Solution Manual for Concrete Structures, HAR/CDR Edition

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SolutionsManualtoaccompanyConcreteStructuresMehdiSetarehRobertDarvas

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CONCRETESTRUCTURESSolutionsManualTableofContentsChapterOne:ReinforcedConcreteTechnology1ChapterTwo:RectangularBeamsandOne-WaySlabs6ChapterThree:SpecialTopicsinFlexure55ChapterFour:ShearinReinforcedConcreteBeams80ChapterFive:Columns114ChapterSix:FloorSystems150ChapterSeven:Foundations,andEarthSupportingWalls158ChapterEight:OverviewofPrestressedConcrete232ChapterNine:MetricSysteminReinforcedConcreteDesignandConstruction2361Study)

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Problem1-1Hydrationisachemicalreactionthatstartswhenwaterisaddedtocement.Ithasthreestages:setting,hardening,andstrengthdevelopment.Thehydrationprocessgeneratesheat.Italsocontinuesthroughoutthelifeofconcretestructuresaslongasthereisfreemoistureavailable.Problem1-2Sinceconcreteisaconstructionmaterialwhichisverystrongincompressionbutweakintension,itisimportantforthestructuraldesignertoknowwhatthecompressioncapacityofconcreteis.Thecompressivestrengthofconcreteismeasuredbyconductinga“cylindertest”.Inthistest,compressionforceisappliedgraduallyonastandard67x12”concretecylinder.Thestressandstrainofthespecimenismeasuredandplotted.Themaximumcompressivestrengthisnotedasf.Problem1-3Theair-entrainingadmixturesarcaddedtoconcretetoincreasetheconcreteresistanceagainstthefreezing/thawingcycles.Asaresult,thisadmixtureimprovesconcretedurability.

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Problem1-4Themodulusofelasticityrelatesthestraintothestressinconcrete.ItcanbedeterminedasaresultofthecylindertestortheuseoftheACIapproximateequation.Problem1-5Themodulusofruptureisthetensilestrengthofconcreteinbending.Problem1-6Deformedbarsareusuallyusedastheprimarysteelreinforcementofstructuralelements.Thebarshaveprotrusionsonthesurfacetoincreasetheirbondagetoconcrete.Weldedwirereinforcementsarethinwiresspacedatcertaindistancesintwoorthogonaldirectionsandfabricatedinlargesheetsorlongrolls.Theweldedwirereinforcementsareusuallyusedwherelargeareasneedtobereinforcedsuchasfloorslabsandwalls.Study

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Problem1-7m2x2ightofbeam=—————-=0.22k/fiweightof7.000ftwy=1.0+0.22=1.22k/fiRol2210)_22My,="issp-k8w=1.22]R=61kR=6.1k2:0"61k}|vo|®||{-6.1kwee©Mowe=15.3fkSketchforProblem1-7Problem1-8w,=145pcff!=3500psiF,=57,000)7;=310003300_53554;1,000J,=7.5y;=7.5v3,500=444psiStudy

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Problem1-9w,=120pcfJ.=3000psiPw=30Ib/ft6-0"w=12S=30Ib/fi121222My,=PPLOPO)yasspopis8484UsingEquation(1-3)forsand-lightweightconcrete:f=fy=0.857.517)=0.85(7.54/3,000)=349psi22S,=YA_SO)sim66Mg=f.S,Mo<My135+1.5P<(sso)12135+1.5P£1,047P<6081bthenP,,,=6081bStudy

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Problem1-10f!=4,000psiw=2.1k/ftAbeamweight=150)8.12=100/b/fi©12712w=2+19_ousp1,0002M,,=we=0.26302f,=7.5f7=7.5{4,000=474psi22S,=bd”302)joi66474(192)My=fS,=——"2=75%-fi=12,000%Mx<M,0.263%"£7.59A<537'Study

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Problem2-1f.=4,000psi1,=60,000psib=14"A,=4#9=4.00in”UsingMethodI:(a)d=28”Mg=?FromFigure2-39,useMethodI:Step1.pte20046100bd14x28Poin=0.0033(fromTable42-4)Pas=0.0207(fromTableA2-3)(Pin=0.0033)<(p=0.0102)<(Dg,=0.0207)OK.Step2.aAsfy__400x60,000oo,0.85/,b 0.85x4,000x14Step3.e=230oyB085Step4.La393=0212<3=0.375—sectionisintension-controlledzoned288$=090Step5.a5.04.Mp=¢Af,(¢-a)-090x400x6028SS)5503.7k—in=458.6k-ftStudy

