Four identical hollow cylindrical columns of mild steel support a mass of $50,000 ext{ kg}$. The in

Question

Four identical hollow cylindrical columns of mild steel support a mass of $50,000	ext{ kg}$. The inner and outer radii of each column are $30	ext{ cm}$ and $60	ext{ cm}$, respectively. Assuming uniform load distribution, what is the compressional strain of each column? (Take $g = 9.8	ext{ m/s}^2$, $Y_{	ext{steel}} = 2.0 	imes 10^{11}	ext{ Pa}$)

Professor PC Thomas Classes · Answer

Load per column $F = \frac{50000 	imes 9.8}{4} = 122500	ext{ N}$. Area $A = \pi(0.6^2 - 0.3^2) = \pi(0.27) \approx 0.848	ext{ m}^2$. Stress $= \frac{122500}{0.848} \approx 1.44 	imes 10^5	ext{ Pa}$. Strain $= \frac{	ext{Stress}}{Y} = \frac{1.44 	imes 10^5}{2 	imes 10^{11}} = 7.2 	imes 10^{-7}$.