Structural Concrete - Hassoun

17.8 Analysis of Two-Way Slabs by the Direct Design Method 653
6. Determine the reinforcement required in the long direction in a table form similar to Example 17.4.
Results will vary slightly from those of Table 17.8.
7. Comparison of results between Examples 17.4 and 17.5 shows that the exterior moment in the
column strip (−143 K ⋅ ft) is greater than that calculated in Example 17.4 (−107 K ⋅ ft) by about
34%, whereas the positive moment (±118.15) is reduced by about 8% (relative to ±128.14). Other
values are almost compatible.
Example 17.6
Design an interior panel of the two-way slab floor system shown in Fig. 17.7. The floor consists of six
panels in each direction, with a panel size of 24 × 20 ft. All panels are supported on 20 × 20 in. columns,
12 ft long. The slabs are supported by beams along the column lines with the cross sections shown in
the figure. The service live load is to be taken as 100 psf, and the service dead load consists of 22 psf of
floor finish in addition to the slab weight. Use normal-weight concrete with f c ′ = 3ksi, f y = 60 ksi, and
the direct design method.
Solution
1. The limitations required by the ACI Code are met. Determine the minimum slab thickness using
Eqs. 17.1 and Eqs. 17.2. The slab thickness has been already calculated in Example 17.2, and
a 7.0 in. slab can be adopted. Generally, the slab thickness on a floor system is controlled by a
corner panel, as the calculations of h min for an exterior panel give greater slab thickness than for
an interior panel.
2. Calculate factored loads:
q D = 22 + weight of slab = 22 + 7 × 150 = 109.5psf
12
q u = 1.2(109.5)+1.6(100) =292 psf
3. The shear stresses in the slab are not critical. The critical section is at a distance d from the face
of the beam. For a 1-ft width:
V u = q u
(
10 − 1 2 beam width − d )
(
= 0.292 10 − 16
2 × 12 − 6 )
= 2.58 K
12
φV c = φ(2λ √ f c ′ )bd = 0.75 × 2 × 1 × √ 3000 × 12 × 6
1000
4. Calculate the total static moments in the long and short directions:
M 0l = q u
8 l 2(l n1 ) 2 = 0.292
8 (20)(22.33)2 = 364.0K⋅ ft
M 0s = q u
8 l 1 (l n2 )2 = 0.292
8 (24)(18.33)2 = 294.3K⋅ ft
5. Calculate the design moments in the long direction: l 1 = 24 ft.
a. Distribution of moments in one panel:
= 6.3K > V u
Negative moment (M n )=0.65M 0l = 0.65 × 364 =−236.6K⋅ ft
Positive moment (M p )=0.35M 0l = 0.35 × 364 = 127.4K⋅ ft