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

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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

dead load+live load

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

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loading on B1

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

loading on B1 as shown

enter image description here

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