Complementary energy based meso-level homogenization for multiscale topology optimization

Abstract

Multiscale topology optimization (MTO) of structures requires the computationally efficient evaluation of the mechanical properties of each mesoscale representative volume element (RVE). This paper presents a novel method that uses a complementary energy (CE) approach to estimate the effective homogenized properties of an RVE, consisting of an embedded structure that is parametrically defined. Visualizing the meso-scale structure as a sweep cross-section makes it possible to quickly and accurately write the CE and then invoke Castigliano’s second theorem to find the force–displacement relationships. A two-dimensional (2D) running example of an anchored torus is employed to motivate the framework in which Lagrange multipliers are used to ensure the physical constraints on the mesostructure. For three-dimensional (3D) RVEs, a Frenet frame representation is used to associate the global forces and moments to a local coordinate system. Two homogenization methods are compared to ensure correctness; the first directly evaluates the constitutive matrix via the kinetic uniform boundary condition (KUBC) approach, while the second is a new finite element based optimization that directly finds the equivalent material properties. The proposed method is compared at the meso-scale against a solid finite element model and a discrete beam model to verify accuracy. Furthermore, a variety of material properties are demonstrated for a 3D example.

Appendix

A running example with several calculations from the proposed CE method are given here to assist the reader with an applied problem and provide results from key calculative steps for verifying work. For the running example, $\xi$ ranges from 0 to $2 \pi$ to create a single loop for the cross section to sweep.

To find the scaling for the running example, a $\zeta$ of 0.01 and $r_t$ of 0.05 is be substituted into Eq. 6 giving:

$$\begin{aligned} \chi _x = \frac{1 - 2(0.05) - 0.01}{1 - (-1)} = 0.445 \end{aligned}$$

(78)

In the running example, the scaling will be equal for every direction, hereon denoted $\chi$. This element is embedded in a square with a unit length that spans between 0 and 1 for both the x and y direction, the resulting corner locations are:

$$\begin{aligned} \textbf{X}_\textbf{c} = \begin{bmatrix} 1 &{} 0 &{} 0 &{} 1 \\ 1 &{} 1 &{} 0 &{} 0 \end{bmatrix} \end{aligned}$$

(79)

Due to the simplicity of the example, we knot the closest points occur at $\vec {\xi }_{\textrm{cl}}=\begin{bmatrix} \pi /4,&3\pi /4,&5\pi /4,&7\pi /4 \end{bmatrix}$. Substituting these values into Eq. 8 results in the following attachment points on the loop:

$$\begin{aligned} \textbf{X}_{\textrm{cl}}&= \begin{bmatrix} 0.445\begin{bmatrix} \frac{\sqrt{2}}{2}&-\frac{\sqrt{2}}{2}&-\frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2} \end{bmatrix} + \begin{bmatrix} 0.5 &{} 0.5 &{} 0.5 &{} 0.5 \end{bmatrix} \\ 0.445\begin{bmatrix} \frac{\sqrt{2}}{2} &{} \frac{\sqrt{2}}{2} &{} -\frac{\sqrt{2}}{2} &{} -\frac{\sqrt{2}}{2} \end{bmatrix} + \begin{bmatrix} 0.5&0.5&0.5&0.5 \end{bmatrix} \end{bmatrix} \end{aligned}$$

(80)

$$\begin{aligned} \textbf{X}_{\textbf{cl}}&= \begin{bmatrix} 0.815 &{} 0.185 &{} 0.185 &{} 0.815 \\ 0.815 &{} 0.815 &{} 0.185 &{} 0.185 \end{bmatrix} \end{aligned}$$

(81)

In Eq. 80 there are positive and negative values since the initial representation was a torus centered at the origin, after scaling and translating all of the values are positive. Once the corners and connection points are defined, the vector which points along each anchor can be found. The anchor vector from Eq. 10 for the first strand, as shown in Fig. 3a, for the running example is:

$$\begin{aligned} \vec {v}^{(1)}_b(\xi )&=\left( \textbf{X}_{\textbf{cl}^{\mathbf {(1)}}}-\textbf{X}_{\textbf{c}}^{\mathbf {(1)}} \right) \xi = \left( \begin{bmatrix} 0.815 \\ 0.815 \end{bmatrix} - \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right) \nonumber \\ \xi&= \begin{bmatrix} -0.185 \xi \\ -0.185 \xi \end{bmatrix} \end{aligned}$$

(82)

With the anchor expressed as a vector, the moment and axial coefficients can be solved. For the first anchor, the moment coefficient vector, $\hat{\textbf{m}}$, defined in Eq. 20 is:

$$\begin{aligned} \hat{\textbf{m}}^{(1)} = \begin{bmatrix} -0.185 \xi + 0.185 \\ 0.185 \xi - 0.185 \\ 1 \end{bmatrix} \end{aligned}$$

