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

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.

Fig.-1

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) = 96 + 102 =198 psf Using Eq.5, with clear span Ln measured face-to-face of columns,

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

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

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.

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.

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.

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

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

Longitudinal distribution of moments

Fig.-2

Fig.-3

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

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

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

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).

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

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.) 2@36 2 @54 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%

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

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

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

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)

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)

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

(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 = 5.50-0.75-0.50= 4.25 in Referring to the fig.5 Vu = 0.198(12)6.65 = 15.8 Kips Fig.-5

(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.