Effects of wall distensibility in hemodynamic simulations of an arteriovenous fistula

Abstract

Arteriovenous fistulae are created surgically to provide adequate access for dialysis patients suffering from end-stage renal disease. It has long been hypothesized that the rapid blood vessel remodeling occurring after fistula creation is in part a process to restore the mechanical stresses to some preferred level, i.e., mechanical homeostasis. The current study presents fluid–structure interaction (FSI) simulations of a patient-specific model of a mature arteriovenous fistula reconstructed from 3D ultrasound scans. The FSI results are compared with previously published data of the same model but with rigid walls. Ultrasound-derived wall motion measurements are also used to validate the FSI simulations of the wall motion. Very large time-averaged shear stresses, 10–15 Pa, are calculated at the fistula anastomosis in the FSI simulations, values which are much larger than what is typically thought to be the normal homeostatic shear stress in the peripheral vasculature. Although this result is systematically lower by as much as 50 % compared to the analogous rigid-walled simulations, the inclusion of distensible vessel walls in hemodynamic simulations does not reduce the high anastomotic shear stresses to “normal” values. Therefore, rigid-walled analyses may be acceptable for identifying high shear regions of arteriovenous fistulae.

Appendices

Derivation of simplified wall motion equations

The first Piola–Kirchhoff stress, Eq. 4, and the the Cauchy stress in the solid, $\mathbf{T}$, are related to each other by the expression

$$\begin{aligned} \mathbf{T} = \frac{1}{J} \, \mathbf{P} \cdot \mathbf{F}^{\mathrm{T}} \end{aligned}$$

(22)

where $\mathbf{F}$ is the deformation gradient tensor and $J$ is the determinant of the deformation gradient tensor, which for an incompressible material is equal to 1. Multiplying Eq. 4 through by the transpose of the deformation gradient, and assuming small deformations, we obtain

$$\begin{aligned} \mathbf{T}&= \big (\mathbf{T}_o + \mathbf{E} \cdot \mathbf{T}_o + \varvec{\Omega } \cdot \mathbf{T}_o + \mathbb C \mathbf{:}\mathbf{E} \big ) \cdot \big (\mathbf{I} + \mathbf{H}^{\mathrm{T}} \big ) + \fancyscript{O}\big (||\mathbf{H}||^2\big ) \nonumber \\&= \mathbf{T}_o \!+\! \mathbf{E}\cdot \mathbf{T}_o \!+\! \mathbf{T}_o\cdot \mathbf{E} \!+\! \mathbb C \mathbf{:}\mathbf{E} \!+\! \varvec{\Omega } \cdot \mathbf{T}_o \!-\! \mathbf{T}_o \cdot \varvec{\Omega } \!+\! \fancyscript{O}\big (||\mathbf{H}||^2\big ) \nonumber \\&= \mathbf{T}_o + \big (\mathbf{I} \, \underline{\otimes } \, \mathbf{T}_o + \mathbf{T}_o \, \overline{\otimes } \, \mathbf{I} + \mathbb C \big )\mathbf{:}\mathbf{E} \nonumber \\&+ \big (\mathbf{I} \, \underline{\otimes } \, \mathbf{T}_o + \mathbf{T}_o \, \overline{\otimes } \, \mathbf{I} \big )\mathbf{:}\,\varvec{\Omega } + \fancyscript{O}\big (||\mathbf{H}||^2\big ) \end{aligned}$$

(23)

and where the tensor product $(\mathbf{M} \overline{\otimes } \mathbf{N})_{ijkl} = M_{il} \, N_{jk}$ for arbitrary second-order tensors $\mathbf{M}$ and $\mathbf{N}$.

For the simplified analysis, we assume that the fistula vessels are thin-walled cylinders undergoing a quasi-static and axisymmetric inflation over the cardiac cycle. The axial strains, $E_{zz}$, can be assumed to be much smaller than the circumferential strains, $E_{\theta \theta }$ since the lengths of the vessels, $\sim$100 mm, are much larger than the vessel radii, $\sim$1 mm. Shear strains and stresses arising from rotations are zero for a simple inflation. Therefore, the circumferential stress, $T_{\theta \theta }$, and radial stress $T_{rr}$ in terms of the strains are

$$\begin{aligned} T_{\theta \theta }&= T_{o,\theta \theta } + \big (2 \cdot T_{o,\theta \theta }+ C_{\theta \theta \theta \theta } \big )E_{\theta \theta } + C_{\theta \theta rr} \, E_{rr}\end{aligned}$$

(24)

$$\begin{aligned} T_{rr}&= T_{o,rr} + \big (2 \cdot T_{o,rr} + C_{rrrr} \big )E_{rr} + C_{rr \theta \theta } \, E_{\theta \theta }. \end{aligned}$$

