+ All Categories
Home > Documents > Foundation Verification Problems

Foundation Verification Problems

Date post: 05-Jan-2016
Category:
Upload: rajat-ramesh
View: 218 times
Download: 0 times
Share this document with a friend
Description:
risa foundation

of 62

Transcript
  • RISAFoundationRapidInteractiveStructuralAnalysisFoundationsVerificationProblems

    26632TowneCentreDrive,Suite210FoothillRanch,California92610(949)9515815(949)9515848(FAX)www.risa.com

  • Copyright2013byRISATechnologies,LLC.Allrightsreserved.Noportionofthecontentsofthispublicationmaybereproducedortransmittedinanymeanswithouttheexpresswritten

    permissionofRISATechnologies,LLC.Wehavedoneourbesttoinsurethatthematerialfoundinthispublicationisbothusefulandaccurate.However,pleasebeawarethaterrorsmayexistinthispublication,andthatRISA

    Technologies,LLCmakesnoguaranteesconcerningaccuracyoftheinformationfoundhereorintheusetowhichitmaybeput.

  • TableofContents

    i

    TableofContents VerificationProblem1:StripFootingDesign..............................................................................................................3VerificationProblem2:SquareSpreadFooting#1...................................................................................................5VerificationProblem3:RectangularSpreadFoot#1..............................................................................................7VerificationProblem4:PileCapShear..........................................................................................................................9VerificationProblem5:EccentricallyLoadedFooting.........................................................................................13VerificationProblem6:CantileverRetainingWall#1..........................................................................................15VerificationProblem7:CantileverRetainingWall#2...........................................................................................17VerificationProblem8:RectangularFooting#2......................................................................................................19VerificationProblem9:SquareFooting#2................................................................................................................21VerificationProblem10:CantileverRetainingWall#3........................................................................................23VerificationProblem11:PileCapDesignExample.................................................................................................25AppendicesAppendixA10:CantileverRetainingWall#3Calculations.......................................................................A10.1AppendixA11:PileCapDesignExampleCalculations................................................................................A11.1

  • VerificationOverview

    1

    VerificationOverview VerificationMethodsWeatRISATechnologiesmaintainalibraryofhundredsoftestproblemsusedtovalidatethecomputationalaspectsofRISAprograms.InthisverificationpackagewewillpresentarepresentativesampleofthesetestproblemsforyourreviewandcompareRISAFoundationtotextbookexampleslistedwithineachproblem.TheinputforthesetestproblemswasformulatedtotestRISAFoundationsperformance,notnecessarilytoshowhowcertainstructuresshouldbemodeledandinsomecasestheinputandassumptionsweuseinthetestproblemsmaynotmatchwhatadesignengineerwoulddoinarealworldapplication.Thedataforeachoftheseverificationproblemsisprovided.ThefileswheretheseRISAFoundationproblemsarelocatedisintheC:\RISA\ExamplesdirectoryandtheyarecalledVerificationProblem1.fnd(2,3,etc).

    VerificationVersionThisdocumentcontainsproblemsthathavebeenverifiedinRISAFoundationversion5.0.2.

  • 2

  • VerificationProblem1:StripFootingDesign

    3

    VerificationProblem1:StripFootingDesignDesignofaWallFootingThisproblemrepresentsatypicaldesignofawallfooting.Thehandverificationofthisproblemcanbetakendirectlyfromthe4theditionofMacgregorandWights,ReinforcedConcreteMechanicsandDesign(Example161,p.802805).

    Description/ProblemStatementA12in.thickconcretewallcarriesservicedeadandliveloadsof10kipsperfootand12.5kipsperfoot,respectively.Theallowablesoilpressure,qa,is5ksfatthelevelofthebaseofthefooting,whichis5ftbelowthefinalgroundsurface.Thewallfootinghasastrengthof3ksiandfy=60ksi.Thedensityofthesoilis120lb/ft3.Notethatthetextdoesnotaccountfortheselfweightofthefooting.Therefore,theRISAmodelhasthedensityoftheconcretematerialsettozero.

    Figure1.1RISAFoundationModelView

    ComparisonComparisonofResults(UnitsSpecifiedIndividually)

    Value RISAFoundation TextValue %DifferenceFactoredNetPressure,qnu(ksf) 6.191 6.19 0

    Vu(k/ft) 7.872 8.513 7.52*Vc(k/ft) 9.613 9.374 2.59Mu(k*ft/ft) 13.455 13.4 0.41

    *Mn(k*ft/ft) 14.268 14.0 1.91Asmin(in^2) 1.451 1.45 0.07

    Table1.1ResultsComparison1ThedetailreportforLC2showsaLoadingDiagramwith6.2ksfonthetoeendand6.18ksfontheheel.Theaverageofthesevaluesisusedintheabovetable.

  • VerificationProblem1:StripFootingDesign

    4

    2ThedetailreportshowsaVuToe=7.88k/ftandaVuHeel=7.86k/ft.Theaverageofthesevaluesisusedintheabovetable.3Thevaluefromthetextisusingad=8.5.RISAFoundationisbeingmoreexactandusingd=1330.5/2=9.75.ThisproducesaVu=(1/12)*(259.75)*6.19=7.87k/ft4Thevaluefromthetextisusingd=9.5whereRISAFoundationisbeingmoreexactandisusingd=9.75.(9.75/9.5)*9.37=9.617k/ft.

    ConclusionInthisexampleitisshownthattheRISAFoundationcalculationsreasonablymatchthetextbookdesignexamplesexceptininstanceswhichareexplainedabove.

  • VerificationProblem2:SquareSpreadFooting#1

    5

    VerificationProblem2:SquareSpreadFooting#1DesignofaSquareSpreadFootingThisproblemrepresentsatypicaldesignofasquarespreadfooting.Thehandverificationofthisproblemcanbetakendirectlyfromthe4theditionofMacgregorandWightsReinforcedConcreteMechanicsandDesign(Example162,p.805810).

    Description/ProblemStatementAsquarespreadfootingsupportsan18in.squarecolumnsupportingservicedeadandliveloadsof400kipsand270kips,respectively.Thecolumnisbuiltof5ksiconcreteandhaseightNo.9longitudinalbarswithfy=60ksi.Thefootinghasconcreteofstrength3ksiandGrade60bars.Thetopofthefootingiscoveredwith6in.offillwithadensityof120lb/ft3anda6in.basementfloor.Thebasementfloorloadingis0.1ksf.Theallowablebearingpressureonthesoilis6ksf.LoadandresistancefactorsaretakenfromACIsections9.2and9.3.

    Figure2.1RISAFoundationModelView

    Solvethemodelandlookatthedetailreportforthefooting.Notethatthetextusesthenetsoilbearingtocalculatethesizeoffooting.ThissizeisuseddirectlyinRISAFoundationandthusthesoiloverburdenandselfweightaresettozero.

  • VerificationProblem2:SquareSpreadFooting#1

    6

    ComparisonComparisonofResults(UnitsSpecifiedIndividually)

    Value RISAFoundation TextValue %DifferenceSoilPressure,qu

    (ksf) 7.311 7.31 11.6VuPunching(k) 804.591 804 0.07

    *VcPunching(k) 0.75*1128.747=846.562 846 0.07VuOneWay(k) 204.254 204 0.12

    *VcOneWay(k) 0.75*411.134=308.352 308 0.11Mu(k*ft) 954.34 954 0.04

    AsRequired(in^2) 7.763 8.41 7.73Table2.1ResultsComparison

    1Toactuallyseethisvalue,checkthe"Service"checkboxforLC2andsolvethemodel.ThenlookatthedetailreportintheSoilBearingsection.Whenviewingtherestoftheresults,uncheckthischeckboxandresolve.2InRISAFoundationtheVcvalueisreportedwithoutthevalue.IftheVcvalueismultipliedbythetextthenthereisagreement.3IfyouuseRISAsvalueofAsRequiredandcalculateanewa,youwillgeta*Mn=954.3k*ft.ThisvalueexceedsMu.TheAsrequiredbythetextisusingabackoftheenvelopecalculationtocomeupwithAsthatisconservativeinthiscase.Whenitcomestothecalculationof*MnRISAisfollowingACI31811Section10.5.3inproviding(4/3)*Asrequired,whereasthetextisnot.

