Quartic formulation of Coulomb 3D frictional contact Olivier Bonnefon, Gilles Daviet

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Quartic formulation of Coulomb 3D frictional contact

Olivier Bonnefon, Gilles Daviet

To cite this version:

Olivier Bonnefon, Gilles Daviet. Quartic formulation of Coulomb 3D frictional contact. [Technical

Report] RT-0400, INRIA. 2011. �inria-00553859�

a p p o r t

t e c h n i q u e

ISSN0249-0803ISRNINRIA/RT--0400--FR+ENG

Modeling, Optimization, and Control of Dynamic Systems

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

Centre de recherche INRIA Grenoble – Rhône-Alpes 655, avenue de l’Europe, 38334 Montbonnot Saint Ismier

Téléphone : +33 4 76 61 52 00 — Télécopie +33 4 76 61 52 52

Quartic formulation of Coulomb 3D frictional contact

Olivier Bonnefon, Gilles Daviet

Theme : Modeling, Optimization, and Control of Dynamic Systems Applied Mathematics, Computation and Simulation

Equipe-Projet BiPoP´

Rapport technique n°0400 — January 10, 2011 — 9 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

Formulation quartique du contact avec frottement de Coulomb tridimensionnel

R´esum´e : Dans ce document, nous reformulons le probl`eme unitaire (1 contact frottant) en la recherche de z´ero d’un polynˆome du quatri`eme degr´e. Si elles existent, cette formulation donne la forme analytique des solutions.

Mots-cl´es : frottement de coulomb, m´ecanique, contact

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 fo- cus 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 fric- tion’s cases.

1.1 The Coulomb friction model

We first define the second order cone, Kµ ⊂ R³, of parameter µ > 0 in the unitary directione∈R³:

Kµ={x∈R³:kx−x.e ek≤µ x.e} (1) Wherea.bis the usual dot product of two vectors.

Coulomb’s frictional law is a relation between the reaction force R ∈ R³ and the relative velocityU ∈R³. In this case, eis 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 andR∈intKµ

– Sliding: R∈∂Kµ and U.e=0 and∃α >0, R− R.e e=−αU (2)

We noteC(µ, e) the set of all (U, R)∈R³×R³satisfying 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 findingU ∈R³andR∈R³subject

to: U =M R+q,

(U, R)∈C(µ, e) (3)

WhereM ∈R^3,3a regular symetric positive matrix andq∈R³. The problem is written in a orthonormal frame (e, et1, et2). We distinguish between the tangen- tial part and the normal part of R. The normal part is defined as RN :=R.e, that is the first coordinate of R.The tangential part is defined asRT ∈R², the coordinates of the vectorR−e.Rin the frame (et1, et2).

2 Enumerative method for solving

It consists in successively looking for a solution in each of the cases defined in (2).

2.1 Take-off case

In this case,Requals to zero. The system (3) leads toU =qwhich is a solution if and only if (q,0)∈C(µ, e) . It consists in checking thate.q >0.

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Quartic formulation of Coulomb 3D frictional contact 4

2.2 Sticking case

Here,U must be equal to zero, and f M is invertible (3) leads toR =−M⁻¹q which is a solution if and only if∅ 6=

R∈R³|M R=−q ∩intKµ. (IfM is invertible, this amounts to checking that −M⁻¹q∈intKµ.)

2.3 Sliding case

Whenµ > 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

It consists in finding α >0 and R ∈∂Kµ such that−α 0

RT

=M R+q.

That is : 

 M+





0 0 0

0 α 0

0 0 α







R+q= 0 (4)

3.1 R

is on a conic

The first line of the system (4) and theR∈∂Kµequation define the intersection between a plan and a cone in R³:

µRN =kRT k

M11

µ kRT k=−q1−M12RT1−M13RT2 (5) That is:

µ²R²_N = (R²_T1+R²_T1)

M11²

µ² (R²_T1+R²_T1) = (−q1−M12RT1−M13RT2)² (6) This means that RT is contained into a conic, whose focus and directrice are:

D:q1+M12RT1+M13RT2= 0 f ocus:O

M11²

µ² Dist(O, RT)²=Dist(D, RT)²(M₁₂² +M₁₃²)

Dist(O,RT) Dist(D,R_T) =^µ

√(M12²+M13²) M11 =e

(7)

The parametric equation is:

RT1=rcos(θ) RT2=rsin(θ) r= _1+ecos(θ^p _−φ)

(8)

With pa simple expression of M11, M12, M13, andφa constant angle between Dand (O, RT1)

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Quartic formulation of Coulomb 3D frictional contact 5

