Design of periodic elastoplastic energy dissipating microstructures

Abstract

The design of periodic elastoplastic microstructures for maximum energy dissipation is carried out using topology optimization. While the topology optimization of elastic microstructures has been performed in numerous studies, microstructural design considering inelastic behavior is relatively untouched due to a number of reasons which are addressed in this study. An RVE-based multiscale model is employed for computational homogenization with periodic boundary constraints, satisfying the Hill-Mandel principle. The plastic anisotropy which may be prevalent in materials fabricated through additive manufacturing processes is considered by modeling the constitutive behavior at the microscale with Hoffman plasticity. Discretization is done using enhanced assumed strain elements to avoid locking from incompressible plastic flow under plane strain conditions and a Lagrange multiplier approach is used to enforce periodic boundary constraints in the discrete system. The design problem is formulated using a density-based parameterization in conjunction with a SIMP-like material interpolation scheme. Attention is devoted to issues such as dependence on initial design and enforcement of microstructural connectivity, and a number of optimized microstructural designs are obtained under different prescribed deformation modes.

Appendices

Appendix A: Sensitivity derivatives for Hoffman model

This appendix presents the derivatives needed for sensitivity analysis using the Hoffman plasticity model which were originally derived in Ref. Zhang et al. (2017). These derivatives are based on the discrete form of the evolution equations which come from the numerical implementation using an elastic predictor/plastic return-mapping algorithm. As this implementation has been detailed in prior studies (De Borst and Feenstra 1990; Schellekens and De Borst 1990; De Souza Neto et al. 2011), it is not presented herein. Interested readers are directed to Appendix A in Zhang et al. (2017) for details on the elastic predictor/plastic return-mapping algorithm.

For the choice of c^k shown in (45), the corresponding local constraint set H^k is

$$ \boldsymbol{H}^{k} = \left[\begin{array}{cc} \boldsymbol{H}^{k}_{1} \\ {\vdots} \\ \boldsymbol{H}^{k}_{n_{ele}} \end{array}\right] = \boldsymbol{0} \quad \text{with} \quad \boldsymbol{H}^{k}_{e} = \left[\begin{array}{llll} \boldsymbol{H}^{k}_{e_{1}} \\ \boldsymbol{H}^{k}_{e_{2}} \\ \boldsymbol{H}^{k}_{e_{3}} \\ \boldsymbol{H}^{k}_{e_{4}} \\ \tilde{\boldsymbol{H}}^{k}_{e} \end{array}\right] $$

(57)

Here,

$$ \tilde{\boldsymbol{H}}^{k}_{e} = \sum\limits_{r = 1}^{n_{ipt}} \boldsymbol{G}^{T}_{e_{r}} \boldsymbol{\sigma}^{k}_{e_{r}} w_{r} = \boldsymbol{0} $$

(58)

and for an elastic step

$$ \boldsymbol{H}^{k}_{e_{r}} = \left\{\begin{array}{llll} \boldsymbol{h}^{k}_{e_{r1}} = \hat{\boldsymbol{\sigma}}^{k}_{e_{r}} - \hat{\boldsymbol{C}}^{e} \left( \hat{\boldsymbol{\varepsilon}}^{k}_{e_{r}} - \hat{\boldsymbol{\varepsilon}}^{p^{k}}_{e_{r}} \right) = \boldsymbol{0} \\ \boldsymbol{h}^{k}_{e_{r2}} = \hat{\boldsymbol{\varepsilon}}^{p^{k}}_{e_{r}} - \hat{\boldsymbol{\varepsilon}} ^{p^{k-1}}_{e_{r}} = \boldsymbol{0} \\ h^{k}_{e_{r3}} = \alpha^{k}_{e_{r}} - \alpha^{k-1}_{e_{r}} = 0 \\ h^{k}_{e_{r4}} = {\Delta}\gamma^{k}_{e_{r}} = 0 \end{array}\right. $$

