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Staticsandstrengthofmaterialsforarchitectureandbuildingconstructionpdf
Tableofcontents:Cover......Page1TableofContents......Page41.Introduction......Page62.Statics......Page203.AnalysisofSelectedDeterminateStructuralSystems......Page1034.LoadTracing......Page2045.StrengthofMaterials......Page2626.Cross-SectionalPropertiesofStructuralMembers......Page3137.BendingandShearinSimpleBeams......Page3478.BendingandShearStressesinBeams......Page3809.ColumnAnalysisandDesign......Page45510.StructuralConnections......Page51311.Structure,Construction,andArchitecture......Page55912.DefinitionofTerms......Page586B......Page588C......Page589F......Page590M......Page591S......Page592W......Page593Z......Page594CitationpreviewStaticsandStrengthofMaterialsOnouyeKaneFourthEditionISBN978-1-29202-707-49781292027074StaticsandStrengthofMaterialsforArchitectureandBuildingConstructionBarryS.OnouyeKevinKaneFourthEditionPearsonNewInternationalEditionStaticsandStrengthofMaterialsforArchitectureandBuildingConstructionBarryS.OnouyeKevinKaneFourthEditionPearsonEducationLimitedEdinburghGateHarlowEssexCM202JEEnglandandAssociatedCompaniesthroughouttheworldVisitusontheWorldWideWebat:www.pearsoned.co.uk©PearsonEducationLimited2014Allrightsreserved.Nopartofthispublicationmaybereproduced,storedinaretrievalsystem,ortransmittedinanyformorbyanymeans,electronic,mechanical,photocopying,recordingorotherwise,withouteitherthepriorwrittenpermissionofthepublisheroralicencepermittingrestrictedcopyingintheUnitedKingdomissuedbytheCopyrightLicensingAgencyLtd,SaffronHouse,6–10KirbyStreet,LondonEC1N8TS.Alltrademarksusedhereinarethepropertyoftheirrespectiveowners.Theuseofanytrademarkinthistextdoesnotvestintheauthororpublisheranytrademarkownershiprightsinsuchtrademarks,nordoestheuseofsuchtrademarksimplyanyaffiliationwithorendorsementofthisbookbysuchowners.ISBN10:1-292-02707-XISBN10:1-269-37450-8ISBN13:978-1-292-02707-4ISBN13:978-1-269-37450-7BritishLibraryCataloguing-in-PublicationDataAcataloguerecordforthisbookisavailablefromtheBritishLibraryPrintedintheUnitedStatesofAmericaPEARSONCUSTOMLIBRARYTableofContents1.IntroductionBarryS.Onouye/KevinKane12.StaticsBarryS.Onouye/KevinKane153.AnalysisofSelectedDeterminateStructuralSystemsBarryS.Onouye/KevinKane974.LoadTracingBarryS.Onouye/KevinKane1995.StrengthofMaterialsBarryS.Onouye/KevinKane2576.Cross-SectionalPropertiesofStructuralMembersBarryS.Onouye/KevinKane3077.BendingandShearinSimpleBeamsBarryS.Onouye/KevinKane3418.BendingandShearStressesinBeamsBarryS.Onouye/KevinKane3759.ColumnAnalysisandDesignBarryS.Onouye/KevinKane44910.StructuralConnectionsBarryS.Onouye/KevinKane50711.Structure,Construction,andArchitectureBarryS.Onouye/KevinKane55312.DefinitionofTermsBarryS.Onouye/KevinKane581Index583IIIIntroduction1DEFINITIONOFSTRUCTUREStructureisdefinedassomethingmadeupofinterdependentpartsinadefinitepatternoforganization(Figures1and2)—aninterrelationofpartsasdeterminedbythegeneralcharacterofthewhole.Structure,particularlyinthenaturalworld,isawayofachievingthemoststrengthfromtheleastmaterialthroughthemostappropriatearrangementofelementswithinaformsuitableforitsintendeduse.Theprimaryfunctionofabuildingstructureistosupportandredirectloadsandforcessafelytotheground.Buildingstructuresareconstantlywithstandingtheforcesofwind,theeffectsofgravity,vibrations,andsometimesevenearthquakes.Figure1Radial,spiralpatternofthespiderweb.Thesubjectofstructureisall-encompassing;everythinghasitsownuniqueform.Acloud,aseashell,atree,agrainofsand,thehumanbody—eachisamiracleofstructuraldesign.Buildings,likeanyotherphysicalentity,requirestructuralframeworkstomaintaintheirexistenceinarecognizablephysicalform.Tostructurealsomeanstobuild—tomakeuseofsolidmaterials(timber,masonry,steel,concrete)insuchawayastoassembleaninterconnectedwholethatcreatesspacesuitabletoaparticularfunctionorfunctionsandtoprotecttheinternalspacefromundesirableexternalelements.Astructure,whetherlargeorsmall,mustbestableanddurable,mustsatisfytheintendedfunction(s)forwhichitwasbuilt,andmustachieveaneconomyorefficiency—thatis,maximumresultswithminimummeans(Figure3).AsstatedinSirIsaacNewton’sPrincipia:Figure2Bowandlatticestructureofthecurrach,anIrishworkboat.Stressesonthehullareevenlydistributedthroughthelongitudinalstringers,whichareheldtogetherbysteam-bentoakribs.Naturedoesnothinginvain,andmoreisinvainwhenlesswillserve;forNatureispleasedwithsimplicity,andaffectsnotthepompofsuperfluouscauses.Figure3MetacarpalbonefromavulturewingandanopenwebsteeltrusswithwebmembersintheconfigurationofaWarrenTruss.FromChapter1ofStaticsandStrengthofMaterialsforArchitectureandBuildingConstruction,FourthEdition,BarryOnouye,KevinKane.Copyright©2012byPearsonEducation,Inc.PublishedbyPearsonPrenticeHall.Allrightsreserved.1Introduction2STRUCTURALDESIGNStructuraldesignisessentiallyaprocessthatinvolvesbalancingbetweenappliedforcesandthematerialsthatresisttheseforces.Structurally,abuildingmustnevercollapseundertheactionofassumedloads,whatevertheymaybe.Furthermore,tolerabledeformationofthestructureoritselementsshouldnotcausematerialdistressorpsychologicalharm.Goodstructuraldesignismorerelatedtocorrectintuitivesensethantosetsofcomplexmathematicalequations.Mathematicsshouldbemerelyaconvenientandvalidatingtoolbywhichthedesignerdeterminesthephysicalsizesandproportionsoftheelementstobeusedintheintendedstructure.Thegeneralprocedureofdesigningastructuralsystem(calledstructuralplanning)consistsofthefollowingphases:■Figure4EiffelTower.■■■■■■Conceivingofthebasicstructuralform.Devisingthegravityandlateralforceresistingstrategy.Roughlyproportioningthecomponentparts.Developingafoundationscheme.Determiningthestructuralmaterialstobeused.Detailedproportioningofthecomponentparts.Devisingaconstructionmethodology.Afteralloftheseparatephaseshavebeenexaminedandmodifiedinaniterativemanner,thestructuralelementswithinthesystemarethencheckedmathematicallybythestructuralconsultanttoensurethesafetyandeconomyofthestructure.Theprocessofconceivingandvisualizingastructureistrulyanart.Figure5NaveofReimsCathedral(constructionbegunin1211).Therearenosetsofrulesonecanfollowinalinearmannertoachieveaso-called“gooddesign.”Theiterativeapproachismostoftenemployedtoarriveatadesignsolution.Nowadays,withthedesignofanylargestructureinvolvingateamofdesignersworkingjointlywithspecialistsandconsultants,thearchitectisrequiredtofunctionasacoordinatorandstillmaintainaleadershiproleevenintheinitialstructuralscheme.Thearchitectneedstohaveabroadgeneralunderstandingofthestructurewithitsvariousproblemsandasufficientunderstandingofthefundamentalprinciplesofstructuralbehaviortoprovideusefulapproximationsofmembersizes.Thestructuralprinciplesinfluencetheformofthebuilding,andalogicalsolution(oftenaneconomicaloneaswell)isalwaysbasedonacorrectinterpretationoftheseprinciples.Aresponsibilityofthebuilder(constructor)istohavetheknowledge,experience,andinventivenesstoresolvecomplexstructuralandconstructionalissueswithoutlosingsightofthespiritofthedesign.Astructureneednotactuallycollapsetobelackinginintegrity.Forexample,astructureindiscriminatelyemployinginappropriatematerialsoranunsuitablesizeandproportionofelementswouldreflectdisorganizationanda2Introductionsenseofchaos.Similarly,astructurecarelesslyoverdesignedwouldlacktruthfulnessandreflectawastefulnessthatseemshighlyquestionableinourcurrentworldsituationofrapidlydiminishingresources.Itcanbesaidthatintheseworks(GothicCathedrals,EiffelTower,FirthofForthBridge),forerunnersofthegreatarchitectureoftomorrow,therelationshipbetweentechnologyandaestheticsthatwefoundinthegreatbuildingsofthepasthasremainedintact.Itseemstomethatthisrelationshipcanbedefinedinthefollowingmanner:theobjectivedataoftheproblem,technologyandstatics(empiricalorscientific),suggestthesolutionsandforms;theaestheticsensitivityofthedesigner,whounderstandstheintrinsicbeautyandvalidity,welcomesthesuggestionandmodelsit,emphasizesit,proportionsit,inapersonalmannerwhichconstitutestheartisticelementinarchitecture.QuotefromPierLuigiNervi,AestheticsandTechnologyinArchitecture,HarvardUniversityPress;Cambridge,Massachusetts,1966.(SeeFigures4and5.)Figure6Tree—asystemofcantilevers.Figure7Beehive—cellularstructure.Figure8Humanbodyandskeleton.3PARALLELSINNATUREThereisafundamental“rightness”inthestructurallycorrectconcept,leadingtoaneconomyofmeans.Twokindsof“economy”arepresentinbuildings.Onesucheconomyisbasedonexpediency,availabilityofmaterials,cost,andconstructability.Theother“inherent”economyisdictatedbythelawsofnature(Figure6).InhiswonderfulbookOnGrowthandForm,D’ArcyWentworthThompsondescribeshowNature,asaresponsetotheactionofforces,createsagreatdiversityofformsfromaninventoryofbasicprinciples.Thompsonsaysthatinshort,theformofanobjectisadiagramofforces;inthissense,atleast,thatfromitwecanjudgeofordeducetheforcesthatareactingorhaveacteduponit;inthisstrictandparticularsense,itisadiagram.Theformasadiagramisanimportantgoverningideaintheapplicationoftheprincipleofoptimization(maximumoutputforminimumenergy).Natureisawonderfulvenuetoobservethisprinciple,becausesurvivalofaspeciesdependsonit.Anexampleofoptimizationisthehoneycombofthebee(Figure7).Thissystem,anarrangementofhexagonalcells,containsthegreatestamountofhoneywiththeleastamountofbeeswaxandisthestructurethatrequirestheleastenergyforthebeestoconstruct.GalileoGalilei(16thcentury),inhisobservationofanimalsandtrees,postulatedthatgrowthwasmaintainedwithinarelativelytightrange—thatproblemswiththeorganismwouldoccurifitweretoosmallortoolarge.InhisDialoguesConcerningTwoNewSciences,Galileohypothesizesthat3Introductionitwouldbeimpossibletobuildupthebonystructuresofmen,horses,orotheranimalssoastoholdtogetherandperformtheirnormalfunctionsiftheseanimalsweretobeincreasedenormouslyinheight;forthisincreaseinheightcanbeaccomplishedonlybyemployingamaterialwhichisharderandstrongerthanusual,orbyenlargingthesizeofthebones,thuschangingtheirshapeuntiltheformandappearanceoftheanimalssuggestmonstrosity....Ifthesizeofabodybediminished,thestrengthofthatbodyisnotdiminishedinthesameproportion;indeed,thesmallerthebodythegreateritsrelativestrength.Thusasmalldogcouldprobablycarryonitsbacktwoorthreedogsofhisownsize;butIbelievethatahorsecouldnotcarryevenoneofhisownsize.Figure9Flyingstructures—abatandOttoLilienthal’shangglider(1896).Economyinstructuredoesnotjustmeanfrugality.Withouttheeconomyofstructure,neitherabirdnoranairplanecouldfly,fortheirsheerweightwouldcrashthemtoearth.Withouteconomyofmaterials,thedeadweightofabridgecouldnotbesupported.Reductionindeadweightofastructureinnatureinvolvestwofactors.Natureusesmaterialsoffibrouscellularstructure(asinmostplantsandanimals)tocreateincrediblestrength-to-weightratios.Ininertgranularmaterialsuchasaneggshell,itisoftenusedwithmaximumeconomyinrelationtotheforcesthatthestructuremustresist.Also,structuralforms(likeapalmleaf,anautilusshell,orahumanskeleton)aredesignedincrosssectionsothattheminimumofmaterialisusedtodevelopthemaximumresistancetoforces(Figure8).Naturecreatesslowlythroughaprocessoftrialanderror.Livingorganismsrespondtoproblemsandachangingenvironmentthroughadaptationsoveralongperiodoftime.Thosethatdonotrespondappropriatelytotheenvironmentalchangessimplyperish.Figure10Theskeletallatticeworkoftheradiolarian(Aulasyrumtriceros)consistsofhexagonalprismsinasphericalform.Figure11BuckminsterFuller’sUnionTankCardome,a384-ft.-diametergeodesicdome.4Historically,humandevelopmentintheareaofstructuralformshasalsobeenslow(Figure9).Forthemostpart,limitedmaterialsandknowledgerestrictedthedevelopmentofnewstructuralelementsorsystems.Evenwithinthelast150yearsorso,newstructuralmaterialsforbuildingshavebeenrelativelyscarce—steel,reinforcedconcrete,prestressedconcrete,compositewoodmaterials,andaluminumalloys.However,thesematerialshavebroughtaboutarevolutioninstructuraldesignandarecurrentlybeingtestedtotheirmateriallimitbyengineersandarchitects.Someengineersbelievethatmostofthesignificantstructuralsystemsareknownand,therefore,thatthefutureliesinthedevelopmentofnewmaterialsandtheexploitationofknownmaterialsinnewways.Advancesinstructuralanalysistechniques,especiallywiththeadventofthecomputer,haveenableddesignerstoexploreverycomplexstructures(Figures10and11)underanarrayofloadingconditionsmuchmorerapidlyandaccuratelythaninthepast.However,thecomputerisstillbeingusedasatooltovalidatetheintentofthedesignerandisnotyetcapableofactual“design.”AIntroductionhumandesigner’sknowledge,creativity,andunderstandingofhowabuildingstructureistobeconfiguredarestillessentialforasuccessfulproject.4LOADSONSTRUCTURESStructuralsystems,asidefromtheirform-definingfunction,essentiallyexisttoresistforcesthatresultfromtwogeneralclassificationsofloads:1.Static.Thisclassificationreferstogravity-typeforces.2.Dynamic.Thisclassificationisduetoinertiaormomentumofthemassofthestructure(likeearthquakes).Themoresuddenthestartingorstoppingofthestructure,thegreatertheforcewillbe.Note:Otherdynamicforcesareproducedbywaveaction,landslides,fallingobjects,shocks,blasts,vibrationfromheavymachinery,andsoon.Alight,steelframebuildingmaybeverystronginresistingstaticforces,butadynamicforcemaycauselargedistortionstooccurbecauseoftheframe’sflexiblenature.Ontheotherhand,aheavilyreinforcedconcretebuildingmaybeasstrongasthesteelbuildingincarryingstaticloadsbuthaveconsiderablestiffnessandsheerdeadweight,whichmayabsorbtheenergyofdynamicforceswithlessdistortion(deformation).Allofthefollowingforcesmustbeconsideredinthedesignofabuildingstructure(Figure12).■DeadLoads.Loadsresultingfromtheself-weightofthebuildingorstructureandofanypermanentlyattachedcomponents,suchaspartitionwalls,flooring,framingelements,andfixedequipment,areclassifiedasdeadloads.Standardweightsofcommonlyusedmaterialsforbuildingareknown,andacompletebuilding’sdeadweightcanbecalculatedwithahighdegreeofcertainty.However,theweightofstructuralelementsmustbeestimatedatthebeginningofthedesignphaseofthestructureandthenrefinedasthedesignprocessproceedstowardcompletion.Asamplingofsomestandardbuildingmaterialweightsusedfortheinitialstructuraldesignprocessis:Figure12Typicalbuildingloads.concrete=150poundspercubicfoot(pcf)timber=35pcfsteel=490pcfbuilt-uproofing=6poundspersquarefoot(psf)half-inchgypsumwallboard=1.8psfplywood,perinchofthickness=3psfsuspendedacousticalceiling=1psfWhenactivatedbyearthquake,staticdeadloadstakeonadynamicnatureintheformofhorizontalinertialforces.Buildingswithheavierdeadloads5Introductiongeneratehigherinertialforcesthatareappliedinahorizontaldirection.■■SnowLoads.Snowloadsrepresentaspecialtypeofliveloadbecauseofthevariabilityinvolved.Localbuildingofficialsorapplicablebuildingcodesprescribethedesignsnowforaspecificgeographicaljurisdiction.Generally,snowloadsaredeterminedfromazonemapreporting50-yearrecurrenceintervalsofanextremesnowdepth.Snowweightscanvaryfromapproximately8pcffordrypowdersnowto12pcfforwetsnow(Figure13).Designloadscanvaryfrom10psfonahorizontalsurfaceto400psfinsomespecificmountainousregions.InmanyareasoftheUnitedStates,designsnowloadscanrangefrom20to40psf.Theaccumulationdepthofthesnowdependsontheslopeoftheroof.Steeperslopeshavesmalleraccumulations.Specialprovisionsmustalsobemadeforpotentialaccumulationofsnowatroofvalleys,parapets,andotherunevenroofconfiguration.DBAuer/EPA/NewscomFigure13LiveLoads.Transientandmovingloadsthatincludeoccupancyloads,furnishings,andstorageareclassifiedasliveloads.Liveloadsareextremelyvariablebynatureandnormallychangeduringastructure’slifetimeasoccupancychanges.Buildingcodesspecifyminimumuniformliveloadsforthedesignofroofandfloorsystemsbasedonahistoryofmanybuildingsandtypesofoccupancyconditions.Thesecodesincorporatesafetyprovisionsforoverloadprotection,allowanceforconstructionloads,andserviceabilityconsiderations,suchasvibrationanddeflectionbehavior.Minimumroofliveloadsincludeallowanceforminorsnowfallandconstructionloads.(SeeTable2foranadditionallistingofcommonliveloadsforbuildings.)Failurefromsnowload.Exceptforabuilding’sdeadload,whichisfixed,theotherforceslistedabovecanvaryinduration,magnitude,andpointofapplication.Abuildingstructuremustneverthelessbedesignedforthesepossibilities.Unfortunately,alargeportionofabuildingstructureexistsforloadsthatwillbepresentatmuchlowermagnitudes—ormayneveroccuratall.Thestructuralefficiencyofabuildingisoftenmeasuredbyitsdeadloadweightincomparisontotheliveloadcarried.Buildingdesignershavealwaysstrivedtoreducetheratioofdeadtoliveload.Newmethodsofdesign,newandlightermaterials,andoldmaterialsusedinnewwayshavecontributedtothedead/liveloadreduction.Thesizeofthestructurehasaninfluenceontheratioofdeadtoliveload.Asmallbridgeoveracreek,forexample,cancarryaheavyvehicle—aliveloadrepresentingalargeportionofthedead/liveloadratio.TheGoldenGateBridgeinSanFrancisco,ontheotherhand,spansalong6Introductiondistance,andthematerialofwhichitiscomposedisusedchieflyincarryingitsownweight.Theliveloadofthevehiculartraffichasarelativelysmalleffectonthebridge’sinternalstresses.Withtheuseofmodernmaterialsandconstructionmethods,itisoftenthesmallerratherthanthelargerbuildingsthatshowahighdead/liveloadratio.Inatraditionalhouse,theliveloadislow,andmuchofthedeadloadnotonlysupportsitselfbutalsoservesasweatherprotectionandspace-definingsystems.Thisrepresentsahighdead/liveloadratio.Incontrast,inalargefactorybuilding,thedeadloadisnearlyallstructurallyeffective,andthedead/liveloadratioislow.Thedead/liveloadratiohasconsiderableinfluenceonthechoiceofstructureandespeciallyonthechoiceofbeamtypes.Asspansincrease,sodothebendingeffectscausedbydeadandliveloads;therefore,morematerialmustbeintroducedintothebeamtoresisttheincreasedbendingeffects.Thisaddedmaterialweightitselfaddsfurtherdeadloadandpronouncedbendingeffectsasspansincrease.Thedead/liveloadrationotonlyincreasesbutmayeventuallybecomeextremelylarge.■WindLoads.Windisessentiallyairinmotionandcreatesaloadingonbuildingsthatisdynamicinnature.Whenbuildingsandstructuresbecomeobstaclesinthepathofwindflow,thewind’skineticenergyisconvertedintopotentialenergyofpressureonvariouspartsofthebuilding.Windpressures,directions,anddurationareconstantlychanging.However,forcalculationpurposes,mostwinddesignassumesastaticforceconditionformoreconventional,lowerrisebuildings.Thefluctuatingpressurecausedbyaconstantlyblowingwindisapproximatedbyameanpressurethatactsonthewindwardside(thesidefacingthewind)andleewardside(thesideoppositethewindwardside)ofthestructure.The“static”ornonvaryingexternalforcesareappliedtothebuildingstructureandsimulatetheactualvaryingwindforces.Figure14Windloadsonastructure.Directwindpressuresdependonseveralvariables:windvelocity,heightofthewindaboveground(windvelocitiesarelowerneartheground),andthenatureofthebuilding’ssurroundings.Windpressureonabuildingvariesasthesquareofthevelocity(inmilesperhour).Thispressureisalsoreferredtoasthestagnationpressure.Buildingsrespondtowindforcesinavarietyofcomplexanddynamicways.Thewindcreatesanegativepressure,orsuction,onboththeleewardsideofthebuildingandonthesidewallsparalleltothewinddirection(Figure14).Upliftpressureoccursonhorizontalorslopingroofsurfaces.Inaddition,thecorners,edges,andeaveoverhangsof7Introductionabuildingaresubjectedtocomplicatedforcesasthewindpassestheseobstructions,causinghigherlocalizedsuctionforcesthangenerallyencounteredonthebuildingasawhole.Windisafluidandactslikeotherfluids—aroughsurfacecausesfrictionandslowsthewindvelocityneartheground.Windspeedsaremeasuredatastandardheightof10meters(33feet)abovetheground,andadjustmentsaremadewhencalculatingwindpressuresathigherelevations.Thewindpressureincreaseswiththeheightofthebuilding.Otherbuildings,trees,andtopographyaffecthowthewindwillstrikethebuilding.Buildingsinvastopenareasaresubjecttolargerwindforcesthanarethoseinshelteredareasorwhereabuildingissurroundedbyotherbuildings.Thesize,shape,andsurfacetextureofthebuildingalsoimpactthedesignwindforces.Resultingwindpressuresaretreatedaslateralloadingonwallsandasdownwardpressureorupliftforces(suction)onroofplanes.■Figure151-storyT~0.1sec.Earthquakeloadsonastructure.4-storyT~0.4sec.15-storyT~1.5sec.Figure16Approximatebuildingperiodsofvibration.8EarthquakeLoads(seismic).Earthquakes,likewind,produceadynamicforceonabuilding.Duringanactualearthquake,therearecontinuousgroundmotionsthatcausethebuildingstructuretovibrate.Thedynamicforcesonthebuildingarearesultoftheviolentshakingofthegroundgeneratedbyseismicshockwavesemanatingfromthecenterofthefault(thefocusorhypocenter)(Figure15).Thepointdirectlyabovethehypocenterontheearth’ssurfaceisknownastheepicenter.Therapidity,magnitude,anddurationoftheseshakesdependontheintensityoftheearthquake.Duringanearthquake,thegroundmasssuddenlymovesbothverticallyandlaterally.Thelateralmovementsareofparticularconcerntobuildingdesigners.Lateralforcesdevelopedinthestructureareafunctionofthebuilding’smass,configuration,buildingtype,height,andgeographiclocation.Theforcefromanearthquakeisinitiallyassumedtodevelopatthebaseofthebuilding;theforcebeingknownasthebaseshear(Vbase).Thisbaseshearisthenredistributedequalandoppositeateachofthefloorlevelswherethemassofthebuildingisassumedconcentrated.Allobjects,includingbuildings,haveanaturalorfundamentalperiodofvibration.Itrepresentsthetimeittakesanobjectorbuildingtovibratethroughonecycleofvibration(orsway)whensubjectedtoanappliedforce.Whenanearthquakegroundmotioncausesabuildingtostartvibrating,thebuildingbeginstodisplace(sway)backandforthatitsnaturalperiodofvibration.Shorter,lowerbuildingshaveveryshortperiodsofvibration(lessthanonesecond),whiletallhighrisescanhaveperiodsofIntroductionvibrationthatlastseveralseconds(Figure16).Fundamentalperiodsareafunctionofabuilding’sheight.Anapproximateestimateofabuilding’speriodisequaltoT=0.1NwhereNrepresentsthenumberofstoriesandTrepresentstheperiodofvibrationinseconds.Thegroundalsovibratesatitsownnaturalperiodofvibration.ManyofthesoilsintheUnitedStateshaveperiodsofvibrationintherangeof0.4to1.5seconds.Shortperiodsaremorecharacteristicofhardsoils(rock),whilesoftground(someclays)mayhaveperiodsofuptotwoseconds.Manycommonbuildingscanhaveperiodswithintherangeofthesupportingsoils,makingitpossibleforthegroundmotiontotransmitatthesamenaturalfrequencyasthatofthebuilding.Thismaycreateaconditionofresonance(wherethevibrationsincreasedramatically),inwhichtheinertialforcesmightbecomeextremelylarge.Inertialforcesdevelopinthestructureduetoitsweight,configuration,buildingtype,andgeographiclocation.Inertialforcesaretheproductofmassandacceleration(Newton’ssecondlaw:F=m*a).Heavy,massivebuildingswillresultinlargerinertialforces;hence,thereisadistinctadvantageinusingalighterweightconstructionwhenseismicconsiderationsareakeypartofthedesignstrategy.Forsometallbuildingsorstructureswithcomplexconfigurationsorunusualmassing,adynamicstructuralanalysisisrequired.Computersareusedtosimulateearthquakesonthebuildingtostudyhowtheforcesaredevelopedandtheresponseofthestructuretotheseforces.Buildingcodesareintendedtosafeguardagainstmajorfailuresandlossoflife;theyarenotexplicitlyfortheprotectionofproperty.5BASICFUNCTIONALREQUIREMENTSTheprincipalfunctionalrequirementsofabuildingstructureare:1.2.3.4.5.6.Stabilityandequilibrium.Strengthandstiffness.Continuityandredundancy.Economy.Functionality.Aesthetics.Primarily,structuraldesignisintendedtomakethebuilding“standup”(Figure17).Inmakingabuilding“standup,”Figure17Stabilityandthestrengthofastructure—thecollapseofaportionoftheUniversityofWashingtonHuskystadiumduringconstruction(1987)duetolackofadequatebracingtoensurestability.Photobyauthor.9Introductiontheprinciplesgoverningthestabilityandequilibriumofbuildingsformthebasisforallstructuralthinking.Strengthandstiffnessofmaterialsareconcernedwiththestabilityofabuilding’scomponentparts(beams,columns,walls),whereasstaticsdealswiththetheoryofgeneralstability.Staticsandstrengthofmaterialsareactuallyintertwined,becausethelawsthatapplytothestabilityofthewholestructurearealsovalidfortheindividualcomponents.Figure18EquilibriumandStability?—sculpturebyRichardByer.Photobyauthor.Thefundamentalconceptofstabilityandequilibriumisconcernedwiththebalancingofforcestoensurethatabuildinganditscomponentswillnotmove(Figure18).Inreality,allstructuresundergosomemovementunderload,butstablestructureshavedeformationsthatremainrelativelysmall.Whenloadsareremovedfromthestructure(oritscomponents),internalforcesrestorethestructuretoitsoriginal,unloadedcondition.Agoodstructureisonethatachievesaconditionofequilibriumwithaminimumofeffort.Strengthofmaterialsrequiresknowledgeaboutbuildingmaterialproperties,membercross-sections,andtheabilityofthematerialtoresistbreaking.Alsoofconcernisthatthestructuralelementsresistexcessivedeflectionand/ordeformation.(a)Continuity—loadsfromtheroofbeamsareredistributedtotheroofcolumnsbelow.Acontinuouspathisprovidedforthecolumnloadstotraveldirectlytothecolumnsbelowandthenontothefoundation.(b)Discontinuityintheverticalelevationcanresultinverylargebeambendingmomentsanddeflection.Structuralefficiencyisenhancedbyaligningcolumnstoprovideadirectpathtothefoundation.Beamsizescanthusbereducedsignificantly.Inthisexample,missingordamagedcolumnscouldalsorepresenthowstructuralframeworkscanhavetheabilitytoredistributeloadstoadjacentmemberswithoutcollapse.Thisisreferredtoasstructuralredundancy.Figure19Examplesofcontinuityandredundancy.10Continuityinastructurereferstoadirect,uninterruptedpathforloadsthroughthebuildingstructure—fromtheroofleveldowntothefoundation.Redundancyistheconceptofprovidingmultipleloadpathsinastructuralframeworksothatonesystemactsasabackuptoanotherintheeventoflocalizedstructuralfailure.Structuralredundancyenablesloadstoseekalternatepathstobypassstructuraldistress.Alackofredundancyisveryhazardouswhendesigningbuildingsinearthquakecountry(Figure19).On9/11,bothoftheWorldTradeCentertowerswereabletowithstandtheimpactofjetlinerscrashingintothemandcontinuestandingforsometime,permittingmanypeopletoevacuate.Thetowersweredesignedwithstructuralredundancy,whichpreventedanevenlargerlossoflife.However,theprocessbywhichthecollapseoftheimpactedstorylevelledtotheprogressivecollapseoftheentirebuildingmayhaveledsomeinvestigatorstohintthataninadequatedegreeofstructuralredundancyexisted.Therequirementsofeconomy,functionality,andaestheticsareusuallynotcoveredinastructurescourseandwillnotbedealtwithinthistext.Strengthofmaterialsistypicallycovereduponcompletionofastaticscourse.Introduction6ARCHITECTURALISSUESAtechnicallyperfectworkcanbeaestheticallyinexpressivebuttheredoesnotexist,eitherinthepastorinthepresent,aworkofarchitecturewhichisacceptedandrecognizedasexcellentfromanaestheticpointofviewwhichisnotalsoexcellentfromthetechnicalpointofview.Goodengineeringseemstobeanecessarythoughnotsufficientconditionforgoodarchitecture.—PierLuigiNerviThegeometryandarrangementoftheload-bearingmembers,theuseofmaterials,andthecraftingofjointsallrepresentopportunitiesforbuildingstoexpressthemselves.Thebestbuildingsarenotdesignedbyarchitectswho,afterresolvingtheformalandspatialissues,simplyaskthestructuralengineertomakesureitdoesnotfalldown.AnHistoricalOverviewItispossibletotracetheevolutionofarchitecturalspaceandformthroughparalleldevelopmentsinstructuralengineeringandmaterialtechnology.Untilthe19thcentury,thishistorywaslargelybasedonstoneconstructionandthecapabilityofthismaterialtoresistcompressiveforces.Lessdurablewoodconstructionwasgenerallyreservedforsmallbuildingsorportionsofbuildings.Figure20Stonehenge.Figure21temple.ConstructionofaGreekperistyleFigure22vault.Stonearch,barrelvault,andgroinNeolithicbuildersuseddrystonetechniques,suchascoursedmasonrywallingandcorbelling,toconstructmonuments,dwellings,tombs,andfortifications.Thesestructuresdemonstrateanunderstandingofthematerialpropertiesofthevariousstonesemployed(Figure20).Timberjoininganddressedstoneworkweremadepossiblebyironandbronzetools.Narrowopeningsinmasonrybuildingwallswereachievedthroughcorbellingandtimberorstonelintels.TheearliestexamplesofvoussoirarchesandvaultsinbothstoneandunfiredbrickconstructionhavebeenfoundinEgyptandGreece(Figure21).ThesematerialsandstructuralinnovationswerefurtherdevelopedandrefinedbytheRomans.TheancientRomanarchitectVitruvius,inhisTenBooks,describedtimbertrusseswithhorizontaltiememberscapableofresistingtheoutwardthrustofslopingrafters.Romanbuildersmanagedtoplacethesemicirculararchatoppiersorcolumns;thelargerspansreducedthenumberofcolumnsrequiredtosupporttheroof.Domesandbarrelandgroinvaultswereimprovedthroughtheuseofmodularfiredbrick,cementmortar,andhydraulicconcrete.TheseinnovationsenabledRomanarchitectstocreateevenlargerunobstructedspaces(Figure22).11IntroductionGradualrefinementsofthistechnologybyRomanesquemasonbuilderseventuallyledtothestructurallydaringandexpressiveGothiccathedrals.Thetall,slendernavewallswithlargestainedglassopenings,whichcharacterizethisarchitecture,aremadepossiblebyimprovementsinconcretefoundationconstruction;thepointedarch,whichreduceslateralforces;flyingarchesandbuttresses,whichresisttheremaininglateralloads;andtheribbedvault,whichreinforcesthegroinandcreatesaframeworkofarchesandcolumns,keepingopaquewallstoaminimum(Figure23).ThemediumofdrawingallowedRenaissancearchitectstoworkonpaper,removedfromconstructionandthesite.Existingtechnicaldevelopmentswereemployedinthesearchforaclassicalidealofbeautyandproportion.Figure23ConstructionofaGothiccathedral.Figure24SportsPalace,reinforcedconcretearena,byPierLuigiNervi.Structuralcastironandlarger,strongersheetsofglassbecameavailableinthelate18thcentury.Thesenewmaterialswerefirstemployedinindustrialandcommercialbuildings,trainsheds,exhibitionhalls,andshoppingarcades.Interiorspacesweretransformedbythedelicatelong-spantrussessupportedontall,slender,hollowcolumns.Theelementsofstructureandcladdingweremoreclearlyarticulated,withdaylightadmittedingreatquantities.Wroughtironand,later,structuralsteelprovidedexcellenttensilestrengthandreplacedbrittlecastiron.ArtNouveauarchitectsexploitedthesculpturalpotentialofironandglass,whilecommercialinterestscapitalizedonthelong-spancapabilitiesofrolledsteelsections.Thetensilepropertiesofsteelwerecombinedwiththehighcompressivestrengthofconcrete,makingacompositesectionwithexcellentweatheringandfire-resistivepropertiesthatcouldbeformedandcastinalmostanyshape(Figure24).Steelandreinforcedconcretestructuralframesenabledbuilderstomaketallerstructureswithmorestories.Thesmallerfloorareadevotedtostructureandthegreaterspatialflexibilityledtothedevelopmentofthemodernskyscraper.Today,pretensionedandposttensionedconcrete,engineeredwoodproducts,tensilefabric,andpneumaticstructuresandotherdevelopmentscontinuetoexpandthearchitecturalandstructuralpossibilities.Therelationshipbetweentheformofarchitecturalspaceandstructureisnotdeterministic.Forexample,thedevelopmentofBuckminsterFuller’sgeodesicdomedidnotimmediatelyresultinaproliferationofdomedchurchesorofficebuildings.Ashistoryhasdemonstrated,vastlydifferentspatialconfigurationshavebeenrealizedwiththesamematerialsandstructuralsystems.Conversely,similarformshavebeengeneratedutilizingverydifferentstructuralsystems.Architectsaswellasbuildersmustdevelopasenseofstructure(Figure25).Creativecollaborationbetweenarchitect,builder,andengineerisnecessarytoachievethehighestlevelofformal,spatial,andstructuralintegration.12IntroductionCriteriafortheSelectionofStructuralSystemsMostbuildingprojectsbeginwithaclientprogramoutliningthefunctionalandspatialrequirementstobeaccommodated.Architectstypicallyinterpretandprioritizethisinformation,coordinatingarchitecturaldesignworkwiththeworkofotherconsultantsontheproject.Thearchitectandstructuralengineermustsatisfyawiderangeoffactorsindeterminingthemostappropriatestructuralsystem.Severalofthesefactorsarediscussedhere.NatureandmagnitudeofloadsTheweightofmostbuildingmaterials(Table1)andtheself-weightofstructuralelements(deadloads)canbecalculatedfromreferencetableslistingthedensitiesofvariousmaterials.Buildingcodesestablishdesignvaluesfortheweightoftheoccupantsandfurnishings—liveloads(Table2)—andothertemporaryloads,suchassnow,wind,andearthquake.Buildinguse/functionSportsfacilities(Figure26)requirelong,clearspanareasfreeofcolumns.Lightwoodframingiswellsuitedtotherelativelysmallroomsandspansfoundinresidentialconstruction.Figure25Foster.HongKongBank,byNormanSiteconditionsTopographyandsoilconditionsoftendeterminethedesignofthefoundationsystem,whichinturninfluencesthewayloadsaretransmittedthoughwallsandcolumns.Lowsoil-bearingcapacitiesorunstableslopesmightsuggestaseriesofpiersloadedbycolumnsinsteadofconventionalspreadfootings.Climaticvariables,suchaswindspeedandsnowfall,affectdesignloads.Significantmovement(thermalexpansionandcontraction)canresultfromextremetemperaturefluctuations.Seismicforces,usedtocalculatebuildingcodedesignloads,varyindifferentpartsofthecountry.BuildingsystemsintegrationAllbuildingsystems(lighting,heating/cooling,ventilation,plumbing,firesprinklers,electrical)havearationalbasisthatgovernstheirarrangement.Itisgenerallymoreelegantandcost-effectivetocoordinatethesesystemstoavoidconflictandcompromiseintheirperformance.Thisisespeciallythecasewherethestructureisexposedanddroppedceilingspacesarenotavailableforductandpiperuns.Figure26SportsPalaceinterior,byPierLuigiNervi(1955).FireresistanceBuildingcodesrequirethatbuildingcomponentsandstructuralsystemsmeetminimumfire-resistancestandards.Thecombustibilityofmaterialsandtheirabilitytocarrydesignloadswhensubjectedtointenseheataretestedtoensurethatbuildingsinvolvedinfirescanbesafelyevacuatedina13IntroductionTable1Selectedbuildingmaterialweights.Assemblylb./ft.2kN/m2Roofs:Three-plyandgravelFive-plyandgravelWoodshinglesAsphaltshinglesCorrugatedmetalPlywoodInsulation—fiberglassbattInsulation—rigid5.56.5221–2.53#/in.0.51.50.260.310.100.100.05–0.120.0057kN/mm0.00250.075Floors:ConcreteplankConcreteslabSteeldeckingw/concreteWoodjoistsHardwoodfloorsCeramictilew/thinsetLightweightconcreteTimberdecking6.512.5#/in.35–452–3.54#/in.158#/in.2.5#/in.0.310.59kN/mm1.68–2.160.10–0.170.19kN/mm0.710.38kN/mm0.08kN/mmWalls:Woodstuds(average)SteelstudsGypsumdrywallPartitions(studsw/drywall)2.543.6#/in.60.0120.200.17kN/mm0.29Table2Selectedliveloadrequirements.*Occupancy/Use(Uniformload)lb./ft.2kN/m2Apartments:PrivatedwellingsCorridorsandpublicrooms401001.924.79Assemblyareas/theaters:FixedseatsStagearea601002.874.7940601.922.87Hotels:PrivateguestroomsCorridors/publicrooms401001.924.79Offices:GeneralfloorareaLobbies/firstfloorcorridor501002.404.794020301.920.961.444080–1001.923.83–4.79401001.924.79Hospitals:PrivateroomsandwardsLaboratories/operatingroomsResidential(private):BasicfloorareaanddecksUninhabitedatticsHabitableattics/sleepingareasSchools:ClassroomsCorridorsStairsandexits:Singlefamily/duplexdwellingsAllotherLoadsareadaptedfromvariousCodesourcesandarelistedhereforillustrativepurposesonly.ConsultthegoverningCodeinyourlocaljurisdictionforactualdesignvalues.*14givenperiodoftime.Woodisnaturallycombustible,butheavytimberconstructionmaintainsmuchofitsstrengthforanextendedperiodoftimeinafire.Steelcanbeweakenedtothepointoffailureunlessprotectedbyfireproofcoverings.Concreteandmasonryareconsideredtobenoncombustibleandarenotsignificantlyweakenedinfires.Thelevelsoffireresistancevaryfromunratedconstructiontofourhoursandarebasedonthetypeofoccupancyandsizeofabuilding.ConstructionvariablesCostandconstructiontimearealmostalwaysrelevantissues.Severalstructuralsystemswilloftenaccommodatetheload,span,andfire-resistancerequirementsforabuilding.Localavailabilityofmaterialsandskilledconstructiontradestypicallyaffectcostandschedule.Theselectedsystemcanberefinedtoachievethemosteconomicalframingarrangementorconstructionmethod.Theuseofheavyequipment,suchascranesorconcretetrucksandpumps,mayberestrictedbyavailabilityorsiteaccess.ArchitecturalformandspaceSocialandculturalfactorsthatinfluencethearchitect’sconceptionofformandspaceextendtotheselectionanduseofappropriatematerials.Wherestructureisexposed,thelocation,scale,hierarchy,anddirectionofframingmemberscontributesignificantlytotheexpressionofthebuilding.Thistext,StaticsandStrengthsofMaterialsforArchitectureandBuildingConstruction,coverstheanalysisofstaticallydeterminatesystemsusingthefundamentalprinciplesoffreebodydiagramsandequationsofequilibrium.Althoughduringrecentyearshaveseenanincredibleemphasisontheuseofcomputerstoanalyzestructuresbymatrixanalysis,itistheauthor’sopinionthataclassicalapproachforabeginningcourseisnecessary.Thistextaimistogivethestudentanunderstandingofphysicalphenomenabeforeembarkingontheapplicationofsophisticatedmathematicalanalysis.Relianceonthecomputer(sometimesthe“blackorwhitebox”)foranswersthatonedoesnotfullyunderstandisariskypropositionatbest.Applicationofthebasicprinciplesofstaticsandstrengthofmaterialswillenablethestudenttogainaclearerand,itishoped,moreintuitivesenseaboutstructure.Statics1CHARACTERISTICSOFAFORCEForceWhatisforce?Forcemaybedefinedastheactionofonebodyonanotherthataffectsthestateofmotionorrestofthebody.Inthelate17thcentury,SirIsaacNewton(Figure1)summarizedtheeffectsofforceinthreebasiclaws:■■■FirstLaw:Anybodyatrestwillremainatrest,andanybodyinmotionwillmoveuniformlyinastraightline,unlessacteduponbyaforce.(Equilibrium)SecondLaw:Thetimerateofchangeofmomentumisequaltotheforceproducingit,andthechangetakesplaceinthedirectioninwhichtheforceisacting.(F=m*a)ThirdLaw:Foreveryforceofaction,thereisareactionthatisequalinmagnitude,oppositeindirection,andhasthesamelineofaction.(Basicconceptofforce.)Newton’sfirstlawinvolvestheprincipleofequilibriumofforces,whichisthebasisofstatics.Thesecondlawformulatesthefoundationforanalysisinvolvingmotionordynamics.Writteninequationform,Newton’ssecondlawmaybestatedasF=m*awhereFrepresentstheresultantunbalancedforceactingonabodyofmassmwitharesultantaccelerationa.Examinationofthissecondlawimpliesthesamemeaningasthefirstlaw,becausethereisnoaccelerationwhentheforceiszeroandthebodyisatrestormoveswithaconstantvelocity.Figure1SirIsaacNewton(1642–1727).BornonChristmasDayin1642,SirIsaacNewtonisviewedbymanyasthegreatestscientificintellectwhoeverlived.Newtonsaidofhimself,“IdonotknowwhatImayappeartotheworld,buttomyselfIseemtohavebeenonlylikeaboyplayingontheseashore,anddivertingmyselfinnowandthenfindingasmootherpebbleoraprettiershellthanordinary,whilstthegreatoceanoftruthlayallundiscoveredbeforeme.”Newton’searlyschoolingfoundhimfascinatedwithdesigningandconstructingmechanicaldevicessuchaswaterclocks,sundials,andkites.Hedisplayednounusualsignsofbeinggifteduntilhislaterteens.Inthe1660s,heattendedCambridgebutwithoutanyparticulardistinction.InhislastundergraduateyearatCambridge,withnomorethanbasicarithmetic,hebegantostudymathematics,primarilyasanautodidact,derivinghisknowledgefromreadingwithlittleornooutsidehelp.Hesoonassimilatedexistingmathematicaltraditionandbegantomovebeyondittodevelopcalculus(independentofLeibniz).Athismother’sfarm,wherehehadretiredtoavoidtheplaguethathadhitLondonin1666,hewatchedanapplefalltothegroundandwonderediftherewasasimilaritybetweentheforcespullingontheappleandthepullonthemooninitsorbitaroundtheEarth.Hebegantolaythefoundationofwhatwaslatertobecometheconceptofuniversalgravitation.Inhisthreelawsofmotion,hecodifiedGalileo’sfindingsandprovidedasynthesisofcelestialandterrestrialmechanics.FromChapter2ofStaticsandStrengthofMaterialsforArchitectureandBuildingConstruction,FourthEdition,BarryOnouye,KevinKane.Copyright©2012byPearsonEducation,Inc.PublishedbyPearsonPrenticeHall.Allrightsreserved.15StaticsThethirdlawintroducesustothebasicconceptofforce.ItstatesthatwheneverabodyAexertsaforceonanotherbodyB,bodyBwillresistwithanequalmagnitudebutintheoppositedirection.Figure2Groundresistanceonabuilding.Forexample,ifabuildingwithaweightWisplacedontheground,wecansaythatthebuildingisexertingadownwardforceofWontheground.However,forthebuildingtoremainstableontheresistinggroundsurfacewithoutsinkingcompletely,thegroundmustresistwithanupwardforceofequalmagnitude.IfthegroundresistedwithaforcelessthanW,whereR<W,thebuildingwouldsettle.Ontheotherhand,ifthegroundexertedanupwardforcegreaterthanW(R>W),thebuildingwouldrise(levitate)(Figure2).CharacteristicsofaForceAforceischaracterizedbyits(a)pointofapplication,(b)magnitude,and(c)direction.Thepointofapplicationdefinesthepointwheretheforceisapplied.Instatics,thepointofapplicationdoesnotimplytheexactmoleculeonwhichaforceisappliedbutalocationthat,ingeneral,describestheoriginofaforce(Figure3).Figure3Ropepullingonaneyebolt.Figure4(a)Ananchordevicewiththreeappliedforces.Inthestudyofforcesandforcesystems,thewordparticlewillbeused,anditshouldbeconsideredasthelocationorpointwheretheforcesareacting.Here,thesizeandshapeofthebodyunderconsiderationwillnotaffectthesolution.Forexample,ifweconsidertheanchorbracketshowninFigure4(a),threeforces—Fl,F2,andF3—areapplied.TheintersectionofthesethreeforcesoccursatpointO;therefore,forallpracticalpurposes,wecanrepresentthesamesystemasthreeforcesappliedonparticleO,asshowninFigure4(b).Magnitudereferstothequantityofforce,anumericalmeasureoftheintensity.Basicunitsofforcethatwillbeusedthroughoutthistextarethepound(lb.or#)andthekilopound(kipork=1,000#).Inmetric(SI)units,forceisexpressedasthenewton(N)orkilonewton(kN)where1kN=1,000N.Thedirectionofaforceisdefinedbyitslineofactionandsense.Thelineofactionrepresentsaninfinitestraightlinealongwhichtheforceisacting.InFigure5,theexternaleffectsontheboxareessentiallythesamewhetherthepersonusesashortorlongcable,providedthepullexertedisalongthesamelineofactionandofequalmagnitude.Figure4(b)Forcediagramoftheanchor.16Ifaforceisappliedsuchthatthelineofactionisneitherverticalnorhorizontal,somereferencesystemmustbeestablished.Mostcommonlyacceptedistheangularsymbolofθ(theta)orφ(phi)todenotethenumberofdegreesthelineofactionoftheforceisinrelationtothehorizontalorverticalaxis,respectively.Onlyone(θorφ)needstobeStaticsindicated.Analternativetoangulardesignationsisasloperelationship.Thesenseoftheforceisindicatedbyanarrowhead.Forexample,inFigure6,thearrowheadgivestheindicationthatapullingforce(tension)isbeingappliedtothebracketatpointO.Byreversingonlythearrowhead(Figure7),wewouldhaveapushingforce(compression)appliedonthebracketwiththesamemagnitude(F10k),pointofapplication(pointO),andlineofaction(θ=22.6°fromthehorizontal).Figure6Threewaysofindicatingdirectionforanangulartensionforce.Figure5Horizontalforceappliedtoabox.Figure7Forceincompression.RigidBodiesPracticallyspeaking,anybodyundertheactionofforcesundergoessomekindofdeformation(changeinshape).Instatics,however,wedealwithabodyofmatter(calledacontinuum)that,theoretically,undergoesnodeformation.Thiswecallarigidbody.WhenaforceofF=10#isappliedtoabox,asshowninFigure8,somedegreeofdeformationwillresult.Thedeformedboxisreferredtoasadeformablebody,whereasinFigure8(b)weseeanundeformedboxcalledtherigidbody.Again,youmustrememberthattherigidbodyisapurelytheoreticalphenomenonbutnecessaryinthestudyofstatics.(a)Original,unloadedbox.(b)Rigidbody(example:stone).Figure8Rigidbody/deformablebody.(c)Deformablebody(example:foam).17StaticsPrincipleofTransmissibilityAnimportantprinciplethatappliestorigidbodiesinparticularistheprincipleoftransmissibility.Thisprinciplestatesthattheexternaleffectsonabody(cart)remainunchangedwhenaforceFlactingatpointAisreplacedbyaforceF2ofequalmagnitudeatpointB,providedthatbothforceshavethesamesenseandlineofaction(Figure9).Figure9Anexampleoftheprincipleoftransmissibility.InFigure9(a),thereactionsRlandR2representthereactionsofthegroundontothecart,opposingtheweightofthecartW.AlthoughinFigure9(b)thepointofapplicationfortheforcechanges(magnitude,sense,andlineofactionremainingconstant),thereactionsRlandR2andalsotheweightofthecartWremainthesame.Theprincipleoftransmissibilityisvalidonlyintermsoftheexternaleffectsonabodyremainingthesame(Figure10);internally,thismaynotbetrue.ExternalandInternalForcesLet’sconsideranexampleofanailbeingwithdrawnfromawoodfloor(Figure11).Figure10Anotherexampleoftheprincipleoftransmissibility.Ifweremovethenailandexaminetheforcesactingonit,wediscoverfrictionalforcesthatdevelopontheembeddedsurfaceofthenailtoresistthewithdrawalforceF(Figure12).Treatingthenailasthebodyunderconsideration,wecanthensaythatforcesFandSareexternalforces.Theyarebeingappliedoutsidetheboundariesofthenail.Externalforcesrepresenttheactionofotherbodiesontherigidbody.18Figure11Withdrawalforceonanail.Let’sconsiderjustaportionofthenailandexaminetheforcesactingonit.InFigure13,thefrictionalforceSplustheforceR(theresistancegeneratedbythenailinternally)resisttheappliedforceF.ThisinternalforceRisresponsibleforkeepingthenailfrompullingapart.Figure12Externalforcesonthenail.Figure13Internalresistingforcesonthenail.StaticsExaminenextacolumn-to-footingarrangementwithanappliedforceF,asillustratedinFigure14.Toappropriatelydistinguishwhichforcesareexternalandwhichareinternal,wemustdefinethesystemweareconsidering.Severalobviouspossibilitiesexisthere:Figure15(a),columnandfootingtakentogetherasasystem;Figure15(b),columnbyitself;andFigure15(c),footingbyitself.InFigure15(a),takingthecolumnandfootingasthesystem,theexternalforcesareF(theappliedforce),Wcol,Wftg,andRsoil.Weightsofbodiesormembersareconsideredasexternalforces,appliedatthecenterofgravityofthemember.Centerofgravity(masscenter)willbediscussedinalatersection.ThereactionorresistancethegroundofferstocounteracttheappliedforcesandweightsisRsoil.Thisreactionoccursonthebaseofthefooting,outsidetheimaginarysystem’sboundary;therefore,itisconsideredseparately.Figure14Acolumnsupportinganexternalload.Whenacolumnisconsideredseparately,asasystembyitself,theexternalforcesbecomeF,Wcol,andRl.TheforcesFandWcolarethesameasinFigure15(b),buttheforceRlisaresultoftheresistancethefootingofferstothecolumnundertheappliedforces(FandWcol)shown.(b)Column.(a)Columnandfooting.Thelastcase,Figure15(c),considersthefootingasasystembyitself.ExternalforcesactingonthefootingareR2,Wftg,andRsoil.TheR2forcerepresentsthereactionthecolumnproducesonthefooting,andWftgandRsoilarethesameasinFigure15(a).(c)Footing.Figure15Differentsystemgroupings.19StaticsNowlet’sexaminetheinternalforcesthatarepresentineachofthethreecasesexaminedabove(Figure16).ExaminationofFigure16(a)showsforcesRlandR2occurringbetweenthecolumnandfooting.Theboundaryofthesystemisstillmaintainedaroundthecolumnandfooting,butbyexaminingtheinteractionthattakesplacebetweenmemberswithinasystem,weinferinternalforces.ForceRlisthereactionofthefootingonthecolumn,whileomR2istheactionofthecolumnonthefooting.FrNewton’sthirdlaw,wecanthensaythatRlandR2areequalandoppositeforces.(a)Relationshipofforcesbetweenthecolumnandfooting.(b)Column.(c)Footing.Figure16Externalandinternalforces.20Internalforcesoccurbetweenbodieswithinasystem,asinFigure16(a).Also,theymayoccurwithinthemembersthemselves,holdingtogethertheparticlesformingtherigidbody,asinFigure16(b)and16(c).ForceR3representstheresistanceofferedbythebuildingmaterial(stone,concrete,orsteel)tokeepthecolumnintact;thisactsinasimilarfashionforthefooting.StaticsTypesofForceSystemsForcesystemsareoftenidentifiedbythetypeortypesofsystemsonwhichtheyact.Theseforcesmaybecollinear,coplanar,orspaceforcesystems.Whenforcesactalongastraightline,theyarecalledcollinear;whentheyarerandomlydistributedinspace,theyarecalledspaceforces.Forcesystemsthatintersectatacommonpointarecalledconcurrent,whileparallelforcesarecalledparallel.Iftheforcesareneitherconcurrentnorparallel,theyfallundertheclassificationofgeneralforcesystems.Concurrentforcesystemscanactonaparticle(point)orarigidbody,whereasparallelandgeneralforcesystemscanactonlyonarigidbodyorasystemofrigidbodies.(SeeFigure17foradiagrammaticrepresentationofthevariousforcesystemarrangements.)OneintelligenthikerobservingthreeotherhikersCollinear—Allforcesactingalongthesamedanglingfromarope.straightline.Figure17(a)Particleorrigidbody.Coplanar—Allforcesactinginthesameplane.Forcesinabuttresssystem.Coplanar,parallel—Allforcesareparallelandactinthesameplane.Abeamsupportedbyaseriesofcolumns.Figure17(b)Rigidbodies.Figure17(c)Rigidbodies.21StaticsLoadsappliedtoarooftruss.Figure17(d)Coplanar,concurrent—Allforcesintersectatacommonpointandlieinthesameplane.Particleorrigidbody.Noncoplanar,parallel—Allforcesareparalleltoeachother,butnotalllieinthesameplane.Columnloadsinaconcretebuilding.Figure17(e)Rigidbodies.Onecomponentofathree-dimensionalspaceframe.Noncoplanar,concurrent—Allforcesintersectatacommonpointbutdonotalllieinthesameplane.Figure17(f)Particleorrigidbodies.Noncoplanar,nonconcurrent—Allforcesareskewed.Arrayofforcesactingsimultaneouslyonahouse.Figure17(g)Rigidbodies.22Statics2VECTORADDITIONCharacteristicsofVectorsAnimportantcharacteristicofvectorsisthattheymustbeaddedaccordingtotheparallelogramlaw.Althoughtheideaoftheparallelogramlawwasknownandusedinsomeformintheearly17thcentury,theproofofitsvaliditywassuppliedmanyyearslaterbySirIsaacNewtonandtheFrenchmathematicianPierreVarignon(1654–1722).Inthecaseofscalarquantities,whereonlymagnitudesareconsidered,theprocessofadditioninvolvesasimplearithmeticalsummation.Vectors,however,havemagnitudeanddirection,thusrequiringaspecialprocedureforcombiningthem.Figure18Cross-sectionthroughagravity-retainingwall.Usingtheparallelogramlaw,wemayaddvectorsgraphicallyorbytrigonometricrelationships.Forexample,twoforcesWandFareactingonaparticle(point),asshowninFigure18;wearetoobtainthevectorsum(resultant).Becausethetwoforcesarenotactingalongthesamelineofaction,asimplearithmeticalsolutionisnotpossible.Thegraphicalmethodoftheparallelogramlawsimplyinvolvestheconstruction,toscale,ofaparallelogramusingforces(vectors)WandFasthelegs.Completetheparallelogram,anddrawinthediagonal.ThediagonalrepresentsthevectoradditionofWandF.AconvenientscaleisusedindrawingWandF,wherebythemagnitudeofRisscaledoffusingthesamescale.Tocompletetherepresentation,theangleθmustbedesignatedfromsomereferenceaxis—inthiscase,thehorizontalaxis(Figure19).AnItalianmathematicianandengineer,GiovanniPoleni(1685–1761),publishedareportin1748onSt.Peter’sdomeusingamethodofillustrationshowninFigure20.Poleni’sthesisoftheabsenceoffrictionisdemonstratedbythewedge-shapedvoussoirswithspheres,whicharearrangedexactlyinaccordancewiththelineofthrust,thussupportingoneanotherinanunstableequilibrium.Inhisreport,PolenireferstoNewtonandhistheoremoftheparallelogramofforcesanddeducesthatthelineofthrustresemblesaninvertedcatenary.Figure19Anotherillustrationoftheparallelogram.Figure20Poleni’suseoftheparallelogramlawindescribingthelinesofforceinanarch.FromGiovanniPoleni,MemorieistorichedellaGranCupoladelTempioVaticano,1748.23StaticsExampleProblems:VectorAddition1Twoforcesareactingonaboltasshown.Determinegraphicallytheresultantofthetwoforcesusingtheparallelogramlawofvectoraddition.1.Drawthe500#and1,200#forcestoscalewiththeirproperdirections.2.Completetheparallelogram.3.Drawthediagonal,startingatthepointoforiginO.4.ScaleoffthemagnitudeofR.5.Scaleofftheangleθfromareferenceaxis.6.Thesense(arrowheaddirection)inthisexamplemovesawayfrompointO.Anothervectoradditionapproach,whichprecededtheparallelogramlawby100yearsorso,isthetriangleruleortip-to-tailmethod(developedthroughproofsbya16thcenturyDutchengineer/mathematician,SimonStevin).Figure21Tip-to-tailmethod.Tip-to-tailsolution.24Tofollowthismethod,constructonlyhalfoftheparallelogram,withthenetresultbeingatriangle.ThesumoftwovectorsAandBmaybefoundbyarrangingtheminatipto-tailsequencewiththetipofAtothetailofBorviceversa.InFigure21(a),twovectorsAandBaretobeaddedbythetip-to-tailmethod.BydrawingthevectorstoscaleandarrangingitsothatthetipofAisattachedtothetailofB,asshowninFigure21(b),theresultantRcanbeobtainedbydrawingalinebeginningatthetailofthefirstvector,A,andendingatthetipofthelastvector,B.Thesequenceofwhichvectorisdrawnfirstisnotimportant.AsshowninFigure21(c),vectorBisdrawnfirst,withthetipofBtouchingthetailofA.TheresultantRobtainedisidenticalinbothcasesformagnitudeandinclinationθ.Again,thesenseoftheresultantmovesfromtheoriginpointOtothetipofthelastvector.NotethatthetriangleshowninFigure21(b)istheupperhalfofaparallelogram,andthetriangleshowninFigure21(c)formsthelowerhalf.Becausetheorderinwhichthevectorsaredrawnisunimportant,whereA+B=B+A,wecanconcludethatthevectoradditioniscommutative.2SolvethesameproblemshowninExampleProblem1,butusethetip-to-tailmethod.StaticsGraphicalAdditionofThreeorMoreVectorsThesumofanynumberofvectorsmaybeobtainedbyapplyingrepeatedlytheparallelogramlaw(ortip-to-tailmethod)tosuccessivepairsofvectorsuntilallofthegivenvectorsarereplacedbyasingleresultantvector.Note:Thegraphicalmethodofvectoradditionrequiresallvectorstobecoplanar.(a)Figure22Parallelogrammethod.(b)(c)AssumethatthreecoplanarforcesA,B,andCareactingatpointO,asshowninFigure22(a),andthattheresultantofallthreeisdesired.InFigure22(b)and22(c),theparallelogramlawisappliedsuccessivelyuntilthefinalresultantforceR¿isobtained.TheadditionofvectorsAandByieldstheintermediateresultantR;RisthenaddedvectoriallytovectorC,resultinginR¿.Asimplersolutionmaybeobtainedbyusingthetip-to-tailmethod,asshowninFigure23.Again,thevectorsaredrawntoscalebutnotnecessarilyinanyparticularsequence.(a)(b)Figure23Illustrationofthetip-to-tailmethod.(c)25StaticsExampleProblems:GraphicalAdditionofThreeorMoreVectors3Twocablessuspendedfromaneyeboltcarry200#and300#loadsasshown.BothforceshavelinesofactionthatintersectatpointO,makingthisaconcurrentforcesystem.Determinetheresultantforcetheeyeboltmustresist.Doagraphicalsolutionusingascaleof1–=100#.Solution:4ThreestructuralmembersA,B,andCofasteeltrussareboltedtoagussetplateasshown.Thelinesofaction(linethroughwhichtheforcepasses)ofallthreemembersintersectatpointO,makingthisaconcurrentforcesystem.Determinegraphically(parallelogramlawortip-totail)theresultantofthethreeforcesonthegussetplate.Useascaleof1mm=400N.Note:Theresultantmustbedenotedbyamagnitudeanddirection.Solution:26Statics5Twoworkersarepullingalargecrateasshown.Iftheresultantforcerequiredtomovethecratealongitsaxislineis120#,determinethetensioneachworkermustexert.Solvegraphically,usingascaleof1–=40#.Solution:Usingtheparallelogramlaw,beginbyconstructingtheresultantforceof120#(horizontallytotheright)toscale.Thesidesoftheparallelogramhaveunknownmagnitudes,butthedirectionsareknown.Closetheparallelogramatthetipoftheresultant(pointm)bydrawinglineA¿paralleltoAandextendingittointersectwithB.ThemagnitudeofBmaynowbedetermined.Similarly,lineB¿maybeconstructedandthemagnitudeofforceAdetermined.Fromscaling,A79#,B53#.ProblemsConstructgraphicalsolutionsusingtheparallelogramlaworthetip-to-tailmethod.1Determinetheresultantofthetwoforcesshown(magnitudeanddirection)actingonthepin.Scale:1–=100#.2Threeforcesareactingontheeyeboltasshown.AllforcesintersectatacommonpointO.Determinetheresultantmagnitudeanddirection.Scale:1mm100N.27Statics3DeterminetheforceFrequiredtocounteractthe600#tensionsothattheresultantforceactsverticallydownthepole.Scale:1–=400#.4ThreeforcesareconcurrentatpointO,andthetensionincableT15,000#withtheslopeasshown.DeterminethemagnitudesnecessaryforT2andT3suchthattheresultantforceof10kactsverticallydowntheaxisofthepole.Scale:1–=2,000#.5Aprecastconcretewallpanelisbeinghoistedintoplaceasshown.Thewallweighs18kNwiththeweightpassingthroughitscenterthroughpointO.DeterminetheforceT2necessaryfortheworkerstoguidethewallintoplace.Scale:1mm100N.28Statics3FORCESYSTEMSResolutionofForcesintoRectangularComponentsAreverseeffectofvectoradditionistheresolutionofavectorintotwoperpendicularcomponents.Componentsofavector(orforce)areusuallyperpendiculartoeachotherandarecalledrectangularcomponents.Thexandyaxesofarectangularcoordinatesystemaremostoftenassumedtobehorizontalandvertical,respectively;however,theymaybechoseninanytwomutuallyperpendiculardirectionsforconvenience(Figure24).AforceFwithadirectionθfromthehorizontalxaxiscanberesolvedintoitsrectangularcomponentsFxandFyasshowninFigure24.BothFxandFyaretrigonometricfunctionsofFandθ,whereFx=Fcosθ;Fy=FsinθIneffect,theforcecomponentsFxandFyformthelegsofaparallelogram,withthediagonalrepresentingtheoriginalforceF.Therefore,byusingthePythagoreantheoremforrighttriangles,F=2F2x+F2yand,FyFytanθ=,or,θ=tan-1abFxFxFigure24Rectangularcomponentsofaforce.ExampleProblems:ResolutionofForcesintoRectangularComponents6Ananchorbracketisacteduponbya1,000#forceatanangleof30°fromthehorizontal.Determinethehorizontalandverticalcomponentsoftheforce.Solution:θ=30°Fx=Fcosθ=Fcos30°Fx=1,000#10.8662=866#Fy=Fsinθ=Fsin30°Fy=1,000#10.502=500#Note:SincethepointofapplicationofforceFisatpointO,thecomponentsFxandFymustalsohavetheirpointsofapplicationatO.29Statics7Aneyeboltonaninclinedceilingsurfacesupportsa100NverticalforceatpointA.ResolvePintoitsxandycomponents,assumingthexaxistobeparallelwiththeinclinedsurface.Solution:Px=Pcos60°=Psin30°Px=100N10.52=50NPy=Psin60°=Pcos30°Py=100N10.8662=86.6NNote:Again,youmustbecarefulinplacingthecomponentforcesPxandPysothattheirpointofapplicationisatA(apulling-typeforce),usingthesameapplicationpointastheoriginalforceP.8Aclotheslinewithamaximumtensionof150#isanchoredtoawallbymeansofaneyescrew.Iftheeyescrewiscapableofcarryingahorizontalpullingforce(withdrawalforce)of40#perinchofpenetration,howmanyinchesLshouldthethreadsbeembeddedintothewall?Solution:Forthisproblem,noticethatthedirectionoftheforceFisgivenintermsofasloperelationship.Thesmall-slopetrianglehasthesameangleθfromthehorizontalastheforceFmakeswiththexaxis.Therefore,wemayconcludethatboththesmall-slopetriangleandthelargetriangle(asshowninthetip-to-taildiagram)aresimilar.30StaticsBysimilartriangles,FyFxF==435‹solvingforthecomponentforces,Fx4=5FFy3sinθ==5Fcosθ=then,orFxF=4544Fx=1F2=1150#2=120#55Fx=Fcosθ,butcosθ=4(perslopetriangle)54Fx=150#ab=120#PCHECKS5Continuingwiththesimilartrianglerelationship,FyF=3533Fy=1F2=1150#2=90#553Fy=Fsinθ=150#ab=90#PCHECKS5Iftheeyescrewiscapableofresistinga40#/in.penetrationinthehorizontaldirection,thelengthofembedmentrequiredmaybecalculatedasFx=120#=140#-in.2*1L2120#=3”embedmentL=40#-in.Awordofcautionconcerningtheequationsforthexandyforcecomponents:Thecomponentsofaforcedependonhowthereferenceangleismeasured,asshowninFigure25.InFigure25(a),thecomponentsFxandFymaybestatedasFx=FcosθFy=FsinθwherethedirectionofforceFisdefinedbytheangleθmeasuredfromthehorizontalxaxis.InthecaseofFigure25(b),thedirectionofFisgivenintermsofanangleφmeasuredfromtheyreferenceaxis.Thisthereforechangesthetrigonometricnotation,whereFx=FsinφFy=FcosφNote:Thereversalofsineandcosinedependsonhowthereferenceangleismeasured.Figure25Forceandcomponents.31StaticsProblemsResolutionofforcesintoxandycomponents.6Determinethexandycomponentsoftheforce,F,shown.7Ifahookcansustainamaximumwithdrawalforceof250Nintheverticaldirection,determinethemaximumtensionTthatcanbeexerted.8Aroofpurlin,supportedbyarafter,mustsupporta300#verticalsnowload.DeterminethecomponentsofP,perpendicularandparalleltotheaxisoftherafter.32StaticsVectorAdditionbytheComponentMethodAswasshowninprevioussections,vectorsmaybeaddedgraphicallybyusingtheparallelogramlaworthemodifiedtip-to-tailmethod.Now,withtheconceptofresolutionofvectorsintotworectangularcomponents,wearereadytobegintheanalyticalapproachtovectoraddition.Thefirststepintheanalyticalapproachinvolvesresolvingeachforceofaforcesystemintoitsrespectivecomponents.Then,theessentialforcecomponentsmaybeaddedalgebraically(asopposedtoagraphicalvectoraddition)toyieldaresultantforce.Forexample,assumewehavethreeforcesA,B,andCactingonaparticleatpointO(Figure26).(a)(b)Figure26Analyticalmethodofvectoraddition.InFigure26(b),eachforceisreplacedbyitsrespectivexandyforcecomponents.AllofthecomponentforcesactingonpointOproducethesameeffectastheoriginalforcesA,B,andC.Thehorizontalandverticalcomponentsmaynowbesummedalgebraically.ItisimportanttonoteherethatalthoughAx,Bx,andCxareactingalongthehorizontalxaxis,theyarenotallactinginthesamedirection.Tokeepthesummationprocesssystematic,itisessentialtoestablishasignconvention(Figure27).Themostcommonlyusedsignconventionforarectangularcoordinatesystemdefinesanyvectoractingtowardtherightasdenotingapositivexdirectionandanyvectoractingupwardasdenotingapositiveydirection.Anythingtotheleftordowndenotesanegativedirection.Figure27Signconventionforforces.InFigure27,aforceFisresolvedintoitsxandycomponents.Forthiscase,theFxcomponentisdirectedtotheright,thereforedenotingapositivexforcecomponent.TheFycomponentispointeddown,representinganegativeyforce.ReturningtotheproblemshowninFigure26,horizontalcomponentswillbesummedalgebraicallysuchthatorRx=-Ax+Bx-CxRx=©FxwhereRdenotesaresultantforce.33StaticsVerticalcomponentsmaybesummedsimilarly,whereorRy=+Ay+By-CyRy=©FyThus,thethreeforceshavebeenreplacedbytworesultantcomponents,RxandRy(Figure28).(a)(b)Figure28FinalresultantRfromRxandRy.Thefinalresultant,orvectorsumofRxandRy,isfoundbythePythagoreantheorem,whereR=21Rx22+1Ry22tanθ=RyRxθ=tan-1a34=©Fy©FxRyRxb;StaticsExampleProblems:VectorAdditionbytheComponentMethod9AnanchordeviceisacteduponbytwoforcesF1andF2asshown.Determinetheresultantforceanalytically.Solution:Step1:F1xF1yF2xF2yResolveeachforceintoitsxandycomponents.====-F1cosα+F1sinα+F2cosβ+F2sinβ1125sinβ=1311212cosβ=13sinα=cosα=F1x=-F1a1221b=-20ka122b=-14.14kF1y=F1a22F2x=F2a1212b=10kab=+9.23k1313F2y=F2aStep2:b=20ka122b=+14.14k55b=10kab=+3.85k1313Resultantalongthehorizontalxaxis.Rx=©Fx=-F1x+F2x=-14.14k+9.23k=-4.91kStep3:Resultantalongtheverticalyaxis.Ry=©Fy=+F1y+F2y=+14.14k+3.85k=+17.99k35StaticsStep4:ResultantofRxandRy.R=2R2x+R2y=2(-4.91)2+(17.99)2=18.65kθ=tan-1aRyRxb=tan-1a17.99b=tan-13.66=74.7°4.9110Ananchoringdeviceissubjectedtothethreeforcesasshown.Determineanalyticallytheresultantforcetheanchormustresist.Solution:Resolveeachforceintoitscomponentparts.-F1x=F1cos25°=125#10.9062=-113#+F1y=F1sin25°=125#10.4232=+53#-F2x=F2a-F2y=F2a122122b=65#ab=65#a122122b=-46#b=-46#+F3x=F3=+100#orbyapplyingtheprincipleoftransmissibility,36StaticsForceF2ismovedalongitslineofactionuntilitbecomesapulling,ratherthanapushing,force.However,theexternaleffectsontheanchordeviceremainthesame.Rx=©Fx=-F1x-F2x+F3xRx=-113#-46#+100#=-59#;Ry=©Fy=+F1y-F2yRy=+53#-46#=+7#cR=21Rx22+1Ry22=21-5922+1722=59.5#tanθ=RyRx=7=0.11959
θ=tan-11.1192=6.8°GraphicalCheck:Scale:1–=40#37Statics11AblockofweightW=500NissupportedbyacableCD,whichinturnissuspendedbycablesACandBC.DeterminetherequiredtensionforcesTCAandTCBsothattheresultantforceatpointCequalszero.Solution:Step1:BecauseallcablesareconcurrentatpointC,isolatepointCandshowallforces.DirectionsforforcesTCD,TCA,andTCBareknown,buttheirmagnitudesareunknown.TCDisequaltotheweightW,whichactsverticallydownward.Step2:Resolveeachforceintoitsxandycomponents.3T5CA4=TCA5TCAx=TCAyTCBx=TCBcos30°=0.866TCBRCBy=TCBsin30°=0.5TCBTCD=W=500NStep3:Writetheresultantcomponentforceequations.Rx=©Fx=TCAx+TCBx=0WesetRx=0tocomplywiththeconditionthatnoresultantforcedevelopsatpointC;thatis,allforcesarebalancedatthispoint.3‹Rx=-TCA+0.866TCB=053T=0.866TCB5CA(1)then,Ry=©Fy=+TCAy+TCBy-500N=04‹Ry=TCA+.5TCB-500N=054T+0.5TCB=500N5CA38(2)StaticsFromwritingtheresultantequationsTCAandTCB,weobtaintwoequations,(1)and(2),containingtwounknowns.Solvingthetwoequationssimultaneously,3T=0.866TCB5CA(1)4T+0.5TCB=500N5CA(2)Fromequation(1),TCA=510.866TCB2=1.44TCB3Substitutingintoequation(2),411.44TCB2+0.5TCB=500N5‹1.15TCB+0.5TCB=500NSolvingforTCB,1.65TCB=500N500NTCB==303N1.65SubstitutingthevalueofTCBbackintoequation(1)or(2),TCA=436.4NGraphicalCheck:BecauseTCDhasaknownmagnitudeanddirection,itwillbeusedasthestarting(orbase)force.Step1:Drawtheforcepolygonusingthetip-to-tailmethod.Step2:DrawforceTCDfirst,toscale.Step3:DrawthelinesofactionforTCAandTCB;theorderdoesnotmatter.ForcesTCAandTCBhaveknowndirectionsbutunknownmagnitudes;therefore,onlytheirlinesofactionaredrawninitially.WeknowthatbecauseR=0,thetipofthelastforcemustendatthetail(theorigin)ofthefirstforce—inthiscase,TCD.Step4:TheintersectionofthetwolinesofactiondeterminesthelimitsforTCBandTCA.Step5:ScaleoffthemagnitudesforTCBandTCA.Step6:TCA=436.4NTCB=303NCHECKSScale:1mm=5N39StaticsProblemsAnalyticalsolutionsusingforgraphically.cecomponents.Check9Threemembersofatrussframeintoasteelgussetplateasshown.AllforcesareconcurrentatpointO.Determinetheresultantofthethreeforcesthatmustbecarriedbythegussetplate.10TwocableswithknowntensionsasshownareattachedtoapoleatpointA.Determinetheresultantforcetowhichthepoleissubjected.Scale:1mm=10N.Note:Thepoleandcablesarecoplanar.11Alaborerishoistingaweight,W=200#,bypullingonaropeasshown.DeterminetheforceFrequiredtoholdtheweightinthepositionshowniftheresultantofFandWactsalongtheaxisoftheboom.40Statics12Oneendofatimberrooftrussissupportedonabrickwallbutnotsecurelyfastened.Thereactionofthebrickwallcanthereforebeonlyvertical.Assumingthatthemaximumcapacityofeithertheinclinedorhorizontalmemberis7kN,determinethemaximummagnitudesofFlandF2sothattheirresultantisverticalthroughthebrickwall.13Theresultantofthreetensionsintheguywiresanchoredatthetopofthetowerisvertical.FindtheunknownbutequaltensionsTinthetwowires.Allthreewiresandthetowerareinthesameverticalplane.41StaticsMomentofaForceThetendencyofaforcetoproducerotationofabodyaboutsomereferenceaxisorpointiscalledthemomentofaforce(seeFigures29aand29b).Quantitatively,themomentMofaforceFaboutapointAisdefinedastheproductofthemagnitudeoftheforceFandperpendiculardistancedfromAtothelineofactionofF.Inequationform,MA=F*d(a)ThesubscriptAdenotesthepointaboutwhichthemomentistaken.(b)Figure29Momentofaforce.Assume,asshowninFigure29(a),thatapersoniscarryingaweightofmagnitudeFatadistanced1fromanarbitrarypointAontheperson’sshoulder.ThepointAhasnosignificanceexcepttoestablishsomereferencepointaboutwhichthemomentscanbemeasured.InFigure29(b)aschematicisshownwiththeforceFappliedonabeamatadistanced1frompointA.ThisisanequivalentrepresentationofthepictorialsketchinFigure29(a),wherethemomentofpointAisMA=F*d1(a)Ifthepersonnowextendshisarmsothattheweightisatdistanced2frompointA,asshowninFigure30(b),theamountofphysicalenergyneededtocarrytheweightisincreased.OnereasonforthisistheincreasedmomentaboutpointAduetotheincreaseddistanced2.ThemomentisnowequaltothatshowninFigure30(a):MA=F*d2(b)Figure30Momentofaforcewithanincreasedmomentarm.42StaticsInmeasuringthedistanced(oftenreferredtoasthemomentarm)betweentheappliedforceandthereferencepoint,itisimportanttonotethatthedistancemustbetheperpendicularmeasurementtothelineofactionoftheforce(Figure31).Amomentofaforceisavectorquantity.Theforceproducingtherotationhasamagnitudeanddirection;therefore,themomentproducedhasamagnitudeandadirection.Unitsusedtodescribethemagnitudeofamomentareexpressedaspound-inch(#-in.,lb.-in.),pound-foot(#-ft.),kip-inch(k-in.),orkip-foot(k-ft.).Thecorrespondingmetric(SI)unitsarenewton-meter(N-m)orkilonewton-meter(kN-m).Directionofamomentisindicatedbythetypeofrotationdeveloped,eitherclockwiserotationorcounterclockwiserotation(Figure32).Figure31Perpendicularmomentarm.Indiscussingforcesinaprevioussection,weestablishedasignconventionwhereforcesactingtotherightorupwardwereconsideredtobepositiveandthosedirectedtotheleftordownwardwereconsideredtobenegative.Likewise,asignconventionshouldbeestablishedformoments.Becauserotationiseitherclockwiseorcounterclockwise,wemayarbitrarilyassignaplus(+)tothecounterclockwiserotationandaminus(-)totheclockwiserotation.Itisperfectlypermissibletoreversethesignconventionifdesired;however,usethesameconventionthroughoutanentireproblem.Aconsistentsignconventionreducesthechancesoferror.Momentscauseabodytohavethetendencytorotate.Ifasystemtriestoresistthisrotationaltendency,bendingortorsionresults.Forexample,ifweexamineacantileverbeamwithoneendsecurelyfixedtoasupport,asinFigure33(a),thebeamitselfgeneratesaresistanceeffecttorotation.Inresistingtherotation,bendingoccurs,whichresultsinadeflection¢,asinFigure33(b).(a)Unloaded.Figure32Sailboatwinch-rotationaboutanaxis.(b)Loaded.Figure33Momentonacantileverbeam.43StaticsNext,considerasituationwheretorsion(twisting)occursbecauseofthesystemtryingtoresistrotationaboutitslongitudinalaxis(Figure34).(a)(b)Figure34Anexampleoftorsiononacantileverbeam.AsshowninFigure34,asteelchannelsectionundereccentricloadingissubjectedtoarotationaleffectcalledtorsion.Momentsaboutacircularshaftorrodareusuallyreferredtoastorque.IfinFigure34(a)thefixedsupportwerereplacedbyahingeorpin,nodistortionofthebeamwouldoccur.Instead,simplerotationatthepinorhingewouldresult.ExampleProblems:MomentofaForce12WhatisthemomentoftheforceFaboutpointA?Solution:TheperpendiculardistancebetweenpointA(ontheheadofthebolt)andthelineofactionofforceFis15inches.Therefore,MA=(-F*d)=(-25#*15–)=-375#-in.Note:Themagnitudeisgiveninpound-inches,andthedirectionisclockwise.44Statics13WhatisMA,withthewrenchinclinedata3in4slope?d;=4115–2=12–5d;istheperpendiculardistancefromAtolineofactionofF.MA=1-F2(d)=-25#112”2=-300#-in.14Theequivalentforcesduetowaterpressureandtheself-weightofthedamareshown.Determinetheresultantmomentatthetoeofthedam(pointA).Isthedamabletoresisttheappliedwaterpressure?Theweightofthedamis36kN.Solution:Theoverturningeffectduetothewaterisclockwise,whiletheweightofthedamhasacounterclockwiserotationaltendencyaboutthe“toe”atA.Therefore,theresultantmomentaboutpointAisMA1=MA=-FW11m2=-20kN11m2=-20kN-m(overturning)MA2=MA=+W(2m)=+36kN(2m)=+72kN-m(stabilizing)becauseMA27MA1Thedamisstable,and,thus,willnotoverturn.45StaticsProblems14Aboxweighing25pounds(assumedconcentratedatitscenterofgravity)isbeingpulledbyahorizontalforceFequalto20pounds.WhatisthemomentaboutpointA?Doestheboxtipover?15Alargewoodbeamweighing800Nissupportedbytwopostsasshown.Ifanunthinkingmanweighing700Nweretowalkontheoverhangportionofthebeam,howfarcanhegofrompointAbeforethebeamtipsover?(Assumethebeamisrestingonthetwosupportswithnophysicalconnection.)16Calculatethemomentthroughthecenterofthepipeduetotheforceexertedbythewrench.17Forthewheelbarrowshown,findthemomentofthe100#weightaboutthecenterofthewheel.Also,determinetheforcePrequiredtoresistthismoment.46Statics18A200#stoneisbeingliftedoffthegroundbyaleveringbar.DeterminethepushPrequiredtokeepthestoneinthepositionshown.Varignon’sTheoremTheFrenchmathematicianPierreVarignondevelopedaveryimportanttheoremofstatics.Itstatesthatthemomentofaforceaboutapoint(axis)isequaltothealgebraicsumofthemomentsofitscomponentsaboutthesamepoint(axis).Thismaybebestillustratedbyanexample(Figure35).(a)Figure35Momentofaforce—Varignon’stheorem.(b)IfweexamineanupturnedcantileveredbeamsubjectedtoaninclinedforceF,asshowninFigure35(a),themomentMAisobtainedbymultiplyingtheappliedforceFbytheperpendiculardistanced(fromAtothelineofactionoftheforce).Findingthedistancedisoftenveryinvolved;therefore,theuseofVarignon’stheorembecomesveryconvenient.TheforceFisresolvedintoitsxandycomponents,asshowninFigure35(b).Theforceisbrokendownintoahorizontalcomponentandaverticalcomponent,andthemomentarmdistancesdxanddyoftherespectivecomponentsareobtained.TheresultingmomentMAisobtainedbyalgebraicallysummingthemomentsaboutpointAgeneratedbyeachofthetwocomponentforces.Inbothcases,themomentsareidenticalinmagnitudeandindirection.47StaticsAproofofVarignon’stheoremmaybeillustratedasinFigure36:F(d)=Fy1dcosθ2+Fx1dsinθ2SubstitutingforFxandFy,F(d)=Fcosθ1dcosθ2+Fsinθ1dsinθ2F(d)=Fdcos2θ+Fdsin2θ=Fd1cos2θ+sin2θ2Butfromtheknowntrigonometricidentity,sin2θ+cos2θ=1F1d2=F1d2‹CHECKSFigure36Varignon’stheorem.ExampleProblems:Varignon’sTheorem15DeterminethemomentMAatthebaseofthebuttressduetotheappliedthrustforceF.UseVarignon’stheorem.ForceFisata5:12slope.Solution:FromVarignon’stheorem,resolveFintoFxandFy;then,MA=+5k115¿2-12k124¿2=+75k-ft.-288k-ft.=-213k-ft.48Statics16A1.5kNforcefromapistonactsontheendofa300mmlever.DeterminethemomentoftheforceabouttheaxlethroughreferencepointO.Solution:Inthiscase,findingtheperpendiculardistancedfortheforceFcanbequitecomplicated;therefore,Varignon’stheoremwillbeutilized.Fy1=Fcos20°Fy1=1.5kN(0.94)=1.41kNFx1=Fsin20°Fx1=1.5kN(0.342)=0.51kNNote:300mm=0.3m.+Mo=+Fx1(0.3m)+Fy1(0)Mo=+(+0.51kN)(0.3m)+0=+0.153kN-m49Statics17Determinethemomentofthe390#forceaboutAby:a.ResolvingtheforceintoxandycomponentsactingatB.b.ResolvingtheforceintoxandycomponentsactingatC.c.DeterminingthemomentduetotheweightWaboutpointA.Solution:a.IsolatepointB,andresolveF=390#intoitsxandycomponents.Fx=1212F=(390#)=360#1313Fy=55F=(390#)=150#1313MA=+Fx110¿2-Fy17.5¿2MA=360#(10’)-150#17.5’2MA=+2,475#-ft.b.Byapplyingtheprincipleoftransmissibility,wecanmoveforceFdowntopointCwithoutreallyalteringtheexternaleffectsonthesystem.Fx=12F=360#13Fy=5F=150#13MA=Fx10¿2+Fy116.5¿2MA=0+150#(16.5¿)=2,475#-ft.c.TheweightW=660#isassumedatthecenterofgravityoftheboom.ThemomentduetoWaboutAcanbeexpressedasMA=-W13.75¿2=-660#13.75¿2=-2,475#-ft.Note:ThemomentsfromforcesFandWarebalancedand,thus,keepboomABfromrotating.Thisconditionwillbereferredtolaterasmomentequilibrium.50StaticsProblems19Thefigureshowstheforcesexertedbywindoneachfloorlevelofasix-storysteelframebuilding.DeterminetheresultantoverturningmomentatthebaseofthebuildingatA.20Determinethemomentofthe1,300#forceappliedattrussjointDaboutpointsBandC.UseVarignon’stheorem.21Araingutterissubjectedtoa30#forceatCasshown.DeterminethemomentdevelopedaboutAandB.22A.Computethemomentofthe1.5kNforceaboutpoint23DeterminetheweightWthatcanbesupportedbytheboomifthemaximumforcethatthecableTcanexertis2,000#.AssumethatnoresultantrotationwilloccuratpointC.(Inotherwords,MC=0.)51StaticsCoupleandMomentofaCoupleAcoupleisdefinedastwoforceshavingthesamemagnitude,parallellinesofaction,butoppositesense(arrowheaddirection).Coupleshavepurerotationaleffectsonabodywithnocapacitytotranslatethebodyintheverticalorhorizontaldirection,becausethesumoftheirhorizontalandverticalcomponentsiszero.Let’sexaminearigidbodyinthex-yplaneacteduponbytwoequal,opposite,andparallelforcesF1andF2(Figure37).Figure37Forcecouplesystem.AssumeapointAontherigidbodyaboutwhichthemomentwillbecalculated.DistancexrepresentstheperpendicularmeasurementfromreferencepointAtotheappliedforceF,anddistheperpendiculardistancebetweenthelinesofactionofF1andF2.MA=+F11x2-F21x+d2whereF1=F2andbecauseF1andF2formacouplesystem,MA=+Fx-Fx-Fd‹MA=-FdThefinalmomentMiscalledthemomentofthecouple.NotethatMisindependentofthelocationofthereferencepointA.MwillhavethesamemagnitudeandthesamerotationalsenseregardlessofthelocationofA(Figure38).MA=-F1d12-F1d22=-F1d1+d22butd=d1+d2MA=-FdFigure38MomentofacoupleaboutA.52Itcanbeconcluded,therefore,thatthemomentMofacoupleisconstant.Itsmagnitudeisequaltotheproduct1F2*1d2ofeitherF(F1orF2)andtheperpendiculardistancedbetweentheirlinesofaction.ThesenseofM(clockwiseorcounterclockwise)isdeterminedbydirectobservation.StaticsExampleProblems:CoupleandMomentofaCouple18Acantileveredbeamissubjectedtotwoequalandoppositeforces.DeterminetheresultantmomentMAatthebeamsupport.Solution:Bydefinition,MA=F1d2MA=+2k15¿2=+10k-ft.Check:MA=-2k110¿2+2k115¿2MA=-20k-ft.+30k-ft.=+10k-ft.Let’sexamineanothercasewherethebeamisextendedanotherfivefeet.MA=+2k15¿2=+10k-ft.Check:MA=-2k(15¿)+2k(20¿)MA=-30k-ft.+40k-ft.=+10k-ft.Weagainnoticethatthemomentofacoupleisaconstantforagivenrigidbody.Thesenseofthemomentisobtainedbydirectobservation(usingyourintuitionandjudgment).53Statics19Aninclinedtrussissubjectedtotwoforcesasshown.DeterminethemomentsatAandBduetothetwoforces.Solution:ByusingVarignon’stheorem,the2,500#inclinedforcecanberesolvedintohorizontalandverticalcomponents.Thetwo1,500#forcesformacouplesystem.Makinguseoftheconceptofcouples,MA=-1,500#18¿2+2,000#14¿2=-12,000#-ft.+8,000#-ft.MA=-4,000#-ft.MB=-1,500#(8¿)+2,000#(10¿)MB=-12,000#-ft.+20,000#-ft.=+8,000#-ft.54StaticsResolutionofaForceintoaForceandCoupleActingatAnotherPointIntheanalysisofsometypesofproblems,itmaybeusefultochangethelocationofanappliedforcetoamoreconvenientpointontherigidbody.Inaprevioussection,wediscussedthepossibilityofmovingaforceFalongitslineofaction(principleoftransmissibility)withoutchangingtheexternaleffectsonthebody,asshowninFigure39.However,wecannotmoveaforceawayfromtheoriginallineofactionwithoutmodifyingtheexternaleffectsontherigidbody,asshowninFigure40.F1=F2(samelineofaction)Figure39Forcemovedalongitslineofaction.