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
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20kN/m$^{3}$

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$^{2}$

Given

1) thickness of brick wall=150mm

=0.15m

H=3m

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

2) thickness of slab=120mm=0.120m

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

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

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

Assume weight of beam=1kN/m

load on beam b$_{1}$

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

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

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

=35.25kN/m

load on beam b$_{2}$

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

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

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

v=115.5kN

m=m max=208.5kN.m

2) Trial section

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

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

=917.4$\times$10$^{3}mm^{3}$

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$\times 10^{3}mm^{3}$

Iz=15521.7$\times10^{4}mm^{4}$

d=D-2tf-2R1

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

=303.2mm

3)selection classification

$\frac{b}{tf}=\frac{bfl2}{tf}=\frac{100}{11.4}$=8.77$\lt$9.4$\epsilon$

$\frac{d}{tw}=\frac{3036}{8}$=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=$\frac{(b_{!}+n_{2})+w\ast fy}{\gamma mo}$

b$_{1}$=75

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

=58.5

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

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

=5.44mm

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

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