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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 Hooke’s law and that plane sections remain plane andperpendiculartotheneutralaxisduringbending.ThelatterissometimesreferredtoasNavier’shypothesis.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:

§ The z -axisisincreasesupward§ Displacementw ispositiveinthedirectionofthe z -axis



Terje Haukaas University of British Columbia www.inrisk.ubc.ca


§ Distributedloadq z ispositivewhenitactsinthedownwarddirection§ Clockwiseshearforceispositive§ Bendingmomentthatimposestensionatthebottomofahorizontalbeam

elementispositive§ Counter-clockwiserotation θ ispositivesothatitcanbeinterpretedasthe

slopeofthedeformedbeamelement§ Tensilestressesandstrainsarepositive,compressionisnegative

EquilibriumThe equilibrium equations are obtained by considering equilibrium in the x -directionfortheinfinitesimalbeamelementinFigure1.Noticethatthedistributedload,q,actsinthedownwarddirection,whilethe z -axisisintheupwarddirection.Thenotationq z isemployedinotherdocumentstoidentifythecasewherepositiveloadactsinthepositive z -direction.

Figure1:Equilibriumforinfinitesimallysmallbeamelement.

Verticalequilibriumyields:

q = !

dV

dx (1)

Momentequilibriumabouttherightmostedgeyields:

V =dM

dx (2)

InEq.(2)itisnotedthat“second-orderterms”areneglected.Thisessentiallymeansthattermswithdx 2(themultiplicationofanexceedinglysmallvaluebyitself)areconsideredapproximatelyequaltozero.

SectionIntegrationIntegrationofaxialstressesoverthecross-section:

M = !" # z d A

A

$ (3)

d x

q

V M M+dM V+dV





wheretheminussignappearsbecauseitiscompressive(negative)stressesinthepositive z -axisdomainthatgivesapositivebendingmoment,i.e.,bendingmomentwithtensionatthebottom.Figure2isintendedtoexplainthisfurther.

Figure2:ThereasonfortheminussigninEq.(3).

MaterialLawThemateriallawthroughoutlinearelastictheoryisHooke’slaw:

! = E "# (4)

Inthecontextoftwo-dimensionaltheoryofelasticity,theuseofEq.(4)impliesa“planestress”materiallaw.Itimpliesthatthereiszerostress,i.e.,aironthesidesofthebeam.Thealternative“planestrain”versionofthetwo-dimensionalHooke’slawismoreappropriateincaseswherethebeamisonlyastripofalongrectangularplatethatissupportedalongthetwolongedges.Inthatcasethestrainisrestrainedinthe y -direction:

! yy =

" yy

E #$ %

" xx

E = 0 & " yy = $ %" xx

(5)

Which,substitutedintothemateriallawinthe x -directionyields:

! xx =

" xx

E

#$

" yy

E

=

" xx

E

#$

$ %" xx( )

E

=

" xx

E

(1#$ 2) & "

xx =

E

1#$

2%!

xx (6)

All the derivations and results in the following are based on the material lawσ xx =E .ε xx from Eq. (4). However, the plain strain version is easily introduced byreplacingtheYoung’smodulus,E ,inanyequationbyE /(1-ν 2).

• Positive tension stress

• Negative z -values

x

z • Negative compression stress

• Positive z -values

Minus sign is cross-section

integral is necessary to get

positive bending moment

M M





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,namelythatstrainis“elongationdividedbyoriginallength”yields:

! =

du

dx

(7)

Next, the axial displacement u is related to the rotation of the cross-section. Inparticular,considertheinfinitesimalrotationdθ oftheinfinitesimallyshortbeamelementinFigure3.Inpassing,itisnotedthatdθ isequaltothecurvature, κ .Under

theassumptionthatplanesectionsremainplaneandperpendiculartotheneutralaxisduringdeformation,eachfibreinthecross-sectionchangelengthproportionaltoitsdistancefromtheneutralaxis.

Figure3:Navier’shypothesisforbeambending.

