Design of energy dissipating elastoplastic structures under cyclic loads using topology optimization

Abstract

A topology optimization approach for designing structures with maximum energy absorption capacity under cyclic loads is proposed. To simulate the Bauschinger effect in materials under cyclic loads, Prager and the Armstrong-Frederick kinematic hardening rules are considered together with the von Mises plasticity in the optimization process. Path-dependent sensitivities are derived analytically using the adjoint method, which are further verified by the central difference method. Effectiveness of the proposed approach is demonstrated on several examples. Results show that the optimized designs with kinematic hardening are remarkably different from the ones obtained with isotropic hardening and are highly dependent on the loading patterns.

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

Numerical implementation of the von Mises plasticity considering mixed isotropic-kinematic hardening discussed in Section 3 is presented in this Appendix. In the context of strain-driven finite element formulation, with the given data at an integration point: ε ^p _m , β _m and α _m at step m, and ε at current step m + 1, the goal is to find the unknown variables: ε ^p, β and α together with the consistent tangent modulus ℂ _T at the current step. The consistent evaluation of the tangent operator 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. The elastic predictor/return-mapping algorithm employed for solving the constitutive model is as follows:

• Step 1: Elastic (trial) step

  Given:

  $$ {{\boldsymbol{\varepsilon}}^p}^{tr}={\boldsymbol{\varepsilon}}_m^p,\ {\boldsymbol{\beta}}^{tr}={\boldsymbol{\beta}}_m,\ {\alpha}^{tr}={\alpha}_m $$

  Evaluate:

  $$ \begin{array}{l}{\boldsymbol{\sigma}}^{tr}=\mathrm{\mathbb{C}}:\left(\boldsymbol{\varepsilon} -{{\boldsymbol{\varepsilon}}^p}^{tr}\right)\hfill \\ {}\mathrm{with}\ \mathrm{\mathbb{C}}=3\kappa {\mathrm{\mathbb{P}}}_{vol}^s+2\mu {\mathrm{\mathbb{P}}}_{dev}^s,\kappa = E/3\left(1-2\nu \right),\mu = E/2\left(1+\nu \right)\hfill \\ {} E\to\ \mathrm{Young}'\mathrm{s}\ \mathrm{Modulus},\nu \to \mathrm{Poisson}'\mathrm{s}\ \mathrm{Ratio}\hfill \end{array} $$

  (A1)

  ${\left[{\mathrm{\mathbb{P}}}_{dev}^s\right]}_{ij kl}=\frac{1}{2}\left({\delta}_{ik}{\delta}_{jl}+{\delta}_{il}{\delta}_{jk}\right)-\frac{1}{3}{\delta}_{ij}{\delta}_{kl}$ and ${\left[{\mathrm{\mathbb{P}}}_{vol}^s\right]}_{ij kl}=\frac{1}{3}{\delta}_{ij}{\delta}_{kl}$

  $$ \begin{array}{l}{\boldsymbol{\eta}}^{tr}={\boldsymbol{s}}^{tr}-{\boldsymbol{\beta}}^{tr}\mathrm{with}{\boldsymbol{s}}^{tr}={\mathrm{\mathbb{P}}}_{dev}^s:{\boldsymbol{\sigma}}^{tr}\hfill \\ {}{\zeta}^{tr}={K}^h{\alpha}^{tr}\hfill \\ {}{\phi}^{tr}\left({\boldsymbol{\sigma}}^{tr},{\boldsymbol{\beta}}^{tr},{\zeta}^{tr}\right)=||{\boldsymbol{\eta}}^{tr}|| - \sqrt{\frac{2}{3}}\left({\sigma}_y+{\zeta}^{tr}\right)\hfill \end{array} $$

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

  $$ {\boldsymbol{\varepsilon}}^p={{\boldsymbol{\varepsilon}}^p}^{tr},\boldsymbol{\beta} ={\boldsymbol{\beta}}^{tr},\alpha ={\alpha}^{tr}\ \mathrm{and}\ \boldsymbol{\sigma} ={\boldsymbol{\sigma}}^{tr} $$

  (A2)
  $$ {\mathrm{\mathbb{C}}}_T=\mathrm{\mathbb{C}} $$

  (A3)

  where ℂ _T is the consistent tangent modulus. 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, the plastic flow is nonzero (γ > 0). Using backward Euler method, the flow rules in Eqs. (9), (10) and (12) or Eq. (13) are discretized as follows

  $$ {\boldsymbol{\varepsilon}}^p={\boldsymbol{\varepsilon}}_m^p+\Delta \gamma \boldsymbol{n} $$

