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# On a hierarchy of effective models for the biomechanics of human compact bone tissue

Grigor Nika

► To cite this version:

Grigor Nika. On a hierarchy of effective models for the biomechanics of human compact bone tissue. IMA Journal of Applied Mathematics, In press, 10.1093/imamat/hxad011. hal-03629864v2

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Submitted on 3 Apr 2023

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Distributed under a Creative Commons Attribution 4.0 International License
On a hierarchy of effective models for
the biomechanics of human compact bone tissue

Grigor Nika*

Mathematics & Computer Science
Karlstad University
Universitetsgatan 2
65188 Karlstad, Sweden
grigor.nika@kau.se

April 3, 2023

**Abstract**

We derive a hierarchy of effective models that can be used to model the biome-
chanics of human compact bone taking into account scale-size effects observed ex-
perimentally. The classification of the effective models depends on the hierarchy of
four characteristic lengths: The size of the heterogeneities, two intrinsic lengths of the
constituents, and the overall characteristic length of the domain. Depending on the
different scale interactions between the size of the heterogeneities, the two intrinsic
lengths of the constituents, and the characteristic length of the domain we obtain ei-
ther an effective Cauchy continuum or an effective Cosserat continuum. The passage
to the limit relies on suitable use of the periodic unfolding operator. Moreover, we
perform numerical simulations to validate our results.

MSC 2020: 74Q05, 35B27, 35J58, 35Q74, 74L15

**Keywords:** non-simple materials, scale-size effects, chirality, compact bone, periodic homogenization

# 1 Introduction

Human compact bone exhibits a rich hierarchical microstructure (see Fig. 1). At the level of the microstructure, we have what is referred to in the medical literature, as osteons [22]. Osteons are porous (often elliptical in nature), cylindrical structures, that themselves possess a microstructure, that give compact bone various properties. As we get older bone loss and bone remodelling is one of the inevitabilities that we face as humans leading to

<footnote>*Email: grigor.nika@kau.se</footnote>
diseases like compromised joints which require total hip arthroplasty (hip replacement). Modern day hip replacement femoral implants are made of solid materials that are considerably stiffer than bone and, moreover, they lack the same (local) kinematic behavior as that of bone. This difference in material properties can lead to a mismatch at the surface of contact between bone and implant which can result in implant failure and a decrease in quality of life for the patient [1]. One way to have better compatibility at the surface of contact between the femoral implant and bone is to perforate the implant in a way that does not compromise its functionality or rigidity. This can be done using topology optimization methods [1] or spinodoid topologies [20] to design the microstructure of the synthetic femoral implant. However, matching the local behavior of bone has remained a challenge since both approaches mentioned above, fall within the classical Cauchy elasticity framework which does not encapsulate the behavior of bone [34].

<figure><img src="image_3.png" /><figcaption>Figure 1: Schematic of a human femur exhibiting the multiple scales. The first two images were generated using data from **BodyParts3D**, © The Database Center for Life Science licensed under CC Attribution-Share Alike 2.1 Japan and the open source software **MeshLab** [4, 3].</figcaption></figure>

A series of experimental works [21, 34, 22] dealing with aspects of micromechanics and biological materials observed non-classical behavior of bone due to microstructural scale-size effects in regions of high strain gradients, e.g. holes or cracks. Specifically, quoting from reference [22] *"strain was redistributed from peak regions to regions which are classically forbidden so that stress concentration was ameliorated for bone"*. Moreover, this "discrepancy" was explained by showing that wet bone follows non-centrosymmetric *Cosserat elasticity* [9, 37, 28, 38, 25, 12, 13, 26, 11, 27, 33]. Furthermore, the above experimental work on compact bone was interpreted within the continuum view in which the forces in the microstructure were averaged. Hence, deriving macroscale equations and boundary conditions that take into account scale-size effects is of great importance. Additionally, understanding the role of the microstructure in bone using Cosserat elasticity is important since it is related with the ability of bone to redistribute stress around "defects" and is consistent with the *"alternate stress pathway of Cosserat elasticity"* (see [34]).

To fix ideas, in this work, we rigorously derive effective models taking into scale-size effects that could be used to study bone macro-samples. Since Cosserat elasticity incorporates scale-size effects, naturally, though intrinsic lengths we obtain a hierarchy of two effective models based on how the aforementioned intrinsic lengths compare with either the "size"
of the heterogeneities or the overall length of the domain. Our starting point is a non-centrosymmetric, heterogeneous, periodic Cosserat elastic solid $\Omega$ with period $\ell$ that is governed by the system of equations,

$$ 
\begin{aligned}
-\operatorname{div} \sigma &= \mathbf{f} && \text{in } \Omega, \\
-\operatorname{div} \mu - \boldsymbol{\epsilon} \cdot \sigma &= \mathbf{g} && \text{in } \Omega,
\end{aligned}
\qquad (1.1) $$

