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Modelling of tear propagation and arrest in fibre-reinforced soft tissue subject to internal pressure

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Abstract

The prediction of soft-tissue failure may yield a better understanding of the pathogenesis of arterial dissection and help to advance diagnostic and therapeutic strategies for the treatment of this and other diseases and injuries involving the tearing of soft tissue, such as aortic dissection. In this paper, we present computational models of tear propagation in fibre-reinforced soft tissue undergoing finite deformation, modelled by a hyperelastic anisotropic constitutive law. We adopt the appropriate energy argument for anisotropic finite strain materials to determine whether a tear can propagate when subject to internal pressure loading. The energy release rate is evaluated with an efficient numerical scheme that makes use of adaptive tear lengths. As an illustration, we present the calculation of the energy release rate for a two-dimensional strip of tissue with a pre-existing tear of length \(a\) under internal pressure \(p\) and show the effect of fibre orientation. This calculation allows us to locate the potential bifurcation to tear propagation in the \((a,p)\) plane. The numerical predictions are verified by analytical solutions for simpler cases. We have identified a scenario of tear arrest, which is observed clinically, when the surrounding connective tissues are accounted for. Finally, the limitations of the models and further directions for applications are discussed.

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Acknowledgments

LW is supported by a China Scholarship Council Studentship and the Fee Waiver Programme at the University of Glasgow.

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Correspondence to X. Y. Luo.

Appendices

Appendix 1: Derivative of Cauchy stress and tangential moduli for a quasi-incompressible material (6)

1.1 Cauchy stress

Follow the standard formulas of the theory of finite elasticity, e.g. see [25]. We firstly calculate the second Piola–Kirchhoff stress,

$$\begin{aligned} {\mathsf {S}} = 2\frac{\partial \varPsi }{\partial {\mathsf {C}}} = 2 \left( \frac{\partial \varPsi _v}{\partial {\mathsf {C}}} + \frac{\partial \varPsi _m}{\partial {\mathsf {C}}} + \frac{\partial \varPsi _f}{\partial {\mathsf {C}}} \right) , \end{aligned}$$
(10)

where

$$\begin{aligned} \frac{\partial \varPsi _v}{\partial {\mathsf {C}}}&= \varPsi _v'(J) \frac{\partial J}{\partial {\mathsf {C}}} = \frac{1}{2} \varPsi _v'(J) J {\mathsf {C}}^{-1} = \frac{K}{2} (J-1) J {\mathsf {C}}^{-1},\nonumber \\ \frac{\partial \varPsi _m}{\partial {\mathsf {C}}}&= \frac{c}{2}\frac{\partial \bar{I}_1}{\partial {\mathsf {C}}} = \frac{c}{2} \left( J^{-2/3}{\mathsf {I}}-\frac{1}{3}\bar{I}_1 {\mathsf {C}}^{-1} \right) ,\nonumber \\ \frac{\partial \varPsi _f}{\partial {\mathsf {C}}}&= \sum \limits _{n=4,6} \psi '(\bar{I}_n) \frac{\partial \bar{I}_n}{\partial {\mathsf {C}}} = \sum \limits _{n=4,6} k_1(\bar{I}_n-1) \exp \left( k_2[\bar{I}_n-1]^2\right) \left( J^{-2/3} {\mathsf {M}}_n -\frac{1}{3}\bar{I}_n {\mathsf {C}}^{-1} \right) . \end{aligned}$$
(11)

Substitute these into (10) to obtain the explicit expression for the second Piola–Kirchhoff stress. Pushing it forward using

$$\begin{aligned} \varvec{\sigma } = J^{-1}{\mathsf {FSF}}^\mathrm{T} \end{aligned}$$

immediately gives the Cauchy stress

$$\begin{aligned} \varvec{\sigma } = K(J-1){\mathsf {I}} + \frac{c}{J}\left( \bar{{\mathsf {b}}}-\frac{1}{3}\bar{I}_1 {\mathsf {I}} \right) +\frac{2}{J} \sum \limits _{n=4,6} k_1 \left( \bar{I}_n-1\right) \exp \left( k_2\left[ \bar{I}_n-1\right] ^2\right) \left( J^{-2/3} {\mathsf {m}}_n-\frac{1}{3}\bar{I}_n {\mathsf {I}} \right) , \end{aligned}$$
(12)

