Quartic formulation of Coulomb 3D frictional contact

Transcription

1 INSTITUT NATIONAL DE RECHERCHE EN INFORMATIQUE ET EN AUTOMATIQUE Quartic formulation of Coulomb 3D frictional contact Olivier Bonnefon Gilles Daviet N 0400 January 10, 2011 Modeling, Optimization, and Control of Dynamic Systems apport technique ISSN ISRN INRIA/RT FR+ENG

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3 Quartic formulation of Coulomb 3D frictional contact Olivier Bonnefon, Gilles Daviet Theme : Modeling, Optimization, and Control of Dynamic Systems Applied Mathematics, Computation and Simulation Équipe-Projet BiPoP Rapport technique n 0400 January 10, pages Abstract: In this paper, we focus on the problem of a single contact with Coulomb friction, which we reduce to a root-finding problem on a degree 4 polynomial. This formulation give us the exact number of solutions, as well as their analytical form when they exist. Key-words: coulomb friction, mechanics, contact Centre de recherche INRIA Grenoble Rhône-Alpes 655, avenue de l Europe, Montbonnot Saint Ismier Téléphone : Télécopie

4 Formulation quartique du contact avec frottement de Coulomb tridimensionnel Résumé : Dans ce document, nous reformulons le problème unitaire1 contact frottant en la recherche de zéro d un polynôme du quatrième degré. Si elles existent, cette formulation donne la forme analytique des solutions. Mots-clés : frottement de coulomb, mécanique, contact

5 Quartic formulation of Coulomb 3D frictional contact 3 1 Introduction Numerical simulation of three dimensional frictionnal contact with impacts has been thoroughly studied in the last two decade[rad05],[bro00],[ab08],[cad09]. In this paper, we use the framework introduced by Moreau [Mor88] and we focus on the solution of the problem for a single contact, which we reduce to a root-finding problem on a degree 4 polynomial. The idea consists in doing an enumerative exploration of the Coulomb friction s cases. 1.1 The Coulomb friction model We first define the second order cone, K µ R 3, of parameter µ > 0 in the unitary direction e R 3 : K µ = {x R 3 : x x.e e µ x.e} 1 Where a.b is the usual dot product of two vectors. Coulomb s frictional law is a relation between the reaction force R R 3 and the relative velocity U R 3. In this case, e is a unitary normal to the surface at the contact point. The relation can be written as a disjonctive formulation: Take-off: R = 0 and U.e 0 Sticking: U = 0 and R intk µ Sliding: R K µ and U.e=0 and α > 0,R R.e e = αu We note Cµ,e the set of all U,R R 3 R 3 satisfying this law. 1.2 Discretised problem The discretization of the mechanical problem is deeply described in[mor88],[ab08]. In the case of a single contact, it consists in finding U R 3 and R R 3 subject to: U = MR+q, 3 U,R Cµ,e Where M R 3,3 a regular symetric positive matrix and q R 3. The problem is written in a orthonormal frame e,e t1,e t2. We distinguish between the tangential part and the normal part of R. The normal part is defined as R N := R.e, that is the first coordinate of R.The tangential part is defined as R T R 2, the coordinates of the vector R e.r in the frame e t1,e t Enumerative method for solving It consists in successively looking for a solution in each of the cases defined in Take-off case In this case, R equals to zero. The system 3 leads to U = q which is a solution if and only if q,0 Cµ,e. It consists in checking that e.q > 0.

6 Quartic formulation of Coulomb 3D frictional contact Sticking case Here, U must be equal to zero, and f M is invertible 3 leads to R = M 1 q which is a solution if and only if { R R 3 MR = q } intk µ. If M is invertible, this amounts to checking that M 1 q intk µ. 2.3 Sliding case When µ > 0, This is the only non-trivial case. The main contribution of this document is the reformulation of this problem as a quartic root-finding problem. In the two following sections, we derive formulations of such polynomials. 3 Quartic polynomial in tanθ 2 0 It consists in finding α > 0 and R K µ such that α RT That is : M R T is on a conic α α = MR + q. R+q = 0 4 The first line ofthe system 4 and the R K µ equation define the intersection between a plan and a cone in R 3 : That is: µr N = R T M 11 µ R T = q 1 M 12 R T1 M 13 R T2 5 µ 2 R 2 N = R2 T1 +R2 T1 M 2 11 µ 2 R 2 T1 +R2 T1 = q 1 M 12 R T1 M 13 R T2 2 6 This means that R T is contained into a conic, whose focus and directrice are: D : q 1 +M 12 R T1 +M 13 R T2 = 0 focus : O M 2 11 µ DistO,R 2 T 2 = DistD,R T 2 M12 2 +M13 2 DistO,R T DistD,R = µ M12 2 +M2 13 T M 11 = e 7 The parametric equation is: R T1 = rcosθ R T2 = rsinθ r = p 1+ecosθ φ 8 With p a simple expression of M 11,M 12,M 13, and φ a constant angle between D and O,R T1

