1) M=150kN.m v=210kN

2) Zpreq=$\frac{m\gamma mo}{fy}$=$\frac{150\times 10^{6}\times 1.1}{250}$

Zpreq=660$\times10^{3}mm^{3}$

select ISLB 350 at 49.5 kg/m having Zp=851.11$\times 10^{3}mm^{3}$

D=350mm bf=165mm tf=11.4mm

tw=7.4mm $\gamma_{1}$=1.6

IZ=13125.3$\times 10^{4}mm^{4}$

d=350-2$\times11.4-2\times$16

d=295.2mm

3) sectional classification

$\frac{b}{tf}=\frac{bf/2}{tf}=\frac{165/2}{11.4}=7.2\lt 9.4\epsilon$

$\frac{d}{tw}=\frac{295.2}{7.4}=39.89\lt 84\epsilon$

seciont is plastic

4) Vd=$\frac{AV\ast fy}{\sqrt{3}\gamma mo}$

=$\frac{(350\times7.4)\ast 250}{\sqrt{3}\times 1.1}$

Vd=339.84$\gt$ V=210kN safe

5) 0.6 Vd=0.6$\times$339.84

=203.91kN

v(210)$\gt$ 0.6Vd=203.91kN

high shear

mdv=md-B(md-nfd)$\leq\frac{1.2\ast ze\ast fy}{\gamma mo}$

B=$(\frac{zv}{vd}-1)^{2}=(\frac{2\times 210}{339.84}-1)^{2}$=0.055

md=$\frac{Bd\ast Zp\ast fy}{\gamma mo}$

=$\frac{1\ast (851.11\times 10^{3})\times 250}{1.1}$

md=193.43$\times 1.^{6}$N.mm

mfd=$\frac{Zd\ast fy}{\gamma mo}$

zfd=zp-$\frac{Aw\ast D}{4}$

=$851.11\times 10^{3}-\frac{(dw\times tw)350}{4}$

=851.11$\times10^{3}-\frac{(295.2\ast 7.4)350}{4}$

=659.968$\times^{3}mm^{3}$

mfc=$\frac{659.968\times 10^{3}\times 250}{1.1}$

=150$\times 10^{6}$Nmm

MDv=193.43$\times 10^{6}$-.0055(193.43$\times 10^{6}---150\times 10^{6})$

mdv=191.05$\times10^{6}N.mm \gt$ m(150kN.m)

Analysis is safe