve l mn.
This loa d wil l induce a failure conditio n i n the weake r laye r (Laye r 2) . However , i f the dept h H i s
relatively larg e the n th e failur e surfac e wil l b e completel y locate d i n Laye r 1 a s show n i n
Fig. 12.16b .
The ultimat e bearing capacitie s o f stri p footings on the surface s of homogeneous thic k bed s
of Layer 1 and Layer 2 may be expressed a s
Layer 1
q\ =
c
\
N
c\
+
-Y\
BN
r\ (12.43 )
Layer 2
1 „. ,
N
y2
(12.44 )
where N
cl
, N . - bearin g capacit y factor s for soil i n Layer 1 for friction angl e 0 j
N
c2
, N y2 = bearing capacit y factor s for soi l i n Layer 2 for frictio n angl e 0
2
For th e footin g founde d a t a dept h ZX , i f th e complet e failur e surfac e lie s withi n th e uppe r
stronger Layer 1 (Fig. 12.16(b) ) an expression for ultimate bearing capacit y o f the upper layer may
be written as
q
u
= v< =
c
i
N
ci
+
Vo
N
q
i
+
^ri
BN
n (12.45 )
If q\ i s muc h greate r tha t q
2
and i f th e dept h H i s insufficien t t o for m a ful l failur e plasti c
condition i n Layer 1 , then the failur e of the footing may be considered du e t o pushing of soil withi n
the boundary ad an d be through the top layer into the weak layer . The resisting forc e fo r punching
may b e assume d t o develo p o n the face s a d and be passin g throug h th e edge s of the footing . The
forces tha t act on these surfaces are (pe r uni t lengt h of footing),
Shallow Foun dat io n I : Ult im at e B earin g Capacit y 523
^ f <*^<*r^<0^
D
f
1
,
Q
. t

1
,
\
,
a b
^^^^^^^^
c
a
i Laye r 1 : stn
\. X i
H
Layer 2: weaker y
2
> 02 '
C
2
(b)
Figure 1 2.1 6 Failur e o f s oi l belo w s t ri p foot in g unde r vert ica l loa d o n s t ron g layer
overl yin g wea k depos i t (aft e r M eyerho f an d Han n a , 1978 )
Adhesive force, C
a
=c
a
H
Frictional force, F, = P si n
(12.46)
where c
a
= unit cohesion, P - passiv e earth pressure per unit length of footing, and < 5 = inclination
of P
p
wit h the norma l (Fi g 12.16(a)) .
The equatio n for the ultimat e bearing capacit y q
u
for the two layer soi l syste m ma y no w be
expressed as
2(C
a
+P sin<5 )
•q =q.+ -
a
- -
p
~ -- yH
"u "b T) ' I
where, q
b
- ultimat e bearing capacit y of Layer 2
The equation for P ma y be written as
(12.47)
H
(12.48)
524 Chapt er 1 2
40
35
30
25
^--J
I 2 0
i
1 5
u
10
Approx. bearing
capacity rati o
0.4
0.2
0
20° 25 ° 30 ° 35 ° 40 ° 45 ° 50 °
0i (deg )
Figure 1 2 .1 7 Coefficien t s of pun chin g s hea r res is t an ce un de r vert ica l loa d (aft e r
M eyerhof an d Han n a , 1978 )
1.0
0.9
§ 0. 7
CU
0.6
0.2 0. 4 0. 6 0. 8
Bearing capacit y rati o q~
L
lq
l
1.0
Figure 1 2.1 8 Plo t o f c
a
/c
1
vers us q
2
lq^ (aft e r M eyerho f an d Han n a , 1978 )
Shal l ow Foundat i o n I : Ul t i mat e Bear i n g Capaci t y 52 5
Substituting for P an d C
a
, the equation for q
u
may be written as
2c
a
H r^H
2 2D
f
^=^
+
-|-
+
-y-
1 +
-^- K
p
tanS-
r
,H
(1249 )
In practice , i t i s convenien t t o us e a coefficien t K
s
o f punchin g shearing resistanc e o n th e
vertical plane through the footing edges s o that
K
s
tan ^= K
p
ta n 8 (12.50 )
Substituting, the equation for q
u
may be written as
2cH ,H
2
2D
Figure 12.1 7 gives the value of K
s
for various values of 0j as afunction of <?
2
/g
r
The variation
i
with q^q
l
i s shown in Fig. 12.18.
Equation (12.45 ) fo r q
t
an d q
b
i n Eq . (12.51 ) ar e fo r stri p footings . Thes e equation s wit h
shape factor s may be written as
-Y,BN
n
s
n
(12.52 )
,2 +\ Y2
BN
Y
2
s
n. (12.53 )
where s
c
, s an d s ar e th e shap e factor s fo r th e correspondin g layer s wit h subscript s 1 and 2
representing layer s 1 and 2 respectively.
Eq. (12.51) can be extended to rectangular foundations by including the corresponding shap e
factors.
The equation for a rectangular footing may be written as
2c H B Y<H
2
2D ,
B
4
u
=<l
b
+ ~1 + Y +-y - l + -~ 'l + -
Case 2 : To p L ayer Dens e Sand and B ottom L ayer Saturate d Sof t Cla y (0
2
= 0 )
The value of q
b
for the bottom laye r fro m Eq . (12.53) may be expressed a s
^b
=C
2
N
c2
S
c2
+
r^
D
f
+H
'> (12-55 )
From Table (12.3), s
c2
= (1+0.2 B/L ) (Meyerhof , 1963 ) and N
c
= 5.14
for 0 = 0. Therefor e
q
b
= 1 + 0.2- 5.Uc
2
+
ri
( D
f
+H)
(12
.
56)
LJ
For C j = 0, q
t
from Eq . (12.52 ) i s
<?, = r^/^V ,!
+
ri
BN
n
s
n (12-57 )
526 Chapt e r 1 2
We may no w writ e an expression for q
u
from Eq . (12.54 ) a s
B
y,H
2
2D
f
B
q = 1 + 0.2— 5.14c
?
+^ -1 + — L i + _K tan0 ,
" L
2
B H L
s l
+ y
l
D
f
<y
]
D
f
N
qlS( il
+~y
l
BN
rl
s
n
(12.58 )
The rati o of q
1
lq
l
ma y be expressed by
C
^ 5.14 c
The valu e of K
s
ma y b e foun d fro m Fig . (12. 17).
Case 3 : Whe n L aye r 1 i s Dense Sand and L ayer 2 i s L oose Sand (c
1
= c
2
= 0 )
Proceeding i n the same wa y as explained earlier the expression for q
u
for a rectangular footing may
be expressed as
q
u
= Y,
f
y.H
2
B 2 D
- ~ (1260)
where q
t
= Y^f^s^ +-Y^BN
n
s
n
(12.61 )
<12
-
62)
Case 4 : L aye r 1 i s Stiff Saturate d Cla y (0
1
= 0 ) an d L ayer 2 i s Saturated Sof t
Clay (0
2
= 0 )
The ultimat e bearing capacit y o f the layered syste m can be give n as
q
u
= 1 + 0.2-5.Uc
2+
1 + |^-
+
y
] Df
<
qt (12
.63)
D
q,= 1 + 0.2- 5.14c,+y,D
/
(12.64 )
L
q
2
_ 5.1 4 c
2
_
(12.65)
Example 1 2 .1 7 .
A rectangular footing of size 3 x 2 m i s founded at a depth o f 1. 5 m i n a clay stratum of ver y stif f
consistency. A cla y laye r o f mediu m consistenc y is locate d a t a dept h o f 1. 5 m ( = H) belo w th e
bottom o f the footing (Fig. Ex. 12.17) . The soi l parameter s o f the two clay layer s ar e a s follows :
Top clay layer : c = 17 5 kN/m
2
Shallow Foun dat io n I : Ult im at e B earin g Capacit y 527
;^:v4^>-i
;
y^v---f-
D
f
= 1. 5m"' '
:
''V'y'i''v^-.'-V'' '•':'*i.;'
:
-
:
''.;
•..' V er y stiff cla y
Soft cla y
Layer 2 c
2
= 40 kN/m
2
y
2
= 17. 0 kN/m
3
Figure Ex . 1 2.17
7
t
= 17. 5 kN/m
3
Bottom layer : c
2
= 40 kN/m
2
y
2
= 17. 0 kN/m
3
Estimate the ultimate bearing capacit y and the allowable bearing pressure on the footing with
a factor of safet y of 3.
Solution
The solutio n t o thi s proble m come s unde r Cas e 4 i n Sectio n 12.14 . W e hav e t o conside r her e
Eqs (12.63) , (12.64 ) an d (12.65) .
The dat a given are:
fl = 2 m,L = 3m, H= 1.5 m (Fig. 12.16a),D
/
= 1.5m , ^ = 17. 5 kN/m
3
.
From Fig. 12.18, for q
2
lq
l
= c
2
/c
l
= 40/175 = 0.23,
the valu e of cjc
l
= 0.83 or c
a
= 0.83^ = 0.83 x17 5 = 145.2 5 kN/m
2
.
From Eq. (12.63 )
B B 2 c H
q = 1 + 0.2— 5.14c ~ + 1 + — -^— + y,D
f
<q,
u
L L B
}
Substituting the known values
2xl 45
-
25x1
-
5
q = 1 + 0.2X- 5.14x40 + 1 + -
y
" 3 3
+n.5xl.5
= 233 + 364 + 26 = 623 kN/m
2
528 Chapt e r 1 2
FromEq. (12.64)
D
q
t
= 1 + 0.2— 5.14c , + y
}
D
f
= l + 0.2x- 5.14x17 5 + 17.5x1.5
3
-1020 + 26-1046 kN/m
2
It i s clea r fro m th e abov e tha t q
u
< q
(
an d a s suc h q
u
i s th e ultimat e bearing capacit y t o b e
considered. Therefor e
kN/m
2
Example 1 2 .1 8
Determine th e ultimat e bearing capacit y of the footing given in Example 12.1 7 i n dense san d with
the bottom laye r being a clay of medium consistency. The parameters of the top layer are :
/! = 18. 5 kN/m
3
, 0 j = 39°
All the other dat a given in Ex. 12.1 7 remain the same. Us e Meyerhof's bearin g capacit y and
shape factors .
Solution
Case 2 of Section 12.1 4 is required t o be considered here .
Given: Top layer: y , = 18. 5 kN/m
3
, 0
1
= 39°, Botto m layer: y
2
= 17.0 kN/m
3
, c
2
= 40 kN/m
2
.
From Table 12. 2 N
yl
( M) = 78.8 fo r 0 j = 39° .
FromEq. (12.59 )
<? 5.14 c 5.14x4 0
q
l
(XSj^flA f j 0. 5x18. 5x2x78. 8
From Fig. 12.1 7 K
s
= 2. 9 for 0 = 39°
Now fro m Eq . (12.58 ) we have
B
y H
2
2 D B
q - 1 + 0.2— 5.14c , + — 1 + —- 1 + — K
L
2
B H L *
7 1 8 S v n 5^ 7x 1 ^ 7
= 1 + 0.2X- 5. 14x40 +
V ;
\ + =——1 + -2.9tan39°+18.5x1. 5
3 2 1. 5 3
-233 + 245 + 28 = 506 kN/m
2
q
u
- 506 kN/m
2
From Eq . (12.58) the limiting value q
t
is
q
t
=
Shallow Foun dat io n I : Ult im at e B earing Capacit y 52 9
where ^ = 18. 5 kN/m
3
, D
f
= 1.5m , B = 2 m.
From Tabl e 12. 2 N
yl
= 78. 8 an d N
ql
= 56.5
B 3 9 2
From Tabl e 12. 3 ^=1 + 0.1^ — =l + 0.1xtan
2
45°+ — x - = l29= s
yl
/_> £ J
Now <?
r
= 18. 5 x 1.5 x56.5 x 1.29 +-x 18.5 x 2x78.8 x 1.29 = 3903 kN/m
2
> q
u
Hence < = 50 6 kN/m
2
1 2 .1 5 B EARI N G CAPACI TY OF FOU NDATI ONS ON TOP OF A
SL OPE
There ar e occasions wher e structures are required to be built on slopes o r near the edges o f slopes .
Since ful l formation s of shear zones under ultimate loading conditions are not possible on the side s
close t o the slope s o r edges, the supporting capacity of soi l on that side get considerabl y reduced .
Meyerhof (1957 ) extende d hi s theorie s t o includ e th e effec t o f slope s o n th e stabilit y o f
foundations.
Figure 12.1 9 shows a section of a foundation wit h the failure surfaces under ultimate loading
condition. The stabilit y of the foundation depends o n the distanc e b of the top edg e o f the slop e
from th e fac e of the foundation.
The for m o f ultimat e bearin g capacit y equatio n fo r a stri p footin g ma y b e expresse d a s
(Meyerhof, 1957 )
1
(12.66)
The uppe r limi t o f th e bearin g capacit y o f a foundatio n in a purel y cohesiv e soi l ma y b e
estimated fro m
q
u
=cN
c
+yD
f
(12.67 )
The resultan t bearing capacit y factor s N
cq
an d N depen d o n th e distanc e b » A 0
an
d th e
DJ B ratio . These bearing capacity factors are given in Figs 12.2 0 and 12.2 1 for stri p foundation in
purely cohesive and cohesionless soil s respectively. It can be seen from th e figures that the bearing
capacity factors increase with an increase of the distance b • Beyond a distance of about 2 to 6 times
the foundatio n widt h B, th e bearin g capacit y i s independen t o f th e inclinatio n of th e slope , an d
becomes the same a s that of a foundation on an extensive horizontal surface.
For a surcharg e ove r th e entir e horizonta l to p surfac e o f a slope , a solutio n o f th e slop e
stability has been obtained on the basis of dimensionless parameters called th e stability number N
s
,
expressed a s
N S
=^H
(12
-
68)
The bearin g capacit y o f a foundatio n o n purel y cohesiv e soi l o f grea t dept h ca n b e
represented b y Eq. (12.67) where the N
c
facto r depends on b a s well as ft, and the stability number
N . Thi s bearin g capacit y factor , whic h i s give n i n th e lowe r part s o f Fig . 12.20 , decreas e
considerably wit h greate r heigh t an d t o a smalle r exten t wit h the inclinatio n o f th e slope . Fo r a
given heigh t an d slop e angle , th e bearin g capacit y facto r increase s wit h a n increas e i n b.
an
d
530 Chapt er 1 2
90°-
Figure 1 2 .1 9 B earin g capacit y o f a s t rip foot in g o n t o p o f a s lope (M eyerhof ,
1957)
beyond a distance of about 2 to 4 times the height of the slope, the bearing capacit y i s independent
of the slope angle. Figur e 12.2 0 shows that the bearing capacity of foundations on top of a slope i s
governed b y foundatio n failur e fo r smal l slop e heigh t (A ^ approachin g infinit y ) an d b y overal l
slope failur e for greate r heights.
The influenc e of groun d wate r and tensio n cracks (i n purel y cohesive soils ) shoul d als o b e
taken int o accoun t i n the study of the overal l stabilit y o f the foundation .
Meyerhof (1957 ) has no t supporte d hi s theor y wit h any practical example s o f failure as any
published dat a wer e no t availabl e for thi s purpose.
B
e
a
r
i
n
g

c
a
p
a
c
i
t
y

f
a
c
t
o
r

N
c
q
*


t
O

U
>

-
P
^

L
f
i
O
N

-
~
J

O
_
X
7-f
30^
"60°
i? / j
<r/
30V
/
/
A!
-30°
Incli
si

10°'
X
X
)
0
PS.'
, .S'
'/
/
°/
/
^°:
^90
natio
ope/
X
•9-- '
''6C
/•
/
^
<f
/
/
90°
^
D
/
/ 0°
/ I
^
f~^'
_9
/
/*
n o f
3
- ' _
X
9
/
'
f
^
^
X"
3° '
/
^
s^-
^
''
s
s
s
^^"^
<*> ***
,
'
,'"
SI
" >
W, =
ope stabi l
factor yv
s
/
c
o
0.5
j
r
/0
.
25
yO. 18
S
- 0 0
ity
Foundation depth/widt h
D
f
/B = 0
D
f
/B=l
0 1 2 3 4 5 6
Distance of foundatio n fro m edg e o f slope b/ B
Figure 12.2 0 B earin g capacit y fact or s fo r s t ri p foun dat io n o n t o p o f s lop e o f
purely cohes iv e m at eria l (M eyerhof , 1957 )
Shallow Foun dat io n I : Ult im at e B earin g Capacit y 531
400
300
200
100
50
25
10
,' "
X
' . - •
A
/
/
II
X
^• 0°
^*
^
30
U
£^
1
iclin
slo
20
"'
atior
pe^

r'
40 V
x I
^^
9
0
^-*-
of
^ * *
**•"**
^-
-
^ —
^-*^
_ — -
F
£
£
L
ii
Ang
„- -
^=
— ~ "
oundation depth/widt h
fit) — U
)
/ D 1
/• //5 — 1
inear inte
itermedia
leof
Ticti<
j
„ - -
40°
^=^
30
*
30
intei
Dn 0
f
_ \ -
^^
rpolation fo r
te depths
nal
0 1 2 3 4 5 6
Distance of foundation fro m edg e of slope b/ B
Figure 1 2 .2 1 B earin g capacit y fact or s fo r s t ri p foun dat io n o n t o p o f s lop e o f
cohes ion les s m at eria l (M eyerhof , 1957 )
Example 1 2 .1 9
A stri p footing i s t o be constructed o n the top of a slope a s per Fig. 12.19 . Th e following data ar e
available:
B = 3m,D
f
= 1.5m , b = 2 m, H= 8 m, p= 30°, y = 18. 5 kN/m
3
, 0 = 0 and c = 75 kN/m
2
,
Determine th e ultimat e bearing capacit y o f the footing.
Solution
PerEq. (12.67) q
u
for 0 = 0 is
From Eq. (12.68)
c 1 5
yH 18.5x 8
= 0.51
and — = — = 0.
From Fig. 12.20 , N =3.4 for N =0.51 , A/ f i = 0.67, an d j8= 30°
D ' C q $ ' c /•" r~
Therefore
a = 75 x 3. 4 + 18. 5 x 1. 5 = 255+28 = 283 kN/m
2
.
532 Chapt e r 1 2
1 2 .1 6 FOU NDATI ON S O N ROCK
Rocks encountere d i n natur e might b e igneous , sedimentar y or metamorphic . Granit e an d basal t
belong t o the firs t group . Granit e is primaril y composed o f feldspar , quart z and mic a possesse s a
massive structure . Basal t i s a dark-colore d fin e graine d rock . Bot h basal t an d granit e unde r
unweathered condition s serv e a s a ver y goo d foundatio n base . Th e mos t importan t rock s tha t
belong t o the second grou p are sandstones, limestones and shales. These rocks are easily weathere d
under hostil e environmenta l conditions and arsuch, th e assessmen t o f bearin g capacit y o f thes e
types require s a littl e care. I n the las t group come gneiss , schist , slat e an d marble . Of thes e rock s
gneiss serve s a s a good bearin g material whereas schis t and slat e posses s poor bearin g quality.
All rock s weathe r unde r hostile environments. The ones tha t ar e close t o the ground surfac e
become weathered mor e tha n the deeper ones . I f a rocky stratu m i s suspected clos e t o the ground
surface, the soundness or otherwise of these rocks mus t be investigated. The quality of the rocks i s
normally designate d b y RQD a s explained i n Chapter 9.
Joints ar e common i n all rock masses . Thi s wor d joint i s used b y geologist s fo r any plane of
structural weakness apar t from fault s withi n the mass. Within the sedimentary rock mas s the joints
are lateral i n extent and form wha t are called bedding planes, and they are uniform throughout any
one be d withi n igneou s roc k mass . Coolin g joint s ar e mor e closel y space d neare r th e groun d
surface wit h wider spacing s at deeper depths . Tension joints and tectonic joint s might be expecte d
to continu e dept h wise . Withi n metamorphi c masses , ope n cleavage , ope n schistos e an d ope n
gneissose plane s ca n b e o f considerabl y furthe r latera l exten t tha n th e beddin g plane s o f th e
sedimentary masses .
Faults an d fissure s happe n i n roc k masse s du e t o interna l disturbances . Th e joint s wit h
fissures an d fault s reduce s th e bearing strengt h of rocky strata .
Since mos t unweathere d intact rocks ar e stronge r an d les s compressibl e tha n concrete , th e
determination o f bearin g pressure s o n suc h material s ma y no t b e necessary . A confine d roc k
possesses greater bearin g strengt h than the rocks expose d a t ground level .
B earing Capacit y o f Rock s
Bearing capacitie s o f rocks ar e ofte n determine d b y crushing a core sampl e i n a testing machine.
Samples use d fo r testing must be fre e fro m crack s an d defects .
In th e roc k formatio n wher e beddin g planes , joints an d othe r plane s o f weaknes s exist , th e
practice tha t i s normall y followe d i s t o classif y th e roc k accordin g t o RQ D (Roc k Qualit y
Designation). Tabl e 9. 2 gives th e classification of the bearing capacit y o f rock accordin g t o RQD.
Peck e t al , (1974) hav e related the RQD to the allowable bearing pressur e q
a
as given in Table 12.5
The RQD fo r use i n Table 12. 5 should be the average withi n a depth belo w foundatio n level
equal t o th e widt h of the foundation , provided th e RQD i s fairl y unifor m within that depth. I f the
upper part of the rock, withi n a depth of about 5/4, is of lower quality, the value of this part should
be used.
Table 1 2. 5 Al l owabl e B earin g Pres s ur e q o n J oin t e d Roc k
RQD
100
90
75
50
25
0
q
g
T on /ft
2
300
200
120
65
30
10
q
g
MP a
29
19
12
6.25
3
0.96
Shallow Foun dat io n I : Ult im at e B earing Capacit y 533
Table 12. 6 Pres um pt iv e allowabl e bearin g pres s ure s on rock s (M Pa ) as
recom m en ded b y various buildin g code s i n USA (aft e r Pec k et al. , 1974 )
Rock t ype Building Code s
B OCA Nat ion a l Un ifor m LO S An g eles
(1968) (1967 ) (1964 ) (1959 )
1. Massive crystalline bedrock,
including granit e diorite, gneiss, 1 0
basalt, har d limeston e and dolomite
2. Foliated rocks such a s schis t or
slate i n soun d conditio n 4
3. Bedded limeston e i n sound condition
sedimentary rocks including hard 2. 5
shales and sandstone s
4. Sof t o r broken bedrock
10 0.2 q*
1.5
0-29.
1.0
0.4
0.3
(excluding shale ) and sof t
limestone
5. Sof t shal e
1.0
0.4
0.2 q
u
0.2 q
u
*q = unconfined compressiv e strength .
Another practic e tha t i s normall y followe d i s t o bas e th e allowabl e pressur e o n th e
unconfined compressive strength, q
u
, of the rock obtained in a laboratory on a rock sample . A factor
of safety of 5 to 8 is used on q
u
to obtain q
a
. Past experience indicates that this method is satisfactory
so long as the rocks i n situ do not possess extensiv e cracks and joints. In such cases a higher factor
of safet y may have to be adopted .
If rock s clos e t o a foundatio n bas e ar e highl y fissured/fractured , the y ca n b e treate d b y
grouting which increases th e bearing capacity of the material .
The bearing capacity of a homogeneous, an d discontinuous rock mass cannot be less than the
unconfined compressiv e strengt h o f th e roc k mas s aroun d th e footin g an d thi s ca n b e take n a s a
lower bound for a rock mas s wit h constant angle of internal friction 0 and unconfined compressive
strength q
ur
. Goodman (1980) suggests the following equation for determining the ultimate bearing
capacity q
u
.
(12.69)
where A f = tan
2
(45° + 0/2), q
ur
= unconfined compressive strengt h of rock.
Recommendations b y Buildin g Code s
Where bedroc k ca n b e reache d b y excavation , th e presumptiv e allowabl e bearin g pressur e i s
specified by Building Codes. Table 12. 7 gives the recommendations of some buildings codes i n the
U.S.
1 2 .1 7 CAS E H I STORY OF FAI L U RE OF TH E TRANSCONA
G RAI N EL EV ATOR
One o f th e bes t know n foundatio n failure s occurre d i n Octobe r 191 3 a t Nort h Transcona ,
Manitoba, Canada . I t wa s ascertaine d late r o n tha t th e failur e occurre d whe n th e foundation
534 Chapt e r 1 2
Figure 1 2.2 2 T h e t ilt ed T r an s con a g rai n el evat o r (Court es y: UMA En g in eerin g
Lt d., M an it oba , Can ada )
Figure 12.2 3 Th e s t raig ht en e d T ran s con a g rai n elevat o r (Court es y : UM A
En g in eerin g Lt d. , M an it oba , Can ada )
Shallow Foun dat io n I : Ult im at e B earin g Capacit y 535
Ground leve l
Organi clay fill
and silt y cal y
V arved brown clay
with horizonta l strati -
fication
Greyish silt y
clay wit h
calcareous sil t
Grey silt y clay wit h
calcareous sil t pocket s
Silty grave l
770
760
750
740
730
720
Refusal
(a) Classification fro m
test borin g
0 1. 0 2. 0 3. 0
Unconfmed compressiv e strengt h (TSF )
(b) V ariation o f unconfmed compressiv e
strength wit h dept h
Figure 12.2 4 Res ult s o f t es t borin g a t sit e o f T ran s con a g rai n elevat o r (Pec k and
B yran t , 1953 )
pressure a t the base wa s about equal t o the calculated ultimat e bearing capacit y o f an underlaying
layer of plastic cla y (Pec k an d Byrant,1953), an d was essentially a shearing failure.
The constructio n o f the sil o starte d i n 191 1 and wa s complete d i n the autumn of 1913 . The
silo i s 77 ft by 19 5 ft in plan and has a capacity o f 1,000,00 0 bushels. I t comprises 6 5 circular bins
and 48 inter-bins. The foundation was a reinforced concrete raf t 2 ft thick and founded at a depth of
12 ft below the groun d surface. The weight of the sil o was 20,000 tons, which was 42. 5 percen t of
the tota l weight , whe n i t wa s filled . Filling th e sil o wit h grai n starte d i n Septembe r 1913 , an d i n
October whe n the silo contained 875,000 bushels, and the pressure on the ground was 94 percent of
the design pressure , a vertical settlement o f 1 ft was noticed. The structur e began t o tilt to the west
and withi n twent y four hour s wa s at a n angl e o f 26.9° fro m th e vertical , th e wes t sid e bein g 2 4 f t
below an d th e eas t sid e 5 f t abov e th e origina l leve l (Szechy , 1961) . Th e structur e tilte d a s a
monolith and there was no damage t o the structure except for a few superficial cracks. Figur e 12.2 2
shows a vie w o f th e tilte d structure . The excellen t qualit y of th e reinforce d concret e structur e is
shown by the fac t that later i t was underpinne d and jacked u p on ne w piers founde d o n rock. Th e
536 Chapt er 1 2
level o f the ne w foundatio n i s 34 ft below th e ground surface . Figure 12.2 3 shows the vie w of the
silo afte r i t was straightene d i n 1916 .
During th e perio d whe n the sil o wa s designed and constructed , soi l mechanic s a s a scienc e
had hardl y begun. The behavior of the foundation unde r imposed load s was not clearly understood.
It was only during the year 195 2 tha t soil investigatio n was carried out close to the silo and the soi l
properties wer e analyze d (Pec k an d Byrant , 1953) . Figur e 12.2 4 give s th e soi l classificatio n and
unconfmed compressiv e strengt h o f th e soi l wit h respec t t o depth . Fro m th e examinatio n o f
undisturbed sample s o f th e clay , i t wa s determine d tha t th e averag e wate r conten t o f successiv e
layers of varved clay increase d wit h their dept h fro m 4 0 percent t o about 60 percent. Th e averag e
unconfmed compressiv e strengt h o f the upper stratu m beneat h th e foundation wa s 1.1 3 tsf, that of
the lower stratum was 0.65 tsf , and the weighted average was 0.93 tsf . The average liqui d limit was
found t o be 10 5 percent; therefor e the plasticity index was 70 percent, whic h indicates that the clay
was highl y colloidal an d plastic. The averag e uni t weight of the soi l was 12 0 lb/ft
3
.
The contact pressur e due to the load from the silo at the time of failure was estimated as equal to
3.06 tsf . The theoretical value s of the ultimate bearing capacit y by various methods ar e as follows.
Methods a., tsf
Terzaghi[Eq. (12.19) ]
Meyerhof [Eq . (12.27) ]
Skempton [Eq . (12.22) ]
3.68
3.30
3.32
The abov e values compar e reasonabl y wel l wit h the actual failur e loa d 3.0 6 tsf . Perloff an d
Baron (1976 ) giv e detail s o f failur e o f the Transcona grai n elevator .
1 2 .1 8 PROB L EM S
12.1 Wha t will be the gross and net allowable bearing pressures o f a sand having 0 = 35° and an
effective uni t weigh t o f 1 8 kN/m
3
unde r the followin g cases: (a ) siz e o f footin g 1 x 1 m
square, (b ) circular footing of 1 m dia. , an d (c) 1 m wide stri p footing.
The footin g i s placed a t a depth of 1 m below th e groun d surface an d th e wate r tabl e i s at
great depth . Us e F^ = 3. Comput e b y Terzaghi' s genera l shea r failur e theory.
1 m :
Sand ••'•:
.• ...;• -.;..• y = 1 8 kN/m
3
12.2 A stri p footing i s founded a t a dept h o f 1. 5 m below th e ground surfac e (Fig . Prob . 12.2) .
The wate r tabl e i s clos e t o groun d leve l an d th e soi l i s cohesionless . Th e footin g i s
supposed t o carry a net saf e loa d o f 400 kN/m
2
wit h F = 3. Given y = 20.85 kN/m
3
and
Shallow Foun dat io n I : Ult im at e B earin g Capacit y 537
0 = 35°, fin d th e require d widt h o f th e footing , usin g Terzaghi' s genera l shea r failur e
criterion.
.' '• ..';:. ' San d
1. 5m • • ' • • ' • ' • •
I i i
• - ' v ' : . - 0 = 35°
; ' ; ; . • ' y
sat
= 20.85 kN/m
3
i i I
Figure Prob . 12. 2
12,3 A t what depth should a footing of size 2 x 3 m be founded to provide a factor of safety of 3
if th e soi l i s stif f cla y having an unconfined compressive strengt h of 12 0 kN/m
2
? The unit
weight of the soi l i s 1 8 kN/m
3
. The ultimat e bearing capacity of the footing is 425 kN/m
2
.
Use Terzaghi's theory . The water table i s close t o the ground surface (Fig. Prob . 12.3) .
D
t
=l
Stiff cla y
q
u
= 120 kN/m
2
y= 1 8 kN/m
3
0 = 0
B xL = 2 x 3m *- |
Figure Prob . 12. 3
12.4 A rectangular footing is founded at a depth of 2 m below the ground surface in a (c - 0 ) soil
having the following properties : porosity n = 40%, G
s
= 2.67, c = 15 kN/m
2
, and 0 = 30° .
The water table is close to the ground surface. If the width of the footing is 3 m, what is the
length required t o carry a gross allowabl e bearing pressure q
a
= 455 kN/m
2
wit h a factor of
safety = 3? Use Terzaghi' s theor y of general shea r failur e (Figure Prob. 12.4) .
538 Chapt er 1 2
V
n 9 m
L )
f
z m
1
#J\
1
1
n = 40%
G, = 2.67
1 c = 1 5 k N/ m
( jj -30
h- B x L = 3 xL - H
Figure Prob . 12.4
12.5 A squar e footin g locate d a t a depth of 5 ft below th e ground surfac e i n a cohesionless soi l
carries a column load o f 13 0 tons. The soi l i s submerged having an effective unit weight of
73 lb/ft
3
an d a n angl e of shearin g resistanc e o f 30°. Determin e th e siz e of th e footin g fo r
F = 3 by Terzaghi' s theor y of general shea r failur e (Fig. Prob . 12.5) .
D
/
= 5 f t ;
i I 1
' San d
11
;: y
b
= 73 lb/ft
3
; • ' 0 = 30 °
|-« B x B ^
Figure Prob . 1 2. 5
12.6 A footin g o f 5 f t diamete r carrie s a saf e loa d (includin g it s sel f weight ) o f 8 0 ton s i n
cohesionless soi l (Fig. Prob. 12.6) . The soi l ha s a n angl e of shearing resistance </ > = 36° an d
an effective unit weigh t of 80 lb/ft
3
. Determin e the dept h o f the foundation for F
s
= 2. 5 by
Terzaghi's genera l shea r fail ur e theory.
£>,=
i I i . •
80 ton
y
h
= 80 lb/ft
3
0 = 36°
5f t
Figure Prob . 1 2. 6
Shallow Foun dat io n I : Ult im at e B earin g Capacit y 539
12.7 I f the ultimat e bearing capacity o f a 4 ft wide stri p footing resting o n the surfac e of a sand
is 5,25 0 lb/ft
2
, wha t wil l b e th e ne t allowabl e pressur e tha t a 1 0 x 1 0 f t squar e footing
resting o n th e surfac e ca n carr y wit h F
S
= 37 Assume tha t th e soi l i s cohesionless . Us e
Terzaghi's theor y of general shea r failure.
12.8 A circular plate of diameter 1.0 5 m was placed o n a sand surface of unit weight 16. 5 kN/m
3
and loade d t o failure . Th e failur e loa d wa s foun d t o giv e a pressur e o f 1,50 0 kN/m
2
.
Determine th e value of the bearing capacity factor N . The angl e of shearing resistanc e of
the san d measure d i n a triaxia l tes t wa s foun d t o b e 39° . Compar e thi s valu e wit h th e
theoretical valu e of N . Use Terzaghi's theor y o f general shea r failure.
12.9 Fin d the net allowable bearing load per foot lengt h of a long wall footing 6 ft wide founded
on a stif f saturate d clay at a depth of 4 ft. The uni t weight of the clay i s 11 0 lb/ft
3
, an d th e
shear strengt h i s 250 0 lb/ft
2
. Assum e th e loa d i s applie d rapidl y suc h tha t undraine d
conditions (0 = 0) prevail. Use F = 3 and Skempton' s metho d (Fig. Prob . 12.9) .
l&%$J 0^-#f$&?i
= 11 0 lb/ft
3
(
vc; c
u
= 2500 lb/ft
2
• v .
t t 1
6f t
Figure Prob. 12. 9
12.10 Th e tota l colum n loa d o f a footin g nea r groun d leve l i s 500 0 kN. Th e subsoi l i s
cohesionless soi l wit h 0=38° and y= 19. 5 kN/m
3
. The footing is to be located a t a depth of
1.50m below ground level. For a footing o f size 3 x 3 m, determine the factor of safety by
Terzaghi's genera l shea r failur e theory i f the wate r tabl e i s at a dept h o f 0. 5 m below th e
base level of the foundation.
.5 m
0.5m
(2 = 5000 kN
• '.;• ;';.• .'.'.."• .' '-: : San d
• i - '( • y = 19. 5 kN/m
3
3m
GWT
Figure Prob . 12.1 0
540 Chapt er 1 2
12.11 Wha t wil l b e th e factor s o f safet y i f th e wate r tabl e i s me t (i ) a t th e bas e leve l o f th e
foundation, an d (ii ) at the ground level i n the case of the footing in Prob. 12.10 , keeping al l
the othe r condition s th e same ? Assum e tha t th e saturate d uni t weigh t o f soi l
7
sat
= 19. 5 kN/m
3
, an d th e soi l abov e th e bas e o f th e foundatio n remains saturate d eve n
under (i ) above .
12.12 I f the factors of safety in Prob. 12.1 1 are other than 3, what should be the size of the footing
for a minimum factor of safet y o f 3 under the wors t condition?
12.13 A footing of siz e 1 0 x 1 0 ft i s founded at a depth o f 5 ft i n a medium stif f cla y soi l having
an unconfmed compressiv e strengt h of 2000 lb/ft
2
. Determine the net safe bearing capacity
of the footing wit h the water table at ground level by Skempton' s method. Assume F
s
= 3.
12.14 I f the average stati c cone penetration resistance, q
c
, in Prob. 12.1 3 is 1 0 t/ft
2
, determine q
na
per Skempton's method. The other conditions remain the same as in Prob. 12.13 . Ignore the
effect o f overburde n pressure .
12.15 Refe r t o Prob . 12.10 . Comput e by Meyerho f theor y (a ) the ultimat e bearing capacity , (b )
the ne t ultimat e bearing capacity , an d (c ) th e facto r o f safet y fo r th e loa d comin g o n th e
column. All the other dat a given in Prob. 12.1 0 remain th e same .
12.16 Refe r t o Prob . 12.10 . Comput e b y Hansen' s metho d (a ) th e ultimat e bearin g capacity ,
(b) the ne t ultimat e bearing capacity , an d (c ) the facto r o f safet y for th e column load . Al l
the othe r dat a remai n th e same . Commen t o n th e result s using the method s o f Terzaghi ,
Meyerhof an d Hansen.
12.17 A rectangular footing of size (Fig . 12.17 ) 1 2 x 24 ft is founded at a depth o f 8 ft below th e
ground surfac e in a ( c - 0 ) soil. The following data are available: (a ) water tabl e at a depth
of 4 f t below groun d level , (b ) c = 600 lb/ft
2
, 0 = 30° , an d 7 = 11 8 lb/ft
3
. Determin e th e
ultimate bearing capacit y by Terzaghi an d Meyerhof's methods .
f
8 f t
M m M
,
c = 600 lb/ft
2
y = 11 8 lb/ft
3
B x L = 12 x 2 4 f t >- |
Figure Prob. 1 2 .1 7
12.18 Refe r t o Prob. 12.1 7 and determine the ultimate bearing capacit y b y Hansen's method . All
the other dat a remai n the same .
12.19 A rectangular footing of size (Fig. Prob. 12.19 ) 1 6 x 24 ft is founded at a depth of 8 ft in a
deep stratum of ( c - 0 ) soil wit h the following parameters :
c = 300 lb/ft
2
, 0 = 30°, E
s
= 15 t/ft
2
, 7 = 10 5 lb/ft
3
, fj. = 0.3.
Estimate th e ultimate bearing capacit y by (a) Terzaghi's method, and (b) V esic's method by
taking int o account the compressibility factors.
Shallow Foundatio n I: Ultimat e B earing Capacit y 541
D
/
= 8f t
UL1
5 x L= 16x 2 4f t
(c - <p) soi l
c = 300 lb/ft
2
^ = 30°
7 = 105 lb/ft
3
^ = 0.3
E, = 75 t/ft
2
Figure Prob. 12.1 9
12.20 A footing of size 1 0 x 1 5 ft (Fig. Prob. 12.20 ) i s placed at a depth of 1 0 ft below the ground
surface i n a deep stratum of saturated clay of sof t t o medium consistency. The unconfme d
compressive strengt h of clay under undrained conditions is 600 lb/ft
2
an d \ J L = 0.5. Assume
7= 9 5 lb/ft
3
an d E
s
= 12 t/ft
2
. Estimat e th e ultimat e bearing capacit y o f th e soi l b y th e
Terzaghi and V esic methods by taking into account the compressibility factors .
M i .
£/;•: 4
U
= 600 lb/ft
2
>••• : V: 7 = 95 lb/ft
3
Figure Prob . 12.2 0
12.21 Figur e Proble m 12.2 1 give s a foundation subjected t o a n eccentri c loa d i n on e direction
with al l the soil parameters. Determine the ultimate bearing capacit y of the footing.
D
f
=Bft
M i l -
e = 1 ft
M
L = 10x2 0 ft
Medium dense sand
7 =110 lb/ft
3
» - x
Figure Prob. 1 2.2 1
542
Chapt er 1 2
12.22 Refe r t o Fig. Prob . 12.22 . Determine th e ultimate bearing capacit y o f the footing i f e
x
= 3 ft
and e = 4 ft . What i s the allowabl e load fo r F = 3?
25f t
*_, i f r
4 f t
~~"~ "TT = 3 f t
- 1
Dense sand
D y = 8 f t
c = 0, 0 = 40
C
x
y = 1 2 U l b / t t
3
Figure Prob . 12.2 2
12.23 Refe r t o Fig . Prob . 12.22 . Comput e th e maximu m an d minimu m contac t pressure s fo r a
column loa d o f Q = 800 tons .
12.24 A rectangular footin g (Fig . Prob . 12.24 ) o f siz e 6 x 8 m i s founded a t a dept h o f 3 m i n a
clay stratum of very stiff consistenc y overlyin g a softer clay stratum at a depth o f 5 m fro m
the ground surface . The soi l parameter s o f the two layer s of soi l are :
Top layer: c , = 200 kN/m
2
, ^ = 18. 5 kN/m
3
Bottom layer : c
2
= 35 kN/m
2
, y
2
= 16. 5 kN/m
3
Estimate th e ultimat e bearing capacity o f the footing .
D
f
=3m
V ery stif f cla y
c, = 200 kN/m
2
y, = 18. 5 kN/m
3
flxL=6x8m
H=2m
Soft cla y
c
2
= 35 kN/m
2
y
2
= 16. 5 kN/m
3
Figure Prob. 12.2 4
Shallow Foun dat io n I : Ult im at e B earin g Capacit y 543
12.25 I f th e to p laye r i n Prob . 12.2 4 i s dens e sand , what is th e ultimat e bearing capacit y o f th e
footing? The soi l parameters of the top layer are:
y
t
= 19. 5 kN/m
3
, 0! = 38°
All th e other dat a given in Prob. 12,2 4 remain the same .
12.26 A rectangular footing of siz e 3 x 8 m i s founded on the to p o f a slope o f cohesive soi l as
given i n Fig. Prob . 12.26 . Determin e the ultimate bearing capacity of the footing.
3 m
b = 2 m~\
^^Nf
- 1 5 m
1 mmI
= 30
(Not to scale )
H=6m
c = 60 kN/m
2
,0 = 0
Y= 17. 5 kN/m
3
Figure Prob . 12.2 6
C H A P T E R 13
S H A L L O W FO UN DA T I ON I I :
S A FE BE A R I NG P R E S S URE A ND
S E T T L E ME N T C A L C UL A T I ON
1 3 .1 I NTRODU CTI O N
Allowable and Saf e Bearing Pressures
The method s o f calculating the ultimat e bearing capacit y of soi l have been discusse d a t lengt h in
Chapter 12 . The theorie s use d i n tha t chapte r ar e base d o n shea r failur e criteria . The y d o no t
indicate the settlement that a footing may undergo under the ultimate loading conditions. From the
known ultimat e bearin g capacit y obtaine d fro m an y on e o f th e theories , th e allowabl e bearin g
pressure can be obtained by applying a suitable factor of safety to the ultimat e value.
When we design a foundation, we must see that the structure is saf e on two counts. They are,
1. Th e supporting soil shoul d be saf e fro m shea r failure due to the loads impose d on it by the
superstructure,
2. Th e settlemen t of the foundation should be withi n permissible limits.
Hence, w e have to deal wit h two types of bearing pressures . The y are ,
1. A pressure tha t is saf e from shea r failure criteria,
2. A pressure that is safe from settlemen t criteria.
For the sake of convenience, let us call the first the allowable bearing pressure and the second
the safe bearing pressure.
In al l ou r design , w e us e onl y th e ne t bearin g pressur e an d a s suc h w e cal l q
na
th e ne t
allowable bearing pressure an d q
s
the net safe bearing pressure. In designing a foundation, we use
545
546 Chapt e r 1 3
the leas t o f th e two bearin g pressures . I n Chapte r 1 2 we learn t tha t q
na
i s obtaine d b y applyin g a
suitable factor of safet y (normall y 3) to the net ultimate bearing capacit y o f soil . In this chapter we
will learn how to obtain q
s
. Even without knowing the values of q
na
and q
s
, it is possible t o say from
experience whic h o f th e tw o value s shoul d be used i n design base d upo n th e compositio n an d
density of soil and the size of the footing. The composition and density of the soil and the size of the
footing decid e th e relative values of q
na
an d q
s
.
The ultimate bearing capacity of footings on sand increases wit h an increase i n the width, and
in the same wa y the settlement of th e footing increases wit h increases i n the width. In other word s
for a give n settlemen t 5
p
the correspondin g uni t soi l pressur e decrease s wit h an increas e i n the
width of the footing. It is therefore, essential to consider tha t settlement wil l be the criterion fo r the
design o f footing s i n san d beyon d a particula r size . Experimenta l evidenc e indicate s tha t fo r
footings smalle r than about 1.2 0 m, the allowable bearing pressure q i s the criterion for the design
of footings , whereas settlemen t i s the criterion fo r footings greate r tha n 1. 2 m width.
The bearing capacit y o f footings on clay is independent of the size of the footings and as such
the uni t bearing pressur e remain s theoreticall y constant in a particular environment. However, th e
settlement o f th e footin g increase s wit h a n increas e i n th e size . I t i s essentia l t o tak e int o
consideration bot h th e shea r failur e and the settlemen t criteri a togethe r t o decide th e saf e bearin g
pressure.
However, footing s on stiff clay, hard clay, and other firm soils generall y requir e no settlement
analysis if the design provide s a mi ni mum factor of safety of 3 on the net ultimate bearing capacit y
of th e soil . Sof t clay , compressibl e silt , an d othe r wea k soil s wil l settl e eve n unde r moderat e
pressure an d therefor e settlemen t analysi s is necessary.
Effect o f Settlemen t o n the Structur e
If the structure as a whole settles uniforml y int o the ground there wil l not be any detrimental effec t
on th e structur e a s such . Th e onl y effec t i t ca n hav e i s o n th e servic e lines , suc h a s wate r an d
sanitary pip e connections , telephon e an d electri c cable s etc . whic h can brea k i f the settlemen t i s
considerable. Suc h unifor m settlement i s possible onl y i f the subsoi l i s homogeneous an d the load
distribution i s uniform . Building s i n Mexic o Cit y hav e undergon e settlement s a s larg e a s 2 m.
However, th e differentia l settlemen t i f i t exceed s th e permissibl e limit s wil l hav e a devastatin g
effect o n th e structure.
According t o experience , th e differentia l settlemen t betwee n part s o f a structur e ma y no t
exceed 7 5 percen t o f th e norma l absolut e settlement . Th e variou s way s b y whic h differentia l
settlements ma y occu r i n a structur e ar e show n i n Fig . 13.1 . Tabl e 13. 1 give s th e absolut e an d
permissible differentia l settlements for various types o f structures .
Foundation settlement s mus t b e estimate d wit h grea t car e fo r buildings , bridges , towers ,
power plant s an d simila r hig h cos t structures . Th e settlement s fo r structure s suc h a s fills ,
earthdams, levees , etc . ca n be estimated wit h a greater margi n of error .
Approaches fo r Determinin g the Ne t Saf e B earing Pressure
Three approache s ma y b e considere d fo r determinin g the ne t saf e bearin g pressur e o f soil . The y
are,
1. Fiel d plat e loa d tests ,
2. Charts ,
3. Empirica l equations.
Shallow Foun dat io n II : Saf e B earin g Pres s ur e an d Set t lem en t Calculat io n 547
Original positio n
of column bas e
Differential settlemen t
(a)
T
H
(c)
t ^ — Relativ e rotation, /?
-Wall o r pane l •
Tension crack s
H
Tension cracks —' I " — Relative deflection, A ^ „ , , . ,
Relative sa g Deflectio n rati o = A/L Relativ e ho g
(b)
Relative rotation,
Figure 13. 1 Defin it ion s o f differen t ial s et t lem en t fo r fram e d an d load-bearin g wall
s t ruct ures (aft e r B urlan d and Wrot h, 1974)
Table 1 3 . 1 aM ax im u m s et t lem en t s an d differen t ial s et t lem en t s o f building s i n cm .
(Aft er M cDon al d an d Sk em pt on , 1955 )
SI. no . Crit erio n Is olat ed foun dat ion s Raf t
1. Angula r distortion 1/30 0
2. Greates t differentia l settlement s
Clays 4- 5
Sands 3-2 5
3. Maximu m Settlement s
Clays 7. 5
Sands 5. 0
1/300
4.5
3.25
10.0
6.25
548 Chapt e r 1 3
Table 1 3 .1 b Perm is s ibl e s et t lem en t s (1955 , U. S. S. R . Buildin g Code )
Sl. n o. Typ e of buildin g Averag e s et t lem en t (cm)
1. Buildin g wi th plai n brickwall s on
continuous an d separat e foundation s wit h
wall lengt h L t o wal l heigh t H
2.
3.
L J H> 2.5
L IH<\ .5
Framed building
Solid reinforce d concret e foundatio n of
blast furnaces, water towers etc.
7.5
10.0
10.0
30
Table 13.1c Perm is s ibl e differen t ia l s et t lem en t (U. S. S. R Buildin g Code , 1955 )
T ype o f s oi l
Sl. n o. T yp e of s t ruct ur e San d an d har d cla y Plas t i c cla y
1. Stee l an d reinforce d concret e structure s 0.002 L 0.002 L
2. Plai n brick wall s in multistory buildings
for L IH < 3 0.0003 L 0.0004 L
L /H > 5 0.0005 L 0.0007 L
3. Wate r towers , silos etc. 0.004 L 0.004 L
4. Slop e o f crane wa y as wel l as track
for bridg e crane trac k 0.003 L 0.003 L
where, L = distance betwee n two columns or part s of structur e that settle differen t amounts , H = Height of
wall.
1 3 .2 FIEL D PL ATE L OAD TEST S
The plat e loa d tes t i s a semi-direc t metho d t o estimat e th e allowabl e bearin g pressur e o f soi l t o
induce a given amount of settlement. Plates, roun d or square, varying in size, from 30 to 60 cm and
thickness of about 2. 5 cm ar e employed for the test .
The loa d o n the plat e i s applied b y making use of a hydraulic jack. Th e reactio n o f the jack
load i s taken by a cross bea m or a steel trus s anchored suitabl y at both the ends. The settlement of
the plate is measured b y a set of three dial gauges of sensitivity 0.02 mm placed 120 ° apart . The dial
gauges ar e fixed to independent support s which remain undisturbe d during the test .
Figure 13.2 a shows the arrangement for a plate load test. The method of performing th e test is
essentially a s follows:
1. Excavat e a pit of size no t less tha n 4 to 5 times the siz e o f the plate. Th e bottom o f the pit
should coincide wit h the leve l of the foundation.
2. I f th e wate r tabl e i s abov e th e leve l o f th e foundation , pump out th e wate r carefull y and
keep i t a t the leve l of the foundation.
3. A suitable siz e of plat e i s selected for the test . Normall y a plate of siz e 30 cm i s used i n
sandy soil s an d a large r siz e i n cla y soils . Th e groun d shoul d b e levelle d an d th e plat e
should be seated ove r the ground.
Shallow Foun dat io n II : Saf e B earin g Pres s ure an d Set t lem en t Calculat io n 54 9
rod -
|k
k^
ns —Channel
IL
Steel girders
g
5IC ^V - ^
""•^X ^-^ |
/ \ (
Anchors ^
1
r_
Extension ^^
pipe ^~^ ^
rt
\
\
—I
13
L_
u
c=
3-1
^^_ Hydraulic
>^ jac k
p
Dial gau£
«h
4
;e
a
i
£55
T^^-^S;
.^^Ita.
J]
£
^
7 N 1 / \
Y
S;
Test plate —/ |^ _ &
p
__^|> ^ Test pit
Section
^na
©
©
©

©
©
©
U1C i
i
i
I
2 Girders
A
i i /
p
\ \
i . t i i i i
)
Test plate
i i
i
1
Support
©
©
©
n
1
©
©
©
4.
Top plan
Figure 13.2 a Plat e load t es t arran g em en t
A seating loa d of about 70 gm/cm
2
is first applie d an d released afte r som e time . A higher
load i s next placed o n the plate and settlement s are recorded b y mean s of the dial gauges .
Observations on every load increment shall be taken until the rate of settlement i s less than
0.25 mm per hour . Load increment s shall be approximatel y one-fift h o f the estimated saf e
bearing capacit y of the soil . The average o f the settlement s recorded by 2 or 3 dial gauges
shall be taken a s the settlement of the plate for each of the load increments .
5. Th e test should continue until a total settlement of 2.5 cm or the settlement at which the soil
fails, whicheve r is earlier, i s obtained. After th e load i s released, the elastic rebound of the
soil shoul d be recorded .
From th e test results , a load-settlement curv e should be plotted a s shown in Fig. 13.2b . The
allowable pressur e on a prototype foundation fo r a n assumed settlement ma y b e found b y making
use o f th e followin g equation s suggeste d b y Terzagh i an d Pec k (1948 ) fo r squar e footing s i n
granular soils .
550
Chapt er 1 3
Plate bearing pressure in kg/cm
2
or T/m
2
i \ q
a
= Net allowable pressure
Figure 1 3 .2 b Load-s et t lem en t curv e o f a plat e-loa d t es t
B
S
f
= S x —-
where 5 , = permissible settlemen t of foundation in mm,
S - settlemen t of plate in mm,
(I S. l b)
B = size of foundation in meters,
b = size of plat e i n meters .
For a plate 1 ft square, Eq. (13.la) ma y be expressed a s
iJ r — 0
f p (13.2)
in whic h S, and 5 ar e expressed i n inches and B i n feet .
The permissibl e settlemen t 5 , fo r a prototyp e foundatio n shoul d b e known . Normall y a
settlement o f 2. 5 cm i s recommended. I n Eqs (13.la) or (13.2) the value s of 5, and b ar e known.
The unknown s ar e 5 an d B. Th e valu e o f S fo r an y assume d siz e B ma y b e foun d fro m th e
equation. Usin g th e plat e loa d settlemen t curv e Fig . 13. 3 th e valu e o f th e bearin g pressur e
corresponding t o th e compute d valu e o f 5 i s found . Thi s bearin g pressur e i s th e saf e bearin g
pressure fo r a give n permissibl e settlemen t 5 ~ Th e principa l shortcomin g o f thi s approac h i s th e
unreliability o f the extrapolation of Eqs (13. la) o r (13.2) .
Since a load tes t i s of short duration, consolidation settlement s canno t be predicted. The tes t
gives th e valu e o f immediat e settlemen t only . If th e underlyin g soil i s sand y i n natur e immediat e
settlement may be taken as the total settlement. If the soil i s a clayey type, the immediate settlement
is onl y a fraction o f th e tota l settlement . Loa d tests , therefore , d o no t hav e muc h significanc e in
clayey soil s t o determine allowabl e pressure on the basis of a settlement criterion.
Shallow Foun dat io n II : Saf e B earin g Pres s ur e an d Set t lem en t Calculat io n 55 1
Pi st p i na H T A •* Foundatio n o f
Flate load Loa d q
n
per uni t area
test
^
ca
/ buildin g
IlJJJJlLLiJ
\y/////////////////^^^^
Stiff cla y
Soft cla y
Pressure bulbs
Figure 13.2 c Plat e load t es t o n n on -hom og en eous s oi l
Plate load test s should be used with caution and the present practice is not to rely too much on
this test . I f th e soi l i s no t homogeneou s t o a grea t depth , plat e loa d test s giv e ver y misleadin g
results.
Assume, a s show n i n Fig . 13.2c , two layer s o f soil . Th e to p laye r i s stif f cla y wherea s th e
bottom laye r i s sof t clay . Th e loa d tes t conducte d nea r th e surfac e o f th e groun d measure s th e
characteristics o f the stif f cla y but does no t indicate the nature of the sof t cla y soi l whic h is below.
The actua l foundation of a building however has a bulb of pressure whic h extends t o a great depth
into th e poo r soi l whic h i s highl y compressible . Her e th e soi l teste d b y th e plat e loa d tes t give s
results which ar e highly on the unsafe side .
A plate loa d tes t i s not recommended i n soils which are not homogeneous a t least t o a depth
equal t o \
l
/2 t o 2 times the widt h of the prototype foundation.
Plate loa d test s shoul d no t be relie d o n t o determine th e ultimat e bearing capacit y of sandy
soils a s the scal e effec t give s very misleading results. However, whe n the test s ar e carried o n clay
soils, the ultimat e bearing capacit y a s determined by the test ma y b e taken a s equal t o that of the
foundation sinc e the bearing capacity of clay is essentially independent of the footing size.
H ousel's (1 9 2 9 ) M etho d o f Determinin g Saf e B earin g Pressur e fro m
Settlement Consideratio n
The metho d suggeste d b y House l fo r determinin g th e saf e bearin g pressur e o n settlemen t
consideration i s based o n the following formula
O = A m + P n C1 3 3 )
^ p p \ ± ~> .~> j
where Q = load applie d o n a given plate, A = contact are a o f plate, P = perimeter o f plate, m = a
constant correspondin g t o the bearing pressure , n - anothe r constan t correspondin g t o perimete r
shear.
Obj ective
To determine th e load (Xan d the siz e of a foundation for a permissible settlemen t 5-. .
Housel suggest s tw o plat e loa d test s wit h plate s o f differen t sizes , sa y B
l
x B^ an d
B
2
x B
2
for this purpose .
552 Chapt e r 1 3
Procedure
1 . Tw o plate loa d test s are to be conducted at the foundation leve l of the prototype as per the
procedure explaine d earlier.
2. Dra w the load-settlemen t curves for each o f the plat e load tests .
3. Selec t th e permissibl e settlement S,. for the foundation.
4. Determin e the loads Q
{
and Q
2
from each of the curves for the given permissible settlemen t
s
f
Now we may writ e the following equations
Q\
=mA
P
\
+np
P
\ (13.4a )
for plat e load tes t 1 .
Q
2
=mA
p2
+nP
p2
(13.4b )
for plat e loa d tes t 2 .
The unknown values of m and n can be found by solving the above Eqs. (13.4a) and (13. 5b).
The equatio n for a prototype foundation may be written as
Q
f
=mA
f
+nP
f
(13.5 )
where A, = area of the foundation, / >,= perimeter o f the foundation.
When A, and P,are known, the size of the foundation can be determined .
Example 1 3 .1
A plat e loa d tes t usin g a plat e o f siz e 3 0 x 3 0 c m wa s carrie d ou t a t th e leve l o f a prototyp e
foundation. Th e soi l a t th e sit e wa s cohesionles s wit h th e wate r tabl e a t grea t depth . Th e plat e
settled b y 1 0 mm a t a load intensit y of 16 0 kN/m
2
. Determine th e settlement of a square footin g of
size 2 x 2 m under the same loa d intensity.
Solution
The settlement of the foundation 5, may be determined fro m Eq. (13. la). ,
=3a24mm
Example 1 3 .2
For Ex. 13. 1 estimate the load intensit y if the permissible settlemen t o f the prototype foundation is
limited t o 40 mm.
Solution
In Ex. 13 . 1, a load intensit y of 16 0 kN/m
2
induces a settlement of 30.24 mm. I f we assume tha t the
load-settlement i s linear withi n a small range, we may writ e
Shallow Foun dat io n II : Saf e B earin g Pres s ur e an d Set t lem en t Calculat io n 55 3
where, q
{
= 160 kN/m
2
, S^ = 30.24 mm, S^ = 40 mm. Substituting the known values
40
q
2
= 160 x —— = 211.64 kN/m
2
Example 13. 3
Two plate load tests were conducted at the level of a prototype foundation in cohesionless soi l close
to each other . The following data are given:
Size of plat e
0.3 x 0. 3 m
0.6 x 0. 6 m
Load applie d
30 kN
90 kN
Settlement recorde d
25 mm
25 mm
If a footing is t o carr y a load o f 1000 kN, determin e the required siz e o f the footin g for th e
same settlement of 25 mm.
Solution
Use Eq. (13.3). For the two plate load test s we may write:
PL Tl: A
pl
= 0.3 x 0.3 = 0.09m
2
; P
pl
= 0.3 x 4 = 1.2m; Q
l
= 30 kN
PL T2: A
p2
=0.6x0. 6 = 0.36m
2
; P
p2
= 0.6 x 4 = 2.4m; Q
2
= 90 kN
Now we have
30 = 0.09m + 1.2n
90 = 0.36m + 2.4n
On solving the equations we have
m = 166.67, and n = 12.5
For prototype foundation, we may write
Q
f
= 1 66.67 A
f
+ 12.5 P
f
Assume the size of the footing as B x B, we have
A
f
= B
2
, P
f
= 4B, an d Q
f
= 1000 kN
Substituting we have
1000 =166.67fl
2
+505
or B
2
+0.35-6 = 0
The solutio n gives B = 2.3 m.
The siz e of the footing = 2.3 x 2. 3 m.
554 Chapt er 1 3
1 3 .3 EFFEC T OF SI Z E OF FOOTI NG S ON SETTL EM EN T
Figure 13.3 a give s typica l load-settlemen t relationship s fo r footing s o f differen t width s o n th e
surface o f a homogeneous san d deposit . I t can b e see n tha t th e ultimat e bearin g capacitie s o f th e
footings pe r uni t area increas e with the increase i n the widths of the footings. However, fo r a given
settlement 5 , such as 25 mm, the soi l pressure i s greater for a footing of intermediate widt h B
b
than
for a large footing with B
C
. The pressures correspondin g t o the three widths intermediate, larg e and
narrow, ar e indicated by point s b, c and a respectively.
The same dat a is used to plot Fig. 13.3 b which shows the pressure per unit area correspondin g
to a given settlement 5j , a s a function o f the widt h o f the footing. The soi l pressur e fo r settlemen t
S
l
increase s fo r increasin g widt h o f th e footing , i f th e footing s ar e relativel y small , reache s a
maximum a t an intermediat e width , and then decreases graduall y with increasing width.
Although th e relation shown in Fig. 13.3 b i s generall y vali d for th e behavior o f footings o n
sand, it is influenced by several factors including the relative density of sand, the depth a t which the
foundation i s established, an d th e position o f the wate r table. Furthermore , th e shap e o f the curve
suggests tha t for narro w footing s smal l variation s i n th e actua l pressure , Fig . 13.3a , ma y lea d t o
large variation i n settlement and in some instances to settlements s o large that the movement would
be considered a bearing capacit y failure . O n the othe r hand , a smal l chang e i n pressur e o n a wide
footing ha s little influence on settlements as small as S
{
, and besides, the value of q
l
correspondin g
to Sj i s far below tha t which produces a bearing capacit y failur e of the wid e footing .
For al l practica l purposes , th e actua l curv e give n i n Fig . 13.3 b ca n b e replace d b y a n
equivalent curv e omn wher e o m i s th e incline d par t an d mn th e horizonta l part . Th e horizonta l
portion of the curve indicates tha t the soi l pressure corresponding t o a settlement S
{
i s independent
of the siz e of the footing. The incline d portion om indicates th e pressur e increasin g wit h width for
the same given settlement S
{
u p to the point m on the curve which is the limi t for a bearing capacit y
Soil pressure, q
Given settlement , S\
Narrow footin g
(a)
(b)
Width o f footing, B
Figure 13. 3 Load-s et t lem en t cur ve s fo r foot in g s o f differen t s ize s
(Peck e t al. , 1974 )
Shallow Foun dat io n II : Saf e B earin g Pres s ur e an d Set t lem en t Calculat io n 55 5
failure. Thi s mean s tha t th e footing s u p t o siz e B
l
i n Fig. 13. 3b shoul d b e checke d fo r bearin g
capacity failur e also whil e selecting a safe bearing pressur e by settlement consideration .
The position of the broken lines omn differs fo r different san d densities o r in other words for
different SP T N values . Th e soi l pressur e tha t produce s a give n settlemen t S
l
o n loos e san d i s
obviously smalle r tha n the soi l pressure that produces th e same settlemen t on a dense sand . Sinc e
N - value increases wit h density of sand, q
s
therefore increases wit h an increase i n the value of N .
13.4 DESIG N CH ARTS FROM SPT V ALUES FOR FOOTINGS ON SAND
The methods suggeste d by Terzaghi et al., (1996) for estimating settlement s an d bearing pressures
of footings founded on sand from SP T values ar e based o n the finding s o f Burland and Burbidge
(1985). The SPT values used are corrected t o a standard energy ratio. The usual symbol N
cor
i s used
in all the cases as the corrected value .
Formulas fo r Settlemen t Calculation s
The followin g formulas were developed fo r computing settlement s fo r squar e footings .
For normall y consolidate d soil s and gravels
cor
For preconsolidated san d and gravel s
(13.6)
for q
s
> p
c
S
c
=B°."-( q
s
-0.67p
c
)
(13
.
7a)
cor
—!± (I3.7b )
N
IA
cor
If the footing is established a t a depth below the ground surface, the removal of the soil above
the bas e leve l make s th e san d belo w th e bas e preconsolidate d b y excavation . Recompressio n i s
assumed fo r bearin g pressure s u p t o preconstruction effectiv e vertica l stres s q'
o
at the bas e o f the
foundation. Thus , for sands normall y consolidated wit h respect t o the original groun d surface and
for value s of q
s
greater tha n q'
o
, we have,
for q
s
> q'
0
S
c
= B
0
'
75
-—( q
s
-Q.61q'
0
)
(1 3 8a)
™ cor
for q
s
<q'
0
S
£
= jfi°-
75
—— q
s
(13.8b )
where
S
c
= settlemen t o f footing , i n mm , a t th e en d o f constructio n an d applicatio n o f
permanent live loa d
B - widt h of footing in m
q
s
= gros s bearin g pressur e o f footin g = QIA, i n kN/m
2
base d o n settlemen t
consideration
Q = tota l load o n the foundation in kN
A = are a of foundation i n m
2
p = preconsolidatio n pressure i n kN/m
2
556 Chapt er 1 3
0.1 1 1 0
Breadth, B( m) — lo g scal e
Figure 13. 4 T hick n es s o f g ran ul a r s oi l ben eat h foun dat io n con t ribut in g t o
s et t lem en t , in t erpret e d fro m s et t lem en t profile s (aft e r B urlan d an d Burbidg e 1985 )
q - effectiv e vertica l pressure at base level
N = averag e correcte d N valu e withi n the dept h o f influenc e Z
;
below th e bas e th e o f
footing
The dept h of influence Z
;
is obtained from
Z^B
0
-
15
(13.9 )
Figure 13. 4 give s th e variatio n o f th e dept h o f influenc e wit h dept h base d o n Eq . (13.9 )
(after Burlan d and Burbidge , 1985) .
The settlemen t of a rectangular footing of size B x L may b e obtaine d from
2
(13.10)
L 1.25( 1/8)
S( L /B> l) = S —= 1 -
c
B L I 5 + 0.25
when th e rati o L IB i s ver y high for a stri p footing, we may writ e
S
c
(strip)
S
r
(square)
= 1.5 6
(13.11)
It may be noted here that the ground water table at the sit e may lie above or within the dept h
of influenc e Z
r
Burlan d an d Burbidg e (1985 ) recommen d n o correctio n fo r th e settlemen t
calculation even if the water table lies within the depth o f influence Z
;
. On the other hand, if for any
reason, th e water table were t o rise int o or above the zone of influence Z
7
after the penetration test s
were conducted, th e actual settlement could be as much as twice the value predicted without taking
the water table int o account.
Shallow Foun dat io n II : Saf e B earin g Pres s ur e an d Set t lem en t Cal cul at io n 557
Chart fo r Estimatin g Allowabl e Soi l Pressur e
Fig. 13. 5 give s a chart for estimating allowable bearing pressur e q
s
(on settlemen t consideration )
corresponding t o a settlement of 1 6 mm fo r differen t value s of T V (corrected). Fro m Eq . (13.6) , an
expression fo r q ma y be written as (for normally consolidated sand )
N
IA yyl.4
1.7fl°-
75
where Q —
1.75
0.75
(13.12a)
(13,12b)
For sand having a preconsolidation pressur e p
c
, Eq. (13.7) may be written as
for q
s
> p
c
q
s
=16Q+Q.61p
c
(13.13a )
for q
s
<p
c
q
s
=3x\ 6Q (13.13b )
If the sand beneath the base of the footing is preconsolidated because excavation has removed
a vertical effectiv e stres s q'
o
, Eq. (13.8) ma y be written as
for q
s
> q'
o
, q
s
=16Q+Q.61q'
o
for q
s
<q'
0
, q
s
(13.14a)
(13.14b)
1 2 3 4 5 6 7 8 9 1 0 2 0 3 0
Width of footing (m )
Figure 13.5 Char t fo r es t im at in g allowabl e s oi l pres s ur e for foot in g o n s an d o n t he
bas is o f res ult s o f s t an dar d pen et rat io n t es t . (T erzag hi , e t al. , 1996 )
558 Chapt e r 1 3
The char t m Fig . 13. 5 give s th e relationship s between B an d Q . The valu e o f q
s
ma y b e
obtained fro m Q for any given width B. The Q to be use d must conform t o Eqs (13.12) , (13.13 )
and (13.14) .
The char t i s constructed for square footings of width B. For rectangular footings, the value of
q
s
shoul d b e reduce d i n accordanc e wi t h Eq . (13.10) . Th e bearin g pressure s determine d b y thi s
procedure correspon d t o a maxi mum settlement of 25 mm a t the end o f construction.
It may be noted her e that the design chart (Fig. 13.5b ) has been develope d b y taking the SPT
values corrected fo r 60 percent of standar d energy ratio.
Example 1 3 . 4
A squar e footin g of siz e 4 x 4 m i s founded at a depth o f 2 m below th e groun d surface i n loose t o
medium dens e sand . The correcte d standar d penetratio n tes t valu e N
cor
= 1 1 . Comput e th e saf e
bearing pressur e q
s
b y usin g th e char t i n Fig . 13.5 . Th e wate r tabl e i s a t th e bas e leve l o f th e
foundation.
Solution
From Fig. 13. 5 Q = 5 for B = 4 m and N
cor
= 11 .
From Eq . (13.12a )
q = 160 = 16x5 = 80 kN/m
2
Example 1 3 . 5
Refer t o Exampl e 13.4 . I f the soi l at the sit e is dense san d wit h N
cor
= 30, estimat e q
s
for B = 4 m.
Solution
From Fig. 13. 5 Q =2 4 fo r B = 4m an d N =30 .
^*~- cor
FromEq. (13.12a)
<7
s
= 16Q = 16 x 24 = 384 kN/m
2
1 3 .5 EM PI RI CA L EQU ATI ON S B ASED ON SPT V AL U ES FOR
FOOTI NG S ON COH ESI ONL ESS SOI L S
Footings o n granular soil s ar e sometimes proportione d usin g empirical relationships . Teng (1969 )
proposed a n equatio n fo r a settlemen t o f 2 5 mm base d o n th e curve s develope d b y Terzaghi an d
Peck (1948) . The modifie d for m of the equation is
(13.15a)
where q - ne t allowable bearin g pressure fo r a settlement of 25 mm in kN/m
2
,
N
cor
= corrected standar d penetration value
R = water tabl e correction facto r (Refe r Section 12.7 )
WZ
F
d
= depth facto r = d + D
f
I B) < 2.0
B = width of footing i n meters ,
D,= depth o f foundation in meters .
Shallow Foun dat io n II : Saf e B earin g Pres s ur e an d Set t lem en t Calculat io n 55 9
Meyerhof (1956 ) propose d th e following equation s which are slightl y different fro m tha t of
Teng
q
s
=\ 2N
cor
R
w2
F
d
fo r 5<1. 2 m (13.15b )
R
w2
F
d
forB> L 2m (13.15c )
where F
d
= ( l + 0.33 D
f
/B) < 1.33.
Experimental result s indicat e tha t th e equation s presente d b y Ten g an d Meyerho f ar e to o
conservative. Bowles ( 1 996) proposes a n approximate increase o f 50 percent ove r that of Meyerhof
which can also be applied t o Teng' s equations . Modified equations of Teng an d Meyerhof are ,
Teng's equatio n (modified),
^=5 3( Af
c o r
- 3) — ±^- R
w2
F
d
(13.16a )
Meyerhof 's equation (modified )
q
s
= 20N
cor
R
w2
F
d
forB<L 2 m (13.16b )
s
c o r R
w2
F
d
forB> l2m
(13
.
16c)
If the tolerabl e settlemen t i s greater tha n 25 mm, the saf e bearing pressur e compute d b y the
above equations can be increased linearl y as,
where q'
s
= net saf e bearin g pressur e fo r a settlement S'mm, q
s
= net saf e bearing pressur e fo r a
settlement o f 25 mm.
13.6 SAF E BEARING PRESSUR E FROM EMPIRICAL EQUATI ONS
B ASED ON CPT V AL U ES FO R FOOTI NG S O N COH ESI ONL ESS SOI L
The static cone penetration test in which a standard cone of 10 cm
2
sectional are a i s pushed into the
soil without the necessity of boring provides a much more accurate and detailed variation in the soil
as discusse d i n Chapte r 9 . Meyerho f (1956 ) suggeste d a se t of empirica l equation s base d o n th e
Terzaghi and Peck curve s (1948). As these equations were also found t o be conservative, modified
forms wit h an increase o f 50 percent over the original values are given below.
q
s
= 3.6q
c
R
w2
kPa fo r B < 1.2 m(13.17a )
( n
2
q
s
=2.lq
c
1 + -R
w2
kPa fo r 5>1. 2 m (13.17b )
V DJ
An approximat e formul a for al l widths
q
s
=2.7q
c
R
w2
kPa (13.17c )
where q
c
is the cone point resistance i n kg/cm
2
and q
s
in kPa.
The abov e equations have been develope d fo r a settlement of 25 mm.
560 Chapt e r 1 3
Meyerhof (1956 ) develope d hi s equation s base d o n th e relationshi p q
c
= 4N
cor
kg/cm
2
fo r
penetration resistanc e i n sand wher e N
cor
i s the corrected SP T value.
Example 1 3 . 6
Refer t o Example 13. 4 and compute q
s
by modified (a) Teng's method , and (b) Meyerhof 's method.
Solution
(a) Teng' s equatio n (modified ) — Eq. (13.16a )
i f D '
where R
w2
= -^1 +- j = 0.5 since D
w2
= 0
F, = \ +—£- = 1 + - = 1. 5 < 2 ,
d
B 4
By substituting
q
s
-53(1 1 -3)1—1 x 0.5 x1.5 - 92 kN/m
2
(b) Meyerhof 's equation (modified ) — Eq. (13.16c)
where R ,=0.5, F , = l + 0. 33x—
f
- = l + 0.33x- =1. 1 65 < 1.33
w2 d
B
4
By substituting
2
<?
y
= 12. 5x11— x0. 5x!. 165-93kN/ m
2
Note: Bot h th e methods giv e the same result .
Example 1 3 . 7
A footing of size 3 x 3 m i s to be constructed a t a site at a depth of 1 .5 m below th e ground surface .
The wate r tabl e i s a t th e bas e o f th e foundation . The averag e stati c con e penetratio n resistanc e
obtained a t th e sit e i s 2 0 kg/cm
2
. Th e soi l i s cohesive . Determin e th e saf e bearin g pressur e fo r a
settlement o f 40 mm.
Solution
UseEq. (13.17b )
Shallow Foun dat io n II : Saf e B earin g Pres s ur e an d Set t lem en t Cal cul at io n 56 1
B
where q
c
= 20 kg/cm
2
, B = 3m,R
w2
= 0.5.
This equation is for 25 mm settlement . By substituting , we have
q
s
= 2.1 x 201 1 + -I x 0.5 = 37.3 kN/m
2
For 40 mm settlement , the valu e of q i s
40
q =37. 3 — =6 0 kN/m
2
*
s
2 5
1 3 .7 FOU NDATI O N SETTL EM EN T
Components o f Total Settlement
The tota l settlemen t of a foundation comprise s thre e part s a s follows
S = S
e
+S
c
+S
s
(13.18 )
where S = tota l settlement
S = elasti c or immediate settlemen t
S
c
= consolidatio n settlemen t
S
s
= secondar y settlemen t
Immediate settlement , S
e
, i s tha t par t o f th e tota l settlement , 5
1
, whic h i s suppose d t o tak e
place during the application of loading. The consolidation settlement is that part which is due to the
expulsion o f por e wate r fro m th e void s an d i s time-dependen t settlement . Secondar y settlemen t
normally start s wit h th e completio n o f th e consolidation . I t means , durin g th e stag e o f thi s
settlement, the pore water pressure is zero and the settlement is only due to the distortion of the soil
skeleton.
Footings founde d i n cohesionles s soil s reac h almos t th e fina l settlement , 5 , durin g th e
construction stag e itsel f du e t o th e hig h permeabilit y o f soil . Th e wate r i n th e void s i s expelle d
simultaneously wit h th e applicatio n o f loa d an d a s suc h th e immediat e an d consolidatio n
settlements i n such soil s ar e rolled int o one.
In cohesive soil s unde r saturated conditions , ther e i s no change i n the wate r conten t during
the stag e o f immediat e settlement . The soi l mas s i s deformed withou t any change i n volume soo n
after th e applicatio n o f th e load . Thi s i s du e t o th e lo w permeabilit y o f th e soil . Wit h th e
advancement of time ther e wil l be gradual expulsion of water under the imposed exces s load. The
time required fo r the complet e expulsio n of water and to reach zer o water pressure may be several
years dependin g upo n the permeabilit y of the soil . Consolidation settlemen t ma y take many years
to reach it s fina l stage . Secondar y settlemen t is supposed t o take place afte r th e completion o f the
consolidation settlement , thoug h in some o f the organi c soil s ther e wil l be overlappin g of the two
settlements t o a certain extent .
Immediate settlement s of cohesive soil s and the total settlement of cohesionless soil s may be
estimated fro m elasti c theory . Th e stresse s an d displacement s depen d o n th e stress-strai n
characteristics o f the underlyin g soil. A realisti c analysi s i s difficul t becaus e thes e characteristic s
are nonlinear . Results from th e theory of elasticity are generally used i n practice, i t being assume d
that th e soi l i s homogeneou s an d isotropi c an d ther e i s a linea r relationshi p betwee n stres s an d
562
Chapt er 1 3
Overburden pressure , p
0
Combined p
0
an d Ap
D
5
= 1. 5t o2 B
0.1 t o 0. 2
Figure 13. 6 Overburde n pres s ur e an d vert ical s t res s dis t ribut io n
strain. A linea r stress-strai n relationshi p i s approximatel y tru e whe n th e stres s level s ar e lo w
relative t o the failur e values . The us e of elastic theor y clearl y involves considerable simplification
of the rea l soil .
Some o f th e result s fro m elasti c theor y requir e knowledg e o f Young' s modulu s ( E
s
), her e
called th e compression o r deformation modulus, E
d
, an d Poisson' s ratio, jU , fo r th e soil .
Seat o f Settlemen t
Footings founde d a t a dept h D, below th e surfac e settl e unde r th e impose d load s du e t o th e
compressibility characteristic s o f th e subsoil . Th e dept h throug h whic h th e soi l i s compresse d
depends upo n th e distributio n of effectiv e vertica l pressur e p'
Q
o f th e overburde n an d th e vertica l
induced stres s A/ ? resulting from th e ne t foundation pressur e q
n
as shown in Fig. 13.6 .
In th e case of deep compressible soils , the lowest level considered i n the settlemen t analysis
is the poin t wher e the vertica l induced stress A/ ? i s of the orde r o f 0. 1 t o 0.2q
n
, where q
n
i s the net
pressure on the foundation from th e superstructure. This dept h works out to about 1. 5 to 2 times the
width o f th e footing . Th e soi l lyin g withi n thi s dept h get s compresse d du e t o th e impose d
foundation pressur e an d causes mor e than 80 percent o f the settlement of the structure . This dept h
D
S
i s called a s the zone of significant stress. If the thickness of this zone is more than 3 m, the steps
to be followed i n the settlemen t analysis are
1. Divid e the zone o f significan t stres s int o layers of thickness not exceeding 3 m,
2. Determin e the effective overburde n pressure p'
o
at the center of each layer ,
3. Determin e the increas e i n vertical stress Ap due t o foundation pressure q a t the cente r of
each laye r along the center line of the footing by th e theor y of elasticity,
4. Determin e th e averag e modulu s of elasticit y an d othe r soi l parameter s fo r eac h o f th e
layers.
1 3 .8 EV AL U ATI O N OF M ODU L U S O F EL ASTI CI TY
The mos t difficul t par t of a settlement analysis is the evaluation of the modulus of elasticity E
s
, that
would confor m t o th e soi l conditio n i n th e field . Ther e ar e tw o method s b y whic h E
s
ca n b e
evaluated. They ar e
Shallow Foun dat io n II : Saf e B earin g Pres s ur e an d Set t lem en t Cal cul at io n 56 3
1. Laborator y method ,
2. Fiel d method .
L aboratory M etho d
For settlement analysis, the values of E
s
a t different depth s below the foundation base are required.
One wa y o f determinin g E
s
i s t o conduc t triaxia l test s o n representativ e undisturbe d sample s
extracted from th e depths required. For cohesive soils, undrained triaxial tests an d for cohesionles s
soils draine d triaxia l test s ar e required . Sinc e i t i s practicall y impossibl e t o obtai n undisturbed
sample o f cohesionles s soils , th e laborator y metho d o f obtainin g E
s
ca n b e rule d out . Eve n wit h
regards t o cohesive soils , ther e will be disturbance to the sampl e at different stage s of handling it,
and a s suc h th e value s o f E
S
obtaine d fro m undraine d triaxial test s d o no t represen t th e actua l
conditions an d normall y give very lo w values . A suggestio n i s t o determin e E
s
ove r th e rang e o f
stress relevan t to the particular problem. Poulo s et al. , (1980) suggest that the undisturbed triaxial
specimen b e given a preliminary preconsolidation under K
Q
conditions with an axial stress equal to
the effectiv e overburde n pressur e a t th e samplin g depth . Thi s procedur e attempt s t o retur n th e
specimen t o its original state of effective stres s i n the ground, assuming that the horizontal effectiv e
stress in the ground was the same as that produced by the laboratory K
Q
condition. Simons and Som
(1970) have shown that triaxial tests on London clay in which specimens wer e brought back to their
original i n situ stresse s gav e elasti c modul i whic h wer e muc h highe r tha n thos e obtaine d fro m
conventional undrained triaxial tests. This has been confirmed by Marsland (1971) who carried out
865 mm diamete r plat e loadin g test s i n 900 mm diamete r bore d hole s i n London clay . Marslan d
found tha t th e averag e modul i determine d fro m th e loadin g test s wer e betwee n 1. 8 to 4. 8 time s
those obtained fro m undraine d triaxial tests. A suggestion t o obtai n the mor e realisti c valu e for E
s
is,
1. Undisturbe d samples obtained from th e field mus t be reconsolidated unde r a stress system
equal t o that in the field (^-condition) ,
2. Sample s mus t be reconsolidate d isotropicall y t o a stres s equa l t o 1/ 2 to 2/ 3 o f the i n situ
vertical stress .
It ma y b e note d her e tha t reconsolidatio n o f disturbe d sensitiv e clay s woul d lea d t o
significant change in the water content and hence a stiffer structur e which would lead to a very high
E,-
Because o f th e man y difficultie s face d i n selectin g a modulu s valu e fro m th e result s o f
laboratory tests , i t has bee n suggeste d tha t a correlation betwee n th e modulu s of elasticit y o f soi l
and th e undraine d shear strengt h may provid e a basis fo r settlemen t calculation. The modulu s E
may b e expressed a s
E
s
= Ac
u
(13.19 )
where the value of A for inorganic stif f cla y varies from about 500 to 150 0 (Bjerrum , 1972 ) an d c
u
is the undrained cohesion. I t may generall y be assumed that highly plastic clays give lower values
for A, and lo w plasticit y give higher values for A. For organic or sof t clay s th e value of A may vary
from 10 0 t o 500 . Th e undraine d cohesio n c
u
ca n b e obtaine d fro m an y on e o f th e fiel d test s
mentioned below an d also discussed i n Chapter 9.
Field methods
Field method s ar e increasingl y use d t o determin e th e soi l strengt h parameters . The y hav e bee n
found t o b e mor e reliabl e tha n th e one s obtaine d fro m laborator y tests . Th e fiel d test s tha t ar e
normally use d for thi s purpose ar e
1. Plat e loa d test s (PLT)
564 Chapt e r 1 3
Table 1 3. 2 Equat ion s fo r com put in g E
s
by m ak in g us e o f SP T an d CP T values (i n
k Pa)
Soil SP T CP T
Sand (normal l y consolidated ) 50 0 ( N
cor
+ 1 5 ) 2 t o 4 q
c
(35000 t o 50000) lo g N
cor
( \ +D
r
2
)q
c
(U.S.S.R Practice )
Sand (saturated ) 25 0 ( N +15 )
Sand (overconsolidated ) - 6 to 30 q
c
Gravelly san d an d grave l 120 0 ( N + 6)
Clayey san d 32 0 ( N
cor
+15 ) 3 to 6 q
c
Silty san d 30 0 ( N
cor
+ 6) 1 to 2 q
c
Soft cla y - 3 to 8 q
c
2. Standar d penetratio n tes t (SPT )
3. Stati c con e penetratio n tes t (CPT )
4. Pressuremete r tes t (PMT )
5. Fla t dilatomete r tes t (DMT )
Plate loa d tests , i f conducte d a t level s a t whic h E
s
i s required , giv e quit e reliabl e value s a s
compared t o laborator y tests . Since thes e test s ar e too expensive t o carr y out , they ar e rarely use d
except i n major projects .
Many investigator s have obtaine d correlations betwee n E
g
an d fiel d test s suc h a s SPT, CP T
and PMT. Th e correlation s betwee n E
S
an d SPT or CPT are applicable mostl y t o cohesionless soil s
and in some cases cohesive soils under undrained conditions. PMT can be used for cohesive soil s to
determine bot h the immediat e and consolidation settlement s together .
Some o f the correlations o f £
y
with N and q
c
are given i n Table 13.2 . Thes e correlations hav e
been collecte d fro m variou s sources.
1 3 .9 M ETH OD S OF COM PU TI NG SETTL EM ENTS
Many methods ar e available for computing elastic (immediate) and consolidation settlements . Onl y
those method s tha t ar e o f practica l interes t ar e discusse d here . The'variou s method s discusse d i n
this chapter ar e the following :
Computation of Elasti c Settlement s
1. Elasti c settlemen t base d o n the theor y o f elasticity
2. Janb u e t al. , (1956) metho d o f determining settlemen t unde r an undrained condition.
3. Schmertmann' s metho d of calculating settlement i n granular soils b y usin g CPT values .
Computation o f Consolidatio n Settlemen t
1. e-\ og p metho d b y makin g use of oedomete r tes t data .
2. Skempton-Bjerru m method.
Shallow Foun dat io n II : Saf e B earin g Pres s ur e an d Set t lem en t Calculat io n 56 5
1 3 .1 0 EL ASTI C SETTL EM EN T B ENEAT H TH E CORNER OF A
U NI FORM L Y L OADE D FL EXI B L E AREA B ASED ON TH E TH EORY OF
EL ASTI CI TY
The ne t elasti c settlement equation for a flexible surface footing may be written as,
c
P
a - >"
2
) ,
S=B-— -I
f
(13.20a )
s
where S
e
= elasti c settlemen t
B = widt h of foundation,
E
s
= modulu s of elasticity of soil ,
fj, = Poisson' s ratio ,
q
n
= ne t foundation pressure,
7, = influenc e factor.
In Eq . (13.20a) , fo r saturate d clays , \ J L - 0.5 , and E
s
i s t o b e obtaine d unde r undraine d
conditions as discussed earlier . Fo r soils othe r than clays, the value of ^ has to be chosen suitably
and th e correspondin g valu e of E
s
ha s t o be determined . Tabl e 13. 3 gives typica l values for /i as
suggested b y Bowles (1996) .
7, is a functio n o f th e L I B rati o o f th e foundation , and th e thicknes s H o f th e compressibl e
layer. Terzaghi has a given a method of calculating 7, from curve s derived by Steinbrenner (1934) ,
for Poisson' s ratio of 0.5, 7, = F
1?
for Poisson' s rati o of zero, 7, = F
7
+ F
2
.
where F
{
an d F
2
are factors whic h depend upo n the ratios of H/B and L IB.
For intermediate values of //, the value of I
f
can be computed by means of interpolation or by
the equation
( l-f,-2f,
2
)F
2
(13.20b)
The value s o f F j an d F
2
ar e give n i n Fig . 13.7a. Th e elasti c settlemen t a t an y poin t N
(Fig. 13.7b ) i s given by
(I - / /
2
)
S
e
a t point N = -S-_ [/ ^ + I
f2
B
2
+ 7
/3
7?
3
+ 7
/4
7?
4
]
(1 3 20c)
Table 13. 3 T ypica l ran g e of val ue s for Pois s on ' s rat i o (B owles , 1996 )
T ype o f s oi l y.
Clay, saturated 0.4-0. 5
Clay, unsaturated 0.1-0. 3
Sandy cla y 0.2-0. 3
Silt 0.3-0.3 5
Sand (dense ) 0.2-0. 4
Coarse (voi d rati o 0. 4 to 0.7) 0.1 5
Fine graine d (voi d rati o = 0.4 t o 0.7) 0.2 5
Rock 0.1-0. 4
566 Chapt er 1 3
V alues of F, _) a n d F
2
( _ _ _
0.1 0. 2 0. 3 0. 4 0. 5 0. 6 0. 7
Figure 1 3 . 7 Set t lem en t due t o loa d o n s urfac e o f el as t i c l aye r (a ) F
1
an d F
2
vers u s
H/B (b) M et hod o f es t im at in g s et t lem en t (Aft e r St ein bren n er , 1934 )
To obtai n th e settlemen t a t th e cente r o f th e loade d area , th e principl e o f superpositio n i s
followed. I n such a case N in Fig. 13.7 b will be at the center o f the area whe n B
{
= B
4
= L
2
= B
3
and
B
2
= L
r
Then the settlement a t the center is equal t o four times the settlement a t any one corner. The
curves i n Fig. 13.7 a ar e based o n the assumption that the modulus of deformation i s constant wit h
depth.
In the case of a rigid foundation , the immediat e settlemen t a t the center i s approximately 0. 8
times tha t obtaine d fo r a flexibl e foundatio n a t th e center . A correctio n facto r i s applie d t o th e
immediate settlemen t t o allow for the depth o f foundation by means o f the depth facto r d~ Fig . 13. 8
gives Fox' s (1948) correctio n curv e for dept h factor . The final elasti c settlemen t i s
(13.21)
where,
"f =
s =
final elastic settlemen t
rigidity factor take n a s equal t o 0. 8 for a highl y rigid foundatio n
depth facto r fro m Fig . 13. 8
settlement fo r a surface flexibl e footing
Bowles (1996 ) has given the influence factor for various shape s o f rigi d and flexible footings
as shown i n Table 13.4 .
Shallow Foun dat io n II : Saf e B earin g Pres s ur e an d Set t lem en t Calculat io n
Table 13. 4 In fluen c e fact o r l
f
(B owles , 1988 )
567
Shape
Circle
Square
Rectangle
L/B = 1. 5
2.0
5.0
10.0
100.0
Flex ible
0.85
0.95
1.20
1.20
1.31
1.83
2.25
2.96
l
f
(aver ag e values )
foot in g Rig i d foot ing
0.88
0.82
1.06
1.06
1.20
1.70
2.10
3.40
Corrected settlemen t for foundation of depth D ,
T~lr-ritli fnr^tn r —
Calculated settlemen t for foundation at surface
Q.50 0.6 0 0.7 0 0.8 0 0.9 0 1. 0
j
Df/^BL
• V
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1 n
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
_ n
1-
9-
L
r
' D,
r
i
m
'//
7TT
y
/
//
^/
r /
1
1
!
!
'/
1
j
\ l l >
//
\ r
I
k
1
I
/
100
25-
/
I//
//
<
— »
<7/
'/<
/
/
'rf
4y
/^
*l
/
-9
- 1
Numbers denote ratio L/B
25
-100
Figure 13. 8 Correct io n curve s for elas t i c s et t lem en t o f flex ibl e rect an g ula r
foun dat ion s a t dept h (Fox , 1948)
568 Chapt er 1 3
1 3 .1 1 JANB U , B JERRU M AND KJAERNSL I ' S M ETH OD OF
DETERM I NI NG EL ASTI C SETTL EM ENT UNDE R U NDRAI NE D
CONDI TI ONS
Probably th e mos t usefu l char t i s tha t give n by Janb u e t al. , (1956 ) a s modifie d b y Christia n an d
Carrier (1978 ) fo r th e cas e o f a constant E
s
wit h respect t o depth . Th e char t (Fig . 13.9 ) provide s
estimates o f th e averag e immediat e settlemen t o f uniforml y loaded , flexibl e strip , rectangular ,
square o r circular footing s on homogeneous isotropi c saturate d clay . The equatio n fo r computing
the settlement ma y be expressed a s
S =
(13.22)
In Eq. (13.20), Poisson' s rati o i s assumed equa l to 0.5. Th e factors fi
Q
an d ^are related t o the
DJ B an d HIB ratio s o f the foundatio n as shown in Fig. 13.9 . V alues of \ J L ^ ar e give n for variou s L I B
ratios. Rigidit y and dept h factor s ar e require d t o be applie d t o Eq . (13.22 ) a s pe r Eq . (13.21) . I n
Fig. 13. 9 th e thicknes s o f compressibl e strat a i s take n a s equa l t o H belo w th e bas e o f th e
foundation wher e a hard stratum i s met with.
Generally, rea l soi l profile s whic h ar e deposite d naturall y consis t o f layer s o f soil s o f
different propertie s underlai n ultimatel y b y a har d stratum . Withi n thes e layers , strengt h an d
moduli generall y increas e wit h depth. The char t give n i n Fig. 13. 9 ma y b e use d fo r th e cas e of E
S
increasing wit h dept h b y replacing th e multilayere d system wit h one hypothetical laye r o n a rigi d
D
1.0
0.9
Incompressible
10
D
f
/B
15 2 0
1000
Figure 13. 9 Fact or s fo r calculat in g t h e averag e im m ediat e s et t lem en t o f a loade d
area (aft e r Chris t ia n an d Carrier , 1978 )
Shallow Foun dat io n II : Saf e B earin g Pres s ur e an d Set t lem en t Calculat io n 56 9
base. Th e dept h o f thi s hypothetica l layer i s successivel y extende d t o incorporat e eac h rea l layer ,
the correspondin g value s o f E
s
bein g ascribe d i n eac h cas e an d settlement s calculated . B y
subtracting th e effec t o f th e hypothetica l layer abov e eac h rea l laye r th e separat e compressio n o f
each laye r may be found an d summed t o give the overal l tota l settlement .
1 3 .1 2 SCH M ERTM ANN' S M ETH O D O F CAL CU L ATI NG
SETTL EM ENT I N G RANU L AR SOI L S B Y USING CR T V AL U ES
It i s normall y taken fo r granted tha t the distribution of vertical strai n under the center o f a footin g
over uniform sand is qualitatively similar to the distribution of the increase i n vertical stress. I f true,
the greatest strai n would occur immediatel y under the footing, which is the position of the greatest
stress increase . Th e detaile d investigation s of Schmertmann (1970), Eggestad , (1963 ) an d others,
indicate that the greatest strai n would occur at a depth equal to half the width for a square or circular
footing. Th e strai n is assumed t o increase fro m a minimum at the base t o a maximum at B/2, then
decrease and reaches zer o at a depth equal to 2B. For strip footings of L/B > 10, the maximum strain
is found t o occur at a depth equal to the width and reaches zer o at a depth equal to 4B. The modified
triangular vertical strain influence factor distribution diagram a s proposed by Schmertmann (1978)
is show n i n Fig. 13.10 . The are a o f thi s diagra m i s relate d t o th e settlement . Th e equatio n (for
square a s well as circular footings) is
IB l
-jj-te (13.23 )
^ s
where, S = tota l settlement ,
q
n
= ne t foundation base pressur e = ( q - q'
Q
),
q = tota l foundatio n pressure,
q'
0
= effectiv e overburden pressur e a t foundation level,
Az = thicknes s o f elemental layer ,
l
z
= vertica l strain influence factor,
Cj = dept h correctio n factor ,
C
2
= cree p factor .
The equations for C
l
and C
2
are
c
i
= 1
~ 0- 5 -7 - (13.24 )
C
2
= l + 0.21og
10 (13>2
5)
where t is time in years for which period settlemen t i s required.
Equation (13.25 ) i s also applicabl e fo r L IB > 10 except tha t the summatio n is from 0 to 4B.
The modulu s of elasticit y t o be used i n Eq. (13.25 ) depends upo n th e typ e of foundation as
follows:
For a square footing ,
E
s
= 2.5q
c
(13.26 )
For a strip footing, L IB > 10,
E = 3 . 5 f l (13.27 )
570 Chapt er 1 3
Rigid foundatio n vertical strai n
influence facto r I
z
0 0. 1 0. 2 0. 3 0. 4 0. 5 0. 6
t
f
l
t
o
N
>
C
o
> 3 B
C*
4B
L
L/B > 1 0
>P e a k / = 0.5 + 0.1.
D
/;
.
•>
I H
- 1,,
m
B/2 for L I B = 1
B for L IB > 10
ill
HI
Depth t o peak /,
Figure 13.1 0 Vert ica l strain Influenc e fact o r diagram s (aft e r Schmertman n e t al. , 1978 )
Fig. 13.1 0 give s th e vertica l strai n influenc e factor /
z
distributio n for bot h squar e an d stri p
foundations i f the ratio L IB > 10. V alues for rectangular foundations for L IB < 10 can be obtained by
interpolation. Th e depth s a t whic h th e maximu m /
z
occur s ma y b e calculate d a s follow s
(Fig 13.10) ,
(13.28)
where p'
Q
= effectiv e overburde n pressur e a t depth s B/ 2 an d B fo r squar e an d stri p
foundations respectively.
Further, / i s equal to 0.1 at the base an d zero a t dept h 2B below th e base for squar e footing;
whereas fo r a stri p foundation it is 0.2 at the base an d zer o a t depth 4B.
V alues of E
5
given in Eqs. (13.26) and (13.27) ar e suggested b y Schmertmann (1978) . Lunne
and Christoffersen (1985 ) proposed th e use of the tangent modulus on the basis of a comprehensive
review of fiel d an d laborator y test s a s follows:
For normall y consolidated sands ,
(13.29) £
5
= 4 4
c
for 9
c
< 10
E
s
= ( 2q
c
+ 20)for\ 0<q
c
<50
E
s
= 12 0 for q
c
> 50
For overconsolidated sands wit h an overconsolidation rati o greater than 2,
(13.30)
(13.31)
(13.32a)
E
s
= 250 for q
c
> 50 (13.32b )
where E
s
an d q
c
are expressed i n MPa .
The cone resistance diagram is divided into layers of approximately constant values of q
c
and
the strain influenc e factor diagram i s placed alongsid e this diagram beneat h the foundation which is
Shallow Foun dat io n II : Saf e B earin g Pres s ur e an d Set t lem en t Calculat io n 571
drawn to the same scale. The settlements of each laye r resulting from the net contact pressure q
n
are
then calculated using the values of E
s
and /
z
appropriate t o each layer. The sum of the settlements in
each laye r i s the n correcte d fo r th e dept h an d cree p factor s usin g Eqs . (13.24 ) an d (13.25 )
respectively.
Example 1 3 .8
Estimate th e immediat e settlemen t o f a concrete footin g 1. 5 x 1. 5 m in size founde d at a depth of
1 m in silty soil whose modulus of elasticity i s 90 kg/cm
2
. The footing is expected t o transmit a unit
pressure o f 200 kN/m
2
.
Solution
Use Eq. (13.20a )
Immediate settlement ,
s
=
E
Assume n = 0.35, /,= 0.82 for a rigid footing.
Given: q = 200 kN/m
2
, B = 1.5 m, E
s
= 90 kg/cm
2
« 9000 kN/m
2
.
By substitutin g the known values, we have
1-035
2
S =200xl . 5 x -
:
-x 0.82 = 0.024 m = 2.4 cm
9000
Example 1 3 . 9
A squar e footing of size 8 x 8 m is founded at a depth of 2 m below th e ground surfac e i n loose to
medium dense san d wit h q
n
= 120 kN/m
2
. Standar d penetration tests conducted a t the site gave the
following correcte d N
6Q
values .
Depth below G.L. (m)
2
4
6
8
"cor
8
8
12
12
Depth below G.L.
10
12
14
16
18
N
cor
11
16
18
17
20
The wate r tabl e i s at the base o f the foundation. Above the wate r tabl e y = 16. 5 kN/m
3
, and
submerged y
b
= 8.5 kN/m
3
.
Compute th e elasti c settlemen t b y Eq . (13.20a) . Us e th e equatio n E
s
= 250 ( N
cor
+ 15) for
computing the modulus of elasticity o f the sand. Assume ] U = 0.3 and the dept h of the compressibl e
layer = 2B= 1 6 m ( = //)•
Solution
For computing the elastic settlement , it is essential t o determine the weighted average valu e ofN
cor
.
The dept h o f th e compressibl e laye r belo w th e bas e o f th e foundatio n i s take n a s equa l t o
16 m ( = H). This dept h ma y be divide d int o three layer s i n suc h a way tha t N
cor
i s approximatel y
constant i n each laye r as given below.
572 Chapt e r 1 3
Layer No .
1
2
3
Depth (m)
From T o
2 5
5 1 1
11 1 8
Thickness
(m)
3
6
7
"cor
9
12
17
The weighte d averag e
9x3 + 12x6 + 17x7
10
^
= 1 3.6 or say 14
I D
From equatio n E
s
= 250 ( N
cor
+ 15) we have
E
s
= 250(14 + 15) = 7250 kN/m
2
The tota l settlemen t o f th e cente r o f th e footin g of siz e 8 x 8 m i s equa l t o fou r time s th e
settlement of a corner o f a footing of size 4 x 4 m.
In the Eq. (13.20a), B = 4 m, q
n
= 120 kN/m
2
, p = 0.3 .
Now fro m Fig. 13.7 , for HIB = 16/ 4 = 4, L IB = 1
F
2
= 0.03 fo r n = 0.5
Now fro m Eq. (13.2 0 b ) T^fo r / * = 0. 3 i s
q-,-2
I-/ / 1-0.3
2
From Eq . (13.20a) we have settlement of a corner o f a footing of size 4 x 4 m as
s =
,
B 7
.
e
" £ ,
7 725
°
With the correction factor , the fina l elasti c settlemen t from Eq . (13.21) is
s
ef
= c
r
d
f
s
e
where C
r
= rigidity factor = 1 for flexibl e footing d, = depth factor
From Fig . 13. 8 fo r
D
f
2 L 4
= 0.5, — = - =1 we have d
r
=0. 85 / * r* A V V \s 1 1 U. V W L* r "
V 4 x 4 B 4 f
Now 5^= 1 x 0.85 x 2.53 = 2.15 cm
The tota l elastic settlemen t of the center of the footing is
S
e
= 4 x 2.15 = 8. 6 cm = 86 mm
Per Tabl e 13.la , th e maximu m permissibl e settlemen t fo r a raf t foundatio n i n san d i s
62.5 mm . Sinc e th e calculated value is higher, the contact pressure q
n
has t o be reduced.
Shallow Foun dat io n II : Saf e B earin g Pres s ur e an d Set t lem en t Calculat io n 57 3
Example 1 3 .1 0
It is proposed t o construct an overhead tan k at a site on a raft foundatio n of size 8 x 1 2 m with the
footing a t a depth of 2 m below ground level. The soi l investigation conducted a t the sit e indicate s
that th e soi l t o a dept h of 20 m i s normall y consolidated insensitiv e inorganic clay wit h th e water
table 2 m belo w groun d level . Stati c con e penetratio n test s wer e conducte d a t th e sit e usin g a
mechanical cone penetrometer . The average value of cone penetration resistance q
c
wa s found t o
be 154 0 kN/m
2
an d th e averag e saturate d uni t weigh t o f th e soi l = 1 8 kN/m
3
. Determin e th e
immediate settlemen t o f th e foundatio n usin g Eq. (13.22) . Th e contac t pressur e q
n
= 10 0 kN/m
2
(= 0.1 MPa). Assume tha t the stratum below 20 m is incompressible .
Solution
Computation of the modulus of elasticit y
Use Eq. (13. 19) with A = 500
where c
u
= the undrained shear strengt h of the soi l
From Eq . (9.14 )
where q
c
= averag e stati c cone penetration resistance = 154 0 kN/m
2
p
o
= averag e tota l overburden pressure = 1 0x 18 = 1 80 kN/m
2
N
k
= 2 0 (assumed)
Therefore c =
154
°~
18
° = 68 kN/m
2
20
E
s
= 500 x 68 = 34,000 kN/m
2
= 34 MPa
Eq. (13. 22)forS
e
i s
_
~
From Fig. 13. 9 for DjE = 2/8 = 0.25, ^
0
= 0.95, for HIB = 16/8 = 2 and UB = 12/8 = 1 .5, ^= 0.6.
Substituting
. 0.95x0.6x0.1x 8
S
e
(average) = -= 0.0134 m = 13.4 mm
From Fig . 13. 8 for D
f
/</BL = 2/ V 8xl2 = 0.2, L / B = 1.5 the dept h factor d
f
= 0.9 4
The corrected settlemen t S
ef
i s
S =0. 94x 1 3.4 = 12. 6 mm
Example 1 3 .1 1
Refer to Example 13.9 . Estimate the elastic settlement by Schmertmann's method by making use of
the relationshi p q
c
= 4 N
cor
kg/cm
2
wher e q
c
= static con e penetratio n valu e i n kg/cm
2
. Assume
settlement i s required a t the end of a period o f 3 years.
574 Chapt er 1 3
5 x L = 8x8 m
y = 16. 5 kN/m
3
Sand
0 0. 1 0.2 0. 3 0. 4 0. 5
Strain influenc e factor , /,
Figure Ex . 1 3 .1 1
0.6 0. 7
Solution
The averag e valu e of for N
cor
eac h laye r given i n Ex. 13. 9 i s given belo w
Layer N o Average
N
Average q
c
kg/cm
2
MP a
9
12
17
36
48
68
3.6
4.8
6.8
The vertica l strai n influenc e factor / wit h respect t o dept h i s calculate d b y makin g use of
Fig. 13.10 .
At the base of the foundation 7 = 0 .1
At dept h B/2 ,
7
; = °-
5 + 0
'\H"
V rO
where q
n
= 12 0 kpa
p'
g
= effectiv e averag e overburden pressure at depth = (2 + B/2) = 6 m below ground level.
= 2 x 16. 5 + 4 x 8 .5 = 67 kN/m
2
.
Shallow Foun dat io n II : Saf e B earin g Pres s ur e an d Set t lem en t Calculat io n 57 5
I
z
(max) = 0.5 +0.1 J— = 0.63
/
z
= 0 a tz = / f= 16 m belo w bas e leve l o f the foundation. The distribution of I
z
i s given in
Fig. Ex. 13.11 . The equation for settlement is
2B I
-^Az
o E
s
where C, = 1-0.5 = 1-0.5
=0
.86
q
n
12 0
C
2
= l + 0.21og^- = l + 0.21og^- =1. 3
where t = 3 years.
The elasti c modulu s E
s
fo r normall y consolidated sand s ma y b e calculate d b y Eq . (13.29) .
E
s
= 4q
c
forq
c
<10MP a
where q
c
i s the averag e fo r eac h layer .
Layer 2 i s divided int o sublayer s 2 a and 2b for computing / . The averag e o f the influence
factors fo r each o f the layers given in Fig. Ex. 13.1 1 are tabulated along wit h the other calculations
Layer No .
1
2a
2b
3
Substituting
Az (cm)
300
100
500
700
q
c
(MPa ) E
s
3.6
4.8
4.8
6.8
; in the equation for settlemen t 5, we
5 = 0.86x1.3x0.12x26.82 = 3.6 cm = 36 mm
(MPa)
14.4
19.2
19.2
27.2
have
I
z
( av)
0.3
0.56
0.50
0.18
Total
^T
6.25
2.92
13.02
4.63
26.82
1 3 .1 3 ESTI M ATI O N OF CONSOL I DATI ON SETTLEMENT BY
U SI NG OEDOM ETE R TES T DAT A
Equations fo r Computing Settlement
Settlement calculation from e-logp curves
A genera l equatio n fo r computin g oedomete r consolidatio n settlemen t ma y b e writte n a s
follows.
Normally consolidated clay s
r, u
C
C , _/
?
0
+ A
P
s
c = //-——log (13.33 )
Po
576 Chapt e r 1 3
Overconsolidated clay s
for p
Q
+A p < p
c
c _ L J
C
s i P Q
+ /
V
O,, — ti1O 2 / 1 7 O / I N
f or / ?
0
< p
c
<p
Q
+ Ap
C,log-^- + C
c
log-^-
(13
_
35)
where C
s
= swell index, and C, = compression inde x
If the thickness of the clay stratu m is more tha n 3 m the stratum has t o be divided int o layer s
of thicknes s les s tha n 3 m . Further , <?
0
i s th e initia l voi d rati o an d p
Q
, th e effectiv e overburde n
pressure corresponding t o the particular layer; Ap is the increase i n the effective stress a t the middle
of the layer due t o foundation loading which is calculated by elastic theory. The compression index ,
and th e swel l inde x may b e the same fo r the entir e depth o r may var y from laye r t o layer .
Settlement calculation from e-p curve
Eq. (13.35 ) ca n b e expresse d i n a different for m a s follows:
S
c
=ZHm
v
kp (13.36 )
where m = coefficien t of volume compressibility
1 3 .1 4 SKEM PTOI M - B JERRU M M ETH OD O F CAL CU L ATI NG
CONSOL I DATI ON SETTL EM EN T (1 9 5 7 )
Calculation o f consolidatio n settlemen t i s base d o n on e dimensiona l tes t result s obtaine d fro m
oedometer test s on representative samples of clay. These tests do not allow any lateral yield during the
test and as such the ratio of the minor to major principal stresses, K
Q
, remains constant . In practice, th e
condition of zer o latera l strain is satisfied only in cases wher e the thickness of the clay layer is small
in compariso n wit h the loade d area . I n man y practical solutions , however , significan t lateral strai n
will occur and the initial pore wate r pressure wil l depend on the in situ stres s condition an d the value
of the pore pressure coefficien t A, which will not be equal to unity as in the case of a one-dimensional
consolidation test . In view of the lateral yield, the ratios of the minor and major principal stresses due
to a given loading conditio n at a given point in a clay layer do not maintai n a constant K
Q
.
The initia l excess por e water pressure at a point P (Fig. 13. 1 1) in the clay layer is given by the
expression
Aw = Acr
3
+ A(Acr, - A<7
3
)
ACT,
where Ao ^ an d Acr
3
ar e th e tota l principa l stress increment s due t o surfac e loading . I t can b e see n
from Eq . (13.37 )
Aw > A<7
3
i f A i s positive
and A w = ACT , ifA = \
Shallow Foun dat io n II: Saf e B earin g Pres s ur e an d Set t lem en t Calculat io n 577
The valu e of A depends o n the type of clay, the stres s level s and the stres s system.
Fig. 13. 1 la present s th e loading condition at a point in a clay laye r below th e central line of
circular footing . Figs. 13.1 1 (b) , (c ) and (d ) show the condition befor e loading , immediatel y afte r
loading an d after consolidation respectively .
By th e one-dimensional method , consolidatio n settlemen t S i s expressed as
(13.38)
By the Skempton-Bejerrum method , consolidation settlemen t i s expressed as
or S=
ACT,
A settlemen t coefficien t ( 3 is used, suc h that S
c
= ( 3S
o
The expressio n fo r ( 3 is
H
T Acr
3
"
+ —-(1-A) <f e
(13.39)
(13.40)
H
^f
h*
/
i
/
/
i
i
^
\
*s
L
o\
71
i
)
\
^u
Arr.
*
t
1
q
n
73 -
Wo
K
0
, ,
-a, - a
L
\
K
>
(b) 1
a'
0
+ Aa
3
a
0
' + Aa, - L
o'
0
+ Aa
3
- AM 1
_ L
a;
r
Aa
l ( o
1

