Static and dynamic modeling of the beams in []

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Titre : Modélisation statique et dynamique des poutres en [...] Date : 25/09/2013 Page : 1/52Responsable : FLÉJOU Jean-Luc Clé : R5.03.40 Révision :

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Static and dynamic modeling of the beams in greatrotations

Summary : This note gives a mechanical formulation of the beams in great displacements and great rotations but with anelastic behavior. The essential difficulty of the analysis of rotations holds so that they are not commutable anddo not constitute a vector space, but a variety.

At any moment, the configuration of a cross-section of beam is defined by the vector - displacement of itscentre of gravity and the vector-rotation of the system of the main axes of inertia compared to a position ofreference. As in classical theory of the beams, the interior efforts are reduced to their resultant and theirmoment on the locus of centres of the sections. The associated deformations are defined.

The linearization of the interior efforts compared to displacements leads to the matrix of usual rigidity, which issymmetrical, and, because of great displacements and rotations, to the geometrical matrix of rigidity, which isunspecified.

The linearization of the inertial forces carries out, for the translatory movement, with the matrix of usual masswhich is symmetrical and, for the rotation movement, with a matrix much more complicated and without anysymmetry.

The diagram of temporal integration is that of Newmark.

This modeling was tested on five problems of reference: three from statics and two of dynamics.

This work was undertaken within the framework of a development of numerical models for the components ofthe lines and the stations. The goal of modeling is the dynamic study of the drivers provided withspacers (forthe lines) or of descents on equipment (for the stations) and subjected to the forces of Laplace resulting fromcurrents of short-circuit.

Warning : The translation process used on this website is a "Machine Translation". It may be imprecise and inaccurate in whole or in part and isprovided as a convenience.Copyright 2021 EDF R&D - Licensed under the terms of the GNU FDL (http://www.gnu.org/copyleft/fdl.html)



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Contents1 Notations.............................................................................................................................................. 4

2 Introduction........................................................................................................................................... 6

3 Kinematics of a beam in finished rotations...........................................................................................7

4 Vector and operator of rotation.............................................................................................................8

4.1 Vector-rotation............................................................................................................................... 8

4.2 Operator of rotation........................................................................................................................ 8

5 Passage des local axes with the axes generals.................................................................................10

6 Interior efforts, deformations and law of behavior..............................................................................11

6.1 Interior efforts............................................................................................................................... 11

6.2 Variation of curve in a point of the locus of centres.....................................................................11

6.3 Virtual work in the beam and vector of the deformations.............................................................12

6.4 Law of behavior........................................................................................................................... 13

7 Elementary inertial forces................................................................................................................... 15

8 Equation of the movement and course of a calculation.....................................................................16

8.1 Equation of the movement not deadened....................................................................................16

8.2 Course of a calculation................................................................................................................16

9 Linearization of the equations of the movement................................................................................17

9.1 Matrices of rigidity........................................................................................................................ 17

9.2 Matrices of inertia........................................................................................................................ 18

9.2.1 Differentiation of the inertia of translation...........................................................................18

9.2.2 Differentiation of the inertia of rotation...............................................................................19

9.2.2.1 Terms coming from the differentiation from............................................................19

9.2.2.2 Terms coming from the differentiation from and.....................................................19

10 Put in work by finite elements..........................................................................................................21

10.1 Matrix of deformation and interior efforts...................................................................................21

10.2 Matrices of rigidity...................................................................................................................... 21

10.3 Inertial forces............................................................................................................................. 22

10.4 Matrix of inertia.......................................................................................................................... 22

10.5 External forces data................................................................................................................... 23

10.6 Linear system of iteration...........................................................................................................23

10.7 Update of displacement, speed and acceleration......................................................................24

10.7.1 Translatory movement......................................................................................................24

10.7.2 Rotation movement..........................................................................................................24

10.8 Update of the vector variation of curve.....................................................................................24

10.9 Initialization before the iterations...............................................................................................25

10.9.1 Translatory movement......................................................................................................25

10.9.2 Rotation movement..........................................................................................................25

11 Schematic organization of a calculation...........................................................................................27




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11.1 Static calculation........................................................................................................................ 27

11.2 Dynamic calculation................................................................................................................... 29

12 Use by Code_Aster........................................................................................................................... 30

13 Digital simulations............................................................................................................................ 31

13.1 Embedded right beam subjected to one moment concentrated in the end (CAS-test SSNL103)

..................................................................................................................................................... 31

13.2 Arc embed-kneecap charged at the top.....................................................................................31

13.3 Arc circular of 45° embedded and subjected in the end to a force perpendicular to its plan.....32

13.4 Movement of an bracket............................................................................................................33

13.5 Setting in rotation of an arm of robot.........................................................................................36

14 Bibliography...................................................................................................................................... 38

15 Description of the versions of the document....................................................................................38

Annexe 1 : Some definitions and results concerning the antisymmetric matrices of order 3................39

Annexe 2 : Treatment of the forces of damping....................................................................................40

Annexe 3 : Algorithm of Newmark in great rotations.............................................................................42

Annexe 4 : Calculation of the differentials of Fréchet............................................................................44

Annexe 5 : Complements on the calculation of the matrices of rigidity.................................................45

Annexe 6 : Principle of the iterative calculation of rotations.................................................................47

Annexe 7 : Need for transport in a space of reference for the vectorial operations relating to the

rotation movement........................................................................................................................ 49

Annexe 8 : Use of the quaternions in modeling of great rotations [bib14] [bib15].................................50




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1 Notations∧ symbol of the vector product.

∧ operator of multiplication by the vector on the left.

q ' derived from q compared to the curvilinear X-coordinate.

q derived from q compared to time.

s curvilinear X-coordinate on the locus of centres of the sections.

u antisymmetric matrix of order 3 partner of axial vector u .

1 matrix unit of order 3.

Df . x directional derivative of f in the direction x

[ dds

1] diagonal matrix DIAG [ dds

,dds

,dds ] .

A , I 1,2 ou 3 surface and moments of inertia compared to main axes 1.2 or 3 of the cross-section.

B matrix of deformation.

C matrix of behavior.

E ,G , Young modulus and rigidity to shearing, density.

e i i=1,3 axes generals of coordinates.

E i s i=1,3 main axes of inertia of the section of X-coordinate s in position of reference.

f s , t linear external force exerted on the beam.

f s ,t force in the beam with the X-coordinate s and at the moment t .

F s , t RT s , t f s , t .

Fext forces external data with the nodes.

F iner ,Fint interior inertial forces and efforts with the nodes.

I tensor of inertia a length unit of beam in deformed position, expressed in theaxes generals.

J tensor of inertia a length unit of beam in position of reference, expressed in theaxes generals.

m s ,t linear external moment exerted on the beam.

m s ,t moment in the beam with the X-coordinate s and at the moment t .

M s , t RT s , t m s , t .

N i function of form relating to the node i .

R s , t operator or matrix, of axes generals, of rotation of the cross-section of X-coordinate s , configuration of reference to that at the moment t .

Ro s rotation making pass from the axes generals to the main axes of inertia of thesection of X-coordinate s in configuration of reference.

R tot s,t R s , t R o s .

SO 3 group of the operators of rotation in space with 3 dimensions.

t i s , t i=1,3 main axes of inertia of the section of X-coordinate s at the moment t .

xo s ,t position, at the moment t , center of the cross-section of X-coordinate s .

xo'−t1 .

E RT .

R tot 00 Rtot

position of the section of X-coordinate s the moment t , d

finished by the vector position x0 center and the vector rotation




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s , t {xo s , t s ,t

s { xo s s } : virtual displacement with the X-coordinate s .

s { xo s s } : correction of displacement to the X-coordinate s .

s , t defining vector, with the X-coordinate s and at the moment t , variation ofcurve compared to the configuration of reference.

X RT .

s , t vector rotation, at the moment t , section of X-coordinate s compared to itsposition of reference.

i−1, in

vector rotation enters the moment i−1 and the iteration n moment i .

in angular velocity of a section of beam calculated with the iteration n moment i .

Q ,Q−1 operator of passage of a vector rotation to the associated quaternion and itsreverse.




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2 IntroductionThe essential difficulty of the mechanics of the beams in great displacements lies in the formulation ofrotations. The rotation of a section compared to a configuration of reference is defined by the vector -rotation ([bib3], [bib4] and [bib5]). The quaternions are used for the update of this vector.In [bib4] and [bib5], the increment of rotation is expressed in the configuration of reference (totalLagrangian diagram). The calculation of the matrices of mass is complicated and cannot becompletely concluded its besides. But finally, all the matrices used are symmetrical.

