Multi-objective concurrent topology optimization of thermoelastic structures composed of homogeneous porous material

Abstract

The present paper studies multi-objective design of lightweight thermoelastic structure composed of homogeneous porous material. The concurrent optimization model is applied to design the topologies of light weight structures and of the material microstructure. The multi-objective optimization formulation attempts to find minimum structural compliance under only mechanical loads and minimum thermal expansion of the surfaces we are interested in under only thermo loads. The proposed optimization model is applied to a sandwich elliptically curved shell structure, an axisymmetric structure and a 3D structure. The advantage of the concurrent optimization model to single scale topology optimization model in improving the multi-objective performances of the thermoelastic structures is investigated. The influences of available material volume fraction and weighting coefficients are also discussed. Numerical examples demonstrate that the porous material is conducive to enhance the multi-objective performance of the thermoelastic structures in some cases, especially when lightweight structure is emphasized. An “optimal” material volume fraction is observed in some numerical examples.

Appendices

Appendix A: Theoretical proof

Proposition

For porous material composed of one solid phase and one void phase, its effective thermal expansion coefficient tensor $\alpha_{\it ij}^H$ equals to that $\alpha_{\it ij}$ of the solid phase, i.e.

$$ \label{eq39} \alpha_{\it ij}^H =\alpha_{\it ij} $$

(39)

Proof

Let $E_{\it ijkl}$ and $E_{\it ijkl}^H$ represent elastic modulus tensor of the solid phase and effective elastic modulus tensor of the porous material respectively. The thermo stress tensor is defined as

$$ \label{eq40} \beta_{\it ij} =E_{\it ijkl} \alpha_{\it kl} $$

(40)

And the effective thermo stress tensor is defined as

$$ \label{eq41} \beta_{\it ij}^H =E_{\it ijkl}^H \alpha_{\it kl}^H $$

(41)

By virtue of homogenization theory, $E_{\it ijkl}^H$ and $\beta_{\it ij}^H$ can be obtained respectively as (Sigmund and Torquato 1997; Rodrigues and Fernandes 1995)

$$ \label{eq42} E_{\it ijkl}^H =\int_Y {\left( {E_{\it ijkl} -E_{ijpm} \frac{\partial X_p^{\it kl} }{\partial y_m }} \right)dY} $$

(42)

$$ \label{eq43} \beta_{\it ij}^H =\int_Y {\left( {\beta_{\it ij} -\beta_{\it kl} \frac{\partial X_k^{\it ij} }{\partial y_l }} \right)dY} $$

(43)

Where Y identifies the design domain of the unit cell occupied with material, and periodic displacements $X^{\it ij}$ are solutions to the equilibrium equations in the unit cell.

Substitute (41) into (43), we can have

$$ \label{eq44} \begin{array}{lll} \beta_{\it ij}^H &=\mu E_{\it ijkl} \alpha_{\it kl} -\displaystyle\int_Y {\left( {E_{klmn} \alpha_{mn} \dfrac{\partial X_k^{\it ij} }{\partial y_l }} \right)dY} \\ &=\mu E_{\it ijkl} \alpha_{\it kl} -\displaystyle\int_Y {\left( {E_{\it mnkl} \alpha_{\it kl} \frac{\partial X_m^{\it ij} }{\partial y_n }} \right)dY} \\ &=\left[ {\mu E_{\it ijkl} -\displaystyle\int_Y {\left( {E_{\it mnkl} \dfrac{\partial X_m^{\it ij} }{\partial y_n }} \right)dY} } \right]\alpha_{\it kl} \end{array} $$

(44)

Since $E_{\it ijpm}=E_{\it pmij}$, (42) can be stated as

$$ \label{eq45} \begin{array}{lll} E_{\it ijkl}^H &=\mu E_{\it ijkl} -\displaystyle\int_Y {\left( {E_{\it pmij} \frac{\partial X_p^{\it kl} }{\partial y_m }} \right)dY} \\ &=\mu E_{\it ijkl} -\displaystyle\int_Y {\left( {E_{\it mnij} \frac{\partial X_m^{\it kl} }{\partial y_n }} \right)dY} \end{array} $$

