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 edge length of an aluminium cube , L = 10cm
so, area of one of the faces of the cube , A = L² = 10 × 10 = 100 cm² or 10^-2 m²
mass is attached to the face of the aluminium cube , m = 100kg
so, applied force = mg = 100 × 9.8 = 980N
given, shear modulus , Y = 25GPa
Y = 25 × 10^9 Pa or 2.5 × 10^10 N/m²
Let ∆L is the vertical deflection of that face .
use formula , Y = FL/A∆L
∆L = FL/AY
= 980 × 0.1/(10^-2 × 2.5 × 10^10)
= 3.92 × 10^-7 m
so, area of one of the faces of the cube , A = L² = 10 × 10 = 100 cm² or 10^-2 m²
mass is attached to the face of the aluminium cube , m = 100kg
so, applied force = mg = 100 × 9.8 = 980N
given, shear modulus , Y = 25GPa
Y = 25 × 10^9 Pa or 2.5 × 10^10 N/m²
Let ∆L is the vertical deflection of that face .
use formula , Y = FL/A∆L
∆L = FL/AY
= 980 × 0.1/(10^-2 × 2.5 × 10^10)
= 3.92 × 10^-7 m
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Edge of the aluminium cube, L = 10 cm = 0.1 m
The mass attached to the cube, m = 100 kg
Shear modulus (η) of aluminium = 25 GPa = 25 × 109 Pa
Shear modulus, η = Shear stress / Shear strain = (F/A) / (L/ΔL)
Where,
F = Applied force = mg = 100 × 9.8 = 980 N
A = Area of one of the faces of the cube = 0.1 × 0.1 = 0.01 m2
ΔL = Vertical deflection of the cube
∴ ΔL = FL / Aη
= 980 × 0.1 / [ 10-2 × (25 × 109) ]
= 3.92 × 10–7 m
The vertical deflection of this face of the cube is 3.92 ×10–7 m.
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