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CE : CIVIL ENGINEERING
Module 8 : Solid Mechanics
INDEX
Pg.
Contents Topics
No.
Introduction 1
Stress and Strain 2
Stress  Strain Relationships and its Constants 8
Free Body Diagrams 10
Mohr’s Circle for principal Strain and Stress 11
Shear Force and Bending Moment Diagrams 13
Notes Bending stress 21
Moment of Inertia of Plane Areas 24
Deflection of Beams 25
Torsion of Circular Shafts 34
Euler Theory of Columns 37
Strain Energy Methods 41
List of Formulae 47
Assignment  1 52
Assignment  2 54
Assignments Assignment  3 57
Assignment  4 60
Assignment  5 63
Test Paper  1 66
Test Paper  2 68
Test Papers Test Paper  3 70
Test Paper  4 73
Test Paper  5 76
Practice Problems 81
SOLUTIONS
Answer Key 90
Assignment
Model Solutions 92
Answer Key 106
Test Paper
Model Solutions 107
Answer Key 121
Practice Problems
Model Solutions 122
 Solid Mechanics

INTRODUCTION
Strength of materials is the scientific area of applied mechanics for the study of the
strength of engineering materials and their mechanical behaviour in general (such as
stress, deformation, strain and stress-strain relations). Strength is considered in terms of
compressive strength, tensile strength, and shear strength, namely the limit states of
compressive stress, tensile stress and shear stress respectively. Certain mechanical
properties are needed to understand the basics.
MECHANICAL PROPERTIES OF MATERIALS
Strength
Strength is the property that enables a metal to resist deformation under load. The
ultimate strength is the maximum strain a material can withstand. Tensile strength is a
measurement of the resistance to being pulled apart when placed in a tension load.
Elasticity
Elasticity is the property of an object or material which causes it to be restored to its
original shape after distortion. It is said to be more elastic if it restores itself more
precisely to its original configuration. A perfectly elastic material is one which returns to
its original shape and size after removal of an applied stress. An inelastic material
becomes permanently deformed following application of a stress.
Isotropic Material
In this type of material, the elastic constants are the same in all directions so if a
specimen is cut from a bulk material, the direction in which it is cut has no affect on the
values. This applies to most metals with no pronounced grain structure.
Orthotropic Material
In this type of material, the elastic constants have different values in the x, y and z
directions so the results obtained in a test depend upon the direction in which the
specimen was cut from the bulk material. This applies to materials with grain structures
such as wood or rolled metals.
Non-Isotropic Material
In this type of material, the elastic constants are unpredictable and the results from any
two tests are never the same. This applies to materials such as glass and other ceramics.
Hardness
Hardness is the property of a material that enables it to resist plastic deformation, usually
by penetration.

Malleability
Malleability is the property of being physically malleable; the property of something that
can be worked or hammered or shaped under pressure without breaking.
Plasticity
Plasticity is the property of a material, such as an adhesive, which permits continuous
and permanent deformation without failure upon the application of a force which exceeds
the yield value of the material.

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Toughness
Toughness is the property that enables a material to withstand shock and to be
deformed without rupturing.
Fatigue
Fatigue strength is the ability of material to resist various kinds of rapidly changing
stresses and is ex- pressed by the magnitude of alternating stress for a specified number
of cycles.

STRESS AND STRAIN

Stress
It is the internal distribution of forces within a body that balance and react to the loads
applied on it.

L
A

L
P
The elongation of a bar is proportional to the tensile force applied on it. This force creates
internal forces in the bar to maintain equilibrium. In discussing the magnitude of internal
forces let us imagine the bar cut into two parts by a cross-section and let us consider the
equilibrium of the lower portion of the bar. At the lower end of this portion the tensile force
P is applied.
On the upper end the forces represent the action of the particles of the upper portion of
the strained bar on the particles of the lower portion. These forces are continuously
distributed over the cross-section. In handling such continuously distributed forces, the
intensity of force, i.e. the force per unit area is very important. The resultant of these
forces will pass through the centroid of the cross-section and will act along the axis of the
bar. Taking into account that the sum of the forces, from the condition of equilibrium,
must be equal to P and denoting the force per unit of cross-sectional area by , we
obtain,
P
=
A
This force per unit area is called unit tensile stress or simply stress.
Stress would either be one  dimensional, two dimensional or three dimensional.