  (A4)
  $$ \boldsymbol{\beta} ={c}_1{\boldsymbol{\beta}}_m+{c}_2\boldsymbol{n} $$

  (A5)
  $$ \alpha ={\alpha}_m+\sqrt{\frac{2}{3}}\Delta \gamma $$

  (A6)

  where the parameters c ₁ and c ₂ depend on the type of kinematic hardening rule. For Prager kinematic hardening

  $$ {c}_1=1,\ {c}_2=\frac{2}{3} H\varDelta \gamma $$

  (A7)

  while for Armstrong-Frederick kinematic hardening

  $$ {c}_1=\frac{1}{1+\varDelta \gamma b},\ {c}_2=\frac{2}{3} H\frac{\varDelta \gamma}{1+\varDelta \gamma b} $$

  (A8)

The stress σ update can be written as

$$ \boldsymbol{\sigma} ={\boldsymbol{\sigma}}^{tr}-2\mu \varDelta \gamma \boldsymbol{n} $$

(A9)

where μ is the shear modulus. The unknown variables are σ, β and Δγ, and the corresponding system of equations are

$$ \begin{array}{l}{\boldsymbol{F}}_1=\boldsymbol{\sigma} - {\boldsymbol{\sigma}}^{tr}+2\mu \varDelta \gamma \boldsymbol{n}=\mathbf{0}\hfill \\ {}{\boldsymbol{F}}_2=\boldsymbol{\beta} - {c}_1{\boldsymbol{\beta}}^{tr} - {c}_2\boldsymbol{n}=\mathbf{0}\hfill \\ {}{F}_3=||\boldsymbol{\eta} || - \left({\sigma}_y+\zeta \right)=\mathbf{0}\hfill \end{array} $$

(A10)

Furthermore, σ and β can be seen as functions of Δγ, which means that the above system of equations can be simplified to one equation which can be used to solve for Δγ. The NR method is used to solve F ₃  =  0 with the Jacobian calculation as follows

$$ \begin{array}{l}\frac{d{ F}_3}{ d\varDelta \gamma} = \frac{d||\boldsymbol{\eta} ||}{ d\varDelta \gamma}-\frac{d\zeta}{ d\alpha}\frac{d\alpha}{ d\varDelta \gamma}\hfill \\ {}\mathrm{with}\ \frac{d||\boldsymbol{\eta} ||}{ d\varDelta \gamma}=-\frac{\partial {a}_1}{\partial \varDelta \gamma}+\frac{1}{2\sqrt{a_2}}\frac{\partial {a}_2}{\partial \varDelta \gamma}\hfill \\ {}\mathrm{where}\ {\boldsymbol{a}}_0\underset{\bar{\mkern6mu}}{\underset{\bar{\mkern6mu}}{\mathrm{def}}}\varDelta {\boldsymbol{s}}^{tr}-{c}_1{\boldsymbol{\beta}}^{tr},\ {a}_1\underset{\bar{\mkern6mu}}{\underset{\bar{\mkern6mu}}{\mathrm{def}}}\varDelta 2\mu \varDelta \gamma +{c}_2\ \mathrm{and}\ {a}_2\underset{\bar{\mkern6mu}}{\underset{\bar{\mkern6mu}}{\mathrm{def}}}{\boldsymbol{a}}_0:{\boldsymbol{a}}_0\hfill \\ {}\mathrm{which}\ \mathrm{gives}\ \frac{\partial {a}_1}{\partial \varDelta \gamma}=2\mu +\frac{\partial {c}_2}{\partial \varDelta \gamma},\ \frac{\partial {a}_2}{\partial \varDelta \gamma}=2{\boldsymbol{a}}_0:\frac{\partial {\boldsymbol{a}}_0}{\partial \varDelta \gamma},\ \frac{\partial {\boldsymbol{a}}_0}{\partial \varDelta \gamma}=-\frac{\partial {c}_1}{\partial \varDelta \gamma}{\boldsymbol{\beta}}^{tr}\hfill \end{array} $$

(A11)

The term $d\alpha / d\varDelta \gamma =\sqrt{2/3}$ can be easily obtained from Eq. (A6). Given the expressions of c ₁ and c ₂ for different hardening rules, it is obvious that