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(b)d=32Mpg=?Repeatsamestepsasinpart(a)a5.04.,My~gap(1-2)090x400xen{32-2246,367.7k—in=530.6k-ft(c)d=36”Mg=?Repeatsamestepsasinpart(a)a5.04NMp=44f,(+-9)-090x400x36-2047,231.7k~in=602.6k—ft(d)d=40Mg=?Repeatsamestepsasinpart(a)a5.04.My=¢4f,(a-2)-0.90a0x60n{0-224)8,095.7k~in=674.6k-ftComparisons:Alinearincreasein“d”leadstoaproportionatelinearincreaseofMg%increaseof%increaseCase“d”overMg(k-ft)overCase(a)Case(a)Study

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UsingMethodII:(a)d=28"Mg=?FromFigure2-40,useMethodII:Step1.pte400=0.0102bd14x28Poin=0.0033(fromTableA2-4)Prax=0.0207(fromTableA2-3)(Prin=0.0033)<(p=0.0102)<(£4,=0.0207)OK.Step2.p=0.0102fy=60,000psi—>TableA2~6b—R=501psifi=4,000psiStep3.22yp,=BER_A9O9CO)_yo12,00012,000®d=32"Mg=?Step1.p=A_4000.00893bd14x320.0033<0.00893<0.0207.".okStep2.p=0.0089Jy=60,000psi—>TableA2—6b—>R=443psif{=4,000psiStep3.Study

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22pyPER_(14)C2(48)012,00012,000(c)d=36"Mg=?Step1.p=20060079bd14x360.0033<0.0079<0.0207.".okStep2.p=0.0079—TableA2-6b—R=397psiStep3.22My=LER(430)(397)_0095-412,00012,000(dd=40Mg=?Step1.pt20050mbd14x400.0033<0.0071<0.0207.".okStep2.p=0.0071—TableA26b—R=359psiStep3.22My=bd"R_(14)(40)(399)_670.1%f12,00012,000Study

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Comparisons:)%increaseof.%increaseCase“d”overMg(k-ft)overCase(a)Case(a)Study

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Problem2-2fl=4,000psif=60,000psid=36"A,=4#9=4.00in"@b=147Mg=?UseMethodI.(FlowchartonFigure2-39)Step1.pds4000.0079bd14x36Poin=0.0033(fromTableA2-4)Pua=0.0207(fromTableA2-3)(Pin=0.0033)<(p=0.0079)<(p,q=0.0207)OK.Step2.Aa=fy__400x60,0000,0.85f,b0.85x4,000x14Step3.=a504ouA085Step4.e3950.1653=0.375—sectionisintension—controlledzoned368¢=0.90’Step5.a5.04.M,=¢Afd-5)=0.90x4.00x60x]36-=-=7,231.7k—in=602.6k—ftStudy

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(b)b=16"Mg=?Repeatingsamestepsasinpart(a):goA00x60,000_,0.85x4,000x16My=¢4f,(4-4)-090xs00036+41)7,300k—in=6083k-fi(c)b=18"Mg=?Repeatingsamestepsasinpart(a):a=4.00x60,000~3.00"0.85x4,000x18Mp=¢4,f,(4-2)090x40036-222).7,352.6k—in=612.7k-fi(db=20"Mpg=?Repeatingsamestepsasinpart(a):oo400x60000oo,0.85x4,000x20Me=¢Af,(+-9)-0.90x4eso36-252).7,394.8kin=6162k-JfiComparisons:Duetotheincreasedwidth,thedepthoftheequivalentstressblockbecomesslightlyless,leadingtoaslightincreaseinMg%increaseof.%increaseCaseb(in)“b”overMg(k-ft)overCase(a)Case(a)Study

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UsingMethodII:(@b=14"Mg=?FromFigure2-40,useMethodII:Step1.pt4000079bd14x36(Pin=0.0033)<(p=0.0079)<(p,,=0.0207)..OK.Step2.0=0.0079fy=60,000psi—TableA2—6b—R=397psiJ.=4,000psiStep3.22My=bd"R_(14)(36)C9)_600.3k-fi12,00012,000(b)b=16"Mg="?Step1.p=2040069bd16x360.0033<0.0069<0.0207.".okStep2.0=0.0069fy=60,000psi—TableA2—6b—R=350psi12=4,000psiStudy

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