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 10¹⁰ N/m²)

Answer 1

Given:

Area under shear, A = 0.5m × 0.5m = 0.25m²

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 × 10¹⁰ N/m²

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 × 10¹⁰ × 0.25 × 1.5 × 10^-3

F = 1.688 × 10⁷ N

Shearing force (F) is 1.688 × 10⁷ N