with $\sigma$ the non-symmetric stress tensor, $\mu$ the couple-stress tensor, $\boldsymbol{\epsilon}$ the Levi-Civita tensor with $\boldsymbol{\epsilon} \cdot \sigma = \sum_{j,k=1}^{3} \boldsymbol{\epsilon}_{ijk}\sigma_{jk}$ for $i \in \{1, 2, 3\}$, $\mathbf{f}$ the body forces, and $\mathbf{g}$ the body couples. Moreover, for $i, j \in \{1, 2, 3\}$, the constitutive relations are given by,

$$ \sigma_{ji} = \sum_{k,\ell=1}^{3} (E_{jik\ell}\gamma_{k\ell} + C_{jik\ell}\kappa_{k\ell}), \qquad (1.2) $$

$$ \mu_{ji} = \sum_{k,\ell=1}^{3} (C_{k\ell ji}\gamma_{k\ell} + M_{jik\ell}\kappa_{k\ell}). \qquad (1.3) $$

Here we have defined the non-symmetric strain tensor and the torsion tensor (or curvature-twist tensor or curvature tensor or curvature) respectively, by,

$$ \gamma_{ji} := \partial_{x_j} u_i - \sum_{k=1}^{3} \epsilon_{kji} \varphi_k \quad \text{and} \quad \kappa_{ji} := \partial_{x_j} \varphi_i, \qquad (1.4) $$

for $i, j \in \{1, 2, 3\}$ where $u$ is the displacement and $\varphi$ is the rotation. Additionally, thermodynamic relations impose the following symmetry on the tensors $E_{jik\ell}$, $M_{jik\ell}$, and $C_{jik\ell}$:

$$ E_{jik\ell} = E_{k\ell ji}, \quad M_{jik\ell} = M_{k\ell ji}, \quad \text{and} \quad C_{jik\ell} = C_{k\ell ji} \qquad (1.5) $$

The tensor $C_{jik\ell}$, expresses the chirality effects [33]. If $C_{jik\ell} \equiv 0$ then the material is called centrosymmetric [33] and the problem has been addressed formally in [15], [14] and rigorously via unfolding in [30]. Regarding non-centrosymmetric material, one can consult [2] as well as recent work in [31] where a cascade of effective models were derived using the multiscale asymptotic expansion technique. The results presented here address rigorous limit passage to some of the models derived in [31]. We address rigorously some of the cases in [31] and not all since certain cases seem to lack coercivity (we suspect that this discrepancy might be attributable to the assumption, made in [31], that chiral effects scale independently of the curvature-twist effects).

Our working technique combines scaling arguments with the periodic homogenization procedure. The unveiling of the structure of the limit problems relies on the use of the periodic unfolding technique. Unfolding has a natural way of bridging the scale between generalized and classical continua [30]. Furthermore, we carry out numerical simulations of the derived effective coefficients to validate our results.
## 2 Dimensional analysis

Let L be the characteristic length of the body Ω and $ℓ$ the characteristic length of the periodic cell. We normalize all the parameters involved, including material parameters, as follows:

$$ \boldsymbol{x}^* = \frac{\boldsymbol{x}}{L}, \quad \boldsymbol{u}^*(\boldsymbol{x}^*) = \frac{\boldsymbol{u}(\boldsymbol{x})}{L}, \quad \boldsymbol{\varphi}^*(\boldsymbol{x}^*) = \boldsymbol{\varphi}(\boldsymbol{x}). \qquad (2.1) $$

In non-centrosymmetric Cosserat media there are two additional and independent length scale parameters that are of importance, namely, the Cosserat intrinsic length $ℓ_{coss}$ of the constituents [14], [15], and the chirality coefficient $ℓ_{chiral}$. The Cosserat intrinsic length and the chirality coefficient are defined as follows,

$$ \mathcal{M} = \mathcal{E} \ell_{\text{coss}}^2, \quad \mathcal{C} = \mathcal{E} \ell_{\text{coss}}^{1/p} \ell_{\text{chiral}}^{1/p'} \quad \text{with} \quad \frac{1}{p} + \frac{1}{p'} = 1, \quad p, p' \in (1, +\infty), \qquad (2.2) $$

where $\mathcal{E} = \max_{z \in Y_\ell} |E_{jikl}(\boldsymbol{z})|$, $\mathcal{C} = \max_{z \in Y_\ell} |C_{jikl}(\boldsymbol{z})|$, $\mathcal{M} = \max_{z \in Y_\ell} |M_{jikl}(\boldsymbol{z})|$ for $i, j, k, l \in \{1, 2, 3\}$, and $Y_\ell = (-\ell/2, \ell/2]^3$ is periodic cell characterizing the body Ω. Additionally, the non-symmetric strain and curvature scale as,

