Structural Concrete - Hassoun

9.6 Distribution of Loads from One-Way Slabs to Supporting Beams 331
The load on column C 3 is one-half the load on column C 4 because it supports loads from slabs on
one side only. Similarly, the loads on columns C 2 and C 1 are
L
Load on C 2 = U S
2 S = load on C 3
( )( L S
Load on C 1 = U S
2 2)
From this analysis, it can be seen that each column carries loads from slabs surrounding the column
and up to the centerline of adjacent slabs: up to L/2 in the long direction and S/2 in the short
direction.
Distribution of loads from two-way slabs to their supporting beams and columns is discussed
in Chapter 17.
Example 9.1
Calculate the design moment strength of a one-way solid slab that has a total depth of h = 7in.andis
reinforced with no. 6 bars spaced at S = 7in.Usef c ′ = 3ksiandf y = 60 ksi.
Solution
1. Determine the effective depth, d:
d = h − 3 in.(cover)−half − bar diameter
4
(See Fig.9.1d)
d = 7 − 3 4 − 6 = 5.875 in.
16
2. Determine the average A s provided per 1-ft width (12 in.) of slab. The area of a no. 6 bar is A b =
0.44 in 2 .
A s = 12A b
= 12(0.44) = 0.754 in. 2 ∕ft
S 7
Areas of bars in slabs are given in Table A.14 in Appendix A.
3. Compare the steel ratio used with ρ max and ρ min .Forf c ′ = 3ksiandf y = 60 ksi, ρ max = 0.01356 and
ρ min = 0.00333, where ρ (used) =0.754∕(12 × 5.875) =0.0107, which is adequate (φ = 0.9).
4. Calculate φ:
a = A s f y ∕(0.85f c ′ b)=0.754(60)∕(0.85 × 3 × 12) =1.48 in.
φM n = 0.9(0.754)(60)(5.875 − 1.48∕2) =209 K ⋅ in. = 17.42 K ⋅ ft
Example 9.2
Determine the allowable uniform live load that can be applied on the slab of the previous example if the
slab span is 16 ft between simple supports and carries a uniform dead load (excluding self-weight) of
100 psf.
Solution
1. The design moment strength of the slab is 17.42 K ft per 1-ft width of slab.
M u = φM n = 17.42 = W u L2
8
The factored uniform load is W u = 0.544 K∕ft 2 = 544 psf.
= W u (16)2
8