Pres Warping en Small

Introduction Assumption I Profile Section Derivation, Example Diff. Eq. Element Design Conclusion

Preliminaries in Warping Torsion

Emanuel [email protected]

Graduiertenkolleg 1462Bauhaus-Universitat Weimar

2. Dezember 2009

Emanuel Bombasaro Preliminaries in Warping Torsion 1 / 42


Part 1

Motivation and Introduction

Assumption

Deduction with the Analogy of Flange Bending (I ProfileSection)

St. Venant Torsion Deduction an a Solid section

Enhance the Approach for Warping Torsion

Examples

Elliptical SectionSpecial Cases of the Elliptical Section



Part 2

Short Recapitulation

Solution for the Differential Equation in Analogy to theBending Theory of Second Order

Practice on a Simple Example

Design of a Structure

Important Remarks when Designing Structures due toWarping Impacts

Conclustion

Information

The slides represent only a minor and basic part of the holelecture, so to reach full comprehension the explanatory notes of thelecturer are indispensable!



St. Venant Torsion

Section can wrap freely, displacement in the x-direction arenot constrained

Torsional moment is not varying much along the elementlength

Section geometry is constant

Section is warping free.

This are very restrictive conditions!

Adhemar Jean Claude Barre de Saint-Venant (* 23. August 1797 in Villiers-en-Bie`re, Seine-et-Marne; 6. January

1886 in St Ouen, Loir-et-Cher) was a french Mathematician and Physicist.



Lemma

Division of the torsional moment in two partsMT = MTp + MTs

primary Torsional MomentMTp = GIT

or St. Venant Torsional Moment. Section can freelywrap and so the displacements in x-direction are notconstrained. x = 0

secondary Torsional MomentMTs = M

results when the warping is constrained and sonormal stresses in x-direction arise. x 6= 0



Appearance Warping Torsional Moment

St. Venant-Trosion (MT = GIT) exist always when a element is

twisted . Warping torsional moment appears additionally;if with twisting of the element warping occurs and

if the twist along the element is not constant



Example

Element with constant,not constrained warpingand so without a warpingtorsional moment



Warping

Warping of a section is only depending on the section geometrywhich means that their exist free warping and restrained warpingsections.Warping free sections are

Circular and Annuluses sections

Section which exists out of two sheets; angles and tees

Square pipes with constant wall thickness

Rectangular pipes only if b/t relation between web and flangeis the same

Warping Free SectionsIf a section is warping free only St. Venant torsionalmoment appears!



Further Assumptions

Further constraints are assumed in the warping torsion approach

With the warping torsion connected secondary sheardeformations are neglected

With the warping torsion connected longitudinal deformationsalong a strait element of the section (web, flange) are linear,BERNOULLI hypothesis.

Section are constant along an element

The solution is based on the linear elasticity, which meanssmall deformations and twist angles.



Reference Note Concerning Secondary Shear Deformations

The neglecting of secondary shear deformations leads at pointswith fixed warping deformations ( = 0) to bigger values for thesecondary torsional moment/ secondary shear stresses. Theinfluence decreases rapidly with the distance from the fixing and isprimary dominate for rectangular pipes.



Derivation Schema for the Diff. Equation, Flange Bending



Differential Equation for Flange Bending

Differential Equation of Flange Bendingx

(EIz

2vx2

)= Qy (x)

with the use ofT = Qy (x) =

MTh

v = (x) h2we obtainEIz,G

h2

4 (x) = MT it follows EI(x) = M

Definitions valid only in this special casewarping moment of second order I, refered to theshear center M, we can expressI = Iz,G

h2

4 Attention! [Lenght6]

The warping moment M [Force Lenght2] can ONLY

IN THIS CASE be computed by M(x) = MG (x)hwith flange moment MG (x).



Basics ForcesThe equilibrium condition for torsional forces leads to

MT =

A

(z xy + y xz) dA (1)y and z coordinates are related to the the shearcenter.

Basic of Geometrical RelationsThe twisting axis of the section is called twist axis.We use =



Relation to Gravity Axis

If the twisting axis (in the shear center) is parallel to the gravityaxis, the movement relative to the gravity axis a rigid body motion.Linear elasticity and small rotation angles are assumed.

