ATSectionofdimension 12cm×12cm×2cm12cm×12cm×2cm and its length 3 meter is used as column with one of its end fixed and the other end free. Take E=2×105 N/mm2E=2×105 N/mm2 Find crippling load.

 ATSectionofdimension 12cm×12cm×2cm$12cm\times 12cm\times 2cm$ and its length 3 meter is used as column with one of its end fixed and the other end free. Take 

E=2×105 N/mm2$E=2\times {10}^{5}N/m{m}^2$ Find crippling load.

Solution

Given:

L=3m=3000mmE=2×105N/mm2$$\begin{gathered} L=3m=3000mm \\ E=2\times {10}^{5}N/m{m}^2 \end{gathered}$$

To find: P = ?

Position of neutral axis (from bottom)y¯=A1y1+A2y2A1+A2=100×20×50+120×20×110(100×20)+(120×20)=82.7mm$$\begin{gathered} Positionofneutralaxis(frombottom) \\ \overline{y}=\frac{A_1{y}_1+A_2{y}_2}{A_1+A_2}=\frac{100\times 20\times 50+120\times 20\times 110}{(100\times 20)+(120\times 20)}=82.7mm \end{gathered}$$

Moment of inertiaIXX=bd312+A1h21+bd312+A2h22=120×20312+120×20(37.30−10)2+20×100312+20×100(82.70−60)2IXX=4.566×106mm4−−−−−−−−−−−−−−−−−−−$$\begin{gathered} Momentofinertia \\ {I}_{XX}=\frac{b{d}^3}{12}+A_1{h}_1^2+\frac{b{d}^3}{12}+A_2{h}_2^2 \\ =\frac{120\times {20}^3}{12}+120\times 20(37.30-10{)}^2+\frac{20\times {100}^3}{12}+20\times 100(82.70-60{)}^2 \\ \underset{\_}{I_{XX}=4.566\times {10}^{6}m{m}^4} \end{gathered}$$

Since one end is fixed and other end is free

Effective length, Le=2L=2×3000=6000mm$L_e=2L=2\times 3000=6000mm$

Crippling load P=π2EIL2e$P=\frac{{\pi }^{2}EI}{L_e^2}$

P=π2×2×105×4.566×106(6000)2=250×103 NP=250 kN