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Copyright © 2007, New Age International (P) Ltd., Publishers
Published by New Age International (P) Ltd., Publishers
All rights reserved.
No part of this ebook may be reproduced in any form, by photostat, microfilm,
xerography, or any other means, or incorporated into any information retrieval
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[email protected]
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NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS
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Visit us at www.newagepublishers.com
ISBN (13) : 9788122425512
Pro¦oco
Mechanical Engineering being core subject of engineering and Technology, is taught to almost all branches
of engineering, throughout the world. The subject covers various topics as evident from the course content,
needs a compact and lucid book covering all the topics in one volume. Keeping this in view the authors
have written this book, basically covering the cent percent syllabi of Mechanical Engineering (TME
102/TME202) of U.P. Technical University, Lucknow (U.P.), India.
From 2004–05 Session UPTU introduced the New Syllabus of Mechanical Engineering which covers
Thermodynamics, Engineering Mechanics and Strength of Material. Weightage of thermodynamics is 40%,
Engineering Mechanics 40% and Strength of Material 20%. Many topics of Thermodynamics and Strength
of Material are deleted from the subject which were included in old syllabus but books available in the
market give these useless topics, which may confuse the students. Other books cover 100% syllabus of this
subject but not covers many important topics which are important from examination point of view. Keeping
in mind this view this book covers 100% syllabus as well as 100% topics of respective chapters.
The examination contains both theoretical and numerical problems. So in this book the reader gets
matter in the form of questions and answers with concept of the chapter as well as concept for numerical
solution in stepwise so they don’t refer any book for Concept and Theory.
This book is written in an objective and lucid manner, focusing to the prescribed syllabi. This book
will definitely help the students and practicising engineers to have the thorough understanding of the
subject.
In the present book most of the problems cover the Tutorial Question bank as well as Examination
Questions of U.P. Technical University, AMIE, and other Universities have been included. Therefore, it is
believed that, it will serve nicely, our nervous students with end semester examination. Critical suggestions
and modifications by the students and professors will be appreciated and accorded
Dr. U.K. Singh
Manish Dwivedi
Feature of book
1. Cover 100% syllabus of TME 101/201.
2. Cover all the examination theory problems as well as numerical problems of thermodynamics, mechanics
and strength of materials.
3. Theory in the form of questions – Answers.
4. Included problems from Question bank provided by UPTU.
5. Provided chapterwise Tutorials sheets.
6. Included Mechanical Engineering Lab manual.
7. No need of any other book for concept point of view.
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lMPCRT/NT CCN\ERSlCN/FCRMÜL/
1. Sine Rule
R
Q
P
O 180–γ
180–α
α
180–β
γ
sin(180 ) sin(180 ) sin(180 )
P Q R
= =
−α −β − γ
2. Important Conversion
1 N = 1 kg X 1 m/sec
2
= 1000 gm X 100 cm/sec
2
g = 9.81 m/sec
2
1 H.P. = 735.5 KW
1 Pascal(Pa) = 1N/m
2
1KPa = 10
3
N/m
2
1MPa = 10
6
N/m
2
1GPa = 10
9
N/m
2
1 bar = 10
5
N/m
2
3. Important Trigonometrical Formulas
1. sin (A + B) = sin A.cos B + cos A.sin B
2. sin (A – B) = sin A.cos B  cos A.sin B
3. cos (A + B) = cos A.cos B – sin A.sin B
4. cos (A – B) = cos A.cos B + sin A.sin B
5. tan (A + B) = (tan A + tan B)/(1 – tan A. tan B)
6. tan (A – B) = (tan A – tan B)/(1 + tan A. tan B)
7. sin2 A = 2sin A.cos A
8. sin
2
A + cos
2
A = 1
9. 1 + tan
2
A = sec
2
A
10. 1 + cot
2
A = cosec
2
A
11. 1 + cosA = 2cos
2
A/2
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CCNTENTS
Preface v
Syllabus
Important Conversion/Formula
Part– A: Thermodynamics (40 Marks)
1. Fundamental concepts, definitions and zeroth law 1
2. First law of thermodynamics 30
3. Second law 50
4. Introduction of I.C. engines 65
5. Properties of steam and thermodynamics cycle 81
Part – B: Engineering Mechanics (40 Marks)
6. Force : Concurrent Force system 104
7. Force : Non Concurrent force system 141
8. Force : Support Reaction 166
9. Friction 190
10. Application of Friction: Belt Friction 216
11. Law of Motion 242
12. Beam 265
13. Trusses 302
Part – C: Strength of Materials (20 Marks)
14. Simple stress and strain 331
15. Compound stress and strains 393
16. Pure bending of beams 409
17. Torsion 432
Appendix
1. Appendix Tutorials Sheets 448
2. Lab Manual 474
3. Previous year question papers (New syllabus) 503
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Fundamental Concepts, Definitions and Zeroth Law / 1
+0)264
1
FÜND/MENT/L CCNCEPTS, DEFlNlTlCNS
/ND ZERCTH L/W
Q. 1: Define thermodynamics. Justify that it is the science to compute energy, exergy and entropy.
(Dec–01, March, 2002, Jan–03)
Sol : Thermodynamics is the science that deals with the conversion of heat into mechanical energy. It is
based upon observations of common experience, which have been formulated into thermodynamic laws.
These laws govern the principles of energy conversion. The applications of the thermodynamic laws and
principles are found in all fields of energy technology, notably in steam and nuclear power plants, internal
combustion engines, gas turbines, air conditioning, refrigeration, gas dynamics, jet
propulsion, compressors,
chemical process plants, and direct energy conversion devices.
Generally thermodynamics contains four laws;
1. Zeroth law: deals with thermal equilibrium and establishes a concept of temperature.
2. The First law: throws light on concept of internal energy.
3. The Second law: indicates the limit of converting heat into work and introduces the principle of
increase of entropy.
4. Third law: defines the absolute zero of entropy.
These laws are based on experimental observations and have no mathematical proof. Like all physical
laws, these laws are based on logical reasoning.
Thermodynamics is the study of energy, energy and entropy.
The whole of heat energy cannot be converted into mechanical energy by a machine. Some portion of
heat at low temperature has to be rejected to the environment.
The portion of heat energy, which is not available for conversion into work, is measured by entropy.
The part of heat, which is available for conversion into work, is called energy.
Thus, thermodynamics is the science, which computes energy, energy and entropy.
Q. 2: State the scope of thermodynamics in thermal engineering.
Sol: Thermal engineering is a very important associate branch of mechanical, chemical, metallurgical,
aerospace, marine, automobile, environmental, textile engineering, energy technology, process engineering
of pharmaceutical, refinery, fertilizer, organic and inorganic chemical plants. Wherever there is combustion,
heating or cooling, exchange of heat for carrying out chemical reactions, conversion of heat into work for
producing mechanical or electrical power; propulsion of rockets, railway engines, ships, etc., application
of thermal engineering is required. Thermodynamics is the basic science of thermal engineering.
Q. 3: Discuss the applications of thermodynamics in the field of energy technology.
2 / Problems and Solutions in Mechanical Engineering with Concept
Sol: Thermodynamics has very wide applications as basis of thermal engineering. Almost all process and
engineering industries, agriculture, transport, commercial and domestic activities use thermal engineering.
But energy technology and power sector are fully dependent on the laws of thermodynamics.
For example:
(i) Central thermal power plants, captive power plants based on coal.
(ii) Nuclear power plants.
(iii) Gas turbine power plants.
(iv) Engines for automobiles, ships, airways, spacecrafts.
(v) Direct energy conversion devices: Fuel cells, thermoionic, thermoelectric engines.
(vi) Air conditioning, heating, cooling, ventilation plants.
(vii) Domestic, commercial and industrial lighting.
(viii) Agricultural, transport and industrial machines.
All the above engines and power consuming plants are designed using laws of thermodynamics.
Q. 4: Explain thermodynamic system, surrounding and universe. Differentiate among open system,
closed system and an isolated system. Give two suitable examples of each system. (Dec. 03)
Or
Define and explain a thermodynamic system. Differentiate between various types of
thermodynamic systems and give examples of each of them. (Feb. 2001)
Or
Define Thermodynamics system, surrounding and universe. (May–03)
Or
Define closed, open and isolated system, give one example of each. (Dec–04)
Sol: In thermodynamics the system is defined as the quantity of matter or region in space upon which the
attention is concentrated for the sake of analysis. These systems are also referred to as thermodynamics
system.
It is bounded by an arbitrary surface called boundary. The boundary may be real or imaginary, may
be at rest or in motion and may change its size or shape.
Everything out side the arbitrary selected boundaries of the system is called surrounding or
environment.
S
urround
in
g
s
S
u
r
r
o
u
n
dings
Boundary
System
Cylinder
Convenient
imaginary
boundary
System
Real
boundary
Piston
Piston
Fig. 1.1 The system Fig. 1.2 The real and imaginary boundaries
The union of the system and surrounding is termed as universe.
Universe = System + Surrounding
Fundamental Concepts, Definitions and Zeroth Law / 3
Types of system
The analysis of thermodynamic processes includes the study of the transfer of mass and energy across the
boundaries of the system. On the basis the system may be classified mainly into three parts.
(1) Open system (2) Closed System (3) Isolated system
(1) Open system
The system which can exchange both the mass and energy (Heat and work) with its surrounding. The mass
within the system may not be constant. The nature of the processes occurring in such system is flow type.
For example
1. Water Pump: Water enters at low level and pumped to a higher level, pump being driven by an
electric motor. The mass (water) and energy (electricity) cross the boundary of the system (pump and motor).
Mass in
Mass Out
Mass may
change
Boundary
free to
move
Heat Transfer
Work Transfer
Fig. 1.3
2.Scooter engine: Air arid petrol enter and burnt gases leave the engine. The engine delivers mechanical
energy to the wheels.
3. Boilers, turbines, heat exchangers. Fluid flow through them and heat or work is taken out or
supplied to them.
Most of the engineering machines and equipment are open systems.
(2) Closed System
The system, which can exchange energy with their surrounding but not the mass. The quantity of matter
thus remains fixed. And the system is described as control mass system.
The physical nature and chemical composition of the mass of the system may change.
Water may evaporate into steam or steam may condense into water. A chemical reaction may occur
between two or more components of the closed system.
For example
1. Car battery, Electric supply takes place from and to the battery but there is no material transfer.
2. Tea kettle, Heat is supplied to the kettle but mass of water remains constant.
Mass may
change
Boundary
free to
move
Heat Transfer
Work Transfer
Fig 1.4
3. Water in a tank
4. Piston – cylinder assembly.
(3) Isolated System
In an Isolated system, neither energy nor masses are allowed to cross the boundary. The system has fixed mass
and energy. No such system physically exists. Universe is the only example, which is perfectly isolated system.
4 / Problems and Solutions in Mechanical Engineering with Concept
Other Special System
1. Adiabatic System: A system with adiabatic walls can only exchange work and not heat with the
surrounding. All adiabatic systems are thermally insulated from their surroundings.
Example is Thermos flask containing a liquid.
2. Homogeneous System: A system, which consists of a single phase, is termed as homogeneous
system. For example, Mi×ture of air and water vapour, water plus nitric acid and octane plus heptanes.
3. Hetrogeneous System: A system, which consists of two or more phase, is termed as heterogeneous
system. For example, Water plus steam, Ice plus water and water plus oil.
Q. 5: Classified each of the following systems into an open or closed systems.
(1) Kitchen refrigerator, (2) Ceiling fan (3) Thermometer in the mouth (4) Air compressor
(5) Pressure Cooker (6) Carburetor (7) Radiator of an automobile.
(1) Kitchen refrigerator: Closed system. No mass flow. Electricity is supplied to compressor motor
and heat is lost to atmosphere.
(2) Ceiling fan: Open system. Air flows through the fan. Electricity is supplied to the fan.
(3) Thermometer in the mouth: Closed system. No mass flow. Heat is supplied from mouth to
thermometer bulb.
(4) Air compressor: Open system. Low pressure air enters and high pressure air leaves the compressor,
electrical energy is supplied to drive the compressor motor.
(5) Pressure Cooker: Closed system. There is no mass exchange (neglecting small steam leakage).
Heat is supplied to the cooker.
(6) Carburetor: Open system. Petrol and air enter and mi×ture of petrol and air leaves the carburetor.
There is no change of energy.
(7) Radiator of an automobile: Open system. Hot water enters and cooled water leaves the radiator.
Heat energy is extracted by air flowing over the outer surface of radiator tubes.
Q. 6: Define Phase.
Sol: A phase is a quantity of matter, which is homogeneous throughout in chemical composition and
physical structure.
If the matter is all gas, all liquid or all solid, it has physical uniformity. Similarly, if chemical composition
does not vary from one part of the system to another, it has chemical uniformity.
Examples of one phase system are a single gas, a single liquid, a mi×ture of gases or a solution of
liquid contained in a vessel.
A system consisting of liquid and gas is a two–phase system.
Water at triple point exists as water, ice and steam simultaneously forms a three–phase system.
Q. 7: Differentiate between macroscopic and microscopic approaches. Which approach is used in
the study of engineering thermodynamics. (Sept. 01; Dec., 03, 04)
Or
Explain the macroscopic and microscopic point of view.Dec–2002
Sol: Thermodynamic studies are undertaken by the following two different approaches.
l. Macroscopic approach–(Macro mean big or total)
2. Microscopic approach–(Micro means small)
The state or condition of the system can be completely described by measured values of pressure,
temperature and volume which are called macroscopic or time–averaged variables. In the classical
Fundamental Concepts, Definitions and Zeroth Law / 5
thermodynamics, macroscopic approach is followed. The results obtained are of sufficient accuracy and
validity.
Statistical thermodynamics adopts microscopic approach. It is based on kinetic theory. The matter
consists of a large number of molecules, which move, randomly in chaotic fashion. At a particular moment,
each molecule has a definite position, velocity and energy. The characteristics change very frequently due
to collision between molecules. The overall behaviour of the matter is predicted by statistically averaging
the behaviour of individual molecules.
Microscopic view helps to gain deeper understanding of the laws of thermodynamics. However, it is
rather complex, cumbersome and time consuming. Engineering thermodynamic analysis is macroscopic and
most of the analysis is made by it.
These approaches are discussed (in a comparative way) below:
Macroscopic approach Microscopic approach
1. In this approach a certain quantity of
matter is considered without taking into
account the events occurring at molecular
level. In other words this approach to
thermodynamics is concerned with gross
or overall behaviour. This is known as
classical thermodynamics.
2. The analysis of macroscopic system
requires simple mathematical formulae.
3. The values of the properties of the system
are their average values. For example,
consider a sample of a gas in a closed
container. The pressure of the gas is the
average value of the pressure exerted by
millions of individual molecules.
Similarly the temperature of this gas is
the average value of transnational kinetic
energies of millions of individual
molecules. these properties like pressure
and temperature can be measured very
easily. The changes in properties can be
felt by our senses.
4. In order to describe a system only a few
properties are needed.
1. The approach considers that the system
is made up of a very large number of
discrete particles known as molecules.
These molecules have different velocities
and energies. The values of these energies
are constantly changing with time. This
approach to thermodynamics, which is
concerned directly with the structure of
the matter, is known as statistical
thermodynamics.
2. The behaviour of the system is found by
using statistical methods, as the number
of molecules is very large. so advanced
statistical and mathematical methods are
needed to explain the changes in the
system.
3. The properties like velocity, momentum,
impulse, kinetic energy, and instruments
cannot easily measure force of impact etc.
that describe the molecule. Our senses
cannot feel them.
4. Large numbers of variables are needed
to describe a system. So the approach is
complicated.
6 / Problems and Solutions in Mechanical Engineering with Concept
Q. 8: Explain the concept of continuum and its relevance in thermodynamics. Define density and
pressure using this concept. (June, 01, March– 02, Jan–03)
Or
Discuss the concept of continuum and its relevance. (Dec–01)
Or
Discuss the concept of continuum and its relevance in engineering thermodynamics. (May–02)
Or
What is the importance of the concept of continuum in engineering thermodynamics. (May–03)
Sol: Even the simplification of matter into molecules, atoms, electrons, and so on, is too complex a picture
for many problems of thermodynamics. Thermodynamics makes no hypotheses about the structure of the
matter of the system. The volumes of the system considered are very large compared to molecular dimensions.
The system is regarded as a continuum. The system is assumed to contain continuous distribution of matter.
There are no voids and cavities. The pressure, temperature, density and other properties are the average
values of action of many molecules and atoms. Such idealization is a must for solving most problems. The
laws and concepts of thermodynamics are independent of structure of matter.
According to this concept there is minimum limit of volume upto which the property remain continuum.
Below this volume, there is sudden change in the value of the property. Such a region is called region of
discrete particles and the region for which the property are maintain is called region of continuum. The
limiting volume up to which continuum properties are maintained is called continuum limit.
For Example: If we measure the density of a substance for a large volume (υ
1
), the value of density is (ρ
1
).
If we go on reducing the volume by δv’, below which the ratio äm/äv deviates from its actual value and
the value of äm/äv is either large or small.
Thus according to this concept the design could be defined as
ρ = lim δv– δv’ [δm / δv]
( ) a
Volume of the system
( ) b
Case II
A
C
B
V
V
Case I
Region of
continnum
Region of
noncontinnum
A
v
e
r
a
g
e
m
a
s
s
d
e
n
s
i
t
y
System
m
=
V
m
Fig 1.5
Q. 9: Define different types of properties?
Sol: For defining any system certain parameters are needed. Properties are those observable characteristics
of the system, which can be used for defining it. For example pressure, temp, volume.
Properties further divided into three parts;
Fundamental Concepts, Definitions and Zeroth Law / 7
Intensive Properties
Intensive properties are those, which have same value for any part of the system or the properties that are
independent of the mass of the system. EX; pressure, temp.
Extensive Properties
EXtensive properties are those, which dependent Upon the mass of the system and do not maintain the same
value for any part of the system. EX; mass, volume, energy, entropy.
Specific Properties
The extensive properties when estimated on the unit mass basis result in intensive property, which is also
known as specific property. EX; sp. Heat, sp. Volume, sp. Enthalpy.
Q. 10: Define density and specific volume.
Sol: DENSITY (ρ)
Density is defined as mass per unit volume;
Density = mass/ volume; ρ = m/v, kg/m
3
P for Hg = 13.6 × 10
3
kg/m
3
ρ for water = 1000 kg/m
3
Specific Volume (ν νν νν)
It is defined as volume occupied by the unit mass of the system. Its unit is m
3
/kg. Specific volume is
reciprocal of density.
ν = v/m; m
3
/kg
Q. 11: Differentiate amongst gauge pressure, atmospheric pressure and absolute pressure. Also give
the value of atmospheric pressure in bar and mm of Hg. (Dec–02)
Sol: While working in a system, the thermodynamic medium exerts a force on boundaries of the vessel in
which it is contained. The vessel may be a container, or an engine cylinder with a piston etc. The exerted
force F per unit area A on a surface, which is normal to the force, is called intensity of pressure or simply
pressure p. Thus
P = F/A= ρ.g.h
It is expressed in Pascal (1 Pa = 1 Nm
2
),
bar (1 bar = 10
5
Pa),
standard atmosphere (1 atm =1.0132 bar),
or technical atmosphere (1 kg/cm
2
or 1 atm).
1 atm means 1 atmospheric absolute.
The pressure is generally represented in following terms.
1. Atmospheric pressure
2. Gauge pressure
3. Vacuum (or vacuum pressure)
4. Absolute pressure
Atmospheric Pressure (P
atm
)
It
is the pressure exerted by atmospheric air on any surface. It is measured by a barometer. Its standard
values are;
1 P
atm
= 760 mm of Hg i.e. column or height of mercury
= ρ.g.h. = 13.6 × 10
3
× 9.81 × 760/1000
8 / Problems and Solutions in Mechanical Engineering with Concept
= 101.325 kN/m
2
= 101.325 kPa
= 1.01325 bar
when the density of mercury is taken as 13.595 kg/m
3
and acceleration due to gravity as 9.8066 m/s
2
Gauge Pressure (P
gauge
)
It is the pressure of a fluid contained in a closed vessel. It is always more than atmospheric pressure. It
is measured by an instrument called pressure gauge (such as Bourden’s pressure gauge). The gauge measures
pressure of the fluid (liquid and gas) flowing through a pipe or duct, boiler etc. irrespective of prevailing
atmospheric pressure.
Vacuum (Or Vacuum pressure) (P
vacc
)
It is the pressure of a fluid, which is always less than atmospheric pressure. Pressure (i.e. vacuum) in a
steam condenser is one such example. It is also measured by a pressure gauge but the gauge reads on
negative side of atmospheric pressure on dial. The vacuum represents a difference between absolute and
atmospheric pressures.
Absolute Pressure (P
abs
)
It is that pressure of a fluid, which is measured with respect to absolute zero pressure as the reference.
Absolute zero pressure can occur only if the molecular momentum is zero, and this condition arises when
there is a perfect vacuum. Absolute pressure of a fluid may be more or less than atmospheric depending
upon, whether the gauge pressure is expressed as absolute pressure or the vacuum pressure.
Inter–relation between different types of pressure representations. It is depicted in Fig. 1.6, which can
be expressed as follows.
p
abs
= p
atm
+ p
gauge
p
abs
= p
atm
– p
vace
1.0132 bar
= 1.0132 bar
= 0 bar
P
gauge
P
atm
P
abs
P
abs
P
vacc
Gauge pressure line
Atmospheric pressure line
Absolute zero pressure line
Fig 1.6 Depiction of atmospheric, gauge, vacuum, and absolute pressures and their interrelationship.
Hydrostatic Pressure
Also called Pressure due to Depth of a Fluid. It is required to determine the pressure exerted by a static
fluid column on a surface, which is drowned under it. Such situations arise in water filled boilers, petrol
or diesel filled tank in IC engines, aviation fuel stored in containers of gas turbines etc.
This pressure is also called ‘hydrostatic pressure’ as it is caused due to static fluid. The hydrostatic pressure
acts equally in all directions on lateral surface of the tank. Above formula holds good for gases also. But
due to a very small value of p (and w), its effect is rarely felt. Hence, it is generally neglected in thermodynamic
calculations. One such tank is shown in Fig. 1.7. It contains a homogeneous liquid of weight density w. The
pressure p exerted by it at a depth h will be given by
Fundamental Concepts, Definitions and Zeroth Law / 9
h
p = wh
Fig 1.7 Pressure under depth of a fluid increases with increase in depth.
Q. 12: Write short notes on State, point function and path function.
STATE
The State of a system is its condition or configuration described in sufficient detail.
State is the condition of the system identified by thermodynamic properties such as pressure, volume,
temperature, etc. The number of properties required to describe a system depends upon the nature of the
system. However each property has a single value at each state. Each state can be represented by a point
on a graph with any two properties as coordinates.
Any operation in which one or more of properties of a system change is called a change of state.
Point Function
A point function is a single valued function that always possesses a single – value is all states. For example
each of the thermodynamics properties has a single – value in equilibrium and other states. These properties
are called point function or state function.
Or
when two properties locate a point on the graph ( coordinates axes) then those properties are called as point
function.
For example pressure, volume, temperature, entropy, enthalpy, internal energy.
Path Function
Those properties, which cannot be located on a graph by a point but are given by the area or show on the
graph.
A path function is different from a point function. It depends on the nature of the process that can
follow different paths between the same states. For example work, heat, heat transfer.
Q. 13: Define thermodynamic process, path, cycle.
Sol: Thermodynamic system undergoes changes due to the energy and mass interactions. Thermodynamic
state of the system changes due to these interactions.
The mode in which the change of state of a system takes place is termed as the PROCESS such as
constant pressure, constant volume process etc. In fig 1.8, process 1–2 & 3–4 is constant pressure process
while 2–3 & 4–1 is constant volume process.
Let us take gas contained in a cylinder and being heated up. The heating of gas in the cylinder shall
result in change in state of gas as it’s pressure, temperature etc. shall increase. However, the mode in which
this change of state in gas takes place during heating shall he constant volume mode and hence the process
shall be called constant volume heating process.
The PATH refers to the series of state changes through which the system passes during a process.
Thus, path refers to the locii of various intermediate states passed through by a system during a process.
CYCLE refers to a typical sequence of processes in such a fashion that the initial and final states are
identical. Thus, a cycle is the one in which the processes occur one after the other so as to finally, land
10 / Problems and Solutions in Mechanical Engineering with Concept
the system at the same state. Thermodynamic path in a cycle is in closed loop
form. After the occurrence of a cyclic process, system shall show no sign of the
processes having occurred. Mathematically, it can be said that the cyclic integral
of any property in a cycle is zero.
1–2 & 3–4 = Constant volume Process
2–3 &4–1 = Constant pressure Process
1–2, 2–3, 3–4 & 4–1 = Path
1–2–3–4–1 = Cycle Fig 1.8
Q. 14: Define thermodynamic equilibrium of a system and state its importance. What are the conditions
required for a system to be in thermodynamic equilibrium? Describe in brief.
(March–02, Dec–03)
Or
What do you known by thermodynamic equilibrium. (Dec–02, Dec–04, may–05, Dec–05)
Sol: Equilibrium is that state of a system in which the state does not undergo any change in itself with
passage of time without the aid of any external agent. Equilibrium state of a system can be examined by
observing whether the change in state of the system occurs or not. If no change in state of system occurs
then the system can be said in equilibrium.
Let us consider a steel glass full of hot milk kept in open atmosphere. It is quite obvious that the heat
from the milk shall be continuously transferred to atmosphere till the temperature of milk, glass and
atmosphere are not alike. During the transfer of heat from milk the temperature of milk could be seen to
decrease continually. Temperature attains some final value and does not change any more. This is the
equilibrium state at which the properties stop showing any change in themselves.
Generally, ensuring the mechanical, thermal, chemical and electrical equilibriums of the system may
ensure thermodynamic equilibrium of a system.
1. Mechanical Equilibrium: When there is no unbalanced force within the system and nor at its
boundaries then the system is said to be in mechanical equilibrium.
For a system to be in mechanical equilibrium there should be no pressure gradient within the system
i.e., equality of pressure for the entire system.
2. Chemical Equilibrium: When there is no chemical reaction taking place in the system it is said
to be in chemical equilibrium.
3. Thermal equilibrium: When there is no temperature gradient within the system, the system is said
to be in thermal equilibrium.
4. Electrical Equilibrium: When there is no electrical potential gradient within a system, the system
is said to be in electrical equilibrium.
When all the conditions of mechanical, chemical thermal, electrical equilibrium are satisfied, the
system is said to be in thermodynamic equilibrium.
Q. 15: What do you mean by reversible and irreversible processes? Give some causes of irreversibility.
(Feb–02, July–02)
Or
Distinguish between reversible and irreversible process (Dec–01, May–02)
Or
Briefly state the important features of reversible and irreversible processes. (Dec–03)
Sol: Thermodynamic system that is capable of restoring its original state by reversing the factors responsible
for occurrence of the process is called reversible system and the thermodynamic process involved is called
reversible process.
p
v
4 1
2 3
Fundamental Concepts, Definitions and Zeroth Law / 11
Thus upon reversal of a process there shall be no trace of the process being occurred, i.e., state changes
during the forward direction of occurrence of a process are exactly similar to the states passed through by
the system during the reversed direction of the process.
1 2 = Reversible process following
equilibrium states
3 4 = Irreversible process following
nonequilibrium states
p
V
2
1
3
4
Fig. 1.9. Reversible and irreversible processes
It is quite obvious that such reversibility can be realised only if the system maintains its thermodynamic
equilibrium throughout the occurrence of process.
Irreversible systems are those, which do not maintain equilibrium during the occurrence of a process.
Various factors responsible for the non–attainment of equilibrium are generally the reasons responsible for
irreversibility Presence of friction, dissipative effects etc.
Q. 16: What do you mean by cyclic and quasi – static process. (March–02, Jan–03, Dec–01, 02, 05)
Or
Define quasi static process. What is its importance in study of thermodynamics. (May–03)
Sol: Thermodynamic equilibrium of a system is very difficult to be realised during the occurrence of a
thermodynamic process. ‘Quasi–static’ consideration is one of the ways to consider the real system as if
it is behaving in thermodynamic equilibr
i
um and thus permitting the thermodynamic study. Actually system
does not attain thermodynamic equilibrium, only certain assumptions make it akin to a system in equilibrium
for the sake of study and analysis.
Quasi–static literally refers to “almost static” and the infinite slowness of the occurrence of a process
is considered as the basic premise for attaining near equilibrium in the system. Here it is considered that
the change in state of a system occurs at infinitely slow pace, thus consuming very large time for completion
of the process. During the dead slow rate of state change the magnitude of change in a state shall also be
infinitely small. This infinitely small change in state when repeatedly undertaken one after the other results
in overall state change but the number of processes required for completion of this state change are infinitely
large. Quasi–static process is presumed to remain in thermodynamic equilibrium just because of infinitesimal
state change taking place during the occurrence of the process. Quasi–static process can be understood from
the following example.
Weight
Lid
Gas
Heating v
p
W
×××××××××××
1 = Initial state
2 = Final state
Intermediate
equilibrium states
Fig 1.9 Quasi static process
12 / Problems and Solutions in Mechanical Engineering with Concept
Let us consider the locating of gas in a container with certain mass ‘W’ kept on the top lid (lid is such
that it does not permit leakage across its interface with vessel wall) of the vessel as shown in Fig. 1.9. After
certain amount of heat being added to the gas it is found that the lid gets raised up. Thermodynamic state
change is shown in figure. The “change in state”, is significant.
During the “change of state” since the states could not be considered to be in equilibrium, hence for
unsteady state of system, thermodynamic analysis could not be extended. Difficulty in thermodynamic
analysis of unsteady state of system lies in the fact that it is not sure about the state of system as it is
continually changing and for analysis one has to start from some definite values.
Let us now assume that the total mass comprises of infinitesimal small masses of ‘w’ such that all ‘w’
masses put together become equal to w. Now let us start heat addition to vessel and as soon as the lifting
of lid is observed put first fraction mass `w’ over the lid so as to counter the lifting and estimate the state
change. During this process it is found that the state change is negligible. Let us further add heat to the
vessel and again put the second fraction mass ‘w’ as soon as the lift is felt so as to counter it. Again the
state change is seen to be negligible. Continue with the above process and at the end it shall be seen that
all fraction masses ‘w’ have been put over the lid, thus amounting to mass ‘w’ kept over the lid of vessel
and the state change occurred is exactly similar to the one which occurred when the mass kept over the lid
was `W’. In this way the equilibrium nature of system can be maintained and the thermodynamic analysis
can be carried out. P–V representation for the series of infinitesimal state changes occurring between states
1 & 2 is also shown in figure 1.9.
Note:
In PV = R
0
T, R
0
= 8314 KJ/Kgk
And in PV = mRT; R = R
0
/M; Where M = Molecular Weight
Q. 17: Convert the following reading of pressure to kPa, assuming that the barometer reads 760mm Hg.
(1) 90cm Hg gauge (2) 40cm Hg vacuum (3) 1.2m H
2
O gauge
Sol: Given that h = 760mm of Hg for P
atm
P
atm
= ρgh = 13.6 × 10
3
× 9.81 × 760/1000 = 101396.16
N/m
2
= 101396.16Pa = 101.39KPa ...(i)
(a) 90cm Hg gauge
P
gauge
= ρgh = 13.6 × 10
3
× 9.81 × 90/100 = 120.07KPa ...(ii)
P
abs
= P
atm
+ P
gauge
= 101.39 + 120.07
P
abs
= 221.46KPa .......ANS
(b) 40cm Hg vacuum
P
vacc
= ρgh = 13.6 × 10
3
× 9.81 × 40/100 = 53.366KPa ...(iii)
P
abs
= P
atm
– P
vacc
= 101.39 – 53.366
P
abs
= 48.02KPa .......ANS
(c)1.2m Water gauge
P
gauge
= ρgh = 1000 × 9.81 × 1.2 = 11.772KPa ...(iv)
P
abs
= P
atm
+ P
gauge
= 101.39 + 11.772
P
abs
= 113.162KPa .......ANS
Fundamental Concepts, Definitions and Zeroth Law / 13
Q. 18: The gas used in a gas engine trial was tested. The pressure of gas supply is 10cm of water
column. Find absolute pressure of the gas if the barometric pressure is 760mm of Hg.
Sol: Given that h = 760mm of Hg for P
atm
P
atm
= ρgh = 13.6 × 10
3
× 9.81 × 760/1000 = 101396.16
N/m
2
= 101.39 × 10
3
N/m
2
...(i)
P
gauge
= ρgh = 1000 × 9.81 × 10/100 = 981 N/m
2
...(ii)
P
abs
= P
atm
+ P
gauge
= 101.39 × 10
3
+ 981
P
abs
= 102.37×10
3
N/m
2
.......ANS
Q. 19: A manometer shows a vacuum of 260 mm Hg. What will be the value of this pressure in N/
m
2
in the form of absolute pressure and what will be absolute pressure (N/m
2
), if the gauge
pressure is 260 mm of Hg. Explain the difference between these two pressures.
Sol: Given that P
Vacc
= 260mm of Hg
P
Vacc
= ρgh = 13.6 × 10
3
× 9.81 × 260/1000
= 34.688 × 10
3
N/m
2
.......ANS ...(i)
P
atm
= ρgh = 13.6 × 10
3
× 9.81 × 760/1000 = 101396.16
N/m
2
= 101.39 × 10
3
N/m
2
...(ii)
P
abs
= P
atm
– P
Vacc
= 101.39 × 10
3
– 34.688 × 10
3
= 66.61 × 10
3
N/m
2
.......ANS
Now if P
gauge
= 260mm of Hg =
P
gauge
= 260mm of Hg = 13.6 × 10
3
× 9.81 × 260/1000 = 34.688 × 10
3
N/m
2
P
abs
= P
atm
+P
gauge
= 101.39 × 10
3
+ 34.688 × 10
3
= 136.07 × 10
3
N/m
2
.......ANS
ANS: P
vacc
= 34.7×10
3
N/m
2
(vacuum), P
abs
= 66.6kPa, 136kpa
Difference is because vacuum pressure is always Negative gauge pressure. Or vacuum in a gauge
pressure below atmospheric pressure and gauge pressure is above atmospheric pressure.
Q. 20: Calculate the height of a column of water equivalent to atmospheric pressure of 1bar if the
water is at 15
0
C. What is the height if the water is replaced by Mercury?
Sol: Given that P = 1bar = 10
5
N/m
2
P
atm
= ρgh , for water equivalent
10
5
= 1000 × 9.81 × h
h = 10.19m .......ANS
P
atm
= ρgh , for Hg
10
5
= 13.6 × 10
3
× 9.81 × h
h = 0.749m .......ANS
ANS: 10.19m, 0.75m
Q. 21: The pressure of a gas in a pipeline is measured with a mercury manometer having one limb
open. The difference in the level of the two limbs is 562mm. Calculate the gas pressure in
terms of bar.
Sol: The difference in the level of the two limbs = P
gauge
P
gauge
= P
abs
– P
atm
14 / Problems and Solutions in Mechanical Engineering with Concept
P
abs
– P
atm
= 562mm of Hg
P
abs
– 101.39 = ρgh = 13.6 × 10
3
× 9.81 × 562/1000 = 75.2 × 10
3
N/m
2
= 75.2 KPa
P
abs
= 101.39 + 75.2 = 176.5kPa
ANS: P = 176.5kPa
Q. 22: Steam at gauge pressure of 1.5Mpa is supplied to a steam turbine, which rejects it to
a condenser at a vacuum of 710mm Hg after expansion. Find the inlet and exhaust
steam pressure in pascal, assuming barometer pressure as 76cm of Hg and density of Hg as
13.6×10
3
kg/m
3
.
Sol: P
gauge
= 1.5 × 10
6
N/m
2
P
vacc
= 710 mm of Hg
P
atm
= 76 cm of Hg = 101.3 × 10
3
N/m
2
P
inlet
= ?
P
inlet
= P
abs
= P
gauge(inlet)
+ P
atm
= 1.5 × 10
6
+ 101.3 × 10
3
P
inlet
= 1.601 × 10
6
Pa .......ANS
Since discharge is at vacuum i.e.;
P
exhaust
= P
abs
= P
atm
– P
vacc
= 101.3 × 10
3
– 13.6 × 10
3
× 9.81 × 710/1000
P
exhaust
= 6.66 × 10
6
Pa .......ANS
ANS: P
inlet
= 1.6×10
6
Pa, P
exhaust
= 6.66×10
3
Pa
Q. 23: A U–tube manometer using mercury shows that the gas pressure inside a tank is 30cm.
Calculate the gauge pressure of the gas inside the vessel. Take g = 9.78m/s
2
, density of mercury
=13,550kg/m
3
. (C.O.–Dec–03)
Sol: Given that P
abs
= 30mm of Hg
P
abs
= ρgh = 13550 × 9.78 × 30/1000 = 39.755 × 10
3
N/m
2
...(i)
P
atm
=ρgh = 13550 × 9.78 × 760/1000 = 100714.44 N/m
2
...(ii)
P
gauge
= P
abs
– P
atm
= 39.755 × 10
3
– 100714.44
= – 60958.74 N/m
2
.......ANS
Q. 24: 12 kg mole of a gas occupies a volume of 603.1 m
3
at temperature of 140°C while its density
is 0.464 kg/m
3
. Find its molecular weight and gas constant and its pressure. (Dec–03–04)
Sol: Given data;
V = 603.1 m
3
T = 140
0
C
ρ = 0.464 kg/m
3
Since PV = nmR
0
T
= 12 Kg – mol
= 12M Kg, M = molecular weight
Since ρ = m/V
0.464 = 12M/603.1
M = 23.32 ...(i)
Now Gas constant R = R
0
/M, Where R
0
= 8314 KJ/kg–mol–k = Universal gas constant
R = 8314/23.32 = 356.52 J/kgk
Fundamental Concepts, Definitions and Zeroth Law / 15
PV = mR
0
T
P = mR
0
T/V, where m in kg, R = 8314 KJ/kg–mol–k
= [(12 × 23.32) × (8314/23.32)(273 + 140)]/ 603.1
P = 68321.04N/m
2
.......ANS
Q. 25: An aerostat balloon is filled with hydrogen. It has a volume of 1000m
3
at constant air
temperature of 27
0
C and pressure of 0.98bar. Determine the load that can be lifted with the
air of aerostat.
Sol: Given that:
V = 100m
3
T = 300K
P = 0.98bar = 0.98 × 10
5
N/m
2
W = mg
PV = mR
0
T
Where m = mass in Kg
R
0
= 8314 KJ/kg/mole K
But in Hydrogen; M = 2
i.e.; R = R
0
/2 = 8314/2 = 4157 KJ/kg.k
0.98 × 10
5
× 1000 = m × 4157 × 300
m = 78.58 kg
W = 78.58 × 9.81 = 770.11 N
....... ANS: 770.11N
Q. 26: What is energy? What are its different forms? (Dec— 02, 03)
Sol: The energy is defined as the capacity of doing work. The energy possessed by a system may be of two
kinds.
1. Stored energy: such as potential energy, internal energy, kinetic energy etc.
2. Transit energy: such as heat, work, flow energy etc.
The stored energy is that which is contained within the system boundaries, but the transit energy
crosses the system boundary. The store energy is a thermodynamic property whereas the transit energy is
not a thermodynamic property as it depends upon the path.
For example, the kinetic energy of steam issuing out from a steam nozzle and impinging upon the
steam turbine blade is an example of stored energy. Similarly, the heat energy produced in combustion
chamber of a gas turbine is transferred beyond the chamber by conduction/ convection and/or radiation, is
an example of transit energy.
Form of Energy
1. Potential energy (PE)
The energy possessed by a body or system by virtue of its position above the datum (ground) level. The
work done is due to its falling on earth’s surface.
Potential energy,PE = Wh = mgh N.m
Where, W = weight of body, N ; m = mass of body, kg
h = distance of fall of body, m
g = acceleration due to gravity, = 9.81 m/s
2
16 / Problems and Solutions in Mechanical Engineering with Concept
2. Kinetic Energy (KE)
The energy possessed by a system by virtue of its motion is called kinetic energy. It means that a system
of mass m kg while moving with a velocity V
1
m/s, does 1/2mV1
2
joules of work before coming to rest.
So in this state of motion, the system is said to have a kinetic energy given as;
K.E. = 1/2mv
1
2
N.m
However, when the mass undergoes a change in its velocity from velocity V
1
to V
2
, the change in kinetic
energy of the system is expressed as;
K.E. = 1/2mv
2
2
– 1/2mv
1
2
3. Internal Energy (U)
It is the energy possessed by a system on account of its configurations, and motion of atoms and molecules.
Unlike the potential energy and kinetic energy of a system, which are visible and can be felt, the internal
energy is invisible form of energy and can only be sensed. In thermodynamics, main interest of study lies
in knowing the change in internal energy than to know its absolute value.
The internal energy of a system is the sum of energies contributed by various configurations and
inherent molecular motions. These contributing energies are
(1) Spin energy: due to clockwise or anticlockwise spin of electrons about their own axes.
(2) Potential energy: due to intermolecular forces (Coulomb and gravitational forces), which keep the
molecules together.
(3) Transitional energy: due to movement of molecules in all directions with all probable velocities
within the system, resulting in kinetic energy acquired by the translatory motion.
(4) Rotational energy: due to rotation of molecules about the centre of mass of the system, resulting
in kinetic energy acquired by rotational motion. Such form of energy invariably exists in diatomic and
polyatomic gases.
(5) Vibrational energy: due to vibration of molecules at high temperatures.
(6) Binding energy: due to force of attraction between various sub–atomic particles and nucleus.
(7) Other forms of energies such as
Electric dipole energy and magnetic dipole energy when the system is subjected to electric and/or
magnetic fields.
High velocity energy when rest mass of the system m
o
changes to variable mass m in accordance with
Eisenstein’s theory of relativity).
The internal energy of a system can increase or decrease during thermodynamic operations.
The internal energy will increase if energy is absorbed and will decrease when energy is evolved.
4. Total Energy
Total energy possessed by a system is the sum of all types of stored energy. Hence it will be given by
E
total
= PE + KE + U = mgh + 1/2mv
2
+ U
It is expressed in the unit of joule (1 J = 1 N m)
Q. 27: State thermodynamic definition of work. Also differentiate between heat and work.
(May02)
HEAT
Sol: Heat is energy transferred across the boundary of a system due to temperature difference between the
system and the surrounding. The heat can be transferred by conduction, convection and radiation. The main
characteristics of heat are:
Fundamental Concepts, Definitions and Zeroth Law / 17
1. Heat flows from a system at a higher temperature to a system at a lower temperature.
2. The heat exists only during transfer into or out of a system.
3. Heat is positive when it flows into the system and negative when it flows out of the system.
4. Heat is a path function.
5. It is not the property of the system because it does not represent an exact differential dQ. It is
therefore represented as δQ.
Heat required to raise the temperature of a body or system, Q = mc (T
2
– T
1
)
Where, m = mass, kg
T
1
, T
2
= Temperatures in °C or K.
c = specific heat, kJ/kg–K.
Specific heat for gases can be specific heat at constant pressure (C
p
) and constant volume (c
v
)
Also; mc = thermal or heat capacity, kJ.
mc = water equivalent, kg.
WORK
The work may be defined as follows:
“Work is defined as the energy transferred (without transfer of mass) across the boundary of a system
because of an intensive property difference other than temperature that exists between the system and
surrounding.”
Pressure difference results in mechanical work and electrical potential difference results in electrical
work.
Or
“Work is said to be done by a system during a given operation if the sole effect of the system on things
external to the system (surroundings) can be reduced to the raising of a weight”.
The work is positive when done by the system and negative if work is done on the system.
Q. 28: Compare between work and heat ? (May–01)
Sol: There are many similarities between heat and work.
1. The heat and work are both transient phenomena. The systems do not possess heat or work. When
a system undergoes a change, heat transfer or work done may occur.
2. The heat and work are boundary phenomena. They are observed at the boundary of the
system.
3. The heat and work represent the energy crossing the boundary of the system.
4. The heat and work are path functions and hence they are inexact differentials.
5. Heat and work are not the properties of the system.
6. Heat transfer is the energy interaction due to temperature difference only. All other energy interactions
may be called work transfer.
7. The magnitude of heat transfer or work transfer depends upon the path followed by the system
during change of state.
Q. 29: What do you understand by flow work? It is different from displacement work? How.
(May–05)
FLOW WORK
Sol: Flow work is the energy possessed by a fluid by virtue of its pressure.
18 / Problems and Solutions in Mechanical Engineering with Concept
X Y
P
X
L
Y
Fig 1.10
Let us consider any two normal sections XX and YY of a pipe line through which a fluid is flowing in
the direction as shown in Fig. 1.10.
Let
L = distance between sections XX and YY
A = cross–sectional area of the pipe line
p = intensity of pressure at section l.
Then, force acting on the volume of fluid of length ‘L’ and
cross–sectional area ‘A’ = p x A.
Work done by this force = p x A x L = p x V,
Where;
V = A x L = volume of the cylinder of fluid between sections XX and YY
Now, energy is the capacity for doing work. It is due to pressure that p x V amount of work has been
done in order to cause flow o£ fluid through a length ‘L‘,
So flow work = p x V mechanical unit
Displacement Work
When a piston moves in a cylinder from position 1 to position 2 with volume changing from V
1
to V
2
, the
amount of work W done by the system is given by W
1–2
=
2
1
.
V
V
p dV
∫
The value of work done is given by the area under the process 1 – 2 on diagram (Fig. 1.11)
1
2
p
1
p
v
p
p
2
V
1
V
2
dV
Fig 1.11 Displacement work
Fundamental Concepts, Definitions and Zeroth Law / 19
Q. 30: Find the work done in different processes?
(1) ISOBARIC PROCESS (PRESSURE CONSTANT)
W
1–2
=
2
1
V
V
p dV
∫
= p (V
2
– V
2
)
P
1 2
V
2
V
1
W – 2
1
p
v
p
1
p
1 2
1
Fig 1.12: Constant pressure process Fig. 1.13: Constant volume process
(2) ISOCHORIC PROCESS (VOLUME CONSTANT)
W
1–2
=
2
1
V
V
p dV
∫
= 0
1 2
( ) V V · 3
(3) ISOTHERMAL PROCESS (T or, PV = const)
W
1–2
=
2
1
V
V
p dV
∫
pV = p
1
V
1
= p
2
V
2
= C.
1 2
2 1
V p
V p
·
p =
1 1
.
p V
V
W
1–2
= p
1
V
1
2
1
V
V
dV
V
∫
= p
1
V
1
ln
2
1
V
V
Fig. 1.14 : Isothermal process
= p
1
V
1
ln
1
2
P
P
(4) POLYTROPIC PROCESS(PV
n
= C)
pV
n
= p
1
1 2 2
n n
V P V C · ·
p =
1 1
( )
n
n
p V
V
W
1–2
=
2
1
V
V
p dV
∫
=
2
2
1 1
1 1
1
1
1
( )
– 1
V
V
n
n
n
V
V
p V
V n
dV p V
V n
− +
·
+
∫
Fig. 1.15 Polytropic process
1
2
p
1
p
pV = C
v
p
2
V
1
V
2
W – 2
1
p
2
2
2
2
n =
n = 1
n = 2
n = 3
2
pV = C
v
20 / Problems and Solutions in Mechanical Engineering with Concept
= ( )
1 1 1–2 1–
2 1
1
n
n
p V
V V
n
−
−
=
1– 1–
2 2 2 1 1 1
1
n n n n
p V V p V V
n
× − ×
−
=
1
1 1 2 2 1 1 2
1
1
1 1
n
n
p V p V p V p
n n p
−
1
¸ _ − 1
· −
1
− −
¸ ,
1
¸ ]
(5) ADIABATIC PROCESS
(PV
γ
= C)
Here δQ or dQ = 0
δQ = dU + dW
0 = dU + dW
dW = dU = – c, dT
dW = pdV [P
1
V
1
γ
= P
2
V
2
γ
= C]
=
2
– 1
2
–
1
1
– 1
v
v
v
v
C V
dV C V dV C
V
γ+
γ
γ
· ·
γ +
∫ ∫
Fig. 1.16 Adiabatic Process
=
1– 1–
1– 1– 2 2 2 1 1 1
2 1
1 1
P V V P V V
C
V V
γ γ γ γ
γ γ
−
1
− ·
¸ ]
− γ −γ
W
1 – 2
=
2 2 1 1 1 1 2 2
– –
1 1
P V P V P V P V
·
γ γ
where γ = C
p
/C
v
Q. 31: Define N.T.P. AND S.T.P.
Sol: Normal Temperature and Pressure (N.T.P.):
The conditions of temperature and pressure at 0°C (273K) and 760 mm of Hg respectively are called
normal temperature and pressure (N.T.P.).
Standard Temperature and Pressure (S.T.P.):
The temperature and pressure of any gas, under standard atmospheric conditions are taken as 15
0
C(288K)
and 760 mm of Hg respectively. Some countries take 25
0
C(298K) as temperature.
Q. 32: Define Enthalpy.
Sol: The enthalpy is the total energy of a gaseous system. It takes into consideration, the internal energy
and pressure, volume effect. Thus, it is defined as:
h = u + Pv
H = U + PV
Where v is sp. volume and V is total volume of m Kg gas.
h is specific enthalpy while H is total enthalpy of m kg gas
u is specific internal energy while U is total internal energy of m kg gas. From ideal gas equation,
Pv = RT
h = u + RT
h = f (T) + RT
PV = C
1
2
V
P
Fundamental Concepts, Definitions and Zeroth Law / 21
Therefore, h is also a function of temperature for perfect gas.
h = f(T)
⇒ dh ∝ dt
⇒ dh = C
p
dT
⇒
2 2
1 1
p
dH mC dT ·
∫ ∫
⇒ H
2
– H
1
= mC
p
(T
2
– T
1
)
Q. 33: Gas from a bottle of compressed helium is used to inflate an inelastic flexible balloon, originally
folded completely flat to a volume of 0.5m
3
. If the barometer reads 760mm of Hg, What is the
amount of work done upon the atmosphere by the balloon? Sketch the systems before and
after the process.
Sol: The displacement work W =
bottle balloon
Pdv Pdv +
∫ ∫
Since the wall of the bottle is rigid i.e.; 0
bottle
Pdv ·
∫
W = ; Pdv
∫
Here P = 760mm Hg = 101.39KN/m
2
dV = 0.5m
3
W = 101.39 × 0.5 KN–m
W = 50.66KJ .......ANS
Q. 34: A piston and cylinder machine containing a fluid system has a stirring device in the cylinder
the piston is frictionless, and is held down against the fluid due to the atmospheric pressure
of 101.325kPa the stirring device is turned 10,000 revolutions with an average torque against
the fluid of 1.275MN. Mean while the piston of 0.6m diameter moves out 0.8m. Find the net
work transfer for the systems.
Sol: Given that
P
atm
= 101.325 × 103 N/m
2
Revolution = 10000
Torque = 1.275 × 10
6
N
Dia = 0.6m
Distance moved = 0.8m
Work transfer = ?
W.D by stirring device W
1
= 2Π × 10000 × 1.275 J = 80.11 KJ ...(i)
This work is done on the system hence it is –ive.
Work done by the system upon surrounding
W
2
= F.dx = P.A.d×
= 101.32 × Π/4 × (0.6)
2
× 0.8
= 22.92 KJ ...(ii)
Net work done = W
1
+ W
2
= –80.11 + 22.92 = –57.21KJ (–ive sign indicates that work is done on the system)
ANS: W
net
= 57.21KJ
22 / Problems and Solutions in Mechanical Engineering with Concept
Q. 35: A mass of 1.5kg of air is compressed in a quasi static process from 0.1Mpa to 0.7Mpa for
which PV = constant. The initial density of air is 1.16kg/m
3
. Find the work done by the piston
to compress the air.
Sol: Given data:
m = 1.5kg
P
1
= 0.1MPa = 0.1 × 10
6
Pa = 10
5
N/m
2
P
2
= 0.7MPa = 0.7 × 10
6
Pa = 7 × 10
5
N/m
2
PV = c or Temp is constant
ρ = 1.16 Kg/m
3
W.D. by the piston = ?
For PV = C;
WD = P
1
V
1
log V
2
/V
1
or P
1
V
1
logP
1
/P
2
ρ = m/V; i.e.; V
1
= m/ρ = 1.5/1.16 = 1.293m
3
...(i)
W
1–2
= P
1
V
1
logP
1
/P
2
= 10
5
× 1.293 loge (10
5
/7 × 10
5
)
= 10
5
× 1.293 × (–1.9459)
= –251606.18J = –251.6 KJ (–ive means WD on the system)
ANS: – 251.6KJ
Q. 36: At a speed of 50km/h, the resistance to motion of a car is 900N. Neglecting losses, calculate
the power of the engine of the car at this speed. Also determine the heat equivalent of work
done per minute by the engine.
Sol: Given data:
V = 50Km/h = 50 × 5/18 = 13.88 m/sec
F = 900N
Power = ?
Q = ?
P = F.V = 900 × 13.88 = 12500 W = 12.5KW ANS
Heat equivalent of W.D. per minute by the engine = power × 1 minute
= 12.5 KJ/sec × 60 sec = 750KJ
ANS: Q =750KJ
Q. 37: An Engine cylinder has a piston of area 0.12m
2
and contains gas at a pressure of 1.5Mpa the
gas expands according to a process, which is represented by a straight line on a pressure
volume diagram. The final pressure is 0.15Mpa. Calculate the work done by the gas on the
piston if the stroke is 0.30m. (Dec–05)
Sol: Work done will be the area under the straight line which is made up of a triangle and a rectangle.
i.e.; WD = Area of Triangle + Area of rectangle
Area of Triangle = ½ × base × height = ½ × AC × AB
AC = base = volume = Area × stroke = 0.12 × 0.30
Height = difference in pressure = P
2
– P
1
= 1.5 – 0.15
= 1.35MPaArea of Triangle
= ½ × (0.12 × 0.30) × 1.35 × 10
6
= 24.3 × 10
3
J = 24.3 KJ ...(i)
Area of rectangle = AC × AD
= (0.12 × 0.30) × 0.15 × 10
6
= 5400 J = 5.4 KJ ...(ii)
B
A
C
2
1.5 MPa
0.15 MPa
P
V
D E
Fundamental Concepts, Definitions and Zeroth Law / 23
W.D. = (1) + (2) = 24.3 + 5.4 = 29.7KJ
ANS 29.7KJ
Q. 38: The variation of pressure with respect to the volume is given by the following equation
p = (3V
2
+ V + 25) NM
2
. Find the work done in the process if initial volume of gas is 3 m
3
and final volume is 6 m
3
.
Sol: P = 3V
2
+ V + 25
Where V
1
= 3m
3
; V
2
= 6m
3
WD =
6
2
2
3
1
(3 25)
V
V
PdV PdV V V dV · · + +
∫ ∫ ∫
= (3V
3
/3 + V
2
/2 + 25V)
6
3
= 277.5J
ANS: 277.5×10
5
N–m
Q. 39: One mole of an ideal gas at 1.0 Mpa and 300K is heated at constant pressure till the volume
is doubled and then it is allowed to expand at constant temperature till the volume is doubled
again. Calculate the work done by the gas. (Dec–01–02)
Sol: Amount of Gas = 1 mole
P
1
= 1.0 MPa
T
1
= 300
0
K Process 1–2: Constant pressure
P
1
V
1
/T
1
= P
2
V
2
/T
2
i.e.; V
1
/T
1
= V
2
/T
2
V
2
= 2 V
1
; i.e.; V
1
/300 = 2V
1
/T
2
T
2
= 600K ...(i)
For 1 mole, R = Universal gas constant
= 8.3143 KJ/kg mole K
= 8314.3 Kg–k
WD =
2
1
∫
Pdv ; Since PV = RT
= PV
2
– PV
1
Fig 1.18
= R(T
2
– T
1
) = 8314.3 (600 – 300) = 2494.29KJ ...(i)
Process 2 – 3: Isothermal process
W
2–3
=
2
1
PdV
∫
= P
2
V
2
lnV
3
/V
2
= RT
2
ln 2V
2
/V
2
= RTln2 = 8314.3 × 600 ln2 = 3457.82KJ
Total WD = WD
1–2
+ WD
2–3
= 2494.29 + 3457.82 = 5952.11 KJ
ANS: 5952.11J
Q. 40: A diesel engine piston which has an area of 45 cm
2
moves 5 cm during part of suction stroke
of 300 cm
3
of fresh air is drawn from the atmosphere. The pressure in the cylinder during
suction stroke is 0.9 × 10
5
N/m
2
and the atmospheric pressure is 1.01325 bar. The difference
between suction pressure and atmospheric pressure is accounted for flow resistance in the
suction pipe and inlet valve. Find the network done during the process. (Dec–01)
Sol: Net work done = work done by free air boundary + work done on the piston
The work done by free air is negative as boundary contracts and work done in the cylinder on the
piston is positive as the boundary expands
P1
P
V
2
3
V1 V2 V3
1 P = C
PV = C
P1
24 / Problems and Solutions in Mechanical Engineering with Concept
Net work done = The displacement work W
=
( ) ( ) +
∫ ∫
bottle balloon
PdV Piston PdV Freeboundary
= [0.9 × 10
5
× 45/(100)
2
× 5/100] + [ – 1.01325 × 10
5
× 300/10
6
]
= – 10.14 Nm .......ANS
Q. 41: Determine the size of a spherical balloon filled with hydrogen at 30
0
C and atmospheric pressure
for lifting 400Kg payload. Atmospheric air is at temperature of 27
0
C and barometer reading
is 75cm of mercury. (May–02)
Sol: Given that:
Hydrogen temperature = 30
0
C = 303K
Load lifting = 400Kg
Atmospheric pressure = 13.6 × 10
3
× 0.75 × 9.81 = 1.00 × 10
5
N/m
2
= 1.00 bar
Atmospheric Temperature = 27
0
C = 300K
The mass that can be lifted due to buoyancy force,
So the mass of air displaced by balloon(m
a
) = Mass of balloon hydrogen gas (m
b
) + load lifted ...(i)
Since PV = mRT; m
a
= P
a
V
a
/RT
a
; R = 8314/29 = 287 KJ/Kgk For Air; 29 = Mol. wt of air
= 1.00 × 10
5
× V/ 287 × 300 = 1.162V Kg ...(ii)
Mass of balloon with hydrogen
m
b
=
PV/RT = 1.00 × 10
5
× V/ (8314/2 × 300) = 0.08V Kg ...(iii)
Putting the values of (ii) and (iii) in equation (i)
1.162V = 0.08V + 400
V = 369.67 m
3
But we know that the volume of a balloon (sphere) = 4/3Πr
3
322 = 4/3Πr
3
r = 4.45 m ......ANS
Q. 42: Manometer measure the pressure of a tank as 250cm of Hg. For the density of Hg 13.6 × 10
3
Kg/m
3
and atmospheric pressure 101KPa, calculate the tank pressure in MPa. (May–01)
Sol: P
abs
= P
atm
+ P
gauge
P
abs
= P
atm
+ ñ.g.h
= 101 × 10
3
+ 13.6 × 10
3
× 9.81 × 250 × 10
–2
= 434.2 × 10
3
N/m
2
= 0.4342 MPa .......ANS
Q. 43: In a cylinder–piston arrangement, 2kg of an ideal gas are expanded adiabatically from a
temperature of 125
0
C to 30
0
C and it is found to perform 152KJ of work during the process
while its enthalpy change is 212.8KJ. Find its specific heats at constant volume and constant
pressure and characteristic gas constant. (May–03)
Sol: Given data:
m = 2Kg
T
1
= 125
0
C
T
2
= 30
0
C
W = 152KJ
H = 212.8KJ
C
P
= ?, C
V
= ?, R = ?
Fundamental Concepts, Definitions and Zeroth Law / 25
We know that during adiabatic process is:
W.D. = P
1
V
1
– P
2
V
2
/γ–1 = mR(T
1
– T
2
)/ γ–1
152 × 10
3
= 2 × R (125 – 30)/(1.4 – 1)
R = 320J/Kg
0
K = 0.32 KJ/Kg
0
K .......ANS
H = mcp dT
212.8 = 2.C
P
.(125 – 30)
C
P
= 1.12 KJ/Kg
o
K .......ANS
C
P
– C
V
= R
C
V
= 0.8 KJ/Kg
o
K .......ANS
Q. 44: Calculate the work done in a piston cylinder arrangement during the expansion process,
where the process is given by the equation:
P = (V
2
+ 6V) bar, The volume changes from 1m
3
to 4m
3
during expansion. (Dec–04)
Sol: P = (V
2
+ 6V) bar
V
1
= 1m
3
; V
2
= 4m
3
WD =
2
1
V
V
PdV PdV ·
∫ ∫
=
4
2
1
( 6 ) V V dV +
∫
= (V
3
/3 + 6V
2
/2)
4
1
= 66J .......ANS
Q. 45: Define and explain Zeroth law of thermodynamics (Dec–01,04)
Or
State the zeroth law of thermodynamics and its applications. Also explain how it is used for
temperature measurement using thermometers. (Dec–00)
Or
State the zeroth law of thermodynamics and its importance as the basis of all temperature
measurement. (Dec–02,05, May–03,04)
Or
Explain with the help of a neat diagram, the zeroth law of thermodynamics. Dec–03
Concept of Temperature
The temperature is a thermal state of a body that describes the degree of hotness or coldness of the body.
If two bodies are brought in contact, heat will flow from hot body at a higher temperature to cold body
at a lower temperature.
Temperature is the thermal potential causing the flow of heat energy.
It is an intensive thermodynamic property independent of size and mass of the system.
The temperature of a body is proportional to the stored molecular energy i.e. the average molecular
kinetic energy of the molecules in a system. (A particular molecule does not have a temperature, it has
energy. The gas as a system has temperature).
Instruments for measuring ordinary temperatures are known as thermometers and those for measuring
high temperatures are known as pyrometers.
Equality of Temperature
Two systems have equal temperature if there are no changes in their properties when they are brought in
thermal contact with each other.
26 / Problems and Solutions in Mechanical Engineering with Concept
Zeroth Law: Statement
When a body A is in thermal equilibrium with a body B, and also separately with a body C, then B and C
will be in thermal equilibrium with each other. This is known as the zeroth law of thermodynamics.
This law forms the basis for all temperature measurement. The thermometer functions as body ‘C’ and
compares the unknown temperature of body ‘A’ with a known temperature of body ‘B’ (reference temperature).
A
B C
Fig. 1.21 Zeroth Law
This law was enunciated by R.H. Fowler in the year 1931. However, since the first and second laws already
existed at that time, it was designated as Zeroth law so that it precedes the first and second laws to
form a logical sequence.
Temperature Measurement Using Thermometers
In order to measure temperature at temperature scale should be devised assigning some arbitrary numbers
to a known definite level of hotness. A thermometer is a measuring device which is used to yield a number
at each of these level. Some material property which varies linearly with hotness is used for the measurement
of temperature. The thermometer will be ideal if it can measure the temperature at all level.
There are different types of thermometer in use, which have their own thermometric property.
1. Constant volume gas thermometer (Pressure P)
2. Constant pressure gas thermometer (Volume V)
3. Electrical Resistance thermometer (Resistance R)
4. Mercury thermometer (Length L)
5. Thermocouple (Electromotive force E)
6. Pyrometer (Intensity of radiation J)
Q. 46: Express the requirement of temperature scale. And how it help to introduce the concept of
temperature and provides a method for its measurement. (Dec–01,04)
Temperature Scales
The temperature of a system is determined by bringing a second body, a thermometer, into contact with the
system and allowing the thermal equilibrium to be reached. The value of the temperature is found by measuring
some temperature dependent property of the thermometer. Any such property is called thermometric property.
To assign numerical values to the thermal state of the system, it is necessary to establish a temperature
scale on which the temperature of system can be read. This requires the selection of basic unit and reference
state. Therefore, the temperature scale is established by assigning numerical values to certain easily
reproducible states. For this purpose it is customary to use the following two fixed points:
(1) Ice Point: It is the equilibrium temperature of ice with air–saturated water at standard Atmospheric
pressure.
(2) Steam Point: The equilibrium temperature of pure water with its own vapour of standard atmospheric
pressure.
Fundamental Concepts, Definitions and Zeroth Law / 27
SCALE ICE POINT STEAM POINT TRIPLE POINT
KELVIN 273.15K 373.15K 273.15K
RANKINE 491.67R 671.67R 491.69R
FAHRENHEIT 32
0
F 212
0
F 32.02
0
F
CENTIGRADE 0
0
C 100
0
C 0.01
0
C
Compansion of references on various scales
373.15.
273.16
273.15
0.00
671.67
491.68
491.67
0.00
211.95
32.02
32.00
–459.67
100
0.01
0.00
–273.15
Normal boiling
point of water
Absolute
zero
Triple point of water
Ice point of water
°F °R °C K
Fig 1.22
Requirement of Temperature Scale
The temperature scale on which the temperature of the system can be read is required to assign the numerical
values to the thermal state of the system. This requires the selection of basic unit & reference state.
Q. 47: Establish a correlation between Centigrade and Fahrenheit temperature scales. (May–01)
Sol: Let the temperature ‘t’ be linear function of property x. (x may be length, resistance volume, pressure
etc.) Then using equation of Line ;
t = A.x + B ...(i)
At Ice Point for Centigrade scale t = 0°, then
0 = A.x
i
+B ...(ii)
At steam point for centigrade scale t = 100
°
, then
100 =A.x
S
+ B ...(iii)
From equation (iii) and (ii), we get
a = 100/(x
s
– x
i
) and b = –100x
i
/(x
s
– x
i
)
Finally general equation becomes in centigrade scale is;
t
0
C = 100x/(x
s
– x
i
) –100x
i
/(x
s
– x
i
)
t
0
C = [(x – x
i
)/ (x
s
– x
i
)]100 ...(iv)
Similarly if Fahrenheit scale is used, then
At Ice Point for Fahrenheit scale t = 32°, then
32 = A.x
i
+ B ...(v)
At steam point for Fahrenheit scale t = 212
°
, then
212 =A.x
S
+ B ...(vi)
From equation (v) and (vi), we get
28 / Problems and Solutions in Mechanical Engineering with Concept
a = 180/(x
s
– x
i
) and b = 32 – 180x
i
/(x
s
– x
i
)
Finally general equation becomes in Fahrenheit scale is;
t
0
F = 180x/(x
s
– x
i
) + 32 – 180x
i
/(x
s
– x
i
)
t
0
F = [(x – x
i
)/ (x
s
– x
i
)]180 + 32 ...(vii)
Similarly if Rankine scale is used, then
At Ice Point for Rankine scale t = 491.67°, then
491.67 = A.x
i
+ B ...(viii)
At steam point Rankine scale t = 671.67
°
, then
671.67 = A.x
S
+ B ...(ix)
From equation (viii) and (ix), we get
a = 180/(x
s
– x
i
) and b = 491.67 – 180x
i
/(x
s
– x
i
)
Finally general equation becomes in Rankine scale is;
t
0
R = 180×/(x
s
– x
i
) + 491.67 – 180x
i
/(x
s
– x
i
)
t
0
R = [(x – x
i
)/ (x
s
– x
i
)] 180 + 491.67 ...(x)
Similarly if Kelvin scale is used, then
At Ice Point for Kelvin scale t = 273.15°, then
273.15 = A.x
i
+ B ...(xi)
At steam point Kelvin scale t = 373.15
°
, then
373.15 = A.x
S
+ B ...(xii)
From equation (xi) and (xii), we get
a = 100/(x
s
– x
i
) and b = 273.15 – 100x
i
/(x
s
– x
i
)
Finally general equation becomes in Kelvin scale is;
t
0
K = 100x/(x
s
– x
i
) + 273.15 – 100x
i
/(x
s
– x
i
)
t
0
K = [(x – x
i
)/ (x
s
– x
i
)] 100 + 273.15 ...(xiii)
Now compare between above four scales:
(x – x
i
)/ (x
s
– x
i
) = C/100 ...(A)
= (F–32)/180 ...(B)
= (R–491.67)/180 ...(C)
= (K – 273.15)/100 ...(D)
Now joining all four values we get the following relation
K = C + 273.15
C = 5/9[F – 32]
= 5/9[R – 491.67]
F = R – 459.67
= 1.8C + 32
Q. 48: Estimate triple point of water in Fahrenheit, Rankine and Kelvin scales. (May–02)
Sol: The point where all three phases are shown of water is known as triple point of water.
Triple point of water T = 273.16
0
K
Let t represent the Celsius temperature then
t = T – 273.15
0
C
Where t is Celsius temperature
0
C and Kelvin temperature T(
0
K)
T
0
F
= 9/5T
0
C
+ 32 = 9/5 × 0.01 + 32 = 32.018
0
F
Fundamental Concepts, Definitions and Zeroth Law / 29
T
0
R
= 9/5T
0
K
= 9/5 × 273.16 = 491.7 R
T
0
C
= 9/5(T
0
K
–32)
T
K
= t
0
C
+ 273.16
T
R
= t
0
F
+ 459.67
T
R
/T
K
= 9/5
Q. 49: During temperature measurement, it is found that a thermometer gives the same temperature
reading in
0
C and in
0
F. Express this temperature value in
0
K. (Dec–02)
Sol: The relation between a particular value C on Celsius scale and F on Fahrenheit sacale is found to be
as mentioned below.
C/100 = (F – 32)/180
As given, since the thermometer gives the same temperature reading say ‘×’ in
0
C and in
0
F, we have
from equation (i)
x/100 = (x – 32) /180
180x = 100(x – 32) = 100x – 3200
x = – 40
0
Value of this temperature in
0
K = 273 + (– 40
0
)
= 233
0
K .......ANS
30 / Problems and Solutions in Mechanical Engineering with Concept
+0)264
2
FlRST L/W CF THERMCD¥N/MlCS
Q. 1: Define first law of thermodynamics?
Sol: The First Law of Thermodynamics states that work and heat are mutually convertible. The present
tendency is to include all forms of energy. The First Law can be stated in many ways:
1. Energy can neither be created nor destroyed; it is always conserved. However, it can change from
one form to another.
2. All energy that goes into a system comes out in some form or the other. Energy does not vanish
and has the ability to be converted into any other form of energy.
3. If the system is carried through a cycle, the summation of work delivered to the surroundings is
equal to summation of heat taken from the surroundings.
4. No machine can produce energy without corresponding expenditure of energy.
5. Total energy of an isolated system in all its form, remain constant
The first law of thermodynamics cannot be proved mathematically. Its validity stems from the fact that
neither it nor any of its corollaries have been violated.
Q. 2: What is the first law for:
(1) A closed system undergoing a cycle
(2) A closed system undergoing a change of state
(1) First Law For a Closed system Undergoing a Change of State
According to first law, when a closed system undergoes a thermodynamic cycle, the net heat transfer
is equal to the network transfer. The cyclic integral of heat transfer is equal to cyclic integral of work
transfer.
. dQ dW ·
∫ ∫
where
∫
stands for cyclic integral (integral around complete cycle), dQ and dW are small elements of heat
and work transfer and have same units.
(2) First Law for a Closed System Undergoing a Change of State
According to first law, when a system undergoes a thermodynamic process (change of state) both heat and
work transfer take place. The net energy transfer is stored within the system and is called stored energy or
total energy of the system.
When a process is executed by a system the change in stored energy of the system is numerically equal
to the net heat interaction minus the net work interaction during the process.
First Law of Thermodynamics / 31
dE = dQ – dW ...(i)
E
2
– E
1
= Q
12
– W
12
Where E is an extensive property and represents the total energy of the system at a given state, i.e.,
E = Total energy
dE = dPE + dKE + dU
If there is no change in PE and KE then, PE = KE = 0
dE = dU, putting in equation (1), we get
dU = dQ – dW
or dQ = dU + dW
This is the first law of thermodynamics for closed system.
Where,
dU = Change in Internal Energy
dW = Work Transfer = PdV
dQ = Heat Transfer = mcdT
{Heat added to the system taken as positive and heat rejected/removal by the system taken as ive}
For a cycle dU = 0; dQ = dW
Q. 3: Define isolated system?
Sol: Total energy of an isolated system, in all its forms, remains constant. i.e., In isolated system there is
no interaction of the system with the surrounding. i.e., for an isolated system, dQ = dW = 0; or, dE = 0,
or E = constant i.e., Energy is constant.
Q. 4: What are the corollaries of first law of thermodynamics?
Sol: The first law of thermodynamics has important corollaries.
Corollary 1 : (First Law for a process).
There exists a property of a closed system, the change in the value of this property during a process
is given by the difference between heat supplied and work done.
dE = dQ – dW
where E is the property of the system and is called total energy which includes internal energy (U), kinetic
energy (KE), potential energy (PE), electrical energy, chemical energy, magnetic energy, etc.
Corollary 2: (Isolated System).
For an isolated system, both heat and work interactions are absent (d Q = 0, d W = 0) and E = constant.
Energy can neither be created nor destroyed, however, it can be converted from one form to another.
Corollary 3 : (PMM  1).
A perpetual motion machine of the first kind is impossible.
Q. 5: State limitations of first law of thermodynamics?
Sol: There are some important limitations of First Law of Thermodynamics.
1. When a closed system undergoes a thermodynamic cycle, the net heat transfer is equal to the net
work transfer. The law does not specify the direction of flow of heat and work nor gives any
condition under which energy transfers can take place.
2. The heat energy and mechanical work are mutually convertible. The mechanical energy can be
fully converted into heat energy but only a part of heat energy can be converted into mechanical
work. Therefore, there is a limitation on the amount of conversion of one form of energy into
another form.
32 / Problems and Solutions in Mechanical Engineering with Concept
Q. 6: Define the following terms:
(1) Specific heat; (2) Joule’s law; (3) Enthalpy
Specific Heat
The sp. Heat of a solid or liquid is usually defined as the heat required to raise unit mass through one degree
temperature rise.
i.e., dQ = mcdT;
dQ = mC
p
dT; For a reversible non flow process at constant pressure;
dQ = mC
v
dT; For a reversible non flow process at constant volume;
C
p
= Heat capacity at constant pressure
C
v
= Heat capacity at constant volume
Joule’s Law
Joules law experiment is based on constant volume process, and it state that the I.E. of a perfect gas is a
function of the absolute temperature only.
i.e., U = f(T)
dU = dQ – dW; It define constant volume i.e dw = 0
dU = dQ; but dQ = mC
v
dT, at constant volume
dU = mC
v
dT; for a perfect gas
Enthalpy
It is the sum of I.E. (U) and pressure – volume product.
H + pv
For unit mass pv =RT
h = C
V
T + RT = (C
V
+ R)T = C
P
T = (dQ)
P
H = mC
P
T
dH = mC
P
dT
Q. 7: What is the relation between two specific heat ?
Sol: dQ = dU + dW; for a perfect gas
dQ at constant pressure
dU at Constant volume; = mC
v
dT = mC
v
(T
2
– T
1
)
dW at constant pressure = PdV = P(V
2
– V
1
) = mR(T
2
– T
1
)
Putting all the values we get
dQ = mC
v
(T
2
– T
1
) + mR(T
2
– T
1
)
dQ = m(C
V
+ R)(T
2
– T
1
)
but dQ = mC
p
(T
2
– T
1
)
mC
p
(T
2
– T
1
) = m(C
V
+ R)(T
2
– T
1
)
C
p
= C
V
+ R; C
p
 C
V
= R ...(i)
Now divided by C
v
; we get
C
p
/C
V
– 1 = R/C
v
; Since C
p
/C
V
= y (gama = 1.41)
y – 1 = R/C
v
;
or C
v
= R/ (y – 1); C
P
= yR/ (y – 1); C
P
>C
V
; y>1
First Law of Thermodynamics / 33
Q. 8: Define the concept of process. How do you classify the process.
Sol: A process is defined as a change in the state or condition of a substance or working medium. For
example, heating or cooling of thermodynamic medium, compression or expansion of a gas, flow of a fluid
from one location to another. In thermodynamics there are two types of process; Flow process and Non
flow process.
Flow Process: The processes in open system permits the transfer of mass to and from the system. Such
process are called flow process. The mass enters the system and leaves after exchanging energy. e.g. I.C.
Engine, Boilers.
NonFlow Process: The process occurring in a closed system where there is no transfer of mass across
the boundary are called non flow process. In such process the energy in the form of heat and work cross
the boundary of the system.
Flow Process
Filling
Emptying
Steady Flow Process
Non Steady
Flow Process
Process
NonFlow Process
V = C
V = C
P = C
P = C
T = C
T = C
PV = C
PV = C
PV = C
x
PV = C
x
U = C
Throtling Process
In steady flow fluid flow at a uniform rate and the flow parameter do not change with time. For
example if the absorption of heat work output, gas flow etc. occur at a uniform rate (Not varying with time),
the flow will be known as steady flow. But if these vary throughout the cycle with time, the flow will be
known as non steady flow process e.g., flow of gas or flow of heat in an engine but if a long interval of
time is chosen as criteria for these flows, the engine will be known to be operating under non – flow
condition.
Q. 9: What is Work done, heat transfer and change in internal energy in free expansion or constant
internal energy process.
P V T
1 1 1
P V T
2 2 2
Insulation
V
P
1
2
A
B
Before expansion After expansion
Gas
Fig 2.1
34 / Problems and Solutions in Mechanical Engineering with Concept
A free expansion process is such a process in which the system expands freely without experience any
resistance. I.E. is constant during state change This process is highly irreversible due to eddy flow of fluid
during the process and there is no heat transfer.
dU = 0; dQ = dW (For reversible process)
dQ = 0; dW = 0; T
1
= T
2
; dU = 0
Q. 10: How do you evaluate mechanical work in different steady flow process?
work done by a steady flow process,
W
1–2
=
2
1
vdp
∫
and work done in a non–flow process,
W
1–2
=
2
1
pdv
∫
1. Constant Volume Process; W
12
= V(P
1
– P
2
)
Steady flow equation
dq = du = dh + d (ke) + d (pv)
Now h = a + pr
Differentiating
dh = du + dtper
= du = pdv = vdp.
1
P P
V
V
1
2
2
Nonflow process
Steady flow process
From First Law of Thermodynamics for a closed system.
dq = du + pdv
db = dg + vdp
∴ dq – = dw = (dq + vdp) + d (ke) + d (pe)
∴ – dw = vdp + d (ke) + d (pe)
if d(ke) = 0 and d (pe) = 0
– dw = vdp
or dw = – vdp
Integrating,
2 2
1 1
dw v vdp · −
∫ ∫
w
1–2
= –
2
1
vdp
∫
First Law of Thermodynamics / 35
2. Constant Pressure process; W
12
= V(P
1
– P
2
) = 0
w
1–2
= –
2
1
vdp
∫
= – v
2
1
dp v ·
∫
1 2
[ ] p p · 3
3. Constant temperature process;
W
1–2
= P
1
V
1
lnP
1
/P
2
= P
1
V
1
lnV
2
/V
1
w
1–2
=
2 2
1 1
1 1
1 1
1 1
"
"
pv p v
p v
vdp dp p v
p v
p
1
1
· −
1
1
¸ ]
∫ ∫
3
= – p
1
v
1
2
1 1
1
dp
p v
p
· −
∫
ln
2
2 2
1
p
p v
p
·
ln
1
2
p
p
= p
1
v
1
ln
2
1
p
v
1 2
2 1
p v
p v
¸ _
·
¸ ,
3
4. Adiabatic Process; W
12
= y(P
1
V
1
– P
2
V
2
)/(y – 1)
pvg =
1 1 2 2
p v p v
γ γ
· · constant
v = v
1
1
t
p
p
γ
¸ _
¸ ,
w
1–2
= –
2
2 2
1
1
1 1
p
vdp v dp
p
γ
¸ _
· −
¸ ,
∫ ∫
w
1–2
= – v
1
2
1
2 – –1 1 1
1
–
1 1 2
1
1
2
– 1
p
p p dp v p
γ
γ γ
γ
· −
+
γ
∫
=
1
1 1
1 1
2 1
–
1
v p
p p
γ
γ − γ −
γ γ
1
−
1
γ −
¸ ]
γ
w
1–2
=
1
γ
γ −
(p
1
v
1
– p
2
v
2
).
5. Polytropic process; W
12
= n(P
1
V
1
– P
2
V
2
)/( n – 1)
w
1–2
=
1
n
n −
(p
1
v
1
– p
2
v
2
).
3
6
/
P
r
o
b
l
e
m
s
a
n
d
S
o
l
u
t
i
o
n
s
i
n
M
e
c
h
a
n
i
c
a
l
E
n
g
i
n
e
e
r
i
n
g
w
i
t
h
C
o
n
c
e
p
t
S.No. PROCESS
. S PVT RELATION WORK DONE dU dQ dH
1. V= C P
1
/T
1
= P
2
/T
2
0 = mC
V
(T
2
– T
1
) = mC
V
(T
2
– T
1
) = mC
P
(T
2
– T
1
)
During Expansion and heating WD and Q is +ive while during Compression and cooling WD and Q is –ive
2. P = C V
1
/T
1
= V
2
/T
2
= P(V
2
– V
1
) = mC
V
(T
2
– T
1
) = mC
P
(T
2
– T
1
) = mC
P
(T
2
– T
1
)
= mR(T
2
– T
1
)
3. T = C P
1
V
1
= P
2
V
2
= P
1
V
1
lnP
1
/P
2
0 Q = W 0
= P
1
V
1
lnV
2
/V
1
= mRT
1
1nV
2
/V
1
4. Pv
γ
= C P
1
V
1
γ
= P
2
V
2
γ
= C = mR (T
2
– T
1
)
/γ–1
= –dW 0 = mCp(T
2
– T
1
)
T
1
/T
2
= (v
2
/v
1
)
γ1
= (P
1
V
1
– P
2
V
2
)/γ–1 = mC
P
(T
2
– T
1
)
= (P
1
/P
2
)
γ1/γ
V
1
/V
2
= (P
2
/P
1
)
1/r
Non – Flow Process
First Law of Thermodynamics / 37
6. Throttling Process
The expansion of a gas through an orifice or partly opened valve is called throttling.
q
1–2
= 0 and w
1–2
= 0
Now h
1
+
2
1
2
V
+gz
1
+ q
1–2
= h
2
+
2
2
2
V
+ gz
2
+ w
1–2
If V
1
= V
2
and z
1
= z
2
, h
1
= h
2
The throttling process is a constant enthalpy process.
If the readings of pressure and temperature of Joule
Thompson porous plug experiment are ploted,
h
1
= h
2
= h
3
= h
4
= h
5
The slope of this constant enthalpy curve is called Joule
Thompson coefficient.
p =
h
dT
dp
1
1
¸ ]
For a perfect gas, p = 0.
Q. 11: Define the following terms:
(1) Control surface
(2) Steam generator
(3) Flow work
(4) Flow Energy
(5) Mass flow rate
Control Surface : A control system has control volume which is separated from its surrounding by
a real or imaginary control surface which is fixed in shape, position and orientation. Matter can continually
flow in and out of control Volume and heat and work can cross the control surface. This is also an open
system.
Steam Generator: The volume of generator is fixed. Water is Supplied. Heat is supplied. Steam
comes out. It is a control system as well as open system.
The flow process can be analysed as a closed system by applying the concept of control volume. The
control surface can be carefully selected and all energies of the system including flow energies can be
considered inside the system. The changes of state of the working substance (mass) need not be considered
during its passage through the system.
PE = force × Distance = (p
1
A
1
).x.
= p
1
V
1
(J)
Now specific volume of working substance is p
1
Fe = p
1
v
1
(J/kg).
Flow Work: The flow work is the energy required to move the working substance against its pressure
It is also called flow or displacement energy .
It a working substance with pressure p, flow through area A, (m
2
) and moves through a distance x.
(m) work required to move the working substance.
Flow work = force X distance = (P.A).x = PV Joule
P
Constant Enthalpy Process
P T
4 4
P T
5 5 P T
1 1
P T
2 2
P T
3 3
38 / Problems and Solutions in Mechanical Engineering with Concept
Control
volume
with
Control
surface
Steam
out
System
Boundary
Water in
Heat
Fig. 2.2 Control volume.
Flow Energy: Flow work analysis is based on the consideration that there is no change in KE, PE,
U. But if these energies are also considered in a flow process, The flow energy per unit mass will be
expressed as
E = F.W + KE + PE + I.E.
E
flow
= PV + V
2
/2 + gZ + U
= (PV + U) + V
2
/2 + gZ
E = h + V
2
/2 + gZ
Mass Flow Rate (m
f
)
In the absence of any mass getting stored the system we can write;
Mass flow rate at inlet = Mass flow rate at outlet
i.e., m
f1
= m
f2
since m
f
= density X volume flow rate = density X Area X velocity = ρ.A.V
ρ
1
.A
1
.V
1
= ρ
2
.A
2
.V
2
or, m
f
= A
1
.V
1
/ν
1
= A
2
.V
2
/ν
2
; Where: ν
1
, ν
2
= specific volume
Q. 13: Derive steady flow energy equation (May05)
Sol: Since the steady flow process is that in which the condition of fluid flow within a control volume do
not vary with time, i.e. the mass flow rate, pressure, volume, work and rate of heat transfer are not the
function of time.
i.e., for steady flow
(dm/dt)
entrance
= (dm/dt)
exit
; i.e, dm/dt = constant
dP/dt = dV/dt = dρ/dt = dE
chemical
= 0
Assumptions
The following conditions must hold good in a steady flow process.
(a) The mass flow rate through the system remains constant.
(b) The rate of heat transfer is constant.
(c) The rate of work transfer is constant.
(d) The state of working:; substance at any point within the system is same at all times.
(e) There is no change in the chemical composition of the system.
If any one condition is not satisfied, the process is called unsteady process.
First Law of Thermodynamics / 39
Let;
A
1
, A
2
= Cross sectional Area at inlet and outlet
ρ
1
,ρ
2
= Density of fluid at inlet and outlet
m
1
,m
2
= Mass flow rate at inlet and outlet
u
1
,u
2
= I.E. of fluid at inlet and outlet
P
1
,P
2
= Pressure of mass at inlet and outlet
ν
1
, ν
2
= Specific volume of fluid at inlet and outlet
V
1
,V
2
= Velocity of fluid at inlet and outlet
Z
1
, Z
2
= Height at which the mass enter and leave
Q = Heat transfer rate
W = Work transfer rate
Consider open system; we have to consider mass balanced as well as energy balance.
2
System Boundary
System
Outlet
Dalum Level
Inlet
A
1
A
2
X
X
2
Z
1
Z
2
Fig 2.3
In the absence of any mass getting stored the system we can write;
Mass flow rate at inlet = Mass flow rate at outlet
i.e., m
f1
= m
f2
since m
f
= density X volume flow rate = density X Area X velocity = ρ.A.V
ρ
1
.A
1
.V
1
= ρ
2
.A
2
.V
2
or, A
1
.V
1
/v
1
= A
2
.V
2
/v
2
; v
1
, v
2
= specific volume
Now total energy of a flow system consist of P.E, K.E., I.E., and flow work
Hence, E = PE + KE + IE + FW
= h + V
2
/2 + gz
Now; Total Energy rate cross boundary as heat and work
= Total energy rate leaving at (2)  Total energy rate leaving at (1)
Q – W= m
f2
[h
2
+ V
2
2
/2 + gZ
2
] m
f1
[h
1
+ V
1
2
/2 + gZ
1
]
For steady flow process m
f
= m
f1
= m
f2
Q – W= m
f
[(h
2
– h
1
) + ½( V
2
2
–V
1
2
) + g(Z
2
–Z
1
)]
For unit mass basis
Q – W
s
= [(h
2
– h
1
) + ½( V
2
2
–V
1
2
) + g(Z
2
–Z
1
)] J/Kgsec
W
s
= Specific heat work
May also written as
dq – dw = dh + dKE + dPE
Or;
h
1
+ V
1
2
/2 + gZ
1
+ q
12
= h
2
+ V
2
2
/2 + gZ
2
+ W
12
40 / Problems and Solutions in Mechanical Engineering with Concept
Q. 14: Write down different cases of steady flow energy equation?
1. Bolter
(kE
2
– kE
1
) = 0, (pE
2
– pE
1
) = 0, w
1–2
= 0
Now, q
1–2
= w
1–2
(h
2
– h
1
) + (kE
2
– kE
1
) + (pE
2
– pE
1
)
q
1–2
= h
2
– h
1
Heat supplied in a boiler increases the enthalpy of the system.
Water in
Boiler
Q
5–2
Steam
out
q
1–2
= w
1–2
(h
2
– h
1
) + (kE
2
– kE
1
) + PE
2
– pE
1
)
– q
1–2
= h
2
– h
1
Heat is lost by the system to the cooling water
q
1–2
= h
1
– h
2
2. Condenser. It is used to condense steam into water.
(kE
2
– kE
1
) = 0 , (pE
2
– pE
1
) = 0
w
1–2
= 0.
q
1–2
– w
1–2
= (h
2
– h
1
) + (kE
2
– kE
1
) + (pE
2
– pE
1
)
– q
1–2
= h
2
– h
1
Heat is lost by the system to the cooling water
q
1–2
= h
1
– h
2
Condense be out
Condenser
Cooling
water out
Cooling
water in
Steam in
1
3. Refrigeration Evaporator. It is used to evaporate refrigerant into vapour.
(kE
2
– kE
1
) = 0, (pE
2
– pE
1
) = 0
w
1–2
= 0
q
1–2
= h
2
– h
1
First Law of Thermodynamics / 41
Evaporator
1 2
Vapour
refrigerant out
Liquid
refrigerant in
Q
1–2
The process is reverse of that of condenser. Heat is supplied by the surrounding to increas3e the
enthalpy of refrigerant.
4. Nozzle. Pressure energy is converted in to kinetic energy
q
1–2
= V, w
1–2
= 0
(pE
2
– pE
1
) = 0
Now, q
1–2
– w
1–2
= (h
2
– h
1
) + (kE
2
– kE
1
) + 0
2 2
2 1
V V
–
2 2
= (h
1
– h
2
)
2
2
V =
2
1
V + 2 (h
1
– h
2
)
V
2
=
2
1 1 2
2 ( ) V h h + −
If V
1
< < V
2
V
2
=
1 2
2 ( ) h h −
Mass flow rate,
m =
1 1 2 2
1 2
A V V A
v v
·
5. Turbine. It is used to produce work.
q
1–2
= 0. (kE
2
– kE
1
) = 0
(pE
2
– pE
1
) = 0
– w
1–2
= (h
2
– h
1
)
w
1–2
= (h
1
– h
2
)
The work is done by the system due to decrease in enthalpy.
6. Rotary Compressor
q
1–2
= 0, (kE
2
– kE
1
) = 0
(pE
2
– pE
1
) = 0
w
1–2
= h
2
– h
1
Work is done to increase enthalpy.
Outlet
2
Nozzle
Intlet
2
Turbine
Steam/gas
W
1–2 Turbine
1
2
Steam gas
Air in
1
Compressor
Air out
2
W
1–2
42 / Problems and Solutions in Mechanical Engineering with Concept
7. Reciprocatin Compressor. It is used to compressor gases.
(kE
2
– kE
1
) = 0, (pE
2
– pE
1
) = 0
q
1–2
– w
1–2
= (h
2
– h
1
) + 0 + 0
– q
1–2
– (–w
1–2
) = h
2
– h
1
w
1–2
= q
1–2
+ (h
2
– h
1
)
Heat is rejected and work is done on the system.
Reciprocating compressor
Cooling of cylinder (by air or water)
q
1–2
w
1–2
2
Air out
1
Air in
Dirrerent Cases of Sfee Sfee
1. Boiler q = h
2
– h
1
2. Condenser q = h
1
– h
2
3. Refrigeration or Evaporator q = h
1
– h
2
4. Nozzle V
2
2
/2 – V
1
2
/2 = h
1
– h
2
5. Turbine W
12
= h
1
– h
2
; WD by the system due to decrease
in enthalpy
6. Rotary compressor W
12
= h
2
– h
1
; WD by the system due to increae in
enthalpy
7. Reciprocating Compressor W
12
= q
12
+ (h
2
– h
1
)
8. Diffuser q – w = (h
2
– h
1
) + ½( V
2
2
–V
1
2
)
Q. 14: 5m
3
of air at 2bar, 27
0
C is compressed up to 6bar pressure following PV
1.3
= constant. It is
subsequently expanded adiabatically to 2 bar. Considering the two processes to be reversible,
determine the net work done, also plot the processes on T – S diagrams. (May – 02)
Sol: V
1
= 5m
3
, P
1
= P
3
= 2bar, P
2
= 6bar, and n = 1.3
V
2
= V
1
(P
1
/P
2
)
1/1.3
= 5(2/6)
1/1.3
= 2.147m
3
Hence work done during process 1 – 2 is W
1–2
= (P
2
V
2
– P
1
V
1
)/(1– n)
= (6 × 10
5
× 2.47 – 2 x 10
5
× 5)/(1–1.3) = – 9.618 × 10
5
J
Similarly to obtain work done during processes 2 – 3, we apply
First Law of Thermodynamics / 43
W
23
= (P
3
V
3
– P
2
V
2
)/(1 – γ); where γ = 1.4
And V
3
= V
2
(P
2
/P
3
)
1/γ
= 2.147(6/2)
1/1.4
= 4.705m
3
P
2
3
1
6 bar
PV = C
1.3
2 bar
V
Fig 2.4
Thus W
23
= (2 × 10
5
× 4.705 – 2 × 10
5
× 2.147)/(1 –1.4) = 8.677 × 10
5
J
Net work done
W
net
= W
12
+ W
23
= – 9.618 × 10
5
+ 8.677 × 10
5
= – 0.9405 × 10
5
J
W
net
= – 94.05 KJ ...ANS
Q. 15: The specific heat at constant pressure of a gas is given by the following relation:
C
p
=0.85+0.00004T+5 x 10T
2
where T is in Kelvin. Calculate the changes in enthalpy and
internal energy of 10 kg of gas when its temperature is raised from 300 K to 2300 K. Take
that the ratio of specific heats to be 1.5. A steel cylinder having a volume of 0.01653 m
3
contains 5.6 kg of ethylene gas C
2
H
4
molecular weight 28. Calculate the temperature to which
the cylinder may be heated without the pressure exceeding 200 bar; given that compressibility
factor Z = 0.605. (Dec0304)
Sol: C
p
= 0.85+0.00004T+5 x 10T
2
dh = m.C
p
.dT
dh = m.
2300
2
2
300
2
(0.85 0.00004T 5 10T )
T
T
dT
·
·
+ + ×
∫
= 10 ×
2300
2
5 2 2 3 3
2 1 2 1 2 1
300
1
0.85 (T – T ) (4 10 / 2) (T – T ) (5 10/ 3) (T – T )
T
T
·
·
1 + × + ×
¸ ]
= 10 x [0.85(2300 – 300) + 4 x 10
5
/2(2300
2
– 300
2
) + 5 x 10/3(2300
3
 300
3
)]
= 2.023 x 10
12
KJ
Change in Enthalpy = 2.023 x 10
12
KJ ...ANS
C
V
= C
P
/γ
du = mC
V
dT
= m.C
P
/γ ⋅ dT
= m/γ.
2300
2
2
300
2
(0.85 0.00004T 5 10T )
T
T
dT
·
·
+ + ×
∫
44 / Problems and Solutions in Mechanical Engineering with Concept
= (10/1.5)
= (10/1.5) × [0.85(2300 – 300) + (4 × 10
5
/2)(2300
2
– 300
2
) + (5 × 10/3)(2300
3
– 300
3
)]
= 1.34 × 10
12
KJ
Change in Internal Energy = 1.34 × 10
12
KJ .......ANS
Now; ν = 0.01653m
3
Pν = ZRT
T = P.V/Z.R = [{200 × 10
5
× 0.01653}/{0.605 × (8.3143 × 10
3
/28) }]
T = 1840.329K .......ANS
Q. 16: An air compressor compresses atmospheric air at 0.1MPa and 27
0
C by 10 times of inlet
pressure. During compression the heat loss to surrounding is estimated to be 5% of compression
work. Air enters compressor with velocity of 40m/sec and leaves with 100m/sec. Inlet and exit
cross section area are 100cm
2
and 20cm
2
respectively. Estimate the temperature of air at exit
from compressor and power input to compressor. (May–02)
Sol: Given that;
At inlet: P
1
= 0.1MPa; T
1
= 27 + 273 = 300K; V
1
= 40m/sec;
A
1
= 100cm
2
At exit: P
2
= 10P
1
= 1.0MPa; V
2
= 100m/sec; A
2
= 20cm
2
Heat lost to surrounding = 5% of compressor work
Since Mass flow rate m
f
= A
1
.V
1
/ν
1
= A
2
.V
2
/ ν
2
;
Where: ν
1
, ν
2
= specific volume ...(i)
(100 × 10
4
× 40)/ ν
1
= (20 × 10
–4
× 100)/ ν
2
...(ii)
or; ν
2
/ν
1
=0.5
Also P
1
ν
1
= RT
1
& P
2
V
2
= RT
2
Or; P
1
ν
1
/T
1
= P
2
ν
2
/ T
2
= R ...(iii)
T
2
/T
1
= (P
2
ν
2
/P
1
ν
1
)
T
2
= T
1
(P
2
ν
2
/P
1
ν
1
) = (10P
1
× 0.5/P
1
) × 300 = 1500K
Also ν
1
= RT
1
/P
1
= {(8.3143 × 10
3
/29) x 300}/(0.1 × 10
6
) = 0.8601 m
3
/kg
From equation (2) m
f
= (100 × 10
4
× 40)/ 0.8601 = 0.465kg/sec
m
f
= 0.465kg/sec .......ANS
Applying SFEE to control volume:
Q – W
S
= m
f
[(h
2
– h
1
) + ½( V
2
2
–V
1
2
) + g(Z
2
–Z
1
)]
Q = 5% of W
S
= 0.05( – W
S
)
– ve sign is inserted because the work is done on the system
–0.05( – W
S
) – W
S
= 0.465[1.005(1500
– 300) + ½( 100
2
–40
2
)/1000]
(Neglecting the change in potential energy)
W
S
= –592.44 KJ/sec .......ANS
–ive sign shows work done on the system
–Power input required to run the compressor is 592.44KW
Q. 17: A steam turbine operating under steady state flow conditions, receives 3600Kg of steam per
hour. The steam enters the turbine at a velocity of 80m/sec, an elevation of 10m and specific
enthalpy of 3276KJ/kg. It leaves the turbine at a velocity of 150m/sec. An elevation of 3m and
First Law of Thermodynamics / 45
a specific enthalpy of 2465 KJ/kg. Heat losses from the turbine to the surroundings amount
to 36MJ/hr. Estimate the power output of the turbine. (May – 01(C.O.))
Sol: Steam flow rate = 3600Kg/hr = 3600/3600 = 1 Kg/sec
Steam velocity at inlet V
1
= 80m/sec
Steam velocity at exit V
2
= 150m/sec
Elevation at inlet Z
1
= 10m
Elevation at exit Z
2
= 3m
Sp. Enthalpy at inlet h
1
= 3276KJ/kg
Sp. Enthalpy at exit h
2
= 2465KJ/kg
Heat losses from the turbine to surrounding Q = 36MJ/hr = 36 x 10
6
/3600 = 10KJ/sec
Turbine operates under steady flow condition, so apply SFEE
For unit mass basis:
Q – W
s
= (h
2
– h
1
) + ½( V
2
2
–V
1
2
) + g(Z
2
–Z
1
) J/Kgsec
– 10 – W
s
= [(2465
– 3276) + (150
2
– 80
2
)/ 2 × 1000 + 9.81(3 –10)/1000]
W
s
= 793 KJ/Kgsec = 793 KW .......ANS
Q. 18: In an isentropic flow through nozzle, air flows at the rate of 600Kg/hr. At inlet to the nozzle,
pressure is 2Mpa and temperature is 127
0
C. The exit pressure is 0.5Mpa. If initial air velocity
is 300m/sec. Determine
(i) Exit velocity of air, and
(ii) Inlet and exit area of the nozzle. (Dec – 01)
Sol:
mf = 600 kg/hr
P = 0.5 MPa
2
P = 2MPa
T = 400 K
C = 300 m/s
1
1
1
2
2
1
1
Fig. 2.5
Rate of flow of air m
f
= 600Kg/hr
Pressure at inlet P
1
= 2MPa
Temperature at inlet T
1
= 127 + 273 = 400K
Pressure at exit P
2
= 0.5MPa
The velocity at inlet V
1
= 300m/sec
Let the velocity at exit = V
2
And the inlet and exit areas be A
1
and A
2
Applying SFEE between section 1 – 1 & section 2 – 2
Q – W
S
= m
f
[(h
2
– h
1
) + ½( V
2
2
–V
1
2
) + g(Z
2
–Z
1
)]
Q = W
S
= 0 and Z
1
= Z
2
46 / Problems and Solutions in Mechanical Engineering with Concept
For air h
2
– h
1
= C
P
(T
2
– T
1
)
0 = C
P
(T
2
– T
1
) + ½( V
2
2
–V
1
2
)
V
2
2
= 2C
P
(T
2
– T
1
) + V
1
2
...(i)
Now T
2
/T
1
= (P
2
/P
1
)
y1/y
For air γ = 1.4
T
2
= 400(0.5/2.0)
1.41/1.4
= 269.18K
from equation 1
V
2
= [2 x 1.005 x 10
3
(400 – 269.18) + (300)
2
]
1/2
V
2
= 594 m/sec .......ANS
Since P
1
ν
1
= RT
1
ν
1
= 8.314 x 400/ 29 × 2000 = 0.05733 m
3
/kg
Also m
f
.ν
1
= A
1
ν
1
A
1
= 600 × 0.05733/3600 × 300 = 31.85mm
2
.......ANS
P
2
ν
2
= RT
2
ν
2
= 8.314 × 269.18/ 29 × 500 = 0.1543 m
3
/kg
Now m
f
.ν
2
= A
2
ν
2
A
2
= 600 × 0.1543/3600 × 594 = 43.29mm
2
.......ANS
Q. 19: 0.5kg/s of a fluid flows in a steady state process. The properties of fluid at entrance are
measured as p
1
= 1.4bar, density = 2.5kg/m
3
, u
1
= 920Kj/kg while at exit the properties are p
2
= 5.6 bar, density = 5 kg/m
3
, u
2
= 720Kj/kg. The velocity at entrance is 200m/sec, while at exit
it is 180m/sec. It rejects 60kw of heat and rises through 60m during the flow. Find the change
of enthalpy and the rate of work done. (May03)
Sol: Given that:
m
f
= 0.5kg/s
P
1
= 1.4bar,
density = 2.5kg/m
3
,
u
1
= 920Kj/kg
P
2
= 5.6 bar,
density = 5 kg/m
3
,
u
2
= 720Kj/kg.
V
1
= 200m/sec
V
2
= 180m/sec
Q = – 60kw
Z
2
– Z
1
= 60m
∆h = ?
W
S
= ?
