Mechanics of Materials 2NdPytel, Kiusalaas材料力学3

MechanicsMechanics ofof MaterialsMaterials 2Nd2Nd Pytel Pytel KiusalaasKiusalaas材料力学材料力学3 3 Any increasein theaxial loadbeyond P cr increasesthe deflectiond maxcatastrophically causing the column to fail On theother hand if theaxialload isdecreased slightlybelow the critical value the oppositee ect ours the columnbees straight The critical load canthus bedefined asthemaximum axial load that a columncan carryand stillremain straight However at the critical load the straightposition of the column is unstablebecause the smallestsideways forcewould causethe columnto deflectlater ally In otherwords the lateralsti ness of the column is zerowhen P P cr b Euler s formulaTheformula for the criticalload of a columnwas derivedin1757by Leon hard Euler the greatSwiss mathematician Euler s analysiswas basedon thedi erential equation of the elastic curved2vdx2 MEI a which weused in the analysisof beamdeflections in Chapter6 Figure10 3 a shows anideal simply supported columnAB subjectedtothe axial load P We assume that thisload is capable ofkeeping the columnin alaterally displacedposition As in the analysisof beams we letx bethedistance measuredalong the column anddenote the lateral deflectionby v The bendingmoment Macting atan arbitrarysection can be obtained fromthe free body diagramin Fig 10 3 b M andv shownon the diagram arepositiveaording to the signconventions introducedinChapter4 Theequilibrium equationSM A 0gives M Pv which uponsubstitutioninto Eq a yieldsd2vdx2 PEIv 0 b Equation b is ahomogeneous linear di erential equationwith constantcoe cients The solution which may be verifiedby directsubstitution isv C1sinffiffiffiffiffiffiPEIrx C2cosf fiffiffiffiffiffiPEIrx c The constantsof integration C1and C2 are determined by theconstraintsimposed by the supports 1 vj x 0 0 which yieldsC2 0 2 vj x L 0 resulting in0 C1sinffiffiffiffiffiffiffiffiffiPL2EIr d Equation d can besatisfied withC1 0 but thissolution isof nointerestbecause itrepresents thetrivial caseP v 0 Other solutionsareffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiff iffiffiffiffiffiPL2 EI p 0 p 2p 3p orP n2p2EIL2 n 0 1 2 3 e FIG 10 3 a Buckling of asimply supported column b free body diagramfor determiningthebending momentM 10 2Critical Load373CopyrightxxCengage Learning All RightsReserved May notbe copied scanned or duplicated in wholeor inpart Due toelectronic rights some thirdparty contentmay besuppressed fromthe eBookand or eChapter s Editorial reviewhas deemedthat anysuppressed contentdoes notmaterially affectthe overalllearning experience Cengage Learningreserves theright toremove additionalcontent atany timeif subsequentrights restrictionsrequire it The casen 0can bediscarded becauseit againyields thetrivial caseP v 0 The criticalload is obtained bysetting n 1 yielding Euler sformula P cr p2EIL2 10 1 This is the smallest value ofP that iscapableof maintainingthelateral dis placement The correspondingequationof the elastiurve called themodeshape isv C1sinpxLas shownin Fig 10 4 a The constantC1is indeterminate implying that themagnitude of the displacementis arbitrary The elastiurves correspondingto n 2and n 3are shownin Figs 10 4 b and c Because thesemode shapesrequire axial loads largerthanP cr they can be realizedonly if the column is bracedat itsmidpoint forn 2 or at its thirdpoints for n 3 The criticalloads ofcolumns withother endsupports can be expressedinterms of the criticalload for a simply supported column Consider forexample the column with built in endsin Fig 10 5 a Its modeshape hasinflectionpoints at the distanceL 4from eachsupport Because the bendingmoment iszero ata pointof inflection due tozero curvature the free bodydiagram in Fig 10 5 b shows that the middlehalf of the column is equiv alent toa simply supported column with an e ective length L e L 2 Therefore the criticalload forthis column isP cr p2EIL2e p2EI L 2 2 4p2EIL2 10 2 This isfour timesthe criticalload fora simply supported column The criticalload fora cantilever column of length Lcan bedeterminedby the same method This columnbehaves as the bottom or top half of asimply supported column as shownin Fig 10 6 Therefore its e ectivelength is L e 2L which resultsin the critical loadP cr p2EIL2e 14p2EIL2 10 3 This isone quarter of the criticalload fora simply supported columnof thesamelength The e ective lengthof thepropped cantilever column in Fig 10 7canbe shownto beapproximately L e 0 7L which is the distancebetween thepointof inflectionand