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Indian standard angle 100$\times$100$\times$6mm is used as strut iin roof truss the length of strut between entrane section at each end 8m cal the strength 1) connect two halts at each end
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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$^{2}$

  1. 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

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