Question

A cube of side 3 cm Is subjected to shearing force. The top of cube is sheared through 0.012 cm with respect to bottom face. If cube is made of a material of shear modulus 2.1 × 10¹⁰Nm^-2 then find 

1. Shearing strain 

2. Shearing stress 

3. Force applied.

Answer 1

1. Shearing strain

= \(\frac {Δl}{l} \) = \(\frac {0.012 cm}{3cm}\)

= 0.004

2. Shearing stress

= Shearing strain × Shear modulus

Shearing stress = 0.004 × 2.1 × 10¹⁰Nm^-1

= 8.4 × 10⁷Nm^-2

3. Shearing stress = \(\frac{S.Force}{Ares}\)

Shearing force = Shearing stress × Area

= 8.4 × 10⁷ × (3 × 10^-1)²

= 7.56 × 10⁴ N

= 75.6 KN.