$$ \gamma^* = \gamma \quad \text{and} \quad \kappa^* = L\kappa. \qquad (2.3) $$

Moreover, we define the scaled stress, couple-stress, and fourth order material tensors as follows,

$$ \sigma^* = \mathcal{E}^{-1}\sigma, \quad \mu^* = (\mathcal{E}L)^{-1}\mu, \qquad (2.4) $$

$$ E^*(\boldsymbol{x}^*) = \mathcal{E}^{-1}E(\boldsymbol{x}), \quad C^*(\boldsymbol{x}^*) = \mathcal{C}^{-1}C(\boldsymbol{x}), \quad M^*(\boldsymbol{x}^*) = \mathcal{M}^{-1}M(\boldsymbol{x}). \qquad (2.5) $$

We remark that the fourth order material tensors $E^*(\boldsymbol{x}^*)$, $C^*(\boldsymbol{x}^*)$, and $M^*(\boldsymbol{x}^*)$ are $Y^*$ periodic where,

$$ Y^* = \frac{\ell}{L}Y, \quad Y := \left(-\frac{1}{2}, \frac{1}{2}\right]^3. \qquad (2.6) $$

Hence, the system of equations in (1.1) scales as:

$$ \begin{aligned} -\operatorname{div} \sigma^* &= \boldsymbol{f}^* && \text{in } \Omega, \\ -\operatorname{div} \mu^* - \boldsymbol{\epsilon} : \sigma^* &= \boldsymbol{g}^* && \text{in } \Omega, \end{aligned} \qquad (2.7) $$

where $\boldsymbol{f}^*$, and $\boldsymbol{g}^*$ are the appropriately scaled body forces and body couples (see [14, Eq. (14), pg. 4589]) and with constitutive laws, for $i, j \in \{1, 2, 3\}$:
$$ \sigma_{ji}^* = \sum_{k,l=1}^{3} \left( \mathbb{E}_{jik\ell}^* \gamma_{k\ell}^* + \left( \frac{\ell_{\text{coss}}}{L} \right)^{1/p} \left( \frac{\ell_{\text{chiral}}}{L} \right)^{1/p'} \mathbf{C}_{ijk\ell}^* \kappa_{k\ell}^* \right), \quad (2.8) $$

$$ \mu_{ji}^* = \sum_{k,l=1}^{3} \left( \left( \frac{\ell_{\text{coss}}}{L} \right)^{1/p} \left( \frac{\ell_{\text{chiral}}}{L} \right)^{1/p'} \mathbf{C}_{k\ell ij}^* \gamma_{k\ell}^* + \left( \frac{\ell_{\text{coss}}}{L} \right)^2 \mathbf{M}_{jik\ell}^* \kappa_{k\ell}^* \right). \quad (2.9) $$

Thus, one can generate an $\varepsilon$ periodic problem by defining the nondimensional number $\varepsilon$ as the ratio of $\ell/L$ and let $\varepsilon \to 0$ to obtain an effective medium. However, different cases ought to be considered depending on how $\ell_{\text{coss}}$ and $\ell_{\text{chiral}}$ scale with $\ell$ and L, respectively. Here we consider the cases,

$$ \ell_{\text{coss}}/L \sim \varepsilon, \quad \ell_{\text{chiral}}/\ell \sim \varepsilon^{p'}, \qquad (\text{HS 1}) $$

$$ \ell_{\text{coss}}/L \sim 1, \quad \ell_{\text{chiral}}/\ell \sim \varepsilon^{p'-1}. \qquad (\text{HS 2}) $$

## 2.1 A remark on scale separation (or lack thereof)

We chose to work with the above scaling, primarily, because of its physical interpretation.

* The (HS 1) scaling indicates that the size of the heterogeneities are comparable to the order of the period. Hence, under this scaling, all scales involved are well separated. This scaling bridges a microscopic generalized continuum model with a macroscopic Cauchy continuum model.

* The (HS 2) scaling indicates that the size of the heterogeneities are comparable to the characteristic length of the overall domain. However, this implies that Cosserat length scale is not well separated. One of the strengths of our approach is that we can still construct an effective model, even though one of the length scales involved is not well separated, which illustrates the power of our approach.

* In contrast, the chirality length scale is chosen to be well separated in either case. This choice was made, primarily, to avoid degenerate cases where we lose coercivity. Hence, it is a purely mathematical assumption. Moreover, our approach deliberately forces the chiral intrinsic length to dependent on the Cosserat intrinsic length (although the Cosserat intrinsic length is independent of the chiral intrinsic length). Again, as we will show in the next section, it has to do with the well-posedness of the model.

