Failure resistant topology optimization of structures using nonlocal elastoplastic-damage model

Abstract

For energy absorbing structures made up of ductile materials, the plastic strain accumulation often leads to early material damage and failure, which can deteriorate the overall structural performance. The goal of this work is to limit this damage in elastoplastic designs using the density-based topology optimization framework such that the optimized structures can absorb energy in a more controllable manner. To this end, an implicit nonlocal coupled elastoplastic damage model is considered for simulating the material damage and softening behavior. The nonlocal effect from the void elements is removed by introducing a scaling scheme for the nonlocal parameters. Path-dependent sensitivity is derived analytically using an adjoint method whose accuracy is further verified by the central difference method. The effectiveness of the proposed method is demonstrated through several numerical examples. It is shown that the load-carrying capacity, ductility, as well as ultimate plastic work dissipation capacity of the optimized design, can be considerably improved by the proposed method.

Appendices

Appendix A: Return-mapping algorithm and consistent tangent modulus

In this appendix, numerical implementation of the nonlocal damage elastoplasticity model given in Section 2 is presented. In the context of the strain-driven finite element formulation, given data at an integration point: $\phantom {\dot {i}\!}\boldsymbol {\varepsilon }_{m}^{p}$, $\phantom {\dot {i}\!}\alpha _{m}$ and $\phantom {\dot {i}\!}{\kappa }_{m}$ at previous step m, and $\phantom {\dot {i}\!}\boldsymbol {\varepsilon }$ and $\phantom {\dot {i}\!}\overline {\alpha }$ at current step $\phantom {\dot {i}\!}m + 1$, the goal is to find the unknown variables: $\phantom {\dot {i}\!}\boldsymbol {\varepsilon }_{m + 1}^{p}$, $\phantom {\dot {i}\!}\alpha _{m + 1}$ and the consistent tangent moduli at the current step. Note that the subscript $\phantom {\dot {i}\!}m + 1$ of the variables at current step is omitted for the sake of clarity, also the step index is put at subscript instead of superscript, and the element number, integration point number are removed for clarity. An elastic predictor/return-mapping algorithm is used to solve the constitutive model as follows.

1.1 Step 1: Elastic trial step

$$ \begin{array}{l} \text{Given}:\quad\quad{\boldsymbol{\varepsilon}^{p^{tr}}}=\boldsymbol{\varepsilon}_{m}^{p},\quad\alpha^{tr}=\alpha_{m}\\ \text{Evaluate}:\\ \quad\quad\boldsymbol{\sigma}^{tr}=\mathbb{C}^{e}:(\boldsymbol{\varepsilon}-\boldsymbol{\varepsilon}^{p^{tr}}),\quad\boldsymbol{s}^{tr}=\mathbb{P}_{dev}^{s}:\boldsymbol{\sigma}^{tr}\\ \quad\quad{\kappa}=\max \{{\kappa}_{m},\overline{\alpha}\},\quad \zeta^{tr}=K^{h}\alpha^{tr}\\ \quad\quad\phi^{tr}(\boldsymbol{\sigma}^{tr},\alpha^{tr},{\kappa})=\|\boldsymbol{s}^{tr}\|-\sqrt{\frac{2}{3}}(1-d({\kappa}))(\sigma_{y}+\zeta^{tr})\ \end{array} $$

(43)

where $\phantom {\dot {i}\!}\mathbb {C}^{e}$ is the fourth-order elasticity tensor and $\phantom {\dot {i}\!}\mathbb {P}_{dev}^{s}$ denotes the fourth-order deviatoric projection tensor; $\phantom {\dot {i}\!}\overline {\alpha }$ is the nonlocal equivalent plastic strain at the integration point.

If $\phantom {\dot {i}\!}\phi ^{tr}\leq 0$, then the current step is an elastic step and the following elastic updates are made

$$ \boldsymbol{\varepsilon}^{p}=\boldsymbol{\varepsilon}^{p^{tr}},\quad\alpha=\alpha^{tr}\quad\text{and}\quad\boldsymbol{\sigma}=\boldsymbol{\sigma}^{tr} $$

(44)

and the consistent algorithmic tangent modulus $\phantom {\dot {i}\!}\boldsymbol {C}_{T}$ is given by

$$ \frac{\partial{\boldsymbol{\sigma}}}{\partial{\boldsymbol{\varepsilon}}}=\mathbb{C}^{e},\quad\frac{\partial{\boldsymbol{\sigma}}}{\partial{\overline{\alpha}}}=\boldsymbol{0},\quad\frac{\partial{\alpha}}{\partial{\boldsymbol{\varepsilon}}}=\boldsymbol{0},\quad\frac{\partial{\alpha}}{\partial{\overline{\alpha}}}= 0 $$

(45)

$$ \boldsymbol{C}_{T}=\large\left[\begin{array}{cc} \frac{\partial{\boldsymbol{\sigma}}}{\partial{\boldsymbol{\varepsilon}}}& \frac{\partial{\boldsymbol{\sigma}}}{\partial{\overline{\alpha}}}\\ \frac{\partial{\alpha}}{\partial{\boldsymbol{\varepsilon}}}& \frac{\partial{\alpha}}{\partial{\overline{\alpha}}} \end{array}\right]=\normalsize\left[\begin{array}{cc} \mathbb{C}^{e}\;&\boldsymbol{0}\\ \boldsymbol{0}\;&0\\ \end{array}\right] $$

(46)

Else if $\phantom {\dot {i}\!}\phi ^{tr}>0$, then there is a plastic flow in this step and the algorithm proceeds to Step 2.

