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Strength of Materials Problem Set for Mechanical Engineering
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GATE/ESE
MECHANICAL ENGINEERING
Strength of Materials
[Student Problem Set]
Table of Contents
Sr. Chapter
Pages
1.
Stress, Strain and Elastic Constants
1 to 32
2.
Principal Stress & Strains - Mohr's Circle
33 to 54
3.
Thermal Stress
55 to 66
4.
Thin Shells
67 to 78
5.
Shear Force and Bending Moment Diagrams
79 to 98
6.
Torsion
7.
Columns
115 to 126
8.
Bending Stresses
127 to 138
9.
Shear Stress in Beams & Combine Loading
139 to 150
99 to 114
10. Theory of Failure
151 to 164
11. Strain Energy
165 to 174
12. Deflection of Beams
175 to 200
1.1 Introduction
Mechanics deals with forces (both internal and
external) and their effects.
NOTE
SOM is also known as solid mechanics, or
mechanics of solids or mechanics of
deformable bodies.
Homogeneous Materials
A material is said to be homogeneous if it
exhibits same properties (elastic properties E,
G, K) at any point in the given direction, i.e.,
for a homogeneous, material properties are
independent of point.
Ex :
A body is said to be a rigid body if the distance
between any two points in the body or on the
body, is invariant.
In engineering mechanics we treat the body
as rigid and we deal only with external forces.
In SOM, the body is treated as a deformable
body and we deal with internal forces.
Aim : The aim of SOM is to develop equations
for stress, strain and to obtain the size by using
mechanical properties.
Isotropic Materials
A material is said to be isotropic, if it exhibits
same elastic properties in any direction at a
given point i.e. for a isotropic material
properties are independent of direction.
Strength of Materials
NOTE
Every homogeneous material need not be
isotropic and similarly every isotropic
material need not be homogeneous.
Fortunately, most of the common
engineering material are both homogeneous
and isotropic.
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Based on the extent of loading :
(a) Point load
(b) Distributed load
If the load acts on a very small area, then it is
a point load.
If the load is distributed over a larger area
then it is a distributed load.
Wood, crystal are anisotropic material i.e.,
these material have different properties in
different directions.
Even if the body is not have homogeneous
and isotropic, it is assumed to be
homogeneous and isotropic.
1.2 Load
Load is an external force or moment experienced
by the member.
1.2.1 Type of Load
Based on the direction of loading
(a) Longitudinal (axial) (parallel to the axis)
(b) Transverse (perpendicular to the axis)
Based on dimensions :
(a) Force Volume
Ex : Buoyancy force, weight, centrifugal
force etc.
(b) Force Area
Ex : Pressure force, drag force, etc.
(c) Force Length
Ex : Surface tension force, cylindrical roller
bearing
Based on variation wrt time :
(a) Static load
(b) Dynamic load
A load is said to be static load if the
magnitude, direction and point of application
(POA) does not change wrt time.
Ex : Self weight
If any of the three (magnitude, direction and
POA) changes wrt time then it is a dynamic
load.
Ex : Crank, connecting rod, piston, gears,
cam and followers, bearings etc.
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Based on load application time :
(a) Gradually applied load : This is the most
idealised type of loading. This loading is also
known as quasi-loading.
In tension/tensile test the material, is
subjected to gradual loading i.e. the load
increases from zero to maximum in an
infinite time.
Stress, Strain & Elastic Constants
1.3 Stress ( )
The internal resistance offered by the material at
a point against the deformation caused due to
external loads. The internal resisting force is due
to intermolecular forces.
(b) Suddenly applied load : For suddenly
applied load ( h 0 ).
Ex : Train moving on a railway track, brake,
clutches etc.
(c) Impact loads : in this type of loading, the
time gap of application of load is small and
the relative velocity exists between loading
and loaded member.
Ex : Charpy test, Izod test, gravity die
forging, hammer blow etc.
Unit :
N
pascal
1.
m2
2. kgf/cm2 9.81104 pascal
1.3.1 Difference Between Stress
And Pressure
S. No.
1.
2.
3.
(d) Shock loads : In shock load, rate of loading
is very high i.e., the time of application of
load is less.
Ex : Bomb blast.
NOTE
SOM basically deals with three S
S
S
S
Stress
Strain
Stability
F
A
4.
5.
6.
Pressure
Pressure is
external normal
force per unit
area.
Pressure is
always normal
to the area
Pressure is a
scalar quantity
Pressure can be
measured.
Due to pressure
there is stress.
At a point, the
pressure is
equal in a all
directions in
static fluid.
Stress
Stress is an
internal resisting
force per unit
area.
Stress need not be
normal to the
area.
Stress is a tensor
of 2nd order.
Stress can not be
measured.
Due to stress no
pressure.
But stress need
not be same in all
directions at a
point.
Strength of Materials
1.3.2 Strength
The maximum stress that a material can resist
without failure is known as strength.
NOTE
Strength depends on material therefore
strength is a material property,
Stress is not a property it depends on load and
area but not on the material.
Stress is developed only when the body is
constrained or restricted.
Stresses are developed only when
deformation or strain is constrained therefore
“Strain is the cause of stress.”
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1.4 Strain ( )
When a force is applied to a body it may result in
change in size or change in shape. This change in
size or shape is known as deformation.
1.4.1 Norman Strain
The extension or contraction of a line segment
per unit length is known as normal strain.
When the material expands or contracts
freely, stress is zero.
1.3.3 TYPES OF STRESSES
L f L0
L0
ve
Lf
L0
L f L0
ve
1
L f L0 (1 )
L0
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NOTE
Strain is dimensionless Sp case of normal
strain :
Stress, Strain & Elastic Constants
NOTE
Normal strains cause change in dimensions
or change in volume. Where as shear strains
cause change in shape.
Strain is a geometrical quantity that is
measured using experimental techniques,
once strain is calculated, stress can be
calculated by using mechanical properties.
Remember
Prismatic bar :
A long straight structural member having
same c/s throughout it’s length is known as a
prismatic bar.
L0 xB xA
L f ( xB U B ) ( xA U A )
L f ( xB xA ) (U B U A )
L f L0
L0
UB U A
xB xA
U
x
1.4.2 Shear Strain ( ) :
1.4.3 Lateral strain :
Every longitudinal strain is associated with
lateral strain. Lateral strain represents normal
strain perpendicular to the direction of loading.
The change in angle that occurs between two
lines segments that were originally
perpendicular to one another is known as
shear strain.
It is expressed in radians.
longitudinal
lateral
dL
(+ ve)
L0
dD
( ve)
D0
1.4.4 Shear strain in x-y plane
Convention :
900 +ve shear strain
900 ve shear strain
2
Strength of Materials
6
1 2 900
Shear strain 90 1 2
tan 1
dv
dx
1 and 2 are very small
tan 2
du
dy
Total shear strain in x-y plane,
1 2
dv du
dx dy
1.5 Tensile Test
This test is conducted on UTM (universal testing
machine) and this machine is used for finding out
tensile strength, compressive strength and shear
strength.
The load is gradually applied from zero to
maximum.
TEST SPECIMEN
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Where A is original area
A
Nominal/conventional/engineering
stress
dL
Where L Original length
L
These stress and strains ate known as engineering
stress-strain,
nominal
stress-strain
or
conventional stress-strain.
Upon performing experiments on different
areas, we get different curve and when these
curves are converted into , the diagram is
independent on dimensions and depends only on
mechanical properties.
Strain
1.5.1 True stress strain diagram :
As the specimen is stretched its cross-sectional
area is reduced and the length between gauge
marks increases if we divided the load with actual
area we get true stress value similarly if we
divide elongation with actual length we get actual
or true strain.
1.5.2 Reason for using
conventional stress-strain
diagram :
Two points A and B are located away from ends
to avoid local effects of grips and to ensure
uniform stress and strain between these marks.
According to ASTM (American society of
testing materials)
All through true stress-strain and conventional
stress strain diagram are different, most
engineering design is done with in the elastic
range, where deformations are very small and
hence the error in using engineering stress strain
diagram is less than 0.1% compared with their
true values.
Lg 5.65 A0
1.5.3 Mechanical properties that
Lg 5d0
can be determined from stress
The testing machine elongates the specimen at a
slow, constant rates until the specimen ruptures
during the test. Continuous readings are taken of
the applied load and elongation. The data are
converted into stress and strain.
strain diagram :
1. Proportionality limit and Hooke’s law :
The stress strain diagram is a straight line
from origin 0 to a point A called
proportionality limit.
This is the result of Hooke’s law i.e.
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E
Where,
E Proportionality constant known as
Young’s modulus or modulus of elasticity
E is a material property
Hooke’s law does not apply over the entire region
of stress strain diagram i.e., its validity ends after
the proportionality, stress is no longer
proportional to strain.
Physical significance of E : It indicates stiffness
of the material i.e. materials with larger E are
more stiffer.
For example : For steel E 200 GPa
ERubber 0.7 GPa
This shows that steel is very much stiffer than
rubber.
E valid only upto proportional limit
1.5.4 Elastic limit
A material is said to be elastic if after being
loaded returns back to it’s original shape when
the load is removed, the elastic limit as its name
suggest is the stress beyond which the material is
no longer elastic.
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Stress, Strain & Elastic Constants
For materials that do not have well defined yield
point, yield stress is determined by offset method
using 0.2% strain.
1.5.6 Ultimate stress :
It is the maximum stress on the stress strain
diagram.
Rupture point/fracture point : It is the point or
stress at which material fractures.
1.5.7 Resilience :
The ability of the material to absorb energy upto
elastic limit is known as resilience.
1.5.8 Toughness :
The ability of the material to absorb energy upto
fracture point is known as toughness.
The elastic limit is larger than proportionality
limit, how ever because of the difficulty in
determining the elastic limit accurately it is
assumed to coincide with proportionality limit.
1.5.5 Yield point :
The point where the stress strain diagram
becomes almost horizontal is called yield point
and the corresponding stress is called as yield
stress or yield strength.
The phenomenon of yielding is unique to
structural steel/mild steel this is because of
carbon bridge atmosphere. Whenever carbon is
in interstitial spaces this phenomenon occurs.
1.5.9 Ductility :
A material that is subjected to large plastic strains
before fracture is known as the ductile material.
Ex : Mildsteel, brass, copper, aluminium etc.
In case of ductile materials, the post elastic strain
> 5%
Strength of Materials
1.5.10 Brittle materials :
Materials that exhibits little or no plastic strain
are known as brittle materials.
Ex : Gray cast iron, ceramics, rubber, glass,
thermosetting plastic etc.
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NOTE
In case of compression as the actual Area
increases the true stress decreases and hence
it is below engineering stress-strain diagram.
1.7 Different types of material
behavior
1.6 TMT (Thermo mechanically
treated bars)
In these bars a layer of martensite is coated to
prevent corrosion.
t 6mm Sheet
Linear elastic
Non-linear
elastic behaviour
t 6 mm Plate
Elasto-plastic
Elasto-plastic
with strain
hardening
Fig. Mild steel under tension
Perfectly plastic
Ideal solid or
rigid solid
Fig. Compression
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Stress, Strain & Elastic Constants
1.9 Effect of increase in % of
NOTE
Brittle materials are weaker in tension
therefore when a tensile load is applied it
fails as shown in figure.
(Brittle materials)
1.8 Loading and unloading
curves
carbon in steel
When carbon percentage is increased in steels :
1. Ductility ( )
2. Ultimate strength
3. Reselience
4. Toughness
NOTE
With decrease in temperature brittleness
increases.
Loading and unloading within elastic limit
Rubber
In case of rubber though complete strain is
recovered loading and unloading curves are not
following the same path, this is because of
heating and this is known as hysteresis.
EC EB EA
True
stress-strain
diagrams
are
important in metal forming because in
Metal forming the material is allowed to
yield.
Rubber has unique behaviour of both
hardening and softening i.e. For the same
strain if the load is less it is in the
softening region.
Strength of Materials
10
1.10 Relationship between
engineering stress and true
stress
The variation of true stress and engineering stress
is in plastic region. Assuming the constancy of
volume in plastic region we have,
V0 VF ,
A0 L0 AF LF
A0 LF
,
AF L0
t
P
A0
P
P A
, t 0
Ainst
A0 Ains
A0 L0 Ainst Linst
A0
L
L L
1
inst 0
Ains
L0
L0
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(At some instant)
Lf
t
L0
Lf
dLinst
ln
Linst
L0
t ln (1 )
1.12 Poisson’s Ratio ( v ) :
When a deformable body is subjected to an axial
load (tensile load), not only does it elongate but
it also contract laterally (transverse).
Similarly when a member is subjected to
compressive loading, it contract in the direction
of force and elongate in lateral direction.
Poisson observed that the ratio of these
strains with in elastic limit is constant. This ratio
is known as Poisson’s ratio i.e., it is defined as
the ratio of lateral strain to the longitudinal strain.
v
Lateralstrain
Longitudinalstrain
t (1 )
( t True stress, engineering stress,
engineering strain)
1.11 Relationship between
engineering strain and true
strain
long
dL
L0
lateral
v
y
x
dD
D0
z
x
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is dimensionless quantity. Generally for
metals it lies between
1 1
to .
4 3
Material
Steel
Al
Concrete
Cork
Rubber
0.3
0.33
0.1
0
0.5
1.13 Shear Stress - Shear
Strain Diagram
Stress, Strain & Elastic Constants
x 0 , y z 0
x 0 , y 0,z 0
x
x
E
y x
z
v x
E
x
E
NOTE
Uni-axial loading produces tri-axial state of
strain.
1.14.2 Bi-axial loading :
With in proportional limit
G
( G Modulus of rigidity or shear modulus.)
Modulus of rigidity represents resistance against
shear deformation.
Larger the rigidity modulus greater is the
resistance and lesser is the shear deformation.
1.14 Types of Loading
1.14.1 Uni-axial loading :
x 0
y 0
z 0
x 0
y 0
z 0
x v y
E
E
y v x
y
E
E
v v y
z x
E
E
v
z ( x y )
E
x
NOTE
Bi-axial loading produces, tri-axial state of
strain.
Strength of Materials
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dV Lbdt bt dL Lt db
1.14.3 Tri-axial loading :
dV dL dt db
V
L t
b
V x y z
x
( y z )
E E
y ( x z )
E E
z ( x y )
E E
y z
V x
(1 2)
E
V
Strain
Load/
Stress
X
x
E
x
y
z
x
y
z
Y
v x
E
v y
y
E
E
v z
E
Z
v z
E
v x
E
v y
E
z
E
NOTE
Volumetric strain ( V ) is known as dilation
Condition for incompressible member :
Case-1 : If x y z 0
Case-2 : If 1 2 0
1
2
Maximum value of Poisson’s ratio :
x v
( y z )
E E
y
E
v
( x z )
E
z v
( x y )
E E
1.15 Volumetric strain (v )
x
,
V x y z
E
x
y
, y x (1 2)
E
E
x
z
E
(1 2) 0
i.e.
x
1 2
1
2
1
2
Volume (V ) Lbt
max
1
2
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1.15.1 Volumetric strain in a
cylindrical bar :
V
1.16 Relationship between
Elastic Constant (E, G & K):
2
DL
4
dV
L (2 D) dD D 2 (dL)
4
4
tan
x
a
L (2 D) dD D 2 (dL)
dV 4
4
V
D2 L
4
Strain in the diagonal
dV
dD dL
2
V
D
L
Sinus
dV
2 D L
V
[ V 2 (Lateral strain) + Longitudinal strain]
1.15.2 Volumetric strain in a
spherical vessel /or sphere :
Volume
Stress, Strain & Elastic Constants
4 3
R
3
4
dV (3R 2 ). dR
3
4
(3R 2 ). dR
dV 3
4 3
V
R
3
dV
dR
3
V
R
V 3 R
V 3 Radial or diametral strain
x
a
A'C '
AC
A'C ' A'C '
AA '
a
x a
1
A'C '
a
2
A'C '
a
2
a
2. 2 a
AC
** AC
2G
2
…. (i)
Strength of Materials
14
AC
E
E
AC
(1 )
E
Physical meaning of K :
…. (i)
(1 )
2G E
E 2G(1 )
Relation Between E and K
Bulk Modulus (K) :
It is defined as the ratio of average stress to
volumetric strain. i.e.
K
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av
v
It represents resistance to dilation
(volumetric strain). Under the action of
loading, greater the bulk modulus, lesser is
the dilation
Relationship between E, G and K :
E 2G(1 )
E
1
2G
E 3K (1 2)
E
1 2
3K
1
E
1
2 3K
…..(i)
…..(ii)
From equation (i) and (ii),
E
1
E
1 1
2G
2 3K
2( E 2G) 3K E
2G
3K
2 E 4G 3K E
2G
3K
6KE 12GK 6KG 2GE
2E(3K G) 18 KG
E 2G(1 ) 3K (1 2)
E
av
v
v
x y z
Type of material
3
x y z
E
(1 2)
3( x y z )
3E
v
3 aV
(1 2)
E
E
3av
(1 2)
v
E 3K (1 2)
9 KG
3K G
(1 2)
Isotropic
Orthotropic
Anisotropic
Number of independent
elastic constant
2
9
21
NOTE
Number of independent elastic constant are
those constant with which the stressstrain diagram can be plotted.
Generally for metals E K G
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1.17 Deformation of axially
Stress, Strain & Elastic Constants
Axial rigidity :
AE is known as axial rigidity.
loaded members
Stiffness
1.17.1 Prismatic bar
P
Axial stiffness
P
AE
PL
L
AE
For a rigid member the axial stiffness is
infinite.
1.17.2 Deflection of a stepped
bar : (Bars in series)
1 2 3
PL1
PL2
PL3
....
A1E1 A2 E2 A3 E3
1.17.3 Elongation of a tapered
bar :
d
dx
E
P
d
E.
A
dx
P dx
d
AE
L
d
0
P dx
AE
PL
AE
Conditions for using this equation :
1. Homogeneous and isotropic
2. Prismatic bar (constant cross section area)
3. Pure axial load
4. Loading is with in proportional limit.
Strength of Materials
16
r2 r1
K
L
y
tan K
x
rx r1 Kx
tan
L
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y Kx
P1
PA1E1
A1E1 A2 E2
P2
PA2 E2
A1E1 A2 E2
Similarly,
L
Pdx
P dx
2
Ax E
E 0 Rx
0
L
P
dx
E 0 (r1 Kx)2
L
P 1
1
.
E (r1 Kx) K 0
1
1
r1 KL r1
r1 KL r2
P
EK
PL
r1 r2 E
4PL
d1 d 2 E
PL
PL
1
A1 E1 ( A1 A2 ) Eeq
PL
PL
A1 E1 A2 E2 ( A1 A2 ) Eeq
Eeq
A1E1 A2 E2
A1 A2
1.17.5 Elongation of a bar due to its
self weight :
1.17.4 Compound bars (In
parallel) :
Specific weight
g
L1 L2 L
d
P P1 P2
1 2
L
PL
PL
P
AE
1
2 1 1 1
A1 E1 A2 E2
P2 A2 E2
PA E
P P1 1 2 2
A1 E1
A E A2 E2
P P1 1 1
A1 E1
0
Axdx
AE
x dx
E
L2
2E
gL2
AE
Weight
Volume
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Practice Questions
Stress, Strain & Elastic Constants
1.5
Common Data Questions 1.1 & 1.2
A hypothetical load elongation curve for a 13
mm diameter tensile specimen with 50 mm
gauge length is as shown in the diagram
below.
1.1
1.2
1.3
1.4
Bar -1 has a diameter d, length L, and elastic
modulus E and subjected to tensile load P,
resulting in an elongation of 1 . Bar -2 has
diameter, 2d, length 2L, an elastic modulus
2E and subjected to tensile load 2P, resulting
in an elongation of 2 . Find the ratio 1 / 2 .
1.6
The axial force diagram for the weightless
beam subjected to the inclined force P = 5 kN
is
The Young’s modulus is
(A) 101 GPa
(B) 148 GPa
(C) 201 GPa
(D) 301 GPa
(A)
The ultimate tensile strength of the material
is
(A) 207 MPa
(B) 247 MPa
(C) 222 MPa
(D) 267 MPa
(B)
The engineering stress - strain diagram of a
mild steel indicates stress values of 300 MPa
and 400 MPa at 10% and 15% strains,
respectively and exhibits necking at a strain
of 27%. The difference in true stress values
between these points is :
(A) 100 MPa
(B) 120 MPa
(C) 130 MPa
(D) 140 MPa
A 9 kN tensile load will be applied to a 50 m
length steel wire E = 200 GPa. The normal
stress in the wire must not exceed 150 MPa
and the increase in the length of the wire
should be at most 25 mm. Which among
these could be the smallest diameter of the
wire so that the wire does not fail ?
(A) 5.75 mm
(B) 7.75 mm
(C) 8.75 mm
(D) 10.7 mm
(C)
(D)
1.7
A two - bar pin - jointed truss is subjected to
a load P as shown in figure. The axial stress
in member 1 is :
Strength of Materials
(A)
(C)
1.8
18
P
A1
(B)
PE2
E1 A1 E2 A2
(D)
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PE1
E1 A1 E2 A2
The maximum P that will not exceed a stress
P
A1 A2
or in Bronze of 100 MPa is
in steel of 140 MPa, in Aluminum of 90 MPa
Two circular rods shown below carry the
same axial load P. The Rod - A has Uniform
cross -section and the Rod - B has non -
(A) 25000 N
(B) 20000 N
Uniform cross -section as shown. The ratio of
(C) 15000 N
(D) 10000 N
elongation of Rod - A to Rod - B is given by
1.9
(A) 1:1
(B) 1:2
(C) 2:1
(D) 3:1
1.11
Two bars of length are joined together to
form a structural system that is fixed at top
and bottom as shown. The bottom part BC of
the structure has a cross sectional area twice
that of the top part AB. Both the bars are
made of the same elastic material. What is the
maximum stress in the top part AB of this
structural system if a load P acts at B as
shown. Neglect the self-weight of the system.
An elastic rod AB is held between two rigid
supports as shown in figure. An axial load P
is applied at a distance of L/3 from the left
end. The support reaction at B is
1.12
1.10
(A) 0
(B) P/3
(C) 2P/3
(D) P
For the cantilever beam as shown in figure
the cross sectional area of the steel,
aluminum and bronze part is 500mm2 ,
400mm2 and 200mm2 respectively.
(A) 2P/A
(B) P/A
(C) P/2A
(D) P/3A
Two wires are connected to a rigid bar as
shown in the figure. The wire on the left is of
steel having a cross sectional area of 0.1cm2
and Young’s modulus of 200,000 MPa. The
wire on the right is made of aluminum having
a cross- sectional area of 0.2cm2 and a
Young’s modulus of 66,667 MPa. If the load
W is to be placed on the rigid bar so as to keep
the bar horizontal, the distance ‘X’ from the
left end (steel wire end) where this weight
should be placed is :
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Stress, Strain & Elastic Constants
1.15
1.16
1.13
1.14
(A) 5.6 cm
(B) 8.57 cm
(C) 9.21 cm
(D) 11.24 cm
A stepped circular shaft , fixed at one end , is
subjected to two axial forces as shown below.
The maximum tensile stress in the shaft is
The tapered rod shown in Fig is subjected to
an axial pull. The ratio of its extension based
on actual calculation to that based on an
average diameter is
(A) 1
(B) 0.75
(C) 0.5
(D) 1.5
The fiber AB has length L and orientation .
If its ends A and B undergo small
displacements u A and vB as shown,
determine the strain in the fiber.
(A) 120 MPa
(B) 210 MPa
(C) 153 MPa
(D) 390 MPa
A three -bar structure made of bars pinned to
each other at G and at the supports B, C and
D is subjected to a load P as shown. All the
bars have the same cross-section, A and
elastic modulus, E. Compute the deflection of
joint G due to the applied load. ( 600 )
uA
v
sin B
L
L
u
v
(B) cos A sin B
L
L
u
v
(C) sin A cos B
L
L
u
v
(D) sin A cos B
L
L
A non - uniform loading on the bar causes a
normal strain which can be expressed as
Ex kx 2 , where k is a constant. Determine
(A) cos
1.17
the displacement of the end B of the bar.
PL
3 AE
4 PL
(C)
3 AE
(A)
PL
2 AE
4 PL
(D)
5 AE
(B)
Strength of Materials
(A)
KL3
3
(B)
Kulkarni Academy
KL2
3
KL2
2
An isotropic body is subjected to a state of
stress given by :
x 10 MPa and
(C) KL2
1.18
20
(D)
xy yx –20 MPa. Assuming G = 0.4 E,
the volumetric strain is
(A)
5
E
(B)
7.5
E
10
15
(D)
E
E
A thin square plate is in a state of pure shear
as shown in Fig. As a result, the angle
(C)
1.19
1.21
between edges AB and AD is no longer 900
but it is now 89.910 . The shear strain at point
A is
A square plate of side 1m and thickness 1cm
is subjected to a tensile stress of 100 MPa and
a shear stress of 50MPa. The Youngs
modulus and the Poisson’s ratio of the
material of the plate are respectively 2 105
MPa and 0.3. The change in volume of this
plate is
(A) 0.2 105 m3
1.22
(A)
1
1000
(B)
1
1000
(D)
2000
1000
An aluminium specimen with an initial gauge
diameter d0 10 mm and a gauge length
(C)
1.20
l0 100mm is subjected to tension test. A
tensile force P = 50 kN is applied at ends of
the specimen as shown resulting in an
elongation of 1mm in the gauge length. The
Poisson’s ratio () of the specimen
is_________ .
Shear modulus of the material G = 25 GPa.
Consider engineering stress – strain
conditions.
(B) 1.5 105 m3
(C) 1105 m3
(D) 0.5 105 m3
At a point ‘O’ on a metal sheet a square
OABC of a side length is drawn. The square
undergoes a small uniform elastic
deformation and deforms to OA*B*C
(dashed lines) as shown in the figure . All
dimensions are in mm and the figure is not to
scale. The normal strains x ,y and shear
strain
xy
developed
respectively are
in
the
square
Kulkarni Academy
21
Stress, Strain & Elastic Constants
(A) – 0.0020, 0.0025 and 0.0020
(B) 0.0020, – 0.0025 and – 0.0020
(A)
1
2(1 )
(B)
(C)
1
(1 )
(D)
(C) 0.0025, – 0.0020 and 0.0020
(D) – 0.0020, 0.0025 and – 0.0020
1.23
An object made of elastic material (Modulus
of Elasticity E and Poisson’s Ratio ) is
restrained from moving in the x - direction by
two rigid walls as shown in the figure : it is
free to move in the other two directions. A
stress of is applied on it in the y- direction
as shown. Find the relationship between the
applied compressive stress and the
compressive strain in the y-direction
1.25
(A) E
(C)
1.24
E
1
(B)
E
1
(D)
E
1 2
1
(1 )
A rod of length L area of cross-section A,
density and modulus of elasticity E hangs
vertically from a roof. The maximum
longitudinal strain in the rod is
(A) 0
(C)
1.26
1
2(1 )
gL
E
(B)
gL
2E
(D)
2gL
E
A cylindrical chamber is filled with an elastic
material of modulus E, and Poisson’s ratio
, as shown. A piston of diameter 2r, is pushed
down on the elastic material by a force F.
Neglecting friction, and assuming that the
chamber and the piston are rigid, the piston
moves a distance of
An elastic material of Young' modulus E and
Poisson's ratio is subjected to a
compressive stress of 1 in the longitudinal
direction. Suitable lateral compressive stress
2 are also applied along the other two
(A)
lateral directions to limit the net strain in each
of the lateral directions to half of the
magnitude that would be under 1 acting
Fh 22
Fh 22
(B)
1
1
r 2 E 1
r 2 E 1
(C)
Fh 22
Fh 22
(D)
1
1
r 2 E 1
r 2 E 1
alone. The magnitude of 2 is
Strength of Materials
22
Common Data Questions 1.27 & 1.28
Kulkarni Academy
1.30
A rod of length L and diameter D is subjected
to a tensile load P. Which of the following is
A 10 mm thick steel rectangular plate of size
100 mm 200 mm is subjected to biaxial
stresses of x 150 MPa, y 200 MPa, as
sufficient to calculate the resulting change in
diameter ?
shown below. The Young’s modulus and
poissons ratio are 200 GPa and 0.3
respectively.
(A) Young’s modulus
(B) Shear modulus
(C) Poisson’s ratio
(D) Both Young’s modulus and shear
modulus
1.31
A metallic rod of 500mm length and 50 mm
diameter, when subjected to a tensile force of
100kN at the ends, experiences an increase in
its length by 0.5 mm and a reduction in its
diameter by 0.015 mm. The Poisson’s ratio of
1.27
the rod material is _________ .
The change in the thickness of the plate is
(A) 2.39 m
(B) 5.25 m
1.32
A rod is subjected to a uni-axial load within
linear elastic limit. When the change in the
(C) 7.12 m
1.28
1.29
(D) 9.16 m
stress is 200 MPa, the change in the strain is
The change in the surface area of the plate is
0.001. If the Poisson’s ratio of the rod is 0.3,
(A) 9.72 mm2
(B) 13.61 mm2
the modulus of rigidity (in GPa ) is ______.
(C) 17.52 mm2
(D) 24.50 mm2
A rod of length L having uniform crosssection area A is subjected to a tensile force
P as shown in the figure below. If the
Young’s modulus of the material varies
linearly from E1 to E2 along the length of the
rod, the normal stress developed at the
section -SS is
(A)
P
A
(B)
P( E1 E2 )
A( E1 E2 )
(C)
PE2
AE1
(D)
PE1
AE2
Kulkarni Academy
A
23
Stress, Strain & Elastic Constants
1.2
Answer Key
(C)
Ultimate tensile strength
1.1
B
1.2
C
1.3
C
1.4
D
1.5
2
1.6
A
1.7
A
1.8
C
1.9
B
1.10
D
1.11
D
1.12
B
1.13
C
1.14
D
1.15
A
1.16
B
1.17
A
1.18
A
Given data :
300MPa , 10%
1.19
C
1.20
0.2732
1.21
A
400MPa , 15%
1.22
D
1.23
D
1.24
B
1.25
B
1.26
A
1.27
B
1.28
D
1.29
A
1.30
D
1.31
0.3
1.32
76.92
E
1.1
3000 9.81 N
221.72MPa
2
(13)2 mm
4
Hence, the correct option is (C).