(59)

while for a plastic step

$$ \boldsymbol{H}^{k}_{e_{r}} = \left\{\begin{array}{llll} \boldsymbol{h}^{k}_{e_{r1}} = \hat{\boldsymbol{\sigma}}^{k}_{e_{r}} - \hat{\boldsymbol{C}}^{e} \left( \hat{\boldsymbol{\varepsilon}}^{k}_{e_{r}} - \hat{\boldsymbol{\varepsilon}}^{p^{k}}_{e_{r}} \right) = \boldsymbol{0} \\ \boldsymbol{h}^{k}_{e_{r2}} = \hat{\boldsymbol{\varepsilon}}^{p^{k}}_{e_{r}} - \hat{\boldsymbol{\varepsilon}} ^{p^{k-1}}_{e_{r}} - {\Delta}\gamma^{k}_{e_{r}} \left( \boldsymbol{P}\hat{\boldsymbol{\sigma}}^{k}_{e_{r}} + \boldsymbol{q} \right) = \boldsymbol{0} \\ h^{k}_{e_{r3}} = \alpha^{k}_{e_{r}} - \alpha^{k-1}_{e_{r}} - \sqrt{\frac{2z^{k}_{e_{r}}}{3}} {\Delta}\gamma^{k}_{e_{r}} = 0 \\ h^{k}_{e_{r4}} = \frac{1}{2} \hat{\boldsymbol{\sigma}}^{k^{T}}_{e_{r}} \boldsymbol{P} \hat{\boldsymbol{\sigma}} ^{k}_{e_{r}} + \boldsymbol{q}^{T} \hat{\boldsymbol{\sigma}}^{k}_{e_{r}} - \zeta^{2}(\alpha^{k}_{e_{r}}) = 0 \end{array}\right. $$

(60)

with $z^{k}_{e_{r}} = \left ( \boldsymbol {P}\hat {\boldsymbol {\sigma }}^{k}_{e_{r}} + \boldsymbol {q} \right )^{T} \boldsymbol {Z} \left ( \boldsymbol {P}\hat {\boldsymbol {\sigma }}^{k}_{e_{r}} + \boldsymbol {q} \right )$. The derivative ∂H^k/∂u^k− 1 = 0 while

$$\begin{array}{@{}rcl@{}} \frac{\partial{\boldsymbol{H}^{k}}}{\partial{\boldsymbol{u}^{k}}} &=& \left[\begin{array}{cc} \frac{\partial{\boldsymbol{H}^{k}_{1}}}{\partial{\boldsymbol{u}^{k}}} \\ {\vdots} \\ \frac{\partial{\boldsymbol{H}^{k}_{n_{ele}}}}{\partial{\boldsymbol{u}^{k}}} \end{array}\right] \text{with } \\ \frac{\partial{\boldsymbol{H}^{k}_{j}}}{\partial{\boldsymbol{u}^{k}}}&=& \overset{n_{ele}}{\underset{e = 1}{\mathcal{A}}} \left( \frac{\partial{\boldsymbol{H}^{k}_{j}}}{\partial{\boldsymbol{u}^{k}_{e}}} \right), \quad j = 1,2,...,n_{ele} \end{array} $$

(61)

The terms $\partial {\boldsymbol {H}^{k}_{j}} / \partial {\boldsymbol {u}^{k}_{e}}$ are nonzero only when j = e and are calculated as