¢16¢2Figure40Forcemovedtoanewlineofaction.ExaminationofFigure40showsthatiftheappliedforceFischangedfrompointAtopointBonthecantileveredbeam,differingdeflectionsatthefreeendresult.Thedeflection¢2(FappliedatpointB)isconsiderablylargerthan¢1(FappliedatpointA).Let’sapplyaforceFatpointBasshowninFigure41(a).TheobjectiveistohaveFmovedtopointAwithoutchangingtheeffectsontherigidbody.TwoforcesFandF¿areappliedatAinFigure41(b)withalineofactionparalleltothatoftheoriginalforceatB.TheadditionoftheequalandoppositeforcesatAdoesnotchangetheeffectontherigidbody.WeobservethattheforcesFatBandF¿atAareequalandoppositeforceswithparallellinesofaction,thusformingacouplesystem.Bydefinition,themomentduetothecoupleisequalto(F)(d)andisaconstantvalueanywhereontherigidbody.ThecoupleMcanthenbeplacedatanyconvenientlocationwiththeremainingforceFatA,asshowninFigure41(c).(a)(b)Theprecedingexamplemaythenbesummarizedasfollows:AnyforceFactingonarigidbodymaybemovedtoanygivenpointA(withaparallellineofaction)providedthatacoupleMisadded.ThemomentMofthecoupleisequaltoFtimestheperpendiculardistancebetweentheoriginallineofactionandthenewlocationA.(c)Figure41Movingaforcetoanotherparallellineofaction.55StaticsExampleProblems:ResolutionofaForceintoaForceandCoupleActingatAnotherPoint20Abentconcretecolumnissubjectedtoadownwardforceof10k.Todesignthecolumn,itisnecessarytohavethecompressiveforceappliedthroughtheaxisofthecolumn.ShowtheequivalentforcesystemwhentheforceismovedfromAtoB.ApplyanequalandoppositepairofforcesatB.Solution:The10kverticalforceupwardatBandthe10kforceatAconstituteacouplesystem.Coupleshaveonlyrotationaltendenciesandcanbeappliedanywhereontherigidbody.Mcouple=-10k15¿2=-50k-ft.Anequivalentrepresentationwiththe10kforceactingatBisaccompaniedbyanappliedmomentof50k-ft.clockwise.Theeffectonthecolumniscompressionplusbending.56Statics21Amajorprecastconcretecolumnsupportsbeamloadsfromtheroofandsecondfloorasshown.Beamsaresupportedbyseatsprojectingfromthecolumns.Loadsfromthebeamsareassumedtobeappliedonefootfromthecolumnaxis.Determinetheequivalentcolumnloadconditionwhenallbeamloadsareshownactingthroughthecolumnaxis.Solution:The12kforceproducesa+12k-ft.momentwhenmovedtothecolumnaxis,whilethe10kforcecounterswitha-10k-ft.clockwisemoment.Theresultantmomentequals+2k-ft.Mroof=+12k-ft.-10k-ft.=+2k-ft.Atthesecond-floorlevel,M2=+20k-ft.-15k-ft.=+5k-ft.TheresultanteffectduetothecolumnloadsatthebaseAequalsFresultant=+22k+35k=+57k(compression)Mresultant=+2k-ft.+5k-ft.=+7k-ft.57StaticsProblems24DeterminetheresultantmomentatsupportpointsAandBduetotheforcesactingonthetrussasshown.Assumethatthe10kforceisactingperpendiculartothetrussslope.25Aladdersupportsapainterweighing150#atmidheight.TheladderissupportedatpointsAandB,developingreactionsasshowninthefree-bodydiagram(seeSection5).AssumingthatreactionforcesRAxandRBxdevelopmagnitudesof25#eachandRAy=150#,determineMA,MB,andMC.26Replacethe90kNbeamloadbyanequivalentforcecouplesystemthroughthecolumncenterline.27An85#forceisappliedtothebentplateasshown.Determineanequivalentforce-couplesystem(a)atAand(b)atB.58StaticsResultantofTwoParallelForcesSupposewewishtorepresentthetwoforcesAandBshownonthegirderinFigure42(a)withasingleresultantforceR,whichproducesanequivalenteffectastheoriginalforces.TheequivalentresultantRmustproducethesametranslationaltendencyasforcesAandBaswellasthesamerotationaleffect,asshowninFigure42(b).Figure42(a)Twoparallelforcesactingonagirder.Becauseforcesbydefinitionhavemagnitude,direction,sense,andapointofapplication,itisnecessarytoestablishtheexactlocationoftheresultantRfromsomegivenreferencepoint.OnlyasinglelocationRwillproduceanequivalenteffectasthegirderwithforcesAandB.Figure42(b)EquivalentresultantforceRforAandB.ThemagnitudeoftheresultantRoftheparallelforcesAandBequalsthealgebraicsummationofAandB,whereR=A+B.Directionoftheforcesmustbeaccountedforbyusingaconvenientsignnotation,suchaspositiveforupwardactingforcesandnegativefordownward-actingforces.LocationoftheresultantRisobtainedbytheprincipleofmoments.59StaticsExampleProblem:ResultantofTwoParallelForces22DeterminethesingleresultantR(magnitudeandlocation)thatwouldproduceanequivalenteffectastheforcesshownonthecombinedfooting.Solution:Magnitudeofresultant:R=-20k-60k=-80kTofindthelocationofR,pickaconvenientreferencepointandcalculatemoments.MA=-60k(12¿)=-720k-ft.ThemomentaboutpointAduetoRmustbeequaltotheMAoftheoriginalforcesystemtomaintainequivalence.‹MA=-R1x2-720k-ft.=-R1x2,butR=80k‹x=-720k-ft.=9¿-80kRmustbelocatedninefeettotherightofpointA.60Statics4EQUILIBRIUMEQUATIONS:TWO-DIMENSIONALEquilibriumEquilibriumrefers,essentially,toastateofrestorbalance.RecallNewton’sfirstlaw,whichstates:Anybodyatrestwillremainatrestandanybodyinmotionwillmoveuniformlyinstraightlinesunlessacteduponbyaforce.Theconceptofabodyorparticleatrestunlessacteduponbysomeforceindicatesaninitialstateofstaticequilibrium,wherebytheneteffectofallforcesonthebodyorparticleiszero.Equilibriumornonmotionissimplyaspecialcaseofmotion(Figures44and45).ThemathematicalrequirementnecessarytoestablishaconditionofequilibriumcanbestatedasRx=©Fx=0Ry=©Fy=0Mi=©M=0;wherei=anypointVarioustypesofproblemsrequiretheselectionofonlyone,andothersmaybeall,oftheequationsofequilibrium.However,foranyoneparticulartypeofproblem,theminimumnumberofequationsofequilibriumnecessarytojustifyastateofbalanceisalsothemaximumnumberofequationsofequilibriumpermitted.Becausevariousforcesystemsrequiredifferingtypesandnumbersofequationsofequilibrium,eachwillbediscussedseparately.Figure44Exampleofequilibrium.Figure43LeonardodaVinci(1452–1519).AlthoughmostpopularlyknownasthepainterofTheLastSupper,theMonaLisa,andhisself-portrait,Leonardowasalsoaninventiveengineerwhoconceivedofdevicesandmachinesthatwerewayaheadofhistime.Hewasthefirsttosolvetheproblemofdefiningforceasavector,heconceptualizedtheideaofforceparallelograms,andherealizedtheneedfordeterminingthephysicalpropertiesofbuildingmaterials.Heisreputedtohaveconstructedthefirstelevator,fortheMilanCathedral.Leonardo’skeensenseofobservationandamazinginsightledhimtothenotionoftheprincipleofinertiaandprecededGalileo(byacentury)inunderstandingthatfallingbodiesaccelerateastheyfall.Aswasthetradition,keencompetitionexistedamongthe“artistarchitect-engineers”ofthedayintheireffortstoattachthemselvestothosewhowouldmostgenerouslysupportthem.Unfortunately,competitionwassofiercethatLeonardofeltitnecessarytoguardhisideaswithgreatsecrecy,writingmuchofhisnotebooksincode.Becauseofthis,andbecausesomanyofLeonardo’sachievementswerevisionaryratherthanactual,hisinfluenceonprofessionaldevelopmenthasgenerallybeenconsideredtobeminimal.Itisunknownwhetherachannelexistedthroughwhichsomeofhisideasmayhaveservedtoinspirehissuccessors,mostnotablyGalileoGalilei.Leonardofiguresprominentlyinthissection’sdiscussionofphysicallaws.Figure45Exampleofnonequilibriumorunbalance.61StaticsCollinearForceSystemAcollinearforcesysteminvolvestheactionofforcesalongthesamelineofaction.Thereisnorestrictiononthedirectionoronthemagnitudeofeachforceaslongasallforcesactalongthesameline.Figure46Tensionforcedevelopedinthewebtocarrytheweightofthespider.InFigure46,aspiderisshownsuspendedfromitsweb.Assumingthatthespideriscurrentlyinastationaryposition,astateofequilibriumexists;therefore,the©Fy=0.ThetensiondevelopedinthewebmustbeequaltotheweightWofthespiderforequilibrium.Anotherexampleofacollinearforcesystemisatug-ofwarinadeadlockedsituationinwhichnomovementistakingplace,asshowninFigure47(a).Ifweassumethattheforceexertedbyeachofthefourparticipantsisalongtheaxisofthehorizontalportionoftherope,asshowninFigure47(b),thenallforcesarecollinear.Inequationform,©Fx=0-F1-F2+F3+F4=0Figure47(a)Tug-of-war(deadlocked).Figure47(b)Detailofrope—collinearforces.ConcurrentForceSystemEquilibriumofaparticleIntheprecedingsection,wediscussedthegraphicalaswellasanalyticalmethodsfordeterminingtheresultantofseveralforcesactingonaparticle.Inmanyproblems,thereexiststheconditioninwhichtheresultantofseveralconcurrentforcesactingonabodyorparticleiszero.Forthesecases,wesaythatthebodyorparticleisinequilibrium.Thedefinitionofthisconditionmaybestatedasfollows:Figure48ConcurrentforcesystematC.Whentheresultantofallconcurrentforcesactingonaparticleiszero,theparticleisinastateofequilibrium.Anexampleofacoplanar,concurrentforcesystemisaweightsuspendedfromtwocables,asshowninFigure48.CableforcesAC,BC,andDCintersectatacommonpointC.62StaticsUsingtheconcurrentpointCastheorigin,aforcediagram(Figure49)oftheforcesatpointCisdrawn.WefoundinSection2thatbyresolvingeachforce(foraseriesofconcurrentforces)intotheprimaryxandycomponents,wecanalgebraicallydeterminetheresultantRxandRyforthesystem.Tojustifyaconditionofequilibriuminacoplanar(two-dimensional),concurrentforcesystem,twoequationsofequilibriumarerequired:Rx=©Fx=0Ry=©Fy=0Thesetwoconditionsmustbesatisfiedbeforeequilibriumisestablished.Notranslationineitherthexorydirectionispermitted.Figure49ForcediagramofconcurrentpointC.Equilibriumofcollinearandcoplanar–concurrentforcesystemsarediscussedlater,undertheheadingEquilibriumofaParticle.Nonconcurrent,CoplanarForceSystemEquilibriumofarigidbodyWewillnowconsidertheequilibriumofarigidbody(arigidbodybeingassumedasasystemconsistingofaninfinitenumberofparticles,suchasbeams,trusses,columns,etc.)underaforcesystemthatconsistsofforcesaswellascouples.Inhisnotes,LeonardodaVinci(1452–1519),asshownFigure43,includednotonlysketchesofinnumerablemachinesandmechanicaldevicesbutalsomanyillustratedtheoreticalrelationshipstoderiveorexplainphysicallaws.Hedealtwiththecenterofgravity,theprincipleoftheinclinedplane,andtheessenceofforce.OnepartofdaVinci’sstudiesincludedtheconceptofstaticequilibrium,asshowninFigure50.Arigidbodyissaidtobeinequilibriumwhentheexternalforcesactingonitformasystemofforcesequivalenttozero.Failuretoprovideequilibriumforasystemmayresultindisastrousconsequences,asshowninFigure51.Mathematically,itmaybestatedas©Fx=0©Fy=0©Mi=0wherei=anypointontherigidbodyFigure50StudiesofstaticequilibriumbyLeonardodaVinci.63StaticsTheseequationsarenecessaryandsufficienttojustifyastateofequilibrium.Becauseonlythreeequationsmaybewrittenforthecoplanarsystem,nomorethanthreeunknownscanbesolved.Alternatesetsofequilibriumequationsmaybewrittenforarigidbody;however,itisrequiredthattherealwaysbethreeequationsofequilibrium.Oneforceequationwithtwomomentequationsorthreemomentequationsrepresentalternatesetsthatarevalid.©Fx=0;©Mi=0;©Mj=0orFigure51Anexampleofnonequilibrium—TacomaNarrowsbridgebeforecollapse.CourtesyoftheUniversityofWashingtonLibraries,SpecialCollections,UW21413.©Fy=0;©Mi=0;©Mj=0or©Mi=0;©Mj=0;©Mk=0Free-BodyDiagramsAnessentialstepinsolvingequilibriumprvolvesthedrawingoffree-bodydiagrams.oblemsin-Thefree-bodydiagram,orFBD,istheessentialkeytomodernmechanics.EverythinginmechanicsisreducedtoforcesinanFBD.Thismethodofsimplificationisveryefficientinreducingapparentlycomplexmechanismsintoconciseforcesystems.What,then,isanFBD?AnFBDisasimplifiedrepresentationofaparticleorrigidbodythatisisolatedfromitssurroundingsandonwhichallappliedforcesandreactionsareshown.Allforcesactingonaparticleorrigidbodymustbeconsideredwhenconstructingthefree-body,andequallyimportant,anyforcenotdirectlyappliedonthebodymustbeexcluded.Acleardecisionmustbemaderegardingthechoiceofthefree-bodytobeused.Forcesthatarenormallyconsideredtobeactingonarigidbodyareasfollows:■■■■■■64Externallyappliedforces.Weightoftherigidbody.Reactionforcesorconstraints.Externallyappliedmoments.Momentreactionsorconstraints.Forcesdevelopedwithinasectionedmember.StaticsFree-BodyDiagramsofParticlesTheFBDofaparticleisrelativelysimple,becauseitonlyshowsconcurrentforcesemanatingfromapoint.Figure52(a)and52(b)giveexamplesofsuchFBDs.Figure52(a)Beambeinghoistedbyacranecable.FBDofconcurrentpointC.Figure52(b)Signsuspendedfromastrutandcable.FBDofconcurrentpointB.65StaticsExampleProblems:EquilibriumofaParticle23TwocablesareusedtosupportaweightW=200#suspendedatC.Usingbothananalyticalaswellasagraphicalmethod,determinethetensiondevelopedincablesCAandCB.AnalyticalSolution:a.DrawanFBDoftheconcurrentpointC.b.Resolveallangularorslopedforcesintotheirrespectivexandycomponents.CAxCAyCBxCBy====-CAcos60°+CAsin60°+CBcos30°+CBsin30°c.FortheparticleatCtobeinequilibrium,©Fx=0and©Fy=0[©Fx=0]-CAx+CBx=0orbysubstituting,-CAcos60°+CBcos30°=0AnalyticalMethod—FBDofConcurrentPointC.(1)[©Fy=0]+CAy+CBy-W=0bysubstituting,+CAsin60°+CBsin30°-200#=0(2)d.Solveequations(1)and(2)simultaneouslytodeterminethedesiredcabletensions.-CA10.52+CB10.8662=0+CA(0.866)+CB(0.5)=200#(1)(2)Rewritingequation(1),CA=+10.8662CB0.5=1.73CBSubstitutingintoequation(2),1.7310.8662CB+0.5CB=200#2CB=+200#;‹CB=+100#CA=1.731100#2=+173#66(1)StaticsPerhapsamoreconvenientmethodofaccountingforforcecomponentsbeforewritingtheequationsofequilibriumistoconstructatable.ForceCACBW=200#FxFy+CBcos30°=+0.866CB0+CBsin60°=+0.5CB-200#-CAcos60°=-0.5CA+CAsin60°=+0.866CAThetwoequationsofequilibriumarethenwrittenbysummingverticallyalloftheforceslistedundertheFxcolumn,andsimilarlyfortheFycolumn.GraphicalSolution(Scale:1–=200#):Inthegraphicalsolution,eitherthetip-to-tailortheparallelogrammethodmaybeemployed.Tip-to-TailMethod:a.BeginthesolutionbyestablishingareferenceoriginpointOonthex-ycoordinateaxis.b.DrawtheweightW=200#toscaleinthegivendownwarddirection.c.TothetipofthefirstforceW,placethetailofthesecondforceCB.DrawCBata30°inclinationfromthehorizontal.d.BecausethemagnitudeofCBisstillunknown,weareunabletoterminatetheforce.Onlythelineofactionoftheforceisknown.e.EquilibriumisestablishedinthegraphicalsolutionwhenthetipofthelastforceclosesonthetailofthefirstforceW.Therefore,thetipofforceCAmustcloseattheoriginpointO.ConstructthelineofactionofforceCAatanangleof60°fromthehorizontal.TheintersectionoflinesCAandCBdefinesthelimitsofeachforce.ThemagnitudesofCBandCAcannowbescaledoff.ParallelogramMethod:a.DrawtheknownforceWtoscale,originatingatreferencepointO.b.ConstructaparallelogramusingtheweightWtorepresenttheresultantforceordiagonaloftheparallelogram.Thelinesofactionofeachcable,CBandCA,aredrawnfromboththetipandthetailofforceW.WherethelinesofCBandCAintersect(thecornersoftheparallelogram),thelimitofeachforceisestablished.67Statics24TwocablesaretiedtogetheratCandloadedasshown.AssumingthatthemaximumpermissibletensioninCAandCBis3kN(thesafecapacityofthecable),determinethemaximumWthatcanbesafelysupported.ForceCACBWFx5-CA134+CB50Fy12CA133+CB5-W+45CA+CB=013545213*CB=CBCA=5525[©Fx=0]-(1)‹CA=2.08CBThisrelationshipiscrucial,becauseittellsusthatforthegivenarrangementofcables,CAwillcarrymorethantwicetheloadinCBforequilibriumtoexist.Becausethemaximumcabletensionisrestrictedto3kN,thisvalueshouldbeassignedtothelargerofthetwocabletensions.‹CA=3kNCB=CA3kN==1.44kN2.082.08Note:Ifthe3kNvaluewereassignedtoCB,thenCAwouldbe6.24kN,whichobviouslyexceedstheallowablecablecapacity.Thisproblemiscompletedbywritingthesecondequationofequilibrium.[©Fy=0]+123CA+CB-W=0135(2)SubstitutingthevaluesforCAandCBintoequation(2),12313kN2+11.44kN2=W135W=2.77kN+0.86kN=3.63kN+GraphicalSolution(Scale:1mm=50N):Usingthetip-to-tailmethod,aforcetriangleisconstructedsuchthatW,CA,andCBformaclosedtriangle.ThetipofthelastforcemustendonthetailofthefirstforceforRx=©Fx=0andRy=©Fy=0WeightWisknowntobeaverticalforcethatclosesattheoriginO.TheonlythingknownaboutCBandCAistheirlinesofaction.Visually,however,itisapparentthatCAmustbethe3kNtensionforcesothatCBdoesnotexceedtheallowabletension.IfCBweredrawnas3kN,CAwouldendupbeingmuchlarger.68Statics25ThetensioninthecableCBmustbeofaspecificmagnitudenecessarytoprovideequilibriumattheconcurrentpointofC.IftheforceintheboomACis4,000#andQis800#,determinetheloadP(vertical)thatcanbesupported.Inaddition,findthetensiondevelopedincableCB.Solvethisproblemanalyticallyaswellasgraphicallyusingascaleof1–=800#.AnalyticalSolution:ForceFxFyQ-Qcos30°=-800#10.8662=-693#-Qsin30°=-800#10.52=-400#-ACcos60°=-4,000#10.52=-2,000#+ACsin60°=+4,000#10.8662=+3,464#ACPCB0-P+CBcos30°=+0.866CBRy=©Fy=0yRx=©Fx=0-CBsin30°=-0.5CBForequilibriumtoexist‹Rx=[©Fx=0]-693#-2,000#+0.866CB=0+693#+2,000#=3,110#0.866Ry=[©Fy=0]-400#+3,464#-P-0.513,110#2=0CB=P=3,464#-400#-0.513,110#2;P=1,509#GraphicalSolution:BeginthegraphicalsolutionbydrawingtheknownforceAC=4,000#(a5–lineat60°fromthehorizontalaxis).Then,tothetipofAC,drawforceQ(1–longand30°fromthehorizontal).ForcePhasaknowndirection(vertical)butanunknownmagnitude.ConstructaverticallinefromthetipofQtorepresentthelineofactionofforceP.Forequilibriumtobeestablished,thelastforceCBmustcloseattheoriginpointC.DrawCBwitha30°inclinationpassingthroughC.TheintersectionofthelinesPandCBdefinesthelimitsoftheforces.MagnitudesofPandCBareobtainedbyscalingtherespectiveforcelines.69Statics26DeterminethetensileforcesinthecablesBA,BC,CD,andCEassumingW=100#.AnalyticalSolution:Becausethisprobleminvolvessolvingfourunknowncableforces,asingleFBDoftheentiresystemwouldbeinappropriate,becauseonlytwoequationsofequilibriumcanbewritten.ThisproblemisbestsolvedbyisolatingthetwoconcurrentpointsBandCandwritingtwodistinctsetsofequilibriumequationstosolveforthefourunknowns.Note:TheFBDofparticleBshowsthatonlytwounknowns,ABandCB,arepresent.IntheFBDofB,thedirectionsofcableforcesBAandBChavebeenpurposelyreversedtoillustratehowforcesassumedinthewrongdirectionarehandled.ForceABCBWFx+AB-CBcos30°0Fy0-CBsin30°-100#[©Fx-0]+AB-CBcos30°=0(1)[©Fy=0]-CBsin30°-100#=0(2)AB=+0.866(CB)+0.5CB=-100#‹CB=-200#ThenegativesignindicatesthattheassumeddirectionforCBisincorrect;CBisactuallyatensionforce.Themagnitudeof200#iscorrecteventhoughthedirectionwasassumedincorrectly.SubstitutingthevalueofCB(includingthenegativesign)intoequation(1),AB=+0.8661-200#2AB=-173.2#ABwasalsoassumedinitiallyasacompressiveforce,butthenegativesignintheresultindicatesthatitshouldbetensile.Note:IntheFBDabove,thedirectionofforceCB(tension)hasbeenchangedtoreflectitscorrectdirection.70StaticsForceCBCDCEFxFy-(200#)cos30°=-173.2#-(200#)sin30°=-100#0+CD34-CE+CE55[©Fy=0]-100#+CE=4CE=0551+100#2=+125#4[©Fx=0]-173.2#+CDCD=+173.2#+3CE=0531+125#25CD=+248#71StaticsProblemsForproblems28through33,drawFBDs.28Thesmallderrickshownontherightconsistsoftwoposts,ABandBC,supportingaweightW=1,000#.FindthereactionsRAandRC.29TwomembersACandBCarepinnedtogetheratCtoprovideaframeforresistinga500Nforceasshown.DeterminetheforcesdevelopedinthetwomembersbyisolatingjointC.30AneyeboltatAisinequilibriumundertheactionofthefourforcesshown.DeterminethemagnitudeanddirectionofP.31Asphereweighing2.5kNhasaradiusr=0.15m.Assumingthetwosupportingsurfacesaresmooth,whatforcesdoesthesphereexertontheinclinedsurfaces?Note:Reactionsonroundobjectsactperpendiculartothesupportingsurfaceandpassthroughthecenteroftheobject.72Statics32Aworkerispositioningaconcretebucket,weighing2,000#,bypullingonaropeattachedtothecrane’scableatA.Theangleofthecableis5°offverticalwhentheworkerpullswithaforceofPata20°anglefromthehorizontal.DeterminetheforcePandthecabletensionAB.33AweightW=200#issupportedbyacablesystemasshown.DetermineallcableforcesandtheforceintheverticalboomBC.73Statics5FREE-BODYDIAGRAMSOFRIGIDBODIESFree-bodydiagramsofrigidbodiesincludeasystemofforcesthatnolongerhaveasinglepointofconcurrency.Forcesarenonconcurrentbutremaincoplanarinatwodimensionalsystem.ThemagnitudesanddirectionsoftheknownexternalforcesshouldbeclearlyindicatedonFBDs.Unknownexternalforces,usuallythesupportreactionsorconstraints,constituteforcesdevelopedontherigidbodytoresisttranslationalandrotationaltendencies.Thetypeofreactionofferedbythesupportdependsontheconstraintcondition.SomeofthemostcommonlyusedsupportconstraintsaresummarizedinTable1.Also,theactualsupportconditionsforrollers,pins,andrigidconnectionsareshowninTable2.Note:IndrawingtheFBDs,theauthorwillplacea“hash”markonforcearrowsthatdenotereactions.Thishelpstodistinguishthereactionforcesfromotherappliedforcesandloads.Table1(a)Supportconditionsforcoplanarstructures.IdealizedSymbolReactionsNumberofUnknownsOneunknownreaction.Reactionisperpendiculartothesurfaceatthepointofcontact.Oneunknownreaction.Thereactionforceactsinthedirectionofthecableorlink.74StaticsTable1(b)Supportsandconnectionsforcoplanarstructures.IdealizedSymbolReactionsNumberofUnknownsOneunknownreaction.Reactionisperpendiculartothesupportingsurface.Twounknownreactions.Unknownreactions.FxandFy.Threeunknownreactions.Unknownreactions.ForcesFx,Fy,andresistingmomentM.75StaticsTable2Connectionandsupportexamples.RollerRockerPinFixed76StaticsMostproblemsdealtwithinthistextwillbeassumedweightlessunlessotherwisespecified.Whenevertherigidbodyweightissignificantinaproblem,onecaneasilyincludeitinthecalculationsbyaddinganotherforcepassingthroughthecentroid(centerofgravity)oftherigidbody.Whenthesenseofthereactingforceormomentisnotapparent,arbitrarilyassignadirectiontoit.Ifyourassumptionhappenstobeincorrect,thecalculatedanswer(s)intheequilibriumequationswillresultinanegativevalue.Themagnitudeofthenumericalanswerisstillcorrect;onlytheassumeddirectionoftheforceormomentiswrong.Ifthenegativeansweristobeusedinfurthercomputations,substituteitintoequationswiththenegativevalue.Itisrecommendedthatnochangesinvectordirectionbeattempteduntilallcomputationsarecompleted.Free-bodydiagramsshouldincludeslopesandcriticaldimensions,becausethesemaybenecessaryincomputingmomentsofforces.Figures53through55showexamplesofsuchFBDs.(a)Pictorialdiagram.(b)Free-bodydiagramofthebeam.Figure53Simplebeamwithtwoconcentratedloads.77Statics(a)Pictorialdiagram.(b)Free-bodydiagram.Figure54Cantileverbeamwithaconcentratedanduniformload.(a)Pictorialdiagram.(b)Free-bodydiagram.Figure55Windloadonapitchedroof.Windloadsonpitchedroofsaregenerallyappliedperpendiculartothewindwardsurface.Anotheranalysiswouldexaminetheupliftforcesontheleewardslope.Purlinsthatrunperpendiculartotheplaneofthetrussaregenerallylocatedatthetrussjointstominimizebendinginthetopchordmember.78StaticsExampleProblems:EquilibriumofRigidBodies27Abeamloadedwitha500#forcehasoneendpinsupportedandtheotherrestingonasmoothsurface.DeterminethesupportreactionsatAandB.Solution:ThefirststepinsolvinganyoftheseequilibriumproblemsistheconstructionofanFBD.DirectionsofAx,Ay,andBarearbitrarilyassumed.a.ThepinsupportatAdevelopstworeactionconstraints:AxandAy.Bothforcesareindependentofeachotherandconstitutetwoseparateunknowns.b.ReactionBfromthesmoothsurfacedevelopsperpendiculartotheinclineofthesurface.c.Becausetheforceequationsofequilibrium(©Fx=0and©Fy=0)areinthexandyreferencecoordinatesystem,forcesthatareinclinedshouldberesolvedintoxandycomponents.Bx=34BandBy=B55Note:TheslopeofreactionforceB(4:3)isthereverseofthesurfaceslope(3:4).BxandByarecomponentsofthereactionforceBandarenotindependentofeachotherasAxandAywere.BywritingBxandByasfunctionsofB,theFBDstillinvolvesonlythreeunknowns,whichcorrespondtothethreeequationsofequilibriumnecessaryforarigidbody.3©Fx=04+Ax-Bx=0ButsinceBx=Axthen,Ax=(1)3B,53B=05+3B53©Fy=04+Ay-500#+By=0Ay=500#-(2)4B53©MA=04=500#(6¿)+By(10¿)=0(3)Note:Themomentequationcanbewrittenaboutanypoint.Normally,thepointchoseniswhereatleastoneoftheunknownsisconcurrent.Thus,theintersectingunknowncanbeexcludedinthemomentequation,becauseithasnomomentarm.79StaticsSolvingequation(3),(10¿)By=+(500#)(6¿)4(10¿)abB=+3,000#-ft.5B=+375#ThepositivesignforthesolutionofBindicatesthattheassumedsenseforBintheFBDwascorrect.Substitutingintoequations(1)and(2),3Ax=+1+375#2=+255#5Ay=500#-41+375#2=+200#5TheassumeddirectionsforAxandAywerecorrect.28DrawanFBDofmemberABD.SolveforsupportreactionsatAandthetensionincableBC.Solution:ThedirectionsforAx,Ay,andBCareallassumed.Verificationwillcomethroughtheequilibriumequations.[©MA=0]+0.707BC(2.5m)-2.4kN(5m)=0(1)BC=2.4kN(5m)=6.79kN0.707(2.5m)‹BCx=0.707(6.79kN)=4.8kN‹BCy=0.707(6.79kN)=4.8kN3gFx=0]+Ax-4.8kN=0(2)Ax=+4.8kN3©Fy=04-Ay+4.8kN-2.4kN=0Ay=+2.4kN80(3)Statics29DeterminethesupportreactionsforthetrussatjointsAandD.Solution:3©MD=04+Ax(20¿)-867#(10¿)-1,000#(20’)=0(1)Ax=+1,434#;Assumeddirectioncorrect3©Fx=04-Dx-500#+Ax=0(2)Dx=1,434#-500#=+934#;AssumeddirectionOK3©Fy=04-Dy-867#-1,000#=0Dy=-1,867#;Assumeddirectionincorrect(3)30DeterminetheresistingmomentMRAatthebaseoftheutilitypoleassumingforcesT1=200#andT2=300#areasshown.WhatarethehorizontalreactionsAxandAyatthebase?(ShowFBD.)Solution:T1x=T1cos15°=200#(0.966)=193#T1y=T1sin15°=200#(0.259)=51.8#T2x=T2cos10°=300#(0.985)=295.4#T2y=T2sin10°=300#(0.174)=52#3©MA=04-MRA+295.4#(30¿)-193#(35¿)(1)+52#(6¿)-51.8#(4¿)=0solvingforMRA:MRA=+2,212#-ft.3©Fx=04-295.4#+193#+Ax=0(2)3©Fy=04-52#-51.8#+Ay=0(3)‹Ax=+102.4#‹Ay=+103.8#81Statics31Acompoundbeamsupportstwoverticalloadsasshown.DeterminethesupportreactionsdevelopedatA,B,andE,andalsotheinternalconstraintforcesatCandD.Solution:TheFBDoftheentirebeamsystemshowsatotalofsixconstraintreactionsdeveloped.BecauseonlythreeequationsofequilibriumareavailableforagivenFBD,allsupportreactionscannotbedetermined.Incasessuchasthese,wherethesystemiscomposedofseveraldistinctelements,themethodofsolutionshouldinvolvedrawingFBDsoftheindividualelements.Note:InternalconstraintforcesatCandDareshownequalandoppositeoneachoftheconnectedelements.SelecttheelementalFBDwiththefewestnumberofunknownforces,andsolvetheequationsofequilibrium.82StaticsFBD(b):3©Fx=04Cx=03©MC=04-3,000#(6¿)+Dy(12¿)=0Dy=+1,500#3©Fy=04+Cy-3,000#+Dy=0Cy=+3,000#-1,500#Cy=+1,500#Cx,Cy,andDyarenowknownforcesforFBDs(a)and(c).FBD(a):3©Fx=04+Ax-Cx=0butCx=0;‹Ax=03©MA=04-2,000#(8¿)+By(12¿)-Cy(20¿)=0By=+2,000#(8¿)+1,500#(20¿)=+3,830#-ft.12¿3©Fy=04+Ay-2,000#+By-Cy=0Ay=+2,000#-3,830#+1,500#=-330#Note:ThenegativeresultforAyindicatesthattheinitialassumptionaboutitsdirectioniswrong.FBD(c):3©Fx=04Ex=03©Fy=04-Dy+Ey=0Ey=+1,500#3©ME=04+Dy(16¿)-MRE=0MRE=+(1,500#)(16¿)=+24,000#-ft.83StaticsProblems34ApoleABleansagainstasmooth,frictionlesswallatB.CalculatetheverticalandhorizontalcomponentsofthereactionsofAandB.35ThegirdershownissupportedbycolumnsatAandB.Twosmallerbeamspushdownwardonthegirderwithaforceof40kNatC,andtwootherbeamspushdownwardwithaforceof50kNatD.FindthereactionsatAandB.36Abridgeoverariverisloadedatthreepanelpoints.DeterminethesupportreactionsatAandB.37Windforcesonthewindwardroofslopeareappliednormallytotheuppertrusschord.DeterminethewallreactionsdevelopedatAandD.84Statics38Aninclinedking-posttrusssupportsaverticalandhorizontalforceatC.DeterminethesupportreactionsdevelopedatAandB.39Athree-spanoverhangbeamsystemisusedtosupporttheroofloadsforanindustrialbuilding.DrawappropriateFBDs,anddeterminethesupportreactionsatA,B,E,andFaswellasthepin(hinge)forcesatCandD.40DeterminethereactionsdevelopedatsupportpointsA,B,C,andD.41SolveforthesupportreactionsatA,B,andC.85Statics6STATICALINDETERMINACYANDIMPROPERCONSTRAINTSIntheanalysisofabeam,truss,orframework,thefirststepusuallyinvolvesthedrawingofanFBD.FromtheFBD,wecandetermine(a)whetherthenecessaryandavailableequationsofequilibriumaresufficienttosatisfythegivenloadconditionsand(b)theunknownsupportforces.Figure56(a)Trusswithhingeandrollersupport.Asanexample,let’sexamineatruss,asshowninFigure56(a),withtwoappliedloads,F1andF2.Ahinge(pin)supportisprovidedatAandarollersupportatB.AnFBDofthistrussshowsthattwosupportforcesaredevelopedatAandonlyaverticalreactionexistsatB,asshowninFigure56(b).Thethreesupportforcesaresufficienttoresisttranslationinboththexandydirectionsaswellasrotationaltendenciesaboutanypoint.Therefore,thethreeequationsofequilibriumaresatisfied,andAx,Ay,andBycanbeeasilydetermined.Incasessuchasthese,thereactionsaresaidtobestaticallydeterminate,andtherigidbodyissaidtobecompletelyconstrained.Figure56(b)FBD—Determinateandconstrained.Nowconsiderthesametruss,asshowninFigure57(a),butwithtwopinsupports.AnFBDofthetrussshowsthatatotaloffoursupportreactionsarepresent:Ax,Ay,Bx,andBy.Thesesupportconstraintsadequatelyresisttranslational(xandy)aswellasrotationaltendenciestosatisfytheprimaryconditionsofequilibrium,Figure57(b).ThreeequationsofequilibriumFourunknownsupportreactionsFigure57(a)Trusswithtwohingedsupports.Figure57(b)FBD—Staticallyindeterminateexternally.86Whenthenumberofunknownsexceedsthenumberofequationsofequilibrium,therigidbodyissaidtobestaticallyindeterminateexternally.Thedegreeofindeterminacyisequaltothedifferencebetweenthenumberofunknownsandthenumberofequationsofequilibrium.AswiththecaseshowninFigure57(b),thetrussconstraintsareindeterminatetothefirstdegree.StaticsTwosupportsareprovidedforthetrussinFigure58(a);botharerollers.TheFBDrevealsthatonlytwoverticalsupportreactionsdevelop.BothAyandByhavethecapabilityforresistingtheverticalforceF1,butnohorizontalreactionisprovidedtoresisthorizontaltranslationcausedbyforceF2.ThreeequationsofequilibriumTwounknownsupportreactionsTheminimumnumberofequilibriumconditionsthatmustbesatisfiedisthree,butbecauseonlytwosupportconstraintsexist,thistrussisunstable(orpartiallyconstrained).(a)Pictorialdiagram.Ageneralizationthatseemsapparentfromthethreepreviousexamplesisthatthenumberofsupportunknownsmustbeequaltothenumberofequationsofequilibriumforarigidbodytobecompletelyconstrainedandstaticallydeterminate.Note,however,thatwhilethisgeneralizationisnecessary,itisnotsufficient.Consider,forexample,thetrussshowninFigure59(a)and59(b),whichissupportedbythreerollersatA,B,andC.ThreeequationsofequilibriumThreeunknownsupportreactionsAlthoughthenumberofunknownsisequaltothenumberofequationsofequilibrium,nosupportcapabilityexiststhatcanrestrainhorizontaltranslation.Theseconstraintsareimproperlyarranged,andthisconditionisreferredtoasimproperlyconstrained.(b)Free-bodydiagram.Figure58Tworollers—partiallyconstrained/unstable.ArearrangementofthethreerollersshowninFigure60(a)and60(b)couldeasilymakethetrussstableaswellasstaticallydeterminate.(a)Pictorialdiagram.(a)Pictorialdiagram.(b)Free-bodydiagram.Figure59Tworollers—partiallyconstrained/unstable.(b)Free-bodydiagram.Figure60Threesupports—stableanddeterminate.87StaticsThreeEquationsofEquilibriumandThreeUnknownSupportReactionsClassificationofStructuresBasedonConstraintsStructureNumberofUnknownsNumberofEquationsStaticalCondition33Staticallydeterminate(calledasimplebeam)53Staticallyindeterminatetotheseconddegree(calledacontinuousbeam)23Unstable33Staticallydeterminate(calledacantilever)33Staticallydeterminate(calledanoverhangbeam)43Staticallyindeterminatetothefirstdegree(calledaproppedbeam)63Staticallyindeterminatetothethirddegree(calledafixed-endedbeam)33Staticallydeterminate33Unstable(improperlyconstrained)63Staticallyindeterminatetothethirddegree1.2.3.4.5.6.7.8.9.10.88StaticsSupplementaryProblemsVectorAddition:Section242DeterminetheresultantofthetwoforcesactingatpointO.Scale:1/2–=100#.43Determinetheresultantofthethreeforcesshownactingontheanchordevice.Usethetip-to-tailgraphicalmethodinyoursolution.FollowasequenceofA-B-C.Scale:50mm=1kNor1mm=20N.44Apileresiststhetensiondevelopedbythreestaycablesforamajortent(membrane)structure.Assumingnobendingisdesiredinthepile(noresultanthorizontalcomponent),whatisthemagnitudeoftheforceSiftheangleθS=30°?Ifthepileresistswithacapacityof500#persquarefootofembeddedsurface,whatistherequiredpenetrationhforthepile?Scale:1–=8k.89StaticsForceSystems:Section345SolvefortheresultantforceatAusingtheanalyticalmethod.46KnowingthatthemagnitudeoftheforcePis500#,determinetheresultantofthethreeforcesappliedatA.47DeterminetheresultantatpointD(whichissupportedbythecrane’smast)ifAD=90kN,BD=45kN,andCD=110kN.MomentofaForce:Section348DeterminetheforceFrequiredatBsuchthattheresultantmomentatCiszero.Showallnecessarycalculations.90Statics49Acrookedcantileverbeamisloadedasshown.DeterminetheresultantmomentatA.50DeterminetheresultantmomentMAatthebaseoftheutilitypoleassumingforcesT1andT2asshown.ResultantofParallelForces:Section351Findthesingleresultantforcethatwouldduplicatetheeffectofthefourparallelforcesshown.Usethereferenceorigingiveninthediagram.52Athree-meter-longhorizontalwoodbeamweighs30Npermeteroflengthandsupportstwoconcentratedverticalloadsasshown.Determinethesingleresultantforcethatwouldduplicatetheeffectofthesethreeparallelforces.Usethereferenceoriginshown.53A16-foot-longhorizontalwoodmemberweighs20lb./ft.andsupportsthreeconcentratedloads.DeterminethemagnitudesofcomponentsAandBthatwouldbeequivalentineffectasthefourparallelforcesactingonthesystem.91StaticsEquilibriumofaParticle:Section454DeterminethetensionincabledevelopedinstrutAC.ABandtheforce55DeterminetheforcedevelopedinmembersACandBCduetoanappliedforceF=2kNatconcurrentjointC.Inyourfinalanswer,indicatewhetherthemembersareintensionorcompression.Solvethisproblemanalyticallyandgraphically.Scale:1mm=20N.56CablesBAandBCareconnectedtoboomDBatconcurrentpointB.DeterminetheforcesincableBAandboomDBusingtheanalyticalmethod.AssumeaconditionofequilibriumexistsatB.57DeterminethetensionsinCAandCBandthemaximumweightWifthemaximumcablecapacityforACandCBis1.8kN.92Statics58DeterminetheweightWrequiredtoproduceatensileforceof1,560#incableAB.Also,determinetheforcesinBC,BE,andCD.EquilibriumofRigidBodies:Section559DeterminethesupportreactionsatAandB.60ConstructtheappropriateFBDs,andsolveforthesupportreactionsatAandC.61Abridgespansacrossarivercarryingtheloadsshown.DeterminethesupportreactionsatAandB.62CalculatethebeamreactionsatsupportsA,C,andD.DrawallappropriateFBDs.93StaticsSummary■■■■Forceisdefinedasapushorapullonabodyandtendstochangethebody’sstateofrestormotion.Also,aforceisavector,characterizedbyitsmagnitude,direction,sense,andpointofapplication.Theprincipleoftransmissibility,asitappliestorigidbodies,statesthataforcemaybemovedanywherealongitslineofactionwithoutchangingtheexternaleffectsonthebody.Twoforces(vectors)addaccordingtotheparallelogramlaw,inwhichthecomponentsformthesidesoftheparallelogramandtheresultantisrepresentedbythediagonal.Areverseofvectoradditionistheresolutionofaforceintotwoperpendicularcomponents,generallythexandyaxes.ForceF,havinganinclinationθwithrespecttothehorizontalxaxis,isresolvedintoxandycomponentsexpressedasFx=FcosθandFy=Fsinθ.TheforceFrepresentsthediagonalofarectangle,andFxandFyaretherespectivesides.FromthePythagoreantheoremforrighttriangles,FyFyF=2F2x+F2yandtanθ=orθ=tan-1abFxFx■AresultantofaseriesofconcurrentforcesisexpressedasR=2R2x+R2yandRx=©FxandRy=©Fy■■■Thedirectionoftheresultantforceisdeterminedbyusingthetrigonometricfunction:Ryθ=tan-1abRxAlgebraicsummationofforcesinthexandydirectionsassumesasignconventioninwhichhorizontalforcesdirectedtotherightareconsideredtobepositive(negativeifdirectedtotheleft)andverticalforcesactingupwardareconsideredtobepositive(negativeifactingdownward).Momentisexpressedasaforcetimestheperpendiculardistancetoareferencepoint:M=F*d.Directionisdeterminedbythetendencyofaforcetoproduceclockwiseorcounterclockwiserotationaboutthereference.■94Varignon’stheoremstatesthatthemomentofaforceaboutapointisequaltothealgebraicsumofthemomentsofitscomponentsaboutthesamepoint.Statics■■■Amomentcoupleistheproductoftwoforceshavingthesamemagnitude,parallellinesofaction,butoppositesense.Couplesproducepurerotationwithnotranslation.Theresultantoftwoparallelforcesmustproduceanequivalenteffect(translationalandrotational)astheoriginalforces.Equilibriumofacoplanar-concurrentforcesystem(equilibriumofaparticle)requirestheapplicationoftwoequations:Rx=©Fx=0andRy=©Fy=0■Coplanarforcesactingonarigidbodymustsatisfythreeconditionsforequilibrium:©Fx=0,©Fy=0;©Mi=0■AnessentialstepinsolvingequilibriumprinvolvestheconstructionofFBDs.oblemsAnswerstoSelectedProblems1R=173lb.;θ=50°fromthehoriz.;φ=40°fromthevertical29AC=768N(compression);BC=672N(tension)3F2=720lb.31A=2.24kN;B=0.67kN5T2=3.6kN6Fx=800lb.;Fy=600lb.33CD=245.6lb.(T);DE=203.4lb.(T);AC=392.9lb.(T);BC=487.7lb.(C)8Px=94.9lb.;Py=285lb.35A=43.33kN;B=46.67kN10R=1,079N;θ=86.8°fromthehorizontalreferenceaxis37Ay=3,463lb.;Dx=3,000lb.;Dy=1,733lb.12F1=6.34kN;F2=7kN3813T=4.14kips;R=–11.3kAx=.705kN;Ay=.293kN;Bx=.295kN;By=.707kN14MA=0.Theboxisjustonthevergeoftippingover.40Ay=240lb.(↓);Bx=0;By=720lb.(↑);Cy=480lb.;Dx=300lb.(←);Dy=80lb.(↓)16MA=–420lb.-in.(clockwise)4118P=10.3lb.19MA=–656kN-m(clockwise)FD=18.9k;Ax=15.2k;Ay=1.3k(↓);BD=17.2k;DC=17.5k21MA=108.8lb.-in.(counterclockwise);MB=–130.6lb.-in.(clockwise)42R=720lb.;θR=72.5°44S=20.5k;R=42.5k;h=78’23W=1,400lb.46R=1,867lb.;θ=22°25MA=MB=MC=048F=137.4lb.27MA=–850lb.-in.;MB=–640lb.-in.50MA=3,990lb.-ft.(counterclockwise)28A=732lb.;C=518lb.52R=40N(↓)atanimaginarylocationwherex=5.4mtotheleftoftheorigin95Statics9654AC=5.36k(C);AB=4.64k(T)56BA=658.2lb.;DB=1,215.2lb.58BC=1,800lb.;BE=1,680lb.;CD=2,037lb.;W=2,520lb.60Ay=1kN;Bx=0;By=0.8kN;Cx=0;Cy=3.5kN;MC=12.9kN-m62Ax=180lb.(→);Ay=52.5lb.(↑);By=187.5lb.;Cy=322lb.(↑);Dx=60lb.(→);Dy=145.5lb.(↑)AnalysisofSelectedDeterminateStructuralSystemsFromChapter3ofStaticsandStrengthofMaterialsforArchitectureandBuildingConstruction,FourthEdition,BarryOnouye,KevinKane.Copyright©2012byPearsonEducation,Inc.PublishedbyPearsonPrenticeHall.Allrightsreserved.97AnalysisofSelectedDeterminateStructuralSystemsFigure1Hingedconnectionofasupportend—RavennaBridge,Seattle.Hingedsupportsfeatureprominentlyinthestructuralsystemsdiscussedinthischapter.ThesymbolforahingedconnectionisshowninTable2.1(b).PhotobyChrisBrown.1EQUILIBRIUMOFAPARTICLESimpleCablesCablesareahighlyefficientstructuralsystemwithavarietyofapplicationssuchas:■■■■Suspensionbridges.Roofstructures.Transmissionlines.Guywires,etc.Cablesorsuspensionstructuresconstituteoneoftheoldestformsofstructuralsystems,asattestedtobytheancientvine-and-bamboobridgesofAsia.Asastructuralsystem,acablesystemislogicalandobjectifiesthelawsofstaticsinvisualterms.Alaypersoncanlookatitandunderstandhowitworks.98AnalysisofSelectedDeterminateStructuralSystemsFigure2Cable-stayedpedestrianbridge.Photobyauthor.Itisasimpleengineeringrealitythatoneofthemosteconomicalwaystospanalargedistanceisthecable.Thisinturnderivesfromtheuniquephysicalfactthatasteelcableintension,poundforpound,isseveraltimesstrongerthansteelinanyotherform.Inthecable-stayedpedestrianbridgeshowninFigure2,thegreatstrengthfactorofsteelintensionisputtoworktoreduceenormouslythedeadloadofthestructurewhileprovidingamplestrengthtosupportthedesignliveload.Cablesarerelativelylight,andunlikebeams,arches,ortrusses,theyhavevirtuallynorigidityorstiffness.Themostcommonmaterialusedforbuildingcablestructures(suspensionbridges,roofstructures,transmissionlines,etc.)ishigh-strengthsteel.Cablesareassumedtobeflexibleelementsbecauseoftheirrelativelysmallcross-sectionaldimensioninrelationtotheirlength.Thebendingcapacityincablesthereforeisusuallyassumedtobenegligible.Cablescancarryloadonlyintension,andtheymustbekeptintensionatalltimesiftheyaretoremainstable.Thestabilityofatensilestructuremustbeachievedthroughacombinationofshape(geometry)andprestress.Beamsandarchessubjecttochangingloadsdevelopbendingmoments,butacablerespondsbychangingitsshapeorconfiguration(Figure3).Figure3Cableshapesrespondtospecificloadconditions.99AnalysisofSelectedDeterminateStructuralSystemsThePrincipalElementsEverypracticalsuspensionsystemmustincludeallofthefollowingprincipalelementsinoneformoranother.VerticalsupportsortowersTheseprovidetheessentialreactionsthatkeepthecablesystemabovetheground.Eachsystemrequiressomekindofsupportingtowers.Thesemaybesimpleverticalorslopingpiersormasts,diagonalstruts,orawall.Ideally,theaxesofthesupportsshouldbisecttheanglebetweenthecablesthatpassoverthem(Figure4).Figure4Cable-supportedroof—BartleHall,KansasCity,Missouri.PhotobyMattBissen.MaincablesThesearetheprimarytensileelements,carryingtheroof(orsometimesfloors)withaminimumofmaterial.Steelusedincablestructureshasbreakingstressesthatexceed200,000psi(poundspersquareinch).AnchoragesAlthoughthemaincablescarrytheirloadsinpuretension,theyareusuallynotvertical,whereasthegravityforcesare.Thisresolutionisaccomplishedbecausesomepartofthestructureprovidesahorizontalforceresistance;thisiscalledtheanchorage.Inthecaseofasuspensionbridge,themaincablesarecarriedovergentlycurvedsaddlesontopofthetowersandondownintomassiveconcreteabutmentsorintobedrock,asshowninFigure5(a).Inthecable-stayedbridgesystemshowninFigure5(b),theverticalcableononesideoftheverticaltowerisbalancedbyanequivalentcableontheotherside.Horizontalthrustisresolvedinthelongitudinalbridge-deckframework.100AnalysisofSelectedDeterminateStructuralSystems(a)Suspensionbridge.(b)Cable-stayedbridge.Figure5Cable-supportedbridges.Forbuildings,theresolutionofhorizontalthrustisusuallyverycumbersome,involvingtremendousmass.Itisusuallybesttousesomepartofthebuildingitselfastheanchorage:afloorthatcanactasabracefromonesidetotheotheror,moresimply,acompressionringifthebuildingformsasmooth,closedcurveinplan.Circularplansarewellsuitedto,andcommonlyusedfor,suspensionroofs(Figure6).Figure6Suspendedroofsystem.101AnalysisofSelectedDeterminateStructuralSystemsStabilizersFigure7Cableflutterinalightweightroof.Figure8“GallopingGertie,”TacomaNarrowsBridge.CourtesyoftheUniversityofWashingtonLibraries,SpecialCollections,UW21422.102Stabilizersarethefourthelementrequiredtopreventcablesfromundergoingextremeshapechangesundervaryingloadconditions.Lightweightroofsystems,suchascablesormembranes,aresusceptibletopronouncedundulationorflutteringwhenacteduponbywindforces(Figure7).Everyformofabuilding,likeeveryphysicallaw,hasitslimitationsorbreakingpoints.Inthebeam,whichresistsloadsinbending,thistakestheformofcrackingorshearing.Inthearch,whoseprimaryloadingisincompression,thisoccursasbucklingorcrushing.Andinthecable,whichresistsloadonlythroughtension,thedestructiveforceisvibration—particularlyflutter,acomplexphenomenonthatbeliesthelightnessofitsname.DavidB.Steinman,oneofthegreatU.S.suspensionbridgeengineers,isolatedandidentifiedthisphenomenonofflutterin1938.Allmaterialsofwhatevernaturehaveanaturalmolecularvibrationorfrequencyrange.Ifanoutsideforceactinguponamaterialcomeswithinthatfrequencyrange,causingthematerialtovibrateinternally,orflutter,avibrationalstatemaybereachedwheretheouterandinnerforcesareintune(calledresonance),andthematerialundergoesdestruction.Evenwithoutreachingresonance,theunevenloadingofoutsideforces,suchaswind,maycauseamaterialtovibratevisiblyupanddown,buildinguprhythmicallytodestruction.Itwasthesealliedforces,plusdesignflaws,thatreducedtheTacomaNarrowsBridgetorubble(Figure8).Inheavy,earthbound,compressivestructures,thenaturalfrequenciesaresolowthatfewexternalforcescanbringthemtoresonance,andsheerweighthastheeffectofcheckingvibrations.Incablestructures,however,thelightandexceedinglystrongmaterialsaresoextremelysensitivetounevenloadingthatvibrationandflutterbecomemajordesignconsiderations.AftertheTacomaNarrowsfailure,alargegroupoftopU.S.engineersandscientists,AnalysisofSelectedDeterminateStructuralSystemsparticularlyintheaerodynamicsfield,thoroughlyinvestigatedandreportedonthephenomenon.DavidSteinmanworkedindependentlyfor17years,devisinganintegralsystemofdampingthatsubtlyoutwitsthephenomenonwithoutasacrificeinweightoreconomy.Withbuildings,theproblemanditssolutionarerelatedtotheroofsurface.Whatistobeusedtospanacrossthecables?Ifitissometypeofmembrane,suchasatentstructure,actingonlyintension,thentheproblemofdynamicinstabilitybeginsimmediatelyandmaybesolvedbypretensioning.If,however,thesurfaceistobewoodplanking,metaldecking,orathinconcreteslab,itisthenrigidandcanresistnormal(perpendicular)forcesthroughbending.Theproblemofflutterandmovementisminimizedforthesurfacebutremainsaconcernforthemaincables(Figure9).Thestabilizingfactorsfortheprimarycablescanbedeadweight,arigidsurfacethatincludesthemaincables,asetofsecondarypretensionedcableswithreversecurvaturefromthemaincables,orrestrainingcables(Figure10).Figure9Figure10(a)Stabilizingtheroofstructure.Increaseofdeadweight.103AnalysisofSelectedDeterminateStructuralSystemsFigure10(b)Stiffeningthroughconstructionasaninvertedarch(orshell).Figure10(c)Spreadingagainstacablewithoppositecurvature.Figure10(d)Tensioningagainstacablewithoppositecurvature.Figure10(e)Fasteningwithtransversecablesanchored.Figure10(f)104Cablenetstructure.AnalysisofSelectedDeterminateStructuralSystemsCableGeometryandCharacteristicsOneoftheadvantagesoftensionstructuresisthesimplicitywithwhichonecanvisualizetheshapeofthetensionelementsunderload.Aperfectlyflexiblecableorstringwilltakeonadifferentshapeforeveryvariationinloading;thisisreferredtoasthefunicularorstringpolygon.Undertheactionofasingleconcentratedload,acableformstwostraightlinesmeetingatthepointofapplicationoftheload;whentwoconcentratedloadsactonthecable,itformsthreestraightlines(polygonform);andsoon.Iftheloadsareuniformlydistributedhorizontallyacrosstheentirespan(suspensionbridge),thecableassumestheshapeofaparabola.Iftheloadsaredistributeduniformlyalongthelengthofthecableratherthanhorizontally,suchasasuspendedchainloadedbyitsownweight,thenthecableassumesthenatural(funicular)shapecalledacatenary,acurveverysimilartoaparabola(Figure11).(a)Simpleconcentratedload—triangle.(b)Severalconcentratedloads—polygon.Changesinshapeduetoasymmetricalloading,suchassnow,overonlyportionsoftheroofareessentiallythesameasthoseduetomovingloads.Inbothcases,thegreatertheliveload/deadloadratio,thegreaterthemovement.Someofthebasiccharacteristicsofacablesystemareinherentinitsgeometry(Figure12).Figure12/Lhr====Cablecharacteristics.(c)Uniformloads(horizontally)—parabola.(d)Uniformloads(alongthecablelength)—catenary.cablespanAC+CB=cablelengthsagh>/=sag-to-spanratioThevaluesofrusuallyaresmall—inmostcases,onthe11orderof10to15.CablelengthLforasingleconcentratedloadcanbedeterminedeasilybyusingthePythagoreantheorem.Saghandcablespan/areusuallygivenorknown.(e)Comparisonofaparabolicandacatenarycurve.Figure11Geometricfunicularforms.105AnalysisofSelectedDeterminateStructuralSystemsCableswithaSingleConcentratedLoadWhenacableissubjectedtoasingleconcentratedloadatmidspan,thecableassumesasymmetricaltriangularshape.Eachhalfofthecabletransmitsanequaltensileforcetothesupport.Figure13FBDofacablewithasingleconcentratedload.InFigure13,assumethat/=24¿andh=3¿andthattheseareknownvalues.ThecablelengthLcanbefoundby:L=AC+CB=/21+4r2=2/2+4h2derivedfromthePythagoreantheorem.Forthisparticulargeometryandloadcondition,theforcethroughoutthecableisthesame.Thetensileforcedevelopedinthecablepassesthroughtheaxis,orlineofaction,ofthecable.Cables,therefore,performsimilarlytorigidtwo-forcemembers.Figure14FBDofconcurrentpointC.IsolatingtheconcurrentpointC,wecandrawtheFBDshowninFigure14.Thesaginacableisimportant,becausewithoutit,loadscannotbetransmittedtothesupport(Figure15).Figure15Cableloadedatmidspan.312TTx=T12.412.4h3¿1r===/24¿8Ty=L=/21+4r2=24121+4ab=24.8¿C8BecausetheTycomponentateachsupportisequal,CgFy=0D2Ty-W=0;Ty=W2Byvirtueoftheslope,Tx=1212WTy=ab=2W332ThehorizontalcomponentTxdevelopedatthesupportisknownasthethrust.106AnalysisofSelectedDeterminateStructuralSystemsExampleProblems:Cables1Inthisexample(Figure16),weassumethatasthesagincreases(rincreases),thetensioninthecabledecreases.Minimizingthesaginacablecauseslargetensileforcestodevelop.Solution:(Extremelylargesag)h=12¿,L=6¿Tx=312T;TyT12.412.4CgFy=0D2Ty-W=0;Ty=Tx=r=W2W33Wab=Ty=121228h12==2/6Figure16FBDofalarge-sagcablewithaloadatmidspan.2Isitpossibletohaveacablesupportaloadandhavezerosag(Figure17)?Solution:ThiscablesystemcannotsupportW,becausenoTycomponentsexistfor[gFy=0].Thetensiondevelopediscompletelyhorizontal,whereFigure17Cablewithnosag.Tx=T,Ty=0Itisimpossibletodesignacablewithzerosag,ortoquoteLordKelvin:Thereisnoforce,howevergreat,Canstretchacord,howeverfine,IntoahorizontallineThatshallbeabsolutelystraight.107AnalysisofSelectedDeterminateStructuralSystems3Determinethetensiondevelopedinthecable,thesupportreactions,andthecablelengthL(Figure18).Figure18(a)Cablewithasingleconcentratedload.Solution:/=10mh=2mh21r===/105Figure18(b)FBDofthecable.2A15AAy=15Ax=1083B
110BBy=110Bx=AnalysisofSelectedDeterminateStructuralSystemsB110m2-45kN14m2=0110B=18110kN=56.9kNC©MA=0D+Bx=3118110kN2=54kN110By=18110kN=18kN1103©Fy=04Ay-45kN+18kN=0Ay=27kNNotethatAyZByduetoasymmetricalloading:Ax=2Ay=54kN1515Ax=154kN2=2715kN22=60.4kNPCriticalTensionA=Note:Ax=Bx;thehorizontalcomponentofthetensionforceisthesameatanypointofthecable,because©Fx=0.Thesupportforcesandcableforcesareofthesamemagnitudebutoppositesense.ThecablelengthisdeterminedasL=212m22+14m22+212m22+16m22L=220m2+240m2=10.79m4Determinethesupportreactions,cabletensions,andelevationsofpointsBandDwithrespecttothesupports(Figure19).Thecableweightisassumedtobenegligible.Figure19Cablewithmultipleconcentratedloads.109AnalysisofSelectedDeterminateStructuralSystemsSolution:CgME=0D-Ay132¿2-Ax18¿2+100#124¿2+150#116¿)+100#18¿2=0=32Ay+8Ax=5,600#-ft.FBD(b)Portionofthecable.(1)FBD(a)Entirecable.CgMC=0D+Ax11¿2-Ay116¿2+100#18¿2=016Ay-Ax=800#-ft.(2)Solvingequations(1)and(2)simultaneously,Ax=400#,Ay=75#;TAB=407#CgFy=0D+75-100#-150#-100#+Ey=0(3)CgFx=0D-400#+Ex=0(4)Ey=+275#Ex=400#KnowingthesupportreactionsatAandE,wecanassumethattheforceincableABisthesameasthesupportforcesatAandthatthecableforceinEDisthesameasthesupportforcesatE.However,thetensioninthecableatthetwoattachmentpoints(AandE)willhavedifferentvalues.Also,becausecablesbehaveastwo-forcemembers,theforcerelationshipsinxandyalsogiveusinformationaboutthesloperelationshipinthecablesegments.110AnalysisofSelectedDeterminateStructuralSystemsForexample,Ayy=8¿Axy=75#18¿2=1.5¿400#yEy8¿=Ex;y=275#18¿2=5.5¿400#NowthattheelevationsofpointsB,C,andDareknown,tensionforcesinBCandCDcanbedetermined.TBCy=.58TBC;TBCx=T8.028.02BC8T=0;TBC=401#8.02BC81401#2=400#=8.02.5=1401#2=25#8.02CgFx=0D-400#+TBCxTBCyTDCx=83.5T;T=T8.73DCDCy8.75DCCgFx=0D-TDCx+400#=0;TDCx=400#3.5TDCy=1400#2=175#88.731400#2=436.5#TDC8Summary:TAB=407#TBC=401#TDC=436.5#TDE=485.4#Note:TABZTDE111AnalysisofSelectedDeterminateStructuralSystemsProblems1Threeequalloadsaresuspendedfromthecableasshown.IfhB=hD=4¿,determinethesupportcomponentsatEandthesagatpointC.2UsingthediagraminProblem1,determinethesagatpointCifthemaximumtensioninthecableis1,200#.3Determinethecabletensionbetweeneachforce,anddeterminetherequiredlengthofthecable,forthesystemshown.4IfthecableinProblem3hadamaximumtensilecapacityof20k,determinetheminimumsagpermittedatC.112AnalysisofSelectedDeterminateStructuralSystems2EQUILIBRIUMOFRIGIDBODIESSimpleBeamswithDistributedLoadsDistributedloads,asthetermimplies,actonarelativelylargearea—toolargetobeconsideredasapointloadwithoutintroducinganappreciableerror.Examplesofdistributedloadsinclude:■■■■Furniture,appliances,people,etc.Windpressureonabuilding.Fluidpressureonaretainingwall.Snowloadonaroof.Pointorconcentratedloadshaveaspecificpointofapplication,whereasdistributedloadsarescatteredoverlargesurfaces.Mostcommonloadconditionsonbuildingstructuresbeginasdistributedloads.(SeeFigures20to24.)Figure20Singleconcentratedload,withanFBDofthesteelbeam.Figure21girder.Multipleconcentratedloads,withanFBDofthewood113AnalysisofSelectedDeterminateStructuralSystemsFigure22UniformlydistributedloadwithanFBDofthesteelbeam.Figure23Uniformlydistributedload,withaconcentratedload,withanFBDofthesteelbeam.Figure24Lineardistributionduetohydrostaticpressure,withanFBDoftheretainingwall.Supportreactionsforbeamsandotherrigidbodiesarecalculatedinthesamemanneremployedforconcentratedloads.Theequationsofequilibrium,gFx=0;gFy=0;andgM=0arestillvalidandnecessary.114AnalysisofSelectedDeterminateStructuralSystemsTocomputethebeamreactions,adistributedloadisreplacedbyanequivalentconcentratedload,whichactsthroughthecenterofgravityofthedistribution,orwhatisreferredtoasthecentroidoftheloadarea.Themagnitudeoftheequivalentconcentratedloadisequaltotheareaundertheloadcurve.Itshouldbenoted,however,thattheconcentratedloadisequivalenttothedistributedloadonlyasfarasexternalforcesareconcerned.Theinternalstresscondition,primarilybending,andthedeformationofthebeamareverymuchaffectedbyachangeofauniformloadtoaconcentratedload.Distributedloads(Figure25)maybethoughtofasaseriesofsmallconcentratedloadsFi.Forthisreason,themagnitudeofthedistributionisequaltothesummationoftheloadseries.Figure25Equivalentloadsystems.Thelocationoftheequivalentconcentratedloadisbasedonthecentroidoftheloadarea.Bygeometricconstruction,thecentroidsoftwoprimaryshapesareshowninFigures26and27.Figure26Centroidofarectangle(Areab*h).Figure27Centroidofatriangle(Area1/2*b*h).115AnalysisofSelectedDeterminateStructuralSystemsTrapezoidalshapesmaybethoughtofastwotriangles,orasarectanglewithatriangle.ThesetwocombinationsareillustratedinFigure28.Figure28(a)Area(1/2bh1)(1/2bh2).Figure28(b)Area(bh1)(1/2b(h2h1)).ExampleProblems:SimpleBeamswithDistributedLoads5DeterminethesupportreactionsatAandB.Solution:a.Calculatethemagnitudeoftheequivalentconcentratedforcefortheuniformdistribution.W=areaunderloadcurveW=1ω212¿=1500#>ft.212¿W=6,000#116AnalysisofSelectedDeterminateStructuralSystemsb.Theequivalentconcentratedforcemustbeplacedatthecentroidoftheloadarea.Becausetheloadareaisrectangularinthiscase,thecentroidislocatedathalfofthedistanceofthedistribution.c.Threeequationsofequilibriumcannowbewritten,andthesupportreactionsAx,Ay,andBycanbedetermined.CgFx=0DAx=0CgMA=0D-4,000#16¿2-6,000#114¿2+By120¿2=0By=+5,400#CgFy=0D+Ay-4,000#-6,000#+By=0Ay=+4,600#6SolveforthesupportreactionsatAandB.Solution:Atrapezoidaldistributionmaybehandledasarectangleplusatriangle,orastwotriangles.Thisproblemwillbesolvedintwowaystoillustratethisconcept.Alternate1:W1=1k>ft.118¿2=18kW2=1>2112¿212k>ft.2=12k117AnalysisofSelectedDeterminateStructuralSystemsCgFx=0DBx=0CgMB=0D+W214¿2+W119¿2-Ay112¿2=0Ay=12k14¿2+18k19¿2Ay=+17.5k12¿CgFy=0D+Ay-18k-12k+By=0By=+18k+12k-17.5k=12.5kAlternate2:W1=6¿11k>ft.2=6kW2=1>2112¿211k>ft.2=6kW3=1>2112¿213k>ft.2=18kCgFx=0DBx=0CgMB=0D+W1115¿2+W218¿2+W314¿2-Ay112¿2=0Ay=6k115¿2+6k18¿2+18k14¿212¿=17.5kCgFy=0D-W1-W2-W3+Ay+By=0By=6k+6k+18k-17.5k=12.5k118AnalysisofSelectedDeterminateStructuralSystemsProblemsFortheproblemsbelow,drawFBDs,anddeterminethesupportreactions.5Thebeamsupportsaroofthatweighs300poundsperfootandsupportsaconcentratedloadof1,200poundsattheoverhangingend.DeterminethesupportreactionsatAandB.6Thecarpenteronthescaffoldweighs800N.TheplankABweighs145N>m.DeterminethereactionsatAandB.7Atimberbeamisusedtosupporttwoconcentratedloadsandadistributedloadoveritscantileveredend.SolveforthepostreactionsatAandB.119AnalysisofSelectedDeterminateStructuralSystemsForProblems8through10,solveforthesupportreactionsatAandB.891011Analyzethereinforcedconcretestairwayandlandingshownbysamplinga1-foot-widestripofthestructure.Concrete(deadload)weighs150poundspercubicfoot,andtheliveload(occupancy)isequaltoabout100psf.120AnalysisofSelectedDeterminateStructuralSystems3PLANETRUSSESFigure29IllustrationofatrussbridgebyPalladio(bottomofPlateIV,BookIII,TheFourBooksofArchitecture).DevelopmentoftheTrussThehistoryofthedevelopmentofthetrussisalegacyoffitsandstarts.TheearliestevidenceoftrusstechnologyappearsinRomanstructures.ThisearlyevidenceoftheexistenceofthetrussisseeninVitruvius’swritingsonRomanbuildings.Inthesewritings,VitruviusclearlypointsoutthepresenceofmoderntrusstechnologyandleavesuswithlittledoubtthattheRomanbuildersunderstoodtheideasembodiedinthedistributionofforcesascarriedbyatriangulararrangementofmembers.ThisconceptoftriangularbracingintrussframeworksbeforetheRomanexampleisstillatopicofdebateforarchitecturalhistorians,largelybecausenophysicalevidenceofearliertrussedwoodenroofstructuresappearstohavesurvived.HistoricalresearchintoAsianbuildingtraditionshasalsobeenunabletoconfirmanyearlierevidenceoftrusstechnology.Afteraperiodofdisuse,Palladio,duringtheRenaissance,revivedtheuseoftrussframeworksandbuiltseveraltimberbridgesexceeding100feetinspan(Figure29).SincePalladio,recordsindicateacontinuedtraditionoftrussconstructionforbridgesandmanyothertypesofstructures.Thefirsttrussesusedwoodastheirmainstructuralmaterial.However,withtheadventoflarge-scaleironsmeltingandtheincreasedspanrequirementsofeverlargerbridges,theinclusionofothermaterialswasinevitable.Ironandthensteelemergedasthedominanttrussmaterialsandgreatlycontributedtothepopularityofthetechnologybyallowinggreaterandgreaterspanningpossibilities.121AnalysisofSelectedDeterminateStructuralSystems©Hulton-DeutschCollection/CORBISTothisday,buildingsandhighwaysaredottedwithtrussesfabricatedfromstructuralsteel.Theintroductionofstructuralsteelasaconstructionmaterialgavethedesigneranidealmediumforuseinthefabricationoftrusses.Steel,amaterialwithhightensileandcompressivestrength,couldbefastenedtogethereasilytoproducestrongconnections.Manufacturedinsectionsofvaryingshapes,cross-sectionalareas,andlengths,steelwastheanswerfortrussconstruction.Figure30TheFirthofForthBridgeinScotland,byBenjaminBaker,representsaninnovativeunderstandingoftrussbehavior.Photographerunknown.Becausethetheoryofthestaticallydeterminatetrussisoneofthesimplestproblemsinstructuralmechanics,andbecausealltheelementsforasolutionwereavailableinthe16thcentury,itissurprisingthatnoseriousattempttowardscientificdesignwasmadebeforethe19thcentury.Theimpetuswasprovidedbytheneedsoftherailways,whoseconstructioncommencedin1821.Theentireproblemofanalysisanddesignwassolvedbetween1830and1860(Figure30).Sincethattime,trussconstructionhasbeenappliedtootherstructures,suchaslong-spanroofs,inadditiontobridges(Figure31).Figure31Trussedarches(haunches)usedintheconstructionoftheblimphangarsduringWorldWarII.NASTillamook,Oregon(1942).PhotographcourtesyofTillamookCountyPioneerMuseum.122AnalysisofSelectedDeterminateStructuralSystemsDefinitionofaTrussAtrussrepresentsastructuralsystemthatdistributesloadstosupportsthroughalineararrangementofvarious-sizedmembersinpatternsofplanartriangles.Thetriangularsubdivisionoftheplanarsystem,Figure32(b),producesgeometricunitsthatarenondeformable(stable).(a)Unstable.Figure32(b)Stable.Aframeandatruss.Anidealpin-connectedtrussisdefinedasastableframeworkconsistingofelementsconnectedonlyattheirends;thus,nomemberiscontinuousthroughajoint.Noelementisrestrainedfromrotationaboutanyaxisperpendiculartotheplaneoftheframeworkandpassingthroughtheendsoftheelements.Theearlytrusseswerejoinedattheirendsbypins,resulting—exceptforsomefriction—inanidealtruss.Hence,underloadsappliedonlyatthejoints,themembersassumeeithertensile(elongation)orcompressive(shortening)forces.Inreality,however,itisimpracticaltojointrussmembersbypins.Thecurrentmethodofjoiningmembersisbybolting,welding,oracombinationofthetwo(Figure33).Figure33(a)Actualtruss.Figure33(b)joints.Idealizedtrusswithpin123AnalysisofSelectedDeterminateStructuralSystemsBecauseasingleboltoraspotweldisusuallynotsufficienttocarrytheloadinatrussmember,groupsofsuchfastenersareneeded.Whenevermorethanoneindividualfastenerisusedatajoint,thejointbecomesasomewhatrigidconnection,withtheabilitytodevelopsomemomentresistance.Hence,themembersofarealtruss,besideselongatingorshortening,alsotendtobend.Bendingstresses,however,areoftensmallincomparisontothoseresultingfromtensionorcompression.Inawell-designedtruss,thesebendingstresses(orso-calledsecondarystresses)arelessthan20%ofthetensileorcompressivestressesandareusuallyignoredinpreliminarydesign.Actualtrussesaremadeofseveraltrussesjoinedtogethertoformaspaceframework,asshowninFigure34.Secondarytrussingorcross-bracingbetweentheprimarytrussesprovidestherequiredstabilityperpendiculartotheplaneofthetruss.Eachtrussisdesignedtocarrythoseloadsthatactinitsplaneand,thus,maybetreatedasatwo-dimensionalstructure.Ingeneral,themembersofatrussareslenderandcansupportlittlebendingduetolateralloads.Allloads,therefore,shouldbeappliedtothevariousjointsandnotdirectlyonthemembersthemselves.Theweightsofthemembersofthetrussarealsoassumedtobeappliedtothejoints;halfoftheweightofeachmemberisappliedtoeachofthetwojointsthememberconnects.Inmostpreliminarytrussanalyses,memberweightsareneglected,becausetheyaresmallincomparisontotheappliedloads.Insummary,apreliminarytrussanalysisassumesthefollowing:Figure34Typicaluseoftrussesinbridgesandbuildings.1241.Membersarelinear.2.Membersarepin-connectedattheends(joints).3.Theweightoftrussmembersisusuallyneglected.4.Loadsareappliedtothetrussatthepinnedjointsonly.5.Secondarystressisneglectedatthejoints.AnalysisofSelectedDeterminateStructuralSystemsThus,eachmemberofthetrussmaybetreatedasatwoforcemember,andtheentiretrussmaybeconsideredasagroupoftwo-forcemembersjoinedbypins.Two-forcemembersareassumedtohavetheirloadsappliedonlyattheendpinorhinge;theresultantforceinthemembermustbealongtheaxisofthemember.Anindividualmembermaybesubjectedtoeithertensileorcompressiveforces.(a)FBDofthetruss.Figure35(b)FBDofjointB.FBDsofatrussandajoint.WhenisolatedasanFBD,atwo-forcememberisinequilibriumundertheactionoftwoforces—oneateachend.Theseendforces,therefore,mustbeequal,opposite,andcollinear,asshowninFigure35(b).Theircommonlineofactionmustpassthroughthecentersofthetwopinsandbecoincidentwiththeaxisofthemember.Whenatwoforcememberiscut,theforcewithinitisknowntoactalongitsaxis.Figure36givesexamplesofbridgetrusses.CommonrooftrussesareshowninFigure37.Figure36Examplesofbridgetrusses.125AnalysisofSelectedDeterminateStructuralSystemsFigure37126Examplesofcommonrooftrusses.AnalysisofSelectedDeterminateStructuralSystemsStabilityandDeterminacyofTrussesAninitialfirststepintheanalysisofatrussisthecalculationofitsexternaldeterminacyorindeterminacy.Iftherearemorereactioncomponentsthanapplicableequationsofequilibrium,thetrussisstaticallyindeterminatewithregardtoreactions—or,asusuallystated,thetrussisstaticallyindeterminateexternally.Iftherearefewerpossiblereactioncomponentsthanapplicableequationsofequilibrium,thestructureisunstableandundergoesexcessivedisplacementundercertainloadapplications.(Figure38).(a)Unstable(horizontally).(b)Unstable(horizontally).(c)Indeterminate(vertically).(d)Indeterminate(horizontally).Figure38Examplesofexternalinstabilityandindeterminacy.Atrusscanalsobedeterminate,indeterminate,orunstableinternallywithrespecttothesystemofbararrangement.Atrusscanbestaticallydeterminateinternallyandstaticallyindeterminateexternally,andthereverseofthiscanalsobetrue.Forthebarstoformastableconfigurationthatcanresistloadsappliedatthejoints,thebarsmustformtriangularfigures.Thesimpleststable,staticallydeterminatetrussconsistsofthreejointsandthreemembers,asshowninFigure39(a).Alargertrusscanbecreatedbyaddingtothesimpletruss.Thisrequirestheadditionofonejointandtwomembers,asshowninFigure39(b).127AnalysisofSelectedDeterminateStructuralSystemsIftheprocessofassemblinglargerandlargertrussescontinues,asshowninFigure39(c),adefiniterelationshipbetweenthenumberofjointsandthenumberofmembersisobserved.Everytimetwonewmembersareadded,thenumberofjointsisincreasedbyone.(a)(b)(c)Figure39Trussdevelopment.Thenumberofbarsasafunctionofthenumberofjointsforanytrussedstructureconsistingofanassemblageofbarsformingtrianglescanbeexpressedasb=2n-3wherebequalsthenumberofmembers(bars)andnequalsthenumberofjoints.Thisequationisusefulindeterminingtheinternaldeterminacyandstabilityofatruss;however,itisnotsufficientbyitself.Visualinspectionandanintuitivesensemustalsobeutilizedinassessingthestabilityofthetruss.128AnalysisofSelectedDeterminateStructuralSystemsFigure40showsexamplesofinternalstability,instability,anddeterminacy.NotethatthetrussshowninFigure40(d)isunstable.Here,theequationaloneisinsufficientinestablishingstability.Thesquarepanelinthecentermakesthistrussunstable.b=21n=122(n)–3=2(12)–3=21(a)Determinate.b=18(b)Indeterminate.b=16(c)Unstable.b=17n=10b=18>2(10)–3=17(Toomanymembers)n=10b=16<2(10)–3=17(Toofewmembers—squarepanelisunstable)n=10b=17=2(10)–3=17(d)Unstable.Figure40Examplesofinternalstabilityanddeterminacy.129AnalysisofSelectedDeterminateStructuralSystemsForceAnalysisbytheMethodofJointsThefirsthistoricalrecordofatrussanalysiswasdonein1847bySquireWhipple,anAmericanbridgebuilder.In1850,D.J.Jourawski,aRussianrailwayengineer,developedthemethodreferredtoastheresolutionofjoints.Thissectionwilldiscussthisjointmethod,whichinvolvesthenow-familiarrepetitiveuseofequationsofequilibrium.Oneofthekeyideasinthejointmethodmaybestatedasfollows:Foratrusstobeinequilibrium,eachpin(joint)ofthetrussmustalsobeinequilibrium.Thefirststepinthedeterminationofthememberforcesinatrussistodeterminealltheexternalforcesactingonthestructure.Aftertheappliedforcesaredetermined,thesupportreactionsarefoundbyapplyingthethreebasicequationsofstaticequilibrium.Next,ajointwithnomorethantwounknownmemberforcesisisolated,andanFBDisconstructed.Becauseeachtrussjointrepresentsatwodimensionalconcurrentforcesystem,onlytwoequationsofequilibriumcanbeutilized:gFx=0andgFy=0Thesetwoequilibriumequationspermitonlytwounknownstobesolved.Progressionthengoesfromjointtojoint,alwaysselectingthenextjointthathasnomorethantwounknownbarforces.130AnalysisofSelectedDeterminateStructuralSystemsExampleProblems:ForceAnalysisbytheMethodofJoints7SolveforthesupportreactionsatAandC,andthendetermineallmemberforces(Figure41).Figure41Methodofjoints.Solution:Step1:ConstructtheFBDoftheentiretruss.Step2:Solveforexternalsupportreactions.CgFx=0D-Ax+2kN=0Ax=2kNCgMA=0D-2kN19m2-2kN112m2+Cy124m2=0Cy=+1.75kNCgFy=0D+Ay-2kN+Cy=0Ay=0.25kNStep3:Isolateajointwithnomorethantwounknownmemberforces.ADandABareassumedastensionforces.Theequationsofequilibriumwillverifytheassumeddirections.131AnalysisofSelectedDeterminateStructuralSystemsStep4:Writeandsolvetheequationsofequilibrium.Assumebothmembersareintension.JointAForceAxFxFy0+.25kN3+AB5-2kNAy4+AB5ABAD+AD0035AB=-0.42kNPassumedwrongdirectionCgFy=0D+0.25kN+AB=045CgFx=0D-2kN+AB+AD=041-0.42kN25AD=+2kN+0.33kN=+2.33kNAD=+2kN-Note:TheFBDofjointArepresentstheforcesappliedtothetheoreticalpinatjointA.Theseforcesareactingequalandoppositeonthecorrespondingmembers,asshowninFigure42.Figure42132FBDofmembersandpin.AnalysisofSelectedDeterminateStructuralSystemsForcesdirectedawayfromapin(joint)aretensionforcesonthepin;conversely,forcespushingtowardapinarecompressionforces.Thesepinforcesshouldbeshownintheoppositedirectiononthemembers.Anassumptionmustbemadeaboutthedirectionofeachmemberforceonthejoint.Intuitionaboutthewaytheforceisappliedisperfectlyacceptable,becausetheequationsofequilibriumwillverifytheinitialassumptionintheend.Apositiveresultfortheequationindicatesacorrectassumption;conversely,anegativeanswermeansthattheinitialassumptionshouldbereversed.Manyengineersprefertoassumeinallcasesthattheforcesfromallunknownmembersareintension.Theresultantsignfromthesolutionoftheequationsofequilibriumwillconformtotheusualsignconventionfortrusses—thatis,plus1+2fortensionandminus1-2forcompression.Step5:Proceedtoanotherjointwithnomorunknownforces.ethantwoJointDCgFx=0D-2.33kN+DC=0DC=+2.33kNCgFy=0D+DB-2kN=0DB=+2kNJointBKnownforcesshouldbeshowninthecorrectdirection.ForceFxFyBD0-2kN3+10.42kN2=+.25kN52kNABBC+2kN4+1.42kN2=+.33kN54+BC503-BC5133AnalysisofSelectedDeterminateStructuralSystems45BC=-2.92kNPwrongdirectionassumedCgFx=0D+2kN+0.33kN+BC=035-2kN+.25kN+1.75kN=0PCHECKSCgFy=0D-2kN+0.25kN-1-2.92kN2=0Asummarydiagram,calledaforcesummationdiagram,shouldbedrawnasalaststep.Forcesummationdiagram.8DeterminethesupportreactionsandallmemberforcesforthetrussshowninFigure43.NotethatBDdoesnotconnectto,butbypasses,memberAC.Figure43Truss.Note:AhingeatbothAandDenablesthewalltofunctionasaforcememberAD,thusmakingthisastableconfiguration.IfarollerwereusedatD,theframeworkwouldbeunstable,resultingincollapse.134AnalysisofSelectedDeterminateStructuralSystemsSolution:FBDofentiretruss.Alltrussmembersareassumedtobetwo-forcememberswhenloadsareappliedatthejointsonly.Memberforcesarethereforeassumedtopassthroughtheaxisofthemember.BecauseonlymemberBDisattachedtothehingesupportatD,wecanassumethatsupportreactionsDxandDymustberelatedbyvirtueoftheslopeofBD,whereDx=D3DandDy=110110SupportreactionsAxandAy,however,havenodistinguishablesloperelationship,becauseAxandAymustserveasasupportfortwomembers,ABandAC.Whenonlyonetwo-forcememberframesintoasupport,wecanestablishasloperelationshipwithrespecttothemember.Therefore,CgMA=0D-20110¿2+Dx18¿2=0Dx=+25kD=Dx110=+25kA110B,andDy=3A25110B3D==+75k110110CgFx=0D-Ax+Dx=0Ax=+25kCgFy=0D-Ay+Dy-20k=0Ay=+75k-20k=+55k135AnalysisofSelectedDeterminateStructuralSystemsIsolateJointCForce20kFx0FyCA-CA0CB--20k3CB3.6CgFy=0D-20k+CB=+36kCgFx=0D-CACA=-31+36k23.6+2CB3.62CB=03.63CB=03.6=-30kJointAForceFxFyAy-25k0-30k0-55k0AB++AxCAAB12AB12OnlyoneequationofequilibriumisnecessarytodetermineforceAB,butthesecondequationmaybewrittenasacheck.CgFy=0D-55k+AB=+A5512BkAB=012CgFx=0D-25k-30k+0=0PCHECKForcesummationdiagram.136A5512Bk12=0AnalysisofSelectedDeterminateStructuralSystems9Thisexamplewillillustratea“shortcut”versionofthemethodofjoints.Theprincipleofequilibriumisstilladheredto;however,separateFBDswillnotbedrawnforeachsuccessivejoint.Instead,theoriginalFBDdrawnforthedeterminationofsupportequationswillbeusedexclusively.Thestepsoutlinedintheearlierexamplesarestillvalidand,thus,willbeemployedinthis“quick”method(Figure44).Figure44Step1:Step2:Quickmethodofjoints.DrawanFBDoftheentiretruss.SolveforthesupportreactionsatAandB.FBD(a)Step3:Isolateajointwithnomorethantwounknowns.Inthismethod,itisunnecessarytodrawanFBDofthejoint.Instead,usetheFBDfromsteps1and2,andrecordyoursolutionsdirectlyonthetruss.Itisusuallyeasiertofindthehorizontalandverticalcomponentsinallthemembersfirstandthendeterminetheactualresultantmemberforcesattheveryend.Itiscommonintrussanalysistousesloperelationshipsinsteadofanglemeasurements.Inthisillustration,diagonalmembershavevertical:horizontalslopesof4:3(3:4:5triangle).Becausealltrussmembersareassumedtobetwo-forcemembers,theverticalandhorizontalforcecomponentsarerelatedbythesloperelationship4:3.Step4:Equationsofequilibriumfortheisolatedjointwillbedonementally,withoutwritingthemout.Occasionally,however,somecomplexgeometrieswillrequireequationstobewrittenoutandsolvedsimultaneously.Let’sbeginatjointA(jointBisalsoapossiblestartingpoint).MembersAC(horizontalandverticalcomponents)andAE(horizontalcomponentonly)areunknown.BecausememberAChasaverticalcomponent,solvegFy=0.Thisresultsinaverticalcomponentof4k(tobalanceAy=4k).Therefore,withACy=4k,andbecauseACxandACyarerelatedbyvirtueoftheirsloperelationship,ACx=FBD(b)3ACy=3k;4137AnalysisofSelectedDeterminateStructuralSystemsNext,solvethehorizontalequilibriumcondition.ReactionAx=6kisdirectedtotheleft;therefore,AEmustbeaforceof9kgoingtotheright.JointAisnowinequilibrium.Recordtheresults(incomponentform)attheoppositeendofeachmember.FBD(c)Step5:Proceedtothenextjointwithnomorunknowns.Let’ssolvejointE.ethantwoThetwounknownsEF(horizontal)andEC(vertical)arereadilysolved,becauseeachrepresentsasingularunknownxandy,respectively.Inthehorizontaldirection,EFmustresistwith9ktotheleft.MemberforceEC=6kupwardtocounterthe6kappliedload.FBD(d)RecordmemberforcesEFandECatjointsFandC,respectively.Continuetoanotherjointwithnomorethantwounknowns.TryjointC.MembersCD(horizontal)andCF(horizontalandvertical)areunknown.Solvetheverticalconditionofequilibriumfirst.CAy=4kisupwardandCE=6kisdownward,leavinganunbalanceof2kinthedowndirection.Therefore,CFymustresistwithaforceof2kupward.Throughthesloperelationship,CFx=FBD(e)FBD(f)1383CF=1.5k4yInthehorizontaldirection,CDmustdeveloparesistanceof7.5ktotheleft.AnalysisofSelectedDeterminateStructuralSystemsStep6:RepeatStep5untilallmemberforcesaredetermined.Thefinalresultsareshownintheforcesummationdiagram.Atthispoint,resultantmemberforcesmaybecomputedbyusingtheknownsloperelationshipforeachmember.FBD(g)Forcesummationdiagram.ProblemsUsingthemethodofjoints,determinetheforceineachmemberofthetrussshown.Summarizetheresultsonaforcesummationdiagram,andindicatewhethereachmemberisintensionorcompression.Youmaywanttotrythe“quick”methodforProblems14through17.1213139AnalysisofSelectedDeterminateStructuralSystems14151617140AnalysisofSelectedDeterminateStructuralSystemsMethodofSectionsIn1862,theGermanengineer,AugustRitter,devisedanotheranalyticalapproachtotrussanalysis,themethodofsections.Rittercutthetrussalonganimaginarylineandreplacedtheinternalforceswithequivalentexternalforces.Bymakingspecificcutsandtakingmomentsaboutconvenientpointsonthetrusssection(cut)FBD,themagnitudeanddirectionofthedesiredmemberforceswereobtained.Themethodofsectionsisparticularlyusefulwhentheanalysisrequiresthesolutionofforcesinafewspecificmembers.Thismethodavoidsthelaborioustaskofajointby-jointanalysisfortheentiretruss.Insomeinstances,thegeometryofthetrussmaynecessitateuseofthemethodofsectionsinconjunctionwiththemethodofjoints.Exampleproblemswillbeusedtoillustratetheprocedureforsolvingspecificmemberforcesbythemethodofsections.ExampleProblems:MethodofSections10SolveformemberforcesBCandBE,asshowninFigure45.Figure45Spacediagramofthetruss.Solution:Step1:DrawanFBDoftheentiretruss.Solveforthesupportreactions.Thismaynotbenecessaryinsomecases,dependingonwhichsectionedFBDisused.FBD(a)Entiretruss.141AnalysisofSelectedDeterminateStructuralSystemsCgMG=0D-10k145¿2-10k160¿2+Ax120¿2=0Ax=+52.5kCgFy=0D+Gy-10k-10k=0Gy=+20kCgFx=0D-Ax+Gx=0Gx=+52.5kStep2:Passanimaginarylinethroughthetruss(thesectioncut).Thesectionlinea-amustnotcutacrossmorethanthreeunknownmembers,oneofwhichisthedesiredmember.Thislinedividesthetrussintotwocompletelyseparatepartsbutdoesnotintersectmorethanthreemembers.EitherofthetwoportionsofthetrussobtainedmaythenbeusedasanFBD.FBD(b)Sectioncutthroughtruss.Squigglelinesindicatethedesiredmemberforces.Note:Theinternalforcesinthememberscutbythesectionlineareshownasexternaldottedlinestoindicatethelineofactionofthesememberforces.Step3:DrawtheFBDofeitherportionofthetruss.EitherFBD(c)orFBD(d)maybeusedforthesolutionofmemberforcesBCandBE.FBD(d)Rightportionofthetruss.142FBD(c)Leftportionofthetruss.AnalysisofSelectedDeterminateStructuralSystemsInFBD(c),jointsCandEareimaginary,andjointsA,B,andGarereal.Themethodofsectionsoperatesontheideaofexternalforceequilibrium.UncutmembersABandBGhaveinternalforcesonlyanddonotaffecttheexternalforceequilibrium.NotethatifFBD(c)ischosenforthesolvingofmemberforcesBCandBE,thesupportreactionsareessentialandmustbesolvedfirst.CutmembersBC,BE,andGEareassumedtoactintension(T),asopposedtocompression(C).SolvingforBC,CgME=0D+Ax120¿2-Gy130¿2-BC120¿2=0¿1BC2120¿2=152.5k2120¿2-120k2130¿2BC=+22.5k1T2MomentsaretakenaboutimaginaryjointEbecausetheothertwounknownforces,BEandGE,intersectatE;therefore,theyneednotbeconsideredintheequationofequilibrium.ThesolutionofBCiscompletelyindependentofBEandGE.SolvingforBE,CgFy=0D+Gy-BEy=0‹BEy=+20k;butBEy=4BE,5thenBE=55BEy=120k2=+25k1T244Anequationinvolvingthesummationofforcesintheverticaldirectionwaschosen,becausetheothertwocutmembers,BCandGE,arehorizontal;therefore,theyareexcludedfromtheequationofequilibrium.Again,BEwassolvedindependentoftheothertwomembers.IfmemberforceGEisdesired,anindependentequationofequilibriumcanbewrittenasfollows:CgMB=0D+Gx120¿2-Gy115¿2+GE120¿2=0GE=-37.5k1C2143AnalysisofSelectedDeterminateStructuralSystemsEquationsofequilibriumbasedonFBD(d)wouldyieldresultsidenticaltothosewithFBD(c).Inthiscase,supportreactionsareunnecessary,becausenoneappearsinFBD(d).MembersCB,EB,andEGarethecutmembers;therefore,theyareshownasforces.JointsBandGareimaginary,becausetheywereremovedbythesectioncut.SolvingforCB,CgME=0D+CB120¿2-10k115¿2-10k130¿2=0CB=+22.5k1T2SolvingforEB,CgFy=0D-10k-10k+EBy=0EBy=+20k55EB=EBy=120k2=+25k1T244SolvingforEG,CgMB=0D-EG120¿2-10k130¿2-10k145¿2=0EG=-37.5k1C211DeterminetheforceinmembersBC,BG,andHG,asshowninFigure46.Figure46144Bowstringorcrescenttruss.AnalysisofSelectedDeterminateStructuralSystemsSolution:Solveforthesupportreactions,andthenpasssectiona-athroughthetruss.FBD(a)Entiretruss.SolvingforHG,CgMB=0D+HG115¿2-17.5k115¿2=0HG=+17.5k1T2FBD(b)Next,solveforBC,3BCBC;BCy=110110CgMG=0D-Ay130¿2+10k115¿2-BCx115¿2BCx=-BCy115¿2=0substituting:FBD(c)3BCBC115¿2+115¿2=-17.5k130¿2+10k115¿2110110BC=-6.25110k1C2145AnalysisofSelectedDeterminateStructuralSystemsThesolutionformemberforceBGcanbesolvedindependentofthealreadyknownmemberforcesHGandBCbymodifyingFBD(b).BG12BGBGy=12BGx=BecauseforcesHGandBCarenotparalleltoeachother,thereexistssomepointAOwheretheyintersect.Ifthisintersectionpointisusedasthereferenceoriginforsummingmoments,HGandBCwouldnotappearintheequationofequilibrium,andBGcouldbesolvedindependently.ThehorizontallengthAOisdeterminedbyusingtheslopeofforceBCasareference.Byutilizingasimilartrianglerelationship,AOisdeterminedtobe30¿.Therefore,CgMo=0D-10k145¿2-BGx115¿2-BGy145¿2+Ay130¿2=0BGBG115¿2+145¿2=17.5k130¿2-10k145¿21212BG=+1.2512k1T2146AnalysisofSelectedDeterminateStructuralSystemsProblems18SolveforAC,BC,andBDusingonlyonesectioncut.UsetherightportionfortheFBD.19SolveforBC,CH,andFH.20SolveforBE,CE,andFJ.21SolveforAB,BH,andHG.Useonlyonesectioncut.147AnalysisofSelectedDeterminateStructuralSystemsDiagonalTensionCountersAnarrangementofwebmemberswithinapanel,whichwascommonlyusedintheearlytrussbridges,consistedofcrosseddiagonals,asshowninFigure47.Figure47Diagonaltensioncounters.Crosseddiagonalsarestillusedinthebracingsystemsofbridgetrusses,buildings,andtowers.Thesecrosseddiagonalsareoftenreferredtoasdiagonaltensioncounters.Strictlyspeaking,suchasystemisfoundtobestaticallyindeterminate,becauseintheequation,b=2n-3;b=6,n=4672142-3Thephysicalnatureofthediagonalsinsomeofthesesystems,however,“breaksdown”intoastaticallydeterminatesystem.Diagonals,whicharelongandslenderandarefabricatedofcross-sectionssuchasroundbars,flatbars,orcables,tendtoberelativelyflexibleincomparisontotheothertrusselements.Becauseoftheirbucklingtendencies,theseslenderdiagonalmembersareincapableofresistingcompressiveforces.Inatrusswithatensioncountersystem,thediagonalslopinginonedirectionissubjecttoatensileforce,whereastheotherslopingmemberwouldbesubjecttoacompressiveforce.Ifthecompressivememberbuckles,theracking(paneldeformationcausedbyshearloading)inthepanelisresistedbythesingle-tensiondiagonal.Thecompressiondiagonalisineffective,andtheanalysisproceedsasifitwerenotpresent.Itis,however,necessarytohaveapairofcountersforagivenpanelinatruss,becauseloadreversalsmayoccurundervaryingloadconditions(e.g.,movingloads,windloads,etc.).148AnalysisofSelectedDeterminateStructuralSystemsDeterminingtheeffectivetensioncountercanbeaccomplishedreadilyusingamodifiedmethodofsectionsapproach.Themodificationinvolvescuttingthroughfourtrussmembers,includingbothcounters,ratherthantheusualthree-membercutrequiredinapplyingthemethodofsections,asindicatedintheprevioussection.ExampleProblem:DiagonalTensionCounters12ForthetrussshowninFigure48,determinetheeffectivetensioncounterandtheresultingloaddevelopedinit.Figure48Trusswithdiagonaltensioncounters.Solution:FBD(a)Sectiona-acuttingthroughfourmembers.Step1:DrawanFBDoftheentiretruss.Step2:SolveforthesupportreactionsatAandD.Step3:Passasectionthroughthetruss,cuttingthecountersandmembersBCandFE.Step4:Isolateoneportionofthecuttruss.149AnalysisofSelectedDeterminateStructuralSystemsFBD(b).Note:Sectioncuta-acutsthroughfourmembers,whichinmostinstancesisnotpermissible.However,becauseoneofthecountersisineffective(equaltozero),onlythreemembersareactuallycut.CountersBEandFCareeffectiveonlyintension;thus,theymustbeshownintensionintheFBD.Theequationofequilibriumnecessarytodeterminewhichcounteriseffectiveisthesummationofforcesintheverticaldirection.ForcesBCandFEarehorizontalforces;therefore,theyarenotconsideredintheequationofequilibrium.orCgFy=0D+Ay-10k-BEy=0(1)+Ay-10k+FCy=0(2)TheunbalancebetweenAy=6.7k1c2andtheapplied10k1T2loadis3.3k1T2.ComponentforceFCyhasthepropersensetoputthehalf-truss,FBD(b),intoequilibrium.Therefore,equation(2)iscorrect.Solving:butFCy=3.3kFC=5FC=+5.55k1T23yCounterBEisineffectiveandassumedtobezero.150AnalysisofSelectedDeterminateStructuralSystemsProblemsDeterminetheeffectivetensioncountersandtheirrespectivemagnitudes.222324151AnalysisofSelectedDeterminateStructuralSystemsZero-ForceMembersManytrussesthatappearcomplicatedcanbesimplifiedconsiderablybyrecognizingmembersthatcarrynoload.Thesemembersareoftenreferredtoaszero-forcemembers.ExampleProblems:Zero-ForceMembers13Let’sexamineatrussloadedasshowninFigure49.Byobservation,canyoupickoutthezero-forcemembers?MembersFC,EG,andHDcarrynoload.TheFBDsofjointsF,E,andHshowwhy.Figure49members.FBDofaparallelchordtrusswithzero-forceSolution:JointFCgFy=0D+FC=0JointECgFy=0D-EG=0JointHCgFy=0D+HD=0152AnalysisofSelectedDeterminateStructuralSystems14VisualinspectionofthetrussshowninFigure50indicatesthatmembersBL,KC,IE,andFHarezero-forcemembers(refertothepreviousproblem).FBDsofjointsKandI,similartothepreviousproblem,wouldshowthatKCandIEarezero,becausegFy=0.Figure50members.FBDofarooftrusswithzero-forceConstructanx-yaxissysteminwhichthexaxisisalongmemberlinesBAandBC.MemberforceBListheonlyonethathasaycomponent.Solution:CgFy=0D-BLy=0‹BL=0ThesameconditionisalsotrueatjointF.Ageneralrulethatcanbeappliedindeterminingzero-forcemembersbyinspectionmaybestatedasfollows:Atanunloadedjointwherethreemembersframetogether,iftwomembersareinastraightline,thethirdisazeroforcemember.Therefore,underfurtherinspection,LC,CJ,JE,andEHarealsofoundtobezero.Remember,zero-forcemembersarenotuselessmembers.Eveniftheycarrynoloadunderagivenloadingcondition,theymaycarryloadsiftheloadingconditionchanges.Also,thesemembersareneededtosupporttheweightofthetrussandmaintainthedesiredshape.JointB(similarforjointF).153AnalysisofSelectedDeterminateStructuralSystemsProblemsIdentifythezero-forcemembersinthetrussesbelow.252627154AnalysisofSelectedDeterminateStructuralSystems©VintageImages/Alamy4PINNEDFRAMES(MULTIFORCEMEMBERS)Figure51Heavytimber-framedbarn.Thisisanexampleofapinnedwoodframe.Photographerunknown.MultiforceMembersInSection3,trusseswerediscussedasstructuresconsistingentirelyofpinjointsandofstraighttwo-forcememberswheretheresultantforcedevelopedwasdirectedalongthemember’saxis.Nowwewillexaminestructuresthatcontainmembersthataremultiforcemembers—thatis,amemberacteduponbythreeormoreforces(Figure51andFigure52).Theseforcesaregenerallynotdirectedalongthemember’saxis;thus,theresultantmemberforcedirectionisunknown.Bendingofthememberistypicallytheoutcome.Pinnedframesarestructurescontainingmultiforcemembersthatareusuallydesignedtosupportanarrayofloadconditions.Somepinnedframesmayalsoincludetwoforcemembers.Forcesappliedtoatrussorpinnedframemustpassthroughthestructuralframeworkand,eventually,worktheirwaytothesupports.Inpinnedframes,forcesactfirstonthemembers,andthenmembersloadtheinternalpinsandjoints.Pinsredirecttheloadstoothermembersand,eventually,tothesupports.ExaminetheFBDsofboththetrussandframeshowninFigures53and54,respectively,andcomparetheloadtransfersthatresultinthemembersandpins.Figure52Exampleofpinnedframes.155AnalysisofSelectedDeterminateStructuralSystemsFigure53(a)two-force.MembersACandBCareFigure53(b)FBDofeachtwo-forcemember.Note:IntheFBDofmemberBC,showninFigure53(b),CxtogetherwithPwillresultinanxcomponentcompatibletoCyandproducearesultantCthatpassesthroughtheaxisofthemember.Figure54(a)Pinnedframewithmultiforcemembers.Figure54(b)FBDofeachmultiforcemember.Resultantsofthecomponentforcesatpinjointsdonotpassthroughtheaxisoftherespectivemember.Thelinesofactionofthepinforcesareunknown.156AnalysisofSelectedDeterminateStructuralSystemsRigidandNonrigidPinnedFramesinRelationtoTheirSupportSomepinnedframesceasetoberigidandstablewhendetachedfromtheirsupports.Thesetypesofframesareusuallyreferredtoasnonrigid,orcollapsible(Figure55).Otherframesremainentirelyrigid(retaintheirgeometry)evenifthesupportsareremoved(Figure56).