(a)
(a) Physical plan e (b) Initial condition s
(c) Immediately afte r loadin g (d ) After consolidatio n
Figure 13.1 1 I n s itu effect ive s t res s e s
578 Chapt er 1 3
Circle
Strip
Normally consolidate d
~ I ~ *
V ery
sensitive
clays
0.2 0.4 0. 6 0. 8
Pore pressur e coefficient A
1.0 1.2
Figure 1 3 .1 2 Set t lem en t coefficien t vers u s pore-pres s ur e coefficien t fo r circula r
an d s t rip foot in g s (Aft e r Sk em pt o n an d Bjerrum , 1957)
Table 13. 5 Value s o f s et t lem en t coefficien t
T ype o f cl a y
V ery sensitiv e clays (sof t al l uvi a l an d marin e clays)
Normally consolidate d clays
Overconsolidated clay s
Heavily Overconsolidate d clays
1.0 t o 1. 2
0.7 t o 1. 0
0.5 t o 0.7
0.2 t o 0. 5
°
r
S
c
=f tS
oc
(13.41)
where f t i s called th e settlement coefficient.
If i t ca n b e assume d tha t m
v
an d A ar e constan t wit h dept h (sub-layer s ca n b e use d i n th e
analysis), the n f t can be expresse d a s
(13.42)
where a =—
dz
(13.43)
Taking Poisson' s rati o ( J i a s 0. 5 fo r a saturate d cla y durin g loadin g unde r undraine d
conditions, th e value of ( 3 depends onl y on the shape of the loaded are a and the thickness o f the clay
layer i n relatio n t o th e dimension s o f th e loade d are a an d thu s f t ca n b e estimate d fro m elasti c
theory.
The valu e of initial exces s por e wate r pressur e (Aw ) should , i n general , correspon d t o th e in
situ stres s conditions . The us e of a value of pore pressur e coefficien t A obtained fro m the result s of
Shallow Foun dat io n II : Saf e B earin g Pres s ur e an d Set t lem en t Calculat io n 57 9
a triaxia l tes t o n a cylindrical cla y specime n i s strictl y applicabl e onl y fo r th e conditio n o f axia l
symmetry, i e., for the case of settlement under the center of a circular footing. However, th e value
of A so obtained wil l serve as a good approximatio n for the case of settlement under the center of a
square footing (using the circular footing of the same area) .
Under a strip footing plane strai n conditions prevail. Scott (1963) has shown that the value of
AM appropriat e i n the case of a strip footing ca n be obtained by using a pore pressure coefficien t A
s
as
A
s
=0.86 6 A+ 0.211 (13.44 )
The coefficien t A
S
replace s A (th e coefficien t fo r th e conditio n o f axia l symmetry ) i n
Eq. (13.42) for the case of a stri p footing, th e expression for a bein g unchanged.
V alues of th e settlemen t coefficien t / 3 for circula r and stri p footings, i n terms o f A and ratio s
H/B, ar e given i n Fi g 13.12 .
Typical values of / 3 are given in Table 13. 5 for variou s types of clay soils .
Example 1 3 .1 2
For th e proble m give n i n Ex . 13.1 0 comput e th e consolidatio n settlemen t b y th e Skempton -
Bjerrum method. The compressible laye r of depth 16m below th e base of the foundation is divided
into four layer s an d th e soi l propertie s o f each laye r ar e given i n Fig. Ex . 13.12 . Th e ne t contac t
pressure q
n
= 100 kN/m
2
.
Solution
From Eq . (13.33) , th e oedometer settlemen t for the entire clay layer syste m ma y be expressed a s
C p + Ap
From Eq . (13.41) , th e consolidation settlemen t as per Skempton-Bjerrum may be expresse d
as
S
c - fi
S
oe
where / 3 = settlemen t coefficient which can be obtaine d fro m Fig . 13.1 2 for variou s values
of A and H/B.
p
o
= effectiv e overburde n pressure a t the middle o f each laye r (Fig . Ex . 13.12 )
C
c
= compressio n inde x of each laye r
//. = thicknes s of i th layer
e
o
- initia l void rati o of each laye r
Ap = th e exces s pressur e a t th e middl e o f eac h laye r obtaine d fro m elasti c theor y
(Chapter 6)
The average por e pressur e coefficient is
„ 0. 9 + 0.75 + 0.70 + 0.45 _ _
A = = 0.7
4
The detail s of the calculations are tabulated below.
580 Chapt er 1 3
G.L.
- Laye r 1
Layer 2
a,
<u
Q 1 0
12
14
16 -
18
Layer 3
Layer 4
fl xL=8x 12 m
G. L.
moist uni t weight
y
m
= 1 7 k N/ m
3
C
c
= 0.16
A = 0. 9
Submerged uni t weigh t y
b
i s
y
b
= (17.00 - 9.81 ) = 7.19 kN/m
3
e
0
= 0.8 4
C
c
= 0.14
A - 0.7 5
, = 7.69 kN/m
3
C =0. 1 1
y
fc
= 8.19kN/m
3
A = 0.70
e
0
= 0.73
C
c
= 0.09
A = 0.45
y
b
= 8.69 kN/m
3
Figure Ex . 1 3 .1 2
Layer No .
1
2
3
4
H. (cm)
400
400
300
500
p
o
(kN/m^ )
48.4
78.1
105.8
139.8
A/? (kN/m
z
75
43
22
14
0.16
0.14
0.11
0.09
6
o
0.93
0.84
0.76
0.73
ltj
b
0.407
0.191
0.082
0.041
Total
4^ (cm)
13.50
5.81
1.54
1.07
21.92
PorH/B = 16/8 = 2, A = 0.7, fro m Fig . 13.1 2 we have 0= 0.8.
The consolidatio n settlement 5
C
is
5 = 0.8 x 21.92 = 17.53 6 cm = 175.36 mm
1 3 .1 5 PROB L EM S
13.1 A plat e loa d tes t wa s conducted i n a medium dens e san d a t a dept h o f 5 f t below groun d
level i n a test pit . The size of the plate used was 1 2 x 1 2 in. The dat a obtained fro m th e test
are plotted i n Fig. Prob . 13. 1 as a load-settlement curve. Determine fro m the curve the net
safe bearin g pressur e fo r footing s o f siz e (a ) 1 0 x 1 0 ft , an d (b ) 1 5 x 1 5 ft . Assume th e
permissible settlement for the foundation i s 25 mm.
Shallow Foundatio n II : Saf e B earin g Pres s ur e an d Set t lem en t Calculat io n 581
Plate bearing pressure, lb/ft
2
2 4 6 8 x l 0
3
0.5
1.0
1.5
Figure Prob . 1 3. 1
13.2 Refe r t o Prob. 13.1 . Determine th e settlement s of the footings given in Prob 13.1 . Assume
the settlemen t o f th e plat e a s equa l t o 0. 5 in . Wha t i s th e ne t bearin g pressur e fro m
Fig. Prob. 13. 1 for the computed settlements of the foundations?
13.3 Fo r Problem 13.2 , determin e th e saf e bearin g pressure o f the footings i f the settlemen t i s
limited t o 2 in.
13.4 Refe r to Prob. 13.1 . If the curve given in Fig. Prob. 13. 1 applies t o a plate test of 12 x 1 2 in.
conducted i n a cla y stratum , determin e th e saf e bearin g pressure s o f th e footing s fo r a
settlement of 2 in.
13.5 Tw o plate load test s wer e conducted i n a c-0 soil as given below.
Size of plates (m)
0.3 x 0.3
0.6 x 0.6
Load k N
40
100
Settlement (mm)
30
30
Determine th e required siz e of a footing to carry a load of 125 0 kN for the same settlement
of 30 mm.
13.6 A rectangular footing of size 4 x 8 m is founded at a depth of 2 m below the ground surface
in dens e san d an d th e wate r tabl e i s a t th e bas e o f th e foundation . N
CQT
= 3 0
(Fig. Prob. 13.6) . Comput e the safe bearing pressur e q usin g the chart given in Fig. 13.5 .
5 x L = 4 x 8 m
I
D
f
=2m
Dense sand N
cor
( av) = 30
Figure Prob. 13. 6
582
Chapt er 1 3
13.7 Refe r to Prob. 13.6 . Comput e q
s
by using modified (a) Teng's formula , and (b) Meyerhof 's
formula.
13.8 Refe r t o Prob . 13.6 . Determin e th e saf e bearin g pressur e base d o n th e stati c con e
penetration tes t valu e based o n the relationship give n i n Eq. (13.7b) for q = 120 kN/m
2
.
13.9 Refe r t o Prob . 13.6 . Estimat e th e immediat e settlemen t o f th e footin g b y usin g
Eq. (13.20a) . The additiona l data availabl e are :
H = 0.30, I
f
= 0.8 2 fo r rigi d footin g and E
s
= 11,00 0 kN/m
2
. Assume q
n
= q
s
a s obtaine d
from Prob . 13.6 .
13.10 Refe r to Prob 13.6. Compute the immediate settlemen t for a flexible footing, given ^ = 0.30
and E
s
= 1 1,000 kN/m
2
. Assume q
n
= q
s
13.11 I f th e footin g given i n Prob . 13. 6 rest s o n normall y consolidated saturate d clay , compute
the immediat e settlemen t usin g Eq. (13.22) . Us e th e followin g relationships.
q
c
= 120 kN/m
2
E
s
= 600c
tt
kN/m
2
Given: y
sat
= 18.5 kN/m
3
,^ = 150 kN/m
2
. Assume tha t th e incompressibl e stratu m lie s
at a t dept h o f 1 0 m belo w th e bas e o f th e foundation.
13.12 A footing of size 6 x 6 m rests i n medium dense san d at a depth of 1 .5 below groun d level.
The contact pressure q
n
= 175 kN/m
2
. The compressible stratu m below the foundation base
is divide d int o thre e layers . Th e correcte d N
cor
value s fo r eac h laye r i s give n i n
Fig. Prob . 13.1 2 wit h othe r dat a . Comput e th e immediat e settlemen t usin g Eq . (13.23) .
Use the relationship q
c
= 400 N
cor
kN/m
2
.
0 1
1 e
2-
4-
6-
8-
10-
10 -
//A>
7sat =
10
15
'"cor o x o m
U 2 U
" * q
n
= 175 kN/m
2
* G.L .
V
19/kN/m
3
1 1 1 1 1 ^ ^ ! : i ^5r n
Layer 1 dens e sand
Layer 2 dens e sand
y
s a t
= 19. 5 kN/m
3
20
Layer 3 dens e sand
Figure Prob. 1 3 .1 2
Shallow Foun dat io n II : Saf e B earin g Pres s ur e an d Set t lem en t Cal cul at io n 583
13.13 I t is proposed t o construct an overhead tank on a raft foundatio n of size 8 x 1 6 m with the
foundation a t a dept h o f 2 m belo w groun d level . Th e subsoi l a t th e sit e i s a stif f
homogeneous clay with the water table at the base of the foundation. The subsoil is divided
into 3 layers an d th e propertie s o f eac h laye r ar e give n i n Fig . Prob . 13.13 . Estimat e th e
consolidation settlemen t by the Skempton-Bjerrum Method .
G.L.
e s
JS
a,
y
m
=18. 5kN/m
3
5 x L= 8 x 1 6m
q
n
= 15 0 kN/m
2
G.L.
D
f
=2m
— Laye r 1
e
0
= 0.8 5
y
sat
= 18. 5 kN/m
3
C
c
= 0.1 8
A = 0.74
Layer 2
y
s at
= 19. 3 kN/m
3
C
c
= 0.16
A = 0.83
Layer 3
e
0
= 0.68
y
sat
= 20.3 kN/m
3
C
c
= 0.13
A = 0.5 8
Figure Prob . 13.1 3
13.14 A footing of siz e 1 0 x 1 0 m i s founde d a t a depth of 2. 5 m below groun d level o n a sand
deposit. Th e wate r tabl e i s at the base o f the foundation. The saturate d uni t weight of soi l
from groun d leve l t o a dept h o f 22. 5 m i s 2 0 kN/m
3
. The compressibl e stratu m of 2 0 m
below th e foundatio n base i s divided int o three layer s wit h corrected SP T value s (/V ) an d
CPT value s ( q
c
} constan t in eac h laye r a s give n below.
Layer N o Depth fro m (m )
foun dat ion leve l
From T o
"
q (av) MP a
1
2
3
0
5
11.0
5
11.0
20.0
20
25
30
8.0
10.0
12.0
Compute th e settlement s b y Schmertmann' s method .
Assume th e ne t contac t pressure a t th e bas e o f th e foundatio n i s equa l t o 7 0 kPa , an d
t- 10 year s
584 Chapt e r 1 3
13.15 A square rigi d footing of size 1 0 x 1 0 m is founded at a depth of 2. 0 m below groun d level .
The typ e o f strat a met a t the sit e i s
Dept h bel o w G . L . (m)
O t o 5
5 t o 7m
Below 7 m
Type o f soi l
Sand
Clay
Sand
The wate r tabl e i s a t th e bas e leve l o f th e foundation . Th e saturate d uni t weigh t o f soi l
above th e foundation base i s 20 kN/m
3
. The coefficien t of volume compressibilit y o f clay,
m
v
, is 0.0001 m
2
/kN, and the coefficient of consolidation c
v
, is 1 m
2
/year. The total contac t
pressure q = 100 kN/m
2
. Water tabl e i s at the bas e leve l o f foundation.
Compute primar y consolidation settlement .
13.16 A circular tank of diameter 3 m is founded at a depth of 1 m below ground surfac e on a 6 m
thick normall y consolidate d clay . Th e wate r tabl e i s a t th e bas e o f th e foundation . Th e
saturated uni t weight of soil i s 19. 5 kN/m
3
, and the in-situ void rati o e
Q
is 1.08 . Laborator y
tests o n representativ e undisturbe d samples o f th e cla y gav e a valu e o f 0. 6 fo r th e por e
pressure coefficien t A an d a valu e o f 0. 2 fo r th e compressio n inde x C
f
. Comput e th e
consolidation settlemen t o f the foundatio n for a total contac t pressur e o f 95 KPa. Us e 2: 1
method fo r computing Ap .
13.17 A raft foundatio n of size 1 0 x 40 m i s founded at a depth o f 3 m below groun d surfac e an d
is uniforml y loaded wit h a net pressure o f 50 kN/m
2
. The subsoi l i s normally consolidate d
saturated cla y t o a dept h o f 2 0 m belo w th e bas e o f th e foundatio n wit h variabl e elasti c
moduli wit h respect t o depth. For the purpose o f analysis , the stratum is divided int o thre e
layers wit h constant modulus as given below:
Layer No
1
2
3
Dept h
From
3
8
18
bel ow g roun d (m )
To
8
18
23
El as t ic M odulu s
E
s
(MPa)
20
25
30
Compute th e immediate settlement s by using Eqs (13.20a). Assume the footing i s flexible.
CH APTER 1 4
SHALLOW FOUNDATION III :
COMBINED FOOTINGS AND MAT FOUNDATIONS
1 4.1 I NTRODU CTI O N
Chapter 1 2 has considere d th e common method s o f transmittin g loads t o subsoi l throug h sprea d
footings carryin g singl e column loads. Thi s chapte r consider s th e following types o f foundations:
1. Cantileve r footings
2. Combine d footing s
3. Ma t foundations
When a column i s nea r o r righ t nex t t o a propert y limit , a squar e o r rectangula r footin g
concentrically loaded unde r the column woul d extend int o the adjoining property. I f the adjoining
property i s a public side walk or alley, local building codes ma y permit such footings to project int o
public property . Bu t whe n th e adjoinin g propert y i s privatel y owned , th e footing s mus t b e
constructed withi n the property. I n such cases, ther e ar e three alternative s whic h are illustrate d i n
Fig. 14. 1 (a). These ar e
1. Cantilever footing. A cantileve r o r stra p footin g normall y comprise s tw o footing s
connected by a beam called a strap. A strap footing is a special cas e of a combined footing .
2. Combined footing. A combined footin g is a long footing supporting two or more column s
in one row.
3. Ma t o r raft foundations. A ma t o r raf t foundatio n i s a large footing , usuall y supportin g
several column s in two or more rows .
The choice between these types depends primarily upon the relative cost. In the majority of cases,
mat foundation s are normall y use d wher e the soi l ha s low bearing capacit y an d wher e th e total are a
occupied by an individual footing is not less than 50 per cent of the loaded are a of the building.
When the distances between the columns and the loads carried by each column are not equal ,
there will be eccentric loading. The effect o f eccentricity i s to increase the base pressure on the side
585
586 Chapt er 1 4
of eccentricit y an d decreas e i t on the opposit e side . Th e effec t o f eccentricity o n the base pressur e
of rigid footing s i s also considere d here .
Mat Foundatio n i n Sand
A foundatio n i s generall y terme d a s a ma t i f th e leas t widt h i s mor e tha n 6 meters . Experienc e
indicates tha t the ultimate bearing capacity of a mat foundation on cohesionless soi l i s much higher
than tha t of individual footings of lesser width . With the increasing widt h of the mat , or increasing
relative densit y of the sand, the ultimate bearing capacity increase s rapidly . Hence, th e danger that
a large ma t may break int o a sand foundation is too remote t o require consideration. O n account of
the larg e siz e o f mat s th e stresse s i n th e underlyin g soi l ar e likel y t o b e relativel y hig h t o a
considerable depth . Therefore , th e influenc e o f loca l loos e pocket s distribute d a t rando m
throughout th e san d i s likel y t o b e abou t th e sam e beneat h al l part s o f th e ma t an d differential
settlements ar e likel y t o b e smalle r tha n thos e o f a sprea d foundatio n designe d fo r th e sam e soi l
Mat
Strap
footin
\ — Combine d footin g
/
Property lin e
(a) Schemati c plan showin g mat,
strap and combined footings
T T T T T T T T T T T T T T I q
(b) Bul b of pressure for vertical stres s for differen t beam s
Figure 14. 1 (a ) Types o f f oot i ngs ; (b ) beams o n compr essi bl e subgr ad e
Shallow Foun dat io n III : Com bin e d Foot in g s an d M at Foun dat io n 58 7
pressure. The methods of calculating the ultimate bearing capacity dealt with in Chapter 1 2 are also
applicable to mat foundations.
Mat Foundatio n i n Clay
The ne t ultimat e bearing capacit y that can be sustaine d by the soi l a t the base of a mat o n a dee p
deposit of clay or plastic silt may be obtained in the same manner as for footings on clay discussed
in Chapter 12 . However, by using the principle of flotation, th e pressure on the base of the mat that
induces settlement can be reduced by increasing the depth of the foundation. A brief discussio n on
the principle of flotation is dealt wit h in this chapter.
Rigid an d Elasti c Foundatio n
The conventiona l method o f desig n o f combine d footing s and ma t foundation s is t o assum e th e
foundation a s infinitel y rigi d an d the contact pressur e i s assumed t o have a planar distribution. In
the case of an elastic foundation, the soil is assumed t o be a truly elastic solid obeying Hooke's law
in al l directions. The design of an elastic foundation requires a knowledge of the subgrade reaction
which i s briefl y discusse d here . However , th e elasti c metho d doe s no t readil y len d itsel f t o
engineering applications because i t is extremely difficul t an d solutions are available for onl y a few
extremely simpl e cases .
1 4.2 SAF E B EARI N G PRESSU RE S FOR M AT FOU NDATI ON S ON
SAND AND CL A Y
M ats o n Sand
Because the differential settlement s o f a mat foundation are less than those o f a spread foundation
designed fo r th e sam e soi l pressure , i t i s reasonabl e t o permi t large r saf e soi l pressure s o n a raf t
foundation. Experienc e has shown that a pressure approximately twice a s great a s that allowed for
individual footings may be used because it does not lead to detrimental differential settlements. The
maximum settlemen t o f a ma t ma y b e abou t 5 0 m m ( 2 in ) instea d o f 2 5 m m a s fo r a sprea d
foundation.
The shap e o f th e curv e i n Fig. 13.3(a ) show s tha t th e ne t soi l pressur e correspondin g t o a
given settlemen t i s practicall y independen t o f th e widt h o f th e footin g o r ma t whe n th e widt h
becomes large . The safe soil pressure fo r design may with sufficient accurac y be taken as twice the
pressure indicate d i n Fig . 13.5 . Pec k e t al. , (1974 ) recommen d th e followin g equatio n fo r
computing net saf e pressure ,
q
s
= 2lN
cor
kPa (14.1 )
for 5 < N
cor
< 50
where N
cor
i s the SPT valu e corrected fo r energy, overburden pressure an d fiel d procedures .
Eq. 14. 1 give s q
s
value s abov e th e wate r table . A correctio n facto r shoul d b e use d fo r th e
presence o f a water tabl e a s explained i n Chapter 12 .
Peck e t al. , (1974) also recommend tha t the q
s
values as given by Eq. 14. 1 may be increase d
somewhat i f bedrock i s encountered at a depth less tha n about one hal f the widt h of the raft .
The value of N to be considered i s the average of the values obtained up to a depth equal to the
least widt h o f th e raft . I f th e averag e valu e o f N afte r correctio n fo r th e influenc e of overburden
pressure an d dilatancy is less tha n about 5, Peck e t al. , say that the sand is generall y considered t o
be too loose for the successful use of a raft foundation . Either the sand should be compacted o r else
the foundation should be established o n piles o r piers.
588 Chapt e r 1 4
The minimu m dept h o f foundatio n recommende d fo r a raf t i s abou t 2. 5 m belo w th e
surrounding ground surface . Experienc e ha s shown tha t i f the surcharg e i s less tha n thi s amount ,
the edges o f the raft settl e appreciably more than the interior because o f a lack of confinement of the
sand.
Safe B earin g Pressures o f M at s o n Clay
The quantit y in Eq. 12.25(b ) is the net bearing capacit y q
m
at the elevation of the base of the raft i n
excess o f tha t exerte d b y th e surroundin g surcharge . Likewise , i n Eq . 12.25(c) , q
na
i s th e ne t
allowable soi l pressure . B y increasin g th e dept h o f excavation , th e pressur e tha t ca n safel y b e
exerted b y th e buildin g i s correspondingl y increased . Thi s aspec t o f th e proble m i s considere d
further i n Section 14.1 0 in floating foundation.
As for footing s o n clay, the factor o f safet y agains t failur e of the soi l beneat h a mat on cla y
should not be les s tha n 3 under normal loads , o r less tha n 2 under the most extreme loads .
The settlemen t o f the mat under the give n loading condition shoul d b e calculated a s per the
procedures explaine d i n Chapte r 13 . The ne t saf e pressur e shoul d b e decide d o n th e basi s o f th e
permissible settlement .
1 4.3 ECCENTRI C L OADI NG
When th e resultant of loads on a footing does not pass throug h the center of the footing, the footing
is subjecte d t o wha t i s calle d eccentric loading. Th e load s o n th e footin g ma y b e vertica l o r
inclined. If the loads ar e inclined i t may be assumed tha t the horizontal component i s resisted b y the
frictional resistanc e offere d by the base of the footing. The vertical componen t i n such a case is the
only facto r fo r th e desig n o f th e footing . Th e effect s o f eccentricit y o n bearin g pressur e o f th e
footings hav e been discusse d i n Chapte r 12 .
14.4 TH E COEFFI CI ENT OF SU B G RADE REACTI ON
The coefficient of subgrade reaction is defined as the ratio between th e pressure agains t the footing
or mat an d the settlement at a given point expressed a s
where &
y
= coefficien t of subgrade reactio n expresse d a s force/length
3
(FZr
3
),
q = pressur e o n the footing or mat at a given point expressed a s force/length
2
(FZr
2
),
S = settlemen t o f th e sam e poin t o f th e footin g o r ma t i n th e correspondin g uni t o f
length.
In other word s the coefficient of subgrade reactio n i s the unit pressure require d t o produce a
unit settlement . In clayey soils, settlement under the load takes place ove r a long period o f time and
the coefficient shoul d be determined o n the basi s o f the final settlement . On purel y granula r soils ,
settlement take s plac e shortl y afte r loa d application . Eq . (14.2 ) i s base d o n tw o simplifyin g
assumptions:
1 . Th e valu e of k^ i s independent of the magnitud e of pressure .
2. Th e valu e of &
s
has th e same valu e for ever y poin t on the surfac e o f the footing .
Both th e assumptions are strictl y not accurate. The valu e of k
s
decreases with the increase o f
the magnitude of the pressur e an d i t is not the same fo r ever y poin t of the surfac e o f the footing a s
the settlemen t of a flexible footing varies from poin t to point . However th e method i s supposed t o
Shallow Foun dat io n III: Com bin ed Foot in g s an d M at Foun dat io n 58 9
give realistic value s for contact pressures an d i s suitabl e for beam or mat design whe n only a low
order of settlement is required.
Factors Affectin g the V alu e o f k
s
Terzaghi (1955 ) discussed th e various factors that affec t th e value of k
s
. A brief descriptio n o f hi s
arguments is given below.
Consider tw o foundatio n beams o f width s B
l
an d B
2
suc h tha t B
2
= nB
{
restin g o n a
compressible subgrad e and each loaded so that the pressure against the footing is uniform and equal
to q for both the beams (Fig . 14 . Ib). Consider th e same points on each beam and , let
>>! = settlemen t o f beam o f widt h B\
y
2
= settlemen t o f beam o f width B
2
q q
Ir — Qt"l H I
7
" —
Hence ** i ~
an a K
s2 ~
y\ ./ 2
If th e beam s ar e restin g o n a subgrad e whos e deformatio n propertie s ar e mor e o r les s
independent of dept h (suc h as a stif f clay ) then i t can be assume d tha t the settlement increase s i n
simple proportion t o the dept h o f the pressure bulb.
Then y
2
= ny
l
k —3—± ?L -k i
and
s2
~ nv ~ v B ~
l
B '
A genera l expressio n fo r k
s
ca n no w b e obtaine d i f w e conside r B
{
a s bein g o f uni t widt h
(Terzaghi use d a uni t width of one foot whic h converted t o metri c unit s may b e take n a s equal t o
0.30 m) .
Hence b y putting B
}
= 0.30 m, k
s
= k
s2
, B = B
2
, w e obtain
k,=Q3-j- (14.4 )
where k
s
i s the coefficient of subgrade reaction o f a long footing of widt h B meter s an d resting on
stiff clay ; k
sl
i s th e coefficien t o f subgrad e reactio n o f a lon g footin g o f widt h 0.3 0 m
(approximately), resting on the same clay. It is to be noted here that the value of k
sl
i s derived fro m
ultimate settlement values , that is, after consolidatio n settlement i s completed .
If th e beam s ar e restin g o n clea n sand , th e fina l settlemen t value s ar e obtaine d almos t
instantaneously. Sinc e th e modulu s o f elasticit y o f san d increase s wit h depth , th e deformatio n
characteristics o f th e san d chang e an d becom e les s compressibl e wit h depth . Becaus e o f thi s
characteristic o f sand , th e lowe r portio n o f the bul b of pressure fo r beam B
2
i s les s compressibl e
than that of the sand enclosed i n the bulb of pressure of beam B
l
.
The settlement value y
2
lies somewhere between y
l
an d ny
r
I t has been shown experimentally
(Terzaghi and Peck, 1948 ) tha t the settlement , y, of a beam o f widt h B resting o n sand i s given by
the expressio n
2B ^
where y
{
= settlement o f a beam of width 0.30 m and subjected t o the same reactive pressure as the
beam of width B meters .
590 Chapt er 1 4
Hence, th e coefficien t of subgrad e reactio n k^ o f a beam o f widt h B meter s ca n b e obtaine d
from th e following equation
5 +0.30
= k..
(14.6)
where k
s{
= coefficient o f subgrad e reaction o f a beam o f widt h 0.3 0 m resting o n th e same sand .
M easurement o f Ar
s1
A valu e fo r £
v l
fo r a particula r subgrad e ca n b e obtaine d b y carryin g ou t plat e loa d tests . Th e
standard size of plate used for this purpose is 0.30 x 0.30 m size. Let k
}
b e the subgrade reaction for
a plat e of siz e 0.30 x 0.30 r n size.
From experiment s i t has bee n foun d tha t &
? 1
~ k
{
fo r san d subgrades , bu t for clay s k
sl
varie s
with th e lengt h of th e beam. Terzaghi (1955) give s the following formula for clays
*5i
=
*i
L + 0.152
( 14.1 a)
where L = lengt h of the beam i n meters an d the widt h of the beam = 0.30 m . For a very long beam
on clay subgrad e w e ma y writ e
Procedure t o Fin d Ar
s
For sand
1. Determin e k^ fro m plat e load tes t o r fro m estimation .
2. Sinc e &s
d
~ k
{
, us e Eq. (14.6 ) t o determine k
s
for san d for any give n width B meter .
For clay
1 . Determin e k
{
fro m plat e load test or from estimation
2. Determin e &
y
, fro m Eq . (14.7a ) a s the lengt h of beam i s known.
3. Determin e k
s
from Eq. (14.4 ) fo r the give n width B meters .
When plat e loa d test s ar e used, k
{
ma y b e foun d b y one of th e two ways ,
1. A bearin g pressur e equa l to not mor e tha n the ultimat e pressur e and the correspondin g
settlement i s used fo r computing k
{
2. Conside r th e bearing pressure corresponding t o a settlement of 1. 3 mm for computing k
r
Estimation o f Ar
1
V alue s
Plate load tests ar e both costl y and time consuming. Generall y a designer requires onl y the values
of the bending moment s and shear forces withi n the foundation. With even a relatively large error in
the estimation o f k
r
moment s and shear force s can be calculated with little error (Terzaghi , 1955) ;
an error of 10 0 per cent in the estimation of k
s
may change the structural behavior of the foundation
by up t o 1 5 per cent only .
Shallow Foun dat io n III : Com bin e d Foot in g s an d M at Foun dat io n 59 1
Table 1 4.1 a /T
I
value s for foun dat ion s o n s an d (M N/m
3
)
Rel at ive den s it y Loos e Mediu m Dens e
SPT Val ues (Un correct ed ) <1 0 10-3 0 >3 0
Soil, dr y o r mois t 1 5 4 5 17 5
Soil submerge d 1 0 3 0 10 0
Table 14.1 b /:
1
value s for foun dat io n o n clay
Consistency
c
u
(kN/m
2
)
*, (MN/m
3
)
Stiff
50-100
25
V ery st i f f
100-200
50
Har d
>200
100
Source: Terzaghi (1955)
In th e absenc e o f plat e loa d tests , estimate d value s of k
l
an d hence k
s
ar e used . The values
suggested b y Terzaghi fo r k\ (converte d int o S.I. units ) are given i n Table 14.1 .
1 4.5 PROPORTI ONI N G OF CANTI L EV ER FOOTI NG
Strap or cantilever footings are designed on the basis of the following assumptions:
1 .Th e strap i s infinitely stiff . I t serves t o transfer the column loads t o the soi l wit h equal and
uniform soi l pressur e under both the footings.
2. Th e strap is a pure flexural membe r and does not take soil reaction. To avoid bearing on the
bottom o f the stra p a few centimeter s of the underlying soil ma y b e loosene d prio r t o the
placement o f concrete .
A stra p footin g i s use d t o connec t a n eccentricall y loade d colum n footin g clos e t o th e
property lin e to an interior column as shown in Fig. 14.2 .
With the above assumptions, the design of a strap footing is a simple procedure. I t starts with
a trial value of e, Fig. 14.2 . Then th e reactions R
l
an d R
2
ar e computed by the principl e of statics .
The tentative footing areas ar e equal to the reactions R
{
an d R
2
divided by the safe bearing pressure
q . With tentative footing sizes, th e value of e is computed. These step s ar e repeated unti l the trial
value of e is identical wit h the fina l one . The shear s an d moment s in the strap ar e determined, an d
the straps designe d t o withstand the shear and moments. The footings are assumed t o be subjected
to unifor m soi l pressur e an d designe d a s simpl e sprea d footings . Unde r th e assumption s give n
above the resultants of the column loads Q
l
an d Q
2
would coincide with the center of gravity of the
two footing areas . Theoretically , th e bearing pressur e woul d be unifor m under both th e footings.
However, i t is possible that sometimes the full desig n liv e load acts upon one of the columns whil e
the other may be subjected t o little live load. In such a case, the full reductio n of column load fro m
<2
2
to R
2
may not be realized. It seems justified the n that in designing the footing under column Q
2
,
only the dead loa d o r dead loa d plus reduced liv e load shoul d be used on column Q
v
The equations for determining the position of the reactions (Fig. 14.2 ) ar e
R2 = 2~(14.8 )
L
R
where R
{
an d R
2
= reactions fo r th e column loads <2 j an d Q
2
respectively, e = distance o f R
{
fro m
Q
V
L
R
= distance between R
{
an d R
r
592 Chapt er 1 4
Property lin e
~\,
x
1
[
i
h-
r
\
i
9,
•*•—
n
—i
nl
<
1
1
f t
p
3.—
*,
Col
, , |f .
-*» •
1
B
l
= 2( e + b
[
/2)
Strap
1
T
/
Strap
7^^ /^cs \ /^/ ^
. / . .
S
h—
r
i y
c
>
£
•^
1 I
ol
/
V ,
's.
Q
1
h <
2
'
\_
H
(
/ —^
' i
T
i
r
- / 7,
:oi2
q
s
Figure 1 4. 2 Prin ciple s o f can t il eve r o r s t ra p foot in g des ig n
1 4.6 DESI G N OF COM B I NED FOOTI NG S B Y RI G I D M ETH OD
(CONV ENTI ONAL M ETH OD)
The rigi d metho d o f design of combined footing s assumes that
1. Th e footin g o r ma t i s infinitel y rigid , an d therefore , th e deflectio n o f th e footin g o r ma t
does no t influenc e th e pressur e distribution,
2. Th e soi l pressur e i s distributed i n a straight line or a plane surface such that the centroi d of
the soi l pressur e coincide s wit h th e lin e o f actio n o f th e resultan t forc e o f al l th e load s
acting o n th e foundation.
Design o f Combine d Footing s
Two o r mor e column s i n a ro w joine d togethe r b y a stif f continuou s footin g for m a combine d
footing a s shown i n Fig. 14.3a . The procedur e o f design for a combined footin g is as follows:
1. Determin e th e total column loads 2<2 = Q
{
+Q-, + Q
3
+ ... an d location of the line of action
of th e resultan t ZQ. I f an y colum n i s subjecte d t o bendin g moment , th e effec t o f th e
moment shoul d be taken int o account.
2. Determin e th e pressur e distributio n q per linea l lengt h of footing.
3. Determin e th e width, B, of the footing.
4. Dra w th e shea r diagra m alon g th e lengt h o f th e footing . B y definition , the shea r a t an y
section alon g th e beam i s equal to the summation of all vertical forces t o the lef t o r right of
the section. Fo r example, the shear at a section immediatel y t o the lef t o f Q
{
i s equal to the
area abed, an d immediatel y t o th e righ t o f Q
{
i s equa l t o ( abed - Q
{
) a s show n i n
Fig. 14.3a .
5. Dra w th e momen t diagra m alon g th e lengt h o f th e footing . B y definitio n th e bendin g
moment a t an y sectio n i s equa l t o th e summatio n o f momen t du e t o al l th e force s an d
reaction t o th e lef t (o r right ) o f th e section . I t i s als o equa l t o th e are a unde r th e shea r
diagram t o the lef t (o r right ) of the section .
6. Desig n th e footing a s a continuous beam t o resist th e shea r an d moment .
7. Desig n th e footing for transvers e bending in the same manne r as for sprea d footings .
Shallow Foun dat io n III : Com bin e d Foot in g s an d M at Foun dat io n 593
Q
(a) Combined footing
T
Q
\ — fl
-
i , ; , , i i ,
i
T
^c
R
Qi
1
1
(b) Trapezoidal combined footin g
Figure 14. 3 Combi ne d o r t r apezoi da l f oot i n g desi g n
It shoul d be noted her e tha t the end column along the property lin e may be connected t o the
interior column by a rectangular or trapezoidal footing. In such a case no strap is required and both
the column s togethe r wil l be a combine d footin g a s shown i n Fig. 14.3b . I t i s necessar y tha t th e
center o f are a o f th e footin g mus t coincide wit h the cente r o f loadin g fo r th e pressur e t o remai n
uniform.
14.7 DESI G N OF M AT FOU NDATI O N B Y RI G I D M ETH OD
In the conventional rigi d method th e mat is assumed t o be infinitely rigi d and the bearing pressure
against th e botto m o f th e ma t follow s a plana r distributio n wher e th e centroi d o f th e bearin g
pressure coincide s wit h the lin e of action of the resultant force of al l loads actin g on the mat . The
procedure o f design i s as follows:
1. Th e column loads of all the columns coming fro m th e superstructur e are calculate d a s per
standard practice . Th e loads includ e live and dead loads .
2. Determin e the line of action of the resultant of all the loads. However, the weight of the mat
is no t include d i n th e structura l desig n o f th e ma t becaus e ever y poin t o f th e ma t i s
supported by the soi l unde r it, causing no flexural stresses .
3. Calculat e the soi l pressur e a t desired location s b y the use of Eq. (12.73a)
594 Chapt e r 1 4
Q.Q
t
e
K
Q,e,
q = —
L
± x± - v
A I 1
y
x
where Q
t
= Z<2 = total loa d on the mat ,
A = total area o f the mat,
x, y = coordinates o f an y give n point on th e ma t wit h respec t t o th e x an d y axe s passin g
through th e centroi d o f the are a o f the mat ,
e
x
, e
v
= eccentricities of the resultant force,
/
v
, I = moments o f inerti a of the ma t wit h respect t o the x and y axes respectively .
4. Th e mat i s treated a s a whole in each o f two perpendicular directions . Thus th e total shea r
force acting on any section cutting across th e entire mat is equal to the arithmetic sum of all
forces and reactions (bearing pressure) to the lef t (o r right) of the section. The tota l bending
moment acting on such a section is equal to the sum of all the moments t o the lef t (o r right)
of th e section .
14.8 DESIG N OF COMBINED FOOTI NG S B Y EL ASTI C LINE M ETH OD
The relationshi p betwee n deflection , y , a t an y poin t o n a n elasti c bea m an d th e correspondin g
bending moment M ma y be expresse d b y th e equation
(14.10)
dx~
The equation s for shear V and reaction q at the same poin t may be expressed a s
(14.11)
(14.12)
where x i s the coordinat e alon g the lengt h of the beam.
From th e basi c assumptio n of a n elastic foundation
where, B = width of footing, k - coefficien t o f subgrade reaction.
Substituting for q, Eq. (14.12 ) may be written as
(14-13)
x
The classica l solution s o f Eq . (14.13 ) bein g o f close d form , ar e no t genera l i n thei r
application. Heteny i (1946 ) develope d equation s fo r a loa d a t an y poin t alon g a beam . Th e
development o f solution s i s base d o n th e concep t tha t th e bea m lie s o n a be d o f elasti c spring s
which i s base d o n Winkler' s hypothesis. A s pe r thi s hypothesis , the reactio n a t an y poin t o n th e
beam depend s onl y on the deflectio n at that point.
Methods ar e als o availabl e fo r solvin g the beam-proble m o n a n elasti c foundatio n b y th e
method o f finit e difference s (Malter , 1958) . Th e finit e elemen t metho d ha s bee n foun d t o b e th e
most efficient of the methods fo r solving beam-elastic foundatio n problem. Compute r programs are
available for solvin g the problem.
Shallow Foun dat io n III : Com bin e d Foot in g s an d M at Foun dat io n 59 5
Since al l th e method s mentione d abov e ar e quit e involved , the y ar e no t deal t wit h here .
Interested reader s ma y refer t o Bowles (1996) .
14.9 DESIG N OF MAT FOU NDATI ON S B Y EL ASTI C PLATE METH OD
Many methods ar e available for the design of mat-foundations. The on e that is very much in use is
the finit e differenc e method . Thi s metho d i s base d o n th e assumptio n that th e subgrad e ca n b e
substituted by a bed o f uniformly distributed coil spring s with a spring constant k
s
whic h i s called
the coefficient of subgrade reaction. The finit e differenc e method uses the fourth order differentia l
equation
q-k w
- -
s

D
where H . = — + -^
+

(14
.,4)
q = subgrade reaction pe r uni t area,
k
s
= coefficient of subgrade reaction,
w = deflection,
Et
3
D = rigidity of th e ma t = -
E = modulus of elasticit y of the material o f the footing,
t = thickness of mat ,
fji = Poisson's ratio .
Eq. (14.14) may be solved by dividing the mat into suitable square grid elements, an d writing
difference equation s for each of the grid points. By solving the simultaneous equations so obtained
the deflection s a t al l th e gri d point s ar e obtained . Th e equation s ca n b e solve d rapidl y wit h an
electronic computer. After the deflections are known, the bending moments are calculated using the
relevant difference equations.
Interested readers may refer to Teng (1969) or Bowles (1996) for a detailed discussio n of the
method.
1 4 .1 0 FL OATI N G FOU NDATI O N
G eneral Consideratio n
A floatingfoundation fo r a building is defined as a foundation in which the weight of the building
is approximatel y equa l t o the ful l weigh t including wate r of the soi l removed fro m th e sit e of the
building. Thi s principl e o f flotatio n ma y b e explaine d wit h reference t o Fig . 14.4 . Fig . 14.4(a )
shows a horizonta l groun d surfac e wit h a horizonta l wate r tabl e a t a dept h d
w
belo w th e ground
surface. Fig . 14.4(b ) show s a n excavatio n mad e i n th e groun d t o a dept h D wher e D > d
w
, and
Fig. 14.4(c ) shows a structur e built in the excavation and completel y fillin g it .
If th e weigh t of th e buildin g is equa l t o the weigh t of the soi l an d wate r remove d fro m th e
excavation, the n i t i s eviden t tha t th e tota l vertica l pressur e i n th e soi l belo w dept h D i n
Fig. 14.4(c ) i s the same a s i n Fig. 14.4(a ) before excavation.
Since th e wate r leve l ha s no t changed , th e neutra l pressur e an d th e effectiv e pressur e ar e
therefore unchanged . Since settlement s ar e caused by a n increas e i n effective vertical pressure , i f
596
Chapt er 1 4
we coul d mov e fro m Fig . 14.4(a ) t o Fig . 14.4(c ) withou t th e intermediat e cas e o f 14.4(b) , th e
building i n Fig. 14.4(c ) woul d not settl e at all.
This is the principle of a floating foundation, an exact balance of weight removed against
weight imposed. The result is zero settlement of the building.
However, i t may b e noted , tha t we cannot jump fro m the stage shown i n Fig. 14.4(a ) t o the
stage i n Fig. 14.4(c ) wit hout passing through stage 14.4(b) . Th e excavation stage o f the building is
the critical stage.
Cases ma y aris e wher e we cannot have a full y floatin g foundation. The foundation s of thi s
type ar e sometime s calle d partly compensated foundations (a s agains t fully compensated o r fully
floating foundations) .
While dealing with floating foundations , we have to consider the following two types of soils.
They are :
Type 1 : Th e foundatio n soils ar e o f suc h a strengt h that shear failur e o f soi l wil l not occur
under th e buildin g loa d bu t th e settlement s an d particularl y differential settlements, wil l be to o
large and will constitute failureo f the structure. A floating foundation is used t o reduce settlements
to an acceptabl e value.
Type 2: The shear strength of the foundation soi l is so low that rupture of the soil would occur
if th e building were t o be founded at ground level. In the absenc e o f a strong layer at a reasonabl e
depth, the building can only be buil t on a floating foundatio n which reduces the shear stresses t o an
acceptable value . Solving this problem solves the settlement problem.
In both the cases, a rigid raft or box type of foundation is required for the floating foundation
[ Fig. I4.4( d)j
(a) (b ) (c )
Balance o f stresses i n foundation excavation
(d) Rigid raft foundation
Figure 14. 4 Prin ciple s o f fl oat in g foun dat ion ; an d a t ypical rig i d raf t foun dat io n
Shallow Foun dat io n III : Com bin e d Foot in g s an d M at Foun dat io n 59 7
Problems t o b e Considered i n the Desig n o f a Floating Foundatio n
The following problems ar e to be considered durin g the design and construction stage of a floating
foundation.
1. Excavation
The excavatio n fo r th e foundatio n has t o b e don e wit h care. Th e side s o f th e excavatio n shoul d
suitably be supporte d b y sheet piling, soldier pile s an d timber or some othe r standar d method .
2. Dewaterin g
Dewatering wil l be necessary whe n excavation has to be taken below the water table level . Care has
to be taken to see that the adjoining structure s are not affected du e to the lowering of the water table.
3. Critica l dept h
In Type 2 foundations the shear strengt h of the soil is low and there is a theoretical limi t to the depth
to whic h a n excavatio n ca n b e made . Terzagh i (1943 ) ha s propose d th e followin g equatio n fo r
computing the critical dept h D
c
,
D=-^
S
for a n excavation whic h is long compared t o its width
where 7 = uni t weight of soil ,
5 = shea r strengt h of soi l = qJ 2,
B = widt h of foundation,
L = lengt h o f foundation.
Skempton (1951) proposes the following equation for D
c
, which is based on actual failures in
excavations
D
c=
N
cJ (14.16 )
or the factor of safet y F
s
agains t bottom failur e for an excavation of dept h D is
F -N
S
M. — 1 T
5 c
rD+
P
where N
c
i s th e bearin g capacit y facto r a s give n b y Skempton , an d p i s th e surcharg e load . Th e
values of N
c
ma y be obtained fro m Fi g 12.13(a) . The above equations may be used to determine the
maximum dept h o f excavation.
4. B otto m heave
Excavation for foundations reduces the pressure in the soil below the founding depth which results
in the heaving of the bottom of the excavation. Any heave which occurs wil l be reversed an d appea r
as settlemen t durin g th e constructio n o f th e foundatio n and th e building . Though heavin g o f th e
bottom of the excavation cannot be avoided i t can be minimized to a certain extent . There ar e three
possible cause s o f heave :
1. Elasti c movemen t o f the soi l as the existing overburden pressure i s removed .
2. A gradual swelling of soi l du e t o the intake of water i f there i s some dela y fo r placing the
foundation o n the excavated botto m o f the foundation.
598 Chapt e r 1 4
3. Plasti c inwar d movement of the surroundin g soil.
The las t movemen t o f th e soi l ca n b e avoide d b y providin g prope r latera l suppor t t o th e
excavated side s o f the trench.
Heaving ca n b e minimize d b y phasin g ou t excavatio n i n narro w trenche s an d placin g th e
foundation soo n afte r excavation . I t ca n b e minimize d b y lowerin g th e wate r tabl e durin g th e
excavation process . Frictio n pile s ca n als o b e use d t o minimiz e th e heave . Th e pile s ar e drive n
either befor e excavatio n commence s o r whe n the excavation i s a t hal f dept h an d th e pil e top s ar e
pushed down t o below foundatio n level. As excavation proceeds , th e soi l start s t o expand bu t thi s
movement is resisted by the upper part of the piles which go into tension. This heave is prevented or
very muc h reduced .
It i s only a practical and pragmatic approach tha t would lead t o a safe and soun d settlement
free floatin g (or partl y floating ) foundation.
Example 1 4 . 1
A bea m o f lengt h 4 m and widt h 0.75 m rest s o n stif f clay . A plat e loa d tes t carried ou t a t the sit e
with th e us e o f a squar e plat e o f siz e 0.3 0 m give s a coefficien t of subgrad e reactio n k
l
equa l t o
25 MN/m
3
. Determine the coefficient of subgrade reaction k
s
for the beam.
Solution
First determine &
s l
fro m Eq . (14.7a ) for a beam o f 0.30 m . wide and lengt h 4 m. Next determine k
s
from Eq . (14.4 ) fo r th e same bea m o f widt h 0.75 m .
, . L + 0.152 4 + 0.152 , - , . „ „
3
k , = k, = 25 = 17. 3 MN/m
3
sl l
1.5 L 1.5x 4
t 03
*
£L=
03x173
=7MN/ m3
B 0.7 5
Example 1 4. 2
A beam o f lengt h 4 m and widt h 0.75 m rests i n dry medium dens e sand . A plate loa d tes t carrie d
out a t th e sam e sit e an d a t th e sam e leve l gav e a coefficien t o f subgrad e reactio n k\ equa l t o
47 MN/m
3
. Determine th e coefficient of subgrade reaction fo r the beam.
Solution
For sand the coefficient of subgrade reaction, &
s l
, for a long beam o f width 0.3 m is the same as that
for a squar e plat e of siz e 0. 3 x 0. 3 m that is k
sl
= k
s
. k
s
no w can be found from Eq . (14.6 ) a s
5 + 0.3
2
0.7 5 + 0.30
2
3
k = k, = 47 = 23 MN/m
3
1
2 B 1. 5
Example 1 4 . 3
The followin g informatio n i s give n fo r proportionin g a cantileve r footin g wit h referenc e t o
Fig. 14.2 .
Column Loads: Q
l
= 1455 kN, Q
2
= 1500 kN .
Size of column: 0. 5 x 0. 5 m.
L = 6.2m,q = 384 kN/m
2
Shallow Foun dat io n III: Com bin ed Foot in g s an d M at Foun dat io n 59 9
It i s required t o determine th e size of the footings for columns 1 and 2.
Solution
Assume th e widt h of the footin g for column 1 = B
{
= 2 m.
First trial
Try e = 0.5 m. Now, L
R
= 6.2 - 0. 5 = 5.7 m.
Reactions
-— =145 5 1 + —=1583k N
L
R
5. 7
/? = a- — = 1500-
1455X
°'
5
=1372k N
2
L
R
5. 7
Size of footings - First tria l
1583
Col. 1 . Are a o f footing A , = = 4.122 sq.m
6 [
38 4
4
1372
Col. 2 . Are a of footing A
0
= = 3.57 sq. m
2
38 4
Try 1. 9 x 1.9 m
Second tria l
B, b . 2 0. 5
New valu e of e =— - = = 0.75 m
2 2 2 2
New L
D
= 6.20-0.75 = 5.45m
075
R, =145 5 1 + — =1655k N
1
5.4 5
2= 15QQ
_ 1455x0.7 5
2
5.4 5
= = 4 3
j
1
38 4
--
=
338
S
q. m o r 1.84 x 1.84 m
2
38 4
Check e = -± —
L
= 1.04 - 0.2 5 = 0.79 « 0.75 m
2 2
Use 2.08 x 2.08 m for Col . 1 and 1.9 0 x 1.9 0 m for Col . 2 .
Note: Rectangula r footings ma y be use d fo r bot h th e columns .
600 Chapt e r 1 4
Example 1 4. 4
Figure Ex . 14. 4 give s a foundation beam wit h the vertica l load s an d momen t actin g thereon. Th e
width of the beam i s 0.70 m and dept h 0.50 m. A uniform load of 1 6 kN/m (including the weight of
the beam ) i s impose d o n th e beam . Dra w (a ) th e bas e pressur e distribution , (b) th e shea r forc e
diagram, an d (c ) the bending moment diagram. The lengt h of the beam i s 8 m.
Solution
The step s t o be followed are:
1 . Determin e th e resultan t vertical forc e R o f th e applie d loading s an d it s eccentricit y wit h
respect to the centers of the beam.
2. Determin e th e maximum and minimum base pressures .
3. Dra w th e shea r an d bending moment diagrams.
R = 320 + 400 + 1 6 x 8 = 848 kN.
Taking the moment about the right hand edge o f the beam, we have,
o2
160 = 2992
2992
or x = -= 3.528 m
848
e
= 4.0 - 3.52 8 = 0.472 r n to the right of center of the beam. Now from Eqs 12.39(a) and (b),
using e
y
= 0,
6x0.472
A L 8x0. 7 8
Convert th e base pressure s pe r uni t area t o load per unit length of beam.
The maximu m vertical load = 0.7 x 205.02 = 143.5 2 kN/m.
The minimum vertical load = 0.7 x 97.83 - 68.4 8 kN/m.
The reactiv e loadin g distribution is given in Fig. Ex . 14.4(b) .
Shear force diagra m
Calculation o f shea r fo r a typica l point suc h a s th e reactio n poin t R
l
(Fig . Ex . 14.4(a) ) i s
explained below .
Consider force s t o the lef t o f R
{
(withou t 320 kN).
Shear forc e V = upwar d shear forc e equa l to the area abed - downwar d force du e to distributed
load on beam ab
68.48 + 77.9 _
1 C
^
1 XT
_ - - 16x 1 = 57. 2 kN
2
Consider t o the right of reaction poin t R
}
(wit h 320 kN).
V = - 32 0 +57.2 = - 262. 8 kN.
In th e sam e awa y th e shea r a t othe r point s ca n b e calculated . Fig . Ex . 14.4(c ) give s th e
complete shea r forc e diagram.
Shallow Foun dat io n III : Com bin e d Foot in g s an d M at Foun dat io n 601
320 kN 400 kN
320 kN
160 kN
(a) Applied loa d
68.48 kN/ m
a b
143.52 kN/ m
77.9 kN/m 134.1 4 kN/m
(b) Base reactio n
57.2 kN
-262.8 kN
277.2 kN
-122.8 kN
(c) Shear forc e diagra m
27.8 kN m 62. 2 kNm
348 kN
1m 6 m 1 m
(d) Bending moment diagra m
Figure Ex . 14. 4
Bending Momen t diagra m
Bending moment a t the reaction poin t R
l
= moment due to force equal t o the area abed + moment
due t o distributed load o n beam ab
= 68.48x- + — x--16x-
2 2 3 2
= 27.8 kN-m
The moment s a t othe r point s ca n b e calculate d i n th e sam e way . Th e complet e momen t
diagram i s given in Fig. Ex. 14.4(d )
602 Chapt e r 14
Example 1 4 . 5
The end column alon g a property lin e i s connected to an interior colum n b y a trapezoidal footing .
The followin g data ar e given with reference t o Fig. 14.3(b) :
Column Loads: Q
l
= 2016 kN, Q
2
= 1560 kN.
Size o f columns: 0.46 x 0.46 m.
L
c
= 5.48 m.
Determine th e dimension s a an d b o f th e trapezoida l footing . Th e ne t allowabl e bearin g
pressure q
m
=190 kPa.
Solution
Determine the center of bearing pressure jc
2
from th e center of Column 1 . Taking moments of all the
loads abou t the cente r of Column 1 , we have
-1560x5.48
1560x5.48
x
_ 239
m
2
357 6
Now jc , = 2.39 + — = 2.62 m
1
2
Point O in Fig. 14.3(b ) i s the center of the are a coincidin g with the center o f pressure .
From th e allowabl e pressure q
a
= 190 kPa, th e area o f the combined footing required is
190
From geometry , th e are a o f the trapezoidal footing (Fig. 14.3(b) ) is
2 2
or ( a + b) = 6.34 m
where, L =L
c
+b
l
= 5.48 + 0.46 = 5.94 m
From th e geometr y of the Fig. (14.3b) , the distance of the center of area x
{
ca n be written in
terms o f a, b and L as
_ L la + b
*
l
~ 3 a + b
2a + b3x , 3x2.6 2
or = —- = 1.32 m
a + bL 5.9 4
but a + b= 6.32 m or b = 6.32 - a . Now substituting for b we have,
6.34
and solving , a = 2.03 m , from which , b = 6.34 - 2.0 3 = 4.31 m.
Shallow Foun dat io n III : Com bin e d Foot in g s an d M at Foun dat io n 60 3
1 4 .1 1 PROB L EM S
14.1 A bea m o f lengt h 6 m an d widt h 0.8 0 m i s founde d o n dens e san d unde r submerge d
conditions. A plate load test with a plate of 0.30 x 0.30 m conducted at the site gave a value
for th e coefficien t o f subgrad e reactio n fo r th e plat e equa l t o 9 5 MN/m
3
. Determin e th e
coefficient o f subgrade reaction fo r the beam.
14.2 I f th e beam i n Pro b 14. 1 i s founde d i n ver y stif f cla y wit h th e valu e fo r k
}
equa l t o
45 MN/m
3
, what is the coefficient of subgrade reaction fo r the beam?
14.3 Proportio n a strap footing given the following data wit h reference t o Fig. 14.2 :
Qj = 580 kN, 0
2
= 900 kN
L
c
= 6.2 m, b
l
= 0.40 m , q
s
= 120 kPa.
14.4 Proportio n a rectangula r combine d footin g give n th e followin g dat a wit h referenc e t o
Fig. 14. 3 (the footing is rectangular instead of trapezoidal) :
Qj = 535 kN, Q
2
= 900 kN, b
{
= 0.40 m,
L = 4.75 m,q =10 0 kPa.
CHAPTER 1 5
DEEP FOUNDATION I :
PILE FOUNDATION
1 5 .1 I NTRODU CTI O N
Shallow foundation s ar e normall y used wher e th e soi l clos e t o th e groun d surfac e an d u p t o th e
zone o f significant stress possesse s sufficien t bearin g strengt h t o carr y th e superstructur e loa d
without causin g distres s t o th e superstructur e due t o settlement . However , wher e th e to p soi l i s
either loos e o r sof t o r of a swelling type the load from th e structure has t o be transferred t o deeper
firm strata .
The structural loads may be transferred to deeper fir m strat a by means of piles. Piles ar e long
slender columns either driven, bored o r cast-in-situ. Drive n piles are made of a variety of materials
such a s concrete , steel , timbe r etc. , wherea s cast-in-situ pile s ar e concret e piles . The y ma y b e
subjected to vertical or lateral loads or a combination of vertical and lateral loads. If the diameter of
a bored-cast-in-situ pil e i s greater tha n about 0.75 m , it i s sometimes calle d a drilled pier , drille d
caisson or drilled shaft . The distinction made between a small diameter bored cast-in-situ pil e (less
than 0.7 5 m ) an d a larger on e i s just fo r th e sak e o f design considerations . Th e desig n o f drilled
piers i s deal t wit h i n Chapte r 17 . This chapte r i s concerne d wit h drive n pile s an d smal l diamete r
bored cast-in-situ pile s only.
1 5 .2 CL ASSI FI CATI O N O F PILES
Piles ma y b e classifie d a s lon g o r shor t i n accordanc e wit h th e L id rati o o f th e pil e (wher e
L = length, d = diameter o f pile) . A shor t pil e behave s a s a rigi d body an d rotates a s a uni t under
lateral loads . Th e loa d transferre d t o the ti p o f the pil e bear s a significant proportion o f the tota l
vertical loa d o n th e top . I n th e cas e o f a long pile , th e lengt h beyond a particular dept h lose s it s
significance under lateral loads, but when subjected to vertical load, the frictional load on the sides
of the pile bears a significant par t to the total load.
605
606 Chapt e r 1 5
Piles ma y furthe r b e classifie d as vertica l pile s o r incline d piles. V ertical piles ar e normally
used t o carr y mainl y vertical loads an d ver y littl e lateral load. When piles ar e inclined at an angl e
to the vertical , they ar e called batter piles or raker piles. Batter piles ar e quite effective fo r taking
lateral loads , bu t whe n use d i n groups , they als o ca n tak e vertica l loads . Th e behavio r o f vertical
and batte r piles subjected t o lateral loads i s dealt wit h i n Chapter 16 .
Types o f Pile s Accordin g t o Thei r Composition
Piles ma y be classified according to thei r composition as
1. Timbe r Piles ,
2. Concret e Piles ,
3. Stee l Piles .
Timber Piles: Timber piles ar e made of tree trunk s with the branches trimmed off. Such piles shal l
be o f soun d qualit y and fre e o f defects . Th e lengt h o f th e pil e ma y b e 1 5 m o r more . I f greate r
lengths ar e required, they may b e spliced. The diamete r o f the piles a t the but t end ma y var y fro m
30 to 40 cm. The diamete r at the ti p end shoul d not be les s tha n 1 5 cm.
Piles entirel y submerge d i n wate r las t lon g withou t deca y provide d marin e borer s ar e no t
present. Whe n a pil e i s subjecte d t o alternat e wetting and dryin g the usefu l lif e i s relativel y shor t
unless treated wit h a wood preservative , usually creosot e a t 250 k g per m
3
for piles i n fres h wate r
and 35 0 kg/m
3
in sea water .
After bein g driven to fi nal depth , al l pile heads, treate d o r untreated, shoul d be sawe d squar e
to sound undamaged wood t o receive the pile cap. But before concret e for the pile cap is poured, the
head of the treated pile s shoul d be protected b y a zinc coat, lead paint or by wrapping the pile heads
with fabri c upon which hot pitch is applied.
Driving of timber piles usually results in the crushing of the fibers on the head (or brooming)
which ca n b e somewha t controlled by using a driving cap, o r ring around the butt.
The usua l maxi mum design loa d pe r pil e does not excee d 25 0 kN. Timber pile s ar e usually
less expensiv e i n places wher e timber i s plentiful .
Concrete Piles. Concrete pile s are either precast o r cast-in-situ piles . Precas t concret e pile s
are cast an d cured i n a casting yard and then transported t o the sit e of work for driving. If the work
is of a ver y big nature , they may b e cast at the sit e also.
Precast pile s ma y b e mad e o f unifor m section s wit h pointe d tips . Tapere d pile s ma y b e
manufactured whe n greate r bearin g resistance i s required. Normall y pile s o f squar e o r octagona l
sections ar e manufacture d since thes e shape s ar e eas y t o cas t i n horizonta l position . Necessar y
reinforcement i s provided t o take care of handling stresses. Pile s ma y als o be prestressed .
Maximum load o n a prestressed concret e pil e i s approximately 2000 kN and on precast pile s
1000 kN . The optimu m load range i s 400 t o 600 kN.
Steel Piles. Steel pile s ar e usuall y rolled H shape s o r pipe piles , //-piles ar e proportioned t o
withstand larg e impac t stresse s during hard driving. Pipe pile s ar e either welde d o r seamless stee l
pipes whic h may b e driven either open-end or closed-end. Pip e pile s ar e ofte n fille d wit h concret e
after driving , although in some cases thi s is not necessary. The optimum load range on steel piles i s
400 t o 1200kN .
15.3 TYPE S OF PILES ACCORDING T O TH E METHOD OF INSTALLATION
According to the method o f construction, ther e ar e three types of piles. The y ar e
1. Drive n piles,
2. Cast-in-situ piles and
3. Drive n an d cast-in-situ piles.
Deep Foun dat io n I : Pil e Foun dat io n 60 7
Driven Pile s
Piles ma y b e o f timber , stee l o r concrete . Whe n th e pile s ar e o f concrete , the y ar e t o be precast .
They ma y b e drive n eithe r verticall y o r a t a n angl e t o th e vertical . Pile s ar e drive n usin g a pil e
hammer. When a pile i s driven int o granular soil, th e soi l s o displaced, equa l t o the volume of the
driven pile, compacts the soil around the sides since the displaced soi l particles enter the soil space s
of th e adjacen t mas s whic h lead s t o densificatio n o f th e mass . Th e pil e tha t compact s th e soi l
adjacent t o it is sometimes calle d a compaction pile. The compaction of the soil mas s around a pile
increases it s bearing capacity.
If a pil e i s drive n int o saturate d silt y o r cohesiv e soil , th e soi l aroun d th e pil e canno t b e
densified becaus e of its poor drainage qualities. Th e displaced soi l particles cannot enter th e voi d
space unless the water in the pores is pushed out. The stresses developed i n the soil mass adjacent to
the pil e du e t o the drivin g of the pil e hav e t o be borne b y the por e wate r only . This result s i n the
development o f pore wate r pressure and a consequent decrease i n the bearing capacit y o f the soil .
The soi l adjacen t t o the piles i s remolded an d loses t o a certain exten t it s structura l strength. The
immediate effect o f driving a pile i n a soil wit h poor drainage qualities is, therefore, t o decrease it s
bearing strength . However , wit h the passag e of time , the remolde d soi l regain s par t of its los t
strength due t o the reorientation of the disturbed particles (whic h is termed thixotrophy} an d due to
consolidation of the mass . The advantage s and disadvantages of driven piles are :
Advantages
1. Pile s can be precast t o the required specifications.
2. Pile s of any size, lengt h and shape can be made in advance and used at the site. As a result,
the progress of the wor k wil l be rapid .
3. A pile driven into granular soil compact s the adjacent soi l mass an d as a result the bearing
capacity of the pil e i s increased .
4. Th e wor k i s nea t an d clean . Th e supervisio n o f wor k a t th e sit e ca n b e reduce d t o a
minimum. The storag e spac e required i s very much less.
5. Drive n piles may convenientl y be used in places wher e it is advisabl e no t to drill holes fo r
fear o f meeting groun d water under pressure.
6. Driven s pile ar e the most favored for works over water such as piles i n wharf structure s or
jetties.
Disadvantages
1. Precas t o r prestresse d concret e pile s mus t be properl y reinforce d t o withstan d handling
stresses durin g transportation and driving.
2. Advanc e planning is required fo r handling and driving.
3. Require s heav y equipment for handling and driving.
4. Sinc e th e exac t lengt h require d a t the sit e canno t b e determine d i n advance , th e metho d
involves cuttin g of f extr a length s o r addin g mor e lengths . Thi s increase s th e cos t o f th e
project.
5. Drive n piles ar e not suitabl e in soils of poor drainage qualities. If the driving of piles i s not
properly phase d an d arranged, ther e is every possibility of heaving of the soil or the liftin g
of the driven piles during the driving of a new pile.
6. Wher e the foundations of adjacent structure s are likely to be affected du e to the vibrations
generated b y the driving of piles, drive n piles shoul d not be used.
608 Chapt e r 15
Cast- in- situ Piles
Cast-in-situ pile s ar e concret e piles . Thes e pile s ar e distinguishe d fro m drille d pier s a s smal l
diameter piles . They ar e constructed by making holes i n the ground t o the required dept h an d then
filling th e hol e wit h concrete . Straigh t bored pile s o r piles wit h one or more bulb s at interval s may
be cast at the site. The latter type are called under-reamed piles. Reinforcement may be used as per
the requirements. Cast-in-situ pile s hav e advantages as well as disadvantages.
Advantages
1. Pile s o f an y siz e and lengt h may b e constructed a t the site .
2. Damag e du e t o drivin g an d handlin g that is common i n precas t pile s i s eliminated i n this
case.
3. Thes e pile s ar e ideall y suite d i n place s wher e vibration s o f an y typ e ar e require d t o b e
avoided t o preserve th e safet y o f the adjoining structure.
4. The y ar e suitabl e i n soil s o f poo r drainag e qualitie s sinc e cast-in-sit u pile s d o no t
si gni fi cant l y di st ur b the surroundin g soil.
Disadvantages
1. Installatio n of cast-in-situ pile s require s careful supervisio n an d qualit y control o f al l th e
materials use d i n the construction.
2. Th e metho d i s quit e cumbersome. I t need s sufficien t storag e spac e fo r al l th e material s
used i n the construction.
3. Th e advantag e of increased bearing capacity due t o compaction i n granular soil that could
be obtaine d by a driven pile i s not produced b y a cast-in-situ pile .
4. Constructio n of piles in holes where there is heavy current of ground water flow o r artesian
pressure i s ver y di ffi cul t .
A straigh t bored pil e is shown in Fig. 15. 1 (a).
Driven and Cast-in-situ Pile s
This typ e ha s th e advantage s an d disadvantage s of bot h th e drive n and th e cast-in-situ piles . Th e
procedure o f installing a driven an d cast-in-situ pil e is as follows:
A stee l shel l i s drive n into the groun d wit h the ai d o f a mandre l inserte d int o the shell . Th e
mandrel i s withdraw n and concret e i s place d i n th e shell . Th e shel l i s mad e o f corrugate d an d
reinforced thi n shee t stee l (mono-tub e piles) o r pipe s (Armc o welde d pipe s o r commo n seamles s
pipes). The piles of this type are called a shell type. The shell-less type is formed b y withdrawing the
shell while the concrete i s being placed. I n both the types of piles the bottom of the shell is closed with
a conical tip which can be separated from the shell. By driving the concrete out of the shell an enlarged
bulb ma y b e forme d i n bot h th e type s o f piles . Franki pile s ar e o f thi s type . The commo n type s of
driven and cast-in-sit u piles are given in Fig. 15.1. In some cases th e shell will be left i n place and the
tube i s concreted. Thi s type of pile is very much used i n piling over water.
1 5 .4 U SE S OF PI L ES
The majo r uses of piles are:
1. T o carry vertica l compression load .
2. T o resist uplif t load .
3. T o resist horizonta l or incline d loads.
Deep Foun dat io n I : Pil e Foun dat io n 609
Fluted stee l
shell fille d
with concret e
Franki pil e
\
Corrugated
steel shel l
filled wit h \
concrete
Bulb
(a) (b ) (c ) (d )
Figure 1 5. 1 T ype s o f cast-in-situ an d driven cast-in-situ con cret e pile s
Normally vertica l pile s ar e use d t o carr y vertica l compressio n load s comin g fro m
superstructures suc h as buildings, bridges etc . The piles ar e used i n groups joined togethe r b y pile
caps. The loads carried b y the piles are transferred to the adjacent soil. If all the loads coming on the
tops o f pile s ar e transferre d t o the tips , suc h pile s ar e calle d end-bearing o r point-bearing piles.
However, i f al l th e loa d i s transferred t o the soi l alon g the lengt h of the pil e suc h piles ar e calle d
friction piles. If , i n th e cours e o f drivin g a pil e int o granula r soils , th e soi l aroun d th e pil e get s
compacted, suc h piles ar e called compaction piles. Fig. 15.2(a ) shows piles used for the foundation
of a multistoried building to carry load s fro m th e superstructure.
Piles ar e also used to resist uplif t loads . Piles used for this purpose ar e called tension piles or
uplift piles or anchor piles. Uplift load s ar e developed du e t o hydrostati c pressure o r overturning
movement as shown in Fig. 15.2(a) .
Piles ar e als o use d t o resist horizonta l o r incline d forces . Batte r pile s ar e normall y use d t o
resist larg e horizontal loads . Fig . 15.2(b ) shows the use of piles t o resist latera l loads .
1 5 .5 SEL ECTI O N O F PIL E
The selectio n o f th e type , lengt h and capacit y i s usuall y mad e fro m estimatio n base d o n th e soi l
conditions and the magnitude of the load. I n large cities, where the soi l conditions ar e well known
and wher e a large numbe r of pil e foundations hav e been constructed, the experience gaine d i n the
past i s extremel y useful . Generall y th e foundatio n design i s mad e o n th e preliminar y estimate d
values. Before the actual construction begins, pile load test s must be conducted to verify th e design
values. The foundation design mus t be revised according t o the test results. The factors that govern
the selectio n o f piles are :
1. Lengt h of pil e i n relation t o the load an d type of soi l
2. Characte r o f structure
3. Availabilit y of material s
4. Typ e of loadin g
5. Factor s causin g deterioratio n
610 Chapt er 1 5
A multi-storied building on pile s
Uplift o r
anchor pile s
^ Uplif t
pile
-^CCompression
pile
Piles use d to resist uplift loads
Figure 1 5 .2 (a ) Prin ciple s o f fl oat in g foun dat ion ; an d a t ypical rig i d raf t foun dat io n
Retaining
wall Bridge pie r
v/JW\>!*K
Batter pile
Figure 1 5 .2 (b ) Pile s us e d t o res is t l at era l load s
6. Eas e o f maintenanc e
7. Estimate d cost s o f type s o f piles , takin g int o account th e initia l cost , lif e expectanc y an d
cost o f maintenanc e
8. Availabilit y of fund s
All the above factor s have t o be largel y analyze d befor e decidin g u p on a particular type .
1 5 .6 I NSTAL L ATI O N OF PI L ES
The method o f installing a pile at a sit e depends upo n the type of pile. The equipment require d fo r thi s
purpose varies. The following types of piles are normally considered for the purpose o f installation
1. Drive n pile s
The pile s tha t come unde r thi s category are,
a. Timbe r piles ,
Deep Foun dat io n I : Pil e Foun dat io n 61 1
b. Stee l piles , //-section and pipe piles ,
c. Precas t concret e o r prestressed concret e piles , eithe r soli d or hollow sections .
2. Drive n cast-in-situ pile s
This involve s driving of a stee l tub e t o th e require d dept h wit h th e en d close d b y a detachabl e
conical tip. The tube is next concreted an d the shell is simultaneously withdrawn. In some cases the
shell wil l not be withdrawn.
3. B ore d cast-in-situ pile s
Boring i s done either by auguring or by percussion drilling. Afte r borin g i s completed, th e bore i s
concreted wit h or without reinforcement.
Pile Drivin g Equipmen t fo r Drive n and Driven Cast-in-situ Pile s
Pile driving equipment contains three parts. They ar e
1. A pile frame,
2. Pilin g winch,
3. Impac t hammers.
Pile Frame
Pile drivin g equipmen t i s require d fo r driven pile s o r drive n cast-in-situ piles . Th e drivin g pil e
frame mus t be such that it can be mounted on a standard tracked crane base machine for mobility on
land site s o r on framed bases fo r mountin g on staging s or pontoons i n offshore construction. Fig .
15.3 give s a typical pile frame for both onshor e an d offshore construction. Both the types must be
capable o f ful l rotatio n an d backwar d o r forward raking. All types of frames consist essentiall y of
leaders, whic h ar e a pai r o f stee l member s extendin g for th e ful l heigh t o f th e frame an d whic h
guide the hamme r an d pil e as i t i s driven into the ground. Where lon g piles hav e t o be driven th e
leaders ca n be extended a t the top by a telescopic boom .
The bas e frame may be mounted on swivel wheel s fitte d wit h self-contained jacking screw s
for levelin g the frame o r i t may be carried o n steel rollers . Th e roller s run on stee l girder s o r long
timbers an d th e frame i s moved alon g b y winchin g from a deadman se t on the rolle r track , o r by
turning the rollers b y a tommy-bar placed i n holes a t the ends of the rollers. Movements parallel t o
the roller s ar e achieved b y winding in a wire rope terminatin g in hooks o n the ends o f rollers; th e
frame the n skid s i n eithe r directio n alon g the rollers. I t i s important t o ensur e tha t th e pil e frame
remains i n its correct position throughou t the driving of a pile.
Piling Winches
Piling winche s ar e mounte d o n th e base . Winche s ma y b e powere d b y steam , diese l o r gasolin e
engines, o r electri c motors . Steam-powere d winche s ar e commonl y use d wher e stea m i s used for
the pilin g hammer. Diese l o r gasoline engines , or electri c motor s (rarely ) ar e used i n conjunction
with drop hammer s o r where compressed ai r is used t o operate th e hammers .
I mpact H ammer s
The impact energy for driving piles may be obtained by any one of the following types of hammers.
They ar e
1. Dro p hammers ,
2. Single-actin g steam hammers ,
3. Double-actin g stea m hammers ,
612
Chapt er 1 5
To crane for lifting .
I Assembl y rest s on pile
A durin g drivin g
Static
r weigh t
Oscillator
u i_r -
Pile
9310mm
(a) The Ackermanns M14-5P pile frame (b ) Diagrammatic sketch of vibratory pil e driver
Figure 15. 3 Pil e drivin g equipm en t an d vibrat or y pil e drive r
4. Diese l hammer ,
5. V ibrator y hammer.
Drop hammers ar e a t present use d fo r smal l jobs. Th e weigh t i s raise d an d allowe d t o fal l
freely o n the top of the pile. The impact drives the pile int o the ground.
In th e cas e o f a single-acting steam hammer stea m o r ai r raise s th e moveabl e par t o f th e
hammer which then drops by gravity alone. The blows in this case ar e much more rapidl y delivered
than fo r a dro p hammer . Th e weight s o f hammer s var y fro m abou t 150 0 t o 10,00 0 k g wit h th e
length of stroke being about 90 cm. In general the ratio of ram weight to pile weight may vary from
0.5 t o 1.0 .
In the case of a double-acting hammer stea m o r air is used to raise the moveabl e part o f the
hammer an d also t o impart additional energy durin g the down stroke . The downwar d acceleratio n
of the ram owin g to gravit y is increased b y the acceleration du e t o steam pressure . Th e weight s of
hammers var y from abou t 350 to 2500 kg . The length of stroke varie s from abou t 20 to 90 cm. The
rate of driving ranges from 30 0 blows per minute for the light types, to 100 blows per minute for the
heaviest types .
Diesel o r internal combustion hammers utiliz e diesel-fuel explosion s t o provide th e impac t
energy to the pile. Diese l hammers have considerable advantag e over steam hammer s because they
are lighter , mor e mobil e an d us e a smaller amount of fuel . Th e weight of the hamme r varie s fro m
about 100 0 t o 2500 kg.
Deep Foun dat io n I : Pil e Foun dat io n 61 3
The advantage of the power-hammer type of driving is that the blows fall i n rapid successio n
(50 to 15 0 blows per minute) keeping the pile in continuous motion. Since the pile i s continuously
moving, th e effect s o f th e blow s ten d t o conver t t o pressur e rathe r tha n impact , thu s reducin g
damage t o the pile.
The vibration method o f driving piles i s now coming int o prominence. Drivin g i s quiet and
does not generate loca l vibrations. V ibration driving utilizes a variable spee d oscillato r attache d to
the top of the pil e (Fig. 15.3(b)) . It consists of two counter rotating eccentric weights which are in
phase twic e per cycl e (180 ° apart ) i n the vertica l direction . Thi s introduce s vibratio n throug h the
pile which can be made to coincide wit h the resonant frequency o f the pile. As a result, a push-pull
effect i s created at the pile tip which breaks up the soil structure allowing eas y pile penetration into
the groun d wit h a relativel y smal l drivin g effort . Pil e drivin g b y th e vibratio n metho d i s quit e
common i n Russia.
Jetting Piles
Water jetting ma y b e use d t o ai d the penetration of a pile int o dense san d or dens e sand y gravel .
Jetting i s ineffectiv e i n fir m t o stif f clay s o r an y soi l containin g much coarse t o stif f cobble s o r
boulders.
Where jetting i s require d fo r pil e penetration a strea m o f wate r i s discharge d nea r th e pil e
point or along the sides of the pile through a pipe 5 to 7.5 cm in diameter. An adequate quantit y of
water is essential fo r jetting. Suitable quantities of water for jetting a 250 to 350 mm pil e are
Fine sand 15-2 5 liters/second ,
Coarse sand 25-4 0 liters/second ,
Sandy gravel s 45-60 0 liters/second .
A pressure of at least 5 kg/cm
2
or more i s required.
PART A—V ERTI CAL LOAD BEARING CAPACITY OF A
SINGLE V ERTI CAL PIL E
1 5 .7 G ENERA L CONSI DERATI ON S
The bearing capacit y o f groups of piles subjecte d to vertical o r vertical and latera l load s depend s
upon the behavior of a single pile. The bearing capacit y o f a single pile depends upo n
1. Type , size and lengt h of pile,
2. Typ e of soil ,
3. Th e method o f installation .
The bearin g capacit y depend s primaril y o n th e metho d o f installatio n an d th e typ e o f soi l
encountered. The bearing capacity of a single pile increases wit h an increase i n the size and length.
The position of the water table also affect s th e bearing capacity .
In orde r to be abl e to design a saf e and economical pil e foundation, we have to analyz e the
interactions betwee n th e pil e an d th e soil , establis h th e mode s o f failur e an d estimat e th e
settlements fro m soi l deformatio n under dead load , servic e loa d etc . Th e desig n shoul d compl y
with the following requirements .
1. I t shoul d ensur e adequat e safet y agains t failure ; th e facto r o f safet y use d depend s o n th e
importance o f th e structur e an d o n th e reliabilit y o f th e soi l parameter s an d th e loadin g
systems use d i n the design .
614 Chapt e r 1 5
2. Th e settlement s shoul d b e compatibl e wit h adequat e behavio r o f th e superstructur e t o
avoid impairing its efficiency .
L oad Transfe r M echanis m
Statement o f the Proble m
Fig. 15.4(a ) give s a singl e pil e o f unifor m diamete r d (circula r o r an y othe r shape ) an d lengt h L
driven int o a homogeneou s mas s o f soi l o f know n physica l properties . A stati c vertica l loa d i s
applied o n the top. I t is required t o determine the ultimat e bearing capacit y Q
u
of the pile.
When the ultimate load applie d on the top of the pile is Q
u
, a part of the load i s transmitted to
the soi l alon g th e lengt h o f th e pil e an d th e balanc e i s transmitte d t o th e pil e base . Th e loa d
transmitted to the soi l along the lengt h of the pile is called the ultimate friction load or skin load Q
f
and tha t transmitte d t o th e base i s called th e base o r point load Q
b
. The tota l ultimat e load Q
u
is
expressed a s the su m of these two, that is,
Q
u
= Q
b
+ Q
f
=q
b
A
b
+f
s
A
s
(15.1 )
where Q
u
= ultimate load applie d o n the top of the pil e
q
b
= ultimate unit bearing capacit y of the pile at the bas e
A
b
= bearing are a o f the base o f the pil e
A
s
= total surfac e area o f pile embedded belo w groun d surfac e
f
s
- uni t ski n friction (ultimate )
L oad Transfer M echanism
Consider th e pil e shown i n Fig. 15.4(b ) is loaded t o failure by gradually increasing the load o n the
top. I f settlemen t o f th e to p o f th e pil e i s measure d a t ever y stag e o f loadin g afte r a n equilibrium
condition i s attained, a load settlemen t curve as shown in Fig. 15.4(c ) can be obtained .
If the pil e i s instrumented, the load distributio n along the pile can be determined a t differen t
stages o f loading an d plotted a s shown in Fig. 15.4(b) .
When a load Q
{
act s on the pile head, the axial load a t ground level i s also Q
r
but at level A
l
(Fig. 15.4(b)) , the axial load i s zero. The total load Q
{
i s distributed a s friction loa d withi n a length
of pile L
{
. The lower section A
{
B o f pile wil l not be affected b y this load. As the load at the top is
increased t o Q
2
, the axial load a t the bottom of the pile is just zero. The tota l load Q
2
is distributed
as friction load alon g the whole length of pile L . The friction loa d distribution curves along the pile
shaft ma y b e as shown in the figure. If the load put on the pile i s greater tha n <2
2
, a part o f this load
is transferred t o the soi l a t the bas e a s point loa d an d the rest i s transferred t o the soi l surrounding
the pile . Wit h th e increas e o f loa d Q o n th e top , bot h th e frictio n an d poin t load s continu e t o
increase. The frictio n loa d attain s an ultimate value £J,at a particular load level , say Q
m
, at the top,
and any further incremen t of load added t o Q
m
will not increase the value of Q~ However , the point
load, Q , still goes on increasing till the soil fail s by punching shear failure . It has been determine d
by V an Wheele (1957 ) that the point load Q increase s linearl y with the elasti c compression o f the
soil a t the base .
The relativ e proportion s o f the load s carrie d b y ski n frictio n an d bas e resistanc e depend o n
the shea r strengt h an d elasticit y of th e soil . Generall y the vertica l movemen t o f th e pil e whic h is
required t o mobiliz e ful l en d resistanc e i s muc h greate r tha n tha t require d t o mobiliz e ful l ski n
friction. Experienc e indicate s tha t i n bore d cast-in-situ pile s ful l frictiona l loa d i s normall y
mobilized a t a settlement equa l to 0.5 to 1 percent of pile diameter and the full bas e load Q
b
at 10 to
20 percent of the diameter. But , if this ultimate load criterion i s applied t o piles of large diameter in
clay, the settlemen t a t the workin g load (wit h a factor o f safet y of 2 on the ultimat e load) ma y b e
excessive. A typica l load-settlemen t relationshi p o f frictio n loa d an d bas e loa d i s show n i n
Deep Foun dat io n I: Pil e Foun dat ion 615
Curves showing the
loads carried by
pile shaf t
_ [ _ ( _ , Bas e or Tip of pile
Q.
(b) Load-transfer curve s
Load (kN)
„() 100 0 200 0 300 0 400 0 500 0
25
I
5 0
<U
2 7 5 *\ * ~ *
100
125
150
Shaft
load
Base
load
(c) Load-settlement curve (d) Load-settlement relationships for
large-diameter bored and cast-in-place piles
(after Tomlinson , 1986)
Figure 15. 4 Loa d t ran s fer m echan is m
Fig. 15.4(d ) (Tomlinson , 1986 ) fo r a large diamete r bore d an d cast-in-situ pil e i n clay. It ma y be
seen fro m thi s figur e tha t th e ful l shaf t resistanc e i s mobilize d a t a settlemen t o f onl y 1 5 mm
whereas th e ful l bas e resistance , an d th e ultimat e resistance o f th e entir e pile , i s mobilize d a t a
settlement of 12 0 mm. The shaf t loa d a t a settlement of 1 5 mm i s only 100 0 k N which is about 25
percent of the base resistance. I f a working load of 2000 kN at a settlement of 1 5 mm is used for the
design, at this working load, the ful l shaf t resistanc e will have been mobilize d wherea s onl y about
50 percent o f the bas e resistanc e ha s been mobilized. Thi s means i f piles ar e designe d t o carr y a
working loa d equa l t o 1/ 3 t o 1/ 2 th e tota l failur e load , ther e i s ever y likelihoo d o f th e shaf t
resistance being fully mobilize d at the working load. This has an important bearin g on the design.
The typ e o f load-settlemen t curv e fo r a pil e depend s o n th e relativ e strengt h value s o f th e
surrounding and underlying soil. Fig. 15. 5 gives the type s of failure (Kezdi, 1975). They are as follows:
616 Chapt er 1 5
(c)
S = Settlement
r
s
= Shear strengt h
Q = load on the pil e
(e)
Figure 1 5 . 5 T ype s o f fail ur e o f piles . Fig ure s (a ) to (e ) in dicat e ho w s t ren g t h o f s oi l
det erm in es t h e t yp e o f fail ure : (a ) buck ling i n ver y wea k s urroun din g soil ; (b ) g en eral
s hear fail ur e i n t he s t ron g lowe r s oil ; (c ) s oil o f un ifor m s t ren g t h; (d ) low s t ren g t h
s oil i n t he lowe r l ayer , s k i n frict io n predom in an t ; (e ) s k in frict io n i n t en s ion
(Kezdi, 1975 )
Fig 15.5(a ) represent s a driven- pile (woode n or reinforce d concrete) , whos e ti p bear s o n a
very har d stratu m (rock) . The soi l aroun d the shaf t i s too wea k t o exer t an y confinin g pressure or
lateral resistance . I n suc h cases , th e pil e fail s lik e a compressed , slende r colum n o f th e sam e
material; afte r a mor e o r les s elasti c compressio n bucklin g occurs . Th e curv e show s a definit e
failure load .
Fig. 15.5(b ) i s th e typ e normall y me t i n practice . Th e pil e penetrate s throug h layer s o f soi l
having low shear strength down to a layer having a high strength and the layer extending sufficientl y
below the tip of the pile. At ultimate load Q
u
, there will be a base general shea r failure at the tip of the
pile, since the upper layer does no t prevent the formation of a failur e surface. The effec t o f the shaf t
friction i s rather less , sinc e the lowe r dens e laye r prevent s th e occurrenc e o f excessive settlements .
Therefore, th e degree o f mobilization of shear stresses along the shaft wil l be low. The load settlement
diagram i s of the shape typical for a shallow footing on dense soil .
Fig. 15.5(c ) shows the case wher e the shear strengt h of the surrounding soil is fairly uniform;
therefore, a punching failure is likely to occur. The load-settlement diagram does not have a vertical
tangent, and there i s no definit e failure load . The loa d wil l be carried b y point resistance a s well as
by ski n friction .
Fig 15.5(d ) is a rare case wher e the lower layer is weaker. In such cases, the load will be carrie d
mainly by shaf t friction , and the point resistance i s almost zero . The load-settlemen t curv e shows a
vertical tangent , which represents th e load when the shaf t frictio n has been full y mobilized .
Deep Foun dat io n I : Pil e Foun dat io n 61 7
Fig. 15.5(e ) is a case when a pull, -Q, acts on the pile. Since the point resistance i s again zero
the same diagram, as in Fig. 15.5(d) , wil l characterize the behavior, but heaving occurs .
Definition o f Failur e L oa d
The method s of determinin g failur e load s base d on load-settlemen t curve s are describe d in
subsequent sections . However , i n th e absenc e o f a loa d settlemen t curve , a failur e loa d ma y b e
defined a s that which causes a settlement equal t o 1 0 percent of the pil e diameter o r widt h (as per
the suggestio n o f Terzaghi ) whic h i s widel y accepte d b y engineers . However , i f thi s criterio n i s
applied t o piles o f large diamete r i n clay an d a nominal factor of safet y o f 2 i s used t o obtai n th e
working load, then the settlement at the working load may be excessive.
Factor o f Safet y
In almos t al l cases where piles ar e acting as structural foundations, the allowabl e loa d i s governe d
solely fro m consideration s of tolerable settlemen t at the working load.
The working load for all pile types in all types of soil may be taken as equal to the sum of the
base resistanc e an d shaf t frictio n divide d b y a suitabl e factor o f safety . A safet y facto r o f 2. 5 i s
normally used. Therefore w e may write
Q -^^(.5.2 )
a
2. 5
In case where the values of Q
b
and Q.can be obtained independently, the allowable load can
be written as
Q =^- + ^-(15.3 )
a
3 1. 5
It is permissible t o take a safety factor equal to 1. 5 for the skin friction because the peak value
of ski n frictio n o n a pil e occur s a t a settlement o f onl y 3- 8 m m (relativel y independen t o f shaf t
diameter an d embedde d lengt h but ma y depen d o n soi l parameters ) wherea s th e bas e resistanc e
requires a greater settlemen t for ful l mobilization.
The least of the allowable loads given by Eqs. (15.2) and (15.3) is taken as the design working
load.
1 5.8 M ETH OD S OF DETERM I NI NG U L TI M ATE L OAD B EARI N G
CAPACI TY OF A SI NG L E V ERTI CA L PIL E
The ultimat e bearin g capacity , Q
u
, o f a singl e vertica l pil e ma y b e determine d b y an y o f th e
following methods .
1. B y the use of stati c bearing capacit y equations .
2. B y the use of SPT and CPT values.
3. B y fiel d loa d tests .
4. B y dynami c method.
The determination of the ultimate point bearing capacity, q
b
, of a deep foundation on the basis
of theory is a very complex on e since there ar e many factors which cannot be accounted for in the
theory. The theor y assume s tha t the soi l i s homogeneous an d isotropi c whic h i s normall y no t the
case. All the theoretical equations are obtained based on plane strain conditions. Only shape factors
are applie d t o tak e car e o f th e three-dimensiona l natur e o f th e problem . Compressibilit y
618 Chapt e r 1 5
characteristics o f the soi l complicat e th e problem further . Experienc e an d judgment ar e therefor e
very essential i n applying any theory to a specific problem. The ski n load Q, depends o n the nature
of th e surfac e o f th e pile , th e metho d o f installatio n o f th e pil e an d th e typ e o f soil . A n exac t
evaluation of QAs a difficult job eve n if the soi l is homogeneous ove r the whol e lengt h of the pile.
The proble m become s al l th e mor e complicate d i f th e pil e passe s throug h soil s o f variabl e
characteristics.
1 5 .9 G ENERA L TH EORY FOR U L TI M ATE B EARI N G CAPACI TY
According t o V esic (1967) , onl y punching shear failure occur s i n deep foundations irrespectiv e of
the density of the soil s o long as the depth-widt h ratio L id i s greater than 4 where L = length of pile
and d = diameter (or width of pile). The types of failure surfaces assumed by different investigators
are shown i n Fig. 15.6 for th e genera l shea r failur e condition. The detaile d experimenta l stud y of
V esic indicate s that the failur e surface s do not revert bac k t o the shaf t a s shown i n Fig. 15.6(b).
The tota l failure load Q
u
may be written as follows
Q = n+ W = Q, +Q, + W (154 )
*-•« x-'u p *-•£ > *-• / p \
i
~
1
-^}
where Q
u
= load a t failure applied t o the pil e
Q
b
= bas e resistance
Qs = shaf t resistanc e
W = weigh t of the pile .
The genera l equatio n for the base resistanc e may be writte n as
Qb= cN
c
+q'
0
N
q
+-ydN
Y
A
b
(15.
5)
where d — width or diamete r of th e shaf t a t base leve l
q'
o
= effective overburde n pressure at the base leve l of the pil e
A
b
= base are a of pil e
c - cohesio n of soil
y = effectiv e uni t weight of soi l
N
c
, N , N = bearing capacit y factors which take int o account the shap e factor .
Cohesionless Soils
For cohesionles s soils ,
term q
o
N fo r deep foundations . Therefore Eq . (15.5) reduce s t o
For cohesionles s soils , c = 0 an d the ter m l/2ydN become s insignifican t in compariso n wit h the
(15.6)
Eq. (15.4 ) may no w be written as
Q
b
-Q
u
+W
p
=q'
0
N
q
A
b +
W
p
+Q
f
(15.7 )
The ne t ultimat e load i n excess o f the overburden pressur e loa d q
o
A
b
is
If we assume , fo r al l practical purposes, W an d q'
0
A
b
ar e roughl y equal fo r straigh t sid e o r
moderately tapere d piles, Eq. (15.8) reduces t o
Deep Foun dat io n I : Pil e Foun dat io n
Q
u
619
\ Qu
L 111 ill , I , , , ,_ M
46
ill
(a) (c)
(b)
Figure 15. 6 Th e s hape s of failur e s urface s at t h e tip s o f pile s as as s um e d by (a)
Terzaghi, (b ) Meyerhof , an d (c ) Vesi c
or (15.9)
where A
s
= surfac e area of the embedded lengt h of the pile
q'
o
= average effective overburde n pressure over the embedded dept h of the pil e
K
s
= averag e latera l earth pressur e coefficien t
8 = angl e of wall friction .
Cohesive Soil s
For cohesive soils such as saturated clays (normally consolidated), we have for </ > = 0, N - 1 and N
= 0. The ultimat e base load fro m Eq . (15.5) is
620
Q
b
=( c
h
N
c
+q'
o
)A
b
The ne t ultimat e base loa d i s
Chapt er 1 5
(15.10)
Therefore, th e ne t ul t i mat e load capacit y of the pile, Q , is
or Q
u
=c
b
N
c
A
b
+A
s
a c
u
(15.12 )
where a = adhesion factor
c
u
= average undraine d shear strengt h of clay alon g th e shaf t
c
b
= undrained shear strengt h of clay at the base leve l
N
C
= bearing capacit y factor
Equations (15.9) and (15.12) ar e used for analyzing the net ultimate load capacit y o f piles i n
cohesionless an d cohesiv e soil s respectively . I n eac h cas e th e followin g type s o f pile s ar e
considered.
1 . Drive n pile s
2. Drive n and cast-in-sit u piles
3. Bore d pile s
1 5 .1 0 U L TI M AT E B EARI NG CAPACI T Y I N COH ESI ONL ESS SOI L S
Effect o f Pil e I nstallatio n o n the V alu e o f the Angl e o f Frictio n
When a pile i s driven into loose san d it s density is increased (Meyerhof , 1959) , and the horizontal
extent of the compacted zon e has a width of about 6 to 8 times the pile diameter. However , in dens e
sand, pil e drivin g decrease s th e relativ e densit y becaus e o f th e dilatanc y o f th e san d an d th e
loosened san d alon g the shaf t ha s a width of about 5 times the pile diameter (Kerisel , 1961) . On the
basis o f fiel d an d mode l tes t results , Kishid a (1967 ) propose d tha t th e angl e o f interna l frictio n
decreases linearl y from a maximum value of 0
2
at the pil e ti p t o a low value of 0
t
a t a distance of
3.5dfrom th e tip where d is the diameter of the pile, 0j is the angle of friction before th e installation
of the pile and 0
2
after th e installation as shown in Fig. 15.7 . Based on the field data, the relationship
between 0
l
an d 0
2
in sands may be written as
(j ), + 4 0
(15.13)
Figure 1 5 . 7 T h e effec t o f drivin g a pil e o n
Deep Foun dat io n I : Pil e Foun dat io n 62 1
An angl e o f 0 , = 0
2
= 40° i n Eq . (15.13 ) mean s n o chang e o f relativ e densit y du e t o pil e
driving. V alue s o f 0
}
ar e obtaine d fro m insitu penetratio n test s (wit h n o correctio n du e t o
overburden pressure , bu t correcte d fo r fiel d procedure ) b y usin g th e relationship s establishe d
between 0 an d SP T o r CP T values . Kishid a (1967 ) ha s suggeste d th e followin g relationshi p
between 0 and the SPT value N
cor
a s
0° =^20N
cor
+l5° (15.14)
However, Tomlinson (1986) is of the opinion that it is unwise to use higher values for 0 du e
to pile driving. His argument is that the sand may not get compacted, a s for example, when piles are
driven into loose sand, the resistance is so low and little compaction i s given to the soil. He suggests
that th e valu e o f 0 use d fo r th e desig n shoul d represent th e i n situ conditio n tha t existed befor e
driving.
With regar d t o driven and cast-in-situ piles , ther e i s no suggestion b y any investigator as t o
what value of 0 should be used for calculating the base resistance. However, it is safer to assume the
insitu 0 value for computing the base resistance.
With regar d t o bore d an d cast-in-situ piles, th e soi l get s loosene d durin g boring. Tomlinson
(1986) suggests that the 0 value for calculating bot h the base and skin resistance should represent the
loose state . However, Poulos et al., (1980) suggest s that for bored piles, the value of 0be take n as
0 = ^ - 3 (15.15 )
where 0 j = angl e of internal friction prio r t o installation of the pile.
1 5 .1 1 CRI TI CA L DEPT H
The ultimat e bearing capacit y Q
u
in cohesionless soil s as per Eq. (15.9) i s
Q
u
= 9'o
N
q
A
b
+
«'
0
Ks
imSA
s (15.16a )
or Q
u
=1
b
\ +fA (15.16b )
Eq. (15.16b) implies that both the point resistance q
b
and the ski n resistance f
s
ar e function s
of the effective overburden pressur e q
o
in cohesionless soils and increase linearly with the depth of
embedment, L , of the pile . However , extensiv e researc h wor k carrie d out by V esi c (1967 ) has
revealed tha t the base an d frictional resistances remai n almos t constan t beyond a certain dept h of
embedment whic h i s a functio n o f 0 . This phenomeno n wa s attribute d t o archin g by V esic. On e
conclusion from th e investigation of V esic is that in cohesionless soils , th e bearing capacit y factor,
N , is not a constant depending on 0 only, but also on the ratio L id (wher e L = length of embedment
of pile, d = diameter o r widt h of pile). In a similar way, the frictional resistance , f
s
, increase s wit h
the L /d rati o an d remain s constan t beyond a particular depth. Let L
C
b e th e depth , whic h ma y b e
called th e critical depth, beyon d which both q
b
and/
5
remain constant . Experiments of V esic have
indicated that L
c
i s a function o f 0 . The L J d rati o as a function o f 0 may be expressed a s follows
(Poulos and Davis, 1980)
For 28° <0< 36.5 °
L J d = 5 + 0.24 (0° -28° ) (15.17a )
For 36.5° < 0 < 42°
L J d = 1+ 2.35(0° - 36.5° ) (15.17b )
The abov e expression s hav e been develope d base d o n the curve given by Poulos an d Davis,
(1980) givin g the relationship between L J d an d 0°.
622 Chapt er 1 5
The Eqs. (15.17 ) indicat e
L J d=5 a t 0 = 28 °
L
c
/d = l a t 0 = 36.5 °
L
c
/d=20 a t 0 = 42 °
The 0 values to be used for obtaining L J d ar e as follows (Poulos and Davis , 1980 )
for drive n piles 0 = 0.7 5 0 j + 10 ° (15.18a )
for bore d piles : 0 = 0 , - 3 ° (15.18b )
where 0 j = angl e o f internal friction prio r t o the installation of the pile .
1 5 .1 2 TOM L I NSON' S SOL U TI ON FOR Q
b
I N SAND
Driven Piles
The theoretica l N facto r i n Eq. (15.9) is a function o f 0. There i s great variatio n i n the values ofN
derived b y differen t investigator s as shown i n Fig. 15.8 . Compariso n o f observed bas e resistance s
of pile s b y Nordlun d (1963 ) an d V esi c (1964 ) hav e show n (Tomlinson , 1986 ) tha t N value s
established b y Berezantsev et al., (1961) which take into account the depth t o width ratio of the pile,
CQ
1000
100
10
1. Terzaghi(1943 )
2. V esi c (1963)
3. Berezantse v (1961)
4. Br i nchHansen(1951 )
5. Skempt onet al (1953 )
6. Caquot-Kerisel(1956 )
I / //
7
7
1. Br i nchHansen(1961 )
8. Meyerhof ( 1953) Bore d Pile s
9. Meyerhof ( 1953) Drive n Piles
10. D e Beer (1945 )
25 3 0 3 5 4 0 4 5
Angle of internal friction 0 °
50
Figure 15. 8 Bearin g capacit y fact or s fo r circula r dee p foun dat ion s (aft e r Kezdi , 1975)
Deep Foun dat io n I : Pil e Foun dat io n
200
623
150
£ 100
50
20 25 3 0 3 5 4 0
Angle of internal frictio n 0°
45
Figure 1 5. 9 B erezan t s ev' s bearin g capacit y fact or , N (aft e r Tom lin s on , 1986 )
most nearl y conform t o practical criteri a o f pil e failure . Berezantsev' s value s of N a s adopted b y
Tomlinson (1986) ar e given in Fig. 15.9 .
It may be see n fro m Fig . 15. 9 that there i s a rapid increas e i n T V fo r high values of 0, giving
thereby high values of base resistance. As a general rul e (Tomlinson, 1986) , the allowable working
load on an isolated pil e driven to virtual refusal, usin g normal driving equipment, in a dense sand or
gravel consistin g predominantl y o f quart z particles , i s give n b y th e allowabl e loa d o n th e pil e
considered a s a structural member rather than by consideration of failure of the supporting soil, or
if the permissibl e workin g stress o n the material o f the pil e i s not exceeded, the n the pil e wil l not
fail.
As pe r Tomlinson , th e maximu m bas e resistanc e q
b
i s normall y limite d t o 1100 0 kN/m
2
(110 t/ft
2
) whateve r might be the penetration depth of the pile .
B ored an d Cast-in-sit u Pile s i n Cohesionless Soil s
Bored pile s ar e formed i n cohesionless soil s b y drilling wit h rigs. The sides o f the holes migh t be
supported by the use of casing pipes. When casing is used, the concrete i s placed i n the drilled hol e
and th e casin g i s graduall y withdrawn . In al l th e case s th e side s an d botto m i f th e hol e wil l b e
loosened a s a resul t o f th e borin g operations , eve n thoug h i t ma y b e initiall y b e i n a dens e o r
medium dens e state . Tomlinso n suggest s tha t the value s of th e parameter s i n Eq. (15.9 ) mus t b e
calculated by assuming that the 0 value will represent the loose condition .
However, whe n piles ar e installed b y rotar y drilling under a bentonite slurr y for stabilizing
the sides , i t ma y b e assume d tha t th e 0 valu e use d t o calculat e bot h th e ski n frictio n an d bas e
resistance wil l correspond t o the undisturbed soil condition (Tomlinson, 1986) .
The assumptio n o f loos e condition s fo r calculating ski n frictio n an d bas e resistanc e mean s
that the ultimate carrying capacit y o f a bored pil e in a cohesionless soi l wil l be considerably lowe r
than that of a pile drive n i n the same soi l type . As per De Beer (1965) , th e base resistance q
b
of a
bored an d cast-in-situ pil e i s about one third of that of a driven pile.
624 Chapt e r 1 5
We may write ,
q
b
(bore d pile ) = (1/3) q
h
(driven pile)
So far as friction loa d is concerned, the frictional paramete r ma y be calculated b y assuming a
value of 0 equal to 28° whic h represents the loose conditio n o f the soil .
The same Eq. (15.9) may be used to compute Q
u
based o n the modifications explained above .
15.13 MEYERH OF' S METHOD OF DETERMINING Q
b
FOR PILES I N SAND
Meyerhof (1976 ) take s int o accoun t th e critical dept h rati o ( L J d) fo r estimatin g th e valu e o f Q
b
,
Fig. 15.1 0 show s th e variatio n of L J d fo r bot h th e bearing capacit y factor s N
c
an d N a s a function
of 0. According t o Meyerhof, the bearing capacity factors increase with L
h
/d an d reach a maximum
value at L J d equa l t o about 0. 5 ( L J d), wher e L
b
i s the actual thickness of the bearing stratum. For
example, i n a homogeneou s soi l (15.6c ) L
b
i s equa l t o L , th e actua l embedde d lengt h o f pile ;
whereas i n Fig. 15.6b, L
h
i s less than L .
1000
10 2 0 3 0
Angle of interna l friction i n degree s
40 4 5
Figure 1 5.1 0 B earin g capacit y fact or s an d crit ical dept h rat io s LJd fo r drive n pile s
(aft er M eyerhof , 1976 )
Deep Foun dat io n I : Pil e Foun dat io n 62 5
As pe r Fig. 15.10, th e valu e of L J d i s abou t 25 for 0 equal t o 45° an d i t decreases wit h a
decrease i n th e angl e o f frictio n 0 . Normally , th e magnitud e o f L
b
ld fo r pile s i s greate r tha n
0.5 ( L J d) s o that maximum values ofN
c
an d T V ma y apply for the calculation of q
b
, the uni t bearing
pressure of the pile. Meyerhof prescribes a limiting value for q
b
, based on his findings on static cone
penetration resistance. The expression fo r the limiting value, q
bl
, is
for dens e sand : q
bl
= 5 0 A^ tan 0 kN/m
2
(15.19a )
for loos e sand : q
bl
= 2 5 N
q
ta n 0 kN/m
2
(15.19b )
where 0 is the angl e of shearing resistance o f the bearing stratum. The limiting q
bl
values given by
Eqs (15.9 a an d b ) remai n practicall y independen t o f th e effectiv e overburde n pressur e an d
groundwater conditions beyond the critical depth.
The equation for base resistanc e i n sand may now be expressed a s
Qb=«o
N
q
A
b*<*u
A
b (15-20 )
where q'
o
= effective overburde n pressur e at the tip of the pile L J d an d N = bearing capacity factor
(Fig. 15.10) .
Eq. (15.20) is applicable onl y for driven piles in sand. For bored cast-in-sit u piles the value of
q
b
i s t o be reduced b y one third to one-half .
Clay Soi l ( 0 = 0 )
The base resistanc e Q
b
for piles i n saturated cla y soi l may be expressed a s
Q
b
=N
c
c
u
A
b
=9c
u
A
b
(15.21 )
where N
c
= 9, and c
u
= undrained shear strengt h of the soi l a t the base leve l of the pile .
1 5 .1 4 V ESI C' S M ETH OD OF DETERM I NI NG Q
b
The unit base resistanc e o f a pile in a ( c - 0 ) soil may be expressed a s (V esic, 1977 )
3
b
=
cN
c
+
3'o
N
*
q
(15.22 )
where c = uni t cohesio n
q'
0
= effectiv e vertica l pressure at the base level of the pil e
N *
c
and N *= bearin g capacit y factors related t o each othe r by the equation
A£= (# ;-i)cot0 (15.23 )
As per V esic, the base resistance is not governed by the vertical ground pressure q'
Q
but by the
mean effective normal groun d stress cr
m
expressed a s
l+2K
o
,
a
m
=
— -^-
3
- 3o (15.24 )
in whic h K
O
= coefficient of earth pressur e for the at rest condition = 1 - si n 0.
Now the bearing capacit y i n Eq. (15.22) may be expressed a s
=<
+
^< (15-25 )
626 Chapt e r 1 5
An equatio n fo r AT
f f
can b e obtaine d fro m Eqs . (15.22), (15.24 ) an d (15.25 ) a s
3/V *
N
°
=
1 + 2K (15.26 )
o
V esic has developed a n expression fo r N *
a
based o n the ultimat e pressure neede d t o expand a
spherical cavit y i n a n infinit e soi l mass a s
where a , = , a~ = I ^I t a n ^ , a-, ~ *""""" r
anc
j y y - tan
2
(45° + <b!2)
1
3- s i n ^
2
\ 2 )
3
( 1 + si n^) ^
According t o V esic
l + /
r
A
£,
(15.28)
l
r
- rigidit y index =
where /
rr
= reduce d rigidit y index for the soil
A = averag e volumetri c strai n i n the plasti c zon e below the pil e poin t
E
s
= modulu s of elasticit y of soi l
G = shea r modulu s of soi l
j. = Poisson' s rati o of soi l
Figures 15. 1 1 (a) and 15. 1 l(b) give plots of A^ versus 0, and N *
c
versus 0fo r variou s values of
l
rr
respectively .
The values of rigidity index can be computed knowing the values of shear modulus G ^ and the
shear strengt h s ( = c + q'
o
tan 0) .
When a n undrained condition exists in the saturated clay soi l or the soil i s cohesionless an d is
in a dense stat e we have A = 0 and in such a case /. = I
rr
.
For 0 = 0 (undrained condition), we have
A7 = 1.33(ln/,
r
+ l) + !+l (15.30 )
The valu e of l
f
depend s upo n the soi l state , (a ) for sand , loose or dense an d (b ) for cla y low,
medium or high plasticity . For preliminary estimates th e following values of l
r
may be used .
Soil type l
r
Sand ( D
r
= 0.5-0.8) 75-15 0
Silt 50-7 5
Clay 150-25 0
Deep Foun dat io n I : Pil e Foun dat io n 627
(a)
1000
CQ
100
10
XXX'/
10 2 0 3 0
Angle of internal friction 0°
40
500
400
200
100
60
40
20
10
50
(b)
1000
100
10
10 2 0 3 0
Angle of internal friction 0
C
40
/,,=
200
100
60
40
20
10
50
Figure 1 5 .1 1 (a ) B earin g capacit y fact or ^ N*
a
(b ) B earin g capacit y fact o r A T
( Veslc, 1977 )
628 Chapt er 1 5
Table 1 5 . 1 B earin g capacit y fact or s A/ * an d N* b y J an bu
V =

0
5
10
20
30
35
40
45
75°
N
\
N
*c
1.00
1.50
2.25
5.29
13.60
23.08
41.37
79.90
5.74
6.25
7.11
11.78
21.82
31.53
48.11
78.90
90°
N
\
N
*c
1.00
1.57
2.47
6.40
18.40
33.30
64.20
134.87
5.74
6.49
8.34
14.83
30.14
46.12
75.31
133.87
105°
N
\
N
*c
1.00
1.64
2.71
7.74
24.90
48.04
99.61
227.68
5.74
7.33
9.70
18.53
41.39
67.18
117.52
226.68
1 5 .1 5 JANB U ' S M ETH OD OF DETERM I NI NG Q
b
The bearing capacit y equatio n of Janbu (1976) is the same a s Eq. (15.22) and i s expressed a s
Q
b
=Wc
+
Vo
N
^
A
b (15-31 )
The shap e o f the failur e surfac e a s assumed b y Janbu i s similar t o that given i n Fig. 15.6(b) .
Janbu' s equation fo r N *i s
F i 1
= ta n 0+ V I + tan
2
0 (15.32)
where i// = angl e a s shown i n Fig. 15.6(b) . This angl e varie s fro m 60 ° i n sof t compressibl e soi l t o
105° i n dens e sand . Th e value s fo r N *
c
use d b y Janb u ar e th e sam e a s thos e give n b y V esi c
(Eq. 15.23) . Tabl e 15. 1 gives the bearing capacit y factor s o f Janbu.
Since Janbu' s bearin g capacity factor N *depends o n the angl e y/ , there ar e two uncertaintie s
involved i n thi s procedure. The y ar e
1. Th e difficult y i n determining th e value s o f i/^fo r differen t situation s a t base level.
2. Th e settlemen t require d a t th e bas e leve l o f th e pil e fo r th e ful l developmen t o f a plasti c
zone.
For ful l bas e loa d Q
b
t o develop , a t leas t a settlemen t o f abou t 1 0 to 2 0 percen t o f th e pil e
diameter i s required whic h i s considerable fo r larger diamete r piles .
1 5 .1 6 COY L E AND CASTEL L O' S METH OD OF ESTIMATING Q
b
I N SAND
Coyle an d Castell o (1981 ) made us e of the result s of 24 ful l scal e pil e loa d test s drive n i n sand fo r
evaluating the bearing capacit y factors. The form of equation used by them is the same as Eq. (15.6 )
which may be expressed a s
Q
b
=Vo
N
q
A
i, (15.33 )
where q' - effectiv e overburden pressur e a t the base leve l of the pile
N *= bearin g capacit y facto r
Coyle an d Castell o collecte d dat a fro m th e instrumented piles , an d separated fro m th e tota l
load th e base loa d an d frictio n load . The total forc e a t the top of the pile was applied b y means o f a
jack. Th e soi l a t the sit e wa s generall y fine sand wit h some percentag e o f silt . The lowes t an d the
Deep Foun dat io n I : Pil e Foun dat io n 629
J O
Bearing capacit y factor N
q
*
20 4 0 6 0 8010 0 20 0 (lo g scale )
Figure 1 5 .1 2 A/ * vers us Lid (aft e r Coyl e an d Cas t ello , 198 1
highest relativ e densitie s wer e 4 0 t o 10 0 percent respectively . Th e pil e diamete r wa s generall y
around 1. 5 f t and pil e penetratio n was about 50 ft . Closed en d stee l pip e wa s use d fo r th e test s i n
some place s an d precast squar e piles or steel H piles wer e used at other places .
The bearing capacity factor N *wa s evaluated with respect t o depth ratio L id i n Fig. 15.1 2 for
various values of </> .
1 5 .1 7 TH E U L TI M ATE SKI N RESI STANC E O F A SI NG L E PIL E I N
COH ESI ONL ESS SOI L
Skin Resistanc e (Straight Shaft )
The ultimat e skin resistance i n a homogeneous soil as per Eq. (15.9) i s expressed a s
In a layered syste m of soi l
then be expressed a s
q
o
,
(15.34a)
K
s
an d 8 vary with respect t o depth. Equation (15.34a ) may
Q=
(15.34b)
where q'
o
, K
S
an d < 5 refer t o thickness dz o f each laye r and P i s the perimete r o f th e pile .
As explaine d i n Sectio n 15.1 0 the effectiv e overburde n pressur e doe s no t increas e linearl y
with depth and reaches a constant value beyond a particular depth L
c
, called the critical dept h which
is a functio n o f 0 . I t i s therefor e natura l t o expec t th e ski n resistanc e f
s
als o t o remai n constant
beyond dept h L . The magnitude of L may be taken as equal to 2Qd.
630 Chapt er 1 5
Table 15. 2 Val ue s o f K
s
an d 8(B rom s , 1966 )
Pile m at erial
Val ues o f K
s
Low D
r
Hig h D
r
Steel
Concrete
Wood
20°
3/40
2/30
0.5
1.0
1.5
1.0
2.0
4.0
3.0
2.5
- 2.0
1.5
l . Q
_l 1 1 1
-
-
^/
1 1 1 1
Driven
piles /
/
i i i i
i i i i _
/:
/ '--
i i i r
1.6
1.2
0.8
0.4
i i
- D
-
-
i i
i i i i
riven pi l
\
\
1
s
1 1 1 1
1 1 1 1
e
V '
i i
-
/ Jacked .
x
^- Bore d
piles
i i i i i i
28 3 3 3 8 4 3

