Capturing the inter-particle force distribution in granular material using LS-DEM

Abstract

Particle shape, as one of the most important physical ingredients of granular materials, can greatly alter the characteristic of inter-particle force distribution which is of vital importance in understanding the mechanical behavior of granular materials. However, currently both experimental and numerical studies remain limited in this regard. In this paper, we for the first time validate the ability of the level set discrete element method (LS-DEM) on capturing the inter-particle force distribution among particles of arbitrary shape. We first present the technical detail of LS-DEM; we then apply LS-DEM to simulate experiments of shearing granular materials composed of arbitrarily shaped particles. The proposed approach directly links experimentally measured material properties to model parameters such as contact stiffness without any calibration. Our results show that LS-DEM is able to not only capture the macro scale response such as stress and deformation, but also to reproduce the particle scale contact information such as the distribution of contact force magnitude, contact orientation and contact friction mobilization. Our work demonstrates the promising potential of LS-DEM on studying the mechanics and physics of natural granular material and on aiding design granular particle shape for novel macro-scale mechanical property.

Appendices

Appendix 1: Resolve the mesh-dependency of current LS-DEM implementation

As mentioned in Sect. 2, in LS-DEM currently force and moment contributions from all penetrating nodes are considered, which in fact will cause LS-DEM to be mesh-dependent in displacement controlled loading condition - the mechanical response of a particle system will depend sensitively on the discretized fineness of each particle’s surface, i.e. how many nodes each particle has. However, such mesh-dependent behavior vanishes for force controlled loading condition. To see this, using both LS-DEM and DEM with exactly the same model parameters we present several numerical tests of two identical frictional disk with radius $R = 15 \,\,\text {mm}$ vertically stacked between two rigid walls with the top wall being moved downward via either displacement controlled ($\Delta$) or force controlled condition (f), as shown in Fig. 12. In either case we output two quantities: the contact force magnitude |F|, and the inter-particle penetration d. For DEM $d = 2R - |\varvec{r}_1 -\varvec{r}_2|$ where $\varvec{r}_1,\varvec{r}_2$ are the centroid position of the two particle respectively; for LS-DEM $d = \sum _zd_z$ where we sum over the penetration $d_z$ of all penetrating nodes. For LS-DEM simulation each disk surface is randomly spatially discretized with either 30, 50 or 70 nodes. As shown in Fig. 13, the mesh-dependence problem appears for displacement controlled loading condition while vanishes for force controlled case. This can be explained by the following: in displacement controlled case the top wall is displaced downward for a certain amount that will lead to larger contact force with denser disk surface discretization, which subsequently leads to increasing |F| and d; while for the force controlled case, external force is already known and is used to compute displacement by enforcing equilibrium − no matter how dense the surface discretization is the total force from all penetrating nodes should always equilibrate the externally prescribed one. Therefore the results for |F| and d all collapse onto those computed from DEM.

Fig. 12

a Displacement controlled and force controlled loading condition with prescribed $\Delta$ and f respectively, and both with output the inter-particle force F and penetration d; b, c Loading curve of input $\Delta$ and f

Fig. 13

Inter-particle force magnitude |F| and penetration d response for displacement controlled case (a, b), and for force controlled case (c, d) from DEM simulation and LS-DEM simulations with 30, 50 or 70 nodes

In order to solve this mesh-dependency problem and further show that LS-DEM converges to DEM with proper modification, two simple approaches are tested: regarding the contributions from all penetrating nodes, we either take average or only consider the one with maximum penetration:

$$\varvec{f}_{n} = \frac{1}{P}\sum _{z=1}^{P}\varvec{f}_{n,z}, \quad \varvec{f}_{t} = \frac{1}{P}\sum _{z=1}^{P}\varvec{f}_{t,z}$$

(18)

or

$$\begin{aligned} \varvec{f}_{n}&= \varvec{f}_{n,z_m}\left| d_{z_m}=\text {max}_{1\le z\le P}\left\{ d_{z}\right\} \right. , \nonumber \\ \varvec{f}_{t}&=\varvec{f}_{t,z_m}^{j,i}\left| d_{z_m}=\text {max}_{1\le z\le P}\left\{ d_{z}\right\} \right. \end{aligned}$$

(19)

As shown in Fig. 14, for displacement controlled loading condition, both approaches resolve the mesh-dependency problem but only the modification of considering maximum penetration can further make LS-DEM converge to DEM: the computed |F| and d from LS-DEM converge to those computed from DEM as the node number N discretizing a grain surface is increased from 30 to 70. We can expect that as N goes larger and larger, LS-DEM will converge to DEM for simulating circular particles with exactly the same model parameters.