Naturally, one could consider different scalings than the one proposed above. We will not address other type of scalings here. Rather we will leave their treatment to future work. Finally, if no confusion arises, henceforth, we will omit the * notation for the sake of simplicity and expediency of presentation.
## 2.2 Scaling of the stress and couple-stress under HS 1

If $\ell_{\text{chiral}}/\ell = \varepsilon^{p'}$ then $\ell_{\text{chiral}}/L = \varepsilon^{p'+1}$. Hence, the stress and couple-stress become,

$$ \sigma_{ij}^{\varepsilon} = \sum_{k, \ell=1}^{3} \left( E_{ijkl} \left( \frac{\mathbf{x}}{\varepsilon} \right) \gamma_{kl}^{\varepsilon} + \varepsilon^2 C_{ijkl} \left( \frac{\mathbf{x}}{\varepsilon} \right) \kappa_{kl}^{\varepsilon} \right), \quad (2.10) $$

$$ \mu_{ij}^{\varepsilon} = \sum_{k, \ell=1}^{3} \left( \varepsilon^2 C_{klij} \left( \frac{\mathbf{x}}{\varepsilon} \right) \kappa_{kl}^{\varepsilon} + \varepsilon^2 M_{ijkl} \left( \frac{\mathbf{x}}{\varepsilon} \right) \gamma_{kl}^{\varepsilon} \right). $$

## 2.3 Scaling of the stress and couple-stress under HS 2

If $\ell_{\text{chiral}}/\ell = \varepsilon^{p'-1}$, then $\ell_{\text{chiral}}/L = \varepsilon^p$. Hence, the stress and couple-stress become,

$$ \sigma_{ij}^{\varepsilon} = \sum_{k, \ell=1}^{3} \left( E_{ijkl} \left( \frac{\mathbf{x}}{\varepsilon} \right) \gamma_{kl}^{\varepsilon} + \varepsilon C_{ijkl} \left( \frac{\mathbf{x}}{\varepsilon} \right) \kappa_{kl}^{\varepsilon} \right), \quad (2.11) $$

$$ \mu_{ij}^{\varepsilon} = \sum_{k, \ell=1}^{3} \left( \varepsilon C_{klij} \left( \frac{\mathbf{x}}{\varepsilon} \right) \kappa_{kl}^{\varepsilon} + M_{ijkl} \left( \frac{\mathbf{x}}{\varepsilon} \right) \gamma_{kl}^{\varepsilon} \right), $$

where $\gamma_{ji}^{\varepsilon}:=\partial_{x_j} u_i^{\varepsilon} - \sum_{k=1}^{3} \epsilon_{kji} \varphi_k^{\varepsilon}$ and $\kappa_{ji}^{\varepsilon}:=\partial_{x_j} \varphi_i^{\varepsilon}$ for $i, j \in \{1, 2, 3\}$.

## 133 Notation

To expedite the presentation of our results, here onwards we will make use of the following notation:

136 - Throughout the article we employ the Einstein summation notation of repeated indices unless otherwise stated.

138 - The third order tensor $\epsilon_{ijk}$ is the Levi-Civita symbol that is equal to 1 if $(i, j, k)$ is an even permutation of $(1, 2, 3)$, $-1$ if it is an odd permutation, and zero if any index is repeated.

141 - Tensor contractions: Let $v$ be a vector, A and B be second order tensors. We define the following contractions of tensor-tensor or tensor-vector contractions:

143 1. A:B = $A_{ij} B_{ij}$ (Frobenius inner product)

144 2. $\epsilon.v = \epsilon_{ijk} v_i$ (contraction of the Levi-Civita tensor with a vector leads to second order tensor)

146 2. $\epsilon:A = \epsilon_{ijk} A_{ij}$ (contraction of the Levi-Civita tensor with a second order tensor leads to a vector)

147
### 3 The microscopic mathematical model

We consider an elastic solid with periodic microstructure of period $\varepsilon$ occupying a region $\Omega$.

The region $\Omega$, that the body occupies, is assumed to be a connected (possibly non-simple),

open set of $\mathbb{R}^3$ with a Lipschitz boundary. $Y = (-1/2, 1/2]^3$ is the unit cube in $\mathbb{R}^3$ and

$\mathbb{Z}^3$ is the set of all vectors with integer components. For every positive $\varepsilon$, let $N_\varepsilon$ be the set of all points $m \in \mathbb{Z}^3$ such that $\varepsilon(m + Y)$ is strictly included in $\Omega$. Let $T \subset \text{interior}(Y)$

be a Lipschitz domain such that $\bar{T} \subset Y$ with $T_m^\varepsilon := \varepsilon(m + T)$ will represent the region

containing one of the material phases (see Fig. 2). Hence, we can define the following

subsets of $\Omega$:

$$ \Omega_{1\varepsilon} := \bigcup_{m \in N_{\varepsilon}} T_m^{\varepsilon}, \quad \Omega_{2\varepsilon} := \Omega \setminus \Omega_{1\varepsilon}, \quad \Omega := \Omega_{1\varepsilon} \cup \Omega_{2\varepsilon}. $$