1.3
(C)
Explanation
(B)
Given Data :
t1 (1 ) 300(1.1) 330MPa
d 13mm , Lg 50mm
t2 400(1 0.15) 460MPa
t2 t1 460 330 130MPa
Hence, the correct option is (C).
1.4
(D)
Given data :
P 9kN 9000 N
L 50m
2000 9.81
(13) 2
4
E
147.816GPa
0.05
50
Hence, the correct option is (B).
Esteel 200GPa
() per 150MPa
() per 25mm
d wire ?
Stress
P
A
Strength of Materials
150 N/mm 2
d2
24
9000
2
(d )
4
Kulkarni Academy
1.6
(A)
9000 4
150
d 8.74mm
If we consider,
d 8.74mm , 25mm .
25
Hence, the correct option is (A).
PL
AE
1.7
(A)
4 9000 50 10
d 2 200 10
d 10.75mm
d 10.75mm
If
induced per
So we select 10.75 mm diameter.
P F1 cos60 F1 cos60
Hence, the correct option is (D).
1.5
F1 P
2
1 Bar
2 Bar
Dia d
dia-2d
Length L
Length 2L
Elastic modulus E
elastic modulus 2E
Tensile load P
Load 2P
1
4PL
d 2 E
2
1
P
A1
Hence, the correct option is (A).
1.8
(C)
4(2 P)(2 L)
(2d )2 (2 E )
4 PL 4 1
1
d 2 E 8 2
2
1
2
1
2
2
Hence, the correct answer is (2).
4 PL
d 2 E
L
L
4P 2 4P
3
B
23
2
(2d ) E d E
A
Kulkarni Academy
25
8PL 1 4 PL
4d 2 E 3 3d 2 E
2 PL
4 PL
B
2
3 d E 3d 2 E
2 PL 1
B 2 A
d E 2
B
A
2
B
Stress, Strain & Elastic Constants
1.10
Given data :
As 500mm2
AAl 400mm2
Abronze 200mm2
( per )steel 140 MPa
A : B 2 :1
( per ) Al 90 MPa
Hence, the correct option is (C).
1.9
(D)
( per )bronze 100 MPa
(B)
Pmax ?
N
5 PS
2
mm
500
Ps 14000 N
140
Compatibility equation
1 2 0
P
400
PAl 36000 N
90
Beam is not to fail so safe condition we select
10000 N.
Hence, the correct option is (D).
1.11
(D)
1 2
1 2
L
2L
( P RA )
3
3
AE
AE
RA 2 P 2 RA
3
3
3
R
2P
3 A
3
3
2P
RA
3
RB P RA
RA
2P
P
RB
3
3
Note : If the no. of static equilibrium equations
are less than the no. of unknowns then it is known
as statically indeterminate member.
Hence, the correct option is (B).
RB P
2P
200
Pb 10000 N
100
AB BC
PL PL
AE 1 AE 2
Strength of Materials
26
Kulkarni Academy
RAB ( P RAB )
A
2A
2RAB P RAB
PS PAl W
2.5PAl PAl W
W 3.5PAl
3RAB P
RAB
M 0 0
P
3
PS x PAl (30 x)
2.499PAl x PAl (30 x)
P
P
AB 3
A 3A
Hence, the correct option is (D).
1.12
(B)
3.499 x 30
30
x
8.5738cm
3.499
Hence, the correct option is (B).
1.13
Given data :
Steel wire
AS 0.1cm
(C)
Aluminium wire
2
ES 200 GPa
AAl 0.2cm2
EAl 66.667 GPa
500 103 N
1
2
(100) 2 mm
4
1 63.66 MPa
300 103
MPa
2
(50)
4
2 152.78 MPa
2
Keep the bar horizontal
steel Al
PS LS
P L
Al Al
AS ES AAl E Al
PS 60
PAl 100
0.1 200 0.2 66.667
PS
2.4999 2.5
PAl
Hence, the correct option is (C).
1.14
(D)
Kulkarni Academy
27
Stress, Strain & Elastic Constants
F1 F3
533.33PL
E
L
L1 L3
cos
F1 cos F2 F3 cos P
... (ii)
1
1
2
PL
AE
2F1 cos F2 P
Hence, the correct option is (A).
1.16
(B)
AE
L
AE cos AE
21
P
L / cos
L
AE cos3 AE
2
P
L
L
600
P
4 PL
5 AE
Hence, the correct option is (D).
1.15
(A)
A ' B ' L2f ( L sin VB )2 ( L cos U A )2
VB2 0
2
U A 0
L2f L2 2VB L sin 2U A L cos
L2f
2
L
Extension based on actual dimension (1 )
4 PL
1
(0.05)(0.15) E
533.33PL
1
E
Average diameter (dm ) d1d2
… (i)
Extension based on average diameter (2 )
2
4 PL
0.08660 E
2
2VB sin 2U A cos
L
L
2V sin 2U A cos
1 B
L
L
L
Lf
(1 x)n 1 nx
Lf
L
0.05 0.15
0.08660 m
1
1
1 2VB sin 2U A cos
2
L
L
VB
U
sin A cos
L
L
Hence, the correct option is (B).
1
2
Strength of Materials
1.17
28
(A)
Kulkarni Academy
1.19
(C)
x kx 2
90 89.91
0.090
180
2000
Hence, the correct option is (C).
x
dx
1.20
Given Data :
d0 10 mm
x dx
P 50 kN
kx2dx
Total displacement
L
0
L0 100 mm
kL3
kx dx
3
2
1 mm (in gauge length)
1
0.01
L 100
Hence, the correct option is (A).
1.18
0.2732
(A)
50 103
636.619 MPa
2
(10)
4
E 63.661 GPa
E 2G(1 )
Given Data :
x 10 MPa
xy yx 20 MPa
G 0.4 E
E 2G (1 )
0.25
v x (1 2 )
E
v
10
(1 0.5)
E
v
5
E
Hence, the correct option is (A).
1.2732 (1 )
0.2732
Hence, the correct answer is 0.2732.
1.21
(A)
Kulkarni Academy
Given Data :
x 100 MPa , 50 MPa
29
Stress, Strain & Elastic Constants
x :
E 2 105 MPa
v 0.3
x x
E
v x
y
E
v x
z
E
V x
v
[1 2v]
V
E
100
[1 0.6]
2 105
V
2 104
V
L f OA ' (1 0.002)2 (0.001)2
L f OA ' 0.9980
x
L f L0
0.0020
L0
y :
V 11102 m3
V 102 m3
V 2 106 m3
0.2 105 m3
Hence, the correct option is (A).
1.22
L0 1 m
L f (1 0.0025)2 (0.003)2
L f 1.00250
(D)
y
L f L0
L0
0.0025
As the angle is more than 900 , shear strain is –
ve.
Hence, the correct option is (D).
1.23
(D)
x 0.0020
y 0.0025
xy
dV dU
dx dy
0.001 0.005
0.002 0.0035
x 0
x 0
y 0
y 0
Strength of Materials
x 0
30
z 0 (Free)
Kulkarni Academy
1.25
(B)
y
x
y
x v
( y z ) 0 ;
E E
y
E
y
y
y
E
y
y
y
E
y
E
x v y
v
( x z )
E
L2
2E
v x
E
x
E
x y
(1 v 2 )
E y
(1 v )
gL2
2E
gL
l 2E
Hence, the correct option is (B).
1.26
(A)
2
E
(1 v 2 )
Hence, the correct option is (D).
1.24
(B)
x 0
x 0
z 0
y 0
y 0
z 0
x z
2 v
[1 2 ]
E E
2 v
v
[1 2 ] 1
E E
2E
v
2 v1 v2 1
2
v
2 (1 v) 1
2
lateral
2
v
1
2(1 v)
Hence, the correct option is (B).
Area r 2
x x ( y z ) 0
E E
x v( y z ) { z x }
x v( y x )
v y x (1 v)
y
y
y
y
E
y
E
y
E
v
( x z )
E
2v
x
E
2v 2 y
(1 v)
x
v y
(1 v)
Kulkarni Academy
31
y
y 2v 2
1
E 1 v
y
F 2v 2
1
2
h
r E 1 v
Z
150
0.3
Z
200 103
200
0.3
5.25 104
3
200 10
F
F
y A r 2
Fh 2v 2
1
r 2 E 1 v
Hence, the correct option is (A).
1.27
Stress, Strain & Elastic Constants
4
Z 5.25 10 Z
5.25 103 mm 5.25 μm.
Hence, the correct option is (B).
1.28
(D)
(B)
Given Data :
x 150 MPa
y 200 MPa
E 200 GPa
v 0.3
Original surface area 100 200 mm2
20000 mm2
x x y
E
E
150
200
x
0.3
3
200 10
200 103
4.5 104
L
x
4.5 104
L
L 100 4.5 104
L 4.5 102 mm 0.045 mm
y
x
E
E
200
0.3 150
7.75 10 4
3
200 10 200 103
b
y
7.75 10 4
b
b 200 7.75 10 4 0.155 mm
Final length and width
y
Z
v v
z
x y
Z
E
E
L f 100.045 mm
bF 200.155 mm
Strength of Materials
32
Final surface area L f b f
20024.50 mm2
Kulkarni Academy
1.31
Given Data :
Change in surface area Final surface area
– Initial surface area
L 500 mm , d 50 mm
P 100 kN ,
20024.50 20000
L 0.5
d 0.015
24.50 mm2
d
v d
L
L
Option the correct option is (D).
1.29
0.3
(A)
0.015
v 50 0.3
0.500
500
Hence, the correct answer is (0.3).
1.32
ss
P
[ b /c stress is independent of material]
A
It depends only on c/s area & applied load].
Hence, the correct option is (A).
1.30
76.92
Given Data :
Change in stress 200 MPa d
Change in strain 0.001 d
E
(D)
d
200 GPa
d
v 0.3
E 2G(1 0.3)
P
v
A
; if E is known then
PL
(For , E is necessary)
AE
d
l
d For , G is necessary.
l
d
L
d
L
For calculating d , both E & G is known to
us.
Hence, the correct option is (D).
E
d
200 GPa
d
200
G G 76.92 GPa
2 1.3
Hence, the correct answer is (76.92).
2.1 Introduction
Case-2 :
For Maximum shear
Expression for normal and shear stress on oblique
plane when a bar is subjected to axial loading :
2 900 , 450
max
n cos2 450
2
n
0
2
NOTE
On a plane of maximum or minimum
normal stress shear stress is zero and these
planes are known as principal planes. i.e. on
principal planes normal stress is maximum
and shear stress is zero.
Failure of Brittle and Ductile materials
Brittle (Weak in tension)
n
s
Pn P cos
cos 2
A
A'
cos
Ductile (Weak in shear)
Ps P sin
sin 2
A
A'
2
cos
Case-1 : n is maximum
When, 00
(n )max cos2
(n )max
s
sin 2 ( 00 )
2
s 0
For mentioning stress completely not only
magnitude and direction are required but the
plane on which it is acting is also required and
hence stress is tensor of 2nd order.
Strength of Materials
Mass density
34
Velocity
acceleration
Stress, strain
moment of
inertia
Tensor of
zero order
Tensor of
first order
Tensor of 2nd
order
Magnitude +
No direction
Magnitude +
direction
Magnitude +
direction + plane
Kulkarni Academy
From force equilibrium equation,
Fx 0
yx dx dz 'yx dx dz
yx yx'
Fy 0
xy dy dz xy' dy dz
xy xy'
Stress representation on 3rd element :
M 00' 0 [Moment equilibrium equation]
yx dx dz dy xy dy dz dx
yx xy
xx
yx
zx
xy
yy
zy
xz xx
yz xy
zz xz
xy
yy
yz
xz xy yx
yz xz zx
zz yz zy
NOTE
To specify stress completely at a point 6
component [3 normal and 3 shear] are required.
Stress matrix is a symmetric matrix.
2D stress element :
xx xy
xy yy
For a 2D stress elements 3, components [2
Normal + 1 Shear] are required.
Plane Stress vs Plane Strain
Plane stress :
When one of the surfaces is not subjected to
any load, then the opposite surface is also not
subjected to any load. Such a loading system
is known as plane stress system.
Kulkarni Academy
35
Principal Stresses, Strains & Mohr’s Circle
Plane strain :
It is a condition in which strain in one of the
primary directions, [Say (z)] is zero, plane strain
can be realized when one of the dimension is very
large.
Example :
Long pipe line, rolling etc.
z
z
( x y )
E E
Plane stress z 0 , z 0
Plane stress
In a plane stress condition stresses in one of
the primary directions (say z) is zero.
To realize plane stress conditions, one
dimension must be very small.
Example : Thin cylinder subjected to internal
pressure.
z
( x y )
E
z may or may not be zero. Therefore, plane
stress condition does not always lead to plane
strain condition.
z
z
( x y )
E E
For plane strain z 0
z ( x y )
z may or may not be zero. Therefore, plane
strain condition does not always lead to plane
stress condition.
Simultaneous occurrence of plane strain and
plane stress condition :
1. 0 (cork)
Pd
Pd
1
2
4t
2t
Eg :
Let,
d
20 (Condition for thin shell)
t
P 5MPa
d
50
t
2 125MPa
1 62.5MPa
3 5MPa (Negligible)
2. x y 0
Strength of Materials
Sign convention :
1. All the outward normal forces are taken as
positive and all inward normal forces are
taken as –ve.
36
Kulkarni Academy
2.2 GENERAL EQUATIONS
FOR PLANE STRESS
TRANSFORMATION
2. Shear stress is taken as +ve if it acts upwards
on the Right face.
x '
3. Clockwise rotations are taken as – ve and
anticlockwise rotations are taken as +ve.
y'
x y
2
x y
2
y
x
cos 2 xy sin 2
2
…..(i)
y
x
cos 2(90 )
2
xy sin 2(90 )
x y
y
x
cos 2 xy sin 2
2
2
…..(ii)
Add equation (i) and (ii),
y '
x ' y ' x y
By using force equilibrium along and to the
inclined plane we have
x '
x y
2
y
x
cos 2 xy sin 2
2
y
x ' y ' x
sin 2 xy cos 2
2
The sum of normal stresses on any two
mutually perpendicular plane is invariant i.e.
it is independent of .
As x ' , y ' depend on angle ' ' let us find out an
angle such that at this angle x ' is maximum.
Kulkarni Academy
37
x y
y
x
cos 2 xy sin 2
2
2
d x '
For maximum x ' ;
0
d
For maximum,
y
d x'
0 x
( sin 2) 2
d
2
x '
xy cos 2 2 0
xy
tan 2
x y
2
xy
y
x
2
Principal Stresses, Strains & Mohr’s Circle
On plane where the normal stress is
maximum or minimum the shear stress is
zero, such plane is known as principal plane
( P ) .
y x y
x ' x
cos 2 xy sin 2
2 2
x y x y
x y 2 2
2
2
2
x y
2
2 xy
xy ( xy )
2
x y
2
2 xy
2
x '
x y
2
y
x
cos 2 xy sin 2
2
x y
2
y
2
x
xy
2
2
On this plane again the shear stress is zero
therefore this is also a principal plane.
x y x y
y 2
2
1 x
2
2
x y
2
2 xy
xy . xy
2
x y
2
2 xy
1
x y
2
y
2
x
xy
2
2
y
x 'y ' x
sin 2 xy cos 2
2
x 'y '
y
y
x
xy
xy x
2
2 0
2
2
x y
x y
2
2
2 xy
2 xy
x y x ' y ' 1 2
Strength of Materials
38
Kulkarni Academy
y
2
x
xy
2
2
max
Observation :
tan 2P
2P is separated by 180 .
2 xy
( x y )
0
tan 2S
P is separated by 90 .
0
Maximum shear stress condition :
d x ' y '
d
0
d x ' y '
y
x
2cos 2 xy (2sin 2) 0
d
2
2 xy
tan 2P tan 2S 1
y
x ' y ' x
sin 2 xy cos 2
2
For x ' y ' maximum
( x y )
2P and 2S are perpendicular (i.e. 900 )
90
450
2
Planes of maximum normal stress and the
P and S are
planes of maximum shear stress are at 450 to
each other.
y
tan 2 x
2
xy
This is the plane on which the shear stress is
maximum.
x '
max
x y x y
2 2
x y
2
2 xy
xy . xy
2
x y
2
2 xy
2
x y
max
2
y
x
cos 2 xy sin 2
2
x y
2
x y
2 xy
x y
2
2 xy
2
y
xy x
2
2
x y
2
2 xy
Kulkarni Academy
max
39
x y
2
Principal Stresses, Strains & Mohr’s Circle
2.3 Graphical representation
avg
Remarks
On the plane of maximum shear stress the normal
stress is equal to average normal stress i.e. On the
planes of maximum shear stress normal stress
need not be zero.
of stress transformation
Mohr’s Circle
Reason’s for using Mohr’s circle :
We can visualize how shear stress and normal
stress component vary with angle.
All the relevant data can be obtained from the
diagram without referring to transformation
equation.
Sign convention :
y x y
x ' x
cos 2 xy sin 2
2 2
1
2
x y
2
x y
2
y
2
x
xy
2
2
y
2
x
xy
2
….. (i)
2
…. (ii)
Now subtracting equation (ii) from equation (i)
we get,
2
…..(i)
y
x
cos 2 xy sin 2
2
2
2
y
x ' y ' x
sin 2 xy cos 2 …...(ii)
2
Adding equation (i) and (ii),
2
2
y
2
1 2 2 x
xy
2
2
x y
x '
2
1 2
max In plane maximum shear stress.
2
x y
2
x '
[ x ' y ' 0]
2
2
2
y
2
x
xy
2
This is the equation of a circle with centre at
y
c x
,0 and radius
2
Strength of Materials
40
y
2
x
xy max
2
We know that
2
y
2
x
xy
2
2
max
Therefore, max Radius of Mohr’s circle
Kulkarni Academy
8. Point 1 on the circle indicates maximum
normal stress and zero shear stress. Therefore
this maximum normal stress is maximum
principal stress.
Similarly point 2 indicates minimum,
normal stress and zero shear stress, this is
minimum principal stress.
Method of constructing Mohr’s circle :
Let the stressed element be as shown on figure,
1. Calculate the center of Mohr’s circle as
x y
, 0
2
2. Locate this point as C, Let the origin be O (0,
0).
3. Take any reference plane in a stressed
element. Let this stressed element be
designated as A.
4. On A, the normal stress is x and the shear
stress is xy (counter clockwise) or this shear
stress tend to rotate the element in CCW
direction. Locate this reference point in the
diagram.
5. With C as centre and CA as radius draw the
circle.
6. Join centre C and reference A, therefore for
this Mohr’s circle CA is the reference line.
7. If we want to calculate stresses on any plane
which is inclined at angle in stressed
element, it should be rotated by 2 on Mohr’s
circle, maintaining the same sense of
rotation.
NOTE
In a stressed element if the rotation is ,
On Mohr’s circle, rotation is 2 [in the
same sence of rotation].
Kulkarni Academy
Special cases of Mohr’s circle :
1. An element subjected to uni-axial tension :
41
Principal Stresses, Strains & Mohr’s Circle
Centre ,0
2
Radius
2
x , y 0 , xy 0
Centre ,0
2
Radius
2
1 , 2 0
3. Pure shear :
x 0 , y 0
Centre = (0, 0)
Radius
2. Uni-axial compression :
x , y 0 , xy 0
1 , 2
Strength of Materials
4. Element subjected to tension
compression of equal magnitude :
42
and
Kulkarni Academy
R = Radius 0
In this case the shear stress is zero on every
plane and hence every plane acts as principle
plane.
PRINCIPAL STRAIN
x , y
Centre (0, 0)
If the material is isotropic it is found that the axis
of principal stresses and principal strain coincide.
Radius
x y
y xy
1,2
x
2
2 2
xy Total shear strain in x-y plane.
5. Element subjected to compression in all
sides :
[Hydrostatic condition]
x , y , 0
Centre (, 0)
2
2
Relationship between maximum principal
stress and maximum principal strain :
1 1 2
…..(i)
E
E
2 2 1
E
E
Multiplying on both sides,
2 2 2 1
…..(ii)
E
E
Adding equation (i) and (ii),
1 2 1 (1 2 )
E
1
E
(1 2 )
(1 2 )
2
E
(2 1 )
(1 2 )
Kulkarni Academy
2.2
2.3
Principal Stresses, Strains & Mohr’s Circle
Practice Questions
P
2.1
43
(C)
A principal plane is one where the shear
stress will be
(A) Maximum
(B) Minimum
(C) Zero
(D) Average of principal stresses
Consider the following statements:
State of stresses at a point when completly
specified, enables one to determine the
1. Principal stresses at the point
2. Maximum shearing stress at the point
3. Stress components on any arbitrary
plane containing the point
Which of these statements are correct ?
(A) 1, 2 and 3
(B) 1 and 3
(C) 2 and 3
(D) 1 and 2
A bar of cross-sectional area A is subjected
to an axial load P. Let be the average shear
stress acting on a plane oriented at an angle
to the axis, as shown. Which one of the
choices below corresponds to the variation of
with respect to 0 .
2
0
0
4
P
2A
0
0
4
P
2A
2.4
0
(C) Are such that the former is inclined at
450 to the latter.
2.5
(D) Need not always exist.
A point in a body is subjected to a biaxial
state of stress, equal in magnitude but
opposite in nature. On a plane inclined at an
angle 450 with respect to x- axis (passing
through the point), the
(A) Shear and normal stresses are zero
(B) Normal stress is maximum and shear
stress is zero
(C) Shear stress is maximum and normal
stress is zero
(D) Shear stress is maximum and normal
stress is non-zero
A Mohr circle with center (0, 0) and radius
xy is shown in the figure
The state of stress is represented by
0
0
4
P
4A
2
P
2A
0
4
2
P
P
0
2A
4A
For an isotropic elastic material the principal
axes of stress and the principal axes of strain
(A) Are independent of each other
(B) Coincide with each other
2
P
A
(B)
2
(D)
2.6
(A)
(A)
Strength of Materials
44
Kulkarni Academy
(B)
2.10
(C)
(D)
2.7
2.8
2.9
2.11
A point in a body is subjected to a bi-axial
state of stress, equal in magnitude but
opposite in nature. On a plane inclined at an
angle 450 with respect to the x- axis (passing
through the point), the
(A) Shear and normal stresses are zero
(B) Normal stress is maximum and shear
stress is zero
(C) Shear stress is maximum and normal
stress is zero
(D) Shear stress is maximum and normal
stress is non-zero
In a piece of material tensile stresses P1 and
P2 act on a mutually perpendicular planes
accompanied by a shear stress 3. The
condition for both the principal stresses to
have the same sign is
2
(A) P1 P2 q
(B) PP
1 2 q
2
2
(C) PP
(D) PP
1 2 q
1 2 q
Which one of the following statements is
true?
(A) In a tensile test on a rod made of ductile
material, failure occurs along a plane
making 450 with respect to the axis of
the rod.
(B) In a tensile test on a rod made of brittle
material, failure occurs along a plane
making 450 with respect to the axis of
the rod.
(C) In a torsion test on a rod made of ductile
material, failure occurs along a plane
making 450 with respect to the axis of
the rod.
(D) In a torsion test on a rod made of brittle
material, failure occurs along a plane
making 00 with respect to the axis of the
rod.
The normal stresses in a two-dimensional
stress state are 80 MPa and 20 MPa
respectively. The normal stress on the plane
of maximum shear stress is
(A) 50 MPa
(B) 30 MPa
(C) 80 MPa
(D) 20 MPa
The principal stresses at a point on a body
subjected to a state of plane stress are 10 MPa
and 20 MPa. The magnitude of shear stress
on a plane in which the normal stress is 12
Mpa is
(A) 3 MPa
2.12
(B) 4 MPa
(C) 5 MPa
(D) 8 MPa
The state of stress at a point in a body is
represented using components of stresses
along X and Y-directions as shown. Which
one of the following represents the state of
stress along X’ and Y’ axes ? (X’-axis is at
450 clockwise with respect to X - axis).
(A)
(B)
Kulkarni Academy
45
Principal Stresses, Strains & Mohr’s Circle
2.14
(C)
The state of stress at a point is as shown
below. Both the normal and shear stresses on
a plane, inclined at an angle of 450 with
horizontal are zero. If x y 200 MPa the
shear stress xy is
(D)
2.13
Mohr’s circle for the state of plane stress at a
point is shown in the figure. Unit of stress is
Mpa and the circle is drawn not to scale.
Which one of the following options (stress
values in MPa ) is true?
2.15
(A) 50 MPa
(B) 70 MPa
(C) 100 MPa
(D) 200 MPa
A wooden block of length 400 mm , width 50
mm and depth 100 mm is subjected to
uniaxial load as shown in the figure .An
inclined plane ABCD is shown which makes
an angle with the XZ plane and the line CD
is parallel to the Z- axis. The normal stress on
the plane ABCD is n1 when 300 and the
normal stress on the plane ABCD is n 2
when 1200 . The value of
(A) A 50, B 10, 1 30, 2 70
(B) A 50, B 20, 1 30, 2 50
(C) A 30, B 30, 1 30, 2 10
(D) A 20, B 10, 1 50, 2 30
n 2
is ______
n1
Strength of Materials
2.16
46
Kulkarni Academy
Which of the following figures represents the
state of stress at the point
The following state of plane stress exists at a
point P in a loaded body. (as shown in figure).
The combination of principal stresses which
corresponds to the given stress state is
max (MPa)
2.17
min (MPa)
(A)
4
0
(B)
4
–5
(C)
3
–3
(D)
5
–5
(A)
(B)
(C)
(D)
Common Data Questions 2.19 & 2.20
At a point in an object subjected to plane
stress conditions, the state of stress is as
shown in the Figure
The shear stress in MPa on the inclined
plane AB shown in the Figure is
2.19
2.18
(A) 5(1 3)
(B) 5(1 3)
(C) 5( 3 1)
(D) 5( 3 1)
The Mohr’s circle at a point is shown in
Figure
2.20
2.21
One of the principal stresses (in MPa ) is
(A) 40
(B) 80
(C) 120
(D) 140
The normal stress on the plane AB (in MPa)
is
(A) 30
(B) 70
(C) 100
(D) 110
The state of plane strain at a point is given by
x 60 10 6 , y 0 and xy 80 10 6 .
The principal strains are
(A) 0, 100 106
(B) 50 106 , 50 106
(C) 20 106 , 80 106
(D) 20 106 , 80 106
Kulkarni Academy
2.22
47
Principal Stresses, Strains & Mohr’s Circle
At a point in a stressed. body, the strains
measured in three directions inclined at
00 , 450 , 900 to x-axis are found to be
500 106 , 400 106
300 106
and
respectively. The shear strain at this point is
2.23
(A) 250 106
(B) 0
(C) 100 106
(D) 400 106
(A) 1 0.001 and 2 0.001
are glued at an inclined section CC. The bars
(B) 1 0.001 and 2 0.001
(C) 1 0.001 and 2 0.001
are subjected to an axial force P as shown in
(D) 1 0.001 and 2 0.001
Two uniform bars of cross -sectional area A
figure. The uniform shear stress developed in
2.26
the glued joint is
For a loaded body representing a two
dimensional
plane
problems,
the
displacement components along x and y at
any point ( x, y) are u x 2 y 2 , v 2 y
respectively. Principal strains at the point
(3,1) in the body are
(A) 1 8, 2 2
(B) 1 6.24, 2 1.76
2.24
(A)
P
sin cos
A
(B)
(C)
P
cos 2
A
(D)
(C) 1 0, 2 3
P 2
sin
A
P
cos
A
(D) 1 5, 2 3
2.27
The state of stress at a point in a loaded body
is given as x 40MPa,
xy 10MPa .
The state of stress at a point is as shown in
the Figure. The magnitude of the maximum
shear stress is
y 60MPa,
The sum of the principal
stresses at that point is
2.25
(A) 20 MPa
(B) 50 MPa
(C) 100 MPa
(D) 110 MPa
A rectangular sheet ABCD of dimensions a
and b along X and Y directions, respectively,
is stretched to a rectangular AB’C’D’, as
shown. The maximum principal strain (1 )
and minimum principal strain (2 ) due to the
stretch are given by
2.28
(A) 9 MPa
(B) 3 MPa
(C) 4.5 MPa
(D) 6 MPa
At a point in a body subjected to plane stress,
the state of stress is as shown in the figure.
One of the principal stresses is 180 MPa.
Find the unknown shear stress (in
MPa).____________
Strength of Materials
48
Kulkarni Academy
2.31
At a point in a body, 1 0.0004 and
2 0.00012 .
If
E 2 105 MPa
and
0.3, the smallest normal stress and the
largest shearing stress are
2.29
State of stress at a point in a strained body is
shown in Figure. Which one of the figure
given below represents correctly the Mohr’s
circle for the state of stress?
2.32
(A)
(B)
(C)
(D)
For
40 MPa and 40 MPa
0 MPa and 40 MPa
80 MPa and 0 MPa
0 MPa and 80 MPa
a point in a body subjected to a plane
stress condition ( x 100 MPa, y 50
MPa and xy yx 25 MPa), the maximum
principal stress in MPa is_____
2.33
(A)
(B)
(C)
(D)
2.30
The principal stresses at a point in twodimensional stress system are 1 and 2
corresponding principal strains are 1 and 2
. If E and denote Young’s modulus and
Poisson’s ratio, respectively, then which one
of the following is correct?
