written 5.8 years ago by |
2) 1 bolt at each end 3) welded at each end
Given ISA=100$\times$100$\times$6
l=3m=3000mm
rvv=19.2mm
off length =0.854$\ast$lenght of strut(0.821)
Leff=KL=1$\times$3000=3000mm
Area=1167mm$^{2}$
- Pd=Ae$\times$fcd $ \ \ \ \ \ $ Ae=1167mm$^{2}$
fcd=$\frac{fy/ymo}{\phi +[\phi^{2}-\lambda^{2}]}^{0.5}$
$\phi=0.5[1+\alpha(\lambda-0.2)+\lambda^{2}$]
$\alpha$=0.49 .. . .. . .class c for angle
$\lambda e=\sqrt{K1+K2\lambda vv^{2}+K3\lambda^{2}\phi}$
$\lambda w=\frac{L/rvv}{\epsilon\sqrt{\frac{\phi^{2}E}{250}}}$& $\lambda \phi=\frac{b1+b2/2t}{\epsilon\sqrt{\frac{\phi^{2}E}{250}}}$
$\lambda $vv=1.73 $\lambda\phi$=0.188
Case I-considering tow bolt at each end
Table 12 IS Pg48
fixed condition
K1=0.2 K2=0.35 K3=20
$\lambda e=\sqrt{0.2+0.35(1.73)^{2}+20(0.188)^{2}}$
$\lambda e$=1.397
$\phi=0.5[1+0.49(1.397-0.2)+(1.397)^{2}$
$\phi$=1.79
fy=250
fcd=79.55mpa =79.55N/mm$^{2}$
Pd=92.84kN
Hindge condition
K1=0.7 K2=0.6 K3=5
$\lambda e=\sqrt{0.7+0.6(1.73)^{2}+5(0.188)^{2}}$
$\lambda e$=1.63
$\phi=0.5[1+0.49(1.637-0.2)+1.63^{2}]$
$\phi$=2.1788
fcd=$\frac{250/1.1}{2.1788[2.1755^{2}-1.63^{2}}]^{0.5}$=62.72Kn
Pd-Ae$\times$fcd
=1167$\times$62.72
Pd=73.19KN
angles is connected by two bolts Pd=73.19KN
Case II Consider one bolt at each end fixed
K1=0.75 K2=0.35 K3=20
$\lambda e=\sqrt{0.75+0.35(1.73)^{2}+20(0.188)^{2}}$
$\lambda$e=1.58
$\phi=0.5[1+0.49(1.58-0.2)+1.58^{2}]$
$\phi$=2.09
fcd=$\frac{250/1.1}{2.09[2.09^{2}-1.58)^{0.5}}$
fcd=65.72
Pd=1167$\times$65.72
Pd=76.69$\times10^{3}$N
Hinde end
K1=1.25
K2=0.5
K3=60
$\lambda e=\sqrt{1.25+0.5(1.75)^{2}+60(0.188)^{2}}$
$\lambda e$=2.206
$\lambda=0.5[1+0.49(2.206-0..2)+2.206^{2}]$
$\phi$=3.424
fcd=$\frac{250/1.1}{3.424+[3.424^{2}-2.206^{2}]^{0.5}}$
fcd=37.67
Pd=1167$\times$37.67
Pd=43.89KN
Whenangle is connected by each end Pd=43.89 KN