(a)
30 3 5 4 0

(b)
25'
0.5 1. 0 1. 5 2. 0
Pile taper angle , a> °
(c)
Figure 1 5 .1 3 Val ue s of K
s
t a n 8 i n s an d as pe r (a ) Poulos an d Davi s 1980, (b)
M eyerhof, 1 976 an d (c) t aper fact o r F
w
(aft er Nordlund , 1 963)
Eq. (15.17) can be used for determining the_critical length L
c
for any given set of values of
and d. (Xcan be calculated from Eq . (15.34) i f A ^ and 8 are known.
Deep Foun dat io n I : Pil e Foun dat io n 63 1
The value s of K
s
an d 5 vary not only with the relative density and pile material but also with
the method o f installation of the pile .
Broms ( 1 966) has related the values of K
s
and 8 to the effective angl e of internal friction 0 of
cohesionless soils for various pile materials and relative densities (D
r
) as shown in Table 15.2 . The
values ar e applicabl e t o drive n piles . As pe r th e presen t stat e o f knowledge , th e maximu m ski n
friction i s limited t o 1 10 kN/m
2
(Tomlinson , 1986) .
Eq. (15.34) ma y also be written as
Q = Pq'
o
j3dz
(1
5.
35)
o
where, / 3 = K
s
ta n 8 .
Poulos and Davis, (1980) hav e given a curve giving the relationship between /3an d 0° which
is applicabl e fo r drive n pile s an d al l type s o f materia l surfaces . Accordin g t o the m ther e i s no t
sufficient evidenc e t o show that /? would vary with the pile material. The relationship between ft and
0 i s give n i n Fig. 15.13(a). Fo r bore d piles , Poulo s e t al , recommen d th e relationshi p give n by
Meyerhof (1976 ) betwee n 0 and 0 (Fig. 15.13(b)) .
Skin Resistanc e o n Tapered Pile s
Nordlund (1963) has shown that even a small tape r of 1 ° on the shaf t give s a four fol d increas e i n
uni t frictio n i n mediu m dens e san d unde r compressio n loading . Base d o n Nordlund' s analysis ,
curves have been developed (Poulo s and Davis, 1980 ) giving a relationship between tape r angl e of
and a taper correction facto r F
w
, which can be used in Eq. (15.35) a s
Q
f
= FvPq'oPte (15.36 )
o
Eq. (15.36 ) give s th e ultimat e ski n loa d fo r tapere d piles . Th e correctio n facto r F
w
ca n b e
obtained fro m Fig. 15.13(c). The value of 0 to be used fo r obtaining F
w
i s as per Eq. (15.18a) for
driven piles .
15.18 SKI N RESISTANCE Q
f
BY COYLE AND CASTELL O METHOD (1981)
For evaluating frictional resistance , Q~ for piles in sand, Coyle and Castello (1981) made use of the
results obtained from 2 4 fiel d test s on piles. The expression for <2,is_th e one given in Eq. (15.34a).
They develope d a chart (Fig. 15.14 ) giving relationships between K
s
an d 0 for various L id ratios .
The angl e of wall friction 8 is assumed equal t o 0.80. The expression fo r Q
f
is
Q
f
=A
s
q'
o
K
s
tanS (15.34a )
where q'
o
= average effectiv e overburde n pressur e an d S = angle of wal l friction = 0.80.
The valu e of K
s
ca n be obtained Fig. 15.14.
1 5 .1 9 STATI C B EARI NG CAPACI T Y OF PILES I N CL AY SOI L
Equation fo r U ltimat e B earin g Capacit y
The stati c ultimate bearing capacit y of piles i n clay as per Eq. (15.12) is
Q
u
= Q
b
+Q
f
= c
b
N
c
A
b
+ac
u
A
s
(15.37 )
632
Chapt er 1 5
Earth pressure coefficien t K
s
j >= 3 0 3 2 3 4 36 °
Figure 15.1 4 Coefficien t K vers u s L/d, 8 = 0.8 0 (aft er Coyl e an d Cast ello, 1981 )
For layered cla y soil s where the cohesive strengt h varies along the shaft , Eq . (15.37) may be
written as
(15.38)
B earing Capacit y Facto r N
c
The valu e o f th e bearin g capacit y facto r N
c
tha t i s generall y accepte d i s 9 whic h i s th e valu e
proposed b y Skempto n (1951 ) fo r circula r foundation s for a L I B rati o greate r tha n 4 . Th e bas e
capacity o f a pile i n clay soi l may no w be expressed as
Q, = 9c,A, ( 15 39)
*--b b b \ L ~J .~J SJ
The value of c
b
may be obtained either from laboratory test s on undisturbed samples or from
the relationships established between c
u
and field penetratio n tests. Eq. (15.39) i s applicable for all
types of pile installations ,
Skin Resistanc e b y a-M etho d
Tomlinson (1986 ) ha s give n som e empirica l correlation s fo r evaluatin g a i n Eq . (15.37 ) fo r
different type s of soil conditions and L id ratios. His procedure require s a great deal of judgment of
the soi l condition s i n th e fiel d an d ma y lea d t o different interpretations b y differen t geotechnica l
engineers. A simplifie d approach fo r suc h problems woul d be needed . Denni s and Olso n (1983b )
Deep Foun dat io n I : Pil e Foun dat io n 633
made us e o f th e informatio n provide d b y Tomlinso n an d develope d a singl e curv e givin g th e
relationship between a and the undrained shear strengt h c
u
of clay as shown in Fig. 15.15 .
This curve can be used to estimate the values of a for piles wit h penetration lengths less than
30 m. As the lengt h of the embedment increase s beyon d 3 0 m, the value of a decreases . Pile s of
such great lengt h experience elasti c shortenin g that results in small shear strai n or slip at great depth
as compared to that at shallow depth . Investigation indicate s that for embedment greater than abou t
50 m the value of a from Fi g 15.15 should be multiplied by a factor 0.56. Fo r embedments between
30 and 50 m, the reduction facto r may be considered t o vary linearly from 1. 0 to 0.56 (Denni s and
Olson, 1983a , b )
Skin Resistanc e b y A- M etho d
V ijayvergiya an d Focht (1972 ) hav e suggested a different approac h fo r computing ski n load QAov
steel-pipe pile s on the basis of examination of load tes t results on such piles. The equation i s of the
form
Q
f
= A( q'
0
+ 2c
M
) A
s
(15.40 )
where A = frictiona l capacit y coefficient ,
q'
o
= mean effectiv e vertical stres s betwee n th e ground surfac e an d pil e tip .
The othe r term s ar e alread y defined . A i s plotte d agains t pil e penetratio n a s show n i n
Fig. 15.16 .
Eq. (15.40) has been found very useful for the design of heavily loaded pip e piles for offshore
structures.
/MV Iethod o r the Effectiv e Stres s M etho d o f Computing Ski n Resistanc e
In thi s method, th e uni t skin friction f
s
i s defined as
f
s
=
K
s
t an Sq'
o
= ( 3~q'
o
(15.41 )
where f t = the ski n factor = K
s
ta n <5 , (15.42a )
K = lateral eart h pressur e coefficient ,
Undrained cohesio n c
u
, kN/m
2
Figure 15.1 5 Adhesio n fact or a. for pile s wit h penetration lengths les s than 50 m in clay.
(Data from Dennis an d Olson 1983 a, b; Stas an d Kulhawy, 1984)
634 Chapt er 1 5
V alue of A
I.2 0. 3 0. 4 0.5
Figure 1 5 .1 6 Frict ion a l capacit y coefficien t A vs pil e pen et rat io n
(Vij ayverg iya an d Focht , 1972 )
8 = angl e o f wall friction,
q'
o
= average effectiv e overburde n pressure.
Burland (1973) discusses the value s to be used for ) 3 and demonstrates tha t a lower limi t for
this factor for normall y consolidated clay can be written as
~ (15.42b )
(15.42c)
(15.42d)
As per Jaky (1944 )
therefore /3 = ( 1 - si n 0') tan 0'
where 0 ' = effectiv e angl e of internal friction.
Since th e concep t o f thi s metho d i s based o n effective stresses, th e cohesio n intercep t o n a
Mohr circle i s equal to zero. For driven piles in stiff overconsolidate d clay, K
s
i s roughly 1. 5 times
greater than K
Q
. For overconsolidate d clays K
O
ma y b e found from th e expression
K
o
= (1 - si n <!)'}TJ R~ ( 15.42e)
where R - overconsolidatio n rati o of clay.
For clays , 0' may be take n i n the range of 20 t o 30 degrees. In suc h a case the value of |3 in
Eq. (15.42d) varies betwee n 0.24 an d 0.29 .
Deep Foun dat io n I : Pil e Foun dat io n 63 5
M eyerhof 's Metho d (1 9 7 6 )
Meyerhof ha s suggested a semi-empirical relationship for estimating skin friction i n clays.
For driven piles:
f
s
= \ .5c
u
tan 0' (15.43 )
For bored piles :
f
s
=c
u
tanp (15.44 )
By utilizing a value of 20° fo r </> ' for th e stif f t o very stif f clays , the expressions reduce t o
For driven piles:
/, = 0-55c, (15.45 )
For bored piles:
f
s
=0.36c
u
(15.46 )
In practice the maximum value of unit friction fo r bored piles i s restricted t o 10 0 kPa.
1 5.20 B EARI N G CAPACI TY OF PILES I N G RANU L AR SOI LS
B ASED ON SPT V AL U E
Meyerhof (1956 ) suggest s the followin g equation s for singl e piles i n granular soils based o n SPT
values.
For displacement piles:
Q
u
= Q
b
+ Q
f
= 40N
cor
( L /d)A
b
+ 2N
cor
A
s
(15.47a )
for //-piles:
Q
u
= 40N
cor
( L /d)A
b
+ N
cor
A
s
(15.47b )
where q
b
= 40 N
cor
( L ld) < 400 N
cor
For bored piles:
Qu=M
N
cor
A
b+°*™cor
A
, (15-48 )
where Q
u
= ultimate total load in kN
N
cor
= average corrected SPT value below pile tip
N
cor
= corrected averag e SPT value along the pile shaf t
A
b
= base are a of pile in m
2
(for H-piles including the soil between the flanges)
A
s
= shaft surfac e area in m
2
In English units Q
u
for a displacement pile is
0
u
(kip) =Q
b
+Q
f
= O.WN
cor
( L /d)A
b
+O.Q4N
cor
A
s
(15.49a )
where A
b
= base area i n ft
2
an d A
s
= surface area i n ft
2
and 0.80^, AA
A
< SN
cor
A
b
( kip) (15.49b )
a
636 Chapt er 1 5
A minimum factor o f safet y of 4 is recommended. Th e allowabl e load Q
a
is
Q ==±
U*4
(15.50)
Example 1 5 . 1
A concret e pil e o f 45 c m diamete r was drive n int o sand o f loos e t o medium densit y t o a dept h of
15m. The following properties ar e known:
(a) Average uni t weight of soil alon g the length of the pile, y = 17.5 kN/m
3
, average 0 = 30° ,
(b) averag e K
s
= 1.0 and 8= 0.750 .
Calculate (a ) th e ultimat e bearin g capacit y o f th e pile , an d (b ) th e allowabl e loa d wit h
F
s
= 2.5. Assume the water table i s at great depth . Use Berezantsev' s method .
Solution
From Eq . (15.9 )
Q
u
=Q
b
+Q
f
=q'
0
A
b
N
q
+ q'
0
AK
s
ta n S
where ' = L = 17.5 x 15 =262.5 kN/m
2
= 131.25 kN/m
2
J
O f\ '
A, = —~ x0. 45
2
= 0.159 m
2
b 4
Q
u
15m
Qf
Sand
y = 17. 5 kN/m
3
0 = 30°
£,= 1.0
Q
b
Figure Ex . 1 5 . 1
Deep Foun dat io n I : Pil e Foun dat io n 63 7
A
s
=3. 14x0. 45x1 5 = 21.195 m
2
S = 0.750 = 0.75x30 = 22.5°
tan 8 =0.4142
L 1 5
From Fig. 15.9, TV fo r — = — —= 33.3 and 0 = 30° i s equal to 16.5.
q
a 0.4 5
Substituting the known values, we have
Q
u
=Q
b
+Q
f
=262.5x0.159x16. 5 + 131.25x21.195x1.0x0.4142
= 689+ 1152 = 1841 k N
Example 1 5 . 2
Assume in Ex. 15. 1 that the water table is at the ground surface and v
t
= 18.5 kN/m
3
. All the other
*"^* Sa l
data remain the same. Calculat e Q
u
and Q
a
.
Solution
Water table at the ground surface y
sat
=18. 5 kN/m
3
r
b
=r
M
~r
w
=l 8- 5 - 9. 8 1 =8.69 kN/m
3
^=8. 69x15 = 130.35 kN/m
2
q'
Q
= -x 130.35 = 65.18 kN/m
2
Substituting the known values
Q
u
= 1 30.35 x 0.159 x 16.5 + 65.18 x 21.195 x 1.0 x 0.4142
= 342 + 572 = 914 kN
914
e
a
= — = 366 kN
Note: I t ma y b e note d her e tha t th e presenc e o f a wate r tabl e a t th e groun d surfac e i n
cohesionless soi l reduces th e ultimate load capacit y of pile by about 50 percent .
Example 1 5. 3
A concrete pil e o f 45 cm diameter i s driven to a depth of 1 6 m through a layered syste m of sandy
soil (c = 0). The following data are available .
Top layer 1 : Thickness = 8 m, y
d
= 16.5 kN/m
3
, e = 0.60 and 0 = 30°.
Layer 2: Thickness = 6 m, y
d
= 15.5 kN/m
3
, e - 0.6 5 and 0 = 35°.
Layer 3: Extends to a great depth, y
d
= 16.00 kN/m
3
, e = 0.65 and 0 = 38°.
638 Chapt er 1 5
Q
T
AL, =
AL
2
=
i
AL
3
=
Layer 1
Sand
= 8 m
Layer 2
Sand
= 6 m
= 2 m
La
y
gr 3
Sand
T
y
d]
= 16.5 kN/m
3
0 = 30°, e = 0.6
y
d2
= 15 . 5 kN/m
3
0 = 35°, e = 0.65
y
d3
= 16. 0 kN/m
3
0 = 38°, e = 0.65
Figure Ex . 1 5. 3
Assume that the value of <5i n all the layers of sand is equal to 0.750. The value of K
S
fo r eac h
layer a s equal t o hal f of the passive eart h pressur e coefficient . Th e water tabl e i s at ground level .
Calculate th e value s o f Q
u
an d Q
a
wit h F
s
= 2. 5 b y th e conventiona l metho d fo r Q
f
an d
Berezantsev's method for Q
b
.
Solution
The soi l is submerged throughout the soil profile. The specific gravit y G^ is required for calculating
' sat'
Y G
(a) Usin g the equation Yd
=
~—
L
' calculat e G^ for each laye r sinc e y
d
, Y
w
and e are known.
Y ( G
(b) Usin g the equation 7
sat
=
:
, calculat e y
sat
for each laye r and then y
b
= y
sat
- y
w
for
each layer .
(c) Fo r a layere d syste m o f soil , th e ultimat e loa d ca n b e determine d b y makin g us e o f
Eq. (15.9). Now
Q
u
=Q
b
+Q
f
=q',N
q
A
b
+ P
L
^ K
s
tan^A L
o
(d) q'
Q
a t the ti p of the pil e i s
Deep Foun dat io n I : Pil e Foun dat io n 63 9
(e) q'
Q
a t th e middl e of each layer i s
4oi
=
- AL, r
M
-
(f) A/ j = 95 for 0 = 38° and — = — = 33.33 fro m Fig. 15.9 .
(g) A
fc
= 0.159 m
2
, P= 1.413m .
- 1 / ^ 1 -
(h) K
s
=-tai\
2
\ 45°+—\ =—K
p
. K
s
fo r each laye r can be calculated .
(i) 8 = 0.75</>. The value s of tan 8 can be calculated for each layer .
The computed values for al l the layers ar e given below i n a tabular form.
Layer no . G
s
1.
2.
3.
2.69
2.61
2.69
From middl e o f layer 3
At th e ti p of pil e
Yt,
k N/ m
3
10.36
9.57
10.05
to ti p of pil e
4'o
=
k N/ m
2
41.44 1. 5
111.59 1.84 5
150.35 2.1 0
= 10.05
160.40 kN/m
2
tan 8 A/ ,
m
0.414 8
0.493 6
0.543 2
<2
W
=(160.4x95x0.15 9 + 1.413(41.44x1.5x0.414x8
+ 1 1 1.59 x 1.845 x 0.493 x 6 + 150.35 x 2.10 x 0.543 x 2)
= 2423 + 1636 = 4059 k N
4059
Q
=
a
2.5 2. 5
Example 1 5.4
If th e pile in Ex. 15. 2 is a bored and cast-in-situ, compute Q
u
and Q
a
. All the other dat a remain the
same. Water table is close t o the ground surface.
Solution
Per Tomlinson (1986), the ultimate bearing capacity of a bored an d cast-in-situ-pile in cohesionles s
soil i s reduce d considerabl y du e t o disturbance of the soil . Pe r Sectio n 15.12 , calculat e the bas e
resistance for a driven pile and take one-third of this as the ultimate base resistance for a bored and
cast-in-situ pile.
For computing 8, take 0 = 28° and K = 1.0 from Tabl e 15. 2 for a concrete pile .
640 Chapt e r 1 5
B ase resistanc e for drive n pile
For 0 = 30°, N
q
= 16.5 fro m Fig . 15.9 .
A
h
= 0.159m
2
q'
Q
= 130.35 kN/m
2
(Fro m Ex. 15.2 )
Q
b
= 130.35 x 0.159 x 16. 5 = 342 kN
For bore d pile
e, =-
Skin load
Q
f
=A
For 0= 28° , 5 = 0.7 5 x 28 = 21°, tan<5 = 0.384
A
s
= 21.195 m
2
(fro m Ex . 15.2 )
^=65.18 kN/ m
2
(Ex . 15.2 )
Substituting the known values,
Q
f
= 21.195 x 65.18 x 1.0 x 0.384 = 530 kN
Therefore, Q
u
= 114 + 530 = 644 k N
644
Example 1 5 . 5
Solve th e proble m give n i n Exampl e 15. 1 b y Meyerho f 's method . Al l th e othe r dat a remai n th e
same.
Solution
Per Table 9.3 , th e sand in-sit u ma y be considered i n a loose stat e for 0 = 30°. Th e correcte d SP T
value N
CQT
=10 .
Point bearing capacit y
From Eq . (15.20)
From Eq . (15.19 b)
q
bl
= 25 N ta n 0 kN/m
2
Now Fro m Fig . 15 . 10 N
q
= 60 for 0 = 30°
q'
o
= y L = 17. 5 x 1 5 = 262.5 kN/m
2
q
b
= 262. 5 x 60 = 15,75 0 kN/m
2
a, , = 2 5 x 60 x ta n 30° = 866 kN/m
2
Deep Foun dat io n I : Pil e Foun dat io n 64 1
Hence th e limitin g value for q
b
= 866 kN/m
2
314
N ow,Q
b
=A
b
q
b
= A
b
q
b
= —x(0. 45)
2
x 866 = 138 kN
Frictional resistanc e
Per Sectio n 15.17 , th e uni t ski n resistanc e f
s
i s assume d t o increas e fro m 0 a t groun d leve l t o a
limiting value of/
5/
a t L
c
=2Qd where L
c
= critical depth and d = diameter. Therefore L
c
= 20 x 0.4 5
= 9 m
Now/
s
, = q'
0
K
s
ta n 8=yL
c
K
s
tan 8
Given: y = 17. 5 kN/m
3
, L
c
= 9 m, K
s
= 1.0 and 8= 22.5 ° m.
Substituting and simplifying we have
f
sl
= 17.5 x 9 x 1. 0 x tan 22. 5 = 65 kN/m
2
The ski n load Q
f
= Q
fl
+ Q
f2
= ^f
sl
PL
c
+Pf
sl
( L -L
c
)
Substituting Q
f
= - x 65 x 3.14 x 0.45 x 9 +3.14 x 0.45 x 65(15 - 9)
= 413 + 551 = 964 kN
The failur e load Q
u
is
Q
u
= Q
b
+ Q
f
=
138
+
964
= U02 kN
with F
c
= 2.5,
= 440 k N
~
a
2. 5
Example 1 5 . 6
Determine th e base load o f the problem i n Example 15. 5 by V esic's method . Assume l
r
= l
rr
= 50.
Determine Q
a
for F
s
= 2.5 using the value of QAn Ex 15.5 .
Solution
From Eq . (15.25) fo r c = 0 we have
From Eq. (15.24 )
l + 2K
Q
, _ 1 + 2(1-shift) ,
m
3 3
^=15x17. 5 = 262.5 kN/m
2
1 + 2(1-sin 30°)
O". = x 262.5 = 175 kN/m
2
642 Chapt e r 15
From Fig . 15. 1 la, N *
a
= 36 for 0 = 30 and l
r
= 50
Substituting
q
b
= 175 x 36 = 6300 kN/m
2
3.14
Q
b
= A
b
q
b
= ~—x (0.45)
2
x 6300 = 1001 kN
Q
u
= Q
b
+ Q
f
= 1001 +964 = 1,965 kN
2.5
Example 1 5 . 7
Determine th e bas e loa d o f th e proble m i n Exampl e 15. 1 b y Janbu' s method . Us e if/ = 9 0
Determine Q
a
for F
S
= 2.5 using the Q, estimated i n Example 15.5 .
Solution
From E q (15.31) , fo r c = 0 we have
For 0 = 30° and y = 90°, we hav e A T * = 18. 4 fro m Tabl e 15.1 . q'
Q
= 262. 5 kN/m
2
as in
Ex. 15.5 .
Therefore q
b
= 262. 5 x 18. 4 = 4830 kN/m
2
Q
b
= A
b
q
b
= 0.159 x 4830 = 768 kN
Q
u
= Q
b
+Q
f
= 76 8 + 96 4 = 173 2 kN
2.5
Example 1 5. 8
Estimate Q
b
, Q*Q
u
and <2
a
by the Coyle and Castello method using the data given in Example 15.1 .
Solution
Base loa d Q
b
from Eq . (15.33)
Vb = 4o
N
*
q
From Fig . 15.12 , N *
q
= 29 for 0 = 30° and L id = 33. 3
q'
o
= 262.5 kN/m
2
a s i n Ex. 15. 5
Therefore q
b
= 262.5 x 29 = 7612 kN/m
2
Qb
=A
b^b =0. 159x7612-121 0 k N
From Eq . (15.34a )
Deep Foundat i o n I : Pil e Foundat i o n 64 3
where A
s
= 3.14 x 0.45 x 15 = 21.2 m
2
q' = -x262.5 = 13 1.25 kN/m
2
v
0 2
S = 0.80 = 0.8x30° =24 °
From Fig. 15.14 , K
s
= 0.35 for </ > = 30° an d L id = 33.3
Therefore Q
f
= 2 1.2x13 1.25x0.35 tan 24° =43 4 k N
Q
u
=Q
b
+Q
f
= 121 0 + 434 = 1644 k N
644
2.5
=658 k N
Example 1 5 . 9
Determine Q
b
, Q
f
Q
u
and Q
a
by using the SPT value for 0 = 30° from Fig. 12.8 .
Solution
From Fig. 12.8, N
cor
= 10 for 0 = 30°. Us e Eq. (15.47a) fo r Q
u
Q
u
=Q
b
+Q
f
=WN
cor
± A
b
+2N
cor
A
s
where ( 2
f t
<G
w
= 400 ^
r
A
6
Given: L = 1 5 m, d = 0.45 m, A
b
= 0.159 m
2
, A
s
= 21.2 m
2
Q, =40x1 0 x — x 0.159 = 2120 k N
* 0.4 5
Q
w
= 400x10x0.1 59 = 636 k N
Since 0^, > <2
W
, use <2
W
Q
f
=2x10x21. 2 = 424 k N
Now <2
M
= 636 + 424 = 1060 k N
Q =1 ^ = 424 kN
a
2.5
Example 1 5.1 0
Compare the values of Q
b
, (Land Q
a
obtained by the different method s in Examples 15.1 , and 15.5
through 15. 9 and make appropriat e comments.
644 Chapt er 1 5
Comparison
The value s obtained b y differen t method s ar e tabulated below.
Met hod
No
1.
2.
3.
4.
5.
6.
Exa mpl e
No
15.1
15.5
15.6
15.7
15.8
15.9
I nves t i gat or
Berezantsev
Meyerhof
V esic
Janbu
Coyle & Castell o
Meyerhof
(Based o n SPT)
a
b
kN
689
138
1001
768
1210
636
Of
kN
1152
964
964
964
434
424
QU
kN
1841
1102
1965
1732
1644
1060
Q
a
( F
s
= 2. 5)
kN
736
440
786
693
658
424
Comments
It may be seen fro m th e table above that there are wide variations in the values of Q
b
and Q,between
the different methods .
Method 1 Tomlinso n (1986 ) recommend s Berezantsev' s metho d fo r computin g Q
b
a s thi s
method conform s t o th e practica l criteri a o f pil e failure . Tomlinso n doe s no t
recommend th e critical depth concept .
Method 2 Meyerhof' s metho d take s int o accoun t th e critica l dept h concept . Eq . (15.19 ) i s
based o n this concept. Th e equation [Eq. (15.20) ] q
b
- q'
Q
N
q
doe s not consider the
critical dept h concept wher e q' = effective overburde n pressur e a t the pile ti p level
of the pile. The value o f Q
b
per thi s equatio n i s
Q
b
= q
b
A
b
= 15,75 0 x 0.159 = 2504 kN
which i s ver y hig h an d thi s i s clos e t o th e valu e o f Q
b
( = 2120 kN) b y th e SP T
method. Howeve r Eq. (15.19 b) gives a limiting value fo r Q
b
=l38 kN (here the sand
is considered loos e for 0 = 30.
QAS compute d b y assuming (Xincreases linearly wit h depth fro m 0 at L = 0 to Q
fl
a t
depth L
c
= 20d an d then remains constan t t o the end of the pile .
Though som e investigator s hav e accepte d th e critica l dept h concep t fo r computin g
Q
b
an d (X , i t i s difficul t t o generaliz e thi s concep t a s applicabl e t o al l type s o f
conditions prevailing in the field .
Method 3 V esic' s method i s based o n many assumption s for determinin g the value s o f /., I
rr
,
<7
m
, N *
a
etc. There are many assumptions i n this method. Are thes e assumptions valid
for th e field conditions? Designers hav e t o answer thi s question .
Method 4 Th e uncertainties involved i n Janbu's method ar e given i n Section 15.1 5 and as such
difficult t o assess the validit y of thi s method.
Method 5 Coyl e an d Castello' s metho d i s based o n ful l scal e fiel d test s o n a number o f driven
piles. Thei r bearin g capacit y factor s var y with depth. Of the first five method s liste d
above, th e value of Q
b
obtained b y them i s much higher than the other fou r method s
whereas the value of QAS ver y much lower. But on the whole the value of Q
u
is lower
than the other methods .
Method 6 Thi s metho d wa s develope d b y Meyerho f base d o n SP T values . Th e Q
b
valu e
(=2120kN) b y thi s metho d i s ver y muc h highe r tha n th e precedin g methods ,
Deep Foun dat io n I : Pil e Foun dat io n 64 5
whereas the value of Q,is the lowest of all the methods. In the table give n above the
limiting value of Q
bl
= 636 kN is considered. I n all the si x methods F
s
= 2.5 has been
taken t o evaluate Q
a
whereas i n method 6, F
s
= 4 i s recommended b y Meyerhof.
Which Metho d t o Use
There ar e wid e variation s in th e value s of Q
b
, <2« Q
u
and Q
a
between th e differen t methods . Th e
relative proportion s o f load s carrie d b y ski n frictio n an d bas e resistanc e als o var y betwee n th e
methods. The orde r of preference of the methods may be listed as follows;
Preference No .
1
2
3
4
5
6
Met hod No .
1
2
5
6
3
4
Name o f th e i nvest i gat o r
Berezantsev for Q
h
Meyerhof
Coyle and Castell o
Meyerhof (SPT)
V esic
Janbu
Example 15 . 11
A concrete pil e 1 8 in. i n diameter and 50 ft long is driven into a homogeneous mas s of clay soil of
medium consistency. The wate r table i s at the ground surface. The uni t cohesion o f the soi l under
undrained conditio n i s 105 0 lb/ft
2
an d th e adhesio n facto r a = 0.75 . Comput e Q
u
an d Q wit h
F, = 2.5.
Solution
Given: L = 50 ft, d = 1.5 ft, c
u
= 105 0 lb/ft
2
, a = 0.75.
From Eq. (15.37), we have
Q = Q
h
+Qf =c
h
N A.+A ac
*~-u *--b *-• / b c b s u
where, c
b
= c
u
= 1050 lb/ft
2
; N
c
=%A
b
= 1.766 f t
2
; A
s
= 235.5 ft
2
Substituting the known values, we have
1050x9x1.766 235.5x0.75x105 0 _
"~ 1000 100 0
= 16.69 + 185.46 = 202.15 kip s
Q
=
a
2.5
Example 1 5 .1 2
A concrete pil e of 45 cm diameter i s driven through a system of layered cohesive soils . The length
of the pile is 16m. The following data are available. The water table is close to the ground surface.
Top laye r 1 : Sof t clay , thicknes s = 8 m , uni t cohesio n c
u
= 3 0 kN/m
2
an d adhesio n
factor a = 0.90.
Layer 2: Mediu m stiff , thicknes s = 6 m, unit cohesion c
u
= 50 kN/m
2
an d a = 0.75.
646 Chapt er 1 5
Layer 3 : Stif f stratu m extend s t o a grea t depth , uni t cohesio n c
u
= 105 kN/m
2
an d
a = 0.50.
Compute Q
u
and Q
a
with F
s
= 2.5.
Solution
Here, th e pil e i s driven through clay soil s o f different consistencies.
The equation s for Q
u
expressed a s (Eq. 15.38 ) yiel d
Here, c
b
= c
u
o f layer 3, P = 1.413 m,A
b
= 0.159 m
2
Substituting the known values, we have
Q
u
=9x105x0. 15 9 + 1.413(0.90x30x8
+ 0. 75x50x6 + 0. 50x105x2)
= 150.25 + 77 1.5 = 92 1.75 kN
= = 5
2.5
Q
T
j
AL, =
1
AL
2
=
1
AL3:
Layer 1
= 8 m
Layer 2
- 6 m
:
2 m Laye r 3
T
Soft cla y
c
u
= 30 kN/m
2
a = 0.90
— 45 c m
Medium stif f cla y
c
u
= 50 kN/m
2
a = 0.75
Stiff cla y c
u
= 105 kN/m
2
' a =0.50
Figure Ex . 1 5 .1 2
Deep Foun dat io n I: Pil e Foun dat ion 647
Example 1 5 .1 3
A precas t concret e pil e o f siz e 1 8 x 1 8 i n i s drive n int o stif f clay . Th e unconfme d compressiv e
strength of the clay is 4.2 kips/ft
2
. Determine th e length of pile required t o carry a safe working load
of 90 kips wit h F
s
= 2.5.
Solution
The equation fo r Q
u
is
we hav e
Q
u
= 2.5 x 90 = 225 kips ,
N
c
= 9, c
u
= 2.1 kips/ft
2
a = 0.48 fro m Fig. 15.15 , c
u
= c
u
= 2.1 kips/ft
2
, A
b
= 2.25 ft
2
Assume th e lengt h o f pil e = L ft
Now,A
5
= 4x 1.5 L = 6L
Substituting the known values, we have
225 = 9 x 2.1 x 2.25 + 0.48 x 2.1 x 6L
or 22 5 = 42.525 + 6.05L
Simplifying, w e have
225-42.525
L =
6.05
= 30. 2 f t
/w\
. 1
18 x 1 8 in.
. Stif f cla y
f c
u
= 2.1 kips/ft
2
1
a = 0.4 8
Figure Ex . 1 5 .1 3
648 Chapt e r 1 5
Example 1 5 .1 4
For the problem give n in Example 15.1 1 determine the ski n friction load b y the A-method. All the
other dat a remain the same. Assume the average uni t weight of the soi l is 1 10 lb/ft
3
. Us e Q
b
given in
Ex. 15.1 1 an d determin e Q
a
for F
S
= 2.5.
Solution
PerEq. (15.40)
Q = A( q" +2c )A
x-'f ^" O U ' S
g j = - x 5 0 x 110 = 2,750 lb/ft
2
Depth = 50 f t = 15.24 m
From Fig . 15.16 , A = 0. 2 fo r dept h L = 15.24 m
0.2(2750 + 2x 1050) x 235.5 „„„, , , .
Now O r =— - -- -= 228.44 kip s
f
100 0
Now Q
u
= Q
b
+ Q
f
= 1 6.69 + 228.44 - 245 kips
Q
a
=- = 98 kips
Example 1 5 .1 5
A reinforced concret e pil e of size 30 x 30 cm and 10m long is driven int o coarse sand extending to
a great depth . The average total uni t weight of the soi l is 1 8 kN/m
3
and the average N
cor
valu e is 15.
Determine the allowable load on the pile by the static formula. Use F^ = 2.5. The water table is close
to the ground surface .
Solution
In thi s exampl e only th e /V -valu e is given. The correspondin g </ > value ca n b e foun d fro m Fig . 12. 8
which i s equal t o 32° .
Now fro m Fig. 15.9 , fo r <j> = 32° , an d — = —- = 33.33, the valu e o f N
a
= 25.
a 0. 3
q
A
b
=0. 3x0. 3 = 0. 09m
2
,
A
v
=10x 4x 0. 3 = 12m
2
£=0. 75x32 = 24°, ta n 8 = 0.445
The relative densit y is loose t o medium dense. Fro m Tabl e 15.2 , we may tak e
^ = l + -(2-l) = 1.33
Now, Q
u
= q'
Q
N
q
A
b
+ q'
Q
K
s
\ .mSA
s
y
b
= r
sat
-y
w
= 18.0-9.81 -8.19 kN/ m
3
Deep Foun dat io n I : Pil e Foun dat io n
649
Q
u
//A\\
J
1
fi*
0.30 x 0.30 m
Medium dens e san d
y
s a t
=18kN/ m
3
^ = 32°
Figure Ex . 1 5 .1 5
q'
Q
= y
b
l = 8.19 x 10 = 81.9 kN/m
2
2 2
Substituting the known values, we have
Q
u
= 81.9 x 25 x 0.09 + 40.95 x 1.33 x 0.445 x 12
= 184 + 291 = 475 k N
475
Example 1 5 .1 6
Determine th e allowable loa d on the pile given in Ex. 15.15 by making use of the SPT approach by
Meyerhof.
Solution
Per Ex. 15.15, N
cor
= 15
Expression fo r Q
u
is (Eq. 15. 47 a)
Q
u
= Q
b
+ Q
f
= 4QN
cor
( L /d)A
b
+2N
cor
A
s
Here, we have to assume N
cor
= N
cor
=1 5
650 Chapt e r 1 5
A
b
= 0.3x 0.3 = 0.09 m
2
, A
s
= 4x0. 3x10= 12 m
2
Substituting, we have
G, =40x 1 5 — x 0.09 = 1800 k N
* 0. 3
Q
bl
= 400N
cor
x A
b
= 400 x 15 x 0.09 = 540 k N < Q
b
Hence, Q
bl
governs .
Q
f
= 2N
cor
A
s
= 2 x 15 x 12 = 360 kN
A minimu m F
y
= 4 i s recommended, thus ,
,gLtg/.
=
54 0
+
360
^"fl A A
Example 1 5 .1 7
Precast concret e pile s 1 6 in. i n diamete r ar e require d t o be drive n for a building foundation . The
design loa d o n a singl e pil e i s 10 0 kips . Determin e th e lengt h o f th e pil e i f th e soi l i s loos e t o
medium dense san d with an average N
cor
valu e of 1 5 along the pile and 21 at the ti p of the pile. The
water tabl e ma y b e taken a t the ground level . The averag e saturate d uni t weight o f soi l i s equal t o
120 lb/ft
3
. Us e th e stati c formul a and F
s
= 2.5.
Solution
It i s require d t o determin e th e lengt h o f a pil e t o carr y a n ultimat e loa d o f
Q
u
= 2.5x 10 0 = 250 kips.
The equatio n fo r Q
u
is
The averag e valu e o f 0 alon g th e pil e an d th e valu e a t th e ti p ma y b e determine d fro m
Fig. 12.8 .
For N
cor
= 15, 0 = 32°; for N
£or
= 21, 0 = 33.5°.
Since th e soi l i s submerged
y
b
=12 0 -62.4 -57.6 l b/ f t
3
Now
l b/ f t
2
314
= 28.8L l b/ f t
2
x(1.33)
2
= 1.39ft
2
A =3. 14xl . 33x L = 4. 176Lft
2
Deep Foundatio n I: Pil e Foundation 651
16 i n
/
Medium dense sand
N
cor
=\ 5
y
s at
=18. 51b/ft
3
Figure Ex . 1 5 .1 7
From Fig . 15.9 , N = 40 for L id = 20 (assumed)
From Tabl e 15.2 , K
f
= 1.33 fo r the lower side of medium dense sand
S = -x33.5 = 25.1°, ta n S= 0.469
4
Now by substituting the known values, we have
(57.6L)x 40x1.39 (28.8L ) x 0.469 x 1.33 x (4.1 76L)
1000 100 0
= 3.203L + 0.075L
2
or Z
2
+ 42.7 1L- 3333 = 0
Solving this equation gives a value of L = 40.2 f t or say 41 ft .
Example 1 5 .1 8
Refer t o the proble m i n Exampl e 15.17 . Determin e directl y th e ultimat e and th e allowabl e load s
using N
cor
. All the other data remain the same.
Solution
UseEq. (15.49a)
Q
u
= Q
b
+ Q
f
= O.
Given: N
cor
= 21, N
cor
= 15, d = 16 in, L = 41 f t
652 Chapt e r 1 5
3.14 16_
=139f t
2
A =314 x
l H
x
41-171.65 ft
2
b
4 1 2
s
1 2
Substituting
Q, = 0.80 x 21 —- x 1.39 = 720 < 8 W A . kip s
*^b 1 TO co r b "
Q
bl
=8x21x1. 3 9 = 234 kips
The limitin g value of Q
h
= 234 kips
Q
f
= Q.Q4N
ajr
A
s
= 0.04 x 15 x171.65 = 103 kips
Q
u
= Q
b
+ Q
f
= 234 + 103 = 337 kips
337
1 5 .2 1 B EARI N G CAPACI T Y OF PI L ES B ASED O N STATI C CON E
PENETRATI ON TEST S (CPT )
M ethods o f Determinin g Pil e Capacity
The con e penetratio n tes t ma y be considere d a s a small scal e pil e loa d test . As suc h the result s of
this test yiel d the necessar y parameter s fo r the design of piles subjecte d t o vertical load . The type s
of stati c cone penetrometer s an d the methods of conducting the tests hav e been discusse d i n detai l
in Chapte r 9 . V ariou s method s fo r usin g CPT result s t o predic t vertica l pil e capacit y hav e bee n
proposed. Th e followin g method s wil l be discussed:
1. V ande r V een's method.
2. Schmertmann' s method.
V ander V een' s M etho d fo r Pile s i n Cohesionles s Soil s
In th e V ande r V een e t al. , (1957 ) method , th e ultimat e end-bearing resistanc e o f a pil e i s taken ,
equal t o th e poin t resistanc e o f th e cone . T o allo w fo r th e variatio n o f con e resistanc e whic h
normally occurs, the method considers average cone resistance over a depth equal to three times the
diameter o f the pil e above th e pile point level and one pil e diameter belo w point leve l as shown in
Fig. 15.17(a) . Experienc e ha s show n tha t i f a safet y facto r o f 2. 5 i s applie d t o th e ultimat e en d
resistance as determined fro m con e resistance, the pile is unlikely to settl e mor e than 1 5 mm under
the workin g loa d (Tomlinson , 1986) . Th e equation s for ultimat e bearing capacit y an d allowabl e
load ma y be writte n as,
pile base resistance , q
b
= q (cone ) (15.51a )
ultimate base capacity , Q
b
= A
b
q (15.51b )
A
b
q
allowable bas e load , Q — (15.51c )
F
s
where, q = average con e resistanc e ove r a dept h 4d a s shown i n Fig. 15.17(a ) an d F
s
= factor o f
safety.
Deep Foun dat io n I : Pil e Foun dat io n 65 3
The ski n frictio n o n th e pil e shaf t i n cohesionles s soil s i s obtaine d fro m th e relationship s
established b y Meyerhof (1956 ) a s follows.
For displacement piles , th e ultimat e skin friction, f
s
, i s given by
f
s
=^( kPa) (15.52a )
and for H-section piles , the ultimat e limiting skin friction i s given by
/, =-^-(kPa) (15.52b )
where q
c
= average cone resistance i n kg/cm
2
over the length of the pile shaft unde r consideration.
Meyerhof state s tha t for straigh t side d displacemen t piles , th e ultimat e unit skin friction, f
s
,
has a maximum value of 107 kPa and for H-sections, a maximum of 54 kPa (calculated o n all faces
of flanges and web). Th e ultimat e skin load i s
Q
f
=A
s
f
s
(15.53a )
The ultimat e load capacit y of a pile i s
Q
u
=Q
b
+Q
f
(15.53b )
The allowabl e loa d i s
(15.53c)
If the working load, Q
a
, obtained for a particular position of pile in Fig. 15.17(a), is less than
that required for the structural designer's loadin g conditions, then the pile must be taken to a greater
depth t o increase the ski n friction f
s
o r the base resistance q
b
.
Schmertmann's M etho d fo r Cohesionles s and Cohesive Soil s
Schmertmann (1978 ) recommend s on e procedur e fo r al l type s o f soi l fo r computin g th e poin t
bearing capacit y o f piles. However , for computing side friction , Schmertman n gives two differen t
approaches, on e for sand and one for clay soils .
Point B earin g Capacit y Q
b
in All Types of Soi l
The metho d suggeste d b y Schmertman n (1978 ) i s simila r t o th e procedure s develope d b y
De Ruiter and Beringen (1979) for sand. The principle of this method i s based o n the one suggested
by V ander V een (1957) and explained earlier. The procedure use d in this case involves determining
a representative con e poin t penetratio n value , q , within a dept h betwee n 0. 7 t o 4d belo w th e ti p
level of the pile and Sd above the tip level as shown in Fig. 15. 17(b) and (c). The value of q ma y be
expressed a s
( q
cl
+q
c2
)/2 + q
3
- £2.
(1 5 54)
where q
cl
= average con e resistanc e belo w th e ti p o f th e pil e ove r a dept h whic h ma y var y
between O.ld and 4d, wher e d = diameter of pile,
q
c2
= minimum cone resistance recorded below the pile tip over the same depth O.ld to 4d,
654 Chapt e r 1 5
<7
c3
= averag e o f the envelope of minimum cone resistance recorded abov e the pile tip to a
height o f Sd.
Now, the uni t point resistance of the pile, q
b
, is
q
b
(pile) = q
p
(cone ) (15.55a )
The ultimat e base resistance, Q
h
, of a pile is
Q
b
= \ q
p
(15.55b )
The allowabl e base load , O i s
*~-a
(15.55c)
M ethod o f Computin g the Average Cone Point Resistanc e q
The metho d o f computing q
cl
, q
c2
and q
c3
wit h respec t t o a typical g
c
-plot shown in Fig. 15.17(b )
and (c ) i s explained below.
Case 1 : Whe n th e con e poin t resistance q
c
belo w th e ti p o f a pil e i s lowe r tha n tha t a t th e ti p
(Fig. 15.17(b) ) wi t hi n dept h 4d.
=
where q
o
, q
b
etc., refe r t o the points o, b etc, on the g
c
-profile, q
c2
= q
c
= minimum value below ti p
within a depth o f 4d a t point c on the ^-profile .
The envelop e o f mi ni mum cone resistanc e above the pil e ti p i s as shown by th e arrow mar k
along (15.17b ) aefghk.
+ / 2 + d + d + /2 + d
where q
a
= q
e
, q
f
=q
g
, q
h
= q
k
.
Case 2: When th e cone resistance q
c
below th e pile tip is greater than that at the tip within a dept h
4d. (Fig . 15. 1 7(c)).
In thi s case q i s foun d withi n a total depth of 0.1 das shown i n Fig. 15.17(c) .
q
c2
= q
o
= minimum valu e at th e pil e ti p itself , q
c3
= average o f th e minimu m value s alon g th e
envelope ocde a s before.
In determining the average q
c
above, the minimum values q
c2
selected unde r Case 1 or 2 are
to be disregarded .
Effect o f Overconsolidatio n Rati o in Sand
Reduction factor s hav e bee n develope d tha t shoul d be applie d t o th e theoretica l en d bearin g of a
pile a s determine d fro m th e CP T i f th e bearin g laye r consist s o f overconsolidate d sand . Th e
problem i n many cases will be to make a reasonable estimate of the Overconsolidation ratio in sand.
In sand s wit h a hig h q
c
, some conservatio n i n thi s respect i s desirable , i n particula r fo r shallo w
foundations. Th e influenc e o f Overconsolidatio n on pil e en d bearin g i s on e o f th e reason s fo r
applying a limiting value t o pil e end bearing, irrespectiv e of th e cone resistance s recorde d i n th e
Deep Foun dat io n I : Pil e Foun dat io n 655
J,
q
p
(average )
4d
(a) V ander V een's metho d
Limits fo r Eq. (15.56)
(b) Resistance belo w pil e tip lower than (c ) Resistance belo w pil e ti p greater tha n
that at pile ti p within depth 4d tha t at pile ti p within 0.75 depth
Figure 1 5.1 7 Pil e capacit y b y us e o f CP T values (a ) Van der Veen ' s met hod , an d
(b,c) Schm ert m an n ' s metho d
bearing layer. A limit pile end bearing of 15 MN/m
2
is generally accepte d (D e Ruiter and Beringen,
1979), although in dense sands cone resistance ma y be greater tha n 50 MN/m
2
. It is unlikely that in
dense normall y consolidated san d ultimate end bearing value s higher than 15 MN/m
2
can occur but
this has not been adequatel y confirme d by load tests .
Design CPT V alues fo r San d and Clay
The applicatio n o f CP T i n evaluatin g th e desig n value s fo r ski n frictio n an d bearin g a s
recommended b y De Ruiter and Beringen (1979 ) is summarized i n Table. 15.3 .
656 Chapt er 1 5
Table 1 5 . 3 Appl icat io n o f CR T i n Pil e Des ig n (Aft e r D e Ruit e r an d B erin g en , 1979 )
Item San d Cl ay Leg en d
Unit Friction
Unit end bearing q
t
Mi ni mum o f
,/; = 012MPa
/
2
= CPT sleeve frictio n
/
3
= <7
r
/300 (compression )
or
/
3
= <?
f
/400 (tension)
mi ni mum o f ( 7 fro m
Fig. 15.17(b ) a n d c
f
s
= t f c
u
, where
of = 1 in N.C. clay
= 0. 5 i n O.C. cla y
q = N c
*p c u
N =9
q
c
- con e resistanc e
below pil e ti p
q
;
= ultimate
resistance o f pil e
U ltimate Ski n L oa d Q
f
i n Cohesionless Soil s
For th e computatio n o f ski n load, £> , Schmertman n (1978 ) present s th e followin g equatio n
— /A + f A
o , J C S J c S
(15.56c)
where K = fjf = correction facto r for/
c
f
s
= uni t pile friction
f
c
= unit sleeve frictio n measure d b y th e friction jacket
z = depth to/
c
value considered fro m groun d surfac e
d = pile diamete r o r width
A = pile-soil contact area per/
c
depth interva l
L = embedded dept h o f pile .
When/
c
doe s no t var y significantl y with depth , Eq . (15.56c ) ca n b e writte n i n a simplifie d
form a s
-Cf
2
c (15.56d)
where f
c
i s th e averag e valu e withi n th e depth s specified . Th e correctio n facto r K i s give n i n
Fig. 15.18(a) .
U ltimate Ski n L oa d Q
f
for Pile s i n Clay Soi l
For pile s i n clay Schmertman n give s the expressio n
(15.57)
where, of = ratio o f pil e t o penetrometer sleev e friction ,
/ = average sleev e friction,
A = pile t o soi l contac t area .
Fig. 15.18(b ) gives value s of a'.
Deep Foun dat io n I : Pil e Foun dat io n 657
K values for stee l pipe pil e
0 1. 0 2. 0 3. 0
K values for concret e pil e
0 1. 0 2. 0
10
3 20
30
40
/
(^t i mber - 1-2 5 A'
1. Mechanical penetrometer
2. Electrical penetrometer
(a)
Penetrometer sleev e friction, f
c
kg/c m
(b)
Figure 1 5.1 8 Pen et rom et e r des ig n curve s (a ) for pil e sid e frict io n i n s an d
(Schm ert m an n , 1978) , an d (b) for pil e s id e frict io n i n clay (Schm ert m an n , 1978 )
Example 1 5 .1 9
A concrete pile of 40 cm diameter is driven into a homogeneous mass of cohesionless soil . The pile
carries a saf e loa d o f 65 0 kN . A stati c con e penetratio n tes t conducte d a t th e sit e indicate s a n
average valu e of q
c
= 40 kg/cm
2
alon g the pil e an d 12 0 kg/cm
2
below th e pil e tip. Comput e th e
length o f the pil e wit h F
s
= 2.5. (Fig . Ex. 15.19 )
658
Chapt er 1 5
Solution
From Eq. (15.5la)
q
b
(pile) = q
p
(cone )
Given, q = 120 kg/cm
2
, therefore,
q
b
=l2Q kg/cm
2
= 12 0 x 10 0 = 1200 0 kN/m
2
Per Sectio n 15.12 , q
b
is restricted t o 11,00 0 kN/m
2
.
Therefore,
314
Q
b
=A
b
q
b
=— x0. 4
2
x 11000 = 1382 kN
Assume th e lengt h of the pil e = L m
The average , q
c
= 40 kg/cm
2
PerEq. (15.52a) ,
f
s
= —- kN/ m
2
= — = 20 kN/ m
2
Now, Q
f
= f
s
A
s
= 20 x 3.14 x 0.4 x L = 25.12L kN
Given Q
a
= 650 kN. Wit h F
v
= 2.5, Q
u
= 650 x 2. 5 = 162 5 kN.
Now,
1625-1382
or L =
25.12
= 9.67 m o r say 1 0 m
The pil e has to be driven to a depth of 10 m to carr y a safe loa d of 650 kN wit h F =2.5.
I!"
q
c
- 40 kg/cm
q
c
= 12 0 kg/cm
2
Qb
Figure Ex . 1 5 .1 9
Deep Foun dat io n I : Pil e Foun dat io n 659
Example 1 5 .2 0
A concret e pil e of size 0.4 x 0.4 m is driven to a depth of 1 2 m into medium dense sand. The water
table i s close t o th e groun d surface . Stati c con e penetratio n test s wer e carrie d ou t a t thi s sit e by
using a n electri c con e penetrometer . The value s of q
c
an d f
c
a s obtaine d fro m th e tes t hav e been
plotted against depth and shown in Fig. Ex. 15.20. Determine the safe load on this pile with F
s
= 2.5
by Schmertmann' s method (Sectio n 15.21) .
Solution
First determine the representative con e penetration value q b y using Eq. (15.54)
Now from Fig. Ex . 15.2 0 an d Eq (15.56a)
1
4d
_ 0.7(7 6 + 85) / 2 + 0.3(85 + 71) / 2 + 0.6(71 + 80) / 2
4x0. 4
= 78 kg/cm
2
= q
d
= the minimum value below the tip of pile withi n 4d dept h = 71 kg/cm
2
.
q
c
, kg/cm
2
f
c
, kg/cm
2
,0 2 0 4 0 6 0 8 0 0 0. 5 1. 0
\ Qu
Square concrete
pile 0.4 x 0.4 m
z= L = 12m
Figure Ex . 1 5.2 0
660 Chapt e r 1 5
FromEq. (15.56b )
_ 0. 4x7 1 + 0.3(71 + 65)72 + 2.1x65 + 0.4(65 + 60)7 2
8x0. 4
= 66 kg/cm
2
= 660 t/m
2
(metric)
From Eq. (15.54)
(78 + 71) 72 + 66
^
700 t
/
m
2
(metric)
U ltimate B as e L oad
Q
b
= q
b
A
b
= q
p
A
b
- 70 0 x 0.4
2
= 1 1 2 t (metric)
Frictional L oad Q
f
From Eq . (15.56d )
l-L
where K = correction facto r from Fig . (15.18a) for electrical penetrometer .
L_ 1 2
For — - T ~ - 3v , K = 0.75 for concrete pile . I t is now necessary t o determine the average
sleeve frictio n f
c
betwee n depths z = 0 and z - 8d , and z = 8d and z = L from the top of pile from/
c
profile give n i n Fig. Ex. 15.20 .
Q =0.75[|x0.34 x 10x4x0. 4x3.2 + 0. 71x10x4x0. 4x8. 8]
= 0.75 [8.7 +99.97] = 81.5 t (metric)
Q =Q
h
+ Q
f
= 11 2 + 81.5= 193. 5 t
*^u *z-b -^ 7
Qh
+
Qr 193 5
Q = -2L
=
_—L_ = 77.4
t
(metric) = 759 kN
•*"^/7 r\ c s*\ C
2.5 2. 5
Example 1 5 .21
A concrete pil e of section 0. 4 x 0. 4 m is driven into normally consolidated cla y t o a depth of 1 0 m.
The wate r tabl e i s a t ground level . A stati c cone penetratio n tes t (CPT ) wa s conducte d a t the sit e
with an electric cone penetrometer . Fig. Ex . 15.2 1 gives a profile of q
c
and/
c
values with respect t o
depth. Determine saf e loads o n th e pil e by the following methods:
(a) a-metho d (b ) A-method , given : y
b
= 8. 5 kN/m
3
an d (c ) Schmertmann' s method . Us e a
factor o f safet y o f 2.5.
Solution
(a) a-method
The a-method requires the undrained shear strengt h of the soil. Sinc e thi s is not given, it has
to b e determine d b y usin g the relation betwee n q
c
and c
u
give n i n Eq. (9.14) .
Deep Foun dat io n I : Pil e Foun dat io n 661
q
c
, kg/cm
5 1 0 1 5
Square concret e
pile 0.4 x 0.4 m
Figure Ex . 1 5.2 1
c
u jy b y neglecting th e overburde n effect ,
YV
k
where N
k
= cone facto r = 20.
It is necessary t o determine the average ~c
u
alon g the pile shaft an d c
b
a t the base leve l of the
pile. Fo r this purpose fin d th e corresponding^ (sleeve friction ) value s from Fig . Ex . 15.21 .
- _
1 + 1 6
_
c
^
Average q
c
alon g th e shaft , Q
c
r <>• • • > k g
Average of q
c
withi n a dept h 3d abov e the bas e and d below the bas e of the pil e (Refe r to
Fig. 15.17a )
15 + 18.5
2
= 17kg/cirr
o c
c = —= 0.43 kg/ c m
2
« 4 3 k N/ m
2
" 2 0
662 Chapt e r 1 5
c , = — = 0. 85kg/ cm
2
= 85 k N/ m
2
.
b
2 0
Ultimate Base L oad, Q
b
FromEq. (15.39 )
Q
b
= 9c
b
A
b
= 9 x 85 x 0.40
2
= 12 2 kN.
Ultimate Friction L oad, Q
f
FromEq. (15.37)
Qf=<*u
A
s
From Fig . 15.1 5 for c
u
= 43 kN/m
2
, a = 70
Q
f
= 0.7 0 x 43 x 1 0 x 4 x 0. 4 = 481.6 k N or say 482 kN
Q
u
= 122 + 482 = 604 kN
604
Q
a
= _= 241.6 kN or say 242 kN
j L, *j
(b) A-method
Base L oad Q
b
In thi s method th e base loa d i s the same a s in (a) above. That i s
Q
b
= 122 kN
Friction L oad
From Fig . 15.1 7 A = 0.25 fo r L = 10 m (= 32. 4 ft) . From Eq . 15.4 0
q =1x10x8. 5 = 42.5 k N/ m
2
H
° 2
f
s
= 0.25(42.5 + 2 x 43) = 32 kN / m
2
Q =f
s
A
s
= 32x10x4x0.4 = 512 kN
Q =12 2 + 512 = 634 kN
*^u
Q = — = 254kN.
a
2. 5
(c) Schmertmann' s Metho d
Base load Q
b
Use Eq. (15.54) for determining the representative value for q
p
. Here, th e minimum value for
q
c
is at point O on the ^-profile i n Fig. Ex. 15.2 1 which is the base level of the pile. Now q
cl
i s the
average q
c
at the bas e an d O.ld below th e base o f the pile, that is,
q
0
+g
e
1 6 + 18.5
2
q
cl
= — -- - -= 1 7.25 kg /cm
2
= The averag e of q
c
withi n a depth &d abov e the base leve l
Deep Foun dat io n I : Pil e Foun dat io n 66 3
q
0
+4
k
1 6 + 11 , , _ . . 2
= — = = 13.5 kg/cm
2
2 2
(1725 + 16)72+13.5
1 C 1
.
2
From Eq. (15.45), V
p
= r
= 15
kg
7001
.
2
FromEq. (15.55a )
^(pile) = q
p
(cone ) = 15 kg/cm
2
« 1500 kN/m
2
Q
fc
=
9ft
A
fc
= 1500 x (0.4)
2
= 240 k N
Friction L oad Q
f
Use Eq. (15.57 )
where ex ' = ratio of pil e t o penetrometer sleeve friction .
From Fig. Ex. 15.21 f
c
=
0+L15
= 0.58 kg/cm
2
« 5 8 kN/m
2
.
From Fig. 16.18b for/
c
= 58 kN/m
2
, a' = 0.70
Q
f
=0.7 0 x 58 x 1 0 x 4 x 0.4 = 650 kN
Q
u
= 240 + 650 = 890 kN
Q =^ ^ = 356kN
a
2. 5
Note: The values given in the examples ar e only illustrative and not factual.
1 5 .2 2 B EARI N G CAPACI TY OF A SI NG L E PIL E B Y L OAD TEST
A pile load test is the most acceptable method to determine the load carrying capacity of a pile. The
load tes t ma y be carried ou t either on a driven pil e or a cast-in-situ pile . Loa d test s ma y be made
either o n a singl e pil e o r a group of piles . Loa d test s o n a pil e grou p ar e ver y costl y an d ma y b e
undertaken only i n very important projects.
Pile load test s on a single pile or a group of piles ar e conducted for the determination of
1. V ertica l load bearing capacity,
2. Uplif t loa d capacity,
3. Latera l loa d capacity.
Generally loa d test s ar e mad e t o determin e th e bearing capacit y an d t o establis h th e load -
settlement relationshi p under a compressive load . The other two types o f tests ma y be carried ou t
only whe n piles are required t o resist larg e uplif t o r lateral forces .
Usually pile foundations are designed wit h an estimated capacity whic h is determined fro m a
thorough stud y of the sit e conditions. At the beginning of construction, load test s ar e made for the
purpose of verifyin g the adequac y of the desig n capacity . If the tes t result s sho w an inadequat e
factor of safety or excessive settlement, the design must be revised before construction is under way.
Load test s may be carried ou t either on
1. A working pile or
2. A test pile.
664 Chapt e r 1 5
A working pile is a pile driven or cast-in-situ along with the other piles t o carry the loads fro m
the superstructure. The maximum test load on such piles should not exceed on e and a half times the
design load .
A test pile i s a pile which does not carry the loads coming from th e structure. The maximum
load tha t can be put on such piles may be about 2
J
/2 times the design load or the load imposed must
be suc h as to give a total settlement not less than one-tenth the pile diameter.
M ethod o f Carryin g Ou t V ertica l Pil e L oa d Tes t
A vertica l pil e load tes t assembl y i s shown in Fig. 15.19(a) . It consists of
1. A n arrangemen t t o take the reaction of the load applied on the pil e head,
2. A hydraulic jack o f sufficien t capacit y to appl y load on the pil e head, an d
3. A set of three dia l gauges t o measure settlement of the pil e head.
L oad Application
A load tes t may be of two types:
1. Continuou s load test.
2. Cycli c loa d test .
In th e cas e o f a continuous load test , continuou s increments of loa d ar e applie d t o th e pil e
head. Settlemen t of the pil e head i s recorded a t each loa d level .
In the case of the cyclic load test, the load is raised t o a particular level , then reduced t o zero,
again raise d t o a higher level an d reduced t o zero . Settlement s ar e recorde d a t each incremen t or
decrement o f load. Cycli c load tests help to separat e frictiona l loa d fro m poin t load.
The tota l elastic recover y or settlement S
e
, is due to
1. Th e tota l plasti c recover y of the pil e material,
2. Elasti c recover y of the soi l a t the ti p of the pile, S
g
The total settlement S due to any load can be separated int o elastic and plastic settlements by
carrying out cycli c load test s as shown in Fig. 15.19(b) .
A pile loaded t o Q
l
give s a total settlement S
r
Whe n thi s load i s reduced t o zero, ther e i s an
elastic recovery which is equal to S
el
. This elastic recovery is due to the elastic compression of the pile
material an d the soil. The net settlement or plastic compression is S
r
The pile is loaded agai n fro m
zero to the next higher load Q
2
and reduced t o zero thereafter. The corresponding settlement s may be
found a s before. The method of loading and unloading may be repeated a s before.
Allowable L oa d fro m Singl e Pil e L oa d Tes t Dat a
There ar e many methods b y whic h allowable loads on a single pile ma y be determined b y making
use of load tes t data. If the ultimate load can be determined from load-settlemen t curves, allowable
loads ar e found by dividing the ultimate load by a suitable factor of safety whic h varies from 2 to 3.
A facto r o f safet y o f 2. 5 i s normall y recommended. A fe w o f th e method s tha t ar e usefu l fo r th e
determination of ultimate or allowabl e loads on a single pile ar e given below:
1. Th e ultimat e load, Q
u
, can be determined as the abscissa o f the point where the curved part
of the load-settlement curve changes to a falling straigh t line. Fig. 15.20(a) .
2. Q
u
i s th e absciss a o f th e poin t o f intersectio n o f th e initia l an d fina l tangent s o f th e
load-settlement curve, Fig. 15.20(b) .
3. Th e allowabl e loa d Q i s 5 0 percen t o f th e ultimat e loa d a t whic h th e tota l settlemen t
amounts t o one-tent h of the diameter of the pil e for unifor m diamete r piles.
Deep Foun dat io n I : Pil e Foun dat io n 665
Anchor rod
Anchor pil e
(a)
//A
I
M^\
t
A C
V T
Pile 1 = befor e loa d
2 = afte r loa d
3 = afte r elasti c recover y
(b)
Figure 1 5.1 9 (a ) Vert ical pil e loa d t es t as s em bly , an d (b ) elas t i c com pres s io n a t
the bas e of th e pil e
4. Th e allowable load Q
a
is sometimes taken as equal to two-thirds of the load which causes a
total settlemen t o f 1 2 mm.
5. Th e allowable load Q
a
is sometimes take n as equal to two-thirds of the load which causes a
net (plastic ) settlement of 6 mm.
If pile groups are loaded t o failure, the ultimate load of the group, Q
u
, may be found by any one
of the first two methods mentioned above for single piles. However, if the groups are subjected to only
666 Chapt er 1 5
Q
(a) (b )
Figure 1 5.2 0 Det erm in at io n o f ult im at e loa d fro m load-s et t lem en t curve s
one and a half-times the design load of the group, the allowable load on the group cannot be found on
the basis of 12 or 6 mm settlement criteri a applicable to single piles. In the case of a group wit h piles
spaced at less than 6 to 8 times the pile diameter, the stress interaction of the adjacent piles affect s th e
settlement considerably. The settlemen t criteria applicable t o pile groups shoul d be the same a s that
applicable t o shallow foundations at design loads .
1 5 .2 3 PIL E B EARI N G CAPACI T Y FROM DY NAM I C PIL E DRI V I NG
FORM U L AS
The resistanc e offere d b y a soi l t o penetratio n o f a pil e durin g driving give s a n indicatio n of it s
bearing capacity . Qualitativel y speaking, a pil e whic h meet s greate r resistanc e durin g drivin g i s
capable o f carryin g a greate r load . A numbe r o f dynami c formula e hav e bee n develope d whic h
equate pil e capacity i n terms of driving energy.
The basi s o f al l these formula e is the simpl e energy relationshi p which may be state d by the
following equation . (Fig. 15.21) .
Wh = Q
u
s
or Q
u
=— (15.58 )
5
where W - weigh t of the driving hammer
h = height of fal l o f hammer
Wh = energy o f hammer blow
Q = ultimate resistance t o penetration
s = pile penetratio n unde r one hammer blow
Q
u
s = resisting energy of the pil e
H iley Formul a
Equation (15.58 ) hold s onl y i f th e syste m i s 10 0 percen t efficient . Sinc e th e drivin g o f a pil e
involves many losses, the energy o f the syste m ma y be writte n as
Deep Foun dat io n I : Pil e Foun dat io n 667
$
x Hammer
t,
h
\ '
J /
Pile^k
cap
s = penetration ^ | f
of pile ^
r
x x x
r
3
x
—W
p
= weight of pile
f t
Q
u
Figure 1 5 .2 1 Basi c en erg y relat ion s hi p
Energy input = Energy used + Energy losses
or Energ y used = Energy input - Energ y losses .
The expressions fo r the various energy terms used are
1 . Energ y used = Q
u
s,
2. Energ y input = T]
h
Wh, wher e r\
h
i s the efficienc y o f the hammer.
3. Th e energy losse s ar e due to the following:
(i) Th e energy loss E
{
du e t o the elastic compressions o f the pil e cap , pil e materia l and
the soi l surrounding the pile. The expression for E
l
ma y be written as
where C j = elasti c compressio n o f the pil e ca p
c
2
= elasti c compression o f the pile
c
3
= elasti c compression o f the soil.
(ii) Th e energ y los s E
2
du e t o the interaction o f the pil e hamme r syste m (impac t of two
bodies). Th e expression fo r E
2
may be written as
i-c
r
2
£
7
= WhW -
2 P
W+W
where W = weigh t of pile
C = coefficien t o f restitution.
668 Chapt e r 1 5
Substituting the variou s expressions i n the energy equation and simplifying, we have
^ rj.Wh l + C
2
R
TTF
(15
'
59
>
w
where R -— —
W
Equation (15.59 ) i s calle d th e Hile y formula . The allowabl e loa d Q
u
ma y b e obtaine d b y
dividing Q
u
by a suitable factor of safety.
If the pile tip rests on rock or relatively impenetrable material, Eq. (15.59) i s not valid. Chellis
(1961) suggest s fo r thi s conditio n tha t th e us e o f W 12 instea d o f W ma y b e mor e correct . Th e
various coefficient s used i n the Eq . (15.59) ar e as given below:
1. Elastic compression c
{
o f cap and pile head
Pile M at eria l Ran g e o f Drivin g Ran g e o f c
1
St res s k g /cm
2
Precast concrete pile wit h
packing inside cap 30-15 0 0.12-0-5 0
Timber pil e without cap 30-15 0 0.05-0.2 0
Steel //-pile 30-15 0 0.04-0.1 6
2. Elastic compression c
2
of pile.
This ma y be computed usin g the equation
c -~
AE
where L = embedde d lengt h of the pile ,
A = averag e cross-sectiona l are a o f the pile ,
E = Young's modulus.
3. Elastic compression c
3
of soil.
The average valu e of c
3
may be taken as 0. 1 (the value ranges from 0.0 for hard soi l to 0.2 for
resilient soils) .
4. Pile-hammer efficiency
Hammer Typ e r\
h
Drop
Single actin g
Double actin g
Diesel
1.00
0.75-0.85
0.85
1.00
5. Coefficient o f restitution C
r
Material C
r
Wood pil e 0.2 5
Compact woo d cushio n on steel pil e 0.3 2
Cast iro n hamme r o n concrete pil e without cap 0.4 0
Cast iro n hamme r o n steel pip e without cushion 0.5 5
Deep Foun dat io n I : Pil e Foun dat io n 66 9
Engineering News Recor d (ENR) Formula
The genera l for m o f th e Engineerin g New s Recor d Formul a fo r th e allowabl e loa d Q
a
ma y b e
obtained fro m Eq . (15.59) by putting
r]
h
= 1 and C
r
= 1 and a factor of safety equal to 6. The formula proposed by A.M. Wellington,
editor o f the Engineering News, i n 1886 , is
Wh .
(15-60)
where Q
a
= allowable load in kg,
W = weigh t of hammer i n kg,
h = height of fal l o f hammer i n cm,
s - fina l penetratio n i n cm per blow (whic h i s termed a s set) . The se t i s take n a s the
average penetration per blow for the last 5 blows of a drop hammer or 20 blows of a
steam hammer,
C = empirical constant ,
= 2. 5 cm for a drop hammer ,
= 0.2 5 c m for single an d double acting hammers .
The equations for the various types o f hammers ma y be written as:
1 . Drop hamme r
Wh
2. Single-acting hamme r
m
(15
-
62)
3. Double-acting hamme r
( W + ap)
a = effective are a o f the piston i n sq. cm,
p = mea n effective stea m pressur e i n kg/cm
2
.
Comments o n the Us e of Dynami c Formula e
1. Detaile d investigation s carrie d ou t b y V esic (1967) o n deep foundation s in granula r soil s
indicate tha t th e Engineerin g New s Recor d Formul a applicabl e t o dro p hammers ,
Eq. (15.61), give s pil e load s a s low as 44 %of the actua l loads . I n orde r t o obtai n bette r
agreement betwee n th e on e compute d an d observe d loads , V esi c suggest s th e followin g
values for the coefficien t C in Eq. (15.60) .
For steel pipe piles, C = 1 cm.
For precast concret e pile s C = 1.5 cm.
2. Th e test s carrie d ou t by V esic in granular soil s indicat e that Hiley's formul a does no t give
consistent results . Th e value s compute d fro m Eq . (15.59 ) ar e sometime s highe r an d
sometimes lowe r than the observed values .
670 Chapt e r 1 5
3. Dynami c formulae in general hav e limited value in pile foundation work mainl y because th e
dynamic resistance o f soil does not represent the static resistance, and because often the results
obtained fro m th e us e o f dynami c equation s ar e o f questionabl e dependability . However ,
engineers prefe r to use the Engineering News Record Formul a because o f its simplicity.
4. Dynami c formulae could be used with more confidence i n freely drainin g materials such as
coarse sand . I f th e pil e i s drive n t o saturate d loos e fin e san d an d silt , ther e i s ever y
possibility o f development of liquefaction whic h reduces the bearing capacit y o f the pile .
5. Dynami c formula e are no t recommende d fo r computin g allowabl e load s o f pile s drive n
into cohesive soils . In cohesive soils, the resistance t o driving increases throug h the sudden
increase i n stress i n pore wate r and decreases becaus e o f the decreased valu e of the internal
friction betwee n soi l and pil e because o f pore water. These two oppositel y directe d force s
do no t len d themselve s t o analytica l treatmen t an d a s suc h th e dynami c penetratio n
resistance t o pile driving has no relationship t o stati c bearing capacity .
There i s anothe r effec t o f pil e drivin g i n cohesiv e soils . Durin g drivin g th e soi l become s
remolded an d th e shea r strengt h o f th e soi l i s reduce d considerably . Thoug h ther e wil l b e a
regaining o f shea r strengt h afte r a laps e o f some day s afte r th e drivin g operation, thi s wil l not be
reflected i n the resistance valu e obtained fro m th e dynamic formulae.
Example 1 5.2 2
A 40 x 40 c m reinforced concret e pil e 20 m long i s driven through loose sand an d then int o dense
gravel t o a fina l se t o f 3 mm/blow, usin g a 3 0 k N single-actin g hamme r wit h a strok e o f 1. 5 m.
Determine th e ultimate driving resistance o f the pile if it is fitted with a helmet, plastic dolly and 50
mm packing on the top of the pile. The weight of the helmet an d dolly is 4 kN. The other details are :
weight o f pil e = 7 4 kN ; weigh t o f hamme r = 3 0 kN ; pil e hamme r efficienc y rj
h
= 0.8 0 an d
coefficient o f restitution C
r
= 0.40.
Use the Hiley formula. The sum of the elastic compression C i s
C = c
l
+c
2
+c
3
= 19.6 mm.
Solution
H iley Formul a
Use Eq. (15.59 )
„ ^ *
" s + C \ + R
W
where n. = 0.80, W = 30 kN, h = 1. 5 m, R = —
p
- = ^——-= 2.6, C = 0.40, s = 0.30 cm .
h
W 3 0
Substituting we have,
0. 8x30x150 l + 0. 4
2
x2. 6
0.3 + 156/2 l
+
2.6
=2813x0. 393=1105kN
1 5.24 B EARI N G CAPACI TY OF PILES FOUNDED O N A ROCK Y B ED
Piles ar e a t times require d t o be driven through weak layer s of soi l unti l the tips meet a hard strat a
for bearing . I f the bearing strat a happens t o be rock, th e piles ar e to be driven t o refusal i n order to
obtain the maximum carrying capacity from th e piles. If the rock i s strong at its surface, the pile will
Deep Foun dat io n I : Pil e Foun dat io n 67 1
refuse furthe r drivin g a t a negligible penetration . I n such cases th e carrying capacit y o f the piles i s
governed by the strengt h of the pile shaf t regarde d a s a column as show n in Fig. 15.6(a) . If the soi l
mass through which the piles are driven happens to be stiff clay s or sands, the piles can be regarded
as being supported on all sides from buckling as a strut. In such cases, the carrying capacity of a pile
is calculated from th e safe load on the material of the pile at the point of minimum cross-section. I n
practice, i t i s necessar y t o limi t th e saf e loa d o n pile s regarde d a s shor t column s becaus e o f th e
likely deviations from th e vertica l and the possibility of damage t o the pile during driving.
If piles ar e driven to weak rocks, working loads a s determined b y the availabl e stres s o n the
material of the pile shaft ma y not be possible. In such cases the frictional resistanc e develope d ove r
the penetration into the rock and the end bearing resistance are required t o be calculated. Tomlinson
(1986) suggest s a n equatio n fo r computin g th e en d bearin g resistanc e o f pile s restin g o n rock y
strata as
<7u
= 2 A
V V (15.64 )
where N ^ = tan
2
(45 + 0/2) ,
q
ur
- unconfme d compressiv e strengt h of the rock.
Boring of a hole in rocky strat a for constructing bored piles may weaken the bearing strat a of
some type s of rock. I n such cases low values of skin friction shoul d be used and normall y may not
be more tha n 20 kN/m
2
(Tomlinson , 1986 ) whe n the boring i s through friabl e chal k o r mud stone .
In th e cas e o f moderatel y wea k t o stron g rock s wher e i t i s possibl e t o obtai n cor e sample s fo r
unconfmed compressio n tests , th e en d bearin g resistanc e ca n b e calculate d b y makin g us e o f
Eq. (15.64) .
1 5 .2 5 UPLIF T RESI STANC E O F PILES
Piles ar e als o used t o resist uplif t loads . Pile s use d for thi s purpose ar e called tensio n piles, uplif t
piles o r ancho r piles . Uplif t force s ar e develope d du e t o hydrostati c pressur e o r overturnin g
moments a s shown in Fig. 15.22 .
Figure 15.2 2 shows a straight edged pil e subjected t o uplift force . The equation for the uplif t
force P
U[
ma y b e writte n as
(15.65)
where, P
ul
= uplift capacit y o f pile ,
W = weight of pile,
/ = unit resisting forc e
A
s
= effective are a o f the embedded lengt h of pile .
Uplift Resistanc e o f Pil e i n Clay
For piles embedde d i n clay, Eq. (15.65 ) ma y written as
(15.66)
where, c
u
= average undraine d shear strengt h of clay along the pil e shaft ,
a, = adhesion facto r (= c
fl
/c
M
),
c
a
= average adhesion .
Figure 15.2 3 gives the relationship between a and c
u
based on pull out test results as collected
by Sow a (1970) . As pe r Sowa , th e value s o f c
a
agre e reasonabl y wel l wit h th e value s fo r pile s
subjected to compression loadings .
672
Chapt er 1 5
fr
Figure 1 5 .2 2 Singl e pil e s ubj ect e d t o uplif t
1.25
0.75
0.50
0.25
Average curv e for concret e pil e
Average curve for al l pil e
25 5 0 7 5 10 0 12 5
Undrained shea r strengt h c
u
, kN/m
150
Figure 1 5 .2 3 Rel at ion s hi p bet wee n adhes io n fact o r a and un drain ed s hea r s t ren g t h
c (Source : Poulo s an d Davis , 1980)
U plift Resistanc e o f Pil e i n Sand
Adequate confirmatory data are not available for evaluating the uplif t resistanc e o f piles embedde d
in cohesionles s soils . Irelan d (1957 ) report s tha t th e averag e ski n frictio n fo r pile s unde r
compression loadin g an d uplif t loadin g ar e equal , bu t dat a collecte d b y Sow a (1970 ) an d Down s
and Chieurzzi (1966) indicat e lower values for upwar d loading a s compared t o downward loadin g
especially fo r cast-in-situ piles . Poulo s an d Davi s (1980 ) sugges t tha t th e ski n frictio n o f upwar d
loading ma y b e take n as two-thirds of the calculated shaft resistanc e fo r downwar d loading .
A safet y factor of 3 i s normal ly assumed for calculating the saf e uplif t loa d fo r bot h pile s i n
clay an d sand .
Deep Foun dat io n I : Pil e Foun dat io n 67 3
Example 1 5 .2 3
A reinforced concret e pil e 30 ft long and 1 5 in. in diameter i s embedded i n a saturated clay of very
stiff consistency . Laborator y test s o n sample s o f undisturbe d soi l gav e a n averag e undraine d
cohesive strengt h c
u
= 2500 lb/ft
2
. Determine the net pullout capacity and the allowable pullout load
with F
s
= 3.
Solution
Given: L = 30 ft, d = 15 in. diameter , c
u
= 2500 lb/ft
2
, F
s
= 3.
From Fig . 15.2 3 cjc
u
= 0.41 fo r c
u
= 2500 x 0.0479 « 120 kN/m
2
for concrete pile .
From Eq . (15.66 )
where a = c/c=0.41, c = 2500 lb/ft
2
A =3.14 x — x 30 = 117.75 ft
2
12
Substituting
„ , , 0.41x2500x117.7 5 , „„„, , .
P, (net ) = -= 120.69 kip s
ul
100 0
P
P
ul
(allowed) = ^4 0 kip s
Example 1 5.2 4
Refer t o Ex . 15.23 . I f th e pil e i s embedde d i n mediu m dens e sand , determin e th e ne t pullou t
capacity an d the net allowabl e pullou t load wit h F
s
= 3.
Given: L = 30 ft, 0 = 38°, ~K
s
= 1.5, an d 8 = 25°, 7 (average) = 1.1 0 lb/ft
3
.
The wate r table i s at great depth . Refe r to Section 15.25 .
Solution
Downward ski n resistance Q
f
where q' =- x 30x1 10 = 1650 lb/ft
2
"o
A =3.14x1.25x3 0 = 117.75 f t
2
N
165 0 x 1.5 t an25°x 117.75
<2/-(down) = = 136 kip s
f
100 0
Based o n the recommendations o f Poulos an d Davi s (1980)
2 2
= — <2/(down) = — x 136 = 91 kip s
674 Chapt er 1 5
kips
PART B—PIL E G ROU P
1 5 .2 6 NUMBE R AND SPACI N G OF PI L ES I N A G ROU P
V ery rarely ar e structures founded on single piles. Normally , there will be a minimum of three piles
under a colum n o r a foundatio n elemen t becaus e o f alignmen t problem s an d inadverten t
eccentricities. Th e spacin g o f piles i n a group depends upo n many factor s suc h as
Pile cap
(a) Singl e pile
Isobar o f a
single pile
Isobar o f
a group
Highly
stressed
zone
(b) Group of piles closel y space d
Q
i r
Pile ca p
(c) Group of piles wit h piles fa r apar t
Figure 1 5.2 4 Pres s ur e is obar s o f (a ) s ingl e pile , (b ) group o f piles , clos el y s paced ,
an d (c ) g roup o f pile s wit h pile s fa r apart .
Deep Foun dat io n I : Pil e Foun dat io n 67 5
1. overlappin g of stresses o f adjacent piles,
2. cos t o f foundation,
3. efficienc y o f the pil e group.
The pressure isobars of a single pile wit h load Q acting on the top are shown in Fig. 15.24(a) .
When pile s ar e placed i n a group, there i s a possibilit y the pressur e isobar s o f adjacent piles will
overlap each other as shown in Fig. 15.24(b) . The soil is highly stressed i n the zones of overlapping
of pressures. Wit h sufficient overlap , either the soil wil l fail o r the pile group will settle excessivel y
since the combined pressur e bulb extends t o a considerable dept h below th e base o f the piles. I t is
possible t o avoi d overla p b y installin g th e pile s furthe r apar t a s show n i n Fig . 15.24(c) . Larg e
spacings are not recommended sometimes , sinc e this would result in a larger pile cap which would
increase th e cost o f the foundation.
The spacing of piles depend s upon the method of installing the piles and the type of soil. The
piles ca n b e drive n pile s o r cast-in-situ piles . Whe n th e pile s ar e drive n ther e wil l b e greate r
overlapping of stresses due to the displacement of soil. If the displacement of soil compacts th e soil
in betwee n the piles as in the case o f loose sand y soils, the piles may be placed a t closer intervals.

10 pile 1 1 pile 1 2 pile
Figure 1 5.2 5 T ypica l arran g em en t s o f pile s i n g roup s
676 Chapt e r 1 5
But i f the piles ar e driven into saturated clay or silt y soils , th e displaced soi l wil l no t compac t th e
soil between th e piles. As a result the soi l between the piles may move upwards and in this proces s
lift th e pil e cap . Greate r spacin g between pile s i s require d i n soil s o f thi s typ e t o avoi d liftin g o f
piles. When piles are cast-in-situ, the soils adjacent to the piles are not stressed t o that extent and as
such smaller spacing s are permitted.
Generally, th e spacin g fo r poin t bearing piles , suc h a s pile s founde d on rock, ca n b e much
less tha n fo r frictio n pile s sinc e th e high-point-bearin g stresse s an d th e superpositio n effec t o f
overlap o f th e poin t stresse s wi l l mos t likel y no t overstres s th e underlyin g materia l no r caus e
excessive settlements .
The minimum allowable spacing of piles is usually stipulated i n building codes. The spacings
for straight uniform diameter piles may vary from 2 to 6 times the diameter of the shaft. For frictio n
piles, the minimum spacing recommended i s 3d where d is the diameter of the pile. For end bearing
piles passing through relatively compressible strata , the spacing of piles shal l not be less than 2.5d.
For end bearing pile s passing through compressible strat a and resting i n stiff clay , the spacing may
be increased t o 3.5d. For compaction piles, the spacing may be Id. Typical arrangement s of piles in
groups ar e shown i n Fig. 15.25 .
1 5 .2 7 PIL E G ROU P EFFI CI ENCY
The spacin g o f pile s i s usuall y predetermine d b y practica l an d economica l considerations . Th e
design o f a pile foundatio n subjected to vertical loads consist s of
1. Th e determination o f the ultimat e load bearing capacit y of the group Q
u
.
2, Determinatio n o f the settlement of the group, S , under an allowable load Q .
a o
The ultimat e loa d o f the grou p is generall y differen t fro m th e su m o f th e ultimat e load s o f
individual piles Q
u
.
The factor
Q
gu
E = ( ' 1^A7 ^
is called group efficienc y whic h depends on parameters such a s type o f soil in which th e piles are
embedded, metho d o f installatio n o f pile s i.e . eithe r drive n o r cast-in-situ piles, an d spacin g o f
piles.
There i s n o acceptabl e "efficiency formula" fo r grou p bearin g capacity . Ther e ar e a fe w
formulae suc h a s th e Converse-Labarr e formul a tha t ar e sometime s use d b y engineers . Thes e
formulae ar e empirica l an d giv e efficiency factor s les s tha n unity. Bu t whe n piles ar e installe d i n
sand, efficienc y factor s greate r tha n unit y ca n b e obtaine d a s show n b y V esi c (1967 ) b y hi s
experimental investigatio n o n group s o f pile s i n sand . Ther e i s no t sufficien t experimenta l
evidence t o determine group efficiency fo r piles embedded i n clay soils .
Efficiency o f Pil e G roup s i n Sand
V esic (1967) carried ou t test s on 4 and 9 pile groups driven into sand under controlled conditions .
Piles with spacings 2, 3, 4, and 6 times the diameter were used in the tests. The tests were conducted
in homogeneous , mediu m dense sand . Hi s finding s ar e give n i n Fig . 15.26 . Th e figur e give s th e
following:
1. Th e efficiencies o f 4 and 9 pile groups whe n the pile caps do not rest on the surface .
2. Th e efficiencies of 4 and 9 pile groups when the pil e caps rest o n the surface .
Deep Foun dat io n I : Pil e Foun dat io n
3.0
677
2.5
tq
56
<u
Q.
§
E
1.0
_DQ
//
//
n
i -
1. Point efficiency—average o f al l test s
2. 4 pile group—total efficienc y
3. 9 pile group—tota l efficiency
4. 9 pile group—total efficiency with cap
5. 4 pile group—total efficiency wit h cap
6. 4 pile group—ski n efficiency
7. 9 pile group—ski n efficiency
1 2 3 4 5 6 7
Pile spacin g i n diameter s
Figure 1 5.2 6 Efficienc y o f pil e group s i n sand (V esic , 1 9 6 7 )
3. Th e skin efficiency o f 4 and 9 pile groups .
4. Th e average poin t efficiency o f all the pile groups .
It may be mentioned here that a pile group with the pile cap resting on the surface takes mor e
load tha n one wit h fre e standin g piles abov e th e surface . In the forme r case , a part o f the load i s
taken by the soi l directl y under the cap and the rest i s taken by the piles. The pil e cap behaves th e
same wa y a s a shallow foundatio n of the same size . Though th e percentage o f loa d take n b y th e
group is quite considerable, building codes have not so far considered th e contribution made by the
cap.
It may be see n from th e Fig. 15.2 6 tha t the overall efficiency o f a four pil e grou p with a cap
resting on the surface increases t o a maximum of about 1. 7 at pile spacings of 3 to 4 pile diameters ,
becoming somewha t lowe r wit h a furthe r increas e i n spacing . A sizabl e par t o f th e increase d
bearing capacit y come s fro m th e caps. I f the loads transmitte d by the caps ar e reduced, the group
efficiency drop s t o a maximum of about 1.3 .
V ery simila r result s ar e indicated fro m test s wit h 9 pil e groups . Sinc e th e test s i n thi s case
were carrie d ou t onl y u p t o a spacin g o f 3 pil e diameters , th e ful l pictur e o f th e curv e i s no t
available. However , i t ma y b e see n tha t th e contributio n o f th e ca p fo r th e bearin g capacit y i s
relatively smaller .
V esic measured th e ski n loads of all the piles. The ski n efficiencies for both th e 4 and 9-pile
groups indicat e an increasing trend. For the 4-pile group the efficiency increase s fro m abou t 1. 8 at
2 pile diameters to a maximum o f about 3 at 5 pile diameters and beyond. I n contrast to this, the
average poin t load efficienc y fo r the groups is about 1.01 . V esi c showe d for th e firs t time that the
678 Chapt e r 15
increasing bearin g capacit y o f a pil e grou p fo r pile s drive n i n san d come s primaril y fro m a n
increase i n ski n loads. The point loads seem t o be virtuall y unaffecte d b y group action.
Pile G rou p Efficienc y Equatio n
There ar e many pile group equations. These equation s ar e to be use d ver y cautiously, and ma y in
many case s b e n o bette r tha n a goo d guess . Th e Converse-Labarr e Formul a i s on e o f th e mos t
widely use d group-efficiency equation s which is expressed a s
6( n - \ }m + ( m- l)n
(15
'
68
>
where m = numbe r of columns of piles i n a group,
n = number of rows,
6 = lan~
l
( d/s ) i n degrees ,
d = diameter of pile,
s = spacing of piles center t o center.
1 5.28 V ERTI CA L B EARI NG CAPACI TY O F PIL E G ROU PS
EMBEDDED I N SANDS AN D G RAV EL S
Driven piles . I f piles ar e driven into loose sand s and gravel, the soi l around the piles t o a radius of
at leas t thre e time s th e pil e diamete r i s compacted . Whe n pile s ar e drive n i n a grou p a t clos e
spacing, the soi l aroun d and between the m becomes highl y compacted. Whe n the group i s loaded ,
the pile s an d th e soi l betwee n the m mov e togethe r a s a unit . Thus , th e pil e grou p act s a s a pie r
foundation havin g a base are a equal to the gross plan area contained by the piles. The efficienc y o f
the pil e grou p wil l b e greate r tha n unit y a s explaine d earlier . I t i s normall y assume d tha t th e
efficiency fall s t o unit y whe n th e spacin g i s increase d t o fiv e o r si x diameters . Sinc e presen t
knowledge i s not sufficient t o evaluate the efficiency fo r different spacin g of piles, i t is conservative
to assume an efficiency facto r of unit y for al l practical purposes . We may, therefore, writ e
Q =nQ (15.69 )
^-gu *^u ^ '
where n - th e number of piles in the group.
The procedure explaine d above is not applicable i f the pil e tips rest on compressible soi l such as
silts or clays. When the pile tips rest on compressible soils , the stresses transferre d t o the compressibl e
soils from the pile group might result in over-stressing or extensive consolidation. The carrying capacity
of pile groups under these conditions is governed b y the shear strength and compressibility o f the soil ,
rather than by the 'efficiency'' o f the group within the sand or gravel stratum.
Bored Pil e G roup s I n Sand And G rave l
Bored pile s ar e cast-in-situ concret e piles . The method of installation involves
1 .Borin g a hole of the required diameter and depth,
2. Pourin g i n concrete.
There wil l alway s be a genera l loosenin g o f th e soi l durin g boring an d the n to o whe n th e
boring has to be done below the water table. Though bentonit e slurr y (sometimes calle d a s drilling
mud) i s use d fo r stabilizin g the side s an d botto m o f th e bores , loosenin g o f th e soi l canno t b e
avoided. Cleaning of the bottom of the bore hol e prior to concreting is always a problem whic h will
never be achieved quit e satisfactorily. Since bored pile s do not compact th e soi l between th e piles ,
Deep Foun dat io n I : Pil e Foun dat io n 679
'TirTTT
Perimeter P
n
Figure 1 5.2 7 B loc k failur e o f a pil e group i n clay soi l
the efficienc y facto r wil l neve r b e greate r tha n unity . However , fo r al l practica l purposes , th e
efficiency ma y be taken a s unity.
Pile G roup s I n Cohesive Soil s
The effec t o f drivin g pile s int o cohesiv e soil s (clay s an d silts ) i s ver y differen t fro m tha t o f
cohesionless soils . It has already been explained that when piles are driven into clay soils, particularly
when the soil is soft an d sensitive, there will be considerable remoldin g of the soil. Besides ther e will
be heaving of the soil between the piles since compaction during driving cannot be achieved in soils of
such low permeability. There i s every possibility of lifting of the pile during this process of heaving of
the soil. Bored pile s are, therefore, preferred t o driven piles i n cohesive soils . In case driven piles are
to be used, the following steps shoul d be favored:
1. Pile s shoul d be space d a t greater distance s apart .
2. Pile s shoul d be driven from th e center of the group towards the edges, and
3. Th e rate of driving of each pil e should be adjusted as to minimize the development of pore
water pressure .
Experimental result s hav e indicate d tha t whe n a pil e grou p installe d i n cohesiv e soil s i s
loaded, i t may fai l b y an y one o f the following ways :
1. Ma y fai l a s a block (calle d bloc k failure).
2. Individua l piles i n the group may fail .
680 Chapt e r 15
When pile s ar e space d a t close r intervals , th e soi l containe d betwee n th e pile s mov e
downward wit h th e pile s an d a t failure , pile s an d soi l mov e togethe r t o giv e th e typica l 'block
failure'. Normall y this type of failure occurs when piles are placed within 2 to 3 pile diameters. For
wider spacings , th e piles fai l individually . Th e efficienc y rati o is les s tha n unit y at closer spacing s
and ma y reac h uni t y at a spacing of about 8 diameters.
The equatio n fo r bloc k failur e ma y b e writte n as (Fig . 15.27) .
Qgu=
cN
c
A
g
+p
g
^ (15.70 )
where c = cohesive strengt h of cla y beneat h th e pil e group ,
c = average cohesiv e strength of clay around the group ,
L = lengt h of pile,
P = perimeter o f pil e group,
o
A = sectiona l are a o f group,
0
N
c
= bearing capacit y factor whic h may be assume d a s 9 for deep foundations.
The bearin g capacit y of a pile group on the basi s o f individual pile failur e may be writte n as
Q =nQ (15.71 )
*^gu ^-u ^ '
where n = numbe r of piles i n the group,
Q
u
= bearing capacit y of an individual pile.
The bearing capacit y of a pile group is normally taken as the smaller of the two given by Eqs.
(15.70) an d (15.71).
Example 1 5 .25
A group o f 9 piles wit h 3 piles i n a row wa s driven int o a sof t cla y extending fro m groun d level t o
a grea t depth . Th e diamete r an d th e lengt h o f th e pile s wer e 3 0 c m an d 10 m respectively . Th e
unconfmed compressiv e strengt h o f th e cla y i s 7 0 kPa . I f th e pile s wer e place d 9 0 c m cente r t o
center, comput e th e allowabl e load o n the pil e group o n the basi s o f a shear failur e criterion fo r a
factor o f safet y of 2.5 .
Solution
The allowabl e loa d o n th e grou p i s t o be calculate d fo r two conditions : (a ) bloc k failur e an d (b )
individual pil e failure . The leas t o f the two gives the allowabl e loa d o n the group.
(a) B loc k failure (Fig. 1 5 .2 7 ) . U s e Eq . (1 5 .7 0 ) ,
Q =cN A+PL c wher e N = 9, c = c = 70/2 = 35 kN/m
2
^-gUc R g C
A =2. 1x2. 1 = 4. 4m
2
, P = 4x 2.1 = 8.4 m, L = 10 m
Q =35x9x4. 4 + 8. 4x10x35 = 4326 kN, Q =
^-gu ^-a
25
(b) I ndividua l pil e failure
Q
u
=Q
b
+ Q
f
=<7^ A + a c A
v
. Assume a= 1.
Now, q
b
= cN
c
= 35 x 9 = 315 kN/m
2
, A
b
= 0.07 m
2
,
Deep Foun dat io n I : Pil e Foun dat io n 68 1
A
s
= 3.14x0.3x10 = 9.42m
2
Substituting, Q
u
= 315 x 0.07 +1 x 35 x 9.42 = 352 kN
3168
Q
gu
= «G
U
= 9 x
352 = 3168 kN
> Qa = ~^~ =
1267 kN
The allowabl e load i s 126 7 kN.
1 5.29 SETTLEMEN T OF PILES AND PIL E GROUPS I N SANDS AN D
G RAV EL S
Normally i t i s not necessar y t o comput e the settlement of a singl e pil e a s thi s settlemen t under a
working load will be within the tolerable limits. However, settlement analysis of a pile group is very
much essential. The total settlement analysis of a pile group does not bear any relationship with that
of a single pile sinc e in a group the settlement is very much affected du e to the interaction stresse s
between piles an d the stressed zon e below the tips of piles .
Settlement analysi s o f singl e pile s b y Poulo s an d Davi s (1980 ) indicate s tha t immediat e
settlement contribute s th e majo r par t o f th e fina l settlemen t (whic h include s th e consolidatio n
settlement for saturate d clay soils ) eve n for piles i n clay. As far a s piles i n san d i s concerned, th e
immediate settlement i s almost equal to the final settlement .
However, it may be noted here that consolidation settlement becomes mor e important for pile
groups in saturated clay soils .
Immediate settlemen t o f a singl e pil e ma y b e compute d b y makin g us e o f semi-empirica l
methods. The method a s suggested by V esic (1977) ha s been discusse d here .
In recen t years , wit h th e advent of computers, mor e sophisticate d method s o f analysi s have
been developed t o predict the settlement and load distribution in a single pile. The following three
methods ar e often used .
1. 'Loa d transfer' metho d which is also called as the 't-z ' method .
2. Elasti c method base d o n Mindlin's (1936) equations for the effect s o f subsurface loadings
within a semi-infinite mass.
3. Th e finit e elemen t method .
This chapter discusses onl y the 't-z' method. The analysis of settlement by the elastic method
is quite complicated an d is beyond the scope of this book. Poulos and Davis, (1980) have discussed
this procedur e i n detail . The finit e elemen t metho d of analysi s of a singl e pil e axiall y loade d ha s
been discusse d by man y investigator s such as Ellison et al. , (1971) , Desa i (1974) , Balaa m et al. ,
(1975), etc . The finit e elemen t approac h i s a generalization of the elasti c approach . Th e powe r of
this method lies in its capability to model complicated conditions and to represent non-linear stress/
strain behavior o f the soi l ove r the whol e zone of the soi l modelled. Us e o f computer programs i s
essential an d th e metho d i s mor e suite d t o researc h o r investigatio n o f particularl y comple x
problems than to general design .
Present knowledg e i s not sufficien t t o evaluat e the settlement s o f piles an d pil e groups . For
most engineerin g structures , th e load s t o b e applie d t o a pil e grou p wil l b e governe d b y
consideration o f consolidation settlemen t rathe r than by bearing capacit y o f the grou p divided by
an arbitrary factor of safety of 2 or 3. It has been found from fiel d observatio n that the settlement of
a pil e grou p is many times the settlemen t of a singl e pile at the corresponding workin g load. Th e
settlement o f a grou p i s affecte d b y th e shap e an d siz e o f th e group , lengt h o f piles , metho d o f
installation of piles and possibl y many other factors.
682 Chapt e r 1 5
Semi-Empirical Formula s and Curve s
V esic (1977 ) propose d a n equation t o determine th e settlemen t of a singl e pile . The equatio n ha s
been developed o n the basis of experimental results he obtained from test s on piles. Tests on piles of
diameters rangin g from 2 to 1 8 inches were carried out in sands of different relativ e densities. Tests
were carried ou t on driven piles, jacked piles, and bored piles (jacked piles are those that are pushed
into the ground by using a jack). The equation for total settlement of a single pile may be expresse d
as
c - c . c ( ] s. 79\
O — O t O r \ L J . l £ )
where S = total settlement,
S = settlement of the pil e tip,
S, = settlemen t du e t o the deformation of the pile shaft .
The equatio n fo r S
p
i s
5 _ = - ^ (15.73 )
The equation fo r 5, is
-^ (15.74 )
where Q = point load ,
d = diameter o f the pile at the base,
q
u
- ultimat e point resistance per unit area ,
D
r
= relative density of the sand,
C
w
= settlemen t coefficient, = 0.0 4 fo r driven pile s
= 0.0 5 fo r jacked pile s
= 0.1 8 for bored piles ,
Q, = friction load ,
L = pile length,
A = cross-sectional are a of the pile,
E = modulus of deformation of the pil e shaft ,
a, — coefficient whic h depends on the distribution of ski n friction along the shaf t an d ca n
be taken equa l t o 0.6 .
Settlement o f pile s canno t b e predicte d accuratel y b y makin g us e o f equation s suc h a s th e
ones give n here. On e shoul d use suc h equations wit h caution. I t i s better t o rel y o n loa d test s for
piles i n sands.
Settlement o f Pil e G roup s i n Sand
The relatio n betwee n th e settlemen t o f a group and a singl e pil e a t corresponding workin g load s
may be expressed a s
(15.75)
where F = group settlement factor,
S = settlement of group,
5 = settlemen t of a single pile.
Deep Foundation I : Pil e Foundatio n 683
s 10
ex
2
O
10 2 0 3 0 4 0 5 0
Relative widt h of the group Bid
60
Figure 1 5.2 8 Curv e s howin g t he relat ion s hi p bet wee n g rou p s et t lem en t rat i o an d
relat ive widt h s o f pil e g roup s i n s an d (Ves ic , 1967)
16
*: 1 2
O
z
6 9 1 2 1 5
Width of pile group (m)
18 21
Figure 15.2 9 Curv e s howin g relat ion s hi p bet wee n F an d pil e g roup widt h
(Sk em pt on , e t al. , 1953 )
V esic (1967) obtained the curve given in Fig. 15.2 8 by plotting F agains t Bid wher e d is the
diameter of the pile and B, the distance between the center to center of the outer piles i n the group
(only squar e pil e groups ar e considered). I t shoul d be remembered her e that the curve i s based on
the result s obtaine d fro m test s o n groups of piles embedde d i n mediu m dens e sand . I t i s possibl e
that groups i n much looser o r much denser deposit s might give somewhat differen t behavior . The
group settlement rati o i s very likel y be affecte d b y the rati o of the pil e point settlement S t o total
pile settlement .
Skempton e t al. , (1953) published curves relating F wit h the widt h of pile groups a s shown
in Fig. 15.29 . These curves can be taken as applying to driven or bored piles. Sinc e the abscissa fo r
the curv e i n Fig . 15.2 9 i s no t expresse d a s a ratio , thi s curv e cannot directl y b e compare d wit h
V esic's curv e give n i n Fig. 15.28 . Accordin g t o Fig. 15.2 9 a pil e grou p 3 m wid e woul d settl e 5
times that of a single test pile.
t-z Metho d
Consider a floatin g vertica l pil e o f lengt h L and diamete r d subjecte d t o a vertica l loa d Q (Fig .
15.30). This loa d wil l be transferred t o the surrounding and underlying soil layer s a s described i n
684 Chapt e r 1 5
Section 15.7 . The pil e loa d wi l l be carried partl y by ski n friction (which wil l be mobilize d throug h
the increasin g settlemen t o n th e mantl e surfac e an d th e compressio n o f th e pil e shaft ) an d partl y
through the pil e tip, in the for m of point resistance a s can be seen fro m Fig. 15.30 . Thus th e load i s
taken as the sum of these components. The distribution of the point resistance i s usual ly considered
as uni form; however , the dist ribut ion of the mantl e friction depends o n many factors . Loa d transfe r
in pile-soi l syste m i s a ver y comple x phenomeno n involving a numbe r o f parameter s whic h ar e
di f f i cul t t o evaluat e i n numeri ca l terms . Yet , som e numerica l assessmen t o f loa d transfe r
characteristics of a pil e soi l syste m i s essential for th e rational desig n o f pil e foundation .
The objectiv e o f a loa d transfe r analysi s i s t o obtai n a load-settlemen t curve . Th e basi c
problem o f loa d transfe r is shown in Fig. 15.30 . Th e followin g are t o be determined :
1. Th e vertica l movements s, of the pile cross-sections a t any depth z under loads acting on the
top,
2. Th e correspondin g pil e loa d Q
7
at dept h z actin g on the pil e section ,
3. Th e vertica l movement of the bas e o f the pil e an d the corresponding poin t stress .
The mobilization of skin shear stres s r at any depth z , from the ground surfac e depends o n the
vertical movemen t o f the pil e cross sectio n a t that level . The relationshi p between th e two may b e
linear o r non-linear . The shea r stres s T , reaches th e maximu m value , r, , a t tha t sectio n whe n th e
vertical movemen t o f th e pil e sectio n i s adequate . I t is , therefore , essentia l t o construc t ( i - s )
curves a t variou s depths z a s required .
There wi l l b e settlemen t o f th e ti p o f pil e afte r th e ful l mobilizatio n o f ski n friction . Th e
movement o f th e tip , s
h
, ma y b e assume d t o b e linea r wit h th e poin t pressur e q . Whe n th e
movement o f the tip is adequate, the point pressure reaches th e maximum pressur e q
b
(ultimate bas e
pressure). I n order t o solve the load-transfer problem , i t is esential t o construct a ( q - s ) curve .
(r - s ) and (q
p
- s ) curves
Coyle an d Rees e (1966) proposed a set of average curve s o f load transfe r base d on laborator y tes t
piles and instrumented field piles . These curves ar e limited t o the case of steel-pipe frictio n pile s i n
clay wit h a n embedde d dept h no t exceedin g 10 0 ft . Coyl e an d Sulaima n (1967 ) hav e als o give n
load transfer curves for piles in sand. These curves ar e meant for specific cases and therefore mean t
to solv e specifi c problems an d as such this approac h canno t be considered a s a general case .
V erbrugge (1981 ) proposed a n elastic-plastic model fo r the (r- s ) and ( q - s ) curves based
on CP T results . The slope s of the elasti c portion o f the curves give n ar e
r 0.22 £
(15.76)
s 2R
V 3.1 25E
2R
(15.77)
The valu e o f elasti c modulu s E
v
o f cohesionles s soil s ma y b e obtaine d b y th e followin g
expressions (V erbrugge) ;
for bore d pile s E
f
= (36 + 2. 2 q
c
) kg/cm
2
(15.78 )
for drive n piles £\. = 3(36 + 2. 2 q
c
) kg/cm
2
; (15.79 )
where q
c
- poin t resistanc e of static cone penetromete r i n kg/cm
2
.
The relationshi p recommende d fo r E^ i s for q
c
> 4 kg/cm
2
. The maximu m value s o f T , and !„
for th e plasti c portio n o f ( r - s ) curve s ar e give n i n Tabl e 15. 4 an d 15. 5 fo r cohesionles s an d
cohesive soil s respectively. In Eqs (15.76) and (15.77) the value of E^ can be obtained by any one of
Deep Foun dat io n I: Pil e Foun dat ion 685
T - s curve
44
3
n - 3
n - 2
n - 1
^p(max)
q
p
- s curve
Qn-2
A^nrr
. 1
Pile subjected to Loa d transfe r mechanism
vertical road
Figure 15.3 0 t- z metho d o f an alys i s o f pil e load-s et t lem en t relat ion s hi p
the know n methods . Th e maximu m valu e o f q ( q
b
) ma y b e obtaine d b y an y on e o f th e know n
methods suc h a s
1. Fro m the relationshi p
q
b
= q'
o
N fo r cohesionless soil s
q
b
= 9c
u
for cohesive soils.
Table 15. 4 Recom m en de d m ax im u m ski n s hea r s t res s r
max
for pile s i n
cohes ion les s s oil s (Aft e r Verbrug g e , 1981 )
T
m ax kN/m
2
Pil e type
0.011 q
c
Drive n concret e piles
0.009 q
c
Drive n steel piles
0.005 q
c
Bore d concrete piles
Limiting values
T =8 0 kN/m
2
for bored piles
T =12 0 kN/m
2
fo r drive n piles
686 Chapt er 1 5
Table 1 5 . 5 Recom m en de d m ax im u m s k i n s hea r s t res s T „ for pile s i n cohes iv e
m ax
s oils (Aft e r Verbrug g e , 1981 )
[Val ues recom m en de d ar e for Dut c h con e pen et rom et er ]
Type o f pi l e
Driven
Bored
Mat er i al
Concrete
Steel
Concrete
Steel
Ra nge o f q
c
, k N/ m
2
q
c
< 375
375 < q
c
< 4500
4500 < q
c
q
c
< 450
45 0<4
c
< 1500
1500«?
c
.
q
c
<600
600 <q
c
< 4500
4500 < q
c
q
c
< 50 0
500 <<?
c
< 1500
1500«?
c
.
max
0.053 q
c
18 + 0.006 q
c
0.01 q
c
0.033 q
c
15
0.01 q
c
0.037 q
c
18 + 0.006 q
c
0.01 q
c
0.03 <?
c
15
0.01 q
c
2. Fro m stati c cone penetratio n tes t result s
3. Fro m pressuremete r tes t results
M ethod o f obtaining load- settlemen t curv e (Fig. 1 5 .3 0 )
The approximat e load-settlemen t curve i s obtained poin t by poin t in the following manner:
1. Divid e th e pil e int o any convenient segment s (possibl y 10 for compute r programmin g and
less for hand calculations) .
2. Assum e a point pressure q les s than the maximum q
b
.
3. Rea d th e corresponding displacemen t s fro m th e ( q - s) curve .
4. Assum e tha t th e loa d i n th e pil e segmen t closes t t o th e poin t (segmen t ri) i s equa l t o th e
point load .
5. Comput e th e compression o f the segment n under that load b y
A
As =
where, Q
p
= q
p
A
b
,
A = cross-sectional are a o f segment ,
E = modulus of elasticit y of the pil e material .
6. Calculat e the settlemen t of the top of segment n by
1. Us e the ( i - s ) curves to read th e friction i
n
on segment n, at displacement s
n
.
8. Calculat e th e load i n pile segment ( n - 1 ) by:
Deep Foun dat io n I : Pil e Foun dat io n 687
where Az
n
= length of segment n,
d
n
= average diameter of pile i n segment n (applies t o tapered piles) .
9. D o 4 throug h 8 up t o th e to p segment . The loa d an d displacemen t a t th e to p o f th e pil e
provide one point on the load-settlement curve.
10. Repea t 1 through 9 for the other assumed values of the point pressure, q .
E ma y also be obtained from th e relationship established between E
s
and the field tests such
as SPT, CPT and PMT. I t may be noted here that the accuracy of the results obtained depends upon
the accurac y with which the values of E
s
simulat e the field conditions.
Example 1 5 .2 6
A concret e pil e o f sectio n 3 0 x 3 0 c m i s drive n int o medium dense san d wit h the wate r tabl e at
ground level. The dept h of embedment of the pile is 18m. Static cone penetration test conducted at
the site gives an average value q
c
= 50 kg/cm
2
. Determine the load transfer curves and then calculate
the settlement. The modulus of elasticity of the pile material E i s 21 x 10
4
kg/cm
2
(Fig. Ex. 15.26) .
Solution
It is first necessar y to draw the ( q - s) and ( T- s) curves (see sectio n 15.29) . The curves can be
constructed b y determining the ratios of q / s and t/ s fro m Eq s (15.77) and (15.76) respectively .
Q
p
_ 3.125 £
s ~ 2R
6 m
6 m
6 m
\
f
f
f
3
Q
'
,
Q
'3
l
2
l
+
S
f
tr,
1
T
5
3
tr
5
1 2 L -
f "
-L „
Sand
- = 18 m
Square pile 30 x 30 cm
E
p
= 21 x 10
4
kg/cm
2
50
N
4 0
S
"&>3 0
j *
^ 20
10
0
i
0.6
0.5
0.4
rs
|0. 3
M
^
(-0.2
0.1
0
=
50kP a
^(max)
=10.96 mm
I I -
4 8 1 2
Settlement, s, mm
ax = 0.55 kg/cm
2
(r - s) curve
s(max)
=1.71 mm
i i i i i
0 0. 5 1. 0 1. 5
Settlement, s, mm
2.0
Figure Ex . 1 5.2 6
688 Chapt e r 1 5
T _ 0.22 E
s ~ 2R
where R = radius or widt h of pil e
The valu e o f E
s
fo r a driven pile may b e determined fro m Eq . (15.78) .
E
s
= 3(3 6 + 2. 2 q
c
) kg/cm
2
= 3(3 6 + 2. 2 x 50) = 438 kg/cm
2
q
p
3.125x43 8 ,
Now — -= -= 45.62 kg/cn r
5 2x1 5
r 0.22x43 8 ,
— = -= 3.21 kg/cnr
^ 2x0.1 5
To construct the ( q - s ) and ( T- s ) curves, we have to know the maximum values of g
p
/s and T.
Given q
b
- q
}
= 50 kg/cm
2
- th e maximum value.
From Tabl e 15. 4 r
max
= 0.01 1 q
r
= 0.01 1 x 50 = 0.55 kg/cm
2
Now th e theoretical maxi mum settlement s for ^(max) = 5 0 kg/cm
2
i s
=1-096 c m = 10.96 m m
The curve ( q
}
- s ) may be drawn as shown in Fig. Ex. 15.26 . Similarl y for T
(max)
= 0.55 kg/cm
2
s = — —= 0.171 cm = 1.7 1 mm.
(max) 3.2 1
Now the (T- s ) curve can be constructed as shown i n Fig. Ex . 15.26 .
Calculation of pile settlement
The variou s step s i n the calculations are
1. Divid e the pil e length 1 8 m int o three equal parts o f 6 m each .
2. T o start wit h assume a base pressur e q - 5 kg/cm
2
.
3. Fro m th e ( q
}
- s} curv e s, = 0.12 cm for q = 5 kg/cm
2
.
4. Assum e tha t a load Q
{
= Q = 5 x 900 = 4500 kg act s axiall y on segmen t 1 .
5. No w th e compressio n As , o f segmen t 1 is
. <2)A L 4500x60 0 . . . . .
As, = — —= --=0.014 cm
1
AE, 30x 30x 2 1x l 0
4
6. Settlemen t of the to p o f segmen t 1 is
s
2
= S, + As, = 0.12 + 0.014 = 0.134 cm .
7. No w from (r- s ) curve Fig. Ex. 15.26, i-0.43 kg/cm
2
.
8. Th e pil e loa d i n segment 2 is
<2
2
= 4 x 30 x 600 x 0.43 + 4500 = 30,960 + 4500 = 35,460 kg
9. No w th e compression of segmen t 2 is
<2
2
AL 35,460x60 0
As
?
= — -= -= 0. 1 13 cm
" AE
p
900x21xl 0
4
10. Settlemen t o f the to p o f segmen t 2 is
Deep Foun dat io n I : Pil e Foun dat io n 689
S
2
= s
2
+ As
2
= 0.134 + 0.113 = 0.247 cm .
11. No w fro m ( T - s) curve , Fig. Ex . 15.2 6 T = 0.55 kg/cm
2
fo r s
3
= 0.247 cm. Thi s i s the
maximum shear stress .
12. No w the pil e load i n segment 3 is
<2
3
= 4 x 30 x 600 x 0.55 + 35,460 = 39,600 + 35,460 = 75,060 kg
13. Th e compression o f segment 3 is
_ Q
3
AL_ 75,060x60 0
AE 900x21xl 0
4
= 0.238 cm
14. Settlemen t o f top o f segment 3 is
S
4
= s
t
= s
2
+&s
2
= 0.247 + 0.238 = 0.485 cm.
15. No w from (T - s ) curve, T
max
= 0.55 kg/cm
2
for s > 0.17 cm.
16. Th e pil e load a t the top of segment 3 is
Q
T
= 4 x 30 x 600 x 0.55 + 75,060 = 39,600 + 75,060 = 114,660 kg
= 11 5 tones (metric)
The total settlemen t s
t
= 0.485 cm = 5 mm.
Total pil e loa d Q
T
= 115 tones.
This yields one point on the load settlement curve for the pile. Other points can be obtained
in the same way by assuming different value s for the base pressure q i n Step 2 above. For
accurate results, the pile shoul d be divided into smaller segments .
1 5 .3 0 SETTL EM EN T OF PIL E G ROU PS I N COH ESI V E SOI LS
The tota l settlement s of pil e groups may be calculated by makin g use of consolidation settlemen t
equations. The problem involve s evaluating the increase in stress A/ ? beneath a pile group when the
group i s subjecte d t o a vertica l load Q . The computation of stresse s depend s o n the typ e of soi l
through which the pile passes. Th e methods of computing the stresses ar e explained below:
Fictitious
footing
\/ \ \
'A' iJJ.i_LLi
Weaker
layer
(a) (b) (c)
Figure 1 5.3 1 Set t lem en t o f pil e g roup s i n clay s oil s
690 Chapt e r 1 5
1. Th e soi l i n th e firs t grou p give n i n Fig . 15.3 1 (a ) i s homogeneou s clay . Th e loa d Q i s
assumed t o act on a fictitious footing at a depth 2/3L from the surfac e and distributed ove r
the sectional are a of the group. The loa d on the pile group acting at this level is assumed t o
spread ou t a t a 2 V er t : 1 Hori z slope . Th e stres s A/ ? a t an y dept h z belo w th e fictitiou s
footing ma y b e foun d a s explained in Chapter 6.
2. I n the second grou p given in (b) of the figure, th e pil e passes through a very weak layer of
depth Lj an d the lowe r portion of lengt h L
2
i s embedded i n a strong layer . I n thi s case, the
load Q i s assumed to act at a depth equal to 2/3 L
7
below the surface of the strong layer and
8 ^
spreads a t a 2 : 1 slope a s before .
3. I n the third case shown in (c) of the figure, th e piles ar e point bearing piles. The load i n this
case i s assumed t o ac t a t the leve l of the fir m stratu m and spreads out a t a 2 : 1 slope.
1 5 .3 1 AL L OWAB L E L OADS ON G ROU PS OF PILES
The basic criterion governin g the design of a pile foundation should be the same as that of a shallow
foundation, tha t is, the settlemen t of the foundation must not exceed some permissibl e value . The
permissible value s o f settlement s assume d fo r shallo w foundation s i n Chapte r 1 3 ar e als o
applicable t o pil e foundations . The allowabl e loa d o n a grou p o f pile s shoul d be th e leas t o f th e
values computed o n the basi s o f the following two criteria.
1. Shea r failure ,
2. Settlement .
Procedures hav e been give n in earlier chapter s a s t o how t o comput e the allowabl e loads o n
the basi s o f a shea r failur e criterion . Th e settlemen t o f pil e group s shoul d no t excee d th e
permissible limit s under these loads .
Example 1 5 .2 7
It is required to construct a pile foundation comprised o f 20 piles arrange d i n 5 columns at distances
of 90 cm center t o center. The diameter and lengths of the piles are 30 cm and 9 m respectively. Th e
bottom o f the pile cap i s located a t a depth of 2.0 m from th e ground surface. The detail s o f the soi l
properties etc . ar e as given below wit h reference to ground level as the datum. The wate r tabl e was
found a t a dept h o f 4 m from groun d level.
Soil propert ie s
0
2
4
12
14
17
2
4
12
14
17
-
Silt, saturated , 7= 1 6 kN/m
3
Clay, saturated , 7 = 19. 2 kN/m
3
Clay, saturated , 7 = 19. 2 kN/m
3
, q
u
= 120 kN/m
2
, e
Q
Clay, 7= 18.2 4 kN/m
3
, q
u
= 90 kN/m
2
, e
Q
= 1.08, C
c
Clay, 7= 2 0 kN/m
3
, q
u
= 180 kN/m
2
, e
Q
= 0.70, C
c
=
Rocky stratu m
= 0.80 ,
= 0.34 .
0.2
C
c
= 0.2 3
Compute the consolidatio n settlement of the pile foundatio n i f the total load impose d o n the
foundation i s 2500 kN.
Deep Foun dat io n I : Pil e Foun dat io n 69 1
Solution
Assume tha t the total loa d 2500 kN acts at a depth (2/ 3 )L = (2/3) x 9 = 6 m from the bottom of the
pile cap on a fictitious footing as shown in Fig. 15 . 31 (a). This fictitious footing is now at a depth of
8 m below ground level . The siz e o f the footing i s 3. 9 x 3. 0 m. Now thre e layer s ar e assume d t o
contribute to the settlement of the foundation. They are : L ayer 1 — from 8 m t o 12 m ( = 4m thick )
below ground level; L ayer 2 —from 1 2 m t o l 4 m = 2m thick ; L ayer 3 —from 1 4 m to 1 7 m = 3 m
thick. The increase in pressure due to the load on the fictitious footing at the centers of each layer is
computed o n th e assumptio n tha t th e loa d i s sprea d a t a n angl e o f 2 vertica l t o 1 horizontal
[Fig. 15.31(a) ] startin g fro m th e edge s o f th e fictitiou s footing. Th e settlemen t i s compute d b y
making use of the equation
C p + Ac
where p
o
= the effective overburden pressur e at the middl e of each layer ,
A/7 = the increas e i n pressure a t the middl e of each laye r
Computation o f p
a
For Layer 1 , p
g
= 2x 1 6 +2x 19. 2 + (10 -4) (19.2 -9.81) = 126.74 kN/m
2
For Layer 2, p
o
= 126.74 + 2(19.2-9.81)+ lx (18.24-9.81 ) = 153.95 kN/m
2
ForLayer3, p
o
= 153.95 + 1(1 8.24 -9.81) + 1.5 x (20.0 -9.81) -177.67 kN/m
2
Computation o f Ap
For Layer 1
Area at 2 m depth below fictitious footing = (3. 9 + 2) x ( 3 + 2) = 29. 5 m
2
2500
Ap = — —= 84.75 kN/m
2
For Layer 2
Area a t 5 m depth below fictitious footin g = (3. 9 + 5) x ( 3 + 5) = 71.2 m
2
2500
A/7 = — —= 35. 1 kN/m
2
For Layer 3
Area a t 7.5 m below fictitiou s footing = (3. 9 + 7.5) x ( 3 + 7.5) = 1 19.7 m
2
=
20. 9 kN/m
2
119.7
Settlement computatio n
, c 4x02 3 126.7 4 + 84.75
Layer 1 S, = -log -= 0.1 13m
l
1 + 0.80 126.7 4
2 x 0. 3 4, 153.9 5 + 35.1
nMn
Layer 2 5
9
= -log -= 0.029 m
3 2
1 + 1.08 153.9 5
692 Chapt er 1 5
T
_
c
3x0. 2 , 177.6 7 + 20.9
n m
_
Layer 3 S
3
= -—— log = 0.017 m
1 + 0.7
Total = 0. 159m- 16cm .
177.67
1 5 .3 2 NEG ATI V E FRI CTI ON
Figure 15.32(a ) shows a single pile and (b) a group of piles passing through a recently constructe d
cohesive soil fill . The soil below the fill had completely consolidated under its overburden pressure.
When the fil l start s consolidating under its own overburden pressure, it develops a drag on the
surface of the pile. This drag on the surface of the pile is called 'negative friction'. Negativ e frictio n

1
Fi l l - H ' Negat i ve - .
• •
V
J:B-
f r i ct i on
• ' • ' •
• : Natura l £\ ^M ••& Frictional ;• .
• Stif f Soi l ' ..^' ^^• ' -*' - ffcicfanr' p • .*! resistance ;
' -• >• . ". N .. •••>••. ''f it ••. •:• •
Point resistance
(a)
:
Fill
^ Natural
.' stiff soi l
i^.V-V'l
• ;i 4*3 ^
Figure 1 5 .3 2 Neg at iv e frict io n o n pile s
Deep Foun dat io n I : Pil e Foun dat io n 69 3
may develop i f the fil l materia l is loose cohesionless soil . Negative friction can also occur when fill
is place d ove r pea t o r a sof t cla y stratu m as shown in Fig. 15.32c . The superimpose d loadin g on
such compressibl e stratu m causes heav y settlement of the fil l wit h consequent drag o n piles .
Negative frictio n may develo p b y lowerin g th e groun d wate r whic h increase s th e effectiv e
stress causin g consolidatio n o f th e soi l wit h resultan t settlemen t an d frictio n force s bein g
developed o n the pile .
Negative frictio n mus t b e allowe d whe n considerin g th e facto r o f safet y o n th e ultimat e
carrying capacity of a pile. The factor of safety, F
s
, where negative friction i s likely to occur may be
written as
Ultimate carrying capacit y of a single pile or group of piles
F =
s
Workin g loa d + Negativ e skin friction loa d
Computation o f Negativ e Frictio n o n a Singl e Pil e
The magnitud e of negative friction F
n
fo r a single pile in a fil l ma y b e taken a s (Fig. 15.32(a)) .
(a) Fo r cohesive soil s
F
n
= PL
n
s. (15.80 )
(b) Fo r cohesionless soil s
(15.81)
where L
n
= length of piles i n the compressible material ,
s = shear strengt h of cohesive soil s i n the fill ,
P = perimeter of pile,
K = earth pressur e coefficien t normall y lie s betwee n th e activ e an d th e passiv e eart h
pressure coefficients,
S = angle of wal l friction which may var y from (|)/ 2 t o ()) .
Negative Frictio n o n Pil e G roup s
When a group of pile s passe s throug h a compressibl e fill , th e negativ e friction, F
n
, on th e group
may b e found by an y of the following methods [Fig . 15.32b] .
(a) F
ng
= nF
n
(15.82 )
(b)
F
n
g
=
sL
n
P
g
+ rL A
g
(15.83 )
where n = number of piles i n the group,
y = uni t weight of soil within the pile group to a depth L
n
,
P = perimeter o f pil e group,
A - sectiona l are a of pile group within the perimeter P ,
o o
s = shear strengt h o f soi l alon g the perimeter o f the group.
Equation (15.82 ) give s th e negativ e friction forces o f th e grou p a s equa l t o th e su m o f th e
friction force s o f al l the singl e piles.
Eq. (15.83 ) assume s th e possibilit y o f bloc k shea r failur e along th e perimete r o f the group
which include s the volume of the soi l yL
n
A enclose d i n the group. The maximu m value obtained
from Eq s (15.82 ) o r (15.83) shoul d be use d in the design.
When th e fi l l i s underlai n by a compressibl e stratu m a s show n i n Fig . 15.32(c) , th e tota l
negative friction may be found as follows:
694 Chapt e r 1 5
F
n
8
= "(
F
nl
+
^(15-84 )
F
ng =
S
\
L
\
P
g
+ S
2
L
2
P
g + XlM*+ ^Mg
- P
g
( s
}
L
{
+ s
2
L
2
) + A
g
( y
]
L
l
+ r
2
L
2
) (15.85 )
where L
}
= depth of fi l l ,
L, = dept h of compressibl e natura l soil,
SP s
?
= shea r strength s of the fil l an d compressibl e soil s respectively,
y
p
7
2
= uni t weight s of f i l l an d compressibl e soil s respectively,
F
nl
= negative friction o f a singl e pile in the fill ,
F
n2
= negative friction o f a singl e pil e i n the compressibl e soil .
The maximu m value of the negative friction obtaine d from Eqs . (15.84 ) o r (15.85) shoul d be
used fo r th e design o f pil e groups.
Example 1 5 .2 8
A squar e pil e grou p simila r t o the one shown i n Fig. 15.2 7 passe s throug h a recentl y constructe d
fill. The dept h o f fil l L
n
- 3 m. The diameter o f the pil e i s 30 cm and the piles ar e space d 9 0 cm
center t o center. I f the soi l i s cohesive wit h q
u
= 60 kN/m
2
, and y = 1 5 kN/m
3
, comput e the negative
frictional loa d o n the pil e group.
Solution
The negativ e frictiona l loa d o n the group i s the maximum of [(Eq s (15.82 ) and (15.83) ]
(a) F = nF , an d (b ) F = sL P + vL A ,
^ > ng n '
v
' ng n g ' n # '
where P = 4 x 3 = 1 2 m, A
g
= 3 x 3 = 9 m
2
, c
u
= 60/2 = 30 kN/rn
2
(a) F
n
= 9 x 3. 14x0. 3x3x30- 763 kN
(b) F
ng
= 3 0 x 3 x 12 + 1 5 x 3 x9 -1485 kN
The negativ e frictiona l loa d o n the grou p = 148 5 kN.
1 5 .3 3 U PL I F T CAPACI T Y O F A PIL E G ROU P
The uplif t capacit y o f a pile group, when the vertica l piles ar e arranged i n a closely space d group s
may no t b e equa l t o th e su m o f th e upl i f t resistance s o f th e individua l piles. Thi s i s because , a t
ultimate loa d conditions , the bloc k o f soi l enclose d b y th e pil e grou p get s lifted . Th e manne r i n
which th e loa d i s transferre d fro m th e pil e t o th e soi l i s quit e complex . A simplifie d wa y o f
calculating th e uplif t capacit y o f a pil e grou p embedde d i n cohesionles s soi l i s show n i n
Fig. 15.33(a) . A sprea d o f loa d o f 1 Horiz : 4 V ert from th e pil e group bas e t o the groun d surfac e
may b e take n a s th e volum e o f th e soi l t o b e lifte d b y th e pil e grou p (Tomlinson , 1977) . Fo r
simplicity i n calculation, the weight of the pil e embedded i n the ground i s assumed t o be equal t o
that of the volume of soi l i t displaces. If the pil e group is partly or ful l y submerged , th e submerge d
weight of soi l below th e wate r table has t o be taken.
In th e cas e o f cohesiv e soil , th e upl i f t resistanc e o f th e bloc k o f soi l i n undraine d shea r
enclosed b y th e pil e group given i n Fig . 15.33(b ) ha s t o b e considered. Th e equatio n for th e total
upl i ft capacit y P
u
of the grou p may b e expresse d b y
P,
u
=2L ( L + B)c
u
+ W (15.86 )
Deep Foun dat io n I : Pil e Foun dat io n 695
t t I t t
Block of soi l
lifted b y piles
(a) Uplif t o f a group of closely-spaced piles in cohesionless soil s
t t I t t
Block of soi l
lifted b y piles
Bx L -
(b) Uplif t o f a group of piles in cohesive soils
Figure 15.3 3 Uplift capacit y o f a pil e g rou p
where L = dept h of the pil e block
L and B = overall lengt h and widt h of the pile group
c
u
= average undrained shear strengt h of soi l aroun d the side s of the group
W = combined weigh t of the block of soi l enclose d b y the pil e group plus the weigh t of
the piles and the pile cap .
A factor o f safety o f 2 may be used i n both cases of piles in sand an d clay .
The uplif t efficienc y E o f a group of piles may be expressed a s
(15.89)
where P
US
= uplift capacit y of a single pil e
n = number of piles i n the group
The efficienc y E
u
varies wit h the method o f installation of the piles , lengt h an d spacing and
the type of soil. The available data indicate that E increase s with the spacing of piles. Meyerhof and
Adams (1968) presente d som e dat a o n uplif t efficienc y o f group s o f two an d fou r mode l circula r
696 Chapt er 1 5
footings i n clay . Th e result s indicat e tha t th e uplif t efficienc y increase s wit h th e spacin g o f th e
footings o r bases an d as the depth of embedment decreases, bu t decreases as the number of footings
or bases i n the group increases. How far the footings would represent the piles i s a debatable point.
For uplif t loadin g on pil e groups in sand, there appears t o be littl e data from ful l scale fiel d tests .
1 5 .3 4 PROB L EM S
15.1
15.2
A 45 cm diameter pipe pile of length 12m with closed end is driven into a cohesionless soil
having 0 = 35°. Th e voi d rati o of the soi l i s 0.48 an d G
s
= 2.65. Th e wate r tabl e i s at the
ground surface. Estimate (a) the ultimate base load Q
b
, (b) the frictional load <2 « and (c) the
allowable load Q
a
wi t h F , - 2.5 .
Use th e Berezantse v metho d fo r estimatin g Q
b
. For estimatin g Q~ us e K
s
= 0.75 an d
5=20°.
~
Refer t o Proble m 15.1 . Comput e Q
b
b y Meyerhof' s method . Determin e Q , usin g th e
critical dept h concept , and Q
a
wit h F
?
= 2.5. Al l the other data give n in Prob. 15. 1 remain
the same .
15.3 Estimat e Q
b
b y V esic' s metho d fo r th e pil e give n i n Prob . 15.1 . Assum e I
r
=I
n
= 60.
Determine Q
a
for F
y
= 2.5 and use ( X obtained i n Prob. 15.1 .
15.4 Fo r Problem 15.1 , estimate the ultimate base resistanc e Q
b
by Janbu' s method . Determin e
Q
a
wit h F
s
= 2.5. Us e Q/ obtained i n Prob. 15.1 . Use (/ / = 90°.
15.5 Fo r Problem 1 5.1, estimate Q
b
, Q,, and Q
a
by Coyle and Castello method. All the data given
remain the same .
15.6 Fo r problem 15.1 , determine Q
b
, and Q
a
by Meyerhof' s metho d usin g the relationship
between N
cor
an d 0 given in Fig 12.8 .
,
18.
//&\ //A$*.
10
>m
1
,
6
1
,
2.1
1
m
m
m
\
1 '
XWs /5'W s X*5C X X^C^ v
c
u
= 30 kN/m
2
0 = 0
y = 18. 0 kN/m
3
/ d = 40 cm
c
u
= 50 kN/m
2
0 = 0
y = 18. 5 kN/m
3
c
u
= 15 0 kN/m
2
0 = 0
= 19.0kN/m
J
Figure Prob . 1 5 . 1 1
Deep Foun dat io n I : Pil e Foun dat io n 69 7
15.7 A concret e pil e 40 c m i n diameter i s drive n int o homogeneous san d extendin g t o a grea t
depth. Estimate the ultimate load bearing capacit y an d the allowabl e load wit h F = 3.0 by
Coyle and Castello's method. Given: L = 1 5 m, 0 = 36°, y= 18. 5 kN/m
3
.
15.8 Refe r t o Prob . 15.7 . Estimat e th e allowabl e loa d b y Meyerho f s metho d usin g th e
relationship between 0 and N
cor
give n in Fig. 12.8 .
15.9 A concrete pil e of 1 5 in. diameter, 40 ft long is driven into a homogeneous's stratu m of clay
with th e wate r tabl e a t groun d level . Th e cla y i s o f mediu m stif f consistenc y wit h th e
undrained shear strengt h c
u
= 600 lb/ft
2
. Comput e Q
b
by Skempton' s metho d an d (^by the
or-method. Determine Q
a
for F
s
= 2.5.
15.10 Refe r to Prob 15.9 . Comput e Q, by the A-method. Determine Q
a
by using Q
b
computed in
Prob. 15.9 . Assume y
sat
= 120 lb/ft
2
.
15.11 A pil e o f 4 0 c m diamete r an d 18. 5 m lon g passe s throug h tw o layer s o f cla y an d i s
embedded i n a third layer. Fig. Prob. 15.11 gives the details of the soil system. Compute Q,
by the a - method an d Q
b
by Skempton' s method . Determine Q
a
for F
5
= 2.5.
15.12 A concret e pil e o f siz e 1 6 x 1 6 in. i s drive n int o a homogeneou s cla y soi l o f mediu m
consistency. The wate r table i s at ground level . The undraine d shear strengt h of the soi l i s
500 lb/ft
2
. Determin e the lengt h of pile required to carry a safe load of 50 kips with F
S
= 3.
Use the a - method.
15.13 Refe r t o Prob. 15.12 . Comput e th e require d lengt h of pil e by th e X-method . All th e othe r
data remai n th e same. Assume y
sat
=120 lb/ft
3
.
15.14 A concrete pil e 50 cm in diameter i s driven into a homogeneous mass of cohesionless soil .
The pile i s required to carry a safe load of 700 kN. A static cone penetration tes t conducted
at the sit e gave an average valu e of q
c
= 35 kg/cm
2
along the pil e and 60 kg/cm
2
below th e
base of the pile. Comput e the lengt h of the pile wit h F
s
=3.
15.15 Refe r to Proble m 15.14 . I f the lengt h of the pil e driven is restricted t o 1 2 m, estimat e th e
ultimate load Q
u
and saf e load Q
a
with F
s
= 3. All the other dat a remai n the same .
15.16 A reinforced concrete pile 20 in. in diameter penetrates 40 ft into a stratum of clay and rests
on a medium dense san d stratum. Estimate the ultimate load.
Given: for sand- 0= 35°, y
sat
= 120 lb/ft
3
for cla y y
sat
= 119 lb/ft
3
, c
u
= 800 lb/ft
2
.
Use (a ) th e cc-metho d fo r computin g th e frictiona l load , (b ) Meyerhof' s metho d fo r
estimating Q
b
. The water table is at ground level.
15.17 A ten-stor y buildin g i s t o b e constructe d a t a sit e wher e th e wate r tabl e i s clos e t o th e
ground surface . The foundatio n o f the building will be supporte d o n 30 cm diameter pip e
piles. The botto m o f the pil e ca p wil l be a t a depth o f 1. 0 m below groun d level . The soi l
investigation a t the sit e an d laborator y test s have provided th e saturate d uni t weights, the
shear strengt h values unde r undrained conditions (average), the corrected SPT values, and
the soil profile of the soil t o a depth of about 40 m. The soil profile and the other details ar e
given below.
Dept h (m )
From
0
6.0
22
30
To
6
22
30
40
Soil
Sand
Med. stif f cla y
sand
stiff cla y
v N
'sat cor
k N/ m
3
19 1 8
18
19.6 2 5
18.5
</ >° c (aver age )
k N/ m
2
33°
60
35°
75
698 Chapt e r 1 5
Determine th e ultimat e bearing capacit y of a singl e pil e fo r length s o f (a ) 15m , an d (b )
25 m below th e bottom o f the cap.
Use a = 0.50 an d K
s
= 1.2. Assume S = 0.8 0.
15.18 Fo r a pil e designe d fo r a n allowabl e loa d o f 40 0 k N drive n b y a stea m hamme r (singl e
acting) wit h a rated energy of 2070 kN-cm, what is the approximate terminal set of the pil e
using the ENR formula?
15.19 A reinforce d concret e pil e o f 4 0 c m diamete r an d 2 5 m lon g i s drive n throug h mediu m
dense san d t o a final set of 2. 5 mm, usin g a 40 kN singl e - acting hammer wit h a stroke of
150 cm. Determin e th e ultimat e driving resistance o f th e pil e i f i t i s fitte d wit h a helmet ,
plastic doll y an d 5 0 mm packin g o n th e to p o f th e pile . The weigh t o f th e helmet , wit h
dolly is 4.5 kN. The other particulars are: weight of pile = 85 kN, weight of hammer 35 kN;
pile hamme r efficienc y TJ
h
= 0.85 ; th e coefficien t restitutio n C
r
= 0.45 . Us e Hiley' s
formula. The sum of elastic compression C = c
l
+ c
2
+ c
3
= 20.1 mm.
15.20 A reinforce d concret e pil e 4 5 f t lon g an d 2 0 in . i n diamete r i s drive n int o a stratu m o f
homogeneous saturate d cla y havin g c
u
= 80 0 lb/ft
2
. Determin e (a ) th e ultimat e loa d
capacity an d th e allowabl e loa d wit h F
s
= 3 ; (b ) th e pullou t capacit y an d th e allowabl e
pullout loa d wit h F
s
= 3. Us e the a-metho d fo r estimatin g th e compressio n load .
15.21 Refe r t o Prob. 15.20 . I f the pil e i s driven t o medium dense sand , estimat e (a ) the ultimate
compression loa d an d th e allowabl e load wit h F
S
= 3, and (b ) the pullou t capacity an d th e
allowable pullout load wit h F
s
= 3. Use th e Coyl e an d Castell o method fo r computin g Q
b
and Q,. The other dat a availabl e are: 0 = 36°, an d 7 = 11 5 lb/ft
3
. Assume the water table is
at a great depth .
15.22 A group of nine friction pile s arranged i n a square pattern is to be proportioned i n a deposi t
of mediu m stif f clay . Assuming that the pile s ar e 30 cm diamete r an d 10 m long , fin d th e
optimum spacin g for the piles. Assume a = 0.8 and c
u
- 5 0 kN/m
2
.
15.23 A group of 9 piles wit h 3 in a row was driven into sand at a site. The diameter and length of
the piles are 30 cm and 12m respectively. The properties of the soil are: 0= 30, e - 0.7 , and
G
s
= 2.64.
If th e spacin g o f th e pile s i s 90 cm, comput e th e allowabl e loa d o n th e pil e grou p o n th e
basis of shear failur e fo r F
s
= 2. 0 wit h respect t o ski n resistance, and F
s
= 2.5 wit h respect
to bas e resistance . Fo r 0 = 30° , assum e N = 22.5 an d N = 19.7 . Th e wate r tabl e i s a t
ground level .
15.24 Nin e RC C pile s of diamete r 3 0 cm eac h ar e drive n i n a squar e patter n a t 90 c m cente r t o
center t o a depth of 1 2 m into a stratum of loose to medium dense sand . The botto m o f the
pile ca p embeddin g al l th e pile s rest s a t a dept h o f 1. 5 m below th e groun d surface . At a
depth o f 1 5 m lie s a clay stratu m of thickness 3 m and below whic h lie s sand y strata . Th e
liquid limi t of the clay i s 45%. The saturate d uni t weights of sand and clay ar e 18. 5 kN/m
3
and 19. 5 kN/m
3
respectively . Th e initia l voi d rati o o f th e cla y i s 0.65 . Calculat e th e
consolidation settlemen t o f th e pil e grou p unde r th e allowabl e load . Th e allowabl e loa d
Q = 120 kN.
^•a
15.25 A squar e pil e group consisting of 1 6 piles o f 40 c m diameter passes through two layer s of
compressible soil s as shown in Fig. 15.32(c) . The thicknesses o f the layer s ar e : Lj = 2. 5 m
and L
2
= 3 m. Th e pile s ar e space d a t 10 0 cm cente r t o center . Th e propertie s o f th e fil l
material are : to p fil l c
u
= 2 5 kN/m
2
; th e botto m fil l (peat) , c
u
= 3 0 kN/m
2
. Assum e
7= 1 4 kN/m
3
fo r bot h th e fil l materials . Comput e th e negativ e frictional load o n th e pil e
group.
CH APTER 1 6
DEEP FOUNDATION II :
BEHAVIOR OF LATERALL Y LOADED
V ERTI CAL AND BATTER PILES
16.1 I NTRODU CTI O N
When a soi l o f lo w bearing capacit y extend s t o a considerable depth , pile s ar e generall y use d t o
transmit vertica l an d latera l load s t o th e surroundin g soi l media . Pile s tha t ar e use d unde r tal l
chimneys, television towers, high rise buildings , high retaining walls , offshore structures , etc. ar e
normally subjecte d t o high lateral loads. These pile s o r pile groups shoul d resis t no t onl y vertica l
movements but also lateral movements . The requirements for a satisfactory foundatio n are,
1. Th e vertica l settlemen t o r th e horizonta l movemen t shoul d no t excee d a n acceptabl e
maximum value ,
2. Ther e must not be failur e by yiel d of the surrounding soil or the pile material .
V ertical piles ar e used i n foundations to take normall y vertical loads an d smal l latera l loads .
When the horizontal load per pile exceeds th e value suitable for vertical piles, batter piles ar e used
in combinatio n wit h vertica l piles . Batte r piles ar e als o calle d inclined piles o r raker piles. The
degree of batter, is the angle made by the pile with the vertical, may up to 30°. I f the lateral load act s
on the pile in the direction of batter, it is called a n in-batter or negative batter pile. If the lateral loa d
acts i n the directio n opposit e t o that of the batter , it i s called a n out-batter o r positive batter pile .
Fig. 16 . la show s the two types of batter piles .
Extensive theoretica l an d experimental investigatio n has been conducte d o n singl e vertica l
piles subjecte d t o latera l load s b y man y investigators . Generalize d solution s fo r laterall y loade d
vertical pile s ar e give n by Matloc k an d Rees e (1960) . Th e effec t o f vertica l load s i n additio n t o
lateral load s ha s bee n evaluate d b y Davisso n (1960 ) i n term s o f non-dimensiona l parameters .
Broms (1964a , 1964b ) an d Poulo s an d Davi s (1980 ) hav e given differen t approache s fo r solvin g
laterally loade d pil e problems . Brom' s metho d i s ingeniou s and i s based primaril y on th e us e of
699
700 Chapt e r 1 6
limiting value s o f soi l resistance . Th e metho d o f Poulo s an d Davi s i s base d o n th e theor y o f
elasticity.
The finit e differenc e method of solving the differentia l equatio n for a laterally loade d pil e is
very much in use wher e computer facilities are available . Reese et al. , (1974) and Matlock (1970 )
have developed th e concept o f ( p-y) curve s for solving laterally loaded pil e problems. This method
is quite popular i n the USA an d i n some othe r countries.
However, the work on batter piles is limited as compared t o vertical piles. Three serie s of tests
on singl e ' in' and ' out ' batter piles subjected to lateral loads have been reporte d b y Matsuo (1939).
They wer e run at three scales. The smal l and medium scale test s wer e conducted using timber pile s
embedded i n san d i n th e laborator y unde r controlle d densit y conditions . Loo s an d Bret h (1949 )
reported a fe w mode l test s i n dr y san d o n vertica l an d batte r piles . Mode l test s t o determin e th e
effect o f batter on pile load capacit y have been reported b y Tschebotarioff (1953) , Yoshimi (1964) ,
and Awad and Petrasovit s (1968). The effect o f batter on deflections has been investigate d by Kubo
(1965) and Awad and Petrasovit s (1968) for mode l piles i n sand.
Full-scale fiel d test s o n singl e vertica l and batte r piles , an d als o group s o f piles , hav e bee n
made fro m tim e t o tim e b y man y investigator s i n th e past . Th e fiel d tes t value s hav e bee n use d
mostly t o check th e theorie s formulate d for the behavior o f vertical pile s only . Murthy an d Subb a
Rao (1995 ) mad e us e o f fiel d an d laborator y dat a an d develope d a ne w approac h fo r solvin g the
laterally loade d pil e problem.
Reliable experimental data on batter piles ar e rather scarce compare d t o that of vertical piles.
Though Kubo (1965) used instrumented model piles to study the deflection behavior of batter piles,
his investigation in this field was quite limited. The wor k of Awad and Petrasovits (1968 ) was based
on non-instrumente d piles an d a s such does not throw muc h light on the behavior o f batter piles .
The autho r (Murthy, 1965) conducted a comprehensive serie s o f model test s on instrumented
piles embedde d i n dry sand . The batter used by the author varied fro m -45° t o +45°. A par t of the
author' s study on the behavior of batter piles, based on his own research work , has been include d in
this chapter .
1 6 .2 WI NKL ER' S H Y POTH ESIS
Most o f th e theoretica l solution s fo r laterall y loade d pile s involv e th e concep t o f modulus o f
subgrade reaction o r otherwis e terme d a s soil modulus whic h i s base d o n Winkler' s assumptio n
that a soil mediu m may be approximated by a series o f closely space d independen t elastic springs .
Fig. 16.1(b ) shows a loaded bea m restin g on a elastic foundation . The reactio n a t any point on the
base of th e bea m i s actuall y a function o f ever y poin t along th e bea m sinc e soi l materia l exhibit s
varying degree s o f continuity . Th e bea m show n i n Fig . 16.1(b ) ca n b e replace d b y a bea m i n
Fig. 16.1(c) . I n thi s figur e th e bea m rest s o n a be d o f elasti c spring s wherei n eac h sprin g i s
independent of the other . According t o Winkler's hypothesis , th e reaction a t any point on the base
of the beam i n Fig. 16.1(c ) depends onl y on the deflection at that point. V esic (1961) has shown that
the error inheren t i n Winkler's hypothesi s is not significant .
The proble m o f a laterall y loaded pil e embedde d i n soi l i s closel y relate d t o the beam o n an
elastic foundation . A beam can be loaded a t one or more points along it s length, whereas i n the case
of piles th e external loads an d moment s ar e applie d at or above th e ground surfac e only.
The natur e of a laterally loaded pile-soi l system is illustrated in Fig. 16.1(d ) for a vertical pile .
The sam e principl e applie s t o batte r piles . A serie s o f nonlinea r spring s represent s th e force -
deformation characteristic s of the soil . The springs attached to the blocks o f different size s indicat e
reaction increasin g wit h deflection and then reaching a yield point, or a limiting value that depends
on depth ; th e tape r o n th e spring s indicates a nonlinear variation of loa d wit h deflection . The ga p
between the pile and the springs indicates the molding away of the soil by repeated loading s and the
Deep Foundatio n II : B ehavio r o f Lat erall y Loade d Vert ica l an d B at t e r Pile s 701
'Out' batter o r
positive batte r pil e
y3=Angle of batte r
(a)
'In' batter or
negative batte r pil e
(b)
_ _ - - Surfac e o f
Surface o f assume d
Bearrf
l ast l c

