1D Model of Precursors to Frictional Stick-Slip Motion Allowing for Robust Comparison with Experiments

Abstract

In this article, we study the dynamic behaviour of 1D spring-block models of friction when the external loading is applied from a side, and not on all blocks like in the classical Burridge–Knopoff-like models. Such a change in the loading yields specific difficulties, both from numerical and physical viewpoints. To address some of these difficulties and clarify the precise role of a series of model parameters, we start with the minimalistic model by Maegawa et al. (Tribol. Lett. 38: 313, 2010) which was proposed to reproduce their experiments about precursors to frictional sliding in the stick-slip regime. By successively adding an (i) internal viscosity, (ii) interfacial stiffness and (iii) initial tangential force distribution at the interface, we manage to (i) avoid the model’s unphysical stress fluctuations, (ii) avoid its unphysical dependence on the spatial resolution and (iii) improve its agreement with the experimental results, respectively. Based on the behaviour of this improved 1D model, we develop an analytical prediction for the length of precursors as a function of the applied tangential load. We also discuss the relationship between the microscopic and macroscopic friction coefficients in the model.

Appendices

Appendices

1.1 Appendix A: Relative Viscous Damping in a Linear Chain of Blocks

If friction forces are ignored, the equation of motion for an infinite chain of blocks connected by springs is given by

$$ m\ddot{u}_n = k(u_{n + 1} - 2u_n + u_{n-1}) + \eta(\dot{u}_{n+1} - 2\dot{u}_n + \dot{u}_{n-1}). $$

(22)

We then assume a solution of the form

$$ u_n(t) = {\rm{e}}^{\zeta_\kappa t} {\rm{e}}^{i\kappa n a}, $$

(23)

where ${\zeta_\kappa \in \mathbb{C}}$ and ${\kappa \in \mathbb{R}.}$ Inserting Eq. 23 into Eq. 22 yields the relation

$$ m \zeta_\kappa^2 = k \left( {\rm{e}}^{i\kappa a} - 2 + {\rm{e}}^{-i\kappa a} \right) + \eta \zeta_\kappa \left( {\rm{e}}^{i\kappa a} - 2 + {\rm{e}}^{-i\kappa a} \right), $$

(24)

which can be simplified to

$$ m \zeta_\kappa^2 + 4 \eta \sin^2 \left( \frac{\kappa a}{2} \right) \zeta_\kappa + 4k \sin^2 \left( \frac{\kappa a}{2} \right) = 0, $$

(25)

since

$$ e^{i\kappa a} - 2 + e^{-i\kappa a} = -4\sin^2 \left( \frac{\kappa a}{2} \right). $$

(26)

The complex parameter $\zeta_\kappa$ is then given by

$$ \zeta_\kappa = \frac{-4 \eta \sin^2 \left( \frac{\kappa a}{2} \right) \pm \sqrt{16 \eta^2\sin^4 \left( \frac{\kappa a}{2} \right) -16 km \sin^2 \left( \frac{\kappa a}{2} \right)}}{2m} $$

(27)

The system is critically damped when Eq. 25 only has one solution for $\zeta_\kappa,$ which occurs when the square root is zero:

$$ \eta^2 \sin^2 \left( \frac{\kappa a}{2} \right) = km \quad \Rightarrow \quad \eta = \frac{\sqrt{km}}{\left| \sin \left( \frac{\kappa a}{2} \right) \right|}. $$

(28)

The oscillations that are to be reduced have a wavelength λ = 2a, i.e. a wave number κ = 2π/λ = π/a. Inserting this into Eq. 28 leads to

$$ \eta_{\rm{c}} = \sqrt{k m}, $$

(29)

which is the value of the damping coefficient η for which waves of wavelength λ = 2a are critically damped. Since the absolute value of $\sin$ is always smaller than one, choosing $\eta = \sqrt{km}$ will cause all other waves to be under-damped.

