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Direct Design Method

Direct Design Method. Design of Two Way Floor System for Slab with Beams. Given data [Problem-1].

clinton-carlson
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Direct Design Method

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  1. Direct Design Method Design of Two Way Floor System for Slab with Beams

  2. Given data [Problem-1] Figure-1 shows a two-way slab floor with a total area of 12,500 sq ft. It is divided into 25 panels with a panel size of 25 ft x 20 ft. Concrete strength is and steel yield strength is fy=40,000 psi. Service live load is to be taken as 120 psf. Story height is 12 ft. The preliminary sizes are follows: slab thickness is 6.5 in.; long beams are 14  28 in. Overall; short beams are 12  24 in. overall; columns are 15  15 in. The four kinds of panels (corner, long-sided edge, short-sided edge, and interior) are numbered 1, 2, 3, and 4 in Fig.-1. Determine the total factored static moment in a loaded span in each of the four equivalent rigid frames whose widths are designated A, B, C and D in Fig.2.

  3. Given data Fig.-1

  4. For two way slab (with beams), the total factored static moment in a loaded span in each of the four equivalent rigid frame whose widths are designated A, B, C and D in Fig.2 Fig.-2

  5. The factored load wu per unit floor area is wu=1.2wD +1.6wL =1.2(6.5)(150/12) +1.6(120) =114 + 204 =318 psf ACI states that Ln shall extend from face to face of columns, capitals or walls. For frame A, For frame B, Mo = 224 ft-kips For frame C, For frame D, Mo = 175 ft-kips

  6. Given data [Problem-2] For the two-way slab with beams design problem-1, Compute the ratio  of the flexural stiffness of the longitudinal beam to that of the slab in the equivalent rigid frame for all the beams around panels 1, 2, 3, and 4 in Fig.-3.

  7. where h=overall beam depth t=overall slab thickness bE=effective width of flange bw=width of slab For the two way slab (with beams), the ratio α of the flexural stiffness of the longitudinal beam to that of the slab in the equivalent rigid frame, for all the beams around panels 1, 2, 3, and 4 in Figure.

  8. (a) B1-B2,Referring to Fig.3, the effective width bE for B1-B2 is the smaller of 14 + 2 (21.5) = 57 and 14 + 8 (6.5) =66 thus bE = 57 in. Using Eq.11 Fig.-3

  9. Fig.-4

  10. In which where h = overall beam depth t =overall slab thickness bE =effective width of flange bw = width of web

  11. (a) B1-B2, Using Eq.9, where Ecb=Ecs

  12. (b) B3-B4,Referring to Fig.3, the effective width bE for B3-B4 is the smaller of 14 + 21.5 =35.5 in. and 14 + 4 (6.5) = 40; thus bE = 35.5 in. Using Eq.11 Using Eq.9, where Ecb=Ecs

  13. (c) B5-B6,Referring to Fig.4, the effective width bE for B5-B6 is the smaller of 12 + 2 (17.5) = 47 in. and 12 + 8 (6.5) = 64 thus bE = 47 in. Using Eq.11 Using Eq.9, where Ecb=Ecs

  14. (d) B7-B8,Referring to Fig.4, the effective width bE for B7-B8 is the smaller of 12 +17.5 = 29.5 in. and 12 + 4 (6.5) = 38; thus bE = 29.5 in. Using Eq.11 Using Eq.9, where Ecb=Ecs The resulting  values for B1 to B8 around panels 1, 2, 3 and 4 are shown in Fig.4. For the design, the  values vary between 3.55 and 13.83; thus the equivalent rigid frames have their substantial portion along or close to the column lines, even though their widths vary from 10 to 25 ft.

  15. Given data [Problem-3] For the two-way slab with beams problem-1, Determine the minimum thickness requirement for deflection control; and compare it with the preliminary thickness of 6.5 in.

  16. The average ratios αm for panels 1, 2, 3, and 4 may be computed from the α values shown in Fig.4; thus

  17. Minimumslabthicknessfordeflectioncontrol Slabs Supported on Beams. Slabssupportedonshallowbeamswhereαm≤0.2. The minimum slab thickness requirements are the same as for slabs without interior beams. Slabssupported on medium stiff beams where 0.2 < αm< 2.0. For this case, The minimum is not be less than 5 in. Slabssupported on very stiff beams where αm > 2.0. For this case, The minimum is not to be less than 3.5 in.

  18. Since the αm values for all four panels are well above 2, Eq.13 applies. The minimum thickness for all panels, using Ln=24 ft, Sn=18.83 ft, and fy=40,000 psi, become If a uniform slab thickness for the entire floor area is to be used, the minimum for deflection control is 6.07 in., which compares well with the 6.5 in. preliminary thickness.

  19. Given data [Problem-4] For the two-way slab with beams problem-1, Show that the six limitations of the direct design method are satisfied.

  20. (7) Limitations of Direct Design Method (1) There is a minimum of three continuous spans in each direction. (2) Panels must be rectangular with the ratio of longer to shorter span center-to-center of supports within a panel not greater than 2.0. (3) The successive span lengths center-to-center of supports in each direction do not differ by more than one-third of the longer span. (4) Columns are not offset more than 10% of the span in the direction of the offset.