In [bib1] with [bib3], the increment of rotation is expressed in the last calculated configuration (updatedLagrangian diagram). It is this diagram which we chose. The calculation of the matrices is completedwithout excessive difficulty but they are not symmetrical.

Unlike [bib3], we expressed speeds and the angular accelerations in the axes generals and not in localaxes. The matrices are thus more complicated, but one avoids the ambiguity which appears with theconnection of two beams not colinéaires.




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3 Kinematics of a beam in finished rotations

e3

s

(a)

P

e2

e1

E2

E1

E3

(b)

P't3

t2

t1xos

Position de référence Position à l'instant t

Figure 3-a: Evolution of a section of beam

Let us follow the evolution of a section of beam of its initial position - or of reference - [fig 3 - has] (A)with its position deformed at the moment t [fig 3-a] (b).

The cross-section of the center P beam in position of reference is located by the curvilinear X-

coordinate s of P on the locus of centres (or neutral fibre). One attaches to this section the

orthonormal trihedron E1 E2 E3 : E1 is the unit tangent of the locus of centres in P ; E2 et E 3are directed along the main axes of inertia of the section.

As in [bib1] with [bib5], one makes the assumption that during the movement the initially right sectionsremain plane and do not change a form.Moment 0 at the moment t :

• P comes in P ' and the position of P ' is defined by the vector xo s ,t ;

• the orthonormal trihedron E1 E2 E3 becomes the orthonormal trihedron t 1t 2 t3 . t2 and t 3 are

always directed along the main axes of inertia of the section and t 1 is always normal unit with

this section. But t 1 is not inevitably tangent with the locus of centres in P ' : in other words,

there can be, in deformed position, a slip due to shearing, as in the model of Timoshenko.

The state of the section at the moment t is thus defined by:

• the vector xo s ,t , which gives the position of the centre of gravity;

• the vector-rotation which makes pass from the trihedron E1 E2 E3 with the trihedron t 1t 2 t3 ,

and which is defined in [§4.1].The whole of these two vectors constitutes the vector s , t .




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4 Vector and operator of rotationAppendix 1 gives preliminary results concerning the antisymmetric matrices of order 3.

4.1 Vector-rotationLet us suppose that, in the system of axes generals P e1 e2 e3 [fig 4.1-a], the point M ' results

from M by the rotation of angle around the axis passing by P and of unit vector u . Let uspose:

=u

it is called vector-rotation making pass from M with M ' .According to the formula of Euler-Rodrigues [bib6] p. 186 and [bib7]:

PM '= PMsin u∧ PM 1−cos [u∧[u∧ PM ] ] .

Figure 4.1-a: Representation of a finished rotation

In general, the vector-rotation of the product of two rotations is not the geometrical sum of vector-rotation components. The case 2D is a particularly simple exception: the vectors - rotation,perpendicular to the plan, are added algebraically.

4.2 Operator of rotationTaking into account [éq An1-3], the preceding equation is written:

PM '=[1sin u1−cos u2 ] PM

The expression between hooks defines itoperator of rotation R making pass from PM with PM ' :

R=1sin u 1−cos u2 . éq 4.2-1

One calls “parameters of Euler” of rotation the four following numbers:

e0=cos

2e1=sin

2u1

e2=sin

2u2 e3=sin

2u3

éq 4.2-2

One has obviously:

e02e1

2e2

2e3

2=1.




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Let us pose:

e={e1

e2

e3}=sin

2u .

By using the relation [éq A1-4], one easily puts the expression [éq 4.2-1] of R in the form:

R=2 e02−112 eeT

e0 e . éq 4.2-3

In addition, substitute sin and cos , to the second member of [éq 4.2-1], by their developmentsin whole series, it comes:

R=1 [− 3

3 !

5

5 !...−1

p−1 2p−1

2p−1 !...] u

[ 2

2 !−

4

4 !

6

6 !...−1

p−1 2p

2p ! ] u2

maybe, while using [éq A1-5] and [éq A1-6], exponential form of the operator of rotation:

R=1 u

u 2

2 !...

u p

p !...

=exp u =exp . éq 4.2-4

It appears on [éq 4.2-4] that when ∥∥ 0 ,

R≈1∧ . éq 4.2-5

R=exp u is obviously not calculated by the development [éq 4.2-4], but by the expression [éq

4.2-1].

Since uT=−u , the transposition of all the terms of the second member of [éq 4.2-4] gives:

[exp ]T=exp − , that is to say: éq 4.2-6

RT=R−1

and:

RRT=1 éq 4.2-7

The operators of rotation, orthogonal according to the equation [éq 4.2-7], form a group compared tothe operation of multiplication - noncommutative in 3D - called group of Dregs and indicated by

SO 3 (Special Orthogonal group).




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5 Passage des local axes with the axes generalsThe components of the vectors are expressed in the axes generals e1 e2 e3 [fig 3-a]. The matrices

of the operators who connect them are thus valid only in these axes. But the mechanics of the beamsis formulated much more simply in the local main axes of inertia t 1t 2 t3 in current configuration. One

is thus brought to make the change of axes of the trihedral general e1 e2 e3 with the local trihedron

t 1t 2 t3 by the product R tot of two rotations:

• rotation R o , invariable, which brings the axes generals e1 e2 e3 on the local axes in position

of reference E1 E2 E3 ;

• rotation R , depend on the time, which brings the trihedron E1 E 2 E3 on the local trihedron

in current deformed position t1 t 2 t3 , that is to say:

R tot=RRo éq 5-1

Being given a vector v , components known in the trihedral general, his components in the localtrihedron are the components in the trihedral general of the vector:

V=R totT v . éq 5-2

One can thus replace calculations relating to vectors expressed in local axes in the current

configuration, by same calculations relating to the same vectors turned of R totT

and expressed of

axes generals. In other words, this rotation R totT allows to replace calculations in local axes of the

current configuration, by same calculations in axes generals.




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6 Interior efforts, deformations and law of behaviorOne places oneself within the framework of the theory of the beams. The efforts and the deformationsare defined by their end cells on the line of the centres of gravity of the sections. Thus, virtual work inthe beam is calculated by a simple curvilinear integral along this line.

6.1 Interior effortsOne calls interior efforts on the section of center P ' and of normal t1 the efforts which on this

section the part exerts of the beam located in the direction of t1 [fig 6.1-a]. These efforts form a

torque of which end cells in P ' are: the resultant f , resulting moment m

f

m

P"

P'

t1

Figure 6.1-a: Section of beam reduced to the locus of centres

If f and m are respectively the force and the moment outsides given per not deformed unit of

length f and m are supposed to be independent of the configuration, i.e. “conservative” or “not-

follower” -, the static balance of the section of beam P ' P ' ' of length ds is written:

∂ f∂ sf=0

∂m∂ s

∂ xo

∂ s∧fm=0 .

éq 6.1-1

6.2 Variation of curve in a point of the locus of centresLocal axes of the section of X-coordinate s result from the axes generals by the relation [éq 5 - 1]:

t i s =R s . Ro s ei . éq 6.2-1

While deriving compared to s :

t i '=R ' RoRRo ' ei ,

maybe, by reversing the relation [éq 6.2-1]:

t i=R ' RTRRo ' R o

T RT t i .

Let us pose:

R ' RT= éq 6.2-2

The matrix because derivation is antisymmetric compared to s of [éq 4.2-7] gives:

T=0 .




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It is checked that the matrix 0 defined by:

o=RRo ' RoT RT

is also antisymmetric.

Therefore, while utilizing axial vectors and 0 :

t i '= o ∧ti

• ds o is the vector rotation which makes pass from t i s with t i sds when the beam

undergoes a rotation uniform R '=0 of its position of reference to its current location. o s characterize curve of the configuration of reference with the X-coordinate s .

• ds is the increase in rotation of t i s with t i sds had with the variation of R along the

beam. It is this vector who characterizes variation of curve between the configuration ofreference and the current deformed configuration.

6.3 Virtual work in the beam and vector of the deformationsThis paragraph is the extension to the three-dimensional case of a approach made in [bib8] on theplane beams.

Configuration of the beam at the moment t is defined [§3] by the field s , t :

s , t ={x0 s , t

s , t } .