Since h
2
– h
1
= ∆U + ∆Pν
h
2
– h
1
= [U
2
– U
1
+ (P
2
/ρ
2
– P
1
/ ρ
1
)]
= [(720 –920) × 10
3
+ (5.6/5 – 1.4/2.5) × 10
5
]
= [–200 × 10
3
+ 0.56 × 10
5
] = – 144KJ/kg
∆H = m
f
× (h
2
– h
1
) = 0.5 × (–144) Kj/kg = –72KJ/sec .......ANS
Now Applying SFEE
First Law of Thermodynamics / 47
– Q – W
S
= m
f
[(h
2
– h
1
) + ½( V
2
2
–V
1
2
) + g(Z
2
–Z
1
)]
60 × 10
3
– W
S
= 0.5[ – 144 × 10
3
+ (180
2
– 100
2
)/2 + 9.81 × 60]
W
S
= 13605.7 W = 136.1KW .......ANS
Q. 20: Carbon dioxide passing through a heat exchanger at a rate of 100kg/hr is cooled down from
800
0
C to 50
0
C. Write the steady flow energy equation. Assuming that the change in pressure,
kinetic and potential energies and flow work interaction are negligible, determine the rate of
heat removal. (Take Cp = 1.08Kj/kgK) (Dec03)
Sol: Given data:
m
f
= 100Kg/hr = 100/3600 Kg/sec = 1/36 Kg/sec
T
1
= 800
0
C
T
2
= 50
0
C
Cp = 1.08Kj/kg–K
Rate of heat removal = Q = ?
Now Applying SFEE
Q – W
S
= m
f
[(h
2
– h
1
) + ½( V
2
2
–V
1
2
) + g(Z
2
– Z
1
)]
Since change in pressure, kinetic and potential energies and flow work interaction are negligible, i.e.;
W
S
= ½( V
2
2
–V
1
2
) = g(Z
2
– Z
1
) = 0
Now
Q = m
f
[(h
2
– h
1
)] = m
f
[C
P
.dT] = (1/36) × 1.08 (800 – 50)
Q = 22.5 KJ/sec .......ANS
Q. 22: A reciprocating air compressor takes in 2m
3
/min of air at 0.11MPa and 20
0
C which it delivers
at 1.5MPa and 111
0
C to an after cooler where the air is cooled at constant pressure to 25
0
C.
The power absorbed by the compressor is 4.15KW. Determine the heat transfer in (a)
Compressor and (b) cooler. C
P
for air is 1.005KJ/KgK.
Sol: ν
1
= 2m
3
/min = 1/30 m
3
/sec
P
1
= 0.11MPa = 0.11 x 10
6
N/m
2
T
1
= 20
0
C
P
2
= 1.5 x 10
6
N/m
2
T
2
= 111
0
C
T
3
= –25
0
C
W = 4.15KW
C
P
= 1.005KJ/kgk
Q
12
= ? and Q
23
= ?
From SFEE
Q – W
S
= m
f
[(h
2
– h
1
) + ½( V
2
2
– V
1
2
) + g(Z
2
– Z
1
)]
There is no data about velocity and elevation so ignoring KE and PE
Q
12
– W
1–2
= m[cp(T
2
– T
1
)] ...(i)
Now P
1
ν
1
= mRT
1
m = (0.11 × 10
6
x 1/30)/(287 × 293) = 0.0436Kg/sec; R= 8314/29 = 287 For Air
From equation (i)
Q
12
– 4.15 × 10
3
= 0.0436[1.005 × 10
3
(111 – 20)]
Q
12
= 8.137KJ/sec .......ANS
48 / Problems and Solutions in Mechanical Engineering with Concept
For process 2 – 3; W
23
= 0
Q
23
– W
23
= m[cp(T
2
– T
1
)]
Q
23
– 0 = 0.0436[1.005 x 10
3
(– 111 + 25)]
Q
23
= – 3.768KJ/sec .......ANS
Q. 22: A centrifugal air compressor delivers 15Kg of air per minute. The inlet and outlet conditions
are
At inlet: Velocity = 5m/sec, enthalpy = 5KJ/kg
At out let: Velocity = 7.5m/sec, enthalpy = 173KJ/kg
Heat loss to cooling water is 756KJ/min find:
(1) The power of motor required to drive the compressor.
(2) Ratio of inlet pipe diameter to outlet pipe diameter when specific volumes of air at inlet
and outlet are 0.5m
3
/kg and 0.15m
3
/kg respectively. Inlet and outlet lines are at the same
level.
Sol: Device: Centrifugal compressor
Mass flow rate m
f
= 15Kg/min
Condition at inlet:
V
1
= 5m/sec; h
1
= 5KJ/kg
Condition at exit:
V
2
= 7.5m/sec; h
3
= 173KJ/kg
Heat loss to cooling water Q = –756 KJ/min
From SFEE
Q – W
S
= m
f
[(h
2
– h
1
) + ½( V
2
2
–V
1
2
) + g(Z
2
–Z
1
)]
–756 – W
S
= 15[(173
– 5) + ½( 7.5
2
–5
2
)/1000 + 0]
W
S
= –3276.23KJ/min =  54.60KJ/sec .......ANS
(ive sign indicate that work done on the system) Thus the power of motor required to drive the
compressor is 54.60KW
Mass flow rate at inlet = Mass flow rate at outlet = 15kg/min = 15/60 kg/sec
Mass flow rate at inlet = m
f1
= A
1
.V
1
/ν
1
15/60 = A
1
× 5/0.5
A
1
= 0.025m
2
Now; Mass flow rate at outlet = m
f2
= A
2
.V
2
/ν
2
15/60 = A
2
× 7.5/0.15
A
2
= 0.005m
2
A
1
/A
2
= 5
Πd
1
2
/Πd
1
2
= 5
d
1
/d
2
= 2.236 .......ANS
Thus the ratio of inlet pipe diameter to outlet pipe diameter is 2.236
Q. 23: 0.8kg/s of air flows through a compressor under steady state condition. The properties of air
at entrance are measured as p
1
= 1bar, velocity 10m/sec, specific volume 0.95m
3
/kg and internal
energy u
1
= 30KJ/kg while at exit the properties are p
2
= 8 bar, velocity 6m/sec, specific
volume 0.2m3/kg and internal energy u
2
= 124KJ/kg. Neglecting the change in potential energy.
Determine the power input and pipe diameter at entry and exit. (May05(C.O.))
2
1
0
W
S
Contr ol volum e
q
R
Com pres sor
Fig 2.6
First Law of Thermodynamics / 49
Sol: Device: Centrifugal compressor
Mass flow rate m
f
= 0.8Kg/sec
Condition at inlet:
P
1
= 1 bar = 1 × 10
5
N/m
2
V
1
= 10m/sec;
u
1
= 30KJ/kg ν
1
= 0.95m
3
/kg
Condition at exit:
P
2
= 8bar = 8 × 10
5
N/m
2
V
2
= 6m/sec;
u
2
= 124KJ/kg
ν
2
= 0.2m
3
/kg
The change in enthalpy is given by
h
2
– h
1
= (u
2
+ P
2
U
2
) – (u
1
+ P
1
U
1
)
= (124 × 10
3
+ 8 × 10
5
× 0.2
– (30 × 10
3
+ 1 × 10
5
× 0.95)
= 159000 J/Kg = 159KJ/kg ...(i)
Heat loss to cooling water
Q = – (dU + dW) = – (U
2
– U
1
) – Ws KJ/sec
Q = – (30 – 124) – W
s
= – 96 – Ws ...(ii)
From Sfee
Q – W
s
= m
f
[(h
2
– h
1
) + 1/2 (v
2
2
– V
1
2
) + g (Z
2
– Z
1
)]
– 96 – Ws – W
s
= 0.8 [159 + 1/2 (6
2
– 10
2
)]
– 96 2W
s
= 0.8 [159 + 1/2 (6
2
– 10
2
)]
– 96 – 2W
s
= 101.6
W
s
= – 98.8KJ/sec .......ANS
(–ive sign indicate that work done on the system)
Thus the power of motor required to drive the compressor is 54.60KW
Mass flow rate at inlet = Mass flow rate at outlet
= 0.8 A
1
× 10/0.95
A
1
= 0.076m
2
Π/4.d
intel
2
= 0.076
d
inlet
= 0.096 m = 96.77 mm .......ANS
Now; Mass flow rate of outlet = m
f2
= A
2
.V
2
/u
2
0.8 = A
2
× 6/0.2
A
2
= 0.0266m
2
Π/4.d
outlet
2
= 0.0266
d
outlet
= 0.03395 m = 33.95 mm .......ANS
Fig 2.7
2
1
0
W
S
Contr ol volum e
q
R
Com pres sor
50 / Problems and Solutions in Mechanical Engineering with Concept
+0)264
3
SECCND L/W CF THERMCD¥N/MlCS
Q. 1: Explain the Essence of Second Law?
Sol: First law deals with conservation and conversion of energy. But fails to state the conditions under
which energy conversion are possible. The second law is directional law which would tell if a particular
process occurs or not and how much heat energy can be converted into work.
Q. 2: Define the following terms:
1. Thermal reservoir,
2. Heat engine,
3. Heat pump (Dec05)
Or
Write down the expression for thermal efficiency of heat engine and coefficient of performance
(COP) of the heat pump and refrigerator. (Dec02,04)
Sol: Thermal Reservoir. A thermal reservoir is the part of environment which
can exchange heat energy with the system. It has sufficiently large capacity and
its temperature is not affected by the quantity of heat transferred to or from it.
The temperature of a heat reservoir remain constant. The changes that do take
place in the thermal reservoir as heat enters or leaves are so slow and so small
that processes within it are quasistatic. The reservoir at high temperature which
supplies heat to the system is called HEAT SOURCE. For example: Boiler
Furnace, Combustion chamber, Nuclear Reactor. The reservoir at low temperature
which receives heat from the system is called HEAT SINK. For example:
Atmospheric Air, Ocean, river.
HEAT ENGINE. A heat engine is such a thermodynamics system that
operates in a cycle in which heat is transferred from heat source to heat
sink. For continuous production of work. Both heat and work interaction
take place across the boundary of the engine. It receive heat Q
1
from a
higher temperature reservoir at T
1
. It converts part of heat Q
1
into
mechanical work W
1
. It reject remaining heat Q
2
into sink at T
2
. There is
a working substance which continuously flow through the engine to ensure
continuous/cyclic operation.
Performance of HP: Measured by thermal efficiency which is the degree of useful conversion of heat
received into work.
Soutce
T
1
Sink
T
2
HE Work. W
Q
1
Q
2
Fig 3.1
Second Law of thermodynamics / 51
η
th
= Net work output/ Total Heat supplied = W/Q
1
= (Q
1
– Q
2
) / Q
1
η ηη ηη
th
= 1 – Q
2
/Q
1
= 1 – T
2
/T
1
; Since Q
1
/Q
2
= T
1
/T
2
Or, Thermal efficiency is defined as the ratio of net work gained
(output) from the system to the heat supplied (input) to the system.
Heat Pump: Heat pump is the reversed heat engine which removes heat
from a body at low temperature and transfer heat to a body at higher
temperature.It receive heat Q
2
from atmosphere at temperature T
2
equal to
atmospheric temperature.
It receive power in the form of work ‘W’ to transfer heat from low
temperature to higher temperature. It supplies heat Q
1
to the space to be heated
at temperature T
1
.
Performance of HP: is measured by coefficient of performance (COP).
Which is the ratio of amount of heat rejected by the system to the mechanical
work received by the system.
(COP)
HP
= Q
1
/W = Q1/(Q
1
– Q
2
) = T
1
/ (T
1
– T
2
)
Refrigerator
The primary function of a heat pump is to transfer heat from a low
temperature system to a high temperature system, this transfer of heat can
be utilized for two different purpose, either heating a high temperature
system or cooling a low temperature system. Depending upon the nature of
use. The heat pump is said to be acting either as a heat pump or as a
refrigerator. If its purpose is to cause heating effect it is called operating
as a H.P. And if it is used to create cold effect, the HP is known to be
operating as a refrigerator.
(COP)
ref
= Heat received/ Work Input
= Q
2
/W = Q
2
/(Q
1
– Q
2
)
(COP)
ref
= Q
2
/(Q
1
– Q
2
) = T
2
/(T
1
– T
2
)
(COP)
HP
= (COP)
ref
+ 1
COP is greater when heating a room than when cooling it.
Q. 3: State and explain the second law of thermodynamics? (Dec02)
Sol: There are many different way to explain second law; such as
1. Kelvin Planck Statement
2. Clausius statement
3. Concept of perpetual motion m/c of second kind
4. Principle of degradation of energy
5. Principle of increase of entropy
Among these the first and second are the basic statements while other concept/principle are derived
from them.
Kelvin Plank is applicable to HE while the clausius statement is applicable to HP.
Fig 3.3
Fig 3.2
Space
being
heated
T > T
1 atm
Surroundings
T = T
1 atm
Work. W
Q
1
Q
2
HP
Sorroundings
T = T
1 atm
Space
being
cooled
T < T
2 atm
Work W
Q
1
Q
2
REF
52 / Problems and Solutions in Mechanical Engineering with Concept
Kelvin Plank Statement
Source
T
1
Sink
T
2
HE W = Q – Q
1 2
Q
1
Q
2
Soutce
T
1
Q
1
HE W =
1
Fig. 3.4 Fig. 3.5
Sol: It is impossible to construct such a H.E. that operates on cyclic process and converts all the heat
supplied to it into an equivalent amount of work. The following conclusions can be made from the statement
1. No cyclic engine can converts whole of heat into equivalent work.
2. There is degradation of energy in a cyclic heat engine as some heat has to be degraded or rejected.
Thus second law of thermodynamics is called the law of degradation of energy.
For satisfactory operation of a heat engine there should be a least two heat reservoirs source and sink.
Clausius statement
It is impossible to construct such a H.P. that operates on cyclic process and allows transfer of heat from
a colder body to a hotter body without the aid of an external agency.
Hot body
T
1
Cold body
T
2
R
e
f
r
i
g
e
r
a
t
i
o
n
r
i
e
a
t
p
u
m
p
Q
1
Q
2
Fig. 3.6
Equivalent of Kalvin Plank and Clausius statement
The Kalvin plank and clausius statements of the second law and are equivalent in all respect. The equivalence
of the statement will be proved by the logic and violation of one statement leads to violation of second
statement and vice versa.
Second Law of thermodynamics / 53
Violation of Clausius statement
A cyclic HP transfer heat from cold reservoir(T
2
) to a hot reservoir(T
1
) with no work input. This violates
clausius statement.
A Cyclic HE operates between the same reservoirs drawing a heat Q
1
and producing W as work. As
HP is supplying Q
1
heat to hot reservoir, the hot reservoir can be eliminated. The HP and HE constitute
a HE operating in cycle and producing work W while exchanging heat with one reservoir(Cold) only. This
violates the KP statement
Violation of KP Statement
A HE produce work ‘W’ by exchanging heat with one reservoir at temperature T
1
only. The KP statement
is violated.
HP
HP
E
E
Q
1
W = 0
Q = 0
2
W = Q – Q
1 2
Q + Q
1 2
W = Q
1
Q
1
Q
1
Q
2
Q
2
Q
1
Hot
rosorvoir,
T
1
T
1
Hot
rosorvoir,
T
2
T
2
Fig. 3.7 Violation of Clascius Statement Fig. 3.8 Violation of KP Statement
H.P. is extracting heat Q
2
from low temperature (T
2
) reservoir and discharging heat to high temperature
(T
1
) reservoir and getting work ‘W’. The HE and HP together constitute a m/c working in a cycle and
producing the sole effect of transmitting heat from a lower temperature to a higher temperature. The
clausius statement is violated.
Q.No4: State and prove the Carnot theorem (May – 02, Dec02)
Sol: Carnot Cycle: Sadi carnot; based on second law of thermodynamics introduced the concepts of
reversibility and cycle in 1824. He show that the temperature of heat source and heat sink are the basis
for determining the thermodynamics efficiency of a reversible cycle. He showed that all such cycles
must reject heat to the sink and efficiency is never 100%. To show a non existing reversible cycle,
Carnot invented his famous but a hypothetical cycle known as Carnot cycle.Carnot cycle consist of two
isothermal and two reversible adiabatic or isentropic operation. The cycle is shown in PV and TS
diagrams
54 / Problems and Solutions in Mechanical Engineering with Concept
P
4
3
2
V
V
1
V
3
V
2
V
4
1
P
4
P
3
P
2
P
1
T = T
1 2
4 3
2 1
T
S = S
1 4
Entropy
S = S
2 3
T = T
3 4
Fig. 3.9 Fig. 3.10
Operation 12: T = C
Q
1
= W
12
= P
1
V
1
lnV
2
/V
1
= mRT
1
lnV
2
/V
1
Operation 23: PV
y
= C
Q = W = 0
Operation 34: T = C
Q
2
= W
34
= P
3
V
3
lnV
4
/V
3
= P
3
V
3
lnV
4
/V
3
= mRT
2
lnV
3
/V
4
Operation 41: PV
y
= C
Q = W = 0Net WD = mRT
1
lnV
2
/V
1
– mRT
2
lnV
3
/V
4
;
Since compression ratio = V
3
/V
4
= V
2
/V
1
, T
2
= T
3
W = mRlnV
3
/V
4
(T
1
– T
3
)
Carnot Theorem
No heat engine operating in a cycle between two given thermal reservoir, with fixed temperature can be
more efficient than a reversible engine operating between the same thermal reservoir.
Thermal efficiency η
th
= Work out/Heat supplied
Thermal efficiency of a reversible engine (η
rev
)
η
rev
= (T
1
– T
2
)/T
1
;
No engine can be more efficient than a reversible carnot engine i.e η
rev
> η
th
Carnot Efficiency
η = (Heat added – Heat rejected) / Heat added = [mRT
2
lnV
2
/V
1
– mRT
4
lnV
3
/V
4
]/ mRT
2
lnV
2
/V
1
η = 1 T
1
/T
2
Condition:
1. If T
1
= T
2
; No work, η = 0
2. Higher the temperature diff, higher the efficiency
3. For same degree increase of source temperature or decrease in sink temperature carnot efficiency
is more sensitive to change in sink temperature.
Q.No5. Explain Clausius inequality (Dec02, 05)
Sol: When ever a closed system undergoes a cyclic process, the cyclic integral
∫
dQ/T is less than zero
(i.e., negative) for an irreversible cyclic process and equal to zero for a reversible cyclic process.
The efficiency of a reversible H.E. operating within the temperature T
1
& T
2
is given by:
Second Law of thermodynamics / 55
η = (Q
1
– Q
2
)/Q
1
= (T
1
– T
2
)/T
1
= 1 – (T
2
/T
1
)
i.e., 1 – Q
2
/Q
1
= 1 – T
2
/T
1
; or
Q
2
/Q
1
= T
2
/T
1
; or; Q
1
/T
1
= Q
2
/T
2
Or; Q
1
/T
1
– (– Q
2
/T
2
) = 0; Since Q
2
is heat rejected so –ive
Q
1
/T
1
+ Q
2
/T
2
= 0;
or
∫
dQ/T = 0 for a reversible engine. .......(i)
Now the efficiency of an irreversible H.E. operating within the same temperature limit T
1
& T
2
is given by
η = (Q
1
– Q
2
)/Q
1
< (T
1
– T
2
)/T
1
i.e., 1 – Q
2
/Q
1
< 1 – T
2
/T
1
;
or; Q
2
/Q
1
< T
2
/T
1
;
or; Q
1
/T
1
< Q
2
/T
2
Or; Q
1
/T
1
– (– Q
2
/T
2
) < 0;
Since Q
2
is heat rejected so –ive
Q
1
/T
1
+ Q
2
/T
2
< 0;
or
∫
dQ/T < 0 for an irreversible engine. ...(ii)
Combine equation (i) and (ii); we get
∫
dQ/T d ≤ ≤≤ ≤≤ 0
The equation for irreversible cyclic process may be written as:
∫
dQ/T + I = 0
I = Amount of irreversibility of a cyclic process.
Q. 6: Heat pump is used for heating the premises in winter and cooling the same during summer
such that temperature inside remains 25ºC. Heat transfer across the walls and roof is found
2MJ per hour per degree temperature difference between interior and exterior. Determine the
minimum power required for operating the pump in winter when outside temperature is 1ºC
and also give the maximum temperature in summer for which the device shall be capable of
maintaining the premises at desired temperature for same power input. (May02)
Sol: Given that:
Temperature inside the room T
1
= 25
0
C
Heat transferred across the wall = 2MJ/hr
0
C
Outside temperature T
2
= 1
0
C
To maintain the room temperature 25
0
C the heat
transferred to the room = Heat transferred across the
walls and roof.
Q
1
= 2 x 10
6
x (25 – 1)/3600 = 1.33 x 10
4
J/sec = 13.33KW
For heat pump
COP= T
1
/(T
1
– T
2
) = 298/(298 – 274) = 12.4167
Also COP = Heat delivered / Net work done = Q
1
/W
net
12.4167 = 1.333X 10
4
/W
net
W
net
= 1073.83 J/sec = 1.074KW
Thus the minimum power required by heat pump =
1.074KW
Again, if the device works as refrigerator (in summer)
Q = 2 MJ/
Hr°C
1
Q = 2 MJ/
Hr°C
1
Q
2
Q
2
W°
W°
T
3
HP
Ref
T = 1 + 273
2
T = 25 + 273
1
T = 25 + 273
4
°
°
°
°
Fig. 3.12 Fig. 3.11
56 / Problems and Solutions in Mechanical Engineering with Concept
Heat transfer Q
1
= {2 x 10
6
/ (60 x 60)} x (T
3
– 298) Watt
Now COP = Q
1
/W
net
= T
4
/(T
3
– T
4
)
[2 x 10
6
x (T
3
– 298)]/[60 x 60 x 1073.83] = 298/(T
3
– 298)
On solving
T
3
= 322K = 49
0
C .......ANS
Q. 7: A reversible heat engine operates between temperature 800
0
C and 500
0
C of thermal reservoir.
Engine drives a generator and a reversed carnot engine using the work output from the heat
engine for each unit equality. Reversed Carnot engine abstracts heat from 500
0
C reservoir
and rejected that to a thermal reservoir at 715
0
C. Determine the heat rejected to the reservoir
by the reversed engine as a fraction of heat supplied from 800
0
C reservoir to the heat engine.
Also determine the heat rejected per hour for the generator output of 300KW. (May01)
Sol: Given that
T
1
= 800
0
C = 1073K
T
2
= 500
0
C = 773K
T
3
= 800
0
C = 988K
η
rev
= (Q
1
– Q
2
)/Q
1
= (T
1
– T
2
)/T
1
= W/Q
1
= (1073 – 773)/ 1073 = W/Q
1
W = 0.28Q
1
...(i)
Now for H.P.
Q
4
/(Q
3
– Q
4
) = T
3
/(T
3
– T
2
)
Q
4
/(W/2) = T
3
/(T
3
– T
2
)
Q
4
= (W/2)[T
3
/(T
3
– T
2
)]
= (0.28Q
1
/2) [T
3
/(T
3
– T
2
)]
Q
4
= (0.28Q
1
/2) [988/(988 – 773)]
= 0.643Q
1
Q
4
= 0.643Q
1
.......ANS Fig 3.13
Now if W/2 = 300; W = 600KW
0.28Q
1
= 600
Q
1
= 2142.8KJ/sec .......ANS
Since Q
1
= W + Q
2
Q
2
= 2142.8 – 600 = 1542.85 KJ/sec
Q
2
= 1542.85 KJ/sec .......ANS
Q. 8: Two identical bodies of constant heat capacity are at the same initial temperature T
1
. A
refrigerator operates between these two bodies until one body is cooled to temperature T
2
. If
the bodies remain at constant pressure and undergo no change of phase, find the minimum
amount of work needed to do is, in terms of T
1
, T
2
and heat capacity. (Dec – 02, May  05)
Sol: For minimum work, the refrigerator has to work on reverse Carnot cycle.
dQ
T
∫
= 0
Let T
f
be the final temperature of the higher temperature body and let ‘C’ be the heat capacity.
C
T
f
T
i
T T
C
T T
α α
+
∫
= 0
(T )
1 (T )
3
HE
W
W
2
2
Generator
T
2
500 + 273 = 773 K
HP
Q
3
Q
2
Q
4
Q
1
(800 + 273) K (715 + 273) K
Second Law of thermodynamics / 57
C log
e
f 2
e
i i
T T
Clog
T T
+ = 0
log
e
f 2
2
i
T T
T
= log
e
1 ⇒
f 2
2
i
T T
T
= 1
T
f
=
2
i
2
T
T
work required (minimum)
= C
T T
f i
T T
i 2
dT – C dT
∫ ∫
= C (T
f
– T
i
) – C (T
i
– T
2
)
= C [T
f
+ T
2
– 2T
i
]
w = C
2
i
2
T
T
2
i
2
2
T
T 2T
T
i
1
+ −
1
1
¸ ]
Q. 9: A reversible heat engine operates between two reservoirs at temperature of 600
0
C and 40
0
C.
The Engine drives a reversible refrigerator which operates between reservoirs at temperature
of 40
0
C and – 20
0
C. The heat transfer to the heat engine is 2000KJ and net work output of
combined engine refrigerator plant is 360KJ. Evaluate the heat transfer to the refrigerator
and the net heat transfer to the reservoir at 40
0
C. (Dec – 05)
Sol : T
1
= 600 + 273 = 873K
T
2
= 40 + 273 = 313K
T
3
= – 20 + 273 = 253K
Heat transfer to engine = 200KJ
Net work output of the plant = 360KJ
Efficiency of heat engine cycle,
η = 1 – T
2
/T
1
= 1 – 313/873 = 0.642
W
1
/Q
1
= 0.642W
1
= 0.642 x 2000 = 1284KJ ...(i)
C.O.P. = T
3
/(T
2
– T
3
) = 253/(313 – 253) = 4.216
Q
4
/W
2
= 4.216 ...(ii)
W
1
– W
2
= 360; W
2
= W
1
– 360
W
2
= 1284 – 360 = 924KJ
From equation (ii)
Q
4
= 4.216 x 924 = 3895.6KJ .......ANS
Q
3
= Q
4
+ W
2
= 3895.6 + 924
Q
3
= 4819.6KJ .......ANS
Q
2
= Q
1
– W
1
= 2000 – 1284
Q
2
= 716KJ .......ANS
Heat rejected to reservoir at 40
0
C = Q
2
+ Q
3
= 716 + 4819.6
Heat rejected to reservoir at 40
0
C = 5535.6KJ .......ANS
Heat transfer to refrigerator, Q
4
= 3895.6KJ .......ANS
Q
4
R E
360KJ
Q
2
Q
3
313K
Q = 2000KJ
1
W
1
W
2
(T )
1
(T )
2
253K 873K
Fig 3.14
58 / Problems and Solutions in Mechanical Engineering with Concept
Q. 10: A cold storage of 100Tonnes of refrigeration capacity runs at 1/4
th
of its carnot COP. Inside
temperature is – 15
0
C and atmospheric temperature is 35
0
C. Determine the power required
to run the plant. Take one tonnes of refrigeration as 3.52KW. (Dec – 03(C.O.))
Sol: Given that T
atm
= 35 + 273 = 308K
T
inside
= 15 + 273 = 258K
COP = T
inside
/( T
atm
 T
inside
) = 258 /(308  258) = 5.16 ...(i)
Again COP = Q/W
5.16 x ¼ = 100 x 3.52/ W
W = 272.87 KW .......ANS
Power required to run the plant is 272.87KW
Q. 11: Define entropy and show that it is a property of system. (Dec05)
Sol: Entropy is a thermodynamics property of a system which can be defined as the amount of heat
contained in a substance and its interaction between two state in a process. Entropy increase with addition
of heat and decrease when heat is removed.
dQ = T.dS; T = Absolute Temperature and dS = Change in entropy.
dS = dQ/ T
TS Diagrams
2
1
dS
∫
=
2
1
dQ/T
∫
The area under TS diagram represent the heat added or
rejected. Entropy is a point functionFrom first las
dQ = dU + dW
T.dS = dU + P.dV
Carnot efficiency η = (T
1
– T
2
)/T
1
= dW/dQ
dW = η.dQ; If T
1
– T
2
= 1; η =1/T
dW = dQ/T = dS; if Temperature difference is one.
dS represents maximum amount of work obtainable per degree
in temperature. Unit of Entropy = KJ/K
Principle of Entropy
From claucius inequality
/ 0 dQ T ≤
∫
Since dS = dQ/T for reversible process and dS > dQ/T for
irreversible process
/ dQ T ≤
∫ ∫
dS; or dQ/T d ≤ dS or dS
e ≥ dQ/T
Change in Entropy During Process
1. V = C PROCESS
dQ = mC
V
dT
or, dQ/T = mC
V
dT/T
dS = mC
V
dT/T; or S
2
– S
1
= mC
V
lnP
2
/P
1
1
2
T
2
T
T
A
ds
S
2
S
1
S
Entrepy
T
1
Fig 3.15
Second Law of thermodynamics / 59
2. P = C; PROCESS
dQ= mC
P
dT
or, dQ/T = mC
P
dT/T
dS = mC
P
dT/T; or S
2
– S
1
= mC
P
lnT
2
/T
1
= mC
P
lnV
2
/V
1
3. T = C; PROCESS
dQ = mRTlnV
2
/V
1
or, dQ/T = (mRT/T)lnV
2
/V
1
or S
2
– S
1
= mRlnV
2
/V
1
= m(C
P
– C
V
) lnV
2
/V
1
= mRlnP
1
/P
2
4. PV
C
= C; PROCESS
dQ = 0; dS = 0
5. PV
n
= C; PROCESS
dQ= [(γ – n)/ (γ – 1)] dW
= [(γ – n)/ (γ – 1)] PdV
dQ/T = [(γ – n)/ (γ – 1)] PdV/T
dS = [(γ – n)/ (γ – 1)] mRdV/V
or S
2
– S
1
= [(γ – n)/ (γ – 1)] mR lnV
2
/V
1
Q. 12: Show that the entropy change in a process when a perfect gas changes from state 1 to state
2 is given by S
2
– S
1
= C
p
lnT
2
/T
1
+ RlnP
1
/P
2
. (May–02, 03)
Using clausius equality for reversible cycle,, we have
rev.
G
T
¸ _
¸ ,
∫
= 0 ...(i)
Let a control mass system undergoes a reversible process from state 1 to state 2 along path A and let
the cycle be completed by returning back through path C, which is also reversible, then
2 2
1 1
A C
G G
T T
¸ _ ¸ _
+
¸ , ¸ ,
∫ ∫
= 0 ...(ii)
Also we can move through path B and C then
2 2
1 1
B C
G G
T T
¸ _ ¸ _
+
¸ , ¸ ,
∫ ∫
= 0
From (ii) and (iii)
2 2
1 1
A B
G G
–
T T
¸ _ ¸ _
¸ , ¸ ,
∫ ∫
= 0
2 2
1 1
A B
G G
T T
¸ _ ¸ _
·
¸ , ¸ ,
∫ ∫
The quantity
G
T
¸ _
¸ ,
∫
is independent of path A and B but depends on end states 1 and 2. Therefore
this is point function and not a path function, and hence a property of the system.
2 2
1 1
rev
G
ds
T
¸ _
·
¸ ,
∫ ∫
where; s is specific entropy.
or S
2
– S
1
=
2
1
rev
G
T
¸ _
¸ ,
∫
60 / Problems and Solutions in Mechanical Engineering with Concept
Also; δq = T. ds (for reversible process)
From first law
δq = du + P dv
h = u + Pv
dh = du +Pdv + vdP
dh – v dP = du + Pdv
Using equations
Tds = du + P dv
2
1
∫
ds =
2 2
1 1
du P
+ dv
T T
∫ ∫
s
2
– s
1
= C
v
2 2
1 1
dT R
+ dv
T v
∫ ∫
s
2
– s
1
= C
v
ln (T
2
/T
1
) + R ln (v
2
/v
1
)
T ds = dh – v dP
2
1
∫
ds =
2 2
1 1
dh v
+ dP
T T
∫ ∫
s
2
– s
1
=
2 2
P
1 1
dT R
C – dP
T P
∫ ∫
s
2
– s
1
= C
p
ln (T
2
/T
1
, – R ln P
2
P
1
)
Q. 13: 5Kg of ice at – 10
0
C is kept in atmosphere which is at 30
0
C. Calculate the change of entropy
of universe when if melts and comes into thermal equilibrium with the atmosphere. Take
latent heat of fusion as 335KJ/kg and sp. Heat of ice is half of that of water. (Dec 05)
Sol: Mass of ice, m = 5Kg
Temperature of ice = 10
0
C = 263K
Temperature of atmosphere = 30
0
C = 303K
Heat absorbed by ice from atmosphere = Heat in solid phase + latent heat + heat in liquid phase
= m
i
C
i
dT + M
i
L
i
+ m
w
C
w
dT
= 5 × 4.187/2 ( 0 + 10) + 5 × 335 + 5 × 4.187 × ( 30 – 0)
= 104.675 + 1675 + 628.05
Q = 2407.725KJ
Entropy change of atmosphere (∆s)
atm
= – Q/T = –2407.725/303
(∆s)
atm
= – 7.946KJ/k
Entropy change of ice(∆s)
ice
= Entropy change as ice gets heated from – 10
0
C to 0
0
C + Entropy change as ice melts at 0
0
C to water
at 0
0
C + Entropy change of water as it gets heated from 0
0
C to 30
0
C
=
dQ/T+ dQ/T
∫ ∫
= m
273 303
ice W
263 273
C .dt/T + L/273 + C .dT/T
1
1
1
1
¸ ]
∫ ∫
= 5[(4.18/2)ln273/263 + 335/273 + 4.18ln303/273]
Second Law of thermodynamics / 61
= 5 x 1.7409 = 8.705KJ
Entropy of universe = Entropy change of atmosphere (∆s)
atm
+ Entropy change of ice(∆s)
ice
= – 7.946KJ/k + 8.705KJ
= 0.7605329KJ/kg .......ANS
Q. 14: 0.05m
3
of air at a pressure of 8bar and 280
0
C expands to eight times its original volume and
the final temperature after expansion is 25
0
C. Calculate change of entropy of air during the
process. Assume C
P
= 1.005KJ/kg – k; C
V
= 0.712KJ/kg – k. (Dec–01)
Sol: V
1
= 0.05m
3
P
1
= 8bar = 800KN/m
2
T
1
= 280
0
C = 553K
V
2
= 8V
1
= 0.4m
3
T
2
= 298K
dS = ?
C
P
= 1.005KJ/kg – k;
C
V
= 0.712KJ/kg – k.
R = C
P
– C
V
= 0.293KJ/kg
P
1
V
1
= mRT
1
m = P
1
V
1
/RT
1
= (800 x 0.05)/(0.293 x 553) = 0.247Kg ...(i)
S
2
– S
1
= mC
V
lnT
2
/T
1
+ mRlnV
2
/V
1
= 0.247 x 0.712 ln(298/553) + 0.247 x 0.293 ln8
= –0.108 + 0.15049
S
2
– S
1
= 0.04174KJ .......ANS
Q. 15: Calculate the change in entropy and heat transfer through cylinder walls, if 0.4m
3
of a gas
at a pressure of 10bar and 200
0
C expands by the law PV
1.35
= Constant. During the process
there is loss of 380KJ of internal energy. (Take C
P
= 1.05KJ/kg k and C
V
= 0.75KJ/kgK)
(May – 01)
Sol: ds = ?
dQ = ?
V
1
= 0.4m
3
P
1
= 10bar = 1000KN/m
2
T
1
= 200
0
C = 473K
PV
1.35
= C
dU = 380KJ
C
P
= 1.05, C
V
= 0.75
Since P
1
V
1
= mRT
m = 1000 × 0.4/[(1.005 – 0.75) × 473]
= 2.82kg ...(i)
dU = mC
V
(T
2
– T
1
)
– 380 = 2.82 × 0.75 (T
2
– 473)
T
2
= 292K ...(ii)
W
12
= mR(T
1
– T
2
)/(n – 1)
= [2.82 × 0.3 (473 – 292)]/(1.35 – 1)
= 437.5KJ ...(iii)
62 / Problems and Solutions in Mechanical Engineering with Concept
y = C
P
/C
V
= 1.05/0.75 = 1.4
Q
1–2
= [(γ – n)W
12
]/(γ – 1)
= [(1.4 – 1.35) x 437.5]/(1.4 – 1)
= 54.69KJ .......ANS
S
2
– S
1
= [(γ – n)mRlnV
2
/V
1
]/(γ –1) ...(iv)
Since in isentropic process T
1
/T
2
= (V
2
/V
1
)
n–1
473/292 = (V
2
/V
1
)
1.35 – 1
V
2
/V
1
= 3.96; Putting in equation 4
S
2
– S
1
= [(1.4 – 1.35) × 2.82 × 0.3 × ln3.96]/( 1.4–1)
S
2
– S
1
= 0.145 KJ/K .......ANS
Q. 16: 5 m
3
of air at 2 bar, 27°C is compressed up to 6 bar pressure following PV
1.3
=C. It is
subsequently expanded adiabatically to 2 bar. Considering the two processes to be reversible,
determine the net work. Also plot the processes on T  S diagram. (Dec–01)
Sol: Given that :
Initial volume of air V
1
= 5 m
3
Initial pressure of air P
1
= 2 bar
Final pressure = 6 bar
Compression : Rev. Polytropic process (PV
n
= C)
Expansion :Rev. adiabatic process (PV
1⋅4
= C)
Now, work done during process (12)
1
W
2
=
1 1 2 2
P V – P V
n –1
also
2
1
V
V
=
1
1
1 3
1
2
2
P
2
V 5
P 6
n
−
¸ _
¸ _
·
¸ ,
¸ ,
= 2⋅148m
3
1
W
2
=
2 100 5 6 100 2 148
962 67 kJ
1 3–1
× × − × × ⋅
· − ⋅
⋅
Now, work done during expansion process (23)
2
W
3
=
2 2 3 3
P V P V
1
−
γ −
3
2
V
V
=
1
2
3
P
P
γ
¸ _
¸ ,
⇒ V
3
= 2⋅148
1
1 4 6
2
−
¸ _
¸ ,
= 4⋅708 m
3
2W
3
=
6 100 2 148 2 100 4 708
1 4 1
× × ⋅ − × × ⋅
− −
= 868 kJ
Net work output = W
12
+ W
23
=
– 96267 + 868 = – 94.67 .......ANS
Fig 3.16
Network output
V
1
V
3
V
2
2 bar
P
6 bar
1
2
PV = C
1.4
PV = C
1.3
Second Law of thermodynamics / 63
–ve sign shows that work input required for compression is more that work output obtained during
expansion.
S
1
3
PV = C
1.3
2
b
a
r
T
2
6
b
a
r
Fig. 3.17
Q. 17: One inventor claims that 2 kg of air supplied to a magic tube at 4 bar and 20°C and two
equal mass streams at 1 bar are produced, one at 20°C and other at 80
0
C. Another inventor
claims that it is also possible to produce equal mass streams, one at 40°C and other at 40°C.
Whose claim is correct and why? Consider that it is an adiabatic system. (Take C
P
air 1.012
kJ/kg K) (Dec–02)
Sol: Given that :
Air supplied to magic tube = 2 kg
Inlet condition at magic tube = 4 bar, 20°C
Exit condition : Two equal mass streams, one at – 20°C and other at 80°C for inventor 1. One at 
40°C and other at 40°C for inventorII.
2 Kg at
4 bar, 20°C
Magic tube
1 Kg at 1 bar,
80°C
1 Kg at 1 bar,
20°C
Fig. 3.18
Assume ambient condition 0
°
C i.e. To = 0°C
This is an irreversible process, the claim will be correct if net entropy of the universe (system
surroundings) increases after the process.
Inventor I :
Total entropy at inlet condition is
S
1
=
m
C
p
ln (T
1
/T
0
) = 2 × 1⋅012 ln
20 273
273
+ ¸ _
¸ ,
= 0⋅143 kJ/K
Total entropy at exit condition is :
S
2
= 1.Cp. ln (T
1
/T
0
) + 1.Cp ln (T
2
/T
0
)
S
2
= 1 × 1⋅012 ln
20 273
273
− + ¸ _
¸ ,
+ 1 × 1⋅012 ln
80 273
273
+ ¸ _
¸ ,
64 / Problems and Solutions in Mechanical Engineering with Concept
S
2
= 0.183
S
2
= S
1
⇒ S
2
– S
1
> 0
Thus the claim of inventor is accepatable
For Inventor 2
S
2
= 1. CP. ln (T
1
/T
0
) + 1. C
p
ln (T
2
/T
0
)
= 1 x 1.012ln[( 40 + 273/273)] + 1 x 1.012ln[( 40 + 273)/273]
= – 0.0219KJ/K
Since S
2
<S
1
S
2
– S
1
< 0
This violates the second law of thermodynamics. Hence the claim of inventor is false. – ANS
Q. 18: 0.25Kg/sec of water is heated from 30
0
C to 60
0
C by hot gases that enter at 180
0
C and leaves
at 80
0
C. Calculate the mass flow rate of gases when its C
P
= 1.08KJ/kgK. Find the entropy
change of water and of hot gases. Take the specific heat of water as 4.186KJ/kgk.
(May – 03)
Sol: Given that
Mass of water m
w
= 0.25Kg/sec
Initial temperature of water T
W1
= 30
0
C
Final temperature of water T
W2
= 60
0
C
Entry Temperature of hot gas = T
g1
= 180
0
C
Exit Temperature of hot gas = T
g2
= 80
0
C
Mass flow rate m
f
=?
Specific heat of gas C
Pg
= 1.08KJ/kgK
Specific heat of water C
W
= 4.186KJ/kgK
Heat gives by the gas = Heat taken by water
m
s
C
Ps
.dT
s
= m
W
C
PW
.dT
W
m
s
× 1.08 × (180 – 100) = 0.25 × 4.186 × (60 – 30)
Mass flow rate of gases = ms = 0.291 Kg/sec .......ANS
Change of Entropy of water = ds
W
= m
W
.C
PW
. ln(T
2
/T
1
)
= 4.186 × 0.25 × ln[(60 + 273)/ (30 + 273)]
Change of Entropy of water = 0.099 KJ/
0
K .......ANS
Change of Entropy of Hot gases = ds
g
= m
g
.C
Pg
. ln(T
2
/T
1
)
= 0.291 × 1.08 × ln [(80 + 273)/ (180 + 273)]
Change of Entropy of Hot gas = 0.0783 KJ/
0
K .......ANS
Introductioon To I.C. Engine / 65
+0)264
4
lNTRCDÜCTlCN TC l.C. ENClNE
Q. 1: What do you mean by I.C. Engine? how are they classified?
Sol.: Internal combustion engine more popularly known as I.C.
engine, is a heat engine which converts the
heat energy released b
y
the combustion of the fuel inside the engine cylinder, into mechanical work. Its
versatile advantages such as high efficiency light weight, compactness, easy starting, adaptability,
comparatively lower cost has made its use as a prime mover universal
Classification of I.C. Engines
IC engines are classified according to:
1. Nature of thermodynamic cycles as:
1. Otto cycle engine;
2. Diesel cycle engine
3. Dual combustion cycle engine
2. Type of the fuel used:
1. Petrol engine
2. Diesel engine.
3. Gas engine
4. Bifuel engine
3. Number of strokes as
1. Four stroke engine
2. Two stroke engine
4. Method of ignition as:
1. Spark ignition engine, known as SI engine
2. Compression ignition engine, known as C.I. engine
5. Number of cylinder as:
1. Single cylinder engine
2. Multi cylinder engine
6. Position of the cylinder as:
1. Horizontal engine
2. Vertical engine.
3. Vee engine
66 / Problems and Solutions in Mechanical Engineering with Concept
4. Inline engine.
5. Opposed cylinder engine
7. Method of cooling as:
1. Air cooled engine
2. Water cooled engine
Q. 2: Differentiate between SI and CI engines. (May–02)
Or
What is C.I. Engine, Why it has more compression ratio compared to S.I. Engine. (May05)
Spark Ignition Engines (S.I. Engine)
It works on otto cycle. In Otto cycle, the energy supply and rejection occur at constant volume process
and the compression and expansion occur isentropically. The engines working on Otto cycle use petrol
as the fuel and incorporate a carburetor for the preparation of mixture of air fuel vapor in correct
proportions for rapid combustion and a spark plug for the ignition of the mixture at the end of compression.
The compression ratio is kept 5 to 10.5. Engine has generally high speed as compared to C.I. engine.
Low maintenance cost but high running cost. These engines are also called spark ignition engines or
simply S.I. Engine.
P
V
2
3
4
1
T
S
1
2
3
4
Fig. 4.1
Compression Ignition Engines (C.I. Engine)
It works on dieses cycle. In diesel engines, the energy addition occurs at constant pressure but energy
rejection at constant volume. Here spark plug is replaced by fuel injector. The compression ratio is from
12 to 25. Engine has generally low speed as compared to S.I. engine. High maintenance cost but low
running cost. These are known as compression ignition engines, (C.I) as the ignition is accomplished by
heat of compression.
P
S
1
2 3
4 T
S
2
1
3
4
Fig. 4.2
The upper limit of compression ratio in S.I. Engine is fixed by anti knock quality of fuel. While in
C.I. Engine upper limit of compression ratio is limited by thermal and mechanical stresses of cylinder
material. That’s way the compression ratio of S.I. engine has more compression ratio as compared to S.I.
Engine.
Introductioon To I.C. Engine / 67
Dual cycle is a combination of the above two cycles, where part or the energy is given a constant
volume and rest at constant pressure.
Q. 3: Define Bore, stroke, compression Ratio, clearance ratio and mean effective pressure.
(Dec–01)
Or
Define clearance volume, mean effective pressure ,Air standard cycle, compression Ratio.
(May–02)
Or
Air standard cycle, Cycle efficiency, mean effective pressure. (May–03)
Bore
The inner diameter of the engine cylinder is known as bore. It can be measured precisely by a vernier
calliper or bore gauge. As the engine cylinder wears out with the passage of time, so the bore diameter
changes to a larger value, hence the piston becomes lose in the cylinder, and power loss occurs. To correct
this problem reboring to the next standard size is done and a new piston is placed. Bore is denoted by the
letter ‘D’. It is usually measured in mm (S.I. units) or inches (metric units). It is used to calculated the
engine capacity (cylinder volume).
Stroke
The distance traveled by the piston from its topmost positions (also called as Top dead centre TDC), to its
bottom most position (or bottom dead centre BDC) is called stroke it will be two times the crank radius.
It is denoted by letter h. Units mm or inches (S.L, Metric). Now we can calculate the swept volume as
follows: (L = 2r)
V
S
=
2
4
D
L
1 π
1
¸ ]
If D is in cm and L is also in cm than the units of V will be cm
3
which is usually written as cubic
centimeter or c.c.
Clearance Volume
The volume above the T.D.C is called as clearance volume, this is provided so as to accommodate engine
valves etc. this is referred as (V
C
).Then total volume of the engine cylinder
V =V
S
+ V
C
Compression Ratio
It is calculated as follows
r
k
=
Total volume
Clearance volume
r
k
=
S C
C
V V
V
+
Mean Effective Pressure (P
m
or P
mef
)
Mean effective pressure is that hypothetical constant pressure which is assumed to be acting on the piston
during its expansion stroke producing the same work output as that from the actual cycle.
68 / Problems and Solutions in Mechanical Engineering with Concept
Mathematically,
P
m
=
1 2
Work Output
Swept volume ( )
net
W
V V
·
−
It can also be shown as
P
m
=
Area of Indicator diagram
constant
Length of diagram
×
The constant depends on the mechanism used to get the indicator diagram and has the units bar/m.
Indicated Mean Effective Pressure (P
im
)
Indicated power of an engine is given by
ip =
60, 000
60, 000
im
im
P L A N K
i p
P
L A N K
×
⇒ ·
Break Mean Effective Pressure (P
bm
)
Similarly, the brake mean effective pressure is given by
P
bm
=
60, 000 b p
L A N K
×
where;
ip = indicated power (kW)
bp = Break Powder (kW)
P
im
= indicated mean effective pressure (N/m
2
)
Pbm = Break mean effective Pressure (N/m
2
)
L = length of the stroke
A = area of the piston (m
2
)
N = number of power strokes
= rpm for 2stroke engines = rpm/2 for 4stroke
K = no. of cylinder.
Q. 4: Write short notes on Indicator diagram and indicated power. (Dec–03)
Sol.: An indicated diagram is a graph between pressure and volume. The former being taken on vertical
axis and the latter on the horizontal axis. This is obtained by an instrument known as indicator. The
indicator diagram are of two types;
(a) Theoretical or hypothetical
(b) Actual.
The theoretical or hypothetical indicator diagram is always longer in size as compared to the actual
one. Since in the former losses are neglected. The ratio of the area of the actual indicator diagram to the
theoretical one is called diagram factor.
Q. 5: Explain the working of any air standard cycle (by drawing it on PV diagram) known to you.
Why is it known as ‘Air standard cycle.’? (Dec–01)
Or
Draw the Diesel cycle on PV coordinates and explain its functioning. (Dec–02)
Or
Show Otto and diesel cycle on PV and TS diagram. (May–03)
Introductioon To I.C. Engine / 69
Or
Stating the assumptions made, describe air standard otto cycle. (Dec–04)
Or
Derive a relation for the air standard efficiency of diesel cycle. Also show the cycle on PV and
TS diagram. (Dec–04)
AIR STANDARD CYCLES
Most of the power plant operates in a thermodynamic cycle i.e. the working fluid undergoes a series of
processes and finally returns to its original state. Hence, in order to compare the efficiencies of various
cycles, a hypothetical efficiency called air standard efficiency is calculated.
If air is used as the working fluid in a thermodynamic cycle, then the cycle is known as “Air Standard
Cycle”.
To simplify the analysis of I.C. engines, air standard cycles are conceived.
Assumptions
1. The working medium is assumed to be a perfect gas and follows the relation
pV = mRT or P = pRT
2. There is no change in the mass of the working medium.
3. All the processes that constitute the cycle are reversible.
4. Heat is added and rejected with external heat reservoirs.
5. The working medium has constant specific heats.
Otto Cycle (1876) (Used S. I. Engines)
This cycle consists of two reversible adiabatic processes and two constant volume processes as shown in
figure on PV and TS diagrams.
The process 12 is reversible adiabatic compression, the process 23 is heat addition at constant volume,
the process 34 is reversible adiabatic expansion and the process 41 is heat rejection at constant volume.
PP
Isentropic
process
O
P
P
Q
1
Q
2
3
2
4
1
T
S
1
2
3
4
Fig. 4.3
The cylinder is assumed to contain air as the working substance and heat is supplied at the end of
compression, and heat is rejected at the end of expansion to the sink and the cycle is repeated.
Process
01 = suction
12 = isentropic compression
23 = heat addition at constant volume
34 = isentropic expansion
4–1 = constant volume heat rejection 10 = exhaust
Heat supplied : Q
1
= mc
v
(T
3
– T
2
)
Heat rejected : Q
2
= mc
v
(T
4
– T
1
)
Efficiency :
η = 1
4 1 2
1 3 2
( )
1
( )
v
v
mc T T Q
Q mc T T
−
· −
−
70 / Problems and Solutions in Mechanical Engineering with Concept
η =1 –
4 1
3 2
T T
T T
−
−
Process 1 – 2 : T
1
V
1
γ – 1 = T
2
V
2
γ – 1
1
1 2
2 1
T V
T V
γ−
¸ _
·
¸ ,
or
1 2
2 1
V T
V T
¸ _
·
¸ ,
Process 3 – 4 : T
3
V
3
γ – 1 = T
4
V
4
γ – 1
1
4 2
3 1
V T
V T
γ−
¸ _
·
¸ ,
also
4 1
3 2
V V
V V
·
⇒
3 3 2 4
4 1 2 1
T T T T
T T T T
· ⇒ ·
⇒
3 4
2 1
1 1
T T
T T
− · −
(subtracting 1 from both sides)
⇒
3 2 4 1
2 1
T T T T
T T
− −
·
⇒
1 1
1 4 1 2
2 3 2 1
1
k
T T T V
T T T V r
γ− γ−
¸ _ ¸ _ −
· · ·
−
¸ , ¸ ,
Substituting in eq. (i) hotto = 1 –
1
1
k
r
γ−
where r
k
= compression ratio.
Diesel Cycle (1892) (Constant Pressure Cycle)
Diesel cycle is also known as the constant pressure cycle because all addition of heat takes place at constant
pressure. The cycle of operation is shown in figure 2.4 (a) and 2.4 (b) on PV and TS diagrams.
PP
V
T
S
2
1
4
3
p C =
V
C
=
Isentropic
Process
Q
1
2
3
4
1
Q
2
Fig. 4.4
The sequence of operations is as follows :
1. The air is compressed isentropically from condition ‘1’ to condition ‘2’.
2. Heat is supplied to the compressed air from external source at constant pressure which is represented
by the process 23.
3. The air expands isentropically until it reaches condition ‘4’.
4. The heat is rejected by the air to the external sink at constant volume until it reaches condition T
and the cycle is repeated.
Introductioon To I.C. Engine / 71
The air standard efficiency of the cycle can be calculated as follows:
Heat supplied: Q
l
= Q
2–3
= mc
p
(T
3
– T
2
)
Heat rejected: Q2 = Q
4–1
= mc
v
(T
4
– T
1
)
η =
4 1 2
1 3 2
( )
1 1
( )
v
p
mc T T Q
Q mc T T
−
− ·
−
η = 1 –
4 1
3 2
( )
( )
T T
T T
−
−
Compression ratio : r
k
= V
1
/V
2
...(i)
Expansion ratio : r
e
= V
4
/V
3
...(ii)
Cut of ratio : r
c
= V
3
/V
2
...(iii)
It is seen that r
k
= r
e
r
c
Process 12 : T
1
V
1
γ – 1 = T
2
V
2
γ – 1
1
2
1
2 1
2 1
( ) 1
k
V
T
T T r
V T
γ−
¸ _
· ⇒ · γ −
¸ ,
Process 23 :
3 3 3 3 2 2
2 3 2 2
c
P V V T P V
r
T T V T
· ⇒ · ·
(As P
2
= P
3
)
Process 34 : T
3
V
3
γ–1 = T
4
V
4
γ–1
1
3 4
3 4
3 4
( ) 1
c
T V
T T r
V T
γ−
¸ _
⇒ · ⇒ · γ −
¸ ,
Substituting η = 1 –
4 1
1 1
4 1
( )
( ) ( )
c k
T T
T r T r
γ− γ−
−
γ −
3
1
T
T
= rc (η)
γ–1
3 2 1
2 1
and ( )
c k
T T
r r
T T
γ−
1
· ·
1
1
¸ ]
3
3
4
T
T
= r
c
γ–1
1
4
1
1
c k
c
k
c
r r T
r
T
r
r
γ−
γ
γ−
⋅
· ·
¸ _
¸ ,
η = 1–
1
2
1
1
c
c
T r
T r
γ
1 −
¸ ]
1 γ −
¸ ]
γ = 1–
1
1
1
( ) 1
c
k c
r
r r
γ
γ−
1 −
¸ ]
γ 1 −
¸ ]
As r
c
> 1, so
1
1
1
c
c
r
r
γ
1 −
1
γ −
1
¸ ]
is also > 1,therefore for the same compression ratio the efficiency of the diesel cycle is less than that of
the otto cycle.
72 / Problems and Solutions in Mechanical Engineering with Concept
Q. 6: Compare otto cycle with Diesel cycle?
Sol.: These two cycles can be compared on the basis of either the same compression ratio or the same
maximum pressure and temperature.
P
PV
= C
V
T
S
V C =
p C = 2
2
1
5
3
4
o
2
3
4
5
1
Fig. 4.5
1  2  3  5 = Otto Cycle,
for the same heat rejection Q
2
the higher the
1  2 4  5 = Diesel Cycles, heat given Q
1
,
the higher is the cycle efficiency.
So from TS diagram for cycle 1  2  3  5, Q
1
is more than that for 1  2  7  5 (area under the curve
represents Q,).
Hence η
Otto
> η
Diesel
For the same heat rejection by both otto and diesel cycles.
Again both can be compared on the basis of same maximum pressure and temperature.
P
V
T
S
2
1
1
2
3
4
2
4
3
2
Fig. 4.6
1  2 – 3 – 4 = Otto Cycle; Here area under the curve
1  2′  3  4 = Diesel Cycle
1  2′  3  4 is more than 1  2 – 3 – 4
So ηdiese l > ηotto; for the same T
max
and P
max
Q. 7: Describe the working of four stroke SI engine. Illustrate using line diagrams.
(May–02, May–03, Dec–03)
Or
Explain the working of a 4 stroke petrol engine. (Dec–02)
Four Stroke Engine
Figure. shows the working of a 4 stroke engine. During the suction stroke only air (in case of diesel engine)
or air with petrol (in case of petrol engine) is drawn into the cylinder by the moving piston.
Introductioon To I.C. Engine / 73
Intel velve
( ) Suction a ( ) Suction b ( ) Suction c ( ) Suction d
Exhauel valve
Spark plug
Fig. 4.7. Cycle of events in a four stroke petrol engine
The charge enters the engine cylinder through the inlet valve which is open. During this stroke, the
exhaust valve is closed. During the compression stroke, the charge is compressed in the clearance space.
On completion of compression, if only air is taken in during the suction stroke, the fuel is injected into the
engine cylinders at the end of compression. The mixture is ignited and the heat generated, while the piston
is nearly stationary, sets up a high pressure. During the power stroke, the piston is forced downward by the
high pressure. This is the important stroke of the cycle. During the exhaust stroke the products of combustion
are swept out through the open exhaust valve while the piston returns. This is the scavenging stroke. All
the burnt gases are completely removed from the engine cylinder and the cylinder is ready to receive the
fresh charge for the new cycle.
Thus, in a 4stroke engine there is one power stroke and three idle strokes. The power stroke supplies
the necessary momentum to keep the engine running.
Q. 8: Describe the working of two stroke SI engine. Illustrate using line diagrams.
(May–03, 04, Dec–05)
Two Stroke Engine
In two stroke engine, instead of valves ports are provided, these are opened and closed by the moving
piston. Through the inlet port, the mixture of air and fuel is taken into the crank case of the engine cylinder
and through the transfer port the mixture enters the engine cylinder from the crank case. The exhaust ports
serve the purpose of exhausting the gases from the engine cylinder. These ports are more than one in
number and are arranged circumferentially.
Spark plug
Exhaust port
Inlet port
Piston
Transfer port
Crank shaft
Crank case
Fig. 4.8
74 / Problems and Solutions in Mechanical Engineering with Concept
A mixture of air fuel enters the cylinder through the transfer ports and drives the burnt gases from the
previous stroke before it. As the piston begins to move upwards fresh charge passes into the engine cylinder.
For the remainder of upward stroke the charge taken in the engine cylinder is compressed after the piston
has covered the transfer and exhaust ports. During the same time mixture of air and fuel is taken into the
crank case. When the piston reaches the end of its stroke, the charge is ignited, which exerts pressure on
top of the piston. During this period, first of all exhaust ports are uncovered by the piston and so the exhaust
gases leave the cylinder. The downward movement of the piston causes the compression of the charge taken
into the crank case of the cylinder. When the piston reaches the end of the downward stroke. The cycle
repeats.
Q. 9: Compare Petrol engine with Diesel engine.?
Sol.: (i) Basic cycle: Petrol Engine work on Otto cycle whereas Diesel Engine work on diesel cycles.
(ii) Induction of fuel: During the suction stroke in petrol engine, the air fuel mixture is sucked in the
cylinder while in diesel engine only air is sucked into the cylinder during its suction stroke.
(iii) Compression Ratio: In petrol engine the compression ratio in the range of 5:1 to 8:1 while in
diesel engine it is in the range of 15:1 to 20:1.
(iv) Thermal efficiency: For same compression ratio, the thermal efficiency of diesel engine is lower
than that of petrol engine.
(v) Ignition: In petrol engine the charge (A/F mixture) is ignited by the spark plug after the compression
of mixture while in diesel engine combustion of fuel due to its high temperature of compressed air.
Two Stroke Engine
In two stroke engine all the four operation i.e. suction, compression, ignition and exhaust are completed
in one revolution of the crank shaft.
Four Stroke Engine
In four stroke engine all the four operation are completed in two revolutions of crank shaft.
Application of 2stroke Engines
2 stroke engine are generally used where low cost, compactness and light weight are the major
considerations
Q. 10: compare the working of 4 stroke and 2 – stroke cycles of internal combustion engines.
(Dec–01, 04)
Sol.: The following are the main differences between a four stroke and two stroke engines.
1. In a four stroke engine, power is developed in every alternate revolution of the crankshaft whereas;
in a two stroke engine power is developed in every revolution of the crankshaft.
2. In a two stroke engine, the torque is more uniform than in the four stroke engine hence a lighter
flywheel is necessary in a two stroke engine, whereas a four stroke engine requires a heavier
flywheel.
3. The suction and the exhaust are opened and closed by mechanical valves in a four stroke engine,
whereas in a two stroke engine, the piston itself opens and closes the ports.
4. In a four stroke engine the charge directly enters into the cylinder whereas in a two stroke engine
the charge first enters the crankcase and then flows into the cylinder.
5. The crankcase of a two stroke engine is a closed pressure tight chamber whereas the crankcase of
a four stroke engine even though closed is not a pressure tight chamber.
Introductioon To I.C. Engine / 75
6. In a four stroke engine the piston drives out the burnt gases during the exhaust stroke, whereas,
in a two stroke engine the high pressure fresh charge scavenges out the burnt gases.
7. The lubricating oil consumption in a two stroke engine is more than in four stroke engine.
8. A two stroke engine produces more noise than a four stroke engine.
9. Since the fuel burns in every revolution of the crankshaft in a two stroke engine the rate of cooling
is more than in a four stroke engine.
10. A valve less two stroke engines runs in either direction, whereas a four stroke engine cannot run
in either direction.
Q. 11: What are the advantage of a two stroke engine over a four stroke engine.?
Sol.: The following are the advantages of a two stroke engine over a four stroke engine:
1. A two stroke engine has twice the number of power stroke than a four strokes engine at the same
speed. Hence theoretically a two stroke engine develops double the power per cubic meter of the
swept volume than the four stroke engine.
2. The weight of the two stroke engine is less than four stroke engine because of the lighter flywheel
due to more uniform torque on the crankshaft.
3. The scavenging is more complete in lowspeed two stroke engines, since exhaust gases are not left
in the clearance volume as in the four stroke engine.
4. Since there are only two strokes in a cycle, the work required to overcome the friction and the
exhaust strokes is saved.
5. Since there are no mechanical valves and the valve gears, the construction of two stroke engine
is simple which reduces its initial cost.
6. A two stroke engine can be easily reversed by a simple reversing gear mechanism.
7. A two stroke engine can be easily started than a four stroke engine:
8. A two stroke engine occupies less space.
9. A lighter foundation will be sufficient for two stroke engine.
10. A two stroke engine has less maintenance cost since it requires only few parts.
Q. 12: What are the disadvantages of two stroke engine?
Sol.: The following are some of the disadvantages of two stroke engine when compared with four stroke
engine:
1. Since the firing takes place in every revolution, the time available for cooling will be less than in
a four stroke engine.
2. Incomplete scavenging results in mixing of exhaust gases with the fresh charge which will dilute
it, hence lesser power output.
3. Since the transfer port is kept open only during a short period, less quantity of the charge will be
admitted into the cylinder which will reduce the power output.
4. Since both the exhaust and the transfer ports are kept open during the same period, there is a
possibility of escaping of the fresh charge through the exhaust port which will reduce the thermal
efficiency.
5. For a given stroke and clearance volume, the effective compression stroke is less in a two stroke
engine than in a four stroke engine.
6. In a crankcase compressed type of two stroke engine, the volume of charge down into the crankcase
is less due to the reduction in the crankcase volume because of rotating parts.
7. A fan scavenged two stroke engine has less mechanical efficiency since some power is required to
run the scavenged fan.
76 / Problems and Solutions in Mechanical Engineering with Concept
8. A two stroke engine needs better cooling arrangement because of high operating temperature.
9. A two stroke engine consumes more lubricating oil.
10. The exhaust in a two stroke engine is noisy due to sudden release of the burnt gases.
Q. 13: Calculate the thermal efficiency and compression ratio for an automobile working on otto
cycle. If the energy generated per cycle is thrice that of rejected during the exhaust. Consider
working fluid as an ideal gas with γ γγ γγ = 1.4 (May–01)
Sol.: Since we have
η
otto
= (Q
1
– Q
2
)/Q
1
Where
Q
1
= Heat supplied
Q
2
= Heat rejected
Given that Q
1
= 3Q
2
η
otto
= (3Q
2
– Q
2
)/3Q
2
= 2/3 = 66.6% ...(i)
We also have;
η
otto
= 1 – 1/(r)
γ – 1
0.667 = 1/(r)
1.4 – 1
r = (3)
1/0.4
= 15.59 .......ANS
Q. 14: A 4 stroke diesel engine has length of 20 cm and diameter of 16 cm. The engine is producing
indicated power of 25 KW when it is running at 2500 RPM. Find the mean effective pressure
of the engine. (May–03)
Sol.: Length or stroke = 20 cm = 0.2 m
Diameter or Bore = 16 cm = 0.16 m
Indicating power = 25 KW
Speed = 2500 RPM
Mean effective pressure = ?
K = 1
Indicated power = P
ip
= (P
mef
.L.A.N.K)/60
Where N = N/2 = 1250 RPM (for four stroke engine)
25 × 10
3
= {P
mef
× 0.2 × (π/4)(0.16)
2
× 1250 × 1}/60
P
mef
= 298.415KN/m
2
.......ANS
Q. 15: A 4 stroke diesel engine has L/D ratio of 1.25. The mean effective pressure is found with the
help of an indicator equal to 0.85MPa. The engine produces indicated power of 35 HP. While
it is running at 2500 RPM. Find the dimension of the engine. (Dec–03)
Sol.: L/D = 1.25
P
mef
= 0.85 MPa = 0.85 × 10
6
N/m
2
P
IP
= 35 HP = 35/1.36 KW (Since 1KW = 1.36HP or 1HP = 1/1.36 KW)
N = 2500 RPM = 1250 RPM for four stroke engine (N = N/2 for four stroke)
Indicated power = P
ip
= (P
mef
.L.A.N.K)/60
(35/1.36) × 10
3
= {0.85 × 10
6
× 1.25D × (π/4)(D)
2
× 1250 × 1}/60
D = 0.11397 m = 113.97 mm
L = 1.25 D = 142.46 mm
D = 113.97 mm, L = 142.46 mm .......ANS
Introductioon To I.C. Engine / 77
Q. 16: An engine of 250 mm bore and 375 mm stroke works on otto cycle. The clearance volume is
0.00263 m
3
. The initial pressure and temperature are 1 bar and 50ºC. If the maximum pressure
is limited to 25 bar. Find
(1) The air standard efficiency of the cycle.
(2) The mean effective pressure for the cycle. (Dec–00)
Sol.: Given that:
Bore diameter d = 250mm
Stroke length L = 375mm
Clearance volume V
C
= 0.00263m
3
Initial pressure P
1
= 1bar
Initial temperature P
3
= 25 bar
We know that, swept volume
V
s
=
2
4 4
d L
π π
⋅ ·
× (0.25)
2
× 0.375 = 0.0184077 m
3
Compression ratio ‘r’ =
0.0184077 0.0263
8
0.00263
c s
c
V V
V
+
+
· ·
∴ The air standard efficiency for Otto cycle is given by
η
otto
= 1–
1 1 4 1
1 1
1
( ) (8) r
γ− ⋅ −
· − · 0.5647 or 56.57%
2
1
T
T
= (r)
γ – 1
= (8)
1⋅4 – 1
= 2.297; T
2
= (50 + 273) × 2.297 = 742.06 K
2 1
1 2
P V
P V
γ
¸ _
·
¸ ,
= (8)
1⋅4
= 18.38; P
2
= 1 × 18.38 = 18.38 bar
Process (2 – 3)
V
2
= V
3
;
3 2
2 3
P P
T T
·
T
3
=
25
18.38
× 742.06 = 1009.38
q
s
= C
p
(T
3
– T
2
) = 1.005 (1009.38 – 742.06) = 268.65 kJ/kg
η
otto
=
s
w
q
; w = q
s
× η
otto
= 268.65 × 0.5647 = 151.70 kJ/kg
Mean effective pressure
2 2
151.70
;
0.021 0.00263
m m
W m
P P
V V
×
· ·
− −
m=
5
1 1
3
1
1 10 0.021
0.287 10 (50 273)
P V
RT
× ×
·
× × +
= 0.02265
P
m
=
151.70 0.02265
0.021 0.00263
×
−
= 187 kPa = 1.87 bar
Fig. 4.9
P
3
2
V
c
PV C =
V
s
V
1
4
78 / Problems and Solutions in Mechanical Engineering with Concept
Q. 17: An Air standard otto cycle has a compression ratio of 8. At the start of compression process
the temperature is 26ºC and the pressure is 1 bar. If the maximum temperature of the cycle
is 1080K. Calculate
(1) Net out put
(2) Thermal efficiency. Take C
V
= 0.718 (Dec–04)
Sol.: Compression Ratio (R
c
) = 8
T
l
=26°C = 26 + 273 = 299K = 1 bar
T
3
=1080 k
(i) Net output = work done per kg of air =
w q δ · δ
∫ ∫
Process (1 – 2) Isentropic compression process
1
2 1
1 2
T V
T V
γ−
¸ _
·
¸ ,
T
2
= T
1
1
2
1
P
P
γ−
¸ _
¸ ,
T
2
= T
1
(R
c
)
γ – 1
2
1
P
Rc
P
¸ _
·
¸ ,
3
T
2
= 299 (8)
1.4 – 1
= 299 (8)
0.4
= 299 × 2.29 = 686.29 K
1
3 4
4 3
;
T V
T V
γ−
¸ _
·
¸ ,
3
4 4 1 1.4 1 0.4
10.30 1080 1080
;
8 (8) 2.29
C
T
T T
R
γ− −
· · · ·
=
471.62 k
Net output = work done per kg of air =
w δ
∫
w δ
∫
= C
v
(T
3
– T
2
) – Cv (T
4
– T
1
)
= 0.718 (1080 – 686.92) – 0.718 (471.62 – 299)
= 0.718 × 393.08 – 0.718 × 172.62 = 282.23 – 123.94
Net Output = 158.28 KJ/Kg .......ANS
(ii) η
thermal
=
work done per kg of air
100
heat suplied per kg of air
w
qs
δ
× ·
∫
q
s
= Cv (T
3
– T
2
) = 0.718 (1080 – 686.29) = 282.23 KJ/kg
η thermal =
158.28
100 100
282.23
s
w
q
δ
× · ×
∫
η η η η η thermal =56.08% .......Ans
Q. 18: A diesel engine operating on Air Standard Diesel cycle operates on 1 kg of air with an initial
pressure of 98kPa and a temperature of 36°C. The pressure at the end of compression is 35
bar and cut off is 6% of the stroke. Determine (i) Thermal efficiency (ii) Mean effective
pressure. (May–05)
P
3
2
1
4
V
PVY C =
Isentropic
Processes
Fig. 4.10
Introductioon To I.C. Engine / 79
Sol.: Given that :
m= 1 kg,
P
1
= 98 kPa = 98 × 10
3
Pa;
T
1
=36°C = 36 + 273 = 309 K,
P
2
= 35 bar = 35 × 10
5
Pa
V
3
– V
2
= 0.06V
S
For air standard cycle P
1
V
1
= mRT
1
98 × 103 × V
1
= 1 × 287 × 309
V
1
= 1.10 m3; V
1
= V
2
+ V
3
= 1.10
As process 12 is adiabtic compression process,
1
2 2
1 1
T P
T P
γ−
γ
¸ _
·
¸ ,
1.4 1
5
1.4
2
3
35 10
309 98 10
T
−
¸ _ ×
·
×
¸ ,
⇒ T
2
= 858.28 K
P
2
V
2
= mRT
2
35 × 10
5
× V
2
= 1 × 287 × 858.28;
V
C
= V
2
= 0.07m
3
V
s
= V
1
= 1.10m
3
However, V
3
– V
2
= 0.06 V
s
V
3
– 0.07 = 0.06 × 1.10; V
3
= 0.136m
3
Compression ratio R
c
=
1
2
1.10
0.07
V
V
·
= 15.71
ρ =
1
2
0.136
0.07
V
V
·
= 1.94
γ
thermal
= 1–
1
1 ( 1)
( ) ( 1)
c
R
γ
γ−
1 ρ −
1
γ ρ −
¸ ]
= 1–
1.4
1.4 1 0.4
1 (1.94) 1 1 253 1
1
(15.71) 1.4(1.94 1) (15.71) 1.4 0.94
−
1 − − 1
· −
1
1
− ×
¸ ]
¸ ]
= 1–
1 153 1
1
3.01 1.32 3.01
1
· −
1
¸ ]
(1.16) = 1 – 0.39 = 0.61
P
mef
is given by =
1
( ) ( 1) ( 1)
( 1) ( 1)
c
c
c
R
P R
R
γ γ
1 γ ρ− − ρ −
⋅
1
− γ −
1
¸ ]
= 98 × 10
3
× 15.71
1.4 1 1.4
1.4(15.71 (1.94 1) (1.94 1)
(15.71 1) (1.4 1)
−
1 − − −
1
− −
¸ ]
= 98 × 103 × 15.71
0.4 1.4
1.4(15.71) (0.94) (1.94 1)
14.71 0.4
1 − −
1
×
¸ ]
P
V
1
4
3 2
Fig. 4.11
80 / Problems and Solutions in Mechanical Engineering with Concept
= 1539580
1.32 3.01 (253 1)
5.88
× − − 1
1
¸ ]
=
3.97 1.53 2.44
1539580
5.88 5.88
− 1 1
·
1 1
¸ ] ¸ ]
= 1539580 × 0.415 Pa = 6389257 Pa = 6389.3 KPa .......ANS
Q. 19: Air enters at 1bar and 230ºC in an engine running on diesel cycle whose compression ratio
is 18. Maximum temperature of cycle is limited to 1500ºC. Compute
(1) Cut off ratio
(2) Heat supplied per kg of air
(3) Cycle efficiency. (Dec–05)
Sol.: Given that:
P
1
= 1bar
T
1
= 230 + 273 = 503K
T
3
= 1500 + 273 = 1773K
Compression ratio r = 18
Since T
2
/T
1
= (r)
γ–1
T
2
= T
1
× (r)
γ–1
= 503(18)
1.41
= 1598.37K
(1) Cut off ratio (ρ) = V
3
/V
2
= T
3
/T
2
T
3
/T
2
= ρ
ρ = 1773/1598.37
ρ ρρ ρρ = 1.109 .......ANS
(2) Heat supplied per kg of air
Q = C
P
(T
3
– T
2
) = 1.005 (1773 – 1598.37)
Q = 175.50 KJ/kg .......ANS
(3) Cycle efficiency
η
diesel
= {1 – 1/[γ(r)
γ–1
]}{ (ρ
γ
– 1)/(ρ – 1)}
η
diesel
= {1 – 1/[1.4(18)
1.4–1
]}{ (1.109
1.4
– 1)/( 1.109 – 1)}
η
diesel
= {1 – 0.225}{ (0.156)/(0.109)}
η
diesel
= 0.678
or η ηη ηη
diesel
= 67.8% .......ANS
P
V
3
2
1
4
Fig. 4.12
Properties of Steam and Thermodynamics Cycle / 81
+0)264
5
PRCPERTlES CF STE/M /ND
THERMCD¥N/MlCS C¥CLE
Q. 1: Discuss the generation of steam at constant pressure. Show various process on temperature
volume diagram. (Dec–04)
Sol.: Steam is a pure substance. Like any other pure substance it can be converted into any of the three
states, i.e., solid, liquid and gas. A system composed of liquid and vapour phases of water is also a pure
substance. Even if some liquid is vaporised or some vapour get condensed during a process, the system will
be chemically homogeneous and unchanged in chemical composition.
Assume that a unit mass of steam is generated starting from solid ice at 10
°
C and 1atm pressure in
a cylinder and piston machine. The distinct regimes of heating are as follows :
Regime (AB) : The heat given to ice increases its temperature from 10°C to 0°C. The volume of ice
also increases with the increase in temperature. Point B shows the saturated solid condition. At B the ice
starts to melt (Fig. 5.1, Fig. 5.3).
Regime (BC): The ice melts into water at constant pressure and temperature. At C the melting 
process ends. There is a sudden decrease in volume at 0
°
C as the ice starts to melt. It is a peculiar property
of water due to hydrogen bonding (Fig. 5.3).
Regime (CD): The temperature of water increases an heating from 0
°
C to 100°C (Fig. 5.1). The
volume of water first decreases with the increase in temperature, reaches to its minimum at 4°C (Fig. 5.3)
and again starts to increase because of thermal expansion.
100ºC
A
0ºC
A
B S L
G
Sensible
vaperization
(Pressure = 1 atm)
S
u
p
e
r
h
e
a
t
i
n
g
G
a
s
L V +
E D
F
of super
heat
Heat
heat
Heat supplied
L
i
q
u
i
d
(Fusion)
Latent
Heat of
fusion
S
o
l
i
d
10ºC
Fig 5.1 Generation of steam at 1 atm pressure.
82 / Problems and Solutions in Mechanical Engineering with Concept
Point D shows the saturated liquid condition.
Regime (DE): The water starts boiling at D. The liquid starts to get converted into vapour. The
boiling ends at point E. Point E shows the saturated vapour condition at 100
°
C and 1 bar.
Regime (EF): It shows the superheating of steam above saturated steam point. The volume of vapour
increases rapidly and it behaves as perfect gas.The difference between the superheated temperature and the
saturation temperature at a given pressure is called degree of superheat.
100ºC
O
H O
2
P = 1 atm
W
a
t
e
r
4ºC
0ºC
10ºC
S L +
For process
Pressure = 1 atm
ABCDEF
C
D E
F
G
a
s
V
T
Fig 5.2 Fig 5.3
Point B, C, D, E are known as saturation states. State B : Saturated solid state.
State C & D : Both saturated liquid states.
State C is for hoar frost and state D is for vaporization. State E : Saturated vapour state.
At saturated state the phase may get changed without change in pressure or temperature.
Q. 2: Write some important term in connection with properties of steam.
Or
Short notes on Dryness fraction measurement. (May03)
sensible Heat of Water or Heat of the Liquid or Enthalpy of Liquid (h
f
)
Sol.: It is the quantity of heat required to raise unit mass of water from 0°C to the saturation temperature
(or boiling point temperature) corresponding to the given pressure of steam generation. In Fig 5.5, ‘h
f
’
indicates enthalpy of liquid in kJ/kg. It is different at different surrounding pressures.
Laten Heat of Vapourisation of Steam (h
fg
) : Or, Latent Heat of Evaporation
It is the quantity of heat required to transform unit mass of water at saturation temperature to unit mass of
steam (dry saturated steam) at the same temperature. It is different at different surrounding pressures.
Saturated Steam
It is that steam which cannot be compressed at constant temperature without partially condensing it. In
Fig. 5.5, condition of steam in the line AB is saturated excepting the point A which indicates water at boiling
point temperature. This water is called saturated water or saturated liquid.
The steam as it is being generated from water can exist in any of the three different states given below.
(1) Wet steam
(2) Dry (or dry saturated) steam
(3) Superheated steam.
Properties of Steam and Thermodynamics Cycle / 83
Amongst these, the superheated state of steam is most useful as it contains maximum enthalpy (heat)
for doing useful work. Dry steam is also widely utilized, but the wet steam is of least utility. Different states
of steam and sequential stages of their evolution are shown in Fig. 5.4 ae. Their corresponding volumes
are also shown therein.
WET SATURATED STEAM Wet steam is a twophase mixture comprising of boiling water particles
and dry steam in equilibrium state. Its formation starts when water is heated beyond its boiling point,
thereby causing start of evaporation.A wet steam may exist in different proportions of water particles and
dry steam. Accordingly, its qualities are also different. Quality of wet steam is expressed in terms of dryness
fraction which is explained below.
V
M
Water at 0ºC
( = 0)
( )
x
a
Evaporation
of water ( > 0)
( )
x
b
Wet steam
( = 0.9 say)
( )
x
c
Dry steam
( = 1)
( )
x
d
Superheated
steam ( = 1)
( )
x
e
V
N
V
N
Fig 5.4: Different states of steam and the stages of their evolution.
Dryness Fraction of Steam
Dryness fraction of steam is a factor used to specify the quality of steam. It is defined as the ratio of weight
of dry steam W
ds
present in a known quantity of wet steam to the total weight of Wet steam W
ws
. It is a unit
less quantity and is generally denoted by x. Thus
·
+
ds
ds ws
W
x
W W
it is evident from the above equation that x = 0 in pure water state because W
dS
= 0. It can also be seen,
in Fig. 5.4a that W
ds
= 0 in water state. But for presence of even a very small amount of dry steam i.e.
W
ds
= 0, x will be greater than zero as shown in Fig. 5.4b. On the other hand for no water particles at all
in a sample of steam, W
ws
= 0. Therefore x can acquire a maximum value of 1. It cannot be more than 1.
The values of dryness fraction for different states of steam are shown in Fig. 5.4, and are as follows.
(i) Wet steam 1 > x > 0
(ii) Dry saturated steam x = 1
(iii) Superheated steam x =1
The dryness fraction of a sample of steam can be found experimentally by means of calorimeters.
Dry (Or Dry Saturated) Steam
A dry saturated steam is a singlephase medium. It does not contain any water particle. It is obtained on
complete evaporation of water at a certain saturation temperature. The saturation temperature differs for
different pressures. It means that if water to be evaporated is at higher pressure, it will evaporate at higher
temperature. As an illustration, the saturation temperatures at different pressures are given below for a
ready reference.
84 / Problems and Solutions in Mechanical Engineering with Concept
p (bar) 0.025 0.30 2.0 9.0 25.0 80.0 150.0 200.0
t
sat
(ºC) 21.094 69.12 120.23 175.35 223.93 294.98 342.11 365.71
T
e
m
p
e
r
a
t
u
r
e
i
n
º
C
A
tsºC
O
tsºC tsºC tsupºC
Enthalpy (H)
KJ/kg
C
B
Fig 5.5
Superheated Steam
When the dry saturated steam is heated further at constant pressure, its temperature risesup above the
saturation temperature. This rise in temperature depends upon the quantity of heat supplied to the dry
steam. The steam so formed is called superheated steam and its temperature is known as superheated
temperature t
sup
°C or T
sup
K. A superheated steam behaves more and more like a perfect gas as its temperature
is raised. Its use has several advantages. These are
(i) It can be expanded considerably (to obtain work) before getting cooled to a lower temperature.
(ii) It offers a higher thermal efficiency for prime movers since its initial temperature is higher.
(iii) Due to high heat content, it has an increased capacity to do work. Therefore, it results in economy
of steam consumption.
In actual practice, the process of superheating is accomplished in a super heater, which is installed near
boiler in a steam (thermal) power plant.
Degree of Super Heat
It is the difference between the temperature of superheated steam and saturation temperature corresponding
to the given pressure.
So, degree of superheated = t
sup
– t
s
Where;
t
sup
= Temperature of superheated steam
t
s
= Saturation temperature corresponding to the given pressure of steam generation.
Super Heat
It is the quantity of heat required to transform unit mass of dry saturated steam to unit mass of superheated
steam at constant pressure so,
Super heat = 1 × C
p
× (t
sup
– t
s
) KJ/Kg
Saturated Water
It is that water whose temperature is equal to the saturation temperature corresponding to the given pressure.
Properties of Steam and Thermodynamics Cycle / 85
Q. 3: How you evaluate the enthalpy of steam, Heat required, specific volume of steam, Internal
energy of steam?
(1) Evaluation the Enthalpy of Steam
Let
h
f
= Heat of the liquid or sensible heat of water in KJ/kg
h
fg
= Latent heat of vapourisation of steam in KJ/kg
t
s
= Saturation temperature in 0ºC corresponding to the given pressure.
t
sup
= Temperature of superheated steam in ºC
x = dryness fraction of wet saturated steam
C
p
= Sp. Heat of superheated steam at constant pressure in KJ/kg.k.
T
e
m
p
.
I
n
C
hr
tsºC
O
xhtg
htg
H
sup
H
dry
H
wes
tsºC tsºC
Fig 5.6
(a) Enthalpy of dry saturated steam
1 kg of water will be first raised to saturation temperature (t
s
) for which h
f
(sensible heat of water) quantity of
heat will be required. Then 1 kg of water at saturation temperature will be transformed into 1 kg of dry saturated
steam for which h
fg
(latent heat of steam) will be required. Hence enthalpy of dry saturated steam is given by
H
dry
(or h
g
) = h
f
+ h
fg
kJ/kg
(b) Enthalpy of wet saturated steam
1 kg of water will be first raised to saturation temperature (t
s
) for which h
f
(sensible heat of water) will be
required. Then ‘x’ kg of water at saturation temperature will be transformed into ‘x’ kg of dry saturated
steam at the same temperature for which x.h
fg
amount of heat will be required. Hence enthalpy of wet
saturated steam is given by
H
wet
= h
f
+ x.h
f
g
kJ/kg
(c) Enthalpy of superheated steam
1 kg of water will be first raised to saturation temperature (t
s
) for which h
f
(sensible heat of water) will be
required. Then, 1kg of water at saturation temperature will be transformed into 1kg dry saturated steam at
the same temperature for which h
fg
(latent heat of steam) will be required. Finally, 1kg dry saturated steam
will be transformed into 1kg superheated steam at the same pressure for which heat required is
1 × C
p
(t
sup
–
t
s
) = Cp (t
sup
– t
s
) kJ
Hence enthalpy of superheated steam is given by
H
sup
= h
f
+ h
fg
+ C
p
(t
sup
– t
s
) kJ/kg
86 / Problems and Solutions in Mechanical Engineering with Concept
(2) Evaluation of heat Required
Heat required to generate steam is different from ‘total heat’ or enthalpy of steam. Heat required to generate
steam means heat required to produce steam from water whose initial temperature is tºC (say) which is
always greater than 0ºC. Total heat or enthalpy of steam means heat required to generate steam from water
whose initial temperature is 0°C. If, however, initial temperature of water is actually 0°C, then of course
heat required to generate steam becomes equal to total heat or enthalpy of steam.
hr
tsºC
Heat in
KJ/kg
tsºC
O
htg
Q
sup
Q
dry
Q
wet
h¢
tsºC
xhtg
Fig 5.7
(a) When steam is dry saturated
heat required to generate steam is given by Q
Dry
=h
f
+ h
fg
 h’ kJ/kg,
where h’ = heat required to raise 1 kg water from 0°C to the given initial temperature (say t°C)
of water
= mst =1 × 4.2 × (t – 0) = 4.2t kJ
[sp. heat of water = 4.2 kJ/kg K].
(b) When steam is wet saturated, heat required to generate steam is given by
Q
wet
= h
f
+ x.h
fg

h’ kJ/kg.
(c) When steam is superheated, heat required is given by
Q
sup
=h
f
+ h
fg
+ C
P
(t
sup
 t
s
)  h’ kJ/kg.
(3) Evaluation of Specific Volume of Steam
Specific volume of steam means volume occupied by unit mass of steam. It is expressed in m
3
/kg. Specific
volume of steam is different at different pressure. Again, corresponding to a given pressure specific volume
of dry saturated steam, wet saturated steam and superheated steam will be different from one another.
(a) Sp. volume of dry saturated steam (V
g
or V
Dry
)
It is the volume occupied by unit mass of dry saturated steam corresponding to the given pressure of steam
generation.
Sp. volume of dry saturated steam corresponding to a given pressure can be found out by experiment.
However, sp. volume of dry saturated steam corresponding to any pressure of steam generation can be
found out directly from steam table. In the steam table, ‘v
g
’ denotes the sp. volume of dry saturated steam
in “m
3
/kg”,
(b) Specific volume of wet saturated steam (V
wet
)
It is the volume occupied by unit mass of wet saturated steam corresponding to the given pressure of steam
generation.
Properties of Steam and Thermodynamics Cycle / 87
Sp. volume of wet saturated steam is given by
V
wet
= volume occupied by ‘x’ kg dry saturated steam + volume occupied by (1 – x) kg. water,
where,
x = dryness fraction of wet saturated steam.
Let v
g
= sp. volume of dry saturated steam in m
3
/kg corresponding to given pressure of wet saturated
steam.
v
f
= sp. volume of water in m
3
/kg corresponding to the given pressure of wet saturated steam.
Then,
V
wet
= x.vg + (1 – x) v
f
m
3
/kg.
Since (1 – x) v
f
is very small compared to x.v
g
, it is neglected.
[Average value of v
f
= 0.001 m
3
/kg upto atmospheric pressure].
So,
V
wet
= xv
g
m
3
/kg.
(c) Specific volume of superheated steam
It is the volume occupied by unit mass of superheated steam corresponding to the given pressure of
superheated steam generation. Superheated steam behaves like a perfect gas. Hence the law
1 1 2 2
1 2
·
PV PV
T T
is applicable to superheated steam. Let
v
g
= sp. volume of dry saturated steam corresponding to given pressure of steam generation is m
3
/kg.
T
S
= absolute saturation temperature corresponding to the given pressure of steam generation.
T
sup
= absolute temperature of superheated steam
P = pressure of steam generation
V
sup
= required specific volume of superheated steam.
Then, in the above formula,
P
1
= P
2
V
1
= v
g,
V
2
= V
sup
T
1
= T
s
T
2
= T
sup
sup
sup
·
g
s
v V
T V
V
sup
= v
g
x T
sup
/T
s
m
3
/kg
(4) Evaluation of Internal Energy of Steam
It is the actual heat energy stored in steam above the freezing point of water.
We know that enthalpy = internal energy + pressure energy
= U + PV,
where
U = internal energy of the fluid
PV = pressure energy of the fluid
P = pressure of the fluid
V = volume of the fluid.
If ‘U’ is in kJ/kg, ‘P’ is in kN/m
2
and ‘V’ is in m
3
/kg,
88 / Problems and Solutions in Mechanical Engineering with Concept
then U + PV denotes specific enthalpy is kJ/kg. [sp. enthalpy means enthalpy per unit mass]
From the above equation, we get
u = enthalpy – PV = H – PV kJ/kg,
where H = enthalpy per unit steam in kJ/kg.
(a) Internal energy of dry saturated steam is given by
U
Dry
= H
Dry
– P.v
g
kJ / kg,
where H
Dry
(or h
g
) = enthalpy of dry saturated steam in kJ/kg.
v
g
= sp. volume of dry saturated steam in m
3
/kg, and
P = pressure of steam generation in kN/m
2
(b) Internal energy of wet saturated steam is given by
u
wet
= H
wet
 P.V
wet
kJ/kg,
where
H
wet
= enthalpy of wet saturated steam in kJ/kg
V
wet
= sp. volume of wet saturated steam in m
3
/kg
(c) Internal energy of superheated steam is given by
u
sup
= H
sup
 P.v
sup
kJ
/
kg,
where
H
sup
= enthalpy of superheated steam in kJ/kg.
v
sup
= sp. volume of superheated steam in m
3
/kg.
Q. 4: Write short notes on Steam table.
Sol.: Steam table provides various physical data regarding properties of saturated water and steam. This
table is very much helpful in solving the problem on properties of steam. It should be noted that the
pressure in this table is absolute pressure.
In this table, various symbols used to indicate various data are as stated below:
(1) ‘P’ indicates absolute pressure in bar
(2) ‘t’ indicates saturation temperature corresponding to any given pressure. This has been often
denoted by ‘t
s
’.
(3) ‘v
f
’ indicates specific volume of water in m
3
/kg corresponding to any given pressure.
(4) ‘v
g
’ indicates specific volume of dry saturated steam corresponding to any given pressure.
(5) ‘h
f
’ indicates heat of the liquid in kJ/kg corresponding to any given pressure.
(6) ‘h
fg
’ indicates latent heat of evaporation in kJ/kg corresponding to any given pressure.
(7) ‘h
g
’ indicates enthalpy of dry saturated steam in kJ/kg corresponding to any given pressure.
(8) ‘S
f
’ indicates entropy of water in kJ/kg.K corresponding to any given pressure.
(9) ‘S
g’
indicates entropy of dry saturated steam in kJ/kg.K corresponding to any given pressure.
(10) “S
fg
” indicates entropy of evaporation corresponding to any pressure. There are two types of steam
tables :
One steam table is on the basis of absolute pressure of steam and another steam table is on the basis
of saturation temperature. Extracts of two types of steam tables are given below.
Properties of Steam and Thermodynamics Cycle / 89
Table 5.1. On the Basis of Pressure
Absolute Saturation Sp. volume in Specific enthalpy in kJ/kg Specific
pressure temperature m3/kg entropy in
(P) bar (t) ºC kJ/kg K
Water Steam Water Latent Steam Water Steam
(vf) (vg) (hf) heat (hfg) (hg) (Sf) (Sg)
1.00 99.63 0.001 1.69 417.5 2258 2675.5 1.303 6.056
1.10 102.3 0.00104 1.59 428.8 2251 2679.8 1.333 5.994
1.20 104.8 0.00104 1.428 439.4 2244 2683.4 1.361 5.937
1.50 111.4 0.00105 1.159 467.1 2226 2693.1 1.434 5.790
Table 5.2. On the Basis of Saturation Temperature
Saturation Absolute Sp. volume in Specific enthalpy in kJ/kg Specific
temperature pressure m3/kg entropy in
(t) in ºC (P) in bar kJ/kg K
Water Steam Water Latent Steam Water Steam
(vf) (vg) (hf) heat (hfg) (hg) (Sf) (Sg)
10 0.0123 0.001 106.4 42.0 2477 2519 0.151 8.749
20 0.0234 0.001 57.8 83.9 2454 2537.9 0.296 8.370
40 0.0738 0.001 19.6 167.5 2407 2574.5 0.572 7.684
Ques No5: Explain Mollier diagram and Show different processes on mollier diagram.?
Sol.: A Mollier diagram is a chart drawn between enthalpy H (on ordinate) and entropy Φ or S (on
abscissa). it is also called HΦ diagram. It depicts properties of water and steam for pressures up to 1000
bar and temperatures up to 800°C. In it the specific volume, specific enthalpy, specific entropy, and dryness
fraction are given in incremental steps for different pressures and temperatures. A Mollier diagram is very
convenient in predicting the states of steam during compression and expansion, during heating and cooling,
and during throttling and isentropic processes directly. It does not involve any detailed calculations as is
required while using the steam tables. Sample of a Mollier chart is shown in Fig. 5.8 for a better understanding.
There is a thick saturation line that indicates ‘dry and saturated state’ of steam. The region below the
saturation line represents steam ‘in wet conditions’ and above the saturation line, the steam is in ‘superheated
state’. The lines of constant dryness fraction and of constant temperature are drawn in wet and superheated
regions respectively. It should be noted that the lines of constant pressure are straight in wet region but
curved in superheated region.
Temperature
Saturation line
x
f
f
H
1
H
2
H
E
n
t
h
a
p
l
y
Wet steam ( < 1) x
Superheated steam ( = 1) x
Dry saturated
steam ( =1 ) x
line
Pressure line
Specific
volume line
Dryness
fraction line
Fig 5.8: A sample Mollier diagram (HΦ chart) showing its details.
90 / Problems and Solutions in Mechanical Engineering with Concept
Q. 6: 10 kg of wet saturated steam at 15 bar pressure is superheated to the temperature of 290°C
at constant pressure. Find the heat required and the total heat of steam. Dryness fraction of
steam is 0.85.
Sol.: From steam table, we obtain the following data:
Absolute pressure (P) Saturation Specific enthalpy kJ/kg
bar temperature (t) ºC Water (hf) Latent heat (hfg)
15 198.3 844.6 1947
Total heat of 10 kg wet saturated steam
=10 × 2499.55 = 24995.5 kJ .......ANS
Total heat of 1 kg superheated steam is given by H
sup
= h
f
+
h
fg
+ Cp(t
sup
– t
s
) kJ
= 844.6 + 1947 + 2.1 × (290 – 198.3) kJ
= 2984.17 kJ .......ANS
Total heat of 10 kg superheated steam =10 × 2984.17 = 29841.7 kJ .......ANS
= h
f
+ h
f
g + C
p
(t
sup
–
t
s
)
–
(h
f
+ xh
fg)
= h
fg
+ C
p
(t
sup
– t
s
) – xh
fg
=1947 + 2.1 x (290 – 198.3) – 0.85 x 1947 kJ = 484.62 kJ
Heat required to convert 10 kg wet saturated steam into 10 kg superheated steam
=10 × 484.62 = 4846.2 kJ .......ANS
Total heat of 1 kg wet saturated steam is given by H
wet
= h
f
+ xh
fg
kJ
= 844.6 + 0.85 × 1947 kJ = 2499.55 kJ
Heat required to convert 1 kg wet saturated steam into 1 kg superheated steam
= H
sup
 H
wet
,
where H
sup
= enthalpy of 1 kg superheated steam = h
f
+ h
fg
+ C
p
(t
sup
– t
s
) kJ
H
wet
= enthalpy of 1 kg wet saturated steam = h
f
+ xh
fg
kJ
Heat required to convert 1 kg wet saturated steam into 1 kg superheated steam
Q. 7: Steam is being generated in a boiler at a pressure of 15.25 bar. Determine the specific enthalpy
when
(i) Steam is dry saturated
(ii) Steam is wet saturated having 0.92 as dryness fraction, and
(iii) Steam is superheated, the temperature of steam being 270°C.
Sol.: Note. Sp. enthalpy means enthalpy per unit mass. From steam table, we get the following data:
Absolute pressure (P) Saturation Specific enthalpy kJ/kg
bar temperature (t) ºC Water (hf) Latent heat (hfg)
15 198.3 844.6 1947
15.55 200.0 852.4 1941
Now 15.55 – 15.25 = 0.30 bar
15.5515 = 0.55bar
For a difference of pressure of 0.55 bar, difference of t (or t
s
) = 200 – 198.0 = 2.0°C
For a difference of pressure of 0.30 bar, difference of
t = (2/0.55) × 0.30 =1.091°C.
Corresponding to 15.25 bar, exact value of
Properties of Steam and Thermodynamics Cycle / 91
t = 200 – 1.091 = 198.909°C
For a difference of pressure of 0.55 bar, difference of h
f
(heat of the liquid) = 852.4 – 844.6 = 7.8 kJ/kg
For a difference of pressure of 0.30 bar, difference of h
f
= (7.8/0.55) × 0.30 = 4.255 kJ/kg.
Corresponding to 15.25 bar, exact value of
h
f
= 852.4 – 4.255 = 848.145 kJ/kg.
Again, for a difference of pressure of 0.55 bar, difference of h
f
g (latent heat of evaporation)
= 1947 – 1941 = 6 kJ/kg.
For a difference of 0.30 bar, difference of
h
fg
= (6/0.55) × 0.30 = 3.273 kJ/kg.
Corresponding to 15.25 bar, exact value of
h
f
g = 1941 + 3.273 = 1944.273 kJ/kg
[Greater the pressure of steam generation, less is the latent heat of evaporation.]
The data calculated above are written in a tabular form as below :
Absolute pressure (P) Saturation Specific enthalpy kJ/kg
bar temperature (t) ºC Water (hf) Latent heat (hfg)
15.25 198.909 848.145 1944.273
(i) When steam is dry saturated, its enthalpy is given by
H
Dry
= h
f
+
h
fg
kJ/kg
= 848.145 + 1944.273 = 2792.418 kJ/kg .......ANS
(ii) When steam is wet saturated, its enthalpy is given by
H
wet
= h
f
+ xh
fg
kJ/kg
= 848.145 + 0.92 × 1944.273 = 2636.876 kJ/kg .......ANS
(iii) When steam is superheated, its enthalpy is given by
H
sup
= h
f
+ h
fg
+ C
p
(t
sup
– t
s
) kJ/kg
= 848.145 + 1944.273 + 2.1 x (270 198.909) = 2941.71 kJ/kg .......ANS
Q. 8: 200 litres of water is required to be heated from 30°C to 100°C by dry saturated steam at 10
bar pressure. Find the mass of steam required to be injected into water. Sp. heat of water is
4.2 kj/kg.K.
Sol.: From steam table, we obtain the following data:
Absolute pressure (P) Saturation Specific enthalpy kJ/kg
bar temperature (t) ºC Water (hf) Latent heat (hfg)
10 1799 702.6 2015
Heat lost by 1 kg dry steam = H
Dry
–
h’ kJ,
where
H
Dry
= enthalpy (or total heat) of 1 kg dry saturated steam
h’ = heat required to raise 1 kg water from 0°C to 100°C
(i.e. h’= total heat of 1 kg water at 100°C)
Now,
H
Dry
= h
f
+ h
fg
= 702.6 + 2015 = 2717.6 kJ/kg
h’ =1 × 4.2 × (100 – 0) = 420kJ/kg
92 / Problems and Solutions in Mechanical Engineering with Concept
Heat lost by 1 kg dry saturated steam = 2717.6 – 420 = 2297.6 kJ
Let m = required mass of steam in kg.
Then, heat lost by m kg dry saturated steam = m × 2297.6 kJ
Now, 200 litres of water has a mass of 200 kg.
Heat gained by 200 kg water
= 200 × 4.2 × (100 – 30) kJ = 58800 kJ
Heat lost by m kg steam = heat gained by 200 kg water
m x 2297.6 = 58800
or, m = 25.592 kg .......ANS
Q. 9: One Kg of steam at 1.5MPa and 400ºC in a piston – cylinder device is cooled at constant
pressure. Determine the final temperature and change in volume. If the cooling continues till
the condensation of two – third of the mass. (May – 01)
Sol.: Given that
Mass of steam m = 1kg
Pressure of steam P1 = 1.5MPa = 15bar
Temperature of steam T
1
= 400ºC
From superheated steam table
At P
1
= 15bar, T
1
= 400ºC
Å
1
= 0.1324m
3
/kg
Å
2
= (2 x 0.1324)/3 = 0.0882m
3
/kg
Change in volume “Å = Å
1
 Å
2
= 0.1324 – 0.0882 = 0.0441m
3
/kg
The steam is wet at 15bar, therefore, the temperature will be 198.32ºC.
Q. 10: A closed metallic boiler drum of capacity 0.24m
3
contain steam at a pressure of 11bar and a
temperature of 200ºC. Calculate the quantity of steam in the vessel. At what pressure in the
vessel will the steam be dry and saturated if the vessel is cooled? (May–01)(C.O.)
