The biomechanics of arterial elastin

Uniaxial mechanical experiments have shown that a neo-Hookean/Gaussian model is suitable to describe the mechanics of arterial elastin networks [Gundiah, N., Ratcliffe, M.B., Pruitt, L.A., 2007. Determination of strain energy function for arterial elastin: Experiments using histology and mechanical tests. J. Biomech. 40, 586–594]. Based on the three-dimensional elastin architecture in arteries, we have proposed an orthotropic material symmetry for arterial elastin consisting of two orthogonally oriented and symmetrically placed families of mechanically equivalent fibers. In this study, we use these results to describe the strain energy function for arterial elastin, with dependence on a reduced subclass of invariants, as W = W ( I 1 , I 4 ) . We use previously published equations for this dependence [Humphrey, J.D., Strumpf, R.K., Yin, F.C.P., 1990a. Determination of a constitutive relation for passive myocardium: I. A new functional form. J. Biomech. Eng. 112, 333–339], in combination with a theoretical guided Rivlin–Saunders framework [Rivlin, R.S., Saunders, D.W., 1951. Large elastic deformations of isotropic materials VII. Experiments on the deformation of rubber. Phil. Trans. R. Soc. A 243, 251–288] and biaxial mechanical experiments, to obtain the form of this dependence. Using mechanical equivalence of elastin in the circumferential and longitudinal directions, we add a term in I 6 to W that is similar to the form in I 4 . We propose a semi-empirical model for arterial elastin given by W = c 0 ( I 1 − 3 ) + c 1 ( I 4 − 1 ) 2 + c 2 ( I 6 − 1 ) 2 , where c 0 , c 1 and c 3 are unknown coefficients. We used the Levenberg–Marquardt algorithm to fit theoretically calculated and experimentally determined stresses from equibiaxial experiments on autoclaved elastin tissues and obtain c 0 = 73.96 ± 22.51 kPa , c 1 = 1.18 ± 1.79 kPa and c 2 = 0.8 ± 1.26 kPa . Thus, the entropic contribution to the strain energy function, represented by c 0 , is a dominant feature of elastin mechanics. Because there are no significant differences in the coefficients corresponding to invariants I 4 and I 6 , we surmise that there is an equal distribution of fibers in the circumferential and axial directions.