Mechanical properties of inclined frictional granular layers

Abstract

We investigate the mechanical properties of inclined frictional granular layers prepared with different protocols by means of DEM numerical simulations. We perform an orthotropic elastic analysis of the stress response to a localized overload at the layer surface for several substrate tilt angles. The distance to the unjamming transition is controlled by the tilt angle $\alpha$ with respect to the critical angle $\alpha _c$. We find that the shear modulus of the system decreases with $\alpha$, but tends to a finite value as $\alpha \rightarrow \alpha _c$. We also study the behaviour of various microscopic quantities with $\alpha$, and show in particular the evolution of the contact orientation with respect to the orthotropic axes and that of the distribution of the friction mobilisation at contact.

Appendix: Orthotropic elastic response

In this Appendix, we detail elastic calculations on a 2D orthotropic slab of finite thickness $h$. Following the notations of Fig. 1, we note $(1,2)$ the orthotropic directions, while $(n,t)$ are the directions respectively normal and tangential to the slab. We note $\tau$ the angle between axes $(1,2)$ and $(n,t)$. For the sake of the computation of the stress profiles in response to a force $\varvec{F}_0$ applied at the free surface, one can switch off gravity, and the mechanical equilibrium of the system writes

$$\begin{aligned} \partial _n \sigma _{nn} + \partial _t \sigma _{tn} = 0 \qquad \text{ and } \qquad \partial _n \sigma _{tn} + \partial _t \sigma _{tt} = 0, \end{aligned}$$

(2)

where $\sigma _{ij}$ is the stress tensor. We define the strain tensor $u_{ij}$ from the displacement field $u_i$ as $u_{ij} = \frac{1}{2} (\partial _i u_j + \partial _j u_i)$. It verifies the compatibility condition:

$$\begin{aligned} \partial _n^2 u_{nn} + \partial _t^2 u_{tt} - 2 \partial _n \partial _t u_{tn} = 0. \end{aligned}$$

(3)

Introducing the two Young moduli $E_1$ and $E_2<E_1$, the shear modulus $G$ and two Poisson coefficients $\nu _{12}$ and $\nu _{21}$, the generalised Hooke’s law relating strain and stress tensors writes, in the orthotropic axes, as follows:

$$\begin{aligned} \left( \begin{array}{c} u_{11}\\ u_{22}\\ u_{12} \end{array}\right) =\left( \begin{array}{c@{\quad }c@{\quad }c} \frac{1}{E_1} &{} -\frac{\nu _{21}}{E_2} &{} 0\\ -\frac{\nu _{12}}{E_1} &{} \frac{1}{E_2} &{} 0\\ 0 &{} 0 &{} \frac{1}{2G} \end{array} \right) \left( \begin{array}{c} \sigma _{11}\\ \sigma _{22}\\ \sigma _{12} \end{array} \right) . \end{aligned}$$

(4)

We call $\mathcal {W}_\dagger$ this $3 \times 3$ compliance matrix. It must be symmetric and these coefficients thus verify $\nu _{12}/E_1 = \nu _{21}/E_2$. Elastic energy is well defined if all moduli $E_1,E_2,G$ are positive and $1-\nu _{12}\nu _{21}>0$. In $(n,t)$ axes, we have

$$\begin{aligned} \left( \begin{array}{c} u_{nn}\\ u_{tt}\\ u_{tn} \end{array} \right) = \mathcal {W}_\tau \left( \begin{array}{c} \sigma _{nn}\\ \sigma _{tt}\\ \sigma _{tn} \end{array} \right) \quad \text{ with } \quad \mathcal {W}_\tau = \mathcal {Q}^{-1} \mathcal {W}_\dagger \mathcal {Q} \end{aligned}$$

(5)

and the rotation matrix

$$\begin{aligned} \mathcal {Q} = \left( \begin{array}{ccc} \cos ^2\tau &{} \sin ^2\tau &{} 2\cos \tau \sin \tau \\ \sin ^2\tau &{} \cos ^2\tau &{} -2\cos \tau \sin \tau \\ -\cos \tau \sin \tau &{} \cos \tau \sin \tau &{} \cos ^2\tau -\sin ^2\tau \end{array} \right) . \end{aligned}$$

(6)

The matrix $\mathcal {W}_\tau$ can be made explicit as follows:

$$\begin{aligned} \mathcal {W}_\tau = \frac{1}{E_2} \left( \begin{array}{c@{\quad }c@{\quad }c} A &{} -C &{} 2D\\ -C &{} B &{} 2F\\ D &{} F &{} H \end{array} \right) , \end{aligned}$$

(7)

with

$$\begin{aligned} A&= T \cos ^4\tau + \sin ^4\tau + 2R\cos ^2\tau \sin ^2\tau , \end{aligned}$$

(8)

$$\begin{aligned} B&= \cos ^4\tau + T \sin ^4\tau + 2R\cos ^2\tau \sin ^2\tau , \end{aligned}$$

(9)

$$\begin{aligned} C&= \nu _{21} + \cos ^2\tau \sin ^2\tau (2R-1-T), \end{aligned}$$