Let us calculate work W linear external forces f and m in following virtual displacement:

s={ xo s s } .

One has obviously:

W=∫s1

s2

f . xom . ds .

The equilibrium equations [éq 6.1-1] make it possible to replace the external forces f and m by the

interior efforts f and m :

W=∫s1

s2 [−∂ f∂ s

. xo−∂m∂ s

. − ∂ xo

∂ s∧f . ]ds ,

maybe, by integration by parts of the first two terms:

W=−[f . xom . ]12∫s1

s2

[ f . xo 'm . '−xo '∧f . ] ds .

If one indicates by W ext the total work of the external forces on the beam - length and in ends - thepreceding equation is written:

W ext=∫s1

s2

f . xom . ds[ f . xom . ]12

=∫s1

s2

[ f . xo '−∧xo ' m . ' ] ds . éq 6.3-1

According to the theorem of virtual work for the continuous mediums, the second member is thevirtual work of the interior efforts, which one notes W int . According to the idea of [feeding-bottle 8],

let us seek to put the coefficients of f et de m in the form of the virtual increase in two vectors.




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Let us make the assumption that the vector xo ' tangent with the locus of centres but not inevitably

length unit, does not differ from the unit vector t1 , normal with the section, that by an infinitely small

of the 1er order (about ). That implies that the beam lengthens little and undergoes a weak slip.Then:

W int=∫s1

s2

[ f . xo '−t1 m . ]ds .

Indeed, in virtual displacement:

• xo ' increases xo ' ;• the unit vector t1 turn of ;

• the vector , defining the variation of curve [§6.2], increases ' .

One thus has:

W int=∫s1

s2

[ f . m . ]ds , éq 6.3-2

with:

{ }={xo '−t1

}. éq 6.3-3

In the equation [éq 6.3-2], { fm} and { } thus seem respectively one torque of efforts or

generalized constraints and it torque of the associated deformations. defines lengthening and the slip; the variation of curve [§6.2] defines.

It is observed that, if must be small, on the other hand it does not have there a limitation for .Most beams enter within this framework.

The relation [éq 6.3.1] is written:

W int ; =∫s1

s2

B . { fm}ds éq 6.3-4

with:

B=[dds

1 xo '∧

0dds

1 ] éq 6.3-5

B is called matrix of deformation.

6.4 Law of behaviorAccording to [§5], the components in local axes of the generalized constraint and deformation are thecomponents of axes generals of the vectors:

{FM}= T { f

m}; {EX}=T { } éq 6.4-1

One supposes that the law of behavior is elastic and that, in local axes, it has the same expression asfor a beam of Timoshenko:

{FM}=DIAG [EA ,GA2 ,GA3 ,GI1 , EI 2 , EI 3 ]{EX} éq 6.4-2

A2 and A3 being two surfaces depending on the size and the form of the section.




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One poses:

C=DIAG [EA ,GA2 , GA3 ,GI 1 ,EI 2 ,EI 3 ] .

C is called matrix of behavior.




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7 Elementary inertial forcesInertial forces applied to an element ds form a torque which admits, in the centre of gravity:

• a general resultant, − Ads x0s ,t ;• one equal moment resulting contrary to the absolute velocity of the elementary kinetic moment

H .

To express the angular velocity, let us proceed as to [§6.2] and derive the relation:t i=R Ei ,

compared to t , by taking account that E i does not depend on time. One obtains:

t i=R RT ti. éq 7-1

Let us pose:

R RT= . éq 7-2

By deriving the relation [éq 4.2-7] compared to t , it is seen that the matrix is antisymmetric. If

one indicates by the axial vector of this matrix, the relation [éq 7.1] is written:

t i s , t =∧t i s , t .

is thus it angular Flight Path Vector section of beam of X-coordinate s at the moment t .The elementary kinetic moment has as an expression:

H=ds I=dsR JRT éq 7-3

where J is the tensor of inertia in the configuration of reference:

J=R0DIAG [ I 1 , I 2 , I 3 ]R0T

While deriving compared to time:

H=ds I ds R JRTR J R

T

But:

R JRT=R RTRJR

T= I=∧I

and:

RJ RT=R JR

TR RT=0

because, according to the equation [éq A1-2]:

R RT=− =0 .

From where moment of the inertial forces of the element:

−H=−ds I −ds∧I

The virtual work of the inertial forces thus has as an expression:

W iner=−∫S1

S2

{ A x0I ∧I }.{

0

}ds éq 7-4




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8 Equation of the movement and course of a calculation8.1 Equation of the movement not deadened

If one adds the inertial forces to the external forces, the weak form of the equations of the movement,in other words the virtual work of the forces not balanced in the beam is written:

W , , ; =W int ; −W iner , , ; −W ext ; . éq 8.1-1

With balance W is null, for all ∂ .

W int ; is given by the equation [éq 6.3-4] where the torque of efforts { fm} , has as an

expression, according to [éq 6.4 - 2]:

{ fm}= CT { } éq 8.1-2

The torque of deformations generalized of the second member of [éq 8.1-2] results from the currentlocation: is given by [éq 6.3-3] and rise from [éq 6.2 - 2].

W iner , , ; is given by the equation [éq 7-4].

W ext ; =∫s1

s 2 { fm}. { xo

}ds + work of the concentrated forces éq 8.1-3

If the external forces given are conservative, i.e. independent of the configuration, W ext does not

depend on .

8.2 Course of a calculation• In the dynamic case, one seeks the fields of displacements, speeds and accelerations in a

discrete succession of moments: t 1 , t2 ... ti−1 , t i ... t n .• In the static case, one splits the total load in increments of load which one adds successively

from zero to reconstitute the truck load. With each stage of loading, named by “urgent” abuse,one calculates the field of displacements.

Knowing the state of the structure at the moment t i−1 , one from of deduced his state at the moment

t i by prediction-correction:• In the static case (STAT_NON_LINE), the prediction consists in calculating the answer of the

structure to i - ème increment of load, by preserving its behavior at the moment t i−1 .

• In the dynamic case, one must initially to initialize fields at the moment t i , by formulas risingfrom the algorithm of temporal integration used: in the operator DYNA_NON_LINE, it is the

algorithm of Newmark [§A3]. Then one applies the increase in load enters t i−1 and t i with thebehavior in initialized situation.

In the predicted state, the equilibrium equation is generally not satisfied and one must to correctdisplacements by iterations resting on linearized equations.




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9 Linearization of the equations of the movementLet us suppose calculated the state of the structure to the iteration n moment i , n=1 correspondsto the phase of prediction. The weak form of the equilibrium equations is, with this iteration [éq 8.1-1]:

W in , i

n , in ; =W int i

n ; −W iner in , i

n , in ; −W ext i

n ; éq 9-1

• If this quantity is rather small, within the meaning of the criterion of stop [bib10], one considersthat this n - ième iteration gives the state of the structure to the moment i .

• If not, one calculates corrections of displacement in1

such as:

L {W [ i

n in1 , i

nin1

., i

n in1

..; ]=

W in , i

n , in ; DW i

n , in , i

n ; .in1=0 .

éq 9-2

DW in , i

n , in ; . i

n1 is the differential of Fréchet of W i

n , in , i

n ; in the

direction in1

[An4].

9.1 Matrices of rigidityThey result from the differentiation of Fréchet from W int ; in the direction . According to

the equations [éq 6.3-4] and [éq 6.4-1]:

W int ; =∫s1

s2

{B } . {FM}ds .

That is to say:

DW int .=∫s1

s2

[D {B } . ] . {FM }ds

∫s1

s2

{B } .[D . {FM}] ds∫s 1

s 2

{B }. [ D{FM}.]ds . éq 9.1-1

However, according to the equation [éq 6.3-5]:

{B }={ xo 'xo '∧ ' }.

Thus [éq A4-2]:

D {B } .={ xo '∧0 }.

In addition [éq A4-5]:

D .=[ Rtot 0

0 Rtot] .

Lastly, according to [§6.4]:

D {FM}=C D T { } .

=CT B−C T[ 00 ]{ },

because [éq 6.3-2] and [éq 6.3-5]:Warning : The translation process used on this website is a "Machine Translation". It may be imprecise and inaccurate in whole or in part and isprovided as a convenience.Copyright 2021 EDF R&D - Licensed under the terms of the GNU FDL (http://www.gnu.org/copyleft/fdl.html)



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D{ }.=B ,

and [éq A4.6]:

D RtotT .=−Rtot

T .

