Question

The edge of an aluminium cube is $10 \, cm$ long. One face of the 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 vertical deflection of this face?

• A. $3.92\times 10^{- 7}m$
• B. $5.98\times 10^{- 7} \, m$
• C. $2.72\times 10^{- 7} \, m$
• D. $4.82\times 10^{- 7} \, m$

Answer 1

Answer: (A)

Given, side of a cube (l) = 10 cm = 0.1 m
​Area its each face (A) = l² = (0.1)² = 0.01 m²
Load (m) = 100 kg
Tangential force acting on one face of the cube,
​F = mg = 100 × 9.8 = 980 N
$\text{Shear stress acting on this face} = \frac{\text{F}}{\text{A}}$
$= \frac{\text{980}}{\text{0.01}} \text{N/m}^{2}$
= 9.8 × 10⁴ N/m²
Shear modulus of aluminium (η)) = 25 GPa
= 25 × 10⁹ N/m²
$\text{Shear modulus} \left(\eta\right) = \frac{\text{Shearing stress}}{\text{Shearing strain}}$
$\text{or shearing strain} \left(\frac{\text{Δ} \text{y}}{\text{L}}\right) = \frac{\text{Shearing stress}}{\text{Shear modulus}}$
$\text{or} \text{Δ} \text{y} = \frac{\text{Shearing stress}}{\text{Shear modulus}} \times \text{L}$
$= \frac{\text{9.8} \times \text{10}^{4}}{\text{25} \times \text{10}^{9}} \times \text{0.1}$
= 0.392 × 10^-5 m
= 3.92 × 10^-7 m