Gyroscopic matrices of the right beams and the discs

Transcription

1 Titre : Matrice gyroscopique es poutres roites et es i[...] Date : 15/07/2014 Page : 1/16 Gyroscopic matrices of the right beams an the iscs Summary: This ocument presents the formulation of the matrices of gyroscopic of the elements beams, voluminal an iscrete amping an stiffness. The beams are only right beams (Elements POU_D_T an POU_D_E). The section is constant over the length an of circular form. The material is homogeneous, isotropic. The iscs are cyliners of cross-section whose axis is confuse with the axis of the beam. The isc is suppose to be ineformable. The assumptions selecte are the following ones: Assumption of Timoshenko: transverse shearing an all the terms of inertia are taken into account. This assumption is to be use for weak twinges (Elements POU_D_T). Assumption of Euler: transverse shearing is neglecte. This assumption is checke for strong twinges (Elements POU_D_E). The number of revolutions clean (along the axis of the beam) can be constant or variable. In Coe_Aster, aopte convention efines the positive irection following the axis of rotation as being the irection trigonometrical usual of rotation.

2 Titre : Matrice gyroscopique es poutres roites et es i[...] Date : 15/07/2014 Page : 2/16 Contents 1 The element beam of constant circular section Definition of the reference marks Characteristics Calculation of the kinetic energy of the beam of Timoshenko Functions of interpolation Calculation of the equilibrium equations The circular isc Calculation of the kinetic energy of the isc Calculation of the equilibrium equations The voluminal element 3D Position moving in the revolving reference mark Expression of the energies kinetic an potential Calculation of the equilibrium equations of the system in rotation Description of the versions... 15

3 Titre : Matrice gyroscopique es poutres roites et es i[...] Date : 15/07/2014 Page : 3/16 1 The element beam of constant circular section A beam is a soli generate by a surface of surface S, of which the geometrical centre of inertia G followe a curve C calle average fibre or neutral fibre. Within the framework of this moeling, only the right beams, with constant an circular section are taken into account. For the stuy of the beams in general, one formulates the following assumptions: The cross-section of the beam is ineformable, Transverse isplacement is uniform on the cross-section. These assumptions make it possible to express isplacements of an unspecifie point of the section, accoring to an increase in isplacement ue to the rotation of the section aroun the transverse axes. The iscretization in exact elements of beam is carrie out on a linear element with two noes an six egrees of freeom by noes. These egrees of freeom break up into three translations u, v, w (isplacements accoring to the irections x, y an z ) an three rotations x, y an z. (aroun the axes x, y an z ). th 1 Z er e 2 X U 1 U 2 v 1 v 2 W 1 W 2 In the case of the right beams, the average line is along the axis x local base, isplacement transverse being thus carrie out in the plan y, z. For the storage of the sizes relate to the egrees of freeom of an element in a vector or an elementary matrix (thus of imension 12 or 12 2 ), one arranges initially the variables for noe 1 then those of noe 2. For each noe, one stores initially the sizes relate to the three translations, then those relate to three rotations. For example, a vector isplacement will be structure in the following way: u 1, v 1, w 1, x1, y1, z1 sommet 1, u, v, w,,, x2 y2 z2 sommet Definition of the reference marks One efines: x is the axis of neutral fibre of the beam, y an z are the main axes of inertia of the section, R 0 is the absolute reference mark relate to a section in the initial configuration, R is the reference mark relate to a section in the eforme configuration,

