Final m2 march 2019

Module 2 : Elastic Properties of Materials
(Elasticity, Bending of Beams &Torsion of Cylinder)
CONTENTS
I. Elasticity
1. Concept of Elasticity, Plasticity, Stress, Strain, Tensile Stress, Shear Stress,
Compressive Stress, Strain Hardening and Strain Softening
2. Hooke’s Law
3. Failure (Fracture/Fatigue)
4. Effect of stress, temperature, annealing and impurity on elasticity
5. Importance of Elasticity
6. Different Elastic Moduli- Young’s Modulus(Y), Bulk Modulus(K), Rigidity
Modulus(n)
7. Poisson’s Ratio(σ)
8. Expression for Young’s Modulus(Y), Bulk Modulus(K), Rigidity Modulus(n)
9. Relation Between the Elastic Constants
 Relation Between Shearing Strain, Elongation Strain and compression Strain
 Relation Between Y, n and σ
 Relation Between K, Y and σ
 Relation Between K, n and Y
 Relation Between K. n and σ
10. Limits of Poison’s
II. Bending of Beams
1. Neutral Surface or Neutral Plane
2. Derivation an Expression for Bending Moment
3. Bending Moment of a Beam With Circular and Rectangular Cross Section
4. Derivation an Expression for Young’s Modulus Using Single Cantilever
III. Torsion of Cylinder
1. Expression For Couple Per Unit Twist of a Solid Cylinder
2. Torsional Pendulum-Expression for Period of Oscillation
3. Numerical Problems
IV. Question Bank

MODULE - 2 ELASTIC PROPERTIES OF MATERIALS
Dr. DIVAKARA S, PROF. & HEAD, DEPT. OF PHYSICS, VVCE, MYSURU Page 2
Books Referred:
1. Engineering Physics By S P Basavaraju
2. Engineering Physics By M N Avadhanulu & P G Kshirsagar
3. Engineering Physics By R K Gaur & S L Gupta
4. Introduction to Mechanics By M K Verma
5. Vibrations & Waves By H J Pain
6. Fundamentals of Mechanics of Materials By Akira Todoroki
7. Theory of Elasticity By Timoshenko
8. Elasticity By J R Barber
9. Elasticity By Robert William Soutas-Little
10. Elasticity in Engineering Mechanics By Arthur P Boresi and Ken Chong
11. Elasticity By Martin H Sadd

I. Elasticity
1. Concept of Elasticity, Plasticity, Stress, Strain, Tensile Stress, Shear Stress,
Compressive Stress, Strain Hardening and Strain Softening
(a) Elasticity : The Property of a body due to which it regains its original shape and size
when the deforming force is removed is called elasticity.
(b) Plasticity : The Property of a body due to which it will not regains its original shape
and size when the deforming force is removed is called plasticity.
(c) Stress : The applied force per unit area developed inside the body is called stress
(d) Strain : The ration of change in dimensions to original dimensions is called strain
(e) Tensile Stress or Longitudinal Strain or Linear Strain
 It is the stretching force acting per unit area of the section of the solid along its
length.
 When the deforming force is applied to one end of a body (whose other end is
fixed) along its length, the applied force produces a linear strain. The applied
force per unit cross sectional area gives the longitudinal or tensile stress.
L
F↓
X a
(f) Tensile Strain or Longitudinal Strain or Linear Strain
 Force is applied at one end along the length keeping the other end fixed, the wire
undergoes a change in length ( )
 The ration of change in length to original length is called Tensile or Longitudinal
strain.

(g) Compressive Stress or Volume Stress
 The uniform pressure (Force/Area) acting normally all over the body is called
Volume Stress or Compressive Stress.
 When the deforming force is applied normally and uniformly to the entire surface
of a body, it produces a volume stress.
(h) Compressive Strain or Volume Strain
 If an uniform force is applied all over the surface of a body, the body undergoes a
change in its volume.
 If is change in volume in an original volume V of the body then,
F
F F
F F
F F F
(f) Shear Stress or Tangential Stress
 It is a force acting tangentially per unit area on the surface of a body.
 If a force is applied tangentially to a free portion of the body with another part
being fixed. Its layer slides one over the other. The body experiences a turning
effect and changes its shape. This is called shearing and the angle through which
the turning take place is shearing angle.
(g) Shear Strain
 The shearing angle itself is a measure of the ration of change in dimensions to
original dimensions
X
L

