LECTURE 25 and 26 SimpleBending TheoryORTheoryof Flexurefor InitiallyStraightBeams (Thenormalstressdueto bendingare calledflexurestresses) Preamble: Whena beamhavingan arbitrary crosssectionis subjectedto a transverse loadsthe beamwill bend.in additionto bendingthe othereffectssuchas twistingand buckling mayoccur,andto investigate a problemthatincludesall the combinedeffectsof bending, twistingandbuckling couldbecomea complicated one.Thus we are interested to investigate the bendingeffectsalone,in ordertodoso,we haveto putcertainconstraints onthe geometry ofthebeamandthe mannerofloading. Assumptions: Theconstraints putonthegeometrywould formthe assumptions: 1. Beamis initially straight, andhasa constantcross-section. 2. Beamis madeofhomogeneous materialandthebeamhasa longitudinal planeof symmetry. 3. Resultantofthe appliedloadsliesintheplaneofsymmetry. 4. Thegeometryofthe overallmemberis suchthatbendingnotbuckling is theprimarycause offailure. 5. Elasticlimitis nowhereexceeded andE is sameintensionandcompression. 6. Planecross sectionsremainsplanebeforeandafterbending. H i" F A------ - 3sly} / c Natural surlaeze: {*2g E Fig"W33 Letusconsiderabeaminitiallyunstressed as showninfig 1(a).Nowthe beamis subjected to a constant bendingmoment(i.e.Zero ShearingForce)alongitslength as wouldbeobtainedbyapplying equalcouplesat eachend.ThebeamwillbendtotheradiusR as shownin Fig1(b) As a resultof thisbending,thetopbers ofthe beamwill be subjected to tensionandthe bottomto compression it is reasonableto suppose,therefore,that some where betweenthe two there are pointsat whichthe stress is zero.The locusof all such pointsis knownas neutralaxis. The radiusof curvature R is then measuredto thisaxis.Forsymmetrical sectionsthe N. A is the axisof symmetry butwhateverthe sectionN. A willalwayspassthroughthe centreof the area or centroid. The aboverestrictionshavebeentakenso as to eliminatethe possibility of twisting of the beam. Concept of pure bending: Loadingrestrictions: As we are awareof the factinternalreactionsdeveloped on anycross-section of a beam mayconsistsof a resultantnormalforce,a resultantshearforceand a resultantcouple.in orderto ensurethatthe bendingeffectsaloneare investigated, we shallputa constraint on the loadingsuchthatthe resultantnormalandthe resultant shearforcesarezeroonanycross-section perpendicularto thelongitudinal axisofthemember, ThatmeansF = 0 sinceW = F= D orM: constant. Thus,thezeroshearforcemeansthatthebendingmomentis constant orthe bendingis sameat everycross-section ofthe beam.Sucha situation maybe visualized or envisaged whenthebeamor someportionofthe beam,as beenloadedonlybypurecouplesat itsends.it mustbe recalledthatthecouplesare assumedto be loadedintheplaneofsymmetry. Planeof Symmetry Fig £2) Whena memberis loadedin such a fashionit is said to be in pure bending.Theexamplesof purebendinghavebeenindicatedin EX1andEX2 as shownbelow: Whena beam is subjectedto pure bendingare loadedbythe couplesat the ends,certaincross-sectiongets deformedand we shall haveto makeoutthe conclusion that, 1. Planesectionsoriginallyperpendiculartolongitudinalaxisofthe beamremainplaneand perpendiculartothe longitudinalaxisevenafterbending, i.e. the cross-sectionA'E', B'F( referFig 1(a)) do notgetwarpedor curved. 2. in the deformedsection,the planes of this cross-sectionhavea common intersectioni.e. any time originallyparallelto the longitudinalaxis of the beambecomesan arc of circle. my Transverse Secxim Mantra! Surlsme We knowthatwhen a beam is underbendingthe bres at the top will be lengthenedwhile at the bottomwill be shortenedprovidedthe bendingmomentM acts at the ends. in betweenthesethereare some bres whichremainunchangedin lengththat is theyare notstrained,that is theydo not carryanystress.The planecontaining suchfibres is calledneutralsurface. Theline ofintersectionbetweenthe neutralsurfaceand the transverseexploratorysection is calledthe neutralaxisNeutralaxis(NA) . Bendin Stresses in Beams or Derivation of Elastic Flexural formula : in order to computethe value of bendingstresses developedin a loaded beam,let us considerthe two cross-sectionsof a beamHEand GF, originallyparallelas shown in g 1(a).whenthe beamis to bendit is assumedthatthese sectionsremain paralleli.e. H'E' and GP , the nal positionof the sections,are still straightlines, theythensubtendsome angle6. ConsidernowfiberABin the material,at adistanceyfrom the NA whenthe beambendsthis will stretchto A'B' Therefore, strain infihre