Fig. 14

Inter-particle force magnitude |F| and penetration d response for the same displacement controlled case from DEM simulation and LS-DEM simulations with 30, 50 or 70 nodes; a, b taking average for all penetrating nodes and c, d considering only the node with maximum penetration

Appendix 2: Boundary condition implemenation

Here we discuss our methodology in estimating the variation of $|\varvec{F}_A|$ and $|\varvec{F}_B|$ as $\theta$ increases. We note first that all quantities mentioned here are experimentally measured. Following the discussion in Sect. 3 (Fig. 6), we assume the forces exerted by the particles and from $F_N$ to the boundary “AD-DC-BC” all act at the center of each bar and the former can be estimated based on the stress state of the granular assembly $\varvec{\sigma }_\theta$. In each configuration with a certain $\theta$ value, we have the following unknown vectors: $\varvec{F}_A$, $\varvec{F}_B$ and $\varvec{F}_h$, see Fig. 6. However, we only end up having three instead of six unknowns due to our experiment setup: $\varvec{F}_A$ and $\varvec{F}_B$ should always be perpendicular to AD and BC, and $\varvec{F}_h$ should always be horizontal. We herein denote them as $F_A$, $F_B$ and $F_h$ as the corresponding signed magnitude: if positive the force is along the assumed direction, otherwise opposite. With force and torque equilibrium we have three equations and can therefore solve for $F_A$, $F_B$ and $F_h$. At a certain configuration with a given $\theta$, we assume that:

$$\varvec{F}_A= F_A\begin{bmatrix} \text {cos}\theta&\text {sin}\theta \end{bmatrix},$$

(20)

$$\varvec{F}_B= F_B\begin{bmatrix} \text {cos}\theta&\text {sin}\theta \end{bmatrix},$$

(21)

$$\varvec{F}_h= F_h\begin{bmatrix} 1&0 \end{bmatrix}$$

(22)

Forces exerted by the particles can be estimated by:

$$\varvec{F}_{pq} = -\left( \varvec{\sigma }_\theta \cdot \varvec{n}_{pq}\right) S_{pq}$$

(23)

where “pq” is one of “ AD”, “DC” and “BC” and $S_{pq}$ is the corresponding arm area. By force and torque equilibrium we must have:

$$\sum _l \varvec{F}_l= \varvec{0}$$

(24)

$$\sum _l \varvec{r}_{lM}\times \varvec{F}_l= \varvec{0}$$

(25)

where $\varvec{F}_l$ stands for all external force exerted to “AD-DC-CB” with $\varvec{r}_{lM}$ being the position vector point from point M (center of bar “DC”) to the location of $\varvec{F}_l$. Combining the above equations and after some algebra we arrive at the following linear system:

$$\varvec{A}\varvec{u} = \varvec{b}$$

(26)

where

$$\begin{aligned} \varvec{A}&= \begin{bmatrix} \text {cos}\theta&\text {cos}\theta&1\\ \text {sin}\theta&\text {sin}\theta&0\\ A_{31}&A_{32}&0 \end{bmatrix},\,\, \varvec{u} = \begin{bmatrix} F_A\\ F_B\\ F_h \end{bmatrix}\,\, \text {and}\nonumber \\ \varvec{b}&= \begin{bmatrix} \tau _\theta S_{DC} \\ \sigma _{y,\theta }S_{DC}+F_N\\ b_3 \end{bmatrix} \end{aligned}$$

(27)

with

$$\begin{aligned} A_{31}&= \frac{1}{2}\text {cos}\theta \left( -\,2h_\theta \text {sec}^2\theta -\varvec{r}_{A2}+\varvec{r}_{B2}\right. \nonumber \\&\quad \left. +\,\text {tan}\theta (\varvec{r}_{A1}-\varvec{r}_{B1})\right) \end{aligned}$$

(28)

$$\begin{aligned} A_{32}&= \frac{1}{2}\text {cos}\theta \left( -\,2h_\theta \text {sec}^2\theta + \varvec{r}_{A2}-\varvec{r}_{B2}\right. \nonumber \\&\quad \left. +\,\text {tan}\theta (\varvec{r}_{B1}-\varvec{r}_{A1})\right) \end{aligned}$$

(29)

$$\begin{aligned} b_3&= S_{DC}\left[ (\varvec{r}_{A1}-\varvec{r}_{B1})\text {sin}\theta \sigma _{y,\theta } +\tau _\theta (\text {cos}\theta +\text {sin}\theta )\right. \nonumber \\&\quad \left. -\,(\varvec{r}_{A2}-\varvec{r}_{B2})\text {cos}\theta \sigma _{x,\theta }\right] \end{aligned}$$

(30)

where $\varvec{r}_{A}$ and $\varvec{r}_{B}$ are locations of the slider A and B which both remain unchanged through the experiment with the subscripts “1” and “2” denote the x and y component respectively. Solving the above linear system can give us the estimation of $F_A$ and $F_B$.