    ConclusionInthisexampleitisshownthattheRISAFoundationcalculationsreasonablymatchthetextbookdesignexamplesexceptininstanceswhichareexplainedabove.

  • VerificationProblem3:RectangularSpreadFoot#1

    7

    VerificationProblem3:RectangularSpreadFoot#1DesignofaRectangularSpreadFootingThisproblemrepresentsatypicaldesignofarectangularspreadfooting.Thehandverificationofthisproblemcanbetakendirectlyfromthe4theditionofMacgregorandWightsReinforcedConcreteMechanicsandDesign(Example163,p.810812).

    Description/ProblemStatementNotethatthetextusesthenetsoilbearingtocalculatethesizeoffooting.ThissizeisuseddirectlyinRISAFoundationandthusthesoiloverburdenandselfweightaresettozero.Thisfootinghasbeendesignedassumingthatthemaximumwidthis9ft.Followingthehandcalculationfromthetextbookthefootingisfoundtobe9wideby138longby32thick.Theexampleassumesthesamenetsoilpressureof7.31ksfforboth162and163.However,(11.17ft)2=124.77ft2and13.666ft*9ft=123ft2.Thus,thesmallerfootinginthisexampleproducesaslightlyhighersoilpressurethanthetext.

    Figure3.1RISAFoundationDetailReportView

    Thetextexampleuses#8barsinonedirectionand#5barsintheotherforthebottomsteel.InRISAFoundationthisisnotpossible,sotwofootingshavebeencreatedtoverifythecalculations.NodeN1isusingthe#8barsandnodeN2isusing#5bars.WhenviewingtheresultsinRISAFoundationusethefootingnodenumbersgiveninTable3.1below.

  • VerificationProblem3:RectangularSpreadFoot#1

    8

    ComparisonComparisonofResults(UnitsSpecifiedIndividually)

    Value RISAFoundation TextValue %DifferenceVuOneWay(k)N1 250.23 247 1.31

    *VcOneWay(k)N1 0.75*331.263=248.45 248 0.18MuLong(k*ft)N1 1234.69 1217 1.45

    AsMinLong(in2)N1 6.221 6.22 0.02AsProvidedLong(in2)N1 10.21in2(13#8bars) 11.1in2(14#8bars)1 8.02

    MuShort(k*ft)N1 712.5 702 1.5AsMinShort(in2)N2 9.446 9.45 0.4

    AsProvidedShort(in2)N29.51in2(31#5bars;25

    arebanded)9.61in2(31#5bars;25

    arebanded) 0Table3.1ResultsComparison

    1InthetextapproximatemethodsareusedtodetermineAsReqd.Wecanseethatthe*Mn=1330k*ft.RISAFoundationisabletoremoveabarandstillproducea*MngreaterthanMu.

    ConclusionInthisexampleitisshownthattheRISAFoundationcalculationsreasonablymatchthetextbookdesignexamples,exceptintheinstancesexplainedabove.

  • VerificationProblem4:PileCapShear

    9

    VerificationProblem4:PileCapShearDesignforDepthofFootingonPilesThisproblemrepresentsthedesignforafootingsupportedonpiles.ThehandverificationofthisproblemcanbetakendirectlyfromPCAsNotesonACI31805BuildingCodeRequirementsforStructuralConcrete(Example22.7,p.2220).

    Description/ProblemStatementFootingSize = 8.5x8.5ColumnSize = 16x16PileDiameter = 12in.fc = 4000psiLoadperPile: PD = 20kips PL = 10kips

    Figure4.1RISAFoundationDetailReportView

    NotethatRISAFoundationwillnotplacetopsteelreinforcementinapilecapunlessthereistensioninthetopfaceofthepilecap.Forthisreasona1kip*ftmomentwasaddedtotheOL1loadcategory.Thisistoforcetopsteel,asthisaffectsthepilepunchingshearchecks.Ifthereisnoreinforcementinthetopthentheprogramconsidersthecapunreinforcedforpunchingshearcalculations.

  • VerificationProblem4:PileCapShear

    10

    ComparisonComparisonofResults(Unitsinkips)

    Value RISAFoundation TextValue %DifferenceOnewayBeamShearCapacity,Vn(kips) 180.629*0.75=135.471 135.4

    0.05

    PedestalPunchingShearCapacity,Vn

    (kips) 320/1.004=318.732 319 0.08CornerPilePunchingShearCapacity,Vn

    (kips) 141.913 217 NA3Table4.1ResultsComparison

    1TheprogramgivesVnexplicitly,sothePhiwasmultipliedinheretogetPhi*Vn.2ThePhi*Vnisnotgivenexplicitly.Theprogramgivesthedemandandthecodecheck,sothecalculationaboveshowswhatPhi*VnisinRISAFoundation.3Acoupleofthingsareoccurringhere.Forone,wearetransformingtheroundpunchingshearperimeterintoanequivalentsquareperimeter.Thus,thiswouldcreateadifference.Second,andmoreimportantly,thepunchingshearcapacityisbasedonthesmallestpossibleshearperimeter,bo.ThePCAnotesexampleassumesthatthepunchingshearperimeterwouldoccurallthewayaroundthepile,asshowninFigure4.2below.

    Figure4.2

    Inreality,however,thecrackwillperpetuatethroughadistancedfromtheedgeofthepile.D/2occursatmidwayalongthecrackandisusedforcalculationpurposes.However,thecrackwouldlooklikethisinanelevationview,asshowninFigure4.3.

  • VerificationProblem4:PileCapShear

    11

    Figure4.3

    BecauseofthisthepunchingshearperimetercannotbetakenasshowninthePCAnotes.Insteadyoureallyonlyhaveapartialperimeterbecauseyouwillbreakoutthecornerbeforeyougetallthewayaround.InRISA,includingthesquareperimeteradjustment,itwouldlookasshowninFigure4.4.

    Figure4.4

  • VerificationProblem4:PileCapShear

    12

    ConclusionInthisexampleitisshownthattheRISAFoundationcalculationsreasonablymatchthetextbookdesignexamples.

  • VerificationProblem5:EccentricallyLoadedFooting

    13

    VerificationProblem5:EccentricallyLoadedFootingFootingUnderBiaxialMomentThisproblemrepresentsthecasewhereafootingmaybesubjectedtoanaxialforceandbiaxialmomentsaboutitsxandyaxes.ThisexamplecomesfromtheDesignofReinforcedConcreteStructures,copyright1985Hassoun(Example13.7,p.409413).

    Description/ProblemStatementA12by24columnofanunsymmetricalshedissubjectedtoanaxialloadPD=220kipsandamomentMD=180kftduetodeadload,andanaxialloadPL=165kipsandamomentML=140kftduetoliveload.Thebaseofthefootingis5ft.belowfinalgradeandtheallowablesoilbearingpressureis5ksf.Thefootinghasstrengthof4ksiandasteelyieldof40ksi.Notethatthetextdoesnotaccountfortheselfweightofthefooting.Therefore,theRISAmodelhasthedensityoftheconcretematerialsettozero.

    Figure5.1RISAFoundationDetailReportView

  • VerificationProblem5:EccentricallyLoadedFooting

    14

    ComparisonComparisonofResults(UnitsSpecifiedIndividually)

    Value RISAFoundation TextValue %DifferenceMethod1SoilPressure,

    qn(ksf) 4.283(87.1/90)*4.42

    =4.2771

    0.07Method1Muxx(k*ft) 687.2 687.4 0.03Method1Muzz(k*ft) 523.11 523.2 0.02Method2SoilPressure

    Max,qn(ksf) 4.43 4.422 0.23Method2SoilPressure

    Min,qn(ksf) 1.973 1.98 0.35Method2Muxx(k*ft) 873.6 873 0.07

    Table5.1ResultsComparison1Thetextbookcalculatesarequiredareaof87.1in^2.Theythenchooseanareaof90in^2.Thus,theirvaluehasbeenadjusted.2Thetextbookexamplehasanerror.Theystatethat3.20+1.22=4.22ksfwhencalculatingqmaxformethod2.Thisshouldbe4.42ksf.