$$\begin{array}{@{}rcl@{}} \frac{\partial{\boldsymbol{H}^{k}_{e}}}{\partial{\boldsymbol{u}^{k}_{e}}} &=& \left[\begin{array}{llllllll} \frac{\partial{\boldsymbol{H}^{k}_{e_{1}}}}{\partial{\boldsymbol{u}^{k}_{e}}} &\frac{\partial{\boldsymbol{H}^{k}_{e_{2}}}}{\partial{\boldsymbol{u}^{k}_{e}}} & \frac{\partial{\boldsymbol{H}^{k}_{e_{3}}}}{\partial{\boldsymbol{u}^{k}_{e}}} &\frac{\partial{\boldsymbol{H}^{k}_{e_{4}}}}{\partial{\boldsymbol{u}^{k}_{e}}} & \frac{\partial{\tilde{\boldsymbol{H}}^{k}_{e}}}{\partial{\boldsymbol{u}^{k}_{e}}} \end{array}\right]^{T} \\ \frac{\partial{\boldsymbol{H}^{k}_{e_{r}}}}{\partial{\boldsymbol{u}^{k}_{e}}} &=& \left[\begin{array}{ccc} -\hat{\boldsymbol{C}}^{e} \boldsymbol{T}_{2} \boldsymbol{B}_{r} \\ \boldsymbol{0} \\ \boldsymbol{0} \\ \boldsymbol{0} \end{array}\right] \quad \text{and} \quad \frac{\partial{\tilde{\boldsymbol{H}}^{k}_{e}}}{\partial{\boldsymbol{u}^{k}_{e}}} = \boldsymbol{0} \end{array} $$

(62)

The derivatives ∂H^k/∂c^k and ∂H^k/∂c^k− 1 have the same structure, i.e.

$$\begin{array}{@{}rcl@{}} &\frac{\partial\boldsymbol{H}^{k}}{\partial\boldsymbol{c}^{k}}=\left[\begin{array}{cccc} \frac{\partial\boldsymbol{H}_{1}^{k}}{\partial\boldsymbol{c}_{1}^{k}} & \boldsymbol{0} & {\dots} & \boldsymbol{0}\\ \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{2}^{k}}{\partial\boldsymbol{c}_{2}^{k}} & {\dots} & \boldsymbol{0}\\ {\vdots} & {\vdots} & {\ddots} & \vdots\\ \boldsymbol{0} & \boldsymbol{0} & {\dots} & \frac{\partial\boldsymbol{H}_{n_{ele}}^{k}}{\partial\boldsymbol{c} _{n_{ele}}^{k}} \end{array}\right]\\ &\frac{\partial\boldsymbol{H}^{k}}{\partial\boldsymbol{c}^{k-1}}=\left[\begin{array}{cccc} \frac{\partial\boldsymbol{H}_{1}^{k}}{\partial\boldsymbol{c}_{1}^{k-1}} & \boldsymbol{0} & {\dots} & \boldsymbol{0}\\ \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{2}^{k}}{\partial\boldsymbol{c}_{2}^{k-1}} & {\dots} & \boldsymbol{0}\\ {\vdots} & {\vdots} & {\ddots} & \vdots\\ \boldsymbol{0} & \boldsymbol{0} & {\dots} & \frac{\partial\boldsymbol{H}_{n_{ele}}^{k}}{\partial\boldsymbol{c} _{n_{ele}}^{k-1}} \end{array}\right] \end{array} $$

(63)

since $\boldsymbol {c}^{k}_{i}$ and $\boldsymbol {c}^{k}_{j}$ are independent and $\boldsymbol {H}^{k}_{i}$ and $\boldsymbol {H}^{k}_{j}$ are uncoupled when i ≠ j. The nonzero sub-matrices have the form

