mechanical structural analysis
TRANSCRIPT
-
7/30/2019 mechanical structural analysis
1/16
This document is part of the notes written by Terje Haukaas and posted at www.inrisk.ubc.ca.The notes are revised without notice and they are provided as is without warranty of any kind.
You are encouraged to submit comments, suggestions, and questions to [email protected].
It is unnecessary to print these notes because they will remain available online.
Euler-Bernoulli Beams
TheEuler-Bernoullibeamtheorywasestablishedaround1750withcontributionsfromLeonardEulerandDanielBernoulli.TheworkbuiltonearlierdevelopmentsbyJacob Bernoulli. However, the beam problem had been addressed even earlier.Galileoattemptedoneformulationbutmisplacedtheneutralaxis.LeonardodaVincialso seems to have addressed the problem of beam bending. The two keyassumptions in the Euler-Bernoulli beam theory are that the material is linearelastic according to Hookes law and that plane sections remain plane andperpendiculartotheneutralaxisduringbending.ThelatterissometimesreferredtoasNaviershypothesis.Incontrast,Timoshenkobeamtheory,whichiscoveredinanotherdocument,relaxestheassumptionthatthesectionsremainperpendicular
totheneutralaxis,thusincludingsheardeformation.Inthefollowing,thegoverningequations are established, followed by the formulation and solution of thedifferential equation. Thereafter, the computation of stresses and cross-sectionconstantsisdescribed.
Anumberofsignconventionsareadoptedinthefollowing:
Thez-axisisincreasesupward Displacementwispositiveinthedirectionofthez-axis
-
7/30/2019 mechanical structural analysis
2/16
Terje Haukaas University of British Columbia www.inrisk.ubc.ca
Euler-Bernoulli Beams Page 2
Distributedloadqzispositivewhenitactsinthedownwarddirection Clockwiseshearforceispositive Bendingmomentthatimposestensionatthebottomofahorizontalbeam
elementispositive Counter-clockwiserotation ispositivesothatitcanbeinterpretedasthe
slopeofthedeformedbeamelement Tensilestressesandstrainsarepositive,compressionisnegative
EquilibriumThe equilibrium equations are obtained by considering equilibrium in the x-directionfortheinfinitesimalbeamelementinFigure1.Noticethatthedistributedload,q,actsinthedownwarddirection,whilethez-axisisintheupwarddirection.Thenotationqzisemployedinotherdocumentstoidentifythecasewherepositiveloadactsinthepositivez-direction.
Figure1:Equilibriumforinfinitesimallysmallbeamelement.
Verticalequilibriumyields:
q =
dV
dx (1)
Momentequilibriumabouttherightmostedgeyields:
V =dM
dx (2)
InEq.(2)itisnotedthatsecond-ordertermsareneglected.Thisessentiallymeansthattermswithdx2(themultiplicationofanexceedinglysmallvaluebyitself)areconsideredapproximatelyequaltozero.
SectionIntegrationIntegrationofaxialstressesoverthecross-section:
M = z dAA
(3)
dx
q
VM M+dMV+dV
-
7/30/2019 mechanical structural analysis
3/16
Terje Haukaas University of British Columbia www.inrisk.ubc.ca
Euler-Bernoulli Beams Page 3
wheretheminussignappearsbecauseitiscompressive(negative)stressesinthepositivez-axisdomainthatgivesapositivebendingmoment,i.e.,bendingmomentwithtensionatthebottom.Figure2isintendedtoexplainthisfurther.
Figure2:ThereasonfortheminussigninEq.(3).
MaterialLawThemateriallawthroughoutlinearelastictheoryisHookeslaw:
= E (4)
Inthecontextoftwo-dimensionaltheoryofelasticity,theuseofEq.(4)impliesaplanestressmateriallaw.Itimpliesthatthereiszerostress,i.e.,aironthesidesofthebeam.Thealternativeplanestrainversionofthetwo-dimensionalHookeslawismoreappropriateincaseswherethebeamisonlyastripofalongrectangularplatethatissupportedalongthetwolongedges.Inthatcasethestrainisrestrainedinthey-direction:
yy =yy
E
xx
E= 0 yy = xx
(5)
Which,substitutedintothemateriallawinthex-directionyields:
xx =
xx
E
yy
E
=
xx
E
xx( )
E
=
xx
E
(12)
xx =
E
12
xx (6)
All the derivations and results in the following are based on the material lawxx=E.xx from Eq. (4). However, the plain strain version is easily introduced byreplacingtheYoungsmodulus,E,inanyequationbyE/(1-2).