(10)

$$\begin{aligned} D&= \cos \tau \sin \tau \left[ (\sin ^2\tau - \cos ^2\tau )R + \cos ^2\tau (1+T) - 1 \right] ,\nonumber \\ \end{aligned}$$

(11)

$$\begin{aligned} F&= \cos \tau \sin \tau \left[ (\cos ^2\tau - \sin ^2\tau )R + \sin ^2\tau (1+T) - 1 \right] ,\nonumber \\ \end{aligned}$$

(12)

$$\begin{aligned} H&= \nu _{21} - 2\cos ^2\tau \sin ^2\tau (2R-1-T) + R, \end{aligned}$$

(13)

and where we have introduced the two dimensionless numbers

$$\begin{aligned} T \!=\! \frac{E_2}{E_1} \!=\! \frac{\nu _{21}}{\nu _{12}}, \quad \text{ and } \quad R = \frac{1}{2} \, E_2 \left( \frac{1}{G} - \frac{\nu _{12}}{E_1} - \frac{\nu _{21}}{E_2} \right) . \qquad \end{aligned}$$

(14)

With the four roots $X_k$ ($k=1, ..., 4$) of the biquadratic equation $X^4+2RX^2+T=0$, that is

$$\begin{aligned} X = \pm \sqrt{-R \pm (R^2-T)^{1/2}}, \end{aligned}$$

(15)

the general solution of the problem can be written as sums of Fourier modes:

$$\begin{aligned} \sigma _{nn} (n,t)&= \sum _{k=1}^4 \,\,\int \limits _{-\infty }^{+\infty } \!\!\!\! b_k(q) \, e^{iqt+iY_k qn} dq, \end{aligned}$$

(16)

$$\begin{aligned} \sigma _{tt} (n,t)&= \sum _{k=1}^4 \,\,\int \limits _{-\infty }^{+\infty } \!\!\!\! b_k(q) \, Y_k^2 \, e^{iqt+iY_k qn} dq, \end{aligned}$$

(17)

$$\begin{aligned} \sigma _{tn} (n,t)&= - \sum _{k=1}^4 \,\,\int \limits _{-\infty }^{+\infty } \!\!\!\! b_k(q) \, Y_k \, e^{iqt+iY_k qn} dq, \end{aligned}$$

(18)

where $Y_k=(X_k-\tan \tau )/(1+X_k\tan \tau )$. The four functions $b_k$ are determined by the boundary conditions at the top and the bottom of the slab.

At the free surface ($n=0$), the overload force imposes two components of the stress:

$$\begin{aligned} \sigma _{nn} = F_0 \cos \theta \, \varDelta (t) \quad \text{ and } \quad \sigma _{tn} = F_0 \sin \theta \, \varDelta (t), \end{aligned}$$

(19)

where $\theta$ is the angle between $\varvec{F}_0$ and the direction of the $n$ axis (see Fig. 1), and where $\varDelta (t)$ is a normalised function which tells how this force is distributed along the surface—e.g. a Dirac or a Gaussian of width $w_F$. We need here its Fourier transform $s(q)$. For the Gaussian case, $s(q) = \frac{1}{2\pi } \exp (-\frac{1}{2} w_F^2 q^2)$. We typically take $w_F \rightarrow 0$ (a $\delta$-peak). These top conditions (19) then give

$$\begin{aligned} \sum _{k=1}^4 b_k = F_0 \cos \theta \, s(q) \quad \text{ and } \quad \sum _{k=1}^4 b_k Y_k = - F_0 \sin \theta \, s(q).\nonumber \\ \end{aligned}$$

(20)

At the bottom of the slab ($n=h$), we impose rigid and rough conditions, i.e. vanishing displacements in both $t$ and $n$ directions: $u_t=u_n=0$. In order to get equations on the functions $b_k$, we must transform these conditions into equations on the stress components. Taking its derivative along $t$, the condition $u_t=0$ gives $u_{tt}=0$, i.e.

$$\begin{aligned} -C \sigma _{nn} + B \sigma _{tt} + 2F \sigma _{tn} = 0, \end{aligned}$$

(21)

leading to

$$\begin{aligned} \sum _{k=1}^4 b_k \left[ -C - 2F Y_k + B Y_k^2 \right] e^{iY_k q h}= 0. \end{aligned}$$

(22)

Similarly, the condition $u_n=0$ gives, after a double derivative along $t$, the relation $2 \partial _t u_{tn} = \partial _n u_{tt}$, leading to

$$\begin{aligned} \sum _{k=1}^4 b_k \left[ 2D + (C-2H) Y_k + 4F Y_k^2 - B Y_k^3 \right] e^{iY_k q h}= 0.\nonumber \\ \end{aligned}$$

(23)

The four linear Eqs. (20, 22, 23) can be inverted, leading to large but analytic expressions for the functions $b_k$. Integrations over $q$ involved in Eqs. 16–18 must, however, be computed numerically. Finally, the stress components, made dimensionless by $F_0/h$, can be plotted for given values of the five parameters $\tau$, $T$, $R$, $\nu _{21}$ and $\theta$, as functions of $t/h$ at a given depth (e.g. $n=h$). We checked that the results are insensitive to the value of $w_F/h$, as long as it remains small.