PCI 6 th 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

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 Pn Pn, Mn fPn, fMn Mn

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.

13. Step 2 – Determine Maximum Moment • 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

14. Step 2 – Determine Maximum Moment • 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

15. Step 2 – Determine Maximum Moment • 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

16. Step 2 – Determine Maximum Moment

17. Step 2 – Determine Maximum Moment Pn Pn, Mn fPn,fMn Mn

18. Step 3 – Determine Mo for Pn= 0 • Same methods used in flexural member design Where: and

19. Step 3 – Determine Mo for Pn= 0 Pn, Mn fPn, fMn Pn Mn

20. Step 3 – Determine Mo for Pn= 0 Pn, Mn fPn, fMn Compression controlled (f = 0.65 or 0.70) Pn Tension controlled (f = 0.9) Mn

21. Step 4 – Additional Points • Select a value of “c” and calculate a = β1c • Determine the value of Acomp from the geometry of the section • Determine the strain in the reinforcement assuming that εc= 0.003 at the compression face of the column. For prestressed reinforcement, add the strain due to the effective prestress εse= fse/Eps

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 fPn and fMn • For compression controlled sections (without spiral reinforcement) the net tensile strain εt in the extreme tension steel has to be less than or equal to that at the balance point • f = 0.65

26. Step 4 – Additional Points Pn, Mn fPn, fMn Pn Mn

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 Pn, Mn fPn, fMn 0.80fPo or 0.85fPo Pn Mn

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 β1 = 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:

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 Pnb = (Acomp-A′s)(0.85f′c) +A′s f′s - As fs = (100.8 - 2)(4.25) + 2 (60)- 2(60) = 419.9 kips f Pnb = 0.65(419.9) = 273 kips

36. Step 2 – Determine Pnb and Mnb • With no prestressing steel Mnb = (Acomp − A′s)(yt − y′)(0.85f′c) + A′sf′s(yt − d′) + Asfs(d − yt) = (100.8 – 2)(10 – 4.20)(4.25) + 2(60)(10 – 2.5) + 2(60)(17.5 – 10) = 2435 + 900 + 900 = 4235 kip-in. fMnb = 0.65(4235 kip-in.) = 2752 kip-in. = 229 kip-ft

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

38. Step 3 – Determine Mo • Strength Reduction Factor

39. 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 Step 4 – Plot 3 Point Interaction

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

42. Slenderness / Secondary Effects

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 + βd 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.