http www.inrisk.ubc.ca process.php file=structural analysis warping torsion

7/27/2019 Http Www.inrisk.ubc.CA Process.php File=STRUCTURAL ANALYSIS Warping Torsion

1/15

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.

Warping Torsion

Inadditiontoshearstresses,somememberscarrytorquebyaxialstresses.Thisiscalled warping torsion. This happens when the cross-section wants towarp, i.e.,displaceaxially,butispreventedfromdoingsoduringtwistingofthebeam.Notallcross-sectionswarp,andeventhosethatwarpdonotcarrytorquebyaxialstressesunless they are axially restrained at some location(s) along the member. Cross-sections that do NOT warp include axisymmetric cross-sections and thin-walledcross-sectionswithstraightpartsthatintersectatonepointthecross-section,suchas X-shaped, T-shaped, and L-shaped cross-sections. For these cross-sections alltorque is carried by shear stresses, i.e., St. Venant torsion, regardless of theboundaryconditions.

WarpingofI-sectionsAsapedagogicalintroductiontowarpingtorsion,considerabeamwithanI-section,such as awide-flange steel beam.When torsion is applied to the beam then theflanges of this cross-section experiences bending in the flange-planes. In otherwords, torsion induces bending about the strong axis of the flanges. When theflanges are fixed at some point, such as in a cantilevered beam with a fully


2/15

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

Warping Torsion Page 2

clampedend,someofthetorqueiscarriedbyaxialstresses.Tounderstandthis,denotethebendingmomentandshearforceineachflangeby MandV,respectively,asshowninFigure1.

Figure1:WarpingofI-section

(zisthelocalaxisforbendingofflange,nottheglobalz-axisofthecross-section).

Thetorquethatthecross-sectioncarriesbybendingintheflangesis:

T = h !V

(1)where h is the distance between the flanges and V is positive shear force inaccordancewiththedocumentonEuler-Bernoullibeamtheory.Thatdocumentalsoprovides the equilibrium equation that relates shear force to bending moment,whichyields:

V =dM

dx ! T = h " dM

dx (2)

Thebeamtheoryalsoprovidetherelationshipbetweenbendingmomentandflangedisplacement,w:

M = EI! d2

w

dx2 " T = h !EI! d

3

w

dx3 (3)

where I is themoment of inertia of one flange about its local strong axis. Next,Figure1isreviewedtodeterminetherelationshipbetweenwand:

w = !"#h

2 $ T = !

h2

2#EI#

d3"

dx3 (4)

which resulted in the differential equation for warping torsion of an I-section.However,thisequationisgenerallywritteninthisformat:

T = !ECw "d

3#

dx3 (5)

whichimpliesthat,forI-sections,thecross-sectionalconstantforwarpingis:

Cw= I!

h2

2 (6)

!

z, w

h!

z, w


3/15



whereitisreiteratedthatI is themomentof inertiaof one flangeabout its localstrongaxis.

CompleteDifferentialEquationforTorsion

As mentioned earlier, when warping is restrained the torque is carried by both

shear stresses, i.e., St. Venant torsion and axial stresses, i.e., warping torsion.Specifically,thetorquefromshearandaxialstressesaresuperimposed,whichleadstothefollowingcompletedifferentialequationfortorsion:

T = GJ!d"

dx# EC

w!

d3"

dx3 (7)

Whenequilibriumwithdistributedtorquealongthebeam, mx,isincluded,i.e.,mx=dT/dx,thenthefulldifferentialequationreads

ECw!

d4"

dx4#GJ!

d2"

dx2= m

x (8)

Solution

ThecharacteristicequationtoobtainthehomogeneoussolutionforthedifferentialequationinEq.(8)reads

!4 "GJ

ECw

#!2= 0 (9)

The roots are 0, 0, (GJ/ECw), and (GJ/ECw). Accordingly, the homogeneoussolutionis

!(x) = C1"e

GJ ECw "

x+ C

2"e

# GJ ECwx+C

3"x + C

4 (10)

whichguidestheselectionofshapefunctionsifanexactstiffnessmatrixwithbothSt.Venantandwarpingtorsionissought.Anotherwayofexpressingthesolutionis:

!(x) = C1 "sinh GJ ECw "x( )+C2 "cosh GJ ECw "x( )+C3 "x +C4 (11)

where the coefficients, Ci, in Eq. (10) are different from those in Eq. (11). Forexample,thehomogeneoussolutionforacantileveredbeamthatisfullyfixedat x=0andsubjectedtoatorque,To,atx=Lhasthesolution

!(x) =1

GJ ECw

"T

o

GJ"

tanh GJ ECw"L( ) " cosh GJ ECw "x( )#1$% &'

#sinhGJ EC

w "x

( )+ GJ EC

w "x

(

)

**

+

,

-- (12)

FromthissolutionthetorquecarriedbySt.Venanttorsioniscomputedby:

TSt.V.