edia
foundatio n
Reactions ar e Closel y
function o f
s
P
ac
f
ed
every point elasti c
along the beam
s
P
nn
g
s
(c)
P.
dUUmH I
JLr| (d)
Figure 1 6. 1 (a ) B at t er piles , (b , c ) Win k ler' s hypot hes i s an d (d ) t he con cep t o f
lat erally loade d pile-s oi l s ys t e m
increasing stiffness o f the soi l is shown by shortening of the springs as the depth below the surface
increases.
16.3 TH E DI FFERENTI AL EQU ATI ON
Compatibility
As state d earlier , th e proble m o f th e laterall y loade d pil e i s simila r t o th e beam-on-elasti c
foundation problem . Th e interactio n between th e soi l an d th e pil e o r th e bea m mus t be treate d
702 Chapt e r 1 6
quantitatively i n th e proble m solution . The tw o condition s tha t mus t b e satisfie d fo r a rationa l
analysis of the proble m are ,
1. Eac h elemen t of the structur e must be i n equilibrium and
2. Compatibilit y mus t b e maintaine d betwee n th e superstructure , th e foundatio n an d th e
supporting soil .
If th e assumptio n i s mad e tha t th e structur e ca n b e maintaine d b y selectin g appropriat e
boundary conditions at the top of the pile, the remaining problem i s to obtain a solution that insures
equilibrium an d compatibilit y of eac h elemen t o f th e pile , takin g int o accoun t th e soi l respons e
along th e pile. Suc h a solution can be made by solvin g the differential equatio n that describes th e
pile behavior .
The Differentia l Equatio n of the Elasti c Curve
The standar d differentia l equation s fo r slope , moment , shea r an d soi l reactio n fo r a bea m o n a n
elastic foundatio n are equall y applicable t o laterall y loaded piles .
The deflection of a point on the elastic curve of a pile is given by y. The *-axis is along the pile
axis and deflection i s measured norma l t o the pile-axis.
The relationship s betwee n y , slope , moment , shea r an d soi l reactio n a t an y poin t o n th e
deflected pil e ma y be writte n as follows.
deflection o f the pil e = y
dy
slope o f the deflected pil e S = — (16.1 )
dx
d
2
y
moment o f pil e M = El—- (16.2 )
dx
2
shear V=EI^-%- (16.3 )
dx*
d
4
y
soil reaction, p - El —- (16.4 )
dx*
where El is the flexural rigidit y of the pile material .
The soi l reactio n p a t any point at a distance x alon g the axi s of the pil e ma y b e expressed a s
p = -E
s
y (16.5 )
where y is the deflection a t point jc, and E
s
is the soil modulus. Eq s (16.4) and (16.5) when combined
gives
s
y = Q (16.6 )
dx*
which i s called th e differential equation for the elastic curve wit h zero axial load .
The key to the solution of laterally loaded pil e problems lie s i n the determination o f the value
of th e modulu s o f subgrad e reactio n (soi l modulus ) wit h respec t t o dept h alon g th e pile .
Fig. 16.2(a ) shows a vertical pile subjected to a lateral load at ground level . The deflecte d positio n
of th e pil e and th e correspondin g soi l reaction curv e ar e als o shown i n the same figure . Th e soi l
modulus E
s
a t any point x below the surface along the pil e as per Eq. (16.5 ) i s
*,=-£(16.7 )
Deep Foundatio n II : B ehavio r o f Lat erall y Loade d Vert ical an d Batte r Pile s
P „„. t. yxxvxxxxv J
"vm c— ^ ^*W5 _ Deflecte d pil e position
Soil reaction curv e
703
(a) Laterall y loaded pil e
o
C/5
Secant
modulus
Depth
Deflection y
(b) Characteristi c shape of a p-y curv e
(c) Form of variation of E
s
wit h depth
Figure 16. 2 Th e con cep t o f (p-y) curves : (a ) a lat erall y loade d pile, (b) charact eris t i c
shape o f a p-ycurve , an d (c) t he for m o f variat io n o f E
s
wit h dept h
As the loa d P
t
at the top of the pil e increase s the deflectio n y and the correspondin g soi l
reaction p increase . A relationshi p between p an d y a t any dept h j c may b e establishe d a s shown in
Fig. 16.2(b) . I t ca n be see n tha t the curve i s strongl y non-linear, changing from a n initia l tangent
modulus E
si
t o an ultimate resistance p
u
. E
S
i s not a constant and changes wit h deflection.
There ar e many factor s tha t influenc e the valu e o f E
s
suc h a s the pil e widt h d, the flexura l
stiffness El, the magnitude of loading P
f
an d the soi l properties .
The variation of E wit h depth for any particular load leve l may be expressed a s
E = n x"
^s
n
h
x (16.8a)
in whic h n
h
i s termed th e coefficient o f soil modulus variation. The valu e of the powe r n depend s
upon the typ e of soi l and th e batter of the pile. Typical curves for the for m of variation of E
s
wit h
depth for value s of n equal to 1/2 , 1 , and 2 are given 16.2(c) . The mos t useful for m of variation of
E i s the linear relationship expressed a s
(16.8b)
which i s normally used by investigators for vertical piles.
704 Chapt er 1 6
Table 16. 1 T ypica l value s of n, fo r cohes iv e soil s (Tak e n from Poulo s an d Davis , 1980 )
Soil typ e
Soft N C cla y
NC organi c cla y
Peat
Loess
n
h
I b/ i n
3
0.6 t o 12.7
1.0 t o 2. 0
0.4 t o 1. 0
0.4 t o 3. 0
0.2
0.1 t o 0. 4
29 t o 40
Reference
Reese an d Matlock , 1956
Davisson an d Prakash , 1963
Peck an d Davisson , 1962
Davisson, 1970
Davisson, 1970
Wilson an d Hi l l s , 1967
Bowles, 1968
Table 16. 1 give s som e typica l value s fo r cohesiv e soil s fo r n
h
an d Fig . 16. 3 give s th e
relationship between n
h
and th e relativ e densit y of sand (Reese , 1975) .
16.4 NON- DI M ENSI ONA L SOL U TI ON S FOR V ERTI CAL PILES
SU B JECTED TO L ATERA L L OADS
Matlock an d Rees e (1960) hav e give n equations for th e determinatio n o f y, S, M, V , and p a t an y
point x alon g th e pil e based o n dimensional analysis. The equations ar e
deflection,
slope,
moment,
shear,
S =
>T
3
~
;
El
P,T
2
El
A +
y
A i
A
s
+
' M,T
2
~
t
El
r
M T
' E
D
_ El
B
M,
T
M.
B.
P
r
., . n — t A + L R
soil reaction, / ' j,
n
p
T
^
2
p
where T i s the relativ e stiffnes s facto r expressed a s
T-
1
(16.9)
(16.10)
(16.11)
(16.12)
(16.13)
(16.14a)
for
For a general cas e
E = n,x
s n
T =
El "+
4
(16.14b)
In Eq s (16.9 ) throug h (16.13) , A an d B ar e th e set s o f non-dimensiona l coefficient s whos e
values are given i n Table 16.2 . The principl e of superposition for the deflection o f a laterally loade d
Deep Foun dat io n II : B ehavio r o f Lat erall y Loade d Vert ica l an d B at t e r Pile s 705
80
70
60
50
40
30
20
10
V ery
loose
Loose
Medium
dense
Dense
V ery
dense
Sand
above
the wate r
table
Sand
below
the water
table
20 4 0 6 0
Relative density, D
r
%
80 100
Figure 16. 3 Variat io n o f n. wit h relat ive den s it y (Rees e , 1975)
pile is shown in Fig. 16.4 . The A and B coefficients ar e given as a function o f the depth coefficient,
Z, expressed a s
Z= -(I6.14c )
The A and B coefficients tend to zero when the depth coefficient Z is equal to or greater than
5 or otherwise the length of the pile is more than 5T. Such piles are called long or flexible piles. The
length of a pile loses it s significance beyond 5T.
Normally w e nee d deflectio n an d slop e a t groun d level . Th e correspondin g equation s fo r
these may be expressed a s
PT
^
El
MT
^
El
(16.15a)
PT
2
MT
S =1.62- — + 1.75—*—
8
El El
(16.15b)
706
Chapt er 1 6
P,
M, M,
Figure 16. 4 Prin cipl e o f s uperpos it io n fo r t h e defl ect io n o f lat erall y loade d pile s
y fo r fixe d head i s
PT
3
8
El
Moment a t ground level for fixe d hea d i s
M
t
= -Q.93[ P
t
T]
(16.16a)
(16.16b)
1 6 .5 p- y CU RV E S FOR TH E SOL U TI ON O F L ATERAL L Y L OADED
PI L ES
Section 16. 4 explains the methods of computing deflection, slope, moment , shea r an d soil reactio n
by making us e of equations develope d by non-dimensional methods . The prediction of the variou s
curves depend s primaril y on th e singl e paramete r n
h
. I f i t i s possibl e t o obtai n th e valu e of n
h
independently fo r each stag e of loading P
r
th e p-y curve s at different depth s alon g the pil e can be
constructed a s follows:
1.
2.
3.
4.
5.
Determine th e value of n
h
for a particular stage o f loading P
t
.
Compute T from Eq . (16.14a ) for the linear variatio n of E
s
wit h depth.
Compute y at specifi c depths x = x
{
,x = x
2
, etc. alon g the pile by making use of Eq. (16.9) ,
where A and B parameters ca n be obtained from Table 16. 2 for various depth coefficients Z.
Compute p b y making use of Eq. (16.13) , sinc e T is known, for each o f the depths x = x^
x = jc
0
, et c .
Since the values of p an d y ar e known at each o f the depths jc
p
x
2
etc. , on e point on the p-y
curve at each o f these depths i s also known.
6. Repea t step s 1 through 5 for different stage s of loading and obtain the value s of p an d y for
each stag e o f loading and plot t o determine p-y curve s at each depth .
The individual p-y curve s obtained by the above procedure a t depths x
{
, x
2
, etc. can be plotted
on a common pai r of axes t o give a family o f curves for the selected depth s below the surface . The
p-y curv e show n i n Fig . 16.2 b i s strongl y non-linear and thi s curve ca n b e predicte d onl y i f th e
Deep Foun dat io n II : B ehavio r o f Lat erall y Loade d Vert ica l an d B at t e r Pile s 707
Table 16. 2 Th e A an d B coef f i ci ents a s obt ai ne d b y Rees e an d Mat l oc k (1956 ) fo r
l ong ver t i ca l pi l e s o n th e assumpt i o n E
s
= n
h
x
Z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.2
1.4
1.6
1.8
2.0
3.0
4.0
5.0
Z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.2
1.4
1.6
1.8
2.0
3.0
4.0
5.0
A
y
2.435
2.273
2.112
1.952
1.796
1.644
1.496
1.353
1.216
1.086
0.962
0.738
0.544
0.381
0.247
0.142
-0.075
-0.050
-0.009
By
1.623
1.453
1.293
1.143
1.003
0.873
0.752
0.642
0.540
0.448
0.364
0.223
0.112
0.029
-0.030
-0.070
-0.089
-0.028
0.000
XL
s
-1.623
-1.618
-1.603
-1.578
-1.545
-1.503
-1.454
-1.397
-1.335
-1.268
-1.197
-1.047
-0.893
-0.741
-0.596
-0.464
-0.040
0.052
0.025
B
s
-1.750
-1.650
-1.550
-1.450
-1.351
-1.253
-1.156
-1.061
-0.968
-0.878
-0.792
-0.629
-0.482
-0.354
-0.245
-0.155
0.057
0.049
0.011
A
m
0.000
0.100
0.198
0.291
0.379
0.459
0.532
0.595
0.649
0.693
0.727
0.767
0.772
0.746
0.696
0.628
0.225
0.000
-0.033
B
m
1.000
1.000
0.999
0.994
0.987
0.976
0.960
0.939
0.914
0.885
0.852
0.775
0.668
0.594
0.498
0.404
0.059
0.042
0.026
A „
V
1.000
0.989
0.966
0.906
0.840
0.764
0.677
0.585
0.489
0.392
0.295
0.109
-0.056
-0.193
-0.298
-0.371
-0.349
-0.016
0.013
B
*
0.000
-0.007
-0.028
-0.058
-0.095
-0.137
-0.181
-0.226
-0.270
-0.312
-0.350
-0.414
-0.456
-0.477
-0.476
-0.456
-0.0213
0.017
0.029
A
p
0.000
-0.227
-0.422
-0.586
-0.718
-0.822
-0.897
-0.947
-0.973
-0.977
-0.962
-0.885
-0.761
-0.609
-0.445
-0.283
0.226
0.201
0.046
B
P
0.000
-0.145
-0.259
-0.343
-0.401
-0.436
-0.451
-0.449
-0.432
-0.403
-0.364
-0.268
-0.157
-0.047
0.054
0.140
0.268
0.112
-0.002
708 Chapt e r 1 6
values o f n
h
are known for each stag e o f loading. Further , th e curve can be extende d unti l the soi l
reaction, /?, reaches a n ultimat e value, p
u
, a t any specifi c dept h x below th e ground surface .
If n
h
values are not known to start with at different stage s of loading, the above method canno t
be followed. Supposing p-y curves can be constructed by some other independent method , then p-y
curves ar e th e startin g point s t o obtai n th e curve s o f deflection , slope , momen t an d shear . Thi s
means we are proceeding i n the reverse direction in the above method. The methods o f constructing
p-y curve s and predicting the non-linear behavior of laterall y loaded pile s ar e beyond the scope of
this book. Thi s metho d ha s been deal t wit h i n detail by Reese (1985) .
Example 1 6 . 1
A steel pip e pil e of 61 cm outside diameter wit h a wall thickness of 2.5 cm is driven int o loose sand
( D
r
= 30%) unde r submerged condition s t o a depth of 20 m. The submerge d uni t weight of the soil
is 8.75 kN/m
3
and the angle of internal friction i s 33°. Th e El value of the pile is 4.35 x 10
11
kg-cm
2
(4.35 x 10
2
MN-m
2
). Comput e the ground line deflection of the pile unde r a lateral loa d o f 268 kN
at ground leve l unde r a free hea d condition using the non-dimensional parameter s o f Matlock an d
Reese. Th e n
h
valu e from Fig . 16. 3 for D
r
= 30%i s 6 MN/m
3
for a submerged condition .
Solution
From Eq . (16.15a )
PT
3
y
= 2.43-^ —for M = 0
*El
FromEq. (16.14a) ,
r-
n
h
where, P
t
- 0.26 8 MN
El = 4.35 x 10
2
MN-m
2
n
h
= 6 MN/m
3
_
4.35 x l O
2
I
- - 2.35 m
6
2.43 x 0.268 x(2.35)
3
n nA
Now v = -~ -— = 0.0194 m = 1.94 cm
yg
4.3 5 x l O
2
Example 1 6 . 2
If the pil e i n Ex. 16. 1 is subjected t o a lateral loa d a t a height 2 m above groun d level , wha t wil l be
the ground line deflection?
Solution
From Eq . (16.15a)
PT
3
MT
2
y = 2.43-^ —+ 1.62
8
El El
Deep Foun dat io n II : B ehavio r o f Lat erall y Loade d Vert ica l an d B at t e r Pile s 70 9
As i n Ex. 16. 1 T = 2.3 5 m, M, = 0.268 x 2 = 0.536 MN- m
2.43 x 0.268 x (2.35)
3
1.6 2 x 0.536 x (2.35)
2
Substituting, y
g
= ^ ^
+
^ ^
= 0.0194 + 0.0110 = 0.0304 m = 3.04 cm.
Example 1 6. 3
If the pile in Ex. 16. 1 i s fixed agains t rotation, calculate the deflection at the ground line.
Solution
UseEq. (16.16a )
_ 0.93P,r
3
y
*~ ~El~
The value s of P
f
Tand E l ar e as given in Ex. 16.1 . Substituting these value s
0.93 x 0.268 x(2.35)
3
4.35 x l O
2
= 0.0075 m = 0.75 cm
1 6.6 B ROM S ' SOL U TI ONS FOR L ATERAL L Y L OADE D PILE S
Broms' (1964a , 1964b ) solution s for laterally loaded pile s deal wit h the following:
1 . Latera l deflections of piles at ground level at working loads
2. Ultimat e lateral resistanc e of piles under lateral loads
Broms' provide d solution s fo r bot h shor t an d lon g pile s installe d i n cohesiv e an d
cohesionless soil s respectively . H e considere d pile s fixe d o r fre e t o rotat e a t th e head . Latera l
deflections a t workin g load s hav e bee n calculate d usin g the concep t o f subgrad e reaction . I t i s
assumed that the deflection increases linearly with the applied loads when the loads applied are less
than one-hal f t o one-third of the ultimate lateral resistance of the pile.
L ateral Deflection s a t Workin g L oad s
Lateral deflections at working loads can be obtained from Fig . 16. 5 for cohesive soi l and Fig. 16. 6
for cohesionles s soil s respectively. For piles i n saturated cohesive soils , th e plot in Fig. 16. 5 gives
the relationships between the dimensionless quantity ( 3L and ( y
o
kdL )IP
t
fo r free-head and restrained
piles, wher e
El = stiffnes s o f pile section
k = coefficien t o f horizontal subgrade reaction
d - widt h or diameter of pile
L = lengt h of pile
A pile i s considered lon g or short on the following condition s
Free-head Pile
Long pil e when ft L > 2.50
Short pile when j B L < 2.5 0
710
Chapt er 1 6
1 2 3 4 5
Dimensionless length , fi L
Figure 16. 5 Chart s fo r calculat in g l at era l defl ect ion a t t h e g roun d s ur fac e of
horizon t ally loade d pil e i n cohes ive s oi l (aft e r B rom s 1964a )
4 6
Dimensionless length, rjL
10
Figure 16. 6 Chart s for cal cul at in g lat eral deflect ion a t t he g roun d s urfac e of
horizon t ally loade d piles i n cohes ion l es s s oil (aft e r B rom s 1964b)
Deep Foundatio n II : B ehavio r o f Lat erall y Loade d Vert ica l an d B at t e r Pile s 711
Fixed-head Pil e
Long pil e whe n ft L > 1. 5
Short pil e when fi L < 1. 5
Tomlinson (1977) suggests that it is sufficientl y accurat e to take the value of k in Eq. (16.17)
as equal t o k\ give n in Table 14.1(b) .
Lateral deflections a t working loads of piles embedded i n cohesionless soil s may be obtained
from Fig . 16. 6 Non-dimensionles s facto r [ v (£7)
3/5
( n
h
)
2/5
] /P
t
L i s plotte d a s a functio n o f r\ L fo r
various value s of e/ L
where y = deflectio n a t ground level
El
1/5
(16.18)
n
h
= coefficient of soi l modulus variation
P
C
= latera l load applie d at or above groun d level
L = lengt h of pile
e = eccentricit y of load.
U ltimate L atera l Resistanc e o f Pile s i n Saturated Cohesiv e Soil s
The ultimat e soil resistance o f piles i n cohesive soil s increases wit h depth from 2c
u
( c
u
= undrained
shear strength) to 8 to 12 c
u
at a depth of three pile diameters (3d) below the surface. Broms (1964a)
suggests a constan t valu e o f 9c
u
belo w a dept h o f l.5d a s th e ultimat e soi l resistance .
Figure 16. 7 give s solutions for shor t pile s an d Fig. 16. 8 for long piles. The solutio n for long piles
0 4 8 1 2 1 6
Embedment length , L id
Figure 16. 7 Ult im at e lat era l res is t an c e o f a s hor t pil e i n cohes ive s oi l relat e d t o
em bedded lengt h (aft e r B rom s (1964a) )
712
Chapt er 1 6
3 4 6 10 2 0 4 0 10 0
Ul t imat e resistance moment,
200 40 0 60 0
Figure 16. 8 Ult im at e l at era l res is t an ce of a lon g pil e i n cohes ive s oi l rel at e d t o
em bedded len g t h (aft e r B rom s ( 1964a) )
200
40
12
Length L id
Figure 16. 9 Ult im at e l at era l res is t an c e o f a s hort pil e i n cohes ion l es s s oi l relat e d t o
em bedded len g t h (aft e r B rom s (1964b) }
involves th e yiel d momen t M, fo r th e pil e section . Th e equation s suggeste d b y Brom s fo r
computing M, are as follows:
Deep Foun dat io n II : B ehavio r o f Lat erall y Loade d Vert ica l an d B at t e r Pile s
1000
713
10 10 0
Ultimate resistance moment , MJ cfyK
1000 10000
Figure 1 6.1 0 Ult im at e lat era l res is t an ce of a lon g pil e i n cohes ion les s s oil relat e d t o
em bedded len g t h (aft e r B rom s (1964b) )
(16.19a)
(16.19b)
For a cylindrical steel pipe sectio n
M
y
=\ 3f
y
Z
For an H-sectio n
M^UfyZ^
where / = yiel d strength of the pile material
Z= sectio n modulu s of the pile section
The ultimat e strengt h o f a reinforce d concret e pil e sectio n ca n b e calculate d i n a simila r
manner.
U ltimate L atera l Resistanc e o f Pile s i n Cohesionless Soil s
The ultimat e lateral resistanc e o f a shor t pile s embedde d i n cohesionles s soi l ca n b e estimate d
making us e o f Fig . 16. 9 an d tha t o f lon g pile s fro m Fig . 16.10 . I n Fig . 16. 9 th e dimensionles s
quantity P
u
/yd
3
K
p
i s plotted against the L id ratio for short piles and in Fig. 16.1 0 P
u
/yd
3
K
p
i s plotted
against M . In both cases th e terms used are
y = effective uni t weight of soi l
Kp = Rankine's passive earth pressure coefficient = tan
2
(45°+0/2)
Example 1 6 . 4
A stee l pip e pil e o f 6 1 c m outsid e diamete r wit h 2. 5 c m wal l thicknes s i s drive n int o saturate d
cohesive soi l to a depth of 20 m. The undrained cohesive strengt h of the soil is 85 kPa. Calculate the
ultimate lateral resistance of the pile by Broms' method with the load applie d a t ground level.
Solution
The pile is considered a s a long pile. Use Fig. 16. 8 to obtain the ultimate lateral resistance P
u
of the
pile.
714 Chapt e r 1 6
M
y
The non-dimensiona l yield moment ~ ,
3
c
u
a
where M
v
= yiel d resistance of the pil e sectio n
1.3 f Z
J y
f = yiel d strengt h o f th e pil e materia l
2800 kg/cm
2
(assumed )
Z= sectio n modulu s = — — [ d
Q
- d
{
]
64 A
I = momen t o f inertia ,
d
g
= outsid e diamete r =6 1 cm,
d
i
= insid e diamete r = 5 6 cm,
R = outsid e radiu s = 30. 5 c m
314
Z = -
:
-[61
4
-56
4
] = 6,452. 6 cm
3
64x30.5
M
y
= 1.3 x 2,800 x 6,452. 6 =23.487 x 10
6
kg-cm .
M 23.48 7 x l Q
6
_
0.85 x61
3
M _
u
From Fig . 16. 8 fo r el d = 0, , , 3 ~ 122 , — ~ 35
dl
P
u
= 35 c
u
d
Q
2
= 35 x 8 5 x 0.61
2
= 1,10 7 k N
Example 1 6 . 5
If the pile give n in Ex. 16. 4 is restrained agains t rotation, calculat e the ultimat e lateral resistanc e P
u
.
Solution
v _
Per Ex. 16. 4 ~ J T~
1 2 2
"
M P
From Fig . 16.8 , fo r —
y
— = 122 , for restrained pil e — ^-50
Therefore p = — x 1,107 = 1,581 kN
" 3 5
Example 1 6. 6
A stee l pip e pil e o f outsid e diamete r 6 1 c m an d insid e diamete r 5 6 c m i s drive n int o a medium
dense san d unde r submerge d conditions . Th e san d ha s a relative densit y o f 60%an d a n angl e o f
internal frictio n o f 38° . Comput e th e ultimat e latera l resistanc e o f th e pil e b y BrorrT s method .
Assume tha t th e yiel d resistanc e o f th e pil e sectio n i s th e sam e a s tha t give n i n E x 16.4 . Th e
submerged uni t weigh t o f th e soi l y
b
=8.7 5 kN/m
3
.
Deep Foun dat io n II : B ehavio r o f Lat erall y Loade d Vert ica l an d B at t e r Pile s 71 5
Solution
From Fig . 16.1 0
Non-dimensional yiel d moment = ^4 ^
where, K
p
= tan
2
(45 + 0/2 ) = tan
2
64 = 4.20,
M
y
= 23.48 7 x 10
6
kg-cm,
Y = 8.7 5 kN/m
3
« 8.75 x 10'
4
kg/cm
3
,
d = 6 1 c m .
Substituting,
M
v
23.48 7 xl 0
6
x! 0
4
. „
= 4o2
8.75 x 61
4
x 4.2
M
y - -"
From Fig . 16.10 , for ,
4
~ ~ 462, for eld - 0 we have
V(
fiv-
Therefore P
u
= 80 yd
3
AT = 80 x 8.7 5 x 0.61
3
x 4. 2 = 667 kN
Example 1 6. 7
If the pile i n Ex. 16. 6 is restrained, wha t is the ultimate lateral resistance o f the pile ?
Solution
M
From Fig . 16.10 , for ,
4
~ -
4&2
, the value
/t** v p
P = 135 Y^ K = 135 x 8.75 x 0.61
3
x 4.2 = 1,12 6 kN.
u I p '
Example 16. 8
Compute the deflection a t ground level by Broms' method for th e pil e give n in Ex. 16.1 .
Solution
FromEq. (16.18 )
1/5 1/ 5
77 = H! L = — =0.42 4
El 4.3 5 x l O
2
r? L = 0. 424x20 = 8.5 .
From Fig . 16.6 , fo r f] L = 8.5, e IL = 0, we have
y£/ )
3/ 5
K)
2 / 5
_
a 2
02P
t
L 0.2x0.268x2 0
y = = — = 0.014 m = 1. 4 cm
y
*( El)
3
'
5
( n. )
2/5
(4.3 5 x 10
2
)
3/5
(6)
2/5
716 Chapt e r 1 6
Example 1 6 . 9
If the pil e given i n Ex. 16. 1 i s only 4 m long, compute the ultimat e lateral resistanc e o f the pil e by
Broms' method .
Solution
Fr omEq. (16.18)
1/5 1/ 5
rj= ^ = ° =0.42 4
El 4.3 5 x l O
2
11 L = 0.424x4=1.696 .
The pil e behaves as an i nfi ni t el y stif f membe r sinc e r\ L < 2.0, L id = 4/0.61 = 6.6 .
From Fig . 16.9 , fo r L id-6.6 , e IL = 0, , we have
P
u
/Y^K
p
= 25.
0 = 33°, y= 8-75 kN/m
3
, d = 61 cm, K = tan
2
(45° + 0/2) = 3.4 .
Now P = 25 yd
3
K = 25 x 8.7 5 x (0.61)
3
x 3. 4 = 16 9 kN
u i p ^ '
If th e san d i s mediu m dense, a s give n i n Ex . 16.6 , the n K = 4.20, an d th e ultimat e latera l
resistance P
u
i s
42
P = — x!69 = 209kN
" 3. 4
As pe r Ex . 16.6 , P
u
fo r a lon g pil e = 66 7 kN , whic h indicate s tha t th e ultimat e latera l
resistance increase s wi t h th e lengt h of th e pil e an d remain s constan t fo r a lon g pile .
1 6 .7 A DI REC T M ETH OD FOR SOL V I NG TH E NON- L I NEAR
B EH AV I OR O F L ATERAL L Y L OADE D FL EXI B L E PI L E PROB L EM S
Key t o the Solutio n
The ke y to the solution of a laterally loaded vertica l pile problem i s the development o f an equation
for n
h
. Th e presen t stat e o f th e ar t doe s no t indicat e an y definit e relationshi p betwee n n
h
, th e
properties of the soil, the pile material, and the lateral loads. However it has been recognize d tha t n
h
depends o n th e relativ e density of soi l for pile s i n san d an d undraine d shear strengt h c for pile s i n
clay. I t is wel l known that the value of n
h
decreases wit h an increase i n the deflection o f the pile. I t
was Palme r e t a l (1948) wh o firs t showe d tha t a change o f widt h d of a pil e wil l have a n effec t on
deflection, momen t and soi l reaction even whil e El i s kept constant for all the widths. The selectio n
of an initial value for n
h
for a particular problem is still difficult an d many times quite arbitrary. Th e
available recommendation s i n this regard (Terzagh i 1955 , an d Reese 1975 ) ar e widel y varying .
The autho r has bee n workin g on thi s problem sinc e a long time (Murthy , 1965) . A n explici t
relationship betwee n n
h
and th e othe r variabl e soi l an d pil e propertie s ha s bee n develope d o n the
principles of dimensiona l analysis (Mur t hy and Subb a Rao, 1995) .
Development o f Expression s fo r n
h
The ter m n
h
may b e expressed a s a function of the following parameter s fo r pile s i n sand an d clay.
Deep Foun dat io n II : B ehavio r o f Lat erall y Loade d Vert ica l an d B at t e r Pile s 71 7
(a) Pile s i n sand
n
h
=f
s
( EI,d,P
e
,Y,® (16.20 )
(b) Pile s i n clay
n
h
=f
c
( EI,d,P
e
,Y,c) (16.21 )
The symbol s use d i n the above expressions have been defined earlier .
In Eqs (16.20) an d (16.21), an equivalent lateral load P
g
a t ground level i s used i n place o f P
t
acting a t a heigh t e abov e groun d level . A n expressio n fo r P
g
ma y b e writte n fro m
Eq. (16.15) as follows.
P = />,( ! + 0.67^) (16.22 )
Now th e equation fo r computing groundline deflection y i s
2.43P T
3
(16
-
23)
Based o n dimensional analysis the following non-dimensiona l groups have been establishe d
for pile s i n sand an d clay.
Piles i n Sand
where C , = correctio n facto r fo r th e angl e o f frictio n 0 . The expressio n fo r C , ha s bee n foun d
separately base d o n a critical stud y of the availabl e data. The expression fo r C\is
C
0
= 3 x 10-
5
(1.316)^° (16.25 )
Fig. 16.1 1 gives a plot of C. versus 0.
Piles i n Clay
The nondimensiona l groups develope d fo r piles i n clay ar e
F
= B -
;
P = — —(16.26 )
In an y lateral load tes t i n the fiel d o r laboratory, th e value s o f El, y, 0 (for sand) an d c (for
clay) are known in advance. From the lateral load tests, the ground line deflection curve P
t
versus y
is known, that is, for any applied loa d P
(
, the corresponding measure d y i s known. The values of T,
n
h
and P
g
ca n be obtained fro m Eq s (16.14a), (16.15) an d (16.22) respectively . C
0
is obtained fro m
Eq. (16.25) for piles i n sand or from Fig. 16. 1 1 . Thus the right hand side of functions F
n
an d F ar e
known a t each loa d level .
A large number of pile test data were analyzed and plots of F
n
versu s F wer e made on log-
log scal e for piles i n sand, Fig. (16.12) and F
n
versu s F fo r piles i n clay, Fig. (16.13). The method
of least squares was used t o determine the linear trend. The equations obtained ar e as given below.
718 Chapt er 1 6
3000 x 10~
2
2000
10 20 3 0
Angle of friction, 0°
40 50
Figure 1 6 .1 1 C . vers us
Piles in Sand
F_= 150. F
Piles in Cla y
F = 125 F.
(16.27)
(16.28)
By substitutin g for F
n
an d F , and simplifying , th e expression s fo r n
h
for pile s i n san d an d
clay ar e obtaine d as
for pile s in sand, n
h
- (16.29)
for pile s i n clay, n , -
pl5
(16.30)
Deep Foundatio n II : B ehavio r o f Lat erall y Loade d Vert ical an d B at t e r Pile s
400
719
10000
Figure 1 6 .1 2 Non dim en s ion a l plo t fo r pile s i n s an d
10000
1000 =
Figure 16.1 3 Non dim en s ion a l plo t fo r pile s i n clay
It can be see n i n the above equations that the numerator s in both cases ar e constants for any
given se t of pil e and soi l properties .
The above two equations can be used to predict the non-linear behavior of piles subjecte d
to lateral load s ver y accurately.
720 Chapt e r 16
Example 1 6.1 0
Solve th e problem i n Exampl e 16. 1 b y the direct method (Murth y and Subb a Rao , 1995) . Th e soi l
is loose sand i n a submerged condition.
Given; El = 4.35 x 1 0
1
' kg- cm
2
= 4.35 x 1 0
5
kN-m
2
d = 6 1 cm, L = 20m, 7 ^ = 8.75 kN/m
3
0 = 33° , P
t
= 268 kN (sinc e e = 0)
Required y a t groun d level
o
Solution
For a pile i n san d fo r th e cas e o f e = 0, us e Eq. (16.29)
n, =
P
e
For 0 = 33°, C
6
= 3 x 10~
5
(1.316)
33
= 0.26 fro m Eq . (16.25 )
150 x0.26 x (8.75)
1
-
5
^4.65 x 10
5
x 0.61 54xl 0
4
54xl 0
4

1 c l l
,
T f
,
= 2,015 kN/ rr r
268
1/5
* 1/ 5
£/
=
415X10 '
2015
Now, usin g Eq. (16.23 )
2.43 x 268 x(2.93)
3
4.35 x l O
5
= 0.0377 m = 3.77 c m
It ma y b e note d tha t the direct metho d give s a greater ground lin e deflection ( = 3.77 cm ) as
compared t o the 1.9 6 c m i n Ex. 16.1 .
Example 1 6 .1 1
Solve the problem i n Exampl e 16. 2 by the direct method. I n thi s case P
t
i s applied a t a height 2 m
above groun d leve l Al l the other dat a remai n th e same .
Solution
From Exampl e 16.1 0
54xl 0
4
For P
e
= P
t
= 268 kN, we have n
h
= 2,015 kN/m
3
, and T = 2.93 m
From Eq. (16.22 )
P =P 1 + 0.67— =26 8 1 + 0.67X— = 391kN
54 x 1 0
4
For P = 391k N, n, = - -= 1,381 kN/m
3
h
Deep Foun dat io n II : B ehavio r o f Lat erall y Loade d Vert ica l an d B at t e r Pile s 72 1
= 3.16m
1,381
As before P =26 8 l + 0.67x— = 382kN
3.16
For P
e
= 382 kN, n
h
= 1,414 kN/m
3
, T = 3.1 4 m
Convergence wil l be reached afte r a few trials . The fina l value s are
P
e
= 387 kN, n
h
= 1718 kN/m
3
, T = 3.02 5 m
Now fro m Eq . (16.23 )
2.43P7
3
2.4 3 x 382 x(3.14)
3
El ~ 4.3 5 x l O
5
= 0.066 m = 6.6 cm
The n
h
valu e fro m th e direc t metho d i s 1,41 4 kN/m
3
wherea s fro m Fig . 16. 3 i t i s
6,000 kN/m
3
. The n
h
from Fig. 16. 3 gives v whic h is 50 percent of the probable valu e and is on the
unsafe side .
Example 1 6 .1 2
Compute the ultimate lateral resistance for the pile given in Example 16. 4 by the direct method. All
the other dat a given in the exampl e remai n th e same .
Given: E l = 4.3 5 x 10
5
kN-m
2
, d = 61 cm, L = 20 m
c
u
= 8 5 kN/m
3
, y
b
= 10 kN/m
3
(assume d for clay)
M
y
= 2,34 9 kN-m; e = 0
Required: The ultimate lateral resistance P
u
.
Solution
Use Eqs (16.30) an d (16.14)
\ 25c
L 5
n= -
" p i-- >
0.2
T
£ /
r =
— (a )
"*
Substituting the known values and simplifyin g
1,600 xl O
5
n
h~ -p& (b )
/
Stepl
]
^
t
P - 1 nnn HV T n , = -
(1000)
1
1,600xlO
5
C A /
,
A 1 X T / 3
Let P = 1 000 kN
n
h rr~ = 5,060 kN/m
J
i^eir, I , UUUKI M ,
h
n^^n^l. 5
435xl Of .
5060
722 Chapt e r 16
For e = 0, from Tabl e 16. 2 and Eq. (16. 1 1) we may writ e
where A
m
= 0.77 (max ) correct t o two decimal places .
For P
t
= 1000 kN , and T= 2.437 m
Af = 0.77 x 100 0 x 2.43 7 = 187 6 kN- m < M .
max y
Step 2
LetP
;
= I S OOk N
n
h
= 275 4 kN/m
3
fro m Eq . (b)
and T = 2.7 5 m from Eq. (a )
Now A/ = 0.7 7 x 1500x2. 7 5 = 3 179 kN-m > M.
ITlaA y
P
U
fo r M , = 234 9 kN-m ca n b e determine d a s
P = 1,000 + (1,500- 1.000) x
(2
'
349
~
1>876)
= l,182kN
(3,179-1,876)
P
u
= 1,100 k N by Brom' s metho d whic h agree s wit h the direct method.
1 6.8 CAS E STU DI ES FO R L ATERAL L Y L OADED V ERTI CAL PI L ES
IN SAN D
Case 1 : M ustang I sland Pil e L oadTes t (Rees e et al. , 1 9 7 4 )
Data:
Pile diameter , d = 2 4 in, steel pip e (drive n pile )
El = 4.85 4 x 10' ° lb-in
2
L = 69f t
e = 1 2 in.
0 = 39 °
Y = 6 6 lb/ft
3
( = 0.0382 lb/in
3
)
M = 7 x 10
6
in-lbs
The soi l wa s fin e silt y san d wit h WT a t ground leve l
Required:
(a) Load-deflectio n curv e ( P
t
vs . y ) and n
h
vs . y
g
curv e
(b) Load-ma x moment curv e (P V s M
max
)
(c) Ultimat e loa d P
u
Solutions:
l50C
||>
r
l5
^fEId
For pile i n sand,
n
h ~ p
e
For 0 = 39°, C
0
= 3x 10~
5
( 1.316)
39
° = 1.34
After subst i t ut i o n and si mpl i fyi ng
1631X10
3
n
h= n (a )
Deep Foun dat io n II : B ehavio r o f Lat eral l y Loade d Vert ica l an d B at t e r Pile s 723
From Eqs( 16.22) and (16.14a) ,
p
e
=
P
t
1 + 0
-
67
7
j_
EI_ 5
n
h
(a) Calculatio n o f G roundline Deflection, y
g
Stepl
Since T is not known t o star t with, assume e = 0, and P
e
= P
t
= 10,000 Ibs
Now, from Eq . (a) , n
h
— = 163 lb/in
3
from Eq . (c),
from Eq . (b) ,
l Ox l O
3
\ _
_ 4.854xlO
10
5
7 = = 49.5 i n
163
P =10x l 0
3
- ~ ~
1 2
49.5
= 11.624xlO
3
Ib s
(b)
(c)
120
100
c
£ 8 0
S?
o
00
Cu
T3
IS
O
60
40
20
EI = 4.854 xl O' °l b- i n
2
,
d = 24 in, e = 12 in,
L = 69 ft , <f> = 39° ,
y = 66 lb/ft
3
1 2 3
Groundline deflection , i n
Reese
P
u
= 102 kip s
Broms
P
u
= 92 kip s
4 8 1 2 x l O
6
Maximum moment , in-l b
(a) P, vs y
g
an d n, , vs >>
g
(b ) P, vs M
max
Figure 16.1 4 M us t an g Is lan d l at era l load t es t
724 Chapt e r 16
Step 2
For P =11. 62xl 0
3
l b,
=
1631x1 0
= 1 4 0 1 b / i n
3
h
1 1.624 x l O
3
As i n Step 1 7=5 1 ins, P
e
= 12.32 x 10
3
Ibs
Step 3
Continue Step 1 and Step 2 unt i l convergence i s reached i n the values of T and P
e
. The fina l value s
obtained fo r P
f
= 1 0 x 10
3
I b ar e T- 51. 6 in, and P
e
= 12.32 x 10
3
Ib s
Step 4
The groun d line deflection may b e obtained from E q (16.23) .
= =
8
El 4.85 4 xl O
10
This deflectio n i s for P
(
= 1 0 x 10
3
Ibs. I n the same wa y th e value s of y ca n b e obtaine d fo r
different stage s o f loadings . Fig . 16.14(a ) gives a plot P, vs. y . Since n
h
is known a t each stag e of
loading, a curve of n
h
vs. y ca n b e plotte d as shown in the same figure .
(b) M aximu m M oment
The calculation s unde r (a ) abov e giv e th e value s o f T fo r variou s load s P
t
. B y makin g us e o f
Eq. (16.11) an d Tabl e 16.2 , momen t distributio n alon g th e pil e fo r variou s load s P ca n b e
calculated. From thes e curve s the maximum moments ma y be obtaine d an d a curve of P
f
vs . M
max
may b e plotte d a s shown in Fig. 16.14b .
(c) Ultimat e L oa d P
u
Figure 16 . 14(b) is a plot of M
max
vs. P
t
. From thi s figure, the value of P
U
i s equal to 10 0 kips for the
ultimate pil e momen t resistanc e o f 7 x 10
6
in-lb . Th e valu e obtaine d b y Broms ' metho d an d b y
computer (Reese , 1986 ) ar e 92 and 10 2 kips respectively
Comments:
Figure 16.14 a give s th e compute d P
t
vs . y curv e b y th e direc t approac h metho d (Murth y an d
Subba Rao 1995 ) an d the observed values . There i s an excellent agreemen t betwee n th e two. In the
same wa y the observed an d th e calculated moments and ultimat e loads agre e well .
Case 2 : Florid a Pil e L oa d Tes t (Davis , 1 9 7 7 )
Data
Pile diameter, d = 5 6 i n steel tub e filled with concret e
El = 132.5 x 10
I 0
l b-i n
2
L = 2 6f t
e - 5 1 ft
0 = 38° ,
Y = 601b/ft
3
M , = 4630ft-kips .
The soi l a t the sit e was medium dense an d wit h water tabl e clos e t o the ground surface .
Required
(a) P
{
vs . y curv e and n
h
vs . y
}
curve
(b) Ultimat e lateral load P
u
Solution
The sam e procedur e a s give n fo r th e Mustan g Islan d loa d tes t ha s bee n followe d fo r
calculating the P
t
vs.y an d n
h
vs. y curves . For getting the ultimat e load P
u
th e P
(
vs . A/
n
Deep Foun dat io n II : B ehavio r o f Lat eral l y Loade d Vert ica l an d B at t e r Pile s 725
100
80
60
•s 4 0
20
\ =84 kips
_ _ (author)
. n
h
vs y
g
Sand
El = 13.25 x 10
! 1
l b-in
2
,
, ,,- • ,-,-. P
u
(Reese ) = 84 kip s
d- 56 in, e - 612 in , " ;_. '
O
. , f
j Kf<\ ^» ° P
H
(Broms) = 84 kips
L = 26 tt, 0 = Jo ,
y = 60 lb/ft
3
1 2
Groundline deflection , i n
0 2 4 6 x l 0
3
Maximum movemen t ft-kip s
(a) (b )
Figure 1 6.1 5 Florid a pil e t es t (Davis , 1977)
is obtained. The valu e of P
U
obtaine d i s equal to 84 kips which is the same a s the ones obtained by
Broms (1964) and Reese (1985) methods. Ther e is a very close agreement betwee n th e compute d
and the observed tes t result s as shown in Fig. 16.15 .
Case 3 : M ode l Pil e Test s i n Sand (M urthy , 1 9 6 5 )
Data
Model pil e test s wer e carrie d ou t t o determin e th e behavio r o f vertica l pile s subjecte d t o latera l
loads. Aluminum alloy tubings, 0.75 i n diameter and 0.035 in wall thickness, were used for the test.
The tes t piles wer e instrumented. Dry clean sand was used for the test at a relative density of 67%.
The othe r detail s ar e given in Fig. 16.16 .
Solution
Fig. 16.1 6 gives the predicted and observed
(a) load-groun d line deflection curve
(b) deflectio n distribution curves along the pil e
(c) momen t and soi l reaction curve s along the pil e
There i s a n excellent agreemen t betwee n th e predicted an d the observe d values . The direct
approach metho d ha s been used .
1 6.9 CAS E STU DI ES FOR L ATERAL L Y L OADE D V ERTI CAL PI L ES
IN CL AY
Case 1 : Pil e loa d test at St . G abrie l (Capazzoli , 1 968 )
Data
Pile diameter, d = 1 0 in, steel pipe filled wit h concret e
726
Chapt er 1 6
Moment, in-l b
0 2 0 4 0 6 0
Deflection, i n
2 4 6 8 x l O "
2
—I I | T | I |
Soil reaction , p, Ib/i n
0 2 4 6
Sand
Murt hy
' = 5.41 x 10
4
lb-in
2
,
d = 0.75 in , e = 0,
L = 30 in, 0 = 40° ,
y = 98 lb/ft
3
/
P, = 20 I b
• Measure d
*J
20
15
5 10
Computed
0 2 4 6 x 10"
2
Ground lin e deflection, i n
Figure 1 6 .1 6 Curve s o f ben din g m om en t , defl ect io n an d s oi l react io n fo r a m odel
pile i n s an d (M urt hy , 1965)
El = 3 8 x 10
8
lb-in
2
L = 115f t
e = 1 2 in .
c = 60 0 lb/ft
2
y = 11 0 lb/ft
3
M
y
= 116f t - ki p s
Water tabl e wa s close t o the ground surface .
Required
(a) P
t
vs. y curv e
(b) the ultimat e latera l load , P
u
Solution
We have ,
(a)
l25c
l5
J EIyd
,1.5 (b)
P
e =
P
,
1 + 0
'
67
7
El
5
(c) T = —
n
h
After substitutin g the known values i n Eq. (a ) and simplifying , we hav e
16045xl 0
3
1.5
Deep Foundatio n II : B ehavio r o f Lat erall y Loade d Vert ica l an d Batte r Pile s 727
(a) Calculation of groundline deflectio n
1. Le t P
e
= P
t
= 500 Ib s
From Eq s (a) and (c), n
h
= 45 lb/in
3
, T = 38.51 in
From Eq . (d) , P
g
= 6044 Ib.
2. Fo r P
e
= 6044 Ib, n
h
= 34 lb/in
3
and T = 41 in
3. Fo r T = 41 in, P
e
= 5980 Ib, and n
h
= 35 lb/in
3
4. Fo r n
h
= 35 lb/in
3
, T = 40.5 in , P
e
=598 8 I b
2A3PJ
3
2.4 3 x 5988 x(40.5)
3
5-
El 38xl 0
8
= 0.25 in
6. Continu e steps 1 through 5 for computing y fo r different load s P
t
. Fig. 16.1 7 gives a plot of
P
t
vs. y whic h agrees ver y wel l wit h the measured values .
(b) Ultimate load P
u
A curve of M
max
vs. P
(
i s given in Fig. 16.1 7 following the procedure give n for the Mustang
Island Test. From thi s curve P
u
= 23 k for M = 116 ft kips. This agrees well with the value s
obtained b y the methods o f Reese (1985 ) and Broms (1964a) .
25
20
C"• * f
t
1 5
of
T3
10
P
u
= 23.00 kips
P
u
(Reese, 1985 ) =21 k
P
u
(Broms) = 24 k
Computed
© Measure d
Clay
£/ =38xl 0
8
l b- i n
2
,
d=10", e=l2",
L= 1 1 5 f t , c = 6001b/ft
2
,
y=1101b/ f t
3
2 4
Groundline deflection, y
g
in
50 100
M,,
150 200 ft-kip s
Maximum moment
Figure 1 6.1 7 St. Gabrie l pil e loa d t es t in cla y
728
Chapt er 1 6
Case 2 : Pil e L oa d Tes t at Ontari o (I smae l and Klym, 1 9 7 7 )
Data
Pile diameter , d - 6 0 in, concrete pil e (Tes t pil e 38)
El = 9 3 x 10
10
lb-in
2
L = 38f t
e = 1 2 in.
c = 20001b/ft
2
Y = 6 0 lb/ft
3
The soi l at the sit e wa s heavily overconsolidate d
Required:
(a)P
r
vs. y curv e
(b) n
h
vs. v curv e
Solution
By substitutin g the known quantities in Eq. (16.30 ) an d simplifying,
68495x10-
n 1. 5
El 5 e
, T= — , an d P
e
= P
t
1 + 0.67 —
2000 T 20 0
1600-•
1200--
800--
400--
0.2 0. 3
Groundline deflection, in
0.4 0.5
Figure 1 6.1 8 On t ari o pil e loa d t es t (38 )
Deep Foun dat io n II : B ehavio r o f Lat erall y Loade d Vert ical an d B at t e r Pile s 729
Follow th e same procedur e a s given for Case 1 to obtain values of y fo r the various loads P
t
.
The loa d deflectio n curve can be obtained fro m th e calculated value s as shown in Fig. 16.18 . The
measured value s ar e als o plotted . I t i s clea r fro m th e curv e tha t ther e i s a ver y clos e agreemen t
between the two. The figur e als o give s the relationship between n
h
and y .
Case 3 : Restraine d Pil e a t the H ea d fo r Offshor e Structur e (M atloc k and
Reese, 1 9 6 1 )
Data
The dat a for the problem ar e taken from Matloc k and Reese (1961) . The pil e i s restrained at
the hea d b y th e structur e on th e to p o f th e pile . Th e pil e considere d i s belo w th e se a bed . Th e
undrained shea r strengt h c an d submerge d uni t weight s ar e obtaine d b y workin g bac k fro m th e
known values of n
h
and T . The other detail s are
Pile diameter, d = 3 3 in, pipe pile
El = 42.3 5 x 10
10
lb-in
2
c = 5001b/ft
2
Y = 4 0 lb/ft
3
P
t
= 150,00 0 Ibs
M -T
(a)
£/ 5
1225 + 1.078 7-
Required
(a) deflection a t the pil e head
(b) moment distributio n diagram
Solution
Substituting the known values in Eq (16.30) an d simplifying,
458 x l O
6
n. -•
Calculations
1. Assum e e = 0,P
g
= P
t
= 150,00 0 Ib
From Eq s (d) and (b) n
h
= 7.9 Ib / in
2
, T= 140 i n
- 140
From Eq . (a)
or
Therefore
12.25 + 1.078x140
M =-0.85 8 P T = Pe
e = 0.858 x 140 = 120 in
P =P
t
1-0.67 — = L5 x l 0
5
1-0.67X —
' T 14 0
= 63,857 I b
33
730 Chapt er 1 6
Now fro m Eq. (d), n
h
= 2.84 lb/in
3
, fro m Eq . (b ) T= 171.6 4 i n
After substitutio n in Eq. (a )
M,
PT
= -0.875 , and e = 0.875 x 171.6 4 = 150. 2 in
P= 1-0.67 X
15
°'
2
x 1.5 x l O
5
=62,20 5 Ibs
171.64
3. Continuin g thi s process for a few more step s ther e wil l be convergence o f value s of n
h
, T and
P
g
. The final values obtained are
n
h
=2.l l b / i n
3
, T= 182.4 in, an d P
e
= 62,246 I b
M, = • ~P
t
e = -150,000 x 150.2 = -22.53 x 10
6
I b - in
2
2A3PJ
3
2.4 3 x 62,246 x(l 82.4)
3
y, =
EI 42.35 x l O
1 0
= 2.17 i n
Moment distributio n along th e pile ma y now be calculated by making us e o f Eq . (16.11)
and Tabl e 16.2 . Please not e that M
t
ha s a negative sign. The moment distribution curve i s given in
Fig. 16.19 . There i s a very close agreemen t between the computed values by direct method and the
Reese an d Matlock method. The deflection and the negative bending moment as obtained by Rees e
and Matloc k ar e
y
m
= 2.307 i n and M
t
= -24.75 x 10
6
lb-in
2
M, Bendin g moment, (10)
6
, in-l b
0 5 -25 -2 0 -1 5 -1 0 - 5
- o Matlock and Reese metho d
Figure 1 6 .1 9 B en din g m om en t dis t ribut io n for a n offs hore pil e s upport e d s t ruct ur e
(M at l ock an d Rees e , 1961)
Deep Foun dat io n II : B ehavio r of Lat eral l y Loaded Vert ical an d B at t e r Pile s 731
16.10 B EH AV I O R OF L ATERAL L Y LOADE D BATTER PILES I N SAND
G eneral Considerations
The earlie r section s deal t wit h th e behavior of long vertica l piles . The autho r has s o far no t come
across any rational approach for predicting the behavior of batter piles subjected to lateral loads. He
has been workin g on thi s problem fo r a long time (Murthy , 1965) . Base d o n the wor k done by the
author and others, a method for predicting the behavior of long batter piles subjected t o lateral load
has no w been developed .
M odel Test s on Piles in Sand (M urthy, 1 9 6 5 )
A series o f seven instrumented model piles were tested i n sand wit h batters varyin g from 0 to ±45°.
Al umi num alloy tubings of 0.75 in outside diameter and 30 in long were used for the tests. Electrica l
resistance gauges were used to measure the flexural strains at intervals along the piles at different loa d
levels. The maximum load applied was 20 Ibs. The pile had a flexural rigidity El = 5.14 x 10
4
lb-in
2
.
The tests were conducted i n dry sand, having a unit weight of 98 lb/ft
3
and angle of friction 0 equal to
40°. Tw o series o f test s wer e conducted-on e serie s wit h loads horizonta l an d th e othe r wit h load s
normal t o the axi s of the pile. The batter s use d wer e 0°, ± 15° , ±30° an d ±45°. Pil e movement s at
ground level were measured wit h sensitive dial gauges. Flexural strains were converted t o moments.
Successive integratio n gave slope s an d deflection s an d successiv e differentiation s gave shear s an d
soil reaction s respectively . A ver y hig h degre e of accurac y was maintaine d throughou t the tests .
Based o n the test results a relationship was established between the n
b
h
values of batter piles and n°.
-30 -15 0°
Batter of pile,
+ 15
C
+ 30°
Figure 16.2 0 Effec t o f bat t e r o n n
b
h
ln°
h
an d n (aft e r M urt hy, 1 965)
7 3 2 Chapt er 1 6
values of vertical piles . Fig . 16.20 gives this relationship between n
b
h
ln°
h
and the angl e of batter /3 . It
is clear from this figure that the ratio increases from a minimum of 0.1 for a positive 30 ° batter pile to
a maximum of 2.2 for a negative 30° batter pile. The values obtained by Kubo (1965) are also shown
in thi s figure. There is close agreemen t betwee n the two.
The othe r importan t facto r i n th e predictio n i s th e valu e o f n i n Eq . (16.8a) . Th e value s
obtained fro m th e experimental tes t results are also give n in Fig. 16.20. The value s of n are equal t o
uni t y fo r vertica l an d negativ e batte r pile s an d increas e linearl y fo r positiv e batte r pile s u p t o a
maximum of 2. 0 at + 30° batter .
In the case of batter piles th e loads an d deflections are considered norma l t o the pil e axi s for
the purpose o f analysis. The corresponding loads and deflections in the horizontal direction ma y be
written a s
P
;
(Hor) =
P
t
( N or)
cos/?
} '
g
(Nor)
cos/3
(16.31)
(16.32)
where P
t
an d y , are norma l t o the pil e axis ; P
f
(Hor) an d y (Hor ) ar e the correspondin g horizonta l
components.
Application o f the U s e o f n
b
h
ln°
h
and n
It is possible no w to predict the non-linear behavior o f laterally loaded batte r piles in the same way
as for vertical pile s by making use of the ratio n
b
h
ln°
h
and the value of n. The validit y of this metho d
is explained b y considering a few case studies .
Case Studies
Case 1 : M ode l Pil e Tes t (M urthy , 1 9 6 5 ) .
Piles o f +15 ° and +30 ° batter s hav e bee n use d her e t o predic t th e P
t
vs . y an d P
{
vs . M
max
relationships. Th e properties o f the pil e and soi l ar e given below.
El = 5. 14 x 10
4
I b in
2
, d = 0.75 in, L = 30 in; e = 0
For 0 = 40°, C. = 1.767 [= 3 x 10'
5
(1.316)*]
150C>
From Eq. (16.29) , n° = *—
1.5
After substitutin g the known values and simplifyin g w e hav e
„ 70 0
Solution: +15 ° batte r pil e
From Fig. 16.20 n
b
h
/n°
h
=OA, n = 1.5
FromEq.
T = l
b
1.5+ 4
-5.33
5.14
Deep Foundatio n II : B ehavio r o f Lat erall y Loade d Vert ica l an d B at t e r Pile s 733
Calculations o f Deflection y
For P
{
= 5 Ibs, n\ = 14 1 lbs/in
3
, n\ = 141 x 0.4 = 56 lb/in
3
and T
b
= 3.5 in
2.43F
3
r
3
y, =
-A- = 0.97 xl 0~
2
in
8
5.1 4 xl O
4
Similarly, y ca n b e calculate d for P
{
= 10, 1 5 and 2 0 Ibs .
The result s ar e plotted i n Fig. 16.21 alon g with th e measured value s ofy . There i s a clos e
agreement between the two.
Calculation o f Maximum Moment , A/
max
For P
t
= 5 Ib, T
b
= 3.5 in, The equation for M i s [Eq. (16. 1 1)]
where A
m
= 0.77 (max) from Tabl e 16.2
By substitutin g and calculating, we have
Similarly M
(max)
can be calculated for other loads. The results are plotted in Fig. 16.21 alon g
with the measured value s of M
(max)
. There i s very close agreemen t between the two.
+ 30°B atte r Pil e
From Fig. 16.20 , n
b
,ln°, = 0.1, and n = 2; T, = — = 4.64 0
n n o h
5.14
0.1667
700
For P
(
= 5 Ibs, n°
h
= 14 1 lbs/in
3
, n\ = 0.1 x 14 1 = 14. 1 lb/in
3
, T
b
= 3.93 in.
For P
t
= 5 Ibs, T
b
= 3.93 in , we have, y
g
= 1.43 x 10~
2
i n
As before, M
(max)
= 0.77 x5 x 3.93 = 1 5 in-lb.
The values ofy an d M
(max)
for other loads can be calculated in the same way. Fig. 16.21 gives
P
t
vs . y an d P
t
vs . M
(max)
alon g wit h measure d values . Ther e i s clos e agreemen t u p t o abou t
-+ 30°
©
Measured
I
4 8 12xl ( T
T
Groundline deflection, y, in
0 4 0 8 0 12 0
Maximum moment , in-lb s
Figure 1 6.2 1 M ode l pile s o f bat t e r + 15 ° an d + 30° (M urt hy , 1965)
734
Chapt er 1 6
P
t
= 10 Ib, an d beyon d thi s load , th e measure d value s ar e greate r tha n th e predicte d b y abou t 2 5
percent whic h is expected since the soi l yield s at a load higher than 1 0 Ib at thi s batter an d ther e i s
a plasti c flow beyond this load.
Case 2 : Arkansa s River Proj ec t (Pile 1 2 ) (Alizade h and Davisson, 1 9 7 0 ) .
Given:
EI = 278.5 x 10
8
l b -in
2
, d = 14 in, e = Q.
0 = 41°, 7=63 lb/ft
3
, j 3=18. 4°(- ve )
From Fig. 16.11 , C
0
= 2.33, fro m Fig . 16.2 0 n
b
h
/n°
h
= 1.7 , n= 1. 0
From Eq. (16.29) , after substitutin g the known values an d simplifying, we have,
i0.2
(a)
n" =
1528 x l O
3
and (b )
T
b
= 39. 8
278.5
Calculation for P. = 12.6
k
From Eq. (a), n\ =\ 2\ lb/in
3
; now n
b
h
- 1. 7 x 12 1 = 206 lb/in
3
From Eq. (b), T
h
- 42.2 7 in
_2.43xl2,600(42.27)
3
_
y
*~ 278. 5xl O
8
M,.
0.083 in
' (max) = °-
77
P
t
T=0.77 x 12. 6 x 3.5 2 = 34 ft-kips .
The value s of y an d M, , for P, = 24.1*, 35.5*, 42.0*, 53.5*, 60 * can b e calculate d i n th e J
g ( ma x ) t
same wa y th e result s ar e plotte d th e Fig. 16.2 2 alon g wit h th e measure d values . Ther e i s a very
close agreement betwee n th e computed and measured value s of y but the computed value s of M
max
200 T
Arkansas River projec t
Pile No 1 2
0.4 0. 8
Ground lin e deflection, i n
1.2
El = 278.5 x 10
8
lb-in
2
, d = 14 in,
= 41° y = 63 lb/ft
3
60
20
— Compute d
© Measured
100 20 0 30 0
Maximum movement ft-kip s
Figure 1 6.2 2 Lat era l l oad t es t -bat t er pil e 12-Ar k an s a s Rive r Proj ect (Al izade h an d
Davis s on , 1970 )
Deep Foundatio n II: B ehavio r o f Lat erall y Loade d Vert ica l an d B at t e r Pile s 735
are higher tha n the measured value s a t higher loads . At a load o f 60 kips, M,. i s higher tha n the
measured b y about 23 %which i s quite reasonable .
Case 3 : Arkansa s River Proj ec t (Pil e 1 3 ) (Alizade h and Davisson, 1 9 7 0 ) .
Given:
El = 288 x 10
8
Ib-ins , d = 14", e = 6 in.
y = 63 lbs/ft
3
, 0 = 41°(C
0
= 2.33)
/3= 18.4 ° (+ve) , n = 1.6, n
b
h
/n°
h
= 0.3
£/ 1.6+ 4 28 8
I L , ~~" £ I ,
b
<
0.1786
(a)
After substitutin g the known values in the equation fo r n°
h
[Eq. (16.29) ] and simplifying, we
have
1597 x l O
3
(b)
Calculations for v fo r P
(
= 141.4k
1. Fro m Eq (b), n\ = 39 lb/in
3
, hence n\ = 0.3 x 39 = 11. 7 lb/in
3
From Eq. (a), T
b
« 48 in.
40 T 10 0
Arkansas Rive r projec t
Pile No 1 3
El = 2.785 x 10' ° lb-in
2
, d = 14 in
0 = 41°,y =
1
0 0. 4 0. 8 1. 2
Ground lin e deflection, i n
100 20 0
Moment ft-kip s
300
Figure 16.2 3 Lat era l loa d t es t -bat t e r pil e 13-Ark an s a s Rive r Proj ec t (Al izade h an d
Davis s on , 1970)
736 Chapt e r 1 6
2 P=P. 1 + 0.67— =41. 4 1 + 0.67X — = 44.86 kip s
z
"
e
' T 4 8
F
3. Fo r P
e
= 44.86 kips, n° = 36 l b / i n
3
, andn * = 11 l b / i n
3
, T
b
~ 48 i n
4. Fina l values : P
e
= 44.86 kips, n
b
h
- 1 1 Ib / in
3
, an d T
b
= 48 in
2. 43PT
3
2.4 3 x 44,860 x(48)
3
nA
.
5
- y
S
= -?T^ = -
!
-r-
1
^ 0. 42m.
El 28 8 x l O
8
6. Follo w Step s 1 t hrough 5 for other loads. Computed an d measured value s of _ y are plotted i n
Fig. 16.2 3 and ther e is a very close agreement between the two. The n
h
values against y
>
are
also plotte d i n th e same figure .
Calculation of Moment Distributio n
The momen t a t any distance x alon g the pil e may be calculated by th e equation
As per the calculations shown above, the value of Twil l be known for any lateral load leve l P .
This mean s [ P
{
T\ wil l b e known . The value s of A an d B ar e function s of th e dept h coefficien t Z
which ca n b e take n fro m Tabl e 16. 2 fo r th e distanc e x( Z = x/T). Th e momen t a t distanc e x wil l be
known fro m th e abov e equation . In the same wa y moment s ma y b e calculate d fo r othe r distances .
The sam e procedur e i s followe d fo r othe r loa d levels . Fig . 16.2 3 give s th e compute d momen t
distribution alon g th e pil e axis. Th e measure d value s o f M ar e shown fo r two loa d level s P
t
= 61.4
and 80. 1 kips . Th e agreemen t betwee n th e measure d an d th e compute d value s is ver y good .
Example 1 6 .1 3
A stee l pip e pil e of 6 1 c m diamete r i s drive n verticall y into a medium dens e san d wit h the wate r
table clos e t o the ground surface. The following data ar e available :
Pile: El = 43. 5 x 10
4
kN-m
2
, L = 2 0 m , th e yiel d momen t M o f th e pil e materia l
= 2349 kN-m .
Soil: Submerge d uni t weigh t y
b
= 8.75 kN/m
3
, 0 = 38° .
Lateral loa d i s applied at ground level ( e = 0)
Determine:
(a) The ultimat e lateral resistance P
u
o f the pil e
(b) The groundl i n e deflection y , at the ultimat e lateral load level .
o
Solution
From Eq . (16.29) th e expression for n
h
is
n, = sinc e P - P , for e - 0 (
a
\
n r > e i \" /
From Eq . (16.25 ) C ^ = 3x 10"
5
(1.326)
38
° = 1.02
Substituting th e known values for n, w e have
150 x 1.02 x(8. 75)
L5
V 43. 5x! 0
4
x 0.61 204xl 0
4
T /
,
n
: = kN/m
J
Deep Foun dat io n II : B ehavio r o f Lat eral l y Loade d Vert ica l an d B at t e r Pile s 73 7
(a) Ultimate latera l loa d P
u
Step 1 :
Assume P
u
= P
(
= 100 0 k N (a )
204 x 10
4
Now fro m Eq . (a ) n, = = 2040 kN/m
3
M h
100 0
i i \ _
Fl
n+4
F J
1+ 4
El
5
FromEq. (16.14a) T = — = — = —
n
h
n
h
n
h
\ _
43<5 x 10
4
5
Substituting and simplyfiin g T = = 2 92 m
2040
The moment equation for e = 0 may be written as (Eq. 16.11 )
Substituting an d simplifyin g we have (where A
OT
(max) = 0.77 )
M
max
= 0.77(1000 x 2.92) = 2248 kN - m
which i s less tha n M
y
= 2349 kN-m.
Step 2:
Try P
t
= 1050 kN .
Following the procedure give n in Step 1
T = 2. 95m for P
t
- 1050k N
NowM
max
= 0.77(1050x2.95) = 2385 k N- m
which i s greater tha n M = 2349 kN-m.
The actua l value P
U
lie s between 100 0 and 105 0 kN whic h can be obtained by proportion as
P = 1000 + (1050 -1000) x
(2349
~
2248)
= 1037 kN
(2385-2248)
(b) Groundline deflection fo r P
u
= 1037 k N
For thi s the value T is required at P
U
= 1037 kN. Following th e same procedure as in Step 1 , we get
T= 2. 29m.
Now from Eq. (16.15a) for e = 0
^^P.T
3
2.4 3 x 1037 x(2.944)
3
rt1
^
0
v
p
= 2.43—— = — = 0.1478 m = 14.78 c m
*El 43. 5 x l O
4
Example 1 6 .1 4
Refer t o Ex . 16.13 . I f th e pip e pil e i s drive n a t a n angl e o f 30 ° t o th e vertical , determin e th e
ultimate latera l resistanc e an d th e correspondin g groundlin e deflection fo r th e loa d applie d (a )
against batter , an d (b ) i n th e directio n o f batter.
In bot h th e case s th e loa d i s applied norma l t o th e pil e axis .
All th e other dat a given in Ex. 16.1 3 remain the same .
738 Chapt e r 16
Solution
From Ex . 16.13 , the expression for n
h
for vertical pile is
204 x l O
4
n, = n°, =
h h
, ,
kN/m
3
+ 30°Batter pil e
From Fig. 16.20 -A_ = 0.1 and n = 2
n
°k
i i j _
£/ "
+4
£ Y
2+ 4
£ 7
6
FromEq. (16. 14b) T=T
b
= — = — = — (c )
2
/i
Determination o f P
U
Stepl
Assume P
e
= P, = 500 kN.
Following the Ste p 1 in Ex. 16.13 , and using Eq. (a ) above
n° = 4,083 kN/m
3
, henc e nf = 4083x0.1« 408 kN/m
3
435xl 0
4 6
Form Eq (c), r = — : = 3.2 m
b
40 8
As before, M
max
= 0.11P
t
T
b
= 0.77 x 500 x 3. 2 = 1232 kN- m < M
y
Step 2
Tr y/ » , = !,000 kN
Proceeding i n the same way as given in Step 1 we have T
b
= 3.59 m, M
max
= 2764 kN-m which
is more than M . The actua l P
u
i s
(2349-1232)
P = 500 + (1000- 500) x^ ^=86 5 k N
(2764-1232)
Step 3
As befor e the correspondin g T
b
for P
U
= 865 kN i s 3. 5 m.
Step 4
The groundline deflection is
+b
2.4 3 x 865 x(3.5)
3
y
+b
= — = 0.2072 m = 20.72 cm
8
43. 5 x l O
4
- 30 °Batter pil e
n~
b
From Fig. 16.20, -2— = 2.2 and n = 1. 0 (d )
n
°h
Deep Foundatio n II : B ehavio r o f Lat erall y Loade d Vert ica l an d Batte r Pile s 73 9
b =
El "
+4
E l
1+ 4
E l
5
^
=
~^
=
~^
n
h
n
h
n
h
Determination of P
u
Step I
Tr yP, = 1000
From Eq. (a ) n°
h
= 2040 kN/m
3
and from Eq . (d ) n~
b
= 2.2x2040 = 4488 kN/m
3
Now fro m Eq . (e), T
b
= 2.5 m
As before M
maY
= 0.77 x 1000 x2. 5 = 1925 kN- m
mSX
which i s less than M = 2349 kN-m
Step 2
Tr yP, = l, 500k N
Proceeding a s in Step \ ,T
b
= 2.1l m , and M
max
= 0.77 x 1500 x 2.71 = 3130 kN-m which is
greater tha n M .
Step 3
The actua l value of P
u
i s therefore
(2349-1925)
P
u
=100 0 + (1500- 1000) x- --= 1253 kN
(2764-1925)
Step 4
Groundline deflectio n
T
b
= 2. 58m for P
u
=l 253 kN
_£2.43x 1 176 x(2.58)
3
n
^
M
Nowy/ = --
A
-— = 0.1202 m =12.0 cm
g
43. 5 x l O
4
The above calculations indicate that the negative batter piles are more resistant to lateral loads
than vertical or positive batter piles. Besides, the groundline deflections of the negative batter piles
are less than the vertical and corresponding positive batter piles .
1 6 .1 1 PROB L EM S
16.1 A reinforced concrete pil e 50 cm square in section i s driven into a medium dense sand to a
depth of 20 m . The san d i s in a submerged state. A lateral load of 50 kN i s applied o n the
pile a t a height o f 5 m above the ground level. Comput e the latera l deflectio n o f the pil e
at ground level. Given: n
h
= 15 MN/m
3
, El = 1 15 x 10
9
kg-cm
2
. The submerged uni t weight
of the soi l i s 8. 75 kN/m
3
.
16.2 I f the pile given in Prob. 16. 1 is full y restraine d at the top, what is the deflection at ground
level?
16.3 I f th e pil e give n i n Prob . 16. 1 is 3 m long , wha t wil l b e th e deflectio n a t groun d leve l
(a) when th e to p o f th e pil e i s free , an d (b ) whe n the to p o f th e pil e i s restrained ? Us e
Broms' method .
740 Chapt e r 16
16.4 Refe r to Prob. 16.1 . Determine the ultimate lateral resistance o f the pile by Broms' method .
Use 0 = 38°. Assume M
y
= 250 kN-m.
16.5 I f the pile given in Prob. 16. 1 is driven into saturated normall y consolidated cla y having an
unconfmed compressiv e strengt h of 70 kPa, wha t woul d be the ultimat e lateral resistanc e
of th e soi l unde r (a ) a free-hea d condition , an d (b ) a fixed-condition ? Make necessar y
assumptions fo r the yiel d strengt h of the material .
16.6 Refe r to Prob. 16.1 . Determine the lateral deflection of the pile at ground level by the direc t
method. Assume m
y
= 250 kN-m and e = 0
16.7 Refe r t o Prob . 16.1 . Determin e th e ultimat e latera l resistanc e o f th e pil e b y th e direc t
method.
16.8 A precas t reinforce d concret e pil e o f 3 0 c m diamete r i s drive n t o a dept h o f 1 0 m i n a
vertical directio n int o a medium dens e san d whic h i s in a semi-dry state . The valu e of the
coefficient o f soil modulu s variation ( n
h
) ma y be assumed a s equal t o 0.8 kg/cm
3
. A lateral
load o f 40 kN is applied a t a height of 3 m above groun d level . Comput e (a ) the deflectio n
at ground level , and (b ) the maximu m bending momen t o n the pil e (assum e E = 2.1 x 10
5
kg/cm
2
).
16.9 Refe r to Prob. 16.8 . Solv e th e problem b y the direc t method . Al l the othe r dat a remai n the
same. Assume 0 = 38° and y = 16. 5 kN/m
3
.
16.10 I f th e pil e i n Prob . 16. 9 i s drive n a t a batte r o f 22.5 ° t o th e vertical , an d latera l loa d i s
applied a t ground level, compute th e normal deflectio n a t ground level , for the cases of the
load actin g i n the direction o f batter and agains t the batter .
CHAPTER 1 7
DEEP FOUNDATION III :
DRILLED PIER FOUNDATIONS
1 7 .1 I NTRODU CTI O N
Chapter 1 5 dealt wit h piles subjected to vertical loads and Chapter 16 with piles subjected to lateral
loads. Drille d pie r foundations , the subjec t matter of thi s chapter, belong t o the same categor y a s
pile foundations. Because pier s and piles serve the same purpose, no sharp deviations can be made
between th e two . The distinction s ar e base d o n the metho d o f installation. A pil e i s installed b y
driving, a pier by excavating. Thus, a foundation unit installed i n a drill-hole ma y als o be called a
bored cast-in-sit u concret e pile . Here, distinction is made between a small diameter pil e and a large
diameter pile. A pile, cast-in-situ, with a diameter less than 0.75 m (or 2.5 ft) is sometimes calle d a
small diamete r pile . A pile greate r tha n thi s siz e i s called a large diamete r bored-cast-in-sit u pile .
The latte r definitio n i s use d i n mos t non-America n countrie s wherea s i n th e USA , suc h large -
diameter bore d pile s ar e calle d drille d piers , drille d shafts , an d sometime s drille d caissons.
Chapter 1 5 deals wit h small diameter bored-cast-in sit u piles i n addition to driven piles .
1 7 .2 TY PE S OF DRILLED PIERS
Drilled pier s ma y b e describe d unde r fou r types . Al l fou r type s ar e simila r i n constructio n
technique, bu t diffe r i n thei r desig n assumption s an d i n th e mechanis m o f loa d transfe r t o th e
surrounding earth mass . These type s are illustrated in Figure 17.1 .
Straight-shaft end-bearing piers develo p thei r suppor t fro m end-bearin g o n stron g soil ,
"hardpan" or rock. The overlyin g soi l i s assumed t o contribut e nothin g t o the suppor t of the loa d
imposed on the pier (Fig. 17. 1 (a)).
741
742
Chapt er 1 7
Q Q
Fill o r
poor
bearing
soil
V
» *
^ 4 *. A^** / £^
Sound rock — ^
(a) Straight -
*\
\
\
\
Roughened o r }
grooved \
sidewall t o , jj
transmit t N
shear ' j j
(c) Straight-sha
sidewall sh<
v
v
v v
' uu
shaft
v ^
V V
H
v
%l
m
ft pi e
jar an
-i
> v V t w i
:'**" * ' * Fil l
Soil \
\
> xiWiiiW !
T: Roc k i
1 . J l\
*rvv:.v» ^«l/ Shea r suppor t — \ ^
-
>
^- En d bearin g > ,
end-bearing pier (b ) Straight-s h
,V ' ' Fi l l V »
^\^\
%
v
\/ /
\/
^ Soi l //
v
v
*V
v *
^
v
^ v
\
^ \ f^ *1
\
\
\
\
^ Roughene d or grooved
I — sidewal l t o transmi t
i i shea r
; N o en d suppor t
/"" (assumed)
"-v^
aft sidewall-shea r pier
I
2
\ r-h-
v V ~
V '
*V >
,' Poo r bearing
, soi l
%>' t» \ w v
5
^
S
*
X
*
!
*^ \\\^^\" ^ V ^\\\NV s^-SV \ N
Soft t o sound A
v
V v
b ^ Goo d bearin g
rock S £
v
^ X
S
oil
^ Shea r suppor t ^V \
|<-fc» ^ ; —•>- End bearing
f M M t t
^- End bearing
' with both ( d) Underreame d (or belled) pie r (30° "bell")
d en d bearin g
(e) Shap e of 45°"bell"(f ) Shape of domed "bell"
Figure 1 7. 1 T ype s o f drille d pier s an d un derream s hape s (Woodwar d e t al. , 1972 )
Straight-shaft side wall friction piers pass through overburden soils that are assumed t o carry
none of the load, an d penetrat e far enough into an assigned bearing stratum to develop design load
capacity b y side wal l friction between the pier and bearing stratu m (Fig. 17.1(b)) .
Combination of straight shaft side wall friction and end bearing piers are of the sam e
construction a s the two mentioned above, but with both side wal l friction an d end bearing assigne d
a role i n carrying the design load. When carried int o rock, thi s pier may be referred t o as a socketed
pier or a "drilled pier wit h rock socket " (Fig. 17.1(c)) .
Belled o r under reamedpiers ar e pier s wit h a botto m bel l o r underrea m (Fig . 17.1(d)). A
greater percentag e o f the imposed load on the pier top i s assumed t o be carried b y the base .
Deep Foun dat io n III : Drille d Pie r Foun dat ion s 74 3
1 7.3 ADV ANTAG E S AND DI SADV ANTAG E S O F DRI L L ED PIER
FOU NDATI ONS
Advantages
1. Pie r o f any lengt h and siz e can be constructed a t the sit e
2. Constructio n equipment is normally mobile and construction can proceed rapidl y
3. Inspectio n o f drilled holes i s possible becaus e o f the larger diameter o f the shaft s
4. V er y large load s ca n b e carrie d b y a singl e drille d pie r foundatio n thus eliminating the
necessity of a pile cap
5. Th e drilled pier i s applicable t o a wide variety of soi l conditions
6. Change s ca n be made i n the design criteria during the progress of a job
7. Groun d vibratio n tha t i s normall y associate d wit h drive n pile s i s absen t i n drille d pie r
construction
8. Bearin g capacit y can be increase d by underreaming the bottom (i n non-caving materials)
Disadvantages
1. Installatio n o f drille d pier s need s a carefu l supervisio n an d qualit y contro l o f al l th e
materials use d i n the construction
2. Th e metho d i s cumbersome. I t needs sufficien t storag e spac e fo r al l th e material s use d i n
the construction
3. Th e advantage of increased bearing capacity due to compaction i n granular soil that could
be obtained i n driven piles i s not there i n drilled pie r construction
4. Constructio n of drilled piers at places wher e there i s a heavy current of ground water flo w
due t o artesian pressure i s very difficul t
1 7.4 M ETH ODS O F CONSTRU CTI ON
Earlier M ethod s
The us e o f drille d pier s fo r foundation s started i n th e Unite d State s durin g th e earl y par t o f th e
twentieth century. The two most common procedures wer e the Chicago and Gow methods shown in
Fig. 17.2 . I n th e Chicag o metho d a circula r pi t wa s excavate d t o a convenien t dept h an d a
cylindrical shel l o f vertica l boards o r stave s wa s placed b y makin g us e o f a n insid e compressio n
ring. Excavation the n continued t o the next board lengt h and a second tier of staves wa s set and the
procedure continued . The tier s coul d be set at a constant diameter or stepped i n about 50 mm. The
Gow method, which used a series of telescopic meta l shells, i s about the same as the current method
of using casing except fo r the telescoping sections reducing the diameter on successive tiers .
M odern M ethod s o f Constructio n
Equipment
There has been a phenomenal growt h in the manufacture and use of heavy duty drilling equipmen t
in th e Unite d State s sinc e th e en d o f Worl d Wa r II . Th e greates t impetu s t o thi s developmen t
occurred i n tw o states , Texa s an d Californi a (Woodwar d e t al. , 1972) . Improvement s i n th e
machines wer e made responding to the needs of contractors. Commercially produced drilling rigs
of sufficien t siz e an d capacit y t o dril l pier hole s come i n a wide variet y of mountings an d driving
arrangements. Mountings are usually truck crane, tractor or skid. Fig. 17. 3 shows a tractor mounted
rig. Drilling machine rating s as presented i n manufacturer' s catalogs an d technical dat a sheet s are
744 Chapt er 1 7
Staves
Compression
ring
< V <
Telescoping
metal
Dig by
hand
(b) The Go w metho d
(a) The Chicag o metho d
Figure 1 7 . 2 Earl y m et hods o f cais s o n con s t ruct io n
usually expresse d a s maximu m hol e diameter , maximu m depth , an d maximu m torqu e a t som e
particular rpm .
Many drille d pie r shaft s throug h soil or sof t roc k ar e drille d wit h the open-heli x auger . Th e
tool ma y b e equippe d wit h a knif e blad e cuttin g edge fo r us e i n mos t homogeneou s soi l o r wit h
hard-surfaced teet h for cutting stif f o r hard soils , stony soils , or soft t o moderately hard rock. These
augers ar e availabl e i n diameter s u p t o 3 m o r more . Fig . 17. 4 show s commerciall y availabl e
models.
Underreaming tool s (o r buckets ) ar e availabl e in a variet y o f designs . Figur e 17. 5 show s a
typical 30° underreamer with blade cutte r for soils that can be cut readily. Most suc h underreaming
tools are limited in size t o a diameter thre e times th e diameter of the shaft .
When roc k become s to o har d t o b e remove d wit h auger-typ e tools , i t i s ofte n necessar y t o
resort t o the us e o f a core barrel . This too l i s a simpl e cylindrical barrel , se t wit h tungsten carbid e
teeth a t th e botto m edge . For har d rock whic h cannot b e cu t readil y wit h th e cor e barre l se t with
hard meta l teeth , a calyx or shot barrel ca n be use d t o cut a core o f rock.
G eneral Constructio n M ethod s o f Drille d Pie r Foundation s
The rotar y drillin g method i s the mos t commo n metho d o f pier constructio n i n th e Unite d States .
The method s of drille d pier construction can be classified i n three categories a s
1. Th e dr y metho d
2. Th e casin g metho d
3. Th e slurr y method
Deep Foun dat io n III : Drille d Pie r Foun dat ion s 74 5
Figure 17. 3 T ract o r m oun t e d hydrauli c drillin g ri g (Court es y: Kell y T ract o r Co ,
USA)
Dry M et ho d o f Con s t ruct io n
The dry method is applicable to soil and rock that are above the water table and that will not cave or
slump whe n th e hol e i s drille d t o it s ful l depth . Th e soi l tha t meet s thi s requiremen t i s a
homogeneous, stif f clay. The first ste p in making the hole is to position the equipment at the desired
location an d t o selec t th e appropriat e drillin g tools . Fig . 17.6(a ) give s th e initia l location . Th e
drilling i s next carried ou t t o it s fil l dept h wit h the spoi l from th e hol e removed simultaneously.
After drillin g is complete, the bottom of the hole is underreamed if required. Fig. 17.6(b ) and
(c) sho w th e nex t step s o f concretin g an d placin g th e reba r cage . Fi g 17.6(d ) show s th e hol e
completely fille d wit h concrete .
746 Chapt e r 1 7
Figure 17. 4 (a ) Sin g le-flig ht aug e r bi t wit h cut t in g blad e for s oils , (b) s in g le-flig ht
aug er bi t wit h hard-m et al cut t in g t eet h for har d s oils , hardpan , an d rock , an d (c) cas t
s t eel heavy-dut y aug e r bit fo r hardpa n an d rock (Source : Woodward e t al. , 1972)
Figure 1 7 . 5 A 30° un derream e r wit h blad e cut t er s fo r s oil s t hat ca n be cut readil y
(Sour ce: Woodwar d e t al . , 1972 )
Deep Foun dat io n III : Drille d Pie r Foun dat ion s 747
Casing M etho d o f Constructio n
The casing method is applicable t o sites where the soil conditions are such that caving or excessive
soil or rock deformation can occur when a hole is drilled. This can happen when the boring i s made
in dr y soil s o r rocks whic h are stable when they are cut but wil l slough soon afterwards . In such a
Competent,
non-caving soil
Surface casing,
if required
^^f^?^^. -
~ soi l
>
&
V
/9K*7re>0«x!W!v/WJ iK\/9s<\/n>\ /?v\
^^—Drop ch"ute
3
(a) (b)
- Competent ,
_ non-cavin g soil
/7!iWWJ«!'W>C'WW^/W\X«<>\
-
Competent,
non-caving soil
^^
» -^
•/'V
.';•
;• • _;
3iKyi>iKxWI)^W\XW\X'iK\
(c) (d)
Figure 17. 6 Dr y m et ho d o f con s t ruct ion : (a ) initiating drilling , (b) s t art in g con cret e
pour, (c ) placin g reba r cag e , an d (d ) com plet ed s haf t (O'Neil l an d Rees e , 1999)
748 Chapt er 1 7
case, th e bor e hol e i s drilled, an d a steel pip e casing i s quickly set to prevent sloughing . Casin g i s
also require d i f drillin g i s require d i n clea n san d belo w th e wate r tabl e underlai n b y a laye r o f
impermeable stone s int o which the drilled shaft wil l penetrate. The casing is removed soo n after the
concrete i s deposited. I n some cases, th e casing may have to be lef t i n place permanently. It may be
noted here that until the casing is inserted, a slurry is used t o maintain the stability of the hole. Afte r
the casin g i s seated , th e slurr y i s baile d ou t an d th e shaf t extende d t o th e require d depth .
Figures 17.7(a ) t o (h ) giv e th e sequenc e o f operations . Withdraw l o f th e casing , i f no t don e
carefully, ma y lea d t o void s or soi l inclusions in the concrete, a s illustrated i n Fig. 17.8 .
' Caving soil ' .-; . •'. . V ' " ' • ' ' " • ' - ' ' • ' • ' ' . " !' .' ,.' :.. . ' - ; " : ' "
-_-_ Cohesiv e soi l .
(a) (b)
_ _ Cohesive soil .
Cohesive soil"_~_~_~_~ .
Cohesive soil.
(c) (d )
Figure 17. 7 Cas in g m et ho d o f con s t ruct ion : (a ) in it iat ing drilling , (b) drillin g wit h
s lurry; (c ) in t roducin g cas in g , (d ) cas in g i s s eale d an d s lurr y i s bein g rem ove d fro m
interior o f casing (continued )
Deep Foun dat io n III : Drille d Pie r Foun dat ion s 749
Slurry Metho d o f Construction
The slurry method of construction involves the use of a prepared slurry to keep the bore hol e stabl e
for th e entire depth of excavation. The soi l conditions for which the slurry displacement metho d is
applicable coul d be an y of the conditions described fo r the casing method. The slurr y method i s a
~ ~ ~ Competent soi l ~ ~ _ ~ J
(e) (f)
Level of
fluid concret e
Drilling flui d
forced fro m
space between
casing and soi l
/W\^>8<\^>B<N^B<N^i8<NX>B<\^>B<NX>9<N
^J
Competent soi l j >
r Caving soi l ;.- > ~ .- >
Competent soi l
/
/*
7$
::: $* *
f !"*;*••
1 J
l;-'l
->i
W
^
-• '• ^i !
*«*
• . «
'•V^".
:
•ft**"'
1
» Wf %
'^• • ,
4*fc
v.^*l
.:-.--^
/WN^SKN^TOv^iKV WNXW
5
[::::::::::
;i:^:;'::..:
v^
»
V
X
£*A
k!4>A
t
>' ; - J
(g)
(h)
Figure 1 7. 7 (con t in ued ) cas in g m et ho d o f con s t ruct ion : (e ) drillin g belo w cas in g ,
(f) un derream in g , (g ) rem ovin g cas in g , an d (h ) com plet ed s haf t (O'Neil l an d
Reese, 1999 )
750
Chapt er 1 7
Casing Casin g Casin g
I bein g 4 bein g f bein g
yA. remove d y^s . remove d /^remove d
z&z
T
conci
'.' *.; ' ,
o
O
U. .
/^\ \
• '. o
• • o
>
.-: U •• ;
X>9<\ /X \
V oid du e t o
/S' archin g
V ,
^
Bearing
^• K
c
o
V
U
/Ws /W \
^Sof t soi l
'*-S
N
squeezing in
Bearing

*

;

V

.

'

*
.
-
)
-

»

C
o
n
c
r
e
t
e


»





"
'
*
(
'
.
*
*
.
'

/
^
r
~
*
~
^

X
^
:
.

'*


>
'
1
xw\
Water flows up
— throug h concret e
causing segregation
/"- Water filled void
Bearing
stratum stratu m stratu m
oo muc h To o l i t t l e V oi d i n water bearing stratum
ete i n casing concret e in casing prio r to placing casing
Figure 17. 8 Pot en t ia l problem s leadin g t o in adequat e s haf t con cret e du e t o
rem oval o f t em por ar y cas in g wit hout car e (D'Appol on ia, e t al. , 1 975)
viable optio n a t any sit e wher e ther e i s a caving soil , and i t coul d b e th e onl y feasibl e optio n i n a
permeable, wate r bearing soi l if it is impossible t o set a casing into a stratum of soil or rock wit h low
permeability. Th e variou s steps i n the construction process are shown in Fig. 17.9. It is essential i n
this method tha t a sufficien t slurr y hea d be availabl e so that the inside pressure i s greater tha n that
from th e GWT o r from the tendency of the soi l t o cave .
Bentonite i s mos t commonl y use d wit h water t o produc e th e slurry . Polymer slurr y i s als o
employed. Som e experimentatio n may be required t o obtain a n optimum percentage fo r a site, but
amounts i n the range o f 4 t o 6 percent b y weight of admixtur e are usuall y adequate.
The bentonit e shoul d be wel l mixe d with water s o that the mixtur e i s not lumpy . The slurr y
should be capable of forming a filter cake on the side of the bore hole. The bore hole is generally not
underreamed for a bell since thi s procedure leaves unconsolidated cuttings on the base an d create s
a possibilit y of trapping slurry betwee n th e concret e bas e an d the bel l roof .
If reinforcing steel i s to be used, the rebar cage is placed i n the slurry as shown in Fig 17.9(b) .
After th e reba r cag e ha s been placed , concret e i s placed wit h a tremie either b y gravit y feed o r by
pumping. If a gravity feed is used, the bottom end of the tremie pipe shoul d be closed with a closure
plate unti l th e bas e o f th e tremi e reache s th e botto m o f th e bor e hole , i n orde r t o preven t
contamination o f th e concret e b y th e slurry . Fillin g o f th e tremi e wit h concrete , followe d b y
subsequent slight lifting o f the tremie, wil l then open th e plate, and concreting proceeds. Car e mus t
be take n tha t the botto m o f the tremi e i s buried in concret e a t least fo r a dept h o f 1. 5 m ( 5 ft). The
sequence o f operation s i s shown in Fi g 17.9(a ) to (d).
Deep Foun dat io n Drilled Pie r Foun dat ion s 751
: Cohesive soi l
• Caving soil
(a) (b)
~ : Cohesive soi l
Caving soi l
. Caving soi l
(c) (d)
Figure 17. 9 Slurr y m et ho d o f con s t ruct io n (a ) drilling t o ful l dept h wit h s lurry;
(b) placin g reba r cag e ; (c ) placin g con cret e ; (d ) com plet e d s haf t (O'Neil l an d
Rees e, 1999 )
1 7 .5 DESI G N CONSI DERATI ON S
The preces s o f the desig n o f a drilled pie r generall y involves the following:
1 . Th e objectives o f selecting drille d pie r foundation s fo r the project .
2. Analysi s of loads comin g o n each pie r foundation element.
3. A detailed soi l investigation and determining the soi l parameter s fo r th e design .
4. Preparatio n o f plan s an d specification s whic h includ e th e method s o f design , tolerabl e
5.
settlement, method s o f construction of piers, etc.
The method o f execution o f the project .
752 Chapt e r 1 7
In genera l the design of a drilled pier may be studie d under the following headings.
1. Allowabl e loads on the pier s based o n ultimat e bearing capacit y theories .
2. Allowabl e loads based on vertica l movement of the piers.
3. Allowabl e loads base d o n lateral bearing capacity of the piers .
In additio n to th e above , th e uplif t capacit y of pier s wit h or withou t underreams ha s t o b e
evaluated.
The followin g types of strat a are considered.
1 . Pier s embedde d i n homogeneous soils, sand or clay.
2. Pier s i n a layered system of soil .
3. Pier s sockete d i n rocks.
It i s better tha t the designe r select shaf t diameter s tha t are multiple s of 15 0 mm ( 6 in) sinc e
these ar e the commonl y availabl e drilling tool diameters.
1 7 .6 L OA D TRANSFE R M ECH ANI SM
Figure 17.10(a ) show s a singl e drille d pie r o f diamete r d, an d lengt h L constructe d i n a
homogeneous mas s o f soi l o f known physical properties . I f thi s pie r i s loade d t o failur e unde r an
ultimate loa d Q
u
, a part o f thi s loa d i s transmitte d to the soi l alon g th e lengt h of th e pie r an d th e
balance i s transmitted t o the pier base. The load transmitte d to the soi l alon g th e pier i s called th e
ultimate frictionload or skin load, Q
f
and tha t transmitted to the base i s the ultimat e base or point
load Q
b
. The total ultimate load, Q
u
, is expressed a s (neglecting the weight of the pier )
where q
b
= ne t ultimat e bearing pressure
A
b
= bas e area
f
si
- uni t ski n resistance (ultimate) of layer i
P. = perimete r of pier i n layer i
Az
(
. = thicknes s of laye r i
N = numbe r of layers
If the pier i s instrumented, the load distribution along the pier ca n be determined a t differen t
stages o f loading . Typical loa d distribution curves plotte d alon g a pier ar e shown i n Fi g 17.10(b )
(O' Neill an d Reese , 1999) . Thes e loa d distributio n curve s ar e simila r t o th e on e show n i n
Fig. 15.5(b) . Since th e load transfe r mechanis m for a pier i s the same a s that for a pile, no furthe r
discussion o n thi s i s necessar y here. However, i t is necessar y t o study i n thi s contex t the effec t o f
settlement o n th e mobilizatio n of sid e shea r an d bas e resistanc e o f a pier . As ma y b e see n fro m
Fig. 17.11 , the maximum values of base and side resistance are not mobilized a t the same value of
displacement. I n some soils , and especiall y i n some brittl e rocks, th e side shea r ma y develop full y
at a smal l valu e o f displacemen t an d the n decreas e wit h furthe r displacemen t whil e th e bas e
resistance i s stil l being mobilize d (O'Neil l an d Reese, 1999) . If the valu e of the sid e resistanc e a t
point A i s added t o the value of the base resistance a t point B, the total resistanc e shown at level D
is overpredicted. O n th e othe r hand , i f the designe r want s t o take advantag e primaril y o f the base
resistance, the side resistance at point C should be added to the base resistance at point B to evaluate
Q . Otherwise, the designer may wis h t o design for the side resistance at point A and disregar d th e
base resistance entirely.
Deep Foun dat io n III : Drille d Pie r Foun dat ion s 753
ia
,
Qt
d
1 1
Q
b
Applied load , kip s
500 100 0 150 0 200 0
(a) (b )
Figure 1 7.1 0 T ypica l s e t o f loa d dis t ribut ion curve s (O'Neil l an d Rees e , 1999 )
Actual
ultimate
resistance
Ultimate sid e
resistance
Ultimate bas e
resistance
Settlement
Figure 1 7.1 1 Con dit io n i n which (Q
b
+ Q
f
) i s not equa l to act ua l ult im at e res is t an ce
754 Chapt er 1 7
1 7 .7 V ERTI CA L B EARI N G CAPACI T Y OF DRI L L ED PI ER S
For th e purpos e o f estimatin g th e ultimat e bearin g capacity , th e subsoi l i s divide d int o layer s
(Fig. 17.12) base d o n j udgment and experience (O' Neil l an d Reese, 1999) . Eac h laye r i s assigne d
one o f four classifications.
1. Cohesiv e soi l [clay s an d plasti c silt s wi t h undraine d shea r strengt h c
u
< 250 kN/m
2
(2.5 t/ft
2
)] .
2. Granula r soi l [cohesionles s geomaterial , suc h a s sand , grave l o r nonplasti c sil t wit h
uncorrected SPT(N ) value s of 50 blows pe r 0.3/ m o r less].
3. Intermediat e geometeria l [cohesiv e geometeria l wit h undrained shear strengt h c
u
betwee n
250 and 2500 kN/m
2
(2. 5 and 25 t sf), o r cohesionless geomaterial wit h SPT(N) value s > 50
blows pe r 0. 3 mj .
4. Roc k [highl y cemente d geomateria l wi t h unconfme d compressiv e strengt h greate r tha n
5000 kN/m
2
(5 0 t sf)J .
The uni t side resistanc e /, (=/
max
) i s computed i n each laye r throug h whic h th e drilled shaf t
passes, an d th e uni t base resistance q
h
(=<7
max
) i s computed fo r the laye r on or i n whic h the base o f
the drille d shaf t i s founded.
The soi l along the whole l ength of the shaf t i s divided int o four layers as shown in Fig. 17.12 .
Effective L engt h fo r Computin g Sid e Resistanc e in Cohesive Soil
O'Neill and Rees e (1999 ) sugges t that the following effective length of pier i s to be considered fo r
computing sid e resistanc e in cohesive soil .
Q
u
/^\ /V3\ /%x$\
Layer 1
t
Layer 2
i ,
Layer 3
!,
Layer 4
d
//T^. /^\//F^ ,
''Qfl
7
1
"2/2 i
l(
Q
fl
*
2
3
' 2/ 4
a
Figure 1 7 .1 2 Idealize d g eom at eria l l ayerin g fo r com put at io n o f com pres s io n loa d
an d res is t an c e (O'Neil l an d Rees e , 1999 )
Deep Foun dat io n Drilled Pie r Foun dat ion s 755
Straight shaft: On e diamete r fro m th e botto m an d 1. 5 m ( 5 feet ) from th e to p ar e t o be exclude d
from th e embedded lengt h of pile for computing side resistance a s shown in Fig. 17.13(a).
Belled shaft: Th e heigh t of the bel l plus the diameter of the shaf t fro m the bottom and 1. 5 m (5 ft )
from th e top ar e t o be excluded as shown in Fig 17.13(b) .
1 7.8 TH E G ENERAL B EARI N G CAPACI T Y EQU ATI ON FOR TH E
B ASE RESI STANC E q
b
( = </
max
)
The equation for the ultimat e base resistance may be expressed a s
,<w
• — vd v d N
/1*
J
V
W
V* * V
9 Y 7 r
(17.2)
where /V
c
, N an d N = bearin g capacit y of factor s for lon g footings
5
C
, s an d s = shap e factors
d
c
, d an d d = dept h factors
q'
0
= effectiv e vertica l pressure a t the bas e leve l of the drilled pie r
7 = effectiv e uni t weight of the soil below the bottom of the drilled shaft t o a
depth = 1. 5 d where d = width or diameter o f pier a t base leve l
c = averag e cohesive strengt h of soi l just below the base .
For deep foundations th e las t ter m i n Eq. (17.2) becomes insignificant and ma y be ignored .
Now Eq. (17.2 ) may be written as
(17.3)
V c + s d ( N — \ }q'
c q q^ q > ^o
/^
L
\
/ww\
d
i
! 5 f t = 1. 5 m
L
.^Xx\S*
• 5' =
E
•s:
i
1
^
"iT
Effective lengt h £ ^
L ' = L -( d+1.5 ) m< g II
UJ
d\
h
d
*j
^?^
1.5 m
d
/ \
(a)
d
b
(b)
Figure 1 7 .1 3 Ex clus io n zon e s for es t im at in g sid e res is t an c e fo r drille d s haft s i n
cohes ive s oil s
756 Chapt e r 1 7
1 7 .9 B EARI N G CAPACI T Y EQU ATI ONS FOR TH E B ASE I N
COH ESI V E SOI L
When the U ndrained Shear Strength, c
u
< 250 kN/m
2
(2 . 5 t/ft
2
)
For 0 = 0, N
q
= 1 and ( N
q
- 1 ) = 0, her e Eq. (17.3) can be written as (V esic, 1972 )
4b =
N
l
c
u (17.4 )
in which
N *=-( \ nI
r
+l) (17.5 )
I
r
- rigidit y index of the soil
Eq. (17.4) i s applicabl e fo r c
u
< 96 kPa and L > 3d (bas e width )
For 0= 0, /. may be expressed a s (O'Neill an d Reese, 1999 )
/r =
3^" (I
7
-
6
)
where E
s
= Young' s modulus of the soi l in undrained loading. Refer to Section 13. 8 for the method s
of evaluating the valu e of E
s
.
Table 17. 1 gives th e value s of I
r
an d N *as a function o f c .
If th e dept h o f bas e ( L ) < 3d (base )
2 L
^(
=
<?
ma x
) = -1 + — N *c
u (17J )
When c
u
> 96 kPa (2000 lb/ft
2
), th e equation for q
b
may be written as
9
b
=*u (17.8 )
for dept h o f base ( = L ) > 3d (base width).
1 7 .1 0 B EARI N G CAPACI T Y EQU ATI O N FOR TH E B AS E I N
G RANU L AR SOI L
V alues N
C
an d N i n Eq. (17.3 ) ar e for stri p footings on the surfac e o f rigi d soil s an d ar e plotted a s
a functio n of 0 i n Fig. 17.14 . V esi c (1977) explaine d tha t durin g bearin g failure , a plasti c failur e
zone develop s beneat h a circula r loade d are a tha t i s accompanie d b y elasti c deformatio n i n th e
surrounding elastic soi l mass. The confinement of the elastic soi l surrounding the plastic soi l has an
effect o n q
b
( = <?
max
). The value s of N
c
an d N ar e therefor e dependen t no t onl y on 0 , but als o o n I
r
.
They mus t be correcte d fo r soi l rigidit y as given below.
Table 1 7 . 1 Val ue s o f l
r
= E
s
/3c
u
an d A/
c
*
', "?
24 kP a (50 0 lb/ft
2
) 5 0 6. 5
48 kP a (1000 lb/ft
2
) 15 0 8. 0
> 96 kPa (2000 lb/ft
2
) 250-30 0 9. 0
Deep Foun dat ion Drilled Pie r Foun dat ion s 757
N
c
(corrected ) = N
c
C
c
N
q
(corrected ) = N
q
C
q
where C
c
and C ar e the correction factors . As per Chen an d Kulhawy (1994)
Eq (17.3) ma y no w be expressed a s
(17.9)
(17.10)
1-C
r - r
q
c q
N
c
tan0
( 1
C
?
=e x p {[-3.8 tan 0] + [(3.07 sin 0)log
10
2/
rr
]/(l + sin 0)} (17.11b )
where 0 i s a n effectiv e angl e o f interna l friction . I
rr
i s th e reduce d rigidit y inde x expresse d a s
[Eq. (15.28) ]
/ =
and / . =
by ignoring cohesion , where ,
(17.12)
(17.13)
CQ
10 2 0 3 0 4 0
Friction angle, 0 (degrees)
50
Figure 1 7.1 4 B earin g capacit y fact or s (Che n an d Kulhawy, 1994)
758 Chapt e r 17
E
d
= drained Young's modulus of the soi l
\ i
d
= drained Poisson' s rati o
A = volumetri c strain wi t hin the plasti c zone during the loading proces s
The expression s for n
d
and A may be writte n as (Chen and Kulhawy , 1994 )
/)
rel
(17.14 )
0.005(1 -0
rel
)g'
0
- - -(17.15 )
where ( j)
rel
= ~^-~-fo r 25 ° < <p° < 45° (17.16 )
45 -25 °
= relative friction angle factor, p
a
= atmospheric pressur e =101 kPa.
Chen an d Kul haw y (1994 ) sugges t that , fo r granula r soils , th e followin g value s ma y b e
considered.
loose soil , E
d
- 100 t o 200p
fl
(17.17 )
medium dense soil , E
d
= 200 t o 500p
a
dense soil , E
d
= 500 t o 1000p
a
The correction factor s C
c
and C indicate d i n Eq. (17.9) nee d be applied onl y if I
n
i s less than
the critical rigidit y inde x (/
r
)
crit
expresse d a s follows
(
7
r )cr =|
e x
P
2
-
85cot 45
°-f (17.18 )
The value s o f critica l rigidit y index ma y b e obtaine d fro m Tabl e 12. 4 fo r pier s circula r o r
square i n section .
If l
rr
> (/
r
)
crit
, the factor s C
c
and C ma y be take n as equal t o unity.
The shape and depth factors in Eq. (17.3) can be evaluated by making use of the relationships
given i n Tabl e 17.2.
Table 17. 2 Shap e an d dept h fact or s (Eq. 17. 3) (Chen an d Kulhawy, 1994)
Fact ors Valu e
N
s 1 + —
N
v-
1 + ta n <f)
sin0)
2
tan'
1

180 d
Deep Foun dat io n III : Drille d Pie r Foun dat ion s 75 9
B ase i n Cohesionless Soi l
The theoretical approach as outlined above is quite complicated and difficult t o apply in practice for
drilled pier s i n granula r soils . Direc t an d simpl e empirica l correlation s hav e been suggeste d b y
O'Neill an d Rees e (1999 ) between SPT T V value an d th e bas e bearing capacit y a s give n below fo r
cohesionless soils.
<2 900k N/ m
2
(17.19a )
<30t sf (17.19b )
where N = SPT valu e < 5 0 blows / 0. 3 m.
B ase i n Cohesionless IG M
Cohesionless IGM' s are characterized by SPT blow counts if more than 50 per 0.3 m. In such cases,
the expression for q
b
is
0.8
<^(=4m ax ) = 0-60 Af
60
^7 q'
0
(17.20 )
tf o
where W
60
= averag e SP T correcte d fo r 6 0 percen t standar d energ y withi n a dept h o f 2 d
(base) belo w th e base . Th e valu e o f A^ i s limite d t o 100 . N o correctio n fo r
overburden pressur e
p
a
= atmospheri c pressure i n the unit s used for q'
o
(=101 kPa i n the S I System)
q'
0
= vertica l effective stres s a t the elevation of the base o f th e drilled shaft .
1 7 .1 1 B EARI N G CAPACI T Y EQU ATI ON S FOR TH E B ASE I N
COH ESI V E I G M OR ROCK (O'NEI L L AND REESE , 1 9 9 9 )
Massive roc k an d cohesive intermediat e materials possess commo n properties . The y posses s lo w
drainage qualitie s under normal loadings but drai n mor e rapidl y unde r large load s tha n cohesiv e
soils. I t is for these reasons undraine d shear strength s ar e used fo r rocks an d IGMs.
If the base of the pier lies in cohesive IGM or rock ( RQD =100 percent) and the depth of socket,
D
v
, in the IGM or rock i s equal to or greater than l.5d, the bearing capacit y may be expressed a s
(17.21)
where q
u
= unconfme d compressiv e strengt h of IGM or rock below th e bas e
For RQD betwee n 70 and 10 0 percent,
For jointed rock o r cohesive IGM
?*(= ^ma x ) = [*°-
5
+( ms°-
5
+s)°-
5
] q
u
(17.23 )
where q
u
i s measured on intact cores fro m withi n 2d (base) below the base o f the drilled pier . In all
the above cases q
b
and q
u
are expressed i n the same units and s and m indicat e the properties of the
rock or IGM mas s that can be estimated from Table s 17. 3 and 17.4 .
760 Chapt er 1 7
Table 1 7. 3 Des cript ion s o f roc k t ype s
Rock t yp e Des cript io n
A Carbonat e rock s wi th well-develope d crysta l cleavag e (eg., dolostone , limestone,
marbl e)
B Li t hi fi e d argi l l aeou s rocks (mudstone, sil t st one , shale , slate )
C Arenaceou s rocks (sandstone , quart zi t e )
D Fine-graine d igneou s rocks (andesite, dolerite , diabase, rhyol i t e)
E Coarse-graine d igneou s an d met amorphi c rock s (amphi bol e , gabbro, gneiss , granite,
norite, quart z-di ori t e )
Table 1 7. 4 Value s o f s an d m(dim en s ion les s ) bas e d o n roc k cl as s ificat io n (Cart e r
an d Kul hawy , 1988 )
Qual i t y o f
rock mas s
Excellent
V ery goo d
Good
Fair
Poor
V ery poor
Joi nt descri pt i o n
and s paci ng
Intact (closed) ;
spacing > 3 m (1 0 ft )
Int erl ocki ng;
spacing o f 1 to 3 m ( 3 t o 1 0 ft)
Slightly weathered ;
spacing o f 1 to 3 m ( 3 t o 1 0 ft )
Moderately weathered;
spacing o f 0. 3 t o 1 m (1 t o 3 ft )
Weathered wi t h gouge (soft material) ;
spacing of 3 0 t o 300 m m ( 1 in . t o 1 ft)
Heavil y weathered ;
spacing o f les s tha n 5 0 m m ( 2 i n.)
s
1
0.1
4x1 Q~
2
10~
4
10~
5
0
V al ue o f m a s f unc t i o n o f roc k t yp e
( A-E) f r o m
A B C D E
7 1 0 1 5 1 7 2 5
3.5 5 7. 5 8. 5 12. 5
0.7 1 1. 5 1. 7 2. 5
0.14 0. 2 0. 3 0.3 4 0. 5
0.04 0.0 5 0.0 8 0.0 9 0.1 3
0.007 0.0 1 0.01 5 0.01 7 0.02 5
1 7 .1 2 TH E U L TI M ATE SKI N RESI STANCE OF COH ESI V E AN D
I NTERM EDI ATE M ATERI AL S
Cohesive Soi l
The proces s o f drilling a borehole fo r a pier in cohesive soil disturbs the natural condition of the soi l
all along th e side t o a certain extent. There i s a reduction in the soi l strengt h not only due t o boring
but als o du e t o stress relie f and the time spent between borin g and concreting. I t is ver y difficul t t o
quantify th e extent of the reduction in strength analytically. In order t o take care o f the disturbance ,
the uni t frictional resistance o n the surfac e of the pier ma y be expressed a s
f
s
=ac
u
(17.24 )
where a — adhesio n facto r
c
u
- undraine d shear strengt h
Relationships hav e bee n develope d betwee n c
u
an d a b y many investigator s base d o n fiel d
load tests . Fi g 17.1 5 give s on e suc h relationshi p in th e for m o f a curv e develope d b y Che n an d
Kulhawy (1994) . Th e curv e has been develope d o n the following assumptions (Fig . 17.15) .
Deep Foun dat io n III : Drille d Pie r Foun dat ion s 761
l . U
0.8
0.6
0.4
0.2
0.0
0
-

