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Lecture # 2

Design of Concrete Structure II. University of Palestine. بسم الله الرحمن الرحيم. Lecture # 2. Instructor:. Eng. Mazen Alshorafa. Design of Concrete Structure II. University of Palestine. Page 1. Direct Design Method [DDM].

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Lecture # 2

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  1. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Lecture # 2 Instructor: Eng. Mazen Alshorafa

  2. Design of Concrete Structure II University of Palestine Page 1 Direct Design Method [DDM] This method (DDM) is restricted to vertical loads only. For lateral loads the Equivalent Frame Method (EFM) should be used. DDM consists of certain steps for distributing moments to slab and beam sections to satisfy safety requirements and most serviceability requirements simultaneously. Limitations of t he Direct Design Method The slab system, to be designed using the DDM, should conform to the following limitations as given by ACI Code 13.6.1: 1. Minimum of three continuous spans in each direction. 2. Panels must be rectangular (L/ S ≤ 2.0). 3. Successive span in each direction must not differ by more than 1/3 the longer span (0.67 ≤[Ln/Ln-1]≤ 1.33). Instructor: Eng. Mazen Alshorafa

  3. Design of Concrete Structure II University of Palestine Page 2 Direct Design Method [DDM] Limitations of t he Direct Design Method (contd.) 4.Columns must not be offset more than 10% of the span in the direction of offset from either axis between centerlines of successive columns. 5. Loads must be due to gravity only and uniformly distributed over the entire panel. The live load must not exceed 2 times the dead load. Instructor: Eng. Mazen Alshorafa

  4. Design of Concrete Structure II University of Palestine Page 3 Direct Design Method [DDM] Limitations of t he Direct Design Method (contd.) 6- For a panel with beams between supports on all sides, the relative stiffness of beams in two perpendicular directions 2 [ ] is not to be less than 0.20 and not greater than 5.0. where α’1 α’2 α’4 l2 α’3 l1 Instructor: Eng. Mazen Alshorafa

  5. Design of Concrete Structure II University of Palestine Page 4 Direct Design Method [DDM] Design Procedure 1- Determination of the total factored static moment [Mo] Total factored static moment for a span is determined in a strip bounded laterally by centerline of panel on each side of centerline of supports, as shown in Figure l1 B A l2 ln1 Instructor: Eng. Mazen Alshorafa

  6. Design of Concrete Structure II University of Palestine Page 5 Direct Design Method [DDM] 1- Determination of the total factored static moment [Mo] (contd.) Total factored static moment for a span is determined in a strip bounded laterally by centerline of panel on each side of centerline of supports, as shown in Figure. B ln1 l1 A l2 Instructor: Eng. Mazen Alshorafa

  7. Design of Concrete Structure II University of Palestine Page 6 Direct Design Method [DDM] 1- Determination of the total factored static moment [Mo] (contd.) Where Wu is the factored load per unit area ln1is clear span in the direction moments are being determined, measured center-to-center of supports and not to be less than 0.65 L1 L2 is taken as the average of adjacent transverse spans. Instructor: Eng. Mazen Alshorafa

  8. Design of Concrete Structure II University of Palestine Page 7 Direct Design Method [DDM] 2- Distribution of Mo to negative and positive moments Instructor: Eng. Mazen Alshorafa

  9. Design of Concrete Structure II University of Palestine Interior span 0.65 Mo 0.35 Mo Page 8 Direct Design Method [DDM] 2- Distribution of Mo to negative and positive moments In an interior span Total factored static moment Mo is distributed in such a way that the negative factored moment at face of support is taken as 0.65 Mo , and positive factored moment at mid span is taken as o 0.35 Mo. To Example Instructor: Eng. Mazen Alshorafa

  10. Design of Concrete Structure II University of Palestine Page 9 Direct Design Method [DDM] 2- Distribution of Mo to negative and positive moments (contd.) In end span Total factored static moment Mo is distributed as shown in the following figures. 1- For exterior edge unrestrained, e.g., supported by masonry wall End span 0.75 Mo 0.63 Mo To Table Instructor: Eng. Mazen Alshorafa

