#### Question

The area of the upper face of a rectangular block is 0.5 m by 0.5 m and the lower face is fixed. The height of the block is 1 cm. A shearing force applied at the top face produces a displacement of 0.015 mm. Find the strain and shearing force.

(Modulus of rigidity: η = 4.5 x 10^{10} N/m^{2})

#### Solution

**Given**:

Area under shear, A = 0.5m × 0.5m = 0.25m^{2}

Height of the block, h = 1cm = 1 × 10^{-2} m,

Displacement of top face = x = 0.015 mm = 15 × 10^{-6} m,

Modulus of rigidity, η = 4.5 × 10^{10} N/m^{2}

**To Find**:

Strain, θ = ?

Shearing force, F = ?

**Formulae:**

θ = x/h

F = ηAθ

**Solution:**

θ = x/h

`theta = (15 xx 10^-6)/10^-2`

`theta = 1.5 xx 10^-3`

**Shearing strain (θ) is 1.5 × 10 ^{-3}**

F = ηAθ

F = 4.5 × 10^{10} × 0.25 × 1.5 × 10^{-3}

F = 1.688 × 10^{7} N

**Shearing force (F) is 1.688 × 10 ^{7} N**