Non-Dimensional Design Procedures for Precast ... - PCI.org

Non-Dimensional Design

Procedures for Precast,

Prestressed Concrete

Hybrid Frames

Rami Hawlleh, Ph.D.ResearcherDepartment of Civil Engineering andMechanicsUniversity of Wisconsin—MilwaukeeCollege of Engineering and AppliedScienceMilwaukee, Wis.

Habib Tabatabai, Ph.D., S.E.Associate Professor

Department of Civil Engineering andMechanics

University of Wisconsin—MilwaukeeCollege of Engineering and Applied

ScienceMilwaukee, Wis

Adeeb Rahman, Ph.D.Associate ProfessorDepartment of Civil Engineering andMechanicsUniversity of Wisconsin—MilwaukeeCollege of Engineering and AppliedScienceMilwaukee, Wis.

Akef Amro, M.S.Structural Engineer

ATKINS Highways & TransportationBristol, U.K.

______

The precast, prestressed concrete hybrid framesystem offers an important feature that allows theelimination of residual drift after an earthquake. Theexisting design process for hybrid frames includesan iterative, step-by-step procedure. The objectiveof this research is to develop a non-iterative, simplified design procedure for precast concrete hybridframes. A set of new non-dimensional parametersand procedures for the design of precast concretehybrid frames has been developed. A fracture criterion for the mild steel reinforcing bars that considers low-cycle fatigue under combined axial andbending strains is proposed. Parametric studieswere performed using the developed non-dimensional procedure. These optimization studies aimedto achieve zero residual drift while the momentcapacity of the section is equal to the applied designmoment. The results of these studies were used togenerate non-dimensional design charts and equations. An example using the new procedures is presented in detail in Appendix B. The results usingconventional procedures and the proposed designcharts and design equations are compared.

The Precast Seismic Structural Systems (PRESSS)research proposed a total of five different seismic structural systems comprising precast concrete

elements.1’2These systems were used in various parts ofthe structural framing in the PRESSS Phase III experimental building that was tested at the University of California atSan Diego.

The system identified as the unbonded post-tensioned con-

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Crete frame with damping (also Called the hybrid frame) performed very well according to the PRESSS evaluation. Thehybrid frame system offers an important feature: It allows theelimination of the residual drift of a structure after an earthquake. The PRESSS report outlined an iterative, step-by-stepdesign procedure for hybrid frames.

Celik and Sritharan proposed a number of modifications tothe original PRESSS procedures.3In this paper, other modifications are also proposed. In some instances, the modifications proposed by Celik and Sritharan and those offered inthis paper address the same issues related to the PRESSS procedures, even though the approach is somewhat different.

HYBRID FRAME CONCEPT DESCRIPTION

Hybrid frames contain precast concrete elements (beamsand columns) that are connected by unbonded post-tensioning steel and partially debonded mild steel reinforcing bars,where both types of steel contribute to the overall momentresistance of the connection (Fig. 1). An important feature ofthis type of connection is the combination of mild steel andpost-tensioning steel. In this “hybrid” connection the mildsteel reinforcement dissipates energy (by yielding in tension and compression) and the post-tensioning steel clampsthe beam against the column. The post-tensioning force actsas a restoring force to bring the frame back to its originalconfiguration after an earthquake and also provides for shearresistance through friction developed at the beam-columninterface.

This combination of reinforcement is intended to create aductile connection and ensure that failure occurs within the

connection. For a given total moment strength, a higher proportion of mild steel reinforcement would result in more apparent damping, lower peak drift, and higher residual drift.

The general design philosophy of the hybrid frame, as presented by Stanton and Nakaki, is to minimize peak drift, maintain zero residual drift, and maximize the moment strengthprovided by the mild steel reinforcement to achieve a ductileconnection.’ Figure 2 shows the relationship between driftand the relative moment strengths of the resisting elements.

A parametric optimization study using non-dimensionalparameters was conducted and is discussed in this paper.

Fig. 2. Drift versus relative strength of resisting elements.’

OBJECTIVES

Following are the objectives of this research:• Develop a set of non-dimensional parameters and

procedures to replace current dimensional parametersin the calculations to design a hybrid frame.

• Perform parametric studies based on these new non-dimensional procedures developed from the PRESSSdesign procedures. These optimization studies weredesigned to achieve zero residual drift while ensuring that the moment capacity of the section was equalto the applied design moment and that other relevantconstraints were satisfied simultaneously.

• Develop and verify design charts and equations for thedesign of hybrid frames. These charts and equationswould allow engineers to design hybrid frames withfewer steps while providing insight into how changing one or more design parameters would affect theoverall design of the frame.

• Develop design criterion for the fracture of mildreinforcing steel based on low-cycle fatigue undercombined axial and bending strains.

HYBRID FRAME:PRESSS DESIGN PROCEDURES

The design equations proposed in the PRESSS report arebased on the following assumptions:’

• The design moments Macs and drifts are known. Theinterface rotation at the design limit state 0des is relatedto the drift ratio based on system geometry and is known.

• The overall dimensions of the frame members are known.• At each interface, the centroid of the post-tensioning

tendon is located at the mid-depth of the beam section,and the beam cross section is constant.

Drift

Peak drift II Residual

Meaps /total beam moment strength at the interface (M+M55)

= moment strength due to the post-tensioning force= moment strength due to mild steel reinforcement

Mild reinforcingbar (A706)(grouted in duct) -

- Reinforcing bar debondedlocally

Post-tensioningtendon (unbonded)

Fiber-reinforcedgrout

Fig. 1. Hybrid frame.

September—October 2006 111

Dimensions and displacements Forces

Fig. 3. Deformations and forces at design drift.1

• The post-tensioning tendon is at incipient yield at thedesign drift.

• The post-tensioning tendon is non-varying and Un-bonded over the entire length of the frame.

• Equal amounts of mild steel reinforcement are used inthe top and bottom of the beam.

• Properties of the materials are known.The PRESSS design equations use deformation-compat

ibility and force-equilibrium relationships to calculate theforces and the resulting moment capacity at the interface between a precast concrete beam and column. Figure 3 showsthe deformations and forces acting on a joint between a precast concrete beam and column in a hybrid-frame connectionsubjected to a design interface rotation of 0d

According to the PRESSS procedures, a total of 17 stepsare required to design the post-tensioning tendon and themild steel reinforcement at the interface. The PRESSS designprocedure is iterative and utilizes a number of dimensionaldesign parameters. A detailed explanation of the PRESSSdesign procedure can be found in the PRESSS reportNo. 3—09)

FRACTURE CRITERIA FOR MILD STEELREINFORCING BARS

The following discussion on cyclic behavior and low-cycle fatigue of mild reinforcing steel is not presented in thePRESSS report but is proposed here as an important consideration in the design of hybrid frames.