2. Hooke’s Law and Stress-Strain Diagram
 Hooke’s Law states that for small deformations, the stress is always proportional
to strain.
Where ‘K’ is the proportionality constant and is the modulus of elasticity
 The figure shows the stress versus strain graph for metallic wire.
 OA is the portion of linearity where stress is directly proportional to strain
(Hooke’s Law is valid only in this region)
 The body continues to exhibit perfect elastic property till B. However, the linear
dependence between stress and strain is not observed in this portion. The point A
and B is called proportional limit and elastic limit respectively.
 At any point between B and D, the body fails to regain its original length if the
applied force is removed but the transverse the dashed line in the figure and the
material is said to be Permanent Set. Such a deformation is called Plastic
Deformation.
 At D, the deformation becomes large enough to pull the molecules far apart so
that the binding force breaks and beyond D the body is no more a single piece. D
is called Fracture Point.
 The force per unit area for which the wire breaks is called Breaking Point. A
highly ductile (Sponge or elastic) metal undergoes large deformation between the
elastic limit and fracture point. Whereas, for a metal having brittle property, the
fracture occurs soon after the elastic limits.
 Between B and D is the plastic range. This is subdivided into Strain Hardening
Region (BX) and Strain Softening Region (XD). X is the peak point of the curve.
X

(a) From B to X, strain hardening takes place in the material with increase in
stress. Strain Harding is defined as the process of making a metal harder by
plastic deformation.
(b) From X to D, strain softening take place as the materials weakens even under
reduced stress. So X is the limit upto which the material stays fit and is
referred to as ultimate strength. The curve will have negative slope soon after
the elastic region indicates that there is softening effect of the material over
this range. This effect is called Strain Softening.
 If a body under stress within its elastic limit, returns to its original form without
any trace of deformation soon after the removal of the deforming force. It is
called a Perfectly Elastic Body.
 If a body deformed under stress, body will not retains its deformed shape and size
even when the stress is decreased or removed is called Plastic Body or Plasticity
(After B).
3. Effect of stress, temperature, annealing and impurity on elasticity
 When certain elastic materials are subjected to continuous stress at elevated
temperature,
(a) Under Constant stress, they begin to undergo creep(Creep is the property due
to which a material under steady stress undergoes deformation continuously)
instead.
(b) At higher temperature which is significantly higher than room temperature,
metals no longer exhibits strain hardening.
 Annealing is a process to make a metal or a glass soft by heating and cooling it
slowly. It improves the elasticity and increases the ductility. It is used to alter the
physical and mechanical properties of metals without changing its shape.
 Addition of Impurities in metals results in increase or decrease of elasticity.
4. Importance of Elasticity
 Pure metals are rarely used in applications because of soft, ductile and low
tensile strength.
 Alloys are generally harder than pure metals. They are produced by bending
different metals. Hence they offer better elastic properties.
 Iron is less elastic than steel. When a tool made of iron is used in the
applications where there is lot of vibrations, a small fracture or cracks are
formed in the tool. In case the tool is made in steel, it springs back to shape
repeatedly as steel is more elastic.

(c) Failure (Fracture/Fatigue)
 A machine is meant for repeated use. Parts of the machines get stressed while
they are loaded. As their use continues, they stressed repeatedly and cracks begin
to form. Formation of cracks results in destruction of continuity. This constitutes
Fracture.
 Fatigue cracks form wherever there are stress concentrations.
 There are two types of fractures namely, brittle fracture and ductile fracture
(a) A brittle fracture occurs due to swift propagation of a crack formulated
suddenly. A completely brittle material suffers fracture almost at the elastic
limit.
(b) A ductile fracture propagates slowly with considerable plastic deformation on
its way. A ductile material redistributes the local stresses accompanied by
plastic deformation so that crack propagates slowly. This slow process is
sustained till ultimate strength is reached beyond which the small microcracks
are generated. Many microcracks join to form central crack leading fracture
called ductile fracture.
5. Different Elastic Moduli
(a) Young’s Modulus (Y)
The ratio of longitudinal stress to longitudinal or linear strain within the elastic
limits is called Young’s Modulus.
(b) Bulk Modulus (K)
The ratio of the compressive stress or volume stress to volume strain without
change in shape of the body within the elastic limit is called the Bulk Modulus.
(c) Rigidity Modulus (n)
The ratio of shearing stress to the shearing strain is called Rigidity Modulus.