(45)

As $E_{\it ijkl}^H =E_{\it klij}^H$, (45) can be written as

$$ \label{eq46} E_{\it klij}^H =\mu E_{\it ijkl} -\int_Y {\left( {E_{\it mnij} \frac{\partial X_m^{\it kl} }{\partial y_n }} \right)dY} $$

(46)

Change index k with i, and l with j, (46) can be stated as

$$ \label{eq47} E_{\it ijkl}^H =\mu E_{\it klij} -\int_Y {\left(E_{\it mnkl} \frac{\partial X_m^{\it ij} }{\partial y_n }\right)dY} $$

(47)

As $E_{\it klij}=E_{\it ijkl}$, from (47), we can have

$$ \label{eq48} E_{\it ijkl}^H =\mu E_{\it ijkl} -\int_Y {\left( {E_{\it mnkl} \frac{\partial X_m^{\it ij} }{\partial y_n }} \right)dY} $$

(48)

Substitute (48) into (44), we can have

$$ \label{eq49} \begin{array}{lll} \beta_{\it ij}^H &=\left[ {\mu E_{\it ijkl} -\int_Y {\left( {E_{\it mnkl} \frac{\partial X_m^{\it ij} }{\partial y_n }} \right)dY} } \right]\alpha_{\it kl} \\ &=E_{\it ijkl}^H \alpha_{\it kl} \end{array} $$

(49)

Since (49) equals to (41), we can have

$$ \label{eq50} \beta_{\it ij}^H =E_{\it ijkl}^H \alpha_{\it kl}^H =E_{\it ijkl}^H \alpha_{\it kl} $$

(50)

Define the compliance tensor $C_{\it ijkl}^H$ as the inverse of $E_{\it ijkl}^H$, and multiply it to the left side of (50), we can obtain the following statement and complete our proof.

$$ \label{eq51} \alpha_{\it kl}^H =\alpha_{\it kl} $$

(51)

It should be mentioned, if the material is composed of three or more phases (at least two solid phases and one void phase), $\alpha _{\it kl}$ could not be extracted from (44) and thus (51) could not be obtained. □

Appendix B: Plane truss example

The stiffness matrix of a plane truss element in local coordinate is

$$ \label{eq52} \overline {\textbf{\emph{K}}}_e =\frac{E_e A_e }{L_e }\left[ {{\begin{array}{@{}rr} 1 \hfill & {-1} \hfill \\ {-1} \hfill & 1 \hfill \\ \end{array} }} \,\right] $$

(52)

Where E _e , A _e and L _e are respectively the elastic modulus, sectional area and length of the truss element. The element stiffness matrix in global coordinate is

$$ \label{eq53} \overline {\textbf{\emph{K}}}_e =({\textbf{\emph{T}}}_e )^{\rm T} {\textbf{\emph{K}}}_e {\textbf{\emph{T}}}_e $$

(53)

Where T _e denotes the transformation matrix, and is expressed as

$$ \label{eq54} {\textbf{\emph{T}}}_e =\left[ {{\begin{array}{@{}cccc@{}} {\cos \theta } \hfill & {\sin \theta } \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & {\cos \theta } \hfill & {\sin \theta } \hfill \\ \end{array} }} \right] $$

(54)

Where θ is the angle between the direction of the truss element and the positive direction of X coordinate axis.

