Given

ISMB 350 bf==140mm tw=8.1mm

ISHB 300 bf=250mm tf=10.6mm

fub=400mpa

fy=250mpa

fu=410mpa

Required stiffness seat angle connection

Step I

length of seat angle bf=140mm

bearing length of seat angle (b)=$\frac{R}{tw\times \frac{fy}{ymo}}$

=$\frac{320}{8.1\times\frac{250}{1.1}}$

b=173.8mm

provide clearance 10mm

Required length of outstandard leg

=173.83+10 =183$\approx$ 200mm

provide seat angle size 200$\times$150$\times$10

Step II

bearing area required for stiffner angle

ABr=$\frac{r}{fy/ymo}$

=$\frac{320\times 10^{3}}{250/1.1}$

=1408mm$^{2}$

Abr each angle=1408/2=704mm$^{2}$

provide 2ISA 90$\times$60$\times$8mm

length of outstanding leg of stiffness angle

$\neq14.6\times$ ta

$\neq14\times(\frac{250}{250})\times8$=112mm

90$\lt$112 Hence ok

Assume distance from on reaction from column fange

R=$\frac{200}{2}$=$\frac{outstanding \ leg \ of \ seat \ angle}{2}$

$(\frac{Vsb}{Vdb})^{2}+(\frac{Tb}{Tdb})^{2}\leq$1.0

shear capacity of bolt=$\frac{fub\times Anb.n}{\sqrt{3}\times1.25}$

Vdsb=65.22kN

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

=2$\times$0.5$\times$8$\times$24$\times$410

Vdpb=80.29kN

least bolt value 65.22kN

No of bolts required per row=$\sqrt{\frac{6m}{n\times p\times bv}}$

=$\sqrt{\frac{6\times(320\times 10^{3}\times 200/2}{2\times 60\times 65.22}}$

=49.5

$\approx$ provide 5 bolts at each row

Dept of stiffen angle=2e+$\sum$p

=2$\times$45+(4$\times$60)

=350mm

Tensile force in critical bolt=$\frac{m.yn}{\sum y^{2}}$

NA=$\frac{h}{7}$

h=D-e\=350-45

=305mm

NA=305/7=43.57

y1=60mm

y2=120mm

y3=180mm

y4=240mm=yn

$\sum$ Fy=2[60+120+180+240]

$\sum$Fy=1200mm

$\sum Fy^{2}=2[60^{2}+120^{2}+280^{2}+240^{2}$]

$\sum Fy^{2}=21600mm^{2}$

m$^{'}$=$\frac{m}{1+\frac{2n}{21}\times\frac{\sum y}{\sum y^{2}}}$

=$\frac{(320\times10^{3})\times(200/2)}{1+(\frac{2\times305}{21}\times \frac{1200}{21600})}$

m${'}$=27.87$\times10^{6}$N.mm

m=$\frac{(27.87\times10^{6})\times 2240}{216000}$

m=30.97$\times 10^{3}$N

Tdb=$\frac{0.9\times fub\times Anb}{1.25}$

=$\frac{0.9\times 400\times 353}{1.25}$

Tdb=101.67kN

Vsb=$\frac{R}{No \ of \ bolts }$

=$\frac{320\times 10^{3}}{10}$=32kN

Vsb=32kN

Put in equation (1)

$(\frac{32\times 10^{3}}{65.22\times 10^{3}})^{2}+(\frac{30.97\times 10^{3}}{101.67\times 10^{3}})^{2}\lt$ 1

0.3$\lt$1.0

hence safe

Unstiffed seat angle connection(welded)

1. length of angle supporting in beam=width of flang of beam (bf)
2. bearing length required at the rout line of beam b=$\frac{R}{tw\times(\frac{fy}{yo})}$ required length of outstanding leg of angle =b+clearance (3 to 15mm)
3. Bearing length on seat angle b1=b-(tf+$\gamma $b)
4. length of seat angle subjected to Bm b1=(b1+clerance )-(ta)-(ra)
5. B.M to be seat angle such to mu=(load on angle )$\times(\frac{b2}{2}) =(\frac{R}{b1})\times (b2)\times(\frac{b2}{2})$design moment (mo)md=(1.2)$\times$(2)\times$(\frac{fy}{\gamma mo})$ design moment $\gt$ ultimate moment md$\gt$mu Hence safe
6. Shear capacity angle=bf$\times ta\times\frac{fu}{\sqrt{3}ymo}$ shera capacity $\gt$ given load then safe
7. Shear strength of beam Vd=$\frac{Aw\times fyw}{\sqrt{3}\times ymo}\gt$given load

8. Design of wield

   1 find p

   2 max size of weld =tp-15 min size of weld

   1. strength of weld=$\frac{fu}{\sqrt{3}ymw}$

9. provide clip angle or cleat angle (80$\times60\times$6mm)at top with 3mm weld