The Stress Relaxation Response of the Porcine Descending Aorta under Combined Normal and Torsional Loadings 23 where ϵ0 and τ0 denote the applied normal strain and shear strain, respectively. We normalized the data by f(t;⃗ p)= σ(t) σ0 , g(t;⃗ p)= τ(t) τ0 . (3) Here, f(t) represents the normalized normal stress and g(t) represents the normalized shear stress. σ0 denotes the initial normal stress at (the normal stress at t = 0+), and τ 0 denotes the initial shear stress. We will model the normalized experimental data using a least-squares regression analysis. We will choose to model the data with f(t;⃗ p)=A+Bexp −t λ t λ n . (4) Equation 4 is a modified version of the Weibull distribution [30]. Other variations of the Weibull distribution have been used to measure the creep and stress relaxation in polymeric materials [8]. In such models, we liken the viscoelastic, microstructural elements to a population of time-dependent mechanical latches, each with their own time to failure. The parameters n and λ are the shape and characteristic time parameters, respectively. Aand B are coefficients. We seek to model the normalized data given in figure 2. We chose the modified Weibull distribution (equation 4) based on the shape of the relaxation data seen in figure 2. We expect an equation of exponential form to fit the data reasonably well. For an initial study, we are seeking to find a nonlinear equation that can be curve fit to the data with a reasonably good coefficient of determination (R-squared value). The vector⃗ pdenotes the material parameters which, in equation 4, are⃗ p = A B λ n . We will model both the normal and shear relaxation data using equation 4. A nonlinear least squares regression produces the best-fit parameters for⃗ p given in Table 1. The experimental data, the modeled equation and the R-squared values are shown in Figure 3. (a) Normalized normal stress over time (b) Normalized shear stress over time Fig. 3 The data and fit using a least-squares regression analysis of the experimental data and equation 4 for the normalized (a) normal and (b) shear stress. We notice the characteristic time parameter, λ, differs between the normal and shear directions. We also note that, even with a nonlinear least-squares regression analysis, the model fails to perfectly capture the data at t =0.1s. Because Figure 3 is on a log-log scale, the difference between the model and data becomes exaggerated compared to a linear scale. Next, we examine the results using linear-log axes to determine the effects of varying the magnitude of compressive strain and shear strain on the relaxation response.
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