(x) =GJ!d"

dx (13)

andthetorquecarriedbywarpingtorsioniscomputedby:


4/15



Twarping(x) = !ECw "d

3#

dx3 (14)

whereTSt.V.(x)+Twarping(x)=Toforall0


5/15



thesedocuments.Althoughthisisashortcoming,thepresentedtheoryissufficientfor many practical applications. This is because many thick-walled cross-sectiontypes are difficult to fully restrain axially. In contrast, it is easier to imagineconnection designs for thin-walled cross-sections that provide sufficient axialrestrainttodeveloptorqueduetoaxialstresses.

Figure2:Beamaxissystemandcorrespondingdisplacements.

Kinematics

Theobjective inthissection istoestablish arelationshipbetweenaxialstrain,x,

andaxialdisplacement, u.Forthispurpose,letthecoordinate sfollowthecentrelineofthecontourofthecross-sectionandlet hdenotethedistancefromthecentreofrotation(ysc,zsc)tothetangentofthecoordinatelines,asshowninFigure2.They-z-axes originate in the centroid of the cross-section, while the shear centrecoordinatesyscandzscarepresentlyunknown.Analogoustotheomissionofsheardeformation in Euler-Bernoulli beam theory, the shear the shear strain xs isneglected in the combined theory of warping torsion and bending. In turn, thisprecludes shear stresses, but as in the beam theory they are recovered later byequilibrium. Furthermore, the shear stresses from St. Venant torsion aresuperimposed so that the total shear strain xs is equal to that from St. Venanttorsion.Moreprecisely,xsisequaltotheshearstrainatr=0fromSt.Venanttheory,

i.e.,atthemid-planeofthecross-sectionprofile.Inpassingitisnotedthataccordingto the St. Venant theory the shear strain at r=0 for open cross-sections is zero.Furthermore,asafundamentalkinematicspostulationit isassumedthatthecross-sectionretainsitsshape.Thisimpliesthats=s=0andthatthedisplacementinthes-directionis:

!v = !v "cos(#)+w "sin(#)+$"h (19)

T, !

y, v

z, w

s, !v

h

r

x, u

Shear centre / centre of twist (ysc,zsc)

Centroid (0, 0)


6/15



wherevandwarethedisplacementsofthecross-sectionandistheanglebetweenthes-axisandthey-axes.ThecontributionstoEq.(19)areillustratedinFigure3.

Figure3:Contributionstothedisplacementalongthe s-axis.

Giventhepreviousassumptionstherelevantkinematicsequationsare !

x= u

,x

(20)

!xs = !v,x + u,s = !xsSt.V. (21)whereitisreemphasizedthat !

xs

StV= 0 foropencross-sections.SubstitutionofEq.

(19)intoEq.(21)yields

!dv

dx"cos(#)+

dw

dx"sin(#)+

d$

dx"h +

du

ds= %

xs

St.V. (22)

Next,therelationshipsbetweenthedifferentials ds,dy,anddzareestablished.WithreferencetoFigure3theyare:

dy

ds= !cos(")

dz

ds= sin(")

(23)

SubstitutionofEq.(23)intoEq.(22)yields

dv

dx!

dy

ds+

dw

dx!

dz

ds+

d"

dx!h +

du

ds= #xs

St.V. (24)

Solvingforduyields

du = !dv

dx"dy!

dw

dx"dz !

d#

dx"h ! $xs

St.V.%&'

()*"ds (25)

Itisdesirabletoexpress !xs

St.V. intermsofd/dxsothatthelattercanbepulledoutsidetheparenthesis.ThisisachievedbyutilizingequationsfromthetheoryofSt.

w

-v

!

s

(ysc,zsc) !v "cos(#)

!

w!sin(")

h

!"h

ds

dy

dz


7/15



Venanttorsion.Foropencross-sections!xs

St.V. vanishes,butingeneralmateriallawyields

!xs

St.V.=

"xs

St.V.