7 Quartic formulation of Coulomb 3D frictional contact The last two lines of the system 4 R T α 0 M 1. + M + R µ 0 α T + q = 0 9 M issymetric,sothereexistsaunitarymatrixv suchthatv MV T d1 0 =. 0 d 2 One can get: Rename: R T V µ M 1. +V R T µ M + α 0 0 α M 1. + d1 +α 0 0 d 2 +α V T VR T +V q = 0 10 R T + q = 0 11 In the plan, V is either a rotation or a symmetry. So R T = VR T is a conic with the same focus and a rotated directrice, which means that there exists φ 1 such that : The equation 11 is : R T1 = rcosθ R T2 = rsinθ r = p 1+ecosθ φ 1 d 1 +α R T1 = q 1 +a 1 R T d 2 +α R T2 = q 2 +a 2 R T The case R T1 = 0 or R T2 = 0 has to be examined. We try to eliminate alpha: that leads to: d 1 RT1 RT2 +α R T1 RT2 = q 1 RT2 +a 1 RT2 R T d 2 RT1 RT2 +α R T1 RT2 = q 2 RT1 +a 2 RT1 R T 14 d 1 d 2 R T1 RT2 = q 1 RT2 + q 2 RT1 +a 1 RT2 a 2 RT1 R T 15 The parametric expression of R T leads to: d 1 d 2 r 2 cosθsinθ = q 1 rsinθ+ q 2 rcosθ+ra 1 rsinθ a 2 rcosθ ie: d 1 d 2 rcosθsinθ = q 1 sinθ+ q 2 cosθ+ra 1 sinθ a 2 cosθ 16 with the expression of r: p d 1 d 2 1+ecosθ φ cosθsinθ = 1 p q 1 sinθ+ q 2 cosθ+ 1+ecosθ φ a 1 1sinθ a 2 cosθ ie: d 1 d 2 pcosθsinθ = 1+ecosθ φ 1 q 1 sinθ+ q 2 cosθ+pa 1 sinθ a 2 cosθ ie: d 1 d 2 pcosθsinθ = pa 1 sinθ a 2 cosθ +1+ecosθcosφ 1 +sinθsinφ 1 q 1 sinθ+ q 2 cosθ ie: 0 = d 1 d 2 pcosθsinθ+p a 1 sinθ+a 2 cosθ +1+ecosθcosφ 1 +esinθsinφ 1 q 1 sinθ q 2 cosθ 17

8 Quartic formulation of Coulomb 3D frictional contact 6 rename: Acosθ 2 +Bsinθ 2 +Csinθcosθ+Dsinθ+Ecosθ = 0 18 with A = e q 2 cosφ 1 B = e q 1 sinφ 1 C = d 1 d 2 p+ecosφ 1 q 1 esinφ 1 q 2 D = q 1 pa 1 E = q 2 +pa 2 Performing the following change of variables: 19 leads to: A 1 t2 2 1+t +B 4t2 2 t = tanθ/2 sinθ = 2t 1+t 2 cosθ = 1 t2 1+t t +C 2t1 t2 2 1+t +D2t+E1 t 2 = 0 2 ie: A1 t Bt 2 +C2t1 t 2 +2Dt1+t 2 +E1 t 2 1+t 2 = 0 ie: P 4 = A E P 3 = 2C +2D P 2 = 4B 2A P 1 = 2C +2D P 0 = A+E 21 Finally, we get 4 possible values for R T, checking the sign of α and R N selects the solutions. 3.3 Case R T12 = 0 From 13, R T1 leads to: From 13, R T2 leads to: R T = R T2 = q1 a 1 R T = 0 ± q1 a 1 22 R T = R T1 = q2 a 2 ± q 2 23 R T = a 2 0 From R T, we have to check the coherence with the equation 12. If it is on the conic, we compute R, and the sign condition of the equation 4 must be checked. 4 Quartic polynomial in α Just like in previous formulation, we want to find α > 0 and R K µ such that 0 U N = 0 and α = MR+q. RT

