Question

A specimen of aluminum having a rectangular cross section 10 mm 12.7 mm (0.4 in. 0.5 in.) is pulled in tension with 35,500 N (8000 lbf) force, producing only elastic deformation. Calculate the resulting strain.

Answer 1

Answer:

$\epsilon = 3.958\times 10^{-3}$

Step-by-step explanation:

The Young's Module of Aluminium is $E = 69\times 10^{9}\,Pa$. The axial stress on the specimen is:

$\sigma = \frac{F}{A_{t}}$

$\sigma = \frac{35500\,N}{(0.01\,m)\cdot (0.013\,m)}$

$\sigma = 2.731\times 10^{8}\,Pa$

The strain is derived of following expression:

$\epsilon = \frac{\sigma}{E}$

$\epsilon = \frac{2.731\times 10^{8}\,Pa}{69\times 10^{9}\,Pa}$

$\epsilon = 3.958\times 10^{-3}$