,
4
1 1 1 1
4
4.^.
» 4+
4
t ^^
4
i i
4
^i ^^ ^
4
i i
1 <
4
r
i i
Design a t
*Interme i
geomater
^
»
i i i i
liate
lal"
^ 4
i i i i
0 1. 0 2. 0 3. 0 4. 0 5 .
Figure 1 7 .1 5 Correlat io n bet wee n a an d cj p
a
f = 0 u p t o 1. 5 m ( = 5 ft) from th e ground level.
f
s
= 0 u p t o a height equal to ( h + d) a s per Fi g 17.1 3
O'Neill an d Reese (1999 ) recommen d th e chart's tren d line given in Fig. 17.1 5 for designing
drilled piers . The suggested relationship s are :
a = 0.55 fo r c
u
I p
a
< 1. 5
and a = 0.55 -0.1
Pa
for 1. 5< c I p
a
<2. 5
(17.25a)
(17.25b)
Cohesive I ntermediat e G eomaterial s
Cohesive IGM' s ar e ver y har d clay-like material s whic h can als o b e considere d a s ver y sof t roc k
(O'Neill and Reese, 1999) . IGM' s are ductile and failure may be sudden at peak load. The value of
f
a
(pleas e not e tha t th e ter m f
a
i s use d instea d o f f
s
fo r ultimat e uni t resistanc e a t infinit e
displacement) depend s upo n th e sid e conditio n o f th e bor e hole , tha t is , whethe r i t i s roug h o r
smooth. For design purposes the side is assumed as smooth. The expression for / ma y be written as
f
a
=a<l
u
(17.26 )
where, q = unconfme d compressiv e strength
/ = th e value of ultimate unit side resistance whic h occurs a t infinite displacement .
Figure 17.1 6 give s a char t for evaluatin g a. Th e char t i s prepared fo r a n effectiv e angl e of
friction between the concrete and the IGM (assuming that the intersurface is drained) and S
t
denote s
the settlement of piers at the top of the socket. Further, the chart involves the use of o
n
lp
a
wher e o
n
is th e norma l effectiv e pressur e agains t th e sid e o f th e borehol e b y th e drille d pie r an d p
a
i s th e
atmospheric pressur e (10 1 kPa).
762 Chapt er 1 7
0.0
^, = 30°
115 <£, „/ <?„< 50 0
5, = 25 mm
Figure 1 7 .1 6 Fact o r a fo r cohes iv e IGM ' s (O'Neil l an d Rees e , 1999)
Table 1 7. 5 Es t im at io n o f E
m
IE
j
bas e d o n ROD (M odifie d aft e r Cart e r an d Kul hawy ,
1988)
ROD (per cen t ) Clos ed j oin t s Open j oin t s
100
70
50
20
1.00
0.70
0.15
0.05
0.60
0.10
0.10
0.05
Note: V alues intermediat e betwee n tabulate d values may b e obtained b y linea r interpolation.
Table 1 7 . 6 f lf
a
bas e d o n EJE, (O'Neil l e t al. , 1996 )
da a in i
1.0
0.5
0.3
0.1
0.05
1.0
0.8
0.7
0.55
0.45
Deep Foun dat io n III: Drille d Pie r Foun dat ion s 763
1.0
0.8
M
-•- Depth = 0 m
-O- Depth = 4 m
-•- Depth = 8 m
-D-Depth = 12m
125 175
Slump (mm)
200 225
Figure 1 7 .1 7 Fact o r M vers u s con cret e slum p (O'Neil l e t al. , 1996 )
O'Neill an d Reese (1999) give the following equation for computing a
n
o
n
= My
c
z
c
(17.27)
where y
c
= th e unit weight of the flui d concret e used for the construction
z
c
= th e dept h of the poin t at which o~
M
i s required
M = a n empirical factor which depends on the fluidit y o f the concrete a s indexed by
the concrete slump
Figure 17.1 7 gives the values of M for various slumps.
The mass modulus of elasticity of the IGM ( E
m
) shoul d be determined before proceeding, in order
to verif y tha t the IGM i s withi n the limits of Fig 17.16 . This requires the average Young's modulus of
intact IGM core (£.) which can be determined in the laboratory. Table 17.5 gives the ratios ofE
m
/E
{
fo r
various values of RQD. V alues QiEJ E
i
les s than unity indicate that soft seams and/or joints exist in the
IGM. These discontinuities reduce the value of f
a
. Th e reduced value off
a
ma y be expressed a s
f = f R
J aa •'a a
where the ratio R
a
=f
aa
/f
a
ca n be determined fro m Tabl e 17.6.
If the socket i s classified as smooth, i t is sufficientl y accurat e to set/^ =/
max
=f
aa
(17.28)
1 7 .1 3 U L TI M AT E SKI N RESI STANCE IN COH ESI ONL ESS SOIL
AND G RAV EL L Y SANDS (O'NEI L L AND REESE , 1 999 }
In Sand s
A general expression fo r total skin resistance i n cohesionless soi l may be written as [Eq. (17.1)]
(17.29)
764 Chapt e r 17
or Q
fi
= />/?.<
;
.Az.
(1730 )
where f
si
= ft.c f '
Ol
(17.31 )
A = K
si
ta n 8 .
<5. = angl e of ski n friction o f th e / t h layer
The followin g equations are provided by O'Neill an d Rees e (1999 ) fo r computing /?..
For SPT N
6Q
(uncorrected ) > 1 5 blows / 0. 3 m
$ = 1.5-0.245[z.]
0
-
5
(17.32 )
For SPT N
6Q
(uncorrected ) < 1 5 blows / 0. 3 m
(17.33)
I n G ravell y Sand s o r G ravel s
For SP T N
6Q
> 15 blows / 0.3 m
fl. =2.0-0.15[z.]°-
75
(17.34 )
In gravell y sands or gravels, use the method for sand s i f yV
60
< 1 5 blows / 0. 3 m.
The definition s of various symbols use d above ar e
/3
;
. = dimensionles s correlation facto r applicabl e t o laye r i . Limite d t o 1. 2 i n sand s
and 1. 8 in gravell y sands and gravel . Minimu m value i s 0.25 i n bot h types o f
soil; f
si
i s limited 200 kN/m
2
(2. 1 tsf )
q'
oi
= vertica l effective stress a t the middl e of each laye r
/V
60
= desig n valu e fo r SP T blo w count , uncorrecte d fo r depth , saturatio n o r fine s
corresponding t o layer /
Z; - vertica l distanc e from th e ground surface , i n meters, t o the middl e of layer i .
The laye r thickness Az
;
. is limited to 9 m.
1 7 .1 4 U L TI M AT E SIDE AND TOTAL RESI STANCE I N ROCK
(O'NEI L L AN D REESE , 1 9 9 9 )
Ultimate Ski n Resistanc e (fo r Smooth Socket)
Rock i s defined as a cohesive geomaterial wit h q
u
> 5 MPa (725 psi). The following equations may
be use d for computin g f
s
( =/
max
) when the pie r is sockete d i n rock. Two methods ar e proposed .
M ethod 1
0.5 0. 5
f
s
( =f
m
J = ^P
a
~ *0.65p
a
A- (17.35 )
* a ' a
where, p
a
= atmospheri c pressur e (= 10 1 kPa)
q
u
= unconfme d compressiv e strengt h of rock mas s
f
c
= 2 8 day compressive cylinde r strength of concrete use d i n the drille d pie r
Deep Foun dat io n III : Drille d Pie r Foun dat ion s 76 5
M ethod 2
0.5
f
s
( =f
m
J = ^P
a
^- (17.36 )
Pa
Carter an d Kulhaw y (1988 ) suggeste d equatio n (17.36) base d o n the analysi s of 25 drilled
shaft socke t test s i n a ver y wide variet y of sof t roc k formations , including sandstone, limestone ,
mudstone, shal e and chalk.
Ultimate Total Resistanc e Q
u
If the base of the drilled pier rests on sound rock, the side resistance can be ignored. In cases where
significant penetratio n of the socket can be made, it is a matter of engineering judgment to decide
whether Q, should be added directl y to Q
b
to obtain the ultimate value Q
u
, When the rock i s brittle
in shear , muc h sid e resistanc e wil l be los t as the settlemen t increase s to the valu e require d to
develop the ful l valu e of q
b
( = q
max
)- I f the rock i s ductile in shear, there is no question that the two
values can be added direcil y (O'Neil l and Reese, 1999) .
1 7 .1 5 ESTI M ATI O N OF SETTL EM ENTS OF DRI L L ED PI ERS AT
WORKI NG L OAD S
O'Neill an d Rees e (1999 ) sugges t th e followin g method s fo r computin g axia l settlement s fo r
isolated drilled piers:
1 . Simple formulas
2. Normalize d load-transfer methods
The tota l settlement S
t
at the pier head at working loads may be expressed a s (V esic, 1977 )
s
t
=S
e
+S
bb
+S
bs
(17.37 )
where, S
e
= elasti c compressio n
S
bb
= settlemen t of the base due to the load transferred to the base
S
bs
= settlemen t of the base due to the load transferred into the soi l along the sides .
The equations for the settlement s are
A
b
E
r_ /~<
bb=
where L = lengt h of the drilled pier
A
b
= bas e cross-sectional are a
E = Young' s modulus of the drilled pier
Q
a
= loa d applie d to the head
<2> = mobilize d sid e resistance whe n Q
a
is applied
766 Chapt er 1 7
Table 1 7 . 7 Val ue s o f C fo r variou s s oil s (Ves ic , 1977)
Soi l C
p
Sand (dense )
Sand (loose )
Clay (st i f f )
Clay (soft )
Silt (dense )
Silt (loose )
0.09
0.18
0.03
0.06
0.09
0.12
Q
bm
= mobilize d base resistance
d = pie r widt h or diamete r
C = soi l factor obtaine d from Tabl e 17. 7
Normalized L oad- Transfe r M ethod s
Reese an d O'Neil l (1988 ) analyze d a serie s o f compression loadin g tes t dat a obtaine d fro m full -
sized drille d pier s i n soil . The y develope d normalize d relation s fo r pier s i n cohesiv e an d
cohesionless soils . Figure s 17.1 8 and 17.1 9 can be used to predict settlement s of piers i n cohesiv e
soils an d Figs. 17.2 0 an d 17.2 1 i n cohesionless soil s includin g soil mixe d wit h gravel.
The boundar y limit s indicated for gravel in Fig. 17.2 0 have been foun d t o be approximatel y
appropriate fo r cemente d fine-graine d desert IGM' s (Wals h et al. , 1995) . Th e rang e o f validit y of
the normalized curves ar e as follows :
).0 0. 2 0. 4 0. 6 0. 8 1. 0 1. 2 1. 4 1. 6 1. 8 2. 0
Settlement ~
Settlement rati o S
R
=
Diameter o f shaf t
Figure 1 7.1 8 Norm alize d sid e loa d t ran s fe r fo r drille d s haf t i n cohes ive s oi l (O'Neil l
an d Rees e , 1999 )
Deep Foun dat io n III : Drille d Pie r Foun dat ion s 767
0 1 3 4 5 6 7
Settlement o f bas e
Diameter o f base
10
Figure 1 7 .1 9 Norm alize d bas e loa d t ran s fe r fo r drille d s haf t i n cohes ive s oi l
(O'Neil l an d Rees e , 1999 )
Range o f result s fo r
deflection-softening respons e
Range o f results fo r
deflection-hardening respons e
Trend line
i.O «
L
0.0 0. 2 0. 4 0. 6 0. 8 1. 0 1. 2 1. 4 1. 6 1. 8 2. 0
S = Settlemen t ~
R
Diamete r o f shaf t
Figure 17.2 0 Norm alize d sid e loa d t ran s fe r fo r drille d s haf t i n cohes ion les s s oi l
(O'Neill an d Rees e , 1999 )
768 Chapt er 1 7
2.Q
1.8
1.6
1.4
1.2
1.0
E 0. 8
D
0.6
0.4
0.2
0.0
Range o f result s
1 2 3 4 5 6 7 8 9 1 0 1 1 1 2
c
Settlemen t o f base _ ,
O/j = %
Diameter o f base
Figure 1 7 .2 1 Norm alize d bas e loa d t ran s fe r fo r drille d s haf t i n cohes ion les s s oi l
(O'Neill an d Reese , 1999 )
Figures 1 7 .1 8 an d 1 7 .1 9
Normalizing factor = shaf t diamete r d
Range of d = 0.4 6 m t o 1.5 3 m
Figures 1 7 .2 0 an d 1 7 .2 1
Normalizing facto r = bas e diamete r
Range o f d = 0.46 m t o 1.53 m
The followin g notations are used i n the figures:
S
D
= Settlemen t rati o = S I d
A a
S
a
= Allowabl e settlement
N
fm
= Normalize d sid e loa d transfe r ratio = QrJ Qr
N
bm
= Normaliz e base loa d transfe r ratio = Q
b
J Q
b
Example 1 7 . 1
A multistory building is t o be constructed i n a stif f t o very stif f clay . The soi l i s homogeneous t o a
great depth . The averag e value of undrained shear strengt h c
u
is 15 0 kN/m
2
. It is proposed t o use a
drilled pier of length 25 m and diameter 1. 5 m. Determine (a ) the ultimate load capacit y of the pier,
and (b ) the allowabl e load o n the pier wit h F
s
= 2.5. (Fig . Ex. 17.1 )
Solution
Base load
When c
u
> 96 kPa (2000 Ib/ft
2
), us e Eq. (17.8 ) fo r computing q
b
. In thi s cas e c
u
> 96 kPa .
a. = 9c = 9 x 15 0 = 135 0 kN/m
2
Deep Foun dat io n III : Drille d Pie r Foun dat ion s 769
Q
, = 25 m
c= 150kN/m
2
Clay
d=1.5m
Figure Ex . 1 7.1
3 14 x 15
2
Base loa d Q
b
= A
b
q
b
= x 1350 =1.76 6 x 135 0 = 2384 kN
Frictional loa d
The uni t ultimate frictional resistance f
s
i s determined using Eq. (17.24 )
fs=
ac
u
From Fig. (17.15) cc= 0.55 for cjp
a
= 150/101 = 1.5
where p
a
i s the atmospheric pressur e =101 kPa
Therefore f
s
= 0.55 x 15 0 = 82. 5 kN/m
2
The effectiv e lengt h of the shaf t fo r computing the frictional load (Fig. 17.13 a) is
L ' = [ L -( d+1.5) ] m = 25- (1. 5+ 1.5 ) = 22m
The effective surfac e area A
s
= ndL ' = 3.14 x 1. 5 x 22 = 103.62 m
2
Therefore Q
f
=f
s
A
s
= 82.5 x 103.6 2 = 8,549 k N
The total ultimat e load i s
Q
u
= Q
f
+ Q
b
= 8,549 + 2,384 = 10,93 3 k N
The allowabl e loa d ma y b e determine d b y applyin g a n overal l facto r o f safet y t o Q
u
.
Normally F
S
= 2.5 i s sufficient .
10933
= 4,373 k N
770 Chapt e r 1 7
Example 1 7 . 2
For the problem give n i n Ex. 17.1 , determine the allowabl e load fo r a settlement of 1 0 mm ( = S ).
All the other dat a remai n the same .
Sol ution
Allowable ski n load
S 1 0
Settlement ratio S
R
= -*-= — x 100 = 0.67%
( 2 l. D X 1 U
Q
f
From Fig . 17.1 8 for S
R
= 0.67%, N
fm
= -^ = 0.95 b y using the trend line.
( 2^ = 0.95 Q
f
= 0.9 5 x 8,549 = 8,122 kN.
Allowable bas e load for S
a
=10 mm
From Fig . 17.1 9 for S
K
= 0.67 %, N, = ^= 0.4
w
A OiH f\
Q
bm
= 0.4 Q
b
= 0.4 x 2,384 = 954 kN.
Now the allowabl e loa d Q
as
based o n settlement consideration i s
Q
as
= Qjm+Q
bm
= 8^12 2 - f 95 4 = 9,076 kN
Q
as
based o n settlement consideration is very much higher than Q
a
(Ex. 17.1 ) and as such Q
a
governs the criteri a for design.
Example 1 7 . 3
Figure Ex . 17. 3 depict s a drille d pie r wit h a belle d bottom . Th e detail s o f th e pil e an d th e soi l
properties ar e give n i n th e figure . Estimat e (a ) th e ultimat e load, an d (b ) the allowabl e loa d wit h
Solution
Based loa d
Use Eq. (17.8 ) fo r computin g q
b
q = 9c = 9x20 0 =1, 800 kN/m
2
Ttd^ 1 14 x 3
2
Base loa d Q
b
=—^xq
b
=- -x 1,800 = 12, 7 17 kN
Frictional loa d
The effective lengt h of shaf t L' = 25 - (2.7 5 + 1.5 ) = 20.75 m
From Eq . (17.24) f
s
=ac
u
For
f« _
=
122. = i o , a = 0.55 fro m Fig . 17.1 5
Pa
10 1
Hence/
y
= 0.55 x 10 0 = 55 kN/m
2
Deep Foun dat io n III : Drille d Pie r Foun dat ion s 771
1.25 m
G
=
/&$^
1.5
Z
25m
1
1
,
l.i
2.75 m —
m
m
t-
^ ^
Clay
c
u
= 100 kN
d= 1. 5m
Clay
w c. . = 200 k l
= 3 m
Figure Ex . 17. 3
Q
f
= PL 'f
s
= 3.14 x 1. 5 x 20.75 x 5 5 = 5,375 k N
Q
u
= Q
b
+ Q
f
= 12,71 7 + 5,375 = 18,09 2 k N
= k N
2.5
Example 1 7. 4
For the problem give n i n Ex. 17.3 , determine the allowabl e loa d Q
as
for a settlement S
a
= 10 mm.
Solution
Skin loa d Q, (mobilized )
Settlement rati o S
R
= — x 100 = 0.67%
1.5 x l O
3
From Fig. 17.1 8 fo r S
p
= 0.67, N , = 0.95 fro m th e trend line.
° A jtn
Therefore Q^ = 0.95 x 5,375 = 5,106 kN
Base loa d Q
bm
(mobilized )
S
R
=
1 0
, x 100 = 0.33%
R
3xl 0
3
From Fig . 17.1 9 for S
R
= 0.33%, N
bm
= 0.3 from th e trend line.
7 7 2 Chapte r 1 7
Hence Q
bm
= 0.3 x 12,71 7 = 3815 k N
Q
as
= Q
frn
+ Q
bm
= 5,106 + 3,815 = 8,921 k N
The facto r of safet y wit h respec t t o Q i s (fro m Ex . 17.3 )
F =
8,921
This i s lo w a s compare d t o th e normall y accepte d valu e o f F
s
= 2.5. Henc e Q
a
rule s th e
design.
Example 1 7 . 5
Figure Ex . 17. 5 show s a straigh t shaf t drille d pie r constructe d i n homogeneou s loos e t o mediu m
dense sand . The pil e and soi l propertie s are :
L = 25m, d= 1. 5m , c = 0,0 =36 ° and y = 17. 5 kN/m
3
Estimate (a ) the ultimate load capacity, and (b) the allowable load wit h F
g
= 2.5. Th e averag e
SPT value N
cor
= 30 for 0 = 36°.
Use (i ) V esic's method, an d (ii ) the O'Neill an d Reese method .
Solution
(i) Vesic's method
FromEq. (17.10) f or e = 0
<?;= 25x1 7.5 = 437.5 kN/m
2
df — 25° 3 6 — 25
From Eq. (17.16) <j)
rd
= *
y
_ ^= -^- = 0.55
FromEq. (17.15)
A
- 0-005(1 - 0-55) x 437.5 ^
101
FromEq. (17.14) ju
d
=0. 1 + 0.3^ =0. 1 + 0.3x0.55 = 0.265
From Eq . (17.17) E
d
= 200p
a
= 200 x 101 =20,200 kN/m
2
_T T M- 7 1 Q N ,
E
d 20,20 0
FromEq. (17.13) / = -2 -= -: -= 25
r
' 2( 1 + 0.265) x 437.5 tan 36°
FromEq. (17.12 ) /
= =
_ _
=
20
"" 1 + A7, 1 + 0.01x25
FromEq. (17. l i b)
C =ex p [-3.8tan36°] + ; ~ °
1 U
= exp-(0.9399) = 0.39 1
3.07 sin 36° Iog
10
2x20
From Fig . 17.14 , N
q
= 30 for 0 = 36°
Deep Foun dat io n III : Drille d Pie r Foun dat ion s 7 7 3
L j — i
'
15m
Sand
0 = 36°
c = 0
y = 17.5 kN/m
3
A^ = 30
d=1.5m
Figure Ex . 1 7. 5
From Table (17.2 ) s
q
= 1+ tan 36° = 1.73
d = l + 2t an36°(l - si n36°)
2
q
18 0
— =1.37 3
1.5
Substituting i n Eq. (a)
q
b
= (30 -1) x 437.5 x 1.73 x 1.373 x 0.391 = 1 1,783 kN/m
2
> 1 1,000 kN/m
2
As per Tomlinson (1986) the computed q
b
should be less than 1 1,000 kN/m
2
.
314
Hence Q
b
= -
1
— x (1.5)
2
x 1 1,000 = 19,429 k N
Skin load Q
f
From Eqs (17.31) and (17.32)
f
s
=fa'
0
. 0 = 1- 5 -0.245z°-
5
, wher e z = - = — = 12. 5 m
Substituting
0 = 1.5 - 0.245 x (1 2.5)°-
5
= 0.63
Hence f
s
= 0.63x437.5 = 275.62 kN/m
2
Per Tomlinson (1986) f
s
shoul d be limite d to 1 10 kN/m
2
. Hence f
s
= 1 10 kN/m
2
Therefore Q
f
= ndL f
s
= 3.14 x 1.5 x 25 x 1 10 = 12,953 kN
774 Chapt e r 1 7
Ultimate load Q
u
= 19,429+12,953 = 32,382 k N
G.=^= 12,953 kN
O' Neill and Reese metho d
This metho d relates q
b
to the SPT N value as per Eq. (17.19a)
q
b
= 51. 5N kN/m
2
= 57.5 x 30 = 1,725 kN/m
2
Q
b
= A
b
q
b
= 1.766 x 1,725 -3,046 k N
The method fo r computing Q, remains the same as above.
Now Q
u
= 3,406 + 12,953 = 15,999
15999 ^ w^v
a
2. 5
Example 1 7 . 6
Compute Q
u
and Q
a
for the pier given in Ex. 17. 5 by the following methods .
1 . Us e th e SP T valu e [Eq. ( 1 5 .48) j fo r bored pile s
2. Us e the Tomlinson method of estimating Q
b
and Table 15. 2 for estimating Q,. Compare the
results of the variou s methods.
Solution
Use of the SPT value [Mey erhof Eq. (15.48)]
q
b
= U3N
cor
= 133x 30 = 3,990 kN/m
2
Q
b
=
3
'
14X(L5)2
x 3,990 = 7,047 k N
f
s
= 0.67 W
cor
= 0.67 x 30 = 20 kN/m
2
Q
f
=3. 14x1. 5x25x2 0 = 2,355 k N
Q
u
= 7,047 + 2,355 = 9,402 k N
_ 9,40 2
a
~ 2. 5 ~ '
Tomlinson Metho d fo r Q
b
For a drive n pile
L 2 5
From Fig . 15. 9 N
o
= 65 for 0 = 36° and — = — * 17
a 1. 5
Hence q
b
= q'
o
N
q
= 437.5 x 65 = 28,438 kN/m
2
Deep Foun dat io n III : Drille d Pie r Foun dat ion s 77 5
For bored pil e
q
b
=-q
b
(driven pile) = -x 28,438 = 9,479 kN/m
2
Q
b
= A
b
q
b
= 1.766 x 9,479 = 16,740 k N
Q
f
fro m Tabl e 15.2
For </ » = 36°, 8 = 0.75 x 36 = 27, and K
s
= 1. 5 (for medium dense sand).
— 43 7 5
f
s
= q'
o
K
s
ta n S= --x 1.5 tan 27° = 167 kN/m
2
As per Tomlinson (1986)/, is limited to 1 10 kN/m
2
. Use/
5
= 1 10 kN/m
2
.
Therefore Q
f
= 3.14 x1.5 x 25 x 1 10 = 12,953 k N
Q
u
= Q
b
+ Q
f
= 16,740 + 12,953 = 29,693 k N
= 1
2.5
Com paris on o f es t im at e d res ults ( F = 2. 5)
Ex a mp l e N o
17.5
17.5
17.6
17.6
Na me o f met ho d
V esic
O' Neill an d Reese, for Q
b
and V esic for Q,
MeyerhofEq. (15.49 )
Tomlinson fo r Q
b
(Fig. 15.9 ) Tabl e 15. 2 for Q
f
Q
b
( k N )
19,429
3,046
7,047
16,740
Q, ( k N)
12,953
12,953
2,355
12,953
Q
u
( k N )
32,382
15,999
9,402
29,693
Q
a
( k N )
12.953
6,400
3,760
11,877
Which method to use
The variatio n i n th e value s o f Q
b
an d Q, ar e ver y larg e betwee n th e methods . Sinc e th e soil s
encountered i n the fiel d ar e generall y heterogeneou s i n character n o theor y hold s wel l fo r al l the
soil conditions . Designer s hav e t o b e practica l an d pragmati c i n th e selectio n o f an y on e o r
combination of the theoretical approache s discusse d earlier .
Example 1 7 . 7
For the problem given in Example 17.5 determine the allowable load for a settlement of 1 0 mm. All
the other dat a remai n the same . Us e (a ) the values of Q , and Q
b
obtained by V esic's method , and
(b) Q
b
from th e O'Neill an d Reese method.
Solution
(a) Vesic' s value s g.and Q
b
Settlement rati o for S = 10 mm i s
d 1. 5 x l O
3
From Fig. 17.20 for S
R
= 0.67%N ^= 0.96 (approx.) using the trend line.
776 Chapt er 1 7
2^= 0.96x0,.= 0.96x12,953 =12,435 k N
From Fig . 17.2 1 for S
R
= 0.67%
N
bm
=0.20 , o r Q
bm
=0.20x19,42 9 = 3,886 k N
Q
as
= 12,435 + 3,886 = 16,321 k N
Shear failur e theor y giv e Q
a
- 12,95 3 k N whic h i s muc h lowe r tha n Q
as
. As suc h Q
a
determines th e criteri a for design.
(b) O'Neill and Rees e Q
b
= 3,046 kN
As above, Q
bm
= 0.20 x 3,046 = 609 kN
Using Qr i n (a ) above ,
Q
as
= 609 + 12,43 5 = 13,04 4 kN
The valu e of Q
as
is closer t o Q
a
(V esic) but much higher than Q
a
calculated by al l the othe r
methods.
Example 1 7 . 8
Figure Ex. 17. 8 shows a drilled pie r penetrating an IGM: clay-shal e t o a depth of 8 m. Joints exist s
within the IGM stratum. The following data ar e available: L
s
= 8 m (= z
c
), d = 1.5 m, q
u
(rock) = 3 x
10
3
kN/m
2
, £. (rock) = 600 x 10
3
kN/m
2
, concret e slump = 17 5 mm, uni t weight of concrete y
c
= 24
kN/m
3
, E
c
(concrete ) = 435 x 10
6
kN/m
2
, and RQD = 70 percent, q
u
(concrete ) = 435 x 10
6
kN/m
2
.
Determine th e ultimate frictional loa d (X(max) .
17m
Soft cla y
d= 1. 5m
Rock
(IGM-clay-shale)
Figure Ex . 1 7. 8
Deep Foun dat io n III: Drille d Pie r Foun dat ion s 77 7
Solution
(a) Determine a i n Eq. (17.26)
f
a
- ocq
u
wher e q
u
= 3 MPa for rock
For th e dept h o f socke t L
s
= $> m, an d slump =17 5 mm
M = 0.76 from Fig. 17.17.
From Eq. (17.27 )
a
n
= My
c
z
c
= 0.76 x 24 x 8 = 146 kN/m
2
p
a
= 101 kN/m
2
, <J
n
/p
a
= 146/101= 1.45
From Fig. 17.16 for q
u
= 3 MPa and &
n
/P
a
= 1-45, w e have a = 0.11
(b) Determinatio n o f f
a
/
fl
= 0.11 x 3 = 0.33 MPa
(c) Determination/^ i n Eq. (17.28)
For RQ D = 70%, EJ E
i
= 0. 1 fro m Tabl e 17. 5 for ope n joints , an d f
aa
/f
a
( = tf
fl
) = 0.5 5 fro m
Table 17.6
/max =faa=°'
55 X
°'
33 =
°"
182 MP &= 182
^^
(d) Ultimate friction loa d Q,
Qf =
PL
faa =
3
-
14
x 1.5 x 8 x 182 = 6,858 k N
Example 1 7 .9
For th e pie r give n i n Ex . 17.8 , determine th e ultimat e bearing capacit y o f th e base . Neglec t th e
frictional resistance . Al l the other dat a remai n the same .
Solution
For RQD between 70 and 10 0 percent
from Eq . (17.22 )
q
b
( = <?
max
) = 4.83(<7
M
)°-
5
MPa = 4.83x(3)°-
5
= 8.37 MPa
Q
b
(max) = —xl . 5
2
x 8.37 = 14.78 MN = 14,780 k N
1 7 .1 6 U PL I F T CAPACI TY OF DRI L L ED PI ERS
Structures subjected t o large overturning moments can produce uplif t load s o n drilled pier s i f they
are use d fo r th e foundation . Th e desig n equatio n fo r uplif t i s simila r t o tha t o f compression .
Figure 17.2 2 shows the forces actin g on the pier under uplift-load Q
u[
. The equation for Q
ul
may be
expressed as
Qul=Qfr
+W
P
=
A
sfr+
W
P
(17-40 )
778
Chapt er 1 7
where, Q
fr
= tota l side resistance for uplif t
W = effectiv e weigh t of the drille d pier
A
S
- surfac e area of the pier
f
r
= frictiona l resistanc e to uplif t
U plift Capacit y o f Singl e Pie r (straigh t edge )
For a drilled pie r i n cohesive soil , th e frictiona l resistanc e may expressed a s (Che n an d Kulhawy,
1994)
f
r
=
= 0.31 + 0.1
Pa
(17.41a)
(17.41b)
where, a = adhesio n factor
c
u
= undraine d shear strengt h of cohesive soi l
p
a
= atmospheri c pressure (101 kPa )
Poulos an d Davis, (1980) sugges t relationships between c
u
and a as given in Fig. 17.23 . The
curves i n the figur e are based o n pul l ou t tes t dat a collected b y Sowa (1970) .
Uplift Resistanc e o f Pier s i n Sand
There ar e n o confirmator y methods availabl e for evaluatin g uplift capacit y o f pier s embedde d i n
cohesionless soils . Poulo s an d Davis, (1980) sugges t that the ski n frictional resistance fo r pul l out
may b e take n as equal to two-thirds of the shaf t resistanc e for downward loading.
U plift Resistanc e o f Pier s i n Roc k
According t o Carter an d Kulhawy (1988), the frictional resistance offered by the surface of the pier
under uplif t loadin g i s almost equal to that for downward loading i f the drilled pie r i s rigid relativ e
Figure 1 7 .2 2 Uplif t for ce s fo r a s t raig ht edg e d pie r
Deep Foun dat io n Drilled Pie r Foun dat ion s 779
to the rock. The effectiv e rigidit y is defined as ( EJ E^dlD)
2
, i n which E
c
and E
m
ar e the Young's
modulus of the drilled pier an d rock mass respectively, d is the socket diamete r and D
S
i s the depth
of the socket . A socket i s rigid when ( EJ E^dlD^
2
> 4. When the effective rigidit y is less tha n 4,
the frictiona l resistanc e f
r
fo r upwar d loadin g ma y b e take n a s equa l t o 0. 7 time s th e valu e fo r
downward loading .
Example 1 7 .1 0
Determine th e uplif t capacit y of the drilled pier given i n Fig. Ex . 17.10 . Neglect th e weight of the
pier.
Solution
FromEq. (17.40 )
From Eq . (17. 4 la) f
r
= ae
u
FromEq. (17.41b )
= 0.31 + 0.1
c = 150 kN/m
2
Clay
Given: L = 25m, d- 1.5m , c =150kN/ m
2
Hence a = 0.31 + 0.17 x
150
101
= 0.5 6
1.5m
25m
f
r
=0.56x15 0 = 84 kN/m
2
Q
ul
=3. 14x1. 5x25x8 4
= 9,891 k N
Figure Ex . 1 7 .1 0
It may be noted her e that/
y
= 82.5 kN/m
2
for downward loadin g and/
r
= 84 kN/m
2
for uplift.
The two values are very close t o each other .
1 7 .1 7 L ATERA L B EARI N G CAPACI T Y OF DRILLED PIER S
It is quite common that drilled piers constructed for bridge foundations and other similar structures
are als o subjecte d t o latera l load s an d overturnin g moments. The method s applicabl e t o piles ar e
applicable t o pier s also . Chapte r 1 6 deals wit h suc h problems . This chapte r deal s wit h one mor e
method as recommended b y O'Neill an d Reese (1999) . Thi s method i s called Characteristi c loa d
method and is described below .
Characteristic L oa d M etho d (Dunca n et a!. , 1994 )
The characteristic load method proceeds b y defining a characteristic or normalizing shear load (P
c
)
and a characteristic or normalizing bending moment (M
c
) as given below.
For clay
M = 3.
ER,
ER,
0.68
0.46
(17.42)
(17.43)
780 Chapt er 1 7
For san d
P =1 .
y'd 0'K
t
0.57
(17.44)
M
c
= 1.33d
3
(£/?,)
ER
{
(17.45)
where
= shaf t diamete r
E = Young's modul us of the shaf t materia l
R
;
= ratio of momen t of inerti a of drilled shaf t t o moment o f inerti a of soli d sectio n ( = 1
for a normal uncracked drilled shaf t without central voids )
c
u
= average valu e of undrained shear strengt h o f the clay i n the top 8 dbelow the groun d
surface
Y ' - averag e effectiv e uni t weigh t of the sand (tota l uni t weigh t abov e th e water table ,
buoyant uni t weight belo w th e wate r table) i n the top %d below th e groun d surfac e
0' = average effectiv e stress frictio n angle for the sand i n the top 8d below groun d surfac e
K = Rankine's passive earth pressur e coefficient = tan
2
(45° + 072 )
In the design method, the moment s and shears are resolved int o groundline values, P
t
and M
(
,
and then divided by the appropriate characteristic load value s [Equations (17.42) throug h (17.45)].
The latera l deflection s a t th e shaf t head , y
;
ar e determine d fro m Figure s 17.2 3 an d 17.24 ,
considering th e condition s o f pile-hea d fixity . Th e valu e o f th e maximu m momen t i n a free - o r
fixed-headed drille d shaf t ca n be determined through the use of figure 17.2 5 i f the onl y load tha t i s
applied i s groun d lin e shear . I f bot h a momen t an d a shea r ar e applied , on e mus t comput e y
t
(combined), an d then solve Eq. (17.46) fo r the "characteristi c length" T (relative stiffnes s factor) .
PT
3
M,T
2
v, (combined) = 2.43 -±— +1.62——
' ' E l El
where / i s the moment o f inerti a of the cross-section of the drilled shaft .
(17.46)
0.045
0.030
o 0.01 5
0.000
0.00
Fixed
Free
0.05 0.1 0
Deflection rati o y I d
0.15
0.015
0.010
0.005
0.000
0.00
1
Fixed
Free^
(b) San d
0.05 0.1 0
Deflection rati o y
p
Id
0.15
Figure 17.2 3 Groundlin e s hear-deflect io n curve s fo r (a ) clay an d (b ) san d
(Dun can e t a!. , 1994 }
Deep Foun dat io n Drilled Pie r Foun dat ion s 781
U.UJ
0.02
0.01
O
nnn
-
-
: /
/
/
/
'
/
(a) Clay
0.00 0.0 5 0.1 0 0 .
Deflection rati o y
m
Id
0.015
0.010
0.005
0.000
(b) Sand
0.00 0.0 5 0.1 0
Deflection rati o y
m
Id
0.15
Figure 17.2 4 Groundlin e moment-deflectio n curves fo r (a ) clay an d (b) san d
(Dun can et al. , 1994)
0.045
0.030
0.015
0.000
Free
(a) Clay
0.020
0.015
a,
o
|0.010
0.005
0.000
Fixed
Free
(b) Sand
0.000 0.00 5 0.01 0 0.01 5
Moment ratio M,/M
C
0.000 0.00 5 0.01 0 0.01 5
Moment ratio M,IM
C
Figure 1 7.2 5 Groun dlin e s hear-m ax im u m momen t curve s for (a ) clay an d (b) san d
(Dun can e t al. , 1994)
The principle o f superposition i s made us e of for computing groun d lin e deflections o f piers
(or piles) subjected t o groundline shears and moments at the pier head. The explanation given here
applies t o a free-head pier . The same principl e applies for a fixed hea d pil e also .
Consider a pier shown in Fig. 17.26(a ) subjected to a shear load P
{
and moment M
t
a t the pile
head a t ground level . The tota l deflectio n y
t
cause d b y th e combine d shea r an d momen t may b e
written as
y
t
=y
p
+
y
m
(n.4? )
where y = deflection du e to shear load P
t
alone wit h M
t
= 0
y
m
= deflection due t o moment M
t
alone wit h P
{
- 0
Again conside r Fig . 17.26(b) . Th e shea r loa d P
t
actin g alon e a t th e pil e hea d cause s a
deflection y (a s above) which is equal to deflection y cause d by an equivalent moment M actin g
alone.
782 Chapt er 1 7
M, M,
(b) (c)
Figure 17.2 6 Principle o f s uperpos it io n fo r computin g groun d lin e deflect io n by
Dun can e t al . , (1999 ) m et ho d fo r a fr ee-head pie r
In the same wa y Fig. 17.26(c ) shows a deflection y
m
cause d b y momen t M
{
a t the pil e head .
An equivalent shea r load P
m
causes the same deflectio n y
m
whic h i s designated her e a s y
mp
. Base d
on the principles explaine d above, groundline deflection a t the pil e head du e t o a combined shea r
load an d momen t ma y b e explaine d a s follows .
1. Us e Fig s 17.2 3 an d 17.2 4 t o comput e groundline deflections v an d y
m
du e t o shea r loa d
and moment respectively.
2. Determin e the groundline momen t M tha t wil l produce th e same deflection a s by a shear
load P (Fig . 17.26(b)) . I n the sam e way , determine a groundlin e shear loa d P
m
, that will
produce th e same deflectio n a s that by the groundline momen t M
r
(Fig. 17.26(c)) .
3. No w th e deflection s cause d b y th e shea r load s P
t
+ P
m
an d tha t cause d b y th e moment s
M
f
+ M
p
ma y b e writte n as follows:
Deep Foundatio n III : Drille d Pie r Foundation s 783
0.2 0.4 0.6 0.8 1.0
1.0
1.5
2.0
Figure 17.2 7 Param et er s Aan d B (M at loc k an d Rees e , 1961
v = V + V J
tp
J
p
J
mp
v = V + V
•'tm •'m Spm
Theoretically y
tp
= y
tm
4. Lastl y th e total deflection y
t
i s obtained as
=
y
'
+ y.)
(17.48)
(17.49)
(17.50)
The distributio n o f momen t alon g a pie r ma y b e determine d usin g Eq . (16.11 ) an d
Table 16. 2 or Fig. 17.27 .
Direct Metho d b y Making Use of n
h
The direct method develope d by Murthy and Subba Ra o (1995) for long laterall y loaded piles has
been explaine d i n Chapte r 16 . Th e applicatio n o f thi s metho d fo r lon g drille d pier s wil l b e
explained wit h a case study.
Example 1 7 .1 1 (O'Neil l and Reese, 1999 )
Refer t o Fig. Ex . 17.11 . Determin e fo r a free-hea d pie r (a ) the groundline deflection , an d (b ) the
maximum bendin g moment . Us e th e Dunca n e t al. , (1994 ) method . Assum e R
f
= 1 i n th e
Eqs (17.44) and (17.45).
Solution
Substituting i n Eqs (17.42) and (17.43)
P =7.34x(0.80)
2
[25xl0
3
x(l)]
0.06
0.68
25xl 0
3
= 17.72MN
784 Chapt er 1 7
M =3. 86(0. 80)
3
[25xl 0
3
x(l )]
0.46
,
25xl 0
3
= 128. 5 MN-m
P, 008 0 M , 0 4
Now -t - = -^-^ = 0.0045 - ^ = -1—= 0.0031
P 17.7 2 M 128. 5
Stepl
From Fig. 17.23a for — = 0.0045
-- = 0.003 or y = 0.003 x 0.8 x10
3
= 2.4 mm
d
From Fig. 17.24a , M,/M
C
= 0.0031
- = 0.006 o r y = 0.006d = 0.006 x 0.8 x 10
3
= 4.8 mm
d
Step 2
From Fig. 17.23(a ) foryjd=0.006 , PJ P
c
= 0.0055
From Fig. 17.24(a) , for y I d = 0.003, M
p
IM
c
= 0.0015.
P, = 80 kN
1,
5 m
V
,
9
//X\ //^\
m
— d = 0. 8m
^^ //> ^\
Clay
c
u
= 60 kPa
El = 52.6 x 10
4
kN-m
2
y = 17. 5 kN/m
3
(assumed)
£= 25x 10
6
kN/m
2
Figure Ex . 1 7 .1 1
Deep Foun dat io n III : Drille d Pie r Foun dat ion s 78 5
Step 3
The shea r loads P
t
and P
m
applie d at ground level, may be expressed a s
P P
-L +_2L = 0.0045 + 0.0055 = 0.01
P P
c c
From Fig . 17.23 ,
^- = 0.013 fo r
P
'
+Pfn
= 0.01
d P
c
or y
tp
=0. 013x(0. 80)xl 0
3
= 10.4 mm
Step 4
In the same wa y as i n Step 3
M, M
—*- + —£- = 0.0031 + 0.0015 = 0.004 6
M
c
M
c
v v + y
,-, ™ i~,r*A • /tin
m
P
m
A A 1 1
From Fig. 17.24 a = — = 0.011
d d
Hence y
tm
= 0.011 x 0. 8 x 10
3
= 8. 8 mm
StepS
From Eq. (17.50 )
y
tp
+
y
[ m
10.4+8. 8
v, = — - -= -= 9.6 mm
1
2 2
Step 6
The maximum moment for the combined shear load and moment at the pier head may be calculated
in th e same way a s explained in Chapter 16. M
(max)
as obtained is
M = 470.5 kN- m
ITlaX
This occur s at a depth of 1. 3 m below ground level.
Example 1 7 .1 2
Solve the problem i n Ex. 17. 1 1 by the direct method.
Given: EI= 52.6 x 10
4
kN-m
2
, d = 80 cm, y= 17.5 kN/m
3
, e = 5 m, L = 9m, c = 60 kN/m
2
and
P
t
= 80 kN.
Solution
Groundline deflectio n
From Eq . (16.30 ) for piers i n clay
n, -
1.5
^
d
786 Chapt e r 17
Substituting and simplifying
125 x 60
1
-
5
V 52.6 x 10
4
x 17.5 x 0.8 80 8 x l O
4
n
h
= — = -—— kN/m
3
1 + — xP
1
'
5 e
0.8
Stepl
Assume P
e
= P
t
= 80 kN,
From Eq. (a), n
h
= 11,285 kN/m
3
and
0.2
A
0 2
El 52. 6 x l O
4
T= — = =2. 16 m
n
h
11,28 5
Step 2
From Eq. (16.22) P=Px 1 + 0.67— =8 0 l + 0.67x — = 204 kN
' T 2.1 6
FromEq. (a), n
h
=2112 kN/m
3
, henc e 7 = 2.86 m
Step 3
Continue the above proces s til l convergenc e is reached. Th e fina l value s are
P
e
= 111 kN, n
h
= 3410 kN/m
3
and T = 2.74 m
For P
e
= 190 kN, w e have n
h
= 8,309 kN/m
3
and T = 2.29 m
Step 4
FromEq. (17.46 )
2.43 x 177 x(2.74)
3
y, = x 1000 = 16.8 mm
1
52. 6 x l O
4
By Dunca n et al , method y
t
= 9.6 mm
Maximum momen t fro m Eq. (16.12 )
M = [ P
t
T] A
m
+( M
t
] B
m
=[ Wx2.W] A
m
+[ WO] B
m
=2l92A
m
+4QQB
m
Dept h xlT = Z A
m
B
m
M^
0 0 1 0
0.4 0.37 9 0.9 9 8 3
0.5 0.4 6 0.9 8 10 1
0.6 0.5 3 0.9 6 11 6
0.7 0.6 0 0.9 4 13 2
0.8 0.6 5 0.9 1 14 2
The maximum bending moment occurs at x/T=0.1 orx = 0
Duncan e t al. , method M(max) = 470.5 kN-m . This occur s a t a
M
2
400
396
392
384
376
364
. 7x2. 74=1. 91
depth o f 1. 3 m.
M ( k N- m )
400
479
493
500
508 (max)
506
m (6.26 ft). As per
Deep Foun dat io n Drilled Pie r Foun dat ion s 7 8 7
250
200 —
£7= 14.1 8 x 10
6
kip-ft
2
= 38°
Circled numeral s show time i n minutes
between observation s
0.10 0.20 0.3 0 0.4 0
Deflection a t elevation of load i n
0.50 0.60
Figure 1 7.2 8 Loa d deflect io n relat ion s hip , Pie r 25 (Davis s o n e t al. , 1969 )
1 7 .1 8 CAS E STU DY O F A DRI L L ED PIER SU B JECTED T O
L ATERAL L OAD S
Lateral loa d tes t wa s performe d o n a circula r drille d pie r b y Davisso n an d Salle y (1969) . Stee l
casing pip e wa s provided fo r th e concret e pier . The detail s o f the pier an d the soi l propertie s are
given i n Fig . 17.28 . Th e pie r wa s instrumente d and subjecte d t o cycli c latera l loads . Th e loa d
deflection curv e as obtained b y Davison et al. , i s shown in the same figure.
Direct method (Murthy and Subba Rao, 1995) has been used here to predict the load displacement
relationship fo r a continuous load increas e b y making use of Eq. (16.29). The predicted curv e i s also
shown i n Fig. 17.28 . There i s an excellent agreement between the predicted and the observed values.
1 7 .1 9 PROB L EM S
17.1 Fig . Prob . 17. 1 show s a drilled pie r of diameter 1.2 5 m constructed fo r the foundation of a
bridge. The soi l investigatio n at the sit e revealed sof t t o medium stif f cla y extending to a
great depth. The other details of th e pier and the soil are given in the figure. Determine (a )
the ultimat e load capacity , an d (b) the allowable loa d fo r F
s
= 2.5. Us e V esic's metho d fo r
base load an d a metho d fo r the skin load.
17.2 Refe r to Prob. 17.1 . Give n d = 3 ft, L = 30 ft, and c
u
= 1050 lb/ft
2
. Determin e th e ultimat e
(a) base loa d capacit y b y V esic' s method , an d (b ) th e frictiona l loa d capacit y b y th e a-
method.
788 Chapt er 1 7
17.3 Fig . Prob . 17. 3 shows a drilled pie r with a belled botto m constructe d fo r the foundation of
a multistory building. The pier passes through two layers of soil. The details of the pier and
the propertie s o f th e soi l ar e give n i n th e figure . Determin e th e allowabl e loa d Q fo r
F
s
= 2.5. Us e (a ) V esic's metho d fo r the base load , an d (b ) the O'Neil l an d Rees e metho d
for ski n load .
Q
Q
T
/W$^
L =
/2*S\
5 m
/
T
//V \\\//9vv \
/ d= 1.25 m
Soft t o medium
stiff cla y
c
u
= 25 kN/m
2
y = 18. 5 kN/m
3
yv
cor
= 4
17.4
17.5
17.6
17.7
V
3C
1C
ft
" ~T~ /
3 f t (
\ \
/
V
Layer 1
^ d = 3 f t
c
u
= 600 lb/ft
2
Clay
Layer 2
v Cla y
N c
u
= 2 100 lb/ft
2
Figure Prob . 1 7. 1
For th e drille d pie r give n i n Fig . Prob . 17.1 ,
determine the working load for a settlement of
10 mm. Al l th e othe r dat a remai n th e same .
Compare th e workin g load wit h th e allowabl e
load Q
a
.
For th e drille d pie r give n i n Pro b 17.2 ,
compute th e workin g load fo r a settlemen t of
0.5 in . an d compar e thi s wit h th e allowabl e
load Q
a
.
If the drilled pie r given i n Fig. Prob . 17. 6 i s to
carry a saf e loa d o f 250 0 kN , determin e th e
length o f th e pie r fo r F
s
- 2.5 . Al l the othe r
data ar e given i n the figure.
Determine th e settlemen t of th e pie r give n i n
Prob. 17. 6 b y th e O'Neil l an d Rees e method .
All the other dat a remai n th e same .
Fig. Prob . 17. 8 depict s a drille d pie r wit h a
belled botto m constructe d i n homogeneou s
clay extending to a great depth. Determine th e
Figure Prob. 17.3
\Q
L = ?
d= 1.25 m
Clay
c= 125 kN/m
2
Figure Prob . 17. 6
Deep Foundatio n III: Drille d Pier Foun dat ion s 789
Q
Q
Clay
c = 15 0 kN/m
2
, = 4 0 f t
2 m
Figure Prob. 17. 8
Loose to
medium dense sand
0 = 34°
y=1151b/ f t
3
Figure Prob. 17.1 0
length of the pier t o carry an allowable load of 3000 kN with a F
g
- 2.5. The other detail s
are given i n the figure .
17.9 Determin e th e settlemen t o f the pie r i n Prob 17. 8 fo r a working loa d o f 3000 kN. All th e
other dat a remai n the same. Us e the length L computed.
17.10 Fig . Prob 17.10 shows a drilled pier. The pier is constructed in homogeneous loose to medium
dense sand . The pier detail s an d the properties o f the soil ar e given i n the figure. Estimate b y
V esic's metho d th e ultimat e load bearing capacity
of the pier .
17.11 Fo r Proble m 17.1 0 determin e th e ultimat e bas e
capacity b y th e O'Neil l an d Rees e method .
Compare thi s valu e wit h th e on e compute d i n
Prob. 17.10 .
17.12 Comput e th e allowabl e loa d fo r th e drille d pie r
given i n Fig . 17.1 0 base d o n th e SP T value . Use
Meyerhof's method .
17.13 Comput e the ultimate base load of the pier in Fig.
Prob. 17.1 0 by Tomlinson' s method .
17.14 A pie r i s installe d i n a rock y stratum . Fig. Prob .
17.14 give s th e detail s o f th e pie r an d th e
properties o f th e roc k materials . Determin e th e
d = 1 . 5 m
Jointed
rock, slightl y
weathered
ultimate frictional loa d <2,(max) .
17.15 Determin e th e ultimat e bas e resistanc e o f th e
drilled pie r i n Prob . 17.14 . Al l th e othe r dat a
remain the same. Wha t i s the allowable loa d wit h
F
s
= 4 by taking into account the frictional loa d Q,
computed i n Prob. 17.14 ?
q
u
(rock) = 2 x 10
3
kN/m
2
£, (rock) = 40 x 10
4
kN/m
2
E
c
(concrete) = 435 x 10
6
kN/m
2
RQD = 50%
q
u
(concrete) = 40 x 10
4
kN/m
2
slump = 175 mm, y
c
= 23.5 kN/m
3
Figure Prob. 17.1 4
790 Chapt er 1 7
17.16 Determin e th e ultimat e point bearing capacity o f th e pie r give n i n Prob . 17.1 4 i f the bas e
rests o n soun d rock wit h RQD = 100%.
Determine th e uplif t capacit y o f th e drille d pie r give n i n Prob . 17.1 . Given : L = 1 5 m, 17.17
d = 1.25 m, and c
u
- 2 5 kN/irr. Negl ec t the wei ght of the pile.
17.18 Th e drille d pie r given in Fig. Prob. 17.1 8 is subjected t o a lateral load of 12 0 kips. The soi l
is homogeneou s clay . Given : d - 6 0 i n. , El = 93 x 10'° lb-in.
2
, L = 38 ft, e- 1 2 in. ,
c - 200 0 I b ft
2
, an d y
h
= 60 lb/ft
3
. Determin e by the Duncan et al. , metho d th e groundl ine
deflection.
P,= 12 0 kip
L = 38 f t
_L
e = 1 2 in.
d = 60 i n
Clay
c
u
= 2000 lb/ft
2
y
b
= 60 lb/ft
3
E/ = 93x 10' ° lb-in.
2
£= 1. 5 x 10
6
l b/ i n.
2
Figure Prob . 1 7 .1 8
CHAPTER 1 8
FOUNDATIONS ON COLLAPSIBLE AND
EXPANSIV E SOIL S
18.1 G ENERA L CONSI DERATI ON S
The structure of soils that experience larg e loss of strength or great increase i n compressibility with
comparatively smal l change s i n stress or deformations i s said t o be metastable (Pec k et al. , 1974) .
Metastable soil s includ e (Peck e t al. , 1974) :
1. Extra-sensitiv e clays such as quick clays,
2. Loos e saturate d sands susceptibl e to liquefaction,
3. Unsaturate d primaril y granula r soil s i n whic h a loos e stat e i s maintaine d b y apparen t
cohesion, cohesio n due t o clays a t the intergranular contact s o r cohesion associated wit h
the accumulation of solubl e salt s as a binder, and
4. Som e saprolite s eithe r above or below the water table in which a high voi d rati o has been
developed a s a resul t o f leachin g tha t has lef t a networ k o f resistant mineral s capabl e of
transmitting stresses aroun d zones i n which weake r mineral s or voids exist .
Footings o n quick clays can be designed by the procedures applicabl e fo r clays as explained
in Chapter 12 . V ery loose sand s shoul d not be used for suppor t of footings. This chapter deal s only
with soils under categories 3 and 4 listed above .
There ar e two types of soils that exhibit volume changes under constant loads wit h changes in
water content. The possibilities are indicated in Fig. 18. 1 which represent th e result of a pair of tests
in a consolidation apparatu s o n identical undisturbe d samples . Curve a represents the e-\ og p curve
for a test starte d at the natural moisture content and to which no water i s permitted access . Curve s
b and c, on the other hand, correspond t o tests on samples to which water is allowed access unde r all
loads unti l equilibrium is reached. If the resulting e-\ og p curve, such as curve b, lies entirely belo w
curve a, th e soil is said to have collapsed. Unde r field conditions , at present overburden pressure/?,
791
792 Chapt er 1 8
P\ Pi
Pressure (lo g scale )
Figure 18. 1 B ehavio r o f s oi l i n doubl e oedom et e r o r paire d con fin e d com pres s io n
t es t (a ) relat ion bet wee n voi d rat i o an d t ot al pres s ur e for s am pl e t o whic h n o wat e r
is added , (b ) rel at ion fo r iden t ica l s am pl e t o whic h wat e r i s al l owe d acces s an d
which ex perien ce s col l aps e, (c ) s am e as (b) for s am pl e t hat ex hibit s s wellin g
(af t er Pec k et al. , 1974)
and voi d rati o e
Q
, the additio n o f wate r a t the commencemen t o f th e test s t o sampl e 1 , causes th e
void rati o to decrease t o e
v
The collapsible settlement S
c
may be expressed a s
S =
(I S. l a)
where H = the thicknes s o f the stratu m in the field .
Soils exhibiting thi s behavior includ e true loess , claye y loose sands i n which the clay serve s
merely a s a binder, loose sand s cemented b y solubl e salts, an d certai n residual soil s suc h as those
derived fro m granite s under conditions of tropical weathering.
On th e othe r hand , i f the additio n o f wate r t o th e secon d sampl e lead s t o curv e c , locate d
entirely abov e a , th e soi l i s sai d t o hav e swelled. A t a give n applie d pressur e p
r
th e voi d rati o
increases to e',, and the corresponding ris e of the ground i s expressed as
S =
(18.Ib)
Soils exhibitin g this behavior t o a marke d degre e ar e usuall y montmorilloniti c clay s wit h
high plasticit y indices.
Foun dat ion s o n Collaps ibl e and Ex pan s ive Soil s 793
PART A—COL L APSI B L E SOI L S
1 8.2 G ENERA L OB SERV ATI ON S
According to Dudley (1970), an d Harden et al., (1973), fou r factor s are needed t o produce collaps e
in a soil structure:
1. A n open , partiall y unstable, unsaturated fabric
2. A high enough net total stres s tha t wil l cause the structure to be metastabl e
3. A bonding or cementing agent that stabilizes the soi l i n the unsaturated condition
4. Th e additio n o f wate r t o th e soi l whic h cause s th e bondin g o r cementin g agen t t o b e
reduced, an d th e interaggregat e o r intergranula r contact s t o fai l i n shear , resultin g i n a
reduction i n total volume of the soi l mass .
Collapsible behavio r of compacted an d cohesive soils depends on the percentage o f fines, the
initial wate r content, the initial dry density and the energy an d the process use d i n compaction .
Current practic e i n geotechnical engineerin g recognizes a n unsaturated soil a s a four phas e
material compose d o f air , water , soi l skeleton , an d contractil e skin . Unde r th e idealization , tw o
phases can flow, that is air and water, and two phases come to equilibrium under imposed loads, tha t
is the soi l skeleto n and contractile skin. Currently, regarding the behavior of compacted collapsin g
soils, geotechnical engineerin g recognized that
1. An y typ e o f soi l compacte d a t dr y o f optimum condition s an d a t a lo w dr y densit y ma y
develop a collapsible fabri c or metastabl e structure (Barden et al., 1973) .
2. A compacted an d metastabl e soi l structur e is supporte d b y microforce s o f shea r strength ,
that i s bonds , tha t ar e highl y dependen t upo n capillar y action . Th e bond s star t losin g
strength wit h the increase of the water content and at a critical degree o f saturation, the soil
structure collapses (Jenning s and Knight 1957; Barde n e t al., 1973) .
Figure 18. 2
Symbols
Major loess
deposits
Reports of collapse
in other type deposits
Locat ion s o f m aj o r loes s depos it s i n t he Unite d St at e s alon g wit h ot he r
s it es o f report e d collaps ibl e s oil s (aft e r Dudley, 1 970)
794 Chapt er 1 8
50
60-
70-
90
100-
110
Soils hav e been
observed t o collaps e
G =2. 6
G, = 2.7
Soils have not
generally bee n
observed t o
collapse
\ I \
10 2 0 3 0 4 0
Liquid limit
50
Figure 18. 3 Collaps ibl e an d n on collaps ibl e loes s (aft e r Holt z an d Hilf , 196 1
3. Th e soi l collaps e progresse s a s th e degre e o f saturatio n increases . Ther e is , however , a
critical degre e o f saturatio n for a give n soi l abov e whic h negligibl e collaps e wil l occu r
regardless of the magnitude of the prewetting overburden pressur e (Jenning s and Burland ,
1962; Housto n et al. , 1989) .
4. Th e collapse o f a soil is associated wit h localized shear failures rather than an overall shea r
failure o f the soi l mass .
5. Durin g wettin g induced collapse, unde r a constant vertical load an d unde r K
o
-oedometer
conditions, a soil specime n undergoe s a n increase i n horizontal stresses .
6. Unde r a triaxial stres s state, th e magnitude of volumetric strai n resultin g fro m a change i n
stress stat e or from wetting , depends on the mean normal total stress an d is independent of
the principal stress ratio.
The geotechnical engineer need s to be able to identify readily the soils that are likely to collapse
and to determine the amount of collapse that may occur. Soils that are likely to collapse ar e loose fills ,
altered windblown sands, hillwash of loose consistency, and decomposed granite s and acid igneous
rocks.
Some soil s at their natural water content will support a heavy load but when water is provided
they underg o a considerabl e reductio n i n volume . Th e amoun t o f collaps e i s a functio n o f th e
relative proportion s o f eac h componen t includin g degree o f saturation , initia l voi d ratio , stres s
history o f the materials, thicknes s of the collapsibl e strata an d the amoun t of added load .
Collapsing soil s of the loessia l typ e ar e found in many part s o f the world . Loes s i s found in
many part s o f th e Unite d States , Centra l Europe , China , Africa , Russia , India , Argentin a an d
elsewhere. Figur e 18. 2 gives the distribution o f collapsible soi l i n the United States .
Foun dat ion s o n Collaps ibl e an d Ex pan s ive Soil s 795
Holtz and Hil f (1961 ) proposed th e us e of the natural dry densit y and liqui d limit as criteria
for predictin g collapse. Figur e 18. 3 show s a plot givin g the relationshi p betwee n liqui d limi t and
dry uni t weight of soil, such that soils that plot above the line shown in the figure are susceptible to
collapse upo n wetting.
18.3 COL L APS E POTENTI A L AND SETTL EM ENT
Collapse Potentia l
A procedure for determining the collapse potentia l of a soil was suggested by Jennings and Knight
(1975). The procedur e i s as follows:
A sample of an undisturbed soil is cut and fit into a consolidometer rin g and loads ar e applied
progressively unti l about 20 0 kPa ( 4 kip/ft
2
) i s reached. At thi s pressur e th e specime n i s floode d
with water for saturation and lef t fo r 24 hours. The consolidation tes t i s carried o n t o it s maximum
loading. The resulting e-log p curv e plotted fro m th e dat a obtained i s shown in Fig. 18.4 .
The collapse potentia l C i s then expressed a s
(18.2a)
in whic h Ae
c
= change i n voi d rati o upon wetting, e
o
= natural void ratio .
The collaps e potentia l is also defined as
C -
C
~
H
(18.2b)
where, A//
c
= change i n the height upo n wetting, H
C
- initia l height .
\
Pressure P J o g p
Figure 18. 4 T ypica l collaps e pot en t ial t es t res ul t
796 Chapt er 1 8
Table 1 8 . 1 Col l aps e pot en t ial val ue s
C//0
0-1
1-5
3-10
10-20
>2 0
Severity o f pr obl e m
No proble m
Moderate troubl e
Trouble
Severe troubl e
V ery sever e troubl e
Jennings an d Knigh t hav e suggeste d som e value s fo r collaps e potentia l a s show n i n
Table 18.1 . These values ar e only qualitativ e t o indicate th e severit y o f the problem.
18.4 COM PU TATI O N OF COL L APSE SETTL EM EN T
The doubl e oedomete r metho d wa s suggeste d b y Jenning s an d Knigh t (1975 ) fo r determinin g a
quantitative measur e o f collapse settlement . The method consist s o f conducting two consolidatio n
tests. Two identical undisturbed soil sample s ar e used i n the tests . The procedur e i s as follows:
1. Inser t tw o identica l undisturbe d samples int o the rings of two oedometers .
Natural moistur e
content curve
Adjusted curv e
(curve 2)
Curve o f sampl e
soaked fo r 24 hr s
1 0. 2 0. 4 0. 6 A > 1. 0
0.
6 1 0 20ton/f r
Figure 18. 5 Doubl e con s olidat io n t es t an d adj us t m en ts fo r n orm all y con s olidat e d
s oil (Cl em en c e an d Fin barr , 1981 )
Foun dat ion s o n Collaps ibl e an d Ex pan s ive Soil s 797
Soil a t natural
moisture content
Adjusted n.m. c
/ curv e
Soaked sampl e
for 2 4 hours
0.1
Figure 18. 6 Doubl e con s olidat io n t es t an d adj us t m en t s fo r overcon s olidat e d soi l
(Clem en ce an d Finbarr, 1981 )
2. Kee p bot h the specimens unde r a pressure of 1 kN/m
2
( = 0.15 lb/in
2
) for a period o f 24
hours.
3. Afte r 2 4 hours, saturate one specimen b y flooding an d keep th e other at it s natural
moisture content
4. Afte r the completion o f 24 hour flooding, continue the consolidation test s for both th e
samples b y doubling the loads. Follo w th e standar d procedur e fo r the consolidation test.
5. Obtai n the necessary dat a from th e two tests, and plot e-log p curve s for bot h the sample s
as shown in Fig. 18. 5 fo r normall y consolidated soil .
6. Follo w th e same procedur e fo r overconsolidated soi l and plot the e-log p curve s as shown
in Fig . 18.6 .
From e-lo g p plots , obtai n th e initial void ratios of the two samples afte r th e first 24 hour of
loading. I t is a fact tha t the two curves do not have the same initial void ratio. The total overburden
pressure p
Q
a t the depth of the sampl e is obtained and plotted on the e-log p curve s in Figs 18. 5 and
18.6. The preconsolidation pressure s p
c
ar e found fro m the soaked curves of Figs 18. 5 and 18. 6 and
plotted.
798 Chapt e r 18
Normally Consolidate d Cas e
For th e cas e i n whic h p
c
/p
Q
i s about uni t y , the soi l i s considered normall y consolidated. I n suc h a
case, compressio n take s plac e alon g the virgi n curve . The straigh t line which is tangentia l to th e
soaked e-lo g p curv e passe s throug h th e poin t ( e
Q
, p
0
) a s shown i n Fig . 18.5 . Through th e poin t
( e
Q
, p
Q
) a curv e i s drawn paralle l t o th e e-lo g p curv e obtained fro m th e sampl e teste d a t natural
moisture content. The settlemen t for any incremen t in pressure A/ ? due t o the foundation load ma y
be expressed i n two parts as
L e H

n c
(18.3a)
where ke
n
= chang e i n void rati o du e to load A p as per the e-log p curv e withou t change i n
moisture content
Ae
c
, = chang e i n voi d ratio at the same loa d Ap with the increas e i n moistur e content
(settlement cause d du e to collapse of the soil structure )
H
c
= thicknes s of soi l stratum susceptible to collapse .
From Eq s (18. 3a) and (18. 3b), the total settlement due t o the collapse of the soi l structur e is
**„+ *e
c
) (18.4 )
Overconsolidated Cas e
In th e cas e o f a n overconsolidate d soi l the rati o p
c
/p
Q
i s greater tha n unity. Draw a curve from the
point ( e
Q
, p
0
) o n the soake d soi l curve parallel to the curve which represents n o change i n moisture
content durin g the consolidation stage . Fo r any load ( p
Q
+ A/?) > p
c
, the settlement o f the foundation
may b e determined b y makin g use of the same Eq. (18.4) . The change s i n voi d ratios /\ e
n
an d Ae
c
are define d i n Fig. 18.6 .
Example 1 8. 1
A footing of size 1 0 x 1 0 ft is founded at a depth of 5 ft below ground level in collapsible soi l of the
loessial type . Th e thicknes s o f th e stratu m susceptibl e t o collaps e i s 3 0 ft . Th e soi l a t th e sit e i s
normally consolidated . In order to determine th e collapse settlement, doubl e oedomete r tests wer e
conducted o n two undisturbed soil sample s a s per the procedur e explained in Section 18.4 . The e-
log p curves of the two samples are given in Fig. 18.5 . The average uni t weight of soil y = 106. 6 lb /
ft
3
an d th e induce d stress A/? , a t the middl e of the stratu m due t o the foundatio n pressure, i s 4,400
lb/ft
2
( = 2.20 t/ft
2
). Estimat e the collapse settlement of the footing under a soaked condition.
Solution
Double consolidatio n test result s of the soi l sample s ar e give n i n Fig. 18.5 . Curv e 1 was obtaine d
with natura l moistur e content . Curv e 3 was obtained fro m th e soake d sampl e afte r 2 4 hours. Th e
virgin curve is drawn in the same wa y a s for a normally loaded cla y soi l (Fig . 7.9a) .
The effectiv e overburden pressure p
0
a t the middl e of the collapsibl e laye r is
p
Q
= 15 x 106. 6 = 1,59 9 lb/ft
2
o r 0. 8 ton/ft
2
Foun dat ion s o n Collaps ibl e an d Ex pan s ive Soils 79 9
A vertical line is drawn in Fig. 18. 5 atp
0
= 0.8 ton/ft
2
. Point A is the intersection of the vertical
line an d th e virgi n curv e givin g th e valu e o f e
Q
= 0.68. p
Q
+ Ap = 0. 8 + 2. 2 = 3. 0 t/ft
2
. A t
(p
0
+ Ap) = 3 ton/ft
2
, w e have (from Fig . 18.5 )
&e
n
= 0.68 - 0.6 2 = 0.06
Ae
c
= 0.62 -0.48 = 0.14
From Eq. (18.3)
A
£
A
1 + 0.68
0.14x30x12 _
n n
.
= -= 30.00 in.
1 + 0.68
Total settlemen t S
c
= 42.86 in .
The tota l settlemen t woul d be reduced i f the thicknes s o f the collapsibl e laye r i s less o r the
foundation pressur e i s less .
Example 18. 2
Refer t o Example 18.1 . Determine th e expected collaps e settlemen t under wetted conditions if the
soil stratu m belo w th e footin g i s overconsolidated . Doubl e oedomete r tes t result s ar e give n i n
Fig. 18.6 . I n this case/?
0
= 0.5 ton/ft
2
, Ap = 2 ton/ft
2
, and/?
c
= 1. 5 ton/ft
2
.
Solution
The virgin curve for the soaked sample can be determined i n the same way as for an overconsolidated
clay (Fig. 7.9b). Doubl e oedometer tes t results are given in Fig. 18.6 . From thi s figure :
e
Q
= 0.6, &e
n
= 0.6 - 0.5 5 = 0.05, Ae
c
= 0.55 - 0.4 8 = 0.07
As i n Ex. 18. 1
^
H
005X30X1 2
1 + 0.6
Total S = 27.00 in .
18.5 FOU NDATI O N DESI G N
Foundation desig n i n collapsibl e soi l i s a ver y difficul t task . The result s fro m laborator y o r fiel d
tests can be used t o predict th e likel y settlemen t tha t may occur unde r sever e conditions . I n many
cases, dee p foundations, such as piles, piers etc, may be used to transmit foundation loads to deeper
bearing strata below the collapsible soi l deposit. In cases wher e it is feasible to support the structure
on shallo w foundation s i n or above th e collapsing soils , the use of continuous stri p footing s ma y
provide a mor e economica l an d safe r foundatio n tha n isolate d footing s (Clemenc e an d Finbarr ,
1981). Differentia l settlement s betwee n column s ca n b e minimized , an d a mor e equitabl e
distribution of stresse s ma y b e achieve d wit h the us e of stri p footing desig n a s shown in Fig. 18. 7
(Clemence an d Finbarr, 1981) .
800 Chapt e r 1 8
Load-bearing
beams
Figure 1 8. 7 Con t in uou s foot in g des ig n wit h l oad-bearin g beam s fo r col l aps ibl e s oi l
(aft er Clem en c e an d Fin barr , 1981 )
18.6 TREATM EN T M ETH OD S FO R COL L APSI B L E SOI LS
On some sites , i t ma y b e feasibl e t o appl y a pretreatment techniqu e eithe r t o stabiliz e th e soi l o r
cause collaps e o f th e soi l deposi t prio r t o constructio n o f a specifi c structure . A grea t variet y o f
treatment method s hav e been use d i n the past . Moistenin g and compaction techniques , wit h eithe r
conventional impact , o r vibrator y roller s ma y be use d fo r shallo w depth s u p t o abou t 1. 5 m. Fo r
deeper depths, vibroflotation , ston e columns , and displacement pile s ma y be tried. Hea t treatmen t
to solidif y th e soi l i n plac e ha s als o bee n use d i n som e countrie s suc h a s Russia . Chemica l
stabilization wit h th e us e o f sodiu m silicat e and injectio n o f carbo n dioxid e hav e bee n suggeste d
(Semkin et al. , 1986) .
Field test s conducte d b y Rollin s et al. , (1990 ) indicat e tha t dynami c compactio n treatmen t
provides th e most effective means of reducing the settlement of collapsible soil s t o tolerable limits .
Prewetting, i n combinatio n wi t h dynami c compaction , offer s th e potentia l fo r increasin g
compaction efficienc y an d uni formi t y , whi l e increasin g vibratio n attenuation. Prewetting wit h a 2
percent solutio n o f sodiu m silicat e provide s cementation tha t reduces th e potentia l fo r settlement .
Prewetting wit h wate r wa s foun d t o b e th e easies t an d leas t costl y treatment , bu t i t prove d t o b e
completely ineffectiv e in reducing collapse potentia l for shallo w foundations . Prewettin g mus t b e
accompanied b y preloading , surcharging or excavation i n order t o be effective.
PART B—EXPANSI V E SOIL S
1 8.7 DI STRI B U TI O N OF EXPANSI V E SOI LS
The proble m o f expansiv e soil s i s widesprea d throughou t world . Th e countrie s tha t ar e facin g
problems wit h expansiv e soil s ar e Aust ral ia, the Unite d States , Canada , China , Israel , India , an d
Egypt. Th e cla y minera l tha t i s mostl y responsibl e fo r expansivenes s belong s t o th e
montmorillonite group . Fig . 18. 8 shows th e distributio n of the montmorillonit e grou p o f mineral s
in th e United States . The majo r concer n wit h expansive soils exist s generall y i n the wester n par t of
the Unite d States . I n th e norther n an d centra l Unite d States , th e expansiv e soi l problem s ar e
primarily relate d t o highly overconsolidated shales . This include s the Dakotas, Montana , Wyomin g
and Colorad o (Chen , 1988) . I n Minneapolis , the expansiv e soi l proble m exist s i n th e Cretaceou s
Foun dat ion s o n Collaps ibl e an d Ex pan s iv e Soil s 801
Figure 18. 8 Gen era l abun dan ce o f m on t m orillon it e i n n ear out cro p bedroc k
form at ion s i n the Unite d St at e s (Chen , 1988 )
deposits alon g th e Mississipp i Rive r an d a shrinkage/swellin g proble m exist s i n th e lacustrin e
deposits i n the Great Lakes Area. I n general , expansiv e soil s ar e not encountered regularl y i n the
eastern part s of the central Unite d States .
In eastern Oklahoma and Texas, th e problems encompass bot h shrinking and swelling. In the
Los Angele s area , th e proble m i s primaril y on e o f desiccate d alluvia l an d colluvia l soils . Th e
weathered volcani c material i n the Denver formation commonly swell s when wetted and is a cause
of major engineerin g problems i n the Denve r area.
The six major natura l hazards are earthquakes, landslides, expansive soils, hurricane, tornad o
and flood . A stud y point s ou t tha t expansive soil s ti e wit h hurrican e wind/stor m surg e fo r secon d
place amon g America' s mos t destructiv e natura l hazard s i n term s o f dolla r losse s t o buildings.
According t o the study, it was projected tha t by the year 2000, losses du e to expansive soi l woul d
exceed 4. 5 billion dollars annuall y (Chen, 1988) .
18.8 G ENERA L CH ARACTERI STI CS O F SWEL L I NG SOI L S
Swelling soils, which are clayey soils, are also called expansive soils. When these soil s are partially
saturated, the y increas e i n volume wit h th e additio n of water . They shrin k greatl y o n dryin g an d
develop crack s o n the surface. These soil s possess a high plasticity index. Black cotton soil s foun d
in many parts of India belong t o this category. Their color varie s from dar k grey t o black. I t is easy
to recognize thes e soil s i n the field during either dry or wet seasons. Shrinkag e cracks ar e visible on
the ground surface during dry seasons. The maximum width of these cracks may be up to 20 mm or
more an d the y trave l dee p int o the ground . A lump of dr y black cotto n soi l require s a hammer t o
break. Durin g rainy seasons , thes e soil s become ver y sticky and very difficul t t o traverse .
Expansive soil s ar e residual soil s whic h ar e the resul t of weathering o f the parent rock. The
depths of these soil s i n some region s ma y be up to 6 m or more. Normall y the water tabl e i s met at
great depths in these regions. As such the soils become we t only during rainy seasons an d are dry or
802 Chapt er 1 8
Increasing moistur e content of soi l
Ground surfac e
D,
D,,< = Unstable zon e
Moisture variatio n
during dry seaso n
Figure 18. 9
Stable zone \
Equilibrium moistur e content (covered area)
Desiccated moistur e content (uncovered natural conditions)
Wet season moistur e conten t (seasona l variation )
Depth of seasonal moistur e content fluctuatio n
Depth of desiccation or unstabl e zone
M ois t ure con t en t variat io n wit h dept h belo w g roun d s urface
(Chen , 1988 )
partially saturate d durin g the dry seasons . I n regions whic h have well-defined, alternately we t and
dry seasons , thes e soil s swel l and shrink in regular cycles. Since moistur e change i n the soil s bring
about sever e movement s o f th e mass , an y structur e buil t o n suc h soil s experience s recurrin g
cracking an d progressiv e damage . I f on e measure s th e wate r conten t o f th e expansiv e soil s wit h
respect t o depth durin g dry and wet seasons , th e variation is similar t o the one shown in Fig. 18.9 .
During dr y seasons , th e natura l wate r conten t i s practicall y zer o o n th e surfac e an d th e
volume of the soi l reaches th e shrinkage limit. The water content increases wit h depth an d reache s
a value w
n
at a depth D , beyond which it remains almost constant. During the wet season the water
content increase s an d reache s a maximum at the surface . The wate r conten t decreases wit h dept h
from a maximum of w
n
at the surfac e to a constant value of w
n
a t almost th e same dept h D
us
. This
indicates tha t the intake of water by the expansive soi l int o its lattice structur e is a maximum at the
surface an d ni l a t dept h D
us
. Thi s mean s tha t the soi l lyin g withi n thi s dept h D
us
i s subjecte d t o
drying an d wettin g an d henc e caus e considerabl e movement s i n th e soil . Th e movement s ar e
considerable close to the ground surfac e an d decrease with depth. The cracks that are developed in
the dry season s clos e du e t o lateral movements during the we t seasons .
Foun dat ion s o n Collaps ibl e an d Ex pan s iv e Soil s 80 3
The zone which lies within the depth D
us
may be called the unstable zone (or active zone ) and
the one below thi s the stable zone. Structures built within thi s unstable zone ar e likely to move up
and dow n accordin g t o season s an d henc e suffe r damag e i f differentia l movement s ar e
considerable.
If a structure is built during the dry season with the foundation lying within the unstable zone,
the bas e o f th e foundatio n experiences a swelling pressure a s th e partiall y saturate d soi l start s
taking i n wate r during the we t season . Thi s swellin g pressure i s due t o constraint s offered b y th e
foundation fo r free swelling . The maximum swelling pressure may be as high as 2 MPa (2 0 tsf). If
the imposed bearin g pressur e o n the foundation by the structure i s less than the swelling pressure ,
the structur e is likel y to be lifte d u p a t least locally which would lead t o cracks i n the structure. If
the imposed bearing pressure is greater than the swelling pressure, there will not be any problem for
the structure . I f o n th e othe r hand , th e structur e is buil t durin g th e we t season , i t wil l definitel y
experience settlemen t as the dry season approaches , whethe r the imposed bearin g pressur e i s high
or low . However , th e impose d bearin g pressur e durin g th e we t seaso n shoul d b e withi n th e
allowable bearing pressure of the soil . The better practice i s to construct a structure during the dry
season and complete it before the wet season.
In covere d area s belo w a buildin g ther e wil l b e ver y littl e chang e i n th e moistur e conten t
except due t o lateral migration of water from uncovere d areas. The moistur e profile i s depicted b y
curve 1 in Fig. 18.9 .
18.9 CL A Y M I NERAL OG Y AN D M ECH ANI SM OF SWEL L I NG
Clays can be divided int o three general group s on the basis o f their crystalline arrangement. They
are:
1. Kaolinit e group
2. Montmorillonit e group (also calle d th e smectit e group )
3. Illit e group .
The kaolinit e group of mineral s ar e the most stabl e of the groups of minerals. The kaolinit e
mineral i s formed b y the stacking o f the crystalline layer s o f about 7 A thick one above th e othe r
with th e bas e o f th e silic a shee t bondin g t o hydroxyl s of th e gibbsit e shee t b y hydroge n bonds .
Since hydroge n bond s ar e comparativel y strong , th e kaolinit e crystal s consist s o f man y shee t
stackings tha t ar e difficul t t o dislodge . The minera l is , therefore , stabl e an d wate r canno t ente r
between the sheet s t o expand the uni t cells .
The structural arrangement of the montmorillonite mineral is composed o f units made of two
silica tetrahedral sheets wit h a central alumina-octahedral sheet . The silica and gibbsit e sheet s ar e
combined i n such a way that the tips of the tetrahedrons of each silica shee t and one of the hydroxyl
layers o f th e octahedra l shee t for m a commo n layer . Th e atoms commo n t o bot h th e silic a and
gibbsite layer s ar e oxyge n instea d o f hydroxyls . The thicknes s o f the silica-gibbsite-silic a uni t is
about 1 0 A. In stacking of these combined unit s one above the other, oxygen layers of each uni t are
adjacent t o oxyge n o f th e neighborin g units , wit h a consequenc e tha t ther e i s a wea k bon d an d
excellent cleavag e betwee n them . Wate r ca n ente r betwee n th e sheet s causin g the m t o expan d
significantly an d thus the structure can break into 10 A thick structural units. The soils containing a
considerable amoun t o f montmorillonit e mineral s wil l exhibi t hig h swellin g an d shrinkag e
characteristics. Th e illit e grou p o f mineral s ha s th e sam e structura l arrangemen t a s th e
montmorillonite group. The presence o f potassium as the bonding material s betwee n unit s makes
the illite minerals swel l less .
804 Chapt e r 1 8
1 8.1 0 DEFINITIO N OF SOM E PARAM ETER S
Expansive soils can be classified o n the basis of certain inheren t characteristic s o f the soil. It is first
necessary t o understand certain basic parameters used i n the classification.
Swelling Potentia l
Swelling potentia l i s define d a s th e percentag e o f swel l o f a laterall y confine d sampl e i n a n
oedometer test which is soaked unde r a surcharge load of 7 kPa ( 1 lb/in
2
) after bein g compacted t o
maximum dr y densit y at optimum moisture content according t o the AASHTO compactio n test .
Swelling Pressure
The swellin g pressure /?
5
, i s defined as th e pressur e required fo r preventin g volume expansio n i n
soil i n contac t wit h water . I t shoul d b e note d her e tha t th e swellin g pressur e measure d i n a
laboratory oedometer i s different fro m tha t in the field. The actual field swelling pressure i s always
less tha n the one measured in the laboratory.
Free Swel l
Free swel l 5, is defined as
V
f
-V.
S
f
=-^—xm (18.5 )
where V
{
= initia l dr y volume of poured soi l
Vr - fina l volum e of poured soi l
According t oHol t z and Gibbs (1956) , 1 0 cm
3
(V . ) of dr y soi l passing thoroug h a No. 40
sieve is poured int o a 100 cm
3
graduated cyl i nder filled wi t h water. The volume of settled soi l
is measure d afte r 2 4 hour s whic h give s th e valu e of V~ Bentonite-cla y i s suppose d t o hav e a
free swel l valu e rangi n g fro m 120 0 t o 200 0 percent . Th e fre e swel l valu e increase s wit h
plasticity index . Holt z and Gibbs suggeste d tha t soil s havin g a free-swell valu e a s low as 100
percent ca n caus e considerabl e damag e t o l i ght l y loaded st ruct ure s and soil s heavin g a fre e
swell valu e belo w 5 0 percen t seldo m exhibi t appreciabl e volum e chang e eve n unde r ligh t
loadings.
1 8.1 1 EV AL U ATI O N O F TH E SWEL L I NG POTENTI AL OF
EXPANSI V E SOI L S B Y SI NG L E I NDEX M ETH OD (CH EN , 1988 )
Simple soi l property tests can be used for the evaluation of the swelling potential of expansive soils
(Chen, 1988) . Such tests ar e easy to perform and should be used as routine tests i n the investigation
of building sites i n those area s having expansive soil. These test s ar e
1. Atterber g limit s tests
2. Linea r shrinkage test s
3. Fre e swell test s
4. Colloi d conten t test s
Atterberg Limits
Holtz and Gibbs (1956) demonstrated that the plasticity index, I
p
, an d the liquid limit, w
/ 5
are usefu l
indices fo r determinin g the swellin g characteristics of mos t clays . Sinc e th e liqui d limi t an d th e
Foun dat ion s o n Collaps ibl e an d Ex pan s ive Soil s 805
Note: Percent swel l measured under 1 psi surcharge for sampl e compacte d o f
optimum wate r content t o maximum density i n standard RASHO test
Clay component : commercial bentonit e
1:1 Commercial illite/bentonit e
6:1 Commercial kaolinite/bentonit e
3:1 Commercial illite/bentonit e
I I
•— Commercia l illit e
1:1 Commercial illite/kaolinit e
Commercial kaolinit e
30 4 0 5 0 6 0 7 0
Percent cla y sizes (fine r tha n 0.002 mm)
80 90 100
Figure 18.1 0 Relat ion s hi p bet wee n percen t ag e o f s wel l an d percen t ag e o f cl ay
s izes fo r ex perim en t a l s oil s (aft e r See d et al. , 1962)
swelling of clays both depend on the amount of water a clay tries to absorb, i t is natural that they are
related. Th e relatio n betwee n th e swellin g potentia l o f clay s an d th e plasticit y inde x ha s bee n
established a s given in Table 18. 2
L inear Shrinkage
The swell potential i s presumed to be related t o the opposite propert y of linear shrinkag e measure d
in a ver y simpl e test . Altmeyer (1955 ) suggeste d the value s given in Table 18. 3 a s a guide t o th e
determination o f potential expansivenes s base d o n shrinkage limit s and linear shrinkage .
Colloid Conten t
There i s a direct relationship between colloid content and swelling potential as shown in Fig . 18.1 0
(Chen, 1988) . For a give n cla y type , th e amoun t o f swel l wil l increas e wit h th e amoun t o f cla y
present i n the soil .
Table 18. 2 Relat io n bet ween s wellin g pot en t ia l an d plas t icity index , /
Plas t icit y in de x l
p
(%) Swellin g pot en t ial
0-15
10-35
20-55
35 and abov e
Low
Medium
High
V ery high
806 Chapt er 1 8
Table 18. 3 Relatio n betwee n swelling potential, shrinkag e limits , an d linear shrinkag e
Shrin k ag e limi t % Linear s hrin k ag e % Deg ree o f ex pan s io n
10-12
> 1 2
>8
5-8
0-5
Critical
Marginal
Non-critical
1 8.1 2 CL ASSI FI CATI O N O F SWEL L I NG SOI L S B Y I NDI REC T
M EASU REM ENT
By utilizin g th e variou s parameter s a s explaine d i n Sectio n 18.11 , th e swellin g potential ca n b e
evaluated withou t resorting to direct measuremen t (Chen, 1988) .
U SB R M etho d
Holtz and Gibbs (1956) developed thi s method which is based on the simultaneous consideration of
several soi l properties . Th e typica l relationship s o f thes e propertie s wit h swellin g potentia l ar e
shown i n Fig. 18.11 . Table 18. 4 has bee n prepare d based o n the curves presented i n Fig. 18.1 1 by
Holtz an d Gibbs (1956).
The relationshi p betwee n the swel l potentia l and the plasticit y inde x can be expresse d as
follows (Chen , 1988 )
(18.6)
where, A = 0.083 8
B = 0.255 8
/ = plasticit y index.
V
o
l
u
m
e

c
h
a
n
g
e

i
n

%
H


K
>

U
>

*
O
O
O

C
T
\
-
£
>


N
>

C
J
f
-"$
1 • /
/• • /
1
i
i i
t '^ i
r *i*
i /
/ '
, 1
.
-;4
^
?
_^ /
:•/
;:/..
• 4
/
/"
0
•i
\
\
f
i
/
1 •
\
\
\ •
v \
\« ^
:
V -
\
\
^•
v
f ,
X
CJJ
£
s
3
T3
OJ
S
O
) 2 0 4 0 0 ' 2 0 4 0 0 8 1 6 2 4
Colloid conten t Plasticit y index Shrinkag e limi t (%)
(%less than 0.001 mm)
(a) (b ) (c )
Figure 18.1 1 Relatio n of volum e chan g e t o (a ) colloid con t en t , (b) plasticity index ,
an d (c ) s hrin k ag e limi t (air-dr y t o s at urat e d con dit io n unde r a load o f 1 Ib pe r s q in )
(Holt z an d Gibbs , 1956 )
Foun dat ion s o n Collaps ibl e an d Ex pan s ive Soil s 807
Table 18. 4 Dat a for m ak in g es t im at e s o f probabl e volum e chan g e s for ex pan s iv e
s oils (Source : Chen , 1988 )
Dat a fro m in de x t es t s *
Colloid con t en t , per -
cen t minu s 0.001 m m
>28
20-13
13-23
< 15
Plas t icit y
in dex
>35
25-41
15-28
<18
Shrin k ag e
limit
<11
7-12
10-16
>15
Probable ex pan s ion ,
percen t t ot a l
vol. chan g e
>30
20-30
10-30
<10
Deg ree o f
ex pan s ion
Very hig h
High
Medium
Low
*Based o n vertica l loading of 1. 0 psi. (afte r Holt z and Gibbs , 1956 )
10
a 5
0
30 35 40 15 2 0 2 5
Plasticity index (%)
1. Holt z an d Gibbs (Surcharg e pressur e 1 psi)
2. Seed , Woodwar d and Lundgren (Surcharge pressure 1 psi)
3. Che n (Surcharg e pressur e 1 psi)
4. Che n (Surcharg e pressure 6.94 psi )
Figure 18.1 2 Relationship s o f volum e chan g e t o plasticit y inde x
(Source: Chen , 1988 )
808
Chapt er 1 8
Figure 18.1 2 shows that with an increase in plasticity index, the increase of swelling potential
is muc h les s tha n predicte d b y Holt z an d Gibbs . Th e curve s give n by Che n (1988) ar e base d o n
thousands of test s performe d ove r a period o f 30 years an d as such are more realistic .
Activity M etho d
Skempton (1953 ) define d activity by the following expression
A = ~^(18.7 )
where / = plasticit y index
C = percentag e of clay size finer than 0.002 m m b y weight.
The activit y method a s proposed b y Seed , Woodward , an d Lundgren, (1962 ) wa s based o n
remolded, artificiall y prepared soil s comprisin g of mixtures of bentonite, illite , kaolinit e and fine
sand i n differen t proportions . The activit y for th e artificiall y prepare d sampl e wa s define d as
activity A =
(18.9)
C-n
where n - 5 for natural soil s and, n = 10 for artificial mixtures .
The propose d classificatio n char t i s show n i n Fig . 18.13 . Thi s metho d appear s t o b e a n
improvement ove r the USER method.
Swelling potential = 25%
Swelling potentia l = 5%
Swelling potentia l = 1.5%
Figure 1 8.1 3
20 3 0 4 0 5 0 6 0 7 0
Percent clay size s (fine r than 0.002 mm)
Cl as s ificat ion char t fo r s wellin g pot en t ia l (aft e r Seed , Woodwar d, an d
Lun dg ren , 1962)
Foun dat ion s o n Collaps ibl e an d Ex pan s ive Soil s 809
The Potentia l V olume Change M etho d (PV C )
A determination of soil volume change was developed b y Lambe under the auspices of the Federa l
Housing Administration (Source: Chen , 1988) . Remolde d sample s wer e specified . The procedur e
is as given below.
The sampl e i s firs t compacte d i n a fixe d rin g consolidomete r wit h a compactio n effec t o f
55,000 ft-l b pe r cu ft . Then a n initial pressure of 200 psi i s applied, an d water added t o the sampl e
which i s partially restraine d b y vertical expansio n b y a proving ring . The proving rin g readin g i s
taken at the end of 2 hours. The reading is converted to pressure and is designated as the swell index.
From Fig . 18.14 , the swel l index can be converted t o potential volume change. Lambe establishe d
the categorie s o f PV C ratin g as shown in Table 18.5 .
The PV C method has been widely used by the Federal Housin g Administration as well as the
Colorado Stat e Highwa y Department (Chen, 1988) .
ouuu
*7nnn
/ \ j\ j\ j
<£H
CT
|400 0
_c
I
2000
1000
200
(
/
^
,
/
^
/
/
. ——
^/
/
— <*
•^
//
/
^
/
/
'/
*/
/
'
/
I
/
/
/
I
1
) 1 2 3 4 5 6 7 8 9 1 0 1 1 1
Non
critical Margina l Critica l V er y Critical
Potential V olume Change (PV C )
Figure 18.1 4 Swel l inde x vers u s potentia l volum e chang e (fro m 'FHA soi l PVC
m et er publicat ion , ' Federa l Housin g Adm in is t rat io n Publicat io n n o . 701) (Source :
Chen , 1988 )
810 Chapt e r 1 8
Table 18. 5 Pot en t ia l volum e chan g e rat in g (PVC)
PV C r at i n g
Less tha n 2
2^
4-6
>6
Cat egor y
Non-critical
Marginal
Critical
V ery critica l
(Source: Chen, 1988)
Figure 18.15(a ) show s a soi l volum e chang e mete r (EL E Internationa l Inc) . Thi s mete r
measures bot h shrinkag e and swellin g of soils , idea l for measurin g swelling of clay soils , and fas t
and easy t o operate.
Expansion I nde x (El)- Che n (1 9 8 8 )
The ASTM Committe e o n Soil and Rock suggeste d th e use of an Expansion Index (El ) as a unified
method t o measur e th e characteristic s o f swellin g soils. I t i s claime d tha t th e El i s a basic inde x
property o f soi l suc h as the liqui d limit, the plastic limit and the plasticity index of the soil .
The sampl e i s sieved throug h a No 4 sieve. Wate r i s added so that the degree of saturation i s
between 49 and 51 percent. The sampl e is then compacted int o a 4 inch diameter mol d in two layers
to give a total compacted dept h of approximates 2 inches. Each laye r i s compacted b y 1 5 blows of
5.5 I b hamme r droppin g fro m a heigh t o f 1 2 inches . Th e prepare d specime n i s allowe d t o
consolidate unde r 1 lb/in
2
pressure for a period of 1 0 minutes, then inundated with water unti l th e
rate o f expansion ceases .
The expansio n index is expressed a s
£/ = — xl OOO
(189 )
h
i
where A/ I = chang e i n thickness of sample , in .
h. = initia l thickness of sample, in .
The classificatio n o f a potentially expansive soi l i s based o n Table 18.6 .
This metho d offer s a simple testing procedure fo r comparing expansiv e soi l characteristics .
Figure 18.15(b ) show s a n ASTMD-829 expansio n inde x tes t apparatu s (ELE Internationa l
Inc). Thi s i s a completely self-containe d apparatu s designed fo r us e i n determining th e expansion
index of soils .
Table 18. 6 Clas s ificat io n o f potentiall y ex pan s ive soi l
Ex pan s ion Index , El Ex pan s io n potential
0-20 V er y low
21-50 Lo w
51-90 Mediu m
91-130 Hig h
> 13 0 V er y hig h
(Source: Chen, 1988)
Foun dat ion s o n Collaps ibl e an d Ex pan s ive Soil s 81 1
Swell I nde x
V ijayvergiya an d Gazzal y (1973 ) suggeste d a simpl e wa y o f identifyin g th e swel l potentia l o f
clays, base d o n the concept o f the swel l index. They define d the swel l index, I
s
, as follows
-
*~~ (18-10 )
where w
n
= natura l moisture content i n percent
\ v
l
= liqui d limit in percen t
The relationshi p betwee n I
s
an d swel l potentia l for a wide rang e o f liqui d limi t i s shown i n
Fig 18.16 . Swell inde x is widel y use d for the design of post-tensioned slab s on expansive soils .
Prediction o f Swelling Potentia l
Plasticity index and shrinkage limit can be used to indicate the swelling characteristics o f expansive
soils. According t o Seed at al., (1962), th e swelling potential is given as a function o f the plasticity
index by the formula
(18.11)
Figure 1 8 .1 5 (a ) Soil volum e chan g e m et er , an d (b) Ex pan s io n in de x t es t apparat us
(Court es y: Soilt es t )
812 Chapt er 1 8
u. /
0.6
0.5
{' 0. 4
<
j
s 0. 3
0.2
0.1
0.0
3
^
Perc
Swe
Percent swell : < 1
Swell pressure: <
^^
1 ^
ent swell : 1 to 4
11 pressure: 0. 3 t o 1.2 5
0.3 ton/sq f
-
ton/sq f t
— L4 - •
Percent swell : 4 to 10
Swell pressure: 1 .25 t o 3 ton/sq f t
^
MPerce
Swell
^~~"1
nt swell : >
pressure: ;
-~
10
> 3 ton/s q f l
t
**—
— • —•
0 4 0 5 0 6 0 7 0 8 (
Liquid limit
Figure 1 8.1 6 Rel at ion s hi p bet wee n s wel l in de x an d liquid limi t fo r ex pan s iv e cl ays
(Source: Chen , 1988)
where S = swellin g potential i n percent
/ = plasticit y index in percent
k = 3. 6 xl O~
5
, a factor for cla y content between 8 and 6 5 percent .
1 8 .1 3 SWEL L I N G PRESSU RE BY DI RECT M EASU REM EN T
ASTM define s swellin g pressur e whic h prevent s th e specime n fro m swellin g o r tha t pressur e
which i s require d t o retur n th e specime n t o it s origina l stat e (voi d ratio , height ) afte r swelling .
Essentially, th e method s o f measurin g swelling pressure ca n b e eithe r stres s controlle d o r strai n
controlled (Chen , 1988) .
In the stress controlle d method , the conventional oedometer i s used. The sample s ar e place d
in th e consolidatio n rin g trimme d t o a height o f 0.7 5 t o 1 inch. Th e sample s ar e subjecte d t o a
vertical pressur e rangin g from 50 0 ps f t o 2000 psf depending upo n the expected fiel d conditions .
On the completion o f consolidation, wate r i s added t o the sample. Whe n th e swelling o f the sampl e
has ceased th e vertical stres s i s increased i n increments until i t has been compresse d t o its origina l
height. Th e stres s require d t o compress th e sampl e t o it s original heigh t i s commonl y terme d the
zero volume change swelling pressure. A typical consolidation curve is shown i n Fig 18.17 .
Foun dat ion s o n Collaps ibl e an d Ex pan s ive Soil s 813
I
I
30
20
'-3 1 0
ta
O
U
-10
9^
^• xj
Placement conditio
Dry densit y
Moisture content
Atterberg limits:
"X
T
F
N
>
\
s
s
\
^
ns
= 76.3 p
\ A. n o
Liquid 1
Plasticit
s
°N
cf
fc
imit = 68%
y index = 17 %
\
N
0.8%expansio n at 10 0 psf
>ressure when wetted
)
100 100 0 1000 0
Applied pressure (psf )
Figure 18.1 7 Typica l s t res s controlle d swell-consolidation curv e
Prediction o f Swelling Pressur e
Komornik et al. , (1969) hav e given an equation for predicting swelling pressure as
\ ogp
s
= 2.132 + 0.0208W, +0.00065^ -0.0269w
n
where p
s
= swellin g pressur e i n kg/cm
2
w
;
= liqui d limi t (%)
w
n
= natura l moisture content (%)
y
d
= dr y densit y of soi l in kg/cm
3
(18.12)
1 8 .1 4 EFFEC T OF I NI TI AL M OI STU RE CONTENT AND I NI TI A L
DRY DENSI T Y ON SWEL L I NG PRESSU RE
The capabilit y o f swellin g decreases wit h a n increas e o f th e initia l water content o f a give n soi l
because its capacit y to absorb wate r decreases wit h the increase of its degree of saturation. It was
found fro m swellin g test s o n black cotto n soi l samples , tha t th e initial wate r conten t ha s a smal l
effect o n swelling pressure unti l i t reaches the shrinkage limit, then its effect increase s (Abouleid ,
1982). This i s depicted i n Fig. 18.18(a) .
The effec t o f initial dry densit y on the swelling percent an d the swelling pressure increase s
with a n increas e o f th e dr y densit y becaus e th e dens e soi l contain s mor e cla y particle s i n a unit
volume an d consequentl y greate r movemen t wil l occu r i n a dens e soi l tha n i n a loose soi l upo n
814 Chapt er 1 8
00
on
18
16
H
1?
10
9
f .
A
0
n

*->
\>
S.L.
I
\
^
\
\
N
V
^
»
0 2 4 6 8 1 0 1 2 1 4 1 6 1 8 2 0 2 2 2 4
Water content %
Black cotton soil
w
. =90, w
p
= 30, w, = 10
Mineralogy: Largely sodium, Montmorillonite
2.0
r
s
,0
j*t
|
C/3
a
L 0
D.
0.0
z
S.I,
1.2 .4 1. 6 1. 8 2. 0 2. 2
Dry density (t/m
3
)
(b)
Figure 18.1 8 (a ) ef f ect o f in it ia l wat e r con t en t o n s wellin g pres s ur e o f blac k cot t o n
s oil, an d (b) ef f ect o f in it ia l dr y den s it y o n s wellin g pres s ur e of blac k cot t o n s oi l
(Source: Abouleid , 1982 )
wetting (Abouleid , 1982) . Th e effec t o f initia l dr y densit y o n swellin g pressur e i s show n i n
Fig. 18.18(b) .
1 8 .1 5 ESTI M ATI N G TH E M AG NI TU DE O F SWEL L I NG
When footing s are buil t i n expansive soil, they experience liftin g du e t o the swellin g or heaving of
the soil . The amoun t of total heave and the rate of heave of the expansive soil on which a structure
is founded are very complex. Th e heave estimat e depend s o n many factors whic h cannot be readil y
determined. Som e o f the major factors that contribute to heaving are:
1. Climati c condition s involvin g precipitation , evaporation , an d transpiratio n affec t th e
moisture i n th e soil . The dept h an d degre e o f desiccatio n affec t th e amoun t of swel l i n a
given soi l horizon.
2. Th e thickness of the expansive soil stratum is another factor. The thickness o f the stratum is
controlled b y the depth t o the water table .
3. Th e dept h t o the water table i s responsible for the change in moisture of the expansive soil
lying abov e th e wate r table. No swellin g of soi l take s plac e whe n i t lie s belo w th e wate r
table.
4. Th e predicted amount of heave depend s o n the nature and degree of desiccation of the soil
immediately afte r constructio n of a foundation.
5. Th e singl e most importan t element controlling the swellin g pressure a s wel l a s the swel l
potential i s th e in-sit u densit y o f th e soil . O n th e completio n o f excavation , th e stres s
condition i n the soi l mas s undergoe s changes , suc h a s the release of stresses due to elastic
Foun dat ion s o n Collaps ibl e an d Ex pan s ive Soil s 81 5
rebound of the soil. If construction proceeds without delay, the structural load compensate s
for th e stres s release .
6. Th e permeabilit y o f the soi l determine s th e rat e o f ingress o f water int o the soi l eithe r by
gravitational flow or diffusion, an d this i n turn determines the rate of heave.
V arious method s hav e bee n propose d t o predic t th e amoun t o f tota l heav e unde r a give n
structural load. The following methods, however , are described here .
1. Th e Department of Navy method (1982 )
2. Th e Sout h African method [als o known as the V an Der Merwe metho d (1964) ]
The Departmen t o f Nav y M etho d
Procedure fo r Estimatin g Tota l Swel l under Structural L oad
1. Obtai n representative undisturbe d samples of soil below the foundation level at intervals of
depth. Th e sample s ar e to be obtaine d during the dry seaso n whe n the moistur e content s
are at their lowest.
2. Loa d specimen s (a t natural moisture content) in a consolidometer unde r a pressure equa l to
the ultimate value of the overburden plus the weight of the structure. Add water to saturat e
the specimen. Measur e th e swell .
3. Comput e th e final swel l i n terms o f percent of original sampl e height .
4. Plo t swel l versu s depth.
5. Comput e th e tota l swel l whic h i s equa l t o th e are a unde r th e percen t swel l versu s dept h
curve.
Procedure fo r Estimating U ndercu t
The procedur e fo r estimating undercut to reduce swel l t o an allowable valu e i s as follows:
1. Fro m the percent swel l versus depth curve, plot the relationship o f total swel l versus dept h
at that height. Tota l swel l at any depth equals area under the curve integrated upwar d fro m
the dept h of zero swell .
2. Fo r a given allowable valu e of swell, read th e amount of undercut necessar y fro m the total
swell versu s depth curve .
V an Der M erwe M etho d (1 9 6 4 )
Probably th e neares t practica l approac h t o th e proble m o f estimatin g swel l i s tha t o f V a n De r
Merwe. Thi s method start s by classifying the swell potential of soil into very high to low categories
as shown in Fig. 18.19 . Then assig n potential expansion (PE) expressed i n in./ft o f thickness base d
on Table 18.7 .
Table 18. 7 Pot en t ia l ex pan s io n
Swell pot en t ia l Pot en t ia l ex pan s io n (PE) in . /ft
V ery hig h 1
High 1/ 2
Medium 1/ 4
Low 0
816 Chapt er 1 8
Reduction factor , F
0.2 0. 4 0. 6 0. 8
Pu
Q -1 6 -
10 2 0 3 0 4 0 5 0 6 0 7 0
Clay fractio n o f whol e sampl e ( %< 2 micron)
(a) (b )
Figure 1 8.1 9 Relat ion s hip s fo r us in g Va n De r M erwe's predict io n m et hod :
(a) pot en t ia l ex pan s iven es s , an d (b) reduct io n fact o r (Va n der M erwe , 1964)
u
-2
-4
-6
-8
-10
-12
-16
-18
on
-ZU
-22
-24
-26
-28
^n
I
-
-
-
-
-
-
/
I
,
/
I
>
/
\
./
'
ix^"
1.0
Procedure fo r Estimatin g Swel l
1 . Assum e th e thicknes s of an expansive soi l laye r or the lowes t leve l o f ground water .
2. Divid e thi s thicknes s ( z) int o several soi l layer s wit h variabl e swel l potential .
-3. Th e tota l expansio n i s expressed as
i=n
A// = A
where A//
e
= total expansio n (in. )
A. =(P£). (AD)
I
. (F).
(18.13)
(18.14)
"
1
(F),. = log" - - ' - - reductio n factor fo r layer / .
z = total thicknes s o f expansive soil laye r (ft)
D
{
= depth t o midpoin t of i t h laye r (ft)
(AD). = thicknes s o f i t h laye r (ft)
Fig. 18.19(b ) gives th e reduction facto r plotte d agains t depth .
Foundations on Collapsibl e an d Ex pan s ive Soil s 817
Outside bric k wal l
Cement concrete
apron wit h ligh t
reinforcement
RCC
foundation
• ;^r - , ~ A • •'„» •„-- • '
/ W\/ / ^\^Sv /#<N s
Granular fil l
Figure 18.20 Foun dat io n i n ex pan s ive s oi l
1 8 .1 6 DESI G N OF FOU NDATI ONS I N SWEL L I NG SOI L S
It i s necessar y t o not e tha t al l part s o f a buildin g wil l no t equall y b e affecte d b y th e swellin g
potential of the soil . Beneath the center of a building where the soi l is protected fro m su n and rain
the moistur e change s ar e smal l an d th e soi l movement s th e least . Beneat h outsid e walls, th e
movement are greater. Damag e t o buildings is greatest on the outside walls due to soil movements.
Three genera l type s of foundations can be considered i n expansive soils. They ar e
1. Structure s that can be kept isolated fro m th e swelling effects o f the soil s
2. Designin g o f foundations that will remain undamage d i n spit e o f swellin g
3. Eliminatio n of swelling potential of soil.
All thre e methods ar e i n us e eithe r singly or i n combination, but th e firs t i s by fa r th e mos t
widespread. Fig. 18.2 0 sho w a typical type of foundation unde r an outside wall . The granula r fill
provided aroun d th e shallow foundatio n mitigates th e effect s o f expansion o f the soils.
1 8 .1 7 DRI L L E D PIER FOU NDATI ON S
Drilled piers ar e commonly used to resist uplift forces caused by the swelling of soils. Drilled piers,
when made with an enlarged base, ar e called, belled piers an d when made without an enlarged bas e
are referred t o as straight-shaft piers.
Woodward, e t al. , (1972 ) commente d o n th e empirica l desig n o f piers : "Man y piers ,
particularly where rock bearing is used, have been designed using strictly empirical consideration s
which are derived fro m regiona l experience". The y further state d that "when surfac e conditions are
well establishe d an d ar e relativel y uniform , an d th e performanc e o f pas t constructions wel l
documented, the design by experience approac h i s usually found t o be satisfactory. "
The principl e of drille d pier s i s t o provid e a relativel y inexpensive wa y o f transferrin g the
structural loads down to stable material or to a stable zone where moisture change s ar e improbable .
818 Chapt e r 1 8
There shoul d b e no direct contac t between th e soi l an d the structure with the exception o f the soil s
supporting th e piers.
Straight- shaft Pier s i n Expansive Soil s
Figure 18.2 1 (a) show s a straight-shaf t drille d pie r embedde d i n expansiv e soil . Th e followin g
notations ar e used .
Lj = lengt h of shaf t i n the unstabl e zone (active zone) affecte d b y wetting .
L
2
= lengt h o f shaf t i n the stabl e zone unaffecte d b y wetting
d = diameter o f shaf t
Q - structura l dead loa d = qA
b
q - uni t dead loa d pressur e and
A
b
= base are a o f pier
When th e soi l in the unstabl e zone takes water during the wet season, th e soi l tries t o expand
which is partially or wholly prevented by the rough surface of the pile shaft of length Lj . As a result
there will be an upward force developed on the surface of the shaft which tries to pull the pile out of
its position. Th e upwar d force ca n be resisted i n the following ways .
1. Th e downwar d dead loa d Q acting on the pier to p
2. Th e resisting forc e provide d b y the shaft lengt h L
2
embedded i n the stabl e zone .
Two approaches fo r solving this problem ma y be considered. The y ar e
1. Th e metho d suggeste d by Che n (1988 )
2. Th e O' Neil l (1988 ) metho d wit h belle d pier .
Two cases may be considered. The y ar e
1. Th e stabilit y o f the pier whe n no downward loa d Q is acting o n the top. For thi s conditio n
a factor o f safet y o f 1. 2 is normall y foun d sufficient.
2. Th e stabilit y o f the pier whe n Q is acting o n the top. For this a value of F
?
= 2.0 is used .
Equations fo r Uplif t Forc e Q
up
Chen (1988 ) suggested th e following equation fo r estimating th e uplif t forc e Q
Qu
P
=7lda
u
P
s
L
i (18.15 )
where d - diamete r o f pier shaf t
a = coefficien t o f uplif t betwee n concret e an d soil = 0.1 5
p
s
= swellin g pressur e
= 10,00 0 psf (480 kN/m
2
) for soi l wit h high degre e of expansio n
= 5,00 0 (240 kN/m
2
) for soi l wit h mediu m degre e of expansio n
The dept h (Lj ) o f the unstable zone (wetting zone) varies wit h the environmental conditions.
According t o Che n (1988 ) th e wettin g zone i s limite d onl y t o th e uppe r 5 fee t o f th e pier . I t i s
possible for the wetting zone to extend beyond 10-1 5 feet i n some countrie s an d limiting the dept h
of unstabl e zon e t o a suc h a lo w valu e of 5 f t ma y lea d t o unsaf e conditions fo r th e stabilit y o f
structures. However , i t is for designers t o decide this dept h L j accordin g to local conditions. With
regards t o swelling pressur e p
s
, i t is unrealistic t o fix any definite valu e of 10,00 0 or 5,000 ps f for
all type s o f expansiv e soil s unde r al l condition s o f wetting . I t i s als o no t definitel y known i f th e
results obtaine d fro m laboratory tests truly represent the in situ swelling pressure. Possibl y on e way
of overcoming thi s complex problem is to relate the upl i f t resistanc e t o undrained cohesive strengt h
Foun dat ion s o n Collaps ibl e an d Ex pan s ive Soil s 81 9
of soi l jus t a s i n th e cas e o f frictio n pier s unde r compressive loading . Equatio n (18.15 ) ma y b e
written as
/") — 'TT/y/ V s* f / 1 O 1 /^ \
^«p ~ i w 1 (18. l u )
where c ^ = adhesion factor between concrete an d soi l unde r a swelling condition
c
u
= unit cohesion under undraine d condition s
It is possible tha t the value of a
?
may be equal to 1. 0 or more accordin g t o the swelling typ e
and environmenta l conditions of th e soil . Local experienc e wil l hel p t o determin e th e valu e of oc
This approac h i s simpl e an d pragmatic.
Resisting Forc e
The length of pier embedded i n the stable zone should be sufficient t o keep the pier being pulled out
of th e groun d wit h a suitabl e factor o f safety . I f L
2
i s the lengt h of th e pier i n th e stabl e zone , th e
resisting forc e Q
R
i s th e frictiona l resistanc e offere d b y th e surfac e o f th e pie r withi n the stabl e
zone. We may writ e
Q
R
=7tiL
2
ac
u
(18.17 )
where a = adhesion facto r under compression loading
c
u
= undrained uni t cohesion o f soil
The value of a ma y be obtained from Fig. 17.15.
Two cases o f stabilit y may be considered :
1. Withou t taking int o account th e dead load Q acting o n the pier top, and using F = 1.2
Q
UP
=f f (18.18a )
2. B y taking int o account the dead load Q and using F
s
= 2. 0
Q
K
(<2
M/7
- 0 = yf (18.18b )
For a given shaf t diameter d equations (18.18a ) and (18.18b) help to determine the length L
2
of the pier i n the stabl e zone. The on e that gives th e maximum length L
2
shoul d be used.
Belled Pier s
Piers wit h a belle d botto m ar e normall y use d whe n larg e uplif t force s hav e t o b e resisted .
Fig. 18.21(b ) shows a belle d pie r wit h al l the forces acting.
The uplif t forc e fo r a belle d pie r i s th e sam e a s tha t applicabl e fo r a straigh t shaft . Th e
resisting forc e equatio n for the pier i n the stabl e zone may be written as (O'Neill , 1988)
Q
Rl
=xdL
2
ac
u
(18.19a )
7T r ~ ~ ir ^
(18.19b)
where
<
N - bearin g capacit y factor
d, = diamete r o f the underream
D
820
Chapt er 1 8
L =
L
2
Q Q
/Ws\//X\\
"
, 1
d
//x6\//x\\//X^ \/XA\
Qu
P
Unstabl e
Unstable
zon e
zone
QR
Stable
zone i i
H
J
/
L_
f
'!
i
t
1
i
i
i
i
/
/
d
\
\
\
\
1
!
1
l l
l l
s
\
'/WS //XS\;
f~
up
L
QR\
O
I I
^
j
L
\
(a) (b )
Figure 18.2 1 Drille d pie r i n ex pan s ive soi l
c = uni t cohesion under undrained condition
7 = uni t weight of soil
The value s of N
C
ar e given i n Table 18. 8 (O'Neill , 1988 )
Two cases o f stabilit y may be written as before.
1. Withou t taking the dead loa d Q and using F - 1. 2
2. B y taking into account the dead load and F
s
= 2.0
Table 18. 8 Val ue s of N.
1.7
2.5
>5.0
Foun dat ion s o n Collaps ibl e an d Ex pan s ive Soil s 821
Dead load Dead load
•3 ^g
1) c ; JD
2r
Reinforcement
for tensio n
Grade bea m
Air space beneath grad e bea m
Swelling
pressure -J
(X
Skin frictio n /
s
kr
Reinforcement
for tensio n
Uplifting pressur e
Resisting forc e
2^? = ^
fc
Bell-pier foundatio n Straight shaft-pie r foundatio n
Figure 18.2 2 Grad e bea m an d pie r s ys t e m (Chen , 1988 )
For a given shaf t diamete r d and base diamete r d
b
, the above equations hel p to determine th e
value of L
T
Th e on e that gives the maximum value fo r L
2
has t o be used i n the design .
Fig. 18.2 2 give s a typica l foundatio n design wit h grad e beam s an d drille d pier s (Chen ,
1988). Th e pier s shoul d be take n sufficientl y belo w the unstabl e zone o f wetting i n order t o resis t
the uplif t forces .
Example 1 8. 3
A footin g founded a t a dept h o f 1 ft below groun d level i n expansiv e soi l wa s subjecte d t o load s
from th e superstructure. Site investigation revealed that the expansive soil extended t o a depth of 8
ft belo w th e bas e o f th e foundation , and th e moistur e content s in the soi l durin g the construction
period wer e a t thei r lowest . I n order t o determine the percent swell , three undisturbe d samples a t
depths of 2,4 and 6 ft were collected an d swell tests were conducted per the procedure describe d i n
Section 18.16 . Fig . Ex . 18.3 a show s th e result s o f th e swel l test s plotte d agains t depth . A lin e
passing throug h th e point s i s drawn . The lin e indicate s tha t th e swel l i s zer o a t 8 f t dept h an d
maximum at a base level equa l t o 3%. Determine (a ) the total swell , and (b) the depth o f undercut
necessary fo r a n allowabl e swel l of 0.03 ft .
Solution
(a) Th e tota l swel l i s equa l t o th e are a unde r th e percen t swel l versu s dept h curv e i n
Fig. Ex . 18.3a .
Total swel l = 1/ 2 x 8 x 3 x 1/10 0 = 0.12 f t
822 Chapt er 1 8
Base of structure Base of structure
®Swell determined from test
1 2
Percent swell
(a) Estimation of total swell
0.04 0.12 0.08
Total swell, ft
(b) Estimation of depth of undercut
0.16
Figure Ex . 1 8. 3
(b) Dept h of undercu t
From th e percen t swel l versu s dept h rel at ionshi p given b y th e curv e i n Fig . 18.3a , tota l
swell a t different depths are calculated and plotted agains t depth i n Fig. 18.3b . For exampl e
the tota l swel l a t depth 2 ft below the foundatio n base i s
Total swel l = 1/ 2 (8 - 2 ) x 2.25 x 1/10 0 = 0.067 ft plotted against depth 2 ft. Similarly total
swell a t other depths can be calculated an d plotted. Poin t B on curve i n Fig. 18.3 b gives the
allowable swel l o f 0.0 3 f t a t a dept h o f 4 f t belo w foundatio n base. Tha t is , th e undercu t
necessary i n clay i s 4 f t which may b e replaced b y a n equivalent thicknes s o f nonswellin g
compacted f i l l .
Example 18. 4
Fig. Ex . 18. 4 shows tha t th e soi l t o a dept h o f 2 0 f t i s a n expansiv e type wit h differen t degree s o f
swelling potential . The soi l mas s t o a 2 0 f t dept h i s divide d int o fou r layer s base d o n th e swel l
potential ratin g give n in Tabl e 18.7 . Calculat e the total swel l per the V an Der Merwe method .
Solution
The procedur e fo r cal cul at i n g the tota l swel l i s explaine d i n Sectio n 18.16 . Th e detail s o f th e
calculated result s ar e tabulated below .
Foun dat ion s o n Collaps ibl e an d Ex pan s iv e Soil s 823
Potential expansio n
F - factor
0.25 0. 5 0.7 5 1. 0 1.5
/z^/rs^?^
Layer 1 5
Layer 2 8
;
Layer 3 2
i
/
Layer 4 5
GWT V i
i/S^^Sx
ft Lo w
ft V er y high
ft Hig h
ft V er y high
Lowest
(a)
The
U
5
u

•B 1 0
o,
u
Q
15
20
Figure 1
det ai l s o f c
1
» /n i
£
*V 0.20
-Wo. 13
^,n\ 7« r
/
F=l o
1
g'
1
(-D/20)
(b)
Ex. 18. 4
al cul at ed result s
Layer No . Thi cknes s P E D F &H
S
( i n .)
AD(ft ) f t
1 5 0 2. 5 0.7 5
2 8 1. 0 9. 0 0.3 5
3 2 0. 5 14. 0 0.2 0
4 5 1. 0 17. 5 0.1 3
0
2.80
0.20
0.65
Total 3.6 5
In the tabl e above D = depth fro m groun d level to the mid-dept h o f the layer considered. F = reduction factor .
Example 18. 5
A drille d pie r [refe r t o Fig. 18.2 1 (a)] wa s constructed i n expansive soil . Th e wate r tabl e wa s not
encountered. The details o f the pier and soi l are :
L = 20 ft, d= 12 in., L
{
= 5ft, L
2
= 15 ft,/?, = 10,000 lb/ft
2
, c
u
= 2089 lb/ft
2
, SPT(N) = 25 blows
per foot ,
Required:
(a) tota l uplif t capacit y Q
u
(b) tota l resisting forc e due to surface frictio n
824 Chapt er 1 8
(c) facto r o f safet y withou t taking int o account th e dead load Q acting o n the top of the pie r
(d) facto r of safet y wit h th e dea d loa d acting on th e top o f the pie r
Assume Q = 10 kips. Calculate Q
u
b y Chen' s metho d (Eq . 18.15) .
Solution
(a) Uplif t forc e Q fro m Eq . (18.15 )
. 3.1 4 x(l )x 0.15x10,000x 5
O = TtdapL , = — = 23.55 kips
up u s l
100 0
(b) Resistin g force Q
R
FromEq. (18.17)
Q
R
= nd( L -L
l
)uc
u
where c
u
= 2089 lb/ft
2
- 100 kN/m
2
—£- = a 1. 0 where p = atmospheric pressur e =101 kPa
Pa 10 1
From Fig . 17.15 , a = 0.55 fo r cjp
a
= 1. 0
Now substitutin g th e known values
Q
R
= 3.14 x 1 x (20 - 5 ) x 0.55 x 2000 = 51,810 I b = 52 kips
Q
z
L
'
y//$\ /A
j
\
1
'2
Unstable
Q
U
p zon e
Stable
zone
.t
Figure Ex . 18. 5
Foun dat ion s o n Collaps ibl e an d Ex pan s ive Soil s 82 5
(c) Facto r o f safet y wit h Q = 0
FromEq. (18.18a)
F = ®R_
=
_ ^L = 2.2 >1.2 required - -OK.
S
Q
UP
23. 5
(d) Facto r o f safet y wit h Q= 1 0 kips
FromEq. (18. 1 8b)
= — = 3.9 > 2.0 required - -OK.
( Q
u
-Q) (23.5-10 ) 13. 5
up
Example 18. 6
Solve Exampl e 18. 5 wit h Lj = 1 0 ft. All the other data remai n the same .
Solution
(a) Uplif t force Q
up
Q
up
= 23.5 x(10/5) = 47.0 kips
where Q = 23.5 kip s for Lj = 5 ft
(b) Resistin g forc e Q
R
Q
R
= 52 x (10/15) = 34.7 kip s
where Q
R
= 52 kips for L
2
= 15 ft
(c) Facto r o f safet y for Q = 0
347
F = ——= 0.74 <1. 2 as required - - not OK .
s
47. 0
M
(d) Facto r o f safet y fo r Q = 10 kips
34 7 3 4 7
F. = -
:
-= — — = 0.94 < 2.0 as require d - - not OK.
s
(47-10 ) 3 7
The abov e calculation s indicat e that if the wettin g zone (unstabl e zone) i s 1 0 ft thic k the
structure wil l no t b e stabl e for L = 20 ft .
Example 1 8. 7
Determine th e length of pier required i n the stable zon e for F
s
=l.2 wher e ( 2 = 0 and F
S
= 2.0 when
2 = 1 0 kips. All the other dat a given in Example 18. 6 remain the same.
Solution
(a) Upliftin g force Q
up
for L, (10 ft) = 47 kip s
(b) Resistin g force for lengt h L
2
in the stabl e zone.
Q
p
= n, a c L , = 3.14 x1 x 0.55 x2000 L, = 3,454 L, lb/ft
2
*^A u U L 2. L
(c) Q = 0, minimum F = 1.2
826 Chapt e r 1 8
or
L 2 =
^
=
3.454 L
2
Q,
v
47
-ooo
solving we have L
2
- 16. 3 ft.
(d) Q = 1 0 kips. Minimum F
s
= 2.0
Q
R
3,45 4 L
2
3,45 4 L
2
F =2. 0 =
(47,000-10,000) 37,00 0
Solving w e have L
2
= 21.4 ft .
The abov e calculations indicate that th e minimu m L
9
= 21. 4 f t or sa y 2 2 f t i s required fo r
the structur e t o be stabl e wit h L
{
= 1 0 ft. The tota l lengt h L = 1 0 + 22 = 32 ft .
Example 18.8
Figure Ex. 18. 8 shows a drilled pier wit h a belled bottom constructed i n expansive soil. Th e water
table i s not encountered. The detail s of the pier and soi l ar e given below:
L
l
- 1 0 ft, L
2
= 10 ft, L
b
= 2.5 ft,d= 1 2 in., d
b
= 3 ft, c
u
= 2000 lb/ft
2
, p^ = 10,000 lb/ft
2
, y= 1 10
lb/ft
3
.
Required
(a) Uplif t forc e Q
up
(b) Resistin g force Q
R
(c) Facto r o f safet y for Q = 0 at the top of the pier
(d) Facto r o f safet y fo r Q = 20 kips at the top o f the pie r
Solution
(a) Uplif t forc e Q
up
As i n Ex. 18.6£
up
= 47ki ps
(b) Resistin g force Q
R
Q
Rl
=ndL
2
ac
u
a = 0.55 a s in Ex. 18. 5
Substituting known values
Q
Rl
= 3.14 x 1 x 1 0 x 0.55 x 2000 = 34540 Ibs = 34.54 kips
where d
b
= 3 ft, c = 2000 lb/ft
2
, N
C
= 7.0 from Table 18. 8 fo r L
2
/d
b
= 10/3 = 3.3 3
Substituting known values
Q
R2
= — [3
2
- (I)
2
] [2000 x 7.0 +110 x 10]
= 6.28 [15,100 ] = 94,828 Ib s = 94.8 kips
Q
R
= Q
Rl
+ Q
R2
= 34.54 + 94.8 = 129. 3 kips
(c) Facto r o f safety for Q - 0
OP 129 3
F -
R
= = 275 > 1 2 --O K A „ X— . / *J -" ^ i*^ *-S A*.
<2
UP
47. 0
Foun dat ion s o n Collaps ibl e an d Ex pan s iv e Soil s 8 2 7
! G
I , = 10ft
= 1 0 ft
Unstable
zone
Q
U
Stable zone
QK
Figure Ex . 18. 8
(d) Facto r o f safet y for Q = 20 kips
h 129. 3
F = •
(47-20)
= 4.79 > 2.0 - - as required OK.
The above calculation s indicat e tha t the design i s over conservative . The lengt h L
2
can be
reduced t o provide a n acceptabl e facto r of safety.
1 8.1 8 EL I M I NATI O N OF SWEL L I N G
The eliminatio n of foundation swelling can be achieved i n two ways. They ar e
1. Providin g a granular bed an d cover below and around the foundation (Fig. 18.19)
2. Chemica l stabilizatio n o f swelling soil s
Figure 18.1 9 gives a typical exampl e o f the firs t type . I n thi s case, th e excavation i s carrie d
out u p t o a depth greate r tha n the widt h of the foundation by abou t 20 t o 30 cms. Freel y drainin g
soil, suc h a s a mixtur e o f san d an d gravel , i s place d an d compacte d u p t o th e bas e leve l o f th e
foundation. A Reinforced concret e footing is constructed at this level. A mixture of sand and gravel
is filled u p loosely over the fill . A reinforced concrete apro n abou t 2 m wide is provided aroun d the
building to prevent moisture directly entering the foundation. A cushion of granular soil s below th e
foundation absorb s the effect o f swelling, and thereby its effect o n the foundation will considerabl y
828 Chapt e r 1 8
be reduced. A foundation o f thi s typ e shoul d b e constructed onl y durin g the dr y seaso n whe n the
soil has shrun k to it s lowest level . Arrangements should be made t o drain awa y the water from th e
granular bas e durin g the rainy seasons .
Chemical stabilizatio n o f swelling soil s by the addition o f lime may be remarkably effectiv e
if th e lime ca n b e mixe d thoroughl y wit h th e soi l an d compacted a t about th e optimu m moistur e
content. The appropriat e percentage usuall y range s from abou t 3 to 8 percent. The lime content is
estimated o n th e basi s o f p H test s an d checke d b y compacting, curin g and testing sample s i n the
laboratory. Th e lim e ha s th e effec t o f reducin g th e plasticit y o f th e soil , an d henc e it s swellin g
potential.
1 8 .1 9 PROB L EM S
18.1 A building was constructe d i n a loessial typ e normall y consolidate d collapsibl e soi l wit h
the foundatio n at a depth of 1 m below groun d level. Th e soi l t o a depth o f 6 m below th e
foundation wa s found to be collapsible on flooding. Th e average overburde n pressure was
56 kN/m
2
. Double consolidomete r test s wer e conducted on two undisturbed samples take n
at a dept h o f 4 m below groun d level , on e wit h it s natural moistur e conten t an d th e othe r
under soaked condition s pe r the procedur e explaine d i n Section 18.4 . Th e followin g dat a
were available .
Applied pressure , kN/m
2
1 0 2 0 4 0 10 0 20 0 40 0 80 0
V oid ratios a t natural
moisture content 0.8 0 0.7 9 0.7 8 0.75 0.72 5 0.6 8 0.6 1
V oid ratios i n the
soaked condition 0.7 5 0.7 1 0.6 6 0.5 8 0.5 1 0.4 3 0.3 2
Plot the e-log p curves an d determine th e collapsible settlement for an increase in pressur e
Ap = 34 kN/m
2
a t the middl e of the collapsible stratum.
18.2 Soi l investigatio n at a site indicated overconsolidated collapsibl e loessia l soi l extending to
a grea t depth . I t i s required t o construc t a footing a t the sit e founde d a t a dept h o f 1. 0 m
below groun d level . Th e sit e i s subjec t t o flooding. The averag e uni t weight of the soi l i s
19.5 kN/m
3
. Two oedometer test s wer e conducte d o n two undisturbe d samples take n a t a
depth o f 5 m from ground level. One test was conducted a t its natural moisture content an d
the other o n a soaked conditio n per the procedure explaine d i n Section 18.4 .
The following test result s ar e available.
Applied pressure , kN/m
2
1 0 2 0 4 0 10 0 20 0 40 0 80 0 200 0
V oid rati o unde r natural
moisture conditio n 0.79 5 0.7 9 0.78 7 0.7 8 0.7 7 0.7 4 0.7 1 0.6 4
V oid rati o unde r
soaked conditio n 0.77 5 0.7 7 0.75 7 0.73 0 0.6 8 0.6 3 0.5 4 0.3 7
The swel l inde x determine d fro m th e rebound curv e of the soaked sample is equal t o 0.08 .
Required:
(a) Plot s o f e-log p curve s for bot h tests.
(b) Determinatio n o f the averag e overburde n pressure a t the middl e of the soi l stratum .
(c) Determinatio n o f the preconsolidation pressur e based on the curve obtaine d from th e
soaked sampl e
(d) Tota l collaps e settlemen t for a n increase i n pressure A/ ? = 710 kN/m
2
?
Foun dat ion s o n Collaps ibl e an d Ex pan s ive Soil s 829
18.3
18.4
A footing for a building is founded 0.5 m below ground level i n an expansive clay stratum
which extends t o a great depth . Swel l test s wer e conducted on three undisturbe d samples
taken a t different depth s and the detail s of the test s ar e given below.
Depth (m)
below GL
1
2
3
Swell %
2.9
1.75
0.63
Required:
(a) Th e total swel l under structural loadings
(b) Dept h of undercut for a n allowabl e swell of 1 cm
Fig. Prob . 18. 4 give s th e profil e o f a n expansiv e soi l wit h varyin g degrees o f swellin g
potential. Calculat e th e total swel l per the V an Der Merwe method .
Potential expansio n ratin g
/XXV X /XXV S //XS \//"/V S
Layer 1
Layer 2
Layer 3
Layer 4
/XA\/X/V N/XXSN/yW , S/*\
6 f t Hig h
4 f t Lo w
8 f t V er y high
6 f t Hig h
Figure Prob . 18. 4
18.5 Fig . Prob. 18. 5 depicts a drilled pier embedded i n expansive soil. The details of the pier and
soil properties ar e given in the figure .
Determine:
(a) Th e tota l uplif t capacity .
(b) Tota l resisting force .
(c) Facto r o f safet y with no load actin g on the top of the pier .
(d) Facto r o f safet y wit h a dead loa d of 10 0 kN on the top of the pier .
Calculate Q
u
b y Chen' s method .
18.6 Solv e problem 18. 5 usin g Eq. (18.16)
18.7 Fig . Prob. 18. 7 shows a drilled pier wit h a belled bottom. All the particulars of the pier and
soil ar e given i n the figure.
Required:
(a) Th e tota l uplif t force .
(b) Th e total resistin g forc e
830
Chapt er 1 8
1
i
\
1
I
'
1
,
L
i
2
i
^_c
g= 10 0 kN
r
— » •
1 Unstabl e
zone

