Answer to Question #122491 in Mechanical Engineering for Ubaid

Question

A circular rod of diameter "d" is 25(mm) and length "L" 2(m) to be stretched outward, with a force "F" of 79(N), The stress(KPa)generated will be

Answer 1

$\sigma=\frac{F}{A}=\frac{F}{\pi d^2/4}=\frac{79}{3.14\cdot 0.025^2/4}=160937Pa\approx161 kPa$