Positive tension stressNegativez-values
x
zNegative compression stressPositivez-values
Minus sign is cross-section
integral is necessary to get
positive bending moment
MM
-
7/30/2019 mechanical structural analysis
4/16
Terje Haukaas University of British Columbia www.inrisk.ubc.ca
Euler-Bernoulli Beams Page 4
KinematicsThe relationship between the axial strain and the transversal displacement of abeamelementissought.Itisfirstrecognizedthatbendingdeformationessentiallyimplies shortening and lengthening of fibres in the cross-section. Fibres on thetensionsideelongate,whilefibresonthecompressionsideshorten.
Thestartingpointfortheconsiderationsistolinktheaxialstraintothechangeoflength of the imaginative fibres that the cross-section is made up of. The sameconsiderationasinkinematicsoftrussmembers,namelythatstrainiselongationdividedbyoriginallengthyields:
=
du
dx
(7)
Next, the axial displacement u is related to the rotation of the cross-section. Inparticular,considertheinfinitesimalrotationdoftheinfinitesimallyshortbeamelementinFigure3.Inpassing,itisnotedthatdisequaltothecurvature, .Under
theassumptionthatplanesectionsremainplaneandperpendiculartotheneutralaxisduringdeformation,eachfibreinthecross-sectionchangelengthproportionaltoitsdistancefromtheneutralaxis.
Figure3:Naviershypothesisforbeambending.
The amount of shortening orelongation depends upon the rotationof the cross-section.AgeometricalconsiderationoftoFigure3showsthattheshorteningandlengthening,i.e.,axialdisplacement,ofeachinfinitesimallyshortfibreis
du = d z (8)
Finally,therotationisrelatedtothetransversaldisplacement.Forthispurpose,
considertwopointsonabeamthatisdxapart,asshowninFigure4.Therelativedisplacement is dw, which is measure positive upwards. Consequently, ageometricalconsiderationofFigure4showsthat:
tan() =
dw
dx (9)
wheretheequationissimplifiedbyassumingthatthedeformationsaresufficientlysmallsothattan().
z
d!
x, u
z, w
-
7/30/2019 mechanical structural analysis
5/16
Terje Haukaas University of British Columbia www.inrisk.ubc.ca
Euler-Bernoulli Beams Page 5
dx
dw
!
!
Figure4:Rotationofthecross-sectionofabeamelement.
Bycombiningequations,thekinematicequationforbeammembersisobtained:
= d2
w
dx2z (10)
This expression implies an approximation of the exact curvature of the beam.Mathematically,curvatureisdefinedas
1
R (11)
whereRistheradiusofcurvatureofthebeam.IntheEuler-Bernoullibeamtheorythatispresentedhere,thecurvatureisapproximatedby
d
dx
d2w
dx2 (12)
Notice that there are two approximation signs. The first alludes to the fact thatdifferentiationiscarriedoutwithrespecttothex-axis.Unlessthedeformationsarenegligiblethisisinaccurate;differentiationshouldbecarriedoutwithrespecttothes-axisthatfollowsthecurvingbeamaxis.ThesecondapproximationisduetoEq.(9).Fromthatequationitisobservedthattheaccurateexpressionforis:
= tan
1 dw
dx
(13)
Ifthisexpressionwas utilizedinthederivationsabovethen thedifferentiationoftheinversetan-functionyields
d
dx=
d2w
dx2
1+dw
dx
2
(14)
-
7/30/2019 mechanical structural analysis
6/16
Terje Haukaas University of British Columbia www.inrisk.ubc.ca
Euler-Bernoulli Beams Page 6