<figure><img src="image_4.png" /><figcaption>Figure 2: Schematic of the heterogeneous Cosserat medium with perfect transmission conditions.</figcaption></figure>

Moreover, we denote by $\Sigma = \partial\Omega$ the boundary of $\Omega$ and $n$ will denote the unit normal

on $\Sigma$ pointing in the outward direction. Hence, the heterogeneous Cosserat continuum is

characterized by the following coupled system:
$$
\begin{align}
-\operatorname{div} \sigma^{\epsilon} &= \boldsymbol{f} && \text{in } \Omega, \nonumber \\
-\operatorname{div} \mu^{\epsilon} - \epsilon \cdot \sigma^{\epsilon} &= \boldsymbol{g} && \text{in } \Omega, \nonumber \\
\boldsymbol{u}^{\epsilon} &= \mathbf{0} && \text{on } \Sigma, \nonumber \\
\varphi^{\epsilon} &= \mathbf{0} && \text{on } \Sigma. \tag{3.1}
\end{align}
$$

Here $\sigma^\epsilon$ is the stress, $\mu^\epsilon$ is the couple-stress, $\boldsymbol{u}^\epsilon$ is the displacement, and $\varphi^\epsilon$ is the rota-
tion. The system of equations (3.1) characterizes the mechanical deformation that the
body undergoes. The equations are fully coupled and the system is closed with homoge-
neous Dirichlet boundary conditions on $\Sigma$. In addition to computing the displacement we
must also compute the rigid rotations which makes for a fully coupled system of partial
differential equations.

167 **Constitutive relations.** The stress and couple-stress are related to the strain and
168 curvature-twist, respectively, though the fourth order material tensors $\mathbf{E}^{\epsilon}(\mathbf{x}) := \mathbf{E}(\frac{\mathbf{x}}{\epsilon})$,
169 $\mathbf{C}^{\epsilon}(\mathbf{x}) := \mathbf{C}(\frac{\mathbf{x}}{\epsilon})$, and $\mathbf{M}^{\epsilon}(\mathbf{x}) := \mathbf{M}(\frac{\mathbf{x}}{\epsilon})$ by the relation (see [33]),

$$
\begin{bmatrix} \sigma^{\epsilon} \\ \mu^{\epsilon} \end{bmatrix} = \begin{bmatrix} \mathbf{E}^{\epsilon} & \mathbf{C}^{\epsilon} \\ \mathbf{C}^{\epsilon \top} & \mathbf{M}^{\epsilon} \end{bmatrix} \begin{bmatrix} \gamma^{\epsilon} \\ \kappa^{\epsilon} \end{bmatrix} \qquad (3.2)
$$

170 The tensors $\mathbf{E}^{\epsilon}$, $\mathbf{M}^{\epsilon}$, and $\mathbf{C}^{\epsilon}$ are linear isotropic tensors in each phase and have the following
171 form,

$$
\begin{align}
\mathbf{E}_{jikl}^{\epsilon}(\mathbf{x}) &:= \mathbf{E}_{jikl}(\frac{\mathbf{x}}{\epsilon}) \nonumber \\
&= \left(\vartheta(\frac{\mathbf{x}}{\epsilon}) + \alpha(\frac{\mathbf{x}}{\epsilon})\right) \delta_{jk} \delta_{il} \tag{3.3} \\
&\quad + \left(\vartheta(\frac{\mathbf{x}}{\epsilon}) - \alpha(\frac{\mathbf{x}}{\epsilon})\right) \delta_{jl} \delta_{ik} + \varrho(\frac{\mathbf{x}}{\epsilon}) \delta_{ij} \delta_{kl}, \nonumber
\end{align}
$$

$$
\begin{align}
\mathbf{M}_{jikl}^{\epsilon}(\mathbf{x}) &:= \mathbf{M}_{jikl}(\frac{\mathbf{x}}{\epsilon}) \nonumber \\
&= (\theta(\frac{\mathbf{x}}{\epsilon}) + \beta(\frac{\mathbf{x}}{\epsilon})) \delta_{jk} \delta_{il} \tag{3.4} \\
&\quad + (\theta(\frac{\mathbf{x}}{\epsilon}) - \beta(\frac{\mathbf{x}}{\epsilon})) \delta_{jl} \delta_{ik} + \rho(\frac{\mathbf{x}}{\epsilon}) \delta_{ij} \delta_{kl}, \nonumber
\end{align}
$$

$$
\begin{align}
\mathbf{C}_{jikl}^{\epsilon}(\mathbf{x}) &:= \mathbf{C}_{jikl}(\frac{\mathbf{x}}{\epsilon}) \nonumber \\
&= c_1(\frac{\mathbf{x}}{\epsilon}) \delta_{kl} \delta_{ij} + c_2(\frac{\mathbf{x}}{\epsilon}) \delta_{ik} \delta_{jl} + c_3(\frac{\mathbf{x}}{\epsilon}) \delta_{il} \delta_{jk}, \tag{3.5}
\end{align}
$$

respectively. The coefficients $\vartheta, \theta, \alpha, \beta, \varrho, \rho$, and $c_1$ for $l = 1, 2, 3$ are material parameters
that are piecewise constant in each phase.
### 3.1 Assumptions