P
Stable
zone

Given:
L, = 3 m, L ~!_ - 1 0 m
d = 40 cm
c
u
= 75 kN/m
2
= 500 kN/m
2
Figure Prob . 18. 5
(c) Facto r o f safet y fo r < 2 = 0
(d) Facto r of safet y for Q = 200 kN .
Use Chen' s method fo r computing Q
18.8 Solv e Prob. 18. 7 by makin g us e of Eq. (18.16) for computin g Q .
Q
d
h
= 1. 2 m
Figure Prob . 18. 7
Given:
L|= 6 m, L ^ - 4 m
L
A
= 0.75 m,d = 0. 4m
c
u
= 75 kN/m
2
y = 17. 5 kN/m
3
p, = 500 kN/m
2
(2 = 200 k N
Foun dat ion s o n Collaps ibl e and Ex pan s ive Soils 831
4 f t
Unstable
zone
Stable zone
Given:
c
u
= 800 lb/ft
2
y = 110 lb/ft
3
p
s
= 10,000 lb/ft
2
Q = 60 kips
Figure Prob. 18. 9
18.9 Refe r to Fig. Prob . 18.9 . Th e following data are available:
Lj = 1 5 ft, L
2
= 1 3 ft, d = 4 ft, ^= 8 ft and L
6
= 6 ft.
All the other dat a ar e given in the figure.
Required:
(a) Th e tota l uplif t forc e
(b) Th e total resisting forc e
(c) Facto r of safety fo r Q = 0
(d) Facto r o f safet y fo r Q = 60 kips.
Use Chen' s metho d fo r computing Q .
18.10 Solv e Prob. 18. 9 using Eq. (18.16) for computing Q
up
.
18.11 I f th e lengt h L
2
i s no t sufficien t i n Prob . 18.10 , determin e th e require d lengt h t o ge t
F. = 3.0.
CHAPTER 1 9
CONCRETE AND MECH ANICALL Y
STABILIZED EARTH RETAINING WALL S
PART A—CONCRETE RETAI NI NG WAL L S
1 9 .1 I NTRODU CTI O N
The commo n type s o f concret e retainin g wall s an d thei r uses wer e discusse d i n Chapte r 11 . The
lateral pressure theories and the methods of calculating the lateral earth pressures wer e described i n
detail i n the same chapter . The two classical eart h pressur e theorie s tha t have been considere d ar e
those of Rankine an d Coulomb . I n thi s chapte r w e ar e intereste d i n th e following :
1. Condition s unde r which the theories o f Rankine and Coulomb ar e applicabl e t o cantileve r
and gravity retaining wall s under the active state.
2. Th e commo n minimu m dimensions use d fo r th e two type s o f retainin g wall s mentione d
above.
3. Us e of chart s fo r the computation of active eart h pressure .
4. Stabilit y of retaining walls.
5. Drainag e provision s fo r retaining walls.
1 9.2 CONDI TI ON S UNDE R WH I CH RANKI NE AND COU L OM B
FORM U L AS ARE APPL I CAB L E TO RETAI NI N G WAL L S UNDE R
ACTI V E STAT E
Conj ugate Failur e Plane s Unde r Activ e Stat e
When a backfil l o f cohesionles s soi l i s unde r a n activ e stat e o f plasti c equilibriu m du e t o th e
stretching o f the soi l mas s a t every poin t in the mass , tw o failur e planes calle d conjugate rupture
833
834
Chapt er 1 9
planes ar e formed . Thes e ar e furthe r designate d a s th e inner failure plane an d th e outer failure
plane as shown i n Fig. 19.1 . Thes e failur e plane s mak e angles o f a. an d «
0
with th e vertical . The
equations fo r thes e angle s may b e written as (for a sloping backfill)
a. = +
s-B
- —
2 2
£-J 3
"°~ 2 2
. sinyt f
where si n s =
(19.2)
„_
when / 7=0 ,
. _ „
- = 45°- — , a
n
=
2 2 °
y i e o
- = 45°- —
2 2
The angl e betwee n th e two failure plane s = 90 - 0 .
Conditions fo r the U s e o f Rankine' s Formul a
1 . Wal l shoul d b e vertica l wit h a smooth pressur e face .
2. Whe n wall s are inclined, it should not come i n the way of the formation o f the outer failure
plane. Figur e 19. 1 shows the formation of failure planes. Sinc e the sloping fac e AB' of the
retaining wal l make s a n angl e a
w
greate r tha n a
o
, the wal l doe s no t interfer e wit h th e
formation o f the outer failure plane. The plasti c stat e exist s withi n wedge ACC'.
The metho d o f calculating the latera l pressure o n AB' i s as follows.
1 . Appl y Rankine' s formul a for the vertical sectio n AB.
2. Combin e P wit h W , the weight of soi l withi n the wedge ABB', to give the resultant P
R
.
Let the resultant P
R
i n this case mak e a n angle 8
r
with the normal t o the face of the wall . Let
the maximum angl e o f wall friction be 8
m
. If 8
r
> 8
m
, the soi l slide s alon g th e face AB'of th e wall.
Outer failur e plane ^
Inner failur e plan e
Figure 1 9. 1 Applicat io n o f Ran k in e' s act iv e con dit io n t o g ravit y wall s
Con cret e an d M echan icall y St abilize d Eart h Ret ain in g Wall s 835
Outer failure plane
Inner failure plane
Figure 19. 2 Latera l eart h pressur e o n can t ilever wall s unde r act ive conditio n
In such an eventuality, the Rankine formula is not recommended bu t the Coulomb formul a may be
used.
Conditions fo r the U s e o f Coulomb' s Formula
1. Th e bac k o f the wal l must be plane o r nearly plane .
2. Coulomb' s formula may be applied unde r all other conditions wher e the surface of the wall
is not smoot h an d wher e the soi l slide s alon g the surface.
In general the following recommendations ma y be made for the application o f the Rankine or
Coulomb formul a without the introduction of significant errors:
1. Us e th e Rankin e formul a for cantilever and counterfor t walls.
2. Us e the Coulomb formul a for soli d an d semisoli d gravit y walls.
In th e cas e o f cantileve r wall s (Fig . 19.2) , P
a
i s th e activ e pressur e actin g o n th e vertica l
section AB passing throug h th e heel o f the wall. The pressure is parallel to the backfill surfac e and
acts at a height H/3 fro m the base of the wall where H is the height of the section AB. The resultant
pressure P
R
is obtained by combining the lateral pressure P
a
with the weight of the soil W
s
between
the section AB an d the wall.
1 9.3 PROPORTI ONI N G OF RETAI NI NG WAL L S
Based on practical experience , retainin g walls can be proportioned initiall y which may be checke d
for stabilit y subsequently. The common dimensions used for the various types of retaining walls are
given below.
G ravity Wall s
A gravit y walls may be proportioned i n terms of its height given in Fig. 19.3(a) . Th e minimu m top
width suggested is 0.30 m . The tentative dimension s fo r a cantilever wal l ar e given i n Fig. 19.3(b )
and thos e fo r a counterfort wal l are given i n Fig. 19.3(c) .
836
Chapt er 1 9
0. 3m to///12
Min. batter
1 : 48
0.3 m min.
H h -
Min. batter
1 :48
\ *-B = 0.5 to 0.7//-*|*
-HO. l t fh—
v
= H/8toH/6
I—-B = 0.4 to 0.7#-» -|
= H/l2toH/lO
(a) Gravity wall (b) Cantilever wall
03 W
0.2 m min
(c) Counterfort wall
Fig. 19. 3 T en t at iv e dim en s ion s fo r ret ain in g wall s
19.4 EART H PRESSU RE CH ARTS FOR RETAI NI NG WAL L S
Charts hav e bee n develope d fo r estimatin g latera l eart h pressure s o n retainin g wall s base d o n
certain assumed soi l properties of the backfill materials . These semi empirical methods represent a
body of valuable experience and summarize much useful information . The charts given in Fig. 19. 4
are meant t o produce a design of retaining walls of height s not greate r tha n 6 m. The chart s have
been develope d for fiv e types of backfill materials give n in Table 19.1 . The charts are applicable to
the following categories of backfil l surfaces . They ar e
1. Th e surfac e of the backfil l is plane and carries n o surcharge
2. Th e surfac e o f the backfil l rise s o n a slope from the cres t o f th e wal l t o a leve l a t some
elevation above the crest .
The char t i s drawn to represent a concrete wal l but it may also be used fo r a reinforced soi l
wall. Al l th e dimension s of th e retainin g wall s ar e give n i n Fig. 19.4. Th e tota l horizonta l an d
vertical pressures on the vertical section of A B of height H ar e expressed a s
P, = -K,H
2
n< • n
(19.3)
Con cret e an d M echan icall y St abilize d Eart h Ret ain in g Wall s 837
Table 19. 1 T ype s o f back fil l fo r ret ain in g wall s
T ype B ack fil l m at eria l
Coarse-grained soi l without admixtur e o f fine soil particles,
very permeabl e (clea n san d or gravel )
Coarse-grained soi l of low permeability du e t o
admixture of particles o f sil t siz e
Residual soi l wit h stones fine silt y sand , an d granula r
materials wit h conspicuous clay content
V ery sof t o r sof t clay , organic silts , or silt y clays
Medium o r stif f cla y
Note:
Numerals o n the curves indicat e
soil types a s describe d
in Table 19. 1
For materials o f type 5
computations of pressur e
may be based o n the value of H
1 meter les s than actual value
10 2 0
V alues of slope angle
40
Figure 19. 4 Char t fo r es t im at in g pres s ur e o f back fil l ag ain s t ret ain in g wall s
s upport in g back fill s wit h a plan e s urface . (T erzag hi , Peck , an d Mesri , 1996 )
H,=0
H
Soil type Soil type 2 Soil type 3
15
10
0.4 0.8 1. 0 0 0. 4 0. 8 1. 0 0
V alues of rati o H
}
IH
0.4 0.8 1. 0
Figure 19. 5 Char t for estimatin g pressur e of backfil l against retaining wall s supporting backfill s with a surface that slopes
upward fro m the crest of the wal l for limited distance an d then becomes horizontal . (Terzagh i et al., 1996 )
Con cret e an d Mechanicall y Stabilize d Eart h Retainin g Wall s 839
Soil type 4 Soil type 5
J V
25
3
3
> 10
5
n
*
~~
Ma;
K.
c slop
= 0
; 3:1
Note:
Numerals o n curve s
indicate the followin g
slopes
No.
1
2
3
4
5
Slope
1.5:1
1.75:1
2:1
3:1
6:1
0 0. 2 0. 4 0. 6 0. 8 1. 0 0 0. 2 0. 4 0. 6 0. 8 1. 0
V alues o f ratio H\ IH V alue s of ratio H
}
/H
Figure 19. 5 Con t in ue d
P
V
=-K
V
H
2
(19.4 )
V alues o f K
h
an d K
v
ar e plotte d agains t slop e angl e / ? in Fig . 19. 4 an d th e rati o HJ H i n
Fig. 19.5 .
19.5 STAB I L I T Y O F RETAI NI NG WAL L S
The stabilit y of retaining walls should be checked for the following conditions:
1. Chec k for sliding
2. Chec k for overturning
3. Chec k for bearing capacity failure
4. Chec k for base shear failur e
The minimu m factors of safet y fo r the stabilit y of the wal l are:
1. Facto r o f safet y against sliding =1.5
2. Facto r of safet y against overturning =2. 0
3. Facto r of safety against bearing capacit y failur e = 3.0
Stability Analysi s
Consider a cantilever wall wit h a sloping backfill for th e purpose of analysis. The same principl e
holds for the other types of walls.
Fig. 19. 6 gives a cantilever wall wit h al l the forces acting on the wal l and the base, where
P
a
= activ e earth pressure acting at a height H/3 ove r the base on section AB
h a > ~^
P = P sin/ 3
v a "
13 = slop e angl e of th e backfill
840 Chapt er 1 9
W
c
w.
F
r
(a) Forces acting o n the wal l
-B-
-Key
(b) Provision o f key t o increas e slidin g resistanc e
Figure 19. 6Chec k fo r sl i di n g
weight of soi l
weight o f wal l including base
the resultant of W
s
and W
c
passive eart h pressur e a t the toe side of the wall .
base slidin g resistance
Check fo r Sliding (Fig . 1 9.6 )
The forc e tha t moves th e wal l = horizontal forc e P
h
Con cret e an d M echan icall y St abilize d Eart h Ret ain in g Wall s 841
The force tha t resist s the movement i s
F Rtan8+P (19.5)
R = tota l vertical force = W
s
+ W
c
+ P
v
,
8 = angl e of wal l frictio n
c
a
= uni t adhesion
If the bottom of the base sla b i s rough, as in the case o f concrete poure d directl y o n soil , th e
coefficient o f friction is equal t o tan 0, 0 being the angl e of internal friction of the soil .
The factor of safet y agains t slidin g is
F =-*->
(19.6)
In case F
s
< 1.5, additiona l facto r of safety can be provided b y constructing on e or two keys
at the base level shown in Fig. 19.6b . The passive pressure P (Fig . 19.6a ) in front of the wall should
not be relied upo n unless i t is certain tha t the soi l wil l always remain fir m and undisturbed.
Check fo r Overturnin g
The forces acting on the wall are shown in Fig. 19.7 . Th e overturning and stabilizing moments ma y be
calculated by taking moments about point O. The factor of safety against overturning is therefore
Sum of moment s that resist overturning _ M
R
Sum of overturning moments M
(19.7a)
Figure 19. 7 Chec k fo r overt urn in g
842 Chapt er 1 9
we may writ e (Fig . 19.7 )
Wl +WI + PB
C C S S V
F =
where F shoul d not be less tha n 2.0.
(19.7b)
Check fo r B earin g Capacit y Failur e (Fig . 1 9 .8 )
In Fig. 19.8 , W
(
is the resultant of W
s
and W
c
. P
R
is the resultant of P
a
and W
f
and P
R
meets th e
base a t m. R i s the resultan t of al l th e vertica l forces actin g a t m wit h an eccentricit y e . Fig. 19. 8
shows th e pressur e distribution at the base wit h a maximum q
t
at the toe and a minimum q
h
at the
heel.
An expression fo r e may be writte n as
B ( M
R
- M
0
)
2 I V
where R = XV = sum of al l vertical force s
(19.8a)
Toe
Figure 19. 8 Stabilit y agains t bearin g capacit y failur e
Con cret e an d M echan icall y St abilize d Eart h Ret ain in g Wal l s 843
The values o f q
t
and q
h
may be calculated b y making us e of the equation s
B
B
B
(19.8b)
(19.8c)
where, q
a
= R/B = allowable bearin g pressure .
Equation (19.8 ) i s vali d fo r e< B/6 . Whe n e = B/6, q
t
= 2q
a
an d q
h
= 0. The bas e widt h B
should b e adjuste d t o satisf y Eq . (19.8 ) . When th e subsoi l belo w th e bas e i s o f a lo w bearin g
capacity, the possible alternativ e is to use a pile foundation.
The ultimate bearing capacit y q
u
may be determined using Eq. (12.27) taking into account the
eccentricity. I t must be ensured that
B ase Failur e o f Foundatio n (Fig . 1 9 .9 )
If th e bas e soi l consist s o f mediu m t o sof t clay , a circula r sli p surfac e failur e ma y develo p a s
shown i n Fig. 19.9 . The most dangerou s slip circle is actually the one that penetrates deepest int o
the sof t material . Th e critica l sli p surfac e mus t be locate d b y trial . Suc h stabilit y problems ma y
be analyze d eithe r b y the metho d o f slices o r an y other metho d discusse d i n Chapter 10 .
Figure 19. 9 St abilit y ag ain s t bas e sli p s urfac e s hear failur e
844 Chapt er 1 9
Drainage Provisio n fo r Retainin g Wall s (Fig. 1 9 .1 0)
The saturatio n of the backfil l o f a ret ai ni ng wal l i s always accompanied by a substantial hydrostatic
pressure on the back of the wall. Saturation of the soil increases the earth pressure by increasing the
uni t weight . I t i s therefor e essentia l t o eliminat e or reduc e por e pressur e b y providin g suitabl e
drainage. Fou r type s of drainage ar e given in Fig. 19.10. The drains collect the water that enters th e
backfill an d this may be disposed of through outlets in the wal l called weep holes. The grade d filte r
material shoul d b e properl y designe d t o prevent clogging by fine materials . The present practice i s
to use geotextile s or geogrids .
The weep hole s are usually mad e by embedding 10 0 mm ( 4 in.) diameter pipe s in the wal l
as shown in Fig. 19.10. The vertica l spacing between horizonta l rows of weep hole s shoul d not
exceed 1. 5 m. Th e horizonta l spacin g i n a give n ro w depend s upo n th e provision s mad e t o
direct th e seepag e wate r toward s the wee p holes .
Percolating
•*- wate r durin g
rain
Percolating
water durin g
Permanently
drained
(c) (d)
Figure 19.1 0 Diag ra m s howin g provis ion s fo r drain ag e o f back fil l behin d ret ainin g
wal l s : (a ) vert ical drain ag e l aye r (b) in clin ed drain ag e layer fo r cohes ion les s back fil l ,
(c) bot t o m drai n t o accel erat e con s olidat ion o f cohes iv e bac k fill , (d ) horizon t al drai n
an d s eal com bin ed wit h in clin ed drain ag e l aye r for cohes iv e back fil l (T erzag h i e t al. ,
1996)
Con cret e an d M echan icall y St abilize d Eart h Ret ain in g Wall s 845
Example 1 9 . 1
Figure Ex . 19.1(a ) show s a sectio n o f a cantileve r wal l wit h dimension s an d force s actin g
thereon. Chec k th e stabilit y o f th e wal l wit h respec t t o (a ) overturning , (b ) sliding , an d (c )
bearing capacity .
Solution
Check fo r Rankine' s conditio n
FromEq. (19.1b)
where sin f =
2 2
si n5 sin!5
c
sin ( j) si n 30 °
ore *31°
_ 90-30 31-1 5
(Xn —
= 0.5176
= 22
C
The oute r failur e lin e AC i s drawn making an angl e 22 ° with the vertical AB. Since thi s line
does no t cut the wall Rankine' s conditio n i s vali d i n this case .
T
// = 0.8 + 7 = 7.8m
/>„
4.75m
F
R
A I ( c - 0) soil
c = 60kN/m
2
( /) - 25 ° y = 19 kN/m
3
Figure Ex . 19. 1 (a)
846
Chapt er 1 9
Rankine active pressure
Height o f wal l = AB = H = 7.8 m (Fig. Ex. 19 . l(a))
where K, = tan
2
(45°-^ / 2) = -
3
substituting
P
a
= - x 18.5 x (7.8)
2
x - = 187.6 kN / m of wall
2 3
P
h
= P
a
cosj3 = 187.6 cos 15°= 181. 2 k N /m
P
v
= P
a
si n 0= 187.6 sin 15° = 48. 6 k N / m
Check fo r overturnin g
The force s actin g o n th e wal l i n Fig . Ex . 19.1(a ) ar e shown . Th e overturnin g an d stabilizin g
moments ma y be calculated b y taking moments abou t point O.
The whol e sectio n i s divide d int o 5 part s a s show n i n th e figure . Le t thes e force s b e
represented by vv
p
vv
2
, . . . vv
5
and the corresponding lever arms as /
p
/
2
, .. . 1
5
. Assume the weight of
concrete y
c
= 24 kN/m
3
. The equation for the resisting moment is
M
R
— Wj /j + w
2
/
9
+ .. . w
5
/
5
The overturnin g moment is
M
0
= .
P
h
3
The detail s o f calculations are tabulated below.
Section
No.
1
2
3
4
5
Area
( m
2
)
1.20
18.75
3.56
3.13
0.78
U n i t wei gh t
k N/ m
3
18.5
18.5
24.0
24.0
24.0
Wei ght
k N/ m
22.2
346.9
85.4
75.1
18.7
P
v
= 48. 6
2, = 596.9
Lever
a r m( m)
3.75
3.25
2.38
1.50
1.17
4.75
Moment
k N- m
83.25
1127.40
203.25
112.65
21.88
230.85
2
W
= 1 , 779.3 = M
R
M
0
= 181. 2x2. 6 = 471.12 kN- m
F =
M
R
_ 1,779. 3
~M~~ 471.1 2
- 3.78 > 2.0 - - OK .
Check fo r sliding (Fig . 19.1a )
The force tha t resists th e movement a s per Eq. (19.5) i s
F
R
=
Ca
B + R tan 5 + P
p
Con cret e an d M echan icall y St abilize d Eart h Ret ain in g Wall s 847
where B = width = 4.75 m
c
a
~
ac
u'
a
~ adhesion factor = 0.55 fro m Fig . 17.1 5
R = total vertical force I
v
= 596.9 kN
For the foundation soil:
S = angle of wal l friction ~ 0 = 25°
FromEq. (11.45c)
where h = 2 m, y= 1 9 kN/m
3
, c = 60 kN/m
2
K = tan
2
(45° + §12) = tan
2
(45° + 25/2) = 2.46
substituting
p = -xl9x2
2
p
2
. 46= 470 kN/m
7 m
0.75 mt
e = 0.183m
B/2 B/2
Figure Ex . 1 9 .Kb)
848 Chapt er 1 9
= 60 x 4.75 + 596. 9 tan 25° + 470 = 285 + 278 + 470 = 103 3 kN/ m
P= 1 8 1.2 kN/ m
P
h
181. 2
1.5 -OK .
Normally th e passive earth pressure P
n
is not considered i n the analysis. By neglecting P
p
, the
factor o f safet y i s
1033- 470
181.2 181. 2
_
Check fo r bearing capacity failur e (Fig . 19.Ib)
From Eq. (19.8 b and c) , the pressures a t the toe and heel o f the retaining wall may be writte n as
R
E
1-
B
where e = eccentricit y o f th e tota l loa d R ( = SV ) actin g o n th e base . Fro m Eq . (19.8a) , th e
eccentricity e may be calculated.
€=
B
2
Now
q
f
=
R 2
596.9 , 6x0.18 3
4.75
.
1 +
4.75
596.9
= 154. 7 kN/m
2
596.9 . 6x0.18 3
n
, ,
1 1 V T /
,
q
h
= 1 — — = 96.6 kN/m
2
4.75 4.75
The ul t i mat e beari n g capaci t y q
u
ma y b e determine d b y Eq . (12.27) . I t ha s t o b e
ensured tha t
where F = 3
Con cret e an d M echan ical l y St abil ize d Eart h Ret ain in g Wall s 849
PART B —M ECH ANI CAL L Y STAB I L I Z ED EARTH
RETAI NI NG WAL L S
1 9.6 G ENERA L CONSI DERATI ONS
Reinforced eart h i s a construction materia l composed o f soil fil l strengthene d b y the inclusion of rods ,
bars, fiber s o r net s whic h interac t wit h th e soi l b y mean s o f frictiona l resistance . Th e concep t o f
strengthening soi l wit h rod s o r fiber s i s no t new . Throughout th e age s attempt s hav e bee n mad e t o
improve the qualit y o f adobe bric k b y addin g straw . Th e present practice i s t o us e thi n meta l strips ,
geotextiles, an d geogrids as reinforcing materials for the construction of reinforced eart h retaining walls.
A ne w er a of retaining wall s wit h reinforced eart h wa s introduce d b y V idal (1969) . Metal strip s wer e
used as reinforcing materia l a s shown in Fig. 19.1 1 (a). Here the metal strips exten d from th e panel bac k
into the soil to serve the dual role of anchoring the facing units and being restrained throug h the frictional
stresses mobilized between the strips and the backfill soil . The backfill soi l creates the lateral pressure
and interact s wit h the strip s t o resis t it . The wall s ar e relativel y flexibl e compare d t o massive gravit y
structures. Thes e flexible wall s offe r many advantages including significan t lower cost per square meter
of exposed surface. The variations in the types effacing units , subsequent t o V idal's introduction o f the
reinforced eart h walls , are many. A few of the types that are currently in use are (Koerner, 1999 )
Figure 1 9 .1 1 (a) Com pon en t part s an d k e y dim en s ion s of rein force d eart h wall
(Vidal, 1969 )
850 Chapter 1 9
1. Facin g panel s wit h meta l stri p reinforcemen t
2. Facin g panels wit h wir e mes h reinforcemen t
3. Soli d panels wit h ti e back anchor s
4. Anchore d gabio n wall s
5. Anchore d cri b wall s
Facing units
Rankine wedge — \
H
As required
Reinforcing strips
"• ;''• : T!
:
• . *. ' • -A-: Selec t fil l ;
;
. >;' • ;. ' _..
Original ground
or other backfil l
1
(sO. 8/ / ) '
(b) Line details of a reinforced earth wal l i n place
(c) Front face o f a reinforced earth wal l unde r construction for a bridge approac h fil l using patented precast
concrete wal l face uni t s
Figure 1 9 .1 Kb ) and (c) Rein force d eart h wall s (B owles , 1996 )
Con cret e an d M echan ical l y St abil ize d Eart h Ret ain in g Wall s 851
6. Geotextil e reinforce d wall s
7. Geogri d reinforce d wall s
In all cases, the soil behind th e wall facing is said to be mechanically stabilized earth (MSE )
and the wal l system i s generally calle d a n MSE wall .
The thre e component s o f a MSE wal l ar e th e facin g unit , the backfil l an d th e reinforcin g
material. Figur e 19.11(b ) show s a sid e vie w o f a wal l wit h meta l stri p reinforcemen t an d
Fig. 19. 1 l(c) th e fron t fac e of a wall under construction (Bowles , 1996) .
Modular concret e blocks , currentl y called segmenta l retainin g wall s (SRWS , Fig . 19.12(a) )
are mos t commo n a s facin g units . Som e o f th e facin g unit s ar e show n i n Fig . 19.12. Mos t
interesting in regard t o SRWS ar e the emerging bloc k systems wit h openings, pouches , o r planting
areas withi n them. These opening s ar e soil-filled and planted wit h vegetation that is indigenous to
the area (Fig . 19.12(b)) . Further possibilities i n the area of reinforced wal l systems coul d be in the
use o f polyme r rope , straps , o r ancho r tie s t o th e facin g i n unit s o r t o geosyntheti c layers , an d
extending the m int o the retained eart h zon e as shown in Fig. 19.12(c) .
A recent study (Koerne r 2000 ) has indicated that geosyntheti c reinforce d wall s ar e the least
expensive of any wal l type and for al l wall height categories (Fig . 19.13) .
1 9 .7 B ACKFI L L AND REI NFORCI N G M ATERI AL S
B ackfill
The backfill , i s limite d t o cohesionless , fre e drainin g materia l (suc h a s sand) , an d thu s th e ke y
properties ar e the densit y and the angl e of internal friction .
Facing
system
(varies)
Block system
with opening s
for vegetation
iK&**¥$r$k
l)\fr^$ffi:™:.fo:
• v'fo .'j
;
?ffij?.
i
yfcX'r
(a) Geosynthetic reinforced wal l (b) Geosynthetic reinforced "live wall "
Polymer ropes
or stra
P
s
Soil anchor
Rock ancho r
(c) Future types of geosynthetic anchorage
Figure 1 9.1 2 Geos yn t het i c us e for rein force d wall s an d bulk heads (Koern er , 2000)
852 Chapt er 1 9
900
800-
700-
|600H
c3
? SCO -
'S
§ 400-
U
300-
200-
1004
1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3
Height of wal l (m)
Figure 1 9 .1 3 M ea n val ue s o f variou s cat eg orie s o f ret ain in g wal l cos t s
(Koern er, 2000)
Reinforcing materia l
The reinforcement s ma y b e strip s or rods o f meta l o r sheet s o f geotextile , wir e grid s o r geogrid s
(grids mad e fro m plastic) .
Geotextile is a permeable geosyntheti c comprised solel y of textiles. Geotextiles ar e used with
foundation soil , rock , eart h o r an y othe r geotechnica l engineering-relate d materia l a s a n integra l
part of a human made project , structure, or system (Koerner, 1999) . AASHT O (M288-96 ) provide s
(Table 19.2 ) geotextil e strengt h requirement s (Koerner, 1999) . Th e tensil e strengt h o f geotextil e
varies wit h the geotextil e designatio n as pe r th e desig n requirements . For example, a wove n slit -
film polypropylen e (weighin g 240 g/m
2
) has a range of 30 to 50 kN/m. The friction angle betwee n
soil an d geotextile s varie s wit h th e typ e o f geotextil e an d th e soil . Tabl e 19. 3 gives value s o f
geotextile frictio n angles (Koerner , 1999) .
The tes t propertie s represen t a n idealize d conditio n an d therefor e resul t i n th e maximu m
possible numerica l value s whe n use d directl y i n design . Mos t laborator y tes t value s canno t
generally b e use d directl y an d mus t b e suitabl y modifie d fo r in-sit u conditions . Fo r problem s
dealing wit h geotextiles th e ultimat e strength T
U
shoul d be reduced b y applyin g certai n reductio n
factors t o obtai n the allowabl e strength T
a
a s follows (Koerner, 1999) .
T =T
I
RF
ID
x RF
CR
x RF
CD
x RF
BD
(19.9)
where
RF
CR
=
RF
BD =
RF
CD =
allowable tensil e strength
ultimate tensil e strengt h
reduction factor for installatio n damag e
reduction facto r for cree p
reduction facto r for biological degradation and
reduction factor for chemica l degradation
Typical value s for reduction factors ar e give n i n Table 19.4.
Table 19. 2 AASHT O M288-9 6 Geotextile strength propert y requirement s
Geotextile Cl assi fi cat i on * tt
Case 1
strengt h
sea m
strength $
strengt h
Puncture strength
strengt h
Test Unit s
methods
ASTM N
D4632
ASTM N
D4632
ASTM N
D4533
ASTM N
D4833
ASTM kP a
D3786
Elongation
< 5 0 %
1400
1200
500
500
3500
Elongation
> 50 %
900
810
350
350
1700
Case 2
Elongation
< 5 0 %
1100
990
400
400
2700
Elongation
> 5 0 %
700
630
250
2505
1300
Case 3
Elongation
< 5 0 %
800
720
300
300
2100
Elongation
>50 %
500
450
180
180
950
measured i n accordance wit h ASTM D4632 . Wove n geotextiles fai l a t elongations (strains)< 50%, whil e nonwovens fai l a t elongation (strains ) > 50%.
Whe n sewnseams are required. Overla p sea m requirement s are applicatio n specific .
require d MARV tea r strengt h for wove n monofilament geotextiles i s 250 N.
854 Chapt er 1 9
Table 19. 3 Pea k s oil -t o-g eot ex t il e frict io n an g le s an d efficien cie s i n s elect e d
cohes ion l es s s oil s *
Geotextile typ e
Woven, monofilamen t
Woven, slit-fil m
Nonwoven, heat-bonde d
Nonwoven, needle-punche d
Concrete san d
(0 = 30° )
26° (8 4 %)
24° (77%)
26° (84 %)
30° (100%)
Rounded san d
(0 = 28° )
-
24° (84 %)
-
26° (92 %)
Silty san d
(0 - 26° )
-
23° (87 %)
-
25° (96 %)
*Number s i n parentheses ar e th e effi ci enci es . V alue s suc h a s thes e shoul d not b e use d i n fina l design . Sit e
specific geotextile s an d soil s mus t b e individuall y tested an d evaluate d i n accordanc e wit h th e particula r
project conditions : saturation , typ e o f liquid , norma l stress , consolidatio n time , shea r rate , displacemen t
amount, an d s o on. (Koerner , 1999)
Table 19. 4 Recom m en de d reduct io n fact o r value s fo r us e in [Eq. (19. 9) ]
Ra nge o f Re duc t i o n Fact ors
Appl i c a t i on
Area
Separation
Cushioning
Unpaved road s
Walls
Embankments
Bearing capacit y
Slope stabilizatio n
Pavement overlay s
Railroads (filter/sep. )
Flexible form s
Silt fence s
I n s t a l l a t i o n
Damage
1.1 t o
1.1 t o
1.1 to
1.1 t o
1.1 t o
1.1 t o
1.1 t o
1.1 t o
1.5 t o
1.1 t o
1.1 to
2
2,
2,
2,
2,
.5
.0
.0
.0
.0
2.0
1.
1,
3,
1.
1.
.5
.5
.0
,5
.5
Cr eep*
1
1
1
2,
2,
.5 t o
.2 t o
.5 t o
. Ot o
. Ot o
2,
1
2
4.
3,
.5
.5
.5
.0
.5
2.0 to 4.0
2.
1
1
1.
1
. Ot o
. Ot o
. Ot o
.5 t o
.5 t o
3.
2,
1.
3.
2,
,0
,0
,5
,0
.5
Che mi c al
Degr adat i on
1.0 t o
1.0 t o
1.0 t o
1.0 to
1.0 t o
1.0 t o
1.0 t o
1.0 t o
1.5 t o
1.0 to
1.0 t o
1.5
2.0
1.5
1.5
1.5
1.5
1.5
1.5
2.0
1.5
1.5
Bi ol ogi cal
Degr adat i on
1.0 to
1.0 t o
1.0 t o
1.0 to
1.0 t o
1.0 to
1.0 to
1.0 to
1.0 t o
1.0 to
1.0 t o
1.2
1.2
1.2
1.3
1.3
1.3
1.3
1.1
1.2
1.1
1.1
*Th e lo w en d o f th e rang e refer s t o application s whic h hav e relativel y shor t servic e lifetime s an d / o r
situations wher e creep deformation s are no t critica l t o the overal l system performance. (Koerner , 1999)
Table 19. 5 Recom m en de d reduct io n fact or value s fo r us e in Eq. (19. 10) fo r
det erm in in g al l owabl e t en s il e s t ren g t h of g eog rid s
Appl i cat i on Are a
Unpaved road s
Paved road s
Embankments
Slopes
Walls
Bearing capacity
DC D C
Hh
/D
hh
C/?
1.1
1.2
1.1
1.1
1.1
1.2
to
to
to
to
to
to
1.6
1.5
1.4
1.4
1.4
1.5
1.5 t o
1.5 t o
2.0 t o
2.0 t o
2.0 t o
2.0 t o
2,
2.
3,
3,
3,
3,
.5
.5
.0
.0
.0
.0
R
f~CD * " B£>
1.0
1.1
1.1
1.1
1.1
1.1
to
to
to
to
t o
to
1.5
1.6
1.4
1.4
1.4
1.6
1.0 t o
1.0 t o
1.0 t o
1.0 t o
1.0 t o
1.0 t o
1.1
1.1
1.2
1.2
1.2
1.2
Con cret e an d M echan icall y St abilize d Eart h Ret ain in g Wall s 85 5
G eogrid
A geogrid is defined a s a geosynthetic materia l consistin g o f connected paralle l sets of tensile rib s
with aperture s o f sufficien t siz e t o allo w strike-throug h o f surroundin g soil , stone , o r othe r
geotechnical materia l (Koerner , 1999) .
Geogrids ar e matri x lik e material s wit h larg e ope n space s calle d apertures , whic h ar e
typically 1 0 t o 10 0 mm betwee n th e ribs , calle d longitudinal an d transverse respectively . Th e
primary functio n o f geogrid s i s clearl y reinforcement . The mas s o f geogrid s range s fro m 20 0 t o
1000 g/m
2
and the open area varie s from 40 to 95 %. It is not practicable to give specific values for
the tensil e strengt h of geogrid s becaus e o f it s wid e variatio n i n density . In suc h cases on e ha s t o
consult manufacturer' s literature for th e strengt h characteristics of thei r products . The allowabl e
tensile strength , T
a
, ma y b e determine d b y applyin g certai n reductio n factor s t o th e ultimat e
strength T
U
a s i n the cas e o f geotextiles. Th e equatio n is
rri _ rri
~
The definitio n o f the variou s terms i n Eq (19.10) i s the same a s i n Eq. (19.9) . However , th e
reduction factor s ar e different . These values ar e given i n Table 19. 5 (Koerner , 1999) .
Metal Strips
Metal reinforcement strips are available in widths ranging from 7 5 to 10 0 mm and thickness on the
order o f 3 t o 5 mm, wit h 1 mm on eac h fac e exclude d fo r corrosion (Bowles , 1996) . The yiel d
strength o f stee l ma y b e take n a s equa l t o abou t 3500 0 lb/in
2
(24 0 MPa ) o r a s pe r an y cod e o f
practice.
19.8 CONSTRU CTI O N DETAIL S
The method o f construction of MSE wall s depends upo n the type effaci ng uni t and reinforcing
material use d i n the system. The facing uni t which is also calle d th e skin can be either flexibl e
or stiff , bu t mus t b e stron g enoug h t o retai n th e backfil l an d allo w fastening s fo r th e
reinforcement t o be attached. Th e facing unit s require onl y a small foundatio n fro m which they
can b e built , generally consistin g of a trench filled wit h mass concret e givin g a footing similar
to thos e use d i n domesti c housing . The segmenta l retaining wall sections o f dry-laid masonr y
blocks, ar e shown i n Fig. 19.12(a) . The block syste m wit h openings fo r vegetation i s shown in
Fig. 19.12(b) .
The construction procedure wit h the use of geotextiles i s explained in Fig. 19 . 14(a). Here, the
geotextile serve both as a reinforcement and also as a facing unit . The procedure i s described belo w
(Koerner, 1985 ) wit h reference t o Fig. 19.14(a) .
1. Star t wit h an adequat e working surface and staging area (Fig . 19.14a) .
2. La y a geotextile shee t o f proper widt h on the ground surface with 4 t o 7 ft at the wal l face
draped ove r a temporary woode n for m (b).
3. Backfil l ove r this sheet wit h soil . Granular soils or soils containing a maximum 30 percent
silt and /or 5 percent cla y ar e customary (c).
4. Constructio n equipmen t mus t work from th e soi l backfil l an d be kept of f the unprotecte d
geotextile. Th e spreadin g equipmen t should be a wide-tracked bulldoze r tha t exert s littl e
pressure agains t th e groun d o n whic h i t rests . Rollin g equipmen t likewis e shoul d b e of
relatively light weight.
856
Chapt er 1 9
Temporary
wooden for m
(a) (b)
/?^\/2xs\/^\/
(c)
C
(e)
(f)
(g)
(h)
Figure 1 9 .1 4 (a ) Gen era l con s t ruct ion procedure s fo r us in g g eot ex t il e s i n fabri c
wal l con s t ruct io n (Koer n er , 1985)
5. Whe n th e firs t laye r ha s bee n folde d ove r th e proces s shoul d b e repeate d fo r th e secon d
layer wit h the temporar y facin g form bein g extende d fro m th e origina l groun d surfac e o r
the wall being steppe d bac k abou t 6 inches s o that the form can be supported from the firs t
layer. I n the latter case, th e support stakes mus t penetrate th e fabric .
6. Thi s proces s i s continued unti l the wal l reaches it s intended height .
7. Fo r protection agains t ultraviolet light and safet y against vandalism the faces of such wall s
must be protected. Bot h shotcret e and gunit e have been use d fo r thi s purpose .
Figure 19.14(b ) shows complet e geotextil e wall s (Koerner, 1999) .
Con cret e an d M echan ical l y St abil ize d Eart h Ret ain in g Wall s 85 7
Figure 1 9 .1 4 (b ) Geot ex t il e wal l s (Koern er , 1999 )
19.9 DESIG N CONSI DERATI ONS FOR A M ECH ANI CAL L Y
STAB I L I Z ED EARTH WALL
The desig n o f a MSE wal l involve s the following steps :
1. Chec k fo r internal stability, addressing reinforcement spacing an d length .
2. Chec k fo r external stabilit y of the wall against overturning, sliding, and foundation failure.
The general considerations for the design are :
1. Selectio n o f backfil l material : granular , freel y drainin g materia l i s normall y specified .
However, wit h the adven t of geogrids, th e use of cohesive soi l i s gaining ground.
2. Backfil l shoul d b e compacte d wit h car e i n orde r t o avoi d damag e t o th e reinforcin g
material.
3. Rankine' s theor y fo r the active stat e i s assumed t o be valid .
4. Th e wal l shoul d be sufficientl y flexibl e for the development of active conditions .
5. Tensio n stresse s ar e considered fo r the reinforcement outside the assumed failur e zone .
6. Wal l failur e wil l occu r in one of three ways
r
-z )
Surcharge
lie
/ '
(90-0)= 45° -0/2
Failure plane
45°
(a) Reinforced earth-wall profil e wit h surcharg e loa d
h- * -H
(b) Latera l pressure distribution diagrams
Figure 1 9.1 5 Prin ciple s of MSE wall desig n
Con cret e an d M echan icall y St abilize d Eart h Ret ain in g Wall s 859
Reinforcement
Figure 1 9.1 6 T ypica l ran g e i n s t ri p rein forcem en t s pacin g fo r rein force d eart h
wal l s (B owl es , 1996 )
a. tensio n in reinforcement s
b. bearin g capacit y failur e
c. slidin g of the whol e wall soil system.
7. Surcharge s ar e allowe d o n th e backfill . Th e surcharge s ma y b e permanen t (suc h a s a
roadway) or temporary.
a. Temporar y surcharge s withi n the reinforcement zone wil l increase th e latera l pressur e
on the facing unit which in turn increases the tension in the reinforcements, but does not
contribute t o reinforcement stability.
b. Permanen t surcharge s withi n the reinforcement zone wil l increase th e latera l pressur e
and tension i n the reinforcement and wil l contribute additional vertica l pressur e fo r the
reinforcement friction.
c. Temporar y o r permanent surcharge s outsid e th e reinforcemen t zon e contribut e latera l
pressure whic h tends to overturn the wall.
8. Th e tota l lengt h L of the reinforcement goes beyon d the failure plane AC by a length L
g
.
Only lengt h L
g
(effective length) i s considere d fo r computin g frictiona l resistance . Th e
length L
R
lyin g withi n th e failur e zon e wil l no t contribut e fo r frictiona l resistanc e
(Fig. 19.15a) .
9. Fo r the propose of design the total length L remains the same for the entire height of wall H.
Designers, however , may use thei r discretion t o curtail th e lengt h at lowe r levels . Typica l
ranges i n reinforcement spacing are given in Fig. 19.16 .
1 9.1 0 DESIG N M ETH O D
The following forces ar e considered :
1. Latera l pressur e o n the wal l due t o backfil l
2. Latera l pressur e du e t o surcharge i f present on the backfil l surface.
860 Chapt e r 1 9
3. Th e vertica l pressur e a t any dept h z on the stri p due t o
a) overburde n pressur e p
o
onl y
b) overburde n pressur e p
o
an d pressure due t o surcharge .
L ateral Pressur e
Pressure due to Overburden
Lateral eart h pressur e du e t o overburden
At dept h z
Pa
= P
OZ
K
A
= yzK
A
(19.11a )
At dept ht f
Pa
=p
oH
K
A
=yHK
A
( 19. lib)
Total active eart h pressur e
p =-vH
2
K. (19.12 )
a 2 ^
Pressure Du e t o Surcharge (a) o f Limite d Width, and (b ) U niforml y Distribute d
(a) From Eq. (11.69 )
/•^
^ =-^-(/?-sin/?cos2a) (19.13a )
n
(b) q
h
= q
s
K
A
(19.13b )
Total latera l pressur e due t o overburden and surcharg e at any dept h z
+q
h
) (19.14 )
V ertical pressur e
V ertical pressur e a t any dept h z due t o overburden only
P
0
=rz (19.15a )
due to surcharge (limited width )
(19.15b)
where th e 2: 1 (2 vertical : 1 horizontal) method i s used for determining Ag at any depth z.
Total vertica l pressur e du e t o overburden and surcharge at any dept h z.
(19.15C)
Reinforcement and Distributio n
Three types o f reinforcements ar e normall y used. They ar e
1 . Meta l strip s
2. Geotextile s
3. Geogrids .
Con cret e an d M echan icall y St abilize d Eart h Ret ain in g Walls 86 1
Galvanized stee l strip s of widths varying from 5 to 100 mm and thickness from 3 to 5 mm are
generally used . Allowance for corrosion i s normally made whil e deciding th e thickness a t the rat e
of 0.001 in. per year an d the lif e spa n is taken as equal t o 50 years. The vertical spacin g ma y rang e
from 2 0 to 15 0 cm ( 8 to 60 in.) and can vary with depth. The horizontal latera l spacin g ma y be on
the order o f 80 to 15 0 cm (30 to 60 in.). The ultimate tensile strengt h may be taken as equal to 240
MPa (35,000 lb/in.
2
). A factor of safety in the range of 1. 5 to 1.67 is normally used t o determine th e
allowable steel strength f
a
.
Figure 19.1 6 depicts a typica l arrangemen t o f meta l reinforcement . Th e propertie s o f
geotextiles an d geogrid s hav e bee n discusse d i n Sectio n 19.7 . However, wit h regards t o spacing ,
only th e vertica l spacin g i s t o b e considered . Manufacturer s provide geotextile s (o r geogrids ) i n
rolls o f various lengths an d widths. The tensil e force per unit width must be determined .
Length o f Reinforcement
From Fig. 19. 15 (a)
L = L
R
+ L
e
= L
R
+L
{
+L
2
(19.16 )
where L
R
= ( H -
z
) ta n (45° - 0/2)
L
e
= effectiv e length of reinforcement outside the failure zon e
Lj = lengt h subjected t o pressure ( p
0
+ Ag) = p
o
L
2
= lengt h subjecte d t o p
o
only .
Strip Tensil e Forc e at any Dept h z
The equation for computing T is
T = p
h
xhxs/stnp = ( KK
A
+q
h
)hxs (19.17a )
The maximum tie force wil l be
T( max)=( yHK
A
+q
hH
)hxs (19.17b )
where p
h
= yzK
A
+q
h
q
h
= latera l pressur e a t depth z due t o surcharg e
<*hH = ^ at depth//
h = vertica l spacin g
s = horizonta l spacin g
where P
a
= l/2yff-K
A
— Rankine' s lateral force
P = latera l forc e du e t o surcharge
(19.18)
Frictional Resistance
In th e cas e o f strip s o f widt h b bot h side s offe r frictiona l resistance . Th e frictiona l resistanc e F
R
offered b y a stri p a t an y dept h z mus t b e greater tha n th e pullou t forc e T by a suitabl e facto r o f
safety. We may writ e
F
R
=2b[ ( p
0
+bq)L
l
+p
0
L
2
] tanS<TF
s
(19.19 )
or F
R
= 2b[ p
0
L
l
+p
o
L
2
] tanS<TF
s
(19.20 )
where F ma y be taken a s equal t o 1.5.
862 Chapt e r 1 9
The friction angle 8 between the strip and the soil may be taken as equal to 0 for a rough strip
surface an d for a smooth surfac e 8 may li e between 1 0 to 25°.
Sectional Are a o f Metal Strips
Normally th e widt h b o f the stri p i s assume d i n the design . Th e thicknes s t has t o be determine d
based o n T (max) and the allowabl e stress f
a
i n the steel . If/ i s the yiel d stress of steel , the n
/v
Normally F^ (steel ) range s fro m 1. 5 to 1.67 . The thicknes s t may b e obtained from
T(max)
t =
(19.22)
The thicknes s o f t is to be increased t o take care of the corrosion effect . The rat e of corrosio n
is normall y take n a s equal t o 0.001 in/y r for a lif e spa n of 50 years .
Spacing o f G eotextil e L ayer s
The tensil e forc e T per unit width of geotextile laye r at any depth z may be obtained from
T = p
h
h = ( yzK
A
+q
h
)h (19.23 )
where q
h
= lateral pressure either du e to a stripload o r due to uniformly distributed surcharge .
The maximu m valu e of th e compute d T shoul d be limite d t o th e allowabl e valu e T
a
a s pe r
Eq. (19.9). As such we may writ e Eq. (19.23) a s
T
a
=TF
s
=( yzK
A
+q
h
)hF
s
(19.24 )
T T
or h =
where F^ = factor of safet y (1. 3 to 1.5 ) when using T
a
.
Equation (19.25 ) i s used for determining the vertica l spacing o f geotextil e layers .
Frictional Resistanc e
The frictional resistance offere d b y a geotextile layer for the pullout force T
a
may be expressed a s
T
a
F
s
(19.26 )
Equation (19.26) expresses frictional resistanc e per unit width and both sides of the sheets ar e
considered.
Design with G eogri d L ayers
A tremendous numbe r of geogrid reinforce d wall s have been constructe d i n the past 1 0 years (Koerner ,
1999). Th e type s of permanent geogrid reinforced wall facings are as follows (Koerner, 1999) :
1. Articulated precast panels ar e discrete precast concrete panel s with inserts for attaching the
geogrid.
2. Full height precast panels ar e concrete panels temporarily supporte d until backfill is complete .
3. Cast-in-place concrete panels ar e often wrap-aroun d wall s that ar e allowe d t o settl e and ,
after 1/ 2 to 2 years, ar e covered wit h a cast-in-place facing panel .
Con cret e an d M echan icall y St abil ize d Eart h Ret ain in g Wall s 863
4. Masonry block facing walls ar e an exploding segment o f the industr y with many differen t
types currently available, all of which have the geogrid embedde d betwee n th e blocks an d
held by pins, nubs, and/or friction.
5. Gabion facings ar e polymer or steel - wire basket s filled wit h stone, having a geogrid hel d
between th e baskets an d fixed wit h rings and/or friction.
The frictiona l resistance offered by a geogrid agains t pullout may be expressed a s (Koerner ,
1999)
(19.27)
where C. = interaction coefficient = 0.75 (ma y vary)
C
r
= coverage rati o = 0.8 (may vary)
All the other notations are already defined. The spacing of geogrid layer s may be obtained fro m
Ph
where p
h
= lateral pressure pe r unit length of wall.
1 9 .1 1 EXTERNA L STAB I L I TY
The MSB wall system consist s of three zones. Thye are
1. Th e reinforce d eart h zone .
2. Th e backfill zone.
(19.28)
Backfill H
(a) Overturning considerations
B
c - c o
Backfill
(b) Sliding consideration s
Wall
Backfill
Foundation soi l
(c) Foundation consideration s
Figure 1 9.1 7 Ex t ern a l s t abilit y con s iderat ion s fo r rein force d eart h wall s
864
Chapt er 1 9
3. Th e foundatio n soil zone .
The reinforced eart h zone is considered a s the wall for checking th e internal stability wherea s
all three zone s ar e considered fo r checking the external stability. The soil s o f the first two zones ar e
placed i n layer s an d compacte d wherea s the foundation soil i s a normal one . The propertie s o f the
soil i n each of the zone s may be the same o r different . However , th e soi l i n the firs t two zone s i s
normally a free drainin g material such as sand .
It i s necessar y t o chec k th e reinforce d eart h wal l (widt h = B) fo r externa l stabilit y whic h
includes overturning, sliding and bearing capacit y failure . These are illustrated i n Fig. 19.17 . Active
earth pressur e o f th e backfi l l actin g o n th e interna l face AB o f th e wal l i s take n i n th e stabilit y
analysis. The resultan t earth thrus t P
a
i s assumed to act horizontall y at a height H/ 3 abov e the bas e
of the wall . The method s o f analysi s are the same a s for concret e retainin g walls.
Example 1 9 . 2
A typica l sectio n o f a retainin g wal l wit h th e backfil l reinforce d wit h meta l strip s i s show n i n
Fig. Ex . 19.2 . Th e followin g data ar e available:
Height H = 9 m; b = 100 mm; t = 5 mm\ f
y
= 240 MPa; F
s
fo r stee l = 1.67 ; F
s
o n soi l frictio n
= 1.5 ; 0=36° ; 7 = 17. 5 kN/m
3
; 5 = 25°;/ ? x s = 1 x 1 m.
Required:
(a) Length s L and L
e
a t varying depths.
(b) Th e larges t tension Tin th e strip.
(
u
• \
2
3
4
9m 5
6
7
3m 8
9
V O
o r
110.5 m
/
' /
/
l l
' ^- - Failure plane /
/i = 1 m .^* ^ /
/'
7
f
Wall / /
/ / Meta l
///\\//^\
Backfill
0 = 36°
y = 17.5kN/m
3
/ / sinp s
Sand /
v
' y = l 7 . 5 k N / m
3
^^^-10 0 mm x 5 mm
^ £ 2 ,\ -
36
o ' ^ —Steppe d
' ^ / remtorce n
' Hkt r i hnt i n
• /
G
- ^
m
/ * Linea r var
/ ,2 1 ,_,. , o t lengt h c
^ ' - i ^ 1-1- 1 • - /
/
/ X
\63° i
\0. 5 m
lent
n (1 )
iation
)f strip s (2 )
Well footin g = 36°
y
= 17.5 kN/m
3
Figure Ex . 19. 2
Con cret e an d M echan icall y St abilized Eart h Retainin g Walls 86 5
(c) Th e allowabl e tension in the strip.
(d) Chec k fo r external stability.
Solution
From Eq. (19.17a), the tension i n a strip at depth z is
T=YzK
A
shforq
h
= Q
where y= 17. 5 kN/m
3
, K
A
= tan
2
(45° - 36/2 ) = 0.26, s = 1m; h= 1 m.
Substituting
T= 17. 5 x 0.26 (1 ) [1] i= 4.55zkN/strip.
F
S
T 1.5x4.55 z
L = -- -= -= 4.14m
e
2yzbtanS 2xl 7. 5xO. l xO. 47x z
This show s tha t th e lengt h L
g
= 4.14 m i s a constan t wit h depth . Fig . Ex . 19. 2 show s th e
positions o f L
g
for stri p numbers 1, 2 .. . 9 . Th e first stri p is located 0. 5 m below the backfill surface
and the 9th at 8.5 m below wit h spacings at 1 m apart. Tension in each of the strips may be obtained
by usin g the equation T = 4.55 z. The tota l tension ^L T a s computed i s
Zr = 184.2 9 kN/m sinc e s = 1 m.
As a check th e total active earth pressure is
p = ~yH
2
K. = -17.5 x9
2
x0. 26 = 184.28 k N/ m = £7
a
2
A
2
The maximum tension is in the 9th strip, that is, at a depth of 8.5 m below the backfill surface.
Hence
T= YZ K
A
sh = 17.5 x 8. 5 x 0.26 x 1 x 1 = 38.68 kN/stri p
The allowabl e tensio n i s
740 v 10
3
where f
a
= = 143.7 x 10
3
k N / m
2
1.67
Substituting T
a
= 143.7 x 10
3
x 0.005 x 0.1 » 72 kN > T- OK .
The tota l lengt h of stri p L at any depth z i s
L = L
R
+ L
e
= ( H-z) ta n (45 - 0/2 ) + 4.14 = 0.51 (9 - z ) + 4.14 m
where H = 9 m.
The lengths as calculated have been shown in Fig. Ex. 19.2 . It is sometimes convenient to use
the same lengt h L with depth or stepped i n two or more blocks or use a linear variation as shown in
the figure.
Check fo r Externa l Stability
Check o f bearing capacit y
It is necessary t o check the base of the wall with the backfill for the bearing capacity per unit length
of the wall . The widt h of the wal l may be taken as equal to 4.5 m (Fig. Ex. 19.2) . The procedur e a s
explained i n Chapte r 1 2 ma y b e followed . Fo r al l practica l purposes , th e shape , depth , an d
inclination factors ma y b e taken a s equal t o 1 .
866 Chapt e r 1 9
Check for sliding resistance
Sliding resistanc e F ^
F =
Driving forc e P
a
where F
K
= W tan 8 =
4
'
5 +8
'
5
x 17.5 x 9 tan 36°
K
2
= 1024 x 0.73 = 744 kN
where 8 = 0 = 36° for the foundation soil , an d W - weigh t of the reinforced wal l
P
a
= 184.2 8 kN
744
F =— = 4>1.5 - - O K
s
184.2 8
Check for overturning
F
M
« F
<™
O
From Fig. Ex. 19. 2 takin g moment s of al l forces abou t O, we hav e
M^=4. 5 x 9x l 7. 5 x — + - x9x (8.5-4.5)(4.5 + -)x 17.5
= 159 5 + 183 7 = 3432 kN-m
M = P x —= 184.28x- = 553 k N-m
0 a
3 3
3432
F = —= 62>2- OK
5
55 3
Example 1 9 .3
A sectio n o f a retainin g wal l wit h a reinforce d backfil l i s show n i n Fig. Ex. 19.3 . The backfil l
surface i s subjected t o a surcharge o f 30 kN/m
2
. Required :
(a) Th e reinforcement distribution .
(b) Th e maximu m tensio n i n the strip .
(c) Chec k fo r external stability.
Given: b = 100 mm, t = 5 mm,/
fl
= 143. 7 MPa, c = 0, 0 = 36°, 8 = 25°, y = 17. 5 kN/m
3
,
s = 0.5 m, and h = 0.5 m.
Solution
FromEq. (19.17a )
where y = 17. 5 kN/m
3
, K
A
= 0.26, A
c
= h x s = (0.5 x 0.5) m
2
FromEq. (19.13a)
2
<7
f rn
q
h
=
n
Con cret e an d M echan icall y St abilize d Eart h Ret ain in g Wall s 867
,
H = A
\
;
1
2
1.75m
3
-*— 4
k 5 m 5
6
7
8
9
" 1 111 * • • " i i n — i
A B
1 i • i ' \ ' i ' i
0.25m / / , ' ' *, ' , , ° /^<S \//S$\
* / /
x
x ' ; \•
7 i
0.5m / / ,' / \
* /
/
' ' 1 \• > T n^^t r. n
i l J L
. i i
/ / ' / \ •
I
2 rJacKl m saii u
-^sK' /
2
^
/ ^" ' » \ r
//>' 0=19.07 ° / \
3
x
//
x
a = 29.74° ^- Failure plan e \
*- L
J ?
=1. 4m -H^ -
L
iI^
475m
-H
Sand /
r
- „- -
/
1
\ * y -17 5 lb/ft
3
27° /1.5 m 0 = 36°
""~^
/
„ 6 = 25° (for the strips')
7
\45° + 0/2 = 63° J
/' \ , • 0.25m*
'/////A t
Figure Ex . 19. 3
Refer t o Fig. E\ . 19. 3 for the definition o f a an d )S.
^ = 30 kN/m
2
The procedur e fo r calculating length L of the stri p for one dept h z = 1.75 m (stri p number 4)
is explained below. The same method i s valid for the other strips .
Strip No. 4. Depth z= 1.75 m
Pa
= YzK
A
= 17. 5 xl.75x 0.2 6 = 7.96 kN/m
2
From Fig . Ex . 19.3 , 0 = 19.07° = 0.3327 radian s
a= 29.74 °
fl = = 3 0 kN/m
2
2x30
3.14
[0. 0.3327- sin 19.07° cos59.5° = 3.19 kN/ m
Figure Ex . 19. 3 shows the surcharge distribution at a 2 (vertical) t o 1 (horizontal) slope . Pe r
the figur e at depth z = 1.75 m, L
l
= 1.475 m from the failure line and L
R
= ( H- z) tan (45° - 0/2 )
= 2.75 ta n (45° - 36°/2 ) = 1.4m from the wall to the failure line. It is now necessary t o determine L
2
(Refer t o Fig . 19.15a) .
868 Chapt e r 1 9
Now T= (7.96 + 3.19) x 0. 5 x 0. 5 = 2.79 kN/strip .
The equation fo r the frictional resistanc e pe r stri p i s
F
R
= 2b ( yz + Aq) L
{
ta n 8 + ( yz L
2
ta n 8) 2b
From th e 2: 1 distribution Ag at z = 1.75 m i s
A? = -£- =-^-=10.9 kN/m
2
B + z1 + 1.75
/ ?
0
= 17.5xl.75 = 30.63 kN/m
2
Hence p
o
= 10.9 + 30.63 = 41.53 kN/m
2
Now equatin g frictiona l resistanc e F
R
t o tension in the stri p wit h F
s
= 1.5, we have
F
R
-1.5 T . Given b - 10 0 mm. Now from Eq. (19.20)
F
R
= 2btanS( p
o
L
l
+ p
o
L
2
) = 1. 5 T
Substituting and taking 8 = 25°, we have
2 x 0.1 x 0.47 [41.53 x 1.475 + 30.63 L
2
] = 1.5x2.7 9
Simplifying
L
2
= -0.546 m -0
Hence L
e
= L
{
+ 0 = 1.47 5 m
L =L
R
+ L
e
=l.4+ 1.47 5 = 2.875 m
L can be calculated i n the same wa y at othe r depths .
Maximum tension T
The maximu m tension i s in stri p numbe r 9 at depth z = 4.25m
Allowable T
a
=f
a
bt = 143.7 x 10
3
x 0. 1 x 0.005 = 71.85 kN
T = ( yzK
A
+q
h
)sh
where yzK
A
= 17
-
50 x 4
-
25 x
°-
26
=
19
-
34 kN/m2
q
h
= 0.89 kN/m
2
from equation fo r q
h
at depth z = 4.25m.
Hence T = (19.34 + 0.89) x 1/ 2 x 1/ 2 = 5.05 kN/stri p < 71.85 k N - OK
Example 1 9. 4 (Koerner , 1 9 9 9 )
Figure Ex . 19. 4 shows a sectio n o f a retainin g wal l wit h geotextil e reinforcement . Th e wal l i s
backfilled wit h a granular soi l havin g 7=18 kN/m
3
and 0 = 34° .
A woven slit-fil m geotextil e wit h warp (machine ) direction ultimat e wide-widt h strengt h of
50 kN/m an d having 8= 24 ° (Tabl e 19.3 ) i s intended t o be used i n its construction .
The orientatio n o f th e geotextil e i s perpendicula r t o th e wal l fac e an d th e edge s ar e t o b e
overlapped t o handl e th e wef t direction . A facto r o f safet y o f 1. 4 i s t o b e use d alon g wit h site -
specific reductio n factor s (Tabl e 19.4) .
Required:
(a) Spacin g o f the individua l layers of geotextile .
(b) Determinatio n o f th e lengt h o f th e fabri c layers .
Con cret e an d Mechanicall y Stabilized Eart h Retainin g Wall s 869
4 m
Layer No.
1.8m
..(
3 - -
4. .
2. 1m
5
"
r
= 6 m
9 - -
1 0 - -
11 - •
12 • •
13 - -
14- -
t * |* t ~ F
Reinforced
earth wal l
y = 1 8 kN/m
3
~
0 = 36°, d = 24°
T
i nv
c
w,
//A\\//AV S ^
Foundation soi l
y = 18. 0 kN/m
3
H
2 m
0 = 34°, d= 25.5°
(a) Geotextile layers
Figure Ex . 19.4
p
a
= 30.24 kN/m
2
(b) Pressure distribution
(c) Chec k the overlap .
(d) Chec k for external stability.
The backfil l surface carries a uniform surcharg e dead load of 1 0 kN/m
2
Solution C7U I U U U I I
(a) The latera l pressur e p
h
a t any depth z is expresse d as
where p
a
= yzK
A
,q
h
= q K
A
, K
A
= tan
2
(45° - 36/2 ) = 0.26
Substituting
p
h
= 18 x 0.26 z + 0.26 x 1 0 = 4.68 z + 2.60
From Eq. (19.9), the allowabl e geotextile strength is
T =T
a U RF
ID
X RF
CR
X RF
CD
X RF
BD
= 50
1
1.2x2.5x1.15x1.1
= 13. 2 kN/ m
870 Chapt e r 1 9
From Eq . (19.17a), the expression for allowable stress i n the geotextile a t any depth z may be
expressed a s
T
h = —-
where h = vertical spacing (lif t thickness)
T
a
= allowable stres s i n the geotextil e
p
h
= lateral eart h pressur e a t depth z
F
s
= factor o f safet y = 1. 4
Now substituting
13.2 13. 2
h =
[4.68(z) + 2.60]1.4 6.55(z ) + 3.64
13 2
At z = 6m , h= ' = 0.307 m o r say 0.30 m
6. 55x6 + 3.64
13 2
At / = 33 m h = -' - -= 0.52 m or say 0.50 m
~ —' •*• ' "M , _ _ . _ _ _ , . . j
6.55x3.3 + 3.64
13.2
At z = 1 3m h = -
:
-= 1.08 m, but use 0.65 m for asuitabl e distribution.
6.55x1.3 + 3.64
The dept h 3. 3 m o r 1. 3 m ar e use d j us t a s a tria l an d erro r proces s t o determin e suitabl e
spacings. Figure Ex. 19. 4 show s th e calculate d spacings o f the geotextiles .
(b) Length of the Fabric Layers
From Eq . (19.26 ) we ma y writ e
L =
s s
=
. 2.60)1. 4
=
e
2 x l 8 z t a n 2 4 ° ~
e
~
From Fig . (19.15 ) th e expression fo r L
R
i s
L
R
=( H- z ) tan(45° - ^/2 ) = ( H-z) tan(45 ° - 36/2 ) = (6.0 - z) (0.509)
The tota l lengt h L i s
Con cret e an d M echan icall y St abilize d Eart h Ret ain in g Wall s 871
The compute d L and suggeste d L ar e give n in a tabular form below.
Layer N o
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Dept h z
(m)
0.65
1.30
1.80
2.30
2.80
3.30
3.60
3.90
4.20
4.50
4.80
5.10
5.40
5.70
6.00
Spaci ng h
( m)
0.65
0.65
0.50
0.50
0.50
0.50
0.30
0.30
0.30
0.30
0.30
0.30
0.30
0.30
0.30
L
e
( m)
0.49
0.38
0.27
0.26
0.25
0.24
0.14
0.14
0.14
0.14
0.14
0.14
0.14
0.14
0.13
L
e
( mi n )
( m)
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
L
R
( m)
2.72
2.39
2.14
1.88
1.63
1.37
1.22
1.07
0.92
0.76
0.61
0.46
0.31
0.15
0.00
L ( c al )
( m)
3.72
3.39
3.14
2.88
2.63
2.37
2.22
2.07
1.92
1.76
1.61
1.46
1.31
1.15
1.00
L (suggest ed)
(m)
4.0
-
-
3.0
-
-
-
-
2.0
-
-
-
-
-
-
It may b e noted her e that the calculated value s of L
g
ar e ver y smal l and a minimum value of
1.0 m shoul d be used .
(c) Chec k fo r the overlap
When th e fabri c layer s ar e lai d perpendicula r t o th e wall , th e adjacen t fabri c shoul d overla p a
length L
g
. The minimum value o f L
o
is 1 .Om. The equation for L
o
ma y be expressed as
L =
/i[4.68(z) + 2.60]1.4
4xl 8(z)t an24°
The maximu m value of L
o
i s at the uppe r layer at z = 0.65. Substituting for z we have
0.65 [4.68(0.65) + 2.60] 1.4
n

c
L
a
= = 0.25 m
4 x 18(0.65) tan 24°
Since thi s value of L
o
calculated is quite low, use L
o
= 1.0m for all the layers .
(d) Check for external stabilit y
The total activ e eart h pressur e P
a
is
P =-yH
2
K
A
=- x 18x6
2
x0.28 = 90.7 kN/ m
a
2
A
2
Resisting momen t M
R
W
{
/ j + W
2
1
2
+ W
3
/
3
+ P1
4
A< ™ ...H ..—. . i n . . i ™ i.. . H i ..-. -
5
Drivin g moment M
o
p
a
(
H
/3)
where W
l
= 6 x 2 x 1 8 = 216 kN and /, = 2/2 = 1m
W
2
= ( 6-2.1) x (3 - 2 ) (18) = 70.2 kN, and 1
2
= 2.5m
W
3
= (6 - 4.2 ) (4 -3) (18 ) = 32.4 kN and /
3
= 3.5m
872 Chapt e r 1 9
_^
o
—Z. I o> 2. — L/IS.
90.7 x (2)
Check for slidin g
Total resisting force F
R
Total drivin g force F
d
F
R
= w
i
+ W
2
+ W
= (216 + 70.2 + 32.4)tan 25.5°
= 318.6x0.477= 15 2 kN
F, = P = 90.7 kN
a n
152
Hence F = -= 1.68 >1.5 -O K
5
90. 7
Check fo r a foundation failur e
Consider the wal l a s a surface foundatio n wit h D
f
=0. Sinc e the foundation soil is cohesionless, w e
may write
Use Terzaghi' s theory. For 0 = 34°, N . = 38, and B = 2m
^ =- x ! 8x 2 x 38 = 684kN/ m
2
The actua l loa d int ensit y o n the base of the backfil l
^(actual) = 1 8 x 6+ 1 0 = 1 18 kN/m
2
684
F
s
= — —= 5. 8 > 3 whi c h i s acceptable
118
Example 1 9 . 5 (Koerner , 1 9 9 9 )
Design a 7m high geogrid-reinforced wal l when the reinforcement vertical maximum spacing mus t
be 1. 0 m. The coverag e rati o i s 0.80 (Refe r t o Fig . Ex. 19.5) . Given: T
u
= 15 6 kN/m, C
r
= 0.80,
C = 0.75. The othe r detail s are given in the figure .
Solution
I nternal Stabilit y
From Eq . (19.14)
K
A
= tan
2
(45° - 0/2) = tan
2
(45° - 32/2 ) = 0.31
p
h
= (18 x z x 0.3 1) + (15 x 0.31) = 5.58z + 4.65
Con cret e an d M echan icall y St abilize d Eart h Ret ain in g Wall s 873
q
s
= 15kN/m
2
= lm
I
I
• = 1 8 kN/m
3
> = 3 2
W
//^^^/( ^///( ^//.if^/y^^//^^
Foundation soi l
5 m
Bearing capacit y
q
u
= 600 kN/m
2
Foundation pressur e
Figure Ex . 19. 5
1. For geogri d vertica l spacing .
Given T
u
= 156 kN/ m
From Eq. (19.10) and Table 19.5 , w e have
T = T
*~n • *• 1 1
1
RF
ID
xR F
CR
x RF
BD
x RF
CD
T =15 6
1
1.2x2.5x1.3x1.0
= 40 kN/m
But us e r
design
= 28.6 kN/ m wit h F
f
= 1. 4 on T
a
From Eq. (19.28 )
T =
design
8 7 4 Chapte r 1 9
5.58z + 4.65
28.6 = h-
or h =
0.8
22.9
5.58z + 4.65
Maximum dept h fo r h = 1m is
229
1.0 = : o r z = 3.27 m
5.58^ + 4.65
Maximum dept h fo r h = 0.5m
229
0.5 = o r z = 7.37 m
5.58z + 4.65
The distributio n o f geogri d layer s i s shown i n Fig. Ex. 19.5.
2. Embedment lengt h o f geogri d layers .
From Eq s (19.27 ) an d (19.24 )
Substituting known value s
2 x 0.75 x 0. 8 x ( L
e
) x 1 8 x ( z) ta n 32 ° = h (5.58^ + 4.65) 1.5
Q- i f r (0-6 2 z +0.516)/z
Simplfymg L
e
=
The equatio n fo r L
R
i s
L
R
=( H- z) tan(45° -<f> 12) = (7 -z) tan(45° - 32/ 2)
= 3.88-0.554(z)
From the above relationships th e spacing o f geogrid layer s and their lengths ar e given below.
Layer
No.
1
2
3
4
5
6
7
8
9
10
11
Dept h
( m)
0.75
1.75
2.75
3.25
3.75
4.25
4.75
5.25
5.75
6.25
6.75
Spaci ng
h (m )
0.75
1.00
1.00
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
L
e
( m)
0.98
0.92
0.81
0.39
0.38
0.37
0.36
0.36
0.36
0.35
0.35
L
e
( mi n )
( m)
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
L
R
( m)
3.46
2.91
2.36
2.08
1.80
1.52
1.25
0.97
0.69
0.42
0.14
L ( c a l )
( m)
4.46
3.91
3.36
3.08
2.80
2.52
2.25
1.97
1.69
1.42
1.14
L ( r e qui r e d )
( m)
5.0
5.0
5.0
5.0
5.0
5.0
5.0
5.0
5.0
5.0
5.0
Con cret e an d M echan icall y St abilize d Eart h Ret ain in g Wall s 87 5
External Stabilit y
(a) Pressure distributio n
P
a
=-yH
2
K
A
= - x! 7x7
2
t an
2
(45° - 30/ 2) = 138.8 kN/m
P
q
=q
s
K
A
H = 15x0.33x7 = 34.7 kN/ m
Total - 173.5 kN/ m
1. Check for sliding (neglectin g effec t of surcharge )
F
R
= WtenS = yxHxL tsn25°= 1 8 x 7 x 5.0 x 0.47 = 293.8 kN/ m
p = p
a
+ p^ = 173.5 kN/ m
F =
293. 8
= L69 >L5 Q K
s
173. 5
2. Check for overturnin g
Resisting moment M
R
= Wx— = 18x 7x 5x — =15 75 k N- m
H H
Overturning moment M
o
- P
a
x — + P
q
x —
or M
0
= 138.8x- + 34.7x- = 445.3 k N- m
0
3 2
F
=
_
=
3.54 > 2.0 O K
5
445. 3
3. Check for bearing capacit y
T, . .
M
o
44
5.3
Eccentncity e = --— = -= 0.63
W + q
s
L 18x7x 5 + 15x5
e = 0.63 <- = -=0.83 Ok
6 6
Effective lengt h = L - 2e = 5- 2 x 0.63 = 3.74m
Bearing pressure = f l 8x 7 + 151 -=189 k N/ m
L J
V 3. 74;
F = — = 3.17 > 3.0 OK
5
18 9
1 9 .1 2 EXAM PL E S OF M EASU RED L ATERAL EARTH PRESSU RES
B ackfill Reinforce d wit h M etal Strip s
Laboratory test s wer e conducte d o n retainin g wall s wit h backfill s reinforced wit h meta l strip s
(Lee et al. , 1973) . Th e wall s were built within a 30 in. x 48 in. x 2 in. wooden box. Skin elements
876 Chapt er 1 9
Later eart h pressur e - psi
0.05 0.1 0.15 0.2 0.25 0.05 0.1 0.15
012
o.
Q
16
17
(a)
Loose san d
ties break
5 = 8 in
L= 1 6 i n
(b)
Dense san d
pull out
5= 8 in
L = 1 6 in
. Tension i s converte d
to equivalent earth pressure
^/&$V<SZ<S^^
_ Rankin e active,
corrected for sil o
effect o f bo x
0 Measure d wit h eart h
pressure gauge s
Figure 19.1 8 T ypica l ex am pl e s o f m eas ure d l at era l eart h pres s ure s j us t prio r t o
wal l fail ur e ( 1 in . = 25. 4 m m ; 1 ps i = 6. 8 9 k N/m
2
) (Le e et al. , 1973 )
were made from 0.012 in aluminum sheet. The strips (ties) used for the tests wer e 0.155 in wide and
0.0005 in thick aluminu m foil . The backfil l consiste d of dry Ottawa No . 90 sand. The smal l wall s
built of these material s i n the laboratory were constructed in much the same wa y as the larger walls
in th e field . Tw o differen t san d densitie s wer e used : loose , correspondin g t o a relativ e density ,
D
r
= 20%, an d mediu m dense , correspondin g t o D
r
= 63%, an d th e correspondin g angle s o f
internal friction were 31° and 44° respectively. SR-4 strain gages wer e used on the ties to determine
tensile stresse s i n the ties during the tests.
Examples o f the typ e of eart h pressur e dat a obtaine d fro m two typica l test s ar e shown i n
Fig. 19.18 . Dat a i n Fig . 19.18(a ) refe r t o a typica l test i n loos e san d wherea s dat a i n 19.18(b )
refer t o tes t i n dens e sand . Th e tie s length s wer e differen t fo r th e tw o tests . Fo r comparison ,
Rankine latera l eart h pressur e variatio n wit h dept h i s als o shown . I t ma y b e see n fro m th e
curve tha t th e measure d value s o f th e eart h pressure s follo w closel y th e theoretica l eart h
pressure variatio n u p t o two third s of the wal l hei ght but fal l of f comparatively t o lowe r value s
in th e lowe r portion .
Field Stud y o f Retainin g Wall s wit h G eogri d Reinforcemen t
Field studie s o f th e behavio r o f geotextil e o r geogri d reinforce d permanen t wal l studie s ar e
fewer i n number . Ber g e t al. , (1986 ) reporte d th e fiel d behavio r o f tw o wall s wit h geogri d
reinforcement. On e wal l i n Tucson, Arizona, 4.6 m high, used a cumulative reduction facto r of
2.6 on ultimat e strengt h for allowabl e strengt h T
a
and a value of 1. 5 as a global factor o f safety.
The secon d wal l wa s i n Li t honi a , Georgia , an d wa s 6 m high . I t use d th e sam e factor s an d
design method . Fig . 19.1 9 present s the resul t s for bot h the wall s shortl y after construction wa s
complete. I t ma y b e note d t ha t th e horizonta l pressure s a t variou s wal l height s ar e
overpredicted fo r eac h wal l , tha t is , th e wal l design s tha t wer e use d appea r t o b e quit e
conservative.
Con cret e an d M echan ical l y St abilize d Eart h Ret ain in g Wal l s 877
<u
I
2
cu
is
M-H
2 4
Lateral pressur e o
h
(kPa)
10 2 0 3 0 4 0
Tolerances
soil weight = 6%
Field measuremen t = 20%
Field
measurements
Rankine
lateral
pressure
°4
Lateral pressur e a
h
(kPa)
10 2 0 3 0 4 0
Tolerances
soil weight = 6%
Field measurement = 20%
Field
measurements
Rankine
lateral
pressure
(a) Results of Tucson, Arizona, wall (b) Results of Lithonia, Georgia, wall
Figure 1 9.1 9 Com paris o n o f m eas ure d s t res s e s t o des ig n s t res s e s fo r t w o g eog ri d
rein forced wal l s (B er g et al. , 1986 )
1 9 .1 3 PROB L EM S
19.1 Fig . Prob . 19. 1 give s a sectio n o f a cantileve r wall . Chec k th e stabilit y o f th e wal l wit h
respect t o (a) overturning, (b) sliding, and (c ) bearing capacity .
= 10°
// = 5.75m
0.5m
0. 75m
Foundation soi l
0 = 20° c = 30 kN/m
2
y = 18. 5 kN/m
3
Figure Prob. 19. 1
18f t
Foundation soi l
y=1201b/ ft
3
0 = 36°
Figure Prob. 19. 3
878 Chapt er 1 9
19.2 Chec k the stabilit y of the wall given in Prob. 19. 1 for the conditio n that the slope is horizonta l
and the foundation soi l is cohesionless with 0 = 30°. Al l the other data remain the same.
19.3 Chec k th e stabilit y o f th e cantileve r wal l give n i n Fig . Prob . 19. 3 fo r (a ) overturning ,
(b) sliding , an d (c) bearing capacit y failure .
19.4 Chec k the stabilit y of the wal l i n Prob. 19. 3 assuming (a ) /3 = 0, and (b ) the foundation soi l
has c = 300 lb/ft
2
, y = 11 5 lb/ft
3
, an d 0 = 26° .
19.5 Fig . Prob . 19. 5 depict s a gravi t y retaining wall. Check th e stabilit y o f th e wal l fo r sliding ,
and overturning .
19.6 Chec k th e stabilit y of the wal l given in Fig. Prob . 19.6 . Al l the dat a ar e given on the figure .
19.7 Chec k the stabilit y o f th e gravit y wal l give n i n Prob. 19. 6 wit h th e foundatio n soi l havin g
properties 0 = 30°, 7 = 11 0 lb/ft
3
an d c = 500 lb/ft
2
. Al l the other dat a remai n th e same .
19.8 Chec k the stabilit y o f the gravit y retaining wal l give n i n Fig. Prob . 19.8 .
19.9 Chec k the stabilit y of the gravity wall given in Prob. 19. 8 for Coulomb's condition. Assum e
5=2/30.
19.10 A typica l sectio n o f a wal l wit h granula r backfil l reinforce d wit h meta l strip s i s give n i n
Fig. Prob . 19.10 . The followin g dat a ar e available .
H=6m,b = 15 mm, t - 5 mm,/
v
= 240 MPa, F
s
fo r steel = 1.75 , F
s
o n soi l frictio n = 1.5 .
The othe r dat a ar e given i n the figure . Spacing : h = 0.6 m, an d s = 1 m.
Required
(a) Lengths o f ti e a t varying depth s
(b) Check for externa l stabilit y
19.11 Solv e th e Prob . 19.1 0 wit h a unifor m surcharg e actin g o n th e backfil l surface . Th e
intensity o f surcharg e i s 2 0 kN/m
2
.
19.12 Figur e Prob. 19.1 2 shows a section o f a MSB wall wit h geotextil e reinforcement .
5 f t
1 m
J i m
35ft
Foundation soil : y - 18. 5 kN/m
3
,
<j> = 20°, c = 60 kN/m
2
Figure Prob. 19. 5
-H 5f t
Not t o scale
y= 118 lb/ft
3
= 35°
y
c
= 150 lb/ft
3
(concrete)
5f t
I
5f t
25f t
Foundation soil: y = 120 lb/ft
3
0 = 36°
Figure Prob. 19. 6
Con cret e an d M echan icall y St abilized Eart h Ret ain in g Walls 879
y=1151b/ft
2
= 35°
y
c
(concrete)
= 1501b/ft
3
:i»
19.13
16ft H
Foundation soil : y = 12 0 lb/ft
3
,0 = 36°
Figure Prob . 19. 8
Required:
(a) Spacing o f the individua l layers of geotextil e
(b) Length o f geotextile i n each laye r
(c) Check fo r external stability
Design a 6 m hig h geogrid-reinforce d wal l (Fig. Prob . 19.13) , wher e th e reinforcemen t
maximum spacin g mus t b e a t 1. 0 m. Th e coverag e rati o C
r
= 0. 8 an d th e interactio n
coefficient C . = 0.75, an d T
a
= 26 kN/m. (r
design
)
Given : Reinforced soi l propertie s : y = 1 8 kN/m
3
0 = 32°
Foundation soi l : y = 17. 5 kN/m
3
0= 34°
Metal strips b = 75 mm, t = 5 mm
6m
> = 0.6m
5= 1 m
Backfill
y = 17. 0 kN/m
3
= 34°
Wall
0 = 36°, c = 0, 6 = 24°, y = 18.0 kN/m
3
Foundation soi l
0 = 36°, y = 17.0 kN/m
3
Figure Prob . 19.1 0
880 Chapt er 1 9
• Geotextile reinforcemen t
c
/
Wall
= 36°, 6 = 24°
Backfill
Granular soi l
y = 17. 5 kN/m
3
0 = 35°
T = 12 kN/m
y=17. 5 kN/m
3
Foundation soil : y = 18. 5 kN/ m, 0 = 36°.
Figure Prob . 1 9 .1 2
/ —Geogri d reinforcemen t
6 m
z
y = 1 8 kN/m
3
Backfill
0 = 32°,
y = 1 8 kN/m
3
= 32°
T
a
= 26 kN/ m
y = 17. 5 kN/m
3
0 = 34°
Figure Prob . 1 9 .1 3
CHAPTER 20
SH EET PIL E WAL L S AND BRACED CUTS
2 0 .1 I NTRODU CTI O N
Sheet pil e wall s ar e retainin g wall s constructe d t o retai n earth , wate r o r an y othe r fil l material .
These wall s are thinner in section as compared t o masonry walls described i n Chapter 19 . Sheet pile
walls ar e generall y use d fo r the following:
1. Wate r fron t structures , for example, i n building wharfs, quays, and pier s
2. Buildin g diversion dams , suc h as cofferdam s
3. Rive r bank protectio n
4. Retainin g th e sides o f cuts made i n eart h
Sheet pile s ma y b e o f timber , reinforce d concret e o r steel . Timbe r pilin g i s use d fo r shor t
spans and to resist ligh t lateral loads. They ar e mostly used for temporary structure s such as braced
sheeting i n cuts . I f use d i n permanent structure s above th e wate r level , the y requir e preservativ e
treatment and eve n then, their span of lif e i s relatively short. Timber shee t pile s ar e joined t o eac h
other b y tongue-and-groov e joint s a s indicate d i n Fig . 20.1. Timbe r pile s ar e no t suitabl e fo r
driving i n soils consisting o f stones a s the stones woul d dislodge the joints .
Groove \ / Tongu e
Figure 20. 1 Tim be r pil e wal l s ect io n
881
882 Chapt er 2 0
o~ ~ o" ~ a~ T 3
D_ _ o _ _Q _ _d b. _o __Q _ .a _O- _ d
Figure 20. 2 Rein force d con cr et e Shee t pil e wal l s ect io n
(a) Straigh t shee t pil ing
(b) Shallo w arch-we b piling
(c) Arch-web pi l i ng
(d) Z-pile
Figure 20. 3 Shee t pil e s ect ion s
Reinforced concret e shee t pile s ar e precas t concret e members , usuall y wit h a tongue-and -
groove joint . Typica l sectio n o f pile s ar e shown i n Fig . 20.2 . Thes e pile s ar e relativel y heav y an d
bulky. They displac e larg e volume s of soli d during driving. This larg e volum e displacemen t o f soi l
tends t o increase th e driving resistance. The design of piles has t o take int o account the large driving
stresses an d suitabl e reinforcement has t o be provided for thi s purpose.
The mos t commo n type s o f pile s use d ar e stee l shee t piles . Stee l pile s posses s severa l
advantages ove r th e othe r types . Some o f the important advantages are :
1. The y ar e resistant to high dri vi ng stresses as developed i n har d or rocky materia l
2. The y ar e lighte r in section
3. The y ma y be used severa l times
Sheet Pil e Wall s an d B race d Cut s 88 3
4. The y ca n be used eithe r below or above wate r and possess longer lif e
5. Suitabl e joints whic h do no t defor m durin g driving can be provided t o have a continuous
wall
6. Th e pil e lengt h can be increased eithe r by welding or bolting
Steel sheet piles are available in the market in several shapes. Some of the typical pile sections are
shown in Fig. 20.3. The archweb and Z-piles are used to resist large bending moments, as in anchored or
cantilever walls . Where th e bendin g moment s ar e less , shallow-arc h pile s wit h corresponding smalle r
section moduli can be used. Straight-web sheet piles are used where the web will be subjected to tension, as
in cellular cofferdams. The ball-and-socket type of joints, Fig. 20. 3 (d), offer less driving resistance than the
thumb-and-finger joints, Fig. 20.3 (c).
20.2 SH EE T PIL E STRU CTU RES
Steel shee t piles may conveniently be used in several civil engineering works. They ma y be used as:
1. Cantileve r shee t pile s
2. Anchore d bulkhead s
3. Brace d sheetin g i n cuts
4. Singl e cel l cofferdams
5. Cellula r cofferdams , circula r type
6. Cellula r cofferdams (diaphragm)
Anchored bulkhead s Fig . 20. 4 (b ) serv e th e sam e purpos e a s retainin g walls . However , i n
contrast t o retaining wall s whos e weigh t always represent a n appreciabl e fractio n of the weight of
the slidin g wedge , bulkhead s consis t o f a singl e ro w o f relativel y ligh t shee t pile s o f whic h th e
lower end s ar e drive n int o the eart h an d th e uppe r end s ar e anchore d b y ti e o r ancho r rods . Th e
anchor rods are held i n place by anchors which are buried in the backfill a t a considerable distanc e
from th e bulkhead.
Anchored bulkhead s ar e widel y use d fo r doc k an d harbo r structures . Thi s constructio n
provides a vertical wall so that ships may tie up alongside, or to serve as a pier structure, which may
jet ou t int o th e water . I n these cases sheetin g ma y be required t o laterall y suppor t a fil l o n which
railway lines , roads o r warehouse s ma y be constructed s o that shi p cargoes ma y be transferre d t o
other areas . The us e o f an anchor ro d tend s t o reduce th e latera l deflection , th e bending moment ,
and th e dept h of the penetration o f the pile.
Cantilever sheet piles depen d fo r their stability on an adequate embedment int o the soi l belo w
the dredge line . Sinc e th e pile s ar e fixe d onl y a t the bottom an d ar e fre e a t the top , the y ar e calle d
cantilever sheet piles. These piles ar e economical onl y for moderate wall heights, sinc e the required
section modulu s increases rapidl y wit h an increas e i n wall height , as the bending momen t increase s
with the cube of the cantilevered height of the wall. The lateral deflection of this type of wall, because
of the cantilever action, wil l be relativel y large. Erosion an d scour in front o f the wall , i.e. , lowerin g
the dredg e line , shoul d be controlled sinc e stabilit y of the wal l depends primaril y o n the develope d
passive pressure in front o f the wall.
20.3 FRE E CANTI L EV ER SH EET PIL E WAL L S
When the height of earth t o be retained b y sheet piling is small , the piling acts as a cantilever. Th e
forces actin g on shee t pil e wall s include:
1. Th e active earth pressure o n the back of the wall which tries to push the wall away from the
backfill
884 Chapt er 20
2. Th e passive pressur e i n front o f the wall below the dredge line. The passive pressur e resist s
the movement s o f the wall
The activ e an d passiv e pressur e distribution s o n th e wal l ar e assume d hydrostatic . I n th e
design o f th e wall , althoug h the Coulom b approac h considerin g wal l frictio n tend s t o b e mor e
realistic, th e Rankine approach (wit h the angl e o f wal l friction 8 = 0) i s normall y used .
The pressur e due to water may be neglected if the water levels on both sides of the wall are the
same. I f the difference in level is considerable, the effect o f the difference on the pressure wil l have
to be considered . Effectiv e uni t weight s of soi l shoul d be considere d i n computing th e activ e an d
passive pressures .
V
Sheet pile
Backfill
\
Anchor rod
(a) Cantilever sheet piles (b) Anchored bulk head
Sheeting
(c) Braced sheeting i n cut s (d) Singl e cel l cofferda m
(e) Cellular cofferda m
Diaphragms Granular fil l
Tie-rods <\
Outer sheet
pile wal l c
Inner sheet
pile wal l
(f) Cellula r cofferdam ,
diaphragm type
(g) Doubl e sheet pile wall s
Figure 20. 4 Us e o f shee t pi l e s
Sheet Pil e Wall s an d B race d Cut s
P
a
885
//AXV/A//,
O'
Pp-Pa
D
(a) (b )
Figure 20. 5 Ex am pl e illus t rat in g eart h pres s ur e o n can t ileve r s hee t pilin g
G eneral Principl e o f Desig n o f Fre e Cantileve r Shee t Pilin g
The actio n of the eart h pressur e agains t cantilever sheet piling can be bes t illustrate d by a simpl e
case shown in Fig. 20. 5 (a) . In thi s case, th e sheet piling is assumed t o be perfectl y rigid. When a
horizontal forc e P i s applie d a t th e to p o f th e piling , th e uppe r portio n o f th e pilin g tilt s i n th e
direction of P and the lowe r portio n move s i n the opposit e direction a s shown by a dashed lin e in
the figure. Thu s the piling rotates abou t a stationary point O'. The portion above O' i s subjected t o
a passive eart h pressure from th e soi l on the lef t sid e of the piles and an active pressure on the right
side of the piling, whereas the lower portion O'g i s subjected to a passive earth pressure on the right
side and a n active pressure o n the lef t sid e of the piling. At point O' th e piling does not move and
therefore i s subjected t o equal and opposit e eart h pressures (at-rest pressure from bot h sides ) wit h
a ne t pressur e equa l t o zero . Th e ne t eart h pressur e (th e differenc e betwee n th e passiv e an d th e
active) i s represente d b y abO'c i n Fig . 20. 5 (b) . Fo r th e purpos e o f design , th e curv e bO'c i s
replaced by a straight line dc. Point d is located at such a location on the line af tha t the sheet piling
is in stati c equilibrium under the action of force P and the earth pressures represente d b y the area s
ade and ecg. The position of point d can be determined by a trial and error method .
This discussion lead s t o the conclusion that cantilever sheet pilin g derives it s stabilit y fro m
passive eart h pressur e o n bot h side s o f th e piling . However , th e distributio n of eart h pressur e i s
different betwee n shee t pilin g i n granula r soil s an d shee t pilin g i n cohesiv e soils . Th e pressur e
distribution i s likel y t o change wit h time for sheet pilings in clay.
20.4 DEPT H OF EM B EDM ENT OF CANTI L EV ER WAL L S I N SANDY
SOI L S
Case 1 : With Water Tabl e a t G rea t Dept h
The activ e pressure actin g on the back of the wal l trie s to move the wal l away from the backfill . If
the dept h of embedment i s adequate the wal l rotates about a point O' situate d above the bottom of
the wall as shown i n Fig. 20. 6 (a) . The types of pressure that act on the wal l when rotation is likely
to take place abou t O' are :
1. Activ e earth pressur e at the back of wal l fro m th e surface of the backfil l down to the point
of rotation, O'. The pressure i s designated as P
al
.
886
Chapt er 20
2. Passiv e eart h pressur e i n fron t o f the wal l from th e poin t of rotation O' t o th e dredg e line .
This pressur e i s designated a s P^ .
3. Activ e eart h pressur e i n fron t o f th e wal l fro m th e poin t o f rotatio n t o th e botto m o f th e
wall. Thi s pressur e i s designated a s P^.
4. Passiv e eart h pressur e a t the back o f wal l from the point of rotation O' t o the bottom o f the
wall. This pressure i s designated a s P
}2
.
The pressure s actin g on the wal l are shown in Fig. 20. 6 (a) .
If the passive an d active pressures ar e algebraically combined, th e resultant pressure distributio n
below the dredge lin e will be as given in Fig. 20.6 (b) . The various notations used are :
D = minimu m depth o f embedment wit h a factor of safet y equal t o 1
K
A
= Rankin e active earth pressur e coefficient
K
p
- Rankin e passive eart h pressur e coefficient
K = K
p
-K
A
p
a
= effectiv e activ e eart h pressur e actin g agains t th e shee t pil e a t th e dredg e lin e
p = effectiv e passive eart h pressur e a t the base o f the pil e wal l and actin g towards th e
backfill = DK
/A\VA\VA\VA\\
O'
• . • • • . - San d r " :, • • . - •
(a) (b )
Figure 20. 6 Pres s ur e dis t ribut ion o n a can t ilever wall .
Sheet Pil e Wall s an d B race d Cut s 88 7
p' = effectiv e passive eart h pressur e a t the base of the sheet pil e wal l acting against the
backfill sid e o f th e wal l = p"
p
+ yKD
o
p" = effectiv e passiv e eart h pressur e a t level of O=
/
yy
Q
K+ jHK
p
Y = effectiv e uni t weight of the soi l assume d th e same belo w an d abov e dredg e lin e
y
Q
= dept h o f poin t O belo w dredg e lin e wher e th e activ e an d passiv e pressure s ar e
equal
y = heigh t o f point of application o f the total activ e pressure P
a
abov e point O
h = heigh t of point G above th e base o f the wal l
D
O
= heigh t of point O above th e base o f the wall
Expression fo r y
0
At poin t O , the passiv e pressur e actin g toward s th e righ t shoul d equa l th e activ e pressur e actin g
towards the left , tha t i s
Therefore, ^
0
(
K
P
~
Expression fo r h
For stati c equilibrium, th e sum of all the forces i n the horizontal direction mus t equal zero . That is
p
a
- \ P
P
V> -y
Q
) + \ ( P
p
+ P'
p
)h = 0
Solving fo r h,
. p
P
(
D
-yJ -i
p
a
h =
Taking moment s o f al l the forces abou t the bottom o f the pile, an d equating t o zero,
P ( D
0
+ p ) - i p
p
xD
0
x-± + ± ( p
p
+ p'
p
)xhx^ = 0
or 6P
a
( D
0
+y)-p
p
Dt+( p
p
+p'
p
)h
2
=0 (20.3 )
Therefore,
Substituting i n Eq. (20.3) fo r p , p' an d h an d simplifying, ,
888 Chapt e r 2 0
D
0
4
+ C,D
0
3
+ C
2
D<2 + C
3
D
0
+ C
4
= 0 (20.4 )
where,
The solutio n of Eq. (20.4 ) give s the dept h D
Q
. The metho d o f trial and error i s generall y adopte d t o
solve this equation. The minimum depth of embedment D with a factor of safety equal to 1 is therefore
D = D
0
+ y
Q
(20.5 )
A minimum factor o f safet y o f 1. 5 to 2 may be obtained by increasing th e minimum depth D
by 2 0 to 40 percent .
Maximum Bendin g M oment
The maximu m momen t o n section AB i n Fig. 20.6(b) occur s a t the point o f zer o shear . This poin t
occurs belo w poin t O in the figure . Let thi s point be represente d b y poin t C at a dept h y
0
belo w
point O. The ne t pressur e (passive ) of th e triangl e OCC' must balanc e th e ne t activ e pressur e P
a
acting abov e th e dredge line . The equatio n for P
a
i s
-
or y
0
= - T r (20.6 )
where 7 = effectiv e unit weight of the soil. If the water table lies above point O, 7 will be equal to y
b
,
the submerge d uni t weight of the soil .
Once the point of zero shear i s known, the magnitude of the maximum bending moment may
be obtained as
M
max =
P
a( y
+
yo^ o^ ^ = ?
a
( j + y
o
) ~ > K (20.7 )
(20.8)
The sectio n modulu s Z
s
o f the shee t pil e may be obtaine d fro m the equatio n
M
^--f^
h
where, / , = allowabl e flexura l stres s of the sheet pile .
Sheet Pil e Wall s an d B race d Cut s 889
Figure 20. 7 Sim plifie d m et ho d o f det erm in in g D fo r can t ileve r s hee t pil e
Simplified M etho d
The solutio n of the fourt h degree_equatio n is quite laborious and the problem ca n b e simplified by
assuming th e passiv e pressur e p' (Fig . 20.6) a s a concentrate d forc e R actin g a t th e foo t o f th e
pile. The simplifie d arrangement i s shown in Fig. 20.7 .
For equilibrium, the moments of the active pressure on the right and passive resistance o n the
left abou t the point of reaction R mus t balance.
=0
Now, P
D
= ^
and
Therefore, K
p
D
3
-K
A
( H + D)
3
- 0
or KD
3
-3HD( H + D)K=0. (20.9)
The solutio n o f Eq. (20.9) gives a value for D which i s at least a guide t o the required depth .
The dept h calculate d shoul d be increased b y at least 20 percent t o provide a factor of safet y and to
allow extr a lengt h to develop th e passive pressure R. An approximate dept h of embedment ma y be
obtained fro m Tabl e 20.1 .
Case 2 : Wit h Water Tabl e Withi n the B ackfil l
Figure 20.8 give s the pressure distribution against the wall with a water table at a depth h
l
below the
ground level . Al l the notation s give n i n Fig. 20. 8 ar e the same a s those give n i n Fig. 20.6 . I n this
case th e soi l abov e th e wate r tabl e ha s a n effectiv e uni t weigh t y an d a saturate d uni t weigh t y
sat
below th e water table. The submerge d uni t weight is
890 Chapt e r 20
Table 2 0 .1 Approx im at e pen et rat io n (D) of s hee t pilin g
Re l a t i ve densi t y
V ery loos e
Loose
Firm
Dense
Dept h, D
2.0 H
1.5 H
1.0 H
0.75 H
Source: Teng , 1969 .
Yb
=
'Ysat ~ YH. )
The activ e pressure at the water table is
Pl=V
h
l
K
A
and p
a
a t the dredge lin e is
P
a
= Y
h
i
K
A
+
Y
fc
h
2
K
A
= (Y/i, + Y^
2
) K
The othe r expressions are
P
P
=
P
P
=
Pa
p
p
( D-y
o
)-2P
a
h = ~~
The fourt h degre e equatio n in terms of D
g
i s
D
o
4
+ C, D^ + C
2
D
0
2
+ C
3
D
o
+ C
4
= 0 (20.10 )
where,
Sheet Pil e Wall s an d B race d Cut s 891
Dredge line
Figure 20. 8 Pres s ur e dis t ribut ion o n a can t ilever wal l wit h a wat er t abl e i n t h e
back fil l
The dept h o f embedment ca n be determined as i n th e previous case an d als o th e maximum
bending moment can be calculated. The depth D computed should be increased b y 20 to 40 percent .
Case 3 : Whe n the Cantileve r i s Free Standin g wit h No B ackfil l (Fig . 2 0 .9 )
The cantileve r i s subjecte d t o a lin e loa d o f P pe r uni t lengt h o f wall . Th e expression s ca n b e
developed o n th e sam e line s explaine d earlie r fo r cantileve r wall s wit h backfill . Th e variou s
expressions ar e
h =
2yDK
where K=( K
p
-K
A
)
The fourt h degre e equatio n i n D i s
D
4
+ C
l
D
2
+ C
2
D + C
3
= 0 (20.11)
8P
where
C
2 = —
12PH
892
Chapt er 20
Figure 20. 9 Fre e s t an din g can t il eve r wit h n o back fil l
4P
2
Equation (20.11) gives the theoretical depth D which should be increased b y 20 to 40 percent .
Point C in Fig. 20. 9 i s the point of zer o shear . Therefore ,
(20.12)
where 7 = effective uni t weigh t of the soi l
Example 2 0 . 1
Determine th e dept h o f embedmen t fo r th e sheet-pilin g show n i n Fig . Ex . 20.l a b y rigorou s
analysis. Determin e als o the minimum section modulus . Assume an allowable flexura l stress/ ^ =
175 MN/m
2
. The soi l has an effective unit weight of 1 7 kN/m
3
and angl e of internal friction of 30° .
Solution
For 0= 30°, K. = tan
2
(45 ° - 0/2 ) = tan
2
30 = -
K — =3 , K = K
a
— K.= 3 — — 2.67.
p
K '
p A
3
The pressur e distributio n along the sheet pile is assumed a s shown i n Fig. Ex . 20. Kb)
p
a
= y HK
A
= 17x 6 x- = 34 k N / m
2
Sheet Pil e Wall s an d B race d Cut s 89 3
From E q (20.1)
P 3 4
= o.75 m.
7( K
p
-K
A
) 17x2.6 7
^=^# + ^0 =1*34x6 + ^x34x0. 75
= 10 2 + 12.7 5 = 1 14.75 kN/meter lengt h of wal l or sa y 1 15 kN/m.
= 17x6x3+ 17xDx2. 67 = 306 + 45.4Z)
p'
p
= yHK
p
+ W
0
( K
p
-K
A
) = 17x6x 3+ 17x0.75x2.67 = 340 k N/ m
2
To find y
1_ H 1 _ 2
= - x 34x 6x (2 + 0.75) + -x34x0. 75x -xO.7 5 =286. 9
2 2 3
^_ 286. 9 286. 9 ^^ ^
Therefore, v = -= -= 2.50 m.
P
a
11 5
Now D
Q
can be found from Eq. (20.4) , namel y
D
0
4
+ qrg + c
2
D
0
2
+ c
3
D
0
+ c
4
= o
17x2.67 '
z
y K 17x2.6 7
(2 x 2.50 x 17 x2.67 + 340) = -189.9
= -310.4
(17x2.1
6x 115x2.50x 340 + 4x (115)
2
-4 — —
( y K )
2
(17X2.67)
2
Substituting for C
p
C
2
, C
3
and C
4
, and simplifying we have
D
0
4
+ 7.49D
0
3
- 20.3D 2 -189.9D
Q
- 310. 4 = 0
This equatio n whe n solved b y the method o f trial and error give s
D
0
* 5. 3m
Depth o f Embedmen t
D = D
0
+ y
Q
= 5.3 + 0.75 = 6.05 m
Increasing D by 40%, w e have
D (design) = 1. 4 x 6.05 = 8.47 m or say 8. 5 m.
894
Chapt er 2 0
H=6m
D = ?
Sand
y = 17kN/m
3
D
(a) (b)
Figure Ex . 20. 1
(c) Section modulu s
From Eq . (20.6) (Th e poin t of zer o shear )
2x115
yKV 17x2.6 7
= 2.25 m
M
'"
= 1 15(2.50 + 2.25) - - (2.25)
3
x 17 x 2.67
6
= 546. 3 - 86. 2 = 460 kN-m/m
From Eq. (20.8 )
Section modulu s
460
/, 17 5 X10
3
-26. 25X10-
2
m
3
/ mof wal l
Example 2 0 . 2
Fig. Ex . 20. 2 show s a free standin g cantilever sheet pil e wit h no backfill drive n int o homogeneou s
sand. The following dat a ar e available:
H = 20 ft , P = 3000 Ib/f t o f wall , 7 = 11 5 lb/ft
3
, 0 = 36° .
Determine: (a ) the dept h o f penetration, D, and (b) the maximum bending momen t M
max
.
Sheet Pil e Wall s an d B race d Cut s 895
Solution
20f t
P = 3000 Ib
y =115 1b / f t
3
0 = 36°
Fig. Ex . 20.2
K
n
= tan
2
45 ° + $.= tan
2
45 ° + — = 3.85
K
A
= — = — = 0.26
A
K
p
3.8 5
K = K
p
- K
A
= 3.85 - 0.2 6 = 3.59
The equation for D i s (Eq 20.11)
where
8P 8x3000
= -58.133
yK 115x3.5 9
12PH 12x3000x2 0
115x3.59
• = -1144
4x3000
2
(l 15x3.59)'
Substituting and simplifying , w e have
D^-58.133 D
2
- 174 4 D-211.2 = 0
From the above equation D =13.5 ft .
From Eq. (20.6)
• = -211.2
12P^
=
\ 2P_
yK \ yK
2x3000
115x3.59
= 3.81f t
896 Chapt e r 20
Fr omEq. (20.12 )
= 3000(20 + 3.81)-
6
115x(3.81)
3
x3.59
6
= 71,430 - 3,80 6 = 67,624 Ib-ft /f t o f wall
M
max
= 67
'
624 ]b
'
ft/ft
°
f Wal1
2 0 .5 DEPT H OF EM B EDM ENT OF CANTI L EV ER WAL L S I N
COH ESI V E SOI L S
Case 1 : When th e B ackfil l is Cohesive Soil
The pressur e distributio n o n a sheet pil e wal l is shown i n Fig. 20.10 .
The activ e pressur e p
a
a t any dept h z ma y b e expresse d a s
where
o~
v
= vertica l pressure, y z
z = dept h fro m th e surfac e o f th e backfill .
The passiv e pressur e p
}
a t any dept h v below th e dredge lin e ma y b e expresse d a s
The activ e pressur e distributio n on th e wal l fro m th e backfil l surfac e t o th e dredg e lin e i s
shown i n Fig. 20.10. The soi l i s supposed t o be in tension up t o a depth o f Z
Q
and the pressure on the
wall i s zero i n thi s zone . Th e ne t pressur e distributio n on the wal l i s shown b y the shade d triangle .
At th e dredge line (a t point A)
(a) Th e activ e pressur e p actin g towards th e lef t i s
p
a
=
Y
HK
A
~2cjK~
A
when 0 = 0 p
a
= yH - 2c = yH - q
u
(20.13a )
where q
u
- unconfi ne d compressiv e strengt h of the clay soi l = 2c.
(b) Th e passiv e pressur e actin g towards th e right at the dredg e lin e is
p
}
-2c sinc e ^ = 0
or P
p
= q
u
The resultan t of th e passive and activ e pressures a t the dredg e lin e i s
P
P
=Pa=Q
u
-( YH-ci
u
) = 1q
u
-YH-p (20.13b )
The resultan t o f th e passiv e and activ e pressures a t any dept h y belo w th e dredge line i s
Sheet Pil e Wal l s an d B race d Cut s 89 7
passive pressure , p = Yy + q
u
active pressure, p
a
= y( H + y) - q
u
The resultant pressur e i s
Pp-Pa = p = ( ry+q
u
)-\ -r( H+y)-q
u
] = 2q
u
-rH (20.14 )
Equations (20.13b ) an d (20.14 ) indicat e tha t th e resultan t pressur e remain s constan t a t
( 2q
u
-yH) a t all depths .
If passive pressur e i s developed o n the backfill side a t the bottom o f the pil e (poin t B) , then
p = Y ( H + D) + q
u
actin g towards th e lef t
p
a
= yD-q
u
actin g toward s the right
The resultant is
p
/
(20.15 )
For stati c equilibrium, the su m of all the horizontal forces mus t be equal zero , tha t is,
P
a
-( 2q
u
-
Y
H}D + ( 2q
u
+2q
u
)h = 0
Simplifying,
P
a
+2q
u
h- 2q
u
D + yHD = 0 , therefore,
D( 2q
u
-yH}-P
a
(20.16)
Also, fo r equilibrium, the sum of the moment s a t any point shoul d be zero. Takin g moment s
about the base ,
h
2
( 2q-yH)D
2
P
a
( y + D) + — ( 2q
u
)-^
L
-^ -= 0 (20.17 )
Substituting for h i n (Eq. 20.17) and simplifying,
C
1
D
2
+C
2
D+C
2
=0 (20.18 )
where C
{
= ( 2q
u
- yH)
_ _
a
The depth computed fro m Eq . (20. 1 8) should be increased b y 20 to 40 percent s o that a factor
of safet y of 1. 5 to 2. 0 may b e obtained. Alternatively the unconfmed compressive strengt h q
u
may
be divided b y a factor o f safety.
898 Chapt er 20
Dredge leve l
Figure 20.1 0 Dept h o f em bedm en t o f a can t il ever wal l i n cohes ive s oi l
Limiting H eight o f Wal l
Equation (20.14) indicates that when ( 2q
u
- y// ) = 0 the resultant pressure is zero. The wall will not
be stable. In order tha t the wall may be stable, the condition that must be satisfie d i s
*-> yH
F
where F = factor of safety .
(20.19)
M aximum B endin g M oment
As pe r Fig . 20.10 , th e maxi mu m bending momen t may occu r withi n the dept h ( D-h) belo w th e
dredge line. Let thi s depth be ;y
0
below the dredge line for zer o shear . We may write,
or y
0
~~
The expressio n fo r maximum bending moment is,
M = P ( v + v ) —
max a
v
^ o
J
' ~
where p = 2q
u
- yH
The sectio n modulu s of the sheet pil e may now be calculated a s before.
(20.20a)
(20.20b)
Sheet Pil e Wall s an d B race d Cut s 899
Case 2 : Whe n the B ackfil l i s Sand wit h Water Tabl e at G rea t Dept h
Figure 20.1 1 give s a cas e wher e th e backfil l i s san d wit h n o wate r tabl e within . The following
relationships may be written as:
Pa_=
p - 2 q
u
- yH = 4c -yH
p' = 2q
u
+ yH = 4c + yH
J _
a o
h =
A second degre e equatio n in D can be developed a s befor e
CjD
2
+ C
2
D + C
3
= 0
where C
l
= ( 2q
u
-yH)
c, = -2P_
(20.21)
(20.22)
H
where y = — , q
u
= 2c
An expression for computing maximum bending moment may be written as
U -f f f + y)-^--*^
max a
v
-
r
- ^ o ^ ^ ,
(20.23)
Sand
J L,
Point of
zero shea r
,
= p
Figure 2 0 .1 1 Shee t pil e wall embedde d i n clay with san d backfill .
900
where
2( 2q
u
-yH}
Chapt er 20
(20.24)
Case 3 : Cantileve r Wal l wit h San d B ackfil l an d Water Tabl e Abov e Dredge
Line [Fig . 2 0 .1 2]
The variou s expression s fo r thi s cas e ma y b e develope d a s i n th e earlie r cases . Th e variou s
relationships may b e writte n as
Pi = ^I
K
A
P
L
= Yhi
K
A
+
p = 2q-YH
K
A
= ( yhj +
Yb
h
2
) K
A
h
_ ( 2c
iu
-( yh
l
+r
b
h
2
)] D-P
a
The expressio n for the second degre e equatio n in D i s
C
}
D
1
+ C
2
D + C
3
= 0
where
(20.25)
(20.26)
(20.27)
Sand y , 0, c = 0
Sand y
sat
, 0, c = 0
Clay /sat , 0 - 0, c
Figure 2 0 .1 2 Can t ileve r wal l wit h s an d back fil l an d wat er t abl e
Sheet Pil e Wall s an d B race d Cut s 901
C
3
= -
Eq (20.27 ) ma y b e solve d fo r D. The dept h compute d shoul d be increase d b y 2 0 t o 40%t o
obtain a factor o f safet y o f 1. 5 to 2.0 .
Case 4 : Free- Standin g Cantileve r Shee t Pil e Wal l Penetrating Cla y
Figure 20.1 3 shows a freestandin g cantileve r wal l penetratin g clay . An expressio n fo r D ca n b e
developed a s before. Th e variou s relationships are given below.
ry _ _ / •
The expressio n fo r D is
Cj D
2
+ C
2
D + C
3
= 0
where C j = 2q
u
C
2
= -2P
(20.28)
The expression fo r h is
ft =
2
^-
f
(20.29)
The maximum momen t ma y be calculated pe r uni t lengt h of wal l by using the expressio n
(20.30)
H
D
C
y
• ° Point of zer o shear
Clay y
sat
,0
Figure 20.1 3 Fre e s t an din g can t ileve r wal l pen et rat in g cla y
902 Chapt er 20
where
~ dept h t o the point of zero shear .
(20.31)
Example 2 0 .3
Solve Exampl e 20.1 , i f th e soi l i s cla y havin g a n unconfine d compressiv e strengt h o f
70 kN/m
2
and a uni t weigh t of 1 7 kN/m
3
. Determin e th e maximum bending moment .
Solution
The pressur e distributio n is assumed as shown i n Fig. Ex . 20.3.
For 0
u
=0, p
a
=yH-q
u
= 17x6-70 = 32 k N/ m
2
Figure Ex . 20.3
1 1
- ^
a
(/ / - z
0
) = - x32
= 2x70- 17x6 =3 8 k N / m
of wall
p
f
=
Sheet Pil e Wall s an d B race d Cut s 90 3
y=^( H-z
0
) = ^(6-4.12) = 0.63 m
For the determination o f h, equate the summation of al l horizontal force s t o zero, thu s
or 30 - 3 8 xD + -(38 + 242)^ = 0
_ 3.8D- 3
Therefore h = -
14
For the determinatio n of D, taking moment s of all the forces abou t the bas e of the wall , we
have
/
o 7
or 3 0 (D + 0.63)- 38 x — + (38 +242)x — = 0
2 6
Substituting for h we have,
O O J~)_1
3D + 1.89-1.9D
2
+4.7 = 0
14
Simplifying, w e have
D
2
-1.57D + 1.35 = 0
Solving D = 2.2 m; Increasing D by 40%, we have D = 1.4(2.2) = 3.1 m.
Maximum bending momen t
From Eq . (20.20 )
-- 2
y
0
= = = 0.79 m
p 3 8
y~= 0.63 m
= 42.6 -1 1.9 = 30.7 kN-m/m of wall
Example 20. 4
Solve Exampl e 20. 1 if the soi l below th e dredge lin e is clay having a cohesion o f 35 kN/m
2
and the
backfill i s sand having an angle of internal friction o f 30°. The uni t weight of both th e soils may be
assumed a s 1 7 kN/m
3
. Determine th e maximum bending moment .
904
Solution
Refer t o Fig. Ex. 20.4
p
a
p = _ x 3 4 x 6 = kN/m o f wal l
p = 2q
u
- y H = 2 x 2 x 35 - 17x6 -38 k N/ m
2
From Eq. (20.22 )
C, D
2
+ C
2
D + C
3
= 0
where C
l
= (2 <y
w
- 7 H ) = (2 x 2 x 35 - 1 7 x 6) = 38 kN/m
2
= p
C
9
= - 2 P . = - 2 x 102 = - 20 4 kN
C
/ >
f l
(
p
a
+ 6 <
?«;y) 102(10 2 + 6 x 7 0 x 2)
X# + 4
M
17x 6 + 70
Substituting an d si mpl i f yi n g
38 D
2
-204 D- 558. 63-0
or D
2
-5. 37 D- 14. 7 =0
Solving th e equations , w e have D ~ 7.37 m
Increasing D b y 40%, we have
D (design ) = 1.4 (7.37) = 10. 3 m
= -558.63
Chapt er 20
Fig. Ex. 20. 4
Sheet Pil e Wall s an d B race d Cut s 905
Maximum Bending Moment
From Eq . (20.23 )
y= — = -= 2m, p = 38 kN/m
2
_ P
a
yH
2
K
2p 2x38
M
max =
= 2.66m
= 475.32-134.44 = 340.9 kN-m/m of wall .
M
max
= 340.9 kN-m/m of wall
Example 20.5
Refer t o Fig . Ex. 20.5. Solve th e proble m i n Ex. 20.4 if th e wate r tabl e i s abov e th e dredg e
line.
Given: h
l
= 2.5 m, y
s a t
= 1 7 kN/m
3
Assume the soil above the wate r table remains saturated. All the other dat a given in Ex. 20.4
remain th e same .
= 30°
y = 1 7 kN/m
3
Clay
c = 35 kN/m
2
= 0
Figure Ex . 20.5
906 Chapt e r 20
Solution
h
l
= 2.5 m, h
2
= 6 - 2. 5 = 3.5 m, y
b
= 17 - 9.8 1 = 7.19 kN/m
3
Pl
= Yh
v
K
A
=llx 2. 5 x 1/ 3 = 14.17 kN/m
2
P
a
=P
l
+ Y
b
h
2
K
A
= 14.1 7 + 7. 19 x 3. 5 x 1/ 3 = 22.56 kN/m
2
1 1
= - x 14.17 x 2.5 -f 14.1 7 x 3.5 +- (22.5 6 -14.17) x 3.5
£, Z*
= 17.71 + 49.6 + 14. 7 =82 kN/m
Determination o f y (Refe r t o Fig. 20.12)
Taking moment s o f al l the forces above dredge lin e about C we have
f 25} 3 5 3 5
82y = 17.71 3. 5 + — +49. 6 x — + 14.7 x —
( . 3 J 2 3
- 76.74 + 86.80 + 17.15 = 180.69
_ 180.6 9 „„ „
y=
__
2
.
20m
From Eq. (20.27) , th e equation for D i s
C
{
D
2
+ C
2
D + C
3
= 0
where C
l
= [2 q
u
-( jh^ + y
b
h
2
)]
= [140 - (1 7 x 2.5 + 7.19 x3.5)] = 72.3
C = -2P=- 2x8 2 = -164
=
(8 2 + 6x7Qx2 . 2 ) x82
3
q
u
+( Yh
l
+r
b
h
2
) 7 0 + 17x2.5 + 7.19x3.5)
Substituting we have,
72.3 D
2
- 1 64 D- 599 = 0
or D
2
- 2.2 7 D- 8.285 = 0
solving we have D ~ 4.23 m
Increasing D by 40%; th e design value is
D (design) = 1.4(4.23 ) = 5.92 m
Example 20. 6
Fig. Exampl e 12. 6 gives a freestandin g shee t pil e penetratin g clay . Determin e th e dept h o f
penetration. Given : H = 5 m, P = 40 kN/m, an d q
u
= 30 kN/m
2
.
Solution
FromEq. (20.13a )
~=2-H=2 = 2x30= 60kN/m
2
Sheet Pil e Wall s an d B race d Cut s 907
From Eq . (20.28) , Th e expressio n fo r D is
Cj D
2
+ C
2
D + C
3
= 0
where C
t
= 2 ^= 60
C
2
= - 2 F = -2 x 40 = - 80
( P + 6q
u
H)P _ (4 0 + 6x30x5)40
3
~ ^
3 0
Substituting an d simplifying
60 D
2
- SOD- 1253 = 0
or D
2
- 1. 33D- 21 =0
Solving D ~ 5. 3 m. Increasing b y 40%we have
D (design) = 1.4(5.3 ) = 7.42 m
= -1253
// = 5m
h = 4.63 m
Clay
0 = 0
c = 1 5 kN/m
2
5. 3m
= 60 kN/m
2
» +—60 kN/m
2
Figure Ex. 20. 6
908 Chapt e r 20
From Eq . (20.29 )
, 2q
u
D-P 2 x 30x 5 . 3m- 4 0
h = — = = 4.63 m
2^ 2 x 3 0
20.6 ANCH ORE D B U L KH EAD: FREE- EART H SU PPOR T M ETH OD-
DEPTH OF EM B EDM ENT OF ANCH ORED SH EET PI L E S I N
G RANU L AR SOI L S
If th e shee t pile s hav e bee n drive n to a shallo w depth , th e deflectio n o f a bulkhea d i s somewha t
similar to that of a vertical elastic beam whose lower end B is simply supported an d the other end is
fixed as shown i n Fig. 20.14 . Bulkheads which satisfy t hi s condition ar e called bulkhead s with free
earth support. Ther e ar e two method s of applying the factor o f safet y i n the desig n o f bulkheads .
1. Comput e the mi ni mu m depth of embedment and increas e the valu e by 20 t o 40 percent t o
give a factor of safet y of 1. 5 to 2.
2. Th e alternativ e method i s t o appl y the facto r o f safet y t o K
p
an d determin e th e dept h o f
embedment.
M ethod 1 : Minimum Dept h o f Embedmen t
The wate r tabl e i s assume d t o be a t a dept h h\ fro m th e surfac e o f th e backfill . Th e ancho r ro d i s
fixed at a height h
2
above th e dredge line . The sheet pil e is held in position by the anchor ro d and the
tension i n the rod i s T . The force s tha t are acting on the shee t pil e are
1. Activ e pressure du e t o the soi l behin d the pile,
2. Passiv e pressur e du e t o the soi l i n front o f the pile, an d
3. Th e tensio n i n the ancho r rod .
The problem i s to determine the mi ni mum dept h of embedment D. The force s tha t are acting
on th e pil e wal l ar e shown i n Fig. 20.15.
The resultan t of th e passiv e an d activ e pressure s actin g belo w th e dredg e lin e i s show n i n
Fig. 20.15 . The distanc e y
Q
t o the poi n t of zer o pressur e i s
The syste m i s i n equilibrium when the sum o f the moment s o f al l the force s abou t an y poin t
is zero. Fo r convenience if the moment s ar e take n abou t the ancho r rod ,
But P
p
=\ y
b
KD%
h
4
=h
3
+y
o
+-D
( >
Therefore,
Sheet Pil e Wall s an d B race d Cut s 909
Deflected /
Dredge shap e o f wal l /
level . ,
//A\V /A\V /A\V /A\V /A\V /A\V /A\V /
Passive
wedge
/A\V /\\V /A\V /A\V /A\V /A\V /A\V /A\V /A
' 4 '
, . **r T" i f s
B
S Ancho r
Active '
wedge /
Figure 20.1 4 Con dit ion s fo r free-eart h s upport o f a n an chore d bulk head
Simplifying th e equation,
>2 + C
3
= 0 (20.32 )
Figure 2 0 .1 5 Dept h o f em bedm en t o f a n an chored bulk head by t h e free-eart h
s upport m et ho d (m et ho d 1 )
910 Chapt e r 20
where C
{
= ——
Y
h
= submerged uni t weigh t of soi l
K - K
p
-K
A
The forc e i n the anchor rod, T
a
, is found b y summing the horizontal forces a s
T
a
=P
a
~P
P
(20.33 )
The mi ni mu m depth of embedment i s
D=D
0
+y
0
(20.34 )
Increase th e dept h D by 20 t o 40%t o give a factor o f safet y o f 1. 5 to 2.0.
M aximum B endin g M omen t
The maximum theoretica l momen t i n this case may be at a point C any depth h
m
below groun d level
which lie s betwee n h
}
an d H wher e th e shea r i s zero. Th e dept h h
m
may b e determine d fro m th e
equation
\ Pi\ -
T
a^(
h
m
-\ ^\ Y
b
( h
m
-h^K
A
=Q(20.35 )
Once h
m
is known the maximum bending moment ca n easil y be calculated.
M ethod 2 : Dept h o f Embedment b y Applying a Facto r o f Safet y t o K
(a) G ranula r Soi l B ot h i n the B ackfil l and B elo w the Dredg e Lin e
The force s tha t ar e acting on the shee t pil e wal l are as shown i n Fig. 20.16 . The maximu m passiv e
pressure tha t can be mobilized i s equal t o the area of triangle ABC show n i n the figure. The passive
pressure tha t ha s t o b e use d i n the computatio n i s the are a o f figur e ABEF (shaded) . Th e triangl e
ABC i s divided b y a vertical line EF suc h that
Area ABC
Area ABEF = - -—— — - P'
Factor o r safet y
p
The widt h o f figur e ABEF an d the poin t of applicatio n o f P' ca n b e calculate d withou t any
difficulty.
Equilibrium o f th e syste m require s tha t th e su m o f al l th e horizonta l force s an d moment s
about an y point , for instance , about the anchor rod , shoul d be equal t o zero.
Hence, P'
p
+ T
a
-P
a
=0 (20.36 )
P
a
y
a
-
ph
*=
Q
(20.37 )
where,
Sheet Pil e Wall s an d B race d Cut s 911
//\\V / \\V /\\V /\\\//\\V /\\\
Figure 20.1 6 Dept h o f em bedm en t b y free-eart h s upport metho d (m et ho d 2)
and F
s
= assumed facto r of safety .
The tensio n i n th e ancho r ro d ma y b e foun d fro m Eq . (20.36 ) an d fro m Eq . (20.37 ) D ca n b e
determined.
(b) Dept h o f Embedmen t whe n the Soi l B elo w Dredge Line i s Cohesive and the
B ackfill G ranular
Figure 20.1 7 show s the pressure distribution .
The surcharg e a t the dredge lin e due t o the backfil l ma y be written as
q = yh
l
+y
b
h
2
= y
e
H (20.38 )
where h
3
= depth of water above th e dredge line , y
e
effectiv e equivalen t unit weight of the soil , an d
The active eart h pressur e actin g towards the lef t a t the dredge lin e is (when 0 = 0)
The passive pressur e actin g towards the right is
The resultan t of the passive an d active earth pressures i s
(20.39)
912 Chapt er 20
-P
Figure 20.1 7 Dept h o f em bedm en t whe n t h e s oi l belo w t h e dredg e lin e i s
cohes ive
The pressur e remains constant with depth. Taking moments o f all the forces abou t the anchor
(20.40)
(20.41)
rod,
where y
a
= the distanc e of the ancho r rod fro m P
a
.
Simplifying Eq . (15.40) ,
where C , = 2h
3
or
The forc e i n the ancho r rod i s given by Eq . (20.33) .
It can be see n fro m Eq. (20.39 ) tha t the wal l will be unstabl e if
2q
u
-q =0
4c - q = 0
For all practical purpose s q - jH - /// , the n Eq. (20.39) ma y be written as
4c - y# = 0
c 1
or
(20,42)
Sheet Pil e Wall s an d B race d Cut s 91 3
Eq. (20.42 ) indicate s that the wal l i s unstable if the ratio clyH i s equal t o 0.25. N
s
i s termed i s
Stability N umber. The stabilit y is a function o f the wall height //, but is relatively independent of the
material use d i n developing q. If the wal l adhesion c
a
is taken int o account the stabilit y number N
s
becomes
(20.43)
At passive failur e ^l + c
a
/c i s approximately equal t o 1.25 .
The stabilit y number for shee t pil e wall s embedded i n cohesive soil s ma y b e written as
1.25c
(20.44)
When th e factor o f safety F = 1 and — = 0.25, N , = 0.30.
y s s
The stabilit y number N
s
require d i n determining the depth o f sheet pil e wall s is therefor e
N
s
= 0.30 x F
s
(20.45 )
The maximum bendin g momen t occur s a s per Eq. (20.35) at depth h
m
which lies between h
l
and H.
20.7 DESIGN CH ARTS FOR ANCH ORED B U L KH EADS I N SAND
Hagerty an d Nofa l ( 1 992) provided a set of design chart s fo r determinin g
1 .Th e dept h o f embedmen t
2. Th e tensil e forc e i n the anchor ro d and
3. Th e maximu m moment i n the sheet pilin g
The chart s ar e applicabl e t o shee t pilin g in san d an d th e analysi s i s base d o n th e free-eart h
support method . Th e assumption s mad e fo r the preparation o f the design chart s are :
1. Fo r active earth pressure , Coulomb' s theor y i s valid
2. Logarithmi c failur e surface below the dredge lin e for the analysis of passive eart h pressure .
3. Th e angl e of friction remain s th e same abov e an d below th e dredge line
4. Th e angl e of wal l friction betwee n the pile an d the soi l i s 0/ 2
The variou s symbol s use d i n the charts ar e the same a s given i n Fig. 20.1 5
where,
h
a
= the dept h o f the ancho r rod below th e backfil l surfac e
h
l
=th e dept h o f the wate r tabl e from th e backfill surfac e
h
2
- dept h of the water above dredg e lin e
H = height of the shee t pil e wal l above the dredge lin e
D = the minimum depth o f embedment require d by the free-eart h suppor t metho d
T
a
= tensile forc e i n the ancho r ro d pe r uni t lengt h of wal l
Hagerty an d Nofal developed th e curves given in Fig. 20. 1 8 on the assumption that the water
table i s a t th e groun d level , tha t i s h
{
= 0. Then the y applie d correctio n factor s fo r /i , > 0. These
correction factor s ar e given i n Fig. 20.19 . The equations for determining D, T
a
and M
(max)
are
914
Chapt er 2 0
0.05
0.2 0. 3
Anchor depth ratio, hJ H
Figure 20.1 8 Gen eral ize d (a ) depth o f em bedm en t , G
d
, (b) an cho r for c e G
(f
an d (c)
m ax im um m om en t G (aft e r Hag ert y an d Nofal , 1992)
Sheet Pil e Wall s an d B race d Cut s 915
1.18
1.16
1.14
e 1.1 0
o
1.08
1.06
1.04
0.0
1.08
1.06
1.04
0.94
0.1 0.2 0.3
0.4
0.3
0.2
0.1
(a)
0.4 0.5
(c)
0.1 0. 2 0. 3 0. 4
Anchor dept h rati o h
a
/H
0.5
Figure 20.1 9 Correct io n fact or s fo r variat io n o f dept h o f wat e r h
r
(a ) dept h
correct ion C
d
, (b) an chor for ce correct io n C
t
an d (c ) momen t correct io n C
m
(aft e r
Hag ert y an d Nofal , 1992 )
916 Chapt e r 20
D = G
d
C
d
H (20.4 6 a)
T
a
= G
t
C
t
j
a
H
2
(20.4 6 b)
M
(max) - G
n
C
n
^ (20.4 6 c)
where,
G
d
= generalize d non-dimensiona l embedment = D/H for ft, =0
G
r
= generalize d non-dimensiona l anchor force = T
a
I ( Y
a
H
2
) fo r h
l
~0
G
m
- generalize d non-dimensiona l moment = M
(max)
/ y
a
(#
3
) for / ij =0
C
rf
, C
r
, C
m
= correctio n factor s for h
{
> 0
Y
fl
= averag e effectiv e uni t weigh t of soi l
- / v h
2
+ v h
2
+ 2v / ? / z V /7
2
' - ' m "l ^ I f e
W
2
Z
' m ' M >V
//7
Y
m
= mois t o r dr y uni t weigh t of soi l above th e wate r tabl e
Y^ = submerge d uni t weigh t of soi l
The theoretical dept h D as calculated by the use of design chart s has t o be increased b y 20 to
40%t o give a factor of safet y of 1. 5 t o 2. 0 respectively.
20.8 M OM EN T REDU CTI ON FOR ANCH ORED SH EET PIL E WAL L S
The desig n o f anchore d shee t pilin g by the free-eart h metho d i s based o n the assumptio n tha t the
piling i s perfectl y rigi d an d th e eart h pressur e distributio n is hydrostatic , obeyin g classica l eart h
pressure theory . I n reality , th e shee t pil in g i s rathe r flexibl e an d th e eart h pressur e differ s
considerably fro m th e hydrostati c distribution.
As suc h th e bendin g moment s M(max ) calculated b y th e latera l eart h pressur e theorie s ar e
higher tha n th e actua l values . Row e (1952 ) suggeste d a procedur e t o reduc e th e calculate d
moments obtaine d b y the/ree earth support method.
Anchored Piling i n G ranular Soil s
Rowe (1952 ) analyzed sheet piling in granular soils and stated that the following significant factors
are required t o be take n i n the design
1. Th e relativ e densit y of the soi l
2. Th e relative flexibilit y o f the pilin g whic h is expressed a s
H
4
p=l09xlO-
6
(20.47a )
El
where,
p = flexibilit y numbe r
H = th e tota l heigh t of the pilin g in m
El = th e modulu s of elasticity and the moment of inerti a of the piling ( MN -m
2
) pe r m of
wall
Eq. (20.47a ) ma y be expresse d i n English units as
H
4
p = —(20.47b )
tA
where, H i s in ft, E i s in lb/in
2
and / i s in in
4
//if-of wal l
Sheet Pil e Wall s an d B race d Cut s 917
Dense san d and gravel Loose sand
1.0
S 0. 6
o
'i 0.4
od
0.2
0
T
= aH
D
_L
-4.0 -3. 5 -3. 0 -2. 5
Logp
-2.0
(a)
0.4
2.0
H
Logp = -2. 6
(working stress)
Logp = -2.0
(yield point of piling)
1.0 1. 5
Stability numbe r
(b)
Figure 20.2 0 Bendin g momen t i n an chored s hee t pilin g by free-eart h s uppor t
m et hod, (a ) in g ran ular s oils , an d (b ) i n cohes ive s oil s (Rowe , 1952)
Anchored Pilin g i n Cohesive Soil s
For anchore d pile s i n cohesive soils , th e most significan t factors are (Rowe, 1957 )
1. Th e stabilit y number
918 Chapt e r 20
(20
-
48)
2. Th e relative height of piling a
where,
H = heigh t of piling above th e dredge lin e in meter s
y = effectiv e uni t weigh t of the soi l above th e dredge lin e = moist uni t weight abov e
water leve l and buoyant unit weigh t below wate r level , kN/m
3
c = th e cohesion o f the soi l below the dredge line , kN/m
2
c
a
= adhesio n betwee n the soi l and the shee t pil e wall , kN/m
2
c
-2- = 1 . 2 5 for design purpose s
c
a = rati o betwee n H and H
M
d
= desig n momen t
Af „ = maximu m theoretical momen t
iTlaX
Fig. 20.20 gives charts for computing design moments for pile walls in granular and cohesive
soils.
Example 20.7
Determine th e dept h o f embedment an d the force i n the ti e rod of the anchored bulkhea d show n in
Fig. Ex . 20.7(a) . Th e backfil l abov e an d belo w th e dredg e lin e i s sand , havin g th e followin g
properties
G^ = 2.67, y
sat
= 1 8 kN/m
3
, j
d
= 13 kN/m
3
and 0 = 30°
Solve th e proble m b y th e free-eart h suppor t method . Assume th e backfil l abov e th e wate r
table remain s dry.
Solution
Assume th e soi l abov e th e water tabl e i s dry
For ^=30° , K
A
=^, £,,=3. 0
and K = K
p
-K
A
-3 — = 2.67
y
b
= y
sat
- y
w
= 18 - 9.8 1 = 8.19 kN/m
3
.
where y
w
= 9.81 kN/m
3
.
The pressur e distribution along the bulkhead is as shown i n Fig. Ex. 20.7(b)
Pj = Y
d
h
l
K
A
= 13x2x- = 8.67 k N/ m
2
at GW level
3
p
a
= p
{
+y
b
h
2
K
A
= 8.67 + 8.19 x 3 x -= 16.86 kN/m
2
at dredge lin e leve l
Sheet Pil e Wall s an d B race d Cut s 919
Pa
16.86
= 0.77 m
Y
b
xK 8.19x2.6 7
1 - , - , 1 , _
= - x 8.67 x 2 + 8.67 x 3 + -( 1 6.86 - 8.67) 3
2 2
+ - x 16.86 x 0.77 = 53.5 kN / m of wall
2
_L
(a) (b )
Figure Ex . 20. 7
To find y , taking moments of areas about 0, we have
53.5xy = -x8. 67x2 - + 3 + 0.77 +8. 67x3(3/ 2 + 0.77)
+ -(16.86- 8.67) x 3(37 3 +0.77) + -xl6.86x-x0.77
2
=122. 6
We have
Now
53.5
, y
a
= 4 + 0.77 -2.3 = 2.47 m
p
p
=- = 10.93 D
2
and it s distance from th e anchor rod i s
h
4
= h
3
+ y
0
+ 2 7 3D
0
= 4 + 0.77 + 2 / 3D
0
= 4.77 + 0.67Z)
0
920 Chapt e r 20
Now, takin g the moment s of the forces abou t the ti e rod, w e have
P
\ S -. , — D \ s h
y — *• ' '/i
53.5 x 2.47 = 10.93D^ x (4.77 + 0.67D
0
)
Simplifying, w e have
D
0
~ 1. 5 m, D = y
Q
+ D
Q
= 0.77 + 1. 5 =2.2 7 m
D (design ) = 1. 4 x 2.2 7 = 3.18 m
For finding the tensio n i n the ancho r rod, w e have
Therefore, T
a
=P
a
-P
p
= 53. 5 -10.93(1.5)
2
= 28.9 kN/ m o f wall for the calculated dept h D
0
.
Example 20 .8
Solve Exampl e 20. 7 b y applying F
v
= 2 t o the passive eart h pressure .
Solution
Refer t o Fig. Ex . 20.8
The followin g equations ma y be written
p' = 1Y
b
KpD
1
. — = -x 8.19 x 3 D
2
x - = 6.14 D
2
P 2 F
x
2 2
p
p
=y
b
K
p
D= 8.1 9 x 3D -24.6D
FG
a
P
n
D-h
BC p D
1 n/ 1
^ or h = D( l-a)
D+h D+h
an -
2
p
2
Area ABEF = ap = ax24.6D
or 6.14 D
2
=a( D + h) x 12. 3 D
Substituting for h = D(1 - a ) and simpl ifying
we have
2 «
2
- 4 a + l = 0
Solving the equation, we get a = 0.3.
Now h = D (1-0.3) = 0.7 D and AG = D - 0.7 D = 0.3 D
Taking moment s o f th e are a ABEF abou t th e bas e o f th e pile , an d assumin g ap = 1 i n
Fig. Ex . 20.8 we have
-(l)x0.3Z> -X0.3Z7 + 0.7D +(l )x0. 7£>x— -
Z, ~j •* —
Sheet Pil e Wall s an d B race d Cut s 921
h
a
=lm
Figure Ex . 20.8
simplifying we have y
p
- 0.44 D
y = 0.44 D
Now h, = h, + (D -y~ ) = 4 + (D - 0.4 4 D) = 4 + 0.56Z)
4 3 •
7
p
/
From th e active eart h pressur e diagra m (Fig . Ex . 20.8) we have
p
l
=y
d
h
l
K
A
=l3x2x-=$.61 kN/m
2
P
a
=P
l
+y
b
( h
2
+ D) K
A
= 8.67 + = 1 6.86 + 2.73D
(3+
2 2
Taking moment s o f active and passive forces abou t the ti e rod, an d simplifying , we hav e
(a) fo r moment s du e t o active forces = 0.89D
3
+ 13.7D
2
+ 66.7 D+ 104
(b) fo r moment s du e to passive forces = 6.14£>
2
(4 + 0.56D) = 24.56D
2
+ 3.44D
3
922 Chapt er 20
Since th e su m o f the moment s abou t th e ancho r rod shoul d be zero , w e have
0.89 D
3
+ 13. 7 D
2
+ 66.7D + 10 4 = 24. 56 D
2
+ 3.44 D
3
Si mpl i fyi ng w e hav e
D
3
+ 4.26 D
2
- 26 . 16 D - 40. 8 = 0
By solvin g the equation we obtai n D = 4.22 m wit h F
S
= 2. 0
Force i n the ancho r ro d
T =P -P'
a a p
where P
a
= 1.36 D
2
+ 16.8 6 D + 47 = 1.3 6 x (4.22)
2
+ 16.8 6 x 4.25 + 47 = 14 3 kN
P'
p
= 6.14 D
2
= 6.14 x (4.22)
2
= 10 9 kN
Therefore T
a
= 143 - 10 9 = 34 kN/m lengt h of wall.
Example 20. 9
Solve Exampl e 20.7 , i f the backfil l i s sand wi t h 0 = 30° and th e soi l belo w th e dredge line i s cla y
having c = 20 kN/m
2
. Fo r bot h th e soils , assum e G
v
= 2.67 .
Solution
The pressur e distributio n along th e bulkhead is as shown i n Fig. Ex . 20.9 .
p, = 8.67 k N/ m
2
a s in Ex. 20.7 , p
a
= 16.86 kN/m
2
= - x 8.67 x 2 + 8.67 x 3 + - ( l 6.86 - 8.67 ) x 3 -47.0 kN/ m
h, = 1 m
h, = 2 m
= 5m
h-, = 4 m
D
= 3 m
c = 0
HHHi VHH
PC: H
Clay
c = 20 kN/m
2
=
2
<-
Figure Ex . 20. 9
Sheet Pi l e Wal l s an d Brace d Cut s 92 3
To determine y
a
, take moment s about the tie rod.
P
a
xy
a
=- x8. 67x 2 - x2- l +8. 67x3x2. 5
+ -(16.86 - 8.67) x 3 x 3 = 104.8
2
V
'
_ _ 104 8 _ 104, 8
Therefore ^ « ~ ~~~ ~ ~
223 m
Now, < ? = Y
d
h
{
+ r A
=
13x 2 + 8.19x 3 = 50.6 kN/m
2
p = 2 x 2 x 2 0- 50.6 = 29.4 kN/m
2
Therefore, P=p*D = 29 AD kN/m
and «
4
Taking moment s of forces abou t the ti e rod, we have
or
47x2.23-29.4D 4 + — = 0
2
or 1.47D
2
+11.76£>-10.48 = 0 o r D = 0. 8 m.
D = 0.8 m is obtained wit h a factor of safet y equal t o one. It shoul d be increased b y 20 to 40
percent t o increas e th e facto r o f safet y fro m 1. 5 t o 2.0. For a facto r o f safet y o f 2 , th e dept h o f
embedment shoul d be at least 1.1 2 m. However the suggested dept h (design ) = 2 m.
Hence P fo r D (design) = 2 m i s
P
p
= 29 Ax 2 = 58.8 kN/ m length of wal l
The tension in the tie rod is
T
a
=P
a
-P
p
=41-59 = -12 kN/ m of wall.
This indicate s tha t the ti e rod wil l not b e i n tension unde r a design dept h o f 2 m. Howeve r
there i s tension for the calculated dept h D = 0.8 m.
Example 20.1 0
Determine the dept h of embedment for the sheet pil e given in Fig. Exampl e 20.7 usin g the design
charts give n in Section 20.7.
Given: H = 5 m, h
l
= 2 m, h
2
~ 3 m, h
a
= 1 m, h
3
= 4 m, 0 = 30°
924 Chapt e r 20
Solution
^ = 1 = 0.2
H 5
From Fig. 20.18
For h
a
IH = 0.2, and 0 = 30° we have
G
d
= 0.26, G
t
= 0.084, G
m
= 0.024
From Fig . 20.1 9 for 0 = 30, hJ H- 0. 2 and / z, / H = 0.4 we have
C
d
= 1.173 , C, = 1.073 , C
m
= 1.03 6
Now fro m Eq . (20.4 6 a )
D=G
d
C
d
H = Q.26x 1 . 1 7 3 x 5 = 1. 52 m
D (design) = 1. 4 x 1.5 2 = 2.13 m.
where y
a
-
H
1
V l
^
V 9 V
^ = 1 1 . 2 7 k N/ m
2
5
2
Substituting an d simplifying
T
a
= 0.084 x 1.073 x 11.2 7 x 5
2
= 25. 4 k N
M =G C Y H
3
max m m 'a
= 0.024 x 1.036 x 11. 27x5
3
= 35. 03kN-m/ mofwal l
The value s of D (design ) and T
a
fro m Ex . 20.7 ar e
D (design ) = 3.18 r n
T
a
= 28.9 kN
The desig n char t give s less by 33%i n the valu e of D (design ) and 12 %in the valu e of T
a
.
Example 2 0 .11
Refer t o Exampl e 20.7 . Determin e fo r the pil e (a) the bending moment M
max
, and (b ) the reduce d
moment b y Rowe' s method .
Solution
Refer t o Fig. Ex. 20.7. Th e followin g dat a ar e available
H = 5 m, h
l
- 2 m, H
2
= 3 m, /z
3
= 4 m
Y
rf
= 1 3 kN/m
3
, Y
6
= 8.19 kN/m
3
and 0 = 30 °
The maximu m bendin g momen t occur s a t a dept h h
w
fro m groun d leve l wher e th e shea r i s
zero. The equatio n whic h gives th e valu e of h i s
-p.h.-T +p.( h -h.) + -v,
~ " 11 a " 1
v
m \' 9 "
Sheet Pil e Wall s an d B race d Cut s 92 5
where p
l
= 8.67 kN/m
2
/ i j = 2 m, T
a
= 28.9 kN/ m
Substituting
-x8.67x2-28.9+8.67(/z -2 ) + -x8.19x-(/z - 2 )
2
= 0
2
m
2 3
m
Simplifying, w e have
^+2.35^-23.5 = 0
or h
m
~ 3.81 m
Taking moment s abou t the point of zero shear ,
M
miiX
=-\ PA ^-f*! +T
a
( h
m
-h
a
)-p
l
( hm
~
2
hl)
~Y
b
K
A
( h
m
-hf
= --x8. 67x2 3.81--X 2 + 28.9(3.81-1)- 8.67
(3
'
81
~
2)
—l8.19x-(3.81-2)
3
2 3 2 6 3
= - 21.4 1 + 81.2 - 14. 2 - 2.70 « 42.8 kN-m/m
From Ex . 20.7
£> (design) = 3. 18m, H = 5m
Therefore # = 8. 18 m
H
4
From Eq . (20.47a ) p = 109 x 10"
6

