Structural Concrete - Hassoun

688 Chapter 17 Design of Two-Way Slabs
It can be seen that the fixed-end moment coefficient, k ′′ = 0.084, is very close to the coefficient
1
= 0.0833 usually used to calculate the fixed-end moments in beams. This is expected because
12
the part of the span that has a variable moment of inertia is very small in flat plates where no
column capital or drop panels are used. In this example, only parts AB and CD, each equal to
0.83 ft, have a higher moment of inertia than I s . In flat plates where the ratio of the span to column
width is high, say, at least 20, the coefficient 0.0833 may be used to calculate approximately the
fixed-end moments. Fixed-end moment (due to q u = 276 psf) = 0.084(0.245)(20)(25) 2 = 256 K ⋅ ft.
The factors K, K s ′,andK′′ can be obtained from tables prepared by Simmonds and Misic [18] to
meet the ACI requirements for the equivalent frame method.
9. Moment distribution can be performed on half the frame due to symmetry. Once the end negative
moments are computed, the positive moments at the center of any span can be obtained by subtracting
the average value of the negative end moments from the simple beam positive moment.
The moment distribution is shown in Fig. 17.39. The final bending moments and shear forces are
shown in Fig. 17.40.
10. Slabs can be designed for the negative moments at the face of the columns as shown in Fig. 17.40.
Example 17.14 SI Units
Use the direct design method to design a typical interior flat slab with drop panels to carry a dead load
of 8.6 kN/m 2 and a live load of 11 kN/m 2 . The floor system consists of six panels in each direction, with
a panel size of 6.0 × 5.4 m. All panels are supported by 0.4-m-diameter columns with 1.0-m-diameter
column capitals. The story height is 3.0 m. Use f c ′ = 28 MPa and f y = 400 MPa.
Solution
1. All the ACI limitations to using the direct design method are met. Determine the minimum slab
thickness, h, using Eqs. 17.1 and Eqs. 17.2. The diameter of the column capital equals 1.0 m.
The equivalent square column section of the same area will have a side of √ πr 2 = √ π(500) 2 =
885 mm or 900 mm.
Clear span (long direction) =6.0 − 0.9 = 5.1m
Clear span (short direction) =5.4 − 0.9 = 4.5m
Because no beams are used α fm = 0, β s = 1.0, and β = 6.0 m/5.4 m = 1.11. From Table 17.1,
minimum slab thickness h = l n /33 = 5100/33 = 155 mm, but because a drop panel is used, h may
be reduced by 10% if drop panels extend a distance of at least l/6 in each direction from the
centerline of support and project below the slab a distance of at least h/4. Therefore, use a slab
thickness h = 0.9 × 155 = 140 mm and a drop panel length and width as follows:
Long direction l 1
3 = 6.0
3 = 2.0m
Short direction l 2
3 = 5.4
3 = 1.8m
The thickness of the drop panel is 1.25h = 1.25 × 140 = 175 mm. Increase drop panel thickness
to 220 mm to provide adequate thickness for punching shear and to avoid the use of a high
percentage of steel reinforcement. All dimensions are shown in Fig. 17.41.
2. Calculate factored loads:
q u = 1.2 × 8.6 + 1.6 × 11 = 28 kN∕m 2