Question (contd.) find out the stress produced at extreme fibre of the column section. Take E=200GPa

Given data:

$ D= 300mm\\ d=220mm\\ l=4m\\ t=40mm\\ Load(P)=100\ kN\\ eccentricity\ e=40\ mm\\ E=200GPa=2\times10^5MPa$

To find: Extreme stress

For a hollow circular column

Area of the column,

$A=\frac{\pi}{4}(D^2-d^2)=\frac{\pi}{4}(300^2-220^2)=32672.56\ mm^2$

Moment of inertai of the column section

$ I=\frac{\pi}{64}(D^4=d^4)=\frac{\pi}{64}(300^4-220^4)=282.62\times10^6mm$

Section Modulus (z) =$\frac{I}{y}=\frac{282.62\times10^6}{150}=1.884\times10^6mm^3$

Since the column is pinned at both ends

Left = L = 4m =4000mm

Extereme Stress:

Maximum bending moment = P.e sec$\left(\frac{1}{2}\sqrt{\frac{p}{EI}}\right)$

Let us determine the angle

$ \frac{L}{2}\sqrt{\frac{P}{EI}}=\frac{4000}{2}\sqrt{\frac{100\times10^3}{2\times10^5\times282.62\times10^6}}=0.08\ radian=4^\circ58'\\ \sec 4^\circ58'=1.003$

maximum BM=$100\times10^3\times40\times1.003$

$\underline{BM=4.01\times10^6N-mm}$

maximum compressive Stress $T_{max}$

$ \sigma_{max}=T_{max}=\frac{P}{A}+\frac{M}{Z}=\frac{1000\times10^3}{32672.56}+\frac{4.01\times10^6}{1.884\times10^6}\\ \underline{ [\sigma_{max}=5.189\ N/mm^2]}$