which reduces to the expression in Eq. (12) when the slope dw/dx is small.However, the curvature expression in Eq. (14) is still approximate because thedifferentiationiscarriedoutwithrespecttothex-axisandnotthebeamaxis.Frommathematics,theexactcurvatureexpressionis:
=
d2w
dx2
1+dw
dx
2
3
2
(15)
DifferentialEquationThegoverningdifferentialequationforbeammembersisobtainedbycombiningtheequationsforequilibrium,sectionintegration,materiallaw,andkinematics:
q = dVdx
= d2
Mdx2
= d2
dx2 z dA
A
=
d2
dx2E z dA
A
= d2
dx2E
d2w
dx2z2 dA
A
= EId
4w
dx4
(16)
wherethemodulusofelasticityisassumedconstantoverthecross-sectionandthemomentofinertiaisdefined:
I=
z
2dA
A (17)InEq.(16)itisassumedthatthecross-sectionishomogeneoussothat Eisconstant.Forcompositecross-sectionsthisassumptionisinvalid,andtherevisedversionofEq.(16)is
q = d4w
dx4Ez2 dA
A
(18)
Oneapproachtoretainingtheoriginaldefinitionof Iistofirstselectareference-valueoftheYoungsmodulusandassumethatallpartsofthecross-sectionhasthatE-value.Next,thewidthofeachpartof thecross-section ismodified ifitsYoungsmodulusisdifferentfromthereferencevalue.Thechangeinwidthisproportionalto the difference inE-value. E.g., ifE is twice the reference value then thewidthshouldbedoubled.IfthisprocedureisfollowedthenEq.(17)remainsvalidforthedeterminationofI.
-
7/30/2019 mechanical structural analysis
7/16
Terje Haukaas University of British Columbia www.inrisk.ubc.ca
Euler-Bernoulli Beams Page 7
GeneralSolutionAlthoughsolvingthedifferentialequationisnotpartoftypicalstructuralanalysisitis instructive to study its solution for simple reference cases. In particular, thesolutionofthedifferentialequationisthestartingpointfortheselectionofshapefunctions in the finite element method. Those shape functions are often
approximate,whilethesolutionofthedifferentialequationrevealstheexactshapewhenthememberdeforms.Thegeneralsolutionofthedifferentialequationrevealswhetherthefiniteelementshapefunctionsareexactornot.Forbeammembers,thegeneralsolutionofthedifferentialequationisobtainedbyintegratingfourtimes:
w(x) =1
24
qz
EIx
4+C1 x
3+C2 x
2+C3 x +C4 (19)
Given a uniform distributed load qz, the displaced shape is a fourth orderpolynomial. Without any distributed load, the displacement shape of a beammember is a third-order polynomial. To obtain the solution for a specific beamproblem,boundaryconditionsarespecified.Toprescribearotation,shearforce,orbending moment, the following equations are useful, obtained by combining thegoverningequationsthatareestablishedearlier:
=dw
dx (20)
M = EId
2w
dx2 (21)
V = EId
3w
dx3 (22)
As an illustration of the solution to the differential equation for beam bending,Figure5showsplotsofw(x),(x),M(x),andV(x)forasimplysupportedbeamwithuniformly distributed load. The illustration ismadewith qz=L=EI=1. In Figure 5,noticethatthedisplacementw(x)isnegative,i.e.,downwardsandthat(x)correctlyshowsthattheslopeisnegativeleftofthemid-span.Furthermore,noticethattheplotsofM(x)andV(x)are identical tothe respectivesection forcediagrams,withoneexception:plottingM(x)yieldsadiagramdrawnonthecompressionside,whileinthesenotesthebendingmomentdiagramsareconsistentlydrawnonthetensionside.
-
7/30/2019 mechanical structural analysis
8/16
Terje Haukaas University of British Columbia www.inrisk.ubc.ca
Euler-Bernoulli Beams Page 8
Figure5:Exampleofresponsefunctionsforbeamelement.