  11. Design of Concrete Structure II University of Palestine Page 10 Direct Design Method [DDM] 2- Distribution of Mo to negative and positive moments (contd.) 2- For slab with beams between all supports, 3A- For slab without beams, e.g., Flat Plate End span Interior beam Edge beam 0.7 Mo 0.16 Mo 0.57 Mo End span 0.7 Mo 0.26 Mo To Example To Table 0.52 Mo Instructor: Eng. Mazen Alshorafa

  12. Design of Concrete Structure II University of Palestine Page 11 Direct Design Method [DDM] 2- Distribution of Mo to negative and positive moments (contd.) 3B- For slab without beams between interior supports but with edge beam, 4- For exterior edge fully restrained, Monolithic concrete wall End span Edge beam 0.7 Mo 0.3 Mo 0.5 Mo End span 0.65 Mo 0.65 Mo To Table 0.35 Mo Instructor: Eng. Mazen Alshorafa

  13. Design of Concrete Structure II University of Palestine Case 1 Case 2 Case 3A Case 3B Case 4 Slab without Beams between Interior supports Slab with Beams between All supports` Exterior Edge Fully Restrained Exterior Edge Unrestrained Without Edge Beams With Edge Beams Interior negative moment 0.70 0.70 0.70 0.65 0.75 0.57 0.52 0.50 0.35 Positive moment 0.63 Exterior negative moment 0.16 0.26 0.30 0.65 0.00 Page 12 Direct Design Method [DDM] 2- Distribution of Mo to negative and positive moments (contd.) Distribution factors applied to static moment Mo for positive and negative moments in end span. Instructor: Eng. Mazen Alshorafa

  14. Design of Concrete Structure II University of Palestine Page 13 Direct Design Method [DDM] 3- Distribution of the positive and negative factored moments to the column and middle strips Columns strip: is a design strip with a width on each side of a column centerline equal to smaller of 0.25 l1 0.25 l2 Column strip includes beams, if any. Middle strip: is a design strip bounded by two column strips. l2A l2B l2A /2 l2A /2 l2B /2 l2B /2 min.(l1 /4, l2A /4) min.(l1 /4, l2B /4) min.(l1 /4, l2A /4) Interior column strip Exterior column strip l1 ½ Middle strip ½ Middle strip ½ Middle strip Instructor: Eng. Mazen Alshorafa

  15. Design of Concrete Structure II University of Palestine l2 / l1 1.0 2.0 0.5 75 75 α1 l2 / l1 = 0 75 α1 l2 / l1 ≥ 1.0 75 45 90 Page 14 Direct Design Method [DDM] 3- Distribution of the positive and negative factored moments to the column and middle strips (contd.) a- Factored moments in column strips For Interior Panel For interior negative moment, column strips have to be proportioned to resist the following portions in percent of the interior negative factored moments with linear interpolation made for intermediate values. α1 is the ratio of flexural stiffness of beam to stiffness of slab in direction l1. To Example Instructor: Eng. Mazen Alshorafa

  16. Design of Concrete Structure II University of Palestine Page 15 Direct Design Method [DDM] a- Factored moments in column strips (contd.) For Exterior Panel For exterior negative moments, column strips should be proportioned to resist the following portions in percent of the exterior negative factored moments with linear interpolation made for intermediate values. & β1 is the ratio of torsional stiffness of edge beam to flexural stiffness of a width of slab equal to beam length l2 / l1 1.0 2.0 0.5 100 100 100 β1 = 0.0 α1 l2 / l1 = 0 β1 ≥ 2.5 75 75 75 100 100 100 β1 = 0.0 α1l2 / l1 ≥ 1.0 β1 ≥ 2.5 75 45 90 To Example Instructor: Eng. Mazen Alshorafa

  17. Design of Concrete Structure II University of Palestine y1 y1 x1 x1 y2 y2 x2 x2 Page 16 Direct Design Method [DDM] a- Factored moments in column strips (contd.) Where x is the shorter overall dimension of rectangular part of cross section y is the longer overall dimension of rectangular part of cross section. The cross section is to be divided into Separate rectangular parts and carrying out the summation given in previous Equation in such away to give the largest value of C. Instructor: Eng. Mazen Alshorafa

