mechanical structural analysis

7/30/2019 mechanical structural analysis

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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.

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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


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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


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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


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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


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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)


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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.


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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.


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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


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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.


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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:


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= 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


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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)


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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


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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.


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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


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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.

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