Structural Concrete - Hassoun

668 Chapter 17 Design of Two-Way Slabs
Figure 17.31
Example 17.10: Shear reinforcement no. 3 at 3.5 in.
4. Calculate shear reinforcement:
φV s =(V u − φV c )=195 − 156.3 = 38.7K φ = 0.75 V s = 51.6K
V s (for one face of critical section) = V s
4 = 51.6 = 12.9K
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Use no. 3 U-stirrups, A v = 0.22 in. 2 (for two legs). The spacing is S = A v f y d/V s = 0.22(60)(7.5)/
12.9 = 7.7 in. Maximum spacing is d/2 = 7.5/2 = 3.75 in.; let s = 3.5 in.
5. Distribution of stirrups: The number of stirrups per one side of column is 43/3.5 = 12.3, or 13
stirrups. Total distance is 13(3.5) = 45.5 in. (Fig. 17.31).
Example 17.11 Flat-Slab Floor System
Using the direct design method, design a typical 24 × 20−ft interior flat-slab panel with drop panels only
(Fig. 17.32). All panels are supported by 20 × 20-in. columns, 12 ft long. The slab carries a uniform
service live load of 80 psf and a service dead load of 24 psf, excluding self-weight. Use f c ′ = 4ksi,and
f y = 60 ksi. (The solution is similar to Example 17.3.)
Solution
1. Determine slab and drop panel thicknesses using Table 17.1.
a. The clear span is 24 − 20 = 22.33 ft. For an exterior panel, minimum h = l 12 n /33 = 8.12 in.,
whereas for an interior panel, minimum h = l n /36 = 7.44 in. Use a slab thickness of 8 in. The
projection below the slab is h∕4 = 8 = 2.0 in.; thus, the drop panel thickness is 10 in.
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b. Extend the drop panels L∕6 = 24 = 4 ft in each direction from the centerline of support in the
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long direction and 20 = 3.33 ft, or 3.5 ft, in the short direction. Thus, the total size of one drop
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panel is 8 × 7 ft (Fig. 17.32).