Contact Mechanics and Elements of Tribology

Transcription

1 Contact Mechanics and Elements of Tribology Lectures 2-3. Mechanical Contact Vladislav A. Yastrebov MINES ParisTech, PSL Research University, Centre des Matériaux, CNRS UMR 7633, Evry, Centre des Matériaux January 24, 206

2 Outline Lecture 2 Balance equations 2 Intuitive notions 3 Formalization of frictionless contact 4 Evidence friction 5 Contact types 6 Analogy with boundary conditions Lecture 3 Flamant, Boussinesq, Cerruti 2 Displacements and tractions 3 Classical elastic problems

3 Boundary value problem in elasticity V.A. Yastrebov 3/45 Reference and current configurations x = X + u Balance equation (strong form) f σ + ρf = 0, x Ω i = v Displacement compatibility u c 2 c ε = = ( u + u ) 2 2 u f Constitutive equation σ = = W (ε) = Boundary conditions Two bodies in contact Dirichlet: u = u 0, x Γ u Neumann: n σ = = t 0, x Γ f

4 Boundary value problem in elasticity V.A. Yastrebov 4/45 Reference and current configurations x = X + u Balance equation (strong form) f σ + ρf = 0, x Ω i = v Displacement compatibility u c 2 c ε = = ( u + u ) 2 2 u f Constitutive equation σ = = W (ε) = Boundary conditions Dirichlet: u = u 0, x Γ u Two bodies in contact Include contact conditions... Neumann: n σ = = t 0, x Γ f

5 Intuitive conditions V.A. Yastrebov 5/45 No penetration Ω (t) Ω 2 (t) = f 2 No adhesion n σ = n 0, x Γ i c u c 2 c 3 No shear stress 2 u f n σ = (I n n) = 0, x Γ i c Two bodies in contact Intuitive contact conditions for frictionless and nonadhesive contact

6 Intuitive conditions V.A. Yastrebov 6/45 No penetration Ω (t) Ω 2 (t) = f 2 No adhesion n σ = n 0, x Γ i c u c 2 c 3 No shear stress 2 u f n σ = (I n n) = 0, x Γ i c Two bodies in contact Intuitive contact conditions for frictionless and nonadhesive contact

7 Gap function Gap function g gap = penetration asymmetric function defined for separation g > 0 contact g = 0 penetration g < 0 governs normal contact Master and slave split Gap function is determined for all slave points with respect to the master surface g>0 non-contact n n n g<0 penetration g=0 contact Gap between a slave point and a master surface

8 Gap function Gap function g gap = penetration asymmetric function defined for separation g > 0 contact g = 0 penetration g < 0 governs normal contact Master and slave split Gap function is determined for all slave points with respect to the master surface Normal gap g n = n [r s ρ(ξ π ) ], n is a unit normal vector, r s slave point, ρ(ξ π ) projection point at master surface g>0 non-contact n n n g<0 penetration g=0 contact Gap between a slave point and a master surface ( n ( r s Definition of the normal gap

9 Frictionless or normal contact conditions V.A. Yastrebov 9/45 No penetration Always non-negative gap g 0 No adhesion Always non-positive contact pressure σ n 0 Complementary condition Either zero gap and non-zero pressure, or non-zero gap and zero pressure g σ n = 0 No shear transfer (automatically) σ t = 0 σ n = (σ n) n = σ : (n n) = = σ t = σ n σ = n n = n σ (I n n ) = = n 0 Scheme explaining normal contact conditions g

10 Frictionless or normal contact conditions V.A. Yastrebov 0/45 No penetration Always non-negative gap g 0 No adhesion Always non-positive contact pressure σ n 0 Complementary condition Either zero gap and non-zero pressure, or non-zero gap and zero pressure g σ n = 0 No shear transfer (automatically) σ t = 0 σ n = (σ n) n = σ : (n n) = = σ t = σ n σ = n n = n σ (I n n ) = = n 0 contact g = 0, n < 0 non-contact g > 0, n = 0 restricted regions Improved scheme explaining normal contact conditions g

11 Frictionless or normal contact conditions V.A. Yastrebov /45 In mechanics: Normal contact conditions Frictionless contact conditions Hertz -Signorini [2] conditions Hertz -Signorini [2] -Moreau [3] conditions also known in optimization theory as Karush [4] -Kuhn [5] -Tucker [6] conditions contact g = 0, n < 0 g 0, σ n 0, gσ n = 0 n 0 non-contact g > 0, n = 0 restricted regions Improved scheme explaining normal contact conditions g Heinrich Rudolf Hertz ( ) a German physicist who first formulated and solved the frictionless contact problem between elastic ellipsoidal bodies. 2 Antonio Signorini ( ) an Italian mathematical physicist who gave a general and rigorous mathematical formulation of contact constraints. 3 Jean Jacques Moreau (923) a French mathematician who formulated a non-convex optimization problem based on these conditions and introduced pseudo-potentials in contact mechanics. 4 William Karush (97 997), 5 Harold William Kuhn (925) American mathematicians, 6 Albert William Tucker ( ) a Canadian mathematician.