It is shown [A5] that:• the sum of the first two integrals of the second member of [éq 9.1-1] can be put in the form:

∫s1

s2

T YT EY ds ;

• the third integral can be written:

∫s1

s2

T BT

CT B ds∫s1

s2

T BTZ ds ,

with:

Y=[dds

1 0

0dds

1

0 1];

E=0 0 −f∧0 0 −m∧

f∧ 0 xo ' f ; éq 9.1-2

Z=[0 Rtot C1 R totT Rtot C1

−1R totT f

v

0 R tot C2 RtotT Rtot C2

−1 R totT m

v ] éq 9.1-3

C1 and C2 being two submatrices of C :

C1=DIAG [EA ,GA2 ,GA3 ] ,C2=DIAG [GI1 ,EI 2 , EI 3 ] .

YT EY is called geometrical matrix of rigidity éq 9.1-4

BT C T matrix of material rigidity is called éq 9.1-5

Lastly, during the differentiation a matrix appears which does not appear in [bib2]:

BT Z that we call complementary matrix éq 9.1-6

9.2 Matrices of inertiaThey result from the differentiation of Fréchet from W iner [éq 7-4] in the direction . More

precisely, one places oneself in the configuration of n - ième iteration of the moment i and one

differentiates in the direction in1

.

9.2.1 Differentiation of the inertia of translation A x0




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One has immediately, according to the equations [éq An4-4] on the one hand and [éq An3.1-4] on theother hand:

D A xo ,in . xo , i

n1= A xo ,i

n1= A t2

xo , in1

éq 9.2.1-1

9.2.2 Differentiation of the inertia of rotation I ∧IAccording to the equation [éq 7-3]:

I=R JRT

J being a constant tensor.

9.2.2.1 Terms coming from the differentiation from IAccording to the equations [éq A4-3] and [éq A4-4], these terms are:

[− I ^I − I

^ I ]i

n

in1 éq 9.2.2.1 - 1

all sizes appearing in the hook being taken in the iteration n moment i .

9.2.2.2 Terms coming from the differentiation from and According to the expressions [éq A3.2-3] and [éq A3.2-4], speed and angular acceleration with theiteration of n the moment i are:

in=Ri

n Rin−1, T

in−1

tRi

n Ri−1T i−1,i

n− i−1, i

n−1

in=Ri

n Rin−1, T

in−1

1 t2 R i

n Ri−1T i−1, i

n− i−1,i

n−1 .

A variation vector-rotation can affect only the sizes relative to this iteration n moment i ,since, in the two preceding relations, the other sizes are fixed. In other words, only are to be

differentiated R in et i−1,i

n, increment of vector-rotation of the moment with i−1 the iteration n

moment i .

However [éq A4-5]:

D Rin . i

n1=

i

n1 Rin.

And, according to [bib3]:

D i−1,in . i

n1=T yi−1,i

n in1 ,

with:

T = 1∥∥

2

T

∥∥

2

tan∥∥2[1− 1

∥∥2

T ]−

2.




4fd7ba1f647a

Terms coming from differentiation from and are thus:

−I , in R i

nR in−1,T

in−1

^−[ i

nI , in−I , i

n in

^ ] R inR i

n−1,T in−1

^

1

t 2 {I , in t [ in I , in−I , i

n i

n ^ ]}

{[−R inR i−1

T i−1, in

− i−1,in−1 ]

^R i

nR i−1T T i−1, i

n }

éq 9.2.2.2 - 1

combination of the three preceding matrices being to multiply by in1

.




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10 Put in work by finite elementsOne gives below the representation in finite elements of the matrices of [§9]. These matrices appear inexpressions to integrate along the beam. One thus calculates them at the points of Gauss. …N i , N j are the values taken, at the point of Gauss considered, by the functions of form relating to the

nodes i , j ….

The matrices of rigidity connect the increase in displacement of the node j with the increase in force

interns with the node i for the element e .

10.1 Matrix of deformation and interior effortsThe matrix of deformation has as a continuous expression [éq 6.3-5]:

B=[dds

1 xo '∧

0dds

1 ] .

In finite elements, the contribution of the displacement of the node i with the deformation at the pointof Gauss considered is obtained by multiplying the 6 components of this displacement by the matrix:

Bih=[N i ' 1 N i xo ' G

0 N i ' 1 ] éq 10.1-1

The superscript h indicate that it is about the discretized shape of the matrix B .

According to [éq 6.3-4]:

F int ie =∫e

BihT { f

m}ds éq 10.1-2

is the interior effort applied to the node i element e and due to the stress field generalized { fm} in

the element. This constraint is calculated according to the equation [éq 8.1-2]. But it should be noticedthat:

• on the one hand [éq 6.3-3]: R totT=Rtot

T xo−e1 ;• in addition, the vector is updated with each iteration, as it is indicated to [§10.8].

10.2 Matrices of rigidityThe expression continues matrix of material rigidity [éq 9.1-5] is:

BTCT B .

One from of deduced, for the finite element e , the matrix connecting the increase in displacement of

the node j with the increase in force interns with the node i :

Smat i je

=∫eBi

hTCT B j

h ds . éq 10.2-1

One calculates numerically CT at the point of current Gauss.

The expression continues geometrical matrix of rigidity [éq 9.1-4] is:

YT E Y




4fd7ba1f647a

where:

Y=[dds

1 0

0dds

1

0 1] ,

and E is given by the equation [éq 9.1-2].

One from of deduced, as for the matrix of material rigidity:

Sgéom i je

=∫eYi

hT E Y jh ds , éq 10.2-2

where:

Y ih=[

N i' 1 0

0 N i' 1

0 N i' 1 ] ,

and where E is calculated numerically at the point of current Gauss.

The expression continues complementary matrix of rigidity [éq 9.1-5] is:

BT Z ,

where Z is given by the equation [éq 9.1-3].

One from of deduced:

Scompl ije

=∫eBi

hT N j Z ds ,

where Z is calculated numerically at the point of current Gauss.

10.3 Inertial forcesAccording to [éq 7.3-1]:

Finer , ie

=−∫e { N i A x0

N i I ∧I }ds éq 10.3-1

is the inertial force of the element e with the node i .

10.4 Matrix of inertiaLet us pose:

I0, iner=[ A1 00 J ] and M0 iner ij

e=∫e N i N j I0 inerds

It is seen, according to [éq 7-4] and [éq 9.2.1-1], that 1

t2Mo iner

e is well the matrix of inertia of the

element e for the translatory movement (diagonal submatrix A1 of I0 iner ). But according to[éq 9.2.2.1 - 1] and [éq 9.2.2.2 - 1], it is not the matrix of inertia of rotation.

Nevertheless, the matrix Mo iner , assembly of Mo inere

, is used to calculate itinitial acceleration

beam when it leaves its position of reference with one worthless initial speed o=0 . Indeed,

one has then, according to [éq 7.3-1]:

Fext t=0 =−Finer t=0 =Mo iner {xo

w }o ,




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since, in position of reference: I=J

Let us call J ' the sum of the two matrices [éq 9.2.2.1 - 1] and [éq 9.2.2.2 - 1] and let us pose:

I rot=[0 00 J ' ] and M rot i j

e=∫e

N i N j Irot ds

Also let us pose:

I0 rot=[0 00 J

] et M0 rot i j

e=∫e

N i N j I0 rot ds

The matrix of complete inertia of an element of beam is obviously:

M inere =

1 t 2

Mo inere −

1 t 2

Mo rote Mrot

e.

10.5 External forces dataAccording to [éq 8.1-3]:

F ext ie =∫e {N i f

N i m}ds éq 10.5-1

is the force applied to the node i element e who is equivalent to the external forces distributed.

10.6 Linear system of iterationBy discretization in finite elements, the equation [éq 9-1] gives:

W h=F int−F iner−Fext . .

In addition, by supposing that the external forces are conservative, one has, according to [§10.2] and[§10.4]:

DW h .=[SmatSgeomScomplMrot−1

t 2Mo rot

1 t 2

Mo iner ] . .

The relation [éq 9-2], in front of being checked by all , thus leads, with the iteration n moment i, with the linear system following in i

n1 :

[Smat, in

Sgeom, in

Scompl, in

Mrot, in−

1

t 2Mo rot

1

bêat t 2Mo iner ]{ xo ,i

n1

in1}

=F ext, i−F int, inF iner,i

n .