(A) E 1
E
[1 2 ]
(B)
1 2
E
[1 2 ]
(C)
1 2
(D) E (1 2 )
A circle of diameter a mm is scribed on an
unstressed aluminum plate of thickness t =
0.1 a mm. Forces acting in the plane of the
plate later cause normal stresses xx and
zz 2 . The elastic modulus is E GPa and
Poisson’s ratio is 0.25. Determine the change
in principal diameters AB and CD of the
original circle.
0.5 a
1.75 a
; CD
(A) AB
E
E
0.5 a
1.75 a
; CD
(B) AB
E
E
0.5 a
1.75 a
; CD
(C) AB
E
E
0.5 a
1.75 a
; CD
(D) AB
E
E
Kulkarni Academy
49
Answer Key
A
2.4
2.1
C
2.2
A
2.3
C
2.4
B
2.5
C
2.6
C
2.7
C
2.8
B
2.9
A
2.10
A
2.11
B
2.12
A
2.13
A
2.14
D
2.15
3
2.16
D
2.17
D
2.18
C
2.19
C
2.20
D
2.21
C
2.22
B
2.23
A
2.24
A
2.25
A
2.26
B
2.27
B
2.28
60
2.29
C
2.30
D
2.31
C
2.32
110.355
2.33
B
2.5
x y
2
y
x
cos 2 xy sin 2
2
x ' 0
0
y
x ' y ' x
sin 2 xy cos 2
2
x ' y '
(1)
2
(C)
(A)
x ' y '
Hence, the correct option is (A).
2.3
(C)
x '
Hence, the correct option is (C).
2.2
(B)
Hence, the correct option is (B).
Explanation
E
2.1
Principal Stresses, Strains & Mohr’s Circle
(C)
Hence, the correct option is (C).
2.6
(C)
Hence, the correct option is (C).
2.7
Hence, the correct option is (C).
P
sin cos
A
900
P
cos sin
A
2.8
00 ; 0 | 450 ;
0
0
(C)
P
| 900 ; 0
2A
450
P
2A
Hence, the correct option is (C).
900
0
(B)
Strength of Materials
50
PP
P P
1 1 2 1 2 q 2 (+ve)
2
2
2
Kulkarni Academy
2.12
(A)
PP
P P
2 1 2 1 2 q 2 (+ve when)
2
2
2
P1 P2 P1 P2
2
q
2 2
2
2
1
[4 P1 P2 ] q 2
4
2
PP
1 2 q
Hence, the correct option is (B).
2.9
(A)
Hence, the correct option is (A).
2.10
(A)
Normal stress on the plane of maximum shear
stress is average normal stress.
x y
80 20
50MPa
2
2
Hence, the correct option is (A).
avg
2.11
(B)
Given data :
1 20MPa , 2 10MPa
1 2
5MPa
2
From Mohr’s circle
Hence, the correct option is (A).
2.13
(A)
Hence, the correct option is (A).
2.14
(D)
( x ')450 0 x y 200MPa
(x ' y ' )450 0 xy ?
Radius
( x ' y ' )450
y
x
2
xy cos 2
0
sin 2
0
0
y x y
( x ' )450 x
cos 2
2 2
xy sin 2 0
200 200
0
0 xy .sin 90 0
2
52 32
4MPa
Hence, the correct option is (B).
xy 200MPa
Hence, the correct option is (D).
Kulkarni Academy
2.15
51
(3)
Principal Stresses, Strains & Mohr’s Circle
44
44
2
(3)
2
2
2
42 32
1,2 5MPa
max 5MPa , min 5MPa
Hence, the correct option is (D).
2.17
n1 ( 300 )
(D)
n 2 ( 1200 )
x
10 103 N
2MPa
50 100 mm2
n 2 cos 2 (90 120)
3
n1 cos 2 (90 30)
Or
x y x y
cos 2
n 2 2 2
2
n1 x y x y
cos 2
2 2
1
2 2
cos 2(90 120) 1 cos 600
2 2
3
0
2 2
1
cos120
cos 2(90 30)
2 2
Hence, the correct answer is (3).
2.16
y
x ' y ' x
sin 2 xy cos 2
2
10 10
0
0
sin 60 10cos 60
2
5 3 5
5[ 3 1]
Hence, the correct option is (D).
2.18
(D)
(C)
Hence, the correct option is (C).
2.19
(C)
Given data :
x 100MPa , y 40 MPa , xy 40 MPa
y
x y
2
x
xy
2
2
2
1,2
1,2 70 303 402
1,2 70 50
1,2
x y
2
y
2
x
xy
2
2
1 120MPa , 2 20 MPa
Hence, the correct option is (C).
Strength of Materials
2.20
52
(D)
( x ' ) AB
450
Kulkarni Academy
2.23
y x y
x
cos 2 xy sin 2
2 2
20
70 cos900 40 sin 900
2
(A)
Hence, the correct option is (A).
2.24
(A)
Hence, the correct option is (A).
2.25
(A)
70 40 110MPa
Hence, the correct option is (D).
2.21
(C)
x 60 106
y 0
xy 80 106
1,2
x y
2
y xy
x
2 2
2
2
0.001a
0.001
a
y 0.001
x
30 106 (30 106 )2 (40 106 ) 2
30 106 50 106
Hence, the correct option is (A).
1 80 106
2 20 10
2.26
6
v 2y
(B)
00 500 106
(B)
u x2 y 2
Hence, the correct option is (C).
2.22
1 2 0.001
du
dx
dv
y
dy
x
x y
450 400 106
900 300 106
y
x
2
2
x y
2
x y
2
x 2 x
2
x 6
y
x
2
x y 800 106
400 106 400 106
xy
xy
2
xy 0
Hence, the correct option is (B).
y 2
v u
x y
0 2y
Vxy
2
1,2
x y
2
y Vxy
x
2
2
2
2
Kulkarni Academy
53
62
62 2
2
2 2
2
Principal Stresses, Strains & Mohr’s Circle
2.29
2
Hence, the correct option is (C).
4 4 1
2.30
4 5
(B)
Hence, the correct option is (B).
1 6.236
2.31
2 4 5 1.76
(C)
1 0.004 , 2 0.00012
Hence, the correct option is (B).
2.27
(C)
E 2 105 MPa , 0.3
(B)
1
E
(1 2 )
(1 2 )
2
E
(2 1 )
(1 2 )
1
2 105
[0.0004 0.3(0.00012)]
(1 0.3)2
1 80MPa
2
y
2
x
xy
2
max
2
max
0 (3)2 3MPa
Hence, the correct option is (B).
2.28
2 105
[0.00012 0.3 0.0004]
(1 0.3)2
2 0MPa
y
x y
2
x
xy
2
2
2
1,2
(Largest)
(Smallest)
1 2 80
40MPa
2
2
Hence, the correct option is (C).
2.32
110.355
Given data :
x 100MPa
(60)
y 50MPa
1 180
2
160
160
2
180
xy
2
2
180 80 80
2
2
xy
xy yx 25MPa
1
x y
2
y
2
x
xy
2
1002 802 2xy
1 75 252 252
xy 60MPa
1 110.355MPa
Hence, the correct answer is 60.
2
Hence, the correct answer is 110.355
Strength of Materials
2.33
(C)
d a mm
t 0.1a mm
AB ?
CD ?
xx
zz 2
E E GPa
0.25
x
x
y z
E E
0.25
2
E
E
0.5
x
E
E
AB 0.5
a
E
AB 0.5a
E
z z y x
E E
2 0.25
()
E
E
CD 1.75
1.75a
CD
a
E
E
x
Hence, the correct option is (B).
54
Kulkarni Academy
3.1 Introduction
Freely expanding x y z 0
When a material is heated (or cooled) stress may
be developed and these stresses are known as
thermal stresses. If the material is free to expand
or contract no stresses are developed.
Material
value
Steel and concrete
12 106 / 0 C
In bar
invar 1.2 106 / 0 C
Aluminium
Al 23 106 / 0 C
Brass
brass 19 106 / 0 C
Copper
cu 16 106 / 0 C
3.2 Coefficient of Thermal
Expansion ( ) :
It represents strain per
temperature difference.
1
1
Unit : 0 or
K
C
L 1
.
L T
L LT
L ' L LT
unit
degree
NOTE
Thermal strains are reversible in Nature.
i.e. when the member is brought back to
the original temperature its original
shape is restored.
If the material is constrained in any
directions, stresses are developed in that
direction.
3.3 Expression for Thermal
Stresses :
Case : 1 Free expansion
L ' L(1 T )
Similarly,
t ' t (1 T )
b ' b (1 T )
L T
t T L t b
b T
0 (Free expansion)
T
Strength of Materials
56
Case : 2 A rod completely constrained in axial
direction.
Kulkarni Academy
Expansion prevented LT a
L T a
P( L a)
AE
L a ; a 0
E ( L T a)
L
Case : 4 A metal cube of side ‘L’ fixed in Xdirection only
dL 1
.
L T
dL L T
(Free expansion)
Expansion prevented L T
dL
PL
AE
L T
PL
AE
th TE (Compressive)
From the equation of thermal stress it can be
observed that the thermal stress is
independent on area and depends on material.
Case : 3 Partially constrained
TE
X
dX
0
L
Y Z
Calculation for Y :
Free expansion L T
Poisson effect due to constrained in X-direction
L 1
.
L T
L LT (Free expansion)
X L
E
Y L T
L
[ TE ]
E
Y LT [1 ]
Kulkarni Academy
57
Thermal Stresses
L T Free expansion
Case : 5 Cube of length ‘L’ constrained in X
and Y direction.
L
(Direct compression)
E
L
(Poisson Y-effect)
E
L
(Poisson Z-effect)
E
X 0
TE
(1 )
Z 0
Case : 7
X
Y
z L T
L L
E
E
z L T
2L TE
E 1
A bar AB fixed at one end A and supported by a
spring at B as shown in figure and this system is
heated by T . Find the stress developed in the
bar.
1
Z TL
1
Case : 6 A cube constrained in all directions.
X 0
Y 0
Z 0
L L L
L T
0
E
E
E
TE
(1 2)
dL
PL FS L
AE AE
Free expansion L T
Net expansion of bar L T
FS L
AE
Strength of Materials
58
Compression of spring
Kulkarni Academy
Case : 2
FS
K
Expansion = Compression
L T
FS L FS
AE K
L T
FS
A
L A
E K
L T
TE
AE
L A
1
E K
KL
R
1
A1
TE
AE
1
KL
R 1 A1
3.4 Composite bar
R
2
A2
Case : 1
R 2 A2
1 A1 2 A2
…..(i)
Total free expansion
L11T L22T
Compression due to R
Cu Compressive
RL1
RL2
A1E1 A2 E2
Steel Tension
As there is NO external force therefore
compressive force in copper should be equal to
tensile force in steel.
PCu PSteel
Cu ACu S AS
…..(i)
Net change in length of copper = Net change in
length of steel
LCu T
PCu L
PL
L S T S
ACu ECu
AS ES
S Cu
(Cu S )T
ES ECu
…..(ii)
Expansion = Compression
L1 L2
L11T L2 2T
E1
E2
…..(ii)
Kulkarni Academy
P
3.1
59
Thermal Stresses
(A) 10 E1 E2
Practice Questions
(B) 20 E1 E2
The strain induced in an unconstrained rod
that is heated uniformly depends
(B) Only on the temperature change
EE
(C) 20 1 2
E1 E2
(C) Neither on the coefficient of expansion
nor on the temperature change
EE
(D) 10 1 2
E1 E2
(A) Only on the coefficient of expansion
(D) Both on the coefficient of expansion and
on the temperature change
3.2
3.4
An intially unstressed rod of length L is fixed
at both ends, as shown in figure. The rod is
heated uniformly, to raise the temperature by
T . The young’s modulus is E, Poisson’s
ratio is v and coefficient of thermal expansion
is . The axial stress in the rod, assuming the
rod to be weightless, is
A cube, made of aluminium, of dimension
0.1m 0.1m 0.1m, rests against a rigid wall
(where normal in the y-directions) as shown
in the figure. Another parallel rigid wall is
located at a clearance of 0.2 mm from the
block. Assuming all contacts to be
frictionless, if the block is heated by
T 1500 C, the normal stress yy induced
in the
block
is
(for
aluminum
6 0
E 70GPa: v 0.3 ; 20 10 / C )
3.3
ET
1 v 1 2v
(A) Zero
(B)
(C) ET
(D) 2ET
Two bars of differently young’s moduli, E1
and E2 , but with the same cross sectional
(A) yy 7 MPa
(B) yy 7 MPa
area, A, and coefficient of thermal expansion,
, are attached together at one end and fixed
at the other as shown in figure.
(C) yy 70MPa
(D) yy 0
The construction of this setup was carried out
at an ambient temperature of 250 Celsius.
The stress in the bars when the temperature is
uniformly increased by 100 Celsius is
3.5
A solid steel cube constrained on all six faces
is heated so that the temperature rises
uniformly by T . If the thermal coefficient
of the material is , Young’s modulus is E
and the Poisson’s ratio is v, the thermal stress
developed in the cube due to heating is
(A)
T E
1 2
(B)
2 T E
1 2
(C)
3 T E
1 2
(D)
T E
3 1 2
Strength of Materials
3.6
60
Determine the temperature rise necessary to
induce buckling in a 1m long circular rod of
diameter 40 mm shown in the figure below.
Assume the rod to be pinned at its ends and
the coefficient of thermal expansion as
Kulkarni Academy
Common Data Questions 3.9 & 3.10
A steel bar of rectangular cross-section is
heated uniformly and the rise in the
temperature is T . The Young’s modulus is
E, the Poisson’s ratio is v and the coefficient
of thermal expansion is . The bar is
completely restrained in the axial direction
and lateral directions.
2.0 106 / 0 C. Assume uniform heating of
the bar.
3.9
3.7
At a temperature of 400 C , a rod tightly fits
between two rigid walls such that the
compressive stress in the rod is 60 MPa.
Given E 200GPa and 20 106 / 0 C ,
find the temperature at which the rod will just
lose contact with the walls.
3.8
A composite system of two metal bars, as
shown below, is made of two dissimilar
materials having areas of cross section A1 and
A2. Young’s moduli E1 and E2 and
coefficients of thermal expansion 1 and 2 .
If the temperature of the system is raised by
T , then the resultant axial force required to
be applied to the rigid end plates to maintain
the same length L is
3.10
The thermal stress developed in the bar along
the axial-direction is
(A)
ET
1 2
(B) ET
(C)
ET
1 2
(D)
Assume that the bar is allowed to deform
freely in the lateral directions, while keeping
the axial direction restrained. The percentage
change in the magnitude of axial thermal
stress for v 0.25 is
(A) 0
(B) 25
(C) 50
(D) 100
Common Data Questions 3.11 & 3.12
Two roads are joined together and the entire
assembly is supported between two rigid
walls, as shown in the figure. The crosssectional area and Young’s modulus for both
the rods are 0.01m2 and 10 GPa, respectively.
The coefficient of thermal expansion for the
two rods are 1 4 106 / 0 C and
2 106 / 0 C
(A)
ET
1 2
respectively.
The
entire
assembly is heated by 1000 C. Neglect the
effect of Poisson’s ratio.
E11 A1 E22 A2 T
(B) 1/ E11 A1 1/ E22 A2 T
1
(C)
E1 E2 1 2 A1 A2 T
(D)
E11 A1 / E22 A2 T
Kulkarni Academy
3.11
3.12
61
The stress in rod 1 (in MPa) is
(A) 4.0
(B) 3.0
(C) 2.5
(D) 1.0
Thermal Stresses
3.14
19 106 per 0C respectively, the temperature
rise necessary to cause all the applied load to
be supported by the steel rods is
Considering the displacement to the right as
positive, the displacement (in mm) of the
interface between the two rods is
(A) 0.2
(B) 0.1
(C) 0.1
(D) 0.2
If the coefficient of thermal expansion for
steel and bronze are 11106 per 0C and
3.15
Common Data Questions 3.13 & 3.14
A rigid block weighing 60 kN is supported by
three rods symmetrically placed as shown in
figure. The lower ends of the rods are
assumed to have been at the same level before
the block is attached. The cross-sectional
areas of the rods and the modulus of elasticity
of the materials of the rods are given as
(A) 100 C
(B) 10.580 C
(C) 150 C
(D) 20.650 C
A composite bar of length ‘L’ is made of a
centrally placed steel plate (50 mm wide 10
mm thick) with two copper plates (each 30
mm wide 5 mm thick) connected rigidly on
each side. If the temperature of the composite
bar is raised by 500C find the stress developed
in each copper plate in MPa
(For Steel : Es 2 105 MPa and
s 12 106 / 0C;
For copper : Ec 1105 MPa and
c 17 106 / 0C )
Esteel 2.11011 Pa
3.16
Ebronze 0.98 10 Pa
11
Asteel 5 104 m2
Abronze 10 104 m2
3.13
The stress in the steel rod is
(A) 48.65 MPa
(B) 52.35 MPa
(C) 60 MPa
(D) 40 MPa
A steel frame as shown in figure is fitted with
an equal length of an aluminum rod at room
temperature (total area of steel = area of
aluminum = 200mm2). When fitted they are
in stress free state.
Given
Ea 70GPa, Es 210 GPa
and
a 25 106 / 0C and s 12.5 106 / 0C,
for a temperature rise of 800 C the load in the
aluminum bar is
Strength of Materials
62
Kulkarni Academy
A
3.17
(A) 21.0 kN
(B) 18.0 kN
(C) 15.8 kN
(D) 10.5 kN
A cantilever rod of length L, area of circular
cross section A and moment of inertia I, is
subjected to temperature change. The rod has
a space for free expansion after which the
free end gets locked and fixed into the wall at
B. Find the expression for increase in
temperature T at which the elastic
Answer Key
3.1
C
3.2
C
3.3
C
3.4
C
3.5
A
3.6
49.340 C
3.7
250 C
3.8
A
3.9
C
3.10
C
3.11
B
3.12
D
3.13
A
3.14
B
3.15
19.23
3.16
D
3.17
A
E
3.1
Explanation
(C)
Hence, the correct option is (C).
3.2
(C)
Hence, the correct option is (C).
3.3
(C)
instability first occurs. Assume E to be
constant with temperature change and as
the coefficient of linear expansion.
T 100 C
1 2
(A) T
1
421
2
L
AL 1 / L
(B) T
1
21
2
L
AL 1 / L
(C) T
1
1
L
4AL
2
1
421
(D) T 2
L
AL
Free expansion :
L (10) L (20) 20 L
Compression :
PL PL
AE1 AE2
Expansion = Compression
20 L
PL 1
1
A E1 E2
EE
20 1 2
E1 E2
Hence, the correct option is (C).
Kulkarni Academy
3.4
63
Thermal Stresses
(C)
TE
(1 2)
L T Free expression
T 150 C ,
0
E 70GPa , 0.3
20 106 / 0 C
L
(Direct compression)
E
L
(Poisson Y - effect)
E
L
(Poisson Z - effect)
E
Hence, the correct option is (A).
0
L T a
P( L a )
AE
3.6
0.1
0.1 20 10 150 0.2 10
70 109
6
3
70MPa (Compressive)
Hence, the correct option is (C).
3.5
(A)
49.340C
When both the ends are pinned the buckling load
2 EI
P 2
L
I MOI
When both the ends are fixed bucking load is
P
42 EI
L2
P
2 EI
L2
P
TE
A
X 0
Y 0
Z 0
2 EI
TE
AL2
(40 103 ) 4
64
20 106 T
3 2
(40 10 ) 1
4
2
L T
L L L
0
E
E
E
T 49.340 C
Hence, the correct answer is 49.340C.
Strength of Materials
3.7
64
250C
Kulkarni Academy
3.9
(C)
Hence, the correct option is (C)
3.10
(C)
Free in lateral direction and axial direction
restrained.
2 TE
C 60MPa (Compressive)
Constrained in all sides
E 200GPa
1
20 106 / 0 C
TE
1 2 TE (– ve means compressive)
60 20 106 200 103 (Tf 400 C)
2 TE
T f 25 C
0
0
Hence, the correct answer is 25 C.
3.8
(A)
TE
if 0.25
(1 2)
% change
2 1
100 50%
2
Hence, the correct option is (C)
3.11
(B)
A1 A2 0.01m2
1 4 106 / 0 C
E1 E2 10GPa
2 106 / 0 C
T 1000 C
L1T
P1 1TA1E1
P2 L
A2 E2
L1 2m
P2 2TA2 E2
L2 1m
L 2T
PL
1
A1E1
P P1 P2
P (1 A1E1 2 A2 E2 ) T
Hence, the correct option is (A).
Free expansion,
L11T L2 2T
RL1 RL2
A1E1 A2 E2
Kulkarni Academy
65
2 4 106 100 1106 100
Thermal Stresses
Given that,
W 60kN ,
R 2
1
3
10 10 0.01 0.01
9 104
ES 2.11011 pa
AS 5 104 m2
R
[3]
0.0110 103
Ebr 0.98 1011 pa
R 0.03MN
Ab 10 104 m2
R 0.03
1
3MPa (Compressive)
A1 0.01
W 2PS Pb
steel bronze
Hence, the correct option is (B)
3.12
…..(i)
PS 0.5
5 104 2.11011
(D)
Pb 1
10 10 0.98 1011
4
PS 2.1428 Pb
60 2 PS
S
Shift = 0.8 0.6 0.2mm
RL1
0.6mm
A1E1
Hence, the correct option is (D)
3.13
(A)
PS
2.1428
PS 24.32kN
Free expansion L11T 0.8mm
Compression due to R
…..(ii)
PS
48.648MPa
AS
Hence, the correct option is (A)
3.14
(B)
S 11106 / 0 C
b 19 106 / 0 C
Lbb L LS S T
b
35.68
35.680MPa
Ab
FS LS
AS ES
T 10.580 C
Hence, the correct option is (B)
Strength of Materials
3.15
19.23
66
Kulkarni Academy
S Al
ES 2 105 MPa , S 12 106 / 0 C
Al S
( Al S )T
EAl ES
EC 1105 MPa , C 17 106 / 0 C
Al 0.0525GPa
Given that,
( Al S )
Load in Al bar Al AAl
0.0525 103 MPa 200
10500 N
10.5kN
PS Pcu Pcu (because load of each copper)
PS 2Pcu
S (10 50) 2 cu (10 300)
Hence, the correct option is (D)
3.17
(A)
S 0.6 cu
S cu
( cu S ) T
ES Ecu
0.6 cu
cu 5 (17 106 12 106 ) 50
5
2 10 110
cu 19.23MPa
Hence, the correct answer is 19.23.
3.16
Free expansion LT
Expansion prevented LT
(D)
Pcr
P( L )
AE
42 EI
( L ) 2
L T
42 EI ( L )
.
( L )2 AE
L T
4 2 I
A( L )
1
4 2 I
T
L
AL(1 / L)
PS PAl
S AS Al AAl
Hence, the correct option is (A).
4.1 Introduction
Circumferential Stress
It is a container which is used for carrying fluid
under pressure.
If
d
20 then such a shell is known as thin
t
shells.
Example : Water bottles, soft drink cans etc.
When a cylinder is subjected to internal
pressure it is subjected to
1.
Circumferential or hoop stress
2.
Longitudinal stress
3.
Radial stress
When a pressure Vessel is subjected to
internal pressure for calculation purposes
Total vertical upward force Prl sin d
0
gauge pressure is taken into account because
externally it is subjected to atmospheric
Total downward force c (2 t l )
pressure.
4.2 Thin cylindrical shells
For static equilibrium
Fupward Fdownward
Prl sin d 2 t l
c
0
Prl [cos cos0] 2ct l
c
Pr Pd
(Circumferential stress)
t
2t
Strength of Materials
Longitudinal Stress : This stress is
developed only when both ends are closed.
68
Kulkarni Academy
1 c
2 l
1 c
Pd
Pd
2
22
2t
4t
1 22
P d 2 l (dt )
4
l
Pd
4t
Whenever a cylindrical shell is subjected to
internal pressure both longitudinal and
circumferential (hoop) stresses are tensile in
Nature i.e. longitudinal stress results is
increase in length and circumferential stress
results in increase in diameter.
Therefore, when a pressure vessel is
subjected to external pressure, the
circumferential and longitudinal stresses are
compressive in Nature.
If a cylinder is subjected to external pressure
length and diameter decreases.
These hoop stress and longitudinal stresses are
principal stresses because these are the maximum
and minimum normal stresses without shear.
Therefore the maximum principal stress
Pd
and the
(1 c ) is hoop stress i.e. 1
2t
minimum principal stress is longitudinal stress is
Pd
equal to
.
4t
4.3 Strain Analysis
1. Change in length :
dL L c
L
E
E
Pd Pd
L
4tE 2tE
L
L
Pd
(1 2)
4tE
L
PdL
(1 2)
4tE
2. Change in diameter :
C
dd
d
C
c L
E
E
C
Pd Pd
2tE 4 Et
C
Pd
[2 ]
4 tE
Change in diameter(dd )
Stressed element
Pd 2
[2 ]
4 tE
Kulkarni Academy
69
3. Change in volume :
Thin Shells
Maximum shear stress in the plane of
1 and r (radial stress)
V 2 c L
Pd
Pd
2
(2 )
(1 2)
4tE
4tE
Pd
V
(5 4)
4tE
(max )1 r
4.4 Maximum shear stress
1 r
2
( r 0 )
1 Pd
2
4t
Maximum shear stress in the plane of
2 and r
(max )2r
2 r
2
2 Pd
2
8t
Absolute maximum shear stress :
Maximum of
Maximum
1 2 1 r 2 r
,
,
2
2
2
Pd Pd Pd
,
,
8t 4t 8t
Absolute max or maximum shear stress
max
1 2 c l
2
2
4.5 Thin spherical shells
1 Pd Pd
2 2t
4t
max
Pd
, Maximum in plane
8t
Shear stress
Pd
8t
This is the maximum in plane shear stress.
When a thin cylinder is subjected to internal
pressure, the radial stress which is equal to
pressure is very small compare to 1 and 2 .
Therefore, in thin cylindrical shells only two
stresses are taken. i.e. 1 and 2 .
P d 2 c (dt )
4
c
Pd
4t
Pd
4t
Strength of Materials
70
For spherical pressure shell
Pd
c l
4t
Kulkarni Academy
Maximum shear stress :
max
1 2
2
Mohr’s Circle for thin spherical pressure shell
max 0 (in a plane, shear stress = 0)
(in the plane of 1 and 2 )
In the plane of 1 and r
Mohr’s circle is a point for thin spherical shell
subjected to internal pressure.
When a thin spherical pressure shell subjected to
internal pressure the principal stresses are :
1 2
Pd
4t
3Pd
V
(1 )
4tE
1 r 1
2
2
max
Pd
8t
Absolute maximum shear stress :
Strain analysis :
Pd
1 c
4t
Pd
2 l
4t
( c l )
d 1 2 c l
E
E
E
E
Pd
(1 )
d c (1 )
4tE
E
Pd
d
(1 )
4tE
V 3 d
max
( r 0 )
2 1 r 2 r
max Max 1
,
,
2
2
2
Pd Pd
Max 0,
,
8 t 8t
(max )absolute
Pd
8t
Maximum shear stress is zero in the plane of l
and c . Whereas the absolute maximum shear
stress is
Pd
.
8t
Kulkarni Academy
71
4.6 Thin cylinder with
Hemispherical ends
1. Same
stress
at
the
junction
(i.e.
circumferential stress).
(c )cyl (c )sp
Pd Pd
2t c 4t s
tc 2ts
…..(i)
2. In order to prevent distortion or cracks at the
junction.
For no distortion the circumferential
strain in the cylinder and sphere must be
same.
(c )cyl
Pd
(2 )
4tc E
(c ) sp
Pd
(1 )
4ts E
(c )cy (c )sp
tc 2 v
ts 1 v
…..(ii)
If these two conditions are to be valid
simultaneously then should be equal to zero
and for metal this is impossible.
From design point of view, spherical pressure
vessels
are
preferred
over
cylindrical
pressure vessel because of uniform lower
circumferential stress.
Thin Shells
Strength of Materials
72
4.5
Practice Questions
P
Kulkarni Academy
A thin cylinder with closed lids is subjected
to internal pressure. State of stress at point x
4.1
A thin walled cylindrical pressure vessel has
is given by.
diameter ‘D’ and length ‘L’. It is closed at
both ends and subjected to internal pressure
‘P’. The principal stresses, if the metal
thickness is ‘t’, are
(A)
pD pD
,
t 2t
(B)
pD pD
,
2t 3t
pD pD
pD pD
(D)
,
,
2t 4t
t 3t
A Cylindrical tank with closed ends is filled
(A)
(C)
4.2
(B)
with compressed air at a pressure of 500 kPa.
The inner radius of the tank is 2 m, and it has
wall thickness of 10 mm. The magnitude of
(C)
(D)
maximum in-plane shear stress (in MPa)
is______.
4.3
A thin walled spherical shell is subjected to
4.6
an internal pressure. If the radius of the shell
200 mm and wall thickness 10mm is filled
is increased by 1% and the thickness is
with a gas at pressure 10MPa. The maximum
reduced by 1% with the internal pressure
shear stress developed in the wall is
remaining the same, the percentage change in
(A) 25 MPa
(B) 50 MPa
(C) 100 MPa
(D) 200 MPa
the circumferential (hoop) stress is
4.4
A closed thin circular cylinder of diameter
(A) 0
(B) 1
(C) 1.08
(D) 2.02
A thin cylindrical tube of inner diameter d,
thickness t is closed at both ends and is
subjected to internal pressure p’. The tube
also carries a torque T. The stresses at any
(C)
Pd Pd T
,
,
2t 4t d 2t
Pd Pd T
(B)
,
,
2t 2t d 2t
(D)
An open ended thin – walled straight pipe is
made of a material that can carry a maximum
shear stress of max . The pipe is of diameter,
d and thickness, t The maximum internal
pressure allowable is given by (neglecting the
normal stress in the radial direction)
point (x , and x ) are
Pd Pd 2T
(A)
,
,
2t 4t d 2t
4.7
Pd Pd 2T
,
,
4t 2t d 2t
t
(A) 4 max
d
(C)
t
2 max
d
t
(B) 2 max
d
t
(D) max
d
Kulkarni Academy
4.8
73
A thin walled cylinder with open ends is
4.12
A cylindrical pressure vessel of 3 m outside
subjected to uniform internal pressure p
diameter is 10.8 m long. The wall thickness
alone. The wall thickness is t, internal radius
is 25 mm and it is subjected to an internal
is r and the Young’s Modulus is E. The
pressure of 800 kPa. Find the change in
increase in radius of the cylinder due to the
diameter and length, given Young’s modulus
internal pressure is
200GPa and Poisson’s ratio 0.3.