12 Contact problem V.A. Yastrebov 2/45 Problem Find such contact pressure p = n σ n 0 = f which being applied at Γ c and Γ 2 c results in x = x 2, x Γ c, x 2 Γ 2 c and evidently Ω (t) Ω 2 (t) = u 2 u c 2 c f Unfortunately, we do not know Γ c in advance, it is also an unknown of the problem. Related problem Suppose that we know p on Γ c Then what is the corresponding displacement field u in Ω i? Two bodies in contact

13 Contact problem V.A. Yastrebov 3/45 Problem Find such contact pressure p = n σ n 0 = f which being applied at Γ c and Γ 2 c results in x = x 2, x Γ c, x 2 Γ 2 c and evidently Ω (t) Ω 2 (t) = u 2 u c 2 c f Unfortunately, we do not know Γ c in advance, it is also an unknown of the problem. Related problem Suppose that we know p on Γ c Then what is the corresponding displacement field u in Ω i? Recall Flamant problem Two bodies in contact

14 Evidence of friction V.A. Yastrebov 4/45 Existence of frictional resistance is evident Independence of the nominal contact area Think about adhesion and introduce a threshold in the interface τ c Globally: - stick: T < T c (N) - slip: T = T c (N) From experiments: - Threshold T c N - Friction coefficient f = T c /N Locally - stick: σ τ < τ c (σ n ) Rectangle on a flat surface - slip: σ τ = f σ n

15 Evidence of friction Existence of frictional resistance is evident Independence of the nominal contact area Think about adhesion and introduce a threshold in the interface τ c Globally: - stick: T < T c (N) - slip: T = T c (N) From experiments: - Threshold T c N - Friction coefficient f = T c /N Locally - stick: σ τ < τ c (σ n ) Rectangle on a flat surface - slip: σ τ = f σ n Torque V.A. Yastrebov 5/45

16 Types of contact V.A. Yastrebov 6/45 Known contact zone conformal geometry flat-to-flat, cylinder in a hole initially non-conformal geometry but huge pressure resulting in full contact Unknown contact zone general case Point and line contact Frictionless conservative, energy minimization problem Frictional path-dependent solution, from the first touch to the current moment Example

17 Types of contact V.A. Yastrebov 7/45 Known contact zone conformal geometry flat-to-flat, cylinder in a hole initially non-conformal geometry but huge pressure resulting in full contact Unknown contact zone general case Point and line contact Frictionless conservative, energy minimization problem Frictional path-dependent solution, from the first touch to the current moment Example

18 Types of contact V.A. Yastrebov 8/45 Known contact zone conformal geometry flat-to-flat, cylinder in a hole initially non-conformal geometry but huge pressure resulting in full contact Unknown contact zone general case Point and line contact Frictionless conservative, energy minimization problem Frictional path-dependent solution, from the first touch to the current moment Example

19 Analogy with boundary conditions V.A. Yastrebov 9/45 Flat geometry Compression of a cylinder Frictionless u z = u 0 Full stick conditions u = u 0 e z Rigid flat indenter u z = u 0

20 Analogy with boundary conditions V.A. Yastrebov 20/45 Flat geometry Compression of a cylinder Frictionless u z = u 0 Full stick conditions u = u 0 e z Rigid flat indenter u z = u 0 Curved geometry Polar/spherical coordinates u r = u 0 If frictionless contact on rigid surface y = f (x) is retained by high pressure (X + u) e y = f ((X + u) e x )

21 Analogy with boundary conditions V.A. Yastrebov 2/45 Flat geometry Compression of a cylinder Frictionless u z = u 0 Full stick conditions u = u 0 e z Rigid flat indenter u z = u 0 Curved geometry Polar/spherical coordinates u r = u 0 If frictionless contact on rigid surface y = f (x) is retained by high pressure (X + u) e y = f ((X + u) e x ) Transition to finite friction From full stick, decrease f by keeping u z = 0 and by replacing in-plane Dirichlet BC by in-plane Neumann BC