  18. Design of Concrete Structure II University of Palestine l2 / l1 1.0 2.0 0.5 60 60 α1 l2 / l1 = 0 60 α1 l2 / l1 ≥ 1.0 75 45 90 Page 17 Direct Design Method [DDM] a- Factored moments in column strips (contd.) For Positive Moments For Positive moments, the column strips have to be proportioned to resist the following portions in percent of the positive factored moments with linear interpolation made for intermediate values. α1 is the ratio of flexural stiffness of beam to stiffness of slab in direction l1. To Example Instructor: Eng. Mazen Alshorafa

  19. Design of Concrete Structure II University of Palestine Page 18 Direct Design Method [DDM] b- Factored moments in middle strips The portion of the negative and positive factored moments not resisted by column strips shall be proportionately assigned to corresponding half middle strips. -ve Mmiddle= negative factored moment–negative column strip moment +ve Mmiddle= positive factored moment–positive column strip moment Each middle strip shall be proportioned to resist the sum of the moments assigned to its two half middle strips. Instructor: Eng. Mazen Alshorafa

  20. Design of Concrete Structure II University of Palestine Page 19 Direct Design Method [DDM] c- Factored moments in beams caused by slab loads Beams between supports are to be designed to resist 85 % of column strip moments if ≥ 1. For values of between 1.0 and zero, proportion of column strip moments resisted by beams is obtained by linear interpolation between 85 and zero. Beams85% of Mc for 85% x of Mc for Slabs Remainder of Mc “column strip moments” In addition to these moments, beams are to be proportioned to resist moments caused by factored loads applied directly on these beams. Instructor: Eng. Mazen Alshorafa

  21. Design of Concrete Structure II University of Palestine Page 20 Direct Design Method [DDM] d- Factored shears in beams caused by slab loads In addition to these shears, beams are to be proportioned to resist shears caused by factored loads applied directly on these beams. Beams with are to be ≥ 1 designed for shear caused by factored loads on tributary areas which are bounded by 45 degree lines drawn from the corners of the panels and the centerlines of the adjacent panels parallel to the long sides as shown. For values of less than 1.0, linear interpolation assuming beams carry no load at less than 1.0, is permitted to be used. Instructor: Eng. Mazen Alshorafa

  22. Design of Concrete Structure II University of Palestine fy (kg/cm2) To Example ln /36 ln /30 ln /33 ln /33 ln /33 ln /36 4200 Page 21 Depth limitations of the ACI Code a- Slab without interior beams According to ACI Code, for slabs without interior beams spanning between the supports and having a ratio of long to short span not greater than 2, the minimum thickness shall be in accordance with the provision of the following table, and shall not be less than the following values: a. Slabs without drop panels -------------------------------- 12.5 cm b. Slabs with drop panels ------------------------------------ 10.0 cm Instructor: Eng. Mazen Alshorafa

  23. Design of Concrete Structure II University of Palestine As specified by ACI Code, for slabs with beams spanning between the supports on all sides, the minimum thickness (hs, min) shall be as follows: a- For (αm ≤ 0.20), b- for (0.20<αm ≤ 2.0), c- for (2.0<αm), α1 α2 α4 α3 Page 22 Depth limitations of the ACI Code b- Slab with beams on all sides To Example Instructor: Eng. Mazen Alshorafa

  24. Design of Concrete Structure II University of Palestine Page 23 Depth limitations of the ACI Code b- Slab with beams on all sides (contd.) d- At discontinuous edges, an edge beam shall be provided with a stiffness ratio α not less than 0.80; otherwise the minimum thickness required by Equations in branch b or c shall be increased by at least 10 % in the panel with a discontinuous edge. The following figure shows dimensions of interior and edge beams that need to be considered in relative stiffness calculations. Instructor: Eng. Mazen Alshorafa

  25. Design of Concrete Structure II University of Palestine Page 24 Depth limitations of the ACI Code b- Slab with beams on all sides (contd.) To Example Instructor: Eng. Mazen Alshorafa

  26. Design of Concrete Structure II University of Palestine Page 22 Minimum extension for reinforcement in slabs without beams Instructor: Eng. Mazen Alshorafa

  27. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Example # 1 Instructor: Eng. Mazen Alshorafa