4 Titre : Matrice gyroscopique es poutres roites et es i[...] Date : 15/07/2014 Page : 4/16 By not consiering torsion, the passage of the reference mark R 0 with the reference mark R be carrie out with the assistance 3 rotations, two following y an z, an a rotation aroun x, note, such as: : number of revolutions clean of the tree 1.2 Characteristics Each element is an isoparametric element beam of circular an constant section. One takes into account transverse shearing in the formulation of this element (right beam of Timoshenko). Notations: x is the axis of neutral fibre of the line of trees, ensity: length of the element: Young moulus: E Fish moule: G= E 2 1 section: interior ray: R i external ray: R e surface: A= R e 2 R i2 polar inertia: I x = 2 R e 4 R i4 inertia of section: I yz =I y =I z = 4 R e 4 R i4 1.3 Calculation of the kinetic energy of the beam of Timoshenko One calculates the kinetic energy of the element beam of Timoshenko by consiering the eformations of membrane an inflection. The expression of the kinetic energy is obtaine while integrating over the length of the element beam: T = 1 2. A ]=[ I with: [ J [ u 2 v 2 ẇ 2 ] x 1 2. R /R0.[ J ]. R/ R0 x x 0 0 ] 0 I y 0 with 0 0 I z I (in m 4 ) That is to say a right beam of axis ox for the not eforme configuration, it is necessary to efine two intermeiate bases to characterize the Flight Path Vector of rotation R/ R0. Passage e la bases B (O, x, y, z ) at the base B 1 (O, x 1, y 1, z 1 ) by a rotation of axis oy of amplitue y x, t such as: y 1 =y Passage e la bases B ( O, x 1, y 1, z 1 ) at the base B 2 ( O, x 2, y 2, z 2 ) by a rotation of axis o z 1 of amplitue z x,t such as: z 2 = z 1 an y 1 =cos z x,t. y 2 sin z x,t. x 2

5 Titre : Matrice gyroscopique es poutres roites et es i[...] Date : 15/07/2014 Page : 5/16 Rotation at the angular velocity t place has along the axis o x 2. Thus the vector of rotation is written: R/ R0 = t. x 2 y x, t. y 1 z x, t. z 1 Since the operator [ J ] element beam is written in the base B 2 who correspons to the eforme position, it is imperative unless changing basic the operator of inertia, to write the Flight Path Vector of rotation R/R0 in the base B 2. R/R0 = t. x 2 y x,t. cos z x,t. y 2 sin z x, t. x 2 z x,t. z 2 By consiering that angles y x, t an z x, t are small, it is legitimate to carry out a evelopment limite to orer 1. The expression of the Flight Path Vector R/R0 becomes then: R/R0 = t y x, t. z x, t. x 2 y x,t. y 2 z x, t. z 2 It remains to evelop the following scalar prouct: 1 2 R /R0. [ J ]. R /R0 x = 1 2 I yz For an element beam of constant section, the expression becomes: T = 1 2 A With: [ u 2 v 2 ẇ 2 ] x 1 2 I yz I x = 2. [ R e 4 R e4 ] [ y 2 z2 ] x 1 2 I x.. 2 I x y. z x [ 2 y z2 ] x 1 2 I x.. 2 I x y. z x I yz =I y =I z = 4. [ R e 4 R e4 ] The various terms of the kinetic energy represent: for the first term, the kinetic energy of translation, for the two following terms, the kinetic energy of rotation, for the fourth term, the gyroscopic term of effect. 1.4 Functions of interpolation For the eformations of membrane (traction an compression), the fiel ux is approache by a linear function of isplacements of noes 1 an 2 of the element beam: u x = N 1 x N 2 x { u 1 u 2} with {N 1 x =1 x N 2 x = x For the eformations of inflection, one uses cubic functions of type moifie Hermit. Degrees of freeom v x, y x, w x, z x are thus interpolate as follows:

6 Titre : Matrice gyroscopique es poutres roites et es i[...] Date : 15/07/2014 Page : 6/16 {v 1 v x = 1 x 2 x 3 x 4 x v { 2 v θ z x= 5 x 6 x 7 x 8 x 1 w x = 1 x 2 x 3 x 4 x {w y1 } w 2 y2 1 y x= 5 x 6 x 7 x 8 x {w y1 } w 2 y2 z1 z2} 1 z1 } v 2 z2 One efines K yz coefficient of shearing in the irections y an z. For an element beam of constant section: K yz = 720.α2 6 with = R e. R i 1 R 2 i 2 R e While noting yz = 12.E. I yz K yz. A.G. 2, functions i 1 x= 1 [ 1 yz 6 5 x= 2 x = [ 1 yz 6 x = 1 1 yz [ 3 x = 1 [ 1 yz 6 7 x = 2. x 3 3. x 2 yz. x 1 yz ].1φ yz. x. [ 1 x ] x 3 4 yz 2 x 2 2 yz 2 3. x 2 4 yz. x 1 yz ] 2. x 3 3. x 2 yz. x ].1 yz. x. [ 1 x ] are as follows efine:. x ]