22 Analogy with boundary conditions II V.A. Yastrebov 22/45 In general Type I: prescribed tractions p(x, y), τ x (x, y), τ y (x, y) Type II: prescribed displacements u(x, y) Type III: tractions and displacements u z (x, y), τ x (x, y), τ y (x, y) or p(x, y), u x (x, y), u y (x, y) Type IV: displacements and relation between tractions u z (x, y), τ x (x, y) = ±fp(x, y)

23 To be continued...

24 Concentrated forces Normal force: in-plane stresses and displacements (plane strain) σ r = 2N π cos(θ) r or σ x = 2N π x 2 y (x 2 +y 2 ) 2, σ y = 2N π y 3 (x 2 +y 2 ) 2, σ xy = 2N π u r = + ν N cos(θ) [2( ν) ln(r) ( 2ν)θ tan(θ)] + C cos(θ) πe xy 2 (x 2 +y 2 ) 2 u θ = + ν N sin(θ) [2( ν) ln(r) 2ν + ( 2ν)( 2θ ctan(θ))] C sin(θ) πe Tangential force σ r = 2T π sin(θ) r or σ x = 2T π x 3 (x 2 +y 2 ) 2, σ y = 2T π xy 2 (x 2 +y 2 ) 2, σ xy = 2T π x 2 y (x 2 +y 2 ) 2 u r = + ν T sin(θ) [2( ν) ln(r) ( 2ν)θctan(θ)] C sin(θ) πe u θ = + ν T cos(θ) [2( ν) ln(r) 2ν + ( 2ν)( + 2θ tan(θ))] + C cos(θ) πe

25 Distributed load Distributed tractions p(x)dx = dn(x), τ(x)dx = dt(x) Use superposition principle for the stress state and for displacements Tractions on the surface σ x (x, y) = 2y π p(s)(x s) 2 ds ((x s) 2 + y 2 ) 2 2 π τ(s)(x s) 3 ds ((x s) 2 + y 2 ) 2 σ y (x, y) = 2y3 π σ xy (x, y) = 2y2 π p(s) ds a ((x s) 2 + y 2 ) 2y2 2 π p(s)(x s) ds ((x s) 2 + y 2 ) 2y 2 π τ(s)(x s) ds ((x s) 2 + y 2 ) 2 τ(s)(x s) 2 ds ((x s) 2 + y 2 ) 2

26 Distributed load Distributed tractions p(x)dx = dn(x), τ(x)dx = dt(x) Use superposition principle for the stress state and for displacements Tractions on the surface σ x (x, y) = 2y π p(s)(x s) 2 ds ((x s) 2 + y 2 ) 2 2 π τ(s)(x s) 3 ds ((x s) 2 + y 2 ) 2 σ y (x, y) = 2y3 π σ xy (x, y) = 2y2 π p(s) ds a ((x s) 2 + y 2 ) 2y2 2 π p(s)(x s) ds ((x s) 2 + y 2 ) 2y 2 π τ(s)(x s) ds ((x s) 2 + y 2 ) 2 τ(s)(x s) 2 ds ((x s) 2 + y 2 ) 2

27 Distributed load Distributed tractions p(x)dx = dn(x), τ(x)dx = dt(x) Use superposition principle for the stress state and for displacements Consider displacements on the surface ( 2ν)( + ν) u x (x, 0) = 2E x p(s) ds x p(s) ds 2( ν2 ) πe Tractions on the surface τ(s) ln x s ds+c

28 Distributed load Distributed tractions p(x)dx = dn(x), τ(x)dx = dt(x) Use superposition principle for the stress state and for displacements Consider displacements on the surface Or rather their derivatives along the surface ( 2ν)( + ν) u x (x, 0) = 2E u x,x (x, 0) = x Near-surface stress state p(s) ds x p(s) ds 2( ν2 ) πe ( 2ν)( + ν) p(x) 2( ν2 ) E πe Tractions on the surface τ(s) x s ds τ(s) ln x s ds+c

29 Distributed load Distributed tractions p(x)dx = dn(x), τ(x)dx = dt(x) Use superposition principle for the stress state and for displacements Consider displacements on the surface Or rather their derivatives along the surface u y (x, 0) = ( 2ν)( + ν) 2E x τ(s) ds x τ(s) ds 2( ν2 ) πe Tractions on the surface p(s) ln x s ds+c 2

