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

An$\gt $Ag required

Tdn=08$\times$1180$\times\frac{410}{1.25}$=309.63KN

block shear

Tdb$_{1}=\frac{Avg.fu}{\sqrt{3}.\gamma m0}+\frac{0.9,Afn.fu}{\gamma m1}$

Tdb$_{2}=\frac{0.9\times Avn.fu}{\sqrt{3}.\gamma m1}+\frac{Afg.fy}{\gamma mo}$

Afg=40$\times$10=400mm$^{2}$

Afn=$(40-\frac{1}{2}\times22)\times10=290mm^{2}$

Tdb$_{1}=\frac{3400\times250}{\sqrt{3}\times1.1}+0.9\times290\times\frac{410}{1.25}$

Tdb$_{1}$=531.74KN

Tdb$_{2}=\frac{0.9 \times 1970\times410}{\sqrt{3}\times1.25}+\frac{400\times250}{1.1}$

Tdb$_{2}=426.66\times10^{3}$

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

=$\frac{1401\times250}{1.1}$

Tdg=318.40KN

Design strength of angle =309.63KN

(least of 318.40,426.66)

Total tensile strength of angle 309.63$\gt$250KN

Hence design is safe