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Design stiffen seat connection for ISMB 350 transmitting factor and reaction 320kN to column section ISHB 300 Use 410 steel and 4.6 grade bolt
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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)=Rtw×fyymo

=3208.1×2501.1

b=173.8mm

provide clearance 10mm

Required length of outstandard leg

=173.83+10 =183 200mm

provide seat angle size 200×150×10

Step II

bearing area required for stiffner angle

ABr=rfy/ymo

=320×103250/1.1

=1408mm2

Abr each angle=1408/2=704mm2

provide 2ISA 90×60×8mm

length of outstanding leg of stiffness angle

14.6× ta

14×(250250)×8=112mm

90<112 Hence ok

Assume distance from on reaction from column fange

R=2002=outstanding leg of seat angle2

(VsbVdb)2+(TbTdb)21.0

shear capacity of bolt=fub×Anb.n3×1.25

Vdsb=65.22kN

bearing capacity of bolt=2×kb×t×d×fu

=2×0.5×8×24×410

Vdpb=80.29kN

least bolt value 65.22kN

No of bolts required per row=6mn×p×bv

=6×(320×103×200/22×60×65.22

=49.5

provide 5 bolts at each row

Dept of stiffen angle=2e+p

=2×45+(4×60)

=350mm

Tensile force in critical bolt=m.yny2

enter image description here

NA=h7

h=D-e\=350-45

=305mm

NA=305/7=43.57

y1=60mm

y2=120mm

y3=180mm

y4=240mm=yn

Fy=2[60+120+180+240]

Fy=1200mm

Fy2=2[602+1202+2802+2402]

Fy2=21600mm2

m=m1+2n21×yy2

=(320×103)×(200/2)1+(2×30521×120021600)

m=27.87×106N.mm

m=(27.87×106)×2240216000

m=30.97×103N

Tdb=0.9×fub×Anb1.25

=0.9×400×3531.25

Tdb=101.67kN

Vsb=RNo of bolts

=320×10310=32kN

Vsb=32kN

Put in equation (1)

(32×10365.22×103)2+(30.97×103101.67×103)2< 1

0.3<1.0

hence safe

enter image description here

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=Rtw×(fyyo) required length of outstanding leg of angle =b+clearance (3 to 15mm)
  3. Bearing length on seat angle b1=b-(tf+γb)
  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 )×(b22)=(Rb1)×(b2)×(b22)design moment (mo)md=(1.2)×(2)\times(fyγmo) design moment > ultimate moment md>mu Hence safe
  6. Shear capacity angle=bf×ta×fu3ymo shera capacity > given load then safe
  7. Shear strength of beam Vd=Aw×fyw3×ymo>given load
  8. Design of wield

    1 find p

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

    1. strength of weld=fu3ymw
  9. provide clip angle or cleat angle (80×60×6mm)at top with 3mm weld

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