x
1D Stress

GATE/CE/SLP/SM/Notes/Pg.2
 Notes on Solid Mechanics

y

x x

y
2D Stress

y

z

x
x

z

y
3D Stress

Direct and Shear Stress

Let us take a prismatic bar acted upon by an axial tension P.
N A


P P

B
Let us take a section AB inclined to the plane of the axis. The internal stresses at the
section balance the external force P applied at the left end. The resultant of the forces
distributed over the cross-section.
AB is equal to P. Taking the area of cross-section of the bar normal to the axis as S and
 be the angle between the x  axis and plane AB, then the stress over the cross-section
 is,
P
 = cos  x cos
S

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P
where x = is the stress on the cross-section normal to the axis of the bar. We can
S
see that the stress  over any inclined cross-section of the bar is smaller than the stress
x over the cross-section normal to the axis of the bar. For  = /2, the stress becomes
zero.
This stress component x, is parallel to the axis of the bar and is resolved into two
components.
The stress component n perpendicular to the cross-section is called the normal or direct
stress.
n =  cos  = x cos 2
The tangential component  is called shearing stress and has the value,

 =  sin  = x cos  sin = x sin2
2
A
n
 P
S 

B
It is seen that the normal stresses n produce extension of the element in the direction of
the normal to the cross-section AB and the shearing stresses produce sliding of section
AB.
From the above equations we see that the maximum normal stress acts on cross-
sections normal to the axis of the bar and we get,
(n)max = x
The maximum shearing stress, acts on cross-sections inclined at 45 to the axis of the
bar, where sin 2  = 1, and has the magnitude
1
()max =  x
2
Although the maximum shearing stress is one half the maximum normal stress, this
stress is sometimes the controlling factor when considering the ultimate strength of
materials which are much weaker in shear than in torsion.
Plane stress is a two-dimensional state of stress in a body. This is a good model when a
flat thin body is loaded in the plane of the body. A small volume element in equilibrium
experiences forces in balance. By specifying some co-ordinates, the forces P and Q can
be resolved normal and perpendicular to the faces of the volume element.

GATE/CE/SLP/SM/Notes/Pg.4
 Notes on Solid Mechanics

The stresses on the element are:

 Normal stresses:
 σx = Px/∆A
 σy = Qy/∆A
 Shear stresses:
 τxy = Py/∆A
 τyx = Qx/∆A

Materials in a stressed component often have direct and shear stresses acting in two or
more directions at the same time. This is a complex stress situation.
Consider a rectangular part of the component material as shown in the figure below.
Stress σx acts on the x plane and σy acts on the y plane. The shear stress acting on the
plane on which σx acts is τx and τy acts on the plane on which σy acts. The shear stresses
are complementary and so must have opposite rotation. We will take clockwise shear to
be positive and anti-clockwise as negative.

ydirection

xdirection

Now consider a plane at angle  to the x plane. The plane is in equilibrium so all the
forces and moments on the plane must add up to zero. The stresses must first be turned
into forces. If the material is 1 m thick normal to the paper then the areas are x and y on
the sides and y/sin or x /cos on the sloping plane.
Consider the balance of forces on the right lower corner. First the forces are resolved
perpendicular and parallel to the plane.

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yy  c
  g
b yy
d xx
a xx
 h
e f

  
a  yy sin  b  y y cos 
c  yy cos  d  yy sin 
e  xx sin  f  xx cos 
g  x x cos  h  xx sin 
All the forces normal to the plane must add up to (y/sin) .


y

y
sin 


Balancing normal forces we have a + c + e + g = (y/sin) , and

All the forces parallel to the plane must add up to (y/sin) 
- f + h + b - d = (y/sin) 
Making the substitutions and conducting algebraic process will yield the following results.
y
  yy sin   yy cos   xx sin   xx cos 
sin 
x x
  y sin2   y sin  cos   x sin2   sin  cos 
y y

