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 × 10 9 N / m2
Modulus of elasticity, $η = \frac{Stress }{ Strain} = \frac{\frac{F} { A} }{ Strain}$

$∴ Strain = \frac{F }{Aη }$

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

= 3.65 × 10–3