Question

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

Answer 1

$SSA=l\times b$

$=19.1\times {10}^{-3}\times 15.2\times {10}^{-3}$

$=2.9\times {10}^{-4}{m}^2$

$\eta =42\times {10}^9\frac{N}{m^2}$

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

$=\frac{\frac{F}{A}}{strain}$

$strain=\frac{F}{A\eta }$

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

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