$$\begin{array}{@{}rcl@{}} &\frac{\partial\boldsymbol{H}^{k}_{e}}{\partial\boldsymbol{c}^{k}_{e}}=\left[\begin{array}{cccccc} \frac{\partial\boldsymbol{H}_{e_{1}}^{k}}{\partial\boldsymbol{c}_{e_{1}}^{k}} & \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{e_{1}}^{k}}{\partial\tilde{\boldsymbol{\alpha}}_{e}^{k}}\\ \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{e_{2}}^{k}}{\partial\boldsymbol{c}_{e_{2}}^{k}} & \boldsymbol{0} & \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{e_{2}}^{k}}{\partial\tilde{\boldsymbol{\alpha}}_{e}^{k}}\\ \boldsymbol{0} & \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{e_{3}}^{k}}{\partial\boldsymbol{c}_{e_{3}}^{k}} & \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{e_{3}}^{k}}{\partial\tilde{\boldsymbol{\alpha}}_{e}^{k}}\\ \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0} &\frac{\partial\boldsymbol{H}_{e_{4}}^{k}}{\partial\boldsymbol{c}_{e_{4}}^{k}} &\frac{\partial\boldsymbol{H}_{e_{4}}^{k}}{\partial\tilde{\boldsymbol{\alpha}}_{e}^{k}} \\ \frac{\partial\tilde{\boldsymbol{H}}_{e}^{k}}{\partial\boldsymbol{c}_{e_{1}}^{k}} & \frac{\partial\tilde{\boldsymbol{H}}_{e}^{k}}{\partial\boldsymbol{c}_{e_{2}}^{k}} &\frac{\partial\tilde{\boldsymbol{H}}_{e}^{k}}{\partial\boldsymbol{c}_{e_{3}}^{k}} & \frac{\partial\tilde{\boldsymbol{H}}_{e}^{k}}{\partial\boldsymbol{c}_{e_{4}}^{k}} & \boldsymbol{0} \end{array}\right]\\ \end{array} $$

$$\begin{array}{@{}rcl@{}} &\frac{\partial\boldsymbol{H}^{k}_{e}}{\partial\boldsymbol{c}^{k-1}_{e}}=\left[\begin{array}{ccccc} \frac{\partial\boldsymbol{H}_{e_{1}}^{k}}{\partial\boldsymbol{c}_{e_{1}}^{k-1}} & \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{e_{1}}^{k}}{\partial\tilde{\boldsymbol{\alpha}}_{e}^{k-1}} \\ \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{e_{2}}^{k}}{\partial\boldsymbol{c}_{e_{2}}^{k-1}} & \boldsymbol{0} & \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{e_{2}}^{k}}{\partial\tilde{\boldsymbol{\alpha}}_{e}^{k-1}} \\ \boldsymbol{0} & \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{e_{3}}^{k}}{\partial\boldsymbol{c}_{e_{3}}^{k-1}} & \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{e_{3}}^{k}}{\partial\tilde{\boldsymbol{\alpha}}_{e}^{k-1}} \\ \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0} &\frac{\partial\boldsymbol{H}_{e_{4}}^{k}}{\partial\boldsymbol{c}_{e_{4}} ^{k-1}} & \frac{\partial\boldsymbol{H}_{e_{4}}^{k}}{\partial\tilde{\boldsymbol{\alpha}}_{e}^{k-1}} \\ \frac{\partial\tilde{\boldsymbol{H}}_{e}^{k}}{\partial\boldsymbol{c}_{e_{1}}^{k-1}} & \frac{\partial\tilde{\boldsymbol{H}}_{e}^{k}}{\partial\boldsymbol{c}_{e_{2}}^{k-1}} &\frac{\partial\tilde{\boldsymbol{H}}_{e}^{k}}{\partial\boldsymbol{c}_{e_{3}}^{k-1}} & \frac{\partial\tilde{\boldsymbol{H}}_{e}^{k}}{\partial\boldsymbol{c}_{e_{4}}^{k-1}} & \boldsymbol{0} \end{array}\right] \end{array} $$

(64)

The terms in $\partial {\boldsymbol {H}^{k}_{e}} / \partial {\boldsymbol {c}_{e_{1}}^{k-1}}$ are the same for both elastic and plastic steps, i.e.

$$ \frac{\partial{\boldsymbol{H}_{e_{r}}^{k}}}{\partial{\boldsymbol{c}_{e_{r}}^{k-1}}} = \left[\begin{array}{cccc} \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0} \\ \boldsymbol{0} & -\boldsymbol{T}_{2} & \boldsymbol{0} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{0} & -1 & 0 \\ \boldsymbol{0} & \boldsymbol{0} & 0 & 0 \end{array}\right], \quad \frac{\partial\boldsymbol{H}_{e_{r}}^{k}}{\partial\tilde{\boldsymbol{\alpha}}_{e}^{k-1}} = \boldsymbol{0}, \quad \frac{\partial\tilde{\boldsymbol{H}}_{e}^{k}}{\partial\boldsymbol{c}_{e_{r}}^{k-1}} = \boldsymbol{0} $$

