Simple single-scale microstructures based on optimal rank-3 laminates

Abstract

With the goal of identifying optimal elastic single-scale microstructures for multiple loading situations, the paper shows that qualified starting guesses, based on knowledge of optimal rank-3 laminates, significantly improves chances of convergence to near optimal designs. Rank-3 laminates, optimal for a given set of anisotropic loading conditions, are approximated on a single scale using a simple mapping approach. We demonstrate that these mapped microstructures perform relatively close to theoretical energy bounds. Microstructures with a performance even closer to the bounds can be obtained by using the approximated rank-3 structures in a further step as starting guesses for inverse homogenization problems. Due to the nonconvex nature of inverse homogenization problems, the starting guesses based on rank-3 laminates outperform classical starting guesses with homogeneous or random material distributions. Furthermore, the obtained single-scale microstructures are relatively simple, which enhances manufacturability. Results, obtained for a wide range of loading cases, indicate that microstructures with performance within 5–8% of the theoretical optima can be guarantied, as long as feature sizes are not limited by minimium size constraints.

Appendix: Reconstructing a rank-3 laminate from moments

In this section, we discuss the method proposed by Lipton (1994) to reconstruct the relative layer contributions p_n and orientations 𝜃_n of a rank-3 laminate from the optimal moments. Our discussion is similar to the practical implementation given by Díaz et al. (1995), and included for completeness.

We can reduce the set of optimal moments from four to three, by rotating the set of moments (m₁, m₂, m₃, m₄) to $(\tilde {m}_{1},\tilde {m}_{2},\tilde {m}_{3},0)$ using a change of reference frame and rotation angle γ such that $\tilde {\theta }_{n} = \theta _{n}+\gamma$ and the following:

$$\begin{array}{@{}rcl@{}} \tilde{m}_{1} &=& \sum\limits_{n = 1}^{N}p_{n} \text{cos}(2\tilde{\theta}_{n}), \quad \tilde{m}_{2} = \sum\limits_{n = 1}^{N}p_{n} \sin(2\tilde{\theta}_{n}), \\ \tilde{m}_{3} &=& \sum\limits_{n = 1}^{N}p_{n} \text{cos}(4\tilde{\theta}_{n}), \quad \tilde{m}_{4} = \sum\limits_{n = 1}^{N}p_{n} \sin(4\tilde{\theta}_{n}). \end{array} $$

(15)

By using the rotated reference frame for the specification of the layer tangents t_n, used in (5), the following relations can be found as follows:

$$\begin{array}{@{}rcl@{}} \tilde{m}_{1} &=& m_{1}\text{cos}(2\gamma) - m_{2}\sin(2\gamma),\\ \tilde{m}_{2} &=& m_{1}\sin(2\gamma) + m_{2}\text{cos}(2\gamma), \\ \tilde{m}_{3} &=& m_{3}\text{cos}(4\gamma) - m_{4}\sin(4\gamma), \\ \tilde{m}_{4} &=& m_{3}\sin(4\gamma) + m_{4}\text{cos}(4\gamma). \end{array} $$

(16)

Hence, we can find γ that ensures $\tilde {m}_{4}= 0$ using the following:

$$ \gamma = \frac{1}{4} \text{arctan}\left( \frac{-m_{4}}{m_{3}}\right). $$

(17)

From (17), it can be seen that γ is periodic every π/4. This means that there are at least four rotated sets $\tilde {\boldsymbol {m}}$ to describe the microstructure. Furthermore, the feasible rotated set of moments $\tilde {\mathcal {M}}$ is bounded by the same constraints as in (18).

$$ \tilde{\mathcal{M}} = \tilde{\boldsymbol{m}} \in \mathbb{R}^{3}, s.t. \left\{\begin{array}{ll} \tilde{m}_{1}^{2} + \tilde{m}_{2}^{2} \leq 1, \\ -1 \leq \tilde{m}_{3} \leq 1, \\ \frac{2\tilde{m}_{1}^{2}}{1+\tilde{m}_{3}}+\frac{2\tilde{m}_{2}^{2}}{1-\tilde{m}_{3}} \leq 1. \end{array}\right. $$

(18)

Feasible set $\tilde {\mathcal {M}}$ is a convex set as can be seen in Fig. 16.

Fig. 16

Convex set $\tilde {\mathcal {M}}$

The boundary of this convex set $\partial \tilde {\mathcal {M}}$ satisfies the following:

$$ \frac{2\tilde{m}_{1}^{2}}{1+\tilde{m}_{3}}+\frac{2\tilde{m}_{2}^{2}}{1-\tilde{m}_{3}} = 1, $$

(19)

while the four corner points also satisfy the following:

$$ \tilde{m}_{1}^{2} + \tilde{m}_{2}^{2} = 1. $$

(20)

It can easily be verified using (15) that the four corner points $\tilde {\boldsymbol {m}} = \left \{1,0,1\right \}$, {− 1, 0, 1}, {0, 1,− 1} and {0,− 1,− 1} correspond to rank-1 laminates, with the corresponding layer directions $\tilde {\theta }_{1}= 0$, π/2, π/4, and − π/4 respectively. Hence, if both (19) and (20) are satisfied, the microstructure is a rank-1 microstructure. Depending on the choice of γ, the unique layer orientation 𝜃₁ can be obtained; furthermore, corresponding p₁ = 1.