El
Assume E = 20.7 x 10
4
MN/m
2
For a section P
z
- 27 , / = 25.2 x 10~
5
m
4
/m
. . 10. 9xl O~
5
x(8. 18)
4
^
o r
^
1r t
.
Substituting p- ; —^ = 9.356 x l O~
3
&
2.0 7 x! 0
5
x 25.2 xl O-
5
logp = log ^y = -2.0287 or say 2.00
Assuming the sand backfil l is loose, w e have from Fig . 20.20 (a )
M
d
— = 0.32 f
or
i
0
g p =_2.0 0
max
Therefore M
d
(design ) = 0.32 x 42.8 = 13. 7 kN-m/m
20.9 ANCHORAG E O F BULKHEADS
Sheet pil e wall s are many times tied t o some kin d of anchors through tie rods t o give them greate r
stability a s shown i n Fig. 20.21. The type s o f anchorage tha t ar e normall y use d ar e als o show n in
the same figure .
926 Chapt er 20
Anchors suc h a s anchor wall s and anchor plate s whic h depend fo r their resistance entirely on
passive eart h pressur e mus t be given such dimensions that the anchor pul l does not exceed a certai n
fraction o f the pul l required t o produce f ai l ur e . Th e rati o betwee n th e tension i n the ancho r T an d
the maxi mu m pul l whi c h the anchor ca n stan d i s called th e factor o f safet y o f the anchor .
The type s o f anchorage s give n i n Fig . 20.21 are :
1. Deadmen, anchor plates, anchor beams etc.: Deadme n ar e shor t concret e block s o r
continuous concret e beam s deri vi n g their resistance fro m passiv e eart h pressure. Thi s typ e
is suitabl e whe n i t can b e installe d below th e leve l o f the origina l groun d surface .
2. Anchor block supported b y battered piles: Fi g (20.21b) shows an anchor block supporte d
by tw o battere d piles . The forc e T
a
exerte d b y th e ti e ro d tend s t o induc e compressio n i n
pile P j an d t ensi o n i n pile P
9
. This typ e i s employed wher e fir m soi l i s a t great depth .
3. Sheet piles: Shor t shee t pile s ar e drive n t o for m a cont inuou s wal l whic h derive s it s
resistance fro m passiv e eart h pressur e i n the same manne r a s deadmen.
4. Existing structures: Th e rod s ca n b e connecte d t o heav y foundat i on s suc h a s buildings ,
crane foundation s etc .
Original
ground Backfi l l
Original
ground
Backfill
Concrete cast against
original soi l
Sand an d grave l
compacted i n layers
(a)
Precast concret e
Final groun d
Ori gi nal
ground
Backfill
Cont i nuous
sheet pile s
(c)
Comp.
Tension
pile
T
a
= anchor pul l
Pairs of shee t pile s drive n t o
greater dept h a t frequen t interval s
as vertica l suppor t
(d)
Figure 2 0 .2 1 T ype s o f an chor ag e : (a ) deadm an ; (b ) brace d piles ; (c ) s hee t piles ; (d)
l arg e s t r uct ur e (aft e r T en g , 1969 )
Sheet Pil e Wall s an d B race d Cut s 92 7
L ocation o f Anchorage
The minimu m distanc e betwee n th e sheet pil e wal l an d th e ancho r block s i s determine d b y th e
failure wedge s of the shee t pil e (unde r free-earth suppor t condition) and deadmen. The anchorag e
does not serve any purpose i f i t is located withi n the failure wedge ABC show n in Fig. 20.22a .
If the failure wedges of the sheet pile and the anchor interfere with each other, the location of
the ancho r a s show n i n Fig. 20.22b reduce s it s capacity . Ful l capacit y o f th e anchorag e wil l be
available i f it is located i n the shaded are a shown in Fig. 20.22c. In thi s case
1 . Th e active sliding wedge of the backfill does not interfere with the passive sliding wedge of
the deadman.
2. Th e deadman i s located below the slope line starting from th e bottom of the sheet pile and
making an angle 0 with the horizontal, 0 being the angle of internal friction o f the soil .
Capacity o f Deadma n (After Teng, 1 969)
A serie s o f deadme n (ancho r beams , ancho r block s o r ancho r plates ) ar e normall y place d a t
intervals parallel t o the shee t pil e walls. These ancho r blocks may be constructed nea r the ground
surface o r at great depths , an d in short lengths or in one continuous beam. The holding capacity of
these anchorages i s discussed below.
Continuous Ancho r Beam Near G round Surface (Teng, 1 969)
If the lengt h o f the beam i s considerabl y greater tha n its depth, it is called^ a continuous deadman.
Fig. 20.23(a) shows a deadman. I f the depth t o the top of the deadman, h , is less tha n about one -
third t o one-hal f o f H (wher e H i s dept h t o th e botto m o f th e deadman) , th e capacit y ma y b e
calculated b y assumin g that th e to p o f th e deadma n extend s t o th e groun d surface . The ultimat e
capacity of a deadman ma y be obtained from (pe r uni t length)
For granular soil ( c = 0 )
or T
u
=^yH
2
( K
p
-K
A
) (20.49 )
For clay soi l ( 0 =0 )
-
=
^ ~ (20-50 )
where q
u
= unconfme d compressiv e strengt h of soil ,
Y = effectiv e unit weight of soil , an d
K
p
, K
A
= Rankine' s active and passive eart h pressure coefficients.
It ma y b e note d her e tha t the activ e eart h pressur e i s assume d t o b e zer o a t a dept h = 2c/y
which i s th e dept h o f th e tensio n cracks. I t i s likel y that the magnitud e an d distributio n o f eart h
pressure may change slowl y with time. For lack of sufficient dat a on this, the design of deadmen i n
cohesive soil s shoul d be made wit h a conservative factor of safety.
928 Chapt er 20
Sliding surface
Sliding surfac e
Ore pil e
Soft cla y
Anchorage subjecte d t o
other horizontal forces
Two sliding wedge s interfer e
with eac h othe r
(b)
Deadman locate d i n this
area ha s ful l capacit y
Figure 20.2 2 Locat io n o f deadm en : (a ) of f ers n o res is t an ce ; (b ) efficien cy g reat l y
im paired; (c ) full capacit y, (aft e r T en g , 1969 )
Short Deadma n Nea r G roun d Surfac e i n G ranular Soi l (Fig . 2 0 .2 3 b )
If the lengt h of a deadman i s shorter than 5h ( h = height of deadman) there will be an end effect with
regards t o th e holdin g capacit y of th e anchor . The equatio n suggested b y Teng fo r computin g th e
ultimate tensile capacity T
u
i s
T -
where
h
h
L
H
P ,
'
height of deadman
depth t o the top o f deadman
length o f deadma n
depth t o the bottom o f the dead ma n fro m th e groun d surfac e
total passive an d active earth pressures per uni t lengt h
(20.51)
Sheet Pil e Wal l s an d B race d Cut s 929
Ground surfac e
Anchor pull
H ^
Deadman
Active
wedge
Granular soi l
Cohesive soi l
Passive
wedge
a Activ e
**" * / wedg e
Deadman
h i '
HH
T
1
Footing
(c)
Figure 20.2 3 Capacit y o f deadm en : (a ) continuous deadmen nea r groun d s urface
(hlH < 1/ 3 ~ 1/2) ; (b) s hort deadm e n n ea r g roun d s urface ; (c ) deadm en a t g reat
dept h belo w g roun d s ur fac e (aft er T en g , 1969 )
K
o
Y
K
0
a, K A
coefficient o f earth pressures at-rest, taken equal to 0. 4
effective uni t weight of soi l
Rankine's coefficient s o f passive and active earth pressure s
angle of internal frictio n
Anchor Capacit y o f Shor t Deadma n i n Cohesive Soi l Near G roun d Surfac e
In cohesive soils , the second ter m of Eq. (20.51) shoul d be replaced by the cohesive resistanc e
T
u
= L (P -P
c
) + q
u
lf- (20.52 )
where q = unconfmed compressiv e strengt h of soil .
Deadman at G rea t Dept h
The ultimat e capacit y o f a deadma n a t grea t dept h belo w th e groun d surfac e a s show n i n
Fig. (20.23c ) i s approximately equal to the bearing capacity of a footing whos e base i s located a t a
depth ( h + ft/2), corresponding t o the mi d height of the deadman (Terzaghi , 1943) .
930 Chapt er 20
U ltimate L atera l Resistanc e o f V ertica l Ancho r Plate s i n Sand
The load-displacemen t behavio r o f horizontall y loade d vertica l ancho r plate s wa s analyze d b y
Ghaly (1997) . H e mad e us e o f 12 8 publishe d fiel d an d laborator y tes t result s an d presente d
equations fo r computing th e following:
1. Ultimat e horizontal resistance T
u
of singl e anchors
2. Horizonta l displacemen t u at any loa d leve l T
The equation s ar e
a
nr
C
A
yAH
t an0 A
T
= 2.2
where
A
H
h
Y
0.3
(20.53)
(20.54)
area o f anchor plat e = h
L
depth fro m the groun d surface t o the bottom o f the plat e
height o f plat e an d L = width
effective uni t weight of the san d
angle o f friction
16
12 - -
General equatio n -
Square plat e
4 8 1 2 1 6 2 0
Geometry factor , —
A
Figure 20.2 4 Relat ion s hi p o f pul l out -capacit y fact o r vers u s g eom et r y fact o r (aft e r
Ghaly, 1997 )
Sheet Pil e Wall s an d B race d Cut s
1.0
931
0.0
0.00 0.0 2 0.0 4 0.0 6 0.0 8
Displacement ratio, u/H
0.10
Figure 20.2 5 Relat ion s hi p of loa d rat i o vers u s dis placem en t rat i o [fro m dat a
report ed b y Da s an d Seel ey, 1975) ] (aft e r Ghaly , 1997)
C
A
= 5. 5 for a rectangular plate
= 5. 4 for a general equation
= 3. 3 for a square plate,
a = 0.3 1 for a rectangular plate
0.28 fo r a general equation
0.39 fo r a square plate
The equation s develope d ar e vali d for relative depth ratio s ( Hlh) < 5. Figs 20.2 4 an d 20.2 5
give relationships in non-dimensional form fo r computing T
u
and u respectively. Non-dimensional
plots for computing T
u
for squar e and rectangular plates are als o given in Fig. 20.24.
2 0 .1 0 B RACE D CU TS
G eneral Consideration s
Shallow excavations can be made without supporting the surrounding material i f there i s adequat e
space t o establish slope s a t which the material can stand. The steepest slope s tha t can be used i n a
given localit y ar e bes t determine d b y experience . Man y building sites exten d t o th e edge s o f th e
property lines. Under these circumstances, the sides of the excavation have to be made vertical and
must usually be supporte d b y bracings.
Common method s o f bracing the side s whe n the dept h of excavation does no t exceed abou t
3 m ar e shown i n Figs 20.26(a ) an d (b) . The practic e i s t o driv e vertical timbe r plank s known as
sheeting alon g th e side s o f th e excavation . The sheetin g i s hel d i n plac e b y mean s o f horizontal
932 Chapt er 20
beams calle d wales that in turn are commonly supported by horizontal struts extendin g from side t o
side o f th e excavation . Th e strut s ar e usual l y o f timbe r fo r width s no t exceedin g abou t 2 m. Fo r
greater width s metal pipe s calle d trench braces ar e commonly used .
When th e excavatio n dept h exceed s abou t 5 t o 6 m, the us e o f vertica l timbe r sheetin g wil l
become uneconomical. According to one procedure, steel sheet piles are used aroun d the boundary
of th e excavation . As th e soi l i s remove d fro m th e enclosure , wale s an d strut s ar e inserted . Th e
wales ar e commonly o f steel and the strut s may be of steel or wood. The proces s continue s until the
Steel
sheet piles
m^
Wale
Hardwood block
(a)
Wedge
Wale
Hardwood bloc k
Ground level
while tieback
is installed
Lagging-
Final groun d
level \
• Grout or concret e
Spacer
Steel
anchor rod
Bell
(c)
Figure 20.2 6 Cros s s ect ion s , t hroug h t ypica l bracin g i n dee p ex cavat ion , (a ) s ide s
ret ain ed b y s t ee l s hee t piles , (b ) s ide s ret ain e d b y H pile s an d lag g in g , (c ) on e o f
s ever al t iebac k s ys t em s fo r s upport in g vert ica l s ide s o f ope n cut . s evera l s et s o f
an chor s m a y b e us ed , a t differ en t el evat ion s (Peck , 1969 )
Sheet Pil e Wall s an d B race d Cut s 93 3
excavation i s complete . I n mos t type s o f soil , i t ma y b e possibl e t o eliminat e shee t pile s an d t o
replace the m wit h a series of//pil e s space d 1. 5 to 2.5 m apart . The //piles, known as soldier piles
or soldier beams, are drive n wit h thei r flange s parallel t o the side s o f the excavatio n a s shown i n
Fig. 20.26(b) . A s th e soi l nex t t o th e pile s i s remove d horizonta l board s know n a s lagging ar e
introduced a s shown i n the figur e an d ar e wedge d agains t th e soi l outsid e th e cut . As th e genera l
depth o f excavatio n advance s fro m on e leve l t o another , wale s an d strut s are inserte d i n the sam e
manner a s for stee l sheeting .
If th e widt h of a deep excavatio n i s to o grea t t o permi t economica l us e o f strut s acros s th e
entire excavation , tiebacks ar e ofte n use d a s a n alternativ e t o cross-bracing s a s show n i n
Fig. 20.26(c) . Incline d hole s ar e drille d int o th e soi l outsid e th e sheetin g o r H piles . Tensil e
reinforcement i s the n inserte d an d concrete d int o th e hole . Eac h tiebac k i s usuall y prestresse d
before th e dept h o f excavation i s increased .
Example 2 0 .1 2
Fig. Ex . 20.12 gives a n anchor plat e fixe d verticall y i n medium dense san d wit h the bottom o f the
plate a t a dept h o f 3 ft below th e groun d surface . The siz e o f the plat e i s 2 x 1 2 ft . Determin e th e
ultimate lateral resistanc e o f the plate. The soi l parameters ar e 7= 11 5 lb/ft
3
, 0 = 38° .
Solution
For al l practica l purpose s i f L /h_ > 5 , th e plat e ma y b e considere d a s a lon g beam . I n thi s cas e
L /h = 12/2 = 6 > 5. If the depth h to the top of the plate is less than about 1/ 3 to 1/ 2 of// (wher e H is
the depth t o the bottom o f the plate), the lateral capacity ma y be calculated usin g Eq. (20.49). In this
case hlH = 1/3, A s suc h
where K = tan
2
45 ° + — = 4.20 4
p
2
K
A
= — — = 0.238
4.204
T = -x 1 15 x3
2
(4.204 -0.238) = 2051 lb/f t
Example 2 0 .1 3
Solve Exampl e 20.1 2 if 0 = 0 and c = 300 lb/ft
2
. Al l the other dat a remai n th e same .
934 Chapt e r 20
Solution
Use Eq. (20.50 )
T =4cH-— = 4x300x3-
2x30(h
= 2035 Ib/f t
Y 11 5
Example 2 0 .1 4
Solve the problem i n Exampl e 20.12 for a plate length of 6 ft. All the other dat a remai n th e same .
Solution
Use Eq . (20.51 ) for a shorter lengt h of plate
where ( P - P
a
) = 2051 Ib/f t fro m Ex . 20.12
K
o
= 1 - si n 0 = 1 - si n 38° =0.384
tan 0 = ta n 38° = 0.78
J K^ = V 4.204 = 2.05, ^K~
A
= V 0.238 - 0.488
substituting
T
u
= 6 x2051 + -x0. 384xl 15(2.05 + 0.488) x 3
3
x 0.78 = 12,306 + 787 = 13,093 I b
Example 2 0 .1 5
Solve Exampl e 20.13 if the plat e length L =6 ft . All the other data remai n th e same .
Solution
Use Eq. (20.52 )
T
u
= L ( P
p
-P
u
)
+
q
u
lP
where P
p
-P
u
= 2035 Ib/f t fro m Ex. 20.13
q
u
= 2 x 300 = 600 Ib/ft
2
Substituting
T
u
= 6 x 2035 + 600 x 3
2
= 12,21 0 + 5,400 = 17,61 0 I b
Example 2 0 .1 6
Solve the problem i n Example 20.14 using Eq. (20.53). Al l the other data remai n th e same .
Solution
Use Eq. (20.53 )
5.4 Y AH H °'
28
tan0 A
where A = 2 x 6 = 1 2 sq ft, H = 3 ft, C
A
= 5.4 and a = 0.28 for a general equatio n
Sheet Pil e Wal l s an d B race d Cut s 935
tan 0 = tan 38° = 0.7 8
Substituting
0.28
5 . 4XH5 X12 X3 3 j
=
0.78 1 2
20.11 LATERA L EARTH PRESSURE DISTRIBUTION ON BRACED-CUTS
Since mos t ope n cut s ar e excavate d i n stage s withi n the boundarie s o f shee t pil e wall s o r wall s
consisting of soldier piles and lagging, and since struts are inserted progressively as the excavation
precedes, th e wall s are likely to deform as shown in Fig. 20.27. Little inward movement can occur
at the top of the cut after the first stru t is inserted. The pattern of deformation differs s o greatly fro m
that require d fo r Rankine' s stat e tha t the distributio n of eart h pressur e associate d wit h retaining
walls i s no t a satisfactor y basi s fo r desig n (Pec k e t al , 1974) . Th e pressure s agains t th e uppe r
portion of the wall s are substantiall y greater than those indicated by the equation.
p
*v
for Rankine' s conditio n
(20.55)
where
p
v
- vertica l pressure,
0 = frictio n angle
Apparent Pressur e Diagrams
Peck (1969) presente d pressure distribution diagrams on braced cuts. These diagrams are based on
a wealt h o f informatio n collected b y actua l measurements in the field . Pec k calle d thes e pressur e
diagrams apparent pressure envelopes whic h represen t fictitiou s pressur e distribution s fo r
estimating strut loads i n a system of loading. Figure 20.28 gives the apparent pressure distribution
diagrams as proposed by Peck .
Deep Cut s i n Sand
The apparen t pressure diagram for san d given in Fig. 20.2 8 wa s developed b y Pec k (1969 ) afte r a
great dea l o f study of actual pressur e measurements on braced cut s used for subways.
Flexible
sheeting
Rigid
sheeting
Deflection
at mud lin e
(a) (b)
y//////////////////////
(c)
Figure 20.2 7 T ypical pat t er n o f deform at io n o f vert ica l wall s (a ) an chored
bulk head, (b ) braced cut , an d (c ) t ieback cu t (Pec k e t al. , 1974)
936 Chapt er 2 0
The pressur e diagra m give n i n Fig. 20.28(b ) is applicable t o both loos e and dense sands. Th e
struts ar e t o be designe d base d o n thi s apparent pressur e distribution . The mos t probabl e valu e of
any individual stru t loa d i s about 25 percent lowe r than the maximum (Peck, 1969) . It may be note d
here tha t this apparen t pressur e distribution diagram is based o n the assumption that the water tabl e
is below th e botto m o f the cut.
The pressur e p
a
i s unifor m wi th respec t t o depth. The expressio n fo r p i s
p
a
= 0.65yHK
A
(20.56 )
where,
K
A
= tan
2
(45 ° - 0/2 )
y = uni t weigh t of san d
Cuts i n Saturated Clay
Peck (1969) developed tw o apparent pressur e diagrams , on e for sof t t o medium cla y an d the othe r
for stif f fissure d clay . H e classifie d thes e clay s on th e basi s o f non-dimensiona l factor s (stabilit y
number N J a s follows .
Stiff Fissure d clay
N = (20.57a)
Soft t o Mediu m clay
N,= —
c
(20.57b)
where y = unit weight of clay, c - undraine d cohesion ( 0 = 0)
Sand
H
0.65 yH tar T (45 ° -0/2)
(b)
(i) Stif f fissure d cla y (ii ) Sof t t o mediu m cla y
yH yH
c ~ c
0.25 H
0.50 H
0.25 H
0.2 yH
to 0. 4 yH
(c)
(ii)
yH-4c
0.25 H
0.75 H
(d)
Figure 20.2 8 Apparen t pres s ur e diag ra m for calculat in g load s i n s t rut s of brace d
cut s : (a ) s k et ch o f wal l o f cut , (b ) diag ram fo r cut s i n dr y o r m ois t s an d , (c ) diag ra m
for cl ay s i f j HIc i s les s t ha n 4 (d ) diag ram fo r cl ay s i f yH/c i s g reat er t ha n 4 wher e c
is t he aver ag e un drain e d s hearin g s t ren g t h o f t h e s oi l (Peck , 1969 )
Sheet Pil e Wall s an d B race d Cut s 937
0 0. 1 0. 2 0. 3 0. 4
V alues of yH
0.5 0 0.1 0. 2 0. 3 0. 4
0.5 H
l.OH
London
16m
A.J
Oslo
4 m
Houston
(a)
• Humbl e Bldg (16m)
A 50 0 Jefferson Bld g (10m)
• On e Shel l Plaza (18m)
(b)
Figure 20.2 9 M ax im u m apparen t pres s ure s for cut s i n s t iff clays : (a ) fis s ured cl ays
in Lon do n an d Os lo , (b) s t iff s lick en s ide d clays i n Hous t o n (Peck , 1969)
The pressur e diagram s fo r thes e tw o type s o f clay s ar e give n i n Fig . 20.28(c ) an d (d )
respectively. The apparent pressure diagram for soft t o medium clay (Fig. 20.28(d)) has been foun d
to b e conservativ e fo r estimatin g load s fo r desig n supports . Fig . 20.28(c) show s th e apparen t
pressure diagra m fo r stiff-fissure d clays . Mos t stif f clay s ar e wea k an d contai n fissures . Lowe r
pressures shoul d be use d onl y when the result s of observations o n simila r cut s i n th e vicinit y so
indicate. Otherwise a lower limit for p
a
= 0.3 yH shoul d be taken. Fig. 20.29 gives a comparison of
measured an d computed pressures distributio n for cuts in London, Oslo an d Houston clays.
Cuts i n Stratified Soil s
It i s very rare t o find unifor m deposit s of sand or clay to a great depth. Many times layer s o f sand
and clays overlying one another the other are found i n nature. Even the simplest of these conditions
does no t len d itsel f t o vigorou s calculation s o f latera l eart h pressure s b y an y o f th e method s
H
Sand
Clay
72
0 = 0
Figure 20.3 0 Cut s i n s t rat ified s oil s
938 Chapt e r 20
available. Base d o n fiel d experience , empirica l o r semi-empirica l procedure s fo r estimatin g
apparent pressure diagrams may be justified. Pec k (1969) propose d th e following unit pressure for
excavations in layered soil s (san d an d clay) with san d overlying as shown i n Fig. 20.30 .
When layer s o f san d an d sof t cla y ar e encountered , th e pressur e distributio n show n i n
Fig. 20.28(d ) ma y b e used i f the unconfme d compressiv e strengt h q
u
i s substituted by th e averag e
q
u
and the unit weight of soil 7 by the average value 7 (Peck , 1969) . The expressions for q
u
and
7 are
— 1
3u = "77 tX i
K
S
h
i
2 tan
0 +
h
2
n
<? J (
20
-
58
)
n
Y= — [ y
l
h
l
+ Y
2
h
2
] (20.59 )
where
H = tota l dept h of excavation
7p 7
2
= uni t weight s of sand and cla y respectivel y
h
l
,h
2
= thicknes s o f sand and clay layer s respectivel y
K
s
= hydrostati c pressur e rati o fo r th e san d layer , ma y b e take n a s equa l t o 1. 0 for
design purpose s
0 = angl e of friction o f sand
n = coefficien t of progressive failur e varie s from 0. 5 to 1. 0 which depends upo n the
creep characteristic s of clay. For Chicago cla y n varies fro m 0.75 t o 1.0 .
q
u
= unconfme d compression strengt h of clay
2 0 .1 2 STAB I L I T Y OF B RACED CU TS I N SATU RATED CL A Y
A braced-cu t ma y fai l a s a uni t du e t o unbalanced external forces o r heaving of th e botto m o f the
excavation. I f the external forces acting on opposite sides of the braced cut are unequal, the stability
of the entir e syste m has t o be analyzed . If soi l on one sid e o f a braced cu t i s removed du e t o some
unnatural forces the stability of the system will be impaired. However, we are concerned her e about
the stabilit y of the bottom of the cut . Two cases may arise . They ar e
1. Heavin g i n cla y soi l
2. Heavin g in cohesionless soi l
H eaving i n Clay Soi l
The dange r of heaving is greater i f the bottom of the cut is soft clay . Even in a soft cla y bottom, two
types of failur e ar e possible. They ar e
Case 1: When th e clay below th e cut i s homogeneous a t least up t o a depth equal 0. 7 B where B is
the widt h of the cut.
Case 2: When a hard stratum is met withi n a depth equal t o 0.7 B.
In th e firs t cas e a ful l plasti c failur e zon e wil l b e forme d an d i n th e secon d cas e thi s i s
restricted a s show n i n Fig . 20.31 . A facto r o f safet y o f 1. 5 i s recommende d fo r determinin g th e
resistance here. Shee t pilin g i s t o be drive n deepe r t o increas e the facto r o f safety . Th e stabilit y
analysis of the bottom o f the cut a s developed b y Terzaghi (1943) is as follows.
Sheet Pil e Wall s an d B race d Cut s 939
Case 1 : Formatio n o f Ful l Plasti c Failur e Z on e B elo w th e B otto m o f Cut .
Figure 20.31 (a) is a vertical section through a long cut of width B and depth //in saturated cohesive
soil ( 0 = 0) . Th e soi l belo w th e botto m o f th e cu t i s unifor m u p t o a considerabl e dept h fo r th e
formation o f a ful l plasti c failur e zone. The undrained cohesive strengt h of soi l i s c. The weigh t of
the block s o f cla y o n eithe r sid e o f th e cu t tend s t o displac e th e underlyin g cla y towar d th e
excavation. I f th e underlyin g cla y experience s a bearin g capacit y failure , th e botto m o f th e
excavation heave s an d the earth pressur e agains t the bracing increases considerably .
The anchorag e load bloc k of soi l a b c d in Fig. 20.31 (a) of widt h B (assumed) a t the level
of the bottom of the cut pe r uni t length may b e expressed a s
(20.60)
(20.61)
Q = yHB-cH=BH
The vertica l pressur e q per uni t length of a horizontal, ba, is
B B
(a)
/x x \
H
'
/
s
£
^D^
\
' >
-» t ^
2fi, ~
= fi
^ t ^
s
/WN /WN /W X
^
1
1'
1
(b)
/////////////////////^^^^
Hard stratu m
Figure 20.3 1 St abilit y o f brace d cut : (a ) heave of bot t o m o f t im bere d cut i n s of t
clay i f n o har d s t rat um in t erfere s wit h fl ow o f clay , (b ) as before, i f cla y res t s at
s hallow dept h belo w bot t o m o f cut o n har d s t rat um (aft e r T erzag hi, 1943 )
940 Chapt e r 20
The bearin g capacit y q
u
per uni t area a t level ab i s
q
u
= N
c
c = 5.1c (20.62 )
where N =5.1
The facto r of safet y against heaving is
p =
q
« =
5
'
7c
q
H y-4 (20.63 )
B
Because o f th e geometrica l condition , i t ha s bee n foun d tha t th e widt h B canno t excee d
0.7 B. Substitutin g this value for 5 ,
F
-= , . ,
(20.64)
Q.1B
This indicate s that the widt h of the failur e sli p is equal to B V2 = 0.7B.
Case 2 : Whe n the Formatio n o f Ful l Plasti c Z on e i s Restricted b y the
Presence o f a H ar d L aye r
If a har d laye r i s locate d a t a dept h D belo w th e botto m o f th e cu t (whic h i s les s tha n 0.75), th e
failure of the bottom occur s as shown in Fig. 20.31(b). The widt h of the stri p which can sink is also
equal t o D.
Replacing 0.75 by D i n Eq. (20.64) , th e factor of safet y is represented b y
5.7c
F =
u
-
H Y- — (20-65 )
D
For a cut in sof t cla y with a constant value of c
u
below the bottom of the cut, D in Eq. (20.65)
becomes large , an d F
s
approaches th e value
F
_ - » _ 5- 7
S
~~J T~~N ^
(20
'
66)
where #
5
= — (20.67 )
C
u
is terme d th e stability number. Th e stabilit y numbe r i s a usefu l indicato r o f potentia l soi l
movements. The soi l movement i s smaller for smalle r values of N
s
.
The analysi s discussed s o far i s for long cuts. For shor t cuts, square , circula r or rectangular,
the factor of safet y against heave ca n b e found i n the same wa y a s for footings .
2 0 .1 3 B JERRU M AN D EID E (1 9 5 6 ) M ETH OD OF ANAL Y SI S
The method o f analysis discussed earlie r give s reliable result s provided the widt h of the braced cu t
is larger than the depth of the excavation and that the braced cut is very long. In the cases wher e the
braced cut s ar e rectangular , square or circular in plan or the depth of excavation exceeds the width
of the cut, the following analysis should be used.
Sheet Pil e Wall s an d B race d Cut s 941
Circular or square - = 1. 0
A-/
1 I
H
q
mm
y£0^S/£0^N
H
0 1
Figure 20.3 2 St abilit y o f bot t o m ex cavat ion (aft e r Bj errum an d Eide , 1956 )
In thi s analysi s th e brace d cu t i s visualize d as a deep footin g whos e dept h an d horizonta l
dimensions are identical t o those at the bottom of the braced cut . This deep footing woul d fail i n an
identical manner t o the bottom braced cu t failed by heave. The theory o f Skempton fo r computin g
N
c
(bearin g capacit y factor ) fo r differen t shape s o f footin g i s mad e us e of . Figur e 20.32 give s
values of N
c
a s a function o f H/B for long, circular or square footings. For rectangular footings, the
value of N
c
ma y be compute d b y the expressio n
N (sq) (20.68)
where
length o f excavatio n
width o f excavatio n
The facto r of safet y fo r botto m heav e may b e expresse d a s
F —
cN .
where
yti + q
= effectiv e uni t weight of the soi l above the bottom of the excavatio n
= unifor m surcharge load (Fig . 20.32 )
Example 20 .1 7
A lon g trenc h i s excavate d i n mediu m dens e san d fo r th e foundatio n of a multistore y building .
The side s o f th e trenc h ar e supporte d wit h sheet pil e walls fixe d i n plac e b y strut s and wale s a s
shown i n Fig. Ex . 20.17. The soi l propertie s are :
7= 18. 5 kN/m
3
, c = 0 and 0 = 38°
Determine: (a ) Th e pressur e distributio n on the wall s wit h respect t o depth .
(b) Stru t loads. The struts are placed horizontally at distances L = 4 m center to center .
(c) Th e maximu m bending momen t fo r determinin g th e pil e wal l section .
(d) Th e maximu m bending moment s for determining the sectio n o f th e wales .
Solution
(a) Fo r a brace d cu t i n san d us e th e apparen t pressur e envelop e give n i n Fig . 20.2 8 b . The
equation for p
a
i s
p
a
= 0.65 y H K
A
= 0.65 x 18. 5 x 8 tan
2
(45 - 38/2 ) = 23 kN/m
2
942 Chapt er 2 0
p
a
= 23 kN/nr T
H
D
(a) Section
—3 m
p
a
= 0.65 yHK
A
(b) Pressur e envelope
-3 m-
8.33 kN
1. 33m 1. 33m i .6 7
C
23 kN \ / 2 3 kN
Section Dfi , " Sectio n B\ E
(c) Shear forc e distribution
•*-! m- ^
38.33 kN
Figure Ex . 20.1 7
Fig. Ex . 20.17b shows th e pressur e envelope ,
(b) Stru t loads
The reactions a t the ends o f strut s A, B and C are represented b y R
A
, R
B
an d R
c
respectivel y
For reaction R
A
, tak e moment s abou t B
fl. x 3 = 4x 2 3x - or R . = — = 61.33 k N
A 2 3
R
B1
= 23 x 4-61. 33 = 30.67 kN
Due t o the symmetr y o f the load distribution ,
R
Bl
= R
B2
= 30.67 kN, and R
A
= R
c
= 61.33 kN.
Now the strut loads ar e (for L - 4 m)
Strut A, P
A
= 61.33 x 4 -245 kN
Strut B, P
B
= ( R
Bl
+ R
B2
) x 4 = 61.34 x 4 -245 kN
Strut C, P
c
= 245 kN
Sheet Pil e Wall s an d B race d Cut s 94 3
(c) Momen t of the pil e wal l section
To determine moment s at different point s it is necessary t o draw a diagram showing the shear
force distribution.
Consider section s DB
{
an d B
2
E of the wall in Fig. Ex. 20. 17(b). The distribution of the shear
forces ar e shown in Fig. 20.17(c ) along wit h the points of zero shear .
The moment s at different point s may be determined as follows
1
M
A
= ~x 1 x23 = 1 1.5 kN- m
A
2
1
M
c
= -x 1 x 23 = 1 1.5 kN- m
1
M
m
= -x 1.33 x 30.67 = 20.4 kN- m
M
n
= ~x 1.33 x 30.67 = 20. 4 kN- m
The maximum moment A/
max
= 20.4 kN-m. A suitable section of sheet pile can be determined
as per standar d practice .
(d) Maximu m moment for wale s
The bendin g moment equation for wales is
RL
2
where R = maximum strut load = 245 kN
L = spacing o f strut s = 4 m
245 x 4
2
M
m a x
= -—= 490 kN-m
A suitabl e sectio n fo r the wales can be determined a s per standar d practice .
Example 2 0 .1 8
Fig. Ex . 20.18a gives the sectio n o f a long braced cut . The side s ar e supported b y stee l shee t pil e
walls with struts and wales. The soi l excavated at the sit e i s stif f cla y wit h the following propertie s
c = 800 lb/ft
2
, 0 = 0, y =11 5 lb/ft
3
Determine: (a ) Th e eart h pressur e distribution envelope.
(b) Stru t loads .
(c) Th e maximum moment of the sheet pil e section .
The strut s are placed 1 2 ft apart cente r t o center horizontally.
Solution
(a) Th e stabilit y numbe r N
s
fro m Eq . (20.57a) i s
=
c 80 0
944 Chapt er 20
D
H —
J flk. ^90^
2S f t Cla v
c - 80 0 lb/ft
2
0 = 0
y = 11 5 lb/ft
3
/^^v /£7
iv
5 f t
A |
7.5 f t
5
7.5 f t
C ,
5 f t
£,
S^ /W \xi?<S ^ /^Ss . /W^ v
^^, ^9S$ s
*U A
R R
R R
R
c
C
p
^\
5
^690 lb/ft
2
^V .
.*——
^690 lb/ft
2
^
~^^ '
p
a
= 863 lb/ft
2
ft
,
ft
*
""
6.25 f t
12 5 f t
6.25 f t
(a) Section of the braced trench
3518 Ib
(b) Pressure envelope
3518 Ib
5 f t —
D A
- 7. 5f t - 7 . 5 f t -
B, n
4. 2f t
1725 I b
(c) Shear force diagra m
5 f t
C E
1725 I b
2848 I b 284 8 Ib
Figure Ex . 20.1 8
The soi l i s stif f fissure d clay . A s suc h th e pressur e envelop e show n i n Fig . 20.28(c ) i s
applicable. Assume p
a
- 0. 3 f H
p
a
= 0.3x 115x2 5 = 863 lb/ft
2
The pressur e envelop e is drawn as shown i n Fig. Ex. 20.18(b) .
(b) Stru t loads
Taking moment s abou t the stru t head B
{
( B )
R
A
x7.5 = -863 x 6.25f — + 625\ + 863 x ^
A
2 I 3 ) 2
= 22 .4 7 x 10
3
+ 16.8 5 x 10
3
= 39.32 x 10
3
R
A
= 5243 lb/f t
R
R
.=-x 863 x 6.25 + 863 x 6.25 -5243 -2848 lb/f t
Dl >~\
Due t o symmetry
R - R = 5243 lb/f t
r{ C
R
B2
= R
Bl
= 2848 lb/f t
Sheet Pil e Wall s an d B race d Cut s 94 5
Strut load s are :
P
A
- 524 3 x 1 2 = 62,916 I b = 62.92 kip s
P
B
= 2 x 284 8 x 1 2 = 68,352 I b = 68.35 kip s
P
c
= 62.92 kip s
(c) Moment s
The shea r forc e diagra m i s shown in Fig. 20.18c for sections DB
{
an d B
2
E
Moment a t A = -x 5 x 690 x - = 2,875 Ib-ft/f t o f wall
2 3
Moment at m = 2848 x 3.3 - 86 3 x 3.3 x — = 4699 Ib-ft/f t
2
Because o f symmetrica l loadin g
Moment a t A = Moment a t C = 2875 Ib-ft/f t o f wall
Moment a t m = Moment a t n = 4699 Ib-ft/f t o f wall
Hence, th e maximu m moment = 4699 Ib-ft/f t o f wall.
The section modulu s and the required shee t pile section ca n be determine d i n the usual way.
2 0 .1 4 PIPIN G FAI L U RES I N SAND CU TS
Sheet pilin g i s used fo r cut s i n san d an d the excavation mus t be dewatere d b y pumpin g fro m th e
bottom o f th e excavation . Sufficien t penetratio n below the botto m o f the cu t mus t be provide d t o
reduce th e amoun t o f seepage an d t o avoi d th e danger of piping.
Piping i s a phenomeno n o f wate r rushin g u p throug h pipe-shape d channel s du e t o larg e
upward seepag e pressure . Whe n pipin g take s place , th e weigh t o f the soi l i s counteracte d b y th e
upward hydraulic pressure an d as such there i s no contact pressure between the grains at the bottom
of the excavation. Therefore, i t offers n o lateral suppor t t o the shee t pilin g an d as a result th e shee t
piling may collapse . Furthe r the soi l wil l become ver y loose an d may not have any bearing power .
It i s therefore , essentia l t o avoi d piping. For furthe r discussion s on piping, se e Chapte r 4 o n Soi l
Permeability an d Seepage . Pipin g ca n b e reduce d b y increasin g th e dept h o f penetration o f shee t
piles below th e bottom o f the cut.
2 0 .1 5 PROB L EM S
20.1 Figur e Prob . 20. 1 shows a cantilever sheet pil e wal l penetrating mediu m dens e san d wit h
the following propertie s o f the soil:
Y= 1151b/ft
3
, 0=38° ,
All the other dat a ar e given i n the figure .
Determine: (a ) th e dept h o f embedmen t fo r design , an d (b ) th e maximu m theoretica l
moment of the sheet pile .
20.2 Figur e Prob . 20. 2 show s a shee t pil e penetrating medium dens e san d wit h th e following
data:
h
{
= 6 ft, h
2
= 18 ft, y
s a t
= 12 0 lb/ft
3
, 0 = 38° ,
Determine: (a ) th e dept h o f embedmen t fo r design , an d (b ) th e maximu m theoretica l
moment o f the shee t pile . Th e sand above the water tabl e i s saturated .
946
Chapt er 2 0
y
s a t
=1201b/ f t
3
Sand
y
s a t
=1201b/ f t
j
^
g
A. _ oo o a : ^ t
0- ^8
M
c = 0
Sand
D = ?
y (moist ) = 17
0 = 34°, c = 0
_ T ^ . ^ _... , _ . ^ _ .
y
sat
= 19. 5 kN/m
3
0 = 34°
c = 0
Figure Prob. 20. 1
20.3
Figure Prob. 20. 2 Figure Prob. 20. 3
Figure Prob . 20. 3 show s a shee t pil e penetratin g loos e t o mediu m dens e san d wit h th e
following data :
/z, = 2 m, h
2
= 4 m, 7 (moist) = 1 7 kN/m
3
, y
sat
= 19. 5 kN/m
3
, 0 = 34° ,
Determine: (a ) th e dept h o f embedment , an d (b ) th e maximu m bendin g momen t o f th e
sheet pile .
20.4 Solv e Proble m 20. 3 fo r the water table at great depth . Assume y = 1 7 kN/m
3
. All the othe r
data remai n th e same .
20.5 Figur e Proble m 20. 5 show s freestandin g cantilever wal l wit h n o backfill . Th e shee t pil e
penetrates mediu m dens e san d wit h the following data :
H = 5 m, P = 20 kN/m, y = 17. 5 kN/m
3
, 0 = 36° .
Determine: (a ) the dept h of embedment , an d (b) the maximum momen t
20.6 Solv e Pro b 20. 5 wit h the following data:
H = 20 ft , P = 3000 Ib/ft , 7=11 5 lb/ft
3
, 0 = 36°
20.7 Figur e Prob . 20. 7 show s a shee t pil e wall penetrating clay soil and the backfil l is also clay.
The followin g data ar e given.
H = 5 m, c = 30 kN/m
2
, y = 17. 5 kN/m
3
Determine: (a ) the dept h of penetration, an d (b) the maximum bending moment .
20.8 Figur e Prob . 20. 8 ha s san d backfil l an d cla y belo w th e dredge line . The propertie s o f the
Backfill are :
0=32°, y= 17. 5 kN/m
3
;
Determine th e dept h o f penetration of pile .
20.9 Solv e Prob . 20. 8 wi t h th e wat er t abl e abov e dredg e l i ne : Gi ve n h
{
= 3 m, h
2
= 3 m,
y
sat
= 18 kN/ m
3
, wher e h
{
= depth of water tabl e below backfil l surface an d h
2
= ( H - h
{
).
The soi l abov e th e wate r table is also saturated . All the other dat a remai n th e same .
20.10 Figur e Prob. 20.10 shows a free-standing shee t pile penetrating clay. The following data are
available:
H = 5 m, P = 50 kN/m, c = 35 kN/m
2
, y = 17. 5 kN/m
3
,
Determine: (a ) the dept h o f embedment , and (b ) the maximum bendin g moment .
Sheet Pil e Wall s an d B race d Cut s 947
,
{
I/
a

z
'
i
i
•*H
)
P = 20 kN/m
H =

7 = 17. 5 kN/m
3
0 = 36°
C = 0 D :
1
5 m
7
Clay
c = 30 kN/m
3
7 =17.5 kN/m
3
Clay
Sand
7 =17.5 kN/m
3
= 32°
Clay
Figure Prob . 20. 5 Figure Prob . 20. 7 Figure Prob . 20. 8
20.11 Solv e Prob. 20.10 wit h th e following data :
H = 15 ft , P = 3000 Ib/ft , c = 300 lb/ft
2
, 7 = 10 0 lb/ft
3
.
20.12 Figur e Prob . 20.1 2 show s a n anchore d shee t pil e wal l fo r whic h th e followin g dat a ar e
given
H = 8 m, h
a
= 1.5 m, h
l
= 3 m, y
sat
= 19.5 kN/m
3
, 0 = 38°
Determine: the forc e i n the ti e rod .
Solve th e proble m b y th e free-eart h suppor t method .
= 5m
D
= 50 kN/m
Clay
c = 35 kN/m
2
7 = 17.5 kN/m
3
h
a
- 1. 5 m
_ T
H = 8 m
Sand
7
s at
= 19. 5 kN/m
3
j » = 3 m
Sand "
7
sat
= 19. 5 kN/m
3
0 = 38°, c = 0
; = 6. 5 m
= 5 m
Sand
Figure Prob . 20.1 0 Figure Prob . 20.1 2
948 Chapt er 20
20.15
20.13 Solv e th e Prob. 20.12 with the following data: H=24 ft, h
l
= 9 ft, h
2
= 15 ft, and h
a
= 6 ft.
Soil properties : 0 = 32°, y
sat
= 12 0 lb/ft
3
. The soi l abov e th e water tabl e i s also saturated .
20.14 Figur e Prob . 20.1 4 give s a n anchore d shee t pil e wal l penetratin g clay . Th e backfil l i s
sand. Th e followin g data ar e given:
H = 24 ft , h = 6 ft , h, = 9 ft ,
' a ' I '
For sand : d) = 36°, y . = 12 0 lb/ft
3
,
' ' Sal
The soi l abov e th e wate r tabl e i s als o saturated .
For clay: 0 = 0, c = 600 lb/ft
2
Determine: (a ) the dept h o f embedment , (b) the force i n the ti e rod, an d (c ) the maximum
moment.
Solve Prob . 20.1 4 wit h the followin g data :
H=8m, h
a
= 1. 5m, h
}
- 3m
For sand : y =19. 5 kN/m
3
, ( b ~36 °
' SU l '
The san d abov e th e WT i s also saturated.
For clay: c = 30 kN/m
2
Solve Prob . 20.1 2 b y the us e of design chart s given i n Section 20.7 .
Refer t o Prob . 20.12 . Determin e fo r th e pil e (a ) th e bendin g momen t M
max
, an d (b ) th e
reduced momen t b y Rowe' s method .
20.18 Figur e Prob . 20.1 8 gives a rigid ancho r plat e fixed verticall y i n mediu m dens e san d wit h
the botto m o f th e plat e a t a dept h o f 6 f t belo w th e groun d surface . Th e heigh t ( h ) an d
length (L ) o f th e plat e ar e 3 f t an d 1 8 f t respectively . Th e soi l propertie s are :
y= 12 0 lb/ft
3
, an d 0= 38°. Determine the ultimate lateral resistance pe r unit length of the
plate.
20.16
20.17
h
a
= 6 ft \
i, = 9 ft
V
= 24 ft
Sand
y
s a t
= 12 0 lb/ft
3
0 = 36°, c = 0
Clay
D
Clay
0 = 0,
c = 600 lb/ft
2
Y= 12 0 lb/f t
L= 1 8 f t
Figure Prob . 20.1 4 Figure Prob . 20.1 8
20.19 Solv e Prob . 20.1 8 for a plate of lengt h = 9 ft. All the other dat a remai n th e same .
20.20 Solv e Prob . 20.1 8 fo r th e plat e i n cla y ( 0 = 0 ) havin g c = 400 lb/ft
2
. Al l th e othe r dat a
remain th e same .
Sheet Pil e Wall s an d B race d Cut s
949
Sand
y
sat
= 18. 5 kN/m
3
0 = 38°, c = 0
Fig. Prob . 20.2 3
20.21 Solv e Prob. 20.20 for a plate of length 6 ft. All the other data remain the same .
20.22 Solv e the prob. 20.19 by Eq (20.53). All the other determine the same .
20.23 Figur e Prob . 20.23 shows a braced cu t in medium dense sand . Given 7= 18. 5 kN/m
3
, c = 0
and 0 = 38°.
(a) Dra w th e pressur e envelope , (b ) determin e th e stru t loads , an d (c ) determin e th e
maximum moment o f the sheet pil e section .
The struts are placed laterall y at 4 m center t o center .
20.24 Figur e Prob . 20.2 4 show s th e sectio n o f a brace d cu t i n clay . Given : c = 650 lb/ft
2
,
7= 11 5 lb/ft
3
.
(a) Dra w the earth pressure envelope, (b) determin e the strut loads, an d (c) determin e the
maximum moment of the sheet pile section.
Assume that the strut s are placed laterall y at 1 2 ft center to center .
Fig. Prob . 20.2 4
CHAPTER 21
SOIL IMPROV EMENT
2 1 .1 I NTRODU CTI O N
General practic e i s t o us e shallo w foundation s for th e foundation s of building s an d othe r suc h
structures, i f th e soi l clos e t o th e groun d surface possesse s sufficien t bearin g capacity . However ,
where th e top soi l i s either loos e o r soft , th e loa d fro m th e superstructur e has t o be transferred t o
deeper fir m strata . I n suc h cases , pil e or pie r foundations are the obvious choice .
There i s als o a thir d metho d whic h ma y i n some case s prov e mor e economica l tha n dee p
foundations o r wher e th e alternat e metho d ma y become inevitabl e du e t o certai n sit e an d othe r
environmental conditions . Thi s thir d metho d come s unde r th e headin g foundation soil
improvement. I n the case of earth dams, there is no other alternative than compacting the remolde d
soil in layers to the required density and moisture content. The soil for the dam wil l be excavated at
the adjoining areas an d transported t o the site. There ar e many methods by which the soil at the site
can b e improved . Soi l improvemen t i s frequentl y terme d soil stabilization, whic h i n it s broades t
sense i s alteratio n o f an y propert y o f a soi l t o improv e it s engineerin g performance . Soi l
improvement
1. Increase s shea r strength
2. Reduce s permeability, an d
3. Reduce s compressibilit y
The method s o f soi l improvement considered i n this chapter ar e
1. Mechanica l compactio n
2. Dynami c compactio n
3. V ibroflotatio n
4. Preloadin g
5. San d and stone columns
951
952 Chapt e r 21
6. Us e of admixture s
7. Injectio n o f suitabl e grout s
8. Us e o f geotextile s
2 1 .2 M ECH ANI CA L COM PACTI O N
Mechanical compactio n i s th e leas t expensiv e o f th e method s an d i s applicabl e i n bot h
cohesionless an d cohesiv e soils . Th e procedur e i s t o remov e firs t th e wea k soi l u p t o th e dept h
required, an d refil l o r replac e th e sam e i n layer s wit h compaction . I f th e soi l excavate d i s
cohesionless o r a sand-sil t cla y mixture , th e sam e ca n b e replace d suitabl y i n layer s an d
compacted. I f the soi l excavate d is a fine sand , silt or soft clay , i t is not advisabl e t o refill th e same
as thes e materials , eve n unde r compaction , ma y no t giv e sufficien t bearin g capacit y fo r th e
foundations. Sometime s i t migh t b e necessar y t o transpor t goo d soi l t o th e sit e fro m a lon g
distance. Th e cos t o f suc h a project ha s t o be studie d carefull y befor e undertakin g the same .
The compaction equipmen t to be used on a project depends upo n the siz e o f the project an d
the availabilit y o f th e compactin g equipment . In project s wher e excavatio n an d replacemen t ar e
confined t o a narrow site , onl y tampers o r surface vibrators may be used. On the other hand, i f the
whole are a o f the project i s to be excavated and replaced i n layers with compaction, suitabl e roller
types o f heav y equipmen t can b e used . Cohesionles s soil s ca n b e compacte d b y usin g vibratory
rollers an d cohesiv e soil s b y sheepsfoo t rollers .
The contro l o f fiel d compactio n i s ver y importan t i n orde r t o obtai n th e desire d soi l
properties. Compactio n o f a soi l i s measure d i n terms o f th e dr y uni t weight o f th e soil . Th e dr y
unit weight , y
d
, ma y b e expresse d a s
1 + w
where,
y
t
= total uni t weight
w = moistur e content
Factors Affecting Compaction
The factor s affectin g compactio n ar e
1. Th e moistur e conten t
2. Th e compactiv e effor t
The compactiv e effor t i s defined as the amount of energy imparted t o the soil . With a soil of
given moistur e content , increasin g th e amoun t o f compactio n result s i n close r packin g o f soi l
particles an d increase d dr y uni t weight . Fo r a particula r compactiv e effort , ther e i s onl y on e
moisture conten t whic h give s th e maximu m dry uni t weight . The moistur e conten t tha t give s th e
maximum dr y uni t weigh t i s calle d th e optimum moisture content. I f th e compactiv e effor t i s
increased, th e maximu m dr y uni t weigh t als o increases , bu t th e optimu m moistur e conten t
decreases. I f al l th e desire d qualitie s o f th e materia l ar e t o b e achieve d i n th e field , suitabl e
procedures shoul d be adopted t o compact the earthfill. The compactive effort t o the soil is imparted
by mechanical roller s or any other compacting device. Whether the soil i n the field has attained the
required maximu m dry uni t weight can be determined by carrying out appropriat e laborator y test s
on the soil . The following tests ar e normally carried ou t i n a laboratory.
1 . Standar d Proctor tes t (AST M Designatio n D-698), an d
2. Modifie d Procto r tes t (AST M Designatio n D-1557)
Soil Im provem en t 953
2 1 .3 L AB ORATOR Y TESTS ON COM PACTI ON
Standard Procto r Compactio n Test
Proctor (1933 ) develope d thi s tes t i n connectio n wit h th e constructio n o f eart h fil l dam s i n
California. Th e standar d siz e o f th e apparatu s use d fo r th e tes t i s give n i n Fi g 21.1 . Tabl e 21. 1
gives th e standar d specification s fo r conductin g th e tes t (AST M designatio n D-698) . Thre e
alternative procedure s ar e provide d base d th e soi l materia l use d fo r the test .
Test Procedur e
A soi l a t a selected wate r content i s placed i n layers int o a mold o f given dimension s (Tabl e 21. 1
and Fig. 21.1) , wit h each laye r compacted b y 25 or 56 blows of a 5.5 I b (2. 5 kg ) hammer droppe d
from a height o f 1 2 in (30 5 mm) , subjectin g the soi l t o a total compactiv e effor t o f abou t 12,37 5
fl-lb/ft
3
(60 0 kNm/m
3
). The resulting dr y uni t weight is determined. Th e procedure i s repeated for
a sufficien t numbe r o f wate r content s t o establis h a relationship betwee n th e dr y uni t weight and
the water content o f the soil . Thi s data , when , plotted, represent s a curvilinear relationshi p known
as th e compactio n curv e o r moisture-densit y curve . Th e value s o f wate r conten t an d standar d
maximum dr y uni t weight ar e determined fro m th e compaction curv e a s shown i n Fig. 21.2 .
It em
T able 21. 1 Specificat io n fo r s t an dar d Proct o r com pact io n t es t
Procedure
A B
1. Diamete r o f mol d
2. Heigh t o f mol d
3. V olum e of mol d
4. Weigh t o f hamme r
5. Heigh t o f dro p
6. No . o f layer s
7. Blow s pe r laye r
8. Energ y o f
compaction
9. Soi l materia l
4 in . (101. 6 mm)
4.584 in . (116.4 3 mm )
0.0333 ft
3
(94 4 cm
3
)
5.5 I b (2. 5 kg )
12.0 in . (304. 8 mm)
3
25
12,375 ft-lb/ft
3
(600 kN-m/m
3
)
Passing No . 4 siev e
(4.75 mm) . Ma y b e use d
if 20 %o r les s retaine d
on No . 4 sier e
4 in . (101. 6 mm )
4.584 in . (116.43 mm)
0.0333 ft
3
(94 4 cm
3
)
5.5 I b (2. 5 kg )
12.0 in . (304. 8 mm )
3
25
12,375 ft-lb/ft
3
(600 kN-m/m
3
)
Passing No 4 siev e (4.7 5 mm) .
Shall b e use d i f 20 %o r mor e
retained o n No . 4 siev e an d
20%o r les s retaine d o n
3/8 i n (9. 5 mm ) siev e
6 in . (152. 4 mm )
4.584 in . (116.4 3 mm )
0.075 ft
3
(212 4 cm
3
)
5.5 I b (2. 5 kg )
12.0 in . (304. 8 mm)
3
56
12,375 ft-lb/ft
3
(600 kN-m/m
3
)
Passing No . 4 siev e
(4.75 mm) . Shal l b e
used i f 20%o r mor e
retained o n 3/ 8 in .
(9.5 m m ) siev e an d
less tha n 30 %retaine d
on 3/ 4 in. (19 mm) sieve
954
Chapt er 2 1
6cm
11. 7cm
101. 6mm
or 152. 4 mm
Collar
Metal mol d
Hammer
Detachable bas e plat e —J
h = 30 or 45 cm W= 2. 5 or 4.5 kg
(a)
5 cm
Figure 2 1 . 1 Proct o r com pact io n apparat us : (a ) diag ram m at ic s k et ch , an d (b)
phot og raph o f m old , an d (c ) aut om at i c s oi l com pact o r (Court es y : Soilt es t )
M odified Procto r Compactio n Test (AST M Designation : D1 5 5 7 )
This tes t metho d cover s laborator y compactio n procedure s use d t o determin e th e relationshi p
between wate r conten t an d dr y uni t weigh t of soil s (compactio n curve ) compacted i n a 4 in . o r 6
in. diamete r mol d wit h a 1 0 I b ( 5 kg ) hamme r droppe d fro m a heigh t o f 1 8 in . (45 7 mm )
producing a compactive effor t o f 56,25 0 ft-lb/ft
3
(2,70 0 kN-m/m
3
). As i n th e cas e o f the standar d
test, th e cod e provide s thre e alternativ e procedures base d o n th e soi l materia l tested . Th e detail s
of th e procedure s ar e give n i n Tabl e 21.2 .
Soil Im provem en t 955
Table 21 . 2 Specificat io n fo r m odifie d Proct o r com pact io n t es t
Item Procedure
B
1.
2.
3.
4.
5.
6.
7.
8.
Mold diamete r
V olume of mold
Weight of hammer
Height of drop
No. of layers
Blows / layer
Energy of compaction
Soil material
4 in. (101.6mm)
0.0333 ft
3
(944 cm
3
)
10 Ib (4.54 kg)
18 in. (457.2mm)
5
25
56,250 f t lb/ft
3
(2700 kN-m/m
3
)
May b e use d if 20%o r
less retained on No. 4
sieve.
4 in. (101.6mm)
0.0333 ft
3
(944 cm
3
)
10 Ib (4.54 kg)
18 in. (457.2mm)
5
25
56,250 ft lb/ft
3
(2700 kN-m/m
3
)
Shall be use d if 20%
or more retained on
No. 4 sieve and
20%or less retained
on the 1/ 8 in. sieve
6 in. (101.6mm)
0.075 ft
3
( 2 124 cm
3
)
101b(4. 54kg)
18 in. (457.2mm)
5
56
56,250 ft lb/ft
3
(2700 kN-m/m
3
)
Shall be used i f more-
than 20%retaine d on
3/8 in. sieve and less-
than 30%retained on
the 3/4 in. sieve
(19 mm)
Test Procedur e
A soi l a t a selecte d wate r conten t i s place d i n fiv e layer s int o a mol d o f give n dimensions , wit h
each layer compacted b y 25 or 56 blows of a 10 Ib (4.54 kg ) hammer dropped fro m a height of 18
in. (45 7 mm) subjectin g th e soi l t o a tota l compactiv e effor t o f abou t 56,25 0 ft-lb/ft
3
(2700 kN-m/m
3
). Th e resultin g dr y uni t weigh t i s determined . Th e procedur e i s repeate d fo r a
sufficient numbe r of water contents to establish a relationship between th e dry unit weight and the
water content for the soil . Thi s data , whe n plotted, represent s a curvilinear relationship known as
the compaction curv e or moisture-dr y uni t weight curve. The valu e of the optimum wate r content
and maximu m dry uni t weight ar e determine d fro m th e compactio n curv e a s shown i n Fig. 21.2 .
Determination o f Z er o Ai r V oid s Lin e
Referring t o Fig. 21.3 , w e have
O W
Degree o f saturation, y
Water content ,
w —
W
Dry weigh t of solids , W
s
= V
S
G
S
y
w
= G
s
y
w
sinc e V
s
= 1
W wG Y
y
Therefore
y
w
r
w
wG
s
V
(21.2)
956 Chapt er 2 1
2.2
2.1
2.0
1.7
1.6
Max. dry uni t wei|
Max. dry
• unit weight
I _ l
90%saturatio n curve
I !
^ 95%saturation curve
Opt. moisture content
100%saturation line
(zero air voids)
8 1 2 1 6 2 0 2 4
Moisture content, w, percent
28 32
Figure 21 . 2 M ois t ure-dr y uni t weig h t relat ion s hi p
or V. =
wG
Dry uni t weight
W G y
s ' w
(21.3)
1 +
In Eq . (21.3) , sinc e G an d y , remain constant fo r a particula r soil , th e dr y uni t weight i s a
function o f wate r conten t fo r an y assume d degre e o f saturation . I f S = 1 , the soi l i s full y saturate d
(zero ai r voids). A curv e giving the relationshi p between y
d
an d w may be drawn by makin g us e of
Eq. (20.3 ) fo r 5 = 1. Curves ma y be drawn for differen t degree s of saturation such as 95, 90, 80 etc
Air
Water
Solids
Figure 2 1 . 3 B loc k diag ra m fo r det erm in in g zer o ai r void s lin e
Soil Im provem en t 95 7
percents. Fig . 21. 2 give s typica l curves for differen t degree s o f saturatio n along wit h moisture-dr y
unit weigh t curves obtaine d b y differen t compactiv e efforts .
Example 2 1 . 1
A procto r compactio n tes t wa s conducte d o n a soi l sample , an d th e followin g observation s wer e
made:
Water content , percent 7. 7 11. 5 14. 6 17. 5 19. 7 21. 2
Mass o f wet soil , g 173 9 191 9 208 1 203 3 198 6 194 8
If the volume of the mold used wa s 950 cm
3
and the specific gravity of soils grains was 2.65 ,
make necessar y calculation s an d draw , (i ) compactio n curv e an d (ii ) 80 %an d 100 %saturatio n
lines.
Solution
From th e know n mas s o f th e we t soi l sampl e an d volum e of th e mold , we t densit y o r we t uni t
weight i s obtained b y the equations,
/
. M Mas s of wet sample in gm , ,
/ 0, (g/ cm
3
) = —= — f 2 — ory
t
= ( k N/ m
3
) = 9.81 xp, ( g/ c m
3
)
V 95 0 cm -
Then fro m th e we t densit y an d corresponding moistur e content , th e dr y densit y o r dr y unit
weight i s obtained from ,
/ >y = —
o r
r
r f
= — d
l + w
d
l + w
Thus fo r eac h observation , th e we t densit y an d the n th e dr y densit y ar e calculate d an d
tabulated a s follows:
Water content , percen t 7. 7 11. 5 14. 6 17. 5 19. 7 21. 2
Mass o f wet sample, g 173 9 191 9 208 1 203 3 198 6 194 8
Wet density, g/cm
3
1.8 3 2.0 2 2.1 9 2.1 4 2.0 9 2.0 5
Dry density , g/cm
3
1.7 0 1.8 1 1.9 1 1.8 2 1.7 5 1.6 9
Dry uni t weight kN/m
3
16. 7 17. 8 18. 7 17. 9 17. 2 16. 6
Hence th e compactio n curve , whic h i s a plo t betwee n th e dr y uni t weigh t an d moistur e
content can be plotted a s shown in the Fig. Ex. 21.1. The curve gives,
Maximum dr y uni t weight , MDD = 18.7 kN/m
3
Optimum moistur e content , OMC = 14.7 percen t
For drawing saturation lines, make us e of Eq. (21.3) , viz. ,
wG
1 +
s
-
where, G ^ = 2.65 , given , S = degree o f saturatio n 80 %an d 100%, w = water content , ma y b e
assumed a s 8%, 12%, 16%, 20%and 24%.
958 Chapt er 21
22
21
20
19
18
17
16
15
5= 100%
10 /
X
15
0MC = 14.7%
Moisture content , %
Fig. Ex . 2 1 . 1
20
Hence fo r eac h valu e of sat urat i o n an d wate r content , find } g , an d t abul at e:
Water content , percentage 8 1 2 1 6 2 0 2 4
7^ kN/m
3
for 5 = 100 %
y
d
kN/m
3
for 5 = 80%
21.45 19.7 3 18.2 6 17. 0 15.6 9
20.55 18.6 1 17.0 0 15.6 4 14.4 9
With thes e calculations , saturatio n l i ne s fo r 100 %an d 80 %ar e plotted , a s show n i n th e
Fig. Ex . 21.1.
Also th e saturation , correspondin g t o MOD = 18. 7 kN/m
3
an d OM C = 14.7 %ca n b e
calculated as ,
G
s
y
w
_ 2.65x9.8 1
18-7 = =
H' GV
S
1 +
0.147x2.65
S
which give s S = 99.7%
Example 2 1 . 2
A smal l cylinde r havi n g volum e o f 60 0 cm
3
i s presse d int o a recentl y compacte d fil l o f
embankment fillin g th e cylinder. The mas s of the soi l i n the cylinder is 1100 g. The dr y mas s o f the
soil i s 910 g . Determine the voi d ratio and the sat ur at i on of the soil . Take th e specifi c gravit y of the
soil grain s a s 2.7.
Solution
Wet densit y of soi l
Water content , w =
1100
600
1100-910 19 0
910
1.83 g / c m
3
or y, - 17.9 9 kN/ m
3
= 0.209 = 20.9 %
Soil Im provem en t 95 9
Y 179 9
Dry uni t weight , v , = -
L L
- = -
:
-= 14.88 kN/ m
3
5 d
l + w 1 + 0.209
From, y. = -- w e have e =
y
d
Substituting an d simplifyin g
e= _ ! =0.7 8
14.88
r
wG
s 0.209x2.7x10 0 ^ ^
From, S e =wG<, o r 5 = --= -= 72.35%
e 0.7 8
2 1 .4 EFFEC T OF COM PACTI ON ON ENG I NEERI NG B EH AV I OR
Effect o f M oistur e Conten t o n Dr y Densit y
The moistur e content affect s th e behavior of the soil . When th e moisture conten t i s low, the soi l i s
stiff an d difficult t o compress. Thus , low unit weight and high air contents ar e obtained (Fig . 21.2) .
As th e moistur e conten t increases , th e wate r act s a s a lubricant , causing th e soi l t o softe n an d
become mor e workable . Thi s result s in a denser mass , highe r uni t weights an d lower ai r content s
under compaction . Th e wate r an d ai r combinatio n ten d t o kee p th e particle s apar t wit h furthe r
compaction, an d preven t an y appreciabl e decreas e i n th e ai r content o f th e tota l voids , however ,
continue to increase wit h moisture conten t and hence the dry uni t weight of the soi l falls .
To the righ t o f th e pea k o f th e dr y uni t weight-moistur e content curv e (Fig . 21.2) , lie s th e
saturation line. The theoretical curv e relating dry density with moisture content wit h no air voids is
approached bu t never reached sinc e it is not possible to expel by compaction al l the air entrapped i n
the voids of the soil .
Effect o f Compactiv e Effor t o n Dr y Uni t Weight
For all types of soil wit h all methods of compaction, increasing the amounts of compaction, tha t is,
the energy applie d pe r uni t weight of soil , result s i n a n increas e i n the maximu m dr y uni t weight
and a corresponding decreas e in the optimum moisture content a s can be see n i n Fig. 21.4 .
Shear Strengt h o f Compacte d Soi l
The shea r strengt h of a soil increase s wit h the amount of compaction applied . The mor e th e soi l i s
compacted, the greater i s the value of cohesion and the angle of shearing resistance. Comparin g the
shearing strengt h wit h the moistur e content fo r a given degree of compaction, i t i s foun d tha t the
greatest shea r strengt h i s attained a t a moistur e content lowe r than the optimum moistur e conten t
for maximu m dr y uni t weight . Fig . 21. 5 show s th e relationshi p betwee n shea r strengt h an d
moisture-dry uni t weight curves for a sandy clay soil . It might be inferred from this that it would be
an advantag e t o carr y ou t compactio n a t th e lowe r valu e o f th e moistur e content . Experiments ,
however, hav e indicate d tha t soil s compacte d i n thi s wa y ten d t o tak e u p moistur e an d become
saturated wit h a consequent los s o f strength.
Effect o f Compactio n o n Structur e
Fig. 21. 6 illustrate s th e effect s o f compactio n o n cla y structur e (Lambe , 1958a) . Structur e (o r
fabric) i s the term used to describe th e arrangement of soil particles an d the electric force s betwee n
adjacent particles .
960 Chapt er 21
120
115
110
105
~ 10 0
95
90
85
1 '
2 3
3 c
4 '
/
*
A
Q
rj
X
x^
rr
X
/
<x
'v
°v
^
^
\-i"
^
^«— —
~ \
^
N
N
N
"^
Blows pe r Hamme r Hamme r
No. Layer s Laye r weight(lb ) dro p (in.)
1 5 5 5 1 0 1 8 (mod. AASHTO)
2 5 2 6 1 0 1 8
3 5 1 2 1 0 1 8(std. AASHTO)
4 3 2 5 5'/ 2 1 2
Note: 6 in . diameter mold used fo r all tests
Fi gure 21. 4Dyna mi c compact i o n
curves f o r a si l t y cl a y ( f r o m Tu r n b u l l ,
1950)
10 1 5 2 0
Water conten t (%)
1.9
1.7
1.5
Unconflned compressiv e strengt h
Moisture-unit weigh t curv e
12 1 4 1 6 1 8
Moisture conten t percen t
2.5
2.0
1.5 >
1.0
20
0.5
Figure 2 1 . 5 Relat ion s hi p
bet ween com pact io n an d
s hear s t ren g t h curve s
T3
T3
High compactiv e
effort
Low
compactive
effort
Molding wate r conten t
Figure 21 .6 Effect s o f
com pact ion o n s t ruct ur e
(from Lam be , 1 958a)
Soil Im provem en t 961
The effect s o f compactio n condition s o n soi l structure , an d thu s o n th e engineerin g
behavior o f th e soil , var y considerabl y wit h soi l typ e an d th e actua l conditions unde r whic h the
behavior i s determined .
At lo w wate r content , W
A
i n Fig . 21.6 , th e repulsiv e force s betwee n particle s ar e smalle r
than th e attractiv e forces , an d a s suc h th e particle s flocculat e i n a disorderl y array . As th e wate r
content increase s beyon d W
A
, the repulsion betwee n particle s increases , permittin g the particles t o
disperse, makin g particles arrang e themselve s i n an orderly way. Beyond W
B
the degree of particl e
parallelism increases , bu t the density decreases. Increasing the compactive effor t a t any given water
content increase s the orientation o f particles an d therefor e give s a higher densit y a s indicated i n
Fig. 21.6 .
Effect o f Compactio n o n Permeabilit y
Fig. 21.7 depict s th e effect o f compaction on the permeability o f a soil. The figure shows the typical
marked decrease in permeability that accompanies a n increase i n molding water content on the dry
side o f th e optimu m wate r content . A minimu m permeabilit y occurs a t wate r content s slightl y
above optimu m moistur e conten t (Lambe , 1958a) , afte r whic h a sligh t increas e i n permeabilit y
occurs. Increasing th e compactive effor t decrease s th e permeability o f the soil .
10-
§10-
io-
9
130
126
60
• 53 12 2
118
Q
*-
Shows chang e i n moisture
x
( , o
(expansion prevented )
11 1 3 1 5
Water content , (%)
17 19
Figure 2 1 . 7 Com pact ion -perm eabilit y t es t s o n Siburu a cla y (fro m Lam be , 1962 )
962
Chapt er 2 1
Dry compacte d or
undisturbed sampl e
Wet compacted o r
remolded sampl e
Stress, natura l scal e
(a) Low-stress consolidation
Dry compacte d o r
undisturbed sampl e
Wet compacted o r
remolded sampl e
Rebound for both sample s
Stress, log scal e
(b) High-stress consolidation
Figure 21 .8 Effec t o f on e-dim en s ion a l com pres s io n o n s t ruct ur e (Lam be , 1958b )
Effect o f Compactio n o n Compressibilit y
Figure 21. 8 illustrates th e differenc e i n compactio n characteristic s betwee n tw o saturate d cla y
samples a t the same density, one compacted o n the dry side of optimum and one compacted o n the
wet sid e (Lambe , 1958b) . A t lo w stresse s th e sampl e compacte d o n th e we t sid e i s mor e
compressible tha n the one compacted o n the dry side. However, at high applied stresse s the sampl e
compacted o n the dryside is more compressibl e tha n th e sampl e compacted o n the wet side .
2 1 .5 FIEL D COM PACTI ON AN D CONTROL
The necessar y compactio n of subgrades of roads, earth fills, an d embankments ma y be obtained by
mechanical means . The equipmen t that ar e normall y used for compaction consist s o f
1. Smoot h whee l roller s
2. Rubbe r tire d roller s
3. Sheepsfoo t roller s
4. V ibrator y rollers
Laboratory test s on the soi l t o be used for construction in the fiel d indicat e the maximum dry
density tha t ca n b e reache d an d th e correspondin g optimu m moistur e conten t unde r specifie d
methods o f compaction . Th e fiel d compactio n metho d shoul d b e s o adjuste d a s t o translat e
Soil Im provem en t 96 3
Figure 2 1 . 9 Sm oot h whee l rolle r (Court es y : Cat erpil l ar , USA)
laboratory conditio n int o practice a s far a s possible. The two important factors tha t ar e necessar y
to achiev e th e objective s i n th e fiel d ar e
1. Th e adjustmen t o f th e natura l moistur e content i n th e soi l t o th e valu e at whic h the fiel d
compaction i s most effective .
2. Th e provision o f compacting equipmen t suitable for the wor k at the site .
The equipment use d fo r compaction ar e briefly described below:
Smooth Wheel Rolle r
There ar e two types of smoot h whee l rollers . On e type has two large wheels , on e i n the rear an d a
similar singl e dru m i n th e front . Thi s typ e i s generall y use d fo r compactin g bas e courses . Th e
equipment weighs from 5 0 to 12 5 kN (Fig.21.9). The other type is the tandem roller normall y used
for compactin g pavin g mixtures. Thi s rolle r ha s larg e singl e drums i n th e fron t an d rea r an d th e
weights of the roller s rang e fro m 1 0 to 200 kN.
964 Chapt e r 21
Figure 2 1 .1 0 Sheeps foo t rolle r (Court es y : Vibr om a x Am eric a In c. )
Rubber Tire d Rolle r
The maximu m weigh t of thi s roller ma y reac h 200 0 kN. The smalle r roller s usuall y have 9 t o 1 1
tires on two axles wit h the tires spaced s o that a complete coverag e i s obtained wit h each pass. The
tire load s o f th e smalle r rolle r ar e i n th e rang e o f 7. 5 k N an d th e tir e pressure s i n th e orde r o f
200 kN/m
2
. The larger roller s have tire loads ranging from 10 0 to 500 kN per tire, and tire pressure s
range from 400 t o 100 0 kN/m
2
.
Sheepsfoot Roller
Sheepsfoot roller s ar e availabl e in drum widths ranging from 12 0 to 18 0 cm and i n drum diameter s
ranging fro m 9 0 t o 18 0 cm. Projection s lik e a sheepsfoo t ar e fixe d o n th e drums . The length s of
these projections rang e from 17. 5 cm to 23 cm. The contact area of the tamping foot ranges from 35
to 56 sq. cm. The loaded weigh t per drum ranges from abou t 30 kN for the smalle r sizes t o 13 0 kN
for th e large r size s (Fig . 21.10) .
V ibratory Roller
The weight s of vibratory rollers range fro m 12 0 to 300 kN. In some units vibration i s produced by
weights place d eccentricall y o n a rotating shaf t i n suc h a manner tha t the force s produce d b y th e
rotating weights are essentially i n a vertical direction. V ibratory rollers ar e effective for compactin g
granular soil s (Fig . 21.11) .
Soil Im provem en t 96 5
Figure 2 1 .1 1 Vibrat or y dru m o n s m oot h whee l rolle r (Court es y : Cat erpillar , USA )
Selection o f Equipmen t fo r Compactio n i n the Fiel d
The choic e o f a roller fo r a given job depend s on the type of soi l t o be compacted an d percentag e
of compactio n t o b e obtained . Th e type s o f roller s tha t ar e recommende d fo r th e soil s normall y
met are :
Type o f soil Typ e of roller recommende d
Cohesive soi l Sheepsfoo t roller , o r Rubbe r tire d rolle r
Cohesionless soil s Rubber-tire d rolle r o r V ibratory roller .
Method o f Compactio n
The firs t approac h t o th e proble m o f compactio n i s t o selec t suitabl e equipment . I f th e
compaction i s require d fo r a n eart h dam, th e numbe r of passe s o f th e rolle r require d t o compac t
the give n soi l t o th e require d densit y a t th e optimu m moistur e conten t ha s t o b e determine d b y
conducting a fiel d tria l tes t a s follows:
The soi l i s wel l mixe d wit h wate r whic h woul d giv e th e optimu m wate r conten t a s
determined i n th e laboratory . I t i s then sprea d ou t i n a layer . The thicknes s o f th e laye r normall y
varies fro m 1 5 to 22.5 cm. The numbe r of passes require d t o obtain the specified densit y has t o be
found b y determining the densit y of the compacted materia l afte r ever y definit e number of passes.
The densit y ma y b e checke d fo r differen t thicknes s i n th e layer . Th e suitabl e thicknes s o f th e
layer and the number of passes require d t o obtain the required densit y wil l have t o be determined .
966 Chapt e r 21
In cohesiv e soils , densitie s o f th e orde r o f 95 percen t o f standar d Procto r ca n b e obtaine d
with practicall y an y o f th e roller s an d tampers ; however , vibrator s ar e no t effectiv e i n cohesiv e
soils. Wher e hig h densitie s ar e require d i n cohesiv e soil s i n th e orde r o f 9 5 percen t o f modifie d
Proctor, rubbe r tired roller s wit h tir e loads i n the order o f 10 0 kN an d tir e pressur e i n the orde r of
600 kN/m
2
ar e effective .
In cohesionles s sand s an d gravels , vibratin g typ e equipmen t i s effectiv e i n producin g
densities u p t o 10 0 percen t o f modifie d Proctor . Wher e densitie s ar e neede d i n exces s o f 10 0
percent of modified Proctor suc h as for base courses for heavy duty air fields and highways, rubber
tired roller s wit h tir e loads o f 13 0 k N an d abov e an d tir e pressure o f 100 0 kN/m
2
ca n b e use d t o
produce densitie s up t o 10 3 to 10 4 percent of modifie d Proctor.
Field Contro l o f Compactio n
M ethods o f Contro l of Densit y
The compaction o f soil in the field mus t be such as to obtain the desired uni t weight at the optimum
moisture content . Th e fiel d enginee r ha s therefor e t o mak e periodi c check s t o se e whethe r th e
compaction i s giving desire d results. The procedure of checking involves :
1. Measuremen t o f the dr y uni t weight , and
2. Measuremen t o f the moisture content.
There ar e many method s for determining the dry uni t weight and/or moistur e content of the
soil in-situ . The importan t methods are :
1. San d con e method ,
2. Rubbe r balloo n method ,
3. Nuclea r method, and
4. Procto r needl e method.
Sand Con e M etho d (ASTM Designatio n D- 1 5 5 6 )
The san d fo r th e san d cone metho d consist s of a sand pourin g jar show n i n Fig. 21.12 . The jar
contains uni f or ml y graded clea n an d dr y sand . A hol e about 1 0 cm i n diamete r i s mad e i n th e
soil t o b e teste d u p t o th e dept h required . Th e weigh t o f soi l remove d fro m th e hol e i s
determined an d it s wate r conten t i s als o determined . Sand i s run int o th e hol e fro m th e jar b y
opening th e valv e abov e th e con e unti l th e hol e an d th e con e belo w th e valv e i s completel y
filled. The valv e i s closed. Th e jar i s calibrated t o give the weigh t of the san d tha t j ust fill s th e
hole, tha t is , th e differenc e i n weigh t of th e jar befor e an d afte r fi l l i n g th e hol e afte r allowin g
for th e weigh t o f san d containe d i n th e con e i s th e weigh t o f san d poure d int o th e hole .
Let
W
s
= weight of dr y san d poured int o the hol e
G
s
= specific gravit y o f sand particles
W = weight of soi l taken out o f the hol e
w = wate r content of th e soi l
V olume of san d i n the hol e = volume of soi l taken out o f the hol e
W
that is, V = —?—( 21 Aa)
Soil Im provem en t
967
Sand-cone
apparatus
3785 cm
3
(1-gal)
II-*— 16 5 mm —H
U— 17 1 mm —*- l
165 mm
171 mm
ASTM dimension s
Mass of sand
to fill cone and
template groov e
Base
template
(a)
Fig. 2 1 .1 2 San d-con e apparat us : (a ) Schem at ic diag ram , an d (b ) Phot og rap h
(21.4b)
' " s
V
The dry unit weight of soil, y
d
-
W WGy
The bulk unit weight of soil, /= — = —
V W
s
r
t
l+w
Rubber Balloo n M etho d (ASTM Designation : D 2 1 6 7 )
The volum e o f a n excavate d hol e i n a give n soi l i s determine d usin g a liquid-fille d calibrate d
cylinder fo r fillin g a thi n rubber membrane . Thi s membran e i s displace d t o fil l th e hole . Th e in -
place unit weight is determined b y dividin g the wet mass of the soi l removed b y the volume of the
hole. The wate r (moisture ) content an d the in-place uni t weight ar e used t o calculat e th e in-place
dry uni t weight . Th e volum e i s rea d directl y o n th e graduate d cylinder . Fig . 21.1 3 show s th e
equipment.
968 Balloon
density apparatus
Chapt er 2 1
Carrying
handle I
Rubber
membrane
Base
tempi at e~~X
!/P%
[— o
40
OQ
— 120
1480
1520
F— 1550
1
1600 )
1
"\S^^v
*« r"**"—— "^
« \
11
r >
II ' ( _
-1
-11
1
II
i
i
1
II
II
III
II
II
Graduated
cylinder
(direct reading)
Hand
pump
Pump
control
valve
(a)
Figure 2 1 .1 3 Rubbe r balloo n den s it y apparat us , (a ) diag ram m at ic s k et ch , an d
(b) a phot og raph
Nuclear M etho d
The moder n instrumen t fo r rapi d an d precis e fiel d measuremen t o f moistur e conten t an d uni t
weight i s th e Nuclea r density/Moistur e meter . Th e measurement s mad e b y th e mete r ar e
non-destructive an d requir e no physica l o r chemica l processin g o f the materia l bein g tested . Th e
instrument may b e used either i n drilled holes or on the surface of the ground. The mai n advantag e
of thi s equipmen t is t ha t a singl e operato r ca n obtai n a n immediat e an d accurat e determinatio n of
the in-situ dr y densit y an d moisture content .
Proctor Needl e M etho d
The Procto r needl e metho d i s one o f th e method s develope d fo r rapi d determinatio n o f moistur e
contents o f soil s in-situ. I t consist s o f a needl e attache d t o a sprin g loade d plunger , th e ste m o f
which i s calibrate d t o rea d th e penetratio n resistanc e o f th e needl e i n lbs/in
2
o r kg/cm
2
. Th e
needle i s supplie d wit h a serie s o f bearing point s s o tha t a wid e rang e o f penetratio n resistance s
can b e measured . Th e bearin g area s tha t ar e normall y provide d ar e 0.05 , 0.1 , 0.25 , 0.5 0 an d
1.0 sq. in. The apparatu s i s shown in Fig. 21.14. A Proctor penetromete r se t is shown i n Fig. 21.1 5
(ASTMD-1558).
Soil Im provem en t 969
T
Figure 2 1 .1 4 Proct o r n eedl e Figur e 21.15 Proct o r pen et rom et e r s e t (Court es y :
Soilt es t )
L aboratory Penetratio n Resistanc e Curv e
A suitabl e needle poin t is selected fo r a soil to be compacted. I f the soi l i s cohesive, a needle wit h
a large r bearing are a i s selected . Fo r cohesionless soils , a needle wit h a smaller bearing are a wil l
be sufficient . Th e soi l sampl e i s compacted i n the mold .
O
O
ao 1 .
&i.
1.5
\
/
\Penetra t
_
r
\
Moisture de
ion resistanc
~N
V \
nsity curve
e curve
\
A
3 6 1 2 1 8 2 "
"
^
O

t
o
-
P
^

O
s

0
O

O

O

C
P
e
r
c
e
n
t
a
g
e

r
e
s
i
s
t
a
n
c
e
,

k
g
/
c
m
2
Moisture content. %
Figure 21 .1 6 Fiel d m et ho d o f det erm in in g wat e r con t en t by Proct o r needl e method
970 Chapt e r 21
The penetromete r wit h a know n bearin g are a o f th e ti p i s force d wit h a gradua l unifor m
push a t a rat e o f abou t 1.2 5 c m pe r se c t o a dept h o f 7. 5 c m int o th e soil . Th e penetratio n
resistance i n kg/cm
2
i s read of f th e calibrate d shaf t o f th e penetrometer . Th e wate r content o f th e
soil an d th e correspondin g dr y densit y ar e als o determined . Th e procedur e i s repeate d fo r th e
same soi l compacte d a t differen t moistur e contents . Curve s givin g th e moisture-densit y an d
penetration resistance-moistur e conten t relationshi p ar e plotte d a s shown i n Fig . 21.16 .
To determine the moisture content i n the field, a sample o f the wet soil is compacted into the
mold unde r the same condition s a s use d i n the laborator y fo r obtaining the penetration resistanc e
curve. Th e Procto r needl e i s force d int o th e soi l an d it s resistanc e i s determined . Th e moistur e
content i s read from the laborator y calibratio n curve.
This metho d i s quit e rapid , an d i s sufficientl y accurat e fo r fine-graine d cohesiv e soils .
However, th e presence of gravel or small stones in the soil makes th e reading on the Proctor needl e
less reliable . I t is not ver y accurat e i n cohesionless sands.
Example 2 1 . 3
The following observations wer e recorded whe n a sand cone test was conducted fo r finding the unit
weight of a natural soil :
Total densit y of san d used i n the tes t = 1. 4 g/cm
3
Mass o f the soi l excavated fro m hol e = 950 g.
Mass o f the san d fillin g th e hol e = 700 g.
Water conten t of the natura l soil = 1 5 percent.
Specific gravit y of the soi l grain s = 2, 7
Calculate: (i ) the wet uni t weight , (ii ) the dry uni t weight , (iii) the voi d ratio, and (iv) the degree of
saturation.
Solution
V olume o f the hole V = — = 500 cm
3
p
1. 4
950
Wet densit y of natural soil , p. - -= 1.9 g/cm
3
or y
t
= 18.64 kN/m
3
y
' 5 0 0 '
p 1. 9
Dry densit y p , = — — = -= 1.65 g/ cm
3
y y Hd
1 + w 1 + 0.15
f~! J "I
x l o r 1.6 5 + 1.65^ = 2.7
w
A . ' w i
l+e l+e
Therefore e -— -'• — = 0.64
1.65
„ vvG
9
0.15x2.7x10 0 ^
m
And S = --= -= 63%
e 0.6 4
Soil Im provem en t 97 1
Example 2 1 . 4
Old records o f a soil compacte d i n the pas t gave compaction wate r content of 15 %and saturation
85%. What might be the dr y densit y of the soil ?
Solution
The specifi c gravity of the soi l grains is not known, but as it varies i n a small range o f 2. 6 to 2.7, it
can suitabl y be assumed. An average valu e of 2.65 i s considered here .
wG
s 0.15x2.6 5
nA
^
Hence e = = = 0.47
S 0.8 5
f O A^
and dr y density p, = —— p - —'• x 1 = 1.8 g/cm
3
o r dry unit weight = 17.66 kN/m
3
d
l + e
w
1 + 0.47
5
Example 2 1 . 5
The followin g data ar e availabl e i n connection wit h the construction of an embankment:
(a) soi l from borro w pit : Natural densit y = 1.7 5 Mg/m
3
, Natura l water conten t = 12%
(b) soi l after compaction: density = 2 Mg/m
3
, wate r content = 18%.
For every 10 0 m
3
of compacted soi l of the embankment, estimate :
(i) th e quantit y of soi l t o be excavated fro m th e borrow pit , an d
(ii) th e amount of wate r t o be adde d
Note: 1 g/cm
3
= 1000 kg/m
3
= 10
3
x 10
3
g/m
3
= 1 Mg/m
3
wher e Mg stands for Megagra m
10
6
g.
Solution
The soi l i s compacted i n the embankment wit h density of 2 Mg/m
3
and wit h 18 %wate r content .
Hence, fo r 10 0 m
3
of soi l
Mass o f compacted we t soi l = 10 0 x 2. 0 = 200 Mg = 200 x 10
3
kg.
200 20 0
Mass of compacted dr y soi l = = = 169. 5 Mg = 169.5 x 10
3
k g
v 3
1 + w 1 + 0.18
6 6
Mass of wet soi l t o be excavated = 169.5(1 + w) = 169.5(1 + 0.12) = 189.84 M g
189 84
V olume of the wet soi l t o be excavated = '• — = 108.48 m
3
1.75
Now, in the natural state, the moistur e present in 169. 5 x 10
3
kg of dr y soi l woul d be
169.5 x 10
3
x 0.12 = 20.34 x 10
3
kg
and the moistur e which the soi l wil l possess durin g compaction i s
169.5 x 10
3
x 0. 1 8 = 30.51 x 10
3
kg
Hence mas s of wate r to be adde d fo r ever y 10 0 m
3
of compacted soi l i s
(30.51 - 20.34 ) 10
3
= 10.17 x 10
3
kg.
972 Chapt er 21
Example 2 1 . 6
A sampl e o f soi l compacte d accordin g t o the standar d Procto r tes t ha s a densit y o f 2.06 g/cm
3
a t
100%compactio n an d a t a n optimum water content of 14%. Wha t i s the dr y uni t weight ? What i s
the dr y uni t weigh t a t zer o air-voids ? I f th e void s becom e fille d wit h wate r wha t woul d b e th e
saturated uni t weight ? Assume G
s
= 2.67 .
Solution
Refer t o Fig. Ex. 21.6. Assume V= total volume = 1 cm
3
. Since wate r content i s 14 %we may write,
—^ = 0.14 o r M =0. 14 M
MS
and since , M
W
+ M
S
= 2.06 g
0.14M + M =1.14 M =2.0 6
s s s
2.06
or M = = 1.80 7 e
1.14
M
w
= 0.14 x 1.807 = 0.253 g.
M, 1.80 7
By definition , p
d
=
v i
The volume of solid s (Fig . Ex. 21.6) i s
1.807
= 1.80 7 g/ cm
j
o r y
d
= 1-807x9.81 = 17.73 kN/n r
V =
2.67
= 0.68 g/cn r
The volume of voids = 1 -0.6 8 = 0.32 cm
3
The volume o f wate r = 0.253 cm
3
The volume of air = 0.320 - 0.25 3 = 0.067 cm
3
If al l th e ai r i s squeeze d ou t o f th e sample s th e dr y densit y a t zer o ai r void s woul d be , b y
definition,
Air
Water
M w
Solids
Figure Ex . 2 1 . 6
Soil Im provem en t 973
1.807
= 1.94 g/cm
3
o r y, = 1.94 x 9.81 = 19.03 kN/m
3
a
0.6 8 +0.253
on the other hand, if the ai r voids also wer e fille d wit h water,
The mas s of water woul d be = 0.32 x 1 = 0.32 g
The saturate d density is
1.807 + 0.32
Aat ~
1
= 2.13 g/cm-
3
o r y
sat
= 2.13 x 9.81 = 20.90 kN/n r
2 1 .6 COM PACTI O N FO R DEEPER L AY ER S OF SOI L
Three types of dynami c compaction for deeper layer s of soi l are discussed here . The y are:
1. V ibroflotation .
2. Droppin g of a heavy weight.
3. Blasting .
V ibroflotation
The V ibroflotatio n techniqu e i s use d fo r compactin g granula r soi l only . Th e vibroflo t i s a
cylindrical tub e containin g wate r jets a t to p an d botto m an d equippe d wit h a rotatin g eccentri c
weight, which develops a horizontal vibratory motion as shown in Fig. 21.17. The vibroflot is sunk
into the soil using the lower jets and is then raised i n successive smal l increments, during which the
surrounding material i s compacted b y the vibration process. The enlarged hole around the vibroflot
is backfilled with suitable granular material. This method is very effective for increasing the density
of a sand deposi t fo r depths u p t o 30 m. Probe spacing s o f compaction hole s shoul d b e o n a grid
pattern of about 2 m to produce relative densities greater than 70 percent over the entire area. I f the
sand i s coarse, th e spacings ma y be somewhat larger.
In sof t cohesiv e soi l and organic soil s the V ibroflotation techniqu e has been use d wit h gravel
as the backfill material. The resulting densified stone column effectively reinforce s softer soils and
acts as a bearing pil e for foundations.
Figure 21 .1 7 Com pact io n b y usin g vibroflo t (B rown , 1977 )
974 Chapt e r 21
Dropping o f a H eavy Weigh t
The repeate d droppin g o f a heav y weigh t o n t o th e groun d surfac e i s on e o f th e simples t o f th e
methods o f compacting loos e soil .
The method , know n a s dee p dynami c compactio n o r dee p dynami c consolidatio n ma y b e
used t o compact cohesionles s o r cohesive soils . The method use s a crane t o lif t a concret e o r steel
block, weighing up to 500 kN and up to heights of 40 to 50 m, from whic h height it is allowed to fal l
freely o n t o th e groun d surface . The weigh t leaves a dee p pi t a t th e surface . Th e proces s i s then
repeated eithe r a t th e same locatio n or sequentiall y ove r othe r part s o f th e are a t o be compacted .
When the required numbe r of repetitions i s completed ove r the entire area, th e compaction a t depth
is completed. Th e soil s near the surface, however , are in a greatly disturbed condition. The top soi l
may then be levelled and compacted, using normal compactiing equipment. The principal claims of
this method are :
1. Dept h o f recompaction ca n reac h u p t o 1 0 to 1 2 m.
2. Al l soil s ca n be compacted .
3. Th e method produce s equa l settlements more quickl y than do static (surcharge type ) loads .
The dept h of recompaction, D, in meters is approximately given by Leonards, e t al., (1980) as
(21.5)
where W = weight o f falling mas s i n metri c tons,
h = height of drop i n meters .
Blasting
Blasting, through the use of buried, time-delayed explosiv e charges, ha s been use d t o densify loose ,
granular soils . Th e sand s an d gravel s mus t b e essentiall y cohesionles s wit h a maximu m o f 1 5
percent o f their particle s passin g th e No. 200 sieve size and 3 percent passin g 0.00 5 mm size . Th e
moisture conditio n o f the soi l i s also importan t for surfac e tension force s i n the partiall y saturate d
state limi t the effectiveness of the technique. Thus the soil , as well as being granular , must be dry or
saturated, whic h require s sometime s prewettin g th e sit e vi a constructio n o f a dik e an d reservoi r
system.
The techniqu e require s carefu l plannin g an d i s use d a t a remot e site . Theoretically , a n
individual charge densifie s the surrounding adjacent soil and soil beneat h th e blast. It should not lif t
the soil situate d abov e th e blast, however, since the upper soi l shoul d provide a surcharge load . The
charge shoul d no t creat e a crate r i n th e soil . Charg e delay s shoul d b e time d t o explod e fro m th e
bottom o f the layer being densifie d upwar d in a uniform manner. The uppermost par t of the stratum
is alway s loosened , bu t thi s ca n b e surface-compacte d b y vibrator y rollers . Experienc e indicate s
that repeated blast s o f smal l charges ar e more effective than a single large charge fo r achieving the
desired results .
2 1 .7 PREL OADI N G
Preloading i s a technique that can successfull y b e use d t o densif y sof t t o ver y sof t cohesiv e soils .
Large-scale construction site s composed o f wea k silt s and clays or organi c material s (particularl y
marine deposits) , sanitar y lan d fills , an d othe r compressibl e soil s ma y ofte n b e stabilize d
effectively an d economicall y b y preloading . Preloadin g compresse s th e soil . Compressio n take s
place whe n the water in the pores o f the soil i s removed whic h amounts to artificial consolidation of
soil i n the field . I n orde r t o remov e th e wate r squeeze d ou t o f the pores an d haste n th e perio d o f
consolidation, horizontal and vertical drains are required to be provided i n the mass. The preload i s
Soil Im provem en t 975
Filter be d
Earthfill
A
H

