Code_Aster. A finite element of cable-pulley

Transcription

1 Titre : Un élément fini de câble-poulie Date : 09/12/2013 Page : 1/14 A finite element of cable-pulley Summary : The pulleys play an important role in the structures of cables such as the air electric lines. It is thus useful to know to model them in one at the same time realistic and powerful way. One presents the finite element formulation of a length of cable passing by a pulley: expressions of the internal forces and the matrix of rigidity. The pulley can be fixed or supported by a flexible structure. Its position on the cable with balance is not known a priori: this position is that for which on both sides the tension of the cable is the same one.

2 Titre : Un élément fini de câble-poulie Date : 09/12/2013 Page : 2/14 Contents 1 Notations Introduction Assumptions and definition of a finite element of cable - pulley Internal forces of a finite element of cable-pulley Matrix of rigidity Matrix of mass Introduction of the element of cable-pulley into Code_Aster Example of application Conclusion Bibliography Description of the versions of the document...11 Annexe 1 Calculation of the matrix of rigidity...12 Annexe 2 Figure of balance of an inextensible cable weighing subjected to a tension given in the end... 13

3 Titre : Un élément fini de câble-poulie Date : 09/12/2013 Page : 3/14 1 Notations A E F surface of the cross-section of the cable. Young modulus. vector of the internal forces of the element. F n force interns with the node n. H horizontal component of the tension [ An1]. I 3 matrix unit of order 3. K l l 0 l i matrix of rigidity of the element. current length of the element. length at rest. initial length. N 3 N 1 euclidian norm of l 2 N 3 N 2 l 2 euclidian norm of l 2. N N i S s T T i u u n w x n current tension of the cable constituting the element. initial tension (prevoltage). arrow of a range of cable [ An1]. length of a range [ An1]. current temperature. initial temperature. vector-displacement of the nodes compared to the initial position. displacement of the node n compared to its initial position. weight per unit of length. vector-position of the node n in initial configuration. thermal dilation coefficient. current deformation compared to the initial configuration.

4 Titre : Un élément fini de câble-poulie Date : 09/12/2013 Page : 4/14 2 Introduction One uses pulleys, during the construction of the air electric lines, for the operation of installation of the cables. The cable in the course of installation [Figure 2-a] is fixed at the one of the supports of stop of the canton, it rests on pulleys placed at the bottom of the insulators of the supports of alignment and it is retained by a force on the level of the second support of stop. While exploiting this force - or by moving its point of application - one adjusts the arrow of the one of the ranges, that which is subjected to constraints of environment. Then one removes the pulleys and one fixes the cable at the insulators. The length of the cable in the various ranges is then fixed and it determines the later behavior of the line under statical stresses (wind, overload of white frost) and in dynamic mode (movement due to the forces of Laplace created by the currents of short-circuit). Figure 2-a: Pose of a cable in a canton with two ranges The finite element of cable-pulley presented here makes it possible to model the operation of installation and thus to calculate, in a natural way, the length of cable in the various ranges. The idea of this finite element came to us some time after the conversation [bib1] and we presented his formulation in [bib2].

5 Titre : Un élément fini de câble-poulie Date : 09/12/2013 Page : 5/14 3 Assumptions and definition of a finite element of cable - pulley N 3 N 1 N 2 Figure 3-a: Length of cable passing by a pulley - reality Let us take a length of cable N 1 N 2 passing by the pulley N 3 [Figure 3-a]. This pulley is not inevitably fixed and can, for example and as it is the case in the example of [ 8], being assembled at the end of a cable. N 3 N 1 N2 Figure 3-b: Length of cable passing by a pulley - modelled The pulley is supposed to be specific [Figure 3-b]. One makes moreover following assumptions: The position of balance of N 3 is not known, but it is necessarily on the section N 1 N 2 deformed starting from its initial position. In modeling of the lines, the horizontal movements are of low amplitude and this assumption is generally not restrictive; Two bits N 3 N 1 and N 3 N 2 are always rectilinear, like elements of cable of the 1 er order.

6 Titre : Un élément fini de câble-poulie Date : 09/12/2013 Page : 6/14 It rises from this assumption that: =x 1 u 1 x 3 u 3 éq 3-1 l 2 =x 2 u 2 x 3 u 3 = T éq 3-2 l 2 = l 2 T l 2 The overall length of the two bits is: in the current location, where the temperature is T : l= l 2 ; in the initial position, indicated by the index i, where the tension is N i and the temperature T i : l i = i l 2 i ; in a not forced position, indicated by index 0, where the temperature is T 0 : l 0 = 0 l 2 0. The pulley is without friction and thus the tension is the same one in the two bits. It rises from this assumption that the deformation is also the same one and one takes for value of this one the measurement of the relative lengthening of the section compared to the initial state: = l l i l 0 éq 3-3 must remain small, so that the section A is regarded as constant. It will be noted that, within the framework of the finite element method, the linear loads do not prevent the tension from being constant of N 1 with N 2. These forces are indeed concentrated with the nodes N 1 and N 2 and onaxis pulley N 3. The relation of behavior is elastic: N =EA [ e α T T i ] N i éq 3-4 One calls finite element of cable-pulley, a length of cable the preceding assumptions. N 1 N 2 N 3 satisfying