1.2 Appendix B: Tangential Force Profiles and Characteristic Length with a Tangential Stiffness of the Interface

An analytical expression for the characteristic length l ₀ can be found. In order to do so, the following assumptions are made: N ≫ 1, l ₀/L ≪ 1 and slow loading compared with the internal dynamics of the system, which enables a static analysis. The system is first placed in a static state with an initial tangential force profile given by τ ⁰ _n , and then loaded slowly from the left. The equilibrium of all non-edge blocks writes

$$ k\left( u_{n+1} - 2u_n + u_{n-1} \right) - k_{\rm{t}} \left( u_n - u_n^{\rm stick} \right) = 0. $$

(30)

We introduce a new variable u _n ′ defined by

$$ u_n = u_n' + u_n^0, $$

(31)

where u ⁰ _n is the initial position of block n. Inserting Eq. 31 into Eq. 30 yields

$$ k\left( u_{n+1}' - 2u_n' + u_{n-1}' \right) - k_{\rm{t}} u_n' + \tau_n^0 - k_{\rm{t}} \left( u_n^0 - u_n^{\rm stick} \right) = 0, $$

(32)

where

$$ \tau_n^0 = k\left( u_{n+1}^0 - 2u_n^0 + u_{n-1}^0 \right). $$

(33)

The two terms τ ⁰ _n and $-k_{\rm{t}} \left( u_n^0 - u_n^{\rm stick} \right)$ cancel in Eq. 32 since the initial state is static, and thus

$$ k\left( u_{n+1}' - 2u_n' + u_{n-1}' \right) - k_{\rm{t}} u_n' = 0. $$

(34)

The above equation can be rewritten to

$$ k a^2 \frac{u_{n+1}' - 2u_n' + u_{n-1}'}{a^2} - k_{\rm{t}} u_n' = 0, $$

(35)

where a = L/(N − 1) is the lattice spacing. Since N ≫ 1, the first term in Eq. 35 can be replaced by the second spatial derivative, and replacing u _n ′ with u′(na) = u′(x) yields

$$ k a^2 \frac{\partial^2 u'(x)}{\partial x^2} - k_{\rm{t}} u'(x) = 0, $$

(36)

which has the general solution

$$ u'(x) = A {\rm{e}}^{x/l_0} + B{\rm{e}}^{-x/l_0}, \quad l_0 = \sqrt{\frac{k}{k_{\rm{t}}}}a. $$

(37)

The tangential force is given by

$$ \tau_n = k \left( u_{n+1} - 2u_n + u_{n-1} \right) $$

(38)

$$ = k \left( u_{n+1}' - 2u_n' + u_{n-1}' \right) + \tau_n^0. $$

(39)

By replacing again finite differences with second-order derivatives,

$$ \tau(x) = ka^2 \frac{\partial^2 u'(x)}{\partial x^2} + \tau^0(x), $$

(40)

the general expression for the shear force profile can be found by using Eq. 37, which yields

$$ \tau(x) = \frac{ka^2 l_0^2}{L^2} \left( A{\rm{e}}^{x/l_0} + B{\rm{e}}^{-x/l_0} \right) + \tau^0(x). $$

(41)

The system is loaded from the left, and at the beginning of an event the tangential force on block 1 is equal to the static friction threshold μ_s p ₁. Provided l ₀/L ≪ 1, the trailing edge will not be affected by the loading. The latter of these two boundary conditions yields

$$ \tau(L) = \frac{ka^2 l_0^2}{L^2} \left( A {\rm{e}}^{L/l_0} + B{\rm{e}}^{-L/l_0} \right) + \tau^0(L) $$

(42)

$$ \approx \frac{ka^2 l_0^2}{L^2} \left( A{\rm{e}}^{L/l_0} \right) + \tau^0(L) $$

(43)

$$ = \tau^0 (L), $$

(44)

i.e. A = 0. The first boundary condition yields

$$ \tau(0) = \frac{ka^2 l_0^2}{L^2} B + \tau^0(L) = \mu_{\rm{s}} p_1, $$

(45)

and the tangential force is therefore given by

$$ \tau(x) = \left( \mu_{\rm{s}} p_1 - \tau^0 (x) \right){\rm{e}}^{-x/l_0} + \tau^0(x). $$

(46)

The characteristic length l ₀ is given by Eq. 37, and inserting for k given by Eq. 4 and a yields

$$ l_0 = \sqrt{\frac{k}{k_{\rm{t}}}}a = \sqrt{\frac{E L S}{(N - 1) k_{\rm{t}}}}, $$

(47)

and hence Eq. 12 for N >> 1.

