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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Start your 14-day free trial to unlock the full solution →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 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 is force per area parallel to the face, shear strain is the angle of distortion (or deflection divided by height), and they are related by the shear modulus :
Where , with being the deflection and the height of the cube (the distance between the fixed and moving faces).
Let’s work it through.
- Identify the shear force. The mass of 100 kg exerts a weight downward. That force acts parallel to the face of the cube (tangential to the fixed wall).
- 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:
- Compute shear stress. Shear stress is force divided by area:
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Recall the shear modulus. Given .
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Find shear strain. From ,
That’s a very small number — aluminium is stiff, so the distortion is tiny. …
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