Sol.: Given that:
Capacity of drum V
1
= 0.24m
3
Pressure of steam P
1
= 11bar
Temperature of steam T
1
= 200ºC
At pressure 11bar from super heated steam table
At 10 bar and T = 200ºC; Å =0.2060m
3
/kg
At 12 bar and T = 200ºC; Å =0.1693 m
3
/kg
Using linear interpolation:
(Å – 0.2060)/(0.1693 – 0.206) = (11 – 10)/(12 – 10)
Å = 0.18765 m
3
/kg
Quantity of steam = V/Å = 0.24/0.18765 = 1.2789 kg
From Saturated steam table
At 11bar; T
sat
= 184.09ºC
2000C > 184.09ºC
i.e. steam is superheated
If the vessel is cooled until the steam becomes dry saturated, its volume will remain the same but its
pressure will change.
From Saturated steam table; corresponding to Å
g
= 0.18765, the pressure is 1122.7KPa .......ANS
Properties of Steam and Thermodynamics Cycle / 93
Q. 11: (a) Steam at 10 bar absolute pressure and 0.95 dry enters a super heater and leaves at
the same pressure at 250°C. Determine the change in entropy per kg of steam. Take
C
ps
= 2.25 kJ/kg K
(b) Find the internal energy of 1 kg of superheated steam at a pressure of 10 bar and 280ºC.
If this steam is expanded to a pressure of 1.6 bar and 0.8 dry, determine the change in internal
energy. Assume specific heat of superheated steam as 2.1 kJ/kgK. (Dec–01)
Sol.: (a) Given that :
P =10bar
x = 0.95
t
sup
= 250°C
From Saturated steam table
t
sat
= 179.9
Now, entropy of steam at the entry of the superheater
s
1
= s
f1
+ x
1
s
fg1
= 2.1386 + 0.95 × 4.4478 = 6.3640 kJ/kg K
entropy of the steam at exit of superheater
s
2
= sgf + C
ps
ln
sup
sat
¸ _
¸ ,
T
T
= 6.5864 + 2.25 ln
250 273
179.9 273
+ ¸ _
+
¸ ,
= 6.9102 kJ/kg K
Change in entropy = s
2
– s
1
= 6.9102 – 6.3640
= 0.5462 kJ/kg K
(b) Given that
State 1 : 10 bar 280°C
State 2 : 1.6 bar, 0.8 dry
Specific heat of superheated steam = 2.1 kJ/kg K
Internal energy at state 1 is :
u
1
=
u
g
+ m.c.(T
1
– T
sat
) = (h
g
 pv
g
) + m.c. (T
1
 T
sat
)
= (2776.2 – 1000 × 0. 19429) + 2.1(280 – 179.88) = 2792.16 kJ/kg
Internal energy at state 2;
u
2
= uf
2
+ xu
f
g
2
= (h
f
– Pv
f
)
2
+ x [h
fg
– P (v
fg
)]
2
= (h
f
– Pv
f
) + x (h
g
– hf) – P (v
g
– v
f
)]
= (h
f
– Pv
f
) + x ((h
g
– Pvg) – (h
f
– P
vf
))
= [475.38 + 160 (0.0010547)] + [(2696.2 – 160 * (1.0911))
– (475.38 – 160 (0.0010547))]
= 475.21 + 0.8 (2521.62 – 475.21)
= 2112.34 kJ/kg
Change in internal energy = 211234  2792.16 =  679.82 kJ/kg .......ANS
ve sign shows the reduction in internal energy.
94 / Problems and Solutions in Mechanical Engineering with Concept
Q. 12: A cylindrical vessel of 5m
3
capacity contains wet steam at 100KPa. The volumes of vapour
and liquid in the vessel are 4.95m
3
and 0.05m
3
respectively. Heat is transferred to the
vessel until the vessel is filled with saturated vapour. Determine the heat transfer during the
process. (Dec–00)
Sol.: Given that:
Volume of vessel V = 5m
3
Pressure of steam P = 100KPa
Volume of vapour Vg = 4.95m
3
Volume of liquid V
f
= 0.05m
3
Since, the vessel is a closed container, so applying first law analysis, we have:
Q
2
–
1
W
2
= U
2
– U
1
1
W
2
= ∫ PdV = 0
1
Q
2
= U
2
– U
1
U
1
=
1 1 1 1
⋅ + ⋅
f f g g
m u m u
m
f
=
1
0.05
47.94
0.001043
· ·
f
f
V
v
(using table B – 2)
m
g
=
1
4.95
1.694
g
g
V
v
·
= 2.922 kg
The final condition of the steam is dry and saturated but its mass remains the same.
The specific volume at the end of heat transfer = v
g2
But v
2
=
5.0
0.0983
(47.94 2.922)
V
m
· ·
+
Now v
2
= v
g2
= 0.0983
The pressure corresponding to v
g
= 0.0983 from saturated steam table is 2030kPa or 2.03 bar.
At 2.03 bar U
2
= u
g2
.
m = (47.94 + 2922) × 2600.5 = 132.26 MJ
1
Q
2
= U
2
– U
1
= 132.26 – 27.33 = 104.93 MJ .......ANS
Q. 13: Water vapour at 90kPa and 150°C enters a subsonic diffuser with a velocity of 150m/s and
leaves the diffuser at 190kPa with a velocity of 55m/s and during the process 1.5 kJ/kg of heat
is lost to surroundings. Determine
(i) The final temperature
(ii) The mass flow rate.
(iii) The exit diameter, assuming the inlet diameter as 10cm and steady flow. (May01)
Sol.: Given that :
Pressure at inlet = 90kPa = P
1
Temperature at inlet = 150
°
C = t
1
Velocity at inlet = 150 m/s = V
1
Pressure at exit = 190kPa = P
2
Velocity at exit = 55 m/s = V
2
Working substancesteam
Type of Process: Flow type
Governing Equation : S.F.E.E.
Fig 5.9
f 10
cm
= 90 kPa
= 150ºC
= 150 m/s
P
T
C
1
1
1
= 190 kPa
= 55 k/s
P
C
2
2
q = 1.5 kJ/kg
1
1
2
2
Properties of Steam and Thermodynamics Cycle / 95
(i) Calculation for Final Temperature
From steady flow energy equation:
Q – W
s
= m
f
[(h
2
– h
1
) + ½(V
2
2
– V
1
2
) + g(z
2
– z
1
)]
W
S
= g(z
2
– z
1
) = 0
(since, there is no shaft and no change in datum level takes place)
Q = m
f
[(h
2
– h
1
)+ ½(V
2
2
– V
1
2
)] ...(i)
since, the working substance is steam the properties of working substance at inlet and exit should be
obtained from steam table.
At stage (1) for P
1
= 90kPa and t
1
= 150°C
t
1
> t
sat
i.e.; superheated vapour
The steam thus behaves as perfect gas.
since y = 1.3 for superheated vapour and R = 8.314/18
kJ/kg K = 0.4619 kJ/kg K
C
p
=
1.3
1 1.3 1
R
¸ _ γ ¸ _
⋅ ·
γ − −
¸ , ¸ ,
× 0.4619 = 2.00 kJ/kg K
From equation (1)
–1.5 = 2 (T
2
– T
1
) +
2 2
(55) (155)
2
¸ _ −
¸ ,
× 10
–3
4.118 = T
2
– T
1
T
2
= 4.118 + 150 = 154.12ºC ......ANS
(ii) Calculation for Mass Flow Rate
Now using ideal gas equation, assuming that superheated vapour behaves as ideal gas:
v
1
=
3
1
3
1
0.4619 (150 273) 10
90 10
RT
P
× + ×
·
×
m
3
/kg
v
1
=2.170 m3/kg
v
2
=
3
2
3
2
0.4619 (154.12 273) 10
90 10
RT
P
× + ×
·
×
m
3
/kg
v
2
= 1.038 m
3
/kg
Mass flow rate can be obtained by using continuity equation
mf ⋅ v = A
1
V
1
= A
2
V
2
m
f
=
2
1 1
1
(0.10) 150
4
2.170
AV
v
π
× ×
· = 0.543 kg/sec .......ANS
(iii) Calculation for Exit Diameter
A
2
=
2
1 1 2
1 2
(0.10) 150 1.038
4
2.170 55
AV v
v V
π
× × ×
× ·
×
A
2
= 0.010246 m
2
d
2
=
2
4 A ×
π
= 0.1142 = 11.42 cm .......ANS
96 / Problems and Solutions in Mechanical Engineering with Concept
Q. 14: A turbine in a steam power plant operating under steady state conditions receives superheated
steam at 3MPa and 350ºC at the rate of 1kg/s and with a velocity of 50m/s at an elevation of
2m above the ground level. The steam leaves the turbine at 10kPa with a quality of 0.95 at
an elevation of 5m above the ground level. The exit velocity of the steam is 120m./s. The
energy losses as heat from the turbine are estimated at 5kJ/s. Estimate the power output of
the turbine. How much error will be introduced, if the kinetic energy and the potential energy
terms are ignored? (Dec–01)
Sol.: Given that; the turbine is running under steady state condition.At
inlet: P
1
= 3MPa; T
1
= 350°C; m
f
= 1 kg/sec; V
1
= 50 m/s; Z
1
=2mAt exit:
P
2
= 10kPa; x = 0.95; Z
2
= 5 m; V
2
= 120 m/s Heat exchanged during
expansion = Q = – 5 kJ/sec From superheat steam table, at 3MPa and
350ºC h
1
= 3115.25 kJ/kgh
2
= h
f2
+ xh
fg2
From saturated steam table; at
10kPa h
2
= 191.81 + 0.95(2392.82) = 2464.99 kJ/kg
From steady flow energy equation:
Q – W
s
= m
f
[(h
2
– h
1
)+ ½(V
2
2
– V
1
2
) + g(z
2
– z
1
)]
– 5 – W
S
= 1x [ (2464.99 – 3115.25) + { (120)
2
– (50)
2
}/2 × 1000
+ 9.8 (5 – 2 )/1000] – W
S
= – 639.28 kJ/sec
W
S
= 639.28 kJ/sec .......ANS
If the changes in potential and kinetic energies are neglected; then
SFEE as; Q – W
s
= m
f
(h
2
– h
1
) – 5 –
1
W
2
= 1 × (2464.99 – 3115.25)
1
W
2
= 645.26 KJ/sec
% Error introduced if the kinetic energy and potential energy terms
are ignored:
% Error = [(W
s
–
1
W
2
)/W
s
] × 100
= [(639.28 – 645.26)/639.28] x 100
= – 0.935%
So Error is = 0.935% .......ANS
Q. 15: 5kg of steam is condensed in a condenser following reversible constant pressure process from
0.75 bar and 150ºC state. At the end of process steam gets completely condensed. Determine
the heat to be removed from steam and change in entropy. Also sketch the process on Ts
diagram and shade the area representing heat removed. (May–02)
Sol.: Given that :
At state 1:P
1
= 0.75 bar and T
1
= 150°C Applying SF’EE
to the control volume Q – W
s
= m
f
[(h
2
– h
1
)+ ½(V
2
2
– V
1
2
)
+ g(z
2
– z
1
)] neglecting the changes in kinetic and potential
energies.i.e.; Ws = ½(V
2
2
– V
1
2
) = g(z
2
– z
1
) = 0 i.e.; Q = m
f
[(h
2
– h
1
)] From super heat steam table at P
1
= 0.75 bar and
T
1
= 150°C We have
(75 – 50)/(100 – 50) = (h
1
– 2780.08)/(2776.38 – 2780.08)
h
1
= 2778.23 KJ/kg
Also, entropy at state (1)
(75 – 50)/(100 – 50) = (s
1
– 7.94)/(7.6133 – 7.94)
s
1
= 7.77665 KJ/kgK
at state (2), the condition is saturated liquid.
Q = 5 kJ/sec
Control volume
Turbine
Steam in
Fig. 5.10
1
2
Steam out
System
boundary
W
O
T
D
R
3 Mpa
10 Mpa
Fig. 5.11
T
s
1
2
Water
in
Water
out
Steam in
Steam out
Control volume
1
2
Fig 5.12
Properties of Steam and Thermodynamics Cycle / 97
0.1 bar
s
Fig. 5.15
1
2
From saturated steam table h
2
= h
f
= 384.36 kJ/kg s
2
= s
f
= 1.2129 kJ/kgK
T
S
Fig. 5.13
2
1
Q = 384.36 – 2778.23 = –2393.87kJ/kgve sign shows that heat is rejected by system Total heat
rejected = 5 × 2393.87 = 11.9693 MJ similarly, total change in entropy
= m(s
2
– s
1
) = 5(l.2129 – 7.7767) = – 32.819 kJ/K .......ANS
Q. 16: In a steam power plant, the steam 0.1 bar and 0.95 dry enters the condenser and leaves as
saturated liquid at 0.1 bar and 45°C. Cooling water enters the condenser in separate steam
at 20°C and leaves at 35
°
C without any loss of its pressure and no phase change. Neglecting
the heat interaction between the condenser and surroundings and changes in kinetic energy
and potential energy, determine the ratio of massflow rate of cooling water to condensing
steam. (Dec–02)
Sol.: Given that:
Inlet condition of steam : Pressure ‘P
1
’ = 0.1 bar dryness fraction x = 0.95 Exit condition of steam
:saturated water at 0.1 bar and 45°C Inlet temperature of cooling water = 20°C Exit temperature of cooling
water = 35°C Applying SFEE to control volume Q – W
s
= m
f
[(h
2
– h
1
)+ ½(V
2
2
– V
1
2
) + g(z
2
– z
1
)]
Water
in
Water
out
Control volume
Steam in
Steam out
1
2
Fig 5.14
neglecting the changes in kinetic and potential energies. And there is no shaft work i.e.; W
s
= 0
Q = m
f
(h
2
– h
1
)
h
1
= h
f1
+ xhf
g1
= 191.81 + 0.95 (2392.82) = 2464.99 kJ/kg
h
2
= h
f2
at 0. 1 bar (since the water is saturated liquid)
h
2
= 191.81 kJ/kg
Q = m
f
(191.81 – 2464.99)
= – 2273.18m
f
kJ/kg ...(i)
ve sign shows the heat rejection.
98 / Problems and Solutions in Mechanical Engineering with Concept
By energy balance;Heat lost by steam = Heat gained by cooling water
Q = m
fw
.C
w
(T
4
– T
3
) = m
fw
x 4.1868(35 – 20) = 62.802m
fw
...(ii)
Equate equation (i) and (ii); We get Q = 2273.18m
f
= 62.802m
fw
m
fw
/m
f
= 36.19 Ratio of mass flow
rate of cooling water to condensing steam = m
fw
/m
f
= 36.19 .......ANS
Q. 17: What do you know about Steam power cycles. And what are the main component of a steam
power plant.?
Sol.: Steam power plant converts heat energy Q from the combustion of a fuel into mechanical work W of
shaft rotation which in turn is used to generate electricity. Such a plant operates on thermodynamic cycle
in a closed loop of processes following one another such that the working fluid of steam and water repeats
cycles continuously. If the first law of thermodynamics is applied to a thermodynamic cycle in which the
working fluid returns to its initial condition, the energy flowing into the fluid during the cycle must be equal
to that flowing out of the cycle.
Q
in
+ W
in
= Q
out
+ W
out
Or, Q
in
– Q
out
= W
out
– W
in
Where;
Q = Rate of Heat transfer
W = Rate of work transfer, i.e. power
Heat and work are mutually convertible. However, although all of a quantity of work energy can be
converted into heat energy (by a friction process), the converse is not true. A quantity of heat cannot all
be converted into work.
Heat flows by virtue of a temperature difference, and which means that in order to flow, two heat
reservoirs must be present; a hot source and a cold sink. During a heat flow from the hot source to the cold
sink, a fraction of the flow may be converted into work energy, and the function of a power plant is to
produce this conversion. However, some heat must flow into the cold sink because of its presence. Thus
the rate of heat transfer Q
out
of the cycle must always be positive and the efficiency of the conversion of
heat energy into work energy can never be 100%. Thermodynamic efficiency for a cycle, nt
h
is a measure
of how Well a cycle converts heat into work.
Engine
Turbine
Boiler
Q
out
Q
out
W
out
Condenser
Heated water
Cold water
Fuei
Air
Pump
Cold sink
Reservoir
Hot Source
Reservoir
Steam
Combustion
products
W
out
Fig 5.16
Properties of Steam and Thermodynamics Cycle / 99
Components of a Steam Power Plant
There are four components of a steam power plant:
1. The boiler: Hotsource reservoir in which combustion gases raise steam.
2. Engine/Turbine: The steam reciprocating engine or turbine to convert a portion of the heat energy
into mechanical work.
3. Condenser: Cold sink into which heat is rejected.
4. Pump: Condensate extraction pump or boiler feed pump to return the condensate back into the
boiler.
Q. 18: Define Carnot vapour Cycle. Draw the carnot vapour cycle on TS diagram and make the
different thermodynamics processes. (Dec–01, Dec–03)
Sol.: It is more convenient to analyze the performance of steam power plants by means of idealized cycles
which are theoretical approximations of the real cycles. The Carnot cycle is an ideal, but nonpractising
cycle giving the maximum possible thermal efficiency for a cycle operating on selected maximum and
minimum temperature ranges.It is made up of four ideal processes: 1  2 : Evaporation of water into
saturated steam within the boiler at the constant maximum cycle temperature T
l
(= T
2
)
2
3
4
1
Turbine
Compressor
Condenser
T T
1 2
=
T T
1 4
=
T K [ ]
4 3
x
3
2
1
0 S S
1 4
= S S
2 3
= s
W
out
W
in
Q
in
Q
out
Fig. 5.17 Carnot Cycle
2  3 : Ideal (i.e., constantentropy) expansion within the steam engine or turbine i.e., S
2
= S
3
.
3  4 : Partial condensation within the condenser at the constant minimum cycle temperature T
3
(= T
4
).
4  1 : Ideal (i.e., constantentropy) compression of very wet steam within the compressor to complete
the cycle, i.e., S
4
= S
1
.
S
2
– S
1
=
in
t
Q
mT
Qin = (m T
1
) (S
2
– S
1
)
Qout = (m T
3
) (S
3
– S
4
)
(S
2
– S
1
) = (S
3
– S
4
)
Qout = (v
1
T
3
) (S
2
– S
1
)
η
s
= 1 –
out 3 2 1 3
in 1 2 1 1
( ) ( )
1 1
( ) ( )
Q mT S S T
Q mT S S T
−
· − · −
−
100 / Problems and Solutions in Mechanical Engineering with Concept
Q. 19: What are the limitations and uses of Carnot vapour cycle.?
Limitations
This equation shows that the wider the temperature range, the more efficient is the cycle.
(a) T
3
: In practice T
3
cannot be reduced below about 300 K (27ºC), corresponding to a condenser pressure
of 0.035 bar. This is due to two tractors:
(i) Condensation of steam requires a bulk supply of cooling water and such a continuous natural
supply below atmospheric temperature of about 15°C is unavailable.
(ii) If condenser is to be of a reasonable size and cost, the temperature difference between the condensing
steam and the cooling water must be at least 10°C.
(b) T
I
: The maximum cycle temperature T
l
is also limited to about 900 K (627°C) by the strength of the
materials available for the highly stressed parts of the plant, such as boiler tubes and turbine blades.
This upper limit is called the metallurgical limit.
(c) Critical Point : In fact the steam Carnot cycle has a maximum cycle temperature of well below this
metallurgical limit owing to the properties of steam; it is limited to the criticalpoint temperature of
374°C (647 K). Hence modern materials cannot be used to their best advantage with this cycle
when steam is the working fluid. Furthermore, because the saturated water and steam curves converge
to the critical point, a plant operating on the carnot cycle with its maximum temperature near the
criticalpoint temperature would have a very large s.s.c., i.e. it would be very large in size and very
expensive.
(d) Compression Process (4  1 : Compressing a very wet steam mixture would require a compressor of
size and cost comparable with the turbine. It Would absorb work comparable with the developed by
the turbine. It would have a short life because of blade erosion and cavitations problem. these reasons
the Carnot cycle is not practical.
Uses of Carnot Cycle
1. It is useful in helping us to appreciate what factors are desirable in the design of a practical cycle;
namely a maximum possible temperature range.
maximum possible heat addition into the cycle at the maximum cycle temperature
a minimum possible work input into the cycle.
2. The Carnot cycle also helps to understand the thermodynamic constraints on the design of cycles. For
example, even if such a plant were practicable and even if the maximum cycle temperature could be
900K the cycle thermal efficiency would be well below 100%. This is called Cartrot lintitation.
η
th
= 1 –
3
1
300
1
900
T
T
· −
= 66.7%
A hypothetical plant operating on such a cycle would have a plant efficiency lower than this owing
to the inefficiencies of the individual plant items.
η
plant
=
η
th
×
η
item 1
×
η
item 2
×
η
item 3
× ...
Q. 20: What is the performance criterion of a steam power plant.?
Sol.: The design of a power plant is determined largely by the consideration of capital cost and
operating cost; the former depends mainly on the plant size and latter is primarily a function of the
overall efficiency of the plant. In general the efficiency can usually be improved, but only by increasing
the capital cost of the plant, hence a suitable compromise must be reached between capital costs and
operating costs.
Properties of Steam and Thermodynamics Cycle / 101
l. Specific steam consumption (S.S.C.). The plant capital cost is mainly dependent upon the size of
the plant components. These sizes will themselves depend on the flow rate of the steam which is passed
through them.
Hence, an indication of the relative capital cost of different steam plant is provided by the mass flow
rate m of the steam required per unit power output, i.e., by the specific steam consumption (s.s.c.) or steam
rate
s.s.c. =
/ 3600 3600
/
m kg s m kg m kg kg
kW kWs kWh kWh W W W W m
· · ·
In M.K.S. system.
1 horsepower hour ≈ 632 k cal
1 kilowatt hour ≈ 860 k cal.
∴ s.s.c. =
632
W
kg/HPhr =
860
W
kg/k Wh
3. Work ratio is defined as the ratio of net plant output to the gross (turbine) output.
Work ratio =
out in 1
out 1
c
W W W W
W W
−
·
Q. 21: Explain Rankine cycle with the help of PV, Ts and Hs diagrams. (May–05)
Or
Write a note on Rankine cycle. (Dec–01, Dec–05)
Sol.: One of the major problems of Carnot cycle is compressing a very wet steam mixture from the
condenser pressure upto the boiler pressure. The problem can be avoided by condensing the steam
completely in the condenser and then compressing the water in a comparatively small feed pump. The
effect of this modification is to make the cycle practical one. Furthermore, far less work is required to
pump a liquid than to compress a vapour and therefore this modification also has the result that the feed
pump’s work is only one or two per cent of the work developed by the turbine. We can therefore neglect
this term in our cycle analysis.
Q
out
Q
in
W
in
W
out
Turbine
Condenser
Boiler
Pump
2
4
3
1
Fig 5.18
The idealized cycle for a simple steam power plant taking into account the above modification is called
the Rankine cycle shown in the figure, Fig. 5.18. It is made up of four practical processes:
102 / Problems and Solutions in Mechanical Engineering with Concept
(a) 1  2 :Heat is added to increase the temperature of the highpressure water up to its saturation value
(process 1 to A). The water is then evaporated at constant temperature and pressure (process A
to 2). Both processes occur within the boiler, but not all of the heat supplied is at the maximum
cycle temperature. Thus, the .mean temperature at which heat is supplied is lower than that in the
equivalent Carnot cycle. Hence, the basic steam cycle thermal efficiency is inherently lower.Applying
the first law of thermodynamics to this process:
(Q – Q ) + (W – W )
in out in out
.
º º
=
( )
fluid final initial
m h h −
0; 0, 0
out in out
Q W W · · ·
∴
2 1
( )
in f
Q m h h · −
(b) 2  3: The high pressure saturated steam is expanded to a low pressure within a reciprocating
engine or a turbine.
If the expansion is ideal (i.e., one of constant entropy), the cycle is called the Rankine cycle.
However, in actual plant friction takes place in the flow of steam through the engine or turbine
which results in the expansion with increasing entropy. Applying first law to this process:
(Q – Q ) + (W – W )
in ou t in out
. . . .
º º º
=
f
m
(h
final
h
mitial
)
in in out
0 0 0. Q W W · · ·
out 3 4
( )
f
Q m h h · −
(c) 3  4:The lowpressure ‘wet steam is completely condensed at constant condenser pressure back
into saturated water. The latent heat of condensation is thereby rejected to the condenser cooling
water which, in turn, rejects this heat to the atmosphere. Applying first law of the thermodynamics,
(Q – Q ) + (W – W )
in ou t in out
. . . .
º º º
=
f
m
(h
final
h
initial
)
(d) 4  1:The low pressure saturated water is pumped back up to the boiler pressure and, in doing so,
it becomes subsaturated. The water then reenters the boiler and begins a new cycle. Applying the
first law:
in out in out final initial
( ) ( ) ( )
f
Q Q W W m h h − + − · −
in out out
0, 0 0 Q Q W · · ·
in 1 4
( ).
f
W m h h · −
However, W
in
can be neglected with reasonable accuracy and we can assume h
1
= h
4
.
The thermal efficiency of the cycle is given by:
2 3
out in out 2 3
th
2 1 2 3 in in
( )
( )
f
f
m h h
W W W h h
m h h h h Q Q
−
− −
η · · · ·
− −
Specific steam consumption is given by:
2 3
3600 kg 3600
. .
kWh /
s s c
h h W m
1
· ·
1
−
¸ ]
kg/kWh
Q. 22: Compare Rankine cycle with Carnot cycle
Sol.: Rankine cycle without superheat : 1 – A – 2 – 3 – 4 – 1.
Rankine cycle with superheat : 1 – A – 2 – 2′ –3′ – 4 – 1.
Carnor cyle without superheat : A – 2 – 3 –4″ – A.
Properties of Steam and Thermodynamics Cycle / 103
Carnot cycle with superheat : A – 2″ – 3′ – 4² – A.
Fig. 5.20
T K [ ]
1
A
4 4² 3 3
X
3
X
3
s
2²
2¢
2
(1) The thermal efficiency of a Rankine cycle is lower than the equivalent Carnot cycle. Temperature
of heat supply to Carnot cycle = T
A
; Mean temperature of heat supply to Rankine
cycle =
1 2 1 2
,
2 2
A
T T T T
T
+ +
>
(2) Carnot cycle needs a compressor to handle wet steam mixture whereas in Rankine cycle, a
small pump is used.
(3) The steam can be easily superheated at constant pressure along 2  2' in a Rankine cycle.
Superheating of steam in a Carnot cycle at constant temperature along A  ?” is accompanied
by a fall of pressure which is difficult to achieve in practice because heat transfer and expansion
process should go side by side. Therfore Rankine cycle is used as ideal cycle for steam power
plants.
104 / Problems and Solutions in Mechanical Engineering with Concept
+0)264
6
FCRCE. CCNCÜRRENT FCRCE S¥STEM
Q. 1: Define Engineering Mechanics
Sol.: Engineering mechanics is that branch of science, which deals the action of the forces on the rigid
bodies. Everywhere we feel the application of Mechanics, such as in railway station, where we seen the
railway bridge, A car moving on the road, or simply we are running on the road. Everywhere we saw the
application of mechanics.
Q. 2: Define matter, particle and body. How does a rigid body differ from an elastic body?
Sol.: Matter is any thing that occupies space, possesses mass offers resistance to any stress, example Iron,
stone, air, Water.
A body of negligible dimension is called a particle. But a particle has mass.
A body consists of a No. of particle, It has definite shape.
A rigid body may be defined as the combination of a large no. of particles, Which occupy fixed
position with respect to another, both before and after applying a load.
Or, A rigid body may be defined as a body, which can retain its shape and size even if subjected to
some external forces. In actual practice, no body is perfectly rigid. But for the shake of simplicity, we take
the bodies as rigid bodies.
An elastic body is that which regain its original shape after removal of the external loads.
The basic difference between a rigid body and an elastic body is that the rigid body don't change its
shape and size before and after application of a force, while an elastic body may change its shape and
size after application of a load, and again regain its shape after removal of the external loads.
Q. 3: Define space, motion.
Sol.: The geometric region occupied by bodies called space.
When a body changes its position with respect to other bodies, then body is called as to be in motion.
Q. 4: Define mass and weight.
Sol.: The properties of matter by which the action of one body can be compared with that of another is
defined as mass.
m = ρ ⋅ ν
Where,
ρ = Density of body and ν = Volume of the body
Weight of a body is the force with which the body is attracted towards the center of the earth.
Q. 5: Define Basic S.I. Units and its derived unit.
Sol.: S.I. stands for “System International Units”. There are three basic quantities in S.I. Systems as concerned
to engineering Mechanics as given below:
Force: Concurrent Force System / 105
Sl.No. Quantity Basic Unit Notation
1 Length Meter m
2 Mass Kilogram kg
3 Time Second s
Meter: It is the distance between two given parallel lines engraved upon the polished surface of a
platinumIridium bar, kept at 00C at the “International Bureau of Weights and Measures” at Serves, near Paris.
Kilogram: It is the mass of a particular cylinder made of Platinum Iridium kept at “International
Bureau of Weights and Measures” at Serves, near Paris.
Second: It is 1/(24 × 60 × 60)th of the mean solar day. A solar day is defined as the time interval
between the instants at which the sun crosses the meridian on two consecutive days.
With the help of these three basic units there are several units are derived as given below.
Sl.No. Derived Unit Notation
1 Area m
2
2 Volume m
3
3 Moment of Inertia m
4
4 Force N
5 Angular Acceleration Rad/sec
2
6 Density kg/m
3
7 Moment of Force N.m
8 Linear moment kg.m/sec
9 Power Watt
10 Pressure/stress Pa(N/m
2
)
11 Mass moment of Inertia kg.m
2
12 Linear Acceleration m/s
2
13 Velocity m/sec
14 Momentum kgm/sec
15 Work Nm or Jule
16 Energy Jule
Q. 6: What do you mean by 1 Newton's? State Newton's law of motion
Sol.: 1Newton: It is magnitude of force, which develops an acceleration of 1 m/s
2
in 1 kg mass of the
body.
The entire subject of rigid body mechanics is based on three fundamental law of motion given by an
American scientist Newton.
Newton’s first law of motion: A particle remains at rest (if originally at rest) or continues to move
in a straight line (If originally in motion) with a constant speed. If the resultant force acting on it is Zero.
Newton's second law of motion: If the resultant force acting on a particle is not zero, then acceleration
of the particle will be proportional to the resultant force and will be in the direction of this force.
F = m.a
Newton's s third law of motion: The force of action and reaction between interacting bodies are
equal in magnitude, opposite in direction and have the same line of action.
106 / Problems and Solutions in Mechanical Engineering with Concept
Q. 7: Differentiate between scalar and Vector quantities. How a vector quantity is represented?
Sol.: A quantity is said to be scalar if it is completely defined by its magnitude alone. Ex: Length, area,
and time. While a quantity is said to be vector if it is completely defined only when its magnitude and
direction are specified. For Ex: Force, velocity, and acceleration.
Vector quantity is represented by its magnitude, direction, point of application. Length of line is its
magnitude, inclination of line is its direction, and in the fig 6.1 point C is called point of application.
H
B
A
Q
T
C
Line of Action
Fig 6.1
Here AC represent the vector acting from A to C
T = Tail of the vector
H = Head of the vector
Q = Direction of the vector
Arrow represents the Sense.
Q. 8: What are the branches of mechanics, differentiate between static's, kinetics and kinematics.
Sol.: Mechanics is mainly divided in to two parts Static's and Dynamics, Dynamics further divided in
kinematics and kinetics
Statics: It deals with the study of behavior of a body at rest under the action of various forces, which
are in equilibrium.
Dynamics: Dynamics is concerned with the study of object in motion
Kinematics: It deals with the motion of the body with out considering the forces acting on it.
Kinetics: It deals with the motion of the body considering the forces acting on it.
Kinematics
Kinetics
Mechanics of Rigid body
Static’s
(Body is at rest)
Dynamics
(Body is in motion)
Fig 6.2
Q. 9: Define force and its type?
Sol.: Sometime we push the wall, then there are no changes in the position of the wall, but no doubt we
apply a force, since the applied force is not sufficient to move the wall, i.e no motion is produced. So this
is clear that a force may not necessarily produce a motion in a body. But it may simply tend to do, So we
can say
The force is the agency, which change or tends to change the state of rest or motion of a body. It is
a vector quantity.
Force: Concurrent Force System / 107
A force is completely defined only when the following four characteristics are specified Magnitude,
Point of application, Line of action and Direction.
OR:
The action of one body on another body is defined as force.
In engineering mechanics, applied forces are broadly divided in to two types. Tensile and compressive
force.
Tensile Forces
A force, which pulls the body, is called as tensile force. Here member AB is a tension member carrying
tensile force P. (see fig 6.3)
Compressive Force
A force, which pushes the body, is called as compressive force. (see fig 6.4)
B
B
P
P
P
P
A
A
Fig 6.3: Tensile Force Fig 6.4: Compressive force
Q. 10: Define line of action of a force?
Sol.: The direction of a force along a straight line through its point of application, in which the force tends
to move a body to which it is applied. This line is called the line of action of the force.
Q. 11: How do you classify the force system?
Sol.: Single force is of two types i.e.; Tensile and compressive. Generally in a body several forces are
acting. When several forces of different magnitude and direction act upon a rigid body, then they are form
a System of Forces, These are
Coplanar
Collinear
Concurrent
Concurrent
Parallel
Parallel
NonCoplanar
Force System
Nonconcurrent
Nonparallel
Nonconcurrent
Nonparallel
Fig 6.5
Coplanar Force System: The forces, whose lines of action lie on the same plane, are known as
coplanar forces.
NonCoplanar Force System: The forces, whose lines of action not lie on the same plane, are known
as noncoplanar force system.
Concurrent Forces: All such forces, which act at one point, are known as concurrent forces.
CoplanarConcurrent System: All such forces whose line of action lies in one plane and they meet
at one point are known as coplanarconcurrent force system.
CoplanarParallel Force System: If lines of action of all the forces are parallel to each other and they
lie in the same plane then the system is called as coplanarparallel forces system.
108 / Problems and Solutions in Mechanical Engineering with Concept
CoplanarCollinear Force System: All such forces whose line of action lies in one plane also lie
along a single line then it is called as coplanarcollinear force system.
Nonconcurrent Coplanar Forces System: All such forces whose line of action lies in one plane but
they do not meet at one point, are known as nonconcurrent coplanar force system.
CONCURRENT FORCE SYSTEM
Q. 12: State and explain the principle of transmissibility of forces?
(Dec00, May01, May(B.P.)01,Dec03)
Sol.: It state that if a force acting at a point on a rigid body, it may be considered to act at any other point
on its line of action, provided this point is rigidly connected with the body. The external effect of the force
on the body remains unchanged. The problems based on concurrent force system (you study in next article)
are solved by application of this principle.
O¢
O¢ O¢
O
O
O
F
F
F
2
= F
F
1
= F
F
1
= F
Fig 6.6 Fig 6.7 Fig 6.8
For example, consider a force ‘F’ acting at point ‘O’ on a rigid body as shown in fig(6.6). On this rigid
body,” There is another point O
1
in the line of action of the force ‘F’ Suppose at this point O
1
two equal
and opposite forces F
1
and F
2
(each equal to F and collinear with F) are applied as shown in fig(6.7).The
force F and F
2
being equal and opposite will cancel each other, leaving a force F
1
at point O
1
as shown
in fig(6.7).But force F
1
is equal to force F.
The original force F acting at point O has been transferred to point O
1
, which is along the line of action
of F without changing the effect of the force on the rigid body. Hence any force acting at a point on a rigid
body can be transmitted to act at any other point along its line of action without changing its effect on the
rigid body. This proves the principle of transmissibility of forces.
Q. 13: What will happen if the equivalent force F and F acting on a rigid body are not in line?
Explain.
Sol.: If the equivalent force of same magnitude 'F' acting on a rigid body are not in line, then no change
of the position of the body, Because the resultant of both two forces is the algebraic sum of the two forces
which is F – F = 0 or F + F = 2F.
Q. 14: What will happen if force is applied to (i) Rigid body (ii) Non Rigid body?
Sol.: (i) Since Rigid body cannot change its shape on application of any force, so on application of force
“It will start moving in the direction of applied force without any deformation.”
(ii) NonRigid body change its shape on application of any force, So on application of force on Non
Rigid body " It will start moving in the direction of applied force with deformation.”
Force: Concurrent Force System / 109
Q. 15: Define the term resultant of a force system? How you find the resultant of coplanar concurrent
force system?
Sol.: Resultant is a single force which produces the same effect as produced by number of forces jointly
in a system. In equilibrium the magnitude of resultant is always zero.
There are many ways to find out the resultant of the force system. But the first thing to see that how
many forces is acting on the body,
1. If only one force act on the body then that force is the resultant.
2. If two forces are acting on the rigid body then there are two methods for finding out the resultant,
i.e. 'Parallelogram law' (Analytical method) and 'triangle law' (Graphical method).
3. If more than two forces are acting on the body then the resultant is finding out by 'method of
resolution' (Analytical method) and 'Polygon law' (Graphical method).
So we can say that there are mainly two type of method for finding the resultant.
1. Analytical Method.
2. Graphical Method
Analytical Method
Graphical Method
Methods for the resultant force
Parallelouram Law of Forces
(For two forces)
Triangle law of Forces
(For two forces)
Polygon law of Forces
(For more than two forces)
Funicular Polygen
Mehod of Resolution
(For more than two forces)
Fig 6.9
Finally; The resultant force, of a given system of forces, may be found out analytically by the following
methods:
(a) Parallelogram law of forces
(b) Method of Resolution.
Q. 16: State and prove parallelogram law of forces
Sol.: This law is used to determine the resultant of two forces acting at a point of a rigid body in a plane
and is inclined to each other at an angle of a.
It state that “If two forces acting simultaneously on a particle, be represented in magnitude and direction
by two adjacent sides of a parallelogram then their resultant may be represented in magnitude and direction
by the diagonal of the parallelogram, which passes through their point of intersection.”
Let two forces P and Q act at a point ‘O’ as shown in fig (6.10).The force P is represented in
magnitude and direction by vector OA, Where as the force Q is represented in magnitude and direction by
vector OB, Angle between two force is ‘a’.The resultant is denoted by vector OC in fig. 6.11. Drop
perpendicular from C on OA.
Let,
P,Q = Forces whose resultant is required to be found out.
θ = Angle which the resultant forces makes with one of the forces
α = Angle between the forces P and Q
110 / Problems and Solutions in Mechanical Engineering with Concept
Now ∠CAD = α :: because OB//CA and OA is common base.
In ∆ACD :: cosα = AD/AC ⇒ AD=ACcosα
:: But AC = Q; i.e., AD = QCosα ...(i)
And sina = CD/AC ⇒ CD = ACsinα
⇒ CD = Q sinα ...(ii)
Now in ∆OCD ⇒ OC
2
= OD
2
+ CD
2
⇒ R
2
= (OA + AD)
2
+ CD
2
= (P + Qcosα)
2
+ (Qsinα)
2
⇒ = P
2
+ Q
2
Cos
2
α + 2PQcosα + Q
2
sinα
R =
2 2
(P + Q + 2PQcos
It is the magnitude of resultant ‘R’
C
D
A
R
a
a
a
0
0
A
P
B
B
Q
Q
Fig 6.10 Fig 6.11
Direction (θ θθ θθ):
in ∆OCD tan θ = CD/OD = Qsinα/(P + Qcosα)
i.e., θ = tan
–1
[Qsinα / (P + Qcosα)]
Conditions
(i) Resultant R is max when the two forces collinear and in the same direction.
i.e., α = 0° ⇒ Rmax = P + Q
(ii) Resultant R is min when the two forces collinear but acting in opposite direction.
i.e., α = 1800 ⇒ Rmin = P– Q
(iii) If a = 900, i.e when the forces act at right angle, then
R = √P
2
+ Q
2
(iv) If the two forces are equal i.e., when P = Q ⇒ R = 2P.cos(θ/2)
Q. 17: A 100N force which makes an angle of 45º with the horizontal xaxis is to be replaced by two
forces, a horizontal force F and a second force of 75N magnitude. Find F.
Sol.: Here 100N force is resultant of 75N and F Newton forces, Draw a Parallelogram with Q = 75N and
P = F Newton
θ = 45° and α is not given.
We know that
tanθ = Qsinα/(P + Qcosα)
tan45° = 75sinα/(F + 75cosα)
since tan45° =1 → 75sinα = F + 75cosα
or, F = 75(sinα – cosα) ...(i)
Now, R = (P
2
+ Q
2
+ 2PQcosα)
1/2
(100)
2
= F
2
+ 75
2
+ 2.F.75.cosα
“
100
45
75
F
Fig 6.12
Force: Concurrent Force System / 111
F
2
+ 150.F.cosa = 4375
F(F + 75cosα + 75cosα) = 4375
F(75sinα + 75cosα) = 4375
F(sinα + cosα) = 58.33 ...(ii)
Value of ‘F’ from eq(i) put in equation(ii), we get
75(sinα – cosα)(sinα + cosα) = 58.33
sin
2
α – cos
2
α = 0.77
– cos
2
α = 0.77 → α = 70.530 ...(iii)
Putting the value of a in equation (i), we get
F = 45.71N .......ANS
Q. 18: Find the magnitude of two forces such that if they act at right angle their resultant is
10 KN, While they act at an angle of 60º, their resultant is 13 KN.
Sol.: Let the two forces be P and Q, and their resultant be ‘R’
Since R =
2 2
( 2 cos ) P Q PQ + + α
Case–1: If α = 90°, than R = (10)
1/2
KN
10 = P
2
+ Q
2
+ 2PQcos90°
10 = P
2
+ Q
2
, cos90° = 0 ...(i)
Case–2: If α = 600, than R = (13)
1/2
KN
13 = P
2
+ Q
2
+ 2PQcos60°
13 = P
2
+ Q
2
+ PQ, cos60° = 0.5 ...(ii)
From equation (i) and (ii)
PQ = 3 ...(iii)
Now (P + Q)
2
= P
2
+ Q
2
+ 2PQ = 10 + 2.3 = 16
P + Q = 4 ...(iv)
(P – Q)
2
= P
2
+ Q
2
– 2PQ = 10 – 2 × 3
P – Q = 2 ...(v)
From equation (v) and (iv)
P = 3KN and Q = 1KN .......ANS
Q. 19: Two forces equal to 2P and P act on a particle. If the first force be doubled and the second
force is increased by 12KN, the direction of their resultant remain unaltered. Find the value
of P.
Sol.: In both cases direction of resultant remain unchanged, so we used the formula,
tanθ = Qsinα/(P + Qcosα)
Case1: P = 2P, Q = P
tanθ = Psinα/(2P + Pcosα) ...(i)
Case2: P = 4P, Q = P + 12
tanθ = (P + 12)sinα/(4P + (P + 12)cosα) ...(ii)
Equate both equations:
Psinα/(2P + Pcosα) = (P + 12)sinα/(4P + (P + 12)cosα)
4P
2
sinα + P
2
sinαcosα + 12Psinαcosα
= 2P
2
sinα + 24Psinα + P
2
sinαcosα + 12Psinαcosα
2P
2
sinα = 24Psinα
P = 12KN .......ANS
112 / Problems and Solutions in Mechanical Engineering with Concept
Q. 20: The angle between the two forces of magnitude 20KN and 15KN is 60º, the 20KN force being
horizontal. Determine the resultant in magnitude and direction if
(i) the forces are pulls
(ii) the 15KN force is push and 20KN force is a pull.
Sol.: Since there are two forces acting on the body, So we use Law of Parallelogram of forces.
Case1: P =20 KN, Q = 15KN, α = 60°
20KN
15KN
60°
R
Fig 6.13
R
2
= P
2
+ Q
2
+ 2PQcosα = 20
2
+ 15
2
+ 2 × 20 × 15cos60°
R = 30.41KN .......ANS
tanθ = Qsinα/(P + Qcosα) = 15sin60°/(20 + 15cos60°)
θ θθ θθ = 25.28º .......ANS
Case2: Now angle between two forces is 120º, P = 20KN, Q = 15KN, α =120°
R
120°
15 KN
20 KN
Fig 6.14
R
2
= P
2
+ Q
2
+ 2PQcosα = 20
2
+ 15
2
+ 2 × 20 × 15cos120°
R = 18.027KN .......ANS
tanθ = Q sinα/(P + Qcosα) = 15sin120°/(20 + 15cos120°)
θ θθ θθ = –46.1º .......ANS
Q. 21: Explain composition of a force. How you make component of a single force?
Sol.: When a force is split into two parts along two directions not at right angles to each other, those parts
are called component of a force. And process is called composition of a force.
In BOAC, angle BOC = angle OCA = β
(Because // lines OB and AC)
Angle CAO = 180  (α + β)
Q
P
A D
R
B C
0
a
b
a
Fig 6.15
Force: Concurrent Force System / 113
Using sine rule in Triangle OCA
OA/sinβ = OC/sin(α + β) = AC/sina → P/sinβ = R/sin(α + β) = Q/sinα
Or we can say that; P = R.sinß/sin(α + β)
Q = R.sinα/ sin(α + β)
Here P and Q are component of the force ‘R’ in any direction.
Q. 22: A 100N force which makes as angle of 45º with the horizontal xaxis is to be replaced by two
forces, a horizontal force F and a second force of 75N magnitude. Find F.
Sol.: given Q = 75N and P = F N
θ = 45º and α is not given.
We know that
Q = R.sina/ sin (α + β)
75 = 100sin45°/sin (45 + β)
on solving, ß = 25.530
P = R.sinβ/sin (α + β)
F = 100sin25.53°/sin (45° + 25.53°)
F = 45.71N .......ANS
Q. 23: What is resolution of a force? Explain principle of resolution.
Sol.: When a force is resolved into two parts along two mutually perpendicular directions, without changing
its effect on the body, the parts along those directions are called resolved parts. And process is called
resolution of a force.
A P cos q
P sin q
0
X
q
P
C B
Y
C
0
q
P
A
Fig 6.17 Fig 6.18
Horizontal Component (∑H) = Pcosθ
Vertical Component (∑V) = Psinθ
A
P sin q
P cos q
0
X
q
P
C B
Y
P
0
q
A
C
Fig 6.19 Fig 6.20
Horizontal Component (∑H) = Psinθ
Vertical Component (∑V) = Pcosθ
Principle of Resolution: It states, “The algebraic sum of the resolved parts of a number of forces in
a given direction is equal to the resolved part of their resultant in the same direction.”
Fig 6.16
F
a
b
75
100
114 / Problems and Solutions in Mechanical Engineering with Concept
Q.No24: What is the method of resolution for finding out the resultant force.
Or
How do you find the resultant of coplanar concurrent force system?
Sol.: The resultant force, of a given system of forces may be found out by the method of resolution as
discussed below:
Let the forces be P
1
, P
2
, P
3
, P
4
, and P
5
acting at ‘o’. Let OX and OY be the two perpendicular
directions. Let the forces make angle a
1
, a
2
, a
3
, a
4
, and a
5
with Ox respectively. Let R be their resultant
and inclined at angle θ. with OX.
Resolved part of ‘R’ along OX = Sum of the resolved parts of P
1
, P
2
, P
3
, P
4
, P
5
along OX.
X
P
1
P
2
P
5
P
4
P
3
Y
0 a
1
a
a
a
3
a
2
Fig 6.21
i.e.,
Resolve all the forces horizontally and find the algebraic sum of all the horizontally components
(i.e., ∑H)
Rcosθ = P
1
cosα
1
+ P
2
cosα
2
+ P
3
cosα
3
+ P
4
cosα
4
+ P
5
cosα
5
= X (Let)
Resolve all the forces vertically and find the algebraic sum of all the vertical components (i.e., ∑V)
Rsin? = P
1
sinα
1
+ P
2
sinα
2
+ P
3
sinα
3
+ P
4
sinα
4
+ P
5
sinα
5
= Y (Let)
The resultant R of the given forces will be given by the equation:
R = √ (∑V)
2
+ (∑H)
2
We get R
2
(sin
2
θ + cos
2
θ) = P
1
2
(Sin
2
α
1
+ cos
2
α
1
) + 
i.e., R
2
= P
1
2
+ P
2
2
+ P
3
2
+ 
And The resultant force will be inclined at an angle ‘θ’with the horizontal, such that
tanθ = ∑V/∑H
NOTE:
1. Some time there is confusion for finding the angle of resultant (θ), The value of the angle θ will
be very depending upon the value of ∑V and ∑H, for this see the sign chart given below, first for
∑H and second for ∑V.
Force: Concurrent Force System / 115
(+ , –)
(+ , + )
(–, –)
(–, +)
Fig 6.22
a. When ∑V is +ive, the resultant makes an angle between 0º and 180º. But when ∑V is –ive, the
resultant makes an angle between 180° and 360°.
b. When ∑H is +ive, the resultant makes an angle between 0º and 90° and 270° to 360°. But when
∑H is ive, the resultant makes an angle between 90° and 270°.
2. Sum of interior angle of a regular Polygon
= (2.n – 4).90°
Where, n = Number of side of the polygon
For Hexagon, n = 6; angle = (6 X 2–4) X 90 = 720°
And each angle = total angle/n = 720/6 = 120°
3. It resultant is horizontal, then θ = 0º
i.e. ∑H = R, ∑V = 0
4. It Resultant is vertical, then θ = 90º; i.e., ∑H = 0, V = R
Q. 25: What are the basic difference between components and resolved parts?
Sol.: 1. When a force is resolved into two parts along two mutually perpendicular directions, the parts along
those directions are called resolved parts. When a force is split into two parts along two directions not at right
angles to each other, those parts are called component of a force. And process is called composition of a force.
2. All resolved parts are components, but all components are not resolved parts.
3. The resolved parts of a force in a given direction do not represent the whole effect of the force in
that direction.
Q. 26: What are the steps for solving the problems when more than two coplanar forces are acting
on a rigid body.
Sol.: The steps are as;
1. Check the Problem for concurrent or Non concurrent
2. Count Total No. of forces acting on the body.
3. First resolved all the forces in horizontal and vertical direction.
4. Make the direction of force away from the body.
5. Take upward forces as positive, down force as negative, Left hand force as negative, and Right
hand force as positive
6. Take sum of all horizontal parts i.e., ∑H
7. Take sum of all vertical parts i.e., ∑V
8. Find the resultant of the force system using,
R = √ (∑V)
2
+ (∑H)
2
9. Find angle of resultant by using tanθ = ∑V/∑H
10. Take care about sign of ∑V and ∑H.
116 / Problems and Solutions in Mechanical Engineering with Concept
Q. 27: A force of 500N is acting at a point making an angle of 60° with the horizontal. Determine
the component of this force along X and Y direction.
500 cos 60°
60°
500 N 500 sin 60°
Fig 6.23
Sol.: The component of 500N force in the X and Y direction is
∑H = Horizontal Component = 500cos60°
∑V = Vertical Component = 500sin60°
∑H = 500cos60°, ∑ ∑∑ ∑∑V = 500sin60° .......ANS
Q. 28: A small block of weight 300N is placed on an inclined plane, which makes an angle 600 with
the horizontal. What is the component of this weight?
(i) Parallel to the inclined plane
(ii) Perpendicular to the inclined plane. As shown in fig(6.24)
CG
Inclined
Plane
Block
W = 300N
W = 300N
60°
300 sin 60°
300 sin 60°
60°
(–, +)
Fig 6.24 Fig 6.25
Sol.: First draw a line perpendicular to inclined plane, and parallel to inclined plane
∑H = Sum of Horizontal Component
= Perpendicular to plane
= 300cos60° = 150N .......ANS
∑V = Sum of Vertical Component
= Parallel to plane
= 300sin600 = 259.81N .......ANS
NOTE: There is no confusion about cosθ and sinθ, the angle ‘θ’ made by which plane, the component
of force on that plane contain cosθ, and other component contain sinθ.
Q. 29: The 100N force is applied to the bracket as shown in fig(6.26). Determine the component of
F in,
(i) the x and y directions
(ii) the x’ and y’ directions
(iii) the x and y’ directions
Force: Concurrent Force System / 117
30°
x
x¢
y¢
y
F N = 100
Fig 6.26
Sol.:
(1) Components in x and y directions
∑H = 100cos500 = 64.2N .......ANS
∑V = 100sin500 = 76.6N .......ANS
(2) Components in x′ and y′ directions
∑H′ = 100cos200 = 93.9N .......ANS
∑V′ = 100sin200 = 34.2N .......ANS
(3) Components in x and y′ directions
∑H = 100cos500 = 64.2N .......ANS
∑V′ = 100sin200 = 34.2N .......ANS
Q. 30: Determine the x and y components of the force exerted on the pin at A as shown in fig (6.27).
B B
A A
T
e
C C
200 mm 200 mm
Cable Cable
300 mm 300 mm
2000 N 2000 N
Fig 6.27 Fig6.28
Sol.: Since there is a single string, so the tension in the string throughout same, Let ‘T’ is the tension in
the string.
At point C, there will be an equal and opposite reaction, so
T = 2000N ...(i)
Now tanθ = 200/300 => θ =33.69°
Horizontal component of T is;
∑H = Tcosθ = 2000cos33.69°
= 1664.3N .......ANS
Vertical component of T is;
∑V = Tsinθ = 2000sin33.69°
= 1109.5N .......ANS
118 / Problems and Solutions in Mechanical Engineering with Concept
Q. 31: Three wires exert the tensions indicated on the ring in fig (6.29). Assuming a concurrent
system, determine the force in a single wire will replace three wires.
Sol.: Single force, which replaces all other forces, is always the resultant of the system, so first resolved
all the forces in horizontal and vertical direction
∑H = Sum of Horizontal Component
= 60 cos 0° + 20 cos 68° + 40 cos 270°
= 67.49N ...(i)
∑V = Sum of Vertical Component
= 60 sin 0° + 20 sin 68° + 40 sin 270°
= –21.46N ...(ii)
Let R be the resultant of coplanar forces
R = (∑H
2
+ ∑V
2
)
1/2
= (67.49
2
+ (–21.46)
2
)
1/2
R = 70.81N .......ANS
θ = tan
–1
(R
V
/R
H
)
= tan
–1
(–21.45/67.49)
θ = –17.63° .......ANS
Angle made by resultant (70.81),–17.63° and lies in forth coordinate.
Q. 32: Four forces of magnitude P, 2P, 5P and 4P are acting at a point. Angles made by these forces
with xaxis are 0°, 75°, 150° and 225° respectively. Find the magnitude and direction of resultant
force.
P
225°
75°
4P
5P
2P
150°
Fig. 6.30
Sol.: first resolved all the forces in horizontal and vertical direction
∑H = Sum of Horizontal Component
= P cos 0° + 2Pcos 75° + 5Pcos 150° + 4Pcos 225°
= –5.628P ...(i)
∑V = Sum of Vertical Component
= Psin 0° + 2Psin 75° + 5Psin 150° + 4Psin 225°
= 1.603P ...(ii)
R = ((–5.628P)
2
+ (1.603P)
2
)
1/2
R = 5.85P .......ANS
θ = tan
–1
(R
V
/R
H
)
20 N
60 N
40 N
68°
Fig 6.29
Force: Concurrent Force System / 119
= tan
–1
(1.603P/–5.628P)
θ = –15.89° .......ANS
Angle made by resultant (5.85P),–15.890 and lies in forth coordinate.
Q. 33: Four coplanar forces are acting at a point. Three forces have magnitude of 20, 50 and 20N
at angles of 45°, 200° and 270° respectively. Fourth force is unknown. Resultant force has
magnitude of 50N and acts along xaxis. Determine the unknown force and its direction from
xaxis.
q
R
P
200°
270°
50
20
20
45°
Fig. 6.31
Sol.: Let unknown force be ‘P’ which makes an angle of ‘θ’ with the xaxis, If R
H
and R
V
be the sum of
horizontal and vertical components of the resultant, and resultant makes an angle of θ’ with the horizontal.
Then;
∑H = Rcosθ = Horizontal component of resultant
∑V = Rsinθ = Vertical component of resultant
Since Resultant make an angle of 00 (Since acts along xaxis) with the Xaxis so
∑H = Rcos 0° = R
∑V = Rsin 0° = 0
i.e., ∑H = R and ∑V = 0
i.e., R = ∑H = 50
∑H = 20cos 45° + 50 cos 200° + Pcosθ + 20cos 270° = 50
On solving Pcosθ = 82.84 ...(i)
As the same,
∑V = 20sin 45° + 50sin 200° + Psinθ + 20sin 270° = 0
On solving Psinθ = 22.95 ...(ii)
Now, square both the equation and add
P
2
cos
2
θ + P
2
sin
2
θ = 22.952 + 82.842
P = 85.96N .......ANS
Let angle made by the unknown force be ?
tanθ = Psinθ/Pcosθ
= 22.95/82.84
θ θθ θθ = 15.48º .......ANS
Angle made by unknown force is 15.48° and lies in first coordinate.
Q. 34: Determine the resultant ‘R’ of the four forces transmitted to the gusset plane if θ θθ θθ = 45° as
shown in fig(6.32).
Sol.: First resolved all the forces in horizontal and vertical direction, Clearly note that the angle measured
by xaxis,
120 / Problems and Solutions in Mechanical Engineering with Concept
∑H = 4000cos 45° + 3000cos 90°+1000cos 0°+5000cos 225°
= 292.8N ...(i)
∑V = 4000sin 45° + 3000sin 90°+ 1000sin 0°+5000sin 225°
= 2292.8N ...(ii)
R
2
= R
H
2
+ R
V
2
R
2
= (292.8)
2
+ (22923.8)
2
R = 2311.5N .......ANS
Let angle made by resultant is θ
tanθ = ∑V /∑H
= 2292.8/292.8
θ θθ θθ = 82.72º .......ANS
Q. 35: Four forces act on bolt as shown in fig (6.33). Determine the resultant of forces on the bolt.
Sol.: First resolved all the forces in vertical and horizontal directions; Let
∑H = Sum of Horizontal components
∑V = Sum of Vertical components
∑H = 150cos 30° + 80cos 110° + 110cos 270° + 100cos 345°
= 199.13N ...(i)
∑V = 150sin 30° + 80sin 110° + 110sin 270° + 100sin 345°
= 14.29N ...(ii)
x
x
y
y
F
2
F
1
F
1
F
4
F
4
F
3
F
3
30°
20°
15°
15°
20°
30°
F
2
80 N
80 N
= 150 N
= 150 N
= 110 N
= 110 N
= 110 N
= 100 N
Fig. 6.33 Fig. 6.34
R = (∑H
2
+ ∑V
2
)
1/2
= {(199.13)
2
+ (14.29)
2
}
1/2
R = 199.6N .......ANS
Let angle made by resultant is θ
tanθ = ∑V/∑H
= 14.29/199.13
θ θθ θθ = 4.11º .......ANS
Q. 36: Determine the resultant of the force acting on a hook as shown in fig (6.35).
Sol.: First resolved all the forces in vertical and horizontal directions Let
∑H = Sum of Horizontal components
∑H = 80cos 25° + 70cos 50° + 50cos 315°
= 152.86N ...(i)
∑V = Sum of Vertical components
300 N
4000 N
q
45°
x
y
5000 N
1000 N
Fig. 6.32
Force: Concurrent Force System / 121
x
y
50 kN
50 sin 315
80 cos 25
50 cos 315
70 cos 50
80 sin 25
70 sin 50
25°
25°
45°
80 kN
70 kN
Fig. 6.35 Fig. 6.36
∑V = 80sin 25° + 70sin 50° + 50sin 315°
= 52.07N ...(ii)
R = (R
H
2
+ R
V
2
)
1/2
= {(152.86)
2
+ (52.07)
2
}
1/2
R = 161.48N .......ANS
Let angle made by resultant is θ
tanθ = ∑V/∑H ⇒ = 52.07/152.86
θ θθ θθ = 18.81° .......ANS
Q. 37: The following forces act at a point:
(i) 20N inclined at 300 towards North of east
(ii) 25N towards North
(iii) 30N towards North west,
(iv) 35N inclined at 400 towards south of west.
Find the magnitude and direction of the resultant force.
Sol.: Resolving all the forces horizontally i.e. along EastWest, line,
∑H = 20cos30°+ 25cos90°+ 30cos135° +35cos220°
= (20 × 0.886) + (25 × 0) + {–30(–0.707) + 35(–0.766) N
= –30.7 N ...(i)
North
25 N
30 N
20 N
30° 45°
40°
West East
South
Fig. 6.37
And now resolving all the forces vertically i.e., along NorthSouth line,
∑V = 20sin 30° + 25sin 90° + 30sin 135° + 35sin 220°
122 / Problems and Solutions in Mechanical Engineering with Concept
= (20 × 0.5) + (25 × 1.00) + (30 × 0.707) + 35 × (– 0.6428)
= 33.7N ...(ii)
We know that the magnitude of the resultant force,
R = √∑H
2
+ ∑V
2
On solving, R = 45.6 N .......ANS
Direction of the resultant force;
tanθ = ∑V /∑H
Since ∑H is –ve and ∑V is +ve, therefore θ lies between 90° and 180°.
Actual θ = 180° – 47° 42’
= 132.18º .......ANS
Q. 38: Determine the resultant of four forces acting on a body shown in fig (6.38).
X
X¢
Y¢
Y
2.24 KN
3 KN
2 KN
3.9 KN
67.38° 60°
26.50°
0
30°
X
Y
3.9 KN
2 KN
12
5
60°
30°
1
2
3 KN
2.24 KN
Fig. 6.38 Fig. 6.39
Sol.: Here 2.24KN makes an angle tan
–1
(1/2) with horizontal. Also 3.9KN makes an angle of tan
–1
(12/5) with horizontal.
Let the resultant R makes an angle θ with xaxis. Resolving all the forces along xaxis, we get,
∑H = 3cos 30° + 2.24cos 153.5° + 2cos 240° + 3.9cos 292.62°
= 1.094KN ...(i)
Similarly resolving all the forces along yaxis, we get
∑V = 3sin 30° + 2.24sin153.5° + 2sin 240° + 3.9sin2 92.62° = –2.83KN = ...(ii)
Resultant R = {(1.094)
2
+ (–2.83)
2
}
1/2
= 3.035KN .......ANS
Angle with horizontal
θ = tan
–1
(–2.83/1.094)
= 68.86º .......ANS
Q. 39: The forces 20N, 30N, 40N, 50N and 60N are acting on one of the angular points of a regular
hexagon, towards the other five angular points, taken in order. Find the magnitude and direction
of the resultant force.
Sol.: In regular hexagon each angle is equal to 120°, and if each angular point is joint together, then each
section makes an angle of 30°.
First resolved all the forces in vertical and horizontal directions Let
Force: Concurrent Force System / 123
60
30
40
50
30
20
30
30
30
F
A
B
C
E
D
Fig. 6.40
∑H = Sum of Horizontal components
∑H = 20cos 0° + 30cos 30° + 40cos 60° + 50cos 90° + 60cos 120°
= 35.98N ...(i)
∑V = Sum of Vertical components
∑V = 20sin 0° + 30sin 30° + 40sin 60° + 50sin 90° + 60sin 120°
= 151.6N ...(ii)
R = (∑H
2
+ ∑V
2
)
1/2
= {(35.98)
2
+ (151.6)
2
}
1/2
R = 155.81N .......ANS
Let angle made by resultant is θ
Tan θ = ∑V /∑H = 151.6/35.98
θ θθ θθ = 76.64° .......ANS
Q. 40: The resultant of four forces, which are acting at a point, is along Yaxis. The magnitudes of
forces F
1
, F
3
, F
4
are 10KN, 20KN and 40KN respectively. The angle made by 10KN, 20KN
and 40KN with Xaxis are 300, 900 and 1200 respectively. Find the magnitude and direction
of force F
2
, if resultant is 72KN.
F
30°
90°
120° F
2
R = 72 KN
F
4
= 10 KN
F
3
= 20 KN
F
1
= 10 KN
X X
Y
Y
Fig. 6.41
Sol.: Given that resultant is along Yaxis that means resultant(R) makes an angle of 90° with the Xaxis,
i.e., horizontal component of R is zero, and Magnitude of resultant is equal to vertical component, Let
∑H = Sum of Horizontal components = 0
∑V = Sum of Vertical components
124 / Problems and Solutions in Mechanical Engineering with Concept
R = (∑H
2
+ ∑V
2
)
1/2
= (0 + ∑V
2
)
1/2
R = ∑V;
Let unknown force be F
2
and makes an angle of Φ with the horizontal Xaxis;
Now resolved all the forces in vertical and horizontal directions;
∑H = 10cos 30° + 20cos 90° + 40cos 120° + F
2
cosΦ
0 = F
2
cosΦ – 11.34
F
2
cosΦ = 11.34 ...(i)
72 = 10sin 30° + 20sin 90° + 40sin 120° + F
2
sin Φ
72 = F
2
sinΦ + 59.64
F
2
sinΦ = 12.36 ...(ii)
Divide equation (ii) by (i), we get
tanΦ = 12.36/11.34
Φ ΦΦ ΦΦ = 47.460 .......ANS
Putting the value of Φ in equation (i) we get
F
2
cos 47.46 = 11.34 ⇒ F
2
= 16.77KN .......ANS
Q. 41: A body is subjected to the three forces as shown in fig 6.42. If possible, determine the direction
θ θθ θθ of the force F so that the resultant is in Xdirection when:
(1) F = 5000N ;
(2) F = 3000N. (Dec(C.O)03)
Sol.: Since Resultant is in X direction, i.e., Vertical component of resultant is zero.
∑V = 0
R = ∑H
Resolve the forces in X and Y direction
∑V = 2000cos 60° + 3000 – Fcosθ = 0
4000–Fcosθ = 0
or, Fcosθ = 4000 ...(i)
Now
(i) If F = 5000
cosθ = 4/5, θ θθ θθ = 36.86° .......ANS
(ii) If F = 3000
cosθ = 4/3, θ θθ θθ = Not possible .......ANS
Q. 42: State the condition necessary for equilibrium of rigid body. What will happen if one of the
conditions is not satisfied?
Sol.: When two or more than two force act on a body (all forces meet at a single point) in such a way that
body remain in state of rest or continue to be in linear motion, than forces are said to be in equilibrium.
According to Newton's law of motion it means that the resultant of all the forces acting on a body in
equilibrium is zero. i.e.,
R = 0,
∑V = 0,
∑H = 0
q
F
X
60°
2000 N
3000 N
Force: Concurrent Force System / 125
When body is in equilibrium, then there are two types of forces applied on the body
Applied forces
None applied forces
Self weight ( = m.g. act vertically downwards)
Contact reaction (Action = reaction
W
NOTE
If the resultant of a number of forces acting on a particle is zero, the particle will be in equilibrium.
Such a set of forces, whose resultant is zero, are called equilibrium forces.
The force, which brings the set of forces in equilibrium, is called an equilibrant. As a matter of fact,
the equilibrant is equal to the resultant force in magnitude, but opposite in nature.
Q. 43: Explain ‘action’ and ‘reaction’ with the help of suitable examples.
Sol.: Two body A and B are in contact at point ‘O’. Body A
Press against the body B. Hence action of body A on the body B is F. Reaction of Body B on body
A is R. From Newton's third law of motion (i.e., action = reaction), both these forces are equal there for
F = R
i.e., Action = Reaction
A
B
F
R
0
Fig 6.43
Or, “Any pressure on a support causes an equal and opposite pressure from the support so that action
and reaction are two equal and opposite forces.”
Q. 44: Describe the different uses of strings. Illustrate the tension in the strings.
Sol.: When a weight is attached to a string then it will be in tension. Various diagrams are shown below
to describe this concept.
T
2
T
1
T
1
W
3
W
2
W
1
W
Tension ‘T’
String not
continous
String
continous
Pulley is
massles
no friction
126 / Problems and Solutions in Mechanical Engineering with Concept
W
W
1
W
2
T
2
T
2
T
1
T
1
T
1
Pulley
Continous string
Motion
Motion
Pulley has
mass
Strings
are
different
T
3
T
2
T
2
T
1
T
1
W
Knot
Friction between
belt & pulley present
Fig 6.44 Different uses of String
Q. 45: What is the principle of equilibrium:
Sol.: Principle of equilibrium may be divided in to three parts;
(1) Two Force Principle: Since Resultant is zero when body is in equilibrium, so if two forces are
acting on the body, then they must be equal, opposite and collinear.
(2) Three Force Principle: As per this principle, if a body in equilibrium is acted upon by three
forces, then the resultant of any two forces must be equal, opposite and collinear with the third force. For
finding out the values of forces generally we apply lamis theorem
(3) Four Force Principle: As per this principle, if four forces act upon a body in equilibrium, then the
resultant of any two forces must be equal, opposite and collinear with the resultant of the other two.
And for finding out the forces we generally apply;
∑H = ∑V = 0, because resultant is zero.
Q. 46: What is free body diagram?
Sol.: An important aid in thinking clearly about problems in mechanics is the free body diagram. In such
a diagram, the body is considered by itself and the effect of the surroundings on the body is shown by forces
and moments. Free body diagrams are also used to show internal forces and moments by cutting away the
unwanted portion of a body.
Force: Concurrent Force System / 127
W
R
C
B
F
String
W
C
B
A
Fig. 6.45
Such a diagram of the body in which the body under consideration is freed from all the contact surface,
and all the forces acting on it.(Reaction) are drawn is called a free body diagram.
Q. 47: Explain lami's theorem?
Sol.: It states that “If three coplanar forces acting at a point be in equilibrium, then each force is proportional
to the sine of the angle between the other two.” Mathematically,
P/sin β = Q/sinγ = R/sinα
a
b
g
R
P
Q
0
Fig 6.46
Q. 48: Explain law of superposition?
Sol.: When two forces are in equilibrium (equal, opposite and collinear), their resultant is zero and their
combined action on a rigid body is equivalent to that of no force at all., Thus
“The action of a given system of forces on a rigid body will in no way be changed if we add to or
subtract from them another system of forces in equilibrium.”, this is called law of superposition.
Q. 49: What are the steps for solving the problems of equilibrium in concurrent force system.
Sol.: The steps are as following:
1. Draw free body diagram of the body.
2. Make the direction of the forces away from the body.
3. Count how many forces are acting on the body.
2. If there is three forces are acting then apply lamis theorem. And solved for unknown forces.
3. If there are more then three forces are acting then first resolved all the forces in horizontal and
vertical direction, Make the direction of the forces away from the body.
4. And then apply equilibrium condition as R
H
= R
V
= 0.
128 / Problems and Solutions in Mechanical Engineering with Concept
Q. 50: Three sphere A, B, C are placed in a groove shown in fig (6.47). The diameter of each sphere
is 100mm. Sketch the free body diagram of B. Assume the weight of spheres A, B, C as 1KN,
2KN and 1KN respectively.
W
B
R
C
R
B
R
A
+ C
+ B
+ A
q
q
150 mm
Fig 6.47 Fig 6.48
Sol.: For θ,
cosθ = 50/100, cosθ = .5 , θ = 60°
FBD of block B is given in fig 9.47
Q. 51: Two cylindrical identical rollers A and B, each of weight W are supported by an inclined plane
and vertical wall as shown in fig 6.49. Assuming all surfaces to be smooth, draw free body
diagrams of
(i) roller A,
(ii) roller B
(iii) Roller A and B taken together.
Sol.: Let us assumed
W = Weight of each roller
R = Radius of each roller
R
A
= Reaction at point A
R
B
= Reaction at point B
R
C
= Reaction at point C
R
D
= Reaction at point D
30°
B
A
C
P
E
D
E
Q
W
W
B
R
D
D
E
Q
W
30°
R
B
C
R
C
Fig 6.49 Fig 6.50 FBD of Roller ‘B’
Force: Concurrent Force System / 129
R
A
A
P
F
R
B
30°
B
R
C
C
E
Q
W
W
30°
R
A
A
R
D
P
F
W
Fig 6.51 FBD of Roller ‘A’ Fig 6.52 FBD of Roller ‘B’ & ‘A’ taken together
Q. 52: Three forces act on a particle ‘O’ as shown in fig(6.53).Determine the value of ‘P’ such that
the resultant of these three forces is horizontal. Find the magnitude and direction of the fourth
force which when acting along with the given three forces, will keep ‘O’ in equilibrium.
P
10°
40°
30°
O
200 N
500 N
Fig 6.53
Sol.: Since resultant(R) is horizontal so the vertical component of resultant is zero, i.e.,
∑V = 0, ∑H = R
∑V = 200sin10° + Psin50° + 500sin150° = 0
On solving, P = –371.68N ...(i)
∑H = 200cos10° + Pcos50° + 500cos150° = 0
Putting the value of ‘P’, we get
∑H = –474.96N ...(ii)
Let Unknown force be ‘Q’ and makes an angle of ? with the horizontal Xaxis. Additional force makes
the system in equilibrium Now,
∑H = Qcosθ –474.96N = 0
i.e., Qcosθ = 474.96N(3)
Since ∑V already zero, Now on addition of force Q, the body be in equilibrium so again ∑V is zero.
∑V = 200sin10° –371.68sin50° + 500sin150° + Qsin θ = 0
But 200sin10° – 371.68sin500 + 500sin1500 = 0 by equation (1)
So, Qsinθ = 0, that means Q = 0 or sinθ = 0,
Q is not zero so sinθ = 0, θ = 0
Putting θ = 0 in equation (iii),
Q = 474.96N, θ θθ θθ = 0º .......ANS
130 / Problems and Solutions in Mechanical Engineering with Concept
Q. 53: An Electric light fixture weighing 15N hangs from a point C, by two strings AC and BC. AC
is inclined at 600 to the horizontal and BC at 450 to the vertical as shown in fig (6.54),
Determine the forces in the strings AC and BC
E
C
C
D
B
F
O
A
T
2
T
2
T
1
T
1
15 N
150°
135°
30°
30°
45°
45°
45°
60°
Fig 6.54 Fig 6.55
Sol.: First draw the F.B.D. of the electric light fixture,
Apply lami's theorem at point ‘C’
T
1
/sin 150° = T
2
/sin 135° = 15/sin75°
T
1
= 15.sin150°/ sin75°
T
1
= 7.76N .......ANS
T
2
= 15.sin135°/ sin75°
T
2
= 10.98N .......ANS
Q. 54: A string ABCD, attached to two fixed points A and D has two equal weight of 1000N attached
to it at B and C. The weights rest with the portions AB and CD inclined at an angle of 300
and 600 respectively, to the vertical as shown in fig(6.56). Find the tension in the portion AB,
BC, CD
D
C
B
A
120°
30° 60°
1000 N
1000 N
Fig 6.56
C
B
A
T
1
T
1
T
2
T
2
C
B
D
120°
120°
60° 60°
30°
1000 N
1000 N
Fig 6.57 Fig 6.58
Sol.: First string ABCD is split in to two parts, and consider the joints B and C separately
Let,
T
1
= Tension in String AB
Force: Concurrent Force System / 131
T
2
= Tension in String BC
T
3
= Tension in String CD
Since at joint B there are three forces are acting. SO Apply lamis theorem at joint B,
T
1
/sin60° = T
2
/sin150° = 1000/sin150°
T
1
= {sin60° × 1000}/sin150°
= 1732N .......ANS
T
2
= {sin150° × 1000}/sin150°
= 1000N .......ANS
Again Apply lamis theorem at joint C,
T
2
/sin120° = T
3
/sin120° = 1000/sin120°
T
3
= {sin120° × 1000}/sin120°
= 1000N .......ANS
Q. 55: A fine light string ABCDE whose extremity A is fixed, has weights W
1
and W
2
attached to it
at B and C. It passes round a small smooth peg at D carrying a weight of 40N at the free end
E as shown in fig(6.59). If in the position of equilibrium, BC is horizontal and AB and CD
makes 150° and 120° with BC, find (i) Tension in the portion AB,BC and CD of the string and
(ii) Magnitude of W
1
and W
2
.
E
D
C B
A
W
2
W
1
40 N
150° 120°
Fig 6.59
C
D
B C
B
A
W
2
W
1
150° 120° 40 N
Fig 6.60 Fig 6.61
Sol.: First string ABCD is split in to two parts, and consider the joints B and C separately
Let,
T
1
= Tension in String AB
T
2
= Tension in String BC
T
3
= Tension in String CD
T
4
= Tension in String DE
T
4
= T
3
= 40N
132 / Problems and Solutions in Mechanical Engineering with Concept
Since at joint B and C three forces are acting on both points. But at B all three forces are unknown
and at point C only two forces are unknown SO Apply lamis theorem first at joint C,
T
2
/sin150° = W
2
/sin120° = 40/sin90°
T
2
= {sin150° × 40}/sin90°
= 20N .......ANS
W
2
= {sin120° × 40}/sin90°
= 34.64N .......ANS
Now for point B, We know the value of T
2
So, Again Apply lamis theorem at joint B,
T
1
/sin90° = W
1
/sin150° = T
2
/sin120°
T
1
= {sin90° × 20}/sin120°
= 23.1N .......ANS
W
1
= {sin150° × 20}/sin120°
= 11.55N .......ANS
Q. 56: Express in terms of θ θθ θθ, β ββ ββ and W the force T necessary to hold the weight in equilibrium as
shown in fig (6.62). Also derive an expression for the reaction of the plane on W. No friction
is assumed between the weight and the plane.
Sol.: Since block is put on the inclined plane, so plane give a vertical reaction on the block say ‘R’. Also
resolved the force ‘T’ and ‘W’ in perpendicular and parallel to plane, now
For equilibrium of the block,
Sum of components parallel to plane = 0, i.e., ∑ H = 0
Tcos β – Wsinθ = 0 ...(i)
Or T = Wsinθ θθ θθ/cosβ ββ ββ .......ANS
Sum of components perpendicular to plane = 0,
i.e., ∑V = 0
R + Tsinβ – Wcosθ = 0
Or R = Wcosθ – Tsinβ ...(ii)
T sin b
T cos b
W cos q
W sin q
W
W
T
T
R
0
b
b
q
Fig 6.62 Fig 6.63
Putting the value of T in equation(ii), We get
R = W{cosθ θθ θθ – sinθ θθ θθ.tanβ ββ ββ} .......ANS
Hence reaction of the plane = R = W{cosθ θθ θθ – sinθ θθ θθ.tanθβ θβ θβ θβ θβ} .......ANS
Q. 57: For the system shown in fig(6.64), find the additional single force required to maintain
equilibrium.
Sol.: Let α and β be the angles as shown in fig. Resolved all the forces horizontal and in vertical direction.
When we add a single force whose magnitude is equal to the resultant of the force system and direction
is opposite the the direction of resultant. Let;
Force: Concurrent Force System / 133
∑H = Sum of horizontal component
∑V = Sum of vertical component
First
∑H = 20cosα + 20cos(360° – β)
∑H = 20cosα – 20cosβ ...(i)
Now
∑V = 20sinα + 20sin(3600 – β)
–50 = –50 + 20sin α + 20sin β ...(ii)
Hence the resultant of the system = R = (∑H
2
+ ∑V
2
)
1/2
Let additional single force be ‘R’ and its magnitude is equal to
R’ = R = [(20cosα αα αα – 20cosβ ββ ββ)
2
+ (–50 + 20sinα αα αα + 20sinβ ββ ββ)
2
]
1/2
.......ANS
This force should act in direction opposite to the direction of force ‘R’.
Q. 58: A lamp of mass 1Kg is hung from the ceiling by a chain and is pulled aside by a horizontal
chord until the chain makes an angle of 600 with ceiling. Find the tensions in chain and chord.
Sol.: Let,
T
chord
= Tension in chord
T
chain
= Tension in chain
9.81 N
9.81 N
T
chord
T
chord
0
0
T
chain
T
chain
60°
150°
120°
90°
Fig 6.65 Fig 6.66
W = weight of lamp = 1 × g = 9.81N
Consider point ‘C’, there are three force acting, so apply lamis
theorem at point ‘C’, as point C is in equilibrium
Tchord/sin150° = T
chain
/sin90° = 9.81/sin120°
T
chord
= 9.81 × sin150° /sin120°
T
chord
= 5.65N .......ANS
T
chain
= 9.81 × sin90° /sin120°
T
chain
= 11.33N .......ANS
Q. 59: A roller shown in fig(6.67) is of mass 150Kg. What force T is necessary to start the roller over
the block A?
Sol.: Let R be the reaction given by the block to the roller, and supposed to act at point A makes an angle
of ? as shown in fig,
For finding the angle θ,
Sinθ = 75/175 = 0.428
θ θθ θθ = 25.37°
Fig 6.64
a
b
20 N
50 N
20 N
134 / Problems and Solutions in Mechanical Engineering with Concept
Apply lami's theorem at ‘A’, Since the body is in equilibrium
T/sin(90° + 25.37°) = 150 × g/ sin(64.63° + 65°)
T = [150 × g × sin(90° + 25.37°)]/ sin(64.63° + 65°)
T = 1726.33N
A
T
R
A
150 × g
25°
75
100
175
1
7
5
= 65°
q
q
= 64.63°
90° – °
90° – 25°
25°
175 mm
100 mm
Fig 6.67 Fig 6.68
Q. 60: Three sphere A, B and C having their diameter 500mm, 500mm and 800mm respectively are
placed in a trench with smooth side walls and floor as shown in fig(6.69).The center to center
distance of spheres A and B is 600mm. The weights of the cylinders A, B and C are 4KN, 4KN
and 8KN respectively. Determine the reactions at P, Q, R and S.
B A
C
S
P
Q R
2 1
0
65°
75°
800 mm
Fig 6.69
( ) a
( ) c ( ) b
R
P
R
S
R
R
R
Q
A
R
1
R
2
C
4 KN
4 KN
65°
75°
a
a
a
b
8 KN
Fig 6.70
Force: Concurrent Force System / 135
Sol.: From triangle ABC in fig 6.69
Cosα = AD/AC = 300/(250 + 400)
Cosα = 62.51°
Consider FBD of sphere C(Fig 6.70(a))
Consider equilibrium of block C
∑H = R
1
Cosα – R
2
Cosα = 0
i.e., R
1
= R
2
...(i)
∑H = R
1
sinα – R
2
sinα – 8 = 0 ⇒ putting R
1
= R
2
∑V = R
1
sinα – R
1
sinα = 8 ⇒ 2R
1
= 8/sina
= R
1
= 8/2sinα =4.509
i.e., R
1
= R
2
= 4.509KN
Consider equilibrium of block A
∑H = R
p
sin75° – R
1
cosα = 0
= > R
P
= R
1
cosα/sin75° = 4.5cos62.51°/sin75°
R
p
= 2.15KN ........ANS
∑V = R
p
cos75° – R
1
sin62.50 + R
Q
– W
A
= 0
R
Q
= 7.44KN .......ANS
Consider equilibrium of block B
∑H = R
S
sin65° – R
2
cosα = 0
=> R
S
= R
2
cosα/sin65° = 4.5cos62.51°/sin65°
R
S
= 2.29KN .......ANS
∑V = R
S
cos65° – R
2
sinα + R
R
– W
B
= 0
∑V = 2.29cos65° – 4.509sin62.5° + R
R
– 4 = 0
R
R
= 7.02KN ........ANS
Q. 61: Determine the magnitude and direction of smallest force P required to start the wheel over
the block. As shown in fig(6.71).
P
10 kN
cm = 60
30°
15
Fig. 6.71
Sol.: Let the reaction of the block be R. The least force P is always perpendicular in the reaction R. When
the wheel is just on the point of movement up, then it loose contact with inclined plane and reaction at this
point becomes zero.
Consider triangle OMP
OM = 60cm
OP = 60–15 = 45cm
MP = {(OM)
2
– (OP)
2
}
1/2
136 / Problems and Solutions in Mechanical Engineering with Concept
M
P
P
R
P
10 kN
45 cm
1
5
3
0
°
0
b
10 kN
90°
90°
1
0
8
.
6
°
4
1
.
4
°
+
3
0
°
Fig. 6.72 Fig. 6.73
= {3600 – 2025}
1/2
= 39.68cm
tan β = MP/OP = 39.68/45, β ββ ββ = 41.400
Using lamis theorem at point O
P/sin108.6° = 10/sin90° = R/sin161.4°
P = (10 × sin108.6°)/sin90° =9.4KN
Hence smallest force P = 9.4KN
Q. 62: A heavy spherical ball of weight W rests in a V shaped trough whose sides are inclined at
angles α αα αα and β ββ ββ to the horizontal. Find the pressure on each side of the trough. If a second ball
of equal weight be placed on the side of inclination α αα αα, so as to rest above the first, find the
pressure of the lower ball on the side of inclination β ββ ββ.
Sol.: Let
R
1
= Reaction of the inclined plane AB on
the sphere or required pressure on AB
R
2
= Reaction of the inclined plane AC on
the sphere or required pressure on AC
The point O is in equilibrium under the action of the
following three forces: W, R
1
, R
2
Case – 1:
Apply lami's theorem at point O
R
1
/sinß = R
2
/sin(180 – α) = W/sin(α + β)
or R
1
= Wsinβ ββ ββ/sin(α αα αα + β ββ ββ) .......ANS
and R
2
= Wsinα αα αα/sin(α αα αα + β ββ ββ) .......ANS Fig 6.74
Case – 2: Let
R
3
= Reaction of the inclined plane AC on the bottom sphere or required pressure on AC
Since the two spheres are equal, the center line O
1
O
2
is parallel to the plane AB.
When the two spheres are considered as a single unit, the action and reaction between them at the
point of contact cancel each other. Considering equilibrium of two spheres taken together and resolving the
forces along the Line O
1
O
2
, we get
R
1
R
2
W
B
C
A
a
a
0
Force: Concurrent Force System / 137
E
1
B
C
a
a
F E
A
M
W
W
O
1
R
3
R¢
1
R¢
2
O
2
b
b
1st ball
2nd ball
Fig. 6.75
R
3
cos{90° – (α + β)} = Wsinα + Wsinα
R
3
sin(α + β) = 2Wsinα
Or, R
3
= 2Wsinα αα αα/sin(α αα αα + β ββ ββ) .......ANS
Q. 63: A right circular roller of weight 5000N rests on a smooth inclined plane and is held in position
by a cord AC as shown in fig 6.76. Find the tension in the cord if there is a horizontal force
of magnitude 1000N acting at C. (May–0203)
F
A
D
R
B
B
C
G
C
E
B
A
W = 5000 N
W = 5000 N
P = 1000 N
P = 1000 N
20°
20°
30°
30°
Fig 6.76 Fig 6.77
R
B
C
B
A
T
5000 N
1000 N
70° 10°
Fig 6.78
Sol.: Let R
B
be the contact reaction at point B. This reaction makes an angle of 20° with the vertical
Yaxis.
Let Tension in string AC is ‘T’, which makes an angle of 100 with the horizontal Xaxis as shown
in fig (6.78).
See fig(6.77)
138 / Problems and Solutions in Mechanical Engineering with Concept
In Triangle EBD
Angle BDE = 20°, Angle BED = 90°,
Angle EBD = 90° – 20° =70°
Since Angle EBD = Angle FBC = 70°,
Now In Triangle FBC
Angle FBC = 70°, Angle CFB = 90°,
Angle FCB = 90° – 70° = 20°
i.e., R
B
makes an angle of 20° with the vertical
Now In Triangle ACF
Angle CAF = 30°, Angle AFC = 90°,
Angle ACF = 90° – 30° = 60°
Now Angle GCB = 90°,
Angle GCA = 90° – 20° – 60° = 10°
i.e., Tension T makes an angle of 10° with the Horizontal
Consider Fig(3), The body is in equilibrium, SO apply condition of equilibrium
R
H
= 0
1000 + R
B
cos70° – Tcos10° = 0
1000 + 0.34RB – 0.985T = 0
R
B
=2.89T – 2941.2 ...(i)
R
V
= 0
R
B
sin70° – 5000 – Tsin10° = 0
0.94R
B
– 5000 – 0.174T = 0 ...(ii)
Putting the value of RB in equation (ii), We get
T = 3060N .......ANS
Q. 64: Fig 6.79, shows a sphere resting in a smooth V shaped groove and subjected to a spring force.
The spring is compressed to a length of 100mm from its free length of 150mm. If the stiffness
of spring is 2N/mm, determine the contact reactions at A and B. (MAY 0203)
A
B
sphere
30° 60°
W = 40 N
k = 2N/mm
100 mm
R
B
R
A
40 N + 100N = (140 N)
60° 30°
0
Fig 6.79 Fig 6.80
Sol.: The spring is compressed from 150mm to 100mm. So it is exiting a compressive force, which is acting
vertically downward on the sphere.
Since,
Spring force(F) = K.x
Force: Concurrent Force System / 139
Given that K = 2N/mm
x = 150 – 100 = 50mm
F = 2 × 50 = 100N ...(i)
Let R
A
and R
B
be the contact reaction at Pont A and B.
Here wt of sphere and F are collinear force, both act down ward so the net force is = 100 + 40, acting
down ward.
Apply lamis theorem at point ‘O’
R
A
/sin(90° + 30°) = R
B
/sin(90° + 60°)
= 140/ sin(180° – 90°)
On solving
R
A
= 121N .......ANS
R
B
= 70N .......ANS
Q. 65: Three sphere A, B and C weighing 200N, 400N and 200N respectively and having radii 400mm,
600mm and 400mm respectively are placed in a trench as shown in fig 6.81. Treating all
contact surfaces as smooth, determine the reactions developed.
C
A
B
D
45°
4
0
0
400
600
600
R
2
R
3
R
6
A
R
1
R
4
R
5
200 N
200 N
400 N
q
4
5
°
45°
45°
45°
45°
4
5
°
Fig 6.81 Fig 6.82
Sol.: From the fig 6.81
Sinα = BD/AB = (600  400)/(400 + 600) = 0.2
α = 11.537°
Referring to FBD of sphere A (Fig a)
R
2
cosα = 200
R
2
= 200/cos11.537° = 204.1 N .......ANS
And R
1
– R
2
sinα = 0
R
1
= 40.8N .......ANS
Referring to the FBD of sphere C [Fig. 6.82(b)],
Sum of forces parallel to inclined plane = 0
R
4
cosα – 200cos45° = 0
R
4
= 144.3 N .......ANS
Sum of forces perpendicular to inclined plane = 0
R
4
cos(45 – α) – R
3
cos45° = 0
R
3
= 170.3N .......ANS
140 / Problems and Solutions in Mechanical Engineering with Concept
Referring to FBD of cylinder B (Fig. 6.82(c)]
∑V = 0
R
6
sin45° – 400 – R
2
cosα – R
4
cos (45 + α) = 0
R
6
sin 45° = 400 + 204.1 cos11.537° + 144.3cos56.537°
R
6
= 961.0 N .......ANS
∑H = 0
R
5
– R
2
sinα – R
4
sin (45 + α) – R
6
cos45° = 0
R
5
= 204.1 sin11.537 + 144.3sin56.537 + 961.0cos45°
R
5
= 840.7 N .......ANS
Force: NonConcurrent Force System / 141
+0)264
7
FCRCE. NCN  CCNCÜRRENT
FCRCE S¥STEM
Q. 1: Define Nonconcurrent force system. Why we find out the position of Resultant in Non
concurrent force system?
Sol.: In Equilibrium of concurrent force system, all forces are meet at a point of a body. But if the forces
acting on the body are not meet at a point, then the force system is called as Nonconcurrent force system.
In concurrent force system we find the resultant and its direction. Because all the forces are meet at one
point so the resultant will also pass through that point, i.e. the position of resultant is already clear. But in non
concurrent force system we find the magnitude, direction and distance of the resultant from any point of the
body because forces are not meet at single point they act on many point of the body, so we don’t know the
exact position of the resultant. For finding out the position of resultant we used the concept of moment.
Q. 2: Define Moment of a Force? What is moment center and moment arm? Also classify the moment.
Sol.: It is the turning effect produced by a force, on the body, on which it acts. The moment of a force is
equal to the product of the force and the perpendicular distance of the point about which the moment is
required, and the line of action of the force.
The force acting on a body causes linear displacement, while moment causes angular displacement.
O
L
F
P
Body
Fig. 7.1
If M = Moment
F = Force acting on the body, and
L = Perpendicular distance between the point about which the moment is required and the line of
action of the force. Then M = F.L
142 / Problems and Solutions in Mechanical Engineering with Concept
The point about which the moment is considered is called Moment Center. And the Perpendicular
distance of the point from the line of action of the force is called moment Arm.
D
1
D
2
1
2
3
Fig. 7.2
The moment is of the two types:
Clockwise moment:
It is the moment of a force, whose effect is to turn or rotate the body, in the clockwise direction. It
takes +ive.
F
r
O
M
Fig. 7.3
Anticlock wise Moment:
It is the moment of a force, whose effect is to turn or rotate the body, in the anticlockwise direction.
It take ive.
F
r O
M
Fig. 7.4
In Fig. 7.2; Moment about Point 1 = F.D
2
(Clock wise)
Moment about Point 2 = F.D
1
(Anti Clock wise)
Moment about Point 3 = 0
i.e. if point lie on the line of action of a force, the moment of the force about that point is zero.
Q. 3: How you represent moment Graphically?
Sol.: Consider a force F represented, in magnitude and direction, by
the line AB. Let ‘O’ be a point about which the moment of this force
is required to be found out.
From ‘O’ draw OC perpendicular to AB. Join OA and OB.
Now moment of the force P about O = F X OC = AB.OC
But AB.OC is equal to twice the area of the triangle ABO.
Thus the moment of a force about any point is geometrically
equal to twice the area of the triangle, whose base is the line
representing the force and whose vertex is the point, About which the
moment is taken.
Mo = 2.Area of Triangle OAB
Unit of moment = Nm
Fig. 7.5
O
A B
F c
Force: NonConcurrent Force System / 143
Q. 4: State Varignon’s theorem. How it can help on determination of moments? In what condition
is it used?
Sol.: Varignon’s theorem also called Law of Moment.
The practical application of varignon’s theorem is to find out the position of the resultant from any
point of the body.
It states “If a number of coplanar forces are acting simultaneously on a particle, the algebraic sum
of the moments of all the forces about any point is equal to the moment of their resultant force about the
same point.”
Proof: Let us consider, for the sake of simplicity, two concurrent forces P and Q represented in
magnitude and direction by AB and AC as shown in fig. 7.6.
Let ‘O’ be the point, about which the moment are taken, through O draw a line OD parallel to the
direction of force P, to meet the line of action of the force Q at C. Now with AB and AC as two adjacent
sides, complete the Parallelogram ABDC as shown in fig. 7.6. Joint the diagonal AD of the parallelogram
and OA and OB. From the parallelogram law of forces, We know that the diagonal AD represents in
magnitude and direction, the resultant of two forces P and Q. Now we see that the moment of the force
P about O: = 2. Area of the triangle AOB ...(i)
O
C
Y
Q
R
P
B
X
D
Fig. 7.6
Similarly, moment of the force Q about O: = 2. Area of the triangle AOC ...(ii)
And moment of the resultant force R about O: = 2.Area of the triangle AOD ...(iii)
But from the geometry of the fig.ure, we find that
Area of triangle AOD = Area of triangle AOC + Area of triangle ACD
But Area of triangle ACD = Area of triangle ABD = Area of triangle AOB
(Because two “AOB and ADB are on the same base AB and between the same // lines)
Now Area of triangle AOD = Area of triangle AOC + Area of triangle AOB
Multiply both side by 2 we get;
2. Area of triangle AOD = 2.Area of triangle AOC + 2. Area of triangle ACD, i.e.
Moment of force R about O = Moment of force P about O + Moment of force Q about O
or,
Where R ⋅ d = ∑M
∑M = Sum of the moment of all forces
d = Distance between the resultant force and the point where moment of all forces are taken.
This principle is extended for any number of forces.
144 / Problems and Solutions in Mechanical Engineering with Concept
Q. 5: How do you find the resultant of Non  coplanar concurrent force system?
Sol.: The resultant of nonconcurrent force system is that force, which will have the same rotational and
translation effect as the given system of forces, It may be a force, a pure moment or a force and a moment.
R = {(∑H)
2
+(∑V)
2
}
1/2
Tan θ = ∑V/∑H
∑M = R ⋅ d
Where,
∑H = Sum of all horizontal component
∑V = Sum of all vertical component
∑M = Sum of the moment of all forces
d = Distance between the resultant force and the point where moment of all forces are taken.
Q. 6: How you find the position of resultant force by moments?
Sol.: First of all, find the magnitude and direction of the resultant force by the method of resolution. Now
equate the moment of the resultant force with the algebraic sum of moments of the given system of forces
about any point or simply using Varignon’s theorem. This may also be found out by equating the sum
of clockwise moments and that of the anticlockwise moments about the point through which the resultant
force will pass.
Q. 7: Explain principle of moment.
Sol.: If there are number of coplanar nonconcurrent forces acted upon a body, then for equilibrium of the
body, the algebraic sum of moment of all these forces about a point lying in the same plane is zero.
i.e. ∑ ∑∑ ∑∑M = 0
Or we can say that,
clock wise moment = Anticlockwise moment
Q. 8: What are the equilibrium conditions for nonconcurrent force system?
Sol.: For Equilibrium of nonconcurrent forces there are three conditions:
1. Sum of all the horizontal forces is equal to zero, i.e
∑H = 0
2. Sum of all the horizontal forces is equal to zero, i.e
∑V = 0
3. Sum of the moment of all the forces about any point is equal to zero, i.e
∑M = 0
If any one of these conditions is not satisfied then the body will not be in equilibrium.
Q. 9: Define equilibrant.
Sol.: The force, which brings the set of forces in equilibrium, is called an equilibrant. As a matter of fact,
the equilibrant is equal to the resultant force in magnitude, but opposite in nature.
Q. 10: What are the cases of equilibrium?
Sol.: As the result of the acting forces, the body may have one of the following states:
(1) The body may move in any one direction:
It means that there is resultant force acting on it. A little consideration will show, that if the body is
to be at rest or in equilibrium, the resultant force causing movement must be zero. or ∑H and ∑V must be
zero.
∑ ∑∑ ∑∑H = 0 and ∑ ∑ ∑ ∑ ∑V = 0
Force: NonConcurrent Force System / 145
(2) The body may rotate about itself without moving:
It means that there is single resultant couple acting on it with no resultant force. A little consideration
will show, that if the body is to be at rest or in equilibrium, the moment of the couple causing rotation must
be zero. or
∑M = 0
(3) The body may move in any one direction, ant at the same time it may also rotate about itself:
It means that there is a resultant force and also resultant couple acting on it. A little consideration will
show, that if the body is to be at rest or in equilibrium, the resultant force causing movement and the
resultant moment of the couple causing rotation must be zero. i.e.
∑ ∑∑ ∑∑H = 0 , ∑ ∑∑ ∑∑V = 0 and ∑ ∑ ∑ ∑ ∑M = 0
(4) The body may be completely at rest:
It means that there is neither a resultant force nor a couple acting on it. A little consideration will show,
that in this case the following condition are already satisfied:
∑ ∑∑ ∑∑H = 0 , ∑ ∑∑ ∑∑V = 0 and ∑ ∑∑ ∑∑M = 0
Q. 11: Determine the resultant of four forces tangent to the circle of radius 3m shown in fig. (7.7).
What will be its location with respect to the center of the circle? (Dec–0304)
80 N
150 N
150
50 N 50
45º 45º
100 N 100
80
O
d
R
45º
Fig. 7.7 Fig. 7.8
Sol: Let resultant be ‘R’ which makes an angle of θ with the horizontal X axis. And at a distance of x from
point ‘O’. Let ∑H and ∑V be the horizontal and vertical component.
∑H = 150 – 100 cos 45º = 79.29N ...(i)
∑V = 50 – 100 sin 45º – 80 = –100.7N ...(ii)
R = {∑H
2
+ ∑V
2
}
1/2
R = {(79.29)
2
+ (100.7)
2
}
1/2
R = 128.17N .......ANS
Calculation For angle θ
tan θ = ∑V/∑H
= –100.71/79.28
θ θθ θθ = –51.78º .......ANS
Calculation For distance ‘d’
According to Varignon’s theorem, R ⋅ d = ∑M
(Taking moment about point ‘O’)
i.e. 128.17 X d = 150 X 3 –50 X 3 + 100 X 3 – 80 X 3
d = 2.808 m .......ANS
146 / Problems and Solutions in Mechanical Engineering with Concept
Q. 12: Determine the moment of the 50N force about the point A, as shown in fig. (7.9).
Sol.: Taking moment about point A,
∑M
A
= 50 cos 150º X 150 –50 sin 150º X 200
(Negative sign because, both moments are anticlockwise)
∑ ∑∑ ∑∑M
A
= –11475.19N –mm .......ANS
Hence moment about A = 11475.19N –mm (Anticlockwise)
200 mm
150 mm
150º
50 N
A
200 mm
150 mm
50 sin 150º
50 cos 150º
A
Fig. 7.9 Fig. 7.10
Q. 13: Determine the resultant of the four forces acting on the plate shown in fig. (7.11)
y
30 N
25 N
x
10
25
0
35 N
5 mm
20
5
30º
Fig. 7.11
Sol.: Let us assume R be the Resultant force is acting at an angle of θ with the horizontal. And ∑H and
∑V be the sum of horizontal and vertical components.
∑H = 25 + 35 cos 30º –30 cos 45º
= 34.09N ...(i)
∑V = 20 + 35 sin 30º +30 sin 45º
= 58.71N ...(ii)
R = {H
2
+ ∑V
2
}
1/2
R = 67.89N .......ANS
For direction of resultant
tan θ = ∑V/∑H
= 58.71/34.09
θ θθ θθ = 59.85º .......ANS
Q. 14: A beam AB (fig. 7.12) is hinged at A and supported at B by a vertical cord, which passes over
two frictionless pulleys C and D. If pulley D carries a vertical load Q, find the position x of
the load P if the beam is to remain in equilibrium in the horizontal position.
L
C
A
x
P
B
D
Q
T T
Q
L
A B
P
x
T Q = /2
Fig. 7.12 Fig. 7.13 Fig. 7.14
Force: NonConcurrent Force System / 147
Sol.: First consider the free body diagram of block Q,
From the fig. 7.13,
2T = Q, T = Q/2
i.e tension in the rope = Q/2
Now consider the F.B.D. of the beam as shown in fig. 7.14, Here two forces are acting force ‘P’ at
a distance ‘X’ from point ‘A’ and T = Q/2 at a distance ‘l’ from point ‘A’
Taking moment about point ‘A’, i.e. ∑M
A
= 0
P X x = Q/2 X l
X =
QL
2P
.......ANS
Q. 15: A uniform wheel of 600 mm diameter, weighing 5KN rests against a rigid rectangular block
of 150mm height as shown in fig. 7.15. Find the least pull, through the center of the wheel,
required just to turn the wheel over the corner A of the block. Also find the reaction of the
block. Take the entire surface to be smooth.