6. Poisson’s Ratio (σ)
 Linear or Longitudinal strain is the ratio of change in length to original length
(x/L). The longitudinal strain produced per unit stress (T) is called Longitudinal
Stress Coefficient.
 Lateral strain (d/D) is the ratio of change in diameter (d) to the original diameter
(D) (by considering the deforming force acts on circular cross section). The lateral
strain produced per unit stress is called Lateral Strain Coefficient.
 The ratio of lateral strain to the longitudinal strain is a constant within the elastic
limits of the body is called Poisson’s Ratio.
7. Expression for Young’s Modulus(Y), Bulk Modulus(K), Rigidity Modulus(n) in
terms of α and β.
(a)
(b)
( )
(c)
( )

8. Relation Between the Elastic Constants
(a) Relation between shearing strain, elongation strain and compression strain
Consider a face of a cube (APSD) whose lower surface (DS) is fixed to a rigid
support. When a deforming force is applied to its upper face along AP, it causes
relative displacements at different parts of the cube (A moves to Aʹ, P to Pʹ, AS to
AʹS, PD to PʹD). Let θ be the angle of shear and diagonals are perpendicular to
each other.
If PX is drawn perpendicular to PʹD and AʹY to AS, then DP=DX and AʹS=YS.
i.e. PʹX is the extension in an original length PD and AY is the contraction in an
original length AS.
√ √
√
√
√
√
√
√
√
(If L is the length of the each side of the cube, then √ )

(b) Relation between Y, n and σ
Consider a face of a cube (APSD) whose lower surface (DS) is fixed to a rigid
support. When a deforming force is applied to its upper face along AP, it causes
relative displacements at different parts of the cube (A moves to Aʹ, P to Pʹ, AS to
AʹS, PD to PʹD). Let θ be the angle of shear and diagonals are perpendicular to
each other.
If PX is drawn perpendicular to PʹD and AʹY to AS, then DP=DX and AʹS=YS.
i.e. PʹX is the extension in an original length PD and AY is the contraction in an
original length AS
Now the shearing strain occurring along AP can be treated as equivalent to a
longitudinal strain along the diagonal DPʹ and an equal lateral strain along the
diagonal AʹS.
√
√
√
√
√ ( )
√
(√ ) ( )
√
( )
( ) ( )
( )
( ) ( )
( )
( )

(c) Relation between K, Y and σ
 Produces an increase in length of in X-direction and the other two
stresses are perpendicular to X-direction, they produce contraction
and .
 Hence a length which was unity along X direction,
Length along X-axis =
Length along Y-axis =
Length along Z-axis =
( )( )( )
Since α and β are very small and their product can be neglected.
( ) ( )
( )( )
, [ ( )( )]
Since the cube is under consideration of unit volume,
Increase in volume [ ( )( )] ( )
If instead of stress T, a pressure P is applied,
Increase in volume ( )
( ) ( )
( ) ( )
( )
( )
 Let 𝑇𝑥 𝑇𝑦 𝑎𝑛𝑑 𝑇𝑧 be the length, breadth and
height of a cube along X, Y and Z-axis
respectively.
 Let α be the longitudinal strain coefficient
along the direction of the forces
(extension of length).
 β be the lateral strain coefficient in a
direction perpendicular to the forces
(Contraction of lengths).