30 Distributed load Distributed tractions p(x)dx = dn(x), τ(x)dx = dt(x) Use superposition principle for the stress state and for displacements Consider displacements on the surface Or rather their derivatives along the surface u y (x, 0) = ( 2ν)( + ν) 2E u y,x (x, 0) = x τ(s) ds x τ(s) ds 2( ν2 ) πe ( 2ν)( + ν) τ(x) 2( ν2 ) E πe Tractions on the surface p(s) ln x s ds+c 2 p(s) x s ds

31 Rigid stamp problem V.A. Yastrebov 3/45 Link displacement derivatives with tractions τ(s) π( 2ν) ds = x s 2( ν) p(x) πe 2( ν 2 ) u x,x(x, 0) p(s) π( 2ν) ds = x s 2( ν) τ(x) πe 2( ν 2 ) u y,x(x, 0) If in contact interface we can prescribe p, u x,x or τ, u y,x, then the problem reduces to The general solution (case a = b): F (x) ds = U(x) x s F (x) = π 2 a 2 x 2 a a2 x 2 U(s) ds C a + x s π a 2 x, C = F (s)ds 2 a

32 Rigid stamp problem V.A. Yastrebov 32/45 Link displacement derivatives with tractions τ(s) π( 2ν) ds = x s 2( ν) p(x) πe 2( ν 2 ) u x,x(x, 0) p(s) π( 2ν) ds = x s 2( ν) τ(x) πe 2( ν 2 ) u y,x(x, 0) If in contact interface we can prescribe p, u x,x or τ, u y,x, then the problem reduces to The general solution (case a = b): F (x) ds = U(x) x s F (x) = π 2 a 2 x 2 a a2 x 2 U(s) ds C a + x s π a 2 x, C = F (s)ds 2 a flat frictionless punch, consider P.V.

33 Three-dimensional problem V.A. Yastrebov 33/45 Analogy to Flamant s problem Potential functions of Boussinesq Boussinesq problem concentrated normal force Cerruti problem concentrated tangential force Displacements decay as r u r (x, y, 0) = 2ν 4πG u z (x, y, 0) = ν 4πG Stress decay as r 2 N x2 + y 2 N x2 + y 2 Superposition principle

34 Why is the sky dark at night? V.A. Yastrebov 34/45

35 Why is the sky dark at night? V.A. Yastrebov 35/45 Olbers paradox or dark night sky paradox Two nominally-flat elastic half-spaces in contact At small scale they are rough with asperity density D Vertical displacement decay u z /r At every asperity, force F Sum up displacements induced by all forces u z 2π R 0 r 0 F r r dr dφ R

36 Classical contact problems Various problems with rigid flat stamps: circular, elliptic, frictionless, full-stick, finite friction Hertz theory normal frictionless contact of elastic solids E i, ν i and z i = A i x 2 + B i y 2 + C i xy, i =, 2 Wedges (coin) and cones Circular inclusion in a conforming hole Steuermann, 939,Goodman, Keer, 965 Frictional indentation z x n Incremental approach Mossakovski, 954 self-similar solution Spence, 968, 975 Adhesive contact Johnson et al, 97, 976 Contact with layered materials (coatings) Elastic-plastic and viscoelastic materials Sliding/rolling of non-conforming bodies K.L. Johnson I.G. Goryacheva Cattaneo, 938,Mindlin, 949,Galin, 953,Goryacheva, 998 Note: u r ( 2ν)/G, so if ( 2ν )/G = ( 2ν 2 )/G 2 tangential tractions do not change normal ones V.L. Popov V.A. Yastrebov 36/45

37 Hertzian contact No friction, no adhesion Two elastic materials E, ν, E 2, ν 2 Effective elastic modulus E = ν2 + ν2 2 E E 2 Two parabolic surfaces z = ax 2 + by2, z 2 = cx dy2 2 Solids of revolution or cylinders R,R 2, effective curvature radius: R = + R R 2 Displacement and contact radius (half-length): δ = a 2 /R Contact pressure: p(r) = p 0 r2 a 2, r < a Original paper by Henrich Hertz On the contact of elastic solids (ENG trans.) (6 pages) His theory, worked out during the Christmas vacation 880 at the age of 23(!), aroused considerable interest... K.L. Johnson V.A. Yastrebov 37/45