P
O
A
150 mm
600 mm
Fig. 7.15
P
150 mm
300 mm
B
O
A
R
5 KN
Fig. 7.16
Sol.: Let P = least pull required just to turn the wheel
Least pull must be applied normal to AO. F.B.D of wheel is shown in fig. 7.16, from the fig.,
sin θ = 150/300, θ = 30º
AB = {(300)
2
– (150)
2
}
1/2
= 260 mm
Now taking moment about point A, considering body is in equilibrium
P X 300 –5 X 260 = 0
P = 4.33 KN .......ANS
148 / Problems and Solutions in Mechanical Engineering with Concept
Calculation for reaction of the block
Let R = Reaction of the block
Since body is in equilibrium, resolving all the force in horizontal direction and equate to zero,
R cos 30º – P sin 30º = 0
R = 2.5KN .......ANS
Q. 16: In the fig. (7.17) assuming clockwise moment as positive, compute the moment of force F =
4.5 KN and of force P = 3.61 KN about points A, B, C and D. Each block is of 1m
2
.
A F
C
B D
P
2
1
Fig. 7.17
Sol.: Here tan θ
1
= 3/4 ⇒ θ
1
= 36.86º
tan θ
2
= 3/2 ⇒ θ
2
= 56.3º
First we find the moment of force F about points A,B,C and D
F = 4.5KN
(1) About point A:
M
A
= –F cos 36.86º X 3 – F sin 36.86º X 1
= –13.50 KN –m .......ANS
(2) About point B:
M
B
= F cos 36.86º X 3 + F sin 36.86º X 4
= 21.59KN–m .......ANS
(3) About point C:
M
C
= F cos 36.86º X 0 – F sin 36.86º X 5
= 13.49 KN–m .......ANS
(4) About point D:
M
D
= F cos 36.86º X 3 – F sin 36.86º X 1
= 8.10KN–m .......ANS
Now we find the moment of force P about points A,B,C and D
P = 3.61KN
(1) About point A:
M
A
= –P cos56.3º X 3 + P sin 56.3º X 2
= 0.002KN–m .......ANS
(2) About point B:
M
B
= P cos 56.3º X 3 – P sin 56.3º X 3
= –7.007KN–m .......ANS
Force: NonConcurrent Force System / 149
(3) About point C:
M
C
= –P cos56.3º X 0 –P sin 56.3º X 4
= –12.0134KN–m .......ANS
(4) About point D:
M
D
= –P cos 56.3º X 3 –P sin 56.3º X 2
= –11.998 KN–m .......ANS
Q. 17: A uniform wheel of 60 cm diameter weighing 1000 N rests against rectangular obstacle
15 cm high. Find the least force required which when acting through center of the wheel will
just turn the wheel over the corner of the block. Find the angle of force with horizontal.
P
min
R
B
B
C
O
W
D
15 cm
15 cm
30 cm
B
Fig. 7.18
Sol.: Let,
P
min
= Least force applied as shown in fig. 7.18
α = Angle of the least force
From triangle OBC, BC = BOsinα
BC = 30sinα
In Triangle BOD, BD = {(BO)
2
– (OD)
2
}
1/2
BD = (30
2
– 15
2
)
1/2
= 25.98
Taking moment of all forces about point B, We get
P
min
X BC – W X BD = 0
P
min
– W X BD/BC
P
min
= 1000 X 25.98 /30sinα
We get minimum value of P when α is maximum and maximum value of α is at 90º
i.e. 1, putting sinα =1
P
min
= 866.02N .......ANS
Q. 18: A system of forces is acting at the corner of a rectangular block as shown in fig. 7.19. Determine
magnitude and direction of resultant.
Sol.: Let R be the resultant of the given system. And ∑H and ∑V be the horizontal and vertical component
of the resultant.
∑H = 25 – 20 = 5KN ...(i)
∑V = –50 – 35 = – 85KN ...(ii)
150 / Problems and Solutions in Mechanical Engineering with Concept
R
2
= ∑H
2
+ ∑V
2
R
2
= (5)
2
+ (85)
2
R = 85.14N .......ANS
Let Resultant makes an angle of ¸ with the horizontal
tan θ = ∑V/∑H
= 85/5
B
C
D
d
A
R
4 m
20 N
35 N
25 N
50 N
Fig. 7.19
θ θθ θθ = – 86.63º .......ANS
Let resultant ‘R’ is at a perpendicular distance ‘d’ from point A,
For finding the position of the resultant i.e. ‘d’, taking moment about point ‘A’., or apply varignon’s
theorem
R.d = 25 X 3 + 35 X 4
d = (75 + 140)/85.14
d = 2.53 m from point A .......ANS
Q. 19: Find the magnitude and direction of resultant of Coplanar forces shown in fig. 7.20.
(Dec–0001)
20 KN
B C
45º
10 2 KN
D
A
10 KN
20 cm
10 KN
2
0
c
m
Fig. 7.20
Sol.: Using the equation of equilibrium,
∑H = –20 + 10 + 10√2 cos45º
∑H = 0 ...(i)
∑V = –10 + 10√2 sin 45º
∑V = 0 ... (ii)
Since ∑H and ∑V both are zero, but in non concurrent forces system, the body is in equilibrium when
∑H = ∑V = ∑M = 0
So first we check the value of ∑M, if it is zero then body is in equilibrium, and if not then that moment
is the resultant.
Taking moment about point A,
Force: NonConcurrent Force System / 151
∑M
C
= 0
= –10 × 20 –10 × 20 = – 400 KN–cm,
Since moment is not zero i.e. Body is not in equilibrium, Hence the answer is
M = 400 KN–cm (Anticlockwise)
Q. 20: Three similar uniform slabs each of length ‘2a’ are resting on the edge of the table as shown
in fig. 7.21. If each slab is overhung by maximum possible amount, find amount by which the
bottom slab is overhanging. (Dec–0001)
2a
2a
A1
A2
A3
a
a
v
a
w
w w
×
x
Fig. 7.21 Fig. 7.22
Sol.: The maximum overhang of top beam is ‘a’,
Now taking moment about point A
2
, considering all load acting on middle beam.
–W(a – X) + W.X = 0
on solving X = a/2 ...(i)
Now taking moment about point A
3
–W(a – Y) + W[Y – (a – X)] + W(X + Y) = 0
–Wa + WY + WY – Wa + WX + WX + WY = 0
3Y – 2a + 2X = 0
Y = a/3 .......ANS
Since bottom beam overhang by a/3 amount
Q. 21: Determine the resultant of force system acting tangential to the circle of radius 1m as shown
in fig. 7.23. Also find its direction and line of action (May–0001)
120 N
50 N
150 N
80 N
0
Fig. 7.23
Sol.: ∑H = 120 – 150 = –30N ...(i)
∑V = 50 – 80 = –30N ...(ii)
R = (∑H
2
+ ∑V
2
)
1/2
R = ((–30)
2
+ (–30)
2
)
1/2
R = 42.43 N .......ANS
152 / Problems and Solutions in Mechanical Engineering with Concept
tan θ = 30/30
θ θθ θθ = 45º .......ANS
Now for finding the position of the resultant Let the perpendicular distance of the resultant from center
‘O’ be ‘d’.
Apply varignon’s theorem, taking moment about point O.
R.d = –80 X 1 + 150 X 1 + 120 X 1 – 50 X 1
42.42 X d = –80 X 1 + 150 X 1 + 120 X 1 – 50 X 1
d = 3.3m .......ANS
Q. 22: A vertical pole is anchored in a cement foundation. Three wires are attached to the pole as
shown in fig. 7.24. If the reaction at the point. A consist of an upward vertical of 5000 N and
a moment of 10,000 Nm as shown, find the tension in wire. (May 0001(B.P.))
B
T1
A
10000 N.m
5000 N
T2
T3
45º
4.5 m
1.5 m
45º
30º
Fig. 7.24
Sol.: Resolve all the forces in horizontal and vertical direction. From the condition of equilibrium
Taking moment about point B, We get
T
3
sin 30º X 4.5 –10000 = 0
T
3
= 4444.44N .......ANS
∑H= 0
T
3
sin 30º + T
2
cos 45º – T
1
sin 60º = 0
2222.22 + 0.707 T
2
– 0.866 T
1
= 0 ...(i)
∑V= 0
T
3
cos 30º + 5000 – T
2
sin 45º – T
1
cos 60º = 0
8849 – 0.707T
2
– 0.5T
1
= 0 ...(ii)
From equation (i) and (ii)
T
1
= 8104.84N and T
2
= 6783.44N ......ANS
Q. 23: A man raises a 10 Kg joist of length 4m by pulling on a rope, Find the tension T in the rope
and reaction at A for the position shown in fig. 7.25. (May 0001(B.P.))
Force: NonConcurrent Force System / 153
T
B
B
T
A
E
C
45º
25º
20º
D
RAH
RAV
45º W = 10 kg
A
45º
25º
Fig. 7.25 Fig. 7.26
Sol.: Apply condition of equilibrium
∑H = 0 R
AH
– T cos 20º = 0 R
AH
= T cos 20º ...(i)
∑V = 0 R
AV
– 10 – T sin 20º = 0 R
AV
= 10 + T sin 20º ...(ii)
Now taking moment about point A
T sin 20º × AC + 10 sin 45º × AE – Tcos20 × BC = 0,
AC = 4 cos 45º = 2.83 m
BC = 4 sin45º = 2.83 m
AE = 2 cos 45º = 1.41 m
T × 0.34 × 2.83 + 10 × 0.71 × 1.41 – T × 0.94 × 2.83 = 0,
0.9622 T + 10.011 – 2.66 T = 0
T = 5.9 Kg .......ANS
Putting the value of T in equation (i) and (ii)
R
AH
= 5.54 Kg .......ANS
R
AV
= 15.89 Kg .......ANS
Q. 24: The 12m boom AB weight 1 KN, the distance of the center of gravity G being 6 m from A.
For the position shown, determine the tension T in the cable and the reaction at B.
(Dec–0304)
A
G
B
15º
2.5 KN
1 KN
HA
A
T
VA
30º
1 KN
2.5 KN
15º
15º
6
m
6
m
B
30º
Fig. 7.27 Fig. 7.28
Sol.: The free body diagram of the boom is shown in fig. 7.28
∑M
A
= 0
T sin 15º X 12 – 2.5 X 12 cos 30º – 1 X 6 cos 30º = 0
T = 10.0382 KN .......ANS
Reaction at B = (2.5
2
+ 10
2
+ 10 X 2.5 X cos 75º)
1/2
R
B
= 10.61 KN .......ANS
154 / Problems and Solutions in Mechanical Engineering with Concept
Q. 25: Define and classified parallel forces?
Sol.: The forces, whose lines of action are parallel to each other, are known as parallel forces. They do not
meet at one point (i.e. Nonconcurrent force). The parallel forces may be broadly classified into the following
two categories, depending their direction.
There are two types of parallel force
1. LIKE PARALLEL FORCES
The forces whose lines of action are parallel to each other and all of them act in the same direction
are known as like parallel forces.
2. UNLIKE PARALLEL FORCES
The forces whose lines of actions are parallel to each other, and all of them do not act in the same
direction are known as unlike parallel forces.
Q. 26: A horizontal line PQRS is 12 m long, where PQ = QR = RS = 4m. Forces of 1000, 1500, 1000
and 500 N act at P, Q, R and S respectively with downward direction. The lines of action of
these make angle of 90º, 60º, 45º and 30º respectively with PS. Find the magnitude, direction
and position of the resultant force.
1000 N
P
Q R S
4 m 4 m 4 m
1500 N 1000 N
500 N
90º
60º 45º
30º
Fig. 7.29
The system of given forces is shown in fig. 7.29
Let R be the resultant of the given system. And R
H
and R
V
be the horizontal and vertical component
of the resultant.
Resolving all the forces horizontally
∑H = –1000 cos 90º – 1500 cos 60º – 1000 cos 45º – 500 cos 30º
∑H = –1890 N ...(i)
Resolving all the forces vertically
∑V = –1000 sin 90º – 1500 sin 60º – 1000 sin 45º – 500 sin 30º
∑V = –3256N ...(ii)
Since, R = √(∑H)
2
+(∑V)
2
R = √(1890)
2
+ (3256)
2
R = 3764N .......ANS
Let θ = Angle makes by the resultant
tan θ = ∑V/∑H
= 3256/1890 ⇒ θ = 59.86º
For position of the resultant
Let, d = Distance between P and the line of action of the resultant force.
Apply varignon’s theorem
R.d = 1000 sin 90º × 0 + 1500 sin 60º
× 4 + 1000 sin 45º × 8 + 500 sin 30º × 12
3256.d = 13852
d = 3.67 m .......ANS
Force: NonConcurrent Force System / 155
Q. 27: Replace the two parallel forces acting on the control lever by a single equivalent force R.
Sol.: Since single equivalent force is resultant.
Let ∑H and ∑V be the horizontal and vertical component of the resultant.
Resolving all the forces horizontally
∑H = 50 – 80 = –30N ...(i)
Since there is no vertical force i.e. the resultant is horizontal. Now for finding out
the point of application of resultant, Let resultant is at a distance of ‘d’ from point ‘O’.
Apply varignon’s theorem, and taking moment about point ‘O’
R.d = 50 × 80 – 80 × 50 = 0
But R = ∑H = –30N, so d = 0
d = 0, means point of application of resultant is ‘O’
Hence an equivalent force 30N acts in –ive xaxis at point ‘O’ which replace the
given force system.
Q. 28: A system of loads acting on a beam is shown in fig. 7.31. Determine the resultant of the loads.
Sol.: Let R be the resultant of the given system. And ∑H and ∑V be the horizontal and vertical component
of the resultant. And resultant makes an angle of θ with the horizontal.
Resolving all the forces horizontally
∑H = 20 cos 60º
∑H = 10KN ...(i)
Resolving all the forces vertically
A B
d
2 m 2 m
20 kN
R
20 kN
30 kN
3 m 2 m
60º
Fig. 7.31
∑V = 20 + 30 + 20 sin 60º
∑V = 67.32KN ...(ii)
Since, R = √(∑H)
2
+ (∑V)
2
⇒ √(10)
2
+ (67.32)
2
R = 68.05KN .......ANS
Let θ = Angle makes by the resultant
tan θ = ∑V/∑H
= 67.32/10 ⇒ θ = 81.55º
For position of the resultant
Let, d = Distance between Point A and the line of action of the resultant force.
Apply varignon’s theorem
R.d = 20 × 2 + 30 × 4 + 20 sin 30º × 7
68.05.d = 281.2
d = 4.132 m .......ANS
Fig. 7.30
50 N
30
50
80 N
O
156 / Problems and Solutions in Mechanical Engineering with Concept
Q. 29: Define couple and Arm of couple?
Sol.: If two equal and opposite parallel forces (i.e. equal and unlike) are acting on a body, they don’t have
any resultant force. That is no single force can replace two equal and opposite forces, whose line of action
are different. Such a set of two equal and opposite forces, whose line of action are different, form a couple.
Thus a couple is unable to produce any translatory motion (motion in a straight line). But a couple
produce rotation in the body on which it acts.
Arm of Couple
The perpendicular distance (d) between the lines of action of the two equal and opposite parallel forces,
is known as arm of couple.
F
F
d
Fig. 7.32
Q. 30: Define different types of couple?
Sol.: There are two types of couples:
1. Clockwise Couple
A couple whose tendency is to rotate the body on which it acts, in a clockwise direction, is known as a
clockwise couple. Such a couple is also called positive couple.
F
F
d
Fig. 7.33
2. Anticlockwise Couple
A couple whose tendency is to rotate the body on which it acts, in a anticlockwise direction, is known as
a anticlockwise couple. Such a couple is also called Negative couple.
F
F
d
Fig. 7.34
Q 31: What is the moment of a couple?
Sol.: The moment of a couple is the product of the force (i.e. one of the forces of the two equal and opposite
parallel forces) and the arm of the couple.
Force: NonConcurrent Force System / 157
Mathematically:
Moment of a couple = F.d Nm or Nmm
The moment of couple may be clockwise or anticlockwise.
The effect of the couple is unchanged if::
1. The couple is shifted to any other position.
2. The couple is rotated by an angle.
3. Any pair of force whose rotation effect is the same replaces the couple.
4. Sum of forces forming couple in any direction is zero.
Q. 32: What are the main characteristics of couple?
Sol.: The main characteristics of a couple
1. The algebraic sum of the forces, consisting the couple, is zero.
2. The algebraic sum of the moment of the forces, constituting the couple, about any point is the same,
and equal to the moment of the couple itself.
3. A couple cannot be balanced by a single force, but can be balanced only by a couple, but of
opposite sense.
4. Any number of coplanar couples can be reduced to a single couple, whose magnitude will be equal
to the algebraic sum of the moments of all the couples.
Q. 33: Define magnitude of a couple.
Sol.: For a system, magnitude of a couple is equal to the algebraic sum of the moment about any point
If the system is reduces to a couple, the resultant force is zero, (i.e. ∑ ∑∑ ∑∑H = ∑ ∑∑ ∑∑V = 0) but ∑M ≠ 0, i.e.
the moment of the force system is the resultant.
Q. 34: A rectangle ABCD has sides AB = CD = 80 mm and BC = DA = 60 mm. Forces of 150 N each
act along AB and CD, and forces of 100 N each act along BC and DA. Make calculations for
the resultant of the force system.
150 N
100 N
D
A
C
B
80 mm
100 N
150 N
60 N
Fig. 7.35
Sol.: Let R be the resultant of the given system. And ∑H and ∑V be the horizontal and vertical component
of the resultant. And resultant makes an angle of θ with the horizontal.
Resolving all the forces horizontally
∑H = 150 – 150
∑H = 0KN ...(i)
Resolving all the forces vertically
∑V = 100 – 100
158 / Problems and Solutions in Mechanical Engineering with Concept
∑V = 0KN ...(ii)
Since ∑H and ∑V both are 0, then resultant of the system is also zero.
But in Nonconcurrent forces system, the resultant of the system may be a force, a couple or a force
and a couple
i.e. in this case if couple is not zero then couple is the resultant of the force system.
For finding Couple, taking moment about any point say point ‘A’.
M
A
= –150 × 60 – 100 × 80, both are anticlockwise
Then, Resultant moment = couple = –17000N–mm .......ANS
Q. 35: A square block of each side 1.5 m is acted upon by a system of forces along its sides as shown
in the adjoining fig.ure. If the system reduces to a couple, determine the magnitude of the
forces P and Q, and the couple.
y
D
Q
P
C
B
x
A
45º
150 N
150 N
300 N
Fig. 7.36
Sol.: If the system is reduces to a couple, the resultant force is zero,
(i.e. ∑ ∑∑ ∑∑H = ∑ ∑∑ ∑∑V = 0) but ∑ ∑∑ ∑∑M ≠ ≠ ≠ ≠ ≠ 0,
i.e. the moment of the force system or couple is the resultant of the force system.
∑H = 150 – 150 cos 45º – P = 0
P = 43.95N .......ANS
∑V = 300 – 150 sin 45º – Q = 0
Q = 193.95N .......ANS
Now moment of couple = Algebraic sum of the moment of forces about any corner, say A
= –300 × 1.5 – 43.95 × 1.5 = –515.925 Nm .......ANS
ive means moment is anticlockwise.
Q. 36: Resolve a force system in to a single force and a couple system. Also explain Equivalent force
couple system.
Or
‘Any system of coplaner forces can be reduced to a force – couple system at an arbitrary
point’. Explain the above statement by assuming a suitable system
Force: NonConcurrent Force System / 159
Sol.: A given force ‘F’ applied to a body at any point A can always is replaced by an equal force applied
at another point B together with a couple which will be equivalent to the original force.
Let us given force F is acting at point ‘A’ as shown in fig. (7.37).
This force is to be replaced at the point ‘B’. Introduce two equal and opposite forces at B, each of
magnitude F and acting parallel to the force at A as shown in fig. (7.38). The force system of fig. (7.38)
is equivalent to the single force acting at A of fig. (7.37). In fig. (7.38) three equal forces are acting. The
two forces i.e. force F at A and the oppositely directed force F at B (i.e. vertically downwards force at B)
from a couple. The moment of this couple is F × x clockwise where x is the perpendicular distance between
the lines of action of forces at A and B. The third force is acting at B in the same direction in which the
force at A is acting.
F
F F
F
M F X = .
A
A
X
B
B
B A
M
F
Fig. 7.37 Fig. 7.38 Fig. 10.39
In fig. (7.39), the couple is shown by curved arrow with symbol M. The force system of fig. (7.39)
is equivalent to fig. (7.38). Or in other words the Fig. (7.39) is equivalent to Fig. (7.37). Hence the given
force F acting at A has been replaced by an equal and parallel force applied at point B in the same direction
together with a couple of moment F × x.
Thus force acting at a point in a rigid body can be replaced by an equal and parallel force at any other
point in the body, and a couple.
Equivalent force System
An equivalent system for a given system of coplanar forces is a combination of a force passing through
a given point and a moment about that point. The force is the resultant of all forces acting on the body.
And the moment is the sum of all the moments about that point.
Hence equivalent system consists of:
1. A single force R passing through the given point, and
2. A single moment (∑M)
Where,
R = the resultant of all force acting on the body
∑M = Sum of all moments of all the forces about point P.
Q. 37: In designing the lifting hook, the forces acting on a horizontal section through B may be
determined by replacing F by a equivalent force at B and a couple. If the couple is 3000 N
mm, determine F. Fig. (7.40).
160 / Problems and Solutions in Mechanical Engineering with Concept
B B F
F
E E
F F
40 mm 40 mm
Fig. 7.40 Fig. 7.41
Sol.: Force ‘F’ is replaced at point B, by a single force ‘F’ and a single couple of magnitude 3000 Nmm.
Now apply two equal and opposite force i.e. ‘F’ at point B. as shown in Fig. 7.41. Now force ‘F’ which
is act at point E and upward force which is act at point B makes a couple of magnitude = Force × distance
= F × 40
But 40F = 3000 i.e.
F = 75 N ........ANS
Q. 38: Two parallel forces are acting at point A and B respectively are equivalent to a force of
100 N acting downwards at C and couple of 200Nm. Find the magnitude and sense of force
F
1
and F
2
shown in Fig. 7.42.
100 N
F
1
F
2
C B
4 m 3 m
A
d
R R
E
C
R
200 Nm
Fig. 7.42 Fig. 7.43
Sol.: The given system is converts to a single force and a single couple at C. Let R be the resultant of
F
1
and F
2
.
R
H
= 0
R
V
= –F
1
–F
2
R
V
= –(F
1
+ F
2
) ...(i)
Since R = (R
2
V
)
1/2
= (F
1
+ F
2
)
Let resultant R act at a distance ‘d’ from the point C.
Now the single force i.e. R is converted in to a single force and a couple at C
Now apply two equal and opposite force i.e. ‘R’ at point C. as shown in Fig. 7.43. Now force ‘R’ which
is act at point E and upward force which is act at point C makes a couple of magnitude = Force × distance
= R × d
But R.d = 200 Nm
And single force R which is downward direction = 100 N (ive for downward)
i.e (F
1
+ F
2
) = 100 or F
1
+ F
2
= 100 ...(ii)
Now taking moment about point C., or apply varignon’s theorem.
R.d = 4F
1
+ 7F
2
, but R.d = 200,
Force: NonConcurrent Force System / 161
4F
1
+ 7F
2
= 200 ...(iii)
solving equation (ii)and (iii)
F
1
= 500/3 N .......ANS
F
2
= –200/3 N .......ANS
Q. 39: A system of parallel forces is acting on a rigid bar as shown in Fig. 7.44. Reduce this system
to
(i) a single force
(ii) A single force and a couple at A
(iii) A single force and a couple at B.
32.5 N
1 m 1 m 1.5 m
3.5 m
150 N 67.5 N 10 N
A
C D B
Fig. 7.44
Sol.: (i) A single force: A single force means just to find out the resultant of the system.
Since there are parallel force i.e resultant is sum of vertical forces,
R = 32.5 –150 + 67.5 –10 = 60
R = (∑V
2
)
1/2
R = 60N (downward) .......ANS
Let d = Distance of resultant from A towards right.
To find out location of resultant apply varignon’s theorem :
R.d = 150 × 1 – 67.5 × 2 + 10 × 3.5
60.d = 150 × 1 – 67.5 × 2 + 10 × 3.5
d = 0.833m
i.e resultant is at a distance of 0.83 m from A.
(ii) A single force and a couple at A: It means the whole system is to convert in to a single force
and a single couple.
Since we convert all forces in to a single force i.e. resultant.
Now apply two equal and opposite force i.e. ‘R’ at point A. Now force ‘R’ which is act at point E and
upward force which is act at point A makes a couple of magnitude,
Magnitude = Force × distance = 60 × 0.833
R = 60 N
0.83 m
A E B
Fig. 7.45
= 49.98 Nm .......ANS and a single force of magnitude = 60N ........ANS
162 / Problems and Solutions in Mechanical Engineering with Concept
(iii) A single force and a couple at B: Since AE = 0.833 m, then BE = 3.5 – 0.833 = 2.67 m
Now, the force R = –60N is moved to the point B, by a single force R = –60N
and a couple of magnitude = R × BE = –60 × 2.67 = 160 Nm
R = 60 N 60 N
0.83 m
3.5 m
A
E B
B A
60 N
M
A
= 60 × 0.833 Nm
Fig. 7.46 Fig. 7.47
B
M
B
R = 60 N 60 N
60 N
60 N
0.83 m 2.667 m
A
E B
Fig. 7.49 Fig. 7.50
Hence single force is 60 N and couple is 160 Nm
Q. 40: The two forces shown in Fig. (7.51), are to be replaced by an equivalent force R applied at
the point P. Locate P by finding its distance x from AB and specify the magnitude of R and
the angle O it makes with the horizontal.
O
A B
P
X
150 mm
180 mm
1500 N
1000 N
50
mm
30º
Fig. 7.51
Sol.: Let us assume the equivalent force R (Resultant force) is acting at an angle of θ with the horizontal.
And ∑H and ∑V be the sum of horizontal and vertical components.
∑H = –1500 cos 30º
= –1299N ...(i)
∑V = 1000 –1500 sin 30º
= 250 N ...(ii)
R = {∑H
2
+ ∑V
2
}
1/2
Force: NonConcurrent Force System / 163
R = 1322.87 N .......ANS
For direction of resultant
tan θ = ∑V/∑H
= 250/–1299
θ = –10.89º .......ANS
Now for finding the position of the resultant, we use Varignon’s theorem,
i.e R × d = ∑M , Take moment about point ‘O’
1322.878 × x = 1500 cos 30º × 180 + 1500 sin 30º × 50 –1000 × 200
on solving x = 53.92 mm .......ANS
Q. 41: Fig. 7.52 shows two vertical forces and a couple of moment 2000 Nm acting on a horizontal
rod, which is fixed at end A.
1. Determine the resultant of the system
2. Determine an equivalent system through A. (May 0001(B.P.))
0.8 m
A
C D B
4000 N
2000 N – M
1 m
1.5 m
2500 N
Fig. 7.52
Sol.: (i) Resultant of the system
∑V = –4000 + 2500 = –1500N
R = (∑V
2
)
1/2
= 1500N (acting downwards)
for finding the position of the resultant, apply varignon’s theorem
i.e Moment of resultant = sum of moment of all the forces about any point.
Let from point be A, and distance of resultant is ‘d’ m from A
R.d = 4000 × 1 + 2000 – 2500 × 2.5
1500 × d = 250 ⇒ d = 0.166 m from point A
(ii) Equivalent system through A
Equivalent system consist of:
1. A single force R passing through the given point, and
2. A single moment (∑M)
Where,
R = the resultant of all force acting on the body
∑M = Sum of all moments of all the forces about point A.
Hence single force is = 1500 N; And couple = 250 Nm
Q. 42: A rigid body is subjected to a system of parallel forces as shown in Fig. 7.53. Reduce this
system to,
(i) A single force system
(ii) A single force moment system at B (May–0102)
164 / Problems and Solutions in Mechanical Engineering with Concept
A B
D C
0.4 m
15 N 60 N 10 N 25 N
0.3 m 0.7 m
Fig. 7.53
Sol.: It is the equivalent force system
R = 15 – 60 + 10 –25 = –60 N
(Acting downward)
Now taking moment about point A, apply varignon’s theorem
R.X = 60 × 0.4 – 10 × 0.7 + 25 × 1.4
60.X = 52, × = 0.867 m
Where X is the distance of resultant from point A.
(1) A single force be 60 N acting downward
(2) Now a force of 60 N = A force of 60 N (down) at B
and anticlockwise moment of 60 × (1.4 – 0.866) = 31.98 Nm at point B.
60 N force and 31.98 Nm moment anticlockwise .......ANS
Q. 43: A rigid bar CD is subjected to a system of parallel forces as shown in Fig. 7.54. Reduce the
given system of force to an equivalent force couple system at F. (Dec–0304)
C E F
D
30 kN
1 m 2 m 2 m
80 kN
60 kN
40 kN
Fig. 7.54
30 kN
1/3m
30 kN
30 KN 10 KNM
K F
F
Fig. 7.55 Fig. 7.56
Sol.: First find the magnitude and point of application of the resultant of the system, Let R be the resultant
of the given system. And ∑H and ∑V be the horizontal and vertical component of the resultant.
∑H = 0, because no horizontal force
∑V = 30 + 60 – 80 – 40
⇒ –30KN (ive indicate down ward force.)
Since, R = √(∑H)
2
+ (∑V)
2
⇒ √(0)
2
+ (–30)
2
R = 30KN (Downwards) .......ANS
For position of the resultant
Let, d = Distance between Point F and the line of action of the resultant force.
Apply varignon’s theorem, take moment about point ‘F’
Force: NonConcurrent Force System / 165
R.d = 30 × 3 – 80 × 2 + 40 × 2
30.d = 10
d = 1/3m
Now it means resultant is acting at a distance of 1/3m from point F. Now the whole system is converted
to a single force i.e. resultant, which is act at a point ‘K’. Now apply two equal and opposite forces at point
F. as shown in Fig. 7.55. Now resultant force which is act at point K and upward force which is act at point
F makes a couple of magnitude = Force × distance
= 30 × 1/3 = 10KN–m (clockwise)
So two force replace by a couple at point F.
Now the system contains a single force of magnitude 30 KN and a couple of magnitude 10 KN–m.
Q. 44: What force and moment is transmitted to the supporting wall at A? (Refer Fig. 7.57)
5 kN/m
10 kN/m
1 m 1.5 m
15 kN 0.5 m
A B
1.5 m
Fig. 7.57
∑H = 0
∑V = – 5 × 1.5 + 15 +
1
2
× 1.5 × 10
= 15 kN
M
A
= 1.5 × 5 × 0.75 – 15 × 2 –
1
2
× 1.5 × 10 × (2.5 – 1.0)
= – 35.625 kNm
A force of 15 KN (vertical) is transmitted to the wall along with an anticlockwise moment of
35.625 kNm.
166 / Problems and Solutions in Mechanical Engineering with Concept
+0)264
8
FCRCE . SÜPPCRT RE/CTlCN
Q. 1: Define a beam. What are the different types of beams and different types of loading?
(Dec–05)
Sol.: A beam may be defined as a structural element which has one dimension (length) considerable larger
compared to the other two direction i.e. breath and depth and is supported at a few points. It is usually
loaded in vertical direction. Due to applied loads reactions develop at supports. The system of forces
consisting of applied loads and reaction keep the beam in equilibrium.
Types of Beam
There are mainly three types of beam:
1. Simply supported beam
2. Over hang beam
3. Cantilever beam
1. Simply Supported Beam : The beam on which the both ends are simply supported, either by point
load or hinged or roller support.
P
1 P
2
R
2
R
1
Fig 8.1
Fig 8.2
Force: Support Reaction / 167
2. Over–Hanging Beam: The beam on which one end or both ends are overhang (or free to air.) are
called overhanging beam.
Fig 8.3
3. Cantilever Beam: If a beam is fixed at one end and is free at the other end, it is called cantilever
beam, In cantilever beam at fixed end, there are three support reaction a horizontal reaction (R
H
), a vertical
reaction(R
V
), and moment(M)
Fig 8.4
Types of Loading
Mainly three types of load acting on any beam;
1. Concentrated load
2. Uniformly distributed load
3. Uniformly varying load
1. Concentrated load (or point load): If a load is acting on a beam over a very small length. It is
called point load.
R
B
B
W
2
L
L
2
L
1
R
A
A
W
1
Fig 8.5
2. Uniformly Distributed Load: For finding reaction, this load may be assumed as total load acting
at the center of gravity of the loading (Middle point).
R
B
R
A
B
A
Fig 8.6
168 / Problems and Solutions in Mechanical Engineering with Concept
Fig 8.7
3. Uniformly Varying Load: In the diagram load varying from Point A to point C. Its intensity is zero
at A and 900N/M at C. Here total load is represented by area of triangle and the centroid of the triangle
represents the center of gravity.
Thus total load =
1
2
⋅ AB ⋅ BC
And C.G. =
1
3
⋅ AB meter from B.
=
2
3
⋅
AB meter from A. Fig 8.8
Q. 2: Explain support reaction? What are the different types of support and their reactions?
Sol.: When a number of forces are acting on a body, and the body is supported on another body, then the
second body exerts a force known as reaction on the first body at the points of contact so that the first body
is in equilibrium. The second body is known as support and the force exerted by the second body on the
first body is known as support reaction.
There are three types of support;
1. Roller support
2. Hinged Support
3. Fixed Support
1. Roller Support: Beams end is supported on rollers. Reaction is at right angle. Roller can be treated
as frictionless. At roller support only one vertical reaction.
90°
R
V
Fig 8.10
Fig 8.9 Fig. 8.10
900 N/m
9m
C
B
A
Force: Support Reaction / 169
2. Hinged (Pin) Support: At a hinged end, a beam cannot move in any direction support will not
develop any resisting moment, but it can develop reaction in any direction.
In hinged support, there are two reaction is acting, one is vertical and another is horizontal. i.e.,
R
H
and R
V
A
R
R
V
R
H
Fig 8.11
3. Fixed Support: At such support the beam end is not free to translate or rotate at fixed end there
are three reaction a horizontal reaction (R
H
), a vertical reaction(R
V
), and moment(M)
R
V
R
MA
A
R
H
Fig 8.12
14.3.5 Rocker Support: Only one reaction i.e., R
H
Q. 3: Determine algebraically the reaction on the beam loaded as shown in fig 8.13. Neglect the
thickness and mass of the beam.
4 m 7 m 4 m
2 m
40KN
80°
60° 45°
30KN
10KN
20KN
Fig 8.13
170 / Problems and Solutions in Mechanical Engineering with Concept
Sol.: Resolved all the forces in horizontal and vertical direction.
Let reaction at hinged i.e., point A is R
AH
and R
AV
, and reaction at roller support is R
BV
Let ∑H & ∑V is the sum of horizontal and vertical component of the
forces ,The supported beam is in equilibrium, hence
∑H = ∑H = 0
∑H = R
AH
– 20cos60° + 30cos45° – 40cos80° = 0
R
AH
= – 4.26 KN ...(i)
∑V = R
AV
– 10 – 20sin60° – 30sin45° – 40sin80° + R
BV
= 0
R
AV
+ R
BV
=87.92 KN ...(ii)
Taking moment about point A
10 × 2 + 20sin60° × 6 + 30sin45° × 13 – 40sin80° × 17 – R
BV
× 17= 0
R
BV
= 62.9 KN ...(iii)
Putting the value of R
BV
in equation (ii)
R
AV
= 25.02 KN
Hence R
AH
= – 4.26KN, R
AV
= 25.02KN, R
BV
= 62.9KN .......ANS
Q. 4: A light rod AD is supported by frictionless pegs at B and C and rests against a frictionless wall
at A as shown in fig 8.14 . A force of 100N is applied at end D. Determine the reaction at A,
B and C.
100 N
100 N
0
.
2
m
0
.
2
m
0
.
2
m
9
0
°
R
C
R
B
R
A
A
C
D
A
B
Fig 8.14 Fig 8.15
Sol.: Since roller support at point B, C, so only vertical reactions are there say R
B
, R
C
. At point A rod is
in contact with the wall that is wall give a contact reaction to the rod say R
A
.
Let rod is inclined at an angle of θ. Rod is in equilibrium position.
∑V = 0
R
B
cosθ – R
C
cosθ + 100cosθ = 0
R
C
– R
B
= 100 ...(i)
Taking moment about point A:
∑M
A
= 100 × 0.6 – RC × 0.4 + RB × 0.2 = 0
2R
C
– R
B
= 300 ...(ii)
Solving equation (i) and (ii)
R
C
= 200 .......ANS
R
B
= 100 .......ANS
∑H = 0
Force: Support Reaction / 171
R
A
+ R
B
sinθ – RCsinθ + 100sinθ = 0
R
A
+ 100sinθ – 200sinθ +100sinθ = 0
RA = 0 .......ANS
Q. 5: Find the reaction at the support as shown in fig 8.16.
5 m
5 m
3 m
3 m
5 kN
5 kN
10 kN
10 kN
10 kN
10 kN
10 kN
10 kN
R
HA
R
VA
R
VB
B A
Fig 8.16 Fig 8.17
Sol.: First draw the FBD of the system as shown in fig 8.17.
Since hinged at point A and Roller at point B. let at point A R
AH
and R
AV
and at point B R
BV
is the
support reaction.
∑H = 0
R
AH
–5 = 0
R
AH
= 5KN .......ANS
∑V = 0
R
AV
+ R
BV
– 10 – 10 –10 = 0
R
AV
+ R
BV
= 30KN ...(i)
Taking moment about point A:
∑M
A
= 10 × 5 + 10 × 10 – 5 × 6 – R
VB
× 5 = 0
R
BV
= 24KN .......ANS
Putting the value of R
BV
in equation (i)
R
AV
= 6KN .......ANS
Q. 6: A fixed crane of 1000Kg mass is to lift 2400Kg crates. It is held in place by a pin at A and
a rocker at B. the C.G. is located at G. Determine the components of reaction at A and B after
drawing the free body diagram.
2400 kg
1000kg
2400 kg
4 m
4 m
2 m
2 m
1.5 m
R
BH
R
AH
R
AV
B
B
G G
A
A
Fig 8.18 Fig 8.19
Sol.: Since two reaction (Vertical and Horizontal) at pin support i.e., R
AH
and R
AV
. And at rocker there will
be only one Horizontal reaction i.e., R
BH
.
172 / Problems and Solutions in Mechanical Engineering with Concept
First draw the FBD of the Jib crane as shown in fig 8.19. The whole system is in equilibrium.
Take moment about point A
∑M
A
= – R
BV
× 1.5 + 1000 × 2 + 2400 × 6 = 0
R
BH
= 10933.3Kg .......ANS
∑H = 0
R
AH
+ R
BH
= 0
i.e., R
AH
= – R
BH
R
AH
= –10933.3Kg .......ANS
∑V = 0
R
AV
– 1000 – 2400 = 0
R
AV
= 3400Kg .......ANS
Q. 7: A square block of 25cm side and weighing 20N is hinged at A and rests on rollers at B as
shown in fig 8.20. It is pulled by a string attached at C and inclined at 300 with the horizontal.
Make calculations for the force P to be applied so that the block gets just lifted off the roller.
30°
30°
45°
W = 20 N
W = 20 N
0.25 m
A A
B B
D D C C
P
P
Fig 8.20 Fig 8.21
Sol.: From the Free body diagram the block is subjected to the following set of forces.
1. Force P
2. Weight of the block W
3. Reaction R
A
at the hinged point
4. When the block is at the state of just being lifted off the roller, reaction R
B
= 0
∑M
A
= 0
– Pcos30° × 0.25 – Psin30° × 0.25 + 20 × 0.125 = 0
– 0.22 P – 0.125P + 2.5 = 0
P = 7.27N .......ANS
Q. 8: Two weights C = 2000N and D = 1000N are located on a horizontal beam AB as shown in the
fig 8.22. Find the distance of weight ‘C’ from support ‘A’ i.e., ‘X’ so that support reaction at
A is twice that at B. (May–00–01)
Sol.: Since given that R
A
= 2R
B
∑H = 0
R
AH
= 0 ...(i)
∑V = 0
Force: Support Reaction / 173
4m
1m
x
B
A
D
C
Fig. 8.22
2000 N
× 1m
1000 N
4m
B
A R
AH
R
A R
B
Fig. 8.23
R
A
+ R
B
– 2000 – 1000 = 0
R
A
+ R
B
= 3000N
But R
A
= 2R
B
i.e., R
B
= 1000N ...(ii)
R
A
= 2000N ...(iii)
Taking moment about point A:
∑M
A
= 2000 × x + 1000 × (x + 1) – R
B
× 4 = 0
2000 × x + 1000 × (x + 1) –1000 × 4 = 0
2000x + 1000x + 1000 – 4000 = 0
3000x = 3000
x = 1m .......ANS
Q. 9: A 500N cylinder, 1 m in diameter is loaded between the cross pieces AE and BD which make
an angle of 60º with each other and are pinned at C. Determine the tension in the horizontal
rope DE assuming that the cross pieces rest on a smooth floor. (Dec–01–02)
Sol.: Consider the equilibrium of the entire system.
C is the pin joint, making the free body diagram of ball and rod separately.
2R
N
cos60° = 500 ...(i)
R
N
= 500KN
R
A
+ R
B
= 500N ...(ii)
Due to symmetry R
A
= R
B
= 250N
CP = 0.5cot30° = 0.866m
174 / Problems and Solutions in Mechanical Engineering with Concept
1.8
1.2
60° 60°
500 N
60°
1
.
8
m
1
.
2
m
500 N
1 m
D
P
T
R
N
R
V
R
H
R
N
R
N
C
B
R
B
B A
C
E
D
Fig 8.24 Fig 8.25 Fig 8.26
Taking moment about point C,
T × 1.8cos30° – R
N
× CP – R
B
× 1.2sin30° = 0
T × 1.8cos300 = R
N
× CP + R
B
× 1.2sin30°
Putting the value of CP, R
N
, and R
B
T = 374N .......ANS
Q. 10: A Force P = 5000N is applied at the centre C of the beam AB of length 5m as shown in the
fig 8.27. Find the reactions at the hinge and roller support. (May–01–02)
Sol.: Hinged at A and Roller at B, FBD of the beam is as shown in fig 14.70
2.5 m
2.5 m
2.5 m
2.5 m
30°
30°
P = 5000 N 5000 N
B
B C
A
R
AH
R
AV
R
BV
A
Fig 8.27 Fig 8.28
∑H= 0
R
AH
–5000cos30° = 0
R
AH
= 4330.127N .......ANS
∑V = 0
R
AV
+ R
BV
–5000sin30° = 0
R
AV
+ R
BV
= 2500N ...(i)
Taking moment about point B:
∑M
B
= R
AV
× 5 – 5000sin30° × 2.5 = 0
R
AV
= 1250N .......ANS
From equation (i)
R
BV
= 1250N .......ANS
Q. 11: The cross section of a block is an equilateral triangle. It is hinged at A and rests on a roller
at B. It is pulled by means of a string attached at C. If the weight of the block is Mg and the
string is horizontal, determine the force P which should be applied through string to just lift
the block off the roller. (Dec–02–03)
Force: Support Reaction / 175
Sol.: When block is just lifted off the roller the reaction at B i.e., R
B
will be zero.
60° 60°
Mg
a
B
A
2a
P
C
Fig. 8.29
For equilibrium, R
A
= R
B
= Mg/2
At this instance, taking moment about ‘A’
P × 3a = Mg.a√3
P = Mg/√ √√ √√3 .......ANS
Q. 12: A beam 8m long is hinged at A and supported on rollers over a smooth surface inclined at 300
to the horizontal at B. The beam is loaded as shown in fig 8.30. Determine the support reaction.
(May–02–03)
2 m 2 m 3 m 1 m
30°
45°
10 kN
8 kN 10 kN
B A
Fig 8.30
30°
10 KN 8 KN
10 KN
B
R
AH
R
AV
R
E
Fig 8.31
8 sin 45°
8 cos 45° 10 10
R
AH
R
AV
R
B
sin 30°
R
B
cos 30°
Fig 8.32
176 / Problems and Solutions in Mechanical Engineering with Concept
Sol.: F.B.D. is as shown in fig 8.32
∑H = 0
R
AH
⋅ + 8cos45° – R
B
sin30° = 0
0.5R
B
– R
AH
⋅ = 5.66 ...(i)
∑V = 0
R
AV
–10 – 8cos45° – 10 + R
B
cos30° = 0
R
AV
+ 0.866R
B
= 25.66 ...(ii)
Taking moment about point A:
∑M
A
= 10 × 2 + 8cos45° × 4 + 10 × 7 – R
B
cos30° × 8= 0
R
B
= 16.3KN .......ANS
From equation (ii)
R
AV
= 11.5KN .......ANS
From equation (1)
R
AH
= 2.5KN .......ANS
Q. 13: Calculate the support reactions for the following. Fig(8.33).
2 m
2 m 2 m 3 m
5 KN
10 KN/M
gap gap
E
D
C
B
A
Fig 8.33
Sol.: First change UDL in to point load.
Resolved all the forces in horizontal and vertical direction. Since roller at B (only one vertical reaction)
and hinged at point A (one vertical and one horizontal reaction).
Let reaction at hinged i.e., point B is R
BH
and R
BV
, and reaction at roller support i.e. point D is R
DV
Let ∑H & ∑V is the sum of horizontal and vertical component of the forces ,The supported beam is
in equilibrium, hence
∑H = ∑V = 0
R
H
= R
BH
= 0
R
BH
= 0 ...(i)
∑V = R
BV
–50 –5 – R
DV
= 0
R
BV
+ R
DV
= 55 ...(ii)
Taking moment about point B
50 × 0.5 – R
BV
× 0 – R
DV
× 5 + 5 × 7 = 0
R
DV
=12 KN .......ANS
Putting the value of R
BV
in equation (ii)
R
BV
= 43KN .......ANS
Hence R
BH
= 0, R
DV
= 12KN, R
BV
= 43KN
Q. 14: Compute the reaction at A and B for the beam subjected to distributed and point loads as
shown in fig (8.34). State what type of beam it is.
Force: Support Reaction / 177
L L L
B
A
W
P N/m
Fig 8.34
L L L
B
R
AV
R
BV
R
AH
W
A
P L L/2
Fig 8.35
Sol.: First change UDL in to point load.
Resolved all the forces in horizontal and vertical direction. Since roller at B (only one vertical reaction)
and hinged at point A (one vertical and one horizontal reaction).
Let reaction at hinged i.e., point A is R
AH
and R
AV
, and reaction at roller support i.e., point B is R
BV
Let ∑H & ∑V is the sum of horizontal and vertical component of the forces ,The supported beam is
in equilibrium, hence Draw the FBD of the diagram as shown in fig 8.35
Since beam is in equilibrium, i.e.,
∑H = 0;
R
AH
= 0 .......ANS
∑V = 0 ; R
AV
+ R
BV
– P.L – W = 0
R
AV
+ R
BV
= P.L + W ...(i)
Taking moment about point A,
P.L × L/2 + W × 2L – RBV × 3L = 0 ...(ii)
R
BV
= P.L/6 + 2W/3 .......ANS
Put the value of R
BV
in equation (i)
R
AV
= 5P.L/6 + W/3 .......ANS
Q. 15: Find the reactions at supports A and B of the loaded beam shown in fig 8.36.
20 kN
30 kN
2 m 2 m 2 m 1 m
4 m
45°
30 kN
A B
Fig 8.36
178 / Problems and Solutions in Mechanical Engineering with Concept
22 3 2
2
20 KN 120 KN
60 cos 45
60 sin 45
A
R
AV
R
BV
B
Fig 8.37
Sol.: First change UDL in to point load.
Resolved all the forces in horizontal and vertical direction. Since roller at A (only one vertical reaction)
and hinged at point B (one vertical and one horizontal reaction).
Let reaction at hinged i.e., point B is R
BH
and R
BV
, and reaction at roller support i.e.. point A is R
AV
Let ∑H & ∑V is the sum of horizontal and vertical component of the forces, The supported beam is
in equilibrium, hence Draw the FBD of the beam as shown in fig 8.37.
Since beam is in equilibrium, i.e.,
∑H = 0;
R
BH
– 60cos45° = 0
R
BH
= 42.42KN .......ANS
∑V = 0;
R
AV
+ R
BV
– 20 –120 – 42.4 = 0
R
AV
+ R
BV
= 182.4KN ...(i)
Taking moment about point A,
20 × 2 + 120 × 4 + 42.4 × 7 – R
BV
× 9 = 0 ...(ii)
R
BV
= 90.7KN .......ANS
Put the value of R
BV
in equation (i)
R
AV
= 91.6KN .......ANS
Hence reaction at support A i.e., R
AV
= 91.6KN
reaction at support B i.e., R
BV
= 90.7KN, R
BH
= 42.4KN
Q. 16: The cantilever is shown in fig (8.38), Determine the reaction when it is loaded..
16 kN/m 20 kN 12 kN
1 m 1 m
2 m
10 kN
R
AH
M
A
R
AV
Fig 8.38
32 KN 20 KN 12 KN 10 KN
1 1 1 1
R
AH
M
A
R
AV
Fig 8.39
Sol.: In a cantilever at fixed end (Point A) there is three reaction i.e., R
AH
, M
A
, R
AV
First draw the FBD of the beam as shown in fig 8.39, Since beam is in equilibrium, i.e.,
Force: Support Reaction / 179
∑H = 0;
R
AH
= 0
R
AH
= 0 .......ANS
∑V = 0;
R
AV
–32 – 20 –12 – 10 = 0
R
AV
= 74KN .......ANS
Taking moment about point A,
– M
A
+ 32 × 1 + 20 × 2 + 12 × 3 + 10 × 4 = 0
M
A
= 148KN–m .......ANS
Hence reaction at support A i.e., R
VA
= 74KN, R
HA
= 0KN, M
A
= 148KN–m
Q. 17: Determine the reactions at A and B of the overhanging beam as shown in fig (8.40).
40 kN m
30 kN
2 m 1 m 2 m 3 m
45°
20 kN/m
A
B
Fig 8.40
R
AH
R
AV
R
BV
40 kN m
30 sin 45
30 cos 45º
1 m 1 m 2 m 3 m
40
A
B
Fig 8.41
Sol.: First change UDL in to point load.
Resolved all the forces in horizontal and vertical direction. Since hinged at point A (one vertical and
one horizontal reaction).
Let reaction at hinged i.e., point A is R
AH
and R
AV
, Let ∑H & ∑V is the sum of horizontal and vertical
component of the forces, The supported beam is in equilibrium, hence Draw the FBD of the beam as shown
in fig 8.42, Since beam is in equilibrium, i.e.,
∑H = 0;
R
AH
= 30cos45° = 21.2KN
R
AH
= 21.21KN .......ANS
∑V = 0;
R
AV
–30sin45 – 40 + R
BV
= 0
R
AV
+ R
BV
= 61.2KN ...(i)
Taking moment about point B,
R
AV
× 6 + 40 – 30sin45 × 1 + 40 × 1 = 0
180 / Problems and Solutions in Mechanical Engineering with Concept
R
AV
= – 9.8 KN .......ANS
Putting the value of R
AV
in equation (i), we get
R
BV
= 71KN .......ANS
Hence reaction at support A i.e., R
AV
= –9.8KN, R
AH
= 21.2KN, R
BV
= 71KN
Q. 18: Find out reactions at the grouted end of the cantilever beam shown in fig 8.42.
10KN/m
100KN/m
5m
3m
3m
5m
5KN/ m
Fig 8.42
100KN/m
7.5m
3m
1.5 1.5
15KN
2.5m
50KN
M
A
R
AV
R
AH
Fig 8.43
Sol.: Draw F.B.D. of the beam as shown in fig 8.43. First change UDL in to point load. Since Point A is
fixed point i.e., there is three reaction are developed, R
AH
, R
AV
, M
A
. Let ∑H & ∑V is the sum of horizontal
and vertical component of the forces, The supported beam is in equilibrium, hence
R = 0,
∑H = ?V =0
∑ ∑∑ ∑∑H = 0; R
AH
= 0 .......ANS
∑ ∑∑ ∑∑V = 0; R
AV
– 50 + 15 = 0, R
AV
= 35KN .......ANS
Now taking moment about point ‘A’
–M
A
+ 50 × 2.5 + 100 – 15 × 14.5 = 0
M
A
= 7.5 KN–m .......ANS
Q. 19: Find the support reaction at A and B in the beam as shown in fig 8.44.
5KN
1m 1m 1m
3m
10KN/m
2KN/m
1KN/m
2m
3KN/m
A
M
P
Q
B
N
Fig 8.44
Force: Support Reaction / 181
5KN
1 1 1 1.5 1.5
6KN
1m
2×2
3
m
10KN–m
X
W
MNQB
W
NPQ
R
AV
R
VB
R
AH
Fig 8.45
Sol.: First draw the FBD of the beam as shown in fig 8.45
In the fig 8.46,
6KN is the point load of UDL
W
MNQB
= Weight of MNQB
= UDL × Distance(MB)
= 1 × 2
= 2KN, act at a point 1m vertically from point B
W
NPQ
= Weight of Triangle NPQ
= 1/2 × MB × (BP – BQ)
= 1/2 × 2 × (3 – 1)
= 2KN and will act at MB/3 = 2/3m from point B
Since hinged at point A and Roller at point B. let at point A R
HA
and R
VA
and at point B R
VB
is the
support reaction, Also beam is in equilibrium under action of coplanar non concurrent force system, therefore:
∑H = 0
R
AH
–W
MNQB
– W
NPQ
= 0
R
AH
– 2 – 2 = 0
R
AH
= 4KN .......ANS
∑V = 0
R
AV
+ R
BV
– 5 – 6 = 0
R
AV
+ R
BV
= 11KN ...(i)
Taking moment about point A:
M
A
= 5 × 1 – 10 + 6 × 4.5 – R
BV
× 6 – W
NPQ
× (2 – 4/3) – W
MNQB
× 1 = 0
5 × 1 – 10 + 6 × 4.5 – R
BV
× 6 – 2 × (2 – 4/3) – 2 × 1 = 0
R
BV
= 3.11KN .......ANS
Putting the value of R
BV
in equation (i)
R
AV
= 7.99KN .......ANS
Q. 20: What force and moment is transmitted to the supporting wall at A in the given cantilever
beam as shown in fig 8.46. (May–02–03)
0.5m
1.5 m
0.5m
5kN/m
15kN
(5 × 1.5)KN 15KN
M
A
R
AH
R
AV
Fig 8.46 Fig 8.47
182 / Problems and Solutions in Mechanical Engineering with Concept
Sol.: Fixed support at A, FBD of the beam is as shown in fig 8.47
∑H = 0
R
AH
= 0
R
AH
= 0 .......ANS
∑V = 0
R
AV
– 7.5 + 15 = 0
R
AV
= –7.5KN .......ANS
–ive sign indicate that we take wrong direction of R
AV
, i.e., Force act vertically downwards.
Taking moment about point A:
∑M = – M
A
+ 7.5 × 0.75 – 15 × 2 = 0
M
A
= 7.5 × 0.75 – 15 × 2
⇒ M
A
= –24.357KN–m .......ANS
–ive sign indicate that we take wrong direction of moment, i.e., moment is clockwise.
Q. 21: Determine the reactions at supports of simply supported beam of 6m span carrying increasing
load of 1500N/m to 4500N/m from one end to other end.
4500N/m
6m A
C
R
A
B
R
B
D
E
4 = 2/3 × 6
6
1
2
1500 × 6
3
(4500 – 1500) ×
R
A
R
B
Fig 8.48 Fig 8.49
Sol.: Since Beam is simply supported i.e., at point A and point B only point load is acting. First change
UDL and UVL in to point load. As shown in fig 8.49. Let ∑H & ∑V is the sum of horizontal and vertical
component of the Resultant forces, The supported beam is in equilibrium, hence resultant force is zero.
Draw the FBD of the beam as shown in fig 8.49,
Divided the diagram ACBE in to two parts A triangle CDE and a rectangle ABCE.
Point load of Triangle CDE =1/2 × CD × DE = 1/2 × 6 × (4.5 –1.5)= 9KN
act at a distance 1/3 of CD (i.e., 2.0m )from point D
Point load of Rectangle ABCD = AB × AC = 6 × 1.5 = 9KN
act at a distance 1/2 of AB (i.e., 3m )from point B
Now apply condition of equilibrium:
∑H = 0;
R
AH
= 0 .......ANS
∑V = 0;
R
A
–1500 × 6 – 3000 × 3 + R
B
= 0
R
A
+ R
B
= 18000 N ...(i)
Now taking moment about point ‘A’
– R
B
× 6 + 9000 × 3 9000 × 4 = 0
R
B
= 10500 Nm .......ANS
Putting the value of R
B
in equation (i)
R
A
= 7500 Nm .......ANS
Force: Support Reaction / 183
Q. 22: Calculate the support reactions for the beam shown in fig (8.50).
3 m 3 m 2m 2m
10kN/m
40kN/m
30 kN
18 kN/m
45°
A
B C D
E
2
50
40 KNm
30 cos 45°
27
2
R
AH
R
AV
2.5
R
DV
Fig 8.50 Fig 8.51
Sol.: Since Beam is overhang. At point A hinge support and point D Roller support is acting. First change
UDL and UVL in to point load. As shown in fig 8.51. let ∑H & ∑V is the sum of horizontal and vertical
component of the Resultant forces, the supported beam is in equilibrium, hence resultant force is zero.
Convert UDL and UVL in point load and draw the FBD of the beam as shown in fig 8.51
∑H = 0;
R
AH
= 30cos45°
R
AH
= 21.21KN .......ANS
∑V = 0;
R
AV
–50 – 30sin450 + R
DV
– 27 = 0
R
AV
+ R
DV
= 98.21 KN ...(i)
Now taking moment about point ‘A’
– R
DV
× 7 + 50 × 2.5 + 40 + 30 sin 450 × 5 + 27 × 9 = 0
R
DV
= 73.4 KN .......ANS
Putting the value of R
DV
in equation (i)
R
AV
= 24.7KN .......ANS
Q. 23: Determine the reactions at supports A and B of the loaded beam as shown in fig 8.52.
10kN/m
10kN/m
20kN/m
1 m 2 m
3 m
E
F
A
C
D
B
H
G
5KN 10KN
2/3 2/3 2.83
0.5
1/3
30KN 15KN
A
E C F G B
R
BH
R
BV
R
AV
Fig 8.52 Fig 8.53
Sol.: First consider the FBD of the diagram 8.52. as shown in fig 8.53. In which Triangle CEA, AED and
FHG shows point load and also rectangle FHDB shows point load.
Point load of Triangle CEA = 1/2 × AC × AE = 1/2 × 1 × 10 = 5KN,
act at a distance 1/3 of AC (i.e., 0.333m )from point A
Point load of Triangle AED = 1/2 × AD × AE = 1/2 × 2 × 10 = 10KN
act at a distance 1/3 of AD (i.e., 0.666m )from point A
Now divided the diagram DBGF in to two parts A triangle FHG and a rectangle FHDB.
Point load of Triangle FHG = 1/2 × FH × HG = 1/2 × 3 × (20 – 10) = 15KN
act at a distance 1/3 of FH (i.e., 1.0m )from point H
Point load of Rectangle FHDB = DB × BH = 3 × 10 = 30KN
act at a distance 1/2 of DB (i.e., 1.5m )from point D
184 / Problems and Solutions in Mechanical Engineering with Concept
At Point A roller support i.e., only vertical reaction (R
AV
), and point B hinged support i.e., a horizontal
reaction (R
BH
) and a vertical reaction (R
BV
). All the point load are shown in fig 8.53
∑H = 0;
R
BH
= 0
R
BH
= 0 .......ANS
∑V = 0;
R
AV
+ R
BV
– 5 – 10 – 30 – 15 = 0
R
AV
+ R
BV
= 60KN ...(i)
Now taking moment about point ‘A’
– 5 × 1/3 + 10 × 0.66 + 30 × 3.5 + 15 × 4 – R
BV
× 5 = 0
R
BV
= 34 KN .......ANS
Putting the value of RDV in equation (1)
R
AV
= 26KN .......ANS
Q. 24: Determine the reactions at the support A, B, C, and D for the arrangement of compound
beams shown in fig 8.54
6kN
1m 1m 1m 1m
1m 1m 1m 1m 1m
6kN
8kN
8kN 10kN 4kN 4kN
A
C
F
B
D
Fig 8.54
10kN
1m
E
B
1m 1m 1m 2m
4kN 6kN 8kN
R
E
R
B
Fig 8.55
6kN
1m
A
F
C D
1m 1m
3m
3m 2m
8kN
R
A
R
F
R
C
R
D
Fig 8.56 Fig 8.57
Sol.: This is the question of multiple beam (i.e., beam on a beam). In this type of question, first consider
the top most beam, then second last beam as, In this problem on point E and F, there are roller support,
and this support give reaction to both up and down beam. Consider FBD of top most beam EB as shown
in fig 8.55
Force: Support Reaction / 185
∑V = 0;
R
E
+ R
B
– 10 – 4 – 6 – 8 = 0
R
E
+ R
B
= 28KN ...(i)
Now taking moment about point ‘E’
10 × 1 + 4 × 2 + 6 × 3 + 8 × 4 – R
B
× 6 = 0
R
B
= 11.33 KN .......ANS
Putting the value of R
B
in equation (i)
R
E
= 16.67 KN .......ANS
Consider FBD of second beam AF as shown in fig 8.56:
∑V = 0;
R
A
+ R
F
– 6 – 8 – R
E
= 0
R
A
+ R
F
= 30.67 KN ...(ii)
Now taking moment about point ‘A’
6 × 1 + 8 × 2 + 16.67 × 3 – RF × 6 = 0
RF = 12 KN .......ANS
Putting the value of R
F
in equation (ii)
R
A
= 18.67 KN .......ANS
Consider FBD of third beam CD as shown in fig 8.57:
∑V = 0;
R
C
+ R
D
– R
F
= 0
R
C
+ R
D
= 12 KN ...(iii)
Now taking moment about point ‘C’
– R
D
× 5 + 12 × 3 = 0
R
D
= 7.2 KN .......ANS
Putting the value of R
D
in equation (iii)
R
C
= 4.8 KN .......ANS
Q. 25: Determine the reactions at A,B and D of system shown in fig 8.58. (Dec–01–02)
3 KN/m
5m
2m
2m 3m 2m 2m
12 KN/m
12 KN/m
A
R
C
R
DV
R
OH
B
C
D
D
Fig 8.58 Fig 8.59
15KN
3.33 m 2m 3m
2.5 m
22.5KN
D R
DH
R
BH
R
BV
R
C
= 21.43 KN
R
A
R
DV
R
C
UDL UVL
Fig 8.60 Fig 8.61
186 / Problems and Solutions in Mechanical Engineering with Concept
Solution: Consider FBD of top most beam as shown in fig 8.59 and 8.60
∑H = 0
R
DH
= 0 ...(i)
∑V = 0
R
C
+ R
DV
– 15 – 22.5 = 0
R
C
+ R
DV
= 37.5KN ...(ii)
Taking moment about point C:
∑M
C
= 15 × 2.5 + 22.5 × 3.33 – R
DV
× 7 = 0
R
DV
= 16.07KN .......ANS
From equation (ii)
R
C
= 21.43KN .......ANS
Consider FBD of bottom beam as shown in fig 8.61
∑H = 0
R
BH
= 0 ...(iii)
∑V = 0
R
A
+ R
BV
– R
C
= 0
R
A
+ R
BV
= 21.43KN ...(iv)
Taking moment about point A:
∑M
A
= R
C
× 2 – R
BV
× 5 = 0
R
BV
= 8.57KN .......ANS
From equation (iv)
R
A
= 12.86KN .......ANS
Q. 26: Determine the reactions at supports A and D in the structure shown in fig–8.62
(Dec–(C.O)–03)
80 KN
0.5 m
3 m 1 m 2 m
A
B
C
D
80 KN
R
AH
R
AV
R
BV
A
3m 1m
B
Fig 8.62 Fig 8.63
R
BV
R
CV
R
DV
R
DH C
B
D
Fig 8.64
Sol.: Since there is composite beam, there fore first consider top most beam,
Let reaction at A is R
AH
and R
AV
Force: Support Reaction / 187
Reaction at B is R
BV
Reaction at C is R
AV
Reaction at D is R
DH
and R
DV
Draw the FBD of Top beam as shown in fig 8.63,
∑H = 0
R
AH
= 0 .......ANS
∑V = 0
R
AV
+ R
BV
– 80 = 0
R
AV
+ R
BV
= 80KN ...(i)
Taking moment about point A:
∑M
A
= 80 × 3 – R
BV
× 4 = 0
R
BV
= 60KN .......ANS
From (i), R
AV
= 20KN .......ANS
Consider the FBD of bottom beam as shown in fig 8.64,
∑H = 0
R
DH
= 0 .......ANS
∑V = 0
R
CV
+ R
DV
– R
BV
= 0
R
CV
+ R
DV
= 60KN ...(ii)
Taking moment about point D:
∑M
D
= –60 × 4.5 + R
CV
× 4 = 0
R
CV
= 67.5KN .......ANS
From (ii), R
DV
= –7.5KN .......ANS
Q. 27: Explain Jib crane Mechanism.
Sol.: Jib crane is used to raise heavy loads. A load W is lifted up
by pulling chain through pulley D as shown in adjacent figure
8.65. Member CD is known as tie, and member AD is known as
jib. Tie is in tension and jib is in compression. AC is vertical
post. Forces in the tie and jib can be calculated. Very often chain
BD and Tie CD are horizontal. Determination of forces for a
given configuration and load is illustrated through numerical
examples.
Q. 28: The frictionless pulley A is supported by two bars AB and AC which are hinged at B and C
to a vertical wall. The flexible cable DG hinged at D goes over the pulley and supports a load
of 20KN at G. The angle between the various members shown in fig 8.66. Determine the forces
in AB and AC. Neglect the size of pulley. (Dec–01–02)
Sol.: Here the system is jib–crane. Hence Member CA is in compression and AB is in tension. As shown
in fig 8.67.
Cable DG goes over the frictionless pulley, so
Tension in AD = Tension in AG
= 20KN
FBD of the system is as shown in fig 8.67
∑H = 0
Fig 8.65
T
I
E
C
H
A
I
N
J
I
B
C
A
B
D
W
188 / Problems and Solutions in Mechanical Engineering with Concept
Psin30° – Tsin60° – 20sin60° = 0
0.5P – 0.866T = 17.32KN ...(i)
3
0
°
3
0
°
20 KN
20 KN
20 KN
3
0
°
3
0
°
6
0
°
6
0
°
C
D
A
T
P
G
A
B
Fig 8.66 Fig 8.67
∑V = 0
Pcos30° + Tcos60° – 20 – 20cos60° = 0
0.866P + 0.5T = 30KN ...(ii)
Multiply by equation (i) by 0.5 , we get
0.25P – 0.433T = 8.66 ...(iii)
Multiply by equation (ii) by 0.866 , we get
0.749P + 0.433T = 25.98 ...(iv)
Add equation (i) and (ii), we get
P = 34.64KN .......ANS
Putting the value of P in equation (i), we get
17.32 – 0.866T = 17.32
T = 0 .......ANS
Q. 29: The lever ABC of a component of a machine is hinged at B, and is subjected to a system of
coplanar forces. Neglecting friction, find the magnitude of the force P to keep the lever in
equilibrium.
Sol.: The lever ABC is in equilibrium under the action of the forces 200KN, 300KN, P and R
B
, where R
B
required reaction of the hinge B on the lever.
Hence the algebraic sum of the moments of above forces about any point in their plane is zero.
Moment of R
B
and B is zero, because the line of action of R
B
passes through B.
Taking moment about B, we get
200 × BE – 300 × CE – P × BF = 0
since CE = BD,
200 × BE – 300 × BD – P × BF = 0
Force: Support Reaction / 189
200 × BCcos30° – 300 × BCsin30° – P × ABsin60° = 0
200 × 12 × cos30° – 300 × 12 × sin30° – P × 10 × sin60° = 0
P = 32.10KN .......ANS
Let
R
BH
= Resolved part of R
B
along a horizontal direction BE
R
BV
= Resolved part of R
B
along a horizontal direction BD
∑H = Algebraic sum of the Resolved parts of the forces along
horizontal direction
∑v = Algebraic sum of the Resolved parts of the forces along
vertical direction
∑H = 300 + R
BH
– Pcos20°
∑H = 300 + R
BH
– 32.1cos20° ...(i)
∑v = 200 + R
BV
– Psin20°
∑v = 200 + R
BV
– 32.1sin20° ...(ii)
Since the lever ABC is in equilibrium
∑H = R
V
= 0, We get
R
BH
= –269.85KN
R
BV
= –189.021KN
R
B
= {(RBH)2 +(RBV)2}1/2
R
B
= {(–269.85)2 + (–189.02)2}1/2
R
B
= 329.45KN .......ANS
Let θ = Angle made by the line of action of R
B
with the horizontal
Then, tanθ = R
BV
/R
BH
= –189.021/–269.835
θ θθ θθ = 35.01° .......ANS
190 / Problems and Solutions in Mechanical Engineering with Concept
+0)264
9
.41+61
Q. 1: Define the term friction?
Sol.: When a body moves or tends to move over another body, a force opposing the motion develops at
the contact surfaces. This force, which opposes the movement or the tendency of movement, is called
frictional force or simply friction. Frictional force always acts parallel to the surface of contact, opposite
to the moving direction and depends upon the roughness of surface.
A frictional force develops when there is a relative motion between a body and a surface on application
of some external force.
A frictional force depends upon the coefficient of friction between the surface and the body which can
be minimized up to a very low value equal to zero (theoretically) by proper polishing the surface.
Q. 2: Explain with the help of neat diagram, the concept of limiting friction.
Sol.: The maximum value of frictional force, which comes into play, when a body just begins to slide over
the surface of the other body, is known as limiting friction. Consider a solid body placed on a horizontal
plane surface.
R
W
block
P F
Fig 9.1
Let
W = Weight of the body acting through C.G. downwards.
R = Normal reaction of body acting through C.G. downwards.
P = Force acting on the body through C.G. and parallel to the horizontal surface.
F = Limiting force of friction
If ‘P’ is small, the body will not move as the force of friction acting on the body in the direction
opposite to 'P' will be more than ‘P’. But if the magnitude of ‘P’ goes on increasing a stage comes, when
the solid body is on the point of motion. At this stage, the force of friction acting on the body is called
‘LIMITING FORCE OF FRICTION (F)’.
R = W; F = P
Friction / 191
If the magnitude of ‘P’ is further increased the body will start moving. The force of friction, acting
on the body is moving, is called KINETIC FRICTION.
Q. 3: Differentiate between;
(a) Static and Kinetic Friction
(b) Sliding and rolling Friction.
Sol.: (a) Static Friction: When the applied force is less than the limiting friction, the body remains at rest
and such frictional force is called static friction and this law is known as law of static friction.
It is the friction experienced by a body, when it is at rest. Or when the body tends to move.
Kinetic (Dynamic) Friction
When the applied force exceeds the limiting friction the body starts moving over the other body and the
friction of resistance experienced by the body while moving. This is known as law of Dynamic or kinetic
friction.
Or
It is the friction experienced by a body when in motion. It is of two type;
1. Sliding Friction
2. Rolling Friction
(b) Sliding Friction: It is the friction experienced by a body, when it slides over another body.
Rolling Friction
It is the friction experienced by a body, when it rolls over the other.
Q. 4: Explain law of coulomb friction? What are the factor affecting the coefficient of friction and
effort to minimize it.
Sol.: Coulomb in 1781 presented certain conclusions which are known as Coulomb's law of friction. These
conclusions are based on experiments on block tending to move on flat surface without rotation. These laws
are applicable at the condition of impending slippage or once slippage has begun. The laws are enunciated
as follows:
1. The total force of friction that can be developed is independent of area of contact.
2. For low relative velocities between sliding bodies, total amount of frictional force is independent
of the velocity. But the force required to start the motion is greater than that necessary to maintain
the motion.
3. The total frictional force that can be developed is proportional to the normal reaction of the surface
of contact.
So, coefficient of friction(µ) is defined as the ratio of the limiting force of friction (F) to the normal
reaction (R) between two bodies.
Thus,
µ = Limiting force of friction/ Normal reaction
= F/R
or, F = µ.R, Generally µ < 1
The factor affecting the coefficient of friction are:
1. The material of the meeting bodies.
2. The roughness/smoothness of the meeting bodies.
3. The temperature of the environment.
192 / Problems and Solutions in Mechanical Engineering with Concept
Efforts to minimize it:
1. Use of proper lubrication can minimize the friction.
2. Proper polishing the surface can minimize it.
Q 5: Define the following terms;
(a) Angle of friction
(b) Angle of Repose
(c) Cone of Friction
Sol.: (a) Angle of Friction (θ θθ θθ)
W
P
A
F = R
S
q
R
Fig 9.2
It is defined as the angle made by the resultant of the normal reaction (R) and the limiting force of
friction (F) with the normal reaction (R).
Let, S = Resultant of the normal reaction (R) and limiting force of friction (F)
θ = Angle between S and R
Tan θ = F/R = µ
Note: The force of friction (F) is always equal to µR
(b) Angle of Repose (α αα αα)
O
W
F
R
=
m
M
o
t
i
o
n
R
q
A
Fig 9.3
It is the max angle of inclined plane on which the body tends to move down the plane due to its own
weight.
Consider the equilibrium of the body when body is just on the point of slide.
Friction / 193
Resolving all the forces parallel and perpendicular to the plane, we have:
µR = W.sina
R = W.cosa
Dividing 1 by 2 we get Tana µ
But µ = tan θ , θ = Angle of friction
i.e., θ = α
The value of angle of repose is the same as the value of limiting angle of friction.
(c) Cone of Friction: When a body is having impending motion
in the direction of P, the frictional force will be the limiting friction
and the resultant reaction R will make limiting friction angle θ with
the normal. If the body is having impending motion in some other
direction, again the resultant reaction makes limiting frictional angle
θ with the normal in that direction. Thus, when the direction of
force P is gradually changed through 360°, the resultant R generates
a right circular cone with semicentral angle equal to θ.
If the resultant reaction is on the surface of this inverted right
circular cone whose semicentral angle is limiting frictional angle
(θ), the motion of body is impending. If the resultant is within this
cone, the body is stationary. This inverted cone with semicentral
angle, equal to limiting frictional angle θ, is called cone of friction.
It is defined as the right circular cone with vertex at the point
of contact of the two bodies (or surfaces), axis in the direction of normal reaction (R) and semivertical
angle equal to angle of friction (θ). Fig (9.4) shows the cone of friction in which,
O = Point of contact between two bodies.
R = Normal reaction and also axis of the cone of friction.
θ = Angle of friction
Q. 6: What are the types of Friction?
Sol.: There are mainly two types of friction,
(i) Dry Friction (ii) Fluid Friction
(i) Dry Friction: Dry friction (also called coulomb friction manifests when the contact surfaces are
dry and there is tendency for relative motion.
Dry friction is further subdivided into:
Sliding Friction
Fiction between two surfaces when one surface slides over another.
Rolling Friction
Friction between two surfaces, which are separated by balls or rollers.
It may be pointed out that rolling friction is always less than sliding friction.
(ii) Fluid Friction: Fluid friction manifests when a lubricating fluid is introduced between the contact
surfaces of two bodies.
If the thickness of the lubricant or oil between the mating surfaces is small, then the friction between
the surfaces is called GREASY OR NONVISCOUS FRICTION. The surfaces absorb the oil and the
contact between them is no more a metaltometal contact. Instead the contact is through thin layer of oil
and that ultimately results is less friction.
O
P
q
Point of
Contact
Cone of
Friction
Axis
R
Fig 9.4
194 / Problems and Solutions in Mechanical Engineering with Concept
When a thick film of lubricant separates the two surfaces, metallic contact is entirely nonexistent. The
friction is due to viscosity of the oil, or the shear resistance between the layers of the oil rubbing against
each other. Obviously then these occurs a great reduction in friction. This frictional force is known as
Viscous Or Fluid Friction.
Q. 7: Explain the laws of solid friction?
Sol.: The friction that exists between two surfaces, which are not lubricated, is known as solid friction. The
two Surfaces may be at rest or one of the surface is moving and other surface is at rest. The following are
the laws of solid friction.
1. The force of friction acts in the opposite direction in which surface is having tendency to move.
2. The force of friction is equal to the force applied to the surfaces, so long as the surface is at rest.
3. When the surface is on the point of motion, the force of friction is maximum and this maximum
frictional force is called the limiting friction force.
4. The limiting frictional force bears a constant ratio to the normal reaction between two surfaces.
5. The limiting frictional force does not depend upon the shape and areas of the surfaces in contact.
6. The ratio between limiting friction and normal reaction is slightly less when the two surfaces are
in motion.
7. The force of friction is independent of the velocity of sliding.
The above laws of solid friction are also called laws of static and dynamic friction or law of friction.
Q. 8: “Friction is both desirable and undesirable” Explain with example
Sol.: Friction is Desirable: A friction is very much desirable to stop the body from its moving condition.
If there is no friction between the contact surfaces, then a body can't be stopped without the application
of external force. In the same time a person can't walk on the ground if there is no friction between the
ground and our legs also no vehicle can run on the ground without the help of friction.
Friction is Undesirable: A friction is undesirable during ice skating or when a block is lifted or put
down on the truck with the help of some inclined plane. If the friction is more between the block and
inclined surface, then a large force is required to push the block on the plane.
Thus friction is desirable or undesirable depending upon the condition and types of work.
Q. 9: A body on contact with a surface is being pulled along it with force increasing from zero. How
does the state of motion of a body change with force. Draw a graph and explain.
Sol.: When an external force is applied on a body and increases
gradually then initially a static friction force acts on the body
which is exactly equal to the applied force and the body will
remain at rest. The graph is a straight line for this range of
force shown by OA on the graph. When the applied force
reaches to a value at which body just starts moving, then the
value of this friction force is known as limiting friction. Further
increase in force will cause the motion of the body and the
friction in this case will be dynamic friction. This dynamic
friction remains constant with further increase in force.
Points to be Remembered
1. If applied force is not able to start motion; frictional force will be equal to applied force.
2. If applied force is able to start motion, and then applied force will be greater than frictional force.
3. The answer will never come in terms of normal reaction.
4. Assuming the body is in limiting equilibrium.
Solved Problems on Horizontal Plane
Force O
Limiting Friction
Dynamic Friction
A
Fig. 9.5
Friction / 195
Q. 10: A body of weight 100N rests on a rough horizontal surface (µ = 0.3) and is acted upon by a
force applied at an angle of 300 to the horizontal. What force is required to just cause the
body to slide over the surface?
Sol.: In the limiting equilibrium, the forces are balanced. That is
∑H = 0;
F = Pcosθ
∑V = 0;
R = W – Psinθ
Also F = µR
P.cosθ = µ(W– P.sinθ)
P.cosθ = µ.W– µ.P.sinθ
µ.P.sinθ + P.cosθ = µ.W
P(µ.sinθ + cosθ) = µ.W
P = µ.W/(Cosθ + µ.sinθ)
= 0.3× 100 / (Cos30° + 0.3sin30°)
= 29.53N .......Ans
Q. 11: A wooden block of weight 50N rests on a horizontal plane. Determine the force required
which is acted at an angle of 150 to just (a) Pull it, and (b) Push it. Take coefficient friction
= 0.4 between the mating surfaces. Comment on the result.
F R = m
W = 50 N
q = 15°
P
1
Motion
R
F R = m
W
P
2
q = 15°
Motion
R
Fig 9.7 Fig 9.8
Sol.: Let P
1
be the force required to just pull the block. In the limiting equilibrium, the forces are balanced.
That gives
∑H = 0; F = P
1
cosθ
∑V = 0; R = W – P
1
sinθ
Also F = µR
µ(W – P
1
sinθ) = P
1
cosθ
or P
1
= µW / (cosθ + µsinθ)
= 0.4 × 50 /( cos 15° + 0.4 sinl5°)
= 18.70N .......Ans
(b) Let P
2
be the force required to just push the block. With reference to the free body diagram
(Fig. 9.8),
F R = m
W = 100 N
q = 30°
P
Motion
R
Fig. 9.6
196 / Problems and Solutions in Mechanical Engineering with Concept
Let us write the equations of equilibrium,
∑H = 0; F = P
2
cosθ
∑V = 0; R = W + P
2
Sinθ
Also F = µR
µ (W + P
2
sinθ) = P
2
cosθ
or P
2
= µW / (cosθ – µsinθ)
= 0.4 × 50 /( cos 15° – 0.4 sinl5°)
= 23.17N .......Ans
Comments. It is easier to pull the block than push it.
Q. 12: A body resting on a rough horizontal plane required a pull of 24N inclined at 30º to the plane
just to move it. It was also found that a push of 30N at 30º to the plane was just enough to
cause motion to impend. Make calculations for the weight of body and the coefficient of
friction.
Sol.: ∑H = 0; F = P
1
cosθ
∑V = 0; R = W – P
1
sinθ
Also F = µR
µ(W – P sinθ) = P
1
cosθ
or P
1
= µ W / (cosθ + µsinθ) ...(i)
With reference to the free body diagram (Fig (9.9) when push is applied)
∑H = 0; F = P
2
cosθ
∑V = 0; R = W + P
2
sinθ
Also F = µR
µ(W + P
2
sin θ) = P
2
cosθ
P
2
= µW/(cosθ – µsinθ) ...(ii)
From expression (i) and (ii),
F R = m
W
q = 30°
P1
=
2
4
N
Motion
R
F R = m
W
q = 30°
P1
=
2
4
N
Motion
R
Fig 9.9 Fig 9.10
P
1
/ P
2
= (cos θ – µsin θ )/ (cos θ + µsin θ)
24/30 = (cos 30°– µ sin 30°)/ (cos 30°+ µsin 30°)
= (0.866 – 0.5µ)/(0.866 + 0.5µ)
0.6928 + 0.4µ = 0.8660.5u
On solving
µ = 0.192 .......Ans
Putting the value of µ in equation (i) we get the value of W
W = 120.25N .......Ans
Friction / 197
Q. 13: A block weighing 5KN is attached to a chord, which passes over a frictionless pulley, and
supports a weight of 2KN. The coefficient of friction between the block and the floor is 0.35.
Determine the value of force P if (i) The motion is impending to the right (ii) The motion is
impending to the left.
3
0
°
2kN
5kN
P
5kN
R
F R = m
3
0
°
P
2kN
5kN
R
F R = m
30°
P
2kN
Fig 9.11 Fig 9.12 Fig 9.13
Sol.: Case1
From the FBD of the block,
∑V = 0 → –5 + R + 2sin30º = 0
R = 4KN
∑H = 0 → –P + 2cos30º – 0.35N = 0
P = 2cos30° – 0.35 × 4 = 0
P = 0.332KN .......Ans
Case2: Since the motion impends to the left, the friction force is directed to the right, from the FBD
of the block:
∑V = 0 → –5 + R + 2sin30° = 0
R = 4KN
∑H = 0 → –P + 2cos30° + 0.35N = 0
P = 2cos30° + 0.35 × 4 = 0
P = 3.132KN .......Ans
Q. 14: A block of 2500N rest on a horizontal plane. The coefficient of friction between block and the
plane is 0.3. The block is pulled by a force of 1000N acting at an angle 30º to the horizontal.
Find the velocity of the block after it moves over a distance of 30m, starting from rest.
F = R
W
30°
1000 N
R
Fig 9.14
Sol.: Here ∑V = 0 but ∑H ≠ 0, Because ∑H is converted into ma
∑V = 0
R + 1000sin30° – W = 0, W = 2500N
R = 2000N ...(i)
198 / Problems and Solutions in Mechanical Engineering with Concept
∑H ≠ 0
∑H = µR – 1000cos30° = 266.02N ...(ii)
Since ∑H ≠ 0
By newtons third law of motion
F = ma
266.02 = (2500/g) × (v
2
– u
2
)/2.s ⇒ v
2
= u
2
+ 2as
u = 0, v
2
= {266.02 × 2 × s × g}/2500
v
2
= {266.02 × 2 × 30 × 9.71}/2500
v = 7.91m/sec .......ANS
Q. 15: Homogeneous cylinder of weight W rests on a horizontal floor in contact with a wall (Fig
12.15). If the coefficient of friction for all contact surfaces be µ, determine the couple M acting
on the cylinder, which will start counter clockwise rotation.
r
w
m
m R
1
m R
2
R
1
W
R
2
Fig 9.15 Fig 9.16
Sol.: ∑H = 0 ⇒ R
1
– µR
2
= 0
R
1
= µR
2
...(i)
∑V = 0 => R
2
+ µR
1
= W ...(ii)
Putting the value of R
1
in equation (ii), we get
R
2
+ µ2R
2
= W
R
2
= W/(1 + µ
2
) ...(ii)
Putting the value of R
2
in equation (i), we get
R
1
= µW/(1 + µ
2
) ...(iv)
Taking moment about point O, We get
M
O
= µR
1
r + µR
2
r
= µr{R
1
+ R
2
}
= µr{(µW/(1 + µ
2
)) + (W/(1 + µ
2
))} = µrW{(1 + µ)/(1 + µ
2
)}
M
O
= µrW(1 + µ)/(1 + µ
2
) .......ANS
Q. 16: A metal box weighing 10KN is pulled along a level surface at uniform speed by applying a
horizontal force of 3500N. If another box of 6KN is put on top of this box, determine the force
required.
Friction / 199
3500
R
W = 10 KN
m R
R
1
P
m R
1
10 + 6 = 16 KN
Fig 9.17 Fig 9.18
Sol.: In first case as shown in fig 12.17
∑H = 0
µR = 3500 ...(i)
∑V = 0
R = W = 10KN = 10000 ...(ii)
Putting the value of R in equation (i)
µ = 0.35 ...(iii)
Now consider second case: as shown in fig 9.18
Now normal reaction is N
1
,
∑H = 0
P – µR
1
= 0
P = µR
1
...(iv)
∑V = 0
R
1
= W = 10KN + 6KN = 16000
R
1
= 16000 ...(v)
Putting the value of R
1
in equation (iv)
P = 0.35 × 16000
P = 5600N .......ANS
Q. 17 Block A weighing 1000N rests over block B which weights 2000N as shown in fig (9.19). Block
A is tied to wall with a horizontal string. If the coefficient of friction between A and B is 1/
4 and between B and floor is 1/3, what should be the value of P to move the block B. If (1)
P is horizontal (2) P is at an angle of 300 with the horizontal.
R
1
m R
1
T
1000 N
P
B
A
W
P
R
2
m R
1
m R
2
R
1
Fig 9.19 Fig 9.20 Fig 9.21
Sol.: (a) When P is horizontal
Consider FBD of block A as shown in fig 12.20.
200 / Problems and Solutions in Mechanical Engineering with Concept
∑V = 0
R
1
= W = 1000
R
1
= 1000 ...(i)
∑H = 0
T = µ
1
R
1
= 1/4 × 1000 = 250
T = 250N ...(ii)
Consider FBD of block B as shown in fig 9.21.
∑V = 0; R
2
– R
1
– W = 0
R
2
= 1000 + 2000 Fig 9.22
R
2
= 3000 N ...(iii)
∑H =0
P = µ
1
R
1
+ µ
2
R
2
= 250 + 1/3 × 3000
P = 1250N .......ANS
(2) When P is inclined at an angle of 30° Consider fig 9.22
∑H = 0
Pcos 30° = µ
1
R
1
+ µ
2
R
2
= 250 + 1/3 × R
2
R
2
= 3(Pcos300 – 250) ...(iv)
∑V = 0
R
2
– R
1
– W + Psin30° = 0
R
2
+ Psin30° = R
1
+ W = 3000 ...(v)
Putting the value of R
2
in equation (v)
3(Pcos30° – 250) + 0.5 × P = 3000
On solving P = 1210.43N .......ANS
Q. 18: Two blocks A and B of weight 4KN and 2KN respectively are in equilibrium position as shown
in fig (9.23). Coefficient of friction for both surfaces are same as 0.25, make calculations for
the force P required to move the block A.
W
b
30°
F R
b b
= m
R
b
30°
A
B
P
Fig 9.23 Fig 9.24
Sol.: Considering equilibrium of block B. Resolving the force along the horizontal and vertical directions:
Tcos30° –µR
b
= 0;
Tcos30° = µR
b
...(i)
R
b
+ Tsin30° – W
b
= 0;
Tsin30° = Wb – Rb ...(ii)
Dividing Equation (i) and (ii), We get
tan30° = (W
b
– R
b
)/µR
b
30°
W
B
P
R
2
m R
1
m R
2
R
1
F
b
F = R
a a
R + W
b a
P
R
a
Fig. 9.25
Friction / 201
0.5773 = (2– R
b
)/0.25R
b
;
0.1443R
b
= 2– R
b
R
b
= 1.748N
F
b
= µR
b
= 0.25 × 1.748 = 0.437N
Considering the equilibrium of block A: Resolving the forces along the horizontal and vertical directions,
F
b
+ µR
a
– P = 0; P = F
b
+ µR
a
R
a
– R
b
– W
a
= 0; R
a
= R
b
+W
a
= 1.748 + 4 = 5.748
P = 0.437 + 0.25 × 5.748
P = 1.874N .......Ans
Q. 19: Determine the force P required to impend the motion of the block B shown in fig (9.26). Take
coefficient of friction = 0.3 for all contact surface.
B
R
B
500 N
P
F R
1
= 0.3
A
F
2
= 0.3 R
B
R
A
R
A
F
A
= 0.3
T
A
300N
400N
500N
300N
A
P
C
B
Fig 9.26 Fig 9.27 (a) Fig 9.27 (b)
Sol.: Consider First FBD of block A Fig 12.27 (a)
∑V = 0 → R
A
= 300N
∑H = 0 → T = 0.3NA
T = 90N
Consider FBD of Block B
∑V = 0 → R
B
= R
A
+ 500
R
B
= 800N
∑H = 0 → P = 0.3N
A
+ 0.3R
B
= 0.3 (300 + 800)
P = 330N .......Ans
Q. 20: Block A of weight 520N rest on the horizontal top of block B having weight 700N as shown
in fig (9.28). Block A is tied to a support C by a cable at 300 horizontally. Coefficient of
friction is 0.4 for all contact surfaces. Determine the minimum value of the horizontal force
P just to move the block B. How much is the tension in the cable then.
W
A
= 520 N
W
B
= 700 N
P
T
30°
B
A
R
1
R
1
m R
1
30
W
W
B
m R
1
m R
2
P
T
R
2
Fig 9.28 Fig 9.29 Fig 9.30
202 / Problems and Solutions in Mechanical Engineering with Concept
Sol.: Consider First FBD of block A Fig 9.29
∑H = 0
µR
1
= Tcos30°
0.4R
1
= 0.866T
R
1
= 2.165T ...(i)
∑V = 0
W = R
1
+ Tsin30°
520 = 2.165T + 0.5T
520 = 2.665T
T = 195.12N ...(ii)
Putting in (i) we get
R
1
= 422.43N ...(iii)
Consider First FBD of block A Fig 9.30
∑V = 0
R
2
= R
1
+ W
B
R
2
= 422.43 + 700
R
2
=1122.43N ...(iv)
∑H = 0
P = µR
1
+ µR
2
P = 0.4(422.43 + 1122.43)
P = 617.9N .......ANS
Q. 21: Explain the different cases of equilibrium of the body on rough inclined plane.
Sol.: If the inclination is less than the angle of friction, the body will remain in equilibrium without any
external force. If the body is to be moved upwards or downwards in this condition an external force is
required. But if the inclination of the plane is more than the angle of friction, the body will not remain in
equilibrium. The body will move downward and an upward external force will be required to keep the body
in equilibrium.
Such problems are solved by resolving the forces along the plane and perpendicular to the planes. The
force of friction (F), which is always equal to µ.R is acting opposite to the direction of motion of the body
CASE 1: magnitude of minimum force ‘p’ which is required to move the body up the plane. When
‘p’ is acted with an angle of ϕ.
m
R
P
O
W
X
A
R
Fig 9.31
Resolving all the forces Parallel to Plane OA:
PcosΦ – µR – W.sinα = 0 ...(i)
Friction / 203
Resolving all the forces Perpendicular to Plane OA:
R + PsinF – W.cosα = 0 ...(i)
Putting value of ‘R’ from (ii) in equation (i) we get
P = W.[(µ.cosα + sinα)/( µ.sinΦ + cosΦ)]
Now putting µ = tanθ, on solving
P = W.[sin(α + θ)/cos(Φ – θ)] ...(iii)
Now P is minimum at cos(Φ – θ) is max
i.e., cos(Φ – θ) = 1 or Φ – θ = 0 i.e., Φ = θ
P
min
= W.sin(θ + Φ)
CASE2: magnitude of force ‘p’ which is required to move the body down the plane. When ‘p’ is acted
with an angle of ϕ.
m
R
P
O
W
X
A
R
Fig 9.32
Resolving all the forces Parallel to Plane OA:
PcosΦ + µR – W.sinα = 0 ...(i)
Resolving all the forces Perpendicular to Plane OA:
R + PsinΦ – W.cosα = 0 ...(ii)
Putting value of ‘R’ from (ii) in equation (i) we get
P = W.[(sinα – µ.cosα)/(cosΦ – µ.sinΦ)]
Now putting µ = tanθ, on solving
P = W.[sin(α – Φ)/cos(Φ + θ)]
CASE3: magnitude of force ‘p’ which is required to move the body down the plane. When ‘p’ is acted
horizontally
m
R
P
O
W
X
A
R
Fig 9.33
Resolving all the forces Parallel to Plane OA:
Pcosα + µR – W.sinα = 0 ...(i)
Resolving all the forces Perpendicular to Plane OA:
204 / Problems and Solutions in Mechanical Engineering with Concept
R – Psinα – W.cosα = 0 ...(ii)
Putting value of ‘R’ from (ii) in equation (i) we get
P = W.[(sinα – µ.cosα)/(cosα + µ.sinα)]
Now putting µ = tanθ, on solving,
P = W.tan(α – θ)
CASE4: magnitude of force ‘p’ which is required to move the body up the plane. When ‘p’ is acted
horizontally
m
R
P
O
W
X
A
R
Fig 9.34
Resolving all the forces Parallel to Plane OA:
Pcosα – µR – W.sinα = 0 ...(i)
Resolving all the forces Perpendicular to Plane OA:
R – Psinα – W.cosα = 0 ...(ii)
Putting value of ‘R’ from (ii) in equation (i) we get
P = W.[(sinα + µ.cosα)/(cosα – µ.sinα)]
Now putting µ = tanθ, on solving,
P = W.tan(α αα αα + θ θθ θθ)
Problems on Rough Inclined Plane
Q. 22: Determine the necessary force P acting parallel to the plane as shown in fig 9.35 to cause
motion to impend. µ = 0.25 and pulley to be smooth.
P
45°
4
5
0
N
1350 N
F
1
R
1
P
x
T
y
450N
R
2
F
2
T
1350
Fig 9.35 Fig 9.36 Fig 9.37
Sol.: Since P is acting downward; the motion too should impend downwards.
Consider first the FBD of 1350N block, as shown in fig (9.37)
∑V = 0
R
2
– W = 0
R
2
= 1350N ...(i)
∑H = 0
Friction / 205
– T + µR
2
= 0
Putting the value of R
2
and µ
T = 0.25(1350)
= 337.5N ...(ii)
Now Consider the FBD of 450N block, as shown in fig (9.36)
∑V = 0
R
1
– 450sin45° = 0
R
1
= 318.2 N ...(i)
∑H = 0
T – P + µR
1
– 450sin45° = 0
Putting the value of R
1
, µ and T we get
P = T + µR
1
– 450sin45° = 0
= 337.5 + 0.25 × 318.2  450sin45°
P = 98.85 N .......ANS
Q. 23: Determine the least value of W in fig(9.38) to keep the system of connected bodies in equilibrium
µ for surface of contact between plane AC and block = 0.28 and that between plane BC and
block = 0.02
R
1
2000N
F R
2 2
= 0.2
y
30°
T
W
2000N
C
60°
30°
T
W
R
2
R
2
F
2
= 0.8
60°
60°
x
y
Fig 9.38 Fig 9.39 Fig 9.40
Sol.: For least value W, the motion of 2000N block should be impending downward.
From FBD of block 2000N as shown in fig 12.39
∑V = 0
R
1
– 2000cos30° = 0
R
1
= 1732.06N ...(i)
∑H = 0
T + µ
1
R
1
– 2000sin30° = 0
T = 2000sin30° – 0.20 × 1732.06, T = 653.6N ...(ii)
Now Consider the FBD of WN block, as shown in fig (9.40)
∑V = 0
R
2
= Wcos60° = 0
R
2
= 0.5W N ...(iii)
∑H = 0
T – µ
2
R
2
– Wsin60° = 0
653.6 = Wsin60° – 0.28 × 0.5W
W
LEAST
= 649.7N .......ANS
206 / Problems and Solutions in Mechanical Engineering with Concept
Q. 24: Block A and B connected by a rigid horizontally bar planed at each end are placed on inclined
planes as shown in fig (9.41). The weight of the block B is 300N. Find the limiting values of
the weight of the block A to just start motion of the system.
60°
45°
W = 300 N
m
A
= 0.25
B
A
// B = 0.3
300N
45°
O C
R
Fig 9.41 Fig 9.42
Sol.: Let W
a
be the weight of block A. Consider the free body diagram of B. As shown in fig 12.42. And
Assume A
B
be the Axis of reference.
∑V = 0;
Rsin45° – µBRcos45° – 300 = 0
On solving, R = 606.09N ...(i)
∑H = 0;
C – RCos45° – µBRsin45° = 0 ...(ii)
Putting the value of R, we get
C = 557.14N ...(iii)
Where C is the reaction imparted by rod.
Consider the free body diagram of block A as shown in fig 9.43
∑H = 0;
C + µARcos60° – Rcos30° = 0 ...(iv)
Putting all the values we get
R = 751.85N ...(v)
∑V = 0;
µ
A
Rsin60° + Rsin60° – W = 0
On solving, W = 538.7N ...(vi)
Hence weight of block A = 538.7N .......ANS
Q. 25: What should be the value of the angle shown in fig 9.44 so that the motion of the 90N block
impends down the plane? The coefficient of friction for the entire surface = 1/3.
q
9
0
N
3
0
N
D
i
r
e
c
t
i
o
n
o
f
M
o
t
i
o
n
3
0
s
i
n
q
T
F
=
R
1
1
m
30 cos q
30N
q
R
1
Fig 9.44 Fig 9.45
60
C
30°
R
m
A
R
W
Fig. 9.43
Friction / 207
Sol.: Consider the equilibrium of block 30N
∑V = 0;
R
1
– 30cosθ = 0,
R
1
= 30cosθ ...(i)
∑H = 0;
T – µR
1
– 30sinθ = 0,
T = 10cosθ + 30sinθ ...(ii)
Consider the equilibrium of block 90N
∑V = 0;
R
2
– R
1
– 90cosθ = 0
R
2
= 120cosθ ...(iii)
∑H = 0;
90sinθ – µR
1
– µR
2
= 0,
90sinθ = 10cosθ + 40sinθ ...(vi)
tanθ = 5/9 i.e., θ = 29.050
Q. 26: A block weighing 200N is in contact with an inclined plane (Inclination = 30º). Will the block
move under its own weight. Determine the minimum force applied (1) parallel (2) perpendicular
to the plane to prevent the motion down the plane. What force P will be required to just cause
the motion up the plane, µ = 0.25?
30°
W = 200N
R
m R
30°
O
W
Fig 9.47 Fig 9.48
m R
m R
m R
P
P
P
W
W
W
30°
30°
30°
R
R R
Fig 9.49 Fig 9.50 Fig 9.51
Sol.: Consider the FBD of block as shown in fig 9.48
From the equilibrium condition
Sum of forces perpendicular to plane = 0
R – Wcos30° = 0;
R = Wcos30° ...(i)
Sum of forces parallel to plane = 0
µR – Wsin30° = 0 ...(ii)
D
i
r
e
c
t
i
o
n
o
f
M
o
t
i
o
n
9
0
s
i
n
q
90N
R
2
90 cos q
F
=
R
1
m
1
F
=
R
2
2
m
q
q
R
1
Fig 9.46
208 / Problems and Solutions in Mechanical Engineering with Concept
Now body will move down only if the value of µR is less than Wsin30°
Now, µR = 0.25 × W(0.866) = 0.2165W ...(iii)
And Wsin30° = 0.5W ...(vi)
Since value of (iv) is less than value of (iii) So the body will move down.
(i) When Force acting parallel to plane as shown in fig 9.49
Frictional force is acting up the plane
∑V = 0;
R = 0.216W ...(v)
∑H = 0;
P = Wsin30° – µR
P = 0.5 × 200 – 0.216 × 200
P = 56.7N .......ANS
(ii) When Force acting perpendicular to plane as shown in fig 12.50
Frictional force is acting up the plane
∑H = 0;
Wsin30° – µR = 0
R = 400 ...(vii)
∑V = 0;
P + R = Wcos30°
P = 0.866 × 200 – 400
P = –226.79N .......ANS
(iii) The force P required to just cause the motion up the plane as shown in fig 12.51. Frictional force
is acting down the plane
Sum of force perpendicular to plane = 0
R = Wcos30
= 172.2N ...(viii)
Sum of force Parallel to plane = 0
P – µR – Wsin30 = 0
P = 0.25 × 172.2 – 200sin30
P = 143.3N .......ANS
Q. 27: A body of weight 50KN rests in limiting equilibrium on a rough plane, whose slope is 30º. The
plane is raised to a slope of 45º; what force, applied to the body parallel to the inclined plane,
will support the body on the plane.
Sol.: Consider When the slope of the plane be 30°
Sum of forces parallel to plane = 0
µR – Wsin30° = 0 ...(i)
Sum of forces perpendicular to plane = 0
R – Wcos30° = 0
R = Wcos30° ...(ii)
Putting the value of R in equation (i) We get
µ = tan30º = 0.577 ...(iii)
Friction / 209
30°
m R
W
30°
R
45°
m R
W
R
P
30°
W
Fig. 9.52 Fig. 9.53 Fig. 9.54
Now consider the case when the slope is 45°, Let Force P required to support the body.
In this case
Sum of forces parallel to plane = 0
P – µR – Wsin45° = 0 ...(iv)
Sum of forces perpendicular to plane = 0
R – Wcos45° = 0
R = Wcos45° ...(v)
Putting the value of R in equation (iv) we get
P = µWcos45° – Wsin45°
P = 15.20KN .......ANS
Q. 28: Force of 200N is required just to move a certain body up an inclined plane of angle 150, the
force being parallel to plane. If angle of indication is made 20º the effort again required
parallel to plane is found 250N. Determine the weight of body and coefficient of friction.
m R
P = 200 N
W
15°
15°
R
m R
P = 200 N
W
20°
R
Fig 9.55 Fig 9.56
Sol.:
Case1
Consider When the slope of the plane be 15º
Sum of forces parallel to plane = 0
P – µR – Wsin15° = 0 ...(i)
Sum of forces perpendicular to plane = 0
R = Wcos15° ...(ii)
Putting the value of (ii) in (i), We get
P = µWcos15° + Wsin15° = 0
200 = 0.96µW + 0.25W ...(iii)
210 / Problems and Solutions in Mechanical Engineering with Concept
Case2
Consider When the slope of the plane be 20º
Sum of forces parallel to plane = 0
P – µR – Wsin20° = 0 ...(iv)
Sum of forces perpendicular to plane = 0
R = Wcos20° ...(v)
Putting the value of (v) in (iv), We get
P = µWcos20° + Wsin20° = 0
250 = 0.939µW + 0.34W ...(vi)
Solved equation (iii) and (vi) we get
W = 623.6N and µ = 0.06 .......ANS
Q. 29: A four wheel drive can as shown in fig (9.57) has mass of 2000Kg with passengers. The
roadway is inclined at an angle with the horizontal. If the coefficient of friction between the
tyres and the road is 0.3, what is the maximum inclination that can climb?
