foundation verification problems
DESCRIPTION
risa foundationTRANSCRIPT
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RISAFoundationRapidInteractiveStructuralAnalysisFoundationsVerificationProblems
26632TowneCentreDrive,Suite210FoothillRanch,California92610(949)9515815(949)9515848(FAX)www.risa.com
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Copyright2013byRISATechnologies,LLC.Allrightsreserved.Noportionofthecontentsofthispublicationmaybereproducedortransmittedinanymeanswithouttheexpresswritten
permissionofRISATechnologies,LLC.Wehavedoneourbesttoinsurethatthematerialfoundinthispublicationisbothusefulandaccurate.However,pleasebeawarethaterrorsmayexistinthispublication,andthatRISA
Technologies,LLCmakesnoguaranteesconcerningaccuracyoftheinformationfoundhereorintheusetowhichitmaybeput.
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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
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VerificationOverview
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VerificationOverview VerificationMethodsWeatRISATechnologiesmaintainalibraryofhundredsoftestproblemsusedtovalidatethecomputationalaspectsofRISAprograms.InthisverificationpackagewewillpresentarepresentativesampleofthesetestproblemsforyourreviewandcompareRISAFoundationtotextbookexampleslistedwithineachproblem.TheinputforthesetestproblemswasformulatedtotestRISAFoundationsperformance,notnecessarilytoshowhowcertainstructuresshouldbemodeledandinsomecasestheinputandassumptionsweuseinthetestproblemsmaynotmatchwhatadesignengineerwoulddoinarealworldapplication.Thedataforeachoftheseverificationproblemsisprovided.ThefileswheretheseRISAFoundationproblemsarelocatedisintheC:\RISA\ExamplesdirectoryandtheyarecalledVerificationProblem1.fnd(2,3,etc).
VerificationVersionThisdocumentcontainsproblemsthathavebeenverifiedinRISAFoundationversion5.0.2.
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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.
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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.
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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.
Figure2.1RISAFoundationModelView
Solvethemodelandlookatthedetailreportforthefooting.Notethatthetextusesthenetsoilbearingtocalculatethesizeoffooting.ThissizeisuseddirectlyinRISAFoundationandthusthesoiloverburdenandselfweightaresettozero.
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VerificationProblem2:SquareSpreadFooting#1
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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.
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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.
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VerificationProblem3:RectangularSpreadFoot#1
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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.
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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
Figure4.1RISAFoundationDetailReportView
NotethatRISAFoundationwillnotplacetopsteelreinforcementinapilecapunlessthereistensioninthetopfaceofthepilecap.Forthisreasona1kip*ftmomentwasaddedtotheOL1loadcategory.Thisistoforcetopsteel,asthisaffectsthepilepunchingshearchecks.Ifthereisnoreinforcementinthetopthentheprogramconsidersthecapunreinforcedforpunchingshearcalculations.
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VerificationProblem4:PileCapShear
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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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VerificationProblem4:PileCapShear
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Figure4.3
BecauseofthisthepunchingshearperimetercannotbetakenasshowninthePCAnotes.Insteadyoureallyonlyhaveapartialperimeterbecauseyouwillbreakoutthecornerbeforeyougetallthewayaround.InRISA,includingthesquareperimeteradjustment,itwouldlookasshowninFigure4.4.
Figure4.4
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VerificationProblem4:PileCapShear
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ConclusionInthisexampleitisshownthattheRISAFoundationcalculationsreasonablymatchthetextbookdesignexamples.
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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.
Figure5.1RISAFoundationDetailReportView
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VerificationProblem5:EccentricallyLoadedFooting
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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.
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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.
Figure6.1RISAFoundationDetailReportView
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VerificationProblem6:CantileverRetainingWall#1
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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.
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VerificationProblem7:CantileverRetainingWall#2
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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.
Figure7.1RISAFoundationDetailReportView
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VerificationProblem7:CantileverRetainingWall#2
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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.
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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
Figure8.1RISAFoundationDetailReportView
Notethattheselfweightandoverburdenwereinputaszeroandtheallowablesoilpressurewasaddeddirectlyas4.50ksf.
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VerificationProblem8:RectangularFooting#2
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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.
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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
Figure9.1RISAFoundationDetailReportView
Notes:
Becausetheexampledoesnotusetheselfweightofthefootinginthecalculationandinsteadjustgivesanaverageweightbetweenthesoilandconcrete,thedensityof
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VerificationProblem9:SquareFooting#2
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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.
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VerificationProblem10:CantileverRetainingWall#3
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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.
Figure10.1RISAFoundationDetailReportView
Note:Theretainingwalliscantileveredandthebaseisnotrestrainedagainstsliding.
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VerificationProblem10:CantileverRetainingWall#3
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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.
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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
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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.
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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