L-1
Elasticity and Plasticity
When the shape or size of a body has been altered by the application of a force or a
system of forces, there is usually some tendency for the body to recover its original shape or size
on the removal of the force. This property of the body by virtue of which it tends to regain its
original shape or size on the removal of deforming force is called elasticity.
The property of the body by virtue of which it tends to retain the altered size and shape
on removal of deforming forces is called plasticity.
1.2
Stress and Strain
Stress is a quantity that characterizes the strength of the forces causing the deformation,
on a “force per unit area” basis. The deforming force per unite area of the body is called stress.
The SI unit of stress is the Pascal (abbreviated Pa, and named for the 17th century French scientist
and philosopher Blaise Pascal). One Pascal equals one Newton per square meter.1 Pascal = 1Pa =
1N/m2. Strain is a quantity which describes the resulting deformation. Strain is the fractional
deformation produced in a body when it is subjected to a set of deforming forces. Strain being
ratio has no units.
There are following three types of stress and strain
(i)
Tensile and compressive stress and strain
(ii) Bulk stress and strain
(iii) Shear stress and strain
1.2
1.3
Physics for Technologists
Tensile and Compressive Stress and Strain
Initial
state
lo

F
F

An object in tension
Tensile stress at the cross section is defined as the ratio of the force F to the
cross
– sectional area A.
Tensile stress =
F
A
The tensile strain of the object is equal to the fractional change in length, which is the
ratio of the elongation  to the original length  o .
Tensile strain =
  o 

o
o
Tensile strain is stretch per unit length
Properties of Matter and Sound
1.3
Initial
state
lo

F
F
l
An object in compression
When forces on the ends of a bar are pushes rather than pulls the bar is in compression
and the stress is a compressive stress. The compressive strain of an object is defined in the same
way as tensile strain, but   has the apposite direction.
1.4
Bulk Stress and Strain
Initial
state
F
F
F
F
F
F
An object under bulk stress
Pressure plays the role of stress in a volume deformation.
The force F per unit area is called the pressure
P
F
A
1.4
Physics for Technologists
Bulk stress =  P , an increase in pressure. The fractional change in volume that is the
ratio of the volume change  V to the original volume Vo, is called as bulk strain.
Bulk (volume) strain =
V
V0
Volume strain is the change in volume per unit volume.
Shear Stress and Strain
h
Initial
state
x
F11
F11
An object under shear stress
Shear strain is defined as the ratio of the displacement x to the transverse dimension h
Shear strain =
1.5
x
h
Hooke’s Law
Hooke’s law states that within the elastic limit, the stress developed is directly
proportional to the strain. The constant of proportionality is the elastic modulus ((or modulus of
elasticity).
Stress
= elastic modulus (Hooke’s law)
Strain
Properties of Matter and Sound
1.5
Plastic range
Stress
Elastic limit
Elastic range
0
Permanent set
Strain
Stress - Strain diagram
If we plot a graph between stress and strain we get a curve as shown in Fig. 1.5 and it is
called stress - strain diagram. It is clear from this graph that Hooke’s law holds good only for
the straight line portion of the curve
1.6
Elastic Moduli
1.6.1
Three kinds of elastic molduli
The coefficient of elasticity or modulus of elasticity indicates how a specimen behaves
when subjected to given stress. This has the same units as stress that is Nm-2 or Pa.
Table 1.1 Three kinds of elastic moduli
Elastic Modulus
Definition
Nature of strain
Tensile stress
Young’s modulus (Y)
Tensile strain
Change of shape and size
Bulk stress
Bulk modulus (B)
Change of size but not shape
Bulk strain
Shear modulus or
Rigidity modulus (S)
Shear stress
Change of shape but not size
Shear strain
1.6
Physics for Technologists
Table 1. 2 Approximate elastic moduli
Young’s Modulus
Bulk Modulus
Shear Modulus
Y(Pa)
B(Pa)
S(Pa)
Aluminium
7.0x1010
7.5x1010
2.5x1010
Brass
9.0 x1010
6.0 x1010
3.5 x1010
Copper
11 x1010
14 x1010
4.4 x1010
Crown Glass
6.0 x1010
5.0 x1010
2.5 x1010
Iron
21 x1010
16 x1010
7.7 x1010
Lead
1.6 x1010
4.1 x1010
0.6 x1010
Nickel
21 x1010
17 x1010
7.8 x1010
Steel
20 x1010
16 x1010
7.5 x1010
Material
1.6.2
Compressibility
The reciprocal of bulk modulus is called compressibility. The unit of compressibility is
same as that of reciprocal pressure, Pa-1
1.6.3
Poisson’s ratio
Within the elastic limits the ratio of the lateral strain to the longitudinal strain is constant
for the material of the body and is known as Poisson’s ratio and is denoted by  .
Worked Example 1.1: A steel rod 2.0m long has a cross sectional area of 0.30cm 2. The rod is
now hung by one end from a support structure and a 550kg milling
machine is hung from the rod’s lower end. The Young’s modulus of
steel is 20x1010Pa. Determine the stress, the strain and the elongation
of the rod.
Stress 
F (550kg)  (9.8m / s 2 )