where \(\bar{{\mathsf {b}}} = J^{-2/3}{\mathsf {FF}}^\mathrm{T}\) and \({\mathsf {m}}_n={\mathsf {FM}}_n{\mathsf {F}}^\mathrm{T}\). In particular,

$$\begin{aligned} {\mathsf {m}}_4=\mathbf {a}_1 \otimes \mathbf {a}_1 \quad \text {and} \quad {\mathsf {m}}_6=\mathbf {a}_2 \otimes \mathbf {a}_2, \end{aligned}$$

where \(\mathbf {a}_i={\mathsf {F}}\mathbf {A}_i~(i=1,2)\) represents the deformed vector of the unit vector \(\mathbf {A}_i\) characterising the orientation of the \(i\hbox {th}\) family of fibres in the reference configuration.

1.2 Tangent moduli

Similarly, the material tangent moduli associated with the increment of the second Piola–Kirchoff stress \({\mathsf {S}}\) and the Green strain tensor \({\mathsf {E}} = \frac{1}{2}({\mathsf {C-I}})\) is derived first:

$$\begin{aligned} \mathfrak {C} = 2 \frac{\partial {\mathsf {S}}}{\partial {\mathsf {C}}} = 2 \left( \frac{\partial {\mathsf {S}}^v}{\partial {\mathsf {C}}} + \frac{\partial {\mathsf {S}}^{m}}{\partial {\mathsf {C}}}+\frac{\partial {\mathsf {S}}^{f}}{\partial {\mathsf {C}}} \right) , \end{aligned}$$
(13)

where \({\mathsf {S}}^x=2 {\partial \varPsi _x}/{\partial {\mathsf {C}}}, x =\{v,m,f\}\). In index notation,

$$\begin{aligned} \frac{\partial S^v_{IJ}}{\partial C_{KL}}&= J C_{IJ}^{-1} \varPsi _v''(J) \frac{\partial J}{\partial C_{KL}} + \varPsi _v'(J) C_{IJ}^{-1} \frac{\partial J}{\partial C_{KL}} + \varPsi _v'(J) J \frac{\partial C_{IJ}^{-1}}{\partial C_{KL}} \nonumber \\&= \frac{1}{2} J C_{IJ}^{-1} C_{KL}^{-1} \left[ (J \varPsi _v''(J) + \varPsi _v'(J) \right] + J \varPsi _v'(J) \frac{\partial C_{IJ}^{-1}}{\partial C_{KL}}, \nonumber \\ \frac{\partial S^{m}_{IJ}}{\partial C_{KL}}&= c \left[ -\frac{1}{3} \left( \frac{\partial \bar{I}_1}{\partial C_{KL}} C_{IJ}^{-1} + \bar{I}_1 \frac{\partial C_{IJ}^{-1}}{\partial C_{KL}} \right) + \frac{\partial J^{-2/3}}{\partial C_{KL}} \delta _{IJ} \right] , \nonumber \\ \frac{\partial S^{f}_{IJ}}{\partial C_{KL}}&= \sum \limits _{n=4,6} 2 \left[ \psi ''(\bar{I}_n) \frac{\partial \bar{I}_n}{\partial C_{IJ}} \frac{\partial \bar{I}_n}{\partial C_{KL}} -\frac{1}{3} \psi '(\bar{I}_n) \left( \frac{\partial \bar{I}_n}{\partial C_{KL}} C_{IJ}^{-1} + \bar{I}_n \frac{\partial C_{IJ}^{-1}}{\partial C_{KL}} + J^{-2/3} C_{KL}^{-1} A_{IJ} \right) \right] . \end{aligned}$$
(14)