1.2 Step 2: Plastic return mapping

In this step, the plastic flow is nonzero (i.e., $\phantom {\dot {i}\!}\gamma >0$). Using backward Euler method, the flow rules in (8) and (9) are discretized as

$$ \boldsymbol{\varepsilon}^{p}=\boldsymbol{\varepsilon}_{m}^{p}+{\Delta}\gamma\boldsymbol{n}\quad\quad \alpha=\alpha_{m}+\sqrt{\frac{2}{3}}{\Delta}\gamma $$

(47)

$$ \boldsymbol{\sigma}=\boldsymbol{\sigma}^{tr}-2\mu{\Delta}\gamma\boldsymbol{n}\quad\text{and}\quad\boldsymbol{s}=\boldsymbol{s}^{tr}-2\mu{\Delta}\gamma\boldsymbol{n} $$

(48)

in which (48)₂ further leads to

$$ \|\boldsymbol{s}\|=\|\boldsymbol{s}^{tr}\|-2\mu{\Delta}\gamma\quad\text{and}\quad\boldsymbol{n}=\boldsymbol{n}^{tr}=\frac{\boldsymbol{s}^{tr}}{\|\boldsymbol{s}^{tr}\|}\ $$

(49)

The yield function can be then expressed as

$$ \phi=\|\boldsymbol{s}^{tr}\|-2\mu{\Delta}\gamma-\sqrt{\frac{2}{3}}(1-d({\kappa}))\left(\sigma_{y}+\zeta(\alpha)\right) $$

(50)

The NR method can then be used for solving (50) for $\phantom {\dot {i}\!}{\Delta }\gamma$ where the Jacobian is given by

$$ \frac{\partial\phi}{\partial{\Delta}\gamma}=-2\mu-\frac{2}{3}(1-d({\kappa}))\frac{\partial{\zeta}}{\partial{\alpha}} $$

(51)

After obtaining $\phantom {\dot {i}\!}{\Delta }\gamma$, the internal state variables $\phantom {\dot {i}\!}\boldsymbol {\varepsilon }^{p}$ and $\phantom {\dot {i}\!}\alpha$ can be calculated through (47) while the stress tensor $\phantom {\dot {i}\!}\boldsymbol {\sigma }$ can be obtained through (48)₁. Next, to calculate the tangent moduli $\phantom {\dot {i}\!}\partial \boldsymbol {\sigma }/\partial \boldsymbol {\varepsilon }$, $\phantom {\dot {i}\!}\partial \boldsymbol {\sigma }/\partial \overline {\alpha }$, $\phantom {\dot {i}\!}\partial {\alpha }/\partial \boldsymbol {\boldsymbol {\varepsilon }}$ and $\phantom {\dot {i}\!}\partial {\alpha }/\partial \overline {\alpha }$, some results due to tensor algebra have to be derived, which are

$$ \begin{array}{l} \displaystyle\frac{\partial{\|\boldsymbol{s}^{tr}\|}}{\partial\boldsymbol{\varepsilon}}= 2\mu\boldsymbol{n}^{tr}\\ \displaystyle\mathbb{B}=\frac{\partial\boldsymbol{n}}{\partial\boldsymbol{\varepsilon}}=\frac{2\mu}{\|\boldsymbol{s}^{tr}\|}(\mathbb{P}_{dev}^{s}-\boldsymbol{n}^{tr}\otimes\boldsymbol{n}^{tr})\\ \displaystyle\frac{\partial{d}}{\partial{\boldsymbol{\varepsilon}}}= 0,\quad\frac{\partial{\|\boldsymbol{s}^{tr}\|}}{\partial{\overline{\alpha}}}= 0 \end{array} $$

(52)

As a result of consistency condition, the total differential of the yield equation gives

$$ d\phi=\frac{\partial\phi}{\partial{\boldsymbol{\varepsilon}}}:d\boldsymbol{\varepsilon}+\frac{\partial\phi}{\partial{\overline{\alpha}}}d\overline{\alpha}= 0 $$

(53)

with

$$ \frac{\partial\phi}{\partial\boldsymbol{\varepsilon}}=\frac{\partial\|\boldsymbol{s}^{tr}\|}{\partial\boldsymbol{\varepsilon}}-2\mu\frac{\partial({\Delta}\gamma)}{\partial\boldsymbol{\varepsilon}}-\frac{2}{3}(1-d)K^{h}\frac{\partial({\Delta}\gamma)}{\partial\boldsymbol{\varepsilon}} $$

(54)

$$\begin{array}{@{}rcl@{}} \frac{\partial\phi}{\partial\overline{\alpha}}&=&\frac{\partial\|\boldsymbol{s}^{tr}\|}{\partial\overline{\alpha}}-2\mu\frac{\partial({\Delta}\gamma)}{\partial\overline{\alpha}}+\sqrt{\frac{2}{3}}\frac{\partial{d}}{\partial{\kappa}}\frac{\partial{\kappa}}{\partial\overline{\alpha}}(\sigma_{y}+\zeta({\Delta}\gamma))\\ &&-\frac{2}{3}(1-d)K^{h}\frac{\partial({\Delta}\gamma)}{\partial\overline{\alpha}} \end{array} $$

(55)

Due to the independency of $\phantom {\dot {i}\!}d\boldsymbol {\varepsilon }$ and $\phantom {\dot {i}\!}d\overline {\alpha }$, we further have $\phantom {\dot {i}\!}\partial \phi /\partial \boldsymbol {\varepsilon }= 0$ and $\phantom {\dot {i}\!}\partial \phi /\partial \overline {\alpha }= 0$, which lead to

$$ \frac{\partial({\Delta}\gamma)}{\partial\boldsymbol{\varepsilon}}=c_{1}\boldsymbol{n}^{tr}\quad\text{with}\quad{c_{1}}=\frac{2\mu}{2\mu+\frac{2}{3}(1-d)K^{h}} $$

(56)

$$ \frac{\partial({\Delta}\gamma)}{\partial\overline{\alpha}}= \left\{\begin{array}{lll} \frac{\sqrt{\frac{2}{3}}(\sigma_{y}+\zeta({\Delta}\gamma))}{2\mu+\frac{2}{3}(1-d)K^{h}}\frac{\partial{d}}{\partial{\kappa}}\quad&&\text{if}\;{\kappa}=\overline{\alpha}\\ 0\quad&&\text{otherwise}\\ \end{array}\right. $$

(57)

With the derivations in (52), (56) and (57), the consistent algorithmic tangent modulus $\phantom {\dot {i}\!}\boldsymbol {C}_{T}$ can be calculated as

$$ \begin{array}{lll} \displaystyle\frac{\partial\boldsymbol{\sigma}}{\partial\boldsymbol{\varepsilon}}=\mathbb{C}^{e}-2\mu{\Delta}\gamma\mathbb{B}-2\mu{c_{1}}\boldsymbol{n}^{tr}\otimes\boldsymbol{n}^{tr}\\ \displaystyle\frac{\partial\boldsymbol{\sigma}}{\partial\overline{\alpha}}=-2\mu\frac{\partial({\Delta}\gamma)}{\partial\overline{\alpha}}\boldsymbol{n}^{tr}\\ \displaystyle\frac{\partial{\alpha}}{\partial\boldsymbol{\varepsilon}}=\sqrt{\frac{2}{3}}c_{1}\boldsymbol{n}^{tr},\quad\frac{\partial\alpha}{\partial\overline{\alpha}}=\sqrt{\frac{2}{3}}\frac{\partial{\Delta}\gamma}{\partial\overline{\alpha}}\\ \end{array} $$