AEl= change in length _ orginallength : A'e' AB AB ElutAEl=CDandCD=C D refertofig1(a)anc|figl(lJ) CD strain = A'El'CD SinceCD and CD are on the neutralaxisand it is assumedthatthe Stresson the neutralaxiszero.Therefore,therewon't be anystrainon the neutralaxis = (R+y]BRB = RB+yBRB :1 RE HE R stress strain Therefore equating the twostrains as obtaineclfromthetworelationsie, However =E whereE=ioung's Modulusofelasticity 0E 0:}: or _=_ E W y R MA Consideranyarbitraryacrosssectionof beam,as shownabovenowthe strainon a bre at a distancey fromthe NA, is givenbythe expression o=Ey R if the shaded strip is of area'c|A' thentheforoe onthe strip is F=o6A=Ey»SA e Mom ent ahouttheneutral axis would he =F_y =%5:262-K The toatl moment for the whole crosssection istherefore equal to M: E 25A=E ERr 25A R25 Nowtheterm2 yzaixis theproperty ofthematerial andis calledas a secondmomentofareaofthecrosssection andis denotedbyasymbol1. Therefore E w1=i combining equation 1 and 2 we get This equation is known as the BendingTheory Equation.Theabove proof has involvedthe assumptionof pure bendingwithout any shear force being present. Thereforethis termedas the pure bendingequation.This equationgivesdistributionofstresses whichare normalto crosssectioni.e. in x-direction. SectionModulus: Fromsimple bendingtheoryequation,the maximumstress obtainedin anycrosssectionis givenas M U rn= T 5" m man: man: Foranygivenallowablestress the maximummomentwhichcan be acceptedbya particularshapeof crosssectionis therefore Second Moment of Area : Takingan analogyfrom the mass momentof inertia,the secondmomentof area is definedas the summationof areas times the distancesquaredfrom a fixedaxis. (This propertyarisedwhile we were drivingbendingtheoryequation).This is also knownas the momentofinertia.Analternativenamegivento this is secondmoment ofarea,because thefirstmoment being thesumofareas timestheirdistance fromagiven axisandthesecond moment being thesquare ofthedistance ml" 3,3dA . Consideranycrosssectionhavingsmall elementofaread Athen bythe denition lX(Mass Momentoflnertia aboutx-axis)= I 3,2,3;-N andly(Mass |\/loment oflnertia abouty-axis)=J' X2CIA Nowthe momentofinertia aboutan axisthroughD and perpendicularto the planeof gure is calledthe polar momentofinertia. (The polarmomentofinertia is also the area momentofinertia). i.e, J=polar momentofinertia = Ir3.:lA = It >8+MA = Ix1dA+Iy1dA Ix + IY orJ=|X+|l,r ......... ..(1) Therelation(1) is knownas the gergendicularaxis theoremand maybe statedas follows: The sum of the Momentof Inertiaabout any two axes in the plane is equal to the momentof inertiaaboutan axis perpendicularto the plane,the three axes being concurrent,i.e,the threeaxesexisttogether. CIRCULAR SECTION : Fora circularx-section,the polarmomentof inertiamaybe computedin the followingmanner Consideranycircularstrip of thicknessEirlocatedat a radius'2. Thanthe areaofthe circularstrip wouldbe dA= Znr.Sr ThsJ=,[r2I:lA Taking the limits ofinterg ration from D to u:lf2 cl 5 J = _[r22:Itr5r o d 5 J2:'I[TL i r 9'rcl4 howeverbyperpenclioularairistheorem J = |:ur+|3r Eiutforthe circular oross-section,the Inand lyare both equal being moment ofinertia about a diameter 1 ldia2 EJ mi Idia_- E fora hollow circularseotionofdiameterDand cl, thehralLiesofJanc|lareclefineclas HED4er) T 3-r[D4 d4] 54 ParallelAxis Theorem: The momentofinertia aboutanyaxis is equalto the momentofinertia abouta parallelaxisthroughthe centroidplus the areatimes the squareofthe distancebetween the axes. 2 H T i. 2: if Z2 is anyaxis in the planeof cross-sectionand XXis a parallelaxisthroughthe centroidG,of the crosssection,then I: =Hy+h)2dz-a bydenition (moment ofinertia about anaxis ZZ) =l[+2yh+h?]ciA = ly3dA+h3ldA +2hlydA Since lydA=o = ly*ciA+h*ldA = ly%iA+h9A 2 = II +.""\h2 II = is (sincecrosssection axesalsopassthroughG) Where A =Total area ofthe section RectangularSection: Fora rectangularxsectionof the beam,the secondmomentof area maybe computedas below: Considertherectangularbeamcross-sectionas shownaboveand an elementof areadA,thickness dy, breadthB locatedat a distanceyfrom the neutralaxis,which bysymmetrypassesthroughthe centreofsection.Thesecondmomentofarea las dened earlierwouldbe INA= [y3dA Thus,forthe rectangularsectionthe secondmomentofareaaboutthe neutralaxis i.e.,an axisthroughthe centreis givenby II 53 r_u|UJ "-CC --| to 0:n|'~3w . ~I-= H.A ofdomed rectangle ofshaded portion .| _EID3 2bcI3 '-V? 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