7 Titre : Un élément fini de câble-poulie Date : 09/12/2013 Page : 7/14 4 Internal forces of a finite element of cable-pulley Let us point out the following definition: one calls internal forces of a finite element of structure the forces which it is necessary to exert in its nodes to maintain it in its current deformed configuration. In the case of a finite element of cable-pulley, the internal forces result immediately from statics. One has indeed [Figure 4-a]: F 1 = N éq 4-1 F 2 = N l 2 l 2 and, to ensure balance: F 3 = F 1 F 2. F 1 and F 2 having even module, F 3 is directed according to the bisectrix of the angle N 1 N 3 N 2. F 3 F 1 F 2 N 3 N 1 N 2 F 1 F 1 F 2 Figure 4-a: Internal forces of an element of cable-pulley F 2 F 3 is applied toaxis pulley. The system F internal forces of the element is thus: 1 F={F F 3} 2. F

8 Titre : Un élément fini de câble-poulie Date : 09/12/2013 Page : 8/14 5 Matrix of rigidity The matrix of rigidity K element is derived from Fréchet of F in the direction of displacement u nodes: F=K u. K is calculated by the following classical formula, used intensively in [bib3], [bib4] and [bib5]: F=lim 0 F u u. éq 5-1 K 11 K=[ N I l 3 K 12 K 11 K 12 N I 1 l 3 1 ] T K 12 K 22 N I l 3 K 22 K T 12 N I 2 l 3 2 K 11 K T 12 N I l 3 K 22 K 12 N I 1 l 3 K 11 K 22 K 12 K T l 2 N I 3 éq 5-2 The detail of calculations is given in [ An1] and the final expression of K is the following one: N is given by [éq 3-4] and [éq 3-3]; K 11 = EA N l 0 K 12 = EA l l 0 l T 2 ; 2 K 22 = EA l 0 N l 2 1 l l l T ; 1 1 l l l T K is symmetrical, because of the symmetry of K 11 and of K 22 and of total symmetry per blocks. But K depends on displacements on N 1,N 2 et N 3 via, l 2 et N : the finite element of cablepulley is thus an element nonlinear.

9 Titre : Un élément fini de câble-poulie Date : 09/12/2013 Page : 9/14 6 Matrix of mass This matrix intervenes obviously only in the dynamic problems. The element of cable - pulley is not used in Code_Aster that for the quasi-static problems of installation of cables [ 7]. Note: One presents nevertheless in [bib2] the example of a dynamic problem comprising a cable-pulley. The matrix of mass of the element N 1 N 2 N 3 is obtained by assembling the coherent matrices of mass of the elements of cable with two nodes N 3 N 1 and N 3 N 2 [bib6] and by adding the specific mass of the pulley. It should be noted that, during a dynamic analysis, this matrix of mass must be updated because the lengths and l 2 vary. 7 Introduction of the element of cable-pulley into Code_Aster The element of cable-pulley is supported by a mesh SEG3. In the order AFFE_MODELE, under the keyword factor AFFE, one must define the arguments of the keywords as follows: keyword GROUP_MA PHENOMENON MODELING argument group of meshs MECHANICAL CABLE_POULIE of cable-pulley The constitutive material must be elastic. In the order AFFE_CARA_ELEM, the cable-pulleys are treated like cables. As the element of cable-pulley is nonlinear [ 5] and that, for the moment, it is used only in statics [ 6], it is accessible only by the operator STAT_NON_LINE. Under the keyword factor BEHAVIOR, the arguments of the keywords are the following: keyword GROUP_MA RELATION DEFORMATION argument group of meshs ELAS GROT_GDEP of cable-pulley Lastly, the force of gravity acting on the nodes N 1 et N 2 of an element of cable-pulley [fig 3 - B] is following because it depends on the length of the bits N 3 N 1 et N 3 N 2. For a structure comprising at least an element of cable-pulley, one must specify this load (following) in STAT_NON_LINE under the keyword factor EXCIT. One finds an example of application in test SSNL100A [V ]. Note: The element of cable-pulley does not make it possible to take into account deformation due to the temperature.

10 Titre : Un élément fini de câble-poulie Date : 09/12/2013 Page : 10/14 8 Example of application This example is that of the installation of a cable with two ranges and imposed tension of adjustment. On [Figure 7-a], O is the anchoring of the cable on the support of stop of left. P 1 is a first pulley placed at the bottom of the lifting chain P 1 C, fixed in C with the support of alignment. P 2 is one second pulley placed on the support of stop of right-hand side. O, P 1 and P 2 are, to simplify, located on horizontal. z O x Q 1 R 1 Q 2 P 2 R2 N p P 1 Q 1 ' C R 1 ' Q 2 ' R 2 ' Figure 7-a: Balance of a cable with two ranges, subjected to a tension of adjustment given. OP 1 = P 1 P 2 =100 m ; w=30 N /m ; E A=5 x10 7 N ; N p =5 000 N In initial position, the cable at rest, supposed in weightlessness, is right: line in feature of axis of [Figure 7-a]. For modeling in finite elements, this line is cut out in: ten elements of cable with two nodes enters O and Q 1 ; an element of cable-pulley Q 1 R 1 P 1 ; nine elements of cable enters R 1 and Q 2 ; an element of cable-pulley Q 2 R 2 P 2. One simultaneously subjects the cable to gravity and the tension of adjustment N p exerted in R 2. The position of balance (line in feature full with [Figure 7-a]) is reached into 11 iterations by the operator STAT_NON_LINE of Code_Aster. The arrows are of m and m, respectively in the range of left and that of right-hand side. For a cable inextensible of a range of 100 m, linear weight 30 N /m and subjected to a tension in the end of N, the theoretical arrow is of m [ An2]. While exploiting the tension of adjustment, one can adjust one of the two arrows to a value fixed in advance.