The amount of shortening orelongation depends upon the rotationof the cross-section.AgeometricalconsiderationoftoFigure3showsthattheshorteningandlengthening,i.e.,axialdisplacement,ofeachinfinitesimallyshortfibreis

du = !d" # z (8)

Finally,therotationθ isrelatedtothetransversaldisplacement.Forthispurpose,

considertwopointsonabeamthatisd x apart,asshowninFigure4.Therelativedisplacement is dw , which is measure positive upwards. Consequently, ageometricalconsiderationofFigure4showsthat:

tan(! ) =

dw

dx"! (9)

wheretheequationissimplifiedbyassumingthatthedeformationsaresufficientlysmallsothattan(θ )≈θ .

z

d!

x, u

z, w





dx

dw

!

!

Figure4:Rotationofthecross-sectionofabeamelement.

Bycombiningequations,thekinematicequationforbeammembersisobtained:

! = "d

2

w

dx2# z (10)

This expression implies an approximation of the exact curvature of the beam.Mathematically,curvatureisdefinedas

! "

1

R (11)

whereRistheradiusofcurvatureofthebeam.IntheEuler-Bernoullibeamtheorythatispresentedhere,thecurvatureisapproximatedby

! "d #

dx"

d 2w

dx2 (12)

Notice that there are two approximation signs. The first alludes to the fact thatdifferentiationiscarriedoutwithrespecttothe x -axis.Unlessthedeformationsarenegligiblethisisinaccurate;differentiationshouldbecarriedoutwithrespecttothe“s-axis”thatfollowsthecurvingbeamaxis.ThesecondapproximationisduetoEq.(9).Fromthatequationitisobservedthattheaccurateexpressionforθ is:

! = tan

"1 dw

dx

# $ %

& ' ( (13)

Ifthisexpressionwas utilizedinthederivationsabovethen thedifferentiationoftheinversetan-functionyields

! "d #

dx=

d 2w

dx2

$ % &

' ( )

1+dw

dx

$ % &

' ( ) 2$

% &'

( )

(14)





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 thedifferentiationiscarriedoutwithrespecttothe x -axisandnotthebeamaxis.Frommathematics,theexactcurvatureexpressionis:

! =

d 2w

dx2

" # $

% & '

1+dw

dx

" # $

% & ' 2"

# $%

& '

3

2

(15)

DifferentialEquationThegoverningdifferentialequationforbeammembersisobtainedbycombiningtheequationsforequilibrium,sectionintegration,materiallaw,andkinematics:

q = ! dV dx

= ! d 2

M dx2

= d 2

dx2" # z d A

A$

=

d 2

dx2E #% # z d A

A

$ = !d 2

dx2E #

d 2w

dx2# z2 d A

A

$

= ! EI d

4w

dx4

(16)

wherethemodulusofelasticityisassumedconstantoverthecross-sectionandthemomentofinertiaisdefined:

I =

z

2d A

A! (17)

GeneralSolutionAlthoughsolvingthedifferentialequationisnotpartoftypicalstructuralanalysisitis instructive to study its solution for simple reference cases. In particular, thesolutionofthedifferentialequationisthestartingpointfortheselectionofshapefunctions in the finite element method. Those shape functions are oftenapproximate,whilethesolutionofthedifferentialequationrevealstheexactshapewhenthememberdeforms.Thegeneralsolutionofthedifferentialequationreveals

whetherthefiniteelementshapefunctionsareexactornot.Forbeammembers,thegeneralsolutionofthedifferentialequationisobtainedbyintegratingfourtimes:

w( x) =1

24!

q z

EI ! x

4+C

1! x

3+C

2! x

2+C

3! x +C

4 (18)

Given a uniform distributed load q z , 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 beam





problem,boundaryconditionsarespecified.Toprescribearotation,shearforce,orbending moment, the following equations are useful, obtained by combining thegoverningequationsthatareestablishedearlier:

! =dw

dx (19)

M = EI d

2w

dx2 (20)

V = EI d

3w

dx3 (21)

As an illustration of the solution to the differential equation for beam bending,Figure5showsplotsofw ( x ),θ ( x ),M ( x ),andV ( x )forasimplysupportedbeamwithuniformlydistributedload.Theillustrationismadewithq z =L=EI =1.

Figure5:Exampleofresponsefunctionsforbeamelement.

With reference to Figure 5, notice that the displacement w ( x ) is negative, i.e.,

downwardsandthatθ ( x )correctlyshowsthattheslopeisnegativeleftofthemid-span. Furthermore, notice that the plots of M ( x ) and V ( x ) are identical to therespective section force diagrams, with one exception: plotting M ( x ) yields adiagramdrawnonthecompressionside,whileinthesenotesthebendingmomentdiagramsareconsistentlydrawnonthetensionside.





