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? (Take modulus of elasticity of copper $140\times {10}^{9}m/se{c}^2$)

Answer 1

Given, the dimensions of a piece of copper are 15.2 mm×19.1 mm. The applied tensile force is 44500 N and the modulus of elasticity of copper is $140\times {10}^{9}N/m^2$ .

$Y=\frac{stress}{strain}$

$strain=\frac{stress}{Y}=\frac{F}{YA}$

$strain=\frac{\left(44500\right)}{(290\times {10}^{-6}\times 140\times {10}^9)}$

$strain=1.09\times {10}^{-3}$