Structural Concrete - Hassoun

332 Chapter 9 One-Way Slabs
2. W u = 1.2D + 1.6L
D = 100 psf + self − weight = 100 + 7 (150) =187.5psf
12
544 = 1.2(187.5)+1.6L L = 200 psf
Example 9.3
Design a 12-ft simply supported slab to carry a uniform dead load (excluding self-weight) of 120 psf
and a uniform live load of 100 psf. Use f c ′ = 3ksi,f y = 60 ksi, λ = 1, and the ACI Code limitations.
Solution
1. Assume a slab thickness. For f y = 60 ksi, the minimum depth to control deflection is L∕20 =
12(12)∕20 = 7 in. Assume a total depth of h = 7 in. and assume d = 6 in. (to be checked later).
2. Calculate factored load: weight of slab = 7 (150) =87.5psf
12
W u = 1.2D + 1.6L = 1.2(87.5 + 120)+1.6(100) =409 psf
For a 1-ft width of slab, M u = W u L 2 ∕8.
M u = 0.409(12)2 = 7.362 K ⋅ ft
8
3. Calculate A s ∶ For M u = 7.362 K ⋅ ft, b = 12 in., and d = 6in., R u = M u ∕bd 2 = 7.362(12,000)∕
(12)(6) 2 = 205 psi. From tables in Appendix A, ρ = 0.0040 M
2
u
= 7.362 K ⋅ ft
5. Calculate the secondary (shrinkage) reinforcement normal to the main steel. For f y = 60 ksi,
ρ min = 0.0018
A sh = ρbh = 0.0018(12)(7) =0.1512 in. 2
Choose no. 4 bars, A b = 0.2 in. 2 , S = 12A b ∕A s = 12(0.2)∕0.1512 = 15.9 in. Use no. 4 bars spaced
at 15 in.
( )
6. Check shear requirements: V u at a distance d from the support is 0.409 12
− 6 = 2.25 K.
2 12
φV c = φ2λ√ f c ′ bd = 0.75(2)(1)( √ 3000)(12 × 6)
= 5.9K
1000
1
2 φV c = 2.95 K > V u
so the shear is adequate.
7. Final section: h = 7 in., main bars = no. 4 spaced at 8 in., and secondary bars = no. 4 spaced
at 15 in.