G (26)

and !xsSt

.

V.

= d" dr where is Prandtls stress function, which is cross-sectiondependent.Forthin-walledcross-sectionswithonecellagoodstressfunctionis

!(s,r) = K "1

2+

r

t

#$%

&'( (27)

Asaresult,theshearstressis

!xs

St.V.="

,r=

K

t (28)

Toexpressthestressintermsofd/dxitisfirstnotedthat

T =GJd!

dx (29)

andthatthetorque,T,canberelatedtothestressbythestress-resultantequation

T = 2 ! "dAA

# (30)

ForthestressfunctioninEq.(27)theintegralevaluates

T = 2 !K!Am (31)

Finally,Eqs.(31)and(29)substitutedintoEq.(28)yieldsthedesiredexpression: !

xs

St.V.=

K

t=

T

t"2 "Am

=

GJ

2 " t"Am

"

d#

dx (32)

SubstitutionintoEq.(25)yields

du = !dv

dx"dy !

dw

dx"dz !

d#

dx" h !

J

2 " t"Am

$%&

'()"ds (33)

Integrationyields

u(y,z)=

u!

dv

dx"y!

dw

dx"z!

d#

dx"$

(34)whereuis theintegrationconstant,whichrepresentsauniformaxialdisplacementofthecross-sectionandhasbeendefinedas:

! " h #J

2 $ t$Am

%&'

()*ds+ (35)


8/15



Itisreemphasizedthatthisexpressionisvalidforcross-sectionswithonecell.Foropencross-sectionstheshearstraininthemid-planeofthecross-sectionprofileiszero,whichimpliesthat

! " hds# (36)

Finally,combiningEq.(20)withEq.(34)yieldsthefinalkinematicsequation:

!x=

du

dx"d2v

dx2#y"

d2w

dx2#z "

d2$

dx2#% (37)

MaterialLaw

Hookeslawprovidestherelationshipbetweenaxialstressandaxialstrain:

!x= E"#

x (38)

SectionIntegration

Integrationofaxialstressoverthecross-sectionyieldstheaxialforce:

N = !xdA

A

" (39)

Integrationofaxialstressmultipliedbydistancefromthecentroidyieldsbendingmoment:

Mz = ! "x #ydAA

$ (40)

Integrationofaxialstressmultipliedbydistancefromthecentroidyieldsbendingmoment:

My = ! "x #z dAA$ (41)

Integrationofaxialstressmultipliedbythepreviouslydefinedquantityisdefinedasthebi-moment:

B ! " #x

$%dAA

& (42)

Equilibrium

Distributedaxial loadalongthebeamis relatedtotheaxialforcebythefollowingequilibriumequation:

qx= !

dN

dx (43)

Distributedloadinz-direction,whichisassumedtoactthroughtheshearcentreoritwillcontributetomx,isrelatedtotheshearforcebythefollowingequilibriumequation:


9/15



qz = !

dVz

dx (44)

Equilibrium also provides the relationship between shear force and bendingmoment:

Vz =

dMy

dx (45)

Thecorrespondingequilibriumequationsintheotherdirectionare:

qy =

dVy

dx (46)

and

Vy =

dMz

dx (47)

Finally,equilibriumfordistributedtorquealongthebeamyields:

mx= !

dT

dx

(48)

DifferentialEquations

SubstitutionofthekinematicsequationinEq.(37)intothemateriallawinEq.(38)yields:

!x = E"du

dx#E"

d2v

dx2"y #E"

d2w

dx2"z #E"

d2$

dx2"% (49)

Substitution of Eq. (49) into the section integration Eqs. (39) to (42) yields thefollowingsetofequations:

N

Mz

My

B

!

"

##

$

##

%

&

##

'

##

= E(

dAA

) * ydAA

) * zdAA

) * +dAA

)

* ydAA

) y2 dAA

) y (z dAA

) y (+dAA

)

* zdAA

) y (z dAA

) z2 dAA

) z (+dAA

)

* +dAA)

y (+dAA)

z (+dAA)

+2 dAA)

,

-

.

.

.

.

.

.

.

.

.

.

/

0

1111111111

du

dx

d2w

dx2

d2v

dx2

d22

dx2

!