  28. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Example # 1 Design the two-way flat plate given the following: interior columns are 40×40cm, exterior columns are 30×30cm, covering materials weigh 2.0 kN/m2, equivalent partition load 0.5 kN/m2 and the live load is 2.5 kN/m2. Use fc’ = 28 MPa, fy = 420 MPa. 7 m 7 m 7 m 7 m 6 m 6 m 6 m Instructor: Page 0 Eng. Mazen Alshorafa

  29. Design of Concrete Structure II University of Palestine Solution 1- Evaluate slab thickness ln1 = 7-0.20-0.15= 6.65 m ln2 = 7-0.20-0.20= 6.60 m ln3 = 7-0.15-0.15= 6.70 m For flat plates with no edge beams, min. slab thickness hmin =6700/30= 223.3 mm taken as 220 mm. 7 m 7 m 7 m From table 7 m 6 m 6 m 6 m Instructor: Page 0 Eng. Mazen Alshorafa

  30. Design of Concrete Structure II University of Palestine Solution 2- Check limitations for slab analysis by the DDM: The first five conditions are satisfied, while the sixth condition does not apply due to the nonexistence of beams 3- Calculate the factored load on the slab Wu = 1.2[0.22 (25)+ 2.0+0.5] +1.6 [ 2.5] = 13.6 kN/m2 7 m 7 m See the limitation of DDM 7 m 7 m 6 m 6 m 6 m Instructor: Page 0 Eng. Mazen Alshorafa

  31. Design of Concrete Structure II University of Palestine Solution 4- Check slab thickness for shear davg =220−20−14=186mm, assuming φ14mm reinforcing bars. For example, take Interior columns **** Punching shear **** bo =4(400+186)= 2344mm= 2.344m Vu =13.6 [ 7(6) - (0.586)2]= 566.53 kN ΦVc is the smallest of βc =long side/short side of the column, βc = 40/40 =1.0 C1+d=586mm d/2 7 m C2+d=586mm d/2 6 m Instructor: Page 0 Eng. Mazen Alshorafa

  32. Design of Concrete Structure II University of Palestine sec 2 d 7 m d sec 1 6 m Solution **** Beam shear **** Section 1-1 Section 2-2 Note (Check slab thickness for corner and edge columns for shear) Instructor: Page 0 Eng. Mazen Alshorafa

  33. Design of Concrete Structure II University of Palestine Solution 5- Calculate the factored static moment Column and middle strips in the short direction Instructor: Page 0 Eng. Mazen Alshorafa

  34. Design of Concrete Structure II University of Palestine Solution Width of intermediate strip = 7000 mm and width of column strip is the smaller of ( l1/ 2) and (l2/ 2), taken as (6/2) = 3 m. Total factored static moment: Clear span for exterior panels = 6.0 – 0.15 – 0.2 = 5.65 m Clear span for interior panels = 6.0 – 0.2 – 0.2 = 5.60 m The larger of the two values will be used in moment calculations. Mo=wu l2 ln2 /8 = 13.6 (7) (5.65)2 / 8 = 379.9 kN.m 6- Distribute the total factored static moment into positive and negative moments End Span Interior Span 0.7 Mo=266 0.65 Mo= 247 0.65 Mo= 247 0.26 Mo=98.8 0.35 Mo= 133 0.52 Mo=197.5 Instructor: Page 0 Eng. Mazen Alshorafa

  35. Design of Concrete Structure II University of Palestine Solution 7- Distribute the positive and negative moments to the column and middle strips For Exterior Negative Moment l2/l1 = 7/6 =1.17, α1 = 0 and βt = 0, there is no beams between supports Mc = 100% MT = 1 * 98.8 = 98.8 kN.m MM = 98.8-98.8 = 0.0 For Interior Negative Moment l2/l1 = 7/6 =1.17 and α1 = 0 Mc =75% MT =0.75*266=199.5 kN.m MM = 266 – 199.5 = 66.5 kN.m For Positive Moments l2/l1 = 7/6 =1.17 and α1 = 0 Mc =60% MT =0.6*197.5=118.5 kN.m MM = 197.5 – 118.5 = 79 kN.m End Span 266.0 98.8 197.5 Total Moment 199.5 98.8 118.5 Column strip Moment 66.5 0.0 79 Middle strip Moment Instructor: Page 0 Eng. Mazen Alshorafa

