Zp=M$\times\frac{8mo}{fy}=3.34\times 10^{6} \times\frac{1.10}{250}$=14696mm$^{3}$

From shear table select ISLC 100 mm having h=100mm, bp=50mm , tf=6.4mm

tw=4mm $\gamma$=6mm, Ixx=164.7$\times 10^{4}$mm Iyy=24.8$\times 10^{4}mm^{4}$, Zxx=32.9$\times 10^{3} mm^{3}$ Zyy=7.3$\times 10^{3}mm^{3}$

Zpz, prov=38.09$\times 10^{3} mm^{3}$

Zpy=Zy$\times$ s

=7.3$\times 10^{3} \times 1.1576$=8.45$\times 10^{3} mm^{3}$

h$_{1}$=74.3mm

sectional classification:

$\frac{b}{tf}=\frac{50}{6.4}=7.8\lt 9.4\epsilon$

$\frac{d}{tw}=\frac{74.3}{4}=18.58\lt 42\epsilon$

section is plastic

check for moment capacity of section

Mdz=Bd$\times ZPz\times\frac{fy}{\gamma y}{\gamma mo}$....(Bb=1)

=1$\times 38.09\times 10^{3}\frac{ 250}{1.10}$

=5654.55$\times 10^{3}$N.mm

=8.66$\times10^{6}$N.mm

Mdy=Bd$\times Zpy\times \frac{gy}{\gamma y}{\gamma mo}=12\times 8.45\times 10^{3}\times\frac{250}{1.10}$

=1.92$\times10^{6}$ N.mm

Using Interaction Equation

$\frac{Mz}{Mdz}+\frac{My}{MDy}\leq$1

$\frac{3.34\times 10^{6}}{8.66\times 10^{6}}+\frac{1.47\times 10^{6}}{1.92\times 10^{6}} \leq$ 1

1.15 $\lt$ unsafe

check for shear capacity (Z directions)

Vd=$\frac{Fy\times h.tw}{\sqrt{3}\times \gamma mo}\frac{250\times 100\times 4}{\sqrt{3}\times 1.10}$=52.49kN

Also

0.6Vd=31.49kN $\gt$ 2.54Kn sfe

Shear capacity in y direction

Vd=$\frac{fy\times bf\times tf}{\sqrt{3}\times \gamma mo}=\frac{250\times (2\times 50)\times 6.4}{\sqrt{3}\times 1.10}$

=83.98$\times 10^{3}$N

Also 0.6Vd=50.39 kN $\gt$ 1.47kN

Check for deflection

$\int$max=$\frac{span}{300}=\frac{4000}{300}=$ 13.33 mm

working $\int=\frac{5}{384}\times \frac{wL^{4}}{EI}=\frac{5}{384}\times \frac{0.845\times 4000^{4}}{2\times 10^{5}\times 164.7\times 10^{4}}$

$\int=8.55mm\lt \int$ max