Microstructural Statistics Informed Boundary Conditions for Statistically Equivalent Representative Volume Elements (SERVEs) of Polydispersed Elastic Composites

Abstract

The statistically equivalent RVE or P-SERVE have been introduced in Swaminathan et al. (J Compos Mater 40(7):583–604, 2006) and Ghosh (Micromechanical analysis and multi-scale modeling using the voronoi cell finite element method. CRC Press/Taylor & Francis, Boca Raton, 2011) as the smallest microstructural volume element in non-uniform microstructures that has effective material properties equivalent to those of the entire microstructure. An important consideration is the application of appropriate boundary conditions for optimal SERVE domains. The exterior statistics-based boundary conditions or ESBCs have been developed in Ghosh and Kubair (J Mech Phys Solids 96:1–24, 2016), Kubair and Ghosh (Int J Solids Struct 112:106–121, 2017), Kubair et al. (J Comput Mech 52(21):2919–2928, 2018), accounting for the statistics of fiber distributions and interactions in the domain exterior to the SERVE. The ESBC-based SERVEs have been validated for effective convergence in evaluating homogenized stiffnesses and optimal domains for micromechanical analysis. Validation is also conducted with an experimentally studied carbon-fiber epoxy-matrix polymer matrix composite (PMC). The performance of the SERVE with ESBCs is compared with other boundary conditions, as well as with the statistical volume elements (SVE). The tests clearly show the significant advantages of the ESBCs in terms of accuracy of the homogenized stiffness and efficiency.

Appendix: Eshelby Tensors for Circular Cylindrical Fibers

For a cylindrical fiber of circular cross-section with a radius a and centroid at x ^I, let r = x −x ^I, x being a generic field point. Let $\rho = \frac {a}{r}$ with ${r}=\lvert \mathbf {x}-{\mathbf {x}}^I\rvert$ and $\theta =\angle (\mathbf {x} - {\mathbf {x}}^{I})$. Then the interior and exterior Eshelby tensors S _ijkl and $\hat {G}_{ijkl}\left (\mathbf {x},{\mathbf {x}}^I \right )$ have been given in [41] as:

$$\displaystyle \begin{aligned} S_{ijkl} = \{ \alpha \}^T \{\Theta_{ijkl} \}(\theta) ~~~\mbox{and}~~~{\hat{G}}_{ijkl}\left(\mathbf{x},{\mathbf{x}}^I \right)= \{\beta \}^T ({r}) \{ \Theta_{ijkl} \}(\theta) {} \end{aligned} $$

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The material-dependent vectors {α} and {β} are:

$$\displaystyle \begin{aligned} \{ \alpha \} =\frac{1}{8 \left( 1-\nu^{M} \right)} \left\{\begin{array}{c} 4 \nu^{M} -1\\ 3-4 \nu^{M}\\ 0\\ 0\\ 0 \end{array}\right\}, ~~\{\beta \}({r})= \frac{\rho^{2}}{8\left(1-\nu^{M}\right)} \left\{\begin{array}{c} -2\left( 1+2 \nu^{M} \right) + 9\rho^2\\ 2- 3 \rho^2\\ 4 ( 1+2\nu^{M} ) -12 \rho^2\\ 4 -12 \rho^2\\ 16-24\rho^2 \end{array}\right\} \end{aligned}$$

The parameter ν ^M is the Poisson’s ratio of the matrix material. The circumference basis tensor is given as:

$$\displaystyle \begin{aligned} \{\Theta_{ijkl} \}(\theta)= \left\{\begin{array}{c} \delta_{ij} \delta_{kl}\\ \delta_{ik} \delta_{jl} + \delta_{il} \delta_{jk}\\ \delta_{ij} n_{k} n_{l}\\ n_{i} n_{j} \delta_{kl}\\ n_{i} n_{j} n_{k} n_{l} \end{array}\right\}, ~~\mbox{where} \left\{\begin{array}{c} n_{1}\\ n_{2}\\ n_{3} \end{array}\right\}= \left\{\begin{array}{c} \cos \theta\\ \sin \theta\\ 1 \end{array}\right\} \end{aligned}$$

For the cylindrical fiber of circular cross-section, the interior and exterior displacement-transfer tensors are given by:

$$\displaystyle \begin{aligned} T_{ijk} \left(\mathbf{x},{\mathbf{x}}^I \right)= \{ \eta \}^T(r) \{\Psi_{ijk} \}(\theta) ~~~\mbox{and}~~~D_{ijk} \left(\mathbf{x},{\mathbf{x}}^I \right)= \{\gamma \}^T (r) \{ \Psi_{ijk} \}(\theta) {} \end{aligned} $$

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where

$$\displaystyle \begin{aligned} \{ \eta \}(r) =a \frac{\rho}{8 \left( 1 -\nu^{M} \right)} \left\{\begin{array}{c} 4 \nu^{M} -1\\ 3 - 4 \nu^{M}\\ 0 \end{array}\right\}, ~~\{\gamma \}(r)= a \frac{\rho}{8 \left( 1 -\nu^{M} \right)} \left\{\begin{array}{c} -2 \left(1-2 \nu^{M} \right) + \rho^2\\ 2 \left(1-2 \nu^{M} \right) + \rho^2\\ 4 \left(1 - \rho^2\right) \end{array}\right\} \end{aligned}$$

and

$$\displaystyle \begin{aligned} \{\Psi_{ijk} \}(\theta)= \left\{\begin{array}{c} n_{i} \delta_{jk}\\ n_{j} \delta_{ik} + n_{k} \delta_{ij}\\ n_{i} n_{j} n_{k} \end{array}\right\} \end{aligned}$$