We frame model (3.1) under the following general assumptions:

• The body force $f$ and body couple $g$ are such that $f \in L^2(\Omega; \mathbb{R}^3)$ and $g \in H(\text{curl}, \Omega)$.

• The fourth order material characterization tensors $E^\epsilon, M^\epsilon, C^\epsilon$ belong in $L^\infty$. Moreover, $E^\epsilon$ and $M^\epsilon$ satisfy the Legendre condition, i.e for any fourth rank tensor $A$,

$$A_{\alpha\beta ij} \xi_i^\alpha \xi_j^\beta \geq \lambda |\xi|^2, \quad \forall \xi \in \mathbb{R}^{3 \times 3} \qquad (3.6)$$

where $\lambda > 0$.

**Existence and uniqueness.** Denote by,

$$\mathcal{J}_{\epsilon}[v, \psi] = \frac{1}{2} \int_{\Omega} [\gamma \ \kappa] \begin{bmatrix} E^{\epsilon} & \hat{C}^{\epsilon} \\ \hat{C}^{\epsilon \top} & \hat{M}^{\epsilon} \end{bmatrix} \begin{bmatrix} \gamma \\ \kappa \end{bmatrix} dx \qquad (3.7)$$

where $\gamma := \nabla v - \varepsilon.\phi$ and $\kappa := \nabla \psi$. Then the total energy of the system described in (3.1),

$$\mathcal{E}_{\epsilon}[v, \psi] = \mathcal{J}_{\epsilon}[v, \psi] - \int_{\Omega} (f \cdot v + g \cdot \psi) dx. \qquad (3.8)$$

The tensors in the matrix with the hat notation are simply place holders. So, under the (HS 1) scaling $\hat{C}^\epsilon \equiv \epsilon^2 C^\epsilon$ and $\hat{M}^\epsilon \equiv \epsilon^2 M^\epsilon$, while under the (HS 2) scaling $\hat{C}^\epsilon \equiv \epsilon C^\epsilon$ and $\hat{M}^\epsilon \equiv M^\epsilon$. Computing the Euler-Lagrange equations and grouping terms by test function (see [30]) one can recover system (3.1) in the sense of distributions.

One can readily verify that the above energy is sequentially weakly lower semicontinuous in $H_\Sigma^1(\Omega, \mathbb{R}^3) \times H_\Sigma^1(\Omega, \mathbb{R}^3)$. Thus, it remains to show that the energy is also coercive under each of the two schemes (HS 1) and (HS 2) and then we can simply rely on the direct method that will guarantee the existence of a unique minimizer.

**Coercivity under the (HS 1) scaling.** Using the Legendre condition for the tensors $E, M$, and Cauchy's inequality with $\delta$ we obtain,

$$\begin{align}
\hat{c} \|\gamma^\epsilon\|_{L^2(\Omega; \mathbb{R}^{3 \times 3})}^2 + \tilde{c} \epsilon^2 \|\kappa^\epsilon\|_{L^2(\Omega; \mathbb{R}^{3 \times 3})}^2 & \nonumber \\
& \leq \mathcal{J}_{\epsilon}[\mathbf{u}^\epsilon, \varphi^\epsilon] - 2\epsilon^2 \int_{\Omega} C_{jikl} \left(\frac{\mathbf{x}}{\epsilon}\right) \gamma^\epsilon : \kappa^\epsilon dx \nonumber \\
& \leq \mathcal{J}_{\epsilon}[\mathbf{u}^\epsilon, \varphi^\epsilon] + 2\epsilon^2 \int_{\Omega} |\gamma^\epsilon| |\kappa^\epsilon| dx \tag{3.9} \\
& \leq \mathcal{J}_{\epsilon}[\mathbf{u}^\epsilon, \varphi^\epsilon] + \frac{\epsilon^2}{2\delta} \|\gamma^\epsilon\|^2 + 2\epsilon^2 \delta \|\kappa^\epsilon\|^2. \nonumber
\end{align}$$

Thus,
$$ (\tilde{c} - 2\delta)\epsilon^2 \|\kappa^\epsilon\|_{L^2(\Omega; \mathbb{R}^{3\times3})}^2 + (\hat{c} - \frac{\epsilon^2}{2\delta}) \|\gamma^\epsilon\|_{L^2(\Omega; \mathbb{R}^{3\times3})}^2 \le \mathcal{J}_\epsilon[\mathbf{u}^\epsilon, \varphi^\epsilon]. \quad (3.10) $$

