Given

ISMB 250

bf=125m tf=12.5m tw=6.9

$\gamma$1=13m

ISHB 200 at 37.92 bf=200mm tf=9mm

tw=6.1mm

R=85kN(factored load)

$\phi$ of bolt=16mm

Anb=157mm$^{2}$

fub=4$\times$100=400N/mm$^{2}$

fy=250N/mm$^{2}$

fu=410N/mm$^{2}$

required no of bolts

shearing capacity of bolt =$\frac{fub\times Anb\times nn}{\sqrt{3}\times\gamma mb}$

=$\frac{400\times 157\times 1}{\sqrt{3}\times 1.25}$

Vdsb=29kN

bearing capacity of bolt=2$\times$K$\times$t$\times$d$\times$fu

Kb=$(\frac{e}{3d0},\frac{p}{3d0}-1, \frac{fyb}{fu},1)$ min value

Vdpb=2$\times0.56\times9\times16\times 410$

Vdps=66.129kN

No .of bolts=$\frac{factored \ load}{bolt \ value}$ (min value Vdsb & Vdpb)

=$\frac{85}{29}$=2.39$\approx$4Nos

Shear capacity of seat angle

=$\frac{bf}{\sqrt{3}\times ymo}$

=$\frac{125\times 10\times 250}{\sqrt{3}\times 1.1}$

=164$\gt$35

Step II length of angle supporting beam(b)=125mm

Step III bearing length of route line of beam(b)

(b)=$\frac{R}{\frac{tw\times fyh}{ymo}}$==$\frac{80}{\frac{6.9\times250}{1.1}}$=54.2mm

b=54.2mm

Step IV length of out standing leg=54.2+10=64.2

Assume 75mm

provide 150$\times75\times$10

b1=b-(tf-$\gamma$1)

=54.2-(12.5+13)

b1=28.7mm

b2=(b1+10)-ta-$\gamma$2

=(28.7+10)-10-10

b2=18.7mm

Step V bending moment of root angle =$\frac{R}{b1}\times b2\times\frac{b2}{2}$

=$\frac{85}{28.7}\times18.7\times \frac{18.7}{2}$

mu=517.83kN/mm

Step VI

moment resisting capacity

md=1.2$\times\frac{250}{1.1}\times \frac{12.5\times10^{3}}{6}$

md=568.18kN.mm

md$\gt$mu hence safe

Step VIII Shear strength of beam

Vd=$\frac{aw\times fyw}{\sqrt{3}\times ymo}$=$\frac{250\times6.9)\times 250}{\sqrt{3}\times1.1}$

Vd=226.35