Cantilever Slab Design Calculation & Procedure
Design of cantilever slab to Eurocode 2
- Span of slab 1.5m
- Variable load 4kN/mm2
- Slab thickness 175mm
- Fck 25N/mm2 fyk 500N/mm2
- Cover to the reinforcements 25mm
- Office building
Slab loading
Self weight = 175x25x10-3 = 4.375kN/mm2
Ultimate load = 1.35gk+1.5qk = 1.35×4.375+1.5×4 = 11.91 kN/mm2
Bending moment
n = 11.91 kN/mm2
Bending moment
M = 11.91*1.5*1.5/2 = 13.4 kNm
Assume T10 bars used for the span
Effective depth = 175-25-5 = 145 mm
Reinforcement
K = M/bd2fck=13.4×10^6/(1000×145^2×25)=0.0255
K’ = 0.60δ-0.18δ2-0.21
No redistribution ,Therefore
δ =1
k’ =0.21
k’>kCompression reinforcement is not required
Z = (d/2)*(1+(1-3.53k)^0.5) ≤ 0.95
d = (145/2)*(1+(1-3.53*0.0255)^0.5) ≤ 0.95*145= 141.66 > 137.75 Therefore
Z = 137.75
As = M/0.87fyk*Z= 13.4*10^6/(0.87*500*137.75) = 224 mm2/m
Provide T10 @ 200mm C/C (As pro. = 393mm2/m
Check for deflection (same method as two way slab)
Allowable span/d eff. = (l/d)*F1*F2*F3
ρ = As req. /bd
For cantilevered slab
K = 0.4
ρ o = (fck ^.5)/1000 = (25^.5)/1000 = 0.005ρ= 224/ (1000*145) = 0.00154
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)0.005/0.00154]+ 3.2*(25^0.5)**[(0.005/0.00154) – 1]^1.5} = 35.69
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 = (500/1.15)(224/393)(4.375+0.3*4) /(1.35*4.375 + 1.5*4)(1/1) = 116.1 N/mm2
F3 = 310/116.1 = 2.67 ≥ 1.5
Hence,
F3 = 1.5
Allowable span/d eff. = 35.69*1*1*1.5 = 53.54
Actual span/ d eff. = 1500/145 = 10.34
Deflection check is ok.