Mander et al. experimentally evaluated the low-cycle fatigue behavior of ASTM A6 15, Grade 40 (280 MPa) deformed reinforcing steel bars subjected to cyclic axial strainamplitudes ranging from yield to 6%. The relationship of

total strain amplitude ç to fatigue life for mild steel reinforcing bars is given by Mander et al.:

LIEç = —-

= 0.0795(2Nf)°8

where the total strain amplitude ç is also equal to

where

E —C.= mat mm

2

(1)

ç= the maximum total strain in a cycleEm. = the minimum total strain in a cycleN = the number of cycles to failureUsing Eq. (1), the number of cycles to failure N can be

calculated:

N0.0018

— (s )2.23

The hysteretic behavior of bars subjected to cyclic deformations can be considered when determining the stress levelin the bar at maximum design strain. The maximum tensilestress fm coffesponding to La can be calculated using thefollowing equation proposed by Mander et al.:

fmaa, TEO = 7.48(2A)°541 (ksi) [51.57(Nf)°541(MPa)]

Substituting for Nf from Eq. (1) yields

Lax. T = 157(Ea)O2 (ksi) [1082(Ea)°2(MPa)}

The over-strength factor in tension tsdtt can be calculated

112 PCI JOURNAL

Step 6

=

= E,, + LIE,, (XE,,,.,

= E,,c,,

j: .tp,d,.x

Step 7

Assume value for r(o.2%-o.25%)

Q =-L, check r 0.45Q

A.?,, (i — — o• 5P ILk.,)

0.5(1_f I11,kJ[’J

.,, .

;, ,,.

.1 , , , ,,,--.

S

r. .

= •p1[

+ ( .—

Step 1

. ..‘

NEstablish material properties (see Table I)for and use Eq. (2), PRESSS, orACT T1.2-03 recommendationsFor c,,,,., use Eq. (5) or other criteria

c;:

r,, , ‘

____________________

,:‘

Step 11 E +i- with k,,1, = 1.0 (PRESSS)

r .:

Step 2Determine Mje, and B,k, using DBD or FBD

Step 10

Step 3= L,,,/Ii,

,,:

0. 5Ii/,,,,,, (i— I3 ‘L...)

AA =—--

x

Step 4

Assume a value for Q) =

‘.

.. 1..

(Initial estimate 0.5)

. ..,,,.,

= 2.4e,,,,,, []M,.

Mth,

‘I,

6,,, = 6,,,,,,, — 6,,,,,,.

1:I::-1:

‘

]4 :1L’ ‘‘-,,‘•“

L>(l—?7,,,., —)e,,,.,

Jig — 6,.,,

Step 50.11

Assume a value for ii,,,,, =

= 0.81 (i,, -

(1, t\. ç.,l)

L,,, (Required) = L,, — 2L,,

Fig. 4. Flowchart for non-dimensional design of hybrid frames.


Assuming ç,,, = 0, thenA —

cJv

(2)

Therefore, using Eq. (2), the value for c = 0.04— s,,,1, = 0.08) andf, = 60 ksi (414 MPa) would be 1.37, whichis close to the 1.35 values suggested by Stanton and Nakaki.Similarly, the value for ç = 0.02 (that is, e,,,,,.,

—

s,,,,, =

0.04) andf, = 60 ksi (414 MPa) would be 1.2.Celik and Sritharan also propose that over-strength factors

be a function of the interface rotation.3However, they use anon-cyclic stress-strain behavior for the mild steel reinforcement to arrive at their over-strength factors. Here, we proposea cyclic approach utilizing the strain amplitude e as an important parameter.

According to Chang and Mander,5 the equivalent number of equi-amplitude cycles of building motion in a designearthquake N,. can be conservatively estimated using the following equation:

N,. =7T3

whereT = the fundamental period of the structure in secondsThe designer should consider Eq. (3) when choosing the

number of design cycles ND, which can be assumed equal tobe a multiplier m times N,., as in Eq. (4):

ND = fiN,.

Considering that the return period for the design earthquakeis significantly larger than the design life of the structure, onemay argue that the multiplier fli should be equal to 1.0. However, one of the stated advantages of the hybrid frame is thatthe structure re-centers itself after an earthquake. Therefore, ifin were set to 1.0, the structure would theoretically be re-centered after the design earthquake occurs, but all of its low-cyclefatigue capacity would be depleted. Therefore, the designershould consider these factors when selecting the value of m.Dutta et al. implicitly suggest that a structure should be capableof sustaining a foreshock, a main shock, and an aftershock.6

Substituting ND from Eq. (4) for Nf in Eq. (1) yields

= 0.024m°457°’5

= 0.048nv°457°’5 (5)

The mild steel reinforcement in the hybrid frame is subjected to strains ranging from approximately zero (sn,,,, = 0) toa maximum design tension strain (e,,,,., = 2ev). Equation (5)can be used to estimate the maximum design strain in the barfor a given period of the structure. This suggested procedureis based on Mander and Panthaki’s work on ASTM A 615bars.4 It should be noted, however, that Mander et al. suggestthat Eq. (1) can be used for all types of steel if the superscripton the(2Nf) factor is changed from -0.448 to -0.5.

Tests on ASTM A 706 and ASTM A 615 bars are beingconducted at the University of Wisconsin—Milwaukee to verify the validity of Mander and Panthaki’ s equation for barsother than ASTM A 615 and to develop a modified equationfor low-cycle fatigue if needed. A new equation based on results of testing the ASTM A 706 and ASTM A 6t5 bars willbe proposed. In the meantime, the previously proposed equa

(3) tions can be used. Alternatively, the designer can select sbased on any other rational criterion.

NON-DIMENSIONAL FORMULATION

The availability of simpler procedures to design hybridframes would help this new technology gain more wide-

(4) spread acceptance. One method would be to develop dimensionless parameters and equations applicable to most cases.The step-by-step PRESSS procedures can be restated in anon-dimensional form as described in the following steps.Figure 4 summarizes the resulting non-dimensional procedure in a flowchart.

Step 1: Determine the Material Properties and RelevantParameters

Table 1 identifies some of the design parameters for thehybrid frame components. The yield strength of the reinforcement is multiplied by the material over-strength factors

(‘‘.,d and for tension and compression, respectively) tocalculate the stress in the reinforcement at a certain designstrain specified by the engineer.

Table 2 lists the mean over-strength factors and their corresponding bar strains as proposed in the PRESSS report.’

