1. Compression Component Design
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 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′

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 = 0Pn
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.
, which occurs when the net tensile strain in the extreme tension steel is equal to fy/Es.

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
determine the stress in the all the elements using strain compatibility:

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 MomentPn
Pn, Mn
fPn, fMn Mn

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 capacity

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

20. Step 3 – Determine Mo for Pn = 0
Compression controlled
(f = 0.65 or 0.70) Pn
Pn, Mn
fPn, fMn
Tension controlled
(f = 0.9)
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 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 = at the compression face of the column. For prestressed reinforcement, add the strain due to the effective prestress εse = fse/Eps
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 εt 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 ≤ εt. For tension controlled sections in which ≥ εt, f = 0.9.
For sections in which εt is between these limits, f = + εt. 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 PointsPn
Pn, Mn
fPn, fMn 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
0.80fPo or 0.85fPo Pn
Pn, Mn
fPn, fMn Mn

29. Example, Find Interaction Diagram for 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:
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
Pnb = (Acomp-A′s)(0.85f′c) +A′s f′s - As fs
= ( )(4.25) + 2 (60)- 2(60)
= 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
= 4235 kip-in.
fMnb = 0.65(4235 kip-in.) = 2752 kip-in. = 229 kip-ft
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

42. Slenderness / Secondary Effects
Sections through of ACI 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 of ACI 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.

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 4-108

52. Figure (page 2-54)
Page 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 , ACI , 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 of ACI 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. Moments without secondary effect
Step 2 – Calculate Final Displacement No Wind, Assumed Un-Cracked Section
Moments without secondary effect

60. Calculate displacement for secondary moments due to Axial load
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 = in.

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

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

64. Step 3 – Check Moments on Panel Without Wind
Conservative Mu at L/2
Mu = (1.069) = 33.9 kip-in
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
Bending Stress
-33psi
Half Panel weight [100psf(15ft)/(8(12))](1.2)
19 psi
Prestress
250 psi
Dead load
10,000 lbs(1.2)/[8 in.(96 in.)]
16 psi
Total Stress Tension Face
252 psi
252 psi in compression panel has not cracked so assumed section properties are valid.
Therefore, the analysis is valid.

67. Axial Load Displacement
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
= in

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

69. Additive wind load (suction) = 30 psf
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. Total initial mid-span bow including eccentricity and wind
Step 5 – Calculate Final Displacement With Wind, Assumed Un-Cracked Section
Total initial mid-span bow including eccentricity and wind
e = = in

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

72. 1st Iteration for P-D Effect
Step 5 – Calculate Final Displacement With Wind, Assumed Un-Cracked Section
1st Iteration for P-D Effect
∆ = 0.028(1.302in.) = in
2nd Iteration
e = in in. = in
∆ = 0.028(1.338 in.) = in
3rd Iteration
e = in in. = in
∆ = 0.028(1.339 in.) = 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
Bending Stress
-292psi
Half Panel weight
19 psi
Prestress
250 psi
Dead load
16 psi
Total Stress Exterior Face
-7 psi(T)
Therefore, the analysis is 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. Points Are inside Curve
Figure (pages 2 – 54)
Points Are inside Curve
Page 2-54

78. Questions?