    ConclusionInthisexampleitisshownthattheRISAFoundationcalculationsreasonablymatchthetextbookdesignexamples.

  • VerificationProblem6:CantileverRetainingWall#1

    15

    VerificationProblem6:CantileverRetainingWall#1DesignofaCantileverRetainingWallThisexamplecomesfromthePrinciplesofFoundationEngineering,3rdEditionbyDas,copyright1995.ThisisexampleA.8onP798.InthisproblemwewillcomparetheserviceabilitychecksforaretainingwallexampletotheoutputfromRISAFoundation.

    Description/ProblemStatementThecrosssectionofacantileverretainingwallisshownbelow.Forthiscase,fy=413.7MN/m2andfc=20.68MN/m2.Notes:

    RISAFoundationusesRankinesmethodtocalculatelateralsoilpressurecoefficients.ThisexampleusesCoulombsmethod.BecauseofthistheKLatToewassetto2.04.

    Thecoefficientoffrictioninthisexampleiscalculatedas:Tan(2/3*)=0.237.Thisisthevalueenteredintheprogram.

    Theultimatebearingpressureisinthisexampleiscalculatedas574.07,sothisisenteredastheallowablebearingintheprogram.

    Figure6.1RISAFoundationDetailReportView

  • VerificationProblem6:CantileverRetainingWall#1

    16

    ComparisonComparisonofResults(UnitsSpecifiedIndividually)

    Value RISAFoundation TextValue %DifferenceMresistAgainst

    Overturning(kNm/m) 1030.034 1044.3(1128.98)1 1.37Moverturning(kNm/m) 379.047 379.25 0.05VresistAgainstSliding

    (kN/m) 147.278433.17 171.39106.67=155.12 5.04

    Vsliding(kN/m) 158.853 158.95 0.06MaxBearingPressure

    (kPa) 199.349 189.23 5.36BearingUC .347

    189.2/574.07=.3293 5.47

    Table6.1ResultsComparison1Thetextbookaccountsfortheslopingouterfaceofthewall,whichRISAFoundationdoesnot.Also,theverticalportionofthesoilforceinthetextisassumedtoactattheedgeoftheheel.InRISAFoundationweassumethisforcetoactattheinsidefaceofthewall.Thesedifferenceswouldequal1128.9811.792.6*28.03=1044.312kNm/m.2Thetextbookassumescohesion.RISAFoundationassumescohesionlesssoil.TheygiveaVresist=111.5+106.7+215=433.17kN/m.The106.7isacohesiontermthatRISAdoesntaccountfor.The215comespassivepressureincludingcohesion.Thecohesionterm=171.39kN/mwhichRISAdoesntaccountfor.AccountingforthesecohesiondifferencesbetweenRISAFoundationandthetextgivesavalue=433.17171.39106.67=155.1kN/m.3ThetextusestheMresisttocalculatethebearingpressure.Becausethisisdifferent,thepressurecalculationisdifferent.

    ConclusionInthisexampleitisshownthattheRISAFoundationcalculationsreasonablymatchthetextbookdesignexamplesafteraccountingfordifferencesincalculationprocedures.

  • VerificationProblem7:CantileverRetainingWall#2

    17

    VerificationProblem7:CantileverRetainingWall#2DesignofReinforcedConcreteCantileverRetainingWallsInthisproblemwewillcomparetheserviceabilitychecksforaretainingwallexampletotheoutputfromRISAFoundation.ThisexamplecomesfromReinforcedConcreteDesign,ThirdEdition,copyright1992bySpiegelandLimbrunner.Thisisdesignexample81onP214.

    Description/ProblemStatementDesignData:unitweightofearthwe=100lb/ft3,allowablesoilpressure=4,000psf,equivalentfluidweightKawe=30100lb/ft3,andsurchargeloadws=400psf.Thedesiredfactorofsafetyagainstoverturningis2.0andagainstslidingis1.5.

    Figure7.1RISAFoundationDetailReportView

  • VerificationProblem7:CantileverRetainingWall#2

    18

    Note:TheshearkeyhasbeenomittedfromtheRISAFoundationmodel,asthiswillaffectthecalculationsforslidingandoverturning.Thetextexampledidnotassumeakeywhenperformingthosecalculations.

    Comparison

    Value RISAFoundation TextValue%

    DifferenceMResist(k*ft) 131.169 131.7 0

    MOverturn(k*ft) 48.6 48.6 0VResist(kips) 10.008 9.855 1.55VSlide(kips) 7.02 7.02 0

    MaxSoilPressure(ksf) 3.101 3.043 1.9

    MuofHeel(k*ft) 46.69 67.65 NA1VuHeel(k*ft) 11.22 20.82 NA1

    VnofHeel(kips) 18.301*(0.85/0.75)=20.742 20.76 0.1AsTop(in2) #7Bars@8"oc #7Bars@8"oc 0

    MuofToe(k*ft) 18.473 20.476 NA3VuofToe(kips) 6.47 13.07 NA4VnofToe(kips)

    17.315*(0.85/0.75)=19.62** 19.64 0.1

    AsBot(in2) #7Bars@16"oc#7Bars@16"

    oc 0MuStemBase(k*ft) 63.4 63.431 0.05VuStemBase(kips) 10.023(LC2) 10.049 0.26VnofStem(kips) 15.281*(0.85/0.75)=17.318 17.391 0.42

    AsStem(in2) #8Bars@9"oc #8Bars@9"oc 0Table7.1ResultsComparison

    1Inthetextexamplethe"relieving"momentduetotheupwardsoilpressureontheheelisnotaccountedfor.ThisisaccountedforinRISA.2Thisvalueisbeingadjustedforthechangeinshearfrom0.85to0.75.3Inthetextexamplethe"relieving"momentduetothedownwardsoilpressureonthetoeisnotaccountedfor.ThisisaccountedforinRISA.4Inthetextexampletheshearlocationistakenasthefaceofwall.InRISAwearecomingoutadistance"d"fromthewallandchecktheshearatthatlocation.

    ConclusionInthisexampleitisshownthattheRISAFoundationcalculationsreasonablymatchthetextbookdesignexample.

  • VerificationProblem8:RectangularFooting#2

    19

    VerificationProblem8:RectangularFooting#2RectangularReinforcedConcreteFootingsThisproblemrepresentsatypicaldesignofarectangularspreadfooting.ThisexamplecomesfromReinforcedConcreteDesign,ThirdEdition,copyright1992bySpiegelandLimbrunner.Thisisdesignexample104onP310.

    Description/ProblemStatementAconcretefooting4ft.belowthefinishedgroundlinesupportsan18in.squaretiedinteriorconcretecolumn.Thetotalfootingthicknessis24in.Onedimensionofthefootingislimitedtoamaximumof7ft.

    ServiceDL =175kipsServiceLL =175kipsfc(footingandcolumn) =3000psiSteelYieldfy =60ksiLongitudinalcolumnsteel =No.8barsSoilDensity =100lb/ft3AllowableSoilPressure =5ksfEffectiveAllowableSoilPressure =4.50ksf

    Figure8.1RISAFoundationDetailReportView

    Notethattheselfweightandoverburdenwereinputaszeroandtheallowablesoilpressurewasaddeddirectlyas4.50ksf.

