Topology optimization of structures with anisotropic plastic materials using enhanced assumed strain elements

Abstract

The focus of this paper is on consistent and accurate adjoint sensitivity analyses for structural topology optimization with anisotropic plastic materials under plane strain conditions. In order to avoid the locking issue, the Enhanced Assumed Strain (EAS) elements are adopted in the finite element discretization, and the anisotropic Hoffman plasticity model, which can simulate the strength differences in tension and compression, is incorporated within the framework of density-based topology optimization. The path-dependent sensitivity analysis is presented wherein the enhanced element parameters are consistently incorporated in the constraints. The objective of topology optimization is to maximize the plastic work. Several numerical examples are presented to show the effectiveness of the proposed framework. The results illustrate that the optimized topologies are highly dependent on the plastic anisotropic material properties.

Appendices

Appendix A: Elastic predictor/return-mapping algorithm and consistent tangent modulus

Numerical implementation of the Hoffman anisotropic model discussed in Section 3 is presented in this Appendix. In the context of strain-driven finite element formulation, for the given data at an integration point: ${\overline{\boldsymbol{\varepsilon}}}_m^p$ and α _m at step m, and $\overline{\boldsymbol{\varepsilon}}$ at current step (i.e. step m + 1), the goal is to find the unknown variables: ${\overline{\boldsymbol{\varepsilon}}}^p$, α together with the consistent tangent moduli ${\overline{\boldsymbol{C}}}_T$ at the current step. The consistent evaluation of the tangent operator ${\overline{\boldsymbol{C}}}_T$ ensures the quadratic convergence of the global NR solver. Note that the subscript (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 employed for solving the constitutive model as follows:

• Step 1: Elastic trial step

Given: ${{\overline{\boldsymbol{\varepsilon}}}^p}^{tr}={\overline{\boldsymbol{\varepsilon}}}_m^p\kern0.5em ,\kern0.75em {\alpha}^{tr}={\alpha}_m$

Evaluate:

$$ \begin{array}{l}{\overline{\boldsymbol{\sigma}}}^{tr}={\overline{\boldsymbol{C}}}^e\left(\overline{\boldsymbol{\varepsilon}}-{{\overline{\boldsymbol{\varepsilon}}}^p}^{tr}\right)\kern0.5em \\ {}{\zeta}^{tr}=1+\left(\frac{K^p}{\sigma_y}\right){\alpha}^{tr}\\ {}{\phi}^{tr}\left({\overline{\boldsymbol{\sigma}}}^{tr},{\zeta}^{tr}\right)=f\left({\overline{\boldsymbol{\sigma}}}^{tr}\right)-{\zeta^{tr}}^2\end{array} $$

(A1)

If ϕ ^tr ≤ 0 then the current step is an elastic step and the following elastic updates are made

$$ {\overline{\boldsymbol{\varepsilon}}}^p={{\overline{\boldsymbol{\varepsilon}}}^p}^{tr}, \alpha ={\alpha}^{tr}\kern1em \mathrm{and}\kern0.5em \overline{\boldsymbol{\sigma}}={\overline{\boldsymbol{\sigma}}}^{tr} $$

(A2)

$$ {\overline{\boldsymbol{C}}}_T={\overline{\boldsymbol{C}}}^e $$

(A3)

where ${\overline{\boldsymbol{C}}}_T$ is the consistent tangent moduli. Else, if ϕ ^tr > 0 then there is a plastic flow in this step and the algorithm proceeds to Step 2.

• Step 2: Plastic return mapping

In this step, plastic flow is nonzero (γ > 0). Using backward Euler method the flow rules in (17) and (18) are discretized as

$$ {\overline{\boldsymbol{\varepsilon}}}^p={\overline{\boldsymbol{\varepsilon}}}_m^p+\Delta \gamma \left(\boldsymbol{P}\overline{\boldsymbol{\sigma}}+\boldsymbol{q}\right) $$

(A4)

$$ \begin{array}{l}\alpha ={\alpha}_m+\sqrt{\frac{2}{3}}\Delta \gamma {z}^{1/2}\\ {}z\overset{\mathrm{def}}{=}{\left(\boldsymbol{P}\overline{\boldsymbol{\sigma}}+\boldsymbol{q}\right)}^T\boldsymbol{Z}\left(\boldsymbol{P}\overline{\boldsymbol{\sigma}}+\boldsymbol{q}\right)\end{array} $$

