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Design a bracket connection to transfer and end reaction of 225 kN factored as shown end reaction from gurder is distance 300mm from column phase design bolted connection fro Tglange with column
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flange steel is 410 and bolt 4.6 grade

Given P=225kN

e=300mm

fub=400

Assume dia bolt=24mm

Step I =find bolt value

Shear capacity of bolt=fub×An×nn3.ymb

=400×353×13×1.25

Vdsb=65.22KN

Step II No of bolts required6mn×p×Bv

m=P.e=225×003=67.5kN.m

n=2

No of bolt required=6×675002×601×65.22

Pitch p=(2.5×24)

=60mm=7.29

No of bolts required=8

Step III Total height of plate

e=1.7×2.6=44.2

e=1.7ds=45mm

H=23+p pitch

=2×45+(87×60)

H=510mm

h=510-45=465mm

Step IV

Interaction equation

(VsbVdb)2+(TbTdb)21.0

Step V

direct shear=ReactionNo. of bolt=2252×8=14.06kN

max tensile force in critical bolt=m×yny2

m=m1+(2h2l×yy2)

675001+(2×46521×yy2)

enter image description here

y1(45+60)-66.48=38.57

y2=38.57+60=98.57

y3=98.57+60=158.87

y4=158.57+60=218.57

y5=218357+60=278.57

y6=278.57+60=358.57

y7=338.57+60=398.57

y=2(38.57+98.57+158.5+218.57+278.57+338.57+398.57)

y=3059.98m

y2=2(38.5572+98.572+158.572+218.572+278.572+338.572+398.572)

y2=878419.83mm

m=675001+(2×46521×3059.98870419.83)

m=58406 kN,,

Te=m×yny2

yn=398.57mm

=58406×398.57870419.83

Te=26.74kN

Tb=0.9×fub×Anb

=0.9× 400× 353

Tb=127.2800kN

(VsbVdb)2+(TbTdb)21

=(14.265.22)+(26.74127.28)1

0.11 hence safe

enter image description here

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