Cross-sectionAnalysisCross-section analysis determines cross-section parameters and stresses in thecross-section.Above,theEuler-Bernoullibeamtheoryisestablishedfora2Dbeam,i.e.,abeamthatbendsinoneplaneaboutoneaxis.However,thederivationsarevalidfora3Dbeamaswell,becausebendingaboutthetwoprincipalaxesofthecross-sectionaredecoupled.Tomaintaintheinclusionof3Dbeamsinthefollowing,both axes of the cross-section, i.e., they-axis and thez-axis, are now consideredsimultaneously. The cross-section parameters in Euler-Bernoulli beam theoryincludethecentroidcoordinates,theshearcentrecoordinates,theorientationoftheprincipal axes, and the moment of inertia, I, defined in Eq. (17). For rectangularcross-sectionsofwidthbandheighththatintegralyields
I= b z2 dzh/2
h/2
=b h
3
12 (23)
Formorecomplicatedcross-sectionsthegeneralprocedureisto1)determinethelocationoftheneutralaxis,2)determinethemomentsofinertiaIyandIzabouttheneutralaxisforselectedaxesyandz,3)determinetheorientationoftheprincipalaxes,ifthosearenottheselectedaxesyandz,and4)re-calculationofthemoments
-
7/30/2019 mechanical structural analysis
9/16
Terje Haukaas University of British Columbia www.inrisk.ubc.ca
Euler-Bernoulli Beams Page 9
of inertia about the principal axes. Each of these items are addressed in thefollowing.
NeutralAxis
For homogeneous cross-sections the neutral axis coincide with the geometrical
centroidofthecross-section.Forcompositecross-sectionsitispossibletoscaletheareaofeachcross-sectionpart(aboutitslocalcentroid)proportionaltotheE-valuerelativetoareferencevalue.Thisissimilartowhatwassuggestedearlierinthescaling of cross-sectionwidth in the computation of the moment of inertia, andimpliesthatthecentroidcoincidewiththeneutralaxis.Thecentroidisdeterminedbythefirstarea-momentsydAandzdAbeingequaltozero.Inpractice,itisusefultofirstselectareferencepointinthecross-section.Letyoandzodenotethedistancefrom the reference point to the centroid. Next, divide the cross-section intoconvenient segmentswith areaAi and letyi andzi denote the distance from thereference point to the centroid of each segment.When denoting the total cross-sectionareabyA,thefollowingexpressionsofthefirstarea-momentsmustbeequal
tozero:
Ayo = Ai yi = 0
Azo = Ai zi = 0 (24)
wherethesumsare takenoverallpartsof the cross-section.Solvingforyoandzoyields:
yo =Ai yiA
zo=
Aiz
iA
(25)
In words, obtain the locationof the neutral axis by summing areamultiplied bydistanceforallpartsofthecross-section,anddividethesumbythetotalarea.
MomentsofInertia
Iy and Iz are computed separately about the neutral axis, and the theoreticalfoundationisprovidedbyEq.(16).Inpracticeitisremainsconvenienttodividethecross-section into parts, each with local moment of inertia denoted by Ii. Thecontributions to the global moment of inertia from each part are summed inaccordancewithSteinersformula:
Iy = Iy,i +zi
2Ai( )
Iz = Iz,i + yi
2Ai( )
(26)
whereyiandziaredistancesfromtheneutralaxisoftheentirecross-sectiontothecentroidofthepart.
