Cantilever Slab Design

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/σ_s ≤ 1.5         

σ_s                                          = (fyk/γ_s)(As,req/As,prov)(SLS Loads / ULS loads)(1/δ)

                                              =  (fyd)(As,req/As,prov)(gk+ Ψ2qk) /(γ_G gk + γ_Q qk)(1/δ)

σ_s                                          = (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.