Picking $\delta > 0$ appropriately and using the smallness of $\epsilon > 0$ we can guarantee that the above expression is always positive. Moreover, by [18, Thm. 3.1] (see also [29], [19]) in $H_\Sigma^1(\Omega, \mathbb{R}^3) \times H_\Sigma^1(\Omega, \mathbb{R}^3)$, the following estimate holds:

$$ \begin{aligned} \|\mathbf{u}^\epsilon\|_{H_\Sigma^1(\Omega; \mathbb{R}^3)}^2 &+ (\|\varphi^\epsilon\|_{L^2(\Omega; \mathbb{R}^3)}^2 + \epsilon^2 \|\nabla\varphi^\epsilon\|_{L^2(\Omega; \mathbb{R}^{3\times3})}^2) \\ &\le c (\|\mathbf{f}\|_{L^2(\Omega; \mathbb{R}^3)}^2 + \|\mathbf{g}\|_{L^2(\Omega; \mathbb{R}^3)}^2), \end{aligned} \quad (3.11) $$

for some generic constant $c$ independent of $\epsilon$.

**Coercivity under the (HS 2) scaling.** Coercivity in this case can be shown in exactly the same way as for (HS 1). We simply write it down and leave the details to the interested reader.

$$ (\tilde{c} - \epsilon) \|\kappa^\epsilon\|_{L^2(\Omega; \mathbb{R}^{3\times3})}^2 + (\hat{c} - \epsilon) \|\gamma^\epsilon\|_{L^2(\Omega; \mathbb{R}^{3\times3})}^2 \le \mathcal{J}_\epsilon[\mathbf{u}^\epsilon, \varphi^\epsilon], \quad (3.12) $$

with the estimate,

$$ \begin{aligned} \|\mathbf{u}^\epsilon\|_{H_\Sigma^1(\Omega; \mathbb{R}^3)}^2 &+ \|\varphi^\epsilon\|_{H_\Sigma^1(\Omega; \mathbb{R}^3)}^2 \\ &\le c (\|\mathbf{f}\|_{L^2(\Omega; \mathbb{R}^3)}^2 + \|\mathbf{g}\|_{L^2(\Omega; \mathbb{R}^3)}^2), \end{aligned} \quad (3.13) $$

for some generic constant $c$ independent of $\epsilon$. Thus, the direct method furnishes a unique minimizer. Furthermore, the results in (3.10) and (3.12) prove that the symmetric “quadratic form” in (3.7) is positive definite. Computation of the Euler-Lagrange equations leads,

$$ \int_{\Omega} [\bar{\gamma} \bar{\kappa}] \begin{bmatrix} \hat{E}^{\epsilon} & \hat{C}^{\epsilon} \\ \hat{C}^{\epsilon\top} & \hat{M}^{\epsilon} \end{bmatrix} \begin{bmatrix} \gamma^{\epsilon} \\ \kappa^{\epsilon} \end{bmatrix} dx = \int_{\Omega} [\mathbf{f} \mathbf{g}] \begin{bmatrix} \mathbf{v} \\ \mathbf{\psi} \end{bmatrix} dx \quad (3.14) $$

with $\bar{\gamma}:=\nabla\mathbf{v}-\epsilon.\psi$ and $\bar{\kappa}:=\nabla\psi$. Grouping terms by test function we obtain the weak form of (3.1) which reads as follows: Find $(\mathbf{u}^\epsilon, \varphi^\epsilon) \in H_\Sigma^1(\Omega; \mathbb{R}^3) \times H_\Sigma^1(\Omega; \mathbb{R}^3)$ such that,

$$ \int_{\Omega} \sigma^{\epsilon}: \nabla \mathbf{v} \, d\mathbf{x} - \int_{\Omega} \mathbf{f} \cdot \mathbf{v} \, d\mathbf{x} = 0 \quad \text{for all } \mathbf{v} \in H_{\Sigma}^{1}(\Omega, \mathbb{R}^{3}), \quad (3.15) $$

$$ \int_{\Omega} \mu^{\epsilon}: \nabla \psi \, d\mathbf{x} - \int_{\Omega} (\epsilon.\psi): \sigma^{\epsilon} \, d\mathbf{x} - \int_{\Omega} \mathbf{g} \cdot \psi \, d\mathbf{x} = 0 \quad \text{for all } \psi \in H_{\Sigma}^{1}(\Omega, \mathbb{R}^{3}). \quad (3.16) $$

Hence, we can recover (3.1) in sense of distributions as usual.
## 4 Periodic unfolding and homogenization

In the next two subsections we recall the definitions and properties of the periodic unfolding operator [6, 10, 7, 5, 8] and present our main results. We will list some of their properties, leaving the interested reader to consult [6, 10, 7] for further details regarding proofs.