(A) Zero
(B)
(C)
pt 2
2 Er
4.13
pr
Et
(D)
A cast iron pipe of 1 m diameter is required
to withstand a 200 m head of water. If the
2
2
4.9
Thin Shells
pr
r
Et
limiting tensile stress of the pipe material is
20 MPa, then the thickness of the pipe will be
A thin circular cylindrical vessel of diameter
d, length l and wall thickness t is subjected to
an internal pressure p. The minimum
principal stress is
(A) 4
pd
, where k is
kt
4.14
(A) 25 mm
(B) 50 mm
(C) 75 mm
(D) 100 mm
Assertion (A): In cylindrical shells with
hemispherical ends, the hemispherical ends
(B) 3
are thicker than the cylindrical section.
(C) 2
(D) 1
Reason (R): The value of Poisson’s ratio for
Common Data Questions 4.10 & 4.11
most metals is 0.3.
A thin pressure vessel of spherical shape has
wall thickness of 10 mm and 1800 mm
diameter.
Its
Young’s
modulus
4.15
is
and 5 mm thickness is subjected to an internal
E 210GPa and Poisson’s ratio 0.3
4.10
pressure of 10 MPa and a torque of 2000 Nm.
Calculate the magnitude of the principal
If the pressure vessel experiences an internal
stresses.
pressure of 0.95 MPa, the approximate
maximum membrane stress is
4.11
A thin cylinder of 100 mm internal diameter
4.16
A compressed air tank having an inner
(A) 35 MPa
(B) 43 MPa
diameter of 480 mm and a wall thickness of
(C) 50 MPa
(D) 57 MPa
8mm is formed by welding two steel
The change in its diameter is
hemispheres. If the allowable shear stress in
the steel is 40 MPa, find the maximum
(A) 2 mm
(B) 1.0 mm
(C) 0.13 mm
(D) 0.25 mm
permissible pressure (in MPa) inside the tank.
Strength of Materials
74
Common Data Questions 4.17 & 4.18
A steel cylindrical pressure vessel has an
inner radius of 1.8 m and a wall thickness of
20 mm.
4.17
For an internal pressure of 800 kPa, the
maximum shear stress for the cylindrical part
of the vassel is
4.18
(A) 16 MPa
(B) 18 MPa
(C) 20 MPa
(D) 0
At which of the following internal pressure
will the cylindrical vessel yield as per the
Tresca criterion if the yield strength of the
material in tension is 320 MPa
4.19
(A) 3.55 MPa
(B) 7.1 MPa
(C) 1.775 MPa
(D) 4.0 MPa
A thin walled cylindrical pressure vessel
having mean radius 100 mm and wall
thickness 5 mm, is subjected to internal
pressure p. If the factor of safety is 2 and the
yield stress in shear is 100 MPa, find the
maximum value of p (in MPa).
4.20
The thin walled cylinder can be supported in
one of two ways as shown. Determine the
state of stress in the wall of the cylinder for
both cases if the piston P causes the internal
pressure to be 450 kPa.
Kulkarni Academy
The wall has a thickness of 6 mm and the
internal diameter of the cylinder is 200 mm.
Kulkarni Academy
A
75
Answer Key
4.4
3
C
4.2
25
4.3
D
4.4
D
4.5
D
4.6
B
4.7
A
4.8
C
4.9
A
4.10
B
4.11
D
4.12
0.5097
4.13
B
4.14
B
4.15
100
4.16
5.33
4.17
B
4.18
B
4.19
10
4.20.
E
4.1
Thin Shells
(a)
L 0 MPa, c 7.5 MPa
(b)
L 3.75 MPa, c 7.5 MPa
(D)
When a cylinder is subjected to internal pressure
and torque then it will be subjected to normal and
shear stresses.
Explanation
(C)
x
Pd
4t
Pd
2t
Hence, the correct option is (C)
4.2
25
(max )in plane
Pd
8t
4.5
500 10 4000
25 106 Pascal
8 10
(D)
3
Hence, the correct option is (D)
4.6
(B)
Hence, the correct answer is 25.
max
(D)
r r
r 1.01r
t t
t 0.99 t
c
2T
d 2t
Hence, the correct option is (D)
25 MPa
4.3
T
2T
2
2 r t d 2t
Pd Pr
4t
2t
c'
Pd 10 200
50MPa
4t
4 10
Hence, the correct option is (B)
4.7
(A)
P(1.01) r
t (0.99) t
c' 1.0202 c
% change
c ' c
100
c
1.0202 1
100 2.02%
1
Hence, the correct option is (D)
L 0 , C 0
max
Pd
0
Pd
2t
2
4t
Strength of Materials
max
Pmax
76
Pd
4t
Kulkarni Academy
4.11
c
t
4 max
d
(C)
Open ended thin cylinder (L 0, C 0)
Increase in radius due to internal pressure.
r C
C
L
r
E
E
r Pr
r tE
Pr 2
r
tE
Pd 2
(1 )
4tE
0.95 (1800)2
[1 0.3] 0.2565mm
4 10 210 103
Hence, the correct option is (D).
0.5097
Cylindrical pressure vessel
D0 3m , L 10.8m
t 25mm , P 800kPa
E 200GPa , 0.3
Hence, the correct option is (C)
4.9
4.12
0
d Pd
(1 )
d
4tE
d
Hence, the correct option is (A)
4.8
(D)
d 3 (2 25) 2.95m
(A)
Minimum principal stress
l
pd
4t
d Pd
(2 )
d
4tE
k 4
So,
Hence, the correct option is (A).
4.10
d
(B)
Thin spherical shell
0.800 (2.95) 2
(2 0.3)
25
3
4
(200 10 )
1000
t 10mm
d 5.9177 104 m 0.59177mm
d 1800mm
L Pd
(1 2)
L 4tE
E 210GPa
0.3
L
Pinternal 0.95MPa
Maximum membrane stress
Pd
4t
42.75MPa 43MPa
Hence, the correct option is (B)
0.8 (2.95)10.8
(1 0.6)
25
3
4
200 10
1000
L 5.0976 104 m
L 0.5097mm
Hence, the correct answer is 0.5097.
Kulkarni Academy
4.13
77
(B)
Thin Shells
4.16
5.33
Given data :
d 1mm
d 480mm , t 8mm
P gh 1000 9.81 200
P gh 1.962MPa
L
Pd
Pd
C
2t
4t
(Both are tensile stress)
allow 40MPa
For sphere max
σlimiting 20MPa
20
1.962 1000
2t
t 24.525 2 50mm
Hence, the correct option is (B).
4.14
40
P 480
88
P 5.33MPa
Hence, the correct answer is 5.33 MPa.
4.17
(B)
Given data :
(B)
d 1.8m 1800mm 2
Hence, the correct option is (B).
4.15
Pd
8t
t 20mm
P 800kPa
100
max
Given data :
d 100mm , t 5mm
P 10MPa , T 2000 N-m
Pd 800 1800 2
8t
8 20
max 18 103 kPa 18MPa
Hence, the correct option is (B).
4.18
(B)
1
Pd
2t
1
10 100
25
max
1 100MPa
max
Pd
4t
Pd
4t
160
P 1800
4 20
2
2 50MPa
Hence, the correct answer is 100 MPa.
P?
y
2
320
160MPa
2
P 7.1MPa
Hence, the correct option is (B).
Strength of Materials
4.19
78
Kulkarni Academy
L 0
10
dmean 200mm
L
Pd
3.75MPa
4t
C
Pd
7.5MPa
2t
rmean 100mm
t 5mm
FOS = 2
allow
100
50MPa
2
Yield stress in shear 100MPa
max
Pd
4t
P 10MPa
Hence, the correct answer is 10MPa.
4.20
Hence, the correct answer is
L 0
C
Pd
2t
450 200
7500kPa 7.5MPa
2 6
(a)
L 0 MPa, c 7.5 MPa
(b)
L 3.75 MPa, c 7.5 MPa
5.1 Introduction
Beam :
A slender long straight members carrying loads
perpendicular to the longitudinal axis of the
member are known as beams. Therefore beams
are transverse loaded members. In case of bars
the load acts along the axis of the member.
Shear force :
It is numerically equal to the algebraic sum of all
transverse component of external forces
(including reaction components) acting on an
isolated segment but in opposite direction to
satisfy equilibrium conditions.
NOTE
It is convenient to consider a beam that
has minimum number of loads.
Sign convention for shear force :
Right face
These beams are known as planer structures
because all the loads act in the plane of figure.
Types of beams :
Left face
+ve shear force
Example 1
–ve shear force
Right face
Left face
Strength of Materials
80
Kulkarni Academy
Bending moment :
It is numerically equal to the algebraic sum of
moments of all forces including reactive forces
and couples on an isolated system.
Sign convention for bending moment
Shear force diagram :
It is a diagram which shows variation of shear
force along the length of the beam.
BMD :
It is a diagram which shows variation of
bending moment along the length of the
beam.
Example 3
Example 2
Example 4
Kulkarni Academy
R 2P ( P P) 0
CW CW CW CCW
PL PL P(3L) 2 P(4 L) 3PL (CCW )
81
Shear Force and Bending Moment Diagrams
Example 6
Resisting moment
3PL(CW)
M x 3PL P( x L)
M C 2PL , M D 2PL
Example 5
RA RB 0 ,
MA 0
C0 RB L 0 ,
RB
Example 7
RA RD 42kN
MA 0
RD 5 28 5 14 2
RD 24kN
RA 18kN
M x 18x 14( x 2)
x5
M C 48kN-m
C0
, RA RB ,
L
Strength of Materials
Mx
82
Kulkarni Academy
Pb
Pab
x
L
L
ab
Pa Pb
L
L
Example 8
14
7
W W
12
6
7
1
RD W W W
6
6
RA
Example 10
RA RB F
MA 0
RB L Fd Fa
RB
F (a d )
L
RA F RB
RA
F (b d )
L
Example 9
Draw SFD for the beam shown in figure
RF Y6W 2W
RA RD W
MD 0
RA RB
WL
2
Kulkarni Academy
83
Shear Force and Bending Moment Diagrams
NOTE
The slope of SFD is equal to negative of
rate of loading.
M0 0
VX
M Vdx wdx.
WL
Wx
2
SF 0 at x
dx
(M dM ) 0
2
dM
v
dx
L
2
BM :
…..(ii)
NOTE
MX
WL
WX
x
2
2
M max
2
Slope of BM diagram = Shear force.
Example 11
L
WL2
at x
2
8
Graphical method for constructing SFD
and BMD :
Fig.
Fy 0
V wdx (V dV ) 0
( w Rate of loading)
dV
Slope of SFD.
dx
dV
w
dx
…..(i)
Strength of Materials
84
Kulkarni Academy
dV
w
dx
dM
V
dx
dM Vdx
w
Negative slope
M2
x2
M1
x1
dM Vdx
dM
V
dx
V 0
Change in B.M = Area of SFD between these two
points.
Effect of point load on SFD and BMD :
dM
0
dx
Observation :
(Maximum B.M)
FY 0
dV
w
dx
V2
X2
V1
X1
dV wd x
dV P
When P acts downwards on the beam dV P
i.e. there is a sudden drop in SFD due to
downward force. Similarly if the SFD rising
suddenly in SFD then it is a indication of upward
point load.
M0 0
M Vdx P
V2 V1 w( X 2 X1 )
Difference between S.F = Area of loading
diagram between these two points.
dx
(M dM ) 0
2
P
V dx dM
2
P
dM V dx
2
dx 0
dM 0
Kulkarni Academy
85
Shear Force and Bending Moment Diagrams
dM
V
dx
dV P
Due to point load the change in B.M is zero but
due to point load the S.F. on the left and right side
is different therefore the slope of the bending
moment diagram changes due to point load.
Downward point load (BMD)
Due to concentrated moment there is a sudden
change in BMD i.e. if there is a clockwise
concentrated moment there is a sudden rise in
bending moment similarly if there is an
Upward point load (BMD)
Effect of concentrated moment on
SFD and BMD :
anticlockwise concentrated moment, there is a
sudden fall in bending moment.
Point of Contraflexure : The point where the
BM change its sign, may be from hogging to
sagging or sagging to hogging is known as point
of contraflexure or point of inflection, at this
point the bending moment changes to zero. i.e. at
the P.O.C. the bending moment is zero.
Fy 0
V (V dV ) 0
dV 0
(V Constant)
Due to concentrated moment there is no
change in shear force.
M0 0
Vdx M M 0 (M dM ) 0
dx 0
dM M 0
BMD
BMD
Strength of Materials
Practice Questions
P
5.1
86
Kulkarni Academy
5.4
For the beam shown, Which of the following
are discontinuous at the mid-span ?
A beam of length 2.0 m is simply supported
at both ends as shown in figure. For an
unknown load X at its mid-point, the beam
experiences a shear force represented by
SFD. The X is
(A) A moment of 1000 Nm
(A) Bending moment only
(B) A concentrated force of 1000 N
(B) Axial force and bending moment only
(C) A moment of 500 Nm
(C) Bending moment and shear force only
(D) Axial force, bending moment and shear
force
5.2
(D) A concentrated force of 500 N
5.5
The beam shown below carries two external
moments. A counterclockwise moment of
magnitude 2M acts at point B and a
clockwise moment of magnitude M acts at
the free end, C. The beam is fixed at A. The
shear force at a section close to the fixed end
is equal to
5.6
5.3
(A) 2M/L
(B) M/L
(C) 0
(D) – M/L
The bending moment at point B on the
cantilever beam in the figure is
(A) – 2.0 PL
(B) 1.0 PL
(C) – 0.5 PL
(D) – 1.5 PL
A simply supported beam with an
overhanging end is loaded as shown below.
The maximum bending moment in the beam
is
A cantilever beam is subjected to a force and
a moment as shown in the figure. The
bending moment at the fixed end is
5.7
(A) 1 PL
(B) 2 PL
(C) 3 PL
(D) 4 PL
(A) 2 kNm
(B) 1 kNm
(C) 0.75 kNm
(D) 0.25 kNm
A beam having both sides overhang is loaded
as shown in figure. To make moments equal
in both supports, the ratio of w and q has to
be
Kulkarni Academy
87
Shear Force and Bending Moment Diagrams
(A)
(B)
5.8
(A) 2 : 3
(B) 4 : 7
(C) 8 : 27
(D) 4 : 9
The shear force and the bending moment
respectively at section A-A for the beam
loaded as shown in the figure is
(C)
(D)
5.11
5.9
5.10
(A) P, 0
(B) 3 P, 2 PL
(C) 2 P, 3 PL
(D) 0, PL
A beam with overhangs carries one point load
acting downwards and the other upward. The
clockwise moment pb is applied at each
support. The bending moment at the midpoint
of the beam is
(A) 0
(B) PL/2
(C) PL
(D) PbL
Which one of the following represents the
correct bending moment diagram of the beam
PQR loaded as shown in the figure?
5.12
A beam ABCD with simple supports at A, B
and D and an internal hinge at C is subjected
to loads as shown in the figure. The
reaction at middle support is given by
(A) 7.5 kN
(B) 10 kN
(C) 20 kN
(D) 22.5 kN
A cantilever beam OP is connected to another
beam PQ with a pin joint aas shown in the
figure. A load of 10 kN is applied at the midpoint of P5. The magnitude of bending
moment (in kN-m) at fixed end O is___.
(A) 2.5
(B) 5
(C) 10
(D) 25
Strength of Materials
5.13
88
Kulkarni Academy
For a simply supported beam as shown in
figure, the bending moment diagram is
shown. The unknown load marked as X is
5.16
(A)
(B)
(C)
(D)
A simply supported beam AB is subjected to
a horizontal force, P, as shown in the figure.
Shear force at section a - a is given by
0
(A) Load of 2000 N inclined at 30 with
horizontal
(B) Vertical load of 1000 N
(C) Moment of 1000 Nm
(D) Load of 2000 N inclined at 300 with
vertical
5.14
If the following sign convention is used for
the shear force in a beam :
5.17
5.18
5.15
(B) 8, – 8
(C) 4, – 8
(D) 4, – 4
A cantilever beam carries the antisymmetric
load shown, where w is the peak intensity of
the distributed load. Qualitatively, the correct
bending moment diagram for this beam is
(B) 3Ph/(2L)
(C) 3Ph/L
(D) PL/h
In a cantilever beam of length 2 m, the shear
force in newton (N) along the length is given
by v( x) 5x 2 , where x is the distance in
meter measured from the fixed end. The
magnitude of the load intensity at the midspan of the beam is
the values of the largest and smallest shear
forces in the beam shown in the Figure are
(A) 8, – 4
(A) Ph/L
(A) 0
(B) 1 N/m
(C) 5 N/m
(D) 10 N/m
Find the maximum bending moment
(magnitude wise) in kN-m for the beam
shown in the figure.__________
Kulkarni Academy
5.19
89
Shear Force and Bending Moment Diagrams
A weightless beam subjected to two point
loads is shown in the figure below :
(A)
The shear force diagram of the beam is
(B)
(A)
(C)
(B)
(D)
(C)
5.22
(D)
5.20
The bending moment diagram of a simplysupported beam loaded by moments M 0 and
2M 0 (shown in figure), is given in
The figure shows the dimensions as well as
loading on a simply supported beam. The
distance (from end A) of the point where
maximum bending moment occurs is
(A)
5.21
(A) 4 m
(B) 6 m
(C) 3.5 m
(D) Zero
The simply supported beam shown below is
subjected to a clockwise moment M at point
A and two counter clockwise moments 2M
and M at points B and C, respectively. Which
one of the following is the correct bending
moment diagram (tensile at bottom is
positive moment) for the beam?
(B)
(C)
(D)
Strength of Materials
5.23
90
Kulkarni Academy
A simply-supported beam of length l is
(B)
subjected to a clock-wise couple moment M 0
at C as shown in figure. The shear force
diagram of the beam is given by
(C)
(D)
(A)
Common Data Questions 5.25 & 5.26
A simply supported beam with overhang in
(B)
one side is loaded as shown in Fig. One of the
diagonals of the square cross-section of the
(C)
beam is kept horizontal.
(D)
5.24
The beam shown below is loaded with a
concentrated clockwise moment of 80 kN-m
at point B. The bending moment diagram (in
5.25
kN-m ) is
5.26
(A)
The shear force at P is
(A) 21.20 kN
(B) 41.40 kN
(C) 61.60 kN
(D) 81.80 kN
The bending moment at S is
(A) 29.23 kNm
(B) 39.23 kNm
(C) 49.22 kNm
(D) 59.23 kNm
Kulkarni Academy
A
91
Shear Force and Bending Moment Diagrams
5.2
Answer Key
(C)
5.1
B
5.2
C
5.3
B
5.4
A
5.5
B
5.6
B
5.7
D
5.8
A
5.9
A
5.10
B
5.11
D
5.12
C
RAH 0 , RAV 0
5.13
C
5.14
A
5.15
C
F 0
5.16
A
5.17
D
5.18
50
SF at a section close to the fixed end = 0
Hence, the correct option is (C)
5.19
C
5.20
C
5.21
A
5.22
D
5.23
C
5.24
D
5.25
B
5.26
C
E
5.1
5.3
(B)
Explanation
(B)
RB 2L Fh
M 0 3PL PL
2PL(CW)
Hence, the correct option is (B).
5.4
(A)
C0
500
L
C0 500 L 500 2 1000 Nm
Hence, the correct option is (B).
Hence, the correct option is (A).
Strength of Materials
5.5
92
Kulkarni Academy
R1 R2 4
(B)
R1 0
MB ?
R2 4kN
Vx 0 2( x 1)
Vx 2( x 1)
VQ 0kN
VR 2kN
VS 4kN
M x R1 x W ( x 1)
2PL(CW) PL(CCW)
PL
W ( x 1) 2
2
MQ 0
Hence, the correct option is (B).
5.6
( x 1)
2
MR
W (2 1)2 W
1kN-m
2
2
MS
2 (3 1)2
40
2
(B)
MS 0
Hence, the correct option is (B).
5.7
M X2
2
W ( x 1)
R2 ( x 2)
2
MA
2
(3 1)2
4(3 2)
2
4
2 4 4 0
2
(D)
WL L qL L
MB
2 4 3 6
W q
8 18
W 4
q 9
Hence, the correct option is (D).
Kulkarni Academy
5.8
93
Shear Force and Bending Moment Diagrams
Pb Pb P(b L) Pb R2 L 0
(A)
Pb PL Pb R2 L 0
R2 P
( BM )MM Px P( x b) Pb
At
x b
L
2
( BM ) MM Pb
PL PL
Pb = 0
2
2
Hence, the correct option is (A).
5.10
(B)
M A P x 2P( x L)
M A Px 2Px 2PL
Px 2PL
x 2L
Mx 0
Hence, the correct option is (A).
5.9
(A)
R1 R2 P P
M X 5 10 x
R1 R2 0
M P 25kN-m
R1 R2
M Q 5kN-m
MA 0
Hence, the correct option is (B).
Strength of Materials
5.11
(D)
94
Kulkarni Academy
5.13
(C)
Falling (CCW)
Moment of 1000 N-m
Hence, the correct option is (C).
5.14
(A)
MA 0
RB
90
22.5kN
4
Hence, the correct option is (D).
5.12
(C)
M 0 10kN-m
Hence, the correct option is (C).
RA RB 16kN
( RB 6) 36 36 0
RB
72
12
6
Largest = 8 kN
Smallest 4kN
Hence, the correct option is (A).
Kulkarni Academy
5.15
(C)
95
Shear Force and Bending Moment Diagrams
5.17
(D)
V ( x) 5 x 2
dV
w
dx
10x w
(Load intensity at mid span) (at x 1m )
(Rate of loading)
w 10 N/m
Hence, the correct option is (D).
Shear force at fixed and free end = 0
5.18
50
Therefore, slope of BMD = 0
dM
V
dX
dM
0
dX
RB RD 60kN
More over SF changes sign at centre therefore
slope of BMD changes suddenly at the centre.
MD 0
Hence, the correct option is (C).
20 7.5 RB 5 40 2 0
5.16
RB 46kN
(A)
RD 14kN
MB 0
R1 L Ph
R1
Ph
L
Hence, the correct option is (A).
At
x0
x 2.5
MA 0
M B 50kN-m Maximum
Strength of Materials
96
Kulkarni Academy
RA RB 12000
RB 8 16000 24000 0
20 x 46( x 2.5)
M C (at x 5.5 ) 20 5.5 46(5.5 2.5)
110 138 28kN-m
MD 0
Hence, the correct answer is 50 kN-m.
5.19
(C)
40000
5000 N
8
R A 7000 N
Maximum BM
At SF = 0
VX 7000 2000 x 0
7000 2000 x
(From end A)
x 3.5m
Hence, the correct option is (C).
RB
5.21
(A)
RA RC 0
MA 0
RC L M 2M M 0
R1 R2 0
Hence, the correct option is (C).
5.20
(C)
RC L 2M
2M
2M
RC
RA
L
L
M X RA x M
2M
xM (x 0)
L
MA M
2M L
MB
M 2M
L 2
L
xL
2
M X RA x M 2M
M X M at
( x L)
Hence, the correct option is (A).
2M
xM
L
Kulkarni Academy
5.22
(D)
97
Shear Force and Bending Moment Diagrams
5.23
(C)
RA RB 0
RB L M 0 0
RA RB 0
MA 0
( RB L M 0 ) 2M 0 0
RB L M 0 0
RB
M0
L
RA
M0
L
M X RA x 2M 0
M
0 x 2M 0
L
At x 0
( M X ) A 2M 0
RB
M0
L
RA
M 0
L
Hence, the correct option is (C).
5.24
(D)
MA 0
80 RC 8 20 12 0
RC
240 80 320
40kN
8
8
RA 20 RC 20 40 20kN
0 x4
M x RA x
20 x
MA 0
At
B ( x L)
M 0
MB
L 2M 0
L
M0
Hence, the correct option is (D).
M B 20 4 80kN
4 x8
M x RA x 80
20 x 80
x4
Strength of Materials
MB 0
M C 20 8 80 80kN-m
MD 0
Hence, the correct option is (D).
5.25
B
MQ 0
RP 4 25 100 2 18.75 0.5 0
RP 41.40kN
Hence, the correct option is (B).
5.26
(C)
RP RQ 118.78
RQ 118.78 RP 77.37kN
M x RP x 100( x 2) 25
M S 41.4 3 100 25
M S 49.22kN-m
Hence, the correct option is (C).
98
Kulkarni Academy
6.1 Introduction
Shaft :
It is a Machine element which rotates about it’s
longitudinal axis and transmit power. The word
torsion means twisting.
Torsion refers to the twisting of the member
when it is loaded by couples that produced
rotation about its longitudinal axis, this couple is
known as twisting couple.
Examples : Motor shaft, turbine shaft, I.C engine
shaft etc.
Assumptions :
1. Material is homogeneous and isotropic.
2. Circular cross-sections remain circular and
are perpendicular to the axis of the shaft i.e.
the cross-section rotate as a rigid body.
3. The distance between cross-sections do not
change, i.e. length of the shaft remains
constant (No Normal strain in the axial
direction).
4. The load is with in elastic limit.
5. The member is weightless i.e. the effect of
bending is negligible.
Torsion equation :
Representation of torque :
For twisting the couple is in the plane of crosssection.
Pure torsion :
A member is said to be in pure torsion if it is
subjected to equal and opposite couples in a plane
perpendicular to longitudinal axis such that the
torque remains constant throughout the length of
the shaft.
A member may not be in pure torsion always
but the portion of the member may be in pure
torsion.
Angle of twist
Shear strain
tan
BB1
L
( is very small)
tan
r
…..(i)
' ' is in radian
L
As the loading is within elastic limit, shear stress
is directly proportional to shear strain.
G
Gr
L
Strength of Materials
G
r
L
100
…..(ii)
Kulkarni Academy
Therefore, r
Variation of shear stress :
Let be the angle of twist at the free end, x be
angle of twist at a distance x from fixed end,
therefore from the figure it is observed that
xL
x
Representation of stressed element
on the surface :
Therefore, L
Constant
L
Therefore at the fixed end the angle of twist
0 and at the free end at “ x L ”, is
maximum.
NOTE
When a shaft is subjected to torsion the
surface element is under pure shear. At
450 on this stressed element there are
normal stresses = Shear stress.
At R Shear strain is
they fail at 450 under the action of tensile
stress.
At r Shear strain is '
Observations :
1. The shear strain f (r ) i.e. with increase in
radius the shear strain increases and it is
maximum at the outer surface.
2. Shear strain is independent of length.
3. The angle of twist is independent of radius
r.
4. The angle of twist f ( L) , i.e. zero at fixed
end and maximum at the free end.
From equation 2 we have
Gr
L
For a given material G is constant, we know
that Constant.
L
As brittle materials are weak in tension
Similarly, when the ductile material is
subjected to torsion and ductile materials
are weak under shear. Therefore they fail
along the cross-section.
Kulkarni Academy
101
Torsion
1. Circle :
dF .2 r dr
dT d F .r
I ZZ I XX IYY
dT .2 r dr. r
D4
64
T .2 r 2 dr
I XX IYY
Gr
L
I ZZ J
…..(ii)
2. Hollow shaft :
Gr
T
2 r 2dr
L
T
G 2
r 2 r dr
L
T
G
J
L
r
T G
J
L
D4
32
2
dA J
…..(iii)
All these conclusions should be written after
equation number (iii)
From equation (ii) and (iii)
J
4
[ D0 D i4]
32
Polar section modulus ( Z P ) :
J
R
Circle / Solid shaft :
ZP
I G
r L L
' ' is in radian
Resisting torque
max
r
R
Polar M.O.I (J) :
This represents the resistance of material to
twisting about polar axis or longitudinal axis. i.e.
if J is more its resistance to twisting will be
more.
J D3 R3
R 16
2
Hollow shaft :
J
D
ZP ; R 0
R
2
ZP
D04 Di4
ZP
16 D0
Strength of Materials
Power transmitted by shaft :
P
W Fx T
t
t
t
102
Kulkarni Academy
T1 T2 T
total AB BC
TL
GJ
T
Torsional stiffness
AE Axial rigidity
EI Flexural rigidity
GJ Torsional rigidity
T GJ
L
P T
Shafts in parallel :
1.
Power transmitted for a given material and at a
given speed.
2N
60
At a given speed P T
P T
T1 T2 T
…..(i)
1 ( 2 ) 0
1 2
T max
J r
R
max
P
T 1L1 T2 L2
G1 J1 G2 J 2
…..(ii)
2. Compound shaft :
16T
D3
max D3
16
P D3 (For a given material)
Shaft in series :
1 2
T T1 T2
Comparison of hollow and solid shaft for
power transmission conditions :
1. Same material
2. Same weight
3. Same length
Kulkarni Academy
103
D0
K
Di
K 1
Torsion
K 4 1 3
Di
16 K
3
D
Ps
16
Ph
K 4 1 3
Di
16 K
Ps K D
Ph K 4 1 Di
Ws Wh
gAs L gAh L
As Ah
2 2
D ( D0 Di2 )
4
4
D2 D02 Di2
D2 Di2 ( K 2 1)
D Di
K 2 1
Ps K ( K 2 1) K 2 1
Ph
( K 2 1)( K 2 1)
Or
T max
J
R
T max
P K K 2 1
Ph
( K 2 1)
Ph
K 2 1
Let K 2
Ps K K 2 1
Ph 1.44 Ps
J
R
J
ZP
R
( max is same because so same material)
T ZP
P ZP
D
K 2 1 (From equation (i))
Di
3
Ps
K
4
[ K 2 1]2
Ph ( K 1)
…..(i)
D0
K
Di
3
Power T
Ph Z Ph
Ps Z Ps
Z Ps
3
D
16
Z Ps
D04 Di4
16 D0
NOTE
For the same material the power
transmitted by hollow shaft is 44% more
than (if K 2 ) the solid shaft.