1  cos 2 sin2 sin2  sin  cos 

  y  y  x  x
2 2 tan  tan 
1  cos 2 sin2 sin2 1  cos 2
  y  y  x  x
2 2 2 2
y cos 2 sin2 sin2  x cos 2
   y  y  x   x
2 2 2 2 2 2

GATE/CE/SLP/SM/Notes/Pg.6
 Notes on Solid Mechanics

The shear stress on both planes are equal so denote them both by xy  x  y . Hence,

 
 x  y 

 x  y  cos 2
 xy sin2 …(1.1)
2 2
Repeating the process for the shear stress we get

 
 x  y  sin2
 xy cos 2 …(1.2)
2
Strain
It is the geometrical expression of deformation caused by the action of stress on a
physical body. Strain therefore expresses itself as a change in size and/or shape. If strain
is equal over all parts of the body, it is referred to as homogeneous strain; otherwise, it is
inhomogeneous strain.
L L

A
extension L
strain  
length L
Strain has no units because it is the ratio of two lengths.

Relationship between Principal Stress & Strain

In any stress system, there are 3 mutually perpendicular planes on which only direct
stress acts and there is no shear stress. These are the principal planes and the stresses
are the principal stresses. These are designates σ1, σ2 and σ3. The corresponding
strains are the principal strains ε1, ε2 and ε3.
1

2

3

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The strain in each direction is given by

1 1
1 = (1v2 v3 ) =  1v(2 3 )
E E
1 1
2 = (2 v1v3 ) =  2 v(13 )
E E
1 1
3 = (3 v1v3 ) =  3 v(12 )
E E

where  = Poisson’s ratio.

 1  2 
 Volumetric strain = 1  2  3   1  2  3   
 E 
Volumetric Strain
Consider a cube of a certain material which is stressed in the x-direction by a
compressive pressure as shown in the figure alongside.
The change in volume is L2 L.
Now consider that the cube is strained by
an equal amount in the y and z-directions
also. With very little error the total change in
volume is 3L2∆L. The original volume is L3.
When a solid object is subjected to a
pressure p such that the volume is reduced,
the volumetric strain is εv = change in
volume/original volume.
In the case of a cube this becomes
L 3L
 v  3L2 3 
L L
v = 3 
 is the equal strain in all three directions.
In case of a beam, volumetric strain is given by
P  1 2 
 vol  
E  1/  
where P = load
 = Poisson’s ratio

STRESS-STRAIN RELATIONSHIPS AND ITS CONSTANTS

Young’s Modulus
For the description of the elastic properties of linear objects like wires, rods, columns
which are either stretched or compressed, a convenient parameter is the ratio of the
stress to the strain, a parameter called the Young's modulus of the material.
Young's modulus can be used to predict the elongation or compression of an object as
long as the stress is less than the yield strength of the material.

GATE/CE/SLP/SM/Notes/Pg.8
 Notes on Solid Mechanics

x - direction

Young’s modulus : y - direction

stress F / A
E 
strain L / L
Hooke’s Law
Hooke's law simply states that the extension of a spring (or any other stretchable object)
is directly proportional to the force acting on it.
Stress  Strain
This law is only true if the elastic limit of the object has not been reached.
If the elastic limit has been reached the object will not return to its original shape and may
eventually break.
Bulk Modulus of Elasticity
The Bulk Modulus of Elasticity is the elastic response to hydrostatic pressure and
equilateral tension or the volumetric response to hydrostatic pressure and equilateral
tension. It is also the property of a material that determines the elastic response to the
application of stress.
Poisson’s Ratio
Poisson's Ratio is the ratio of transverse strain to corresponding axial strain on a material
stressed along one axis.
Consider a piece of material in 2-dimensions. The stress in the y direction is y and there
is no stress in the xdirection. When it is stretched in the ydirection, it causes the
material to get thinner in all the other directions at right angles to it. This means that a
negative strain is produced in the xdirection. For elastic materials, it is found that the
applied strain (εy) is always directly proportional to the induced strain (εx) and its ratio is
called Poisson’s Ratio.
x
 
y
The strain produced in the xdirection is εx = - νεy
If stress is applied in xdirection then the resulting strain in the ydirection would similarly
be εy = - νεx
The resulting strain in any one direction is the sum of the strains due to the direct force
and the induced strain from the other direct force.
  y
 x  x  y  x  
E E E
1
 x    x  y  ...(1A)
E
1
Similarly  y    y  x  …(1B)
E

