Question

A specimen of aluminum having a rectangular cross section 9.6 mm × 12.9 mm (0.3780 in. × 0.5079 in.) is pulled in tension with 35600 N (8003 lbf) force, producing only elastic deformation. The elastic modulus for aluminum is 69 GPa (or 10 × 106 psi). Calculate the resulting strain.

Answer 1

Answer:

$\epsilon=4.16\times 10^{-3}$

Explanation:

It is given that,

Dimension of specimen of aluminium, 9.6 mm × 12.9 mm

Area of cross section of aluminium specimen, $A=9.6\times 12.9=123.84\times 10^{-6}\ mm^2$

$A=123.84\times 10^{-6}\ m^2$

Tension acting on object, T = 35600 N

The elastic modulus for aluminum is, $E=69\ GPa=69\times 10^9\ Pa$

The stress acting on material is proportional to the strain. Its formula is given by :

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

$\sigma$ is the stress

$\epsilon=\dfrac{F}{EA}$

$\epsilon=\dfrac{35600}{69\times 10^9\times 123.84\times 10^{-6}}$

$\epsilon=4.16\times 10^{-3}$

Hence, this is the required solution.