Strength of Materials
6.2
6.3
Kulkarni Academy
Practice Questions
P
6.1
104
In a solid circular bar of diameter D, a
D
concentric hole is made of diameter . The
2
ratio of the torque carried by the hollow bar
to that of the solid bar in order to develop the
same magnitude of shear stress (maximum)
will be
8
15
(A)
(B)
9
16
1
5
(C)
(D)
2
6
A circular shaft of linear elastic material is
subjected to a pure torque T. The maximum
shear stress developed is . The maximum
tensile stress developed in the shaft is
3
(A)
(B)
2
4
(C)
(D) 2
A stepped shaft of uniform material with
shear modulus of 105 MPa is shown in figure.
The lengths and polar moments of inertia
I
p
are indicated in the figure. A torque of
6.5
6.6
6.7
5000 N-M is applied at the free end. The twist
in radians at the free end is
6.4
(A) 0.3
(B) 0.5
(C) 0.2
(D) 0.7
For a stepped shaft shown in figure, the ratio
T
of the torque at A to that at C, i.e, A is
TC
6.8
(A) 2
(B) 4
(C) 8
(D) 16
A fixed beam Ps is made up of two dissimilar
materials and is subjected to a torque T at Q
T
as shown in figure. The ratio of p is
Ts
(A) 1
(B) 3
(C) 2
(D) 4
Solid shaft A of diameter d and length l is
subjected to a torque T. Another shaft B of
same material and length but of half the
diameter is also subjected to the same torque
T. The ratio of the angles of twist of shaft B
to that of shaft A is
(A) 32
(B) 16
(C) 8
(D) 4
A circular rod of length L and torsional
rigidity GJ is fixed at one end and free at the
other end. If a twisting moment T is applied
L
at a distance of
from the fixed end. The
2
angle of twist at free end will be
(A)
TL
2GJ
(B)
TL
GJ
(C)
2TL
GJ
(D)
3TL
GJ
A hollow shaft d0 2d1 where d 0 and d1
are the outer and inner diameters respectively
needs to transmit 20KW power at 3000 rpm.
If the maximum permissible shear stress is 30
MPa, d 0 is
(A) 11.29 mm
(B) 22.58 mm
(C) 33.87 mm
(D) 45.16 mm
Kulkarni Academy
6.9
105
A stepped shaft PQR is fixed at both the ends
as shown in figure. A torque T is applied at
point 6. The polar moments of inertia of the
shaft PQ and QR are J1 and J2 respectively. G
is the modulus of rigidity. The angle of twist
at point Q due to torque T, is given by
(radians)
Torsion
6.13
6.14
(A)
2TL
GJ1
(B)
TL
GJ 2
2 1 TL
TL
2
(C) (D)
J1 2 J 2 G
J1 J 2 G
6.10
6.11
A long shaft of diameter d is subjected to
twisting moment T at its ends. The maximum
normal stress acting at its cross section is
equal to
16T
(A) Zero
(B)
d 3
32T
64T
(C)
(D)
3
d
d 3
Polar MOI
I p , in cm4, of a rectangular
section having width, b 2cm and depth,
6.12
d 6cm is in a hollow circular shaft of outer
diameter 20 mm and thickness 2 mm,
subjected to a torque of 92.7 N-m will be
(A) 59 MPa and 47.2 MPa
(B) 100 MPa and 80 MPa
(C) 118 MPa and 160 MPa
(D) 200 MPa and 160 MPa
A hollow shaft of 1 m length is designed to
transmit a power of 30 KW at 700 rpm. The
maximum permissible angle of twist is 10.
The inner diameter of the shaft is 0.7 times
the outer diameter. The modulus of rigidity is
80 GPa. The outside diameter (in mm) of the
shaft is______
6.15
Two shafts A and B are made of the same
material. The diameter of shaft B twice that
of shaft A. The ratio of power which can be
transmitted by shaft A to that of shaft B is (if
maximum shear stress remains the same)
(A)
1
2
(B)
1
4
(C)
1
8
(D)
1
16
A torque of 1 N-m is transmitted through a
steeped shaft as shown in the figure. The
torsional stiffness of individual sections of
lengths MN, NO and OP are 20 Nm/rad, 30
Nm/rad and 60 Nm/rad respectively. The
angular deflection between the ends M and P
of the shaft is
(A) 0.05rad
(B) 0.1rad
(C) 0.5rad
(D) 1rad
A solid shaft of diameter d and L is fixed at
both ends. A torque T0 is applied at a
L
from the left end as shown in the
4
figure. the maximum shear stress in the shaft
is
distance,
(A)
16T0
d 3
(B)
12T0
d 3
(C)
8T0
d 3
(D)
4T0
d 3
Strength of Materials
6.16
106
Two solid circular shafts of radii R1 and R2
Kulkarni Academy
6.20
A stepped circular shaft made of steel is
are subjected to same torque. The maximum
rigidly fixed at two supports A and C as
shear stresses developed in the two shaft are
shown in figure. A torque of 680 Nm is
1 and 2 . If
R1
2, then 2 is_____
R2
1
applied on the shaft at point B. The diameter
of portion AB is twice that of portion BC. The
magnitudes of torque reactions at supports A
Common Data Questions 6.17 & 6.18
and C respectively are
A solid Circular steel shaft of 50 mm
diameter fixed at one end, is subjected to
torques as shown. The shear modulus of the
materials is 80 GPa
(A) 640 Nm, 40 Nm
(B) 40 Nm, 640 Nm
(C) 340 Nm, 340 Nm
6.17
6.21
the length PQ is
6.18
6.19
(D) 544 Nm, 136 Nm
The maximum shear stress due to torsion in
Two solid shafts A and B are made of the
(A) 15.75 MPa
(B) 21.22 MPa
same material. Shaft A is of 50 mm diameter
(C) 30.56 MPa
(D) 51.21 MPa
and shaft B is of 100 mm diameter. The
The rotation of the free end S due to the
strength of shaft B is
torsion is
(A) 2 times as that of shaft A
(A) 0.250
(B) 0.580
(B) 4 times as that of shaft A
(C) 1.220
(D) 1.250
(C) 6 times as that of shaft A
If the diameter of a thin hollow homogeneous
elastic tube is doubled while retaining
thickness, within elastic limit, the ratio of the
(D) 8 times as that of shaft A
6.22
A composite shaft is made of a steel tube with
an inner brass core perfectly bonded together
maximum allowable torque to weight would
as shown. The shaft is fixed at one end and
(A) Remain the same
subjected to a torque of 2T at the other end.
(B) Nearly double
Shear modulus of steel is G and that of brass
(C) Become nearly four fold
(D) Become nearly eight-fold
is
G
.
2
Kulkarni Academy
107
The outer radius of the steel tube is R 2r
and radius of the inner brass core is r. The
magnitude of shear stress at the interface
(point x) and in the steel tube is closest to
6.23
(A)
0.041T
r3
(B)
0.082T
r3
(C)
0.16T
r3
(D)
0.41T
r3
A composite circular shaft is comprised of a
steel core surrounded by an aluminium
annulus perfectly bonded to each other as
shown in the figure. If it is subjected to pure
torque, which one of the following statements
is TRUE?
(A) Only shear stress is continuous across
the steel-aluminum interface
(B) Only shear strain is continuous across
the steel-aluminum interface
(C) Both Shear stress and shear strain are
continuous across the steel-aluminum
interface
(D) Both Shear stress and shear strain are
discontinuous
across
aluminum interface.
the
steel-
Torsion
Strength of Materials
A
108
Kulkarni Academy
6.2
Answer Key
6.1
B
6.2
C
6.3
D
6.4
D
6.5
A
6.6
B
6.7
A
6.8
B
6.9
D
6.10
A
6.11
B
6.12
44.52
6.13
C
6.14
B
6.15
B
6.16
8
6.17
C
6.18
B
6.19
B
6.20
A
6.21
D
6.22
B
6.23
B
(C)
X 'Y '
X Y X Y
2
2
cos 2 xy sin 2
X 'Y ' xy sin 900
xy
Maximum tensile stress (at 450 )
Hence, the correct option is (C).
E
Explanation
6.3
(D)
Given data :
6.1
(B)
G 105 MPa ,
D04 Di4
Th Z Ph 16 Di
3
Ts Z Ps
D
16
D
D4
2
D4
T 5000 N-m
T1 T2 T 5000 N-m
() free end 1 2
T .L TL TL 1 1
GJ1 GJ 2 G J1 J 2
4
Th
1
15
1
Ts
16 16
Th 15
Ts 16
Hence, the correct option is (B).
5000 102 N-m 1m
105 106 N/m 2
1
1
24 (102 )4 10 (102 )4
5000
0.1416 108
11
10
() free end 0.708 radian
Hence, the correct option is (D).
Kulkarni Academy
6.4
109
Torsion
(D)
B
A
AB BC
TA L
TC L
GJ AB GJ BC
B
A
TA J AB
TC J BC
4
(d )
JA
32
4
JB
d
32 2
B
24 16
A
(2d ) 4
TA 32
TC
(d ) 4
32
TA
16
TC
Hence, the correct option is (B).
6.7
Hence, the correct option is (D).
6.5
TL
GJ B
TL
GJ A
I
A
IB
(A)
(A)
PQ QS
L
2
TP L TS (3L)
GJ
GJ
TP GPQ 3 4 106 3
1
TS
GQS
12 106
No torque in BC portion it is only rotating
TP
1
TS
Torque diagram
Hence, the correct option is (A).
6.8
Hence, the correct option is (A).
Same material GA GB
Diameter
(B)
Hollow shaft
D0
K 2
Di
(B)
Length
Torque
L
Constant
L
B C
TP TS T
6.6
TL
;
2GJ
Shaft A
d
l
T
Shaft B
d
2
l
T
P 20kW
N 3000rpm
(max ) per 30MPa
P
2NT
60
Strength of Materials
2 3000 T
60
T 63.66 N-m
20 103
max
Kulkarni Academy
6.10
(A)
16 TD0
( D04 Di4 )
16 63.66 103 N-mm D0
1
D04 1
16
15
30 D03 16 63.66 103
16
D0 22.58mm
30
Hence, the correct option is (B).
6.9
110
(D)
16T
d 3
But in this problem he is asking maximum
normal stress acting at its cross-section. So, this
is the case of pure torsion (i.e. pure shear),
Maximum normal stress at its cross section =
Zero.
y x y
x ' y ' x
cos 2 xy sin 2
2
2
max
(Not its cross-section)
xy
x ' y '
at 450
Hence, the correct option is (A).
1 2
TP TR T
TP (2 L) TR ( L)
GJ1
GJ 2
TP
J
1
TR 2 J 2
JT
TP 1 R
2J2
J1TR
TR T
2J2
J
TR 1 1 T
2J2
T
TR
J1
1 2 J
2
TL
TL
R
GJ 2
J1
1 2 J GJ 2
2
2
TL
G J1 2 J 2
Hence, the correct option is (D).
6.11
(B)
D0 20mm
Di 16mm
T 92.7 N-m
Kulkarni Academy
outer
111
16TD0
( D04 Di4 )
6.13
16 92.7 10 20
[204 164 ]
3
outer 99.957 MPa 100MPa
outer inner
20
16
2
2
100 inner
10
8
(C)
Same material
Shaft A
G
d
Hence, the correct option is (B).
PA 1
PB 8
Hence, the correct option is (C).
6.14
(B)
44.52
T
T
K
K
Given that :
Hollow shaft
L 1m , P 30kW , N 700 rpm
max 10
rad
180
mp mn n0 0 p
1 1 1
20 30 60
Di 0.7 D0
K 1.4285
G 80GPa
2NT
P
60
mp 0.1rad
Hence, the correct option is (B).
6.15
(B)
30 103 60
409.25 N-m
2 700
TL
( T 409.25N-m )
GJ
T
409.25 103 (N-mm) 1000 mm
1800 80 103 N-mm2 ( D 4 D 4 )
0
i
32
Di4 (0.7 D0 )4
409.25 106
1800 80 103 0.7599 D 4
0
32
D0 44.52mm
Hence, the correct answer is 44.52.
Shaft B
G
2d
3
d
PA Z A
1
32
3
PB Z B
(2d )3 2
32
[Similar rule]
inner 80MPa
6.12
Torsion
Shafts in parallel
T1 T2 T0
1 2
L
3L
T2
4
4
GJ
GJ
T1
3
T2
T1
Strength of Materials
T1 3T2
112
Kulkarni Academy
6.18
T1 T2 T0
S () PQ ()QR
3T2 T2 T0
T1
(750 250) 103 500
80 103 (50)4
32
0.010185 radian
16T1 16 3T0
d 3
4d 3
max
12T0
d 3
Hence, the correct option is (B).
L
[TPQ TQR ]
GJ
3T0
4
max
6.16
TL
TL
GJ PQ GJ QR
T0
4
T2
(B)
0.58360 (in degree)
Hence, the correct option is (B).
6.19
(B)
8
16T
3
1 d13 d 2
2 16T d1
d 23
R1
2
R2
1 1 1
2 2 8
3
2
8
1
d
J dt
2
J A r 2
2
J (2dt )(d )2
J A r 2
2
Hence, the correct answer is 8.
6.17
J 2rt.r 2 2r 3t
(C)
d
( dt )
T
2
X max
W 1 g ( dt ) L d
2
d
T
X max
W 1 g L 2
TPQ 500 250 750 N-m
16T 16 750 103
N-mm
d 3
(50)3
(max ) PQ 30.557 30.56MPa
Hence, the correct option is (C).
…..(i)
( (2d )t ) (d ) 2
T
Y max
W 2 g ( (2d )t ) L d
.d
…...(ii)
g L
From equation (i) and (ii)
X 1
Y 2
(Nearly double)
Y 2X
Hence, the correct option is (B).
Kulkarni Academy
6.20
113
(A)
Torsion
6.22
(B)
TA TC 680 N-m
AB BC
TA L
T L
C
GJ AB GJ BC
(2d )3
TA 32
TC
( d )3
32
2T TB TS
B S
TA
16
TC
TB L
TL
S
GB J B GS J S
TA 16TC
TA TC 680
TB
TS
4
G r
G ((2r )4 r 4 )
.
2 2
2
16TC TC 680
TC
680
40 N-m
17
TA 680 40 640 Nm
TA 640 Nm
TC 40 Nm
Hence, the correct option is (A).
6.21
TB
TS
4
r
[15r 4 ]
2
4
4TB
2 TS
4
r
15 r 4
TS 30TB
2T
(D)
Shaft A
d 50mm
TS
TS
30
TS 1.9354 T
(steel ) X
Shaft B
d ' 100mm= 2d
T d3
TA
d3
1
3
TB (2d ) 8
TB 8TA
Hence, the correct option is (D).
(S ) X
TS r
J
1.935T . r
(15 r 4 )
2
0.082T
r3
Hence, the correct option is (B).
Strength of Materials
6.23
114
(B)
G
C
r
L
G
(at interface r is constant)
GAl Gsteel
So at interface shear stress is not continuous.
r Constant.
G
G
r
(as r is constant, is constant at interface)
Hence, the correct option is (B).
Golden sentence :
As G is different for different material at the
interface therefore is also different at the
interface, therefore shear stress is not continuous
at the interface. i.e. at the interface will be
different in different material.
We know that r
Constant.
L
is constant and r is same at the interface
L
therefore is same at interface.
i.e. shear strain is continuous at the interface not
the shear stress.
Kulkarni Academy
7.1 Introduction
Column :
A vertical member of a structure which is under
compression.
Strut :
It is relatively smaller in size compare d to
column but it may be inclined also. This term is
generally used in reference to trusses.
Strut also carries compressive load.
Boom :
It is a compression member used in cranes.
Classification of columns :
1. Short column
L 8 d Fails by crushing
tan
2. Intermediate column
8 d L 30 d
Fails by both crushing and buckling
3. Long columns
L 30 d Fails by buckling
In the design of long columns Euler’s theory
is used and in the design of intermediate column
Ranking theory is used.
2
L
Px P tan
( is small)
Total disturbing force 2P tan 2P
Case : 1
If disturbing force > Restoring force
2P K
Buckling :
Due to the sudden loss of stiffness, the
column deflects laterally and this is known as
buckling. It is elastic instability.
Ideal column :
A column is said to be an ideal column if it is
perfectly straight before loading and made of
homogeneous material and load passes through
c.g. of cross section.
2
L
L
2
2P
2
K
L
P
KL
4
(Unstable)
Case : 2
If restoring force > Disturbing force
P
KL
(Stable equilibrium)
4
Strength of Materials
Effective length :
116
Kulkarni Academy
Where,
n 0,1,2,3,.....
When end moments are applied, the column
deflects
laterally with
the
maximum
displacement max , which is proportional to M 0 .
Now suppose that we gradually apply the axial
load ‘P’. While at the same time decreasing end
moments so that max does not change.
When end moments become zero max is
maintained only by axial load. The axial load
required to hold the column in its deflected
position without any moment is called critical
load or buckling load.
Effective length ( Le ) : Unsupported length
between points of zero moment is known as
effective length.
Euler’s theory of column :
1. The column is an ideal column.
2. The column fails by buckling alone.
3. The failure occurs only by buckling i.e.
buckling load is less than crushing load and
hence in Euler theory stress is always less
than yield stress.
By using the equation
EId 2 y
M
dx 2
We can calculate the critical load.
P
n 2 2 EI
L2e
When n 0, P 0 , this is trivial solution
(Meaning less) / Unimportant solution. Therefore
let is calculate the min value of P at which the
column buckles, this load is known as critical
load.
According to Euler theory, the buckling load or
critical load is given by
Pcr
2 EI
L2e
Le Effective length
n indicates the number of times the curvature
changes.
n 2; Le
L
2
Kulkarni Academy
n 3; Le
117
L
3
Columns
Slenderness ratio ( ) :
2 EI least
Pcr
L2e
2
Ileast AKleast
Pcr
2 E
A L 2
e
K
least
2 E
c 2 ;
Slenderness ratio
Le
K least
For mild steel
y PL 250MPa
E 200 103 MPa
For applying Euler’s equation cr y
2 E
y
2
bd 3
db3
, Iy
Ix
12
12
Ix I y
Pcr
2 EI Least
L2e
NOTE
Critical load does not depend on strength
of ve material it depends only on E and
dimensions.
Two
dimensionally
identical columns one of high strength
steel and other of ordinary steel will
buckle under the same critical load
because they have same value of E.
The critical load which we derived has
meaning only when the stress is less than
proportionality limit or yield or elastic
limit.
2 200 103
250
2
89.9
90
Limitation of Euler’s theory of
column :
It is not valid for short columns because in short
column crushing occurs before buckling for the
validity of Euler’s theory, buckling should occur
before crushing. Therefore, if Euler’s theory is to
be applied for mild steel the slenderness ratio
must be greater than 90.
Slenderness ratio helps us in classifying the
column as
Short 30
Intermediate 30 80
Long 80
Strength of Materials
Effective
length
118
of
column
different support :
with
Kulkarni Academy
Case : 3
Both ends fixed
Case : 1
Both end hinged
Le
L
2
Pcr
42 EI
L2
Case : 4
One end fixed and other end pinned.
Pcr
2 EI
L2
Case : 2 One end fixed & other end free
Pcr
Pcr
2 EI
4 L2
22 EI
L2
A Fails by crushing
2 E
cr 2
Kulkarni Academy
119
Columns
1. Analysis :
Design of intermediate column :
Case : 1
Intermediate column fails by both crushing and
buckling Rankine’s formula is used in the design
of intermediate column. It is an empirical
relation.
1 1 1
P Pc Pe
P Rankine’s crippling load
P
Pc Pe
Pc Pe
2 EI
( Le1 )2
Pc c . A
2 EI
Pcr2
( Le2 )2
2 EI
Pe 2
Le
Pcr2 Pcr1
P
Pcr1
Le2 Le1
For design point of view, we consider more
effective length.
Case : 2
Pc
P
1 c
Pe
c . A
A
1 2c
EA
2
P
c . A
;
1 2
c
2 E
Rankine constant depends upon the material.
Le2 Le1
So effective length
2L
3
1
Mild steel
7500
1
Cast iron
1600
1
Wood
750
Strength of Materials
P
120
Kulkarni Academy
Q.5
Practice Questions
The maximum compressive load that can be
applied on a hinged-hinged column of cross-
Q.1
The Euler’s buckling load of a column fixed
section 20mm 10mm and length 2000mm
at both the ends is P. If one of the ends is
is (allowable compressive stress 250MPa:
made free, the buckling load shall change to
E 210GPa )
(A)
P
16
P
(C)
4
Q.2
(B)
P
8
P
(D)
2
Q.6
(A) 0.86kN
(B) 3.45kN
(C) 25kN
(D) 50kN
The buckling load of a slender column
clamped at both the ends is 4000 N. The
The Euler buckling load in an axially loaded
column is subjected to an axial compression.
slender column
During the course of service, one of the ends
(A) Increases with increase in slenderness
gets detached from the clamp and becomes
ratio
free end. The absolute percentage change in
(B) Decrease with increase in slenderness
the buckling load due to the change in the end
ratio
condition is
(C) Is not affected by slenderness ratio
(D) None of the above
Q.3
Q.4
A column has a rectangular cross-section of
Q.7
(A) 50.00
(B) 75.00
(C) 83.25
(D) 93.75
A rigid bar AB is hinged at B through a
10mm 20mm and a length of 1m. The
torsional spring with spring constant kt . For
slenderness ratio of the column is close to
small rotations of the bar AB about B, the
(A) 200
(B) 346
critical load Pcr is given by
(C) 477
(D) 1000
For a slender steel column of circular cross
section, the critical buckling load is pcr . If
the diameter of the column is doubled
(keeping other material and geometrical
parameters same), then the critical buckling
load of the column is
P
(A) cr
16
(C) 2 Pcr
(B) 8Pcr
(D) 16 Pcr
(A)
k1
2L
(B)
k1
L
(C)
2k1
L
(D)
4k1
L
Kulkarni Academy
Q.8
121
Find the maximum force P (in kN) that can
be applied to the planar structure ABC so as
to prevent buckling in any of the members.
Consider buckling only in the plane of the
Columns
Q.10 Two massless rigid bars, each of length
a 0.5m, are connected by a rotational
spring having stiffness k 1000 N.m/rad.
Find the buckling load P (in kN)
structure Joint B is a pin connection. Use
E 200 GPa
for
both
members. The
diameter of member AB is 10 mm and the
diameter of member BC is 15 mm.
Q.11 The column fails by, P 430kN
Yield stress = 250 MPa
E 200GPa
L 1m
1 2.198 10 7 m4
A 1.662 103 m2
Q.9
A rigid bar compressed by a vertical force P
and connected by a horizontal spring is as
shown in figure. The buckling load for the
column is
(A) KL
KL
(C)
2
(B) 2KL
(D) 4KL
(A) Buckling only
(B) Yielding only
(C) Buckling and yielding simultaneously
(D) Will not fail
Q.12 The rod PQ of length L and with flexural
rigidity EI is hinged at both ends. For what
minimum force F is it expected to buckle?
(A)
2 EI
L2
(B)
(C)
2 EI
2 L2
(D)
22 EI
L2
2 EI
2 L2
Strength of Materials
122
Kulkarni Academy
Q.13 An axially loaded column is made of a
Q.15 A rigid bar of length L is hinged at the lower
material whose yield stress is 2 108 Pa. If
end and is loaded by a compressive force P at
the Young’s modulus of the material is 200
the upper end. At the mid length of this bar
GPa, from the plot of the compressive stress
two springs of spring constant ‘k’ is
slenderness ratio curve shown in the figure,
transversely attached. The critical load for the
the Euler buckling load formula is valid for:
stability of its equilibrium is given by
(A) All regions of the curve
(B) For region ABCD
(C) For region OCDE
(D) For region DEFG
(A) 2.0kL
(B) 1.5kL
(C) 1.0kL
(D) 0.5kL
Q.16 The figure below shows 4 long columns with
different support conditions but the same
flexural rigidity EI. Let P, Q, R and S be the
Q.14 A column of length L hinged at two ends is as
shown in figure. If the column is prevented
values of their critical buckling load as shown
below the respective columns.
from bending in the form of one lobe by a
restraint at its midpoint, the Euler crippling
load is given as
2 EI
(A)
L2
22 EI
(B)
L2
42 EI
(C)
L2
2 EI
(D)
4 L2
P, Q, R and S can be arranged in increasing
order as
(A) Q, S, R, P
(B) Q, P, R, S
(C) Q, R, P, S
(D) Q, R, S, P
Kulkarni Academy
A
123
I AK 2
Answer Key
7.1
A
7.2
B
7.3
B
7.4
D
7.5
A
7.6
D
7.7
B
7.8
1.37
7.9
A
7.10
4
7.11
B
7.12
C
7.13
D
7.14
C
7.15
D
7.16
D
E
7.1
Columns
Explanation
I
20 103
A
12 (10 20)
K 2.88mm
L
SR e
Kleast
K
1000
346.41
2.88
Hence, the correct options is (B).
7.4
SR
(D)
(B)
Pcr
Column fixed at both end
42 EI
P
L2
If one of the ends is made free.
2 EI
P'
4 L2
P
16
P'
P
P'
16
2 EI least
Pcr X (constant)
L2e
Pcr I
Pcr1 X d 4 Pcr
64
Pcr2 X (2d ) 4
64
4
Pcr1
d
Pcr2 (2d ) 4
Hence, the correct options is (A).
7.2
(B)
Pcr
2 EI 2 EA
2
L2e
1
If is increases,
2
Pcr decreases.
Pcr
Hence, the correct options is (B).
7.3
(B)
2 EI least
L2e
Pcr2 16 Pcr
Hence, the correct options is (D).
7.5
Pcr
(A)
2 EI least
L2
(Both ends are hinged Le L )
Le 2000
692.82
2 20 103 (20 10) SR
Pcr
K
2.88
(692.82)2
20 103
Pcr 863.59 N
K
12 (20 10)
Pcr 0.863kN
K 2.88mm
Hence, the correct options is (A).
Strength of Materials
7.6
124
(D)
Kulkarni Academy
7.8
1.37
Slender column clamped at both the ends.
Pcr 4000 N
42 EI
L2
If one end free :
Pcr '
2 EI
4 L2
Pcr '
4000
250
16
% change
DAB 10mm , DBC 15mm
4000 250
100 93.75%
4000
(Decrease in load)
P
2 EI
Pcr 2
L
2
Hence, the correct options is (D).
7.7
(B)
2 200 103 (10) 4
P
64
Pcr
4
(1000)
2
P 1.370kN
P
Pcr
2
2 200 103
(15)4
64
(2000) 2
P Kt
L
tan
L
PL Kt
Kt Pcr L
Pcr
Kt
L
Hence, the correct options is (B).
P 1.734kN
To prevent buckling, in any member, we consider
min of (1.370, 1.734)
1.37 kN
Hence, the correct answer is 1.37.
Kulkarni Academy
7.9
(A)
125
Columns
7.11
(B)
P 430kN
y 250MPa
E 200GPa
L 1m
I 2.198 107 m4
A 1.662 103 m2
Pcr
M0 0
P K ( L)
2 200 103 2.198 107 106
1
Pcr 433kN 430kN (i.e. not buckle)
Pcr
P KL
Hence, the correct options is (A).
7.10
2 EI
L2e
Check for crushing
Pcrushing c A
4
250 106 1.622 103 415.5kN
As the actual load is more than the crushing load,
so it is failed by crushing or yielding.
Hence, the correct options is (B).
7.12
tan
a
a
a
P Kt 2
P(a) Kt (2)
2 Kt 2000
4kN
a
0.5
Hence, the correct answer is 4.
P
(C)
P cos 45 F
P 2F
Both ends hinged
2 EI
2F 2
L
2
EI
2 EI
P 2 2F 2
L
L
2
EI
F
2 L2
Hence, the correct options is (C).
Strength of Materials
7.13
126
Kulkarni Academy
( K1 K2 )L P
(D)
P ( K1 K2 ) L
y 2 108 Pa
E 200GPa
2 E
y
2
99.9
100
Euler buckling load formula is valid for
Region DEFG
Hence, the correct options is (D).
7.14
(C)
P 2 K
P
L
2 2
KL
2
P 0.5 KL
Hence, the correct options is (D).
7.16
Le
L
2
Pcr
2 EI
L2e
Pcr
42 EI
L2e
(D)
Hence, the correct options is (C).
7.15
(D)
P
1
L2e
Order of critical buckling load.
QRSP
Hence, the correct options is (D).
8.1 Introduction
Assumptions:
1. Material is homogeneous and isotropic.
2. The beam is initially straight and unstressed.
3. Elastic limit is not exceeded.
4. Young’s modulus is same in compression
and tension.
5. The plain section of beam remains plain
before and after bending.
Pure bending/simple bending:
A member is said to be in pure bending when
Fig. A deformed beam loaded transversely to its
axis
NOTE
It’s not necessary for the complete beam
to undergo pure bending part of the
beam may also undergo pure bending.
it is subjected to equal and opposite couples in
the plane along the longitudinal axis.
For pure bending the shear force is zero.
SFD
Pure bending in BC region because in BC
Region Shear force is equal to ZERO.