W mg =
q
0
.2
5
m
0.5 m
1 m
R
2
g
q
m R
1
1
W
2
m R
2
4
0.25m
R
2
0.5m
1m
Fig 9.57 Fig 9.58
Sol.: Let the maximum value for inclination is θ for body to remain stationary.
Let 0.25m distance is the distance between the inclined surface and C.G. Now,
Sum of forces parallel to plane = 0
Wsinθ = µ(R
1
+ R
2
) ...(i)
Sum of forces perpendicular to plane = 0
R
1
+ R
2
= Wcosθ ...(ii)
Putting the value of (ii) in (i), We get
Wsinθ = µ(Wcosθ)
Or µ = tanθ
θ = tan
–1
(0.3)
θ θθ θθ = 16.69º .......ANS
Q. 30: A weight 500N just starts moving down a rough inclined plane supported by force 200N acting
parallel to the plane and it is at the point of moving up the plane when pulled by a force of
300N parallel to the plane. Find the inclination of the plane and the coefficient of friction
between the inclined plane and the weight.
Sol.: In first case body is moving down the plane, so frictional force is acting up the plane
Let θ be the angle of inclination and µ be the coefficient of friction.
Friction / 211
q
500
m R
200N
R
q
500
m R
300
R
Fig. 9.59 Fig. 9.60
Sum of forces parallel to plane = 0
200 + µR = 500sinθ ...(i)
Sum of forces perpendicular to plane = 0
R = 500cosθ ...(ii)
Putting the value of (ii) in equation (i)
200 + 500µcos? = 500sinθ ...(iii)
Now 300N is the force when applied to block, it move in upward direction. Hence in this case frictional
force acts downward.
Sum of forces perpendicular to plane = 0
R = 500cosθ ...(vi)
Sum of forces parallel to plane = 0
300 = µR + 500sinθ ...(v)
Putting the value of (iv) in equation (v)
300 = 500µcosθ + 500sinθ ...(vi)
Adding equation (iii) and (vi), We get
Sinθ = 1/2; or θ θθ θθ = 30° ........ANS
Putting the value in any equation we get
µ = 0.115 .......ANS
Q. 31: What is ladder friction? How many forces are acting on a ladder?
Sol.: A ladder is an arrangement used for climbing on the walls It essentially consists of two long uprights
of wood or iron and connected by a number of cross bars. These cross bars are called rungs and provide
steps for climbing. Fig 9.61. shows a ladder AB with its end A resting on the ground and end B leaning
against a wall. The ladder is acted upon by the following set of forces:
F R
a a
= m
A
q
W
R
a
B
R
b
F R
b b
= m
Fig. 9.61
212 / Problems and Solutions in Mechanical Engineering with Concept
(1) Weight W acting downwards at its mid point.
(2) Normal reaction Rh and friction force F
h
= µR
h
at the end B leaning against the wall. Since the
ladder has a tendency to slip downwards, the friction force will be acting upwards. If the wall is
smooth (µ = 0), the friction force will be zero.
(3) Normal reaction R
a
and friction force F
a
=µR
a
at the end A resting on the floor. Since the ladder,
upon slipping, tends to move away from the wall, the direction of friction force will be towards
the wall.
Applying equilibrium conditions, the algebraic sum of the horizontal and vertical component of forces
would be zero.
Problems on Equilibrium of The Body on Ladder
Q. 32: A ladder 5m long rests on a horizontal ground and leans against a smooth vertical wall at an
angle 70° with the horizontal. The weight of the ladder is 900N and acts at its middle. The
ladder is at the point of sliding, when a man weighing 750N stands 1.5m from the bottom of
the ladder. Calculate coefficient of friction between the ladder and the floor.
Sol.: Forces acting on the ladder is shown in fig 9.62
Resolving all the forces vertically,
RV = R – 900 – 750 = 0
R = 1650N ...(i)
Now taking moment about point B,
R × 5sin20° – F
r
× 5cos20°
– 900 × 2.5sin20° – 750 × 3.5sin20° = 0
Since F
r
= µR, and R = 1650N; F
r
= 1650µ
Putting, the value of R and F
r
µ = 0.127 .......ANS Fig. 9.62
Q. 33: A uniform ladder of length 13m and weighing 250N is placed against a smooth vertical wall
with its lower end 5m from the wall. The coefficient of friction between the ladder and floor
is 0.3. Show that the ladder will remain in equilibrium in this position. What is the frictional
force acting on the ladder at the point of contact between the ladder and the floor?
Sol.: Since the ladder is placed against a smooth vertical wall, therefore there will be no friction at the point
of contact between the ladder and wall. Resolving all the force horizontally and vertically.
∑H = 0, F
r
– R
2
= 0 ...(i)
∑V = 0, R
1
– 250 ...(ii)
From the geometry of the figure, BC = 12m
Taking moment about point B,
R
1
× 5 – F
r
× 12 – 250 × 2.5 = 0
F
r
= 52N .......ANS
For equilibrium of the ladder, Maximum force of friction available
at the point of contact between the ladder and the floor = µR
= 0.3 × 250 = 75N .......ANS
Thus we see that the amount of the force of friction available at
the point of contact (75N) is more than force of friction required for
equilibrium (52N). Therefore, the ladder will remain in equilibrium in
this position.
1
3
m
Fr
B
R
2
C
R
1
250N
5m
Fig. 9.63
70°
1
.
5
m
Fr
B
C
R
900N
750N
5m
Friction / 213
Q. 34: A uniform ladder of 7m rests against a vertical wall with which it makes an angle of 45º, the
coefficient of friction between the ladder and the wall is 0.4 and that between ladder and the
floor is 0.5. If a man, whose weight is one half of that of the ladder, ascends it, how high will
it be when the ladder slips?
Sol.: Let,
X = Distance between A and the man, when the ladder is at the point of slipping.
W = Weight of the ladder
Weight of man = W/2 = 0.5W
Fr
1
= 0.5R
1
...(i)
Fr
2
= 0.4R
2
...(ii)
Resolving the forces vertically
R
1
+ Fr
2
– W – 0.5W = 0
R
1
+ 0.4R
2
= 1.5W ...(iii)
Resolving the forces Horizontally
R
2
– Fr
1
= 0; R
2
= 0.5R
1
...(iv)
Solving equation (iii) and (iv), we get
R
2
= 0.625W, Fr
2
= 0.25W
Now taking moment about point A,
W × 3.5cos45° + 0.5W × xcos45° – R
2
× 7sin45° – Fr
2
× 7cos45°
Putting the value of R
2
and F
r2
, we get
X = 5.25m .......ANS
WEDGE FRICTION
Q. 35: Explain how a wedge is used for raising heavy loads. Also gives principle.
Sol.: Principle of wedge : A wedge is small piece of material with two of their opposite faces not parallel.
To lift a block of weight W, it is pushed by a horizontal force P which lifts the block by imparting a reaction
on the block in a direction 1
r
to meeting surface which is always greater than the total downward force
applied by block W This will cause a resultant force acting in a upward direction on block and it moves
up.
P
wedge
W
Fig 9.65
Q. 36: Two wedge blocks A and B are employed to raise a load of 2000 N resting on another block
C by the application of force P as shown in Fig. 9.66. Neglecting weights of the wedge blocks
and assuming coefficient of friction µ = 0.25 for all the surfaces, determine the value of P for
impending upward motion of the block C.
Sol.: The block C, under the action of forces P on blocks A and B, tends to move upward. Hence the
frictional forces will act downward. What holds good for block A, the same will hold good for block B.
tan ϕ = µ = 0.25 (given),
45°
Fr
1
B R
2
0.5W
W
x
G
7m
Fr
2
Fig. 9.64
214 / Problems and Solutions in Mechanical Engineering with Concept
where ϕ is the angle of friction
ϕ = 14° Refer Fig. 9.66
Consider the equilibrium of block C : Refer Fig. 9.67
B
A
P P
2000 N
15° 15°
C F
A
R
A
R
B
N
B
B
N
A
A
C
f
F
B
2000 N
M
o
t
i
o
n
M
o
t
i
o
n
Fig 9.66 Fig 9.67
It is acted upon by the following forces :
(i) Load 2000 N,
(ii) Total reaction R
A
offered by wedge block A, and
(iii) Total reaction R
B
offered by wedge block B.
Using Lami's theorem, we get
2000
sin 50° sin (180 29 ) sin (180 29 )
A B
R R
= =
° − ° ° − °
2000
sin 58 sin 29 sin 29
A B
R R
= =
° ° °
R
A
= R
B
2000 sin 29
sin 58
× °
= =
°
1143 N
Refer Fig. 9.69 Fig 9.68
Consider equilibrium of block A:
It is acted upon by the following forces :
(i) Force P,
(ii) R
A
(from block C), and
(iii) Total reaction R offered by horizontal surface.
29°
14°
R
R
N
f
F
P
P
Motion
15°
R
A R
A
Fig 9.69
R
A
R
B
15° 15°
15° 15°
29° 29°
14° 14°
2000 N
Friction / 215
Using Lami’s theorem, we have
[ ] sin 180 (29 14 ) sin (90 14 )
A
R
P
=
° − ° + ° ° + °
sin 137 sin 104
A
R
P
=
° °
P =
1143 sin 137
( 1143 )
sin 104
A
R N
× °
=
°
3
=
1143 0.682
0.97
×
= 803 N
Hence P = 803 N .......ANS
216 / Problems and Solutions in Mechanical Engineering with Concept
+0)264
10
/PPLlC/TlCN CF FRlCTlCN. 8ELT FRlCTlCN
Q. 1: What is belt? How many types of belt are used for power transmission?
Sol: The power or rotary motion from one shaft to another at a considerable distance is usually transmitted
by means of flat belts, Vee belts or ropes, running over the pulley. But the pulleys contain some friction.
Types of Belts
Important types of belts are:
1. Flat belt 2. V belt 3. Circular Belt
Circular belt
Pulley
VBelt
Pulley
Rectangular belt
(a) (b) (c)
Fig 10.1
Flat Belt
The flat belt is mostly used in the factories and workshops. Where a moderate amount of power is to be
transmitted, from one pulley to another, when the two pulleys are not more than 10m apart.
VBelt
The Vbelt is mostly used where a great amount of power is to be transmitted, from one pulley to another,
when the two pulleys are very near to each other.
Circular Belt or Rope
The circular belt or rope is mostly used where a great amount of power is to be transmitted from one pulley
to another, when the two pulleys are more than 5m apart.
Application of Friction: Belt Friction / 217
Q. 2: Explain how many types of belt drive used for power transmission? Also derive their velocity
ratio.
Sol: There are three types of belt drive:
(1) Open belt drive
(2) Cross belt drive
(3) Compound belt drive
(1) Open Belt Drive
When the shafts are arranged in parallel and rotating in the same direction, open belt drive is obtained.
In the diagram 10.2, pulley 'A' is called as driver pulley because it is attached with the rotating shaft.
Driven or
Follower
Pulley
Drive Pulley
Slack Side
Tight Side
Totating Shaft
A
B
+ +
Fig 10.2
Velocity Ratio (V.R.) for Open Belt Drive
D
x
A
+
α
α
α
N
M E
H
F
B
C K
Fig 10.3
Consider a simple belt drive (i.e., one driver and one follower) as shown in fig 10.3.
218 / Problems and Solutions in Mechanical Engineering with Concept
Let
D
1
= Diameter of the driver
N
1
= Speed of the driver in R.P.M.
D
2
, N
2
= Corresponding values for the follower
Length of the belt,
that passes over the driver, in one minute = Π.D
1
.N
1
Similarly,
Length of the belt,
That passes over the follower, in one minute = Π.D
2
.N
2
Since the length of belt, that passes over the driver in one minute is equal to the length of belt that
passes over the follower in one minute, therefore:
Π.D
1
.N
1
= Π.D
2
.N
2
Or, velocity ratio = N
2
/N
1
= D
1
/D
2
If thickness of belt 't' is given then
V.R = N
2
/N
1
= (D
1
+ t)/(D
2
+ t)
(2) Cross Belt Drive
When the shafts are rotating in opposite direction, cross belt drive is obtained.
A
G
K
+
C
N
D
x
F
B
H
α
M
E
Fig 10.4
In the diagram 13.4, pulley 'A' is called as driver pulley because it is attached with the rotating shaft.
Velocity ratio is same as for open belt
V.R. = N
2
/N
1
= D
1
/D
2
If thickness of belt 't' is given then
V.R = N
2
/N
1
= (D
1
+ t)/(D
2
+ t)
(3) Compound Belt Drive
When a number of pulleys are used to transmit power from one shaft to another then a compound belt drive
is obtained.
Application of Friction: Belt Friction / 219
3
1
2
1
4
2
3
4
+
+
+
Driver Pulley
Follower or
Driven Pulley
Fig 10.5
Velocity Ratio for Compound Belt Drive
Speed of last follower = Product of diameter of driver(odd dia)
Speed of first driver Product of diameter of follower(even dia)
N
4
/N
1
= (D1.D
3
)/(D
2
.D
4
)
Q. 3: What is slip of the belt? How slip of belt affect the velocity ratio?
Sol: When the driver pulley rotates, it carries the belt, due to a firm grip between its surface and the belt.
The firm between the pulley and the belt is obtained by friction. This firm grip is known as frictional grip.
But sometimes the frictional grip is not sufficient. This may cause some forward motion of the driver pulley
without carrying the belt with it. This means that there is a relative motion between the driver pulley and
the belt. The difference between the linear speeds of the pulley rim and the belt is a measure of slip.
Generally, the slip is expressed as a percentage. In some cases, the belt moves faster in the forward
direction, without carrying the driver pulley with it. Hence in case of driven pulley, the forward motion of
the belt is more than that of driver pulley.
Slip of belt is generally expressed in percentage(%).
Let v = Velocity of belt, passing over the driver pulley/min
N
1
= Speed in R.P.M. of driver
N
2
= Speed in R.P.M. of follower
S
1
= Slip between driver and belt in percentage
S
2
= Slip between follower and belt in percentage
The peripheral velocity of the driver pulley
= ω
1
.r
1
=
1 1 1
1
2 N ,
× (D /2)
60 60
Π
= ...(i)
Now due to Slip between the driver pulley and the belt, the velocity of belt passing over the driver
pulley will decrease
Velocity of belt =
1 1 1 1 1 1 1
.N .D (, ,
60 60 100 60
s Π Π
× = (1–s
1
/100) ...(ii)
220 / Problems and Solutions in Mechanical Engineering with Concept
Now with this velocity the belt pass over the driven pulley,
Now
Velocity of Driven = Velocity of Belt  Velocity of belt X (S
2
/100)
1 1
.N .D
60
Π
(1–s
1
/100) –
1 1
.N .D
60
Π
(1–s
1
/100)(s
2
/100)
1 1
.N .D
60
Π
(1–s
1
/100)(1–s
2
/100) ...(iii)
But velocity of driven =
2 2
.N .D
60
Π
...(iv)
Equate the equation (iii) and (iv)
1 1
.N .D
60
Π
(1–s
1
/100)(1–s
2
/100) =
2 2
.N .D
60
Π
N
2
D
2
= N
1
D
1
(1–s
1
/100–s
2
/100 + s
1
.s
2
/10,000)
= N
1
D
1
[1(s
1
+s
2
)/100], Neglecting s
1
.s
2
/10,000, since very small
If s
1
+ s
2
= S = Total slip in %
N
2
/N
1
= D
1
/D
2
[1–S/100]
This formula is used when total slip in % is given in the problem
NOTE: If Slip and thickness both are given then, Velocity ratio is,
V.R = N
2
/N
1
=
1
2
(D +t)
(D +t)
[1–s/100]
Q. 4: Write down different relations used in belt drive.
Sol: Let:
D
1
= Diameter of the driver
N
1
= Speed of the driver in R.P.M.
D
2
= Diameter of the driven or Follower
N
2
= Speed of the driven or follower in R.P.M.
R
1
= Radius of the driver
R
2
= Radius of the driven or Follower
t = Belt thickness (if given)
X = Distance between the centers of two pulleys
α = Angle of lap (Generally less than 10º)
θ = Angle of contact (Generally greater than 150º)
(always express in radian.)
µ = Coefficient of friction
s = Total slip in percentage(%)
L = Total length of belt
Formula For Open Belt Drive
V.R. V.R = N
2
/N
1
Thickness is considered V.R = N
2
/N
1
=
1
2
(D +t)
(D +t)
Application of Friction: Belt Friction / 221
Slip is considered V.R = N
2
/N
1
=
1
2
D
D
[1–s/100]
Slip and thickness both are
considered V.R = N
2
/N
1
=
1
2
(D +t)
(D +t)
[1–s/100]
Angle of contact θ = Π – 2α
Angle of lap Sinα = (r
1
–r
2
)/X
Length of belt L = Π (r
1
+ r
2
) +
2
1 2
(r – r )
X
+ 2X
Q. 5: Prove that the ratio of belt tension is given by the T
1
/T
2
= e
µθ µθ µθ µθ µθ
B
T
2
Driven
Pulley
T
1
F
N
R
P
M
2
F = R µ
(T + T) δ
α
α
δθ
θ
δθ
Fig 10.6
Let T
1
= Tension in the belt on the tight side
T
2
= Tension in the belt on the slack side
θ = Angle of contact
µ = Coefficient of friction between the belt and pulley.
α = Angle of Lap
Consider a driven or follower pulley. Belt remains in contact with EBF. Let T
1
and T
2
are the tensions
in the tight side and slack side.
Angle EBF called as angle of contact = Π.–2α
Consider a driven or follower pulley.
Belt remains in contact with NPM. Let T
1
and T
2
are the tensions in the tight side and slack side.
Let T be the tension at point M & (T + δT) be the tension at point N. Let d? be the angle of contact
of the element MN. Consider equilibrium in horizontal Reaction be 'R' and vertical reaction be µR.
Since the whole system is in equilibrium, i.e.,
∑V = 0;
Tsin (90 – δθ/2) + µR  (T + δT)sin(90 – δθ/2) = 0
Tcos (δθ/2) + µR = (T + δT) cos (δθ/2)
Tcos (δθ/2) + µR = Tcos(δθ/2) + δTcos(δθ/2)
µR = δTcos(δθ/2)
222 / Problems and Solutions in Mechanical Engineering with Concept
Since δθ/2 is very small & cos0° = 1, So cos(δθ/2) = 1
µR = δT ...(i)
∑H = 0;
R–Tcos(90 – δθ/2)–(T + δT)cos(90 – δθ/2) = 0
R = Tsin(δθ/2) + (T + δT)sin(δθ/2)
Since δθ/2 is very small So sin(δθ/2) = δθ/2
R = T(δθ/2) + T(δθ/2) + δT(δθ/2)
R = T.δθ + δT(δθ/2)
Since δT(δθ/2) is very small So δT(δθ/2) = 0
R = T.δθ ...(ii)
Putting the value of (ii) in equation (i)
µ.T.δθ = δT
or, δT/T = µ.δθ
Integrating both side:
T 0
1
T 0
2
T/T = , δ µ δθ
∫ ∫
Where θ = Total angle of contact
ln(T
1
/T
2
) = µ.θ
or, T
1
/T
2
= e
µ.θ θθ θθ
Ratio of belt tension = T
1
/T
2
= e
µθ
Belt ratio is also represent as 2.3log(T
1
/T
2
) = µ.θ
Note that θ is in radian
In this formula the main important thing is Angle of contact(θ)
For Open belt drive:
Angle of contact (θ) for larger pulley = Π + 2α
Angle of contact (θ) for smaller pulley = Π – 2α
For cross belt drive:
Angle of contact (θ) for larger pulley = Π + 2α
Angle of contact (θ) for smaller pulley = Π + 2α
(i.e. for both the pulley, it is same)
But for solving the problems, We always take the Angle of contact (θ) for smaller pulley
Hence,
Angle of contact (θ) = Π – 2α – for open belt
Angle of contact (θ) = Π + 2α – for cross belt
Q. 6: Explain how you evaluate power transmitted by the belt.
Sol: Let T
1
= Tension in the tight side of the belt
T
2
= Tension in the slack side of the belt
V = Velocity of the belt in m/sec.
= πDN/60 m/sec, D is in meter and N is in RPM
P = Maximum power transmitted by belt drive
The effective tension or force acting at the circumference of the driven pulley is the difference between
the two tensions (i.e., T
1
– T
2
)
Application of Friction: Belt Friction / 223
Effective driving force = (T
1
–T
2
)
Work done per second = Force X Velocity
= F X V N.m
= (T
1
–T
2
)X V N.m
Power Transmitted = (T
1
–T
2
).V/1000 Kw
(Here T
1
& T
2
are in newton and V is in m/sec)
Note:
1. Torque exerted on the driving pulley = (T
1
–T
2
).R
1
Where R
1
= radius of driving pulley
2. Torque exerted on the driven pulley = (T
1
–T
2
).R
2
Where R
2
= radius of driven pulley
Q. 7: What is initial tension in the belt?
Sol: The tension in the belt which is passing over the two pulleys (i.e driver and follower) when the pulleys
are stationary is known as initial tension in the belt.
When power is transmitted from one shaft to another shaft with the help of the belt, passing over the
two pulleys, which are keyed, to the driver and driven shafts, there should be firm grip between the pulleys
and belt. When the pulleys are stationary, this firm grip is increased, by tightening the two ends of the belt.
Hence the belt is subjected to some tension. This tension is known as initial tension in the belt.
Let To = initial tension in the belt
T
1
= Tension in the tight side
T
2
= Tension in the slack side
To = (T
1
+ T
2
)/2
Q. 8: With the help of a belt an engine running at 200rpm drives a line shaft. The Diameter of the
pulley on the engine is 80cm and the diameter of the pulley on the line shaft is 40cm. A 100cm
diameter pulley on the line shaft drives a 20cm diameter pulley keyed to a dynamo shaft. Find
the speed of the dynamo shaft when: (1) There is no slip (2) There is a slip of 2.5% at each
drive.
100 cm
80 cm
40 cm
20
cm
Dynamo
Shaft
Engine
Shaft
Line
Shaft
Fig 10.7
Sol:
Dia. of driver pulley (D
1
) = 80cm
Dia. of follower pulley (D
2
) = 40cm
Dia. of driver pulley (D
3
) = 100cm
Dia. of follower pulley (D
4
) = 20cm
Slip on each drive, s
1
= s
2
= 2.5
224 / Problems and Solutions in Mechanical Engineering with Concept
Let N
4
= Speed of the dynamo shaft
(i) When there is no slip
Using equation
N
4
/N
1
= (D
1
.D
3
)/(D
2
.D
4
)
N
4
= N
1
X (D
1
.D
3
)/(D
2
.D
4
)
= [(80 X 100) X 200]/(40 X 20)
N
4
= 2000RPM .......ANS
(ii) When there is a slip of 2.5% at each drive
In this case we will have the equation of:
N
4
/N
1
= [(D
1
.D
3
)/(D
2
.D
4
)][1–s
1
/100][1–s
2
/100]
Putting all the values, we get
N
4
= N1 X [(D
1
.D
3
)/(D
2
.D
4
)][1– s
1
/100][1– s
2
/100]
N
4
= 200 X [(80 X 100)/(40 X 20)][1 – 2.5/100][1– 2.5/100]
N
4
= 1901.25R.P.M. .......ANS
Q. 9: Find the length of belt necessary to drive a pulley of 500mm diameter running parallel at a
distance of 12m from the driving pulley of diameter 1600m.
Sol: Given Data
Dia. of driven pulley (D
2
) = 500mm = 0.5m
Radius of driven pulley (r
2
) = 0.25m
Centre distance (X) = 12m
Dia. of driver pulley (D
1
) = 1600mm = 1.6m
Radius of driver pulley (r
1
) =0.8m
Since there is no mention about type of belt(Open or cross type)
So we find out for both the cases.
(i) Length of the belt if it is open
WE know that: L = Π (r
1
+ r
2
) +
2
1 2
(r – r )
X
+ 2X
Putting all the value
L = Π (0.8 + 0.25 ) +
2
(0.8 – 0.25)
12
+ 2 X 12
L = 27.32m .......ANS
(ii) Length of the belt if it is cross
WE know that: L = Π (r
1
+ r
2
) +
2
1 2
(r – r )
X
+ 2 X
Putting all the value
L = Π (0.8 + 0.25 )+
2
(0.8 – 0.25)
12
+ 2 X 12
L = 27.39m .......ANS
Q. 10: Find the speed of shaft driven with the belt by an engine running at 600RPM. The thickness
of belt is 2cm, diameter of engine pulley is 100cm and that of shaft is 62cm.
Application of Friction: Belt Friction / 225
Sol: Given that
Speed of driven shaft (N
2
) = ?
Thickness of belt (t) = 2cm
Diameter of driver shaft (D
2
) = 100cm
Diameter of driven shaft (D
1
) = 62cm
Speed of driver shaft (N
1
) = 600rpm
Since we know that,
V.R = N
2
/N
1
=
1
2
(D +t)
(D +t)
N
2
= N
1
X [(D
1
+ t)/(D
2
+ t)]
Putting all the value,
N
2
= 600 X [(62 + 2)/(100 + 2)]
N
2
= 376.47RPM .......ANS
Q. 11: A belt drives a pulley of 200mm diameter such that the ratio of tensions in the tight side and
slack side is 1.2. If the maximum tension in the belt is not to exceed 240KN. Find the safe
power transmitted by the pulley at a speed of 60rpm.
Sol: Given that,
D
1
= Diameter of the driver = 200mm = 0.2m
T
1
/T
2
= 1.2
Since between T
1
and T
2
, T
1
is always greater than T
2
,
Hence T
1
= 240KN
N
1
= Speed of the driver in R.P.M. = 60PRM
P = ?
We know that
T
1
/T
2
= 1.2
T
2
= T
1
/1.2 =240/1.2 = 200KN ...(i)
V = Velocity of the belt in m/sec.
= πDN/60 m/sec, D is in meter and N is in RPM
= (3.14 X 0.2 X 60)/60 = 0.628 m/sec (ii)
P = (T
1
 T
2
) X V
P = (240  200) X 0.628
P = 25.13KW .......ANS
Q. 12: Find the power transmitted by cross type belt drive connecting two pulley of 45.0cm and
20.0cm diameter, which are 1.95m apart. The maximum permissible tension in the belt is 1KN,
coefficient of friction is 0.20 and speed of larger pulley is 100rpm.
Sol: Given that
D
1
= Diameter of the driver = 45cm = 0.45m
R
1
= Radius of the driver = 0.225m
D
2
= Diameter of the driven = 20cm = 0.2m
R
2
= Radius of the driven = 0.1m
X = Distance between the centers of two pulleys = 1.95m
T
1
= Maximum permissible tension = 1000N
µ = Coefficient of friction = 0.20
N
1
= Speed of the driver(Larger pulley) in R.P.M. = 100RPM
226 / Problems and Solutions in Mechanical Engineering with Concept
Since we know that,
Power Transmitted = (T
1
–T
2
).V/1000 Kw ...(i)
Tension is in KN and V is in m/sec
First ve find the velocity of the belt,
V = Velocity of the belt in m/sec.
Here we take diameter and RPM of larger pulley
= πDN/60 m/sec, D is in meter and N is in RPM
= (3.14 X0.45 X 100) /60
= 2.36m/sec ...(ii)
Now Ratio of belt tension, T
1
/T
2
= e
µ.θ
...(iii)
Here we don't know the value of θ, For θ, first find the value of α, by the formula,
Angle of Lap for cross belt α = sin
–1
(r
1
+ r
2
)/X
= sin
–1
(0.225 + 0.1)/1.95
= 9.59° ...(iv)
Now Angle of contact (θ) = Π + 2α  for cross belt
θ = Π + 2 X 9.59°
= 199.19°
= 199.19°(Π/180°) = 3.47rad ...(v)
Now putting all the value in equation (iii)
We get
1000/T
2
= e
(0.2)(3.47)
T
2
= 498.9 N ...(vi)
Using equation (i), we get
P = [(1000 – 498.9) X 2.36 ]/1000
P = 1.18KW .......ANS
Q. 13: A flat belt is used to transmit a torque from pulley A to pulley B as shown in fig 7.8. The
radius of each pulley is 50mm and the coefficient of friction is 0.3. Determine the largest
torque that can be transmitted if the allowable belt tension is 3KN.
Sol: Radius of each pulley = 50mm,
R
1
= R
2
= 50mm
R
1
= Radius of the driver = 50mm
R
2
= Radius of the driven = 50mm
θ = Angle of contact(In radian) = 1800 = p,
µ = Coefficient of friction = 0.3
B
A
200 mm
Fig 13.8
T
1
= Allowable tension = 3KN,
Application of Friction: Belt Friction / 227
T
1
always greater than T
2
Using the relation T
1
/T
2
= e
µθ
Putting all the value,
3/T
2
= e
(0.3)(π)
On solving T
2
= 1.169KN .......ANS
Since Radius of both pulley is same;
So, Torque exerted on both pulley is same and
= (T
1
–T
2
).R
1
= (T
1
–T
2
).R
2
Putting all the value we get,
(3 – 1.169) X 50 = 91.55 KNmm .......ANS
Q. 14: An open belt drive connects two pulleys 120cm and 50cm diameter on parallel shafts 4m
apart. The maximum tension in the belt is 1855.3N. The coefficient of friction is 0.3. The
driver pulley of diameter 120cm runs at 200rpm. Calculate (i) The power transmitted
(ii) Torque on each of the two shafts.
Sol: Given data:
D
1
= Diameter of the driver = 120cm = 1.2m
R
1
= Radius of the driver = 0.6m
N
1
= Speed of the driver in R.P.M. = 200RPM
D
2
= Diameter of the driven or Follower = 50cm = 0.5m
R
2
= Radius of the driven or Follower = 0.25m
X = Distance between the centers of two pulleys = 4m
µ = Coefficient of friction = 0.3
T
1
= Tension in the tight side of the belt = 1855.3N
Calculation for power transmitting:
Let
P = Maximum power transmitted by belt drive
= (T
1
–T
2
).V/1000 KW ...(i)
Where,
T
2
= Tension in the slack side of the belt
V = Velocity of the belt in m/sec.
= πDN/60 m/sec, D is in meter and N is in RPM ...(ii)
For T
2
,
We use the relation Ratio of belt tension = T
1
/T
2
= e
µθ
...(iii)
But angle of contact is not given,
let
θ = Angle of contact and, θ = Angle of lap
for open belt, Angle of contact (θ) = Π – 2α ...(iv)
Sinα = (r
1
– r
2
)/X = (0.6 – 0.25)/4
α = 5.02° ...(v)
Using the relation (iii), θ = Π – 2α = 180 – 2 X 5.02 = 169.96°
= 169.96° X Π/180 = 2.97 rad ...(iv)
Now using the relation (iii)
228 / Problems and Solutions in Mechanical Engineering with Concept
1855.3/T
2
= e
(0.3)(2.967)
T
2
= 761.8N ...(vii)
For finding the velocity, using the relation (ii)
V = (3.14 X 1.2 X 200)/60 = 12.56 m/sec ...(viii)
For finding the Power, using the relation (i)
P = (1855.3 – 761.8) X 12.56
P = 13.73 KW .......ANS
We know that,
1. Torque exerted on the driving pulley = (T
1
– T
2
).R
1
= (1855.3 – 761.8) X 0.6
= 656.1Nm .......ANS
2. Torque exerted on the driven pulley = (T
1
– T
2
).R
2
= (1855.3 – 761.8) X 0.25
= 273.4.1Nm .......ANS
Q. 15: Find the power transmitted by a belt running over a pulley of 600mm diameter at 200r.p.m.
The coefficient of friction between the pulleys is 0.25; angle of lap 160º and maximum tension
in the belt is 2.5KN.
Sol: Given data
D
1
= Diameter of the driver = 600mm = 0.6m
N
1
= Speed of the driver in R.P.M. = 200RPM
µ = Coefficient of friction = 0.25
θ = Angle of contact = 160º
= 1600 X (π/180) = 2.79rad
(Angle of lap is always less than 10º, so it is angle of contact which is always greater than 150º, always
in radian)
T
1
= Maximum Tension = 2.5KN
Let
T
2
= Tension in the slack side of the belt
V = Velocity of the belt in m/sec.
= πDN/60 m/sec, D is in meter and N is in RPM
P = Power transmitted by belt drive
We know that
Power Transmitted = (T
1
– T
2
).V KW, T
1
& T
2
in KN
Here T
2
and V is unknown
Calculation for V
V = πDN/60 m/sec, D is in meter and N is in RPM
Putting all the value,
V = (3.14 X 0.6 X 200)/60 = 6.28m/sec ...(i)
Calculation for T
2
We also know that,
Ratio of belt tension, T
1
/T
2
= e
µθ
Putting all the value,
Application of Friction: Belt Friction / 229
2.5/ T
2
= e
(0.25 × 2.79)
T
2
= 1.24KN ...(ii)
Now, P = (2.5 – 1.24) × 6.28
P = 7.92 KW .......ANS
Q. 16: An open belt runs between two pulleys 400mm and 150mm diameter and their centers are
1000mm apart. If coefficient of friction for larger pulley is 0.3, then what should be the value
of coefficient of friction for smaller pulley, so that the slipping is about to take place at both
the pulley at the same time?
Sol: Given data
D
1
= 400mm, R
1
= 200mm
D
2
= 150mm, R
2
= 775mm
X = 1000mm
µ
1
= 0.3
µ
2
= ?
Sinα = (r
1
– r
2
)/X = (200 – 75)/1000
α = 7.18° = 7.18° × Π/180°
α = 0.1256 rad ...(i)
We know that
For Open belt drive:
Angle of contact (θ) for larger pulley = Π + 2α
Angle of contact (θ) for smaller pulley = Π – 2α
Since, Ratio of belt tension = T
1
/T
2
= e
µθ
It is equal for both the pulley, i.e.,
(T
1
/T
2
)larger pulley = (T
1
/T
2
)smaller pulley
or, e
µ
1
θ
1 = e
µ
2
θ
2 , or µ
1
θ
1
= µ
2
θ
2
putting all the value, we get,
(0.3)(Π + 2α) = (µ
2
)(Π – 2α)
(0.3)(Π + 2 × 0.1256) = (µ
2
)(Π – 2 X 0.1256)
on solving, µ
2
= 0.352 .......ANS
Q. 17: A belt supports two weights W
1
and W
2
over a pulley as shown in fig 7.9. If W
1
= 1000N, find
the minimum weight W
2
to keep W
1
in equilibrium. Assume that the pulley is locked and µ
= 0.25.
A
B
T
2
T
1
0
β
W
1
= 1000N
W
2
Fig 10.9
230 / Problems and Solutions in Mechanical Engineering with Concept
Sol : Let the tensions in the belt be T
1
and T
2
as shown, since the weight W
2
just checks the tendency of
weight W
1
to move down, tension on the side of W
1
is larger.
That is, T
1
>T
2
µ = 0.25, θ = Π, W
1
= 1000N
Using the relation Ratio of belt tension
= T
1
/T
2
= eµθ
W
1
/T
2
= e
(0.25)(Π)
On solving, T
2
= W
2
= 456N .......ANS
Q. 18: An open belt running over two pulleys 24cm and 60cm diameters. Connects two parallel shaft
3m apart and transmits 3.75KW from the smaller pulley that rotates at 300RPM, µ = 0.3, and
the safe working tension in 100N/cm width. Determine
(i) Minimum width of the belt.
(ii) Initial belt tension.
(iii) Length of the belt required.
Sol: Given that,
D
1
= 60cm
D
2
= 24cm
N
2
= 300rpm
µ = 0.3
X = 3m = 300cm
P = 3.75KW
Safe Tension = Maximum tension = 100N/cm width = 100b N b = width of belt
T
max
= 100b ...(i)
Let θ = Angle of contact
Sinα = (r
1
– r
2
)/X = (30 – 12)/300 ; α = 3.45°, ...(ii)
θ = Π – 2α = (180 – 2 X 3.45) = 173.1°
= (173.1°)X Π/180 = 3.02rad ...(iii)
Now,
Using the relation, Ratio of belt tension = T
1
/T
2
= e
µθ
= e
(0.3)(3.02)
T
1
= 2.474T
2
...(iv)
Now,
V = πDN/60 m/sec, D is in meter and N is in RPM
= 3.14 X (0.24)(300)/60 = 3.77m/sec ...(v)
Power Transmitted (P) = (T
1
– T
2
).v/1000 Kw
3.75 = (T
1
– T
2
)X 3.77/1000
T
1
– T
2
= 994.7N ...(vi)
From relation (iv) and (v), we get:
T
1
= 1669.5N ...(vii)
T
2
= 674.8N ...(viii)
(i) For width of the belt
But T
1
= T
max
= 100b; 1669.5 = 100b; b = 16.7cm .......ANS
(ii) For initial tension in the belt
Let To = initial tension in the belt
Application of Friction: Belt Friction / 231
To = (T
1
+ T
2
)/2
= (1669.5 + 674.8)/2
To = 1172.15N .......ANS
(iii) For length of belt
L = Π(r
1
+ r
2
) +
2
1 2
(r – r )
X
+ 2X
Putting all the value, we get
L = 7.33m .......ANS
Q. 19: Determine the minimum value of weight W required to cause motion of a block, which rests
on a horizontal plane. The block weighs 300N and the coefficient of friction between the block
and plane is 0.6. Angle of warp over the pulley is 90º and the coefficient of friction between
the pulley and rope is 0.3.
Block
Polley
Polley
T
1
T
1
T
2
T
2
W = 300 N
F = R µ
R
W W
Fig 10.10 Fig 10.11 Fig 10.12
Sol: Since the weight W impend vertical motion in the down ward direction, the tension in the two sides
of the pulley will be as shown in fig 10.11
Given date:
T
1
= W, µ = 0.3, θ = 90° = π/2 rad
Using the relation of Ratio of belt tension, T
1
/T
2
= e
µ.θ
W/T
2
= e
(0.3).(p/2)
= 1.6
W = 1.6 × T
2
...(i)
Considering the equilibrium of block:
∑V = 0
R = 300N ...(ii)
∑H = 0
T
2
= µR = 0.3 × 300 = 180N ...(iii)
Equating equation (i) and (iii), we get
W = 1.6 × 180
W = 288N .......ANS
Q. 20: A horizontal drum of a belt drive carries the belt over a semicircle around it. It is rotated anti
clockwise to transmit a torque of 300Nm. If the coefficient of friction between the belt and
rope is 0.3, calculate the tension in the limbs 1 and 2 of the belt shown in figure, and the
reaction on the bearing. The drum has a mass of 20Kg and the belt is assumed to be mass less.
(May–0102)
232 / Problems and Solutions in Mechanical Engineering with Concept
0.5 m
1 2
Fig 10.13
Sol: Given data:
Torque(t) = 300Nm
Coff. of friction(µ) = 0.3
Diameter of Drum (D) = 1m, R = 0.5m
Mass of drum(m) = 20Kg.
Since angle of contact = π rad
Torque = (T
1
– T
2
).R
300 = (T
1
– T
2
) X 0.5
T
1
– T
2
= 600N ...(i)
And, T
1
/T
2
= e
µθ
T
1
/T
2
= e
(0.3)π
T
1
= 2.566T
2
...(ii)
Solving (i) and (ii)
We get,
T
1
= 983.14N .......ANS
T
2
= 383.14N .......ANS
Now reaction on bearing is opposite to the mass of the body, and it is equal to
R = T
1
+ T
2
+ mg
R = 983.14 + 383.14 + 20 X 9.81
R = 1562.484N .......ANS
Q. 21: A belt is stretched over two identical pulleys of diameter D meter. The initial tension in the
belt throughout is 2.4KN when the pulleys are at rest. In using these pulleys and belt to
transmit torque, it is found that the increase in tension on one side is equal to the decrease
on the other side. Find the maximum torque that can be transmitted by the belt drive, given
that the coefficient of friction between belt and pulley is 0.30. (Dec–0203)
B
D
A
T
2
T
1
Fig 10.14
Application of Friction: Belt Friction / 233
Sol: Given data:
Diameter of both pulley = D
Initial tension in belt (T
O
) =2.4KN
Torque = ?
Coefficient of friction (µ) = 0.3
Since dia of both pulley are same, i.e., Angle of contact = π
T
O
= (T
1
+ T
2
)/2
T
1
+ T
2
= 4.8KN ...(i)
Now, Ratio of belt tension = T
1
/T
2
= e
µθ
T
1
/T
2
= e
(0.3)π
T
1
= 2.566T
2
...(ii)
Putting the value of (ii) in equation (i), We get
T
1
= 3.46KN .......ANS
T
2
= 1.35KN .......ANS
Now, Maximum torque transmitted by the pulley = (T
1
– T
2
)D/2
(Since radius of both pulley are same)
Torque = (3.46 – 1.35)D/2 = 1.055D KN–m
Torque = 1.055D KNm .......ANS
Q. 22: A belt is running over a pulley of 1.5m diameters at 250RPM. The angle of contact is 120º
and the coefficient of friction is 0.30. If the maximum tension in the belt is 400N, find the
power transmitted by the belt. (Nov–03 C.O.)
Sol: Given data
Diameter of pulley(D) = 1.5m
Speed of the driver(N) = 250RPM
Angle of contact(?) = 1200 = 1200 X (π/180º) = 2.09 rad
Coefficient of friction(µ) = 0.3
Maximum tension(Tmax) = 400N = T
1
Power (P) = ?
Since P = (T
1
– T
2
) X V Watt ...(i)
T1 is given, and for finding the value of T
2
, using the formula
Ratio of belt tension = T
1
/T
2
= e
µθ
400/T
2
= e
(0.3)(2.09)
T
2
= 213.4N ...(ii)
Now We know that V = πDN/60 m/sec
V = [3.14 X 1.5 X 250]/60 = 19.64m/sec ...(iii)
Now putting all the value in equation (i)
P = (400 – 213.4) X 19.64 watt
P = 3663.88Watt or 3.66KW ......ANS
Q. 23: Explain the concept of centrifugal tension in any belt drive. What are the main consideration
for taking maximum tension?
Sol: We know that the belt continuously runs over both the pulleys. In the tight side and slack side of the
belt tension is increased due to presence of centrifugal Tension in the belt. At lower speeds the centrifugal
tension may be ignored but at higher speed its effect is considered.
234 / Problems and Solutions in Mechanical Engineering with Concept
The tension caused in the running belt by the centrifugal force is known as centrifugal tension. When
ever a particle of mass 'm' is rotated in a circular path of radius 'r' at a uniform velocity 'v', a centrifugal
force is acting radially outward and its magnitude is equal to
2
mv
r
.
i.e., Fc = mv
2
/r
The centrifugal tension in the belt can be calculated by considering the forces acting on an elemental
length of the belt(i.e length MN) subtending an angle δθ at he center as shown in the fig 10.14.
Let
v = Velocity of belt in m/s
r = Radius of pulley over which belt run.
M = Mass of elemental length of belt.
m = Mass of the belt per meter length
T
1
= Tight side tension
T
c
= Centrifugal tension acting at M and N tangentially
F
c
= Centrifugal force acting radially outwards
The centrifugal force R acting radially outwards is balanced by the components of Tc acting radially
inwards. Now elemental length of belt
MN = r. δθ
Mass of the belt MN = Mass per meter length X Length of MN
M = m X r X δθ
Centrifugal force = F
c
= M X v
2
/r = m.r.δθ.v
2
/r
Now resolving the force horizontally, we get
T
c
.sinδθ/2 + T
c
.sinδθ/2 = F
c
Or 2T
c
.sinδθ/2 = m.r.δθ.v
2
/r
At the angle δθ is very small, hence = sinδθ/2 = δθ/2
Then the above equation becomes as
2T
c
.δθ/2 = m.r.δθ.v
2
/r
or T
c
= m.v
2
Important Consideration
1. From the above equation, it is clear that centrifugal tension is independent of T
1
and T
2
. It depends
upon the velocity of the belt. For lower belt speed (i.e., Belt speed less than 10m/s) the centrifugal tension
is very small and may be neglected.
2. When centrifugal tension is to be taken into consideration then total tension on tight side and slack
side of the belt is given by
For tight side = T
1
+ Tc
For slack side = T
2
+ Tc
3. Maximum tension(Tm) in the belt is equal to maximum safe stress in the belt multiplied by cross
sectional area of the belt.
T
m
= σ (b.t)
Where
σ = Maximum safe stress in the belt
b = Width of belt and
Application of Friction: Belt Friction / 235
t = Thickness of belt
T
m
= T
1
+ T
c
 if centrifugal tension is to be considered
= T
1
 if centrifugal tension is to be neglected
Q. 24: Derive the formula for maximum power transmitted by a belt when centrifugal tension in to
account.
Sol: Let T
1
= Tension on tight side
T
2
= Tension on slack side
v = Linear velocity of belt
Then the power transmitted is given by the equation
P = (T
1
–T
2
). V ...(i)
But we know that T
1
/T
2
= e
µθ
Or we can say that T
2
= T
1
/ e
µθ
Putting the value of T
2
in equation (i)
P = (T
1
 T
1
/ e
µθ
).v = T
1
(1 – 1/ e
µθ
). V ...(ii)
Let (1–1/ e
µθ
) = K , K = any constant
Then the above equation is P = T
1
.K. V or KT
1
V ...(iii)
Let T
max
= Maximum tension in the belt
T
c
= Centrifugal tension which is equal to m.v
2
Then T
max
= T
1
+ T
c
T
1
= T
max
– T
c
Putting this value in the equation (iii)
P = K(T
max
– T
c
).V
= K(T
max
– m.V
2
).V
= K(T
max
.v – m.V
3
)
The power transmitted will be maximum if d(P)/dv = 0
Hence differentiating equation w.r.t. V and equating to zero for maximum power, we get
d(P)/dv = K(T
max
– 3.m.V
2
)=0
T
max
– 3mV
2
=0
T
max
= 3mV
2
V = (T
max
/3m)
1/2
...(iv)
Equation (iv) gives the velocity of the belt at which maximum power is transmitted.
From equation (iv) T
max
= 3Tc ...(v)
Hence when the power transmitted is maximum, centrifugal tension would be 1/3rd of the maximum
tension.
We also know that Tmax = T
1
+ T
c
= T
1
+ T
max
/3 ...(vi)
T
1
= T
max
 T
max
/3
= 2/3.T
max
...(vii)
Hence condition for the transmission of maximum power are:
T
c
= 1/3 T
max
, and T
1
= 2/3T
max
...(viii)
NOTE: Net driving tension in the belt = (T
1
– T
2
)
236 / Problems and Solutions in Mechanical Engineering with Concept
STEPS FOR SOLVING THE PROBLEM FOR FINDING THE POWER
1. Use the formula stress (σ) = force (Maximum Tension)/Area
Where; Area = b.t i.e., Tmax = σ.b.t
2. Unit mass (m) = ρ.b.t.L
Where;
ρ = Density of a material
b = Width of Belt
t = Belt thickness
L = Unit length
Take L = 1m, if b and t are in meter
Take L = 100cm, if b and t are in cm
Take L = 1000mm, if b and t are in mm
3. Calculate V using V = πDN/60 m/sec (if not given)
4. T
C
= mV
2
, For finding T
C
5. T
max
= T
1
+ T
c
, for finding T
1
6. For T
2
, Using relation Ratio of belt tension = T
1
/T
2
= e
µθ
7. Power Transmitted = (T
1
–T
2
).V/1000 Kw
Steps for Solving the Problem for Finding the Maximum Power
1. Use the formula stress (σ) = force (Maximum Tension)/Area
Where; Area = b.t i.e. Tmax = σ.b.t
2. Unit mass (m) = ρ.b.t.L
Where
ρ = Density of a material
b = Width of Belt
t = Belt thickness
L = Unit length
Take L = 1m, if b and t are in meter
Take L = 100cm, if b and t are in cm
Take L = 1000mm, if b and t are in mm
3. T
C
=1/3 Tmax = mV
2
, For finding T
C
and velocity (If not given)
We don't Calculate Velocity using V = πDN/60 m/sec (if not given)
5. T
max
= T
1
+ T
c
, for finding T
1
6. For T
2
, Using relation Ratio of belt tension = T
1
/T
2
= e
µθ
7. Maximum Power Transmitted = (T
1
– T
2
).v/1000 Kw
Initial Tension in The Belt
Let To = initial tension in the belt
T
1
= Tension in the tight side
T
2
= Tension in the slack side
T
C
= Centrifugal Tension in the belt
T
o
= (T
1
+ T
2
)/2 + T
C
Application of Friction: Belt Friction / 237
Q. 25: A belt 100mm wide and 8.0mm thick are transmitting power at a belt speed of 160m/minute.
The angle of lap for smaller pulley is 165º and coefficient of friction is 0.3. The maximum
permissible stress in belt is 2MN/m
2
and mass of the belt is 0.9Kg/m. find the power transmitted
and the initial tension in the belt.
Sol.: Given data
Width of belt(b) = 100mm
Thickness of belt(t) = 8mm
Velocity of belt(V) = 160m/min = 2.66m/sec
Angle of contact(?) = 165° = 165° X Π/180 = 2.88rad
Coefficient of friction(µ) = 0.3
Maximum permissible stress(f) = 2 X 10
6
N/m
2
= 2N/mm
2
Mass of the belt material(m) = 0.9 Kg/m
Power = ?
Initial tension (To) = ?
We know that, T
max
= σ.b.t
= 2 X 100 X 8 = 1600N ...(i)
Since m and velocity (V) is given, then
Using the formula, T
C
= mV
2
, For finding T
C
= 0.9(2.66)
2
= 6.4 N ...(ii)
Using the formula, T
max
= T
1
+ T
c
, for finding T
1
1600 = T
1
+ 6.4
T
1
= 1593.6N ...(iii)
Now, For T
2
, Using relation Ratio of belt tension = T
1
/T
2
= e
µθ
1593.6/T
2
= e
(0.3)(2.88)
T
2
= 671.69 N ...(iv)
Now Power Transmitted = (T
1
–T
2
).v/1000 Kw
P = (1593.6 – 671.69).2.66/1000 Kw
P = 2.45KW .......ANS
Let To = initial tension in the belt
T
o
= (T
1
+ T
2
)/2 + T
C
T
o
= (1593.6 + 671.69)/2 + 6.4
T
o
= 1139.045N .......ANS
Q. 26: A belt embraces the shorter pulley by an angle of 165º and runs at a speed of 1700 m/min,
Dimensions of the belt are Width = 20cm and thickness = 8mm. Its density is 1gm/cm
3
. Determine
the maximum power that can be transmitted at the above speed, if the maximum permissible
stress in the belt is not to exceed 250N/cm
2
and µ = 0.25.
Sol: Given date:
Angle of contact(θ) =165° = 165° X Π/180 = 2.88rad
Velocity of belt(V) = 1700m/min = 28.33m/sec
Width of belt(b) = 20cm
Thickness of belt(t) = 8mm 0.8cm
238 / Problems and Solutions in Mechanical Engineering with Concept
density of belt = 1gm/cm
3
Maximum permissible stress(f) = 250 N/cm
2
Coefficient of friction(µ) = 0.25
Maximum Power = ?
We know that, T
max
= σ.b.t
= 250 X 20 X 0.8 = 4000N ...(i)
Since Unit mass (m) = ρ.b.t.L
= 1/1000 X 20 X 0.8 X 100 = 1.6Kg .. (ii)
Since velocity(V) is given, So we don't find the velocity using formula T
C
=1/3 Tmax = mV
2
, then
Using the formula, T
C
= mV
2
, For finding T
C
= 1.6(28.33)
2
= 1284 N ...(iii)
Using the formula, T
max
= T
1
+ T
c
, for finding T
1
4000 = T
1
+ 1284
T
1
= 2716N ...(iv)
Now, For T
2
, Using relation Ratio of belt tension = T
1
/T
2
= e
µθ
2716/T
2
= e
(0.25)(2.88)
T
2
= 1321 N ...(v)
Now Maximum Power Transmitted = (T
1
–T
2
).V/1000 KW
P = (2716 – 1321) X 28.33/1000 KW
P = 39.52KW .......ANS
Q. 27: A belt of density 1gm/cm
3
has a maximum permissible stress of 250N/cm
2
. Determine the
maximum power that can be transmitted by a belt of 20cm X 1.2cm if the ratio of the tight
side to slack side tension is 2.
Sol: Given date
Density of belt = 1gm/cm
3
= 1/1000 Kg/cm
3
Maximum permissible stress(f) = 250 N/cm
2
Width of belt(b) = 20cm
Thickness of belt(t) = 8mm 0.8cm
Ratio of tension (T
1
/T
2
) = 2
Maximum Power = ?
We know that, Tmax = σ.b.t
= 250 X 20 X 1.2 = 6000N ...(i)
Since Unit mass (m) = σ.b.t.L
= 1/1000 X 20 X 1.2 X 100 = 2.4Kg ...(ii)
Since velocity(V) is not given, So we find the velocity using formula T
C
=1/3 Tmax = mV
2
, for
maximum power
Using the formula, 1/3 T
max
= mV
2
V = (T
max
/3m)
1/2
V = (6000/3 X 2.4)
1/2
V = 28.86 m/sec ...(iii)
Using the formula, T
C
= mV
2
, For finding T
C
Application of Friction: Belt Friction / 239
= 2.4(28.86)
2
= 1998.96N ...(iv)
Using the formula, T
max
= T
1
+ T
c
, for finding T
1
6000 = T
1
+ 1998.96
T
1
= 4001N ...(v)
Now, For T
2
, Using relation Ratio of belt tension = T
1
/T
2
= e
µθ
= 2
4001/T
2
= 2
T
2
= 2000.5 N ...(vi)
Now Maximum Power Transmitted = (T
1
–T
2
).V/1000 KW
P = (4001  2000.5) X 28.86/1000 KW
P = 57.73KW .......ANS
Q. 28: What is Vbelt. Drive the expression of Ratio in belt tension for Vbelt
Sol: The power from one shaft to another shaft is also transmitted with the help of Vbelt drive and rope
drive. Fig shows a Vbelt with a grooved pulley.
VBelt
Pulley
VBelt
R
2
R
N R
N
VGrooved
Pulley
α
α
(a) (b)
Fig 10.15
Sol: Let
R
N
= Normal reaction between belt and sides with a grooved pulley.
2α = Angle of groove
µ = Coefficient of friction between belt and pulley.
R = Total reaction in the plane of groove.
Resolving the forces vertically, we get
R = R
N
sin α + R
N
sin α
= 2R
N
sin α
R
N
= (R/2) cosec α ...(i)
Frictional resistance = µR
N
+ µR
N
= 2µR
N
= 2µ(R/2)cosec α
= µR cosec α = R ⋅ µcosec α
Since in flat belt frictional resistance is equal to µR, and in case of Vbelt µcoseca X R
So,
Ratio of Tension in VBelt:: T
1
/T
2
= e
µ.θ θθ θθ.cosec α αα αα
240 / Problems and Solutions in Mechanical Engineering with Concept
Q. 29: What do you mean by rope drive.
Sol: The ropes are generally circular in section. Ropedrive is mostly used when the distance between the
driving shaft and driven shaft is large. Frictional grip in ropedrive is more than that in Vbelt drive.
The ratio of tensions in this case will also be same as in case of Vbelt. Hence ratio of tension will
be as:
Ratio of Tension in Rope Drive:: T
1
/T
2
= e
µ.θ.coseca
Q. 30: The maximum allowable tension, in a Vbelt of groove angle of 30º, is 2500N. The angle of
lap is140º and the coefficient of friction between the belt and the material of the pulley is 0.15.
If the belt is running at 2m/sec, Determine:
(i) Net driving tension (ii) Power transmitted by the pulley, Neglect effect of centrifugal tension.
Sol: Given data
Angle of groove(2α) = 30º, α = 15º
Max. Tension(T
max
) = 2500N
Angle of lap(contact) (θ) = 140º = 140º X (Π/180º) = 2.44 rad
Coefficient of friction (µ) = 0.15
Speed of belt(V) = 2m/sec
We know that,
T
max
= T
1
=2500N
(T
C
is neglected, since belt speed is less than 10m/sec)
Ratio of Tension in VBelt:: T
1
/T
2
= e
µ.θ.cosec α
2500/T
2
= e
(0.15).(2.44).cosec15
T
2
= 2500/4.11
T
2
= 607.85N ...(i)
(i) Net driving tension = (T
1
–T
2
)
= 2500 – 607.85 = 1892.2N .......ANS
(iii)Power transmitted = (T1 – T2)X V W
= (2500 – 607.85) X 2 = 3784.3Watt .......ANS
Q. 31: A pulley used to transmit power by means of ropes, has a diameter of 3.6m and has 15 groove
of 45º angle. The angle of contact is 170º and the coefficient of friction between the ropes and
the groove side is 0.28. The maximum possible tension in the ropes is 960N and the mass of
the rope is 1.5Kg per m length. What is the speed of the pulley in rpm and the power transmitted
if the condition of maximum power prevails?
Sol: Given data
Dia. Of pulley(D) = 3.6m
Number of groove(or ropes) = 15
Angle of groove(2a) = 45°, α = 22.50º
Angle of contact(θ) = 170° = 1700 X (Π/180°) = 2.97 rad
Coefficient of friction(µ) = 0.28
Max. Tension(Tmax) = 960N
Mass of rope(m) = 1.5Kg per m length
For maximum power:
T
c
= 1/3Tm
= 1/3 X 960 = 320N ...(i)
Application of Friction: Belt Friction / 241
T
m
= T
1
+ T
C
960 = T
1
+ 320
T
1
= 640N ...(ii)
Now T
c
= (1/3)T
m
= mV
2
V = (T
m
/3m)1/2
= [960/(3 X 1.5)]
1/2
= 14.6m/sec ...(iiii)
Since V = πDN/60 = 14.6,
N = 77.45R.P.M. .......ANS
Now, Ratio of Tension in VBelt:: T
1
/T
2
= e
µ.θ.cosec α
640/T
2
= e
(0.28).(2.97).cosec22.5
T
2
= 73.08N ...(iv)
Maximum power transmitted(P) = (T
1
–T
2
).v/1000 Kw
P = [(640 – 73.08) X 14.6]/1000 KW
P = 8.277KW
Total maximum power transmitted = Power of one rope X No. of rope
P = 8.277 X 15 = 124.16KW .......ANS
242 / Problems and Solutions in Mechanical Engineering with Concept
+0)264
11
L/WS CF MCTlCN
Q. 1 : Define Kinetics. What is plane motion?
Sol : Kinetics of that branch of mechanics, which deals with the force system, which produces acceleration,
and resulting motion of bodies.
PLANE MOTION: The motion of rigid body, in which all particles of the body remain at a constant
distance from a fixed reference plane, is known as plane motion.
Q. 2 : Define the following terms: Matter, Particle, Body, Rigid body, Mass, Weight and Momentum?
Sol : Matter: Matter is any thing that occupies space, possesses mass offers resistance to any stress,
example Iron, stone, air, Water.
Particle: A body of negligible dimension is called a particle. But a particle has mass.
Body: A body consists of a No. of particle, It has definite shape.
Rigid body: A rigid body may be defined as the combination of a large no. of particles, Which occupy
fixed position with respect to another, both before and after applying a load.
A rigid body may be defined as a body, which can retain its shape and size even if subjected to some
external forces. In actual practice, no body is perfectly rigid. But for the shake of simplicity, we take the
bodies as rigid bodies.
Mass: The properties of matter by which the action of one body can be compared with that of another
is defined as mass.
m = ρ.v
Where,
ρ = Density of body
V = Volume of the body
Weight: Weight of a body is the force with which the body is attracted towards the center of the earth.
Momentum : It is the total motion possessed by a body. It is a vector quantity. It can be expressed
as,
Momentum(M) = mass of the body(m) × Velocity(V) Kgm/sec
Q. 3 : Define different Newton’s law of motion.
Sol.: The entire system of Dynamics is based on three laws of motion, which are the basis assumptions,
and were formulated by Newton.
First Law
A particle remains at rest (if originally at rest) or continues to move in a straight line (If originally in
motion) with a constant speed. If the resultant force acting on it is Zero.
Laws of Motion / 243
It is also called the law of inertia, and consists of the following two parts:
A body at rest has a tendency to remain at rest. It is called inertia of rest.
A body in motion has a tendency to preserve its motion. It is called inertia of motion.
Second Law
The rate of change of momentum is directly proportional to the external force applied on the body and take
place, in the same direction in which the force acts.
Let a body of mass ‘m’ is moving with a velocity ‘u’ along a straight line. It is acted upon a force ‘F’
and the velocity of the body becomes ‘v’ in time ‘t’ then.
Initial momentum = m.u
Initial momentum = m.v
Change in momentum = m(vu)
Rate of change of momentum = change of momentum / Time
= m(vu)/t
but v = u + a.t
a = (vu)/t
i.e Rate of change of momentum = m.a
But according to second law F proportional to m.a
i.e. F = k.m.a Where K = constant.
Unit of force
1N = 1 kgm/sec
2
= 10
5
dyne = 1 grm.cm/sec
2
Third Law
The force of action and reaction between interacting bodies are equal in magnitude, opposite in
direction and have the same line of action.
Q. 4 : A car of mass 400kg is moving with a velocity of 20m/sec. A force of 200N acts on it for 2
minutes. Find the velocity of the vehicle:
(1) When the force acts in the direction of motion.
(2) When the force acts in the opposite direction of the motion.
Sol :
m = 400Kg, u = 20m/sec, F = 200N, t = 2min = 120sec, v =?
Since F = ma
200 = 400 X a
a = 0.5m/sec
2
...(i)
(1) Velocity of car after 120sec, When the force acts in the direction of motion.
v = u + at
= 20 + 0.5 X 120
v = 80m/sec ........ANS
(2) Velocity of car after 120sec, When the force acts in the opposite direction of motion.
v = u  at
= 20  0.5 X 120
v = 40m/sec .......ANS
ve sign indicate that the body is moving in the reverse direction
244 / Problems and Solutions in Mechanical Engineering with Concept
Q. 5 : A body of mass 25kg falls on the ground from a height of 19.6m. The body penetrates into the
ground. Find the distance through which the body will penetrates into the ground, if the
resistance by the ground to penetrate is constant and equal to 4998N. Take g = 9.8m/sec
2
.
Sol : Given that:
m = 25Kg, h = 19.6m, s = ?, F
r
= 4998N, g = 9.8m/sec
2
Let us first consider the motion of the body from a height of 19.6m to the ground surface,
Initial velocity = u = 0,
Let final velocity of the body when it reaches to the ground = v,
Using the equation, v
2
= u
2
+ 2gh
v
2
= (0)
2
+ 2 X 9.8 X 19.6
v = 19.6m/sec ...(i)
When the body is penetrating in to the ground, the resistance to penetration is acting in the upward
direction. (Resistance is always acting in the opposite direction of motion of body.) But the weight of the
body is acting in the downward direction.
Weight of the body = mg = 25 X 9.8 = 245N ...(ii)
Upward resistance to penetrate = 4998N
Net force acting in the upward direction = F
F = F
r
– mg
= 4998 – 245 = 4753N ...(iii)
Using F = ma, 4753 = 25 X a
a = 190.12 m/sec
2
...(iv)
Now, calculation for distance to penetrate
Consider the motion of the body from the ground to the point of penetration in to ground.
Let the distance of penetration = s,
Final velocity = v,
Initial velocity = u = 19.6m/sec,
Retardation a = 190.12m/sec
2
Using the relation, v
2
= u
2
– 2as
(0)
2
= (19.6)
2
– 2 X 190.12 X S
S = 1.01m .......ANS
Q. 6 : A man of mass 637N dives vertically downwards into a swimming pool from a tower of height
19.6m. He was found to go down in water by 2m and then started rising. Find the average
resistance of the water. Neglect the resistance of air.
Sol: Given that:
W= 637N, h = 19.6m, S = 2m, g = 9.8m/sec
2
Let, F
r
= Average resistance
Initial velocity of man u = 0,
V
2
= u
2
+ 2gh
= 0 + 2 X 9.8 X 19.6
V= 19.6 m/sec ...(i)
Now distance traveled in water = 2m,v = 0, u = 19.6m/sec now apply
V
2
= u
2
– 2as
0 = 19.6
2
– 2a X 2
Laws of Motion / 245
a = 96.04m/sec
2
...(ii)
Since, net force acting on the man in the upward direction = F
r
– W
But the net force acting on the man must be equal to the product of mass and retardation.
F
r
– W= ma
F
r
– 637 = (637/g) X 96.04
F
r
= 6879.6N .......ANS
Q. 7 : A bullet of mass 81gm and moving with a velocity of 300m/sec is fired into a log of wood and
it penetrates to a depth of 10cm. If the bullet moving with the same velocity were fired into
a similar piece of wood 5cm thick, with what velocity would it emerge? Find also the force
of resistance assuming it to be uniform.
Sol: Given that
m = 81gm = 0.081Kg, u = 300m/sec, s = 10cm = 0.1m, v = 0
As the force of resistance is acting in the opposite direction of motion of bullet, hence force of
resistance will produce retardation on the bullet, Apply, V
2
= u
2
– 2as
0 = 300
2
– 2a(0.1)
a = 450000m/sec
2
...(i)
Let F is the force of resistance offered by wood to the bullet.
Using equation, F = ma,
F = 0.081 X 450000
F = 36450N .......ANS
Let v = velocity of bullet with which the bullet emerges from the piece of wood of 5cm thick,
U = 300m/sec, a = 450000m/sec
2
, s = 0.05m
Using equation, V
2
= u
2
– 2as
V
2
= 300
2
– 2 X 450000 X 0.05
V = 212.132m/sec .......ANS
Q. 8 : A particle of mass 1kg moves in a straight line under the influence of a force, which increases
linearly with the time at the rate of 60N per sec. At time t = 0 the initial force may be taken
as 50N. Determine the acceleration and velocity of the particle 4sec after it started from rest
at the origin.
Sol: As the force varies linearly with time,
F = mt + C
Differentiate the equation with time,
dF/dt = m = 60(given)
i.e., m = 60 ...(i)
Given that, at t = 0, F = 50N,
50 = 60 X 0 + C
C = 50 ...(ii)
Now the equation becomes,
F = 60t + 50 ...(iii)
Since, F = ma, m = 1Kg
F = ma = 1.a = 60t +50
At t = 4 sec,
a = 60 X 4 + 50 = 290
246 / Problems and Solutions in Mechanical Engineering with Concept
a = 290m/sec
2
.......ANS
also, a = dv/dt
a = dv/dt = 60t + 50
Integration both side for the interval of time 0 to 4sec.
V = ∫
+
4
0
) 50 60 ( dt t
V = (60t
2
+ 50t), limit are 0 to 4
V = 30(4)
2
+ 50 X 4
V = 680m/sec .......ANS
Q. 9 : Determine the acceleration of a railway wagon moving on a railway track if fraction force exerted
by wagon weighing 50KN is 2000N and the frictional resistance is 5N per KN of wagon’s weight.
Sol: Let a be the acceleration of the wagon
Mass (m) = W/g = (50 X 1000/9.81)
Friction force F
r
= 5 X 50 = 250N ...(i)
Net force = F  F
r
= ma
2000 – 250 = (50 X 1000/9.81)a
a = 0.3438m/sec
2
.......ANS
Q.10: A straight link AB 40cm long has, at a given instant, its end B moving along line OX at 0.8m/
s and acceleration at 4m/sec
2
and the other end A moving along OY, as shown in fig 11.1. Find
the velocity and acceleration of the end A and of mid point C of the link when inclined at 30
0
with OX.
Sol: Let the length of link is L = 40cm and AD = Y, OB = X
X
2
+ Y
2
= 1 ...(i)
Diff with respect to time, and ive sign is taken for down word motion of A, when B is moving in +ive
direction, we get
2Xdx/dt – 2Ydy/dt = 0XV
B
– YV
A
= 0 ...(ii)
V
A
= (X/Y)V
B
= (Lcosθ/Lsinθ)V
B
= V
B
/tanθV
A
= 0.8/tan30
0
= 1.38m/sec ...(iii)
V
A
= 1.38m/sec .......ANS
Y
Y
X
A
D
C
B
X
90º
Fig 11.1
Again differentiating equation (2), we get
Xd
2
x/dt
2
+ (dx/dt)
2
– yd
2
y/dt
2
– (dy/dt)
2
= 0
X.a
B
+ (V
B
)
2
– Y.a
A
– (V
A
)
2
= 0
0.4cos30
0
X 0.4 – (0.8)
2
– 0.4sin30
0
X a
A
– (1.38)
2
= 0
1.38 + 0.64 – 0.2a
A
– 1.9 = 0
a
A
= 0.6.6m/sec
2
.......ANS
Laws of Motion / 247
Q.11 : A 20KN automobile is moving at a speed of 70Kmph when the brakes are fully applied
causing all four wheels to skid. Determine the time required to stop the automobile.
(1) on concrete road for which µ µµ µµ = 0.75
(2) On ice for which µ µµ µµ = 0.08
Sol: Given data: W = 20KN, u = 70Kmphr = 19.44m/sec, v = 0, t = ?
Consider FBD of the car as shown in fig 11.2
∑V = 0, R = W ...(i)
∑H = 0, Fr = 0 Fr = µR ...(ii)
Here net force is the frictional force
i.e. F = F
r
ma = µR = µmga = µg ...(iii)
F
r
R
W
Fig 11.2
(1) on concrete road for which µ = 0.75
a = µg = 0.75 X 9.81 = 7.3575
a = 7.35 m/sec
2
...(iv)
Using the relation v = u – at
0 = 19.44 – 7.35t
t = 2.64 seconds .......ANS
(1) On ice for which µ = 0.08
a = µg = 0.08 X 9.81 = 0.7848
a = 0.7848m/sec
2
...(v)
Using the relation v = u – at
0 = 19.44 – 0.7848t
t = 24.77 seconds .......ANS
Q. 12: Write different equation of motion on inclined plane for the following cases.
(a) Motion on inclined plane when surface is smooth.
(b) Motion on inclined plane when surface is rough.
Sol: CASE: 1 WHEN SURFACE SMOOTH
q
q
W cos q
W
R
W
s
i
n
q
Fig 11.3
248 / Problems and Solutions in Mechanical Engineering with Concept
Fig 11.3 shows a body of weight W, sliding down on a smooth inclined plane.
Let,
θ = Angle made by inclined plane with horizontal
a = Acceleration of the body
m= Mass of the body = W/g
Since surface is smooth i.e. frictional force is zero. Hence the force acting on the body are its own
weight W and reaction R of the plane.
The resolved part of W perpendicular to the plane is Wcos θ, which is balanced by R, while the resolved
part parallel to the plane is Wsin θ, which produced the acceleration down the plane.
Net force acting on the body down the plane
F = W.sin θ, but F = m.a
m.a = m.g.sinθ
i.e. a = g.sin θ (For body move down due to self weight.)
and, a = g.sin θ (For body move up due to some external force)
CASE: 2 WHEN ROUGH SURFACE
q
q
W cos q
W
R
W
s
i
n
q
F
=
R
1
Fig 11.4
Fig 11.4 shows a body of weight W, sliding down on a rough inclined plane.
Let,
θ = Angle made by inclined plane with horizontal
a = Acceleration of the body
m = Mass of the body = W/g
µ = Coefficient of friction
F
r
= Force of friction
when body tends to move down:
R = w.cosθ
F
r
= µ.R = µ.W.cosθ
Net force acting on the body F = W.sinθ  µ.W.cosθ
i.e. m.a = W.sinθ  µ.W.cosθ
Put m = W/g we get
a = g.[sinθ  µ.cosθ] (when body tends to move down)
a = –g.[sinθ  µ.cosθ] (when body tends to move up)
Q. 13 :A train of mass 200KN has a frictional resistance of 5N per KN. Speed of the train, at the top
of an inclined of 1 in 80 is 45 Km/hr. Find the speed of the train after running down the incline
for 1Km.
Sol: Given data,
Mass m = 200KN, Frictional resistance F
r
= 5N/KN, sinθ = 1/80 = 0.0125,
Laws of Motion / 249
Initial velocity u = 45Km/hr = 12.5m/sec, s = 1km = 1000m
Total frictional resistance = 5 X 200 = 1000N = 1KN ...(i)
Force responsible for sliding = Wsinθ = 200 X 0.0125 = 2.5KN
Now, Net force, F = F – F
r
= ma
2.5 – 1 = (200/9.81)a
a = 0.0735m/sec
2
...(ii)
Apply the equation, v
2
= u
2
+ 2as
v
2
= 0 + 2 X 0.0735 X 1000
v = 12.1 m/sec .......ANS
Q.13: A train of wagons is first pulled on a level track from A to B and then up a 5% upgrade as
shown in fig (11.5). At some point C, the least wagon gets detached from the train, when it
was traveling with a velocity of 36Km.p.h. If the detached wagon has a mass of 5KN and the
track resistance is 10N per KN, find the distance through which the wagon will travel before
coming to rest. Take g = 9.8m/sec
2
.
Level track
A
B
C
5
%
u
p
g
rad
e
Fig 11.5
Sol: Given that, Grade = 5% or sinθ = 5% = 0.05, u = 36Km.p.h. = 10m/sec,
W = 5KN, V = 0, F
r
= 10N/KN
Let s = Distance traveled by wagon before coming to rest
Total track resistance F
r
= 10 X 5 = 50N ...(i)
Resistance due to upgrade = msinθ = 5 X 0.05 = 0.25KN = 250N ...(ii)
Total resistance to wagon = Net force = 50 + 250 = 300N
But, F = ma, 300 = (5000/9.81)a
a = 0.588m/sec
2
...(iii)
Apply the equation, v
2
= u
2
 2as
0 = (10)
2
– 2 X 0.588 X s
s = 85 m .......ANS
Q.14: Write equation of motion of lift when move up and when move down.
Pulley
T
Force
W
Lift
T
Cable supporting
the Lift
W
Lift
T
W
Lift
Fig 11.6 Fig 11.7 Lift is moving upward Fig 11.8 Lift is moving downward
250 / Problems and Solutions in Mechanical Engineering with Concept
Let,
W = Weight carried by the lift
m = Mass carried by lift = W/g
a = Uniform acceleration
T = Tension in cable supporting the lift, also called Reaction of the lift
For UP MOTION
Net force in upward direction = TW
Also Net Force = m.a
i.e. TW = m.a ...(i)
FOR DOWN MOTION
Net force = W – T
Also Net Force = m.a
i.e.
W – T = m.a ...(ii)
Note: In the above cases, we have taken weight or mass carried by the lift only. We have assumed that the
weight carried by the lift includes weight of the lift also. But sometimes the example contains weight of
the lift and weight carried by the lift separately. In such a case, the weight carried by the lift or weight of
the operator etc, will exert a pressure on the floor of the lift. Whereas tension in the cable will be given
by the algebraic sum of the weight of the lift and weight carried by the lift.
Q.15: An elevator cage of a mineshaft, weighing 8KN when, is lifted or lowered by means of a wire
rope. Once a man weighing 600N, entered it and lowered with uniform acceleration such that
when a distance of 187.5m was covered, the velocity of cage was 25m/sec. Determine the
tension in the rope and the force exerted by the man on the floor of the cage.
Sol: Given data;
Weight of empty lift W
L
= 8KN = 8000N
Weight of man W
m
= 600N
Distance covered by lift s = 187.5m
Velocity of lift after 187.5m v = 25m/sec
Tension in rope T = ?
Force exerted on the man F
m
=?
Apply the relation v
2
= u
2
+ 2as, for finding acceleration
(25)
2
= 0 + 2a(187.5)a = 1.67m/sec
2
...(i)
Cage moves down only when W
L
+ W
m
>T
Net accelerating force = (W
L
+ W
m
) T
Using the relation F = ma, we get (W
L
+ W
m
) T = ma = [(W
L
+ W
m
)/g]a(8000 + 600) – T = [(8000
+ 600)/9.81] X 1.67
T = 7135.98N .......ANS
Calculation for force exerted by the manConsider only the weight of the man,
F
m
– W
m
= maF
m
– 600 = (600/9.81) X 1.67F
m
= 714.37N
Laws of Motion / 251
Lift
Moves
Down
T
W + W
L m
a
Fig 11.9
Since Newton’s third law i.e The force of action and reaction between interacting bodies are equal in
magnitude, opposite in direction and have the same line of action.
i.e., Force exerted by the man = F = 714.37N ........ANS
Q.16: An elevator weight 2500N and is moving vertically downward with a constant acceleration.
(1) Write the equation for the elevator cable tension.
(2) Starting from rest it travels a distance of 35m during an interval of 10 sec. Find the cable
tension during this time.
(3) Neglect all other resistance to motion. What are the limits of cable tension.
Sol: Given data;
Weight of elevator W
E
= 2500N
Initial velocity u = 0
Distance traveled s = 35m
Time t = 10sec
(1) Since elevator is moving down
Net acceleration force in the down ward direction
= W
E
– T = (2500 – T)N ...(i)
The net accelerating force produces acceleration ‘a’ in the down ward direction.
Using the relation, F = ma
2500 – T = (2500/9.81)a
T = 2500 – (2500/9.81)a ........ANS
Hence the above equation represents the general equation for the elevator cable tension when the
elevator is moving downward.
(2) Using relation,
s = ut +
1
2
at
2
= 35 = 0 X 10 + 1/2 X a (10)
2
...(ii)
∴ a = 0.7 m/sec
2
Substituting this value of a in the equation of cable tension
T = 2500 – (2500/9.81) X 0.7T = 2321.61N .......ANS
(3) T = 2500 – (2500/9.81)a
Limit of cable tension is depends upon the value of a, which varies from 0 to g i.e. 9.81m/sec
2
At a = 0, T = 2500
252 / Problems and Solutions in Mechanical Engineering with Concept
i.e elevator freely down
At a = 9.81, T = 0
i.e elevator is at the top and stationary.
Hence Limits are 0 to 2500N .......ANS
Elecator
Moves
Down
T
W
E
a
Fig 11.10
Q. 17: A vertical lift of total mass 500Kg acquires an upward velocity of 2m/sec over a distance of
3m of motion with constant acceleration, starting from rest. Calculate the tension in the cable
supporting the lift. If the lift while stopping moves with a constant deceleration and comes to
rest in 2sec, calculate the force transmitted by a man of mass 75kg on the floor of the lift
during the interval.
Sol: Given data,
Mass of lift M
L
= 500Kg
Final Velocity v = 2m/sec
Distance covered s = 3m
Initial velocity u = 0
Cable tension T = ?
Apply the relation v
2
= u
2
+ 2as
2
2
= 0 + 2a X 3a = 2/3 m/sec
2
...(i)
Since lift moves up, T > M
L
X gNet accelerating force = T – M
L
g, and it is equal to,T – M
L
g = maT
– 500 X 9.81 = 500 X 2/3
T = 5238.5N .......ANS
Let force transmitted by man of mass of 75Kg, is FF – mg = ma For finding the acceleration, using
the relation v = u + at0 = 2 + a X 2
Lift
Moves
Down
T
W
L
a
Fig 11.11
Laws of Motion / 253
a = –1 m/sec
2
...(ii)
Putting the value in equation, F – mg = ma
F – 75 X 9.81 = 75(–1)
F = 660.75N .......ANS
Q. 18: An elevator weight 5000N is ascending with an acceleration of 3m/sec
2
. During this ascent its
operator whose weight is 700N is standing on the scale placed on the floor. What is the scale
reading? What will be the total tension in the cable of the elevator during this motion?
Sol: Given data, W
E
= 5000N, a = 3m/sec
2
, W
O
= 700N,
Let R = Reaction offered by floor on operator. This is also equal to the reading of scale.
T = total tension in the cable
Operator
5000 N
T
R
Cable
E
l
e
v
a
t
o
r
Fig 11.12
Net upward force on operator
= Reaction offered by floor on operator – Weight of operator
= R – 700
But, Net force = ma
R – 700 = (700/9.81)X 3
R = 914.28N .......ANS
Now for finding the total tension in the cable, Total weight of
elevator is considered.
Net upward force on elevator and operator
= Total tension in the cable – Total weight of elevator and operator
= T – 5700
But net force = mass X acceleration
T – 5700 = (5700/9.81) X 3
T = 7445N .......ANS
Q.19: Analyse the motion of connected bodies, which is connected by a pulleys.
Sol: Fig 11.13 shows a light and inextensible string passing over a smooth and weightless pulley. Two
bodies of weights W
1
and W
2
are attached to the two ends of the string.
Let W
1
>W
2
, the weight W
1
will move downwards, whereas smaller weight W
2
will move upwards. For
an inextensible string, the upward acceleration of the weight W
2
will be equal to the downward acceleration
of the weight W
1
.
As the string is light and inextensible and passing over a smooth pulley, the tension of the string will
be same on both sides of the pulley.
Consider the Motion of weight W
1
(Down motion)
W
1
T = m
1
.a ...(i)
254 / Problems and Solutions in Mechanical Engineering with Concept
Consider the Motion of weight W
2
(up motion)
Inextensiable
liqut string
Smooth Puley
T
W
2
W
1
Fig 11.13
TW
2
= m
2
.a ...(ii)
Solved both the equation for finding the value of Tension (T) or acceleration (a)
Q.20: Two bodies weighing 300N and 450N are hung to the two ends of a rope passing over an ideal
pulley as shown in fig (11.14). With what acceleration will the heavier body come down? What
is the tension in the string?
Sol: Since string is light, inextensible and frictionless, so the tension in the string on both side is equal to
T, let acceleration of both the block is ‘a’.
450
450
450
ma
T
300
Fig 11.14
Let 450N block moves down,
Consider the motion of 450N block,
Apply the equation, F = ma
450 – T = (450/9.81) a
450 – T = 45.87a ...(i)
Consider the motion of 300N block,
Apply the equation, F = ma
T  300 = (300/9.81) a
T  300 = 30.58a ...(ii)
Add equation (1) and (2)
150 = 76.45a
a = 1.962m/sec
2
.......ANS
Putting the value of a in equation (i), we get
T = 360N .......ANS
Q.21: Find the tension in the string and accelerations of blocks A and B weighing 200N and 50N
respectively, connected by a string and frictionless and weightless pulleys as shown in
fig 11.15.
Laws of Motion / 255
Sol: Given Data,
Weight of block A = 200N
Weight of block B = 50N
As the pulley is smooth, the tension in the string will be same throughout
Let, T = Tension in the string a = Acceleration of block B
Then acceleration of block A will be equal to half the acceleration of block B.
Acceleration of block
A = a/2 ...(i)
As the weight of block is more than the weight of block B, the block A will move downwards whereas
the block B will move upwards.
T
200 N
50 N
Fig 11.15
Consider the motion of block B,
Net force = T – 50 ...(ii)
Since Net force, F = ma
T – 50 = (50/9.81) a
T – 50 = 5.1a ...(iii)
Consider the motion of block A,
Net force = 200– 2T ...(iv)
Since Net force, F = ma
200– 2T = (200/9.81)(a/2)
200– 2T = 10.19a
100 – T = 5.1a ...(v)
Add equation (3) and (5)
50 = 10.19a
a = 4.9m/sec
2
.......ANS
Putting the value of a in equation (5) we get
T = 75N .......ANS
Q.22: The system of particles shown in fig 11.16 is initially at rest. Find the value of force F
that should be applied so that the system acquires a velocity of 6m/sec after moving 5m.
(Nov–03(C.O.))
Sol: Given data,
Initial velocity u = 0
256 / Problems and Solutions in Mechanical Engineering with Concept
Final velocity v = 6m/sec
Distance traveled s = 5m
For finding acceleration, using the relation, v
2
= 4
2
+ 2as
∴ a = 3.6 m/sec
2
6
2
= 0 + 2a X 5 ... (i)
Apply the relation F = ma,
B
F
A
String
100 N
100 N
Fig 11.16
Let T = Tension in the string, same for both side
Using the relation F = ma, for block A
T – 100 = ma
T – 100 = (100/9.81) X 3.6 ...(ii)
Using the relation F = ma, for block B
100 + F – T = ma
100 + F – T = (100/9.81) X 3.6 ...(iii)
Add equation (2) and (3), we get
F = 2[(100/9.81) X 3.6]
F = 73.5N .......ANS
Q.23: A system of weight connected by string passing over pulleys A and B is shown in fig. Find the
acceleration of the three weights. Assume weightless string and ideal condition for pulleys.
Sol: As the strings are weightless and ideal conditions prevail, hence the tensions in the string passing over
pulley A will be same. The tensions in the string passing over pulley B will also be same. But the tensions
in the strings passing over pulley A and over pulley B will be different as shown in fig 11.17.
Let T
1
= Tension in the string passing over pulley A
T
2
= Tension in the string passing over pulley B
One end of the string passing over pulley A is connected to a weight 15N, and the other end is
connected to pulley B. As the weight 15N is more than the weights (6 + 4 = 10N), hence weight 15N will
move downwards, whereas pulley B will move upwards. The acceleration of the weight 15N and of the
pulley B will be same.
Let, a = Acceleration of block 15N in downward directiona
1
= Acceleration of 6N downward with
respect to pulley B.
Then acceleration of weight of 4N with respect to pulley B = a
1
in the upward direction.
Laws of Motion / 257
Pulley A
4 N
6 N
T
2
T
2
T
2
T
2
Pulley B
15 N
T
1
T
1
Fig 11.17
Absolute acceleration of weight 4N,
= Acceleration of 4N w.r.t. pulley B + Acceleration of pulley B.
= a
1
+ a (upward)
(as both acceleration are in upward direction, total acceleration will be sum of the two accelerations)
Absolute acceleration of weight 6N,
= Acceleration of 6 w.r.t. pulley B + Acceleration of pulley B.
= a
1
– a (downward)
(As a
1
is acting downward whereas a is acting upward. Hence total acceleration in the downward
direction)
Consider the motion of weight 15N
Net downward force = 15 – T
1
Using F = ma,
15 – T
1
= (15/9.81)a ...(1)
Consider the motion of weight 4N
Net downward force = T
2
 4
Using F = ma,
T
2
– 4 = (4/9.81)(a + a
1
) ...(2)
Consider the motion of weight 6N
Net downward force = 6  T
2
Using F = ma,
6 – T
2
= (6/9.81)(a
1
– a) ...(3)
Consider the motion of pulley B,
T
1
= 2T
2
...(4)
Adding equation (2) and (3)
2 = (4/9.81)(a + a
1
) + (6/9.81)(a
1
– a)
9.81 = 5a
1
– a ...(5)
Multiply equation (2) by 2 and put the value of equation (4), we get
258 / Problems and Solutions in Mechanical Engineering with Concept
T
1
– 8 = (8/9.81)(a
1
+ a) ...(6)
Adding equation (1) and (6), we get
15 – 8 = (15/9.81)a + (8/9.81)(a
1
+ a)
23a + 8a
1
= 7 X 9.81 ...(7)
Multiply equation (5) by 23 and add with equation (7), we get
a
1
= 2.39m/sec
2
.......ANS
Putting the value of a
1
in equation (5), we get
a = 2.15m/sec
2
.......ANS
Acceleration of weight 15N = a = 2.15m/sec
2
.......ANS
Acceleration of weight 6N = a = 0.24m/sec
2
.......ANS
Acceleration of weight 4N = a = 4.54m/sec
2
.......ANS
Q.24: A cord runs over two pulleys A and B with fixed axles, and carries a movable pulleys ‘c’ if
P = 40N, P
1
= 20N, P
2
= 30N and the cord lies in the vertical plane. Determine the acceleration
of pulley ‘C. Neglect the friction and weight of the pulley.
Sol: a = a
1
+ a
2
...(1)
For pulley A, Apply F = ma,
T – 20 = (20/10) a
1
, take g = 10m/sec
2
T – 20 = 2a
1
...(2)
For pulley C, 40 – 2T = (40/10)a40 – 2T = 4a ...(3)
For pulley B, T – 30 = (30/10) a
2
T – 30 = 3a
2 ...
(4)
P = 30N
2
P = 20N
1
P = 40N
C
a
2
a
1
B
A
– a
2
a
1
2
(
(
Fig 11.18
From equation (2) and (4)
2a
1
– 3a
2
= 10 ...(5)
Equation (3) can be rewritten as
40 – 2T = 4(a
1
+ a
2
) ...(6)
Now (6) + 2 (4)
40 – 2T + 2T – 2 X 30 = 4(a
1
+ a
2
) + 6a
2
–20 = 4a
1
+ 10a
2
...(7)
Solving equation (5) and (7), we get
a
1
= 5/4 m/sec
2
.......ANS
a
2
= 5/2m/sec
2
.......ANS
Acceleration of ‘C’ = a = a
1
+ a
2
= 5/4 – 5/2 = –1.25m/sec
2
(downward) .......ANS
Laws of Motion / 259
Q.25: Analyse the motion of two bodies connected by a string when one body is lying on a horizontal
surface and other is hanging free for the following cases.
1. The horizontal surface is smooth and the string is passing over a smooth pulley.
2. The horizontal surface is rough and string is passing over a smooth pulley.
3. The horizontal surface is rough and string is passing over a rough pulley.
Sol: CASE1: THE HORIZONTAL SURFACE IS SMOOTH AND THE STRING IS PASSING OVER A
SMOOTH PULLEY:
Fig shows the two weights W
1
and W
2
connected by a light inextensible string, passing over a smooth
pulley. The weight W
2
is placed on a smooth horizontal surface, whereas the weight W
1
is hanging free.The
weight W
1
is moving downwards, whereas the weight W
2
is moving on smooth horizontal surface. The
velocity and acceleration of W
1
will be same as that of W
2
.As the string is light and inextensible and
passing over a smooth pulley, the tensions of the string will be same on both sides of the pulley.
W
2
W
2
W
1
T
Smooth
horizontal surface
T
Pulley
Fig 11.19
For W
1
block: Move down
W
1
– T = (W
1
/g).a ...(1)
For W
2
block
T = (W
2
/g).a ...(2)
(Since W act vertically and T act Horizontally & w.cos90 = 0)
Solve both the equation for the value of ‘T’ and ‘a’.
CASE2: THE HORIZONTAL SURFACE IS ROUGH AND STRING IS PASSING OVER A SMOOTH
PULLEY.
Fig shows the two weights W
1
and W
2
connected by a light inextensible string, passing over a smooth
pulley. The weight W
2
is placed on a rough horizontal surface, whereas the weight W
1
is hanging free.
Hence in this case force of friction will be acting on the weight W
2
in the opposite direction of the motion
of weight W
2
.
Let, µ = Coefficient of friction between weight W
2
and horizontal surface. Force of friction = µR
2
=
µW
2
Motion of W
1
(Down Motion)
W
1
– T = (W
1
/g).a ...(1)
260 / Problems and Solutions in Mechanical Engineering with Concept
W
2
W
2
Rough
surface
Frictional
force
( W )
2
R
2
W
1
T
T
Fig 11.20
Motion of W
2
T – µ.W
2
= (W
2
/g).a ...(2)
Solve the equations for Tension ‘T’ and Acceleration ‘a’
CASE3: THE HORIZONTAL SURFACE IS ROUGH AND STRING IS PASSING OVER A ROUGH
PULLEY.
Fig shows the two weights W
1
and W
2
connected by a string, passing over a rough pulley. The weight
W
2
is placed on a rough horizontal surface, whereas the weight W
1
is hanging free. Hence in this case force
of friction will be acting on the weight W
2
in the opposite direction of the motion. As the string is passing
over a rough pulley. The tension on both side of the string will not be same.
Let, µ
1
= Coefficient of Friction between Weight W
2
and Horizontal plane µ
2
= Coefficient of Friction
between String and pulley T
1
= Tension in the string to which weight W
1
is attached
W
2
W
2
Rough
surface
Frictional
force
( W )
2
R
2
W
1
T
T
Fig 11.21
T
2
= Tension in the string to which weight W
2
is attached
Force of friction = µ
1
R
2
= µ
1
W
2
Consider block W
1
W
1
– T
1
= (W
1
/g).a ...(1)
Consider block W
2
T
2
– µ
2
.W
2
= (W
2
/g).a ...(2)
Another equation is, T
1
/T
2
= e
µ.θ
...(3)
Solve all three equation for the value of ‘a’, ‘T
1
’ and ’T
2
’
Q.26: Two bodies of weight 10N and 1.5N are connected to the two ends of a light inextensible
String, passing over a smooth pulley. The weight 10N is placed on a rough horizontal surface
while the weight of 1.5N is hanging vertically in air. Initially the friction between the weight
Laws of Motion / 261
10N and the table is just sufficient to prevent motion. If an additional weight of 0.5N is added
to the weight 1.5N, determine
(i) The acceleration of the two weight.
(ii) Tension in the string after adding additional weight of 0.5N to the weight 1.5N
Sol: Initially when W
1
= 1.5N, then the body is in equilibrium. i.e. both in rest or a = 0,
Then consider block W
1
R
V
= 0; T = W
1
= 1.5N ...(1)
Consider block W
2
R
V
= 0; R = W
2
= ...(2)
F
r
– T = 0; F
r
= T = 1.5N ...(3)
But, Fr = µR = µW
2
; µW
2
= 1.5; µ X 10 = 1.5, µ = 0.15 ...(4)
Now when Weight W
1
= 2.0N, body moves down Now the tension on both side be T
1
Consider block W
1
W
1
– T
1
= ma2 – T
1
= (2/g)a ...(5)
Consider block W
2
W
2
W
2
Frictional force
( , W )
1 2
R
2
W
1
T
1
T
1
T
Fig 11.22
T
1
– F
r
= ma
T
1
– µW
2
= (10/g)a
T
1
– 1.5 = (10/g)a ...(6)
Solve the equation (5) and (6) for T
1
and a, we get
T
1
= 1.916N, a = 0.408m/sec
2
.......ANS
Q.27: Two blocks shown in fig 11.23, have masses A = 20N and B = 10N and the coefficient of
friction between the block A and the horizontal plane, µ µµ µµ = 0.25. If the system is released from
rest, and the block B falls through a vertical distance of 1m, what is the velocity acquired by
it? Neglect the friction in the pulley and the extension of the string.
Sol: Let T = Tension on both sides of the string.
a = Acceleration of the blocks
µ = 0.25 Consider the motion of block B,
W
B
– T = ma ...(1)
10 – T =
10
2
a
⋅
262 / Problems and Solutions in Mechanical Engineering with Concept
20 kg
10 kg
B
A
Fig 11.23
Consider the motion of block A,
T – µW
A
= ma
T – 0.25 X 20 = (20/g)a ...(2)
Add equation (1) and (2)
10 – 5 = (30/g)a
a = 1.63m/sec
2
...(3)
Now using the relation, v
2
= u
2
+ 2as
v
2
= 0 + 2X 1.63 X 1
v = 1.81m/sec .......ANS
Q.28: Analyse the motion of two bodies connected by a string one of which is hanging free and other
lying on a smooth inclined plane.
Sol.: Consider two bodies of weight W
1
and W
2
respectively connected by a light inextensible string as
shown in fig 11.24
Let the body W
1
hang free and the W
2
be places on an inclined smooth plane.
W
1
will move downwards and the body W
2
will move upwards along the
inclined surface. A little consideration will show that the velocity and acceleration of the body W
1
will
be same as that of W
2
. Since the string is inextensible, therefore tension in both the string will also be equal.
Consider the motion of W
1
W
1
– T = (W
1
/g)a ...(1)
Consider the motion of W
1
m
2
T
m
1
Fig 11.24
T – W
2
sin ± = (W
1
/g)a ...(2)
Solve the equations for ‘T’ and ‘a’
Q.29: Analyse the motion of two bodies connected by a string one of which is hanging free and other
lying on a rough inclined plane.
Sol.: Consider two bodies of weight W
1
and W
2
respectively connected by a light inextensible string as
shown in fig 11.25.
Let the body W
1
hang free and the W
2
be places on an inclined rough plane. W
1
will move downwards
and the body W
2
will move upwards along the inclined surface.
Consider the motion of W
1
W
1
– T = (W
1
/g)a ...(1)
Consider the motion of W
1
T – W
2
sin α – µW
1
cos α = (W
1
/g)a ...(2)
Laws of Motion / 263
Solve the equations for ‘T’ and ‘a’
= 0.2
T
10 kg
30º
15 kg
T
T
Fig 11.25
Q.30: Determine the resulting motion of the body A, assuming the pulleys to be smooth and weightless
as shown in fig 11.26. If the system starts from rest, determine the velocity of the body A after
10 seconds.
Sol.: Given data:
Mass of Block A = 10Kg
Mass of Block B = 15Kg
Angle of inclination α = 30
0
Coefficient of friction m = 0.2
Consider the motion of block B,
The acceleration of block B will be half the acceleration of the block A i.e. a/2,
M
1
g – 2T = m
1
(a/2)
× = 0 2
T
6
4
T
T
15 kg
30°
10 kg
Fig 11.26
15 X 9.81 – 2T = 15 (a/2)
147.15 – 2T = 7.5a ...(1)
Consider the motion of block B,
T – W
2
sin α – µW
1
cos α = (W
1
/g)a
T  m
2
g sin α – 0.2m
2
gcos α = m
2
a
T – 10 X 9.81sin 30
0
– 0.2 X 10 X 9.81cos30
0
= 10a
T – 66.04 = 10a ...(2)
Adding equation (1) with 2 X equation (2)
147.15 – 2T + 2T – 132.08 = 7.5a + 20a
a = 0.54 m/sec
2
.......ANS
Now velocity of the block after 10 sec,
Apply v = u + at
V = 0 + 0.54 X 10
V = 5.4m/sec .......ANS
264 / Problems and Solutions in Mechanical Engineering with Concept
Q.31: In the fig 11.27, the coefficient of friction is 0.2 between the rope and the fixed pulley, and
between other surface of contact, m = 0.3. Determine the minimum weight W to prevent the
downward motion of the 100N body.
4
tan = 33
3
4
1
0
0
N
W
cos = 0.8
sin = 0.6
RN
1
0.3 × RN
1
T
1
w cos
4 sin
Fig 11.27 Fig 11.28
T
1
T
2
RN
2
T
2
100 cos + RN
1
0.3 RN
2
0.3 RN
1
100 N
100 sin
Fig 11.29 Fig 11.30
Sol.: From the given fig tanα = 3/4,
cosα = 4/5 & sinα = 3/5,
Consider equilibrium of block W
R
V
= 0; R
2
= Wcosα ...(1)
R
H
= 0; T
1
= µR
2
+ Wsinα ...(2)
Putting the value of equation(1) in (2)
T
1
= µWcos α + Wsinα
= 0.3 X W(4/5) + W(3/5)
T
1
= 0.84W ...(3)
For pulley; T
2
/T
1
= e
µ
1
θ
T
2
= T
1
X e
µ
1
θ
= 0.84We
(0.2 X )
T
2
= 1.574W ...(4)
Consider equilibrium of block 100N
R
V
= 0; R
1
= 100cos α + R
2
...(5)
R
1
= 100cosα + Wcosα
= 100(4/5) + W(4/5)
R
1
= 80 + 0.8W ...(6)
R
H
= 0;
T
2
= 100sinα –µR
1
µR
2
T
2
= 100(3/5) –0.3[(80 + 0.8W) – W(4/5)]
1.574W = 60 – 24 – 0.24W – 0.24W
W= 17.53N .......ANS
Beam / 265
+0)264
12
*)
Q.1: How you define a Beam, and about Shear force & bending moment diagrams?
Sol.: A beam is a structural member whose longitudinal dimensions (width) is large compared to the
transverse dimension (depth). The beam is supported along its length and is acted by a system of loads at
right angles to its axis. Due to external loads and couples, shear force and bending moment develop at ant
section of the beams. For the design of beam, information about the shear force and bending moment is
desired.
Shear Force (S.F.)
The algebraic sum of all the vertical forces at any section of a beam to the right or left of the section is
known as shear force.
Bending Moment (B.M.)
The algebraic sum of all the moment of all the forces acting to the right or left of the section is known as
bending Moment.
Shear Force (S.F.) and Bending Moment (B.M.) Diagrams
A S.F. diagram is one, which shows the variation of the shear force along the length of the beam. And a
bending moment diagram is one, which shows the variation of the bending moment along the length of the
beam.
Before drawing the shear force and bending moment diagrams, we must know the different types of
beam, load and support.
Q.2: How many types of load are acting on a beam?
A beam is normally horizontal and the loads acting on the beams are generally vertical. The following are
the important types of load acting on a beam.
A B C D
E G
Couple
Varying load
udl
Point load
H
F
Fig 12.1 Various type of load acting on beam
266 / Problems and Solutions in Mechanical Engineering with Concept
Concentrated or Point Load
A concentrated load is one, which is considered to act at a point, although in practical it must really be
distributed over a small area.
Uniformly Distributed Load (UDL)
A UDL is one which is spread over a beam in such a manner that rate of loading 'w' is uniform along the
length (i.e. each unit length is loaded to the same rate). The rate of loading is expressed as w N/m run. For
solving problems, the total UDL is converted into a point load, acting at the center of UDL.
Uniformly Varying Load (UVL)
A UVL is one which is spread over a beam in such a manner that rate of loading varies from point to point
along the beam, in which load is zero at one end and increase uniformly to the other end. Such load is
known as triangular load. For solving problems the total load is equal to the area of the triangle and this
total load is assumed to be acting at the C.G. of the triangle i.e. at a distance of 2/3rd of total length of
beam from left end.
Q.3: What sign convention is used for solving the problems of beam?
Although different sign conventions many be used, most of the engineers use the following sign conventions
for shear forces and bending moment.
(i) The shear force that tends to move left portion upward relative to the right portion shall be called
as positive shear force.
Fig 12.2
(ii) The bending moment that is trying to sag (Concave upward) the beam shall be taken as positive
bending moment. If left portion is considered positive bending moment comes out to be clockwise
moment.
Fig 12.3
To decide the sign of moment due to a force about a section, assume the beam is held tightly at that
section and observe the deflected shape. Then looking at the shape sign can be assigned.
The shear force and bending moment vary along the length of the beam and this variation is represented
graphically. The plots are known as shear force and bending moment diagrams. In these diagrams, the
abscissa indicates the position of section along the beam, and the ordinate represents the value of SF and
BM respectively. These plots help to determine the maximum value of each of these quantities.
