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?

#### Solution 1

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/(triangleL))`

Where

*F* = Applied force = *m*g = 100 × 9.8 = 980 N

*A* = Area of one of the faces of the cube = 0.1 × 0.1 = 0.01 m^{2}

Δ*L* = Vertical deflection of the cube

`:.triangleL = FL/(Aη)`

`= (980xx0.1)/(10^(-2)xx(25xx10^9))`

= 3.92 × 10^{–7} m

The vertical deflection of this face of the cube is 3.92 ×10^{–7} m.

#### Solution 2

Here, side of cube, L = 10 cm =10/100= 0.1 m

∴ Area of each face, A = (0.1)^{2} = 0.01 m^{2}

Tangential force acting on the face,

F = 100 kg = 100 x 9.8 = 980 N

Shear Modulus, η = 25 Gpa = `25 xx 10^9 Nm^(-2)`

Since shear modulus is given as

`eta = "Tangential stress"/"Shearing strain"`

`:."Shearing strain" = "Tangential stress"/"Shear modulus"`

`= F/(Aeta) = 980/(0.01xx25xx10^9) = 3.92 xx 10^(-6)`

Now `"Lateral Strain"/"Side of cube" = "Shearing strain"`

:.Lateral Strain = Shearing stain x Side of the cube

`= 3.92 xx 10^(-6) xx 0.1`

`=3.92 xx 10^(-7) m = 4 xx 10^(-7) m`