"

#####

$

###

##

%

&

#####

'

###

##

(50)

where symmetry is observed. Under certain condition described shortly, theequationsbecomedecoupledandreducesto:

N = EA !du

dx (51)


10/15



My = EIyd

2w

dx2 (52)

Mz = EIzd

2v

dx2 (53)

B = !ECw

d2"

dx2 (54)

where the diagonal components of the matrix in Eq. (50) have been named asfollows:

A = dAA

! (55)

Iz = y2dA

A

! (56)

Iy = z2 dA

A! (57)

Cw= !2 dA

A

" (58)

For the system of equations in Eq. (50) to be decoupled, the six off-diagonalelements of the coefficient matrix must be zero. These six conditions form animportantpartofthecross-sectionanalysis.Infact,theydeterminethefollowingsixunknownsofthecross-section:

1. yo=y-coordinateofthecentroid

2. zo=z-coordinateofthecentroid3. o=orientationoftheprincipalaxes4. C

=normalizingconstantforthe-diagram

5. ysc=y-coordinateoftheshearcentre6. zsc=z-coordinateoftheshearcentre

Specifically,thecoordinatesofthecentroidofthecross-sectionaredeterminedbythetwoequations:

y dAA

! = z dAA

! = 0 (59)

Theorientationoftheprincipalaxesaredeterminedbythecondition: y !z dA

A

" = 0 (60)

Thenormalizationofthe-diagramiscarriedouttosatisfythecondition:

! dAA

" = 0 (61)


11/15



Thefollowingequationsaresatisfiedifthe-diagramisestablishedwithrespecttotheshearcentre,whichimpliesthattheshearcentrecoordinatesaredeterminedbytheseequations:

y !" dAA

# = z !" dAA

# = 0 (62)

Adding equilibriumwith external forces yields the finaldifferential equations foraxialdeformationandbending

qx = !EA "d2uo

dx2

(63)

qz = EIyd

4wD

dx4

(64)

qy = EIzd

4vD

dx

4 (65)

The derivation of the differential equation that combines St. Venant torsion andwarpingtorsionstartswiththedefinitionofthestressresultant:

T = !xs " t"hdAA

# = qs "hdsA

# = qs d$A

# = qs "$[ ]% & $dqsA

# (66)

whereqsistheshearflowandtheboundaryterm[ qs]fromintegrationbypartsiszero.Becauseshearstrainsareomittedfromthewarpingtheoryitisnecessarytoemploy equilibrium to recover the shear flow. With reference to Figure 4,equilibriumyields:

d!x "ds " t+ d#xs "dx " t= 0 $ d!

x

dx" t+ d

#xs

ds" t= 0 $ dqs

ds= % d

!x

dx" t (67)

Figure4:Equilibriumtorecovershearstresses.

!x+ d!

x

!x

dx

ds

!xs+ d!

xs

!xs

x

s

t


12/15



SubstitutionofEq.(67)intoEq.(66)yields:

T = ! "dqsA

# = " $d%x

dx$ t ds

A

# = " $d%x

dxdA

A

# =d

dx" $%x dA

A

# =dB

dx (68)

Adding the torque carried by shear stresses, i.e.,T=GJ(d/dx) and employing Eq.

(54)yields:

T = GJ!d"

dx# EC

w

d3"

dx3 (69)

Adding equilibrium with distributed torque from Eq. (48) yields the completedifferentialequationforSt.Venanttorsionandwarpingtorsion:

mx= EC

w

d4!

dx4"GJ#

d2!

dx2 (70)

ThesolutiontothisdifferentialequationwaspresentedinEq.(10).

Final-diagram

Eqs.(61)and(62)providetheconditionsthatthefinal -diagrammustsatisfy.Inparticular,thefinal-diagrammustbenormalizedanditmustbedrawnaboutthepoint(ysc,zsc),i.e.,theshearcentreofthecross-section.Inordertoarriveatthefinaldiagram,atrialdiagramisfirstdrawn.Thereafter,thetrialdiagramismodifiedtoobtainthefinalone.Forthispurposeitisusefultoestablishanexpressionthatrelatesthe-diagramdrawnaboutanarbitrarypointQtoanotherdiagramdrawnabout another pointSC.Sincethe-values consist of accumulateddouble sectorareas it is of interest to study a generic infinitesimal contribution. To this end,Figure5identifiesbygray-shadingthe-diagramcontributionsforaninfinitesimal

lengthdsofthecross-section.Thedoubleareaofthesector thatoriginatesin(ysc,zsc)is

d!sc = dy " z #zsc( )# dz " y# ysc( ) (71)