We note some useful differentials:

$$\begin{aligned}&\frac{\partial C_{IJ}^{-1}}{\partial C_{KL}} = -\frac{1}{2}\left( C_{IK}^{-1} C_{JL}^{-1} + C_{IL}^{-1} C_{JK}^{-1}\right) ,\\&\frac{\partial \bar{I}_1}{\partial C_{IJ}} = - \frac{1}{3} \bar{I}_1 C_{IJ}^{-1} + J^{-2/3} \delta _{IJ}, \qquad \varPsi _v'(J) = K(J-1), \qquad \varPsi _v''(J) = K,\\&\frac{\partial \bar{I}_n}{\partial C_{IJ}} = -\frac{1}{3} \bar{I}_n C_{IJ}^{-1} + J^{-2/3} A_{IJ}, \quad n =4,6,\\&\psi '(\bar{I}_n) = k_1 (\bar{I}_n - 1) \exp \left( k_2 [\bar{I}_n-1]^2\right) ,\\&\psi ''(\bar{I}_n) = k_1 \exp \left( k_2[\bar{I}_n-1]^2\right) \left[ 1+2k_2(\bar{I}_n-1)^2\right] . \end{aligned}$$

Substituting (14) into (13) gives the explicit expression for the material tangent moduli. Pushing it forward gives the spatial tangent moduli required by a user-provided material model in FEAP. Its components are as follows:

$$\begin{aligned} \mathfrak {c}_{ijkl}&= \frac{1}{J} F_{iI}F_{jJ}F_{kK}F_{lL}\mathfrak {C}_{IJKL} \nonumber \\&= \delta _{ij} \delta _{kl} \left[ J \varPsi _v''(J)+ \varPsi _v'(J) \right] - \varPsi _v'(J) (\delta _{ik} \delta _{jl} + \delta _{il} \delta _{jk})\nonumber \\&- \frac{2}{3}\frac{c}{J} \left[ -\frac{1}{3}\bar{I}_1 \delta _{kl} \delta _{ij} + \delta _{ij} \bar{b}_{kl} + \delta _{kl}\bar{b}_{ij} - \frac{\bar{I}_1}{2} (\delta _{ik}\delta _{jl} + \delta _{il}\delta _{jk}) \right] \nonumber \\&+\frac{4}{J} \sum \limits _{n=4,6} \left\{ \psi ''(\bar{I}_n) \left( -\frac{1}{3}\bar{I}_n \delta _{ij} + J^{-2/3} m_{nij} \right) \left( -\frac{1}{3}\bar{I}_n \delta _{kl} + J^{-2/3} m_{nkl}\right) \right. \nonumber \\&\left. - \frac{1}{3} \psi '(\bar{I}_n) \left[ -\frac{1}{3} \bar{I}_n \delta _{ij} \delta _{kl} - \frac{\bar{I}_n}{2}(\delta _{ik}\delta _{jl} + \delta _{il}\delta _{jk}) + J^{-2/3} (m_{nij} \delta _{kl} + \delta _{ij} m_{nkl}) \right] \right\} . \end{aligned}$$
(15)

Finally, transforming (12) and (15) into the corresponding matrix form gives all formulas for the user subroutine for the HGO material model.

Appendix 2: Verification of material model for simple cases

We verify our model on the basis of comparisons with analytical results for a plane strain problem for a unit-square sample of fibre-reinforced material. For simplicity, both families of fibres have the same orientation, along the \(x\)-axis.

Firstly, we stretch the block along the \(x\)-axis with a stretch ratio \(\lambda _x\). Plane strain and incompressibility ensure that the deformation gradient can be written as

$$\begin{aligned} {\mathsf {F}} =\left( \begin{array}{ccc} \lambda _x &{}\quad 0 &{}\quad 0 \\ 0 &{} \quad 1/\lambda _x &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1 \end{array}\right) . \end{aligned}$$
(16)

Substituting (16) into (4) gives the corresponding strain energy (Fig. 14). For an incompressible material (4) we derive the Cauchy stress

$$\begin{aligned} \varvec{\sigma } = -\mathfrak {L}{\mathsf {I}} + c~{{\mathsf {b}}} + \sum \limits _{n=4,6}2\psi '(I_n){\mathsf {m}}_n, \end{aligned}$$
(17)

where \(\mathfrak {L}\) is the Lagrange multiplier. Without loss of generality, we consider a material with both families of fibres along the horizontal direction, \(\mathbf {A}_1=\mathbf {A}_2=[1,0,0]^T\). Substituting (16) into (17) gives the Cauchy stress

$$\begin{aligned} \varvec{\sigma } = -\mathfrak {L}{\mathsf {I}} + c {\left( \begin{array}{ccc} \lambda _x^2 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 1/{\lambda }_x^2 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1\\ \end{array}\right) } + 2\psi '(I_4) {\left( \begin{array}{ccc} \lambda _x^2 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 \end{array}\right) }. \end{aligned}$$
(18)

