$$ \newcommand{\partd}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\partdd}[2]{\frac{\partial^{2} #1}{\partial {#2}^{2}}} \newcommand{\ddt}{\frac{d}{dt}} \newcommand{\Int}{\int\limits} \newcommand{\D}{\displaystyle} \newcommand{\ie}{\textit{i.e. }} \newcommand{\dA}{\; \mbox{dA}} \newcommand{\dz}{\; \mbox{dz}} \newcommand{\tr}{\mathrm{tr}} \renewcommand{\eqref}[1]{Eq.~(\ref{#1})} \newcommand{\reqs}[2]{\req{#1} and \reqand{#2}} \newcommand{\rthreeeqs}[3]{Eqs.~(\ref{#1}), (\ref{#2}), and (\ref{#3})} $$
Cardiovascular biomechanics

 

 

 

4.4 Anisotropic Materials

Notation for anisotropic materials. Angular momentum \( \Rightarrow \boldsymbol{T} = \boldsymbol{T}^T \). Green's strain \( \boldsymbol{E} \equiv \boldsymbol{E}^T \Leftarrow \boldsymbol{E \equiv H + H^T + H^T H} \). Only 6 distinct values for \( \boldsymbol{T} \) and \( \boldsymbol{E} \). Special notation for coordinate stresses and strains $$ \begin{equation} T = \left [ % \begin{array}{ccc} T_1 & T_6 & T_5 \\ \vdots & T_2 & T_4 \\ \vdots & \vdots & T_3 \end{array} \right ] % \quad \quad E = \left [ % \begin{array}{ccc} E_1 & E_6 & E_5 \\ \vdots & E_2 & E_4 \\ \vdots & \vdots & E_3 \end{array} \right ] % \tag{4.119} \end{equation} $$

Anisotropic constitutive equation A fully anisotropic, linearly elastic constitutive equation $$ \begin{equation} T_\alpha = S_{\alpha\, \beta} E_\beta, \qquad \{\alpha, \beta \} \in \{1 \ldots 6 \} \tag{4.120} \end{equation} $$

\( T_\alpha \) and \( E_\beta \) are \( 6 \times 1 \) vector matrices, whereas \( S_{\alpha \, \beta} \) is a \( 6 \times 6 \) elasticity or stiffness matrix. We will show \( S=S^T \) for a hyperelastic material, i.e., only 21 independent stiffnesses for full anisotropy.

Alternative formulation $$ \begin{equation} E_\alpha = K_{\alpha\, \beta} T_\beta, \qquad K=S^{-1} \tag{4.121} \end{equation} $$ \( K_{\alpha \, \beta} \) is a \( 6 \times 6 \) compliance or flexibility matrix

Stiffness matrix for Hookean materials.

Isotropic, linearly elastic material $$ \begin{equation} S = \frac{\eta}{2(1+\nu)(1-2\nu)} \; \left [ \begin{array}{cccccc} 2(1-\nu) & 2\nu & 2\nu & 0 & 0 & 0 \\ 2\nu & 2(1-\nu) & 2\nu & 0 & 0 & 0 \\ 2\nu & 2\nu & 2(1-\nu) & 0 & 0 & 0 \\ 0 & 0 & 0 & (1-2\nu) & 0 & 0 \\ 0 & 0 & 0 & 0 & (1-2\nu) & 0 \\ 0 & 0 & 0 & 0 & 0 & (1-2\nu) \end{array} \right ] \tag{4.122} \end{equation} $$

Note: \( S=S^T \)

Symmetry of stiffness matrix. Hyperelastic anisotropic materials

Hyperelastic material $$ \begin{align} \boldsymbol{T} & = \partd{\phi}{\boldsymbol{E}} \Leftrightarrow T_{ij} = \partd{\phi}{E_{ij}} \tag{4.123}\\ & \Leftrightarrow T_\alpha = \partd{\phi}{E_\alpha} \tag{4.124} \end{align} $$ The stress tensor \( \boldsymbol{T} \). Elastic energy \( \phi \) per unit volume.