Thedoubleareaofthesectorthatoriginatesin(yQ,zQ)is

d!Q = dy " z #zQ( )# dz " y# yQ( ) (72)

Thedifferencebetweenthetwo-diagramcontributionsinFigure5is

d!sc " d!Q = ysc " yQ( ) #dz " zsc "zQ( ) #dy (73)

Integrationofinfinitesimalcontributionsyieldtheexpressionforthedifferencein-diagrams:

!sc "!Q = ysc " yQ( ) #z " zsc "zQ( ) #y +C (74)

whereCistheintegrationconstant.Rearrangingandrenamingsctoyields:

! =!Q + ysc " yQ( ) #z " zsc "zQ( ) #y+C (75)


13/15



whichistheexpressionforthefinal-diagramonceC,ysc,andzscareknown.

Figure5:Relationshipbetweentwo-diagrams.

The integration constant, C, normalizes the -diagram and is determined byrequiringtheintegralofthe-diagraminEq.(75)tobezeroaccordingtoEq.(61):

!dAA"

= !Q dAA"

+ CdAA"

= 0 # C= $

!Q dA

A

"

A

(76)

wheretermscancelintheintegrationbecause yandzoriginateatthecentroidofthe cross-section. The shear centre coordinates ysc and zsc are determined byrequiringtheintegralsinEq.(62)ofthe-diagraminEq.(75)tovanish:

y !"dAA

# = y !"Q dAA

# $ zsc $zQ( ) ! y2dA

A

# = 0 % zsc = zQ +y !"Q dA

A

#Iz

(77)

z !"dAA#=

z !"Q dA

A#+

ysc $ yQ

( ) ! z2

dAA#

= 0

%ysc

=

yQ $

z !"Q dAA

#Iy (78)

where terms cancel in the integration becausey andz are principal axes of thecross-section. In summary, the following procedure is suggested for thedeterminationofC,ysc,zsc,andultimatelythe-diagram:

1. Selectanarbitrarypoint,Q,inthecross-section

(ysc,zsc)

(yQ,zQ) ds

dz

dy

(y,z)


14/15



2. DrawthetrialQ-diagramaboutQ,i.e.,gathercontributionsaccordingtoEq.(35)alongcross-sectionpartsbyaclockwiseradarsweepabout Q,whileensuring that theQ-diagram is continuous (contributions that are madeclockwise are positive, while those that are made by a counter-clockwisesweeptoensurecontinuousdiagramarenegative)

3. DetermineCbyEq.(76)4. DetermineyscbyEq.(77)5. DeterminezscbyEq.(78)6. Determinefinal-diagrambyEq.(75)

Thereafter,thecross-sectionalconstantforwarpingtorsion,Cw,iscomputedbyEq.(58)andthestressesaredeterminedaccordingtothefollowingequations.

Stresses

Analogous totheEuler-Bernoulli beamtheory,axialstressesare obtainedbyfirstcombiningkinematics fromEq. (37)withmaterial law fromEq. (38),followedbysubstitutionof thefulldifferentialequationswithoutexternalequilibrium,i.e.,Eqs.(51)to(54),whichyields:

!x =N

A"Mz

Iz#y+

My

Iy#z +

B

Cw#$ (79)

Itisobservedthattheomegadiagramshowsthedistributionofaxialstressesinthecross-section due to torsion when warping is restrained. The shear flow isrecoveredbyequilibrium,whichwaspresentedinEq.(67):

qs = t!"xs = t!d"xs0

s

# = $ t!d%x

dxds

0

s

#

= $ t!dN

dx!1

A$dMz

dx!1

Iz!y+

dMy

dx!1

Iy!z +

dB

dx!1

Cw!&'

()

*+,dA

0

s

#

= $dN

dx!1

A!As +

Vy

Iz!Sy $

Vz

Iy!Sz $

dB

dx!1

Cw!S&

(80)

wherethefollowingdefinitionsaremade:

As= dA

0

s

! = t ds0

s

! (81)

Sy = ydA0

s

! = t"yds0

s

! (82)

Sz= zdA

0

s

! = t"z ds0

s

! (83)


15/15



S! = !dA0

s

" = t#!ds0

s

" (84)

http www.inrisk.ubc.ca process.php file=structural analysis warping torsion

Documents