Direct Design Method Design of Two Way floor

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. 2

Fig. -1 3

The dead load for a 5. 5 in slab is w. D=(5. 5/12)(150)=69 psf The factored load per unit area is wu =1. 2 w. D +1. 6 w. L =1. 2(69) +1. 6(60) = 96 + 102 =198 psf Using Eq. 5, with clear span Ln measured face-to-face of columns, 4

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

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 6

Table-1 Minimum thickness of slab without interior beams WITHOUT DROP PANELS f*y EXTERIOR PANELS (ksi) α=0 α 0. 8 WITH DROP PANELS INTERIOR EXTERIOR PANELS α=0 α 0. 8 INTERIOR 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. 8

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. 9

(b) Total factored static moment M 0 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) Mo for C = 46. 3 ft-kips Mo for D = 23. 1 ft-kips =29. 1 ft-kips 10

(c) 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 11

Longitudinal distribution of moments

Fig. -2 13

Fig. -3 14

Table-1: Longitudinal moments (ft-kips) for the flat plate FRAME A B C D M 0 58. 2 29. 1 46. 3 23. 1 Mneg at exterior support, 0. 26 Mo 15. 1 Mpos in exterior span, 0. 52 Mo 30. 3 15. 1 24. 1 12. 0 Mneg at first interior support, 0. 70 Mo 40. 7 20. 4 32. 4 16. 2 Mneg at typical interior support, 0. 65 Mo 37. 8 18. 9 30. 1 15. 0 Mpos in typical interior span, 0. 35 Mo 20. 4 10. 2 16. 2 8. 1 7. 6 12. 0 6. 0 15

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

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 c 1. Fig. -4 17

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). 18

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

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. ) 72 36 3 Half middle strip width (in. ) 2@36 36 2 @54 54 4 C(in 4) from previous calculations 474 363 5 Is(in. 4) in βt 2, 000 2, 500 6 βt= Ecb. C/(2 Ecs. Is) 0. 118 0. 073 7 α 1 0 0 8 L 2/L 1 0. 80 1. 25 9 α 1 L 2/L 1 0 0 10 Exterior negative moment, percent to column strip 98. 8% 99. 3% 11 Positive moment, percent to column strip 60% 60% 12 Interior negative moment, percent to 75% column strip 75% 75% 20

Table-2: Percentage of longitudinal moment in column strip ASPECT RATIO L 2/L 1 Negative moment at α 1 L 2/L 1 = 0 exterior support α 1 L 2/L 1 > 1. 0 0. 5 1. 0 2. 0 βt=0 100 100 βt 2. 5 75 75 75 βt = 0 100 100 βt> 2. 5 90 75 45 α 1 L 2/L 1 =0 60 60 60 α 1 L 2/L 1 > 1. 0 90 75 45 Negative moment at α 1 L 2/L 1 =0 75 75 75 interior support α 1 L 2/L 1 > 1. 0 90 75 45 Positive moment

Table-3: Factored moments in a typical column strip and middle strip EXTERIOR SPAN Line number Moments at critical section (ft-kips) 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 Negative Positive moment INTERIOR SPAN Negative Positive moment Negative moment 22

Table-3: Factored moments in a typical column strip and middle strip EXTERIOR SPAN Line number Moments at critical section (ft-kips) 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 Negative Positive moment INTERIOR SPAN Negative Positive moment Negative moment 23

Table-4: Design of reinforcement in column strip EXTERIOR SPAN INTERIOR SPAN LINE NUMBER ITEM 1 Moment, Table-2, line 3 (ftkips) 2 Width b of drop or strip (in. ) 3 Effective depth d (in. ) 4 Mu/Ø (ft-kips) 5 Rn(psi)= Mu/(Øbd 2) 6 ρ, Eq. or Table A. 5 a 7 As = ρbd 8 As =0. 002 bt* 9 N=larger of (7) or(8)/0. 31 10 N=width of strip/(2 t) 11 N required, larger of (9) or (10) NEGATIV E MOMENT POSITIV E MOMEN T NEGATIV E MOMENT 24

Table-5: Design of reinforcement in Middle Strip EXTERIOR SPAN LINE NUMBER ITEM 1 Moment, Table 3, line 4 (ftkips) 2 Width b of strip (in. ) 3 Effective depth d (in. ) 4 Mu/Ø (ft-kips) 5 Rn(psi)= Mu/(Øbd 2) 6 ρ 7 As = ρbd 8 As =0. 002 bt 9 N=larger of (7) or(8)/0. 31* 10 N=width of strip/(2 t) 11 N required, larger of (9) or (10) NEGATIV E MOMENT POSITIV E MOMEN T INTERIOR SPAN NEGATIV E MOMENT POSITIV E MOMEN T NEGATIV E MOMENT 25

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 26

(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 Referring to the fig. 5 Avg d = 5. 50 -0. 75 -0. 50= 4. 25 in Vu = 0. 198(12)6. 65 = 15. 8 Kips Fig. -5 27

(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. 28