Neutral Surface and Neutral Axis:
Strength of Materials
128
Kulkarni Academy
As the loading is within proportionality limit
Hooke’s law can be apply i.e.
E
y
E
R
Neutral surface: It is the surface which
experiences neither tension nor compression,
E
y R
There is only one neutral surface.
Neutral axis: It is the line of intersection of
cross-section with the neutral surface, there can
be any no. of infinite neutral axis.
or
…. (i)
y
E R
Bending stress
E Young’s modulus
y Distance from NA
R Radius of curvature
Equilibrium: As there is no external force
applied in axial direction therefore net force in
axial direction must be zero.
Undeformed element
Deformed element
Normal strain in ab
a ' b ' ab
ab
( R y)d Rd
Rd
y
R
From this equation it is observed that the normal
strain is directly proportional to the distance from
the neutral axis (y).
In the above equation ve sign shows that for +
ve bending moment (sagging) there will be
compressive stresses when y is + ve i.e., above
the neutral axis. Similarly, when y is ve (below
NA) strain is +ve i.e., tensile.
Force Equilibrium:
Ey
R
dF dA
dF
E
ydA
R
Faxial
E
ydA 0
R
ydA A y
A 0, y 0
y 0 shows that the centroidal axis coincides
with NA therefore in pure bending centroidal
axis always coincides with neutral axis.
Kulkarni Academy
129
Moment Equilibrium:
Ey
R
Bending Stresses
MR E
I
R
…. (ii)
From equation (i) and (ii),
dF dA
E
dF ydA
R
MR E
y
I
R
Moment dF y
Where M R = Moment of resistance
Total moment ( M ) dF y
Above Equation is known as Euler Bernoulli
bending equation.
M
E
ydA y
R
Economical sections:
In a beam having rectangular or circular section
the fibers near neutral axis are under stressed
(less) compared with those at the top or bottom.
The face that the large portion of the crosssection is thus under stressed makes it inefficient
for resisting flexure or bending.
M
indicates that if the
I
y
area of the beam of rectangular cross-section is
rearranged (redistributed) so as to maintain same
depth and same area the MOI would be greatly
increased resulting in greater moment carrying
capacity.
The expression
y dA
2
is the second moment of area about NA
and this is known as area moment of inertia and
it is designated by I.
y dA I
2
M
zz
E
I
R
Let M R be the resisting moment
M M R
M R
MR
EI
R
EI
R
This moment resisting capacity is due to
placing more material at greater distance from
the NA.
Fig. 1
Fig. 2
I Fig 2 I Fig.1
In order to obtain the maximum resistance to
bending it is advisable to use section which have
large area away from the NA and hence Isections and T-sections are preferable.
Strength of Materials
130
Kulkarni Academy
Beams of uniform strength: A beam is said to
be beam of uniform strength if the maximum
bending stress is same at each and every section.
Case I: Cantilever beam subjected to moment
‘M’ at the free end.
Section modulus (Z):
I
is known as section modulus, if section
ymax
modulus (z) is more, max will be less and the
resistance to the bending is more and hence
chances of bending failure will be less.
NOTE
In case of axial loading c/s area is
considered where as in the case of bending
section modulus is considered.
My
I
Mymax
max
I
M
I
Z ; max
Z
ymax
M max Z
max
Mymax
I
max
6M
bd 2
d
M
2
bd 3
12
As max is independent of x therefore max is
constant at each and every section, therefore, this
is the beam of constant strength.
Generally beams are subjected to transverse
loading and hence BM changes along the length
of the beam so it is not a beam of constant
strength, to make such a beam a beam of constant
strength, two techniques are followed.
(a) Varying width, keeping the depth constant
(b) Varying depth, keeping the width constant.
Kulkarni Academy
131
(max )1-1
max
2-2
6 Px
bx d 2
6 PL
bL d 2
Bending Stresses
Let the depth at the fixed end be ‘d’.
dL d
dx d
x
L
Case 1: Varying width, keeping depth
dx x
constant
For a uniform strength beam
max 1-1 max
2-2
6 Px 6 PL
bx d 2 bL d 2
x
bx bL
L
Let width at fixed end bL b
x
bx b
L
M
E 2
y dA
R
Case II: Keeping the width constant, varying
depth.
max 11 max 22
6 Px 6 PL
bd x2 bd L2
dx dL
x
L
Important observation:
Strength of Materials
P
8.1
Practice Questions
132
Kulkarni Academy
8.5
In the flexure theory of beams, the neutral
axis has the following characteristic.
(A) Always passes through the centroid of
the cross section
A homogenous prismatic simply supported
beam is subjected to a point load F. The load
can be placed anywhere along the span of the
beam. The maximum flexural stress
developed in the beam is
(B) Always remains straight after bending
(C) Always lies at the mid height of crosssection
(D) Longitudinal stress is maximum along
the axis.
8.2
In a prismatic beam under the action of pure
bending.
(A) Both the shear stress and shear strain are
zero
(B) Shear stress is zero and shear strain is
non zero.
(C) Shear stress is non -zero and shear strain
is zero
8.6
(D) Both shear stress and shear strain are
non-zero
8.3
A steel wire of diameter 5mm is bent around
a cylindrical drum of radius 0.5m. The steel
wire has modulus of elasticity of 200 GPa.
Find the bending moment in the wire in N-m.
8.7
8.4
A steel beam replaced by a corresponding
aluminium beam of same cross-sectional
shape and dimensions, and is subjected to
same loading. The maximum bending stress
will
(A) Be unaltered
(B) Increase
(C) Decrease
(D) Vary in proportion to their modulus of
elasticity
(A)
3FL
2 BD 2
(B)
3FL
4 BD 2
(C)
2 FL
3BD 2
(D)
4 FL
3BD 2
A hollow circular shaft of inside diameter 10
mm and outside diameter 20 mm is subjected
to a pure symmetric - bending moment of
200 N-m. The magnitude of bending stress at
a point in the plane of loading. Which is at a
distance of 5mm from the neutral axis, is
(A) 0
(B) 68.8 MPa
(C) 135.8 MPa
(D) 271.6 MPa
A test is conducted on a beam loaded by end
couples. The fibers at layer CD are found to
lengthen by 0.03 mm and fibers at layer AB
shorten by 0.09 mm in 20 mm gauge length
as shown in the figure. Taking
E 2 105 N/mm2 , the flexural stress at the
top fibre in N/mm2 is__________.
Kulkarni Academy
8.8
133
A beam of flexural rigidity 8 104 Nm2 is
subjected to four point bending as shown in
figure. The radius of curvature of the portion
BC of the beam is
Bending Stresses
8.12
An unspecified pure bending moment is used
to bend an aluminium rod of radius 2.5 mm
elastically into a circular ring of radius 2 m.
If the same bending moment is used to bend
elastically a copper rod of radius 2 mm, the
radius of the resulting ring (in m) is
(elastic modulus of aluminium is 70 GPa and
elastic modulus of copper is 120 GPa )
8.9
8.10
8.11
(A) 800 m
(B) 600 m
(C) 1000 m
(D) continuously variable
A structural steel beam has an unsymmetrical
I- cross -section. The overall depth of the
beam is 200 mm. The flange stresses at the
top and bottom are 120 N/mm2 and 80
N/mm2 , respectively. The depth of the
neutral axis from the top of the beam will be
(A) 120 mm
(B) 100 mm
(C) 80 mm
(D) 60 mm
A square beam laid flat is then roatated in
such a way that one of its diagonal becomes
horizontal. How is its moment capacity
affected ?
(A) Increases by 41.4%
(B) Increases by 29.27 %
(C) Decreases by 29.27%
(D) Decreases by 41.4 %
A thin steel ruler having its cross-section of
0.0625 cm 2.5 cm bent by couples applied
at its ends so that its length l equal to 25 cm
, When bent, as a circular arc, subtends a
central
angle
Take
600 .
E 2 106 kg/cm2 . The maximum stress
induced in the ruler and the magnitude is
(A) 2618 kg/cm2
(B) 2512 kg/cm2
(C) 2406 kg/cm2
(D) 2301 kg/cm2
8.13
(A) 0.702
(B) 1.404
(C) 1.755
(D) 2.808
The structure shown below is of rectangular
cross-section and carries a load of 10kN at its
free end E. Maximum bending stress (in
MPa) developed in the beam due the external
load is _____________.
The depth of the beam is 300mm and the
width is 150mm.
8.14
In a beam of uniform strength the extreme
fibers at every cross-section are stressed to
the maximum allowable stress. Consider a
solid circular beam of uniform strength
subjected to bending moment. In this beam,
the diameter of the cross-section at any
section is proportional
(A) To cube root of the bending moment at
that section
(B) To the square root of the bending
moment at that section
(C) To the bending moment at that section
(D) Inversely to the bending moment at that
section
Strength of Materials
8.15
134
A cantilever beam of T-section, shown in
figure is carrying a couple moment M 0 at the
free end. Maximum magnitude of bending
stress will occur at
(A) Bottom fibre throughout
(B) Top fibre at fixed end only
(C) Bottom fibre at fixed end
(D) Top fibre throughout
Common Data Questions 8.16 & 8.17
A massless beam has a loading pattern as
shown in the figure. The beam is of
rectangular cross-section with a width of 30
mm and height of 100mm
8.16
The maximum bending moment occurs at
(A) Location B
(B) 2675 mm to the right of A
(C) 2500 mm to the right of A
(D) 3225 mm to the right of A
8.17
The maximum magnitude of bending stress
(in MPa) is given by
(A) 60.0
(B) 67.5
(C) 200.0
(D) 225.0
Kulkarni Academy
Kulkarni Academy
A
135
Answer Key
Bending Stresses
8.4
8.1
A
8.2
A
8.3
12.210
8.4
A
8.5
A
8.6
C
8.7
900
8.8
A
8.9
A
8.10
C
8.11
A
8.12
B
8.13
17.77
8.14
A
8.15
A
8.16
C
8.17
B
(A)
Hence, the correct option is (A).
8.5
(A)
Pab Fx( L x)
L
L
dM
M max ,
0
dx
F
( L 2 x) 0
L
L
x
2
M
E
8.1
Explanation
For,
(A)
Hence, the correct option is (A).
8.2
(A)
VAy
pure bending shear force V 0
Ib
0
G; 0
Hence, the correct option is (A).
8.3
12.210 Nm
L
L
L FL
2
2
M max
L
4
FL
2 3FL
My
max
4 3
BD
I
2 BD 2
12
Hence, the correct option is (A).
F
8.6
E 200 GPa
5
R ' 500 502.5mm
2
I wire (5)4
64
M E
y I
R
(5) 4
64
M
N-mm
502.5
M 12210.79 N-mm
200 103
M 12.210 N-m
Hence, the correct answer is 12.210 N-m.
(C)
M 200 N-m
My
y 5mm
I
200 5
1000 (0.0204 0.0104 )
64
135.8122 MPa
Hence, the correct option is (C).
Strength of Materials
8.7
900
136
Kulkarni Academy
8.8
(A)
BC region is pure bending region, so bending
moment in BC region is constant and SF 0
Pa 100 1 100 N
M E
y I
R
EI 8 104
800 m
M
100
Hence, the correct option is (A).
R
CD 0.03 mm
AB 0.09 mm
8.9
(A)
LCD LAB 20 mm
0.03
1.5 103
20
0.09
AB
4.5 103
20
From similar rule
CD
AB
x
100 x
1.5 103 4.5 103
x
100 x
3x 100 x
x 25mm
topfibre ?
x?
My
I
y
CD
120
x
80 200 x
3
x
2 200 x
600 3x 2 x
5x 600
x
600
120mm
5
Hence, the correct option is (A).
8.10
(C)
CD CD E
1.5 103 2 105 N/mm2 300 MPa
top 300
(From similar rule)
75
25
topfibre 900 MPa
Hence, the correct Answer is 900 MPa.
I
a4
a3
Z1
12
6
Kulkarni Academy
137
Bending Stresses
a4
a3
Z 2 12
a
6 2
2
Z1 Z 2
0.0625
0.03125 cm
2
2 10 0.03125
25 3
ymax
2617.99kg/cm2
Moment capacity affect
a3
a3
Z Z2
6 6 2
1
a3
Z1
6
1
1
2 0.2928
1
by 29.28%
Hence, the correct option is (C).
8.11
Hence, the correct option is (A).
8.12
(B)
Aluminium rod R 2.5 mm
Circular ring radius 2 m
EAl 70 GPa
Copper rod, radius 2 mm
RRing ? , Ecu 120 GPa
Al M E
y
I
R
(A)
EI
EI
M
R Al R Cu
70
(2.5)4 120 (2)4
64
64
2
( RCu ) Ring
( RCu ) Ring 1.404 m
Hence, the correct option is (B).
8.13
E 2 106 kg/cm2 , 600
L R
M E
y I
R
Ey
R
17.77
M 0 50 10 40 kN-m
d 300 mm, b 150 mm
40 103 N-m 150 103
150 3003
12
17.77 MPa
Hence, the correct answer is 17.77 MPa.
Strength of Materials
8.14
138
(A)
Kulkarni Academy
8.16
(C)
M E
y I
R
For uniform strength beam, maximum bending
stress is same at each and every section.
My
I
d 4
I
64
M
d
y
2
d
32
3
M d
M
RA RB 6000 N
M A 0
RB 4 6000 3
3
RB 4500 N
RA 1500 N
[ Constant]
VX RA 3000( x 2)
d3M
Hence, the correct option is (A).
8.15
1500 3000 x 6000 0
(A)
7500 3000x
x
7500
2.5m 2500 mm from A,
3000
SF 0 , BM-Max
M x RA x 3000
( x 2)2
2
At x 2.5 m or 2500 mm (to the right of A)
bending moment is max.
Hence, the correct option is (C).
8.17
As bending moment is same throughout the
beam, so max bending stress depend on y value,
from above fig.
y1 y2
So maximum magnitude of bending stress will
occur at bottom fiber throughout.
Hence, the correct option is (A).
(B)
M 1500 2.5 3000
(0.5)2
2
M 3750 375 3375 N-m
max
3375 103 (N-mm) 50
67.5 MPa
30 1003
12
Hence, the correct option is (B).
9.1 Introduction
Assumptions :
The material is isotropic and homogeneous.
The loading is within elastic limit i.e. Hooks
law is valid.
The shear stress is assumed to be uniform
along the width.
F b A
My
(b dy )
I
Net unbalance force
Due to transverse loading there is a transverse
shear stress and this is accompanied by the
shear in perpendicular direction known as
complementary shear (along longitudinal
direction)
( M dM ) y bdy My(b dy)
I
I
dMyb dy
I
Strength of Materials
140
Kulkarni Academy
d
y
2
y
y
2
h
Y
d y 1d
y
4 2 2 2
d
1 d
V b y y
VAy
2
2 2
bd 3
Ib
b
12
dM
y b dy b dx
I
( dM varies with x not with y so it is constant).
dM
I
h
(bdy). y b dx
y
6V d 2
3 y2
bd 4
At the N.A, y 0
bdy dA
dA
dM
Ay b dx
I
dM 1
Ay
dx Ib
dM
V = Shear force
dx
Shear stress at a distance y from N.A
A Area of the cross-section above y
y Distance of c.g. of A from N.A
b Width of the beam
VAy
Ib
max
6V d 2
.
bd 3 4
max
3V
2 bd
3
max avg
2
Shear stress distribution in rectangular crosssection beam :
NOTE
The shear stress distribution in the beam
along the depth varies parabolically.
Kulkarni Academy
Triangular cross-section beam :
141
Shear Stress in Beam and Combined Loading
In case of bending the bending stresses are
more away from the N.A and hence in case of
bending I-sections are preferred.
Circular cross-section beam :
I-section
NOTE
VAY
Ib
1
Width
1 Shear stress in flange at the junction
For shear stress distribution along the
depth check width (b) at the junction if
width is increases, shear stress is
decreases suddenly or vice versa.
Combined loading :
A force of 15000 N is applied to the edge of the
member as shown in figure. Neglect the weight
of the member and determine the state of stress
at points B and C.
2 Shear stress in the web at the junction
1 b
2 B
NOTE
About 80-90% of the shear load is taken
by web whereas only 10 to 20% of the shear
load is taken by flanges, therefore for
resisting shear load circular sections are
preferred because the area is more near the
centre.
15000 50 750000 N-mm
Bending stress
Strength of Materials
142
Direct stress
Kulkarni Academy
8 0.2
My
375kPa
0.8 0.43
I
12
Stress due to moment of 8 kNm
375kPa
[AD (Tension) , BC (Compression)]
(B ) 11.25 3.75
7.5MPa (+) (Tensile)
C 3.75 11.25
15MPa (Compressive)
Direct (Compressive) stress :
P 15000
3.75MPa ( ve)
A 100 40
My 750000 50
b
11.25MPa
40 1003
I
12
Example 1
A Rectangular block of negligible weight is
subjected to a vertical force of 40 kN which
is applied to its corner. Determine the stress
distribution acting on a section through
ABCD.
Sol.
Let us shift the force from R to T,
My
I
16 0.4
375kPa
0.4 0.83
12
Kulkarni Academy
143
Stress due to moment of 16 kNm
375kPa [AB (+ ve), CD (– ve)]
Direct stress due to load ‘P’ at c.g.
P
40
125kPa ( ve )
A 0.8 0.4
Shear Stress in Beam and Combined Loading
Sol.
Direct stress
P
(Compressive)
bh
Stress at A due to bending
My
I
h
2 6 Pey (+ ve)
3
bh
bh 2
12
Pey .
Total stress at A,
P 6 Pey
0
bh bh2
ey
A 125 375 375 625kPa
B 125 375 375 125kPa
C 125 375 375 875kPa
h
Condition for No tension
6
Similarly
ex
b
6
[Most critical]
D 125 375 375 125kPa
Example 2
A rectangular block of negligible weight and
is subjected to vertical force ‘P’ as shown in
figure. Determine the range of values for the
eccentricity e y of the load along y-axis so that
it does not cause any tensile stress in the
block.
If the load (compressive) is within EFGH, there
will not be any tensile stress in the cross-section,
so that the stress is always compressive in nature.
In the design of pillars (concrete, structures),
it should be seen so that the load is with in the
shaded region EFGH. Because, if the load is
outside this region tensile stresses are setup and
the concrete is weak in tension and it may fail.
Strength of Materials
P
9.1
144
Kulkarni Academy
9.5
Practice Questions
A simply supported beam having a
rectangular cross-section of depth d is
subjected to a vertical concentrated load p at
the mid-span. The maximum shear stress in a
section occurs at
(A) d/2 from the top of the cross-section
(B) d/3 from the top of the cross-section
(C) 2d/3 from the top of the cross-section
9.6
(D) Top of the cross-section
9.2
The transverse shear stress acting in a beam
of rectangular cross-section, subjected to a
transverse shear load, is
A simply supported beam of span L is
subjected to a concentrated load P at
midspan. The cross-section of the beam is
rectangular. The ratio of the shear stress at
1
the depth of the beam to the shear stress
4th
at the centre of the cross-section is
3
1
(A)
(B)
8
2
3
(C)
(D) 2
4
A rectangular cross-section beam of width
w 0.25m and depth d 0.4m is subjected
to a bending moment M 200 N-m and a
uniform axial load of P 200 N as shown.
Measured from the centroidal axis of the
beam, normal stress will be zero at a distance
of
(A) Variablewith maximum at the bottom of
the beam
(B) Variable with maximum at the top of the
beam
(C) Uniform
(D) Variable with maximum on the neutral
axis
9.3
9.4
Consider a simply supported beam of length
50h, with a rectangular cross-section of
depth, h, and width, 2h. The beam carries a
vertical point load, P at its mid-point. The
ratio of the maximum shear stress to the
maximum bending stress in the beam is
(A) 0.02
(B) 0.10
(C) 0.05
(D) 0.01
9.7
(A) y 15mm
(B) y 13.3mm
(C) y 15mm
(D) y 10mm
An L-shaped bar of square cross-section with
sides. b, is loaded as shown in the figure. If
the value of the stress component xx at point
P is zero, the distance a of the force F from
the x-axis should be
A vertical pole, cantilevered at the bottom,
has a solid circular cross-section of diameter
d 49.21mm. It is loaded by a horizontal
force P 6675 N at the top end. The
maximum shear stress in the pole is
(A) 4.25 N/mm
2
(C) 4.68 N/mm2
(B) 5.68 N/mm
2
(D) 7.50 N/mm2
(A) b
(C)
b
3
b
2
b
(D)
6
(B)
Kulkarni Academy
9.8
145
A cantilever beam having cross-sectional
area 0.1 m2 and moment of inertia
1.33 103 m4 as shown in figure is subjected
to uniform tension of 200 N and a couple of
200 N-m at the free end.
Shear Stress in Beam and Combined Loading
9.10
For the component loaded with a force F as
shown in the figure. The axial stress at the
corner point P is
The state of stress at point P, 20 mm above
the neutral axis is
(A)
(B)
(C)
9.9
(D)
Two 50mm diameter solid rods are rigidly
connected together at right angles and loaded
as shown. Use P 1000 kN. At point A
located at the top of the cross-section at the
fixed end, the magnitude of bending stress
and shear stress are
(A) 256MPa, =512MPa
(B) 512MPa, =256MPa
(C) 512MPa, =128MPa
(D) 128MPa, =512MPa
(A)
F 3L b
4b3
(B)
F 3L b
4b3
(C)
F 3L 4b
4b3
(D)
F 3L 2b
4b3
Common Data Questions 9.11 & 9.12
A machine frame shown in the figure below
is subjected to a horizontal force of 600 N
parallel to z-direction.
Strength of Materials
9.11
9.12
146
The normal and shear stresses in MPa at point
P are respectively
(A) 67.9 and 56.6
(B) 56.6 and 67.9
(C) 67.9 and 0.0
(D) 0.0 and 56.6
The maximum principal stress in MPa and
the orientation of the corresponding principal
plane in degrees are respectively
(A) 32.0 and 29.52
(B) 100.0 and 60.48
(C) 32.0 and 60.48
(D) 100.0 and 29.52
9.13
A 40 mm diameter rotor shaft of a helicopter
transmits a torque T 0.16 kN-m and a
tensile force P 24 kN. The maximum
tensile stress (in MPa) induced in the shaft
is____. Use the value of 3.1416.
Kulkarni Academy
Kulkarni Academy
A
147
Answer Key
9.1
A
9.2
D
9.3
D
9.4
C
9.5
C
9.6
B
9.7
D
9.8
C
9.9
C
9.10
D
9.11
C
9.12
D
9.13
80
E
9.1
Shear Stress in Beam and Combined Loading
6
max
6
1
16
0.01
600
(b ) max
600 100
16
Hence, the correct option is (D).
9.4
(C)
Explanation
(A)
4
max avg
3
(D)
Hence, the correct option is (D).
9.3
4P
3A
4
6675
3 (49.21)2
4
4.6794 N/mm2
(max )circular
Hence, the correct option is (A).
9.2
circle
(D)
max
Hence, the correct option is (C).
9.5
max
(C)
6V h 2
(2h) h3 4
P
6P
22
8h 16h 2
P (50h) h
.
My
4
2
(b )max
h3
I
2h
12
2
50 Ph 12
8 2. h 4
600 P
(b )max
16 h2
6
…..(i)
…..(ii)
2
6V d 2 d
() d
bd 3 4 4
y
4
6V d 2
() y 0
.
bd 3 4
4 1
() d 1 1
y
3
4
4 16 16
1
1
() y 0
4
4
4
Hence, the correct option is (C).
Strength of Materials
9.6
148
Kulkarni Academy
Due to F
(B)
Tensile stress
F
b2
Due to moment
200
2kN
0.25 0.4
200 0.2
b (max)
30kPa
0.25 0.43
12
28
32
y 0.4 y
(From similar triangle rule)
11.2 28 y 32 y
y 0.1866
x 0.2 y
0.01333m
13.3mm
nd
II method :
200
200 y 12
0
0.25 0.4 0.25 0.43
y 13.3mm
Hence, the correct option is (B).
9.7
b
2
MY
4
b
I
12
6Fa
3
b
The value of stress component at point P = 0
F 6 Fa
3
So,
b2
b
b
a
6
Hence, the correct option is (D).
b
tensile
Fa.
9.8
(C)
200
2kPa
0.1
My
(bending ) P
I
200 0.02
3.007 kPa (Compression)
1.33 103
Net stress at P 2 3.007 1.007kPa
direct
(Compressive stress)
(D)
Hence, the correct option is (C).
Kulkarni Academy
9.9
149
Shear Stress in Beam and Combined Loading
Bending stress
MY F ( L b) b
b
b4 4
I
2
12
12 F ( L b)
b
16 b3
(C)
(total ) P
=
F 12 F ( L b)
4b2
16 b3
4 Fb 12FL 12Fb
16 b3
12 FL 8Fb F (3L 2b)
16 b3
4 b3
Ans.
Hence, the correct option is (D).
9.11
(C)
M P 2kN-m
16T 16 1000 103
d3
(50 103 )3
16T
128GPa
d3
Ans.
My 1000 103 2 25 103
b
I
(50 103 ) 4
64
512 109 Pa 512GPa
Ans.
Hence, the correct option is (C).
9.10
(D)
Shear stress
16T
16 300
3
56.6MPa
d
(30 103 )3
Bending stress
My
b
I
M P 0.3N-m
600 0.3 180 N-m
Direct stress
d
P
F
2
A 4b
180 15 103
b
67.9 MPa
(30 103 ) 4
64
Hence, the correct option is (C).
Strength of Materials
9.12
150
(D)
direct
max
x y
2
y
2
x
2
2
67.9
67.9
2
(56.6)
2
2
max 99.95MPa
tan 2P
tan 2P
Ans.
2 xy
16T 16 0.16103 103
d3
(40)3
16T
40MPa
d3
max
y
x y
2
x
2
2
max
60
60
(40)2
2
2
2
x y
2 56.6
29.520
67.9
24103 N
2
(40) 2 mm
4
415MPa 60MPa
2
max
Kulkarni Academy
Ans.
2
max 30 302 402
max 30 50 80MPa
Hence, the correct answer is 80.
Hence, the correct option is (D).
9.13
80
T 0.16 kN-m
P 24 kN
10.1 Introduction
The main aim of machine design is to size the
component. The component must be designed in
such a way that it should not fail and if at all it
falls it should fail safe.
Failure is non ability of the component to
perform its function. (failure does not
necessarily) mean separation or fracture of
component. Theories of failure provide a
relationship between strength of the component
subjected to combined loading with that obtained
in uniaxial test
Two modes of failure are considered
(i) Yielding (Ductile) y
(ii) Fracture (Brittle) 𝜎𝑢𝑡
Different theory of failures
(i) Maximum principal stress theory (Rankine’s
theory)
(ii) Maximum principal strain theory (St. Venant
theory)
(iii)Maximum shear stress theory (Tresca / Guest
theory)
(iv) Maximum strain energy theory (Haigh’s
theory)
(v) Maximum distortion energy / max shear
strain theory (Hencky-von mises theory)
(i) Maximum Principal Stress Theory
(Rankine’s theory):
According to this theory failure occurs when
the maximum principal stress in a loaded
member reaches yield stress (for ductile
materials) or ultimate stress (for brittle
materials).
1, 2
x y
1 y
Or
1 ut
2
y
2
x
xy
2
2
Strength of Materials
For no failure
1 y
Or
152
2 y
1 y
2 y
1 ut
2 ut
1 uc 2 uc
max 0.57 yt
From experiments (In general yt yc )
If
y 350 MPa
Kulkarni Academy
When a member is subjected to pure torsion
though the maximum principal stress is less than
yield stress the maximum shear stress is not less
than the permissible maximum shear stress (as
per experiments). It may fail by shear as ductile
9material are weak in shear therefore this theory
is not used for ductile materials.
(Maximum principal stress theory is applicable
for brittle material. This theory considered only
maximum principal stress and disregards other
stresses).
(ii) Maximum Principal Strain Theory:
According to this theory for no failure
maximum principal statin in a loaded
member must be less than strain under
yielding conditions when a member is
subjected to uniaxial loading.
max 200 MPa
{ 1 350 maximum principal stress is less than
y it is safe.}
max
It fails in shear
x 0
y 0
x y
1, 2
2
Principal strain 1
y
2
x
xy
2
1 2
, 2 2 1
E
E
E
E
xy
yt
yt
E
For no failure
1 y
y
350
300 MPa
2
1 yt ;
1 u2
yt
E
E
2 yt ,
2 1
yt
E
E
Kulkarni Academy
153
Theory of failure
1 2 y
Principal strain theory is bounded by four lines
and the shape is Rhombus.
() y
1 2 yt
(1 ) y
2 1 yt
1 1 yt
𝜎2 − 𝜇𝜎1 = 𝜎𝑦𝑡
1 2 yc
2 1 yc
(1 )
For most engineering materials 0.3
0.77 y
Limitations of this theory:
In the first and third quadrant though stresses
are greater than yield stress still it is safe
according to this theory but as per maximum
stress theory it is not safe.
y
0.77 350
270 not safe since {max 200}
This theory gives wrong result when a member is
subjected to pure torsion and hence this theory is
not used in practice.
(iii)Maximum shear stress theory (Tresca
/Guest):
According to this theory, failure of the
component occurs when the maximum shear
stress in the actual loaded component exceeds
yield shear stress.
The results of maximum principal strain
theory are not safe when a member is subjected
to like stresses. The results are in good agreement
when members are subjected to unlike stresses
(second 4 fourth quadrant).
1, 2
x y
2
1 2
2
For uniaxial loading
max
x yt
y 0
[max 0.57 y ]
max 200 MPa
y 350 MPa
xy 0
C yt , 0
2
y
2
x
xy
2
2
Strength of Materials
154
Kulkarni Academy
0
2
R yt
0
2
2
R
yt
2
According to maximum shear stress theory
max 0.5 y and hence it is safe (from
experiment max 0.57 y ).
