Example 1: Design of One-Way Slab

Design interior one-way slabs for the situations shown. Concrete weight = 150 lb/ft3 , fy=60,000 psi, and fc’= 4000psi. Do not use the ACI Code’s minimum thickness for deflections. Steel percentages are given in the figures. The only dead load is the weight of the slab. assume ρ=0.01425

example12bone2bway-1986564

Taking 12 in strip and assuming weight of 12 in strip=0.095 k/ft

ultimate load=1.2*DL+1.6*LL
ultimate load=1.2*0.095+1.6*0.15=0.3525k/ft

Mu=(w*L^2)/8=25.38 k-ft=304,560 lb-in

Mu/b*d^2=ϕ*fy*ρ*(1-ρ*fy/1.7*fc’)
assuming ϕ=0.90
b*d^2=304,560/0.90*60000*0.01425*(1-.01425*60000/1.7*4000)

b*d^2=485.17, b=12
d=6.00 in, assuming h=7.50 in

checking weight=7.5*12*150/(144*1000)=0.094 k/ft   ok

As=ρ*b*d=0.01425*12*6.0=1.026 in2, using #8@9 in, As for the strip=0.79*12/9=1.141 so ok

checking if Ф=0.9 valid or no β1=0.85 for fc’=4000 psi or less a=1.141*60,000/(0.85*4000*12)=1.68in εt=0.003*(6.0)/1.97-0.003=0.0061>0.005 so section is tension control Ф=0.9  is ok checking if ρ>ρmin

ρmin=3*√ fc’/fy and not less than 200/fy

ρmin=3*√ 4000/60000 and not less than 200/60000
ρmin=0.00316 and 0.00333     ok

ФMn=0.9*1.141*60,000*(6-1.97/2)= 308,994.21lb-in>Mu   ok