PCI 6th Edition

1. PCI 6th Edition Compression Component Design

2. Presentation Outline Interaction diagrams Columns example Second order effects Prestress wall panel example

3. Compression Members Proportioned on the basis of strength design. Stresses under service conditions, particularly during handling and erection (especially of wall panels) must also be considered

4. Design Basis The procedures are based on Chapter 10 of the ACI Code Recommendations of the PCI Committee on Prestressed Concrete Columns Recommendations of the PCI Committee on Sandwich Wall Panel Columns

5. Design Process The capacity determined by constructing a capacity interaction curve. Points on this curve are calculated using the compatibility of strains and solving the equations of equilibrium as prescribed in Chapter 10 of the Code (ACI).

6. Reinforcement ACI 318-02 waives the minimum vertical reinforcement requirements for compression members if the concrete is prestressed to at least an average of 225 psi after all losses In addition, the PCI Recommended Practice permits the elimination of lateral ties if: Compression-controlled section Non-prestressed reinforcement is not considered in the calculation of Pn Non-prestressed reinforcement which is added for tension (e.g., for handling) is not considered in the calculation of Pn The nominal capacity is multiplied by 0.85

7. Development Length Mild Reinforcement and prestressed development length can play a significant role in capacity Additional Mild steel or special termination anchorages may be required Mechanical bar termination methods Threaded ends Anchored to end plates

8. Interaction Diagrams Separate curves X, Y for none rectangular cross sections Most architectural precast column sections are not rectangular, therefore it is necessary to calculate the actual centroid of the compression area Instead of using a/2, as for a rectangular cross section, it is necessary to calculate the actual centroid of the compression area which is indicated as y' Instead of using a/2, as for a rectangular cross section, it is necessary to calculate the actual centroid of the compression area which is indicated as y'

9. Interaction Diagram Steps Step 1 – Determine Po pure axial capacity Step 2 – Determine maximum moment Step 3 – Determine Mo for Pn = 0 Step 4 – Determine additional points Step 5 – Calculate the maximum factored axial resistance specified by the Code as: 0.80fPo for tied columns 0.85fPo for spiral columns

10. Step 1 – Determine Po for Mn = 0

11. Step 1 – Determine Po for Mn = 0

12. Step 2 – Determine Maximum Moment For members with non-prestressed reinforcement, this is the balance point For symmetrical prestressed members, it is sufficiently precise to assume that the point occurs when the compression block, a, is one-half the member depth. , which occurs when the net tensile strain in the extreme tension steel is equal to fy/Es. , which occurs when the net tensile strain in the extreme tension steel is equal to fy/Es.

13. Neutral Axis Location, c Where: fy = the yield strength of extreme tension steel Es = Modulus of elasticity of extreme tension steel d = depth to the extreme tension steel from the compression face of the member Step 2 – Determine Maximum Moment

14. Determine the force in steel using strain compatibility Where: ds = Depth of steel es = Strain of steel Es = Modulus of elasticity of reinforcing steel fs = Force in steel Step 2 – Determine Maximum Moment determine the stress in the all the elements using strain compatibility: determine the stress in the all the elements using strain compatibility:

15. Maximum Axial Force Where: Acomp = Compression area A’s = Area of non prestressed compression reinforcing A’ps = Area of compression prestressing reinforcing As = area of reinforcing at reinforcement level y’ = distance from top of c.g. to Acomp Step 2 – Determine Maximum Moment

16. Step 2 – Determine Maximum Moment

17. Step 2 – Determine Maximum Moment

18. Step 3 – Determine Mo for Pn = 0 Same methods used in flexural member design Where: and This is normally done by neglecting the reinforcement above the neutral axis and determining the moment capacityThis is normally done by neglecting the reinforcement above the neutral axis and determining the moment capacity

19. Step 3 – Determine Mo for Pn = 0 Normally done by neglecting the reinforcement above the neutral axis and determining the moment capacityNormally done by neglecting the reinforcement above the neutral axis and determining the moment capacity

20. Step 3 – Determine Mo for Pn = 0 This is a 3 point approximation of the interaction diagram If the applied axial load and moment are within area, there is no need to go further since this is a conservative approximation This is a 3 point approximation of the interaction diagram If the applied axial load and moment are within area, there is no need to go further since this is a conservative approximation

