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Question: A cube of aluminium of sides 0.1 m is subjected to a shearing force of 100 N. The top face of the cu...

A cube of aluminium of sides 0.1 m is subjected to a shearing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be

A

0.02

B

0.1

C

0.005

D

0.002

Answer

0.002

Explanation

Solution

Shearing strain φ=xL=0.02cm0.1m=0.002\varphi = \frac{x}{L} = \frac{0.02cm}{0.1m} = 0.002