1.0Modulus of Elasticity

Young's modulus, or the modulus of elasticity, gauges how stiff a material is by comparing stress (force per unit area) to strain (deformation) within the material’s elastic range. A higher modulus means the material is stiffer, while a lower value indicates greater flexibility. This property is crucial in engineering and materials science as it helps predict how materials will behave when subjected to stress.

2.0Modulus of Elasticity

• The modulus of elasticity, or elastic modulus, quantifies substance stiffness or its ability to resist deformation.
• It measures the relationship between stress (force applied per unit area) and strain (change in length or shape) within a material's elastic range.
•  More is the value of Modulus of Elasticity, more is the Elasticity of material.
•  It means more elastic material will have more tendency to regain its shape under  elastic limit deformation (not permanent deformation).

3.0Young's Modulus of Elasticity

• Within the elastic limit, the ratio of longitudinal stress to longitudinal strain is called Young's modulus of elasticity.

$Y=\frac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}}=\frac{F/A}{\Delta L/L}$

$Y=\frac{FL}{A\Delta L}$

Units of Y: $N/m^2$

Dimensions of Y : $[M^1{L}^{-1}{T}^{-2}]$

$Y=\frac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}}=\frac{F/A}{\Delta L/L}\Rightarrow F=\frac{YA}{L}\Delta L$

Note:For Spring F=kx

For Wire :

$F=\frac{YA}{L}\Delta L$

$kx=\frac{YA}{L}\Delta L\Rightarrow k=\frac{YA}{L}$

4.0Increment of Length Due to Own Weight

A cable   of mass M and length L suspended vertically experiences varying tension, stress, and strain at different points:

 (1) Maximum stress happens at the top.

 (2) Minimum stress is at the lower end.

Consider a dx element of rope at x distance from lower end than tension, $T=\left(\frac{x}{L}\right)Mg$

$\text{Stress}=\frac{T}{A}=\left(\frac{x}{L}\right)\frac{Mg}{A}$

Let increase in length of dx is dy then strain=$\frac{dy}{dx}=\frac{\text{Stress}}{Y}=\frac{Mg}{YA}\frac{x}{L}$

So, Young modulus of elasticity $Y=\frac{\text{Stress}}{\text{Strain}}$

For full length of rope

$\frac{Mg}{LA}{\int }_0^{L}xdx=Y{\int }_0^{\Delta L}dy\Rightarrow \frac{Mg}{LA}\frac{L^2}{2}=Y\Delta L\Rightarrow \Delta l=\frac{MgL}{2AY}$

5.0Bulk Modulus of Elasticity

• It is defined as the ratio of the volume stress to the volume strain

$B=\frac{\text{Pressure}}{\text{Volume Strain}}$

The stress being the normal force applied per unit area and is equal to the pressure applied (p).

$B=\frac{-p}{\frac{\Delta V}{V}}=-V\frac{\Delta p}{\Delta V}$

• Negative sign shows that increase in pressure (p) causes decrease in volume (V).
• The bulk modulus is commonly defined as the ratio of the change in pressure to the resulting change in volume.

$B=-V\frac{\Delta p}{\Delta V}=-V\frac{dp}{dV}$

6.0Compressibility

•  The opposite of the Bulk modulus of elasticity is referred to as compressibility.

$N^{-1}{m}^{2}orPasca{l}^{-1}(P{a}^{-1})$

• Unit of Compressibility in Sl is
• Bulk modulus of solids is about fifty times that of liquids, and for gases it is $10^{-8}$
• times of solids.

$B_{Solids}>B_{Liquids}>B_{Gases}$

• In thermodynamics we will study:

Isothermal bulk modulus of elasticity of gas B = P(pressure of gas)

Adiabatic bulk modulus of elasticity of gas

$B=\gamma \times P,where\gamma =\frac{C_p}{C_v}$

7.0Modulus of Rigidity

• The modulus of rigidity of a material is characterized as the ratio of shearing stress to shearing strain, provided the material remains within its elastic limit.

$\eta =\frac{\text{Shearing Stress}}{\text{Shearing Strain}}=\left(\frac{\frac{{}^{F}Tangential}{A}}{\theta }\right)=\frac{{}^{F}Tangential}{A\theta }$

Note: Angle of shear 'Ф' is always taken in radians

8.0Solved Examples

1. The graph shows the extension of a wire of length 1m suspended from a roof at one end and with a load W connected to the other end. If the cross-sectional area of the wire is $1m{m}^2$ then Calculate the Young's modulus of the material of the wire.

Solution:

$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta L/L}=\frac{FL}{A\Delta L}$

$Y=\frac{L}{A}\times (\text{Slope})=\frac{1}{10^{-6}}\times \frac{10}{(4.0\times 10^{-3})}(2\times 10^{10},{\text{N/m}}^2)$

2. Calculate the force required to increase the length of a steel wire of cross-sectional area $10^{-6}{m}^2$ by 0.5%. Given: Y(for steel) = $2\times 10^{11}{\text{N/m}}^2$

Solution:

$\frac{\Delta L}{L}\times 100=0.5\%=5\times 10^{-3}$

$F=\frac{YA}{L}\Delta L=2\times 10^{11}\times 10^{-6}\times 5\times 10^{-3}=10^{3}N$