Table 1. Material Properties and Relevant Parameters for Hybrid Frame Components

Material Property Symbol

Concrete and interface grout compressive strengths

Post-tensioning tendon initial strain minus losses, initial stress minus losses,. . . . c,f.,E,f,ands,.

modulus of elasticity, yield stress, and yield strain ‘ p p P3

Mild steel reinforcement yield stress, modulus of elasticity, minimum tensile. . . . f,E,f,ande.

strength, and maximum design axial strain ‘y S .u

Mild steel reinforcement over-strength factors (tension and compression) A5d,.S and 2,’,,,.,

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Table 2. Over-Strength Factors Corresponding to Different Strains for ASTM A 7061

System State Strain 2, (Tension) ,i’ (Compression)

First yield 0.002 1.0 1.0

Design 0.04 1.35 1.0

Maximum credible 0.08 1.5 1.0

The design over-strength factors for tension and compression, as suggested by the PRESSS report, are = 1.35 and

= 1.0, respectively.Section 3.3.3 of ACE Tl.2-03, Special Hybrid Moment

Frames Composed of Discretely Jointed Precast and Post-Tensioned Concrete Members, specifies that, in the absenceof test data on stress-strain properties of the mild steel reinforcement,7 the tensile strength of the reinforcement shouldbe taken as the specified minimum tensile strength f andthe reinforcement’s compressive strength must be taken as1.25ff. However, a more rational procedure based on inelastic cyclic behavior of reinforcing steel bars (as shown previously) is desirable when calculating the reinforcing steel’sover-strength factors in tension and compression as well asthe maximum permissible steel strain.

Step 2: Obtain Mde. and °de.

Design loads can be obtained using either force-baseddesign (FBD) or the preferred displacement-based design(DBD) methods.8 Interface rotations °drs can be calculatedfrom the drift ratio using the geometry of the system.

Step 3: Determine the Non-Dimensional ParameterRelated to Beam Geometry

Step 5: Estimate Neutral Axis Parameter at DesignDrift

Assume an initial value for i (relative position of neutralaxis from the compression face at the design drift). The following was suggested in the PRESSS report:’

0.1

=

An approximate value of=

may be used in lieu

of the PRESSS suggestion as discussed in a following sectionin this paper. The initial estimate of the neutral axis location will be adjusted, as necessary, during the iterative designprocedure.

Step 6: Calculate Strain in the Prestressing Tendon at 0dey

The following proposed procedure is slightly differentfrom the PRESSS procedure in that the strain (and not thestress) in the post-tensioning tendon is checked to ensure thatthe tendon does not yield. From Fig. 3, the axial deformationof the post-tensioning steel 4 is

where

L= _in

Jig

= Y2 L for spans of equal length along the entirelength of frame

= (mge ospas) for spans of unequal length= span length (center-to-center of columns)

Step 4: Determine Proportion of Moment CapacityContributed by Post-Tensioning and DeformedReinforcement

Assume a value for w for the ratio of post-tensioning moment to design moment:

Mp,des

Mde.,

where

L1 = 0deyhg(057lde)

hg = the height of the beam and the grout padThe corresponding change in post-tensioning strain z1E is

A 0Ae =—e-=—-(0.5— ) (6)de

The post-tensioning steel should not yield and, therefore,its strain s should be kept at or below yield:

8 =E.+L1E ae (7)p p p py

where= the effective post-tensioning strain after losses

a = a factor introduced to afford the designer the flexibility to control and limit strain in the post-tensioning tendon to any desired level below yield

= post-tensioning tendon yield strainA good initial assumption is w = 0.5.


The a parameter proposed here was not included in the original PRESSS procedures. The ability to maintain the strain inthe post-tensioning tendon below yield (a 1) is crucial forthe stability of the entire frame. Therefore, further research onthe choice of an appropriate value of a is warranted.

The post-tensioning steel stress fp(Ies can then be obtainedfrom the stress-strain relationship for Grade 270 (1860 MPa)prestressing strands as follows:9

For e, e= 0.0086

whereE = the modulus of elasticity of post-tensioning steelCelik and Sritharan also propose that the calculation often-

don stress be based on a calculation of tendon strain,3 usinga stress-strain equation for strands. However, they do not explicitly limit the post-tensioning strain to a level below yield.

Also, the following non-dimensional parameters must bedefined:

qJ3Cpies

I:

IC,

whereb9 = the width of the grout pad (same as width of beam)Fed,, = compressive force in concrete corresponding

to de,

Fcaes = Fpdes ÷ F,dC, — F,’des

where Fp,aes, Fsde,, and F,d, refer to tension force in thepost-tensioning steel, tension force in the mild steel reinforcement, and compression force in the mild steel reinforcement, respectively (Fig. 3).

whereA = the cross-sectional area of the post-tensioningA, = the cross-sectional area of the mild steel

reinforcement (same in top and bottom)Define the following non-dimensional parameters:

A

bhgg

where= compressive strength of concrete and grout pad at

the interface

fp,d, = post-tensioning steel stress at design drift

f,3, = mild steel reinforcement yield stress

Step 7: Calculate Relative Locations of CompressionForce and Neutral Axis

The depth of the equivalent stress block ades, the relativeposition of the compressive force as a fraction of overallheight ad,, and the relative position of neutral axis as a fraction of overall height can be calculated as follows:

a=

des 0•85.4bg

f,, = Ee (8) Fp,aes = APfPd,,

Fsdes = A, .as,desf,y

FSd,, = A,25’desfsy

A

— C

p

where

f = post-tensioning tendon effective stressAccording to Section 18.4.2 of ACT 3 18-02, Building Code

Requirementfor Structural Concrete,’°

T 0.45Q

Substituting for FCdC, (from the previous equations) intoEq. (9) yields:

üdes = [+(ASdCS_ Ad)]

— adC,

fldes— /3 h

Ig

aa — des

des 2hg(9)

116 PCI JOURNAL

Therefore,

rides [+(des_ sdes)] (10)

In Eq. (10), the first term in the bracket, P, is a much largernumber than the second term. Therefore, the following approximation (within 5%) can be written

l.07P1’des = 0.85fl

To solve for rides using Eq. (10), an initial estimate for Fmust be made. Values of [‘range from approximately 0.0020to 0.0025. An equation for calculating x is derived belowusing the following non-dimensional design relationships:

= 0.5(1—/317)

Ahf 0)p g p,des

Md = (l——0.5/31u7d)

Ahaf 1—COS g 5,

= s,des (i__o.sp15)

0.5(1/3lrid)(±-

In this process, x is first calculated using the previous equation and the rides assumed in Step 5. Then, a new value for riis calculated using Eq. (10). The newly calculated value of

is then compared with the previous value in Step 5, andSteps 6 and 7 are repeated until the Tides values converge.

Step 8: Assess Moment Strength of Section at 0des

Calculating the moments of the post-tensioning force andthe deformed mild steel reinforcement forces about the centroid of the concrete compression force yields

M =F h(0.5—cz )p,des p,des g des

MSdes = ,ag(±des)

‘s’,des = ,deshlg (© — Udes)

The total moment strength Mcapbea,s is

M =M +M ÷Mcap beam p,des s,des sties

which should be greater than or equal to the design moment

M Mcap beam des (11)

Therefore, the following non-dimensional relationship canbe written

0.5(1-1ri5,5 (1_

_ 0.5111dCV) ÷

‘,des (0.5ri)]0.5(1_)

(12)

If Eq. (12) is not satisfied, then the value of F in Step7 must be modified and the steps must be repeated untilEq. (12) is satisfied.

