A T-Section of dimension $12cm\times12cm\times2cm$ and its length 3 meter is used as column with one of its end fixed and the other end free. Take $E=2\times10^5\ N/mm^2$ Find crippling load.

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written 7.1 years ago by

teamques10 ★ 69k

• modified 7.0 years ago

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Given:

$L=3m=3000mm\\ E=2\times10^5 N/mm^2$

To find: P = ?

$Position\ of\ neutral\ axis\ (from\ bottom)\\ \bar{y}=\frac{A_1y_1+A_2y_2}{A_1+A_2}=\frac{100\times20\times50+120\times20\times110}{(100\times20) + (120\times20)}=82.7mm$

$Moment\ of\ inertia\\ I_{XX}=\frac{bd^3}{12}+A_1h_1^2+\frac{bd^3}{12}+A_2h_2^2\\=\frac{120\times20^3}{12}+120\times20(37.30-10)^2+\frac{20\times100^3}{12}+20\times100(82.70-60)^2\\ \underline{I_{XX}=4.566\times10^6mm^4}$

Since one end is fixed and other end is free

Effective length, $L_e=2L=2\times3000=6000mm$

Crippling load $P=\frac{\pi^2EI}{L_e^2}$

$P=\frac{\pi^2\times2\times10^5\times4.566\times10^6}{(6000)^2}=250\times10^3\ N\\ \boxed{P=250\ kN}$

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