Explicit parametric solutions of lattice structures with proper generalized decomposition (PGD)

Abstract

Architectured materials (or metamaterials) are constituted by a unit-cell with a complex structural design repeated periodically forming a bulk material with emergent mechanical properties. One may obtain specific macro-scale (or bulk) properties in the resulting architectured material by properly designing the unit-cell. Typically, this is stated as an optimal design problem in which the parameters describing the shape and mechanical properties of the unit-cell are selected in order to produce the desired bulk characteristics. This is especially pertinent due to the ease manufacturing of these complex structures with 3D printers. The proper generalized decomposition provides explicit parametic solutions of parametric PDEs. Here, the same ideas are used to obtain parametric solutions of the algebraic equations arising from lattice structural models. Once the explicit parametric solution is available, the optimal design problem is a simple post-process. The same strategy is applied in the numerical illustrations, first to a unit-cell (and then homogenized with periodicity conditions), and in a second phase to the complete structure of a lattice material specimen.

Appendices

Appendix A. Proper generalized decomposition algorithms

Appendix B. Analytical solutions

Given the unit-cell global stiffness matrix built by Bernoulli beam elements, $\mathbf {K}(\varvec{\mu })$, the equations

$$\begin{aligned} \mathbf {K}(\varvec{\mu }) \, \mathbf {U}(\varvec{\mu })&= \mathbf {0} , \\ \mathbf {C}\,\mathbf {U}(\varvec{\mu })&=\mathbf {Q}(\varvec{\mu }) , \end{aligned}$$

contain equilibrium and periodicity conditions, given a specific load case in the unit-cell. If the constraints are applied using a direct method [11], then the system unknowns can be split into released and constrained such as

$$\begin{aligned} {[}\begin{array}{ll} \mathbf {C}_{\mathrm {R}}&\mathbf {C}_{\mathrm {C}} \end{array}] \; \left\{ \begin{array}{l} \mathbf {U}_{\mathrm {R}}(\varvec{\mu }) \\ \mathbf {U}_{\mathrm {C}}(\varvec{\mu }) \end{array}\right\} - \{ \mathbf {Q}(\varvec{\mu }) \} = \{ \varvec{0} \}, \end{aligned}$$

in a way that the constrained unknowns can be solved in terms of the released

(41)

The system can be solved for the released unknowns only:

$$\begin{aligned} \mathbf {K}_{\mathrm {R}}(\varvec{\mu }) \, \mathbf {U}_{\mathrm {R}}(\varvec{\mu }) = \mathbf {F}_{\mathrm {R}}(\varvec{(}\mu )), \end{aligned}$$

(42)

where

$$\begin{aligned} \begin{aligned} \mathbf {K}_{\mathrm {R}}(\varvec{\mu })&= \varvec{\varphi }^{\mathrm {T}}\,\mathbf {K}(\varvec{\mu })\,\varvec{\varphi }, \\ \mathbf {F}(\varvec{\mu })_{\mathrm {R}}&= -\varvec{\varphi }^{\mathrm {T}}\,\mathbf {K}(\varvec{\mu })\,\mathbf {Q}_{\mathrm {0}}(\varvec{\mu }). \end{aligned} \end{aligned}$$

While $\forall \varvec{\mu } \in \mathcal {D}=I_{1}\times I_{2}\times \ldots \times I_{\texttt {n}_{\texttt {p}}}$, $\mathbf {K}(\varvec{\mu }) \in \mathbb {R}^{({\texttt {n}_{\texttt {dof}}}\times {\texttt {n}_{\texttt {dof}}})}$, $\mathbf {K}_{\mathrm {R}}(\varvec{\mu }) \in \mathbb {R}^{({n_{\mathrm {R}}}\times {n_{\mathrm {R}}})}$. In the unit-cell problem, ${\texttt {n}_{\texttt {dof}}}=24$ is reduced to ${n_{\mathrm {R}}}=13$. Now the system can be solved symbolically with the parameters $\varvec{\mu } = [ t \, \, a \,\, b\, \, \alpha ]^{\textsf {T}}$.