Step 9: Evaluate Restoring Properties of Beam

After reaching the design drift corresponding to 9des andthen returning to zero drift, the restoring moment provided bythe post-tensioning tendon should be greater than the corresponding moment due to the mild steel reinforcement. This isto ensure that the frame re-centers after an earthquake. Bothsets of mild steel reinforcement (top and bottom) would be incompression at that stage. Thus, the forces, depth of equivalentstress block, location of compression force and neutral axis,and the moments at zero drift can be calculated as follows:

where

F =AA,, fsO s sties sy

= A2,,f

F =Af.p0 pp’

F =F -F -F,cO pO sO sO

F = force in the mild tension reinforcement at zero driftF,,0 = force in the mild compression reinforcement at zero

driftF = force in the post-tensioning tendon at zero driftF = force in the concrete at zero drift

where

Fa = sO

0•85fbg

aa =—--

2hg

a0ho =

a0 = the depth of the equivalent stress block at zero drift

rio = the relative position of neutral axis at zero drift

M =Fh(0.5—a)pO pO g 0

M,,0 = F,ohg(1__a0)


M =F h (—a0)sO sO g

reinforcing bar s,5,,,5. Therefore,

whereM,,0 = moment due to post-tensioning taken about the

centroid of the compression force in the concrete atzero drift

M50 = moment due to mild tension reinforcement takenabout the centroid of the compression force in theconcrete at zero drift

M50 = moment due to mild compression reinforcementtaken about the centroid of the compression forcein the concrete at zero drift

To ensure self-centering of the frame after a seismic event,Eq. (13) must be statisfied.

M M +M,p0 sO sO

One may argue that the use of the equivalent stress blockin this case would not be appropriate. However, the a0 termcancels out in the derivation. Therefore, the choice and appropriateness of using the equivalent stress block would notaffect this provision (as given in the following). The previousrequirement for self-centering can be written in the followingsimple and non-dimensional form:

x2)d,K

K=

If Eq. (14) were not satisfied, then w (in Step 4) must bemodified, and Steps 5 to 9 must be repeated.

Step 10: Calculate Steel Areas and Required UnbondedLength of Mild Steel Reinforcing Bar

L55 = the length of mild steel reinforcement that is physically debonded at each interface location

A potential failure mode of the hybrid frame involves thefracturing of the mild steel reinforcing bars. Tests of hybrid-frame connections by Stone et al. indicated fracturingof some bars during cyclic loading.’1Thus, the prediction offracturing of mild steel reinforcing bars is a very importantconcern and should be considered in design. As discussedpreviously, the PRESSS procedures do not specifically ad-

(13) dress this issue.Section 3.3.2 of ACI T1.2-03 recommends that ASTM

A706 reinforcing bars be used in hybrid frames and specifiesa maximum design strain that is 2% less than the strainat the minimum elongation indicated in ASTM A706 for agiven bar size.7 For a No. 8 (25M) bar, this limit would be10%. This limit could be as high as 12% for other bar sizes.Such high strain limits, however, would result in a very limited number of cycles to failure based on the published low-cycle fatigue test results.

(14) Priestley et al. recommended that the maximum strain inthe mild steel reinforcement under cyclic loading be limited to 75% of ultimate strain and 6% (absolute strain) in theabsence of better information)2Equation (5) can be used todetermine Esm based on low-cycle fatigue.

As interface rotation occurs in hybrid frames, the mild reinforcing steel is subjected to axial and bending strains. Thebending strains can be as much as 10% of the axial strains.’3The PRESSS procedures can be modified to include the effect of maximum bending strain Ebfl, corresponding to O.The bar’s bending strain Ebmax at 0d, can be estimated usingthe following equation:

The areas of mild steel reinforcement and post-tensioningtendons can be calculated using the equations shown in Step7. Thus,

and

A=P 0•5hgfpj5(i

— Ii’1a)

AA5 = —s

According to the PRESSS design procedures, the elongation of the mild tension reinforcement A5 at 0des can be calculated as

A = Odashg(171ds)

The strain in the mild steel deformed bar should be equalto or less than the maximum design strain in the mild steel

= the diameter of the barA more detailed equation for calculating Ebm is given in

the research report for this project.’3Growth in the unbonded length of the bar is expected to

occur under high cyclic strain. Raynor et al. studied the bondcharacteristics of bars grouted in light-gauge metal ducts.’4A non-uniform strain distribution occurs at each end of theunbonded region of the bar that penetrates into the grouted(bonded) region of the bar. According to Raynor et al., theelongation due to strain penetration is equivalent to the peakstrain acting uniformly over an equivalent additional unbonded length on each end. The additional equivalent debondedlength L10 can be calculated using the following equation:

0.8l(J -j)db

(f;d)’

where

AL —— (15)

max

where

(d0g =2.4r I—I

b,max smax h I\ gj

where

118 PCI JOURNAL

where= the maximum tensile strength of the steel reinforcing

barNew research may provide more accurate estimates for L,,.

Thus, the total unbonded bar length L, is

where

L1 = L + 2L

L, = the unbonded length of the mild steel reinforcementat each interface

The maximum total strain that the bar is subjected to is thesum of axial and bending strains. A conservative approachhere is proposed to ensure that the total strain is less thanEs.rnax, which is the maximum design strain that the bar can besubjected to based on low-cycle fatigue tests.

The apparent maximum axial design strain 8b is then calculated.

6 =6 —6sb gnus, brnax

To ensure satisfactory performance, the calculated axialstrain in the bar at O,,, must be less than e,,,. Because the totalunbonded length is L,,, (not L,,), Eq. (15) can be rearranged:13

AL —‘

6sh

h eg sb

L (Required) = L, — 2Lua

Step 11: Confine Compression Region

The common design assumption that plane sections remain plane would not be valid at the end region of the beam.1Therefore, the concrete strains cannot be calculated from thecurvature within the plastic hinge length.

Stanton and Nakaki recommended a plastic hinge length ltaken as a function of the compression zone depth such that’

where

1ph = kpi;T1des

hg

kPh = 1 (without experimental validation)The concrete compression strain at the compression face of

the beam e can then be calculated

0des(l7deghlg)

=

ph ph

The compression region should be confined if the calculated concrete strain is greater than the concrete’s strain capacity ( 0.003).

PARAMETRIC STUDIES

The above dimensionless parameters were varied in anextensive parametric study (approximately 1200 trials) involving an optimization process. The objective of the optimization program was to ensure that the moment capacityof the section is greater than MdCS and to satisfy additionalconstraints discussed in the following paragraphs.