As ductile materials are weak in shear and this
theory takes care of shear failure therefore this
theory is predominantly used for ductile
materials.
For no failure
actual xy
(iv) Maximum strain energy theory:
According to this theory failure occurs when
the total strain energy in the actual
component exceeds strain energy under yield
conditions.
max
1 2 2 3 1 3
;
;
2
2
2
For biaxial 3 0
max
1 2 2 1
;
;
2
2
2
1 2 yt
2
2
2 yt
2
2
1 yt
2
2
1 2 y
2 y
1 y
Actual loading
1, 2
1 2 yt
2 yt
1 yt
x y
2
y
2
x
xy
2
2
1
1
1
Total S.E density 1 1 2 2 3 3
2
2
2
1
1
( 2 3 )
E E
2
2
(1 3 )
E E
3
3
(2 1 )
E E
Kulkarni Academy
155
Theory of failure
Uniaxial loading
y
y
E
1
S .E y y
2
S .E
2y
2E
1
SE [1 1 2 2 3 3 ]
2
1 ( 2 3 )
1
E
1
(1 3 )
2 2
2
E
3 (2 1 )
3
E
1
(1 2 3 ) 2 (2 1 3 )
1 E
E
3
2
(3 2 1 )
E
1 2
SE
[1 22 32 2 (12 23 31 )]
2E
According to this theory for no failure the actual
total S.E must be less than strain energy under
yield conditions.
2y
1 2
2
2
[1 2 3 2 (12 23 31 )]
2E
2E
1
2
2y
1 2
2
[1 2 212 ]
2E
2E
12 22 212 2y
2 ()2 2()() 2y
2
For biaxial case (3 0)
212
2
1
2
2
2
y
Limiting condition,
12 22 212 2y {equation of ellipse}
2y
2(1 )
y
2(1 )
y
2 1.3
0.62 y
Strength of Materials
156
Kulkarni Academy
From experiments 0.57 y but as per total
strain energy theory max 0.62 y . It is not safe
under shear.
(V) Maximum shear strain energy or
maximum distortion energy theory or Hencky
Von-Mises theory.
x
v
v
v
E
E
y x z
1 1D v
y x y z
2 2D v
3 x
3 3D v
3
1D , 2 D and 3D responsible for change in
v responsible for change in volume.
1D (2 D 3 D )
E
E
2 D
2 D (1D 3 D )
E
E
3 D
3 D (2 D 1D )
E
E
3 𝜎𝑣2
𝑈𝑣 = 2
1
𝑈𝑣 = 6𝐸 (𝜎1 + 𝜎2 + 𝜎3 )2 (1 − 2𝜇)
Volumetric strain energy
1
1
1
U 1 1 2 2 3 3
2
2
2
in shape hence volumetric strain = zero.
1
2 D 3D
v 1D
(1 2) 0
E
1
3
2
E
E
2
2
2
1 1 2 3 212
U
2 E 223 231
1D 2 D 3D 0
1 1D v
Ud U Uv
2 2D v
1 2 3
v average
3
(1 − 2𝜇)
2
1D , 2 D and 𝜎3𝐷 Responsible only for change
1 2 3 1D 2 D 3D 3v
𝐸
3 1 2 3
Uv
(1 2)
2E
3
∈𝑣 =∈1𝐷 +∈2𝐷 +∈3𝐷 = 0
3 3D v
v
(1 2)
E
1
U v v v
2
1 3
v v (1 2)
2
E
shape.
1D
v
(1 2)
E
U d distortion strain energy
2
2
3
1 1 2 3 212
2 E 223 231
1
(1 2 3 ) 2 (1 2)
6E
Kulkarni Academy
Ud
157
Theory of failure
1 2
[1 22 33 12 23 31 ]
3E
2
2
2
2
2
2
1 1 1 2 2 3 3
Ud
6 E 212 223 231
Ud
1
[(1 2 )2 (2 3 )2 (1 3 )2 ]
6E
For uniaxial yielding
For Torsion:
1 y
2 3 0
Ud
1 2
y
3E
(yield distortion strain energy U d )
According to this theory failure of component
occurs when the distortional strain energy in the
actual component exceeds yield distortional
strain energy under uniaxial loading.
1 max
For biaxial loading (3 0)
2 max
2max (max )2 (max )(max ) 2y
1 2
Ud
(1 22 12 )
3E
32max 2y
For no failure, U d (U d ) yielding
max
1 2
1 2
(1 22 12 )
y
3E
3E
12 22 12 2y
*
Results obtained from experiments are in
complete agreement with the results of this
theory and hence this theory gives accurate
result.
*
Predominantly this theory is used for ductile
materials.
Maximum principal stress theory:
12 22 12 2y
12 22 12 is working stress sunder
distortion strain energy.
3
max 0.57 y
For limiting condition,
y 12 22 12
y
1 y
Strength of Materials
158
Maximum shear stress theory:
2 2 1 y
1
,
,
2
2
2
2
Maximum distortion energy theory:
12 22 12 2y
Theories
max value
Maximum principal stress
yt
Maximum principal strain
0.77 yt
Maximum shear stress
0.5 yt
Maximum strain energy
0.62 yt
Maximum distortion energy
0.57 yt
{Taking 0.3 }
*
Maximum principal stress theory is generally
used for brittle materials and maximum shear
stress and distortion energy theory is used for
ductile materials.
Comparison of Various theory of failure
Distortion energy theory
Maximum shear stress theory
Maximum principal stress theory.
Kulkarni Academy
Element subjected to combined bending and
torsion
x
M
My
I
d
2
4
d
34
x
32M
d 3
…. (i)
T
J r
T
4 d
d
32
2
16T
d 3
1
…. (ii)
x y
2
y
2
x
2
2
2
1 32M
1 32M 16T
3
3
3
2 d
2 d d
2
NOTE: The safest theory under static loading
is maximum shear stress theory i.e. it is the most
conservative theory.
Equivalent Bending Moment ( M e ) :
It is the single moment which produces same
maximum stress as that under the combination of
bending and torsion.
16M
16M 16T
3
3
3
d
d d
1
16
M M 2 T2
3
d
2
2
Kulkarni Academy
e
159
Theory of failure
From equation (i) and (ii),
32M e
d 3
16Te 16
3 M 2 T2
3
d
d
e 1
1
Me M M 2 T 2
2
Te M 2 T 2
Factor of safety:
Equivalent torque Te :
It is such a single torque when acting alone
produces same maximum shear stress under the
combined bending and torsion.
It is the ratio of maximum stress to the
working stress. This factor is provided for
reserved strength, more is the FOS greater is the
reserved strength, more are the dimensions and
less economical.
FOS varies from condition to condition. In case
of shock and impact loads large FOS is used.
x
32M
d 3
max
y
2
x
xy
2
2
16T
3
d
1, 2
1 2
2
max
x y
2
y
2
x
xy
2
2
2
max
32M 1 16T
3
3
d 2 d
max
16
M 2 T2
3
d
max
16Te
d 3
… (i)
… (ii)
Strength of Materials
P
Practice Questions
Which theory of failure is used for aluminum
compounds under steady loading?
(A) Principal stress theory
(B) Principal strain theory
(C) Strain energy theory
(D) Maximum shear stress theory
10.2 An element at the critical section of a
component is in a bi-axial state of stress with
the two principal stresses being 360 MPa and
140 MPa. The maximum working stress
according to distortion energy theory is
(A) 220 MPa
(B) 110 MPa
(C) 314 MPa
(D) 330 MPa
10.3 Match List-I (Theory of failure) with Lit-II
(Predicted ratio of shear stress to direct stress
at yield condition
List-I
(A) Maximum shear stress theory
(B) Maximum distortion energy theory
(C) Maximum principal stress theory
(D) Maximum principal strain theory
List-II
1. 1.0
2. 0.577
3. 0.62
4. 0.5
(A) A-1, B-2, C-4, D-3
(B) A-4, B-3, C-1, D-2
(C) A-1, B-3, C-4, D-2
(D) A-4, B-2, C-1, D-3
10.4 A circular solid shaft is subjected to a
bending moment of 400 kN-m and a trusting
moment of 300 kN-m. On the basis of the
maximum principal stress theory, the direct
stress is and according to the maximum
shear stress theory, the shear stress is . The
ratio is
is
1
3
(A)
(B)
5
9
9
11
(C)
(D)
5
6
160
Kulkarni Academy
10.5
10.1
A machine elements is subjected to the
following bi-axial state of stress
x 80 MPa, y 20 MPa, xy 40 MPa .
If the shear stress of the material is 100 MPa,
the factor of safety as per Tresca’s maximum
shear stress theory is
10.6
(A) 1
(B) 2
(C) 2.5
(D) 3
Which one of the following is NOT correct?
(A) Intermediate principal stress is ignored
when applying the maximum principal
stress theory
(B) The maximum shear stress theory gives
the most accurate results amongst all the
failure theories
(C) As per the maximum stress energy
theory, failure occurs when the strain
energy exceeds a critical value
(D) As per the maximum distortion energy
theory, failure occurs, when the
distortion energy exceeds a critical value
10.7
A thin walled spherical vessel (1 m inner
diameter and 10 mm wall thickness) is made
of a material with y 500 MPa in both
tension and compression. The internal
pressure Py at yield, based on the Van –
Mises yield criterion is
10.8
(A) 500 MPa
(B) 250 MPa
(C) 100 MPa
(D) 20 MPa
A thin rectangular plate made of isotropic
material which satisfies Von – Mises
distortion energy failure criterion as yield
strength of 200 MPa under uniaxial tension.
As shown in the figure, if it is loaded with
uniform tension of 150 MPa along the xdirection. The maximum uniform tensile
stress that can be applied along the ydirection before the plate starts yielding is
about
Kulkarni Academy
161
Theory of failure
10.11 A specimen of steel has yield strength of 700
MPa. The specimen is subjected to a state of
plane stress with 1 2 500 MPa . The
factor of safety according to Van-Mises
theory of failure is________________
10.12 A batch of aluminum alloy yields in uniaxial
tension at the stress of 330 MN/m2. If this
material is subjected to the following state of
10.9
(A) 227 MPa
(B) 77 MPa
stress: x 140 MPa, y 70MPa
(C) 87 MPa
(D) 114 MPa
xy X MPa . The value of x that would
A circular shaft of diameter d, is fixed at one
result is yield according to the Von – Mises
end and subjected to an axial force P and a
failure criterion is _____________.
torque T, at the other end. The torque T is
equal to
Pd
. The tensile yield stress of the
8
shaft material is y , a point on the surface of
the shaft will yield according to the Tresca
10.13 The state of plane stress at a point in a body
is shown in the figure. The allowable shear
stress of the material is 200 MPa. According
to the maximum shear stress theory of failure
the maximum permissible value of (in
MPa) is__________
yield criterion if P is equal to
d 2
(A) y
4 2
d 2
(B) y
4
(C) y d 2
(D) y ( d 2 )
10.10 A closed thin walled cylindrical steel
pressure vessel of wall thickness t = 1 mm is
subjected to internal pressure. The maximum
value of P (in KPa) that the wall can with
stand based on the maximum shear stress
10.14 A steel cylindrical pressure vessel ahs an
inner radius of 1.8 m and wall thickness of 20
mm. At which of the following internal
failure theory is
( ( y 200MPa and mean radius of the
cylinder, r = 1 m )
pressures will the cylindrical vessel yield as
per the Tresca criterion if the yield strength
of the material in tension is 320 MPa
(A) 100
(B) 20
(C) 300
(D) 400
(A) 3.55 MPa
(C) 1.775 MPa
(B) 7.1 MPa
(D) 4 MPa
Strength of Materials
A
162
Answer Key
Kulkarni Academy
10.5
(B)
10.1
D
10.2
C
10.3
D
x 80 , y 20 , xy 40 , max 50
10.4
C
10.5
B
10.6
B
1,2 50 302 402
10.7
D
10.8
A
10.9
A
10.10
200
10.11
1.4
10.12
157.7
10.13
200
10.14
A
E
Explanation
50 50
1,2 100,0
FOS
Hence, the correct option is (B).
10.6
10.1
100
2
50
(B)
Hence, the correct option is (B).
(D)
Hence, the correct option is (D).
10.7
(D)
Sphere
10.2
(C)
d0 1m , t 10mm , y 500MPa
12 22 12 (360)2 (140)2
PD P 103 100 P
1 2 h
4t
4 10
4
(360 140) 2w
Von 12 22 12 (500)2
w 314.32MPa
100 P
500
4
Hence, the correct option is (C).
10.3
P 20MPa
(D)
Hence, the correct option is (D).
Hence, the correct option is (D).
10.8
10.4
(C)
(A)
y 200 , x 150 1
M 400kNm
T 300kNm
16
(M M 2 T 2 )
3
d
16
3 M 2 T2
d
400 500 900 9
500
500 5
Hence, the correct option is (C).
xy 0 , y ? 2
12 22 12 y
1502 2y 150 y 200
22500 + 𝜎𝑦2 − 150𝜎𝑦 = 40000
𝜎𝑦2 − 150𝜎𝑦 = 17500
𝜎𝑦2 − 150𝜎𝑦 − 17500 = 0
y 227.07 MPa
Hence, the correct option is (A).
Kulkarni Academy
10.9
163
(A)
Theory of failure
10.11 1.4
1 2 500MPa , y 700
Pd
8
𝜎𝑦 = 𝜎𝑦
T
Working stress
16𝑇
𝜏𝑚𝑎𝑥 =
𝜋𝑑3
2𝑃
𝜏𝑚𝑎𝑥 =
12 22 12 700
16 𝑃𝑑
= 𝜋𝑑3
8
500 Working stress
𝜋𝑑2
𝜎𝑥 =
𝜎1 , 𝜎2 =
700
1.4
500
Hence, the correct answer is 1.4.
FOS
𝑃
4𝑃
=
𝐴 𝜋𝑑 2
2𝑃
2𝑃
𝜋𝑑
2𝑃
𝜋𝑑
√(
2 ±
10.12 (A)
y 330MPa , x 140
2
2) + (
2𝑃
𝜋𝑑
2)
2
y 70 ,
xy X MPa
𝜎1 , 𝜎2 = 𝜋𝑑2 (1 ± √2)
1, 2 35 (105)2 X 2
𝜏𝑚𝑎𝑥 =
1 35 (105)2 X 2
𝜎𝑦
2
𝜎1 −𝜎2
𝜏𝑚𝑎𝑥 =
2 35 (105)2 X 2
2
𝑃=
𝜋𝑑 2
𝜎12 + 𝜎22 − 𝜎1 𝜎2 = (330)2
𝜎𝑦
4√2
Hence, the correct option is (A).
2(352 (105)2 X 2 ) (352 1052 X 2 ) 3302
352 3 1052 3 X 2 3302 X 157.7MPa
10.10 200
Hence, the correct answer is 157.7 MPa.
t 1mm
10.13 200
d 2m
y max 200MPa
Maximum shear stress theory, y
Pd 2 P P
Pd
, 2
103
2t
4t 2
2t
2P
1
P 103
2t
1
max
1 2 1 2
, ,
2
2 2
= 𝜏𝑚𝑎𝑥 =
𝑃×103
2
=
𝜎𝑦
2
{𝜎𝑦 =200MPa
P 200kPa
Tresca
1 2 1 2
, ,
, 1 , 2
2
2 2
2 2
Note: As 1 and 2 are of opposite nature
max max
therefore absolute max. shear stress will be
1 2
. Hence, the correct answer is 200.
2
max 200
Hence, the correct answer is 200.
Strength of Materials
10.14 (A)
di 3.6m , t 20mm
y 320
y
1
y
2
160
Pd 3.6 P 103
2t
2 20
1 90 P, 2 45 P
max
1
45 P 160
2
P 3.55MPa
Hence, the correct option is (A).
164
Kulkarni Academy
11.1 Introduction
It is defined as the energy which is stored with in
the material. When work has been done on the
material. Here it is assume that the material
within elastic limit. If there is no energy loss due
to heat, the complete energy is recovered.
Let us assume that the load is applied
gradually from 0 to F therefore the average load
0 F F
is
2
2
F
Work
2
Strain energy due to axial loading :
(Neglect self weight of the bar)
SE
P2 L
2 AE
SE
Resilience
Volume
SE
P2 L 1
Volume 2 AE AL
P2
2
2 A2 E 2 E
Strain energy density
SE
1
Volume 2
NOTE
Strain energy is always +ve because if the load is
the compressive load, is ve and is also
ve and hence the product of and is +ve.
Strain Energy under axial loading with self
weight :
1
SE F
2
1
PL
SE P
2
AE
Strength of Materials
166
Kulkarni Academy
F P Ax
Fdx
AE
1
SE F
2
1
L
( P Ax)2 dx
2
SE
AE
0
1
SE M
2
1
L
SE M
2
R
SE
1 2
2 A2 L3
P
L
PAL2
2 AE
3
SE
M 2L
2 EI
Example 1
Find the strain energy in a cantilever beam with
point load ‘P’ at it’s free end.
Sol .
Strain energy due to torsion :
L
1
SE T
2
1
TL
T
2
GJ
SE
T 2L
2GJ
Strain energy due to bending :
SE
0
L
SE
0
SE
M 2 dx
2 EI
P 2 x 2 dx
2 EI
P 2 L3
6 EI
Case 2 : Cantilever beam with UDL ‘W’
M x Wx
x
2
2
L R
L
R
M E
I
R
EI
R
M
Wx 2
dx
L
2
SE
2 EI
0
L
W2
SE
x 4 dx
8EI 0
SE
W 2 L5
40 EI
Kulkarni Academy
Example 2
Calculate strain energy in AB portion
Sol.
RA RB P
RB L P 2L
RB 2 P
RA P
M x Px
L
P 2 x 2 dx
SE
2 EI
0
SE
P 2 L3
6 EI
167
Strain Energy
Strength of Materials
Kulkarni Academy
11.4
Practice Questions
P
11.1
168
A stepped steel shaft is subjected to a
clockwise torque of 10 Nm at its free end.
A member having length L, cross-sectional
Shear modulus of steel is 80GPa. The strain
area A and modulus of elasticity E is
energy stored in the shaft is
subjected to an axial load W. The strain
energy stored in this member is
11.2
(A)
WL2
AE
(B)
WL2
2 AE
(C)
W 2 L2
2 AE
(D)
W 2L
2 AE
What is the ratio of the strain energy in bar X
(A) 1.73 Nmm
(B) 2.52 Nmm
to that in bar Y when the material of the two
(C) 3.46 Nmm
(D) 4.12 Nmm
bars is the same? The cross-sectional areas
11.5
are as indicated over the indicated lengths.
Two shafts of the same material and equal
length are subjected to the same torque. The
diameter of the first shaft is twice that of the
second. The ratio of the strain energy of the
first shaft to that of the second shaft is
11.6
(A) 16:1
(B) 1:16
(C) 1:2
(D) 2:1
Consider a simply supported beam loaded
either by a uniformly distributed transverse
(A)
(C)
11.3
1
3
(B)
4
3
(D)
2
3
load or by a concentrated transverse load
1
6
bending stress in both cases is the same. The
applied at the centre such that the maximum
ratio of the strain energy for the two cases is
A simply supported beam of span ‘L’ is
subjected to a concentrated load W at mid-
(A)
4
5
(B)
(C)
8
5
(D) 1
span. The strain Energy due to bending in the
beam would be
W 2 L3
(A)
48 EI
W 2 L3
(B)
96 EI
W 2 L3
(C)
24 EI
W 2L
(D)
96 EI
11.7
5
8
A cylindrical steel bar of uniform crosssectional area is subjected to an axial tensile
force P and a torque T. Assuming linear
elastic deformation of the bar, the internal
Kulkarni Academy
strain
20P
energy
2
169
stored
in
the
bar
is
11.10 For a bar of circular cross-section and length
8T 2 106 N-m. The axial extension
L the ratio of the torsional to the axial strain
of the bar for P 10 N and T 16 N-m is
11.8
Strain Energy
energy
is
K
times
T
2
/ P2 ,
(A) 256 m
(B) 400 m
T torsional moment, P = axial force ,
(C) 2000 m
(D) 2048 m
where K is
A uniform rod of length l, cross-section area
A and modulus of elasticity E is held rigidly
at both ends as shown in figure. An axial load
P is applied at mid-length of the rod. The
elastic strain energy stored in the rod is
2 1 v
1 v
(B)
2
r
3r 2
1 2v
2v
(C)
(D)
2
2r
r2
(where r radius of gyration)
(A)
11.11 An L-shaped elastic member with flexural
rigidity EI is loaded as shown below: Total
strain energy in the member due to bending
is:
(A)
P2L
2 AE
P2L
(C)
4 AE
11.9
(B)
P2L
16 AE
P2L
(D)
8 AE
In the given pin jointed truss figure the strain
energy stored in the horizontal bar is K times
P2L
, where K is
AE
(A) P2b2 b / 3 a / 2EI
(B) P2b2 a / 3 b / 2EI
(C) P2a 2 b / 3 a / 3EI
(D) P2a 2 a / 3 b / 2EI
11.12 A cantilever beam is subjected to following
three different loading conditions:
(a)
(A) 2.0
(B) 1.5
(C) 0.5
(D)
1
2
(b)
Strength of Materials
170
(c)
(A) A concentrated load P at its free end
(B) A couple M 0 at its free end and
(C) Both loads acting simultaneously
The flexural rigidity of the beam may be
assumed as EI. The strain energy due to
bending when both loads act simultaneously
(A) Can be determined by applying the
principle of superposition and the strain
energy is
P 2 L3 M 02 L
6 EI 2 EI
(B) Can be determined by applying the
principle of superposition and the strain
energy is
P 2 L2 M 0 L3
6 EI
2 EI
(C) Cannot be determined by applying the
principle of superposition and the strain
energy is
P 2 L3 M 02 L PM 0 L2
6 EI 2 EI
2 EI
(D) Cannot be determined by applying the
principle of superposition and the strain
energy is
P 2 L2 M 0 L3 PM 0 L2
6 EI
2 EI
2 EI
Kulkarni Academy
Kulkarni Academy
A
171
Answer Key
11.3
11.1
D
11.2
B
11.3
B
11.4
A
11.5
B
11.6
C
11.7
B
11.8
D
11.9
C
11.10
A
11.11
A
11.12
C
E
Strain Energy
(B)
Mx
Explanation
W
x
2
L /2
11.1
SE 2
(D)
0
L /2
SE 2
0
SE
W 2L
2 AE
L /2
(B)
P2 L
( SE ) x
2 AE
P2 L P2 L
( SE )Y 2 2
A
2 E 2 AE
2
P2 L 1 P2 L
2 AE 2 2 AE
W 2X 2
dx
8EI
W2 X3
SE
4 EI 3 0
(SE due to axial loading)
Hence, the correct option is (D).
11.2
M x2 dx
2 EI
3 P2 L
2 2 AE
P2 L
2
( SE ) X
2 AE
( SE )Y 3 P 2 L 3
2 2 AE
Hence, the correct option is (B).
SE
W 2 L3
4 3 8EI
SE
W 2 L3
96 EI
Hence, the correct option is (B).
11.4
(A)
T 2L
T 2L
SE
2GJ AB 2GJ BC
(10 103 )2 100
(10 103 )2 100
2 80 103 (50) 4 2 80 103 (25) 4
32
32
SE 1.73 N-mm
Hence, the correct option is (A).
Strength of Materials
11.5
172
Kulkarni Academy
(B)
SE1
1st
2nd
d1 2d
d2 d
T 2L
2G (2d ) 4
32
SE2
T 2L
2G (d ) 4
32
WL2 PL
8
4
P
SE1 1
SE2 16
SE2
Hence, the correct option is (B).
11.6
WL
2
(C)
P 2 L3
96 EI
W 2 L5
SE2
384 EI
1
SE1 240
1
SE2
384
SE1 8
SE2 5
Hence, the correct option is (C).
11.7
(B)
WL
Wx 2
Mx
x
2
2
L
SE
0
M x2 dx
2 EI
W 2 L2 x 2 W 2 x 4 W 2 Lx3
dx
L
4
4
2
SE
2 EI
0
SE
1 W 2 L5 W 2 L5 W 2 L5
2 EI 12
20
8
SE
W 2 L5
2 EI
SE1
1 1 1
12 20 8
W 2 L5
240 EI
SE (20 P2 8T 2 ) 106 Nm
P 10 N
T 16 Nm
SE
P2 L T 2 L
2 AE 2GJ
SE 20 106 P2 8 106 T 2
Kulkarni Academy
173
Equate :
Strain Energy
11.9
20 106
L
2 AE
20 2 106
(C)
L
AE
PL
10 20 2 106
AE
400 106 m
106 m m
400 m
R 2P
Hence, the correct option is (B).
11.8
Q R cos 450
(D)
Q 2P
SE
L
L
( P RA )
2
2
AE
AE
QP
P2 L
2 AE
SE 0.5
RA
1
2
P2 L
AE
Hence, the correct option is (C).
11.10 (A)
RA P RA
RA
P
2
( SE ) AC
( SE ) BC
RB
P
2
P2 L
P2 L
2
4 2 AE 16 AE
T2
( SE )Torsion
K 2
( SE )Axial
P
P2 L
16 AE
T 2L
2
2GJ K T
2
P2 L
P
2 AE
T 2 AE
T2
K
2
2
P GJ
P
( SE ) AB
P2 L
P2 L
P2 L
16 AE 16 AE 8 AE
Hence, the correct option is (D).
K 0.5
Strength of Materials
174
I 2G(1 v)
K
K 2 G (2 I )
Kulkarni Academy
11.12 (C)
1 v
K2
K r (Radius of gyration)
K
*
1 v
K 2
r
SE1
P 2 L3
6 EI
SE2
M 02 L
2 EI
Hence, the correct option is (A).
11.11 (A)
M XX Px M 0
( Px M 0 )2 dx
2 EI
0
L
Strain energy due to bending
SE
L
SE
1
( P 2 x 2 M 02 2 PM 0 x) dx
2 EI 0
1
SE
2 EI
( S .E ) AB
( Pb)2 a
2 EI
SE
1
2 EI
L
2 x3
x2
2
P
M
x
2
PM
0
0
3
2 0
2 L3
2
2
P 3 M 0 L PM 0 L
( S .E ) BC
( Px)2 dx
2 EI
0
P 2 L3 M 02 L PM 0 L2
SE
6 EI 2 EI
2 EI
( S .E ) BC
P 2b3
6 EI
Hence, the correct option is (C).
b
( S .E )due to bending
( S .E )bending
P 2b2a P 2b3
2 EI
6 EI
P 2b2 b
a
2 EI 3
Hence, the correct option is (A).
12.1 Introduction
The vertical shift of longitudinal axis after
loading is known as deflection. Deflection is
basically due to bending. The deflection diagram
of the longitudinal axis that passes through
centroidal axis is known as elastic curve.
Methods of finding deflection
1. Double Integration method
2. Macaulay’s method
3. Strain energy method
4. Moment area method
12.2 Double Integration Method
dx R d
…(i)
(Centroidal axis after loading)
dy
dx
is small tan
tan
dy
dx
1 d
R dx
Assumptions:
(i) Loading is within elastic limit.
(ii) The material is homogeneous and isotropic.
(iii)Slopes and deflections are very small.
…(ii)
1 d dy
R dx dx
1 d2y
R dx 2
M E
I
R
(by theory of bending)
{
Strength of Materials
M 1
EI R
M d2y
EI dx 2
d2y
M
dx 2
Due to transverse load (shear load) there is a
deflection.
As the cross-section of beam is very small
compared to length this deflection due to
transverse load is negligible
and
hence
deflection due to bending is taken into
consideration.
Conditions for applying double integration
method:
(i) The beam should be prismatic. (cross section
is same throughout)
(ii) The bending moment equation does not
change along the length.
EI
176
Kulkarni Academy
Integrate,
EI
dy
x2
P C1
dx
2
Integrate,
EI y
P x3
C1 x C2
2 3
Px3
EI y
C1 x C2
6
At the fixed end ( x L)
y 0,
dy
0
dx
EI (0)
𝐶1 = −
EI (0)
PL2
C1
2
𝑃𝐿2
2
PL3 PL2
L C2
6 2
PL3 PL3
0
C2
6
2
C2
PL3 PL3
6
2
C2
PL3
3
EI ( y )
Px3 PL2
PL3
x
6 2
3
2
2
dy Px PL
EI
2
2
dx
Case-I:
Cantilever with a point load at the free end
dy Px 2 PL2
EI
dx
2
2
3
Px PL2 x PL3
EI y
6
2
3
At the free end ( x 0)
dy
PL2
dx
2
2
dy
PL
dx
2 EI
PL3
EI y
3
3
PL
y
3EI
EI
EI
𝐸𝐼
d2y
M
dx 2
𝑑2𝑦
𝑑𝑥 2
= 𝑃𝑥
…(i)
…(ii)
Kulkarni Academy
177
2
dy PL
Slope at the free end
dx 2 EI
Deflection at the free end ( y )
Deflection of Beam
Case-III : Cantilever with a moment M 0
PL3
3EI
Case-II : Cantilever with a point load not at the
free end but some where
EI
d2y
M0
dx 2
EI
dy
M 0 x C1
dx
EI y
M 0 x2
C1 x C2
2
At the fixed end ( x L),
dy
0
dx
EI (0) M 0 L C1
C1 M 0 L
Pa 3
y1
3EI
Pa 2
2 EI
M 0 x2
EI ( y )
C1 x C2
2
At the fixed end x L, y 0
EI (0)
0
M 0 L2
( M 0 L) L C2
2
M 0 L2
( M 0 L) L C2
2
M 0 L2
C2 M 0 L
2
2
C2
tan
y2
b
y2
b
y2 b
Pa 2
y2 b
2 EI
Deflection at the free end (𝑦3 ) y1 y2
Pa3 Pa 2b
y3
3EI 2 EI
M 0 L2
2
M 0 x2
M 0 L2
EI ( y)
M 0 Lx
2
2
dy
EI M 0 x M 0 L
dx
At the free end ( x 0)
dy M L
Slope 0
dx EI
Deflection ( y )
M 0 L2
2 EI
Strength of Materials
Case-IV :
Cantilever with uniformly distributed load
throughout the length
178
Kulkarni Academy
At the fixed end x L, y 0
EI (0)
WL4 WL3
, L C2
24 6
C2
WL4 WL4 3WL4
6
24
24
C2
WL4
8
dy Wx3 WL3
EI
dx
6
6
Wx 4 WL3
WL4
EI y
x
24
6
8
At the free end ( x 0)
3
dy WL
Slope
dx 6 EI
M XX
x
Wx
2
M XX
W x2
2
EI
d2y
M
dx 2
EI
dy Wx 2
dx
2
EI
dy W x3
C1
dx 2 3
EI
dy Wx3
C1
dx
6
W x4
EIy
C1 x C2
6 4
W x4
EIy
C1 x C2
24
dy
At the fixed end x L ,
0
dx
WL3
EI (0)
C1
6
C1
WL3
6
Deflection ( y )
WL4
8EI
Case V:
Cantilever beam subjected to uniformly varying load
with zero intensity at free end and maximum
intensity at fixed end:
Kulkarni Academy
Load
Load
179
Deflection of Beam
Slope at the free end ( x 0)
x Wx
2 L
dy C1
dx EI
2
Wx
2L
dy WL3
dy WL3
dx 24 EI
dx 24 EI
Case-VI :
Simply supported beam of length L and load P at
the centre
Wx 2 x
2L 3
Wx3
M XX
6L
2
d y
EI 2 M
dx
d 2 y Wx3
EI 2
dx
6L
dy W x 4
EI
C1
dx 6 L 4
dy Wx 4
EI
C1
dx 24 L
W x5
EIy
C1 x C2
24 L 5
W x5
EIy
C1 x C2
120 L
M XX
At the fixed end, ( x L) ; slope
Calculate the deflection at the centre.
