Question

A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

Answer 1

Length of the piece of copper, l = 19.1 mm = 19.1 × 10^–3 m

Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10^–3 m

Area of the copper piece:

A = l × b

= 19.1 × 10^–3 × 15.2 × 10^–3

= 2.9 × 10^–4 m²

Tension force applied on the piece of copper, F = 44500 N

Modulus of elasticity of copper, η = 42 × 10⁹ N/m²

Modulus of elasticity, η = ${\displaystyle \frac{\text{Stress}}{\text{Strain}}=\frac{\frac{F}{A}}{\text{Strain}}}$

${\displaystyle \therefore \text{Strain}=\frac{F}{A\eta }}$

${\displaystyle =\frac{44500}{2.9\times {10}^{-4}\times 42\times {10}^9}}$

${\displaystyle =3.65\times {10}^{-3}}$

Answer 2

A = 15.2 x 19.2 x 10^-6 m²; F  = 44500 N; η = 42 x 10⁹ Nm^-2

Strain =  ${\displaystyle \frac{\text{Stress}}{\text{modulus of elasticity}}=\frac{\text{F/A}}{\eta }}$

${\displaystyle =\frac{F}{A\eta }=\frac{44500}{(15.2\times 19.2\times {10}^{-6})\times 42\times {10}^9}=3.65\times {10}^{-3}}$