Based on Eq. (12), the objective function is

0.5 (i- + [s,e.s (1_

- 0.5fl,nd,,) +

0.5(l_ ‘

&,des ( —0.5fl,Jde5)] — 1 des)= 0

CL)

Microsoft® Excel’s “solver” feature was used to find theoptimum solution by changing ijde,’ cv, and T subjected to thefollowing constraints:

• To achieve zero residual drift: x 225d,SK

• To ensure that the post-tensioning steel would notyield: e,,

• To meet ACI 3 18-02 Section 18.4.2 requirements: T0.45Q

The a parameter was assumed to be equal to 1.0 followingthe PRESSS procedures. However, a lesser value for a maybe warranted pending further studies. Table 3 lists the rangeof parameter values used in the parametric studies.

RESUIJS FROM PARAMETRIC STUDIES

Based on the results of the parametric studies,13 a set ofcharts with non-dimensional parameters was generated(Fig. 5—10) and corresponding design equations were recommended. One can calculate the required areas of post-tensioning tendon A and mild steel reinforcement A, by utilizingFig. 5,7, 8, and 10.

It can be observed from Fig. 5 that the required area ofpost-tensioning tendon A increases as ede, increases for agiven q, Mde,, and e,,s,,. On the other hand, A decreases asEs,,ax increases for a given , M,5, and °de, (Fig. 8). The reverse is true for the area of the mild steel reinforcement A,.

The stress in the post-tensioning tendon fpd. at a specifieddesign drift 0de, increases as increases. It also increaseswhen the stress in the post-tensioning tendonf1, increases asshown in Fig. 16.

The designer can also calculate the optimum ratio of thepost-tensioning moment to the total moment strength cvfrom Fig. 6 and 9. The relative area of the post-tensioningtendon to the beam’s cross section F can be calculated fromFig. 13—15. The relative location of the neutral axis ‘ldes can


Table 3. Material Properties and Relevant Parameters for Hybrid Frame Components

Parameter Minimum Maximum

L

0(%) 0.5 4

.

8s,mas (%) 1 8

Ic . . ,. 0.5 0.65J1u

f.,ksi 4 8

5Although AC! allows a ratio of 0.70, a lower value is used here to ensure that the tendon stress would not yield at the design limit state.

Note: 1 ksi = 6.895 MPa.

be calculated from Fig. 14. Figure 14 shows that 7Jde. asso- Mdciated with an optimum design increases as the compres- O.O4

= A h iatE 4%sines

sive strength of the concrete f increases (or j3 decreases). P g piles

It should be emphasized that Fig. 14 reflects a condition inwhich all parameters of design have been modified to arrive

Mdes ate =4%at an optimum solution. Thus, for the different points shown04

= A h fin Fig. 14, the various design parameters (such as T) are not . g piles

held constant. For example, Fig. 13 indicates that F increasesas f’, increases.

Because the curves in the figures show smooth variations, L1.For 0approximate design equations that are functions of 0,r5,55,

80

, and can be derived from the design charts using curvefitting. These equations can be used in lieu of the charts. The

Mproposed non-iterative design equations are as follows: cv = p,des (5.160 +0.04 des

des

Mp,des

= 0.89w -0.037 (16)0,04 cmiii

=9O04

= (—6.70 + O9)27+003)

Ahf desp g p,des

Mdes= 125004es,,(ii0037 (17)

Ahf Mg g p,des des —

A hfs s.des g p,des

M (22.230 + 1 7)(3.5Os+0.02)

desdes =0.882a7 £-0.034

(18)A .Z h f 0.04 s,rnax

s,des g p,deswhere

°des =interface rotation at design limit state expressed inradiansA

F = = y,v1v (in percent) (19)bhgg L

2. For 0 = —-> 8h

g

0.11fldesp

Mco004 = p.des

= (3.790 +o47)0’9Od

where M des

des

= the ratio of post-tensioning moment to total momentstrength corresponding to = 4°7

120 PCI JOURNAL

(l)

n

CDS

I3

.

)

0)

CD CM

?•1

CD CM CD

M/A

s,de

shgf

)-1

-.

CD01

001

001

001

0 01

0

Mp,

pes/

MdS

00

pp

pp

op

..

P01

0101

CY1

0101

01CD

01-

1’3

C.)

401

Mdes

/ Ai,h

Cfpd

es0

00

00

00

PD

CC

01r\)

.C

)01

CDF\

).

0)

0)

01

0

0

01 I\)

0 I’.)

01

0

01

fL

0101

C))

01

0

C))

Cs)

C))

Cs)

0

F)

E)(

fC

.)C.

)Is

)Cs

)-

01C)

)C)

)01

I) C))

.I.)

(C)

Cs)

Cs)

--

001

01

bi

Fig. 8. [MdIAP hCfPd,S] versus for °smax = 2%.

Fig. 9. [MP,dJMd] versus q for 6d = 2%.

Fig. 10. [MdjA2a hgfy] versus for O = 2%.

Mdes/4hg fp,des VS

values of[Ode, = 2%]Es,nas(°7O)0.96

--- 1.50.94

—‘— 2.5

-—-A--— 20.92c4 0.9

--*-- 40.88

—6

- 0.86

0.84

0.82

0.8

0 5 10 15 20 25= L, I h5

CO= ‘p,des “des” b

[Odes = 2%]values of0.57,max

0.55 -—

- 1.5

-—— 20.53

--30.51

50.49

------ 60.47

0.45

0 5 10 15 20 25= L / h5

2.05

[Mdes /(A5hg2s,des.fsy1vs

—. 2.001.95

1.90

1.85

1.80

1.75

1.70

1.65

values of

s (%)s,,flax

—s----- 1.5

—k--— 2----x---- 2.5

-—--—- 3

-.- 3.5

5

— 6—.-—-- 7-------8

0 5 10 15 20= L / h

25

122 PCI JOURNAL

F

i0 CD o C

) CD

CD Co CD

ppp

-M

01

C01

ppp

()

C)

-

C01

0

-I.’

CD °1 C,,

C,, CD p cM

01

Fpp

pp

pp

pp

I’J

r.3

N)

N)

N)

N)

C-

N)

C)

.01

0

?1 CD Co

Ct

°1 Co cM CD CD 9 C’

cM

II (J

“.3

0 “3 01

0 01

IIII

IIII

00-)

C’L

M

CMC

OCo

CO

N)

0-

N)

01

t-

N)

.Cp

01CY

1

N)

C,)

1’ides VSj31

Fig. 14. ‘i versus J91.

Fig. 15.fPdJfP versus forf,,1 = O.65f

Fig. 16. versus for = 2%.

124

0.18

0,17

0.16

0.15

0.14

0.13

0.120.6 0.65 0.7 0.75 0.8 0.85 0.9

J3

.t,’o,desUjy vs

[f = 0.65 f] Od(%)

1.00 —.e—0.5

0.95 ---— 1

0.90 —A— 1.5

0.85

V.0k)

0.75

40.70

0 5 10 15 20 25

= L / hg

fp,desU,,y VS

= 2%]1.00

0.95

0.90

0.85

0.80

0.75

0.70

0.65

0.60

—.---f. = O.5f

O.55

—A— 4’= O.6f1,= O.654

0 510=LIh915

20 25

PCI JOURNAL

M= des=(810des

+ 0.94)19)Ahf

p g p,des

M004

— des=(13650d,,,

+ l.65)200A hf

s ,.des g p,de

A“ for [‘=5 ksi (34.5 MPa) andf =0.65f

‘1,, pu

gg

Ycef =(20des+ 0.21 5)q(3 80+0 02) (in percent)

where= factor that accounts for different values off

if= 0.208f — 0.0355

where= factor that accounts for different values off/f

v =—l.82---+2.l8I

ipu

DESIGN PROCEDURES

In addition to the iterative steps shown in Fig. 4, the following steps can also be followed when using the proposednew design charts or equations.

