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 …