thesimple support This valueyields for the criticalloadP cr p2EIL2e p2EI 0 7L 2A2p2EIL2 10 4 which istwice the criticalloadfora simply supported column FIG 10 4First threebucklingmode shapesof a simply supportedcolumn FIG 10 5Buckling of a columnwith built in ends 374CHAPTER10Columns CopyrightxxCengage Learning All RightsReserved May notbe copied scanned or duplicated in wholeor inpart Due toelectronic rights some thirdparty contentmay besuppressed fromthe eBookand or eChapter s Editorial reviewhas deemedthat anysuppressed contentdoes notmaterially affectthe overalllearning experience Cengage Learningreserves theright toremove additionalcontent atany timeif subsequentrights restrictionsrequire it 10 3Discussion ofCritical LoadsIn the previoussection we discoveredthat the critical or buckling load of acolumn isP cr p2EIL2e 10 5 where thee ective lengthL eof the column isdetermined by the typesof endsupports For a simply supportedcolumn we haveL e L Equation 10 5 shows thatP crdoes notdepend on the strength of thematerialbut onlyonthemodulus of elasticity and the dimensionsof thecolumn Thus two dimensionallyidentical slendercolumns one ofhigh strength steel and theother ofordinary steel will buckleunder thesamecritical loadbecause theyhave thesame modulusof elasticity The criticalload obtainedfrom Eq 10 5 is physicallymeaningfulonly if the stress at bucklingdoes notexceed the proportional limit Thestress in the columnjust beforeit bucklesmay befound bysubstitutingI Ar2into Eq 10 5 where Ais the cross sectional areaand r is the leastradius of gyration of the cross section 1This substitutionyieldss cr P crA p2E L e r 2 10 6 1Here we areusing rto denotethe radius ofgyrationto conformto AISCnotation Do notconfusethis rwith theradius of a circle FIG 10 6Buckling of a cantilevercolumn FIG 10 7Buckling of a proppedcantilevercolumn 10 3Discussion ofCritical Loads375CopyrightxxCengage Learning All RightsReserved May notbe copied scanned or duplicated in wholeor inpart Due toelectronic rights some thirdparty contentmay besuppressed fromthe eBookand or eChapter s Editorial reviewhas deemedthat anysuppressed contentdoes notmaterially affectthe overalllearning experience Cengage Learningreserves theright toremove additionalcontent atany timeif subsequentrights restrictionsrequire it where s cr iscalled the critical stressand theratio L e risknown asthe slen derness ratioof the column Thus P crshould beinterpreted asthe maximumsustainableload onlyif s cr100 as illustratedbytheplot inFig 10 8 The plotalso showsthat thecrit icalstressrapidly decreasesasthe slenderness ratioincreases It must bepointed outthat Fig 10 8shows the stressatfailure not theworking stress Therefore it isnecessary todivide thecritical stressby asuitable factor ofsafety toobtain theallowable stress The factor of safetyshould allowforunavoidable imperfectionsalways presentin areal column such asmanu facturing flawsand eentricityof loading A columnalways tendsto bucklein thedirection thato ers theleastresistance tobending For thisreason buckling oursabout theaxis thatyieldsthe largestslenderness ratio L e r which isusually theaxis ofleastmoment of inertia of the cross section FIG 10 8Critical stress versus slenderness ratio forstructural steel columns Forslenderness ratioslessthan100 thecritical stress is not meaningful 376CHAPTER10Columns CopyrightxxCengage Learning All RightsReserved May notbe copied scanned or duplicated in wholeor inpart Due toelectronic rights some thirdparty contentmay besuppressed fromthe eBookand or eChapter s Editorial reviewhas deemedthat anysuppressed contentdoes notmaterially affectthe overalllearning experience Cengage Learningreserves theright toremove additionalcontent atany timeif subsequentrights restrictionsrequire it Sample Problem10 1Select thelightest W shape that canbeused as a steelcolumn7m longto supportanaxial load of450kN with a factor of safety of3 Use s pl 200MPa andE 200GPa Assume that the column is 1 simplysupported and 2 a propped cantilever SolutionMultiplying thegiven designload bythe factor of safety we getPcr 450 3 1350kN forthe minimum allowable criticalload The selectedsection must be abletocarry thisload withoutbuckling orcrushing The crushingcriterion isPcr A6750mm2 The lightestsection thatmeets boththe bucklingand thecrushing criteriaistheW200 59section Answerwhich has Imin 20 4 106mm4and A 7550mm2 1377CopyrightxxCengage Learning All RightsReserved May notbe copied scanned or duplicated in wholeor inpart Due toelectronic