Elasticity
Date: Aug 17 2026
Elasticity
The property by virtue of which if a material is deformed under the application of external force, after removing the external force, the internal restoring force bring back the body to its original configuration is known as elasticity.
Stress
If a body is deformed by an external force, the body can get its original configuration by the help of the internal restoring force after the removal of external, one.
This restoring force per unit area is known as stress.
\[
\text{Stress}=\frac{\text{restoring force}}{\text{area}}=\frac{F}{A}
\]
\[
=\frac{N}{m^2}
\]
\[
= \left[\frac{N}{m^2}\right]
= \left[\frac{MLT^{-2}}{L^2}\right]
\]
\[
=\left[ML^{-1}T^{-2}\right]
\]
Unit: \(N.m^{-2}\)
Strain
If an elastic body is deformed by an application of external force, the ratio of deformation to its original config. is known as strain.
Let it is unit less and dimensionless quantity.
Let,
- deformation be \(L_2\)
- original config. be \(L_1\)
\[
\text{Strain} =
\frac{\text{change in config.}}
{\text{original config.}}
\]
Longitudinal strain
\[
\text{Longitudinal strain} =
\frac{\text{change in length }(\Delta L)}
{\text{original length }(L)}
\]
Volumetric Strain
\[
\text{Volumetric strain} =
\frac{\text{change in volume }(\Delta V)}
{\text{original volume }(V)}
\]
Shearing Strain
\[
\text{Shearing strain} =
\text{ Angular Change } \space \theta
\]
Hook's Law
It states that under elastic limits, stress of a body is directly proportional to its strain.
i.e.
\[
\boxed{\text{Stress}\propto\text{strain}}
\]
or,
\[
\boxed{\text{stress}=(\text{constant})\times\text{strain}}
\]
The constant of proportionality is known as co-efficient of elasticity or modulus of elastic elasticity.
\[
\boxed{
\text{modulus of elasticity} =
\frac{\text{stress}}{\text{strain}}
}
\]
Unit and Dimension of Modulus of Elasticity
The unit of and dimension of modulus of elasticity is, similar to that of stress.
i.e.
\[
N.m^{-2}
\]
\[
[ML^{-1}T^{-2}]
\]
respectively.
Types of Modulus of Elasticity
1. Young's Modulus (for longitudinal strain)
\[
\boxed{
\text{Young's Modulus} =
\frac{\text{Normal stress}}
{\text{longi. strain}}
}
\]
or
\[
\boxed{
Y=
\frac{F/A}{\Delta L/L}
}
\]
2. Bulk Modulus (for volumetric strain)
\[
\boxed{
\text{Bulk Modulus} =
\frac{\text{Normal stress}}
{\text{Volumetric strain}}
}
\]
\[
\boxed{
K\text{ or }B=
\frac{F/A}{\Delta V/V}
}
\]
3. Shearing Modulus (for shearing strain)
\[
\boxed{
\text{Shearing Modulus} =
\frac{\text{Shearing stress}}
{\text{Shearing strain}}
}
\]
or
\[
\boxed{
\eta=
\frac{F/A}{\theta}
}
\]
Example
Q. A string of 60 cm long is stretched by 2 cm by the application of load 200 gm. What will be the length when a load of 500 gm is applied?
Soln.
Given,
Original length of string \(=60cm\)
By load of 200gm
If load \(=200gm\)
then,
\[
\text{Str. stretched}=2cm
\]
By using Hook's law,
\[
\frac{E_1}{W_1} =
\frac{E_2}{W_2}
\]
1)
\[
\frac{2}{200} =
\frac{E_2}{500}
\]
\[
E_2 =
\frac{2\times500}{200}
\]
\[
E_2=5cm
\]
Hence,
the new length will be
\[
=5+60cm
\]
\[
=65cm
\]
Now
\[
L=60cm=\frac{60}{100}=0.6m
\]
1st case
\[
mg=\frac{200}{1000}\times9.8 =
0.2\times9.8
\]
\[
\Delta L=2cm =
\frac{2}{100}m =
2\times10^{-2} =
0.02m
\]
Let \(A\) be the area of cross of section of the string.
