A column of hollow circular section with external diamter 300mm and thickness 40mm id 4m long. It is pinned at both the ends. The column carries a load of 100 kN at an eccentricity of 40mm,...(contd.)

 A column of hollow circular section with external diamter 300mm and thickness 40mm id 4m long. It is pinned at both the ends. The column carries a load of 100 kN at an eccentricity of 40mm,...(contd.)

Solution

Given data:

D=300mmd=220mml=4mt=40mmLoad(P)=100 kNeccentricity e=40 mmE=200GPa=2×105MPa$$\begin{gathered} D=300mm \\ d=220mm \\ l=4m \\ t=40mm \\ Load(P)=100kN \\ eccentricitye=40mm \\ E=200GPa=2\times {10}^{5}MPa \end{gathered}$$

To find: Extreme stress

For a hollow circular column

Area of the column,

A=π4(D2−d2)=π4(3002−2202)=32672.56 mm2$A=\frac{\pi }{4}(D^2-d^2)=\frac{\pi }{4}({300}^2-{220}^2)=32672.56m{m}^2$

Moment of inertai of the column section

I=π64(D4=d4)=π64(3004−2204)=282.62×106mm$I=\frac{\pi }{64}(D^4=d^4)=\frac{\pi }{64}({300}^4-{220}^4)=282.62\times {10}^{6}mm$

Section Modulus (z) =Iy=282.62×106150=1.884×106mm3$\frac{I}{y}=\frac{282.62\times {10}^6}{150}=1.884\times {10}^{6}m{m}^3$

Since the column is pinned at both ends

Left = L = 4m =4000mm

Extereme Stress:

Maximum bending moment = P.e sec(12pEI−−−√)$\left(\frac{1}{2}\sqrt{\frac{p}{EI}}\right)$

Let us determine the angle

L2PEI−−−√=40002100×1032×105×282.62×106−−−−−−−−−−−−√=0.08 radian=4∘58′sec4∘58′=1.003$$\begin{gathered} \frac{L}{2}\sqrt{\frac{P}{EI}}=\frac{4000}{2}\sqrt{\frac{100\times {10}^3}{2\times {10}^5\times 282.62\times {10}^6}}=0.08radian=4^{\circ }{58}^{\prime } \\ \sec 4^{\circ }{58}^{\prime }=1.003 \end{gathered}$$

maximum BM=100×103×40×1.003$100\times {10}^3\times 40\times 1.003$

BM=4.01×106N−mm−−−−−−−−−−−−−−−−−−−−−$\underset{\_}{BM=4.01\times {10}^{6}N-mm}$

maximum compressive Stress Tmax$T_{max}$

σmax=Tmax=PA+MZ=1000×10332672.56+4.01×1061.884×106[σmax=5.189 N/mm2]−−−−−−−−−−−−−−−−−−$$\begin{gathered} {\sigma }_{max}=T_{max}=\frac{P}{A}+\frac{M}{Z}=\frac{1000\times {10}^3}{32672.56}+\frac{4.01\times {10}^6}{1.884\times {10}^6} \\ \underset{\_}{[{\sigma }_{max}=5.189N/m{m}^2]} \end{gathered}$$