éq 10.6-1

In Code_Aster, elementary matrices Mo inere , who are independent of displacement, are assembled

only once to constitute the total matrix Mo iner , who is in particular used to calculate initial

acceleration [§10.4].

Three matrices of elementary rigidity Smate , Sgeom

eand Scompl

e, the matrix of inertia of rotation, Mrot

e

who depend all the four on displacement, and inertia of corrective rotation stamps it −1

t2Mo rot

e ,

who is invariable, with each iteration, combined then are assembled to constitute a pseudonym

stamps total rigidity Kin .

The linear system [éq 10.6-1] thus becomes:




4fd7ba1f647a

[ Kin

1 t 2

Mo iner]{ xo ,in1

in1}=Fext ,i−F int, i

nFiner, i

n . éq 10.6-2

In the case of a static problem, the preceding system is simplified in:

[K in ] { xo , i

n1

in1}=Fext, i−Fint, i

n éq 10.6-3

where Kin

is the assembly of the only matrices of rigidity to the iteration n “moment” i [§8.2]:

Smat, in

Sgeom, in

Scompl, in .

10.7 Update of displacement, speed and accelerationThe treatment of the translatory movement is classical; that of the rotation movement is doneusing the quaternions [A8].

10.7.1 Translatory movementOne applies the formulas [éq A3.1-3] and [éq A3.1-4].

10.7.2 Rotation movementThe sizes to be updated are:• on the one hand, vector-rotation, speed and angular acceleration;• in addition, for later calculations, the matrix of rotation and the increment of the vector-rotation of

the moment i−1 with the current iteration of the moment i [§A6].

The update of vector-rotations rests on the following property of the quaternions [§A8.4]: “thequaternion of the product of two rotations is equal to the product of the quaternions of componentrotations”.

Thus while posing [§An8.6]:

x o , x =Q in1

xo ,x =Q in ,

it comes:

in1=Q−1 [ xo , x ° xo ,x ] . éq 10.7.2-1

In addition, according to the equation [éq An6-2], if:

yo , y =Q i−1,in ,

then:

i−1,in1=Q−1 [ xo , x ° yo , y ]. éq 10.7.2-2

The update of the matrix of rotation is immediate [§4.2]:

R in1=exp i

n1 ,

who is calculated according to [éq 4.2-1].

Finally speed and the angular acceleration are updated by the relations [éq A3.2-3] and [éq A3.2 - 4].

10.8 Update of the vector variation of curveThe vector , which defines the deformation of rotation [§6.2], must be calculated only at the pointsof Gauss. In Code_Aster, it is treated by means of computer like a “internal variable”.

According to [éq 6.2-2]:




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in=

dds Ri

n .R inT .

And, according to [éq 4.2-6]:

in1=

dds [exp i

n1 Rin ]R i

nT exp − in1 .

That is to say:

in1=

dds [exp i

n1 ] exp − in1

exp in1 i

n exp − in1 .

In the preceding equation, the matrix of the first member is antisymmetric by construction; the secondmatrix of the second member is obviously antisymmetric; thus the first matrix of the second memberis too.One shows in [bib2] Appendix B, that the axial vector of this last matrix is:

=sin∥ i

n1∥

∥ in1∥

in1 '1−sin∥ i

n1∥

∥ in1∥ i

n1 . in1 '

∥ in1∥

in1

∥ in1∥

12 sin

12∥ i

n1∥

12∥ i

n1∥

2

in1∧ i

n1 ' .

Thus:

in1=AXIAL [exp i

n1 in exp− i

n1 ] .

10.9 Initialization before the iterationsIn the dynamic case, if the loading is constant in time, the iterations cannot start at the moment i whoif one initializes some of the fields of displacement, speed and acceleration with values different fromthose from the moment i−1 . These initializations are done as follows.

10.9.1 Translatory movementxo ,i

o=xo , i−1.

Then, according to the equation [éq An3.1-1]:

xo ,io=−

1 t

xo , i−12−1

2xo ,i−1 .

According to the equation [éq An3.1-2]:

xo ,io= xo , i−1 t [ 1− xo ,i−1 xo ,i

o ] .

10.9.2 Rotation movementOne takes expressions similar to the preceding ones:

io= i−1 éq 10.9.2-1

io=−

1 t

i−12−1

2 i−1 éq 10.9.2-2




4fd7ba1f647a

io=i−1 t [ 1− i−1 i

o ]. éq 10.9.2-3

The second members of [éq 10.9.2-2] and [éq 10.9.2-3] have a direction because all the vectors whichappear in it find in the tangent vector space with SO 3 in R i−1 .

Like consequences of the equation [éq 10.9.2-1]:

R io=Ri−1

and:

i−1, io=0 .

One sees that at the first moment i=1 , initializations require the knowledge of initial acceleration

{xo

}o , whose calculation is indicated to [§10.4].




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11 Schematic organization of a calculationThis paragraph shows how are articulated the concepts presented to [§10] in unfolding of acalculation.

11.1 Static calculation




4fd7ba1f647a




4fd7ba1f647a

11.2 Dynamic calculation




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12 Use by Code_AsterThis paragraph indicates how the beams in great displacements in the orders intervene ofCode_Aster.Order Keyword factor Keyword ArgumentAFFE_MODELE AFFE PHENOMENON ‘MECHANICAL’

MODELING ‘POU_D_T_GD’AFFE_CARA_ELEM BEAM SECTION ‘GENERAL’

‘RIGHT-ANGLED‘CIRCLE’

STAT_NON_LINEandDYNA_NON_LINE

BEHAVIOR RELATIONDEFORMATION

‘ELAS_POUTRE_GD’‘GROT_GDEP’




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13 Digital simulationsOne gives five digital simulations below resting on the formulation presented in this note. The threefirst relate to static problems, the two last on dynamic problems.

13.1 Embedded right beam subjected to one moment concentrated in theend (CAS-test SSNL103)That is to say M this moment. The beam is the seat only one constant moment and the equation [éq

6.4-2] shows that the variation of curve X is also constant. The beam thus becomes deformed incircle of radius:

r=E I 3

∥M∥,

in a plan perpendicular to the vector moment.

The figure [fig 13.1-a] shows the deformations of a beam length unit, of which E I 3=2 and subjected

to the moments , 2 and 4

The beam is cut out in 10 finite elements of the 1er order. One applies from the start the final momentand convergence is reached in 3 iterations.

Figure 13.1-a: Beam subjected to one moment in the end

13.2 Arc embed-kneecap charged at the topThe figure [fig 13.2-a] shows the deformations of an arc of 215° of opening, embedded on the right,kneecap on the left and subjected to an increasing force concentrated at the top. The solution of thisproblem is given in [bib11] for an initial ray of 100 and the following characteristics of beam:

EA=GA=5×107 and EI=106

The arc is modelled by 40 elements of the 1er order. One made grow the force up to 890 by eightincrements of 100, an increment of 50 and four increments of 10. Beyond it appears buckling, i.e.displacement continues to grow under a force which decrease brutally. The algorithm describes heredoes not make it possible to take into account such a phenomenon, and diverges for a force from 900.Da Deppo, in [bib11], locates the force criticizes to 897.

There is the table of comparative results according to:




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Force Vertical displacementof the point of

application

Horizontaldisplacement of thepoint of application

Our calculations 890 – 110.5 – 60.2Da Deppo 897 – 113.7 – 61.2

F = 400

F = 700

F = 890 Figure 13.2-a: Arc embed-kneecap charged at the top (Da Deppo)

13.3 Arc circular of 45° embedded and subjected in the end to a forceperpendicular to its planThe problem is three-dimensional. He was proposed in [bib12]. The figure [fig 13.3-a] shows threesuccessive configurations of the beam of initial ray 100, square section and characteristics:

EA=107 ,GA1=GA2=5×106 , GI1=EI 2=EI 3=8 .333×105 .




4fd7ba1f647a

50

40

30

20

10

0

10

20

30

Z

X

Y5040302010 60 70

P = 600 lb

P = 300 lb

A (29.3 70.7 0.0)

CONFIGURATION

CONFIGURATION

A (22.5 59.2 39.5)

A (15.9 47.2 53.4)

INITIALE

FINALE

Figure 13.3-a: Three configurations of the beam (extracted from [bib12])

We modelled the beam by 8 elements of the 1er order. The force grows by increments of 20. Thereare the following comparative results, for the coordinates of the point of application of the force, in theconfigurations of the figure [fig 13.3-a].