7 Titre : Matrice gyroscopique es poutres roites et es i[...] Date : 15/07/2014 Page : 7/16 4 x= [ 1 yz 8 x= 1 1 yz [ x 3 2 yz 2 x 2 yz 2. x ] 3. x 2 2 yz. x ] Note: In the case of elements beams of Euler (Elements POU_D_E) the term yz is null. The vector of the egrees of freeom of the element beam is efine by: q = u 1 v 1 w 1 x1 y1 z1 u 2 v 2 w 2 x2 y2 z2 One poses: u = u 1 u 2 v = v 1 z1 v 2 z2 w = w 1 y1 w 2 y2 By replacing the preceing approximations in the expression of the kinetic energy, one obtains: T = 1 2 u [ M 1 ] { u} 1 2 ẇ [ M 2 ][ M 4 ] { ẇ } 1 2 v [ M 3 ][ M 5 ] { v}. v [ M 6 ] { w } I x. 2 With: [ M 1]= [ M 2 ]= [ M 3 ]= [ M 4 ]=. A.{ N 1 x } N 2 x. N 1 x N 2 x. x. A.{ 1 x 2 x 3 x 4 x }. 1 x 2 x 3 x 4 x.x 1 x. A.{ 2 x 3 x }. 1 x 2 x 3 x 4 x. x 4 x. I yz.{ 5 x 6 x 7 x 8 x }. 5 x 6 x 7 x 8 x.x

8 Titre : Matrice gyroscopique es poutres roites et es i[...] Date : 15/07/2014 Page : 8/16 [ M 5]= [ M 6 ]= 5 x. I yz.{ 6 x 7 x 8 x 5 x. I x.{ 6 x 7 x 8 x }. 5 x 6 x 7 x 8 x. x }. 5 x 6 x 7 x 8 x.x 1.5 Calculation of the equilibrium equations The equations of agrange for the kinetic energy of the beam are written in the following form: t T q i T =0 with q = u v w q i { t That is to say: T t δ u T δu = [ M 1 ] {δ ü} T δ v T δv = [ M 3 ][ M 5 ] {δ v } φ.[ M 6 ] { δ ẇ } φ.[ M 6 ] { δ w } T t δ ẇ T These equations can be put in the form: [ M ] q [ C gyro ] q [ K ][ K gyro ] q = 0 δw = [ M 2][ M 4] { δ ẅ } φ.[ M 6 ] T { δ v } The gyroscopic matrix of amping [C gyro ] system is mae up starting from the matrix [ M 6 ] an of its transpose. It is antisymmetric, an its contribution must be multiplie by the angular velocity. While noting: = yz ρ. I x [ ] [ M 6 ]=

9 Titre : Matrice gyroscopique es poutres roites et es i[...] Date : 15/07/2014 Page : 9/16 [ C gyro]=. I x [ ] ike the matrix [C gyro ] is antisymmetric, only the higher triangle is represente. ( ) mean that the egree of freeom is not concerne with the gyroscopic matrices. The gyroscopic matrix of stiffness [ K gyro ] system is mae up starting from the matrix [ M 6 ]. Its contribution must be multiplie by the angular acceleration. [ K gyro ]=. I x ] [ The full matrix [ K gyro ] is fille in entirety (triangles higher an inferior). Recall: with q = u 1 v 1 w 1 θ x1 θ y1 θ z1 u 2 v 2 w 2 θ x2 θ y2 θ z2 in the case of elements beams of Euler (Elements POU_D_E) the term yz is null. 2 The circular isc The objective of this chapter is to characterize the gyroscopic matrices of an infinitely rigi circular isc, subjecte at a number of constant or variable revolutions. The characteristics of the isc are the following ones: axis of the isc confuse with the axis of neutral fibre of the beam (axis x ) centre of gravity of the isc: C