Procedures When Using Design Charts

1. Obtain the applied moment Mdes and design drift 0des

2. From the estimated fundamental period of the structureT, calculate N using Eq. (3) and E,,m, using Eq. (5).

L3. Specify

4. Calculatef, by using Fig. 15 and 16.5. Calculate A by using Fig. 5 and 8.6. Calculate A, using Fig. 7 and 10.7. Optional step: Calculate Tusing Fig. 11 to 13.8. Calculate L,, using the procedures discussed in Step 10.

Procedures When Using Design Equations

1. Obtain the applied moment Md,, and design drift °des

2. From the estimated fundamental period of the structureT, calculate N, using Eq. (3) and e,,,,, using Eq. (5).

L3. Specify

0.114. Calculate 1”des =

5. Calculate Ae,, using Eq. (6).6. Check whether E ae using Eq. (7).7. Calculate fpd. using Eq. (8).8. Calculate A1, using Eq. (17).9. Calculate A, using Eq. (18).10. Optional step: Calculate Tusing Eq. (19).11. Calculate L,, using the procedures discussed in Step 10.

CONCLUSIONS

The following conclusions are made:• The PRESSS procedures to design hybrid frames

can be modified into a non-dimensional form. Thenon-dimensional procedure, though iterative, can besolved with optimization routines for a wide range ofparameters.

• The solution of the non-dimensional formulation canbe summarized into non-dimensional charts and equations. A set of such design charts and equations wasdeveloped and is presented in this paper. These toolsare designed to enable engineers to perform the designof hybrid frames in fewer numbers of steps while providing insight into the effects of various parameters onthe design.

• The limit state of bar fracture resulting from low-cyclefatigue should be considered in design. The maximumdesign strain in the mild steel reinforcement should belimited based on a low-cycle fatigue criterion. Suggested design equations are provided in this paper.

• The stress in the post-tensioning tendon should becalculated based on tendon strains. In all cases, the post-tensioning strains should be less than the yield strain.Although the PRESSS procedures assume stresses to beat incipient yield, it is recommended that the strain in thetendon be limited to as, where a is a value less than 1.

ACKNOWLEDGMENTS

This research work was supported by the PCI Daniel P.Jenny Research Fellowship for 2003—2004. The supportof PCI is greatly appreciated. The research team thanks Industry Advisory Panel members S. K. Ghosh, Ned Cleland, Roger Becker, Fattah Shaikh, and Sri Sritharan. Theauthors are grateful to John B. Mander for his valuable advice and input. The authors also express their gratitude to thePCI Journal reviewers and editor for their valuable suggestions.

REFERENCES

1. Stanton, J. F., and S. D. Nakaki. 2002. Design Guidelines forPrecast Concrete Seismic Structural Systems. PRESSS ReportNo. 0 1/03-09, PCI, Chicago, IL, and University of Washington


Report No. SM 02-02, Seattle, WA (February).2. Priestley, M. J., S. Sritharan, J. R. Conley, and S. Pampanin.

1999. Preliminary Results and Conclusions from the PRESSSFive-Story Precast Concrete Test Building. PCI Journal, V. 44,No. 6 (November—December): pp. 42—67.

3. Celik, 0., and S. Sritharan. 2004. An Evaluation of Seismic Design Guidelines Proposed for Precast Concrete Hybrid FrameSystems. ISU-ERI-Ames Report ERI-04425. Iowa State University, Ames, IA (January).

4. Mander, J. B., F. D. Panthaki. 1994. Low-Cycle Fatigue Behavior of Reinforcing Steel. Journal ofMaterials in Civil Engineering, V. 6, No. 4: pp. 453—468.

5. Chang, G. A., and J. B. Mander. 1994. Seismic Energy BasedFatigue Damage Analysis of Bridge Columns: Part II—Evaluation of Seismic Demand. Technical Report NCEER-94-0013, Buffalo, NY.

6. Dutta, A., J. B. Mander, and T. Kokorina. 1999. Retrofit forControl and Repairability of Damage. Earthquake Spectra, V.15, No. 4 (November): pp. 657—679.

7. ACT Innovation Task Group 1 and Collaborators and ACI Committee 374. 2003. Special Hybrid Moment Frames Composedof Discretely Jointed Precast and Post-Tensioned ConcreteMembers and Commentary. TI .2-03 and Ti .2R-03. FarmingtonHills, MI: ACT.

8. Priestley, M. J. 2002. Direct Displacement-Based Design ofPrecast/Prestressed Concrete Buildings. PCI Journal, V. 47,No. 6 (November—December): pp. 66—79.

9. PCI Industry Handbook Committee. 2004. PCI Design Handbook: Precast and Prestressed Concrete. Sixth Ed. MNL 120-04. Chicago, IL: PCI.

10. ACT Committee 318. 2002. Building Code Requirements forStructural Concrete (ACI 318—02) and Commentary (ACI318R—02). Farmington Hills, MI: ACT.

11. Stone, W. C., G. S. Cheok, and J. F. Stanton. 1995. Performanceof Hybrid Moment-Resisting Precast Beam-Column ConcreteConnections. ACI Journal, V. 92, No. 2 (March—April): pp.229—249.

12. Priestley, M. J., F. Seible, and G. M. Calvi. 1996. Seismic Design and Retrofit of Bridges. Hoboken, NJ: John Wiley & SonsInc.

13. Rahman, A., H. Tabatabai, and R. Hawileh. 2004. Simplified Design Procedures for Precast Hybrid Frame Buildings.Daniel P. Jen.j flcsearch Fellowship report, PCI, Chicago, IL(September).

14. Raynor, D. J., D. E. Lehman, and J. F. Stanton. 2002. Bond-SlipResponse of Reinforcing Bars Grouted in Ducts. ACI StructuralJournal, V. 99, No. 5 (November—December): pp. 568—577.

