Anelastic reorganisation of fibre-reinforced biological tissues

Abstract

In this work, we contribute to the study of the structural reorganisation of biological tissues in response to mechanical stimuli. We specialise our investigation to a class of hydrated soft tissues, whose internal structure features reinforcing fibres. These are oriented statistically within the tissue, and their pattern of orientation is such that, at each material point, the tissue is anisotropic. From its natural, stress-free state, the tissue can be distorted anelastically into a global reference configuration, and then deformed under the action of external mechanical loads. The anelastic distortions are responsible for changing irreversibly the internal structure of the tissue, which, in the present context, occurs through both the rearrangement of the bonds among the tissue cells and the deformation-driven reorientation of the fibres. The anelastic strains, in addition, are assumed to model the onset and evolution of microcracks in the tissue, which may be triggered by the mechanical loads applied to the tissue in the case of traumatic events, or diseases. For our purposes, we formulate an anisotropic model of remodelling and we consider a fully isotropic model of structural reorganisation for comparison, with the aim to study if, how, and to what extent the evolution of anelastic distortions is influenced by the tissue’s anisotropy.

Appendix: Fourth-order tensors

The notation adopted in the following is taken from [17]. Let $[T{\mathscr {B}}]^{1}_{~1}$, $[T{\mathscr {B}}]_{1}^{~1}$, $[T{\mathscr {B}}]^{2}_{0}$, and $[T{\mathscr {B}}]^{0}_{2}$ denote the spaces of all second-order tensors which, as bilinear maps, read

$$\begin{aligned}&{\varvec{A}}:T^{\star }{\mathscr {B}}\times T{\mathscr {B}}\rightarrow {\mathbb {R}}, \end{aligned}$$

(55a)

$$\begin{aligned}&{\varvec{B}}:T{\mathscr {B}}\times T^{\star }{\mathscr {B}}\rightarrow {\mathbb {R}}, \end{aligned}$$

(55b)

$$\begin{aligned}&{\varvec{T}}:T^{\star }{\mathscr {B}}\times T^{\star }{\mathscr {B}}\rightarrow {\mathbb {R}}, \end{aligned}$$

(55c)

$$\begin{aligned}&{\varvec{Q}}:T{\mathscr {B}}\times T{\mathscr {B}}\rightarrow {\mathbb {R}}, \end{aligned}$$

(55d)

respectively. Let also $([T{\mathscr {B}}]^{2}_{0},\mathrm {sym})$ and $([T{\mathscr {B}}]_{2}^{0},\mathrm {sym})$ be, respectively, the subspaces of $[T{\mathscr {B}}]^{2}_{0}$ and $[T{\mathscr {B}}]^{0}_{2}$ of all symmetric, second-order tensors. The elements of $[T{\mathscr {B}}]^{1}_{~1}$ and $[T{\mathscr {B}}]_{1}^{~1}$ can be written as linear maps from $T{\mathscr {B}}$ into itself, and from $T^{\star }{\mathscr {B}}$ into itself, respectively, while the elements of $[T{\mathscr {B}}]^{2}_{0}$, and $[T{\mathscr {B}}]^{0}_{2}$ can be written as linear maps from $T^{\star }{\mathscr {B}}$ into $T{\mathscr {B}}$, and from $T{\mathscr {B}}$ into $T^{\star }{\mathscr {B}}$, respectively.

Let us also consider the spaces $[T{\mathscr {B}}]^{2}_{~2}$ and $[T{\mathscr {B}}]_{2}^{~2}$ of all fourth-order tensors of the type

$$\begin{aligned}&{\mathbb {T}}\in [T{\mathscr {B}}]^{2}_{~2},~{\mathbb {T}}:T^{\star }{\mathscr {B}}\times T^{\star }{\mathscr {B}}\times T{\mathscr {B}}\times T{\mathscr {B}}\rightarrow {\mathbb {R}}, \\&{\mathbb {Q}}\in [T{\mathscr {B}}]_{2}^{~2},~{\mathbb {Q}}:T{\mathscr {B}}\times T{\mathscr {B}}\times T^{\star }{\mathscr {B}}\times T^{\star }{\mathscr {B}}\rightarrow {\mathbb {R}}. \end{aligned}$$

