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S K Mondal’s
Strength of Materials
Contents
Chapter – 1: Stress and Strain
Chapter - 2 : Principal Stress and Strain
Chapter - 3 : Moment of Inertia and Centroid
Chapter - 4 : Bending Moment and Shear Force Diagram
Chapter - 5 : Deflection of Beam
Chapter - 6 : Bending Stress in Beam
Chapter - 7 : Shear Stress in Beam
Chapter - 8 : Fixed and Continuous Beam
Chapter - 9 : Torsion
Chapter-10 : Thin Cylinder
Chapter-11 : Thick Cylinder
Chapter-12 : Spring
Chapter-13 : Theories of Column
Chapter-14 : Strain Energy Method
Chapter-15 : Theories of Failure
Chapter-16 : Riveted and Welded Joint
Er. S K Mondal
IES Officer (Railway), GATE topper, NTPC ET-2003 batch, 12 years teaching
experienced, Author of Hydro Power Familiarization (NTPC Ltd)
Page 1 of 429

Note
“Asked Objective Questions” is the total collection of questions from:20 yrs IES (2010-1992) [Engineering Service Examination]
21 yrs. GATE (2011-1992)
and 14 yrs. IAS (Prelim.) [Civil Service Preliminary]
Copyright © 2007 S K Mondal
Every effort has been made to see that there are no errors (typographical or otherwise) in the
material presented. However, it is still possible that there are a few errors (serious or
otherwise). I would be thankful to the readers if they are brought to my attention at the
following e-mail address: swapan_mondal_01@yahoo.co.in
S K Mondal
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1.
Stress and Strain
Theory at a Glance (for IES, GATE, PSU)
1.1 Stress (ı)
When a material is subjected to an external force, a resisting force is set up within the component.
The internal resistance force per unit area acting on a material or intensity of the forces distributed
over a given section is called the stress at a point.
x
It uses original cross section area of the specimen and also known as engineering stress or
conventional stress.
Therefore, T
x
P
A
P is expressed in Newton (N) and A, original area, in square meters (m2), the stress ǔ will be
expresses in N/ m2. This unit is called Pascal (Pa).
x
As Pascal is a small quantity, in practice, multiples of this unit is used.
1 kPa = 103 Pa = 103 N/ m2
1 MPa =
106
Pa =
106 N/
m2
(kPa = Kilo Pascal)
=1
N/mm2
1 GPa = 109 Pa = 109 N/ m2
(MPa = Mega Pascal)
(GPa = Giga Pascal)
Let us take an example: A rod 10 mm q 10 mm cross-section is carrying an axial tensile load 10
kN. In this rod the tensile stress developed is given by
Tt
x
P
10 kN
10q103 N
100N/mm2 100MPa
2
A 10 mm q10 mm
100 mm
The resultant of the internal forces for an axially loaded member is
normal to a section cut perpendicular to the member axis.
x
The force intensity on the shown section is defined as the normal stress.
%F
P
T lim
and Tavg
%Al 0 %A
A
x
Tensile stress (ıt)
If ǔ > 0 the stress is tensile. i.e. The fibres of the component
tend to elongate due to the external force. A member
subjected to an external force tensile P and tensile stress
distribution due to the force is shown in the given figure.
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Chapter-1
x
Stress and Strain
S K Mondal’s
Compressive stress (ıc)
If ǔ < 0 the stress is compressive. i.e. The fibres of the
component tend to shorten due to the external force. A
member subjected to an external compressive force P and
compressive stress distribution due to the force is shown in
the given figure.
x
Shear stress ( U )
When forces are transmitted from one part of a body to other, the stresses
developed in a plane parallel to the applied force are the shear stress. Shear
stress acts parallel to plane of interest. Forces P is applied
transversely to the member AB as shown. The corresponding
internal forces act in the plane of section C and are called shearing
forces. The corresponding average shear stress U
P
Area
1.2 Strain (İ)
The displacement per unit length (dimensionless) is
known as strain.
x
Tensile strain ( F t)
The elongation per unit length as shown in the
figure is known as tensile strain.
ǆt = ƦL/ Lo
It is engineering strain or conventional strain.
Here we divide the elongation to original length
not actual length (Lo + % L)
Let us take an example: A rod 100 mm in original length. When we apply an axial tensile load 10
kN the final length of the rod after application of the load is 100.1 mm. So in this rod tensile strain is
developed and is given by
Ft
x
%L L Lo 100.1mm 100 mm 0.1mm
0.001 (Dimensionless)Tensile
Lo
Lo
100 mm
100 mm
Compressive strain ( F c)
If the applied force is compressive then the reduction of length per unit length is known
as compressive strain. It is negative. Then İc = (-ǻL)/ Lo
Let us take an example: A rod 100 mm in original length. When we apply an axial compressive
load 10 kN the final length of the rod after application of the load is 99 mm. So in this rod a
compressive strain is developed and is given by
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