Aspecimen of copper having a rectangular cross section 15.2 mm x19.1 mm (0.6 in. x 0.75 in.) is pulled in tension with 44500 n(1 lbf) force, producing only elastic deformation. calculate the resulting strain.

The elastic deformation is 0.00131.

Explanation:

Given that,

Force F = 44500 N

Cross section

We Calculate the stress

Using formula of stress

Where, F = force

A = area of cross section

Put the value into the formula

We need to calculate the strain

Using formula of strain

Where,

=stress

Y = young modulus of copper

Put the value into the formula

Hence, The elastic deformation is 0.00131.

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.