flange steel is 410 and bolt 4.6 grade

Given P=225kN

e=300mm

fub=400

Assume dia bolt=24mm

Step I =find bolt value

Shear capacity of bolt=$\frac{fub\times An\times nn}{\sqrt{3}.ymb}$

=$\frac{400\times 353\times 1}{\sqrt{3}\times 1.25}$

Vdsb=65.22KN

Step II No of bolts required$\sqrt{\frac{6m}{n\times p\times Bv}}$

m=P.e=225$\times$003=67.5kN.m

n=2

No of bolt required=$\sqrt{\frac{6\times 67500}{2\times 601\times65.22}}$

Pitch p=(2.5$\times$24)

=60mm=7.29

No of bolts required=8

Step III Total height of plate

e=1.7$\times$2.6=44.2

e=1.7ds=45mm

H=23+$\sum$p pitch

=2$\times 45+(87\times 60$)

H=510mm

h=510-45=465mm

Step IV

Interaction equation

$(\frac{Vsb}{Vdb})^{2}+(\frac{Tb}{Tdb})^{2} \leq 1.0$

Step V

direct shear=$\frac{Reaction}{No. \ of \ bolt}=\frac{225}{2\times 8}$=14.06kN

max tensile force in critical bolt=$\frac{m^{'}\times yn}{\sum y^{2}}$

m$^{'}=\frac{m}{1+(\frac{2h}{2l}\times \frac{\sum y}{\sum y^{2}})}$

$\frac{67500}{1+(\frac{2\times465}{21}\times\frac{\sum y}{\sum y^{2}})}$

y1(45+60)-66.48=38.57

y2=38.57+60=98.57

y3=98.57+60=158.87

y4=158.57+60=218.57

y5=218357+60=278.57

y6=278.57+60=358.57

y7=338.57+60=398.57

$\sum$y=2(38.57+98.57+158.5+218.57+278.57+338.57+398.57)

$\sum$y=3059.98m

$\sum y^{2}=2(38.557^{2}+98.57^{2}+158.57^{2}+218.57^{2}+278.57^{2}+338.57^{2}+398.57^{2})$

$\sum y^{2}$=878419.83mm

m${'}=\frac{67500}{1+(\frac{2\times 465}{21}\times\frac{3059.98}{870419.83})}$

m$^{'}$=58406 kN,,

Te=$\frac{m^{'}\times yn}{\sum y^{2}}$

yn=398.57mm

=$\frac{58406\times398.57}{870419.83}$

Te=26.74kN

Tb=0.9$\times fub \times Anb$

=0.9$\times $ 400$\times $ 353

Tb=127.2800kN

$(\frac{Vsb}{Vdb})^{2}+(\frac{Tb}{Tdb})^{2}\leq$1

=$(\frac{14.2}{65.22})+(\frac{26.74}{127.28})\leq 1$

0.1$\leq$1 hence safe