21. Step 4 – Additional Points Select a value of “c” and calculate a = ß1c Determine the value of Acomp from the geometry of the section Determine the strain in the reinforcement assuming that ec = 0.003 at the compression face of the column. For prestressed reinforcement, add the strain due to the effective prestress ese = fse/Eps In this step, use the same method used in step 2 In this step, use the same method used in step 2

22. Step 4 – Additional Points Determine the stress in the reinforcement. For non-prestressed reinforcement

23. Step 4 – Additional Points For prestressed reinforcement, the stress is determined from a nonlinear stress-strain relationship

24. Step 4 – Additional Points If the maximum factored moment occurs near the end of a prestressed element, where the strand is not fully developed, an appropriate reduction in the value of fps should be made

25. Step 4 – Additional Points Calculate f Pn and f Mn For compression controlled sections (without spiral reinforcement) the net tensile strain et in the extreme tension steel has to be less than or equal to that at the balance point f = 0.65 For Grade 60 reinforcement and for prestressed steel, this occurs when 0.002 = et. For tension controlled sections in which 0.005 = et, f = 0.9. For sections in which et is between these limits, 0.48 83 f = + et. For each point plotted on the nominal strength curve, multiply Pn and Mn by f to obtain the design strength curve. For Grade 60 reinforcement and for prestressed steel, this occurs when 0.002 = et. For tension controlled sections in which 0.005 = et, f = 0.9. For sections in which et is between these limits, 0.48 83 f = + et. For each point plotted on the nominal strength curve, multiply Pn and Mn by f to obtain the design strength curve.

26. Step 4 – Additional Points

27. Step 5 – Calculate the Maximum Factored Axial Pmax – = 0.80fPo for tied columns = 0.85fPo for spiral columns

28. Step 5 – Calculate the Maximum Factored Axial

29. Example, Find Interaction Diagramfor a Precast Column Given: Column cross section shown Concrete: f'c = 5000 psi Reinforcement: Grade 60 fy = 60,000 psi Es = 29,000 ksi Problem: Construct interaction curve for bending about x-x axis

30. Solution Steps Initial Step: Determine Column Parameters Step 1 – Determine Po from Strain Diagram Step 2 – Determine Pnb and Mnb Step 3 – Determine Mo Step 4 – Plot and add points as required Step 5 – Calculate maximum design load

31. Determine Column Parameters ß1 = 0.85 – 0.05 = 0.80 d = 20 – 2.5 = 17.5 in d' = 2.5 in 0.85f'c = 0.85(5) = 4.25 ksi Ag = 12(20) = 240 in2 As = As' = 2.00 in2 yt = 10 in

32. Step 1 – Determine Po From Strain Diagram With no prestressing steel, the equation reduces to: Assume all steel is at yield.Assume all steel is at yield.

33. Step 2 – Determine Pnb and Mnb From Strain Diagram determine Steel Stress

34. Step 2 – Determine Pnb and Mnb Determine Compression Area

35. Step 2 – Determine Pnb and Mnb With no prestressing steel

36. Step 2 – Determine Pnb and Mnb With no prestressing steel y' = a/2 = 4.20 in. y' = a/2 = 4.20 in.

37. Step 3 – Determine Mo Conservative solution neglecting compressive reinforcement

38. Step 3 – Determine Mo Strength Reduction Factor

39. Step 4 – Plot 3 Point Interaction From the previous 3 steps, 3 points have been determined. From these 3 points, a conservative 3 point approximation can be determined. Add additional points as required

40. Step 5 – Calculate Maximum Design Load Pmax= 0.80 fPo = 0.80 (808 kips) = 646 kips

41. Wall or Column Effective Width is the Least of The center-to-center distance between loads 0.4 times the actual height of the wall 6 times the wall thickness on either side The Recommended Practice specifies that the portion of a wall considered as effective for supporting concentrated loads or for determining the effects of slenderness shall be the least of the following: The width of the loaded portion plus six times the wall thickness on either side. Ribbed Panels - The width of the rib plus six times the thickness of the wall between ribs or on either side of the rib The Recommended Practice specifies that the portion of a wall considered as effective for supporting concentrated loads or for determining the effects of slenderness shall be the least of the following: The width of the loaded portion plus six times the wall thickness on either side. Ribbed Panels - The width of the rib plus six times the thickness of the wall between ribs or on either side of the rib