9 Quartic formulation of Coulomb 3D frictional contact 7 That is : 0 = M α 0 R+q α R T = µr N Let s decompose M: M := m 0 m 01 m 02 m 01 m 02 If m 00 = 0, since M is positive-semidefinite, m 01 = m 02 = 0. Then : M T with if q N 0, there is no solution to the problem if q N = 0, we can chose an arbitrary α and take any solution of the linear system on R T then set R N = 1 µ R T. We suppose now that m 00 > 0. We can rewrite 24 as: 0 = [ m 00,m 01,m 02 R+q ] N α 0 0 = M T + R 0 α T + R T = µr N m01 m 02 R N +q T and then, substituting the first line into the second one : 0 = m 00,m 01,m 02 R+q N 0 = ΛαR T +w 25 R T = µr N Λα := λ1 +α λ 12 λ 12 λ 2 +α α 0 = M T + 0 α 1 m01 m m 00 m 01,m and To study Λα, we define: w := q T q N m01 m 00 m 02 L := m 00 Λ0 = m 00 M T m01 m 02 m 01,m 02 Since M is positive definite, we have i j m ij m ii m jj. That means that i = 0,1L ii = m 00 m i+1i+1 m 2 0i+1 0. We conclude that trl 0. Moreover, deriving the determinant of L, we remark that. detl = m 00 detm 0

10 Quartic formulation of Coulomb 3D frictional contact 8 L is therefore a 2 2 symmetric matrix with a positive trace and determinant, which means L is positive-semidefinite. So is Λ since m 00 > 0, and α > 0, Λα is positive-definite. We get 1 R T = det Λα λ2 +α λ 12 λ 12 λ 1 +α w 26 Now, let s return to equation 25 and combine its first line into the last one, then square the result: RT1 2 +R2 T2 = µ2 m 2 m 01 R T1 +m 02 R T2 +q N 2 00 Multiplying both sides by detλα 2 : 0 = detλαr T1 2 +detλαr T2 2 µ2 m 2 [m 01 detλαr T1 +m 02 detλαr T2 +detλαq N ] Now remember from 26 that detλαr Ti are linear functions in α, and detλα is a degree 2 polynomial in α. The expression in 27 is therefore a degree 4 polynomial a quartic in α, whose roots in R + are the solution of our frictionnal contact problem. 5 Numerical example We consider the following problem, which is derived from a??? system: M = e e and q = The formulation, using t = tanθ 2, described in the section 3, leads to the following polynomial: and µ = 0.6 Pt = t t t t The formulation, using α, described in 4, leads to the following polynomial: Pα = α α α α Both of thems yield the solution: R = However, we found that in this particular case we could reach the machine precision using the formulation based on t = tanθ 2 of the section 3, while the worse-conditioned polynomial of section 4 gave us an error of Precision measurements were done using the Alart-Curnier [AC91] function.

11 Quartic formulation of Coulomb 3D frictional contact 9 6 Conclusion We presented two distinct approaches for expressing the sliding case of Coulomb s frictional law as a degree 4 polynomial, from which we can deduce the analytic solutions. While in itself restricted to systems with a single contact, this method can be embedded as a local solver in a Gauss-Seidel like algorithm [JAJ98], allowing us to deal with real-world problems. These quartic formulations are implemented in the Siconos Platform [AP07], [AB08], and are used as well in the MECHE framework for simulating fibrous materials. References [AB08] V. Acary and B. Brogliato. Numerical Methods for Nonsmooth Dynamical Systems: Applications in Mechanics and Electronics, volume 35 of Lecture Notes in Applied and Computational Mechanics. Springer Verlag, [AC91] P. Alart and A. Curnier. A mixed formulation for frictional contact problems prone to Newton like solution methods. Comput. Methods Appl. Mech. Eng., 923: , [AP07] V. Acary and F. Pérignon. An introduction to Siconos. Technical Report TR-0340, INRIA, [Bro00] B. Brogliato, editor. Impact in Mechanical systems: Analysis and Modelling, volume 551 of Lecture Notes in Physics. Springer, [Cad09] F. Cadoux. Méthodes d optimisation pour la dynamique non-régulière. PhD thesis, Université Joseph Fourier, November [JAJ98] Franck Jourdan, Pierre Alart, and Michel Jean. A Gauss-Seidel like algorithm to solve frictional contact problems. Comput. Methods Appl. Mech. Eng., :31 47, [Mor88] J.-J. Moreau. Unilateral contact and dry friction in finite freedom dynamics. Nonsmooth mechanics and applications, CISM Courses Lect. 302, , [RAD05] M. Renouf, V. Acary, and G. Dumont. 3d frictional contact and impact multibody dynamics. a comparison of algorithms suitable for real-time applications. In ECCOMAS, 2005.

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