Q: The edge of an aluminium cube is 10 cm long. One face of cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the displacement of upper face relative to lower face.
Sol: Here; L = 10 cm = 10-1 m
A = L2 = 10-2 m2
F= 100 kg wt = 100 × 9.8 N = 9.8 × 102 N.
η = 25 GPa = 25 × 109 Pa =25 × 109 N/m2
$\large \eta = \frac{Shearing \; Stress}{Shearing \; Strain}$
$\large \eta = \frac{F/A}{\Delta L/L}$
$\large \Delta L = \frac{F L }{A \eta}$
$\large \frac{9.8 \times 10^2 \times 10^{-1}}{10^{-2}\times 25 \times 10^9} $
= 0.4 × 10-6 m = 4 × 10-7 m.