PCI
th
6
Edition
Compression Component Design
Presentation Outline
•
•
•
•
Interaction diagrams
Columns example
Second order effects
Prestress wall panel example
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
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
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).
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
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
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
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
Step 1 – Determine Po for Mn = 0
Step 1 – Determine Po for Mn = 0
1200
1000
800
Pn
600
400
Pn, Mn
fPn, fMn
200
0
0
10
20
Mn
30
40
Step 2 – Determine Maximum Moment
• For members with nonprestressed
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.
Step 2 – Determine Maximum Moment
• Neutral Axis Location, c
0.003
c
fy
0.003 
d
Ey
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
• Determine the force in steel using strain compatibility
 c  ds

fs  
 0.003  E s
 c

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
• Maximum Axial Force
Pn   A comp  A 's  A 'ps    0.85  f 'c    fs  A s
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
Mn  Pn  e
  A comp  A 's  A 'ps    y t  y '    0.85  f 'c 
  fs  A s   y t  ds 
Step 2 – Determine Maximum Moment
1200
1000
800
Pn
600
400
200
Pn, Mn
fPn, fMn
0
0
10
20
Mn
30
40
Step 3 – Determine Mo for Pn = 0
• Same methods used in flexural member design
Where:
a

Mn  A ps  fps   dps    A s  fy
2

and
a
A ps  fps  A s  fy  A 's  f 'y
0.85  f 'c  b
a

  dps  
2

Step 3 – Determine Mo for Pn = 0
1200
1000
800
Pn
600
400
Pn, Mn
fPn, fMn
200
0
0
10
20
Mn
30
40
Step 3 – Determine Mo for Pn = 0
1200
Compression
controlled
(f = 0.65 or 0.70)
1000
800
Pn
600
400
Pn, Mn
fPn, fMn
200
Tension controlled
(f = 0.9)
0
0
10
20
Mn
30
40
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
Step 4 – Additional Points
• Determine the stress in the reinforcement.
For non-prestressed reinforcement
fs  Es  es  fy
Step 4 – Additional Points
• For prestressed reinforcement, the stress is
determined from a nonlinear stress-strain relationship
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
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
Step 4 – Additional Points
1200
1000
800
Pn
600
400
Pn, Mn
fPn, fMn
200
0
0
10
20
Mn
30
40
Step 5 – Calculate the Maximum Factored Axial
Pmax –
= 0.80fPo for tied columns
= 0.85fPo for spiral columns
0.8  f  Po  0.8  f  0.85  f`c (A g  A`s  A s  A ps  A`ps ) 
A
ps
 A`ps  fse  0.003  Eps    A ps  A`ps  fy
Step 5 – Calculate the Maximum Factored Axial
0.80fPo or
0.85fPo
1200
1000
800
Pn
600
400
Pn, Mn
fPn, fMn
200
0
0
10
20
Mn
30
40
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
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
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
Step 1 – Determine Po From Strain Diagram
With no prestressing steel, the equation
reduces to:
Po  Pn  0.85  f 'c   A - A 's - As    A 's  As  f y
 0.85  5  240 - 2 - 2    2  2  60   1243 kips
Step 2 – Determine Pnb and Mnb
• From Strain Diagram determine Steel
Stress
 0.003

f 's  e 's  E s  
c

d
'

  Es
 c

 0.003


10.5  2.5   29, 000   66.3ksi  60ksi
 10.5

f 's  60 ksi
Step 2 – Determine Pnb and Mnb
• Determine Compression Area
A comp = a  b = 0.8 10.5 12 
= 100.8 in2
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
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
Step 3 – Determine Mo
• Conservative solution neglecting
compressive reinforcement
a 
A s  fy
0.85  f 'c  b

2.0  60 
4.25 12 
 2.35in
2.35 
 a

Mo  A s fy  d -   2.0   60  17  1959kip - in

2 
 2

Mo  163.3kip  ft
Step 3 – Determine Mo
• Strength Reduction Factor
a
2.35
c 

 2.94in
1
0.8
0.003
0.003
et 
d  c  
17.5  2.94   0.0149
c
2.94
e t  0.005  f  0.9
fMo  0.9 163.3   147kip  ft
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
Step 5 – Calculate Maximum Design Load
Pmax= 0.80 fPo = 0.80 (808 kips)
= 646 kips
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
Slenderness / Secondary Effects
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
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
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
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
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
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
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 secondorder deflections
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.
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
 
