What is the minimum diameter of a brass rod if it is to support a 400 N load without exceeding the elastic limit? Assume that the stress for the elastic limit is 379 MPa.

To find the minimum diameter, and hence minimum cross-sectional area, we assume that the force$F=400N$ brings us to the elastic limit. Then from the stress,

$\frac{F}{A}=379\times {10}^{6}Pa,$ we get

$A=\frac{400N}{379\times {10}^{6}Pa}=1.0554\times {10}^{-6}{m}^2.$

Then, $A=\frac{\pi D^2}{4}$

$D^2=\frac{4A}{\pi }=\frac{4\left(1.0554\times {10}^{-6}{m}^2\right)}{\pi }=1.344\times {10}^{-6}$

and $D=\sqrt{1.344\times {10}^{-6}{m}^2}=1.16\times {10}^{-3}m=1.16mm$