Question

A cylindrical bar of 20mm diameter and 1 m length is subjected to a tensile test. Its longitudinal strain is 4 times that of its lateral strain. If the modulus of elasticity is 2 $\times {10}^5$ N/mm${}^2$, then its modulus of rigidity will be

• A. $8\times {10}^{6}N/m{m}^2$
• B. $8\times {10}^{5}N/m{m}^2$
• C. $0.8\times {10}^{4}N/m{m}^2$
• D. $0.8\times {10}^{5}N/m{m}^2$

Answer 1

The correct option is D $0.8\times {10}^{5}N/m{m}^2$

Poisson's ration $=\frac{\text{Lateral strain}}{\text{Longitudinal strain}}=0.25$
$G=\frac{E}{2(1+\mu )}=\frac{2\times {10}^5}{2.5}$
$=0.8\times {10}^{5}N/m{m}^2$