Answer to Question #172504 in Mechanical Engineering for isaac

Question

) A bar of 20 mm diameter is tested in tension. When a load of 38 kN was applied

extension of 0.12 mm was measured over a length of 200 mm and contraction in

diameter was 0.004 mm. If Young’s modulus is 2 × 105 N/mm2

then the value of

modulus of rigidity is??

Answer 1

Diameter of bar is 20 mm, Load = 38 kN, extension =0.12 mm, length=200 mm,

contraction diameter=0.004 mm, $E= 2 \times 10^5$ ,

lateral strain= $\frac{\delta d}{d}=\frac{0.004}{20}=0.0002$ ,

longitudinal strain=$\frac{\delta l}{l}=\frac{.12}{200}=0.0006$

poisson ratio=$\frac{lateral strain}{longitudinal strain}= \frac{0.0002}{0.0006}=0.333$

We know that relation between Modulus of rigidity and Young's modulus

$E= 2G(1+\nu)$

$2\times 10 ^5 = 2G(1+0.333)$

G= $0.75\times 10 ^5= 7.5 \times 10 ^4 \frac{N}{mm^2}$