An element of $[T{\mathscr {B}}]^{2}_{~2}$ can also be represented as a linear map from $[T{\mathscr {B}}]^{2}_{0}$ into $[T{\mathscr {B}}]^{2}_{0}$. Analogously, an element of $[T{\mathscr {B}}]_{2}^{~2}$ can be represented as a linear map from $[T{\mathscr {B}}]^{0}_{2}$ into $[T{\mathscr {B}}]^{0}_{2}$. For instance, the fourth-order tensor

$$\begin{aligned}&{\mathbb {I}}:[T{\mathscr {B}}]^{2}_{0}\rightarrow ([T{\mathscr {B}}]^{2}_{0},\mathrm {sym}), \nonumber \\&{\mathbb {I}}=\tfrac{1}{2}\left( {\varvec{I}}\,{\underline{\otimes }}\,{\varvec{I}}+{\varvec{I}}\,{\overline{\otimes }}\,{\varvec{I}}\right) , \end{aligned}$$

(56)

where ${\varvec{I}}:T{\mathscr {B}}\rightarrow T{\mathscr {B}}$ is the identity tensor in $T{\mathscr {B}}$, returns the symmetric part of the element of $[T{\mathscr {B}}]^{2}_{0}$ to which it is applied. Given two tensors ${\varvec{A}},{\varvec{D}}\in [T{\mathscr {B}}]^{1}_{~1}$, the representation of the tensor products ${\varvec{A}}{\underline{\otimes }}{\varvec{D}}$ and ${\varvec{A}}{\overline{\otimes }}{\varvec{D}}$ in index notation reads $[{\varvec{A}}{\underline{\otimes }}{\varvec{D}}]^{AB}_{~~\;MN}=A^{A}_{~M}D^{B}_{~N}$ and $[{\varvec{A}}{\overline{\otimes }}{\varvec{D}}]^{AB}_{~~\;MN}=A^{A}_{~N}D^{B}_{~M}$ [9]. Accordingly, in index notation, ${\mathbb {I}}$ is represented by the expression

$$\begin{aligned} {\mathbb {I}}^{AB}_{~~~MN}=\tfrac{1}{2}\left( \delta ^{A}_{~M}\delta ^{B}_{~N}+ \delta ^{A}_{~N}\delta ^{B}_{~M}\right) . \end{aligned}$$

(57)

Thus, for every ${\varvec{T}}\in [T{\mathscr {B}}]^{2}_{0}$, it holds that

$$\begin{aligned} {\mathbb {I}}:{\varvec{T}}=\tfrac{1}{2}\left( {\varvec{T}}+{\varvec{T}}^{ \mathrm {T}}\right) =\mathrm {sym}({\varvec{T}}), \end{aligned}$$

(58)

where the symbol “ : ” stands for “double contraction”. In index notation, it reads $({\mathbb {I}}:{\varvec{T}})^{AB}={\mathbb {I}}^{AB}_{~~~MN}T^{MN}=[\mathrm {sym}({\varvec{T}})]^{AB}$. By definition, ${\mathbb {I}}$ is the identity fourth-order tensor over the space $([T{\mathscr {B}}]^{2}_{0},\mathrm {sym})$. From here on, we consider only the restrictions of the fourth-order tensors of $[T{\mathscr {B}}]^{2}_{0}$ onto $([T{\mathscr {B}}]^{2}_{0},\mathrm {sym})$.

For every ${\varvec{T}}\in ([T{\mathscr {B}}]^{2}_{0},\mathrm {sym})$, the fourth-order tensor

$$\begin{aligned}&{\mathbb {K}}^{*}:([T{\mathscr {B}}]^{2}_{0},\mathrm {sym})\rightarrow ([T{\mathscr {B}}]^{2}_{0},\mathrm {sym}), \nonumber \\&{\mathbb {K}}^{*}=\tfrac{1}{3}{\varvec{C}}^{-1}\otimes {\varvec{C}} \end{aligned}$$

(59)

extracts the spherical part of ${\varvec{T}}$ with respect to the metric ${\varvec{C}}$, i.e.,

$$\begin{aligned} {\mathbb {K}}^{*}:{\varvec{T}}=\tfrac{1}{3}\mathrm {tr}({\varvec{C}}{ \varvec{T}}){\varvec{C}}^{-1}. \end{aligned}$$

(60)