Beam / 267
Sagging
Hogging
+ ve BM – ve BM
Fig 12.4
Q. 4: What is the relation between load intensity, shear force and bending moment?
x dx
F
A
F
B
F
dx
w kN/m
M
M dM +
F dF +
Fig 12.5
Sol.: Consider a beam subjected to any type of transverse load of the general form shown in fig 12.5. Isolate
from the beam an element of length dx at a distance x from left end and draw its free body diagram as
shown in fig 12.5. Since the element is of extremely small length, the loading over the beam can be
considered to be uniform and equal to w KN/m. The element is subject to shear force F on its left hand
side. Further, the bending moment M acts on the left side of the element and it changes to (M + dM) on
the right side.
Taking moment about point C on the right side,
∑M
C
= 0
M – (M + dM) + F X dx – (W X dx) X dx/2 = 0
The UDL is considered to be acting at its C.G.
dM = Fdx – [W(dx)
2
]/2 = 0
The last term consists of the product of two differentials and can be neglected
DM = Fdx, or
F = dM/dx
Thus the shear force is equal to the rate of change of bending moment with respect to x.
Apply the condition ∑V = 0 for equilibrium, we obtain
F – Wdx – (F + dF) = 0
Or W = dF/dx
That is the intensity of loading is equal to rate of change of bending moment with respect to x.
F = dM/dx
and W = dF/dx = dM
2
/dx
2
Q.5: Define the nature of shear force and bending moment under load variation.
Sol.: The nature of SF and BM variation under twoload region is given in the table below
BETWEEN TWO POINTS, IF S.F.D B.M.D
No load Constant Linear
UDL Inclined Linear Parabolic
UVL Parabolic Cubic
268 / Problems and Solutions in Mechanical Engineering with Concept
Q.6: Define point of contraflexure or point of inflexion. Also define the point of zero shear force?
Sol.: The points (other than the extreme ends of a beam) in a beam at which B.M. is zero, are called points
of contraflexure or inflexion.
The point at which we get zero shear force, we get the maximum bending moment of that section/beam
at that point.
Q.7: How can you draw a shear force and bending moment diagram.
Sol.: In these diagrams, the shear force or bending moment are represented by ordinates whereas the length
of the beam represents abscissa. The following are the important points for drawing shear force and bending
moment diagrams:
1. Consider the left or right side of the portion of the section.
2. Add the forces (including reaction) normal to the beam on one of the portion. If right portion of the
section is chosen, a force on the right portion acting downwards is positive while force acting upwards
is negative.
3. If the left portion of the section is chosen, a force on the left portion acting upwards is positive while
force acting downwards is negative.
4. The +ive value of shear force and bending moment are plotted above the base line, and ive value below
the base line.
5. The S.F. diagram will increase or decrease suddenly i.e. by a vertical straight line at a section where
there is a vertical point load.
6. In drawing S.F. and B.M. diagrams no scale is to be chosen, but diagrams should be proportionate
sketches.
7. For drawing S.F. and B.M. diagrams, the reaction of the right end support of a beam need not be
determined. If however, reactions are wanted specifically, both the reactions are to be determined.
8. The Shear force between any two vertical loads will remain constant. Hence the S.F. diagram will be
horizontal. The B.M. diagram will be inclined between these two loads.
9. For UDL S.F. diagram will be inclined straight line and the B.M. diagram will be curve.
10. The bending moment at the two supports of a simply supported beam and at the free end of a cantilever
will be zero.
11. The B.M. is maximum at the section where S.F. changes its sign.
12. In case of overhanging beam, the maximum B.M. will be least possible when +ive max. B.M. is equal
to the ive max. B.M.
13. If not otherwise mentioned specifically, selfweight of the beam is to be neglected.
14. Section line is draw between that points on which load acts.
Numerical Problems Based on Simply supported beam
Q.8: Draw the SF and BM diagram for the simply supported beam loaded as shown in fig 12.6.
2 KN 2 KN
1 m 1 m 1 m
R
A
R
B
B A
C D E
4 KN 2 KN
Fig 12.6
Beam / 269
A
X
1
X
2
C D E
B
R
B
R
A
2KN
4 4
2
+ ve
2
–2
–2
– 4
S.F.D 6
4
0 0
4
B.M.D.
– 4
0
4KN 2KN
X
3
X
4
Fig 12.7
Sol.: Let reaction at support A and B be, R
A
and R
B
First find the support reaction
For that,
∑V= 0
R
A
+ R
B
– 2 – 4 2 = 0, R
A
+ R
B
= 8 ...(1)
Taking moment about point A,
∑M
A
= 0
2 X 1 + 4 X 2 + 2 X 3 – R
B
X 4 = 0
R
B
= 4KN ...(2)
From equation (1), R
A
= 4KN ...(3)
Calculation for the Shear force Diagram
Draw the section line, here total 4 section line, which break the
load R
A
and 2KN(Between Point A and C),
2KN and 4KN(Between Point C and D),
4KN and 2KN (Between Point D and E) and
2KN and RB(Between Point E and B)
Consider left portion of the beam
Consider section 11
Force on left of section 11 is R
A
SF
1–1
= 4KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
A
= SF
C
= 4KN ...(4)
Consider section 22
Forces on left of section 22 is R
A
& 2KN
SF
2–2
= 4 – 2 = 2KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
C
= SF
D
= 2KN ...(5)
270 / Problems and Solutions in Mechanical Engineering with Concept
Consider section 33
Forces on left of section 33 is R
A
, 2KN, 4KN
SF
3–3
= 4 – 2 – 4 = –2KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
D
= SF
E
= –2KN ...(6)
Consider section 44
Forces on left of section 44 is R
A
, 2KN, 4KN, 2KN
SF
4–4
= 4 – 2 – 4 – 2 = – 4KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
E
= SF
B
= –4KN ...(7)
Plot the SFD with the help of above shear force values.
Calculation for the Bending moment Diagram
Let
Distance of section 11 from point A is X
1
Distance of section 22 from point A is X
2
Distance of section 33 from point A is X
3
Distance of section 44 from point A is X
4
Consider left portion of the beam
Consider section 11, taking moment about section 11
BM
1–1
= 4.X
1
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X
1
= 0 to X
1
= 1
At X
1
= 0
BM
A
= 0 ...(8)
At X
1
= 1
BM
C
= 4 ...(9)
i.e. inclined line 0 to 4
Consider section 22,taking moment about section 22
BM
2–2
= 4.X
2
– 2.(X
2
– 1)
= 2.X
2
+ 2
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of Bending moment at both nearest point of the section is varies with
X
2
= 1 to X
2
= 2
At X
2
= 1
BM
C
= 4 ...(10)
At X
2
= 2
BMD = 6 ...(11)
i.e. inclined line 4 to 6
Consider section 33,taking moment about section 33
BM
3–3
= 4.X
3
– 2.(X
3
– 1) – 4.(X
3
– 2)
= –2.X
3
+ 10
It is Equation of straight line (Y = mX + C), inclined linear.
Beam / 271
Inclined linear means value of Bending moment at both nearest point of the section is varies with
X
3
= 2 to X
3
= 3
At X
3
= 2
BM
D
= 6 ...(12)
At X
3
= 3
BM
E
= 4 ...(13)
i.e. inclined line 6 to 4
Consider section 44, taking moment about section 44
BM
44
= 4.X
4
– 2.(X
4
– 1) – 4.(X
4
– 2)  2.(X
4
– 3)
= 4.X
4
+ 16
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of Bending moment at both nearest point of the section is varies with
X
4
= 3 to X
4
= 4
At X
4
= 3; BM
E
= 4 ...(14)
At X
4
= 4; BM
B
= 0 ...(15)
i.e. inclined line 4 to 0
Plot the BMD with the help of above bending moment values.
Q.9: Draw the SF and BM diagram for the simply supported beam loaded as shown in fig. 12.8.
1 m 1 m 1 m 1 m 2 m
R
A
A C D E F B
1 KN 1 KN 2 KN/m
R
B
Fig 12.8
S.F.D.
3
2
2
0
0
0
3
5
6
5
3
0
B.M.D.
–2
–2
–3 –3
A
C D
1KN 1KN
2KN/m
E F B
R
A
R
B
X
1
X
2
X
3
X
4
X
5
Fig 12.9.
272 / Problems and Solutions in Mechanical Engineering with Concept
Sol.: Let reaction at support A and B be, R
A
and R
B
First find the support reaction.
For finding the support reaction, convert UDL in to point load and equal to 2 X 2 = 4KN, acting at
mid point of UDL i.e. 3m from point A.
For that,
∑V = 0
R
A
+ R
B
– 1 – 4 – 1 = 0,
R
A
+ R
B
= 6 ...(1)
Taking moment about point A,
∑M
A
= 0
1 X 1 + 4 X 3 + 1 X 5 – R
B
X 6 = 0
R
B
= 3KN ...(2)
From equation (1), R
A
= 3KN ...(3)
Calculation for the Shear force Diagram
Draw the section line, here total 5section line, which break the
load R
A
and 1KN (Between Point A and C),
1KN and starting of UDL (Between Point C and D),
end point of UDL and 1KN (Between Point E and F) and
1KN and R
B
(Between Point F and B)
Let
Distance of section 11 from point A is X
1
Distance of section 22 from point A is X
2
Distance of section 33 from point A is X
3
Distance of section 44 from point A is X
4
Distance of section 55 from point A is X
5
Consider left portion of the beam
Consider section 11
Force on left of section 11 is R
A
SF
1–1
= 3KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
A
= SF
C
= 3KN ...(4)
Consider section 22
Forces on left of section 22 is R
A
& 1KN
SF
2–2
= 3 – 1 = 2KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
C
= SF
D
= 2KN ...(5)
Consider section 33
Forces on left of section 33 is R
A
, 1KN and UDL (from point D to the section line i.e. UDL on total
distance of (X
3
 2)
SF
3–3
= 3  1  2(X
3
 2) = 6  2X3 KN (Equation of straight line)
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of S.F. at both nearest point of the section is varies with X
3
= 2 to
X
3
= 4
Beam / 273
At X
3
= 2
SF
D
= 2 ...(6)
At X
3
= 4
SF
E
= –2 ...(7)
i.e. inclined line 2 to 2
Since here shear force changes the sign so at any point shear force will be zero and at that point
bending moment is maximum.
For finding the position of zero shear force equate the shear force equation to zero, i.e.
6 – 2X
3
= 0; X
3
= 3m, i.e. at 3m from point A bending moment is maximum.
Consider section 44
Forces on left of section 44 is R
A
, 1KN, 4KN
SF
4–4
= 3 – 1 – 4 = – 2KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
E
= SF
F
= 2KN ...(8)
Consider section 55
Forces on left of section 55 is RA, 1KN, 4KN, 1KN
SF
55
= 3 – 1 – 4 – 1 = –3KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
E
= SF
B
= –3KN ...(9)
Plot the SFD with the help of above shear force values.
Calculation for the Bending moment Diagram
Consider left portion of the beam
Consider section 11, taking moment about section 11
BM
1–1
= 3.X1
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X
1
= 0 to X
1
= 1
At X
1
= 0
BM
A
= 0 ...(10)
At X
1
= 1
BM
C
= 3 ...(11)
i.e. inclined line 0 to 3
Consider section 22,taking moment about section 22
BM
22
= 3.X
2
– 1.(X
2
– 1)
= 2.X
2
+ 1
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X
2
= 1 to X
2
= 2
At X
2
= 1
BM
C
= 3 ...(12)
At X
2
= 2
BM
D
= 5 ...(13)
274 / Problems and Solutions in Mechanical Engineering with Concept
i.e. inclined line 3 to 5
Consider section 33,taking moment about section 33
BM
33
= 3.X
3
– 1.(X
3
– 1) – 2.(X
3
– 2)[(X
3
– 2)/2]
= 2.X
3
+ 1 – (X
3
– 2)
2
It is Equation of Parabola (Y = mX2 + C),
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the
section is varies with X
3
= 2 to X
3
= 4
At X
3
= 2
BM
D
= 5 ...(14)
At X
3
= 4
BM
E
= 5 ...(15)
But B.M. is maximum at X
3
= 3, which lies between X
3
= 2 to X
3
= 4
So we also find the value of BM at X
3
= 3
At X
3
= 3
BM
max
= 6 ...(16)
i.e. curve makes with in 5 to 6 to 5 region.
Consider section 44, taking moment about section 44
BM
44
= 3.X
4
– 1.(X
4
– 1) – 4.(X
4
– 3)
= –2.X
4
+ 13
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X
4
= 4 to X
4
= 5
At X
4
= 4
BM
E
= 5 ...(17)
At X
4
= 5
BM
F
= 3 ...(18)
i.e. inclined line 5 to 3
Consider section 55,taking moment about section 55
BM
55
= 3.X
5
– 1.(X
5
– 1)  4.(X
5
– 3)  1.(X
5
– 5)
= – 3.X
5
+ 18
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X
5
= 5 to X
5
= 6
At X
5
= 5
BM
E
= 3 ...(19)
At X
4
= 6
BM
F
= 0 ...(20)
i.e. inclined line 3 to 0
Plot the BM
D
with the help of above bending moment values.
Beam / 275
Q.10: Draw the SF and BM diagram for the simply supported beam loaded as shown in fig. 12.10
R
A
A
E D C B
20 KN
30 KN
2 KN/m
1.5 m 1m 1m
0.5 m
R
B
Fig. 12.10
Sol.: Let reaction at support A and B be, R
A
and R
B
First find the support reaction. For finding the support
reaction, convert UDL in to point load and equal to 20 X 1.5 = 30KN, acting at mid point of UDL i.e.
0.75m from point A.
For that,
∑V = 0
R
A
+ R
B
 30  20 = 0, R
A
+ R
B
= 50 ...(1)
Taking moment about point A,
∑M
A
= 0
30 X 0.75 + 30 + 20 X 3 – R
B
X 4 = 0
R
B
= 28.125 KN ...(2)
From equation (1), R
A
= 21.875KN ...(3)
R
A
A B
0.75
21.875
–8.175
11.8
1.03125
36.25
28.125
6.25
0
0
+ve
–ve
–28.125 –28.125
0.75
30
2 × 2 = 4
0.5 1 1
R
B
Fig. 12.11
Calculation for the Shear force Diagram
Draw the section line, here total 4section line, which break the
load R
A
and UDL (Between Point A and E),
30KN/m and 20KN (Between Point E and D),
30KN/M and 20KN (Between Point D and C) and
20KN and RB (Between Point C and B)
Let
Distance of section 11 from point A is X
1
Distance of section 22 from point A is X
2
Distance of section 33 from point A is X
3
276 / Problems and Solutions in Mechanical Engineering with Concept
Distance of section 44 from point A is X
4
Consider left portion of the beam
Consider section 11
Force on left of section 11 is R
A
and UDL (from point A to the section line i.e. UDL on total distance
of X
1
SF
1–1
= 21.875 20X
1
KN (Equation of straight line)
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of shear force at both nearest point of the section is varies with X
1
= 0
to X
1
= 1.5
At X
1
= 0
SF
A
= 21.875 ...(4)
At X
1
= 1.5
SF
E
= –8.125 ...(5)
i.e. inclined line 21.875 to – 8.125
Since here shear force changes the sign so at any point shear force will be zero and at that point
bending moment is maximum.
For finding the position of zero shear force equate the shear force equation to zero, i.e.
21.875 20X
1
= 0; X
1
= 1.09375m, i.e. at 1.09375m from point A bending moment is maximum.
Consider section 22
Forces on left of section 22 is RA & 30KN
SF
22
= 21.875 – 30 = – 8.125KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
E
= SF
D
= – 8.125KN ...(6)
Consider section 33
Forces on left of section 33 is R
A
& 30KN, since forces are equal that of section 22, so the value
of shear force at section 33 will be equal that of section 22
SF
33
= 21.875 – 30 = – 8.125KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
D
= SF
C
= – 8.125KN ...(7)
Consider section 44
Forces on left of section 44 is R
A
, 30KN, 20KN
SF
44
= 21.875 – 30 20 = 28.125KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
C
= SF
B
= –28.125KN ...(8)
Plot the SFD with the help of above shear force values.
Calculation for the Bending moment Diagram
Consider left portion of the beam
Consider section 11, taking moment about section 11
BM
11
= 21.875X
1
20X
1
(X
1
/2)
It is Equation of Parabola (Y = mX
2
+ C),
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the
section is varies with X
1
= 0 to X
1
= 1.5
Beam / 277
At X
1
= 0
BM
A
= 0 ...(9)
At X
1
= 1.5
BM
C
= 10.3125 ...(10)
But B.M. is maximum at X
1
= 1.09, which lies between X
1
= 0 to X
1
= 1.5
So we also find the value of BM at X
1
= 1.09
At X
1
= 1.09
BM
max
= 11.8 ...(11)
i.e. curve makes with in 0 to 11.8 to 10.3125 region.
Consider section 22,taking moment about section 22
BM
22
= 21.875X
2
– 30(X
2
– 0.75)
= –8.125.X
2
+ 22.5
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X
2
= 1.5 to X
2
= 2
At X
2
= 1.5
BM
E
= 10.3125 ...(12)
At X
2
= 2
BM
D
= 6.25 ...(13)
i.e. inclined line 10.3125 to 6.25
Consider section 33, taking moment about section 33
BM
33
= 21.875X
3
– 30(X
3
– 0.75) + 30
= –8.125.X
2
+ 52.5
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X
3
= 2 to X
3
= 3
At X
3
= 2
BM
D
= 36.25 ...(14)
At X
3
= 3
BM
C
= 28.125 ...(15)
Consider section 44, taking moment about section 44
BM
44
= 21.875X
4
– 30(X
4
– 0.75) + 30 – 20 (X
4
– 3)
= –28.125.X
4
+ 112.5
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X
4
= 3 to X
4
= 4
At X
4
= 3
BM
C
= 28.125 ...(16)
At X
4
= 4
BM
B
= 0 ...(17)
i.e. inclined line 28.125 to 0
Plot the BM
D
with the help of above bending moment values.
278 / Problems and Solutions in Mechanical Engineering with Concept
Q.11: Determine the SF and BM diagrams for the simply supported beam shown in fig 12.12. Also
find the maximum bending moment.
17.86
17.8 15.76
Pan cubic
S.F.D
Parabolic
0
21.1
1.1
1.1
18.9
18.9
20KN 20KN
R
AH
R
AV
R
DH
R
DV
D C B
A
22 2 2
20KN
10 KN/m
Fig 12.12
Sol.: Since hinged at point A and D, suppose reaction at support A and D be, R
AH
, R
AV
and R
DH
, R
DV
first
find the support reaction. For finding the support reaction, convert UDL and UVL in to point load and,
Point load of UDL equal to 10 X 2 = 20KN, acting at mid point of UDL i.e. 1m from point A.
Point load of UVL equal to 1/2 X 20 X 2 = 20KN, acting at a distance 1/3 of total distance i.e.
1/3m from point D.
For that,
∑V = 0
R
AV
+ R
DV
– 20 – 20 = 0, R
A
+ R
B
= 40 ...(1)
Taking moment about point A,
∑M
A
= 0
20 X 1 + 20 X 5.33 – R
DV
X 6 = 0
R
DV
= 21.1 KN ...(2)
From equation (1), R
AV
= 18.9KN ...(3)
Calculation for the Shear force Diagram
Draw the section line, here total 3section line, which break the
load R
AV
and UDL (Between Point A and B),
No load (Between Point B and C) and
UVL (Between Point C and D).
Let
Distance of section 11 from point A is X
1
Distance of section 22 from point A is X
2
Distance of section 33 from point A is X
3
Consider left portion of the beam
Consider section 11
Beam / 279
Force on left of section 11 is R
AV
and UDL (from point A to the section line i.e. UDL on total distance
of X
1
SF
11
= 18.9 10X
1
KN (Equation of straight line)
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of shear force at both nearest point of the section is varies with X1 = 0
to X
1
= 2
At X = 0
SF
A
= 18.9 ...(4)
At X
1
= 2
SF
B
= –1.1 ...(5)
i.e. inclined line 18.9 to  1.1
Since here shear force changes the sign so at any point shear force will be zero and at that point
bending moment is maximum.
For finding the position of zero shear force equate the shear force equation to zero, i.e.
18.9 –10X
1
= 0; X
1
= 1.89m, i.e. at 1.89m from point A bending moment is maximum.
Consider section 22
Forces on left of section 22 is R
AV
& 20KN
SF
22
= 18.9 – 20 = – 1.1KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
B
= SF
C
= – 1.1KN ...(6)
Consider section 33
Forces on left of section 33 is R
AV
& 20KN and UVL of 20KN/m over (X
3
– 4) m length,
First calculate the total load of UVL over length of (X
3
– 4)
Consider triangle CDE and CGF
DE/GF = CD/CG
Since DE = 20
20/GF = 2/(X
3
– 4)
GF = 10(X
3
– 4)
Now load of triangle CGF = 1/2 X CG X GF = 1/2 X (X
3
– 4) X 10(X
3
– 4)
F
C
(M – 4)
G
2m
D
E
Fig 12.13
= 5(X
3
– 4)
2
, at a distance of (X
3
– 4)/3 from G ...(7)
SF
33
= 18.9 – 20 –5(X
3
 4)
2
= – 1.1 –5(X
3
– 4)
2
(Parabola)
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the
section is varies with X
3
= 4 to X
3
= 6
280 / Problems and Solutions in Mechanical Engineering with Concept
At X
3
= 4
SF
C
= –1.1KN ...(8)
SF
D
= –21.1KN ...(9)
Calculation for the Bending moment Diagram
Consider left portion of the beam
Consider section 11, taking moment about section 11
BM
1–1
= 18.9X
1
–10X
1
.X
1
/2
= 18.9X
1
–5 ⋅ X
1
2
It is Equation of Parabola (Y = mX
2
+ C),
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the
section is varies with X
1
= 0 to X
1
= 2
At X
1
= 0
BM
A
= 0 ...(10)
At X
1
= 2
BM
B
= 17.8 ...(11)
But B.M. is maximum at X
1
= 1.89, which lies between X
1
= 0 to X
1
= 2
So we also find the value of BM at X
1
= 1.89
At X
1
= 1.89
BM
max
= 17.86 ...(12)
i.e. curve makes with in 0 to 17.86 to 17.8 region.
Consider section 22,taking moment about section 22
BM
22
= 18.9X
2
– 20(X
2
– 1)
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X
2
= 2 to X
2
= 4
At X
2
= 2
BM
B
= 17.8 ...(13)
At X
2
= 4
BM
C
= 15.76 ...(14)
i.e. inclined line 17.8 to 15.76
Consider section 33,taking moment about section 33
BM
33
= 18.9X
3
– 20(X
3
– 1) – 5(X
3
– 4)
2
. (X
3
– 4)/3
It is cubic Equation which varies with X
3
= 4 to X
3
= 6
At X
3
= 4
BM
C
= 15.76 ...(15)
At X
3
= 6
BM
D
= 0 ...(16)
Plot the BM
D
with the help of above bending moment values.
Q.12: Draw the SF and BM diagrams for a simply supported beam 5m long carrying a load of 200N
through a bracket welded to the beam loaded as shown in fig 12.14
Sol.:
Beam / 281
2000N
0.5m
C D
A B
3m 2m
Fig 12.14
The diagram is of force couple system, let us apply at C two equal and opposite forces each equal and
parallel to 2000N. Now the vertically upward load of 2000N at C and vertically downward load of 2000N
at D forms an anticlockwise couple at C whose moment is 2000 X 0.5 = 1000Nm
And we are left with a vertically downward load of 2000N acting at C.
Let reaction at support A and B be, R
A
and R
B
first find the support reaction.
Taking moment about point A;
2000 X 3 – 1000 – R
B
X 5 = 0
R
B
= 1000N ...(1)
R
V
= 0, R
A
+ R
B
– 2000 = 0
R
A
= 1000N ...(2)
C
A
R
A
R
B
B
1000 Nm
1000 N
1000 N
S.F.D.
B.M.D.
3000 N – M
1000 N – M
2000 N
3m 2m
Fig 12.15
Calculation for the Shear force Diagram
Draw the section line, here total 2 section line, which break the load
R
A
and 2000N (Between Point A and C),
2000N and R
B
(Between Point C and B).
Let
Distance of section 11 from point A is X
1
Distance of section 22 from point A is X
2
Consider left portion of the beam
Consider section 11
Force on left of section 11 is R
A
SF
11
= 1000N (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
282 / Problems and Solutions in Mechanical Engineering with Concept
SF
A
= SF
C
= 1000N ...(3)
Consider section 22
Forces on left of section 22 is R
A
& 2000N
SF
22
= 1000 – 2000 = –1000 (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
C
= SF
B
= –1000N ...(4)
Plot the SFD with the help of above shear force values.
Calculation for the bending moment Diagram
Consider section 11, taking moment about section 11
BM
11
= 1000.X
1
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X
1
= 0 to X
1
= 3
At X
1
= 0
BM
A
= 0 ...(5)
At X
1
= 3
BM
C
= 3000 ...(6)
i.e. inclined line 0 to 3000
Consider section 22,taking moment about section 22
BM
22
= 1000.X
2
– 2000.(X
2
– 3) – 1000
= –1000.X
2
+ 5000
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of Bending moment at both nearest point of the section is varies with
X
2
= 3 to X
2
= 5
At X
2
= 3
BM
C
= 2000 ...(7)
At X
2
= 5
BM
B
= 0 ...(8)
i.e. inclined line 2000 to 0
Plot the BMD with the help of above bending moment values.
The SFD and BMD is shown in fig (12.15).
Q.13: A simply supported beam 6m long is subjected to a triangular load of 6000N as shown in fig
12.16 below. Draw the S.F. and B.M. diagrams for the beam.
Sol.:
Beam / 283
C
A
–3000Nm
3m
(a) S.F.D.
(a) B.M.D.
–3000Nm
R
A
R
B
E
B
F D
6m
6000N
6000N
Fig 12.16
Let
Suppose reaction at support A and B be, R
A
and R
B
first find the support reaction.
Due to symmetry, R
A
= R
B
= 6000/2 = 3000N ...(1)
Calculation for the Shear force Diagram
Draw the section line, here total 2section line, which break the point A,D and Point D,B
Let
Distance of section 11 from point A is X
1
Distance of section 22 from point A is X
2
Consider left portion of the beam
Consider section 11
Forces on left of section 11 is R
A
and UVL of 6000N/m over X
1
m length,
Since Total load = 6000 = 1/2 X AB X CD
1/2 X 6 X CD = 6000, CD = 2000N ...(2)
First calculate the total load of UVL over length of X
1
Consider triangle ADC and AFE
DC/EF = AD/AF
Since DC = 2000
2000/EF = 3/X
1
EF = (2000X
1
)/3
Now load of triangle AEF = 1/2 X EF × AF
= (1/2 X 2000X
1
)/3 × (X
1
)
=
2
1
1000
3
X ⋅
a distance of X
1
/3 from F ...(3)
SF11 = 3000 – (1000X12)/3 (Parabola)
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the
section is varies with X
1
= 0 to X
1
= 3
At X
1
= 0
SF
A
= 3000N ...(4)
284 / Problems and Solutions in Mechanical Engineering with Concept
At X
1
= 3
SF
D
= 0 ...(5)
Consider section 22
Forces on left of section 22 is R
A
and UVL of 2000N/m(At CD) and UVL over (X
2
– 3) m length,
First calculate the total load of UVL over length of (X
2
– 3)
Consider triangle CDB and BGH
DC/GH = DB/BG
Since DC = 2000
2000/GH = 3/(6  X
2
)
GH = 2000(6X
2
)/3
Now load of triangle BGH = 1/2 X GH X BG
= [1/2 X 2000(6X
2
)/3] X (6X
2
)
= 1000(6 – X
2
)2/3, at a distance of X
1
/3 from F ...(6)
Load of CDB = 1/2 X 3 X 2000 = 3000
Now load of CDGH = load of CDB  load of BGH
= 3000 – 1000(6 – X
2
)
2
/3 ...(7)
SF
22
= 3000 – 3000 – [3000 – 1000(6 – X
2
)
2
/3] (Parabola)
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the
section is varies with X
2
= 3 to X
2
= 6
At X
2
= 3
SF
A
= 0 ...(8)
At X
2
= 6
SF
D
= –3000N ...(9)
Plot the SFD with the help of above value as shown in fig.
Since SF change its sign at X
2
= 3, that means at a distance of 3m from point A bending moment is
maximum.
Calculation for the Bending moment Diagram
Consider section 11
BM
11
= 3000X
1
– [(1000X
1
2
)/3]X
1
/3 (Cubic)
Cubic means a parabolic curve is formed, value of bending moment at both nearest point of the section
is varies with X
1
= 0 to X
1
= 3
At X
1
= 0
BM
A
= 0 ...(10)
At X
1
= 3
BM
D
= 6000 ...(11)
Consider section 22
Point of CG of any trapezium is = h/3[(b + 2a)/(a + b)]
i.e. Distance of C.G of the trapezium CDGH is given by,
= 1/3 X DG X [(GH + 2CD)/(GH + CD)]
= 1/3.(X23).{[2000(6X2)/3] + 2 X 2000)}/{[ 2000(6X2)/3]+[ 2000]}
= {(X2 – 3)(12 – X2)}/{3(9 – X2)} ...(12)
BM
22
= 3000X
2
3000(X
2
2)[30001000(6 – X
2
)
2
/3]{+ (X
2
– 3)(12 – X
2
)}/{3(9 – X
2
)}
(Equation of Parabola)
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the
section is varies with X
2
= 3 to X
2
= 6
At X
2
= 3
BM
A
= 6000 ...(13)
Beam / 285
At X
2
= 6
BM
D
= 0N ...(14)
Plot the BMD with the help of above value.
Note: We also solve the problem by considering right hand side of the portion, example as given
below.
Q.14: A simply supported beam carries distributed load varying uniformly from 125N/m at one
end to 250N/m at the other. Draw the SF and BM diagram and determine the maximum
B.M.
C
1
5
0
k
m
D
F
P
5m
4.75m
–25.7N
H
E
D
135 Nm
750 N
Fig 12.17
Sol.: Total load = Area of the load diagram ABEC
= Rectangle ABED + Triangle DEC
= (AB X BE) + (1/2 X DE X DC) = (9 X 125) + [1/2 X 9 X (250125)]
= 1125N + 562.5N ...(1)
Centroid of the load of 1125N(rectangular load) is at a distance of 9/2 = 4.5m from AD and the
centroid of the load of 562.5N (Triangular load) is at a distance of 1/3 X DE = 1/3 X 9 = 3m from point
A.
Let support reaction at A and B be R
A
and R
B
. For finding the support reaction,
Taking moment about point A,
1125 X 4.5 + 562.5 X 3  R
B
X 9 = 0
R
B
= 750N ...(2)
Now, R
V
= 0
R
A
+ R
B
= 1125 + 562.5 = 1687.5
R
A
= 937.5N ...(3)
Calculation for the Shear force Diagram
Draw the section line, here total 1section line, which break the point A and B
Let
Distance of section 11 from point B is X
Consider right portion of the beam
Consider section 11
Forces on right of section 11 is R
B
and Load of PBEF and Load of EFH
286 / Problems and Solutions in Mechanical Engineering with Concept
SF11 = RB  load on the area PBEF  load on the area EFH
= RB  X.125  1/2.X.FH
In the equiangular triangles DEC and FEH
DC/DE = FH/FE or, 125/9 = FH/X
FH = 125X/9
S.F. between B and A = 750  125X  125X
2
/18 (Equation of Parabola)
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the
section is varies with X = 0 to X = 9
At X = 0
SF
B
= 750N ...(4)
At X = 9
SF
A
= –937.5N ...(5)
Since the value of SF changes its sign, which is between the point A and B we get max. BM
For the point of zero shear,
750 – 125X – 125X
2
/18 = 0
On solving we get, X = 4.75m
That is BM is max. at X = 4.75 from point B
Calculation for the Bending moment Diagram
Consider section 11
BM
1–1
= 750X – PB.BE.X/2 – 1/2.FE.FH.1/3.FE
= 750X – X.125.(X/2) – 1/2.X.(125X/9)(X/3)
= 750x – 125x2/2 – 125X2/54 (Equation of Parabola)
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the
section is varies with X = 0 to X = 9
At X = 0
BM
B
= 0 ...(6)
At X = 4.75
BM
max
= 1904Nm ...(7)
At X = 9
BM
A
= 0 ...(8)
Numerical Problems Based on Cantilever Beam
Q.15: Draw the SF and BM diagram for the beam as shown in fig 12.18. Also indicate the principal
values on the diagrams.
1m
2kN 3kN 3kN
2m 2m
Fig 12.18
Sol.: Let reaction at support A be R
AV
, R
AH
and M(anti clock wise), First find the support reaction
For that,
∑V = 0
R
AV
–2 –3 –3 = 0, R
AV
= 8 ...(1)
Beam / 287
Taking moment about point A,
∑M
A
= 0
–M + 2 X 1 + 3 X 3 + 3 X 5 = 0
M = 26KNm ...(2)
∑H = 0
R
AH
= 0 ...(3)
1 2
2
8 8
S.F.D
B.M.D
6
6
3
–26
–18
–6
3
+ve
–ve
3 3
2
B
A
C D
Fig 12.19
Calculation for the Shear force Diagram
Draw the section line, here total 3 section line, which break the
load R
AV
and 2KN(Between Point A and B),
2KN and 3KN(Between Point B and C),
3KN and 3KN (Between Point C and D).
Consider left portion of the beam
Consider section 11
Force on left of section 11 is R
AV
SF
11
= 8KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
A
= SF
B
= 8KN ...(4)
Consider section 22
Forces on left of section 22 is R
AV
& 2KN
SF
22
= 8 – 2 = 6KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
B
= SF
C
= 6KN ...(5)
Consider section 33
Forces on left of section 33 is R
A
, 2KN, 3KN
SF
33
= 8 – 2 – 3 = 3KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
288 / Problems and Solutions in Mechanical Engineering with Concept
SF
C
= SF
D
= 3KN ...(6)
Calculation for the Bending moment Diagram
Let
Distance of section 11 from point A is X
1
Distance of section 22 from point A is X
2
Distance of section 33 from point A is X
3
Consider section 11, taking moment about section 11
BM
11
= 8.X
1
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X
1
= 0
to X
1
= 1
At X
1
= 0
BM
A
= 0 ...(8)
At X
1
= 1
BM
B
= 8 ...(9)
i.e. inclined line 0 to 8
Consider section 22,taking moment about section 22
BM
22
= 8.X
2
– 2.(X
2
– 1)
= 6.X
2
+ 2
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of Bending moment at both nearest point of the section is varies with
X
2
= 1 to X
2
= 3
At X
2
= 1
BM
B
= 8 ...(10)
At X
2
= 3
BM
C
= 20 ...(11)
i.e. inclined line 8 to 20
Consider section 33, taking moment about section 33
BM
33
= 8.X
3
– 2.(X
3
– 1) – 3.(X
3
– 3)
= 3.X
3
+ 11
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of Bending moment at both nearest point of the section is varies with
X
3
= 3 to X
3
= 5
At X
3
= 3
BM
C
= 20 ...(12)
At X
3
= 5
BM
D
= 26 ...(13)
i.e. inclined line 20 to 26
Plot the BM
D
with the help of above bending moment values.
The SF
D
and BM
D
is shown in fig 12.19.
Beam / 289
Q.16: A cantilever is shown in fig 12.20. Draw the BMD and SFD. What is the reaction at supports?
Sol.:
B.M.D
Curve
– 60
– 40
A
A
14
M
10 10
C
C
B
B
S.F.D
10
R
AH
R
AV
4
C
2 m 4 m
B
A
2KN/m
Fig 12.20
Let reaction at support A be R
AV
, R
AH
and M (anti clock wise), First find the support reaction
For that,
∑V = 0
R
AV
–4 –10 = 0, R
AV
= 14 ...(1)
Taking moment about point A,
∑M
A
= 0
–M + 4 X 1 + 10 X 6 = 0
M = 64KNm ...(2)
∑H = 0
R
AH
= 0 ...(3)
Calculation for the Shear force Diagram
Draw the section line, here total 2 section line, which break the load R
AV
and UDL (Between Point
A and B), point B and 10KN(Between Point B and C).
Let
Distance of section 11 from point A is X
1
Distance of section 22 from point A is X
2
Consider left portion of the beam
Consider section 11
Force on left of section 11 is R
AV
and UDL from point A to section line
SF
11
= 14 2X1 KN (Equation of straight line)
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of shear force at both nearest point of the section is varies with X1 = 0
to X
1
= 2
At X
1
= 0
SF
A
= 14 ...(4)
290 / Problems and Solutions in Mechanical Engineering with Concept
At X
1
= 2
SF
B
= 10 ...(5)
i.e. inclined line 14 to 10
Consider section 22
Forces on left of section 22 is R
AV
& 4KN
SF
22
= 14 – 4 = 10KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
B
= SFC = 10KN ...(5)
Calculation for the Bending moment Diagram
Consider section 11, taking moment about section 11
BM
11
= –64 + 14.X
1
–2.X1(X
1
/2) (Equation of Parabola)
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the
section is varies with X
1
= 0 to X
1
= 2
At X
1
= 0
BM
A
= –64 ...(8)
At X
1
= 2
BM
B
= –40 ...(9)
i.e. parabolic line 64 to 40
Consider section 22,taking moment about section 22
BM
22
= –64 + 14.X
2
– 4.(X
2
– 1)
= –60 + 10X
2
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of Bending moment at both nearest point of the section is varies with
X
2
= 2 to X
2
= 6
At X
2
= 2
BM
B
= –40 ...(10)
At X
2
= 6
BM
C
= 0 ...(11)
i.e. inclined line –40 to 0
Plot the BMD with the help of above bending moment values.
The SFD and BMD is shown in fig (12.20).
Q.17: Fig 12.21 shows vertical forces 20KN, 40KN and UDL of 20KN/m in 3m lengths. Find the
resultant force of the system and draw the shear force and B.M. diagram. (Dec01)
A B
C
20kN
2m
3m
40kN 20KN/m
Fig 12.21
Sol.: Total force acting are 20KN, 40KN and 60KN (UDL),
Hence resultant of the system =
2 2
( H) ( V) ∑ + ∑
∑H = 0 and ∑V = 20 + 40 + 60 = 120KN
R = 120KN .......ANS
Beam / 291
Here total twosection line, which cut AB, AC
Distance of section 11 from point A is X
1
Distance of section 22 from point A is X
2
Consider left portion of the beam
S.F. Calculations:
S.F.
11
= – 20 – 20.X
1
(Equation of inclined line)
At X
1
= 0
SF
A
= –20KN
At X
1
= 1
SF
B
= –40KN
S.F.
22
= –20 – 40 – 20X
2
At X
2
= 1
SF
B
= –80KN
At X
2
= 3
SF
C
= –120KN
Plot the SFD with the help of above value
A B
C
20kN
2m 1m
20KN
40KN
80KN
120KN
30KN
230KN
B.M.D.
S.F.D
40kN 20KN/m
Mc–230KNm
He = 0
V = 120 KN
C
Fig 12.22
B.M. Calculations:
B.M.
1–1
= –20X
1
– 20.X1(X
1
/2) (Equation of Parabola)
At X
1
= 0
BM
A
= 0
At X
1
= 1
BM
B
= –30KNm
292 / Problems and Solutions in Mechanical Engineering with Concept
BM
22
= –20X2 – 40(X
2
1) – 20X2(X
2
/2)
At X
2
= 1
BM
B
= –30KNm
At X
2
= 3
SF
C
= –230KNm
Plot the BMD with the help of above value
Numerical Problems Based on Overhanging Beam
Q.18: Draw the SF diagram for the simply supported beam loaded as shown in fig 12.23.
2
–2.5
– 8
2
0
+ve
–ve
S.F.D.
+ve
3.5
3.5
free body diagrum
D
2K
R
B
B 10 KN
2.5 0.5
C
2 2 R
A
A
2.5 m
.5 m
2 m
2 KN
B
2KN/m
5.5KN
A
Fig 12.23
Sol.: Let reaction at support A and B be, R
A
and R
B
First find the support reaction. For finding the support
reaction, convert UDL in to point load and equal to 2 X 5 = 10KN, acting at mid point of UDL i.e. 2.5m
from point A.
For that,
∑V = 0
R
A
+ R
B
 5.5  10 2 = 0, R
A
+ R
B
= 17.5 ...(1)
Taking moment about point A,
∑M
A
= 0
10 X 2.5 + 5.5 X 2  R
B
X 5 + 2 X 7 = 0
R
B
= 10 KN ...(2)
From equation (1), R
A
= 7.5 KN ...(3)
Calculation for the Shear force Diagram
Draw the section line, here total 3section line, which break the
load R
A
, 5.5KN (Between Point A and E),
5.5KN and UDL (Between Point E and B),
Point B and 2KN (Between Point B and C).
Let
Distance of section 11 from point A is X
1
Distance of section 22 from point A is X
2
Distance of section 33 from point A is X
3
Consider left portion of the beam
Consider section 11
Beam / 293
Force on left of section 11 is R
A
and UDL (from point A to the section line i.e. UDL on total distance
of X
1
SF
11
= 7.5 2X
1
KN (Equation of straight line)
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of shear force at both nearest point of the section is varies with X
1
= 0
to X
1
= 2
At X
1
= 0
SF
A
= 7.5 ...(4)
At X
1
= 2
SF
E
= 3.5 ...(5)
i.e. inclined line 7.5 to 3.5
Consider section 22
Forces on left of section 22 is RA, 5.5KN and UDL on X
2
length
SF
22
= 7.5  5.5  2X
2
= 2  2X
2
(Equation of straight line)
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of shear force at both nearest point of the section is varies with X
2
= 2
to X
2
= 5
At X
2
= 2
SF
E
= –2 ...(4)
At X
2
= 5
SF
B
= –8 ...(5)
i.e. inclined line 2 to 8
Since here shear force changes the sign so at any point shear force will be zero and at that point
bending moment is maximum.
For finding the position of zero shear force equate the shear force equation to zero, i.e.
2  2X
2
; X
2
= 1m, i.e. at 1m from point A bending moment is maximum.
Consider section 33
Forces on left of section 33 is R
A
, 5.5KN and 10KN and R
B
SF
33
= 7.5  5.5 10 +10 = 2KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SF
B
= SF
C
= 2KN ...(7)
Plot the SFD with the help of above shear force values.
Q.19: Draw the shear force diagram of the beam shown in fig 12.24.
Sol.: First find the support reaction, for that
Convert UDL in to point load, Let reaction at C be R
CH
and R
CV
, and at point D be R
DV
.
R
V
= 0
R
CV
+ R
DV
= 1 X 3 + 2R
CV
+ R
DV
= 5KN ...(1)
Taking moment about point C,
3 X 0.5 + 2 X 5 – R
DV
X 4 = 0
R
DV
= 2.875KN ...(2)
From equation (1)
R
CV
= 2.125KN
Calculation for SFD
Here total 4 section line
SF
11
= 1X
1
(inclined line)
294 / Problems and Solutions in Mechanical Engineering with Concept
At X
1
= 0
SF
A
= 0
At X
1
= 1; SF
C
= 1
2m 2m 2m 2m
A
E
C D
B
1 KN/m
1.125 KN
2 KN
–1 KN
–875KN
2 KN
Fig 12.24
SF
22
= 1X
2
 R
CV
(inclined line)
At X
2
= 1
SF
C
= –1.125
At X
2
= 3
SF
E
= 0.875
SF
33
= 3RCV (Constant line)
At X
3
= 3
SF
C
= 0.875
At X
3
= 5
SF
D
= 0.875
SF
44
= 3 R
CV
 R
DV
(Constant line)
At X
4
= 5
S
FD
= 2
At X
4
= 7
SF
B
= –2
Q.20: Find the value of X and draw the bending moment diagram for the beam shown below
12.25. Given that R
A
= 1000 N & R
B
= 4000 N. (May–01)
AA
C
X
B
2m 1m
D
R = 4000N
B
R = 4000N
A
1000N 2000N/m
Fig 12.25
Sol.: For finding the Value of X, For that first draw the FBD, Taking moment about point A
UDL = 2000 X 2 = 4000 acting at a distance of (X + 1) from point A.
Beam / 295
M
A
= 4000 ⋅ (X + 1)  R
B
⋅ (2 + X) + 1000 ⋅ (X + 3) = 0
4000 + 4000X  8000  4000X  1000X 3000 = 0
1000X = 1000
X = 1m
Calculation for Banding Moment diagram
Here total threesection line, which cut AC, CB and BD
Distance of section 11 from point A is X
1
Distance of section 22 from point A is X
2
Distance of section 33 from point A is X
3
Consider left portion of the beam
Consider section 11, taking moment about section 11
BM
11
= 1000.X
1
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X
1
= 0 to X
1
= 1
At X
1
= 0
BM
A
= 0 ...(8)
At X
1
= 1
BM
C
= 1000 ...(9)
i.e. inclined line 0 to 1000 (Inclined line)
Consider section 22, taking moment about section 22
BM
22
= 1000.X
2
2000(X
2
1)
It is Equation of parabola (Y = mX
2
+ C),
Parabola means value of bending moment at both nearest point of the section is varies with X
2
= 1
to X
2
= 3 and make a curve
At X
2
= 1
BM
C
= 1000 ...(8)
At X
2
= 3
BM
B
= –1000 ...(9)
i.e. Curve between 1000 to 1000
Consider section 33, taking moment about section 33
BM
33
= 1000.X
3
–4000(X
3
– 2) + 4000(X
3
– 3)
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X
3
= 3 to X
3
= 4
At X
3
= 3
BM
B
= –1000 ...(8)
At X
3
= 4
BM
B
= 0 ...(9)
i.e. Curve between 1000 to 0
Plot the BMD with the help of above value, BMD is show in fig 12.26.
296 / Problems and Solutions in Mechanical Engineering with Concept
A
1m 2m
K
4000N
1000N–m
1000N–m
1m
C B
1000N
1000N
00 00
Fig 12.26
Q.21: Draw the SFD and BMD for the beam shown in the figure 12.27.
10kN
1.5m 2m 1.5m
10kN
A C D B
Fig 12.27
Sol.: Let Support reaction at A and B be Ra and Rb; and the diagram is symmetrical about y axis so the
both reactions are equal; i.e.
R
a
= R
b
= 10KN
S.F. Calculation
S.F
11
= +10 KN
S.F
A
= S.F
C
= 10
S.F
22
= 10  10 = 0 KN
S.F
C
= S.F
D
= 0
F.B.D.
S.B.D.
10kN
10kN
15kN
15kN
+ve
0
B.M.D.
0
0
+ve
–ve
A
C D
B
10kN
R = 10 kN
A
R = 10 kN
B
1.5m 1.5m 2m
10kN
Fig 12.28
Beam / 297
S.F
33
= 10 – 10 – 10 = – 10 KN
S.F
D
= S.F
B
= – 10
B.M. Calculation
B.M
11
= 10.x1 (Linear)
B.M
A
= 0
B.M
C
= 15KN
B.M
22
= 10.x2 – 10(X
2
– 1.5) (Linear)
B.M
C
= 15 KN
B.M
D
= 15KN
B.M
33
= 10.X
3
– 10(X
3
– 1.5) – 10(X
3
– 3.5) (Linear)
B.M
D
= 15 KN
B.M
B
= 0
Draw the SFD and BMD with the help of above values as shown in fig 12.28.
Note: The B.M. is zero at the point where shear force is zero. And the region where shear force is zero;
the bending moment is constant as shown in fig.
Load Diagram and BM
D
from the Given SF
D
Q.22: The shear force diagram of simply supported beam is given below in the fig 12.29.
Calculate the support reactions of the beam and also draw bending moment diagram of the
beam. (May–01(C.O.))
3.5 KN
H
K
J
E
B
A
1.5 KN
3.5 KN
Slope = 1
Slope = 0
Slope = 0
Slope = 0
F
G
D
C
(+)
(–)
Fig 12.29
Sol.: For the given SF
D
, First we draw the load diagram, and then with the help of load diagram we draw
the BM
D
.
As the slope in SFD is zero. So it indicates that the beam is only subjected to point loads. Let R
A
and
R
B
be the support reaction at A and B and the load R
C
, R
D
and R
E
in down ward direction at point C, D
and E respectively.
Here the graph of SFD moves from AFGCDHEJKB
Consider two points continuously,
Consider AF
Load moves from A to F,
Load intensity at A = R
A
= Last load  first load = 3.5  0 = 3.5KN
i.e. R
A
= 3.5 KN ...(1)
Consider FG
298 / Problems and Solutions in Mechanical Engineering with Concept
Load moves from F to G,
Load intensity = Last load  first load = 3.5  3.5 = 0
i.e. No load between F to G ...(2)
Consider GC
Load moves from G to C,
Load intensity at C = R
C
= Last load  first load = 3.5 – 1.5 = –2KN
i.e. R
C
= –2 KN ...(3)
Consider CD
Load moves from C to D,
Load intensity = Last load – first load = 1.5 – 1.5 = 0
i.e. No load between C to D ...(4)
Consider DH
Load moves from D to H,
Load intensity at D = R
D
= Last load  first load = 1.5 – 1.5 = –3KN
i.e. R
D
= –3 KN ...(5)
Load moves from H to E,
Load intensity = Last load  first load = 1.5 (1.5) = 0
i.e. No load between H to G ...(6)
Consider EJ
Load moves from E to J,
Load intensity at E = R
E
= Last load  first load = –1.5 (–3.5) = –2KN
i.e. R
E
= –2 KN ...(7)
Load moves from J to K,
Load intensity = Last load  first load = –3.5 – (–3.5) = 0
i.e. No load between J to K ...(8)
Consider KB
Load moves from K to B,
Load intensity at B = R
B
= Last load  first load = 0 –(–3.5) = 3.5KN
i.e. R
B
= 3.5 KN ...(9)
Now load diagram is given in fig 12.30
2m
7KNm
10KNm
7KNm B.M.D.
W – 2KN
1
R –3.5KN
A
R –3.5KN
B
W – 2KN
2
W – 2KN
3
A
C D E
B
2m 2m 2m
Fig 12.30
Beam / 299
Now Calculation for BMD
Taking moment about any point gives the value of BM at that point.
Consider left portion of the beam
Taking moment about point A i.e. M
A
= BM
A
= 0
Taking moment about point C, M
C
= BM
C
= 3.5 X 2 = 7KNm
Taking moment about point D, M
D
= BM
D
= 3.5 X 4 – 2 X 2 = 10 KNm
Taking moment about point E, M
E
= BM
E
= 3.5 X 6 – 2 X 4 3 X 2
= 7 KNm
Taking moment about point B, M
B
= BM
B
= 3.5 X 8 – 2 X 6 – 3 X 4 – 2 X 2
= 0 KNm
Draw the BMD with the help of above value.
Q.23: The shear force diagram of simply supported beam is given below in the fig. Calculate the
support reactions of the beam and also draw bending moment diagram of the beam.
(Dec–01)
4kN
2m 2m 2m 2m
4
2kN
2kN
(+)
(–)
Fig 12.31
Sol.: For the given SFD, First we draw the load diagram, and then with the help of load diagram we draw
the BMD.
Let R
A
and R
B
be the support reaction at A and B
Here the graph of SFD moves from AFGCDEHJB
Consider two points continuously,
Consider AF
Load moves from A to F,
Load intensity at A = R
A
= Last load  first load = 4 – 0 = 4KN
i.e. R
A
= 4 KN ...(1)
Consider FG
Load moves from F to G,
Load intensity = Last load – first load = 2 – 4 = –2KN
Since inclined line in BMD indicate that UDL on the beam
Udl = Total Load/Total distance = –2/2 = 1KN/m
(ive means UDL act downward)
i.e. UDL of 1KN/m between F to G ...(2)
Consider GC
Load moves from G to C,
Load intensity at C = R
C
= Last load – first load = 0 – 2 = 2KN
i.e. R
C
= –2 KN ...(3)
Consider CD
300 / Problems and Solutions in Mechanical Engineering with Concept
Load moves from C to D,
Load intensity = Last load – first load = 0 – 0 = 0
i.e. No load between C to D ...(4)
Consider DE
Load moves from D to E,
Load intensity at D = Last load – first load = 0 – 0 = 0
i.e. No load between D to E ...(8)
Load moves from E to H,
Load intensity = Last load  first load = –2 –0 = 2KN
i.e. R
E
= –2KN ...(6)
Consider HJ
Load moves from H to J,
Load intensity = Last load – first load = –1.5 –(–3.5) = –2KN
i.e. R
E
= –2 KN ...(7)
Load moves from J to K,
Load intensity = Last load – first load = –4 – (–2) = –2
Since inclined line in BMD indicate that UDL on the beam
Udl = Total Load/Total distance = –2/2 = –1KN/m
(ive means UDL act downward)
i.e. UDL of 1KN/m between H to I ...(2)
Consider JB
Load moves from J to B,
Load intensity at B = R
B
= Last load  first load = 0 – (–4) = 4KN
i.e. R
B
= 4KN ...(9)
Now load diagram is given in fig
Now Calculation for BMD
Here total threesection line, which cut AC, CD, DB
Distance of section 11 from point A is X
1
Distance of section 22 from point A is X
2
Distance of section 33 from point A is X
3
Consider left portion of the beam
(+)
4KN
A
C D E
B
2KN
v
3
v
2
v
1
r
(–ve)
2KN
2m
w KN/m
1
RA–4KN
6KNm 6KNm 6KNm
2m 2m 2m
Fig 12.32
Beam / 301
B.M. Calculations:
B.M. at A = 0KN.m
B.M. at C = 4 X 2 – 1 X 2 = 6KN.m
B.M. at D = 4 X 4 – 1 X 2 X (2/2 + 2) – 2 X 2 = 6KN.m
B.M. at E = 4 X 6 – 1 X 2 X (2/2 + 4) – 2 X 4 = 6KN.m
B.M. at B = 4 X 8 – 1 X 2 X (2/2 + 6) – 2 X 6 – 2 X 2 – 1 X 2 X (2/2) = 0KN.m
Loading Giagram and SF
D
from the given BM
D
Q.24: The bending moment diagram (BM
D
) of a simple supported beam is given as shown in
fig 12.33. Calculate the support reactions of the beam. (Dec–00)
Sol.:
AA
C D
5 kNM
7 kNM
1m 1m 1m
B
Fig 12.33
Linear variation of bending moment in the section AC, CD and DB indicate that there is no load on
the beam in these sections. Change in the slope of the bending moment at point C and D is indicate that
there must be concentrated vertical loads at these points.
Let point load acting at A, B, C, D are RA, RB, P, Q respectively.
Consider three section line of the beam which cut the line AC, CD and DB respectively. Since the
value of moment at all the section is the last value of the BM at that section.
Consider Section 11, Taking moment from point C,
M
C
= 7 = RA X 1
R
A
= 7KN ...(1)
Consider Section 22, Taking moment from point D,
M
D
= 5 = R
A
X 2 – P X 1
5 = 7 X 2 – P
P = 9KN ...(2)
Consider Section 33, Taking moment from point B,
M
B
= 0 = R
A
X 3 – P X 2 – Q X 1
0 = 7 X 3 – 9 X 2 Q
Q = 3KN ...(3)
NOW R
A
+ R
B
= P + Q
R
B
= 5KN
R
A
= 7KN and R
B
= 5KN .......ANS
302 / Problems and Solutions in Mechanical Engineering with Concept
CHAPTER
13
TRUSS TRUSS TRUSS TRUSS TRUSS
Q. 1: What are truss? When can the trusses be rigid trusses? State the condition followed by simple
truss?
Sol.: A structure made up of several bars (or members) riveted or
welded together is known as frame or truss. The member are welded
or riveted together at their joints, yet for calculation purpose the
joints are assumed to be hinged or pinjoint. We determine the forces
in the members of a perfect frame, when it is subject to some external
load.
Rigid Truss: A truss is said to be rigid in nature when there is
no deformation on application of any external force.
Condition followed by simple truss: The truss which follows
the law n = 2j – 3. is known as simple truss. Wheren = Number of link
or memberj = Number of jointsA triangular frame is the simplest truss.
Q. 2: Define and explain the term: (a) Perfect frame (b) Imperfect frame (c) Deficient frame
(d) Redundant frame.
Perfect Frame
The frame, which is composed of such members, which are just sufficient to keep the frame in equilibrium,
when the frame is supporting an external load, is known as perfect frame. Hence for a perfect frame, the
number of joints and number of members are given as:
n = 2j – 3
Imperfect Fram
An Imperfect frame is one which does not satisfies the relation between the numbers of members and
number of joints given by the equation n = 2j – 3.
This means that number of member in an imperfect frame will be either more or less than (2j3)
It may be a deficient frame or a redundant frame.
Deficient Frame
If the numbers of member in a frame are less than (2j3), then the frame is known as deficient frame.
A
B
C
30º 60º
R
B
R
C
Fig. 13.1
Truss / 303
Redundant Frame
If the numbers of member in a frame are more than (2j3), then the frame is known as redundant
frame.
Q. 3: What are the assumptions made in the analysis of a simple truss?
Sol.: The assumptions made in finding out the forces in a frame are,
(1) The frame is a perfect frame.
(2) The frame carries load at the joints.
(3) All the members are pinjoint. It means members will have only axial force and there will be no
moment due to pin, because at a pin moment becomes zero.
(4) Load is applied at joints only.
(5) Each joint of the truss is in equilibrium, hence the whole frame or truss is also in equilibrium.
(6) The weight of the members of the truss is negligible.
(7) There is no deflection in the members on application of load.
(8) Stresses induced on application of force in the members is negligible.
Q. 4: How can you evaluate the reaction of support of a frame?
Sol.: The frames are generally supported on a roller support or on a hinged support. The reactions at the
supports of a frame are determined by the conditions of equilibrium (i.e. sum of horizontal forces and
vertical forces is zero). The external load on the frame and the reactions at the supports must form a system
of equilibrium.
There are three conditions of equilibrium.
1. ∑V = 0 (i.e. Algebraic sum of all the forces in a vertical direction must be equal to zero.)
2. ∑H = 0 (i.e. Algebraic sum of all the forces in a horizontal direction must be equal to zero.)
3. ∑M = 0 (i.e. Algebraic sum of moment of all the forces about a point must be equal to zero.)
Q. 5: How can you define the nature of force in a member of truss?
Sol.: We know that whenever force is applied on a cross section or beam along its axis, it either tries to
compress it or elongate it. If applied force tries to compress the member force is known as compressive
force as shown in fig (13.2). If force applied on member tries to elongate it, force is known as tensile force
shown in fig. 13.3.
Fig. 13.2
Fig. 13.4 Fig. 13.5
Fig. 13.3
F F
member
F F
member
F F
Axis
If a compressive force is applied on the member as in fig. 13.2, the member will always try to resist
this force & a force equal in magnitude but opposite to direction of applied force will be induced in it as
shown in fig. 13.4, Similarly induced force in member shown in fig. 13.3 will be as shown in fig. 13.5
From above we can conclude that if induced force in a member of loaded truss is like the fig. 13.4
we will say nature of applied force on the member is compressive. If Nature of induced force in a member
of truss like shown in fig. 13.5, then we can say that Nature of force applied on the member is tensile.
304 / Problems and Solutions in Mechanical Engineering with Concept
Q. 6: Explain, why roller support are used in case of steel trusses of bridges?
Sol.: In bridges most of time only external force perpendicular to links acts and the roller support gives
the reaction to link, hence it is quite suitable to use roller support in case of steel trusses of bridges
Q. 7: Where do you find trusses in use? What are the various methods of analysis of trusses? What
is basically found when analysis of a system is done?
Sol.: The main use of truss are:
1. The trusses are used to support slopping roofs.
2. Brick trusses are used in bridges to support deck etc.
Analysis of a frame consists of,
(a) Determinations of the reactions at the supports.
(b) Determination of the forces in the members of the frame.
The forces in the members of the frame are determined by the condition that every joint should be in
equilibrium. And so, the force acting at every joint should form a system in equilibrium. A frame is
analyzed by the following methods,
1. Method of joint.
2. Method of section.
3. Graphical method.
When analysis is done, we are basically calculation the forces acting at each joint by which we can
predict the nature of force acting at the link after solving our basic equation of equilibrium.
Q. 8: How you can find the force in the member of truss by using method of joint? What are the
steps involved in method of joint ?
Sol.: In this method, after determining the reactions at the supports, the equilibrium of every joint is
considered. This means the sum of all the vertical forces as well as horizontal forces acting on a joint is
equal to zero. The joint should be selected in such a way that at any time there are only two members, in
which the forces are unknown.
The force in the member will be compressive if the member pushes the joint to which it is connected
whereas the force in the member will be tensile if the member pulls the joint to which it is connected.
Steps for Method of Joint
To find out force in member of the truss by this method, following three Steps are followed.
Step1: Calculate reaction at the support.
Step2: Make the direction of force in the entire member; you make the entire member as tensile. If
on solving the problems, any value of force comes to negative that means the assumed direction is wrong,
and that force is compressive.
Step3: Select a joint where only two members is unknown.
1 First select that joint on which three or less then three forces are acting. Then apply lami’s theorem
on that joint.
Step4: Draw free body diagram of selected joint since whole truss is in equilibrium therefore the
selected joint will be in equilibrium and it must satisfy the equilibrium conditions of coplanar concurrent
force system.
∑V = 0 and ∑H = 0
Step5: Now select that joint on which four forces, five forces etc are acting. On that joint apply
resolution of forces method.
Note: If three forces act at a joint and two of them are along the same straight line, then for the
equilibrium of the joint, the third force should be equal to zero.
Truss / 305
Q. 9: Find the forces in the members AB, BC, AC of the truss
shown in fig 13.6. C.O. Dec 0405
Sol.: First determine the reaction R
B
and R
C
. The line of action of
load 20KN acting at A is vertical. This load is at a distance of AB
cos 60
º
, from the point B.Now let us find the distance AB,The triangle
ABC is a right angle triangle with angle BAC = 90
º
. Hence AB will
be equal to CB cos60º. AB = 5 X cos 60º = 2.5m Now the distance
of line of action of 20KN from B is = AB cos 60º = 1.25m
Now, taking the moment about point B, we get
R
C
X 5 – 20 X 1.25 = 0
R
C
= 5KN ...(i)
R
B
= 15KN ...(ii)
Let the forces in the member AC, AB and BC is in tension.
Now let us consider the equilibrium of the various joints.
A
B C
R
B
= 15 KN
60º
T
AB
T
BC
C
T
BC
R
C
= 5 KN
T
AC
B
A
30º
Fig 13.7(a) Fig 13 .7(b)
Joint B:
Consider FBD of joint B as shown in fig 13.7(a)
Let,
T
AB
= Force in the member AB
T
BC
= Force in the member BC
Direction of both the forces is taken away from point B. Since three forces are acting at joint B. So
apply lami’s theorem at B.
T
AB
/sin270º = T
BC
/sin30º = R
B
/sin60º
T
AB
/sin270º = T
BC
/sin30º = 15/sin60º
On solving
T
AB
= –17.32KN ...(iii)
T
AB
= 17.32KN (Compressive) .......ANS
T
BC
= 8.66KN ...(iv)
T
BC
= 8.66KN (Tensile) .......ANS
Joint C Fig 13.7(b)
Consider FBD of joint C as shown in fig 13.7 (b)
Let,
T
BC
= Force in the member BC
T
AC
= Force in the member AC
20 KN
5m
Fig. 13.6
60º
B
A
C
R
B
R
C
30º
306 / Problems and Solutions in Mechanical Engineering with Concept
Direction of both the forces is taken away from point C. Since three forces are acting at joint C. So
apply lami’s theorem at C.
T
BC
/sin60º = T
AC
/sin270º = R
C
/sin30º
T
BC
/sin60º = T
AC
/sin270º = 5/sin30º
On solving
T
AC
= –10KN ...(v)
T
AC
= 10KN (Compressive) .......ANS
MEMBER FORCE TYPE
AB 17.32KN COMPRESSIVE
AC 8.66KN TENSILE
BC 10KN COMPRESSIVE
Q. 10: Determine the reaction and the forces in each member of a simple triangle truss supporting
two loads as shown in fig 13.8.
Sol.: The reaction at the hinged support (end A) can have two components acting in the horizontal and
vertical directions. Since the applied loads are vertical, the horizontal component of reaction at A is zero
and there will be only vertical reaction R
A
, Roller support (end C) is frictionless and provides a reaction
R
C
at right angles to the roller base. Let the forces in the entire member is tensile. First calculate the
distance of different loads from point A.
Distance of Line of action of 4KN,
from point A = AF = AE cos 60º = 2X 0.5 = 1m
Distance of Line of action of 2KN,
from point A = AG = AB + BG = AB + BD cos 60º
= 2 + 2 x 0.5 = 3m
Taking moment about point A,
R
C
X 4 – 2 x 3 + 4 x 1 = 0R
C
= 2.5KN ...(i)
R
A
= 4 + 2 – 2.5 = 3.5 KN ...(ii)
4 kN
E
A
B
C
D
60º
2 m
2 m
60º
60º
60º
2 kN 2 kN
4 kN
E
A
R
A
R
C
C
B
D
60º 60º 60º 60º
2 kN
2 3
4
5 6
7
1
Fig 13.8 Fig 13.9
Joint A:
Consider FBD of joint A as shown in fig 13.10 Let, T
AE
= Force in the member AET
AB
= Force in the
member AB Direction of both the forces (T
AE
& T
AB
) is taken away from point A. Since three forces are
Truss / 307
acting at joint A. So apply lami’s theorem at B.T
AE
/sin270º = T
AB
/sin30º = R
A
/sin60º
T
AE
/sin 270º = T
AB
/sin30º = 3.5/sin60º
On solving
T
AE
= – 4.04KN ...(iii)
T
AE
= 4.04KN (Compressive) .......ANS
T
AB
= 2.02KN ...(iv)
T
AB
= 2.02KN (Tensile) .......ANS
Joint C:
Consider FBD of joint C as shown in fig 13.11 Let, T
BC
= Force in the member
BCT
DC
= Force in the member DCDirection of both the forces(T
BC
& T
DC
) is taken
away from point C. Since three forces are acting at joint C. So apply lami’s theorem
at C.T
BC
/sin30º = T
DC
/sin 270º = R
C
/sin 60º
T
BC
/sin30º = T
DC
/sin 270º = 2.5/sin 60º
On solving
T
BC
= 1.44KN ...(v)
T
BC
= 1.44KN (Tensile) .......ANS
T
DC
= 2.88KN ...(iv)
T
DC
= 2.88KN (Compressive) .......ANS
Joint B:
Consider FBD of joint B as shown in fig 13.12 Since,
T
AB
= 2.02KN(T) T
BC
= 1.44KN(T) Let,T
BE
= Force in the member
BET
DB
= Force in the member DB Direction of both the forces
(T
BE
& T
DB
) is taken away from point B. Since four forces are acting at
joint B. So apply resolution of forces as equilibrium at B.
R
H
= 0
– T
AB
+ T
BC
–T
BE
cos 60º + T
BD
cos 60º = 0
– 2.02 + 1.44 – 0.5T
BE
+ 0.5T
BD
= 0
T
BE
– T
BD
= 1.16 ...(vii)
R
V
= 0
T
BE
sin 60º + T
BD
sin 60º = 0
T
BE
= – T
BD
...(viii)
Value of equation (viii) put in equation (vii), we get
i.e –2 T
BE
= 1.16, or
T
BE
= 0.58KN ...(ix)
T
BE
= 0.58KN (Compression) .......ANS
T
BD
= 0.58KN (Tensile) .......ANS
Joint D:
Consider FBD of joint D as shown in fig 13.13 Since, T
CD
= 2.88KN(C) T
BD
= 0.58KN(T) Let,T
ED
= Force in the member ED Direction of forces T
CD
, T
BD
&
T
ED
is taken away from point D. Since four forces are acting at joint D. So apply
resolution of forces as equilibrium at D.
R
H
= 0
– T
ED
– T
BD
cos 60º + T
CD
cos 60º = 0
– T
ED
– 0.58 X 0.5 + (2.88) X 0.5 = 0
T
ED
= – 1.73KN .......ANS
T
AE
T
AB
60º
E
B A
R
A
Fig. 13.10
Fig. 13.11
T
AE
T
AB
60º
D
C
B
R
C
= 2.5 kN
Fig. 13.12
E
B
60º
T
AB
= 2.02 kN T
BC
= 1.44 kN
60º
D
T
BE
T
DB
Fig. 13.13
T
CD
T
ED
B C
D E
2kN
T
DB
308 / Problems and Solutions in Mechanical Engineering with Concept
Member Force Member Force
AE 4.04KN(C) BE 0.58KN(C)
AB 2.02KN(T) BD 0.58KN(T)
BC 1.44KN(T) DE 1.73KN(C)
CD 2.88KN(C)
Q. 11: Determine the forces in all the members of the truss loaded and supported as shown in fig
13.14.
Sol.: The reaction at the supports can be determined by considering equilibrium of the entire truss. Since
both the external loads are vertical, only the vertical component of the reaction at the hinged ends A need
to be considered. Since the triangle AEC is a right angle triangle, with angle AEC = 90º. Then,
AE = AC cos 60º = 5 X 0.5 = 2.5m
CE = AC sin60º = 5 X 0.866 = 4.33m
Since triangle ABE is an equilateral triangle and therefore,
AB = BC = AE = 2.5m
Distance of line of action of force 10KN from joint A,
A
E
D
B
C
60º
10 kN
12 kN
60º
30º
60º
5 m
Fig 13.14
AF = AE cos 60º = 2.5 X 0.5 = 1.25m
Again, the triangle BDC is a right angle triangle with angle BDC = 90º.
Also, BC = AC – AB = 5 – 2.5 = 2.5m
BD = BC cos 60º = 2.5 X 0.5 = 1.25m
Distance of line of action of force 12KN from joint A,
AG = AB + BG = AB + BD cos 60º = 2.5 + 1.25 X 0.5 = 3.125m
Taking moment about end A, We get
R
C
X 5 = 12 X 3.125 + 10 X 1.25 = 50
R
C
= 10KN ...(i)
∑V = 0, R
C
+ R
A
= 10 + 12 = 22KN
R
A
= 12KN ...(ii)
Joint A:
Consider FBD of joint A as shown in fig 13.15 Let, T
AE
= Force in the member AET
AB
= Force in the
member AB Direction of both the forces (T
AE
& T
AB
) is taken away from point A. Since three forces are
Truss / 309
acting at joint A. So apply lami’s theorem at A.T
AE
/sin 270º = T
AB
/sin 30º = R
A
/sin 60º
T
AE
/
sin 270º = T
AB
/sin 30º = 12/sin 60º
T
AE
= –13.85KN ...(iii)
T
AE
= 13.85KN (Compression) .......ANS
T
AB
= 6.92KN ...(iv)
T
AB
= 6.92KN (Tension) .......ANS
Joint C:
Consider FBD of joint C as shown in fig 13.16 Let, T
BC
= Force in the member
BCT
CD
= Force in the member CD Direction of both the forces (T
BC
& T
CD
) is taken
away from point C. Since three forces are acting at joint C. So apply lami’s theorem at
C.
T
BC
/sin 60º = T
CD
/sin 270º = R
C
/sin 30º
T
BC
/sin 60º = T
CD
/sin 270º = 10/sin 30º
T
BC
= 17.32KN ...(v)
T
BC
= 17.32KN (Tension) .......ANS
T
CD
= 20KN ...(vi)
T
CD
= 20KN (compression) .......ANS
Joint B:
Consider FBD of joint B as shown in fig 13.17
Since, T
AB
= 6.92KN
T
BC
= 17.32KN
Let, T
BD
= Force in the member BD
T
EB
= Force in the member EB
Direction of both the forces (T
BD
& T
EB
) is taken away from point B.
Since four forces are acting at joint B. So apply resolution of forces at
joint B.
R
H
= 0
–T
AB
+ T
BC
– T
EB
cos 60º + T
BD
cos 60º = 0
– 6.92 + 17.32 – 0.5 T
EB
+ 0.5 T
BD
= 0
T
BD
– T
EB
= – 20.8KN ...(vii)
R
V
= 0
T
EB
sin 60º + T
BD
sin 60º = 0
T
BD
= – T
EB ...
(viii)
T
BD
= 10.4KN ...(ix)
T
BD
= 10.4KN (Tension) .......ANS
T
EB
= 10.4KN ...(x)
T
EB
= 10.4KN (compression) .......ANS
Joint D:
Consider FBD of joint D as shown in fig 13.18
Since, T
CD
= 20KN
Let, T
ED
= Force in the member ED
Fig. 13.15
T
AE
T
AB
60º
A
R
A
Fig. 13.16
T
CD
T
BC
C
30º
R
A
Fig. 13.17
B
60º 60º
T
BE
T
AB
T
BC
T
DB
Fig. 13.18
12 kN
D
T
ED
T
DE
T
CD
310 / Problems and Solutions in Mechanical Engineering with Concept
Direction of the force (T
ED
) is taken away from point D. Since four forces are acting at joint D. So
apply resolution of forces at joint D.
Resolve all the forces along EDC, we get
T
ED
+ 12cos 60º + T
CD
= 0
T
ED
+ 6 – 20 = 0
T
ED
= 14KN ...(xi)
T
ED
= 14KN (Tension) .......ANS
Member AE AB BC CD BD BE DE
Force in KN 13.85 6.92 17.32 20 10.4 10.4 14
Nature
C = Compression C T T C T C T
T = Tension
Q. 12: A truss is as shown in fig 13.19. Find out force on each member and its nature.
Sol.: First we calculate the support reaction, Draw FBD as shown in fig 13.20
R
H
= 0,
R
AH
– R
BV
cos 60º = 0 ...(i)
R
V
= 0, R
AV
+ R
BV
sin 60º
– 24 – 7 – 7 – 8 = 0 ...(ii)
Taking moment about point
B, R
AV
X 6 – 24 X 3 – 7 X 6 – 8 X 3 = 0 ...(iii)
R
AV
= 23KN ...(iv)
Value of (iv) putting in equation (ii)
We get, R
BV
= 26.6KN ...(v)
Value of (v) putting in equation (i)
We get, R
AH
=13.3KN ...(vi)
Joint E, Consider FBD as shown in fig 13.21 From article 8.8.2,
T
ED
= 0, ...(vii)
T
ED
= 0 .......ANS
And, T
AE
= –7KN ...(viii)
T
AE
= 7KN (compression) .......ANS
q q
q q
E
7 KN 24 KN
C
B
4 m
30º
8 KN
3 m
A
6 m
7 KN
7 KN
E
T
ED
T
CD
T
DB
T
AE
T
AD
R
AH
R
AV
T
AF
30º
A
F
60º
T
BF
R
BV
24 KN
D
7 KN
C
Fig 13.19 Fig 13.20
Truss / 311
Joint C:
Consider FBD as shown in fig 13.22
T
CD
= 0, ...(ix)
T
CD
= 0 .......ANS
And, T
BC
= – 7KN ...(x)
T
BC
= 7KN (compression) .......ANS
Note: Since for perfect frame the condition n = 2j – 3 is necessary to satisfied.
Here Point F is not a joint, if we take F as a joint then,
Number of joint(j) = 6 and No. of member (n) = 7
n = 2j – 3, 7 = 2 X 6 – 3
≠ 9 i.e
i.e if F is not a joint, then j = 5
7 = 2 X 5 – 3
= 7, i.e. F is not a joint. But at joint F a force of 8KN is acting. Which
will effect on joint A and B, Since 8KN is acting at the middle point of AB,
So half of its magnitude will equally effect on joint A and B. i.e. 4KN each
acting on joint A and B downwards
Joint D:
Consider FBD of joint D as shown in fig 13.23
Since, T
CD
= T
ED
= 0KN
Let,T
BD
= Force in the member BD
T
AD
= Force in the member AD
Direction of the force (T
AD
) & (T
BD
) is taken away from point D. Since five forces are acting at joint
D. So apply resolution of forces at joint D.
Resolve all the forces, we get
R
H
= – T
AD
cos θ + cos θ = 0
T
AD
= T
BD
...(xi)
R
V
= –T
AD
sin θ – T
BD
sin θ – 24 = 0
(T
AD
+ T
BD
) = –24/ sin θ
sin θ = 4/5
or, 2T
AD
= 2T
BD
= – 24/(4/5)
T
AD
= T
BD
= – 15KN ...(xii)
T
AD
= 15KN (compression) .......ANS
T
BD
= 15KN (compression) .......ANS
Joint A:
Consider FBD of joint A as shown in fig 13.24
Let, T
AB
= Force in the member AB
Since, T
AD
= –15KN T
AE
= –7KN
Direction of the force (T
AD
) & (T
AE
) & (T
AB
) is taken away from point A. Since
five forces are acting at joint A. So apply resolution of forces at joint A.
Resolve all the forces, we get
R
H
= R
AH
+ T
AB
+ T
AD
cos θ = 0
Fig. 13.21 Fig. 13.22
7 KN 7 KN
E C T
ED
T
CD
T
AE
T
BC
Fig. 13.23
24 KN
D
T
ED
T
AD
T
DE
T
CD
Fig. 13.24
T
AE
T
AD
T
AF
R
AV
R
AH
A
312 / Problems and Solutions in Mechanical Engineering with Concept
= 13.3 + T
AB
– 15 cos θ = 0
13.3 + T
AB
–15(3/5) = 0
T
AB
= – 4.3KN ...(xii)
T
AB
= 4.3KN (compression) ...ANS
Member AB BC CD DE EA AD DB
Force in KN 4.3 7 0 0 7 15 154
Nature
C = Compression
T = Tension C C — — C C CT
Q. 13: A truss is shown in fig(13.25). Find forces in all the members of the truss and indicate whether
it is tension or compression. (Dec–0001)
Sol.: Let the reaction at joint A and E are R
AV
and R
EV
. First we calculate
the support reaction,
R
H
= 0, R
AH
= 0 ...(i)
R
V
= 0, R
AV
+ R
EV
– 10 – 15 – 20 – 10 = 0
R
AV
+ R
EV
= 55 ...(ii)
Taking moment about point A and equating to zero; we get
15 X 3 + 10 X 3 + 20 X 6 – R
EV
X 6 = 0 ...(iii)
R
EV
= 32.5 KN ...(iv)
Value of (iv) putting in equation (ii)
We get, R
BV
= 22.5 KN ...(v)
Consider FBD of Joint B, as shown in fig 13.26
∑H = 0; T
BC
= 0
∑V = 0; T
BA
+ 10 = 0; T
BA
= –10KN(C)
Consider FBD of Joint F, as shown in fig 13.27
∑H = 0; T
FC
= 0
∑V = 0; T
FE
+ 20 = 0; T
FE
= –20KN(C)
Consider FBD of Joint A, as shown in fig 13.28
∑H = 0; T
AD
+ T
AC
cos 45 = 0
T
AD
= –T
AC
cos 45 ...(i)
∑V = 0; T
CA
sin 45
+ 22.5 – 10 = 0
T
CA
= –17.67KN(C)
Putting this value in equation (i), we get
T
AD
= 12.5KN(T)
Consider FBD of Joint D, as shown in fig 8.29
∑H = 0;
– T
AD
+ T
DE
= 0
T
AD
= T
DE
= 12.5KN (T)
∑V = 0; T
DC
– 10
= 0
T
DC
= 10KN (T)
Fig. 13.25
10 KN
10 KN
B
3 m 3 m
3 m
C
D
A E
F
15 KN 20 KN
Fig. 13.26
10 KN
T
AB
T
BC
Fig. 13.27
20 KN
T
CF
T
EF
Fig. 13.28
T
AB
45º
22.5
T
AC
T
AD
Fig. 13.29
T
CD
T
AD
T
ED
10 KN
Truss / 313
Consider FBD of Joint E, as shown in fig 8.30
∑H = 0;
– T
ED
– T
ECn
cos 45 = 0
– 12.5 – T
EC
cos 45 = 0
T
EC
= – 17.67KN(C)
∑V = 0; T
FE
+ T
EC
sin 45
+ 32.5 = 0
T
FE
+ (–17.67) sin 45
+ 32.5 = 0
T
FE
= – 20KN(C)
Forces in all the members can be shown as in fig below.
Fig. 13.31
10 KN
10 KN
B
C
6 KN
12.5 KN(T)
1
0
K
N
(
C
)
1
2
.
5
2
K
N
(
C
)
1
2
.
5
2
(
C
)
2
0
K
N
(
C
)
6 KN
D
A E
F
15 KN 20 KN
Q. 14: Find out the axial forces in all the members of a truss with loading as shown in fig 13.32.
(May–02 (C.O.))
Sol.: For Equilibrium
∑H = 0; R
AH
= 15KN
∑V = 0; R
AV
+ R
BV
= 0
∑M
B
= 0; R
AV
X 4 + 10
X 4 + 5 X 8
= 0
R
AV
= – 20 KN
And R
BV
= 20KN
Consider Joint A as shown in fig 13.33
H = 0; T
AB
= 15KN(T)
∑V = 0; T
AF
= 20KN(T)
Fig. 13.33
Fig. 13.32
T
AF
15
A
T
AB
20
A
E
F
D
B
C
4 m
4 m
4 m
10 KN
5 KN
Fig. 13.30
T
CE
45º
30.5
T
EF
T
ED
314 / Problems and Solutions in Mechanical Engineering with Concept
Consider Joint B as shown in fig 13.34
∑H = 0;
– T
AB
– T
BF
cos 45 = 0
T
BF
= – 15/cos 45 = – 21.21KN
T
BF
= – 21.21KN(C)
∑V = 0;
T
BC
+ T
BF
sin 45 + 20 = 0
T
BC
– 21.21
sin 45 + 20 = 0
T
BC
= – 5KN(C)
Consider Joint F as shown in fig 13.35
∑H = 0;
T
FC
+ T
BF
cos 45 + 10 = 0
T
FC
– 21.21 cos 45 + 10 = 0
T
FC
= 5KN(T)
∑V = 0;
T
FE
– T
FA
– T
BF
sin 45 = 0
T
FE
– 20 + 21.21
sin 45 = 0
T
FE
= 5KN(T)
Consider Joint C as shown in fig 13.36
∑H = 0;
– T
FC
– T
CE
cos 45 = 0
– 5 – T
CE
cos 45 = 0
T
CE
= – 7.071KN(C)
∑V = 0;
T
CD
– T
CB
+ T
CE
sin 45 = 0
T
CD
+ 5 – 7.071
sin 45 + 20 = 0
T
CD
= 0
Consider Joint D as shown in fig 13.37
∑H = 0;
T
ED
= 0
S.No. Member Force (KN) Nature
1. AB 15 T
2. AD 20 T
3. BD 21.21 C
4. BC 5 C
5. DC 5 T
6. DE 5 T
7. CE 7.071 C
8. CF 0 —
9. FE 0 —
Fig. 13.34
T
BF
45º
B
T
BC
T
AB
Fig. 13.35
45º
– 21.21
10 KN
T
CF
F
T
AF
T
EF
Fig. 13.36
T
CE
45º
C
T
CD
T
CF
= 5
F
SC
= – 5
Fig. 13.37
D
T
CD
T
ED
Truss / 315
Q. 15: Determine the magnitude and nature of forces in the various members of the truss shown in
figure 13.38. (C.O. August0506)
Sol.: For Equilibrium
∑H = 0; R
BH
= 0
∑V = 0; R
AV
+ R
BV
= 200
∑M
B
= 0; R
AV
X 6 – 50 X 6 – 100 X 3 = 0
R
AV
= 100 KN
And R
BV
= 100KN
C
B
F D
A
50 kN 100 kN 50 kN
E
2 m 3 m
3 m
Fig. 13.38
Consider joint A; fig 13.39
∑H = 0; T
AC
= 0
∑V = 0; R
AV
+ T
AD
= 0 T
AD
= – 100 KN(C)
Fig. 13.39 Fig. 13.40 Fig. 13.41 Fig. 13.42 Fig. 13.43
T
AD
T
AC
T
FB
50 kN
50 kN
100 kN
E
T
DE
T
EF T
DE
T
EF
T
CE
T
CF
T
FB
F
T
AD
T
DC
D
T
BC
R
BV
T
AV
A
3 m
Consider joint B; As shown in fig 13.40
∑H = 0; T
BC
= 0
∑V = 0; R
BV
+ T
FB
= 0
T
FB
= – 100 KN(C)
Consider joint D; As shown in fig 13.41
∑V = 0; – T
AD
– T
DC
sin 45
– 50 = 0
T
DC
= 70.71KN (T)
∑H = 0; T
DE
+ T
DC
cos 45
= 0
T
DE
= – 50KN (C)
Consider joint F; As shown in fig 13.42
∑V = 0; – T
FB
– T
FC
sin 45
–
50= 0
T
FC
= 70.71KN (T)
∑H = 0; – T
FE
–
T
FC
cos 45
= 0
T
FE
= – 50KN (C)
316 / Problems and Solutions in Mechanical Engineering with Concept
Consider joint E; as shown in fig 13.43
∑V = 0; – T
EC
– 100 = 0
T
EC
= – 100KN (C)
∑H = 0; – T
ED
+ T
EF
= 0
T
ED
= – 50KN (C)
S.No. Member Force(KN) Nature
1. AC 0 —
2. AD 100 C
3. BC 0 —
4. FB 100 C
5. DC 70.71 T
6. DE 50 C
7. FC 70.71 T
8. FE 50 C
9. EC 100 C
10. ED 50 C
Problems on Cantilever Truss
In case of cantilever trusses, it is not necessary to determine the support reactions. The forces in the
members of cantilever truss can be obtained by starting the calculations from the free end of the cantilever.
Q. 16: Determine the forces in all the member of a cantilever truss shown in fig 13.44.
Sol.: From triangle ACE, we have
3 m
4 m
E
D
B C A
q q
q
3 m
2000 N 2000 N
Fig. 13.44
tan θ = AE/AC = 4/6 = 0.66 ...(i)
Also, EC = 2 2
4 6 +
= 7.21m ...(ii)
cos θ = AC/EC = 6/7.21 = 0.8321 ... (iii)
sin θ = AE/CE = 4/7.21 = 0.5548 ...(iv)
Truss / 317
3 m
4 m
E
D
T
ED
T
AD
T
AB
T
BC
T
DE
T
CD
B C A
q q
q
3 m
2000 N 2000 N
Fig 13.45
Joint C:
Consider FBD of joint C as shown in fig 13.46;
Since three forces are acting, so apply lami,s theorem at joint C.
T
BC
/sin(90 – θ) = T
CD
/sin270 = 2000/sin θ
T
BC
/cos θ = T
CD
/sin 270 = 2000/sin θ
T
BC
= 2000/tan θ = 2000/0.66 = 3000.3N ...(v)
T
BC
= 3000.3N (Tensile) .......ANS
T
CD
= – 2000/sin θ = 2000/0.55 = 3604.9N ...(vi)
T
CD
= 3604.9N (Compressive) .......ANS
Joint B:
Consider FBD of joint B as shown in fig 13.47
Since, T
BC
= 3000.3N
Let, T
AB
= Force in the member AB
T
DB
= Force in the member DB
Since four forces are acting at joint B, So apply resolution of forces at
joint B
R
H
= T
AB
– T
BC
= 0, T
AB
= T
BC
= 3000.03 = T
AB
T
AB
= 3000.03 ...(vii)
T
AB
= 3000.03N (Tensile) .......ANS
R
V
= – T
DB
– 2000 = 0
T
DB
= 2000N ...(viii)
T
DB
= 2000N (compressive) .......ANS
Joint D:
Consider FBD of joint D as shown in fig 13.48
Since, T
DB
= – 2000N
T
CD
= 3604.9N
Let, T
AD
= Force in the member AD
T
DE
= Force in the member DE
Since four forces are acting at joint D, So apply resolution of forces at
joint D.
Fig. 13.46
T
BC
2000 N
T
CD
C
q
Fig. 13.47
2000 N
B
T
AB
T
BC
T
DE
Fig. 13.48
D
T
CD
T
ED
T
AD
T
DE
q
318 / Problems and Solutions in Mechanical Engineering with Concept
R
V
= 2000 + T
CD
sin θ + T
AD
sin θ – T
ED
sin θ = 0
2000 + 3604.9 X 0.55 + T
AD
x 0.55 – T
ED
x 0.55 = 0
T
AD
– T
ED
= 7241.26N ...(ix)
R
H
= T
CD
cos θ – T
AD
cos θ – T
ED
cos θ = 0
= 3604.9 = T
AD
+ T
ED
T
AD
+ T
ED
= 3604.9 ...(x)
Solving equation (ix) and (x), we get
T
ED
= 55423.1N ... (xi)
T
ED
= 5542.31N (Tensile) .......ANS
T
AD
= 1818.18N ...(xii)
T
AD
= 1818.18N (compressive) .......ANS
Member AB BC CD DE DB AD
Force in N 3000.03 3000.03 3604.9 5542.31 2000 1818.18
Nature
C = Compression T T C T C C
T = Tension
Q. 17: Determine the forces in the various members of the cantilever truss loaded and supported as
shown in fig. 13.49.
D
1 m
15 kN
1 m
2 m 2 m
B
E
C
q q
q
Sol.: BC =
2 2
(2 1 ) + = 2.23m
sin θ = 1/(2.23) = 0.447
sin θ = 2/(2.23) = 0.894
D
1 m
15 kN
1 m
T
AD
T
DB
T
CD
T
DB
T
AB
T
BC
2 m 2 m
B
E
C
q q
q
Fig 13.50
Truss / 319
Let
T
CD
= Force in the member CD
T
CB
= Force in the member CB
T
DB
= Force in the member DB
T
AB
= Force in the member AB
T
AD
= Force in the member AD
T
BD
= Force in the member BD
Consider Joint C:
Consider FBD of joint C as shown in fig 13.51.
There are three forces are acting so apply lami’s theorem at joint C
T
CD
/sin(90 – θ) = T
BC
/sin 270 = 15/sin θ
T
CD
= 30KN (Tensile) .......ANS
T
BC
= – 33.56 ...(i)
T
BC
= 33.56 (Compressive) .......ANS
Consider Joint B:
Consider FBD of joint B as shown in fig 13.52.
There are three forces are acting so apply lami’s theorem at joint B
T
AB
/sin(90 – θ) = T
BC
/sin90 = T
DB
/sin(180 + θ)
T
4
= –30KN ...(ii)
T
AB
= – 30KN(Compressive) .......ANS
T
DB
= 15 ...(iii)
T
DB
= 15(Tensile) .......ANS
Consider Joint D:
Consider FBD of joint D as shown in fig 13.53.
There are four forces are acting so apply resolution of forces at joint D
R
H
= 0, T
CD
– T
AD
cos θ – T
ED
cos θ = 0
30 – (T
AD
+ T
ED
) cos θ = 0
T
AD
+ T
ED
= 30/cos θ = 30/0/89 = 33.56 ...(iv)
R
V
= 0
T
ED
sin θ  T
AD
sin θ – T
DB
= 0
(T
ED
– T
AD
) sin θ = 15
T
ED
– T
AD
= 15/sin θ
T
ED
– T
AD
= 33.56 ...(v)
Solve equation (iv) and (v), we get
T
AD
= 0 ...(vi)
T
AD
= 0 .......ANS
T
ED
= 33.56 ...(vii)
T
ED
= 33.56 (Tensile) .......ANS
Member CD BC BD BA AD DE
Force in kN 30 33.56 15 30 0 33.56
Nature
C = Compression T C T C — T
T = Tension
Fig. 13.51
15 kN
q
T
BC
T
CD
C
T
AB
T
DE
T
BC
B
Fig. 13.52
Fig. 13.53
T
ED
T
CD
T
DE
D
T
AD
q
q
Truss / 321
Consider joint A:
Fig. 13.58
12 kN
T
OA
T
AB
T
AC
T
AD
∑V = 0;
– 12 – T
AC
– T
AD
. sin 45º = 0
– 12 – 0 – T
AD
. sin 45º = 0
T
AD
= – 12/sin 45º
T
AD
= – 16.97KN (Compressive)
∑H = 0;
T
AB
– T
OA
– T
AD
. cos 45º = 0
16 – T
OA
+ 16.97 . cos 45º = 0
T
OA
= 28KN
T
OA
= 28KN (Tensile)
Consider joint D:
∑V = 0;
T
OD
+ T
AD
. cos θ
1
+ T
CD .
sin θ = 0
T
OD
– 16.97 cos 45º – 17.9 . sin 26.56º = 0
T
OD
= 20KN
T
OD
= 20KN (Tensile)
Member Force Nature (T/C)
BC 17.9 C
AB 16 T
CD 17.9 C
AC 0 —
AD 16.97 C
AO 28 T
OD 20 T
Q. 19: Define method of section? How can you evaluate the problems with the help of method of
section?
Sol.: This method is the powerful method of determining the forces in desired members directly, without
determining the forces in the previous members. Thus this method is quick. Both the method, i.e. method
of joint and method of sections can be applied for the analysis of truss simultaneously. For member near
to supports can be analyzed with the method of joints and for members remote from supports can be quickly
analyzed with the help of method of section.
In this method a section line is passed through the member, in which forces are to be determined in
such a way that not more than three members are cut. Any of the cut part is then considered for equilibrium
under the action of internal forces developed in the cut members and external forces on the cut part of the
Fig. 13.59
T
OD
D
T
AD
R
D
T
CD
θ
θ
1
322 / Problems and Solutions in Mechanical Engineering with Concept
truss. The conditions of equilibrium are applied to the cut part of the truss under consideration. As three
equations are available, therefore, three unknown forces in the three members can be determined. Unknown
forces in the members can be assumed to act in any direction. If the magnitude of a force comes out to be
positive then the assumed direction is correct. If magnitude of a force is negative than reverse the direction
of that force.
Steps Involved for Method of Section
The various steps involved are:
(1) First find the support reaction using equilibrium conditions.
(2) The truss is split into two parts by passing an imaginary section.
(3) The imaginary section has to be such that it does not cut more than three members in which the
forces are to be determined.
(4) Make the direction of forces only in the member which is cut by the section line.
(5) The condition of equilibrium are applied for the one part of the truss and the unknown force in
the member is determined.
(6) While considering equilibrium, the nature of force in any member is chosen arbitrarily to be tensile
or compressive.
If the magnitude of a particular force comes out positive, the assumption in respect of its direction is
correct. However, if the magnitude of the forces comes out negative, the actual direction of the force is
positive to that what has been assumed.
The method of section is particularly convenient when the forces in a few members of the frame is
required to be worked out.
Q. 20: A cantilever truss is loaded and supported as shown in fig 13.60. Find the value of P, which
would produce an axial force of magnitude 3KN in the member AC.
Sol.: Let us assume that the forces is find out in the member AC, DC and DF
Let T
1
= Force in the member AC
T
2
= Force in the member DC
T
3
= Force in the member DF
A
B
C
P
P
3 m
3 m
D
3 m 1.5 m
60º 60º
60º 60º
60º
60º
F
E
2 m
Fig 13.60
Draw a section line, which cut the member AC, DC, and DF.
Consider right portion of the truss, because Force P is in the right portion.
Taking moment about point D,
∑M
D
= 0
– T
1
X AB + P X (AC – AD) + P X (AE – AD) = 0
3 X 2 + P X 1.5 + P X 4.5 = 0
P = 1KN .......ANS
Truss / 323
Q. 21: Find the forces in members BC, BE, FE of the truss shown in fig 13.61, using method of
section. (May–04)
B
D
C
20 kN
3 m
A
3 m 3 m 3 m
30 kN
Fig 13.61
Sol.: First find the support reaction which can be determined by considering equilibrium of the truss.
∑V = 0
R
A
+ R
D
= 50 ...(i)
Taking moment about point A,
∑M
A
= 0
– R
D
X 9 + 20 X 6 + 30 x 3 = 0
R
D
= 23.33KN ...(ii)
Now, from equation (i); we get
R
A
= 26.67KN ...(iii)
Let draw a section line 11 which cut the member BC, BE, FE, and divides the truss in two parts RHS
and LHS as shown in fig 13.62. Make the direction of forces only in those members which cut by the
section line.
LHS
3 m 3 m 3 m
RHS T
BE
T
EF
30 kN
B
A
F E
D
20 kN
C
T
BC
1
1
Fig 13.62
Choose any one part of them, Since both parts are separately in equilibrium. Let we choose right hand
side portion (as shown in fig 13.63). And the Right hand parts of truss is in equilibrium under the action
of following forces,
324 / Problems and Solutions in Mechanical Engineering with Concept
T
EF
T
BE
RHS
D
E
20 kN
T
BC
1
1
Fig 13.63
1. Reaction R
D
= 23.33KN
2. 20KN load at joint C
3. Force T
BC
in member BC (From C to B)
4. Force T
BE
in member BE (From E to B)
5. Force T
FE
in member FE (From E to F)
All three forces are assumed to be tensile.
Now we take moment of all these five forces only from any point of the truss for getting the answers
quickly
Taking moment about point E, of all the five forces given above
∑M
E
= 0
(Moment of Force T
BE
,T
EF
and 20KN about point E is zero, since point E lies on the line of action
of that forces)
– R
D
x ED + T
EF
x 0 + T
BE
x 0 – T
BC
x CE + 20 x 0 = 0
– R
D
x 3 – T
BC
x 3 = 0
T
BC
= – 23.33KN ... (iii)
T
BC
= 23.33KN (Compressive) .......ANS
Taking moment about point B, of all the five forces given above
∑M
B
= 0
(Moment of Force T
BE
, T
BC
force about point B is zero)
– R
D
x FD + T
BC
X 0 + T
BE
X 0 + T
FE
X CE + 20 X BC = 0
– R
D
X FD + T
FE
X CE + 20 X BC = 0
–23.33 X 6 + T
FE
X 3 + 20 X 3 = 0
T
FE
= 26.66KN ...(iii)
T
FE
= 26.66KN (Tensile) .......ANS
Taking moment about point F, of all the five forces given above
∑M
F
= 0
(Moment of Force T
FE
about point B is zero)
– R
D
x FD – T
BC
x EC – T
BE
cos45º
x FE + T
FE
x 0 + 20 x FE = 0
–23.33 x 6 + 23.33 x 3 – T
BE
cos 45º
x 3 + 20 x 3 = 0
T
BE
= 4.71KN ...(iii)
T
BE
= 4.71KN (Tensile) .......ANS
Truss / 325
Q. 22: Determine the support reaction and nature and magnitude of forces in members BC and EF
of the diagonal truss shown in fig 13.64. (May–01, (C.O.))
40 KN
2 m 2 m
2 m
2 m
45
45
B C D A
E F
10 KN
Fig 13.64
Sol.: First find the support reaction which can be determined by considering equilibrium of the truss.
Let R
AH
& R
AV
be the support reaction at hinged support A and R
DV
be the support reaction at roller
support D.
∑H = 0
R
AH
+ 10 = 0
R
AH
= – 10 KN
∑V = 0
R
AV
+ R
DV
= 40 ...(i)
Taking moment about point A,
∑M
A
= 0
40 x 2 – 10 x 2 – R
DV
x 6 = 0
R
DV
= 10KN
From equation (i); R
AV
= 30KN
Let draw a section line 11 which cut the member BC, EC, FE, and divides the truss in two parts RHS
and LHS as shown in fig 13.65. Make the direction of forces only in those members which cut by the section
line. i.e. in BC, EF and EC, Since the question ask the forces in the member BC and EF, but by draw a
section line member EC is also cut by the section line, so we consider the force in the member EC.
40 KN
LHS RHS
T
EF
T
CE
T
BC
2 m
1
1
45
45
B C D A
E F
10 KN
Fig 13.65
Choose any one part of them, Since both parts are separately in equilibrium. Let we choose right hand
side portion (as shown in fig 13.66). And the Right hand parts of truss is in equilibrium under the action
of following forces,
326 / Problems and Solutions in Mechanical Engineering with Concept
C D
45
T
BC
1
1
T
CE
T
EF
F
10 KN
RHS
Fig 13.66
1. Reaction R
DV
= 10KN at the joint D
2. 10KN load at joint F
3. Force T
BC
in member BC
4. Force T
CE
in member CE
5. Force T
FE
in member FE
All three forces are assumed to be tensile.
Now we take moment of all these five forces only from any point of the truss, for getting the answers
quickly
Taking moment about point C, of all the five forces given above
∑M
C
= 0
(Moment of Force T
BC
,T
CE
about point C is zero, since point C lies on the line of action of that forces)
– R
D
x CD + T
EF
x CF – 10x CF = 0
–10 x 2 + T
EF
x 2 – 10x 2 = 0
T
EF
= 20KN ...(iii)
T
EF
= 20KN (Tensile) .......ANS
Taking moment about point E, of all the five forces given above
∑M
E
= 0
(Moment of Force T
EF
,T
EC
and 10KN about point E is zero, since point E lies on the line of action
of that forces)
– R
D
x BD – T
BC
x CF = 0
–10 x 4 – T
BC
x 2 = 0
T
BC
= –20KN ...(iii)
T
BC
= 20KN(Compressive) .......ANS
Q. 23: Determine the forces in the members BC and BD of a cantilever truss shown in the
figure 13.67. (May–04(C.O.))
3 m
2 m
A
E
D
C
2 m
1000 N 1000 N
Fig 13.67
Truss / 327
Sol.: In this problem; If we draw a section line which cut the member BC, BD, AD, ED, then the member
BC and BD cut by this line, but this section line cut four members, so we don’t use this section line. Since
a section line cut maximum three members.
There is no single section line which cut the maximum three member and also cut the member BC and
BD.
This problem is done in two steps
(1) Draw a section line which cut the member BC and DC. Select any one section and find the value
of BC.
(2) Draw a new diagram, draw another section line which cut the member AB, BD and CD, and find
the value of the member BD.
STEP–1
Draw a section line which cut the member BC and CD, as shown in fig 13.68.
3 m
2 m
A
E
D
C
1000 N 1000 N
RHS LHS
B
1
1
T
CD
Fig 13.68
Consider Right hand side portion of the truss as shown in fig 13.69
Taking moment about point D
∑M
D
= 0
1000 x 2 – T
BC
x BD = 0
Consider Triangle CAE and CDB, they are similar
BD/AE = BC/AC
BD/3 = 2/4
BD = 1.5m
1000 x 2 – T
BC
x 1.5 = 0
T
BC
= 1333.33KN ...(iii)
T
BC
= 1333.33KN (Tensile) .......ANS
STEP–2:
Draw a section line which cut the member AB, AD and BD, as shown in fig 13.70.
328 / Problems and Solutions in Mechanical Engineering with Concept
3 m
A
E
D
C
θ
1000 N 1000 N
T
AB
T
DE
B
T
CD
Fig 13.70
Consider Right hand side portion of the truss as shown in fig 13.71
Taking moment about point C
∑M
C
= 0
{Moment of force T
AB
, T
CD
and 1000N acting at C is zero}
–1000 x BD – T
BD
x BC = 0
–1000 x 2 – T
BD
x 2 = 0
T
BD
= – 1000KN ...(iii)
T
BC
= 1000KN (compressive) .......ANS
Q. 24: Find the axial forces in the members CE, DE, CD and BD of the truss shown in fig 13.72.
(May–04(C.O.))
