written 5.7 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=$\frac{fub\times An\times nn}{\sqrt{3}.ymb}$
=$\frac{400\times 353\times 1}{\sqrt{3}\times 1.25}$
Vdsb=65.22KN
Step II No of bolts required$\sqrt{\frac{6m}{n\times p\times Bv}}$
m=P.e=225$\times$003=67.5kN.m
n=2
No of bolt required=$\sqrt{\frac{6\times 67500}{2\times 601\times65.22}}$
Pitch p=(2.5$\times$24)
=60mm=7.29
No of bolts required=8
Step III Total height of plate
e=1.7$\times$2.6=44.2
e=1.7ds=45mm
H=23+$\sum$p pitch
=2$\times 45+(87\times 60$)
H=510mm
h=510-45=465mm
Step IV
Interaction equation
$(\frac{Vsb}{Vdb})^{2}+(\frac{Tb}{Tdb})^{2} \leq 1.0$
Step V
direct shear=$\frac{Reaction}{No. \ of \ bolt}=\frac{225}{2\times 8}$=14.06kN
max tensile force in critical bolt=$\frac{m^{'}\times yn}{\sum y^{2}}$
m$^{'}=\frac{m}{1+(\frac{2h}{2l}\times \frac{\sum y}{\sum y^{2}})}$
$\frac{67500}{1+(\frac{2\times465}{21}\times\frac{\sum y}{\sum y^{2}})}$
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
$\sum$y=2(38.57+98.57+158.5+218.57+278.57+338.57+398.57)
$\sum$y=3059.98m
$\sum y^{2}=2(38.557^{2}+98.57^{2}+158.57^{2}+218.57^{2}+278.57^{2}+338.57^{2}+398.57^{2})$
$\sum y^{2}$=878419.83mm
m${'}=\frac{67500}{1+(\frac{2\times 465}{21}\times\frac{3059.98}{870419.83})}$
m$^{'}$=58406 kN,,
Te=$\frac{m^{'}\times yn}{\sum y^{2}}$
yn=398.57mm
=$\frac{58406\times398.57}{870419.83}$
Te=26.74kN
Tb=0.9$\times fub \times Anb$
=0.9$\times $ 400$\times $ 353
Tb=127.2800kN
$(\frac{Vsb}{Vdb})^{2}+(\frac{Tb}{Tdb})^{2}\leq$1
=$(\frac{14.2}{65.22})+(\frac{26.74}{127.28})\leq 1$
0.1$\leq$1 hence safe