PCI 6 th Edition Compression Component Design. Presentation Outline Interaction diagrams Columns example Second order effects Prestress wall panel example

PCI 6th EditionPCI 6th Edition

Compression Component Design

Presentation OutlinePresentation Outline

• Interaction diagrams

• Columns example

• Second order effects

• Prestress wall panel example

Compression MembersCompression 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 BasisDesign 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 ProcessDesign 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).

ReinforcementReinforcement

• 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 LengthDevelopment 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 DiagramsInteraction 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 StepsInteraction Diagram Steps

Step 1 – Determine Po pure axial capacityStep 2 – Determine maximum moment

Step 3 – Determine Mo for Pn = 0Step 4 – Determine additional points Step 5 – Calculate the maximum factored axial

resistance specified by the Code as:• 0.80Po for tied columns• 0.85Po for spiral columns

Step 1 – Determine Po for Mn = 0Step 1 – Determine Po for Mn = 0

Pn

Mn

Pn, Mn

Pn, Mn

Step 1 – Determine Po for Mn = 0Step 1 – Determine Po for Mn = 0

0

200

400

600

800

1000

1200

0 10 20 30 40

Step 2 – Determine Maximum MomentStep 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.

• Neutral Axis Location, c

Where:

fy = the yield strength of extreme tension steel

Es = Modulus of elasticity of extreme tension steeld = depth to the extreme tension steel from the

compression face of the member

c 0.003

0.003 fy

Ey

d


• Determine the force in steel using strain compatibility

Where:

ds = Depth of steel

sStrain of steel

Es = Modulus of elasticity of reinforcing steel

fs = Force in steel

ss s

c df 0.003 E

c


• Maximum Axial Force

Where:Acomp = Compression area A’s = Area of non prestressed compression

reinforcingA’ps = Area of compression prestressing reinforcingAs = area of reinforcing at reinforcement levely’ = distance from top of c.g. to Acomp

n comp s ps c s sP A A' A' 0.85 f ' f A


n n

comp s ps t c

s s t s

M P e

A A' A' y y' 0.85 f '

f A y d


Pn

Mn

Pn, Mn

Pn, Mn


0

200

400

600

800

1000

1200

0 10 20 30 40

Step 3 – Determine Mo for Pn = 0Step 3 – Determine Mo for Pn = 0

• Same methods used in flexural member design

Where:

and

a

Apsf

ps A

sf

y A'

sf '

y

0.85f 'cb

n ps ps ps s y ps

a aM A f d A f d

2 2


Pn

Mn

Pn, Mn

Pn, Mn

0

200

400

600

800

1000

1200

0 10 20 30 40

0

200

400

600

800

1000

1200

0 10 20 30 40

Pn

Mn

Pn, Mn

Pn, Mn

Compression controlled

( = 0.65 or 0.70)

Tension controlled ( = 0.9)


Step 4 – Additional PointsStep 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


• Determine the stress in the reinforcement. For non-prestressed reinforcement

fsE

s

sf

y


• For prestressed reinforcement, the stress is determined from a nonlinear stress-strain relationship


• 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


• Calculate Pn and 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

= 0.65


Pn

Mn

Pn, Mn

Pn, Mn

0

200

400

600

800

1000

1200

0 10 20 30 40

Step 5 – Calculate the Maximum Factored AxialStep 5 – Calculate the Maximum Factored Axial

Pmax –

= 0.80Po for tied columns

= 0.85Po for spiral columns

o c g s s ps ps

ps ps se ps ps ps y

0.8 P 0.8 0.85 f` (A A A A A )

A A f 0.003 E A A f

0

200

400

600

800

1000

1200

0 10 20 30 40

Step 5 – Calculate the Maximum Factored AxialStep 5 – Calculate the Maximum Factored Axial

Pn

Mn

0.80fPo or 0.85fPo

Pn, Mn

Pn, Mn

Example, Find Interaction Diagramfor a Precast Column

Example, Find Interaction Diagramfor a Precast Column

Given:Column cross section shown

Concrete: f′c = 5000 psiReinforcement: Grade 60

fy = 60,000 psi

Es = 29,000 ksiProblem:

Construct interaction curve for bending about x-x axis

Solution StepsSolution 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 ParametersDetermine 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 DiagramStep 1 – Determine Po From Strain Diagram

With no prestressing steel, the equation reduces to:

0.85 ' - ' - '

0.85 5 240- 2- 2 2 2 60 1243

o n c s s s s yP P f A A A A A f

kips

Step 2 – Determine Pnb and MnbStep 2 – Determine Pnb and Mnb

• From Strain Diagram determine Steel Stress

s s s s

s

0.003f ' ' E c d' E

c

0.003 10.5 2.5 29,000 66.3ksi 60ksi

10.5

f ' 60 ksi


• Determine Compression Area

comp

2

A = a b = 0.8 10.5 12

= 100.8 in


• 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

Pnb = 0.65(419.9) = 273 kips


• With no prestressing steelMnb = (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.