(65)

The terms $\partial \boldsymbol {H}_{e_{r}}^{k} / \partial \tilde {\boldsymbol {\alpha }}_{e}^{k}$ and $\partial \tilde {\boldsymbol {H}}_{e}^{k} / \partial \boldsymbol {c}_{e_{r}}^{k}$ in $\partial \boldsymbol {H}^{k}_{e} / \partial \boldsymbol {c}^{k}_{e}$ are

$$ \frac{\partial\boldsymbol{H}_{e_{r}}^{k}}{\partial\tilde{\boldsymbol{\alpha}}_{e}^{k}} = \left[\begin{array}{cccc} -\hat{\boldsymbol{C}}^{e} \boldsymbol{T}_{2} \boldsymbol{G}_{e_{r}} \\ \boldsymbol{0} \\ 0 \\ 0 \end{array}\right], \quad \frac{\partial\tilde{\boldsymbol{H}}_{e}^{k}}{\partial\boldsymbol{c}_{e_{r}}^{k}} = \left[\begin{array}{llll} w_{r} \boldsymbol{G}_{e_{r}}^{T} & \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0} \end{array}\right] $$

(66)

regardless of whether the step is elastic or plastic. Finally, the term $\partial \boldsymbol {H}_{e_{r}}^{k} / \partial \boldsymbol {c}_{e_{r}}^{k}$ at an elastic step is

$$ \frac{\partial\boldsymbol{H}_{e_{r}}^{k}}{\partial\boldsymbol{c}_{e_{r}}^{k}} = \left[\begin{array}{cccc} \boldsymbol{T}_{1} & \hat{\boldsymbol{C}}^{e} \boldsymbol{T}_{2} & \boldsymbol{0} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{T}_{2} & \boldsymbol{0} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{0} & 1 & 0 \\ \boldsymbol{0} & \boldsymbol{0} & 0 & 1 \end{array}\right] $$

(67)

and at a plastic step is

$$\begin{array}{@{}rcl@{}} &&{}\frac{\partial\boldsymbol{H}_{e_{r}}^{k}}{\partial\boldsymbol{c}_{e_{r}}^{k}} = \left[\begin{array}{llll} \boldsymbol{M}_{11} & \boldsymbol{M}_{12} & \boldsymbol{M}_{13} & \boldsymbol{M}_{14} \\ \boldsymbol{M}_{21} & \boldsymbol{M}_{22} & \boldsymbol{M}_{23} & \boldsymbol{M}_{24} \\ \boldsymbol{M}_{31} & \boldsymbol{M}_{32} & M_{33} & M_{34} \\ \boldsymbol{M}_{41} & \boldsymbol{M}_{42} & M_{43} & M_{44} \end{array}\right]\\ &&\boldsymbol{{}M}_{11} = \boldsymbol{T}_{1}, \quad \boldsymbol{M}_{12} = \hat{\boldsymbol{C}}^{e} \boldsymbol{T}_{2}, \quad \boldsymbol{M}_{13} = \boldsymbol{0}, \quad \boldsymbol{M}_{14} = \boldsymbol{0} \\ &&\boldsymbol{{}M}_{21} = -{\Delta}\gamma^{k}_{e_{r}} \boldsymbol{P} \boldsymbol{T}_{1}, \boldsymbol{M}_{22} = \boldsymbol{T}_{2}, \boldsymbol{M}_{23} = \boldsymbol{0}, \boldsymbol{M}_{24} = -\left( \boldsymbol{P}\hat{\boldsymbol{\sigma}}^{k}_{e_{r}} + \boldsymbol{q} \right) \\ &&\boldsymbol{{}M}_{31} = -\sqrt{\frac{1}{6z^{k}_{e_{r}}}} {\Delta}\gamma^{k}_{e_{r}} \frac{dz^{k}_{e_{r}}}{d\hat{\boldsymbol{\sigma}}^{k}_{e_{r}}} \boldsymbol{T}_{1}, \boldsymbol{M}_{32} = \boldsymbol{0}, M_{33} = 1, M_{34} = -\sqrt{\frac{2z^{k}_{e_{r}}}{3}} \\ &&\boldsymbol{{}M}_{41} = \hat{\boldsymbol{\sigma}}^{k^{T}}_{e_{r}} \boldsymbol{P} \boldsymbol{T}_{1} + \boldsymbol{q}^{T} \boldsymbol{T}_{1}, \boldsymbol{\!M}_{42} = \boldsymbol{0}, \!M_{43} = \!-2\zeta \left( \alpha^{k}_{e_{r}} \right) \!\frac{d\zeta}{d\alpha^{k}_{e_{r}}}, M_{44} = 0\\ \end{array} $$

