long-hand design of a two-way post-tensioned floor system

1S. K. Ghosh Associates Inc.

www.skghoshassociates.com

Long-Hand Design of a Two-WayPost-Tensioned Floor System

Dr Bijan O AalamiProfessor Emeritus,

San Francisco State UniversityPrincipal, ADAPT Corporation; [email protected]

www.adaptsoft.com 1

View of a Two-Way Floor System Reinforced with Unbonded

Tendons

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View of a Two-Way Floor System Reinforced with Grouted Tendons

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Views of the Design Floor Featuring Highly Irregular Support Layout

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1. GEOMETRY AND STRUCTURAL SYSTEM

2. MATERIAL PROPERTIES3. LOADS4. DESIGN PARAMETERS5. ACTIONS DUE TO DEAD

AND LIVE LOADS6. POST-TENSIONING7. CODE CHECK FOR

SERVICEABILITY8. CODE CHECK FOR

STRENGTH9. CODE CHECK FOR INITIAL

CONDITION10.DETAILING

10 STEPS OF DESIGN

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STEP 1GEOMETRY AND STRUCTURAL SYSTEM

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7


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Span TributaryWidthft.

Depthin

lin4

Ybin.

Ytin.

1Left 26.25 9.50 22506 4.75 4.752Middle 30.75 9.50 26364 4.75 4.752Right 30.75 17.50 61248 11.70 5.803Left 34.75 17.50 65132 11.80 5.70

3Middle 34.75 9.50 29794 4.75 4.753Right 34.75 17.50 104268 10.78 6.72

4Left 34.00 17.50 103287 10.75 6.754Middle 34.00 9.50 29151 4.75 4.754Right 34.00 17.50 182219 8.75 8.75

Cantilever 34.00 17.50 182219 8.75 8.75

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STEP 2MATERIAL PROPERTIES

2.1Concretef c=5000psi

Weight=150pcf

ElasticModulus=57000f c=4,030.50ksiCreepCoefficient=2

2.2Nonprestressed(Passive)Reinforcement:fy =60ksi

ElasticModulus=29,000ksi

2.3Prestressing:Material lowrelaxation,sevenwireASTM416

Stranddiameter=in(nominal).

Strandarea=0.153in.2

ElasticModulus=28,000ksi

Ultimatestrengthofstrand(fpu)=270ksi

Contd

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Unbonded system hardware

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Unbonded system




View of a Floor System Constructed withan Unbonded System

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Unbonded system

An example of a grouted system hardware with flat duct

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Grouted (bonded) system




Example of a Floor System Reinforcedwith Grouted Post-Tensioning System

Grouted (bonded) system

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S. K. Ghosh Associates Inc.


PT systems

UnbondedSystemAngularcoefficientoffriction() =0.07Wobblecoefficientoffriction(K)=0.001rad/ftAnchorset(wedgedrawin)=0.25inchStressingforce=80%ofspecifiedultimate

strength

Effectivestressafteralllosses=175ksi

BondedSystemUsesheetmetalorplasticflatductshousingup

tofivestrands

Angularcoefficientoffriction() =0.2Wobblecoefficientoffriction(K)=0.001rad/ftAnchorset(wedgedrawin)=0.25inch


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STEP 3LOADS

3.1SelfweightSlab=(9.5/12)*150pcf=118.75psf

3.2SuperimposedDeadLoadFixturesandfinishes=7.00psf

Partitions=20.00psf

TotalSuperimposedDeadLoad=27.00psf

Totaldeadload=SW+SDL=145.75psf

DeadloadondesignstripSpan1DL=145.75*26.25/1000=3.826klf

Span2DL=145.75*30.75/1000=4.482klf

Span3DL=145.75*34.75/1000=5.065klf

Span4DL=145.75*34.00/1000=4.956klf

Addeddeadloadduetocolumndrop,droppanelandthickenedoverhangiscalculatedandadded.

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STEP 3LOADS

3.3LiveLoad50psf,reducibleperIBCTheliveloadisreducedbasedontheareaitcovers.ThefollowingrelationshipfromIBCisused:

R=0.08(A 150)

R=reductionfactor,nottoexceed40%

A=tributaryinsquarefeet.