(a)Withsupports.Figure55CaseI:Nonrigidwithoutsupports.(a)Withsupports.Figure56(b)Supportsremoved.(b)Supportsremoved.CaseII:Rigidwithoutsupports.InCaseI(nonrigid),thepinnedframecannotjustlybeconsideredasarigidbody,becausetheframecollapsesunderloadingwheneitherorbothofthetwopinsupportsareremoved.MembersACandBCshouldbetreatedastwoseparate,distinctrigidparts(Figure57).Ontheotherhand,theframeinCaseII(rigid)maintainsitsgeometrywiththepinandrollersupportsremoved.StabilityismaintainedbecausetheintroductionofmemberDEformsatriangular,trusslikeconfiguration.Theentirepinnedframeiseasilytransportableandcanbeconsideredasarigidbody(Figure58).157AnalysisofSelectedDeterminateStructuralSystemsFigure57CaseI:Twodistinctrigidparts.Figure58CaseII:Entireframeasarigidbody.Note:Althoughastructuremaybeconsideredasnonrigidwithoutitssupports,anFBDoftheentirenonrigidframecanbedrawnforequilibriumcalculations.158AnalysisofSelectedDeterminateStructuralSystemsProcedureforAnalysisofaPinnedFrame1.DrawanFBDoftheentireframe.2.Solvefortheexternalreactions.a.Iftheframeisstaticallydeterminateexternally(threeequationsofequilibrium,threeunknownreactions),allreactionscanbesolved(Figure59).Figure59■■Pinnedframe.Threeunknowns:Ax,Ay,andByThreeequationsofequilibriumNote:Directionfortheexternalreactionsorpinreactionsmaybeassumed.Awrongassumptionwillyieldthecorrectmagnitudebuthaveanegativesign.b.Iftheframeisstaticallyindeterminateexternally,reactionsmustbesolvedthroughothermeans(Figure60).Figure60■■FBDofthepinnedframe.Byrecognizingtwo-forcemembers.Bywritinganotherequationofequilibriumfortheextraunknown.ThisisaccomplishedbyseparatingtheframeintoitsrigidpartsanddrawingFBDsofeachpart[Figure61(c)].159AnalysisofSelectedDeterminateStructuralSystems(a)Pictorialdiagram.(b)FBD—entireframe.(c)FBDofthecomponents.Figure61FBDoftheframeandcomponents.Forexample,fromtheFBDoftheentireframeshowninFigure61(b),gMB=0EquationincludesAy.SolveforAy.gMA=0SolveforBy.gFx=0EquationintermsofAxandBx.FromtheFBDoftheleftcomponentofFigure61(c),gMC=0SolveforAx.GobacktothegFx=0equationandsolveforBx.3.Tosolveforinternalpinforces,dismembertheframeintoitscomponentparts,anddrawFBDsofeachpart(Figure62).160AnalysisofSelectedDeterminateStructuralSystemsFigure62FBDof(a)entireframeand(b)components.Note:Equalandoppositeforcesatpins(Newton’sthirdlaw).4.Calculatetheinternalpinforcesbywritingequilibriumequationsforthecomponentparts.5.TheFBDsofthecomponentpartsdonotincludeanFBDofthepinitself.Pinsinframesareconsideredasintegralwithoneofthemembers.Whenthepinsconnectthreeormoremembers,orwhenapinconnectsasupporttotwoormoremembers,itbecomesveryimportanttoassignthepintoaparticularmember.Forexample,seeFigure63.Figure63PinassignedtomemberAB.6.Whenloadsareappliedatajoint,theloadmaybearbitrarilyassignedtoeithermember.Note,however,thattheappliedloadmustappearonlyonce.Or,aloadcanbeassignedtothepin,andthepinmayormaynotbeassignedtoamember.Then,asolutionforthepincouldbedoneasaseparateFBD(Figure64).Figure64FBDofjointC.161AnalysisofSelectedDeterminateStructuralSystemsExampleProblems:AnalysisofaPinnedFrame15ForthepinnedframesshowninFigure65,determinethereactionsatAandBandthepinreactionsatC.Figure65(a)Pinnedframes.Solution:Step1:DrawanFBDoftheentireframe.Solveforasmanyexternalreactionsaspossible.Externally:■■Figure65(b)FBDofentirepinnedframe.FourunknownsThreeequationsofequilibriumCgMB=0D-Ay18¿2+400#14¿2+Ax16¿2-260#18¿2=0CgFx=0D-Ax+260#+Bx=0CgFy=0D+Ay-400#+150#+By=0BecausenosupportreactionscanbesolvedusingtheFBDoftheentireframe,goontostep2.162AnalysisofSelectedDeterminateStructuralSystemsStep2:Breaktheframeupintoitscomponentparts.FBD(a)FBD(b)FBDsofcomponentparts.FromFBD(a),CgMc=0D-Ay18¿2+400#14¿2=0Ay=+200#AssumptionOKGoingbacktothe3gMB=04equationfortheentireframe,Ax=200#18¿2+260#18¿2-400#14¿2=+346.7#6¿Bx=Ax-260#=347#-260#=+87#By=400#-Ay-150#=400#-200#-150#=+50#FromFBD(a),CgFx=0D-Ax+Cx=0Cx=+347#CgFy=0D+Ay-400#+Cy=0Cy=+200#orfromFBD(b),CgFx=0D-Cy+260#+Bx=0Cx=260#+87#=+347#PCHECKSCgFy=0D+150#-Cy+By=0Cy=150#+50#=+200#PCHECKS163AnalysisofSelectedDeterminateStructuralSystems16ForthepinnedframeshowninFigure66,determinethesupportreactionsatEandFandthepinreactionsatA,B,andC.Figure66Pinnedframe.Solution:FBDofentireframe.Staticallydeterminateexternally:■■ThreeunknownsThreeequationsofequilibriumCgME=0D+Fy15.33m2-3kN14m2=0Fy=2.25kNCgFy=0D+Ey-3kN+2.25kN=0Ey=+0.75kNCgFx=0DEx=0164AnalysisofSelectedDeterminateStructuralSystemsFBD(a)FBD(b)FBD(c)ComponentFBDs.FromFBD(b),CgMC=0D+By12.67m2-3kN11.33m2=0By=+1.5kNCgFy=0D-By+Cy-3kN=0Cy=3kN+1.5kN=+4.5kNFromFBD(a),CgMB=0DAx=0CgFx=0DBx=0CgFy=0D-Ay+By+Ey=0Ay=+1.5kN+0.75kN=+2.25kNFromFBD(c),CgFx=0D+Ax-Cx=0,butAx=0,thereforeCheck:Cx=0CgFy=0D+2.25kN-4.5kN+2.25kN=0165AnalysisofSelectedDeterminateStructuralSystems5THREE-HINGEDARCHESArchesArchesareastructuraltypesuitableforspanninglongdistances.Thearchmaybevisualizedasa“cable”turnedupsidedown,developingcompressivestressesofthesamemagnitudeasthetensilestressesinthecable.Forcesdevelopedwithinanarchareprimarilycompressive,withrelativelysmallbendingmoments.Theabsenceoflargebendingmomentsmadethearchesofantiquityideallysuitedformasonryconstruction.Contemporaryarchesmaybeconstructedasthree-hinged,two-hinged,orhingeless(fixed)(Figure67).Figure67166Contemporaryarchtypes.AnalysisofSelectedDeterminateStructuralSystemsAnarchisastructuralunitsupportedbyverticalaswellashorizontalreactions,asshowninFigure67.Thehorizontalreactionsmustbecapableofresistingthrustforces,orthearchtendstoflattenoutunderload.Archesincapableofresistinghorizontalthrustdegradeintoatypeofcurvedbeam,asshowninFigure68.Figure68Curvedbeam.Curvedbeamstendtohavelargebendingmoments,similartostraightbeams,andpossessnoneofthecharacteristicsofatruearch.Regardlessoftheshapetakenbyanarch,thesystemshouldprovideanunbrokencompressivepaththroughthearchtominimizeitsbending.Theefficiencyofanarchisdeterminedbythegeometryofitscurvedformrelativetotheloadintendedtobesupported(funicularshape).SomeofthemoreefficientgeometricformsforspecificloadingconditionsareshowninFigure69.Figure69Archshapes.167AnalysisofSelectedDeterminateStructuralSystemsThree-hingedarches(Figure70)arestaticallydeterminatesystemsandwillbetheonlycontemporaryarchtypestudiedinthesubsequentexamples.Figure70Three-hingedarch.Examinationofthethree-hingedarchshowninFigure70revealstworeactioncomponentsateachsupport,withatotaloffourunknowns.Threeequationsofequilibriumfromstatics,plusanadditionalequationofequilibriumofzeromomentattheinternal(crown)hinge,makethisintoastaticallydeterminatesystem.ThearchinFigure70issolvedbytakingmomentsatonesupportendtoobtaintheverticalreactioncomponentatthesecondsupport.Forexample,CgMA=0DsolvedirectlyforByBecausebothsupportsareatthesamelevelforthisexample,thehorizontalreactioncomponentBxpassesthroughA,notaffectingthemomentequation.WhenByissolved,theotherverticalreaction,Ay,maybeobtainedbyCgFy=0DsolveforAyFigure71FBDofeacharchsection.Solve:CgMC=0DofeitherFBD168AnalysisofSelectedDeterminateStructuralSystemsThehorizontalreactioncomponentsareobtainedbytakingmomentsatthecrownhingeC,asshowninFigure71.TheonlyunknownappearingineitherequationisthehorizontalreactioncomponentatAorB.Theotherhorizontalcomponentisfoundbywriting3gFx=04solveforremaininghorizontalreactionThethree-hingedarchhasbeenusedforbothbuildingsandbridges(sometimesintheformofatrussedarch,asshowninFigure72).Three-hingedarchescanundergosettlementsofsupportswithoutinducinglargebendingmomentsinthestructure.Thisistheprimaryadvantageofthethree-hingedarchovertheotherstaticallyindeterminatesystems.Figure72Three-hingedtimberarch,Lillehammer,Norway,OlympicHockeyArena.Archesrequirefoundationscapableofresistinglargethrustsatthesupports.Inarchesforbuildings,itispossibletocarrythethrustbytyingthesupportstogetherwithsteelrods,cable,steelsections,buttresses,foundationabutments,orspeciallydesignedfloors.Goodrockfoundationsprovideidealsupportsforresistinghorizontalthrust(Figure73).Continuousarches.Archwithabutments.Figure73Tiedarch.Methodsofthrustresolution.169AnalysisofSelectedDeterminateStructuralSystemsExampleProblems:Three-HingedArch17DeterminethesupportreactionsatAandBandtheinternalpinforcesatC.Thisexampleutilizesathreehingedarchwithsupportsatdifferentelevations,asshowninFigure74.Figure74Three-hingedarch.FBD(a)Solution:Ax=AA,Ay=1212Cx=CC,Cy=1212FromFBD(a),CgMB=0D+1,000#130¿2+4,000#140¿2+Ax120¿2-Ay1120¿2=0A=+1,90012#andAx=+1,900#,Ay=+1,900#Note:AxandAyresultina1:1relationshipidenticaltothatoftheimaginaryslopealongthelineconnectingAandC.170AnalysisofSelectedDeterminateStructuralSystemsFromFBD(b),CgFx=0D+Ax-Cx=0Cx=+1,900#CgFy=0D+Ay-Cy=0Cy=+1,900#FBD(b)FromFBD(c),CgFx=0D+Cx-1,000#-Bx=0Bx=+1,900#-1,000#=+900#CgFy=0D-4,000#+Cy+By=0By=+4,000#-1,900#=+2,100#FBD(c)Check:UsingFBD(a)again,CgFx=0D+Ax-1,000#-Bx=0+1,900#-1,000#-900#=0PCHECKS171AnalysisofSelectedDeterminateStructuralSystems18Asteelgabledframe(assembledasathree-hingedarch)issubjectedtowindforcesasshowninFigure75.DeterminethesupportreactionsatAandBandtheinternalpinforcesatC.Figure75Three-hingedgabledframe.Solution:FBDofframe(a).DeterminetheequivalentwindforcesF1,F2,andF3.F1=250#>ft.120¿2=5,000#F2=100#>ft.125.3¿2=2,530#F3=100#>ft.125.3¿2=2,530#172AnalysisofSelectedDeterminateStructuralSystemsFBDofframe(b)F3ResolveforcesF2andintoxandycomponentforces:F2x=F3x=812.53k2=0.8k25.3F2y=F3y-2412.53k2=2.4k25.3SolvefortheverticalsupportreactionsAyandBy.CgMA=0D-F1110¿2+F2x124¿2+F2y112¿2-F3x124¿2+F3y136¿2-By148¿2=0-5k110¿2+.8k124¿2+2.4k112¿2-.8k124¿2+2.4k136¿2-By148¿2=0‹By=+1.36kATBCgFy=0D-Ay+2.4k+2.4k-1.36k=0‹Ay=3.44kATBSeparatetheframeatthecrownjointC,anddrawanFBDoftheleftorrighthalf.173AnalysisofSelectedDeterminateStructuralSystemsFBD(c)FBD(d)UsingFBD(c),CgMC=0D-2.4k112¿2-0.8k14¿2+5k118¿2+3.44k124¿2-Ax128¿2=028Ax=140.56‹Ax=5.02k1;2FortheinternalpinforcesatC,CgFx=0D+5k-5.02k-0.8k+Cx=0‹Cx=+0.82kCgFy=0D-3.44k+2.4k+Cy=0‹Cy=+1.04kGoingbacktoFBD(b)oftheentireframe,solveforBx:CgFx=0D+5k-5.02k-0.8k+0.8k-Bx=0‹Bx=-0.02kThenegativesignintheresultforBxindicatesthattheoriginaldirectionassumedintheFBDwasincorrect.‹Bx=0.02k1:2174AnalysisofSelectedDeterminateStructuralSystemsProblemsDetermineallsupportandpinforcesforthemultiforcememberdiagramslistedbelow.282930175AnalysisofSelectedDeterminateStructuralSystems313233176AnalysisofSelectedDeterminateStructuralSystems6RETAININGWALLSAsthenameimplies,retainingwallsareusedtoholdback(retain)solidorothergranularmaterialtomaintainadifferenceingroundelevation.Adamisaretainingwallusedtoresistthelateralpressureofwaterorotherfluids.Therearethreegeneraltypesofretainingwalls:(a)thegravitywall(Figure76),(b)thereinforcedconcretecantileverretainingwall(Figure77),and(c)thereinforcedconcretecantileverretainingwallwithcounterforts(Figure78).Gravityretainingwallsaregenerallybuiltofplainconcreteormasonry.Heighthisgenerallylessthanfourfeet(1.3m).Agravitywalldependsonitsmasstogiveitstabilityagainstthehorizontalforcesfromthesoil.Slidingresistance(friction)isdevelopedbetweentheconcreteandsoilatthebase.Somemajordamsareconstructedasgravitywallsystems,butunderstandably,thebasedimensionsareimmense.Figure76Gravityretainingwall.Figure77wall.ReinforcedcantileverretainingFigure78Counterfortwall.Reinforcedconcretecantileverretainingwallsarethemostfrequentlyusedtypeofretainingwall,withaneffectivenessuptoaheight(h)ofabout20to25feet(6to7.6m).Stabilityofthiswalltypeisachievedbytheweightofthestructureandtheweightofthesoilontheheeloftheslabbase.Sometimesashearkeyisincludedatthebottomoftheslabbasetoincreasethewall’sresistancetosliding.Retainingwallsshouldhavetheirfoundationswellbelowthefrostline,andadequatedrainage(weep)holesnearthebottomofthewallshouldbeprovidedtopermitthewateraccumulationbehindthewalltoescape.Astheheightofaretainingwallincreases,thebendingmomentinthecantileverwallincreases,requiringmorethickness.Theadditionofcounterforts(verticaltriangular-shapedcross-walls)providestheadditionaldepthatthebasetoabsorbthelargebendingstresses.Counterfortwallsbehavelikeone-wayslabsthatspanhorizontallybetweenthecounterforts.Counterfortsarecalledbuttresseswhenthissameconfigurationisusedfortheretainedearththatisontheflatsideofthewall.Saturatedloosesandorgravel,granularsoil,ormudcausepressuresagainstretainingwallsinamannersimilartotruefluids(liquids)byexertingahorizontalpressure.Intrueliquids,likewater,thehorizontalpressureandtheverticalpressurearethesameatagivendepth.However,insoil,thehorizontalpressureislessthantheverticalpressure,withtheratiodependentonthephysical177AnalysisofSelectedDeterminateStructuralSystemspropertiesofthesoil.Soilpressure,aswithliquids,increasesproportionatelywithitsdepthbelowgrade(Figure79).Lateralpressureincreaseslinearlyfromzeroatthetoptoamaximumatthebottomofthefooting.p=ω¿*hwherep=themagnitudeoftheearthpressureinpsfor1kN>m22Figure79FBDofagravityretainingwall.ω¿=the“equivalent”fluidweight(density)ofthesoilinpoundspercubicfeet.Valuesrangefromaminimumof30pcf(forwell-graded,cleangravelsandmixes)to60pcf(forclayeysands).SIvaluesare4.7to9.4kN>m3.h=soildepthinfeet1m2.Therefore,P=1*1pmax*h2*1ft.or1m2whereP=thelateralforce(pounds,kips,N,orkN)basedontheareaofthepressuredistributionactingona1-footwide(1-m-wide)stripofwall.pmax=themaximumpressureatdepthh(psforkN/m2)Equivalentfluidpressureagainstaretainingwallmaycreateconditionsofinstability.Retainingwallsaresusceptibletothreefailuremodes:(a)sliding—whenthefrictionatthefootingbaseisinsufficienttoresistsliding;(b)overturningaboutthetoe—whenthelateralforceproducesanoverturningmomentgreaterthanthestabilizingmomentfromthewall’sweight,slabbaseweight,andthesoilmassabovetheheel;and(c)excessivebearingpressureatthetoe—whenthecombinationoftheverticaldownwardforceandthecompressionatthetoecausedbythehorizontalforceexceedstheallowablebearingpressureofthesoil.Thepressuredistributionunderthebase(Figure80)dependsuponthelocationandmagnitudeoftheresultant(verticalandhorizontalforces)forceasitpassesthroughthefootingbase.Figure80footing.178BearingpressureunderthewallAnalysisofacantileverretainingwallrequiresthattheequilibriumsummationofmomentsaboutthetoeisstable;thatis,theweightofthewallplusthebackfillontheheelexceedstheoverturningmomentoftheactivesoilpressurebyafactorofatleast1.5(asafetyfactorimposedbybuildingcodes).Onceastableconfigurationisachieved,thesoilpressuredistributiononthefootingmustbecalculatedtoensurethatthebearingpressuresarewithinallowablelimitsforthesoilonsite.AnalysisofSelectedDeterminateStructuralSystemsBuildingcodesrequirethatasafetyfactorof1.5alsobeprovidedtopreventslidingfailure.Inadditiontogeneralinstabilityissues,individualcomponentsofaretainingwall(wallthickness,basesize,basethickness,andreinforcementsteelquantityandlocation)mustbedesignedtoresistthebendingmomentsandshearforcesinducedbythesoilpressure.Manyofthefactorsconcernedwithslidinggobeyondthescopeofstatics;therefore,problemsinthissectionwillbelimitedtoinvestigatingawall’sresistancetooverturningandbearingpressurebeneaththetoe.ExampleProblems:OverturningStability19Asmallgravityretainingwall(Figure81)isusedtoaccommodatea412footdropinelevation.Determinethewall’sfactorofsafetyagainstoverturning.γconcrete=150ω¿=40##ft.3andtheequivalentfluiddensityft.3Solution:Analyzetheretainingwallbyassuminga1foottributarylengthofthewallasarepresentationoftheentirewall(Figure82).W1=1#11.5ft.214.5ft.211ft.2a1503b=506#2ft.W2=11ft.214.5ft.211ft.2a150Pmax=ω¿*h=a40#ft.3#ft.3b=675#b*14.5ft.2=180#Figure81Smallgravitydam.ft.211#1p2*1h2*11ft.2=a1802b14.5ft.211ft.22max2ft.=405#P3=TheoverturningmomentMOTMaboutthetoeatAisMOTM=P3*1.5ft.=1405#2*11.5ft.2=608#-ft.Thestabilizingmomentorrightingmoment,MRMisequaltoMRM=W1*11ft.2+W2*12ft.2MRM=1506#211ft.2+1675#212ft.2=1,856#-ft.ThefactorofsafetyagainstoverturningatAisSF=11,856#-ft.2MRM=3.0571.5=MOTM1608#-ft.2∴Thewallisstableagainstoverturning.Figure82Overturningandstabilizingforcesactingonthedam.179AnalysisofSelectedDeterminateStructuralSystemsForceP3=405#representsthehorizontalslidingforcethatmustberesistedatthebaseofthefootingthroughfrictionalresistance.20DeterminetheoverturningstabilityofthereinforcedcantileverretainingwallshowninFigure83.Whatistheslidingforcethatneedstoberesistedatthebaseofthefooting?γconc.=150pcf;γsoil=110pcfω¿=35pcf1equivalentfluiddensity2Solution:Assumea1-foot-widestripofwallasrepresentativeoftheentirewall(Figure84).W1=11ft.2110.67ft.211ft.2a150(wall)Figure83Cantileverretainingwall.W2=11.33ft.217ft.211ft.2a150(base)#ft.3#ft.3W3=14.5ft.2110.67ft.211ft.2a110(soil)b=1,600#b=1,397##ft.3b=5,282#EvaluatingthemomentsaboutthetoeatA,pmax=ω¿*h=a35#ft.3b*112ft.2=420#ft.211#P4=ab*1pmax2*h=ab*a4202b*112ft.222ft.=2,520#MOTM=P4*14ft.2=12,520#2*14ft.2=10,080#-ft.Therightingmoment,orthestabilizingmoment,aboutAisequaltoFigure84Cantileverretainingwallwithoverturningandstabilizingforces.MRM=W112ft.2+W213.5ft.2+W314.75ft.2=11,600#212ft.2+11,397#213.5ft.2+15,282#214.75ft.2MRM=33,180#-ft.ThesafetyfactoragainstoverturningiscomputedasSF=MRM33,180#-ft.==3.2971.5MOTM10,080#-ft.Theretainingwallisstableagainstoverturningaboutthetoe.Frictionalresistancedevelopedbetweenthebottomofthefootingandthesoilmustbeafactorof1.5(safetyfactor)thehorizontalfluidforceP4=2,520#,or3,780#.180AnalysisofSelectedDeterminateStructuralSystemsRetainingwalldesign,initsproportioningandelementsizing,mustensurethatthebearingpressureunderthefooting(atthetoe)remainsbelowtheallowablelimitforthesoilinvolved.Thepressuredistributionvaries,dependingonthelocationandmagnitudeoftheresultant(vectorsumofthehorizontalandverticalforces)asitpassesthroughthebaseofthefooting(Figures85through87).Notallresultantforcesarelocatedbeneaththewallsection.However,itisgenerallymoreeconomicaliftheresultantislocatedwithinthemiddlethirdofthebase(Figure85),becausethenthebearingpressureswillbeincompressionthroughoutthebase.Iftheresultantislocatedrightontheedgeofthemiddlethird,wherex=3a,thepressuredistributionresultsinatriangle,asshowninFigure86.Whentheresultantforceislocatedoutsidethemiddlethird,wherex63a,thepressurewouldindicatetensionatorneartheheel.Tensioncannotdevelopbetweensoilandaconcretefootingthatrestsonit.Therefore,thepressuredistributionshowninFigure87results,withtheimplicationthataslightliftingoffthesoiloccursattheheel.HistoricalexamplesexistofsomecathedralsbuiltduringtheMiddleAgesthatsufferedcatastrophicfailurewhentheresultantforcespassingthroughthestonewallsandfootingsfelloutsideofthemiddlethirdofthebase.Itisgenerallygoodpracticetoproportiontheretainingwalltohavetheresultantfallwithinthemiddlethird.Thispracticehelpstoreducethemagnitudeofthemaximumpressure,anditwillminimizethevariationbetweenthemaximumandminimumpressures.Figure85Trapezoidalpressuredistribution.Figure86Resultantatthethirdpoint—atriangularpressuredistribution.Figure87Tensionpossibleattheheel.ForcediagraminFigure85:Pmax=W14a-6x2a2Pmin=Wa216x-2a2181AnalysisofSelectedDeterminateStructuralSystemsForcediagraminFigure86:Pmax=2WaPmin=0ForcediagraminFigure87:Pmax2W3xPmin=0Ifawallweretobeconstructedonhighlycompressiblesoil,suchassometypesofclay,apressuredistributionasfoundinFigures86and87wouldresult.Thelargersettlementofthetoe,ascomparedtotheheel,wouldresultinatiltingofthewall.Foundationsconstructedoncompressiblesoilsshouldhaveresultantsfallingatornearthecenterofthefooting.Resultantscanfalloutsidethemiddlethirdifthefoundationisconstructedonveryincompressiblesoil,suchaswell-compactedgravelorrock.21ThebearingpressurewillbecheckedforthegravityretainingwallfoundinExampleProblem19.Assumethattheallowablebearingpressureis2,000psf.Figure88Forcesontheretainingwall.WallweightsandthehorizontalfluidforceP3areasshowninFigure88.ThelocationoftheresultantverticalforceWtotalisobtainedbywritingamomentequationinwhichMA=W1*11ft.2+W2*12ft.2MA=1506#211ft.2+1675#212ft.2=1,856#-ft.FromFigure89,amomentequationduetothewallweightcanalsobewrittenasMA=1Wtotal2*1b2whereWtotal=W1+W2=1506#2+1675#2=1,181#Equatingthetwoequations,MA=Wtotal*1b2=11,181#2*1b2=1,856#-ft.‹b=1.57ft.TheresultantforceRoftheverticalandhorizontalcomponentsisequalto:Figure89182Totalwallweightatitscentroid.R=4AW2total+P23B=4A1,1812+4052B=1,249#AnalysisofSelectedDeterminateStructuralSystemsLocatingthepointwheretheresultantintersectsthebaseofthefootingutilizesVarignon’stheorem,inwhichthemomentcausedbytheverticalandhorizontalforcesaboutthetoeatAisthesameasthemomentresultingfromRy(whichisequaltoWtotal)timesthedistancexfrompointA(Figure90).MRM-MOTMWtotalx=Valuesfortheoverturningmoment1MOTM2andtherightingmoment1MRM2areobtainedfromworkdoneinExampleProblem19.11,856#-ft.2-(608#-ft.)x=(1,181#)=1.06ft.Figure90Resultantwithinthemiddlethird.Thedimensionx=1.06ft.fallswithinthemiddlethirdofthebasedimension.2aa…x=1.06ft.…33Becausetheresultantiswithinthemiddlethirdofthebasedimension,theequationsforthemaximumandminimumbearingpressuresarepmax=Wa214a-6x2=pmax=a189pmin=W2a#ft.211,181#212.522b110-6.362=68816x-2a2=a=257#14*2.5ft.-6*1.06ft.21181#2.52#ft.2b16*1.06-2*2.52ft.2Themaximumbearingpressureiswellwithintheallowablebearingstresslimitof2,000psf.183AnalysisofSelectedDeterminateStructuralSystems22PerformacheckofthebearingpressurebeneaththewallfootinginExampleProblem20.Assumeanallowablebearingpressure3,000psf.Solution:ThetotaldownwardforceWtotalanditslocationcanbefoundasfollows(Figure91):Wtotal=W1+W2+W3Wtotal=11,600#2+11,400#2+15,280#2=8,280#Figure91Retainingwallforces.MA=1W1*2ft.2+1W2*3.5ft.2+1W3*4.75ft.2MA=11,600#*2ft.2+11,400#*3.5ft.2+15,280#*4.75ft.2=33,180#-ft.Also,MA=Wtotal*b.Equatingbothequations,MA=18,280#2*1b2=33,180#-ft.‹b=4.0ft.TheresultantforceRiscomputedasR=21W2total+P242=218,28022+12,52022=8,655#UsingthemomentvaluesobtainedinthesolutionofExampleProblem20,thedistancexfromthetoeatAcanbesolved.133,180#-ft.2-110,080#-ft.2MRM-MOTMx==Wtotal8,280#=2.8ft.a2a=2.33ft.6x=2.8ft.6=6.67ft.33Figure92thebase.ResultantforcepassingthroughTheresultantiswithinthemiddlethirdofthebase,asshowninFigure92;therefore,theentirebearingpressurewillbeincompression.Atrapezoidaldistributionresults,inwhich8,28014a6x2=ab14*7ft.-6*2.8ft.2a272=1,893psf=1893psf63000psf;‹OK.pmax=pmax184WAnalysisofSelectedDeterminateStructuralSystemsW16x-2a2a28,280=a2b16*2.8-2*72=473psf7pmin=23Eight-inchnominalconcretemasonry(CMU)blocksareusedtoretainsoilasshowninFigure93.Determinethewall’sstabilityagainstoverturning,thencheckthebearingpressureunderthefootingbaseatthetoe(A).Assumethattheallowablebearingpressureforthesoilislimitedto2,500psf.Assume:γCMU=125#-ft.3;#γconc=1503;equivalentdesignfluidpressureft#ω¿=403ftFigure93ReinforcedCMUretainingwall.Figure94third.ResultantforceoutsidethemiddleSolution:Assumea1-foot-widewidthofwallasrepresentativeoftheentirewall(Figure94).W1=WCMU=11ft.2(3.33ft.)a8#ft.ba1253b=279#12ft10#ft.b(2.33ft.)a1503b=291#12ft=W1+W2=279#+291#=570#W2=Wftg=11ft.2aWtotalThehorizontalforceonthewallduetothefluidpressureisequaltopmax=a40#ft3b(4.17ft.)=167psf111p21h211ft.2=1167#-ft.2214.17ft.211ft.22max2=348#P=ThelocationoftheresultantweightWtotalisdeterminedbysummingmomentsaboutthetoeatA.MA=W1*12ft.2+W2*11.17ft.2MA=1279#212ft.2+1291#211.17ft.2=899#-ft.MA=Wtotal*1b2=1570#2*1b2=899#-ft.‹b=1.58ft.1.6ft.TheresultantforceontheretainingwallisR=215702+34822=668#TheoverturningmomentcausedbythehorizontalforceaboutthetoeatAisMOTM=1348#2*a4.17ft.b=484#-ft.3185AnalysisofSelectedDeterminateStructuralSystemsTherightingorstabilizingmomentduetothetotalweightoftheretainingwalliscalculatedasMRM=Wtotal*b=1570#2*11.6ft.2=912#-ft.Checkingthesafetyfactoragainstoverturning:SF=MRM912#-ft.=1.971.5=MOTM484#-ft.‹OKResultantforceRintersectsthefootingbaseatx=1912-4842#-ft.MRM-MOTM=0.75ft.=Wtotal570#a=0.78ft.7x=0.75ft.3∴Theresultantfallsoutsidethemiddlethirdsectionofthefootingbase.pmax=‹OK1863157022W=507psf62,500psf=3x310.752AnalysisofSelectedDeterminateStructuralSystemsSupplementaryProblemsDistributedLoads:Section2Determinethesupportreactionsfortheproblemsbelow.DrawallappropriateFBDs.34353637187AnalysisofSelectedDeterminateStructuralSystemsTrusses—MethodofJoints:Section3Usingthemethodofjoints,determinetheforceineachmember.Summarizeyourresultsinaforcesummationdiagram.383940188AnalysisofSelectedDeterminateStructuralSystems4142Trusses—MethodofSections:Section343SolveforFG,DG,andAB.189AnalysisofSelectedDeterminateStructuralSystems19044SolveforCD,HD,andHG.45SolveforGB,HB,BE,andHE.46SolveforAB,BC,andDE.AnalysisofSelectedDeterminateStructuralSystemsTrusses—DiagonalTensionCounters:Section3Determinetheeffectivetensioncountersusingthemethodofsections.4748MethodofMembers:Sections4and5Determinethesupportreactionsandallinternalpinforces.49191AnalysisofSelectedDeterminateStructuralSystems50515253192AnalysisofSelectedDeterminateStructuralSystems54555657193AnalysisofSelectedDeterminateStructuralSystemsRetainingWalls:Section6Assumeanallowablebearingpressureof3,000psfforallproblemsbelow.58Agravityretainingwallasshownissubjectedtoalateralsoilpressureasaresultofanequivalentfluiddensityof35pcf.Calculatetheresultanthorizontalpressureagainstthewallandthewall’sfactorofsafetyagainstoverturning.Assumethatconcretehasadensityof150pcf.Checkthebearingpressureunderthefooting.59Ahighgravitywallwithaheightoffivefeet,sixinchesisusedtocontainasoilwithanequivalentfluiddensityof30pcf.Ifconcretehasadensityof150pcf,determinethewall’sfactorofsafetyagainstoverturning.Doesthebearingpressureremainwithintheallowablelimit?60AnL-shapedcantileverretainingwallconstructedofreinforcedconcrete,withadensityof150pcf,iseightfeettallfromthebasetothetopofthesoil.Assumingasoildensityof120pcfandanequivalentfluiddensityof40pcf,
determinetheoverturningstabilityofthewall.CheckthebearingpressurebeneaththetoeatpointA.61UsingthesamewallconfigurationasinProblem60,determinethewallstabilityifthewallstemisprojectedtotheoppositesidewithnosoilaboveit.Checkthebearingpressure.194AnalysisofSelectedDeterminateStructuralSystems62Determinetheoverturningstabilityofthecantileverretainingwallshown.Theequivalentfluiddensityis5.5kN/m3,soildensityis18kN/m3,andtheconcreteweighs23.5kN/m3.Doesthebearingpressurebeneaththetoeremainwithintheallowablebearingpressurelimit?63Checkontheoverturningstabilityofthewallshowniftheequivalentfluidpressureis40pcfandthesoildensityequals115pcf.Useaconcretedensityof150pcf.Evaluatethebearingpressuredevelopedatthebaseofthefooting.195AnalysisofSelectedDeterminateStructuralSystemsSummary■■■■■Tosolveforthesupportreactionsofrigidbodies(beams,trusses,andframes)withdistributedloads,anequivalentconcentratedload(s)issubstitutedthroughthecentroid(center)oftheloadarea.Themagnitudeoftheequivalentconcentratedloadisequaltotheareaunderthedistributedloadcurve.Planartrussesconsistofanarrangementoflinearelementsjoinedbyfrictionlesspins,formingapatternoftriangles.Thetrussisviewedasarigidbodywithindividualmemberssubjectedtotensionorcompressionforceswhenexternalloadsareappliedatthetrussjoints.Membersareconsideredastwo-forcemembers.Forceanalysisofplanartrussescanbecarriedoutusingthemethodofjointsorthemethodofsections.BothmethodsrequiretheuseofFBDsandtheequationsofequilibrium.Pinnedframescontainmembersthataremultiforcemembers—thatis,amemberacteduponbythreeormoreforces,generallyresultinginthememberexperiencingbendingmoment(s).MultipleFBDswithseveralsetsofequilibriumequationsarerequiredtodeterminethesupportreactionsandinternalpinforcesoftheframe.Threehingedarchesareanalyzedinamannersimilartopinnedframes.AnswerstoSelectedProblems1Ex=1,125lb.;Ey=450lb.;hc=5.33’163A=BA=13.27k(59.1kN);CB=12.15k(54.1kN);DC=12.28k(54.6kN);E=ED=13.03k(58kN)185A=750lb.(↑);B=5,850lb.(↑)6A=731N;B=598N8Ax=0;Ay=350lb.;By=1,550lb.10Ax=0;Ay=1,140lb.;By=360lb.11A=1,900lb.;E=1,900lb.13AB=.577kN(C);BC=.577kN(C);CD=1.732kN(C);BE=.577kN(T);EC=.577kN(C);AE=.289kN(T);ED=.866kN(T)14196AB=5.75k(T);BC=5k(T);BE=2.42k(T);BD=.21k(T);CD=7.07k(C);DE=8.56k(C)20222425272830AB=12kN(C);BC=3kN(C);CD=4kN(C);DE=0;EF=3kN(T);CE=5kN(T);BE=12kN(C);BF=15kN(T)AC=20.1k(T);BC=2.24k(C);BD=16k(C)BE=500lb.(C);CE=25025lb.(T);FJ=4,000lb.(C)EH=3.41k(T);HC=.34k(T);BI=2.84k(T)DB=1.2k(T);EA=4.7k(T)GH,GF,EF,FC,CD,andCBBM,MC,FO,OG,GK,GJ,andJH,EO,OKAx=455lb.(→);Ay=67lb.(↓);Bx=455lb.(←);By=417lb.(↑);Cx=455lb.;Cy=267lb.Ax=171.6kN;Ay=153.4kN;Bx=171.6kN;By=161.6kN;Cx=171.6kN;Cy=18.4kNAnalysisofSelectedDeterminateStructuralSystems3233Ax=10kN(→);Ay=2kN(↑);Bx=10kN(←);By=8kN(↑);Cx=10kN;Cy=8kNAx=4k(←);Ay=1k(↑);Cx=4k(→);Cy=3k(↑);BD=6k;Ex=4k;Ey=4k;Bx=4k;By=4k47DG=HG=25k(T);DF=54.2k(C);EG=38.4k(T)49Ax=693lb.(→);Ay=400lb.(↑);MA=3,144lb.-ft(clockwise);Cx=107lb.;Cy=400lb.;Bx=By=800lb.;Dx=Dy=800lb.51Ay=320lb.;Bx=0;By=80lb.Cx=0;Cy=280lb.;MC=1,280lb.-ft.35Ax=676lb.;Bx=308lb.36Ay=15k;Bx=0;By=5k;Cx=3k;Cy=10.5k;Dy=8.5k53Ax=0;Ay=333.3lb.;Dy=166.7lb.;Cx=44.4lb.;Cy=222.2lb.;ABx=44.4lb.;ABy=55.5lb.Ax=Ay=10k;Ex=10k;Ey=0;AB=1022k;BC=10.54k;BE=0;CD=16.67k;BD=13.33k;ED=10k55Ax=2,714lb.(←);Ay=286lb.(↑);Cx=1,726lb.(←);Cy=2,114lb.(↑);Bx=286lb.;By=2,714lb.56Ax=.33kN(←);Ay=4.26kN(↑);Dx=3.33kN(→);Dy=3.26kN(↓);Bx=3.33kN;By=4.26kN58P=157.5lb.,S.F.=1.43<1.50,pmax=2,000psf60MOTM=3,420lb.-ft.,MRM=8,025lb.-ft.,S.F.=2.35>1.50,pmax=2,530psf62P=83.2kN,MOTM=152kN-m,MRM=457kN-m,S.F.=3.0>1.5,pmax=164.4kN/m23739414345A=9,600lb.;B=3,200lb.Ax=200lb.;Ay=150lb.;Fx=200lb.;AB=200lb.(T);BC=200lb.(T);AD=150lb.(T);DF=150lb.(T);FE=250lb.(C);EC=250lb.(C);BD=BE=DE=0Ay=10k(↑);Cx=0;Cy=5k(↑);DG=8.33k(C);AB=4.8k(T);FG=1.87k(T)BG=2,700lb.(C);HE=1,875lb.(T);HB=1,179lb.(T)197198LoadTracingFigure1Loadtracingcanbevisualizedintheexposedlight-frameconstructionofanapartmentcomplexthroughtherooftrusses,studwalls(bearingandpartition),floorjoists,andfoundation.PhotocourtesyoftheSouthernForestProductsAssociation.1LOADTRACINGEarlyinthestructuraldesignphaseofaproject,aninitialassumptionismadebythedesigneraboutthepathacrosswhichforcesmusttravelastheymovethroughoutthestructuretothefoundation(ground).Loads(forces)travelalongloadpaths,andtheanalysismethodisoftenreferredtoasloadtracing(Figure1).FromChapter4ofStaticsandStrengthofMaterialsforArchitectureandBuildingConstruction,FourthEdition,BarryOnouye,KevinKane.Copyright©2012byPearsonEducation,Inc.PublishedbyPearsonPrenticeHall.Allrightsreserved.199LoadTracingEngineersoftenviewstructuresasinterdependentmechanismsbywhichloadsaredistributedtotheirindividualmembers,suchasroofsheathing,floorslabs,rafters,joists(aregular,relativelycloselyspacedseriesofsecondarybeams),beams,andcolumns(Figure2).Thestructuraldesignermakesjudgmentsregardingtheamountofloadassignedtoeachmemberandthemannerinwhichloadstravelthroughoutthestructure(loadpath).Loadtracinginvolvesthesystematicprocessofdeterminingloadsandsupportreactionsofindividualstructuralmembersastheyinturnaffecttheloadingofotherstructuralelements(Figure3).SimpledeterminatestructurescanbethoroughlyanalyzedusingFBDsinconjunctionwiththebasicequationsofequilibriumstudiedpreviously.Usually,theprocessbeginswiththeveryuppermostmemberorlevel,tracingloadsdownward,layerbylayer,untilthelastaffectedmemberunderinvestigationissolved.Inotherwords,startfromtheuppermostroofelement,andworkyourwaydownthroughthestructureuntilyoureachthefoundation.Figure2Loadpathsthroughasimplebuilding.Figure3200Loadsandloadpaths.LoadTracingLoadPathsIngeneral,theshortertheloadpathtoitsfoundationandthefewerelementsinvolvedindoingso,thegreatertheeconomyandefficiencyofthestructure.Themostefficientloadpathsalsoinvolvetheuniqueandinherentstrengthsofthestructuralmaterialsused—tensioninsteel,compressioninconcrete,andsoon.Bending,however,isarelativelyinefficientwaytoresistloads,andasaresult,beamsbecomerelativelylargeasloadsandspansincrease.SketchesofstructuralmembersintheformofFBDsareusedextensivelytoclarifytheforceconditionsofindividualelementsaswellasotherinterconnectedmembers.SimpledeterminatestructurescanbethoroughlyanalyzedusingFBDsinconjunctionwiththebasicequationsofequilibriumstudiedpreviously.Aslongaseachelementisdeterminate,theequationsofequilibriumaresufficienttodetermineallsupportingreactions.Loadtracingrequiresaninitialassessmentofthegeneralstructuralframeworktodeterminewheretheanalysisshouldbegin(Figure4).Figure4path.Decking-beam-girder-columnloadFigure5(a)FBDofthedecking.Figure5(b)FBDofthebeam.Figure5(c)FBDofthegirder.Figure5(d)FBDofthecolumn.Eachtimetheloadpathisredirected,asupportconditioniscreated,andtheloadsandreactionsateachtransfermustbeanalyzed(usingFBDs)andsolved(usingequations)(Figure5).201LoadTracingTributaryAreaLoadsuniformlydistributedoveranareaofrooforfloorareassignedtoindividualmembers(rafters,joists,beams,girders)basedontheconceptofdistributivearea,tributaryarea,orcontributoryarea.Thisconcepttypicallyconsiderstheareathatamembermustsupportasbeinghalfwaybetweentheadjacentsimilarmembers.Asectionofawoodfloorframingsystem(Figure6)willbeusedtoillustratethisconcept.Assumethatthegeneralloadovertheentiredeckareaisauniform50#>ft.2.■■■ThetributarywidthcontributingtotheloadonbeamBis21thedistance(plankspan)betweenAandBplus12thedistancebetweenBandC.ThetributarywidthofloadonbeamB=2+2=4¿.ThesameistrueforbeamC.Similarly,thetributarywidthforedgebeamsAandD=2¿.Beamloadsresultingfromauniformlyappliedloadconditionaredeterminedbymultiplyingtheloadinpoundspersquarefoot1#>ft.22bythetributarywidthofload:ω=1#>ft.22*1tributarywidth2TheloadoneachbeammaybeexpressedasωA,D=150#>ft.2*2¿2=100#>ft.Figure6Woodfloorsystemwithdeckingandbeams.FBDofbeamsAandD.ωB,C=150#>ft.2*4¿2=200#>ft.FBDofbeamsBandC.Loadtracinginvolvesthesystematicprocessofdeterminingloadsandsupportreactionsofindividualstructuralmembersastheyinturnaffecttheloadingofotherstructuralelements.202LoadTracingFramingDesignCriteria:DirectionofSpanArchitecturalcharacterThestructuralframing,ifexposed,cancontributesignificantlytothearchitecturalexpressionofbuildings.Shortjoistsloadingrelativelylongbeamsyieldshallowjoistsanddeepbeams.TheindividualstructuralbaysaremoreclearlyexpressedinFigure7.StructuralefficiencyandeconomyConsiderationsshouldincludethematerialsselectedforthestructuralsystem,thespancapability,andtheavailabilityofmaterialandskilledlabor.Standardsectionsandrepetitivespacingofuniformmembersaregenerallymoreeconomical.(a)Long,lightlyloadedjoistsbearingonshorterbeamscreateamoreuniformstructuraldepth.Spacecanbeconservedifthejoistsandbeamsareflushframed.MechanicalandelectricalsystemrequirementsThelocationanddirectionofmechanicalsystemsshouldbecoordinatedwiththeintendedstructuralsystem.Layeringthestructuralsystemprovidesspaceforductsandpipestocrossstructuralmembers,eliminatingtheneedtocutopeningsinthebeams.Flushorbuttframingsavesspaceinsituationswherefloor-to-floordimensionsarelimitedbyheightrestrictions.(b)Shortjoistsloadingrelativelylongbeamsyieldshallowjoistsanddeepbeams.Theindividualstructuralbaysaremoreclearlyexpressed.OpeningsforstairsandverticalpenetrationsMostframingsystemsaccommodateopenings,butgenerally,itismoreeconomicaltomakeopeningsparalleltothedominantspanningdirection.Additionalheadersandconnectionscreatepointloadsonmembersthataretypicallydesignedforlight,uniformloads,thusincreasingtheirsize.(e)Three-levelframingsystem.Photobyauthor.(c)Loadscanbereducedonselectedbeamsbyintroducingintermediatebeams.(d)Thespancapabilityofthedeckingmaterialcontrolsthespacingofthejoists,whereasbeamspacingiscontrolledbytheallowablejoistspan.Figure7Variousframingoptions.203LoadTracingLoadPaths:PitchedRoofSystemsSingle-levelframingRaftersandceilingjoistscombinetoformasimpletrussspanningfromwalltowall.Inadditiontothetrussaction(rafterspickupcompressionforces,andceilingjoistsdeveloptensiontoresistthehorizontalthrust),raftersexperiencebendingduetotheuniformloadalongtheirlength,asshowninFigures8(a)and8(b).(a)(b)Double-levelframingRoofjoistsorbeamsaresupportedbyaridgebeamatoneendandabearingwallorheaderbeamattheother.Noceilingtiesareused,becausethisarrangementdoesnotdevelopahorizontalthrust(asinthepreviousexample).Noticethateachlevelofstructuralframingspansinadirectionperpendiculartothenextlayer,asshowninFigures8(c)and8(d).(c)(d)Three-levelframingTheloadpathsequenceinthisarrangementstartswiththeloadstransferredfromthesheathing(decking)ontothepurlins,whichdistributeconcentratedloadsontotheroofbeams,whichinturntransmitloadtotheridgebeamatoneendandabearingwallorwallbeamattheother.Columnsorwallframingsupporttheridgebeamateitherend,asshowninFigures8(e)and8(f).(e)Figure8204Framingforpitchedroofs.(f)LoadTracingConstruction:PitchedRoofSystemsSingle-levelframingAcommonroofsystemforresidentialstructuresisarafter/ceilingjoistarrangement.Loadsontotheroofareinitiallysupportedbythesheathing(plywoodorotherstructuralpanelsorskipsheathing,usually1–*4–boardsspacedsomedistanceapart),whichinturnloadstherafters(Figure9).(a)FBD—rafters.Figure9(b)Typicallight-framestructure.PhotocourtesyoftheSouthernForestProductsAssociation.Single-levelframing.Double-levelframingAnothercommonroofframingarrangementinvolvesroofjoistsorbeamsthataresupportedbyaridgebeamatoneendandabearingwallorheaderbeamattheother.Theridgebeammustbesupportedateachendbyacolumnorbearingwall(Figure10).(a)FBD—rafters.Figure10(b)FBD—ridgebeam.(c)Post-and-beamconstruction.Photobyauthor.Double-levelframing.Three-levelframingAmethodusedtoachieveaheavierbeamappearanceisspacingtheroofbeams(ratherthantherafters)fartherapart,typically4to12feeto.c.Perpendiculartotheroofbeamsarepurlins,spacedfromonefoot,sixinchestofourfeeto.c.,supportingsheathing,decking,orametalroof.Inboththetwo-andthree-levelframingsystems,theceilingplanecanfollowtheroofslope(Figure11).(a)FBD—purlins.Figure11Three-levelframing.(b)FBD—roofbeams.(c)FBD—ridgebeam.205LoadTracingLoadPaths:WallSystemsFigure12Uniformwallloadfromaslab.Abearingwallisaverticalsupportsystemthattransmitscompressiveforcesthroughthewallplaneandtothefoundation.Uniformcompressiveforcesalongthelengthofthewallresultinarelativelyuniformdistributionofforce.Concentratedloadsordisruptionsinthestructuralcontinuityofthewall,suchaslargewindowordooropenings,willresultinanonuniformdistributionofcompressiveforcesonthefooting.Bearingwallsystemscanbeconstructedwithmasonry,cast-in-placeconcrete,sitecasttilt-upconcrete,orstuds(woodorlight-gaugemetalframing).Uniformslabloadsaredistributedalongthetopofthebearingwallasω.Amasonryorconcretewallfootingwillberequiredtosupportωplustheadditionalwallweight.Theloadω2=1ω1+wallweight2andremainsauniformload(Figure12).UniformdistributionFigure13Uniformwallloadfromraftersandjoists.Rooforfloorjoists(intypicallight-woodframing)arespaced16or24inchesoncenter.Thisregular,closespacingisassumedasauniformloadalongthetopofthewall.Ifnoopeningsdisrupttheloadpathfromthetopofthewall,auniformloadwillresultontopofthefooting(Figure13).NonuniformdistributionConcentratedloadsdevelopatthetopofawallwhenbeamsarespacedatwideintervals.Dependingonthewallmaterial,theconcentratedloaddistributesalonganangleof45°to60°asitmovesdownthewall.Theresultingfootingloadwillbenonuniform,withthelargestforcesdirectlyundertheappliedload(Figure14).Figure14Concentratedloadsfromwidelyspacedbeams.206LoadTracing“Archingaction”overopeningOpeningsinwallsalsoredirecttheloadstoeithersideoftheopening.Thenaturalstiffnessofaconcretewallundercompressionproducesan“archingaction”thatcontributestothelateraldistributionoftheloads(Figure15).Figure15Archingoverwallopenings.Figure16Studwallwithawindowopening.OpeninginastudwallStudwalls(woodandmetal)aregenerallyidealizedasmonolithicwalls(exceptforopenings)whenloadeduniformlyfromabove.Openingsrequiretheuseofheaders(beams)thatredirecttheloadstoeitherside.Concentratedloadsfromtheheaderreactionsmustbesupportedbyabuildupofstudsresemblingacolumn(Figure16).Concentratedloads—pilastersInspecialcaseswheretheconcentratedloadsareverylarge,wallsmayneedtobereinforcedwithpilastersdirectlyunderthebeam.Pilastersareessentiallycolumnsandcarrythelargeconcentratedloadsdirectlytothefooting.Thewallsbetweenthepilastersarenowconsideredasnonbearingwallsexceptforcarryingtheirownweight(Figure17).Figure17Pilasterssupportingconcentratedbeamloads.207LoadTracingLoadPaths:RoofandFloorSystemsOne-levelframingAlthoughitisnotacommonframingsystem,relativelylong-spanningdeckingmaterialsmaytransmitrooforfloorloadsdirectlytobearingwalls(Figure18).Figure18floor).One-levelframing(roofandFramingplan.Two-levelframingThisisaverycommonfloorsystemthatusesjoiststosupportadeck.Thedeckingislaidperpendiculartothejoistframing.Spandistancesbetweenbearingwallsandbeamsaffectthesizeandspacingofthejoists(Figure19).Figure19Two-levelframing.Framingplan.Three-levelframingWhenbearingwallsarereplacedbybeams(girdersortrusses)spanningbetweencolumns,theframinginvolvesthreelevels.Joistloadsaresupportedbymajorbeams,whichtransmittheirreactionstogirders,trusses,orcolumns.Eachlevelofframingisarrangedperpendiculartotheleveldirectlyaboveit(Figure20).Figure20208Three-levelframing.Framingplan.LoadTracingLoadPaths:RoofandFloorSystemsOne-levelframingPrecasthollow-coreconcreteplanksorheavy-timberplankdeckingcanbeusedtospanbetweencloselyspacedbearingwallsorbeams.Spacingofthesupports(thedistancebetweenbearingwalls)isbasedonthespancapabilityoftheconcreteplanksortimberdecking(Figure21).FBD—plank.Figure21One-levelframing.Two-levelframingEfficientstructuralsectionsinwoodandsteeljoistsallowrelativelylongspansbetweenbearingwalls.Lighterdeckmaterials,suchasplywoodpanels,canbeusedtospanbetweenthecloselyspacedjoists(Figure22).(a)FBD—decking.Figure22(b)FBD—joists.(c)Light-framedjoist-beamassembly.PhotobyChrisBrown.Two-levelframing.Three-levelframingBuildingsrequiringlargeopenfloorareas,freeofbearingwallsandwithaminimumnumberofcolumns,typicallyrelyonthelongspancapabilityofjoistssupportedbytrussesorgirders.Thespacingoftheprimarystructureandthelayeringofthesecondarystructuralmembersestablishregularbaysthatsubdividethespace(Figure23).(a)FBD—decking.(c)FBD—beams.Figure23(b)FBD—joists.(d)FBD—girder.(e)Joist-beamtruss;threelevel-framing.PhotobyChrisBrown.Three-levelframing.209LoadTracingLoadPaths:FoundationSystemsThefoundationsystemforaparticularstructureorbuildingdependsonthesizeofthebuilding,theuseofthestructure,thesubsurfaceconditionsatthesite,andthecostofthefoundationsystemtobeused.Alargebuildingwithheavyloadscanoftenbesupportedonrelativelyinexpensiveshallowfootingsifthesubsurfacesoilsaredenseandstable.However,thesamebuildingconstructedatasitecontainingsoftsoilsorexpansiveclaymayrequirepileorcaissonfoundations.Foundationsaregenerallysubdividedintotwomajorcategories:(a)shallowfoundationsand(b)deepfoundations.ShallowfoundationsFigure24Spreadfooting.Shallowfoundationsessentiallyobtaintheirsupportonsoilorrockjustbelowthebottomofthestructureindirectbearing.Verticalloadsaretransmittedfromwallsorcolumnstoafootingthatdistributestheloadoveranarealargeenoughthattheallowableload-carryingcapacityofthesoilisnotexceededand/orsettlementisminimized.Shallowfoundationsareofthreebasictypes:(a)spreadfootings—individualcolumnfooting,(b)continuousstripfootings—supportingabearingwall,and(c)matfoundations—coveringtheentireplanareaofthebuilding.Spreadfooting.Thisfootingtypeisusuallysquareorsometimescircularinplanandisgenerallysimpleandeconomicalformoderatetohighsoil-bearingcapacities.Thepurposeofthistypeoffootingistodistributetheloadoveralargeareaofsoil.Pedestalandfootingarereinforcedwithsteel(Figure24).Figure25Wallfooting.Wallfooting.Wallfootingsareoneofthemostcommonfootingtypesandsupportrelativelyuniformbearingwallloadsthroughacontinuousfoundationwall.Thewallfootingwidthremainsconstantthroughoutitslengthifnolargeconcentratedloadsoccur(Figure25).Matorraftfoundations.Matfoundationsareusedwhensoilbearingisrelativelylowortheloadsareheavyinrelationtosoil-bearingcapacities.Thisfoundationtypeisessentiallyonelargefootingundertheentirebuilding,thusdistributingtheloadovertheentiremat.Amatiscalledaraftfoundationwhenitisplaceddeepenoughinthesoilthatthesoilremovedduringexcavationequalsmostorallofthebuilding’sweight(Figure26).Figure26210Matorraftfoundation.LoadTracingDeepfoundationsThefunctionofadeepfoundationistocarrybuildingloadsbeneathalayerofunsatisfactorysoiltoasatisfactorybearingstratum.Deepfoundationsaregenerallypiles,piers,orcaissonsinstalledinavarietyofways.Normally,thereisnodifferencebetweenadrilledcaissonandadrilledpier,andmostoften,onlyamodestdifferenceindiameterexistsbetweenthem.Piles,themostcommondeep-foundationsystem,aredrivenintotheearthusingpile-drivinghammerspoweredbydrophammers,compressedair,ordieselengines.Buildingloadsaredistributedtothesoilincontactwiththesurfaceareaofthepilethroughskinfriction(frictionpiles),indirectbearing(bearingpiles)atthebottomofthepileonasoundstratumofearthorrock,oracombinationofskinfrictionanddirectbearing.Figure27Pilefoundations.Figure28Pilecapononepilegroup.Figure29wall.GradebeamsupportingabearingPilefoundations.Timberpilesarenormallyusedasfric-tionpiles,whereasconcreteandsteelpilesaregenerallyusedasbearingpiles.Whenbearingpilesmustbedriventogreatdepthstoreachsuitablebearing,acombinationofsteelandconcreteisused.Hollowsteelshellsaredrivenintothegroundtoapredeterminedbearingpoint,andthenthecasingsarefilledwithconcrete(Figure27).Pilecaps.Individualbuildingcolumnsaregenerallysupportedbyagroup(cluster)ofpiles.Athickreinforcedcapispouredontopofthepilegroupanddistributesthecolumnloadtoallthepilesinthecluster(Figure28).Gradebeams.Pilesorpierssupportingbearingwallsaregenerallyspacedatregularintervalsandconnectedwithacontinuousreinforcedconcretegradebeam.Thegradebeamisintendedtotransfertheloadsfromthebuildingwalltothepiles(Figure29).211LoadTracingFigure30Light-frameconstruction—adrugstoreinQuincy,Washington.PhotobyPhilLust.ExampleProblems:LoadTracingTheexampleproblemsthatfollowwillillustratetheloadtracingmethodologyasitappliestoavarietyofstructuralframeworksandarrangements.Notethatmostoftheexamplesillustratedarewood-framestructures,suchasthoseshowninFigures1and30.Woodframingistheonestructuralmaterialtypethatgenerallyresultsinadeterminateframingsystem,whereassteeland,particularly,cast-inplaceconcreteareoftendesignedtocapitalizeontheadvantagesofindeterminacythroughtheuseofrigidconnectionsand/orcontinuity.212LoadTracing1Inthesingle-bay,post-and-beamdeckillustrated,plankstypicallyareavailableinnominalwidthsoffourorsixinches,butforthepurposesofanalysis,itispermissibletoassumeaunitwidthequaltoonefoot.Determinetheplank,beam,andcolumnreactions.Solution:Loadonthedeck1liveload,orLL2=60psf=8psfDeckweight1deadload,orDL2Totalload1LL+DL2=68psfPLANKREACTIONLookingatthedeckinelevation,theloadωisdeterminedbymultiplyingthepoundspersquarefootloadbythetributarywidthoftheplank.Therefore:ω=68#>ft.211¿2=68#>ft.BEAMREACTIONTheplanksloadthebeamswithaloadof68#perfootoftheplankspan.Halfoftheplankloadistransferredtoeachbeam.Thebeamsareloadedbytheplankswithaloadof272#perfootofthebeamspan.R=68#>ft.218¿2ωL==272#1Beamreaction222Inaddition,thebeamhasaself-weightequivalentto10#/ft.COLUMNREACTIONHalfofeachbeamloadistransferredtothecolumnateachcornerofthedeck.Thecolumnsareloadedbythebeamswithloadsof1,692#ateachcolumn.Assumeeachcolumnhasaself-weightof100#.1272+102#>ft.112¿2ωL=22=1,692#1Columnreaction2R=GROUNDREACTIONTheloadateachcolumnisresistedbyanequivalentgroundreactionof1,792#.213LoadTracing2ThisproblemrepresentsanexpansionofExampleProblem1,wherethedeckinghasanadditionalsixfeettospanandthebeamsareextendedanotherbay.Loadsonthestructuralsystemremainthesame.Determinetheloadsdevelopedineachcolumnsupport.Assumethatcolumnsarelocatedatgrids1-A,2-A,3-A,1-B,2-B,3-B,1-C,2-C,and3-C.Solution:DeckDLDeckLLTotalload=8psf=60psf=68psfBeamself-weight=10#>ft.Columnself-weight=100#PLANKREACTIONω=68#>ft.211¿2=68#>ft.BEAMREACTIONFirst,analyzetheplanksthatspansixfeetbetweengridlinesAandB.R=68#>ft.216¿2ωL==204#1Beamreaction222Next,analyzetheplankloadsandbeamsupportfortheeightfootspanbetweengridsBandC.R=21468#>ft.218¿2ωL==272#1Beamreaction222LoadTracingCOLUMNREACTIONAllofthebeamcasesbelowrepresentuniformlyloadedconditionswithsimplysupportedends,whichresultinreactionsthatareR=ωL>2.Theresultingreactionsofthebeamsrepresenttheloadspresentineachcolumn.First,analyzethebeamsalonggridlineA:Next,analyzethebeamsalonggridlineB:Then,analyzethebeamsalonggridlineC:215LoadTracingCOLUMNLOADSANDREACTIONSTheperimetercolumnsalonggridlines1and3receivehalfoftheloadofeachbeam.Theinteriorcolumnsalonggridline2receiveloadsfromtwobeams,whichareaddedtogethertocalculatethecolumnloads.ColumnB-2FBDofColumnB-2216LoadTracingExampleProblem2—AlternateMethodAnothertechniquemaybeemployedindeterminingthebeamreactionswithoutgoingthroughananalysisoftheplanks.Thismaybeaccomplishedbyevaluatingthetributarywidthsofloadforeachbeamanddirectlycalculatingtheωforeachbeam.Forexample,inthefollowingfigure,thetributarywidthofloadassignedtothebeamsalonggridlineAisthreefeet.Therefore:ω=68#>ft.213¿2+10#>ft.=214#>ft.Thisωvaluecorrespondstotheresultobtainedinthepreviousmethod.AndalonggridlineC,withtributarywidth=4¿,ω=68#>ft.214¿2+10#>ft.=282#>ft.QCHECKSSimilarly,forbeamsalonggridlineB:Tributarywidth=3¿+4¿=7¿‹ω=68#>ft.217¿2+10#>ft.=486#>ft.QCHECKS217LoadTracing3Asteel-framedfloorforanofficebuildingwasdesignedtosupportaloadconditionasfollows:Loads:LiveLoadDeadLoads:ConcreteSteeldeckingMechanicalequipmentSuspendedceilingSteelbeamsSteelgirders=50psf======150#>ft.35psf10psf5psf25#>ft.35#>ft.UsingappropriateFBDs,determinethereactionforcesforbeamsB-1,B-2,andB-3andforgirderG-1.Solution:Loads:4–bA150#>ft.3B12in.>ft.=50psf1slab2+5psf1decking2+10psf1mechanicaleq.2+5psf1ceiling2TotalDL=70psfDL+LL=70psf+50psf=120psfDeadLoads(DL)=a218LoadTracingBeamB-1:(Tributarywidthofloadis6¿)ω1=120#>ft.216¿2+25#>ft.=745#>ft.BeamB-2:(Tributarywidthofloadis6¿+6¿=12¿)ω2=120#>ft.2112¿2+25#>ft.=1,465#>ft.BeamB-3:Thisbeamhastwodifferentloadconditionsduetothechangingtributarywidthcreatedbytheopening.For12feetofspan,ω3=120#>ft.2112¿2+25#>ft.=1,465#>ft.Forsixfeetofspan,ω4=120#>ft.216¿2+25#>ft.=745#>ft.219LoadTracing3©Ma=04-1745#>ft.218¿214¿2-11,465#>ft.2112¿2114¿2+By120¿2=0‹By=13,498#3©Fy=04-1745#>ft.218¿2-11465#>ft.2112¿2+13,498#+Ay=0‹Ay=10,042#GirderG-1:GirderG-1supportsreactionsfrombeamsB-2andB-3.BeamB-1sendsitsreactiondirectlytothecolumnandcausesnoloadtoappearingirderG-1.3©Ma=04-14,650#112¿2-13,498#124¿2-135#>ft.*36¿2118¿2+By136¿2=0‹By=14,512#3©Fy=04-14,650#-13,498#+14,512#+Ay=0‹Ay=14,896#220LoadTracing4Inthisexample,theloadtracewillinvolvetheframingforasmalldeckadditiontoaresidence.Oncepostreactionshavebeendetermined,apreliminaryfootingsizewillbedesignedassumingthesoilcapacityof3,000psfisknownfromageotechnicalinvestigation.Loads:LiveLoad=60#>ft.2DeadLoads:Decking=5#>ft.2Beams=5#>ft.Girder=10#>ft.γconcrete=150#>ft.3(density)Forthisload-traceproblem,wewillinvestigatethefollowing:1.DrawanFBDofthetypicalbeamwithitsloadconditionshown.2.DrawanFBDofthegirderwithitsloadconditionsshown.3.Determinetheloadineachpost.4.Determinethesizexofthecriticalpierfooting(accountfortheweightoftheconcrete).221LoadTracingSolution:1.Beam(typicalinterior):DeadLoad:5psf14¿2Beamwt.ωDL=20#>ft.=5#>ft.=25#>ft.LiveLoad:60psf14¿2=240#>ft.=265#>ft.ωDL+LL2.Beam(typicalexterior):DeadLoad:5psf12¿2Beamwt.ωDL=10#>ft.=5#>ft.=15#>ft.LiveLoad:60psf12¿2=120#>ft.=135#>ft.ωDL+LL3/4.Girderandpost:3©MB=04-3,066#14¿2-3,066#18¿2-1,562#112¿2-110#>ft.2112¿216¿2+Ay18¿2=0‹Ay=7,032#222LoadTracing3,066#-3,066#-3,066#-1,562#2-110#>ft.2112¿2+By1rightside2+7,032#=0‹ByR=2,315#3©Fy=04-ThetotalreactionatpostBisthesumofthereactions:ByR+ByL=4,630#5.Criticalfooting:Thesoiliscapableofresistingatotalbearingpressureof3,000#>ft.2Note:pressure=loadP;q=areaABysettingq=3,000#>ft.2(allowablecapacityofthesoil),weneedtodeducttheweightofthefootingitselftodeterminethefooting’scapacitytoresistappliedloadfromabove.Therefore,qnet=q-footingweight1aspressuremeasuredinpsf2Footingweightcanbesolvedbyconvertingthedensityofconcrete1γconcrete=150#>ft.32intoequivalentpoundspersquarefootunitsbymultiplying:footingwt.1psf2=1γconcrete21thicknessofconcreteinfeet2‹footingwt.=1150#>ft.32110–>12in.>ft.2=125#>ft.2Theremainingsoilcapacitytoresistpointloadsisexpressedasqnet=3,000#>ft.2-125#>ft.2=2,875#>ft.2Becausepressure=loadareaPP=2Ax7,032#P==2.45ft.2‹x2=qnet2,875#>ft.2qnet=‹x=1.57¿=1¿7–squarefooting1theoreticalsize2Practicalsizemaybex=2¿0–223LoadTracing5Calculatetheloadtraceforaslopedroofstructure.Roofing=5psfRoofSheathing=3psfRafters=4#/ft.}AlongrafterSnowload(SL):40psfonhorizontalprojectionofrafter*Beam:16#>ft.*Snowloadsarenormallygivenasaloadonthehorizontalprojectionofaroof.Solution:1.Rafteranalysis:Roofing:Sheathing:Rafter:5psf3psf18psf2124–>12in.>ft.2=16#>ft.+4#>ft.20#>ft.1Alongrafterlength2Adjusttohorizontalprojection:13120#>ft.2=21.7#>ft.1224–=80#>ft.12in.>ft.ω=21.7#>ft.+80#>ft.=101.7#>ft.SL=40#>ft.2*Note:Rafterhorizontallyprojectedforeaseofcomputation.224LoadTracingUsingFBDsketchesandcomputations,determinetheloadconditionon(a)beam,(b)column,and(c)wall(typicalstud).1.Beamanalysis:Thereactionfromatypicalroofjoistontopofthebeamandstudwallis813.6#.However,becausetheroofjoistsoccureverytwofeet,theequivalentloadωisequalto813.6#=406.8#>ft.+16#>ft.1beamweight22¿=422.8#>ft.ω=2.Columnanalysis:ThecolumnloadiscomputedasP=422.8#>ft.124¿2ωL==5,074#22Note:Thisequationsimplydividesthetotalloadontherafterinhalf,becausetherafterissymmetricallyloaded.3.Studwall:Tributarywalllengthperstudis16–=16–=1.33¿12in.>ft.225LoadTracing6Asimplelight-framedwoodbuildingissubjectedtotheloadconditionsasspecified.UsingFBDsandequationsofequilibrium,tracetheloadsthroughthebuildingforthefollowingelements:1.Determinetheequivalent(horizontallyprojected)loadontherafters.2.Determinetheloadperfootonthebearingwall.3.Determinetheloadontheridgebeam.4.Determinethecolumnloads.5.Determinetheminimumwidthofthecontinuousfoundation.6.Determinethesizeoftheinteriorfootings.LoadConditions:SoilBearingPressureFlooringSubfloorJoistsLL1occupancy2SLWalls=2,000psf=2psf=5psf=4psf=40psf=25psf=7psfAlongRafterLength:RoofingSheathingRaftersCeiling====5psf3psf2psf2psfRidgebeamspans16feetfromcolumnsupporttocolumnsupport.226LoadTracingSolution:1.Rafters:SL=25psf:raftersspaced16–oncenter.16¿ωSL=125#>ft.22ab=33.3#>ft.121Horizontalprojection2RoofDL=12psf16¿ωDL=112#>ft.22ab=16#>ft.121Alongrafterlength213ω¿DL=ab116#>ft.2=17.3#>ft.121Equivalentloadhorizontallyprojected2ωtotal=ωSL+ω¿DLωtotal=33.3#>ft.+17.3#>ft.=50.6#>ft.Thereactionateachraftersupportmaybedeterminedusingequilibriumequations.Whenuniformloadsonasimplysupportedmemberarepresent,asimpleformulamaybeusedwhereR=150.6#>ft.2114¿2ωL==354#22227LoadTracing2.Bearingwall:Thereactionoftherafterontothebearingwallis354#every16inches.Aconversionshouldbedonetoexpresstheloadatthetopofthewallinpoundsperlinealfoot.ω=1354#>16–2a12¿b=266#>ft.16Astripofwall1footwideandsevenfeet,sixinchestallweighsωwall=7#>ft.217.5¿2=52.5#>ft.ω=266#>ft.+52.5#>ft.=318.5#>ft.3.Ridgebeam:Rafterreactionsareequalto354#per16inches,or266#/ft.Becausetheridgebeamisrequiredtosupportraftersfrombothsides,ω=21266#>ft.2=532#>ft.Note:Theridgebeamsaretreatedastwosimplespanbeams,each16feetinlength.ExteriorcolumnssupportingtheridgebeamcarryPext=1532#>ft.218¿2=4,256#Note:Theeightfeetrepresentsthetributarybeamlengththatissupportedbytheexteriorcolumns.Interiorcolumnssupportatributarybeamlengthof16feet;therefore,Pint=1532#>ft.2116¿2=8,512#228LoadTracing4.Floorjoists:Joistsarespaced16inchesoncenter,whichalsorepresentsthetributarywidthofloadassignedtoeachjoist.Loads:DL+LL=11psf+40psf=51psf1FloorJoistLoadsandReactions2ωD+L=51#>ft.2a16¿b=68#>ft.12FoundationreactionsmaybeobtainedbyFoundation=168#>ft.217¿5tributarylength62=476#>16–ThecentralfloorbeamsupportsafloorjoistreactionequaltoBeam=168#>ft.2114¿5tributarylength62=952#>16–ConversionofthefloorjoistreactionsintoloadperfootresultsinFoundation:ω=1476#>16in.2aBeam:ω=1952#>16in.2a12¿b=357#1612¿b=714#165.Continuousfoundation:Thestemwallmeasureseightinchesthickandtwofeettall.Thefootingbaseiseightinchesthickandxwide.Loadsfromtheroof,wall,andfloorarecombinedasatotalloadontopofthefoundationstem:ωtotal=318.5#>ft.+357#>ft.=675.5#>ft.Thefoundationwallstemaddsadditionalloadonthefootingequalto8¿Stemweight=ab12¿21150#>ft.32=200#>ft.12Becausethefootingwidthxisunknown,theweightofthefootingbasemustbecomputedintermsofpoundspersquarefoot.Footingweight=a8¿b1150#>ft.32=100#>ft.212229LoadTracingIndeterminingthefootingwidth,examineaunitlength11¿2offoundationasarepresentationoftheentirelength.q=allowablesoilbearingpressure=2,000psfqnet=q-footingweight=2,000psf-100psf=1,900psfThisvalueofqnetrepresentstheresistanceofthesoilavailabletosafelysupporttheloadsofroof,walls,floor,andfoundationstem.ωtotal=675.5#>ft.+200#>ft.=875.5#>ft.TheminimumrequiredresistanceareaofthefootingperunitlengthisA=11¿21x2ωωqnet=;x=xqnet875.5#>ft.x==0.46¿L6–1,900#>ft.2Notethatafootingbaseofsixincheswouldbelessthanthethicknessofthestemwall.Aminimumfootingwidthforaone-story,light-framedassemblyshouldbe12inches.Ifa12inchbasewidthisprovided,theactualpressureonthesoilwillbe875.5#>ft.ω=x=1¿1¿=875.5#>ft.26qnetActualpressure=‹OK.6.Interiorspreadfootings:Adeterminationofindividualpostloadsisnecessarybeforefootingsizescanbecomputed.Assumethespreadfootingshaveathicknessofeightinchesandq=2,000psf:8¿ωfooting=ab1150#>ft.32=100#>ft.212qnet=q-footingwt.=2,000psf-100psf=1,900psfCriticalcentercolumn:14,224#P==7.5ft.2qnet1,900#>ft.2‹x=2.74¿L2¿9–squareA=x2=Otherfooting:x2=5,712#=3.00ft.21,900#>ft.2‹x=1.73¿L1¿9–square230LoadTracingProblemsIneachoftheload-tracingproblemsbelow,constructaseriesofFBDsandshowthepropagationofloadsthroughthevariousstructuralelements.1Determinethecolumnloadsassuming:DL1decking,flooring,etc.2=10psfLL1occupancy2=40psfTotal50psfBeamB-2spansbetweengirderG-3andcolumnB-2andbeamB-3spansbetweengirderG-3andcolumnC-2.2Loads:Roof:DL=10psfLL=25psf1snowhorizontallyprojected2Ceiling:DL=5psfLL=10psfBearingwalls:DL=10psf12ndand3rdfloors2Floors:DL=20psf12ndand3rdfloors2LL=40psf12ndand3rdfloors21.Determinetheequivalent(horizontallyprojected)loadontheraftersspacedattwofeet,zeroinches2’0”oncenter.2.Determinetheloadperfootonthebearingwalls.3.Determinetheloadingandbeamreactionsforeachofthesteelwide-flangebeams.231LoadTracing3Tracetheloadsthroughthefollowingelementsinthisstructure.Occupancyliveloadis40psf,withafloordeadloadof5psf.1.2.3.4.5.6.7.8.Rafters.Studwalls.Roofbeam.Columns(interiorandexterior).Floorjoist.Floorbeam.Loadontopofcontinuousfootings.Criticalinteriorfootingload.4DrawFBDsandshowloadconditionsforB-1,G-1,interiorcolumn,B-2,andG-2.Loads:SLRoofingandjoists=25psf=10psfTrussjoistInsul.,mech.,elec.BeamsB-1andB-2GirdersG-1andG-2====1deck22323psf5psf15#>ft.50#>ft.LoadTracing5RoofLoads:SLShakesPlywoodInsulationJoistsRidgebeam======20psf5psf2psf5psf4#>ft.40#>ft.1.Sketchtheloadanditsmagnitudeactingonthe34-foot-longridgebeam.2.WhatistheforceincolumnsAandB,whichsupporttheridgebeam?6Showgraphically(FBDsforeachelement)theloadtrace(loadcondition)for:1.2.3.4.5.6.7.8.9.Rafter(s).Roofbeam.Exteriorstudwall(s).Interiorcolumns.Floorjoist(s).Floorbeam.Floorpost.Exteriorfoundationwidth(s)(adequacy?).Sizeofcriticalpierfooting.LoadConditions:SL1horiz.proj.2Finishfloor>subfloorJoistsInsulationOccupancy1LL2Bearingwallsγconcrete1densityofconcrete2=======RoofingSheathingRaftersCeilingInsulationSoilbearingpressure=8psf=2psf=3psf=3psf=2psf=2,000psf30psf5psf3psf2psf40psf10psf150#>ft.3AlongRafterLength:233LoadTracing7Fortheillustratedhippedroof,evaluatetheloadconditionson:1.2.3.4.5.Typicaljackrafter.Hiprafter.Ceilingjoist.BeamsB-1,B-2,andB-3.Interiorcolumn.RoofLiveLoads:SL=25psfRoofDeadLoads:Roofing=6psfPlywoodroofsheathing=1.5psfJoistframing=4#>ft.CeilingLoads:DL=7psfLL=20psf234LoadTracing2LATERALSTABILITYLOADTRACINGThestructure,beitlargeorsmall,mustbestableandlasting,mustsatisfytheneedsforwhichitwasbuilt,andmustachievethemaximumresultwithminimummeans.Theseconditions:stability,durability,function,andmaximumresultswithminimummeans—orincurrentterms,economicefficiency—canbefoundtoanextentinallconstructionfromthemudhuttothemostmagnificentbuilding.Theycanbesummedupinthephrase“buildingcorrectly,”whichseemstomemoresuitablethanthemorespecific:“goodtechnicalconstruction.”Itiseasytoseethateachofthesecharacteristics,whichatfirstseemonlytechnicalandobjective,hasasubjective—andIwouldaddpsychological—componentwhichrelatesittotheaestheticandexpressiveappearanceofthecompletedwork.Stableresistancetoloadsandexternalforcescanbeachievedeitherbymeansofstructuresthatthebeholdercanimmediatelyoreasilyperceiveorbymeansoftechnicalartificesandunseenstructures.Itisevidentthateachapproachcausesadifferentpsychologicalreactionwhichinfluencestheexpressionofabuilding.Noonecouldfeelasenseoftranquilaestheticenjoymentinaspacewhosewallsorwhoseroofgavethevisualsensationofbeingonthevergeofcollapse,evenifinreality,becauseofunseenstructuralelements,theywereperfectlysafe.Similarly,anapparentinstabilitymight,undercertaincircumstances,createafeelingofparticularaesthetic—thoughanti-architectural—expression.Figure31Examplesoflateralinstability.Thetopphotoisofanoldcarportwithkneebracesaddedabittoolate.TheothertwophotographsareofwoodenstructuresdamagedintheKobe,Japan,earthquakeof1995.Photosbyauthors.Thusonecanseethateventhemosttechnicalandbasicqualityofconstruction,thatofstability,can,throughthedifferentbuildingmethodsemployedtoassureit,contributegreatlytotheachievementofadeterminedanddesiredarchitecturalexpression.—PierLuigiNervi,AestheticsandTechnologyinBuilding,HarvardUniversityPress,Cambridge,Mass.,1966.Pages2and3.Section1onloadtracingfollowedthepathofloadsthroughastructuralframeworktothefoundation.Deadandliveloadsonthestructureweregravity-inducedandassumedtobeactinginavertical,downwarddirection.Eachjoist,beam,girder,column,andthelikecouldbe235LoadTracinganalyzedusingappropriateFBDsalongwithequationsofequilibrium(inthecaseofstaticallydeterminatesystems).Althoughtheconditionsofequilibriumneedtobesatisfiedforeachelementormemberinthestructuralframework,itisnotasufficientconditiontoensurethegeometricstabilityofthewholestructure(seeFigure31).Stabilitycanbeproblematicforasinglestructuralelement,suchasanoverloadedbeam(Figure32[a])orabuckledcolumn(Figure32[b]),butsometimesanentirestructuralassemblymaybecomeunstableundercertainloadconditions.Geometricstabilityreferstoaconfigurationalpropertythatpreservesthegeometryofastructurethroughitselementsbeingstrategicallyarrangedandinteractingtogethertoresistloads.Figure32(a)Excessivebeamdeflection.Figure32(b)Columnbuckling.Allbuildingstructuresrequireacertainset(s)ofelements,referredtoasabracingsystem(Figure33),whichprovidestherequisitestabilityfortheentirestructuralgeometry.Decisionsaboutthetypeandlocationofthebracingsystemtobeuseddirectlyaffecttheorganizationalplanofthebuildingand,ultimately,itsfinalappearance.Figure33Windactingparalleltotheshearwalls.Figure34Windactingperpendiculartotheshearwalls.236Aprimaryconcerninthedesignofanystructureistoprovidesufficientstabilitytoresistcollapseandalsopreventexcessivelateraldeformation(racking;seeFigure34),whichmayresultinthecrackingofbrittlesurfacesandglass.Everybuildingshouldbeadequatelystiffenedagainsthorizontalforcescomingfromtwoperpendiculardirections.Windandseismicforcesonbuildingsareassumedtoacthorizontally(laterally)andmustberesistedincombinationwithgravityloads.Forexample,whenwindforcespushlaterallyonthesideofaone-story,wood-framedbuilding,thesehorizontalforcesaretransmittedbythesheathingorcladdingtoverticalframingelements(wallstuds),whichinturntransmittheloadstotheroofandfloor.Thehorizontalplanes(roofandfloor)mustbesupportedagainstlateralmovement.Forcesabsorbedbythefloorplanearesentdirectlyintothesupportingfoundationsystem,whereastheroofplane(referredtoastheroofdiaphragm)mustbesupportedbythewallsalignedparalleltothewinddirection.Intypicallight-wood-framedstructures,theselateralforce-resistingwallsarecalledshearwalls.Theuseofverticalwallframingandhorizontaldiaphragmsisthemostcommonsysteminwood-frameLoadTracingbuildings,becausetheroofsheathingcanbedesignedeconomicallytofunctionasbothaverticalload-carryingandlateralload-carryingelement.Roofandfloordiaphragmsmustbecapableoftransmittingtheappliedlateralforcestotheshearwallsthroughtheirplanarstrength;alternately,bracingmustbeprovidedinthehorizontalplane.Loadstransferredfromtheroofdiaphragmtothewallplanearethenchanneledtothefoundation.Let’sreturntotheearlierdiagramofthesimpleroofstructureshowninFigure35(a),whichissupportedbythetwoparallel(N-S)wallsandwhichhastwononstructuralwalls(glass)ontheothertwoparallelplanes.Atributaryslice(Figure35[b])throughtheframeworkrevealsarectilineararrangementthatissimplified,foranalysispurposes,asabeamsupportedbytwoposts.(a)Arrangementoflateralelements.Figure35(b)Idealizedframe.Structurewithtwoparallelshearwalls.Ifweassumetheconstructiontobeofwood,thebeamandpostconnectionsareappropriatelyassumedassimplepins,andthebasesupportsfunctionashinges.Thissimplerectilineargeometrywithfourhinges(Figure36)isFigure36Simpleframewithfourhinges.237LoadTracinginherentlyunstableandrequirestheadditionofotherstructuralelementstopreventlateralcollapsefromhorizontallyappliedloadsorunevenverticalloading.Thereareseveralwaysofachievingstabilityandcounteractingtherackingoftheframeunderverticaland/orhorizontalloading.Notethateachsolutionhasobviousarchitecturalimplications,andselectionofthebracingsystemmustbemadeforreasonsbeyondsimplybeingthemost“efficientstructurally.”DiagonalTrussMemberFigure37Diagonaltrussmember.Asimplewayofprovidinglateralstabilityistointroduceasimplediagonalmemberconnectingtwodiagonallyoppositecorners.Ineffect,averticaltrussiscreated,andstabilityisachievedthroughtriangulation.Ifasinglediagonalmemberisused,itmustbecapableofresistingbothtensionandcompressionforces,becauselateralloadsareassumedtooccurineitherdirection.Memberssubjectedtocompressionhaveatendencyto“buckle”(suddenlossofmemberstability)whentooslender(smallcross-sectionaldimensionwithrespecttolength);therefore,themembersneedtobeproportionedsimilarlytotrussmembersincompression(Figure37).X-BracingMembersFigure38Diagonaltensioncounters.Anotherstrategyinvolvestheuseoftwosmaller,crosssectionallydimensionedX-bracingmembers.TheseX-bracesarealsoknownasdiagonaltensioncounters(discussedinSection3.3),whereonlyonecounteriseffectiveinresistingadirectionallateralload(Figure38).Knee-BracingFigure39Knee-bracing.Acommonlyusedarrangementincarportsandelevatedwooddecksisknee-bracing.Thisstiffeningmethodtriangulatesthebeam-columnconnectiontoprovideadegreeofrigidityatthejoint.Thelargertheknee-bracesare,themoreeffectivetheirabilitytocontrolracking.Bracingisusuallyplacedascloseto45°aspossiblebutwillsometimesrangebetween30°and60°.Knee-bracesdeveloptensionandcompressionforces(liketrussmembers)dependingonthelateralforcedirection(Figure39).GussetPlatesFigure40238Rigidjointconnection.Largegussetplatesateachbeam-columnconnectioncanalsoprovidetherequiredrigiditytostabilizetheframe.However,inboththeknee-braceandgusset-platearrangements,somemovementwillstilloccurbecauseofthepinconnectionsatthebaseofthecolumns.Modifyingthebaseintoamorerigidconnectioncancertainlyaddtotheoverallrigidityoftheframe.Rigidconnectionsinducebendingmomentsinthebeamsandcolumns(Figure40).LoadTracingRigidBaseConditionColumnsplacedatsomedepthintothegroundandsetinconcretecanprovidearigidbasecondition.Resistancetolateralloadscomesthroughthecolumnsactingaslargeverticalcantileversandthehorizontalbeamtransferringloadsbetweenthecolumns(Figure41).Figure41Polestructure—columnswithrigidbases.CombinationKnee-BraceandRigidColumnBaseWhenknee-bracesareusedinconjunctionwithrigidcolumnbases,allconnectionsoftheframearenowrigid,andlateralloadsareresistedthroughthebendingresistanceofferedbythebeamandcolumns.Thelateraldisplacementswouldbelessthaninthethreepreviousexamples(Figure42).Figure42Rigidbaseandknee-bracing.RigidBeam/ColumnJointsMoment-resistingframesconsistoffloororroofmembersinplanewith,andconnectedto,columnmemberswithrigidorsemirigidjoints.Thestrengthandstiffnessofaframeisproportionaltothebeamandcolumnsizesandisinverselyproportionaltothecolumn’sunsupportedheightandspacing.Amoment-resistingframemaybeinternallylocatedwithinthebuilding,oritmaybeintheplaneoftheexteriorwalls.Momentframesrequireconsiderablylargerbeamsandcolumns,especiallyatthelowerlevelsoftallstructures.Allelementsinamomentframeareactuallybeam-columnssubjectedtocombinedstresses(bendingandtensionorcompression).Structuralsteelbeamsandcolumnsmaybeconnectedtogethertodevelopmomentframeactionbymeansofwelding,high-strengthbolting,oracombinationofthetwo(Figures43and44).Figure43Steelorconcreteframe(rigidjoints).239LoadTracingCast-in-placeconcrete,orprecastconcretewithcast-inplacejoints,providestherigidorsemirigidmonolithicjointsrequired.Framesmayconsistofbeamsandcolumns,flatslabsandcolumns,andslabswithbearingwalls.Theinherentcontinuitythatoccursinthemonolithiccastingofconcreteprovidesanaturallyoccurringmoment-resistantconnectionand,thus,enablesmemberstohavecantileverswithverysimpledetailingofthereinforcingsteel.