STEP–1:
First draw a section line which cut the member CE, CD and BD as shown in fig 13.73
Consider RHS portion of the truss as shown in fig 13.74.
Here Member AB = BC = CD = BE
Fig. 1373 Fig. 1372
E
C
D
1 kN
45º 45º
A B
E
C
LHS
RHS
T
CD
T
DE
T
CE
D
1 kN
45º 45º
A B
Taking moment about point C
∑M
C
= 0{Moment of force T
CE
, T
CD
is zero}
1 x CD + T
BD
cos45º x BC = 0
T
BD
=1/ cos 45º = –1.414KN
T
BD
=1.414KN (compressive) .......ANS
T
AB
B
C
1000 N
Fig. 1371
1000 N
T
DE
T
CD
Truss / 329
Taking moment about point E
∑M
E
= 0
{Moment of force T
CE
, 1KN is zero}
T
CD
x ED + T
DB
cos 45º x ED= 0
T
CD
= T
DB
cos 45º T
CD
= – 1KN
T
CD
= 1KN (Tensile) .......ANS
Taking moment about point B
∑M
B
= 0
{Moment of force T
DB
, is zero}
– T
CD
x CB + (1 + T
CE
cos 45º) x CD – T
CE
sin 45º x (ED + CB) = 0
CB = CD = ED
– 1 + 1 + T
CE
cos 45º – 2 T
CE
sin 45º = 0
T
CE
= 0 .......ANS
STEP–2:
Draw another section line which cut the member CE and ED Select RHS portion of the truss;
There are only two forces on the RHS portion
Taking moment about point C We get
T
ED
= 0 .......ANS
Q. 25: A pin jointed cantilever frame is hinged to a vertical wall at A and E, and is loaded as shown
in fig 13.75. Determine the forces in the member CD, CG and FG.
A B C H
D
G
F
E
Fig 13.75
Sol.: First find the angle HDG
Let Angle HDG = θ
tan θ = GH/HD = 2/2 = 1
θ = 45º
A B
2 kN 2 kN
T
CD
T
FG
T
CG
C
F
K
E
θ
1
θ
D
4 kN
LHS
RHS
Fig 13.76
T
CE
T
DE
T
CD
RHS
E
D
1 kN
330 / Problems and Solutions in Mechanical Engineering with Concept
Draw a section line which cut the member CD, CG and FG, Consider RHS portion of the truss as
shown in fig 13.76.
Taking Moment about point G, we get
∑M
G
= 0
{Moment of force T
FG
and T
CG
is zero}
– T
CD
x 2 + 2 x 2 = 0
T
CD
= 2KN (Tensile) .......ANS
Since Angle HDG = DGH = HCG = HGC = 45º
Now for angle GEK; tan¸ = 2/8 = ¼
Angle GEK = 14º
Angle EGK = 76º
Now resolved force T
FG
and T
CG
as
T
FG
sin 76º
T
FG
cos 76º
T
CG
sin 45º
T
CG
cos 45º
Taking Moment about point C, we get
∑M
C
= 0
{Moment of force T
CD
is zero }
– T
CG
cos 45º x 2 + T
CG
sin 45º x 2 – T
FG
sin 76º x 2 – T
FG
sin 76º x 2 + 2 x 4 = 0
– 2 T
FG
(sin 76º
+ cos 76º) + 8 = 0
T
FG
= 3.29KN (Tensile) .......ANS
Taking Moment about point E, we get
∑M
E
= 0
{Moment of force T
FG
is zero}
– T
CD
x 4 + 2 x 10 – T
CG
cos 45º x 8 – T
CG
sin 45º x 2 = 0
– 2 x 4 + 20 10 T
CG
cos 45º = 0
T
CG
= 1.69KN (Tensile) .......ANS
Simple Stress and Strain / 331
+0)264
14
SlMPLE STRESS /ND STR/lN
Q. 1: Differentiate between strength of material and engineering mechanics.
Sol. : Three fundamental areas of mechanics of solids are statics, dynamics and strength of materials.
Strength of materials is basically a branch of `Solid Mechanics'. The other important branch of solid
mechanics is Engineering Mechanics: statics and dynamics. Whereas `Engineering Mechanics' deals with
mechanical behaviour of rigid (nondeformable) solids subjected to external loads, the 'Strength of Materials'
deals with mechanical behaviour of nonrigid (deformable) solids under applied external loads. It is also
known by other names such as Mechanics of Solids, Mechanics of Materials, and Mechanics of Deformable
Solids. Summarily, the studies of solid mechanics can be grouped as follows.
Mechanics
Solid Mechanics
(of rigid bodies)
Engineering Mechanics
or
Applied Mechanics
or
Mechanics of Rigid Bodies
or
Mechanics of Nondefonnable Solids
(of nonrigid bodies)
Strength of Materials
or
Mechanics of Solids
or
Mechanics of Materials
or
Mechanics of Deformable Solids
Fluid Mechanics
Fig. 14.1
Since none of the known materials are rigid, therefore the studies of Engineering Mechanics are based
on theoretical aspects; but because all known materials are deformable, the studies of strength of materials
are based on realistic concepts and practical footings. The study of Strength of Materials helps the design
engineer to select a material of known strength at minimum expenditure.
Studies of Strength of Materials are applicable to almost all types of machine and structural components, all
varieties of materials and all shapes and crosssections of components. There are numerous variety of components,
each behaving differently under different loading conditions. These components may be made of high strength steel,
low strength plastic, ductile aluminium, brittle cast iron, flexiable copper strip, or stiff tungsten.
Q. 2: What is the scope of strength of materials?
Sol. : Strength of materials is the science which deals with the relations between externally applied loads
and their internal effects on bodies.
The bodies are not assumed to be rigid, and the deformation, however small are of major interest.
332 / Problems and Solutions in Mechanical Engineering with Concept
Or, we can say that, When an external force act on a body. The body tends to undergoes some
deformation. Due to cohesion between the molecules, the body resists deformation. This resistance by
which material of the body oppose the deformation is known as strength of material, with in a certain limit
(in the elastic stage). The resistance offered by the materials is proportional to the deformation brought out
on the material by the external force.
So we conclude that the subject of strength of materials is basically a study of
(i) The behaviour of materials under various types of load and moment.
(ii) The action of forces and their effects on structural and machine elements such as angle iron,
circular bars and beams etc.
Certain assumption are made for analysis the problems of strength of materials such as:
(i) The material of the body is homogeneous and isotropic,
(ii) There are no internal stresses present in the material before the application of loads.
Q. 3: What is load?
Sol. : A load may be defined as the combined effect of external forces acting on a body. The load is applied
on the body whereas stress is induced in the material of the body. The loads may be classified as
1. Tensile load
2. Compressive load
3. Torsional load or Twisting load
4. Bending load
5. Shearing loads
Q. 4: Define stress and its type.
Sol. : When a body is acted upon by some load or external force, it undergoes deformation (i.e., change
in shape or dimension) which increases gradually. During deformation, the material of the body resists the
tendency of the load to deform the body, and when the load influence is taken over by the internal resistance
of the material of the body, it becomes stable. The internal resistance which the body offers to meet with
the load is called stress.
Or, The force of resistance per unit area, offered by a body against deformation is known as stress.
Stress can be considered either as total stress or unit stress. Total stress represent the total resistance to an
external effect and is expressed in N,KN etc. Unit stress represents the resistance developed by a unit area
of cross section, and is expressed in KN/m
2
.
If the external load is applied in one direction only, the stress developed is called simple stress Whereas
If the external loads are applied in more than one direction, the stress developed is called compound stress.
Normal stress (σ) = P/A N/m
2
1 Pascal(Pa) = 1 N/m
2
1KPa = 10
3
N/m
2
1MPa = 10
6
N/m
2
1GPa = 10
9
N/m
2
Generally stress are divided in to three group as:
( ) 1Dstress a
( ) 2Dstress b
( ) 3Dstress c
s
y
s
y
s
x
s
x
s
x
s y
s y
s
z
s
z
s
s
Fig 14.2 One, Two and Three dimensional stress
Simple Stress and Strain / 333
But also the various types of stresses may be classified as:
1. Simple or direct stress (Tension, Compression, Shear)
2. Indirect stress (Bending, Torsion)
3. Combined Stress (Combination of 1 & 2)
(a) Tensile Stress
The stress induced in a body, when subjected to two equal and opposite pulls as shown in fig (14.3 (a))
as a result of which there is an increase in length, is known as tensile stress.
Let,
P = Pull (or force) acting on the body,
A = Cross  sectional area of the body,
σ = Stress induced in the body
Fig (a), shows a bar subjected to a tensile force P at its ends. Consider a section xx, which divides
the bar into two parts. The part left to the section xx, will be in equilibrium if P = Resisting force (R).
This is show in Fig (b), Similarly the part right to the section xx, will be in equilibrium if P = Resisting
force as shown in Fig (c), This resisting force per unit area is known as stress or intensity of stress.Tensile
stress (σ) = Resisting force (R)/Cross sectional area
σ σσ σσ
t
= P/A N/m
2
.
Tensile stress
P
P
P
R R
( ) a
( ) b
( ) c
( ) d
Resisting
Force (R)
Resisting
Force (R)
X
P
P
P
P P
Fig 14.3
(b) Compressive Stress
The stress induced in a body, when subjected to two equal and opposite pushs as shown in fig (14.4 (a))
as a result of which there is an decrease in length, is known as tensile stress.
Let, an axial push P is acting on a body of cross sectional area A. Then compressive stress(σ
c
) is given
by;
σ
c
= Resisting force (R)/Cross sectional area (A)
σ
c
= P/A N/m
2
.
334 / Problems and Solutions in Mechanical Engineering with Concept
Compressive streas
P
P
P
R R
( ) a
( ) b
( ) c
( ) d
Resisting
Force (R)
Resisting
Force (R)
X
P
P
P
P P
Fig 14.4
Q. 5: Define strain and its type.
Sol. : STRAIN(e) :When a body is subjected to some external force, there is some change of dimension
of the body. The ratio of change in dimension of the body to the original dimension is known as strain.
Or, The strain (e) is the deformation produced by stress. Strain is dimensionless.
There are mainly four type of strain
1. tensile strain
2. Compressive strain
3. Volumetric strain
4. Shear strain
Tensile Strain
When a tensile load acts on a body then there will be a decrease in crosssectional area and an increase
in length of the body. The ratio of the increase in length to the original length is known as tensile strain.
e
t
= δ δδ δδL/L
L
P
L
P
Fig 14.5
The above strain which is caused in the direction of application of load is called longitudinal strain.
Another term lateral strain is strain in the direction perpendicular to the application of load i.e., δD/D
Compressive Strain
When a compressive load acts on a body then there will be an increase in crosssectional area and decrease
in length of the body. The ratio of the decrease in length to the original length is known as compressive
strain.
e
c
= δ δδ δδL/L
Simple Stress and Strain / 335
L
P
L
P
Fig 14.6
Q. 6: What do you mean by Elastic Limit?
Sol. : When an external force acts on a body, the body tends to undergo some deformation. If the external
force is removed and the body comes back to its original shape and size (which means the deformation
disappears completely), the body is known as elastic body. This property, by virtue of which certain materials
return back to their original position after the removal of the external force, is called elasticity.
The body will regain its previous shape and size only when the deformation caused by the external
force, is within a certain limit. Thus there is a limiting value of force upto and within which, the deformation
completely disappears on the removal of the force. The value of stress corresponding to this limiting force
is known as the elastic limit of the material.
lf the external force is so large that the stress exceeds the elastic limit, the material loses to some extent
its property of elasticity. If now the force is removed, the material will not return to its origin shape and
size and there will be a residual deformation in the material.
Q. 7: State Hook's law.
Sol. : It states that when a material is loaded within elastic limit, the stress is proportional to the strain
produced by the stress. This means the ratio of the stress to the corresponding strain is a constant within
the elastic limit. This constant is known as Modulus of Elasticity or Modulus of Rigidity.
Stress /strain = constant
The constant is known as elastic constant
Normal stress/ Normal strain = Young's modulus or Modulus of elasticity (E)
Shear stress/ Shear strain = Shear modulus or Modulus of Rigidity (G)
Direct stress/ Volumetric strain = Bulk modulus (K)
Q. 8: What do you mean by Young's Modulus or Modulus of elasticity?
Sol. : It is the ratio between tensile stress and tensile strain or compressive stress and compressive strain.
It is denoted by E. It is the same as modulus of elasticity
E = σ σσ σσ/e [σ σσ σσ
t
/e
t
or σ σσ σσ
c
/e
c
]
S.No. Material Young's Modulus(E)
1 Mild steel 2.1 × 10
5
N/mm
2
2 Cast Iron 1.3 × 10
5
N/mm
2
3 Aluminium 0.7 × 10
5
N/mm
2
4 Copper 1.0 × 10
5
N/mm
2
5 Timber 0.1 × 10
5
N/mm
2
The % error in calculation of Young's modulus is: [(E
1
– E
2
)/E
1
] × 100
Q. 9: What is the difference between
(a) Nominal stress and true stress
(b) Nominal strain and true strain?
336 / Problems and Solutions in Mechanical Engineering with Concept
(a) Nominal Stress and True Stress
Nominal stress or engineering stress is the ratio of force per initial cross sectional area (original area
of crosssection).
Nominal stress =
0
Force
initial area of crosssection
P
A
·
True stress is the ratio of force per actual (instantaneous) crosssectional area taking lateral strain into
consideration.
True stress =
Force
Actual area of crosssection
P
A
·
(b) Nominal Strain and True Strain
Nominal Strain is the ratio of change in length per initial length.
Nominal strain =
Change in length
Initial length
L
L
∆
·
True strain is the ratio of change in length per actual length (instantaneous length) taking longitudinal
strain into consideration.
Q. 10: A load of 5 KN is to be raised with the help of a steel wire. Find the diameter of steel wire,
if the maximum stress is not to exceed 100 MNm
2
. (UPTUQUESTION BANK)
Sol.: Given data:
P = 5 KN = 5000N
σ = 100MN/m
2
= 100N/mm
2
Let D be the diameter of the wire
We know that, σ = P/A
σ = P/ (Π/4 × D
2
)
100 = 5000/ (Π/4 × D
2
)
D = 7.28mm .......ANS
Q. 11: A circular rod of diameter 20 m and 500 m long is subjected to tensile force of 45kN. The
modulus of elasticity for steel may be taken as 200 kN/m
2
. Find stress, strain and elongation
of bar due to applied load. (UPTUQUESTION BANK)
Sol.: Given data:
D = 20m
L = 500m
P = 45KN = 45000N
E = 200KN/m
2
= (200 × 1000 N/mm
2
= 200000 N/m
2
Using the relation; σ = P/A = P/(Π/4 × D
2
)
σ = 45000/(Π/4 × 20
2
)
σ σσ σσ = 143.24 N/m
2
.......ANS
E = σ/e
200000 = 143.24 /e
e = 0.000716 .......ANS
Now, e = dL
A
/L
0.000716 = dL
A
/500
dL
A
= 0.36 .......ANS
Simple Stress and Strain / 337
Q. 12: A rod 100 cm long and of 2 cm x 2 cm crosssection is subjected to a pull of 1000 kg force.
If the modulus of elasticity of the materials 2.0 x 106 kg/cm
2
, determine the elongation of the
rod. (UPTUQUESTION BANK)
Sol.: Given data:
A = 2 × 2 = 4cm
L = 100cm
P = 1000kg = 1000 × 9.81 = 9810N
E = 2.0 × 106 kg/cm
2
= 9.81 × 2.0 × 106 kg/cm
2
= 19.62 × 106 N/cm
2
Using the relation; σ = P/A
σ = 9810/4
σ σσ σσ = 2452.5 N/cm2 .......ANS
E = σ/e
19.62 x 10
6
= 2452.5/e
e = 0.000125 .......ANS
Now, e = dL
A
/L
0.000125 = dL
A
/100
dL = 0.0125 .......ANS
Q. 13: A hollow castiron cylinder 4 m long, 300 mm outer diameter, and thickness of metal 50 mm
is subjected to a central load on the top when standing straight. The stress produced is 75000
kN/m
2
. Assume Young's modulus for cast iron as 1.5 x 108 KN/m
2
find
(i) magnitude of the load,
(ii) longitudinal strain produced and
(iii) total decrease in length.
Sol.: Outer diameter, D = 300 mm = 0.3 m Thickness, t = 50 mm = 0.05 m
Length, L = 4 m
Stress produced, σ = 75000 kN/m
2
E = 1.5 x 108 kN/m
2
Here diameter of the cylinder, d = D – 2t = 0.3 – 2 × 0.05 = 0.2 m
(i) Magnitude of the load P:
Using the relation, σ =P/A
or P = σ × A = 75000 × Π/4 (D
2
– d
2
)
= 75000 × Π/4 (0.32 – 0.22)
or P = 2945.2 kN .......ANS
(ii) Longitudinal strain produced, e :
Using the relation,
Strain, (e) = stress/E = 75000/1.5 x 108 = 0.0005 .......ANS
(iii) Total decrease in length, dL:
Using the relation,
Strain = change in length/original length = dL
A
/L
0.0005 = dL
A
/4
dL
A
= 0.0005 × 4m = 0.002m=2mm
Hence decrease in length = 2 mm .......ANS
338 / Problems and Solutions in Mechanical Engineering with Concept
Q. 14: A steel wire 2 m long and 3 mm in diameter is extended by 0.75 mm when a weight P is
suspended from the wire. If the same weight is suspended from a brass wire, 2.5 m long and
2 mm in diameter, it is elongated by 4.64 mm. Determine the modulus of elasticity of brass if
that of steel be 2.0 x 10
5
N/mm
2
. (UPTUQUESTION BANK)
Sol.: Given: L
S
= 2 m,
δs = 3 mm,
δL
S
= 0.75 mm;
Es = 2.0 × 105 N/mm
2
;
L
b
=2.5m;d
b
= 2mm;
δL
b
=4.64m.
Modulus of elasticity of brass, E
b
:
From Hooke's law, we know that;
E = σ/e
= (P/A)/(δL
A
/L) = P.L/A. δL
A
or, P = δL
A
.A.E/L
where,
δL = extension,
L = length,
A = crosssectional area,
and E = modulus of elasticity.
Case I : For steel wire:
P = δLs.As.Es/Ls
or P = [0.75 × (Π/4 × 32) × 2.0× 10
5
]/2000 ...(i)
Case II : For bass wire
P = δLb.Ab.Eb/L
b
or P = [4.64 × (Π/4 × 22) × Eb]/2500 ...(ii)
Equating equation (i) and (ii), we get
[0.75 × (Π/4 x32) × 2.0x10
5
× 2.0 × 10
5
]/2000 = P = [4.64 × (Π/4 × 22) × E
b
]/2500
E
b
= 0.909 × 10
5
N/mm
2
.......ANS
Q. 15: The wire working on a railway signal is 5mm in diameter and 300m long. If the movement
at the signal end is to be 25cm, make calculations for the movement which must be given to
the end of the wire at the signal box. Assume a pull of 2500N on the wire and take modulus
of elasticity for the wire material as 2 × 10
5
N/mm
2
.
Sol.: Given data:
P = 2500N
D = 5 mm
L = 300m = 300 × 1000 mm
D
m
= 25cm
E = 2 × 10
5
N/mm
2
.
We know that, σ = P/A
σ = P/ (Π/4 × D
2
)
σ = 2500/ (Π/4 × 5
2
)
σ = 127.32 N/mm
2
.
Simple Stress and Strain / 339
e = σ/E = 127.32/2 × 10
5
e = 0.0006366
Since e = δL/L
δL = e.L = 0.0006366 × 300 × 1000 = 190.98mm = 19.098 cm
Total movement which need to be given at the signal box end = 25 + 19.098 = 44.098 cm .......ANS
Q. 16: Draw stressstrain diagrams, for structural steel and cast iron and briefly explain the various
salient points on them. (May–01, May–03)
Or;
Draw a stress strain diagram for a ductile material and show the elastic limit, yield point and
ultimate strength. Explain any one of these three. (May–03(CO))
Or;
Draw stressstrain diagram for a ductile material under tension. (Dec–04)
Or;
Draw the stress strain diagram for aluminium and cast iron. (May–05)
Or;
Explain the stressstrain diagram for a ductile and brittle material under tension on common
axes single diagram. (May–05(CO))
Or
Define Ductile behaviour of a metal (Dec–00)
Sol.: The relation between stress and strain is generally shown by plotting a stressstrain (σe) diagram.
Stress is plotted on ordinate (vertical axis) and strain on abscissa (horizontal axis). Such diagrams are most
common in strength of materials for understanding the behaviour of materials. Stressstrain diagrams are
drawn for different loadings. Therefore they are called
Tensile stressstrain diagram
Compressive stressstrain diagram
Shear stressstrain. diagram
StressStrain Curves (Tension)
When a bar or specimen is subjected to a gradually increasing axial tensile load, the stresses and strains
can be found out for number of loading conditions and a curve is plotted upto the point at which the
specimen fails. giving what is known as stressstrain curve. Such curves differ in shape for various materials.
Broadly speaking the curves can be divided into two categories.
(a) Stressstrain carves for ductile materials : A material is said to be ductile in nature, if it elongates
appreciably before fracture. One such material is mild steel. The shape of stressstrain diagram for the mild
steel is shown in Fig. 14.7.
A mild steel specimen of either circular crosssection (rod) or rectangular section (flat bar) is pulled
until it breaks. The extensions of the bar are measured at every load increments. The stresses are calculated
based on the original cross sectional area and strains by dividing the extensions by gauge length. When the
specimen of a mild steel is loaded gradually in tension, increasing tensile load, in tension testing machine.
The initial portion from O to A is linear where strain linearly varies with stress. The line is called line of
proportionality and is known as proportionality limit. The stress corresponding to the point is called “Limit
of Proportionality”. Hook's law obeys in this part, the slope of the line gives, 'modulus of elasticity'.
340 / Problems and Solutions in Mechanical Engineering with Concept
Further increase in load increases extension rapidly and the stress strain diagram becomes curved. At
B, the material reaches its 'elastic limit' indicating the end of the elastic zone and entry into plastic zone.
In most cases A and B coincide. if load is removed the material returns to its original dimensions.
Beyond the elastic limit, the material enters into the plastic zone and removal of load does not return
the specimen to its original dimensions, thus subjecting the specimen to permanent deformation. On further
loading the curve reaches the point `C' called the upper yield point at which sudden extension takes place
which is known as ductile extension where the strain increases at constant stress. This is identified by the
horizontal portion of the diagram. Point C gives 'yield stress'. beyond which the load decreases with increase
in strain upto C' known as lower yield point.
After the lower yield point has been crossed, the stress again starts increasing, till the stress reaches
the maximum value at point `D'. The increase in load causes non linear extension upto point D. The point
D known as 'ultimate point' or 'maximum point'. This point gives the 'ultimate strength' or maximum load
of the bar. The stress corresponding to this highest point `D' of the stress strain diagram is called the
ultimate stress.
Strain (E)
Typical stressstain diagram for a ductile material
A Limit of proportionatity
B Elastic limit
C Upper yield point
C Lower yield point
D Ultimate point
F Rupture point
¢
s
s
s
s
s
S
t
r
e
s
s
(
)
s
0
A
B
C
C
F
Fig. 14.7
After reaching the point D, if the bar is strained further, a local reduction in the cross section occurs
in the gauge length (i.e., formation of neck). At this neck stress increases with decrease in area at constant
load, till failure take place. Point F is called 'rupture point.Note that all stresses are based on original area
of cross section in drawing the curve of Fig 14.7.
1. Yield strength =
0
Load at yield point
A
where (original area ) A
0
=
2
0
0
D
A
π
2. Ultimate strength =
max
0 0
Ultimate load
P
A A
·
3. % Elongation =
0
0
100
F
L L
L
−
×
Simple Stress and Strain / 341
where L
F
= Final length of specimen
L
0
= Original length of specimen
4. % Reduction in area =
0
0
100
F
A A
A
−
×
where A
F
= Final area of cross section
A
0
= Original area of cross section
5. Young’s modulus of elasticity, E =
Stress at any point with in elastic limit
Strainat that point
From the figure clastic limit is upto point B.
(b) Stress strain curves for brittle materials : Materials which show very small elongation before
they fracture are called brittle materials. The shape of curve for a high carbon steel is shown in
Fig. 14.8 and is typical of many brittle materials such as G.I,
concrete and high strength light alloys. For most brittle
materials the permanent elongation (i.e., increase in length) is
less than 10%.
StressStrain Curves (Compression)
For ductile materials stress strain curves in compression are identical
to those in tension at least upto the yield point for all practical purposes.
Since tests in tension are simple to make, the results derived from
tensile curves are relied upon for ductile materials in compression.
Brittle materials have compression stress strain curves usually of
the same form as the tension test but the stresses at various points (Limit of proportionality, ultimate etc)
are generally considerably different.
Q. 17: Define the following terms:
(1) limit of proportionality
(2) yield stress and ultimate stress
(3) working stress and factor of safety.
Sol.: (1) Limit of proportionality: Limit of proportionality is the stress at which the stress  strain diagram
ceases to be a straight line i.e, that stress at which extension ceases to be proportional to the corresponding
stresses.
(2) Yield stress and ultimate stress Yield stress : Yield stress is defined as the lowest stress at which
extension of the test piece increases without increase in load. It is the stress corresponding to the yield
point. For ductile material yield point is well defined whereas for brittle material it is obtained by offset
method. It is also called yield strength.
Yield Stress = Lowest stress = Yield Point Load/ Cross sectional Area
Ultimate stress : Ultimate stress or Ultimate strength corresponds to the highest point of the stress
strain curve. It is the ratio of maximum load to the original area of crosssection. At this ultimate point,
lateral strain gets localized resulting into the formation of neck.
Ultimate stress = Heighest value of stress =
Maximum Load
Original Cross sectional Area
Breaking point
Line of proportionaley
0
S
t
r
e
s
s
(
)
s
Stress ( ) s
Fig. 14.8
342 / Problems and Solutions in Mechanical Engineering with Concept
(3) Working stress and Factor of Safety Working Stress: Working stress is the safe stress taken
within the elastic range of the material. For brittle materials, it is taken equal to the ultimate strength
divided by suitable factor of safety. However, for materials possessing well defined yield point, it is equal
to yield stress divided by a factor of safety. It is the stress which accounts all sorts of uncertainties.
Working stress =
Ultimate strength
Factors of safety
for brittle materials
=
Yield strength
Factors of safety
for ductile materials
It is also called allowable stress, permissible stress, actual stress and safe stress.
Factor of Safety : Factor of safety is a number used to determine the working stress. It is fixed based
on the experimental works on the material. It accounts all uncertainties such as, material defects, unforeseen
loads, manufacturing defects, unskilled workmanship, temperature effects etc. Factor of safety is a
dimensionless number. It is fixed based on experimental works on each materials. It is defined as the
ratio of ultimate stress to working stress for brittle materials or yield stress working stress for ductile
materials.
Q. 18: Define how material can be classified?
Sol.: Materials are commonly classified as:
(1) Homogeneous and isotropic material: A homogeneous material implies that the elastic properties
such as modulus of elasticity and Poisson's ratio of the material are same everywhere in the material system.
Isotropic means that these properties are not directional characteristics, i.e., an isotropic material has same
elastic properties in all directions at any one point of the body:
(2) Rigid and linearly elastic material: A rigid material is one which has no strain regardless of the
applied stress. A linearly elastic material is one in which the strain is proportional to the stress.
S
t
r
e
s
s
S
t
r
e
s
s
Strain
a ( )
Strain
b ( )
Fig. 14.9. (a) Rigid and (b) linearly elastic material
(3) Plastic material and rigidplastic material: For a plastic material, there is definite stress at which
plastic deformation starts. A rigidplastic material is one in which elastic and timedependent deformations
are neglected. The deformation remains even after release of stress (load).
Simple Stress and Strain / 343
Stain
a ( )
Stain
a ( )
Stain
b ( )
S
t
r
e
s
s
S
t
r
e
s
s
Fig. 14.10 (a) Plastic mid (b) rigidplastic material
(4) Ductile mid brittle material: A material which can undergo 'large permanent' deformation in
tension, i.e., it can be drawn into wires is termed as ductile. A material which can be only slightly deformed
without rupture is termed as brittle.
Ductility of a material is measured by the percentage elongation of the specimen or the percentage
reduction in crosssectional area of the specimen when failure occurs. If L is the original length and L’ is
the final length, then
% increase in length =
100
L L
L
′ −
×
The length l' is measured by putting together two portions of the fractured specimen. Likewise if A is the
original area of crosssection and A’ is the minimum cross sectional area at fracture, then
% age reduction in area =
100
A A
A
′ −
×
A brittle material like cast iron or concrete has very little elongation and very little reduction in cross
sectional area. A ductile material like steel or aluminium has large reduction in area and increase in elongation.
An arbitrary percentage elongation of 5% is frequently taken as the dividing line between these two classes
of material.
Q. 19: The following observations were made during a tensile test on a mild steel specimen 40 mm
in diameter and 200 mm long. Elongation with 40 kN load (within limit of proportionality),
δ δδ δδL = 0.0304 mm
Yield load =161 KN
Maximum load = 242 KN
Length of specimen at fracture = 249 mm
Determine:
(i) Young's modulus of elasticity
(ii) Yield point stress
(iii) Ultimate stress
(iv) Percentage elongation.
Sol.: (i) Young's modulus of elasticity E :
Stress, σ = P/A
= 40/[Π/4(0.04)
2
] = 3.18 × 10
4
kN/m
2
344 / Problems and Solutions in Mechanical Engineering with Concept
Strain, e = δL/L = 0.0304/200 = 0.000152
E = stress/ strain = 3.18 × 10
4
/0.000152
= 2.09 × 10
8
kN/m
2
.......ANS
(ii) Yield point stress:
Yield point stress = yield point load/ Cross sectional area
= 161/[Π/4(0.04)
2
]
= 12.8 × 10
4
kN/m
2
.......ANS
(iii) Ultimate stress:
Ultimate stress = maximum load/ Cross sectional area
= 242/[Π/4(0.04)2]
= 19.2 × 10
4
kN/m
2
.......ANS
(iv) Percentage elongation:
Percentage elongation = (length of specimen at fracture  original length)/ Original length
= (249–200)/200
= 0.245 = 24.5% .......ANS
Q. 20: The following data was recorded during tensile test made on a standard tensile test specimen:
Original diameter and gauge length =25 mm and 80 mm;
Minimum diameter at fracture =15 mm;
Distance between gauge points at fracture = 95 mm;
Load at yield point and at fracture = 50 kN and 65 kN;
Maximum load that specimen could take = 86 kN.
Make calculations for
(a) Yield strength, ultimate tensile strength and breaking strength
(b) Percentage elongation and percentage reduction in area after fracture
(c) Nominal and true stress and fracture.
Sol.: Given data:
Original diameter =25 mm
gauge length = 80 mm;
minimum diameter at fracture =15 mm
distance between gauge points at fracture = 95 mm
load at yield point and at fracture = 50 kN
load at fracture = 65 kN;
maximum load that specimen could take = 86 kN.
Original Area Ao = Π/4 (25)
2
= 490.87mm
2
Final Area A
f
= Π/4 (15)
2
= 176.72mm
2
(a) Yield Strength = Yield Load / Original Cross sectional Area
= (50 × 10
3
)/490.87 = 101.86 N/mm
2
.......ANS
Ultimate tensile Strength Maximum Load / Original Cross sectional Area
= (86 × 10
3
)/490.87 = 175.2 N/mm
2
.......ANS
Breaking Strength = fracture Load / Original Cross sectional Area
= (65 × 103)/ 490.87 = 132.42 N/mm
2
.......ANS
Simple Stress and Strain / 345
(b) Percentage elongation = (distance between gauge points at fracture  gauge length)/ gauge length
= [(95 – 80)/80] × 100 = 18.75% .......ANS
percentage reduction in area after fracture = [(Original Area – Final Area)/ Original Area] × 100
= [(490.87 – 176.72)/ 490.87] × 100 = 64% .......ANS
(c) Nominal Stress = Load at fracture / Original Area = (65 × 1000)/ 490.87
= 132.42 N/mm
2
.......ANS
True Stress = Load at fracture / Final Area = (65 × 1000)/ 176.72
= 367.8 N/mm
2
.......ANS
Q. 21: Find the change in length of circular bar of uniform taper.
Sol.: The stress at any cross section can be found by dividing the load by the area of cross section and
extension can be found by integrating extensions of a small length over whole of the length of the bar. We
shall consider the following cases of variable cross section:
Consider a circular bar that tapers uniformly from diameter d1 at the bigger end to diameter d2 at the
smaller end, and subjected to axial tensile load P as shown in fig 14.11.
Let us consider a small strip of length dx at a distance x from the bigger end.
Diameter of the elementary strip:
dx = d
1
– [(d
1
– d
2
)x]/L
= d
1
– kx; where k = (d
1
– d
2
)/L
x
L
dx
P
d
1 d
2
Elementary Strip
Fig 14.11
Crosssectional area of the strip,
A
x
=
2
4 4
x
d
π π
·
(d
1
– kx)
2
Stress in the strip,
σ
x
=
2
2
1
1
4
( )
( )
4
x
P P P
A d kx
c kx
· ·
π
π −
−
Strain in the strip
ε
x
=
2
1
4
( )
x
P
E d kx E
σ
·
π −
Elongation of the strip
346 / Problems and Solutions in Mechanical Engineering with Concept
δl
x
= ε
x
dx = 2
1
4
( )
Pdx
d kx E π −
The total elongation of this tapering bar can be worked out by integrating the above expression
between the limits x = 0 to x = L
δl =
2 2
1 1
0 0
4 4
( ) ( )
L L
Pdx P dx
d kx E E d kx
·
π − π −
∫ ∫
=
1
1
1
0 0
( )
4 4 1
( 1) ( )
L L
d kx
P P
E k EK d kx
−
1 1 −
·
1 1
π − × − π −
1 1
¸ ] ¸ ]
Putting the value of k = (d
1
– d
2
) /l in the above expression, we obtain
δl =
1 2
1 2 1
1
4 1 1
( )
( )
PL
d d l
E d d d
d
l
1
1
− 1
−
π −
1
−
1
¸ ]
=
1 2 2 1
4 1 1
( )
PL
E d d d d
1
−
1
π −
1
¸ ]
=
1 2
1 2 1 2 1 2
4 4
( )
d d
PL PL
E d d d d E d d
−
× ·
π − π
If the bar is of uniform diameter d throughout its length, then
δL = 4.P.L/(Π.E.d
2
)
= P.L/[(Πd
2
/4).E] = P.l/A.E.; Which is same as last article
Q. 22: A conical bar tapers uniformly from a diameter of 4 cm to 1.5 cm in a length of 40 cm. If an
axial force of 80 kN is applied at each end, determine the elongation of the bar. Take
E = 2 × 10
5
N/mm
2
(UPTU QUESTION BANK)
Sol.: Given that; P = 80 × 10
3
N,
E = 200GPa = 2 × 10
5
N/mm2, dL = 40mm, d
2
= 15mm, L = 400mm
Since;
δL =
1 2
4 PL
Ed d π
Putting all the value, we get
δL = [4 × 80 × 10
3
× 400]/[Π(2 × 10
5
) × (40 × 15)]
= 0.3397mm .......ANS
Q. 23: If the Tension test bar is found to taper from (D + a) cm diameter to (D – a)cm diameter, prove
that the error involved in using the mean diameter to calculate Young's modulus is (10a/D)
2
.
Sol.: Larger dia d
1
= D + a (mm)
Smaller dia d
2
= D – a (mm)
Let,
P = load applied on bar (tensile) N
Simple Stress and Strain / 347
L = length of bar (mm)
E
1
= Young's modulus using uniform cross section (N/mm
2
)
E
2
= Young's modulus using uniform cross section (N/mm
2
)
dL = Extension in length of bar (mm)
Using the relation of extension for tapering cross section, we have
δL = 4.P.L/(Π.E.d
1
.d
2
) = 4.P.L/[Π.E
1
.(D + a).(D – a)] = 4.P.L/[Π.E
1
.(D
2
– a
2
)]
E
1
= 4.P.L/[Π. dL.(D
2
– a
2
)] ...(i)
Now using the relation of extension for uniform cross  section; we have
δL = 4.P.L/(Π.E.d
2
) = 4.P.L/(Π.E
2
.D
2
)
E
2
= 4.P.L/[Π. δL.D
2
] ...(ii)
The % error in calculation of Young's modulus is: [(E
1
– E
2
)/E
1
] × 100
Hence proved
Q. 24: A bar of length 25mm has varying cross section. It carries a load of 14KN. Find the extension
if the cross section is given by (6 + x
2
/100) mm
2
where x is the distance from one end in cm.
Take E = 200 GN/m
2
. (Neglect weight of bar)
Sol.: Consider a small element of length dx at distance x from the small element. Due to tensile load applied
at the ends, the element length dx elongates by a small amount ∆x, and
∆x = Pdx/AE = Pdx/[(6 + x
2
/100)]
The total elongation of the bar is then worked out by integrating the above identity between the limits
x = 0 to x = L
δL =
25
2 2
0 0
100 1
600
6
100
L
P P
dx
E x x
E
·
+ ¸ _
+
¸ ,
∫ ∫
Since,
2 2
1 dx
a x a
·
+
∫
tan
–1
(x/a)
Recalling ;
Here, a = (600)
1/2
δL = (100P/E) × (1/(600)
1/2
)tan
–1
25
1/ 2
0
/(600) x 1
¸ ]
Putting P = 14KN and E = 2 × 10
5
N/mm
2
; we get
We get;
δ δδ δδL = 0.227 mm .......ANS
Q. 25: A steel bar AB of uniform thickness 2 cm, tapers uniformly from 1.5 cm to 7.5 cm in a length
of 50 cm. From first principles determine the elongation of plate; if an, axial tensile force of
100 kN is applied on it. [E = 2 x 105 N/mm
2
]
Sol.: Consider a small element of length dx of the plate, at a distance x from the larger end. Then at this
section,
Width w
x
= 100 – (100 – 50) 100
400 8
x x ¸ _
· −
¸ ,
mm
348 / Problems and Solutions in Mechanical Engineering with Concept
Crosssection area A
x
= thickness × width = 10 100
8
x ¸ _
−
¸ ,
mm
2
Stress σ
x
=
3 3
50 10 5 10
10 (100 / 8) (100 / 8)
x
P
A x x
× ×
· ·
− −
N/mm
2
Strain ε
x
=
3
5
5 10 1
(100 / 8) (2 10 ) 40 (100 / 8)
x
E x x
σ
×
· ·
− × × −
10 mm
50 mm
100 mm
50 kN
50 kN
400 mm
dx
x
Fig 14.12
∴ Elongation of the elementary length,
δL
x
= ε
x
× dx =
40 (100 / 8)
dx
x −
The total change in length of the plate can be worked out by integrating the above identity between
the limits x = 0 and x = 400 mm. That is:
δL =
400 400
0 0
1
40 (100 / 8) 40 (100 / 8)
dx dx
x x
·
− −
∫ ∫
=
400
400
0
0
log (100 / 8)
1
0.2 log (100 / 8)
40 1/ 8
e
e
x
x
− 1
1 · − −
1
¸ ]
−
¸ ]
= – 0.2 [ log
e
50 – log
e
100] = 0.2 (log
e
100 – log
e
50)
= 0.2 log
e
100
50
¸ _
¸ ,
= 0.2 × 0.693 = 0.1386 mm
Q. 26: Find ratio of upper end area to lower end area of a bar of uniform strength.
Sol: Figure 14.13 shows a bar subjected to an external tensile load P. If the bar had been of uniform cross
section, the tensile stress intensity at any section would be constant only if the selfweight of the member
is ignored. If the weight of the member is also considered, the intensity of stress increases for sections at
higher level. It is possible to maintain a uniform stress of all the sections by increasing the area from the
lower end to the upper end. Let the areas of the upper and lower ends be A
1
and A
2
respectively. Let A be
the area at a distance x and A + dA at a distance x + dx from the lower end. Let the weight per unit volume
be w Making a force balance for the element ABCD.
Simple Stress and Strain / 349
A
1
A
2
A + dA
D
A
dx
C
A
P
Fig 14.13
σ (A + dA) = σA + w Adx
or, σdA = w Adx
or
dA
A
=
w
σ
dx
Assuming w and s to be uniform,
ln A =
w
σ
x + C
1
At x = 0, A = A
2
= C
1
= ln A
2
ln
2
A
A
=
wx
σ
A = A
2
e
ax/σ
x = L, A = A
1
1
2
A
A
= e
wL/σ
Q. 27: A vertical bar fixed at the upper end, and of uniform strength carries an axial load of 12KN.
The bar is 2.4m long having a weight per unit volume of 0.0001N/mm
3
. If the area of the bar
at the lower end is 520mm
2
, find the area of the bar at the upper end.
Sol.: The bar is of uniform strength, so the stress will remain the same everywhere.
σ = P/A
2
; = 12000/520 = 23.08N/mm
2
Now
1
2
A
A
= e
wL/σ
0.0001
520 2400
23.08
x e x
−
= 520e
0.0104
= 525.44mm
2
.......ANS
Q. 28: Find increase in length of a bar of uniform section due to self weight.
Sol.: Consider a bar of crosssectional area A and length l hanging freely under its own weight (Fig. 14.14).
Let attention be focused on a small element of length dy at distance x from the lower end. If 'ω' is the
specific weight (weight per unit volume) of the bar material, then total tension at section mn equals weight
of the bar for length y and is given by
P = ωAy
350 / Problems and Solutions in Mechanical Engineering with Concept
As a result of this load, the elemental length dx elongates by a small
amount Ax, and
∆x =
Pdy wAy w
dy
AE AE E
· ·
The total change in length of the bar due to self weight is worked out by
integrating the above expression between the limits y = a and y = l. Therefore,
δx =
1
1
2 2
0 0
2 2
w w y w l
y dy
E E E
1
· ·
1
¸ ]
∫
If W is the total weight of the bar (W= wAl), then w = W/Al. In that case, total
extension of the bar
δx =
2
2 2
w l wl
Al E AE
¸ _
·
¸ ,
, W = Total weight
Thus total extension of the bar due to self weight is equal to the extension that would be produced if
onehalf of the weight of the bar is applied at its end.
Q. 29: An aerial copper wire (E =1 x 10
5
N/mm
2
) 40 m long has cross sectional area of 80 mm
2
and
weighs 0.6 N per meter run. If the wire is suspended vertically, calculate
(a) the elongation of wire due to self weight,
(b) the total elongation when a weight of 200 N is attached to its lower end, and
(c) the maximum weight which this wire can support at its lower end if the limiting value of
stress is 65 N/mm
2
.
Sol.: (a) Weight of the wire W = 0.6 × 40 = 24 N
The elongation due to self weight is,
δL = ωL
2
/2 E; where ω is the specific weight (weight per unit volume)
In terms of total weight W = ωAL
δL = WL/2E = 24 × (40 × 10
3
)/2 × 80 × (1 × 10
5
)
= 0.06 mm .......ANS
(b) Extension due to weight P attached at the lower end,
δL = PL/2E
= 200 × (40 × 10
3
)/80 × (1 × 10
5
)
= 1.0 mm
Total elongation of the wire = 0.06 + 1.0 = 1.06 mm
(c) Maximum limiting stress = 65 N/ mm
2
Stress due to self weight equals that produced by a load of half its weight applied at the end. That is
Stress due to self weight = (W/2)/A = (24/2)/80 = 0.15 N/ mm
2
Remaining stress = 65 – 0.15 = 64.85 N/mm2
Maximum weight which the wire can support = 64.85 × 80 = 5188 N
Q. 30: A rectangular bar of uniform crosssection 4 cm × 2.5 cm and of length 2 m is hanging
vertically from a rigid support. lt is subjected to axial tensile loading of 10KN. Take the
dy
m n
x
y
Fig 14.14
Simple Stress and Strain / 351
density of steel as 7850 kg/m
3
. And E = 200 GN/m
2
. Find the maximum stress and the elongation
of the bar. (Dec–03)
Sol.: Data given,cross sectional area A = 4 cm. × 2.Scm.
length L = 2m
axial load
P= I0KN
density of steel ρ = 7850 Kg/m
3
E= 200 GN/m
2
Since δL = PL/AE + ρL
2
/2E
= (10 × 10
3
× 2)/(4 × 2.5 × 10
–4
× 200 × 10
9
)
+ (7850 × 9.81 × 2
2
)/(2 × 200 × 10
9
)
= 0.0001 m Fig 14.15
= 0.1 mm
stress (σ
max
) = E × strain
= 200 × 10
9
× δL/L =200 × 10
9
× (0.0001/2) = 10.08 mpa .......ANS
Q. 31: Determine the elongation due to self weight of a conical bar.
Sol.: Consider a conical bar ABC of length L with its diameter d fixed rigidly at AB. Let attention be
focussed on a small element of length dy at distance y from the lower end point C. Total tension P at section
DE equals weight of the bar for length y and is given by
P = w ×
2
1
3 4
ds y
π ¸ _
¸ ,
where ω is the specific weight of the bar material and ds = DE is the diameter of the elementary strip.
From the similarity of triangles ABC and DEC,
AB
DE
=
L
y
or DE = AB
y y
d
L L
·
P = w ×
2 2
2
1
3 4
d y
y
L
1 π
1
¸ ]
= w
2
3
2
2
d
y
L L
π
As a result of this load, the elemental length dy elongates by a small amount Dy and
∆y =
2
2
12
Pdy d
w
AE L
π
· y
3
dy +
2 2
2
4
d y
L
π
E
=
3
w
E
ydy
The total change in length of the conical bar due to selfweight is worked out by integrating the above
expression between the limits y = 0 to y = L
δL =
2 2
2
0
0
3 3 2 6 6
L
L
w w y wL g
y dy L
E E E E
1 ρ
· · ·
1
¸ ]
∫
Where ρ is the mass density of the bar material
2 m
10 kN
A B
C
D
y
dy
E
Fig 14.16
352 / Problems and Solutions in Mechanical Engineering with Concept
Q. 32: Determine the elongation due to self weight of a tapering rod.
Sol.: Consider a tapering rod hung vertically and firmly fixed at the top position. The rod is of length L
and it tapers uniformly from diameter d
l
to d
2
, Let the sides AC and BD meet at point E when produced.
Extension δL in the length L of the rod ABDE due to self weight is
δL = extension of conical rod AEB due to self weight – extension due to conical segment CEB due
to self weight – extension in rod length L due to weight of segment CED
=
2 2
1 2
( ) 4
6 6
wL w L L PL
E E Ed d
′ ′ −
· ·
π
where P is the weight of segment CED and w is the specific weight of bar material
P = w
2
2
1
( )
4 3
d L L
π 1
′ × × −
1
¸ ]
Substituting this value of tensile load P in the above expression, we get
δl =
2 2
2
1
( ) ( )
6 6 3
d
wL w L L w L L L
E E E d
′ − −
· −
From the geometrical configuration
cot θ =
2 1
/ 2 / 2
( )
d d
L L L
·
−
=
2 1
d d
L L L
·
−
This gives : L′ =
1
1 2
d L
d d −
and L′ – L =
2
1 2
d L
d d −
δL =
1 2 2 2
1 2 1 2 1 2 1
6 6 6
d L d L d L d
w w w
L
E d d E d d E d d d
¸ _ ¸ _ ¸ _
− − ×
− − −
¸ , ¸ , ¸ ,
=
2 2 2
2
1 2 2
2 2
1 2 1 2 1 2 1
3 2( ) 2( ) 2( )
d d d
wL
E d d d d d d d
1
− −
1
− − −
1
¸ ]
=
3 2 2 2
1 1 2 2 1 2
2
1 1 2
2 ( )
3 2 ( )
d d d d d d
wL
E d d d
1 − − −
1
−
1
¸ ]
=
3 3 2 2
1 2 1 2
2
1 1 2
2 3
6 ( )
d d d d
wL
E d d d
1 + −
1
−
1
¸ ]
If the rod is conical, i.e., d
2
= 0
δL =
2
6
wL
E
which is the same expression as derived earlier
A E
E
P
P –1
l
d
2
d
1
q
q
Fig. 14.17
Simple Stress and Strain / 353
Q. 33: A vertical rod of 4 m long is rigidly fixed at upper end and carries an axial tensile load of
50kN force. Calculate total extension of the bar if the rod topers uniformly from a diameter
of 50 mm at top to 30 mm at bottom. Take density of material as 1 × 10
5
kg/m
3
and E = 210
GN/m
2
. (UPTU QUESTION BANK)
Sol.: Extension in the rod due to external load
=
1 2
4 PL
Ed d π
E = 2.1 × 10
5
N/mm
2
= [4 × (50 × 1000) × (4 × 1000)]/[Π × (2.1 × 10
5
) × (50 × 30)]
= 0.8088mm
Extension in the rod due to self weight
=
3 3 2 2
1 2 1 2
2
1 1 2
2 3
6 ( )
d d d d
wL
E d d d
1 + −
1
−
1
¸ ]
where; ω = ρ.g; ρ = 1 × 10
5
Kg/m
3
= 1 × 10
–4
Kg/m
3
=
4 2 3 3 2
5 2
(1 10 9.81) (4 1000) 50 2(30) 3 50 30
6 2.1 10 50 (50 30)
−
1 × × × + − × ×
1
× × −
¸ ]
= 0.0274mm
Total extension in the bar = 0.8088 + 0.0274 = 0.8362mm .......ANS
Q. 34: Explain the principle of superposition.
Sol.: A machine member is subjected to a number of forces acting on its outer edges as well as at some
intermediate sections along its length. The forces are then split up and their effects are considered on
individual sections. The resulting deformation is then given by the algebraic sum of the deformation of the
individual sections. This is the principle of superposition which may be stated as
“The resultant elongation due to several loads acting on a body is the algebraic sum of the elongations
caused by individual loads”
Or
“The total elongation in any stepped bar due to a load is the algebraic sum of elongations in individual
parts of the bar”.
Mathematically
δL =
i n
i
i L
L
·
·
δ
∑
Q. 35: How you evaluate the elongation of a bar of varying cross section?
Sol.: Consider a bar made up of different lengths and having different crosssections as shown in Fig. 14.18.
A , E
1 1
A , E
2 2
A , E
3 3
P
P
L
1
l
2
l
3
Fig 14.18
354 / Problems and Solutions in Mechanical Engineering with Concept
For such a bar, the following conditions apply:
(i) Each section is subjected to the same external pull or push
(ii) Total change in length is equal to the sum of changes of individual lengths
That is:
P
1
= P
2
= P
3
= P as ∑H = 0 & ∑V = 0
and δL = δL
1
+ δL
2
+ δL
3
=
3 3 1 1 2 2
1 2 3
L L L
E E E
σ σ σ
+ +
=
2 3 1 1 2 2
1 1 2 2 3 3
P L P L P L
A E A E A E
+ +
If the bar segments are made of same material, then In that case
E
1
= E
2
= E
3
= E.
δL =
3 1 2
1 2 3
L L L
P
E A A A
1
+ +
1
1
¸ ]
Q. 36: A steel bar is 900 mm long; its two ends are 40 mm and 30 mm in diameter and the length
of each rod is 200 mm. The middle portion of the bar is 15 mm in diameter and 500 mm long.
If the bar is subjected to an axial tensile load of 15 kN, find its total extension. Take E = 200
GN/ m
2
(G stands for giga and 1G = 10
9
)
200 mm 200 mm
2
1
500 mm
1
5
m
m
d
i
a
4
0
m
m
d
i
a
3
0
m
m
d
i
a
Fig 14.19
Sol.: Refer Fig. 14.19
Load, P = 15kN
Area, A
1
= (Π/4) × 40
2
= 1256.6 mm
2
= 0.001256 m
2
Area, A
2
= (Π/4) × 15
2
= 176.7 mm
2
= 0.0001767 m
2
Area, A
3
= (Π/4) × 30
2
= 706.8 mm
2
= 0.0007068 m
2
Lengths: L
1
=200mm = 0.2m, L
2
= 500mm = 0.5m and L
3
=200mm = 0.2m
Total extension of the bar:
Let δL
1
, δL
2
and δL
3
, be the extensions in the parts 1, 2 and 3 of the steel bar respectively. Then,
δL
1
=
3 1 2
2 3
1 2 3
/ .
, , or
/ .
PL PL PL
P A P L PL
L L E L
A E A E A E e L L A AE
σ 1
δ · δ · · · · δ ·
1
δ δ
¸ ]
3
Total extension of the bar,
Simple Stress and Strain / 355
δL = δL
1
+ δL
2
δL
3
=
3 3 1 2 1 2
1 2 3 1 2 3
PL L PL PL L L
P
A E A E A E E A A A
1
+ + · + +
1
1
¸ ]
=
3
9
15 10 0.20 0.50 0.20
200 10 0.001256 0.0001767 0.0007068
× 1
+ +
1
×
¸ ]
= 0.0002454 m = 0.2454 ram
Hence total extension of the steel bar = 0.2454 mm .......ANS
Q. 37: A member ABCD is subjected to point loads P
1
, P
2
, P
3
and P
4
as shown in Fig. 14.20
1600 mm
2
0.75 m 1 m 1.2 m
625 mm
2
900 mm
2
A B C D
P
4
P
3
P
2
P
1
Fig 14.20
Calculate the force P
3
,necessary for equilibrium if P
1
= 120 kN, P
2
= 220 kN and P
4
= 160 kN.
Determine also the net change in length of the member. Take E = 200 GN/m
2
.
(UPTU QUESTION BANK)
Sol.: Modulus of elasticity E = 200 GN/ m
2
= 2 × 10
5
N/mm
2
.
Considering equilibrium of forces along the axis of the member.
P
1
+ P
3
= P
2
+ P
4
;
120 + P
3
= 220 + 160
Force P
3
= 220 + 160 – 120 = 260 kN
The forces acting on each segment of the member
are shown in the free body diagrams shown below:
Let δL
1
, δL
2
and δL
3
, be the extensions in the
parts 1, 2 and 3 of the steel bar respectively. Then,
δL
1
=
3 1 2
2 3
1 2 3
, ,
PL pL PL
L L
A E A E A E
δ · δ ·
Since Tension in AB and CD but compression in
BC, So,
Total extension of the bar,
δL = δL
1
– δL
2
+ δL
3
δL =
3 1 2
1 2 3
PL PL PL
A E A E A E
− −
Extension of segment AB = [(120 × 10
3
) × (0.75 × 10
3
)]/[1600 × (2 × 10
5
)] = 0.28125 mm
Compression of segment BC = [(100 ×10
3
) × (1 × 10
3
)]/[625 × (2 × 10
5
)] = 0.8 mm
Extension of segment CD = [(160 ×10
3
) × (1.2 × 10
3
)]/[900 × (2 × 10
5
)] = 1.0667 mm
Net change in length of the member = δl =0.28125 – 0.8 + 1.0667 = 0.54795 mm (increase) .......ANS
120 kN
220120
= 100 kN
A
120 kN
100 kN
160 kN 160 kN
B
B
C
C D
Fig 14.21
356 / Problems and Solutions in Mechanical Engineering with Concept
Q. 38: The bar shown in Fig. 14.22 is subjected to an axial pull of 150kN. Determine diameter of the
middle portion if stress there is limited to 125N/mm
2
. Proceed to determine the length of this
middle portion if total extension of the bar is specified as 0.15 mm. Take modulus of elasticity
of bar material E = 2 × 10
5
N/mm
2
.
Sol.: Each of the segment of this composite bar is subjected to axial pull P =150 kN.
Axial Stress in the middle portion σ
2
= Axial pull/Area = 150 × 10
3
/[(Π/4).(d
2
2
)]
L
2
150 kN 150 kN
d mm
3
= 50 d mm
3
= 50 d mm
1
= 50
300 mm
d
2
Fig 14.22
Since stress is limited to 125 N/mm
2
, in the middle portion
125 = 150 × 10
3
/[(Π/4).(d
2
2
)]
d
2
2
= 1528.66 mm
Diameter of middle portion d
2
= 39.1mm .......ANS
(ii) Stress in the end portions, σ
1
= σ
3
= 150 × 10
3
/[(Π/4).(50
2
)] = 76.43 N/m
2
Total change in length of the bar,
= change in length of end portions + change in length of mid portion
δL = δL
1
+ δL
2
+ δL
3
= σ
1
L
1
/E + σ
2
L
2
/E + σ
3
L
3
/E; Since E is same for all portions
= σ
1
(L
1
+ L
3
)/E + σ
2
L
2
/E
L
1
+ L
3
= 300 – L
2
Now putting all the values;
0.15 = [76.43(300 – L
2
)]/2 × 10
5
+ 125L
2
/2 × 10
5
L
2
= 145.58 mm .......ANS
Q. 39: A steel tie rod 50mm in diameter and 2.5m long is subjected to a pull of 100 KN. To what length
the rod should be bored centrally so that the total extraction will increase by 15% under the
same pull, the bore being 25mm diameter? For steel modulus of elasticity is 2 × 10
5
N/mm
2
.
Sol.: Diameter of the steel tie rod = 50 mm = 0.05 m
Length of the steel rod, L = 2.5 m
Magnitude of the pull, P = 100 kN
Diameter of the bore = 25 mm = 0.025 m
Modulus of elasticity, E = 200 × 10
9
N/m
2
x
25 mm dia
L m =2.5
( ) a
( ) b
100 kN 100 kN 5
0
m
m
Fig. 14.23
Simple Stress and Strain / 357
Let length of the bore be ‘x’.
Stress in the solid rod σ = P/A
= {(100 × 1000)/[(Π/4)(0.05)2]} = 50.92 × 10
6
N/m
2
Elongation of the rod δL = σL/E
= (50.92 × 10
6
× 2.5) / (200 × 10
9
)
= 0.000636m = 0.636mm
Elongation after the rod is bored = 1.15 × 0.636 = 0.731mm
Area of the reduction section = (Π/4) (0.05
2
– 0.025
2
) = 0.001472m
2
Stress in the reduced section σ
b
= (100 × 1000)/0.001472m
2
= 67.93 × 10
6
N/m
2
Elongation of the rod
= σ(2.5 – x)/E + σ
b
.x/E = 0.731 × 10
–3
= [50.92 × 10
6
(2.5 – x)]/(200 × 10
9
) + (67.93 × 10
6
.×)/(200 × 10
9
) = 0.731 × 10
–3
x = 1.12m
Hence length of the bore = 1.12m .......ANS
Q. 40: A square bar of 25 mm side is held between two rigid plates and loaded by an axial pull equal
to 300kN as shown in Fig. 14.24. Determine the reactions at end A and C and elongation of
the portion AB. Take E = 2 × 10
5
N/mm
2
.
R
c R
c
R
a R
a
P
250 mm
400 mm
C
C D
B
B
A
A
Fig 14.24 Fig 14.25
Sol.: Cross section area of the bar A = 25 × 25 mm
2
Since the bar is held between rigid support at the ends, the following observations need to be made:
(1) Portion AB will be subjected to tension and portion BC will be under compression
(2) Since each ends are fixed and rigid and therefore total Elongation;
δL
ab
– δL
bc
= 0; Elongation in portion AB equals shortening in portion BC. i.e., δL
ab
= δL
bc
(3) Sum of reactions equals the applied axial pull i.e., P = R
a
+ R
c
Apply condition (2), we get
[P
ab
× L
ab
]/A
ab
.E = [P
bc
× L
bc
]/A
bc
.E
(R
a
× 400)/(625 × 2 × 10
5
) = (R
c
× 250)/(625 × 2 × 10
5
)
R
c
= 1.6R
a
...(i)
Now apply condition (3) i.e., P = R
a
+ R
c
300 × 10
3
= R
a
+ 1.6 R
a
R
a
= 1.154 × 10
5
N; R
c
= 1.846 × 10
5
N .......ANS
Q. 41: A rod ABCD rigidly fixed at the ends A and D is subjected to two equal and opposite forces
P = 25 kN at B and C as shown in the fig 14.26 given below: Make calculations for the axial
stresses in each section of the rod.
358 / Problems and Solutions in Mechanical Engineering with Concept
P
A
B C D
P
400 mm
2
400 mm
250 mm
2
250
mm
250
mm
250 mm
2
Fig 14.26
Sol.: The following observations need to be made.
(i) Due to symmetrical geometry and load, reaction at each of the fixed ends will be same both in
magnitude and direction. That is P
a
= P
d
= P
1
(say).
(ii) Segments AB and CD are in tension and the segment BC is in compression.
(iii) End supports are rigid and therefore total change in length of the rod is zero.
The forces acting on each segment will be as shown in Fig. 14.27
Using the relation δL = PL/AE
δL
ab
= (P
1
× 250)/( 250 × E) = P
1
/E .......Extension
δL
bc
= (P – P
1
) × 400)/( 400 × E) = (P – P
1
)/E .......Compression
δL
cd
= (P
1
× 250)/( 250 × E) = P
1
/E .......Extension
Since net change in length = 0 i.e.,
δL
ab
– δL
bc
+ δL
cd
= 0
P
1
/E – (P – P
1
)/E + P
1
/E= 0
Or, P
1
– P + P
1
+ P
1
= 0;
or, P
1
= P/3 = 25/3 KN
And; P – P
1
= 50/3KN
A
P – P
1
P – P
1
B
B C
C D
P
1
P
1
P
1
P
1
Fig 14.27
Now Stress in segment AB and CD
= 25 × 10
3
/3 × 250 = 33.33 N/mm
2
(tensile)
Stress in segment
BC = (50 × 10
3
)/(3 × 400) = 41.67 N/mm
2
(compressive)
Q. 42: A steel bar is subjected to loads as shown in fig. 14.28. Determine the change in length of the
bar ABCD of 18 cm diameter. E = 180 kN/mm
2
. (May05(C.O.))
Sol.: Ref fig 14.28
Since d = 180mm
E = 180 × 10
3
N/mm
2
Simple Stress and Strain / 359
300
mm
310
mm
310
mm
50 kN 30 kN
40 kN
60 kN
A B C D
Fig. 14.28
A B
B C
C D
60 kN
20 kN 20 kN
50 kN
50 kN
60 kN
Fig. 14.29
L
AB
= 300mm
L
BC
= 310mm
L
CD
= 310mm
From the fig 14.29
Load on portion AB = P
AB
= 50 × 10
3
N
Load on portion BC = P
BC
= 20 × 10
3
N
Load on portion CD = P
CD
= 60 × 10
3
N
Area of portion AB = Area of portion BC = Area of portion CD = A = Πd
2
/4
= Π(180)
2
/4 = 25446.9 mm
2
Using the relation δl = Pl/AE
δL
ab
= (50 × 10
3
× 300)/( 25446.9 × 180 × 10
3
) = 0.0033mm  Compression
δL
bc
= (20 × 10
3
× 310)/( 25446.9 × 180 × 10
3
) = 0.0012mm  Compression
δL
cd
= (60 × 10
3
× 310)/( 25446.9 × 180 × 10
3
) = 0.0041mm  Compression
Since net change in length = – δL
ab
– δL
bc
– δL
cd
= – 0.0033 – 0.0012 – 0.0041
= – 0.00856 mm
Decrease in length = 0.00856mm .......ANS
Q. 43: Two prismatic bars are rigidly fastened together and support a vertical load of 45kN as shown
in Fig. 14.30. The upper bar is steel having mass density 7750 kg/m
3
, length 10 m and cross
sectional area 65 cm
2
. The lower bar is brass having mass density 9000 kg/m
3
, length 6 m and
crosssectional area 50 cm
2
. Determine the maximum stress in each material. For steel
ES = 200 GN/m
2
and for brass Eb =100 GN/m
2
.
Sol.: Refer Fig. 14.30. The maximum stress in the brass bar occurs at junction BB, and this stress is caused
by the combined effect of 45kN load together with the weight of brass bar.
360 / Problems and Solutions in Mechanical Engineering with Concept
A
Steel
bar
Brass
bar
P kN = 45
6 m
10 m
B B
A
Fig. 14.30
Weight of brass bar, W
b
= ρ
b
gV
h
= 9000 × 9.81 × (6 × 50 × 10
–4
) = 2648.7 N
Stress at section BB, σ
b
= (P + W
b
)/A
b
= (45000 + 4648.7)/50 × 10
–4
= 9529740 N/m
2
= 9.53 MN/m
2
The maximum stress in the steel bar occurs at section A–A. Here the entire weight of steel and brass
bars and 45kN load gives rise to normal stress.
Weight of steel bar, W
S
= p
s
gV
s
=7750 × 9.81 × (10 × 65 × 10
–4
) = 4941.79 N
Stress at section AA ; σ
s
= (P + W
h
+ W
s
)/A
s
= (45000 + 2648.7 + 4941.79)/ 65 × 10
–4
= 8090845 = 8.09 MN/m
2
.......ANS
Q. 44: For the bar shown in Fig. 14.31, calculate the reaction produced by the lower support on the
bar. Take E = 200 GN/m
2
. Find also the stresses in the bars.
Sol.: Let R
1
= reaction at the upper support;
R
2
= reaction at the lower support when the bar touches it.
If the bar MN finally rests on the lower support,
we have
R
I
+ R
2
= 55kN = 55000
N For bar LM, the total force = R
1
= 55000 – R
2
(tensile)
For bar MN, the total force =R
2
(compressive)
δL
1
=extension of LM = [(55000 – R
2
) × 1.2]/[(110 × 10
–6
)
× 200 × 10
9
]
δL
2
= contraction of MN = [R
2
× 2.4]/[(220 × 10
–6
) × 200 × 10
9
]
In order that N rests on the lower support, we have from
compatibility equation
δL
l
– δL
2
= 1.2/1000 = 0.0012 m
Or,
[(55000 – R
2
) × 1.2]/[(110 × 10
–6
) × 200 × 10
9
]
– [R
2
× 2.4]/[(220 × 10
–6
) × 200 × 10
9
] = 0.0012
on solving;
R
2
= 16500N or, 16.5 KN
.......ANS
R
1
= 5516.5 = 38.5 KN .......ANS
Stress in LM = R
1
/A
1
= 38.5/110 × 10
–6
= 0.350 × 10
6
kN/m
2
= 350 MN/m
2
.......ANS
Stress in MN = R
2
/A
2
= 16.5/220 × 10
–6
= 0.075 × 10
6
kN/m
2
= 75 MN/m
2
.......ANS
2
.
4
m
R
2
N
1.2 mm
55 kN
220 mm
2
A =
2
110 mm
2
L
A =
1
M
R
1
1
.
2
E
Fig. 14.31
Simple Stress and Strain / 361
Q. 45: A 700 mm length of aluminium alloy bar is suspended from the ceiling so as to provide a
clearance of 0.3 mm between it and a 250 mm length of steel bar as shown in Fig. 14.32. A
al
= 1250 mm
2
, E
al
= 70 GN/m
2
, As = 2500 mm
2
, Es = 210 GN/m
2
. Determine the stress in the
aluminium and in the steel due to a 300kN load applied 500 mm from the ceiling.
2
5
0
m
m
0
.
3
m
m
P
Steel
N
3
0
0
k
N
Q
200
mm
5
0
0
m
m
Aluminium
L
M
Fig. 14.32
Sol.: On application of load of 300kN at Q, the portion LQ will move forward and come in contact with
N so that QM and NP will both be under compression. LQ will elongate, while QM and NP will contact
and the net elongation will be equal to gap of 0.3 mm between M and N.
Let σ
1
= tensile stress in LQ
σ
2
= compressive stress in QM
σ
3
= compressive stress in NP
Elongation of LQ = (σ
1
× 0.5)/70 × 10
9
Contraction of QM = (σ
2
× 0.2) / 70 × 10
9
Contraction of NP = (σ
3
× 0.25)/ 210 × 10
9
But force in QM = force in NP
σ
2
× 1250 × 10
–6
= σ
3
× 2500 × 10
–6
We get σ
3
= σ
2
/2
So the Contraction of NP = (σ
2
× 0.25)/(2 × 210 × 10
9
)
Net elongation = δL
LQ
– δL
QM
– δL
NP
= 0.0003
(σ
1
× 0.5)/70 × 10
9
– (σ
2
× 0.2) / 70 × 10
9
– (σ
2
× 0.25)/(2 × 210 × 10
9
) = 0.0003
on solving we get;
3σ
1
– 1.45σ
2
= 2 × 210 × 10
9
× 0.0003 ...(i)
Tensile force in LQ + compressive force in QM = 300000
1250 × 10
–6
× σ
1
+ 1250 × 10
–6
× σ
2
= 300000
We get; σ
1
+ σ
2
= 2.4 × 10
8
N/m
2
...(ii)
From equation (i) and (ii) we get
σ σσ σσ
1
= 1.065 × 10
8
N/m
2
= 106.5 MN/m
2
(tensile) .......ANS
σ σσ σσ
2
= 1.335 × 10
8
N/m
2
= 133.5 MN/m
2
(compressive) .......ANS
σ σσ σσ
3
= σ σσ σσ
2
/2 = 0.667 × 10
8
N/m
2
= 66.7 MN/m
2
(compressive) .......ANS
362 / Problems and Solutions in Mechanical Engineering with Concept
Q. 46: Two parallel steel wires 6 m long, 10 mm diameter are hung
vertically 70 mm apart and support a horizontal bar at their
lower ends. When a load of 9kN is attached to one of the wires,
it is observed that the bar is 24° to the horizontal. Find ‘E’ for
wire.
Sol.: Two wires LM and ST made of steel, each 6 m long and 10 mm
diameter are fixed at the supports and a load of 9kN is applied on wire ST.
Let the inclination of the bar after the application of the load be θ.
The extension in the length of steel wire ST,
δL =70 ⋅ tan θ = 70 × tan 24°
= 70 × 0.4452 = 31.166 mm = 0.031166 m
Strain in the wire,
e = δL/L = 0.031166/6 = 0.005194
and stress in the wire
σ = P/A = 9000/[(Π/4)(10/1000)
2
]
= 11.46 × 10
7
N/m
2
Young's modulus E= σ/e
= 11.46 × 107/0.005194 =2.2 × 10
10
N/m
2
= 22 GN/m
2
.......ANS
Ques No47: A rigid beam AB 2.4m. long is hinged at A and supported as shown in fig14.34. by two
steel wires CD and EF. CD is 6m long and 12mm in diameter and EF is 3m long and 3mm in
diameter. If a load of 2250N is applied at B find the stress in each wire E = 2 × 10
5
N/mm
2
.
Sol.: Let CD = L
1
= 6m
EF = L
2
= 3m
Tension in CD = T
1
N
Tension in EF = T
2
N
For the equilibrium of the beam, taking moment about
the hinge A.
T
1
× 0.60 + T
2
× 1.80 = 2250 × 24
T
1
+ 3T
2
= 9000 ...(i)
Since the beam is rigid it will remain straight
Let extension of CD = δ
1
= DD
1
Extension of EF = δ
2
= FF
1
As the wires extend, the rigid beam takes the position
AD
1
FB
1
δ
1
/δ
2
= AD/AF = 0.60/1.80 = 1/3
δ
2
= 3δ
1
...(ii)
But δ
1
= T
1
L
1
/A
1
E
and δ
2
= T
2
L
2
/A
2
E
i.e.; T
2
L
2
/A
2
E = 3. T
1
L
1
/A
1
E
T
1
= 1/3{(A
1
/A
2
)(L
2
/L
1
)}T
2
T
1
= 1/3{(12/3)
2
(3/6)}T
2
T
1
= 8/3T
2
...(iii)
Fig. 14.33
M
T
T
9 kN
q
6
m
70 mm
S L
2
1
B
1
F
1
F B
2250 N
B F
1.8 m
A
A
0.6 m
L m
2
= 3
T
2
L m
1
= 6
D
1
D
D
T
1
C
Fig. 14.34
Simple Stress and Strain / 363
Solve equation (i) and (iii) we get
T
1
= 4233N and T
2
= 1589N .......ANS
Now
Stress in the wire CD = 4233/ (Π/4)12
2
= 37 N/mm
2
Stress in the wire EF = 1589/ (Π/4)3
2
= 225 N/mm
2
Q. 48: How you determine the stress in composite bar? What is Modular Ratio?
Sol.: It becomes necessary to have a compound tie or strut (column)
where two or more material elements are fastened together to
prevent their uneven straining. The salient features of such a
composite system are:System extends (or contracts) as one unit
when subjected to tensile (or compressive) load. This implies that
deformation (extension or contraction) of each element is same.o
Strain, i.e., deformation per unit length of each element is same.o
Total external load on the system equals the sum of loads carried
by the different materials comprising the composite system. Fig
14.35
Consider a composite bar subjected to load P and fixed at
the top as shown in Fig. 14.35. Total load is shared by the two
bars and as such
P = P
1
+ P
2
= σ
1
A
1
+ σ
2
A
2
...(i)
Further elongation in two bars are same, i.e., δL
1
= δL
2
= δL
3
...(ii)
If strain in the bar are equal e
1
= e
2
; σ
1
/E
1
= σ
2
/E
2
or, σ
1
/ σ
2
= E
1
/E
2
(when length is same)
This ratio E
1
/E
2
is called Modular ratio.
Modular Ratio: Modular ratio is the ratio of moduli of elasticity of two materials. 1t is denoted by µ.
Modular ratio, µ =
1
2
Young's Modulas of Material 1
Young's Modulas of Material 2
E
E
·
Q. 49: Two copper rods one steel rod lie in a vertical plane and together support a load of 50kN as
shown in Fig. 14.36. Each rod is 25 mm in diameter, length of steel rod is 3 m and length of
each copper rod is 2m. If modulus of elasticity of steel is twice that of copper, make calculations
for the stress induced in each rod. It may be presumed that each rod deforms by the same
amount.
Sol.: Each rod deforms by the same amount and accordingly i.e. δL
S
= δL
E
or
L L
s s c c
E E
s c
σ ⋅ σ ⋅
·
or σ
s
=
s c
C
c s
E L
E L
⋅ σ ⋅
σ
s
= 2 × (2/3) σ
c
= 1.33 σ
c
Division of load between the steel and copper rods is as follows : total load = load carried by steel
rod + rod carried by two copper rods
50 × 10
3
= σ
s
A
s
+ 2 σ
c
A
c
= 1.33 σ
c
× (Π/4)(25)
2
+ 2σ
c
× (Π/4)(25)
2
= 1633.78σ
c
P
Bar 2 Bar 1
Fig. 14.35
364 / Problems and Solutions in Mechanical Engineering with Concept
Steel
2 m
3 m
2 m
Copper Copper
W KN – 50
Fig. 14.36
σ
C
= (50 × 10
3
)/1633.78 = 30.60 N/mm
2
.......ANS
σ
S
= 1.33 σ
C
=1.33 × 30.60 = 40.7 N/mm
2
.......ANS
Q. 50: A load of 100 kg is supported upon the rods A and C each of 10 mm diameter and another
rod B of 15 mm diameter as shown in figure 14.37. Find stresses in rods A, B and C.
(May01)
15 cm
EA
KN/mm
= 210
2
P
B
E KN/mm = 110
2
C
B
P
C
A
P
A
20 cm
KN/mm
2
E = 210
100 Kg
C B A
Load of Kg 100
2
0
c
m
1
5
c
m
E KN/mm = 210
2
E KN/mm = 110
2
E KN/mm = 210
2
1 3 2
Fig. 14.37 Fig. 14.38
Sol.: Given data
m = 100Kg
Diameter of rod A and C = 10mm
Diameter of rod B = 15mm
Area of rod ‘A’ = Area of rod ‘C’
= A
A
= π/4.D
2
= π/4.10
2
= 78.54 mm
2
Area of rod ‘B’
= A
B
= π/4.D
B
2
= π/4.15
2
= 176.71 mm
2
Since; P = P
A
+ P
B
+ P
C
100 × 9.81 = σ
A
× A
A
+ σ
B
× A
B
+ σ
C
× A
C
981 = 78.54 σ
A
+ 176.71σ
B
+ 78.54 σ
C
...(i)
The deflection or shorten in length in each rod will be same so,
δL
A
= δL
B
= δL
C
e
A
.L
A
= e
B
.L
B
= e
C
.L
C
Simple Stress and Strain / 365
(σ
A
/E
A
) × L
A
= (σ
B
/E
B
) × L
B
= (σ
C
/E
C
) × L
C
(σ
A
/210) × 200 = (σ
B
/110) × 350 = (σ
C
/210) × 200
σ
A
= σ
C
= 3.34 σ
B
...(ii)
Putting the value of (ii) in equation (i); we get
981 = 78.54 × 3.34σ
B
+ 176.71σ
B
+ 78.54 × 3.34 σ
B
981 = 701.5σ
B
σ σσ σσ
B
= 1.34 N/mm
2
.......ANS
σ σσ σσ
A
= σ σσ σσ
C
= 4.67 N/mm
2
.......ANS
Q. 51: A beam weighing 50 N is held in horizontal position by three wires. The outer wires are of
brass of 1.8 mm dia and attached to each end of the beam. The central wire is of steel of 0.9
mm diameter and attached to the middle of the beam. The beam is rigid and the wires are of
the same length and unstressed before the beam is attached. Determine the stress induced in
each of the wire. Take Young's modulus for brass as 80 GN/m
2
and for steel as 200 GN/m
2
.
Fig 14.39
B
r
a
s
s
w
i
r
e
B
r
a
s
s
w
i
r
e
S
t
e
e
l
w
i
r
e
Beam
P N = 50
Fig. 14.39
Sol.: If P
b
denotes load taken by each brass wire and P
s
denotes load taken by steel wire, then
Total load
P = 2P
b
+ P
s
= 2σ
b
A
b
+ σ
s
A
s
...(i)
As the beam is horizontal, all the wire extend by the same amount. Further since each wire is of same
length, the wires would experience the same amount of strain, thus
e
s
= e
b
σ
s/
E
s
= σ
b/
E
b
σ
s
= (E
s
× σ
b
)/E
b
= (200 × σ
b
)/80 = 2.5σ
b
...(iii)
Putting the value of equation (ii) in equation (i)
50 = 2σ
b
(Π/4)(1.8)
2
+ 2.5σ
b
(Π/4)(0.9)
2
50 = 6.678 σ
b
σ σσ σσ
b
= 7.49 N/mm
2
.......ANS
σ σσ σσ
s
= 2.5σ σσ σσ
b
= 18.71 N/mm
2
.......ANS
Q.52: A concrete column 300mm x 300mm in section, is reinforced by 10 longitudinal 20mm diameter
round steel bars. The column carries a compressive load of 450KN. Find load carried
and compressive stress produced in the steel bars and concrete. Take E
s
= 200GN/m
2
and
E
c
= 15GN/m
2
.
Sol.: Cross sectional Area of column = 300 × 300 = 90000mm
2
366 / Problems and Solutions in Mechanical Engineering with Concept
Area of steel bars As = 10 × (Π/4)(20)
2
= 3141.59mm
2
Area of concrete = 90000 – 3141.59 = 86858.4 mm
2
Each component ( concrete and steel bars) shorten by the same amount under the compressive load,
and therefore
Strain in concrete = strain in steel
σ
c
/Ec = σ
s
/E
s
Where σ
c
, σ
s
are stress induced in concrete and steel respectively
σ
s
= (E
s
σ
c
)/E
c
= (200 × 10
9
/15 × 10
9
) σ
c
= 13.33 σ
c
Further,
Total load on column = Load carried by steel + load carried by concrete
P = σ
s
.A
s
+ σ
c
.A
c
450 × 10
3
= 13.33σ
c
× 3141.59 + σ
c
. 86858.4
= 128735.8 σ
c
σ σσ σσ
c
= 3.59 N/mm
2
.......ANS
σ σσ σσ
s
= 13.33 × 3.59 = 46.95 N/mm
2
.......ANS
Load carried by concrete Pc = σ
cAc
= 3.59 × 86858.4 = 311821.656 N = 311.82KN .......ANS
Load carried by steel P
s
= σ
sAs
= 46.95 × 3141.59 = 147497.65 N = 147.49 KN .......ANS
Q. 53: A load of 300kN is applied on a short concrete column 250 mm × 250 mm. The column is
reinforced by steel bars of total area 5600 mm
2
. If the modulus of elasticity of steel is 15 times
that of concrete, find the stresses in concrete and steel.
If the stress in the concrete should not exceed 4 N/mm
2
, find the area of the steel required so
that the column may support a load of 600 kN. [U.P.T.U. Feb., 2001]
Sol.: do your self.
Q. 54: A solid steel cylinder 500 mm long and 70 mm diameter is placed inside an aluminium cylinder
having 75 mm inside diameter and 100 mm outside diameter. The aluminium cylinder is 0.16
mm. longer than the steel cylinder. An axial load of 500kN is applied to the bar and cylinder
through rigid cover plates as shown in Fig. 14.40. Find the stresses developed in the steel
cylinder and aluminium tube. Assume for steel, E = 220 GN/m
2
and for Al E = 70 GN/m
2
Sol.: Since the aluminium cylinder is 0.16 mm longer than the steel cylinder, the load required to compress
this cylinder by 0.16 mm will be found as follows :
E = stress/ strain = P.L/A.δL
Or P = E.A.δL/L
= 70 × 10
9
× π/4 (0.1
2
 0.075
2
) × 0.00016/0.50016 = 76944 N
When the aluminium cylinder is compressed by its extra length 0.16 mm, the load then shared by both
aluminium as well as steel cylinder will be,
500000 – 76944 = 423056 N
Let e
s
= strain in steel cylinder
e
a
= strain in aluminium cylinder
σ
s
= stress produced in steel cylinder
σ
a
= stress produced in aluminium cylinder
E
s
= 220 GN/m
2
E
a
= 70 GN/m
2
Simple Stress and Strain / 367
As both the cylinders are of the same length and are compressed by the same amount
e
s
= e
a
σ
s
/E
s
= σ
a
/E
a
or; σ
s
= E
s
/E
a
.σ
a
= (220 × 10
9
/70 × 10
9
).
σ
a
= (22/7). σ
a
Also P
s
+ P
a
= P
or; σs A
s
+ σ
a
⋅ A
a
= 423056
(22/7). σ
a
A
s
+ σ
a
A
a
= 423056 ...(i)
A
s
= π/4 (0.07
2
) = 0.002199 m
2
A
a
= π/4 (0.12 – 0.075
2
) = 0.003436 m
2
Putting the value of A
s
and A
a
in equation (i) we get
σ
a
= 27.24 × 10
6
N/m
2
= 27.24 MN/m
2
.......ANS
and σ
S
= 22/7 × 27.24 = 85.61 MN/m
2
Stress in the aluminium cylinder due to load 76944 N
= 76944/ π/4 (0.12  0.0752) Fig 14.40
= 23.39 × 10
9
N/m
2
= 22.39 MN/m
2
Total stress in aluminium cylinder
= 27.24 + 22.39 = 49.63 MN/m
2
.......ANS
andstress in steel cylinder = 85.61 MN/m
2
.......ANS
Ques No55: A steel rod 20 mm diameter passes centrally through a steel tube 25 mm internal
diameter and 40 mm external diameter. The tube is 750 mm long and is closed by rigid
washers of negligible thickness which are fastened by nuts threaded on the rod. The nuts are
tightened until the compressive load on the tube is 20kN. Calculate the stresses in the tube and
the rod. Find the increase in these stresses when one nut is tightened by one quarter of a turn
relative to the other. There are 0.4 threads per mm length. Take E = 200 GN/m
2
.
Steel rod
750
2
0
4
0
2
5
Washer
Steel tube
Nut
Fig 14.41
Sol.: Area of steel rod = A
sr
= π/4 (0.02
2
) = 0.0003142 m
2
Area of steel tube = A
st
= π/4 (0.04
2
– 0.025
2
) = 0.000766 m
2
Load is equally applied on both steel tube and steel rod.
i.e.; σ
st1
.A
st
= σ
sr1
.A
sr
= 2000
Since compression in steel tube and tension in steel rod.
0.0003142 σ
sr1
= 0.000766 σ
st1
= 2000
σ
sr1
= 63.6 MN/m
2
(T)
σ
st1
= 26.1 MN/m
2
(C)
Aluminium
cylinder
Steel
cylinder
Rigid plate
kN 500
500 kN
75 mm
5
0
0
m
m
0.16 mm
100 mm
70 mm
Rigid plate
368 / Problems and Solutions in Mechanical Engineering with Concept
Now when nut is tightened by one quarter, then let σ
sr2
and σ
st2
be the additional stresses produced
in the rod and tube respectively.
Distance traveled by nut = ¼(1/0.4) = 0.625mm = 0.000625m = δL
Total
δL
Total
= δL
Rod
+ δL
tube
= (σ
sr2
/E).L
sr
+ (σ
st2
/E).L
st
; L
st
= L
sr
0.000625 = L/E(σ
sr2
+ σ
st2
) ...(i)
Again load are equal
σ
st2
.A
st
= σ
sr2
.A
sr
σ
sr2
= σ
st2
.A
st
/A
sr
= σ
st2
.0.000766/0.0003142 = 2.44σ
st2
...(ii)
This value put in equation (i) we get
0.000625 = (0.75/200 × 109)(2.44 σ
st2
+ σ
st2
)
σ
st2
= 48.48MN/m
2
σ
sr2
= 118.19MN/m
2
These are increase in stress
Q. 56: Explain the concept of temperature stress.
Sol.: When the temperature of a material changes, there will be corresponding changes in its dimensions.
When a member is free to expand or contract due to the rise or fall of temperature, no stress will be induced
in the member. But if the natural changes in length due to rise or fall of temperature be prevented, stress
will be offered.
If prevented, then stress is induced, which offers strain, which is given by
e = δL/L = α.δt
δL = L.α.δt
σ = e.E = α.δt.E
where
L is the length of the member,
α is coefficient of thermal expansion and
δt is change in temperature.
L
( ) a
( ) b
L
L
Fig 14.42
Case1: If the bar is free to expand; Then no stress induced, only expansion in terms of δL
δL = L.α.δt
Case2: If the bar is rigidly fixed at both end to prevent expansion, or the grip do not yield or
Expansion is prevented; then stress is induced
δL = Ll.α.δt
e = δL/L = α.δt
σ = e.E = α.δt.E
Case3: Some grip is provided for expansion (Yield), same as in railway track
δL = L.α.δt – yield
e = δL/L = (α.δt.L – yield)/L
σ = e.E = {(α.δt.L – yield)/L}E
Simple Stress and Strain / 369
Q. 57: Two parallel walls 6 m apart, are stayed together by a steel rod 20 mm diameter, passing
through metal plates and nuts at each end. The nuts are tightened, when the rod is at a
temperature of 100°C. Determine the stress in the rod, when the temperature falls down to
20°C, if
(1) The ends do not yield.
(2) The ends yield by 1mm.
Take E = 2.0 × 10
5
N/mm
2
, α αα αα
s
= 12 × 10
–6
/°C. (UPTU QB)
Sol.: Given data;
L = 6m = 6000mm
d = 20mm
T
1
= 100°C
T
2
= 20°C
α
s
= 12 × 10
–6
/°C
E = 2 × 10
5
N/mm
2
(1) When the ends do not yield
Thermal stress = σ = αE∆T = 12 × 10
–6
× 2 × 10
5
× (100 – 20)
σ σσ σσ = 192 N/mm
2
.......ANS
(2) The ends yield by 1mm.
σ = E(α.L.∆T – δL)/L
= {(12 × 10
–6
× 80 × 6000  1)/6000 }× 2 × 10
5
= 158.67 N/mm
2
.......ANS
Ques No58: A copper rod 15 mm diameter, 0.8 m long is heated through 50°C. What is its expansion
when free to expand? Suppose the expansion is prevented by gripping it at both ends, find the
stress, its nature and the force applied by the grips, when:
(i) The grips do not yield.
(ii) One grip yields back by 0.5 mm.
Take α αα αα
c
= 18.5 × 10
6
/°C and E
C
= 1.25 × 10
5
N/mm
2
(Feb–01)
Sol.: Given data;
L = 0.8m = 800mm
d = 15mm
∆T = 500C
α
C
= 18.5 × 10
–6
/°C
E
C
= 1.25 × 10
5
N/mm
2
(1) Expansion when free to expand
δL = α.∆T.L = 18.5 × 10
–6
× 50 × 800 = 0.74mm .......ANS
(2) When the ends do not yield
Thermal stress = σ = αE∆T = 18.5 × 10
–6
× 1.25 × 105 × 50
σ σσ σσ = 115.63 N/mm
2
.......ANS
P = σA = 115.63 × π/4(15)
2
= 20.43KN(Compressive) .......ANS
(Since gripping is provided so the force is compressive)
(3) The ends yield by 0.5mm.
σ = (α.L.∆T – ∆L).E/L
= {(18.5 × 10
–6
× 800 × 50 – 0.5)/800 } × 1.25 × 105 = 37.5 N/mm
2
.......ANS
P = σ
A
= 37.5 × π/4(15)
2
= 6.63KN(Compressive) .......ANS
370 / Problems and Solutions in Mechanical Engineering with Concept
Q. 59: A steam pipe is 30 m long at a temperature of 15ºC. Steam at 180°C is passed through the
pipe. Calculate the increase in length when the pipe is free to expand. What stress is induced
in the material if the expansion is prevented ?
(E = 200 GN/m
2
, α αα αα = 0.000012 per °C) (May03 (C.O.))
Sol.: Given data;
L = 30m = 30000mm
T
1
= 15°C
T
2
= 180°C
α
s
= 12 × 106 /0C
E = 200 × GN/m
2
= 2 × 10
5
N/mm
2
(1) When free to expand
δL = ∆.L.∆T = 12 × 10
–6
× 30000 × (180 – 15)
δ δδ δδL = 59.4 mm .......ANS
(2) When expansion is prevented
σ = Eα.∆T
= 2 × 10
5
× 12 × 10
–6
× 165 = 396 N/mm
2
.......ANS
Q. 60: A steel rod 2.5 m long is secured between two walls. If the load on the rod is zero at 20°C,
compute the stress when the temperature drops to 20°C. The crosssectional area of the rod
is 1200 mm
2
, α αα αα = 11.7 µm/(m°C), and E = 200GPa, assuming
(a) that the walls are rigid and
(b) that the walls spring together a total distance of 0.5 mm as the temperature drops.
Sol.: (a) Let the rod be disconnected from the right wall. Temperature deformations can then freely occur.
A temperature drop causes the contraction represented by δ
T
in Fig. 14.43. To reattach the rod to the wall
will evidently require a pull P to produce the load deformation δ
P
so that δ
T
= δ
P
. Thus
α∆tL =
PL L
AE E
σ
·
σ = E.α.∆t = 200 × 10
9
× 11.7 × 10
–6
× 40 = 93.6 × 106 N/mm
2
= 93.6 MPa .......ANS
It may be noted that the stress is independent of the length of the rod.
(b) When the walls spring together, the free temperature contraction is equal to the sum of the load
deformation and the yield of the walls.
T
T
P
P
Yield
( ) B
( ) Nonrigid walls b
( ) Rigid walls a
( ) a
P
P
Fig 14.43
Hence δ
T
= δ
P
+ yield
Now,
α.L.∆t = σ.L/E + yield
Simple Stress and Strain / 371
11.7 × 10
–6
× 2.5 × 40 = (σ × 2.5)/(200 × 10
9
) + 0.5 × 10
–3
we get; σ σσ σσ = 53.6 MPa .......ANS
Thus the yield of the walls reduces the stress considerably.
Q. 61: A circular bar of length 400 mm and tapering uniformly from 50 mm to 25 mm diameter is
held between rigid supports at the ends. Calculate the maximum and minimum stress developed
in the bar when the temperature is raised by 30°C. Take E = 2 × 10
5
N/mm
2
and
α αα αα =1.2 × 10
–5
per°C.
Sol.: Increase in length due to temperature rise, = L.α.∆t = 400 × (1.2 × 10
–5
) × 30 = 0.144 mm)
This elongation sets up a compressive reaction P at the supports and the corresponding shortening in
length is given by;
= 4P.L/4.d
1
.d
2
.E = (4P × 400)/ (4 × 50 × 25 × 2 × 10
5
) = 1.6 × 10
–6
P
From compatibility condition; 1.6 × 10
–6
P = 0.144; P = 0.09 × 106 N
Maximum stress smax = P/A
min
= 0.09 × 10
6
/{π/4 (25)
2
} = 183.44 N/mm
2
.......ANS
Minimum stress smin = P/A
max
= 0.09 × 10
6
/{π/4 (50)
2
} = 45.86 N/mm
2
.......ANS
Q. 62: Explain the effect of temperature change in a composite bar. What is compatibility condition ?
Sol.: Consider temperature rise of a composite bar consisting of two members; one of steel and other of
brass rigidly fastened to each other. If allowed to expand freely;
expansion of brass bar: AB = Lα
b
∆t
expansion of steel bar: AC = Lα
s
∆t
Since coefficient of thermal expansion of brass is greater than that of steel, expansion of brass will be
more.But the bars are fastened together and accordingly both will expand to the same final position represented
by DD with net expansion of composite system AD equal to dl. To attain this position, brass bar is pushed
back and the steel bar is pulled.
AA
D
D
C
Steel
Brass
B
A
l l
Fig 14.44
Obviously compressive stress will be induced in brass bar and tensile stress will be developed in steel
bar. Under equilibrium state;
compressive force in brass = tensile force is steel
σ
c
A
c
= σ
s
A
s
Corresponding to brass rod:
Reduction in elongation,
DB = AB – AD = Lα
b
σ
t
– δL
Strain e
b
= (Lα
b
∆t – δL)/L = αb∆t – e;
Where e = δL/L, is the actual strain of the composite system
Corresponding to steel rod:
extra elongation, CD = AD – AC = δL – Lα
s
∆t
Strain e
s
= (δL – Lα
s
∆t)/L = e – α
b
∆t
372 / Problems and Solutions in Mechanical Engineering with Concept
Adding e
b
and e
s
, we get
e
b
+ e
s
= (δ
b
– δ
s
) ∆t,
It is also called as compatibility condition.
It may be pointed out that the nature of the stresses in the bars will get reversed if there is reduction
in the temperature of the composite system.
Q. 63: A steel tube with 2.4 cm external diameter and 1.8 cm internal diameter encloses a copper rod
1.5 cm diameter to which it is rigidly joined at each end. If at a temperature of 10°C, there
is no longitudinal stress, calculate the stresses in the rod and tube when the temperature is
raised to 200°C. E
S
= 210,000 N/mm
2
, α αα αα
s
= 11 × 10
6
/°C, E
C
= 100,000 N/mm
2
,
α αα αα
C
= 18 × 10
6
/°C [U.P.T.U. Feb–01]
Sol.: Given that:
E
S
= 210,000 N/mm
2
,
E
C
= 100,000 N/mm
2
α
s
= 11 × 10
–6
/°C,
α
C
= 18 × 10
–6
/°C
∆T = 200°C
Apply compatibility condition:
e
C
+ e
S
= (α
C
– α
S
) ∆t,
e
C
+ e
S
= (18 – 11) × 10
6
× 200
e
C
+ e
S
= 0.0014 ...(i)
From the equilibrium condition;
Compressive force on copper = Tensile force on steel
P
B
= P
C
e
C
.A
C
.E
C
= e
S
.A
S
.E
S
e
C
= e
S
[(A
S
/A
C
)(E
S
/E
C
)]
= e
S
[{(π/4)(2.4
2
– 1.8
2
)/(π/4)(1.5
2
)}(210/100)]
e
C
= 2.35e
S
...(ii)
Substituting the value of equation(ii) in equation (i)
2.35e
S
+ e
S
= 0.0014
e
S
= 0.000418 ...(iii)
e
C
= 0.000982 ...(iv)
Stress in steel tube δ
S
= e
S
.E
S
= 0.000418 × 210,000
σ σσ σσ
S
= 87.7 N/mm
2
.......ANS
Stress in copper tube σ
C
= e
C
.E
C
= 0.000982 × 100,000
σ σσ σσ
C
= 98.2 N/mm
2
.......ANS
Q. 64: A steel bar is placed between two copper bars each having the same area and length as the
steel bar at 15°C. At this stage they are rigidly connected together at both ends. When the
temperature is raised to 315°C the length of the bar increases by 1.50 mm. Determine the
original length and the final stresses in the bars. Take, E
S
= 2.1 × 10
5
MPa, E
C
. = 1 × 10
5
N/mm
2
, α αα αα
S
= 0.000012 per °C, α αα αα
C,.
= 0.0000175 per °C. (UPTU QUESTION BANK)
Simple Stress and Strain / 373
Sol.: When the composite system is in equilibrium,
Tensile force in steel bar = compressive force in two copper bars
σ
s
A
s
= 2σ
C
A
C
; σ
s
= 2σ
C
;
Since A
s
= A
C
Using the relation
e
C
– e
S
= (α
C
– α
S
)∆t
σ
C
/E
C
+ σ
S
/E
S
= (1.75 × 10
–5
– 1.2 × 10
–5
) × (250 – 20)
σ
C
/E
C
+ σ
S
/E
S
= 0.001265
Substituting the value of
σ
s
= 2σ
C
; we get
σ
C
/1 × 10
5
+ 2σ
C
/2 × 10
5
= 0.001265
σ σσ σσ
C
= 63.25 N/mm
2
.......ANS
σ σσ σσ
S
= 126.5 N/mm
2
.......ANS
Further, e
C
= α
C
.∆t – e;
where e = δL/L, actual strain of the composite system
e = δL/L = α
C
.∆t – e
C
= α
C
.∆
t
– σ
C
/E
C
= 1.75 × 10
–5
(250 – 20) – 63.25/1 × 10
5
= 0.0033925
Original length of bar, L = δL/0.0033925 = 1.25/0.0033925 = 368.46 mm = 0.3685m .......ANS
Q.65: A flat bar of aluminium alloy 25 mm wide and 5 mm thick is placed between two steel bars
each 25 mm wide and 10 mm thick to form a composite bar 25 mm × 25 mm as shown in Fig.
14.45. The three bars are fastened together at their ends when the temperature is 15°C. Find
the stress in each of the material when the temperature of the whole assembly is raised to
55°C. If at the new temperature a compressive load of 30kN is applied to the composite bar
what are the final stresses in steel and alloy ?
Take E
S
= 200 GN/m
2
, E
al
= 200/3 GN/m
2
α αα αα
s
= 1.2 × 10
–5
per°C, α αα αα
al
= 2.3 × 10
–5
per°C.
Sol.: Refer Fig. 14.45.
Area of aluminium,
A
al
= 25 × 5 = 125 mm
2
= 125 × 10
–6
m
2
Area of steel,
A
S
= 2 × 25 × 10 = 500 mm
2
or 500 × 10
–6
m
2
(i) Stresses due to rise of temperature:
If the two members had been free to expand,
Free expansion of steel = α
s
.∆t.L
s
Free expansion of aluminium = σ
al
.∆t.L
al
But since the members are fastened to each other at the ends, Fig 14.45
final expansion of each member would be the same.
Let this expansion be δ. The free expansion of aluminium is greater than δ while the free expansion
of steel is less than δ. Hence the steel is subjected to tensile stress while aluminium is subjected to compressive
stress.
Let σS. and σ
al
be the stresses in steel and aluminium respectively.
The whole system will be in equilibrium when
Total tension (pull) in steel = total compression (push) in aluminium
2
5
m
m
25 mm
5 mm 10 mm 10 mm
S
t
e
e
l
S
t
e
e
l
A
l
u
m
i
n
i
u
m
374 / Problems and Solutions in Mechanical Engineering with Concept
σ
S
.A
S
= σ
al
.A
al
or σ
S
× 500 × 10
–6
= σ
al
× 125 × 10
–6
σ
s
=
4
al
σ
= 0.25 σ
al
Final increase in length of steel = final increase in length of aluminium
σ
s
.∆t.L
s
+ σ
S
.L
S
/E
S
= α
al
.∆t.La
l
– s
al
.L
al
/E
al
α
s
.∆t + σ
S
/E
S
= α
al
.∆t – σ
al
/E
al
; Since L
S
= L
AL
But ∆t = 55 – 15 = 40°C
1.2 × 10
–5
× 40 + 0.25σ
al
/(200 × 10
9
) = 2.3 × 10
–5
× 40 – σ
al
/(200/3) × 10
9
or; 1.2 × 10
–5
× 40 × 200 × 109 + 0.25σ
al
= 2.3 × 10
–5
× 40 × 200 × 10
9
– 3σal
σ σσ σσ
al
= 27.07 MN/m
2
(Compressive) .......ANS
(ii) Stresses due to external compressive load 30kN :
Let σ
S1
and sal1 be the stresses due to external loading in steel and aluminium respectively.
Strain in steel = e
S
Strain in aluminium = e
al
σ
S1
/E
S
= σ
al1
/E
Al
σ
S1
= E
S
. σ
al1
/ E
Al
= (200. σ
al1
)/(200/3) = 3.σ
al1
But, load on steel + load on aluminium = total load
σ
S1
.A
S
+ σ
al1
.A
al
= 30 × 1000
or; 3.σ
al1
× 500 × 10
–6
+ σ
al1
× 125 × 10
–6
= 3000
σ
al1
= 18.46 MN/m
2
(compressive)
σS1 = 3.σ
al1
= 55.38 MN/m
2
(compressive)
Final stress:
Stress in aluminium = σ σσ σσ
al
+ σ σσ σσ
al1
= 27.07 + 18.46 = 45.53 MN/m
2
(Compressive) .......ANS
Stress in steel= σ σσ σσ
S
+ σ σσ σσ
S1
= –6.76 + 55.38 = 48.62 MN/m
2
(Compressive) .......ANS
Q. 66: A steel rod 40mm in diameter is enclosed by a copper tube of external diameter 50mm and
internal diameter 40 mm. A pin 25mm in diameter is fitted transversely to the assembly at
each end so as to secure the rod and the tube. If the temperature of the assembly is raised by
60°C, find
(i) the stresses in the steel rod and the copper tube, and
(ii) the shear in the pin.
Take E
S
= 2 × 10
5
N/mm
2
, E
C
= 1 × 10
5
N/mm
2
, α αα αα
s
= 1.2 × 10
–5
per °C and ac =1.6 × 10
–5
per°C.
Steel rod
Pin, mm dia 25
Copper tube
Pin
4
0
m
m
5
0
m
m
Fig 14.46
Simple Stress and Strain / 375
Sol.: Area of steel rod; A
S
= π/4 × 40
2
= 400π mm
2
Area of copper tube A
C
= π/4 × (50
2
– 40
2
) = 225π mm
2
Let σ
S
. and σ
C
be the stresses produced in steel and copper respectively due to change in temperature.