  • VerificationProblem8:RectangularFooting#2

    20

    ComparisonComparisonofResults(UnitsSpecifiedIndividually)

    Value RISAFoundation TextValue %DifferenceFactoredSoilPressure,qu(ksf) 6.7391 6.74 0.01ShearDemand,Vutwoway(k) 474.921 475 0.02ShearCapacity,Vntwoway(k)

    *666.031=566.13(=0.85)2 566 0.02

    ShearDemand,Vuoneway(k) 157.246 157.1 0.09ShearStrength,Vnoneway(k)

    *184.035=156.43(=0.85)2 156.4 0.17

    BendingMoment,Mulongdirection(k*ft) 589.67 590 0.06BendingMoment,Mushortdirection(k*ft) 293.05 293 0.02

    Asrequiredlongdirection(in2) 6.884 6.9 0.23Asrequiredshortdirection(in2) 3.303

    4.4/(4/3)=3.33 0.09

    AsrequiredT&S(in2) 5.962 5.96 0.03FootingBearingStrength(in2)

    *1652.4=1156.68(=0.70)4 1157 0.03

    FactoredBearingLoad,Pu(k) 542.5 542.5 0.00Table8.1ResultsComparison

    1Toactuallyseethisvalue,checkthe"Service"checkboxforLC2andsolvethemodel.ThenlookatthedetailreportintheSoilBearingsection.Whenviewingtherestoftheresults,uncheckthischeckboxandresolve.2InRISAFoundationtheVcvalueisreportedwithoutthevalue.IftheVcvalueismultipliedbythetextthenthereisgoodagreement.3Inthetexttheyaremultiplyingby4/3*Asrequiredastheirvalue.RISAFoundationwilldothisaswellwhenactuallyreinforcingthefooting,however,wealsoreporttheAsrequireditself.4InRISAFoundationtheBcvalueisreportedwithoutthevalue.IftheBcvalueismultipliedbythetextthenthereisgoodagreement.

    ConclusionInthisexampleitisshownthattheRISAFoundationcalculationsreasonablymatchthetextbookdesignexample.

  • VerificationProblem9:SquareFooting#2

    21

    VerificationProblem9:SquareFooting#2DesignforBaseArea,Depth,andReinforcementofFootingThisproblemrepresentsatypicaldesignofasquarespreadfootingThehandcalculationcomparisonofthisexamplecomesfromthePCANotesfortheACI31805Example22.1,22.2and22.3(allinoneproblem)onpage227.

    Description/ProblemStatementServiceDeadLoad =350kipsServiceLiveLoad =275kipsServiceSurcharge =100psfWeightofSoilandConcreteaboveFootingBase =130lb/ft3NetAllowableSoilPressure =3.75ksf

    Figure9.1RISAFoundationDetailReportView

    Notes:

    Becausetheexampledoesnotusetheselfweightofthefootinginthecalculationandinsteadjustgivesanaverageweightbetweenthesoilandconcrete,thedensityof

  • VerificationProblem9:SquareFooting#2

    22

    concretehasbeensetto0.TheOverburdenhasalsobeensettozero.Thus,theallowablesoilpressureissimplyaddeddirectlyas3.75ksf.

    ThedfootvalueforfootingsinRISAFoundation=footingthicknessbottomcover1*db.Theexamplesusead=28,thusthebottomcoverissetto4.

    ComparisonComparisonofResults(UnitsSpecifiedIndividually)

    Value RISAFoundation TextValue %DifferenceEx22.1:qs(ksf) 5.0891 5.1 0.22Ex22.2ShearDemand,Vuoneway(k) 242.564 243 0.18

    Ex22.2ShearCapacity,Vnoneway(k)*478.5=358.868

    (=0.75)2 359 0.04Ex22.2ShearDemand,Vutwoway(k) 778.014 780 0.25

    ShearCapacity,Vntwoway(k)*1082=811.593(

    =0.75)2 812 0.05Ex22.2BendingMoment,Mu(k*ft) 1190.77 1193 0.12Ex22.3Asrequired(in2) 9.704 9.6 1.08

    Table9.1ResultsComparison1Toactuallyseethisvalue,checkthe"Service"checkboxforLC2andsolvethemodel.ThenlookatthedetailreportintheSoilBearingsection.Whenviewingtherestoftheresults,uncheckthischeckboxandresolve.2RISAFoundationpresentstheVcvaluewithout.WhenyoumultiplyVcbyyougetagreement.

    ConclusionInthisexampleitisshownthattheRISAFoundationcalculationsreasonablymatchthePCANotesdesignexamples.

  • VerificationProblem10:CantileverRetainingWall#3

    23

    VerificationProblem10:CantileverRetainingWall#3DesignofaCantileverRetainingWallInthisexamplewehaveanonslopingbackfilledretainingwallwithaloadsurchargeandawatertablepresent.Thewallandfootingarenotpouredmonolithically.Footingdowelsoccuratbothfacesofthewallandareofthesamesizeandspacingasthewallreinforcement.Aloadcombinationof1.0*DL+1.0*LL+1.0*HLisusedfortheserviceLCandaloadcombinationof1.2*DL+1.6*LL+1.6*HLisusedforthestrengthLC.InthisexampleRISAFoundationsvaluesarecomparedtothevaluesobtainedfromahandcalculationdoneforsoilpressures,stabilityandalldesignaspectsofthewall.ThishandcalculationislocatedinAppendixA10

    Description/ProblemStatementThisproblemcomesfromahandcalculationverification.Itistestingallresultsforretainingwallstability,soilpressurecalculationsandreinforcementdesign.

    Figure10.1RISAFoundationDetailReportView

    Note:Theretainingwalliscantileveredandthebaseisnotrestrainedagainstsliding.

  • VerificationProblem10:CantileverRetainingWall#3

    24

    ComparisonThissectionisthetabularcomparisonoftheRISAFoundationanswersandthesummaryfromthedetailedvalidationresults.

    ComparisonofResults(UnitsSpecifiedIndividually)1,2Value RISAFoundation HandCalculation %Difference

    LateralEarthPressures NAKLatHeel 0.307 0.307 0KLatHeelSat 0.333 0.333 0KLatToe 3.255 3.255 0

    StabilityChecks OverturningSFMin/SF 0.659 0.659 0SlidingSFMin/SF 1.176 1.176 0

    WallDesign UCMaxInt 1.664 1.678 0.834ShearUCMax 0.624 0.627 0.478DowelShearUCMax 0.455 0.455 0

    FootingSoilPressures qmax(ft)* 5.6 5.603 0.054LsoilLength(ft)* 9.09 9.090 0

    FootingDesign ShearUCHeel 0.746 0.746 0MomentUCHeel 0.967 0.967 0ShearUCToe 0.597 0.597 0MomentUCToe 0.63 0.630 0

    Table10.1ResultsComparison

    1Notethatthevaluesshownherecanbeseengraphicallybylookingatthedetailreportforloadcombination2.2SeeAppendixA10foranindepthhandcalculation.

    ConclusionInthisexampleitisshownthattheRISAFoundationcalculationsreasonablymatchthehandcalculateddesignexample.

  • VerificationProblem11:PileCapDesignExample

    25

    VerificationProblem11:PileCapDesignExampleDesignofaPileCapInthisexamplewehaveapilecapwith12HP14x102pilesprovidingsupport.Thepileshavean85kipcompressioncapacity,a12kiptensioncapacityanda14kipshearcapacity.Thepilecapis42"thickwitha6"pileembedmentandmadefrom4ksilightweightconcrete.Aloadcombinationof1.0*DL+1.0*LLisusedfortheserviceLCandaloadcombinationof1.2*DL+1.6*LLisusedforthestrengthLC.

    Description/ProblemStatementInthisexampleRISAFoundationsvaluesarecomparedtothevaluesobtainedfromahandcalculationdoneforallaspectsofthepilecap.ThishandcalculationislocatedinAppendixA11.

    Figure11.1RISAFoundationModelView

  • VerificationProblem11:PileCapDesignExample

    26

    ComparisonThissectionisthetabularcomparisonoftheRISAFoundationanswersandthesummaryfromthedetailedvalidationresults.