FORCE X Y Z 300 Our calculations 22.3 58.9 40.1

BATHE-BOLOURCHI 22.5 59.2 39.5600 Our calculations 15.7 47.3 53.4

BATHE-BOLOURCHI 15.9 47.2 53.4

13.4 Movement of an bracketIt is about with a three-dimensional dynamic problem dealt in [bib3].An bracket consists of a post and a cross-piece length 10 [fig 13.4-a] (A). The foot of the post isembedded and one applies to the connection a not-following force, perpendicular to the plan of thebracket at rest [fig 13.4-a] (b).




4fd7ba1f647a

10

F

z

y

x

10

(a)

F

50

0 1 2t

(b)

Figure 13.4-a: Bracket subjected to a dynamic force perpendicular to its plan

The characteristics of the elements provided by [bib3] are the following ones:

EA=G A1=G A2=106

G I 1=E I 2=E I 3=103

A=1 I 2= I 3=10 ; I 1=20

It is noticed that these data do not make it possible to identify a material and a section of beam,because one has at the same time:

EI 2

I 2

=102=E

et E A A

=106=E

One can thus deal with this problem only by imposing, by program, a characteristic: one chose toimpose the product EA .

The post and the cross-piece are modelled each one by 4 elements of the 1er order and the durationof the analysis comprises 120 pas equal of 0.25.

The figures [fig 13.4-b] and [fig 13.4-c] give the evolution of the component according to xdisplacement respectively for the connection and the end of the cross-piece. In cartridge, onereproduced the corresponding curves given in [bib3] for two modelings.




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24.0 26.0 28.0 30.0 32.0 temps2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 22.0–4.0

–3.0

–2.0

–1.0

0.0

1.0

2.0

3.0

4.0

5.0

6.0déplacement

100

Displacem

ent

8060402000

–20–40–60–80

–1000.0 30 60 90 120 150 180 210 240 270 300

Time

ABCD

Figure 13.4-b: Bracket of SIMO - Displacement of the connection

perpendicular to the initial plan.

10.00

8.757.50

6.255.00

3.752.501.250.00

1.252.50

3.755.00

6.257.50

8.7510.00

2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 22.0 24.0 26.0 28.0 30.0 32.0

déplacement

Time

temps

100

Displacem

ent

8060402000

–20–40–60–80

–1000.0 30 60 90 120150180 210 240270

ABCD

300

Figure 13.4-c: Bracket of SIMO - Displacement of the end

perpendicular to the initial plan.




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13.5 Setting in rotation of an arm of robotAn arm of robot OA is put moving in the plan e1e2 by a rotation t imposed on its axis 0 [fig

13.5-a]. One wants to calculate the displacement of the end A in a system of axes e1’ e2

’ involved

in rotation t .

Figure 13.5-a: Arm of robot subjected to an imposed rotation.

Characteristics of material:

EA=2.8107

GA2=107

EI=1.4104

A=1.2I =610-4

The step of time evolves of 0.05 at the beginning of the analysis with 0.001 with the end.

The figures [fig 13.5-b] and [fig 13.5-c] give the evolution of following displacement e1 ' and

according to the perpendicular direction. One reproduced, in cartridge, the corresponding curve of[bib3]. When the number of revolutions becomes constant, the arm undergoes a permanentlengthening due to the centrifugal force and it is subjected to an oscillation of inflection of lowamplitude.




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0.00125

0.00000

0.00125

0.00250

0.00375

0.00500

0.00625

0.00750

0.00875

0.01000

0.01125

0.01250

0.01375

0.01500

0.01625

0.01750

0.01875

1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.011.012.013.014.015.016.017.018.0

0.0040

0.0000-0.0008

-0.0056

-0.0104

-0.0152

-0.0200

0.0 6.0 12.0 18.0 24.0 30.0Time

Tip D

ispla

cem

ent

u2 (

L, I)

0.02000

5.14 x 10 -4

~

Figure 13.5-b: Following displacement e1 '

0.04

0.00

- 0.04

- 0.08

- 0.12

- 0.16

- 0.20

- 0.24

- 0.28

- 0.32

- 0.36

- 0.40

- 0.44

- 0.48

- 0.52

- 0.56

- 0.60

1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.012.0 13.014.0 15.0 16.017.018.0

0.10

0.00

0.06

-0.22

-0.38

-0.54

-0.700.0 6.0 12.0 18.0 24.0 30.0

Time

Tip

Dis

pla

cem

ent

u 2 (

L, I

)

Figure 13.5-c: Following displacement e2 '




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14 Bibliography1) J.C. SIMO : With finite strain beam formulation. The three-dimensional dynamic problem.

Share I. Comput. Meth. appl. Mech. Engng. 49,55-70 (1985).

2) J.C. SIMO and L. CONSIDERING-QUOC : With three-dimensional finite-strain rod model.Share II: computational aspects. Comput. Meth. appl. Mech. Engng. 58,79-116 (1986).

3) J.C. SIMO and L. CONSIDERING-QUOC : One the dynamics in space of rods undergoingbroad motions. With geometrically exact approach. Comput. Meth. appl. Mech. Engng.66,125-161 (1988).

4) A. CARDONA and Mr. GERADIN : With beam finite element nonlinear theory with finiterotations. Int. J. Numer. Meth. Engng. 26,2403-2438 (1988).

5) A. CARDONA : Year integrated approach to mechanism analysis. Thesis, University of Liege(1989).

6) J. DUKE and D. BELLET : Mechanics of the real solids. Elasticity. Cepadues-editions (2E

edition, 1984).

7) H. CHENG and K.C. GUPTA : Year historical note one finite rotations. Newspaper of AppliedMechanics 56,139-145 (1989).

8) E. REISSNER : One one-dimensional finite-strain beam theory: the planes problem.Newspaper of Applied Mathematics and Physics 23,795-804 (1972).

9) H. HUTS : Course of General mechanical engineering. Dunod (1961).

10) Mr. AUFAURE : Dynamic nonlinear algorithm of Code_Aster. Note HI - 75/95/044/A.

11) D.A. DADEPPO and R. SCHMIDT : Instability of clamped-hinged circular arches subjected todoes not have load. Trans. ASME 97 (3) (1975).

12) K.J. BATHE and S. BOLOURCHI : Broad displacement analysis of three - dimensional beamstructures. Int. J. Numer. Meth. Engng. 14,961-986 (1979).

13) K.J. BATHE : Finite element procedures in engineering analysis. Prentice-hall (1982).

14) D. HENROTIN : Report of internship at the Institute Montefiore (University of Liege). Privatecommunication.

15) P. of CASTELJAU : Quaternions. Hermes (1987).

15 Description of the versions of the documentVersionAster

Author (S) Organization (S)

Description of the modifications

3 M.AufaureEDF- R&D/MMN

Initial text




4fd7ba1f647a

Annexe 1 : Some definitions and results concerning theantisymmetric matrices of order 3Any vector u of order 3 and components ux , uy , uz one can associate the antisymmetric matrix uof order 3 following:

u=[0 −uz uy

uz 0 −ux

−uy ux 0 ] éq A1-1

Conversely, any antisymmetric matrix of order 3 can be written in the form [A1-1] and one can thusassociate a vector to him u . This vector is called it axial vector matrix.

One sees without difficulty that: u u=0 éq A1-2

in other words the axial vector is clean vector of antisymmetric associated for the eigenvalue 0.

• Whatever the vector v :

u v=u∧v éq A1-3

u v=vuT−u .v 1. éq A1-4

• If u is unit:

u2=uuT (symmetrical matrix)

u3=−u (antisymmetric matrix)

From where:

u2p= −1

p−1u2 (symmetrical matrix) éq A1-5

u2p−1=−1

p−1u (antisymmetric matrix) éq A1-6




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Annexe 2 : Treatment of the forces of dampingThese forces are not taken into account currently in Code_Aster.

A2.1 End assumptions and cells in the center of the sectionOne makes the assumption that the element of volume dv bordering a point M interior with thebeam is subjected to a force of damping being composed of two parts:• for the translatory movement of the section of X-coordinate s which belongs M :

d f 1=−1 xo s , t dv ;

• for the angular rotation movement velocity around the center P' section:

d f 2=−2 s , t ∧ P ' M dv .