(a)Steelmoment-resistingjoint.(b)Concretemomentconnection.Figure44concrete.Jointconnectionsbecomequitecomplexandlaborintensiveforathree-dimensionalrigidframeintwodirections.Thistranslatesintohighercost,soalternativelateralresistancesystems,suchasbracedframesorshearwallsinoneofthedirections,arepreferred.Ifweexaminetheideaofabeamontwocolumnsandimaginearooftrussspanningbetweentwocolumns,againtheissueoflateralstabilitymustberesolved.Althoughtrussesarestableconfigurationsasaresultoftriangulation,atrusssupportedontwopin-connectedcolumnsisunstable(Figure45).RigidconnectionsinsteelandFigure45Geometricallyunstabletrussroofstructure.Becauseknee-braceshelptodeveloprigidityatthecornerconnectionsforthebeamandcolumns,asimilararrangementcanbeprovidedwiththetrusstodevelopresistancetoracking.Knee-bracesattachtocontinuouscolumns,therebydevelopinglateralresistancethroughthecolumns’bendingcapacity(Figure46).(a)Pratttruss.Figure46240(b)Finktruss.Knee-bracedstructurewithrooftrusses.LoadTracingFigure47monitor.ModifiedFinktrusswithsideshedsandThecontinuouscolumnfromthegroundthroughtheentiredepthofthetrussprovidesaverystiffbend(rigidconnection)toresistlateralloads(Figure47).Columnsmustbedesignedtoresistthepotentiallylargebendingmomentsthatdevelop.Manyresidentialandsmalltomidscalecommercialbuildingsdependonthewalls(bearingandnonbearing)ofthestructuretodevelopthenecessaryresistancetolateralforces(Figure48).Thistypeoflateralrestraint,referredtoearlierasashearwall,dependsontheverticalcantilevercapacityofthewall.Thespanofthecantileverisequaltotheheightofthewall.Figure48Explodedviewofalight-framedwoodbuildingshowingthevariouslateralresistingcomponents.241LoadTracingShearwalls(a)Wideshearwall.Ashearwallactsasaverticalcantileverbeamtoprovidelateralsupport(seeFigures48,49,52,and54through56)andexperiencesbothshearandbendingstressesanddeformations.Thelateralresistanceofferedbytheshearwalldependsontherelativerigiditiesofthewallandthehorizontaldiaphragms.Shearwallsforsteelframedbuildingsarerelativelysolidmassesusuallymadeofreinforcedconcreteforavarietyofbuildingscalesandconfigurations.(b)Narrowshearwall.Whenconcretewallsareusedasfire-resistingcompartmentwalls,stairandelevatorcores,andverticalserviceshafts,itisreasonabletoutilizethemforstiffeningthebuildingagainstlateralloads.Acommonstructuralstrategyistouseabracedorrigidframeincombinationwithconcretewalls.Thebeamsofthesteelframemustbeconnectedtotheconcretecoresorshearwallstotransmittheverticalandhorizontalforcestothem.Carefuldetailingoftheframingelementstotheshearwallisessential.Figure49Shearwallproportions.(a)Diagonaltensioncounters.InFigure49,thewidthoftheshearwalldisrelativelylargecomparedwiththeheighth;therefore,sheardeformationreplacesbendingasthesignificantissue.Commonlyusedmaterialsforshearwallsareconcrete,concreteblocks,bricks,andwoodsheathingproducts,suchasplywood,orientedstrandboard(OSB),andwaferboards.MultipleBaysThusfar,thediscussionofframestabilityfromlateralloadshasbeenlimitedtosingle-bay(panel)frames;however,mostbuildingscontainmultiplebaysinthehorizontalandverticaldirections.Theprinciplesthatapplytosingle-bayframesalsoholdtrueformultiple-framestructures.(b)Diagonaltrussbrace.Remember,singlediagonalsmustbecapableoftensionorcompression.Thelengthofdiagonalscanbecomecriticalwhensubjectedtocompressionduetobuckling.Bracingdiagonalsshouldbekeptasshortaspossible.IntheexamplesshowninFigure50,itisquitepossiblethatonlyonepanelneedstobebracedfortheentireframetobestabilized.Itisrarelynecessaryforeverypaneltobebracedtoachievestability.(c)Shearwall.Figure50Bracingsystemsinmultiple-baystructures.242LoadTracingMultistory,MultibayStructuresMultistory,multibaystructuresalsousethesamebracingprinciples.However,asthestructuresbecomemuchtaller(heightgreaterthanthreetimestheleastbuildingdimension),onlycertaintypesofbracingsystemsandmaterialsofconstructionremainpracticalfromastructuraland/oreconomicstandpoint.Knee-braces,althoughappropriateforsmaller,one-ortwo-storystructures,arenotnearlyaseffectiveforlargerstructures.Thehorizontalforcecomponentoftheknee-braceontothecolumnproducessignificantbendingmoments,whichrequirelargercolumnsizes.Largerdiagonalbracesthatgoacrossanentirepanelfromoppositediagonalpointsaremuchmoreeffectivestructurally.(a)X-bracing.Diagonals,X-bracing,andK-trussingonmultistoryframesessentiallyformverticalcantilevertrussesthattransmitlateralloadstothefoundation(Figure51).Thesebracingtechniquesaregenerallylimitedtotheexteriorwallplanesofthebuildingtopermitmoreflexibilityfortheinteriorspaces.Reinforcedconcrete(ormasonry)andbracedsteelframingusedforstairwellsandelevatorsareoftenutilizedaspartofthelateralforcestrategy.(b)Eccentricbracedframe.Combinationsofbracing,shearwall,and/orrigidframesareusedinmanybuildings(Figure52).Larger,multistorybuildingscontainutility/servicecores,whichincludeelevators,stairs,ducts,andplumbingchases,thatarestrategicallyplacedtomeetfunctionalandstructuralcriteria.Becausethesecoresaregenerallysolidtomeetfireproofingrequirements,theycanfunctionasexcellentlateralresistingelements,eitherinisolationoraspartofalargeroverallstrategy.(c)K-trussing.(d)Shearwall.Figure52Combinationoflateralresistingsystems—steelframewithacentralshearwallcore.(e)Rigidframe.Figure51systems.Typesofmultistorybracing243LoadTracingRooforfloordiaphragmstocollectanddistributelateralloadstotheverticalresistingplanesHorizontalroof/floorbracingShearwallRigidframeDiagonalorX-bracingFigure53Bracingforathree-dimensionalframework.(X-bracing,trussdiagonals,knee-braces,shearwalls,andrigidbeam-columnconnectionscouldbeusedtostabilizeanyoftheseplanes.)Three-DimensionalFramesFigure53remindsusthatbuildingsare,indeed,three-dimensionalframeworksandnotplanar,two-dimensionalframes.Alloftheframeexamplesillustratedpreviouslyassumethattounderstandthewhole,onlyarepresentativeportionofanentirestructureneedstobeexamined.Eachplanarframerepresentsjustoneofseveral(ormany)framesthatconstitutethestructure.Itisimportanttonote,however,thatafundamentalrequirementofgeometricstabilityforathree-dimensionalstructureisitsabilitytoresistloadsfromthreeorthogonaldirections.Athree-dimensionalframecanbestabilizedbyuseofbracingelementsorshearwallsinalimitednumberofpanelsintheverticalandhorizontalplanes.Inmultistorystructures,thesebracingsystemsmustbeprovidedateachandeverystorylevel.Thetransverseexteriorwallsofabuildingtransmitthewindforcestotheroofandfloors,whichinturndirectthemtotheutility/servicecores,shearwalls,orbracedframes.Inmostcases,theroofandfloorsystemsformhorizontaldiaphragms,whichcanperformthisfunction.Inwood-framedbuildingsorbuildingswithwoodroofandfloorsystems,therooforfloorsheathingisdesignedandconnectedtothesupportingframingmemberstofunctionasahorizontaldiaphragmcapableoftransferringlateralloadtotheshearwalls.Inbuildingswithconcreteroofandfloorslabs,theslabsarealsodesignedtofunctionasdiaphragms.Itisunlikelythatthehorizontalsystemusedinonedirectionofloadingwillbedifferentfromthehorizontalsystemusedintheotherdirection.Ifthewoodsheathingorreinforcedconcreteslabisdesignedtofunctionasahorizontaldiaphragmforlateralforcesinonedirection,itprobablycanbedesignedtofunctionasadiaphragmforforcesappliedintheotherdirection.Occasionally,whentherooforfloorsheathingistoolightorflexibleand,thus,isunabletosustaindiaphragmforces,thehorizontalframeworkmustbedesignedtoincorporatebracingsimilartothebracedwallsorshearwalls.Horizontalbracingmayconsistoftensioncounters,trusses,orstiffpanelsinstrategiclocations(seeFigure53).244LoadTracingBracingConfigurationsOncetheroofplane(orfloors)hasbeenconfiguredtofunctionasadiaphragm,aminimumrequirementforstabilizingtheroofisthreebraced(orshear)wallsthatareneitherallparallelnorconcurrentatacommonpoint.Thearrangementofthewallsinrelationshiptooneanotheriscrucialinresistingloadsfrommultipledirections(Figure54).Morethanthreebraced(shear)wallsareusuallyprovided,thusincreasingthestructuralstiffnessoftheframeworkinresistinglateraldisplacements(sheardeformation).Braced(shear)wallsshouldbelocatedstrategicallythroughoutastructuretominimizethepotentialoftorsionaldisplacementsandmoments.Acommonsolutionistohavetwoshearwallsparalleltooneanother(areasonabledistanceapart)andathirdwall(orperhapsmore)perpendiculartotheothertwo.Figure54Variousshearwallarrangements—somestable,othersunstable.245LoadTracingMultistoryStructuresInmultistorystructures,lateralloads(fromwindorearthquakeforces)aredistributedtoeachofthefloor(diaphragm)levels.Atanygivenfloorlevel,theremustbearequisitenumberofbraced(shear)wallstotransferthecumulativelateralforcesfromthediaphragmsabove.Eachstorylevelissimilartothesimplestructuresexaminedpreviously,inwhichthediaphragmloadwastransferredfromtheupperlevel(roof)tothelowerlevel(ground).Multistorystructuresaregenerallybracedwithaminimumoffourbracedplanesperstory,witheachwallbeingpositionedtominimizetorsionalmomentsanddisplacement(Figures55through57).Althoughitisoftendesirabletopositionthebracedwallsinthesamepositionateachfloorlevel,itisnotalwaysnecessary.Thetransferofshearthroughanyonelevelmaybeexaminedasanisolatedproblem.(a)Figure55(b)Plandiagram.Shearwallsatthecentralcirculationcore.(a)Figure56Shearwallsattheexteriorcorners.(a)Figure57246(b)Plandiagram.(b)Plandiagram.Rigidframesatendbays(canalsocomprisetheentireskeleton).LoadTracingExampleProblems:LateralStability/DiaphragmsandShearwalls7Anindustrialbuildingwithaplandimensionof30feet*30feetandaheightof10feetissubjectedtoawindloadof20psf.Twobracedexteriorwallsparalleltothewinddirectionareusedtoresistthehorizontaldiaphragmforceintheroof.Assumingdiagonaltensioncountersintwoofthethreebays,determinethemagnitudeofforcedevelopedineachdiagonal.Solution:Diagonaltensioncountersarealwaysinpairs,becauseonecounterwillbuckleasaresultofcompressionloading.Themethodofsectionswasusedindeterminingtheeffectivecounter.ω=20psf*5¿=100#>ft.1100#>ft.2130ft.2ωL==1,500#V=22Inthiscase,countersAFandCHareeffectiveintension,whereasmembersBEandDGareassumedaszeroforcemembers.Utilizingthemethodofsections,drawanFBDoftheframeabovesectioncuta-a.AF12CHCHx=CHy=12AFx=AFy=OnlythexcomponentsofmembersAFandCHarecapableofresistingthe1,500#lateralforce.AssumingAFx=CHx,then3©Fx=04AFx+CHx=1,500#AFx=CHx=750#750=1,061#1T2AF=CH=12Completingtheanalysisusingthemethodofjointsfortrusses:247LoadTracing8Asimplecarportisframedusingknee-bracedframesspacedatfivefeetoncenter.Assumingwindpressureat20psf,analyzeatypicalinteriorframe.Solution:Eachinteriorknee-bracedframeisrequiredtoresistaloadappliedtoatributarywallareaof20squarefeetat20psf.F=p*A=20#>ft.2*20ft.2=400#NoticethatintheFBDoftheframe,atotaloffoursupportreactionsdevelopatAandB.Becauseonlythreeequationsofequilibriumarepermitted,allsupportreactionscannotbesolvedunlessanassumption(s)ismadeabouttheframeoritsloaddistributioncharacteristics.Inthiscase,onlyoneassumptionisnecessary,becausetheexternalsupportconditionisindeterminatetothefirstdegree.AssumeAx=Bx.Then,3©Fx=04Ax+Bx=400#‹Ax=Bx=200#Writingtheotherequationsofequilibrium,weget3©MA=04-400#18¿2+By110¿2=0By=320#3©Fy=04-Ay+By=0Ay=320#Oncethesupportreactionshavebeendetermined,passaverticalsectiona-athroughoneknee-brace.Isolatetheleftcolumn,anddrawanFBD.3©MD=04+CEx12¿2-200#18¿2=0‹CEx=800#CE=80012#CEy=800#3©Fx=04+400#-Ax+CEx-DEx=0248LoadTracingSubstitutingforAxandCEx,DEx=400#-200#+800#DEx=+1,000#3©Fy=04-DEy+CEy-Ay=0DEy=800#-320#DEy=+480#Inasimilarmanner,isolatetherightcolumnusingsectioncutb-b.3©MG=04-200#18¿2+FHx12¿2=0FHx=+800#FH=80012#FHy=800#3©Fy=04+FGy-FHy+By=0FGy=800#-320#=480#3©Fx=04-FGx+FHx-Bx=0FGx=800#-200#=600#9Inthissection,ananalysiswillbeperformedonasimplerectangular,one-storywoodbuilding,illustratingtheloadpropagationandtransferthatoccursinaroofdiaphragmandshearwallresistivesystem.Wood-framedroofsandwallsarestiffenedconsiderablybytheuseofplywoodorOSB/waferboardsheathingactingasdiaphragmsandshearwalls.Windforce(assumedasauniformpressureonthewindwardfrontwall)isinitiallydistributedtotheroofandfloordiaphragm(orfoundation).Conventionalwoodstudframinginthewallsfunctionsasverticalbeamsanddistributeshalfofthewindloadtotheroofandtheotherhalftothefloorconstruction.Pressuresonthewindwardwallandsuctionontheleewardwallconvertintouniformlydistributedloadsωalongthewindwardandleewardboundaryedgesoftheroofdiaphragm.Often,theloadsonthewindwardandleewardedgesarecombinedasoneloaddistributionalongthewindwardedge.249LoadTracingThe12wallheightdimensionrepresentsthetributarywalldimensionthatloadstheroofdiaphragm.Uniformloadωcarriestheunitsofpoundsperlinealfoot,thesameasauniformloadonabeam.ω=windpressurep*12thewallheightInfact,roofdiaphragmsareessentiallytreatedasflat,deepbeamsspanningfromwallsupporttowallsupport.ShearwallsrepresentsupportsfortheroofdiaphragmwithresultingreactionsV,whereV=ωL1pounds22Theintensityoftheshearreactionisexpressedaslowercasev,wherev=V1pounds>ft.2dShearloadVisappliedtothetopedgeoftheshearwall.WallequilibriumisestablishedbydevelopinganequalandoppositeshearreactionV¿atthefoundation,accompaniedbyatension(T)andcompression(C)coupleatthewalledgestocounteracttheoverturningmomentcreatedbyV.ThetensionTisnormallyreferredtoasthetie-downforce.Becausewindcannotbeassumedtoactinaprescribeddirection,anotheranalysisisrequiredwiththewindpressureappliedtotheendwalls,perpendicularindirectiontotheearlieranalysis.Eachwallisdesignedwiththerequisitewallsheathingthickness,nailsize,andspacingtoreflecttheresultsofthewindanalysis.Detailsofthisdesignprocedurearecoveredmorethoroughlyinsubsequentcoursesontimberdesign.Anexcellentdiscussion,withexamplesofshearwallsinactualstructures,canbefoundontheInternetathttp://www.mcvicker.com/vwall/dti.htm.250LoadTracing10AbeachcabinontheWashingtoncoast(100mphwindvelocity)isrequiredtoresistawindpressureof35psf.Assumingwood-frameconstruction,thecabinutilizesaroofdiaphragmandfourexteriorshearwallsforitslateralresistingstrategy.Drawanexplodedviewofthebuilding,andperformalateralloadtraceintheN-Sdirection.ShowthemagnitudeofshearVandintensityofshearvfortheroofandcriticalshearwall.Also,determinethetheoreticaltie-downforcenecessarytoestablishequilibriumoftheshearwall.Notethatthedeadweightofthewallcanbeusedtoaidinthestabilizingofthewall.251LoadTracingSolution:ω=35psf*7.5¿=262.5#>ft.Examiningtheroofdiaphragmasadeepbeamspanning42feetbetweenshearwalls:V=262.5#>ft.142¿2ωL==5,513#22AnFBDoftheshearwallshowsashearV¿developingatthebase(foundation)toequilibratetheshearVatthetopofthewall.Inadditiontoequilibriuminthehorizontaldirection,rotationalequilibriummustbemaintainedbythedevelopmentofaforcecoupleTandCattheedgesofthesolidportionofwall.v=V>shearwalllength=5,513#>15¿=368#>ft.W=deadloadofthewallW=10psf*15¿*15¿=2,250#Tie-downforceTisdeterminedbywritingamomentequationofequilibrium.SummingmomentsaboutpointA,3©MA=04-V115¿2+W115¿>22+T115¿2=015T=5,513#115¿2-2,250#17.5¿2182,695#ft.2-116,875#-ft.2T=15T=4,390#252LoadTracingProblems8Determinetheforcesineachofthemembers,includingtheeffectivetensioncounters.Assumethelateralforcetoberesistedequallybyeachtensioncounter.9DeterminethereactionforcesAandBandallothermemberforces.Unlikethepreviousproblem,knee-braceelementsarecapableofcarryingbothtensionandcompressionforces.AssumeAx=Bx.10Atwo-storywarehouseissubjectedtolateralforcesasshown.Determinetheeffectivetensioncountersandforcesinallothermembers.Assumetheeffectivediagonaltensioncountersatthelowerlevelshareequallyinresistingthehorizontalforces.11Abarnstructureissubjectedtowindvelocitieswiththeequivalentof20psfpressureontheverticalprojection(includingtheroof).Analyzethediaphragmandshearwallforcesassumingthewindhittingthelongdimensionofthebarn.UseexplodedviewFBDsintracingtheloads.Determinetheshearreactionandhold-downforcesatthebaseofthewalls.Note:Thedeadloadofthewallsmaybeusedinhelpingtostabilizeagainstrotationduetooverturningmoments.253LoadTracing12Asmallgarageutilizespremanufacturedtrussesspacedattwofeet,zeroinches2’0”oncenter.Onewallofthegaragehasa22footopening,framedwithalargegluelaminatedgirder.Tracetheloadsfromtheroof,anddeterminetheloadateachendoftheheader.Also,sizetheconcretefootingthatsupportstheheaderpost(assumethefootingissquare,10–thick).Summary■■■254Loadtracinginvolvesthesystematicprocessofdeterminingloadandsupportreactionsofindividualstructuralmembersastheyinturnaffecttheloadingofotherstructuralmembers.SimpledeterminatestructurescanbethoroughlyanalyzedusingFBDsinconjunctionwiththebasicequationsofequilibriumforrigidbodies.Loadsuniformlydistributedoveranareaofrooforfloorareassignedtoindividualmembersbasedontheconceptoftributary(contributory)area.Memberloadareasareassumedtobehalfwaybetweentheadjacentsimilarmembers.Allbuildingstructuresrequireacertainset(s)ofelements(bracingsystem)thatprovidesthenecessarylateralstabilityfortheentirestructuralgeometry.Windandseismicforcesonbuildingsareassumedtoacthorizontally(laterally)andmustberesistedincombinationwithgravityloads.LoadTracingAnswerstoSelectedProblems1ω=50psf×5’=250plfB1:Reaction(R)=1,250lb.B2:R=1,250lb.;B3:R=1,250lb.G1:R=1,250lb.;G2:R=3,750lb.G3:R=1,250/2,500lb.Col.Loads:A1=3,750lb.;D1=3,750lb.;B2=3,750lb.;C2=3,750lb.;A3=5,000lb.;D3=5,000lb.2ωsnow=50plf(horizontalprojection)ωDL=20plf(alongtherafterlength)ωtotal=70.6plf(equiv.horiz.proj.)Wallreaction=388lb.per2’Ridgebeamload=777lb.per2’Ceilingloadω=30plfThird-floorlevel:Loadattopofleftwall=269plfLoadattopofinteriorwall=165plfLoadattopofrightwall=284plfSecond-floorlevel:Loadattopofleftwall=649plfLoadattopofinteriorwall=905plfLoadattopofrightwall=724plfLoadonexterior‘W’beam=1,072plfLoadoninteriorbeam=1,706plf34Roofrafters(joists)=ω=66plf.Reactionatfrontwall=764lb.per2’Reactiononroofbeam=820lb.per2’Reactionatbackwall=396lb.per2’Loadatbaseoffrontwall=446plfLoadatbaseofbackwall=262plfRoofbeamload=410plfFloorjoistreactions:Reactionatfrontwall=672lb.per2’Reactionatbackwall=576lb.per2’Reactiononfloorbm.=1,248lb.per2’Topoffrontwallfooting=782plfTopofbackwallfooting=550plfCriticalinterfootingload=8,664lb.BeamB1:ω=335plf;wall/beamreaction=4,020lb.Girdercarriesconcentratedloadsof8,040lb.every8’oncenterplusthegirderweightof50plfCriticalcolumnload=42kipsBeamB2:ω=255plf;Wall/G2reaction=2,040lb.Girdersupportsaload=2,040lb.concentratedevery6’o.c.,andω=566plffromtrussjoists.5Criticalroofjoist:ωSL=26.7plf;ωDL=20plf;andtheequivalenttotalloadonthehorizontalprojectionoftheroofjoist=51.7plf.Rafterreaction=439lb.per16”.Theridgebeamsupportsatriangularloaddistributionwithapeakvalueof659plf(12ft.totherightofcolumnA).ColumnAload=3,140lb.;ColumnBload=9,420lb.6Rafterloads:ωSL=60plf(horiz.);ωDL=36plfωTotal=98plf(equiv.horiz.proj.)Roofbeamload=759plfTopofleftwall=343plfTopofrightwall=396plfFloorjoistload=66.7plfFloorbeamload=750plfwithcolumnloadsspacedat10’o.c.directlyoverfoundationposts.Topofleftcontinuousfoundationwallload=823plfTopofrightcontinuousfoundationload=926plfRequiredpierfootingsize=2’–10”sq.8AH=0;AG=500lb.(T);BG=400lb.(C);CF=0;CE=500lb.(T);DE=400lb.(C)10GK=5.66k(T);AG=CE=7.07k(T);IJ=JK=4k(C);KL=0;IH=JG=LE=0;KF=4k(C);HG=6k(C);GF=FE=5k(C);HA=0;BG=1k(C);FC=4k(C);ED=5k(C).11Totalloadatroofdiaphragmlevel=8,000lb.or200plfalongthe40ft.edgeoftheroof.Second-floordiaphragmload=8,000lb.or200plf.Shearatthetopofthesecond-floorwalls=4,000lb.;v=200plfthrough
thesecond-storywalls.Shearatthetopofthefirst-floorwalls,V=8,000lb.,andthewallshear,v=400plfThetie-downforceT=5,000lb.255256StrengthofMaterialsIntroductionStaticsisessentiallyforceanalysis:thedeterminationofthetotalinternalforcesproducedinmembersofastructuralframeworkbyexternallyappliedloads.Staticsinitselfisnotthedesignofanymember,butitisafirststepleadingtostructuraldesign.Theprimaryobjectiveforacourseofstudyinstrength(mechanics)ofmaterialsisthedevelopmentoftherelationshipbetweentheloadsappliedtoanonrigidbodyandtheresultinginternalforcesanddeformationsinducedinthebody.Theseinternalforces,togetherwithpredeterminedallowableunitstresses(usuallyexpressedinpoundspersquareinch),arethenusedtodeterminethesizeofastructuralelementrequiredtosafelyresisttheexternallyappliedloads.Thisformsthebasisofstructuraldesign.InhisbookDialoguesConcerningTwoNewSciences(1638),GalileoGalilei(Figure1)madereferencetothestrengthofbeamsandthepropertiesofstructuralmaterials.Hebecameoneoftheearlyscholarswhofosteredthedevelopmentofstrengthofmaterialsasanareaofstudy.1STRESSANDSTRAINItishopedthatthestudyofthestrengthofmaterialswillenablethereadertodevelopalogicalrationalefortheselectionandinvestigationofstructuralmembers.Thesubjectmattercoveredinthischapterestablishesthemethodologyforthesolutionofthreegeneraltypesofproblems:1.Design.Givenacertainfunctiontoperform(thesupportingofaroofsystemoverasportsarenaorthefloorsforamultistoryofficebuilding),ofwhatmaterialsshouldthestructurebeconstructed,andwhatshouldbethesizesandproportionsofthevariouselements?Thisconstitutesstructuraldesign,wherethereisoftennosinglesolutiontoagivenproblem,asthereisinstatics.Figure1GalileoGalilei(1564–1642).Galileo,pushedbyhismathematicianfathertostudymedicine,waspurposelykeptfromthestudyofmathematics.Fatetookaturn,however,andGalileoaccidentallyattendedalectureongeometry.Hepursuedthesubjectfurther,whicheventuallyledhimtotheworksofArchimedes.Galileopleaded,andreluctantly,hisfatherconcededandpermittedhimtopursuethestudyofmathematicsandphysics.Galileo’sfundamentalcontributiontosciencewashisemphasisondirectobservationandexperimentationratherthanonblindfaithintheauthorityofancientscientists.Hisliterarytalentenabledhimtodescribehistheoriesandpresenthisquantitativemethodinanexquisitemanner.Galileoisregardedasthefounderofmodernphysicalscience,andhisdiscoveriesandthepublicationofhisbookMechanicsservedasthebasisforthethreelawsofmotionpropoundedbyIsaacNewtonacenturylater.Galileoisperhapsbestknownforhisviewsonfree-fallingbodies.Legendhasitthathesimultaneouslydroppedtwocannonballs,one10timesheavierthantheother,fromtheLeaningTowerofPisa,bothbeingseenandheardtotouchthegroundatthesametime.Thisexperimenthasnotbeensubstantiated,butotherexperimentsactuallyperformedbyGalileoweresufficienttocastdoubtonAristotelianphysics.FromChapter5ofStaticsandStrengthofMaterialsforArchitectureandBuildingConstruction,FourthEdition,BarryOnouye,KevinKane.Copyright©2012byPearsonEducation,Inc.PublishedbyPearsonPrenticeHall.Allrightsreserved.257StrengthofMaterials2.Analysis.Giventhecompleteddesign,isitadequate?Thatis,doesitperformthefunctioneconomicallyandwithoutexcessivedeformation?Whatisthemarginofsafetyallowedineachmember?Thiswecallstructuralanalysis.3.Rating.Givenacompletedstructure,whatisitsactualload-carryingcapacity?Thestructuremayhavebeendesignedforsomepurposeotherthantheoneforwhichitisnowtobeused.Isthestructureoritsmembersadequatefortheproposednewuse?Thisiscalledaratingproblem.Becausethecompletescopeoftheseproblemsisobviouslytoocomprehensiveforcoverageinasingletext,thistextwillberestrictedtothestudyofindividualmembersandsimplestructuralframeworks.Subsequent,moreadvancedbooksonstructureswillconsidertheentirestructureandwillprovideessentialbackgroundformorethoroughanalysisanddesign.StructuralLoadClassificationLoadsappliedtostructuralelementsmaybeofvarioustypesandsources.Theirdefinitionsaregivenbelowsothattheterminologywillbeclearlyunderstood.Loadsclassifiedwithrespecttotime(Figure2)1.Staticload.Agraduallyappliedloadforwhichequilibriumisreachedinarelativelyshorttime.Liveoroccupancyloadsareconsideredtobestaticallyapplied.2.Sustainedload.Aloadthatisconstantoveralongperiodoftime,suchasthestructureweight(deadload)ormaterialand/orgoodsstoredinawarehouse.Thistypeofloadistreatedinthesamemannerasastaticload.Figure2258Loadsbasedontime.3.Impactload.Aloadthatisrapidlyapplied(anenergyload).Vibrationnormallyresultsfromanimpactload,andequilibriumisnotestablisheduntilthevibrationiseliminated,usuallybynaturaldampingforces.StrengthofMaterialsLoadsclassifiedwithrespecttotheareaoverwhichtheloadisapplied1.Concentratedload.Aloadorforcethatisappliedatapoint.Anyloadthatisappliedtoarelativelysmallareacomparedwiththesizeoftheloadedmemberisassumedtobeaconcentratedload.2.Distributedload.Aloaddistributedalongalengthoroveranarea.Thedistributionmaybeuniformornonuniform.Loadsclassifiedwithrespecttothelocationandmethodofapplication1.Centricload.Aloadinwhichtheresultantconcentratedloadpassesthroughthecentroid(geometricalcenter)oftheresistingcross-section.Iftheresultantconcentratedforcepassesthroughthecentroidsofallresistingsections,theloadingiscalledaxial.ForcePinFigure3hasalineofactionthatpassesthroughthecentroidofthecolumnaswellasthefooting;therefore,loadPisaxial.2.Bendingorflexuralload.Aloadinwhichtheloadsareappliedtransverselytothelongitudinalaxisofthemember.Theappliedloadmayincludecouplesthatlieinplanesparalleltotheaxisofthemember.Amembersubjectedtobendingloadsdeflectsalongitslength.Figure4illustratesabeamsubjectedtoflexuralloadingconsistingofaconcentratedload,auniformlydistributedload,andacouple.Figure3Centricloads.Figure4Bending(flexural)loadsonabeam.3.Torsionalload.Aloadthatsubjectsamembertocouplesormomentsthattwistthememberspirally(Figures5and6).4.Combinedloading.Acombinationoftwoormoreofthepreviouslydefinedtypesofloading.Figure5Torsiononaspandrelbeam.Figure6Torsiononacantileverbeam(eccentricloading).259StrengthofMaterialsConceptofStressStress,likepressure,isatermusedtodescribetheintensityofaforce—thequantityofforcethatactsonaunitofarea.Force,instructuraldesign,haslittlesignificanceuntilsomethingisknownabouttheresistingmaterial,crosssectionalproperties,andsizeoftheelementresistingtheforce(Figure7).Theunitstress,ortheaveragevalueoftheaxialstress,mayberepresentedmathematicallyasf=σ=axialforceP=AperpendicularresistingareawhereP=appliedforceorload(axial);unitsareexpressedas#,kips(k),N,orkNA=resistingcross-sectionalareaperpendiculartotheloaddirection;unitsareexpressedasin.2,ft.2,m2,ormm2Figure7Twocolumnswiththesameload,differentstress.f=σ1sigma2=thesymbol(s)representingunitstress(normal);unitsareexpressedas#/in.2,k/in.2,k/ft.2,andpascal(N/m2)orN/mm2ExampleProblems:Stress1Assumetwoshortconcretecolumns,eachsupportingacompressiveloadof300,000#.Findthestress.Column1hasadiameterof10inchesColumn2hasadiameterof25inches25”-diametercolumn.10”-diametercolumn.A=πA102B4stress=2602=78.5in.300,000#force==3,820#>inresistingarea78.5in.2A=πA252B4stress==491in.2300,000#2491in.=611#>in.2}PLAINCONCRETEStrengthofMaterialsSolution:Column1wouldprobablyapproachacriticalstresslevelinthisexample,whereasColumn2is,perhaps,overdesignedfora300,000#load.TheinferenceinExampleProblem1isthateveryportionoftheareasupportsanequalshareoftheload(i.e.,thestressisassumedtobeuniformthroughoutthecrosssection).Inelementarystudiesofthestrengthofmaterials,theunitstressonanycross-sectionofanaxiallyloaded,two-forcememberisconsideredtobeuniformlydistributedunlessotherwisenoted.NormalstressAstresscanbeclassifiedaccordingtotheinternalreactionthatproducesit.AsshowninFigures8and9,axialtensileorcompressiveforcesproducetensileorcompressivestress,respectively.Thistypeofstressisclassifiedasanormalstress,becausethestressedsurfaceisnormal(perpendicular)totheloaddirection.Thestressedareaa-a,isperpendiculartotheload.Innormalcompressivestress,fc=PAFigure8Normalcompressivestressacrosssectiona-a.whereP=appliedloadA=resistingsurfacenormal(perpendicular)toPInnormaltensilestress,ft=PAFigure9Normaltensilestressthroughsectiona-a.261StrengthofMaterialsShearstressShearstress,thesecondclassificationofstress,iscausedbyatangentialforceinwhichthestressedareaisaplaneparalleltothedirectionoftheappliedload.(Figures10through12).Averageshearstressmayberepresentedmathematicallyasfv=τ=axialforceP=AparallelresistingareawhereFigure10blocks.ShearstressbetweentwogluedP=appliedload(#ork,NorkN)A=cross-sectionalareaparalleltoloaddirection(in.2,m2,mm2)fvorτ=averageunitshearstress(psiorksi,N/mm2)(a)Twosteelplatesboltedusingonebolt.(b)Elevationshowingtheboltinshear.fv=averageshearstressthroughboltcross-sectionA=boltcross-sectionalareafv=(c)Shearforcesactingonthebolt.Figure11(d)FBDofhalfofabolt.Aboltedconnection—singleshear.FBDofmiddlesectionoftheboltinshear.262PAStrengthofMaterialsfv=P2A(twoshearplanes)Free-bodydiagramofmiddlesectionoftheboltinshear.Figure12shear.AboltedconnectionindoubleBearingstressThethirdfundamentaltypeofstress,bearingstress(Figure13),isactuallyatypeofnormalstress,butitrepresentstheintensityofforcebetweenabodyandanotherbody(i.e.,thecontactbetweenbeamandcolumn,columnandfooting,footingandground).Thestressedsurfaceisperpendiculartothedirectionoftheappliedload,thesameasnormalstress.Liketheprevioustwostresses,theaveragebearingstressisdefinedintermsofaforceperunitareafp=PAwherefp=unit-bearingstress(psi,ksi,orpsf;N/mm2orN/m2)P=appliedload(#ork,NorkN)A=bearingcontactarea(in.2orft.2,mm2orm2)Boththecolumnandfootingmaybeassumedtobeseparatestructuralmembers,andthebearingsurfaceisthecontactareabetweenthem.Therealsoexistsabearingsurfacebetweenthefootingandtheground.Figure13Bearingstress—post/footing/ground.263StrengthofMaterialsIntheprecedingthreestressclassifications,thebasicequationofstressmaybewritteninthreedifferentways,dependingontheconditionbeingevaluated.1.f=P>A(Basicequation;usedforanalysispurposesinwhichtheload,membersize,andmaterialareknown.)2.P=f*A(Usedinevaluatingorcheckingthecapacityofamemberwhenthematerialandmembersizeareknown.)3.A=P>f(Designversionofthestressequation;membersizecanbedeterminediftheloadandmaterial’sallowablestresscapabilityareknown.)TorsionalstressThefourthtypeofstressiscalledtorsionalstress(Figure14).Membersintorsionaresubjectedtotwistingactionalongtheirlongitudinalaxescausedbyamomentcoupleoreccentricload(seeFigures4and5).Oneofthemostcommonexamplesofabuildingmembersubjectedtotorsionalmomentsisaspandrel(edge)beam.Mostbuildingmemberssubjectedtotorsionaleffectsarealsoexperiencingeitherbending,shear,tensile,and/orcompressivestresses;therefore,itisrelativelyuncommontodesignspecificallyfortorsion.Ontheotherhand,designsinvolvingmachineryandmotorswithshaftsareextremelysensitivetothestressesresultingfromtorsion.Figure14264Membersubjectedtotorsion.StrengthofMaterialsExampleProblems:Stress2Atypicalmethodoftemporarilysecuringasteelbeamontoacolumnisbyusingaseatanglewithboltsthroughthecolumnflange.Two12–-diameterboltsareusedtofastentheseatangletothecolumn.TheboltsmustcarrythebeamloadofP=5kinsingleshear.Determinetheaverageshearstressdevelopedinthebolts.Solution:PA3.1410.5–22πD2A=2*=2*=0.393in.244fv=τ=↓(twobolts)fv=τ=5k0.393in.2=12.72ksi3Inatypicalfloorsupport,ashorttimberpostiscappedwithasteelchanneltoprovidealargerbearingareaforthejoists.Thejoistsare4–*12–roughcut.Thesteelbaseplateisprovidedtoincreasethebearingareaontheconcretefooting.Theloadtransmittedfromeachfloorjoistis5.0k.Findthefollowing:a.Theminimumlengthofchannelrequiredtosupportthejoistsifthemaximumallowablebearingstressperpendiculartothegrainis400psi.b.Theminimumsizeofpostrequiredtosupporttheloadifthemaximumstressallowedincompressionparalleltothegrainis1,200psi.c.Thesizeofbaseplaterequirediftheallowablebearingonconcreteis450psi.d.Thefootingsizeifallowablesoilfp=2,000psf.265StrengthofMaterialsSolutiona.fp=PAExamineonejoist;P=5,000#fallowable=400#>in.25,000#PA===12.5in.2f400#>in.2A=4–*L>24–*L>2=12.5in.2‹L=6.25in.P;P=2*5.0k=10kAfallowable=1,200#>in.210,000#PArequired===8.33in.2,L=1Af1,200#>in.2b.fc=Minimumsizerequirementofsquarepostis2.89–*2.89–.Practicaltouseatleasta4–*4–post.10,000#P=22.2in.2=f450#>in.2Minimumbaseplate=4.72–squareUseatleasta5–*5–squareplate.c.Arequired=10,000#P==5ft.2;x=2.24¿f2,000#>ft.2Usea2¿-3–*2¿-3–footingsize.d.Arequired=4Apieceofstandardsteelpipeisusedasastructuralsteelcolumnandsupportsanaxialloadof38,000#.Iftheallowableunitstressinthecolumnis12,000psi,whatsizepipeshouldbeused?Solution:Fallowable=12,000#>in.238,000#PPfc;Arequired===3.17in.2Af12,000#>in.2SeetheTableA8intheAppendix.Usea4–-diameterstandardweightpipe.1Area=3.17in.22266StrengthofMaterials5Atimberrooftrussissubjectedtoroofloadsasshown.Becausetimberlengthsarerelativelyrestrictive,itisnecessarytoprovideagluedspliceonthebottomchord.Determinethetensileforceinthebottomchordmember(splice).Assumingthemembersandspliceplateare6–deepandthegluehasashearcapacityof25#/in.2(withalotofsafetyfactor),determinetherequiredlengthLofthesplice.P=2,000#@spliceSolution:fv=τ=P;AArequired=Pfallowable=2,000#25#>in.2=80in.2Eachsideplateprovideshalfoftheresistance.‹Arequired=40in.21persideplate2L=3LA=6–*240in.2A==13.3in.L=33in.6Thefigureshownisaconnectionofthelowerjointonasingle-membertruss.Ifthereactionis10,000#andthemembersare8–*10–,determinetheshearingstressdevelopedonthehorizontalplanea-b-c.Solution:C©Fy=0D-P1y+10,000=0P1y=10,000#P1xP1y10,000#==20,000#sin30°.5=P1cos30°=20k1.8662=17.32k1horiz.thrust2P1=fv=τ=P1x17,320#==216.5#>in.2A8–*10–267StrengthofMaterialsProblems1DeterminethetensilestressdevelopedinmemberABduetoaloadofP=500#atD.ThememberABis12–thickand2”wide.2A10¿*20¿hotelmarqueehangsfromtworodsinclinedatanangleof30°.Thedeadloadandsnowloadonthemarqueeaddupto100psf.DesignthetworodsoutofA36steelthathasanallowabletensilestress:Ft=22,000psi1allowablestress23Ashortsteelcolumnsupportsanaxialcompressiveloadof120,000#andisweldedtoasteelbaseplaterestingonaconcretefooting.a.SelectthelightestW8(wide-flange)sectiontouseiftheunitstressisnottoexceed13,500psi.b.Determinethesizeofthebaseplate(square)requirediftheallowablebearingonconcreteis450psi.c.Calculatetherequiredsizeoffooting(square)iftheallowablesoilpressureisequalto3,000psf.Neglectweightsofcolumn,baseplate,andfooting.268StrengthofMaterials4Assumingadensityof120#/ft.3formasonrybrickwork,determinethemaximumheightofabrickwalliftheallowablecompressivestressislimitedto150#/in.2andthebrickis(a)fourincheswideand(b)sixincheswide.5Theaccompanyingfigureshowspartofacommontypeofrooftruss,constructedmainlyoftimberandsteelrods.Determine:a.Theaveragecompressivestressinthe8–*8–diagonalmemberiftheloadinitis20k.b.Thetensilestressinthe34–diameterthreadedsteelrodiftheloadinitis4k.c.Thebearingstressbetweenthetimberandthe4–*4–squaresteelwasheriftheholeinitis78–diameter.d.Thebearingstressbetweenthebrickwallcolumnandthe8–*10–timberiftheloadinthecolumnis15k.e.ThelengthLrequiredtokeepthedashedportionofthe8–*10–memberfromshearingoffduetothehorizontalthrustof16kagainstthesteelshoe.TheFv=120psiallowable.6TheturnbucklesinthediagramshownaretighteneduntilthecompressionblockDBexertsaforceof10,000#onthebeamatB.MemberDBisahollowshaftwithaninnerdiameterof1.0inchandouterdiameterof2inches.RodsADandCDeachhavecross-sectionalareasof1.0in.2.PinChasadiameterof0.75inch.Determine:a.TheaxialstressinBD.b.TheaxialstressinCD.c.TheshearingstressinpinC.269StrengthofMaterialsDeformationandStrain(a)Sheetofrubber—unloaded.Mostmaterialsofconstructiondeformundertheactionofloads.Whenthesizeorshapeofabodyisaltered,thechangeinanydirectionistermeddeformationandgiventhesymbolδ(delta).Strain,whichisgiventhesymbolε(epsilon)orγ(gamma),isdefinedasthedeformationperunitlength.Thedeformationorstrainmaybetheresultofachangeoftemperatureorstress.Considerapieceofrubberbeingstretched:(b)Sheetofrubber—underload.Figure15Deformationofasheetofrubber.L=OriginallengthW=OriginalwidthW¿=NewwidthδL=Longitudinalchangeinlength(deformation)W-W¿=δt=TransversechangeinlengthInFigure15,therubbertendstoelongateinthedirectionoftheappliedloadwitharesultantdeformationδ;correspondingly,acontractionofthewidthoccurs.Thisdeformationbehavioristypicalofmostmaterials,becauseallsolidsdeformtosomeextentunderappliedloads.Notruly“rigidbodies”existinstructuraldesign.Strainresultingfromachangeinstressisdefinedmathematicallyasε=δLwhereε=unitstrain(in./in.)δ=totaldeformation(in.)L=originallength(in.)Memberssubjectedtoashearstressundergoadeformationthatresultsinachangeinshape.Ratherthananelongationorshortening,shearingstresscausesanangulardeformationofthebody.ThesquareshowninFigure16becomesaparallelogramwhenacteduponbyshearstresses.Shearingstrain,representedbyγ,isγ=Figure16270Sheardeformation.δs=tanφφLWhentheangleφissmall,tanφφ,whereφistheangleexpressedinradians.StrengthofMaterialsExampleProblems:DeformationandStrain7AconcretetestcylinderisloadedwithP=100kandaresultingshorteningof0.036inch.Determinetheunitstraindevelopedintheconcrete.Solution:ε=δ0.036–==0.003in.>in.L12–Notethatthevalueofunitstrainisobtainedbydividingalengthbyalength.Theresultissimplyaratio.8Atrusstierodhasdimensionsasshown.Uponloading,itisfoundthatanelongationof0.400inchoccurredineachtierodassembly.Iftheunitstrainontherodportionequaled0.0026,whatwastheunitstrainonthetwoendclevises?Solution:δrod;δrod=εL=0.0026*120–=0.312–LTotalδ=0.400–-0.312–=0.088–δe0.088–ε===0.00183L48–1L=2¿ateachend=4¿=48–2ε=9ThemidpointCofacabledropstoC¿whenaweightWissuspendedfromit.Findthestraininthecable.Solution:δLδ=Deformation=ChangeinlengthofthecableL=Originallengthofthecableor12cableBC=Oldlength(beforeloading)BC¿=Newlength(afterloading)‹δ=BC-BC¿=2122+12-12¿=12.04¿-12¿=0.04¿=0.48–0.48–=0.0033in.>in.orleaveitunitlessε=12*12ε=271StrengthofMaterialsProblems7Duringthetestofaspecimeninatensiletestingmachine,itisfoundthatthespecimenelongates0.0024inchbetweentwopunchmarksthatareinitiallytwoinchesapart.Evaluatethestrain.8Areinforcedconcretecolumnis12feetlong,andunderload,itshortens18inches–.Determineitsaverageunitstrain.9Aconcretetestcylindereight8–talland4–indiameterissubjectedtoacompressiveloadthatresultsinastrainof0.003in./in.Determinetheshorteningthatdevelopsasaresultofthisloading.10A500-foot-longsteelcableisloadedintensionandregistersanaverageunitstrainof0.005.Determinethetotalelongationduetothisload.272StrengthofMaterials2ELASTICITY,STRENGTH,ANDDEFORMATIONRelationshipBetweenStressandStrainAwidevarietyofmaterialsarepresentlyusedinarchitecturalstructures:stone,brick,concrete,steel,timber,aluminum,plastics,etc.Allhaveessentialpropertiesthatmakethemapplicableforagivenpurposeinastructure.Thecriterionforselection,ataverybasiclevel,isthematerial’sabilitytowithstandforceswithoutexcessivedeformationsoractualfailures.Onemajorconsiderationinanystructuraldesignisdeflection(deformation).Deformationinstructurescannotincreaseindefinitely,anditshoulddisappearaftertheappliedloadisremoved.Elasticityisamaterialpropertyinwhichdeformationsdisappearwiththedisappearanceoftheload(Figure17).(a)Elasticbehavior.Allstructuralmaterialsareelastictosomeextent.Asloadsareappliedanddeformationsresult,thedeformationswillvanishastheloadisremovedaslongasacertainlimitisnotexceeded.Thislimitiscalledtheelasticlimit.Withintheelasticlimit,nopermanentdeformationsresultfromtheapplicationandremovaloftheload.Ifthislimitofloadingisexceeded,however,apermanentdeformationresults.Thebehaviorofthematerialisthentermedplasticorinelastic.Insomematerials,whentheelasticlimitisexceeded,themolecularbondswithinthematerialareunabletoreform,thuscausingcracksorseparationofthematerial.Suchmaterialsaretermedbrittle.Castiron,high-carbonsteel,andceramicsareconsideredtobebrittle;low-carbonsteel,aluminum,copper,andgoldexhibitpropertiesofductility,whichisameasureofplasticity.(b)Linearlyelasticbehavior.Materialsthathavemolecularbondsre-formingafterexceedingtheelasticlimitwillresultinpermanentdeformations;however,thematerialstillremainsinonepiecewithoutanysignificantlossinstrength.Thistypeofmaterialbehavioristermedductile.DuctilematerialsgivewarningofimpendingfailurBrittlematerialsdonot.e.Oneofthemostimportantdiscoveriesinthescienceofmechanicsofmaterialswasundoubtedlythatpertainingtotheelasticcharacterofmaterials.Thisdiscoveryin1678byRobertHooke,anEnglishscientist,mathematicallyrelatesstresstostrain.Therelationship,knownasHooke’slaw,statesthatinelasticmaterials,stressandstrainareproportional.Hookeobservedthisstress-strainrelationshipbyexperimentallyloadingvariousmaterialsintensionandthenmeasuringthesubsequentdeformations.AlthoughHooke’sinitialexperiment,techniques,andtestingequipmenthaveimproved,therelationshipbetweenstressandstrainandthedeterminationofelasticandplasticpropertiesofmaterialsstilluseHooke’sbasicconcept.(c)Plasticbehavior(permanentdeformation).Figure17behavior.Examplesofelasticandplastic273StrengthofMaterialsToday,universaltestingmachines,similartotheoneshowninFigure18,areemployedtoapplypreciseloadsatpreciseratestostandardizedtensileandcompressivespecimens(Figure19).Thetensiletestisthemostcommontestappliedtomaterials.Avarietyofdevicesformeasuringandrecordingstrainordeformationcanbeattachedtothetestspecimentoobtaindataforplottingstress-straindiagrams(orload-deformationcurves).(a)Beforeloading.Thestress-straincurvesobtainedfromtensionorcompressiontestsconductedonvariousmaterialsrevealseveralcharacteristicpatterns(Figure20).Ductilerolledsteels,suchasordinary,low-carbonstructuralsteel,stretchconsiderablyafterfollowingastraight-linevariationofstressandstrain.Forsteelsalloyedwithincreasingamountsofcarbonandotherstrengtheningmaterials,suchaschromium,nickel,silicon,manganese,andsoforth,thetendencytoproducesuchanintermediatestretchingpointbecomesincreasinglyremote.Thestressstraincurvesforheavilyalloyedsteelsaregenerallystraighttoapointashortdistancefromtherupturepoint.(b)Athighstresslevel.Figure18Universaltestingmachine.PhotocourtesyofMTSSystemsCorporation.(c)Atrupture—notethereducedcross-section.Figure19Steelspecimens—original,withload,andatfailure.274StrengthofMaterialsIncontrasttosuchstraight-line,stress-straincurvesarethoseobtainedformaterialssuchascastiron,brass,concrete,wood,andsoforth,whichareoftencurvedthroughoutmostoftheirlength.Figure20Stress-straindiagramforvariousmaterials.Thestress-straincurveforlow-carbon(lessthan0.30%carbon)steel(Figure21)willformthebasisfortheensuingremarksconcerningseveralfamiliarstrengthvalues.Thisdiagramplotsstrainalongtheabscissaandstressalongtheordinate.Thestressisdefinedastheloadinpoundsorkipsdividedbytheoriginalcross-sectionalareaofthespecimen.Asthetestproceeds,withlargerloadsbeingappliedataspecifiedrate,theactualcross-sectionalareaofthespecimendecreases.Athighstresses,thisreductioninareabecomesappreciable.Thestressbasedontheinitialareaisnotthetruestress,butitisgenerallyused(andiscalledtheindicatedstress).Thecalculatedstressinload-carryingmembersisalmostuniversallybasedonthisoriginalarea.Thestrainusedistheelongationofaunitlengthofthetestspecimentakenoverthegaugelengthoftwoinches.Figure21Stress-straindiagramforlowcarbonstructuralsteel.275StrengthofMaterialsUsinganexaggeratedscaleonthestress-straindataformildsteel,asshowninFigure22,thesignificantpointsonthecurvearedefinedasfollows:1.Proportionallimit.Theproportionallimitisthatstressbeyondwhichtheratioofstresstostrainnolongerremainsconstant.ItisthegreateststressthatamaterialiscapableofdevelopingwithoutdeviationfromHooke’slawofstressstrainproportionality.2.Elasticlimit.Locatedclosetotheproportionallimit,yetofentirelydifferentmeaning,istheelasticlimit.Theelasticlimitisthatmaximumunitstressthatcanbedevelopedinamaterialwithoutcausingapermanentset(deformation).Aspecimenstressedtoapointbelowitselasticlimitwillassumeitsoriginaldimensionswhentheloadisreleased.Ifthestressshouldexceeditselasticlimit,thespecimenwilldeformplasticallyandwillnolongerattainitsoriginaldimensionswhenunloaded.Itisthensaidtohaveincurredapermanentset.Figure22Stress-straindiagramformildsteel(A36)withkeypointshighlighted.3.Yieldpoint.Whentheloadonthetestspecimenisincreasedbeyondtheelasticlimit,astresslevelisreachedwherethematerialcontinuestoelongatewithoutanincreaseofload.Thispoint,calledtheyieldpoint,isdefinedasthestressatwhichamarkedincreaseinstrainoccurswithoutaconcurrentincreaseinappliedstress.Aftertheinitialyielding(upperyieldpoint)isreached,theforceresistingdeformationdecreasesduetotheyieldingofthematerial.Thevalueofstressafterinitialyielding(theloweryieldpoint)isusually276StrengthofMaterialstakentobethetruematerialcharacteristictobeusedasthebasisforthedeterminationofallowablestress(fordesignpurposes).Manymaterialsdonotexhibitwell-definedyieldpoints,andtheyieldstrengthisdefinedasthestressatwhichthematerialexhibitsaspecifiedlimitingpermanentset.Thespecifiedset(oroffset)mostcommonlyusedis0.2%,whichcorrespondstoastrainof0.002in./in.(Figure23).Whenatestspecimenisstressedbeyonditselasticlimitandthenhasitsloadreleased,aplotofthedatashowsthatduringtheload-reducingstage,thestress-straincurveparallelstheinitialportionofthecurve.Thehorizontalinterceptalongthexaxisisthepermanentset.Suchaloadcycledoesnotnecessarilydamageamaterialeveniftheimposedstressexceedstheelasticlimit.Ductilitymaybelowered,butthehardness(abilityofamaterialtoresistindentation)andelasticstresslimitofthematerialwillgenerallyincrease.Figure23Stress-straindiagramshowingpermanentset.4.Ultimatestrength.Theultimatestrengthofamaterialisdefinedasthestressobtainedbydividingthemaximumloadreachedbeforethespecimenbreaksbytheoriginalcross-sectionalarea.Theultimatestrength(oftencalledthetensilestrength)ofthematerialissometimesusedasabasisforestablishingtheallowabledesignstressesforamaterial.5.Rupturestrength(breakingstrength,fracturestrength).Inaductilematerial,rupturedoesnotusuallyoccurattheultimateunitstress.Aftertheultimateunitstresshasbeenreached,thematerialwillgenerally“neckdown,”asshowninFigure24(b),anditsrapidlyincreasingelongationwillbeaccompaniedbyadecreasein277StrengthofMaterialsload.Thisdecreasebecomesmorerapidastherupturepointisapproached.Therupturestrength,obtainedbydividingtheloadatrupturebytheoriginalarea(indicatedrupture),haslittleornovalueindesign.Amorecorrectevaluationofthevariationofstressfollowingattainmentoftheultimateunitstressisobtainedbydividingtheloadsbythesimultaneouslyoccurringdecreasingareas(truerupturestrength).6.Elongation.Elongation(Figure24)isameasureoftheabilityofamaterialtoundergodeformationwithoutrupture.Percentageelongation,definedbytheequationbelow,isameasureoftheductilityofamaterial.Ductilityisadesirableandnecessaryproperty,andamembermustpossessittopreventfailureduetolocaloverstressing.(a)Unloaded.Figure24loading.(b)Underload.ElongationofspecimenunderawhereLf-Lob*100%=%ofelongationLoLo=originalspecimenlengthLf=lengthofspecimenatrupture(fracture)7.Reductionofarea.Astheloadonthematerialundergoingtestingisincreased,theoriginalcross-sectionalareadecreasesuntilitisataminimumattheinstantoffracture.Itiscustomarytoexpressthisreductioninareaastheratio(asapercentage)ofthechangeinareatotheoriginalspecimencross-sectionalarea.(SeeFigure24.)%reductioninarea=awhereAo-Afb*100%AoAf=reducedareaatfailureAo=originalcross-sectionalareaThefailedspecimenexhibitsalocaldecreaseindiameterknownasneckingdownintheregionwherefailureoccurs.Itisverydifficulttodeterminetheonsetofneckingdownandtodifferentiateitfromtheuniformdecreaseindiameterofthespecimen.Failureinastructuralsteeltestspecimencommenceswhenthematerialissufficientlyreducedincross-sectionalareaandmolecularbondswithinthematerialbeginbreakingdown.Eventually,theultimatetensilestrengthofthematerialisreached,andactualseparationofthematerialstarts,withshearfailureoccurringattheperipheryofthespecimenat45°angles.Then,completeseparationresultsinthecenterportion278StrengthofMaterialsinwhichtheplaneoffailureisnormaltothedirectionofthetensileforce.Thepercentagereductioninareacanbeusedasameasureofductility.Brittlematerialsexhibitalmostnoreductioninarea,whereasductilematerialsexhibitahighpercentagereductioninarea.Table1showstheaveragestrengthvaluesforselectedengineeringmaterials.Table1Averagestrengthvaluesforselectedengineeringmaterials.YieldStressorProportionalLimit(ksi)MaterialsTensionCompressionShearUltimateStrength(ksi)TensionCompressionModulusofElasticity(ksi)ShearTens.orComp.ShearSteel:A-36StructuralA572Grade50*(alsoA992)3650365022305865404529,00030,00012,00012,000Iron:MalleableWrought2530253012185058403824,00025,00011,00010,000AluminumAlloy:6061-T6Rolled/extruded6063-T6Extrudedtubes3525352520143830241910,00010,0003,8003,800OtherMetals:Brass:70%Cu,30%ZincBronze,castheattreated255515375575485614,00012,0006,0005,0001.00.80.751,6001,6001,200Timber,AirDry:YellowPineDouglasFirSpruce6.25.44.0Concrete:Concrete:1:2:4mix,28days8.46.85.03.03,000*WithspecialrequirementsperAISCTechnicalBulletin#3.279StrengthofMaterials3OTHERMATERIALPROPERTIESCompressionTestsThecompressiontestisusedprimarilytotestbrittlematerials,suchascastironandconcrete(Figure25).Theuniversaltestingmachineisusedforthispurpose,anddataaretakeninamannersimilartothatdiscussedforthetensiontest.Theresultsofthecompressiontestgenerallydefineanelasticrange,aproportionallimit,andayieldstrength.Inthecompressiontest,thecross-sectionalareaofthespecimenincreasesasloadsincrease,whichproducesacontinuouslyrisingstress-straincurve.Figure25Stress-straindiagramforcastiron(compressiontest).CreepThedeformationthatmoststructuralmaterialsundergowhenstressedtotheirallowablelimitatroomtemperatureisacompleteddeformationandwillnotincreasenomatterhowlongthestressisapplied.Atsomehighertemperature,however,thesesamematerialswillrevealacontinuingdeformationorcreepthat,ifpermittedtocontinue,willultimatelyleadtoexcessivedisplacementsorrupture(Figure26).Figure26Strainwithrespecttotime(creep).CyclicStress(Fatigue)Memberssubjectedtorepeatedconditionsofloadingorunloading,ortorepeatedstressreversals,willfailatastressconsiderablylowerthantheultimatestressobtainedinasimpletensiontest.Failuresthatoccurasaresultofthistypeofrepeatedloadingareknownasfatiguefailures.Atheoryoffatiguefailureassumesasuddenchangeinshapeoftheloadedmember.Inconsistenciesinthematerialcauselocalizedstressestodevelopthatarefargreaterthantheaveragestressinthematerial.Theselocalizedstressesexceedthematerial’syieldstrengthandcausepermanentdeformationstooccurlocally.Repeatedpermanentdeformationsinasmallareaeventuallycausehairlinecrackstodevelop.Thiscrackingprocesscontinuesuntiltheaveragestressontheresistingareareachestheultimatestrengthofthematerial.280StrengthofMaterialsPoisson’sRatioThecross-sectionalreduction(neckingdown)duringasteeltensiletesthasadefiniterelationshiptotheincreaseinlength(elongation)experiencedbythespecimen.Whenamaterialisloadedinonedirection,itwillundergostrainsbothperpendicularaswellasparalleltothedirectionoftheload(seeFigure15).TheratioofthelateralorperpendicularstraintothelongitudinaloraxialstrainiscalledPoisson’sratio.Poisson’sratiovariesfrom0.2to0.4formostmetals.Moststeelshavevaluesintherangeof0.283to0.292.Thesymbolµ(mu)isusedforPoisson’sratio,whichisgivenbytheequationµ=εlateralεlongitudinalAllowableWorkingStress—FactorofSafetyAworkingstressorallowablestress(Table2)isdefinedasthemaximumstresspermittedinadesigncomputation.Itisthestressderivedfromtheresultsofmanytestsandtheaccumulatedexperienceobtainedfrommanyyearsoffirsthandobservationintheperformanceofmembersinactualservice.Thefactorofsafetymaybedefinedastheratioofafailureproducingloadtotheestimatedactualload.Theratiomaybecalculatedastheultimatestress(oryield-pointstress)totheallowableworkingstress.Loadresistancefactordesign(LRFD),alsoknownaslimitstateorthestrengthmethod,representsanotherphilosophyusedinthedesignofconcreteandsteeland,morerecently,timber.ModulusofElasticity(Young’sModulus)In1678,SirRobertHookeobservedthatwhenrolledmaterialsweresubjectedtoequalincrementsofstress,theysufferedequalincrementsofstrain—inotherwords,stressisproportionaltostrain.TheratioformedbydividingaunitstressbyitscorrespondingvalueofstrainwassuggestedbyThomasYoungin1807asameansofevaluatingtherelativestiffnessofvariousmaterials(Figure27).ThisratioiscalledYoung’smodulus,orthemodulusofelasticity,anditistheslopeofthestraight-lineportionofthestressstraindiagram:E=fεAlthoughnoportraitofRobertHookesurvivesandhisnameissomewhatobscuretoday,hewasperhapsthegreatestexperimentalscientistofthe17thcentury.Anearthy,albeitcantankerous,fellow,Hookewasoccupiedwithanenormousnumberofpracticalproblems.AmongHooke’sinventionsthatarestillinusetodayaretheuniversaljointusedincarsandtheirisdiaphragm,whichwasusedinmostearlycameras.Hookeexperimentedinawidevarietyoffields,rangingfromphysics,astronomy,chemistry,geology,andbiologytomicroscopy.Heiscreditedwiththediscoveryofoneofthemostimportantlawsinthescienceofmechanicspertainingtotheelasticcharacterofmaterials.Byexperimentingwiththebehaviorofelasticbodies,especiallyspiralsprings,Hookesawclearlythatnotonlydosolidsresistmechanicalloadsbypushingback,theyalsochangeshapeinresponsetotheloads.Itisthischangeinshapethatenablesthesolidtodothepushingback.Unfortunately,Hooke’spersonalityhinderedtheapplicationofhistheoriesonelasticity.Hewasreputedtobeamostnastyandargumentativeindividual,andhereportedlyusedhispowerintheRoyalSocietyagainstthoseheperceivedashisenemies.IsaacNewtonhadthemisfortuneofheadingthatlist.SofiercewasthehatredthatgrewbetweenthesetwomenthatNewton,wholivedfor25yearsafterHooke’sdeath,devotedagooddealoftimetodenigratingHooke’smemoryandtheimportanceofappliedscience.Asaresult,subjectssuchasstructuressufferedinpopularity,andHooke’sworkwasnotmuchfollowedorexploited,forsomeyearsafterNewton’sdeath.281StrengthofMaterialswhereE=modulusofelasticity(ksiorpsi,N/m2pascalorN/mm2)f=stress(ksiorpsi,N/m2pascalorN/mm2)ε=strain(in./in.,mm/mm)Thisratioofstresstostrainremainsconstantforallsteelsandmanyotherstructuralmaterialswithintheirusefulrange.Figure27SirThomasYoung(1773–1829).YoungwasaprodigyinhisinfancywhopurportedlycouldreadattheageoftwoandhadreadtheentireBibletwicebeforetheageoffour.AtCambridge,hisincredibleabilitiesearnedhimthenickname“PhenomenonYoung.”Hematuredintoanadultprodigyandwasknowledgeablein12languagesandcouldplayavarietyofmusicalinstruments.Asaphysician,hewasinterestedinsenseperception.Fromtheeyeandthentolightitself,itfelltoYoungtodemonstratethewavenatureoflight.Turningmoreandmoretophysics,heintroducedtheconceptofenergyinitsmodernformin1807.Inthesameyear,hesuggestedtheratioformedbydividingaunitstressbyitscorrespondingvalueofstrain(Young’smodulus)asameansofevaluatingthestiffnessofvariousmaterials.HewasalsoanaccomplishedEgyptologist,instrumentalindecipheringtheRosettaStone,thekeytoEgyptianhieroglyphics.AppointedtotheChairofNaturalPhilosophyattheRoyalInstitutioninLondon,Youngwasexpectedtodeliverscientificlecturestopopularaudiences.Youngtookthismissionseriouslyandlaunchedintoaseriesoflecturesabouttheelasticityofvariousstructures,withmanyusefulandnovelobservationsonthebehaviorsofwallsandarches.UnfortunatelyforYoung,helackedtheflairandoratoryskillofhiscolleagueHumphreyDavy,andhemightaswellhavebeenrecitinghieroglyphicsasfarashisaudienceswereconcerned.Disappointed,Youngresignedhischairandreturnedtomedicalpractice.Generally,ahighmodulusofelasticityisdesirable,becauseEisoftenreferredtoasastiffnessfactor.MaterialsexhibitinghighEvaluesaremoreresistanttodeformationand,inthecaseofbeams,suffermuchlessdeflectionunderload.NoteinFigure28thatofthethreematerialsshown,thesteelspecimenhasamuchsteeperslopeintheelasticrangethanaluminumorwoodand,therefore,willbemuchmoreresistanttodeformation.TheYoung’smodulusequationmayalsobewritteninaveryusefulexpandedformwheneverthestressanddeformationarecausedbyaxialloads.f=E=P>Aδ>L‹δ=whereδPLAE=====Figure28282δP;ε=AL=PLδA¿PL1Elasticequation2AEdeformation(in.,mm)appliedaxialload(#ork,NorkN)lengthofmember(in.,mm)cross-sectionalareaofmember(in.2,mm2)modulusofelasticityofmaterial(#/in.2ork/in.2,N/m2pascalorN/mm2)E—ratioofstresstostrainforvariousmaterials.StrengthofMaterialsToughnessTheareaunderthestress-straincurve(Figure29)isameasureoftheworkrequiredtocauseafracture.Thisabilityofamaterialtoabsorbenergyuptofractureisalsousedbydesignersasacharacteristicpropertyofamaterialandiscalledtoughness.Toughnessmaybeimportantinapplicationswherestressesintheplasticrangeofthematerialmaybeapproachedbuttheresultingpermanentsetisnotcritical—andissometimesevendesirable.Anexamplewouldbedie-formingsheetmetalforautomobilebodies.Thestress-straindiagramsindicatethatlow-carbon(mild)steelsaremuch“tougher”thanhigh-carbon(higherstrength)steels.Thisconceptissometimesinoppositiontotheinstinctoftheengineertospecifytheuseofa“stronger”steelwhenastructurefailsorseemsindangeroffailing.Thiscouldbeamistakeinlargerstructures,becauseeveninmildsteel,muchofthestrengthisnotreallybeingused.Failureofastructuremaybecontrolledbythebrittlenessofthematerial,notbyitsstrength.Figure29Toughness—areaunderthestressstraindiagram.283StrengthofMaterialsTable2Allowablestressesforselectedengineeringmaterials.UnitWeight(density)(pfc)ModulesofElasticityE(ksi)Metals:A36SteelFy=36ksiA572Grade50/A992A572SteelFy=65ksiAluminumIron(cast)49049049016545029,00030,00030,00010,00015,00022ksi30ksi39ksi16ksi5ksi22ksi30ksi39ksi16ksi20ksiBrittleMaterials:ConcreteStoneMasonryBrickMasonry1501651203,0001,0001,500100psi10psi20psi1,350psi100psi300psi35351,7001,600650psi700psi1,050psi1,000psi625psi625psi1,450psi1,300psi95psi85psi35351,6001,600625psi1,050psi1,000psi975psi565psi440psi1,400psi1,550psi90psi110psi30301,4001,300800psi600psi1,050psi850psi405psi405psi1,150psi1,000psi75psi70psi3537451,8001,8002,0001,100psi1,850psi2,000psi1,650psi2,460psi2,900psi650psi750psi650psi2,400psi2,600psi2,900psi165psi285psi290psiMaterialsWood:Doug-FirLarchNorth*•Joist&Rafters(No.2)•Beams&Posts(No.1)SouthernPine*•Joists&Rafters(No.2)•Beams&Posts(D-No.1)Hem-Fir*•Joists&Rafters(No.2)•Beams&Posts(No.1)WoodProducts:Glu-LamBeamsMicro-LamBeamsParallamBeams*Averagedstressvaluesfordesign.284AllowableAllowableAllowableAllowableAllowableTensionAxialCompressBendingShearStressCompressBearingStressStressFcFc⊥FbFvFt22ksi30ksi39ksi16ksi5ksi14.5ksi20ksi26ksi10ksi7.5ksi100psi10psi30psiStrengthofMaterialsExampleProblems:MaterialProperties10Thefollowingdatawereobtainedduringatensiletestonamildsteelspecimenhavinganinitialdiameterof0.505inch.Atfailure,thereduceddiameterofthespecimenwas0.305inch.Plotthedata,anddeterminethefollowing:a.b.c.d.e.f.g.Modulusofelasticity.Proportionallimit.Ultimatestrength.Percentagereductioninarea.Percentageelongation.Indicatedstrengthatrupture.Truerupturestrength.AxialLoad(#)0Stressf(#/in.2)01,6403,1408,20015,7006,00030,0004,5807,4408,0007,9807,9008,0408,0408,0609,46012,00013,26013,58013,46013,2209,860Elongationper2flLength(in.)022,90037,20040,00039,90039,50040,20040,20040,30047,30060,00066,30067,90067,30066,10049,300.00050.0010.0015.0020.0025.0030.00375.00500.00624.00938.0125.050.125.225.325.475.535.625Strain>L(in./in.)0.00025.00050.00075.00100.00125.00150.001875.00250.00312.00469.00625.0250.0625.1125.1625.2375.2675.3125Solution:Initialdiameter=0.505–Lorig=2.00–Aorig=0.20in.2Reduceddiameteratfailure=0.305–Afail=0.073in.2a.Modulusofelasticity:130,000#>in.22¢f==30,000,000#>in.2E=0.0010¢ε285StrengthofMaterialsb.Proportionallimitisthelastpointofthelinearpartofthediagram.‹fprop=30,000psic.Ultimatestrengthistheabsolutehigheststressmagnitude.‹fult=67,900psid.PercentageareareductionequalsAorig-Afail0.20-0.073*100%=63.5%=Aorig0.20e.PercentageelongationequalsδtotalLoriginal=0.625–*100%=31.25%2.00–f.Indicatedstrengthofrupture:frupture=49,300psig.Truerupturestrength:frupture=9,860#0.073in.2=135,068psiPlotofdataforsteeltensiletest.286StrengthofMaterials11Aone-inch-diametermanganesebronzetestspecimenwassubjectedtoanaxialtensileload,andthefollowingdatawereobtained.Gaugelength10inchesFinalgaugelength12.25inchesLoadatproportionallimit18,500#Elongationatproportionallimit0.016inchMaximumload55,000#Loadatrupture42,000#Diameteratrupture0.845inchFindthefollowing:a.b.c.d.e.f.g.Proportionallimit.Modulusofelasticity.Ultimatestrength.Percentageelongation.Percentagereductioninarea.Indicatedrupturestrength.Truerupturestrength.Solution:Aorig.=0.785in.2;Arupture=0.560in.2a.fprop.=εprop.=b.E=E=18,500#0.785in.20.016in.2=0.0016in.>in.10in.¢f=¢εfprop.εprop.23,567#>in.20.0016in.>in.c.fult==23,567psi=14,729,375psi55,000#max.load==70,100psiAorig.0.785in.2d.%elongation=2.25–*100%=22.5%10–e.%reductionofarea=f.frupture=g.frupture=0.785in.2-0.560in.20.785in.2*100%=28.7%loadatrupture=53,503psi1indicated2loadatrupture=75,000psi1true2AoriginalArupture287StrengthofMaterials12Determinetheallowableloadcapacityofthecolumnshown,assumingthattheallowablecompressivestress1Fc=600psi2accountsforthebucklingpotential.RefertothestandardwoodtablesinTableA1oftheAppendixforinformationoncross-sectionalproperties.Solution:11Pallow=Fc*A=600psi*a3–*3–b=7,350#22WhatisthedeformationforPallow?δ=7,350#110¿*12in.>ft.2PL=0.045–=AE112.25in.2211.6*106#>in.2213A600-foot-longroofcablecannotbepermittedtostretchmorethanthreefeetwhenloaded,ortheroofgeometrywillbeaffectedtoodrastically.IfEs=29*103ksiandtheloadis1,500k,determinetherequiredcablediameterneededtoavoidexcessiveelongationoroverstress.Ft=100ksi(allowabletensilestress).Solution:1,500k1600¿2PLPL=10.34in.2;A==AEδE3¿129*103k>in.22δ=4110.34in.22πD2;D==3.63–4DπA=(diameterbasedonthedeformationrequirement)Sectioncutthroughacableroofstructure.f=1,500kPP;A===15in.2Pgoverns2Aft100k>in.D=D4115in.22π=4.37–Pgoverns(diameterbasedontensilestress)288StrengthofMaterials14Aheavychandelierweighing1,500#issuspendedfromtheroofofatheaterlobby.Thesteelpipefromwhichithangsis20feetlong.Determinethesizeofpipenecessarytocarrythechandeliersafely.UseA36steel.Whatistheresultingelongationofthepipe?Again,we’relookingforarequiredsize;therefore,thisisadesignproblem.Solution:ft=PAArequired=PftForA36steel,Ft=22,000psiArequired=1,500#22,000#>in.2=0.0682in.2RefertotheTableA6intheAppendix.A12-inch-diameterstandardweightpipehasthefollowingcross-sectionproperties:NominalDiameterOuterDiameterInnerThicknessWallFootWeightperAreaDiameter0.5–84–0.622–0.109–0.85#/ft.0.25in.2Arequired=0.0682in.260.25in.2OKTodetermineelongation,δ=1,500#120¿*12in.>ft.2PL=0.05–=AE10.25in.22129,000,000#>in.22289StrengthofMaterials15A15-passengerelevatorina15-storybuildingis1–steelrope.raisedbyusingaone-inch-diameterAssumingthatthecitycoderequiresafactorofsafetyof11againsttheultimatestrengthoftherope,checktheadequacyoftheropeanditselongation.SolutionOne-inch1–-diameterrope:NetresistingareaUltimatestrengthRopewt.Ropelength====0.523in.227tons=54k2.0#>ft.14storiesofropeplus10¿topulleyLoadsonrope:15passengers@150#ea.15-passengerelevatorcabRopelength=14*10¿perstory+10¿1topulley2=150¿*2.0#>ft.=2,250#=1,250#=300#P=3,800#Ultimatestrength=54,000#safestrength=(working)ultimatestrengthsafetyfactor=54,000#=4,909#11Pactual=3,800#64,909#allowableOKδ=29013.8k21150¿*12in.>ft.2PL=0.45–=AE10.523in.22129,000k>in.22StrengthofMaterialsProblems11Two4–-wideby12-feet-highbrickwallsofagaragesupportaroofslab20feetwide.Theroofweighs100psfandcarriesasnowloadof30psf.Checkthecompressivestressatthebaseofthewallassumingbrick(forthisproblem)hasacapacityof125psi.Brickweighs120#/ft.3Hint:Analyzeatypicalonefootstripofwall.12Asteelwire300feetlongand81inchindiameterweighs0.042#/ft.Ifthewireissuspendedverticallyfromitsupperend,calculate(a)themaximumtensilestressduetoitsownweightand(b)themaximumweightWthatcanbesafelysupportedassumingasafetyfactorofthreeandanultimatetensilestressof65ksi.13Astructuralsteelrod112inchesindiameterand25feetlong,insupportingabalcony,carriesatensileloadof29k;E=29*103ksi.a.Findthetotalelongationδoftherod.b.Whatdiameterdisnecessaryifthetotalδislimitedto0.1inch?14A100-foot-longsurveyor’ssteeltapewithacrosssectionalareaof0.006squareinchmustbestretchedwithapullof16#wheninuse.IfthemodulusofelasticityofthissteelisE=30,000ksi,(a)whatisthetotalelongationδinthe100foottapeand(b)whatunittensilestressisproducedbythepull?15Theendsofthelaminated-woodroofarchshownaretiedtogetherwithahorizontalsteelrod90feet,“10inches”long,whichmustwithstandatotalloadof60k.Twoturnbucklesareused.Allthreadedrodendsareupset.a.DeterminetherequireddiameterDoftherodifthemaximumallowablestressis20ksi.b.Iftheunstressedlengthoftherodis90feet,“10inches”andtherearefourthreadsperinchontheupsetends,howmanyturnsofoneturnbucklewillbringitbacktoitsunstressedlengthafterithaselongatedunderfullallowabletensilestress?E=29*103ksi.291StrengthofMaterialsStressConcentrationInourinitialdiscussiononnormalstress(Figure30),wemadetheassumptionthatifaloadisappliedcentrically(throughtheaxisofthemember),thestressdevelopedonthenormalplanecouldbeassumedtobeuniform(Figure31).Formostcasesofnormalstress,thisisapracticalassumptiontomakeforastaticloadcondition.If,however,thegeometryofthememberischangedtoincludediscontinuitiesorchangingcross-sections,stresscannolongerbeassumedtobeuniformacrossthesurface.Figure30292Normalstress.Stresstrajectories,alsocalledisostaticlines,connectpointsofequalprincipalstressandrepresentstresspathsthroughamember.Thisconceptprovidesavisualpictureofthestressdistributionofamemberorstructureundervariousloadingconditions.Stresstrajectoriesarenormallydrawnatequalincrementsofstresstodenoteauniformstress.Acrowdingofthestresstrajectorylinesindicatesstressconcentration,orhighstress,justascontourlinesonatopographicmapindicatesteepgrades(Figures32and33).Figure31Uniformstressdistribution.AFrenchmathematiciannamedBarredeSaint-Venant(1797–1886)observedthatlocalizeddistortionsoccurredinareasofdiscontinuityandthatstressconcentrationsdeveloped,causinganunevendistributionofstressacrossthestressedsurface.However,theselocalizedeffectsdisappearedatsomedistancefromsuchlocations.ThisisknownasSaint-Venant’sprinciple.Figure32Nonuniformstressdistribution.Figure33Columnundercompressiveload.StrengthofMaterialsLoadconcentrations,reentrantcorners,notches,openings,andotherdiscontinuitieswillcausestressconcentrations.These,however,donotnecessarilyproducestructuralfailures,evenifthemaximumstressexceedstheallowableworkingstress.Forexample,instructuralsteel,extremestressconditionsmayberelieved,becausesteelhasatendencytoyield(give),thuscausingaredistributionofsomeofthestressacrossmoreofthecross-section.Thisredistributionofstressenablesthegreaterpartofthestructuralmembertobewithinthepermissiblestressrange.Astressconcentrationinconcreteisaseriousmatter.Excessivetensilestresses,eventhoughlocalized,causecrackstoappearintheconcrete.Overaperiodoftime,thecracksbecomemorepronouncedbecauseofthehighstressconcentrationattheendofthecracks.Crackinginreinforcedconcretecanbeminimizedbyplacingthereinforcingsteelacrosspotentialcracklines.Timberbehavesinmuchthesameway,ascracksappearalongthegrain(Figure34).Figure34(flexure).StresstrajectoriesinabeamFigure35Stresstrajectoriesaroundahole.Inthepast,photoelasticity(theshiningofapolarizedlightonatransparentmaterial)wasoftenusedtoproducestresspatternsforvariousstructuralmembersunderloading.Today,computermodelingandanalysissoftwarearecapableofgeneratingcolorfulstresscontourmapping,visuallyrepresentingthestressintensityforbothindividualmembersandthestructureasawhole.Often,structuralelementswillhavediscontinuities(holesinabeamformechanicalducts,windowopeningsinwalls)thatinterruptthestresspaths(calledstresstrajectories).Thestressatthediscontinuitymaybeconsiderablygreaterthantheaveragestressduetocentricloading;thus,thereexistsastressconcentrationatthediscontinuity(Figure35).Stressconcentrationsareusuallynotsignificantlycriticalinthecaseofstaticloadingofductilematerial,becausethematerialwillyieldinelasticallyinthehigh-stressareasandredistributionofstressresults.Equilibriumisestablished,andnoharmisdone.However,incasesofdynamicorimpactloading,orofstaticloadingofbrittlematerial,stressconcentrationsbecomeverycriticalandcannotbeignored.Stressredistributiondoesnotresulttosuchanextentthatequilibriumismaintained.293StrengthofMaterialsTorsionalStressCharles-AugustindeCoulomb,aFrenchengineerofthe18thcentury,wasthefirsttoexplaintorsioninasolidorhollowcircularshaft.Heexperimentallydevelopedarelationshipbetweentheappliedtorque(T)andtheresultingdeformation(angleoftwist)ofcircularrods.FromthedistortionoftherodshowninFigure36,itisclearthatshearingstressesfvmustexist.Inanelasticmaterial,suchassteel,thestressesincreaseinmagnitudeproportionatelytothedistancefromthecenterofthecircularcrosssectionandflowcircularlyaroundthearea.CoulombderivedbymeansofequilibriumconceptstherelationshipT=Figure36Circularcrosssectionintorsion.πr3fv2whereT=externallyappliedtorsionalmoment(torque)πr2=cross-sectionalareaoftherod(shaft)r=radiusoftherodfv=internalshearstressonthetransverseplaneofftherodHollow,circularcross-sections(pipes)offerthegreatesttorqueresistanceperunitvolumeofmaterial,becausemateriallocatednearthecenterisatalowstressleveland,thus,islesseffective.Noncircularcross-sectionedmembers,suchasrectangularorI-shapedbeams,developacompletelydifferentdistributionofshearstresswhensubjectedtotorsion.Figure37294Rectangularbarintorsion.AfullcenturyafterCoulomb,BarredeSaint-Venantdevelopedthetheorytoexplainthedifferencebetweencirculartorsionandnoncirculartorsion.Circumferentiallinesofacircularrodremainintheiroriginalcross-sectionalplaneundertorsionalforces(seeFigure36),whereasthecorrespondinglinesofarectangularbarwarpoutoftheiroriginalplane(Figure37).ThiswarpingsignificantlyaltersthesimplelinearstressdistributionassumedbyCoulomb.StrengthofMaterials4THERMALEFFECTSMoststructuralmaterialsincreaseinvolumewhensubjectedtoheatandcontractwhencooled.Wheneveradesignpreventsthechangeinlengthofamembersubjectedtotemperaturevariation,internalstressesdevelop.Sometimesthesethermalstressesmaybesufficientlyhightoexceedtheelasticlimitandcauseseriousdamage.Free,unrestrainedmembersexperiencenostresschangeswithtemperaturechanges,butdimensionalchangeresults.Forexample,itiscommonpracticetoprovideexpansionjointsbetweensidewalkpavementstoallowmovementduringhotsummerdays.Preventionofexpansiononahotdaywouldundoubtedlyresultinseverebucklingofthepavement.Thedimensionalchangeduetotemperaturechangesisusuallydescribedintermsofthechangeinalineardimension.Thechangeinlengthofastructuralmember¢Lisdirectlyproportionaltoboththetemperaturechange¢TandtheoriginallengthofthememberLo.Thermalsensitivity,calledthecoefficientoflinearexpansion1α2,hasbeendeterminedforallengineeringmaterials(Table3).Carefulmeasurementshaveshownthattheratioofstrainεtotemperaturechange¢Tisaconstant:δ>Lεstrain==temperaturechange¢T¢Tα=Solvingthisequationforthedeformation,δ=αL¢TwhereαL¢Tδ====coefficientofthermalexpansionoriginallengthofmember(in.)changeintemperature1°F2totalchangeinlength(in.)Table3Linearcoefficientsofthermalexpansion(contraction).MaterialCoefficients(␣)[in./in./ºF]Wood3.0*10-6Concrete6.0*10-6GlassCastironSteelWroughtironCopper4.4*10-66.1*10-66.5*10-66.7*10-69.3*10-6Bronze10.0*10-6Aluminum12.8*10-6Brass10.4*10-6295StrengthofMaterialsOfperhapsevengreaterimportanceinengineeringdesignarethestressesdevelopedbyrestrainingthefreeexpansionandcontractionofmemberssubjectedtotemperaturevariations.Tocalculatethesetemperaturestresses,itisusefultodeterminefirstthefreeexpansionorcontractionofthememberinvolvedandthentheforceandunitstressdevelopedinforcingthemembertoattainitsoriginallength.Theproblemfromthispointonisexactlythesameasthosesolvedintheearlierportionsofthischapterdealingwithaxialstresses,strains,anddeformations.TheamountofstressdevelopedbyrestoringabartoitsoriginallengthLisf=εE=δαL¢TEE==α¢TELL‹f=α¢TEExampleProblems:ThermalEffects16Asurveyor’ssteeltapemeasuresexactly100feetbetweentheendmarkingswhenthetemperatureis70°F.Whatisthetotalerrorthatresultswhenmeasuringatraverse(routeofasurvey)of5,000feetwhenthetemperatureofthetapeis30°F?Solution:δ=αL¢Tαs=6.5*10-6¢T=70°F-30°F=40°Fδ=16.5*10-6in.>in.>°F21100¿*12in.>ft.2140°F2=0.312in.1per100¿2Totallengthofthetraverse=5,000¿5,000¿=50tapelengths100¿>measurement‹δtotal=50*0.312–=15.6–=1.3¿296StrengthofMaterials17AW18×35wide-flangebeam(seeTableA3intheAppendix)isusedasasupportbeamforabridge.Ifatemperaturechange(rise)of40°Foccurs,determinethedeformationthatresults.Solution:δ=αL¢T=16.5*10-6in.>in.>°F2140¿*12in.>ft.2140°2=1.25*10-1in.=0.125–Whatwouldhappenifbothendsofthebeamweresolidlyimbeddedinconcretefoundationsonbothsides?Assumeitisinitiallyunstressedaxially.Solution:Becausethesupportrestrictsanydeformationfromoccurring,stresseswillbeinduced.¢f=εE=αL¢TE=α¢TELf=16.5*10-6in.>in.>°F2140°F2129*106k>in.32=7,540#>in.2297StrengthofMaterials18Computethethermalstressinducedduetoatemperaturechangeof100°F,assumingthatthebeamisabletoexpandfreelyfor14inch.Solutionδ=αL¢T=16.5*10-6>°F21100°F2150¿*12in.>ft.2=0.39–Restraineddeformation:δ¿=0.39–-0.25–=0.14–δ¿Lε¿=andf=ε¿E‹¢f=ε¿E=0.14–δ¿1E2=129*106psi2=6,767psiL600–19Asteelwide-flangebeam(W8×31)isusedtobracetwoshoringwallsasshown.Ifthewallsmove0.01inchoutwardwhenthebeamissubjectedtoa100°Ftemperaturechange,determinethestressinthebeam.SolutionThetotalelongationofthebeamifitisunrestrainedisδ=αL¢T=16.5*10-6>°F2172–21100°F2=0.0468–Becausethewallsmove0.01inch,thebeammustbecompressedbyaforcecausingadeformationofδ¿=0.0468–-0.01–=0.0368–Therefore,f=εE=298δ¿1E2L=0.0368–129*106psi272–=14,822psiStrengthofMaterialsProblems16Alongconcretebearingwallhasverticalexpansionjointsplacedevery40feet.Determinetherequiredwidth20°Fandjustofthegapinajointifitiswideopenatbarelyclosedat80°F.Assumeα=6*10-6>°F.17Analuminumcurtainwallpanel12feethighisattachedtolargeconcretecolumns(topandbottom)whenthetemperatureis65°F.Noprovisionismadefordifferentialthermalmovementvertically.Becauseofinsulationbetweenthem,thesunheatsupthewallpanelto120°Fbutthecolumntoonly80°F.Determinetheconsequentcompressivestressinthecurtainwall.18Thesteelrailsofacontinuous,straightrailroadtrackareeach60feetlongandarelaidwithspacesbetweentheirendsof0.25inchat70°F.a.Atwhattemperaturewilltherailstouchendtoend?b.Whatcompressivestresswillbeproducedintherailsifthetemperaturerisesto150°F?299StrengthofMaterials5STATICALLYINDETERMINATEMEMBERS(AXIALLYLOADED)Intheworkthusfar,ithasalwaysbeenpossibletofindtheinternalforcesinanymemberofastructurebymeansoftheequationsofequilibrium;thatis,thestructureswerestaticallydeterminate.If,inanystructure,thenumberofunknownforcesanddistancesexceedsthenumberofindependentequationsofequilibriumthatareapplicable,thestructureissaidtobestaticallyindeterminate.Areinforcedconcretecolumn,asshowninFigure38,isanexampleofastaticallyindeterminatesysteminwhichitisnecessarytowriteanadditionalequation(s)involvingthegeometryofthedeformationsinthemembersofthestructuretosupplementtheequationsofequilibrium.Thefollowingoutlineisaprocedurethatmightbehelpfulintheanalysisofsomeproblemsinvolvingaxiallyloaded,staticallyindeterminatemembers.1.DrawanFDB.2.Notethenumberofunknownsinvolved(magnitudesandpositions).3.RecognizethetypeofforcesystemontheFBD,andnotethenumberofindependentequationsofequilibriumavailableforthissystem.Figure38Reinforcedconcretecolumnswithlongitudinalsteelandconcretesharingthecompressiveload.Photobyauthor.4.Ifthenumberofunknownsexceedsthenumberofequilibriumequations,adeformationequationmustbewrittenforeachextraunknown.5.Whenthenumberofindependentequilibriumequationsanddeformationequationsequalsthenumberofunknowns,theequationscanbesolvedsimultaneously.Deformationsandforcesmustberelatedtosolvetheequationssimultaneously.Itisrecommendedthatadisplacementdiagrambedrawnshowingdeformationstoassistinobtainingthecorrectdeformationequation.300StrengthofMaterialsExampleProblems:StaticallyIndeterminateMembers(AxiallyLoaded)20Ashort,built-upcolumnmadeupofarough-cut4–*4–timbermember,reinforcedwithsteelsideplates,mustsupportanaxialloadof20k.DeterminethestressthatdevelopsinthetimberandinthesteelplatesifEtimber=1.76*103ksi;Esteel=29*103ksiTherigidplatesatthetopandbottomallowthetimberandsteelsideplatestodeformuniformlyundertheactionofP.Solution:Fromtheconditionofequilibrium:ft=Stressdevelopedinthetimberfs=StressdevelopedinthesteelC©Fy=0D-20k+fsAs+ftAt=0whereAs=2*14–*4–=2in.2At=4–*4–=16in.22in.21fs2+16in.21ft2=20kObviously,theequationofequilibriumaloneisinsufficienttosolveforthetwounknowns.Therefore,anotherequation,involvingadeformationrelationship,mustbewritten.δ=δt=δsBecauseboththetimberandthesteelmembersarequallength,Lt=Ls;then,ε=δtδs=;εt=εsLtLsbutE=‹ftEtft=eof=fεfsEsorε=fE;ft=fs1.76*103ksi29*103ksiEtEs1fs2=0.0611fs2Substitutingintotheequilibriumequation,2in.21fs2+16in.210.061fs2=20k2fs+0.976fs=20ksi;2.976fs=20ksifs=6.72ksift=0.06116.72ksi2=0.41ksi301StrengthofMaterials21Ashortconcretecolumnmeasuring12squareinchesisreinforcedwithfour#8bars1As=4*0.79in.2=3.14in.22andsupportsanaxialloadof250k.Steelbearingplatesareusedtopandbottomtoensureequaldeformationsofsteelandconcrete.CalculatethestressdevelopedineachmaterialifEc=3*106psiandEs=29*106psiSolutionFromequilibrium,C©Fy=0D-250k+fsAs+fcAc=0As=3.14in.2Ac=112–*12–2-3.14in.2141in.23.14fs+141fc=250kFromthedeformationrelationship,δs=δc;Ls=Lc‹δsδc=LLandεs=εcBecauseE=andfsEs=fεfcEcfs=fc29*1031fc2Es==9.67fcEc3*103Substitutingintotheequilibriumequation,3.1419.76fc2+141fc=25030.4fc+141fc=250171.4fc=250fc=1.46ksi‹fs=9.6711.462ksifs=14.1ksi302StrengthofMaterialsProblems19Thecolumnshownconsistsofanumberofsteelreinforcingrodsimbeddedinaconcretecylinder.Determinethecross-sectionalareaoftheconcreteandthesteelandthestressesineachifthecolumndeforms0.01inchwhilecarryinga100kload.Esteel=29*106psiEconcrete=3*106psi20Ashort,hollowsteeltube1L=30–2isfilledwithconcrete.Theoutsidedimensionofthetubeis12.75inches,andthewallthicknessis0.375inch.TheassemblyiscompressedbyanaxialforceP=180kappliedtoinfinitelyrigidcoverplates.Determinethestressdevelopedineachmaterialandtheshorteninginthecolumn.Econcrete=3*103ksiEsteel=29*103ksi21Twosteelplates14inchthickbyeightincheswideareplacedoneithersideofablockofoakfourinchesthickbyeightincheswide.IfP=50,000#isappliedinthecenteroftherigidtopplate,determinethefollowing:a.Stressdevelopedinthesteelandoak.b.DeformationresultingfromtheappliedloadP.Esteel=30*106psi;Eoak=2*106psi22Asurveyor’stapeofhardsteelhavingacrosssectionalareaof0.004squareinchisexactly100feetlongwhenapullof15#isexertedonitat80°F.a.Usingthesamepullonit,howlongwillthetapebeat40°F?b.WhatpullPisrequiredtomaintainitslengthof100feetat40°F?Es=29*103ksi303StrengthofMaterialsSummary■Stressistheintensityofaforce—thequantityofforcethatactsonaunitarea.Mathematically,stressiswrittenasf=■■axialforceP=AresistingforceTwobasicstressesarenormal(tensileorcompressive)andshearingstress.Tensilestressestendtoelongateamember,whereascompressivestressesshortentheelement.Bothareformsofnormalstresswheretheresistingarea(themember’scross-sectionalarea)isperpendiculartotheloaddirection.fcompression=fc=■■■PAdeformationδ=LmemberlengthElasticityisamaterialpropertyinwhichdeformationsdisappearwiththeremovaloftheload.Allstructuralmaterialsareelastictosomeextent.Ifloadsremainbelowtheelasticlimitofthematerial,nopermanentdeformationresultsfromtheapplicationandremovaloftheload.Whenloadsexceedtheelasticlimitinamateriallikesteel,apermanentdeformationresults,andthebehavioriscalledplasticorinelastic.Structuralsteelsexhibitanumberofkeypointswhenseeninagraphdepictingstressversusstrain.■■■■304PAStrainisameasureofdeformationperunitlength.Axialstrain(duetotension,compression,orthermalchange)isdefinedbytheequation:ε=■ftension=ft=Shearstressesdevelopoveranareathatisparalleltothedirectionoftheappliedforces.Theshearstressequationisidenticalwiththatfornormalstressexceptfortheresistingareaused.fshear=fc=■PAElasticlimit.Themaximumstressthatcanbedevelopedwithoutcausingpermanentdeformation.Yieldpoint.Thestressatwhichasignificantstrainoccurswithoutaconcurrentincreaseinstress.Thisstressvalueisusedindeterminingtheallowablestress(safe,workinglevels)usedinthedesignofstructuralsteelmembers.Ultimatestress.Thehighestlevelofstressattainedbeforefailure.Rupture.Thestresslevelatthetimeofphysicalfailure.StrengthofMaterials■Themeasureofelasticityofamaterialisreferredtoasthemodulusofelasticity,orYoung’smodulus.E=fε=stress1intheelasticrange2strainThisratio(E)representstheslopeoftheelasticportionofthestress-straindiagram.HighEvaluesaregenerallydesirable,becausethematerialismoreresistanttodeformation.Themodulusofelasticityisameasureofmaterialstiffness.■■Stressconcentrationsoccurwhenamember’sgeometryischangedtoincludediscontinuitiesorchangingcrosssections.Unrestrainedstructuralmemberswillexperiencedimensionalchangesduetotemperaturechange.δ=αL¢T=1coefficientofthermalexpansion2*1length2*1changeintemperature2■Ifastructuralmemberisfullyrestrainedandnodeformationcanoccurduetothermalchange,internalstresseswillresult.Internalstressiscomputedasf=α¢TE=1coefficientofthermalexpansion2*1changeintemperature2*1modulusofelasticity2AnswerstoSelectedProblems12ft=1,060.7psi15(a)Areq’d.=3in.2;D=1.95”(b)δ=0.75”Oneturnontheturnbuckle=12–Numberofturnsrequired=1.517f=6,140psi18(a)∆T=53.4°F;Tfinal=123.4°F;(b)f=4,994psi19fs=2.42ksi;fc=0.25ksi;As=25.8in.2;Ac=150.8in.220fs=6.87ksi;fc=0.71ksi;δ=0.0071”21(a)fo=.543ksi;fs=8.15ksi;TAB=10k;Areq’d.=0.46in.2;2fora1316–dia.rod;A=0.5185in.4h=180’5(a)fc=312.5psi;(b)ft=13,245psi;(c)fbrg=259.7psi;(d)f=156.3psi;(e)L=16.7in.7ε=0.0012in./in.9δ=0.0264in.11fbearing=P/A=1,780lb./48in.2=37.1psi<125psi∴OK13(a)δ=PL/AE=0.17”(b)δs=0.002”(b)Areq’d.=3in.2;D=1.95”≈2”rod305306Cross-SectionalPropertiesofStructuralMembersFromChapter6ofStaticsandStrengthofMaterialsforArchitectureandBuildingConstruction,FourthEdition,BarryOnouye,KevinKane.Copyright©2012byPearsonEducation,Inc.PublishedbyPearsonPrenticeHall.Allrightsreserved.307Cross-SectionalPropertiesofStructuralMembersIntroductionBeamdesignrequirestheknowledgeofmaterialstrengths(allowablestresses),criticalshearandmomentvalues,andinformationabouttheircross-section.Theshapeandproportionofabeamcross-sectioniscriticalinkeepingbendingandshearstresseswithinallowablelimitsandmoderatingtheamountofdeflectionthatwillresultfromtheloads.Whydoesa2–*8–joiststandingonedgedeflectlesswhenloadedatmidspanthanthesame2–*8–usedasaplank?Columnswithimproperlyconfiguredcross-sectionsmaybehighlysusceptibletobucklingunderrelativelymoderateloads.Arecircularpipecolumns,then,betteratsupportingaxialloadsthancolumnswithacruciformcross-section?(Figure1)Itwillbenecessarytocalculatetwocross-sectionalpropertiescrucialtothedesignofbeamsandcolumns.Thetwopropertiesarethecentroidandthemomentofinertia.1CENTEROFGRAVITY—CENTROIDSCenterofgravity,orcenterofmass,referstomassesorweightsandcanbethoughtofasasinglepointatwhichtheweightcouldbeheldandbeinbalanceinalldirections.Iftheweightorobjectwerehomogeneous,thecenterofgravityandthecentroidwouldcoincide.Inthecaseofahammerwithawoodenhandle,itscenterofgravitywouldbeclosetotheheavysteelhead,andthecentroid,whichisfoundbyignoringweightandconsideringonlyvolume,wouldbeclosertothemiddleofthehandle.Becauseofvaryingdensities,thecenterofgravityandthecentroiddonotcoincide.Figure1Relativeresistanceoffourbeamcross-sections(withthesamecross-sectionalareas)tobending
