Question

A cast iron column has an internal diameter of $200mm$. What should be the minimum external diameter (in m) so that it may carry a load of $1.6millionN$ without the stress exceeding $90N/m{m}^2$? (round off your answer to nearest integer)

Answer 1

Here breaking stress ,$\sigma =\frac{F}{A}=\frac{F}{\pi (R^2-r^2)}$ where $r$ is the internal radius and $R$ be the external radius.
Now, $R^2-r^2=\frac{F}{\pi \sigma }$
$R^2=\frac{F}{\pi \sigma }+r^2=\frac{1.6\times {10}^6}{3.14\times 90\times {10}^6}+(0.1{)}^2=0.0156$
$R=0.125m=125mm$
Thus minimum external diameter is $2R=2\left(125\right)=250mm=0.250m=0$ ; (It is $0$ in nearest integer)