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?

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Answer 1

ANSWER :–

Vertical deflection = $4 \times {10}^{ - 7} m$

EXPLANATION :–

GIVEN :–

• The edge of an aluminium cube is 10 cm.

• A mass of 100 kg is then attached to the opposite face of the cube.

• The shear modulus of aluminium is 25 GPa.

TO FIND :–

Vertical deflection.

SOLUTION :–

Shear modulus :–

The ratio of shear stress and shear strain is called Shear modulus .

• it's denoted by G .

• Formula of G = (F/A)/tan(θ)

• Now let's put value –

=> G = (F/A)/(x/L) [ tan(θ) = x/L ]

=> x = (F×L)/(G×A)

=> x = $\frac{100 \times g \times 10 \times {(10)}^{ - 2} }{ {(10 \times {10}^{ - 2} )}^{2} \times 25 \times {10}^{9} }$

=> x = $\frac{100 \times 10 \times 10 \times {(10)}^{ - 2} }{ {(10 \times {10}^{ - 2}) }^{2} \times 25 \times {10}^{9} }$

=> x = $\frac{100}{100 \times {10}^{- 4} \times 25 \times {10}^{9} }$

=> x = $\frac{1}{ {10}^{- 4} \times 25 \times {10}^{9} }$

=> x = $4 \times {10}^{ - 7} m$

Answer 2

Explanation:

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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 × 10^9 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 m^2

ΔL = Vertical deflection of the cube

∴ ΔL = FL / Aη

= 980 × 0.1 / [ 10^-2 × (25 × 10^9) ]

= 3.92 × 10^–7 m

The vertical deflection of this face of the cube ==> 3.92 ×10^–7 m