Normorethan

R=23.1(1+D/L0)=23.1(1+145.75/50)=90.43%>40%use40%

D=deadload;L0 =applicableliveload.

Span1:Reduction=0.08*(30*26.25 150)=51%

>40%max;use40%

Liveload=(1.00.40)*50.00psf=30.00psf

=30psf*26.25/1000=0.788klf

Spans2tospan4areworkedoutsimilarly21

STEP 3LOADS

Liveloadisgenerallyskipped(patterned),inordertomaximizethedesignvalues.However,fortwowayfloorsystems,ACI31811doesnotrequireliveloadskipping,providedtheratiooflivetodeadloaddoesnotexceed0.75.Inthisexample,asinmostconcretefloorsystemsforresidentialandofficebuildings,theratiooflivetodeadloadislessthan0.75.Hence,theliveloadwillnotbeskipped.(ACI31811,Section13.7.6)

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STEP 4DESIGN PARAMETERS

4.1ApplicableCodesACI3182011;IBC2012

4.2CovertoRebarandPrestressingTendonsUnbondedsystemMinimumrebarcover=0.75;topandbottomCovertotendon=0.75topandbottomCGS: allspans=1.00in.

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STEP4DESIGNPARAMETERS

BondedsystemMinimumrebarcover=0.75in;topandbottomForposttensioningtendons:Covertoduct=0.75in.Distancetocentroidofstrand=0.75+0.4+0.1

=1.25in.Where,0.4inishalfductdiameterandz=0.1inCGSforallspans=1.25in.

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4.3AllowableStressesInACI318/IBCtheallowablestressesfortwowaysystemsandonewaysystemsaredifferent.Thevaluesstatedarefortwowaysystems.Thesevaluesmaynotbeexceeded.UsingACI318,twowaysystemsaredeemedtobeessentiallycrackfreewheninservice.Inservicecracking,ifanyisnotofdesignsignificance.

Asinglerepresentativehypotheticalstressvalueisusedforeachdesignsection.

.

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4.3AllowableStresses.Allowablestressesinconcretearethesameforbondedandunbounded PTsystems

For sustained loadconditionCompression=0.45*fc=0.45*5000=2,250psi

Tension=6*f c=424psi For total loadconditionCompression=0.60*fc=3,000psi

Tension=6*f c=424psi Forinitial condition:Compression=0.60*fci=0.6*3750=2,250psi

Tension=3*f c=184psi

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4.4CrackWidthLimitationandControl.None required in ACI 318/IBC for two-way systems

InEC2designisbasedonauserdefineddesigncrackwidth.

4.5AllowableDeflection

Visual effects(L/240) Damagetononstructuralmembers(L/480) Malfunction

Longtermfactor2

Usecamber,whereneededandpractical

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STEP 5ACTIONS DUE TO DEAD AND LIVE

Usein houseframeprogram,recognizingthatitmayapproximatethepositionofdropsbymodelingthemalignedwithcentroidofslab.

Geometry of Analysis Frame

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STEP 5ACTIONS DUE TO DEAD AND LIVE LOADS

FOS=FaceofSupport

Span Location MD kft ML kft MD+ML kftSpan#1 LeftFOS* 204.25 42.04 246.29

Midspan 158.52 32.63 190.50RightFOS 264.42 54.42 318.84

Span#2 LeftFOS 296.00 60.92 356.92Midspan 185.09 38.09 223.18RightFOS 393.92 80.96 474.88



Cantilever LeftFOS 9.40 1.91 11.31

Forhandcalculationusevaluesatfaceofsupport andmidspan,recognizingthat(i)maximumvaluesdonotnecessarilyoccuratmidspan,andthemaximaofdifferentloadcasesdonotnecessarilycoincide

Actions Due to Dead Load, and Live Load

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STEP 5ACTIONS DUE TO DEAD AND LIVE LOADS

.

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STEP 6POST-TENSIONING

Selection of PT force and profileTwo entry values must be made to initiate the

computations. Select precompression and % of DL to balance

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.

