Question

A piece of copper having a rectangular cross-section of $15.2mm\times 19.1mm$ is pulled in tension with $44,500N$ force, producing only elastic deformation. Calculate the resulting strain? (Modulus of elasticity of copper, $Y=42\times {10}^{9}N{m}^{-2}$)

Answer 1

Length of the piece of copper, $l=19.1mm$
Breadth of the piece of copper, $b=15.2mm$
Area of the copper piece: $A=lb$
$A=19.1\times {10}^{-3}\times 15.2\times {10}^{-3}$

$=2.9\times {10}^{-4}{m}^2$
Tension force applied on the piece of copper, $F=44500N$

Modulus of elasticity of copper, $Y=42\times {10}^{9}N{m}^{-2}$

Modulus of elasticity = Stress / Strain =$\left(F/A\right)$ / Strain
Strain = $F/\left(YA\right)$
= $44500/(2.9\times {10}^{-4}\times 42\times {10}^9$)
= $3.65\times {10}^{-3}$