15. Collins, J. A. 1993. Failure ofMaterials in Mechanical Design:Second Edition. Columbus, OH: Ohio State University.

APPENDIX A: NOTATION

a0 = depth of compression stress block in grout at zero driftA0 = area of post-tensioning steelA = area of mild steel reinforcement in each face of the beamb5 = width of girder (same as width of grout pad) at the

beam-column interfacedb = diameter of mild steel reinforcing barE0 = modulus of elasticity of post-tensioning steelE = modulus of elasticity of mild steel reinforcementf compressive strength of concrete and grout pad at the

interface

f = compressive strength of grout inside duct

= maximum tensile stress at maximum tensile strain= post-tensioning steel stress at design drift

post-tensioning tendon effective stress= post-tensioning tendon yield stress= mild steel reinforcement yield stress= mild steel reinforcement ultimate stress= concrete compressive force at beam-column interface= post-tensioning force in post-tensioning tendon at design

drift= force in tension mild steel reinforcement at design drift= force in compression mild steel reinforcement at the design

driftF,,0 = post-tensioning force in post-tensioning tendon at zero

driftF,,, = force in tension mild steel reinforcement at zero driftF,, = force in compression mild steel reinforcement at zero

drift= height of girder (equal to height of grout pad) at beam-

column interface= overall height of the building in feet= plastic hinge length factor= plastic hinge length

= V2 L for spans of equal length along the entire frame

= (average length of spans)/2 for spans of unequal lengthL = span length (center-to-center of columns)

= unbonded length of mild steel reinforcement at eachinterface

= additional unbonded length at each end of the initialunbonded length of the mild steel reinforcing bar due tohigh cyclic strain

L, total unbonded length of mild steel reinforcing barincluding the effect of L

M,.,, = total moment strength (A4,0,, + A4j= moment strength due to post-tensioning force

M,.,,,,, = moment strength due to mild steel reinforcementequivalent number of equ i-amplitude cycles of buildingmotion in a design earthquake

ND = number of design cyclesNf = number of cycles to failure

Q =ratioofftof01T fundamental period of the structure in secondsad = distance from center of equivalent compression block to

beam compression face divided by beam height= O.5f3,i

i3 = factor defined in ACT 318—02 (Section 102.7.3)10

x = ratio of A0 to A

4 axial deformation of post-tensioning steel= tensile axial deformation of mild steel reinforcement= compressive axial deformation of mild steel reinforcement= concrete compression strain at the compression face of

the beam= maximum total strain in a cycle

c,,,,, = minimum total strain in a cycle= post-tensioning tendon effective strain= post-tensioning tendon yield strain

a.,,, = apparent maximum axial design strain in the mild steelreinforcing bar

8,,,,,,, = maximum design strain in the mild steel reinforcing bar= ratio of L0,, to h,,= ratio off, to f’= Md,,,, divided by (A,, h,,f,,d,,,) when e,,,,,,,,,, is equal to 4%= Twhen f’,, is equal to 5 ksi (35 MPa) andf,,, is equal to

0.65.1,,,,T = ratio of A,, to cross-sectional area of beam

f,,=.rf,,.d

fp,

fpy

.1,,”F,,d,,,

h,kr,,

Iph

126 PCI JOURNAL

distance from neutral axis to beam compression facedivided by beam height

K = ratio off, tof,= mild steel reinforcement tension over-strength factor= mild steel reinforcement compression over-strength factor= factor that accounts for different values off/f, when

determining Tv = factor that accounts for different values of f’ when

determining F= interface rotation at design limit state= value of cv when is equal to 4%= Md divided by (A25dh5fPI) when is equal to 4%

ratio offpdS., to f= distance from center of the mild steel reinforcing bar to

nearest face divided by height of the beam

APPENDIX B: NUMERICAL DESIGN EXAMPLE

In order to validate, verify, and illustrate the effectivenessof the developed design charts and simplified equations, fivedesign examples were presented in the research report for thisstudy.’3 In this paper, only one example is shown. This example includes four separate design procedures: the originalPRESSS procedures, the modified non-dimensional procedures, the developed non-dimensional charts, and proposedequations.

For the sake of brevity, the calculations based on the original PRESSS procedures are not shown. A comparison anddiscussion of various solutions is given. It should be notedthat the values of 1. and )L’sde, recommended in ACT Ti .2-03 are used in this example.

The design parameters in this example are identical to thosecontained in the PRESSS design example presented by S. K.Ghosh at the March 2002 PCI Seismic Seminar in Chicago,

Table Bi. Material Properties and Design Information

Parameter Value Comments

f., ksi 5 Compressive strength of grout pad and concrete

J,’ ksi 5.5 Compressive strength of grout inside duct

ksi 243 Yield strength of Gr. 270 strand (= 0.9,f,)

L’ ksi 270 Minimum strength of Gr. 270 strand

f,,,, ksi 175.5 Initial stress in strand

E, ksi 29,000 Modulus of elasticity of mild steel reinforcement

/., ksi 60 Yield strength of Gr. 60 steel bars

0.055 Cover factor (Fig. 3)

L, ft 45 Tendon length center-to-center of columns (all spans are of equal length)

L,,,,, ft 22.5 L12

E,,, ksi 28,500 Modulus of elasticity of Gr. 270 strand

/3 0.8 /3 0.85 — 0.05 f— 4)

0.046 Maximum design strain in the mild reinforcing steel

) 1.4 Tension over-strength factor7

) 1.25 Compression over-strength factor7

e, 0.00207 fIE,

C,,. 0.0085 f7JE,,

s 0.00616 f,/E,,

1v1,,,,, ft-k 1235 Obtained using displacement-based design

6,,., % 1.96 Interface rotation

Note: I ksi = 6.895 MPa; 1 ft = 0.3048 m; 1 kip = 4.45 kN.


Table 82. Results Compared Using the Four Methods

Method A, in? A5, in.2 L5, in. Remarks

PRESSS procedures 1.96 2.31 14.5 Bending strain and grout effects are not included

Non-dimensional procedures 1.98 2.30 12.5 Bending strain and grout effects are included

Charts 1.9 2.30 12.5 Bending strain and grout effects are included

Design equations 1.9 2.38 12.5 Bending strain and grout effects are included

Note: PRESSS = Precast Seismic Structural Systems; I in. = 25.4 mm; I in.2 = 0.000645 m2.

Ill. However, the calculation steps are somewhat revised toachieve an optimized solution.

Seismic data:• Location: Charleston, S.C.• Site class: D• Building type: five-story office• Three equal bays; span length = 45 ft (13.7 m)• hg = 42 in. (1070 mm), bg = 22.5 in. (570 mm)• Moment-resisting system: hybrid frameTable Ri shows the material properties and other informa

tion used in the design procedures. Table B2 compares theresults from the design procedures.

SOLUTION USING MODIFIEDNON-DIMENSIONAL DESIGN PROCEDURES

BASED ON PRESSS

Step 1: Determine Material Properties and RelevantParameters

See Table B 1 for a list of information used in this example.Please note that the calculations fore5,,,.. utilize the proposedlow-cycle fatigue criterion for the mild reinforcing steel andare not included in the original PRESSS procedures.