(58)

$$ \begin{array}{ll} \boldsymbol{C}_{T}&=\large\left[\begin{array}{ll} \frac{\partial\boldsymbol{\sigma}}{\partial\boldsymbol{\varepsilon}}&\frac{\partial\boldsymbol{\sigma}}{\partial\overline{\alpha}}\\ \frac{\partial\alpha}{\partial\boldsymbol{\varepsilon}}&\frac{\partial\alpha}{\partial\overline{\alpha}} \end{array}\right]\\ &=\normalsize\left[\begin{array}{cc} \mathbb{C}^{e}-2\mu{\Delta}\gamma\mathbb{B}-2\mu{c_{1}}\boldsymbol{n}^{tr}\otimes\boldsymbol{n}^{tr}\;&-2\mu\frac{\partial({\Delta}\gamma)}{\partial\overline{\alpha}}\boldsymbol{n}^{tr}\\ \sqrt{\frac{2}{3}}c_{1}\boldsymbol{n}^{tr}\;&\sqrt{\frac{2}{3}}\frac{\partial{\Delta}\gamma}{\partial\overline{\alpha}}\\ \end{array}\right] \end{array} $$

(59)

The above consistent evaluation of the tangent operator $\phantom {\dot {i}\!}\boldsymbol {C}_{T}$ ensures the quadratic convergence of the global NR solver.

Appendix B: Explicit derivatives for the adjoint sensitivity analysis

This Appendix presents the complete derivations of the explicit derivatives of F, $\phantom {\dot {i}\!}\boldsymbol {R}^{k}$ and $\phantom {\dot {i}\!}\boldsymbol {H}^{k}$ needed in (36).

2.1 Derivatives of objective and constraint functions

According to (28), three target functions (f₀, $\phantom {\dot {i}\!}f_{1}$ and $\phantom {\dot {i}\!}f_{2}$) need sensitivity calculation. The sensitivity of the volume fraction objective function $\phantom {\dot {i}\!}f_{0}$ is straightforward to calculate and is given as

$$ \frac{df_{0}}{d\boldsymbol{\rho}}=\frac{1}{V_{0}}\boldsymbol{V} $$

(60)

where $\phantom {\dot {i}\!}\boldsymbol {V}$ is the element volume vector that collects all the element volume $\phantom {\dot {i}\!}V_{e}$.

For the plastic work constraint $\phantom {\dot {i}\!}F=f_{1}$ defined in (28)₂, since $\phantom {\dot {i}\!}W^{p}$ only depends on $\phantom {\dot {i}\!}\boldsymbol {v}^{k}$ and $\phantom {\dot {i}\!}\boldsymbol {v}^{k-1}$ as indicated by (29), it implies

$$ \frac{\partial{f_{1}}}{\partial\boldsymbol{\rho}}=\boldsymbol{0},\quad\frac{\partial{f_{1}}}{\partial\boldsymbol{U}^{k}}=\boldsymbol{0} $$

(61)

The only non-zero derivatives $\phantom {\dot {i}\!}\partial {f_{1}}/\partial {\boldsymbol {v}}^{k}$ are arranged as

$$ \begin{array}{l} \frac{\partial{f_{1}}}{\partial\boldsymbol{v}^{k}}=\left[\begin{array}{l} \frac{\partial{f_{1}}}{\partial\boldsymbol{v}_{1}^{k}}\quad \frac{\partial{f_{1}}}{\partial\boldsymbol{v}_{2}^{k}}\quad\dots\quad \frac{\partial{f_{1}}}{\partial\boldsymbol{v}_{n_{ele}}^{k}} \end{array}\right]\quad\text{with}\\ \frac{\partial{f_{1}}}{\partial\boldsymbol{v}_{e}^{k}}=\left[\begin{array}{l} \frac{\partial{f_{1}}}{\partial\boldsymbol{v}_{e_{1}}^{k}}\quad \frac{\partial{f_{1}}}{\partial\boldsymbol{v}_{e_{2}}^{k}}\quad \frac{\partial{f_{1}}}{\partial\boldsymbol{v}_{e_{3}}^{k}}\quad \frac{\partial{f_{1}}}{\partial\boldsymbol{v}_{e_{4}}^{k}} \end{array}\right] \end{array} $$

(62)

while the expression of term $\phantom {\dot {i}\!}\partial {f_{1}}/\partial \boldsymbol {v}_{e_{r}}^{k}$ is distinguished from $\phantom {\dot {i}\!}k=n$ and $\phantom {\dot {i}\!}k\neq n$, which reads

$$ \begin{array}{l} \frac{\partial{f_{1}}}{\partial{\boldsymbol{v}_{e_{r}}^{n}}}=-\frac{1}{\overline{W}^{p}}\left[\begin{array}{lllllllll} \frac{1}{2}\left(\boldsymbol{\varepsilon}_{e_{r}}^{p^{n}}-\boldsymbol{\varepsilon}_{e_{r}}^{p^{n-1}}\right)w_{r}&&\frac{1}{2}\left(\boldsymbol{\sigma}_{e_{r}}^{n}+\boldsymbol{\sigma}_{e_{r}}^{{n-1}}\right)w_{r}&&0&&0&&0 \end{array}\right]\\ \frac{\partial{f_{1}}}{\partial{\boldsymbol{v}_{e_{r}}^{k}}}=-\frac{1}{\overline{W}^{p}}\left[\begin{array}{lllllllll} \frac{1}{2}\left(\boldsymbol{\varepsilon}_{e_{r}}^{p^{k + 1}}-\boldsymbol{\varepsilon}_{e_{r}}^{p^{k-1}}\right)w_{r}&&\frac{1}{2}\left(\boldsymbol{\sigma}_{e_{r}}^{k-1}-\boldsymbol{\sigma}_{e_{r}}^{{k + 1}}\right)w_{r}&&0&&0&&0 \end{array}\right] \end{array} $$

(63)