!• '>• '.'• • '.• <• :
N
'•'^\
,
r
/XS\//kS\/
t'?.$?£
f
*
• xx\\//xv\//
v •:,•:*;.;•?•'*
•"xvs //^s\/x
J£\
AN //AN
Load for m fil l
1 1 I 1 I ,
t • .•:•''.•••.•.•.•: ; v^r... J-
[Xs- .>-• • *. • -: v-t
j
1 ... .
^
»
_
Impermeable be d
(a) V ertical sectio n (b) Section of singl e
drain
2R
I I I I
\ / \ » ]/' x x J
1
x x
v
j/' x x;( ' / "\ )
'' V -f; V ^i V^ V ^; V
N|
,
\x x
v
;/ x x
v
;/ x x
v
;/ x x
v
|; x x /
x
- _ -
x
i
x
- _ -
x /
i
s
- _ -
x
^- - - ' ' i "
1
- - ^' '
(c) plan o f san d drain s
Figure 2 1 .1 8 Con s olidat io n o f soi l usin g s an d drain s
generally i n the for m o f a n imposed eart h fil l whic h must be lef t i n place long enoug h t o induc e
consolidation. The process o f consolidation can be checked by providing suitable settlement plates
and piezometers. The greater the surcharge load, shorter the time for consolidation. Thi s is a case of
three-dimension consolidation.
Two types of vertical drains considered ar e
1. Cylindrica l sand drains
2. Wic k (prefabricated vertical ) drains
Sand Drain s
V ertical and horizontal and drains are normally used for consolidating ver y soft clay , silt and other
compressible materials . Th e arrangemen t of sand drains shown in Fig. 21.18 is explained below :
1. I t consist s o f a series o f vertica l sand drains or piles. Normall y medium to coars e san d i s
used.
2. Th e diameter o f the drains ar e generally not less than 30 cm and the drains ar e placed i n a
square grid pattern at distances of 2 to 3 meters apart . Economy requires a careful stud y of
the effec t o f spacing the sand drains on the rate of consolidation.
3. Dept h of the vertical drain s should extend up to the thickness of the compressible stratum.
976 Chapt e r 21
4. A horizonta l blanket of fre e drainin g sand shoul d be place d o n th e to p o f th e stratu m and
the thicknes s of thi s may b e u p t o a meter , and
5. Soi l surcharg e i n th e for m o f a n embankmen t is constructed on to p o f th e san d blanke t i n
stages.
The heigh t o f surcharg e shoul d be s o controlle d a s t o kee p th e developmen t o f por e wate r
pressure i n th e compressibl e strat a a t a lo w level . Rapi d loadin g ma y induc e hig h por e wate r
pressures resultin g i n th e failur e o f th e stratu m by rupture . The latera l displacemen t o f th e soi l
may shea r of f th e san d drains and bloc k the drainage path.
The applicatio n of surcharge squeezes ou t water i n radial directions t o the nearest san d drai n
and als o i n the vertica l direction to the sand blanket . The dashe d line s show n i n Fig. 21.18(b ) are
drawn midwa y between th e drains . The plane s passin g through these line s ma y b e considere d a s
impermeable membrane s an d al l the water wi t hin a block has to flow t o the drain at the center. The
problem o f computin g th e rat e o f radia l drainag e ca n b e simplifie d withou t appreciabl e erro r b y
assuming tha t eac h bloc k ca n be replaced b y a cylinder of radius R suc h that
n R
2
= L ?
where L is the side length o f the prismati c block.
The relatio n betwee n th e tim e t an d degre e o f consolidatio n U
z
% i s determine d b y th e
equation
U
z
% = 100/(T)
wherein,
T = ^4r(21.6 )
H
2
If th e botto m o f th e compressibl e laye r i s impermeable , the n H i s th e ful l thicknes s o f th e
layer.
For radial drainage , Renduli c (1935) has shown that the relation betwee n th e time t and the
degree of consolidation U
r
% can b e expressed a s
U
r
%=WOf( T) (21.7 )
wherein,
is th e tim e factor . Th e relatio n betwee n th e degre e o f consolidatio n U% an d th e tim e facto r T
r
depends o n the valu e of the rati o Rlr. The relation between T
r
and U% fo r ratios o f Rlr equal to 1 ,
10 and 10 0 in Fig. 21.19 are expressed b y curves C
;
, C
10
and C
wo
respectively.
I nstallation o f V ertica l Sand Drain s
The san d drains are installed as follows
1. A casin g pip e o f th e require d diamete r wit h th e botto m close d wit h a loose-fit-con e i s
driven u p t o the required depth,
2. Th e cone i s slightl y separated fro m th e casing by driving a mandrel int o the casing, an d
3. Th e sand of the required gradation is poured int o the pipe for a short dept h an d at the same
time the pipe is pulled up in steps. As the pipe is pulled up, the sand i s forced out of the pipe
by applyin g pressure on t o the surface of the sand. The procedur e i s repeated til l the hole s
is completely fi l l e d wit h sand.
Soil Im provem en t 977
0.6
Time factor T
r
0.8 1.0 1.2
Figure 2 1 .1 9 7 "vers u s U
The sand drains may also be installed by jetting a hole in the soil or by driving an open casing
into the soil , washin g the soi l ou t of the casing, and filling th e hol e wit h sand afterwards .
Sand drain s hav e bee n use d extensivel y i n man y part s o f th e worl d fo r stabilizin g soil s fo r
port development work s and for foundations of structures in reclaimed area s o n the sea coasts. I t is
possible tha t san d drain s ma y no t functio n satisfactoril y i f th e soi l surroundin g th e wel l get s
remolded. Thi s conditio n i s referre d t o a s smear. Thoug h theorie s hav e bee n develope d b y
considering differen t thicknes s o f smea r an d differen t permeability , i t i s doubtfu l whethe r suc h
theories ar e of any practical us e since i t would be very difficult t o evaluate the qualit y of the smea r
in th e field .
Wick (Prefabricate d V ertical ) Drain s
Geocomposites use d a s drainag e medi a hav e completel y take n ove r certai n geotechnica l
application areas . Wic k drains , usuall y consistin g o f plasti c flute d o r nubbe d core s tha t ar e
surrounded by a geotextile filter , have considerable tensil e strength. Wick drains do not require any
sand to transmit flow. Mos t syntheti c drains are of a strip shape. The stri p drains are generall y 100
mm wid e an d 2 t o 6 mm thick . Fig . 21.2 0 show s typical cor e shape s o f stri p drain s (Hausmann ,
1990).
Wick drain s ar e installe d by using a hollow lance. The wic k drai n i s threaded int o a hollow
lance, whic h is pushed (or driven) through the soi l layer , which collapses aroun d it . At the ground
surface th e ends o f the wic k drains (typicall y at 1 to 2 m spacing) are interconnected b y a granular
soil drainag e laye r o r geocomposit e shee t drai n layer . Ther e ar e a numbe r o f commerciall y
available wic k drai n manufacturer s and installatio n contractors wh o provid e informatio n o n th e
current products, styles , properties , an d estimated cost s (Koerner , 1999) .
With regards t o determining wick drain spacings , th e initial focal point i s on the time for the
consolidation o f the subsoi l t o occur. Generall y the time for 90%consolidation (/
90
) i s desired. I n
order to estimate th e time t, it is first necessar y t o estimate a n equivalent sand drain diameter for the
wick drai n used. The equations suggeste d b y Koerner (1999) ar e
(21.9a)
978 Chapt er 2 1
Filter
sleeve
Core
Fluted PV C
or paper
V arious shape s
of cores with
nonwoven
geotextile
filter sleeve s
V\/\/\/\/\/\/\/\/\7\l
• IHIIIUM+M
iin rm n n n rm n j i
Figure 2 1 .2 0 T ypica l cor e s hape s o f s t ri p drain s (Haus m an n , 1990 )
d.. = (21.9b)
where d
sd
= equivalen t san d drai n diamete r
d
v
= equivalen t void circl e diamete r
b,t = widt h and thickness o f the wick drain
n
s
- porosit y o f sand drai n
V oid are a o f wick drain V oid are a o f wick drain
- -
total cros s sectiona l are a of strip b x t
It may b e noted here that equivalent sand drain diameters for various commercially availabl e
wick drain s var y from 3 0 to 50 mm (Korner , 1999) .
The equatio n for estimating the time t for consolidation i s (Koerner , 1999)
8c,
1-U
(21.10)
where t = tim e fo r consolidatio n
c
h
= coefficien t of consolidation of soi l for horizonta l flo w
Soil Im provem en t 97 9
d = equivalen t diameter o f stri p drai n
_ circumferenc e
n
D = spher e o f influence of the stri p drain;
a) for a triangular pattern, D = 1.05 x spacin g D
t
b) fo r a square pattern, D = 1.13 x spacing D
?
D
(
- distanc e between drain s in triangular spacing and
D
s
= distanc e fo r squar e pattern
U = averag e degre e o f consolidation
Advantages o f Usin g Wic k Drain s (Koerner , 1 9 9 9 )
1. Th e analyti c procedur e i s availabl e and straightforwar d in it s use .
2. Tensil e strengt h i s definitely afforde d t o the soft soi l b y the installation of the wic k drains.
3. Ther e i s only nominal resistance t o the flow o f wate r if it enters th e wic k drain .
4. Constructio n equipment i s generally small .
5. Installatio n i s simple, straightforwar d and economic .
Example 2 1 . 7
What is the equivalent sand drain diameter of a wick drain measuring 96 mm wide and 2.9 mm thick that
is 92%void in its cross section ? Use an estimated san d porosity of 0.3 for typical sand in a sand drain.
Solution
Total are a o f wick drai n = b x t = 96 x 2. 9 = 279 mm
2
V oid area of wick drain = n
d
x b x t - 0.9 2 x 279 = 257 mm
2
The equivalent circle diamete r (Eq . 21.9b ) is
Ubtn
d
|4x25 7
d = , = , = 18.1 mm
v
V n V 3.1 4
The equivalent sand drai n diameter (Eq . 21.19a) is
Example 2 1 . 8
Calculate th e times required fo r 50, 70 and 90%consolidation o f a saturated claye y sil t soi l using
wick drains a t various triangular spacings. The wick drains measure 10 0 x 4 mm and the soi l has a
c
h
= 6.5 1Q-
6
m
2
/min.
Solution
In the simplifie d formula the equivalent diameter d of a strip drai n i s
circumference 10 0 + 100 + 4 + 4 ^ _
d = = 66.2 m m
n 3.1 4
980 Chapt e r 21
Using Eq. (21.10 )
D
2
D
t = i n —-0.75 I n
8c, d
substituting the known values
ln-°—0.75 i n .
8(6.5 x! 0~
6
) 0.006 2 1- U
The time s require d fo r the various degrees o f consolidation ar e tabulate d below fo r assume d
theoretical spacing s o f wick drains.
Wick drai n
s pacin g s
2. 1
1.8
1.5
1.2
0.9
0.6
0.3
T im e i n days fo r variou s
deg rees o f con s olidat io n (U)
50%
110
77
49
29
14
4. 8
0.6
70%
192
133
86
50
24
8.4
1.1
90%
367
254
164
95
46
16
2.1
For the triangular pattern, the spacing D
t
i s
D=^~
' 1.0 5
2 1 .8 SAN D COM PACTI O N PILE S AND STONE COL U M N S
Sand Compactio n Pile s
Sand compactio n pile s consist s o f drivin g a hollo w stee l pip e wit h th e botto m close d wit h a
collapsible plat e down to the required depth ; filling it with sand, and withdrawing the pipe whil e air
pressure i s directe d agains t th e san d insid e it . The botto m plat e open s durin g withdrawa l and the
sand backfill s the void s create d earlie r durin g the driving of th e pipe. Th e in-sit u soi l i s densified
while th e pip e i s bein g withdrawn , an d th e san d backfil l prevent s th e soi l surroundin g th e
compaction pip e fro m collapsing as the pipe i s withdrawn. The maximu m limit s on the amount of
fines tha t can be present ar e 1 5 percent passing the No. 200 sieve (0.075 mm ) and 3 percent passin g
0.005 mm. The distance betwee n the piles may have to be planned according t o the sit e conditions.
Stone Column s
The method describe d fo r installing sand compaction pile s or the vibroflot describe d earlier ca n be
used t o construct stone columns . The size of the stones used fo r this purpose rang e from about 6 to
40 mm . Ston e column s hav e particula r applicatio n i n sof t inorganic , cohesiv e soil s an d ar e
generally inserte d o n a volume displacement basis.
Soil Im provem en t 98 1
The diamete r o f the pipe use d eithe r for the construction o f sand drains o r sand compactio n
piles can be increased accordin g t o the requirements. Stones ar e placed i n the pipe instead of sand,
and the technique of constructing stone columns remains the same a s that for san d piles .
Stone column s ar e place d 1 t o 3 m apar t ove r th e whol e area . Ther e i s n o theoretica l
procedure fo r predictin g th e combine d improvemen t obtained , s o i t i s usua l t o assum e th e
foundation load s ar e carrie d onl y b y th e severa l ston e column s wit h n o contributio n fro m th e
intermediate ground (Bowles, 1996) .
Bowles (1996 ) give s a n approximat e formul a fo r th e allowabl e bearin g capacit y o f ston e
columns a s
1a=-jj
L
( *
C+
^ (21.11 )
where K
p
= tan
2
(45° + 072),
0'= drained angl e of friction o f stone,
c - eithe r draine d cohesio n (suggeste d for large areas ) or the undrained shear strengt h c ,
G
r
' = effectiv e radia l stres s a s measure d b y a pressuremete r (bu t ma y us e 2 c i f
pressuremeter dat a ar e no t available),
F
s
= factor o f safety , 1. 5 to 2.0 .
The tota l allowabl e loa d o n a stone column of average cross-sectio n are a A
C
i s
Stone column s should extend through soft cla y to firm strat a t o control settlements . There i s
no en d bearin g i n Eq . (21.11 ) becaus e th e principa l loa d carryin g mechanis m i s loca l perimete r
shear.
Settlement i s usuall y th e principa l concer n wit h ston e column s sinc e bearin g capacit y i s
usually quit e adequat e (Bowles , 1996) . Ther e i s n o metho d currentl y availabl e t o comput e
settlement o n a theoretica l basis .
Stone column s ar e no t applicabl e t o thic k deposit s o f pea t o r highl y organi c silt s o r clay s
(Bowles, 1996) . Ston e column s can b e use d i n loos e san d deposit s t o increas e th e density.
2 1 .9 SOI L STAB I L I Z ATI ON B Y TH E U SE OF ADM I XTU RES
The physica l properties o f soil s ca n ofte n economicall y b e improve d b y th e us e o f admixtures .
Some o f the mor e widel y used admixtures include lime, portland cement an d asphalt . The proces s
of soi l stabilizatio n firs t involve s mixin g wit h th e soi l a suitabl e additiv e whic h change s it s
property an d the n compactin g th e admixtur e suitably . This metho d i s applicabl e onl y for soil s in
shallow foundation s or th e bas e course s o f roads , airfiel d pavements, etc .
Soil-lime Stabilizatio n
Lime stabilizatio n improve s th e strength , stiffnes s an d durabilit y o f fin e graine d materials . I n
addition, lime i s sometimes use d t o improve the properties o f the fine grained fraction of granular
soils. Lime ha s bee n use d a s a stabilizer for soil s i n the bas e course s o f pavement systems , under
concrete foundations, on embankment slope s an d canal linings.
Adding lime to soils produces a maximum density under a higher optimum moisture content
than i n the untreated soil. Moreover , lime produces a decrease i n plasticity index.
Lime stabilizatio n ha s bee n extensivel y use d t o decreas e swellin g potentia l an d swellin g
pressures i n clays. Ordinaril y th e strengt h of wet cla y i s improved whe n a proper amoun t of lime
982 Chapt e r 21
is added. Th e improvemen t i n strength is partly due t o the decrease in plastic properties of the clay
and partl y t o th e pozzolani c reactio n o f lime wit h soil , whic h produces a cemente d materia l tha t
increases i n strengt h wit h time. Lime-treated soils , i n general , have greate r strengt h and a higher
modulus of elasticit y than untreated soils.
Recommended percentage s o f lime for soil stabilization vary from 2 to 10 percent. For coarse
soils suc h as clayey gravels , sandy soils wit h less than 50 per cent silt-clay fraction, the per cent of
lime varies from 2 to 5, whereas for soils wi t h more than 50 percent silt-clay fraction, the percent of
lime lies between 5 and 10 . Lime i s also use d wit h fl y ash . The fl y ash may var y from 1 0 to 20 per
cent, an d the percent o f lime may li e between 3 and 7.
Soil-Cement Stabilizatio n
Soil-cement i s the reaction product of an intimate mixture of pulverized soil and measured amount s
of portlan d cemen t an d water , compacte d t o hig h density . As th e cemen t hydrates , th e mixtur e
becomes a hard, durabl e structura l material. Hardened soil-cemen t ha s the capacity t o bridge ove r
local wea k point s i n a subgrade. When properl y made, i t does not soften when exposed t o wetting
and drying , or freezing an d thawing cycles.
Portland cemen t an d soi l mixed at the proper moistur e content has been use d increasingl y in
recent year s t o stabiliz e soil s i n specia l situations . Probabl y th e mai n us e ha s bee n t o buil d
stabilized base s unde r concret e pavement s for highway s and airfields . Soi l cemen t mixture s ar e
also use d t o provid e wav e protectio n o n eart h dams . Ther e ar e thre e categorie s o f soil-cemen t
(Mitchell an d Freitag , 1959) . The y are :
1. Norma l soil-cemen t usual l y contain s 5 t o 1 4 percen t cemen t b y weigh t an d i s use d
generally fo r stabilizin g low plasticit y soil s an d sand y soils .
2. Plasti c soil-cemen t has enough water to produce a wet consistency simila r t o mortar . Thi s
material i s suitabl e for use as water proof cana l linings and for erosion protectio n o n stee p
slopes wher e roa d buildin g equipment may no t b e used .
3. Cement-modifie d soi l i s a mi x tha t generall y contain s les s tha n 5 percen t cemen t b y
volume. Thi s form s a les s rigi d syste m tha n eithe r o f th e othe r types , bu t improve s th e
engineering properties o f th e soi l an d reduce s th e abi l i t y o f th e soi l t o expand by drawing
in water .
The cemen t requiremen t depends on the gradation of the soil . A wel l graded soi l containing
gravel, coars e san d an d fin e san d wit h o r withou t smal l amount s o f sil t o r cla y wil l requir e
5 percen t o r les s cemen t b y weight . Poorly grade d sand s wit h minima l amount of sil t wil l require
about 9 percent b y weight. The remaining sandy soils wil l generall y requir e 7 percent. Non-plasti c
or moderately plasti c silt y soil s generally require about 1 0 percent, and plasti c clay soils requir e 13
percent o r more .
Bituminous Soi l Stabilization
Bituminous material s suc h a s asphalts , tars , an d pitche s ar e use d i n variou s consistencie s t o
improve th e engineerin g propertie s o f soils . Mixe d wit h cohesiv e soils , bituminou s material s
improve th e bearin g capacit y an d soi l strengt h a t lo w moistur e content . Th e purpos e o f
incorporating bitumen into such soils is to water proof the m as a means t o maintain a low moistur e
content. Bituminou s materials adde d t o san d ac t a s a cementin g agen t an d produce s a stronger ,
more coheren t mass . Th e amoun t o f bi t ume n adde d varie s fro m 4 t o 7 percent fo r cohesiv e
materials an d 4 t o 1 0 percent fo r sand y materials . Th e primar y us e o f bituminou s materials i s i n
road constructio n wher e i t may b e th e primar y ingredient for th e surfac e cours e o r be use d i n th e
subsurface an d base courses fo r stabilizing soils.
Soil Im provem en t 98 3
2 1 .1 0 SOI L STAB I L I Z ATI ON B Y I NJECTI ON O F SU I TAB L E G ROU T S
Grouting i s a proces s whereb y flui d lik e materials , eithe r i n suspension , o r solutio n form , ar e
injected int o the subsurfac e soi l o r rock .
The purpos e o f injectin g a grout ma y b e any one o r mor e o f the following :
1. T o decrease permeability .
2. T o increase shea r strength .
3. T o decrease compressibility .
Suspension-type grout s includ e soil , cement , lime , asphal t emulsion , etc. , whil e th e solutio n typ e
grouts include a wide variety of chemicals. Grouting proves especially effectiv e i n the following cases:
1. Whe n th e foundation has t o be constructe d below th e ground water table . Th e deepe r th e
foundation, th e longe r th e time neede d fo r construction , an d therefore , th e mor e benefi t
gained fro m groutin g a s compare d wit h dewatering .
2. Whe n ther e i s difficul t acces s t o the foundatio n level . Thi s i s ver y ofte n th e cas e i n cit y
work, i n tunne l shafts, sewers , an d subwa y construction.
3. Whe n th e geometri c dimension s o f th e foundatio n ar e complicate d an d involve s man y
boundaries an d contac t zones .
4. Whe n th e adjacen t structure s requir e tha t th e soi l o f th e foundatio n strat a shoul d no t b e
excavated (extensio n o f existin g foundations into deepe r layers) .
Grouting ha s bee n extensivel y used primaril y t o control groun d wate r flow unde r eart h an d
masonry dams , wher e roc k groutin g i s used . Sinc e th e process fill s soi l void s wit h some type of
stabilizing materia l groutin g i s als o use d t o increas e soi l strengt h an d preven t excessiv e
settlement.
Many different materials hav e been injecte d int o soils t o produce change s i n the engineering
properties of the soil . I n one method a casing is driven and injection i s made unde r pressure t o the
soil a t th e botto m o f th e hol e a s th e casin g i s withdrawn . In anothe r method , a groutin g hol e i s
drilled and at each level i n which injection is desired, the drill is withdrawn and a collar i s placed a t
the top of the area t o be grouted and grout is forced int o the soil unde r pressure. Another method is
to perforate th e casing i n the area t o be grouted an d leave the casing permanentl y i n the soil .
Penetration groutin g ma y involv e portland cement o r fine graine d soil s suc h as bentonite or
other materials of a paniculate nature. These material s penetrate only a short distance throug h most
soils an d ar e primaril y useful i n ver y coarse sands or gravels. V iscous fluids, such as a solution of
sodium silicate , ma y b e use d t o penetrat e fin e graine d soils . Som e o f thes e solution s for m gel s
that restric t permeabilit y an d improv e compressibilit y an d strengt h properties .
Displacement groutin g usuall y consist s o f usin g a grou t lik e portlan d cemen t an d san d
mixture whic h whe n forced int o the soi l displace s an d compacts th e surroundin g materia l abou t a
central cor e o f grout . Injectio n o f lime i s sometime s use d t o produc e lense s i n th e soi l tha t wil l
block th e flo w o f wate r an d reduc e compressibilit y an d expansio n propertie s o f th e soil . Th e
lenses ar e produce d b y hydrauli c fracturing o f th e soil .
The injectio n an d groutin g method s ar e generall y expensiv e compare d wit h othe r
stabilization technique s and are primarily used under special situation s as mentioned earlier . Fo r a
detailed stud y on injections, readers ma y refer t o Caron e t al. , (1975) .
2 1 .1 1 PROB L EM S
21.1 Differentiate : (i ) Compactio n an d consolidation , an d (ii ) Standar d Procto r an d modifie d
Proctor tests .
9 8 4 Chapte r 21
21.2 Dra w a n ideal 'compactio n curve' and discuss the effect of moisture o n the dry unit weight
of soil .
21.3 Explain : (i ) th e uni t , i n whi c h th e compactio n i s measured , (ii ) 9 5 percen t o f Procto r
density, (i i i ) zero air-void s line , an d (iv ) effec t o f compactio n o n th e shea r strengt h o f
soil.
21.4 Wha t ar e the types of rollers used for compacting different types o f soil s i n the field? How
do you decide the compactive effort require d for compacting the soil t o a desired densit y in
the field?
21.5 Wha t ar e th e method s adopte d fo r measurin g the densit y o f the compacte d soil ? Briefl y
describe th e on e whic h wi l l sui t al l type s o f soils .
21.6 A soi l havin g a specifi c gravit y of solid s G = 2.75 , i s subjecte d t o Procto r compactio n
test i n a mol d o f volum e V = 945 cm
3
. The observation s recorde d ar e a s follows :
Observation numbe r 1 2 3 4 5
Mass o f we t sample , g 138 9 176 7 182 4 178 4 170 1
Water content , percentage 7. 5 12. 1 17. 5 21. 0 25. 1
What ar e th e value s o f maximu m dr y uni t weigh t an d th e optimu m moistur e content ?
Draw 100 %saturatio n line.
21.7 A fiel d densit y test wa s conducte d b y san d con e method . Th e observatio n dat a ar e give n
below:
(a) Mass of jar wit h cone and sand (before use) = 4950 g, (b) mass of jar wit h cone and sand
(after use ) = 228 0 g , (c ) mas s o f soi l fro m th e hol e = 292 5 g , (d ) dr y densit y o f
sand = 1.4 8 g/cm
3
, (e ) water content of the wet soi l = 12%. Determine th e dr y uni t weight
of compacte d soil .
21.8 I f a claye y sampl e i s saturate d a t a wate r conten t o f 30%, wha t i s it s density ? Assum e a
value for specifi c gravit y of solids .
21.9 A soil i n a borrow pi t is at a dry density of 1. 7 Mg/m
3
wit h a water content o f 12%. If a soil
mass o f 200 0 cubi c mete r volum e i s excavate d fro m th e pi t an d compacte d i n a n
embankment wit h a porosit y o f 0.32, calculat e th e volume of the embankmen t whic h ca n
be constructed ou t of thi s material. Assume G
s
= 2.70.
21.10 I n a Proctor compactio n test , for one observation , the mas s o f th e we t sampl e i s missing .
The oven dr y mas s o f thi s sample wa s 180 0 g. The volume of the mol d use d wa s 950 cm
3
.
If th e saturatio n o f thi s sampl e wa s 8 0 percent , determin e (i ) th e moistur e content , an d
(ii) the total uni t weight of the sample . Assume G
s
= 2.70.
21.11. A field-compacte d sampl e o f a sand y loa m wa s foun d t o hav e a we t densit y o f
2.176 Mg/m
3
at a water content of 10%. The maximu m dry densit y of the soi l obtaine d i n
a standard Proctor tes t was 2.0 Mg/m
3
. Assume G
s
= 2.65. Comput e p
d
, S, n and the percen t
of compaction o f th e fiel d sample .
21.12 A proposed eart h embankment is required to be compacted t o 95%of standar d Procto r dry
density. Tests o n the material to be used for the embankment giv e p
max
= 1.984 Mg/m
3
at an
opt i mum wate r content of 12%. The borro w pi t material i n it s natural condition ha s a voi d
ratio o f 0.60 . I f G = 2.65 , wha t is th e mi ni mu m volume of th e borro w require d t o mak e
1 cu. m of acceptabl e compacted fill ?
21.13 Th e followin g dat a wer e obtaine d fro m a fiel d densit y tes t o n a compacte d fil l o f sand y
clay. Laborator y moistur e densit y test s o n th e fil l materia l indicate d a maximu m dr y
Soil Im provem en t 98 5
density o f 1.9 2 Mg/m
3
a t a n optimu m wate r conten t o f 11%. Wha t wa s th e percen t
compaction o f the fill ? Was the fil l wate r content above or below optimum.
Mass of the moist soil removed from th e test hole = 103 8 g
Mass o f the soi l afte r ove n drying = 914 g
V olume of the test hol e = 478.55 cm
3
21.14 A fiel d density test performed by sand-cone method gave the following data.
Mass of the soi l removed + pan = 1590 g
Mass of the pan = 12 5 g
V olume of the test hole = 750 cm
3
Water content information
Mass of the wet soi l + pan = 404.9 g
Mass of the dr y soi l + pan = 365.9 g
Mass of the pan = 122. 0 g
Compute: p
d
, y
d
, an d the water content of the soil . Assume G
5
= 2.67
APPENDIX A
SI UNITS IN GEOTECHNICAL ENGINEERING
Introduction
There has always been some confusion with regards t o the system of units to be used in engineering
practices an d othe r commercia l transactions . FP S (Foot-pound-second ) an d MK S (Meter -
Kilogram-second) systems are stil l i n use in many parts of the world. Sometimes a mixture of two
or mor e system s are in vogu e makin g the confusio n all the greater . Thoug h the SI (Le Syste m
International d'Unites o r the International System of Units) units was first conceived an d adopte d
in th e yea r 196 0 a t the Elevent h General Conferenc e of Weight s and Measure s hel d i n Paris , th e
adoption o f thi s coherent an d systematicall y constituted system i s stil l slo w becaus e o f th e pas t
association wit h the FPS system. The condition s are no w gradually changing and possibl y i n the
near futur e th e S I syste m wil l be th e onl y syste m of us e i n al l academi c institution s in th e worl d
over. It is therefore essential t o understand the basic philosophy of the SI units.
The Basic s o f the SI System
The S I syste m i s a full y coheren t an d rationalize d system. I t consist s o f si x basi c unit s and two
supplementary units , and several derive d units . (Table A. I)
Table A.1 B as i c unit s o f in t eres t i n g eot echn ical en g in eerin g
1.
2.
3.
4.
5.
Quant i t y
Length
Mass
Time
Electric curren t
Thermodynamic
temperature
Uni t
Meter
Kilogram
Second
Ampere
Kelvin
SI symbo l
m
kg
S
A
K
987
988 Appen dix A
Supplementary U nit s
The supplementar y unit s include the radian an d steradian, th e unit s o f plan e an d soli d angles ,
respectively.
Derived Unit s
The derive d unit s used by geotechnica l engineer s are tabulated in Table A.2.
Prefixes ar e used to indicate multiples an d submultiples o f the basic and derived units as given
below.
Factor
10
6
10
3
io-
3
io~
6
Quant i t y
acceleration
area
density
force
pressure
stress
moment or torque
uni t weigh t
frequency
volume
volume
work (energy )
Prefix Symbo l
mega M
kilo k
milli m
micro ( i
Table A. 2
Uni t
meter per second squared
square meter
kilogram pe r cubi c meter
newton
pascal
pascal
newton-meter
newton pe r cubi c meter
hertz
cubic meter
liter
joule
Derived uni t s
SI symbo l
-
-
-
N
Pa
Pa
N-m
N/m
3
Hz
m
3
L
J
For mul a
m/sec
2
m
2
kg/m
3
kg-m/s
2
N/m
2
N/m
2
kg-m
2
/s
2
kg/s
2
m
2
cycle/sec
-
10-
3
m
3
N-m
M ass
Mass i s a measure o f the amount of matter an object contains . The mas s remain s th e same eve n if
the object' s temperatur e an d it s locatio n change . Kilogram , kg , i s th e uni t used t o measur e th e
quantity of mass containe d i n an object. Sometimes Mg ( megagram) an d gram (g) are als o used as
a measur e o f mass i n an object.
Time
Although the second (s ) is the basic SI time unit, minutes (min), hours (h), days (d) etc. may be used
as and wher e required.
Force
As pe r Newton' s secon d la w o f motion , force , F , i s expresse d a s F = Ma , where , M = mas s
expressed i n kg, and a is acceleration i n units of m/sec
2
. If the acceleration i s g, the standar d value
of which is 9.80665 m/sec
2
~ 9.81 m/s
2
, the force F wil l be replaced b y W, th e weight of the body.
Now the above equatio n may be written as W = Mg.
SI Unit s i n Geot echn ical En g in eerin g 98 9
The correc t uni t to express the weigh t W, of an object is the newton sinc e the weigh t is the
gravitational force that causes a downward acceleration o f the object .
N ewton, N , is defined as the force that causes a 1 kg mass to accelerate 1 m/s
2
. k g - m
or 1 N = 1—
E
-r—
s
2
Since, a newton, is too small a unit for engineering usage, multiples of newtons expressed a s
kilonewton, kN , an d meganewton, MN , ar e used. Some of the usefu l relationship s ar e
1 kilonewton, k N = 10
3
newton = 1000 N
meganewton, MN = 10
6
newton = 10
3
kN = 1000 k N
Stress an d Pressur e
The uni t of stress and pressure i n SI units is the pascal ( Pa) which is equal to 1 newton per squar e
meter (N/m
2
). Since a pascal is too small a unit, multiples of pascals are used as prefixes to expres s
the uni t o f stres s an d pressure . I n engineerin g practice kilopascal s o r megapascal s ar e normall y
used. For example ,
1 kilopascal = 1 kPa = 1 kN/m
2
= 1000 N/m
2
1 megapascal = 1 MPa = 1 MN/m
2
= 1000 kN/m
2
Density
Density is defined as mass per unit volume. In the SI system of units, mass i s expressed i n kg/m
3
. In
many cases, i t may be more convenient to express density in megagrams per cubic meter or in gm per
cubic centimeter. The relationships may be expressed as
1 g/cm
3
= 1000 kg/m
3
= 10
6
g/m
3
= 1 Mg/m
3
It may be noted her e that the density of water, p^ i s exactly 1.00 g/cm
3
at 4 °C, and the variation is
relatively smal l ove r th e rang e o f temperatures i n ordinar y engineering practice . I t i s sufficientl y
accurate t o write
p
w
= 1.00 g/cm
3
= 10
3
kg/m
3
= 1 Mg/m
3
Unit weight
Unit weigh t i s stil l th e commo n measuremen t i n geotechnica l engineerin g practice . Th e
relationship between uni t weight, 7 an d density p, may be expressed a s 7= pg.
For example, if the densit y of water , p
w
= 1000 kg/m
3
, then
,..,„, = ,000 4x 9*1 i= 9810 4-!|
m
j
s
2
m
3
s
2
Iro r n N
Since, 1 N = 1 -£—, y =9810— - = 9.81 kN/m
3
s
2
m
j
990
Appen dix A
Table A.3 Con vers io n fact or s
To conver t
Length
V olume
Force
Stress
Unit weight
Moment
Moment of inertia
Section modulus
Hydraulic
conductivity
Coefficient
of consolidatio n
SI t o FPS
From
m
m
cm
mm
m
2
m
2
cm
2
mm'
m
3
m
3
cm
3
N
kN
kN
kN
N/m
2
kN/m
2
kN/m
2
kN/m
2
kN/m
2
kN/m
3
kN/m
3
N-m
N-m
mm
4
m
4
mm
3
m
3
m/min
cm/min
m/sec
cm/sec
cm
2
/sec
m
2
/year
cm
2
/sec
To
ft
in
in
in
ft
2
in
2
in
2
in
2
ft
3
in
3
in
3
Ib
Ib
kip
US ton
lb/ft
2
lb/ft
2
US ton/ft
2
kip/ft
2
lb/in
2
lb/ft
3
lb/in
3
Ib-ft
Ib-in
in
4
in
4
in
3
in
3
ft/min
ft/min
ft/sec
in/sec
in
2
/sec
inV sec
ft
2
/sec
Multiply by
3.281
39.37
0.3937
0.03937
10.764
1550
0.155
0.155x 10~
2
35.32
61,023.4
0.06102
0.2248
224.8
0.2248
0.1124
20.885 xlO-
3
20.885
0.01044
20.885 x 10-
3
0.145
6.361
0.003682
0.7375
8.851
2.402 x KT
6
2.402 x 10
6
6.102 x 1Q-
5
6.102 x 10
4
3.281
0.03281
3.281
0.3937
0.155
4.915 x 10-
5
1.0764 x lO'
3
FPS t o SI
From
ft
in
in
in
ft
2
in
2
in
2
in
2
ft
3
in
3
in
3
Ib
Ib
kip
US ton
lb/ft
2
lb/ft
2
US ton/ft
2
kip/ft
2
lb/in
2
lb/ft
3
lb/in
3
Ib-ft
Ib-in
in
4
in
4
in
3
in
3
ft/min
ft/min
ft/sec
in/sec
in
2
/sec
in
2
/sec
ft
2
/sec
To
m
m
cm
mm
m
2
m
2
cm
2
mm
2
m
3
m
3
cm
3
N
kN
kN
kN
N/m
2
kN/m
2
kN/m
2
kN/m
2
kN/m
2
kN/m
3
kN/m
3
N-m
N-m
mm
4
m
4
mm
3
m
3
m/min
cm/min
m/sec
cm/sec
cm
2
/sec
m
2
/year
cm
2
/sec
Multiply by
0.3048
0.0254
2.54
25.4
929.03 xlO^
6.452x10^
6.452
645.16
28. 317xlO-
3
1 6.387 x 10~
6
16.387
4.448
4.448 x 10'
3
4.448
8.896
47.88
0.04788
95.76
47.88
6.895
0.1572
271.43
1.3558
0.11298
0.4162 x 10
6
0.4162 x KT
6
0.16387 x 10
5
0.16387 x 10-
4
0.3048
30.48
0.3048
2.54
6.452
20.346 x 10
3
929.03
SI Unit s in Geot echn ica l En g in eerin g 991
Table A.4 Con vers io n fact or s —g en er al
To convert fro m
Angstrom units
Microns
US gallon (gal )
Pounds
Tons (short o r US tons )
Tons (metric )
kips/ft
2
Pounds/in
3
Poise
millipoise
ft/mm
ft/year
cm/sec
To
inches
feet
millimeters
centimeters
meters
inches
cm
3
m
3
ft
3
liters
dynes
grams
kilograms
kilograms
pounds
kips
grams
kilograms
pounds
kips
tons (short or US tons)
lbs/in
2
lbs/ft
2
US tons/ft
2
kg/cm
2
metric ton/ft
2
gms/cm
3
kg/m
3
lbs/ft
3
kN-sec/m
2
poise
kN-sec/m
2
gm-sec/cm
2
ft/day
ft/year
ft/min
m/min
ft/min
ft/year
Multiply by
3.9370079 10~
9
3.28084 x 10-' °
1 xio-
7
1 x io~
8
1 x io-
10
3.9370079 x 1(T
5
3785
3.785 x io-
3
0.133680
3.785
4.44822 x io
5
453.59243
0.45359243
907.1874
2000
2
1 x IO
6
1000
2204.6223
2.2046223
1.1023112
6.94445
1000
0.5000
0.488244
4.88244
27.6799
27679.905
1728
io-
4
io-
3
io-
7
icr
6
1440
5256 x IO
2
1.9025 x 1Q-
6
0.600
1.9685
1034643.6
APPENDIX B
SLOPE STABILITY CHARTS AND TABLES
As per Eq. ( 10.43), the factor o f safet y F
s
i s defined a s
F
s
= m-nr
u
where, m, n = stability coefficients, an d r
u
= pore pressure ratio. The value s of m and n ma y b e
obtained from Figs . B. I t o B.I4
3:1 4: 1
Slope cot/?
5:1 3:1 4: 1
Slope cot /3
Figure B . 1 St abilit y coefficien t s m an d n for c'lyH = 0 (B is ho p and
M org en s t ern , 1960 )
993
994 Appen dix B
= >'40 °
= J >40 °
2:1 3: 1 4: 1 5: 1 2: 1 3: 1 4: 1 5: 1
Slope cot f t Slop e co t /?
Figure B . 2 St abilit y coefficien t s for c'lj H = 0. 02 5 an d n
d
= 1.0 0
(B is hop an d M org en s t ern , 1960)
2:1 3: 1 4: 1 5: 1
40° 5
2:1 3: 1 4: 1 5: 1
Figure B . 3 St abilit y coefficien t s man d /? fo r c'lyH = 0. 025 an d n
d
= 1. 2 5
(B is hop an d M org en s t ern , 1960)
Slope St abilit y Chart s 995
3:1 4: 1 5: 1 2: 1 3: 1 4: 1
cot/J cot/ 3
Figure B. 4 St abilit y coefficien t s m an d n for c'/yH = 0.0 5 an d n
d
= 1.0 0
(B is hop an d M org en s t ern , 1960 )
2:1 3: 1 4: 1 5: 1 2: 1 3: 1 4: 1 5: 1
cot/? cot/3
Figure B . 5 St abilit y coefficien t s m an d n for c'lyH = 0. 0 5 an d n
d
= 1. 2 5
(Bishop an d M org en s t ern , 1960 )
996 Appen dix B
40°
30 4
5:1 2:1 5:1
Figure B. 6 St abilit y coefficien t s m an d n for c'lyH = 0.0 5 an d n
d
= 1. 5 0
(B is hop an d M org en s t ern , 1960 )
3 4
cot/?
40°
35°
5
30"
4
25° n
20° 3
40°
35°
30°
25°
20°
3 4
cot/?
Figure B . 7 St abilit y coefficien t s m an d /? fo r c'lyH = 0. 07 5 t o e circl es
(O'Con n or an d M it chell, 1977)
Slope St abilit y Chart s 997
0
40°
35°
30°
25°
20°
3 4 5 "2 3 4 5
cot/3 cot/ 3
Figure B. 8 St abilit y coefficien ts m an d n for c'lyH = 0.075 an d n
d
= 1.0 0
(O'Con n or e t al. , 1977)
0
3 4 5
cot/?
Figure B. 9 St abilit y coefficien ts m an d n for c'lyH = 0. 075 an d n
d
= 1. 2 5
(O'Con n or an d Mitchell, 1977)
998 Appen dix B
2 = ~
3 4
cot 3
35° 5
30°
4
25° n
20°
3
40°
35°
30°
25°
20°
3 4
cot/3
Figure B .1 0 St abilit y coefficien t s m an d A? fo r c'lyH = 0. 07 5 an d n
d
= 1. 5 0
(O'Con n or an d M it chell, 1977 )
3 4
cot/?
40°
35
C
30°
25°
20°
3 4
cot/3
40°
35°
30°
25°
20°
Figure B .1 1 St abilit y coefficien t s m an d n for c'lj H = 0. 10 0 t o e circl e s
(O'Con n or an d M it chell, 1977 )
Slope St abilit y Chart s 999
40°
7
35°
30°
25°
90°
/u
n
2 3 4 5
cot/3
Figure B .1 2 St abilit y coefficien t s m an d n for c'/yH = 0.100 an d n
d
= 1.0 0
(O'Con n or an d M it chell, 1977 )
40°
35°
30°
25°
20°
2 3 4 5
cot/3
Figure B .1 3 St abilit y coefficien t s m an d n for c'lyH = 0.100 an d n
d
= 1. 2 5
(O'Con n or an d Mitchell, 1977)
1000 Appen dix B
40°
35°
30°
25°
20°
5 2 3 4 5
cot/3
Figure B .1 4 St abilit y coefficien t s man d n for c'lyH = 0. 10 0 an d n
d
= 1. 5 0
(O'Con n or an d M it chell , 1977 )
Slope St abilit y Chart s 1001
Bishop and M orgenster n (1 9 6 0 ) Stabilit y Coefficient s ar e Presente d i n
Tabular For m
F - m -n.r
c'_
Table B 1 St abilit y coefficien t s m an d n for ~ 7 7 ~ °
yh
Stability coefficient s fo r earth slopes
0'
10.0
12.5
15.0
17.5
20.0
22.5
25.0
27.5
30.0
32.5
35.0
37.5
40.0
Slope 2:1
m
0.353
0.443
0.536
0.631
0.728
0.828
0.933
1.041
1.155
1.274
1.400
1.535
1.678
n
0.441
0.554
0.670
0.789
0.910
1.035
.166
.301
.444
.593
.750
1.919
2.098
Slope 3:1
m
0.529
0.665
0.804
0.946
1.092
1.243
1.399
1.562
1.732
1.911
2.101
2.302
2.517
n
0.588
0.739
0.893
1.051
1.213
1.381
1.554
1.736
1.924
2.123
2.334
2.588
2.797
Slope 4:1
m
0.705
0.887
1.072
1.261
1.456
1.657
1.865
2.082
2.309
2.548
2.801
3.069
3.356
n
0.749
0.943
1.139
1.340
1.547
1.761
1.982
2.213
2.454
2.708
2.977
3.261
3.566
Slope 5: 1
m
0.882
1.109
1.340
1.577
1.820
2.071
2.332
2.603
2.887
3.185
3.501
3.837
4.196
n
0.917
1.153
1.393
1.639
1.892
2.153
2.424
2.706
3.001
3.311
3.639
3.989
4.362
Table B 2 St abilit y coefficien t s man d n for ~TJ- 0-02 5
anc
|
n
^ = < |
<t> '
10.0
12.5
15.0
17.5
20.0
22.5
25.0
27.5
30.0
32.5
35.0
37.5
40.0
Slope 2: 1
m
0.678
0.790
0.901
1.012
1.124
1.239
1.356
1.478
1.606
1.739
1.880
2.030
2.190
n
0.534
0.655
0.776
0.898
1.022
1.150
1.282
1.421
1.567
1.721
1.885
2.060
2.247
Slope 3:1
m
0.906
1.066
1.224
1.380
1.542
1.705
1.875
2.050
2.235
2.431
2.635
2.855
3.090
n
0.683
0.849
1.014
1.179
1.347
1.518
1.696
1.882
2.078
2.285
2.505
2.741
2.933
Slope 4:1
m
1.130
1.337
1.544
1.751
1.962
2.177
2.400
2.631
2.873
3.127
3.396
3.681
3.984
n
0.846
1.061
1.273
1.485
1.698
4.916
2.141
2.375
2.622
2.883
3.160
3.458
3.778
Slope 5: 1
m
1.365
1.620
1.868
5.121
2.380
2.646
2.921
3.207
3.508
3.823
4.156
4.510
4.885
n
1.031
1.282
1.534
1.789
2.050
2.317
2.596
2.886
3.191
3.511
3.849
4.209
4.592
1002 Appen dix B
Table B 3 Stabilit y coefficien t s m an d n for ~ 7 7 ~ °-
025
an d n
d
= 1. 2 5
0'
10.0
12.5
15.0
17.5
20.0
22.5
25.0
27.5
30.0
32.5
35.0
37.5
40.0
Slope 2: 1
m
0.737
0.878
1.019
1.162
1.309
1.461
1.619
1.783
1.956
2.139
2.331
2.536
2.753
n
0.614
0.759
0.907
1.059
1.216
1.379
1.547
1.728
1.915
2. 112
2.321
2.541
2.775
Slope 3: 1
m
0.901
.076
.253
.433
.618
.808
2.007
2.213
2.431
2.659
2.901
3.158
3.431
n
0.726
0.908
1.093
1.282
1.478
1.680
1.891
2.111
2.342
2.588
2.841
3.112
3.399
Slope 4:1
m
1.085
1.299
1.515
1.736
1.961
2.194
2.437
2.689
2.953
3.231
3.524
3.835
4.164
n
0.867
1.089
1.312.
1.541
1.775
2.017
2.269
2.531
2.806
3.095
3.400
3.723
4.064
Slope 5: 1
m
1.285
1.543
1.803
2.065
2.344
2.610
2.897
3.196
3.511
3.841
4.191
4.563
4.988
n
1.014
1.278
1.545
1.814
2.090
2.373
2.669
2.976
3.299
3.638
3.998
4.379
4.784
|_,
Table B 4 St abilit y coefficien t s m an d n for ~T
= 0>05
an d n
d
= 1.0 0
0'
10.0
12.5
15.0
17.5
20.0
22.5
25.0
27.5
30.0
32.5
35.0
37.5
40.0
Slope 2: 1
m
0.913
1.030
1.145
1.262
1.380
1.500
1.624
1.753
1.888
2.029
2.178
2.336
2.505
n
0.563
0.690
0.816
0.942
1.071
1.202
1.338
1.480
1.630
1.789
1.958
2.138
2.332
Slope 3: 1
m
1.181
1.343
1.506
1.671
1.840
2.014
2.193
2.380
2.574
2.777
2.990
3.215
3.451
n
0.717
0.878
1.043
1.212
1.387
1.568
1.757
1.952
2.157
2.370
2.592
2.826
3.071
Slope 4: 1
m
1.469
1.688
1.904
2.117
2.333
2.551
2.778
3.013
3.261
3.523
3.803
4.103
4.425
n
0.910
1.136
1.353
1.565
1.776
1.989
2.211
2.444
2.693
2.961
3.253
3.574
3.926
Slope 5: 1
m
1.733
1.995
2.256
2.517
2.783
3.055
3.336
3.628
3.934
4.256
4.597
4.959
5.344
n
1.069
1.316
1.576
1.825
2.091
2.365
2.651
2.948
3.259
3.585
3.927
4.288
4.668
Slope St abilit y Chart s 1003
rv rvc
Table B 5 St abilit y coefficien t s m and n for ~
=
°-
05
an d n = 1. 2 5
0'
10.0
12.5
15.0
17.5
20.0
22.5
25.0
27.5
30.0
32.5
35.0
37.5
40.0
Slope 2: 1
m
0.919
1.065
1.211
1.359
1.509
1.663
1.822
1.988
2.161
2.343
2.535
2.738
2.953
n
0.633
0.792
0.950
1.108
1.266
1.428
1.595
1.769
1.950
2.141
2.344
2.560
2.791
Slope 3: 1
m
1.119
1.294
1.471
1.650
1.834
2.024
2.222
2.428
2.645
2.873
3.114
3.370
3.642
n
0.766
0.941
1.119
1.303
1.493
1.690
1.897
2.113
2.342
2.583
2.839
3.111
3.400
Slope 4:1
m
1.344
1.563
1.782
2.004
2.230
2.463
2.705
2.957
3.221
3.500
3.795
4.109
4.442
n
0.886
1.112
1.338
1 .56 7
1.799
2.038
2.287
2.546
2.819
3.107
3.413
3.740
4.090
Slope 5:1
m
1.594
1.850
2.109
2.373
2.643
2.921
3.211
3.513
3.829
4.161
4.511
4.881
5.273
n
1.042
1.300
1.562
1.831
2.107
2.392
2.690
2.999
3.324
3.665
4.025
4.405
4.806
_ _ _
Table B 6 St abilit y coefficien t s m an d n for 777
=
°-
05
an d n
d
= 1. 5 0
<t> '
10.0
12.5
15.0
17.5
20.0
22.5
25.0
27.5
30.0
32.5
35.0
37.5
40.0
Slope 2: 1
m
1.022
1.202
1.383
1.565
1.752
1.943
2.143
2.350
2.568
2.798
3.041
3.299
3.574
n
0.751
0.936
1.122
1.309
1.501
1.698
1.903
2.117
2.342
2.580
2.832
3.102
3.389
Slope 3: 1
m
1.170
1.376
1.583
1.795
2.011
2.234
2.467
2.709
2.964
3.232
3.515
3.817
4.136
n
0.828
1.043
1.260
1.480
1.705
1.937
2.179
2.431
2.696
2.975
3.269
3.583
3.915
Slope 4:1
m
1.343
1.589
1.835
2.084
2.337
2.597
2.867
3.148
3.443
3.753
4.082
4.431
4.803
n
0.974
1.227
1.480
1.734
1.993
2.258
2.534
2.820
3.120
3.436
3.771
4.128
4.507
Slope 5:1
m
1.547
1.829
2.112
2.398
2.690
2.990
3.302
3.626
3.967
4.326
4.707
5.112
5.543
n
1.108
1.399
1.690
1.983
2.280
2.585
2.902
3.231
3.577
3.940
4.325
4.735
5.171
1004
Appen dix B
Extension o f the B isho p and M orgenster n Stabilit y Coefficient s (O'Conno r
and Mitchell , 1 9 7 7 )
Table B 7 St abil it y coefficien t s m and n for ~ 7 7 - 0.07 5
an c
j t
oe c
j
r c
|
e
yn
<t> '
20
25
30
35
40
Slope 2: 1
m
1.593
1.853
2.133
2.433
2.773
n
1. 158
1.430
1.730
2.058
2.430
Slope 3: 1
m
2.055
2.426
2.826
3.253
3.737
n
1.516
1.888
2.288
2.730
3.231
Slope 4:1
m
2.498
2.980
3.496
4.055
4.680
n
1.903
2.361
2.888
3.445
4.061
Slope 5: 1
m
2.934
3.520
4.150
4.846
5.609
n
2.301
2.861
3.461
4.159
4.918
Table B 8 St abilit y coefficien t s m an d n for 77 7
= 0
-°75
an c
j = 1 .00
0'
20
25
30
35
40
Slope 2: 1
m
1.610
1.872
2.142
2.443
2.772
n
1.100
1.386
1.686
2.030
2.386
Slope 3: 1
m
2.141
2.502
2.884
3.306
3.775
n
1.443
1.815
2.201
2.659
3.145
Slope 4: 1
m
2.664
3.126
3.623
4.177
4.785
n
1.801
2.259
2.758
3.331
3.945
Slope 5: 1
m
3.173
3.742
4.357
5.024
5.776
n
2.130
2.715
3.331
4.001
4.759
Table B 9 St abilit y coefficien t s m an d n fo r
— = 0.075
an d n
d
=
0'
20
25
30
35
40
Slope 2: 1
m
1.688
2.004
2.352
2.782
3.154
n
1.285
1.641
2.015
2.385
2.841
Slope 3: 1
m
2.071
2.469
2.888
3.357
3.889
n
1.543
1.975
2.385
2.870
3.428
Slope 4: 1
m
2.492
2.792
3.499
4.079
4.729
n
1.815
2.315
2.857
3.457
4.128
Slope 5: 1
m
2.954
3.523
4.149
4.831
5.063
n
2.173
2.730
3.357
4.043
4.830
Table B 1 0 St abilit y coefficien t s m and n for ~rj - 0.075
an c
j
n
^
=
i 50
0'
20
25
30
35
40
Slope 2: 1
m
1.918
2.308
2.735
3.211
3.742
n
1.514
1.914
2.355
2.854
3.397
Slope 3: 1
m
2.199
2.660
3.158
3.708
4.332
n
1.728
2.200
2.714
3.285
3.926
Slope 4:1
m
2.548
3.083
3.659
4.302
5.026
n
1.985
2.530
3.128
3.786
4.527
Slope 5: 1
m
2.931
3.552
4.218
4.961
5.788
n
2.272
2.915
3.585
4.343
5.185
Slope St abilit y Chart s 1005
Table B 1 1 St abilit y coefficien t s m an d n for ~ 7 7 ~ 0-100
an c
j
t oe c
j
r c
|
e
yh
0'
20
25
30
35
40
Slope 2: 1
m
1.804
2.076
2.362
2.673
3.012
n
1.201
1.488
1.786
2.130
2.486
Slope 3: 1
m
2.286
2.665
3.076
3.518
4.008
n
1.588
1.945
2.359
2.803
3.303
Slope 4:1
m
2.748
3.246
3.770
4.339
4.984
n
1.974
2.459
2.961
3.518
4.173
Slope 5: 1
m
3.190
3.796
4.442
5.146
5.923
n
2.361
2.959
3.576
4.249
5.019
Table B 1 2 St abilit y coefficien t s m and nfor ~~ °-
100
and n
d
= 1 .00
0'
20
25
30
35
40
S l o pe 2:1
m
1.841
2.102
2.378
2.692
3.025
n
1.143
1.430
1.714
2.086
2.445
S l o pe 3:1
m
2.421
2.785
3.183
3.612
4.103
n
1.472
1.845
2.258
2.715
3.230
S l o pe 4: 1
m
2.982
3.458
3.973
4.516
5.144
n
1.815
2.303
2.830
3.359
4.001
S l o pe 5:1
m
3.549
4.131
4.751
5.426
6.187
n
2.157
2.743
3.372
4.059
4.831
Table B1 3 St abilit y coefficien t s m an d nfor 777 - °-
100
an d n
d
= 1. 2 5
<t> '
20
25
30
35
40
Slope 2: 1
m
1.874
2.197
2.540
2.922
3.345
n
1.301
1.642
2.000
2.415
2.855
Slope 3:1
m
2.283
2.681
3.112
3.588
4.119
n
1.558
1.972
2.415
2.914
3.457
Slope 4: 1
m
2.751
3.233
3.753
4.333
4.987
n
1.843
2.330
2.858
3.458
4.142
Slope 5: 1
m
3.253
3.833
4.451
5.141
5.921
n
2.158
2.758
3:372
4.072
4.872
Table B1 4 St abilit y coefficien t s m an d n for — - 0.100
anc
|
n
^ _ 1 59
0'
20
25
30
35
40
Slope 2:1
m
2.079
2.477
2.908
3.385
3.924
n
1.528
1.942
2.385
2.884
3.441
Slope 3: 1
m
2.387
2.852
3.349
3.900
4.524
n
1.742
2.215
2.728
3.300
3.941
Slope 4:1
m
2.768
3.297
3.881
4.520
5.247
n
2.014
2.542
3.143
3.800
4.542
Slope 5:1
m
3.158
3.796
4.468
5.211
6.040
n
2.285
2.927
3.614
4.372
5.200
R E FE R E N C E S
Abduljauwad, S.N. , Al-Sulaimani , G.J. , Basunbul , A., and Al-Buraim, I . (1998). "Laborator y an d
Field Studie s o f Respons e o f Structure s t o Heav e o f Expansiv e Clay, " Geotechnique, 4 8
No. 1 .
Abouleid, A.F . (1982) . "Measuremen t o f Swellin g an d Collapsibl e Soi l Properties, " Foundation
Engineering, Edite d b y Georg e Pilot . Presse s d e l e col e national e des Pont s e t chaussees ,
Paris, France .
Alizadeh, M, and Davisson, M.T. (1970) . "Lateral Loa d Test on Piles-Arkansas River Project," J .
ofS.M.F.D.,SM5,VoL 96.
Alpan, I. (1967). "Th e Empirica l Evaluation of the coefficient K
0
an d K
or
Soi l and Foundation," /.
Soc. Soil Mech. Found. Eng.
Altmeyer, W.T . (1955) . "Discussio n o f Engineerin g propertie s o f expansiv e clays, " Proc. ASCE,
V ol. 81, No. SM2 .
Amar, S., Jezequel, J.F. (1972). "Essai s en place et en laboratoire sus sols coberents Comparaison s
des resultats, " Bulletin d e L iaison de s L aboratories des Ponts et Chaussees, No. 58.
American Association o f State Highway and Transportation Official s (1982) . AASHTO Materials ,
Part I , Specifications, Washington, D.C.
American Societ y for Testing and Materials (1994). Annual Book of ASTM Standards , V ol. 04.08,
Philadelphia, Pa.
American Societ y o f Civi l Engineers , Ne w Yor k (1972) . "Performanc e o f Eart h an d Earth -
supported Structures, " V ol . 1 , Par t 2 , Proc. o f th e specialty Conf. Purdu e University ,
Lafayette, Indiana .
Anderson, P. (1956). Substructure Analysis an d Design, Secon d Ed. , The Rolan d Pres s Co. , New
York.
Applied Technolog y Council , (1978) . "Tentativ e Provision s fo r th e Developin g o f Seismi c
Regulations for Buildings, " Publication ATC-3-06.
1007
1008 Refer en ce s
Atterberg, A. (1911) . "Ube r di e Physikaliseh e Bodenuntersuchun g Un d Ube r di e Plastizita t de r
Tone," Int. Mitt. Fu r Bodenkunde, V ol . 1 , Berlin.
Awad, A. and Petrasovits, G. (1968). "Consideration o n the Bearing Capacit y of V ertical and Batter
Piles Subjecte d t o Forces Actin g in Different Directions, " Proc. 3r d Budapest Conf. S M and
FE, Budapest.
Azzouz, A.S. , Kriezek , R.J. , an d Corotis , R.B . (1976) . "Regressio n Analysi s o f soi l
compressibility," Soils Found., Tokyo, V ol . 16, No. 2 .
Baguelin, E , Jezequel , J.F. , an d Shields , D.H . (1978 ) "The Pressuremeter an d Foundation
Engineering, Trans Tech. Publications, Clausthal, Germany.
Balaam, N.P. , Poulos , H.G. , an d Booker , J.R. (1975) . "Finit e Elemen t Analysi s o f th e Effect s of
Installation o n Pil e Load-settlemen t Behaviour," Geot. Eng., V ol . 6, No. 1 .
Barden, L. , McGown , A. , an d Collins , K. (1973) . "Th e Collaps e Mechanis m i n Partl y Saturate d
Soil," Engineering Geology, V ol . 7.
Baver, L.D. (1940) . Soil Physics, John Wiley and Sons , Ne w York.
Bazaraa, A.R. , (1967) . "Us e o f Standar d Penetratio n Tes t fo r Estimatin g Settlement s o f Shallo w
Foundations o n Sand, " Ph.D. Thesis, Uni v o f Illinois , U.S.A.
Bell, A.L. (1915) . "Th e Latera l Pressur e an d Resistanc e o f Cla y an d Supportin g Powe r o f Cla y
Foundations," Century o f Soil Mechanics, ICF , London .
Berezantsev, V .G. , Khristoforov , V , an d Golubkov , V . (1961) . Loa d Bearin g Capacit y an d
Deformation o f Piled Foundations , Proc. 5t h Int. Conf. S M and FE, V ol . 2.
Berg, R.R. , Bonaparte , R. , Anderson, R.P. , an d Chouery , V .E. (1986) . "Desig n Constructio n an d
Performance o f Tw o Tensa r Geogri d Reinforce d Walls " Proc. 3r d Intl. Conf. Geotextiles,
V ienna: Austrian Society o f Engineers .
Bergado, D.T. , Anderson , L.R. , Miura , N. , Balasubramaniam , A.S . (1996) . Soft Ground
Improvement, Publishe d by ASCE Press , ASCE, Ne w York.
Biot, M.A. , an d Clingan , F.M . (1941) . "Genera l Theor y o f Three-Dimensiona l Consolidation, "
J . Applied Physics, V ol. 12.
Bishop, A.W. (1954) . "Th e us e o f Por e Pressur e coefficient s i n Practice, " Geotechnique, V ol. 4,
No. 4, London .
Bishop, A.W . (1955) . "Th e Us e o f Sli p Circl e i n th e Stabilit y Analysi s o f Eart h Slopes, "
Geotechnique, V ol. 5, No. I .
Bishop, A.W, (1960-61) . "Th e Measuremen t o f Por e Pressur e i n The Triaxia l Tests, " Proc. Conf.
Pore Pressure an d Suction i n Soils, Butterworths, London.
Bishop, A.W. , an d Bjerrum , L . (1960) . "Th e Relevanc e o f th e Triaxia l Tes t t o th e Solutio n o f
Stability Problems, " Proc. Res. Conf. Shear Strength o f Soils, ASCE.
Bishop, A.W. , an d Henkel , D.J . ( \ 962).The Measurement o f Soil Properties i n th e Triaxial Test,
Second Ed. , Edwar d Arnold , London.
Bishop, A.W. , an d Morgenstern , N . (1960) . "Stabilit y Coefficient s fo r Eart h Slopes, "
Geotechnique, V ol. 10, No. 4 , London .
Bjerrum, L. (1972) . "Embankment s o n Sof t Ground, " ASCE Conf. o n Performance o f Earth an d
Earth-supported Structures, Purdue University.
Referen ces 100 9
Bjerrum, L. (1973) . "Problem s o f soi l mechanic s an d constructio n o n sof t days, " Proc. 8t h Int.
Conf. o n Soil Mech. and Found. Eng., Moscow, 3 .
Bjerrum, L., and Hide, O. (1956). "Stabilit y of Strutted Excavations in Clay," Geotechnique, V ol. 6,
No. 1 .
Bjerrum, L. , Casagrande , A. , Peck , R.B. , an d Skempton , A.W. , eds . (1960) . From Theory t o
Practice i n Soil Mechanics, Selection s fro m th e writing s of Kar l Terzaghi, John Wiley and
Sons.
Bjerrum, L., and Simons, N.E. (1960). "Compariso n o f Shear Strength Characteristics of Normally
Consolidated Clay, " Proc. Res. Conf. Shear Strength Cohesive Soils, ASCE .
Boussinesq, J.V . (1885). Application de s Potentiels a L 'Etude de L 'Equilibre e t due mouvement des
solides Elastiques, Gauthier-V illard , Paris.
Bowles, I.E. (1992) . Engineering Properties o f Soils and their Measurement, McGra w Hill , New
York.
Bowles, J.E. (1996) . Foundation Analysis an d Design, McGraw-Hill , New York.
Briaud, J.L., Smith , T.D., an d Tucker, L.M. (1985). "A Pressuremeter Method for Laterally Loaded
Piles," Int. Conf. ofSM an d FE, Sa n Francisco .
Broms, B.B . (1964a) . "Latera l Resistanc e o f Pile s i n Cohesiv e Soils, " J SMFD, ASCE , V ol . 90,
SM2.
Broms, B.B. (1964b). "Latera l Resistanc e of Piles i n Cohesionless Soils, " J SMFD, ASCE , V ol. 90,
SM3.
Broms, B.B . (1966) . "Method s o f Calculatin g th e Ultimat e Bearin g Capacit y o f Piles— A
Summary," Soils-Soils, No . 18-19 : 21-32 .
Broms, B.B. (1987) . "Soi l Improvement Methods i n Southeast Asia for Sof t Soil, " Proc. 8ARC on
SM an d FE, Kyoto , V ol. 2.
Broms, B.B . (1995) . "Fabri c Reinforce d Soil, " Developments i n Deep Foundations an d Ground
Improvement Schemes, Balasubramanium , et al. , (eds) , Balkem a Rotterdam, ISBN 905410
5933.
Brown, J.D. , an d Meyerhof , G.G . (1969) . "Experimenta l stud y of Bearin g Capacit y i n Layere d
Clays," 7t h ICSFME, V ol . 2.
Brown, R.E. (1977) . "Dril l Ro d Influenc e o n Standar d Penetration Test, " J GED, ASCE, V ol . 103,
SM3.
Brown, R.E . (1977) . "V ibroflotatio n Compactio n of Cohesioinles s Soils, " J . of Geotech. Engg.
Div., ASCE, V ol . 103, No . GT12 .
Burland, J.B. (1973) . "Shaf t Frictio n o f Piles i n Clay-A Simpl e Fundamental Approach," Ground
Engineering, V ol.6 , No. 3.
Burland, J.B., and Burbidge, M.C. (1985) . "Settlement of Foundations on Sand and Gravel," Proc.
ICE, V ol. 78.
Burland, J.B. , an d Wroth , C.P . (1974) . "Allowabl e an d Differentia l Settlemen t o f Structure s
Including Damag e an d Soi l structur e Interaction, " Proceedings, Conf . o n Settlemen t o f
Structures, Cambridge University , England.
1010 Refer en ce s
Burmister, D.M . (1943) . "Th e Theor y o f Stresse s an d Displacement s i n Laye r System s an d
Application t o Design o f Airport Runways," Proc. Highway Res. Board, V ol . 23.
Button, S.J . (1953) . "Th e Bearin g Capacit y o f Footing s o n a two-Laye r Cohesiv e Sub-soil, " 3r d
7C£MF£, V ol. 1 .
Capazzoli, L . (1968) . "Tes t Pil e Progra m a t St . Gabriel , Louisiana , Louis, " / . Capazzoli an d
Associates.
Caquot, A., and Kerisel , J. (1948). Tables for th e Calculation o f Passive Pressure, Active Pressure,
and Bearing Capacity o f Foundations (Translate d b y M.A. Bee , London) , Gauthier-V illars
Paris.
Caquot, A., and Kerisel , J. (1956) . Traite de mechanique des sols, 2n d ed. , Gauthie r V illars, Paris.
Caron, P.C. , an d Thomas, F.B . (1975) . "Injection, " Foundation Engineering Hand book, Edited by
Winterkorn an d Fang, V an Nastrand Reinhold Company, New York.
Carrillo, N. (1942) . "Simpl e Tw o and Three-Dimensional Cases in the Theory o f Consolidation of
Soils," J . Math, Phys. V ol . 21.
Carter, J.P. , an d Kulhawy , F.H . (1988) . "Analysi s an d Desig n o f Drille d Shaf t Foundation s
Socketed int o Rock, " Fina l Repor t Projec t 1493-4 , EPRI EL-5918 , Geotechnica l Group ,
Cornell University , Ithaca.
Casagrande, A . (1931) . Th e Hydrometer Method fo r Mechanical Analysis o f Soils an d other
Granular Materials, Cambridge , Massachusetts.
Casagrande, A . (1932) . "Researc h o f Atterberg Limits of Soils, " Public Roads, V ol . 13 , No. 8 .
Casagrande, A . (1936) . "Th e determinatio n o f th e Preconsolidatio n Loa d an d it s Practica l
Significance," Proc. First Int. Conf. Soil Mech. Found Eng, V ol. 3.
Casagrande, A. (1937). "Seepag e through dams," /. N . Engl. Waterworks Assoc. Ll(2) .
Casagrande, A. (1958-59). "Review o f Past and Current Works on Electro-osmotic Stabilisatio n of
Soils," Harvard Soil Mech., Serie s No. 45. Harvar d Univ. , Cambridge, Mass .
Casagrande, A . (1967) . "Classificatio n and Identificatio n o f Soils, " Proc. ASCE, V ol . 73, No . 6 ,
Part 1 , New York .
Casagrande, A. , an d Fadum, R.E . (1940) . "Note s o n Soi l Testin g for Engineerin g Purposes, " Soil
Mechanics Series, Graduat e School o f Engineering, Harvard University , Cambridge. M.A. ,
V ol. 8.
Casagrande, L. (1932 ) "Naeherungsmethode n zu r Bestimmur g von Art' und Meng e de r sickerung
durch geschuettet e Daemme, " Thesis, Technich e Hochschule , V ienna.
Chellis, R.D. (1961) . Pile Foundations, McGra w Hill , New York.
Chen, F.H . (1988) . Foundations o n Expansive Soils. Elsevie r Scienc e Publishin g Company Inc. ,
New York.
Chen, Y.J., and Kulhawy , F.H. (1994) . "Cas e Histor y Evaluation of th e Behavio r of Drille d Shaf t
Under Axia l an d Latera l Loading, " Fina l Report , Projec t 1493-04 , EPR I TR-104601 ,
Geotechnical Group , Cornell University, Ithacka.
Christian, J.T. , an d Carrier , W.D. (1978) . "Janbu , Bjerrum , an d Kjaernsli' s Chart Reinterpreted, "
Canadian Geotechnical J ournal, V ol . 15.
Referen ces 101 1
Clemence, S.P. , and Finbarr, A.O. (1981). "Desig n Consideration s fo r Collapsibl e Soils, " /. of the
Geotech. Eng. Div., ASCE, V ol . 107, No . GTS.
Coulomb, C.A . (1776) . "Essa i Su r Un e Applicatio n de s regie s de s Maximi s e t Minimu m a
Queiques Problemes d e Statique Relatifs a L 'Architecture'," Mem . Acad. Ro y Pres . Diver s
Savants, Paris, V ol . 3, Paris .
Coyle, H.M. , an d Reese , L.C. (1966) . "Loa d Transfe r for Axiall y Loade d Pile s i n Clay, " J . of the
S.M. an d F . Div., ASCE, V ol . 92, SM2 .
Coyle, H.M. , an d Sulaiman , I.H. (1967) . "Ski n Frictio n for Stee l Pile s i n Sand, " J SMFD, ASCE ,
V ol. 93 , SM 6.
Coyle, H.M. , an d Castello , R.R . (1981) . "Ne w Desig n Correlation s fo r Pile s i n Sand, " J . o f th e
Geotech. Eng. Div., ASCE, V ol . 107, No . GT17 .
Culmann, C. (1875). Di e Graphische Statik, Zurich, Meyer and Zeller .
D'Appolonia, E. , Ellison , R.D. , an d D'Appolonia , DJ . (1975) . "Drille d Piers, " Foundation
Engineering Handbook, Edited : H.F . Winterkor n an d H.Y . Fang , V a n Nostran d Reinhol d
Company, Ne w York.
Darcy, H. (1856) . L es Fontaines Publiques de la Ville de Dijon, Paris .
Das, B.M. , an d Seeley , G.R . (1975) . "Load-Displacemen t Relationship s fo r V ertica l Ancho r
Plates," J . of th e Geotech. Engg. Div., ASCE, V ol. 101, GT 7.
Davis, L.H. (1977) . "Tubula r Stee l Foundation, " Tes t Repor t RD-1517 , Florida Powe r an d Light
company, Miami , Florida .
Davis, M.C.R. , an d Schlosser , F . (1997) . "Groun d Improvemen t Geosystems, " Proc. Third Int.
Conf. o n Ground Improvement Geosystems, London .
Davisson, M.T . (1960) . "Behavio r o f Flexibl e V ertica l Pile s Subjecte d t o Moments , Shea r an d
Axial Load, " Ph.D. Thesis , Univ . of Illinois, Urbana, U.S.A.
Davisson, M.T. , an d Salley , J.R. (1969) . "Latera l Loa d Test s o n Drille d Piers, " Performanc e o f
Deep Foundations , ASTM ST P 444, AST M
De Beer , E.E . (1965) . "Bearin g Capacit y and Settlement of Shallo w Foundations on Sand, " Proc.
of Symp held a t Duke University.
De Ruiter , J. , an d Beringen , F.L. (1979) . "Pil e Foundation s fo r Larg e Nort h Se a Structures, "
Marine Geotechnology, V ol. 3, No. 3.
Dennis, N.D. , an d Olson , R.E . (1983a). "Axia l capacit y of stee l Pip e Pile s i n sand, " Proc. ASCE
Conf. o n Geotech. Practice i n Offshore Eng, Austin, Texas.
Dennis, N.D. , and Olson, R.E. , (1983b) . "Axial Capacit y of Steel Pip e Piles i n Clay, " Proc. ASCE
Conf. o n Geotech. Practice i n Offshore Eng., Austin, Texas.
Department o f th e Navy , Nava l Facilitie s Engineerin g Command . (1982) . Design Manual,
NAV FACDM-71.
Desai, C.S. (1974) . "Numerica l Design-Analysi s for Piles i n Sands," J . Geotech. Eng. Div., ASCE,
V ol. 100 , No . GT6 .
Douglas, B.J . (1984) . Th e Electric Cone Penetrometer Test: A User's Guide t o Contracting fo r
Services, Quality Assurance, Data Analysis, Th e Eart h Technolog y Corporation , Lon g
Beach, California, U.S.A.
1012 Referen ce s
Douglas, B.J., an d Olsen, R.S . (1981) . "Soil Classification Using the Electric Con e Penetrometer ,
Cone Penetratio n Testing and Experience, " ASCE Fall Convention.
Downs, D.I. , an d Chieurzzi , R. (1966). "Transmissio n Towe r Foundations, " J . Power Dm, ASCE ,
V ol. 92, PO2 .
Dudley, J.H. (1970). "Revie w of Collapsible Soils," J . ofS.M. an d F. Div., ASCE, V ol. 96, No. SMS.
Duncan, J. M., Evans, L. T, Jr. , and Ooi P. S. K. (1994). "Lateral Loa d Analysi s of Single Piles and
Drilled Shafts, " /. ofGeotech. Engg. ASCE, V o l 120 , No. 6.
Dunn, I.S. , Anderson, I.R. , and Kiefer , F.W. (1980). Fundamentals o f Geotechnical Analysis, John
Wiley an d Sons .
Dupuit, J . (1863) . Etudes Theoriques e t practiques su r le Mouvement de s eaux dans les Canaux
Decouverts et a trovers les Terrains Permeables, Dunod, Paris .
Eggestad, A . (1963) . "Deformatio n Measuremen t Below a Model Footin g o n th e Surfac e o f Dr y
Sand," Proc. European Conf. Soil Mech., Wiesbaden , 1 .
Ellison, R.D. , D' Appolonia , E. , an d Thiers , G.R . (1971) . "Load-Deformatio n Mechanis m fo r
Bored Piles, " J SMFD, ASCE , V ol . 97, SM4 .
Fadum, R.E . (1948) . "Influenc e V alue s fo r Estimatin g Stresse s i n Elasti c Foundations, " Proc.
Second Int. Conf. SM an d FE , V ol . 3.
Fellenius, W . (1927) . "Erdstatische Berechnungen" (revise d edition , 1939) , W . Ernstu , Sons ,
Berlin.
Fellenius, W. (1947). Erdstatische Berechnungen mit Reibung an d Kohasion an d unter Annahme
keissy Undrischer Gleitf lachen, Wilhel m Ernt and Sohn , 1 1 Ed. , Berlin.
Forchheimer, P . (1930). General Hydraulics, Thir d Ed. , Leipzig.
Foster, C.R. , an d Ashlvin, R.G. (1954) . "Stresse s an d deflection s induce d b y a unifor m circula r
load," Proceedings, HRB .
Fox, E.N. (1948) . "Th e mea n Elasti c Settlement s o f Uniforml y Loade d Are a a t Dept h Belo w th e
Ground Surface, " 2n d ICSMEF, V ol . 1 .
Frohlick, O.K. (1955) . "Genera l Theor y o f Stabilit y of Slopes, " Geotechnique, V ol. 5.
Ghaly, A.M. (1997). "Load-Displacement Predictio n for Horizontally Loaded V ertical Plates," J . of
Geotechnical an d Environmental Engineering, ASCE, V ol . 123, No. 1 .
Gibbs, H.J. , an d Holtz , W.G. (1957) . "Researc h o n Determining The Densit y o f Sand s b y Spoo n
Penetration Testing," 4t h ICSMEF, V ol . 1 .
Gibson, R.E. (1953). "Experimental Determination of the True Cohesion and True angle of Internal
Friction i n Clays, " Proc. Third Int. Conf. Soil Mech. Found Eng., Zurich , V ol. 1.
Gibson, R.E. , an d Lo , K.Y . (1961) . A Theory o f Consolidation fo r Soils Exhibiting Secondary
Compression, Norwegian Geotechnical Institute, Publication No. 41.
Gilboy, G . (1934) . "Mechanci s o f Hydrauli c Fil l Dams, " J ournal Boston Society o f Civil
Engineers, Boston.
Glesser, S.M . (1953) . L ateral L oad Tests o n Vertical Fixed-Head an d Free-Head Pile, ASTM,
STP 154 .
Goodman, R.E. (1980) . Introduction t o Rock Mechanics, Wiley , New York.
Referen ces 101 3
Grim, R.E. (1959). "Physico-Chemica l Propertie s o f Soils," ASCEJ . Soil Mech., 85 , SM2, 1-17 .
Grim, R.E. (1962). Applied Clay Mineralogy, McGra w Hill.
Hagerty, D.J. , an d Nofal , M.M . (1992) . "Desig n Aids : Anchored Bulkhead s i n Sand, " Canadian
Geotech. J ournal, V ol . 29, No. 5.
Hansen, J.B . (1970) . " A Revise d an d Extende d Formul a fo r Bearin g Capacity, " Danish
Geotechnical Institute Bui. No. 28, Copenhagen.
Hanzawa, H. , an d Kishida , T . (1982) . "Determinatio n o f i n situ undraine d strengt h o f sof t cla y
deposits," Soils and Foundations, 22 , No. 2.
Harr, M.E. (1962) . Ground Water and Seepage, McGra w Hill , New York.
Harr, M.E. (1966) . Foundations o f Theoretical Soil Mechanics, McGra w Hill , New York.
Hausmann, M.R . (1990) . Engineering Principles o f Ground Modification, McGra w Hill , Ne w
York.
Hazen, A. (1893) . "Some Physical Propertie s o f Sands an d gravel s with special referenc e t o thei r
use i n filtration, " 24th Annual Report, Massachusetts , State Boar d o f Health.
Hazen, A . (1911) . "Discussio n o f Dam s an d San d Foundations, " b y A.C. Koenig , Tran s ASCE,
V ol. 73.
Hazen, A. (1930) . Hydraulics, Thir d Ed. , Leipzig,
Henkel, D.J . (1960) . "Th e Shea r Strengt h of Saturate d Remoulded Clays, " Proc. Re. Conf. Shear
Strength Cohesive Soils, ASCE .
Hetenyi, M. (1946) . Beams o n Elastic Foundations, The Univ. of Michigan Press , Ann Arbor, MI.
Highter, W.H. , an d Anders, J.C . (1985) . "Dimensionin g Footing s Subjecte d t o Eccentri c Loads, "
J . o f Geotech. Eng., ASCE, V ol . Ill, No. GT5 .
Roll, D.L. (1940) . "Stress Transmission o n Earths," Proc. Highway Res. Board, V ol . 20.
Holl, D.L. (1941). "Plan e Strai n Distribution of Stress in Elastic Media, " Iowa Engg. Expts Station,
Iowa.
Holtz, R.D., and Kovacs, W.D. (1981). An Introduction t o Geotechnical Engineering, Prentic e Hall
Inc., New Jersey .
Holtz, W.G. an d Gibbs , HJ . (1956) . "Engineerin g Propertie s o f Expansive Clays. " Tran s ASCE .
121.
Holtz, W. G. an d Hilf, J . W. (1961). "Settlemen t o f Soil Foundations Due t o Saturation, " Proc. 5t h
Internation o n SMFE, Paris , V ol 1 .
Hough, B.K. (1957). Basic Soil Engineering, Ronal d Press, Ne w York.
Housel, W.S. (1929). " A Practical Metho d for the Selection of Foundations Based o n Fundamental
Research i n Soil Mechanics, " Research Bulletin, No . 13 , University of Michigan, AnnArbor.
Houston, W.N. , an d Houston , S.L . (1989) . "State-of-the-Practic e Mitigatio n Measure s fo r
Collapsible Soi l Sites, " Proceedings, Foundation Engineering: Current Principles an d
Practices, ASCE, V ol . 1.
Hrennikoff, A . (1949) . "Analysi s o f Pile Foundation s wit h Batter Piles, " Proc. ASCE, V ol . 75.
Huntington, W.C. (1957) . Earth Pressures and Retaining Walls, John Wiley and Sons , Ne w York.
1014 Refer en ce s
Hvorslev, M.J. (1949). Surface Exploration an d Sampling o f Soils for Civil Engineering Purposes.
Water ways Experimental Station, Engineering Foundations, Ne w York.
Hvorslev, M.J . (1960) . "Physica l Component s o f Th e Shea r Strengt h o f Saturate d Clays, " Proc.
Res. Conf. o n Shear Strength o f Cohesive Soils, ASCE, Boulder , Colorado.
Ireland, H.O. (1957) . "Pullin g Tests o n Pile s in Sand, " Proc. 4t h Int. Conf. S M and FE, V ol . 2.
Ismael, N.F. , an d Klym , T.W . (1977) . "Behavio r o f Rigi d Pier s i n Layere d cohesiv e Soils" . / .
Geotech. Eng. Div., ASCE. V ol . 104, No . GTS.
Jakosky, J.J . (1950) . Exploration Geophysics, Trij a Publishin g Company , California . U.S.A .
Jaky, J. (1944). "The Coefficien t of Earth Pressur e a t Rest," /. Soc. Hungarian Arch. Eng.
Janbu, N . (1976) . "Stati c Bearin g Capacit y o f Frictio n Piles, " Proc. 6t h European Conf. o n Soil
Mech. and F . Engg.Vol. 1-2 .
Janbu, N. , Bjerrum , L. , an d Kjaernsli , B . (1956) . "V eilednin g V e d losnin g a v fundamentering-
soppgaver," Pub. No. 16 , Norwegian Geotechnical Institute.
Jarquio, R. (1981) . "Tota l Latera l Surcharg e Pressur e Du e t o Stri p Load, " /. Geotech. Eng. Div.,
ASCE, 107 , 10 .
Jennings, J.E., an d Knight , K. (1957). "An Additional Settlement of Foundation Due t o a Collapse
Structure o f Sand y Subsoil s o n Wetting, " Proc. 4t h Int. Conf o n S.M. an d F . Eng., Paris ,
V ol. 1 .
Jennings, J.E. , an d Knight , K. (1975) . "A Guid e t o Construction on o r wit h Materials Exhibiting
Additional Settlement s Du e t o 'Collapse ' o f Grai n Structure, " Proc. Sixth Regional
Conference fo r Africa o n S. M and F . Engineering, Johannesburg.
Jumikis, A.R. (1962) . Soil Mechanics, D . V an Nostrand Co. Inc. , New York.
Kamon, M. , an d Bergado , D.T . (1991) . "Groun d Improvemen t Techniques, " Proc. 9 AR C o n
SMFE, Bangkok , V ol. 2.
Karlsson, R. , V iberg , L. (1967) . "Rati o c Ip i n relatio n t o liqui d limi t an d plasticit y inde x wit h
special referenc e t o Swedis h clays, " Proc. Geotech. Conf, Olso , Norway , V ol. 1.
Kenney, T.C. , Lau , D. , an d Ofoegbu , G.I . (1984) . "Permeabilit y o f Compacte d Granula r
Materials," Canadian Geotech. J .,2\ No . 4.
Kerisel, J . (1961) . "Fondation s Profonde s e n Milie u Sableux, " Proc. 5t h Int. Conf. S M and FE ,
V ol. 2.
Kezdi, A. (1957) . "Th e Bearin g Capacit y o f Pile s an d Pil e Groups, " Proc. 4t h Int. Conf. S M and
FE, V ol . 2.
Kezdi, A. (1965) . "Foundation s Profonde s e n Milie s Sableux, " Proc. 5t h Int. Conf. S M an d FE ,
V ol. 2 .
Kezdi, A. (1965) . "Genera l Repor t o n Dee p Foundations, " Proc. 6t h Int. Conf. o n SM an d FE ,
Montreal.
Kezdi, A . (1975) . "Pil e Foundations, " Foundations Engineering Handbook, Edite d b y H.F .
Winterkorn an d H . Y. Fang, V an Nostrand, New York.
Kishida, H . (1967) . "Ultimat e Bearin g Capacit y o f Pile s Drive n int o Loos e Sand, " Soil an d
Foundations, V ol . 7, No . 3 .
Referen ces 101 5
Koerner, R.M . (1985) . Construction an d Geotechnical Methods i n Foundation Engineering,
McGraw Hill , New York.
Koerner, R.M. (1999). Design with Geosynthetics. Fourt h Edition, Prentice Hall, New Jersey.
Koerner, R.M . (2000) . "Emergin g an d Futur e Development s o f Selecte d Geosyntheti c
Applications" Thirty-Second Terzaghi Lecture, Drexel University, Philadelphia.
Kogler, F, an d Scheidig, A. (1962). Soil Mechanics b y A. R. Jumikis, D. V an Nostrand.
Komornik, A., and David, D. (1969) . "Prediction o f Swelling Pressure o f Clays," J SMFD, ASCE ,
V ol. 95, SMI.
Kovacs, W.D., and Salomone (1982) . "SPT Hammer Energy Measurement, " J GED, ASCE, GT4.
Kozeny, J.S. (1931) . "Groun d wasserbewegun g bei frei m Spiegel , Fluss an d Kanalversicheringi, "
Wasserkraft an d Wassenvitrschaft, No . 3 .
Kubo, J . (1965) . "Experimenta l Stud y o f th e Behaviou r of Laterall y Loade d Piles, " Proc. 6t h
Int.Conf. S M and FE , V ol . 2.
Ladd, C.C. , an d Foott , R. (1974) . "New Desig n Procedure fo r Stabilit y of Sof t Clays, " J . Geotech
Eng. Div., ASCE, V ol. 100, No. GT7 .
Ladd, C.C., Foott, R., Ishihara, K., Schlosser, F, an d Poulos, H.G. (1977). "Stress Deformation and
strength Characteristics, " Proceedings 9t h Int.Conf. S M and FE, Tokyo, V ol. 2.
Lambe, T.W. (1958a). "The Structur e of Compacted Clay, " J . ofS.M. & F. Dm, ASCE , V ol. 84, No.
SM2.
Lambe, T.W. (1958b). "Th e Engineerin g Behavior of Compacted Clay, " J . Soil Mech, Found, Div.,
ASCE, V ol. 84, No. SM2 .
Lambe, T.W. (1962) . "Soi l Stabilization, " Chapte r 4 o f Foundation Engineering, G.A . Leonard s
(ed.), McGraw Hill , New York.
Lambe, T.W. , and Whitman, R.V. (1969). Soil Mechanics, John Wiley and Sons Inc.
Lee, K.L. , an d Singh , A. (1971) . "Relativ e Densit y an d Relativ e Compaction, " J . Soil Mech.,
Found, Div., ASCE, V ol. 97, SM 7.
Lee, K.L. , Adams, B.D. , an d V agneron, J.J. (1973) . "Reinforce d Eart h Retainin g Walls," J ou. of
SMFD., ASCE , V ol . 99, No . S M 10.
Leonards, G.A . (1962). Foundation Engineering, McGra w Hill .
Leonards, G.A. , Cutter, W.A., and Holtz, R.D. (1980). "Dynami c Compaction of Granular Soils,"
J ou. of Geotech. Eng. Div., ASCE, V ol. 106, No. GT1 .
Liao, S.S. , an d Whitman, R.V . (1986). "Overburden Correction Factor s fo r SP T i n Sand, " J GED,
V ol. 112 , No. 3 .
Littlejohn, G.S. , and Bruce, D.A. (1975). "Rock Anchors: State of the Art, Part I - Design, " Ground
Engineering, V ol. 8, No. 3.
Loos, W. , and Breth , H. (1949) . Modellversuche uber Biege Beanspruch - ungen Von Pfahlen an d
Spunwenden, De r Bauingeniur , V ol. 28.
Louden, A.G. (1952). "Th e Computatio n o f Permeability fro m Soi l Tests," Geotech., 3, No. 4.
Lowe, J. (1974). "New Concept s i n Consolidation and Settlement Analysis," J . Geotech. Eng. Div.,
ASCE, V ol . 100 .
1016 Referen ce s
Lunne, T., and Christoffersen , H.P., (1985). Interpretation o f Cone Penetrometer data for Offshore
Sands, Norwegia n Geotechnical Institute, Pub. No. 156 .
Lunne, T. , an d Kelven , A . (1981) . "Rol e o f CP T i n Nort h Se a Foundatio n Engineering, "
Symposium o n Cone Penetration Testing an d Experience, Geotech . Eng . Dev . ASCE , St .
Louis.
Malter, H . (1958) . "Numerica l solution s fo r Beam s o n Elasti c Foundations, " J . Soil Mech an d
Found Dm, LXXXIV , No. ST2 , Par t 1 , ASCE.
Marchetti, S. (1980) . "I n situ Tes t b y Flat Dilatometer," J . of Geotech. Eng. D/v. , ASCE, V ol . 106,
No. GT3 .
Marsland, A. (1971) . "Larg e i n situ test s t o measur e the propertie s o f stif f fissur e clay, " Proc. 1s t
Australian Conf. a n Geomech, Melbourne, 1.
Matlock, H . (1970) . "Correlation s fo r Desig n o f Laterall y Loaded Pile s i n Sof t Clay, " Proc. 2nd
Offshore Tech. Conf., Houston , V ol. 1.
Matlock, H., and Reese, L.C. (1960) . "Generalized Solution s for Laterally Loaded Piles," J SMFD,
ASCE, V ol . 86, N. SMS , Part 1 .
Matlock, H. , an d Reese , L.C . (1961) . "Foundatio n Analysi s o f Offshor e Supporte d Structures, "
Proc. 5th Int. Conf. SM and FE, V ol . 2.
Matsuo, H. (1939) . Test s o n th e Latera l Resistanc e of Pile s (i n Japanese) , Researc h Inst . o f Civ.
Bugg, Min, of Home Affairs, Repor t No. 46.
McDonald, D.H, . and Skempton, A.W. (eds) (1955). "A Survey of Comparison between calculated
and observe d settlement s of structure s on clay, " Conf. o n Correlation o f Calculated an d
Observed Stresses and Displacements, Institutio n o f Civi l Engineers , London.
Means, R.E. , Parcher, J.V . (1965). Physical Properties o f Soils. Prentic e Hal l of India, New Delhi .
Menard, L. (1957). An Apparatus for Measuring the Strength of Soils in Place, M.Sc. Thesis, Univ.
Illinois, Urbana, U.S.A.
Menard, L . (1976) . "Interpretatio n an d Application of pressuremete r Tes t Result s t o Foundatio n
Design, Soils " Soils, No. 26.
Mesri, G. (1973) . "Coefficien t o f Secondar y Compression, " J . Soil Mech. Found. D/v. , ASCE, V ol.
99, SMI .
Mesri, G. , an d Godlewski , P.M . (1977) . "Tim e an d stress-compressibilit y interrelationship, "
J . Geotech. Eng., ASCE, 103 , No. 5 .
Mesri, G., and Choi , Y.K. (1985). "Settlemen t analysis of embankments on sof t clays, " J . Geotech.
Eng., ASCE, III , No. 4.
Mesri, G. , an d Choi , Y.K. (1985b) . "Th e Uniquenes s o f th e End-of-Primar y (EOP ) V oi d ratio -
effective Stres s relationship, " Proc. llth Int. Conf. o n Soil Mech. an d Found Engg., Sa n
Francisco, No. 2 .
Mesri, G. , an d Feng , T.W . (1986) . "Stress-strain-strai n rat e relatio n fo r th e compressibilit y of
sensitive natura l clays, " Discussion. Geotech., 36, No. 2 .
Mesri, G. , Feng , T.W. , AH , S. , an d Hayat , T.M. (1994) . "Permeablit y Characteristic s o f Sof t
Clays," Proc. 13th Int. Conf. O n Soil Mech and Found. Eng., New Delhi .
Referen ces 101 7
Meyerhof, G.G . (1951) . "Th e Ultimat e Bearin g Capacit y o f Foundation, " Geotechnique, V ol. 2,
No. 4.
Meyerhof, G.G . (1953) . "Th e Bearin g Capacit y o f Foundation s Unde r Eccentri c an d Incline d
Loads," 3rd ICSMFE, V ol . 1 .
Meyerhof, G.G. (1956) . "Penetratio n Test s and Bearing Capacity of Cohesionless Soils, " J SMFD,
ASCE,V o. 82, SMI.
Meyerhof, G.G. (1957) . "Discussion s o n Sand Densit y by Spoon Penetration, " 4t h ICSMFE, V ol .
3.
Meyerhof, G.G. (1957). "Th e Ultimat e Bearing Capacity of Foundations on Slopes," 4th ICSMFC,
V ol. 1 , London.
Meyerhof, G.G . (1959) . "Compactio n o f Sand s an d Bearin g Capacit y o f Piles, " J SMFD, ASCE ,
V ol. 85, SM6 .
Meyerhof, G.G. (1963) . Some Recent Research on Bearing Capacity o f Foundation, CGJ Ottawa,
V ol. 1 .
Meyerhof, G.G. (1965) . "Shallo w Foundations, " J SMFD, ASCE , V ol . 91, SM2 .
Meyerhof, G.G. (1974) . "Ultimat e Bearin g Capacity of Footings o n Sand Layer Overlyin g Clay, "
Canadian Geotech. J ., V ol. II, No. 2.
Meyerhof, G.G . (1975) . "Penetratio n Testin g Outsid e Europe, " Genera l Report , Proc. o f th e
European Symp. o n Penetration Testing, V ol. 2, Stockholm.
Meyerhof, G.G . (1976) . "Bearin g Capacit y an d Settlemen t o f Pil e Foundations, " J GED, ASCE ,
V ol. 102 , GT 3.
Meyerhof, G.G., an d Adams J.I. (1968). "Th e Ultimat e Uplift Capacit y of Foundations," Canadian
Geotech. J ., V ol . 5, No. 4.
Meyerhof, G.G. , and Hanna, A.M. (1978). "Ultimat e Bearin g Capacity of Foundations on Layered
Soil Under Inclined Load," Canadian Geotech. J ., V ol. 15, No. 4.
Mindlin, R.D. (1936) . "Force s at a point in the Interior of a Semi-Infinite Solid," Physics, V ol . 7.
Mindlin, R.D. (1939) . "Stres s distribution around a Tunnel," Trans ASCE. Am. Soc. Civil Engrs.
V ol. 104 .
Mitchell, J.K. , an d Freitag , D.R. (1959). " A Review and Evaluation of Soil-Cemen t Pavements, "
J . ofSM an d FD, ASCE, V ol . 85, No. SM6 .
Mohr, O . (1900) . Di e Elastizitatsgrenz e Und Bruse h eine s Material s (Th e elasti c limi t an d th e
failure o f a Material), Zeitschrif t V eneins Deuesche Ingenieure , V ol. 44.
Mononobe, N. (1929) . "On the Determinatio n of Eart h Pressure s Durin g Earthquakes "
Proceedings World Engineering Conference, V ol. 9.
Morgenstern, N.R . (1963) . "Stabilit y Chart s fo r Eart h slope s Durin g Rapi d Drawdown, "
Geotechnique, V ol . 13, No. 2 .
Morrison, C.S. , an d Rees e L.C . (1986) . " A Latera l loa d tes t o f a ful l scal e pil e grou p i n sand. "
Geotechnical Engineerin g center , Burea u o f Engineerin g Research , Universit y o f Texas ,
Austin.
Morrison, E.E . Jr. , an d Ebeling , R.M . (1995) . "Limi t Equilibriu m Computatio n o f Dynami c
Passive Eart h Pressure, " Canadian Geotech. J ., V ol. 32, No. 3.
1018 ' Referen ce s
Murphy, V .A . (1960) . "Th e effec t o f groun d wate r characteristic s o n th e sismi c desig n o f
structures," Proce. 2nd World Conf. o n Earthquake Engineering, Tokyo, Japan.
Murthy, V .N.S , an d Subb a Rao , K.S . (1995) . "Predictio n o f Nonlinea r Behaviou r o f Laterall y
Loaded Long Piles, " Foundation Engineer, V ol . 1, No. 2, New Delhi .
Murthy, V .N.S . (1965) . "Behaviou r o f Batte r Pile s Subjecte d t o Latera l Loads, " Ph.D . Thesis ,
Indian Institut e of Technology, Kharagpur , India.
Murthy, V .N.S. (1982). "Repor t o n Soil Investigation for the construction of cooling water system, "
Part I I of Farakha Supe r thermal Power Project , Nationa l Thermal Powe r Corporation , Ne w
Delhi.
Muskat, M. (1946) . Th e Flow o f homogeneous fluid s through porous media, McGraw Hil l Co.
Nagaraj, T.S. , an d Murthy , B.R.S . (1985) . "Predictio n o f th e Preconsolidatio n Pressur e an d
Recompression Inde x of Soils, " Geotech. Testing J ournal, V ol . 8. No. 4 .
Nelson, J.D. , Miller , D.J . (1992) . Expansive Soils: Problems an d Practice i n Foundation an d
Pavement Engineering, John Wiley & Sons, Inc. , New York.
Newmark, N.M. (1942) . "Influenc e Chart s fo r Computatio n o f Stresses i n Elastic Soils," Univ . of
Illinois Expt . Sin. , Bulletin No . 338 .
Nordlund, R.L. (1963) . "Bearin g Capacit y of Piles i n Cohesionless Soils, " J SMFD, ASCE , V ol . 89,
SM3.
Norris, G.M. , an d Holtz , R.D . (1981) . "Con e penetratio n Testin g an d Experience, " Proc. ASCE
N ational Convention, St . Louis, Missouri , Published by ASCE, NY.
O' Connor , M.J. , an d Mitchell , R.J. (1977) . "An Extension o f the Bisho p and Morgenster n Slop e
Stability Charts, " Canadian Geotech. J . V ol. 14.
O' Neill , M.W. , an d Reese , L.C . (1999) . "Drille d Shafts : Constructio n Procedure s an d Desig n
Methods" Repor t No . FHWA-IF-99-025 , Federa l Highwa y Administratio n Offic e o f
Infrastructure/Office o f Bridge Technology, HIBT, Washington D.C.
O'Neill, M.W. , Townsend, F. C., Hassan, K. H., Buttler A., and Chan, P. S. (1996). "Loa d Transfer
for Intermediat e Geomaterials, " Publicatio n No . FHWA-RD-95-171 , Federa l Highwa y
Administration, Office o f Enggniering and Highwa y Operations R & D, McLean, V . A.
O'Neill, M.W . (1988) . "Specia l Topics i n Foundations," Proc. Geotech. Eng. Div., ASCE National
Convention, Nashville.
Okabe, S . (1926) . "Genera l Theor y o f Eart h Pressures, " / . o f th e J ap. Soc. o f Civil Engineers,
Tokyo, V ol . 12 , No. 1 .
Osterberg, J.O . (1952) . "Ne w Pisto n Type Soi l Sampler, " Eng. N ews. Rec., 148 .
Osterberg, J.O . (1957) . "Influenc e V alue s fo r V ertica l Stresse s i n Semi-Infinit e Mas s du e t o
Embankment Loading, " Proc. Fourth Int. Conf. Conf. SM an d FE , V ol . 1 .
Ostwald, W . (1919) . A Handbook o f Colloid Chemistry, Trans , b y M.H . Fisher , Philadelphia ,
P. Balkiston's So n an d Co .
Palmer, L.A. , an d Thompson, J.B . (1948) . "Th e Eart h Pressur e an d Deflectio n alon g Embedde d
Lengths of Piles Subjecte d t o Lateral Thrusts, " Proc. 2nd Int. Conf. S M and FE, Rottterdam,
V o. 5 .
Referen ces 101 9
Pauker, H.E. (1889). "An explanatory report on the project of a sea-battery" (in Russian), J ournal
of th e Ministry o f Ways an d Communications, St . Petersburg,
Peck, R.B . (1969) . "Dee p excavation s an d Tunnelin g i n Sof t Ground, " Proc., 7t h Int. Conf. o n
SMFE, Mexic o city.
Peck, R.B. , and Byrant , KG. (1953) . "Th e Bearin g Capacit y Failur e o f th e Transcon a Elevator, "
Geotechnique, V ol . 3, No. 5.
Peck, R.B. , Hanson, W.E., and Thornburn, T.H., (1974) . Foundation Engineering, Joh n Wiley and
Sons Inc. , New York.
Perloff, W.H. , and Baron , W. (1976). Soil Mechanics, Principles and Applications, Joh n Wiley &
Sons, Ne w York.
Petterson, K.E . (1955) . "Th e Earl y Histor y o f Circula r Slidin g Surfaces, " Geotechnique. Th e
Institution of Engineers, London, V ol. 5.
Poncelet, J.V . (1840) . "Memoir e su r l a stabilit e de s reretment s e t d e leur s fondation, " Not e
additionelle sur les relations analytiques qui lient entre elles la poussee et la butee de la terre.
Memorial d e 1'officie r d u genie, Paris, V ol. 13.
Poulos, H.G . (1974) . Elastic Solutions for Soil and Rock Mechanics, Joh n Wile y & Sons , Ne w
York.
Poulos, H.G. , an d Davis, E.H. (1980) . Pile Foundation Analysis and Design, John Wiley & Sons,
New York.
Prakash, S. , and Saran , S . (1966) . "Stati c an d Dynami c Earth Pressur e Behin d Retaining Walls,"
Proc. 3rd Symp. o n Earthquake Engineering, Roorke , India, V ol. 1.
Prandtl, L. (1921) . "Uber di e Eindringungsflstigkei t plastische r Baustoff e un d di e Festigkei t vo n
Schneiden," Zeit. angew. Math., 1 , No. 1 .
Proctor, R.R. (1933). "Fou r Articles on the Design and Construction of Rolled Earth-Dams, " Eng.
N ews Record, V ol. 3.
Rankine, W.J.M. (1857). "On the Stability of Loose Earth Dams," Phil . Trans. Royal Soc., V ol. 147,
London.
Reese, L.C . (1975) . "Laterally Loade d Piles, " Proc. of the Seminar Series, Design, Construction
and Performanc e o f Dee p Foundations : Geotech . Grou p an d Continuin g Educatio n
Committee, San Francisco Section , ASCE, Berkeley .
Reese, L.C. (1985). "Behavio r o f Piles and Pile Groups under Lateral Load," Technical Report No.
FHWA/RD-85/106., Federa l Highwa y Administration. Office o f Engineerin g an d Highway
operations, Research an d Development, Washington D.C.
Reese, L.C. , and Matlock, H. (1956). "Non-dimensiona l solutions for Laterally Loaded Pile s wit h
Soil Modulus assumed Proportional to Depth," Proc. 8th Texas Conf. S.M. and F.E. Spec. Pub
29, Burea u of Eng. Res. , Univ. of Texas, Austin.
Reese, L.C. , Cox , W.R. , an d Koop , F.D . (1974) . "Analysi s o f Laterall y Loade d Pile s i n Sand, "
Proc. 6th Offshore Tech. Conf. Houston , Texas.
Reese, L.C. , an d Welch , R.C . (1975) . "Latera l Loading s o f Dee p Foundations , i n Stif f Clay, "
J GED, ASCE, V ol. 101, GT 7.
Reese, L.C. , an d O'Neill , M.W. (1988). "Fiel d Load Test s of Drilled Shafts, " Proc., International
Seminar on Deep Foundations and Auger Piles, V an Impe (ed.), Balkema, Rotterdam.
1020 Refer en ce s
Rendulic, L . (1935) . "De r Hydrodynamisch e spannungsaugleic h i n zentra l entwasserte n
Tonzylindern," Wasserwirtsch. u. Technik, V ol. 2.
Reynolds, O . (1883) . "A n Experimenta l investigatio n o f th e circumstance s whic h determin e
whether th e motion of water shall be Direct or Sinuous and the Law of Resistance in Parallel
Channel," Trans . Roya l Soc . V ol . 174, London .
Richards, R. , an d Elms , D.G. (1979) . "Seismi c Behavio r o f Gravit y Retaining Walls," J ou. o f the
Geotech. Eng. Div., ASCE, V ol . 105, No . GT4 .
Robertson, P.K. , and Campanella, R.G. (1983a). "Interpretation o f Cone Penetratio n Tests," Part I -
Sand, CBJ , Ottawa , V ol. 20, No . 4.
Robertson, P.K. , and Companella, R.G. (1983b). "SPT-CPT Correlations," J GED, ASCE , V ol. 109.
Rowe, P.W. (1952) "Anchore d Shee t Pil e Walls," Proc. Inst. of Civ. Engrs., V ol. 1 , Par t 1 .
Sanglerat, G. (1972). The Penetrometer and Soil Exploration, Elsevier Publishing Co., Amsterdam.
Saran, S. , an d Prakash , S . (1968) . "Dimensionles s parameter s fo r Stati c an d Dynami c Eart h
Pressure fo r Retainin g Walls," India n Geotechnical Journal , V ol. 7, No. 3 .
Scheidig, A. (1934). De r Loss (Loess) , Dresden .
Schmertmann, J.H . (1955) . "Th e Undisturbe d Consolidatio n Behavio r o f Clay, " Trans . ASCE ,
No. 120 .
Schmertmann, J.H . (1970) . "Stati c Con e t o Comput e Stati c Settlemen t Ove r Sand, " J SMFD,
ASCE, V ol . 96, S M 3.
Schmertmann, J.H. (1978) . Guidelines for Cone Penetration Test: Performance an d Design. U.S .
Dept. o f Transportation, Washington , D.C.
Schmertmann, J.H . (1986) . "Suggeste d Metho d fo r Performin g th e Fla t Dilatomete r Test, "
Geotech. Testing J ournal, ASTM, V ol . 9, No. 2.
Schmertmann, J.M. (1975) . "Measuremen t o f i n situ shea r strength, " Proc. ASCE Specially Conf.
on I n Situ Measurement o f Soil Properties, Raleigh , 2.
Scott, R.F . (1963) . Principles o f Soil Mechanics, Addison-Wesel y Publishin g Co. , Inc . London, .
Seed, H.B. , Woodward , R.J. , an d Lundgren , R. (1962) . "Predictio n o f Swellin g Potentia l fo r
Compacted Clays, " J ournal ASCE, SMFD, V ol . 88, No. S M 3.
Seed, H.B. , an d Whitmann , R.V . (1970) . "Desig n o f Eart h Retainin g Structure s fo r Dynami c
Loads," ASCE , Spec . Conf . Latera l stresse s i n th e groun d an d desig n o f eart h retainin g
structures.
Semkin, V .V ., Ermoshin, V .M., and Okishev, N.D. (1986). "Chemical Stabilizatio n of Loess Soil s in
Uzbekistan," Soi l Mechanic s an d Foundatio n Engineerin g (trans , fro m Russian) , V ol. 23 ,
No. 5.
Sichardt, W. (1930). "Grundwasserabsenkung be i Fundierarbeiten, " Berlin, Juliu s Springer .
Simons, N.E . (1960) . "Comprehensiv e Investigatio n o f th e Shea r Strengt h o f a n Undisturbe d
Drammen Clay, " Proc. Res. Conf. Shear Strength Cohesive Soils, ASCE.
Simons, N.E., and Som, N.N . (1970). Settlemen t of Structures on Clay wit h Particular emphasis on
London Cla y Industr y Research Institute , Assoc. Repor t 22.
Skempton, A.W . (1944) . "Note s o n th e compressibilitie s o f clays, " Q.J . Geological. Soc,
London, C (C: Parts 1 & 2).
Referen ces 102 1
Skempton, A.W . (1951) , "Th e Bearin g Capacit y o f Clays, " Proc. Building Research Congress,
V ol. 1 , London.
Skempton, A.W. (1953). "The Colloidal Activity of Clays," Proceedings, 3rd Int. Conf. SMandFE,
London, V ol. 1.
Skempton, A.W. (1954), "Th e Por e Pressure Coefficient s A and B," Geotechnique, V ol. 4.
Skempton, A.W. (1957) , "Th e plannin g an d desig n o f th e ne w Hongkon g airpor t Discussion, "
Proc. Int. of Civil Engineers (London) , V ol. 7.
Skempton, A.W. (1959). "Cast-in-situ Bored Piles in London Clay, " Geotech, V ol . 9.
Skempton, A.W., and Northey, R.D. (1952). "The Sensitivit y of Clays, " Geotechnique, V ol . HI.
Skempton, A.W., and Northey, R.D. (1954)., "Sensitivity of Clays," Geotechnique, V ol . 3, London.
Skempton, A.W. , an d Bjerrum , L . (1957) . " A Contributio n t o th e Settlemen t Analysi s o f
Foundations on Clay, " Geotechnique 7 , London.
Skempton, A.W., Yassin, A.A., and Gibson, R.E. (1953) . "Theori e de l a force Portaut e de s pieuse
dans l e sable, " Annales de 1, " Institut Technique, Batiment 6.
Smith, R.E. , and Wahls, H.E. (1969). "Consolidatio n unde r Constant Rate of Strain" J . Soil Mech.
Found. Div., ASCE, V ol . 95, SM2 .
Sowa, V .A. (1970). "Pullin g Capacit y of Concrete cast-in-situ Bore d Pile," Can. Geotech. J ., V ol. 7.
Spangler, M.G . (1938) . "Horizonta l Pressure s o n Retainin g Walls du e t o Concentrate d Surfac e
Loads," IOWA State Universit y Engineering Experiment Station, Bulletin, No. 140 .
Spencer, E . (1967) . " A Method o f Analysis of th e Stabilit y of Embankments , Assuming Paralle l
Inter-slice Forces, " Geotechnique, V ol . 17, No. 1 .
Stas, C.V . , an d Kulhawy , F.H . (1984) . "Critical Evaluatio n of Desig n Method s fo r Foundation s
under Axial Uplift an d Compression Loading," EPRI Report EL-3771, Palo Alto, California.
Steinbrenner, W . (1934) . "Tafel n zu r Setzungsberechnung, " V ol . 1. , No . 4 , Schriftenreih e de r
strasse 1 , Strasse.
Stokes, G.G. (1856). "On the effect o f the Internal Friction of Fludis on the Motion of Pendulium"
Trans. Cambridge Philosophica l Society , V ol. 9, Part 2.
Szechy, C. (1961). Foundation Failures, Concrete Publications Limited, London .
Tavenas, F. and Leroueil , S . (1987) . "Laborator y an d Stress-strain-time behaviour of sof t clays, "
Proc. Int. Symp o n Geotech. Eng. of soft soils, Mexico City, 2.
Taylor, D.W. (1937) . "Stabilit y of Earth Slopes, " J . of Boston Soc. of Civil Engineers, V ol . 24.
Taylor, D.W. (1948) . Fundamentals o f Soil Mechanics, Joh n Wiley and Sons , Ne w York.
Teng, W.C. (1969) . Foundation Design, Prentice Hall , Englewood Cliffs , NJ .
Terzaghi, K. (1925). "Erdbaumechanik. " Franz, Deuticke , V ienna.
Terzaghi, K. (1943). Theoretical Soil Mechanics, Wile y & Sons, New York.
Terzaghi, K . (1951) . "Th e Influenc e of Moder n Soi l Studie s o n th e Desig n an d Constructio n of
Foundations" Bldg . Research Congr. , London, 1951 , Div . 1, Part III .
1022 Refer en ce s
Terzaghi, K . (1955) . "Evaluatio n o f th e Co-efficien t o f Sub-grad e Reaction, " Geotechnique,
Institute o f Engineers , V ol. 5, No. 4 , London .
Terzaghi, K. , an d Frohlich , O.K . (1936) . Theori e de r setzun g vo n Tonschichte n (Theor y o f
Settlement o f Clay Layers) , Leipzig, Deuticke.
Terzaghi, K. , an d Peck , R.B . (1948). Soil Mechanics i n Engineering Practice, Wiley, New York.
Terzaghi, K. , and Peck, R.B . (1967). Soil Mechanics i n Engineering Practice, John Wiley & Sons,
N.Y.
Terzaghi, K. , Peck , R.B. , an d Mesri , G . (1996) . Soil Mechanics i n Engineering Practice, Joh n
Wiley & Sons , Inc. , Third Edition, New York.
Timoshenko, S.P. , an d Goodier , J.N . (1970) . Theory o f Elasticity, Thir d Ed. , McGra w Hill , New
York.
Tomlinson, M.J . (1977) . Pile Design an d Construction Practice, V iewpoint Publications, London .
Tomlinson, M.J . (1986) . Foundation Design an d Construction, 5th Ed., New York, John Wiley and
Sons.
Tschebotarioff, G.P . (1953) . "Th e Resistanc e o f Latera l Loadin g o f Singl e Pile s an d o f Pil e
Group." AST M specia l Publicatio n No. 154 , V ol . 38.
Tschebotarioff, G.P . (1958) . Soil Mechanics Foundation and Earth Structures, McGraw Hill , New
York.
Tsytovich, N . (1986) . Soil Mechanics, Mi r Publishers , Moscow.
Turnbull, W.J . (1950) . "Compactio n an d Strengt h Test s o n Soil, " Presente d a t Annua l Meeting ,
ASCE (January) .
U.S. Arm y Enginee r Wate r Way s Experiment s Station . (1957) . Th e Unifie d Soi l Classificatio n
system Technica l memo No.3-357, Corps o f Engineers, V icksburg, U.S.A.
U.S. Corp s o f Engineers. (1953) . Technica l Memo 3-360 , Filter Experiment s an d Design Criteria ,
U.S Waterways Experimenta l Station, V icksburg, Miss.
U.S. Departmen t o f the Navy. (1971). "Desig n Manua l - Soi l Machanics , Foundations , and Earth
Structures," NAV F AC DM-7, Washington, D.C.
V an Der Merwe , D.H . (1964) . "Th e Predictio n o f Heave fro m th e Plasticit y Inde x an d Percentag e
clay Fractio n o f Soils, " Civil Engineer i n South Africa, V ol . 6, No . 6.
V an Wheele, A.F . (1957) . " A Method o f separatin g th e Bearin g Capacit y o f a Tes t Pil e int o Ski n
Friction an d Poin t Resistance, " Proc. 4th Int. Conf. S M and FE, V ol . 2.
V ander V een , C. , an d Boersma , L . (1957) . "Th e Bearin g Capacit y o f a Pil e Predetermine d b y a
Cone Penetratio n Test, " Proc. 4th Int. Conf SM . an d F.E., V ol 2 .
V erbrugge, J.C . (1981) . "Evaluatio n d u Tassemen t de s Pieur e a Parti r d e 1 ' Essa i d e Penetratio n
Statique," Revue Francaise de Geotechnique, No. 15 .
V esic, A.S . (1956) . "Contributio n a L' Extud e de s Foundation s Su r Pieu x V erticau x e t Inclines, "
Annales de s Travaux Public s de Belgique, No.6.
V esic, A.S. (1961) . "Bendin g o f Bea m Restin g on Isotropi c Elasti c Solid, " J ou. Eng. Mechs. Div.,
ASCE, V ol . 87, No. EM2 .
Referen ces 102 3
V esic, A.S. (1963). "Bearin g Capacit y of Deep Foundations in Sand," Highway Research Record ,
No. 39, Highway Research Board , Washington D.C.
V esic, A.S. (1963) . "Discussion-Sessio n III, " Proc. of th e 1st. Int. Conf. o n Structural Design o f
Asphalt Pavements, Universit y of Michigau.
V esic, A.S. (1964). "Investigatio n of Bearing Capacity of Piles i n Sand," Proc. N o. Amer. Conf. o n
Deep Foundations, Mexico city, V ol 1 .
V esic, A.S. (1965). "Ultimate Loads and Settlements of Deep Foundations," Proc. of a Symp hel d at
Duke University.
V esic, A.S. (1967) . " A Study of Bearing Capacity of Dee p Foundations, " Final Report , School of
Civil Engg. , Georgia Inst . Tech., Atlanta, U.S.A.
V esic, A.S . (1969) . "Effec t o f Scal e an d Compressibilit y o n Bearin g Capacit y o f Surfac e
Foundations," Discussions, Proceedings, 7t h Int. Conf. SM and FE. Mexico City, V olume III.
V esic, A.S. (1970) . "Test s on Instrumente d Piles - Ogeechee Rive r Site, " J SMFD, ASCE , V ol.96,
No SM2 .
V esic, A.S. (1972). "Expansio n o f cavities in Infinit e Soi l Mass, " J .S.M.F.D., ASCE , V ol . 98.
V esic, A.S. (1973). "Analysis of Ultimate Loads of Shallow Foundations," J SMFD, ASCE , V ol. 99,
SMI.
V esic, A.S. (1975) . "Bearin g Capacit y o f Shallo w Foundations, " Foundation Engineering Hand
Book, V an Nostrand Reinhol d Book Co. , N.Y.
V esic, A.S . (1977) . Design o f Pile Foundations, Synthesi s o f Highwa y Practic e 42 , Res . Bd. ,
Washington D.C .
V idal, H. (1969). "Th e Principl e of Reinforced Earth," HRR No. 282.
V ijayvergiya, V .N. , and Focht , J.A. Jr., (1972). A N ew way to Predict the Capacity of Piles i n Clay,
4th Annual Offshore Tech Conf., Houston , V ol. 2.
V ijayvergiya, V .N., and Ghazzaly, O.I. (1973). "Prediction of Swelling Potential of Natural Clays,"
Third International Research and Engineering Conference on Expansive Clays.
Walsh, K.D., Houston, S.L., an d Houston, W.N. (1995). "Development o f t-z Curves for Cemented
Fine-Grained Soi l Deposits, " J ournal o f Geotechnical Engineering, ASCE , V ol . 121.
Westergaard, H.M. (1917) . "Th e Resistanc e of a group of Piles," /. Western Society o f Engineers,
V ol. 22 .
Westergaard, H.M . (1926) . Stresse s i n Concret e Pavemen t Compute d b y Theoretica l Analysi s
Public Road, V ol. 7, No. 12 , Washington.
Westergaard, H.M. (1938). "A Problem of Elasticity Suggested by Problem in Soil Mechanics, Sof t
Material Reinforce d b y Numerou s Stron g Horizonta l Sheets, " i n Contribution t o th e
Mechanic o f Solids, Stephe n Timoshenko 60t h Anniversary V ol., Macmillan, New York.
Winterkorn, H.F., and Hsai-Yang Fang. (1975). Foundation Engineering Hand Book, V an Nostrand
Reinhold Company, New York.
Woodward, Jr. , R.J. , Gardner , W.S. , an d Greer , D.M. (1972) . Drilled Pier Foundations, McGra w
Hill, New York.
Yoshimi, Y . (1964) . "Pile s i n Cohesionles s Soil s Subjecte d t o Obliqu e Pull, " J SMFD, ASCE ,
V ol. 90, SM6 .
INDEX
Activity 5 7
Adsorbed wate r 15 , 1 6
Angle of obliquit y 25 3
Angle of wal l friction 254 , 42 1
Anisotropic soi l 11 6
Apparent cohesio n 255 , 30 0
Aquifer 9 7
confined 99 , 10 0
unconfined 97 , 9 8
Atterberg limit s 4 6
flow curv e 4 8
liquid limi t 46-5 0
plastic limi t 4 9
shrinkage limi t 5 0
B
Base exchang e 16 , 1 7
Beaming capacity , shallo w foundatio n 48 1
based on CPT 51 8
based on SPT 518 , 51 9
bearing capacit y factor s 489^1-90 , 493 -
494, 50 4
case history , Transcona 533-53 4
depth factor s 50 5
design chart s 555 , 55 8
effect o f compressibilit y 50 9
effect o f eccentri c loadin g 515 , 588
effect o f size of footings 55 4
effect o f water table 494-49 6
empirical correlation s 558-55 9
equation, Terzaghi 48 9
footings o n stratifie d deposit s 521-52 6
foundation o n rock 53 2
foundations o n slop e 52 9
general equatio n 50 3
gross allowabl e 484 , 49 3
load inclinatio n factor s 50 5
net allowabl e 484 , 493, 545
net ultimat e 484 , 493
plate loa d test s 54 8
safe bearing pressur e 48 5
safety facto r 484 , 49 3
seat of settlement 56 2
settlement chart s 55 5
settlement computatio n 561-57 1
settlement differentia l 54 7
settlement permissibl e 54 8
settlements (max ) 54 7
shape factors 50 5
ultimate 483-484 , 489 , 491
Boiling conditio n 14 8
Boring o f hole s
auger metho d 31 8
rotary drillin g 32 0
1025
1026
I ndex
wash boring 31 9
Boussinesq poin t loa d solutio n 3 , 17 4
Capillary wate r 149 , 154 , 15 6
contact angl e 15 0
pressure 14 9
rise i n soi l 14 9
siphoning 15 4
surface tensio n 149 , 15 0
Classical eart h pressur e theor y 2
Coulomb's theor y 2
Rankine' theor y 2
Classification o f soil s 6 9
AASHTO 7 0
textural 6 9
USCS 7 3
Clay minera l 1 1
composition 1 1
formation 1 2
Ulite 11 , 1 4
Kaolinite 11 , 1 2
Montmorillonite 11 , 1 4
structure 11-1 5
Clays
high sensitivit y 21 9
low t o medium sensitivit y 21 9
normally consolidate d 217 , 22 0
overconsolidated 217 , 22 0
Coefficient o f friction 25 4
at rest eart h pressur e 42 2
compressibility 22 2
consolidation 236 , 24 0
earth pressure , activ e 42 7
earth pressure , passiv e 42 8
volume compressibilit y 22 2
Collapse potentia l 79 5
Collapse settlemen t 79 6
Compression 20 9
immediate 20 9
primary 20 9
secondary 20 9
Compression inde x 219 , 223 , 22 4
Conjugate confoca l parabola s 12 7
Consistency inde x 5 5
Consistency limit s 3 , 4 5
Consolidation 207 , 20 8
degree of consolidation 23 8
one-dimensional 209 , 210 , 233
settlement 20 9
test 21 3
time factor 23 6
Consolidometer 21 2
Coulomb's eart h pressur e 45 2
coefficient fo r active 45 4
coefficient fo r at-rest 42 2
coefficient fo r passive 45 6
for activ e stat e 45 2
for passiv e state 45 5
Critical hydrauli c gradient 14 8
Curved surfaces of failure 46 2
earth pressur e coefficien t 46 6
for passiv e stat e 46 2
D
Darcy's law 8 9
Degree o f consolidation 23 8
Density 2 1
Diffused doubl e laye r 1 5
Dilatometer tes t 34 9
Discharge velocit y 9 1
Drilled pie r foundatio n 74 1
design consideration s 75 1
estimation o f vertica l settlemen t 76 5
lateral bearin g capacit y 77 9
methods o f construction 74 3
types 74 1
uplift capacit y 77 7
vertical bearin g capacit y 75 4
vertical bearin g capacit y equatio n 75 5
vertical loa d transfe r 75 2
vertical ultimat e ski n resistanc e 760 , 763 ,
764
Effective diamete r 15 4
Effective stres s 144 , 27 4
Electrical resistivit y metho d 35 4
Embankment loadin g 19 1
Expansion inde x 81 0
Factor o f safet y wit h respect t o cohesion 36 8
safety wit h respect t o heave 13 2
with respec t t o height 36 8
with respec t t o shearing strengt h 36 8
Floating foundatio n 59 5
Flow ne t constructio n 11 6
Flow valu e 26 3
In dex 1027
Free swell 80 4
Geophysical exploratio n 35 2
Grain siz e distribution 4 3
coefficient o f curvature 4 4
gap grade d 4 3
uniformity coefficien t 4 3
uniformly grade d 4 3
well graded 4 3
H
Hydraulic conductivit y 90 , 9 1
by bore-hol e test s 10 1
by constan t hea d 9 2
by pumpin g test 9 7
empirical correlation s 10 3
falling hea d metho d 9 3
for stratifie d layers 10 2
of rocks 11 2
Hydraulic gradien t 87 , 14 7
critical 14 8
Hydrometer analysi s 35 , 38 , 3 9
I
Isobar 19 8
Laminar flow 8 8
Laplace equatio n 11 4
Lateral eart h pressur e 41 9
active 42 0
at rest 42 0
passive 42 0
Leaning Tower of Pisa 2
Linear shrinkage 5 6
Liquid Limi t
by Casagrand e metho d 4 7
by fal l con e method 4 9
by one-poin t metho d 4 8
Liquidity inde x 5 4
M
Meniscus 39 , 41 , 15 2
Meniscus correction 4 1
Mohr circl e of stress 264-26 6
diagram 265 , 269, 27 0
Mohr-coulomb failur e theor y 268 , 26 9
N
Newmark's influence char t 188 , 19 0
influence valu e 18 9
o
Oedometer 21 2
Origin o f planes 266 , 27 1
Overconsolidation rati o 30 6
Percent fine r 4 0
Permeability tes t 92-10 1
Phase relationship s 19-2 5
Pile grou p 67 4
allowable load s 69 0
bearing capacit y 67 8
efficiency 67 6
negative frictio n 69 1
number and spacing 67 4
settlement 680 , 68 9
uplift capacit y 69 4
Piles batte r laterall y loade d 73 1
Piles, vertica l 60 5
classification 60 5
driven 60 7
installation 61 0
selection 60 9
types 60 6
Piles, vertica l loa d capacit y 61 3
adhesion facto r a 63 3
jS-method 63 3
Coyle and Castello' s method 62 8
bearing capacit y o n rock 67 0
general theor y 61 8
Janbu's method 62 8
A-method 63 3
load capacit y b y CPT 65 2
load capacit y b y load tes t 66 3
load capacit y b y SPT 63 5
load capacit y fro m dynami c formula 66 6
load transfer 61 4
methods o f determining 61 7
Meyerhof's metho d 62 4
settlement 68 0
static capacit y i n clays 63 1
t-zmethod 683
Tomlinson's solutio n 62 2
1028
I ndex
ultimate ski n resistanc e 62 9
uplift resistanc e 67 1
V esic's method 62 5
Piles vertical loaded laterall y 699
Broms' solution s 70 9
case studies 72 2
coefficient o f soi l modulu s 70 3
differential equatio n 70 1
direct metho d 71 6
Matlock an d Reese metho d 70 4
non-dimensional solution s 70 4
p-y curve s 70 6
Winkler's hypothesis 70 0
Piping failur e 131 , 94 5
Plastic limi t 4 9
Plasticity char t 59, 75
Plasticity inde x 5 3
Pocket penetromete r 30 4
Pole 26 6
Pore pressur e parameters 29 8
Pore water pressures 14 4
Porosity 2 1
Preconsolidation pressur e 21 8
Pressure bul b 19 8
Pressuremeter 34 3
Pressuremeter modulu s 34 6
Principal plane s 260 , 26 3
Principal stresse s 260 , 263 , 27 5
Proctor test 95 2
modified 95 4
standard 95 3
Pumping tes t 9 7
Q
Quick san d conditio n 14 8
Radius of influence 9 9
Rankine's eart h pressur e 42 5
Relative densit y 24 , 4 4
Reynolds Number 8 9
Rock classificatio n 5
minerals 5 , 6
weathering 7
Rock qualit y designation 326 , 53 2
Secondary compressio n 22 4
coefficient 22 4
compression inde x 22 5
settlement 22 4
Seepage 11 4
determination 12 0
flow net 114-116 , 12 7
Laplace equatio n 11 4
line locatio n 13 0
loss 12 8
pressure 122 , 123 , 14 7
Seepage velocit y 9 1
Seismic refraction method 35 3
Settlement
consolidation settlemen t 219-22 3
secondary compressio n 22 4
Skempton's formul a 22 3
Settlement rate 24 2
Shear test s 27 6
consolidated-drained 27 7
consolidated-undrained 27 7
unconsolidated-undrained 27 7
Shrinkage limi t 5 0
Shrinkage rati o 5 5
Sieve analysi s 3 3
Significant dept h 19 9
Soil classificatio n 10 , 33 9
Soil particl e 9
size an d shap e 9 , 3 2
size distribution 4 3
specific gravit y 2 2
specific surfac e 9
structure 17 , 1 8
Soil permeabilit y 8 7
Soils
aeolin 8
alluvial 8
classification 6 9
colluvial 8
glacial 8
identification 6 8
inorganic 8
lacustrine 8
organic 8
residual 8
transported 8
Specific gravit y 2 2
correction 4 2
Stability analysi s of finit e slopes
Bishop and Morgenster n metho d 40 4
Bishop's metho d 40 0
Culmann metho d 37 6
Friction-circle metho d 38 2
Index 1029
( f)
u
= 0 method 38 0
Morgenstern metho d 40 5
slices method , conventiona l 39 3
Spencer metho d 40 9
Taylor's stabilit y numbe r 38 9
Standard penetratio n tes t 322 , 327
standardization 32 7
Static cone penetratio n tes t 33 2
Stokes' law 3 6
Stress, effectiv e 143-14 4
pore water 143-14 4
Suction pressure 14 9
Surface tensio n 149-150 , 15 5
Swell inde x 223 , 811
Swelling potentia l 804 , 812-81 3
pressure 80 4
Taj Maha l 2
Taylor's stabilit y number 389 , 39 0
Thixotropy 5 9
Torvane shear test 30 2
Toughness inde x 5 4
Transcona grai n elevato r 533 , 536
Turbulent flow 8 8
u
Unconfined aquife r 9 7
Unconfined compressiv e strengt h 5 8
related t o consistency 5 8
Uplift pressur e 12 3
V
V ane shear test 30 0
V elocity
discharge 90 , 9 1
seepage 90 , 9 1
V oid ratio 2 1
V olumetric Shrinkage 5 6
w
Water content 2 1
Westergaard's poin t load formul a 17 5
Zero air void line 95 5

Sponsor Documents

Or use your account on DocShare.tips

Hide

Forgot your password?

Or register your new account on DocShare.tips

Hide

Lost your password? Please enter your email address. You will receive a link to create a new password.

Back to log-in

Close