 1.8  10 8 Pa
5
2
A
3.0  10 m
Strain 
 Stress 1.8  10 8 Pa


 9.0  10 4
o
Y
20  1010 Pa
Elongation =
 = (strain) x  o = (9.0x10-4) (2.0m)
= 0.0018m = 1.8mm
Properties of Matter and Sound
1.7
Worked Example 1.2: The bulk modulus of water is 2.1Gpa. Calculate the volume
contraction of 100mL of water when subjected to a pressure of
1.5Mpa.
Bulk modulus B 
V 
P
(V / Vo
Vo(P) (100mL)  (1.5  10 6 Pa)

B
2.1  10 9 Pa
=
0.071mL
Worked Example 1.3: A box – shaped piece of gelatin dessert has a top area of 15cm 2 and a
height of 3cm. When a shearing force of 0.50N is applied to the upper
surface, the upper surface displaces 4mm relative to the bottom
surface. What are the shearing stress, the shearing strain, and the
shear modulus for the gelatin?
Shear stress 
tan gential force
0.50 N

 333Pa
area of face
15  10  4 m 2
Shear strain 
displacement 0.4cm

 0.1333
height
3cm
ShearModulus 
stress
strain
1.7

333Pa
 2.5kPa
0.133
Twisting Couple on a Cylinder (or Wire)
The twisting of a structural member about its longitudinal axis by two equal and opposite
torques is expressed through a certain angle. The stress seen in this situation is not tensile or
compressive, it is said to be shearing or shear stress. The strain in this case is measured by an
angle in unit of radians
Let us consider a cylindrical rod of length l and radius r with its upper end fixed. Let a
twisting couple-be applied to the lower end of the rod in a plane perpendicular to its length and
let the rod twist through an angle θ (radians). While the rod is twisted restoring couple acts in
the opposite direction and in the position of equilibrium, the twisting couple is equal and
opposite to the restoring couple. To calculate this couple, let us consider the solid cylinder to be
made up of a larger number of concentric thin walled cylinders. Let us consider one such hollow
cylinder of radius x, and radial thickness dx. When the rod is twisted through an angle θ, the
angle through which the rim of the cylinder is sheared is ф.
1.8
Physics for Technologists
Fig. 1. 6 Twisting couple on a cylinder
i.e. BB’=l ф
Also BB’=x θ
:. ф =

x
l
(1)
From this it is clear that with x, ф varies. ф has the maximum value when x is the
greatest. i.e., the strain is maximum on the outermost part of the cylinder and minimum on the
innermost. In other words, the shearing stress is not uniform through out the material.
If N is the rigidity modulus,
N
shearing stress F

angle of shear

Hence
F =N ф=
Nx 

(2)
Now the face area of the hollow cylinder
= 2πxdx
:. Total shearing force on this area
= 2πxdx. Nx θ / l
= 2πN . θ. x2.dx
l
Therefore, moment of this force about the axis 00’
= 2πN θ x2dx . x
l
(3)
Properties of Matter and Sound
1.9
= 2πN θ x3dx
l
(4)
Total twisting couple of the cylinder can be obtained by integrating this expression
between limits x = 0 and x = r.
r

:. Total twisting couple on the cylinder =
2N

l
0
2N

l

r

0
x 3 dx
(5)
r
2N  x 4 
x dx 
 
  4 0
3
(6)
N r 4
2N  r 4 

 
l 4
2l
(7)
If θ = l radian, we have
Twisting couple per unit twist C= πNr4/2 
(8)
This twisting couple per unit twist of the wire is called the torsional rigidity or modulus
of torsion of the cylinder of wire. It is evident form this relation that the couple required is
proportional to the fourth power of the radius.
Note: Hollow Cylinder
For a hollow cylinder of the same length l and of inner radius r1 and outer radius r2,
Twisting couple of the cylinder C 
r2

