Find the maximum and minimum Stress developed at the base of a column loaded as shownin fig. Also draw stress distribution. Take $E=2\times10^5 N/mm^2$

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written 7.1 years ago by

teamques10 ★ 69k

• modified 7.0 years ago

Given data:

$L=6m\\ D=300mm \\ d=250 mm\\ P=150\ kN\\ e=450 mm\\ t=25 mm\\ E=2\times10^5\ N/mm^2$

To find: Maximum & minimum stresses

$A=\frac{\pi}{4}(300^2-250^2)=\frac{\pi}{4}(27500)\\ \underline{A=21598.4\ mm^2}\\ \underline{E=2\times10^5\ N/mm^2}\\ I_{YY}\frac{\pi}{64}(300^4-250^4)=\frac{\pi}{64}(300^4-25^4)\\ I_{YY}=205.86\times10^6mm^4\\ \sigma_d=\frac{P}{A}=\frac{150\times10^3}{21598.4}\\ \underline{\sigma_d=6.95\ N/mm^2}\\ \sigma_b=\frac{M}{I}\times y=\frac{P\times e\times\bar{x}}{I_{YY}}=\frac{150\times10^3\times450\times150}{205.86\times10^6}\\ \underline{\sigma_b=49.18\ N/mm^2}\\ \sigma_{max}=\sigma_d+\sigma_b\\ =6.95+49.18\\ \underline{\sigma_{max}=56.13\ N/mm^2}\\ \sigma_{min}=\sigma_d-\sigma_b\\ =6.95-49.18\\ \underline{\sigma_{min}=-42.23\ N/mm^2}$

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