Design of Simply Supported Slab (one way spanning)

Slab design Example  01 Design of simply supported slab (one way spanning)

•          Span of slab 4.5m
•          Variable load 4kN/mm2
•          Floor finishes and Ceiling load 1kN/mm2
•          Fck 25N/mm2     fyk  500N/mm2
•     Office building
•     No re-distribution of moments (δ = 1)

Assume slab thickness 200mm

                d              = 200-25-5

                                                 = 170mm

Slab loading

Self weight                          = 200x25x10-3

                                = 5kN/m2

Total permanent load    = 1+5

= 6kN/m2

Ultimate load                    = 1.35gk+1.5qk

= 1.35x5+1.5x6

Bending Moment            = wl2/8

= (1.35x5+1.5x6)x4.5^2/8

= 31.9kNm

Reinforcement

K                                             = M/bd2fck

= 31.9x10^6/1000x170^2x25

= 0.044

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

= (170/2)*(1+(1-3.53*0.049)^0.5) ≤ 0.95*170

= 162.3 ≤ 161.5

Therefore

Z                                              = 161.5

As                                           = M/0.87fykz

= 35.69*10^6/(0.87*500*161.5)

                                                = 508 mm2

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                                             = 508/ (1000*170)

                                                = 0.00299

Ρ0 > Ρ

Then

l/d                                          = {11+[1.5*(fck^0.5) ρ o/ ρ ]+ 3.2*(fck^0.5)*[( ρ0/ ρ)-1]^1.5}

                                                ={11+[1.5*(25^0.5)0.005/0.00299]+ 3.2*(25^0.5)*[(0.005/0.00299)-1]^1.5}

                                                = 32.36

Normal slab

F1                                           = 1

Span is less than 7m

F2                                           = 1

Steel ratio

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)(508/525)(5+0.3*6) /(_1.35*5 + 1.5*6)(1/1)

                                                = 151.85 N/mm2
                                               
F3                                           = 310/151.85

                                                = 2.04 ≥ 1.5

Hence,

F3                                = 1.5                                                

Allowable span/d eff.    = 32.36*1*1*1.5

                                                = 48.54

Actual span/ d eff.           = 4500/170

                                                = 26.47

Deflection check is  OK