The sequential addition and migration method to generate representative volume elements for the homogenization of short fiber reinforced plastics

Abstract

We present an algorithm for generating volume elements of short fiber reinforced plastic microstructures for prescribed fourth order fiber orientation tensor, fiber aspect ratio and solid volume fraction. The algorithm inserts fibers randomly into an existing microstructure, and removes the resulting overlap systematically based on a gradient descent method. In contrast to existing methods, large fiber aspect ratios (up to 150) and large volume fractions (60 vol% for isotropic orientation and aspect ratio of 33) can be reached. We study the effective linear elastic properties of the resulting microstructures, depending on fiber orientation, volume fraction as well as aspect ratio, and examine the size of a corresponding representative volume element.

Appendix A

The Appendix contains details how to compute the differentials in Eqs. (5) and (8).

1.1 Differentiating f in (5)

We use the notation of Sect. 2.2. For each pair $i\ne j$ the vector $k_{ij}$ is determined from the minimization problem

$$\begin{aligned} \frac{1}{2} \Vert k_{ij}\Vert ^2\longrightarrow \min _{s,t} \end{aligned}$$

with the constraints $s,t\in [-1,1]$ and

$$\begin{aligned} k_{ij}=x_i - x_j + s\frac{L}{2} p_i - t\frac{L}{2} p_j. \end{aligned}$$

Introducing the dual variables $\alpha ,\beta ,\gamma$ and $\delta$ the corresponding KKT system reads

$$\begin{aligned} 0= & {} \left\langle k_{ij}, \frac{L}{2} p_i \right\rangle -\alpha +\beta ,\nonumber \\ 0= & {} \left\langle k_{ij}, -\frac{L}{2} p_j \right\rangle -\gamma +\delta ,\nonumber \\ 0\le & {} 1+s,\quad (1+s)\alpha =0, \quad \alpha \ge 0,\nonumber \\ 0\le & {} 1-s,\quad (1-s)\beta =0, \quad \beta \ge 0,\nonumber \\ 0\le & {} 1+t,\quad (1+t)\gamma =0, \quad \gamma \ge 0,\nonumber \\ 0\le & {} 1-t,\quad (1-t)\delta =0, \quad \delta \ge 0, \end{aligned}$$

(12)

We examine how $\tfrac{1}{2} \Vert k_{ij}\Vert ^2$ changes with $x_i$, i.e.

$$\begin{aligned} \frac{\partial \left( \tfrac{1}{2}\Vert k_{ij}\Vert ^2\right) }{\partial x_i} [\xi ]= & {} \left\langle k_{ij}, \xi + \frac{\partial s}{\partial x_i}[\xi ]\frac{L}{2}p_i -\frac{\partial t}{\partial x_i}[\xi ]\frac{L}{2}p_j\right\rangle \nonumber \\= & {} \langle k_{ij}, \xi \rangle + \frac{\partial s}{\partial x_i}[\xi ] \left\langle k_{ij},\frac{L}{2}p_i\right\rangle \nonumber \\&+\,\frac{\partial t}{\partial x_i}[\xi ]\left\langle k_{ij},-\frac{L}{2}p_j\right\rangle \nonumber \\= & {} \langle k_{ij}, \xi \rangle + \frac{\partial s}{\partial x_i}[\xi ] (\alpha -\beta ) \nonumber \\&+\,\frac{\partial t}{\partial x_i}[\xi ](\gamma -\delta ) \end{aligned}$$

(13)

where we have used the KKT system (12) in the last line. (Notice that we use $\langle u,v\rangle$ and $u\cdot v$ interchangeably to denote the Euclidean scalar product.)

First we analyze the term $\frac{\partial s}{\partial x_i}[\xi ] \alpha$. We will show that it is identically zero. Suppose that $\alpha >0$. Otherwise there would be nothing to show. Consider the complementary slackness $(1+s)\alpha =0$. As $\alpha >0$, we have $s=-1$. Differentiating the complementary slackness yields

$$\begin{aligned} 0=\frac{\partial s}{\partial x_i}[\xi ] \alpha + (1+s)\frac{\partial \alpha }{\partial x_i}[\xi ] =\frac{\partial s}{\partial x_i}[\xi ] \alpha , \end{aligned}$$

where we have used that $s=-1$. Thus, the term $\frac{\partial s}{\partial x_i}[\xi ] \alpha$ vanishes. Argumenting analogously we see that the other three terms $\frac{\partial s}{\partial x_i}[\xi ]\beta$, $\frac{\partial t}{\partial x_i}[\xi ]\gamma$ and $\frac{\partial t}{\partial x_i}[\xi ]\delta$ in (13) vanish. Thus, we conclude

$$\begin{aligned} \frac{\partial \left( \tfrac{1}{2}\Vert k_{ij}\Vert ^2\right) }{\partial x_i} [\xi ]=\langle k_{ij}, \xi \rangle . \end{aligned}$$