rights some thirdparty contentmay besuppressed fromthe eBookand or eChapter s Editorial reviewhas deemedthat anysuppressed contentdoes notmaterially affectthe overalllearning experience Cengage Learningreserves theright toremove additionalcontent atany timeif subsequentrights restrictionsrequire it Problems10 1A simplysupported steelcolumn is8ft longand has a squarecross section ofside lengthb If the column is to supporta22 kip axialload determine thesmallestvalue of b that would preventbuckling Use E 29 106psi for steel 10 2Solve Prob 10 1if the column ismade ofwood for whichE 1 6 106psi 10 3A40 mm by80 mm timber 2 2m long is used as a column with built inends If E 10GPa ands pl 30MPa determine thelargest axialload that can becarriedwith a factorof safety of2 10 4An aluminumtube oflength8m is used as a simplysupported columncarryinga1 2 kN axialload If theouter diameter of thetube is50mm putethe innerdiameter that would provide a factorof safety of2against buckling UseE 70GPa foraluminum 10 5An aluminumcolumn6ft longhasasolid rectangularcross section3 4in by2in The column is securedat eachend with a boltparallel tothe3 4 in directionas shownin thefigure Thus the endscan rotateabout thez axis butnot aboutthey axis Find thelargest allowableaxialloadusing E 10 3 106psi s pl 6000psi and a factorof safety of2 10 6Two C310 45channels arelaced togetheras shownto form a sectionwithequal momentsofinertiaaboutthex and y axes the lacingdoes notcontribute tothebending sti ness If thissection is used asasimplysupportedcolumn determine a the shortestlength for which the column wouldfail bybuckling and b thelargest allowableaxialloadthatcanbe supportedby a12 m long column with afactorof safetyof2 5 Use E 200GPa ands pl 240MPa 10 7Both members of the truss areW16 67sections Determine thelargest loadWthatcanbe safelycarried bythe truss Use E 29 106psi s pl 36 103psi andafactorof safetyof1 6 Assume thatcross bracing not shownin thefigure prevents deflectionof joint B outof the plane of thetruss 10 8A W section is used asasimplysupportedcolumn8m long Select thelightestshape thancan carryan axialload of270kN with afactorofsafetyof2 5 Use E 200GPa andspl 200MPa FIG P10 5FIG P10 615ftCA BW20ftFIG P10 7378CHAPTER10Columns CopyrightxxCengage Learning All RightsReserved May notbe copied scanned or duplicated in wholeor inpart Due toelectronic rights some thirdparty contentmay besuppressed fromthe eBookand or eChapter s Editorial reviewhas deemedthat anysuppressed contentdoes notmaterially affectthe overalllearning experience Cengage Learningreserves theright toremove additionalcontent atany timeif subsequentrights restrictionsrequire it 10 9Select thelightest W section fora40 ft long columnwithbuilt in endsthatcan carryan axialloadof150kips withafactorofsafetyof2 Assume thatE 29 106psi andspl 30 103psi 10 10The twomembersof the pin jointed woodstructure ABChave identicalsquarecross sectionsof dimensionsb b A1600 lb verticalload actsat B Determine thesmallestvalue ofbthatwouldprovideafactorofsafetyof2 5againstbuckling Use E 1 5 106psi forwood andassumethatjointBis bracedso thatitcan moveonly in theplaneof thestructure 10 11The5 m long woodcolumnis built in atitsbase andstayed bytwo cablesatthe top The turnbucklesin thecables areturned untilthe tensileforce ineachcable isT Determine thevalue ofT thatwould causethecolumnto buckle UseE 10GPa forwood 10 12The L76 76 12 7angle section isused asacantilevercolumnoflengthL Find themaximum allowablevalue ofL if thecolumnis notto bucklewhen the12 kNaxial loadis applied Use E 200GPa 10 13The24 ft long steelcolumnisan S8 23section thatisbuiltinatbothends The midpointof thecolumnisbraced bytwo cablesthat preventdisplacementin thex direction Determine thecritical valueoftheaxialloadP Use E 29 106psi forsteel 2ft1 5ft3ft16001b1 0ftFIG P10 10FIG P10 11FIG P10 12FIG P10 13Problems379CopyrightxxCengage Learning All RightsReserved May notbe copied scanned or duplicated in wholeor inpart Due toelectronic rights some thirdparty contentmay besuppressed fromthe eBookand or eChapter s Editorial reviewhas deemedthat anysuppressed contentdoes notmaterially affectthe overalllearning experience Cengage Learningreserves theright toremove additionalcontent atany timeif subsequentrights restrictionsrequire it 10 14The solidaluminum barof circularcrosssection is fittedsnugly betweentwoimmovable walls Determine thetemperature increasethatwouldcausethebarto buckle Use E 10 3 106psi anda 12 8 10 6 F foraluminum 10 4Design Formulasfor Intermediate