• For Prager linear hardening:

  $$ \frac{\partial {c}_1}{\partial \varDelta \gamma}=0,\ \frac{\partial {c}_2}{\partial \varDelta \gamma}=\frac{2}{3} H $$

  (A12)

• For Armstrong-Frederick nonlinear hardening:

  $$ \frac{\partial {c}_1}{\partial \varDelta \gamma}=-\frac{b}{{\left(1+\varDelta \gamma b\right)}^2},\ \frac{\partial {c}_2}{\partial \varDelta \gamma}=\frac{2}{3} H\left(\frac{1}{1+\varDelta \gamma b}-\frac{\varDelta \gamma b}{{\left(1+\varDelta \gamma b\right)}^2}\right) $$

With the Jacobian, Δγ can be obtained iteratively by solving Eq. (A10)₃. Then a ₀ can be computed, followed by which n, σ and β are sequentially determined.

Next, to compute the consistent algorithmic tangent modulus ℂ _T (Kiran and Khandelwal 2014a), the stress tensor is decomposed as follows

$$ \boldsymbol{\sigma} =\boldsymbol{p}+\boldsymbol{s} $$

(A13)

Then the tangent modulus can be expressed as

$$ {\mathrm{\mathbb{C}}}_T=\frac{d\boldsymbol{\sigma}}{d\boldsymbol{\varepsilon}}=\frac{d\boldsymbol{\sigma}}{d{\boldsymbol{\varepsilon}}^{e^{tr}}}=\frac{d\boldsymbol{p}}{d{\boldsymbol{\varepsilon}}^{e^{tr}}}+\frac{d\boldsymbol{s}}{d{\boldsymbol{\varepsilon}}^{e^{tr}}} $$

(A14)

where

$$ \frac{d\boldsymbol{p}}{d{\boldsymbol{\varepsilon}}^{e^{tr}}}=3\kappa {\mathrm{\mathbb{P}}}_{vol}^s $$

(A15)

$$ \frac{d\boldsymbol{s}}{d{\boldsymbol{\varepsilon}}^{e^{tr}}}=2\mu {\mathrm{\mathbb{P}}}_{d ev}^s-2\mu \varDelta \gamma \frac{d\boldsymbol{n}}{d{\boldsymbol{\varepsilon}}^{e^{tr}}} - 2\mu \boldsymbol{n}\otimes \frac{d\varDelta \gamma}{d{\boldsymbol{\varepsilon}}^{e^{tr}}} $$

(A16)

The term $d\varDelta \gamma / d{\boldsymbol{\varepsilon}}^{e^{tr}}$ can be obtained by differentiating Eq. (A10)₃ as

$$ \frac{ d\varDelta \gamma}{d{\boldsymbol{\varepsilon}}^{e^{tr}}}=\frac{\frac{1}{\sqrt{a_2}}2\mu \left({\mathrm{\mathbb{P}}}_{d ev}^s:{\boldsymbol{a}}_0\right)}{\frac{\partial {a}_1}{\partial \varDelta \gamma}+\frac{1}{\sqrt{a_2}}\frac{\partial {c}_1}{\partial \Delta \gamma}{\boldsymbol{\beta}}^{tr}:{\boldsymbol{a}}_0+\frac{ d\zeta}{d\Delta \gamma}} $$

(A17)

and the term $d\boldsymbol{n}/ d{\boldsymbol{\varepsilon}}^{e^{tr}}$ can be obtained as

$$ \begin{array}{l}\frac{d\boldsymbol{n}}{d{\boldsymbol{\varepsilon}}^{e^{tr}}}=\frac{1}{||\boldsymbol{\eta} ||+{a}_1}\mathrm{\mathbb{Z}with}\hfill \\ {}\mathrm{\mathbb{Z}}=2\mu {\mathrm{\mathbb{P}}}_{d ev}^s-\frac{\partial {c}_1}{\partial \varDelta \gamma}{\boldsymbol{\beta}}^{tr}\otimes \frac{d\varDelta \gamma}{d{\boldsymbol{\varepsilon}}^{e^{tr}}}-\boldsymbol{n}\otimes \left[\frac{1}{\sqrt{a_2}}{\boldsymbol{a}}_0:\left(2\mu {\mathrm{\mathbb{P}}}_{d ev}^s-\frac{\partial {c}_1}{\partial \Delta \gamma}{\boldsymbol{\beta}}^{tr}\otimes \frac{d\varDelta \gamma}{d{\boldsymbol{\varepsilon}}^{e^{tr}}}\right)\right]\hfill \end{array} $$

(A18)