Nonlocal orientational distribution of contact forces in granular samples containing elongated particles

Abstract

In this paper we introduce the branch tensor as an internal variable able to account for the structural anisotropy of a granular sample. The distribution of averaged contact forces is assumed to depend not only on the macroscopic stress and the local orientation, but also on the value of the fabric tensor. In contrast to previous work, including the fabric tensor has the crucial advantage that accounts for all relative positions between interacting particles, through the average value of the branch tensor. Based on a classical representation result, we propose an identification procedure that uses information obtained from both isotropic and anisotropic configurations.

Appendix. Technical details of the result in (12)

This result is obtained in three steps using a classical representation result for isotropic functions due to Wang [15].

1.1 (a) General form for f

Using the general result in [15] we find that the general form of an isotropic function is a linear combination of vector-valued invariants below

$$ {\user2{n}},\quad{\varvec{\Upsigma}}{\user2{n}},\quad {\user2{H}}{\user2{n}},\quad{\varvec{\Upsigma}}{\user2{H}}{\user2{n}}, \quad {\user2{H}}{\varvec{\Upsigma}}{\user2{n}}, \quad{\varvec{\Upsigma}}^2{\user2{n}},\quad {\user2{H}}^2{\user2{n}} $$

(35)

with coefficients depending on the following combinations

$$ \begin{aligned} \,&\hbox{tr}({\user2{H}}),\; {\user2{n}}\cdot{\varvec{\Upsigma}}{\user2{n}},\; {\user2{n}}\cdot{\user2{H}}{\user2{n}},\; \hbox{tr}({\varvec{\Upsigma}}),\; \hbox{tr}({\user2{H}}{\varvec{\Upsigma}}),\;{\varvec{\Upsigma}}{\user2{n}}\cdot{\user2{H}}{\user2{n}}\\ \,&\hbox{tr}({\varvec{\Upsigma}}^2),\ \hbox{tr}({\varvec{\Upsigma}}^3),\; \hbox{tr}({\user2{H}}^2),\ \hbox{tr}({\user2{H}}^3),\; \hbox{tr}({\user2{H}}^2{\varvec{\Upsigma}}),\\ \,&\hbox{tr}({\user2{H}}{\varvec{\Upsigma}}^2),\; \hbox{tr}({\user2{H}}^2{\varvec{\Upsigma}}^2), {\user2{n}}\cdot{\varvec{\Upsigma}}^2{\user2{n}},\; {\user2{n}}\cdot{\user2{H}}^2{\user2{n}}. \end{aligned} $$

(36)

By inspection in (36) only the first line contains terms at most linear in ${\varvec{\Upsigma}}$ and H. Moreover, products between scalar-valued and vector-valued invariants lead to a general form for f as

$${{\user2{f}}}({\varvec{\Upsigma}},{\user2{H}},{\user2{n}})=\sum_{i= 1}^{19}a_i{{\user2{f}}}_i({\varvec{\Upsigma},{\user2{H}}},{\user2{n}}) $$

(37)

where a _i  ∈ $\mathbb{R}$ and the list of vector-valued invariants f _i is given by:

$$\begin{array}{lll}{{\user2{f}}}_{1}&={\user2{n}},&{{\user2{f}}}_{10}={\user2{H}}{\user2{n}}, \\{{\user2{f}}}_2&=\hbox{tr}({\varvec{\Upsigma}}){\user2{n}},&{{\user2{f}}}_{11}=\hbox{tr}({\varvec{\Upsigma}}){\user2{H}}{\user2{n}}, \\{{\user2{f}}}_3&= \hbox{tr}({\varvec{\Upsigma}})\hbox{tr}({\user2{H}}){\user2{n}}, &{{\user2{f}}}_{12}=({\user2{n}}\cdot{\varvec{\Upsigma}}{\user2{n}}){\user2{H}}{\user2{n}}, \\{{\user2{f}}}_4&= \hbox{tr}({\varvec{\Upsigma}})({\user2{n}}\cdot{\user2{H}}{\user2{n}}){\user2{n}},& {{\user2{f}}}_{13}={\varvec{\Upsigma}}{\user2{n}}, \\{{\user2{f}}}_5&=({\user2{n}}\cdot{\varvec{\Upsigma}}{\user2{n}}){\user2{n}},&{{\user2{f}}}_{14}=\hbox{tr}({\user2{H}}){\varvec{\Upsigma}}{\user2{n}}, \\{{\user2{f}}}_6&=\hbox{tr}({\user2{H}})({\user2{n}}\cdot{\varvec{\Upsigma}}{\user2{n}}){ \user2{n}}, & {{\user2{f}}}_{15}=({\user2{n}}\cdot{\user2{H}}{\user2{n}}){\varvec{\Upsigma}}{\user2{n}}, \\{{\user2{f}}}_7&=({\user2{n}}\cdot{\varvec{\Upsigma}}{\user2{n}})({\user2{n}}\cdot{{\mathbf{H}}}{\user2{n}}){\user2{n}},&{{\user2{f}}}_{16}={\varvec{\Upsigma}} {\user2{H}}{\user2{n}}, \\{{\user2{f}}}_8&=\hbox{tr}({\varvec{\Upsigma}}{\user2{H}}){\user2{n}}, & {{\user2{f}}}_{17}={\user2{H}}{\varvec{\Upsigma}}{\user2{n}},\\ {{\user2{f}}}_9&=({\user2{H}}{\user2{n}}\cdot{\varvec{\Upsigma}}{\user2{n}}){\user2{n}}, &{{\user2{f}}}_{18}=\hbox{tr}({\user2{H}}){\user2{n}}, \\&&{{\user2{f}}}_{19}= ({\user2{H}}{\user2{n}}\cdot{\user2{n}}){\user2{n}}. \end{array} $$