Fig. 25

Scaled generalized displacement solution case XX

Fig. 26

Scaled generalized displacement solution case YY

With this, the parametric solution for case XX results:

$$\begin{aligned} \mathbf {U}_{\mathrm {R}}(\varvec{\mu }) = \left\{ \begin{array}{l} u_2 \\ v_2 \\ \theta _2 \\ u_3 \\ v_3 \\ \theta _3 \\ u_4 \\ v_4 \\ \theta _4 \\ u_5 \\ v_5 \\ \theta _5 \\ \theta _6 \end{array} \right\} = \left\{ \begin{array}{c} \frac{b(a \cos (\alpha )-b) \left( \left( a^2-4 t^2\right) \sin ^2(\alpha )-a^2\right) }{a^2 (a+2b)-2b \left( a^2-4 t^2\right) \sin ^2(\alpha )} \\ 0 \\ 0 \\ -\frac{(a \cos (\alpha )-b) \left( a^2 (a+b)-b\left( a^2-4 t^2\right) \sin ^2(\alpha )\right) }{a^2 (a+2b)-2b \left( a^2-4 t^2\right) \sin ^2(\alpha )} \\ 0 \\ 0 \\ -\frac{(a \cos (\alpha )-b) \left( a^2 (a+3b)-3b \left( a^2-4 t^2\right) \sin ^2(\alpha )\right) }{a^2 (a+2b)-2b \left( a^2-4 t^2\right) \sin ^2(\alpha )} \\ 0 \\ 0 \\ -\frac{(a \cos (\alpha )-b) \left( a^2 (2 a+3b)-3b \left( a^2-4 t^2\right) \sin ^2(\alpha )\right) }{a^2 (a+2b)-2b \left( a^2-4 t^2\right) \sin ^2(\alpha )} \\ 0 \\ 0 \\ 0 \\ \end{array} \right\} . \end{aligned}$$

Figure 25 shows a scaled analytical solution for case XX, evaluated at parameters $b=1$, $t=\frac{1}{40}$, $a=0.5$, $\alpha =60^{\circ }$ as an example.

The solution for load case YY becomes:

$$\begin{aligned} \mathbf {U}_{\mathrm {R}}(\varvec{\mu }) = \left\{ \begin{array}{l} u_2 \\ v_2 \\ \theta _2 \\ \theta _3 \\ u_4 \\ v_4 \\ \theta _4 \\ u_5 \\ v_5 \\ \theta _5 \\ u_6 \\ v_6 \\ \theta _6 \end{array} \right\} = \left\{ \begin{array}{c} -\frac{2 a b \left( a^2-4 t^2\right) \cos (\alpha ) \sin ^2(\alpha )}{a^3 + a^2 b + 4 b t^2 + b \left( a^2-4 t^2\right) \cos (2\alpha )} \\ a \sin (\alpha ) \\ 0 \\ 0 \\ -\frac{2 a b \left( a^2-4 t^2\right) \cos (\alpha ) \sin ^2(\alpha )}{a^3 + a^2 b + 4 b t^2 + b \left( a^2-4 t^2\right) \cos (2\alpha )} \\ 2 a \sin (\alpha ) \\ 0 \\ 0 \\ a \sin (\alpha ) \\ 0 \\ -\frac{a b \left( a^2-4 t^2\right) \cos (\alpha ) \sin ^2(\alpha )}{a^3 + a^2 b + 4 b t^2 + b \left( a^2-4 t^2\right) \cos (2\alpha )} \\ a \sin (\alpha ) \\ 0 \\ \end{array} \right\} . \end{aligned}$$

Figure 26 shows a scaled analytical solution for case YY, evaluated at parameters $b=1$, $t=\frac{1}{40}$, $a=0.5$, $\alpha =60^{\circ }$.

Finally, for case XY:

Figure 27 shows a scaled analytical solution for case XY, evaluated at parameters $b=1$, $t=\frac{1}{40}$, $a=0.5$, $\alpha =60^{\circ }$.