Since the surfaces with normal directions parallel to the \(y\)-axis are traction free, we have

$$\begin{aligned} 0=\sigma _{yy}=-\mathfrak {L}+1/\lambda _x^2, \end{aligned}$$
(19)

and thus \(\mathfrak {L}=1/\lambda _x^2\). Substituting into (18) we have

$$\begin{aligned} \sigma _{xx} = -1/\lambda _x^2 + c \lambda _x^2 + 2 \psi '(\lambda _x^2) \lambda _x^2. \end{aligned}$$
(20)

This analytical response is shown in Fig. 15.

Fig. 14
figure 14

The stored energy calculated by the computational model agrees with the analytical expression for the energy in a simple tension experiment

Fig. 15
figure 15

Verification of model for HGO material against analytical results for principal longitudinal stress \( \sigma _{xx} \) vs. stretch \( \lambda _x \)

We now compare the analytical with the numerical results. In the computations, the penalty parameter \(K\) in (6) is chosen to be \(10^5\), at which value or greater the numerical results agree with analytical predictions for both the energy (Fig. 14) and stress (Fig. 15).

Appendix 3: A simple beam model for ERR

Inequality (2) is known as the Griffith criterion when applied to linear elastic problems. Consider a beam of constant Young’s modulus \(E\) and second moment of area \(J\). The beam is bonded to a surface except for a region \(0\le x\le a\), where \(x\) measures the length along the beam from one end. The deflection of the beam is \(w(x)\), and the boundary conditions are

$$\begin{aligned} w(x)=0 \quad \text { for } x\ge a,\quad w''(0)=0,\ w'''(0)=0. \end{aligned}$$
(21)

The equation satisfied by \(w(x)\) depends on the loading experienced by the beam. We take a general function \(F(x,w)\) so that

$$\begin{aligned} EJw''''(x)=F(x,w). \end{aligned}$$
(22)

Different choices of \(F(x,w)\) give different external boundary conditions, e.g. in what follows we simulate the effect of the constraint of the surrounding connective tissue. In particular, we are interested in the calculation of the mechanical energy

$$\begin{aligned} \varPi (a)&=\frac{1}{2}\int _{0}^\infty EJ\left( w''(x)\right) ^2\,\mathrm{d}x+\int _{0}^\infty f(x,w)\,\mathrm{d}x, \end{aligned}$$
(23)

where \(F(x,w)=-\partial f/\partial w\) and \(G=-\hbox {d} \varPi / \hbox {d}a\). For a given value of \(G_c\), (23) enables us to use (2) to determine whether a tear of length \(a\) can propagate.

We now use this simple beam model to explore the type of phenomena we obtained from the numerical experiments. To simulate the boundary condition, we set \(F(x,w)=p\), a constant. Solving the ordinary differential beam equation (22) gives

$$\begin{aligned} w(x)= {\left\{ \begin{array}{ll} \dfrac{p}{24EJ}\left( x^4-4a^3x+3a^4\right) &{}\quad x<a, \\ 0 &{}\quad x>a, \end{array}\right. } \end{aligned}$$
(24)

and therefore, substituting into (23), we find that

$$\begin{aligned} \varPi (a)=-\frac{p^2a^5}{40EJ}. \end{aligned}$$
(25)

The energy release rate is

$$\begin{aligned} G = -\frac{\mathrm{d}\varPi }{\mathrm{d}a} = \frac{p^2a^4}{8EJ}. \end{aligned}$$
(26)

\(G\) is a monotonically increasing function of \(a\) and \(p\), and therefore an increase in either the length of the unbonded region (the tear) or the pressure results in the propagation of the tear being energetically favourable.