38 Hertzian contact V.A. Yastrebov 38/45 No friction, no adhesion Two elastic materials E, ν, E 2, ν 2 Effective elastic modulus E = ν2 + ν2 2 E E 2 Two parabolic surfaces z = ax 2 + by2, z 2 = cx dy2 2 Solids of revolution or cylinders R,R 2, effective curvature radius: R = + R R 2 Displacement and contact radius (half-length): δ = a 2 /R Contact pressure: p(r) = p 0 r2 a 2, r < a Geometries resolved in the framework of Hertz theory

39 Hertzian contact V.A. Yastrebov 39/45 No friction, no adhesion Two elastic materials E, ν, E 2, ν 2 Effective elastic modulus E = ν2 + ν2 2 E E 2 Two parabolic surfaces z = ax 2 + by2, z 2 = cx dy2 2 Solids of revolution or cylinders R,R 2, effective curvature radius: R = + R R 2 Displacement and contact radius (half-length): δ = a 2 /R Contact pressure: p(r) = p 0 r2 a 2, r < a Line contact (cylinders): ( 4PR ) /2 a = πe p 0 = 2P πa Solids of revolution: ( 3PR ) /3 a = 4E p 0 = 3P 2πa 2

40 Westergaard solution V.A. Yastrebov 40/45 No friction, no adhesion Two elastic materials E, ν, E 2, ν 2 E Interface profile y = ( cos(2πx/λ) Full contact pressure: p = πe /λ Contact length: 2a/λ = (2/π) arcsin( p/p ), where p is the applied pressure. Contact pressure distribution: 2 p cos(πx/λ) p(x, a) = sin 2 (πa/λ) sin 2 (πa/λ sin 2 (πx/λ) Illustration from K.L. Johnson (985) [] Westergaard, ASME J Appl Mech (939)

41 Westergaard solution: 3D extension V.A. Yastrebov 4/45 Surface z = (2 cos(2πx/λ) cos(2πy/λ) Full contact pressure: p = 2πE /λ Contact area for a λ: ( ) πa 2 2/3 3 p = π λ 2 8πp Non-contact area for b λ: πb 2 = 3 ( pp ) λ 2 2π Bi-wavy surface [2] [] Johnson, Greenwood, Higginson, Int J Mech Sci (985) [2] Yastrebov, Anciaux, Molinari, Tribol Lett 56 (204)

42 Westergaard solution: 3D extension Surface z = (2 cos(2πx/λ) cos(2πy/λ) Full contact pressure: p = 2πE /λ Contact area for a λ:!2/3 3p πa2 = π λ2 8πp Non-contact area for b λ:! p πb2 3 = λ2 2π p Contact area evolution[2] [] Johnson, Greenwood, Higginson, Int J Mech Sci (985) [2] Yastrebov, Anciaux, Molinari, Tribol Lett 56 (204) V.A. Yastrebov 42/45

43 Westergaard solution: 3D extension Surface z = (2 cos(2πx/λ) cos(2πy/λ) Full contact pressure: p = 2πE /λ Contact area for a λ: ( ) πa 2 2/3 3 p = π λ 2 8πp Non-contact area for b λ: πb 2 = 3 ( pp ) λ 2 2π Contact area fraction, A' (a) (b) Eq. (2) Eq. (3) Numerical results [Krithivasan & Jackson, 2007] Experimental results [Johnson et al., 985] Numerical results [Johnson et al., 985] Approximated curve [Johnson et al., 985] Numerical results [Johnson et al., 985] Current numerical results 0.8 Eq. (3) [] Johnson, Greenwood, Higginson, Int J Mech Sci (985) [2] Yastrebov, Anciaux, Molinari, Tribol Lett 56 (204) Contact area fraction, A' Eq. (2) Normalized mean contact pressure, p' /A' Contact area fraction, A' Normalized pressure, p' Contact area evolution [2] V.A. Yastrebov 43/45

44 Westergaard solution: 3D extension 2 Surface Contact area for a λ:!2/3 3p πa2 = π λ2 8πp Non-contact area for b λ:! p πb2 3 = λ2 2π p [] Johnson, Greenwood, Higginson, Int J Mech Sci (985) [2] Yastrebov, Anciaux, Molinari, Tribol Lett 56 (204).5 Normalized pressure, p/p* Full contact pressure: p = 2πE /λ y x (a) x 2+y2 / 2.5 Normalized pressure, p/p* z = (2 cos(2πx/λ) cos(2πy/λ) (b) y x Numerical results Fit Westergaard's solution Fit cosine wave x/ Pressure distribution along y = x and y = 0[2] V.A. Yastrebov 44/45

45 Thank you for your attention!