Minimumaverageprecompression=150psiMaximumaverageprecompression=300psiTargetbalancedloading=60%oftotaldeadload,

upto80%wherebeneficial

Effectivestressinprestressingstrand:Forunbondedtendons:fse=175ksiForbondedtendons:fse=160ksi

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.

Forunbonded tendons

Forcepertendon=175ksi*0.153in2 =26.77kips/tendonUsemultiplesof26.77kipswhenselectingtheposttensioningforcesfordesign.

Forbondedtendons

Forcepertendons=160ksi*0.153in2 =24.48kips/tendonUsemultiplesof24.48kipswhenselectingtheposttensioningforcesfordesign.

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Selection of design parameters Select average precompression 150 psi Target to balance 60% of DL

Selection of PT force and profile Assume simple parabola mapped within

the bounds of top and bottom covers

Force diagram of simple parabola

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Calculation of balanced loads;adjustment of % of DL balanced


Select critical span

Select max drape

Calculate %of DL balanced (%DL)

%DL < 50%?

Reduce P/AReduce drape

Yes

Increase P/A

NoYes

No

No Yes

Yes No

Assume P/A =150psi[1MPa]

P /A 80%?

P/A>125psi [0.8MPa]?

Go to next span

1

2

F121_ACI_PT_2_way_082012

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Calculation of balanced loads;adjustment of % of DL balanced


Select max drape using tendons from critical span

Calculate %of DL balanced (%DL)

Move to next span

Reduce P/A or tendons to%DL balanced ~ 60% ;

P/A >= 125 psi [0.8 MPa]

YesNo

No Yes

Is it practical to reduce P/A or tendons?

%DL > 80%?

Rais e tendon to reach%DL ~ 60%

Exit after last span

F121_ACI_2-way_PT_force_082012

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Effective force per strand 26.77 k Average precompression 150 psi

Calculate number of strands for each span


Unbonded Tendon

Span1area=26.25*12*9.50=2992.50in2

Span1force=150psi*2992.50in2/1000=448.88kips

Numberoftendons=448.88/26.77=16.77;say17

Span Tributaryft.

Thicknessin.

Areain2 Forcekips.

Tendonsrequired

Tendonsselected

1 26.25 9.50 2992.50 448.88 17 20

2 30.75 9.50 3505.50 525.83 20 20

3 34.75 9.50 3961.50 594.23 23 23

4 34.00 9.50 3876.00 581.40 22 23

Cant.

34.00 17.50 7140.00 1071.00 41 23

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Calculate balanced loads Wb


Span 1

a=4.75 1.00=3.75b=8.50 1.00=7.50L=30.00c={[3.75/7.50]0.5/[1+(3.75/7.50)0.5]}*30=12.43

Wb/tendon=2P*a/c2 =26.77*(2*3.75/12)/12.432= 0.108klf/tendon

For20tendonsWb =0.108*20=2.160klf%DLBalanced=(2.160/3.826)*100=56%

OK(lessthan60%target,butacceptable)

Reaction,left=2.160klf*12.43=26.85kReaction,right=2.160klf*17.57=37.95k

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Span2;balancedloads;continuoustendons


a=7.5L=32.75Wb/tendon=8*P*a/L2

=(8*26.77*7.5/12)/32.752=0.125klf

For20tendonsWb =0.125*20=2.500klf%DLBalanced=2.500/4.482=56%OK

Reaction;left =2.500klf*16.38=40.95kReaction;right=2.500klf*16.38=40.95k

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Span 2; terminated tendons


a=3.75c=0.20*32.75=6.55

Wb=(3*26.77*2*3.75/12)/6.552 =1.170klfForceatdeadend =1.170klf*6.55=7.66k

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Span 2; Moments at change in centroid


M=P*shiftincentroid=P*(YtLeft YtRight)=23*26.77k*(4.75 5.80)/12=53.87kft

Momentatrightsupportcenterline:M=23*26.77*(5.805.70)/12=5.13kft

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Span3;symmetricaltendonspan


a=7.5;L=34.75

Wb/tendon=8*P*a/L2=(8*26.77*7.5/12)/34.752=0.111klf

For23tendons,Wb=0.111*23=2.553klf

%DLBalanced=2.553/5.065=50% 60%OK

Reactionatleft=2.553klf*17.38=44.37kReactionatright=2.553klf*17.38=44.37k

Momentatfaceofleftcolumndrop:

