A model for decoding the life cycle of granular avalanches in a rotating drum

Abstract

Granular materials can behave as harmless sand dunes or as devastating landslides. A granular avalanche marks the transition between these distinct solid-like and fluid-like states. The solid-like state is typically described using plasticity models from critical state theory. In the fluid regime, granular flow is commonly captured using a visco-plastic model. However, due to our limited understanding of the mechanism governing the solid–fluid-like transition, characterizing the material behavior throughout the life cycle of an avalanche remains an open challenge. Here, we employ laboratory experiments of transient avalanches spontaneously generated by a rotating drum. We report measurements of dilatancy and grain kinematics before, during, and after each avalanche. Those measurements are directly incorporated into a rate-dependent plasticity model that quantitatively predicts the granular flow measured in experiments. Furthermore, we find that dilatancy in the solid-like state controls the triggering of granular avalanches and therefore plays a key role in the solid–fluid-like transition. With the proposed approach, we demonstrate that the life cycle of a laboratory avalanche, from triggering to run out, can be fully explained. Our results represent an important step toward a unified understanding of the physical phenomena associated with transitional behavior in granular media.

Appendices

Appendices

Appendix 1: Determination of the governing equation of the flow

In this appendix, we show how the governing equation of the granular flow (5) is derived from the linear momentum balance equation:

$$\frac{a}{g} = \cos \theta \left( {\tan \theta - \mu } \right).$$

(5)

Let us consider a granular flow along an inclined infinite slope. The position in the flow direction and in the flow depth direction is labeled x and y, respectively. The free surface of the flow makes an angle θ with the horizontal.

The local statement of linear momentum balance expressed in full gives:

$$\frac{{\partial \sigma_{xx} }}{\partial x} + \frac{{\partial \sigma_{xy} }}{\partial y} + \rho g\sin \theta = \rho a_{x} ,$$

(6a)

$$\frac{{\partial \sigma_{yx} }}{\partial x} + \frac{{\partial \sigma_{yy} }}{\partial y} + \rho g \cos \theta = \rho a_{y} .$$

(6b)

Exploiting the infinite character of the problem, we assume that the stress components do not vary in the x-direction, such that: $\frac{{\partial \sigma_{xx} }}{\partial x} = \frac{{\partial \sigma_{yx} }}{\partial x} = 0$. Hence, by writing $\sigma_{yy} = \sigma$ and $\sigma_{xy} = \tau ,$ we obtain:

$$\frac{\partial \tau }{\partial y} + \rho g\sin \theta = \rho a_{x} ,$$

(7a)

$$\frac{\partial \sigma }{\partial y} + \rho g\cos \theta = \rho a_{y} .$$

(7b)

The definition of dilatancy β is given by:

$$\beta = \frac{{{\text{d}}\varepsilon_{yy} }}{{{\text{d}}\varepsilon_{xy} }} = \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{v_{y} }}{{v_{x} }} \Rightarrow v_{y} = \beta v_{x} .$$

(8)

We know that in the fluid-like state the granular material reaches critical state so that β = 0. Therefore, we must have v _y  = 0 and a _y  = 0. Equation (7b) can be rewritten as:

$$\frac{\partial \sigma }{\partial y} = - \rho g\cos \theta .$$

(9)

Recalling Coulomb friction law and using the previous expression, we have:

$$\frac{\partial \tau }{\partial y} = \mu \frac{\partial \sigma }{\partial y} = - \mu \rho g\cos \theta .$$

(10)

Combining Eq. (7a) with Eq. (10) further gives:

$$- \mu g\cos \theta + g\sin \theta = a_{x} .$$

(11)

By denoting the surface acceleration in the flow direction a _x by a, we finally obtain the governing equation of the granular flow from (11):

$$\frac{a}{g} = \cos \theta \left( {\tan \theta - \mu } \right).$$

(12)

Appendix 2: Determination of the difference between the angle of avalanche θ _A and angle of repose θ _R

Knowing that in the solid-like state the surface acceleration a = 0 and the mobilized friction $\mu = \beta + \mu_{1}$, Eq. (1) can be expressed as:

$$\sin \theta = \beta \cos \theta + \mu_{l} \cos \theta .$$

(13)

Multiplying each side of Eq. (13) by cos θ _R, we have:

$$\sin \theta \cos \theta_{R} = \beta \cos \theta \cos \theta_{R} + \mu_{l} \cos \theta \cos \theta_{R} .$$

(14)

Using the formula sin (x − y) = sin x cos y − cos x sin y, we obtain:

$$\sin \left( {\theta - \theta_{R} } \right) + \cos \theta \sin \theta_{R} = \beta \cos \theta \cos \theta_{R} + \mu_{l} \cos \theta \cos \theta_{R} .$$

(15)

Moreover, we defined the parameter μ _l to be $\mu_{\text{l}} = \tan \theta_{R} ,$ such that:

$$\sin \left( {\theta - \theta_{R} } \right) + \cos \theta \sin \theta_{R} = \beta \cos \theta \cos \theta_{R} + \cos \theta \sin \theta_{R}$$

(16)

$$\Leftrightarrow \sin \left( {\theta - \theta_{R} } \right) = \beta \cos \theta \cos \theta_{R}$$

(17)

$$\Leftrightarrow \tan \left( {\theta - \theta_{R} } \right) = \beta \frac{{\cos \theta \cos \theta_{R} }}{{\cos \left( {\theta - \theta_{R} } \right)}}.$$

(18)

Once the inclination angle is equal to the angle of avalanche θ _A, the dilatancy reaches its peak value β*. Therefore, the previous equation can be rewritten as:

$$\tan \Delta \theta = \beta^{*} \frac{{\cos \theta_{A} \cos \theta_{R} }}{\cos \Delta \theta }.$$

(19)