written 6.1 years ago by |
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)2≤1.0
shear capacity of bolt=fub×Anb.n√3×1.25
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.yn∑y2
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×∑y∑y2
=(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
Unstiffed seat angle connection(welded)
- length of angle supporting in beam=width of flang of beam (bf)
- bearing length required at the rout line of beam b=Rtw×(fyyo) required length of outstanding leg of angle =b+clearance (3 to 15mm)
- Bearing length on seat angle b1=b-(tf+γb)
- length of seat angle subjected to Bm b1=(b1+clerance )-(ta)-(ra)
- 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
- Shear capacity angle=bf×ta×fu√3ymo shera capacity > given load then safe
- Shear strength of beam Vd=Aw×fyw√3×ymo>given load
Design of wield
1 find p
2 max size of weld =tp-15 min size of weld
- strength of weld=fu√3ymw
provide clip angle or cleat angle (80×60×6mm)at top with 3mm weld