Slab design Example 02 Design of simply supported slab (Two way spanning)
- Span of slab 4.5m & 6.3m
- Variable load 5 kN/mm2
- Slab thickness 200 mm
- Fck 25 N/mm2 fyk 500 N/mm2
- Cover to the reinforcements 25mm
- Office building
- No Moment re-distribution
Ly/Lx =6.3/4.5
=1.4
Therefore slab is two way
Slab loading
Self weight = 200x25x10-3
= 5kN/mm2
Ultimate load = 1.35gk+1.5qk
= 1.35x5+1.5x5
n = 14.25 kN/mm2
Bending moment coefficients for mid span
αsx = 0.099
αsy = 0.051
Bending moment in short span
Msx = αsx*n*(lx)^2
= 0.099*14.25*4.5^2
= 28.57 kN/mm2
Bending moments at long span
Msy = αsy*n*(lx)^2
= 0.051*14.25*4.5^2
= 14.72 kN/mm2
Assume T12 bars used for short span and T10 bars for long span
Effective depth at short span
= 200-25-6
= 169mm
Effective depth at long span
= 200-25-12-5
= 158mm
Reinforcement for short span
K = Msx/bd2fck
= 28.57x10^6/1000x169^2x25
= 0.04
K’ = 0.60δ-0.18δ2-0.21
No redistribution ,Therefore
δ = 1
k’ = 0.21
k’>k
Compression reinforcement is not required
Z = (d/2)*(1+(1-3.53k)^0.5) ≤ 0.95d
= (169/2)*(1+(1-3.53*0.04)^0.5) ≤ 0.95*169
= 162.81 ≤ 160.55
Therefore
Z = 160.55
As = Msx/0.87fyk*Z
= 28.57*10^6/(0.87*500*160.55)
= 409 mm2/m
Provide T10 @ 150mm C/C (As pro. = 523mm2/m)
Check for deflection
Allowable span/d eff. = (l/d)*F1*F2*F3
ρ = As req. /bd
For simply supported slab
K = 1
ρ o = (fck ^.5)/1000
= (25 ^.5)/1000
= 0.005
ρ = 445/ (1000*169)
= 0.00263
Ρ0 > Ρ
Then
l/d = K{11+[1.5*(fck^0.5) ρ o/ ρ ]+ 3.2*(fck^0.5)*[( ρ0/ ρ)-1]^1.5}
= K{11+[1.5*(25^0.5)0.005/0.00263]+ 3.2*(25^0.5)*
*[(0.005/0.00263)- 1]^1.5}
= 38.95
Normal slab
F1 = 1
Span is less than 7m
F2 = 1
F3 = 310/σs ≤
1.5
σs = (fyk/γs)(As,req/As,prov)(SLS Loads / ULS loads)(1/δ)
= (fyd)(As,req/As,prov)(gk+ Ψ2qk) /(γG gk + γQ qk)(1/δ)
σs = (fyk/γs)(As,req/As,prov)(SLS Loads / ULS loads)(1/δ)
= (fyd)(As,req/As,prov)(gk+ Ψ2qk) /(γG gk + γQ qk)(1/δ)
= (500/1.15)(409/523)(5+0.3*5)
/(1.35*5 + 1.5*5)(1/1)
= 155.09 N/mm2F3 = 310/155.09
= 2.0 ≥ 1.5
Hence,
F3 = 1.5
Allowable span/d eff. = 38.95*1*1*1.5
= 58.43
Actual span/ d eff. = 4500/169
= 26.63
Deflection check is ok
Reinforcement for long span
K = Msy/bd2fck
= 14.72x10^6/1000x158^2x25
= 0.024
K’ = 0.60δ-0.18δ2-0.21
No redistribution, Therefore
δ = 1
k’ = 0.21
k’>k
Compression reinforcement is not required
Z = (d/2)*(1+(1-3.53k)^0.5) ≤ 0.95d
= (158/2)*(1+(1-3.53*0.024)^0.5) ≤ 0.95*158
= 154.58 ≤ 150.1
Therefore
Z = 150.1
As = Msx/0.87fyk*Z
= 14.72*10^6/ (0.87*460*150.1)
= 245 mm2/m
Provide T10 @ 200mm C/C (As provided 293mm2/m)
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