GATE/CE/SLP/SM/Notes/Pg.9
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Converting Strain into Stress

We have already derived
 v2
1 = 1
E
2 v1
2 =
E
Rearrange to make 2 the subject.
2 = 2E + v1
Substitute this into the first formula
( Ev1 )v1
2 = 2
E
Rearrange
 E 
1 =  (1v2 )
 1v 2 
If we do the same but make 2 the subject of the formula we get
 E 
2 =  (2 v1 )
 1v 2 
Relationship between the elastic constants
When the material is compressed by a pressure p the stress is equal to -p because it is
compressive. The bulk modulus is then
p
K
v
E
K=
3  1  2 
This shows the relation between E, K and ν.
The relation between E, G and  is given by
E
G=
2(1  )
where E = Young’s modulus, G = Bulk modulus,  = Poisson’s ratio

FREE BODY DIAGRAMS

Free-body diagrams are diagrams used to show the relative magnitude and direction of
all forces acting upon an object in a given situation.

Fnorm

Fapp where,
Fgrav : Gravitational force.
Ffrict Fnorm : Normal force acting due to reaction
from ground.
Fapp : Applied Force
Fgrav Ffrict : Frictional Force between object
and ground.

GATE/CE/SLP/SM/Notes/Pg.10
 Notes on Solid Mechanics

Example :
For a loading of a beam as given below, the reaction at support A is
500 Nm 200N

A B

1m 1m

4m
RA RB
Solution :
RA, RB are the reactions.
Now doing vertical force balance,
RA + RB = 200 N
Doing moment balance about B,
RA  4 + 500 = 200  2
RA =  25 N acting downwards.

Example :
In the above problem, the reaction at support B is ?
Solution :
As calculated before,
RA =  25 N
 RB =  200  25  N  225N 

MOHR’S CIRCLE FOR PRINCIPAL STRAIN AND STRESS

Determining principal stresses and planes
The stresses are a maximum or minimum on the principal planes so using the maximum
or minimum theory we have
d 
   x  y  sin2  2 cos 2  0
d
2 cos 2   x  y  sin2
2
tan2 
x  y
There are two solutions to this equation giving answers less than 360 degrees and they
differ by 90 degrees. From this, the angle of principal plane may be found. From this the
principal stresses are calculated and are found to be

  y    y   42xy
2
x x
max  1  
2 2

  y     y   42xy
2
x x
min  2  
2 2

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Similarly the shear stress is found to be

  y   42xy
2
x 1  2
max  
2 2

  y   42xy
2
 x    2 
min    1 
2  2 

Mohr's Circle illustrates principal strain and strain transformations via a graphical format,

x
x  y Note: A rotation, from a
 c= reference stress state in the
2
y real plane  corresponds to a
rotation of 2 from the
1
reference points in the Mohr’s
circle plane.
y, xy

xy 2
0 2 x 1

y c
2
xy

(x, xy)

2

Fig. Mohr’s Circle

 x  y 
Centre of Mohr’s circle is placed at  ,0 
 2 

2
 x  y  2
The radius of circle = r =    xy
 2 
Principal stresses have values 1, 2 = centre  radius
The maximum shear stress = radius

GATE/CE/SLP/SM/Notes/Pg.12
 Notes on Solid Mechanics

The normal strains are equal to the principal strains when the element is aligned with the
principal directions, and the shear strain is equal to the maximum shear strain when the
element is rotated 45° away from the principal directions.

As the element is rotated away from the principal (or maximum strain) directions, the
normal and shear strain components will always lie on Mohr's Circle.
To establish Mohr's Circle, we first recall the strain transformation formulas for plane
strain at a given location,
  x  y  x  y
 x    cos 2 xy sin2
2 2

    x  y sin2 cos 2
 x y 
2
xy

Using basic trigonometric relations to combine the two above equations we have,
2 2
  x  y  2   x  y  2
  x     xy    xy
 2   2 
This equation is an equation for a circle. To make this more apparent, we can rewrite it
as,
  Avg   xy2 R2
2
x

where,
2
 x  y   x  y 
 Avg  , R    xy
2

2  2 

The circle is centered at the average strain value Avg, and has a radius R equal to the
maximum shear strain,
The Mohr's Circle for plane stress can also be obtained from similar procedures.