stressanddeflection.308Centroidusuallyreferstothecentersoflines,areas,andvolumes.Thecentroidofcross-sectionalareas(ofbeamsandcolumns)willbeusedlaterasthereferenceoriginforcomputingothersectionproperties.Cross-SectionalPropertiesofStructuralMembersThemethodoflocatingthecenterofgravityofamassoranareaisbasedonthemethodofdeterminingtheresultantsofparallelforcesystems.Ifanareaormassisdividedintoalargenumberofsmall,equalareas,eachofwhichisrepresentedbyavector(weight)actingatitscentroid,theresultantvectoroftheentireareawouldactthroughthecenterofgravityofthetotalmassarea.Thecenterofgravityofamassoranareaisthetheoreticalpointatwhichtheentiremassorareacanbeconsideredtobeconcentrated.Todeveloptheequationsnecessaryforcalculatingthecentroidalaxesofanarea,considerasimple,squareplateofuniformthickness(Figure2).x=centroidalxdistance=2¿fromthereferenceyaxisy=centroidalydistance=2¿fromthereferencexaxisItmaybeobviousthatthecentroidislocatedatx=2¿andy=2¿,butamethodologymaybenecessarytohandleodd-shapedorcomplicatedareas.Figure2Centroidoftheentireplate.Figure3Platedividedintofourcomponents.Thefirststep,usingthemethoddescribedbelow,istodividetheareaintosmallerincrementscalledcomponents(Figure3).Eachcomponenthasitsownweightandcentroid,withallweightsdirectedperpendiculartothesurfacearea(x-yplane).Themagnitudeoftheresultantisequaltothealgebraicsumofthecomponentweights.W=©¢WwhereW=totalweightoftheplate¢W=componentweightCentroidsareobtainedbytakingmomentsaboutthexandyaxes,respectively.TheprincipleinvolvedmaybestatedasThemomentofanareaaboutanaxisequalsthealgebraicsumofthemomentsofitscomponentareasaboutthesameaxis.UsingthediagramsinFigures4and5,writethemomentequations:and©My:xW=W1x2+W2x1+W3x2+W4x1©Mx:yW=W1y1+W2y1+W3y2+W4y2309Cross-SectionalPropertiesofStructuralMembersx=©1x¢W2y=©1y¢W2WWIftheplateisdividedintoaninfinitenumberofelementalpieces,thecentroidalexpressionsmaybewrittenincalculusformasx=y=Figure4Elevationsofthewholeplate.冮xdWW冮ydWWAssumingthattheplateisofuniformthicknesst,W=γtAwhereWγtA====totalweightoftheplatedensityoftheplatematerialplatethicknesssurfaceareaoftheplateCorrespondingly,forthecomponentparts(areas)oftheplatewithuniformthicknesst,¢W=γt¢Awhere¢W=weightofcomponentplatearea¢A=surfaceareaofcomponentFigure5310Elevationsofthequarteredplate.IfwereturntothemomentequationswrittenaboveandsubstitutethevaluesγtAforWandγt¢Afor¢W,wefindthatiftheplateishomogeneousandofconstantthickness,γtcancelsoutoftheequation(s).Cross-SectionalPropertiesofStructuralMembersTheresultingmomentequationwouldthenbewrittenas©My:xW=x2¢A1+x1¢A2+x2¢A3+x1¢A4©Mx:yW=y1¢A1+y1¢A2+y2¢A3+y2¢A4x=©1x¢A2y=©1y¢A2AAwhereA=©¢AThemomentofanareaisdefinedastheproductoftheareamultipliedbytheperpendiculardistancefromthemomentaxistothecentroidofthearea.ThecentroidsofsomeofthemorecommonareashavebeenderivedandareshowninTable1.Thederivationaboveshowshowthecentroidofanareacanbelocatedbydividinganareaintoelementalareasandsummingthemomentsoftheelementalareasaboutanaxis.Infindingthecentroidofamorecomplexarea(i.e.,acompositearea),asimilarmethodologyisused.Compositeormorecomplexareasarefirstdividedintosimplergeometricshapes(asshowninTable1)withknowncentroids.Areferenceoriginischosen(usuallythelowerleftcorner)toestablishthereferencexandyaxes.Then,momentsofareaaresummedaboutthereferencexandyaxes,respectively.Thecentroidlocatesthenewreferenceoriginforsubsequentcomputationsofothercross-sectionalproperties(momentofinertiaandradiusofgyration).311Cross-SectionalPropertiesofStructuralMembersTable1Centroidsofsimpleareas.Shape312DrawingAreaxyRectangleb/2h/2bhTriangleb/3h/3bh/2Semicircle04r/3ππr2/2QuarterCircle4r/3π4r/3ππr2/4ParabolicSegment5b/82h/52bh/3ComplementofaParabolicSegment3b/43h/10bh/3Cross-SectionalPropertiesofStructuralMembersExampleProblems:Centroids1Determinethecentroidalxandydistancesfortheareashown.Usethelowerleftcornerofthetrapezoidasthereferenceorigin.Solution:Selectaconvenientreferenceorigin.Itisusuallyadvantageoustoselectareferenceoriginsuchthatxandydistancesaremeasuredinthepositivexandydirectionstoavoiddealingwithnegativesigns.Dividethetrapezoidintosimplergeometricshapes:arectangleandatriangle.Ascompositeareasbecomemorecomplex,itmaybeconvenienttouseatabularformatforcentroidalaxiscalculations.ComponentArea(≤A)xx≤Ayy≤A9–(3–)26–81in.34–54in.34.5–121.5in.31.5–40.5in.3=13.5in.29–(3–)=27in.2A=©¢A=40.5in.2©x¢A=202.5in.3©y⌬A=94.5in.3Thecentroidaldistancesxandyfromthereferenceoriginarex=y=©x¢A202.5in.3=5–=A40.5in.2©y¢AA=94.5in.340.5in.2=2.33–313Cross-SectionalPropertiesofStructuralMembers2FindthecentroidoftheL-shapedareashown.Usethereferenceoriginshown.Solution:Again,thecompositeareawillbesectionedintotwosimplerectangles.ThesolutionbelowisbasedonthediagraminFigure(a).(a)ComponentArea(¢A)xx¢Ayy¢A2–(6–)=12in.21–12in.35–60in.32–(6–)=12in.23–36in.31–12in.3A=©¢A=24in.2314(b)©x¢A=48in.3©y¢A=72in.3Cross-SectionalPropertiesofStructuralMembersThecentroidalaxisfromthereferenceoriginisx=y=48in.3©x¢A==2–A24in.2©y¢AA=72in.324in.2=3–3Determinethecenterofgravity(centroid)ofthetrianglewiththecutoutshown.Thisparticularproblemwillutilizetheconceptofnegativeareasinthesolution.Solution:Thereferenceoriginisagainlocatedinthelowerleftcornerforconvenience(avoidingthenegativedistances).ComponentArea(¢A)xx¢Ayy¢A12–(9–)=54in.224–216in.33–162in.3–3–(5–)=–15in.33.5––52.5in.31.5––22.5in.3©¢A=39in.2x=y=©x¢A=163.5in.3©y¢A=139.5in.3©x¢A163.5in.3=4.2–=A39in.2©y¢AA=139.5in.339in.2=3.6–315Cross-SectionalPropertiesofStructuralMembers4Findthecentroidofthebuilt-upsteelsectioncomposedofaW12*87(wideflange)witha12¿¿*14–coverplateweldedtothetopflange.SeeTableA3intheAppendixforinformationaboutthewideflange.Solution:W12*87:d=12.53–bf=12.13–A=25.6in.2ComponentArea(¢A)xx¢Ayy¢A7in.20–012.78–89.5in.325.6in.20–06.26–160.3in.3A=©¢A=32.6in.2©y¢A=249.8in.3©x¢A=0x=y=©x¢A0==0A32.6in.2©y¢AA=249.8in.332.6in.2=7.7–Thecentroid,orcenterofgravity(CG),nowbecomesthenewreferenceoriginforevaluatingothercross-sectionalproperties.316Cross-SectionalPropertiesofStructuralMembersProblemsFindthecentroidofthefollowingcross-sectionsandplanes.123317Cross-SectionalPropertiesofStructuralMembers45318Cross-SectionalPropertiesofStructuralMembers2MOMENTOFINERTIAOFANAREAThemomentofinertia,orsecond-momentasitissometimescalled,isamathematicalexpressionusedinthestudyofthestrengthofbeamsandcolumns.Itmeasurestheeffectofthecross-sectionalshapeofabeamonthebeam’sresistancetobendingstressanddeflection.Instabilityorbucklingofslendercolumnsisalsoaffectedbythemomentofinertiaofitscross-section.Themomentofinertia,orI-value,isashapefactorthatquantifiestherelativelocationofmaterialinacross-sectionintermsofeffectiveness.AbeamsectionwithalargemomentofinertiawillhavesmallerstressesanddeflectionsunderagivenloadthanonewithalesserI-value.Along,slendercolumnwillnotbeassusceptibletobucklinglaterallyifthemomentofinertiaofitscross-sectionissufficient.Momentofinertiaisameasureofcross-sectionalstiffness,whereasthemodulusofelasticityEisameasureofmaterialstiffness.Area=b*h1in.22Perimeter=2b+2h1in.2Figure6Rectangularbeamcross-section.Theconceptofmomentofinertiaisvitaltounderstandingthebehaviorofmoststructures,anditisunfortunatethisconcepthasnoaccuratephysicalanalogyordescription.ThedimensionalunitformomentofinertiaIisinchestothefourthpower(in.4).Assumingtherectangularcross-sectionshowninFigure6,aphysicaldescriptioncanbegivenasfollows:area=b*h1in.22perimeter=2b+2h1in.2However,themomentofinertiaforthecross-sectionisI=bh34in.12forarectangularcross-section.Figure7(a)2–*6–joist.Figure7(b)6–*2–plank.Thesecondmomentofanarea(areatimesadistancesquared)isquiteabstractanddifficulttovisualizeasaphysicalproperty.Ifweconsidertwoprismaticbeamsmadeofthesamematerialbutwithdifferentcross-sections,thebeamwhosecross-sectionalareahadthegreatermomentofinertiawouldhavethegreaterresistancetobending.Tohaveagreatermomentofinertiadoesnotnecessarilyimply,however,agreatercross-sectionalarea.Orientationofacross-sectionwithrespecttoitsbendingaxisiscrucialinobtainingalargemomentofinertia.A2–*6–rectangularcross-sectionisusedasajoistinFigure7(a)andasaplankinFigure7(b).Fromexperience,itisalreadyknownthatthejoistismuchmoreresistanttobendingthantheplank.Likemanystructuralelements,therectanglehasastrongaxis(orientation)andaweakaxis.Itisfarmoreefficienttoloadacross-sectionsothatbendingoccursaboutitsstrongaxis.319Cross-SectionalPropertiesofStructuralMembersFigure8(a)Wide-flangeshape.Itmayhelptounderstandtheconceptofmomentofinertiaifwedrawananalogybaseduponrealinertiaduetomotionandmass.ImaginethetwoshapesshowninFigure8tobecutoutofa12-inchsteelplateandplacedandspunaboutaxisx-x.Thetwoshapeshaveequalareas,buttheoneinFigure8(a)hasamuchhighermomentofinertiaIx-xwithrespecttotheaxisofspin.Itwouldbemuchhardertostartitspinning,andoncemoving,itwouldbemuchhardertostop.Thesameprincipleisinvolvedwhenafigureskaterspinsonice.Witharmsheldclosein,theskaterwillrotaterapidly;witharmsoutstretched(creatingincreasedresistancetospinand/ormoreinertia),theskaterslowsdown.Inourdiscussionaboutthemethodoffindingthecenterofgravityforareas,eachareawassubdividedintosmallelementalareasthatweremultipliedbytheirrespectiveperpendiculardistancesfromthereferenceaxis.Theprocedureissomewhatsimilarfordeterminingthemomentofinertia.Themomentofinertiaaboutthesameaxiswouldrequire,however,thateachelementalareabemultipliedbythesquareoftherespectiveperpendiculardistancesfromthereferenceaxis.Figure8(b)Cruciformshape.Amomentofinertiavaluecanbecomputedforanyshapewithrespecttoanyreferenceaxis(Figure9).Supposewewantedtofindthemomentofinertiaofanirregulararea,asshowninFigure9,aboutthexaxis.WewouldfirstconsidertheareatoconsistofmanyinfinitelysmallareasdA(wheredAisamuchsmallerareathan¢A).ConsideringthedAshown,itsmomentofinertiaaboutthexaxiswouldbey2dA.However,thisproductisonlyaminuteportionofthewholemomentofinertia.EachdAmakingupthearea,whenmultipliedbythesquareofitscorrespondingmomentarmyandaddedtogether,willgivethemomentofinertiaoftheentireareaaboutthexaxis.Themomentofinertiaofanareaaboutagivenaxisisdefinedasthesumoftheproductsofalltheelementalareasandthesquareoftheirrespectivedistancestothataxis.Therefore,thefollowingtwoequationsfromFigure9areFigure9area.MomentofinertiaofanirregularIx=Iy=冮A0冮A0y2dAx2dAThedifficultyofthisintegrationprocessislargelydependentontheequationoftheoutlineoftheareaanditslimits.Whentheintegrationbecomestoodifficultoramore320Cross-SectionalPropertiesofStructuralMemberssimple,approximatesolutionispermissible,themomentofinertiamaybeexpressedinfinitetermsbyIx=©y2¢AIy=©x2¢AThesolutionbecomeslessaccurateasthesizeofincreased.¢AisAnI-valuehastheunitsoflengthtothefourthpower,becausethedistancefromthereferenceaxisissquared.Forthisreason,themomentofinertiaissometimescalledthesecondmomentofanarea.Moreimportantly,itmeansthatelementsorareasthatarerelativelyfarawayfromtheaxiswillcontributesubstantiallymoretoanI-valuethanthosethatarecloseby.ExampleProblems:MomentofInertiabyApproximation5Toillustratetheapproximatemethodofmomentofinertiacalculation,wewillexaminea4–*8–beamcrosssection.ThemomentofinertiaIisalwayscomputedrelativetoareferencepointoraxis.Themostuseful,andoften-used,referenceaxisisthecentroidalaxis.(Nowyouseethereasonforfindingthecentroidofanarea.)InthefirstapproximatecalculationforIx,let’sassumethecross-sectionaboveisdividedintotwoelementalareas.Eachelementalareameasures4–*4–,andthecomponentcentroidalydistanceistwoinches.Beamcrosssection14–*8–2.A1=16in.2;y1=2–A2=16in.2;y2=2–Ix=©y2¢A=116in.2212–22+116in.2212–22=128in.4Thesmallertheelementsselected,themoreaccuratetheapproximationwillbe.Therefore,let’sexaminethesamecross-sectionandsubdivideitintofourequalelements.Ixbasedontwoelementalareas.321Cross-SectionalPropertiesofStructuralMembersA1=8in.2;y1=+3–A2=8in.2;y2=+1–A3=8in.2;y3=-1–A4=8in.2;y4=-3–Ix=©y2¢A=18in.221+3–22+18in.221+1–22+18in.221-1–22+18in.221-3–22Ix=72in.4+8in.4+8in.4+72in.4=160in.4Ixbasedonfourelementalareas.NoticethatthevalueofIxisdifferentfromthatinthepreviouscalculation.Ifthecross-sectionissubdividedintoevensmallerareas,Ixwillapproachanexactvalue.TocomparethetwocomputationsabovewiththeexactvalueofIxforarectangularcross-section,therectanglewillbesubdividedintoinfinitelysmallareasdA.Ix=冮ydA;Ix=冮Ix=bh3-h3bh3h3bh3cab-abd=c+d=322388122dA=bdyy2dA=冮+h>2-h>2y2bdy=b冮+h>2-h>2y2dy=by33`+h>2-h>2Substitutingb=4–andh=8–,Iexact=14–218–2312=171in.4Themomentofinertiacanbedeterminedabouttheyaxisinthesamemanneraswasdoneforthexaxis.IxbasedonintegrationofareasdA.322Momentsofinertiacalculatedbytheintegrationmethod(exact)forsomebasicgeometricshapesareshowninTable2.Cross-SectionalPropertiesofStructuralMembersTable2Momentsofinertiaforsimplegeometricshapes.ShapeMomentofInertia(Ix)Ix=bh312Ix=bh336Ix=πd4πr4=464Ix=π1D4-d42Ix=r4a64π8b=0.11r489π323Cross-SectionalPropertiesofStructuralMembers6DeterminethevalueofIaboutthexaxis(centroidal).Solution:Thisexamplewillbesolvedusingthenegativeareamethod.SolidVoidsIsolid=bh312bh31216–216–23Iopenings=2*Isolid=12=108in.412–214–23=21.3in.412Ix=108in.4-21.3in.4=86.7in.4Iopenings=2*Note:Ixisanexactvaluehere,becausethecomputationswerebasedonanexactequationforrectangles.324Cross-SectionalPropertiesofStructuralMembersMomentofInertiabyIntegration:Ixc=冮ydA2wheredA=bdy1b=base2yvariesfrom0→+2–whileb=2”0→–2–andfrom:+2–→+3–whileb=6”–2–→–3–}Ixc=}冮ybdy2ATreatbasaconstantwithinitsdefinedboundarylines.Ixc=Ixc=IxcIxc冮+2122y2dy+-22y3+2`6y3`冮+3+3+2162y2dy+6y3`-2冮-2-3162y2dy3-23+23-32=18-3-842+2127-82+21-8-3-27423=10.67+38+38=86.67in.4++325Cross-SectionalPropertiesofStructuralMembers3(a)PrecastconcreteT-section.MOMENTOFINERTIAOFCOMPOSITEAREASInsteelandconcreteconstruction,thecross-sectionsusuallyemployedforbeamsandcolumnsarenotlikethesimplegeometricshapesshowninTable2.Moststructuralshapesareacompositeoftwoormoreofthesimpleshapescombinedinconfigurationstoproducestructuralefficiencyorconstructionexpediency.Wecalltheseshapescompositeareas(Figure10).Instructuraldesign,themomentofinertiaaboutthecentroidalaxisofthecross-sectionisanimportantsectionproperty.Becausemomentsofinertiacanbecomputedwithrespecttoanyreferenceaxis,astandardreferencewasnecessaryindevelopingconsistencywhencomparingthestiffnessofonecross-sectionrelativetoanother.(b)Steelwide-flangesection.(c)Precastconcreteplank.Theparallelaxistheoremprovidesasimplewaytocomputethemomentofinertiaofashapeaboutanyaxisparalleltothecentroidalaxis(Figure11).Theprincipleoftheparallelaxistheoremmaybestatedasfollows:Themomentofinertiaofanareawithrespecttoanyaxisnotthroughitscentroidisequaltothemomentofinertiaofthatareawithrespecttoitsownparallelcentroidalaxisplustheproductoftheareaandthesquareofthedistancebetweenthetwoaxes.TheparallelaxistheoremexpressedinequationformisIx=©Ixc+©Ady2whichisthetransferformulaaboutthemajorxaxisandwhereIx=momentofinertiaofthetotalcross-sectionaboutthemajorcentroidalxaxis1in.42(d)Steelchannelsection.Figure10beams.ColumncompositeshapesforIxc=momentofinertiaofthecomponentareaaboutitsowncentroidalxaxis1in.42A=areaofthecomponent1in.22dy=perpendiculardistancebetweenthemajorcentroidalxaxisandtheparallelaxisthatpassesthroughthecentroidofthecomponent(in.)ThetransferformulaforthemajoryaxisisexpressedasIy=©Iyc+©Adx2Figure11326Transferareaaboutaxesxandy.Cross-SectionalPropertiesofStructuralMembersExampleProblems:MomentofInertiaofCompositeAreas7Determinethemomentofinertiatroidalxaxis.Ixaboutthecen-Note:ThisisidenticaltoExampleProblem6.Solution:Insteadofthenegativeareamethod,thetransferformulawillbeused,employingtheparallelaxistheorem.Thecross-sectionwillbedividedintothreecomponentsasshown(twoflangesandtheweb).Eachcomponentofthecompositeareahasitsowncentroidandisdenotedasxc.Themajorcentroidalaxisoftheentirecross-sectionisX.NotethatthemajorcentroidalaxisXcoincideswiththecomponentaxisxc2.Ascompositeareasbecomemorecomplex,itmaybeadvisabletouseatabularformattominimizeerrorandconfusion.ComponentIxc1in.426–11–23bh3=1212=0.5in.42–14–2312=10.67in.4=0.5in.4©Ixc=11.67in.4A1in.22dy1in.2Ady21in.426in.22.5–37.5in.48in.20–06in.2–2.5–37.5in.4©Ady2=75in.4ThemomentofinertiaoftheentirecompositeareaisIX=©Ixc+©Ady2=11.67in.4+75in.4=86.67in.4Theconceptofthetransferformulainvolvestheadditionalinertiarequiredtomergeortransferanaxisfromonelocationtoanother.ThetermAdy2representstheadditionalinertiadevelopedaboutthemajoraxisXduetocomponentsland2.327Cross-SectionalPropertiesofStructuralMembers8Determinethemomentofinertiaaboutthecentroidalxandyaxesforthecompositeareashown.Solution:Usuallythefirststepindeterminingthemomentofinertiaaboutthemajororcentroidalaxisrequiresthedeterminationofthecentroid(particularlyinunsymmetricalcross-sections).Componentxx¢A10in.20–04.5–45in.316in.20–01–∆AA=©¢A=26in.2©x¢A=0x=0y=328©y¢AA=61in.326in.2=2.35–yy¢A16in.3©y¢A=61in.3Cross-SectionalPropertiesofStructuralMembersdy1=y1-y=4.5–-2.35–=2.15–dy2=y-y2=2.35–-1–=1.35–ComponentA10in.216in.2dyAdy22.15–46.2in.41.35–29.2in.4Ixc2–15–2312=20.8in.48–12–2312=5.3in.4©Ixc=26.1in.4©Ady2=75.4in.4Iyc5–12–2312=3.3in.42–18–2312=85.3in.4©Iyc=88.6in.4dxAdx20–00–0©Adx2=0Ix=©Ixc+©Ady2=26.1in.4+75.4in.4=101.5in.4Iy=©Iyc+©Adx2=88.6in.4+0in.4=88.6in.4329Cross-SectionalPropertiesofStructuralMembers9FindIxandIyfortheL-shapedcross-sectionshown.Solution:Itmaybeconvenienttosetupatableforsolvingthecentroidaswellasthemomentofinertia.ComponentAreaxAxyAy8in.20.5–4in.34–32in.38in.23–24in.31–8in.3©A=16in.2x=ComponentAIxcdy28in.316in.2©Ax=28in.3Ady212=42.67in.412=2.67in.416in.21.5–1218in.4=0.67in.4dxAdx21.25–12.5in.41.25–12.5in.42–14–231.5–18in.412=10.67in.4©Ixc=45.34in.4;©Ady2=36in.4;©Iyc=11.34in.4©Adx2=25in.4Ix=©Ixc+©Ady2;Ix=81.34in.4Iy=©Iyc+©Adx2;Iy=36.34in.4330=2.5–8–11–234–12–238in.240in.3Iyc1–18–238in.2y==1.75–;©Ay=40in.3Cross-SectionalPropertiesofStructuralMembersProblemsDetermineIxandIyforthecross-sectionsinProblems6through8andIxforProblems9through11.678331Cross-SectionalPropertiesofStructuralMembers9AllmembersareS4S(surfacedonfoursides).SeeTableA1intheAppendix.2*4S4S;1.5–*3.5–;A=5.25in.22*12S4S;1.5–*11.25–;A=16.88in.21011SeethesteeltablesintheAppendixTableA3.332Cross-SectionalPropertiesofStructuralMembersExampleProblems10Abuilt-upsteelbeamusesaW18×97onitssidewitha1–*32–verticalplateanda2–*16–horizontalplate.Assumingtheverticalplateiscenteredonthewideflange’sweb,calculateIxandIyaboutthemajorcentroidalaxesofthecross-section.Solution:ComponentAreay2–+32–+28.5in.2Ay0.532–2976.7in.3=34.27–1–×32–=32in.22–+16–=18–2–×16–=32in.21–©A=92.5in.2y=©y¢AA=1,584.6in.392.5in.2576in.332in.3©Ay=1,585in.3=17.1–333Cross-SectionalPropertiesofStructuralMembersComponentAIxc(in.4)dyAdy2Iyc(in.4)dxAdx228.5in.220117.2–8,431in.41,7500–00–00–032in.432in.41"(32")312=2,7310.9–16"(2")312=10.716.1–©Ixc=2,942.7in.426in.48,295in.4©Ady2=16,752in.432"(1")312=2.72"(16")312=682.7©Iyc=2,435.4in.4©Adx2=0Ix=©Ixc+©Ady2=2,942.7in.4+16,752in.4=19,694.7in.4Iy=©Iyc+©Adx2=2,435.4in.4+0in.4=2,435.4in.4334Cross-SectionalPropertiesofStructuralMembers11Aheavilyloadedfloorsystemusesacompositesteelsectionasshown.AC15×40channelsectionisattachedtoIythetopflangeoftheW18×50.DetermineIxandaboutthemajorcentroidalaxesusingthecross-sectionalpropertiesgiveninthesteeltablesforstandardrolledshapes(seeAppendixTablesA3andA4).Solution:Determinethemajorcentroidalaxesasthefirststep.ComponentAreayAy11.8in.2d+tw–x=18+0.52–0.78=17.74–209.3in.314.7in.2d18–==9–22132.3in.3©A=26.5in.2y=©y¢AAComponent=341.6in.326.5in.2©Ay=341.6in.3=12.9–A11.8in.214.7in.2Ixcdyy1–y9.23=(17.7–12.9)=4.8”800_y–y2=(12.9–9)=3.9”©Ixc=809in.4Ady2IycdxAdx2272in.4349in.40”0224in.440.1in.40”0©Ady2=496in.4©Iyc=389in.4©Adx2=0Ix=©Ixc+©Ady2=809in.4+496in.4=1,305in.4Iy=©Iyc+©Adx2=389in.4+0in.4=389in.4335Cross-SectionalPropertiesofStructuralMembersProblemsDeterminethemomentsofinertiaforthecompositeareasusingthestandardrolledsectionsshownbelow.121314Abuilt-upboxcolumnismadebyweldingtwo43–*16–platestotheflangesoftwoC15×50.Forstructuralreasons,itisnecessarytohaveIxequaltoIy.FindthedistanceWrequiredtoachievethis.336Cross-SectionalPropertiesofStructuralMembers4RADIUSOFGYRATIONInthestudyofcolumnbehavior,wewillbeusingthetermradiusofgyration(r).Theradiusofgyrationexpressestherelationshipbetweentheareaofacross-sectionandacentroidalmomentofinertia.Itisashapefactorthatmeasuresacolumn’sresistancetobucklingaboutanaxis.AssumethataW14×90steelcolumn(Figure12)isaxiallyloadeduntilitfailsinabucklingmode(Figure13).Anexaminationofthebuckledcolumnwillrevealthatfailureoccursabouttheyaxis.Ameasureofthecolumn’sabilitytoresistbucklingisitsradiusofgyration(r)value.FortheW14×90,theradiusofgyrationvaluesaboutthexandyaxesareW14*90:rx=6.14–Figure12column.Axiallyloadedwide-flangeFigure13(y)axis.ColumnbucklingabouttheweakFigure14RadiusofgyrationforA1andA2.ry=3.70–Thelargerthervalue,themoreresistanceofferedagainstbuckling.Theradiusofgyrationofacross-section(area)isdefinedasthatdistancefromitsmomentofinertiaaxisatwhichtheentireareacouldbeconsideredasbeingconcentrated(likeablackholeinspace)withoutchangingitsmomentofinertia(Figure14).IfandthenA1=A2Ix1=Ix2Ix=Ar2‹rx2=IxAandrx=ry=IxCAIyDAForallstandardrolledshapesinsteel,theradiusofgyrationvaluesaregiveninthesteelsectionpropertiestableintheAppendix(TablesA3throughA6).337Cross-SectionalPropertiesofStructuralMembersExampleProblems:RadiusofGyration12Usingthecross-sectionshowninExampleProblem11,wefoundthatIx=1,305in.4;A=26.5in.2Iy=389in.4Theradiiofgyrationforthetwocentrcomputedasrx=oidalaxesareIx1,305in.4==7.02–AAD26.5in.2Iyry=DA389in.4D26.5in.2==3.83–13Twoidenticalsquarecross-sectionsareorientedasshowninFigure15.WhichofthesehasthelargerIvalue?Findtheradiusofgyrationrforeach.Solution:a1a23bh3a4==121212Figure(a):Ix=Figure(b):Madeupoffourtriangles‹Ix=Figure15(a)Squarecross-section(a)x(a).Ix=b=0.707a;410.707a*30.707a43236Ix=41bh32h=0.707a;+4a36+41Ady22dy=h0.707a=3310.707a2*0.707a*0.707abab23a4a43a4a4+==36183612Bothcross-sectionshaveequalradiiofgyration.rx=Figure15(b)Squarecross-sectionrotated45°.338a412Ixa2a1a====AASa2C122A3213Cross-SectionalPropertiesofStructuralMembersSummary■■■Centerofgravity(C.G.),orcenterofmass,referstomassesorweightsidealizedasasinglepointatwhichtheweightcouldbeheldandbeinbalanceinalldirections.Centroidreferstothegeometriccenteroflines,areas,andvolumes.Beamandcolumndesignutilizesthecentroidofcross-sectionalareasforcomputingothersectionproperties.Compositeareasaremadeupofsimplerareas.1A=©¢A2■Themomentofaareaaboutanaxisequalsthealgebraicsumofthemomentsofitscomponentareasaboutthesameaxis.x=■■©x¢A;Ay=©y¢AAThemomentofanareaisdefinedastheproductoftheareamultipliedbytheperpendiculardistancefromthemomentaxistothecentroidofthearea.Momentofinertia(I),orsecondmoment,isafactorthatquantifiestheeffectofthebeamorcolumn’scrosssectionalshapeinresistingbending,deflection,andbuckling.Themomentofinertiaisameasureofcrosssectionalstiffness.Ix=©y2¢A;Iy=©x2¢A;■Whencross-sectionalshapesarenotsimplegeometricshapesbutacompositeoftwoormoresimpleshapescombined,theparallelaxistheoremisusedfordeterminingthemomentofinertiaaboutanyaxisparalleltothecentroidalaxis.Ix=©Ixc+©Ady2;Iy=©Iyc+©Adx2■Theradiusofgyrationexpressestherelationshipbetweentheareaofacross-sectionandacentroidalmomentofinertia.Thisshapefactorisusedinmeasuringacolumn’sslendernessanditsresistancetobuckling.rx=IyIx;ry=AADAAnswerstoSelectedProblems1x=5.33”;y=5.67”3x=7.6’;y=5.3’4x=0;y=9.4”6y=2.0”;Ix=17.4in.4x=0.99”;Iy=6.2in.489y=5.74”;Ix=561in.411y=10.4”;Ix=1,518in.412x=–.036”;y=7.0”,Ix=110.4in.4;Iy=35.7in.414Ix=2,299in.4;W=15.8”Ix=1,787in.4;Iy=987in.4339340BendingandShearinSimpleBeamsFromChapter7ofStaticsandStrengthofMaterialsforArchitectureandBuildingConstruction,FourthEdition,BarryOnouye,KevinKane.Copyright©2012byPearsonEducation,Inc.PublishedbyPearsonPrenticeHall.Allrightsreserved.341BendingandShearinSimpleBeamsIntroductionAbeamisalong,slenderstructuralmemberthatresistsloadsthatareusuallyappliedtransverse(perpendicular)toitslongitudinalaxis.Thesetransverseforcescausethebeamtobendintheplaneoftheappliedloads,andinternalstressesdevelopinthematerialasitresiststheseloads.Beamsareprobablythemostcommontypeofstructuralmemberusedintheroofandfloorsofabuildingofanysize,aswellasforbridgesandotherstructuralapplications.Notallbeamsneedtobehorizontal;theymaybeverticaloronaslant.Inaddition,theymayhaveeitherone,two,orseveralreactions.1CLASSIFICATIONOFBEAMSANDLOADSThedesignofabeamentailsthedeterminationofsize,shape,andmaterialbasedonthebendingstress,shearstress,anddeflectionduetotheappliedloads(Figure1).(a)Pictorialdiagramofaloadedbeam.(b)FBDofthebeam.Figure1342Steelbeamwithloadsandsupportreactions.BendingandShearinSimpleBeamsBeamsareoftenclassifiedaccordingtotheirsupportconditions.Figure2illustratessixmajorbeamclassifications.(a)Simplysupported:twosupports.(b)Continuous:threeormoresupports.(c)Cantilever:oneendsupportedrigidly.(d)Overhang:twosupports—oneorbothsupportsnotlocatedattheend.(e)Propped:twosupports—oneendisfixed.(f)Restrainedorfixed:bothsupportsarefixed,allowingnorotationattherestrainedends.Figure2Classificationbasedonsupportconditions.343BendingandShearinSimpleBeamsTypesofConnectionsActualsupportandconnectionconditionsforbeamsandcolumnsareidealizedasrollers,hinges(pins),orfixed.Figure3illustratesexamplesofcommonsupport/connectionconditionsfoundinpractice.(a)Beamsupportedbyaneoprenepad.(b)Beamsupportedbyaconcreteorsteelcylinder.Examplesofrollersupports(a,b).Horizontaldisplacementandrotationarepermitted;maybeduetoloadsorthermalconditions.(c)Timberbeam-columnconnectionwithT-plate.(e)Typicalpin-connectedcolumnbase.Figure3(d)Steelbeamconnectedtoasteelgirder.(f)Trussjoint—threesteelangleswithgussetplate.Classificationbasedonconnectiontypes(continuesonnextpage).Examplesofhingeorpinsupports(c,d,e,f).Allowsacertainamountofrotationattheconnection.344BendingandShearinSimpleBeams(g)Reinforcedconcretefloor-wallconnection.(h)Steelstrapweldedtoagussetplate.(i)Timberpolestructure—embeddedbase.(j)Beam-columnmomentconnection.Examplesoffixedsupport(g,h,i,j).Norotationattheconnection.Figure3Classificationbasedonconnectiontypes(Con’t.)345BendingandShearinSimpleBeamsFigure4illustratesthefourfundamentaltypesofloadsthatcanactonabeam.(a)Concentratedload.(b)Uniformlydistributedload.(c)Nonuniformlydistributedload.(d)Puremoment.Figure4346Classificationbasedonloadtype.BendingandShearinSimpleBeams2SHEARANDBENDINGMOMENTWhenabeamissubjectedtoanyoftheloadingspreviouslydiscussed,eithersinglyorinanycombination,thebeammustresisttheseloadsandremaininequilibrium.Forthebeamtoremaininequilibrium,aninternalforcesystemmustexistwithinthebeamtoresisttheappliedforcesandmoments.Stressesanddeflectionsinbeamsarefunctionsoftheinternalreactions,forces,andmoments.Forthisreason,itisconvenientto“map”theseinternalforcesandtoconstructdiagramsthatgiveacompletepictureofthemagnitudesanddirectionsoftheforcesandmomentsthatactthroughoutthebeamlength.Thesediagramsarereferredtoasload,shear(V),andmoment(M)diagrams(Figure5).■■■TheloaddiagramshowninFigure5isforapointloadatthefreeendofacantileverbeam.(TheloaddiagramisessentiallytheFBDofthebeam.)ThesheardiagramshowninFigure5isagraphofthetransverseshearalongthebeamlength.ThemomentdiagramshowninFigure5isagraphofthebendingmomentalongthebeamlength.Asheardiagramisagraphinwhichtheabscissa(horizontalreferenceaxis)representsdistancesalongthebeamlengthandtheordinates(verticalmeasurementsfromtheabscissa)representthetransverseshearatthecorrespondingbeamsections.Amomentdiagramisagraphinwhichtheabscissarepresentsdistancesalongthebeamandtheordinatesrepresentthebendingmomentatthecorrespondingsections.Figure5Load,shear(V),andmoment(M)diagrams.Shearandmomentdiagramscanbedrawnbycalculatingvaluesofshearandmomentatvarioussectionsalongthebeamandplottingenoughpointstoobtainasmoothcurve.Suchaprocedureisrathertime-consuming,however,andalthoughitmaybedesirableforgraphicalsolutionsofcertainstructuralproblems,twomorerapidmethodswillbedevelopedinSections3and5.347BendingandShearinSimpleBeamsAsignconventionisnecessaryforshearandmomentdiagramsiftheresultsobtainedfromtheirusearetobeinterpretedconvenientlyandreliably.Bydefinition,theshearatasectionisconsideredtobepositivewhentheportionofthebeamtotheleftofthesectioncut(forahorizontalbeam)tendstobeintheuppositionwithrespecttotheportiontotherightofthesectioncut,asshowninFigure6.(+)Shear.(+)Shear.(–)Shear.Figure6(–)Shear.Signconventionforshear.Alsobydefinition,thebendingmomentinahorizontalbeamispositiveatsectionsforwhichthetopfibersofthebeamareincompressionandthebottomfibersareintension,asshowninFigure7.(+)Moment.(+)Moment.Figure7348Signconventionformoment.(–)Moment.(–)Moment.BendingandShearinSimpleBeamsPositivemomentgeneratesacurvaturethattendstoholdwater(concave-upwardcurvature),whereasnegativemomentcausesacurvaturethatshedswater(concavedownwardcurvature).Thisconventionisastandardoneformathematicsandisuniversallyaccepted.Becausetheconventionisrelatedtotheprobabledeflectedshapeofthebeamforaprescribedloadingcondition,itmaybehelpfultosketchintuitivelythebeam’sdeflectedshapetoassistindeterminingtheappropriatemomentsigns(Figure8).Figure8Deflectedshapeduetoloadsonoverhangbeam.TheoverhangbeamshowninFigure8exhibitsachangingcurvaturethatresultsinnegativetopositivetonegativemoments.Theimplicationhereisthatthebeamspanincludesoneormoretransversesectionswherethebendingmomentiszerotoaccommodatetherequiredsignchange.Suchasection,termedtheinflectionpointorpointofinflection,isalmostalwayspresentinoverhangandmultiplespanbeams.Animportantfeatureofthesignconventionusedforshearandmomentdiagramsisthatitdiffersfromtheconventionsusedinstatics.Whenusingtheequationsofequilibrium,forcesdirectedupandtotherightarepositive,andcounterclockwisemomenttendenciesarepositive.Thenewsignconventionisusedspecificallyforplottingtheshearandmomentdiagrams.Makesureyoudonotconfusethetwoconventions.349BendingandShearinSimpleBeams3EQUILIBRIUMMETHODFORSHEARANDMOMENTDIAGRAMSOnebasicmethodusedinobtainingshearandmomentdiagramsisreferredtoastheequilibriummethod.SpecificvaluesofVandMaredeterminedfromstaticsequationsthatarevalidforappropriatesectionsofthemember.Intheexplanationsthatfollow,weshallassumethatthememberisabeamacteduponbydownwardloads,butthemembercouldbeturnedatanyangle.AconvenientarrangementforconstructingshearandmomentdiagramsistodrawanFBDoftheentirebeamandthenconstructshearandmomentdiagramsdirectlybelow.Unlesstheloadisuniformlydistributedorvariesaccordingtoaknownequationalongtheentirebeam,nosingleelementaryexpressioncanbewrittenfortheshearormomentthatappliestotheentirelengthofthebeam.Instead,itisnecessarytodividethebeamintointervalsboundedbyabruptchangesintheloading.Anoriginshouldbeselected(differentoriginsmaybeusedfordifferentintervals),andpositivedirectionsshouldbeindicatedforthecoordinateaxes.BecauseVandMvaryasafunctionofxalongthebeamlength,equationsforVandMcanbeobtainedfromFBDsofportionsofthebeam(seeExampleProblem1).Completeshearandmomentdiagramsshouldindicatevaluesofshearandmomentateachsectionwheretheyaremaximumpositiveandmaximumnegative.Sectionswheretheshearand/ormomentarezeroshouldalsobelocated.350BendingandShearinSimpleBeamsExampleProblems:EquilibriumMethodforShearandMomentDiagrams1Usingtheequilibriummethod,drawtheshearandmomentdiagramsforasimplysupportedbeamwithasingleconcentratedload.Solution:SolveforexternalreactionsatAandB.CutthebeamthroughsectionD-D.DrawanFBDforeachhalfofthebeam.ExaminesegmentADfromtheFBDcutthroughD.3©Fy=04V=4k3©MD=04M=4k1x2FBDofentirebeam(oftenreferredtoastheloaddiagram).Note:ShearVisaconstantbetweenAandC.Themomentvariesasafunctionofx(linearly)betweenAandC.@x=0,M=0@x=6¿,M=24k-ft.ExaminesegmentAEfromtheFBDcutthroughE.3©Fy=04V=10k-4k=6k3©M=04+M+10k1x-6¿2-4k1x2=0M=60k-ft.-6xFBDofbeamsectionscutthroughD.@x=6¿,M=24k-ft.@x=10¿,M=0Note:ShearV=6kremainsconstantbetweenCandB.Themomentvarieslinearly,decreasingasxincreasesfromCtoB.FBDofbeamsectionscutthroughE.351BendingandShearinSimpleBeamsLoaddiagram(FBD).Shear(V)diagram.ShearconstantAtoC(positive).ShearconstantCtoB(negative).Vmax=6k1-2shearMoment(M)diagram.FBDofbeamsectionscutthroughE.Momentsareallpositive.MomentincreaseslinearlyfromAtoC1x=0tox=6¿2.MomentdecreaseslinearlyfromCtoB1x=6¿tox=10¿2.352BendingandShearinSimpleBeams2Drawshearandmomentdiagramsforanoverhangbeamloadedasshown.DeterminethecriticalVmaxandMmaxlocationsandmagnitudes.DrawanFBD.Solveforexternalreactions.Basedonintuition,sketchthedeflectedshapeofthebeamtoassistindeterminingthesignsformoment.Loadedbeam.Loaddiagram(FBD).Cutsectionsa,b,andcbetweenloadsandreactions.Solution:TofindVcritical,examinesection(a)leftandrightofconcentratedloadsandsection(b)atthebeginningandendofdistributedloads.FBDatsectioncuta-a.Sectiona-a,x=0tox=10¿3©Fy=04V=10k1-2shearJustrightofA,V=10kAtx=10¿,V=10kFBDatsectioncutb-b.Sectionb-b,x=10¿tox=20¿JustrightofB,3©Fy=04V=10k1+2shear1constant2JustleftofC,atx=20¿,3©Fy=04andV=10kFBDatsectioncutc-c.Sectionc-c,x=20¿tox=30¿JustrightofC,3©Fy=04V=10k1+2shearJustleftofD,atx=30¿,3©Fy=04andV=20k-10k-2k>ft.1x-202¿V=50k-2x353BendingandShearinSimpleBeamsMmaxoccursatplaceswhereV=0orVchangessign.Thisoccurstwice,atBandbetweenCandD.ForMmaxatB:Examineasectioncutjusttotheleftorrightoftheconcentratedload.M=110k2110¿2=100k-ft.1-2momentForMmaxbetweenCandD:Examinetheequationofsectioncutc-c.3©Fy=04-10k+20k-2k>ft.1x-20¿2-V=0-10k+20k-2x+40-V=0‹V=50-2xBut,MmaxoccursatV=0.‹0=50-2x,x=25¿3©Mc=04atx=25¿+10k125¿2-20k115¿2+2k>ft.15¿212.5¿2+M=0Mmax=25k-ft.1+2momentNote:BeamswithoneoverhangenddeveloptwopossibleMmaxvalues.‹Mcritical=100k-ft.atB1-2momentConstructtheresultingshearandmomentdiagrams.Loaddiagram.Sheardiagram.Momentdiagram.354BendingandShearinSimpleBeamsProblemsConstructshearandmomentdiagramsusingtheequilibriummethod.IndicatethemagnitudesofVmaxandMmax.1234355BendingandShearinSimpleBeams4RELATIONSHIPBETWEENLOAD,TRANSVERSESHEAR,ANDBENDINGMOMENTTheconstructionofshearandmomentdiagramsbytheequilibriummethodisquitetime-consuming,particularlywhenalargenumberofsectioncutsmustbeconsidered.Themathematicalrelationshipsbetweenloads,shears,andmomentscanbeusedtosimplifytheconstructionofsuchdiagrams.TheserelationshipscanbeobtainedbyexamininganFBDofanelementallengthofabeam,asshowninFigure9.Figure9(a)Beamwithageneralizedload.Inthisexample,wewillassumeasimplysupportedbeamloadedwithavaryingdistributedload.Detachasmall(elemental)lengthofthebeambetweensections①and②.DrawanFBDofthebeamsegmentwithanelementallengthofthebeamsegment¢x.V=Shearattheleft;①V+¢V=Shearattheright;②¢V=Changeinshearbetweensections①and②Thebeamelementmustbeinequilibrium,andtheequation3©Fy=04gives3©Fy=04+V-ω1¢x2-1V+¢V2=0+V-ω¢x-V-¢V=0¢V=-ω¢x¢V=-ω¢xNote:Thenegativesignrepresentsanegativeslopeforthisparticularloadcondition.Figure9(b)beam.356AnelementalsectionoftheIncalculus,theaboveexpressiontakesthefollowingform:dV=ωdxBendingandShearinSimpleBeamsTheprecedingequationindicatesthatatanysectioninthebeam,theslopeofthesheardiagramisequaltotheintensityoftheloading.Ifweexaminetheshearonthebeambetweenpointsx1andx2(Figure10),weobtainV=ω¢xbut¢V=V2-V1and¢x=x2-x1‹V2-V1=ω1x2-x12WhendV>dx=ωisknownasafunctionofx,theequationcanbeintegratedbetweendefinitelimitsasfollows:冮v2v1dV=冮x2x1ωdx∫V2-V1=ω1x2-x12(Note:Sameequationasabove.)Thatis,thechangeinshearbetweensectionsatx1andx2isequaltotheareaundertheloaddiagrambetweenthetwosections.Figure10and②.Sectionofbeambetweenpoints①Anotherequationofequilibriumaboutpoint0forFigure9(a)canbewrittenas3©M0=04-V1¢x2-M+1M+¢M2+ω1¢x2a¢xb=02-V¢x-M+M+¢M+ω¢x2=02If¢xisasmallvalue,thesquareof¢xbecomesnegligible.‹¢M=V¢xordM=VdxdM=VdxTheprecedingequationindicatesthatatanysectioninthebeam,theslopeofthemomentdiagramisequaltotheshear.Again,examiningthebeambetweenpoints①and②ofFigure10:¢M=M2-Ml;¢x=x2-xl‹M2-Ml=V1x2-xl2or冮M2M1dM=冮x2x1VdxM2-M1=V1x2-x12357BendingandShearinSimpleBeams5SEMIGRAPHICALMETHODFORLOAD,SHEAR,ANDMOMENTDIAGRAMSThetwoexpressionsdevelopedintheprevioussectioncanbeusedtodrawshearandmomentdiagramsandcomputevaluesofshearandmomentatvarioussectionsalongthebeamasneeded.Thismethodisoftenreferredtoasthesemigraphicalmethod.Beforeillustratingthesemigraphicalmethodforshearandmomentdiagrams,itmightbehelpfultoseetherelationshipthatexistsbetweenallofthediagrams(Figure11).Theexampleshownisforasimplysupportedbeamwithauniformloadovertheentirespan.Alsonecessarybeforeattemptingthesemigraphicalmethodisanunderstandingofbasiccurvesandcurverelationships(Figures12and13).EId4ydVd2MLoad:ω===dxdx2dx4whereE=modulusofelasticityI=momentofinertiaShear:V=Moment:M=d3ydM=EI3dxdxd2ydx2d2yEIMdθ=2=EIdxdxSlope:Figure11Relationshipofload,shear,moment,slope,anddeflectiondiagrams.358θ=Deflection:ydydxBendingandShearinSimpleBeamsZero-degreecurvey=cc=constantFirst-degreecurveStraightline—maybeuniformlyincreasingordecreasing.Slope=¢y¢xy=cxc=constantSecond-degreecurveParabolic—increasing(approachingverticality)ordecreasing(approachinghorizontality).y=kx2+cThird-degreecurveGenerallysteeperthanasecond-degreecurve.y=kx3+k¿x2+ÁFigure12Basiccurves.359BendingandShearinSimpleBeamsZero-degreecurveAzero-degreecurvemayrepresentauniformlydistributedloadortheareaofasheardiagram.x=anypointxalongthebeam.First-degreecurveAfirst-degreecurvemayrepresenttriangularloading,theareaunderthesheardiagramforauniformload,ortheareaunderthemomentdiagramforaconcentratedload.Second-degreecurveAsecond-degreecurveusuallyrepresentstheareaofasheardiagramduetoatriangularloaddistribution,oritcouldrepresentthemomentdiagramforauniformloaddistribution.Figure13360Basiccurvesandtheirproperties.BendingandShearinSimpleBeamsLoad,Shear,andMomentDiagrams(SemigraphicalMethod)Generalconsiderationsfordrawingshearandmomentdiagrams:1.Whenallloadsandreactionsareknown,theshearandmomentattheendsofthebeamcanbedeterminedbyinspection.2.Atasimplysupportedorpinnedend,theshearmustequaltheendreaction,andthemomentmustbezero.3.Bothshearandmomentarezeroatafreeendofabeam(cantileverbeamoroverhangbeam).4.Atabuilt-inorfixed-endbeam,thereactionsareequaltotheshearandmomentvalues.5.Load,shear,andmomentdiagramsareusuallydrawninadefinitesequence,withtheloaddiagramontop,thesheardiagramdirectlybeneaththeloaddiagram,andthemomentdiagrambelowthesheardiagram.6.Whenpositivedirectionsarechosenasupwardandtotheright,auniformlydistributedloadactingdownwardwillgiveanegativeslopeinthesheardiagram,andapositivedistributedload(oneactingupward)willresultinapositiveslope.7.Aconcentratedforceproducesanabruptchangeinshear.8.Thechangeinshearbetweenanytwosectionsisgivenbytheareaundertheloaddiagrambetweenthesametwosections:1V2-V12=ωave1x2-x12.9.Thechangeofshearataconcentratedforceisequaltotheconcentratedforce.10.Theslopeatanypointonthemomentdiagramisgivenbytheshearatthecorrespondingpointonthesheardiagram:Apositiveshearrepresentsapositiveslope,andanegativeshearrepresentsanegativeslope.11.Therateofincreaseordecreaseinthemomentdiagramslopeisdeterminedbytheincreasingordecreasingareasinthesheardiagram.12.Thechangeinmomentbetweenanytwosectionsisgivenbytheareaunderthesheardiagrambetweencorrespondingsections:1M2-M12=Vave1x2-x1213.Amomentcoupleappliedtoabeamwillcausethemomenttochangeabruptlybyanamountequaltothemomentofthecouple.361BendingandShearinSimpleBeamsExampleProblems:ShearandMomentDiagrams3BeamABCisloadedwithasingleconcentratedloadasshown.Constructtheshearandmomentdiagrams.Solution:Load,shear,andmomentdiagramscomeinadefiniteorderbecauseoftheirmathematicalrelationships(seeFigure11).DrawanFBDofthebeam,andsolvefortheexternalsupportreactions.ThisFBDistheloaddiagram.Byinspection,theshearatendAis+4k.BetweenAandC,thereisnoloadshownontheloaddiagram.Therefore,ω=0V2-V1=ω1x2-x12‹V2-V1=0ThereisnochangeinshearbetweenAandC(theshearisconstant).AtC,the10kconcentratedloadcausesanabruptchangeinshearfrom+4kto-6k.Thetotalshearchangeequalsthemagnitudeoftheconcentratedload.BetweenCandB,noloadexists;therefore,thereisnoshearchange.Theshearremainsaconstant-6k.AtsupportB,anupward6kforcereturnsthesheartozero.Thereisnoresultantshearattheveryendofthebeam.Themomentatpinandrollersupportsiszero;pinsandrollershavenocapacitytoresistmoment.Thechangeinmomentbetweenanytwopointsonabeamequalstheareaundertheshearcurvebetweenthesametwopoints:M2-M1=V1x2-x12BetweenAandC,theareaundertheshearcurveistheareaofarectangle:area=6¿*4k=24k-ft.362BendingandShearinSimpleBeamsBecausetheshearareaispositive,thechangeinmomentwilloccuralongapositive,increasingcurve.Thechangeinmomentisuniform(linearlyincreasing).Sheardiagram:Momentdiagram0°curve1°curve1+2area1+2slopeFromCtoB,theareaofthesheardiagramisArea=4¿*6k=24k-ft.ThechangeinmomentfromCtoBis24k-ft.Sheararea:Momentdiagram0°curve1°curve1-2area1-2slopeBecausetheshearareaisnegative,theslopeofthemomentcurveisnegative.ThemomentatBshouldgobacktozero,becausenomomentcapabilityexistsattherollersupport.363BendingandShearinSimpleBeams4ConstructshearandmomentdiagramsforthesimplysupportedbeamABC,whichissubjectedtoapartialuniformload.Solution:DrawanFBDofthebeam,andsolvefortheexternalreactions.Thisistheloaddiagram.Byinspection,weseethatatA,thereactionof15kistheshear.Theshearattheendreactionpointisequaltothereactionitself.BetweenAandBisadownward(-)uniformloadof2k/ft.ThechangeinshearbetweenAandBequalstheareaundertheloaddiagrambetweenAandB.Area=2k/ft.(10ft.)=20k.Therefore,shearchangesfrom+15kto-5k.VgoestozeroatsomedistancefromA.2k>ft.1x2-x12;‹15k=2k>ft.1x2;x=7.5¿uu15k:06V2-V1=ωxBetweenBandC,noloadexistsonthebeam,sonochangeinshearoccurs.ShearisconstantbetweenBandC.Themomentatthepinandrolleriszero.ComputetheareaunderthesheardiagrambetweenAandx.area=A12B17.5ft.2115k2=56.25k-ft.ThechangeinmomentbetweenAandxequals56.25k-ft.,andbecausetheshearareais(+),theslopeofmomentcurveis(+).Afirst-degreeshearcurveresultsinaseconddegreemomentcurve.Theshearcurveispositive,butbecausetheareaisdecreasing,thecorrespondingmomentslopeispositivebutdecreasing.5Drawtheshearandmomentdiagramsforthepartialuniformloadonacantileverbeam.Solution:Solvefortheexternalreactions.Theloaddiagramhasauniformload,whichisazerodegreecurve.Becausetheforceisactingdownward,itconstitutesanegativearea,thusproducinganegativeslopeinthesheardiagram.Theresultingshearareaisdrawnasafirst-degreecurvewithanegativeslope.NoloadexistsbetweenBandCintheloaddiagram;therefore,nochangeinshearresultsbetweenBandC.TheareaunderthesheardiagrambetweenAandBequals50k-ft.Becausetheshearareaisnegative,itproducesamomentwithanegativeslope.AstheshearareaincreasesfromAtoB(becomingmorenegative),themomentcurvedevelopsanincreasing(steeper)negativeslope.TheshearareaisuniformbetweenBandC;therefore,itproducesafirst-degreecurveinthemomentdiagram.Theshearareaisstillnegative;therefore,themomentdiagramisdrawnwithanegativefirst-degreeslope.364BendingandShearinSimpleBeams6Drawtheload,shear,andmomentdiagramsfortheillustratedsingleoverhangbeamwithauniformandconcentratedload.(Note:SingleoverhangsdeveloptwopointsofpossibleMmax.)Solution:Solveforthesupportreactions.Then,usingtheloaddiagram,workfromtheleftendtotherightendofthebeam.Thesheardiagramisafirst-degreecurvewithanegativeslopebetweenAandBandcrossesthezeroaxissixfeettotherightofsupportA.Attheconcentrated1,200#load,anabruptchangeinVresults.Thesheardiagramcontinueswithafirst-degreenegativeslopebetweenBandC,andagain,aconcentratedreactionforceatCcausesanabruptchangeofV.FromCtoD,thesheardiagramchangeslinearlyfrom0.8ktozero.Themomentdiagramdevelopstwopeakpoints,atadistancesixfeettotherightofsupportω(whereV=0)andalsoatreactionsupportω(wherethesheardiagramcrossesthezeroaxis).Thefirst-degreecurvesofthesheardiagramgeneratesecond-degreecurvesinthemomentdiagram.MomentsareatboththehingeatAandthefreeoverhangendatD.NotethatthemomentattherollersupportCisnotzero,becausethebeamcontinuesonasanoverhang.7Foracantileverbeamwithanupturnedend,drawtheload,shear,andmomentdiagrams.Solution:Determinethesupportreactions.Then,movethehorizontal4kforceatCtoalignwiththebeamaxisA-B-C.Becausethe4kforceismovedtoanewlineofaction,amomentM=8k-ft.mustbeaddedtopointC.Thesheardiagramisverysimpleinthisexample.Theleftsupportpushesupwithaforceof2kandremainsconstantuntilB,becausenootherloadsarepresentbetweenAandB.AtB,a2kdownward-actingforcebringsVbacktozero,andVremainszeroallthewaytoC(noverticalloadsoccurbetweenBandC).ThemomentdiagramstartswithamomentattheleftendbecauseofthepresenceofthesupportmomentM=18k-ft.ImaginethebeamcurvatureindeterminingwhetherM=18k-ft.isplottedinthepositiveornegativedirection.Becausethecurvatureduetobendingresultsintensiononthetopsurfaceofthebeam,thesignconventionsaysthisisanegativemomentcondition.BetweenAandB,themomentremainsnegative,butwithapositiveslopeofthefirstdegree.ThereisnochangeinmomentbetweenBandC;therefore,themagnituderemains-8k-ft.,whichcorrespondstotheappliedmomentatC.365BendingandShearinSimpleBeams8Drawtheload,shear,andmomentdiagramsforanoverhangbeamwithatriangularanduniformload.Solution:ItisnecessarytodeterminethedistancexinthesheardiagramwhereV=0.FromAtoD,theshearchangesfrom+3.67ktozero.‹V2-V1=+3.67kω1x2-x12=areaundertheloaddiagrambetweenAandD.Ifwestudytheloaddiagram,wefindthatωvariesfromAtoB.Therefore,ωmustbeafunctionofthedistancex.Usingsimilartriangles,6k>ft.ω=x9¿‹ω=2x3TheareaundertheloaddiagrambetweenAtoDequals12x2ω1x2-x12=ab(x)axb=233V2-V1=ω1x2-x12Equating:3.67k=x2;x=3.32ft.3ThechangeinmomentfromAtoDisfoundbyM2-M1=V1x2-x12(areaunderthesheardiagram)22Area=abbh=ab13.32¿213.67k2=8.12k-ft.33366BendingandShearinSimpleBeamsThecalculatedchangeinmomentbetweenDandB(shearareaDBE)cannotbecalculatedastheareaofaspandrel,becausethereisnozeroslopeanywherealongcurveDE.Instead,theareamustbedeterminedbyusingtheconceptofcalculatingtheareaofthespandrelABEandsubtractingthesectionADB.areaABE1spandrel2=A=1319¿2127k2=81k-ft.totalareaofADB=3.67k*19¿2=33k-ft.area1AD2=2313.32¿213.67k2=8.12k-ft.area1DB2=33k-ft.-8.12k-ft.=24.87k-ft.areaDBE=81k-ft.-24.87k-ft.=56.13k-ft.BetweenDandB,themomentchangesbyM2-M1=56.13k-ft.-8.12k-ft.=48k-ft.ThechangeinmomentbetweenBandCisequaltothetriangularsheararea:area1BC2=A12B14¿2124k2=48k-ft.Thepositiveshearareaproducesadecreasingseconddegreecurve.367BendingandShearinSimpleBeams9Thediagramshowsabearingloadonaspreadfooting.Drawtheload,shear,andmomentdiagramsofthefigureshown.Solution:Atypicalspreadfootingsupportingacolumnloaddevelopsuniqueshearandmomentdiagrams.BetweenAandB,theshearchangesfromzeroto5k(whichistheareaundertheupward-actingloadofthesoilbearingfromAtoB).Becausetheloadenvelopeispositive,theslopeofthesheardiagramispositive.ThecolumnloadatBcausesanabruptchangetooccurinthesheardiagram.ApositiveloadenvelopebetweenBandCagaingeneratesafirst-degreepositiveslopetozero.Themomentattheleftendofthefootingiszeroandincreasespositivelytoamagnitudeof5k-ft.atB.AdecreasingnegativeslopeisgeneratedbetweenBandCasthecurvediminishestozeroatC.M2-M1=V1x2-x12Areaoftriangle:112¿215k2=5k-ft.2(sheardiagram)368BendingandShearinSimpleBeams10Acompoundbeamwithinternalhingesisloadedasshown.Solution:Theproblemillustratedhererepresentsacompoundbeam,whichisessentiallyseveralsimplerbeamslinkedtogetherbyhinges.BeamCDrepresentsasimplebeamwithtwopinsupports,andbeamsABCandDEFaresingleoverhangbeams.InsolvingforthesupportreactionsatA,B,E,andF,detacheachbeamsection,anddrawanindividualFBDforeach.Oncesupportreactionshavebeendetermined,reassemblethebeamintoitsoriginalcondition,andbeginconstructionoftheshearandmomentdiagrams.Remember,thehingesatCandDhavenomomentcapability1M=02.Thesheardiagramcrossesthezeroaxisinfiveplaces;therefore,fivepeakpointsdevelopinthemomentdiagram.MomentismostcriticalatthesupportpointsBandE.369BendingandShearinSimpleBeamsProblemsConstructtheload,shear,andmomentdiagramsforthefollowingbeamconditionsusingthesemigraphicalapproach.5678370BendingandShearinSimpleBeams9101112371BendingandShearinSimpleBeams131415372BendingandShearinSimpleBeamsMostcommerciallyavailablestructuralsoftwarewillvisuallyandnumericallydisplayload,shear,andmomentdiagramsforbeamandframesystems.Figure14isanexampleoftheshearandmomentdiagramsgeneratedforarigidframewithfillerbeams.Figure14Shear,moment,anddeflectiondiagramsforaframeandbeams.ImageisreprintedwithpermissionandcourtesyofIntegratedEngineeringSoftware,Inc.,Bozeman,Montana.Free,downloadablesoftwarecapableofgeneratingshear,moment,anddeflectiondiagramscanbefoundontheInternetat:ece313/beams/beams.html**Note:Thisvendorofferseducationaluseofasubstantialstructuralsoftwarepackageifarrangedthroughyourprofessor.373BendingandShearinSimpleBeamsSummary■■■■■■Stressesanddeflectionsinbeamsarefunctionsoftheinternalreactions,forces,andmoments.Therefore,itisconvenientto“map”theseinternalforcesandshow,inadiagrammaticform,theshearsandmomentsthatactthroughoutthelengthofthebeam.Thesediagramsarereferredtoasload,shear(V),andmoment(M)diagrams.Aloaddiagramdepictstheexternalforcesactingonthebeam,includingthesupportreactions.ThisisessentiallyanFBDofthebeam.Thesheardiagramisagraphofthetransverseshearalongthebeamlength.Themomentdiagramisagraphofthebendingmomentsalongthelengthofthebeam.Onemethodofobtainingtheshearandmomentdiagramsistheequilibriummethod.SpecificvaluesofVandMaredeterminedfromequationsofequilibriumthatarevalidforappropriatesectionsofthebeam.Mathematicalrelationshipsthatexistbetweenloads,shears,andmomentscanbeusedtosimplifytheconstructionofshearandmomentdiagrams.Twoequationsareusedingeneratingthesemigraphicalmethodfordrawingtheshearandmomentdiagrams.dV=ω;dx冮V2V1dV=冮x2x1ωdx;V2-V1=ω1x2-x12Thechangeinshearbetweensectionsx1andx2isequaltotheareaundertheloaddiagrambetweenthetwosections.dM=V;dx冮M2M1dM=冮x2x1Vdx;M2-M1=V1x2-x12Thechangeinmomentbetweensectionsatx1andx2isequaltotheareaunderthesheardiagrambetweenthetwosections.AnswerstoSelectedProblems3741Vmax=10k;Mmax=50k-ft.8Vmax=±360lb.;Mmax=+720lb.-ft.2Vmax=–20k;Mmax=–200k-ft.9Vmax=+3k;Mmax=–15k-ft.3Vmax=+15k;Mmax=–50k-ft.10Vmax=–9.2k;Mmax=+28.8k-ft.4Vmax=±20k;Mmax=+100k-ft.11Vmax=–1,080lb.;Mmax=–3,360lb.-ft.5Vmax=+4k;Mmax=±10k-ft.12Vmax=–18k;Mmax=+31.2k-ft.6Vmax=–45k;Mmax=+337.5k-ft.13Vmax=–38k;Mmax=–96k-ft.7Vmax=+10.5k;Mmax=+27.6k-ft.14Vmax=+7.5k;Mmax=+8.33k-ft.BendingandShearStressesinBeamsIntroductionOneoftheearlieststudiesconcernedwiththestrengthanddeflectionofbeamswascarriedoutbyGalileoGalilei.Galileowasthefirsttodiscussthebendingstrengthofabeam.Hethusbecamethefounderofanentirelynewbranchofscience:thetheoryofthestrengthofmaterials,whichhasplayedavitalpartinmodernengineeringscience.Galileostartedwiththeobservationofacantileverbeam(Figure1)subjectedtoaloadatthefreeend.Heequatedthestaticalmomentsoftheexternalloadwiththatoftheresultantofthetensilestressesinthebeam(assumedtobeuniformlydistributedovertheentirecross-sectionofthebeam,asshowninFigure2)inrelationtotheaxisofrotation(assumedtobelocatedattheloweredgeoftheembeddedcross-section).Galileoconcludedthatthebendingstrengthofabeamwasnotdirectlyproportionaltoitswidthbut,instead,wasproportionaltothesquareofitsheight.However,ashebasedhispropositionmerelyonconsiderationsofstaticsanddidnotyetintroducethenotionofelasticity—anideapropoundedbyRobertHookeahalf-centurylater—Galileohaderredintheevaluationofthemagnitudeofthebendingstrengthinrelationtothetensilestrengthofthebeam.Figure1Cantileverbeamloadedatthefreeend.FromGalileoGalilei,DiscorsieDemostrazioniMatematiche,Leyden,1638.DrawingbasedontheillustrationinSchweizerischeBauzeitung,Vol.119.Expressedinmodernterms,themomentofresistance(inertia)oftherectangularbeamwould,accordingtoGalileo,amounttobh34whichisthreetimesasgreatasthecorrectvalue:3bh12Figure2FlexureaccordingtoGalileo.RedrawnfromanillustrationinSchweizerischeBauzeitung,Vol.116.About50yearsafterGalileo’sobservations,aFrenchphysicist,EdmeMariotte(1620–1684),stillretainingtheconceptofthefulcrumatthecompressionsurfaceoftheFromChapter8ofStaticsandStrengthofMaterialsforArchitectureandBuildingConstruction,FourthEdition,BarryOnouye,KevinKane.Copyright©2012byPearsonEducation,Inc.PublishedbyPearsonPrenticeHall.Allrightsreserved.375BendingandShearStressesinBeamsbeambutreasoningthattheextensionsofthelongitudinalelementsofthebeam(sometimescalledfibers)wouldbeproportionaltothedistancefromthefulcrum,suggestedthatthetensilestressdistributionwasasshowninFigure3.Figure3FlexureaccordingtoMariotte.Mariottelaterrejectedtheconceptofthefulcrumandobservedthatpartofthebeamonthecompressionsidewassubjectedtocompressivestresshavingatriangulardistribution.However,hisexpressionfortheultimateloadwasstillbasedonhisoriginalconcept.TwocenturiesafterGalileoGalilei’sinitialbeamtheory,Charles-AugustindeCoulomb(1736–1806)andLouisMarie-HenriNavier(1785–1836)finallysucceededinfindingthecorrectanswer.In1773,CoulombpublishedapaperthatdiscardedthefulcrumconceptandproposedthetriangulardistributionshowninFigure4,inwhichboththetensileandcompressivestresseshavethesamelineardistribution.Figure4FlexureaccordingtoCoulomb.1FLEXURALSTRAINTheaccuracyofCoulomb’stheorycanbedemonstratedbyexaminingasimplysupportedbeamsubjectedtobending.Thebeamisassumed(a)tobeinitiallystraightandofconstantcross-section,(b)tobeelasticandhaveequalmoduliofelasticityintensionandcompression,and(c)tobehomogeneous—thatis,ofthesamematerialthroughout.Itisfurtherassumedthataplanesectionbeforebendingremainsaplaneafterbending[Figure5(a)and5(b)].Figure5(a)Beamelevationbeforeloading.Beamcross-section.Forthistobestrictlytrue,thebeamneedstobebentonlywithcouples(noshearontransverseplanes).Thebeammustbeproportionedsothatitwillnotbuckle,andtheloadsmustbeappliedsothatnotwisting(torsion)occurs.Figure5(b)376Beambendingunderload.Examiningaportionofthebentbeambetweensectionsa-aandb-b(Figure6),oneobservesthatatsomedistanceCabovethebottomofthebeam,thelongitudinalelements(fibers)undergonochangeinlength.BendingandShearStressesinBeamsThecurvedsurface(①-②)formedbytheseelementsisreferredtoastheneutralsurface,andtheintersectionofthissurfacewithanycross-sectioniscalledtheneutralaxisofthecross-section.Theneutralaxiscorrespondstothecentroidalaxisofacross-section.Allelements(fibers)ononesideoftheneutralsurfacearecompressed,andthoseontheoppositesideareintension.ForthesimplebeamshowninFigure5,theportionofthebeamabovetheneutralsurfaceexperiencescompression,whereasthelowerportionisundergoingtensilestressing.Theassumptionismadethatalllongitudinalelements(fibers)havethesameinitiallengthbeforeloading.ReferringagaintoFigure6,δyδc=ycorδy=Figure6Beamsectionafterloading.yδccwhereδy=deformationdevelopedalongfiberslocatedadistanceybelowneutralsurfaceδc=deformationatthebottomsurfaceofthebeamlocatedadistancecbelowtheneutralsurfaceBecauseallelementshadthesameinitiallength¢x,thestrainofanyelementcanbedeterminedbydividingthedeformationbythelengthoftheelement;therefore,thestrainbecomesεy=yεccwhichindicatesthatthestrainofanyfiberisdirectlyproportionaltothedistanceofthefiberfromtheneutralsurface.Withthepremisethatthelongitudinalstrainsareproportionaltothedistancefromtheneutralsurfaceaccepted,theassumptionisnowmadethatHooke’slawapplies,whichrestrictsstressestomagnitudeswithintheproportionallimitofthematerial.Then,theequationbecomesεyfyfc==yEyyEcc377BendingandShearStressesinBeamsThefinalresult,ifEc=Ey(constant),isfyfc=ycwhichverifiesCoulomb’sconclusion.RedrawingthediagramshowninFigure6toincludethecompressiveaswellasthetensiledeformationsduetobendingstress,wecanuseHooke’slawtoexplainthestressvariationsoccurringonthecross-section(Figure7).Figure7Deformedsectiononbeamduetobending.Thedeformationattheneutralaxisiszeroafterbending;therefore,thestressattheneutralaxisiszero.Atthetopfiber,themaximumshortening(compressivedeformation)occursfromthedevelopmentofmaximumcompressivestresses.Conversely,themaximumtensilestressoccursatthebottomfibers,resultinginamaximumelongationdeformation(Figure8).2FLEXURAL(BENDING)STRESSEQUATIONFigure8Bendingstressesonsectionb-b.Consideraportionofabeamthatissubjectedtopurebendingonlybycouples(designatedbyM)ateachend,asshowninFigure9.Becausethebeamisinequilibrium,themomentsateachendwillbenumericallyequalbutofoppositesense.Duetothemomentcouples,thebeamisbentfromitsoriginalstraightpositiontothecurved(deformed)shapeindicatedinFigure9.Elevationofbeaminbending.Figure9Beamcurvatureduetobendingmoment.378Beamcrosssection.BendingandShearStressesinBeamsBecauseofthisbendingaction,wefindthatthelengthsoftheupperpartsofthebeamdecrease,whilethebottompartsofthebeamlengthen.Thisactionhastheeffectofplacingtheupperportionofthebeamincompressionandthelowerportionofthebeamintension.Anequationmustbeobtainedthatwillrelatebendingstresstotheexternalmomentandthegeometricpropertiesofthebeam.ThiscanbedonebyexaminingasegmentofthebeamwhoseinternalforcesystematanygiventransversesectionisamomentM,asshowninFigure10,whereFigure10Bendingstressesonabeamcross-section.cc=distancefromneutralaxis(N.A.)totheextremecompressivefiberct=distancefromN.A.totheextremetensilefibery=distancefromN.A.tosomearea¢A¢A=smallstripofareaonthebeamcross-sectionIfwedenotetheelementofareaatanydistanceyfromtheneutralaxis(seeFigure10)as¢Aandthestressonitasf,therequirementforequilibriumofforcesyields3©Fx=04©ft¢A+©fc¢A=0whereft=tensilestressbelowtheN.A.fc=compressivestressabovetheN.A.Therefore,©ft¢A=Ftand©fc¢A=Fc379BendingandShearStressesinBeamsIfeachfy¢Aismultipliedbyitsydistanceaboveorbelowtheneutralaxis,M=©fyy¢AwhereM=internalbendingmoment(Figure11)Figure11Bendingstressatanydistancefromtheneutralaxis.ButrememberingtherelationshipdevelopedbyCoulomb,fyfc=ycandfy=yfccSubstitutingthefyrelationshipintothemomentequation:yfcM=©fyy¢A=©fcy¢A=©y2¢AccWedevelopedtherelationshipformomentofinertia,whereI=©y2¢A1momentofinertia2‹M=(a)Rectangularcrosssection.c1=c2;ftop=fbottomfcIcorfb=Mc;flexureformulaIwhere(b)Unsymmetricalcrosssection.c1>c2;ftop>fbottomFigure12Distancestotheextremefiberforbeamcross-sections.380fb=bendingstressattheextremefiber,toporbottomc=distancefromtheN.A.totheextremefiberI=momentofinertiaofthecross-sectionaboutitscentroidalaxis(orN.A.)M=momentatsomepointalongthebeamlengthNote:Bendingstressfisdirectlyproportionaltothevaluec;therefore,thelargestbendingstressonacross-sectionisobtainedbyselectingthelargestcvalueforunsymmetricalcross-sections(Figure12).BendingandShearStressesinBeamsExampleProblems:BendingStress1A4*12S4SDouglasfirbeamisloadedandsupportedasshown.a.Calculatethemaximumbendingstressdevelopedinthebeam.b.DeterminethemagnitudeofthebendingstressdevelopedthreefeettotheleftofsupportB.Solution:Themaximumbendingstressdevelopedinthebeamoccurswherethebendingmomentislargest.Todeterminethemaximummoment,plottheshear(V)andmoment(M)diagrams.VatthreefeettotheleftofB:V=w11.6¿2V=1.6k4*12S4SA=39.4in.2(a)Mmax=8.58k-ft.fb=(b)18.58k-ft.*12in.>ft.215.63–2Mc==1.4ksiIx415.3in.4MomentatthreefeettotheleftofB:M=8.58k-ft.-1211.6¿211.6k2M=8.58k-ft.-1.28k-ft.M=7.3k-ft.f=17,300#-ft.*12in.>ft.215.63–2415.3in.4=1,188psi381BendingandShearStressesinBeams2Abeammustspanadistanceof12feetandcarryauniformlydistributedloadof120#/ft.Determinewhichcross-sectionwouldbetheleaststressed:(a),(b),or(c).Solution:(a)Plank.A=20in.2;Ix=6.7in.4;c=1–fmax=12,160#-ft.*12in.>ft.211–2Mc=3,869#>in.2=I6.7in.4(b)Rectangularbeam.A=20in.2;Ix=41.7in.4;c=2.5–fmax=Mc12,160#-ft.*12in.>ft.212.5–2=1,554#>in.2I141.7in.42(c)I-beam.A=20in.2;Ix=154.7in.4;c=4–fmax=38212,160#-ft.*12in.>ft.214–2Mc==670#>in.24I154.7in.BendingandShearStressesinBeams3AW8*28steelbeamisloadedandsupportedasshown.Determinethemaximumbendingstress.Solution:V=w*8¿=400#>ft.*8¿=3,200#M=3.2k110¿2=32k-ft.fmax=Mc1usingMmaxfromthemomentdiagram)II=98.0in.4c=d8.06–==4.03–22M=32k-ft.*12in.>ft.=384k-in.f=1384k-in.214.03–298.0in.4Fallow=22ksi=15.8ksi‹OKWhatisthebendingstressthatwouldresultifthesteelbeamwerereplacedbya6*16S4SSouthernPineNo.1beam?(ObtainsectionpropertiesfromtimbertablesinAppendixTableA1.)Solution:f=McII=1,707in.4c=f=15.5–d==7.75–221384,000#-in.217.75–21,707in.4=1,743psiFallow=1,550psi‹NotGood383BendingandShearStressesinBeams4Determinethemaximumtensileandcomprbendingstressesinthebeamshown.essiveSolution:ComponentAyAy6in.24–24in.34in.20.5–2in.3©A=10in.2;©Ay=26in.3©Ayy==©AComponent26in.310in.2=2.6–Ixc(in.4)dy(in.)Ay2(in.4)181.411.762.117.640.33©Ixc=18.33in.4;©Ady2=29.4in.4Ix=18.33in.4+29.4in.4=47.73in.4fb=(top)111,200#-ft.*12in.>ft.214.4–2Mct=I47.73in.4=12,390#>in.2f=b(bottom)111,200#-ft.*12in.>ft.212.6–2Mcc=I47.73in.4=7,321#>in.2384BendingandShearStressesinBeamsSectionModulusThemajorityofthestructuralshapesusedinpractice(structuralsteel,timber,aluminum,etc.)arestandardshapesthatarenormallyavailableinindustry.Crosssectionalpropertiessuchasarea(A),momentofinertia(I),anddimensionalsize(depthandwidth)forstandardshapesareusuallylistedinhandbooksandtables.Asameansofexpandingthebasicflexureequationintoadesignform,thetwosectionpropertiesIandcarecombinedasI>c,whichiscalledthesectionmodulus.fb=McM=II>cSectionmodulus:S=I>cTherefore,fb=McM=ISwhereS=sectionmodulus,usuallyaboutthexaxis1in.32M=bendingmomentinthebeam,usuallyMmaxBecauseIandcofstandardsectionsareknown,theirsectionmoduli(S)arealsolistedinhandbooks.Fornonstandardsectionsandforregulargeometricshapes,thesectionmodulusmaybeobtainedbycalculatingthemomentofinertiaIoftheareaandthendividingIbyc,thedistancefromtheneutralaxistotheextreme
