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?

Answer 1

Given

• Edge of cube (length): \( L = 10 \,\text{cm} = 0.10 \,\text{m} \)
• Mass attached: \( m = 100 \,\text{kg} \)
• Shear modulus of aluminium: \( G = 25 \,\text{GPa} = 25 \times 10^{9} \,\text{Pa} \)

1. Force due to the attached mass

Tangential (shearing) force:

\( F = mg = 100 \times 9.8 = 980 \,\text{N} \)

2. Area of the face being sheared

The face is a square of side \( L = 0.10 \,\text{m} \), so:

\( A = L^{2} = (0.10)^{2} = 0.010 \,\text{m}^{2} \)

3. Use shear modulus relation

Shear modulus:

\( G = \dfrac{\text{shear stress}}{\text{shear strain}} = \dfrac{F/A}{x/L} = \dfrac{F L}{A x} \)

Solve for vertical deflection \( x \):

\( x = \dfrac{F L}{A G} \)

4. Substitute values

\( x = \dfrac{980 \times 0.10}{0.010 \times 25 \times 10^{9}} \) \( = \dfrac{98}{0.25 \times 10^{9}} = \dfrac{98}{2.5 \times 10^{8}} \approx 3.92 \times 10^{-7} \,\text{m} \)

Vertical deflection of the free face: \( x \approx 3.9 \times 10^{-7} \,\text{m} \) (about \( 4 \times 10^{-7} \,\text{m} \)).