    ComparisonofResults(UnitsSpecifiedIndividually)1,2Value RISAFoundation HandCalculation %Difference

    FlexuralChecks Muxx(kft) 1432.03 1438 0.42Muzz(kft) 937.13 932.8 0.46Asminx(in^2) 13.835 13.835 0Asminz(in^2) 10.13 10.13 0Asflexxbot(in^2) 20.588 20.588 0Asflexzbot(in^2) 15.075 15.075 0UCMx 0.755 0.753 0.27UCMz 0.445 0.488 8.81

    PunchingShearChecks PedestalPunchingUC 0.719 0.719 0Pile4PunchingCapacity

    (kips)220.284 220.284 0

    Pile4PunchingUC 0.399 0.399 0OneWayShearChecks ShearCapacityVcx(kips) 1186.972 1187 0ShearCapacityVcz(kips) 585.931 591.221 0.89

    PedestalShearCapacities Vc(kips) 48.952 48.952 0Vs(kips) 50.658 50.658 0

    Table11.1ResultsComparison

    1Notethatthevaluesshownherecanbeseengraphicallybylookingatthedetailreportforthepilecap.2SeeAppendixA11foranindepthhandcalculation.

    ConclusionInthisexampleitisshownthattheRISAFoundationcalculationsreasonablymatchthehandcalculateddesignexample.

  • AppendixA10CantileverRetainingWall#3Calculations_________________________________________________________________________Inthisexamplewehaveanonslopingbackilledretainingwallwithaloadsurchargeandawatertablepresent.Herewewillcalculateallsoilpressures,designallaspectsoftheretainingwallandcheckforoverturningandsliding.

    InputParameters

    Theretainingwalliscantileveredandthebaseisnotrestrainedagainstsliding.Thewallandfootingarenotpouredmonolithically.Footingdowelsoccuratbothfacesofthewallandareofthesamesizeandspacingasthewallreinforcement.

    Inthisexamplewewillusealoadcombinationof1.0*DL+1.0*LL+1.0*HLfortheserviceLCandaloadcombinationof1.2*DL+1.6*LL+1.6*HLforthestrengthLC.

    DLFactor 1.2 LLFactor 1.6 HLFactor 1.6

    Geometry

    Hwall 16 ft Hwall=Hsoil Ltoe 3.5 ft Wkey 18 in

    Hwater 6 ft Lheel 5.5 ft Dkey 18 in

    twall 18 in tfoot 18 in Lkey 4.5 ft

    Lwall 10 ft Totallengthofwall

    Lfoot =++Ltoe Lheel tfoot 10.5 ft Overalllengthofthefooting

    Offsetkey =+Lkey Wkey Ltoe twall 1 ft Thekeyoffsetfromtheinteriorfaceofwallandtheinteriorfaceofkey.

    Materials

    conc .15 kip

    ft3

    fc 4 ksi fy 60 ksi

    A10.1

  • Soil

    0.5 Coefoffrictionw/soil 0 backillangle Htoesoil 2 ft

    Soilallow 5 ksf q 500 psf surcharge

    w 62.4 pcf

    m 115 pcf

    m 32 deg

    s 125 pcf

    s 30 deg

    SF 1.5 Thisisthesafetyfactorrequiredforbothslidingandoverturning.

    Note:Themoistsoilpropertiesarealsousedforthetoesoil.

    A10.2

  • WallReinforcingProperties

    dbinside 0.75 in s 8 in spacingofverticalbarsdbhoriz 0.5 in swallhoriz 10 in spacingofhorizontalbarsdboutside 0.5 in

    Numfaces 2 Twofacesofreinforcement

    Asinside = dbinside

    2

    40.442 in

    2 #6barsinterior.

    Asoutside = dboutside

    2

    40.196 in

    2 #4barsexterior

    Ashoriz =2 dbhoriz

    2

    40.393 in

    2 #4barshorizontaleachface

    coverinside 2 in coveroutside 1 in

    Theouterbarsareinthehorizontaldirection.

    A10.3

  • FootingReinforcingProperties

    dbtop 0.75 in stop 8 in

    dbbot 0.75 in sbot 8 in

    dblong 0.5 in slong 16 in

    Astop = dbtop

    2

    40.442 in

    2 #6barsat8"spacingtop

    Asbot = dbbot

    2

    40.442 in

    2 #6barsat8"spacingbot

    Aslong =2 dblong

    2

    40.393 in

    2 #4barsat16"spacinglongitudinaleachface

    covertop 2 in coverbot 3 in

    A10.4

  • Calculations

    ThissectionbreaksdownallofthecalculationsthatoccurwithinRISAFoundationforretainingwalldesign.

    ForceCalculationsForOverturning,SlidingandWallDesign

    LateralEarthPressureCoeficients

    = 0 =m 32 deg =s 30 deg

    Kam =cos (())

    cos (())

    ((cos (())))

    2

    cos m2

    +cos (())

    ((cos (())))2

    cos m2

    0.307

    Kpm =cos (())

    +cos (())

    ((cos (())))

    2

    cos m2

    cos (())

    ((cos (())))2

    cos m2

    3.255

    Kas =cos (())

    cos (())

    ((cos (())))

    2

    cos s2

    +cos (())

    ((cos (())))2

    cos s2

    0.333

    LateralPressureCalculations(Service)

    P1 =Kam q 153.629 psf

    P2 =Kam +q Hwall Hwater m 506.977 psf

    P3 =Kas +q Hwall Hwater m 550 psf

    P4 =++P3 Kas Hwater s w Hwater w 1.05 10

    3 psf

    P5 =++P4 Kas tfoot s w tfoot w 1.175 10

    3 psf

    P6 =++P5 Kas Dkey s w Dkey w 1.299 10

    3 psf

    A10.5

  • P7 =Htoesoil m Kpm 748.555 psf

    P8 = +Htoesoil tfoot m Kpm 1.31 10

    3 psf

    P9 = ++Htoesoil tfoot Dkey m Kpm 1.871 10

    3 psf

    A10.6

  • LateralResultantForceLocationsforOverturning

    H1 = +Hwall tfoot 0.5 8.75 ft

    H2 =++ Hwall Hwater1

    3

    Hwater tfoot 10.833 ft

    H3 = +Hwater tfoot 0.5 3.75 ft

    H4 = +Hwater tfoot1

    3

    2.5 ft

    H5 =1

    3 +Htoesoil tfoot 1.167 ft

    A10.7

  • LateralForceSummationsforOverturning,SlidingandWallDesign

    LF1 =P1 +Hwall tfoot 2.689 kip

    ftThisvaluechangesforall3calculations.

    LF1slide =+LF1 P1 Dkey 2.919 kip

    ft

    LF1wall =P1 Hwall 2.458 kip

    ft

    Thisisthesamevalueforall3calculations.LF2 =

    1

    2 P2 P1 Hwall Hwater 1.767

    kip

    ft

    LF3 = P3 P1 +Hwater tfoot 2.973 kip

    ftThisvaluechangesforall3calculations.

    LF3Slide =+LF3 P3 P1 Dkey 3.567 kip

    ft

    LF3wall = P3 P1 Hwater 2.378 kip

    ft

    LF4 = P5 P31

    2

    +Hwater tfoot 2.342

    kip

    ftThisvaluechangesforall3calculations.

    LF4slide = P6 P31

    2

    ++Hwater tfoot Dkey 3.372

    kip

    ft

    LF4wall = P4 P31

    2

    Hwater 1.499

    kip

    ft

    LF5 =1

    2P8 +Htoesoil tfoot 2.292

    kip

    ftThisvaluechangesforall3calculations.

    LF5slide =1

    2P9 ++Htoesoil tfoot Dkey 4.678

    kip

    ft

    LF5wall =1

    2P7 Htoesoil 0.749

    kip

    ft

    A10.8

  • VerticalForceCalculations(ServiceandStrength)

    w1 =Hwall twall conc 3.6 kip

    ft w1f =DLFactor w1 4.32 kip

    ft

    w2 =tfoot ++twall Ltoe Lheel conc 2.363 kip

    ftw2f =DLFactor w2 2.835

    kip

    ftw3 =Wkey Dkey conc 0.338

    kip

    ft w3f =DLFactor w3 0.405 kip

    ft

    w4 =Ltoe Htoesoil m 0.805 kip

    ftw4f =DLFactor w4 0.966

    kip

    ft

    qtotal =q Lheel 2.75 kip

    ftqtotalf =LLFactor qtotal 4.4

    kip

    ft

    w5 =Lheel Hwall Hwater m 6.325 kip

    ftw5f =DLFactor w5 7.59

    kip

    ft

    w6 =Lheel Hwater s 4.125 kip

    ftw6f =DLFactor w6 4.95

    kip

    ft

    A10.9

  • VerticalForceCentroids

    D1 =+Ltoe twall

    24.25 ft

    D2 =Lfoot

    25.25 ft

    D3 =+Lkey Wkey

    25.25 ft

    D4 =Ltoe

    21.75 ft

    D5 =++Ltoe twall Lheel

    27.75 ft

    D6 =D5 7.75 ft

    StabilityChecks

    OverturningThischeckistakenfromthebaseofthetoeofthefooting.

