DESIGN A SIMPLY SUPPORTED R.C.
C SLAB FOR AN OFFICE FOR THE FOLLOWING DATA
DATA
CLEAR SPAN 4 m 4000 mm
WALL THICKNESS 0.23 m 230 mm
LIVE LOAD 4 kN/m^2
FLOOR FINISH 0.6 kN/m^2
fck 20 N/mm^2
fy 415 N/mm^2
clear cover 20 mm
Dia of the bar is 10 mm
STEP:1 THICKNESS OF SLAB
Assume the efffective depth = d =(span/25)
d 160 mm 0.16 m
Adopting a clear cover of 20mm and using 10 mm diameter bars total depth is
effective cover d' 25 mm
therefore
D 185 mm 0.185 mm
Breadth of a slab b 1000 mm 1m
STEP:2 EFFECTIVE SPAN
THE LEAST IS CONSIDERED
1) (clear span+effective depth) = 4160 mm
4.16 m
2) (clear span +center to center support) = 4230 mm
4.23 m
therefore
effective length = 4160 mm
4.16 m
STEP:3 LOAD CALUCATION
Self weight of a slab = (25*b*D) = 4.625 kN/m^2
Live load of slab = 4 kN/m^2
Floor finish = 1.5 kN/m^2
Total load = 10.125 kN/m^2
Ultimate load = (1.5*Wu) = 15.19 kN/m^2
STEP:4 CALCULATION BENDING MOMENTS AND SHEAR FORCE
Mu =(Wu*l^2)/8 =(0.125*15.19*4.16^2) 32.85 kN /m
Vu =(Wu*l/2) =(0.5*15.19*4.16) 31.59 kn
31.59*10^3 N
STEP:4 CHECK FOR DEPTH
Mulimit =(0.138 fck b d^2)
32.85*10^6 =(0.138*20*1000*d^2)
= dreq 109.097 mm
STEP:5 LIMITING MOMENT OF RESISTANCE
Mulimit =(0.138*fck*b*d^2)
Mulimit 70656000 N/m
Mulimit 70.65 Kn/m
Mu<Mulimit =section is under reinforced
STEP:6 TENSION REINFORCEMENT
Mu (0.87 * fy*Ast*d (1-(Ast fy/b d fck))
32.85*10^6 = =((0.87*415*Ast*d*(1-((Ast*415/1000*160*20))
Ast req = 531 mm^2
Ast pro = 216 mm^2
using 10mm dia bars , the spacing of the bars is computed as
Sv = ((1000*ast/Ast))
Sv = 147.83427
Adopt a spacing of 140mm . Alternate bars are bent up at supports
STEP:7 DISTRIBUTION BARS
Ast =0.12%*b*D 222 mm
provide 8mm diameter bars @ 230 mm center
STEP:8 CHECK FOR SHEAR STRESS
τv = (Vu/bd) = 0.197 N/mm^2
Pt ((100*Ast/((b*d)) = 0.1659
PERMESIBLE SHEAR STREES IN SLAB (REFER TABLE -19 OF IS :456)IS COMPUTED AS
K*τc=(1.23*0.293)=0.36N/mm^2>τv
HENCE THE SLAB IS SAFE IN SHEAR
STEP:9 CHECK FOR DEFLECTION CONTROL
(L/d)max = (L/d)max *Kt * Kc *Kf
= (20*1.4*1*1) = 29
(L/d)pro = =(4160/160) = 26
HENCE , THE DEFLECTION CRITERION IS SATISFIED
Design a singly reinforced concrete beam to suit the following data
Data
clear span L 3000 mm
width of supports b 200 mm
live load LL 6 kN/m
M20 fck 20 N/mm^2
fe415 fy 415 n/mm^2
Step :1 Calucation of dimensions of a beam
Effective depth d 150 mm
Adopt d 160 mm
D 200 mm
b 200 mm
Effective span 3160 mm 3.16 m
Hence
l 3.16 m
Step :2 CALUCATION OF LOAD
Self weight DL (25*b*D) 1 kN/m
Live load LL 6 kN/m
Total load W 7 kN/m
Ultimate load Wu (1.5*W) 10.5 kN/m
Step:3 ULTIMATE MOMENTS AND SHEAR FORCES
Mu =(0.125 Wu L^2) 13.1 Kn-m
Vu =(0.5*Wu*L) 16.59 Kn
Step:4 TENSION REINFORCEMENT
Mulimit 0.138 fck bd^2 14131200 N-mm
14.13 Kn-m
SECTION IS UNDER REINFORCED
Step:5 CHECK FOR REINFORCEMENT
Mu=0.87 Fy Ast d (1-(Ast Fy/Fck bd))
Mu 57768 0.99935156
Ast 227.02
Astmin ((0.85*b*d)/fy) 65.54216867
Ast=227.02mm^2>Astmin
Provide 3 bars of 12mm diameter (Ast=339 mm^2)
And 2 hanger bars of 10mm dia on compression side
Ast provided 339.12 mm^2
Step:6 CHECK FOR SHEAR STRESS
τv (Vu/(bd)) 0.52 N/mm^2
Pt (100*Ast/(bd)) 1.06
REFER TABLE 19OF IS 456:2000 FOR τc
BY INTERPOLATION
1 0.62
1.06 ?