The element nodal load by temperature load in local coordinate is

$$ \begin{array}{lll} \label{eq55} {\overline {\textbf{\emph{F}}}_e}^{^\mathit{Tem}} &=& \int_{\Omega_e } {{\textbf{\emph{B}}}_e^T {\textbf{\emph{D}}}_e \boldsymbol{\varepsilon}_e^0 d\Omega_e }\\ &=&\int_0^{L_e } {\left[{{\begin{array}{@{}c@{}} {-1 \mathord{\left/ {\vphantom {1 {L_e }}} \right. \kern-\nulldelimiterspace} {L_e }} \hfill \\ {1 \mathord{\left/ {\vphantom {1 {L_e }}} \right. \kern-\nulldelimiterspace} {L_e }} \hfill \\ \end{array} }} \right]E_e \gamma \Delta T_e (A_e dl)} \\ &=& E_e A_e \gamma \Delta T_e \left[ {{\begin{array}{@{}r@{}} {-1} \hfill \\ 1 \hfill \\ \end{array} }} \right] \end{array} $$

(55)

Where γ is the thermal expansion coefficient of the material and ΔT _e is the temperature variation of the truss member. The element nodal load by temperature load in global coordinate is

$$ \label{eq56} {\textbf{\emph{F}}}_e^\mathit{Tem} ={\textbf{\emph{T}}}_e {\textbf{\emph{F}}}_e^\mathit{Tem} $$

(56)

Now, let’s consider our plane truss example in Fig. 7. We fix A ₁ and find an optimum A ₂ (0 ≤ A ₂ ≤ 2A ₁) which minimizes obj. And we set the initial value of A ₂ to A ₁/2. Since the structure has only one degree of nodal freedom, after simple derivation, we can easily obtain the structural stiffness matrix, the nodal load by thermal load and mechanical load as

$$\begin{array}{rll} \label{eq57} {\textbf{\emph{K}}} &=& \left[ {\frac{E(A_1 +A_2 )}{2\sqrt 2 L}} \right],\\ {\textbf{\emph{F}}}^\mathit{Tem} &=& \left[ {\frac{\sqrt 2 }{2}E\gamma \Delta T(A_1 -A_2 )} \right],\quad {\textbf{\emph{F}}}=\left[ {-F} \right] \end{array} $$

(57)

Solving equation KU ^M = F ^M, we can obtain U ^M and then

$$ \label{eq58} f_c ={\left({\textbf{\emph{F}}}^M\right)^{\rm T} {\textbf{\emph{U}}}^M} \mathord{\left/ {\vphantom {{(F^M)^{\rm T}U^M} {(F^M)^{\rm T}U_0^M }}} \right. \kern-\nulldelimiterspace} {\left({\textbf{\emph{F}}}^M\right)^{\rm T} {\textbf{\emph{U}}}_0^M }=\frac{3A_1 }{2(A_1 +A_2 )} $$

(58)

Solving equation KU $^\mathit{Tem}$ = F $^\mathit{Tem}$, we can obtain U $^\mathit{Tem}$ and then

$$ \label{eq59} \begin{array}{lll} f_u &={\sum\limits_{k=1}^S {\left( {U_{\lambda (k)}^\mathit{Tem} } \right)^2} } \mathord{\left/ {\vphantom {{\sum\limits_{k=1}^S {\left( {U_{\lambda (k)}^\mathit{Tem} } \right)^2} } {\sum\limits_{k=1}^S {\left( {\left( {U_{\lambda (k)}^\mathit{Tem} } \right)_0 } \right)^2} }}} \right. \kern-\nulldelimiterspace} {\sum\limits_{k=1}^S {\left( {\left( {U_{\lambda (k)}^\mathit{Tem} } \right)_0 } \right)^2} }\\ &=\frac{9(A_1 -A_2 )^2}{(A_1 +A_2 )^2} \end{array} $$

(59)

Since A ₁ is fixed and 0 ≤ A ₂ ≤ 2A ₁, to minimize f _c , A ₂ should satisfy A ₂ = 2 A ₁. And to minimize f _u , A ₂ should satisfy A ₂ = A ₁. Therefore, to minimize obj (ω ∈ [0, 1]), the optimum A ₂ ∈ (A ₁, 2A ₁). This means A ₂ is not “the larger the better” in the sense of minimizing the multi-objective performance of the thermoelastic truss structure.