    MR1 =+++w1 D1 w2 D2 w3 D3 w4 D4 30.884 kip ft

    ft

    MR2 =++w5 D5 +w6 qtotal D6 LF5 H5 104.975 kip ft

    ft

    MR =+MR1 MR2 135.858 kip ft

    ft

    MOT =+++H1 LF1 H2 LF2 H3 LF3 H4 LF4 59.667 kip ft

    ft

    OSF =MR

    MOT2.277

    UCOT =SF

    OSF0.659

    Thisretainingwallpassestheoverturningcheckbecauseithasgreaterthana1.5safetyfactor.

    A10.10

  • Sliding

    Thischeckistakenfromthebottomofkeyelevation.

    FSlide =+++LF1slide LF2 LF3Slide LF4slide 11.625 kip

    ft

    R =++++++w1 w2 w3 w4 w5 w6 qtotal 20.305 kip

    ftTotalverticalforce

    FFriction =R 10.153 kip

    ft

    LF8 =1

    2P9 ++Htoesoil tfoot Dkey 4.678

    kip

    ft

    Theforcesresistingslidingareduetobothfrictionandpassivepressureonthetoesideofthefooting.

    FResist =+FFriction LF5slide 14.831 kip

    ft

    SafetyFactorSliding =FResist

    FSlide1.276

    UCSliding =SF

    SafetyFactorSliding1.176

    Thisretainingwallfailstheslidingcheckbecauseithaslessthana1.5safetyfactor.

    A10.11

  • DesigningtheWallStemThewallstemwaspouredseparatelyfromthefooting.Wherethewallispouredthefootinghasnotbeenintentionallyroughened.Footingdowelsoccuratbothfacesofthewallandareofthesamesizeandspacingasthewallreinforcement.

    =Lwall 10 ft =Hwall 16 ft =twall 1.5 ft

    =Asinside 0.442 in2 #6barsinterior. =coverinside 2 in

    =Asoutside 0.196 in2

    #4barsexterior =coveroutside 1 in=Ashoriz 0.393 in

    2 #4barshorizontaleachface=s 8 in

    =swallhoriz 10 in=Numfaces 2

    Theouterbarsareinthehorizontaldirection.

    AxialandBendingDesign(perfoot)Thesearethecentroidheightsofeachportionofload.

    H1wall =Hwall

    28 ft

    H2wall =+Hwater Hwall Hwater

    39.333 ft

    H3wall =Hwater

    23 ft

    H4wall =Hwater

    32 ft

    H5wall =Htoesoil

    30.667 ft

    A10.12

  • Pu 0 kip

    Mwalls +++LF1wall H1wall LF2 H2wall LF3wall H3wall LF4wall H4wall LF5wall H5wall

    =Mwalls 45.787 kip ft

    ft

    =HLFactor 1.6

    Mwallf =HLFactor Mwalls 73.26 kip ft

    ft

    dcant =twall coverinside dbhoriz dbinside

    215.125 in

    dprime =++coveroutside dbhoriz dboutside

    21.75 in

    awall =Asinside fy

    0.85 fc s0.975 in aprime =

    Asoutside fy

    0.85 fc s0.433 in

    Mnwall

    Asinside fy

    dcant awall

    2

    1212

    s

    =Mnwall 48.501 kip ft

    ftThisisthemomentcapacityinthewallnotconsideringcompressionreinforcement

    A10.13

  • wall 0.9

    Note:Theprogramtakesintoaccountcompressionreinforcementaswell,sotheprogramreportedvalueisalittlelarger(44.029).

    PhiMnwall =wall Mnwall 43.651 kip ft

    ft

    BendingInteraction =Mwallf

    PhiMnwall1.678

    ReinforcementProvidedChecks(forentirewall)

    HorizontalReinforcement

    BarsHoriz1 =Numfaces Hwall

    swallhoriz38.4

    BarsHoriz =roundBarsHoriz1 38 Thetotalnumberofhorizontalbarsinthewall.

    Asprovh =BarsHoriz Ashoriz

    27.461 in

    2 Asprovided(H)

    rhoprovh =Asprovh

    12 Hwall twall1.799 10

    4 RhoProvided(H)

    rhominh .002 Rhomin(H)

    Asminh =rhominh Hwall twall 6.912 in2 Asmin(H)

    InsideFaceVerticalReinforcement

    BarsVertInt1 =Lwall

    s15

    BarsVertInt =roundBarsVertInt1 15 Thetotalnumberofinteriorverticalbarsinthewall.

    Asprovint =BarsVertInt Asinside 6.627 in2 IntAsProvided(V)

    rhoprovint =Asprovint

    Lwall twall 122.557 10

    4 IntrhoProvided(V)

    A10.14

  • OutsideFaceVerticalReinforcement

    BarsVertExt1 =Lwall 12

    s180

    BarsVertExt =roundBarsVertExt1 180 Thetotalnumberofexteriorverticalbarsinthewall.

    Asprovext =BarsVertExt Asoutside 35.343 in2 ExtAsProvided(V)

    rhoprovext =Asprovext

    Lwall twall 120.001 ExtrhoProvided(V)

    TotalVerticalReinforcement

    rhominv .0015 rhomin(V)

    Asminv =rhominv Lwall 12 twall 38.88 in2 Asmin(V)

    ShearDesignConcretecheck:

    Vwallds1 =++LF1wall Hwall dcant

    HwallLF3wall

    Hwater dcant

    Hwater

    LF2 5.91 kip

    ft

    Vwallds2 =P4 P3

    2

    Hwater dcant

    2

    Hwater

    P7

    2 Htoesoil dcant

    2

    Htoesoil0.833

    kip

    ft

    Fortheconcretecheckweareusingtheshearforceatadistancedfromthebase.

    Vwallds =+Vwallds1 Vwallds2 6.743 kip

    ft fc 4000 lbf

    2

    in4

    Vwalldf =HLFactor Vwallds 10.788 kip

    ft v 0.75

    Vc =2 fc dcant 2.2958 10

    4 lbf

    ft

    A10.15

  • PhiVcwall =v Vc 1.7219 10

    4 lbf

    ft

    ShearConcInteraction =Vwalldf

    PhiVcwall0.627

    SteelCheck(shearfriction)Inthisexamplethewallisnotpouredmonolithicallywiththefooting.AllcodereferencesarepertheACI31811.

    Vwallbases =+++LF1wall LF2 LF3wall LF4wall LF5wall 7.353 kip

    ft

    Vwallbasef =HLFactor Vwallbases 11.765 kip

    ft

    HereweareusingtheAsofthewallreinforcing,asthedowelsfromthefoundationmatchthewallr/f.Avf =

    +Asinside Asoutside 12 in

    ft

    s0.957

    in2

    ft

    =fy 60 ksi

    conc 0.6 Thisassumesthatthesurfaceofthefootingwherethewallispouredisnotintentionallyroughened.