(d) Relation between K, n and Y
The relation between K, Y and σ is,
( )
( )
The relation between Y, n and σ is, ( )
( )
( )
(e) Relation between K, n and σ
The relation between Y, n and σ is, ( )
The relation between K, Y and σ is, ( )
( )
(f) Limiting Values of Poison’s Poisson’s Ratio (σ)
We know the relation, ( ) ( )
( ) ( )
 If σ is positive then left side of the equation is positive. If left side of the
equation is positive then right side should also be positive when σ is not more
than ½. Therefore, σ takes the value less than 0.5.
 If σ is given negative value then its right side will become positive. If right
side of the equation is positive then left side should be positive when σ is not
more than -1.
 Thus the values of σ lie between -1 to 0.5. However, negative value of σ
means an elongation of the body accompanied by lateral expansion which is
not observed in practice.
 The limiting value for σ is taken between 0 and 0.5.

II. Bending of Beams
 A homogeneous body of uniform cross section whose length is large compared
to its other dimensions is called beam.
 Different types of beams,
 Applications : Beams are used in the,
 Fabrication of trolley ways
 Chassis as truck beds
 Elevators
 Construction of platform and bridges
 Girders in buildings and bridges
1. Neutral Surface or Neutral Plane
 Neutral surface is a layer of uniform beam which does not undergo any change
in its dimensions when the beam is subjected to bending within its elastic limit.
 If a load is attached to the free end of the beam, the beam bends. A filament like
upper layer of the beam (AB) will be elongated to and the lower layer (EF)
of the beam will be contracted to . But there will be always the particular
layer whose length will not change (CD). Such a layer is called Neutral Surface
and the axis or a line is called Neutral Axis.
 Neutral Axis is a longitudinal line along which the neutral surface is intercepted
by any longitudinal plane considered in the plane of bending.

2. Derivation an Expression for Bending Moment
 The moment of applied couple (two parallel forces that are equal in magnitude
and opposite in direction) subjected to which the beam undergoes bending
longitudinally is called the Bending Moment.
 Consider a long uniform beam MN whose one end is fixed at M. The beam can
be made up of a number of parallel layers like AB, CD, EF, etc..
 If now a load is attached to the beam at N, the beam bends. The successive layer
now is strained. A layer AB above the neutral surface (CD) will be elongated to
and the one like EF below the neutral surface will be contracted to .
 Let R be the radius of the circle to which the neutral surface forms a part
CD=Rθ, where θ is the common angle.
[( ) ]
Where F be the force acting on the beam and a is the area of the layer AB.
Moment of force about the neutral axis=F X its distance from the neutral axis
∑ ∑
∑

3. Bending Moment of a Beam With Circular and Rectangular Cross Section
(i) Bending moment for a beam of rectangular cross-section
( )
Where b and d are the breadth and thickness of the beam respectively.
(ii) Bending Moment for a beam of circular cross-section
( )
is the radius of the beam
4. Derive an Expression for Young’s Modulus Using Single Cantilever
Consider a uniform beam of length L fixed at M. Let a load W act on the beam at N.
as a result beam bends. Consider a point P on the free beam at a distance x from the
fixed end ( ). Let be its position after the beam is bent.
Bending Moment = Force X Perpendicular Distance
( )
( )
If y is the depression of the point P then, (Approximately)
Where R is the radius of the circle to which the bent beam becomes a part.
( )
( )
( )
( ) ( )

Integrating both sides, ∫ ( ) ∫ ( )
( ) * + ( )
Where is the constant of integration. But ( ) is the slope of the tangent drawn to
the bent beam at a distance ‘x’ from the fixed end. ( )
Equ. (1) becomes,
And ( ) * +
* +
Integrating both the sides, ∫ ∫ * +
( )
Where is the constant of integration. When x=0, there is no depression at M.
Hence y=0 at x=0 then
( )
At the loaded end (N),
( )
Where b and d are the breadth and thickness of the rectangular beam respectively
𝒀
𝟒𝒎𝒈𝑳 𝟑
𝒚 𝟎 𝒃𝒅 𝟑

III. Torsion of Cylinder
1. Expression for Couple Per Unit Twist (Torsion) of a Solid Cylinder
 Consider a long cylindrical rod of length ‘L’ and radius ‘R’ rigidly fixed at its
upper end. Let be its axis.
 The cylindrical rod made of thin concentric, hollow cylindrical layers each of the
thickness ’dr’. ‘r’ be the radius of an one concentric layer.
 If the rod is now twisted at its lower end then the concentric layer slide one over
the other. The movement will be zero at the fixed end ‘X’, and it gradually
increases along the downward direction and a point B at its bottom shifts to .
 is the angle of shear. Since ϕ is small, we have . Also
, then arc length .
( )
The above equation for one layer of the cylinder.
𝐵𝑋𝐵 𝜙
𝐵𝑂 𝐵 𝜃
𝐵𝑂 𝐵 𝑂 𝑟