42. Slenderness / Secondary Effects Sections 10.10 through 10.13 of ACI 318-02 contain provisions for evaluating slenderness effects (buckling) of columns. Additional recommendations are given in the Recommended Practice. The term “slenderness effects,” can be described as the moments in a member produced when the line of action of the axial force is not coincident with the displaced centroid of the member. These moments, which are not accounted for in the primary analysis, are thus termed “secondary moments.” These secondary moments arise from changes in the geometry of the structure, and may be caused by one or more of the following: Sections 10.10 through 10.13 of ACI 318-02 contain provisions for evaluating slenderness effects (buckling) of columns. Additional recommendations are given in the Recommended Practice. The term “slenderness effects,” can be described as the moments in a member produced when the line of action of the axial force is not coincident with the displaced centroid of the member. These moments, which are not accounted for in the primary analysis, are thus termed “secondary moments.” These secondary moments arise from changes in the geometry of the structure, and may be caused by one or more of the following:

43. Causes of Slenderness Effects Relative displacement of the ends of the member due to: Lateral or unbalanced vertical loads in an unbraced frame, usually labeled “translation” or “sidesway.” Manufacturing and erection tolerances

44. Causes of Slenderness Effects Deflections away from the end of the member due to: End moment due to eccentricity of the axial load. End moments due to frame action continuity, fixity or partial fixity of the ends Applied lateral loads, such as wind Thermal bowing from differential temperature Manufacturing tolerances Bowing due to prestressing

45. Calculation of Secondary Effects ACI allows the use of an approximate procedure termed “Moment Magnification.” Prestressed compression members usually have less than the minimum 1% vertical reinforcement and higher methods must be used The PCI Recommended Practice suggests ways to modify the Code equations used in Moment Magnification, but the second-order, or “P-?” analysis is preferred

46. Second-Order (P-?) Analysis Elastic type analysis using factored loads. Deflections are usually only a concern under service load, the deflections calculated for this purpose are to avoid a stability failure The logic is to provide the same safety factor as for strength design

47. Second-Order (P-?) Analysis Iterative approach Lateral deflection is calculated, and the moments caused by the axial load acting at that deflection are accumulated Convergence is typical after three or four iterations If increase in deflection is not negligible the member may be approaching stability failure

48. Second-Order (P-?) Analysis Cracking needs to be taken into account in the deflection calculations The stiffness used in the second order analysis should represent the stiffness of the members immediately before failure May involve iterations within iterations Approximations of cracked section properties are usually satisfactory

49. Second-Order (P-?) Analysis Section 10.11.1 of ACI 318-02 has cracked member properties for different member types for use in second-order analysis of frames Lower bound of what can be expected for equivalent moments of inertia of cracked members and include a stiffness reduction factor fK to account for variability of second-order deflections

50. Second-Order (P-?) Analysis Effects of creep should also be included. The most common method is to divide the stiffness (EI) by the factor 1 + ßd as specified in the ACI moment magnification method A good review of second-order analysis, along with an extensive bibliography and an outline of a complete program, is contained in Ref. 24.

51. Example, Second Order Analysis of Uncracked Member Given: An 8 in. thick, 8 ft wide prestressed wall panel as shown. Loading assumptions are as follows: Axial load eccentricity = 1 in (at one end) Assume midspan bowing = 1.0 in outward. Wind load = 30 psf Pinned top and bottom Concrete: f`c = 5000 psi Ec = 4300 ksi Example page 4-108Example page 4-108

52. Figure 2.7.5 (page 2-54) Page 2-54 Figure 2.7.5Page 2-54 Figure 2.7.5

53. Assumptions Panel has 250 psi compressive stress due to prestressing after losses Simple span Maximum moment occur at L/2

54. Solution Steps Step 1 – Check P-Critical Step 2 – Calculate Final Displacement without wind assuming un-cracked section Step 3 – Check moments on panel including secondary moments without wind Step 4 – Check for cracking without wind Step 5 – Calculate Final Displacement with wind assuming un-cracked section Step 6 – Check moments on panel including secondary moments with wind Step 7 – Check for cracking with wind Step 8 – Compare results against interaction diagram