(68)

The derivative with respect to the element density variables is

$$ \frac{\partial\boldsymbol{H}^{k}}{\partial\boldsymbol{\rho}} = \left[\begin{array}{ccccc} \frac{\partial\boldsymbol{H}_{1}^{k}}{\partial\rho_{1}} & \boldsymbol{0} & {\dots} & \boldsymbol{0} \\ \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{2}^{k}}{\partial\rho_{2}} & {\dots} & \boldsymbol{0} \\ {\vdots} & {\vdots} & {\ddots} & {\vdots} \\ \boldsymbol{0} & \boldsymbol{0} & {\dots} & \frac{\partial\boldsymbol{H}_{n_{ele}}^{k}}{\partial\rho _{n_{ele}}} \end{array}\right] $$

(69)

The nonzero entries $\partial \boldsymbol {H}_{e}^{k} / \partial \rho _{e}$ are

$$\begin{array}{@{}rcl@{}} &&\frac{\partial\boldsymbol{H}_{e}^{k}}{\partial\rho_{e}} = \left[\begin{array}{ccccc} \frac{\partial\boldsymbol{H}_{e_{1}}^{k}}{\partial\rho_{e}} &\frac{\partial\boldsymbol{H}_{e_{2}}^{k}}{\partial\rho_{e}} & \frac{\partial\boldsymbol{H}_{e_{3}}^{k}}{\partial\rho_{e}} & \frac{\partial\boldsymbol{H}_{e_{4}}^{k}}{\partial\rho_{e}} & \frac{\partial\tilde{\boldsymbol{H}}_{e}^{k}}{\partial\rho_{e}} \end{array}\right] \\ &&\frac{\partial\tilde{\boldsymbol{H}}_{e}^{k}}{\partial\rho_{e}} = \boldsymbol{0} \end{array} $$

(70)

For an elastic step, $\partial \boldsymbol {H}_{e_{r}}^{k} / \partial \rho _{e}$ is

$$ \frac{\partial\boldsymbol{H}_{e_{r}}^{k}}{\partial\rho_{e}} = \left[\begin{array}{cccc} -\frac{\partial{\hat{\boldsymbol{C}}^{e}}}{\partial{\rho_{e}}} \left( \hat{\boldsymbol{\varepsilon}}^{k}_{e_{r}} - \hat{\boldsymbol{\varepsilon}}^{p^{k}}_{e_{r}} \right) \\ \boldsymbol{0} \\ 0 \\ 0 \end{array}\right] $$

