Question

A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm (0.60 in. × 0.75 in.) is pulled in tension with 44,500 N (10,000 lbf) force, producing only elastic deformation. Calculate the resulting strain.

Answer 1

Answer:

0.00119

Explanation:

F = Force = 44500 N

A = Area = $0.0152\times 0.0191\ m^2$

E = Young's modulus of copper = $128\times 10^{9}\ Pa$

Stress is given by

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

Strain is given by

$\epsilon=\frac{\sigma}{E}\\\Rightarrow \eplison=\frac{\frac{F}{A}}{E}\\\Rightarrow \epsilon=\frac{\frac{44500}{0.0152\times 0.0191}}{128\times 10^{9}}\\\Rightarrow \epsilon=0.00119$

The resulting strain is 0.00119