An ultrasound-based approach for the characterization of fluid–structure interaction of large arterial vessels

Abstract

An ultrasound-based approach to characterize the fluid–structure interaction in large arterial vessels is presented. The ultrasound-based data are fed into a new dynamic model accounting for a two-dimensional (2D) stress state, which in turn provides a better estimate of the material elasticity under dynamic loading. In order to validate the semi-empirical model, a compliant, synthetic vessel was subjected to a range of pulsatile and steady flow profiles. Ultrasound imaging was used to capture the flow field through the compliant vessel and its change in diameter over time. Internal pressure was extracted from ultrasound image velocimetry using spatial integration of the Navier–Stokes equation, and used to find the pressure–area relationship. Two constitutive laws describing a one-dimensional expansion of a cylindrical vessel, the Laplace law and one from Olufsen (Am J Physiol-Heart Circul Physiol 276(1):H257–H268, 1999), were also used to estimate the instantaneous elastic modulus. A uniaxial tensile test of the vessel material was performed to provide validation criteria. Under steady flow, the Laplace law predicted the elasticity of the vessel material with 255% error and the results from Olufsen (Am J Physiol-Heart Circul Physiol 276(1):H257–H268, 1999) had an error of 99%. In contrast, our developed 2D stress model predicted the elasticity with less than 10% error. The Laplace law and the Olufsen (Am J Physiol-Heart Circul Physiol 276(1):H257–H268, 1999) model were revealed to be flow-dependent such that the trend of the resultant elastic modulus varied for each pulsatile flow case. However, the 2D stress model showed no flow dependency, presenting consistent elasticity results across all test cases.

Graphic abstract

Appendix

Summing the forces on the element in Fig. 5b results in the following two equations:

$$\begin{aligned} F_{x}= & {} -p\,\mathrm{d}A\sin {\theta } + (T + \mathrm{d}T)\cos (\theta + \mathrm{d}\theta ) - T\cos (\theta ), \end{aligned}$$

(7)

$$\begin{aligned} F_{y}= & {} -p\,\mathrm{d}A\cos {\theta }\sin (\phi ) + (T + \mathrm{d}T)\sin (\theta + \mathrm{d}\theta )\sin (\phi )\nonumber \\&- T\sin (\theta )\sin (\phi ) - \sigma _{C}h\,\mathrm{d}s\sin (\phi )\,d\phi , \end{aligned}$$

(8)

where all variables are a function of time. Moving forward, we assume quasi-steady conditions and neglect the acceleration (i.e. $F_{x} = 0$, $F_{y} = 0$) of the elastic vessel wall so that the pressure forces are balanced with the tension forces. On the right-hand side, p and h can be obtained from the ultrasound data and we can resolve $\mathrm{d}A$ and $\mathrm{d}s$ as the follows:

$$\begin{aligned} \mathrm{d}A = r\,d\phi \,\mathrm{d}s, \quad \mathrm{d}s = \sqrt{1 + \left( \frac{\mathrm{d}y}{\mathrm{d}x}\right) ^2}\mathrm{d}x \end{aligned}$$

where r and $\frac{\mathrm{d}y}{\mathrm{d}x}$ can also be found from the ultrasound data. Focusing on Eq. (7), we can simplify further by expanding $\cos (\theta + \mathrm{d}\theta )$, and with the small angle approximation we set $\cos (\mathrm{d}\theta )$ equal to 1 and $\sin (\mathrm{d}\theta )$ equal to $\mathrm{d}\theta$. Equation (7) then becomes

$$\begin{aligned}&-p\,\mathrm{d}A\sin {\theta } + T\cos (\theta ) - T\sin (\theta )\,\mathrm{d}\theta +\mathrm{d}T\cos (\theta )\nonumber \\&\qquad - \mathrm{d}T\sin (\theta )\mathrm{d}\theta - T\cos (\theta ) = 0. \end{aligned}$$

(9)

Equation (9) can be further reduced by acknowledging that the term $\mathrm{d}T\sin (\theta )\,\mathrm{d}\theta$ is small and may be neglected due to the multiplication of two differential quantities. In addition, $T\cos (\theta )$ cancels out, and we are left with

$$\begin{aligned} -p\,\mathrm{d}A\sin {\theta } - T\sin (\theta )\,\mathrm{d}\theta + \mathrm{d}T\cos (\theta ) = 0. \end{aligned}$$

(10)

We recognize that the latter two terms on the left-hand side in Eq. (10) are the result of a product derivative, and then, we can also substitute for $\mathrm{d}A$ and T. Following these substitutions, we can remove $\sigma _{L}h\,d\phi$ from the derivative as these terms are considered constant along x, while r and $\cos (\theta )$ are a function of x, yielding

$$\begin{aligned} -p\,\sin {\theta }r\,d\phi \,\mathrm{d}s + \sigma _{L}h\,d\phi \,\,d(r\cos (\theta )) = 0. \end{aligned}$$

(11)

Now we rearrange and integrate over $\phi$ from 0 to $\pi$ and substitute $\mathrm{d}s$ to obtain:

$$\begin{aligned} p\,\pi \sin {\theta }r\sqrt{1 + \left( \frac{\,\mathrm{d}y}{\,\mathrm{d}x}\right) ^2}\,\mathrm{d}x = \sigma _{L}h\pi \,d(r\cos (\theta )). \end{aligned}$$

(12)

Finally, we perform another integration over x to obtain:

$$\begin{aligned} \int _{x_{1}}^{x_{2}}p\,\sin (\theta )r\sqrt{1 + \left( \frac{\mathrm{d}y}{\mathrm{d}x}\right) ^2}\,\mathrm{d}x = \sigma _{L}h\,r\cos (\theta ) \left| _{x_{1}}^{x_2}\right. . \end{aligned}$$

(13)

We now move on to the y-direction forces, represented by Eq. (8), to which we can immediately apply the trigonometric identity presented before and simplify the products of the resulting derivative for the terms involving the force of tension in a similar manner to that done for the x-direction. We will also substitute for tension here, and we obtain the following:

$$\begin{aligned}&-p\,r\,d\phi \,\mathrm{d}s\cos {\theta }\sin (\phi ) +\sigma _{L}h\,d\phi \,\,d(r\sin (\theta ))\sin (\phi ) \nonumber \\&\qquad -\sigma _{C}h\,\mathrm{d}s\sin (\phi )\,d\phi = 0. \end{aligned}$$

(14)

Once again, we integrate over $\phi$ from 0 to $\pi$ to obtain:

$$\begin{aligned} -2p\,r\,\mathrm{d}s\cos {\theta } + 2\sigma _{L}h\,\,d(r\sin (\theta )) - 2\sigma _{C}h\,\mathrm{d}s = 0, \end{aligned}$$

(15)

and rearranging this equation and performing another integral over x results in

$$\begin{aligned} \int _{x_{1}}^{x_{2}} p\,r\,\mathrm{d}s\cos {\theta } = \sigma _{L}h\,r\sin (\theta ) \left| _{x_{1}}^{x_2} - \sigma _{C} \int _{x_{1}}^{x_{2}} h\,\mathrm{d}s. \right. \end{aligned}$$

(16)

From Eq. (13) it is possible to isolate for the longitudinal stress (Eq. (4)), and from Eq. (16) we can isolate for the circumferential stress (Eq. (5)).