K  L 1 360

 156
r
8 1
12
Figure 2.7.5 (page 2-54)
Assumptions
• Panel has 250 psi compressive stress
due to prestressing after losses
• Simple span
• Maximum moment occur at L/2
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
Step 1 – Check P-Critical
2
Pc =
  EIeff
l
2
Where:
EIeff – Effective Stiffness
Step 1 – Calculate Effective Stiffness
EIeff =
0.70  Ec  Ig
1  d
Where:
1+d = Accounts for sustained loads
Step 1 – Calculate Effective Stiffness
bh3
96(83 )
Ig 

 4096 in4
12
12
1.2 10 
1.2DL
d 

 0.556
Pu
1.2(10)  1.6  6 
EIeff 
0.70  Ec  Ig
1  d

0.70  4300  4096 
1.56
 7.90  106 kip  in2
Step 1 – Check P-Critical
2
Pc =
  EIeff
l
2

2
6
2
  7.9  10 kips  in
2
360 in
Pc  Pu
 602 kips
Step 2 – Calculate Final Displacement
No Wind, Assumed Un-Cracked Section
• Moments without secondary effect
Step 2 – Calculate Final Displacement
No Wind, Assumed Un-Cracked Section
• Calculate displacement for secondary
moments due to Axial load
2

21.6 1 3602


Pu  e  l
i 

 0.0221 in
6
16  E c  Ig 16 7.90  10

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.
Step 2 – Calculate Final Displacement
No Wind, Assumed Un-Cracked Section
• Additional Mid-Span displacement due
to P- effect
2

21.6  e   3602
Pu  e  l
i 

6
8  EIeff
8 7.90  10


  0.044
e
Step 2 – Calculate Final Displacement
No Wind, Assumed Un-Cracked Section
• 1st Iteration for P- 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)
Step 3 – Check Moments on
Panel Without Wind
• Conservative Mu at L/2
Mu = 10.8 + 21.6(1.069) = 33.9 kip-in
Step 4 – Check for Cracking Without Wind
• Tension Stress at exterior face
Mu  y 33.9  4 

 33 psi (Ten)
I
4096
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.)]
Total Stress Tension Face
16 psi
252 psi
Therefore, the analysis is valid.
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
Step 5 – Calculate Final Displacement
With Wind, Assumed Un-Cracked Section
Determine Effective Stiffness for Wind case
When considering wind, βd = 0
EIeff =
0.70  Ec  Ig
1  d



0.70 4300 4096in4
1.0
 1.23  107 kip  in2
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
w=
5  wu  l 4
384  EIeff





 0.28in
384 1.23  10 kip  in 
5 192 / 12 3602
7
2
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
Step 5 – Calculate Final Displacement
With Wind, Assumed Un-Cracked Section
• Additional mid-span displacement due
to P- effect
 i=
Pu  e  l 2
8  Ec  Ig

  
21.6 e 3602

7
2
8 1.23  10 kip  in


 0.028 e
Step 5 – Calculate Final Displacement
With Wind, Assumed Un-Cracked Section
• 1st Iteration for P- 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)
Step 6 – Check Moments on Panel With Wind
Pu  e
wl2
Mu 
+ Pu   +
2
8
0.192 3602
21.6 1
Mu 
+ 21.6 1.339 +
 299 kip  in
2
8



 
Step 7 – Check for Cracking With Wind

Muy  y
I

   292 psi(Tension)
229 4
4096
Step 7 – Check for Cracking With Wind
Bending Stress
-292psi
Half Panel weight
19 psi
Prestress
Dead load
Total Stress Exterior Face
250 psi
16 psi
-7 psi(T)
Therefore, the analysis is valid.
Step 8 – Check Interaction
• No Wind
21.6 kips
Pu =
+ 1.2  (15ft)  0.1 kips = 4.50 kips
ft
ft
8 ft
33.9in.-kips
Mu =
= 0.35 kip-ft
ft
[12 in. (8ft)]
ft
• With Wind
Mu =
299 in.-kips
= 3.1 kip-ft
ft
[12 in. (8ft)]
ft
Figure 2.7.5 (pages 2 – 54)
Points Are inside Curve
Questions?