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

`\epsilon=4.16* 10^(-3)\`

Step-by-step 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* 12.9=123.84* 10^(-6)\ mm^2`

`A=123.84* 10^(-6)\ m^2`

Tension acting on object, T = 35600 N

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

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

`\epsilon=(\sigma)/(E)`

`\sigma` is the stress

`\epsilon=(F)/(EA)`

`\epsilon=(35600)/(69* 10^9* 123.84* 10^(-6))`

`\epsilon=4.16* 10^(-3)`

Hence, this is the required solution.