Steady Sliding Frictional Contact Problems in Linear Elasticity

Abstract

The existence and uniqueness of an equilibrium solution to frictional contact problems involving a class of moving rigid obstacles is studied. At low friction coefficient values, the steady sliding frictional contact problem is uniquely solvable, thanks to the Lions-Stampacchia theorem on variational inequalities associated with a nonsymmetric coercive bilinear form. It is proved that the coerciveness of the bilinear form can be lost at some positive critical value of the friction coefficient, depending only on the geometry and the elastic properties of the body. An example presented here, shows that infinitely many solutions can be obtained when the friction coefficient is larger than the critical value. This result is paving the road towards a theory of jamming in terms of bifurcation in variational inequality. The particular case where the elastic body is an isotropic half-space is studied. The corresponding value of the critical friction coefficient is proved to be infinite in this case. In the particular case of the frictionless situation, our analysis incidentally unifies the approaches developed by Lions-Stampacchia (variational inequalities) and Hertz (harmonic analysis on the half-space) to contact problems in linear elasticity.

Appendices

Appendix A: The Lions-Stampacchia Theorem

The material presented in this appendix is reproduced from [10].

Let H be an arbitrary Hilbert space, the scalar product and norm of which are denoted by (⋅ ,⋅) and ∥⋅∥. A bilinear form a:H×H→ℝ is said to be:

• continuous, if there exists a real constant C such that:

  $$\forall u,v\in H,\quad\bigl| a(u,v) \bigr| \leq C \| u \|\| v \|,$$

• coercive, if there exists a real constant α>0 such that:

  $$\forall u\in H,\quad a(u,u) \geq\alpha\| u \|^2.$$

Theorem 6

(Lions-Stampacchia)

Let a(⋅ ,⋅) be a coercive, continuous bilinear form on a Hilbert space H, and let K be a nonempty closed convex subset of H. Given an arbitrary φ∈H′, there exists uniquely u∈K such that:

$$ \forall v\in K,\quad a(u,v-u) \geq\langle\varphi, v-u\rangle.$$

(17)

In addition, if a is symmetric, then u is characterized by being the unique minimizer of the functional:

$$\frac{1}{2}a(v,v) -\langle\varphi,v \rangle,$$

over K.

Proof

Based on the Riesz-Fréchet representation theorem, there exists uniquely f∈H such that:

$$\forall v\in H,\quad\langle\varphi,v \rangle= (f,v).$$

Similarly, given an arbitrary u∈H, the mapping v↦a(u,v) is a continuous linear form, and there exists Au∈H such that:

$$\forall v\in H,\quad a(u,v) = (Au,v).$$

Obviously, A:H→H is a linear mapping such that:

The variational inequality (17) amounts to finding u∈K such that:

$$\forall v\in H,\quad(Au,v-u)\geq(f,v-u).$$

Arbitrarily taking ρ>0, the latter inequality can be rewritten:

$$\forall v\in H,\quad(\rho f-\rho Au + u - u,v-u)\leq0,$$

or equivalently:

$$u = \text{proj} [\rho f-\rho Au + u,K ] .$$

The original problem therefore reduces to that of finding a fixed point of the mapping:

$$Sv = \text{proj} [\rho f-\rho Av + v,K ].$$

Therefore, the claim will follow from Banach fixed point theorem provided that the mapping S is a strict contraction, that is, there exists L<1 such that:

$$\forall v_{1},v_{2}\in K,\quad\| Sv_{1}-Sv_{2}\| \leq L \| v_{1}-v_{2} \|,$$

for some appropriate value of ρ. But, in view of the contraction property of the projection onto K, we obtain:

and the constant between brackets in the last term is made to be smaller than 1 by taking¹ ρ in ]0,2α/C ²[. It should now be clear that the desired conclusion follows from Banach’s fixed point theorem.

In the particular case where a is symmetric, then a defines a new scalar product on H the associated norm of which is equivalent to that of H. Based on the Riesz-Fréchet representation theorem, there exists uniquely g∈H such that:

$$\forall v\in H,\quad\langle\varphi,v \rangle= a(g,v).$$

The variational inequality (17) can therefore be rewritten equivalently:

$$\forall v\in H,\quad a(g-u,v-u) \leq0,$$

which states that u is simply the projection of g onto K. Thus, u achieves the minimum:

$$\min_{v\in K} a(g-v , g-v),$$

or equivalently:

$$\min_{v\in K} \bigl[a(v , v) - 2 a(g,v) \bigr],$$

or equivalently:

$$\min_{v\in K} \biggl[\frac{1}{2}a(v , v) - \langle\varphi, v\rangle\biggr].$$

 □

Appendix B: Moving Line Load on the Surface of a Half-Space

The material presented in this appendix is adapted from the book [4].

