Page: flooring system for industrial shed . Design the beams SB MB and the connections between them USe I-sections for beam and provide cover plates if necessary.Top flanges of beams are at same level
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and embedded in concrete of 200mm thick RCC skab. The parapet wall of 230mm thickness and 1.2m in height is provided on all peripheral beams

used calculation

For given Slab pane

ly/lex=5/3=1.666$\lt$ 2

hence is it is two way slab

The load is transfer as shown in fig

self weight is Rcc slab=0.15$\times$25=375kbN/m$^{2}$

uve load=4kn/m$^{2}$

Total floor load=7.75kn/m$^{2}$

Tributary area of slab to transfer load =1/2$\times4\times$=10m$^{2}$

floor load on beam B$_{3}$ per m run

=$\frac{7.75\times 10}{5}$=15.5mm

load of brick wall perm run =0.15$\times3.50\times$20=10.5kN/m

Total load per m run on Beam B$_{3}$

15.5+15.5=26kN/m

Assuming self weight of beam=$\frac{Total \ imposed \ load}{300}=\frac{26 \times 4.6}{3.20}$ =0.398 say 0.75$\times$ kN/m

self weight =0.75kN/m

Total working load in Beam B$_{3}$=28+0.75=26.75kN/m

wl/2=66.875

VB=66.875kN

point load on the center of beam AB=66..875km end reaction of beam B1 wall load on Beam B1=10.5kN/m

Total imposed load w=66.825+(10.5$\times$5)=119.325 kN

self weight =$\frac{115.375\times 1}{300}$=0.397 say 0.75kN/m

Total load on Beam B$_{1}$=10..5+0.75=11.25kN

max Bm=m=$\frac{66.875\times 5}{2}+\frac{11.25\times 5^{2}}{8}$=202.33kN.m

factored shear force Vd=1.5$\times 61.562=$92.343kN

Factored BbM=Md=1.5$\times 202.33$=303.495 kN.m

Assuming plastic section

(Zp)req=Md/fy/$\gamma$ ms=$\frac{303.495}{250/1.1}$=1.33

(Zp)req=1.32

(Ze)req=$\frac{1.33\times 1000}{1.14}$=1.166m$^{3}$

section ISLB 450

check for shear Vdr=$\frac{fy\times tw\times g}{\gamma mo\times \sqrt{3}}$=$\frac{250\times8.6\times 450}{1.1 \times \sqrt{3}}$=507.80kM $\gt$ Vd also Vd/Vdr $\lt$ 0.6 =0.18$\lt$0.6

sallwable =L/300=500/300=16.66mm

$\int$max=$\frac{5wl^{4}}{384 eI}=\frac{5\times 7.75\times 5000^{4}}{384\times 2\times 10^{5}\times275.36\times 10^{6}}$

$\int$max=11.45 mm

AS $\int$ max $\lt \int allowable$

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