(25)

In the membrane state, the radial component of the stress is negligible such that $T_{rr}\approx 0$ and $T_{o,rr}\approx 0$ and the radial strains can be eliminated from the above equations such that

$$\begin{aligned} T_{\theta \theta }&= T_{o,\theta \theta } + \big (2\cdot T_{o,\theta \theta } + \overline{C}_{\theta \theta \theta \theta } \big )E_{\theta \theta } \end{aligned}$$

(26)

where $\overline{C}_{\theta \theta \theta \theta } = C_{\theta \theta \theta \theta } - \frac{C_{rr \theta \theta } C_{\theta \theta r r}}{C_{rrrr}}$. One can also invoke the law of Laplace for a thin-walled cylinder which relates the internal pressure within the vessel to the stress, in which case

$$\begin{aligned} T_{\theta \theta }&= \frac{P \, R_o}{h_o} \end{aligned}$$

(27)

$$\begin{aligned} T_{rr}&= -\frac{P}{2} \end{aligned}$$

(28)

which is a statically determinate state of stress for the cylinder. The circumferential strain for a long cylinder in terms of the radial deformations is simply $E_{\theta \theta } = \eta _r / R_o$. Thus, combining Eq. 26 with Eq. 27, and writing the strains in terms of displacements, one obtains

$$\begin{aligned} \frac{(P-P_o) \, R_o}{h_o} = \bigg (2\frac{P_o \, R_o}{h_o} + 4\mu _s \bigg )\frac{\eta _r}{R_o} \end{aligned}$$

(29)

which is the expression previously given in Sect. 2.1.

Calculation of wall shear stress

The total wall viscous shear stress traction acting on the solid wall with unit normal in the current configuration, $\mathbf{n}_s$, is given as

$$\begin{aligned} \varvec{\tau }(\mathbf{x},t) = \mu \Big ( \big (\nabla \mathbf{u} \big ) + \big (\nabla \mathbf{u} \big )^{\mathrm{T}} \Big )\cdot \mathbf{n}_s \end{aligned}$$

(30)

where $\mu$ is the fluid viscosity. Given that the displacements are small, the unit normal in the current configuration can be approximated with the unit normal in the reference configuration, $\mathbf{N}_s$, such that

$$\begin{aligned} \mathbf{n}_s = \mathbf{N}_s + \fancyscript{O}\big (||\mathbf{H}||\big ). \end{aligned}$$

The instantaneous wall shear stress is computed as the absolute value of the wall shear stress vector at position $\mathbf{x}$ and time $t$ such that

$$\begin{aligned} \tau (\mathbf{x},t) = \big (\tau _{s}^{2}(\mathbf{x},t) + \tau _{m}^{2} (\mathbf{x},t) \big )^{1/2} \end{aligned}$$

(31)

where $\tau _s$ and $\tau _m$ are the two components of the wall shear stress vector which are perpendicular to the wall-normal vector. The wall-normal component of the wall shear stress is neglected. The time-averaged wall shear stress, TAWSS, over $n$ number of cardiac cycles is computed by

$$\begin{aligned} \text {TAWSS}(\mathbf{x}) = \frac{1}{n\cdot T} \int \limits _{0}^{n\cdot T} \tau (\mathbf{x},t) \, \mathrm{d}t \end{aligned}$$

(32)

where $T$ is the period of the cardiac cycle.

Furthermore, we define a “wall shear stress duty factor,” $\textit{DF}(\mathbf{x})$, which quantifies the fraction of the cardiac cycle for which the wall shear stress is above a certain stress threshold as

$$\begin{aligned} \textit{DF}(\mathbf{x}) = \frac{1}{n\cdot T} \int \limits _{0}^{n\cdot T} \phi (\mathbf{x},t) \, \mathrm{d}t \end{aligned}$$

(33)

where

$$\begin{aligned} \phi (\mathbf{x},t) = {\left\{ \begin{array}{ll} 1 &{} \quad \text {if}\quad \,\tau (\mathbf{x},t) \ge \tau _o \\ 0 &{} \quad \text {if}\quad \,\tau (\mathbf{x},t) < \tau _o \end{array}\right. } \end{aligned}$$

(34)

and where $\tau _o$ is some shear stress threshold. The “highly stressed lumen area,” $A_{\tau }$, is then defined as

$$\begin{aligned} A_{\tau } = \int \limits _{A} \textit{DF}(\mathbf{x}) \, \mathrm{d}A \end{aligned}$$

(35)

where $A$ is the luminal surface area. This is an arbitrary yet simple measure of high shear acting on the vessels. Since the duty factor can only range from 0 to 1, the stressed area is weighted by the length of time the wall shear is above the given threshold.