It may be noted that steel is in tension and copper is in compression. For equilibrium of the assembly,
Total tension (pull) in steel = total compression (push) in copper
σ
S
.A
S
= σ
C
.A
C
or σ
S
× 400π = σ
C
× 225π
σS = (9/16) σ
C
Actual expansion of steel = Actual compression of copper
α
s
.∆t.L
s
+ σ
S
.L
S
/E
S
= α
C
.∆t.L
C
– σ
C
.L
C
/E
C
αs.∆t + (9/16)σ
C
/E
S
= αC.∆t  σ
C
/E
C
;
Since L
S
= L
AL
and σS = (9/16)σ
C
But ∆t = 600C
1.2 × 10
–5
× 60 + (9/16)σ
C
/(2 × 10
5
) = 1.6 × 10
–
5 × 60 – σ
C
/1 × 10
5
72 + (9/32) σ
C
= 96 – σs
C
σ σσ σσ
C
= 18.286 N/m
2
(Compressive) .......ANS
σ σσ σσ
S
= 9/16 × 18.286 = 10.286N/m
2
(tensile) .......ANS
Q. 67: The composite bar consisting of steel and aluminium components as shown in Fig. 14.47 is
connected to two grips at the ends at a temperature of 60°C. Find the stresses in the two rods
when the temperature fall to 20°C
(i) if the ends do not yield,
(ii) if the ends yield by 0.25 mm.
Take E
S
= 2 × 10
5
and E
a
= 0.7 × 10
5
N/mm
2
, α αα αα
s
= 1.17 × 10
–5
and α αα αα
a
= 2.34 × 10
–5
per°C. The
areas of steel and aluminium bars are 250 mm
2
and 375 mm
2
respectively.
Aluminium
400 mm
250 mm
2
375 mm
2
800 mm
Fig 14.47
Sol.: A
a
/A
S
= 375/250 = 1.5
Free contraction of the composite bar
= α
s
.∆t.L
s
+ αa.∆t.La
= 1.17 × 10
–5
× 40 × 800 + 2.34 × 10
–5
× 40 × 400
= 0.7488 mm
When the contraction is prevented, partially tensile stresses are induced in the rods.
A
S
.σ
S
= A
a
.σ
a
σ
S
= (375/250).σ
a
= 1.5.σ
a
376 / Problems and Solutions in Mechanical Engineering with Concept
(1) When the rod do not yield, contraction prevented in steel and aluminimu = 0.7488 mm.
σ
S
/E
S
. L
S
+ σ
a
/E
a
. L
a
= 0.7488
(1.5.σ
a
/2 × 10
–5
) × 800 + (σ
a
/0.70 × 10
–5
) × 400 = 0.7488
11.7143 × 10
–3
σ
a
= 0.7488
σ σσ σσ
a
= 63.92 N/mm
2
.......ANS
σ σσ σσ
S
= 1.5 × 63.92 = 95.88 N/mm
2
.......ANS
(ii) When the ends yield by 0.25 mm
Contraction prevented = 0.7488 – 0.25 = 0.4988mm
11.7142 × 10
–3
.σ
a
= 0.4988
σ σσ σσ
a
= 42.58 N/mm
2
.......ANS
σ σσ σσ
S
= 1.5 × 42.58 = 63.87 N/mm
2
.......ANS
Q. 68: Explain strain energy and Resilience. (Dec–01; May–05(C.O.), May–02)
Sol.: When an external force acts on an elastic material and deforms it, internal resistance is developed in
the material due to cohesion between the molecules comprising the material. The internal resistance does
some work which is stored within the material as energy and this strain energy within elastic limit is known
as resilience.
Whatever energy is absorbed during loading, same energy is recovered during unloading and the
material springs back to its original dimension. Machine members like helical, spiral and leaf springs
possess this property of resilience.
A body may be subjected to following types of loads:
(1) Gradually applied load
(2) Suddenly applied load
(3) Falling or impact loads
(A) Gradually Applied Loads:
Load applied to a bar starts from zero and increases linearly
until the bar is fully loaded. When the load is within elastic
limit, the plot of load (stress) versus deformation (strain)
is linear (Fig. 14.48).
Work done = average load × deformation
= (1/2)P.δL = (½) (σA) × (σL/E)
= (½) (σ
2
/E)(AL) = (σ
2
/2E) × Volume
Work done = (σ
2
/2E) × Volume
The strain energy U stored in the bar equals the work
done and therefore,
U= (σ
2
/2E) × volume
σ = P/A
The maximum strain energy absorbed by a body upto its elastic limit is termed as Proof Resilience and
this proof resilience per unit volume is called Modulus of Resilience.
Proof resilience = (σ
e
2
/2E) × volume
where se is the stress at elastic limit.
Modulus of resilience = σ
e
2
/2E
l
Elongation
P
L
o
a
d
Elastic limit
stress
s
s
e
Fig. 14.48
Simple Stress and Strain / 377
(B) Suddenly Applied Load:
Load is applied suddenly and this remains constant throughout the process
of deformation. Accordingly the plot between load and elongation will be
parallel to xaxis.
Work done = area of shaded portion = P.δL = P(σ
su
.L/E)
The subscript su refers to suddenly applied load.
The work done equals the strain energy given by
(σ
su
2
/2E)(AL)
(σ
su
2
/2E)(AL) = p(σ
su
.L/E)
or, σ
su
= 2P/A = 2 × stress due to gradually applied load
Thus the instantaneous stress induced in a member due to suddenly applied load is twice that when
the load is applied gradually.
(C) Impact Loads
Impact loading occurs when a weight is droped on a member from some height. The kinetic energy of the
falling weight is utilized in deforming the member. Refer Fig. 14.50, a rod of crosssectional area A and
length l is fixed at one end and has a collar at the outer end. A weight W, slides freely on the rod and is
dropped on the collar through height. The falling weight causes impact load and that leads to extension δl
i
,
and tensile stress σL
i
, (subscript i refers to impact load).
Now
external work done = energy stored in the rod
W(h + δL
i
) = (σ
i
2
/2E) × volume = (σ
i
2
/2E) × AL
W[h + (σ
i
.L)/L] = (σ
i
2
/2E) × AL
Or, (AL/2E)σ
i
2
– (WL/E)σ
i
– Wh = 0
Solution of this quadratic equation gives
2
2
2
4 ( )
2 2
2
2
WL WL AL
Wh
E E E W W WhE
AL
A A AL
E
¸ _ ¸ _
t − × × −
¸ , ¸ ,
· t +
×
Negative sign is inadmissible as the stress cannot be compressive when
the bar gets elongated.
s
i
=
2 2
1 1 1
W W h AE W h AE
A A WL A WL
1
+ + · + +
1
1
¸ ]
Following relations are worth noting:
(i) If δl
i
is neglected as compared to h, (if δL × 1000 < < < h), then
Wh =
2
i
E
σ
⋅ AL and σ
i
=
2W hE
Al
(ii) If h = 0, then
W =
2
2
i
L
E E
σ
σ⋅
· ⋅ AL and σ
i
=
2W
A
Note: Strain energy becomes smaller and smaller as cross sectional area of the bar is increased over
more and more of its length.
Elongation
P
L
o
a
d
l
Fig 14.49
L
i
L
h
Collar
Rod
W
Fig 14.50
378 / Problems and Solutions in Mechanical Engineering with Concept
Now strain energy for any types of load = U =
2
1
2
AL
E
σ
⋅
where σ = P/A for qradually applied load
= 2 P/A for suddenly applied load
=
2
1 1
W hAE
A W L
1
+ +
1
⋅
¸ ]
for impact load
Q. 69: Three bars of equal length and having, crosssectional areas in ratio l:2:4, are all subjected
to equal load. Compare their strain energy. (Dec03)
Sol.: Let A, 2A, 4A be the areas.
Loads are equal = P
Each have equal length = L
Now σ
1
= P/A; σ
2
= P/2A; σ
3
= P/4A;
Since strain energy = U = σ
2
.Vol/2E
Strain energy in first bar = U
1
= σ
1
2
.Vol/2E = (1/2E)(P/A)
2
.AL = P
2
.L/2AE
Strain energy in second bar = U
2
= σ
2
2
.Vol/2E = (1/2E)(P/2A)
2
.2AL = P
2
.L/4.AE
Strain energy in third bar = U
3
= σ
3
2
.Vol/2E = (1/2E)(P/4A)
2
.4AL = P
2
.L/8AE
Now;
U
1
: U
2
: U
3
= 1 : ½ : ¼
Q. 70: A 1 m long steel rod of rectangular section 80 mm × 40 mm is subjected to an axial tensile
load of 200kN. Find the strain energy and maximum stress produced in it for the following
cases when load is applied gradually and when load falls through a height of 100 mm. Take
E = 2 × l0
5
N/mm
2
. (May2005)
Sol.: When gradually applied load
σ = P/A
= (200 × 1000)/(80 × 40) = 62.5 N/mm
2
Strain energy = σ
2
× volume / 2E
= (62.5
2
× 80 × 40 × 1000) / ( 2 × 2 × 10
5
)
= 31250 Nmm .......ANS
When load falls through a height of 100mm
σ =
2
2 W W EWh
A A AL
¸ _
+ +
¸ ,
=
2
5
200 1000 200 1000 2 2 10 100 200 1000
80 40 80 40 80 40 1000
¸ _ × × × × × × ×
+ +
× × × ×
¸ ,
= 62.5 +
2 5
(625) 25 10 + ×
= 62.5 + 3906.25 2500000 + = 62.5 + 1582.37
= 1644.87 N/mm
2
.......ANS
U =
2
5
1 1644 87
2 2 10
⋅
× ×
×
80 × 40 × 1000
U = 21644778.5 Nmm .......ANS
Simple Stress and Strain / 379
Q. 71: A bar of 1.2 cm diameter gets stretched by 0.3 cm under a steady load of 8KN. What stress
would be produced in the bar by a weight of 0.8KN. Which falls through 8 cm before
commencing the stretching of the rod, which is initially unstressed. Take E = 200GN/m
2
.
(UPTUQB)
Sol.: Crosssectional area of the bar = A = π/4.d
2
= π/4(1.2/1000)
2
= 0.0001131 m
2
Steady load = 8 kN
Elongation under steady load, δL = 0.3 cm = 0.003 m
Falling load = 0.8 kN
Distance through which the weight falls, h = 8 cm = 0.08 m
Modulus of elasticity, E = 200 GN/m
2
Instantaneous stress produced due to the falling load, σ
i
= ?
In order to first find length of the bar, using the following relation, we have
δL = WL/AE or; L = δL.A.E/W
L =
9
0.003 0.0001131 200 10
8 1000
× × ×
×
= 8.48 m
Now to calculate instantaneous stress si due to falling load 0.8 kN using the following relation, we have
σ
i
=
9
0.8 1000 2 0.08 0.0001131 200 10 2
1 1 1
0.0001131 0.8 1000 8.84
W hAE
A WL
1 1 × × × × ×
+ + · × +
1 1
× ×
1 ¸ ] ¸ ]
= 7073386.3 (1 + 23.119)
σ σσ σσ
i
= 170.6 × 10
6
N/m
2
or 170.6 MN/m
2
. .......ANS
Q. 72: A bar 3m long and 5 cm diameter hands vertically and has a collar securely attached at the
lower end. Find the maximum stress induced when;
(i) a weight 2.5KN falls from 12 cm on the collar
(ii) a weight of 25KN falls from 1 cm on the collar Take E = 2.0 × 105 N/mm
2
. (UPTUQB)
Sol.: Crosssectional area of the bar = A = π/4.d
2
= π/4(50)
2
= 1962.5 mm
2
(i)Instantaneous elongation of bar ;
δL = WL/AE = (2500 × 3000)/(1962.5 × 2 × 10
5
) = 0.01911mm
This elongation is very small as compared to 120mm height of fall and as such can be neglected.
Accordingly stress induced in the bar can be worked out by using the relation. σ
1
2WhE
A L
·
⋅
=
5
2 2500 120 2 10
1962.5 3000
× × × ×
×
= 142.77N/mm
2
.......ANS
(ii) Instantaneous elongation
δL = WL/AE = (25000 × 3000)/(1962.5 × 2 × 10
5
) = 0.1911mm
This elongation is comparable to 10 mm height of fall. Further the falling weight is large and hence
extension of the bar cannot be neglected. Accordingly stress induced in the bar is worked out from
the relation
σ
i
=
2
1 1
W hAE
A WL
1
+ +
1
1
¸ ]
380 / Problems and Solutions in Mechanical Engineering with Concept
=
5
25000 2 1962.5 2 10 10
1 1
1962.5 25000 3000
1
× × × ×
× + + 1
×
1
¸ ]
= 143.68 N/mm
2
.......ANS
Q. 73: A load of 100 N falls by gravity a vertical distance of 300 cm When it is suddenly stopped by
a collar at the end of a vertical rod of length 6m and diameter 2 cm. The top of the bar is
rigidly fixed to a ceiling. calculate the maximum stress and the strain induced in the bar. Table
E = 1.96 × 10
7
N/cm
2
. (UPTUQB)
Sol.: Given data:
Weight of the object = 100N
Height of fall = 300 cm
Length of vertical rod = 6m = 600cm
Diameter of rod = 2cm
E = 1.96 × 10
7
N/cm
2
Crosssectional area of the rod
= A = π/4.d
2
= π/4.(2)
2
= 3.142 cm
2
Maximum stress induced in the bar;
σ
i
=
2
1 1
W hAE
A WL
1
+ +
1
1
¸ ]
=
7
100 2 300 3.142 1.96 10
1 1
3.142 100 600
1
× × × ×
× + + 1
×
1
¸ ]
= 25007.945 N/cm
2
.......ANS
Strain induced in the bar due to impact e
i
;
We know that
ei = σ
i
/E = 25007.945/(1.96 × 10
7
) = 0.001276
ei = 0.001276 .......ANS
Q. 74: A steel specimen 1.5cm
2
in cross section stretches 0.05mm over 5 cm gauge length under an
axial load of 30 KN. Calculate the strain energy stored in the specimen at this point. If the
load at the elastic limit for specimen is 50KN, Calculate the elongation at the elastic limit.
Sol.: Cross sectional area of specimen A = 1.5 cm
2
= 1.5 × 10
–4
m
2
Increase in length over 5cm gauge length δL = 0.05mm = 0.05 × 10
–3
m
Axial load W = 30KN
Load at elastic limit = 50KN
Strain energy stored in the specimen
U = σ
2
A.L/2E = ½. W. δL = ½ × (30 × 1000) × 50 × 10
–3
U = 0.75 Nm or J .......ANS
Also E = W.L/A.δL
= {(30 × 1000) × (5/100)}/{(1.5 × 10
–4
) × (0.05 × 10
–3
)}
= 200 × 10
9
= 200 GN/m
2
Elongation at elastic limit, δL
δL = W.L/A.E = {(50 × 1000) × (5/100)}/{(1.5 × 10
–4
) × (200 × 10
9
)}
δ δδ δδL = 0.0000833 m = 0.0833 mm .......ANS
Simple Stress and Strain / 381
Q. 75: A wagon weighing 35KN is attached to a wire rope and moving down an incline plane at speed
of 3.6Km/hr when the rope jams and the wagon is suddenly brought to rest. If the length of the
rope is 60 meters at the time of sudden stoppage, calculate the maximum instantaneous stress
and maximum instantaneous elongation produced. Diameter of rope = 30mm. E = 200GN/m
2
.
Sol.: Weight of the wagon W = 35KN
Speed of the wagon, v = 3.6 km/hr = 1m/sec
Diameter of the rope, d = 30mm = 0.03m
Length of the rope at the time of sudden stoppage, L = 60m
Maximum instantaneous stress σ
i
;
The kinetic energy of the wagon = ½.mv
2
= Strain energy
= ½. (35 × 1000 /9.81) × 12 = 1783 Nm or J ...(i)
This energy is to be absorbed by the rope at a stress σ
i
Now, strain energy stored = σ
i
2
.A.L/2E
= {σ
i
2
× π/4 × (0.032) × 60}/(2 × 200 × 109) = 0.0106 σ
i
2
/1011 ...(ii)
Equating equation (i) and (ii)
σ σσ σσ
i
= 129.69 × 10
6
N/m
2
= 129.69 MN/m
2
.......ANS
Maximum instantaneous elongation of the rope, δL
Using the relation,
δL = σ
i
.L/E = ( 129.69 × 10
6
× 60)/(200 × 10
9
)
δ δδ δδL = 389.07 × 10
–4
m = 38.9 mm .......ANS
Q. 76: A steel wire 2.5 mm diameter is firmly held in clamp from which it hangs vertically. An anvil
the weight of which may be neglected, is secured to the wire 1.8 m below clamp. The wire is
to be tested allowing a weight bored to slide over the wire to drop freely from 1 m above the
anvil. Calculate the weight required to stress the wire to 1000MN/m
2
assuming the wire to be
elastic upto this stress. Take: E = 210 GN/m
2
.
Sol.: Diameter of the steel wire, d = 2.5mm = 2.5 × 10
–3
m
Height of fall, h = 1m
Length of the wire, L = 1.8m
Instantaneous stress produced σ
i
= 1000 MN/m
2
E = 210 GN/m
2
Weight required to stress the wire, W;
Instantaneous extension,
δL = σ
i
.L/E = (1000 × 10
6
× 1.8)/(210 × 10
9
) = 0.00857 m
Equating the loss of potential energy to strain energy stored by the wire, we have
W(h + δL) = σ
i
2
.A.L/2E
W (1 + 0.00857) = {(1000 × 10
6
)
2
× π/4(2.5 × 10
–3
)
2
× 1.8}/{2 × 210 × 10
9
}
W = 20.85N .......ANS
Q. 77: Explain the concept of Complementary shear stress. (May–05(C.O.), Dec05)
Sol.: It states that a set of shear stresses across a plain is always accompanied by a set of balancing shear
stresses across the plane and normal to it.
Shear force on face AB = τ.AB
Shear force on face CD = τ.CD ...(i)
These parallel and equal forces form a couple.
The moment of couple = τ.AB.AD or τ.CD.AD ...(ii)
382 / Problems and Solutions in Mechanical Engineering with Concept
C C
D D
B B A A
t
t t
t
t
( ) a ( ) b
Fig. 14.51
For equilibrium they must be a restoring couple.
Shear force on face AD or BC = τ
l
.AD or τ
l
.BC ...(iii)
They also form a couple as
= τ
l
.AD.AB or τ
l
.BC.AB ...(iv)
The moment of two couple must be equal;
τ.AB.CD = τ
l
.AD.AB or τ = τ
l
τ
l
is called complementary shear stress
Q. 78: Explain longitudinal strain and lateral strain. What is the relation between longitudinal and
lateral strain.
Sol.: Longitudinal strain is the longitudinal deformation expressed as a dimensionless constant and is
defined as the ratio of change in length to the initial length.
e = δL/L
Lateral strain is the lateral deformation expressed as a dimensionless constant and is defined as the
ratio of change in lateral dimension to the initial lateral dimension
Lateral strain = Change in lateral dimension/Original lateral dimension
= Change in diameter/Original diameter; For circular bar
= Change in width or depth/Original width or depth; For rectangular bar
LATERAL STRAIN = POISSON'S RATIO X LONGITUDINAL STRAIN
Q. 79: What do you meant by shear stress and shear strain?
Sol.: Stress and strain produced by a force tangential to the surface of a body are known as shear stress
and shear strain
Shear Stress
Shear stress exists between two parts of a body in contact, when the two parts exert equal and opposite
force on each other laterally in a direction tangential to their surface of contact.
Figure 14.52 shows a section of rivet subjected to equal and opposite forces P causing sliding of the
particles one over the other.
From figure it is clear that the resisting force of the rivet must be equal to P. Hence, shearing stress
τ is given by
τ = total tangential force/(Surface Area)
Simple Stress and Strain / 383
P =P¢
P¢
P¢
P
P
P
P
P
P
Fig. 14.52
= P/A ; A = Π.d.t
The tensile stress and compressive stress are also known as “direct stresses" and shearing stress as
"tangential stress”.
The common examples of a system involving shear stress are riveted and welded joint, towing device,
punching operation etc.
Shear Strain
In case of a shearing load, a shear strain will be produced
which is measured by the angle through which the body
distorts.In Fig. 14.53 is shown a rectangular block LMNP
fixed at one face and subjected to force F. After application
of force, it distorts through an angle Φ and occupies new
position LM’N’P. The shear strain (e
s
) is given by
e
s
= NN’/NP = tanΦ
= Φ (radians) ...... since Φ is very small.
The above result has been obtained by assuming NN’ equal
to arc (as NN’ is small) drawn with centre P and radius
PN.
Q. 80: A steel punch can be worked to a compressive stress
of 800 N/mm
2
. Find the least diameter of hole which
can be punched through a steel plate 10 mm thick if
its ultimate shear strength is 350 N/mm
2
.
(UPTU QUESTION BANK)
Sol.: Let d be the diameter of hole in mm
Area being sheared = π.d.t = πd × 10 = 10πd mm
2
Force required to punch the hole = Ultimate shear
strength x area sheared
= 350 × 10πd = 3500πd (N)
Cross sectional area of the hole = (π/4)d
2
mm
2
P L
N
F
N ¢ M M ¢
Fig 14.53
Punch
Plate
P
t
d
Fig. 14.54
384 / Problems and Solutions in Mechanical Engineering with Concept
Compressive stress on the punch
σ
C
= 3500πd/{(π/4)d
2
} = 14000/d
But σ
C
is limited to 800 N/mm
2
and therefore
800 = 14000/d
d = 17.5mm .......ANS
Q. 81: Write short notes on:
(a) Modulus of Rigidity or Shear Modulus (G)
(b) Hydrostatic stress
(c) Volumetric strain (e
V
)
(d) Bulk Modulus or Volume modulus of elasticity (K)
(e) Poisson's ratio (µ)
(a) Modulus of Rigidity or Shear Modulus
It is the ratio between shear stress(τ) and shear strain(e
s
). It is denoted by G. It is the same as Shear modulus
of elasticity
G = τ/e
s
(b) Hydrostatic Stress
When a body is immersed in a fluid to a large depth, the body gets subjected to equal external pressure
at all points of the body. This external pressure is compressive in nature and is called hydrostatic stress.
(c) Volumetric Strain
The hydrostatic stress cause change in volume of the body, and this change of volume per unit is called
volumetric strain e
v
.
Or, It is defined as the ratio between change is volume and original volume of the body, and is denoted
by e
v
,
e
v
= change in volume/ original volume = δV/V
e
v
= e
x
+ e
y
+ e
z
i.e., Volumetric strain equals the sum of the linear normal strain in x, y and z direction.
(d) Bulk Modulus or Volume Modulus of Elasticity
It may be defined as the ratio of normal stress(on each face of a solid cube) to volumetric strain. It is
denoted by K. It is the same as Volume modulus of elasticity. K is a measure of the resistance of a material
to change of volume without change of shape or form.
K = Hydrostatic pressure / Volumetric strain.
n
V
K
e
σ
·
(e) Poisson’s Ratio (µ)
If a body is subjected to a load, its length changes; ratio of this change in length to the original length
is known as linear or primary strain. Due to this load, the dimensions of the body change; in all directions
at right angles to its line of application the strains thus produced are called lateral or secondary or
transverse strains and are of nature opposite to that of primary strains. For example, if the load is tensile,
there will be an increase in length and a corresponding decrease in crosssectional area of the body (Fig.
14.55). In this case, linear or primary strain will be tensile and secondary or lateral or transverse strain
compressive.
Simple Stress and Strain / 385
Initial transverse
dimension
Final transverse
dimension
Longitudinal
elongation
Fig. 14.55
Poisson’s ratio is the ratio of lateral strain to the longitudinal strain. It is an elastic constant having
the value always less than 1. It is denoted by ‘µ’ ( l/m)
Poisson’s Ratio (µ) = Lateral Strain / Longitudinal Strain; always less than 1.
Sl. No. Material Poisson’s ratio
1. Aluminium 0.330
2. Brass 0.340
3. Bronze 0.350
4. Cast iron 0.270
5. Concrete 0.200
6. Copper 0.355
7. Steel 0.288
8. Stainless steel 0.305
9. Wrought iron 0.278
Q. 82. A steel bar 2 m long, 20 mm wide and 10 mm thick is subjected to a pull of 20KN in the
direction of its length. Find the changes in length, breadth and thickness. Take E = 2 × 10
5
N/mm
2
and poisons ratio 0.30. (UPTU QUESTION BANK)
Sol.: Longitudinal strain = δL/L = stress/ modulus of elasticity
= (P/A)/E = P/AE = 20 × 10
3
/{(20 × 10) × (2 × 10
5
)} = 0.5 × 10
–3
Change in length δL = longitudinal strain x original length
= (0.5 × 10
–3
) × (2 × 10
3
) =1.0 mm (increase)
Lateral strain = Poisson's ratio × longitudinal strain =0.3 × (0.5 × 10
–3
)=0.15 × 10
–3
The lateral strain equals δb/b and δt/t
Change in breadth δb = b × lateral strain = 20 × (0.15 × 10
–3
)
= 3 × 10
–3
mm (decrease) .......ANS
Change in thickness δt = t × lateral strain = 10 × (0.15 × 10
–3
)
= 1.5 × 10
–3
mm (decrease) .......ANS
Q. 83: A bar of steel 25 cm long, of rectangular crosssection 25 mm by 50 mm is subjected to a
uniform tensile stress of 200 N/mm
2
along its length. Find the changes in dimensions.
E = 205,000 N/mm
2
Poisson's ratio = 0.3. (UPTU QUESTION BANK)
Sol.: do your self
386 / Problems and Solutions in Mechanical Engineering with Concept
Q. 84: A 500 mm long bar has rectangular crosssection 20 mm × 40 mm. The bar is subjected to:
(i) 40 kN tensile force on 20 mm × 40 mm face.
(ii) 200kN compressive force on 20 mm × 500 mm face
(iii) 300kN tensile force on 40 mm × 500 mm face.
Find the change in dimensions and volume, if E = 2 × 10
5
N/mm
2
and poisson ratio = 0.3.
[U.P.T.U. March–02]
200 KN
40
40 KN
40 KN
2
0
300 KN 200 KN
500
Fig 14.56
Sol.: E = 2 × 10
5
N/mm
2
,
1
m
= µ = ⋅3
σ
x
=
3
40 10
20 40
x
x
P
A
×
·
×
= 50 N/mm
2
(tensile stress)
σ
y
=
3
200 10
20 500
y
y
P
A
×
·
×
= 20 N/mm
2
(compressive stress)
= – 20 N/mm
2
(tensile stress)
σ
z
=
3
300 10
40 500
z
z
P
A
×
·
×
= 15 N/mm
2
(tensile stress)
e
x
=
y
x
z
E mE mE
σ
σ
− −
=
5 5 5
50 20 0 3 15 0 3
2 10 2 10 2 10
× ⋅ × ⋅
+ −
× × ×
e
x
= 000257
e
y
=
y
x z
E m E mE
σ
σ σ
− −
×
Simple Stress and Strain / 387
=
5 5
20 50 0 3
2 10 2 10
− × ⋅
−
× ×
e
y
= – 0001975
e
z
=
y
x z
E mE mE
σ
σ σ
− −
=
5 5 5
15 50 0 3 20 0.3
2 10 2 10 2 10
× ⋅ ×
− +
× × ×
e
z
= ⋅00003
Change in length in x direction ∆l = l × e
x
= 500 × ⋅000257
∆lx = ⋅1285 mm
in y direction ∆b = e
y
× b
= – ⋅000197 × 90
= – ⋅0076 mm (decreases)
in z direction ∆w = e
z
× w
= ⋅00003 × 20
eυ = e
x
+ e
y
+ e
z
= ⋅00009
∆v = υ × e
υ
= 20 × 40 × 500 × ⋅00009
= 36 mm
3
Q. 85: A 2m long rectangular bar of 7.5 cm × 5 cm is subjected to an axial tensile load of 1000kN.
Bar gets elongated by 2mm in length and decreases in width by 10 × 10
–6
m. Determine the
modulus of elasticity E and Poisson's ratio of the material of bar. (Dec–2002)
d cm = 5
0 = 75 cm 2 m
1000 kN
1000 kN
Bar
Fig. 14.57
Sol.: Given:
L = 2m;
B = 7.5cm = 0.075m;
D = 5cm = 0.05m
P = 1000kN
δL = 2mm = 0.002m
δb = 10 × 10
–6
m.
Longitudinal strain e
L
= e
t
= δL/L = 0.002/2 = 0.001
Lateral strain = δb/b = 10 × 10
–6
/0.075 = 0.000133
Tensile stress (along the length) σ
t
= P/A = (1000 × 1000)/(0.075 × 0.05) = 0.267 × 10
9
N/m
2
Modulus of elasticity, E = σ
t
/e
t
= 0.267 × 10
9
/0.001 = 267 × 10
9
N/m
2
.......ANS
Poisson’s ratio = Lateral strain/ Longitudinal strain = (δb/b)/(δL/L) = 0.000133/0.001
388 / Problems and Solutions in Mechanical Engineering with Concept
Q. 86: Prove that E = 3K (1 – 2µ).
Sol.: Consider a cubical element subjected to volumetric stress
σ which acts simultaneously along the mutually perpendicular
x, y and zdirection.
The resultant strains along the three directions can be
worked out by taking the effect of individual stresses.
Strain in the xdirection,
e
x
= strain in xdirection due to σ
x
– strain in xdirection
due to σ
Y
– strain in xdirection due to
σ
Z
= σ
x
/E – µ.σ
y
/E – µ.σ
z
/E ...(i)
But σ
x
= σ
y
= σ
z
= σ
e
x
=
(1 2 )
E E E E
σ σ σ σ
−µ −µ · − µ
Likewise e
y
=
(1 2 )
E
σ
− µ
and e
z
=
(1 2 )
E
σ
− µ
Volumetric strain
e
x
= e
x
+ e
y
+ e
z
=
3
(1 2 )
E
σ
− µ
Now, bulk modulus
K=
volumetric stress
volumetric strain
=
3
3(1 2 )
(1 2 )
E
E
σ
·
σ
− µ
− µ
or, E = 3K (1 – 2µ)
E= 3K (1 – 2µ) ...(i)
Q. 87: Derive the relation E = 2C (1 + 1/m) where; E =Young's modulus, C = modulus of rigidity
1/m = Poisson's ratio. (Dec–2004, May–2005)
Sol.: Consider a cubic element ABCD fixed at the bottom face and
subjected to shearing force at the top face. The block experiences
the following effects due to this shearing load:
shearing stress t is induced at the faces DC and AB.
complimentary shearing stress of the same magnitude is
set up on the faces AD and BC.
The block distorts to a new configuration ABC’D’.
The diagonal AC elongates (tension) and diagonal BD
shortens (compression).
Longitudinal strain in diagonal AC
=
AC AC AC AE EC
AC AC AC
′ ′ ′ − −
· ·
...(i)
where CE is perpendicular from C onto AC′
Fig 14.58
x
z
s
x
s
x
s
x
s
z
s
y
s
y
y
B A
t
t
t
C¢ C D¢ D
F
E
Fig 14.58
Simple Stress and Strain / 389
Since extension CC′ is small, ∠ACB can be assumed to be equal ∠ACB which is 45°. Therefore
EC′ = CC′ cos 45° =
2
CC′
Longitudinal strain =
tan
2 2 2 2 2 2
CC CC CC
BC AC BC
′ ′ ′ φ φ
· · · ·
×
...(ii)
Where, Φ = CC’/BC represents the shear strain
In terms of shear stress t and modulus of rigidity C, shear strain = τ/C
longitudinal strain of diagonal AC = τ/2C ...(iii)
The strain in diagonal AC is also given by
= strain due to tensile stress in AC – strain due to compressive stress in BD
=
(1 )
E E E
τ τ τ ¸ _
− − µ · + µ
¸ ,
...(v)
From equation (iv) and (v), we get
=
2C E
τ τ
·
(1 + µ)
or E = 2C (1 + µ) ...(vi)
Q. 88: What is the relation between elastic constant E, C and K.?
Sol.: With reference to the relations (1) and (6) derived above,
E = 2C(1 + µ) = 3K(1–2µ)
To eliminate µ from these two expressions for E, we have
µ =
1
2
E
C
−
and 3 1 2 1
2
E
E k
C
1 ¸ _
· − −
1
¸ , ¸ ]
or E = 3K
3
1 2 3 3 9
E E KE
k K
C C C
1 ¸ _ 1
− − · − · −
1 1
¸ , ¸ ] ¸ ]
or E +
3KE
C
=
3
9 ; 9
C K
K E K
C
+ ¸ _
·
¸ ,
or E =
9
3
KC
C K +
E = 2C (1 + µ) = 3K (1 – 2µ) =
9
3
KC
C K +
Q. 89: A circular rod of 100 mm diameter and 500 m long is subjected to a tensile force of 1000KN.
Determine the modulus of rigidity, bulk modulus and change in volume if poisons ratio
= 0.3 and Young's Modulus = 2 × 10
5
N/mm
2
. (UPTU QUESTION BANK)
Sol.: Modulus of rigidity
G = E/2.(1 + µ) = 2 × 10
5
/ 2(1 + 0.3) = 0.769 × 10
5
N/mm
2
.......ANS
Bulk modulus
K = E/3(1 – 2µ) = 2 × 105/3(1 – 2 × 0.3) = 1.667 × 10
5
N/mm
2
.......ANS
390 / Problems and Solutions in Mechanical Engineering with Concept
Normal stress σ = P/A = 1000 × 10
3
/ π/4(100)
2
= 127.388 N/mm
2
Linear (Longitudinal ) strain = δL/L = Normal stress/ Young's modulus
= 127.388/ 2 × 10
5
= 0.000637
Diametral (Lateral) strain
= δd/d = µ. δL/L = 0.3 × 0.000637 = 0.0001911
Now volume of a circular rod
= V = π/4.d
2
.L
Upon differentiation
δV = π/4[ 2.d.δd.L + d
2
.δL]
Volumetric strain
δV/V = π/4[ 2.d.δd.L + d
2
.δL]/ π/4.d
2
.L = 2δd/d + δL/L
Substituting the value of δd/d and δL/L as calculated above, we have
δV/V = 2 (0.0001911) + 0.000637 = 0.0002548
The ive sign with δd/d stems from the fact that whereas the length increases with tensile force, there
is decrease in diameter.
Change in volume
δV = 0.0002548 [π/4(100)
2
× 500] = 1000.09 mm
3
.......ANS
Q. 90: What do you know about properties of a metal?
Sol.: Different materials posses different properties in varying degree and therefore behave in different
ways under given conditions. These properties includes mechanical properties, electrical properties, thermal
properties, chemical properties,magnetic properties and physical properties. We are basically interested in
knowing as to how a particular material will behave under applied load i.e. in knowing the mechanical
properties.
Q 91: What is mechanical properties of material? Define strength.
Sol.: Those characteristics of the materials which describe their behaviour under external loads are known
as Mechanical Properties. The most important and useful mechanical properties are:
Strength:
It is the resistance offered by a material when subjected to external loading. So Stronger the material
the greater the load it can withstand. Depending upon the type of load applied the strength can be tensile,
compressive,shear or torsional.
Proportional Limit
Elastic Limit
Breaking Point
Plastic Zone
Elastic Zone
Unit Elongation
U
n
i
t
L
o
a
d
Ultimate Strength Yield Point
Fig 14.59 A Typical StressStrain Curve
Simple Stress and Strain / 391
The stress at the elastic limit is known as yield Strength.
And the maximum stress before the fracture is called ultimate strength. While in tension the ultimate
strength of the material represents it tenacity.
Q. 92: Write short notes on:
Sol.: Elasticity, stiffness, Plasticity, Malleability, Ductility, Brittleness, Toughness
Elasticity: Elasticity of a material is its power of coming back to its original position after deformation
when the stress or load is removed.Elasticity is a tensile property of its material.
Proportional Limit: It is the maximum stress under which a material will maintain a perfectly uniform
rate of strain to stress.
Elastic Limit: The greatest stress that a material can endure without taking up some permanent set is
called elastic limit.
Stiffness: It is the property of a material due to which it is capable of resisting deflection or elastic
deformation under applied loads.also called rigidity.
The degree of stiffness of a material is indicated by the young’s modulus. The steel beam is stiffer or
more rigid than aluminium beam.
Plasticity: The plasticity of a material is its ability to change some degree of permanent deformation
without failure. This property is widely used in several mechanical processes like forming, shaping, extruding,
rolling etc. Due to this properties various metal can be transformed into different products of required shape
and size. This conversion into desired shape and size is effected either by the application of pressure , heat
or both. Plasticity increase with increase in temp.
Malleability: Malleability of a material is its ability to be flattened into their sheets without creaking
by hot or cold working. Aluminum, copper, tin lead steel etc are malleable metals.
Ductility: Ductility is that property of a material, which enables it to draw out into thin wire. Mild
steel is a ductile material. The percent elongation and the reduction in area in tension is often used as
empirical measures of ductility.
Brittleness: The brittleness of a material is the property of breaking without much permanent distortion.
There are many materials, which break or fail before much deformation take place. Such materials are
brittle e.g. glass, cast iron. Therefore a nonductile material is said to be a brittle material. Usually the
tensile strength of brittle materials is only a fraction of their compressive strength. A brittle material should
not be considered as lacking in strength. It only shows the lack of plasticity.
Toughness: Toughness is a measure of the amount of energy a material can absorb before actual
fracture or failure takes place. The toughness of a material is its ability to withstand both plastic and elastic
deformation. “The work or energy a material absorbs is called modulus of toughness”
For Ex: If a load is suddenly applied to a piece of mild steel and then to a piece of glass the mild steel
will absorb much more energy before failure occurs. Thus mild steel is said to be much tougher than a glass.
Q. 93: Write short notes on: Hardness, Impact Strength
Sol.: Hardness: Hardness is defined in terms of the ability of a material to resist screeching, abrasion,
cutting, indentation or penetration. Many methods are now in use for determining the hardness of a material.
They are Brinell, Rockwell and Vickers.
Hardness of a metal does not directly related to the hardenability of the metal. Hardenability is indicative
of the degree of hardness that the metal can acquire through the hardening process. i.e., heating or quenching.
Impact Strength: It can be defined as the resistance of the material to fracture under impact loading,
i.e under quickly applied dynamic loads.Two standard tests are normally used to determine this property.
1. The IZOD impact test.
2. The CHARPY test.
392 / Problems and Solutions in Mechanical Engineering with Concept
Q. 94: What is fatigue; how it is related to creep?
Sol.: Fatigue : Failure of a material under repeated stress is known as fatigue and the maximum stress that
a metal can withstand without failure for a specific large number of cycle of stress is called Fatigue limit.
Creep: The slow and progressive deformation of a material with time at constant stress is called creep.
There are three stages of creep. In the first one,the material elongates rapidly but at a decreasing rate. In
the second stage,the rate of elongation is constant. In the third stage, the rate of elongation increases rapidly
until the material fails. The stress for a specifid rate of strain at a constant temperature is called creep
strength.
Creep Curve and Creep Testing: Creep Test is carried out at high temp. A creep curve is a plot of
elongation of a tensile specimen versus time, For a given temp. and under constant stress. Tests are carried
out for a period of a few days to many years.
Primary
Fracture
Tertiary
Creep
Secondary Creep
Instantaneous elongation
Time
S
t
r
a
i
n
Creep
Fig 14.60
Creep curve shows four stages of elongation:
1. Instantaneous elongation on application of load.
2. Primary creep:Work hardening decreases and recovery is slow.
3. Secondary creep:Rate of work hardening and recovery processes are equal.
4. Tertiary creep: Grain boundary cracks.Necking reduces the cross sectional area of the test specimen.
Compound Stress and Strain / 393
+0)264
15
CCMPCÜND STRESS /ND STR/lN
Q. 1: Define Compound Stress.
Sol.: Simple stresses mean only tensile stress or compressive stress or only shear stress. Tensile and
compressive stresses act on a plane normal to the line of action of these stresses, and shear stress acts on
a plane parallel to the line of action of this stress. But when a plane in a strained body is oblique to the
applied external force, this plane may be subjected to tensile or compressive stress and shear stress. i.e.;
Such a system or a plane in which direct or normal stresses and shear stresses act simultaneously are
called compound stress or complex stress.
Q. 2: Define the concept of plane stress.
Sol.: In a cubical element of a strained material is acted on by stresses
acting on only two pairs of parallel planes and the third pair of parallel
planes is free from any stress, it is said that the element is under the
action of plane stresses. So, plane stress condition can be called two
dimensional stress condition. Let a cubical element ABCD taken from
a strained body be subjected to normal stresses σ
1
and σ
2
and shear
stress ¹.
Planes AD and BC are subjected to normal stress σ
1
and shear
stress¹, and planes AB and CD are subjected to normal stress σ
2
and
shear stress¹. But no stress acts on the third pair (front face and rear face
of ABCD) of parallel planes. Hence it is said that the cubical element
ABCD is under the action of plane stresses.
Q. 3: Explain Principal Planes and Principal Stresses. (Dec–00)
Sol.: When an element in a strained body is under the action of plane stresses, it is found that there exist
two mutually perpendicular planes of the element on which normal stresses
are maximum and minimum and no shear stress acts on these planes. These
planes are called principal planes.
From the figure 15.2; The principle planes can be determined by
2
tan 2
xy
x y
τ
θ ·
σ − σ
Since shear stress on these plane are zero, therefore they are also
called the shear free plane.
The maximum and minimum principal stresses acting on principal
planes are called principal stress.
A
f
1
f
1
f
2
f
2
q
B
q
q
C D
Fig. 15.1
Fig. 15.2
y
y
xy
xy
xy
x
x
394 / Problems and Solutions in Mechanical Engineering with Concept
The principal stress having maximum value is called “major principal stress” and the principal stress
having minimum value is called “minor principal stress”.
2 2
1,2
1
( ) ( ) (4 )
2
x y x y xy
1
σ · σ + σ t σ −σ τ
1
¸ ]
Major principal stress =
2 2
1
1
( ) ( ) (4 )
2
x y x y xy
1
σ · σ + σ t σ −σ τ
1
¸ ]
Minor principal stress =
2 2
2
1
( ) ( ) (4 )
2
x y x y xy
1
σ · σ + σ t σ − σ τ
1
¸ ]
Where σ
x
, σ
y
be the direct stresses in x and y direction and σ
xy
be the shear stress normal to
plane xy.
Resultant stress is given by the equation
2 2
r n t
σ · σ + τ
Q. 4: Derive the equation for principal stresses and principal planes for an element subjected to
compound stresses.
Sol.: For the state of stress shown in fig. the normal stress and shear stress on any oblique plane inclined
at an angle θ can be determined by,
y
y
x
x
E
n
( ) AE
y
( ) AB
x
(BE)
( ) AE
E
E B
Fig 15.3
2 2
x y x y
n
σ + σ σ −σ ¸ _ ¸ _
σ · +
¸ , ¸ ,
cos 2θ + τ
xy
sin 2θ
and
2
x y
σ − σ ¸ _
τ ·
¸ ,
sin 2θ − τ
xy
= cos 2θ
Since the principal plane should carry only normal stress, shear stress acting on it is to be zero.
i.e.,
2
x y
σ − σ ¸ _
τ ·
¸ ,
sin 2θ − τ
xy
= cos 2θ
( )
2
x y
σ −σ 1
1
1
¸ ]
sin 2θ · τ
xy
= cos 2θ
or
tan 2θ =
2
xy
x y
1 τ
1
σ − σ
1
¸ ]
Compound Stress and Strain / 395
This gives one of the principal planes, other principal plane shall be perpendicular to this plane.
Substituting the value of θ in σ
n
expression by rearranging as below.
sin 2θ =
2 2
2
( ) 4
xy
x y xy
t τ
σ − σ + τ
and cos 2θ =
2 2
( )
( ) 4
x y
x y xy
t σ − σ
σ − σ + τ
2
2 2 2 2 2
( ) ( ) ( ) 2
2 2
( ) 4 ( ) 4
x y x y x y xy
n
x y xy x y xy
1
σ − σ σ − σ t σ −σ τ
1
σ · + +
1
σ − σ + τ σ − σ + τ
1
¸ ]
2 2
2 2
( ) ( ) 4
2
2 ( ) 4
x y x y xy
x y xy
σ + σ σ + σ t τ
· t
× σ + σ + τ
2 2
1,2
1
( ) ( ) 4
2
x y x y xy
1
σ · σ −σ t σ · σ + τ
1
¸ ]
It gives the values of principal stresses. The largest one is called major and
smaller is called minor principal stress.
Q. 5: Obtain the expression for maximum shearing stress and maximum shearing planes.
Sol.: Condition for maximum shearing stress, :
0
d
d
τ
·
θ
sin 2 cos 2 0
2
x y
xy
d
d
¹ ¹ σ − σ ¸ _
¹ ¹
θ − τ θ · ' ;
θ
¹ ¹ ¸ , ¹ )
or cos 2 sin 2
2
x y
xy
σ − σ
θ · τ θ
or
tan 2
2
x y
xy
¸ _ σ − σ
θ · −
τ
¸ ,
This expression give the value of a maximum shearing plane. Another plane which also carries maximum
shearing stress is normal to this plane. Substituting the value of θ in τ expression with the following
rearrangement.
2 2
( )
sin 2
( ) 4
x y
x y xy
t σ − σ
θ ·
σ − σ + τ
2 2
2
cos 2
( ) 4
xy
x y xy
τ
θ ·
σ − σ + τ
max
2
x y
σ − σ ¸ _
τ ·
¸ ,
sin 2θ – τ
xy
cos 2θ
max
2 2 2 2
( ) 2
2
( ) 4 ( ) 4
x y x y xy
x y xy x y xy
σ − σ t σ −σ τ ¸ _
τ · − τ
σ − σ + τ σ − σ + τ
¸ ,
Fig. 15.4
± ( – )
x y
± 2
xy
x
y
x
y
–
)
+
4
2
2
Fig. 15.5
±
(
–
)
x
y
± 2
xy
x
y
x
y
–
)
+
4
]
2
2
396 / Problems and Solutions in Mechanical Engineering with Concept
2 2
2 2
( ) 4
( ) 4
x y xy
x y xy
σ − σ + τ
·t
σ − σ + τ
2 2
max
1
( ) 4
2
x y xy
τ ·t σ − σ + τ
or
max
1
( )
2
x y
τ ·t σ − σ
Q. 6: Prove that the plane inclined at 45º to the plane carrying the greatest normal stress carries
the maximum shear stress.?
Sol.: The principal planes are determined by,
2
tan 2
xy
x y
τ
θ ·
σ − σ
and maximum shearing planes by,
tan 2
2
x y
xy
σ −σ
′ θ ·
− τ
Multiply both expressions,
tan 2θ . tan 2θ′ = – 1
When the products of slopes of two lines = – 1;then the two line will be orthogonal,
2θ′ = 2θ + 90°
θ′ = θ + 45°
i.e., Maximum shearing planes are always inclined at 45° with principal planes.
Q. 7: Prove the when stress are unequal and alike
(i) Normal stress
¸ _ ¸ _
¸ , ¸ ,
o +o o  o
o ÷ + cos 20
2 2
x y x y
n
(ii) Shear stress
)
1
= (o +o sin20
2
x y
τ
Sol.: By drawing the F.B.D. of wedge BCE, considering unit thickness of element.
Fig. 15.6 Fig. 15.7
y
y
x
x
E
B
D
A
C
y
( ) EC
x
(BE)
x
(BE)
( ) BE
E
B
C
Let σ
n
and τ be normal and shear stresses on the plane BE, Applying Equilibrium conditions to the
wedge BCE:
Compound Stress and Strain / 397
∑Fx = 0
σ
x
(BC) – σ
n
(BE) cos θ − τ (BE) cos (90 – θ) = 0
σ
x
(BC) – σ
n
(BE) cos θ − τ (BE) sin θ = 0 ...(i)
∑Fy = 0
σ
y
(EC) – σ
n
(BE) sin θ + τ (BE) sin (90 – θ) = 0
σ
x
EC – σ
n
(BE) sin θ + τ (BE) cos θ = 0 ...(ii)
Dividing equations (i) and (ii) by BE and replacing
cos
BC
BE
· θ
and
sin
EC
BE
· θ
σ
x
cos θ – σ
n
cos θ – τ sin θ = 0 ...(iii)
σ
y
sin θ – σ
n
sin θ + τ cos θ = 0 ...(iv)
Multiplying equation (iii) by cos θ, (iv) by sin θ and adding
σ
x
cos
2
θ – σ
n
cos
2
θ + σ
y
sin
2
θ − σ
n
sin
2
θ = 0
σ
n
= σ
x
cos
2
θ + σ
y
sin
2
θ
Putting cos
2
θ =
1 cos
2
+ θ
and sin
2
θ =
1 cos 2
2
− θ
1 cos 2 1 cos 2
2 2
n x y
+ θ − θ 1 1
σ · σ + σ
1 1
¸ ] ¸ ]
cos 2
2 2
x y s y
n
σ − σ σ −σ ¸ _ ¸ _
σ · + θ
¸ , ¸ ,
Multiplying equation (iii) by sin θ and (iv) by cos θ and then subtracting.
σ
x
sin θ cos θ – τ
sin
2
θ − σ
y
sin
θ cos
θ – τ
cos
2
θ = 0 [3 (sin
2
θ + cos
2
θ) = τ]
or (σ
x
– σ
y
) sin
θ cos
θ –
τ ·
0
or τ = (σ
x
– σ
y
) sin
θ cos
θ
τ =
1
( )sin 2
2
x y
σ − σ θ
.
Q. 8: Derive the expression for normal and shear stress on a plane AE inclined at an angle B with
AB subjected to direct stresses of compressive nature (both) of σ σσ σσ
x
and σ σσ σσ
y
on two mutually
perpendicular stresses as shown in fig 15.8.
y
y
x
x
E
D C
B A
n
( ) AE
y
( ) AB
x
(BE)
( ) AE
E
E B
Fig 15.8
Sol.: Consider the unit thickness of the element. Applying equations of equilibrium to the free body diagram
of wedge ABE.
Let σ
n
and τ be the normal and shear stresses on the plane AE.
398 / Problems and Solutions in Mechanical Engineering with Concept
∑Fx = 0
σ
x
(BE) – τ (AE) cos θ − σ
n
(AE) cos (90 – θ) = 0
σ
x
(BE) + τ (AE) cos θ + σ
n
(AE) sin θ = 0 ...(i)
∑Fy = 0
σ
y
(AB) + τ (AE) sin θ − σ
n
(AE) sin (90 – θ) = 0
σ
y
(AB) + τ (AE) sin θ − σ
n
(AE) cos θ = 0 ...(ii)
Dividing equations (i) and (ii) by AE and replacing
sin θ =
BE
AE
and cos θ =
AB
AE
– σ
x
sin θ + τ cos θ + σ
n
sin θ = 0 ...(iii)
– σ
x
cos θ + τ sin θ – σ
n
cos θ = 0 ...(iv)
Multiplying equation (iii) by sin θ, (iv) by cos θ and subtracting
– σ
x
sin
2
θ + σ
n
sin
2
θ – σ
y
cos
2
θ + σ
n
cos
2
θ = 0
σ
n
= σ
x
sin
2
θ + σ
y
cos
2
θ
Putting cos
2
θ =
1 cos 2
2
+ θ
and sin
2
θ =
1 cos 2
2
− θ
σ
n
=
2 2
x y x y
σ + σ σ −σ ¸ _ ¸ _
+
¸ , ¸ ,
cos 2θ
Also multiplying (iii) by cos θ and 4 by sin θ and then adding.
− σ
x
sin θ cos θ + τ cos
2
θ + σ
y
cos θ sin θ + τ sin
2
θ = 0
τ = (σ
x
+ σ
y
) sin θ cos θ
or τ =
2
x y
σ + σ ¸ _
¸ ,
sin 2θ
Q. 9: Obtain the expression for normal and tangential stresses on a plane BE inclined at an angle
θ θθ θθ with BC subjected to compound stresses as shown in fig 15.9.
y
y
x
x
E
B
D
A
C
Fig 15.9
Sol.: Consider the unit thickness of the element. Applying equations of equilibrium to the free body diagram
of wedge BCE.
Compound Stress and Strain / 399
Let σ
n
and τ be the normal and shear stresses on the plane BE,
∑Fx = 0
σ
x
(BC) – τ
xy
(EC) – τ (EB) cos (90 – θ) − σ
n
(EB) cos θ = 0
∑Fy = 0
σ
x
(EC) + τ
xy
(BC) + τ (EB) sin (90 – θ) − σ
n
(EB) sin θ = 0
Dividing both equations by BE replacing
sin θ =
EC
BE
and cos θ =
BC
BE
σ
x
cos θ + τ
xy
sin θ – τ sin θ − σ
n
cos θ = 0 ...(i)
σ
y
sin θ + τ
xy
cos θ + τ cos θ − σ
n
sin θ = 0 ...(ii)
Multiplying equation (i) and cos θ, (ii) by sin θ and then adding,
σ
x
cos
2
θ + τ
xy
sin θ cos θ – σ
n
cos
2
θ + σ
y
sin
2
θ + τ
xy
cos θ sin θ − σ
n
sin
2
θ = 0
σ
x
cos
2
θ + σ
y
sin
2
θ + 2τ
xy
sin θ cos θ · σ
n
or σ
n
= σ
x
cos
2
θ + σ
y
sin
2
θ + τ
xy
sin 2θ
Putting cos
2
θ =
1 cos 2
2
+ θ
and sin
2
θ =
1 cos 2
2
− θ
σ
n
=
1 cos 2 1 cos 2
2 2
x y
+ θ − θ 1 1
σ + σ
1 1
¸ ] ¸ ]
+ τ
xy
sin 2θ
i.e., σ
n
=
2 2
x y x y
y
σ + σ σ − σ ¸ _ 1
+ σ 1
1
¸ , ¸ ]
cos 2θ + τ
xy
sin 2θ
Multiplying equation (i) by sin θ, (ii) by cos θ and then subtracting (ii) from (i)
(σ
x
cos θ sin θ + τ
xy
sin
2
θ – τ
sin
2
θ) (σ
y
sin θ cos θ + τ
xy
cos
2
θ + τ
cos
2
θ = 0
(σ
x
– σ
y
) cos θ sin θ + τ
xy
(sin
2
θ − cos
2
θ) − τ = 0
τ =
2
x y
σ + σ
sin 2θ – τ
xy
cos
θ (cos
2
θ – sin
2
θ = cos 2θ
Q. 10: Find the principal stresses for the state of stress given below (May–02 (C.O.))
50 Mpa
100 Mpa
Fig 15.11
Sol.: Given that
σ
x
= 100MPa
σ
y
= 0
τ
xy
= 50MPa
Since we known that; the principal stresses are given by
σ
1 or 2
=
2 2
1
( ) ( ) (4 )
2
x y x y xy
1
σ + σ t σ −σ + τ
1
¸ ]
Fig. 15.10
y
( ) EC
x
(BE)
x
(BE)
( ) BE
E
B
C
400 / Problems and Solutions in Mechanical Engineering with Concept
σ
1, 2
= 1/2 [(100 + 0) t {(100 – 0)
2
+ 4 × (50)
2
}
1/2
]
σ
1,2
= 50 ± 70.71
σ
1
= 50 + 70.71 = 120.71MN/m
2
.......ANS
σ
2
= 50 – 70.71 = – 20.71MN/m
2
.......ANS
Q. 11: Determine σ σσ σσ
n
and σ σσ σσ
t
for a plane at θ θ θ θ θ = 25º, for the element shown in figure. (Dec–03 (C.O.))
1
= 25º
1
= 80 MN/m
2
Fig 15.12
Sol.: Given that:
θ =25º
σ
1
= σ
x
= 80MN/m
2
σ
y
= 0
Since we know that
(i) Normal stress
σ
y
=
cos 2
2 2
x y x y
σ + σ σ −σ ¸ _ ¸ _
+ θ
¸ , ¸ ,
= (80 + 0)/2 + ½(80 – 0) cos50º
= 40 + 40 cos 50º
= 65.711MN/m
2
.......ANS
(ii) Shear stress
τ =
1
( )sin 2
2
x y
σ − σ θ
= ½ (80 – 0) sin 50º
= 40 sin 50º
= 30.642MN/m
2
........ANS
Q. 12: In an elastic material, the direct stresses of 120 MN/m
2
and 90 MN/m
2
are applied at a certain
point on planes at right angles to each other in tension and compressive respectively.
Estimate the shear stress to which material could be subjected, if the maximum principal
stress is 150 MN/m
2
. Also find the magnitude of other principal stress and its inclination to
120 MN/m
2
. (May–01 (C.O.))
Sol.: Given that
σ
x
= 120MN/m
2
(tensile i.e + ive)
σ
y
= 90MN/m
2
(Compressive i.e. – ive)
τ
xy
= ?
σ
1
= 150MN/m
2
Since we have
Compound Stress and Strain / 401
σ
1
=
2 2
1
( ) ( ) (4 )
2
x y x y xy
1
σ − σ + σ − σ + τ
1
¸ ]
150 = ½[(120 – 90) + {(120 – (– 90))
2
+ 4(τ
xy
)
2
}
1/2
]
τ ττ ττ
xy
= 84.85MN/m
2
........ANS
Now the magnitude of other principal stress σ
2
σ
2
=
2 2
1
( ) ( ) (4 )
2
x y x y xy
1
σ − σ − σ −σ + τ
1
¸ ]
σ
2
= ½[(120 – 90) – {(120 – (– 90))
2
+ 4(84.85)
2
}
1/2
]
σ σσ σσ
2
= – 120MN/m
2
.......ANS
The direction of principal planes is:
tan 2θ =
2
xy
x y
τ
σ −σ
tan 2θ = (2 × 84.85)/(120 – (– 90))
2θ = 38.94º or 2218.94º
θ θθ θθ = 19.47º or 109.47º .......ANS
Q. 13: A load carrying member is subjected to the following stress condition;
Tensile stress σ σσ σσ
x
= 400MPa;
Tensile stress σ σσ σσ
y
= – 300MPa;
Shear stress τ ττ ττ
xy
= 200MPa (Clock wise);
Obtain
(1) Principal stresses and their plane
(2) Maximum shearing stress and its plane. (Dec–00 (C.O.))
Sol.: Since Principal stresses are given as:
Major Principle stress = σ
1
=
2 2
1
( ) ( ) (4 )
2
x y x y xy
1
σ + σ + σ − σ + τ
1
¸ ]
Minor Principle stress = σ
2
=
2 2
1
( ) ( ) (4 )
2
x y x y xy
1
σ + σ − σ − σ + τ
1
¸ ]
or;
σ
1 or 2
=
2 2
1
( ) ( ) (4 )
2
x y x y xy
1
σ + σ t σ − σ + τ
1
¸ ]
Given that;
σ
x
= 400MPa;
σ
y
= – 300MPa;
τ
xy
= 200MPa (Clock wise);
σ
1,2
= ½ [(400 – 300) ± {(400 + 300)
2
+ 4 x (200)
2
}
1/2
]
σ
1
= 453.11MPa .......ANS
σ
2
= – 353.11MPa .......ANS
The direction of principal planes is:
tan 2θ =
2
xy
x y
τ
σ −σ
tan 2θ = (2 x 200)/(400 – (– 300))
402 / Problems and Solutions in Mechanical Engineering with Concept
2θ = 29.04º or 209.04º
θ θθ θθ = 14.52º or 104.52º .......ANS
Since Maximum Shear stress is at θ = 45º
τ =
1
2
(σ
1
– σ
2
) sin 2θ
= ½ (453.11 + 353.11) sin 90º
= 403.11MPa .......ANS
Now plane of maximum shear
θ
s
= θ
p
+ 45º
θ
s
= 14.52º + 45º or 104.52º
+ 45º
θ θθ θθ
s
= 59.52º or 149.52º .......ANS
Q. 14: A piece of steel plate is subjected to perpendicular stresses of 50 N/mm
2
tensile and
50N/mm
2
compressive as in the fig 5.13. Calculate the normal and shear/stresses at a plane
making 45º. May–03 (C.O.))
50 N/mm
2
50 N/mm
2
50 N/mm
2
45º
45º
50 N/mm
2
Fig 15.13
Sol.: Given that:
σ
x
= 50 N/mm
2
σ
y
= – 50 N/mm
2
θ = 45º
τ = 0
Normal stress is given as:
σ
n
=
2 2
x y x y
σ + σ σ −σ ¸ _ ¸ _
+
¸ , ¸ ,
cos 2θ + τ
xy
sin 2θ
σ
n
= ½ (50 – 50) + ½ (50 – (–50)) cos 90º + 0 x sin 90º
σ σσ σσ
n
= 0 .......ANS
Shear stress at a plane is given by the following equation:
τ =
2
x y
σ −σ ¸ _
¸ ,
sin 2θ – τ
xy
cos 2θ
τ = ½ (50 – (–50)) sin 90º – 0 x cos90º
τ ττ ττ = 50N/m
2
.......ANS
Compound Stress and Strain / 403
Q. 15: The state of stress at a point in a loaded component principal stresses is found to be as
given below : σ σσ σσ
x
=50 GN/m
2
; σ σσ σσ
y
=150 GN/m
2
; τ ττ ττ
xy
=100 GN/m
2
;Determine the principal stresses and
maximum shearing stress. Find the orientations of the planes on which they act. (Dec–03 (C.O.))
Sol.: Principal stresses is given by the equation:
σ
1,2
=
2 2
1
( ) ( ) (4 )
2
x y x y xy
1
σ + σ t σ − σ + τ
1
¸ ]
σ
1,2
= ½ [(50 + 150) ± {(50 – 150)
2
+ 4 x (100)
2
}
1/2
]
σ
1,2
= 100 ± 111.8
σ
1
= 211.8 GN/m
2
.......ANS
σ
2
= – 11.8GN/m
2
.......ANS
Now Maximum shear stress is given by the equation:
t =
1 2
1
( ) sin2
2
σ + σ θ
(Since Maximum Shear stress is at θ = 45º)
= ½ (211.8 + 11.8) sin 90º
= 111.8GN/m
2
.......ANS
The orientation of the planes on which they act is given by the equation:
2θ =
2
xy
x y
τ
σ −σ
tan 2θ = (2 × 100)/(50 – 150))
2θ = – 63.43º
θ = – 31.72º
Major principal plane = θ θθ θθ = – 31.72º .......ANS
Miner principal plane = θ θ θ θ θ + 90º = 58.28º .......ANS
Q. 16: A plane element is subjected to following stresses σ σσ σσ
x
= 120KN/m
2
(tensile), σ σσ σσ
y
= 40 KN/m
2
(Compressive) and τ ττ ττ
xy
= 50KN/m
2
(counter clockwise on the plane perpendicular to xaxis)
find
(1) Principle stress and their direction
(2) Maximum shearing stress and its directions.
(3) Also, find the resultant stress on a plane inclined 40º with the xaxis. (May–05 (C.O.))
Sol.: Given that:
σ
x
= 120KN/m
2
σ
y
= – 40 KN/m
2
τ
xy
= – 50KN/m
2
(i) Calculation for Principle stress and their direction
Principal stresses is given by the equation:
σ
1,2
=
2 2
1
( ) ( ) (4 )
2
x y x y xy
1
σ + σ t σ − σ + τ
1
¸ ]
σ
1,2
= ½ [(120 – 40) ± {(120 + 40)
2
+ 4 x (–50)
2
}
1/2
]
σ
1,2
= 40 ± 94.34
σ σσ σσ
1
= 134.34KN/m
2
.......ANS
σ σσ σσ
2
= – 54.34KN/m
2
.......ANS
404 / Problems and Solutions in Mechanical Engineering with Concept
The Direction of the plane is given by the equation:
tan 2θ =
2
xy
x y
τ
σ −σ
tan 2θ = (2 x (–50))/(120 + 40))
2θ = – 32º
θ = – 16º
Major principal plane = θ θθ θθ = – 16º .......ANS
Miner principal plane = θ θθ θθ + 90º = 74º .......ANS
(ii) Maximum shear stress is given by the equation:
τ =
1
2
(σ
1
– σ
2
) sin 2θ
(Since Maximum Shear stress is at θ = 45º)
= ½ (134.34 + 54.34) sin 90º
= 94.34KN/m
2
.......ANS
(iii) Resultant stress on a plane inclined at 40º with xaxis
Since
σ
n
=
2 2
x y x y
σ + σ σ −σ ¸ _ ¸ _
+
¸ , ¸ ,
cos 2θ + τ
xy
sin 2θ
σ
n
= ½ (120 – 40 ) + ½ (120 + 40) cos 80º – 50 sin 80º
= 40 + 13.89 – 49.24
= 4.65 KN/m
2
τ =
2
x y
σ −σ
sin 2θ – τ
xy
cos 2θ
τ
t
= ½ (120 + 40) sin 80º + 50 cos 80º
τ
t
= 87.47KN/m
2
Resultant stress is given by the equation
σ
r
=
2 2
n t
σ + τ
σ
r
= (4.65
2
+ 87.47
2
)
1/2
σ σσ σσ
r
= 87.6 KN/m
2
.......ANS
Q. 17: Using Mohr’s circle, derive expression for normal and tangential stresses on a diagonal plane
of a material subjected to pure shear. Also state and explain mohr’s theorem for slope and
deflection. (Dec–00, May–01 (C.O.))
Sol.: Mohr circle is a graphical method to find the stress system on any inclined plane through the body.
It is a circle drawn for the compound stress system. The centre of the circle has the coordinate
, 0
2
x y
σ − σ ¸ _
¸ ,
and radius of circle is
2
2
2
x y
σ − σ ¸ _
+ τ
¸ ,
by drawing the mohr’s circle of stress the following three systems
can be determined.
Compound Stress and Strain / 405
(a) The normal stress, shear stress, and resultant stress on any plane.
(b) Principal stresses and principal planes.
(c) Maximum shearing stresses and their planes along with the associated normal stress.
Mohr’s circle can also be drawn for compound strain system.
1. Mohr’s circle of stress with reference to two mutually perpendicular principal stresses acting on
a body consider both dike stresses.
y
y
( ) a ( ) b
x
x
r
n
x
O
C
P
2
Q A B
Fig 15.14
Both principal stresses may be considered as (a) Tensile and (b) Compressive.
Let us for tensile
i.e.; σ
x
> σ
y
Steps:
l. Mark OA = S
x
and OB = S
y
, along xaxis (on +ve side if tensile and ve side if compressive).
2. Draw the circle with BA as diameter, called mohr’s circle of stress.
3. To obtain stress on any plane ¸ as shown in fig. 15.14 (a) measure angle ACP = 2¸ in counterclock
wise direction.
4. The normal stress on the plane is S
n
= OQ
Shear stress is τ = PQ
Resultant stress = σ
r
= OP
And Angle POQ = ϕ is known as angle of obliquity.
2. Mohr’s stress circle for a two dimensional compound stress condition shown in fig.
Let for a tensile
σ
x
> σ
y
Steps:
l. Mark OA = σx and OB = σy along x axis
[on +ve side if tensile and –ve side if compessive.]
2. Mark AC = τ
xy
and BD = τ
xy
Fig. 15.15
y
y
xy
xy
xy
xy
x
x
406 / Problems and Solutions in Mechanical Engineering with Concept
D
B Q
O
S
C
P
2
2
xy
mex
mex
y
n
x
1
Fig 15.16
3. Join C and D which bisects AB at E the centre of Mohr’s circle.
4. With E as centre, either EC or ED as a radius draw the circle called Mohr’s circle of stress.
5. Point P and Q at which the circle cuts S axis gives principal planes OP = σ
1
, and OQ = σ
2
gives
the two principal stresses.
6. ∠CEP = 2θ
1
and ∠CEQ = 2θ
2
is measured in anticlockwise direction.
θ
1
= (1/2) ∠CEP and θ
1
= (1/2) ∠CEQ indicates principal planes.
7. ER = EC = τ
max
is the maximum shearing stress.
θ′
1
= (1/2) ∠CER and θ′
2
= (1/2) ∠CES
is measured anticlockwise direction gives maximum shearing planes.
8. To obtain stress on any plane ‘θ’ measure ∠CEX = 2θ in anticlockwise direction.
Normal stress on the plane is,
σ
n
= OY
Shear Stress, τ = XY
Resultant Stress is σ
r
= OX
And
∠XOY = ϕ is known as angle of obliquity.
Q. 18: A uniform steel bar of 2 cm × 2 cm area of crosssection is subjected to an axial pull of 40000
kg. Calculate the intensity of “normal stress, shear stress and resultant stress on a plane
normal to which is inclined at 30° to the axis of the bar. Solve the problem graphically by
drawing Mohr Circle. (Dec–01 (C.O.))
Sol.: Given that :
Load applied ‘P’= 4000 × 9.8 = 39.2 kN
σ
x
= 39.2/(2 × 10
–2
)
2
= 98MN/m
2
P = 4000 kg
P
30º
30º
B
A
10 cm 2 cm
2
c
m
Fig 15.17
Compound Stress and Strain / 407
Steps to draw Mohr’s circle.
Step1: Take origin ‘O’ and draw a horizontal line OX.
Step2: Cut off OA equal to σ
x
by taking scale
1 mm = 1MN/m
2
O Q
P
X
C
60º
4
3
m
m
73.5 cm
8
5
m
m
Fig 15.18
Step 3: Bisect OA at C
Step 4: With C as center and radius CA draw a circle.
Step 5: At C draw a line CP at an angle 2θ with OX meeting the circle. At P (θ is angle made by
oblique plane with minor principle stress, here zero).
Step 6: Through P draw perpendicular to OX, it intersect OX at Q, join OP. Measure OQ, PQ and
OQ as σ, τ and σ
r
. respectively. Therefore;
Normal stress on the plane σ = OQ = 73.5 × 1 = 73.5MN/m
2
Tangential or shear stress on the plane τ = PQ = 43 × 1 = 43 MN/m
2
And Resultant stress σ
r
= OP = 85 × 1 = 85 MN/m
2
.
Q. 19: The stresses on two mutually perpendicular planes are 40N/mm
2
(Tensile) and 20N/mm
2
(Tensile). The shear stress across these planes is 10N/mm
2
. Determine by Mohr’s circle method
the magnitude and direction of resultant stress on a plane making an angle 30º
with the plane
of first stress.
Sol.: For the given stress system, the mohr’s circle has been drawn and this depicts as shown in fig.
y
0
B
P
E
A
x
F
Q C
2
Fig 15.19
408 / Problems and Solutions in Mechanical Engineering with Concept
OA = σx = 40N/mm
2
OB = σy = 20N/mm
2
AE = BF = τ = 10N/mm
2
2θ = 120º
Scale 1cm = 5N/mm
2
From Measurement:
σ
n
= OQ = 27N/mm
2
σ
t
= PQ = 13.5N/mm
2
σ
r
= OP = 30N/mm
2
And ϕ = 27º
Q. 20: At a point in a stressed body, the principal stresses are σ σσ σσ
x
= 80 kN/m
2
(tensile) and σ σσ σσ
y
= 40
kN/m
2
(Compressive). Determine normal and tangential stresses on planes whose normal are
at 30° arid 120° with xaxis using Mohr’s stress circle Method.
y
= 40 kN/m
2
B
( ) a ( ) b
P
P
A
x Q C
2
4
0
º
6
0
º
Q
y
= 40 kN/m
2
x
= 80 kN/m
2
x
= 80 kN/m
2
Fig. 15.20
Sol.:
Steps:
1. Taking origin at O draw the axes.
2. Cut off OA = σ
x
= 80 kN/m
2
and OB = –σ
y
= –40 kN/m
2
. Let us choose a suitable scale, say
1 mm = 2 kN/m
2
3. Bisect AB at C.
4. With C as centre and CA as radius, draw a circle, which is Mohr’s circle.
5. At C draw a straight line CP making an angle 2θ = 60º with CX.
6. From P draw perpendicular to OX to intersect at Q.
7. Then OQ = σ and PQ = τ
It is found σ = 50 kN/m
2
and τ ττ ττ = 52kN/m
2
.......ANS
Similarly, at C draw a straight line, CP making an angle 2θ = 240° with CX. From P′ draw perpendicular
to OX to intersect at Q′. Then
OQ′ ′′ ′′ = σ σσ σσ = –10kN/m
2
and P′ ′′ ′′Q′ ′′ ′′= –52 kN/m
2
.......ANS
Pure Bending of Beam / 409
+0)264
16
PÜRE 8ENDlNC CF 8E/M
Q. 1: Explain the concept of centre of gravity and centroid.
Sol.: A point may be found out in a body through which the resultant of all such parallel forces acts. This
point through which the whole weight of the body acts irrespective of the position of the body is known
as centre of gravity. Every body has one and only one centre of gravity.
The plane fig like rectangle, circle, triangle etc., have only areas, but no mass. The centre of area of
such figure is known as centroid.
The method of finding out the centroid of a fig is same as that of finding out the C.G. of a body.
X = ∑A
i
. x
i
/A
Y = ∑A
i
. y
i
/A
C.G. = (X,Y)
Q. 2: Explain the following terms:
(i) Area moment of inertia
(ii) Theorem of perpendicular axis,
(iii) Theorem of parallel axis.
(iv) Radius of Gyration
(v) Axis of symmetry. (May–02, Dec–01 (C.O.))
(i) Area Moment of Inertia
Moment of a force about a point is the product of the force (F) and the perpendicular distance (d)
between the point and the line of action of the force i.e. F.d. This moment is also called first moment of
force.
If this moment is again multiply by perpendicular distance (d) between the point and the line of action
of the force i.e.; F.d
2
. This quantity is called moment of moment of a force or second moment of force
or force moment of inertia. If we take area instead of force it is called Area Moment of inertia.
Unit of area moment of inertia (A.d
2
) = m
4
For rectangular body: M.I. about XX axis; I
GXX
= bd
3
/12
For rectangular body: M.I. about YY axis; I
Gyy
= db
3
/12
For circular body: I
GXX
= I
Gyy
= πD
4
/64
For hollow circular body: I
GXX
= I
Gyy
= π(D
4
– d
4
)/64
(ii) Theorem of perpendicular axis
It states “If I
XX
and I
YY
be the M.I. of a plane section about two mutually perpendicular axes meeting at
410 / Problems and Solutions in Mechanical Engineering with Concept
a point. The M.I. I
ZZ
perpendicular to the plane and passing through the intersection of XX and YY is given
by the relation”
I
ZZ
= I
XX
+ I
YY
I
ZZ
also called as Polar moment of inertia.
(iii) Theorem of parallel axis
It states “If the M.I. of a plane area about an axis passing through its C.G. be denoted by I
G
. The M.I. of
the area about any other axis AB, parallel to first and a distance ‘h’ from the C.G. is given by ”
I
AB
= I
G
+ a.h
2
Where;
I
AB
= M.I. of the area about an axis AB
I
G
= M.I. of the area about its C.G.
a = Area of the section
h = Distance between C.G. of the section and the axis AB
This formula is reduced to;
I
XX
= I
G
+ a.h
2
; h = distance from x – axis i.e.; Y – y
I
YY
= I
G
+ a.h
2
; h = distance from y – axis i.e.; X – x
(iv) Radius of Gyration (K)
The radius of gyration of a given lamina about a given axis is that distance from the given axis at which
all elemental parts of the lamina would have to be placed so as not to alter the moment of inertia about
the given axis.
K = (I/A)
1/2
K
xx
= (I
xx
/A)
1/2
K
yy
= (I
yy
/A)
1/2
Where;
K
xx
= Radius of gyration of the area from xx axis
K
yy
= Radius of gyration of the area from yy axis
(v) Axis of symmetry
If in a diagram half part of the diagram is mirror image of next half part, then there is a symmetry in the
diagram. The axis at which the symmetry is create, called axis of symmetry.
C.G. of the body will lies on axis of symmetry.
If symmetrical about Y axis, then X = 0
If symmetrical about X axis, then Y = 0
Q. 3: What is bending stress ?
The bending moment at a section tends to bend or deflect the beam and the internal stresses resist its
bending. The process of bending stops when every cross section sets up full resistance to the bending
moment. The resistance offered by the internal stresses to the bending is called the bending stress.
Q. 4: Write down the different assumptions in simple theory of bending. (Dec–05 (C.O.))
The following assumptions are made in the theory of simple bending:
1. The material of the beam is homogeneous (i.e.; uniform in density, strength etc.) and isotropic
(i.e.; possesses same elastic property in all directions.)
2. The cross section of the beam remains plane even after bending.
Pure Bending of Beam / 411
3. The beam in initially straight and unstressed.
4. The stresses in the beam are within the elastic limit of its material.
5. The value of Young’s modulus of the material of the beam in tension is the same as that in
compression.
6. Every layer of the beam material is free to expand or contract longitudinally and laterally.
7. The radius of curvature of the beam is very large compared to the cross section dimensions of the
beam.
8. The resultant force perpendicular to any cross section of the beam is zero.
Q. 5: What is simple bending or pure bending of beam? (Dec–01, 04, 05 (C.O.))
If portion of a beam is subjected to constant bending moment only and no shear force acts on that portion
as shown in the Fig. 16.1, that portion of the beam is said to be under simple bending or pure bending.
A
A
S.F. Diagram
B.M. Diagram
B C
a
a
D
W W
W
B
+ve
+ve
ve
C
C
D
W
Wa Wa
R = W
D
R = W
A
Fig 16.1
A simply supported beam loaded symmetrically as shown in the figure, will be subjected to a constant
bending moment over the length BC and on this length shear force is nil. So the portion BC is said to be
under simple bending.
Q. 6: Define the following:
(1) Nature of bending stress
(2) Neutral layer and Neutral axis
(3) Nature of Distribution of bending stress.
(1) Nature of Bending Stress
If a beam is not loaded, it will not bend as shown in Fig. 16.2 (a). But if a beam is loaded, it will bend
as shown in Fig. 16.2 (b), whatever may be the nature of load and number of loads. Due to bending of the
beam, its upper layers are compressed and the lower layers are stretched. Therefore, longitudinal compressive
stresses are induced in the upper layers and longitudinal tensile stresses are induced in the lower layers.
These stresses are bending stress. In case of cantilevers, reverse will happen, i.e., tensile stresses will be
induced in the upper layers and compressive stresses will be induced in the lower layers.
412 / Problems and Solutions in Mechanical Engineering with Concept
( ) Neutral
layer
a
Y
Y
N A
Neutral
axis
Fig. 16.2
(2) Neutral Layer and Neutral Axis
In a beam or cantilever, there is one layer which retains its original length even after bending. So in this
layer neither tensile stress nor compressive stress is set up. This layer is called neutral layer.
Neutral axis is the line of intersection of the neutral layer with any normal section of the beam. If a
cutting plane YY is passed across the length of the beam, the line NA becomes the line of intersection of
the neutral layer and the normal section of the beam. Therefore, NA is the neutral axis. Since no bending
stress is set up in the neutral layer, the same is true to neutral axis, i.e., no bending stress is set tip in neutral
axis. It will be proved that the neutral axis passes through the C.G. of the section of the beam or cantilever.
(3) Nature of Distribution of Bending Stress
N A
Neutral axis
f
c
F
t
Fig 16.3 Nature of stress distribution in the section of a beam
It will be proved that the bending stress at any layer of the section of a beam varies directly as the
distance of the layer from the neutral axis. The bending stress is maximum at a layer whose distance from
the neutral axis is maximum. The bending stress is gradually reduced as the neutral axis more and more
becomes nearer, it becomes zero at the neutral axis, and then again the bending stress increases in the
reverse direction (see Fig. 16.3) as the distance of the laver from the neutral axis N – A is increased. The
arrows indicating the magnitudes of the bending stress at different layers of a section above and below the
neutral axis have been given in opposite directions just to show the difference in nature of stresses in these
areas.
Pure Bending of Beam / 413
Q. 7: Differentiate between direct stress and bending stress.
Direct tensile and compressive stress is set up due to load applied parallel to the length of the object, and
direct shear stress is set up in the section which is parallel to the line of action of the shear load. But
bending stress is set up due to load at right angles to the length of the object subjected to bending.
In case of direct stress, nature and intensity of stress is the same at any layer in the section of the object
subjected to direct stress, but in case of bending stress nature of stress is opposite on opposite sides of the
neutral axis, and intensity of stress is different at different layers of the section of the object subjected to
bending.
In case of direct stress, intensity of stress is the same in a section taken through any point of the object.
But in case of bending stress, intensity of stress is different at the same layer of the section taken through
different points of the object.
Q. 8: Explain Moment of resistance?
Sol.: Two equal and unlike parallel forces whose lines of action are not the same, form a couple. The
resultant compressive force (P
C
) due to compressive stresses on one side of the neutral layer (or neutral
axis) is equal to the resultant tensile force (P
t
) due to tensile stresses on the other side of it. So these two
resultant forces form a couple, and moment of this couple is equal and opposite to the bending moment
at the section where the couple acts. This moment is called moment of resistance (M.R.).
Neutral
layer
P
t
P
c
Fig 16.4
So, moment of resistance at any section of a beam is defined as the moment of the couple, formed by
the longitudinal internal forces of opposite nature and of equal magnitude, set up at that section on either
side of the neutral axis due to bending.
In magnitude moment of resistance of any section of a beam is equal to the bending moment at that
section of it.
Q. 9: Derive the bending equation i.e.; M/I = σ σσ σσ/y = E/R. (Dec–04)
Sol.: With reference to Fig. 16.5 (a), let us consider any two normal sections AB and CD of a beam at a
small distance δL apart (i.e., AC = BD = δL). Let AB and CD intersect the neutral layer at M and N
respectively.
Let;
M = bending moment acting on the beam
θ = Angle subtended at the centre by the arc.
R = Radius of curvature of the neutral layer M′N′.
At any distance ‘y’ from the neutral layer MN, let us consider a layer EF.
Fig. 16.5 (b) shows the beam due to sagging bending moment. After bending, A′B′, C′D′, M′N′ and
E′F′ represent the final positions of AB, CD, MN and EF respectively.
When produced, A′B′ and C′D′ intersect each other at O subtending an angle θ radian at O, which is
the centre of curvature.
414 / Problems and Solutions in Mechanical Engineering with Concept
Since δL is very small, arcs A′C′, M′N′, E′F′ and B′D′ may be taken as circular.
Now, strain in the layer EF due to bending is given by e = (E′F′ – EF)/EF = (E′F′ – MN)/MN
Since MN is the neutral layer, MN = M′N′
e =
( ) E F MN R y R y y
MN R R R
′ ′ ′ ′ − + θ− θ θ
= = =
′ ′ θ θ
...(i)
Let; σ = stress set up in the layer EF due to bending
E = Young’s modulus of the material of the beam.
Then E =
e
σ
or, e =
e
σ
...(ii)
Equate equation (i) and (ii); we get
y
R E
σ
=
Neutral
layer
AA
M
E
B D
Before bending
( ) a
F
N
C
Neutral
layer
After bending
( ) b
A
M
E
y
B D
F
N
C
Fig. 16.5
or, σ σσ σσ/y = E/R ...(iii)
N A
dy
y
Neutral axis
Fig 16.6
With reference to Fig. 16.6.
At the distance ‘y’, let us consider an elementary strip of very small thickness dy. We have already
assumed that ‘σ’ is the bending stress in this strip.
Let dA = area of this elementary strip.
Then, force developed in this strip = σ.dA.
Then, elementary moment of resistance due to this elementary force is given by dM = f.dA.y
Total moment of resistance due to all such elementary forces is given by
dM dA y = σ× ×
∫ ∫
or, M =
dA y σ× ×
∫
...(iv)
Pure Bending of Beam / 415
From Eq. (iii), we get
σ =
.
E
y
R
×
Putting this value of f in Eq. (iv), we get
M =
2
E E
y dA y dA y
R R
× × × = ×
∫ ∫
But
2
1 dA y ⋅ =
∫
where I = Moment of inertia of the whole area about the neutral axis NA.
M = (E/R) . I
M/I = E/R
Thus; M/I = σ/y = E/R
Where;
M= Bending moment
I = Moment of Inertia about the axis of bending i.e; I
xx
y = Distance of the layer at which bending stress is consider
(We take always the maximum value of y, i.e., distance of extreme fibre from N.A.)
E = Modulus of elasticity of the beam material.
R = Radius of curvature
Q. 10: What is section modulus (Z)? What is the value of Bending moment in terms of section modulus?
(Dec–01, May–02)
Sol.: Section modulus is the ratio of M.I. about the neutral axis divided by the outer most point from the
neutral axis.
Z = I/y
max
.
For Circular Shaft (Z) = I/y = (πD
4
/64)/D/2 =(πD
3
/32)
For Hollow Shaft (Z) = I/y = {π(D
4
– d
4
)/64}/D/2 =(π/32)( D
4
– d
4
)/D
For Rectangular section (Z) = I/y = (bd
3
/12)/d/2 =bd
2
/6
Section modulus represent the strength of the section of the beam.
Since; M = σ.I/y = σ.Z
Stress at the outer fiber will be maximum. i.e.;
M = σ σσ σσ
max
.(I/y
max
) = σ σσ σσ
max
.Z
Q. 11: What is the relation between maximum tensile stress and maximum compressive stress in any
section of a beam?
Sol.: For generalization, let us assume inverted angle section of a beam as shown below:
N
Neutral
axis
D
A
y
c
Y
t
Fig 16.7
416 / Problems and Solutions in Mechanical Engineering with Concept
In case of a beam, maximum compressive stress will be set up in the topmost layer, and maximum
tensile stress will be set up in the bottom most layer due to bending.
Let
σ
c
= maximum compressive stress
σ
t
= maximum tensile stress
y
C
= distance of the topmost layer from the neutral axis NA
y
t
= distance of the bottommost layer from the neutral axis NA.
Then, according to bending equation, we get
M/I = σ/y
where
σ = maximum bending stress,
y = distance of the layer at which maximum bending stress occurs, the distance being measured from
the neutral axis,
M = maximum moment of resistance
= maximum B.M.
I = Moment of inertia (M.I.) of the section of the beam about the neutral axis.
M/I = σ
c
/y
c
Also; M/I = σ
t
/y
t
Equate both we get;
σ
c
/y
c
= σ
t
/y
t
or; σ
c
/σ
t
= y
c
/y
t
Q. 12: Write down the basic formula for maximum bending moment in some ideal cases.
Sol.:
Berm together with Maximum B.M. Section when
nature of load maximum B.M. occurs
1. WL It occurs at the fixed end.
l
W
Cantilever loaded with
one point load wat the
free end.
2.
2
2
WL
It occurs at the fixed end.
where
W = total value of U.D.L.
20 × l
Cantilever loaded with
U.D.L. over the enitre
length.
Contd...
w/unit length
l
Pure Bending of Beam / 417
Berm together with Maximum B.M. Section when
nature of load maximum B.M. occurs
3.
4
WL
If occurs at the midspan.
where
Boam loaded with one W = point load placed at
point load at the midspan. the midspan.
4.
2
8
WL
If occurs at the midspan.
where
W = total value of U.D.L.
= 20 × l.
Q. 13: Find out the M.I. of T section as shown in fig 16.8 about XX and YY axis through the C.G.
of the section.
150 mm
150 mm
50 mm
50 mm
2
Fig 16.8
Sol.: Since diagram is symmetrical about y axis i.e. X = 0
A
1
= 150 × 50 = 7500 mm
2
A
2
= 50 × 150 = 7500 mm
2
y
1
= (150 + 50/2) = 175 mm
y
2
= 150/2 = 75 mm
Y = (A
1
y
1
+ A
2
y
2
)/(A
1
+ A
2
)
= (7500 × 175 + 7500 × 75)/(7500 + 7500) = 125 mm
C.G. = (0,125)
Moment of inertia (M.I.) about xx axis = I
XX
= I
XX1
+ I
XX2
I
XX1
= I
GXX1
+ A
1
h
1
2
= (bd
3
/12)
1
+ A
1
(Y – y
1
)
2
= 150 × 50
3
/12 + 150 × 50 (125 – 175)
2
= 20.31 × 10
6
mm
4
...(i)
I
XX2
= I
GXX2
+ A
2
h
2
2
= (bd
3
/12)
2
+ A
2
(Y – y
2
)
2
= 50 × 150
3
/12 + 50 × 150 (125 – 75)
2
= 32.8125 x 10
6
mm
4
...(ii)
I
XX
= I
XX1
+ I
XX2
= 20.31 × 10
6
+ 32.8125 × 10
6
= 53.125 × 10
6
mm
4
...(iii)
Moment of inertia (M.I.) about yy axis = I
yy
= I
yy1
+ I
yy2
Since X = 0 i.e.; X
1
= X
2
= 0
1/2 1/2
W
w/unit length
l
418 / Problems and Solutions in Mechanical Engineering with Concept
I
yy1
= I
Gyy1
+ A
1
h
1
2
= (db
3
/12)
1
+ A
1
(X – X
1
)
2
= (db
3
/12)
1
= 50 × 150
3
/12
= 14 × 10
6
mm
4
...(iv)
I
yy2
= I
Gyy2
+ A
2
h
2
2
= (db
3
/12)
2
= 150 × 50
3
/12
= 1.5 x 10
6
mm
4
...(v)
I
yy
= I
yy1
+ I
yy2
= 14 × 10
6
+ 1.5 × 10
6
= 15.5 × 10
6
mm
4
...(vi)
I
XX
= 53.125 × 10
6
mm
4
; I
yy
= 15.5 × 10
6
mm
4
.......ANS
Q. 14: Find the greatest and least moment of inertia of an inverted T–section shown in fig 16.9.
1
2
15 cm
Flange
Web
5 cm
20 cm
5 cm
Fig 16.9
Sol.: Since diagram is symmetrical about y axis i.e. X = 0, x
1
= x
2
= 0
A
1
= 5 × 20 = 100cm
2
A
2
= 15 × 5 = 75cm
2
y
1
= (5 + 20/2) = 15cm
y
2
= 5/2 = 2.5cm
Y = (A
1
y
1
+ A
2
y
2
)/(A
1
+ A
2
)
= (100 × 15 + 75 × 2.5)/(100 + 75) = 9.64 cm
C.G. = (0, 9.64)
Moment of inertia (M.I.) about xx axis = I
XX
= I
XX1
+ I
XX2
I
XX1
= I
GXX1
+ A
1
h
1
2
= (bd
3
/12)
1
+ A
1
(Y – y
1
)
2
= 5 × 20
3
/12 + 5 × 20 (9.64 – 15)
2
= 6206.09 cm
4
...(i)
I
XX2
= I
GXX2
+ A
2
h
2
2
= (bd
3
/12)
2
+ A
2
(Y – y
2
)
2
= 15 × 5
3
/12 + 15 × 5 (9.64 – 2.5)
2
= 3979.72 cm
4
...(ii)
I
XX
= I
XX1
+ I
XX2
= 6206.09 + 3979.72 = 10186.01 cm
4
...(iii)
Moment of inertia (M.I.) about yy axis = I
yy
= I
yy1
+ I
yy2
; h = 0
Since X = 0 i.e.; X
1
= X
2
= 0
I
yy1
= I
Gyy1
+ A
1
h
1
2
= (db
3
/12)
1
+ A
1
(X – X
1
)
2
= (db
3
/12)
1
I
yy2
= I
Gyy2
+ A
2
h
2
2
= (db
3
/12)
2
I
yy
= I
yy1
+ I
yy2
= 20 × 5
3
/12 + 5 × 15
3
/12 = 1614.25 cm
4
...(iv)
Greatest Moment of inertia = I
XX
= 10186.01 cm
4
.......ANS
Pure Bending of Beam / 419
Q. 15: An I section has the following dimensions Top flange = 8 cm × 2 cm; Bottom flange = 12 cm
× 2 cm; Web = 12 cm × 2 cm; Over all depth of the section = 16 cm. Determine the MI of the
I section about two centroidal axis.
8 cm
12 cm
1
3
2
2 cm
2 cm
12 cm
2 cm
Fig 16.10
Sol.: Since diagram is symmetrical about y axis i.e. X = 0, x
1
= x
2
= 0
A
1
= 8 × 2 = 16 cm
2
A
2
= 12 × 2 = 24 cm
2
A
3
= 12 × 2 = 24 cm
2
y
1
= (2 + 12 + 2/2) = 15 cm
y
2
= (2 + 12/2) = 8 cm
y
2
= 2/2 = 1 cm
Y = (A
1
y
1
+ A
2
y
2
+ A
3
y
3
)/(A
1
+ A
2
+ A
3
)
= (16 × 15 + 24 × 8 + 24 × 1)/(16 + 24 + 24) = 7.125cm
C.G. = (0, 7.125)
Moment of inertia (M.I.) about xx axis = I
XX
= I
XX1
+ I
XX2
I
XX1
= I
GXX1
+ A
1
h
1
2
= (bd
3
/12)
1
+ A
1
(Y – y
1
)
2
= 8 × 2
3
/12 + 16 (7.125 – 15)
2
= 997.58 cm
4
...(i)
I
XX2
= I
GXX2
+ A
2
h
2
2
= (bd
3
/12)
2
+ A
2
(Y – y
2
)
2
= 2 × 12
3
/12 + 24 (7.125 – 8)
2
= 306.375 cm
4
...(ii)
I
XX3
= I
GXX3
+ A
3
h
3
2
= (bd
3
/12)
3
+ A
2
(Y – y
3
)
2
= 12 × 2
3
/12 + 24 (7.125 – 1)
2
= 908.375 cm
4
...(iii)
I
XX
= I
XX1
+ I
XX2
+ I
XX3
= 997.58 + 306.375 + 908.375 = 2212.33 cm
4
...(iv)
Moment of inertia (M.I.) about yy axis = I
yy
= I
yy1
+ I
yy2
; h = 0
Since X = 0 i.e.; X
1
= X
2
= 0
I
yy
= I
yy1
+ I
yy2
+ I
yy3
= (db
3
/12)
1
+ (db
3
/12)
2
+ (db
3
/12)
3
= 2 × 8
3
/12 + 12 × 2
3
/12 + 2 × 12
3
/12 = 381.33 cm
4
...(iv)
I
XX
= 2212.33 cm
4
; I
yy
= 381.33 cm
4
.......ANS
420 / Problems and Solutions in Mechanical Engineering with Concept
Q. 16: Determine the MI of an unequal angle section 15cm × 10cm × 1.5cm with longer leg vertical
and flange upwards.
10 cm
2
15 cm
15 cm
1.5
cm
Fig 16.11
Sol.: A
1
= 10 × 1.5 = 15 cm
2
A
2
= 15 × 1.5 = 22.5 cm
2
y
1
= (13.5 + 1.5/2) = 14.25 cm
y
2
= 13.5/2 = 6.75 cm
x
1
= (10/2) = 5 cm
x
2
= 1.5/2 = 0.75 cm
X = (A
1
x
1
+ A
2
x
2
)/(A
1
+ A
2
)
= (15 × 5 + 22.5 × 0.75)/(15 + 22.5) = 2.56 cm
Y = (A
1
y
1
+ A
2
y
2
)/(A
1
+ A
2
)
= (15 × 14.25 + 22.5 × 6.75)/(15 + 22.5) = 9.94 cm
C.G. = (2.56,9.94)
Moment of inertia (M.I.) about xx axis = I
XX
= I
XX1
+ I
XX2
I
XX
= (bd
3
/12)
1
+ A
1
(Y – y
1
)
2
+ (bd
3
/12)
2
+ A
2
(Y – y
2
)
2
= 10 × 1.5
3
/12 + 15 (9.94 – 14.25)
2
+ 1.5 × 13.5
3
/12 + 22.5 (9.94 – 6.75)
2
= 795.07 cm
4
.......(i)
Moment of inertia (M.I.) about yy axis = I
yy
= I
yy1
+ I
yy2
I
yy
= I
yy1
+ I
yy2
= [(db
3
/12)
1
+ A
1
(X – X
1
)
2
] + [(db
3
/12)
2
+ A
1
(X – X
2
)
2
]
= 1.5 × 10
3
/12 + 15 (2.56 – 5)
2
+ 13.5 × 1.5
3
/12 + 22.5 (2.56 – 0.75)
2
= 284.44 cm
4
...(ii)
I
XX
= 795.07 cm
4
; I
yy
= 284.44 cm
4
.......ANS
Q. 17: A beam made of C.I. having a section of 50mm external diameter and 25 mm internal diameter
is supported at two points 4 m apart. The beam carries a concentrated load of 100N at its
centre. Find the maximum bending stress induced in the beam. (Dec–01 (C.O.))
Pure Bending of Beam / 421
Sol.:
100 N
+ 25
+ 50
4 m = 400 mm
Fig 16.12
This problem is the case of simply supported beam with load at its mid point, and in this case maximum
bending moment = M = WL/4
= (100 × 4)/4 = 100 Nm = 100 × 10
3
N.mm ...(i)
Let I = Moment of inertia =
64
π
(d
0
4
– d
i
4
) = π/64 (50
4
– 25
4
) = 287621.4 mm
4
...(ii)
y = d/2 = 50/2 = 25mm ...(iii)
We know that
Maximum stress during Bending moment (at y = 25mm) = M.y/I
= (100 × 10
3
× 25)/287621.4 = 8.69 N/mm
2
Maximum bending stress = 8.69 N/mm
2
.......ANS
Q. 18: A steel bar 10 cm wide and 8 mm thick is subjected to bending moment. The radius of neutral
surface is 100 cm. Determine maximum and minimum bending stress in the beam. (May–02)
Sol.:
100 mm
8 mm
100 mm
y
y
Fig 16.13
Assume for steel bar E = 2 × 10
5
N/mm
2
y
max
= 4mm
R = 1000mm
σ
max
= E.y
max
/R = (2 × 10
5
× 4 )/1000
We get maximum bending moment at lower most fiber, Because for a simply supported beam tensile
stress (+ive value) is at lower most fiber, while compressive stress is at top most fiber (–ive value).
σ σσ σσ
max
= 800 N/mm
2
.......ANS
y
min
= – 4mm
R = 1000mm
∑
min
= E.y
min
/R = (2 × 10
5
x – 4 )/1000
σ σσ σσ
max
= –800 N/mm
2
.......ANS
422 / Problems and Solutions in Mechanical Engineering with Concept
Q. 19: A simply supported rectangular beam with symmetrical section 200mm in depth has moment
of inertia of 2.26 x 10
5
m
4
about its neutral axis. Determine the longest span over which the
beam would carry a uniformly distributed load of 4KN/m run such that the stress due to
bending does not exceed 125 MN/m
2
. (May–03)
Sol.: Given data:
Depth d = 200mm = 0.2m
I = Moment of inertia = 2.26 × 10
5
m
4
UDL = 4KN/m
Bending stress σ = 125 MN/m
2
= 125 × 10
6
N/m
2
Span = ?
Since we know that Maximum bending moment for a simply supported beam with UDL on its entire
span is given by = WL
2
/8
i.e; M = WL
2
/8 ...(i)
Now from bending equation M/I = σ/y
max
y
max
= d/2 = 0.2/2 = 0.1m
M = σ.I/y
max
= [(125 × 10
6
) × (2.26 × 10
5
)]/ 0.1 = 28250 Nm ...(ii)
Substituting this value in equation (i); we get
28250 = (4 × 10
3
)L
2
/8
L = 7.52m .......ANS
Q. 20: A rectangular beam 300 mm deep is simply supported over a span of 5 m. What uniformly
distributed load per meter the beam may carry? If the bending stress is not to exceed
130N/mm
2
. Take I = 8.5 × 10
6
mm
4
. (Sep–(C.O.)03)
Sol.: Given data:
σ = 130 N/mm
2
I = 8.5 × 10
6
mm
4
y = d/2 = 300/2 = 150mm
L = 5m = 5000mm
Let UDL = W N/m
Maximum bending moment for a simply supported beam with UDL on its entire span is given by
= WL
2
/8
i.e; M = WL
2
/8 ...(i)
Now from bending equation M/I = σ/y
max
M = Ã.I/y
max
= [(130) × (8.5 × 10
6
)]/ 150 = 7366666.67 Nmm ...(ii)
Substituting this value in equation (i); we get
7366666.67 = W(5000)
2
/8
W = 2.357 N/mm = 2357.3 N/m .......ANS
Q. 21: A rectangular beam of 200 mm
in width and 400 mm in depth is simply supported over a span
of 4m and carries a distributed load of 10 KN/m. Determine maximum bending stress in the
beam. (Dec –03)
Sol.:
200 m
400 m
10KN/m
4 m
A B
Fig 16.14
Pure Bending of Beam / 423
Given data:
b = 200 mm = 0.2 m
d = 400 mm = 0.4 m
L = 4 m
W = 10KN/m
σ
max
= ?
We know that σ
max
/y = M/I; σ
max
= y.M/I
Here;
y = d/2 = 0.2 m
M = WL
2
/8 = (10 × 10
3
× 4
2
)/8 = 20000 Nm
I = bd
3
/12 = (0.2 × 0.4
3
)/12 = 0.001066 m
4
Putting all the value we get;
σ
max
= (0.2 × 20000)/0.001066 = 3750000 N/m
2
= 3.75MPa .......ANS
Q. 22: A wooden beam of rectangular cross section is subjected to a bending moment of 5KNm. If
the depth of the section is to be twice the breadth and stress in wood is not to exceed
60N/cm
2
. Find the dimension of the cross section of the beam. (Dec–05)
Sol.: Since We know that σ/y = M/I;
Where
σ = 60 N/cm
2
= 60 × 10
4
N/m
2
d = 2b
y = d/2
M = 5KNm = 5 × 10
3
Nm
I = bd
3
/12; for rectangular cross section
Substituting all the values we get
60 × 10
4
/ (d/2) = 5 × 10
3
/(bd
3
/12)
d = 2b;
60 × 10
4
/ (2b/2) = 5 × 10
3
/(b.(2b)
3
/12)
b = 0.232 m = 23.2 cm
d = 2b = 46.4 cm
b = 23.2 cm, d = 46.4 cm .......ANS
Q. 23: Find the dimension of the strongest rectangular beam that can be cut out of a log of 25 mm
diameter.
Sol.: b
2
+ d
2
= 25
2
d
2
= 25
2
– b
2
Since M/I = σ/y; M = σ(I/y) = σ.Z
M will be maximum when Z will be maximum
Z = I/y = (bd
3
/12)/(d/2) = bd
2
/6 = b.(25
2
– b
2
)/6
The value of Z maximum at dZ/db = 0;
i.e.; d/db[25
2
b/6 – b
3
/6] = 0
25
2
/6 – 3b
2
/6 = 0b2 = 625/3;
b = 14.43 mm .......ANS
d = 20.41 mm .......ANS
Fig. 16.15
d
b
25
424 / Problems and Solutions in Mechanical Engineering with Concept
Q. 24: In previous question specify the safe maximum Spain for the simply supported beam of
rectangular section when it is to carry a UDL of 2.5KN/m and bending stress are limited to
10MN/m
2
.
Sol.: Since M/I = σ/y; M = σ(I/y) = σ(bd
3
/12)/(d/2) = σ(bd
2
)/6
M = {(10 × 10
6
) × 14.43 × 20.41
2
× 10
–6
}/6 = 10KN–m ...(i)
Since Maximum bending moment for a simply supported beam with UDL on its entire span is given by
M = WL
2
/8
10 = 2.5.L
2
/8;
L = 5.66 m .......ANS
Q. 25: A beam having I – section is shown in fig is subjected to a bending moment of 500 Nm at its
Neutral axis. Find maximum st