11 Titre : Un élément fini de câble-poulie Date : 09/12/2013 Page : 11/14 9 Conclusion The finite element of cable-pulley presented in this note, of a very simple mechanical formulation, has performances comparable to those of an element of ordinary cable. It is very convenient and even essential for a realistic modeling of the air electric lines. It should find other applications, in particular in Robotics. 10 Bibliography 1) J.L. LILIEN : Private communication. 2) Mr. AUFAURE : With finite element of cable passing through has pulley. Computers & Structures 46, (1993). 3) J.C. SIMO, L. CONSIDERING-QUOC : With three-dimensional finite-strain rod model. Share II: computational aspects. Comput. Meth. appl. Mech. Engng 58, (1986). 4) A.CARDONA, Mr. GERADIN : With beam finite element nonlinear theory with finite rotations. Int. J. Numer. Meth. Engng. 26, (1988). 5) A. CARDONA : Year integrated approach to mechanism analysis. Thesis, University of Liege, Belgium (1989). 6) Mr. AUFAURE : Modeling of the cables in Code_Aster, in the course of drafting. 7) H. MAX IRVINE : Cable Structures. MIT Close (1981). 11 Description of the versions of the document Author (S) Aster Organization (S) 3 MR. AUFAURE (EDF/IMA/MMN) 10,1 J.M.PROIX (EDF- R&D/AMA) Description of the modifications Initial text Modification of GREEN in GROT_GDEP

12 Titre : Un élément fini de câble-poulie Date : 09/12/2013 Page : 12/14 Annexe 1Calculation of the matrix of rigidity One shows here how applies the formula [éq 5-1] to the calculation of the first three lines of the matrix K [éq 5-2], those which are relating to the force F 1. The other lines are obtained either by permutation of the indices, or by summation of two preceding lines. The first lines are thus obtained by calculating the derivative: lim e 0 d de F 1 u ed u éq An1-1 and by putting in factor the vector u. According to the relations [éq 4-1], [éq 3-4] and [éq 3-3], one a: ={ F 1 u u EA [ u d u l 2 u u l i ] i} T T l i N u u 0 u u with, according to [éq 3-1]: and, according to [éq 3-2]: Consequently: u u =x 1 u 1 u 1 x 3 u 3 u 3 lim 0 u u = T u u u u. lim 0 and by permutation of indices 1 and 2: Finally: lim 0 lim 0 d d u u = u 1 u 3 d d u u = 1 l T u 1 u u 1 u 3 d d l 2 u u = 1 l 2 u l T 2 u u 2 u 3 d d u u = 1 3 u l T 1 u u 1 u 3 While carrying the preceding expressions in [éq An1-1] and by putting in factor the vector u, one easily obtains the first three lines of K.

13 Titre : Un élément fini de câble-poulie Date : 09/12/2013 Page : 13/14 Annexe 2Figure of balance of an inextensible cable weighing subjected to a tension given in the end Let us take a cable [An2-a Figure] whose end is fixed at the point O and of which the other end P is subjected to the tension N p. O and P are on same horizontal and distant ones of s. The linear weight is w. The arrow is sought S. z O (0, 0) x S P s, 0 N p An2-a figure: Heavy cable in balance One finds in [bib7], p 6, the following well-known formulas: figure of balance of the cable: z x = H w [ cosh w H s 2 x ws cosh 2H ] ; éq An2-1 tension: N x = H cosh w H s 2 x. éq An2-2 H is the horizontal, constant tension along the cable since the external force distributed - the weight - is vertical. H is calculated by [éq An2-2] written in P : cosh ws 2H 2N p ws ws 2H =0. ws 2H is thus root of the transcendent equation: cosh X = 2N p ws X.

14 Titre : Un élément fini de câble-poulie Date : 09/12/2013 Page : 14/14 This equation has two roots [An2-b Figure] if: with: and: 2N p ws p 0, p 0 =sinh x 0 x 0 =cotanh x 0 x 0 0. cosh X 1 2 N p w s w s 2 H X An2-b figure: Calculation of ws 2H The smallest root, which corresponds to the greatest tension of the cable, is only useful. The other root corresponds to a sag of the considerable cable, of about size of its range. ws 2H being calculated, the arrow results from [éq An2-1]: S= s 2 2H ws ws cosh 2H 1.