The deviatoric part of ${\varvec{T}}$ with respect to the metric ${\varvec{C}}$ is obtained by substracting ${\mathbb {K}}^{*}:{\varvec{T}}$ to ${\varvec{T}}$. This operation can be represented by the application of the fourth-order tensor

$$\begin{aligned}&{\mathbb {M}}^{*}:([T{\mathscr {B}}]^{2}_{0},\mathrm {sym})\rightarrow ([T{\mathscr {B}}]^{2}_{0},\mathrm {sym}) \nonumber \\&{\mathbb {M}}^{*}={\mathbb {I}}-{\mathbb {K}}^{*}, \end{aligned}$$

(61)

to ${\varvec{T}}$ i.e.,

$$\begin{aligned} {\mathbb {M}}^{*}:{\varvec{T}}=({\mathbb {I}}-{\mathbb {K}}^{*}): {\varvec{T}}={\varvec{T}}-\tfrac{1}{3}\mathrm {tr}({\varvec{C}} {\varvec{T}}){\varvec{C}}^{-1}. \end{aligned}$$

(62)

Clearly, it holds that $\mathrm {tr}\left[ {\varvec{C}}\left( {\mathbb {M}}^{*}:{\varvec{T}}\right) \right] =0$. We remark that, by their own definition, ${\mathbb {K}}^{*}$ and ${\mathbb {M}}^{*}$ constitute the partition of unity, i.e., ${\mathbb {I}}={\mathbb {K}}^{*}+{\mathbb {M}}^{*}$.

In analogous manner, we introduce the identity fourth-order tensor over the space $([T{\mathscr {B}}]^{0}_{2},\mathrm {sym})$, i.e.,

$$\begin{aligned}&{\mathbb {I}}^{\mathrm {T}}:([T{\mathscr {B}}]^{0}_{2},\mathrm {sym}) \rightarrow ([T{\mathscr {B}}]^{0}_{2},\mathrm {sym}), \nonumber \\&{\mathbb {I}}^{\mathrm {T}}=\tfrac{1}{2}\left( {\varvec{I}}^{\mathrm {T}} \,{\underline{\otimes }}\,{\varvec{I}}^{\mathrm {T}}+{\varvec{I}}^{\mathrm {T}} \,{\overline{\otimes }}\,{\varvec{I}}^{\mathrm {T}}\right) , \end{aligned}$$

(63)

where ${\varvec{I}}^{\mathrm {T}}:T^{\star }{\mathscr {B}}\rightarrow T^{\star }{\mathscr {B}}$ is the identity tensor in $T^{\star }{\mathscr {B}}$. For every ${\varvec{Q}}\in ([T{\mathscr {B}}]^{0}_{2},\mathrm {sym})$ it holds that

$$\begin{aligned} {\mathbb {I}}^{\mathrm {T}}:{\varvec{Q}}=\tfrac{1}{2}\left( {\varvec{Q}}+{\varvec{Q}}^{\mathrm {T}}\right) \equiv {\varvec{Q}}. \end{aligned}$$

(64)

The spherical and the deviatoric parts of ${\varvec{Q}}$ with respect to the inverse metric ${\varvec{C}}^{-1}$ are extracted by employing the fourth-order tensors

$$\begin{aligned}&{\mathbb {K}}^{*\mathrm {T}}:([T{\mathscr {B}}]^{0}_{2},\mathrm {sym}) \rightarrow ([T{\mathscr {B}}]^{0}_{2},\mathrm {sym}), \nonumber \\&{\mathbb {K}}^{*\mathrm {T}}=\tfrac{1}{3}{\varvec{C}}\otimes {\varvec{C}}^{-1}, \end{aligned}$$

(65)

and

$$\begin{aligned}&{\mathbb {M}}^{*\mathrm {T}}:([T{\mathscr {B}}]^{0}_{2}, \mathrm {sym})\rightarrow ([T{\mathscr {B}}]^{0}_{2},\mathrm {sym}), \nonumber \\&{\mathbb {M}}^{*\mathrm {T}}={\mathbb {I}}^{\mathrm {T}}-{ \mathbb {K}}^{*\mathrm {T}}, \end{aligned}$$

(66)

respectively, which are such that

$$\begin{aligned}&{\mathbb {K}}^{*\mathrm {T}}:{\varvec{Q}}=\tfrac{1}{3}\mathrm {tr}({\varvec{C}}^{-1}{\varvec{Q}}){\varvec{C}}, \end{aligned}$$

(67)

$$\begin{aligned}&{\mathbb {M}}^{*\mathrm {T}}:{\varvec{Q}}=({\mathbb {I}}^{\mathrm {T}}-{\mathbb {K}}^{*\mathrm {T}}):{\varvec{Q}}={\varvec{Q}}-\tfrac{1}{3}\mathrm {tr}({\varvec{C}}^{-1}{\varvec{Q}}){\varvec{C}}. \end{aligned}$$

(68)

In this case, it holds that $\mathrm {tr}\left[ {\varvec{C}}^{-1}\left( {\mathbb {M}}^{*\mathrm {T}}:{\varvec{Q}}\right) \right] =0$.

