Given

load=250KN

Design bolted connection

1) Shearing strength of bolt $\frac{fub\times(nn.Anb)}{\sqrt{3}.\gamma mb}$

=$\frac{400\times(1\times245)}{\sqrt{3}\times1.25}$

Vdsb=45.26KN

strength of bolt=45.26KN

No of bolt =$\frac{250}{45.26}=5.52\sim$7 Nos

Tdg=$\frac{Ag.\times fy}{\gamma mo}$

Ag=$\frac{Tdg\times \gamma mo}{fy}$

=$\frac{250\times1.1}{250}$

=1100mm$^{2}$

provide 90$\times60\times10mm$ Aprov=1401mm$^{2}$

e=1.7$\times$22=40mm

e=2.5$\times$d=2.5$\times$20=50mm

Assume gauge distance g=50mm

Tdn=$0.9\times Anc\times\frac{fu}{\gamma ml}+BAgo\frac{fy}{\gamma mo}$

or

Tdn=$\gamma\times An\frac{fu}{\gamma m1}$

Anc=$(90-\frac{10}{2})\times10-(22\times10)$

=630mm$^{2}$

Ago=$(60-\frac{10}{2})\times10=550mm^{2}$ …