Question

Explain the different types of modulus of elasticity.

Answer 1

From Hooke’s law, the stress in a body is proportional to the corresponding strain, provided the deformation is very small. Here we shall define the elastic modulus of a given material. There are three types of elastic modulus.

1. Young’s modulus
2. Rigidity modulus (or Shear modulus)
3. Bulk modulus

Young’s Modulus: When a wire is stretched or compressed, then the ratio between tensile stress (or compressive stress) and tensile strain (or compressive strain) is defined as Young’s modules.

Young modulus of a material = ${\displaystyle \frac{\text{Tensile stress or compressive stress}}{\text{Tensile strain or compressive strain}}}$

Y = ${\displaystyle \frac{{\sigma }_{\text{t}}}{{\epsilon }_{\text{t}}}}$ or Y = ${\displaystyle \frac{{\sigma }_{\text{c}}}{{\epsilon }_{\text{c}}}}$

S.I. unit of Young modulus is Nm⁻² or pascal.

Bulk modulus: Bulk modulus is defined as the ratio of volume stress to the volume strain.

Bulk modulus, K = ${\displaystyle \frac{\text{Normal (Perpendicular) stress or Pressure}}{\text{Volume strain}}}$

The normal stress or pressure is σ_n = ${\displaystyle \frac{{\text{F}}_{\text{n}}}{\Delta \text{A}}=\Delta \text{P}}$

The volume strain is ε_v = ${\displaystyle \frac{\Delta \text{V}}{\text{V}}}$

Therefore, Bulk modulus is K = ${\displaystyle -\frac{{\sigma }_{\text{n}}}{{\epsilon }_{\text{v}}}=-\frac{\Delta \text{P}}{\frac{\Delta \text{V}}{\text{V}}}}$

The negative sign indicates when pressure is applied to the body, its volume decreases. Further, the equation implies that a material can be easily compressed if it has a small value of bulk modulus. In other words, bulk modulus measures the resistance of solids to change in their volume. For example, we know that gases can be easily compressed than solids, which means, gas has a small value of bulk modulus compared to solids. The S.I. unit of K is the same as that of pressure i.e., Nm^–2 or Pa (pascal).

The rigidity modulus or shear modulus: The rigidity modulus is defined as the ratio of the shearing stress to shearing strain,

η_R = ${\displaystyle \frac{\text{Shearing stress}}{\text{Angle of shear or shearing strain}}}$

The shearing stress is σ_s = ${\displaystyle \frac{\text{Tangential force}}{\text{Area over which it is applied}}=\frac{{\text{F}}_{\text{t}}}{\Delta \text{A}}}$

The angle of shear or shearing strain ε_s = ${\displaystyle \frac{\text{x}}{\text{h}}}$ = θ

Therefore, Rigidity modulus is η_R = ${\displaystyle \frac{{\sigma }_{\text{s}}}{{\epsilon }_{\text{s}}}=\frac{\frac{{\text{F}}_{\text{t}}}{\Delta \text{A}}}{\frac{\text{x}}{\text{h}}}=\frac{\frac{{\text{F}}_{\text{t}}}{\Delta \text{A}}}{\theta }}$

Further, the above implies, that a material can be easily twisted if it has a small value of rigidity modulus. For example, consider a wire, when it is twisted through an angle θ, a restoring torque is developed, that is

${\displaystyle \tau \propto \theta }$

This means that for a larger torque, the wire will twist by a larger amount (angle of shear θ is large). Since rigidity modulus is inversely proportional to the angle of shear, the modulus of rigidity is small. The S.I. unit of ηR is the same as that of pressure i.e., Nm^–2 or pascal.