Let us consider an isotropic homogeneous linearly elastic half-space. Some orthonormal Cartesian coordinate system (x,y,z) is chosen so that the half-space is defined by: z>0. Since only the plane strain situation will be considered here, it is convenient to take:

$$\varOmega= \bigl\{ (x,z)\in\mathbb{R}^2 ; z>0 \bigr\}$$

to denote a corresponding two-dimensional “slice” of the half-space. The Young’s modulus is denoted by E, the Poisson ratio by ν∈]−1,1/2[ and the volumic mass by ρ. We take c _p >c _s to denote the velocities of the pressure and the shear waves in the isotropic elastic medium:

$$c_{p}^2 = \frac{1-\nu}{(1+\nu)(1-2\nu)}\frac{E}{\rho},\qquad c_{s}^2 = \frac{1}{2(1+\nu)}\frac{E}{\rho}.$$

Let us consider the situation where some homogeneous force (F _x ,F _z ) concentrated along the moving line x=wt, z=0 of the boundary is applied to the half-space which is assumed to be free of body forces. A steady displacement field is sought, that is, a displacement field which is constant in the moving frame. The variables x and z will be also used in the moving frame. As usual, u is taken to denote the displacement, ε(u) to denote the associated linearized strain tensor, and σ(u) to denote the corresponding Cauchy stress tensor.

Only the subsonic situation:

$$0<w< c_{s}<c_{p}$$

will be considered here. We therefore set:

The function R(w) has one and only one root on the interval ]0,c _s [. This root, which is the velocity of the Rayleigh waves, is denoted by c _r . In what follows, the specific value w=c _r is excluded.

Theorem 7

Let (F _x ,F _z )∈ℝ² be taken arbitrarily. All the tempered distributions $\mathbf{u}\in\mathcal{S}' (\overline{\varOmega};\mathbb{R}^{2} )$ such that:

($C_{\mathrm{c}}^{\infty}(\overline{\varOmega};\mathbb{R}^{2} )$ stands, as usual, for the space of C ^∞ test-functions compactly supported in the closed half-space $\overline {\varOmega}$) are given by:

where D _x , D _z , Ω, Σ are four arbitrary real constants, and $U_{xx}^{0}$, $U_{xz}^{0}$, $U_{zx}^{0}$, $U_{zz}^{0}$ are the four functions in C ^∞(Ω;ℝ) defined by:

The corresponding Cauchy stress field is given by the three functions in C ^∞(Ω;ℝ) defined by:

In particular, we can prescribe the supplementary condition: lim_∞ σ(u)=0, which imposes Σ=0 and determines σ(u) uniquely. The arbitrariness still remaining in u because of the three constants D _x , D _z , Ω can be interpreted as being some arbitrary (linearized) rigid displacement.

Proof

Perform a Fourier transform with respect to x, then solve the ordinary differential equation with respect to z, and finally, perform the inverse Fourier transform. □

The explicit knowledge of the fundamental solution makes it possible to solve by convolution the Neumann problem with arbitrary, compactly supported, surface traction distributions (t _x ,t _z ):

where the supplementary condition: lim_∞ σ(u)=0 is imposed. The stress field is therefore uniquely determined. It is in C ^∞(Ω;ℝ³), but it is generally not square-integrable. No energy can therefore be associated with that solution. It can also be seen that the displacement field is infinite at infinity, and we cannot superimpose any conditions such as: lim_∞ u=0, in order to set the arbitrary rigid displacement. The only possible procedure here consists in superimposing the condition:

$$\mathbf{u} = O \bigl(\log\bigl(x^2+z^2\bigr) \bigr),\quad\text{as } x^2+z^2 \rightarrow\infty,$$

which entails Ω=0, and will always be assumed from now on. In this case, the surface displacement (u _x ,u _z ) must be:

To eliminate the arbitrary constants D _x , D _z , it is convenient to take a derivative with respect to x. Taking $\operatorname{pv} 1/x$ (for “principal value”) to denote the distributional derivative of log|x|, we obtain the explicit final form of the Neumann-Dirichlet operator:

Appendix C: Point Load on the Surface of a Half-Space

Let us consider an isotropic homogeneous linearly elastic half-space. Some orthonormal Cartesian coordinate system (x,y,z) is chosen so that the half-space Ω is defined by: z>0. The Young’s modulus is denoted by E, and the Poisson ratio by ν∈]−1,1/2[.

Theorem 8

Let (F _x ,F _z )∈ℝ² be taken arbitrarily. All the tempered distributions $\mathbf{u}\in\mathcal{S}' (\overline{\varOmega};\mathbb{R}^{3} )$ such that:

$$\forall\boldsymbol{\varphi} \in C_{\mathrm{c}}^\infty\bigl(\overline{\varOmega};\mathbb{R}^3 \bigr), \quad\bigl\langle\sigma_{ij}(\mathbf{u}),\varepsilon_{ji}(\boldsymbol{\varphi}) \bigr\rangle_{\mathcal{S}',\mathcal{S}}= F_{x} \varphi_{x}(0,0,0) + F_{z}\varphi_{z}(0,0,0) ,$$