r1
 2N  3

 x dx
 l 
2N

l
(9)
r2

x 3 dx
(10)
r1
r2
2N  x 4 

 
l  4 r
1

N
2l
(11)
(r24  r1 )
If θ = 1 radian, twisting couple per unit twist C=
4
N 4 4
(r2 – r1 )
2l
(12)
(13)
1.10 Physics for Technologists
Worked Example 1.4: A wire of length 1 meter and diameter 1 mm is fixed at one end and a
couple is applied at the other end so that the wire twists by π/2 radians.
Calculate the moment of the couple required if rigidity modulus of the
material = 2.8 ×1010 N/m2.
Rigidity modulus of the material N = 2.8x1010 N/m2
Angle twisted by wire θ =

radians
2
Length of the wire l = 1 meter
Diameter of the wire d = 1 mm = 1 x 10 -3 meter
Radius of the wire r 
Required couple C 

d 1  10 3

 0.5  10 3 meters
2
2
 N r4
2l
3.14  (2.8  1010 )  (0.5  3.14)  (0.5  10 3 ) 4
2 1
= 4.3 x10-3 N-m.
Worked Example 1.5: A wire of length l m and diameter 1mm is clamped at one of its ends.
Calculate the couple required to twist the other end by 90o. Given
N = 2.8 × 1010 N/m2.
The torque required to twist the free end of a clamped wire of length 
through θ radian will be

N r 4
2l

For θ = 90o = π/2 radian , C=
 2 Nr 4
4
N= 2.8x10-10 N/m2, l=1m
r = 5mm = 0.0005 m
:.C 
2  2.8 1010  (5 104 )4
  4.32×10-3 Nm
4
Properties of Matter and Sound
Worked Example 1.6:
1.11
An iron wire of length l m and radius 0.5 mm elongates by 0.32 mm
when stretched by a force of 49 N, and twists through 0.4 radian when
equal and opposite torques of 3x10-3 N-m are applied at its ends.
Calculate elastic constant for iron.
Y
Stress F / r 2
Fl

Strain
x/l  r2x
Force F = 49 N, radius r = 5x10-4 m and elongation x = 32x10-5 m and l =
1m
Y 
49 1
  (5  10 )  (32  10 5 )
4 2
= 19.5x1010N/m2.
Worked Example 1.7: Two solid cylinders of the same material having length  and 2  , and
radii r and 2r joined coaxially. Under a couple applied between the
free ends, the shorter cylinder shows a twist of 30o. Calculate the twist
of the longer cylinder.
If  is the couple, and it produces twist θ in the shorter cylinder and
twist θ’ in their longer cylinder. Then
 
 Nr 4
2l
 
 N (2r ) 4
2(2l )
'
:. θ’ = θ/8 = 30o / 8 = 3.75o.
Worked Example 1.8: One end of wire of 4 mm radius and 100 cm in length is twisted
through 60o. Calculate the angle of shear on its surface.
If θ is the twist at the free end and ф is the angle of shear, then ф = r θ /l
As  = 1m, r=4x10-3m and θ = 60o
:. ф = 4x10-3 x60 = 0.24°.
Worked Example 1.9: The restoring couple per unit twist in a solid cylinder of radius 5.0 cm
is 10-1 N-m. Find the restoring couple per unit twist in a hollow
cylinder of the same material, mass and length but the internal radius
is 12cm.
The restoring couple per unit twist in a solid cylinder of length of l is
given by
C = πN r4 / 2l
1.12 Physics for Technologists
Here r=0.05m and C=10-1N-m.

N
2l

C
10 1

r4
(0.05) 4
If r1 and r2 are the internal and external radii of the hollow cylinder of
the same length and mass and of same material, then
M = π ( r22-r21)   = π r2   or r22-r21 = r2
Here r1 = 0.12mm and r = 0.05m
r22 = r2 + r12 = (0.05)2 + (0.12)2 = (0.13)2
:. r2 = 0.13m.
The restoring couple per unit twist for this hollow cylinder is
C
 N (r 4 2  r 41 )
2l

10 1
(0.13) 4  (0312) 4 
(0.5) 4
= 1.25x10-4 N.m.
Exercise Problem1.1
A wire of length 1 metre and diameter 1mm is clamped at one of its
ends. Calculate the couple required to twist the other end by 90o. Given
the modulus of rigidity is 298GPa.
Hint :
  90 o 
Twisting couple =

2
radians
Nr 4
 4.313×10-3 Nm
2l
Exercise Problem 1.2. Given a 2m length and 1mm diameter of a steel wire suspender
vertically down keeping the top end fixed. What is the tangential force
required to twist the wire at the force end by 1 radian?. Rigidity
modulus of the wire = 80 GPa.
Hint : Couple required =
Nr 4
2
 tangential force × diameter
Hence, tangential force = couple required  diameter = 3.928 N