Four identical hollow cylindrical columns of mild steel support a big structure of mass $50\times {10}^3 \mathrm{kg}$. The inner and outer radii of each column are 50 cm and 100 cm respectively. Assuming uniform local distribution, calculate the compression strain of each column. [use $\mathrm{Y}=2.0\times {10}^{11} \mathrm{Pa}, \mathrm{g}=9.8 \mathrm{m}/{\mathrm{s}}^2$] 
 

Given,

Mass of big structure M = $50\times {10}^3\mathrm{kg}$

$\mathrm{r} = 30 \mathrm{cm} = 0.3 \mathrm{m} , \mathrm{Outer} \mathrm{radius} \mathrm{R} = 60 \mathrm{cm} = 0.6 \mathrm{m}$

$\mathrm{Young}’\mathrm{s} \mathrm{modulus} \mathrm{Y}=2\times {10}^{11}\mathrm{Pa}$

$\mathrm{Total} \mathrm{force} \mathrm{exerted}, \mathrm{F}=\mathrm{Mg}=50000\times 9.8\mathrm{N}$

Stress equals the force applied to a single column.

$=50000\times \frac{9.8}{4}=122500\mathrm{N}$

$\mathrm{Young}\text{'}\mathrm{s} \mathrm{modulus}, \mathrm{Y}=\frac{\mathrm{stress}}{\mathrm{strain}}$

$\Rightarrow \mathrm{strain}=\frac{{\displaystyle \frac{\mathrm{F}}{\mathrm{A}}}}{\mathrm{Y}}$

$\mathrm{Area}=\mathrm{A}=({\mathrm{R}}^2-{\mathrm{r}}^2)=\mathrm{\pi }\left({\left(0.6\right)}^2-{(0.3)}^2\right)$

$\Rightarrow \mathrm{strain}=122500/[\mathrm{\pi }\left({\left(0.6\right)}^2-{(0.3)}^2\right)\times 2\times {10}^{11}]=7.22\times {10}^{-7}$