written 6.1 years ago by |
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×nn√3.ymb
=400×353×1√3×1.25
Vdsb=65.22KN
Step II No of bolts required√6mn×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)2≤1.0
Step V
direct shear=ReactionNo. of bolt=2252×8=14.06kN
max tensile force in critical bolt=m′×yn∑y2
m′=m1+(2h2l×∑y∑y2)
675001+(2×46521×∑y∑y2)
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′×yn∑y2
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)2≤1
=(14.265.22)+(26.74127.28)≤1
0.1≤1 hence safe