For the damage constraint $\phantom {\dot {i}\!}F=f_{2}$ defined in (28)₃, since $\phantom {\dot {i}\!}D_{max}(\boldsymbol {x})$ only depends on $\phantom {\dot {i}\!}\boldsymbol {\rho }$ and $\phantom {\dot {i}\!}\boldsymbol {v}^{k}$ as indicated by (30) and (31), implies $\phantom {\dot {i}\!}\partial {f_{2}}/\partial \boldsymbol {U}^{k}=\boldsymbol {0}$. The non-zero derivative $\phantom {\dot {i}\!}\partial {f_{2}}/\partial \boldsymbol {\rho }$ is arranged as

$$ \begin{array}{ll} \frac{\partial{f_{2}}}{\partial\boldsymbol{\rho}}=&\left[ \frac{\partial{f_{2}}}{\partial\rho_{1}}\quad \frac{\partial{f_{2}}}{\partial\rho_{2}}\quad\dots\quad \frac{\partial{f_{2}}}{\partial\rho_{n_{ele}}} \right]\quad\text{with}\\ \frac{\partial{f_{2}}}{\partial\rho_{e}}=&\displaystyle\left({\sum}_{e = 1}^{n_{ele}}\left({\sum}_{r = 1}^{n_{ipt}}\left({\rho_{e}^{q}}{d_{e_{r}}^{n}}\right)^{p_{a}}\right)\right)^{\frac{1}{p_{a}}-1}\times\\ &\displaystyle{\sum}_{r = 1}^{n_{ipt}}\left(({\rho_{e}^{q}}d_{e_{r}}^{n})^{p_{a}-1}\left(q\rho_{e}^{q-1}d_{e_{r}}^{n}+{\rho_{e}^{q}}\frac{\partial{d_{e_{r}}^{n}}}{\partial\rho_{e}}\right)\right) \end{array} $$

(64)

where $\phantom {\dot {i}\!}\partial {d}_{e_{r}}^{n}/\partial \rho _{e}$ can be derived based on (6) using chain rule, which reads

$$ \frac{\partial{d}_{e_{r}}^{n}}{\partial\rho_{e}}=\frac{\partial{d}_{e_{r}}^{n}}{\partial{\kappa}_{th_{e}}}\frac{\partial{\kappa}_{th_{e}}}{\partial\rho_{e}}+\frac{\partial{d}_{e_{r}}^{n}}{\partial{\Gamma}}\frac{\partial{\Gamma}}{\partial{\kappa}_{th_{e}}}\frac{\partial{\kappa}_{th_{e}}}{\partial\rho_{e}}+\frac{\partial{d}_{e_{r}}^{n}}{\partial\beta_{e}}\frac{\partial\beta_{e}}{\partial\rho_{e}} $$

(65)

in which the derivatives $\phantom {\dot {i}\!}\partial {d}_{e_{r}}^{n}/\partial {\kappa }_{th_{e}}$, $\phantom {\dot {i}\!}{\partial {d}_{e_{r}}^{n}}/{\partial {\Gamma }}$, $\phantom {\dot {i}\!}{\partial {\Gamma }}/{\partial {\kappa }_{th_{e}}}$ and $\phantom {\dot {i}\!}{\partial {d}_{e_{r}}^{n}}/{\partial \beta _{e}}$ can be straightforwardly obtained from (6) and (7), and are omitted here. The derivatives of damage parameters $\phantom {\dot {i}\!}{\partial {\kappa }_{th_{e}}}/{\partial \rho _{e}}$ and $\phantom {\dot {i}\!}{\partial \beta _{e}}/{\partial \rho _{e}}$ can be calculated by the material interpolation shown in (21) and (22) as

$$ \begin{array}{l} \frac{\partial{\kappa}_{th_{e}}}{\partial\rho_{e}}=p_{4}({\kappa}_{th}-{\kappa}_{th_{min}})\rho_{e}^{(p_{4}-1)}\\ \frac{\partial\beta_{e}}{\partial\rho_{e}}=-p_{5}(\beta_{v}-\beta)\rho_{e}^{(p_{5}-1)}\\ \end{array} $$

(66)

The other non-zero derivative $\phantom {\dot {i}\!}\partial {f_{2}}/\partial \boldsymbol {v}^{k}$ is given by

$$ \begin{array}{l} \frac{\partial{f_{2}}}{\partial\boldsymbol{v}^{k}}=\left[ \frac{\partial{f_{2}}}{\partial\boldsymbol{v}_{1}^{k}}\quad \frac{\partial{f_{2}}}{\partial\boldsymbol{v}_{2}^{k}}\quad\dots\quad \frac{\partial{f_{2}}}{\partial\boldsymbol{v}_{n_{ele}}^{k}} \right]\quad\text{with}\\ \frac{\partial{f_{2}}}{\partial\boldsymbol{v}_{e}^{k}}=\left[ \frac{\partial{f_{2}}}{\partial\boldsymbol{v}_{e_{1}}^{k}}\quad \frac{\partial{f_{2}}}{\partial\boldsymbol{v}_{e_{2}}^{k}}\quad \frac{\partial{f_{2}}}{\partial\boldsymbol{v}_{e_{3}}^{k}}\quad \frac{\partial{f_{2}}}{\partial\boldsymbol{v}_{e_{4}}^{k}} \right] \end{array} $$

(67)

where the term $\phantom {\dot {i}\!}{\partial {f_{2}}}/{\partial \boldsymbol {v}_{e_{r}}^{k}}$ is distinguished from $\phantom {\dot {i}\!}k=n$ and $\phantom {\dot {i}\!}k\neq n$, which is expressed as

$$ \begin{array}{ll} \frac{\partial{f_{2}}}{\partial\boldsymbol{v}_{e_{r}}^{k}}&=\displaystyle\rho_{e}^{qp_{a}}\left(d_{e_{r}}^{k}\right)^{p_{a}-1}\left({\sum}_{e = 1}^{n_{ele}}\left({\sum}_{r = 1}^{n_{ipt}}\left({\rho_{e}^{q}}{d_{e_{r}}^{k}}\right)^{p_{a}}\right)\right)^{\frac{1}{p_{a}}-1}\times\\ &\quad \left[\boldsymbol{0} \quad \boldsymbol{0} \quad 0 \quad 0 \quad \frac{\partial{d}_{e_{r}}^{k}}{\partial{\kappa}_{e_{r}}^{k}}\right]\quad(k=n)\\ \frac{\partial{f_{2}}}{\partial\boldsymbol{v}_{e_{r}}^{k}}&=\boldsymbol{0}\quad(k\neq n)\\ \end{array} $$

