written 5.9 years ago by
teamques10
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modified 5.9 years ago
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A] Design of slab S1
ly=5m
ln=4m
lyln=54=1.25<2
∴ Two way continuous slab
1) Depth Calculation
dread=lxl2×M.F
= 400026×1.4
= 109.89mm
dreq≈120mm
Overall Depth
D=d+c.c+ϕ2
= 120+20+102
D = 145 mm
2) Load Calculation
a) s/w of slab = 25×D
= 25×0.145=3.625kN/m2
b) Live load = 3kN/m2
c) Floor Finish = 1.5kN/m2w=8.124kN/m2
wu=1.5×8.125
wu=12.18kN.m
3) Calculation of B.M Coefficient
lyln=1.25
αx=0.075 and αy=0.056
Mux=αx×wu×l2n
= 0.075×12.18×42
Mux=14.616kN.m
Muy=0.056×12.18×42
Muy=10.91kN.m
4) Check for depth
d=√14.616×10620×1000×0.138
d=72.72mm<120mm
∴ safe in depth
5) Calculation of steel
a) Astx = ?
Astx=0.5×20×1000×120415[1−√1−4.6×14.616×10620×1000×1202]
Astx=359.91mm2
Astmin=0.12100×1000×145
Astmin=174mm2
Astx>Astmin
Use Astx=359.91mm2
use 10 mm ϕ bar
Spacing=π4×102359.91×1000
= 218.22
≈200mm
Provide 10 mm ϕ bar @ 200 mm c/c
b) Asty = ?
Muy=10.91kN.m
Astx=0.5×20×1000×120415[1−√1−4.6×10.91×10620×1000×1202]
Asty=263.98mm2
Asty>Astmin
use 10 mm ϕ bar
Spacing=π4×102263.98×1000
= 297.73
≈275mm
Provide 10 mm ϕ bar @ 275 mm c/c
6) Check for shear
1) Vux=0.076×12.18×4=3.65kN
2) Vuy=0.056×12.18×4=2.72kN
Vuc=kτuc bd
k=1.3ford lt150mm
Astp=π4×102200×1000=392.7mm2
%pt=392.71000×120×100
τuc=0.396
Vuc=kτucbd
1.3×0.396×1000×120
Vuc=61.77kN>3.656kN
∴Safe in shear
7) Check for deflection
Fs=0.58fyAstreqAstd
= 0.58×415×359.91392.7
Fs=220.6
≈240N/mm2
%pt=0.327

M.F = 1.5
d=400026×1.6=102.56<120mm
∴ Safe in deflection
7) Check for development length
Ld≤1.3M1V+lo
M1=14.616kN.m
V=3.65kN
lo=12dord
lo=120mm
∴RHS=5325.66 mm
LHS=0.87fy4τbd
= 0.87×4×1.2×1.8
= 470.11mn
∴RHS>LHS
Safe in development length
B) Beam design B1

1) Load Calculation
a) Load transformed by Slab S1
= wulx2[1−13β2]
= 12.18×42[1−13×1.252]
= 19.16kW/m
b) Wall load = 1.5×l×b×h
1.5×20×0.23×3.2
= 22.08kN/h
C) S/W of beam = 10% (a+b)
= 10% (19.16 + 22.08)
= 4.124 kN/m
= wu=45.364kN/n

VA=VB=45.36×52=113.41

midspan AB
Mu=141.76kN.m

bf=lo6+6Df+bw
0.7×50006+6×145+230
1683.33mm
D=500010
D=500mm
d=500−25=475mm
Assume x≤Df
Mur=0.36Fckxubf(d−0.42xl)
=0.36×20×145×1683.33(475−0.42×145)
Mur=727.73>14176kN
Assumption is correct
Ast1=0.5×20×230×475415[1−√1−4.6×141.76×10620×1000×4702]
Ast=1027.54mm2
Use 20 mm ϕ bar
No. of Bar = 1027.54pi4×202
= 3.27
≈4Nos
Provide 4 -20 mm ϕ bars
Design of shear reinforcement
VuD=113.41kN
DVuc=τucbd
Ast=4×π4×202=1256.64mm2
%pt=Astbd×100
%pt=1.16
τuc0.652
Vuc=τucbd
= 0.652×230×475
Vuc=71.23kN
Vus=VuD−Vuc
= 113.41−71.23
Vus=42.18kN
Vus=Vusv=42.18kN
Sv=0.87FyAsvdVusv
Sv=0.87×415×2×π4×82×47542.18×103
= 402.82
≈300mm
∴ 2L-8mm ϕ 300 mm c/c
