written 5.9 years ago by
teamques10
★ 69k
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modified 5.9 years ago
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Solution:
ly=9m
lx=4m
lylx=94=2.25>2
One way continuous slab
Given:
Live load = 3KN/m2
Partition load = 1KN/m2
F.F = 0.6KN/m2
M20 Fe415
beam = 250 mm
brick wall = 230 mm
1] Calculation of depth
dreq=lxld×MF=400026×1.4=109.89∼120mmD=120+20+102=145mm
2] Load calculation
i) Dead load
a) s/w of slab =25×D=25×0.145=3.625KN/m2Partition load=1KN/m2Floor finish=0.6KN/m2----------------5.225KN/m2WuDL=1.5×5.225=7.84KN/m
ii) Live load
LL=3KN/m2
WULL=4.5KN/m2

D.L+112−116116−112
L.L+110−19112−19
Mu1=112×7.84×42+110×4.5×42=17.65KNmMuB=−110×7.84×42−19×4.5×42=−20.54KNmMu2=116×7.84×42+112×4.5×42=13.84KNmMuc=−112×7.84×42+(−19)×4.5×42=−18.45KNm
4] Check for depth
mmmax=Rumax bd2d=√20.54×1000.138×20×1000d=86.26mm<120mm∴Safe in depth
5] Calculation of steel
a) Main steel
1) Ast1=0.5×20×1000×120415[1−√1−46×176.5×10620×1000×1202]=441.24mm2Astmin=174mm2Use commϕSpacing=π/4×102441.24×1000=177.99mm∼150mmUse 10mm ϕ @ 150 mm
2) AstB=521.30mm2
Provide 10mm ϕ 125 mm c/c
3) Ast2=339.53mm2
Provide 10mm ϕ @ 200 mm c/c
4) Astc=463.14mm2
Provide 10mm ϕ @ 150 mm c/c
b) Distribution steel
Astmin=174mm2
Provide 8 mm ϕ @ 275 mm c/c
Astp=π/4×102150×1000=523.6mm2
6] Check for deflection Page no 38 IS456
Fs=0.58×415×441.2953.6=202.83N/mm2∼24N/mm2Pt%=523.61000×120×100=0.436%MF=1.3d=400026×1.3=118.34mm<120mm∴Safe in deflection
7] Check for shear
At B
VuB=αDL×WUD×lx+αuWULL×lx=0.6×7.84×4+0.6×7.84×4=37.63KNVuD=VuB=37.63KNVuc=k τUC bdk=1.3 (for d≤150)Pt%=Astbd×100=521.301000×120×100=0.430.25−0.360.43−10.50−0.48τuc=0.4120N/mm2Vuc=k τuc bd=1.3×0.421×1000×120=65.67KN>VUD=37.63KN∴Safe in shear
8] Check for development length
Ld=0.87fyϕ4τbd=0.87×415×104×1.2×1.6=470.11mmM1=17.652=8.825KNVA=21.9KNlo=12ϕ or d=120mm=644.03mmLd<13M1V+ls
