Question

A cast iron column has an internal diameter of $200\,mm$ . What should be the minimum external diameter (in m) so that it may carry a load of $1.6\, \times {10^6}\,N$ without stress exceeding $90\,N\,m{m^{ - 2}}$ .(round off your answer to the nearest integer)

Answer 1

Stress is defined as the ratio of the applied force often known as load to the area on which it is applied.In this question, since we are asked to find the minimum external diameter, we shall calculate the breaking stress using the above-mentioned formula.We shall plug in the values in the formula to get the minimum external diameter.

Formula used:
We shall use the formula of stress given as
$\sigma = \dfrac{F}{A}$
where $\sigma$ is the stress, $F$ is the force (or load) and $A$ is the area on which the load is applied.
Also, the area of the cast iron column would be given by
$A = \pi ({R^2} - {r^2})$
where $R$ is the external radius and $r$ is the internal radius.

Complete step by step answer:
At the breaking stress, the value of the external diameter will be minimum.
Given that $F = 1.6 \times {10^6}\,N$, $\sigma = 90\,Nm{m^{ - 2}}$ and $r = \dfrac{{200}}{2} = 100\,mm$ .
We know that $\sigma = \dfrac{F}{A}$ and $A = \pi ({R^2} - {r^2})$ .
Substituting in the stress equation,
$\sigma = \dfrac{F}{{\pi ({R^2} - {r^2})}}$
Now plugging in the known values,
$90 = \dfrac{{1.6 \times {{10}^6}}}{{\pi ({R^2} - {{100}^2})}}$
Further solving the equation,
\Rightarrow 5 = \dfrac{{{{10}^5}}}{{\pi ({R^2} - {{100}^2})}} \\\
\Rightarrow \dfrac{{5\pi }}{{{{10}^5}}} = \dfrac{1}{{({R^2} - {{100}^2})}} \\\
Rearranging the terms we get,
${R^2} - {100^2} = \dfrac{{{{10}^5}}}{{5\pi }} \\\$
\Rightarrow {100^2} + \dfrac{{{{10}^5}}}{{5\pi }} = {R^2} \\\
Now solving the equation, we get,
${R^2} = 0.0156$
$\Rightarrow R = 125\,mm$
Converting to m,
$R = 0.125\,m$
Now since $R = 0.125\,mm$ the diameter will be equal to $2R = 2 \times 0.125\, = 0.250\,m$.

Hence the value of minimum external diameter is $0.250\,m$.Rounding off to the nearest integer, we get the value as zero.

Note: In the formula of stress which is given by $\sigma = \dfrac{F}{A}$ in which the given force is the perpendicular force better known as thrust. In cases where the applied load is inclined, we take the perpendicular component of the force to calculate the stress.Everywhere else when not mentioned, we take the load to be acting in a perpendicular direction.