    Vn =Avf fy conc 113.056 1

    mkip Equation1125

    PerSection11.6.6fymustbetaken

  • Vn1 =0.2 fc Ac 172.8 kip

    ft

    Vn2 = +480 psi 0.08 fc Ac 172.8 kip

    ftVn4 =0.2 fc Ac 172.8

    kip

    ft

    Vn3 =1600 psi Ac 345.6 kip

    ftVn5 =800 psi Ac 172.8

    kip

    ft

    Vnrough min ,,,Vn Vn1 Vn2 Vn3 Vnsmooth =min ,,Vn Vn4 Vn5 34.459 kip

    ft

    SteelConcInteraction =Vwallbasef

    v Vnsmooth0.455

    DesigningtheFooting

    SoilPressureCalculation(forFootingDesign)

    MOTS =HLFactor +++H1 LF1 H2 LF2 H3 LF3 H4 LF4 95.467 kip ft

    ft

    MRS1 DLFactor +++++w1 D1 w2 D2 w3 D3 w4 D4 w5 D5 w6 D6

    MRS2 +LLFactor qtotal D6 HLFactor LF5 H5

    MRS =+MRS1 MRS2 172.625 kip ft

    ft

    RS =+DLFactor +++++w1 w2 w3 w4 w5 w6 LLFactor qtotal 25.466 kip

    ft

    xRS =MRS MOTS

    RS3.03 ft

    e1S =Lfoot

    2xRS 2.22 ft =

    Lfoot

    61.75 ft

    LbasesoilS =3 xRS 9.09 ft

    A10.17

  • qmaxS =if

    else

  • DesignoftheHeel(Shear)

    =covertop 2 in

    =dbtop 0.75 in

    =dblong 0.5 in

    =stop 8 in

    =slong 16 in

    =Astop 0.442 in2

    =Aslong 0.393 in2

    dheel =tfoot covertop dbtop

    215.625 in

    Becausethefootingwilltendtoshearoffasshownhere,theshearcheckshouldoccuratthefaceofwall.

    A10.19

  • =qtotalf 4.4 kip

    ft

    =w5f 7.59 kip

    ft

    =w6f 4.95 kip

    ft

    Vuheel1 =+++w5f w6f qtotalf DLFactor conc tfoot Lheel 18.425 kip

    ft

    LsoilheelS =LbasesoilS Ltoe twall 4.09 ft

    qmaxheelS =qmaxS LbasesoilS Ltoe twall

    LbasesoilS2.521 ksf

    Vuheel2 =1

    2LsoilheelS qmaxheelS 5.155

    kip

    ft

    Vuheel =Vuheel1 Vuheel2 13.27 kip

    ft

    Vuheel1isthetotaldownwardshearforceontheheel.Vuheel2isthetotalupwardshearforceontheheel.Becausethenetforceisdownward,thelocationoftheshearingisconirmed.

    A10.20

  • fc 4000 lbf

    2

    in4

    Vcheel =2 fc dheel 23.717 kip

    ft

    PhiVcheel =v Vcheel 17.788 kip

    ft

    ShearheelInteraction =Vuheel

    PhiVcheel0.746

    DesignoftheHeel(Moment)

    Muheel =Vuheel1 Lheel

    2Vuheel2

    1

    3LsoilheelS 43.642

    kip ft

    ftfc 4 ksi

    aheel =

    12 in

    stopAstop fy

    0.85 12 in fc0.975 in Astop1 =

    Astop

    1 ft0.442

    in2

    ft

    Thereinforcementspacingisat8"oc,sothemomentcapacityisnormalizedtobeperfoot.

    Mnheel =12 in

    stopAstop1 fy

    dheel aheel

    2

    50.157 kip ft

    ft

    =wall 0.9

    PhiMnheel =wall Mnheel 45.142 kip ft

    ft

    BendheelInteraction =Muheel

    PhiMnheel0.967

    A10.21

  • DesignoftheToe(Shear)=coverbot 3 in

    =dbbot 0.75 in

    =dblong 0.5 in

    =sbot 8 in

    =slong 16 in

    =Asbot 0.442 in2

    =Aslong 0.393 in2

    dtoe =tfoot coverbot dbbot

    214.625 in

    Becausethefootingwilltendtoshearoffasshownabove,theshearcheckshouldoccuratadistancedfromthefaceofwall.

    A10.22

  • qtoedS = +LbasesoilS Ltoe dtoe qmaxS

    LbasesoilS4.197 ksf

    VutoeOT = Ltoe dtoe

    +qtoedS 1

    2 qmaxS qtoedS

    11.179 kip

    ft

    VutoeR = +w4f DLFactor conc tfoot Ltoe

    Ltoe dtoe

    Ltoe

    1.246 kip

    ft

    fc 4000 lbf

    2

    in4

    Vutoe =VutoeOT VutoeR 9.933 kip

    ftVctoe =2 fc dtoe 22.199

    kip

    ft

    PhiVctoe =v Vctoe 16.649 kip

    ft

    SheartoeInteraction =Vutoe

    PhiVctoe0.597

    A10.23

  • DesignoftheToe(Moment)

    qtoefaceS = LbasesoilS Ltoe qmaxS

    LbasesoilS3.446 ksf

    MutoeOS +Ltoe qtoefaceSLtoe

    2

    1

    2Ltoe qmaxS qtoefaceS

    2

    3Ltoe

    =MutoeOS 29.916 kip ft

    ft

    VutoeRbend =VutoeR Ltoe

    Ltoe dtoe1.911

    kip

    ft

    MutoeR =VutoeRbendLtoe

    2

    3.344 kip ft

    ft

    Mutoe =MutoeOS MutoeR 26.571 kip ft

    ftfc 4 ksi

    Asbot1 =Asbot

    1 ft0.442

    in2

    ftatoe =

    12 in

    sbotAsbot fy

    0.85 12 in fc0.975 in

    Mntoe =12 in

    sbotAsbot1 fy

    dtoe atoe

    2

    46.844 kip ft

    ft=wall 0.9

    PhiMntoe =wall Mntoe 42.16 kip ft

    ft

    BendtoeInteraction =Mutoe

    PhiMntoe0.63

    A10.24

  • AppendixA11PileCapDesignCalculations_______________________________________________________________________________Inthisexamplewehaveapilecapwith12HP14x102pilesprovidingsupport.Thepileshavean85kipcompressioncapacity,a12kiptensioncapacityanda14kipshearcapacity.Thepilecapis42"thickwitha6"pileembedmentandmadefrom4ksilightweightconcrete.Aloadcombinationof1.0*DL+1.0*LLis usedfortheserviceLCandaloadcombinationof1.2*DL+1.6*LLisusedforthestrengthLC.

    Geometry,MaterialsandCriteriaLcap 183 in Wcap 134 in tcap 42 in embed 6 in

    fy 60 ksi fc 4 ksi 0.75 conc .11 kip

    ft3

    Hped 24 in N 12 NumberofPilesLped 24 in dpile 14 in SideDimensionofPileWped 24 in dbar 0.75 in Diameterofreinforcement

    lx 49 in Distancefromc/lofpedestaltoc/lofpilesinthexdirection.l1z 24.5 in Distancefromc/lofpedestaltoc/lof1stpilesinthezdirection.l2z 73.5 in Distancefromc/lofpedestaltoc/lof2ndpilesinthezdirection.

    wx =lx Wped

    237 in Distancefrompilescentroidtofaceofpedestalinx

    direction.

    w1z =l1z Wped

    212.5 in Distancefrom1stpilescentroidtofaceofpedestalin

    zdirection.

    A11.1

  • Distancefrom2ndpilescentroidtofaceofpedestalinzdirection.w2z =l2z

    Wped

    261.5 in

    EffectiveDepthCalculations(forbending)

    c 1.5 in Cover(topandbottom)

    d =tcap embed c dbar 33.75 in Distancefromthetopofcaptocentroidofbottomreinforcementdtop =tcap embed 36 in Distancefromthetopofcaptothetopofthepiles

    AppliedLoadsPd 250 kip Vx 20 kip

    Pl 350 kip Vz 40 kip

    Mx =Vz

    +Hped tcap

    2

    150 kip ft Mz =Vx

    +Hped tcap

    2

    75 kip ft

    wped =Hped Lped Wped conc 0.88 kip wcap =Lcap Wcap tcap conc 65.562 kip

    Ptot =+++Pd Pl wped wcap 666.442 kip

    PileForces(Service)WewillassumetheindividualpileforcesarecorrectandusetheRISAFoundationoutput.