At this point, it seems worth mentioning the parametric analytical solutions have the size of a unit-cell reduced by its periodic constraints. For a full structure done by repetitions of unit-cells, or a pattern with irregularities, the cost of obtaining such expressions becomes computationally unfeasible.

Finally, the components of the effective elasticity matrix are calculated from (37) and are shown in Eqs. (43)–(48). It is worth mentioning that these results are affected by a factor of E, the Young’s modulus of the constituent material, which is here omitted for the sake of clarity:

$$\begin{aligned}&\mathrm {C}^{\mathrm{eff}}_{11}(\varvec{\mu }) = -\frac{t \csc (\alpha ) (a \cos (\alpha )-b) \left( \left( a^2-4 t^2\right) \cos (2 \alpha )+a^2+4 t^2\right) }{a \left( a^3+b \left( a^2-4 t^2\right) \cos (2 \alpha )+a^2 b+4 t^2 b\right) } \end{aligned}$$

(43)

$$\begin{aligned}&\mathrm {C}^{\mathrm{eff}}_{22}(\varvec{\mu }) = -\frac{t \sin (\alpha ) \left( a^3-a \left( a^2-4 t^2\right) \cos (2 \alpha )+4 a t^2+16 t^2 b\right) }{(a \cos (\alpha )-b) \left( a^3+b\left( a^2-4 t^2\right) \cos (2 \alpha )+a^2 b+4 t^2b\right) } \end{aligned}$$

(44)

$$\begin{aligned}&\mathrm {C}^{\mathrm{eff}}_{33}(\varvec{\mu })= -\frac{16 t^3 \sin (\alpha ) (a \cos (\alpha )-b)}{-b^2\left( a^2+5 a b-4 t^2\right) \cos (2 \alpha )+8 a^2 t^2+a^2 b^2-16\, a b\, t^2 \cos (\alpha )+5 a b^3+4 t^2 b^2} \end{aligned}$$

(45)

$$\begin{aligned}&\mathrm {C}^{\mathrm{eff}}_{12}(\varvec{\mu }) = \frac{t \left( 4 t^2-a^2\right) \sin (2 \alpha )}{a^3+b\left( a^2-4 t^2\right) \cos (2 \alpha )+a^2b+4 t^2b} \end{aligned}$$

(46)

$$\begin{aligned}&\mathrm {C}^{\mathrm{eff}}_{13}(\varvec{\mu }) = 0 \end{aligned}$$

(47)

$$\begin{aligned}&\mathrm {C}^{\mathrm{eff}}_{23}(\varvec{\mu }) = 0 \end{aligned}$$

(48)

Using (43), (45) and (46) the orthotropic Poisson’s ratios can be calculated as follows, which assumes a state of plane stress ([6]):

$$\begin{aligned} \nu _{12}^{\mathrm{eff}}(\varvec{\mu })&= \frac{\mathrm {C}^{\mathrm{eff}}_{12}(\varvec{\mu })}{\mathrm {C}^{\mathrm{eff}}_{22}(\varvec{\mu })} = \frac{2 \left( a^2-4 t^2\right) \cos (\alpha ) (a \cos (\alpha )-b)}{a^3-a \left( a^2-4 t^2\right) \cos (2 \alpha )+4 a\, t^2+16 t^2 b} \end{aligned}$$

(49)

$$\begin{aligned} \nu _{21}^{\mathrm{eff}}(\varvec{\mu })&= \frac{\mathrm {C}^{\mathrm{eff}}_{12}(\varvec{\mu })}{\mathrm {C}^{\mathrm{eff}}_{11}(\varvec{\mu })} = \frac{a \left( a^2-4 t^2\right) \sin (\alpha ) \sin (2 \alpha )}{(a \cos (\alpha )-b) \left( \left( a^2-4 t^2\right) \cos (2 \alpha )+a^2+4 t^2\right) } \end{aligned}$$

(50)

Fig. 27

Scaled generalized displacement solution case XY