and \(Y\) as the i.e. the co-efficient of elasticity of the material.
\[
Y=
\frac{mg\times g\times L}{A(\Delta L)}
\]
\[
= \frac{0.2\times9.8\times0.6}
{A(0.02)}
\qquad -(1)
\]
In 2nd case the applied load is 500gm
Thus,
\[
Y = \frac{0.5\times9.8\times0.6}
{A(\Delta L_2)}
\qquad -(2)
\]
\(Y\) remains same in both situation.
So,
\[
(1)=(2)
\]
\[
\frac{0.2\times9.8\times0.6}
{A(0.02)} =
\frac{0.5\times9.8\times0.6}
{A(\Delta L_2)}
\]
i)
\[
\frac{0.2}{0.02} =
\frac{0.5}{\Delta L_2}
\]
\[
\Delta L_2 =
\frac{0.02\times0.5}{0.2}
\]
\[
=5\times10^{-2}
\]
\[
=0.05m
\]
\[
=5cm
\]
Hence,
\[
\text{new }L=L+\Delta L_2
\]
\[
=60+5
\]
\[
\boxed{=65cm}
\]
Question
Q. The Young's Modulus of steel \(2\times10^{11}N.m^{-2}\)
Q.
The Young's Modulus for steel \(2\times10^{11}N.m^{-2}\) is the inter atomic spacing of material is \(2.8\). Find the increasing inter atomic spacing for the stress of \(10^9N.m^{-2} Å\)
Soln.
Force
\[
Y=2\times10^{11}N.m^{-2}
\]
\[
L=2.8 Å =
2.8\times10^{-10}m
\]
\[
=28\times10^{-11}m
\]
Stress
\[
= \frac{F}{A} =
10^9N.m^{-2}
\]
We know that
\[
Y=
\frac{\text{Stress}}{\text{Strain}}
\]
or,
\[
\text{Strain} =
\frac{\text{Stress}}{Y}
\]
\[
\frac{\Delta L}{L} =
\frac{\text{Stress}}{Y}
\]
\[
= \Delta L =
\frac{\text{Stress}}{Y}\times L
\]
\[
\frac{10^9\times28\times10^{-11}}
{2\times10^{11}}
\]
\[
\Delta L =
14\times10^{9-11-11}
\]
\[
= 14\times10^{9-22}
\]
\[
= 14\times10^{-13}m
\]
\[
= 0.014 Å
\]
Poisson's Ratio
\[
\text{Poisson's Ratio} =
\frac{\text{Lateral strain}}
{\text{Longitudinal strain}}
\]
or
\[
\sigma =
\frac{-\Delta D/D}{\Delta L/L}
\]
\[
0<\sigma<1
\]
The Poisson's Ratio always stays between \(0\) & \(1\).
Relation b/w Stress and Strain
- \(A\) = Elastic limit
- \(B\) = Yield point
OA:
Linear portion where Hook law is valid.
BC:
The plastic region where Hook's law is not obeyed.
Elastic limit:
The maximum point of over stress \(\propto\) strain.
Yield Point
The maximum limit of elasticity.
Question
Calculate the percentage of increase in aluminium wire of \(2.2mm\) diameter stressed by a load of \(100gm\).
Youngs modulus of the wire is
\[
12.5\times10^{10}N.m^{-2}
\]
Given:
Diameter of wire \(=2.2mm\)
Load \(=100gm\)
Youngs Modulus
\[
=12.5\times10^{10}N.m^{-2}
\]
Soln.
To find:
- increase in length \(\%=?\)
Area of cross section of wire is
\[
\frac{\pi d^2}{4}
\]
\[
= \frac{22}{7}\times\frac{11}{10}\times\frac{11}{10}\times\frac14
\]
\[
= \frac{1331}{350}
\]