(A5)

The stress $\overline{\boldsymbol{\sigma}}$ updates can be written as

$$ \overline{\boldsymbol{\sigma}}={\overline{\boldsymbol{C}}}^e{\overline{\boldsymbol{\varepsilon}}}^{e^{tr}}-\varDelta \gamma {\overline{\boldsymbol{C}}}^e\left(\boldsymbol{P}\overline{\boldsymbol{\sigma}}+\boldsymbol{q}\right) $$

(A6)

The unknown variables are selected as $\overline{\boldsymbol{\sigma}}$, α and Δγ, and the corresponding system of equations are

$$ \begin{array}{l}{\boldsymbol{F}}_1=\overline{\boldsymbol{\sigma}}-{\overline{\boldsymbol{C}}}^e{\overline{\boldsymbol{\varepsilon}}}^{e^{tr}}+\varDelta \gamma {\overline{\boldsymbol{C}}}^e\left(\boldsymbol{P}\overline{\boldsymbol{\sigma}}+\boldsymbol{q}\right)=0\\ {}{F}_2=\alpha -{\alpha}_m-\sqrt{\frac{2}{3}}\varDelta \gamma {z}^{1/2}=0\\ {}{F}_3=\frac{1}{2}{\overline{\boldsymbol{\sigma}}}^T\boldsymbol{P}\overline{\boldsymbol{\sigma}}+{\boldsymbol{q}}^T\overline{\boldsymbol{\sigma}}-{\zeta}^2\left(\alpha \right)=0\end{array} $$

(A7)

Furthermore, $\overline{\boldsymbol{\sigma}}$ and α can be seen as functions of Δγ, which means the above system of equations can be reduced to one equation which can be used to solve for Δγ. The Newton Raphson method is used for solving F ₃ = 0 with the Jacobian calculated through

$$ \begin{array}{l}\frac{d{F}_3}{d\varDelta \gamma } = {\overline{\boldsymbol{\sigma}}}^T\boldsymbol{P}\frac{d\overline{\boldsymbol{\sigma}}}{d\varDelta \gamma }+{\boldsymbol{q}}^T\frac{d\overline{\boldsymbol{\sigma}}}{d\varDelta \gamma }-2\zeta \frac{d\zeta }{d\alpha}\frac{d\alpha }{d\varDelta \gamma}\\ {}\mathrm{with}\ \frac{d\overline{\boldsymbol{\sigma}}}{d\varDelta \gamma }=-{\left[\boldsymbol{I}+\varDelta \gamma {\overline{\boldsymbol{C}}}^e\boldsymbol{P}\right]}^{-1}{\overline{\boldsymbol{C}}}^e\left(\boldsymbol{P}\overline{\boldsymbol{\sigma}}+\boldsymbol{q}\right)\\ {}\mathrm{and}\kern0.5em \frac{d\alpha }{d\varDelta \gamma }=\sqrt{\frac{2}{3}}{z}^{\frac{1}{2}}+\sqrt{\frac{1}{6}}{z}^{-\frac{1}{2}}\varDelta \gamma \frac{dz}{d\overline{\boldsymbol{\sigma}}}\ \frac{d\overline{\boldsymbol{\sigma}}}{d\varDelta \gamma }=0\end{array} $$

(A8)

which are obtained by taking the differential of the equations F ₁ = 0 and F ₂ = 0. After calculating Δγ, $\overline{\boldsymbol{\sigma}}$ is updated by

$$ \overline{\boldsymbol{\sigma}}={\left(\boldsymbol{I}+\varDelta \gamma {\overline{\boldsymbol{C}}}^e\boldsymbol{P}\right)}^{-1}\left({\overline{\boldsymbol{\sigma}}}^{tr}-\varDelta \gamma {\overline{\boldsymbol{C}}}^e\boldsymbol{q}\right) $$

(A9)

and α is updated using (A5) and ${\overline{\boldsymbol{\varepsilon}}}^p$ is updated using (A4). Next, to compute the consistent algorithmic tangent modulus ${\overline{\boldsymbol{C}}}_T$, i.e. $\frac{d\overline{\boldsymbol{\sigma}}}{d\overline{\boldsymbol{\varepsilon}}}=\frac{d\overline{\boldsymbol{\sigma}}}{d{\overline{\boldsymbol{\varepsilon}}}^{e^{tr}}}$, the total differential of the system of (A7) is carried out considering variation of total strain $\overline{\boldsymbol{\varepsilon}}$, i.e.

