Question

A copper wire of circular cross section (diameter = 2 mm) was subjected to bending moment which shaped the wire into a semi -circle of radius 1m. The maximum bending stress developed in the wire is _____MPa. $E_{Cu}=140\text{Gpa})$

• A. 140

Answer 1

The correct option is A 140
By flexure formula
$\frac{\sigma }{y}=\frac{M}{I}=\frac{E}{R}$
$\sigma =\frac{E}{R}y$
Given, E = 140 GPa $=140\times {10}^{3}N/m{m}^2$
$y=\left(\frac{\text{diameter of wire}}{2}\right)=\left(\frac{2}{2}\right)=1mm$
R =1 m = 1000 m
$\sigma =\frac{140\times {10}^3\times 1}{1000}=140N/m{m}^2$
$=140\text{MPa}$