-
7/30/2019 mechanical structural analysis
10/16
Terje Haukaas University of British Columbia www.inrisk.ubc.ca
Euler-Bernoulli Beams Page 10
PrincipalAxes
Supposetheyandzaxeswereselectedarbitrary,originatingattheneutralaxis,andthat it is unknown whether they represent the principal axes. To determine theorientationoftheprincipalaxesrelativetotheselectedaxes,first,withthehelpofEq.(10),considerthestrainatalocationinthecross-section:
= o d
2v
dx2y
d2w
dx2z (27)
Next,themateriallawinEq.(4)providesthestress:
= Eo Ed2v
dx2y E
d2w
dx2z (28)
IntegrationofaxialstressyieldsthebendingmomentsMyandMzabouttheyandzaxes,respectively.Firstconsiderthebendingmomentaboutthey-axis:
My=
zdA
A=
E
ozdA
A+
Ed2v
dx2
yz
dA
A+
Ed2w
dx2z
2dA
A (29)The first integralvanishesbecausezoriginatesattheneutralaxis,whilethelastterm is theordinary bendingmoment from Eq. (21).As a result,Eq. (29) canberewrittenas:
My = EIyz d
2v
dx2+EI
d2w
dx2 (30)
wheretheproductofinertia,Iyz,hasbeendefinedas:
Iyz = y z dAA
(31)
Similarly, Iyz appears in the expression for Mz. It is possible and relativelystraightforward to establish formulas for Iyz and compute stresses in term of Iyz.However,itispreferredtorotatetheaxissystemsothatIyziszero,whichimpliesarotationof theoriginalaxestocoincidewiththeprincipal axes.Knowledgeoftheprincipal axes is beneficial because bending about the two principal axes aredecoupledandtheelementaryformulasforbeambendingremainvalid.
AxialStress
Forbendingaboutoneaxis,aconvenientapproachforobtainingtheaxialstressintermsofthebendingmomentistocombinemateriallawandkinematicsequations,whichyields:
= Ed
2w
dx2z (32)
Then substitute the differential equation without equilibrium equations, i.e., Eq.(21),toobtain:
-
7/30/2019 mechanical structural analysis
11/16
Terje Haukaas University of British Columbia www.inrisk.ubc.ca
Euler-Bernoulli Beams Page 11
= M
Iz (33)
Inthecontextofthe in-planebendingpresenterearlier,it isnotedthata positivebendingmoment,i.e.,tensionatthebottom,correctlyyieldsnegativestressesatthetop,i.e.,compression,wherezispositive.ThisisthereasonfortheminussigninEq.(33), which also correctly gives positive tension stresses at the bottom when apositivemomentactsonthecross-section.Insummary,thebeamtheorypresentedinthisdocumentconsistsofthegoverningequationsshowninFigure6.
Figure6:GoverningequationsinEuler-Bernoullibeamtheory.
ShearStressandShearCentreforOpenCross-sections
Whenapproachingshearstressesinbending,ananomalyinEuler-Bernoullibeamtheory is first noted. The theory is based on the assumption that plane sectionsremainplaneandperpendiculartotheneutralaxis.Inotherwords,theonlystrainthat takes place is the axial shortening or elongation of the fibres in the cross-
section.Effectively,thispreventsshearstrain.Withnoshearstrainthereisnoshearstress,whichaddsuptozeroshearforce.Inotherwords,theshearforceisnotpartofthetheory.Thisisananomaly,becauseshear forcewilldevelopeveninsimplebeamsthataresubjectedtotransversalload.Theanomalyiscustomarilyaddressedbyrecoveringtheshearforcebyequilibriumoncethebendingmomentiscomputed.Infact,accordingtoEq.(2)theshearforceisequaltothederivativeofthebendingmoment;thisistheequationthatrecoverstheshearforce.Anotherdocumenton
q
= d
2w
dx2z
= E
= M
IzM = z dA
A
q = EId4w
dx4
q = dV
dx
V=dM
dx
M = EId2w
dx2
w
M
-
7/30/2019 mechanical structural analysis
12/16
Terje Haukaas University of British Columbia www.inrisk.ubc.ca
Euler-Bernoulli Beams Page 12
Timoshenkobeamtheorydescribesanapproachtofurtherextendthebeamtheorytoincludedeformationduetoshearforces.
Toobtainexpressionsfortheshearstress,,intermsoftheshearforce,V,considertheinfinitesimallyshortbeamelementinFigure7.Furthermore,consideracutinthecross-sectionandletqsdenotetheshearflowatthatlocation.
Figure7:Shearflowbyequilibriumofinfinitesimalbeamelement.
