Foundation Verification Problems

RISAFoundationRapidInteractiveStructuralAnalysisFoundationsVerificationProblems

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TableofContents

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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

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

VerificationVersionThisdocumentcontainsproblemsthathavebeenverifiedinRISAFoundationversion5.0.2.

VerificationProblem1:StripFootingDesign

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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

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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

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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.


Solvethemodelandlookatthedetailreportforthefooting.Notethatthetextusesthenetsoilbearingtocalculatethesizeoffooting.ThissizeisuseddirectlyinRISAFoundationandthusthesoiloverburdenandselfweightaresettozero.

VerificationProblem2:SquareSpreadFooting#1

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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

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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

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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

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

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


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


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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.


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Figure4.3

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

Figure4.4


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ConclusionInthisexampleitisshownthattheRISAFoundationcalculationsreasonablymatchthetextbookdesignexamples.

VerificationProblem5:EccentricallyLoadedFooting

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

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


VerificationProblem5:EccentricallyLoadedFooting

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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

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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.



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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.


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

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



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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

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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


Notethattheselfweightandoverburdenwereinputaszeroandtheallowablesoilpressurewasaddeddirectlyas4.50ksf.

VerificationProblem8:RectangularFooting#2

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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

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

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


Notes:

Becausetheexampledoesnotusetheselfweightofthefootinginthecalculationandinsteadjustgivesanaverageweightbetweenthesoilandconcrete,thedensityof

VerificationProblem9:SquareFooting#2

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concretehasbeensetto0.TheOverburdenhasalsobeensettozero.Thus,theallowablesoilpressureissimplyaddeddirectlyas3.75ksf.

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


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.


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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.


Note:Theretainingwalliscantileveredandthebaseisnotrestrainedagainstsliding.


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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

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

Description/ProblemStatementInthisexampleRISAFoundationsvaluesarecomparedtothevaluesobtainedfromahandcalculationdoneforallaspectsofthepilecap.ThishandcalculationislocatedinAppendixA11.


VerificationProblem11:PileCapDesignExample

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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


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


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


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


kip

ft

qtotal =q Lheel 2.75 kip

ftqtotalf =LLFactor qtotal 4.4

kip

ft

w5 =Lheel Hwall Hwater m 6.325 kip


kip

ft

w6 =Lheel Hwater s 4.125 kip


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





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

Date post:	05-Jan-2016
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