two channel back to back * talce steel fe 410

Given

Pu=1100 kN

=1100$\times 10^{3}N$

leff=KL

=1$\times$9000(both end hinged)

Step I Pu=Area $\ast$ fcd

1100$\times10^{3}=Area\ast$150 Assume fcd=150

Area=733.33mm$^{2}$

Area of each channel =$\frac{7333.330}{2}=3666.67mm^{2}$

provide ISMC 300

Aprov =4564mm$^{2}$

Total provide=2$\times$4564

=9128mm$^{2}$

$\gamma$xx=118.1mm

$\lambda$act=$\frac{9000}{118.1}$

$\lambda$act=76.20

$\lambda$e=1.1$\times \lambda$act

=1.1$\times$76.20

=$\lambda$e=83.83

80 …