Theshearflowistheforceperunitlengthofthebeamthatensuresequilibriumwiththeaxialstresses,whicharegreaterononesidethantheotherduetodM:
qs dx = ddAAs
=dM
Iz dA
As
(34)
whereAsisthecross-sectionalareaoutsidethecut.GiventhatV=dM/dx,thisyields
qs =V
IS (35)
where
S= zdAAs
(36)
Theshearstressiscalculatedbydistributingtheshearflowoverthethickness,t,ofthecross-sectionattheparticularlocation:
= V SI t
(37)
For example, for a rectangular cross-section themaximum shear stress is at theneutralaxis,withvalueequalto
=3
2
V
A
(38)
dx
VM
M+dM
V+dV
Axial stresses
Shear stresses
(shear flow)
-
7/30/2019 mechanical structural analysis
13/16
Terje Haukaas University of British Columbia www.inrisk.ubc.ca
Euler-Bernoulli Beams Page 13
Theshearcentreofacross-section,sometimescalledthecentreoftwist,isthepointwheretheresultantoftheshearforcemustacttoavoidrotationofthecross-section.Thecoordinatesoftheshearcentrearedenotedbyyscandzsc,andthereareseveraltechniques to determine them. The simplest case is double-symmetric cross-sections;forthesecross-sectionstheshearcentrecoincidewiththecentroid.Infact,
ifacross-sectionhasanaxisofsymmetrythentheshearcentreislocatedonthisaxis.Forgeneralcross-sections,oneapproachtodetermineyscandzscisdescribedinthedocumentonwarpingtorsion,wheretheomegadiagramisutilized.However,asomewhatsimplerapproach,whentheconsiderationofwarpingtorsionisoffthetable,isofferedhere.Theprincipleissimple;bydefinition,themomentoftheshearflowabouttheshearcentremustbezero.Thisleadstothefollowingproceduretodetermine the coordinatesofthe shearcentre,providedyandzaretheprincipalaxesthroughthecentroidofthecross-section:
1. Selectanarbitrarypointas trialshear centre,andletyscandzscdenotethecoordinatesoftheshearcentrerelativetothecentroid;inotherwords,let yscandzscdenotethedistancesfromthecentroidtotheshearcentre
2. InaccordancewithEq.(35),determinetheshearflowinthecross-sectionduetoashearforceinthez-direction
3. WritetheequationthatexpressesthemomentoftheshearflowinItem2aboutthetrialshearcentre;ingeneral,bothyscandzscwillappearinthisexpression
4. SimilartoItem2,determinetheshearflowinthecross-sectionduetoashearforceinthey-direction
5. SimilartoItem3,writetheequationthatexpressesthemomentoftheshearflowinItem4aboutthetrialshearcentre
6. SettheequationsfromItems3and5bothequaltozeroandsolvethesetwoequationsforthetwounknownsyscandzsc
Onlyonemomentequation isneededfor single-symmetric cross-sections; inthatcasestheproceduresimplifiesto:
1. Selectanarbitrarypointalong thesymmetryaxisastrial shearcentre,andletedenotethedistancefromthecentroidtothatpoint
2. InaccordancewithEq.(35),determinetheshearflowinthecross-sectionduetoashearforceinthedirectionperpendiculartotheaxisofsymmetry
3. WritetheequationthatexpressesthemomentoftheshearflowinItem2aboutthetrialshearcentre;ewillappearinthisexpression
4. SettheequationfromItems3equaltozeroandsolvefore
ShearStressandShearCentreforClosedCross-sectionsThe determinationof shearstressandshear centre for closed cross-sections, i.e.,cross-sectionswithcells,canbeapproachedintwoways.Inthefirstapproach,theshearcentrecoordinatesarefirstdeterminedusingtheomegadiagramforthin-walled cross-sections, as described in the document on warping torsion. In thefollowing, suppose the cross-section has one cell. In this approach, a cut isintroducedtoyieldanopencross-section.Acoordinate soriginatesatthecutand
-
7/30/2019 mechanical structural analysis
14/16
Terje Haukaas University of British Columbia www.inrisk.ubc.ca
Euler-Bernoulli Beams Page 14
traces the cross-section around the cell. The unknown shear flow at the cut isdenotedqo,andtheshearflowatallotherlocationsisdeterminedrelativeto qoinaccordancewithEq.(35),sothat
q(s) = qo +V
IS(s) (39)
Onceqisdeterminedatalllocationsoftheopenedcross-section,themomentoftheshearflowabouttheknownshearcentreiscomputedas
T = q hds = qo hds +V
IShds (40)
wheretheintegralsaremadearoundthecell,startingats=0,andh(s)isthedistancefromtheshearcentretothetangentlineofthecontourofthecross-sectionats.Bydefinitionthemoment,T,abouttheshearcentremustbezero,andsolvingfor qoyields
qo= V
I Shds hds
= V2 Am I
Shds (41)
wherethelastequalityisobtainedbyrecognizingthattheintegralofharoundthecross-section is twice the cell area,Am. Having the value of qo, the shear flow isdeterminedatotherlocationsinaccordancewithEq.(39).
