Design of cantilevre slab to Eurocode 2

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Slab design Example 03 Design of cantilever slab

•          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

= 1.35x4.375+1.5x4

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

=13.4x10^6/(1000x145^2x25)

=0.0255

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

= (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

                                                = 0.005

ρ                                             = 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)

                                                = (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