10 Titre : Matrice gyroscopique es poutres roites et es i[...] Date : 15/07/2014 Page : 10/16 interior ray: R i external ray: R e thickness: h presumely uniform ensity: Deuce values: mass of the isc: M = h R e 2 R i 2 moment of inertia mass/axes y or z calculate in the centre of gravity C : I yz = M R 2 e3. R 2 i h 2 mass moment of inertia compare to the axis x calculate in the centre of gravity C : I x = M 2 R 2 er 2 i Note: Axes C x, C y an C z being main axes of inertia of the isc, proucts of inertia I xy, I yz an I xz are worthless. The symmetry of the isc compare to the axes C y an C z impose: I yz =I y =I z The isplacement of the center of the isc is given by: u.xv. yw.z One notes: R / R0 : the Flight Path Vector of rotation of the isc x. R/ R0 = t : number of revolutions clean 2.1 Calculation of the kinetic energy of the isc One calculates the kinetic energy of the isc by applying the formula of Huygens: T = 1 2 M. V C, D/ R R/ R0. [ J ]. R/ R0 T = 1 2 M. u 2 v 2 ẇ R/ R0. [ J ]. R /R0 with: [ J ]=[ I x 0 0 ] 0 I y 0 with I yz =I y =I z 0 0 I z By eveloping the preceing expression, one obtains: T = 1 2. u 2 v 2 ẇ I y z. y 2 z I x. 2 2 y.. z The various terms of the kinetic energy represent: for the first term, the kinetic energy of translation, for the secon term, the kinetic energy of rotation, for the term 1 2 I x. 2, the clean energy of rotation, an for the term I x. y.. z, the gyroscopic effect. 2.2 Calculation of the equilibrium equations

11 Titre : Matrice gyroscopique es poutres roites et es i[...] Date : 15/07/2014 Page : 11/16 The equations of agrange are use to formulate the ynamic balance of the isc. In this case, typical case the eformation energy is worthless (infinitely rigi isc) an no external effort is consiere, one thus has: t T q i element isc. T q i =0 with q = u v w y z : vector of the egrees of freeom of the One oes not take account of the egree of freeom because it is consiere that the number of revolutions clean is impose an thus known. The following equations then are obtaine: { T t y t T u T =M.ü δu t T v T =M. v δv t T ẇ T δw =M. ẅ T t z T y = I yz. y I x.. z I x.. z T z =I yz. z I x.. y These equations can be put in the form: [ M ] q [ C gyro ] q [ K ][ K gyro ] q = 0 The gyroscopic matrix of amping of the isc is obtaine as from the moment of inertia I x. It is antisymmetric, an its contribution must be multiplie by the clean angular velocity [C gyro ]= I.[ ] x I x 0 with u v w x y z vector of the egrees of freeom of the element isc an such as: x = The inent correspons to the egree of freeom of rotation along the axis of the beam an leas obviously to worthless terms. The gyroscopic matrix of stiffness of the isc is also obtaine as from the moment of inertia I x. Its contribution must be multiplie by the clean angular acceleration.

12 Titre : Matrice gyroscopique es poutres roites et es i[...] Date : 15/07/2014 Page : 12/ [ K gyro ]=.[ I x The voluminal element 3D ] The setting in equation of a mobile structure can be one, either in the fixe Galilean reference mark (OXYZ), or in the reference mark of inertia (O' X' Y' Z') attache with the structure. For a ynamic analysis, the choice of the reference mark of inertia as frame of reference makes it possible to simplify the formulation of the ifferential equations governing their ynamic behavior. This choice makes it possible to characterize the effect Coriolis (gyroscopic effect seen of the rotor) which is ae to amping. Also, two other effects of type apparent stiffness are ae to the matrix of stiffness: it is the centrifugal effect an the gyroscopic effect associate with the variation spee. 3.1 Position moving in the revolving reference mark In any general information, it will be consiere that mobile reference frame, whose origin is characterize by a translation st= t s x, s y, s y compare to the reference mark of inertia, turns at the angular velocity = t x, y, z aroun an unspecifie axis passing by the origin. It is important to note that the components number of revolutions are measure in reference mark corotationnel. One will note ' the projection this number of revolutions in the fixe reference mark,