SHEAR FORCE AND BENDING MOMENT DIAGRAMS

Types of Beams and Loads :

Point Loads
A point load is shown as a single arrow and acts at a point.

Uniform Loads
Uniform loads are shown as a series of arrows and has a value of w N/m. For any given
length x meters, the total load is wx Newton and this is assumed to act at the centre of
that length.

Beams
Here, we discuss the simplest types of beams having a vertical plane of symmetry
through the longitudinal axis. It is assumed that all applied forces are vertical and act in
the plane of symmetry so that bending occurs in the same plane.

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P
a b


R1 R2
The above figure represents a beam with simply supported ends. Points of support A and
B are hinged so that the ends of the beam can rotate freely during bending.
P
a
B
A
M1 R1
The above figure represents a cantilever beam. The end A of this beam is built into the
wall and cannot rotate during bending, while the end B is totally free.
A C
B

The above figure represents a beam with an overhanging end. This beam is hinged to an
immovable support at the end A and rests on a movable support at C.
All the above three cases represent statically determinate beams since the reactions at
the supports produced by a given load can be determined from the equations of statics.
For the simply supported beam carrying a vertical load P, we see that the reaction R2 at
the end B must be vertical, since this end is free to move horizontally. Then from the
equations of statics, X = 0, it follows that reaction R1 is also vertical. The magnitudes of
R1 and R2 are then determined from the equations of moments. Equating to zero the sum
of the moments of all forces with respect to point B, we obtain
R1  Pb  0
from which,
Pb
R1 

Similarly, by considering the moments with respect to the point A, we obtain,
Pa
R2 

Bending Moment and Shearing Force :
Consider a simply supported beam acted upon by vertical forces P1, P2 and P3 as shown
below. Assume that the beam has an axial plane of symmetry and that the loads act in
this plane. Then from symmetry, we conclude that bending must also occur in this same
plane.

GATE/CE/SLP/SM/Notes/Pg.14
 Notes on Solid Mechanics

i) To investigate the stresses produced in a beam during bending, let us imagine that
the beam AB is cut in two parts by a crosssection ab taken at any distance x from
the left support, and that the portion of the beam to the right is removed.
R1 R2
P2
a
P1 P3

x
b (a)


a
c2 P2

c1 P1

R1 x
(b)
b

V
M
(c)

In discussing the equilibrium of the remaining left hand portion of the beam, we
consider not only the external forces such as loads P1, P2 and reaction R1, but also
the internal forces which are distributed over the cross-section ab and which
represent the action of the right portion of the beam on the left portion. These internal
forces must be of such a magnitude as to equilibrate the above mentioned external
forces P1, P2 and R1.
From statics, we know that a system of parallel forces can be replaced by one force
equal to the algebraic sum of the given forces together with a couple. Thus, the
balancing force here is,
V = R1  P1  P2
and the magnitude of the couple is
M  R1x  P1  x  C1   P2  x  C2 
The force V, which is equal to the algebraic sum of the external forces to the left of
the cross-section ab, is called the shearing force at the cross-section ab. The couple
M, which is equal to the algebraic sum of the moments of the external forces to the
left of the cross-section ab with respect to the centroid of this cross-section is called
the bending moment at the cross-section ab.

GATE/CE/SLP/SM/Notes/Pg.15
Vidyalankar : GATE – CE

ii) Let us now consider a distributed load rather than a number of concentrated forces
acting on a beam, then denoting the load per unit length by q, the reactions in this
case are,
q
R1  R2 
2

R1 R2
x a
q C

b
A  B

a
q
R1 C
x b

Let us investigate the stresses distributed over a cross-section ab by considering the

equilibrium of the left portion of the beam, as shown in the above figure. The external
forces acting on this portion of the beam are the reaction R1 and the load uniformly
distributed along the length x. This latter load has a resultant equal to qx. The
algebraic sum of all forces to the left of the cross-section ab is R1  qx. The moment
of the distributed load about point C is evidently equal to,
x qx 2
qx  
2 2
We obtain for the algebraic sum of the moments the expression,
qx 2
R1x 
2
All the forces acting on the left portion of the beam can now be replaced by one force
acting in the plane of the cross-section ab and equal to
 