To consider the effect of surrounding connective tissues, we set \(F(x,w)=p-kw\), where the constant \(k\) is the stiffness per unit length of the springs, as shown in Fig. 16. Consequently, \(f=-pw+kw^2/2\), and the solution for \(w(x)\) is

$$\begin{aligned} w(x)=\frac{p}{k}+W(x), \end{aligned}$$
(27)

where \(W(x)\) satisfies

$$\begin{aligned} W''''+4\lambda ^4W(x)=0,\quad \lambda ^4=\frac{k}{4EJ}. \end{aligned}$$
(28)

This is solved to give

$$\begin{aligned} W(x)=e^{-\lambda x}\left[ A\cos (\lambda x)+B\sin (\lambda x)\right] +e^{\lambda x}\left[ C\cos (\lambda x)+D\sin (\lambda x)\right] , \end{aligned}$$
(29)

with \(A, B, C\) and \(D\) chosen to satisfy the boundary conditions. Non-dimensionalising the deflection with \(p/k\) and \(x\), with \(\left( 4EJ/k\right) ^{1/4}\), leads to the canonical problem

$$\begin{aligned} \frac{\mathrm{d}^4y}{\mathrm{d}s^4}=4-4y, \end{aligned}$$
(30)

with boundary conditions \(y''(0)=y'''(0)=0\) and \(y(\alpha )=y'(\alpha )=0\), where \(\alpha =a/l\). The solution to this problem is

$$\begin{aligned} y(s) = 1+\frac{\left( \sin s \cosh s+\cos s\sinh s\right) \left( \cos \alpha \sinh \alpha -\sin \alpha \cosh \alpha \right) -2\cos s\cosh s\cos \alpha \cosh \alpha }{\cos ^2\alpha +\cosh ^2\alpha }. \end{aligned}$$
(31)

The mechanical energy is

$$\begin{aligned} \varPi =\left( \frac{p}{k}\right) ^2EJ\frac{l}{l^4}\int _0^\alpha \frac{1}{2}\left( y''(s)\right) ^2\,\mathrm{d}s+\frac{1}{2}k\left( \frac{p}{k}\right) ^2l\int _0^\alpha y(s)^2\,\mathrm{d}s-\frac{p^2}{k}l\int _0^\alpha y(s)\,\mathrm{d}s. \end{aligned}$$
(32)

This expression simplifies to

$$\begin{aligned} \varPi =\frac{p^2l}{k}\left[ \int _0^\alpha \frac{1}{8}\left( y''(s)^2+4y(s)^2\right) -y(s)\,\mathrm{d}s\right] , \end{aligned}$$
(33)

and then we obtain the ERR

$$\begin{aligned} G = -\dfrac{\hbox {d}\varPi }{\hbox {d}a} = - \dfrac{p^2}{k}\left\{ \dfrac{1}{8}\left[ y''(\alpha )^2 + 4 y(\alpha )^2 \right] -y(\alpha ) \right\} . \end{aligned}$$
(34)
Fig. 16
figure 16

Effect of connective tissue using beam model. The semi-infinite beam, of constant Young’s modulus \(E\) and second moment of area \(J\), is bonded to a surface except for a region \(0<x<a\), which represents the tear. The spring bed represents the surrounding connective tissues

We display the curve of \(G(a)\) for a set of typical parameters in Fig. 17. When subject to a constant pressure, \(G(a)\) is not a monotonically increasing function of \(a\), and propagation arrest occurs. This is qualitatively similar to what is seen in Fig. 12 in the numerical simulations for a strip of fibre-reinforced tissues subject to finite strain.

Fig. 17
figure 17

Tear arrest is also demonstrated by the beam model when connective tissue is present. The ERR \(G\) is no longer a monotonic function of \(a\), and for a given critical value \(G_c\), a tear of length \(a\), where \(a_1 < a < a_2\) or \(a > a_3\), will propagate. However, the tear arrests when \(a_2 < a < a_3\)

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Wang, L., Roper, S.M., Luo, X.Y. et al. Modelling of tear propagation and arrest in fibre-reinforced soft tissue subject to internal pressure. J Eng Math 95, 249–265 (2015). https://doi.org/10.1007/s10665-014-9757-7

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