Note :- consier minimum size of ISA 50$\times 50\times$6mm for truss

used min size of bolt d=16m

do-18mm e=30mm, P=40mm

a) Design at Tie member: consider 2 ISA B/B of 10mm the G.P

1) Design of bolted connection Volsb=$\frac{Fub\times Anb\times Nb}{\sqrt {3}\gamma mb}$=$\frac{2\times 157\times400}{\sqrt{3}\times1.25}$

=58.01$\times10^{3}$n

Kb=[$\frac{e}{3do},\frac{p}{3do}-0.25, \frac{Fub}{Fu},1]$min

=[0.56,0.49,0.98,1]min

Kb=0.49

Vdpb=2.5$\times5b\times d\times t\times\frac{fu}{\gamma mb}$min

=2.5$\times0.49\times19\times10\times\frac{400}{1.25}$

=62.72$\times10^{3}$N

B.V =58.01kN(mm)

No of bolt required=$\frac{23.39}{85.01}$=0.4=2

=2bolt(minimum)

provide 2-16 mm $\phi$ bolt in member L$_{1}L_{2}$ as end connection

2) Design L$_{1}$L$_{2}$ for Tension

Area required =$\frac{1.2\times Pu\times Fmo}{FY}$=1.2$\times23.39$$\times 10^{3}\times$2.52 Area=123.5mm$^{2}$ from steel table select z ISA 60$\times$60$\times$6mm having A prav=1368mm$^{2}$ $\gamma$xx=18.2mm

$\gamma$yy=28.5mm

Take $\gamma$min=18.2mm

A net=1368-[2$\times18\times4]$=1152mm$^{2}$

Tdg=Ag$\times\frac{ fy}{\gamma mo}$=1368$\times\frac{250}{1.10}$

=310.91$\times10^{3}$ N

Tdn==$\alpha\times An\times \frac{F u}{\gamma m_{1}}=0.6\times 1152\times\frac{410}{1.25}$

=226.71$\times10^{3}$N

Block shear

Avg=70$\times$6=420mm$^{2}$

Avm=[70-1.5$\times18]\times$6

=258mm$^{2}$

Atg=25$\times$6=150mm$^{2}$

Atn=(25-0.5$\times18)\times6=96mm^{2}$

Tdb$_{1}=2\times[\frac{AVg\times fy}{\sqrt{3}\times \gamma mo}+\frac{0.9 Atn\times fy}{\gamma mo}]$

=167$\times 10^{3}$N

Tdb$_{2}$=2$\times[\frac{0.9\times Avn\times Fu}{\sqrt{3}\times \gamma m}+\frac{Afg\times Fy}{\gamma mo}]$

=156.13$\times10^{3}$N

Tensile capacity=156.13kN$\gt$ 23.39kN safe in Tension