Integrated on the volume of beam length ds , these forces admit for end cells in P' :• the force:

d f=−1 A xo s , t ds ;

• moment:

d m=−2ds∫section P ' M∧[ s , t ∧ P ' M ]d=−2

Is ,t ds

A2.2 Elementary forces of dampingThe virtual work of the forces of damping is:

W amor=−∫S1

S2 {1 A x0

2

I}{ x0

}ds

If damping into account is taken, W amor must be cut off to the second member from the equation [éq9-1].

It results from this, as with [§10.3] for the inertial forces, that the contribution of the element e by the

strength of damping with the node i is:

Famor ie

=−∫e {N i1 A x0

N i

2

I}ds

If damping into account is taken, these forces must be added to the second member of the equation[éq 10.6 - 1].

A2.3 Matrices of dampingAccording to the relations [éq A4-3] and [éq A4-7] giving the differentials of Fréchet of xo and of in the direction , and given that ([éq A4-5] and [éq A4-6]):

D I . =[−I^I ]

DW amor .=−∫S1

S2

.[1 A1 0

02

I ] ds−∫S 1

S 2

.[0 0

0−2

I

^]ds




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One from of deduced the matrices from damping of the element e :

• except for the sign, relation between the increase speed to the node j and the increase in force

of damping to the node i :

Camor i je

=∫e N i N j [1 A1 0

02

I ]ds

• except for the sign, relation between the increase in displacement to the node j and the increase

in force of damping to the node i :

Samor i je

=∫e N i N j [0 0

0−2

I

^]ds

In the hook of the first member of the equation [éq 10.6.1], it is necessary to add:

• the matrix

tCamor ;

• the matrix Samor .

The matrix Camore . is symmetrical, but the matrix Samor

e . is antisymmetric.




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Annexe 3 : Algorithm of Newmark in great rotationsA3.1 Classical diagram of Newmark in translation

The state of the locus of centres (displacement, speed and acceleration) is supposed to be known atthe moment i−1 The algorithm of Newmark [bib10] and [bib13], rests on the following developments

of displacement and speed, with the iteration n1 moment i :

xo ,in1=xo ,i−1 t xo , i−1

t2

2[ 1−2 xo ,i−12 xo ,i

n1 ] éq A3.1-1

xo ,in1=xo ,i−1 t [ 1− xo ,i−1 xo , i

n1 ] éq A3.1-2

and are the parameters of Newmark which, in the case of the “rule of the trapezoid”, are worth

14

and 12

: the hooks in the equations [éq A3.1-1] and [éq A3.1-2] are then the arithmetic mean of

xo ,i−1 and of xo ,in1

Récrivant each one of these relations to the iteration n moment i and cutting off memberwith member from the preceding relations, it comes:

xo ,in1=xo ,i

n 1

t 2 xo ,in1−xo ,i

n and xo ,in1=xo ,i

n

txo ,i

n1−xo ,i

n If one poses:

xo ,in1=xo , i

n xo ,i

n1 , éq A3.1-3

one thus has:

xo ,in1=xo ,i

n

1 t 2

xo ,in1

and xo ,in1=xo ,i

n

t xo , i

n1éq A3.1-4

A3.2 Diagram of Newmark adapted to great rotationsTo unify calculations, one would like to be able to write, in rotation, the following similar relations:

in1=i

n

t i−1, i

n1 − i−1, in and i

n1=in

1 t2 i−1,i

n1 − i−1,in

where i−1, in

and i−1, in1

are respectively the vectors increment of rotation of the section considered,

between the moment i−1 and iterations n and n1 moment i [A6].

But these relations are not exact because one cannot combine a speed (or an acceleration) angularwith the iteration ( n1 ) moment i with the homologous size with the iteration n and of vector-

rotation counted starting from the configuration at the moment i−1 [A7]. One can combine only afterhaving brought by rotation (one says “transported”) all the sizes in the same configuration. Onechooses to simplify the configuration of reference.

Thus let us pose:

in1=Ri

n1,Ti

n1 in=Ri

n , Ti

n éq A3.2-1

Ain1=R i

n1,Ti

n1 Ain=Ri

n ,Ti

n éq A3.2-2

i−1,in1=Ri−1

Ti−1,i

n1 i−1,in=R i−1

T i−1,i

n

The algorithm of Newmark in rotation results in the following relations:• for the operator of rotation, [éq A6-1]:

R in1=exp i

n1R in ;

• for the increment of vector-rotation since the moment i−1 , [éq A6-2]:

exp i−1,in1 =exp i

n1 exp i−1,in ;




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• for the angular velocity:

in1= i

n

t i−1, i

n1 − i−1, in ;

• for the angular acceleration:

Ain1=Ai

n1

t 2 i−1,in1 − i−1, i

n .

The two preceding relations give, by opposite transport on the configuration n1i and taking into

account the relations [éq A3.2-1] and [éq A3.2-2]:

in1=R i

n1 Rin ,Ti

n

tRi

n1 R i−1T i−1, i

n1− i−1,i

n éq A3.2-3

in1=Ri

n1 Rin ,Ti

n

1 t 2

Rin1 Ri−1

T i−1, in1− i−1, i

n . éq A3.2-4




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Annexe 4 : Calculation of the differentials of FréchetOne will have to use thereafter the identity following, known as hook of Dregs, of which thedemonstration is immediate. If A and B are two antisymmetric matrices of the 3E order, of axial

vectors A and B , then one has, for any vector v :

A B− B A v=A∧B ∧v éq A4-1

In other words: is the axial vector of the antisymmetric matrix A B− B A .

Differential of Fréchet of the function is called f in the direction x , quantity:

D f x . x=grad f . x=lim 0

dd

f x x .

It is the principal part of the increase in f corresponding to the increase x variable x .Here the catalogue of the differentials of Fréchet intervening in this note.

D xo ' . xo=lim 0

dd

xo xo '= xo '= xo ' éq A4-2

D xo. xo= xo éq A4-3

D xo. xo= xo éq A4-4

D R .=lim0

dd

[exp R ]= R éq A4-5

D RT .=lim 0

dd

[RT exp − ]=−RT éq A4-6

because, according to the equation [éq 4.2-6]:

[exp ]T=exp − .

=R RT .

However:

D R .=lim 0

dd

[exp R ]∙=

R R .

Therefore, according to [éq A4-6]:

D .=D R . RTR D RT . =

− .

Maybe, by using it hook of Dregs [éq A4-1]: D .= −∧ éq A4-7




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Annexe 5 : Complements on the calculation of the matrices ofrigidityThis appendix develops calculations of [§9.1].

A5.1 Geometrical matrix of rigidity

[D {B } . ] . {FM} {B }. [D . ]{FM}= xo '∧ . f xo 'xo '∧ . f ' . m .

By rearranging the terms and by using the vectorial identity:

a . b∧c =c . a∧b ,

The second preceding member is written:

− xo ' . f∧− ' .m∧ . f∧ xo ' . xo ' f =Y .E Y .

A5.2 Matrices of material and complementary rigidityAs it is shown with [§9.1]:

D{FM}.=CT B−CT 0

0 { }.

However:

{ }=C−1

T { fm}.

Consequently:

{B} . D {FM }. ={B} .CT B éq A5.2-1

−{B } .CT 00 C±1

T { f

m}.

The term:

{B} .CT B

conduit immediately with matrix of material rigidity.

In addition:

00 C−1

T { f

m}=−{ Rtot C1−1R tot

T f ∧

R tot C2−1 Rtot

T m ∧

}.

The second term of the equation [éq A5.2-1] is thus written:

B .[0 Rtot C1 R totT R tot C1

±1R totT f

∧

0 Rtot C2 R totT Rtot C2

±1R totT m

∧ ] .

While indicating by Z the matrix between hooks of the expression which precedes, Bt Z is the

complementary matrix of rigidity.




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Annexe 6 : Principle of the iterative calculation of rotationsThis appendix visualizes the operation of exponentiation, defines the vectors increment of rotationwhich intervene in appendix 3 and gives the relation between them.

A6-a figure: Representation of the operators intervening in

the rotation of the sections. : symbol of exponential projection

The curved surface of the figure [fig. A6-a] represents the unit SO 3 operators of rotation R . One

appeared tangent spaces in SO 3 in R i−1 , rotation calculated at the moment i−1 , and in, n R in

- ième iteration in the calculation of rotation at the moment i .