$$ \begin{array}{l}d{\boldsymbol{F}}_1=\frac{\partial {\boldsymbol{F}}_1}{\partial \overline{\boldsymbol{\sigma}}}d\overline{\boldsymbol{\sigma}}+\frac{\partial {\boldsymbol{F}}_1}{\partial \alpha }d\alpha +\frac{\partial {\boldsymbol{F}}_1}{\partial \Delta \gamma }d\left(\Delta \gamma \right)+\frac{\partial {\boldsymbol{F}}_1}{\partial {\overline{\boldsymbol{\varepsilon}}}^{e^{tr}}}d{\overline{\boldsymbol{\varepsilon}}}^{e^{tr}}=0\\ {}d{F}_2=\frac{\partial {F}_2}{\partial \overline{\boldsymbol{\sigma}}}d\overline{\boldsymbol{\sigma}}+\frac{\partial {F}_2}{\partial \alpha }d\alpha +\frac{\partial {F}_2}{\partial \Delta \gamma }d\left(\Delta \gamma \right)+\frac{\partial {F}_2}{\partial {\overline{\boldsymbol{\varepsilon}}}^{e^{tr}}}d{\overline{\boldsymbol{\varepsilon}}}^{e^{tr}}=0\\ {}d{F}_3=\frac{\partial {F}_3}{\partial \overline{\boldsymbol{\sigma}}}d\overline{\boldsymbol{\sigma}}+\frac{\partial {F}_3}{\partial \alpha }d\alpha +\frac{\partial {F}_3}{\partial \Delta \gamma }d\left(\Delta \gamma \right)+\frac{\partial {F}_3}{\partial {\overline{\boldsymbol{\varepsilon}}}^{e^{tr}}}d{\overline{\boldsymbol{\varepsilon}}}^{e^{tr}}=0\end{array} $$

(A10)

Where

$$ \begin{array}{l}{\boldsymbol{A}}_{11}=\frac{\partial {\boldsymbol{F}}_1}{\partial \overline{\boldsymbol{\sigma}}}={\boldsymbol{I}}_6+\varDelta \gamma {\overline{\boldsymbol{C}}}^e\boldsymbol{P}, \kern0.75em {\boldsymbol{A}}_{12}=\frac{\partial {\boldsymbol{F}}_1}{\partial \alpha }=0, \kern0.75em {\boldsymbol{A}}_{13}=\frac{\partial {\boldsymbol{F}}_1}{\partial \Delta \gamma }={\overline{\boldsymbol{C}}}^e\left(\boldsymbol{P}\overline{\boldsymbol{\sigma}}+\boldsymbol{q}\right),\\ {}{\boldsymbol{A}}_{21}=\frac{\partial {F}_2}{\partial \overline{\boldsymbol{\sigma}}}=-\sqrt{\frac{2}{3}}\varDelta \gamma \frac{1}{2}\frac{1}{z^{1/2}}\ \frac{dz}{d\overline{\boldsymbol{\sigma}}}\kern0.5em ,\kern0.75em {A}_{22}=\frac{\partial {F}_2}{\partial \alpha }=1, \kern0.75em {A}_{23}=\frac{\partial {F}_2}{\partial \Delta \gamma }=-\sqrt{\frac{2}{3}}{z}^{1/2},\\ {}{\boldsymbol{A}}_{31}=\frac{\partial {F}_3}{\partial \overline{\boldsymbol{\sigma}}}={\overline{\boldsymbol{\sigma}}}^T\boldsymbol{P}+{\boldsymbol{q}}^T, \kern0.75em {A}_{32}=\frac{\partial {F}_3}{\partial \alpha }=-2\zeta \frac{d\zeta }{d\alpha }, \kern0.75em {A}_{33}=\frac{\partial {F}_3}{\partial \Delta \gamma }=0,\kern0.75em \\ {}\frac{\partial {\boldsymbol{F}}_1}{\partial {\overline{\boldsymbol{\varepsilon}}}^{e^{tr}}}=-{\overline{\boldsymbol{C}}}^e\kern0.5em ,\kern1em \frac{\partial {F}_2}{\partial {\overline{\boldsymbol{\varepsilon}}}^{e^{tr}}}=0, \kern0.75em \frac{\partial {F}_3}{\partial {\overline{\boldsymbol{\varepsilon}}}^{e^{tr}}}=0\end{array} $$

