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

Answer:

The resulting strain is $1.39\times 10^{-3}$.

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 :

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

E is elastic modulus of Cu is are of cross section

$\epsilon=\dfrac{44500}{15.2\times 19.1\times 10^{-6}\times 110\times 10^9}\\\\\epsilon=1.39\times 10^{-3}$

So, the resulting strain is $1.39\times 10^{-3}$.