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Exercises · 8.6

Q.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?

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The vertical deflection is found using shear strain = shear stress / shear modulus. The shear force is the weight (100 kg × 9.8 m/s²), the area is the face area (0.01 m²), and the height is the cube edge (0.1 m). The deflection comes out to 4×10−74 \times 10^{-7} m.

This is a problem about shear deformation, not tensile or compressive strain. When you fix one face of a cube to a wall and hang a weight from the opposite face, the weight pulls that face downward. The cube doesn't stretch lengthwise — instead, its shape distorts: the top face shifts sideways relative to the bottom face. That sideways shift is the vertical deflection we need.

The key idea: shear stress τ\tau is force per area parallel to the face, shear strain γ\gamma is the angle of distortion (or deflection divided by height), and they are related by the shear modulus GG:

τ=G⋅γ\tau = G \cdot \gamma

Where γ=Δxh\gamma = \frac{\Delta x}{h}, with Δx\Delta x being the deflection and hh the height of the cube (the distance between the fixed and moving faces).

Let’s work it through.

  1. Identify the shear force. The mass of 100 kg exerts a weight F=mgF = mg downward. That force acts parallel to the face of the cube (tangential to the fixed wall).

F=100×9.8=980 NF = 100 \times 9.8 = 980 \ \text{N}

  1. Find the area of the face where the force is applied. The cube edge is 10 cm = 0.1 m, so each face has area:

A=(0.1)2=0.01 m2A = (0.1)^2 = 0.01 \ \text{m}^2

  1. Compute shear stress. Shear stress is force divided by area:

τ=FA=9800.01=98 000 Pa=9.8×104 Pa\tau = \frac{F}{A} = \frac{980}{0.01} = 98\,000 \ \text{Pa} = 9.8 \times 10^4 \ \text{Pa}

  1. Recall the shear modulus. Given G=25 GPa=25×109 PaG = 25 \ \text{GPa} = 25 \times 10^9 \ \text{Pa}.

  2. Find shear strain. From τ=Gγ\tau = G \gamma,

γ=τG=9.8×10425×109=3.92×10−6\gamma = \frac{\tau}{G} = \frac{9.8 \times 10^4}{25 \times 10^9} = 3.92 \times 10^{-6}

That’s a very small number — aluminium is stiff, so the distortion is tiny. …

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