Question

Q3 (50 pts): Design the concrete slab shown below for bending moment: Wu= 0.13 k/ft including self weight 12' fe-4,000 psi fy=60 ksi

Answer 1

Complete design of slab as per ACI is given below.

Given, L= fc' = fy= 12 ft 4 ksi 60 ksi b1 = 0.85 1. Thickness as per ACI specification hmin = L/20 (for SSB h = 7 in 2. Loads Wu = 0.13 k/ft * Mu= wl2/8 Mu= 2.34 k-ft Mu= 28.08 k-in 3. Required resistance Mn = Mu/ Mn = Mu/0.9 Mn = 31 k-in 4. As,bar = 0.2 in2 Dimension (Assume #4 bars d= h - cover - 0.500 d= h - 0.75 - (0.5*0.5) d = 6.00 in 5. Main reinforcement Steel reinforcement ratio =P Pmin = 0.0018 As.min = Pmin * b*h As.min = 0.0018 * 12 *h As,min= 0.1512 in2 Pmax= Ptc = 0.85 * 1 * (fc/fy) * (3/8) Pmax= 0.018 As,tc = Pmax * b*d As.tc= 1.30 in 2

Actual steel required Mn= As * fy* (0.9d) Therefore, As = Mn/(fy *0.9d) As = 0.10 in 2 As min > AS Therefore, Provide As = 0.15 in2 Assuming #5 bar Sreq = 12 * (As,bar/As) S= 15.87 in Provide S= 15.00 in Maximum spacing = min (18,3h) Smax= 18 Min 21 Therefore, Provide #4 bars with spacing 15 in Steel area per ft As = As,bar *12/S As = 0.16 in2 * Find actual moment capacity c= (As*fy) / (0.85*fc'*b1*b) 0.28 in C= Mn= As * fy* (d- (b1*c/2)) Mn = 42 k-in 31 Safe 6. Temperature and shrinkage steel As.req = 0.0018 * b*h As = 0.1512 Assuming #4 bars (As,bar) = 0.2 S= 15 in 7. Final design values Main steel = #4 bars @ 15" spacing Distribution steel = #4 bars @ 15" spacing