V  R1  qx  q   x 
2 
together with a couple equal to,
qx 2 qx
M  R1x     x
2 2

GATE/CE/SLP/SM/Notes/Pg.16
 Notes on Solid Mechanics

Sign Conventions :
The bending moment and the shearing force at a cross-section have the following
sign conventions.
M

(+) ()
Bending Moment

(+) ()
Shearing Force

Relation between Bending Moment and Shearing Force :

Consider an element of a beam cut out by two adjacent cross-sections ab and a1b1 which
are at a distance shearing force at the cross-section ab, the action of the left portion of
the beam on the element is represent by the force V and the couple M as shown in the
figure below.
a a1

V
V
M dx
M + dM

b b1
If no forces act on the beam between cross-sections ab and a1b1 the shearing forces at
these two sections are equal. Regarding the bending moment, there is an increase which
equals the moment of the couple represented by the two equal and opposite forces V,
i.e.,
dM  Vdx
and
dM
V
dx
Similarly, for a distributed load of intensity q acting between the cross-sections ab and
a1b1, the total load acting on the element is q dx.
Then,
dV = q dx
which gives,
dV
 q
dx

GATE/CE/SLP/SM/Notes/Pg.17
Vidyalankar : GATE – CE

Bending Moment and Shear Force Diagrams :

A representation in which the abscissa indicate the position of the cross-section and the
ordinate represents the value of the bending moment or shearing force which acts at this
cross-section, is called Bending Moment and Shearing Force Diagrams.
i) Consider a simply supported beam with a single concentrated load P.
The reactions in this case are,
Pb Pa
R1  and R2 
 
R1 P R2
a b
m

x n

Taking a cross-section mn to the left of P, it can be concluded that at such a cross-
section,
Pb Pb
V and M  x
 
For a cross-section to the right of the load, we obtain
Pb Pb
V  P and M  x  Px  a ,
 
x always being the distance from the left end of the beam.
a
c
Pb
(+)


Pa
()

c b
Shear Force Diagram
c1

Pab
(+)


a1 b1
Bending Moment Diagram

The shearing force and bending moment diagrams are drawn as shown above.
Note that the shearing force diagram consists of two rectangles of equal areas. Also,
the rectangles are in opposite directions.
In the case of a simply supported beam, the moments at the end vanish.

GATE/CE/SLP/SM/Notes/Pg.18
 Notes on Solid Mechanics

ii) If several loads act on a beam, the beam is divided into several portions and
expressions for V and M are established for each portion, as shown below.

R1 a3 b3 R2

a2 P3
P2
P1
a1

a c
P1
d
c
R1

b
P2

R2
e
d P3
e
Shear Force Diagram

R1a1P1(a2  a1)
R1a1
R2a3

a b

Bending Moment Diagram

iii) Consider the case of a uniformly distributed load, as shown below :

a b

GATE/CE/SLP/SM/Notes/Pg.19
Vidyalankar : GATE – CE

a
q
2
c
q
2
Shear Force Diagram
b
b1

q 2
8

a1 c1
Bending Moment Diagram
Now at the cross-section ab at a distance x from the left support,
 
V  q   x  and
2 
qx
M   x
2
We see here that the shearing force diagram consists of an inclined straight line.
Also, the bending moment here is a parabolic curve with its vertical axis at the middle
of the span of the beam. The maximum value is obtained at the middle of the span,
giving,
q 2
Mmax 
8
If concentrated loads and distributed loads act on the beam simultaneously, it is
advantageous to draw the diagrams separately for each kind of loading and obtain
the total values of V or M at any cross-section by summing up the corresponding
ordinates of the two partial diagrams.