Thesecondapproachdeterminesbothshearflowandshearcentre.Thisapproachexplicitlyrecognizesthatthedeterminationofshearflowinaclosedcross-sectionisastaticallyindeterminateproblem.Inotherwords,equilibriumequationsaloneare insufficient todetermine the sought forces. From this viewpoint, each cell is
associated with one redundant. Similar to the flexibility method in fundamentalstructural analysis, the solution approach involves removing the capacity of thestructure to carry the sought forces, i.e., to introduce cuts, and then toenforcecompatibility equations that are solved for the unknown forces. In the following,considerathin-walledcross-sectionwithonecell,andintroduceonecuttomakeitanopencross-section.
-
7/30/2019 mechanical structural analysis
15/16
Terje Haukaas University of British Columbia www.inrisk.ubc.ca
Euler-Bernoulli Beams Page 15
Figure8:Twocontributionstoshearstrain.
Thesoughtgapisthediscrepancyinu-displacementoneachsideofthecut.Tofindthisu-displacement, all infinitesimal contributions,du,areintegratedalongthe s-axisdefinedearlier.Toobtaintheexpressionforduitisusefultoconsiderthetotalshear strain of an infinitesimal plane element in the cross-section. Figure 8illustratesthattherearetwocontributionsinthiskinematicsconsideration:
= 1+
2=
du
ds+d
dxh (42)
whereistherotationofthecross-section.Solvingforduyields
du = ds d
dxh ds (43)
Materiallawprovidesthefollowingexpressionfortheshearstrainintermsoftheshearstressandthustheshearflow:
=
G=
q
G t (44)
where t is the thickness of the cross-section-wall, which may vary around thecircumference.SubstitutionofEq.(44)intoEq.(43)yields
du =q
G tds
d
dxh ds (45)
Integrationaroundthecellyieldsthetotalgapopeningatthecut:
u =q
G tds
d
dxhds =
q
G tds
d
dx hds =
q
G tds
d
dx2 Am (46)
h
dx
!1
d!
ds
!2
du
-
7/30/2019 mechanical structural analysis
16/16
Terje Haukaas University of British Columbia www.inrisk.ubc.ca
Euler-Bernoulli Beams Page 16
Settingu=0yieldsthesoughtcompatibilityequation:
q
G tds =
d
dx2 Am (47)
Next,itisunderstoodthattherotationofthecross-sectionmustbezeroiftheshear
forceisappliedattheshearcentre.Settingd/dx=0inEq.(47)yields:
q
G tds
2Am= 0 (48)
whichmeansthatthefollowingconditionmustbeenforced:
q
G tds = 0 (49)
Asdoneearlier,theunknownshearflowatthecutisdenoted qo,sothattheshearflowatotherlocationsisqo+VS/I.SubstitutionintoEq.(49)yields
qo
G tds +
VS
IG tds = 0 (50)
Solvingforqoyields:
qo =
V
I
S
G tds
1
G tds
(51)
Whenthematerialishomogeneous,theexpressionsimplifiesto:
q
o =V
I
Stds
1
tds
(52)
Havingqoyieldstheshearflowatanylocationbecause
q(s) = qo +V
IS(s) (53)
and having the shear flow yields the shear centre coordinates because the totalmomentoftheshearflowabouttheshearcentremustbezero.