Note that in a 3D situation, the exponential decay of the tangential stress with x would be replaced by a power law [9, 18].

1.3 Appendix C: Derivation of the Prediction of Precursor Lengths in Our Improved Model

We start with Eq. 16 and use the assumed tangential force profile in Eq. 21, shown in Fig. 14. Again, we go to the limit $N \to \infty,$ resulting in the substitution

$$ \sum_{n=1}^N \tau_n \to \frac{N}{L} \int\limits_0^L \tau(x) {\hbox {d}} x , \qquad n \to xN/L. $$

(48)

The tangential force after a precursor of length L _p is then given by

$$ F_{\rm{T}} = \frac{N}{L} \left[ \int\limits_0^{L_{\rm{p}}} \tau(x) {\hbox {d}} x + \int\limits_{L_{\rm{p}}}^{L} \tau(x) {\hbox {d}} x \right] $$

(49)

$$ = \frac{N}{L} \left[ \int\limits_0^{L_{\rm{p}}} \mu_{\rm{k}} p(x) {\hbox {d}} x + \int\limits_{L_{\rm{p}}}^{L} \left(\alpha p(L_{\rm{p}}) - \tau^0 (x) \right) {\rm{e}}^{-\frac{x - L_{\rm{p}}}{l_0}} + \tau^0(x) {\hbox {d}} x \right]. $$

(50)

We limit ourselves to predicting the precursors in Fig. 13b, i.e. using a tangential interfacial stiffness and linear initial tangential forces as depicted in Fig. 13a, but with θ = 0. The normal and initial tangential force are then given by

$$ p(x) = p = F_{\rm{N}}/N ={ \hbox {constant}}, $$

(51)

$$ \tau^0 (x) = \beta p \frac{2(x - L/2)}{L}, $$

(52)

where the parameter β determines the slope in the initial tangential force profile. Inserting Eqs. 51 and 52 into Eq. 50 yields

$$ F_{\rm{T}} = \frac{N}{L} \left[ \int\limits_0^{L_{\rm{p}}} \mu_{\rm{k}} p {\hbox {d}} x + \right. \left. \int\limits_{L_{\rm{p}}}^{L} \left(\alpha p - \beta p \frac{2(x - L/2)}{L} \right) {\rm{e}}^{-\frac{x - L_{\rm{p}}}{l_0}} + \beta p \frac{2(x - L/2)}{L} {\hbox {d}} x \right]. $$

(53)

The above integrals can be calculated easily, and the result is the tangential load F _T as a function of the precursor length L _p:

$$ F_{\rm{T}}(L_{\rm{p}}) = F_{\rm{N}} \left[ \mu_{\rm{k}} \frac{L_{\rm{p}}}{L} + 2\beta \frac{l_0^2}{L^2} \left({\rm{e}}^{-\frac{L-L_{\rm{p}}}{l_0}} - 1 \right) + \beta \frac{\left(L-L_{\rm{p}}\right) L_{\rm{p}}}{L^2} + \right.\\ \left. \frac{l_0}{L} \left( \beta \left(1+{\rm{e}}^{-\frac{L-L_{\rm{p}}}{l_0}} - 2\frac{L_{\rm{p}}}{L} \right) + \alpha \left(1 - {\rm{e}}^{-\frac{L-L_{\rm{p}}}{l_0}}\right) \right) \right]. $$

(54)

We observe that again F _T(L) =  μ_k F _N.