(71)

and for a plastic step it is

$$ \frac{\partial\boldsymbol{H}_{e_{r}}^{k}}{\partial\rho_{e}} = \left[\begin{array}{cccc} -\frac{\partial{\hat{\boldsymbol{C}}^{e}}}{\partial{\rho_{e}}} \left( \hat{\boldsymbol{\varepsilon}}^{k}_{e_{r}} - \hat{\boldsymbol{\varepsilon}}^{p^{k}}_{e_{r}} \right) \\ -{\Delta}\gamma^{k}_{e_{r}} \left( \frac{\partial\boldsymbol{P}}{\partial\rho_{e}} \hat{\boldsymbol{\sigma}}^{k}_{e_{r}} + \frac{\partial\boldsymbol{q}}{\partial\rho_{e}} \right) \\ -\sqrt{\frac{1}{6z^{k}_{e_{r}}}} \frac{\partial z^{k}_{e_{r}}}{\partial\rho_{e}} {\Delta} \gamma^{k}_{e_{r}} \\ \frac{1}{2} \hat{\boldsymbol{\sigma}}^{k^{T}}_{e_{r}} \frac{\partial\boldsymbol{P}}{\partial\rho_{e}} + \frac{\partial\boldsymbol{q}^{T}}{\partial\rho_{e}} \hat{\boldsymbol{\sigma}}^{k}_{e_{r}} - 2\zeta \frac{\partial\zeta}{\partial\rho_{e}} \end{array}\right] $$

(72)

The derivatives $\partial {\hat {\boldsymbol {C}}^{e}} / \partial {\rho _{e}}$, ∂P/∂ρ_e, ∂q/∂ρ_e, $\partial z^{k}_{e_{r}} / \partial \rho _{e}$ and ∂ζ/∂ρ_e come from the material interpolation.

Appendix B: Sensitivity verification

The sensitivity analysis outlined in Section 3.3 is verified in this Appendix by comparing the values computed using the analytical adjoint method with those obtained numerically using the central difference method (CDM). The CDM calculates the sensitivity of a response function F with respect to a design variable x_e as

$$ \frac{\partial F(\boldsymbol{x})}{\partial x_{e}} \approx \frac{F(\boldsymbol{x} + {\Delta}\boldsymbol{h}) - F(\boldsymbol{x} - {\Delta}\boldsymbol{h})} {2{\Delta} h} $$

(73)

where x is the full vector of design variables and Δh is a vector with entries of zero except at the index corresponding to x_e, where the entry is the perturbation value Δh. The perturbation value for CDM used in the following verification study is Δh = 10^− 5.

The example used for sensitivity verification is a square RUC domain discretized into a 10 × 10 mesh of plane strain Q1/E4 EAS elements with a thickness of 1. The discretized domain and element numbering are shown in Fig. 15a. The size of the domain is set to 1000 by 1000 and the elastic material parameters are E = 2500 and ν = 0.38. The hardening coefficient K^h = 125 and is normalized by the yield stress σ_y = 20.

Fig. 15

Square RUC domain with element numbering and density distribution

The density distribution used for this example is shown in Fig. 15b, where the density values within the square inclusion are ρ_in = 0.2 while the outer density values are ρ_out = 0.7. The penalization parameters p₀ and p₁ are set to 3 and 2.5, respectively. No filter is utilized so that in this case ∂F/∂x = ∂F/∂ρ. The RUC is subject to a simple shear deformation by prescribing the macroscopic strain tensor as $\hat {\overline {\boldsymbol {\varepsilon }}} = \left [\begin {array}{llll} 0 \quad 0 \quad 0.2 \end {array}\right ]$. Two verification cases are run using this example. In case 1, the scaling factors χ and ω are set to 1 and the angle θ = 0, resulting in the von Mises model without rotation. In case 2, ω = 4, χ = 2 and θ = π/4. Figure 16 shows the sensitivity values calculated using the two cases of material parameters. From here it can be seen that the sensitivity values match closely, verifying the correct implementation of the path-dependent adjoint sensitivity analysis from Section 3.3.

Fig. 16

Comparison of sensitivity values obtained using the adjoint method and CDM with Δh = 10^− 5 for two cases of Hoffman plasticity under simple shear