Question

A cube of aluminium of each side 4 cm is subjected to a tangential (shearing) force. The top of the cube is sheared through 0.012 cm w.r.t. the bottom face. Find (a) shearing strain, (b) shearing stress, (c) shearing force. (Given η = 2.08 η 10¹¹ dyne cm^-2)

Answer 1

Given :

Length of each side, L = 4 cm

Length displacement, x = 0.012 cm

η = 2.08 × 10¹¹ dyne cm^-2

(a) Shearing strain, θ = \(\frac{x}{L}\) = \(\frac{0.012}{4}\) = 0.003

(b) Using relation, η = \(\frac{F}{θA'}\)

Shearing stress = \(\frac{F}{A}\) = ηθ

= 2.08 × 10¹¹ × 3 × 10⁻³

= 6.24 × 10⁸ dyne cm^-2.

(c) Shearing force, F = Shearing stress × area

= 6.24 × 10⁸ × L²

= 6.24 × 10⁸ × 4²

= 9.984 × 10⁹ dyne.