(68)

Based on (6), the following derivative is obtained

$$ \frac{\partial{d}_{e_{r}}^{k}}{\partial{\kappa}_{e_{r}}^{k}}\,=\,\beta_{e}{e}^{-\beta_{e}\left({\kappa}_{e_{r}}^{k}-{\kappa}_{th_{e}}\right)}{\Gamma}\left({\kappa}_{e_{r}}^{k}\right)+\left(1\,-\,e^{-\beta_{e}({\kappa}_{e_{r}}^{k}-{\kappa}_{th_{e}})}\right)\frac{\partial{\Gamma}}{\partial{\kappa}}|_{{\kappa}={\kappa}_{e_{r}}^{k}} $$

(69)

where $\phantom {\dot {i}\!}{\partial {\Gamma }}/{\partial {\kappa }}$ can be obtained from (7).

2.2 Derivatives of $\boldsymbol {R}^{\text {\textit {k}}}$

According to (38), the derivative of $\phantom {\dot {i}\!}\boldsymbol {R}^{k}$ with respect to the solution variable $\phantom {\dot {i}\!}\boldsymbol {U}^{k}$ reads

$$ \frac{\partial\boldsymbol{R}^{k}}{\partial\boldsymbol{U}^{k}}=\overset{n_{ele}}{\underset{e = 1}{\mathscr{A}}}\left(\frac{\partial\boldsymbol{R}^{e^{k}}}{\partial\boldsymbol{U}_{e}^{k}}\right) $$

(70)

with

$$\begin{array}{@{}rcl@{}} \begin{array}{ll} \frac{\partial\boldsymbol{R}^{k}}{\partial\boldsymbol{U}_{e}^{k}}&=\left[\begin{array}{ll} \frac{\partial\boldsymbol{R}_{u}^{e^{k}}}{\partial\boldsymbol{u}_{e}^{k}}&\frac{\partial\boldsymbol{R}_{u}^{e^{k}}}{\partial\overline{\boldsymbol{\alpha}}_{e}^{k}}\\ \frac{\partial\boldsymbol{R}_{\overline{\alpha}}^{e^{k}}}{\partial\boldsymbol{u}_{e}^{k}}&\frac{\partial\boldsymbol{R}_{\overline{\alpha}}^{e^{k}}}{\partial\overline{\boldsymbol{\alpha}}_{e}^{k}} \end{array}\right]\\ &=\left[\begin{array}{lc} \boldsymbol{0}&\boldsymbol{0}\\\boldsymbol{0}&\sum\limits_{r = 1}^{n_{ipt}}\left(\boldsymbol{N}_{\overline{\alpha}_{r}}^{T}\boldsymbol{N}_{\overline{\alpha}_{r}}+(\rho_{e}^{p_{6}}\ell)^{2}\boldsymbol{B}_{\overline{\alpha}_{r}}^{T}\boldsymbol{B}_{\overline{\alpha}_{r}}\right)w_{r} \end{array}\right] \end{array} \end{array} $$

(71)

while the derivative $\phantom {\dot {i}\!}\partial \boldsymbol {R}^{k}/\partial \boldsymbol {U}^{k-1}=\boldsymbol {0}$. The derivative of $\phantom {\dot {i}\!}\boldsymbol {R}^{k}$ with respect to the auxiliary variable $\phantom {\dot {i}\!}\boldsymbol {v}^{k}$ is calculated as

$$ \begin{array}{l} \frac{\partial\boldsymbol{R}^{k}}{\partial\boldsymbol{v}^{k}}= \overset{n_{ele}}{\underset{e = 1}{\mathscr{A}}}\left(\frac{\partial\boldsymbol{R}_{e}^{k}}{\partial\boldsymbol{v}^{k}}\right)\quad\text{with}\\ \frac{\partial\boldsymbol{R}^{e^{k}}}{\partial\boldsymbol{v}^{k}}=\left[ \frac{\partial\boldsymbol{R}^{e^{k}}}{\partial\boldsymbol{v}_{1}^{k}}\quad\dots\quad\frac{\partial\boldsymbol{R}^{e^{k}}}{\partial\boldsymbol{v}_{n_{ele}}^{k}}\right] \end{array} $$

(72)

where the derivatives $\phantom {\dot {i}\!}\partial \boldsymbol {R}^{e^{k}}/\partial \boldsymbol {v}_{j}^{k}=\boldsymbol {0}$ for $\phantom {\dot {i}\!}j\neq e$. The only non-zero derivative $\phantom {\dot {i}\!}\partial \boldsymbol {R}^{e^{k}}/\partial \boldsymbol {v}_{e}^{k}$ is calculated as

$$ \begin{array}{ll} \frac{\partial\boldsymbol{R}^{e^{k}}}{\partial\boldsymbol{v}_{e}^{k}}&=\left[ \frac{\partial\boldsymbol{R}^{e^{k}}}{\partial\boldsymbol{v}_{e_{1}}^{k}}\,\, \frac{\partial\boldsymbol{R}^{e^{k}}}{\partial\boldsymbol{v}_{e_{2}}^{k}}\,\, \frac{\partial\boldsymbol{R}^{e^{k}}}{\partial\boldsymbol{v}_{e_{3}}^{k}}\,\, \frac{\partial\boldsymbol{R}^{e^{k}}}{\partial\boldsymbol{v}_{e_{4}}^{k}}\right]\quad\text{with}\\ \frac{\partial\boldsymbol{R}^{e^{k}}}{\partial\boldsymbol{v}_{e_{r}}^{k}}&=\left[ \frac{\partial\boldsymbol{R}^{e^{k}}}{\partial\boldsymbol{\sigma}_{e_{r}}^{k}}\,\, \frac{\partial\boldsymbol{R}^{e^{k}}}{\partial\boldsymbol{\varepsilon}_{e_{r}}^{p^{k}}}\,\, \frac{\partial\boldsymbol{R}^{e^{k}}}{\partial\alpha_{e_{r}}^{k}}\,\, \frac{\partial\boldsymbol{R}^{e^{k}}}{\partial{\Delta}\gamma_{e_{r}}^{k}}\,\, \frac{\partial\boldsymbol{R}^{e^{k}}}{\partial{\kappa}_{e_{r}}^{k}}\right]\\ &=\left[\begin{array}{ccccc} w_{r}\boldsymbol{B}_{u_{r}}^{T} &\quad \boldsymbol{0} &\quad \boldsymbol{0} &\quad \boldsymbol{0} &\quad \boldsymbol{0}\\ \boldsymbol{0} &\quad \boldsymbol{0} &\quad -w_{r}\rho_{e}^{p_{7}}\boldsymbol{N}_{\overline{\alpha}_{r}}^{T} &\quad \boldsymbol{0} &\quad \boldsymbol{0} \end{array}\right] \end{array} $$