(A11)

where $\frac{dz}{d\overline{\boldsymbol{\sigma}}}=2{\left(\boldsymbol{P}\overline{\boldsymbol{\sigma}}+\boldsymbol{q}\right)}^T\boldsymbol{Z}\boldsymbol{P}$ and $\frac{d\zeta }{d\alpha }=\left(\frac{K^p}{\sigma_y}\right)$.

Equation (A10) can be written in matrix–vector form as

$$ \left[\begin{array}{ccc}\hfill {\boldsymbol{A}}_{11}\hfill & \hfill 0\hfill & \hfill {\boldsymbol{A}}_{13}\hfill \\ {}\hfill {\boldsymbol{A}}_{21}\hfill & \hfill 1\hfill & \hfill {A}_{23}\hfill \\ {}\hfill {\boldsymbol{A}}_{31}\hfill & \hfill {A}_{32}\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{c}\hfill d\overline{\boldsymbol{\sigma}}\hfill \\ {}\hfill d\alpha \hfill \\ {}\hfill d\left(\Delta \gamma \right)\hfill \end{array}\right]=\left[\begin{array}{c}\hfill {\overline{\boldsymbol{C}}}^ed{\overline{\boldsymbol{\varepsilon}}}^{e^{tr}}\hfill \\ {}\hfill 0\hfill \\ {}\hfill 0\hfill \end{array}\right] $$

(A12)

With simple manipulations, the tangent moduli can be obtained as

$$ {\overline{\boldsymbol{C}}}_T=\frac{d\overline{\boldsymbol{\sigma}}}{d{\overline{\boldsymbol{\varepsilon}}}^{e^{tr}}} = {\left({\boldsymbol{A}}_{11}-\frac{{\boldsymbol{A}}_{13}}{A_{23}}\left({\boldsymbol{A}}_{21}-\frac{1}{A_{32}}{\boldsymbol{A}}_{31}\right)\right)}^{-1}{\overline{\boldsymbol{C}}}^e $$

(A13)

Appendix B: Calculation of the terms $\partial {\overline{\boldsymbol{C}}}^e/\partial {\rho}_e$, ∂P/∂ρ _e , ∂q/∂ρ _e , ∂z/∂ρ _e and ∂ζ/∂ρ _e

• a. $\partial {\overline{\boldsymbol{C}}}^e/\partial {\rho}_e$

$$ \frac{\partial {\overline{\boldsymbol{C}}}^e}{\partial {\rho}_e}=\left[\begin{array}{cccccc}\hfill \frac{\partial {C}_{11}}{\partial {\rho}_e}\hfill & \hfill \frac{\partial {C}_{12}}{\partial {\rho}_e}\hfill & \hfill \frac{\partial {C}_{13}}{\partial {\rho}_e}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill \frac{\partial {C}_{12}}{\partial {\rho}_e}\hfill & \hfill \frac{\partial {C}_{22}}{\partial {\rho}_e}\hfill & \hfill \frac{\partial {C}_{23}}{\partial {\rho}_e}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill \frac{\partial {C}_{13}}{\partial {\rho}_e}\hfill & \hfill \frac{\partial {C}_{23}}{\partial {\rho}_e}\hfill & \hfill \frac{\partial {C}_{33}}{\partial {\rho}_e}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{\partial {C}_{44}}{\partial {\rho}_e}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{\partial {C}_{55}}{\partial {\rho}_e}\hfill & \hfill 0\hfill \\ {}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{\partial {C}_{66}}{\partial {\rho}_e}\hfill \end{array}\right] $$

