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Design laterally supported beam of effective span 6m for the following data 1) max Bm=150kN.m 2) max shear force V=210kN grade of steel fe 410 required design laterally supported beam
1 Answer
written 5.2 years ago by |
1) M=150kN.m v=210kN
2) Zpreq=$\frac{m\gamma mo}{fy}$=$\frac{150\times 10^{6}\times 1.1}{250}$
Zpreq=660$\times10^{3}mm^{3}$
select ISLB 350 at 49.5 kg/m having Zp=851.11$\times 10^{3}mm^{3}$
D=350mm bf=165mm tf=11.4mm
tw=7.4mm $\gamma_{1}$=1.6
IZ=13125.3$\times 10^{4}mm^{4}$
d=350-2$\times11.4-2\times$16
d=295.2mm
3) sectional classification
$\frac{b}{tf}=\frac{bf/2}{tf}=\frac{165/2}{11.4}=7.2\lt 9.4\epsilon$
$\frac{d}{tw}=\frac{295.2}{7.4}=39.89\lt 84\epsilon$
seciont is plastic
4) Vd=$\frac{AV\ast fy}{\sqrt{3}\gamma mo}$
=$\frac{(350\times7.4)\ast 250}{\sqrt{3}\times 1.1}$
Vd=339.84$\gt$ V=210kN safe
5) 0.6 Vd=0.6$\times$339.84 …