∫
* +
( )
2. Torsional Pendulum-Expression for Period of Oscillation
 A set up in which a rigid body is suspended by a wire
clamped to a support and the body executes to and
from turning motions with the wire as its axis is
called Torsional Pendulum and the oscillations are
called Torsional Oscillations.
 The time period of oscillations for torsional pendulum,
√
Where ‘I’ is the moment of inertia of the rigid body about the axis through the
wire and ‘C’ is the couple per unit twist for the wire.
 Applications
 It is possible to evaluate moment of inertia of regular bodies but to find the
moment of inertia of irregular bodies, we have to use torsional pendulum
method only.
 Rigidity modulus of the wire can be found.
𝑪𝒐𝒖𝒑𝒍𝒆 𝒑𝒆𝒓 𝒖𝒏𝒊𝒕 𝒕𝒘𝒊𝒔𝒕 𝑪 (
𝝅𝒏𝑹 𝟒
𝟐𝑳
)

3. Numerical Problems
Formulas at a glance


 ( ) ( )
 ( ) ( ) * +
 ( ) ( )

 ( )

 √
01 Calculate the force required to produce an extension of 1mm in steel wire of length 2m and
diameter 1mm (Given: Young’s Modulus of the steel, Y=2 X 1011
N/m2
).
02 Calculate the extension produced in a wire of length 2m and radius 0.013cm due to a force of
14.7N is applied along its length (Given: Young’s Modulus of the steel, Y=2 X 1011
N/m2
). Jan 19
03 Calculate the torque required (twisting couple) to twist a wire of length 1.5m, radius 0.0425cm,
through an angle (π/45) radian. If the value of Rigidity modulus of its material is 8.3 X 1010
N/m2
.
Jan 2019
Sol :
F=?
X=1mm
D=1mm R=0.5mm
L=2m, Y=2 X 1011
N/m2
Y=2 X 1011
N/m2
𝑌
𝐹𝐿
𝑎𝑥
𝐹𝐿
( 𝜋𝑅 ) 𝑥
𝐹
𝑌𝜋𝑥𝑅
𝐿
𝐹
𝑋 11 𝑋 𝜋 𝑋 ( 𝑋 − ) 𝑋 ( 5 𝑋 − )
𝟕𝟖 𝟓𝟒𝑵
𝜏
𝜋𝑛𝑅
𝐿
𝑋 𝜃
𝜏
𝜋 𝑋 8 𝑋 𝑋 ( 𝑋 )
𝐿
𝑋
𝜋
𝟏 𝟗𝟖 𝑿 𝟏𝟎 𝟒
𝑵𝒎
Sol :
F=14.7N
X=?
R=0.013cm
L=2m
Y=2 X 1011
N/m2
𝑌
𝐹𝐿
𝑎𝑥
𝐹𝐿
( 𝜋𝑅 ) 𝑥
𝑥
𝐹𝐿
𝑌( 𝜋𝑅 )
𝑥
7 𝑋
𝑋 11 𝑋 𝜋 𝑋 ( 𝑋 − )
𝟐 𝟕𝟕 𝑿 𝟏𝟎 𝟑
𝑚
Sol :
L=1.5m
R=0.0425cm
θ=π/45
n=8.3 X 1010
N/m2
τ=?