(B1)

$$ \begin{array}{l}\frac{\partial {C}_{11}}{\partial {\rho}_e}=\frac{1-{\nu}_{23}{\nu}_{32}}{\overline{\nu}}\frac{\partial {E}_1}{\partial {\rho}_e};\kern0.5em {C}_{22}=\frac{1-{\nu}_{31}{\nu}_{13}}{\overline{\nu}}\frac{\partial {E}_2}{\partial {\rho}_e};\kern0.75em {C}_{33}=\frac{1-{\nu}_{12}{\nu}_{21}}{\overline{\nu}}\frac{\partial {E}_3}{\partial {\rho}_e}\\ {}{C}_{12}=\frac{\nu_{12}+{\nu}_{32}{\nu}_{13}}{\overline{\nu}}\frac{\partial {E}_2}{\partial {\rho}_e};\kern0.5em {C}_{13}=\frac{\nu_{13}+{\nu}_{12}{\nu}_{23}}{\overline{\nu}}\frac{\partial {E}_3}{\partial {\rho}_e};\kern0.75em {C}_{23}=\frac{\nu_{23}+{\nu}_{21}{\nu}_{13}}{\overline{\nu}}\frac{\partial {E}_3}{\partial {\rho}_e}\\ {}{C}_{44}=\frac{\partial {G}_{12}}{\partial {\rho}_e};\ {C}_{55}=\frac{\partial {G}_{23}}{\partial {\rho}_e};\ {C}_{66}=\frac{\partial {G}_{31}}{\partial {\rho}_e}\\ {}\overline{\nu}=1-{\nu}_{12}{\nu}_{21}-{\nu}_{23}{\nu}_{32}-{\nu}_{31}{\nu}_{13}-2{\nu}_{21}{\nu}_{32}{\nu}_{13}\\ {}\frac{\partial {E}_i}{\partial {\rho}_e}=p\left({E}_{i,0}-{E}_{i, min}\right){\rho}_e^{p-1}, \kern0.5em i=1,2,3\\ {}\frac{\partial {G}_{ij}}{\partial {\rho}_e}=p\left({G}_{ij,0}-{G}_{ij, min}\right){\rho}_e^{p-1}, \kern1em \left(i,j\right)=\left(1,2\right),\ \left(2,3\right),\ \left(3,1\right)\end{array} $$

(B2)

• b. ∂P/∂ρ _e , ∂q/∂ρ _e and ∂z/∂ρ _e

By defining ${a}_i=\frac{1}{{\overline{\sigma}}_{ii}^t{\overline{\sigma}}_{ii}^c}$, (i = 1, 2, 3), and using chain rule

$$ \frac{d{a}_i}{d{\rho}_e}=-\frac{1}{{\left({\overline{\sigma}}_{ii}^t{\overline{\sigma}}_{ii}^c\right)}^2}\left(\frac{d{\overline{\sigma}}_{ii}^t}{d{\rho}_e}{\overline{\sigma}}_{ii}^c+\frac{d{\overline{\sigma}}_{ii}^c}{d{\rho}_e}{\overline{\sigma}}_{ii}^t\right) $$

(B3)

So the derivatives of P and q with respect to density ρ _e are calculated as

$$ \begin{array}{l}\frac{\partial \boldsymbol{P}}{\partial {\rho}_e}=2\left[\begin{array}{cccccc}\hfill \frac{\partial \left({c}_1+{c}_3\right)}{\partial {\rho}_e}\hfill & \hfill -\frac{\partial {c}_1}{\partial {\rho}_e}\hfill & \hfill -\frac{\partial {c}_3}{\partial {\rho}_e}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill -\frac{\partial {c}_1}{\partial {\rho}_e}\hfill & \hfill \frac{\partial \left({c}_2+{c}_1\right)}{\partial {\rho}_e}\hfill & \hfill -\frac{\partial {c}_2}{\partial {\rho}_e}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill -\frac{\partial {c}_3}{\partial {\rho}_e}\hfill & \hfill -\frac{\partial {c}_2}{\partial {\rho}_e}\hfill & \hfill \frac{\partial \left({c}_3+{c}_2\right)}{\partial {\rho}_e}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{\partial {c}_4}{\partial {\rho}_e}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{\partial {c}_5}{\partial {\rho}_e}\hfill & \hfill 0\hfill \\ {}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{\partial {c}_6}{\partial {\rho}_e}\hfill \end{array}\right]\\ {}\frac{\partial \boldsymbol{q}}{\partial {\rho}_e}={\left[\begin{array}{cccccc}\hfill \frac{\partial {c}_7}{\partial {\rho}_e}\hfill & \hfill \frac{\partial {c}_8}{\partial {\rho}_e}\hfill & \hfill \frac{\partial {c}_9}{\partial {\rho}_e}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]}^T\\ {}\frac{\partial z}{\partial {\rho}_e}=2{\left(\boldsymbol{P}\overline{\boldsymbol{\sigma}}+\boldsymbol{q}\right)}^T\boldsymbol{Z}\left(\frac{\partial \boldsymbol{P}}{\partial {\rho}_e}\overline{\boldsymbol{\sigma}}+\frac{\partial \boldsymbol{q}}{\partial {\rho}_e}\right)\end{array} $$