  36. Design of Concrete Structure II University of Palestine 185.25 61.25 247.0 185.25 61.25 247.0 133.0 79.8 53.2 Solution 7- Distribute the positive and negative moments to the column and middle strips (contd.) For Interior Negative Moment l2/l1 = 7/6 =1.17 and α1 = 0 Mc =75% MT =0.75*247=185.25 kN.m MM = 247-185.25 = 61.75 kN.m For Positive Moments l2/l1 = 7/6 =1.17 and α1 = 0 Mc =60% MT =0.6*133=79.8 kN.m MM = 133-79.8 = 53.2 kN.m Interior Span Total Moment Column strip Moment Middle strip Moment Instructor: Page 0 Eng. Mazen Alshorafa

  37. Design of Concrete Structure II University of Palestine 199.5 199.5 185.3 185.3 98.8 98.8 79.8 118.5 118.5 Column strip Moment 66.5 66.5 61.3 61.3 0.0 0.0 53.2 79.0 79.0 Middle strip Moment Solution 7- Distribute the positive and negative moments to the column and middle strips (contd.) Instructor: Page 0 Eng. Mazen Alshorafa

  38. Design of Concrete Structure II University of Palestine 1443 0.00259 98.80 6Φ12 mm 1739 8Φ12 mm 118.5 0.00312 2997 14Φ12 mm 199.5 0.00537 6Φ12 mm 1159 0.00208 79.8 Solution 8- Design the reinforcement a - Column strip reinforcement Design sections at maximum +ve and -ve moments as rectangular sections where, d = 186 mm, and b = 3000 mm. Use fc ′ = 28 MPa , and fy= 420 MPa (mm2) 8Φ12 mm 8Φ12 mm 8Φ12+5Φ12 mm 5Φ12 mm Instructor: Page 0 Eng. Mazen Alshorafa

  39. Design of Concrete Structure II University of Palestine - 0.00 0.00 - 1584 7Φ12 mm 79.0 0.0018 1584 ----------- 0.0018 66.5 7Φ12 mm 1584 0.0018 53.2 بسم الله الرحمن الرحيم Solution 8- Design the reinforcement (contd.) b - Half middle strip reinforcement Design sections at maximum +ve and -ve moments as rectangular sections where, d = 186 mm, and b = 4000 mm. Use fc ′ = 28 MPa , and fy= 420 MPa (mm2) - 8Φ12 mm 8Φ12+8Φ12 mm 8Φ12 mm Instructor: Page 0 Eng. Mazen Alshorafa

  40. Design of Concrete Structure II University of Palestine 8Φ12 8Φ12 8Φ12 7Φ12 7Φ12 7Φ12 6Φ12 6Φ12 14Φ12 14Φ12 8Φ12 8Φ12 6Φ12 8Φ12 8Φ12 5Φ12 8Φ12 8Φ12 8Φ12 7Φ12 7Φ12 7Φ12 بسم الله الرحمن الرحيم Solution 9- Design the reinforcement (contd.) Instructor: Page 0 Eng. Mazen Alshorafa

  41. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Solution 5- Calculate the factored static moment Column and middle strips in the long direction Instructor: Page 0 Eng. Mazen Alshorafa

  42. Design of Concrete Structure II University of Palestine Solution Width of intermediate strip = 6000 mm and width of column strip is the smaller of ( l1/ 2) and (l2/ 2), taken as (6/2) = 3 m. Total factored static moment: Clear span for exterior panels = 7.0 – 0.15 – 0.2 = 6.65 m Clear span for interior panels = 7.0 – 0.2 – 0.2 = 6.60 m The larger of the two values will be used in moment calculations. Mo=wu l2 ln2 /8 = 13.6 (6) (6.65)2 / 8 = 451.1 kN.m 6- Distribute the total factored static moment into positive and negative moments End Span Interior Span 0.7 Mo=315.8 0.65 Mo= 293.2 0.65 Mo= 293.2 0.26 Mo=117.3 0.35 Mo= 157.9 0.52 Mo=234.6 Instructor: Page 0 Eng. Mazen Alshorafa