Cross-sectionParametersThe only cross-section constant in fundamental 2D beam theory is the cross-sectionalmomentofinertia,I ,definedinEq.(17).Intheformula, z isthedistancefrom the neutral axis of the cross-section. For simple cross-sections, mostprominentlytherectangularonewithwidthbandheighth,theintegralisevaluated

analytically:

I = b ! z2d z

"h/2

h/2

# =

b !h3

12 (22)

Formorecomplicatedcross-sectionsthefollowingproceduremaybehelpful:

1. Determinethelocationoftheneutralaxis.Forhomogeneouscross-sections,theneutralaxispassesthroughthecentroidofthecross-section.Inotherwords,whenthecoordinate z has itsorigin at the centroid then the staticmoment∫ z d Aiszero.Inpractice,firstselectareferenceaxisinthecross-sectionthatisparalleltotheneutralaxiasandlet z odenoteitsdistancetothe

trueneutralaxis.Furthermore,let z idenotethedistancefromthearbitrarilyselected axis to the centroid of each sub-area, Ai that the cross-sectionconsistsof.Thedistancetothetrueneutralaxisisdeterminedfrom:

A· zo= A

i· z

i

All parts of thecross-section

! " zo=

Ai· z

i! A

(23)

2. Determine the local moment of inertia, I i, of each sub-area of the cross-sectionaboutthelocalcentroidaxisofthatpart.

3. Addcontributionstotheglobalmomentofinertiafromeachcross-section

partaccordingtoSteiner’sformula: I = I

i+ z

i

2· A

i

All parts of thecross-section

! (24)

where z iisthedistancefromtheneutralaxisoftheentirecross-sectiontothecentroidofthepart.

PrincipalAxes

Thisdocumentdescribesbendingaboutoneaxis,i.e.,bendingof2Dbeams.Withthemost common cross-sections it is straightforward to understand which cross-sectionaxisthebeamwillbendaround.Forexample,adouble-symmetriccross-

section bends around the two symmetry-axes. However, additional analysis isnecessaryforgeneralasymmetriccross-sectionstodeterminethe“principalaxes,”i.e., the bending axes of the cross-section. As a starting point, suppose two axesdirections y and z are arbitrary selected; they originate in the centroid but theygenerallydonotcoincidewiththeprincipalaxes.Thetotalaxialstrainatalocationofthecross-sectionisexpressedinthefollowingextendedversionofEq.(10):





! = ! o "d

2v

dx2# y "

d 2w

dx2# z (25)

Fromthisstrain,themateriallawinEq.(4)providesthestress,andthestressisintegrated byEq. (3) toobtain the following expression for the bendingmoment

aboutthey-axis:

M = ! E "# o " z d A A

$ + E "d

2v

dx2" y " z d A A

$ + E "d

2w

dx2" z

2d A

A

$ (26)

Thefirstintegralvanishesbecause z originatesatthecentroid,whilethelasttermistheordinarybendingmomentfromEq.(20).Asaresult,Eq.(26)isrewrittenas:

M = EI yz !

d 2v

dx2+ EI !

d 2w

dx2 (27)

wherethe“productofinertia,”I yz ,hasbeendefinedas:

I yz = y ! z d A A" (28)

ItisrelativelystraightforwardtoestablishformulasforI yz andcomputestressesintermofI yz ,etc.However,itisalsoalwayspossibletorotatetheaxissystemsothatI yz is zero; then the axis system is referred to as the principal axes. This isadvantageousbecausetheelementaryformulasforbeambendingremainvalid.

(Materialtobeaddedhere.)

Stresses

AlthoughtheEuler-Bernoullibeamtheoryisformulatedintermsofaxialstress,σ ,beam bending involves both axial and shear stresses. The axial stress is directlyrelatedtothebendingmoment,whiletheshearstressisdirectlyrelatedtotheshearforceasdescribedshortly.Onewayofobtaininganexpressionforaxialstressintermsofthebendingmomentistocombinemateriallawandkinematicsequations,whichyields:

! = " E #d

2w

dx2# z (29)

Then substitute the differential equation without equilibrium equations, i.e., Eq.(20),toobtain:

! = "

M

I # z (30)

It is noted that a positive bendingmoment, i.e., tension at the bottom, correctlyyieldsnegativestressesatthetop,i.e.,compression,where z ispositive.Thisisthereason for theminussign inEq.(30),whichalso correctly givespositive tensionstresses at the bottom when a positive moment acts on the cross-section. In





summary, the beam theory presented in this documentconsists of the governingequationsshowninFigure6.

