Four identical hollow cylindrical columns of mild steel support a big structure of mass 50 * 10³ kg. The inner and outer radii of each column are 50cm and 100cm respectively. Assuming uniform local distribution, calculate the compression strain of each column.

[Use Y = 2.0 * 10¹¹ Pa, g = 9.8 m/s²]

Load of mass will be equally distributed among the four colours so force on each columns will be 125 * 10³ N.

Cross section area of the column = $\pi \left[{\left(1\right)}^2-{\left(0.5\right)}^2\right]=2.355m^2$

Using young's modulus :  $\epsilon =\frac{\sigma }{Y}=\frac{F}{AY}=\frac{125*10^3}{2.355*2*10^{11}}=2.65*10^{-7}$