Mx =

A

(zxy + yxz ) dA (2)

with help of the coordinative relations y = y yM andz = z zM we obtain

Mx = MT (3)



Twisted Element Fiber When twisting an element due to torsionalaction a strait element fiber AB is turned into a helixAB . This leads to a deformation in w and vdirection but also in a displacement in u direction,which causes the section to be warped.If this warping is constrained this leads to normalstresses x and this case warping torsion has to beconsidered.

Small Deformations When the problem is reduced to smalldeformations and twisting angles, helixes can beconsidered to be strait lines. So the relation betweenrotation and twisting of the section can beexpressed = 4x .



Derivation at a Solid Section Element

The derivation is performed at a general solid section, we use thepole coordinate system

y = r cos und z = r sin.The nodes P and Q are inthe twisted configuration P

and Q . The displacementsresults to

v = r cos ( + ) r cos, w = r sin ( + ) r sinafter linearization we getv = xz , w = xy



Constitutive Relations

y =v

y, z =

w

z, yz = 2yz =

v

z+w

y(4)

it results y = 0, z = 0, yz = 0. Which shows that under puretorsion action on an element no distortion occurs. With x = 0,y = 0, z = 0 results x =

ux = 0 and so u=u(y,z).

Following approach u(y , z) = (y , z) is used. The constitutiverelations

xy = G(

y z), xz = G

(

z+ y

)(5)

2

y2+2

z2= 4 = 0 (6)

(y , z) is call warping function which fulfills the LAPLACEdifferential equation.



Solving the Warping Function

Instead of matching to the boundary conditions, it is moreadvantageous to fit the PRANDTL torsion function (y , z). Sothe shear stresses xy and xz can be expressed

xy =

z, xz =

y(7)

What we see is that the PRANDTL torsion function is a stressfunction. With the help of the constitutive relations we obtain

z= G

(

y z),

y= G

(

z+ y

)(8)



PRANDTL Torsion Function

PRANDTL Torsion FunctionThe partial z derivation of the first equation andpartial y derivation of the second equation andsumming up leads to

4 = 2G (9)So we see that the PRANDTL torsion function fulfilsthe POSSION differential equation.

Boundary ConditionsThe surface of the element under torsional action isfree of stresses, so with the condition on the figureon slide 16 xzxy =

dzdy we obtain

xz dy +xy dz = y

dy +

zdz = d = 0 (10)



Inserting

Inserting (7) in (2) results

MT =

A

[

y(y) +

z(z)

]dA + 2

AdA (11)

With the help of the GAUSS integration method on the first termof the equation we get

MT =

C (yny + znz ) dC + 2

AdA (12)

when looking at solid sections, along C is zero, it follows

MT = 2

AdA (13)



Inserting

Inserting (5) in (2) results

MT = G

A

(y2 + z2 + y

z z

y

)dA (14)

The Integral

IT =

A

(y2 + z2 + y

z z

y

)dA (15)

which is call torsional second moment. IT is a section value. GITis called torsion stiffness. For the warping free element it results

MT = MTp = GIT (16)



Warping

with the help of equation (7) and (8) we find

u

y= G

(

y z),u

z= G

(

z+ y

)(17)

and we find the separable approach function

u(x , y , z) = (x)M(y , z) (18)

the unit warping function M is relative to thesection shear centre

In previous slides we used u(y , z) = (y , z) as approachfunction. (In the literature mostly M is used.)



Warping Stressis , Warping shear flow T

The warping stress results in the case of restrained warpingwith the help of eq. (18)

x = = Eu

x= EM , (19)

because the constraint x = 0 can not be fulfilled any more.

The warping shear flow T can be found

T = E

AM(s)dA = ES

(20)

by evaluating the equilibrium conditions d t ds + dT dx = 0.S is in analogy called warping first order moment.



