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. Assume elastic modulus of Cu to be 110GPa. (Points: 5).

Answer 1

The resulting strain is
`1.39* 10^(-3)`.

Step-by-step explanation:

A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm

Force, F = 44,500 N

Th elastic modulus of Cu to be 110 GPa

The resulting strain is given by the formula as follows :

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

E is elastic modulus of Cu is are of cross section

`\epsilon=(44500)/(15.2* 19.1* 10^(-6)* 110* 10^9)\\\\\epsilon=1.39* 10^(-3)`

So, the resulting strain is
`1.39* 10^(-3)`.