  43. Design of Concrete Structure II University of Palestine Solution 7- Distribute the positive and negative moments to the column and middle strips For Exterior Negative Moment l2/l1 = 6/7 =0.86, α1 = 0 and βt = 0, there is no beams between supports Mc = 100% MT = 1 * 117.3 = 117.3 kN.m MM = 98.8-98.8 = 0.0 For Interior Negative Moment l2/l1 = 6/7 =0.86 and α1 = 0 Mc =75% MT =0.75*315.8=236.9 kN.m MM = 315.8 – 236.9 = 78.9 kN.m For Positive Moments l2/l1 = 6/7 =0.86 and α1 = 0 Mc =60% MT =0.6*234.6=140.8 kN.m MM = 234.6 – 140.8 = 93.8 kN.m End Span 315.8 117.3 234.6 Total Moment 236.9 117.3 140.8 Column strip Moment 78.9 0.0 93.8 Middle strip Moment Instructor: Page 0 Eng. Mazen Alshorafa

  44. Design of Concrete Structure II University of Palestine 293.2 219.9 73.3 219.9 293.2 73.3 157.9 94.7 63.2 Solution 7- Distribute the positive and negative moments to the column and middle strips (contd.) For Interior Negative Moment l2/l1 = 6/7 =0.86 and α1 = 0 Mc =75% MT =0.75*293.2=219.9 kN.m MM = 293.2-219.9 = 73.3 kN.m For Positive Moments l2/l1 = 6/7 =0.86 and α1 = 0 Mc =60% MT =0.6*157.9=94.7 kN.m MM = 157.9-94.7 = 63.2 kN.m Interior Span Total Moment Column strip Moment Middle strip Moment Instructor: Page 0 Eng. Mazen Alshorafa

  45. Design of Concrete Structure II University of Palestine 236.9 199.5 219.9 219.9 219.9 219.9 236.9 117.3 98.8 117.3 94.7 94.7 140.8 118.5 140.8 Column strip Moment Solution 7- Distribute the positive and negative moments to the column and middle strips (contd.) 78.9 78.9 73.3 73.3 73.3 73.3 0.0 0.0 63.2 63.2 93.8 93.8 Instructor: Page 0 Eng. Mazen Alshorafa

  46. Design of Concrete Structure II University of Palestine (mm2) 1721 117.3 0.00309 7Φ12 mm 9Φ12 mm 2080 10Φ12 mm 9Φ12 mm 140.8 0.00373 3599 16Φ12 mm 9Φ12+7Φ12 mm 236.9 0.00645 1381 6Φ12 mm 7Φ12 mm 94.7 0.00248 3324 16Φ12 mm 7Φ12+ 7Φ12 mm 219.9 0.00597 Solution 8- Design the reinforcement a - Column strip reinforcement Design sections at maximum +ve and -ve moments as rectangular sections where, d = 186 mm, and b = 3000 mm. Use fc ′ = 28 MPa , and fy= 420 MPa Instructor: Page 0 Eng. Mazen Alshorafa

  47. Design of Concrete Structure II University of Palestine (mm2) - - - - - 1368 7Φ12 mm 6Φ12 mm 93.8 0.00245 1188 - 6Φ12+6Φ12 mm 78.9 0.0018 1188 5Φ12 mm 6Φ12 mm 63.2 0.0018 1188 - 6Φ12+ 6Φ12 mm 73.3 0.0018 Solution 8- Design the reinforcement b - Half middle strip reinforcement Design sections at maximum +ve and -ve moments as rectangular sections where, d = 186 mm, and b = 3000 mm. Use fc ′ = 28 MPa , and fy= 420 MPa Instructor: Page 0 Eng. Mazen Alshorafa

  48. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Example # 2 Instructor: Eng. Mazen Alshorafa

  49. Design of Concrete Structure II University of Palestine 6 m 6 m 6 m 7.5 m 7.5 m 7.5 m بسم الله الرحمن الرحيم Example # 1 For the two-way solid slab with beams on all column lines. All columns are 30cm×30cm, all beams are 30×60cm, covering materials weigh 2.0 kN/m2, and the live load is 3.0 kN/m2. Use fc’ = 28 MPa, fy = 420 MPa. Required : Evaluate slab thickness and moments acting on any of the internal beams, using the DDM. Instructor: Page 0 Eng. Mazen Alshorafa

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