(73)

while the derivative of $\phantom {\dot {i}\!}\boldsymbol {R}^{k}$ with respect to $\phantom {\dot {i}\!}\boldsymbol {v}^{k-1}$ is zero, i.e., $\phantom {\dot {i}\!}\partial \boldsymbol {R}^{k}/\partial \boldsymbol {v}^{k-1}=\boldsymbol {0}$. Finally, the derivative of $\phantom {\dot {i}\!}\boldsymbol {R}^{k}$ with respect to $\phantom {\dot {i}\!}\boldsymbol {\rho }$ is arranged as

$$ \begin{array}{l} \frac{\partial\boldsymbol{R}^{k}}{\partial\boldsymbol{\rho}}\,=\, \overset{n_{ele}}{\underset{e = 1}{\mathscr{A}}}\left(\frac{\partial\boldsymbol{R}^{e^{k}}}{\partial\rho_{e}}\right)\quad\text{with}\\ \frac{\partial\boldsymbol{R}^{e^{k}}}{\partial\rho_{e}}\,=\,\left[\begin{array}{c} \boldsymbol{0}\\ 2p_{6}\ell^{2}\rho_{e}^{2p_{6}-1}\sum\limits_{r = 1}^{n_{ipt}}w_{r}\boldsymbol{B}_{\overline{\alpha}_{r}}^{T}\boldsymbol{B}_{\overline{\alpha}_{r}}\overline{\boldsymbol{\alpha}}_{e}^{k}-p_{7}\rho_{e}^{p_{7}-1}\sum\limits_{r = 1}^{n_{ipt}}w_{r}\boldsymbol{N}_{\overline{\alpha}_{r}}^{T}\alpha_{e_{r}}^{k} \end{array}\right] \end{array} $$

(74)

2.3 Derivatives of $\boldsymbol {H}^{\text {\textit {k}}}$

As indicated by (41) and (42), $\phantom {\dot {i}\!}\boldsymbol {H}^{k}$ only depends on the solution field $\phantom {\dot {i}\!}\boldsymbol {U}^{k}$ at current step. Thus, $\phantom {\dot {i}\!}\partial \boldsymbol {H}^{k}/\partial \boldsymbol {U}^{k-1}=\boldsymbol {0}$ and the expressions for non-zero derivatives $\phantom {\dot {i}\!}\partial \boldsymbol {H}^{k}/\partial \boldsymbol {U}^{k}$ are the same for elastic and plastic steps, which can be written as

$$ \frac{\partial\boldsymbol{H}^{k}}{\partial\boldsymbol{U}^{k}}=\left[\begin{array}{c} \frac{\partial\boldsymbol{H}_{1}^{k}}{\partial\boldsymbol{U}^{k}}\\ \vdots\\ \frac{\partial\boldsymbol{H}_{n_{ele}}^{k}}{\partial\boldsymbol{U}^{k}}\\ \end{array}\right]\quad \text{with}\quad \frac{\partial\boldsymbol{H}_{j}^{k}}{\partial\boldsymbol{U}^{k}}=\overset{n_{ele}}{\underset{e = 1}{\mathscr{A}}}\left(\frac{\partial\boldsymbol{H}_{j}^{k}}{\partial\boldsymbol{U}_{e}^{k}}\right)\\ $$

(75)

It can be seen that $\phantom {\dot {i}\!}\partial \boldsymbol {H}_{j}^{k}/\partial \boldsymbol {U}_{e}^{k}=\boldsymbol {0}$ for $\phantom {\dot {i}\!}j\neq e$, so the non-zero term $\phantom {\dot {i}\!}\partial \boldsymbol {H}_{e}^{k}/\partial \boldsymbol {U}_{e}^{k}$ is calculated as

$$ \begin{array}{l} \frac{\partial\boldsymbol{H}_{e}^{k}}{\partial\boldsymbol{U}_{e}^{k}}=\left[\begin{array}{c} \frac{\partial\boldsymbol{H}_{e_{1}}^{k}}{\partial\boldsymbol{U}_{e}^{k}}\\ \frac{\partial\boldsymbol{H}_{e_{2}}^{k}}{\partial\boldsymbol{U}_{e}^{k}}\\ \frac{\partial\boldsymbol{H}_{e_{3}}^{k}}{\partial\boldsymbol{U}_{e}^{k}}\\ \frac{\partial\boldsymbol{H}_{e_{4}}^{k}}{\partial\boldsymbol{U}_{e}^{k}}\\ \end{array}\right]=\left[\begin{array}{cc} \frac{\partial\boldsymbol{H}_{e_{1}}^{k}}{\partial\boldsymbol{u}_{e}^{k}}&\frac{\partial\boldsymbol{H}_{e_{1}}^{k}}{\partial\overline{\boldsymbol{\alpha}}_{e}^{k}}\\ \frac{\partial\boldsymbol{H}_{e_{2}}^{k}}{\partial\boldsymbol{u}_{e}^{k}}&\frac{\partial\boldsymbol{H}_{e_{2}}^{k}}{\partial\overline{\boldsymbol{\alpha}}_{e}^{k}}\\ \frac{\partial\boldsymbol{H}_{e_{3}}^{k}}{\partial\boldsymbol{u}_{e}^{k}}&\frac{\partial\boldsymbol{H}_{e_{3}}^{k}}{\partial\overline{\boldsymbol{\alpha}}_{e}^{k}}\\ \frac{\partial\boldsymbol{H}_{e_{4}}^{k}}{\partial\boldsymbol{u}_{e}^{k}}&\frac{\partial\boldsymbol{H}_{e_{4}}^{k}}{\partial\overline{\boldsymbol{\alpha}}_{e}^{k}}\\ \end{array}\right]\quad \text{with}\\ \frac{\partial\boldsymbol{H}_{e_{r}}^{k}}{\partial\boldsymbol{u}_{e}^{k}}=\left[\begin{array}{c} -\mathbb{C}^{e}:\frac{\partial\boldsymbol{\varepsilon}_{e_{r}}^{k}}{\partial\boldsymbol{u}_{e}^{k}}\\ \boldsymbol{0}\\\boldsymbol{0}\\\boldsymbol{0}\\\boldsymbol{0}\\ \end{array}\right]\quad\text{and}\quad\frac{\partial\boldsymbol{H}_{e_{r}}^{k}}{\partial\overline{\boldsymbol{\alpha}}_{e}^{k}}=\left[\begin{array}{c} \boldsymbol{0}\\\boldsymbol{0}\\\boldsymbol{0}\\\boldsymbol{0}\\\boldsymbol{z}_{1} \end{array}\right]\\ \text{where}\quad\boldsymbol{z}_{1}=\left\{\begin{array}{ll} \boldsymbol{0},& \,\,\text{if}{\;\;}{\kappa}_{e_{r}}^{k-1}>\boldsymbol{N}_{\overline{\alpha}_{r}}\overline{\boldsymbol{\alpha}}^{e}\\ -\boldsymbol{N}_{\overline{\alpha}_{r}},&\,\,\text{otherwise} \end{array}\right. \end{array} $$

