Given

w=40kN (Service load)

leff=40(of beam)

Required - Design laterally supported beam

factored load=40$\times$1.5=60kN

w=$\frac{60}{4(leff)}$=15kN/m

Step I determine max factor shear face (v) and Bm(m) for given loading and support condition

v=$\frac{wl}{2}=\frac{15\times4}{2}$=30kN

m=$\frac{wl^{2}}{8}=\frac{15\times4^{2}}{8}$=30kN.m

Step II Trial section

zpreq=$\frac{m.req}{fy}$

=$\frac{(30\times10^{6})\times1.1}{250}$

Zpreq=132$\times10^{3}mm^{3}$

ISMB175 zp=166.08$\times10^{3}mm^{3}$

h=175 bf=90mm tf=8.6

tw=5.5 r=10

IZ=12.75$\times10^{4}mm^{4}$

Ze=Zxx=145.4$\times 10^{3}mm^{3}$

Step III check section IS pg 18

$\frac{b}{tf}=\frac{bf/2}{tf}=\frac{90/2}{8.6}$=5.23

$\epsilon=\sqrt{\frac{250}{fy}}$=1

$\frac{b}{tf}$=5.23$\lt$9..4$\epsilon$

$\lt$9.4$\times$ 1

$\lt$ 9.4

$\frac{d}{tw}=\frac{D-2tf-2R_{1}}{tw}$=$\frac{137.5}{5.5}$=25

$\frac{d}{tw}=25\lt 84\epsilon$ secton is plastic

Step IV Design shear strenght Vd

Vd=$\frac{Vn}{\gamma mo}=\frac{Av\ast fy}{\sqrt{3}.\gamma mo}$

Av=hf$\ast$tw

=175$\ast$ 5.5=962.5

=$\frac{962.5\times 250}{\sqrt{3}\times1.1}$

Vd=126.30$\times 10^{3}$N Vd=126.30$\times10^{3}N\gt V=30\times 10^{3}$N

Analysis is sage

Step V Design bearing strength md

0.6Vd=75.78$\times$10$^{3}$N V$\lt$ 0.6Vd--low shear

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

Bb=1 plastic section

fy=250

zp=166.08$\times10^{3}mm^{3}$

md=$\frac{1\times250\times 166.08\times 10^{3}}{11}$

md=37.75kN.M

md$\lt\frac{1.2\times ze\times fy}{\gamma mo}$

$\lt 1.2\times 145.\times 4\times10^{3}\times \frac{250}{1.1}$

md$\leq$39.65kN.m

Analysis is safe

Step VI web buckling

fwb=(b1+n1)$\ast $tw$\ast $ fcd

b1=75 Assume

N1=$\frac{D}{2}=\frac{175}{2}$=87.5

fcd=$\frac{Kl}{\gamma min}=\frac{0.7\times d}{t/\sqrt{12}}$ d=D-2tf-2R$_{1}$=137.5

=$\frac{0.7\times137.5}{5.5/\sqrt{12}}$

$\frac{Kl}{\gamma min}=\lambda$=60.62

60$ \ \ \ \ \ $ 168

60.72 $ \ $ fcd

70 $ \ \ \ \ \ \\ $ 152

fcd=167

fwb=(75+87.5)$\ast$5.5$\ast$167--pg 67

fwb=149.256$\times10^{3}\gt$ V

2) $\frac{d}{tw}\lt$ 67.8

$\frac{137.5}{5.5}\lt$67

25$\lt$67

shear buckling check is not required

2) Web crippling

fwx=$\frac{[(b_{1}+n_{2})+tw]+fyw}{\gamma mo}$

N$_{2}$=2.5[tf+$\gamma_{1}$]

=2.5[8.6+10]

n$_{2}$=46.5

fwc=$\frac{(75\times46.5)\ast 5.5\ast250}{1.1}$

Fwc=151.88$\times10^{3}\gt V=30\times10^{3}$kN

3) Deflection

$\int act \lt \int$limit

$\int act=\frac{5 wl^{4}}{384EI}$

permissible =$\frac{span}{300}$

$\frac{4000}{300}$

$\frac{\frac{5}{384}\times(\frac{15}{1.5})\ast(1000)^{4}}{(2\times10^{5}\ast(1274\times10^{4})}$=13.33

=13.08

1308$\lt \int$ perm=13.33

Analysis is sage

Select ISMB 175 of beam section laterally supported through out