Question

A circular rod of 100 mm diameter and 500 mm length is subjected to a tensile force of 1000 kN. Determined the modulus of rigidity (G) if E = 2 × 10⁵ N/mm² and Poisson’s ratio = 0.3.

1.

0.335 × 105 N/mm2

2.

0.555 × 105 N/mm2

3.

0.7692 × 105 N/mm2

4.

0.2256 × 105 N/mm2

Answer 1

Correct Answer - Option 3 :

0.7692 × 105 N/mm2

Concept:

Strain acting on a body due to stress is given by - 

σ = ϵ × E

where E = Young's Modulus of Elasticity.

The relation between Young's Modulus of Elasticity (E) and Modulus of Rigidity (G) is given by - 

E = 2G(1 + μ)

where μ = Poisson's ratio. 

Calculation:

Given:

d = 100 mm, l = 500 mm, P = 1000 kN ⇒ 10⁶ N, E = 2 × 105 N/mm2 and Poisson’s ratio = 0.3

E = 2G(1 + μ)

2 × 10⁵ = 2 × G × (1 + 0.3)

∴ G = 0.7692 × 10⁵ N/mm²