(B4)

With

$$ \begin{array}{l}\frac{d{c}_1}{d{\rho}_e}=0.5\left(\frac{d{a}_1}{d{\rho}_e}+\frac{d{a}_2}{d{\rho}_e}-\frac{d{a}_3}{d{\rho}_e}\right)\\ {}\frac{d{c}_2}{d{\rho}_e}=0.5\left(-\frac{d{a}_1}{d{\rho}_e}+\frac{d{a}_2}{d{\rho}_e}+\frac{d{a}_3}{d{\rho}_e}\right)\\ {}\frac{d{c}_3}{d{\rho}_e}=0.5\left(\frac{d{a}_1}{d{\rho}_e}-\frac{d{a}_2}{d{\rho}_e}+\frac{d{a}_3}{d{\rho}_e}\right)\\ {}\frac{d{c}_4}{d{\rho}_e}=-2\frac{1}{{\left({\overline{\sigma}}_{12}^0\right)}^3}\frac{d{\overline{\sigma}}_{12}^0}{d{\rho}_e}\\ {}\frac{d{c}_5}{d{\rho}_e}=-2\frac{1}{{\left({\overline{\sigma}}_{23}^0\right)}^3}\frac{d{\overline{\sigma}}_{23}^0}{d{\rho}_e}\\ {}\frac{d{c}_6}{d{\rho}_e}=-2\frac{1}{{\left({\overline{\sigma}}_{13}^0\right)}^3}\frac{d{\overline{\sigma}}_{13}^0}{d{\rho}_e}\\ {}\frac{d{c}_7}{d{\rho}_e}={a}_1\left(\frac{d{\overline{\sigma}}_{11}^c}{d{\rho}_e}-\frac{d{\overline{\sigma}}_{11}^t}{d{\rho}_e}\right)+\frac{d{a}_1}{d{\rho}_e}\left({\overline{\sigma}}_{11}^c-{\overline{\sigma}}_{11}^t\right)\\ {}\frac{d{c}_8}{d{\rho}_e}={a}_2\left(\frac{d{\overline{\sigma}}_{22}^c}{d{\rho}_e}-\frac{d{\overline{\sigma}}_{22}^t}{d{\rho}_e}\right)+\frac{d{a}_2}{d{\rho}_e}\left({\overline{\sigma}}_{22}^c-{\overline{\sigma}}_{22}^t\right)\\ {}\frac{d{c}_9}{d{\rho}_e}={a}_3\left(\frac{d{\overline{\sigma}}_{33}^c}{d{\rho}_e}-\frac{d{\overline{\sigma}}_{33}^t}{d{\rho}_e}\right)+\frac{d{a}_3}{d{\rho}_e}\left({\overline{\sigma}}_{33}^c-{\overline{\sigma}}_{33}^t\right)\end{array} $$

(B5)

where the expressions of $\frac{d{\overline{\sigma}}_{ii}^c}{d{\rho}_e}$, $\frac{d{\overline{\sigma}}_{ii}^t}{d{\rho}_e}$ (i = 1, 2, 3) and $\frac{d{\overline{\sigma}}_{ij}^0}{d{\rho}_e}$ ((i, j) = (1, 2), (2, 3), (1, 3)) can be calculated from (20).

• c. ∂ζ/∂ρ _e

$$ \frac{\partial \zeta }{\partial {\rho}_e}=\frac{1}{\sigma_y}\left(\frac{d{K}^p}{d{\rho}_e}-\frac{K^p}{\sigma_y}\frac{d{\sigma}_y}{d{\rho}_e}\right)\alpha $$

(B6)

where $\frac{d{K}^p}{d{\rho}_e}$ and $\frac{d{\sigma}_y}{d{\rho}_e}$ can be obtained from (20).