Question (contd.) carries a load of 80 kN ata n eccentricity of 25mm from the column axis. Take E = 200GPa

Subject : Structural Analysis 1

Topic : Struts

Difficulty : High

Given data:

$ D=200mm\\ d=160mm\\ L=6m \\ t=20mm \\ P=80\ kN\\ eccentricity\ e=25mm\\ E=200GPa=2\times10^5MPa$

To find: $\underline{Extreme Stress}$

$Area\ of\ column,\ A=\frac{\pi}{4}(D^2-d^2)\\ = \frac{\pi}{4}(200^2=160^2)\\\underline{A=11309.73 mm^2}\\ I=\frac{\pi}{64}(D^4-d^4)=\frac{\pi}{64}(200^4-160^4)=46.36\times10^6mm\\ z=\frac{I}{y}=\frac{46.36\times10^6}{100}=463.6\times10^3mm^3 $

Since the column is fixed at both ends

$ L_{eff}=L=6000mm\\ L_{eff}=\frac{L}{2}=3000mm$

$\underline{Extreme\ Stresses:}$

Maximum bending moment =

p.e. sec$\left( \frac{L}{2}\sqrt{\frac{P}{EI}}\right) $

Determine the angle $\frac{L}{2}\sqrt{\frac{P}{EI}}$

$ \frac{L}{2}\sqrt{\frac{P}{EI}}=\frac{6000}{2}\left[ \sqrt{\frac{80\times10^3}{2\times10^5\times46.38\times10^6}}\right]=0.27 rad =16^\circ12'\\ \sec 16^\circ12'=1.041$

Maximum bending moment

= $ 80\times10^3\times25\times1.041=\underline{2.08\times10^6\ Nmm}$

Maximum Compresive stress

$ \sigma_{max}=\frac{P}{A}+\frac{M}{Z}=\frac{80\times10^3}{11309.73}+\frac{2.08\times10^6}{4.636\times10^6}\\ \underline{\sigma_{max}=7.52\ N/mm^2}\\ \sigma_{min}=\frac{P}{A}-\frac{M}{Z}=\frac{80\times10^3}{11309.73}-\frac{2.08\times10^6}{4.636\times10^6}\\ \underline{\sigma_{min}=6.62\ N/mm^2}$