M=P*ShiftinCentroid

=23*26.77k*(5.70 4.75)/12=48.74kftMomentatfaceofrightdroppanel:M=23*26.77*(4.75 6.72)/12=101.08kft

Momentatcenterlineofrightsupport:

M=23*26.77*(6.72 6.75)/12=1.54kft44

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Span 4; unsymmetrical tendon span


a=4.75 1.00=3.75b=8.50 1.00=7.50L=34.50c={[3.75/7.50]0.5/[1+(3.75/7.50)0.5]}*34.5=14.29Wb/tendon=26.77kips*(2*3.75/12)/14.292

=0.082klf/tendon

For23tendons,Wb =0.082*23=1.886klf%DLBalanced=(1.886/4.956)*100=38%

Reaction,left=1.886klf*20.21=38.12kReaction,right=1.886klf*14.29=26.95kMomentatleftdroppanelface:M=23*26.77*(6.75 4.75)/12=102.62kft

Momentatrightcantileverface:M=23*26.77*(4.75 8.75)/12=205.24kft

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Cantilever; Balanced Loads


Tendonishorizontalandstraight.Nodeadloadisbalanced.Momentduetodeadendanchoredawayfromcentroid:M=23*26.77*(8.75 4.75)/12=205.24kftM=205.24kft

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Summary of balanced loads


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Validationofbalancedloadsthroughequilibriumcheck


Forces =26.85+(30*2.16) 37.95 40.95+(32.75*2.50)(6.55*1.17)+7.6640.9544.37+(34.75*2.553) 44.3738.12+(34.5*1.886) 26.95

=0.05k 0OK

M3rdSupport =26.85*62.75+(37.95+40.95)32.757.66*6.55 2.16*30*47.75 (2.50*32.752 /2)+(2.553*34.752/2)+(1.170*6.552/2)+1.886*34.5*52(44.37+38.12)*34.75 6.95*69.2553.87+5.13+48.74101.08 1.54+102.62 205.42+205.42=0.96kft 0,OK

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Computationofactionsfrombalancedloads


Usingtheinhouseframeprogram,andbalancedloadsasappliedforces,determinethemomentsatfaceofsupportsandmidspans

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ACI 318-11 requirements for serviceability Load combinations Stress check Minimum reinforcement Deflection check.

Load combination Total load condition

1.00DL+1.00LL+1.00PT Sustained load condition

1.00DL+0.30LL+1.00PT Stress check

STEP 7CODE CHECK FOR SERVICEABILITY

=(MD +ML +MPT)/S+P/AS=I/Yc;I=secondmomentofareaof;Yc =distancetofarthesttensionfiber

ConsiderlocationsA,BandCascriticallocationsforstresscheck 50

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Stress check at point A face of support

STEP 7CODE CHECK FOR SERVICEABILITY

Stop =103287/6.75=15302in3

Sbot=103287/10.75=9608in3

A=34*12*9.5+144*8=5028in3

P/A =615.71*1000/5028=122psi

Totalloadcombination: =(MD +ML +MPT)/S+P/AMD +ML +MPT =(615.08 124.0+321.67)

=417.41kft

Bottomfiber: =417.41*12*1000/9608 122

=643psiCompression40%maxLiveload=(1.0-0.40)*50.00psf=30.00psf =30psf*26.25/1000=0.788klf

Spans2tospan4:By inspectionthemaximumreductionof40%maybeusedLiveload=(1.0-0.40)*50.00psf=30.00psfSpan2LL=30psf*30.75/1000=0.923klfSpan3LL=30psf*34.75/1000=1.043klfSpan4LL=30psf*34.0/1000=1.020klfCantilever:Reduction=0.08*(34*2.5150)=0%

Liveload=50psf=50psf*34.0/1000=1.700klfLL/DLratio=50/145.75=0.34

long-hand design of a two-way post-tensioned floor system

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