Warping Torsional Moment and Warping Moment

For a general section we obtain with equation (20) andM =

s0 r

Mt (s)ds

MTs =

T(x , s)r

Mt (s)ds = E

A

[M(s)

]2dA (21)

the integral is defined as warping second order moment I, so wecan express the warping moment in analogue to the resultingstresses

M =

A

MdA = E

A

[M]2

dA = EI (22)



Synopsis

Returning to the Lemma

MT = MTp + MTs (23)

Primary Part; St. Venant Torsion

MTp = GIT (24)

Secondary Part; Warping Torsional Moment

MTs = EI (25)Differential Equation for Torsional Load Action

MT = GITEI GIT(x)EI(x) = mT (x)

(26)



Example Elliptic Section

Shape Functiony2

a2+ z

2

b2 1 = 0

Stress Fucntion = a2b2

a2+b2

(y2

a2+ z

2

b2 1)

G

MT = 2

A dA =a3b3pia2+b2

G comination with the stress function

results = MTabpi(

y2

a2+ z

2

b2 1)

and so

xy = 2MTab3pi z und xz = 2MTa3bpiy . IT = a3b3pi

a2+b2

The warping function results to = a2b2a2+b2

yz



Special Case; Circular Section

with a = b = R, y = r cosz = r sin we obtainxy = xz = 2MTR4pi r . IT = R

4pi2

The warping function results to = 0 and so obviously nowarping!



Special Case; Rectangular Section

We replace b with the half of the narrow side of the section b andwe look at the limit b/a 0

so [(b)2 z2]G, obviouslyxy 2Gz , xz 0 and with L = 2a und b = 2b we obtainIT Lb33 .The warping function results to yz and obviously warpingexists and the section values has to be corrected!IT = 1Lb

3 und max = 2MTLb2

. are depending on the relation between length and hight L/b ofthe section.



Differential Equation for Warping Torsion Function (withoutsecondary shear deformations)

EI(x) GIT(x) = mT (x) (27)

(x). . . torsion twist angle EI. . . warping stiffnesGIT . . . torsional stiffness (St. Venant) mT (x). . . uniform torsionalload

Differential Equation for Second Order Bending (withoutshear deformations)

EIw (x) N II w (x) = q(x) (28)

w(x). . . bending ordinate EI . . . bending stiffnessN II . . . longitudinal force q(x). . . uniform distributed load



Warping Torsion Bending Theory 2nd OrderTerms

+ Twist Angle w Bending Ordinate

Twist w Angel of Bending Ordi-nate

M Warping Moment M Bending Moment

MT + Torsional Moment R Transversal Load

MTpPrimary Torsional Mo-ment (St. Venant)

N II w 2nd Order part of R

MTsSecondary Torsional Mo-ment

Q Shear Force

mT +Sectional Torsional Mo-ment

q Sectional Uniformal Load

MeT +Concentrated TorsionalMoment

P Concentrated Load

EI Warping Stiffness EI Bending Stiffness

GITTorsional Stiffness (St.Venant)

N II Longitudinal Force



Warping Torsion Bending Theory 2nd OrderEquations

= MTp/GIT w = N II w /N II (Identitie) = M/EI w = M/EI

MT = MTp + MTs R = NII w + Q

M T = mT R = qM = MTs M

= QDifferential Equation

EI GIT = mT EIw N II w = q

Boundary ConditonsCradle BearingM = 0, = 0

M = 0, w = 0

Fixing = 0, = 0 w = 0, w

= 0

Free EndM = 0, MT = 0

M = 0, R = 0

Fixed HeadstockMT = 0,

= 0 R = 0, w = 0



Application on Elements, Example

m = 10kNm/ml = 10.0mE = 21 000kN/cm2

= 0.3

Sections

I Profile Rectangular Pipe

h = 30cmb = 20cm

t = 1.5cms = 1.0cm



0 1 2 3 4 5 6 7 8 9 1020

10

0

10

20

x

MTs

[kN

m], M

[kN

m2]

Torsionsmomente

M

0 1 2 3 4 5 6 7 8 9 1040

20

0

20

40

MTp

[kN

m]

0 1 2 3 4 5 6 7 8 9 100

100

200

[]

Torsionsdrehwinkel und Verdrillung

x0 1 2 3 4 5 6 7 8 9 10

1

0

1

[]

Figure: Condition Diagrams I Profile

0 1 2 3 4 5 6 7 8 9 100

0.5

1x 105

x

MTs

[kNm

], Mt [k

Nm2 ]

Torsional Moment

Mt

0 1 2 3 4 5 6 7 8 9 1050

0

50

MTp

[kNm

]

0 1 2 3 4 5 6 7 8 9 100

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

[!