M xx
dy
0
dx
WL4
C1
24 L
WL3
C1
24
At the fixed end, x L, y 0
EI (0)
EI (0)
WL5 WL3
L C2
120 L 24
WL4 WL4
24 120
WL4
C2
30
Deflection at the free end ( x 0 )
C
WL4
WL4
y 2
y
EI 30 EI
30 EI
C2
Px
2
EI
d2y
M
dx 2
EI
d 2 y Px
dx 2
2
EI
dy P x 2
C1
dx 2 2
EI
dy Px 2
C1
dx
4
EI ( y )
Px3
C1 x C2
43
Px3
EI ( y )
C1 x C2
12
At x 0 , y (deflection) = 0
EI (0)
P(0)3
C1 (0) C2
12
C2 0
At x
L dy
, 0
2 dx
Strength of Materials
180
Kulkarni Academy
EI
d2y
M
dx 2
2
PL
EI (0) C1
42
C1
PL2
16
EI
d 2 y WLx Wx 2
dx 2
2
2
EIy
Px3 PL2 x
0
12
16
EI
dy WL x 2 W x3
C1
dx
2 2 2 3
Px3 PL2 x
EIy
12
16
EI
dy WL 3 W 3
x x C1
dx
4
6
L
ycentre x
2
EI y
WL x3 W x 4
C1 x C2
4 3 6 4
EI y
WL 3 W 4
x x C1 x C2
12
24
3
P L PL2 L
EI y
12 2
16 2
y
PL3
48 EI
Case-VII:
Uniformly distributed load on simply supported
beam
At x 0, y 0
x
L dy
, 0
2 dx
At x 0, y 0
EI (0) C2
C2 0
At x
L dy
,
0
2 dx
2
EI (0)
0
M XX
WL
x
x Wx
2
2
WLx Wx 2
2
2
3
L
C1
2
WL3 WL3
C1
16
48
C1
M XX
WL L W
4 2
6
WL3 WL3 WL3 3WL3
2WL3
48
16
48
48
C1
WL3
24
EI ( y)
WL x3 Wx 4
C1 x C2
12
24
Kulkarni Academy
181
Deflection of Beam
d2y
EI 2 M 0
dx
WL x3 Wx 4 WL3
EI ( y)
x
12
24
24
L
x
2
ycentre ymax
3
EI ( ymax )
4
EI
WL L W L WL L
12 2 24 2
24 2
EI y M 0
3
WL4 WL4 WL4
96 384 48
4WL4 WL4 8WL4
384
ycentre
5WL4
(Maximum deflection)
384 EI
4
5WL
384 EI
Case-VIII:
Simple supported beam with end moments Mo
x2
C1 x C2
2
At x 0, y 0
L
EI (0) M 0 C1
2
ycentre
dy
M 0 x C1
dx
At x
M L
L dy
, 0 C1 0
2 dx
2
EI (0) 0 0 C2
C2 0
M 0 x2 M 0 L
EI ( y )
x
2
2
𝐸𝐼(𝑦) =
𝑀0 𝑥 2
2
ymax ycentre
−
𝑀0 𝐿
2
𝑥
L
x
2
EI ( ymax )
M 0 L2 M 0 L L
2 4
2 2
EI ( ymax )
M 0 L2 M 0 L2
8
4
ymax
M 0 L2
8 EI
( ve sign shows deflection(y) in downward
direction)
R1 R2 0
M R1 0
R2 ( L) M 0 M 0 0
R2 0 , R1 0
M XX M 0
ymax
M 0 L2
8 EI
12.3 Macaulay’s Method
It is a slight modification of double integration
method and this method can be used when the
bending moment equation VARIES ALONG
THE LENGTH.
Strength of Materials
182
Example 1
Sol.
Kulkarni Academy
Example 2
M XX R1 x P1 x a P2 x b
Sol.
M XX R1 x M1 M 2
EI
d2y
M XX
dx 2
EI
d2y
R1 x P1 x a P2 x b
dx 2
M XX R1 x M1 x a 0 M 2 x b 0
dy
x 2 P1 x a 3
EI
R1
dx
2
6
P x b 3
2
C1 x C2
6
Boundary conditions :
At x 0, y 0
EI
d2y
R1 x M1 x a 0 M 2 x b 0
dx 2
dy R1 x 2
EI
M1 x a M 2 x b C1
dx
2
EI y
x L, y 0
Note :
The term x a must be integrated as
x a 2
x2
ax .
but not
2
2
P 0 a 3 P2 0 b 3
EI (0) 0 1
0 C2
6
6
0 0 0 0 0 C2
C2 0
R1L P1 L a
6
6
P2 L b 3
C1L 0
6
P L a 3 P2 L b 3 R1L3
C1L 1
6
6
6
P
P
R L3
C1L 1 [ L a 3 ] 2 [ L b 3 ] 1
6
6
6
EI (0)
C1
3
R1 x3
x a 2
M1
6
2
x b 2
M2
C1 x C2
2
Boundary conditions :
At x 0, y 0
x L, y 0
0 a 2
EI (0) 0 M 1
2
0 b 2
M2
0 C2
2
0 0 0 0 0 C2
3
P1
P
R L2
[ L a 3 ] 2 [ L b 3 ] 1
6L
6
6
C2 0
R1L3
L a 2
EI (0)
M1
6
2
L b 2
M2
C1L 0
2
L b 2
L a 2 R1L3
C1L M 2
M1
2
2
6
C1
M 2 L b 2 M1 L a 2 R1L2
L
2
L
2
6
Kulkarni Academy
183
12.4 Strain Energy Method
(i) Castigliano’s first theorem
The partial derivative of total strain energy
with respect to load gives deflection of that
point under the load in the direction of load.
U
P
If the deflection is to be calculated where
there is no load, introduce an imaginary load at
that point.
Calculate the total strain energy and
differentiate partially with that load and finally
put the imaginary load zero.
Example 3
Calculate deflection at the free end
Deflection of Beam
Sol.
{At the free end no load present so introduce
a imaginary load P and at last put imaginary
load zero.}
M XX M 0 Px
(P is imaginary load)
( M 0 Px)2 dx
0 2EI
L
L
1
( M 02 P 2 x 2 2 PxM 0 ) dx
2 EI 0
L
1 2
P 2 x3
x2
U
M
x
2
P
M0
0
2 EI
3
2
0
2
P 2 L3
M
L
PM 0 L2
0
3
U
1
2 EI
U
1
P 2 EI
L3
0
2 P M 0 L2 1
3
Put P 0 (Imaginary load)
Sol.
(ii) Castigliano’s second theorem
M XX Px
L
L
M 2 dx
P 2 x 2 dx
0 2EI 0 2EI
Slope at any point is equal to the partial
derivative of total strain energy with respect to
moment at that point i.e.
L
P 2 x3
2 EI 3 0
P 2 L3
6 EI
U
M 0 L2
2 EI
P 2 L3
6 EI
U
M
If there is no moment introduced imaginary
moment and calculate the total strain energy and
differentiate partially with moment and put the
moment zero to get this load at that point.
U
L3
PL3
2P
P 6 EI
3EI
Example 4
Calculate deflection at the free end
M 2L
U
2 EI
U 2ML
ML
M 2 EI
EI
Strength of Materials
184
Kulkarni Academy
( Pa)2 b P 2b3
2GJ
6 EI
U U AB U BC
U BC
There is no moment present in given problem so
introduce an imaginary moment M at free end in
clockwise direction.
Bending moment at section X-X
M XX Px M
U
U 2 Pa3 2 Pa 2b 2 Pb3
P 6 EI
2GJ
6 EI
( Px M ) 2 dx
0 2EI
L
P 3 3 Pa 2b
(a b )
3
GJ
L
1
( P 2 x 2 M 2 2 PMx) dx
2 EI 0
L
1 P 2 x3
2 PMx 2
2
M x
2 EI 3
2 0
1 P 2 x3
M 2 x PML2
2 EI 3
U
1
[0 2ML PL2 ]
M 2 EI
Put M 0
1
[ PL2 ]
2 EI
P 2 a 3 P 2 a 2 b P 2 b3
6 EI
2GJ
6 EI
Example 6
Find the deflection at point A under the load P.
Sol.
PL2
2 EI
Example 5
Fine the deflection at point A under the load.
M B PR sin
/2
U
0
( PR sin ) 2 Rd
2 EI
P 2 R3
U
2 EI
Sol.
U U AB U BC
a
U AB
( Px)2 dx
2 EI
0
𝑃2 𝑎3
P 2 x3
=
6𝐸𝐼
6 EI 0
a
sin 2
U
sin
2
d
0
1 cos 2
2
P 2 R3
U
2 EI
2
/2
/2
0
3 /2
P R
4 EI
1 cos 2
d
2
(1 cos 2) d
0
Kulkarni Academy
P 2 R3
U
4 EI
P 2 R3
U
4 EI
185
Deflection of Beam
1 0
/2
sin 2
2 0
1
2 2 sin 2 2
P 2 R3
U
0
4 EI 2
P 2 R3
U
8EI
U 2 PR3 PR3
P
8EI
4 EI
3
PR
4 EI
ML
EI
ML
2
EI
2 0
Example 8
12.5 Moment Area Method
M XX Px
1 0
2 1 Area
dx Rd
1 d
R dx
M E
1 M
I
R
R EI
1 M d
R EI dx
M
d
dx
EI
The difference in slope between any two points
M
is equal to area of
diagram between those two
EI
points.
Example 7
d 2 1
M
L
EI
1 PL PL2
L
2 EI 2 EI
PL2
2 0
2 EI
2
PL
2
2 EI
Example 9
Strength of Materials
186
2 1 Area
1 L PL
2 2 4 EI
2 0
Kulkarni Academy
12.5.1 Moment area method
for deflection
PL2
16EI
PL2
2
16EI
t A/ B : Tangential deviation of A with respect to B
t B / A : Tangential deviation of B with respect to A
Example 10
Example 11
2 1 Area of
M
diagram between (i) and
EI
(ii),
2 0
M 2L M L
2 EI 3 EI 3
2
ML ML
3EI 3EI
2
2ML
3EI
tB / A yB Deflection at the free end.
ML
EI
Moment of the area about B
ML L
EI 2
Area
ML2
2 EI
Kulkarni Academy
187
Deflection of Beam
Example 12
t A/ B ycentre
A
1 L PL
PL2
2 2 4 EI 16 EI
PL2 L
16 EI 3
PL3
48 EI
tB / A y free end
Example 14
Find the deflection at the free end.
1
PL PL2
Area L
2
EI 2 EI
Moment of area about B
PL2 2 L PL3
2 EI 3 3EI
Example 13
Sol.
tB / A y free end
y free end A1 x1 A2 x2
ML L ML 2 L
3EI 6 3EI 3
ML2 2ML2 5ML2
18EI 9 EI
18 EI
Strength of Materials
12.2
A steel beam of breadth 120 mm and height
750 mm is loaded as shown in the figure.
A cantilever beam of length L, with uniform
cross-section and flexural rigidity, EI is
loaded uniformly by a vertical load, w per
unit length. The maximum vertical deflection
of the beam is given by
(A)
wL4
8 EI
(B)
wL4
16 EI
(C)
wL4
4 EI
(D)
wL4
24 EI
wL4
48EI
(B)
4
(C)
5wL
384 EI
Assume Esteel 200 GPa.
12.4
A simply supported beam of span L and
flexural rigidity EI carries a uniformly
distributed load w/unit length. The deflection
at the mid span of the beam is
(A)
12.3
Kulkarni Academy
Common Data Questions 12.4 & 12.5
Practice Questions
P
12.1
188
12.5
5wL4
96 EI
The beam is subjected to a maximum bending
moment of
(A) 3375kN-m
(B) 4750 kN-m
(C) 6750 kN-m
(D) 8750 kN-m
The value of the maximum deflection of the
beam is
(A) 93.75 mm
(B) 83.75 mm
(C) 73.75 mm
(D) 63.75 mm
4
(D)
3wL
16 EI
12.6
The flexural rigidity (EI) of a cantilever beam
is assumed to be constant over the length of
the beam shown in figure. If a load P and
PL
beanding moment
are applied at the free
2
end of the beam then the value of the slope at
the free end is
12.7
(A)
1 PL2
2 EI
(B)
PL2
EI
(C)
3 PL2
2 EI
(D)
5 PL2
2 EI
For the beam shown, if the maximum
deflection occurs at a distance x from support
P, which one of the following is TRUE?
(A) 0 x a
(B) x a
(C) a x L
(D) x L
A cantilever beam of span L is acted on by a
concentrated load P at the free end. To have
the same rotation at the free end under the
action of a concentrated moment M 0 at the
free end, the value of
M0
will be
P
(A) L
(B) 1.5L
(C) 1.75L
(D) 2L
Kulkarni Academy
12.8
12.9
189
The slope and deflection at the free end of a
variable cross section cantilever beam
subjected to a bending moment at the free end
as shown in the figure is
(A)
2ML 5ML2
,
3EI 18EI
(B)
ML ML2
,
EI 2 EI
(C)
ML ML2
,
1.5EI 3EI
(D)
ML ML2
,
3EI 3EI
A beam simply supported at the ends carries
a uniformly distributed load at the mid span,
If the span is doubled, the deflection at the
mid span will become
(A) 2 times
(B) 4 times
(C) 8 times
(D) 16 times
12.10 A beam is fixed at the left end and supported
by a spring at the other end. The length of the
beam is L and its flexural rigidity is EI. The
3EI
spring constant of the spring is k 3 , A
L
vertical downward load P is applied at the
right end. The deflection of the point under
the load P is
Deflection of Beam
12.11 A free end of a cantilever is attached to a
spring having a spring constant K as shown
in the figure. Assuming that the spring is undeformed prior to the application of the load
P, the deflection at the end C (Spring end)
after the load is applied is.
(A)
PL3
3EI KL3
(B)
PL3
3EI KL3
(C)
PL3
3EI 6 KL3
(D)
PL3
3EI 6 KL3
Common Data Questions 12.12 & 12.13
A cantilever beam of flexural rigidity
EI 81MN-m2 is loaded as shown in Fig:
12.12 The bending moment at the fixed end is
(A) 0
(B) 900 kN-m
(C) 1800 kN-m
(D) 8100 kN-m
12.13 The deflection at the free end is
(A) 0
(B) 75 mm
(C) 150 mm
(D) 300 mm
12.14 Two identical cantilever beams are supported
as shown, with their free ends in contact
through a rigid roller. After the load P is
applied, the free ends will have
(A) Equal deflections but not equal slopes
3
3
(A)
PL
9 EI
(B)
2 PL
9 EI
(B) Equal slopes but not equal deflections
(C) Equal slopes as well as equal deflections
3
3
(C)
PL
6 EI
(D)
5 PL
9 EI
(D) Neither equal
deflections
slopes
nor
equal
Strength of Materials
Common Data Questions 12.15 & 12.16
A cantilever beam of unknown material
(which is homogenous, linearly elastic and
isotropic) and an unknown cross-section
(which is uniform and symmetric) is given in
the figure. The stiffness of the end spring is
k 2000 N/m and end load P 1000 N;
length of the beam L 1m.
12.15 If the deflection at the free end (Under load
P, with end moment M 0 ) is measured as
5mm, the flexural rigidity EI for the beam
is (in Nm2)
(A) 66, 666
(B) 66, 000
(C) 67, 3000
(D) 64, 000
12.16 The value of the additional end moment M
(in N.m) required to obtain an upward
deflection of 1 mm at the free end, is
(moment is positive in counter clockwise
direction)
(A) 533.33
(B) 533.33
(C) 528
(D) 528
190
Kulkarni Academy
12.17 Neglecting the axial compression of member
AB, the deflection of point C in the direction
of the load is
(A)
2PL3
Ea 4
(B)
4PL3
Ea 4
(C)
8PL3
Ea 4
(D)
16PL3
Ea 4
12.18 The maximum bending stress in the frame is
(A)
3PL
a3
(B)
6PL
a3
(C)
9PL
a3
(D)
12PL
a3
12.19 Group I contains beams with different types
of supports and loading conditions. The
beams have the same flexural rigidity EI and
span L. Group II contains the maximum
deflections. Match the beam from Group I
with the maximum deflection given in Group
II.
Group-I
P.
Common Data Questions 12.17 & 12.18
A frame ABC is shown in the figure
Members AB and BC both have a length of
L, and Young’s Modulus E. Members AB
and BC both have a square cross-section of
side a. A load P is applied at point C as shown
in the figure.
Q.
R.
S.
Kulkarni Academy
191
Group-II
1.
1 FL3
3 EI
2.
1 FL3
8 EI
3.
1 FL3
48 EI
4.
5 FL3
384 EI
(A) P 3, Q 4, R 1, S 2
(B) P 3, Q 2, R 1, S 4
(C) P 4, Q 3, R 1, S 2
(D) P 3, Q 4, R 2, S 1
12.20 The simply supported beam is constructed by
welding a rigid beam to a deformable one
(with flexural rigidity EI), as shown in figure.
The deflection under the load P is
(A)
PL3
6 EI
(B)
PL3
8EI
(C)
PL3
12 EI
(D)
PL3
3EI
12.21 Consider a cantilever beam, having
negligible mass and uniform flexural rigidity,
with length 0.01 m. The frequency of
vibration of the beam, with a 0.5 kg mass
attached at the free tip, is 100 Hz. The
flexural rigidity (in N.m2) of the beam
is______.
Deflection of Beam
12.22 The area moment of inertia about the neutral
axis of a cross section at a distance x
measured from the free end is (EI = Flexural
rigidity)
bxt 3
bxt 3
(A)
(B)
12l
6l
3
bxt
xt 3
(C)
(D)
12l
24l
12.23 The maximum deflection of the beam is
12Pl 3
24Pl 3
(A)
(B)
Ebt 3
Ebt 3
8Pl 3
6Pl 3
(C)
(D)
Ebt 3
Ebt 3
12.24 The vertical deflection at the free end of the
cantilever beam is shown in figure is
1400
1400
(B)
3EI
EI
100
200
(C)
(D)
EI
EI
12.25 A frame is subjected to a load P as shown in
the figure. The frame has a constant flexural
rigidity EI. The effect of axial load is
neglected. The deflection at point A due to
the applied load P is
(A)
Common Data Questions 12.22 & 12.23
(A)
1PL3
3EI
(B)
2 PL3
3EI
(C)
PL4
EI
(D)
4 PL3
3EI
Strength of Materials
192
Kulkarni Academy
12.26 A frame of two arms of equal length L is
12.28 A force P is applied at a distance x from the
shown in the adjacent figure. The flexural
end of the beam as shown in the figure. What
rigidity of each arm of the frame is EI. The
would be the value of x so that the
vertical deflection at the point of application
displacement at ‘A’ is equal to zero?
of load P is
(A)
PL3
3EI
(B)
2 PL3
3EI
(C)
PL3
EI
(D)
4 PL3
3EI
12.27 The horizontal displacement at D of the
frame shown in figure is (neglect axial strain
energy and assume EI to be constant
throughout)
(A)
6P
EI
(B)
9P
EI
(C)
45P
EI
(D)
729P
EI
(A) 0.5 L
(B) 0.25 L
(C) 0.33 L
(D) 0.66 L
Kulkarni Academy
A
193
Deflection of Beam
PL2 PL2
1 2
2 EI 2 EI
Answer Key
12.1
A
12.2
C
12.3
B
12.4
A
12.5
A
12.6
C
12.7
B
12.8
A
12.9
D
12.10
B
12.11
B
12.12
A
12.13
C
12.14
A
12.15
B
12.16
A
12.17
D
12.18
B
12.19
A
12.20
C
12.21
0.0658
12.22
B
12.23
D
12.24
B
12.25
D
12.26
B
12.27
C
12.28
C
PL2
EI
M XX Px M
( Px M ) 2 dx
2 EI
0
L
U
L
1
P 2 x 2 dx M 2dx 2 PMxdx
2 EI 0
L
E
12.1
1 P 2 x3
2 PMx 2
2
U
M
x
2 EI 3
2 0
Explanation
(A)
U
1 P 2 L3
M 2 L PML2
2 EI 3
U
1
[0 2ML PL2 ]
M 2 EI
1
2 EI
Hence, the correct option is (A).
12.2
PL
2
2 L 2 PL
PL2
EI
Hence, the correct option is (B).
(C)
12.4
(A)
Deflection is maximum at the center
Max. deflection =
5wL4
384 EI
Hence, the correct option is (C).
12.3
(B)
b 120 mm , d 750 mm
E 200 103 MPa , L 15m
PL2
Slope due to P 1
2 EI
ML PL L PL2
Slope due to M 2
EI
2 EI 2 EI
M max
120 kN / m (15)2 m2
8
M max 3375kNm
Hence, the correct option is (A).
Strength of Materials
12.5
194
(A)
ymax ycentre
12.8
5WL4
384 EI
120 750
bd
12
12
9
I 4.218 10 mm4
l 15m 15000 mm
I
3
(A)
Hence, the correct option is (A).
12.9
(D)
3
5 120 103
(15000) 4
384
103
200 103 4.218 109
ycentre 93.75mm
Hence, the correct option is (A).
12.6
Kulkarni Academy
(C)
ymid
5WL4
384 EI
ymid L4
ymid (2L)4
ymid 16L4
Hence, the correct option is (D).
12.10 (B)
aL
axL
In case of simply supported beam with a point
load not at the center the maximum deflection
occurs at the mid point of load and center of the
beam.
Hence, the correct option is (C).
12.7
(B)
PL3
3EI KL3
Given, K
3EI
L3
PL3
3EI
3EI 3 L3
L
PL3
6 EI
Hence, the correct option is (B).
12.11 (B)
PL2
2
2 EI PL M 0 L
EI
M 0 L 2 EI
EI
M0 L
P
2
M0
0.5 L
P
Hence, the correct option is (B).
Kulkarni Academy
195
PL3
y1
3EI
y2
Deflection of Beam
12.13 (C)
FS L3
3EI
Net deflection y1 y2
Deflection of spring
PL3 100 103 N 93 m3
3EI
3 81106 Nm 2
FS
F
K S
K
y1
y1 y2
y1 0.3m 300mm
PL3 FS L3 FS
3EI 3EI
K
ML2 900 103 92
y2
2 EI
2 81106
PL3 FS L3 FS
3EI 3EI K
PL3 FS KL3
1
3EI K 3EI
3
12.14 (A)
3EI KL
PL
3EI
3EI
3
ynet y2 y1 450 300 150mm
Hence, the correct option is (C).
KL
PL
1
3EI
3EI
3
y2 0.45m 450mm
3
PL3
3EI KL3
Hence, the correct option is (B).
12.12 (A)
As the roller is always in contact with both the
rollers. Therefore, the deflection is same but
slopes are not same.
Hence, the correct option is (A).
12.15 (B)
K 2000 N / m
P 1000 N
L 1m
EI 81MN-m2
81106 Nm2
Hence, the correct option is (A).
y1
PL3
3EI
FS L3
y2
3EI
Strength of Materials
196
FS
FS
K
FS K
Kulkarni Academy
12.17 (D)
K
2000 5 103 10
Spring deflection y1 y2
FS PL3 FS L3
K 3EI 3EI
L3
5 103
( P FS )
3EI
13
5 103
(1000 10)
3EI
1
5 103
(990)
3EI
990
EI
3 5 103
EI 66000 Nm2
U U AB U BC
a a3 a 4
I
12
12
L
( Px)2 dx
U BC
2 EI
0
Hence, the correct option is (B).
12.16 (A)
FS
FS K
K
U BC
P 2 L3
6 EI
U AB
P 2 L2 L
2 EI
U AB
P 2 L3
2 EI
P 2 L3 P 2 L3
U
2 EI 6 EI
3
2000 N / m 110 m 2 N
U 2 PL3 2 PL3
P 2 EI 6 EI
PL3 PL3 PL3 1
1
EI 3EI EI 3
PL3 FS L3 ML2
1103
3EI 3EI 2 EI
1000 1
2 1
M 1
1103
3 66000 3 66000 2 66000
M 533.33 Nm
Hence, the correct option is (A).
3
3
2
4 PL3 4 PL3 12
3EI
3E a 4
16PL3
Ea 4
Hence, the correct option is (D).
Kulkarni Academy
12.18 (B)
197
Deflection of Beam
12.20 (C)
Note :
I
I
a a3
12
Rigid member cannot deform and hence it will
not absorb any strain energy. Only deformable
member absorbs strain energy.
Px
2
M
4
a
12
L
M max PL
U
0
P 2 x 2 dx
4 2 EI
L
P2 x2
U
24 EI 0
M
I
y
My
I
max
M y
max max
I
max
a
PL
2
a4
12
max
U
P 2 L3
24 EI
U 2 PL3
P 24 EI
PL3
12 EI
Hence, the correct option is (C).
12.21 0.0658
L 0.01m
m 0.5kg
f 100 Hz
EI ?
6PL
a3
Hence, the correct option is (B).
2f f
12.19 (A)
Hence, the correct option is (A).
K
m
1
2
f
1 K
2 m
Strength of Materials
100
1 K
2 0.5
K 197392 N/m
PL3
P 3EI
K 3
3EI
L
K
3EI
L3
197392
198
Kulkarni Academy
12.23 (D)
M XX Px
L
( Px) 2 dx
0 2EI
L
U
0
P 2 x 2 dx 12 L
2 E bxt 3
L
6P2 L
xdx
Ebt 3 0
U
3 EI
(0.01)3
EI 0.0658 Nm2
Hence, the correct answer is 0.0658.
L
6P2 L x2
6 P 2 L L2
U
Ebt 3 2 0 Ebt 3 2
U
3P 2 L3
Ebt 3
U 3.2 PL3
P
Ebt 3
12.22 (B)
6 PL3
Ebt 3
Hence, the correct option is (D).
12.24 (B)
b bx
L x
bx
a 2 m , b 1m , L 3m , P 100 103 N
bx
L
Pa 3
y1
3EI
bx t 3
I
12
I
bxt 3
12 L
Hence, the correct option is (B).
Pa 2
2 EI
y
tan 2 y2 b tan
b
y2 b
Pa 2
y2 b
2 EI
Kulkarni Academy
199
y y1 y2
P 2 L3 P 2 L3
U
6 EI 6 EI
Pa3 Pa 2b
3EI 2 EI
y
Deflection of Beam
100 103 23 100 103 22 1
y
3EI
2 EI
1400
y
3EI
Hence, the correct option is (B).
12.25 (D)
P 2 L3
U
3EI
U 2 PL3
P 3EI
2 PL3
3EI
Hence, the correct option is (B).
2 3
U BC
P L
6 EI
U AB
P 2 L2 L
2 EI
12.27 (C)
U U AB U BC
P 2 L3 P 2 L3
U
2 EI 6 EI
U
2 P 2 L3
3EI
U 2.2 PL
P
3EI
3
4 PL3
3EI
Hence, the correct option is (D).
12.26 (B)
U CD
P 2 L3
6 EI
U BC
P 2 L2 L
2 EI
U BC
P 2 L3
2 EI
M ( Px PL)
( Px PL)2 dx
2 EI
0
L
U AB
L
U AB
MB 0
RA ( L) PL 0
RA P
U U AB U BC
1
( P 2 x 2 P 2 L2 2 P 2 xL) dx
2 EI 0
L
U AB
1 P 2 x3
x2
2 2
2
P L x 2P L
2 EI 3
2 0
U AB
1 P 2 L3
P 2 L2 L P 2 LL2
2 EI 3
Strength of Materials
U AB
P 2 L3
6 EI
U U AB U BC UCD
U
P 2 L3 P 2 L3 P 2 L3
6 EI 2 EI 6 EI
U
5P 2 L3
6 EI
U 10 PL3 5PL3
P
6 EI
3EI
5P(3)3
3EI
45P
EI
Hence, the correct option is (C).
12.28 (C)
PL3 P( L x) L2
3EI
2 EI
L Lx
3
2
2L 3L 3x
3x L
x 0.33 L
Hence, the correct option is (C).
200
Kulkarni Academy
0
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