13 Titre : Matrice gyroscopique es poutres roites et es i[...] Date : 15/07/2014 Page : 13/16 IE = t R ' where R t is the matrix of basic change, forme by the cosine irectors of the basic vectors of the reference mark of inertia, expresse in the fixe reference frame. The position of a point P in the reference frame of inertia (OXYZ), note y, has then as an expression: y=st R t[ xux,t ] where x is the initial position of the point P in the inertial system an u is the vector isplacement resulting from the ynamic eformation of the structure at a moment given T. Absolute velocity ẏ point P is efine as being the erivative first compare to the time of the vector y. She is written: ẏ=ṡṙ xur u where ṡ is the spee of traverse of the O' origin of the turning reference mark, written in the fixe reference mark. It is shown that the erivative Ṙ matrix of basic change can be also written as being the vector prouct between the vector of rotation in the fixe reference mark ' an D stamps ite basic change R or then as being the prouct of the matrix R an of the antisymmetric matrix who inclues the three components of the tensor of rotation in the reference mark of inertia : 0 z y Ṙ=' R=R with =[ ] z 0 x y x 0 While inserting this efinition in the expression absolute velocity, it results from it that: ẏ=ṡ' R xur u Absolute acceleration ÿ point P is efine as being the erivative secon compare to the time of the Flight Path Vector y. She is written: ÿ= s ' R xu' [ ' ' R xu]2 ' R ur ü with s the acceleration of translation of the O' origin of the turning reference mark, written in the fixe reference mark an ' is the instantaneous acceleration of rotation efine by the relation = t R '. By concern of clearness an without losing the general information, one will suppose in the continuation of the ocument that the O' origin of the turning reference mark is fixe, IE. ṡ= s=0. After pre-multiplication by the reverse transformation t R, one then obtains the absolute spee an acceleration expresse in the mobile reference frame: t R ẏ= u R xu

14 Titre : Matrice gyroscopique es poutres roites et es i[...] Date : 15/07/2014 Page : 14/16 t R ÿ=ü xu [ xu]2 u In this expression, which represents the theorem of the composition of accelerations of a material point, the terms are recognize: of relative acceleration ü, which contributes to the matrix of mass; of acceleration of training xu [ xu] (nap of the effect Euler ue to acceleration of rotation an of the effect of softening centrifugal, which contributes to the matrix of stiffness); of complementary acceleration or Coriolis 2 u, who contributes to the matrix of amping. 3.2 Expression of the energies kinetic an potential In its general form, the kinetic energy is obtaine starting from the absolute velocity y as follows: T= 1 2 ẏ t ẏ The evelopment of the terms gives the following expression accoring to isplacement u : T= 1 2 u t u u t u x t 2 x u t 2 u u t 2 x u t x The potential energy of the system (internal eformation energy an work of the external forces) has as a classical expression: U= 1 2 t u t f u t t where is the vector associate with the tensor with the eformations, is the matrix of behavior an where f an t are, respectively, the vectors of the voluminal an surface forces external. One approaches the vector isplacement by the finite element metho. With this intention, one will use the classical functions of form escribe in the ocument [R ]. Displacement is written then in the shape of the prouct of a matrix of interpolation of isplacements, note B, an of a vector of the generalize coorinates q. The energies kinetic an potential of the eformable boy are written then accoring to the structural matrices as follows: T= 1 2 qt [ M ] q 1 2 qt [ G ] q 1 2 qt [ N ] q q t 1 2 x t 2 x B t 2 x q t B t x U= 1 2 qt [ K ] q q t B t f q t B t t The first three terms of the kinetic energy highlight the matrices:

15 Titre : Matrice gyroscopique es poutres roites et es i[...] Date : 15/07/2014 Page : 15/16 of mass [ M ]= B t B ; of Coriolis [G ]=2 B t B ; of centrifugal acceleration [ N ]= B t 2 B. The first term of the potential energy highlights the matrix of stiffness [ K ]= B t B. 3.3 Calculation of the equilibrium equations of the system in rotation By isregaring a possible classical function of issipation, characterize by the matrix of amping, the equations of agrange for the energies kinetic an potential of the soli are written as follows: t T q T q U q =0 While inserting the expressions of energies in the equations of agrange, one fins: t [ [ M] q1 2 [ G ] q1 2 [G] q[ N ] q t [ B t t ]=0 B t x B t 2 x [ K ] q B t f ] By clarifying the temporal erivative an by taking account of the erivative of the matrix of transformation in the equations of agrange, one obtains, after simplification an rearrangement of the terms, the following matric form: [ M ] q [G ] q [ K ][ P ][ N ] q = r [ P ] is the matrix of angular acceleration, efine by [ P ]= 1 2 [Ġ]= B t B, with: =[ 0 z y z 0 x y x 0 ] r is the vector joining together the terms of the member of right-han sie. He gathers the external excitations B t f B t t an centrifugal prestressing B t x 2 x ). 4 Description of the versions Aster Author (S) Organization Description of the (S) moifications 9.4 E. BOYERE, X. RAUD EDF/R & D AMA Initial text 9,8 Mr. Torkhani Correction of shells

16 Titre : Matrice gyroscopique es poutres roites et es i[...] Date : 15/07/2014 Page : 16/16 EDF/R & D AMA