Since $\tilde {\mathcal {M}}$ is a convex set, each point $\tilde {\boldsymbol {m}}$ can be described as a combination of a corner point $\tilde {\boldsymbol {a}}$ and a point $\tilde {\boldsymbol {b}}$ on $\partial \tilde {\mathcal {M}}$ as follows:

$$ \tilde{\boldsymbol{m}} = \alpha\tilde{\boldsymbol{a}} +(1-\alpha)\tilde{\boldsymbol{b}}. $$

(21)

Since $\tilde {\boldsymbol {a}}$ corresponds to a rank-1 laminate, point $\tilde {\boldsymbol {b}}$ on boundary $\partial \tilde {\mathcal {M}}$ has to correspond to a rank-2 laminate. Hence,

$$ \frac{2\tilde{b}_{1}^{2}}{1+\tilde{b}_{3}}+\frac{2\tilde{b}_{2}^{2}}{1-\tilde{b}_{3}} = 1. $$

(22)

Furthermore, we have the following:

$$ \alpha = \frac{\tilde{m}_{1}-\tilde{b}_{1}}{\tilde{a}_{1}-\tilde{b}_{1}} = \frac{\tilde{m}_{2}-\tilde{b}_{2}}{\tilde{a}_{2}-\tilde{b}_{2}} =\frac{\tilde{m}_{3}-\tilde{b}_{3}}{\tilde{a}_{3}-\tilde{b}_{3}}. $$

(23)

If we take one of the corner points, e.g., $\tilde {\boldsymbol {a}} = \left \{1,0,1\right \}$, we can solve for $\tilde {\boldsymbol {b}}$ and α, using the Equations above. We know that the rank-2 laminate can be described using two relative layer contributions $p_{1}^{\tilde {b}}$ and $p_{2}^{\tilde {b}}$, and two angles $\theta _{1}^{\tilde {b}}$ and $\theta _{2}^{\tilde {b}}$, such as the following:

$$\begin{array}{@{}rcl@{}} p_{1}^{\tilde{b}} & +& p_{2}^{\tilde{b}} = 1, \\ \tilde{b}_{1} & =& p_{1}^{\tilde{b}} \text{cos}(2\theta_{1}^{\tilde{b}}) + p_{2}^{\tilde{b}} \text{cos}(2\theta_{2}^{\tilde{b}}),\\ \tilde{b}_{2} & =& p_{1}^{\tilde{b}} \sin(2\theta_{1}^{\tilde{b}}) + p_{2}^{\tilde{b}} \sin(2\theta_{2}^{\tilde{b}}), \\ \tilde{b}_{3} & =& p_{1}^{\tilde{b}} \text{cos}(4\theta_{1}^{\tilde{b}}) + p_{2}^{\tilde{b}} \text{cos}(4\theta_{2}^{\tilde{b}}). \end{array} $$

(24)

This is system of four equations can be solved for the four unknowns. To do so, one can describe $\tilde {\boldsymbol {b}}$ in terms of two angles, 0 ≤ t ≤ 2π and 0 ≤ β ≤ π/2 (Lipton 1994).

$$ \left\{\tilde{b}_{1},\tilde{b}_{2},\tilde{b}_{3}\right\} = \left\{\text{cos}(\beta)\text{cos}(t),\sin(\beta)\sin(t),\text{cos}(2\beta) \right\}. $$

(25)

The corresponding solution for the rank-2 laminate can then be written as follows:

$$\begin{array}{@{}rcl@{}} s & =& \text{cos}^{2}(2\beta)-2\text{cos}(2\beta)\text{cos}(2t)+ 1, \\ \delta & =& \text{arctan}\left( \frac{\sqrt{s}}{1-\text{cos}^{2}(2\beta)} \right), \\ p_{1}^{\tilde{b}} & =& \frac{1}{2}\left( 1 + \frac{2\text{cos}(2\beta)\sin(2t)}{\sqrt{s}} \right), \\ p_{2}^{\tilde{b}} & =& 1 - p_{1}^{\tilde{b}}, \\ \theta_{1}^{\tilde{b}} & =& \text{arctan}\left( \frac{p_{2}^{\tilde{b}}\sin(2\delta)-\sin(\beta)\sin(t)}{-(p_{1}^{\tilde{b}}+p_{2}^{\tilde{b}}\text{cos}(2\delta)+\text{cos}(\beta)\text{cos}(t))} \right), \\ \theta_{2}^{\tilde{b}} & =&\theta_{1}^{\tilde{b}} + \delta. \end{array} $$

(26)

The corresponding rank-3 laminate in global frame of reference can thus be written as follows:

$$\begin{array}{@{}rcl@{}} p_{1} & =& \alpha, \quad \theta_{1} = -\gamma,\\ p_{2} & =& (1-\alpha)p_{1}^{\tilde{b}}, \quad \theta_{2} = \theta_{1}^{\tilde{b}}-\gamma,\\ p_{3} & =& (1-\alpha)p_{2}^{\tilde{b}}, \quad \theta_{3} = \theta_{2}^{\tilde{b}}-\gamma. \end{array} $$

(27)