55. Step 1 – Check P-Critical Where: EIeff – Effective Stiffness Note: The bending stiffness EI is multiplied by 0.70 per Section 10.11.1, ACI 318-02, for uncracked wall sections. This includes a stiffness reduction factor fK to cover the variability of computed deflections. Once the moments are established, the f factor from Section 9.3.2.2 of ACI 318-02 is used to determine the strength of the cross sectionNote: The bending stiffness EI is multiplied by 0.70 per Section 10.11.1, ACI 318-02, for uncracked wall sections. This includes a stiffness reduction factor fK to cover the variability of computed deflections. Once the moments are established, the f factor from Section 9.3.2.2 of ACI 318-02 is used to determine the strength of the cross section

56. Step 1 – Calculate Effective Stiffness Where: 1+bd = Accounts for sustained loads

57. Step 1 – Calculate Effective Stiffness

58. Step 1 – Check P-Critical

59. Step 2 – Calculate Final Displacement No Wind, Assumed Un-Cracked Section Moments without secondary effect

60. Step 2 – Calculate Final Displacement No Wind, Assumed Un-Cracked Section Calculate displacement for secondary moments due to Axial load

61. Step 2 – Calculate Final Displacement No Wind, Assumed Un-Cracked Section Total mid-span displacement = Axial load displacement + initial mid-span bow = 1.0 in. + 0.0221 = 1.022 in.

62. Step 2 – Calculate Final Displacement No Wind, Assumed Un-Cracked Section Additional Mid-Span displacement due to P-D effect

63. Step 2 – Calculate Final Displacement No Wind, Assumed Un-Cracked Section 1st Iteration for P-D Effect ? = 0.044(1.022 in) = 0.045 in 2nd Iteration e = 1.022 in. + 0.045in. = 1.067 in ? = 0.044(1.067 in.) = 0.047 in 3rd Iteration e = 1.022 in. + 0.047 in.= 1.069 in ? = 0.044(1.069 in.) = 0.047 in (convergence)

64. Step 3 – Check Moments onPanel Without Wind Conservative Mu at L/2 Mu = 10.8 + 21.6(1.069) = 33.9 kip-in The maximum moment does not occur at midheight but can be assumed to act very close to it.The maximum moment does not occur at midheight but can be assumed to act very close to it.

65. Step 4 – Check for Cracking Without Wind Tension Stress at exterior face

66. Step 4 – Check for Cracking Without Wind 252 psi in compression panel has not cracked so assumed section properties are valid.252 psi in compression panel has not cracked so assumed section properties are valid.

67. Step 5 – Calculate Final Displacement With Wind, Assumed Un-Cracked Section Axial Load Displacement Deflection due to Pue is the same as the previous case = 0.0221 in

68. Step 5 – Calculate Final Displacement With Wind, Assumed Un-Cracked Section Determine Effective Stiffness for Wind case When considering wind, ßd = 0

69. Step 5 – Calculate Final Displacement With Wind, Assumed Un-Cracked Section Additive wind load (suction) = 30 psf wu = 30 psf (8ft)(0.8) = 192 lb/ft

70. Step 5 – Calculate Final Displacement With Wind, Assumed Un-Cracked Section Total initial mid-span bow including eccentricity and wind e = 1.0 + 0.022 + 0.28 = 1.302 in

71. Step 5 – Calculate Final Displacement With Wind, Assumed Un-Cracked Section Additional mid-span displacement due to P-D effect

72. Step 5 – Calculate Final Displacement With Wind, Assumed Un-Cracked Section 1st Iteration for P-D Effect ? = 0.028(1.302in.) = 0.036 in 2nd Iteration e = 1.302 in. + 0.036 in. = 1.338 in ? = 0.028(1.338 in.) = 0.037 in 3rd Iteration e = 1.302 in. + 0.037 in. = 1.339 in ? = 0.028(1.339 in.) = 0.037 in. (convergence)

73. Step 6 – Check Moments on Panel With Wind

74. Step 7 – Check for Cracking With Wind

75. Step 7 – Check for Cracking With Wind 7 psi in tension panel has not cracked so assumed section properties are valid.7 psi in tension panel has not cracked so assumed section properties are valid.

76. Step 8 – Check Interaction No Wind With Wind

77. Figure 2.7.5 (pages 2 – 54) Page 2-54Page 2-54

78. Questions?