design the connection

GIven

ISLB=500 at 950N/m $\frac{750}{10}$=75kg/m

D=500, tf=14.1, tw=3.22 --- steel table

ISHB 300 at 630N/m $\frac{630}{10}$=63kg/m

D=300, tf=10.6 tw=9.4 steel table Anb=2245

Assume ISA =100$\times$100$\times$8

connecting ISA to the web of beam

Step I

Shear capacity of bolt=$\frac{fub\times(nm Anb)}{\sqrt{3}\times ymb}$

=$\frac{400\times(2\times 245)}{\sqrt{3}\times1.25}$

vdsb=90.52kN

Step II bearing strength of bolt =2$\times$kb$\times$d$\times$t$\times$fu

kB=$\frac{e}{3do}$e=1.7$\times$do

e=1.7$\times$22=37.4$\approx$40

$\frac{e}{3do}=\frac{40}{3\times 22}$=0.60

P=2.5$\times$d

p=2.5$\times$20=50mm

$\frac{p}{3do}$-0.25=

$\frac{50}{3\times22}$-0.25=0.51

$\frac{fub}{fu}=\frac{400}{410}$=0.97

min value pf kb=0.51

Vdpb=2$\times$kb$\times$d$\times$9twx$\times$fu

=2$\times$ 0.51$\times$ 20$\times$8$\times$410

Vdpb=976.94

bolt value=min value of vdsb and vdpb=66.941

Step III No of bolt=$\frac{load}{bolt / value}$

$\frac{150}{76.94}$=3.89$\approx$4 No

=1.94$\approx$2 bolt

provide 2 no of 20 mm bolt edge distance 40mm

pitch distance =50mm

length of angle=2e+$\sum$p

=2$\times$40+50\=130mm

connecting ISA with column flange

Shear capacity of bolt Vdsb=$\frac{fub\times(nn.Anb)}{\sqrt{3}\times ymb}$

=$\frac{400\times(1\times 245)}{\sqrt{3}\times1.25}$

vdsb=45.26KNI

shearing capacity of bolt=2kb$\times$ d$\times$ tw$\times$ fu

$\frac{e}{3do}=\frac{40}{3times 22}$=0.56

$\frac{p}{3do}-0.25=\frac{50}{3\times 22}$=0.25=0.51

$\frac{fub}{fu}=\frac{400}{410}$=0.97

Vdpb=2$\times$0.51$\times$20$\times $8$\times$410

Vdpsb=66.91kN

No of bolt=$\frac{load}{bolt \ value}\frac{150}{45.26}$=3.31$\approx$ 4No

length of angle=2e+$\sum$ p

=2$\times$40+50$\times$3

=230m

provide 4 no of 20mm bolt edge distnace $\lt$ 10mm pitch distance 50 mm

Check of thickness of angle

Vd=$\frac{Av\times fym}{\sqrt{3}\times ymo}$

=$\frac{2(100\times8)\times 250}{\sqrt{3}\times 1.1}$

vd=209.94kN Connection is sage

Vd$\gt$ end reaction (150) connection is safe