That is to say i−1 the vector rotation corresponding to. R i−1 There exists a vector increment of

rotation i−1,in

such as:

R in=exp i−1,i

n Ri−1.

If the equilibrium equation of the beam is not satisfied, at the moment i , by R in

, a correction is

sought in1 de i−1,i

n . in1

is obtained by linearization, while replacing in the equilibrium

equation, according to equivalence [éq 4.2-5]:

exp in1R i

n par 1 in1 Ri

n .

The second of the preceding expressions is not an operator of rotation, but is in tangent space with

SO 3 in R in

[fig A6-a]. Having calculated in1 ,R i

n1 from of deduced while projecting by exp

on SO 3 :

R in1=exp i

n1R in . éq A6-1




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The increment of swing angle i−1,in1 making pass from R i−1 with R i

n1 is such as:

R in1=exp i−1,i

n1 Ri−1.

Vector-rotation not being additive, one does not have :

i−1,in1= i

n1 i−1,i

n ,

but one has :

exp i−1,in1 =exp i

n1 exp i−1,in éq A6-2

With [§10.7.2], one solves the preceding equation compared to i−1,in1

by using the properties of the

quaternions.

Increments of swing angle i−1,in et i−1,i

n1 are used to calculate the corrections speed and

acceleration of ni with n1i .




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Annexe 7 : Need for transport in a space of reference forthe vectorial operations relating to the rotationmovement

The sizes kinematics, speeds and accelerations, in a given configuration are in the tangent vectorspace with SO 3 at the point defined by the rotation of the configuration compared to the position ofreference. One can combine the sizes relative to two distinct configurations only after havingtransported them in the same vector space taken for reference. The example which follows will makeunderstand this need.

Let us examine the angular velocity of a gyroscope at three moments t 1 , t2 and t 3 [fig A7-a]. Let us

suppose that one passes from configuration 1 to configuration 2 by the rotation of angle − around

e1 , and of configuration 1 with configuration 3 by the rotation of angle −/2 around the same

axis. In 1, the angular velocity is carried by the axis of the gyro and has as components 0, 0, in

the axes generals e1 e2 e3 . In 2, the angular velocity is also carried by the axis and has as general

components 0, 0,– 3 . One wants to determine the general components speed into 3, knowingthat this speed is the average speeds into 1 and 2.

That the angular velocity in 3 is the average angular velocities in 1 and 2 does not mean that it is theaverage in the axes generals, but in the axes related to the gyro. In the example, since the gyro turnsaround its axis, in 1 with speed , in 2 with speed 3 , then in 3 it turns around this axis with speed

2 . Therefore, in 3, the general components angular velocity are 0,2 ,0 .

Taking into account [§5], one gets the preceding result in “transporting” the angular Flight Path Vectorof configuration 2 on configuration 1, taken for reference, i.e. while making turn this vector of the

angle around e1 . Its general components are then 0,0,3 . One makes the average of this

vector and the angular Flight Path Vector into 1, to obtain the vector of components 0,0,2 . One

“transports” finally this last vector on configuration 3 by making it turn of −/2 around e1 , and one

leads well to the angular Flight Path Vector of general components 0, 2 ,0 .

A7-a figure: Evolution angular velocity of a gyroscope.




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Annexe 8 : Use of the quaternions in modeling of greatrotations [bib14] [bib15]

A8.1 DefinitionsOne quaternion, noted q or x o , x , is defined by the whole of a scalar xo and of a vector with

three dimensions x x1 x2 x3 :

xo , x =xox1 x2x3 , ,

, and being three numbers satisfying the relations:

2=

2=

2=−1 ;

=− = ;=−= ; =−= .

Combined quaternionCombined, noted ∗ , from a quaternion is obtained by changing the sign of the vectorial part:

x o ,x ∗= xo ,−x .

Purely vectorial quaternion

A quaternion is known as purely vectorial when its scalar part is worthless and that it is thus of the

form: 0, x . A quaternion v is purely vectorial if and only if:

v v ∗=0 .

A8.2 Elements of algebra of the quaternionsMultiplicationBy applying the definition, one shows immediately that the multiplication, noted ° , of two quaternionsis:

xo , x ° yo ,y = xo yo−x . y , xo yyo xx∧y .

It is checked that this multiplication is:• associative:

[ xo , x yo , y ]° zo , z= xo , x ° zo , z yo , y ° zo , z ;

• noncommutative because:

x o , x ° yo , y −yo , y ° xo , x =0,2x∧y .

Normalizes of a quaternion, noted ∥.∥ :

∥ xo , x ∥2= xo , x ° xo , x

∗= x0

2 x12 x2

2 x32 .

One quaternion is unit if its standard is equal to the unit. It is checked that:

[q1 ° q2 ]∗

=q2 ∗° q1

∗.

A8.3 Representation of a rotation by a quaternionIf v1 is a purely vectorial quaternion and u a unit quaternion, then the quaternion v2 such as:

v2 = u ° v1 ° u ∗

éq A8.3-1

is purely vectorial and has even standard that v1 . Indeed:




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v2 v2 ∗=u ° v1 ° u

∗u ° v1

∗ ° u ∗

= u ° [ v1 v1∗]° u

∗= 0 ;

v2 ° v2 ∗=u ° v1 ° u

∗° u ° v1

∗° u

∗=∥v1 ∥

2.

If one poses:

v1 =0,x ;v2 =0,y ,

it is seen that, by the relation [éq A8.3-1], the unit quaternion u an orthogonal transformation of the

vector defines x on the vector y . This transformation is the rotation whose matrix is defined by [éqA8.5-1].

opposite rotation is defined by the quaternion u *

, because:

u ∗° v2 ° u = u

∗° u ° v1 ° u

∗° u =v1 .

A8.4 Successive rotationsLet us subject the purely vectorial quaternion v1 two successive rotations defined by the unit

quaternions u1 et u2 :

v2 =u1 ° v1 ° u1 ∗

and:

v3 = u2 ° v2 ° u2 ∗=u2 ° u1 ° v1 °[ u2 ° u1 ]

∗ éq A8.4-1

One sees on the last member of [éq A8.4-1] that the unit quaternion defining the product of tworotations is equal to the product of the quaternions of these rotations.

A8.5 Matric expressionsThat is to say q a quaternion:

q q = xo ,x .

Let us pose:

q ={xo

x },

q is the formed vector by the four components of the quaternion.

Let us define two matrices built on q :

Aq =[ xo −xT

x x o1x ] et Bq =[ xo −xT

x xo 1−x] .

One checks without sorrow, according to the rule of multiplication, that:

q1 ° q2 =Aq1 q2 =B

q2 q1 .

Now let us take the rotation defined by the unit quaternion u =eo ,e . If:

v2 = u v1 u ∗

.




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Then:

v2 =Au

Bu T v1 .

In addition:

A u

Bu T=[1 0T

0 R ] ,

with:

R=eeTeo

2 12 eo ee2

,

maybe, by using the relation [éq A1-4]:

R=2eo2−112 eeT

eo e éq A8.5-1

By bringing the preceding relation closer to the equation [éq 4.2-3], one sees that the components ofthe unit quaternion defining a rotation are not other than the parameters of Euler of this rotation.

A8.6 Passage d' a vector-rotation with the associated quaternion and viceversa

The operator Q who makes pass from a vector rotation in the associated quaternion is definedby the relations [éq 4.2 - 2]:

eo=cos 12∥∥ éq A8.6-1

e=sin 12∥∥ ∥∥ éq A8.6-2

The operator Q−1 is less simple because the reverse of a trigonomitric function does not have a

single determination.

But one notices on [éq 10.7.2-1] and [éq 10.7.2-2] that Q−1 is used to calculate the vector - rotation

n1 deduced from vector-rotation

n by correction n1 . In general:

∥n1∥<<∥ n

∥ .

We thus adopted the following strategy: among all the determinations of the vector n1 one takes

that to which the module is closest to:

∥n1

n∥ .

In the case plan, this strategy is rigorous [§ 4.1], but in the three-dimensional case it is not it because:

n1

n≠

n1 .From [éq A8.6-1], one draws:

12∥

n1∥=±cos−1 eo 2k .

The adopted strategy leads to only one determination of 12∥ n1∥ .

[éq A8.6-2] gives then:

n1=2

12∥ n1∥

sin 12∥

n1∥

e .


Documents

Static and dynamic modeling of the beams in []