A structurally based viscoelastic model for passive myocardium in finite deformation

Abstract

This paper discusses the finite-deformation viscoelastic modeling for passive myocardium tissue. The formulations established can also be applied to model other fiber-reinforced soft tissue. Based on the morphological structure of the myocardium, a specific free-energy function is constructed to reflect its orthotropicity. After deriving the stress–strain relationships in the simple shear deformation, a genetic algorithm is used to optimally estimate the material parameters of the myocardial constitutive equation. The results show that the proposed myocardial model can well fit the shear experimental data. To validate the viscoelastic model, it is used to predict the creep and the dynamic responses of a cylindrical model of the left ventricle. Upon comparing the results calculated by the proven myocardial elastic model with those by the viscoelastic model, the merits of the latter are discussed.

Appendix: Stress for the cylindrical model

In this appendix, the formulation of the deviatoric Cauchy stress, ${\hat{\sigma }}=\mathbf {F}^{\mathrm {T}}{\hat{\mathbf {S}}}\mathbf {F}$, appearing in the analysis of the cylindrical model is explicitly presented. The first ingredients for obtaining ${\hat{\sigma }}$ are the deformation invariants involved, which can be expressed as

$$\begin{aligned}&I_1 = r_i^2+r^2 \left( \frac{1}{R^2}+v^2\right) +\lambda ^2 \\&I_{4f}= \frac{r^2 \cos (\beta )^2}{R^2}-\frac{2 r^2 v \cos (\beta ) \sin (\beta )}{R}\nonumber \\&~~~~~~~~~~+\left( r^2 v^2+\lambda ^2\right) \sin (\beta )^2 \\&I_{4s} = r_i^2 \\&I_{8fs} = 0 \end{aligned}$$

Hence, the terms related to the invariant $I_{8fs}$ cause no effect on the stress.

To make the presentation tractable, the stress components are treated separately:

• (1) for the component associated with the first invariant,

  $$\begin{aligned} (\varvec{\sigma })_1= & {} \mathbf {F}(\mathbb {P}:\mathbf {I}) \mathbf {F}^{\mathrm {T}}\\ {}= & {} cof_{1R}\mathbf {e}_R\mathbf {e}_R +cof_{1\varTheta }\mathbf {e}_{\varTheta }\mathbf {e}_{\varTheta }\\&+\,cof_{1Z}\mathbf {e}_Z\mathbf {e}_Z+cof_{1Z\varTheta }\mathbf {e} _Z\mathbf {e}_{\varTheta } \end{aligned}$$

  in which the three coefficients are explicitly expressed as

  $$\begin{aligned}&cof_{1R} = \frac{1}{3} \left[ 2 r_i^2+r^2 \left( -\frac{1}{R^2} -v^2\right) -\lambda ^2\right] \\&cof_{1\varTheta } = \frac{2 r^2 \left( 1+R^2 v^2\right) -R^2 \left( r_i^2+\lambda ^2\right) }{3 R^2} \\&cof_{1Z} = \frac{1}{3} \left[ -r_i^2+r^2 \left( -\frac{1}{R^2}-v^2\right) +2 \lambda ^2\right] \\&cof_{1Z\varTheta }=r v \lambda \end{aligned}$$

• (2) for the component in the fiber direction associated with the invariant $I_{4f}$,

  $$\begin{aligned} (\varvec{\sigma })_{4f}= & {} \mathbf {F}(\mathbb {P}: \mathbf {f}_0\otimes \mathbf {f}_0)\mathbf {F}^{\mathrm {T}}\\= & {} cof_{4fR}\mathbf {e}_R\mathbf {e}_R +cof_{4f\varTheta }\mathbf {e}_{\varTheta }\mathbf {e}_{\varTheta } +cof_{4fZ}\mathbf {e}_Z\mathbf {e}_Z \end{aligned}$$

  whose coefficients are

  $$\begin{aligned} cof_{4fR}= & {} -\frac{1}{3 R^2}\Big (r^2 \cos (\beta )^2+\,R \sin (\beta ) [-2 r^2 v \cos (\beta )\\\nonumber&+R \left( r^2 v^2+\lambda ^2\right) \sin (\beta )]\Big ) \\ cof_{4f\varTheta }= & {} \frac{1}{3 R^2}\Big (2 r^2 \cos (\beta )^2-4 r^2 R v \cos (\beta ) \sin (\beta )\\ \nonumber&+\,R^2 \left( 2 r^2 v^2-\lambda ^2\right) \sin (\beta )^2\Big )\\ cof_{4fZ}= & {} \frac{1}{3 R^2}\Big (-r^2 \cos (\beta )^2+R^2 \left( -r^2 v^2+2 \lambda ^2\right) \\&\times \sin (\beta )^2+r^2 R v \sin (2 \beta )\Big )\\ cof_{4fZ\varTheta }= & {} \frac{r \lambda \sin (\beta ) [-\cos (\beta )+\,R v \sin (\beta )]}{R} \end{aligned}$$

• (3) for the component in the sheet direction associated with the invariant $I_{4s}$,

  $$\begin{aligned} (\varvec{\sigma })_{4s}= & {} \mathbf {F}(\mathbb {P}:\mathbf {s}_0 \otimes \mathbf {s}_0)\mathbf {F}^{\mathrm {T}}\\= & {} cof_{4sR}\mathbf {e}_R\mathbf {e}_R +cof_{4s\varTheta }\mathbf {e}_{\varTheta }\mathbf {e}_{\varTheta }+cof_{4sZ} \mathbf {e}_Z\mathbf {e}_Z \end{aligned}$$

  whose coefficients are

  $$\begin{aligned} cof_{4sR}= & {} \frac{2 r_i^2}{3}\\ cof_{4s\varTheta }= & {} -\frac{r_i^2}{3} \\ cof_{4sZ}= & {} -\frac{r_i^2}{3}. \end{aligned}$$