Figure6:GoverningequationsinEuler-Bernoullibeamtheory.

ShearStressandShearCentreforOpenCross-sections

When approaching shear stresses from bending, an anomaly in Euler-Bernoullibeam theory is first noted. The theory is based on the assumption that planesections remain plane and perpendicular to the neutral axis. In otherwords, the

onlystrainthattakesplaceistheaxialshorteningorelongationofthefibresinthecross-section.Effectively,thispreventsshearstrain.Withnoshearstrainthereisnoshearstress,whichaddsuptozeroshearforce.Inotherwords,theshear forceisnotpartofthetheory.Thisisananomaly,becauseshearforcewilldevelopeveninsimple beams that are subjected to transversal load. Theanomaly is customarilyaddressedbyrecoveringtheshearforcebyequilibriumoncethebendingmomentiscomputed.Infact,accordingtoEq.(2)theshear forceisequaltothederivativeofthe bending moment; this is the equation that recovers the shear force. AnotherdocumentonTimoshenkobeamtheorydescribesanapproachtofurtherextendthebeamtheorytoincludedeformationduetoshearforces.

Toobtainexpressionsfortheshearstress,τ ,intermsoftheshearforce,V,considertheinfinitesimallyshortbeamelementinFigure7.Furthermore,considera“cut”inthecross-sectionandletqsdenotethe“shearflow”atthatlocation.

! "

q

! = "

d 2w

dx2# z

! = E "#

! = "

M

I # z

M = !" # z d A

A

$

q = ! EI d

4w

dx4

q = !dV

dx

V =dM

dx

M = EI d

2w

dx2

w

M





Figure7:Shearflowbyequilibriumofinfinitesimalbeamelement.

Theshearflowistheforceperunitlengthofthebeamthatensuresequilibriumwiththeaxialstresses,whicharegreaterononesidethantheotherduetodM :

qs !dx = d " dA As

# =

dM

I ! z dA

As

# (31)

where Asisthecross-sectionalareaoutsidethecut.GiventhatV=dM/dx ,thisyields

qs =

V

I !S (32)

where

S = zdA

As

! (33)

Theshearstressiscalculatedbydistributingtheshearflowoverthethickness,t ,ofthecross-sectionattheparticularlocation:

! =V "S

I " t (34)

For example, for a rectangular cross-section themaximum shear stress is at theneutralaxis,withvalueequalto

! =

3

2 "

V

A (35)

Theshearcentreofacross-section,sometimescalledthecentreoftwist,isthepointwheretheresultantoftheshearforcemustacttoavoidrotationofthecross-section.Thecoordinatesoftheshearcentrearedenotedby y scand z sc,andthereareseveraltechniques to determine them. The simplest case is double-symmetric cross-sections;forthesecross-sectionstheshearcentrecoincidewiththecentroid.Infact,ifacross-sectionhasanaxisofsymmetrythentheshearcentreislocatedonthis

d x

V M

M+dM

V+dV

Axial stresses

Shear stresses

(“shear flow”)





axis.Forgeneralcross-sections,oneapproachtodetermine y scand z scisdescribedinthedocumentonwarpingtorsion,wherethe“omegadiagram”isutilized.However,asomewhatsimplerapproach,whentheconsiderationofwarpingtorsionisoffthetable,isofferedhere.Theprincipleissimple;bydefinition,themomentoftheshearflowabouttheshearcentremustbezero.Thisleadstothefollowingprocedureto

determine the coordinatesofthe shearcentre,provided y and z aretheprincipalaxesthroughthecentroidofthecross-section:

1. Selectanarbitrarypointas trialshear centre,andlet y scand z scdenotethecoordinatesoftheshearcentrerelativetothecentroid;inotherwords,let y scand z scdenotethedistancesfromthecentroidtotheshearcentre