    Ppile1 54.1593 kip Ppile8 59.2103 kipPu1 76.1068 kip Pu8 84.1884 kip

    Ppile2 56.6083 kip Ppile9 49.5675 kipPu2 80.0252 kip Pu9 68.7599 kip

    Ppile3 59.0573 kip Ppile10 52.0164 kipPu3 83.9435 kip Pu10 72.6782 kip

    Ppile4 61.5062 kip Ppile11 54.4654 kipPu4 87.8619 kip Pu11 76.5966 kip

    Ppile5 51.8634 kip Ppile12 56.9144 kipPu5 72.4333 kip Pu12 80.515 kip

    Ppile6 54.3124 kip Pu6 76.3517 kip

    Ppile7 56.7613 kip Pu7 80.2701 kip

    A11.2

  • PileCapFlexuralDesignForthelexuraldesignwearesimplytakingtheworstcasemomentateitherfaceofthepedestalandcheckingagainstthat.TodothisIsimplycomparethepileforcesforeachsideofthepedesalandtaketheworstcaseforces.

    wucapresistx =1.2

    Wcap Wped

    2

    Lcap tcap conc 32.292 kip

    wucapresistz =1.2

    Lcap Lped

    2

    Wcap tcap conc 34.178 kip

    Mux + ++Pu3 Pu7 Pu11 w1z ++Pu4 Pu8 Pu12 w2z wucapresistx Lcap Lped

    4

    =Mux 1.438 10

    3 kip ft

    Muz = +++Pu1 Pu2 Pu3 Pu4 wx wucapresistz Wcap Wped

    4932.815 kip ft

    Herearethecalculationsforminimumsteelforbothtemperatureandshrinkageandlexure.

    Asminx =.0018 Lcap tcap 13.835 in2

    Asminz =.0018 Wcap tcap 10.13 in2

    Asflexxbot =

    200 lbf

    in2

    Lcap d

    fy20.588 in

    2Asflexzbot =

    200 lbf

    in2

    Wcap d

    fy15.075 in

    2

    Asreqdxbot 6.226 in2

    ValuesgivenintheprogramAsprovxbot 12.812 in

    2

    ax =Asprovxbot fy

    0.85 Lcap fc1.235 in

    PhiMnx =0.9 Asprovxbot fyd ax

    2

    1.91 103 kip ft

    UCMx =Mux

    PhiMnx0.753

    Asreqdzbot 9.609 in2

    ValuesgivenintheprogramAsprovzbot 14.137 in

    2

    A11.3

  • az =Asprovzbot fy

    0.85 Wcap fc1.862 in

    PhiMnz =0.9 Asprovzbot fyd az

    2

    2.088 103 kip ft

    UCMz =Muz

    PhiMnx0.488

    InthexdirectiontheAsreqd(andeven4/3Asreqd)islessthantheminimumtemperatureandshrinkagesteel,theprogramusesthatminimum.Inthezdirectionthe4/3*Asreq'disgreaterthantheAsS&T,thusweuse9.609*4/3=12.812in^2.

    PedestalPunchingShearCheck

    db =d 33.75 in Effectivedepthofslabforpedestalpunching.

    L1 =+Wped db 57.75 in

    Sidedimensionsfortheshearperimeter.L2 =+Lped db 57.75 in

    Pupileped =+++++++++Pu1 Pu2 Pu3 Pu4 Pu5 Pu8 Pu9 Pu10 Pu11 Pu12 783.109 kip

    Thisvaluerepresentsthesumofthefactoredaxialforcesinpilesoutsideofthepedestalpunchingshearperimeter.

    wucapped =1.2 Wcap Lcap L1 L2 tcap conc 67.975 kip

    Thisistheselfweightofthepilecapthatisoutsideofthepedestalpunchingshearperimeter.

    Pupunch =Pupileped wucapped 715.134 kip

    Muxped =1.6 Mx 240 kip ft

    Forceinthepedestal.Muzped =1.6 Mz 120 kip ft

    bo =2 +L1 L2 231 in Punchingshearperimeter.

    c1 =L1

    228.875 in Thisisthedistancefromcentroidtoextremeiber.

    A11.4

  • Ac =bo db 7.796 10

    3 in2 Acistheperimeterareaoftheshearcone.

    Jc =++db +Wped db

    3

    6

    +Wped db db3

    6

    db +Lped db +Lped db2

    2

    4.704 106 in

    4

    Jcisthepolarmomentofinertiaandthisequationcanbefoundinthecommentarytosection11.11.7.2oftheACI31811.

    0.4

    umax =++Pupunch

    Ac

    Muxped c1

    Jc

    Muzped c1

    Jc0.102 ksi

    Thisisthecriticalpunchingshearstress,combiningtheaxialandmomentforcestransmittedthroughthepedestal.Punchingequationscanbefoundinthecommentarytosection11.11.7.2oftheACI31811.Notethatherewearecombiningthestressesduetothemomentstogettheworstcasestressatacornerofthepedestalpunchingshearperimeter.

    fc 4000 lbf

    2

    in4

    PhiVcpunch =0.75 4 fc bo db 1.479 10

    3 kip

    PhiVny = PhiVcpunch

    bo db0.142 ksi

    Punchcodecheck =umax

    PhiVny0.719

    PilePunchingShearCheckHerewewilldoapunchingshearcheckforpile4,theworstcaseone.TheprogramlooksateachpileandcalculatesapunchingshearperimeterforInterior,EdgeandCornerscenariosandchoosesthesmallestvalueforthecheck.Forroundpiles,wecalculateanequivalentsquaredimensionsuchthattheperimeterofbothareequal.

    =dpile 14 in dtoppunch =tcap embed 36 in

    Lpile =++11 in dpile dtoppunch

    243 in

    Becausethereisnotopreinforcementinthepilecap,theslabisconsideredunreinforcedforpilepunching.BecauseofthisourPhifactorisnow0.55andweessentiallytake2/3oftheoriginalstrength(thus4goesto8/3).Theratioof2/3*(0.55/0.75)is0.4888.Intheprogramweuseablanket50%reduction.

    A11.5

  • 0.55 bo1 =2 Lpile 86 in

    PhiVcpunch = 8

    3fc bo1 dtop 215.389 kip Ifweweretocalculateitexactly.

    PhiVcpunch2 =0.75 4 fc bo1 dtop

    2220.284 kip Thisisthevaluethe

    programreports.

    =Pu4 87.862 kip

    Puratio =Pu4

    PhiVcpunch20.399

    OneWayShearCheck

    =w1z 12.5 in =wx 37 in

    =d 33.75 in =d 33.75 in

    Becauseinthexdirectionw>d,thecriticallocationisatadistancedfromthepedestal.Thismeansthatweneedtocalculatetheweightofthepilecapresistingtheshearatthislocation.

    wucapresistxshear =

    Wcap Wped

    2dLcap tcap conc 10.397 kip

    Vux =+++Pu1 Pu2 Pu3 Pu4 wucapresistxshear 317.54 kip

    Becauseinthezdirectionwwz,thereforethecriticallocationforshearatthefaceofthepedestal.

    Asprovidedz 12.8122 in2

    Asprovidedx =Asminx 13.835 in2

    provz =Asprovidedz

    Wcap d0.002833 provx =

    Asprovidedx

    Lcap d0.00224

    A11.6

  • Shearstrengthinthexdirectiondz >wz,thereforethecriticallocationforshearatthefaceofthepedestalandCRSIDesignHandbookequation132onP.1326isused.

    =Mux

    Vuz d1.114 Mu/Vu*dmustbelessthanorequalto1.0,souse1.0.

    MVratio 1

    cx =d

    w1z

    (( 3.5 2.5 MVratio))

    +1.9 fc 2500 lbf

    in2

    provz MVratio

    262.46 psi

    cmax =10 fc 632.456 psi

    Vc_x =cx Wcap d 1.187 10

    3 kip

    Shearstrengthinthezdirectiondz


Recommended