1. 786 Design of Two Way floor system for Flat Plate Slab
Direct Design Method
Design of Two Way floor system for
Flat Plate Slab

2. Given data [Problem-1]
Figure-1 shows a flat plate floor with a total area of 4500 sq ft. It is divided into 25 panels with a panel size of 1512 ft. Concrete strength is and steel yield strength is fy= 50,000 psi. Service live load is 60 psf. Story height is 9 ft. All columns are rectangular, 12 in. in the long direction and 10 in. in the short direction. Preliminary slab thickness is set at 5.5 in. No edge beams are used along the exterior edges of the floor. Compute the total factored static moment in the long and short directions of a typical panel in the flat plate design as shown in Fig.-1.

3. Fig.-1

4. The dead load for a 5.5 in slab is
wD=(5.5/12)(150)=69 psf
The factored load per unit area is
wu =1.2wD +1.6wL
=1.2(69) +1.6(60)
= =198 psf
Using Eq.5, with clear span Ln measured face-to-face of columns,

5. Given data [Problem-2]
Review the slab thickness and other nominal requirements for the dimensions in the flat plate design example.

6. Minimum slab thickness
For fy=50 ksi, for a flat plate which inherently has α = 0, and Ln=15-1=14 ft, from Table-1,
min t=linear interpolation between fy=40 ksi and fy=60 ksi

7. Table-1 Minimum thickness of slab without interior beams
WITHOUT DROP PANELS
WITH DROP PANELS
f*y
EXTERIOR PANELS
INTERIOR
(ksi)
α = 0
α  0.8
PANELS 40 60 75
*For fy between 40 and 60 ksi, min. t is to be obtained by linear interpolation.

8. A table value seems appropriate and entirely within the accuracy of engineering knowledge regarding deflection. The 5.5 in. slab thickness used for all panel satisfies the ACI-Table minimum and exceeds the nominal minimum of 5 in. for slabs without drop panel and without interior beams.

9. Given data [Problem-3]
For the flat plate design problem-1, compute the longitudinal moments in frames A, B, C and D as shown in Figs.1 & 2.

10. Total factored static moment M0
The total factored static moment Mo from the results of previously found; thus
Mo for A = 58.2 ft-kips
Mo for B = 0.5(58.2) =29.1ft-kips
Mo for C = 46.3 ft-kips
Mo for D = 23.1 ft-kips

11. Longitudinal moments in the Frame.
The longitudinal moments in frames A, B, C, D are computed using Case-3 of Fig.22 for the exterior span and Fig.19 for the interior span. The computations are shown in Table-1 and the results are summarized in Fig.2&3.
Longitudinal Moments (ft-kips) for the flat plate

12. Longitudinal distribution of moments

13. Fig.-2

14. Fig.-3

15. Table-1: Longitudinal moments (ft-kips) for the flat plate
FRAME A B C D M0
58.2
29.1
46.3
23.1
Mneg at exterior support, 0.26Mo
15.1
7.6
12.0
6.0
Mpos in exterior span, 0.52Mo
30.3
24.1
Mneg at first interior support, 0.70Mo
40.7
20.4
32.4
16.2
Mneg at typical interior support, 0.65Mo
37.8
18.9
30.1
15.0
Mpos in typical interior span, 0.35Mo
10.2
8.1

16. Given data [Problem-4]
For the flat plate design problem-1, Compute the torsional constant C for short and long beam

17. Since no actual edge beams are used, the torsional members is, according to Fig.4, equal to the slab thickness t by the column width c1.
Fig.-4

18. Given data [Problem-5]
Divide the 5 critical moments in each of the equivalent rigid frames A, B, C, and D, as shown in Fig.2&3, into two parts: one for the half column strip (for frames B and D) or the full column strip (for frames A and C), and the other for half middle strip (for frames B and D) or the two half middle strips on each side of the column line (for frames A and C).

19. The percentages of the longitudinal moments going into the column strip width are shown in lines 10 to 12 of Table-2.

20. Table-2 Transverse Distribution of Longitudinal Moment
LINE
NUMBER
EQUIVALENT RIGID FRAME A B C D 1
Total transverse width (in.)
144 72
180 90 2
Column strip width (in.) 36 3
Half middle strip width (in.) 54 4
C(in4) from previous calculations
474
363 5
Is(in.4) in βt
2,000
2,500 6
βt= EcbC/(2EcsIs)
0.118
0.073 7 α1 8
L2/L1
0.80
1.25 9
α1 L2/L1 10
Exterior negative moment, percent to column strip
98.8%
99.3% 11
Positive moment, percent to column strip
60% 12
Interior negative moment, percent to column strip
75%

21. Table-2:Percentage of longitudinal moment in column strip
ASPECT RATIO L2/L1
0.5
1.0
2.0
Negative moment at α1L2/L1
= βt=0
100
exterior support
βt2.5 75
α1L2/L1
> βt = 0
βt> 2.5 90 45
Positive moment α1L2/L1
= 0 60
> 1.0
Negative moment at α1L2/L1
interior support α1L2/L1

22. Table-3: Factored moments in a typical column strip and middle strip
EXTERIOR SPAN
INTERIOR SPAN
Line number
Moments at critical section (ft-kips)
Negative moment
Positive moment 1
Total M in column and middle strips
(frame A) 2
Percentage to column strip
(Table-2) 3
Moment in column strip 4
Moment in middle strip

23. Table-3: Factored moments in a typical column strip and middle strip
EXTERIOR SPAN
INTERIOR SPAN
Line number
Moments at critical section (ft-kips)
Negative moment
Positive moment 1
Total M in column and middle strips
(frame C) 2
Percentage to column strip
(Table-2) 3
Moment in column strip 4
Moment in middle strip

24. Table-4: Design of reinforcement in column strip
EXTERIOR SPAN
INTERIOR SPAN
LINE
NUMBER
ITEM
NEGATIVE MOMENT
POSITIVE MOMENT 1
Moment, Table-2, line 3 (ft-kips) 2
Width b of drop or strip (in.) 3
Effective depth d (in.) 4
Mu/Ø (ft-kips) 5
Rn(psi)= Mu/(Øbd2) 6
ρ, Eq. or Table A.5a 7
As = ρbd 8
As =0.002bt* 9
N=larger of (7) or(8)/0.31 10
N=width of strip/(2t) 11
N required, larger of (9) or (10)

25. Table-5: Design of reinforcement in Middle Strip
EXTERIOR SPAN
INTERIOR SPAN
LINE
NUMBER
ITEM
NEGATIVE MOMENT
POSITIVE MOMENT 1
Moment, Table 3, line 4 (ft-kips) 2
Width b of strip (in.) 3
Effective depth d (in.) 4
Mu/Ø (ft-kips) 5
Rn(psi)= Mu/(Øbd2) 6 ρ 7
As = ρbd 8
As =0.002bt 9
N=larger of (7) or(8)/0.31* 10
N=width of strip/(2t) 11
N required, larger of (9) or (10)

26. Given data [Problem-6]
Investigate the shear strength in wide-beam and two-way actions in the flat plate slab system for an interior column with no bending moment to be transferred . Note that

27. (a)Wide beam action. Assuming ¾ in clear cover and #4 bars, the average effective depth when bars in two directions are in contact is Avg d = = 4.25 in
Referring to the fig Vu = 0.198(12)6.65 = 15.8 Kips
Fig.-5

28. (b)Two- way action, Referring to Fig.5
The perimeter of the critical section at d/2 around the column is
Shear reinforcement is not required at this interior location.