2. InaccordancewithEq.(32),determinetheshearflowinthecross-sectionduetoashearforceinthe z -direction

3. WritetheequationthatexpressesthemomentoftheshearflowinItem2aboutthetrialshearcentre;ingeneral,both y scand z scwillappearinthisexpression

4. SimilartoItem2,determinetheshearflowinthecross-sectionduetoashearforceinthe y -direction

5. SimilartoItem3,writetheequationthatexpressesthemomentoftheshearflowinItem4aboutthetrialshearcentre

6. SettheequationsfromItems3and5bothequaltozeroandsolvethesetwoequationsforthetwounknowns y scand z sc

Onlyonemomentequation isneededfor single-symmetric cross-sections; inthatcasestheproceduresimplifiesto:

1. Selectanarbitrarypointalong thesymmetryaxisastrial shearcentre,andletedenotethedistancefromthecentroidtothatpoint

2. InaccordancewithEq.(32),determinetheshearflowinthecross-sectionduetoashearforceinthedirection perpendicular totheaxisofsymmetry

3. WritetheequationthatexpressesthemomentoftheshearflowinItem2aboutthetrialshearcentre;ewillappearinthisexpression

4. SettheequationfromItems3equaltozeroandsolvefore

ShearStressandShearCentreforClosedCross-sections

The determinationof shearstressandshear centre for closed cross-sections, i.e.,cross-sectionswithcell,canbeapproachedintwoways.Inthefirstapproach,theshear centre coordinates are first determined, using the “omega diagram” asdescribedinthedocumentonwarpingtorsion.Next,a“cut”inthecross-sectionismadetoyieldanopencross-section;thenewcoordinate s,whichtracesthecross-sectionaroundthecell,originatesatthecut.Theunknownshearflowatthecutisdenotedqo,andtheshearflowatallotherlocationsaredeterminedrelativetothisvalueinaccordancewithEq.(32):

q = qo+V

I !S (36)





Onceqisdeterminedatalllocationsofthecross-section,themomentoftheshearflowabouttheshearcentreiscomputed:

T = q !hds!" = qo !hds!" +V

I !S !hds!" (37)

wheretheintegralsaremadearoundthecell,andh(s)isthedistancefromtheshearcentre to the tangent line of the cross-section at s. By definition the moment,T ,abouttheshearcentremustbezero,andsolvingforqoyields:

qo= !

V

I "

S "hds!# hds!#

= !V

2 " Am " I " S "hds!# (38)

wherethelastequalityexpressesthattheintegralof haroundthecross-sectionistwicethecellarea, Am.Havingthevalueofqo,theshearflowisdeterminedatotherlocationsbyEq.(36).Thesecondapproachtodeterminetheshearflowandshearcentreinclosedcross-sectionsentailsdeterminingtheshearcentreattheend.Also

this approach introduces a cut in the cross-section, but now compatibility isconsideredinsteadofmomentequilibrium.Tothisend,considerFigure8.

Figure8:Twocontributionstoshearstrain.

Figure 8 illustrates the two following contributions to shear strain in aninfinitesimalelementofthecross-section:

! = ! 1+ !

2=

du

ds+d "

dx#h (39)

whereφ istherotationofthecross-section.Materiallawstatesthat:

h

dx

! 1

d !

ds

! 2

du





! ="

G=

q

G # t (40)

andcombinationofEqs.(39)and(40)yields

du =q

G!t !ds "

d #

dx!h !ds (41)

Integrationaroundthecellyieldsthetotal“gapopening”duetothecut:

u =q

G ! t ds!" #

d $

dx!hds!" (42)

Toensurecompatibility,thisgapopeningmustbezero:u=0.Fromthatequationitisofinteresttosolvefor d φ /dxbecausethecross-sectionshouldnotrotateduetoshearforce:

d !

dx=

q

G " t ds!#

hds!# =

q

G " t ds!#

2 Am

= 0

(43)

Eq.(36)providestheexpressionfortheshearflow,q,whichyields:

qo

G ! t ds!" +

V !S

I !G ! t ds!" = 0 (44)

Solvingforqoyields:

qo = !

V

I "

S

G " t ds!#

1

G " t ds!

#

(45)

Whenthematerialishomogeneous,theexpressionsimplifiesto:

q

o =!V

I "

S

t ds!#

1

t ds!#

(46)

Havingthevalueofqo,theshearflowisdeterminedatotherlocationsbyEq.(36).

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