Structural Concrete - Hassoun

336 Chapter 9 One-Way Slabs
• If the live load on the ribbed slab is less than 3 kN/m 2 (60 psf) and the span of ribs exceeds
5 m (17 ft), a secondary transverse rib should be provided at midspan (its direction is perpendicular
to the direction of main ribs) and reinforced with the same amount of steel as
the main ribs. Its top reinforcement shall not be less than half of the main reinforcement in
the tension zone. These transverse ribs act as floor stiffeners.
• If the live load exceeds 3 kN/m 2 (60 psf) and the span of ribs varies between 4 and 7 m
(13 and 23 ft), one traverse rib must be provided, as indicated before. If the span exceeds 7 m
(23 ft), at least two transverse ribs at one-third span must be provided with reinforcement,
as explained before.
Example 9.6
Design an interior rib of a concrete joist floor system with the following description: Span of rib = 20 ft
(simply supported), dead load (excluding own weight) = 16 psf, live load = 85 psf, f c ′ = 4ksi, and
f y = 60 ksi.
Solution
1. Design of the slab: Assume a top slab thickness of 2 in. that is fixed to ribs that have a clear spacing
of 20 in. No fillers are used. The self-weight of the slab is 2 × 150 = 25 psf.
12
Total DL = 16 + 25 = 41 psf
U = 1.2D + 1.6L = 1.2 × 41 + 1.6 × 85 = 185 psf
M u = UL2 (Slabisassumedfixedtoribs.)
12
= 0.185 ( ) 20 2
= 0.043 K ⋅ ft = 0.514 K ⋅ in.
12 12
Considering that the moment in slab will be carried by plain concrete only, the allowable flexural
tensile strength is f t = 5 √ f c ′ , with a capacity reduction factor φ = 0.55, f t = 5 √ 4000 = 316 psi.
Flexural tensile strength = Mc = φf
I t
where
I = bh3
12 = 12(2)3 = 8in. 4 c = h 12
2 = 2 2 = 1in.
I
M = φf t
c = 0.55 × 0.316 × 8 = 1.39 K ⋅ in.
1
This value is greater than M u = 0.514 K ⋅ in., and the slab is adequate. For shrinkage reinforcement,
A s = 0.0018 × 12 × 2 = 0.043 in. 2 Use no. 3 bars spaced at 12 in. laid transverse to the
direction of the ribs. Welded wire fabric may be economically used for this low amount of steel
reinforcement. Use similar shrinkage reinforcement no. 3 bars spaced at 12 in. laid parallel to the
direction of ribs, one bar on top of each rib and one bar in the slab between ribs.
2. Calculate moment in a typical rib:
Minimum depth = L 20 × 12
= = 12 in.
20 20
The total depth of rib and slab is 10 + 2 = 12 in. Assume a rib width of 4 in. at the lower end that
tapers to 6 in. at the level of the slab (Fig. 9.8). The average width is 5 in. Note that the increase
in the rib width using removable forms has a ratio of about 1 horizontal to 12 vertical.
Weight of rib = 5
12 × 10 × 150 = 52 lb∕ft
12