Mnb = 0.65(4235 kip-in.) = 2752 kip-in. = 229 kip-ft

Step 3 – Determine MoStep 3 – Determine Mo

• Conservative solution neglecting compressive reinforcement

s y

c

A f 2.0 60a 2.35in

0.85 f ' b 4.25 12

o s y

o

a 2.35M A f d- 2.0 60 17- 1959kip- in

2 2

M 163.3kip ft

Step 3 – Determine MoStep 3 – Determine Mo

• Strength Reduction Factor

1

t

t

a 2.35c 2.94in

0.8

0.003 0.003d c 17.5 2.94 0.0149

c 2.940.005 0.9

oM 0.9 163.3 147kip ft

Step 4 – Plot 3 Point InteractionStep 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 LoadStep 5 – Calculate Maximum Design Load

Pmax= 0.80 Po = 0.80 (808 kips)

= 646 kips

Wall or ColumnWall 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 EffectsSlenderness / Secondary Effects

Causes of Slenderness EffectsCauses 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 EffectsCauses 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 EffectsCalculation 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-∆) AnalysisSecond-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


• 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


• 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


• 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 K to account for variability of second-order deflections


• 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

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 Lr

1 360 8 1

12

156

Figure 2.7.5 (page 2-54)Figure 2.7.5 (page 2-54)

AssumptionsAssumptions

• Panel has 250 psi compressive stress due to prestressing after losses

• Simple span

• Maximum moment occur at L/2

Solution StepsSolution Steps

Step 1 – Check P-CriticalStep 2 – Calculate Final Displacement without wind

assuming un-cracked sectionStep 3 – Check moments on panel including secondary

moments without windStep 4 – Check for cracking without windStep 5 – Calculate Final Displacement with wind

assuming un-cracked sectionStep 6 – Check moments on panel including secondary

moments with windStep 7 – Check for cracking with windStep 8 – Compare results against interaction diagram

Step 1 – Check P-CriticalStep 1 – Check P-Critical

Where:

EIeff – Effective Stiffness

P

c=2 EI

eff

l 2

Step 1 – Calculate Effective StiffnessStep 1 – Calculate Effective Stiffness

Where:

1+d = Accounts for sustained loads

EI

eff=

0.70EcI

g

1 d

Step 1 – Calculate Effective StiffnessStep 1 – Calculate Effective Stiffness

c g 6 2eff

d

0.70 E I 0.70 4300 4096EI 7.90 10 kip in

1 1.56

3 34

g

du

bh 96(8 )I 4096 in

12 121.2 101.2DL

0.556P 1.2(10) 1.6 6

Step 1 – Check P-CriticalStep 1 – Check P-Critical

Pc=2 EI

eff

l 22 7.9106 kipsin2

360 in2602 kips

PcP

u

Step 2 – Calculate Final Displacement No Wind, Assumed Un-Cracked Section


• Moments without secondary effect



• Calculate displacement for secondary moments due to Axial load

22u

i 6c g

21.6 1 360P e l0.0221 in

16 E I 16 7.90 10



• Total mid-span displacement = Axial load displacement + initial mid-span bow

= 1.0 in. + 0.0221 = 1.022 in.



• Additional Mid-Span displacement due to P- effect

22u

i 6eff

21.6 e 360P e l0.044 e

8 EI 8 7.90 10



• 1st Iteration for P- Effect∆ = 0.044(1.022 in) = 0.045 in

• 2nd Iteratione = 1.022 in. + 0.045in. = 1.067 in∆ = 0.044(1.067 in.) = 0.047 in

• 3rd Iteratione = 1.022 in. + 0.047 in.= 1.069 in∆ = 0.044(1.069 in.) = 0.047 in (convergence)

Step 3 – Check Moments onPanel Without Wind

Step 3 – Check Moments onPanel Without Wind

• Conservative Mu at L/2

Mu = 10.8 + 21.6(1.069) = 33.9 kip-in

Step 4 – Check for Cracking Without WindStep 4 – Check for Cracking Without Wind

• Tension Stress at exterior face

u33.9 4M y

33 psi (Ten)I 4096

Step 4 – Check for Cracking Without WindStep 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

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



Determine Effective Stiffness for Wind case

When considering wind, βd = 0

EI

eff=

0.70EcI

g

1 d

0.70 4300 4096in4

1.01.23107 kipin2



Additive wind load (suction) = 30 psf

wu = 30 psf (8ft)(0.8) = 192 lb/ft

w=

5wul 4

384EIeff

5 192 / 12 3602

384 1.23107 kipin2 0.28in



Total initial mid-span bow including eccentricity and wind

e = 1.0 + 0.022 + 0.28 = 1.302 in



• Additional mid-span displacement due to P- effect

i=

Puel 2

8EcI

g

21.6 e 3602

8 1.23107 kipin2 0.028 e



• 1st Iteration for P- Effect∆ = 0.028(1.302in.) = 0.036 in

• 2nd Iteratione = 1.302 in. + 0.036 in. = 1.338 in∆ = 0.028(1.338 in.) = 0.037 in

• 3rd Iteratione = 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 WindStep 6 – Check Moments on Panel With Wind

Mu

Pue

2 + P

u +

wl 2

8

Mu

21.6 12

+ 21.6 1.339 + 0.192 3602

8299 kipin

Step 7 – Check for Cracking With WindStep 7 – Check for Cracking With Wind

Muyy

I

229 4 4096

292 psi(Tension)

Step 7 – Check for Cracking With WindStep 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.

Step 8 – Check InteractionStep 8 – Check Interaction

• No Wind

• With Wind

Pu =

21.6 kips

8 ft + 1.2(15ft)0.1 kips

ft = 4.50 kips

ft

Mu =

33.9in.-kips

[12in.ft

(8ft)] = 0.35 kip-ft

ft

Mu =

299 in.-kips

[12in.ft

(8ft)] = 3.1 kip-ft

ft

Figure 2.7.5 (pages 2 – 54)Figure 2.7.5 (pages 2 – 54)

Points Are inside Curve

Questions?Questions?

Documents

PCI 6 th Edition Compression Component Design. Presentation Outline Interaction diagrams Columns example Second order effects Prestress wall panel example