04 Calculate the angular twist of a wire of length 0.3m and radius 0.2mm when a torque of 5 X 10-4
Nm is applied. Rigidity modulus of its material is 8.3 X 1010
N/m2
.
05 An increment in length by 1mm was observed in a gold wire of diameter 0.3mm, when it was
subjected to a longitudinal force of 2N and a twist of 0.1rad was observed in the same wire when
its one end is subjected to a torque of 7.9 X 10-7
Nm, while its other end was fixed. Calculate the
value of Poisson’s ratio for gold.
06 A rod of cross section of area 1cm X 1cm is rigidly planted into the earth vertically. A string
which can withstand a maximum tension of 2kg is tied to the upper end of the rod and pulled
horizontally. If the length of the rod from the ground level is 2m. Calculate the distance through
which its upper end is displaced just before the string snaps (Young’s Modulus of steel, Y=2 X
1011
N/m2
).
07 The free end of a single cantilever depressed 10mm under a certain load. Calculate the
depression under the same load for another cantilever of the same material but 2 times in length,
2 times in width and 3 times in thickness. Assume both the cantilevers to be having rectangular
cross section.
Sol :
L=0.3m
R=0.2mm
θ=?
n=8.3 X 1010
N/m2
τ= 5 X 10-4
Nm
𝐶
𝜏
𝜃
𝜃
𝜏
𝐶
𝝉
(
𝜋𝑛𝑅
𝐿
)
𝟓 𝑿 𝟏𝟎 𝟒
𝝅 𝑿 8 𝑋 𝑿( 𝑋 )
𝟐 𝑿 𝟎 𝟑
𝟎 𝟕𝟓𝒓𝒂𝒅
Sol :
x=1mm
d=0.3mm R=0.15mm
F=2N
θ=0.1rad
τ= 7.9 X 10-7
Nm
σ=?
𝑌
𝐹𝐿
𝑎𝑥
𝐹𝐿
( 𝜋𝑅 ) 𝑥
𝑋 𝐿
𝜋 𝑋 ( 𝑋 ) 𝑋 𝑋
8 𝑋 𝐿 𝑁 𝑚
𝐶
𝜋𝑛𝑅
𝐿
𝑛
𝐿𝐶
𝜋𝑅
𝐿(𝜏 𝜃)
𝜋𝑅
𝑋 𝐿 𝑋 7 9 𝑋 7
𝜋 𝑋 ( 𝑋 ) 𝑋
9 9 𝑋 9
𝐿 𝑁 𝑚
𝑌
𝑛
( 𝜎) 𝜎
𝑌
𝑛
8 𝑋 𝐿
9 9 𝑋 𝐿
𝟎 𝟒𝟐𝟒
Sol :
b=1cm and
d=1cm
m=2kg
L=2m
Y= 2 X 1011
N/m2
y0=?
𝑦
𝑚𝑔𝐿
𝑌𝑏𝑑
𝑋 𝑋 9 8 𝑋 ( )
𝑋 𝑋 𝑋( )
𝟎 𝟑𝟏𝟒𝒎
Sol : First Cantilever:𝐿𝑒𝑛𝑔𝑡 𝐿 𝐵𝑟𝑒𝑎𝑑𝑡 𝑏 𝑇 𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑑 𝑌𝑜𝑢𝑛𝑔 𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠
𝑌 𝐿𝑜𝑎𝑑 𝑊 𝑎𝑛𝑑 ( 𝑦 ) 𝑚𝑚
Second Cantilever:
𝐿𝑒𝑛𝑔𝑡 ( 𝐿 ) 𝐿 𝐵𝑟𝑒𝑎𝑑𝑡 ( 𝑏 ) 𝑏 𝑇 𝑖𝑐𝑘𝑛𝑒𝑠𝑠( 𝑑 ) 𝑑 𝑌𝑜𝑢𝑛𝑔 𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠( 𝑌 )
𝑌 𝐿𝑜𝑎𝑑 ( 𝑊 ) 𝑊 𝑎𝑛𝑑 ( 𝑦 ) ?