  21. (7) Limitations of Direct Design Method (5) The load is due to gravity only and is uniformly distributed over an entire panel, and the service live load does not exceed two times the service dead load. (6) The relative stiffness ratio of L21/α1 to L22/α2 must lie between 0.2 and 5.0, where  is the ratio of the flexural stiffness of the included beam to that of the slab.

  22. The first four limitations are satisfied by inspection. For the fifth limitation, For the sixth limitation, referring to Fig.3&4 and taking L1 and L2 in the long and short directions, respectively, poll Panel 1,

  23. Panel 2, Panel 3, Panel 4

  24. Given data [Problem-5] For the two-way slab with beams problem-1, Determine the longitudinal moments in frames A, B, C, and D. as shown in Fig.2&5.

  25. (a) Check the six limitations for the direct design method. These limitations have been checked previously. (b) Total Factored Static Moment Mo The total factored static moments M0 for the equivalent rigid frames A, B, C, and D have been computed previously; they are • M0 (frame A) = 448 ft-kips • M0 (frame B) = 224 ft-kips • M0 (frame C) = 349 ft-kips • M0 (frame D) = 175 ft-kips

  26. Fig.-5

  27. Fig.-6

  28. Longitudinal Moments in the frames The longitudinal moments in frames A, B, C, and D are computed using Case-2 of Fig.21(DDM) for the exterior span and Fig.19(DDM) for the interior span. The computations are as shown below, and the results are summarized in Fig.-5&6.

  29. For Frame A:M0=448 ft-kips Mneg at exterior support = 0.16(448) =72 ft-kips Mpos in exterior span = 0.57(448) =255 ft-kips Mneg at first interior support = 0.70(448) =313 ft-kips Mneg at typical interior support = 0.65(448) =291 ft-kips Mpos in typical interior span = 0.35(448) =157 ft-kips

  30. For Frame B:M0=224 ft-kips Mneg at exterior support = 0.16(224) =36 ft-kips Mpos in exterior span = 0.57(224) =128 ft-kips Mneg at first interior support = 0.70(224) =157 ft-kips Mneg at typical interior support = 0.65(224) =146 ft-kips Mpos in typical interior span = 0.35(224) =78 ft-kips

  31. For Frame C:M0=349 ft-kips Mneg at exterior support = 0.16(349) =56 ft-kips Mpos in exterior span = 0.57(349) =199 ft-kips Mneg at first interior support = 0.70(349) =244 ft-kips Mneg at tupical interior support = 0.65(349) =227 ft-kips Mpos in typical interior span = 0.35(349) =122 ft-kips

  32. For Frame D:M0=175 ft-kips Mneg at exterior support = 0.16(175) =28 ft-kips Mpos in exterior span = 0.57(175) =100 ft-kips Mneg at first interior support = 0.70(175) =123 ft-kips Mneg at tupical interior support = 0.65(175) =114 ft-kips Mpos in typical interior span = 0.35(175) =61 ft-kips

  33. Given data [Problem-6] For the two-way slab with beams problem-1, Compute the tensional constant C for the edge and interior beams in the short and long directions

  34. The torsional constant C equals, where x = shorter dimension of a component rectangle y= longer dimension of a component rectangle and the component rectangles should be taken in such a way that the largest value of C is obtained.

  35. Each cross-section shown in Fig.7 may be divided into component rectangles in two different ways and the larger value of C is to be used. Fig.-7

  36. For long direction,

  37. For long direction,

  38. For short direction,

  39. For short direction,

  40. Given data [Problem-7] For the two-way slab with beams as described in problem-1, Distribute the longitudinal moments computed for frames A, B, C, and D [Fig.5&6] into three parts-- for the longitudinal beam, for the column strip slab, and for the middle strip slab.

  41. (a) Negative moment at the face of exterior support. For Frame A,

  42. (a) Negative moment at the face of exterior support. Table-1 shows the linear interpolation for obtaining the column strip percentages from the prescribed limits of Table-2(DDM). The total moment of 72 ft-kips is divided into three parts, 92.6% to column strip (of which 85% goes to the beam and 15% to the slab since and 7.4% to the middle strip slab. The results are shown in Table-2

  43. Table-:Percentage of longitudinal moment in column strip

  44. Table-1: Linear interpolation for column strip percentage of exterior negative moment-frame A

  45. (b) Negative moments at exterior face of first interior support and at face of typical interior support. Frame A Using the prescribed values in Table-2(DDM), the proportion of moment going to the column strip is determined to be 81% by linear interpolation.

  46. Table-2: Transverse Distribution of Longitudinal Moments -72 -313 -291 -291 +255 +157 Moments in A

  47. (a) Negative moment at the face of exterior support. Frame B,

  48. (b) Negative moments at exterior face of first interior support and at face of typical interior support. Frame B, The proportion of moment is 81% for column strip, the same as for frame A.

  49. Table-3: Transverse distribution of longitudinal moments -36 -157 -146 -146 +128 +78 Moments in B

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