]

Torsional Twist Angle and Twisting

x0 1 2 3 4 5 6 7 8 9 10

4

3

2

1

0

1

2

3

4x 103

v [

]

Figure: Condition Diagrams Rectangular PipeEmanuel Bombasaro Preliminaries in Warping Torsion 33 / 42


Results

I Profile Rectangular Pipe

M = 19.81kNm2 M = 0.00kNm

2

MTs = 14.52kNm MTs = 0.00kNmMTp = 35.48kNm MTp = 50.00kNm = 111.87 = 0.43

= 0.66 = 3 103



Figure: Result taken from Fiedrich + Lochner



0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.5

1

1.5

x

MTs

[kNm

], Mt [k

Nm2 ]

Torsional Moments

Mt

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 51.5

1

0.5

0

MTp

[kNm

]

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 52

1

0

[

]

Torsional Twist Angle and Twisting v

x0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

0.02

0.01

0

v [

]

Figure: System, Diagrams I Profile Section, Cantilever



Forces and Stresses [kN/cm2]

Node a b 1 2 3

N [kN] 100.00 100.00 x ,N 1.13 1.13 1.13My [kNm] 10.00 10.00 x ,My 1.43 1.43 0.00Mz [kNm] 15.00 15.00 x ,Mz 5.00 0.00 0.00M[kNm

2] 0.10 1.50 x ,M 5.13 0.00 0.00MTp[kNm] 0.00 1.03 Tp 2.87 2.87 1.91MTs [kNm] 0.00 1.03 Ts 0.00 0.17 0.00 [] 0.00 1.49 [] 0.00 0.02v 13.63 5.85 3.50

x =N

A+

MyIy

z +MzIz

y +MIR ; = Tp + Ts (29)

v =2x + 3 2 (30)



= 1.0 takes into account secondary shear deformations



Conclusion

If a section is not warping free both the secondary torsionalmoment and the normal stresses due to the warping momenthas to be taken into account.

In addition the secondary shear deformations and sheardeformations may have to be taken into account.

When performing numerical computations, analyzations has tobe done with caution, because solving the differential equationcan lead to wrong results and so structural failure.



Cradle Bearing in Steel Structures



Handbook of Continuum Mechanics: General Concepts -Thermoelasticity, J. Salencon, Springer 2001

Festigkeitslehre, 2. Auflage, H. Mang, G. Hofstetter, Springer2000

Baustatik Theorie I. und II. Ordnung, 4. Auflage, H. RubinK-J. Schneider, Werner Verlag 2002

Schub und Torsion in geraden Staben, 3. Auflage, W. Francke,H. Friedmann, Vieweg 2005

Mechanik der festen Korper, 2. Auflage, H. Parkus, Springer2005

Vorlesungen uber Stahlbau. Grundlagen, 2. Auflage, Kh. Roik,Ernst & Sohn 1983


TitleMotivation and IntroductionMotivation and Introduction

AssumpionBasics in Warping TorsionWarpingFurther Assumptions

Derivation with the Help of Flange BendingDerivation Schema for the Diff. Equation, Flange Bending

Derivation of Fundamental Relations on a Solid SectionPreliminariesDerivation at a Solid Section ElementSolving the Warping FunctionWarping MomentExample Elliptic SectionCircular SectionRectangular Section

Solution in Analogy to the Second Order Bending TheoryEquationsAnalogy

Application on Elements and DesignApplication on ElementsI Profile with Eccentric Horizontal Force, CantileverCondition Diagrams for Special ExamplesDesign Tables

ConclusionConclusionExample; Bearings in Steel StructuresReference

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