ColumnsInthe previoussection we showedthat ifa columnis su ciently slender asmeasured bythe slenderness ratioL e r buckling oursatastress thatisbelow theproportionallimit At theother extreme we haveshort columns where thelateraldis placements play a negligiblerole inthe failuremechanism Therefore thesecolumns failwhen P A reachesthe yield stress ofthematerial Various designformulas havebeen proposedfor columnsof inter mediate length which bridgethe gapbetween shortand long columns These formulasare primarilyempirical innature being derivedfrom theresultsof extensivetest programs Material propertiesplayamajor roleinthe failureof intermediatecolumns Hence di erent designformulasfor di erent materialscanbefound invarious engineeringhandbooks anddesigncodes a Tangent modulus theoryConsider a column made of materialthat hasno distinctyield point such asaluminum Suppose thatthe pressive stress s cr justprior tobucklingexceeds theproportionallimitspl as indicatedinthe pressivestress strain diagraminFig 10 9 Any additionalincrement ofstrain d wouldresult inthe stress increment ds E td where E tisthe slope ofthediagramat thepoint s s cr Thus thebendingsti ness ofthecolumnat bucklingisdetermined by E t which issmaller thanthe elasticmodulus E To aountforthe reducedsti ness the tangent modulus theory replaces EbyE t inEq 10 6 resulting inthe followingexpression forthecriticalstress s cr P crA p2E t L e r 2 10 7 Although Eq 10 7 aounts forthe nonlinearityofthestress strain dia gram beyondtheproportionallimit its theoreticalbasis issomewhatweak Therefore this equationshould beviewed asan empiricalformula However the resultsobtainedfromEq 10 7 are insatisfactory agreementwithexperimental results Because the slopeofthestress strain diagram called the tangentmod ulus is notconstant beyondtheproportionallimit the evaluationof thecriticalload fromEq 10 7 is notstraightforward The di culty isthat theslopeE tatthecriticalstress isnotknown beforehand after all its valueFIG 10 9If s cr spl the tangentmodulustheoryreplacesE inEuler sformula withEt theslopeof thestress strain curveatthepoints scr FIG P10 14380CHAPTER10Columns CopyrightxxCengage Learning All RightsReserved May notbe copied scanned or duplicated in wholeor inpart Due toelectronic rights some thirdparty contentmay besuppressed fromthe eBookand or eChapter s Editorial reviewhas deemedthat anysuppressed contentdoes notmaterially affectthe overalllearning experience Cengage Learningreserves theright toremove additionalcontent atany timeif subsequentrights restrictionsrequire it depends onthecriticalstress Therefore Eq 10 7 mustbesolved fors crandEtsimultaneously bytrial and error Figure10 10shows theplot ofthecriticalstress obtainedfrom tangentmodulustheory againstthe slenderness ratio foraluminum columns As canbeseen thetangent modulustheorysmoothly connectsthe curvesfor shortandlong columns b AISC specificationsforsteelcolumnsThe AmericanInstitute ofSteel ConstructionAISC specifieswhat isknownas theLoad andResistance FactorDesign LRFD method forputingthe pressivestrengths ofsteelcolumns 2The LRFDmethod di erentiates betweenslender andnonslender sec tions The wallsof nonslender sections arethick enoughto allowthe stresstobee fullyplastic withoutbuckling locally no wrinklingoftheflangesor theweb On theother hand slendersectionsmay developlocalizedbuckling inthe inelasticzone whichhasan adversee ect onthe bucklingstrength ofthecolumn Slender sectionshave specialdesign proceduresthatresult incolumns thatare notas economicalas nonslenderdesigns There fore the use of slendersections isdiscouraged A structuralsection isdefined tobe nonslenderif itmeets thecriteriashown inFig 10 11 Most ofstandard structuralsections listedin AppendixBfall intothe nonslendercategory Aording tothe LRFDspecifications the nominal buckling stressofacolumnwithanonslender sectionis givenbythe following s nom 0 658s yp scr s ypif Le r 4 71ffiffiffiffiffiffiffiffiffiffififfiE s ypp0 877scrif Le r 4 71ffiffiffiffiffiffiffiffiffiffififfiE s ypp 10 8 Tangent modulustheory FIG 10 10Critical stressversus slendernessratio foraluminum columns 2AISC alsoallows theuseofan oldermethod known asallowable strengthdesign ASD 10 4Design Formulasfor IntermediateColumns381CopyrightxxCengage Learning All RightsReserved May notbe copied scanned or duplicated in wholeor inpart Due toelectronic rights some thirdparty contentmay besuppressed fromthe eBookand or eChapter s Editorial reviewhas deemedthat anysuppressed contentdoes notmaterially a