(83)

and the axial force coefficient vector that was defined in Eq. 22 is:

$$\begin{aligned} \hat{\textbf{p}}^{(1)} = \begin{bmatrix} -0.707&-0.707&0 \end{bmatrix} \end{aligned}$$

(84)

After defining the moment and axial coefficient vectors, the total complementary energy is found by integrating the entire length by setting $\xi _1 = 1$. The total CE for the first anchor is:

$$\begin{aligned} \hat{\textbf{A}}^{(1)} = \begin{bmatrix} 1.305\text {e}{-}{4} & -1.048\text {e}{-}{4} & -9.519\text {e}{-}{4}\\ -1.048\text {e}{-}{4} & 1.305\text {e}{-}{4} & 9.519\text {e}{-}{4}\\ -9.519\text {e}{-}{4} & 9.519\text {e}{-}{4} & 1.027\text {e}{-}{2} \end{bmatrix} \end{aligned}$$

(85)

The resulting assembled $\hat{A}$ matrix from Eq. 25 is detailed in Table 8. It should be observed that the magnitude of the values of each anchor are the same but there are differences in the signs due to the starting and end points of the anchors. The symmetry of the torus causes the magnitudes of the values to be equal.

Table 8 Assembled $\hat{A}$ matrix

With the CE contribution analyzed for the anchors it is appropriate to transition to the solving the loop by first finding the vectors pointing from the loop segments to the splice. For the second segment, the vector pointing to the splice, labelled $\vec {v}_l^{(0)}$ in Fig. 3d, is:

$$\begin{aligned} \vec {v}_l^{(0)}= & {} \begin{bmatrix} \chi x(0) + x_{\textrm{shift}} - (\chi x(\xi )^T + x_{\textrm{shift}}) \\ \chi y(0) + y_{\textrm{shift}} - (\chi y(\xi )^T + y_{\textrm{shift}}) \\ 1 \end{bmatrix} \end{aligned}$$

(86)

$$\begin{aligned} \vec {v}_l^{(0)}= & {} \begin{bmatrix} \chi (x(0) - x(\xi )) \\ \chi (y(0) - y(\xi )) \\ 1 \end{bmatrix} \end{aligned}$$

(87)

$$\begin{aligned} \vec {v}_l^{(0)}= & {} \begin{bmatrix} 0.445(1-\cos (\xi )) \\ 0.445(-\sin (\xi )) \\ 1 \end{bmatrix} \end{aligned}$$

(88)

and the vector pointing at the first loop-anchor interaction point, $\vec {v}_l^{(1)}$ in Fig. 3d, is:

$$\begin{aligned} \vec {v}_1^{(1)}= & {} \begin{bmatrix} \chi \left( x\left( \frac{\pi }{4}\right) - x(\xi )\right) \\ \chi \left( y\left( \frac{\pi }{4}\right) - y(\xi )\right) \\ 1 \end{bmatrix} \end{aligned}$$

(89)

$$\begin{aligned} \vec {v}_1^{(1)}= & {} \begin{bmatrix} \chi \left( \frac{\sqrt{2}}{2} - x(\xi )\right) \\ \chi \left( \frac{\sqrt{2}}{2} - y(\xi )\right) \\ 1 \end{bmatrix} \end{aligned}$$

(90)

For the running example, the values of $\frac{\mathrm{{d}}x}{\mathrm{{d}}\xi }$ and $\frac{\mathrm{{d}}y}{\mathrm{{d}}\xi }$ are:

$$\begin{aligned} \frac{dx}{\mathrm{{d}}\xi } = \frac{d}{\mathrm{{d}}\xi } \chi \cos (\xi ) = -\chi \sin (\xi ) \end{aligned}$$

(91)

and

$$\begin{aligned} \frac{\mathrm{{d}}y}{\mathrm{{d}}\xi } = \frac{d}{\mathrm{{d}}\xi } \chi \sin (\xi ) = \chi \cos (\xi ) \end{aligned}$$

(92)

For the axial contributions, the scaling will not factor into the final results in the running example since the values are normalized.

For simplicity the start point for the running example is $\xi _0 = 0$.

Tables 9, 10, and 11 demonstrate three different states of the A matrix for the running example. Table 9 is the integration from the splice to the first encountered interaction point. The size of the matrix that is non-zero is the first $3\times 3$ since only the internal forces at the splice effect the energy. Table 10 shows the integration from anchor 1 to anchor 2 to demonstrate how the matrix encompasses additional forces as the integration traverses the loop. The upper left $6\times 6$ matrix are densely filled since both the internal forces and forces at the first interaction point effect the energy. Lastly, Table 11 shows the entire A matrix with all of the integrations summed, the Lagrange multiplier constraints, and the equilibrium equations.

Table 9 Integration results for the A matrix for running example between loop splice and interaction point 1

Table 10 Integration results for the A matrix for running example between loop interaction points 1 and 2

Table 11 Assembled A matrix for running example