Finally, we introduce the fourth-order tensor

$$\begin{aligned}&{\mathbb {I}}^{\sharp *}:([T{\mathscr {B}}]^{0}_{2},\mathrm {sym})\rightarrow ([T{\mathscr {B}}]^{2}_{0},\mathrm {sym}), \nonumber \\&{\mathbb {I}}^{\sharp *}=\tfrac{1}{2}\left( {\varvec{C}}^{-1}\,{\underline{\otimes }}\,{\varvec{C}}^{-1}+{\varvec{C}}^{-1}\,{\overline{\otimes }}\,{\varvec{C}}^{-1} \right) . \end{aligned}$$

(69)

For every ${\varvec{Q}}\in ([T{\mathscr {B}}]^{0}_{2},\mathrm {sym})$, it holds that

$$\begin{aligned}&{\mathbb {I}}^{\sharp *}:{\varvec{Q}}={\varvec{C}}^{-1}{\varvec{Q}}{\varvec{C}}^{-1}. \end{aligned}$$

(70)

In index notation, Eq. (70) implies $({\mathbb {I}}^{\sharp *}:{\varvec{Q}})^{AB}=({\varvec{C}}^{-1})^{AM}Q_{MN}({\varvec{C}}^{-1})^{NB}$, which means that ${\mathbb {I}}^{\sharp *}$ raises the indices of ${\varvec{Q}}$ through the inverse metric tensor ${\varvec{C}}^{-1}$ rather than through ${\varvec{G}}^{-1}$, the latter being the inverse of the metric tensor ${\varvec{G}}$ in the undeformed configuration. In analogy with ${\mathbb {K}}^{*}$ and ${\mathbb {M}}^{*}$, we also consider the fourth-order tensors

$$\begin{aligned}&{\mathbb {K}}^{\sharp *}:([T{\mathscr {B}}]^{0}_{2},\mathrm {sym})\rightarrow ([T{\mathscr {B}}]^{2}_{0},\mathrm {sym}), \nonumber \\&{\mathbb {K}}^{\sharp *}=\tfrac{1}{3}{\varvec{C}}^{-1}\,{\otimes }\,{\varvec{C}}^{-1}, \end{aligned}$$

(71a)

$$\begin{aligned}&{\mathbb {M}}^{\sharp *}:([T{\mathscr {B}}]^{0}_{2},\mathrm {sym})\rightarrow ([T{\mathscr {B}}]^{2}_{0},\mathrm {sym}), \nonumber \\&{\mathbb {M}}^{\sharp *}={\mathbb {I}}^{\sharp *}-{\mathbb {K}}^{\sharp *}. \end{aligned}$$

(71b)

For every ${\varvec{Q}}\in ([T{\mathscr {B}}]^{0}_{2},\mathrm {sym})$, we obtain

$$\begin{aligned}&{\mathbb {K}}^{\sharp *}:{\varvec{Q}}=\tfrac{1}{3}\mathrm {tr}({\varvec{C}}^{-1}{\varvec{Q}}){\varvec{C}}^{-1}, \end{aligned}$$

(72a)

$$\begin{aligned}&{\mathbb {M}}^{\sharp *}:{\varvec{Q}}={\varvec{C}}^{-1}{\varvec{Q}}{\varvec{C}}^{-1}-\tfrac{1}{3}\mathrm {tr}({\varvec{C}}^{-1}{\varvec{Q}}){\varvec{C}}^{-1}. \end{aligned}$$

(72b)

Note that the second-order tensor ${\mathbb {M}}^{\sharp *}:{\varvec{Q}}$ is deviatoric in the sense that $\mathrm {tr}[{\varvec{C}}({\mathbb {M}}^{\sharp *}:{\varvec{Q}})]=0$.