(76)

For the derivatives $\phantom {\dot {i}\!}\partial \boldsymbol {H}^{k}/\partial \boldsymbol {v}^{k}$ and $\phantom {\dot {i}\!}\partial \boldsymbol {H}^{k}/\partial \boldsymbol {v}^{k-1}$, the structures of the derivative matrices are

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

(77)

This is because $\phantom {\dot {i}\!}\boldsymbol {v}_{i}^{k}$ and $\phantom {\dot {i}\!}\boldsymbol {v}_{j}^{k}$ are independent and $\phantom {\dot {i}\!}\boldsymbol {H}_{i}^{k}$ and $\phantom {\dot {i}\!}\boldsymbol {H}_{j}^{k}$ are uncoupled given $\phantom {\dot {i}\!}i\neq j$. As a result, only submatrices lying on the diagonal are non-zeros. Moreover, the non-zero submatrices share the same structure as

$$ \begin{array}{l} \frac{\partial\boldsymbol{H}_{e}^{k}}{\partial\boldsymbol{v}_{e}^{k}}=\left[\begin{array}{cccc} \frac{\partial\boldsymbol{H}_{e_{1}}^{k}}{\partial\boldsymbol{v}_{e_{1}}^{k}} & \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0}\\ \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{e_{2}}^{k}}{\partial\boldsymbol{v}_{e_{2}}^{k}} & \boldsymbol{0} & \boldsymbol{0}\\ \boldsymbol{0} & \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{e_{3}}^{k}}{\partial\boldsymbol{v}_{e_{3}}^{k}} & \boldsymbol{0}\\ \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{e_{4}}^{k}}{\partial\boldsymbol{v}_{e_{4}}^{k}}\\ \end{array}\right]\\ \frac{\partial\boldsymbol{H}_{e}^{k}}{\partial\boldsymbol{v}_{e}^{k-1}}=\left[\begin{array}{cccc} \frac{\partial\boldsymbol{H}_{e_{1}}^{k}}{\partial\boldsymbol{v}_{e_{1}}^{k-1}} & \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0}\\ \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{e_{2}}^{k}}{\partial\boldsymbol{v}_{e_{2}}^{k-1}} & \boldsymbol{0} & \boldsymbol{0}\\ \boldsymbol{0} & \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{e_{3}}^{k}}{\partial\boldsymbol{v}_{e_{3}}^{k-1}} & \boldsymbol{0}\\ \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{e_{4}}^{k}}{\partial\boldsymbol{v}_{e_{4}}^{k-1}}\\ \end{array}\right] \end{array} $$

(78)

For $\phantom {\dot {i}\!}\partial \boldsymbol {H}_{e}^{k}/\partial \boldsymbol {v}_{e}^{k-1}$, both elastic step and plastic step give

$$ \begin{array}{l} \frac{\partial\boldsymbol{H}_{e_{r}}^{k}}{\partial\boldsymbol{v}_{e_{r}}^{k-1}}=\left[\begin{array}{ccccc} \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0}\\ \boldsymbol{0} & -\mathbb{I}_{4}^{s} & \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0}\\ \boldsymbol{0} & \boldsymbol{0} & -1 & 0 & 0\\ \boldsymbol{0} & \boldsymbol{0} & 0 & 0 & 0\\ \boldsymbol{0} & \boldsymbol{0} & 0 & 0 & z_{2}\\ \end{array}\right]\quad \text{with}\\ z_{2}=\left\{\begin{array}{ll} -1,\quad&\qquad\text{if}\;\;{\kappa}_{e_{r}}^{k-1}>\boldsymbol{N}_{\overline{\alpha}_{r}}\overline{\boldsymbol{\alpha}}^{e}\\ 0,\quad&\qquad\text{otherwise} \end{array}\right. \end{array} $$

(79)

while for $\phantom {\dot {i}\!}\partial \boldsymbol {H}_{e}^{k}/\partial \boldsymbol {v}_{e}^{k}$, elastic step and plastic step have to be distinguished. For elastic step, it is

$$ \frac{\partial\boldsymbol{H}_{e_{r}}^{k}}{\partial\boldsymbol{v}_{e_{r}}^{k}}=\left[\begin{array}{ccccc} \mathbb{I}_{4}^{s} &\quad \mathbb{C}^{e} &\quad \boldsymbol{0} &\quad \boldsymbol{0} &\quad {\boldsymbol{0}}\\ \boldsymbol{0} &\quad \mathbb{I}_{4}^{s} &\quad \boldsymbol{0} &\quad \boldsymbol{0} &\quad \boldsymbol{0}\\ \boldsymbol{0} &\quad \boldsymbol{0} &\quad 1 &\quad 0 &\quad 0\\ \boldsymbol{0} &\quad \boldsymbol{0} &\quad 0 &\quad 1 &\quad 0\\ \boldsymbol{0} &\quad \boldsymbol{0} &\quad 0 &\quad 0 &\quad 0\\ \end{array}\right] $$