Linearly anisotropic material $$ \begin{equation} T_\alpha = S_{\alpha \beta} E_\beta \tag{4.125} \end{equation} $$

Hyperelastic linearly anisotropic material $$ \begin{align} \partd{T_\alpha}{E_\beta} &= S_{\alpha \beta} = \frac{\partial^2\phi}{\partial E_\alpha \partial E_\beta} \equiv \frac{\partial^2\phi}{\partial E_\beta \partial E_\alpha} = S_{\beta \alpha} \tag{4.126}\\ & \Rightarrow S_{\alpha \beta} = S_{\beta \alpha} \Leftrightarrow S=S^T \tag{4.127} \end{align} $$ $$ \begin{equation} \Rightarrow S_{\alpha \beta} = S_{\beta \alpha} \Leftrightarrow S=S^T \tag{4.128} \end{equation} $$ Independent stiffnesses reduces from 36 to 21.

Materials with one plane of symmetry:

Figure 32: Materials with one plane of symmetry.

Structure symmetric with respect to a plane through the particle. The mirror image of the structure is identical to the structure itself. The number of stiffnesses is reduced from 21 to 13.

The fig a) shows a plane of symmetry normal to \( \boldsymbol{e}_3 \). The fig b) shows \( 180^\circ \) rotation about the \( \boldsymbol{e}_3 \)-axis A state of strain \( E \) will produce the \( \Rightarrow \bar{S} = S \) The number of stiffnesses is reduced from 21 to 13.

\( \pi \)-rotation about the \( \boldsymbol{e}_3 \)-axis $$ \begin{equation} T = \left [ % \begin{array}{rrr} T_1 & T_6 & T_5 \\ \vdots & T_2 & T_4 \\ \vdots & \vdots & T_3 \end{array} \right ] % \quad \quad E = \left [ % \begin{array}{rrr} E_1 & E_6 & E_5 \\ \vdots & E_2 & E_4 \\ \vdots & \vdots & E_3 \end{array} \right ] % \tag{4.129} \end{equation} $$

Transformation matrix $$ \begin{equation} Q = % \left [ % \begin{array}{rrr} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right ]% ,\quad Q_{\pi} \left [ % \begin{array}{rrr} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array} \right ]% \tag{4.130} \end{equation} $$ $$ \begin{equation} \bar{T} = Q^T T Q= \left [ % \begin{array}{rrr} T_1 & T_6 & -T_5 \\ \vdots & T_2 & -T_4 \\ \vdots & \vdots & T_3 \end{array} \right ] % ,\quad \bar{E} = Q^T E Q = \left [ % \begin{array}{rrr} E_1 & E_6 & -E_5 \\ \vdots & E_2 & -E_4 \\ \vdots & \vdots & E_3 \end{array} \right ] % \tag{4.131} \end{equation} $$ Plane symmetric anisotropic constitutive equations $$ \begin{align} T_{\alpha} = S_{\alpha \, \beta} E_{\beta}, \qquad \bar{T}_{\alpha} = \bar{S}_{\alpha \, \beta} \bar{E}_{\beta} \tag{4.132} \end{align} $$ Plane symmetry \( \Rightarrow \bar{S}_{\alpha \, \beta} = S_{\alpha \, \beta} \). Notation: $\lambda$= 4 and 5, and \( \rho, \; \gamma \) = \( 1 \ldots 3 \) and 6. From the constitutive equations $$ \begin{align} T_{\lambda} &= S_{\lambda \rho} E_{\rho} + S_{\lambda 4} E_{4} + S_{\lambda 5} E_{5} \tag{4.133}\\ -T_{\lambda} &= S_{\lambda \rho} E_{\rho} + S_{\lambda 4} (-E_{4}) + S_{\lambda 5} (-E_{5}) \tag{4.134}\\ T_{\gamma} &= S_{\gamma \rho} E_{\rho} + S_{\gamma 4} E_{4} + S_{\gamma 5} E_{5} \tag{4.135}\\ T_{\gamma} &= S_{\gamma \rho} E_{\rho} + S_{\gamma 4} (-E_{4}) + S_{\gamma 5} (-E_{5}) \tag{4.136} \end{align} $$ By comparison \( S_{\lambda \, \rho} = 0 \) and \( S_{\gamma \lambda} = 0 \).