Example :
A simply supported beam of length L, cross-section A, carrying a uniformly distributed
load of W will have maximum bending moment of:
Solution :

A B

GATE/CE/SLP/SM/Notes/Pg.20
 Notes on Solid Mechanics

Due to symmetry, the maximum bending moment will occur at the center. Now, taking
a cross sect ion at the center, and doing a moment balance about point A, we have,
M = (W/L x L/2) x L/4 = WL/8, where the first term is the net force due to the
distributed load and the second term is the distance from A at which it is acting.
The bending moment at any distance x from point A is :
C

W/L = q

M
A
x
W/2
Taking moment about C, we get,

W  W x W Wx 2 L
M =  x   x  x , 0x
 2   L 2 2 2L 2
 Bending moment diagram looks like :

WL
8
x
O L

W
The relationship between M and  q is as follows :
L
W Wx 2 qL qx 2
M x  x
2 2L 2 2
dM qL
  qx  = shear stress at any section x.
dx 2
d2 M
&  q  0
dx 2
Thus distributed load q is second differential of moment M.

GATE/CE/SLP/SM/Notes/Pg.21
Vidyalankar : GATE – CE

ASSIGNMENT  1
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each

1. In case of non-dilatent materials, the maximum value of Poisson’s ratio is:

(A) 0.25 (B) 0.33
(C) 0.47 (D) 0.5
2. If S is the elongation produced in a bar of length L, diameter D and crosssectional
area A, by a direct load of W, then Young's modulus may be given by the following:
(A) E = WL/SA (B) S = WE/A
(C) W = SDE/A (D) A = WSL/E
3. The numerical value of Young’s modulus of elasticity in ascending order of glass,
aluminium, copper, wrought iron and tungsten are given in:
(A) Tungsten, wrought iron, copper, aluminium, glass
(B) wrought iron, copper, aluminium, glass, tungsten
(C) copper, aluminium, glass, tungsten, wrought iron
(D) glass, aluminium, copper, wrought iron, tungsten
4. Strain is:
(A) The change in metals
(B) The change in shape and direction in metals
(C) The result of load application on a body
(D) The change of form produced in a piece by the action of load
5. Tensile stress is:
(A) Stress due to any force
(B) Stress due to change in length under a load
(C) Stress caused by varying load
(D) Stress measured by the ratio of the increase or decrease in length of the
unloaded piece under tensile force.
6. A material capable of absorbing large amounts of energy is known as:
(A) Ductile (B) Shock proof
(C) Resilient (D) Hard
Q7 to Q18 carry two marks each

7. The ratio of bulk modulus to shear modulus for Poisson’s ratio 0.25 is:
(A) 3/2 (B) 5/3
(C) 1 (D) 3/5
8. A copper bar having crosssectional area of 1000mm2 is subjected to the forces
shown in figure below:

GATE/CE/SLP/SM/Assign/Pg.52
 Assignment on Solid Mechanics

Which point does not change its position after loading?

(A) B only (B) C only
(C) B and C only (D) D only
9. A cylindrical bar of L meters length deforms by  mm. The strain in the bar will be:
(A) / L (B) 0.1  / L
(C) 0.01  / L (D) 0.001  / L
10. Materials exhibiting time bound behaviour are known as:
(A) isentropic (B) reactive
(C) fissile (D) visco elastic
11. If Young’s modulus of elasticity is determined for mild steel in tension and
compression, the two values will have a ratio (Et/Ec) of:
(A) 1 (B) 1.2
(C) 1.4 (D) 0.8
12. Poisson’s ratio for cast iron is:
(A) 0.27 (B) 0.31
(C) 0.33 (D) 0.36
13. A body having similar properties throughout its volume is said to be:
(A) Homogenous (B) Isotropic
(C) Isentropic (D) Continuous
14. A beam of uniform strength is one in which:
(A) The cross-section remains same throughout
(B) The bending moment is same at every section
(C) The stiffness is same at every section
(D) The bending stress is same at every section
15. It is desired to punch a hole of 20mm diameter in a plate 20mm thick. If the shear
stress of mild steel is 30kg/mm2 the force necessary for punching would be
approximately in the range:
(A) 10-15 tonnes (B) 15-20 tonnes
(C) 20-25 tonnes (D) 35-45 tonnes
16. A solid uniform metal bar of diameter D and length L is hanging vertically from its
upper end. If W is the total weight of the bar and E is the Young’s modulus of
elasticity, the elongation of the bar due to self weight would be:
(A) WL/2AE (B) 2WL/AE
(C) WL/AE (D) 4WL/AE
17. The relation between modulus of elasticity E, modulus of elasticity in shear G and
Poisson’s ratio v is:
(A) E = Gv (B) E = G(v + 1)
(C) E = 2G(v + 1) (D) E = 4G(1 + 2v)
18. The relationship between modulus of elasticity E and Bulk modulus of elasticity K
is:
(A) K = E/(1  2v) (B) K = E/(3  2v)
(C) K = E/(3(1  v) (D) K = E/3(1  2v)