(80)

while for plastic step it is

$$ \frac{\partial\boldsymbol{H}_{e_{r}}^{k}}{\partial\boldsymbol{v}_{e_{r}}^{k}}\,=\,\left[\begin{array}{ccccc} \mathbb{I}_{4}^{s} & \mathbb{C}^{e} & \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0}\\ \!-{\Delta}\gamma_{e_{r}}^{k}\mathbb{A} & \mathbb{I}_{4}^{s} & \boldsymbol{0} & -\boldsymbol{n}_{e_{r}}^{k} & \boldsymbol{0}\\ \boldsymbol{0} & \boldsymbol{0} & 1 & -\sqrt{\frac{2}{3}} & 0\\ \boldsymbol{n}_{e_{r}}^{k} & \boldsymbol{0} & \!-\sqrt{\frac{2}{3}}(1\!-d_{e_{r}}^{k}){K_{e}^{h}} & 0 & \sqrt{\frac{2}{3}}(\sigma_{y_{e}}\,+\,\zeta_{e_{r}}^{k})\frac{\partial{d}_{e_{r}}^{k}}{\partial{\kappa}_{e_{r}}^{k}}\\ \boldsymbol{0} & \boldsymbol{0} & 0 & 0 & 1\\ \end{array}\right] $$

(81)

where $\phantom {\dot {i}\!}{\partial {d}_{e_{r}}^{k}}/{\partial {\kappa }_{e_{r}}^{k}}$ can be calculated using (69) and

$$ \mathbb{A}=\frac{\partial\boldsymbol{n}_{e_{r}}^{k}}{\partial\boldsymbol{\sigma}_{e_{r}}^{k}}=\frac{1}{\|\boldsymbol{s}_{e_{r}}^{k}\|}(\mathbb{P}_{dev}^{s}-\boldsymbol{n}_{e_{r}}^{k}\otimes\boldsymbol{n}_{e_{r}}^{k}) $$

(82)

Finally, since $\phantom {\dot {i}\!}\boldsymbol {H}_{e}^{k}$ only depends on $\phantom {\dot {i}\!}{\rho }_{e}$, the derivative $\phantom {\dot {i}\!}\partial \boldsymbol {H}^{k}/\partial \boldsymbol {\rho }$ is calculated as

$$ \frac{\partial\boldsymbol{H}^{k}}{\partial\boldsymbol{\rho}}\,=\,\left[\begin{array}{cccc} \frac{\partial\boldsymbol{H}_{1}^{k}}{\partial\rho_{1}} & \boldsymbol{0} & {\dots} & \boldsymbol{0}\\ \boldsymbol{0} & \frac{\partial\boldsymbol{H}_{2}^{k}}{\partial\rho_{2}} & {\dots} & \boldsymbol{0}\\ {\vdots} & {\vdots} & {\ddots} & \vdots\\ \boldsymbol{0} & \boldsymbol{0} & {\dots} & \frac{\partial\boldsymbol{H}_{n_{ele}}^{k}}{\partial\rho_{n_{ele}}}\\ \end{array}\right] \mathrm{\!with} \frac{\partial\boldsymbol{H}_{e}^{k}}{\partial\rho_{e}}\,=\,\left[\begin{array}{c} \frac{\partial\boldsymbol{H}_{e_{1}}^{k}}{\partial\rho_{e}}\\ \frac{\partial\boldsymbol{H}_{e_{2}}^{k}}{\partial\rho_{e}}\\ \frac{\partial\boldsymbol{H}_{e_{3}}^{k}}{\partial\rho_{e}}\\ \frac{\partial\boldsymbol{H}_{e_{4}}^{k}}{\partial\rho_{e}}\\ \end{array}\right] $$

(83)

where the calculation of the term $\phantom {\dot {i}\!}\partial \boldsymbol {H}_{e_{r}}^{k}/\partial {\rho }_{e}$ is different for elastic and plastic steps. For elastic step, it is

$$ \frac{\partial\boldsymbol{H}_{e_{r}}^{k}}{\partial\rho_{e}}=\left[\begin{array}{c} -\frac{\partial\mathbb{C}^{e}}{\partial\rho_{e}}:\boldsymbol{\varepsilon}_{e_{r}}^{e^{k}}\\ \boldsymbol{0}\\ 0\\ 0\\ 0\\ \end{array}\right] $$

(84)

while for plastic step, it is

$$ \frac{\partial\boldsymbol{H}_{e_{r}}^{k}}{\partial\rho_{e}}\,=\,\small\left[\begin{array}{c} -\frac{\partial\mathbb{C}^{e}}{\partial\rho_{e}}:\boldsymbol{\varepsilon}_{e_{r}}^{e^{k}}\\ \boldsymbol{0}\\ 0\\ \!\!-\sqrt{\frac{2}{3}}(1\,-\,d_{e_{r}}^{k})\left(\frac{\partial\sigma_{y_{e}}}{\partial\rho_{e}}\,+\,\frac{\partial\zeta_{e_{r}}^{k}}{\partial\rho_{e}}\right)\,+\,\sqrt{\frac{2}{3}}\frac{\partial d_{e_{r}}^{k}}{\partial\rho_{e}}(\sigma_{y_{e}}\,+\,\zeta_{e_{r}}^{k})\\ 0\\ \end{array}\right] $$

(85)

Here $\phantom {\dot {i}\!}\partial d_{e_{r}}^{k}/\partial \rho _{e}$ can be again calculated by (69). With the material parameters interpolation presented in Section 4.1, following derivatives complete the calculations shown in (85)

$$ \begin{array}{l} \frac{\partial\mathbb{C}^{e}}{\partial\rho_{e}}=p_{1}(E-E_{min})\rho_{e}^{(p_{1}-1)}\mathbb{C}_{0}\\ \frac{\partial\sigma_{y_{e}}}{\partial\rho_{e}}=p_{2}(\sigma_{y}-\sigma_{y_{min}})\rho_{e}^{(p_{2}-1)}\\ \frac{\partial\zeta_{e_{r}}^{k}}{\partial\rho_{e}}=\alpha_{e_{r}}^{k}{p_{3}}(K^{h}-K_{min}^{h})\rho_{e}^{(p_{3}-1)} \end{array} $$

(86)

where $\phantom {\dot {i}\!}\mathbb {C}_{0}$ is the isotropic elasticity tensor evaluated with $\phantom {\dot {i}\!}E = 1$.