fiber.Insymmetricalsections,chasonlyonevalue,butinunsymmetricalsections,cwillhavetwovalues,asshowninFigure12(b).Intheanalysisanddesignofbeams,however,weareusuallyinterestedinthemaximumstressthatoccursintheextremefiber.Inallsuchproblems,thegreatestvalueofcmustbeused.Ifwerewritethebasicflexureequationintoadesignform,Srequired=MFbwhereFb=allowablebendingstress(ksiorpsi)M=maximumbendingmomentinthebeam(k-in.or#-in.)theusefulnessofthesectionmodulusbecomesquiteapparent,becauseonlyoneunknownexistsratherthantwo(Iandc).385BendingandShearStressesinBeamsExampleProblems:SectionModulus5TwoC10*15.3steelchannelsareplacedbacktobacktoforma10-inch-deepbeam.DeterminethepermissiblePifFb=22ksi.AssumeA36gradesteel.Solution:Ix=67.4in.4*2=134.8in.4Mmax=12152152+1P>22152Mmax=12.5+2.5P=112.5k-ft.+2.5P2*12in.>ft.f=MMc=ISM=FbSS=2*13.5in.3=27in.3EquatingbothMmaxequations,M=22ksi*27in.3=594k-in.112.5k-ft.+2.5P2112in.>ft.2=594k-in.Dividingbothsidesoftheequationby12in./ft.,12.5k-ft.+2.5¿1P2=49.5k-ft.2.5P=37kP=14.8k6Atimberfloorsystemutilizing2*10S4Sjoistsspansalengthof14feet(simplysupported).Thefloorcarriesaloadof50psf(deadloadplusliveloadDL+LL).Atwhatspacingshouldthejoistsbeplaced?AssumeDouglasfir-LarchNo.21Fb=1,450psi2.Solution:Basedontheallowablestresscriteria,f=MMc=ISMmax=S*fb=121.4in.3211.45k>in.32=31k-in.M=38631k-in=2.58k-ft.12in.>ft.BendingandShearStressesinBeamsBasedonthebendingmomentdiagram,Mmax=ωL28Therefore,ω=8ML2SubstitutingforMobtainedpreviously,ω=but312.58kft2114¿22=0.105k>ft.=105#>ft.ω=#>ft.2*tributarywidth1joistspacings2105#>ft.ωs===2.1¿50psf50#>ft.2s=25–spacingUse24–oncenter(o.c.)spacing.Note:Spacingismorepracticalforplywoodsubflooring,basedonafourfootmoduleofthesheet.7Designtheroofandsecond-floorbeamsifFb=1,550psi(SouthernPineNo.1).Solution:Loadconditions:Roof:SL+DL(roof)=200#/ft.Walls:400#concentratedloadonbeamsatsecondfloorRailing:100#concentratedloadonbeamoverhangSecondFloor:DL+LL=300#/ft.(alsoondeck)387BendingandShearStressesinBeamsRoofbeamdesign:Mmax=3,600#-ft.Srequired=13.6k-ft.2112in.>ft.21.55k>in.2FromtimbertableintheAS=30.7in.3B=27.9in.3Appendix:Use4*8S4S.Second-floorbeamdesign:Mmax=9,112#-ft.MSrequired=Fb19.112k-ft.2112in.>ft.2Srequired==70.5in.321.55k>in.FromtimbertableintheAS=73.8in.3B388Appendix:Use4*12S4S.BendingandShearStressesinBeamsProblems1Acantileverbeamhasaspanofninefeetwithaconcentratedloadof2,000#atitsunsupportedend.IfaW8*18isused1Fb=22ksi2,isitsafe?2Thesingleoverhangbeamusesa4*12S4SDouglasfir–LarchNo.1member.Determinethemaximumbendingstrengthdeveloped.Isitsafelydesigned?1Fb=1,300psi23A16-foot-longsingleoverhangbeamisloadedasshown.AssumingaW8*35,determinethemaximumbendingstressdeveloped.(A992steel,Fb=30ksi)4Abeamasshownsupportsafloorandpartitionwherethefloorloadisassumedtobeuniformlydistributed(500#/ft.)andthepartitioncontributesa1,000#concentratedload.SelectthelightestW8steelsectionifFb=22ksi.389BendingandShearStressesinBeams5AW8*18floorbeamsupportsaconcreteslabandamachineweighing2,400#.DrawshearVandmomentMdiagrams,anddeterminetheadequacyofthebeambasedonbendingstress.(FbforA992steelis30ksi.)6Alintelbeamoveradoorwayopening10feetwidesupportsatriangularloadasshown.AssumingthelintelbeamtobeaW8*15(A36steel),determinethebendingstressdeveloped.Whatsizetimberbeam,eightinchnominalwidth,couldbeusedifFb=1,600psi?7Glu-laminatedbeamsareusedtosupporttheroofandpulleyloadatawarehouse.Beamsspan24feetplusaneightfootoverhangovertheloadingarea.Determinethebendingstressadequacyofthebeam.PropertiesofGlu-lam:bhSIFb390=====6.75–12–162in.3974in.42,400psiBendingandShearStressesinBeams8AW8*28steelbeamisloadedandsupportedasshown.Determinethemaximumbendingstressdevelopedatthewall.Whatisthebendingstressatapointfourfeettotherightofthefreeendofthebeam?(ConstructtheshearVandmomentMdiagrams.)9Selectthelightest14-inch-nominal-depthWbeamtocarrytheloadshown.AssumeA992steel.1Fb=30ksi.210AW18*40(A36)beamisusedtosupportthreeconcentratedloadsofmagnitudeP.DeterminethemaximumpermissibleP.DrawshearVandmomentMdiagramsasanaid.391BendingandShearStressesinBeams3SHEARINGSTRESS—LONGITUDINALANDTRANSVERSEFigure13Transverseshearofabeam(seeSection7.2).(a)Beamwithnoload.Inadditiontotheinternalbendingmomentpresentinbeams,asecondimportantfactortobeconsideredwhendeterminingthestrengthofbeamsisshear.AninternalshearforceVgenerallyispresentand,insomecases,maygovernthedesignofbeams.Manymaterials(e.g.,wood)areprimarilyweakinshear;forthisreason,theloadthatcanbesupportedmaydependontheabilityofthematerial(beam)toresistshearingforces.Becausebeamsarenormallyhorizontalandthecrosssectionsuponwhichbendingstressesareinvestigatedarevertical,theseshearingstressesinbeamsaregenerallyreferredtoasvertical(transverse)andhorizontal(longitudinal).Transverseshearaction(Figure13)isapureshearingconditionandoccursevenwherethereisnobendingofthebeam.However,beamsdobend,andwhentheydo,fibersononesideoftheneutralaxisareplacedincompressionandthoseontheothersideareplacedintension.Ineffect,thefibersoneithersideoftheneutralsurfacetendtoslipindirectionsoppositetooneanother.Theexistenceofhorizontal(longitudinal)shearingstressesinabentbeamcanreadilybevisualizedbybendingadeckofcards.Theslidingofonesurfaceoveranother,whichisplainlyvisible,isashearingactionthat,ifprevented,willsetuphorizontalshearingstressesonthosesurfaces(Figure14).(b)Longitudinalshear.Figure14beams.LongitudinalshearstressesinIfoneconstructsabeambystackingone4–*4–memberontopofanotherwithoutfasteningthemtogetherandthenloadsthisbeaminadirectionnormaltothebeamlength,theresultingdeformationwillappearsomewhatlikethatshowninFigure15(a).Thefactthatasolidbeamdoesnotexhibitthisrelativemovementoflongitudinalelements,asshowninFigure15(b),indicatesthepresenceofshearingstressesonlongitudinalplanes.TheevaluationoftheseshearingstresseswillnowbestudiedbymeansofFBDsandtheequilibriumapproach.(a)Two4–*4–sunfastened(largedeflection).(b)Solidsection(smallerdeflectionunderload).Figure15392Theeffectofshearingstresses.BendingandShearStressesinBeamsRelationshipBetweenTransverseandLongitudinalShearingStressWepreviouslydevelopedamethodofplottingsheardiagramsbasedonbeamsexperiencingtransverseshearingaction.Thissectionwillnowshowthatatanypointinadeflectedbeam,theverticalandhorizontalshearingstressesareequal.Therefore,theshearVdiagramisarepresentationofbothtransverseandlongitudinalshearalongthebeam.ConsiderasimplysupportedbeamasshowninFigure16(a).Whenasectiona-aispassedthroughthebeam,ashearforceV,representingthesumtotalofallunittransverseshearingstressesonthecutsection,developsasshowninFigure16(b).Ifwenowisolateasmall,squareelementofthisbeam,thefollowingrelationshipdevelops:(a)Simplysupportedbeam.(b)Transverseshearforce(V).V=©fvAwherefv=unitshearingstressA=cross-sectionalareaofbeamRemovingthesmallelementalsquarefromthebeam,wedrawanFBDshowingtheforcesactingonit(Figure17).shearstressalongsectioncuta-aAssumethat(c)Transverseshearingstress.Figure16Transverseshearstressatsectiona-a.¢y=¢xandthattheelementalsquareisverysmall.fv=transverseshearstressForequilibriumvertically,(a)Elementalsquare.3©Fy=04fv1=fv21formsamomentcouple2Toplacetheelementalsquareinrotationalequilibrium,summomentsaboutpointO.But3©Mo=04fv11¢x2=fv31¢y2¢x=¢yTherefore,(b)Transverseshearstresses.fv1=fv2=fv3=fv4Shearsfv3andfv4formacounterclockwisecouple.Fromtheprecedingexample,wecanconcludethatftransverse=flongitudinalatagivenpointalongthebeamlength.(c)Longitudinalshearstresses.Figure17Shearstressesonaunitstressblock.393BendingandShearStressesinBeams4DEVELOPMENTOFTHEGENERALSHEARSTRESSEQUATIONToarriveatarelationshipfortheshearingstress,considerthebeamshowninFigure18.Atsectiona-athemomentisMa,andatsectionb-b,anincrementaldistancetotheright,themomentisMb.FromthemomentdiagraminFigure18,weseethatMb7Ma(a)Beamsectionbetweensectionsaandb.Figure18Shear(V)andmoment(M)diagramofabeamunderload.(b)Bendingstressesonthebeamsectiona-b.Figure19Bendingstressonabeamsection.Therefore,Fca6FcbandFta6FtbIsolatingasmallsectionofincrementalbeam(betweensectionsa-aandb-bshowninFigures19aand19b)abovetheneutralsurface,Figure20showsthedistributionoftensileandcompressivebendingstresses.Intheelementcdef,theforcesC1andC2aretheresultantsofthecompressivestressesthatactonthetransverseplanescdandef.ShearforceVonplanedfisrequiredforhorizontalequilibrium.C27C1Figure20FBDoftheupperportionofthebeambetweensectionsa-b.394©Fx=0;C1+V-C2=0‹V=C2-C1=1fv21b21¢x2BendingandShearStressesinBeamsExaminethecross-sectionofthisisolatedbeamsegment(Figure21).¢A=smallincrementofareay=distancefromtheN.A.totheareaAFromtheflexureformula,fy=MyITheforceagainstthearea¢AequalsFigure21Upperportionofthebeamcross-section.My¢A¢Afy=IButifwesumallofthe¢AvaluesintheshadedcrosssectionshowninFigure21,cAreaoftheshadedd=©¢AcrosssectionTotalforceontheshadedCcross-sectionalareaatS=C2sectionb-bC2=©Mby¢AI=Mb©¢AyIwhereMb=internalbendingmomentatsectionb-b,obtainedfromthemomentdiagram(Figure18)I=momentofinertiaoftheentirebeamcross-section,aconstantFigure22Beamsegmentbetweena-b.©¢Ay=Sumofallthe¢AvaluesthatcomposetheshadedareatimestherespectiveydistancefromtheN.A.©¢Ay=¢Aywherey=distancefromtheN.A.tothecentroidoftheshadedcross-sectionAyisnormallyreferredtoasthestaticalorfirstmoment.ThesymbolQwillbeusedtorepresentthevalueAy:Q=Ay395BendingandShearStressesinBeamsNext,substituting:C2=MbQMaQ;andC1=IIButV=C2-C1Therefore,V=MaQMbQQ=1Mb-Ma2IIIForbeamsofconstantcross-section:Q=constant,I=constantLookingagainatthesectionofthebeambetweensectiona-aandb-bshowninFigure22,whereMb7Ma,theconditionofverticalandmomentequilibriummustbeestablished:3©Fy=04Va=Vb=VT1transverse23©Mo=04+Mb-Ma-VT¢x=0Mb-Ma=VT¢xSubstituting(Mb-Ma)intotheearlierequation,Vlongitudinal=Q1V¢x2ITwhereVL=shearforceactingonthelongitudinalbeamsurface1area=b¢x2fv=shearstress(longitudinal)=fv=fv=396Vb¢xQ1VT¢x2Ib¢xVshearareaBendingandShearStressesinBeamsSimplifying,theresultingequationrepresentsthegeneralshearformula:fv=VQIbwherefv=unitshearingstress;transverseorlongitudinalV=shearinthebeamatagivenpointalongthebeamlength,usuallyobtainedfromthesheardiagramQ=Ay=firstmomentA=areaaboveorbelowthelevelatwhichtheshearstressisdesiredy=distancefromthebeamcross-section’sN.A.tothecentroidoftheareaaboveorbelowthedesiredplanewhereshearstressisbeingexaminedIx=momentofinertiaoftheentirebeamcross-sectionb=widthofthebeamattheplanewheretheshearstressisbeingexamined397BendingandShearStressesinBeamsExampleProblems:ShearStress8Calculatethemaximumbendingandshearstressforthebeamshown.Solution:IxcAdyAdy24in.412in.22–48in.436in.412in.22–48in.4Component©Ady2=96in.4©Ixc=40in.4(a)Sectionabovetheneutralaxis.Ix=136in.4ComponentAyAy12in.22–24in.32in.21>2–1in.3©A=14in.2(b)Sectionbelowtheneutralaxis.y=25in.314in.2fbmax=398©Ay=Q=25in.3=1.79–15,000#-ft.*12in.>ft.215–2Mc=2,200#>in.2=4I136in.BendingandShearStressesinBeamsFrom(a):Vmax=1,000#1fromsheardiagram2Q=©Ay=Ay=14in.211.79–2=25in.3Ix=136in.41fortheentirecross-section2b=2–‹fv=1,000#A25in.3BVQ=92psi=Ib136in.4(2in.)From(b):V=1,000#Q=Ay=25in.3Ix=136in.4b=2–‹fv=1,000#A25in.3B136in.412in.2=92psiNote:Picktheeasierhalfofthecross-sectionincalculatingQ=Ay.(c)Shearplanebetweenflangeandweb.Whatshearstressdevelopsatthebaseoftheflange?(Thiscalculationwouldgiveanindicationastowhatkindofshearstressmustberesistedifglue,nails,oranyotherfasteningdeviceisusedtojointheflangetothestem.)V=1,000#I=136in.4b=2–or6–1butasmallerbgivesalargerfv‹Useb=2–2Q=Ay=112in.2212–2=24in.3‹fv=1,000#A24in.3B136in.412in.2=88.3psi(d)Shearplaneabovetheshearplane.399BendingandShearStressesinBeams9Determinethemaximumshearstressdevelopedonthebeamcross-sectionshownbelow.Solution:IxcComponent216214231221122312Beamcross-section.=64in.4=288in.4Ix=©Ixc=352in.4Twolocationswillbeexaminedtodeterminethemaximumshearstress.Oneshearplanewillbethroughtheneutralaxis(normallythecriticallocation),andtheotherwillbewherethesectionnecksdowntotwoinches.ComponentAyAy8in.24–32in.328in.21–28in.3©Ay=60in.3Q=Ay=©Ay=60in.36,000#A60in.3BVQ=73.1psifv==Ib352in.4114in.2(atN.A.)400BendingandShearStressesinBeamsComponentAyAy8in.24–32in.3©Ay=32in.3Q=©Ay=32in.36,000#A32in.3BVQfv==272.7psi=Ib352in.412in.2‹fvmax=272.7psiThisiswherelongitudinalshearfailurewouldprobablyoccur.10Forthebeamcross-sectionshown,determinethelongitudinalshearstressthatdevelopsattheneutralaxisandatoneinchincrementsabovetheneutralaxis.UseVmaxforyourcalculations.Solution:Forthegeneralshearstressequation,adeterminationmustbemadefortheVmaxandthecross-sectionalpropertiesI,b,andQ.TheVmaxvalueismostconvenientlyobtaineddirectlyfromthesheardiagram.BecausethemomentofinertiaIisconstantforagivencross-section,itmaybecalculatedas14–218–23bh3==171in.4Ix=1212Thewidthofthecross-section(shearplane)isalsoconstant;therefore,b=4–.Beamcross-section.401BendingandShearStressesinBeamsThevaluesofQ=Ayforeachofthefourshearplanes,includingtheneutralaxisatwhichthehorizontal(longitudinal)shearisdesired,areshowninFigure23(a)through23(d)andaretabulatedasfollows:Ix=171in.4,b=4–Q=Ay=116in.2212–2=32in.315,800#2A32in.3BVQfv==271psi=Ib171in.414in.2Figure23(a)axis(N.A.)Shearstressesattheneutral(N.A.)Ix=171in.4,b=4–Q=Ay=14–213–212.5–2=30in.315,800#2A30in.3BVQ=254psifv==Ib171in.414in.211–aboveN.A.2Figure23(b)theN.A.Shearstresses1–aboveIx=171in.4,b=4–Q=Ay=18in.2213–2=24in.315,800#2A24in.3BVQ=204psi=fv=Ib171in.414in.212–aboveN.A.2Ix=171in.4,b=4–Q=Ay=14in.2213.5–2=14in.315,800#2A14in.3BVQ=119psi=fv=Ib171in.414in.2402Figure23(c)Shearstresses2–abovetheN.AFigure23(d)theN.A.Shearstresses3–above13–aboveN.A.2Plottingtheshearstressvaluesonagraphadjacenttothebeamcross-section,weobtainaparaboliccurveasshowninFigure24.Hadthevaluesofshearingstressbeenobtainedforthecorrespondingpointsbelowtheneutralaxis,weshouldhavefoundcorrespondingmagnitudes.Bycompletingthecurve,itwillbenotedthatthemaximumvalueofhorizontalshearingstressoccursattheneutralplane(surface),whereAyisamaximumandthebendingstressesareequaltozero.BendingandShearStressesinBeamsBeamcross-section.Figure24Graphofstressintensitiesatvariouslocationsonthebeamcross-section.Shearstressdistributiononarectangularcross-section.Becauseofitsfrequentuseindesign,anexpressionforthemaximumhorizontalshearingstressoccurringinsolidrectangularbeams(primarilytimberbeams)maybederived(Figure25).Shearplaneismaximumattheneutralaxis,asfoundinFigure24.fv=VQ;IbIx=b=b;y=bh3;12A=b*hbh=22h4Q=AyFigure25Beamcross-section.Therefore,fv=Vaabhh*b243bhb1b212=12Vbh2238bh=3V2bhHowever,bh=areaoftheentirebeamcross-section.Simplifying,fvmax=(N.A.)3V1.5V=2AAforsolidrectangularcross-sectionswherefvmax=maximumshearingstressattheN.A.(N.A.)V=maximumshearontheloadedbeamA=cross-sectionalareaofthebeam403BendingandShearStressesinBeamsFromtheequationjustdeveloped,wefindthatthemaximum(design)shearstressforarectangularbeamis50%largerthantheaverageshearvalue(Figure26).CrossSection.Figure26Shearstressgraphofarectangularcrosssection.Shearstressdistribution—keypoints.11Abuilt-upplywoodboxbeamwith2*4S4Stopandbottomflangesisheldtogetherbynails.Determinethepitch(spacing)ofthenailsifthebeamsupportsauniformloadof200#/ft.alongthe26footspan.Assumethenailshaveashearcapacityof80#each.Built-upplywoodboxbeamcrosssection.Solution:Constructthesheardiagramtoobtainthecriticalshearconditionanditslocation.Notethattheconditionofsheariscriticalatthesupportsandthattheshearintensitydecreasesasyouapproachthecenterlineofthebeam.ThiswouldindicatethatthenailspacingPvariesfromthesupporttomidspan.Nailsarecloselyspacedatthesupport,butincreasingspacingoccurstowardmidspan,followingthesheardiagram.fv=Ix=404VQIb14.5–2118–2312-13.5–2115–2312=1,202.6in.4BendingandShearStressesinBeamsQ=Ay=A5.25in.2B18.25–2=43.3in.3Shearforce=fv*AvwhereAv=shearareaAssumethefollowing:F=capacityoftwonails(oneeachside)attheflange,representingtwoshearsurfacesF=fv*b*p=‹F=p*VQ*bpIbFIxVQp=IVQAtthemaximumshearlocation(support)whereV=2,600#,12nails*80#>nail2A1,202.6in.4Bp==1.71–12,600#2A43.3in.3BCheckingthespacingrequirementpatdifferentlocationsalongthebeam,weobtainagraphicalplot(likethesheardiagram)ofthespacingrequirements.Atthesupport,V0=2,600#,p0=1.36–V2=2,200#;p2=V4=1,800#;p4=V6=1,400#;p6=V8=1,000#;p8=V10=600#;p10=12*80#>nail2A1,202.6in.4B12,200#2A43.3in.3B=2.02–11,800#2A43.3in.3B=2.47–11,400#2A43.3in.3B=3.17–11,000#2A43.3in.3B=4.44–1600#2A43.3in.3B=7.04–12*80#>nail2A1,202.6in.4B12*80#>nail2A1,202.6in.4B12*80#>nail2A1,202.6in.4B12*80#>nail2A1,202.6in.4BInpracticalnailspacing,incrementsofhalfaninchoroneinchshouldbeused.405BendingandShearStressesinBeamsShearingStressVariationsinBeamsBeamsmustbedesignedtosafelywithstandthemaximumstressesduetobendingandshear.Thevariationoftensileandcompressivebendingstressesoveracross-sectionalareawasdiscussedinSection2.Asinbendingstress,shearstressalsovariesonacross-section,asillustratedforarectangularcross-sectioninFigure26.Withfewexceptions,themaximumshearingstressgenerallyoccursattheneutralaxis.(a)Rectangularbeam.Shearingstressvariationoverthecross-sectionofaT-beam,I-beam,andwide-flangesectionisillustratedinFigure27(b),Figure27(c),andFigure28.ThedashedcurveinFigure27(c)indicateswhatthestressvariationwouldbeifthebeamareahadremainedrectangularwithaconstantwidthb.ThisvariationwouldbesimilartothatshowninFigure27(a).Thesuddenincreaseinshearstressattheundersideofthetopflangecomesfromthechangeofthewidthfrombtotin(b)T-beam.fv=VQlbAsimilarchangeoccursattheflange-to-webtransitionofaT-beam,asshowninFigure27(b),buthere,thecurvebelowtheneutralaxisfollowstheusualpatternforarectangularbeam.(c)I-beam.Figure27Variationsofshearingstress.Uponexaminationoftheshearstressdistributionforawide-flangesection,wefindthatmostoftheshearisresistedbytheweb,andverylittleresistanceisofferedbytheflanges.Theoppositeistrueinthecaseforflexuralstresses—theflangesresistmostofthebendingstress,andthewebofferslittleresistancetobending(Figure28).ThecalculationoftheexactmaximumstressmagnitudeusingVQ>Ibcanbecomedifficultbecauseofthepresenceoffillets(rounding)wheretheflangejoinstheweb.AhighlevelofaccuracyisevenhardertoachieveinchannelsorstandardI-shapesthathaveslopingflangesurfaces.Accordingly,theAmericanInstituteofSteelConstruction(AISC)recommendstheuseofamuchsimpler,approximateformulaforthecommonsteelshapes:Wide-flangesection.Shearstressdistribution.Figure28Shearstressdistributionforawide-flangesection.fvaverage=whereVtwdV=shearforced=beamdepthtw=webthicknessThisformulagivestheaverageunitshearingstressfortheweboverthefulldepthofthebeam,ignoringthecontributionoftheflange(Figure29).406BendingandShearStressesinBeamsWebsresistapproximately90%ofthetotalshearforstructuralshapes,asshowninFigure29.Incontrast,flangesresist90%ofthebendingstresses.Dependingontheparticularsteelshape,theaverageshearstressformulafvaverage=Vtwdcanbeasmuchas20%inerrorinthenonconservativedirection.Thismeansthatwhenashearingstresscomputedfromthisequationgetswithin20%ofthemaximumallowableshearstress,theactualmaximumstress1VQ>Ib2mightexceedtheallowablestressbyasmallamount.Wide-flangeshape.Standardshape.Figure29(a)Figure29(b)Fortunately,thislowlevelofaccuracyisseldomaproblemfortworeasons:1.Structuralsteelsareverystronginshear.2.Mostbeamsandgirdersinbuildings,unlikethoseinsomemachines,havelowshearingstresses.Highshearingstressmaybepresentinshort-span,heavilyloadedbeamsoriflargeconcentratedloadsareappliedadjacenttoasupport.Indeterminingthesizeofasteelbeam,flexuralstressesordeflectionwillusuallygovern.Whenshearingstressesdobecomeexcessive,steelbeamsdonotfailbyrippingalongtheneutralaxis,asmightoccurintimberbeams.Rather,itisthecompressionbucklingoftherelativelythinwebthatconstitutesashearfailure(Figure30).TheAISChasprovidedseveraldesignformulasfordeterminingwhenextrabearingareamustbeprovidedatconcentratedloadsorwhenwebstiffenersareneededtopreventsuchfailures(Figure31).Channelshape.Figure29(c)Figure30Webbucklinginsteelbeams.Figure31Webstiffeners.407BendingandShearStressesinBeamsExampleProblems:ShearingStress12AnAmericanStandardS12×31.8beamresistsashearV=12katthesupports.Determinetheaveragewebshearstress.Fv=14.5ksi(A36steel).Solution:fvaverage=V=12kVtwdtw=0.35–d=12–fvaverage=12k=2.86ksi6Fv=14.5ksi10.35–2112–2‹OK13An(A36)W12*50beamisloadedasshown.Calculatethecriticalfvaverage.Solution:Fv=14.5ksi1A36steel2V=35ktw=0.37–d=12.19–fvaverage=‹OK408V35k=7.76ksi6Fv=14.5ksi=twd10.37–2112.19–2BendingandShearStressesinBeamsProblems11Twosteelplates(A572,Fy=50ksi)areweldedtogethertoformaninvertedT-beam.Determinethemaximumbendingstressdeveloped.Alsodeterminethemaximumshearstressattheneutralaxisofthecross-sectionandattheintersectionwherethestemjoinstheflange.12AlogofdiameterDisavailabletobeusedasabeamcarryingauniformlydistributedloadof400#/ft.overalengthof32feet.DeterminetherequireddiameterDnecessaryifFb=1,200psiandFv=100psi.13The20-footbeamshownhasacross-sectionbuiltupfroma1–*10–steelplateweldedontothetopofaW8*31section.DeterminethemaximumloadwthebeamcansustainwhenthesteelsectionreachesamaximumallowablebendingstressofFb=22ksi.Forthewcalculated,determinetheshearstressfvdevelopedbetweentheplateandthetopflangesurface(usetheflangewidthforb).409BendingandShearStressesinBeams14Alintelbeam12feetlongisusedincarryingtheimposedloadsoveradoorwayopening.Assumingthatabuilt-upboxbeamisusedwitha12inchoveralldepthasshown,determinethemaximumbendingstressandshearstressdeveloped.15Thecross-sectionoftherough-cuttimberbeamshownisloadedwithwoversixfeetofthespan.DeterminethemaximumvalueofwiftheallowablebendingstressisFb=1,600psiandtheallowableshearstressisFv=85psi.16A4*12S4Sbeamcarriestwoconcentratedloadsasshown.AssumingFb=1,600psiandFv=85psi,determine:a.ThemaximumpermissibleloadP.b.ThebendingandshearstressfourfeettotherightofsupportA.410BendingandShearStressesinBeams17Thebeamshownisconstructedbyweldingcoverplatestotwochannelsections.Whatmaximumuniformlydistributedloadcanthisbeamsupportona20footspanifFb=22ksi?Checktheshearstresswheretheplateattachestothechannelflange.18SelectthelightestWsectionsteelbeambasedonthebendingcondition.Checkfvaverageforthebeamselected.Fb=22ksiFv=14.5ksifA36Steel19Aplankisbeingusedtosupportatriangularloadasshown.Assumingtheplankmeasures12incheswide,determinetherequiredplankthicknessifFb=1,200psiandFv=100psi.20Abuilt-upplywoodboxbeamwith2–*4–blockingtopandbottomisheldtogetherbynailsalongthetopandbottomchords.Determinethepitch(spacing)ofthenailsifthebeamsupportsa5kconcentratedloadatmidspan.Thenailsarecapableofresisting80#eachinshear.411BendingandShearStressesinBeams5DEFLECTIONINBEAMSAsdiscussedinearliersectionsofthischapter,thedesignofbeamsforaparticularloadandsupportconditionrequiresthestrengthinvestigationofbendingstressandshearstress.Quitefrequently,however,thedesignofabeamisgovernedbyitspermissibledeflection.Indesign,deformation(calleddeflectioninbeams)oftensharesanequivalentimportancewithstrengthconsiderations,especiallyinlong-spanstructures.Deflection,astiffnessrequirement,representsachangeintheverticalpositionofabeamduetotheappliedloads.Loadmagnitude,beam-spanlength,momentofinertiaofthebeamcross-section,andthebeam’smodulusofelasticityareallfactorsintheamountofdeflectionthatresults.Generally,theamountofallowabledeflection,orpermissibledeflection,islimitedbybuildingcodesorbypracticalconsiderations,suchasminimizingplastercrackinginceilingsurfacesorreducingthespringinessofafloor.Wood,asastructuralmaterial,islessstiff(hasalowerE-value)thansteelorconcrete;hence,deflectionisalwaysaconcern.Detrimentaleffectsfromlargedeflectionscanincludenailpoppingingypsumceilings,crackingofhorizontalplastersurfaces,andvisiblesaggingofceilingsandfloors.Insomedesignsituations,primarilylongerspans,awoodmembersatisfyingthestrengthrequirementswillnotnecessarilysatisfydeflectioncriteria.Steelbeams,althoughstrongerrelativetowood,stillneedtobecheckedfordeflection.Particularcaremustbegiveninlong-spansituationsbecauseofthelikelihoodofobjectionablesagorpondingofwater.Pondingispotentiallyoneofthemostdangerousconditionsforflatroofs.Itoccurswhenaflatroofdeflectsenoughtopreventnormalwaterrunoff.Instead,somewatercollectsinthemidspan,andwiththeaddedweightofaccumulatedwater,theroofdeflectsalittlemore,allowingevenmorewatertocollect,whichinturncausestherooftodeflectmore.Thisprogressivecyclecontinuesuntilstructuraldamageorcollapseoccurs.Buildingcodesrequirethatallroofsbedesignedwithsufficientslopetoensuredrainageafterlong-termdeflectionorthatroofsbedesignedtosupportmaximumroofloads,includingthepossibleeffectsofponding.TheallowabledeflectionlimitsforbeamsaregiveninTable1.TheselimitsarebasedontheAmericanInstituteofTimberConstruction(AITC),AmericanInstituteofSteelConstruction(AISC),andUniformBuildingCode(UBC)standards.412BendingandShearStressesinBeamsTable1Recommendedallowabledeflectionlimits.LiveLoadDeadandLiveLoadsIndustriall/180l/120CommercialandinstitutionalWithoutplasterceilingWithplasterceilingl/240l/360l/180l/240Ordinaryusage*l/360l/240UseClassificationRoofBeams:FloorBeams:*Ordinaryusagereferstofloorsintendedforconstructioninwhichwalkingcomfortandtheminimizingofplastercrackingareprimaryconsiderations.Thecalculationofactualbeamdeflectionsisoftenapproachedfromamathematicalviewpointthatrequiresthesolutionofasecond-orderdifferentialequationsubjecttotheloadingandthetypeofendsupportsofthebeam.Thismethodismathematicallystraightforward,butitpresentsformidableproblemsassociatedwithevaluationoftheproperboundaryconditionsaswellasinthemathematicsrequiredtoobtainthesolution.Therearemanywaystoapproachtheproblemofbeamdeflections:themoment-areamethod,conjugatebeam,doubleintegration,andformulas.Thissectionwilldealexclusivelywiththeuseofestablisheddeflectionformulasfoundinstandardhandbooks,suchastheAISCmanual,timberdesignmanuals,andthelike.Today,deflectionsareautomaticallycalculatedformostbeamdesignsdoneonacomputer.Theintentofthissectionistopresentafewfundamentalconceptsdealingwithdeflectionanditsroleinbeamdesignratherthantoexplorethemanysophisticatedmathematicaltechniquesthatmaybeemployedinobtainingdeflectionvalues.Anunderstandingofthebasicsofdeflectionwillenabletheuserofcomputersoftwaretobetterunderstandtheresultsobtained.413BendingandShearStressesinBeamsTheElasticCurve—RadiusofCurvatureofaBeamWhenabeamdeflects,theneutralsurfaceofthebeamassumesacurvedposition,whichisknownastheelasticcurve(Figure32).Figure32(b)Afterloading.Figure32(a)Beforeloading.Itisassumedfrombeamtheorythatplanesaandb[Figure32(a)],whichwereparallelbeforeloading,willremaindqplaneafterbendingsoastoincludeasmallangle[Figure32(b)].Ifthecurvatureissmall,wemayassumethatRa=Rb=R,theradiusofcurvatureoftheneutralsurface(elasticcurve).Thelengthofthesegmentbetweensectionsaandbisdesignatedas¢x.IfdθisverysmallandRisverylarge,then¢x=Rdθ,becauseforverysmallanglessinθ=tanθ=θ.FromFigure33,¢x¿=1R+c2dθbythesamereasoning.Length:ab=¢xa¿b¿=¢x¿Thetotalelongationthatthebottomfiberundergoesisδc=¢x¿-¢xSubstitutingthevaluesgivenabove,wegetδc=1R+c2dθ-Rdθ=Rdθ+cdθ-Rdθ‹δc=cdθFromε=δ>L,wegetFigure33andb-b.414Sectionofbeambetweena-aεc=c1dθ2c¢x¿-¢x==R¢xR1dθ2BendingandShearStressesinBeamsSimilarly,itcanbeshownthattheunitstrainatanydistanceyfromtheneutralsurfacecanbewrittenasεy=yRWeknowthatfyMyMy;soεy=andfy=εy=EIEIEquatingthetwoexpressionsforεy,wegetMyEI=yRorR=MEIIor=MREIwhereR=radiusofcurvatureM=bendingmomentatsectionwhereRisdesiredE=modulusofelasticityI=momentofinertiaofthebeamcross-sectionBecausetheflexureformulawasusedtoobtainthisrelationship,itwillbevalidonlyforthosemembersthatmeettheassumptionsmadeinthederivationoftheflexureformula.EandIwillusuallybeconstantsforagivenbeam.Theradiusofcurvatureequationaboveisconsideredasabasicequationinthedevelopmentofdeflectionformulas.DeflectionFormulasManyloadingpatternsandsupportconditionsoccursofrequentlyinconstructionthatreferencemanuals(e.g.,AISC,AITC,etc.)andengineeringhandbookstabulatetheappropriateformulasfortheirdeflections.AfewofthemorecommoncasesareshowninTable2.Moreoftenthannot,therequireddeflectionvaluesinabeamdesignsituationcanbedeterminedviatheseformulas,andonedoesnotneedtoresorttodeflectiontheory.Evenwhentheactualloadingsituationdoesnotmatchoneofthetabulatedcases,itissufficientlyaccurateformostdesignsituationstoapproximatethemaximumdeflectionbyusingoneormoreoftheformulas.Computedactualdeflectionsmustbecomparedagainsttheallowabledeflectionspermittedbythebuildingcodes.¢actual…¢allowable415BendingandShearStressesinBeamsTable2Commoncasesofbeamloadinganddeflection.BeamLoadandSupportActualDeflection*¢max=5ωL4384EI¢max=PL348EI¢max=PL323PL3=648EI28.2EI¢max=PL320.1EI(atthecenterline)¢max=ωL4384EI(atthecenterline)(atthecenterline)(a)Uniformload,simplespan(atthecenterline)(b)Concentratedloadatmidspan(atthecenterline)(c)Twoequalconcentratedloadsatthirdpoints(d)Threeequalconcentratedloadsatquarterpoints(e)Uniformloadbothendsfixed416BendingandShearStressesinBeamsTable2Continued.BeamLoadandSupportActualDeflection*¢max=ωL48EI(atthefreeend)¢max=PL33EI(atthefreeend)(f)Cantileverwithuniformload(g)Cantileverwithconcentratedloadattheend*Becausespanlengthofbeamsisusuallygiveninfeetanddeflectionsareininches,aconversionfactormustbeincludedinallofthedeflectionformulasabove.Multiplyeachdeflectionequationbyconversionfactor=CF=112in.>ft.23=1,728in.3>ft.3417BendingandShearStressesinBeamsExampleProblems:DeflectioninBeams14UsingDF-LNo.1,designthesimplysupportedfloorbeamshowntomeetbending,shear,andmomentcriteria.1>46b>h61>2¢allow1DL+LL2=L>240;¢allow1LL2=L>360Fb=1,300psi;Fv=85psi;E=1.6*106psiSolution:BendingMmax=200#>ft.120¿22ωL2==10,000#-ft.88Srequired=ShearVmax=10k-ft.*12.in.>ft.Mmax==92.3in.3Fb1.3*103k>in.2200#>ft.120¿2ωL==2,000#22Arequired=1.5*12,000#21.5Vmax==35.3in.2Fv85#>in.2Deflection(Allowable)¢allow1DL+LL2=or¢allow(LL)=20¿*12in.>ft.L==1–24024020¿*12in.>ft.L==0.67–360360NotethatthecalculatedSrequiredandArequiredvaluesdonotaccountforthebeam’sownweight.Try6*12S4S.1A=63.25in.2;Sx=121.23in.3;Ix=697.07in.42Economical(efficient)beamsusuallyhavewidth-to-depth1b>h2ratiosof146b>h612.Checktheeffectofthebeam’sweightasitaffectsthebendingandshearstresscondition.BendingSadd.=418MaddFbBendingandShearStressesinBeamswhereMadd.=additionalbendingmomentduetothebeam’sweight.ωbeamL28Madd.=Conversionforwooddensityof35pcf(DouglasfirandSouthernPine)topoundsperlinealfoot(plf)ofbeamisωbeam=0.252*cross-sectionalareaofbeamωbeam=0.252*63.25=16plf‹Madd.=Sadd.=16#>ft.120¿228=800#-ft.800#ft.*12in.>ft.2Madd.==7.4in.3Fb1,300psi‹Stotal=92.3in.3+Sadd.=92.3in.3+7.4in.3Stotal=99.7in.36121.2in.3‹OKShearVadd.=additionalsheardevelopedduetothebeam¿sweight‹Vadd.=Aadd.=16#>ft.120¿2ωbeamL==160#221.5Vadd.1.5*160#==2.8in.2Fv85psi‹Atotal=35.3in.2+Aadd.=35.3in.2+2.8in.2Atotal=38.1in.2663.25in.2‹OKDeflection(Actual)¢actual=51100#>ft.2120¿2411,728in.3>ft.325ωLLL4=0.32–=384EI38411.6*106psi21697.1in.42¢actual1LL2=0.32–6¢allow1LL2=0.67–¢actual=51216#>ft.2120¿2411,728in.3>ft.325ωtotalL4=0.7–=384EI38411.6*106psi21697.1in.42Note:ωtotal=216#/ft.includesthebeamweight.¢actual1DL+LL2=0.7–6¢actual1DL+LL2=1–‹OKUse:6*12S4S.419BendingandShearStressesinBeams15DesignaSouthernPineNo.1beamtocarrytheloadsshown(roofbeam,noplaster).Assumethebeamissupportedateachendbyaneightinchblockwall.Fb=1,550psi;Fv=110psi;E=1.6*106psiSolution:BendingSrequired=12.8k-ft.*12in.>ft.Mmax==99.1in.3Fb1.55ksiArequired=1.5*12,750#21.5Vmax==37.5in.2Fv110psiShearDeflection(Allowable)¢allow=15¿*12in.>ftL==0.75–240240Try6*12S4S.AA=63.3in.2;Sx=121in.3;Ix=697in.4BωbeamL0.252*63.3=16#>ft.BendingMadd.16#>ft.115¿22ωbeamL2==450#-ft.=88Sadd.=450#-ft.*12in.>ft.Madd.=3.5in.3=Fb1,550psiStotal=99.1in.3+3.5in.3=102.5in.36121in.3‹OKNote:Sadd.isusuallyapproximately2%to5%ofSrequired.ShearVadd.=16#>ft.115¿2ωbeamL==120#22Aadd.=1.5Vadd.1.5*120#==1.6in.2Fv110psiAtotal=37.5in.2+1.6in.2=39.1in.2663.3in.2‹OK420BendingandShearStressesinBeamsActualDeflectionUsingsuperposition(thecombination,orsuperimposing,ofoneloadconditionontoanother),¢actual=5ωLLL423PL3+[emailprotected]centerline2384EI648EI2312,000#2A15¿B311,7282+384A1.6*106BA697in.4B684A1.6*106BA697in.4B=0.12–+0.35–=0.47–60.75–‹OK‹¢actual=¢actual51100+162A15¿B411,7282Checkthebearingstressbetweenthebeamandtheblockwallsupport.fp=2,870#P==65.2psiAbearing44in.2TheallowablebearingstressperpendiculartothegrainforSouthernPineNo.1is:Fc=440psi‹OKUse6*12S4S.421BendingandShearStressesinBeams16Asteelbeam(A572Grade50)isloadedasshown.Assumingadeflectionrequirementof¢total=L>240andadepthrestrictionof18inchesnominal,selectthemosteconomicalsection.Fb=30ksi;Fv=20ksi;E=30*103ksiSolution:Vmax=24kMmax=238k-ft.Steelbeamsareusuallydesignedforbending.Onceatrialsectionhasbeenselected,shearanddeflectionarechecked.Bending238k-ft.*12in.>ft.M=95.2in.3=Fb30ksiSreq’d=TryW18*55.1Sx=98.3in.3;Ix=890in.4;tw=0.39–;d=18.11–2Madd.=Sadd.=55#>ft.128¿22ωbeamL2==5,390#-ft.885.39k-ft.*12in.>ft.Madd.=2.2in.3=Fb30ksiStotal=95.2in.3+2.2in.3=97.4in.3698.3in.3‹OKShearCheck55#>ft.128¿2ωbeamL==770#22Vadd.=fvaverage=Vmax24,000#+770#=3,510psi=twd10.39–2118.11–2fvaverage=3,510psi620,000psi‹OKDeflectionCheck¢allow=¢actual=PL35ωL4+48EI384EI20k128¿231,728UseW18*55.422511.06k>ft.2128¿241,72848A30*103B(890)13842A30*103B18902=0.59–+0.55–=1.14–61.4–‹OK¢actual=¢actual28¿*12in.>ft.L==1.4–240240+BendingandShearStressesinBeams17Apartialplanofanofficebuildingisshown.AllstructuralsteelisA36steel.DesignatypicalinteriorbeamB1,andrestricttheliveloaddeflectionto¢LL6L>360.Limitdepthto14inches.Alsodesignthespandrelbeam,restrictingitstotaldeflectionto¢LL6L>240.Limitdepthto18inches.Loads:Concretefloor:150pcf2.5psf1–finishwoodfloor:Suspendedfire-resistantceiling:3.0psfLL:70psf*Curtainwall:400#/ft.*OccupancyforofficebuildingwithmoveablepartitionsA36steel:Fb=22ksiFv=14.5ksiE=29*103ksiPartialfloorplan(officebuilding).Solution:BeamB1Design5–concretefloor:1–finishwoodfloor:62.5psf2.5psfSuspendedceiling:3.0psfTotalDL68psfTotalLL=70psf*8¿(tributarywidth)=560#/ft.TotalDL+LL=138psf*8¿(trib.width)=1,104#/ft.Sectiona-a.Mmax=1.1k>ft.128¿22ωL2==108k-ft.88Srequired=108k-ft.*12in.>ft.Mmax==59in.3Fb22ksiTryW14*43(S=62.7in.3;I=428in.4).NocheckisnecessaryforSadd.Sectionb-b.423BendingandShearStressesinBeamsDeflection28¿*12in.>ft.L==0.93–36022ksi510.56k>ft.2128¿2A1,728in.3>ft.3B5ωLLL4===0.62–384EI384A29*103ksiBA428in.4B=0.62–6¢allow(LL)=0.93–‹OK¢allow(LL)=¢actual(LL)¢actual(LL)UseW14*43.BeamReactionontoSpandrelBeamR=11.1+0.043k>ft.2128¿2ωtotalL==16.1k22SpandrelDesignCurtainwall=400plfSpanlength=32¿¢allow(D+L)=32¿*12in.>ft.L==1.6–240240Mmax=309k-ft.Srequired=309k-ft.*12in.>ft.22ksi=168.5in.3TryW18*97ASx=188in.3;Ix=1,750in.4B.Madd.=Sadd.=0.097k>ft.132¿228=12.4k-ft.12.4k-ft.*12in.>ft.22ksi=6.8in.3Stotal=168.5+6.8=175.3in.36188in.3‹OKDeflection¢total(DL+LL)=¢total(DL+LL)=5ωL4PL3+384EI20.1EI51.52132¿2411,7282384A29*103B11,7502¢actual=0.23–+0.91–=1.14–6+16.1132¿2311,728220.1A29*103B11,7502L=1.6–‹OK240ShearCheckfvaverage=V30.6k+1.6k=3.2ksi614.5ksi‹OK=twh10.535–2118.59–2UseW18*97.424BendingandShearStressesinBeams18ThesundeckistobeframedusingHem-FirNo.2gradetimber.Joistsarespacedattwofeeto.c.withaspanlengthof10feet.Oneendofthejoistissupportedbyaconcretefoundationwallandtheotherendbyabeam.Thesupportingbeamisactuallymadeoftwobeams,splicedatthecenterofthespan.Joistsandbeamsaretobeconsideredassimplysupported.Loads:2–plankdeck=5psfLL=60psfForJoists:¢total6L>240Fb=1,150psiFramingplanofsundeck.Fv=75psiE=1.4*106psiForBeams:¢total6L>240Fb=1,000psiFv=75psiE=1.4*106psiSolution:JoistDesignDL—2–deck=5psfLL=60psfTotalDL+LL=65psfElevationA-A.ωDL+LL=65psf*2¿=130#>ft.Vmax=Mmax=130#>ft.110¿2ωL==650#22130#>ft.110¿22ωL2==1,625#-ft.881.51650#21.5V==13in.22Fv75#>in.#-ft.*12in.>ft.1,625M===17in.3Fb1,150psiArequired=SrequiredTry2*10.ElevationB-B.1A=13.88in.2;Sx=21.4in.3;Ix=98.9in.4;ωbeam=3.5#>ft.2Note:Hem-Firhasadensityof30pcf;hence,theconversionfactorisωbeam=0.22*cross-sectionalareaofthebeam.425BendingandShearStressesinBeamsBendingandShear3.5#>ft.110¿2Vadd.=21.5A17.5in.3BAadd.=75psi=17.5#=0.35in.2Atotal=13.4in.2‹OK3.5#-ft.110¿22Madd.=Sadd.=8=43.75#-ft.43.75#-ft.*12in.>ft.1,150psi=0.5in.3Stotal=17.5in.3‹OKDeflectionSectionC-C.¢allow(DL+LL)=¢actual(DL+LL)10¿*12in.>ftL==0.5–24024051130+3.52110¿2411,72825ωL4==384EI38411.4*1062198.92¢actual(DL+LL)=0.22–60.5–‹OKUse2*10joistsattwofeeto.c.(equivalentto2psf)BeamDesignBecausejoistsarespaceduniformlyandoccuratarelativelyclosespacing,assumeloadstobeuniformlydistributedonthebeam.Beamsplicedetail.Loads:2–plankdeck=5psf2*10@2¿o.c.=2psfLL=60psfω=67psf*5¿=335#>ft.Vmax=Mmax=Beamsupport.426335#>ft.18¿2ωL==1,340#22335#>ft.18¿22ωL2==2,680#-ft.88BendingandShearStressesinBeamsArequired=Srequired=1.511,340#21.5V==26.8in.2Fv75psi2,680#-ft.*12in.>ft.M=32.2in.3=Fb1,000psiTry4*10.1A=32.38in.2;Sx=49.91in.3;Ix=230.84in.4;ωbeam=8#>ft.2Aadd.andSadd.shouldnotbecriticalhere.¢allow=¢actual=8¿*12in.>ft.L==0.4–24024051335#>ft.218¿2411,72825ωL4=384EI38411.4*10621231in.42¢actual=0.1–60.4–‹OKUse4*10S4Sbeam.427BendingandShearStressesinBeamsProblems21AssumingA36steel,selectthemosteconomicalW8section.Checktheshearstress,anddeterminethedeflectionatthefreeend.Fb=22ksiFv=14.5ksiE=29*103ksi22DesignaDouglasfir–LarchNo.1beamtosupporttheloadshown.Fb=1,300psiFv=85psiE=1.6*106psi¢allow1LL2=L>36023Designthebeamshownassumingtheloadsareduetodeadandliveloads.¢allow1LL+DL2=L>240AssumingthebeamisSouthernPineNo.1supportedatbothendsbygirdersasshown,calculateandcheckthebearingstressdevelopedbetweenthe6×_beamanda6*12girder.Fb=1,550psi;Fv=110psi;E=1.6*106psi;Fc=410psi24DesignB1andSB1assumingA36steel.Maximumdepthforeachisrestrictedto16inches–nominal.LL40psfConcrete150pcfCurtainwallonspandrelbeamSB1300plfSuspendedplasterceiling5psfMetaldeck4psf¢LL6L>360forB1;¢DL+LL6L>240forSB1428BendingandShearStressesinBeams6LATERALBUCKLINGINBEAMSInthepreviousdiscussiononbeams,itwasimpliedthatmakingabeamasdeep(largeIx)aspossiblewasgenerallyadvantageous,becausetheIxandSxvaluesaremaximized.Thereare,however,limitsonhowdeepabeamshouldbewhenusedinthecontextofthebuilding.Whenasimplysupportedbeamissubjectedtoaload,thetopflangeorsurfaceisincompression,whilethebottomflangeorsurfaceisintension.Atthecompressionsideofthebeam,thereisatendencyforittobuckle(deflectsideways),justasacolumncanbuckleunderaxialloading.Inacantileveroroverhangbeam,thebuckling,orsidesway,willdevelopduetothecompressiononthebottomsurfaceofthebeam(Figure34).Verynarrow,deepbeamsareparticularlysusceptibletolateralbuckling,evenatrelativelylowstresslevels.(a)Simplysupportedbeam.Toresistthetendencyofabeamtodisplacelaterally,eitherthecompressionsurfaceneedstobebracedbyotherframingmembersorthebeamneedstobereproportionedtoprovidealargerIy.Thevastmajorityofbeams,suchasfloorandroofbeamsinbuildings,arelaterallysupportedbythefloororroofstructuresattachedtoandsupportedbythem.Steeldeckingweldedtothebeams,beamswiththetopflangeembeddedintheconcreteslab,orcompositeconstruction(steelbeamsmechanicallylockedtothesteeldeckingandconcreteslab)areexamplesoflateralsupportforsteelbeams.(b)Cantileverbeam.Figure34Lateralbucklinginbeams.Woodframingtypicallyemployscontinuoussupportalongthetopcompressionsurfacethroughsheathingnailedatarelativelyclosespacingandsolidblockingtoproviderestraintagainstrotationattheends.Dependingonthespanofthewoodbeam,bridgingorsolidblockingisprovidedatintervalstoresistlateralbuckling.Someroofbeamsthatsupportrelativelylightweightroofsheathingarenotconsideredtobelaterallysupported.Figure35(a)Typicalwoodfloorjoistwithcontinuousnailing.Figure35(b)Concreteslab/beamcastmonolithically.429BendingandShearStressesinBeamsCertainbeamsareinherentlystableagainstanylateralbucklingtendencybyvirtueoftheircross-sectionalshapes.Forexample,arectangularbeamwithalargewidth-to-depthratio(IyandIxarerelativelyclose)andloadedintheverticalplaneshouldhavenolateralstabilityproblem(Figure35).Awide-flangebeamhavingacompressionflangethatisbothwideandthicksoastoprovidearesistancetobendinginahorizontalplane(relativelylargeIy)willalsohaveconsiderableresistancetobuckling(Figure36).Figure35(c)steelbeam.CompositeconcreteslabwithTheproblemoflateralinstabilityinunbracedsteelbeams(Wshapes)isamplified,becausethecross-sectionaldimensionsaresuchthatrelativelyslenderelementsarestressedincompression.Slenderelementshavelargewidth-to-thicknessratios,andtheseelementsareparticularlysusceptibletobuckling.Abeamthatisnotlaterallystiffincross-sectionmustbebracedeverysooftenalongitscompressivesidetodevelopitsfullmomentcapacity.Sectionsnotadequatelybracedorlaterallysupportedbysecondarymembers(Figure34)couldfailprematurely.Figure35(d)framing.Timberbeamwithjoist2–*12–Joist:Ix=178in.4Iy=3.2in.4Ix>Iy=55.6(a)Poorlateralresistance.6–*10–Beam:Ix=393in.4Iy=132in.4Ix>Iy=3.3(b)Goodlateralresistance.W14*22Beam:Ix=199in.4Iy=7in.4Ix>Iy=28.4(c)Poorlateralresistance.W14*82Beam:Ix=882in.4Iy=148in.4Ix>Iy=6.0(d)Goodlateralresistance.Figure36InSection2,thedesignofsteelbeamsassumedanallowablebendingstressofFb=0.6Fy,whereFb=22ksiforA36steel.Steelbeamslaterallysupportedalongtheircompressionflanges,meetingthespecificrequirementsofFb=0.66Fy,wheretheAISC,mayuseanallowableFb=24ksiforA36steel.Whentheunsupportedlengthsofthecompressionflangesbecomelarge,allowablebendingstressesmaybereducedbelowtheFb=0.6Fylevel.430BendingandShearStressesinBeamsForthepurposesofpreliminarysizingofsteelbeamsinarchitecturalpractice,andinparticularforthistext,theallowablebendingstresswillbetakenasFb=0.60FyInthecaseoftimberbeams,thedimensionsofthecrosssectionsaresuchthatthedepth-to-widthratiosarerelativelysmall.Acommonmethodofdealingwiththelateralstabilityissueistofollowrulesofthumbthathavedevelopedovertime.Theserulesapplytosawnlumberbeamsandjoists/rafters(Table3).Thebeamdepth-to-widthratiosarebasedonnominaldimensions.Table3Lateralbracingrequirementsfortimberbeams.BeamDepth/WidthRatioTypeofLateralBracingRequired2to1None3to1Theendsofthebeamshouldbeheldinposition5to1Holdcompressionedgeinline(continuously)6to1Diagonalbridgingshouldbeused7to1BothedgesofthebeamshouldbeheldinlineExample431BendingandShearStressesinBeams7INTRODUCTIONTOLOADRESISTANCEFACTORDESIGN(LRFD)Allofthepreviousanalysisanddesignproblemsintroducedinthistexthavebeenbasedontheallowablestressesofamemberdesignedintimberorsteel.Formerlycalledtheworkingstressorserviceloadmethod,theallowablestressdesign,orASD,methodhasbeentheclassicalapproachusedformanyyearstodesignstructuresofsteel,timber,andconcrete.However,thismethodologyisslowlygivingwaytowhatisknownasthestrengthmethod.Inconcretedesign,becauseofthecomplexitiesofanalyzingcompositesectionsusingtheworkingstressmethod,theACI318concretespecificationshaveemployedthestrengthbasedmethodsincethe1970s.Whendesigninginsteelandtimber,achoiceofdesignphilosophiesneedstobemade—namely,ASDorthestrength-basedmethod,currentlyreferredtoastheloadandresistancefactordesignmethod,orLRFD.AlthoughLRFDisrelativelynewtotimberdesign,itisnowincludedwithASDinthecurrenteditionsoftheNationalDesignSpecificationforWoodConstructionManual.Inthe13theditionoftheAISC’sSteelConstructionManual,eitherdesignprocessisallowed.DesigningsteelstructuresusingASDisrelativelysimpleandhasaproventrackrecordinprovidingabasisforsafeandreliabledesign.TheASDphilosophyisbasedonkeepingthememberstressesbelowaspecifiedpercentageoftheyieldstressinthesteel.Adequacyoftheproposeddesignisevaluatedbasedonspecifiedlimitssetontheallowablestress,stability,anddeformation.Amarginofsafety,obtainedbydividingtheallowablestressintothefailurestress,generallytheyieldstressforsteel,constitutesthefactorofsafety.Structuralelementsareproportionedsuchthattheactualcomputedstressesbasedonexpectedloads(bothdeadloadandliveload)arelessthantheallowablestresses,wellwithintheelasticstressrange.Thebehaviorofthestructureduetoanoverloadorfailureisnotconsidered.Alsoknownasstrengthdesign,limitstatedesign,orultimateloaddesign,LRFDisusedtodesignastructuretowithstandthemostcriticalcombinationoffactoredloadsappliedtothemember.Limitstateisaconditioninwhichastructureorastructuralcomponentisnolongerfitoruseful.Astructuralmembercanhaveseverallimitstates.Strengthlimitstatesconcernsafetyandrelatetomaximum432BendingandShearStressesinBeamsload-carryingcapacity,whichmayrefertomomentcapacity,shearcapacity,buckling,orplastichingeformation.Serviceabilitylimitstatesrelatetoperformanceundernormalconditions(unacceptableelasticdeformationordrift,unacceptablevibrationandpermanentdeformation).Essentially,ASDcomparesactualandallowablestresses,whileLRFDcomparesstrengthtoactualstrength.Loadandresistancefactordesignforsteelwasinitiallyproposedin1978andformallyadoptedbytheAISCin1986.Initially,however,theshifttoLRFDwasnotuniversallyembracedbytheengineeringprofession,eventhoughalmostalluniversitiesshiftedtoteachingthestrengthmethodwithin10yearsofitsintroduction.TheLRFDmethodisbelievedtohaveanumberofadvantagesovertheallowablestressmethod:■■■LRFDaccountsforthenonlinearnatureofthestress-straindiagramformaterialssubjectedtohighstresslevels.Itprovidesabetterrepresentationofthebehaviorofsteel-compositemembers(Figures37and38)andofmemberssubjectedtolargeseismicloads.Thedeadloadsactingonastructuremaybedeterminedwithsomedegreeofaccuracy,whileliveloadsarelesspredictable.TheLRFDmethodaccountsforthisbyspecifyinghigherloadfactorsforliveloads.BecausetheASDmethodhasnosimilarprovisionavailable,theresultisamoreconservativedesignforstructureswithahighratioofdeadloadtoliveloadandaninadequatesafetyfactorforstructureswithahighratioofliveloadtodeadload.LRFDaccountsforthevariabilityinnominalstrengthofdifferenttypesofmembersbyapplyingdifferentvaluesfortheresistancefactortothenominalstrength.IntheASDmethod,achievingauniformreliabilityfordifferentmembersisnotpossible.IntheLRFDmethod,probabilitytheoryisusedtoestablishanacceptablemarginofsafetybasedonthevariabilityofanticipatedloadsandmemberstrength.Failureresultswhenthenominalstrengthofamemberisincapableofresistingtheappliedforcesonthemember.Steelbeamwithasolidslabandshearstudsforcompositeaction.Figure37Compositebeam.EncasedsteelcolumnReinforcedcolumnwithencasedsteelConcretefilledtubularorpipecolumnFigure38Compositecolumns.ThethirdeditionoftheAISC’sManualofSteelConstruction:LoadandResistanceFactorDesignrequiresthatallsteelstructuresandmembersbeproportionedsothatnostrengthlimitstateisexceededwhensubjectedtoallrequiredfactoredloadcombinations.Inotherwords,thedesignstrengthofamemberhastobegreaterthanorequaltotherequiredresistance.Therequiredresistanceistheeffect(moment,shear,axialtensionorcompression,etc.)causedbythehighestfactoredloadcombination.433BendingandShearStressesinBeamsThegeneralequationoftheLRFDspecificationis©γiQi…φRnwhereγ=loadfactorforthetypeofload(deadload,liveload,wind,earthquake,etc.)Rn=NominalcapacityφRn=designstrengthf=P/A(axialstress,ksi)60RnφRn50RuptureQi=nominalloadeffectγi=loadfactorcorrespondingtoQi©γiQi=requiredresistanceorstrengthFy=36ksi(yield)40φ=resistancefactorcorrespondingtoRn30Fa=22ksi(0.6Fy)20Necking10PlasticrangeElasticrange0i=typeofload(deadload,liveload,wind,etc.)Fult.=58ksi.001.015Strainhardeningrange.040.240Unitstrain,ε(in./in.)-(nottoscale)Figure39Stress-straindiagramformild(A36)steel.Rn=nominalresistanceorstrength(ultimatecapacity;force,moment,shear,orstress)φRn=designstrengthNominalstrengthisthecapacityofastructuralcomponenttoresisttheeffectsofloads,basedonmaterialstrengths(yieldstrengthorultimatestrength)(Figure39)obtainedthroughlaboratoryandfieldtestingorderivedfromformulasusingacceptedprinciplesofstructuralmechanics.Therequiredultimatestrengthofamember1©γiQi2consistsofthemostcriticalcombinationsoffactoredloadsappliedtothemember.Factoredloadsconsistofworkingloads,orserviceloadsmultipliedbytheappropriateloadfactorstoaccountfortheinherentuncertaintiesintheloads.Therequiredstrength©γQisdefinedbysixloadcombinations.Aloadcombinationmayhaveuptothreeseparateparts,whichcanconsistof(a)thefactoreddeadloadeffect,(b)the50-yearmaximumliveloadeffect,and(c)apossibleliveloadeffectlabeledasanarbitrary-point-in-time,orAPT,value.SuchAPTvaluestypicallyareonlyafractionoftheirdesignvalues,becausethisreflectsasmallprobabilitythattheywouldoccursimultaneouslywiththeother50-yearmaximumliveloadadfullgravityload.Nominalloadsthatmustbeconsideredindesignincludethefollowing:D=deadload;includestheweightofstructuralelementsandotherpermanentelementssupportedbythestructure,suchaspermanentpartitionsL=liveloadduetooccupancyandmovableequipmentLr=roofliveload434BendingandShearStressesinBeamsR=rainload;initialrainwaterloadoricewaterloadexclusiveofthepondingcontributionS=snowloadW=windloadE=earthquakeloadThesixloadconditionsfortherequiredstrength©γQareasfollows:■■■■■■©γQ=1.4D1duringconstructionwherethepredominantloadisdeadload2©γQ=1.2D+1.6L+0.51LrorSorR21whenthemaximumvaluesofoccupancyliveloadsgoverntheloadingcondition2©γQ=1.2D=1.61LrorSorR2+10.5L*or0.8W)(whenthemaximumvaluesofroofliveload,rainwater,orsnowgovernstheloadingcondition)©γQ=1.2D+1.3W+0.5L*+0.51LrorSorR2(whensubjectedtomaximumvaluesofwindload,increasingtheeffectsofdeadload)©γQ=1.2D;1.0E+0.5L*+0.2S(whensubjectedtomaximumvaluesofseismicload,increasingtheeffectsofdeadload)©γQ=0.9D;11.3Wor1.0E2(whensubjectedtomaximumvaluesofwindorseismicload,opposingtheeffectsofdeadload)*Replace0.5Lwith1.0Lforgarages,placesofpublicassembly,andareaswhereL>100#/ft.2.Thedesignstrengthofamemberconsistsofthetheoreticalultimate,ornominal,strengthofthememberRnmultipliedbytheappropriateresistancefactorφ,whichaccountsforthevariabilityinthenominalstrength.Resistancefactorφforthevariousstressconditionsisasfollows:φb=0.90beams1flexure2φv=0.90beams1shear2φc=0.85compressionmembersφt=0.90tensionmembers1yieldingstate2φt=0.75tensionmembers1fracturestate2435BendingandShearStressesinBeamsExampleProblems:LRFDMethod19TypicalW1855floorbeamsofanofficebuildingarespacedat12feetcentertocentersupportingasuperimposeddeadloadof90#>ft.2andaliveloadof50#>ft.2.Determinethegoverningloadcombinationandthecorrespondingfactoredload.Solution:DL:D=90#>ft.2*12.0¿+55#>ft.=1,135#>ft.LL:L=50#>ft.2*12.0¿=600#>ft.Twoloadcombinationsapplytothisfloorframing,becausethereisnorooflive,snow,rain,orwindload,orearthquake,toconsider.1.4D=1.4*1,135#>ft.=1,589#>ft.1DLonly21.2D+1.6L=1.2*1,135#>ft.+1.6*600#>ft.=2,322#>ft.Themaximumfactoredloadof2,322#/ft.governsthedesignforthefloorbeams.20Aroofdesignsupportsadeadloadof65#>ft.2andasnowloadof40#>ft.2Inaddition,awindpressureof20#>ft.2(upliftordownward)mustbeaccommodated.Determinethegoverningloadcondition.Solution:Sixloadcombinationsmustbeconsideredinthisexample.1.4D=1.4*65#>ft.2=91#>ft.21.2D+1.6L+0.5S=1.2*65#>ft.2+0+0.5*40#>ft.2=98#>ft.21.2D+1.6S+0.8W=1.2*65#>ft.2+1.6*40#>ft.2+0.8*20#ft.2=158#>ft.21.2D+1.3W+0.5S=1.2*65#>ft.2+1.3*20#>ft.2+0.5*40#>ft.2=124#>ft.2436BendingandShearStressesinBeams1.2D;1.0E+0.2S=1.2*65#>ft.2+0+0.2*40#>ft.2=86#>ft.20.9D;1.3W=0.9*65#>ft.2+1.3*20#>ft.2=84.5#>ft.2Inthisexample,thegoverningloadcombinationfortheroofdesignis1.2D+1.6S+0.8W=158#>ft.2StandardStructuralSteelsShapesandSectionsManygradesofsteelandavarietyofstrengthsareusedinbuildingdesign,butcertainsectionsmaybeavailableonlyinspecificgrades.Forexample,A992isofferedwithFy=50ksibutprimarilyinWshapes.A36steelhasaloweryieldandultimatetensilestrengthbutistheprimarysourceforM,S,C,L,andMCshapesaswellasforplates,rods,andbars.HPpilesaregenerallyavailableinA572Grade50andA36Grade36.WhenhigherstrengthsarerequiredforWsections,A572Grade60orGrade65arespecified.ElasticrangeFlexuralStressesDesign,usingtheLRFDphilosophy,assumesthatabeamhastheabilitytodevelopyieldstressesoveritsentirecross-section(Figure40),whichrepresentsthestrengthlimitstate.Theplasticdesignconcepttakesadvantageoftheductilepropertyofamaterial,whichischaracterizedbyitsunrestrictedplasticflow(Figure41).Thisabilityofthebeamtodevelopahighercapacitywhileallfibersthroughoutthecross-sectionareyieldingdependsupontheindividualelementsofthatbeam(theflangesandtheweb)remainingstable.Abeamthathasthisstabilityisreferredtoasbeingcompact(Figure42),anditcanattainitsplasticmomentcapacity1Mp2andexhibitnolocalbuckling.Thisplasticmomentcapacityisattainedwhenallofthebeam’sfibershaveyieldedandis,onaverage,10%to12%greaterthanthemomentneededtoproducefirstyield.Whenabeamreachesitsplasticmomentcapacity,itPlasticrangeFyyieldpointF(stress)Flexuralstress,orbendingstress,istheprimaryconcerninthedesignofbeams.Flexuralmemberstrengthislimitedbylocalbucklingofacross-sectionalelement(thebeam’sweborflange),lateral-torsionalbucklingoftheentiremember,orthedevelopmentofaplastichingeataparticularcross-section.εy=yieldstrainεy(strain)Beyondtheyieldpoint,asignificantdeformationoccurswithoutanincreaseofstress.Thebehaviorofthesteelisinelasticorplastic.Actualfailuredoesnotoccuruntilexcessivedeformationresultsorruptureofthematerial.Figure40Idealizedstress-straindiagramforductilesteel.437BendingandShearStressesinBeamsCf<Fyd/6d2d/3Cf<FyTTC=T=fb(bxd/2)(1/2)2M=Cx(2d/3)=fb(bd/6)ElasticmomentBendingStressintheElasticRangeNofibershaveyieldedandM<Myieldd/6tod/4dCf=FyC<2d/3Tf=FyTOutermostfibersonlyhaveyieldedM>Mydd/4Elastic/PlasticMomentCf>FyCd/2f>FyTTC=T=f(bxd/2)AllfibershaveundergoneyieldingM>Mplastic2Mp=C(d/2)=Fy(bd/4)PlasticResistingMomentFigure41Beamstressesfromelastic,partialyielding,andfullyielding.bfbftwhcdtfFigure42438dtwhctfDefinitionsforcomputingbeamcompactness.BendingandShearStressesinBeamshasyieldedeveryfiberwithinitscross-sectionanddevelopsaplastichinge(Figure43).Thetermplastichingereflectsthefactthatthebeamhasnomorerotationalcapacityandwillmostlikelycollapseifnomomentredistributionoccurs(Figure44).Theplastichingeconditionrepresentstheabsolutelimitofusefulnessofthecross-section.Onlybeamsthatarecompact(resistanttolocalbuckling)andadequatelybraced(topreventlateral-torsionalbuckling)canattainthisupperlimitofflexuralstrength.TheAISCcriteriafordeterminingthecompactnessofabeamcross-section(seeFigure42)topreventlocalbucklingisexpressedasplastichingeCantileverBeamOnehingeformsSimpleBeamOnehingeformsFixed-EndBeamThreehingesformbf65…2tf2Fy(checksforcompactnessoftheflangesofWandotherIshapesandchannels)wherebf=flangewidth(in.)tf=flangethickness(in.)Fy=minimumyieldstress(ksi)andashc640…tw2Fy(checksforcompactnessoftheweb)wherehc=heightoftheweb(in.)tw=webthickness(in.)ForA36steelwithFy=36ksi,hcbf…10.8and…1072tftwForA572andA992withFy=50ksi,hcbf…9.19and…90.52tftwFigure43hinges.InstabilityresultingfromplasticωLC2LωL242ωL82−ωL12(a)Momentdistribution-elasticrangeω+LCLMp2ωL8MpMp(b)Plasticmomentachievedatthreelocations-supportsandmid-span.ω+(c)Plastichingesformatthesupportsfirst,thenasadditionalloadsareapplied,thethirdplastichingeresultsatmid-span.Figure44Plastichingesandmomentredistribution.439BendingandShearStressesinBeamsExampleProblems:Compactness21CheckthecompactnessofaW10*45ifitismadewithA992steel1Fy=50ksi2.BracingofferedbyothercrossmembersConcretefloorslabSteeldeckShearstudsSteelfloorbeamCompositebeamaction-aslabwithshearstudsthatstabilizethecompressionflangeofthebeamFigure45Bracedcompressionflanges.Solution:FromTableA3intheAppendix,bf=8.020–;tf=0.620–tw=0.350–;hc=10.10–-210.620–2=8.76–Checkingthecompactnessoftheflange,bf>2tf=8.020–>210.620–2=6.4769.191flangeiscompact2Checkforwebcompactness,hc>tw=8.76–>0.35–=25.0690.51webiscompact2Loadcausinglateral-torsionalbucklingofthecompressionflangeUnbracedbeamDisplacementandrotationofthebeamcross-sectionFigure46Lateral-torsionalbucklingofthecompressionflange.LateralSupportoftheCompressionFlangeInacompactrolledbeambendingaboutitsstrongaxis(x-x),thecontrollingfactoraffectingthecapacityofthebeamisthelateralsupportofthecompressionflange(Figure45).Fullsupportofthecompressionflangeismorelikelytoproduceaplastichingeasitslimitstate.Inadequatebracingofthecompressionflangecanresultincolumn-typeinstabilityinbeams,atypeofbehaviorthatisreferredtoaslateral-torsionalbuckling.Theslendercompressionflangebeginstobuckleoutofplane,andthebeamundergoesatorsionalcomponentcausedbythedownwardforcesalongthetopflange(Figure46).Figure45demonstratestwocommonwaysinwhichbracingofthebucklingplanecanbeachieved,thusallowingthecompressionflangetobestabilizedandincreasethebeam’sabilitytoresistmoreload.Again,referringtoFigure41,therelationshipsbetweenmomentandmaximum(extremefiber)stressforagivencross-sectionatvariousstagesofloadingareasfollows:Intheelasticrange,beforetheyieldpoint,M=S1fb2Atinitialyielding,My=S1Fy2Atfullplastification(i.e.,plastichinge),Mp=Z1Fy2whereM=bendingmomentduetotheappliedloads(k-in.)My=bendingmomentatinitialyielding(k-in.)Mp=plasticmoment(k-in.)S=elasticsectionmodulus(in.3)440BendingandShearStressesinBeamsZ=plasticsectionmodulus(in.3)fb=bendingstressintheelasticrange(ksi)Fy=minimumyieldstress(ksi)RecallfromSection2thattheelasticsectionmodulus,ortheelasticsectionmodulus,S=I/c,whereI=momentofinertiaofthecross-sectionaboutitscentroidalaxis(in.4).Usingarectangularbeamcross-sectionasanexample,theelasticsectionisgivenbytheequationS=bh2>61in.32whereastheformulafortheplasticsectionmodulusisZ=bh2>4whereb=widthoftherectangularbeamcross-section(in.)h=heightordepthofthebeam(in.)AcomparisonoftheplasticmomenttothemomentatinitialyieldshowstheratioMp>My=1Fy21bh2>42>1Fy21bh2>62=1.5TheratioMp>Myindicates,intermsoftheyieldmoment,themagnitudeofthemomentcausingfullyieldingofthecross-section.Thisratioisindependentofthematerialpropertiesofthemember;itdependssolelyonthecrosssectionalpropertiesZandS.TheratioZ>Sisdefinedastheshapefactor.Shapefactorsofwide-flangedbeamsvarybetween1.10and1.18,withtheaveragebeing1.14.Shapefactorsforothercross-sectionalshapesare1.70foraroundbarand1.50forarectangle.Asindicatedearlier,thefullyieldingofacross-sectionofamemberresultsinaplastichingeifnoinstabilityispresent.Forasimplysupportedbeam(seeFigure43),loadshigherthantheultimateload(correspondingtoMp)willcausetheplastichingetorotateand,inturn,causefailureofthebeam.Intheexampleofafixed-endbeamoflengthL,supportingauniformlydistributedloadw,thegreatestmomentthatdevelopsiswL2>12attheends.Theelasticstrengthistheloadmagnitudethatinitiatesyieldingattheseends.Uponanincreaseofloadintensity,plastichingesformattheendsofthebeam,whilethemomentincreasesatthecenterportiontomaintainequilibriumwiththeload.Thisredistributionofmomentcontinuesuntilyieldingresultsatthecenterwiththeformationofathirdplastichingeinthebeamatmidspan.A“mechanism”isthenformed,andthebeamcantakenoadditionalload.Theresultisinstabilityandthebeamfails.441BendingandShearStressesinBeamsDesignforBendingThemostbasicbeamdesignrequirementintheLRFDmethodisthatthedesignmomentcapacityφbMnmustbegreaterthanorequaltotherequired(factored)flexuralstrengthMu.Therequiredflexuralstrengthisgenerallyreferredtoastheultimatemoment.φbMnÚMuwhereφb=0.90forbendingMn=nominalmomentcapacityMu=required(factored)momentAsdiscussedpreviously,bendingstrengthisaffectedbytheunbracedlengthofthecompressionflangeandthecompactnessofthemember.Forlaterallybracedcompactsections,wheretheplasticmomentisreachedbeforelocalbucklingoccurs,designisbasedonthelimitstateofyielding.Therefore,Mn=Mp=FyZxThedesignformoftheequationfordeterminingtherequiredplasticsectionmodulusZxfromthemaximumfactoredloadmomentMu(k-ft.)isZx=12MuφbFyIssuesoflateral-torsionalbuckling,flangelocalbuckling,andweblocalbucklingarebeyondthescopeandobjectivesofthistextandwillnotbecovered.ShearStressesAlthoughflexuralstrengthusuallycontrolstheselectionofrolledbeams,shearstrengthshouldbechecked.Shearmaybecriticalincasesofshort-spanmembers,especiallythosesupportinglargeconcentratedloads.Forrolledshapes,withanaxisofsymmetryintheplaneofloading,theflangecapacitytoresistshearisnegligible,andnowebstiffenersarerequired.TheAISCLRFDspecificationequationsmaybesimplifiedasfollows:Forh418…,tw2FyaγiRi=Vu…φvVn=0.910.6FywAw2442BendingandShearStressesinBeamswhereh=thecleardistancebetweenfillets(forrolledshapes)tw=webthickness(in.)d=overalldepthoftherolledsection(in.)Vu=maximumshearfromfactoredloadsφv=resistancefactorforshear(0.9)Vn=nominalshear(ultimatecapacity)Fyw=yieldstrengthofthesteelinthewebAw=twd=areaofthewebOtherAISCLRFDformulasaregivenforcaseswhereh418…tw2Fybuttheywillnotbeconsideredinthistext.ExampleProblems:MomentCapacity22DeterminethemomentcapacityofanASTM992W18*40floorbeamwithFy=50ksi.Assumethatthebeamisfullybracedlaterallyalongthecompressionflange(seeFigure47)withacompositeconcretedecksecurelyanchoredwithshearstuds.Solution:Acheckmustfirstbemadetoverifyitscompactness.FortheW18*40,d=17.90–;bf=6.015–;tf=0.525tw=0.315;h=15.5–;Zx=78.4in.4bf6.015–==5.73…9.19for50ksisteel2tf2*0.525–andhc15.50–==49.2…90.5tw0.315TheW18*40sectioniscompact,andthemomentcapacitywillbeequaltotheplasticmoment.Mn=Mp=FyZx=A50k>in.2BA78.4in.4B=3,920k-in.Mn=3,920>12=327k-ft.Compositebeamwithmetaldeckinglaidperpendiculartothebeamspan.Shearstudsareweldedtothebeamflangeanddeckingtocreatethecompositeactionwiththeconcreteslab.Figure47Compositebeamexample.443BendingandShearStressesinBeams'2412'1B-G12B-12'12'3BG223Asimplysupportedbeam(B-2)supportsauniformlydistributedloadandisbracedatsixfootintervals(seeFigure48)inadditiontoacompositeslabthatissecuredtothecompressionflangeofthebeamviashearstuds.AssumethatthebeamismadeofASTMA992gradewithFy=50ksi.SelectthelightestWshapetosafelysupportthefollowingloads:4B-DL=100psf;LL=75psfBeamsarespacedat12feeto.c.Bracing@6’o.c.Solution:ωDL=100#>ft.2*12¿=1,200#>ft.Figure48Isometricviewofpartialsteelframingarrangement.ωLL=75#>ft.2*12¿=900#>ft.Becausethisexamplerepresentsafloorloadcondition,thetwofactoredloadconsiderationsthatapplyare1.4D=1.4*1,200#>ft.=1680#>ft.1forDLonly21.2D+1.6L=1.211,200#>ft.2=1.61900#>ft.2=2,880#>ft.Thefactoredloadof2,880#/ft.governsthedesignforthefloorbeams.ω=2.88k/ft.ωL/2L=24’Mu=ωL/212.88k-ft.2124¿22ωL2==207.4k-ft.88Vu=
12.88k-ft.2124¿2ωL==34.6k22Zx=121207.4k-ft.212Mu==55.3in.32φbFy10.902A50k>in.BFromTablesA11(a)andA11(b)intheAppendix,selectthelightesttrialsize.TryW18*35.Zx=66.5in.3,d=17.7in.,h=15.5in.bf=6.00in.,tf=0.425in.,tw=0.30in.)Mp=FyZx=A50k-in.2Ba66.5in.3b=277k-ft.12in.>ft.φMp=10.9021277k-ft.2=249k-ft.444BendingandShearStressesinBeamsThisvaluecanalsobefounddirectlyinTableA11(b)intheAppendix.Next,checkthecompactnessofthebeamsection.bf6.00–=7.0669.19(theflangeiscompact)=2tf210.425–2h15.5–=51.7690.5(thewebiscompact)=tw0.300–Shearcheck:418h…tw2FythenuseVu…φvVn=1.9021.6Fytwd2h15.5–418418=51.76===59tw0.300–2Fy250Therefore,φvVn=10.90210.62150210.30*17.7–2=143.4kVu=34.6k6φvVn=143.4k‹thebeamisOKinshearTheW18*35isadequateandthelightestsectiontouseasafloorbeam.Summary■■■■Beamssubjectedtotransverseloadsdevelopinternalmoments,whichresultinbendingaction.Partofthebeamcross-sectionissubjectedtocompressivestresses,whiletheremainingportionofthecross-sectionisintension.Thetransitionofstressfromcompressiontotensionoccursattheneutralaxis,whichcorrespondstothecentroidalaxisofthebeam’scross-section.Flexuralstressesaredirectlyproportionaltothebeam’sbendingmomentandinverselyproportionaltothebeam’smomentofinertiarelativetotheneutralaxis.Bendingstressesvaryfromzeroattheneutralaxistoamaximumatthetopand>orbottomfibers.Generally,inbeamdesign,themaximumbendingstressisused,andtheequationisexpressedasfb=1moment2*1distancefromN.A.toextremefiber2Mc=I1momentofinertia2445BendingandShearStressesinBeams■Themajorityofthestructuralshapesusedinpracticearestandardshapes,normallyavailableinindustry.Crosssectionalpropertiesareknownandavailableintablesandhandbooks.Therefore,thebasicflexureequationcanbesimplifiedbysubstitutingatermcalledthesectionmodulus(S),whereS=■Substitutingthesectionmodulusintotheflexureformula,theequationbecomesfb=■IcMMMc==II>CSRewritingtheequationintoadesignform,appropriatebeamssizescanbeselectedfromtables.Srequired=■■Inadditiontobendingmoment,beamsaresubjectedtotransverse(vertical)andlongitudinal(horizontal)shears.Atanygivenpointalongthelengthofthebeam,thetransverseshearandlongitudinalshearareequal.Therefore,theshear(V)diagramcanbeusedintheanalysisordesignofbeamsforshear.Thegeneralshearstressatanypointonabeamisexpressedasfv=■maximumbendingmomentMmax.=Fballowablebendingstress1shearforce2*1firstmomentAy2VQ=lb1momentofinertia2*1widthofshearplane2Shearingstressinwide-flange,channel,andT-sectionsinsteelaregenerallycarriedbytheweb.Verylittleshearresistanceisofferedbytheflange;therefore,thegeneralshearstressequationissimplifiedasfvaverage=1shearforce2V=twd1webthickeness2*1beamdepth2Thisformulagivestheaverageunitshearingstressfortheweboverthefulldepthofthebeam.Structuralsteelisveryresistantinshear,therefore,theaverageshearstressformulaisgenerallysufficientasacheck.■Forsolidrectangularcross-sectionsintimber,thegeneralshearstressformulasimplifiestofvmaximum=4461.5*1shearforce21.5V=A1cross-sectionalarea2BendingandShearStressesinBeams■■■■Themaximumshearstressinarectangularbeamis50%largerthantheaveragestressvalue.Timberbeamsareoftencriticalinshear,becausetheallowableshearstressforwoodisverylow.Thethirdrequirementforbeamdesignisdeflection.Astiffnessrequirement,deflectionrepresentsachangeinverticalpositionofabeamduetotheappliedloads.Generally,theamountofallowableorpermissibledeflectionislimitedbybuildingcodes.Deflectionsareafunctionofloadmagnitude,beamspanlength,themomentofinertiaofthebeamcrosssection,andthebeam’smodulusofelasticity.Actualbeamdeflectionsmustbecomparedagainsttheallowabledeflectionlimitedbycode.¢actual6¢allowableAnswerstoSelectedProblems1ƒ=M/S=14.2ksi<22ksi2Mmax=4.28k-ft.;fb=696psi<Fb=1,300psi4Mmax=17.2k-ft.;Smin.=9.38in.3UseW8×13(Sx=9.91in.3)6(a)Mmax=16.67k-ft.;f=17ksi(b)Sreq’d.=125in.3;use8×12S4S8Mmax=30.4k-ft.;fmax=15ksi;at4’fromthefreeend,f=1.88ksi10Mmax=125.4k-ft.;P=15.7k11Vmax=8.75k;Mmax=43.75k-ft.Ix=112.6in.4;fb=27.5ksi;attheN.A.:fv=1.37ksi;attheweb/flange:fv=1.20ksi12Vmax=6,400lb.;Mmax=51,200lb.-ft.Basedonbending:Radius=8.67”;basedonshear:Radius=5.2”;use18”-diameterlog14Vmax=1,800lb.;Mmax=7,200lb.-ft.;Ix=469.9in.4;fb=1,170psi;fv=196psi16(a)Basedonbending:P=985lb.Basedonshear:P=1,340lb.Bendinggovernsthedesign.(b)At4’fromtheleftsupport,V=1,315lb.;M=5,250lb.-ft.;fb=854psi;fv=50.2psi.18Mmax=32k-ft.;Sreq’d.=17.5in.3;Fromsteeltables:UseW12×19(Sx=21.3in.3)orW10×19(Sx=18.8in.3)Averagewebshearfv=5.14ksi20Ix=1,965in.4;Q=72in.3;V=2,500lb.p=FI/VQ=1.75”(spacingofnails)22Mmax=22,400lb.-ft.;Sreq’d.=207in.3;Vmax=5,600lb.;Areq’d.=98.8in.2;8×16S4S;∆allow=L/360=0.53”∆LL=0.16”<0.53”;∴8×16S4SOK23Vmax=2,000lb.;Areq’d.=27.3in.2;Mmax=12,000lb.-ft.;Sreq’d.=92.9in.3;Fora4×14S4S∆allow=L/240=0.8”∆actual=0.48”<0.8”∴OKfbearing=109psi.Use4×14S4S24Mmax=48,000lb.-ft;Sreq’d.=26.2in.3;Try:W14×22;∆allow=0.73”;∆LL=0.29”<.73”∴OKUseW14×22forbeamB1.ForSB1:Mmax=95.6k-ft.;Sreq’d.=52.1in.3;TryW16×36;∆allow=1.2”(DL+LL);∆actual=0.78”(DL+LL);fv(ave)=2.7ksi<Fv=14.5ksi;UseW16×36forSB1.447448ColumnAnalysisandDesignFromChapter9ofStaticsandStrengthofMaterialsforArchitectureandBuildingConstruction,FourthEdition,BarryOnouye,KevinKane.Copyright©2012byPearsonEducation,Inc.PublishedbyPearsonPrenticeHall.Allrightsreserved.449ColumnAnalysisandDesignIntroductionColumnsareessentiallyverticalmembersresponsibleforsupportingcompressiveloadsfromroofsandfloorsandtransmittingtheverticalforcestothefoundationsandsubsoil.Thestructuralworkperformedbythecolumnissomewhatsimplerthanthatofthebeam,becausetheappliedloadsareinthesameverticalorientation.Althoughcolumnsarenormallyconsideredtobeverticalelements,theycanactuallybepositionedinanyorientation.Columnsaredefinedbytheirlengthdimensionbetweensupportendsandcanbeveryshort(e.g.,footingpiers)orverylong(e.g.,bridgeandfreewaypiers).Theyareusedasmajorelementsintrusses,buildingframes,andsubstructuresupportsforbridges.Loadsaretypicallyappliedatmemberends,producingaxialcompressivestresses.(a)Coveredwalkway.PhotobyMattBissen.(b)Compressionstrutsinabiplane.PhotobyChrisBrown.Figure1450Examplesofcompressionmembers.Commontermsusedtoidentifycolumnelementsincludestuds,struts,posts,piers,piles,andshafts,asshowninFigure1.Virtuallyeverycommonconstructionmaterial,includingsteel,timber,concrete(reinforcedandprestressed),andmasonry,isusedforcolumnconstruction.Eachmaterialpossessescharacteristics(materialandproduction)thatpresentopportunitiesandlimitationsontheshapesofcross-sectionsandprofileschosen.Columnsaremajorstructuralcomponentsthatsignificantlyaffectthebuilding’soverallperformanceandstabilityand,thus,aredesignedwithlargersafetyfactorsthanotherstructuralcomponents.Failureofajoistorbeammaybelocalizedandmaynotseverelyaffectthebuilding’sintegrity;however,failureofastrategiccolumnmaybecatastrophicforalargeareaofthestructure.Safetyfactorsforcolumnsadjustfortheuncertaintiesofmaterialirregularities,supportfixityatthecolumnends,andtakeintoconsiderationconstructioninaccuracies,workmanship,andunavoidableeccentric(off-axis)loading.ColumnAnalysisandDesign1SHORTANDLONGCOLUMNS—MODESOFFAILUREThelargeslabsofstoneusedatStonehengewereextremelymassiveandtendedtobestabilizedbytheirownweight.MassivestonecolumnscontinuedtobeusedinGreekandRomanstructures,butwiththedevelopmentofwroughtiron,castiron,steel,andreinforcedconcrete,columnsbegantotakeonmuchmoreslenderproportions.Columnslendernessgreatlyinfluencesacolumn’sabilitytocarryload.Becauseacolumnisacompressionmember,itwouldbereasonabletoassumethatonewouldfailduetocrushingorexcessiveshorteningoncethestresslevelexceededtheelastic(yieldpoint)limitofthematerial.However,formostcolumns,failureoccursatalowerlevelthanthecolumn’smaterialstrength,becausemostcolumnsarerelativelyslender(longinrelationtotheirlateraldimension)andfailduetobuckling(lateralinstability).Bucklingisthesudden,uncontrolled,lateraldisplacementofacolumn,atwhichpointnoadditionalloadcanbesupported.Thesidewaysdeflectionorbucklewilleventuallyfailinbendingifloadsareincreased.Veryshort,stoutcolumnsfailbycrushingduetomaterialfailure;long,slendercolumnsfailbybuckling—afunctionofthecolumn’sdimensionsanditsmodulusofelasticity(Figure2).Figure2(a)Crushing:Shortcolumn—exceedmaterialstrength.ShortColumnsStresscomputationsforshortcolumnsareverysimpleandrelyonabasicstressequation.Iftheloadandcolumnsizeareknown,theactualcompressivestressmaybecomputedasfa=Pactual…FaAwherefaAPactualFa====actualcompressivestress(psiorksi)cross-sectionalareaofcolumn(in.2)actualloadonthecolumn(#ork)allowablecompressivestresspercodes(psiorksi)Thisstressequationcanbeeasilyrewrittenintoadesignformwhendeterminingtherequisiteshortcolumnsizeiftheloadandallowablematerialstrengthareknown:Arequired=Figure2(b)instability.Buckling:Longcolumn—elasticPactualFawhereArequired=minimumcross-sectionalareaofthecolumn451ColumnAnalysisandDesignLongColumns—EulerBucklingThebucklingphenomenoninslendercolumnsisduetotheinevitableeccentricitiesinloadingandthelikelihoodofirregularitiesinamaterial’sresistancetocompression.Bucklingcouldbeavoided(theoretically)iftheloadsappliedwereabsolutelyaxialandthecolumnmaterialwastotallyhomogeneouswithnoimperfections.Obviously,thisisnotpossible;hence,bucklingisafactoflifeforanyslendercolumn.Figure3LeonhardEuler(1707–1783).Knownasoneofthemostprolificmathematiciansofalltime,Eulerwroteprofuselyoneverybranchofthesubject.Hislearnedpaperswerestillbeingpublished40yearsafterhisdeath.Progressively,hebegantoreplacethegeometricmethodsofproofusedbyGalileoandNewtonwithalgebraicmethods.Hecontributedconsiderablytothescienceofmechanics.Hisdiscoveryinvolvingthebucklingofthinstrutsandpanelsresultedfromthetestingofhisinvention,calledthe“calculusofvariation,”tosolveaprobleminvolvingcolumnsbucklingundertheirownweight.Itwasnecessarytousethecalculusofvariationtosolvethishypotheticalproblembecausetheconceptsofstressandstrainwerenotinventeduntilmuchlater.Eulerandhisfamily,Swiss-Germanbyorigin,weresupportedincomfortalternatelybytherulersofRussiaandPrussia.DuringastayinRussia,hechallengedthevisitingFrenchphilosopherandatheist,Diderot,toadebateonatheism.TothegreatamusementofCatherinetheGreatandothersofthecourt,EuleradvancedhisownargumentinfavorofGodintheformofasimpleandcompletelya+bn=x,irrelevantequation:“Sir,nhenceGodexists.”AllmathematicswasbeyondpoorDiderot,andhewasleftspeechless.Assumingthathehadbeenshownproof,whichheclearlydidnotunderstand,andfeelingafool,DiderotleftRussia.452Theloadcapacityofaslendercolumnisdirectlydependentonthedimensionandshapeofthecolumnaswellasonthestiffnessofthematerial(E).Itisindependentofthestrengthofthematerial(yieldstress).Thebucklingbehaviorofslendercolumns,withintheirelasticlimit,wasfirstinvestigatedbyaSwissmathematiciannamedLeonhardEuler(1707–1783)(Figure3).Euler’sequationpresentstherelationshipbetweentheloadthatcausesbucklingofapinnedendcolumnandthestiffnesspropertiesofthecolumn.ThecriticalbucklingloadcanbedeterminedbytheequationPcritical=π2EIminL2wherePcritical=criticalaxialloadthatcausesbucklinginthecolumn(#ork)E=modulusofelasticityofthecolumnmaterial(psiorksi)Imin=smallestmomentofinertiaofthecolumncross-section(in.4)L=columnlengthbetweenpinnedends(in.)Notethatasthecolumnlengthbecomesverylong,thecriticalloadbecomesverysmall,approachingzeroasalimit.Conversely,veryshortcolumnlengthsrequireextremelylargeloadstocausethemembertobuckle.Highloadsresultinhighstresses,whichcausecrushingratherthanbuckling.ColumnAnalysisandDesignTheEulerequationdemonstratesthesusceptibilityofthecolumntobucklingasafunctionofthecolumnlengthsquared,thestiffnessofthematerialused(E),andthecross-sectionalstiffnessasmeasuredbythemomentofinertia(I)(Figure4).Figure4(a)Stableequilibrium:Longcolumn(Plessthanthecriticalload)—columnstiffnesskeepsthememberinastateofstableequilibrium.Figure4(b)Neutralequilibrium:Longcolumn1P=Pcrit.2—thecolumnloadequalsthecriticalbucklingload;thememberisinastateofneutralequilibrium.Figure4(c)Unstableequilibrium:Longcolumn1P7Pcrit.2—thememberbucklessuddenly,changingtoastateofinstability.453ColumnAnalysisandDesignTounderstandthephenomenonofbucklingfurther,let’sexamineaslendercolumnwithaslightinitialbowtoitbeforeloading(Figure5).Figure5(a)Columnwithaslightbow‘e’fromvertical.BecausetheloadPisoffset(eccentrictothecentralaxisofthecolumn),amomentM=P*eresultsinbendingstressesbeingpresentinadditiontothecompressivestressf=P>A.Iftheloadisincreased,additionalmomentresultsinbendingthecolumnfurtherand,thus,inalargereccentricityordisplacement.ThismomentM¿=P¿*¢resultsinanincreasedbendingthatcausesmoredisplacement,thuscreatinganevenlargermoment(P-¢effect).Aprogressivebendingmomentanddisplacementcontinuesuntilthestabilityofthecolumniscompromised.Thecriticalloadatwhichthelimitofthecolumn’sabilitytoresistuncontrolled,progressivedisplacementhasbeenreachedisreferredtoasEuler’scriticalbucklingload.ItisimportanttonoteagainthattheEulerequation,whichcontainsnosafetyfactors,isvalidonlyforlong,slendercolumnsthatfailduetobucklingandinwhichstressesarewellwithintheelasticlimitofthematerial.Shortcolumnstendtofailbycrushingatveryhighstresslevels,wellbeyondtheelasticrangeofthecolumnmaterial.SlendernessRatiosFigure5(b)TheoffsetloadPproducesamomentM=P*e.Thegeometricpropertyofacross-sectioniscalledtheradiusofgyration.Thisdimensionalpropertyisbeingrecalledhereinconnectionwiththedesignofcolumns.AnotherusefulformoftheEulerequationcanbedevelopedbysubstitutingtheradiusofgyrationforthemomentofinertia,inwhichIAAI=Ar2r=wherer=radiusofgyrationofthecolumncross-section(in.)I=least(minimum)momentofinertia(in.4)A=cross-sectionalareaofthecolumn(in.2)ThecriticalstressdevelopedinalongcolumnatbucklingcanbeexpressedasfcriticalFigure5(c)P¿7P:Increasedloadwithanincreaseddisplacement¢7e.454π2E1Ar22Pcriticalπ2E===AAL21L>r22Columnsectionswithhighrvaluesaremoreresistanttobuckling(Figure6).Becausetheradiusofgyrationisderivedfromthemomentofinertia,wecandeducethatcross-sectionalconfigurationiscriticalingeneratinghigherrvalues.ColumnAnalysisandDesignTheL/rtermintheequationaboveisknownastheslendernessratio.Thecriticalbucklingstressofacolumndependsinverselyonthesquareoftheslendernessratio.Highslendernessratiosmeanlowercriticalstresses(Figure7)thatwillcausebuckling;conversely,lowerslendernessratiosresultinhighercriticalstress(butstillwithintheelasticrangeofthematerial).Theslendernessratioisaprimaryindicatorofthemodeoffailureonemightexpectforacolumnunderload.Asacomparisonofsteelcolumnsectionsoftenfoundinbuildings,notethedifferenceinrmin.valuesforthreesectionsshowninFigure8.Allthreesectionshaverelativelyequalcross-sectionalareasbutverydifferentradiiofgyrationaboutthecriticalbucklingaxis.Ifallthreecolumnswereassumedas15feetinlengthandpinconnectedatbothends,thecorrespondingslendernessratiosarequitedifferentindeed.Ingeneral,themostefficientcolumnsectionsforaxialloadsarethosewithalmostequalrxandryvalues.Circularpipesectionsandsquaretubesarethemosteffectiveshapes,becausetheradiiofgyrationaboutbothaxesarethesame1rx=ry2.Forthisreason,thesetypesofsectionsareoftenusedascolumnsforlighttomoderateloads.However,theyarenotnecessarilyappropriateforheavyloadsandwheremanybeamconnectionsmustbemade.Thepracticalconsiderationsandadvantagesofmakingstructuralconnectionstoeasilyaccessiblewide-flangeshapesoftenoutweighthepurestructuraladvantagesofclosedcross-sectionalshapes(e.g.,tubesandpipes).Specialwide-flangesectionsarespecificallymanufacturedtoproviderelativelysymmetricalcolumns(rx>ryratiosapproaching1.0)withlargeload-carryingcapability.Mostofthesecolumnsectionshavedepthandflangewidthsapproximatelyequal(“boxy”configuration)andaregenerallyinthe10,12,and14inchnominaldepthcategory.Figure8areas.Figure6axis.ColumnbucklingaboutitsweakFigure7ratio.BucklingstressversusslendernessComparisonofsteelcross-sectionswithequivalent455ColumnAnalysisandDesignExampleProblems:ShortandLongColumns—ModesofFailure1Determinethecriticalbucklingloadfora3–φstandardweightsteelpipecolumnthatis16feettallandpinconnected.AssumethatE=29*106psi.Solution:FromtheEulerbucklingequation,Pcritical=π2EIL2Theleast(smallest)momentofinertiaisnormallyusedintheEulerequationtoproducethecriticalbucklingload.Inthisexample,however,Ix=Iyforacircularpipe:I=3.02in.41Thereisnoweakaxisforbuckling.2Pcritical=13.1422129*106psi2A3.02in.4B116¿*12in.>ft.22=23,424#Theaccompanyingcriticalstresscanbeevaluatedasfcritical=Pcritical23,424#==10,504psiA2.23in.2Thiscolumnbucklesatarelativelylowstresslevel.Fcompression=22ksi2Determinethecriticalbucklingstressfora30-foot-longW12*65steelcolumn.Assumesimplepinconnectionsatthetopandbottom.Fy=36ksi1A36steel2E=29*103ksiSolution:fcritical=π2E1L>r22ForaW12*65,rx=5.28–,ry=3.02–ComputetheslendernessratioL>rforeachofthetwoaxes.456ColumnAnalysisandDesignSubstitutethelargerofthetwovaluesintotheEulerequation,becauseitwillyieldthemorecriticalstressvalue30¿*12in.>ft.L==68.2.rx5.28–30¿*12in.>ft.L==119.2;Governsry3.02–(producesasmallerstressvalueatbuckling)fcritical=π2A29*103ksiB1119.222=20.1ksiTheuseofL>rxwouldclearlyyieldamuchlargerstressvalue.Thisindicatesthatthecolumnwouldbuckleabouttheyaxisunderamuchsmallerloadthanwouldberequiredtomakeitbuckletheotherway.Inpracticalterms,thismeansthatincaseofoverload,thecolumnwouldnotbeabletoreachthecriticalloadnecessarytomakeitbuckleaboutitsstrongaxis;itwouldfailatalowerloadvaluebybucklingaboutitsweakaxis.Therefore,incomputingcriticalloadandstressvalues,alwaysusethegreaterL>rvalue.457ColumnAnalysisandDesign2ENDSUPPORTCONDITIONSANDLATERALBRACINGInthepreviousanalysisofEuler’sequation,eachcolumnwasassumedtohavepinnedendsinwhichthememberendswerefreetorotate(butnottranslate)inanydirection.If,therefore,aloadisappliedverticallyuntilthecolumnbuckles,itwilldosoinonesmoothcurve(seeFigure8).Thelengthofthiscurveisreferredtoastheeffectivelength,orthebuckledlength.Inpractice,however,thisisnotalwaysthecase,andthelengththatisfreetobuckleisgreatlyinfluencedbyitsendsupportconditions.Figure9Effectivecolumnlengthversusactuallength.Theassumptionofpinnedendsisanimportantone,becauseachangeofendconditionsimposedonsuchacolumnmayhaveamarkedeffectuponitsload-carryingcapacity.Ifacolumnissolidlyconnectedatthetopandbottom,itisunlikelytobuckleunderthesameloadassumedforapinnedendcolumn.Restrainingtheendsofacolumnfrombothtranslationandafree-rotationconditiongenerallyincreasestheload-carryingcapacityofacolumn.Allowingtranslationaswellasrotationattheendsofacolumngenerallyreducesitsload-carryingcapacity.Columndesignfomulasgenerallyassumeaconditioninwhichbothendsarefixedintranslationbutfreetorotate(pinconnected).Whenotherconditionsexist,theloadcarryingcapacityisincreasedordecreased,sotheallowablecompressivestressmustbeincreasedordecreasedortheslendernessratioincreased.Forexample,insteelcolumns,afactorKisusedasamultiplierforconvertingtheactuallengthtoaneffectivebucklinglengthbasedonendconditions(Figure9).ThetheoreticalKvalueslistedinFigure10arelessconservativethantheactualvaluesoftenusedinstructuraldesignpractice.CaseA:BothEndsPinned—Structureadequatelybracedagainstlateral(windandearthquake)forces.Le=LK=1.0Pcritical=π2EIL2Examples:Timbercolumnnailedtopandbottom.Figure10(a)CaseA—effectivebucklinglength,bothendspinned.458Steelcolumnwithsimpleclipangleconnectiontopandbottom.ColumnAnalysisandDesignCaseB:BothEndsFixed—Structureadequatelybracedagainstlateralforces.Le=0.5LK=0.5Pcritical=π2EI10.5L22=4π2EIL2Examples:Concretecolumnrigidlyconnected(monolithicallycast)tolargebeamstopandbottom.Steelcolumnrigidlyconnected(welded)tolarsteelbeamstopandbottom.geFigure10(b)CaseB—effectivebucklinglength,bothendsfixed.CaseC:OneEndPinnedandOneEndFixed—Structureadequatelybracedagainstlateralforces.Le=0.707LK=0.7Pcritical=π2EI10.7L22=2π2EIL2Examples:Concretecolumnrigidlyconnectedtoconcreteslabatthebaseandattachedtolight-gaugeroofingatthetop.Figure10(c)CaseC—effectivebucklinglength,oneendpinnedandoneendfixed.CaseD:OneEndFreeandOneEndFixed—Lateraltranslationpossible(developseccentriccolumnload).Le=2.0LK=2.0Pcritical=π2EI12L22=124πEI2LExamples:Watertankmountedonasimplepipecolumn.Flagpoleanalogy.Figure10(d)CaseD—effectivebucklinglength,oneendfreeandoneendfixed.459ColumnAnalysisandDesignCaseE:BothEndsFixedwithSomeLateralTranslationLe=1.0LK=1.0Pcritical=Figure10(e)CaseE—effectivebucklinglength,bothendsfixedwithsomelateraltranslation.π2EI(L)2Examples:Columnsinarelativelyflexiblerigidframestructure(concreteorsteel).CaseF:BasePinned,TopFixedwithSomeLateralTranslationLe=2.0LK=2.0Pcritical=Figure10(f)CaseF—effectivebucklinglength,basepinned,topfixedwithsomelateraltranslation.460π2EI12L22=124πEI2LExamples:Steelcolumnwitharigidconnectiontoabeamaboveandasimplepinconnectionatthebase.Thereissomeflexibilityinthestructure,allowingcolumnloadstobepositionedeccentrically.ColumnAnalysisandDesignIntermediateLateralBracingIntheprevioussection,wefoundthattheselectionofthetypeofendconnectionuseddirectlyinfluencedthebucklingcapacityofacolumn.Fixedconnectionsseemtobeanobvioussolutiontominimizingcolumnsizes;however,thecostassociatedwithachievingrigidconnectionsishigh,andsuchconnectionsaredifficulttomake.Also,timbercolumnsaregenerallyassumedtobepinconnected,becausethematerialstrengthgenerallyprecludestheconstructionoftruerigidjoints.Whatothermethodsaretheretoachieveanincreaseincolumncapacitywithoutspecifyinglargercolumnsizes?Acommonstrategyusedtoincreasetheeffectivenessofacolumnistointroducelateralbracingabouttheweakaxisofbuckling(Figure11).Infillwallpanels,windowheaders,girtsforcurtainwalls,andothersystemsprovidelateralbracingpotentialsthatcanbeusedtoreducethebucklinglengthofthecolumn.Bracingprovidedinoneplanedoesnot,however,provideresistancetobucklingintheperpendicularplane(Figure12).Columnsmustbecheckedinbothdirectionstodeterminethecriticalslendernessratiotobeusedinanalysisordesign.Figure11Wide-flangecolumnsectionbracedabouttheweakaxisofbuckling.Veryslendersectionscanbeusedforcolumnsiftheyareadequatelybracedorstiffenedagainstbucklingbetweenfloors,asseeninFigures13(a)–(d).Lateralrestraintprovidedbythebracingisforbucklingresistanceabouttheyaxisonly.Thecolumnisstillsusceptibletobucklinginonesmoothcurveaboutthexaxis.Slendernessratiosmustbecalculatedforbothaxestodeterminewhichdirectiongoverns.Figure12RectangulartimbercolumnbracedabouttheYaxisbutfreetobuckleabouttheXaxis.461ColumnAnalysisandDesignP1=π2EIP3=L2π2EIA13LB2=9P1(c)Third-pointbracing.