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

0.02

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0.1

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0.005

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0.002

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Solution

The correct option is D 0.002
Shearing strain $ϕ = \frac{x}{L} = \frac{0.02 c m}{0.1 m} = 0.002$

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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

The correct option is D 0.002Shearing strain ϕ=xL=0.02cm0.1m=0.002

The correct option is D 0.002
Shearing strain $ϕ = \frac{x}{L} = \frac{0.02 c m}{0.1 m} = 0.002$