GATE/CE/SLP/SM/Assign/Pg.53
Vidyalankar : GATE – CE

TEST PAPER  1
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each
1. The value of Poisson’s ratio for cork is taken as:
(A) Zero (B) 0.1
(C) 0.2 (D) 0.3
2. The phenomenon under which the strain of a material varies under constant stress is
known as:
(A) Creep (B) Strain hardening
(C) Fatigue failure (D) Hysteresis
3. The approximate value of allowable stress for carbon steel under static
loading(kg/cm2 ) is:
(A) 500-1000 (B) 4000-7500
(C) 10000-15000 (D) 15000-20000
4. Within elastic limits the greatest amount of strain energy per unit volume that a
material can absorb is known as:
(A) Shock proof energy (B) Resilience
(C) Proof resilience (D) Impact energy limit
5. A high strength steel band-saw of 90 mm width and 0.5 mm thickness runs over
a pulley of 500 mm diameter. Assuming E = 200 GPa, the maximum flexural
stress developed would be
(A) 300 MPa (B) 100 MPa
(C) 200 MPa (D) 150 MPa

Q6 to Q13 carry two marks each

6. The volumetric strain of a body having modulus of elasticity E and subjected to a

load P, and having Poisson’s ratio 0.25 is:
(A) P/E (B) P/2E
(C) 2P/E (D) P/4E
7. A bar of uniform crosssection of 400 mm 2 is loaded a shown in the figure. The
stress at section 11 is

(A) 50N / mm2 (B) 100N / mm2

(C) 25N / mm2 (D) 200N / mm2
8. In the simple bending theory, one of the assumptions usually made is that the
plane sections before bending remain plane after bending. This assumption
implies that:
(A) Stress is uniform in the beam cross-section
(B) Strain is uniform in the beam cross-section
(C) Stress is proportional to the distance from the neutral axis
(D) Strain is proportional to the distance from the neutral axis

GATE/CE/SLP/SM/Test Paper/Pg.66
 Test Paper on Solid Mechanics

9. Identify the true statement about true stress-strain method:

(A) There is no such phenomenon like true stress or true strain
(B) True stress is load per unit area (actual) and similarly true strain is
determined under actual conditions
(C) This method can be used for compression tests as well
(D) It is more sensitive to changes in mechanical conditions
10. A simply supported beam of span L which carries over its full span a uniform load
of W, will have zero shear force at:
(A) ends
(B) mid span
(C) both at ends as well as mid span
(D) L/4 from either ends
11. A uniform steel rope 400mts long is hung vertically. The weight of the steel is
7.8 kg/cm2 and E = 2100 kg/mm2 . The ratio of elongation of first 150mts to that of
300mts would be:
(A) 16:1 (B) 4:1
(C) 2:1 (D) 1:4
12. If the value of Poisson’s ratio is zero, then it means that:
(A) The material is rigid
(B) The material is perfectly plastic
(C) There is no longitudinal strain in the material
(D) The longitudinal strain in the material is infinite
13. The number of elastic constants for a completely anisotropic elastic material
which follows Hook’s law is:
(A) 2 (B) 3
(C) 4 (D) 21
Q14(a) & (b) carry two marks each

Linked Answer Question

14(a). For a simply supported loading as
given, the distance x at a which
the shear force is 0 is
kL
(A) L/3
(B) L
L A B
(C)
3 x
(D) L/2
L
14(b). For the above problem, the reaction at support B is :
KL2 KL2
(A) (B)
2 3
KL2 KL2
(C) (D)
6 8



GATE/CE/SLP/SM/Test Paper/Pg.67