(a)Nobracing.P2=2πEIA12LB2P4==4P1π2EIA23LB2=9P41(b)Bracedatmidpoint.Figure13462Effectivebucklinglengthsforvariouslateralsupportconditions.(d)Asymmetricalbracing.ColumnAnalysisandDesignExampleProblems:EndSupportConditionsandLateralBracing3Determinethecriticalbucklingloadfora48S4SDouglasfircolumnthatis18feetlongandbracedatmidheightagainsttheweakdirectionofbuckling.E=1.3*106psi.(b)Weakaxisbuckling.(a)Strongaxisbuckling.Solution:4*8S4SAIx=111.2in.4;Iy=25.9in.4;A=25.38in.2BrxIx=2.1–AAryIy=1.01–AALoadcausingbucklingaboutthexaxis:3.14211.3*106psi2A111.2in.4Bπ2EIx=Pcritical=Lx2118¿*12in.>ft.22=30,550#Loadcausingbucklingabouttheyaxis:Pcritical=π2EIyLy2=3.142A1.3*106psiBA25.9in.4B19¿*12in.>ft.22=28,460#Becausetheloadrequiredtocausethemembertobuckleabouttheweakeryaxisislessthantheloadthatisassociatedwithbucklingaboutthestrongerxaxis,thecriticalbucklingloadfortheentirecolumnis28.46k.Inthiscase,thememberwillbuckleinthedirectionoftheleastdimension.Whencolumnsareactuallytested,adifferenceisusuallyfoundbetweenactualbucklingloadsandtheoreticalpredictions.Thisisparticularlytrueforcolumnsnearthetransitionbetweenshort-andlong-columnbehavior.Theresultisthatbucklingloadsareoftenslightlylowerthanpredicted,particularlynearthetransitionzone,wherefailureisoftenpartlyelasticandpartlyinelastic(crushing).463ColumnAnalysisandDesign4AW840steelcolumnsupportstrussesframedintoitsweb,whichservetofixtheweakaxisandlightbeamsthatattachtotheflange,simulatingapinconnectionaboutthestrongaxis.Ifthebaseconnectionisassumedasapin,determinethecriticalbucklingloadthecolumniscapableofsupporting.Solution:W8*40AA=11.7in.2;rx=3.53–;Ix=146in.4;ry=2.04–;Iy=49.1in.4BThefirststepistodeterminethecriticalaxisforbuckling(i.e.,whichonehasthelargerKL>r).Topofcolumnframing.Deeptrussesmaybeconsideredasrigidsupportsforbucklingabouttheyaxis.Lightbeamssimulatepinsupportsaboutthexaxis.Le=KL=0.7134¿2=23.8¿23.8¿*12in.>ft.KL==140ry2.04–WeakAxisLe=L;K=1.0;KL=37¿137¿*12in.>ft.2KL==125.8rx3.53–Columnbaseconnection.TheW8*40(A36)steelcolumnisconnectedatthebaseviaasmallbaseplatewithtwoanchorbolts.Thisconnectionisassumedasapinthatallowsrotationintwodirectionsbutnotranslation.TheweakaxisforthiscolumniscriticalbecauseKLKL7ryrxPcritical=fcritical464π2EIyL2e=π2EIy1KL22=3.142A29*103ksiBA49.1in.4B123.8*12in.>ft.22=172.1kPcritical172.1k==14.7ksi=A11.7in.2StrongAxisColumnAnalysisandDesign5Arectangularsteeltubeisusedasa36footcolumn.Ithaspinnedends,anditsweakaxisisbracedatmidheightbyamasonryinfillwallasshown.Determinethecolumn’scriticalbucklingload.E=29*103ksi.Solution:4–*2–*14–rectangulartube1A=2.59in.2;Ix=4.69in.4;rx=1.35–;Iy=1.54in.4;ry=0.77–2Again,thefirststepinthesolutionmustinvolvethedeterminationofthecriticalslendernessratio.WeakAxisK=0.7KL=0.7*18¿=12.6¿=151.2–KL151.2–=196.4=ry0.77–StrongAxisK=1.0KL=1.0*36¿=36¿=432–KL432–==320;Governsrx1.35–WeakaxisKLKL7,bucklingismorecriticalabouttherxrystrongaxis.BecausePcritical=fcritical=π2EIx1KLx22=3.142A29*103ksiBA4.69in.4B=7.19k1432–22Pcritical7.19k==2.78ksiA2.59in.2Strongaxis465ColumnAnalysisandDesign6Determinethebucklingloadcapacityofa2*4stud12feethighifblockingisprovidedatmidheight.AssumeE=1.2*106psi.Solution:2*4S4SAA=5.25in.2;Ix=5.36in.4;Iy=0.984in.4Brx=ry=Ix5.36==1.01–AAA5.25Iy0.984==0.433–AAA5.25WeakAxisL=12¿K=0.5KL=0.5*12¿=6¿=72–KL72–==166.3ry0.433–StrongAxisL=12¿K=1.0KL=1.0*12¿=12¿=144–KL144–=142.6=rx1.01–Weakaxis.TheweakaxisgovernsbecausePcritical=fcritical=Strongaxis.466π2EI1KL22=KLKL7ryrx3.142A1.2*106psi210.984in.4B=2,246#172–22Pcritical2,246#==428psiA5.25in.2ColumnAnalysisandDesignProblems1AW8*31steelcolumn20feetlongispinsupportedatbothends.Determinethecriticalbucklingloadandstressdevelopedinthecolumn.E=29*103ksi.2Two312–φstandardpipesectionsarestrappedtogethertoformacolumnasshown.Ifthecolumnispinconnectedatthesupportsandis24feethigh,determinethecriticalaxialloadwhenbucklingoccurs.E=29*106psi.3DeterminethemaximumcriticallengthofaW10*54columnsupportinganaxialloadof250k.E=29*103ksi.4Aneight-inch-diametertimberpoleisfixedintoalargeconcretefootingatgradeandiscompletelypinconnectedatitsupperend.Howhighcanthepolebeandstilljustsupportaloadof25k?E=1.0*106psi.5Determinethecriticalbucklingloadandstressofan8–*6–*38–rectangularstructuraltubeusedasacolumn38feetlong,pinconnectedtopandbottom.6Determinethecriticalbucklingloadandstressforthecolumnshown.467ColumnAnalysisandDesign3AXIALLYLOADEDSTEELCOLUMNSMuchofthediscussionthusfarhasbeenlimitedtoveryshortcolumnsthatcrushand,ontheotherendofthescale,longandslendercolumnsthatbuckle.Somewhereinbetweenthesetwoextremesliesazonewherea“short”columntransitionsintoa“long”column.Euler’sbucklingequationassumesthatthecriticalbucklingstressremainswithintheproportionallimitofthematerialsothatthemodulusofelasticityEremainsvalid.SubstitutingtheproportionallimitvalueFproportional=31,000psiforA36steel(closetotheFy=36,000psi)intoEuler’sequation,theminimumslendernessrationecessaryforelasticbehaviorisfoundtobefcritical=Pcriticalπ2E=;thenA1L>r22/π2129,000,0002π2E==96=r31,000BBP>AColumns(A36steel)withslendernessratiosbelowL>r…96generallyexhibitcharacteristicsofinelasticbucklingorcrushing.TheupperlimitofKL>rforsteelcolumnsdependsongoodjudgmentandsafedesignandisusuallyestablishedbythebuildingcode.StructuralsteelcolumnsarelimitedtoaslendernessratioequaltoKL…200rInreality,columnsdonottransitionabruptlyfromshorttolong,orviceversa.Atransitionzoneexistsbetweenthetwoextremesandisnormallyreferredtoastheintermediatecolumnrange.Intermediatecolumnsfailbyacombinationofcrushing(oryielding)andbuckling(Figure14).Figure14Columnclassificationbasedonslenderness.468ColumnAnalysisandDesignTheinitiallyflatportionofthecurve(intheshort-columnrange)indicatesmaterialyieldingwithnobucklingtakingplace.Onthefarendofthecurve1K/>r71202,thecompressivestressesarerelativelylow,andbucklingisthemodeoffailure.Intheintermediate-columnrange1406K/>r61202,failurehasaspectsofbothyieldingandbuckling.Theload-carryingabilityofintermediate-lengthcolumnsisinfluencedbyboththestrengthandelasticpropertiesofthecolumnmaterial.Empiricaldesignformulas,basedonextensivetestingandresearch,havebeendevelopedtocoverthedesignofcolumnswithinthelimitsofeachcolumncategory.Since1961,theAISChasadoptedasetofcolumndesignformulasthatincorporatetheuseofavariablefactorofsafety,dependingonslenderness,fordeterminingallowablecompressivestress.AISCformulasrecognizeonlytwoslendernesscategories:short/intermediateandlong(Figure15).SlendercolumnsaredefinedasthosehavingaKL>rexceedingavaluecalledCc,inwhich:Cc=2π2EBFywhere:E=modulusofelasticityFy=yieldstressofthesteelMildsteel(A36)withFy=36ksihasCc=126.1;highstrengthsteelwithFy=50ksihasaCc=107.0.TheCcvaluerepresentsthetheoreticaldemarcationlinebetweeninelasticandelasticbehavior.Figure15formulas.AllowablecompressivestressbasedonAISC469ColumnAnalysisandDesignTheAISCallowablestress1Fa2formulaforshort/intermediatecolumns1KL>r6Cc2isexpressedasFa=c1-1K/>r222C2cd1Fy231K/>r21K/>r235+38Cc8C3c(AISCEq.E2–1)whereK/>r=thelargesteffectiveslendernessratioofanyunbracedlengthofcolumnFa=allowablecompressivestress(psiorksi)WhenaxiallyloadedcompressionmembershaveaK/>r7Cc,theallowablestressiscomputedasFa=12π2E231K/>r22(AISCEq.E2–2)Notethatthetwoprecedingequationsrepresentactualdesignequationsthatcanbeusedtosizecompressionelements.Theseequationsappearratherdaunting,especiallyE2–1.Fortunately,theAISCManualofSteelConstructionhasdevelopedadesigntableforK/>rfrom1to200withtherespectiveallowablestressFa.NocomputationsusingE2–1andE2–2arenecessary,becausetheequationshavebeenusedingeneratingthesetables(Tables1and2).Instructuralwork,pinnedendsareoftenassumedeveniftheendsofsteelcolumnsaretypicallyrestrainedtosomedegreeatthebottombybeingweldedtoabaseplate,whichinturnisanchor-boltedtoaconcretefooting.Steelpipecolumnsgenerallyhaveplatesweldedateachendandthenboltedtootherpartsofastructure.Suchrestraints,however,varygreatlyandaredifficulttoevaluate.Thus,designersrarelytakeadvantageoftherestrainttoincreasetheallowablestress,whichthereforeaddstothesafetyfactorofthedesign.470ColumnAnalysisandDesignOntheotherhand,testshaveindicatedthatinthecaseoffixed-endconditions,the“theoretical”K=0.5valuesaresomewhatnonconservativewhendesigningsteelcolumns.Becausetruejointfixityisrarelypossible,theAISCrecommendstheuseofrecommendedK-values(Figure16).Figure16AISC-recommendeddesignvaluesforK.ReproducedwiththepermissionoftheAmericanInstituteofSteelConstruction,Chicago,Illinois;fromtheManualofSteelConstruction:AllowableStressDesign,Ninthed.,Secondrev.(1995).Allexamplesdiscussedinthischapterassumethatthecolumnsarepartofabracedbuildingsystem.Sideswayisminimizedthroughtheuseofaseparatebracingsystem(bracedframeorshearwalls),andtheK-valuesforthebracedcolumnsneednotexceed1.0.Inunbracedbuildings,suchasthoseutilizingrigidframes,sideswaycanresultineffectivecolumnlengthsgreaterthantheactualcolumnlength1K71.02.Amuchmoreinvolvedanalysisisrequiredforcolumnswithsideswayand,thus,willnotbediscussedinthistext.471ColumnAnalysisandDesignAnalysisofSteelColumnsColumnanalysisimpliesthedeterminationoftheallowablecompressivestressFaonagivencolumnoritsallowableloadcapacityPallowable(Figure17).Asimpleanalysisprocedureisoutlinedbelow.Given:Columnlength,supportconditions,gradeofsteel1Fy2,appliedload,andcolumnsize.Required:Checktheadequacyofthecolumn.Inotherwords,isPactual6PallowableFigure17Analysisofsteelcolumns.Procedure:a.CalculateK/>rmin.;thelargestK/>rgoverns.b.EntertheappropriateAISCTable(Table1or2onthefollowingpages).c.PickouttherespectiveFa.d.Compute:Pallowable=Fa*AwhereA=cross-sectionalareaofthecolumn(in.2)Fa=allowablecompressivestress(ksi)e.Checkthecolumnadequacy.IfPallowable7PactualthenOK.IfPallowable6Pactualthenoverstressed.472ColumnAnalysisandDesignTable1ReproducedwiththepermissionoftheAmericanInstituteofSteelConstruction,Chicago,Illinois;fromtheManualofSteelConstruction:AllowableStressDesign,Ninthed.,Secondrev.(1995).473ColumnAnalysisandDesignTable2ReproducedwiththepermissionoftheAmericanInstituteofSteelConstruction,Chicago,Illinois;fromtheManualofSteelConstruction:AllowableStressDesign,Ninthed.,Secondrev.(1995).474ColumnAnalysisandDesignExampleProblems:AxiallyLoadedSteelColumns7AW12*531Fy=36ksi2columnistobeusedasaprimarysupportinabuilding.Iftheunbracedheightofthecolumnis16feetwithbothendsassumedaspinconnected,computetheallowableloadonthecolumn.Solution:EntertheAISCsectionpropertiestable(AppendixTableA3),andextractthedatafortheW12*53column.W12*53:A=15.6in.2;rx=5.23in.,ry=2.48in.Becausethecolumnisassumedtobepinconnectedatbothendsandforbothdirections(axes)ofbuckling,theleastradiusofgyration1ry2willyieldthemorecritical(larger)slendernessratio.Thecriticalslendernessratioisthencomputedas11.02116¿*12in.>ft.2K/=77.42=ry2.48–TodeterminetheallowablecompressivestressFa,enterTable1withthecriticalslendernessratio.K/>r77KL>r=77.42:Interpolating78Fa15.69ksi15.64ksi15.58ksiTheallowablecapacityoftheW12*53iscomputedas:Pallowable=Fa*A=15.64k>in.2*15.6in.2=244k475ColumnAnalysisandDesign8A24-foot-tallsteelcolumn(W14*82)withFy=50ksihaspinsatbothends.Itsweakaxisisbracedatmidheight,butthecolumnisfreetobucklethefull24feetinthestrongdirection.Determinethesafeloadcapacityforthiscolumn.Solution:PropertiesoftheW14*82:A=24.1in.2rx=6.05–ry=2.48–Weakaxisbuckling.Strongaxisbuckling.Computetheslendernessratioaboutbothaxestodeterminethecriticaldirectionforbuckling.24¿*12in.>ft.K/=47.6=rx6.05–12¿*12in.>ft.K/==58.1ry2.48–Thelargerslendernessratiogoverns;therefore,theweakaxis(y)bucklingisusedindeterminingFa.FromTable2,InterpolatingforK/>r=58.1;Fa=23.04ksi‹Pallowable=Fa*A=23.04k>in.2*24.1in.2=555k476ColumnAnalysisandDesign9A4–φstandardweightsteelpipe1Fy=50ksi2supportsaroofframingsystemasshown.Thetimberbeamto-columnconnectionisconsideredtobeapin,whilethebaseofthecolumnisrigidlyembeddedintotheconcrete.Iftheloadfromtheroofis35k,isthecolumnadequate?Solution:Four-inch”-diameterstandardweightpipe:A=3.17in.2r=1.51in.AlthoughthetheoreticalKis0.7forthesupportconditionshown,theAISC-recommendedvalueforuseindesignisK=0.80K/>r=0.80*118¿*12in.>ft.21.51–=114.43UsingTable2,Fa=11.4ksiPa=11.4k>in.2*3.17in.2=36.14k735kThecolumnisadequate.477ColumnAnalysisandDesignProblems7Determinetheallowableloadcapacity1Pallowable2foranFy=36ksisteelcolumn,W12*65,whenL=18¿anda.thebaseandtoparebothfixed.b.thebaseisfixedandthetopispinned.c.boththetopandbottomarepinned.8TwoC12*20.7channelsectionsareweldedtogethertoformaclosedboxsection.IfL=20¿andthetopandbottomarepinned,determinetheallowableaxialloadcapacityPa.AssumeFy=36ksi.9Anangle5–*312–*12–isusedasacompressionmemberinatruss.IfL=7¿,determinetheallowableaxialloadforanFy=36ksi.10DeterminethemaximumallowableheightofanA36column(5–φstandardsteelpipe)iftheappliedloadis60k.Assumethetoptobepinconnectedandthebasetobefixed.11Atwo-story,continuousW12*106columnofA36steelsupportsaroofloadof200kandanintermediate(secondfloor)loadof300k.Assumethetopandbottomhavepinconnections.Isthecolumnsectionshownadequate?Note:Assumethesecond-floorloadtobeappliedatthetopofthecolumn—thiswillresultinasomewhatconservativeanswer.Theconceptofintermediateloadsismuchmorecomplicatedandwillnotbediscussedfurtherinthistext.478ColumnAnalysisandDesignDesignofSteelColumnsThedesignofaxiallyloadedsteelcolumns(inotherwords,theselectionofanappropriatecolumnsize)isusuallyaccomplishedbyusingspecializedcolumntables,suchasthosecontainedinthenintheditionoftheAISC’sManualofSteelConstruction:AllowableStressDesign.Structuraldesignvariesfromanalysisinthateachproblemhasseveralpossibleanswers.Forexample,theselectionofacolumnsizeisjustlydependentonstrengthandsafetyrequirements,butotherissues(architecturalaswellasconstructional)mayinfluencethefinalselection.BecausetheAISCColumnDesignTablesareassumednottobeavailable(itwouldrequirethepurchaseoftheAISCmanual),steelcolumndesignwillinvolveaniterative,trial-and-errorprocess.Thismethodologyappearstobelongandtedious,butinfact,veryfewcyclesareusuallynecessarytozoominonasolution.Anearlierdiscussionofefficientcolumncross-sectionsforaxialloads(seeFigure8)suggestedtheuseofcircularor“boxier”wide-flangemembers.Alongwithspatialandconstructionalconcerns,relativemaximumorminimumsizesmayalreadybespecifiedbythearchitect,thuslimitingthearrayofchoicesthatwouldotherwisebeavailable.Thisinnowaylimitsthedesignpossibilities;infact,ithelpsguidethestructuraldesignandchoicesmadebytheengineer.Smaller-scalesteelstructuresmayuse8and10inchnominalsizewide-flangecolumns,whilelargerbuildingswithheavierloadswilloftenuse12and14inchnominalsizes.Thesesectionsarethe“boxier”orsquaresizes,withthedepthandflangewidthofthecross-sectionbeingapproximatelyequal.Onetrial-and-errorproceduremaybeoutlinedasfollows(Figure18):Given:Columnlength,supportconditions,gradeofsteel1Fy2,appliedload1Pactual2.Required:Columnsizetosafelysupporttheload.Procedure:a.Guessatasize.Butwheredoesonebegin?Ifitwereasmaller-scalebuilding,tryasquareW8orW10inthemiddleoftheweightgrouping.Asimilarestimateusinglargersectionsisappropriateforheavierloads.b.Oncethetrialsizehasbeenselected,crosssectionalpropertiesareknown.Computethecriticalslendernessratio,takingintoaccounttheendconditionsandintermediatebracing.479ColumnAnalysisandDesignc.UsingthelargerK/>rvalue,enterTable1or2.ObtaintherespectiveFa.d.CalculatethePallowable=Fa*Aofthetrialsection.e.ChecktoseeifPallowable7Pactual.IfPallowable6Pactual,thenthecolumnisoverstressed,andalargersectionshouldbeselectednext.Ifthetrialsectionismuchtoostrong,cycleagainwithasmallersize.Onewaytochecktherelativeefficiencyofthecross-sectionistoexamineitspercentageofstresslevel.%ofstress=Figure18Designofsteelcolumns.Pactual*100%PallowableApercentageofstressinthe90%to100%levelisveryefficient.f.Repeatthisprocessuntilanadequatebutefficientsectionisobtained.Note:Steps(b)through(e)areessentiallytheprocedureusedpreviouslyintheanalysisofsteelcolumns.480ColumnAnalysisandDesignExampleProblems:DesignofSteelColumns10SelectthemosteconomicalW12shapecolumn18feetinheighttosupportanaxialloadof600kusingA572Grade50steel.Assumethatthecolumnishingedatthetopbutfixedatthebase.Solution:Asafirstguessinthistrial-and-errorprocess,tryW12*96(aboutthemiddleoftheavailable“boxier”sections).W12*961A=28.2in.2,rx=5.44–,ry=3.09–2Compute1K/>r2critical:10.802118¿*12in.>ft.2K/==31.8rx5.44–10.802118¿*12in.>ft.2K/==55.9ry3.09Thelargerslendernessiscritical;therefore,useK/>r=55.9EnterTable2,andobtaintherespectiveFa:Fa=23.41ksiPallowable=Fa*APallowable=23.41k>in.2*28.2in.2=660k7600k%stress=Pactual600k*100%=*100%=91%Pallowable660kThisselectionisquiteefficientandstillabitunderstressed.Therefore,useW12*96.481ColumnAnalysisandDesign11SelectthemosteconomicalW8shapecolumn,16feetlong,withP=180k.Assumethatlateralbracingisprovidedatmidheightintheweakaxisofbucklingandthatthetopandbottomarepinconnected.Fy=36ksi.Solution:Again,weneedtobeginbyguessingatasizeandthencheckingtheadequacyoftheselection.TryW8*35.W8*35AA=10.3in.2;rx=3.51–;ry=2.03–BDeterminethecriticalslendernessratio:Weakaxis.Strongaxis.10.802118¿*12in.>ft.2K/=55.9=ry3.0910.802118¿*12in.>ft.2K/==31.8rx5.44–Bucklingaboutthestrongaxisismorecriticalinthisexamplebecauseofthelateralbracingprovidedfortheweakaxis.Therefore,theFavalueisobtainedfromTable1basedonK/>r=54.7.Fa=17.93ksiPallowable=Fa*A=117.93ksi2*A10.3in.2B=184.7kPallowable=184.7k7Pactual=180k180k*100%=97.5%184.7k(veryefficientselection)%stress=Therefore,useW8*35.482ColumnAnalysisandDesign12Afour-storybuildinghasastructuralsteelbeamgirder-columnframingsystem.Columnsarespacedat20feeto.c.inonedirectionand30feeto.c.intheperpendiculardirection.Aninteriorcolumnsupportsatypicaltributaryfloorareaof600squarefeet.Forapreliminarydesign,findaneconomicalW10orW12sectionforaninteriorfirst-floorcolumn.Assumethatthecolumnshaveunsupportedlengthsof14feetandK=1.0.Fy=36ksi.RoofLoads:DL80psfSL30psfFloorLoads:DL100psfLL70psfSolution:Totalroofloads:DL⫹SL80⫹301110psf2*A600ft.2B=66,000#=66kTotalfloorloads:DL⫹LL100⫹701170psf2*A600ft.2B=102,000#=102kperfloorTheloadatthetopoftheinteriorfirst-floorcolumnisaresultoftheroofplusthreefloorloads.Totalloadonthefirst-floorcolumn=PactualPactual=66k1roof2+31102k21floors2=372kTryW10*60.W10*601A=17.6in.2;ry=2.57¿2Theassumptionbeingmadeisthattheyaxisisthecriticalbucklingdirection,becausenoweakaxisbracingisprovided.FloorPlan.11.02114¿*12in.>ft.2K/=65.4=ry2.57–‹Fa=16.9ksiPa=16.9ksi*17.6in.2=297.4k6372kTherefore,thiscolumnsectionisoverstressed.Selectalargersection.483ColumnAnalysisandDesignTryW10*77.W10*771A=22.6in.2,ry=2.60–211.02114¿*12in.>ft.2K/=64.6=ry2.60–‹Fa=16.98ksiPa=16.98ksi*22.6in.2=383.7kPa=383.7k7372k‹OK%stress=372*100%=97%383.7UseW10*77.AcolumndesignusingaW12sectioncanbecarriedoutusinganidenticalprocedure.TheresultingW12sizewouldbeW12*721Pa=377k2Bothofthesesectionsareadequateforstressandefficientinmaterial.However,theW12*72ismoreeconomical,becauseitis5poundsperfootlessinweight.Thefinaldecisiononselectionwillundoubtedlyinvolveissuesconcerningdimensionalcoordinationandconstruction.484ColumnAnalysisandDesignProblemsNote:ThefollowingcolumndesignproblemsassumepinnedendstopandbottomandFy=36ksi.12Selectthemosteconomicalsteelpipecolumn(standardweight)tosupportaloadof30kandalengthof20feet.13WhatisthemosteconomicalW8columnforProblem12?14Selectanappropriatesteelcolumnsection24feetlong,bracedatmidheightabouttheweakaxis,thatsupportsaloadof350k.UseaW14section.(SeeExampleProblem11.)15Asix-storybuildinghasastructuralsteelbeamcolumnframethatisappropriatelyfireproofed.Thecolumnsarespaced20feeto.c.inonedirectionand25feeto.c.intheperpendiculardirection.Atypicalinteriorcolumnsupportsatributaryfloorareaof500squarefeet.Thegoverningbuildingcodespecifiesthattheframemustbedesignedtowithstandthedeadweightofthestructureplusaroofsnowloadof40psfandaliveloadoneachfloorof125psf.Thedeadweightoftheroofisestimatedtobe80psf,andeachflooris100psf.Theunsupportedlengthoftheground-floorcolumnis20feet,andthecolumnsattheotherfloorlevelsare16feet.Designatypicalinteriorthird-floorcolumnandthefirst-floorcolumnusingthemosteconomicalW12sectionateachlevel.485ColumnAnalysisandDesign4AXIALLYLOADEDWOODCOLUMNSWoodcolumnsarecommonlyfoundsupportingbeamsandgirders,whichinturnsupporttributaryareasofroofandfloorloads.Otherstructuralmembers,suchasbridgepiers,compressionchordsofatruss,orthestudsinaloadbearingwall,subjectedtocompressionarealsodesignedusingthesamemethodsutilizedforbuildingcolumns.AsdiscussedinSection1,longcolumnswilltendtobuckleundercriticalload,whileshortcolumnswillfailbythecrushingoffibers.Forwoodcolumns,theratioofthecolumnlengthtoitswidthisjustasimportantasitisforsteelcolumns.However,inwoodcolumns,theslendernessratioisdefinedasthelaterallyunsupportedlengthininchesdividedbytheleast(minimum)dimensionofthecolumn(Figure19).Figure19columns.Slendernessratioofwoodslendernessratio=Ldmin.=Ld1whered16d2Woodcolumnsarerestrictedtoamaximumslendernessratio:L…50dKL…200usedwhichisapproximatelythesameasthermin.forsteelcolumns.AlargerL>dratioindicatesagreaterinstabilityandatendencyforthecolumntobuckleunderloweraxialload.Figure20RecommendeddesignKecoefficientvaluestoaccountforendfixity.486TheeffectivelengthofsteelcolumnswasdeterminedbyapplyingaKfactor(seeFigure16)totheunsupportedlengthofthecolumntoadjustforendfixity.Similareffectivelengthfactors,calledKeinwoodcolumns,areusedtoadjustforthevariousendconditions.Infact,therecommendeddesignKevaluesareidenticaltothoseofsteelcolumnsexceptforacolumnwithapinnedbaseandrigidtopconnection,susceptibletosomesidesway(Figure20).ColumnAnalysisandDesignMostwoodconstructionisdetailedsuchthattranslation(sidesway)isrestrainedbuttheendsofthecolumnarefreetorotate(i.e.,pinconnection).TheKevalueisgenerallytakenas1.0,andtheeffectivelengthisequaltotheactualunsupportedlength.Evenifsomefixitymayexistinthetoporbottomconnection,itisdifficulttoevaluatethedegreeoffixitytoassumeindesign.Therefore,Ke=1.0isanacceptableassumptionthatisusuallyabitconservativeinsomeconnectionconditions.Woodcolumnscanbesolidmembersofrectangular,round[Figure21(a)],orothershapes,orspacedcolumnsbuiltupfromtwoormoreindividualsolidmembersseparatedbyblocking[Figure21(b)].Becausethemajorityofallwoodcolumnsinbuildingsaresolidrectangularsections,theanalysisanddesignmethodsexaminedinthissectionwillbelimitedtothesetypes.Amorethoroughtreatmentforthedesignofwoodelementsisusuallycoveredinadvancedstructurescourses.Figure21(a)Anexampleofapolecolumn.Figure21(b)Anexampleofaspacedcolumn.In1992,theAmericanForestandPaperAssociation,initsNationalDesignSpecificationforWoodConstruction(NDS–91),approvedanewstandardandincorporatedanewmethodologyandequationsforthedesignofwoodelementsandconnections.Previouscategorizingofwoodcolumnsintotheshort-,intermediate-,orlong-columnrangeresultedinthreedifferentequationsforeachrespectiveslendernessrange.TheNDS–91nowutilizesasingleequation,providingacontinuouscurveovertheentirerangeofslendernessratios(Figure22).Figure22Ylinencolumncurve—allowablestressversusslendernessratio.487ColumnAnalysisandDesignTheallowablecompressivestressforanaxiallyloadedwoodcolumnofknownsizeisexpressedasfc=wherefcPAFc¿====P…Fc¿Aactualcompressivestressparalleltograinaxialcompressiveforceinthemembercross-sectionalareaofthecolumnallowablecompressivestressparalleltograinToobtaintheallowablecompressivestressFc¿,manyadjustmentstothetabulatedbasestressarenecessary.TheNDS–91definestheFc¿asFc¿=Fc1CD21CM21Ct21CF21Cp2where:Fc¿=allowablecompressivestressparalleltograinFc=tabulatedcompressivestressparalleltograin(foundinbuildingcodetables,NDStables,andwooddesignhandbooks)CD=loaddurationfactor(definedlaterinthissection)CM=wetservicefactor(accountsformoisturecontentinthewood)=1.0fordryserviceconditionsasinmostcoveredstructures,dryserviceconditiondefinedas:moisturecontent…19%forsawnlumbermoisturecontent…16%forglu-lamsCt=temperaturefactor(usuallytakenas1.0fornormaltemperatureconditions)CF=sizefactor(anadjustmentbasedonmembersizesused)Cp=Columnstabilityfactor(accountsforbucklingandisdirectlyaffectedbytheslendernessratio)Becausetheobjectiveinthistextistoanalyzeanddesignstructuralelementsinapreliminaryway(ratherthanthefullcomplementofequationsandchecksperformedbyastructuralengineer),theprecedingallowablecompressivestressequationwillbesimplifiedasfollows:Fc¿=Fc*CpwhereFc*=Fc1CD21CM21Ct21CF2FcCD(forpreliminarycolumndesign)ThissimplificationassumesCM,Ct,andCFareallequalto1.0,whichisgenerallythecaseforamajorityofwoodcolumns.488ColumnAnalysisandDesignNowawordabouttheloaddurationfactorCD.Woodhasauniquestructuralpropertyinwhichitcansupporthigherstressesiftheappliedloadsareforashortperiodoftime.Alltabulatedstressvaluescontainedinbuildingcodes,NDS,orwooddesignmanualsapplyto“normal”loaddurationanddryserviceconditions.TheCDvalueadjuststabulatedstressestoallowablevaluesbasedontheduration(time)ofloading.“Normal”durationistakenas10years,andCD=1.0.Short-durationloadingfromwind,earthquake,snow,orimpactallowsCDvalueshigherthan1.0butlessthan2.0(Figure23).ThecolumnstabilityfactorCpmultipliedbyFcessentiallydefinesthecolumncurve(equation)asshowninFigure22.Thisequation,originallydevelopedbyYlinen,explainsthebehaviorofwoodcolumnsastheinteractionoftwomodesoffailure:bucklingandcrushing.Cp=1+1FcE>Fc*22c-Cc1+1FcE>Fc*22c2d-Figure23factors.MadisoncurveforloaddurationFcE>F*ccwhereFcE=EulercriticalbucklingstressforcolumnsKcEE¿KcEE=21/e>d21/e>d22Fc*FcCDE¿=modulusofelasticityassociatedwiththeaxisofcolumnbucklingc=bucklingandcrushinginteractionfactorforcolumns=0.9forglu-lamcolumns=0.8forsawnlumbercolumnsKcE=0.30forvisuallygradedlumber=0.418forglu-lamsThecolumnstabilityfactorCpisdirectlyaffectedbytheEulerbucklingstressvalueFcE,whichinturnisinverselyproportionaltothesquareofacolumn’sslendernessratio.Atable,tosimplifythecomputationsforpreliminarycolumnanalysis/design,couldbecreatedbyinputtingslendernessratiosbetween1and50,whichresultsinFcEvaluesforsawnandglu-lammembers.Then,ifvariousFcEvaluesweredividedbyFc*,generatingratiosofa1FcE>Fc*2,acomputercouldeasilycalculatecorrespondingCpvalues.Table3wascreatedforthispurposeandeliminatesthenecessityoflaboriouscomputationsforCp.489ColumnAnalysisandDesignTable3TabledevelopedandpermissionforusegrantedbyProfessorEdLebert,DepartmentofArchitecture,UniversityofWashington.490ColumnAnalysisandDesignAnalysisofWoodColumnsAsimpleprocedurecanbeadoptedforcheckingtheadequacyorcapacityofwoodencolumns(Figure24).Thismethodologyisforapproximateanalysisandassumesthesimplificationsdiscussedintheprevioussection.Given:Columnsize,columnlength,gradeandspeciesoflumber,andendconditions.Required:Theallowablecapacityofacolumnortheadequacyofagivencolumn.Procedure:a.Calculatethe1/e>d2min.b.ObtainF¿c(allowablecompressivestress)whereFc¿=Fc1CD21CM21Ct21CF21Cp2orFc¿=Fc*CpCompute:FcE=Figure24columns.CheckingthecapacityofwoodenKcEE1/e>d22KcE=0.31sawnlumber2KcE=0.4181glu-lams2c=0.81sawnlumber2c=0.91glu-lams2c.ComputeFc*FcCDFcEd.CalculatetheratioFc*e.EnterTable3;obtainrespectiveCpf.Calculate:Fc¿=Fc*Cp‹Pallowable=Fc¿*A…PactualwhereA=cross-sectionalareaofthecolumn491ColumnAnalysisandDesignExampleProblems:AnalysisofWoodColumns13A6*8DouglasfirNo.1postsupportsaroofloadof20k.Checktheadequacyofthecolumnassumingpinsupportconditionsatthetopandbottom.UseFc=1,000psiandE=1.6*106psi.Solution:6*8S4SDouglasfirNo.1AA=41.25in.2B12¿*12in.>ft./e==26.2d5.5–FcE=0.3E1/e>d22=0.3A1.6*106B126.222=699psiFc*FcCDLoaddurationfactorforsnowisCD=1.15115%increaseinstressabove“normal”condition2‹Fc*=11,000psi211.152=1,150psiThecolumnstabilityfactorCpcanbeobtainedfromTable3byenteringtheratio699psiFcE==0.61Fc*1,150psi‹Cp=0.506Fc¿=Fc*Cp=11,150psi2*1.5062=582psiThen,Pallowable=Pa=Fc¿*A=1582psi2*A41.25in.2B=24,000#Pa=24k7Pactual=20kThecolumnisadequate.492ColumnAnalysisandDesign14An18-foot-tall6*8SouthernPinecolumnsupportsaroofload(deadloadplusaseven-dayliveload)equalto16k.Theweakaxisofbucklingisbracedatapointninefeet,sixinchesfromthebottomsupport.Determinetheadequacyofthecolumn.Solution:6*8S4SSouthernPinePost:1A=41.25in.2;Fc=975psi;E=1.6*106psi2Checktheslendernessratioabouttheweakaxis:19.5¿*12in.>ft.2/e==20.7d5.5–Theslendernessratioaboutthestrongaxisis118¿*12in.>ft.2/e==28.8d7.5–Thisvaluegoverns.FcE=0.3E1/e>d22=0.3A1.6*106psiB128.822=579psiFc*FcCDwhereCD=1.25for7-day-durationload‹Fc*=975psi*1.25=1,219psi579psiFcE==0.47Fc*1,219psiFromTable3,Cp=0.412.‹Fc¿=Fc*Cp=1,219psi*0.412=502psiPa=Fc¿*A=1502psi2*A41.25in.2B=20,700#Pa=20.7k7Pactual=16kips‹OK493ColumnAnalysisandDesign15An11-foot-tallDouglasfirglu-lamcolumnisusedtosupportaroofload(deadloadplussnow)asshown.Apartial-heightwallbracesthe518–direction,andkneebracesfromthebeamsupportthesixinchface.Determinethecapacityofthecolumn.Solution:518–*6–glu-lampost:1A=30.8in.2;Fc=1650psi;E=1.8*106psi2Bucklingintheplaneofthe6–dimension:8¿*12in.>ft./e==16PGovernsd6–Inthe518–direction:8¿*12in.>ft./e==16PGovernsd6–Comparingthebucklingconditioninbothdirections,the6–directionismorecriticaland,therefore,governs.0.418*A1.8*106psiB0.418E=FcE=1/e>d2211622=2,939psiFc*FcCD=1,650psi*11.152=1,898psi2,939psiFcE=1.55=Fc1,898psiFromTable3,Cp=0.883.‹Fc¿=Fc*Cp=1,898psi*10.8832=1,581psiPa=Fc¿*A=1,581psi*A30.8in.2B=48,700#494ColumnAnalysisandDesignDesignofWoodColumnsColumndesigninwoodisatrial-and-errorprocess(Figure25).Startbymakingaquickestimateonsize(tryoutyourintuition),andcheckouttheadequacyorinadequacybyfollowingtheanalysisproceduregivenintheprevioussection.Axiallyloadedwoodcolumnswithoutmidheightbracingaregenerallysquareincross-sectionor,insomecases,justslightlyrectangular.Fortunately,therearefewerpossiblewoodsectionstochoosefromcomparedwiththewidearrayofsizesavailableinsteel.Onedesignprocedureusingthetrial-and-errormethodcouldbethefollowing:Given:Columnlength,columnload,gradeandspeciesoflumbertobeused,andendconditions.Required:Aneconomicalcolumnsize.Figure25Columndesigninwood.Procedure:a.Guessatatrialsize;unlessthecolumn’sweakaxisisbraced,trytoselectasquareoralmostsquarecross-section.b.Followthesamestepsusedintheanalysisprocedureoftheprevioussection.c.IfPallowableÚPactual,thenOK.d.IfPallowable…Pactual,pickalargersizeandcyclethroughtheanalysisprocedureagain.495ColumnAnalysisandDesignExampleProblems:DesignofWoodColumns16A22-foot-tallglu-lamcolumnisrequiredtosupportaroofload(includingsnow)of40k.Assuming834–inonedimension(tomatchthebeamwidthabove),determinetheminimumcolumnsizeifthetopandbottomarepinsupported.Selectfromthefollowingsizes:834–*9–AA=78.75in.2B834–*101012–AA=91.88in.2B834–*12–1A=105.0in.22Solution:Glu-lamcolumn:1Fc=1,650psi,E=1.8*106psi2Try834–*1012–834–*1012–AA=91.9in.2B/edmin.FcE==22*12=30.26508.750.418E1/e>d22=(max.slendernessratio)0.418A1.8*106B130.222=825psiFc*FcCD=1,650psi11.152=1,898psiFcE825==0.43Fc*1,898FromTable3,Cp=0.403Fc¿=Fc*Cp=1,8981.4032=765psiPa=Fc¿*A=765psi191.9in.22=70,300#740,000#Cycleagain,tryingasmaller,moreeconomicalsection.Try834–*9–.834–*9–1A=78.8in.22Becausethecriticaldimensionisstill834–,thevaluesforFcE,F*c,andF¿callremainthesameasintrial1.Theonlychangethataffectsthecapabilityofthecolumnistheavailablecross-sectionalarea.‹Pa=Fc¿A=1765psi2178.8in.22=60,300#Pa=60.3k740kUse834–*9–glu-lamsection.496ColumnAnalysisandDesignProblems16A66S4SSouthernPine(DenseNo.1)columnisusedtosupportheadersthatcarryloadsfromroofjoists.Determinethecapacityofthecolumnassumingpinconnectionsatthetopandbottom.Assumeaseven-daydurationroofliveload.17An8*8S4Sfirst-floorcolumnsupportsaloadofP1=20kfromroofandfloorsaboveandanadditionalsecond-floorloadP2=12k.Determinetheadequacyofthecolumnassuminga“normal”loadduration.634–*1012–18Determinetheaxialloadcapacityofaglu-lamcolumnAA=70.88in.2Bassuminglateralbracingabouttheweakaxisatthemidheightlevel.Assumepinconnectionstopandbottominbothdirectionsofbuckling1Fc=1,650psi;E=1.8*106psi2.497ColumnAnalysisandDesign19Aninteriorbearingwallinthebasementofaresidenceutilizes24S4Sstudsspacedat16incheso.c.tosupportthefloorloadabove.Sheathingisprovidedonbothsidesofthewallandservestopreventbucklingabouttheweakaxisofthemember.Determinethepermissibleloadω(inpoundsperlinealfoot)assumingHem-Fir(joist/planks).Then,usingtheωvaluecomputed,determinethebearingstressthatdevelopsbetweenthestudandsoleplate.20A48S4SDouglasfircolumnsupportsaroofbeamasshown.Thebeam-columnconnectionisconsideredtobeapin.Thelowerendispinnedforbucklingaboutthestrongaxisbutisfixedintheweakaxisbythepartialwallthatmeasures2feetinheight.DeterminethetributaryareathatcanbesupportedbythiscolumnifDL=20psfandSL=30psf.21Determinetheminimumsizecolumn(SouthenPineDenseNo.1)requiredtosupportanaxialloadofP=25kassuminganeffectivecolumnlength/e=16¿.22Aninteriorglu-lamcolumnsupportsaroofloadofP=15k.Thetotalcolumnheightis24feet,butkneebracingfromthebeamsreducestheunsupportedheightto18feet.Assuming634–inondimension,determinetheminimumsizerequired.UseFc=1,650psiandE=1.8*106psi.Selectasizefromthefollowing:634–*712–AA=50.63in.2B634–*9–AA=60.75in.2B634–*1012–AA=70.88in.2B498ColumnAnalysisandDesign5COLUMNSSUBJECTEDTOCOMBINEDLOADINGORECCENTRICITYSofar,theprevioussectionshaveassumedcompressionmemberssubjectedtoconcentricloading(loadsactingthroughthecentroidofthecolumnsection).Thestudyofaxiallyloadedcolumns(Figure26)wasessentialtotheunderstandingoftheprimaryissueofslendernessanditsrelationshiptofailuremodesinvolvingcrushingandbuckling.Inpractice,however,concentricloadingisrarelythecase.Thissectionwillintroducetheideaofeccentricity(Figure27)and/orsideloading(Figure28)andtheireffectoncolumnbehavior.Figure26column.Concentrically(axially)loadedFigure27Eccentricallyloadedcolumn.Figure28sideload.ColumnwithcompressionplusManycolumnsaresubjectedtobendingincombinationwithaxialcompressionloads.Nonuniformbearing,misalignmentoftheframing,oreventhecrookednessofamemberwillcausealoadtomissthecentroidofthecolumncross-section.Compressionmemberscarryingbendingmomentduetoeccentricityorsideloadinginadditiontocompressionarereferredtoasbeamcolumns(Figure29).Anassumptionthatwasmadeforaxiallyloadedcolumnswastherelativeuniformityofthestressdistributionoverthecross-sectionalarea,asshowninFigure30(a).Bendingstress,whichinvolvestensionandcompressionstresses,mustbeaddedalgebraicallytothecompressivestressesfromgravityloading.Ifabeamisveryflexibleandthecolumnisveryrigid,theeccentricityeffectwillbesmall,becausemostofthebendingstresswillbeabsorbedbythebeam.Relativelysmalleccentricitiesalterthefinalstressdistribution,butthecross-sectionwillremainincompression,althoughnonuniform,asshowninFigure30(b).Ontheotherhand,ifarigidbeamisconnectedtoalessrigidcolumn,aconsiderablylargeeccentricitywillbetransmittedtothecolumn.Whenlargeeccentricitiesexist,tensilestressesmaydevelopoverpartofthecross-section,asshowninFigure30(c).Thetensionstressesthatdevelopedinmasonryconstructionofthepastwereformerlyofgreatconcern,buttheyhavelittlesignificanceforthebuildingsystemsandmaterialsusedtoday.Timber,steel,prestressedconcrete,andreinforcedconcreteallpossessgoodtensilecapability.ThemasonryconstructionoftheGothiccathedralsrequiredakeenawarenessoftheresultantforce(fromtheflyingbuttressandtheverticalpier)remainingwithinazone(themiddlethird)ofthecross-sectiontoavoiddevelopingtensilestresses.Theareawithinthethirdpointofeachfaceofthepieriscalledthekernarea(Figure31).499ColumnAnalysisandDesign(a)Framedbeam(shear)connection.e=Eccentricity;M=P*e(b)Momentconnection(rigidframe).M=Momentduetobeambending(c)Timberbeam–columnconnection.e=d>2=eccentricity;M=P*e(d)Upperchordofatruss—compressionplusbending.ω/2M=8Figure29Examplesofbeam-columnaction.(a)Axiallyloaded—uniformcompressivestress.Figure30500(b)Smalleccentricity—linearlyvaryingstress.Stressdistributionforeccentricallyloadedrectangularcolumns.(c)Largeeccentricity—tensilestressonpartofcross-section.ColumnAnalysisandDesignBeamcolumnsareevaluatedusinganinteractionequationthatincorporatesthebendingstresswiththecompressivestress.ThegeneralinteractionequationisexpressedasfaFa+fbFb…1.(interactionequation)wherefa=P>A[theactualcompressive(axial)stress]Fa=allowablecompressivestress[basedonK/>r(steel)or/e>d(timber)]MMcfb==(actualbendingstress)ISM=P*eforeccentricallyloadedmembersM=bendingmomentduetosideloadorrigidframeactionFb=allowablebendingstress(valuesfromtables)Figure31Kernareasfortwocross-sections.Ifamemberissubjectedtoaxialcompressionandbendingaboutboththexandyaxes,theinteractionformulaisadaptedtoincorporatethebiaxialbending(Figure32).Therefore,themostgeneralizedformoftheequationisfaFa+where:fbx=fby=fbxFbx+fbyFby…1.0(forbiaxialbending)M=actualbendingstressaboutthexaxisSxM=actualbendingstressabouttheyaxisSyWhenbothbendingandaxialforcesactonamember,themagnitudeoftheaxialstressisexpressedasacertainpercentageoftheallowableaxialstress,andthebendingstresswillbeacertainfractionoftheallowablebendingstress.Thesumofthesetwopercentagescannotexceedunity(100%stress).TheinteractioncurveshowninFigure33illustratesthetheoreticalcombiningoftheaxialcompressiveandbendingstresses.Figure32Exampleofbiaxialbendingduetotwoeccentricloads.501ColumnAnalysisandDesignFigure33Interactioncurveforcompressionandbending.Bendingmomentsincolumns,whethertheyresultfromlateralforces,appliedmoments,oreccentricityoftheendloads,cancauseamembertodeflectlaterally,resultinginadditionalbendingmomentduetotheP-¢effect(seeFigure5).InFigure33,aslendercolumndeflectsanamount¢duetoasideload.However,thelateraldisplacementgeneratesaneccentricityfortheloadP,whichresultsinthecreationofadditionalmomentatthemidheightofthecolumnequaltoP*¢(knownasasecondorderbendingmomentandalsoasamomentmagnification).SlendercolumnsareparticularlysensitivetothisP-¢effectandmustbeaccountedforintheinteractionequation(Figure34).TheAISC(steel)andNDS(timber)manualshaveintroducedamagnificationfactortoincorporatetheP-¢effect,whichresultsfromtheinitialbendingmoment.AgeneralizedrepresentationforbothsteelandwoodisfaFa+fb1magnificationfactor2Fb…1.0Theactualanalysis/designequationforsteelasspecifiedbytheAISCisexpressedasCMyfbyfaCMxfbx++fafaFaa1bFbxa1bFF¿exF¿eybywhere1=magnificationfactortoaccountforP-¢1-fanFe¿Figure34502P-¢effectonaslendercolumn.Fe¿=12π2E231K/>r22=Euler’sformulawithasafetyfactorCM=modificationfactor,whichaccountsforloadingandendconditionsColumnAnalysisandDesignAnequivalentequationspecifiedbytheNDSforanalyzinganddesigningwoodcolumnssubjectedtocompressionanduniaxialbendingisfcfc¿+wherefcFc¿fbxf¿bx3-1fc>FcEx24…1.0=equivalentofthesteelratioforcompressionfc=P=actualcompressivestressAFc¿=FcCDCMCtCFCp=allowablecompressivestressfbx=M=bendingstressaboutthexaxisSxFbx¿=allowablebendingstressaboutthexaxis1C1-fc>FcExD=magnificationfactorforP-¢effectAnalyzinganddesigningbeamcolumnsusingtheAISCandNDSequationsaremoreappropriatelydoneinfollow-upcoursesdealingspecificallywithsteelandwooddesign.Oversimplificationoftheprecedingequationsdoesnotnecessarilyresultinappropriateapproximations,evenforpreliminarydesignpurposes.Thistextdoesnotincludeproblemsinvolvingtheuseoftheinteractionequation.503ColumnAnalysisandDesignSummary■■fa=Columnslendernessgreatlyinfluencesacolumn’sabilitytosupportload.Short,stoutcolumnsfailbycrushingduetomaterialfailure,whilelong,slendercolumnsfailbybuckling.Buckling(elasticinstability)isthesudden,uncontrolledlateraldisplacementofacolumn.Shortcolumnsareevaluatedbasedonthebasicstressequation:1actualload2Pactual=…Fa1allowablecompressivestress2A(cross-sectionalarea)■Pcritical=π2EIminL2=ThecriticalbucklingloadofaslendercolumnwaspredictedbyLeonhardEuler.π21modulusofelectricity21smallestmomentofintertia21columnlengthbetweenpinnedends2Asthecolumnlengthincreases,thecriticalloaddecreases.Euler’sequationcontainsnosafetyfactor.Pcriticalrepresentstheactualfailureload.■Thelengthofacolumnbetweenpinnedendsdividedbythecolumn’sleastradiusofgyrationisknownastheslendernessratio.slendernessratio=Lrmin.Highslendernessratiosmeanlowercriticalstressesthatresultinbuckling.Columnsectionswithhighradiiofgyrationraremoreresistanttobuckling.■■Euler’sbucklingequationassumespinnedends.Ifacolumnsupportatthetopand/orbottomisrestrainedfromrotation,theload-carryingcapacityofacolumnincreases.TheslendernessratioofacolumnisadjustedbyaKfactorto
incorporatetheendsupportconditions.slendernessratio=KLrwhereKisthemultiplierforconvertingtheactuallengthtoaneffectivebucklinglengthbasedonendconditions.■504Acolumn’seffectivenesscanbeincreasedbytheintroductionoflateralbracingabouttheweakaxis.Lateralbracingreducestheeffectivebucklinglengthofacolumn,therebyresultinginahigherloadcapacity.ColumnAnalysisandDesign■■■■Structuralsteelcolumnsareclassifiedasshort,intermediate,andlong.ThetwoAISCallowablecompressivestressformulasrecognizeonlytwoslendernesscategories—short/intermediateandlong.Shortcolumnsareexpressedasafunctionofslendernessandtheyieldstrengthofthesteel.ThelongcolumnformulaisamodifiedversionoftheEulerbucklingformulawithasafetyfactorincorporated.Theslendernessofrectangulartimbercolumnsisevaluatedbyslendernessratio=■■■Ldmin.=1columnlength21minimumcolumndimension2Mostwoodconstructionassumesthattheendsarepinnedandfreetorotate1K=12.TheanalysisanddesignofwoodcolumnsgenerallyfollowtheprovisionsestablishedbytheNationalDesignSpecificationforWoodConstructionmanual.Manycolumnsusedinactualbuildingsaresubjectedtobendingincombinationwithaxialcompressiveloads.ThegeneralconditionofstressisexpressedasfaFa+fbFb…1.01interactionformula21actualcompressivestress21allowablecompressivestress2+1actualbendingstress21allowablebendingstress2…1.0AnswerstoSelectedProblems1Pcritical=184.2k;fcritical=20.2ksi3L=28.6’5KL/ry=193.2;Pcr=73.64k;fcr=7.7ksi7(a)KL/ry=46.5;Pa=356k(b)KL/ry=57.2;Pa=339k(c)KL/ry=71.5;Pa=311k9KL/rz=111.25;Fa=11.5ksi;Pa=46k11Weakaxis:KL/ry=54;Pa=561.3k;Strongaxis:KL/rx=57;Pa=553k13W8×24;KL/r=149;Pa=47.6k15Roofload=60k;floorload=112.5kPactual=397.5k(3rd-floorcolumn)W12×79(Pa=397.6k)Pactual=622.5k.(1st-floorcolumn)W12×136(Pa=630k)16Le/d=30.5;Fc=459.5psi;Pa=13.9k18Strongaxiscontrolsthedesign;Le/d=25.15;Fc=1,021psi;Pa=72.4k20Weakaxis:Le/d=21.9(governs)Strongaxis:Le/d=16.6;Pa=17.54k;Atrib.=350.8ft.221Use8×8;Pa=32k>25k505506StructuralConnectionsFromChapter10ofStaticsandStrengthofMaterialsforArchitectureandBuildingConstruction,FourthEdition,BarryOnouye,KevinKane.Copyright©2012byPearsonEducation,Inc.PublishedbyPearsonPrenticeHall.Allrightsreserved.507StructuralConnectionsIntroductionAnexaminationwasmadeoftheloadpathsthroughastructuralassemblage,beginningwiththeroof,floors,walls,andcolumnsandeventuallytracingtheloadstothefoundation.ForcesandsupportreactionswereobtainedforeachstructuralelementthroughtheuseofFBDsandtheequationsofequilibrium.Later,theconstructionofload,shear,andmomentdiagramsenabledustodeterminethecriticalshearsandmomentsusedinthedesignofbendingmembers,whichincludejoists,rafters,beams,andgirders.Beamandgirderreactionswerethenusedinthedesignoftimberandsteelcolumns.Figure1SirHumphreyDavy(1778–1829).Thecolleagueand,insomeways,therivalofThomasYoung,DavywasappointedProfessorofChemistryattheRoyalInstituteattheageof24inthesameyearthatYoungwasappointed.Davy,incontrasttoYoung,exhibitedadynamicandcaptivatingpublicpersonalityinhislectures,whichgarneredbothmoneyandpublicityfortheInstitute.Davywasextremelyprolificinhiswork,isolatinghimselfandconductingextensiveresearch.Hediscoveredtheexhilaratingandanestheticeffectsofnitrousoxide(laughinggas)aswellaspioneeringinthefieldofelectricalengineering.Oneofthemajorcontributionstothemodernworldofconstructionwashisdevelopmentofarcweldingintheearly1800s.DavyremainedattheInstituteforthedurationofhisprofessionalcareerandprospered.HeeventuallywasknightedandbecameSirHumphrey.HealsoservedaspresidentoftheRoyalSociety.508Individualstructuralmembersweredesignedasisolatedelements,assumingthatthetransferofloadsfromonemembertothenextwasassured.Inreality,however,allstructuralassemblagesareinterconnected,andindividualmembersworkincombinationwithothermembersthroughphysicalconnections.Structuralmembersmustbejoinedinsuchawaythatpermitsthesafetransferofloadsfromonemembertoanother.Manyofthestructuralfailuresinbuildingsandbridgesoccurintheconnectionofmembersandnotinthemembersthemselves.Thetechniquesusedbycarpenters,welders,andironworkersinmakingaspecificjointcancreatesituationsthatarenotamenabletomathematicalcomputations.Animproperlyinstalledbolt,undersizedorimproperstrengthratingofthebolts,boltholesthataretoolarge,weldsmadewithimproperpenetrationoftheweldmaterial,impropertighteningofabolt,andsooncanallleadtounsatisfactorystructuralattachments.Toprecludesuchbadjoints—oratleastlimitthenumberofthemthatoccur—variousstandardcodesareineffect.Timberdesign,includingitsconnections,isoverseenbytheAITCandtheNDSbytheAmericanForestandPaperAssociation.Insteeldesign,theAISC’sManualofSteelConstructionandthespecificationsoftheAmericanWeldingSociety(AWS)arethemostcomprehensiveandmostwidelyusedinindustry.StructuralConnections1STEELBOLTEDCONNECTIONSBoltsusedinbuildingconnectionsaregenerallysubjectedtoforcesthatcausetension,shear,oracombinationofthetwo.Typicalconnectionssubjectingfastenerstoaxialsheararesplicesusedintrusses,beam-lapsplices,andgussets(Figure2).Bracketsandwebshearsplicesaretypicaleccentricshearconnections(Figures3and4).Boltsintensionarecommoninhangerconnections(Figure5).Typicalbeam-to-columnmomentconnectionsareexamplesofcombinedtensionandshearforces(Figure6).Figure2Rooftrusswithatensionsplice—axialshear.Figure3shear.Bracketconnection,millbuilding—eccentricFigure4Beamsplice—eccentricshear.509StructuralConnectionsTheconnectionshowninFigure6(a)isnolongerrecommendedbecauseofitsdifficultyinfabricationandexpense.Weldinghassimplifiedthemomentconnectionconsiderably,asshowninFigure6(b).Manymomentconnectionstodayutilizeacombinationofshopweldingwithfieldboltingtosimplifytheerectionprocessaswellaskeepcostdown.Figure5Hangerconnection—boltsintension.(a)Boltedtee-stubconnection—moment.Figure6510Boltsinshearandtension.(b)Typicalmomentconnectionwithboltsandwelds.(c)Kneebraceconnectionintensionandshear.StructuralConnectionsFailureofBoltedConnectionsThechoiceofsuitabledesignstressesforboltedjointsisnotasimpletask.Considerationmustbegivenbothtothepossiblemodesoffailureofagivenjointandtothebehaviorofmaterialsundersuchloading.Inaddition,themethodsoffabricatingaparticularjointmayinducelatentlocalstressesorphysicalconditionsthatcouldcausethejointtofail.However,thedesignofaboltedjointinmanycasesisacomparativelysimplematter,andthecomputationsaregenerallynotveryinvolved.Indesigningaboltedjointproperly,onemustanticipateandcontrolthemaximumstressesdevelopedatcriticalsections.Becauseitistobeexpectedthatfailurewilloccuratoneofthesecriticalsections,anyknowledgeastowherethesesectionsmaybelocatedprovidesinformationforasuccessfuldesign.Severalcommoncriticalstressconditionsdevelopinalljoints,eachofwhichiscapableofproducingfailure.ShowninFigures7through12arethefivebasictypesoffailure:1.Shearofthebolt(Figures7and8).2.Bearingfailure(crushing)oftheconnectedmembersagainstthebolt(Figure9).3.TensionfailureoftheconnectedmembermaterialFigure7Shearfailure—singleshear(one(Figure10).shearsurface).4.Endtear-outoftheconnectedmember(Figure11).5.Blockshear,aconditionthatcanoccurwhenthetopflangeofthebeamiscopedtointersectwithagirder(Figure12).Boltedjointsareusuallycategorizedbythetypeandcomplexityofthejoint.Themostcommonlyencounteredboltedjointsarethelapjointandthebuttjoint,shownschematicallyinFigures7and8.ShearfailureTheresistingshearstressdevelopedisonlyanaverageshearstress.Shearstressesarenotuniformontheboltcross-sectionbutareassumedsofordesign.Singleshear:fv=P…Fvn*A(ave)where:fvPAFvn=====averageshearstress(psiorksi)loadontheplateorconnection(#ofkips)cross-sectionalareaofboltAin.2BAllowableshearstressofbolt(psiorksi)numberofbolts511StructuralConnectionsDoubleshear:fv=(ave)P…Fv2A*nwherefvP2AFvn=====averageshearstress(ksiorpsi)loadsontheplateorconnection(#orkips)twocross-sectionalareasperboltAin.2BAllowableshearstress(ksiorpsi)numberofboltsFigure8Doubleshear—twoshearplanestoresist.Bearingfailure(Figure9)Bearingfailureinvolvestheconnectingmembersandnottheboltitself.ThecontactsurfaceoftheboltpushingontheplateistakenasAp=n*t*dwhereAptdnFigure9512Bearingfailure.====bearingareaofplateAin.2Bplatethickness(in.)boltdiameter(in.)numberofboltsinbearingStructuralConnectionsTheaveragebearingstressdevelopedbetweentheboltandplatecanbeexpressedasP…FpApfp=(ave)wherefp=averagebearingstress(psiorksi)ApPFpFpFu=====bearingareaAin.2Bappliedload(#orkips)allowablebearingstress(psiorksi)1.2Fuultimatestress(psiorksi)TensilefailureTheshadedareainFigure10representstheamountofmaterialleftacrosstheteartoresisttheappliedloadafterdeductingthevoidleftbythebolthole:Anet=1b-n*D2twhereAnet=netplatearearesistingtensionAin.2Bb=platewidth(in.)D=diameterofastandardsizebolthole.AD=boltdiameter+161–Bt=platethickness(in.)n=numberofboltsinarowTheaveragenettensionintheplateacrossthefirstlineofboltsisexpressedasft=(net)P…FtAnetFigure10Tensilefailure.513StructuralConnectionswhereft=averagetensionstressacrossthenetareaofthemember(psiorksi)P=appliedload(#orkips)Ft=allowabletensilestressacrossthenetareaFt=0.5FuFu=ultimatetensilestress(psiorksi)Endtear-out(Figure11)andblockshear(Figure12)arepossiblefailuremodeswhenhigh,allowablefastenervaluesareusedinconjunctionwithrelativelythinmaterial.However,failureislesslikelytooccurbecauseofthedesignspecificationsrequiringampleedgeandenddistancesandminimumboltspacing.Figure11Endtear-out.Figure13illustrates,forpreliminarydesignpurposes,theminimumpitchandedgedistancesspecifiedbytheAISCtoavoidthesefailuremodes.TheAISCspecifiesanabsoluteminimumpitchandgage(center-to-center)spacingof223timestheboltdiameter(d),with3dbeingthepreferredspacing.Insomecases,adimensionofthreeinchesisusedforallsizesofboltsuptooneinchindiameter.Shearandbearingfailures,theresultofexcessivestressesonthebolts,areavoidedbyprovidingasufficientnumberoffastenerstokeepthestresseswithintheallowablelimits.Figure12Blockshear—tensionfailureattheendconnectionsalongtheperimeterofagroupofbolts.Figure13514Minimumpitchandedgedistance.StructuralConnectionsDesignStressesforBoltsBoltinghasreplacedrivetingasameansofstructurallyconnectingmembers,becauseitisquieter,simpler,andcanbeperformedmorequicklywithsmallercrews,resultinginlowerlaborcosts.Possiblythemostcommonconstructionmethodologytodayistoutilizeshopweldingincombinationwithfieldbolting.Threeprimarytypesofboltsareusedinsteelconstructiontoday.Commonbolts,alsoknownasroughboltsorunfinishedbolts,aredesignatedasAmericanSocietyofTestingMaterials(ASTM)A307andaresuitableforuseinlightsteel-framestructureswherevibrationandimpactdonothavetobeconsidered.Also,unfinishedboltsarenotpermittedundersomebuildingcodesintheconstructionofbuildingsexceedingcertainheightlimits.Wheretheseconditionsexist,nutsmaybecomeloose,impairingthestrengthoftheconnection.Thetwoothertypesofbolts,designatedasASTMA325andA490,arehigh-strengthboltsarethemostwidelyusedfastenersforstructuralsteelconnectionsmadeinthefield(Figure14).High-strengthboltsareidentifiedonthetopoftheheadwiththelegendA325orA490andthemanufacturer’smark.Boltsareavailableinavarietyofsizes,from58–to112–indiameter.However,themosttypicallyusedsizesinbuildingconstructionare34–and78–.Larger-diameterboltsgenerallyrequirespecialequipmentaswellasincreasedspacingandedgedistancesforproperboltplacement.AllowableAISCshearcapacitiesarelistedinTable1,andAISCallowablebearingvaluesarecontainedinTable2.Figure15showstwoplatesheldtogetherbyahighstrengthboltwithanutandtwowashers.Whenahighstrengthboltistightened,averyhightensileforceTdevelopsinthebolt,thustightlyclampingtheconnectedpartstogether.ItistheresultingfrictionforceSthatresiststheappliedload;unlikeunfinishedbolts,thereisnoactualshearorbearingstress.WhenthejointloadPexceedsthemagnitudeS,slipoccurs.Ifthereisslipbetweentheplates,theedgesoftheplatesarebroughtincontactwiththeshankofthebolt,therebyproducingbearingandshearingstressesinthebolt.Ifthereisnoslip,theloadPistransmittedfromoneplatetotheotherbythefrictionalresistance.Figure14ASTMA490tension-controlboltwiththeflutedtipthattwistsoffatthepropertightness.AvailableinA325andA490forhigh-strengthbolting.Imagereprintedwithpermission,courtesyoftheLeJeuneBoltCompany.Figure15High-strengthboltinsingleshear.Theoretically,thebolt’sshankandthesurfacesoftheplatesthroughwhichitpassesarenotincontactatall,becausetheholesarepunchedslightlylargerthanthebolt.Evenwhenaconnectionofthistypeissubjectedtovibration,thehighresidualtensilestresspreventsloosening.Mechanicallyfastenedconnectionsthattransmitloadbymeansofshearintheirfastenersarecategorizedaseitherslip-critical(SC)orbearing-type(NorX).Slip-criticalconnections,alsoformerlyreferredtoasfriction-typeconnections,dependuponsufficientlyhighclampingforcetopreventslipoftheconnectedpartsunderanticipatedserviceconditions.515StructuralConnectionsBearing-typeconnectionsarebasedonthecontact(bearing)betweenthebolt(s)andthesidesoftheholestotransfertheloadfromoneconnectedelementtoanother.Beforetensionisappliedtohigh-strengthboltsthroughtightening,thejointsurfacesadjacenttoboltheads,nuts,orwashersmustbefreeofscale,burrs,dirt,andotherforeignmaterial.Figure16Tension-controlwrench.Imagereprintedwithpermission.CourtesyoftheLeJeuneBoltCompany.Tighteningofhigh-strengthboltsinslip-criticalconnectionsisaccomplishedbyspecialtorqueorimpactwrenchestoatensionequalto70%ofthespecifiedminimumtensilestrengthofthebolt.TheResearchCouncilonStructuralConnectionsprovidesfourmethodsofcontrollingbolttension:(a)thecalibratedwrench,(b)thetwistoff-typetension-controlbolt,(c)thedirecttensionindicator,and(d)theturn-of-nutmethod,whichrequiresspecifiedadditionalnutrotationaftertheboltsare“snugtight.”Snugtightisdefinedasthefulleffortofanironworkerwithanordinaryspudwrenchthatbringstheconnectedpliesintofirmcontact.Whenboltsarenotfullytensioned,theconnectionsaredesignedasbearing-typeconnections,whereasmallamountofmovement(slippage)isexpected(Figures16and17).ThehighclampingforceproducedbytheproperlytightenedA325andA490boltsissufficienttoassurethatslippagebetweentheconnectedpartswillnotoccuratfullallowablestressinslip-criticalconnectionsandprobablywillnotoccuratworkingloadsinbearing-typeconnections.Theallowablestressinbearing-typeconnectionsisbasedonafactorofsafetyof2.0ormore,whichoveralongperiodofobservationhasbeenfoundtobeadequate.Thissafetyfactorishigherthanthatusedforthedesignoftheconnectedmember.TheallowableshearstressesforA325andA490boltsusedinstandardroundholesareasfollows:Figure17“Snugging”theboltsandnutswithatension-controlwrench.Imagereprintedwithpermission.CourtesyoftheLeJeuneBoltCompany.A325-SCFv=17ksi(slip-criticalconnection,standardroundhole).A325-NFv=21ksi(bearing-typeconnectionwiththreadsincludedinshearplane).A325-XFv=30ksi(bearing-typeconnectionwiththreadsexcludedfromshearplane).A490-SCFv=22ksi(slip-criticalconnection,standardroundhole).A490-NFv=28ksi(bearing-typeconnectionwiththreadsincludedinshearplane).A490-XFv=40ksi(bearing-typeconnectionwiththreadsexcludedfromshearplane).SeeTable1fortheallowableAISCshearstressofotherbolts.516StructuralConnectionsTheefficiencyofthreadedfastenersinresistingshearinbearing-typeconnectionsisreducedwhentheboltthreadsextendintotheshearplane(s)betweentheconnectedplates(Figures18and19).Allowableshearstressvaluesremainthesameforslip-criticalconnectionsthathaveboltthreadsincludedintheshearplanes.ASTMA325andA490high-strengthboltsarepurposefullyproportionedandproducedsothatthethreadscanbeexcludedfromtheshearplaneswhenitisdesirable.Severaltypesofholesaretypicallyusedforboltedconnections(Figure20):(a)standardroundholes,(b)oversizedholes,(c)short-slottedholes,and(d)long-slottedholes.Figure18plane.ThreadsexcludedfromtheshearFigure19threads.ShearplanepassingthroughFigure20Typicallyusedholetypes.1Standardroundholesaremade16–largerthanthediameterofthebolt.Insteel-to-steelstructuralconnections,standardroundholescanbeusedinmanyjointapplicationsand,insomecases,arepreferred.Forexample,standardroundholesarecommonlyusedingirderandbeamconnectionstocolumnsasameansofcontrollingthecenterto-centerdimensionbetweencolumnsandtofacilitatethealignmentofthecolumntoaplumbposition.Oversizedandslottedholesaretypicallyusedinthefieldtoreducethefit-upandassemblytimeduringerection.3–largerOversizedholeshavenominaldiametersupto167thanbolts8–andlessindiameter,whileboltswitha1–diameterwillhaveahole14–larger.Oversizeholesarepermittedinslip-criticalconnectionsonly.1–widerthantheboltdiameterandShort-slottedholesare16havealengththatdoesnotexceedtheoversizeholedi1–.Thistypeofholecanbeusedmensionsbymorethan16ineitherbearing-typeorslip-criticalconnections,butifusedinbearing-typeconnections,theslotsmustbeperpendiculartothedirectionofload.1–widerthantheboltdiameterandLong-slottedholesare16havealengthnotexceeding212timestheboltdiameter.Thistypeofconnectionmaybeusedinslip-criticalconnectionswithoutregardfortheloaddirection.However,inbearing-typeconnections,theloadmustbeperpendiculartotheslotdirection.Slottedholesareparticularlyusefulwheresomeamountoffieldadjustmentisnecessary.Long-slottedholescanonlybeusedinoneoftheconnectedmembersatajoint;theothermembermustuseastandardroundholeorbewelded.Allowableshearstressvaluesforboltsinslip-criticalconnectionsremainunchangedfortheoversizedandshortslottedholes.However,allowableshearstressvaluesdecreasewhenusinglong-slottedholesforloadsappliedperpendiculartotheslot,andanevenlargerreductionoccursforloadsappliedparalleltotheslot.517StructuralConnectionsTable1Copyright©AmericanInstituteofSteelConstruction,Inc.Reprintedwithpermission.Allrightsreserved.518StructuralConnectionsTable2Copyright©AmericanInstituteofSteelConstruction,Inc.Reprintedwithpermission.Allrightsreserved.519StructuralConnectionsExampleProblems1DeterminetheallowableloadcapacityoftheconnectionshowninFigure21if78–φA307unfinishedboltsareusedwithstandard-sizeroundholes.AssumetheplatesareA36steel.Solution:Figure21Typicalbuttsplice.Threepossiblefailuremodescouldoccurinthistypicalbuttsplicecondition.Shear,bearing,andnettensionwillbecheckedtodeterminethecriticalconditionthatgovernstheconnectioncapacity.Shear—DoubleShearTakeasectioncutthroughtheconnectionatthebuttsplice,anddrawanFBDforone-halfoftheassembly.Theboltspassthroughthreeplatesandarethussubjectedtodoubleshear.ThegeneralequationfordeterminingtheshearcapacityofthisconnectionisPv=Fv*AvBoltsindoubleshear.whereFv=10ksi(seeTable1)Av=2bolts*2doubleshear*aπ*d2b472ab8Av=2*2*Jπ*K=2.41in.24Pv=Fv*Av=10k>in.2*24.1in.2=24.1kAnotherwayofobtainingthesameresultbutminimizingsomeofthecomputationsistouseTable.1,wheretheactualloadcapacitiesofcommonlyusedboltsizesandgradesaregivenforbothsingleanddoubleshear.‹Pv=2bolts*12k>bolt=24kBearing(indoubleshear)Unfinishedboltsarecheckedforbearinginwhich,assumingstandardroundholes,theallowablestressistakenasFp=1.2FuwhereFu=58ksiforA36steelBearingstressonplate.520Fp=1.2*158ksi2=69.6ksiThecenterplateiscritical.StructuralConnections‹Pp=2bolts*A38–*78–B*69.6k>in.2=45.7kOr,usingtheAISCallowablebearinginTable2,Pp=2bolts*22.8k>bolt=45.6kRememberthatbearingfailureisintheplatematerialbeingconnectedandnotintheconnector(bolt).NetTension—AtConnectionNettensionresultsinthetearingoftheplateduetoinsufficientmaterial(cross-section)toresistthetensionstress.Thenumberandplacementofboltsinarowacrosstheconnectiongreatlyinfluencesthesusceptibilityoftheplatetonettensionfailure.Ft=0.5Fu=0.5158ksi229ksiAnet=A38–B*A312–-1516–2B=0.96in.2Pt=0.96in.2*29k>in.=27.8kTensiononthegrossareaoftheplate(inaregionbeyondtheconnection):Nettensionstress.Agross=A38–B*A312–B=1.31in.2Pt=Ft*AgrossandFt=0.6Fy=0.6136ksi2=22ksi‹Pt=1.31in.2*22k>in.2=28.2k(gros)Becausetheshearcheckresultedinthesmallestallowablevalue,itgovernsthecapacityoftheconnection.Tensionstressonthegrossareaofbar.‹Pallowable=24k521StructuralConnections2ThebuttspliceshowninFigure22usestwo8*38–platesto“sandwich”inthe8*12–platesbeingjoined.Four78–φA325-SCboltsareusedonbothsidesofthesplice.AssumingA36steelandstandardroundholes,determinetheallowablecapacityoftheconnection.Solution:Figure22Buttspliceconnection.Shear,bearing,andnettensionwillbecheckedtodeterminethecriticalconditionthatgovernsthecapacityoftheconnection.Shear:UsingtheAISCallowableshearinTable1,Pv=20.4k>bolt*4bolts=81.6k1doubleshear2Bearing:UsetheAISCallowablebearingvaluefoundinTable2.Thethinnermaterialwiththelargestproportionalloadgoverns;therefore,the12–centerplategoverns.Assumetheboltsareata3dspacing,centertocenter.SectioncutA–A.Pb=30.5k>bolt*4bolts=122kTension:Thecenterplateiscritical,becauseitsthicknessislessthanthecombinedthicknessofthetwoouterplates.Holediameter=1boltdiameter2+=Pt=Ft*Anet1516–116–=78–+116–whereFt=0.5Fu=0.5158ksi2=29ksiPt=29k>in.2*3.06in.2=88.7kThemaximumconnectioncapacityisgovernedbyshear.Pallow=81.6k522StructuralConnections3AsimpletrussconnectionisaccomplishedusingA325-Nboltsinstandardroundholes.DeterminethesizeoftheboltsrequiredfortheloadconditionshowninFigure23.Figure23Typicaltrussconnectionwithgussetplate.Solution:Eachtrussmemberwillbeexaminedindividuallytodeterminetheminimumnumberofboltsrequired.Shearandbearingwillbecheckedineachdesign.However,nonettensioncomputationwillbemade,becausethedoubleanglesandgussetplatehavelargecross-sectionalareas.DiagonalMembersAandBShear:Doubleshear,twobolts:30k=15k>bolt2boltsUsingTable1,2-34–φA325-N1Pv=2*18.6k>bolt=37.2k2Bearing:The38–gussetplateiscriticalinbearing.Abolt=15k>boltFp=15k69.6k>in.2=0.216in.2Bearingarea:Ap=d*t=0.216in.2d=0.216in.2=0.576in.3–.8523StructuralConnectionsTwo58–φboltsarenecessaryforbearing.Sheargovernsthedesign;therefore,usetwo34–φbolts.HorizontalMemberCTheunbalancedloadis30kinthehorizontaldirection.P=30kThedesignloadisthesameasformembersAandB;therefore,two34–φA325-Nboltsarerequired.However,itisusefulinpracticetoprovideanoddnumberoffastenerssuchthattheintersectionofthelinesofforcefromthetwodiagonalsoccurinthecenterofthehorizontalmembers’boltpattern.Thistendstoreducethepossibilityofundueeccentricityattheconnection.Also,itisadvisabletomaintainthesameboltsizethroughoutaconnectiontominimizeerrorsresultingfromboltsubstitution.Therefore,usethree34–A325-Nbolts.4Forthethree-rowboltedbuttjointshowninFigure24,determinetheloadthatcanbecarriedbasedonshear,bearing,andtension.AssumeA325-SCboltsinstandardroundholes.Figure24Atypicalboltpattern.Solution:Shear:Six34–φboltsindoubleshear(seeTable1):Pv=6bolts*15k>bolt=90kBearing:Thecenterplateiscriticalinbearing.UsingTable2,Pp=6bolts*26.1k>bolt=156.6kNetTension:Thetensioncapabilityoftheplate(center)willbecheckedacrossthethreerowsofbolts.Thisparticulartypeofboltarrangementissometimesusedtoreducethepossibilityofnettensionfailure.Theideaistohavetheloadtransferthroughtherowsofbolts,diminishingtheforceprogressivelyforeachsubsequentrowofbolts.524StructuralConnectionsSection1Ft=29ksiHolediameterD=34–+116–=0.8125–1–*19–-0.813–2=4.1in.22(acrossonebolt)Anet=Pt1=Ft*Anet=29k>in.2*4.1in.2Section2Theboltacrosssection1reducesthetotaltensileloadoccurringatsection2.Therefore,thetensilecapacityofsection2willincludetheshearcontributionofoneboltfromsection1.1–*19–-2*0.813–2=3.69in.221acrosstwobolts2Anet=Pt2=129k>in.2*3.69in.22+15k(1boltinshear)=92kSection3Thissectionwillincludetheshearcontributionofboltsinsections1and2.Thenetareaoftheplateacrosstherowofboltsatsection3isAnet=1–*19–-3*0.813–2=3.28in.22Pt3=29k>in.2*3.28in.2+3115k2=140.1kBasedonexaminingtheconditionsofshear,bearing,andnettensionacrossthreedifferentsections,thecapacityoftheconnectionisgovernedbyshear:Pallow=Pv=90k525StructuralConnectionsProblems1DeterminetheallowableloadPpermittedforthisdoubleshearjointconnectionassumingA36steelandA325-SCboltsinstandardroundholes.2Theverticalsteelbarshownis38–thickandmustbedesignedtowithstandatensileloadP=28k.TwoA325-Xboltswillbeused.AssumingA36materialandstandardroundholes,calculatethefollowing:a.Therequireddiameterdofthebolts.b.TherequiredwidthWofthebar.3Aconnectionofthetypeshownusesthree34–φA325-X(STD)boltsintheupperconnectionandtwo78–φA325-X(STD)boltsinthethreeinchbar.WhatisthemaximumloadPthatthisconnectioncansupport?Note:Thisisanacademicexercise.Generally,itisnotadvisabletousedifferent-sizedboltsinthesameconnection.4Determinethenumberofboltsnecessaryforeachmemberframingintothetrussjointshown.Boltsare34–φA325-X(NSL),andmembersareA36steel.526StructuralConnections5Determinethecapacityofthisbuttsplicebasedonshear,bearing,andnettension.TheplatesaremadeofA36steel,andthefourboltsoneachsideofthespliceareA325SCwithstandardroundholes.6Asuspensionbridgeoverariverusesasystemoflinkedbarsconnectedasshownforthemainsuspensionsystem.AssumingA36steelandA490-Xbolts,determinethemaximumloadPthatthesystemcancarry.Checkforshear,bearing,andnettensionattheboltandtensionofthemember.StandardFramedBeamConnectionsStandardAISCtablesareavailabletocoverthedesignofavastmajorityoftypicalstructuralconnectionswherefillerbeamsframeintogirdersorgirdersframeintocolumns(Figures25and26).Thistypeofstandardshearconnectionconsistsoftwoclipanglesplacedbacktobackoneithersideofthebeamorgirderweb.Whenabeamframesintoagirdersuchthattheuppersurfacesofthetopflangesareatthesameelevation,thetermflushtopisused.Toaccomplishthis,itisnecessarytocutawayaportionoftheupperflangeasshownintheillustrationontherightinFigure25.Thisisknownascoping,orblocking,andforeconomy,itshouldbeavoidedwheneverpossible.Figure25Typicalbeam-girdershearconnection.527StructuralConnectionsFramedbeamconnectionsaregenerallydesignedforshear,bearing,andwebtear-outorblockshear(forbeamswiththetopflangecoped).AsampleAISCtableisshowninTable3foruseindesigningslip-criticalandbearingtypeconnectionsbasedonshearcapacityforstandardsizeholes.OtherAISCtables(notincludedinthistext)areusedtocheckthebearingandwebtear-outcapacities.Figure26Figure27angles.528Beamcross-sectionwithclipStandardframedbeam-columnconnection.Table3hasprovisionsforbolttype,boltsize,holetype,numberofbolts(usingathreeinchpitchdimension),anglethickness,andlength.High-strengthbolts,ineitherslip-criticalorbearing-typeconnections,assumeadoubleshearconditionthroughthebeamwebandasingleshearattachmenttothecolumnflangeorgirderweb.Clipanglethicknessandlengtharedependentonthefastenersize,themagnitudeoftheappliedload,andspacelimitationwithinthebeam’sflanges.Anglesmustbeabletofitbetweentheclearanceoftopandbottomflangefillets(Figure27).Anglelengthsaregenerallyatleastone-halfofthebeam’sdepthtoprovidesomeresistancetoendrotationatthebeam’send.StructuralConnectionsTable3Copyright©AmericanInstituteofSteelConstruction,Inc.Reprintedwithpermission.Allrightsreserved.529StructuralConnectionsTable3Continued.Copyright©AmericanInstituteofSteelConstruction,Inc.Reprintedwithpermission.Allrightsreserved.530StructuralConnectionsExampleProblems5UsingtheAISCframedbeamconnectionboltshearinTable3,determinetheshearadequacyoftheconnectionshowninFigure28.Whatthicknessandanglelengtharerequired?Endbeamreaction=60k.Figure28Typicalbeam-columnconnection.Solution:Enteraboltdiameterof34–φ(fromTable3),A325-Ntypefasteners,andn=4boltsindoubleshearthroughthebeamweb.Theshearallowable=74.2k760k.Theconnectionisadequateinshear.5Anglesare16–thickandhavealengthofL=1112–.BecausetheanglesarelessthanthecleardimensionTbetweentheflangefillets,thereshouldnotbeaproblemoffit.6DeterminethenumberofboltsrequiredfortheconnectioninFigure29basedonsheariftheendreactionis120k.Whatistherequiredanglethicknessandlength?Doestheanglefitwithintheflanges?Solution:FromTable3,thisconnectionrequiresfive78–φA325-Nboltsthroughthebeamwebandten78–φboltsthroughthegirderweb.Figure29Typicalbeam-girdershearconnection.Shearcapacity=126k7120k.Anglesare38–thickand1412–long.BecauseT=15.5–isgreaterthanL=14.5–,theconnectionanglesshouldfitadequatelybetweentheflanges.531StructuralConnectionsProblems7Abeam-to-girderconnectionisboltedusingtwoclipanglesandfiveA490-Xboltsasshown.Thebeamreactionisequalto210kips.AssumingA36steelandthreeinchboltspacing,determinetheboltdiameterrequired,theclipanglethickness,andtheanglelength.8Astandardbeam-columnframedconnectionusesA36steelwith34–φA325–SCboltsatthreeinchspacing.Fortheconnectionshown,determinethefollowing:a.Themaximumallowableshearcapacityfortheconnection.b.Thenumberofboltsrequired.c.ThelengthLoftheclipangle.532StructuralConnections2WELDEDCONNECTIONSWelding,asordinarilyconsideredforstructuraluse,maybedefinedasamethodofjoiningmetalsbyfusionwithouttheapplicationofpressure.Themetalatthejoint,togetherwithadditionalmetalsuppliedintheformoffillermetal(fromanelectrode),ismelted,formingasmallpoolorcrater.Uponcooling,theweldandbasemetalformacontinuousandalmosthomogeneousjoint.ManyweldingprocessesarerecognizedbytheAWS,butforstructuralsteelusedinbuildingconstruction,arcweldingisthemethodgenerallyused.Forthistext,thetermweldingreferstoarcwelding,inwhichthefusionprocessoccursbythegenerationofheatfromanelectricarc.ArcweldingwasfirstmadepossiblebythediscoveryoftheelectricarcbySirHumphreyDavy(seeFigure1atthebeginningofthechapter)earlyinthe19thcentury.Healsodevelopedthemethodologyofstartingandmaintaininganelectricarc.Figure30Theweldingcircuit.Figure31Shieldedmetalarcwelding.Electricarcweldingrequiresapowersourceconnectedinacircuitthatincludesagroundcabletothepiecebeingweldedand,ontheelectrodecable,theelectrodeholderandelectrode(Figure30).Asustainedarcisformedbetweentheworktobeweldedandtheelectrodeinagap,completingtheelectricalcircuit.Theresistancefromtheairorgasinthegaptransformstheelectricalenergyintoheatatextremelyhightemperatures(approximately6,500°Fattheelectrodetip).Intenseheatisgeneratedbythearcinwhichthebasemetalandelectrodefillermetalliquefy(atatemperatureinexcessof3,000°F)intoapool(calledacrater).Asthemoltenmetalcoolsandsolidifies,themetalsarejoinedintoametallurgicallysolid,homogeneouspiece.Shieldedmetalarcwelding(Figure31)isusedtocontroltheoxidationofthemoltenpooltopreventporosityinthemetal(causingembrittlement)andtocontrolthemeltingoftherodformoreeffectivepenetrativepower.Shieldinginmanualarcweldingisgenerallyaccomplishedthroughtheuseofachemicalcoatingontheelectrode,whileautomaticweldingfrequentlyusesapowderedfluxtosubmergethearcandprotectthemoltenmetalfromair.Inrecentyears,greatadvanceshavebeenmadeinautomaticandsemiautomaticweldingprocessessothatmanualweldingtodayisgenerallylimitedtoshortweldsandfieldwelding(weldingdoneatthesite).Averycommonpracticeformakingastructuralconnectionistoshopweldaconnectingdevice—clipangles,bearingplates,andsoon—toonememberandthenfastenthroughboltingtoaconnectingmemberinthefield.Insomeinstances,suchasinmoment-resistingconnections,fullyweldedjointsareareasonableoption.Becauseweldedmemberscanbeattachedtogetherformomentcapabilitywithoutusingconnectingplatesorangles(Figures32and33),theweldedconnectionisusuallysimpler,ismorecompact,andrequiresasmallercrew.Figure32Boltedtee-stubmomentconnection—notrecommendedfornewconstruction.533StructuralConnectionsHolesforboltsareavoided;therefore,thegrosssectionratherthanthenetsectionisusedtodeterminethecrosssectionalareaofmembersintension.Figure33Typicalweldedmomentconnection.Onoccasion,problemswillariseintheapplicationofweldingtostructuralconnections.Theselectionofawrongelectrode,useofanimproperamperage/voltagesettingontheweldingmachine,toorapidacoolingrateoftheweld,anddevelopmentofinternalstressesfromdifferentialcoolingaresomeofthefactorsaffectingproperweldments.Inthepast,theseconsiderationswereprimarilytheconcernofthewelder.However,withtheinceptionofbettermethodsandstandardization,muchofthisresponsibilityhasnowbeenshiftedfromtheweldertotheAWSweldingcode.Seriouslyflawedworkhasbeensubstantiallyeliminatedbyrequiringeachweldertopassrigidqualificationtestsandsubmithisorherworktothecarefulscrutinyofatrainedinspector.Tofurthertestthesafetyofweldedjoints,ultrasonictestingandmagneticparticleinspectionareoccasionallyusedtolocateinternalflaws.Designersofstructuralconnections,whetherboltedorwelded,shouldalwaysbeawareoftheactualconditionsduringtheerectionproceduretofacilitatethatprocessandtoprovideforaneconomicalsolution.Awidevarietyofconnectiontypesandcombinationsarepossible,andanexperienceddesignerismostdesirableindeterminingapracticalandeconomicalconnection.TypesofWeldedJointsTherearevarioustypesofweldedjointsincommonuse.Theselectionoftheappropriatetypeisafunctionofthemagnitudeoftheloadatthejoint,thedirectionoftheappliedload,theconfigurationofthejoint,thedifficultyofthejointpreparation,andthecostoferection.Filletwelds(Figure34)andgroove(butt)welds(Figure35)arethetwomostcommonlyusedweldtypesinbuildingconstruction.Figure35534Typicalgroove(butt)welds.Figure34Typicalfilletwelds.StructuralConnectionsOnoccasion,plugandslotwelds(Figure36)areusedforspecialcircumstances.Thefollowingdiscussionwillbelimitedtoload-carryingfilletandgroovewelds.CommonsymbolsusedfordesignatingthetypeofweldisshowninTable4.AppropriateweldsymbolsareindicatedforfilletandgrooveweldsinFigures34and35.Filletweldsandgrooveweldsdifferprimarilyfromthemannerinwhichthestresstransfertakesplace.Grooveweldsarenormallyindirecttensionorcompression(Figure37),whereasfilletweldsaregenerallysubjectedtoshearaswellastensionorcompression(Figure38).Figure36Figure37(b)Full-orpartial-penetrationgrooveweldsdevelopfullcompressivecapabilityofthesection.Thestrengthofafull-penetrationgrooveweldisproportionaltoitscross-sectionalareaandthestrengthofthefillermetal.BecausethefillermetalfromtheelectrodeexceedsthestrengthoftypicalA36steelbasemetal,thegrooveweldisstrongerthanthebasematerialinshear,tension,andcompression.Thestrengthforfull-penetrationgrooveweldsisconservativelyassumedtobeequaltothatofthebasematerial.Inotherwords,agrooveweldofthesamecross-sectionastheconnectedmembersisassumedtobe100%efficientintransferringstress.Ifagrooveweldweremadewithanincompletepenetration,itsstrengthwouldhavetobereducedinaccordancewiththeweldingcodeused.Grooveweldsaregenerallyusedforstructuralassembliesinwhichfull-strengthweldsaremandatory.Theyrequirerelativelylargeamountsofweldmetalandcansometimesexperienceproblemsduringtheweldingprocess.Grooveweldsalsorequirethecuttingofstructuralmemberstomoreorlessexactlengthsfortheendstobutt,andtheynecessitateextensiveedgepreparation.Asaresult,grooveweldsaremoreexpensivetoproducethanfilletwelds.Plugandslotwelds.Figure37(a)Full-penetrationgroovewelddevelopsthefulltensilecapabilityoftheplate.Figure38(a)Filletweldsresistinshear.Figure38(b)Filletweldresiststensionthroughshearacrossthroat.535StructuralConnectionsTable4Copyright©AmericanInstituteofSteelConstruction,Inc.Reprintedwithpermission.Allrightsreserved.536StructuralConnectionsThefilletweldisoneofthemostcommonlyusedwelds.Itistheweldbywhichsteelfabricatorsjoinplatematerialtomakebuilt-upbeamsandgirdersand,morefrequently,tojoinbeamstocolumnsortogirders.Eventhoughgrooveweldspossessgreaterstrengththanfilletwelds,moststructuralconnectionsarejoinedbyfilletwelding.Filletweldsallowforgreaterfit-uptolerancesandgenerallyrequirenoedgepreparationbeforewelding.Theultimatestrengthofafilletweldisdependentuponthedirectionoftheappliedload,whichisparallel(longitudinal)ortransversetotheweld.Experimentshaveshownthattheendsofafilletweld,lyingparalleltothelineofactionoftheload,carryhigherunitstressesthanthemidportionoftheweld,asillustratedinFigure39.Also,whenanendweldiscombinedwithlongitudinalwelds,theunitstressintheend(transverse)weldwillbeapproximately30%higherthanthoseintheside(longitudinal)welds;however,thisfactisnotrecognizedbymostdesignspecifications.Figure39Longitudinalandtransversestressesinfilletwelds.Infilletwelds,withatheoreticallytriangularcrosssection,thecriticalstressisassumedtobeactingontheminimumthroatarea,regardlessofthedirectionoftheappliedload.Thethroatofafilletweld(Figure40)ismeasuredfromtheroot(insidevertexofthetriangle)tothetheoreticalfaceoftheweld.ThethroatisequaltotheproductofthetheoreticalthroatTandtheweldlength.Shear,bending,andaxialforcesallcauseshearstresses(acrossthethroat)infilletwelds.Filletweldsaregenerallyspecifiedwithequallegs,andthelengthoftheselegsisconvenientlyusedtorepresentthesizeoftheweld.TheeffectivethroatTthicknessofanequal-leg45°filletweldisconsideredasT=0.707*weldsizeThecompatibleandmostcommonlyusedelectrodesforweldingA36steelaretheE60XXandE70XX,wheretheprefixEdenoteselectrodeandthefirsttwodigitsindicatetheultimatetensilestrengthinthousandsofpoundspersquareinch.Forexample,anE70XXelectrodehasanultimatetensilecapacityof70ksi.Thenext-to-lastdigitindicatestheweldposition(Figure41)inwhichtheelectrodeiscapableofmakingsatisfactorywelds.Forexample,E701XE702XFigure40Partsofafilletweld.AllpositionsFlatpositionandhorizontalfilletsFilletweldstrengthisbasedontheallowableshearstressfortheweldmetalacrosstheeffectivethroatarea.TheAISCspecificationslimittheallowableshearstressontheeffectiveareato30%ofthenominaltensilestrengthoftheweldmetal.Therefore,forA36steelandE60XXandE70XXelectrodes,Fv=0.30*60ksi=18ksi1E60XX2Fv=0.30*70ksi=21ksi1E70XX2537StructuralConnectionsWeldstrengthsperinchofweldforanysizeequal-legweldcanbefoundbymultiplyingtheweldsizeby0.707timestheallowableshearstress.Throatdimension:T=0.707*weldsizeThroatarea:Athroat=T*lengthofweldFora1–weld:Athroat=T*1–weldstrengthperinch=throatarea1At2*allowableshearstress1Fv2Table5isincludedforquickercomputationsinvolvingfilletwelds.Table5AllowablestrengthoffilletweldsperinchofweldFigure41position.TypesofweldsbasedonweldWeldSize(in.)E70XX(k/in.)316–14–2.7838–5.5712–7.4234–11.133.71516–4.64716–6.4958–9.27Inadditiontofilletweldstrengthbasedonthesizeandlengthoftheweld,otherweldingcodeprovisionsareaddressedfullyintheAISC’sManualofSteelConstructionAllowableStressDesignandthestructuralcodeoftheAWS.Someoftheothercodeitemsarethefollowing:■538Themaximumsizeofafilletweldappliedtoasquareedgeofaplateorsection14–ormorein1thicknessshouldbe16–lessthanthenominalthicknessoftheedge.Alongedgesofmateriallessthan14–thick,themaximumsizemaybeequaltothethicknessofthematerial.StructuralConnections■Theminimumsizeofafilletweldisdependentonthethickeroftwomembersbeingweldedbutcannotexceedthethicknessofthethinnermember.Theminimumsizeoffilletweldsis18–formaterialwith3–foramaterialthicknessathicknessof14–orless,16over14–to12–,14–formaterialthicknessover12–to34–,5–formaterialthicknessover34–.and16■■■■Theminimumeffectivelengthoffilletweldsshouldbefourtimesthenominalsize,orelsetheweldsizeistobetakenas14–ofitseffectivelength.Iftwoormoreweldsareparalleltoeachother,thelengthmustbeatleastequaltotheperpendiculardistancebetweenthem(Figure42).Figure42welds.MinimumlengthforparallelFigure43Endreturnsforfilletwelds.Theminimumlengthofintermittentfilletweldsshouldbenotlessthanfourtimestheweldsize,withaminimumof112–.Sideorendfilletweldsterminatingattheendsorsidesshouldbereturned,ifpractical,aroundthecornersforadistancenotlessthantwotimesthenominalweldsize(Figure43).Addedstrengthisgiventofilletweldswithendreturns.Costconcernsalsodictatetoalargeextentthesizeofthefilletweldtobeused.Weldmetalvolumehasadirectcorrelationtothelaborcostsinvolvedindepositingtheweld.Themosteconomicalweldminimizesweldmetalvolumeand,atthesametime,reducestheheatinputandtheassociatedshrinkageanddistortionofthejoint.Minimizingtheweldmetalalsominimizesthepotentialforweld5–orless,defects.Filletweldsizesshouldbekeptto165becausethe16–weldisthelargestweldsizethatcanbedepositedinonepasswiththeshieldedmetalarcprocessinthehorizontalandflatpositions.Largerfilletweldsgenerallyrequiretwoormorepasses.Inpractice,itisalsoadvisabletomaintainthesame-sizefilletthroughouttheconnection.Achangeoffilletsizenecessitatesachangeofweldingrodsand,therefore,slowstheworkandmaycauseerrors.539StructuralConnectionsExampleProblems7DeterminethecapacityoftheconnectioninFigure44assumingA36steelwithE70XXelectrodes.Solution:Capacityofweld:5Fora16–filletweld,S=3.98k>in.Weldlength=22–Weldcapacity=22–*3.98k>in.