Calculate the estimated fundamental period of the structure

T=0.03(h)4=0.03(65.5) = 0.691 sec.

where= the overall height of the building in feet

The number of equi-amplitude cycles associated with oneearthquake can be calculated:

N = 7T = 7(0.69l) = 7.92 cycles

The number of cycles to failure should ideally be higherthan the number of cycles associated with a single earthquake. The choice of the number of cycles is up to the designer. Here, based on similar assumptions made by others,we assume that ND = mN with m = 1.0. Further research onthe proper value of m is needed.

Enter the value of ND for Nf in the following equation:

=2 0.046.s,max a

It should be emphasized that this procedure for estimatinge5,, is a recommendation, and the designer can use other appropriate means of selecting

f.des = = 1.4 (60 ksi) = 84 ksi (580 MPa)

Jde.s = sdesfsy = 1.25 (60 ksi) = 75 ksi (517 MPa)

Step 2: Obtain Mdes and 0de.

See Table B].

Step 3: Determine Non-Dimensional Parameter Relatedto Beam Geometry

22.5xl2=643

h 42g

Step 4: Determine Proportion of Moment CapacityContributed by Post-tensioning and DeformedReinforcement

Assume w = 0.5 18In this case, the assumption is based on an optimum solu

tion calculated using the Excel solver feature. If an optimumsolution is not available, an iterative approach must be used.

Step 5: Estimate Neutral Axis Parameter Ts1dm at DesignDrift

Assume

0.11 0.1117 =—=--———=0.138des /3 0.80

Step 6: Calculate Strain in Prestressing Tendon at 0des

This step is a modified form of the PRESS S procedure to ensure that the post-tensioning steel wifi not deform beyond yield.

& =-(o.s_i ) 0.0196(050138)00011p des 6.43

S =0.0795(2Nf)°448 = 0.023

where

128 PCI JOURNAL

e,,1=0.0616 0.29=

+ = 0.0073 0.5(1 —0.8x 0.134)+ [1.4(1 —0.055--0.5x 0.8 x 0.134)0.859

Check whether e,, ae1,,+1.25(0.055 —0.5 x 0.8 x 0.134)1= 0.869Let a =

0.0073 <0.0085 OK= = 207.0 ksi (1427 MPa)

0.5(1 —0.8x0.134)=0.862 OK

Let 0.518

60=0.29

207.0 Step 9: Evaluate the Restoring Properties of the Beam

,J,,207.0414 K=--=—-—=0.342j s.o 175.5

= — — = 12.0x = 0.859 = 2x 1.25x0.342 = 0.855 OK

Step 10: Calculate Steel Areas and Required UnbondedLength of Mild Steel Reinforcing Bar

Step 7: Calculate Relative Locations of CompressionForce and Neutral Axis

wMA =

0.5h f10.0021 g PdS(l Ide)

In this case, the assumption is based on an optimum solution calculated using the Excel solver feature. If an optimumsolution is not available, an iterative approach must be used. A

= 0.518x 1235x 12= l.98in.2 (1276 mm2)

I’ 0.5(l—0.8x0.134)x42x207

= = —— = 0.0285f. 175.5

A 1.98A=—-=—————=230 in.2 (1486mm2)Check whether

x 0.8591=0.0021 0.45 Q = 0.45 x 0.0285 = 0.0128 OK

To calculate the unbonded length of reinforcing steel, theA2 (1—

— O.SPlnd ) — maximum design tensile strain must first be determined basedx

= s,deson low-cycle fatigue considerations.

o.s(i _/31des)[_ 1— 0.046

0.29 xl .4(1 — 0.055 — 0.5x 0.8 x 0.1375)

o.s(1_o.8o.i37s)1_L_i= 0.859

8b,rnx=2.

bJnaxl h J‘ gj

0.518

db = 1 in. (No. 8 [25M1 ASTM A706 bars)

[ + (ses — s,des)] == 2.4 x 0.046(1= 0.00263

17des= 0.85/3

0.00214i.4+

12.0(1.4_1.25)]=o.134

42

0.85x0.8[ 0.859

tsb = sniax —8b,,,ax

7?des = 0.134 0.1375 OK—No need to iterate.= 0.046—0.00263 = 0.043

Step 8: Assess the Moment Strength of the Section at °des

L (1—d —)od = (1—0.134—0.055)0.0196=0.369

A0.5(1

— Pfld ) + [.dns.(i

— — 0.5/3117d)hg

sb0.043

and0.5(1

— 1317des)


L 0.8 i(f— f 0.81 f —

3. Check whether e a

=I su sj’) l s,des sy

dh (%)15

— (fd)15 Leta= I0.81(1.4x60—60)

= =1.51 =‘0.0073<0.0085 OK(5.5)1.5

4. Ca1culatef,For h5 = 42 in. (1067 mm)

L,, 42 x 0.369 = 15.5 in. (394 mm) I = E r = 207 ksi (1430 MPa)p,des p pL5,, = L,,, — = 15.5 —2 x 1.51 = 12.5 in. (318 mm)

L5.For =—‘-8

Step 11: Confine Compression Region hg

= 1.0 (based on PRESSS) M

_________

0.037des•‘250.04s,,naxAhf

— O.— 0.0 196 p g

=0.0196k 1.0

04 =(679des+ 0.9)p276’,°°3ph

e> 0.003 (The solution provides proper confinement toprevent spalling.) = (—6.7 x 0.0196 + 0.9) x 6.43(27500l9003) = 0.897

SOLUTION USING PROPOSED DESIGN CHARTSMd =11.1

A h L25x0.04600371]xO.897=0.897

1. Go to Fig. 15 and determine: g p,des

=0.87j;,ies=0.87x243 =A= 1235x12

= 19 in2(l230mm2)0.897x42x 207

=211.4lksi (1458 MPa)

L2. Go to Fig. 5 and determine: 6. For = 8

h

A’! Mddes

= 0.885 0.882&J sAhf A.? h 004 sums

p g p,dus s s,des gip,des

1235x12=A = = 1.9in.2(1230mm2)

0.885 X 42 x 211.41 0.04 = (22.230d + I.7)Ø_(35002)

3. Go to Fig. 7 and determine: = (22.23 x 0.0196+1.7) x 6.43_(35x00196+002)

—1.8

______________

= —1.79Ah f M

des= [0.882 x 0.046_0034 ]x 1.8 = 1.76

AA hgfpdesS s,des

1235x12=2.35 jfl.21 (1520mm2)

______

A =1.79x42x 1.4x60 1235x12

= 2.38 (1535 mm2)A =

______

1.76x42x1.4x60

SOLUTION USING DESIGN EQUATIONSTable B2 compares the results of the various methods.

1. Calculate 2ldes

0.11fldus =

= 0.14Use thirteen 112-in.-diameter (13 mm) post-

2. Calculate i tensioning strands (Ap,p,oyj4e4= 1.9 in.2 [1230 mm2]),along with three No. 8 (25M) bars (Au,prove4 = 2.4 in.2

AAs = —fl- = 0.001 [1550 mm2]).

Pu

130 PCI JOURNAL

Non-Dimensional Design Procedures for Precast ... - PCI.org

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