(38)

1.2 (b) Restrictions due to consistency relation

In (37) the a _i are restricted by the consistency condition (8); identification for symmetric part gives

$$ \begin{aligned} &a_1=0,\\ &a_5=-4a_2\\ &a_9=-4a_8-\frac{1}{2}a_7-a_6-a_4-4a_3,\\ &a_{12}=-4a_{11}-\frac{1}{2} a_7-a_6-3a_4-4a_3,\\ &a_{13} =1+2a_2,\\ &a_{15} =-4a_{14}-4a_3-a_4-3a_6-\frac{1}{2} a_7,\\ &a_{17} =-a_{16}+2a_{14}+2a_{11}+2a_8+\frac{1}{2}a_7 \\ &\quad + 3a_6+3a_4+8a_3,\\ &a_{18} = \frac{1}{2} a_{10},\\ &a_{19} =-2a_{10}. \end{aligned} $$

(39)

and identification of the skew-symmetric part leads to

$$ a_{16}=2a_{14}+a_8+\frac{1}{4}a_7+2a_6+a_4+4a_3. $$

(40)

We shall denote by g _i the factors in f of a _i . Relations (39) and (40) reduce the number of coefficients a _i to nine, for i ∈ I = {2,3,4,6,7,8,10,11,14}.

1.3 (c) Linear independence of remaining invariants

Using g _i we are lead to

$$ {{\user2{f}}}={\varvec{\Upsigma}}{\user2{n}}+\sum_{i\in I} a_i{\user2{g}}_i. $$

(41)

The following results are straightforward

• If

  $$ a_2{\user2{g}}_2+a_3{\user2{g}}_3+ a_4{\user2{g}}_4+a_6{\user2{g}}_6+a_7{\user2{g}}_7+a_{10}{\user2{g}} _{10}={\user2{0}} $$

  for any ${\varvec{\Upsigma}}$ and any H then a ₂ = a ₃ = a ₄ = a ₆ = a ₇ = a ₁₀ = 0.

• We have for any ${\varvec{\Upsigma}}$ and H:

  $$ -{\user2{g}}_3+{\user2{g}}_4+{\user2{g}}_6+{\user2{g}}_8= {\user2{0}},\quad {\user2{g}}_{\user2{4}}={\user2{g}}_{11},\quad {\user2{g}}_6={\user2{g}}_{14}. $$

• One can easily check that

  $$ {\user2{I}}_2:={\user2{g}}_2=2{\varvec{\Upsigma}}{\user2{n}}+ (\hbox{tr}{\varvec{\Upsigma}}){\user2{n}}- 4({\user2{n}}\cdot({\varvec{\Upsigma}}{\user2{n}})){\user2{n}}, $$

  and

  $$ {\user2{g}}_{10}= 2{\user2{H}}{\user2{n}}+(\hbox{tr}{\user2{H}}){\user2{n}}- 4({\user2{n}}\cdot({\user2{H}}{\user2{n}})){\user2{n}}, $$

  while the four remaining invariants can be combined to give

  $$ \begin{aligned} {\user2{I}}_3&:={\user2{g}}_3-4{\user2{g}}_4,\quad {\user2{I}}_4:={\user2{g}}_3- 4{\user2{g}}_6, \\ {\user2{I}}_5&:={\user2{g}}_3-8{\user2{g}}_7,\quad {\user2{I}}_6:={\user2{g}}_3-16{\user2{g}}_7, \end{aligned} $$

  thus leading finally to (12)–(17).