=87.6kFigure44Platefilletweldedonthreesides.Capacityofplate:Ft=0.6Fy=22ksiallowplatecapacity=38–*6–*22k>in.2=49.5k‹platecapacitygoverns,Pallow=49.5kTheweldsizeusedisobviouslytoostrong.Towhatsize,then,cantheweldbereducedsothattheweldstrengthismorecompatibletotheplatecapacity?TomaketheweldcapacityLplatecapacity,22–*1weldcapacityperin.2=49.5kweldcapacityperinch=49.5k=2.25k>in.22in.3–weld1S=2.78k>in.2.FromTable5,use16Minimumsizefillet=316–basedona38–thickplate.8Determinethesizeandlengthoflongitudinalfilletweldsthatwilldevelopthestrengthofthesmallerplate(Figure45).AssumeA36steelwithE70XXelectrodes.Solution:platecapacity=4–*38–*22k>in.2=33kFigure45Parallelfilletwelds.Maximumweldsize(limitedbytheplatethickness):weldsize=38–-116–=516–Note:Thisisgood,becausethisweldsizecanbedepositedinonepass.Allowableweldstrength:S=4.64k>in.totalweldlengthrequired=54033k=7.1in.4.64k>in.StructuralConnectionsRoundingupwardtothenearest14–,useatotalof714–of55516–weldor16–*38–oneachside.Note:TheAWSspecifiesthattheweldlengthoneachsideoftheplateforparallelweldsshouldnotbelessthantheperpendiculardistancebetweenthewelds.Thisrequirementistoensurefulldevelopmentoftheplatecapacity.‹Use4–minimumofweldlengthoneachsideoftheplate.Asmallerweldsizecanbetriedbecausemoreweldlengthisrequired.Try:14–weldwithS=3.71k>in.totalweldlengthrequired=33k=8.9in.3.71k>in.‹Use14–*412–weldoneachside.9Determinethecapacityofthefull-penetrationgrooveweldshowninFigure46.AssumeA36steelwithE70XXelectrodes.Solution:Full-penetrationgrooveweldscarrythefullcapacityoftheplate.‹Pt=38–*4–*22k>in.2=33kFigure46weld.Transversefull-penetrationgroove541StructuralConnectionsEccentricityinWeldedJointsOneofthemostcommonexamplesofeccentricallyloadedweldedjointsisthatofastructuralangleweldedtoagussetplateasshowninFigure47.TheloadPintheangleispresumedtoactalongitscentroidalaxis.Consequently,becauseanangleisanasymmetricalcross-section,theweldsmarkedL1andL2aremadeunequalinlengthsothattheirstresseswillbeproportionalinaccordancewiththedistributedareaoftheangle.Figure47Angleweldedtoagussetplate.WritingtheequationsofequilibriumfortheappliedforceandweldresistanceR1andR2,3©Fx=04R1+R2=PorS1L1+S2L2=PIfthestrengthoftheweld(Table5)isdefinedasS,3©M0=04R1*a=R2*borS1L1a=S2L2bandiftheweldsizeisconstant,S1=S2whereL1=542L2baStructuralConnectionsExampleProblem516–anglelargegussetplateby41–filletDeterminethelengthsL1and10AnL3*2*(A36steel)isattachedtoaweldssshowninFigure48.L2tosupportatensileloadof30k.Assumetheangletobesubjectedtorepeatedstressvariations(minimizeeccentricity).Figure48Weldedconnectionofangleironandgussetplate.3©Fx=04R1+R2=30k3©M0=04R111.98–2=R211.02–2R1=1.02R2=0.515R21.98Substituting:FDBoftheangleironwelds.0.515R2+R2=30kR2=19.8kandR1=10.2kFor14–filletwelds(E70electrode),Weldstrength:S=4.64k>in.R1=S*L1=3.71k>in.1L12Use:14–*234–weld.R2=S*L2=3.71k>in.1L22Use:14–*538–weld.Checkthetensilecapacityoftheangle:Pt=A*Ft=1.46in.2*22k>in.2=32.1k730k‹OK543StructuralConnectionsProblemsAssumeineachproblemthatthebasemetalisA36steelandthattheelectrodesareE70XX.9Determinethemaximumload-carryingcapacityofthislapjoint.10Determinetheshearcapacityofthefilletweldshown.11WhatlengthLisrequiredtodevelopthefullcapacityoftheplate?12Determinethecapacityofthe14–filletweldconnectionshown.Whatwouldthecapacityoftheconnectionbeifafull-penetrationgrooveweldwereusedinstead?13Computethelengthandsizeofthefilletweldneededtodevelopthefulltensilestrengthoftheangle.Useafull-transversefilletweldontheendandbalancedweldsonthesides(forminimizingeccentricity).544StructuralConnections3COMMONFRAMINGDETAILSINSTEELStructuralanalysistheoryassumesthatconnectionsbetweenbeamsandcolumnsorbetweencolumnsandfoundationsarerigid(fixed),allowingnorelativerotationbetweenconnectedelementsortruepins,hingedwithnomomentresistance.Inreality,structuraldesigndealswithconnectionsthatfitneitheroftheseassumptionsfully.Mostconnectionsexhibitsomedegreeofmomentresistanceandavaryingdegreeofrotationaljointresistance.Thethreebasiclateralresistingsystemsforsteel-framedstructuresaretherigidframe(Figure49),thebracedframe(Figure50),andtheshearwallsystem.Rigidframesystemsrequiretheuseofrigid,momentresistingconnections,whilebracedframesaregenerallydesignedaspinconnections.Materialandlaborcostsaregenerallymuchlowerinpinconnectionsascomparedtorigid,momentconnections.©Aardvark/AlamyThetruebehavioroftheconnection(i.e.,momentandrotationalcharacteristics)affectsthestrengthandstabilityoftheindividualconnectedelementsandthestabilityoftheentirebuildingstructure.Properdetailingoftheconnectionstoprovideacontinuousloadpaththroughtheinterconnectedelementsisessentialinassuringthattheresistingforceswilldevelopinthephysicalconnectiontoprovideoverallstructuralstability.Considerationofalateralresistingstrategyforastructuralframeworkshouldoccurintheearlyplanningstagesofaproject.Selectionofalateralresistingsystemgreatlyinfluencesthedesignofindividualmembersandtheirconnections.Figure49stability.ExposedrigidframeforlateralExamplesofConnectionDetails©Caro/AlamyItisparticularlyuseful,whenstudyingstructuralsteelframing,toexaminesomeofthestandardortypicaldetailsofconnectionsusedinthepreparationofstructuraldrawings.TheaccompanyingdrawingsinFigures51throughFigure55showstandarddetails,inaverybasicandgeneralform,thatarefrequentlyemployedindetaildrawingsofsteelbuildings.SelecteddetailsarereferencedtothehypotheticalstructureshowninFigure51.Theexamplestructureassumesalateralstrategythatutilizesarigidframesystemworkinginconjunctionwithacentrallylocated,concreteshearcore.Lateralstabilityisachievedthroughtheuseofabracedframe-shearwallsystemintheperpendiculardirection.Toillustratethevarioustypesofconnectionsemployedwiththedifferenttypesofbracing,acompositebracedframehasbeencreated.Thisisnotatypeofbracedframesystemthatonewouldnormallyseeinanactualbuildingframe.Figure50Concentricbracinginlowertwofloorsandeccentricbracinginuppertwostories.545StructuralConnectionsFigure51Steel-framedbuildingutilizingarigidframewithcoreshearwallsinthelongitudinaldirectionandabracedframewithcoreshearwallsinthetransversedirection.546StructuralConnectionsRigidFramesRigidframesareconstructedwithbeamandcolumnsrigidlyattachedusingmoment-resistingconnections.Therigidframederivesitsstrengthtoresistgravityandlateralloadsfromthemomentinteractionbetweenbeamsandcolumns(asshowninFigure52).Beam-columnconnectionsmaintaintheirrelative90°orientationtoeachotherunderload,eveniftheconnectionassemblyrotatesasaunit.Rigidframe-beam/columnandsupportRigidbeam/columnconnectionandrigidbases.Allmembersshareinresistingthelateralforcethroughbending.Two-hingedframe-pinnedbaseMostframesassumepinconnectionsatthebasesincefootingsaresusceptibletosomedegreeofrotation.Figure52Rigidframewithbeam-columninteraction.Columns,beams,girders,andjointsareresponsiblefortransferringthehorizontal,vertical,androtational(moment)forcesthroughouttherigidframe.Becausebendingmomentsaresharedbybothbeamsandcolumns,membersizesaregenerallyheavierthanwouldbefoundinbracedframesystems.Stabilityofthestructureismaintainedbytheusuallyhighstiffness1Ix2requiredinthecolumnandbeamorgirder.Therefore,membersaretypicallyorientedtotakeadvantageofitsstrongaxis.Beam-columnconnectionsoftenconsistofashearconnectionforgravityloadsactingincombinationwithfieldweldedbeamflangesformomentresistance(Figure53).Three-hingedframeThree-hingedframe;onehingeatthebeam’smid-spanThree-hingedframepitchedroofRigidframewithapitchedroof.Largebendingmomentsdevelopatthehaunch(pitched-rafterandcolumnjoint).Figure53Examplesofsingle-bay,single-storyrigidframes.Comparedwithbracedframesorshearwallstructures,rigidframeshavetheadvantageofprovidingunobstructedspace.However,underhighseismicloading,thelargedeformationsencounteredmaycausedistresstothearchitecturalfinishes.Rigidframesystemsareeffectiveforlow-tomedium-risebuildings.547StructuralConnectionsFigure54Commonrigidconnectiondetailsforbeam-column,columnsplice,andcolumn-foundation.548StructuralConnectionsBracedFramesBracedframesfoundinusetodayincludeX-bracing,K-bracing(VeeorinvertedVee),andeccentricbracing.Twobasiccategoriesofbracedframesaretheconcentricallyandeccentricallybracedframes.Diagonalbracing,X-bracing,andK-orVee-bracingareclassifiedasconcentricallybraced.Themembersofaconcentricallybracedframeactasaverticaltrusssystem,anddiagonalmembersaregenerallyassumedtoactprimarilyintensionandsometimescompression.DiagonaltensionX-bracingistypicallyanalyzedashavingonlytensionforces.Thisdesignassumptionutilizesonlyone-halfofthememberstoresistlateralloads,whiletheadjacentmemberwithinthesamepanelisassumedtobenegligibleinresistingcompressivestress.K-bracingisusedindesigncircumstanceswhenaccessthroughthebracingplaneisrequired.TheinvertedVee(K-bracing)allowsclearancefordoorways,corridors,androoms.Bracingmembersineccentricallybracedframes,asshowninFigure55(a)and55(b),areconnectedtothebeamsoastoformashort“linkbeam”betweenthebraceandthecolumnorbetweentwoopposingbraces.Linkbeamsactasa“fuse”topreventotherelementsintheframefrombeingoverstressed.Inlow-to-moderategroundshaking,aneccentricallybracedframeperformsasabracedframeratherthanasamomentframe.Therefore,thestructureexperiencessmallerlateraldisplacements,minorarchitecturaldamage,andnostructuraldamage.Inmajorearthquakes,thelinkbeam(Figures56and57)isspecificallydesignedtoyield,therebyabsorbinglargequantitiesofseismicenergyandpreventingbucklingoftheotherbracingmembers.Bracedframesaremorecosteffectivewhencomparedtorigidframes.∆PlastichingesFigure55(a)brace.DiagonaleccentricFigure55(b)EccentricK-brace.Figure56(a)Detailofaneccentricallybracedframeconnection.EndstiffenersIntermediatestiffenersBeamFigure57Linkbeamrotation.BracesFigure56(b)Linkbeamdetail.549StructuralConnectionsAsummaryofthevariouslateralresistingsystemsinsteelisshowninFigure58.Figure58Lateralresistingstrategyusingaconcentricallyandeccentricallybracedframe.550StructuralConnectionsSummary■Fivebasictypesoffailurethatcancauseacriticalstressconditionatajointare■■■■■■Therearethreeprimarytypesofboltsusedinsteelconstructioncurrently:■■■■■■■■■Shearofthebolt.Bearingfailureoftheconnectedmembersagainstthebolt.Tensionfailureoftheconnectedmembermaterial.Endtear-outoftheconnectedmember.Blockshear.ASTMA307unfinishedbolt—usedinlightsteelframestructureswherevibrationandimpactarenotcritical.ASTMA325andA490high-strengthbolts—themostwidelyusedfastenersofsteelconstruction.Mechanicallyfastenedhigh-strengthconnectionsthattransmitloadbymeansofshearintheirfastenersarecategorizedaseitherslip-critical(SC)orbearing-type(NorX).Slip-criticalconnectionsdependuponsufficientlyhighclampingforcetopreventslipoftheconnectedpartsunderserviceconditions.Bearing-typeconnectionsarebasedonthecontact(bearing)betweenthebolt(s)andthesidesoftheholestotransferloadfromoneconnectedmembertoanother.StandardAISCtablesareusedtocoverthedesignofavastmajorityoftypicalstructuralconnectionswherefillerbeamsframeintogirdersorgirdersintocolumns.Thevarioustypesofweldedjointscommonlyuseafilletweldoragroove(butt)weld.Filletweldsresistloadinshear.Thecriticalshearstressisassumedtobeactingontheminimumthroatareaoftheweld.Thethroatofafilletweldismeasuredfromtheroottothetheoreticalfaceoftheweld.Thestrengthofafull-penetrationgrooveweldisproportionaltoitscross-sectionalareaandthestrengthofthefillermetal.Ingeneral,agrooveweldisassumedequaltothestrengthofthebasematerialbeingwelded.551StructuralConnectionsAnswerstoSelectedProblems13552(a)Pv=90.1k(b)634–A325-SCShear:Pv=20.8k(governs)Bearing:Pp=65.2kNettension:Pt=71.9kPlatetension:P=66kGroupA:Shear:Pv=79.5kBearing:Pp=78.3kNettension:Pt=115.5kGroupB:Shear:Pv=72.2kBearing:Pp=61kNettension:Pt=29.9k(governs)Platecapacityintension:P=33k4MemberA:3bolts;MemberB:3bolts;MemberC:2bolts:MemberD:4bolts6Shear:Pv=119kBearing:Pp=95.7kNettension:Pt=117.8k(governs)(c)L=1712–9Platecapacity:Pt=49.5k5–weldsize:Using16Weldcapacity=55.7kLengthofweld=12”115Assumingthemaximumweldsizeof16–isused:L=5.6”12Filletweld:P=59.4kFull-penetrationgrooveweld:Pt=79k133Usingtheminimumsizefilletweldof16–:L1=8.01”L2=18.64”Structure,Construction,andArchitectureFromChapter11ofStaticsandStrengthofMaterialsforArchitectureandBuildingConstruction,FourthEdition,BarryOnouye,KevinKane.Copyright©2012byPearsonEducation,Inc.PublishedbyPearsonPrenticeHall.Allrightsreserved.553Structure,Construction,andArchitectureBuildingCaseStudy:REIFlagshipStore,Seattle,WAArchitects:MithunPartners,Inc.StructuralEngineers:RSP-EQEGeneralContractor:GallLandauYoung554Figure1Eastelevation,designsketch.CourtesyofMithunPartners,Inc.Figure2Eastelevation,constructiondrawing.CourtesyofMithunPartners,Inc.Figure3Eastelevation,photographofthecompletedbuilding.PhotographbyRobertPisano.Structure,Construction,andArchitectureIntroductionItisdifficulttoseparatepreciselythecontributionsoftheengineers,architects,andcontractorstothesuccessofabuildingproject.Theprocessofdesignandconstructionvarieswitheachowner,site,design,andconstructionteam.Mostbuildingprojectsbeginwithaclientprogramoutliningthefunctionalandspatialrequirements,whichistheninterpretedandprioritizedbythearchitect,whocoordinatesarchitecturaldesignworkwiththeworkofotherconsultantsontheproject.Thearchitectandstructuralengineermustsatisfyawiderangeoffactorsindeterminingthemostappropriatestructuralsystem.1INITIATIONOFPROJECT—PREDESIGNREIOutgrowsCapitolHillStoreandSiteRecreationEquipment,Inc.,orREI,hasbeenmakingandsellingoutdoorgearandclothinginSeattlesince1938.ThedesignoftheREIflagshipstorebeganwithananalysisofREI’sformerconvertedwarehousebuildingandsiteinSeattle’sCapitolHillneighborhood.Aprogramdocumentwaspreparedthatoutlinedfunctionalandspacerequirementsforthestore.Theneedforadditionalretailspace,structuralimprovementsforseismicsafety,handicappedaccessibility(elevatorsinparticular),expandedloading,andparkingwouldrequiredifficultandcostlyrenovationoftheexistingbuildingaswellasacquisitionofadditionalsitearea.Afterevaluatingseverallocations,asiteintheCascadeneighborhoodwasselectedandpurchasedfortherelocationofthestore.Figure4Theoriginalstorewashousedinaconvertedwarehousewithexposedheavytimberframing.Therugged,functionalcharacterofthebuildingwasconsistentwiththenatureofREIproducts.PhotocourtesyofMithunPartners,Inc.Thegeneralcontractor(GallLandauYoung)wasengagedbeforetheselectionofthearchitectandengineeringconsultants.Thecollaborationofthedesignteamandgeneralcontractorhelpedensurethatissuesofconstructability—schedule,cost,andavailabilityofmaterials—wereconsideredfromtheverybeginningofthedesignprocess.Theselectedsitewasafullcityblock—approximately90,400squarefeet—andwasoccupiedbyseveralexistingbuildingsandpavedparkingareas.Thesize,configuration,andpoorconditionoftheexistingbuildingsmadethemunsuitableforadaptationandrenovation.Theneighborhoodwasavariedmixofindustrial,commercial,andresidentialbuildingsjustoutsidethedowntowncoreandadjacenttoanoisyinterstatefreeway.Soilconditionsweretypicalforthearea:compactedglacialtill,withnounusualgroundwaterproblems.Figure5Siteanalysisdiagramshowingadjacentbuildingsandroadsinrelationtosolarorientation.SketchcourtesyofMithunPartners,Inc.Figure6Earlysiteplandiagramsstudyingbuildingandopenspaceplacementonthesite.SketchcourtesyofMithunPartners,Inc.555Structure,Construction,andArchitecture2DESIGNPROCESSTheOwner’sProgramandRequirementsREImemberswerefondoftheoldwarehousestore,anditwasdecidedthatthenewstoreshouldreflectthefeelandcharacteroftheformerstore.Inadditiontothe98,000squarefeetofretailsalesspace,a250-seatmeetingroom,a100-seatdelicatessen,administrativeoffices,arentalrepairshop,andamultistoryrock-climbingstructureweredesired.Inaddition,cityzoningregulationsrequired160,000squarefeetofparkingfor467cars,loadingdocksforlargetrucks,andsomesmalllandscapedareas.Becauseoftheircommitmenttoenvironmentalquality,REIrequestedthatthebuildingrespondtotheregionandclimate,conservingbothenergyandmaterials.Byexposingthestructuralelementsandmechanicalsystems,materialandlabor(money)thatwouldbeneededtoconcealandfinishthespaceswouldbeconserved.Expressingthestructuralandmechanicalsystemscontributedsignificantlytothefunctional,no-nonsensecharacterofthebuilding.Italsorequiredagreaterdesignandcoordinationefforttomakepresentablethesystemsthataretypicallyhidden.Figure8Designmassingsketches,option4.Thebulkyparkingstructureandretailblockdominatethesite.Aninterioratriumisrequiredtoadmitdaylightintothelowerfloors.SketchcourtesyofMithunPartners,Inc.Figure9Designmassingsketches,option5.Parkingistuckedbelowthebuilding,leavingmoreexterioropenspaceforlandscaping.Splittingtheprogramintoseparateblockscreatessmallerbuildingsthatbetterfittheexistingneighborhoodscale.SketchcourtesyofMithunPartners,Inc.556Buildingcoderequirementslimittheareaandheightofbuildingsbasedontheiroccupancyoruseandtheirtypeofconstruction.Generally,largerbuildingsareallowedifmoreexpensive,fire-resistiveconstructionisemployed.Afteranalyzingthesite,program,andbuildingcoderequirements,severaloptionsweredevelopedandevaluated.Figure7Photographofmassingmodel,option5,usedtostudytheexteriormassingandformofthebuilding.PhotocourtesyofMithunPartners,Inc.Structure,Construction,andArchitectureOption4splittheprojectintotwoparts,aneight-levelparkinggarageandathree-levelretailbuildingonthesouthpartofthesite.Thelargeparkingstructurewithcarsontherooftopandbulkyretailbuildingwereinconsistentwiththescaleoftheexistingneighborhoodandwouldhaverequiredvisitorstoenterthebuildingviaelevators.Inaddition,thelargeretailfloorsrequiredacostlyatriumtobringadequatenaturallighttotheinterior.Thisalternativewasabandonedinfavorofoption5.Theparkingstructurewaslargelyscreenedfromviewbyplacingmostoftheparkingbelowgradeundertheretailbuildingontheuphillpartofthesite.Byworkingwiththeexistingslope,excavationwaslimited,andalarger,moreusableoutdoorspaceforlandscapingwasprovidedonthesunnysouthportionofthesite.Breakingtheretailbuildingintotwoparts,atwo-levelwarehouse-likebuildingalongthefreewayedgeprotectedtheinterioroftheblockfromtrafficnoise,andafour-storyconcretebuildingatthenorthwestcornerbroketheprojectintosmallercomponentsmoreappropriatetothecharacteroftheneighborhood.Alarge,welcomingentryporchwascreatedonthepedestrian-orientedwestsideofthebuilding,anda85-foot-tallglassenclosureforthe65-foot-highclimbingpinnaclewasaddedtothesoutheastcorner,visiblefromthefreeway.Figure10Schematicdesignsketchofoption5asseenfromEastlakeAvenue.SketchcourtesyofMithunPartners,Inc.Figure11Presentationmodelshowingtheentryspaceatthesouthwestcorner.Mostofthelarge-scaledecisionsaboutthebuildingformandsiteplanhavebeenestablishedatthispoint.PhotocourtesyofMithunPartners,Inc.557Structure,Construction,andArchitecture3SCHEMATICDESIGNFigure12(a)Buildingsection—schematicdesignsketch.Figure12(b)Buildingsection—designdevelopmentdrawing.Furtherdevelopmentandrefinementsofthespacesandstructuralsystemresultedinachangefrombowstringtrussestoglu-lambeamsattheYaleStreetportion(attheleftsideofthedrawings).DrawingscourtesyofMithunPartners,Inc.Theselecteddesignwasfurtherrefinedandtestedtomorecarefullyfittheprogramtothesite.Circulation(themovementofpeopleandcarsinandoutofthesite)andbuildinghorizontallyaswellasvertically,wereworkedoutinplansandsectiondrawings.Thebelow-gradeparkinglevelsprovidedintegralconcreteretainingwallsandasubstantialfoundationforthebuilding.Themainsalesareasweredesignedaslargeopenspacestorecallthecharacteroftheoriginalwarehousestore.Studiesforbringingnaturallightintothemainsalesareasuggestedthattheroofbeslopedtoalloweastlighttoenterhighabovethewallinthemornings,whenheatgainwouldbelessproblematic.Theroofslopealsocreatedalarger-scaleprotectivewallonthefreewaysideofthebuildingandalower,pedestrian-scalewallonthelandscapedentryside.Theentryporchprotectedthewestsidefromprevailingrainsandthelateafternoonsun.Rainwaterfromthelargeshedroofwouldbecollectedonthewestsideofthebuildingandusedtosupplementandrechargeawaterfallwithinthelandscapedcourtyard.Mechanicalequipmentforheatingandcoolingthebuildingwaslocatedcentrallyontherooftoptodistributetheconditionedairmoreeconomically.Figure13Aphotographoftheentryelevator/stairtowerandfrontporchillustrateshowearlydesignideasshowninFigure11werefinallydesignedandbuilt.PhotographbyRobertPisano.558Structure,Construction,andArchitectureThearchitectsresearchedtheSeattleandUniformBuildingCoderequirementsanddeterminedthatbasedonoccupancy/use,floorarea,andheight,theprojectshouldbeseparatedintothreebuildings(Figure14).Althoughthethreebuildingswerecontiguousspatially,theywouldbeseparatedbyfire-resistivewallsorfloors.Eachbuildingwasadifferentconstructiontype,withspecificfire-resistivestandardsforstructuralelementsandwalls.■■■Building1.Athree-levelparkinggaragelocatedprimarilybelowgradeandusingthemostfireresistiveconstruction(TypeI).Structuralelementsmustbenoncombustible:steel,iron,concrete,ormasonry.Building2.Atwo-levelretailstructurelocatedabovetheparkinglevelsandcontainingthemainsalesareas.ConstructiontypeisIII-N(ornonrated),whichallowsstructuralelementstobeofanymaterial,includingwood.Building3.Atwo-levelstructureabovetheparkinggaragecontainingsomeretailandstorage,asmallauditorium,arestaurant,andofficespace.ConstructionisTypeIII–1hour,requiringstructuralelementstobeprotectedbyaone-hourfireratedassembly.Sprayed-onfireproofingorgypsumwallboardarecommonlyusedtoprotectstructuralelements.ThecityofSeattlealsoallowsforheavy-timberconstructiontoqualifyasTypeIII–1hour.Figure14Buildingcodediagram.Costestimatesandconstructionfeasibilitystudieswereprovidedbythebuildingcontractoratcriticalpointsthroughoutthedesignprocess.Figure16Photographofparkinggarageposttensionedconcreteslabatthedropsupportandcolumn(alsoillustratedinFigure19).PhotocourtesyofMithunPartners,Inc.Figure15Board-formedreinforcedconcreteisusedforthefour-hourfire-resistiveexteriorbearingwallsatBuilding3(locatedatthenorthwestcornerofthesite).Thetextureoftherough-sawnformboardshelpsestablishanindustrialcharacterandprovidesadurableweatherenclosure.PhotocourtesyofMithunPartners,Inc.559Structure,Construction,andArchitecture4DESIGNDEVELOPMENTANDCONSTRUCTIONDOCUMENTSFigure17Partialeast-westbuildingsectionshowingthestructuralbaysofthesalesfloorsalignedabovetheparkinglevels.DrawingcourtesyofMithunPartners,Inc.Inthecourseofschematicdesign,theorganizationandapproximatespacingofthestructuralbayswerecoordinatedwiththeparkinglayout.Columns,bearingwalls,andshearwallswerecarefullylocatedtousespaceontheparkinglevelsefficiently.Becausetheretailsalesfloorscouldbearrangedmoreflexibly,theverticalstructureatthoselevelscouldalignwithcolumnsandbearingwallsattheparkinglevelsbelowwithaminimumoftransferbeamsorunusablespace.Thethreestructuralbayshelpedorganizespacesonthesalesfloors—locatingcirculationandstairsinthecentertwo-storybay,withdisplayareasinthesidebays.Whiletheverticalloadtraceanalysisgenerallyproceedsfromthetopdown(beginningwiththeroofloadsandendingatthefoundation),thelayoutofthestructuralbayswasdevelopedfromthebottomup,helpingtokeeptheverticalloadpathasdirectaspossible.Thearchitectsandstructuralengineersbasedtheselectionofmaterialsforstructuralelementsandenclosureassembliesonseveralcriteria:■Figure18Plandiagramofparkingspacerequirementsanddimensionsofstructuralbays.DrawingcourtesyofMithunPartners,Inc.■■■■■Buildingcoderequirementsforfireresistance.Structuralproperties,performance,andefficiency.Resistancetoweatheringordecay;longevity.Appearance,aesthetics,orcharacter.Costandavailability,includingskilledconstructionlabor.Resourceefficiency.ParkingGarageandBuilding3Reinforcedconcreteisaneconomicalanddurablematerialandwasthereforeusedforthe
parkinggarage.Locatedbelowground,concretewasusedforretainingwallsonthefreewayside.Theconcreteceiling/floorprovidedahorizontalfireseparationandsatisfiedtheTypeIfireresistiveconstructionrequirementforstructuralelements.Posttensionedreinforcedconcreteslabswereusedfortheirefficientspan-to-depthratio.Theslabswereonly612–thick,spanning30feetinsomelocations.Droppanelsatcolumn-slabconnectionsreinforcedtheslabagainst“punchingshear”stressesandprovidedadditionaldepthforbendingstressesoverthecolumns.Byreducingtheverticaldimension(depth)oftheparkingstructure,costsweresavedonexcavation,shoring,andverticalstructure.Figure19Structuraldetailsofposttensionedconcreteslabatcolumnanddetailofconventionallyreinforcedcolumnatfooting.DrawingscourtesyofMithunPartners,Inc.560Conventionallyreinforcedconcretewasusedforcolumns,bearingwalls,shearwalls,retainingwalls,andfootings.Allconcreteutilizedrecycledflyashinthemix,awasteby-productofcoal-generatingpowerplants.Structure,Construction,andArchitectureFigure20Planatthebasementparkinglevel.DrawingcourtesyofMithunPartners,Inc.12124a121224OPENTOBELOW56311OPEN1123456789101112FreestandingClimbingRockViewingBalconyApparelFootwearaboottesttrailCustomerServiceChildren'sPlayAreaTravelAssistanceArtGalleryMeetingRoomCafe´RepairShopStockrooms781092NDFLOORPLANFigure21Planatupperretaillevel.Thepathofmovementthroughthebuildingandthearrangementofsalesareasrelateto,butarenotdeterminedby,thestructuralgrid.DrawingcourtesyofMithunPartners,Inc.561Structure,Construction,andArchitectureFigure22Structuralplanatparkinglevel.Concretefootings,columns,andshearwallsarearrangedinaregularspacingtoaccommodatecarparkingandmaneuvering.DrawingcourtesyofMithunPartners,Inc.Figure23Roofframingplan.Thestructuralgridestablishedattheparkinglevelisapparentattheroofframing.Theshadedareasattheroofedgeindicateshearpanelsintheroofdiaphragm.DrawingcourtesyofMithunPartners,Inc.562Structure,Construction,andArchitectureRetailSalesFloorsSincetherewerenofire-resistiverequirementsforstructuralelementsinBuilding2,housingtheretailsalesfloors,woodwasselectedasaneconomicalstructuralmaterialthatwouldalsoevokethewarmercharacteroftheoriginalstore.Thebaysandspanningrequirementsoftheprimarystructurefortheroofandfloorwereeffectivelyestablishedbythecolumnlocationsintheparkinglevelsbelow.Theuseofheavytimbers,commoninearlyAmericanwarehouseconstruction,wasconsidered.Today,however,theavailabilityofqualitytimbersismostlylimitedtorecycledmaterialfromoldandabandonedstructures,andcostsarerelativelyhigh.Steelbeamsortrussesandglulaminatedtimbersaremorecost-effectiveandcouldspantherequiredbaysefficiently,butthesewerenotasappropriateinscaleorcharacter.Ultimately,trussesmadewithglu-lamchordsandcompressionmembersaswellassteeltensionmembersandconnectionsweredeveloped.Thetrussesspannedeast-westadistanceof45feet.Thearchitects,contractors,andstructuralengineersdiscussedtheideaofusingaverycommonandeconomicalindustrialroofsystem.Followingthesediscussions,thestructuralengineersdevelopedamodified“Berkeley”system.Partlyprefabricatedonsite,58–plywooddeckingwasnailedto2–*6–joistsspaced24incheso.c.Joistswereeightfeetlongandspannedbetweenglu-lambeams.The4¿*8¿modulardimensionstookadvantageofthedimensionsoftheplywoodsheathing.The318–*18–glu-lampurlinsspanned25feet,6inchesbetweenthetrusses.Theeightfootspacingoftheglu-lambeamsestablishedthespacingoftrusspanelpoints.Becauseoftheslopingtopchord,thetrusspanelschangedproportion,andthediagonaltensionmemberswerenotactingasefficientlyastheycould.Tomaintainaconsistentappearanceandconstructiondetail,thetensionrodsweresizedfortheworstcaseandremainedthesamediameter,evenastheloadsvaried.Figure24Diagramofthemodified“Berkeley”systemroofstructure.Figure25Constructiondrawing—elevationoftherooftrusses.Glu-lambeamsframeontothetrussatpanelpointseightfeeto.c.DrawingcourtesyofMithunPartners,Inc.563Structure,Construction,andArchitectureFigure26Photographofretailsalesareasandtwo-storycentralcirculationbay.Theroofandfloortrusseseasilyspanbetweencolumns,allowingmoreflexiblearrangementsofdisplayfixtures.Mostoftheinteriorwallsarenonstructural,soeventheroomscanbereconfiguredovertimetoaccommodatechangesinthebuilding’suse.Four-logcolumnsareboltedtogethertosupportthestairsandextendtothesteelroofdeck,wheretheyaresupportedlaterallybyangledbraces.Eventhoughfewerorsmallerlogscouldadequatelysupportthestair,severallargelogswereusedtokeepthesupportvisuallyinscalewiththeheightofthespace.PhotographbyRobertPisano.564Structure,Construction,andArchitectureFigure27Constructiondrawingsdetailingtheconnectionsatrooftrussjoints.Asteelgussetplate,drilledtoaccepttheclevispin/tensionrod,isweldedtoabentplate,whichinturnisboltedthroughtheglu-lamtrussmember.Anothersteelplateontheoppositesideactsasalargewasher.DrawingscourtesyofMithunPartners,Inc.Thefloorwasframedinmuchthesamewayastheroof,butbecauseofthegreaterliveloads(75psf),thefloorsheathingwas118–-thickplywoodon3*8S4Sjoistsspaced32incheso.c.Thesespannedeightfeetbetween518–*21–glu-lambeams,whichinturnweresupportedbyfloortrussesatpanelpoints.Figure28Photographoftheconnectiondetailatarooftruss.PhotographcourtesyofMithunPartners,Inc.565Structure,Construction,andArchitectureFigure29Rooftopmechanicalequipmentinthecenterbayissupportedbyareinforcedconcreteslabcastonsteelformdeck.Thesteeldeckingisweldedtoaseriesofsteelpurlinsthataresupportedbysteelgirdersspanningeast-west.Loadsfromthesegirdersandtherooftrussesaretransferredtosteelcolumnslocatedabovetheconcretecolumnsintheparkinggarage.Inadditiontosupportinggravityloads,thesesteelcolumnsandgirdersformmoment-resistingframesthatprovidelateralstabilityintheeast-westdirection.PhotocourtesyofMithunPartners,Inc.566Structure,Construction,andArchitectureLateralLoadsSeattleisinarelativelyhighseismiczone.Inaddition,thelargewallareasofthestoreaccumulatedsignificantwindloads.Forcesintheeast-westdirectionwereresistedbyaseriesofductilesteelmoment-resistingframes,whichweretiedtotheroofandfloordiaphragmsatthetrusses.Shearwallsorcross-bracedframeswouldhaverestrictednorth-southmovementthroughthebuilding,dividingthespaceintoaseriesofsmallbays.Steelmomentframeshadseveraldistinctadvantagesoverconcrete:partialprefabrication,easierandfastererection,connectioncompatibilitywiththewoodandsteeltrusses,lighterweight,andsmallersize.Figure30Diagramoftheeast-westlateralloadbracingsystemconsistingofductilesteelmoment-resistingframes.Plywoodroofsheathingonglu-lampurlinsHeatingductSteeldeckingonsteelbeamsGlu-lamatflooredgeDuctilesteelmomentframeFigure31Sevenductilesteelmoment-resistingframesformthetwo-storycenterbayoftheretailbuilding.Theframesareanchoredtoconcretecolumnsthatextendthroughtheparkingstructuretoconcretefootings.PhotographbyRobertPisano.567Structure,Construction,andArchitectureFigure32Constructiondetailsatmoment-resistingcolumn-beamconnection.Steelstiffenerplatesandanglesareaddedtoresistlocalbendingofthecolumnattheflanges.CourtesyofMithunPartners,Inc.Figure34Constructiondrawingofatypicalductileframe.CourtesyofMithunPartners,Inc.Figure33Photographofmoment-resistingcolumn-beamconnectionatfirst-floortruss.Additionalsteelplateswereweldedtothetopandbottomflangesofthebeams,preventingstructuralfailureinanearthquake.StudyoftheNorthridge,CA,earthquakeledtothedevelopmentofthisdetail,whichexceedscurrentseismiccoderequirementsinSeattle.PhotocourtesyofMithunPartners,Inc.568Structure,Construction,andArchitectureNorth-southlateralloadswereresistedthroughplywoodroofandfloordiaphragms,whichwerecollectedanddirectedintosteeldiagonalbracedframesintheexteriorwalls.Thediagonalcross-braceswereveryeffective,andonlytwobaysontheeastandwestwallswererequired.Thesebraceswereexposedontheexterioroftheeastelevation,andwindowswerelocatedtheretofurtherrevealtheirpresence.Theexposedportionofthediagonalbracesontheeastelevationweredesignedtoappearsymmetricaltopandbottom,matchingthebracingontheclimbingpinnacletower.Behindthemetalsiding,thediagonalbracesextendedtothefloorframinganddiaphragm.Figure35Diagramofthediagonalbracingsystemattheeastandwestwallsoftheretailbuilding.DrawingscourtesyofMithunPartners,Inc.Figure36Photographofthediagonalbracingsystemmeetingthebuildingenvelope.Thestructuralbaysarefurtherdefinedbysettingthewallback,usingalighter-colorsiding,andplacingwindowsbehindthebracedframe.PhotocourtesyofMithunPartners,Inc.Figure37Photographofthediagonalbracesattheeastelevation.Thebracesdonotalignwiththesecondfloordiaphragm;instead,theyextendtothefloorconcealedbehindtheexteriorsiding.PhotographbyRobertPisano.569Structure,Construction,andArchitectureFigure38Constructiondrawings—elevationofatypicaldiagonalbracedframe.DrawingscourtesyofMithunPartners,Inc.Figure39Constructiondrawingsatdiagonalbracedframe.Theconcentricsectionofthesteelpipeefficientlyresistsaxialtensionandcompressionloads.DrawingscourtesyofMithunPartners,Inc.570Structure,Construction,andArchitecture5INTEGRATIONOFBUILDINGSYSTEMSAllbuildingsystems—lighting,heatingandcooling,ventilation,plumbing,firesprinklers,andelectrical—havearationalbasisthatgovernstheirarrangement.Itisgenerallymoreelegantandcost-effectivetocoordinatethesesystems,thusavoidingconflictandcompromiseintheirperformance.Thisisespeciallythecasewherestructureisexposedanddroppedceilingspacesarenotavailabletoconcealductandpiperuns.Afterthespacesandsystemshadbeenroughlyarrangedinschematicdesign,thearchitectsandengineersworkedthroughseveralgenerationsofplansandsectionstorefinethesizeandlocationofsystemcomponentsandresolveanyconflictsbetweensystems.Figure41Mechanicalsupplyandreturnairducts,lightingfixtures,andthefiresprinklersystemarecoordinatedwiththestructuralelementsandspatialrequirements.Theopenwebsoftheroofandfloortrussesallowspacefortheseothersystemstorunperpendiculartotheframing.PhotocourtesyofMithunPartners,Inc.Figure40Roofdrainsarelocatedoneithersideoftheglu-lamroofbeamattheentryporch.Steeldownspoutsaresupportedbythecolumninthewall.Photobyauthor.571Structure,Construction,andArchitectureFigure42Sectionatrooftopmechanicalequipmentshowsstructuralframeinrelationtoairhandlerandductwork.DrawingcourtesyofMithunPartners,Inc.Figure43Themiddlebayoftheretailsalesbuildingprovidesacentraldistributionlocationfortheventilationducts.PhotocourtesyofMithunPartners,Inc.572Structure,Construction,andArchitectureGlu-lambeamateaveprovideslateralsupportforroofedgeandtopofwallframing.Roofandfloortrussesbeardirectlyonsteelcolumns.Glu-lamledgerbeamsupportsporchroofbeams.Aluminumwindowsystemissupportedverticallybymetalwallstudsandhorizontallybyglu-lambeams.Photographoftheentrywallongridline6underconstruction.Steelcolumnsandglu-lambeamssupportthelight-gaugesteelstudframing.PhotocourtesyofMithunPartners,Inc.Figure44Constructionsectiondrawingthroughtheentryporchlookingnorth.Unliketraditionalload-bearingwallconstruction,manynewerbuildingsseparatethesupportingstructureandenclosure.Consideredasasystem,theenclosingelementsprovidethermalandweatherproofingprotection.Thealuminum-framedwindowsystemissupportedbyglu-lambeamsandbracedverticallybycolumns.Thesesupportconnectionsmustbecarefullydetailedtoaccommodatethedeflectionanddisplacementoftheprimarystructureunderloadwithoutloadingtheenclosingelements.Differentialthermalexpansionofthestructuralframeandskinmustalsobeconsidered.Anelevationofthesecondarywallframing(insetdrawing)wasprovidedbythestructuralengineerstodescribethelocation,size,andconnectionofbeamstocolumnsandwallframing.DrawingscourtesyofMithunPartners,Inc.573Structure,Construction,andArchitectureFigure45Thisviewoftheentryporch(lookingsouth)showsthewallandroofstructuredescribedinthedrawingsshowninFigure44.Becausethewallsarenotloadbearing,windowsanddoorscanbelocatedanywhereexceptatcolumns.Columnsandglu-lambeamsarerequiredtosupporttheactualweightofthewallaswellasthewindandseismic(lateral)loads.Thesesecondarystructuralmembersareprotectedfromweatherandtemperaturebyinsulationandmetalwallpanels.PhotographbyRobertPisano.574Structure,Construction,andArchitectureDetailsandConnectionsConnectionsmustbecarefullydesignedtotransferloadspredictably,particularlywhentheloadpathisredirectedorchanged.Thestructuralaction(tension,compression,bending,torsion,shear)ofeachassemblyultimatelydeterminesthebehaviorofthestructuralframework.IntheREIbuilding,pinorhingejointswouldbeusedtopreventthetransferofbendingmomentsataconnectionbyaccommodatingcontrolledmovementorrotation.Attheentrystairtower,constructionmaterialswouldbeclearlyidentifiable,witheachconnectionvisibleanddirectlycontributingtotheruggedutilitariancharacterofthebuilding.(a)Figure46Theshedroofframingattheentrystairtower(a)issupportedbyasmallwideflangebeamandtwosteelpipecolumns(b)triangulatedina“V.”Acast-in-placeconcretewallcarriesthecolumnloadstothefoundationandground.(b)Figure47(a)Entrytowerunderconstructionshowstheglu-lambeamssupportedonsteelwide-flangebeamsandcolumns.Steelreinforcingbarsfortheconcretestairscantileverfromtheconcretesupportingwall.Figure47(b)Byexaminingtheroofsupportatthesouthendoftheentrystairtower,onecanseehowthetheoreticalprinciplesofstaticsandstrengthofmaterialsareapplied.Thesectioniscutthroughtheroofbehindthesteelpipe“V.”The4–*6–purlinsrunperpendiculartotwopairsofglu-lambeams,whichareshownatthesectioncut.PhotographsanddrawingsonthispagearecourtesyofMithunPartners,Inc.575Structure,Construction,andArchitectureBentsteelplateconnectorswereweldedtothetoptothesteelwide-flangebeamtoaccepttheslopingglu-lambeams.The58*6–lagscrewsthroughthebottombearingplateintothebeamsandtwothroughboltsatthesideplatescompletedtheconnection.AshortW6*20sectionweldedtothetopofthebeamactedasaspacerandconnector.ThesewerepredrilledandweldedtotheW10*steelbeaminthefabricationshop.Boltholesintheglu-lamsandpurlinswereshopdrilledtospeedonsiteerection.Thefour-inch-diametersteelpipeactedasbothcolumnandlateralbrace.Asteelkerfplate,cuttotherequiredangle,wasweldedinaslotcutinthepipe.Thiscommonsteelconnectionenabledincreasedweldingatthejoint,andprovideda34–-thicksteelbearingplateforthebeam.Theprefabricatedbeamanddiagonalsupportwerefieldbolted.Slottedboltholesallowedsomeassemblytoleranceandthermalmovementofthebeam.Figure48TheW10*steelbeamhasstiffenerplatesweldedtothewebandflangestopreventlocalfailureasloadsandstressesincreaseattheconnection.Photobyauthor.Carefulpreparationandreviewofshopdrawingsensuredthattheassemblywouldactasintendedandthatthepartswouldfittogetherefficientlyonsite.Inconstruction,importantdetailsconsidertheproperties,shape,andphysicaldimensionsofthematerialstobejoinedaswellasthetoolsandtradespeopleneededtoassemblethem.Figure49The4–*6–woodroofpurlinsaresecuredateachendtothepairsofglu-lambeamsbysteelboltsandsteelsideplateconnectors.Theseconnectorspreventthepurlinsfromslidingandoverturning.DrawingscourtesyofMithunPartners,Inc.576Structure,Construction,andArchitecture6CONSTRUCTIONSEQUENCEBuildingproceedsfromthegroundup.Sitework,demolition,clearing,excavation,shoring,andgradingpreparedthesiteforfoundations.Whileconcretefoundationsandsubstructurewerebeingcompletedonsite,manycomponentsofthestructuralframewereprefabricatedinshopsandtruckedtothesiteasneeded.Thisoverlappingofconstructionsavedtimeandmoney.Largestationarytoolsandequipmentallowedclosertolerancesinshopfabrication—cutting,drilling,andwelding.InSeattle,rainywinterweathermadeshopfabricationevenmoreattractive.Somepartsofthestructure,liketheglu-lamandsteeltrusses,wereprefabricatedandsetinplacebycrane.Thesteelmomentframesweretoolargetotrucktothesite,sotheywerepartlyprefabricated.Erectionboltswereusedtosecurememberstemporarilywhilesiteweldingwascompleted.Figure50Sitecastingisusedtoachievemonolithic,moment-resistingjointsinconcrete.Thegeneralcontractorisresponsiblefortheengineeringofformworkandscaffoldingrequiredforlargeandcomplexpours.PhotocourtesyofMithunPartners,Inc.577Structure,Construction,andArchitectureFigure51Erectionsequenceoftheclimbingtowerbracedframes.Portionsofthefourframeswereshopfabricatedandshippedtothesite,wheretheywereassembledandwelded,eachontopoftheother.Theframes,cornercolumns,andcornerstrutswerethenerectedinoneweekendsothatthemobilecranewouldnotinterferewithfreewaycommuters.PhotoscourtesyofMithunPartners,Inc.578Structure,Construction,andArchitecture7CONCLUSIONTheREIbuilding,likemostsuccessfulbuildingprojects,involvedtheeffortofmanydesignandconstructionprofessionalsworkingtogetherinthespiritofcollaboration.Theearlydecisiontoexposeandexpressthestructuralsystemrequiredcollaborationfromtheearliestdesignsketches.Thearchitects,structuralengineers,andbuilders,whilenotalwaysagreeingonthesamecourseofaction,neverthelessunderstoodandappreciatedoneanother’sexpertise.Thiscreativeconflict,managedconstructively,resultedinaprojectgreaterthanthesumoftheindividuals.Thestructuralengineersforthisprojectdidnotsimplycalculateloads,stresses,anddeflectionstoensurethatthearchitect’sdesignwouldstandup.Atthesametime,thearchitectsdidnotsimplyconceiveofasculpturalorspatialformwithoutregardforbasicprinciplesofstructuralsoundness,order,andstability.Structuraldesignisrarely,ifever,entirelyrationalorpure.Engineersrelyonprofessionalexperienceandintuitionaswellastheanalyticalmethodsandformulasdevelopedoverthecourseofengineeringhistory.Thestructuralsystemasawholemustbeconfigured.Assumptionsaboutprobableloadsandloadpathsmustbemadeandtestedbycalculationtoensurethestabilityofastructure.Eachmemberandconnectionmustbedesignedtoactasintended,resistingloadsandstressconcentrations.Theeffectsoftimeandweathermustbeanticipatedtoaccommodatemovementandpreventcorrosionorprematurefailure.Afterabuildingisfinallyconstructedandloaded,itisverydifficulttodetermineifitisbehavingexactlyasdesigned.Loadsmaynotbedistributedorconcentratedasanticipated,andundersomeconditions,structuralmembersmaybeactingincompressionratherthantension.Therefore,buildingcodesandgoodengineeringpracticeprovideadequatesafetyfactorsandredundancy.TheREIbuildingdemonstratesthatgoodstructuraldesignandarchitecturearepossiblewithoutoverlyexpressive,exotic,oruntestedconfigurationsandmaterials.Itispossibletoutilizetime-testedengineeringprinciplesandmaterialsinanimaginativeandappropriatewaytoachieveasuccessfulbuilding.Figure52Photographoftheclimbingpinnacleandenclosingtower.PhotocourtesyofMithunPartners,Inc.579Structure,Construction,andArchitectureFigure53580Oneoftwointeriorstairstructures.PhotographbyRobertPisano.DefinitionofTermsMeasurementameasureoflengthU.S.Unitsinch(in.or′′)Metric(S.I.)millimeter(mm)ameasureofareasquareinches(in.)squaremillimeters(mm2)pound(lb.or#)newton(N)feet(ft.or′)meter(m)2squarefeet(ft.2)poundmass(lbm)ameasureofmassameasureofforceameasureofstress(force/area)squaremeters(m2)kilogram(kg)kilopound=1,000lb.(k)psi(lb./in.or#/in.)22moment(force×distance)ksi(k/in.2)psf(lb./ft.2or#/ft.2)pound-feet(lb.-ft.or#-ft.)aloaddistributedoverlengthω(lb./ft.,#/ft.,orplf)ameasureofpressuredensity(weight/volume)kip-feet(k-ft.)γ(lb./ft.or#/ft.)33kilonewton=1,000N(kN)pascal(N/m2)kilopascal=1,000Panewton-meter(N-m)kilonewton-meter(kN-m)ω(kN/m)γ(kN/m3)force=(mass)×(acceleration);accelerationduetogravity:32.17ft./sec.2=9.807m/sec.2Conversions1m=39.37in.1ft.=0.3048m1m2=10.76ft.21ft.2=92.9×10-3m21kg=2.205lb.-mass1lbm=0.4536kg1kN=224.8lb.-force1lb.=4.448N1kPa=20.89lb./ft.21lb./ft.2=47.88Pa1MPa=145lb./in.21lb./in.2=6.895kPa1kN/m=68.52lb./ft.1lb./ft=14.59N/m1kg/m=0.672lbm/ft.1lbm/ft.=1.488kg/mPrefixSymbolFactorgiga-G109or1,000,000,000mega-M106or1,000,000kilo-k103or1,000milli-m10-3or0.001FromtheDefinitionofTermsofStaticsandStrengthofMaterialsforArchitectureandBuildingConstruction,FourthEdition,BarryOnouye,KevinKane.Copyright©2012byPearsonEducation,Inc.PublishedbyPearsonPrenticeHall.Allrightsreserved.581582IndexPagereferencesfollowedby"f"indicateillustratedfiguresorphotographs;followedby"t"indicatesatable.AAbscissa,275,347Abutments,100Accelerationground,8-9,15-16Accident-preventionprogramdevelopmentof,12Accountingconsistencydefined,325Accurate,319,415ACI,432Additionalpaymentcontractor,554-555Airbarriersrigid,212,534Aircompressorscapacity,210,532types,6-7,210,534Airconditioningmethods,6-7,534Airleakagerate,534Airpollutionbuildings,6-8,573construction,6-7,533-534,570vehicular,7Air-handlingequipmentandsystemsconfigurations,9,240,326,579AISC,279,406-407,469-472,514-516Allowablebendingstress,385,501Allowablesoilpressure,268Allowablestress,264,386,470-471,516Allowablestressdesign,432,471,538Allowablestresses,284,308,432-433Alphabettypesillustrated,244,537roman,121text,258,432,488,533Aluminumroofing,5windows,574Aluminumalloyallowablestressesfor,284strengthof,4,272-274AmericanInstituteofSteelConstruction(AISC)columndesignformulas,469filletweldsand,535recommendedK-values,471AmericanInstituteofTimberConstruction(AITC),412AmericanWeldingSociety(AWS)structuralcode,538Amortizationscheduleschedule,14,555Analyticalmethodofvectoraddition,33Anchorrodsgrades,437uplift,436Anchorages,100Anglehorizontal,30-31,100-101,267steel,99-100,265-266,458-459,527,575-577vertical,30,100,206Anglesright,527Answerstoselectedproblems,95,196,255,305,339,374,447,505,552Appliances,113Arcwelding,508Archesanalysis,98-100three-hinged,166two-hinged,166Archimedes,257Architecturaldetailsmembersin,238-239,549stairs,14,203,560wall,204,569Architecturalmodelslandscaping,556Architecturehistoryof,121Areagross,521Asphaltshingles,14Asphaltshinglesapplication,14Assumptions,415,545,579Axiallyloadedsteelcolumnsanalysisof,14,98-101,217,300,458exampleproblems,26,107,212,260,321,362,381,463,520steelcolumndesign,479BBalloonframeconstructionbracing,9,121,236-238,431,461-462,545,567floorjoists,199posts,72,237,450studs,14,206,433,450,573Barsizes,520supports,46Basewood,18,236-238,291,486Baseplatesholes,177,515-517,576shear,8,177,240,262-265,398-400,515-517,575stiffened,236,461Baseshear,8Batchingequipmentasphalt,14Beamlaminated,390precast,57,209BeamcolumnsBendingandshearstressesinbeams,375-397classificationof,21,262,342deflectionin,412designof,6,236,308,342,385,454,508gradebeams,211Beam-columnsinteractionequation,501-503trusschord,84Beamsbendingstress,308,342,380-381,499bendingstresses,242,378-379,560blockshear,528box,404cable-supported,100cantilever,195,241-243,343,375,575centroids,115,309Beam(s)continuous,8,88,210-211,299,343,533Beamscoped,528deflections,319,347,412-413designprocedure,250holesin,293,576lateralbracing,431,458lateralbuckling,429-430lateral-torsionalbuckling,437loads,10,57,99-100,199-207,257-258,308,342,376,460,508,560location,9,115,203,319,400,572-573longitudinal,100,342,376-377,537momentofinertia,308,358,380neutralaxis,377-380plastichinge,433reinforcedconcrete,211,293,345,499,559-560sectionmodulus,385-386serviceability,433shearstresses,308,392-393,537Beam(s)spandrel,259,423-424Beamsstability,9-10,99,238,431-432,545,566steelbeams,218,407,459,563timberbeams,403torsion,294,376,575transverse,294,342,376,537tributaryarea,202,498unbracedlength,442vibrations,9Beams,reinforcedconcretebalanced,100,544failureof,283,441,450,551rectangular,29,117,249,294,319,375,461shearin,397,551strengthrequirements,412T-beam,406bearingbeams,10,196,200,284,367-369,406-408,498,527-528,558bolts,265,510-517,576-577connections,203,497,510-529,560plates,262,515,576stiffeners,407stress,183,262,406,498,511-513Bearingcapacityallowable,68,178,203,257,407,451,511-517ultimate,285,432,513-514Bearingcapacityequationsextended,14,53introductionto,432Bearingstress,183,265,421,498,513belowgrade,178,557Bendingandshearinsimplebeamsclassificationofbeamsandloads,342equilibriummethodforshearandmomentdiagrams,350-351shearandbendingmoment,347Bendingandshearstressesinbeamsdeflectioninbeams,412flexural(bending)stressequation,378flexuralstrain,376Bendingstrength,375Bendingstress,308,342,380-381,499Biaxialbending,501Billingsinexcessofcostsandprofitcalculation,7,127,321-322,399,579changesin,77,105,350,564Blockshear,511Blocksissue,242Boltholes,508,576Boltscommon,509,576doubleshear,520edgedistance,514framedbeamconnections,527-528holetypes,517machine,534singleshear,265,511snugtight,516spacing,514-515unfinished,515Bondpatterns:brick,273Bracedframes,240,545,567Braces(bracing)diagonal,148-149,238,431,549,563Bracingbracingconfigurations,245concentricbracing,545583diagonalbracing,549,569K-bracing,549knee-bracing,238-239x-bracingmembers,238Bracingsystemsdescriptionofmultistory,243-244,556Brass,275Brickmasonryterms,450Brick(s)building,10-11,290constructionof,11-12Bridgingmetal,428woodframing,429Brittlematerials,273Brittleness,283Bronze,11,279Bucklingcompression,101-102,242,407,450-452,549Euler,452-454flangelocal,442weblocal,442Buildingcode,14,61,290,412,468,534-535,556industry,446orientation,555paper,376,487Buildingcasestudybuildingcoderequirementsand,559conclusion,579constructionsequence,577designdevelopmentandconstructiondocuments,560designprocess,5,555-556integrationofbuildingsystems,571schematicdesign,557Buildingcodesapplicationof,14,533-534model,556-557special,6,23,515-516Buildingmaterialscost,3,210,510,555Buttjoint,511CCablestructurescable-stayed,99-101catenary,105forcesin,7,64,111,244,300lengths,122,430,452reactions,64,100,200-202Cablesuspensionsystemsanchorages,100cablegeometryandcharacteristics,105cables,62,98-100cableswithasingleconcentratedload,106stabilizers,102verticalsupports,100Cable-stayed,99-101Caissons,211Cantileverretainingwallsheel,177-178stem,194,229-230toe,178-185Capillaryrisetimerateof,15Carpetandcarpettileflooringcharacteristicsof,15-16,105,468Cashflowaverage,293,437complex,7size,6Ceilingjoist,203,429suspended,5,423Ceilings,suspendedacoustical,5Cells,3Centerpunchescenter,10,19,115,206,259,309,387,514,560Centricloads,259Centroidscenterofgravityand,308ofsimpleareas,312Changeorderdealingwith,296,413internal,115,294-295,369Channels,386584Chords,411,486,563Circularcurvecompound,369geometry,105-106radius,294,414-415,454reverse,43Cladding,exteriorwallstone,11,181Classificationofbeamsandloadssupportconditionsand,458Claysurface,211Claysoil(s)compacted,182,555Clearing,577Clevis,565ClimaticconditionsUnitedStates,6Coal,560Coefficientofthermalexpansion,295Collinearforcesystemsequilibriumand,88Columnanalysisanddesignaxiallyloadedsteelcolumns,468axiallyloadedwoodcolumns,486combinedloadingoreccentricity,499endsupportconditionsandlateralbracing,458modesoffailure,451,511Columnsbeamcolumns,499buckling,242,308,433,451-454coderequirementsfor,560composite,12,326-327,432-433,545continuouscolumns,240criticalbucklingload,452-454criticalbucklingstress,468design,9,56,260,308,432-434,449,508,559-560eccentricloads,501effectivelength,458Eulerbuckling,489footings,13,210,547,560forms,12generally,11,203,432,458,509,560largeeccentricity,499longcolumns,486radiusofgyration,454shortcolumns,451slender,12,238,319,451-452slendernessratio,458splices,509steelcolumns,458,508,566tied,567timbercolumns,461Combinedloadingcolumnsand,240,508,573-574Combinedshearplustensionwelds,508Commercialbuildingsdetails,545,560elevations,8,170structuraldrawings,545commercialplangroupsarchitectural,11-13,555mechanical,556structural,11-13,123,209,412,483,554-556Commitment,556Compositebeamsponding,412practicalconsiderations,412shoring,560Compression,17,101-102,201,265,348,375-377,450-452,535,563Compressionbuckling,407Computeranalysestrusses,123-125Computer-aideddesignanddrafting(CADD)profiles,450scales,242Computer-aideddraftinganddesign(CADD)historicaloverview,11resources,3software,293,373,413systemcomponents,571three-dimensionalspace,22Concentricbracing,545Concentricloading,499Concreteblock,269,557cast-in-place,206,575cast-in-placeconcrete,206,575cellular,4column,19-20,201,259-261,326,344-345,429,449-452,559-560compressivestrength,12compressivestrengthof,12fireresistance,13-14,560floorslab,440footing,19-20,180,206-207,259,450,560foundation,2,182,200-201,425,558foundationwall,210,425girders,202,428,566high-strength,99,239incompression,102,238,265,429-430,499intension,99,272-273,429making,12,429modulusofelasticity,279,412,451-452pavement,295propertiesof,11-12,273,390,452pumps,14reinforcedconcretecantileverretainingwalls,177Roman,11,121,451slab,14,103,206,390,459,559-560strength,1,99,239,259-261,389,449,544-545,575testing,272-274,434,452type,5-6,74,103,210,261,346,413,461,545,556weightof,4-5,177,213,412,574Concretefloors,beam-supportedbeamandgirder,508Concretejoistsreinforced,4-5,120,207,272,345,433,450-451,559-560steeland,12,211-212,281,326,432,544Concretemasonryunits(C.M.U.s)lintel,390piers,11,100,211,450Concretemasonryunits(CMUs)shapes,99,320,430,450Concrete,precastarchitecturalandstructural,12curtainwalls,461Concrete,reinforcedprestressed,4,450Concreteslab-on-groundfibrous,4joints,11,196,239-240,295,461,533-534,565Concreteslabsflat,177,240,412,469posttensioned,12,559-560ribbed,12Conditionsweather,7,579working,281,432,516,579Conductivityelectricalthermal,13Configurationscross-sectional,325-327Connectiondouble,263,520welded,268,345,459,533,565-566Connectionsfailureofboltedconnections,511framedbeam,527-528friction-typeconnections,515moment,75,239-240,345,454,509-510,566pin,76,123,238,344,455-456,545,565rigidjoints,239,461simple,204,343-345,456,511Connectors,576Consolidationofsoilprimary,209secondary,209settlement,182,210Constructiondrainageafter,412modern,7,121,257,375,508quality,235,556team,2,555Constructioncompanycharacteristics,15-16,105,235,450Constructiondocuments(CD)set,14,103,236,432,469,577stage,14ConstructionjobsintheUnitedStates,9Constructionsurveyscontrol,238,515-516Constructionvolumesshrinkage,539Continuouscolumns,240Contourlinesnatural,105values,68,105-106,308Contractor(general/prime)framing,13-14,545,555tile,14Contributionmarginratio,431Contributoryarea,202Controllimitslower,239,412,515upper,468Conversion,228-229,417Coordination,484,556Coordinator,2Coping,527Corners,7,67,238,293,539Costactual,4,534deterministic,12operating,14other,3-4,461,535,555Costsdirect,516,560Counterfortwall,177Couple,52-53,250,264,361,393Coverrequirementsslabs,244,560Cranetower,578-579Crawlspacesinsulation,299Creep,280Crew,533Criticalbucklingload,452-454Criticalload,452-454Crosssectionslevel,10,168,203-204,261,397,451Crowding,292Currentperiodnetincomeuse,14,23-24,124,236,289,490,526Curtainwalls,glassfiber-reinforcedconcrete(GFRC)componentsof,29,179,247,577Curtainwalls,glass-aluminumdetailsof,560Curtainwalls,precastconcrete(PC)advantagesof,212Curvesdata,274length,275Cutandfillsiteplan,555DDatacurve,275-277Deadloadsexample,2-3,23,107,204,260,350-351,386,456,517onabeam,250,259,346,446Decisionmakingobjective,55Deckingcorrugated,14roof,102-103,200,429,562-567Deepfoundationdesignandmethodsdesignconsiderations,102Deepfoundationscaissons,211Deflection,6,43,236,273,308,342,375,451,573Deflectioninbeamsdeflectionformulas,413Deformationloadsand,1,235,433Deformationandstrainconceptof,10,15-16,121,202,260,319-320,367,375-376,478Degrees,16Designlumber,431,488-489truss,1,121,267,478,509Designer,3-4,122,199-200,534Detaildrawingssections,571standards,559Detailssection,249,560Diagonalbridging,431Diagonaltensioncounterssupplementaryproblems,89Diagramsmatrix,14Diaphragms,236-237,567Differentvalues,110,433Dimensionalcoordination,484Dimensioningsystems,indrawingstolerances,577Discovery,273,452,533Distancemeasurementrecording,274slope,45,205,363-365traditional,7Distributedloads,113,249,353Distributivearea,202Dome,4,23Doubleshear,520Drawingglass,12stonework,11Drawingboards/tablescovers,14Drawingfilesorganization,560Drawingsassembly,559detail,560floor,560-561large-scale,557presentation,557site,557-559Dressed,11Drift,433Drilledpiersreinforcing,575Droppanels,560Dryunitweightmaximum,3,41,178,265,398,478-479,538Drywallpartitions,14Ductilematerials,273Ductility,273Durability,235EEarthpressureactive,178atrest,15lateral,8,177-178,247,437Earthquakeloadsbracedframesanddescribed,11,295,309,574Earthquakesepicenter,8intensity,8magnitude,8waves,8Eccentricconnectionsbolted,263,509-511,564-565Eccentricloads,501Economyframingdesigncriteriaandstructuraldesignand,4,479,579Edgedistance,514Effectivelength,458,539EiffelTower,2-3Elasticlimitsstressand,273-274,308,412Electricaldesignandwiringaccessibility,555mechanicalequipment,218,558specification,432,487Electricalresistancesystemselements,4-5,99-100,202,293,319,376,508,556Electronicdistancemeasurement(EDM)accuracyof,376precise,274Elevationsfour,309Engineer,12-13,23-24,130,283,479Engineeredwood,12Entry,557-558Epicenter,8Equilibriumparticle,16,98rigidbody,22,157-158two-forcemembers,125Equilibriumofrigidbodiessimplebeamswithdistributedloads,113Equipmentadequate,516,557available,13,433,515,571desired,556erection,567poor,555Erection,510,567Errormultiple,113Estimatefinal,479Euler’scriticalbucklingload,454Euler’sequation,452Evidence,121Excavationgeneral,577Excavation(s)open,209,556Exteriorfinishmaterialsplywood,5,209,569woodshingles,14Exteriorinsulationandfinishsystem(EIFS)basics,413categoriesof,549Externalforces,7,18-19,102,235,374FFactorofsafetyagainstoverturning,179-180Factoredloads,432Failure:formwork,577Failureofboltedconnectionsbearingfailure,511-512endtear-out,511shearfailure,278,401,511tensilefailure,513Fallprotectionsystems,5-7,16,180-181Family,14,452Fatigue,280Fields,281Filletwelds,534-535Finishgrade,422,557Firesprinklers,13,571Firealarmsystemscoderequirements,556Fireblockinglocations,555requirement,422Fireprotectionequipmentandsystemsdesigncriteria,203Firesafetyprogramsresponse,3,281Fire-ratedroofs,14FirthofForthBridge,3,122Flangelocalbuckling,442Flatslabs,240Flat-roofconstructionoverhangs,7Flexuralstrain,376Flexure,293,375-376Flexureformula,380Flooropenings,11-12,203Floorssheathing,244,498,563Flownetsupliftforces,8FluorescentlightsourcesFrequency,102Flux,533Flyash,560Flyingarches,12Footingspread,210,368Footingscombined,12,229individual,211loadson,227locationof,182post,230settlementof,182585sizeof,14,210soiland,181trapezoidal,181Forceanalysisbymethodofjointsofpinnedframes,155Forcesystemsconcurrentforcesystems,21ForcesystemsVarignon’sTheorem,47-48,183Forcedwarm-airsystemsductwork,572Forcesexternalandinternal,18lineofaction,15-18,106,259,537momentofaforce,42-44parallelogram,23-27Forcesresultantoftwoparallelforces,59-60resultants,156,394Forcestypesofforcesystems,21Varignon’stheorem,47-49,183vector,23-24,181Foundationwallsbasement,561thickness,5,177,230Foundationsdeep,210-211earthpressure,178gradebeam,211mat,210raft,210shallow,210Foundations,deeppiles,211,450Framessidesway,471Framingfloors,14,244-245,545,556hours,14Framingdesigncriteria,203Framingplan,208,425,562Free-bodydiagrams,64-65Free-bodydiagramsofrigidbodies,74Freeway,450,555Frictionlossescompressedair,211Friction-typeconnections,515Fullyrestrained,305GGalileo,3,61,257,376,452GalileoGalilei,3,61,257,375Generalconditions,4,178,201,433,504,551Generalledgeraccounts,433Generaloverheadbudgetestimate,9incremental,394Generalshear,394Geographicinformationsystem(GIS)direction,7,15-17,235-236,476,545,566-567point,8,16-18,106,211,308-309,380,557Geographicinformationsystems(GIS)history,6,121Geometricconstructionreduction,6,517Geometry,11,99,236-237,257Girts,461Glacialtill,555Gothiccathedrals,3,499Grade,211,279,386,467,557-559Gradebeams,211Grading,577Granularsoiluniform,14,210well-graded,178Graphicalvectoraddition,33Gravityloads,236,547,566Gravityretainingwalls,177Gridlines,214Groove(butt)welds,534Grossprofitdefinedmargin,258,433Ground,1,15-16,100,199,257,485,533,560Groundresistance,16Groundwatertablepositionof,412586Gussetplates,238Gypsumwallboard,5,559HHammersdiesel,211pile-driving,211soft,210Head,44,308,515Header,204-205Heatgaincalculationprocedure,320,534Heatingandcoolingsystemsdefinitions,438Heating,ventilating,andair-conditioning(HVAC)protectionof,9Heating-coolingplanfirst-floor,483,568Heavytimberconstruction,14Hinges,237,344,439,549Hooke,Robert,273,375Hooke’slaw,273,377-378Horizontaldrainagefabric,12Horizontalline,107Horizontalmillingmachinescontrols,203Housing,563Humanerrorfactorsoverload,6,432Hydrostaticpressure,114IImageanalysisclassification,5,258,468Importancefactorice,320Inelasticbuckling,468Informationamountof,308In-placedensitydeterminationof,137,277,328,465Inspectionwork,365,533-534Inspector,534Integrationofbuildingsystemsdetailsandconnections,575Intensityearthquake,8Interestratefactors,534Interiorplanningentry,557-558Intermediatecolumnrange,468Intermediatestiffeners,549Internalforces,10,18,141-143,257,347Interpolatingcontourlinesmathematical,61Ironcast,12,273,451wrought,12,279,451Isometricdrawingaxes,311JJointscomponent,137-138expansion,295-296movement,10,238,295,575-576slip,551thermalmovement,299,576Joistheader,207Joists.SeealsoBeamsbridging,429hanger,509tail,30Judgment,53,468Jurisdiction,6Just-in-time(JIT)manufacturingrequirementsof,9-10,121,430,563KKfactor,486K-brace,549Kern,499Kernarea,499Kernareas,501Kneebrace,510Knee-bracing,238-239K-valuesforwoodcolumns,486Llaboravailability,203,555Lagscrews,576Landscaping,556-557Landslides,5Lateralbucklinginbeams,406-408intrusses,450LateralforcesEarthquake,5,235,434,458,568Lateralinstability,235,430,451Lateralstabilityloadtracingcombinationknee-braceandrigidcolumnbase,239diagonaltensioncounters,148-149,238gussetplates,238rigidbeam/columnjoints,239three-dimensionalframes,244LeadershipinEnergyandEnvironmentalDesign(LEED)ExistingBuildings,555Levelline,274-275,431rod,294Lightconstructionsteelframe,5,242,551Lightcontrolstransmission,98-99Lightingdesignconstructiondrawings,570Limitstates,432-433Lineofactionforceand,43Lineweightsvarying,7,308Linescurved,100dimension,99,295linesofconstructionarrowhead,24sectioncut,142,249,520,575Loadcombinations,433-434Loadfactors,433-434Loadresistancestructuralfailures,293Loadtracinglateralstabilityloadtracing,235tributaryareas,486Loadsanticipated,433,515,579beam,7,21,102,201,259,308,342-348,375-377,450,508-509,571ceilingjoists,204combinations,116,243,433-434deadloads(DL),218dead/liveloadratio,6-7determining,2,128,200,304,309,349,392,451gravity,1-2,23,100,235-236,308-309,434,499,547,566headers,203,461intended,1-2,203,413,576liveloadreduction,6nonuniformdistributionof,206path,7,201,560-561rafters,11,200,431,508rain,436rooflive,6,234,434-436,497rooftrusses,125-126,199,563staticloads,5tributarywidth,213,423uniformslabloads,206windloads,7,78,148,567Low-carbonsteel,273Lumbermoisturecontent,488MManganese,274Manual,AISC,413,470ManualofSteelConstruction,433,470-471,508Mariotte,Edme,375mark,74,515Masonrycourseof,560mortar,11weights,14,268Masonryconstructionothermaterials,121Mass,5,15,101,235,309,581Materialorder,413stored,258Materialpropertiesallowableworkingstress,281compressiontests,274cyclicstress(fatigue),280modulusofelasticity(Young’smodulus),281Poisson’sratio,281safetyfactor,178-179,290,433,470,516torsionalstress,294Materialsdamaged,10solid,1,242-243,281,392Measurementarea,581weight,42,581Measurementsdistance,294mechanicaldrawings,558Mechanicalequipment,218,558Metaldeck,428Metalroofingaluminum,4,573cooling,13,533-534Meter,43,581Meters,8,581Metricchangesstructuralsteel,304Mildsteel,276,469Minimumflexuraltensionsteelareaone-wayslabs,177rectangularbeams,403Modulardraftingprinciples,10Modulusofelasticity(Young’smodulus),281-282Molten,533Momentarm,42-43,320Momentmagnification,502Momentofinertiagrosssection,534Momentsbending,10,43,99,238-243,259,308,346-353,375-379,501-503,508,575couple,53,252,264,393rotational,43-45,252,393,547Motion,7-9,15,257,320Multiforcemembers,155-156NNAHBResearchCenterhomeprogramarchitect,12-13,555foundations,182Nailspopping,412Naturalperiodofvibration,8-9Neckingdown,278Negativemoment,365Netarea,514Neutralaxis,377-380Nickel,274Noiseandvibrationloss,9-10,238,273restraint,241,429,461source,437,533nonbearing,207Nonferrousmetalsbrass,279bronze,279zinc,279nonstructural,237,564Normalstressminimum,3-4,183,265-266,439,454,514Nuts,515-516OOccupancyload,6,119-120,233,433-436,560OccupationalSafetyandHealthAct(OSHAct)coverage,258OccupationalSafetyandHealthAdministration(OSHA)inspector,534Officeproject,559space,12,559-560Openshop,556Ordinate,275Orientedstrandboard,242OSB,242Overhang,46,343,387Overload,6,432,457Overturning,45,178-180,250,576Owner’sprogram,556Owningandoperatingcostsrepair,561PP6,405Pad,344Parallelaxistheorem,326-327Parallelogramofforces,23Particle-sizedistributioncurves,359-360Path,7,167,201,545,560-561Pavement,295Penetrationresistancevaluesfor,13,433Performancebondformof,12Perimeter,216,319,514Photographs,235,575Pierformed,12Pilecap,211Pilecaps,211Pilefoundations,211Pilesgroup,211hammers,211Pinnedframesforceanalysisof,196multiforcemembers,155-156Pinnedjoints,124Pipingequipmentandsystemsthermalexpansion,13,295,573Plainconcrete,177,260Planetrussesstabilityanddeterminacyof,127zero-forcemembers,152-154Plank-and-beamconstructionlimitations,102,450Plants,4,560Plastic,273,433-434,549Plasticdesignofcontinuousbeamsplasticmoment,437-443plasticsectionmodulus,441-442Plasticflow,437PlasticlimitPlasticity,273plastics,273Plategirders,weldedbendingstrength,442Plates,238,262,336,409,470,515,568Platformframeconstructionfloortrusses,564-565Plumbingarrangement,13-14,571Plywoodpanels,205,562-563Pneumaticstructures,12Pointofinflection,349Polygon,39Porosity,533Positioning,73Positivemoment,349Posttensioned,12,559-560Powercapability,533Prestress,99Processaveragedescribing,23Projectbased,6,556Projectdeliverdesignphase,5,199Propertydamage,277Pumps,14Purlins,78,204,563Pythagoreantheorem,29,105-106QQualityfunctiondeployment(QFD)structureof,1Qualityimprovementstructurefor,563RRadians,270Raftership,234jack,234Reactions,18,100,200-202,342,508Reduction,4,275,517ReimsCathedral,2Reinforcedconcretereinforcingsteel,240,293Reinforcedconcretecantileverretainingwalls,177Reinforcingbar,294diameter,576number,303-304Reinforcingsteel,240,293Relationships,23,110,356,440Reliability,433Residentialplanninginteriorspaces,12siteanalysis,555Resistancefactor,281,432-435Resolutionofjoints,130Resonance,9,102Respect,109,238,258-259,319,348Rest,15Retainingstructuresretainingwalls,177-178,558Retainingwallsfactorofsafetyof,290tiltingof,182Returnonsales,572Ridgebeam,204Rigidbodiesforcesand,16,257,374free-bodydiagramsof,65Rigidbody,17-18,157-158Rigidconnection,124,241,460,548Rigidframes,243,471,547Rigidjoints,239,461Rolledshapes,335,442-443Roller,76,134,344Roofpitch,411ridge,226Rooftrussesmodified,33,149,241,505,563prefabricated,563Roofingbuilt-up,5Roofsdecking,14,203-205Roof(s)inverted,104liveload,6-7,99,433-436,485snowload,6,32,113,268,435-436,485Roofs,low-slopedecks,14membranes,102Rulesofthumb,431Runoff,412R-valuesofglass,12SSafetycrane,578factorof,178-180,281,432,469-470,516steelconstruction,432-433,470-471,515Safetyandhealthteamcontractors,555Saint-Venant’sprinciple,292Schematicdesign,557Schools,14Scope,179,258,442Sectionmodulus,385-386587sectionsfull,291,438,488half,150,351,375importance,412removed,10,144,210subdivisionof,123Seepageforcesvelocity,7-8,15Semigraphicalmethodforload,shear,andmomentdiagramsbasiccurvesand,358generalconsiderationsfordrawing,361Sensitivity,3,295Serviceloads,434Shaft,44,269Shakes,8,233Shallowstructuralfoundationspreadfootings,13,230Shearandmomentdiagramstypical,124,206,265,344,423,483,509-510,555Sheardiagramsbeamsand,239,308,342,407,537,575-576Shearstrengththeorystresspaths,292-293Shearstressequations,114,254,300,309,349-350,386,470,508longitudinalandtransverse,537Shearstudeconomy,9,527Shoring,298,560Shortcolumns,451Shrinkage,539SIsystemofunitsloadsin,22,100Sidesway,429,471Sidewalk,295Signconventionsforshearandmomentdiagrams,348-350Silicon,274Siteanalysiscirculation,246,558neighborhood,555-557Sketchroofsheathing,234Slendernessratio,456Slip-criticalconnections,515-517Slotwelds,535Snow,6,32,105,224,268,435-436,485Snowload,6,32,113,268,435-436,485Society,281,508Softwarelimits,412-413Soilexpansive,210foundationsystems,210granular,177Soleplate,498SouthfaceEnergyandEnvironmentalResourceCenterpartners,554-563Spaceforcesystems,21Spantablesfloorandroofbeams,429Splices,509Sprinklers,13,571Stairswidthof,202Standardframedbeamconnectionsclipanglesand,532Staticequilibrium,61,130Staticloads,5Staticsstaticalindeterminacyand,86Stationpointlocating,183,309Steelbearingplates,302,533tensilepropertiesof,12workers,27,508Steelboltedconnectionsdesignstressesforbolts,515SteelframeconstructionRvalues,454roofframing,205,477,562Steelframingdetailsbracedframes,240,545,567examplesofconnectiondetails,545rigidframes,243,471,547588shearwalls,236-237,471,546,560Steel,light-gaugefloorframing,202,436,569framingmembers,14,244,429wallframing,204,573Steel,structuralboltsandwelds,510low-carbon(mild),283Steinman,DavidB.,103Steps,137,480Stiffnessdeflectionand,10,413Stone,exteriorcladdingwiththin,14,103,452Stonehenge,11,451Stonework,11Stormdrainagesystemgutter,51Strainhardening,434Strawbaleconstructionload-bearing,11,573Strengthofmaterialsmaterialproperties,10-11,280,441staticallyindeterminatemembers(axiallyloaded),300-301stressandstrain,257,452thermaleffects,295-296Stressaxial,259-261,308,433-434,450,537compression,535normal,260-261,392,488Stressandstrainbearingstress,183,265,421,498,513deformationandstrain,270-271normalstress,261shearstress,262,342,394,511-512Stress(es)neutral,378-380,453Stress/strainrelationshipstress-straincurves,274-275Strike,8Stringers,1Structuralconnectionssteelboltedconnections,509weldedconnections,533Structuraldesigneconomyand,201Structuraldrawings,545Structuralmembersstressesin,166,283,384-386,515Structuralmodelsexteriorwalls,239,569Structuralsteelcompositecolumns,433density,178-179,269tensilestrength,12,277-278,437,516yieldstrength,277,434,505Structuralsystemsbeamswithdistributedloads,113new,3-4,128,258,432,487,533,556Structuralsystemspinnedframes,155planetrusses,121Structuralsystemsthree-hingedarches,166Suction,7-8,249Superposition,421Supportconstraints,74Surfaceplatesgradesof,437Suspendedceiling,218,423Suspension,98-102,527Three-hingedarches,166Tileceramic,14Tilt-upwalls,reinforced-concreteaesthetics,3,560Timber,1,41,119,209,265,344-345,383,450,508,559Timberconstruction,14,412,559Timberpiles,211Topography,8Torque,44,294,516Torsionangleoftwist,294warping,294Torsionalstressstressandstrainand,273Totalqualitysuccessof,555Transversebeamssupport,11,84-85,99,200-202,342-343,381,458,508,559Travel,10,199-200,561Triangulation,238Tributaryarea,202,498Trucks,14,556Truestress,275Truss:bridge,84,130,486,527TrussesFink,240-241methodofjoints,196,247methodofsections,149,247staticallyindeterminate,127Trusses,woodspecifying,433Truth,15Two-storywoodlightframebuilding,step-bystepprocessfloorsheathing,244,565Typeofconstruction,556typesofsurveysrange,6,178,432,454-455TWallcurtain,299,423-424openingsin,203Wallsinfill,461tilt-up,206Walls,masonrybearinginreinforcedconcrete,293Warping,294Watermolecule,16resistance,16,177Webbuckling,407Webstiffeners,407Webs,407,571WeightTables,13,288,332,383,470,527-528Telephonesystemstypesofsystems,21Temporarysheds,12Tensilestrength,12,277-278,375,516Tensilestress,261,376,514-515Tensionmembersdeformations,10,242,257,378,547Tensionrods,563Text,10,16,258,431-432,471,528Thermalexpansion,13,295,573Third-angleprojectionplanes,244,263Three-dimensionalspace,22UUltimateload,376Ultrasonictesting,534Uniformlydistributedload,114,259,360-361,382Unitsofforce,16Universaltestingmachines,274VValuescaleuses,522Varignon,Pierre,23Vault,11-12Vectoradditionscharacteristicsofvectors,23componentmethod,33Verticalbandsawswelding,239,411Verticalline,69Verticalsupports,100Vibrationsnaturalfrequency,9resonance,9,102Views,257Visualinspection,128Visualize,105,319Voidratiocritical,432,454-455,559Wofconcrete,12,223Weldedconnectionseccentricityinweldedjoints,542typesofweldedjoints,534Weldedjointsfilletwelds,534-535groove(butt)welds,534Weldinginspection,534shieldedmetalarc,533Weldingandcuttingarcwelding,508Weldsfillet,534-535groove,534-535slot,535Wellpointsinstalled,508Wide-flangesection,326,406Wide-flangesections,455Windowsframe,568-570Woodengineered,12girder,113,201,428Woodlightframe(WLF)constructioncontemporary,166evolutionof,11Woodshinglesmaterial,122,203,276,392,504,559-560Worktypeof,14,300,535Wrenchescombination,515torque,516Wroughtiron,12,295,451YYieldpoint,276,437,451Yieldstrength,277,434,505Yieldstress,279,432,452Young,Thomas,281-282,508Young’smodulus,281-282ZZinc,279Zoningregulations,556589590
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