08 A solid lead sphere of radius 10.3 m is subjected to a normal pressure of 10N/m2
acting all over
the surface. Determine the change in its volume. (Given: Bulk Modulus for lead, K=4.58 X
1010
N/m2
).
09 A water column of length 1m is held in a cylinder, between its base and a tightly fitted piston.
Calculate the distance through which the piston would move if the water column is subjected to a
pressure of 205 X 1014
N. (Given: Bulk Modulus for water, K=2.05 X 109
N/m2
).
10 A brass bar of length 1m, 0.01m2
in section is clamped firmly in a horizontal position at one end.
A weight of 1kg is applied at other end. What depression would be produced (Given: Young’s
Modulus for brass, Y=9.78 X 1010
N/m2
).
11 A wire of length 2m and radius 2mm is fixed to the center of a wheel. A torque of magnitude
0.0395Nm is applied to twist the wire. Find the rigidity modulus of the wire if the angular twist is
0.038rad,
Sol :
R=10.3m
P=10N/m2
K=4.58 X 1010
N/m2
V'=?
𝐾
𝑃𝑉
𝑉
𝑉
𝑃𝑉
𝐾
𝑃
𝐾
( 𝜋𝑅 )
𝑋 𝑋 𝜋 𝑋 ( )
8 𝑋 𝑋
𝟗 𝟗𝟗 𝑿 𝟏𝟎 𝟕
𝒎 𝟑
𝐿 𝑚
Sol :
P=205 X 1014
N
K=2.05 X 109
N/m2
Δx=?
𝜋𝑟 ∆𝑥
𝐾
𝑃𝑉
𝑉
𝑃( 𝜋𝑟 𝐿)
( 𝜋𝑟 𝑥)
𝑃𝐿
𝑥
𝑉
𝑃𝐿
𝐾
𝑋 𝑋
𝑋
𝟏𝟎 𝟕
𝒎
( 𝑦 )
𝑊 𝐿
𝑌 𝑏 𝑑
𝑎𝑛𝑑 ( 𝑦 )
𝑊 𝐿
𝑌 𝑏 𝑑
𝑊 ( 𝐿)
𝑌 ( 𝑏) ( 𝑑)
( 𝑦 )
( 𝑦 )
𝑊 ( 𝐿)
𝑌 ( 𝑏) ( 𝑑)
𝑊 𝐿
𝑌 𝑏 𝑑
( )
𝑋 ( ) 8
( 𝑦 )
8
𝑋( 𝑦 )
8
𝑋 𝑋 𝟏 𝟒𝟖 𝑿 𝟏𝟎 𝟑
𝒎
Sol :
L=1m
A=0.01m2
b=0.1m & d=0.1m
m=1kg
Y= 9.78 X 1010
N/m2
y0=?
𝑦
𝑚𝑔𝐿
𝑌𝑏𝑑
𝑋 𝑋 9 8 𝑋 ( )
9 78 𝑋 𝑋 𝑋( )
𝟒 𝟎𝟏 𝑿 𝟏𝟎 𝟔
𝒎
Sol :
L=2m, R=2mm
θ=0.038rad
n=?
τ= 0.0395Nm
𝜏
𝜋𝑛𝑅
𝐿
𝑋 𝜃 𝑛
𝐿𝜏
𝜋 𝜃 𝑅
𝑋 𝑋 9
𝜋 𝑋 8 𝑋 ( 𝑋 )
𝟖 𝟐𝟕 𝑿 𝟏𝟎 𝟏𝟎
𝑵 𝒎 𝟐

IV. Question Bank
Module 2: Elastic Properties of Materials
Q. No. Question Bank
01 Explain the terms Elasticity and Plasticity.
02 Explain the different types of stresses and strains.
03
Define Young’s Modulus, Bulk Modulus and Rigidity Modulus and mention the
expression for the same.
04 Mention the expression for Y, K and n in terms of α and β.
05 State and explain Hooke’s law through the stress-strain curve.
06 Explain the difference between strain hardening and strain softening.
07 Explain the effect of stress, temperature, annealing and impurity on elasticity.
08 Explain the importance of elasticity in engineering applications.
09 Explain the fracture in elasticity.
10 Derive the relation between shearing strain, elongation strain and compression strain.
11 Derive the relation between Y, n and σ.
12 Derive the relation between K. Y and σ.
13 Derive the relation between K, n and Y.
14 Derive the relation between K, n and σ.
15 State Poission’s Ratio and explain the limiting value of it.
16 Describe Neutral surface and neutral plane with the help of suitable diagram.
17 Derive an expression for bending moment or moment of force.
18 Derive an expression for the Young’s modulus of the material of a single cantilever.
19 Derive an expression for couple per unit twist (Torsion) of a solid cylinder.
20 Write a note on Torsional Pendulum and mention its applications.
21 Problems on elastic properties of materials.

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