Using this previous result, the differentiating the objective function

$$\begin{aligned} f(x_1,\ldots ,x_N,p_1,\ldots ,p_N) = \frac{1}{2} \sum _{i<j} \delta _{ij}^2 \end{aligned}$$

with $\delta _{ij}=\max \{0,D-\Vert k_{ij}\Vert \}$ in (5) proceeds as follows:

$$\begin{aligned} \frac{\partial f}{\partial x_i}[\xi ]= & {} \frac{1}{2} \sum _{i<j} \frac{\partial (\delta _{ij}^2)}{\partial x_i}[\xi ]\nonumber \\= & {} -\sum _{i<j} \delta _{ij}\frac{\partial /\partial x_i (\tfrac{1}{2} \Vert k_{ij}\Vert ^2)[\xi ] }{\Vert k_{ij}\Vert }\nonumber \\= & {} -\sum _{i<j} \delta _{ij}\left\langle \frac{k_{ij}}{\Vert k_{ij}\Vert }, \xi \right\rangle , \end{aligned}$$

leading to the first line in (6).

Differentiating w.r.t. $p_i$ proceeds similarly. Notice that in contrast to (13) the product sL / 2 appears

$$\begin{aligned} \frac{\partial \left( \tfrac{1}{2}\Vert k_{ij}\Vert ^2\right) }{\partial p_i} [\eta ]= & {} \langle k_{ij}, sL/2\xi \rangle +\, \frac{\partial s}{\partial p_i}[\eta ] (\alpha -\beta ) \nonumber \\&- \frac{\partial t}{\partial p_i}[\eta ] (\gamma - \delta ). \end{aligned}$$

However, this is the only difference. Denoting the s-parameter for the i-j-collision by $s_{ij}$ similar arguments as for the $x_i$-derivative permit us to conclude

$$\begin{aligned} \frac{\partial f}{\partial p_i}[\eta ] = -\sum _{i<j} \delta _{ij}\left\langle \frac{\tfrac{L}{2}s_{ij}k_{ij}}{\Vert k_{ij}\Vert }, \eta \right\rangle , \end{aligned}$$

representing the second line in (6).

1.2 Computing the gradient of the orientation term

We wish to differentiate the function

$$\begin{aligned} h(p_1,\ldots p_N)=\frac{1}{8} \Vert \mathbb {A}-\mathbb {A}_N\Vert ^2 \end{aligned}$$

with $\mathbb {A}_N=\sum _{i=1}^N p_i\otimes p_i\otimes p_i\otimes p_i$ at $p_i$ in direction $\eta$ orthogonal to $p_i$. Then,

$$\begin{aligned} \frac{\partial h}{\partial p_i}[\eta ]= & {} -\frac{1}{4} (\mathbb {A}-\mathbb {A}_N)\!\cdot \!\cdot \!\cdot \!\cdot (\eta \otimes p_i\otimes p_i \otimes p_i\nonumber \\&+\,p_i \otimes \eta \otimes p_i \otimes p_i + p_i\otimes p_i \otimes \eta \otimes p_i\nonumber \\&+\, p_i \otimes p_i\otimes p_i \otimes \eta ). \end{aligned}$$

Taking into account the full symmetry of $\mathbb {A}$ and $\mathbb {A}_N$ w.r.t. the permutation of indices the latter expression can be simplified to

$$\begin{aligned} \frac{\partial h}{\partial p_i}[\eta ]= & {} -(\mathbb {A}-\mathbb {A}_N)\!\cdot \!\cdot \!\cdot \!\cdot \eta \otimes p_i\otimes p_i \otimes p_i\nonumber \\= & {} -\left[ (\mathbb {A}-\mathbb {A}_N)\!\cdot \!\cdot \!\cdot p_i\otimes p_i \otimes p_i\right] \cdot \eta , \end{aligned}$$

proving (9).