3.2 The last two lines of the system (4)

kRT k µ M˜1.+

M˜ +

α 0

0 α

RT + ˜q= 0 (9) M˜ is symetric, so there exists a unitary matrixV such thatVM V˜ ^T =

d1 0 0 d2

. One can get:

kRT k

µ VM˜1.+V

M˜ +

α 0 0 α

V^TV RT +Vq˜= 0 (10) Rename:

kR¯T k µ M¯1.+

d1+α 0 0 d2+α

RT + ¯q= 0 (11) In the plan,V is either a rotation or a symmetry. So ¯RT =V RT is a conic with the same focus and a rotated directrice, which means that there existsφ1 such that :

R¯T1=rcos(θ) R¯T2=rsin(θ) r= _1+ecos(θ^p _−φ1)

(12) The equation (11) is :

(d1+α) ¯RT1 =−q¯1+a1kRT k

(d2+α) ¯RT2 =−q¯2+a2kRT k (13) The case ( ¯RT1= 0 or ¯RT2= 0) has to be examined. We try to eliminatealpha:

d1R¯T1R¯T2+αR¯T1R¯T2=−q¯1R¯T2+a1R¯T2kRT k

d2R¯T1R¯T2+αR¯T1R¯T2=−q¯2R¯T1+a2R¯T1kRT k (14) that leads to:

(d1−d2) ¯RT1R¯T2=−q¯1R¯T2+ ¯q2R¯T1+ (a1R¯T2−a2R¯T1)kRT k (15) The parametric expression of ¯RT leads to:

(d1−d2)r²cos(θ)sin(θ) =−q¯1rsin(θ) + ¯q2rcos(θ) +r(a1rsin(θ)−a2rcos(θ)) ie: (d1−d2)rcos(θ)sin(θ) =−q¯1sin(θ) + ¯q2cos(θ) +r(a1sin(θ)−a2cos(θ))

(16) with the expression of r:

(d1−d2)_1+ecos(θ^p _−φ1)cos(θ)sin(θ) =

−q¯1sin(θ) + ¯q2cos(θ) +_1+ecos(θ^p _−φ1)(a1sin(θ)−a2cos(θ)) ie: (d1−d2)pcos(θ)sin(θ) =

(1 +ecos(θ−φ1))(−q¯1sin(θ) + ¯q2cos(θ)) +p(a1sin(θ)−a2cos(θ)) ie: (d1−d2)pcos(θ)sin(θ) =p(a1sin(θ)−a2cos(θ))

+(1 +e(cos(θ)cos(φ1) +sin(θ)sin(φ1)))(−q¯1sin(θ) + ¯q2cos(θ)) ie: 0 = (d1−d2)pcos(θ)sin(θ) +p(−a1sin(θ) +a2cos(θ))

+(1 +ecos(θ)cos(φ1) +esin(θ)sin(φ1))(¯q1sin(θ)−q¯2cos(θ)) (17)

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Quartic formulation of Coulomb 3D frictional contact 6

rename:

Acos(θ)²+Bsin(θ)²+Csin(θ)cos(θ) +Dsin(θ) +Ecos(θ) = 0 (18) with A=−e¯q2cos(φ1)

B=e¯q1sin(φ1)

C= (d1−d2)p+ecos(φ1)¯q1−esin(φ1)¯q2

D= ¯q1−pa1

E=−q¯2+pa2

(19)

Performing the following change of variables:

t=tan(θ/2) sin(θ) =_1+t^2t2

cos(θ) = ¹⁻_1+t^t22

(20)

leads to:

A⁽¹⁻_1+t^t22⁾² +B_1+t^4t22 +C^2t(1−_1+t2^t2)+D2t+E(1−t²) = 0

ie: A(1−t²)²+ 4Bt²+C2t(1−t²) + 2Dt(1 +t²) +E(1−t²)(1 +t²) = 0 ie: P4=A−E P3=−2C+ 2D P2= 4B−2A

P1= 2C+ 2D P0=A+E

(21) Finally, we get 4 possible values forRT, checking the sign ofαandRN selects the solutions.

3.3 Case R

= 0

From (13),RT1 leads to:

kRT k=|R¯T2|=_a^q¯1₁ R¯T =

0

±_a^q¯11

(22)

From (13),RT2leads to:

kRT k=|R¯T1|=_a^q¯2₂ R¯T =

±a^q¯22

0

(23)

From ¯RT, 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 findα >0 andR∈∂Kµsuch that UN = 0 and−α

0 RT

=M R+q.

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Quartic formulation of Coulomb 3D frictional contact 7

That is :











0 =



 M+





0 0 0

0 α 0

0 0 α







R+q kRT k = µRN

(24)

Let’s decomposeM:

M :=





m0 m01 m02

m01 MT

m02





Ifm00= 0, sinceM is positive-semidefinite,m01=m02= 0. Then :

• ifqN 6= 0, there is no solution to the problem

• ifqN = 0, we can chose an arbitraryαand take any solution of the linear system onRT then setRN =_µ¹ kRT k.