1.25 0.67
τc 0.24
0.05
-0.62
τc 0.632 N/mm^2
NOMINAL SHEAR REINFORCEMENT IS PROVIDED OF 2 NOS OF 8mm DIA
Asv (3.14*D^2)/4 50.24 mm^2
((0.87*Fy*Asv)) 227
Sv
0.4*b
Sv (0.75*d) 120
SV 300 300
PROVIDE THE LEAST SPACING 120mm
THEREFORE ,PROVIDE 2L- #8mm DIA OF 120mm SPACING C/C
STEP:7 CHECK FOR DEFLECTION
Pt 1.06
fs 0.58*fy(Ast req/Ast pro) 161.1352
K1 1.2
K2 1
K3 1
(L/d)max 24
(L/d)provided 19.75 (L/d)max > (L/d)provided
HENCE DESIGN IS SAFE
DESIGN A SIMPLY SUPPORTED R.C.C SLAB FOR AN OFFICE FOR THE FOLLOWING DATA
DATA
lx = 4m 4000 mm
ly = 5m 5000 mm
fck = 20 N/mm^2
fy = 415 N/mm^2
live load = 4 kN/m^2
STEP:1 CHECK THE ROOM IS ONE WAY OR TWO WAY SLAB
(ly/lx) = 1.25 < 2
Design for Two way slab
STEP:2 DEPTH OF SLAB
Depth = (lx/25) 160 mm
Breadth = 1000 mm 1m
Adopt the overall depth as = 170 mm 0.17 m
effective cover = (clear cover +(dia/2))
assume clear cover = 20 mm
dia of the bar as = 10 mm
Therefore
effective cover = 25 mm
effective depth = 145 mm 0.145 m
STEP:3 EFFECTIVE SPAN
Effective span = (Clear span +effective depth
= 4145 mm
= 4.145 m
STEP:4 CALUCATION OF LOADS
SELF WEIGHT OF SLAB = (25*b*D)
= 4.25 kN/m^2
live load = 4 kN/m^2
floor finish = 0.6 kN/m^2
Total W = 8.85 kN/m^2
Ultimate load Wu = 13.275 kN/m^2
STEP:4 CALCULATION OF ULTIMATE DESIGN MOMENTS AND SHEAR FORCE
Refer table9.2 of IS code of 456:2000 read out the moment co-efficients for (ly/lx)=1.25
(ly/lx) 1.25 αx 0.076 αy 0.056
Mux = (αx*Wu*lx) = 17.33 kN-m
Muy = (αy*Wu*lx) = 12.77 kN-m
shear force
(0.5*Wu*lx) = 27.51 kN
STEP:5 CHECK FOR DEPTH
Mu = (0.138*fck*b*d^2)
d = (Mu*10^6/(0.138*fck*b))^(1/2)
dreq = 79.24905 mm
dprov = 145 mm dreq < dpro
Hence the design is safe
STEP:6 REINFORCEMENT CALCULATION
@ shorter span
Mu x = 0.87*fy*Astx*fy*d*(1-(Astx*fy/b*d*fck))
Astx = 302 mm^2
area of one bar = 78.5 mm^2
sv = (area of one bar*1000/(Astx)) = 259.9338 mm
Therefore, = 255 mm
Provide 10 mm dia bars of 255 mm c/c distance @ shorter span
@ longer span
Muy = 0.87*fy*Asty*fy*d*(1-(Asty*fy/b*d*fck))
Asty = 237 mm^2
sv = (area of one bar*1000/(Astx)) = 331.2236 mm
Therefore, = 300 mm
Provide 10 mm dia bars of 300 mm c/c distance @ longer span
Minimum reinforcement
(Ast)min = (0.12%*b*D) = 204 mm^2
Astx > Asty > (Ast)min
Hence the design is safe
STEP:7 CHECK FOR SHEAR STRESS
τv = (Vu/bd) = 0.190
Pt = (100*Astx/b*d) = 0.208276 N/mm^2
Refer Table -19 of IS 456:2000 and read out the permisible shear stress as
τc*k = (1.26*0.32) = 0.4 N/mm^2
τc > τv
Hence slab is safe against shear
STEP:8 CHECK FOR DEFLECTION
(l/d)max = 20*Kt*Kc*Kf ( for simply supported )
Pt = 0.208 N/mm^2
Refer IS 456:2000 code book of fig 7.2 read out Kt value
Therefore,
Kt = 1.7 Kc&Kf = 1
(l/d)max = 34
(lx/d) = 28.6 < (l/d)max
Hence the deflection control is satisfied
13106100
0.999352
361.05
36323.51
190.5873
