Elastic Modulus>Modulus of Rigidity

Shear Stress:

When a tangential force F is applied parallel to the surface of a block, shear stress is produced.

• Force is parallel to the cross-section (unlike tensile stress, where force is perpendicular)
• Unit: N/m² or Pa

Shear Strain:

When the block is subjected to shear stress, the top surface displaces relative to the bottom surface by a small distance Δl. The angle of deformation θ is called shear strain.

• Unit: Radians (dimensionless ratio)
• Valid for small displacements

How Shear Deformation Works

When two equal and opposite forces (forming a couple) are applied to the top and bottom surfaces of a block:

• The upper surface shifts horizontally by a distance Δl
• The corner angles change by a small amount θ
• The block gets distorted but maintains its volume

This is the same effect as fixing the bottom of a block and pushing only the top surface.

Problem: Calculate the modulus of rigidity of a metal cube of side 40 cm subjected to a shearing force of 2000 N. The upper surface is displaced by 0.5 cm with respect to the bottom.

Given Data:

• Side of cube, l = 40 cm = 0.40 m
• Shearing force, F = 2000 N = 2 × 10³ N
• Displacement, Δl = 0.5 cm = 0.005 m
• Area, A = l² = (0.40)² = 0.16 m²

To Find:

• Modulus of rigidity, η

Solution:

1. Calculate shear strain (θ)

• θ = \[\frac{\Delta l}{l}=\frac{0.005}{0.40}\] = 0.0125

2. Apply the modulus of rigidity formula

• η = \[\frac{F}{A\cdot\theta}\]

3. Substitute the values

• η = \[\frac{2.0\times10^3\mathrm{~N}}{(0.16\mathrm{~m}^2)\times(0.0125)}\]

4. Calculate

• η = \[\frac{2000}{0.002}=1.0\times10^6\mathrm{~N/m^2}\]

Answer: The modulus of rigidity of the metal = 1.0 × 10⁶ N/m² or 1.0 × 10⁶ Pa