to back in opposite site of gusset plate 2)Angle place same side of gusset

Angle are tack bolted & provided wotj min of two bolts

Given P=178KN

l=3.05m

leff=0 8$\times$3.08

=2464mm

consider fed=150mpa

Pd=Ae$\times$fcd

178$\times10^{3}=Ae\times150$

back to back connection Ae=1186.67mm$^{2}$ Area required

Area of each angle=$\frac{1186.67}{2}$=56.33mm$^{2}$

Assume fcd=90mpa

Pd=Ae$\times$fcd

178$\times10^{3}$=Ae$\times$90

Ae=1977.78mm$^{2}$

Area of each angle = $\frac{1977.78}{2}=988.88mm^{2}$

select Anlge=$\times$80$\times$50$\times$10

A=1202mm$^{2}$

Area provided =1202$\times$2=2404mm$^{2}$

$\gamma$xx=$\sqrt{\frac{$Ixx}{A}}$ $\gamma uu$=$\sqrt{\frac{Iuu}{A}}$

Iuu=22.10$\times10^{4}mm^{4}$--steel table

Iyy$^{1}$=Iyy1+Iyy2

=2(Iyy+Ah$^{2})$

=$2(22.10\times10^{4}+1202\times(\frac{10}{2}+13.2)^{2})$

Iyy=$1.23\times10^{6}mm^{6}$

$\gamma yy=\sqrt{\frac{1.23\times10^{6}}{2404}}$

$\gamma yy$=22.63mm

$\lambda=\frac{kL}{\gamma yy}=\frac{2464}{22.69}=108.56$

100 $\ \ \ \ \ \ $ 107

108 $\ \ \ \ \ \ \ $ fcd

110 $\ \ \ \ \ \ \ \ $ 94.6

$\frac{110-100}{110-106.56}=\frac{94.6-107}{94.6-fcd}$

$\frac{10}{1.44}=\frac{-12.4}{94.6-fcd}$

-17.856=946-10fc

fcd=96.38 N/mm$^{2}$

Pd=A$\times$fcd

=2404$\times$96.38

Pd=231.71 KN =178 KN hence safe

Case second

angle connect same side of gusset plate

Iyy=Iyy1+Iyy2

=2(22.1$\times10^{4}+1202(17.2)^{2})$

Iyy=860$\times87\times10^{3}mm^{4}$

$\gamma yy=\sqrt{\frac{860.87\times10^{3}}{2404}}$=18.92mm

$\lambda=\frac{2464}{18.92}$=130.20

130 743 $\frac{140=130}{140-130.20}=\frac{66.2-743}{66.2-fcd}$

fcd=74.13N/mm$^{2}$

Pd=2404$\times$74.13

=178.22KN$\gt$178 hence safe