Page: following fig show typical framing plan of steel building design beam b$_{1},b_{2}$ laterally supported use following data 1) beam support brick wall of 150mm thick and 3m high having unit weight


2) thickness of Rcc slab is 120mm and top flang embedded in concrete having unit weight 25kN//m$^{3}$

3) L.L =3kN/m$^{2}$ and F.F.L=0.75kN/m$^{2}$

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1) thickness of brick wall=150mm



$\int$ m=20kN/m$^{3}$

2) thickness of slab=120mm=0.120m

$\int c$=25kN/M$^{3}$

3) L.L =3kN/m$^{3}$


Assume weight of beam=1kN/m

load on beam b$_{1}$

Slab load

1) concrete load=25$\times$2$\times$0.12=6kN/m

2) F.F.L=0.75$\times$2=1.5kN/m

3) L.L=3$\times$2=6kN/m

Total slab load=13.5kN/m

factored slab load=13.5$\times1.5$=20.25kN/m

Wall load

Wall load=20$\times$3$\times0.15$ kN/m=6kN/m

factored load=9$\times$1.5=13.5kN/m

weight of beam $b_{1}$=1kN/m

factored weight of beam=1.5kN/m

Total load on beam b$_{1}$=20.25+13.5+1.5


load on beam b1

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load on beam b$_{2}$

load on beam b$_{2}$=13.5+1.5=15kN/m

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1) V=$\frac{wl}{2}+\frac{2w}{2}$

=$\frac{15\times 6}{2}+\frac{2\times 10.5}{2}$


m=m max=208.5kN.m

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2) Trial section

Zreq=$\frac{m.\gamma ma}{fy}$

=$\frac{208.5\times 10^{6}\times 1.1}{250}$


select ISWB 350 having Zp= 995.49$\times 10^{3}mm^{3}$

D=350 bf=200mm tf=114.4 mm

tw=8mm $\gamma_{1}$=12mm

Ze=887$\times 10^{3}mm^{3}$



=350-(2$\times 11.4)-(2\times 12)$


3)selection classification


$\frac{d}{tw}=\frac{3036}{8}$=37.9$\lt$ 846

section is plastic

4) Vd=$\frac{Av\times fy}{\sqrt{3}\times \gamma mo}$

=$\frac{(350\times 8)\times 250)}{\sqrt{3}\times 1.1}$

Vd=367.41$\gt$ 113.5 safe

5) 0.6Vd=0.6$\times$ 367.41

=220.44v(115.5)$\lt$ Vd(220.44) low shear

Md=$\frac{Bd\ast ZP\ast fy}{\gamma mo}$ $\frac{1\times (99.549\times 10^{3})\times 250}{1.1}$

=226.25kN $\gt$ 208.5kN.m safe

$\lt$ 1.2$\times ze\ast \frac{fy}{\gamma mo}$

$\lt(1.2\times 887\times 10^{3})\ast \frac{250}{1.1}$

$\lt$ 241.91kN.m

(Ze=887$\times10^{3}mm^{3})$ safe

6) 1)$\frac{d}{tw}$=37.9$\lt $ 67$\epsilon$ no need to check shear buckling

2) web crippling

fwc=$\frac{(b_{!}+n_{2})+w\ast fy}{\gamma mo}$


n$_{2}$=2.5(tf+$\gamma _{!})$ =2.5(11.4+12)


fwc=$\frac{(75+58.5)\ast 8\times 250}{1.1}$

=242.75kN $\gt$ 115.5kN safe

3) Deflection

Permissible =$\frac{span}{300}=\frac{6000}{300}$=20mm

$\int$act =$\frac{5}{384}$ $\frac{wl^{4}}{EI}+\frac{w_{2}\times a(3l^{2}-4a^{2})}{24EI}$

=$\frac{5(15/1.5)\times 6^{4}}{384(2\times 10^{5})\times (15521.7\times 10^{4})}$


$\int $ act $\lt$ permissible safe again deflection

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modified 9 weeks ago  • written 3 months ago by gravatar for stanzaa37 stanzaa3720
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