($C_{\mathrm{c}}^{\infty}(\overline{\varOmega};\mathbb{R}^{3} )$ stands, as usual, for the space of C ^∞ test-functions compactly supported in the closed half-space $\overline {\varOmega}$) are given by:

where D _x , D _y , D _z , Ω _x , Ω _y , Ω _z are six arbitrary real constants, and $U_{xx}^{0}$, $U_{xz}^{0}$, $U_{yx}^{0}$, $U_{yz}^{0}$, $U_{zx}^{0}$, $U_{zz}^{0}$ are the six functions in C ^∞(Ω;ℝ) defined by:

where r ²=x ²+y ²+z ². The corresponding Cauchy stress field is then given by the six functions in C ^∞(Ω;ℝ) defined by:

The stress field σ(u) is therefore uniquely determined and square-integrable. The displacement field u is determined only up to an arbitrary (linearized) rigid displacement, which can be fixed by prescribing the supplementary condition: lim_∞ u=0, which imposes D _x =D _y =D _z =Ω _x =Ω _y =Ω _z =0 and determines u uniquely.

Proof

Perform a Fourier transform with respect to x and y, then solve the ordinary differential equation with respect to z, and finally, perform the inverse Fourier transform. □

The fundamental solution given by Theorem 8 yields an explicit form of the Neumann-Dirichlet operator for the isotropic homogeneous linearly elastic three-dimensional half-space.

Corollary 2

Let t _x ,t _z be arbitrary compactly supported distributions in the variable (x,y)∈ℝ². Let us consider the Neumann problem in the isotropic homogeneous elastic three-dimensional half-space involving:

• (compactly supported) surface tractions with components (t _x ,t _z ) on the boundary,

• no body forces,

• the following condition at infinity: lim_∞ u=0.

Then, the surface displacement (u _x (x,y),u _y (x,y),u _z (x,y)) is given by:

where the convolution products are understood in the sense of distributions.

Proposition 2

For f∈L ¹(ℝ²;ℝ)∩L ²(ℝ²;ℝ), we adopt the following definition of the Fourier transform:

$$\mathcal{F}[f] \stackrel{\mathrm{def}}{=} \frac{1}{2\pi}\int _{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(x,y)e^{ix\xi+iy\eta}\,\mathrm{d}x\,\mathrm{d}y = \hat{f}(\xi,\eta).$$

Then, we have:

$$\everymath{\displaystyle}\begin{array}{l@{\quad}l@{\qquad}l@{\quad}l}1. &\mathcal{F} \biggl[\frac{1}{\sqrt{x^2+y^2}} \biggr] = \frac{1}{\sqrt{\xi^2+\eta^2}},&3. & \mathcal{F} \biggl[\frac{xy}{(x^2+y^2)^{3/2}} \biggr] =\frac{-\xi\eta}{(\xi^2+\eta^2)^{3/2}},\\\noalign{\vspace{5pt}}2. &\mathcal{F} \biggl[\frac{x}{x^2+y^2} \biggr] = \frac{i\xi}{\xi ^2+\eta^2},&4. & \mathcal{F} \biggl[\frac{x^2}{(x^2+y^2)^{3/2}} \biggr] =\frac{\eta^2}{(\xi^2+\eta^2)^{3/2}}.\end{array}$$

Proof

• 1. This is formulae 3.754.2 and 6.671.14 in [6],

• 2. use formula 3.723.4 in [6],

• 3. this is formulae 3.754.3 and 6.691 in [6],

• 4. this is formulae 3.773.6 and 6.699.12 in [6].  □

Let f∈H ^−1/2(]−1,1[²;ℝ) be arbitrary. The extension of f by zero outside ]−1,1[² is in H ^−1/2(ℝ²;ℝ), and still denoting it by f, the expression:

$$\biggl(\int_{\mathbb{R}^2} \frac{|\hat{f}(\xi,\eta)|^2}{\sqrt{1+\xi ^2+\eta^2}}\,\mathrm{d}\xi\,\mathrm{d}\eta\biggr)^{1/2}$$

defines a norm and this norm is equivalent to that of H ^−1/2(]−1,1[²). Adapting the proof of Theorem 3 in [1], it can actually be seen that the same conclusion holds true with the expression:

$$\biggl(\int_{\mathbb{R}^2} \frac{|\hat{f}(\xi,\eta)|^2}{\sqrt{\xi^2+\eta ^2}}\,\mathrm{d}\xi\,\mathrm{d}\eta\biggr)^{1/2}.$$

Hence, in view of Proposition 2, the convolution products:

$$f(x,y)\stackrel{x,y}{*}\frac{1}{\sqrt{x^2+y^2}},\qquad f(x,y)\stackrel{x,y}{*}\frac{x}{x^2+y^2},$$

are in $H_{\mathrm{loc}}^{1/2}$. The proof of the following proposition is quite straightforward and has therefore been omitted.

Proposition 3

For arbitrary f,g∈H ^−1/2(]−1,1[²;ℝ), we define the bilinear forms:

which are continuous on H ^−1/2(]−1,1[²;ℝ). The bilinear form a(⋅ ,⋅) is symmetric and defines a scalar product inducing a norm which is equivalent to that of H ^−1/2(]−1,1[²;ℝ). The bilinear form $\tilde {a}(\cdot\,,\cdot)$ is skew-symmetric.