We suppose now thatm00>0. We can rewrite (24) as:











0 = (m00, m01, m02)·R+qN

0 =

MT+

α 0 0 α

RT+

m01

m02

RN+qT

kRT k = µRN

and then, substituting the first line into the second one :







0 = (m00, m01, m02)·R+qN

0 = Λ(α)RT +w kRT k = µRN

(25)

with Λ(α) :=

λ1+α λ12

λ12 λ2+α

=MT +

α 0 0 α

− 1 m00

m01

m02

(m01, m02) and

w:=qT − qN

m00

m01

m02

To study Λ(α), we define:

L:=m00Λ(0) =m00MT − m01

m02

(m01, m02)

SinceM is positive definite, we have∀i6=j|mij| ≤ √miimjj. That means that ∀i= 0,1Lii=m00m(i+1)(i+1)−m²_0(i+1)≥0. We conclude thattrL≥0.

Moreover, deriving the determinant ofL, we remark that detL=m00detM ≥0

.

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Quartic formulation of Coulomb 3D frictional contact 8

Lis therefore a 2×2 symmetric matrix with a positive trace and determinant, which means L is positive-semidefinite. So is Λ (sincem00 >0), and ∀α >0, Λ(α) is positive-definite.

We get

RT =− 1 det Λ(α)

λ2+α −λ12

−λ12 λ1+α

w (26)

Now, let’s return to equation (25) and combine its first line into the last one, then square the result:

R²_T1+R²_T2= µ²

m²₀₀(m01RT1+m02RT2+qN)² Multiplying both sides by (det Λ(α))²:

0 = (det Λ(α)RT1)²+ (det Λ(α)RT2)²

− µ²

m²₀₀[m01det Λ(α)RT1+m02det Λ(α)RT2+ det Λ(α)qN]² (27) Now remember from (26) that det Λ(α)RT i are linear functions in α, and det Λ(α) is a degree 2 polynomial inα.

The expression in (27) is therefore a degree 4 polynomial (a quartic) inα, 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 =





0.01344 −9.421e−07 0.001486

−9.421e−07 0.1061 0.0001733 0.001486 0.0001733 0.001442



 andq=





−0.1458

−0.2484

−0.1515



 andµ= 0.6 The formulation, usingt = ^tan(θ)₂ , described in the section 3, leads to the

following polynomial:

P(t) =−1.349578×10⁻¹t⁴−1.879711t³−6.600166×10⁻²t²+8.851747×10⁻¹t+1.349783×10⁻¹ The formulation, usingα, described in 4, leads to the following polynomial:

P(α) =−7.65275α⁴−1.62238α³−7.13689×10⁻²α²+7.43684×10⁻⁴α+3.70595×10⁻⁵ Both of thems yield the solution:

R=





10.2059 1.93189 5.8108





However, we found that in this particular case we could reach the machine precision using the formulation based on t= ^tan(θ)₂ of the section 3, while the worse-conditioned polynomial of section 4 gave us an error of 10⁻⁸. Precision measurements were done using the Alart-Curnier [AC91] function.

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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], al- lowing 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 Dy- namical Systems: Applications in Mechanics and Electronics, vol- ume 35 of Lecture Notes in Applied and Computational Mechanics.

Springer Verlag, 2008.

[AC91] P. Alart and A. Curnier. A mixed formulation for frictional contact problems prone to Newton like solution methods. Comput. Methods Appl. Mech. Eng., 92(3):353–375, 1991.

[AP07] V. Acary and F. P´erignon. An introduction to Siconos. Technical Re- port TR-0340, INRIA, http://hal.inria.fr/inria-00162911/en/, 2007.

[Bro00] B. Brogliato, editor. Impact in Mechanical systems: Analysis and Modelling, volume 551 ofLecture Notes in Physics. Springer, 2000.

[Cad09] F. Cadoux.M´ethodes d’optimisation pour la dynamique non-r´eguli`ere.

PhD thesis, Universit´e Joseph Fourier, November 2009.

[JAJ98] Franck Jourdan, Pierre Alart, and Michel Jean. A Gauss-Seidel like algorithm to solve frictional contact problems.Comput. Methods Appl.

Mech. Eng., 155(1-2):31–47, 1998.

[Mor88] J.-J. Moreau. Unilateral contact and dry friction in finite freedom dynamics. Nonsmooth mechanics and applications, CISM Courses Lect. 302, 1-82 (1988)., 1988.

[RAD05] M. Renouf, V. Acary, and G. Dumont. 3d frictional contact and impact multibody dynamics. a comparison of algorithms suitable for real-time applications. InECCOMAS, 2005.

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