### 4.1 The periodic unfolding operator

We define the following domain decompositions (see [6, 10, 7, 8]):

$$ K_{\varepsilon}^{-} := \{ \ell \in \mathbb{Z}^{d} \mid \varepsilon(\ell + Y) \subset \bar{\Omega} \}, \quad \Omega_{\varepsilon}^{-} := \text{int} \left( \cup_{\ell \in K_{\varepsilon}^{-}} \varepsilon(\ell + Y) \right), \quad \Lambda_{\varepsilon}^{-} := \Omega \setminus \Omega_{\varepsilon}^{-}. \quad (4.1) $$

<figure><img src="image_5.png" /><figcaption>Figure 3: Unfolding operator on a periodic grid</figcaption></figure>

Let $[z]_Y = ([z_1], \dots, [z_d])$ denote the integer part of $z \in \mathbb{R}^3$ and denote by $\{z\}_Y$ the difference $z - [z]_Y$ which belongs to $Y$. Regarding our multiscale problem that depends on a small length parameter $\varepsilon > 0$, we can decompose any $x \in \mathbb{R}^3$ using the maps $[\cdot]_Y : \mathbb{R}^3 \to \mathbb{Z}^3$ and $\{\cdot\}_Y : \mathbb{R}^3 \to Y$ the following way (see Fig. 3 (right)),

$$ \mathbf{x} = \varepsilon \left( \left[ \frac{\mathbf{x}}{\varepsilon} \right]_Y + \left\{ \frac{\mathbf{x}}{\varepsilon} \right\}_Y \right). \quad (4.2) $$

For any Lebesgue measurable function $\varphi$ on $\Omega$ we define the periodic unfolding operator by,

$$ \mathcal{T}_{\varepsilon}(\varphi)(\mathbf{x}, \mathbf{y}) = \begin{cases} \varphi \left( \varepsilon \left[ \frac{\mathbf{x}}{\varepsilon} \right]_Y + \varepsilon \mathbf{y} \right) & \text{for a.e. } (\mathbf{x}, \mathbf{y}) \in \Omega_{\varepsilon}^{-} \times Y \\ 0 & \text{for a.e. } (\mathbf{x}, \mathbf{y}) \in \Lambda_{\varepsilon}^{-} \times Y. \end{cases} \quad (4.3) $$

**Proposition 4.1.** For any $p \in [1, +\infty)$ the unfolding operator $\mathcal{T}_{\varepsilon} : L^p(\Omega) \to L^p(\Omega \times Y)$ is linear, continuous, and has the following properties:

I. $\mathcal{T}_{\varepsilon}(\varphi \psi) = \mathcal{T}_{\varepsilon}(\varphi) \mathcal{T}_{\varepsilon}(\psi)$ for every pair of Lebesgue measurable functions $\varphi, \psi$ on $\Omega$
II. For every $\varphi \in L^1(\Omega)$ we have,

$$ \frac{1}{|Y|} \int_{\Omega \times Y} \mathcal{T}_{\varepsilon}(\varphi)(\mathbf{x}, \mathbf{y}) \, d\mathbf{x} \, d\mathbf{y} = \int_{\Omega_{\varepsilon}^{-}} \varphi(\mathbf{x}) \, d\mathbf{x} = \int_{\Omega_{\varepsilon}} \varphi(\mathbf{x}) \, d\mathbf{x} - \int_{\Lambda_{\varepsilon}^{-}} \varphi(\mathbf{x}) \, d\mathbf{x} \quad (4.4) $$

III. $\|\mathcal{T}_{\varepsilon}(\varphi)\|_{L^p(\Omega \times Y)} \le |Y|^{1/p} \|\varphi\|_{L^p(\Omega)}$ for every $\varphi \in L^p(\Omega)$

IV. $\mathcal{T}_{\varepsilon}(\varphi) \to \varphi$ strongly in $L^p(\Omega \times Y)$ for $\varphi \in L^p(\Omega)$ as $\varepsilon \to 0$

V. If $\{\varphi_{\varepsilon}\}_{\varepsilon}$ is a sequence in $L^p(\Omega)$ such that $\varphi_{\varepsilon} \to \varphi$ strongly in $L^p(\Omega)$, then $\mathcal{T}_{\varepsilon}(\varphi_{\varepsilon}) \to \varphi$ strongly in $L^p(\Omega \times Y)$

VI. If $\varphi \in L^p(Y)$ is $Y$-periodic and $\varphi_{\varepsilon}(\mathbf{x}) = \varphi(\frac{\mathbf{x}}{\varepsilon})$ then $\mathcal{T}_{\varepsilon}(\varphi_{\varepsilon}) \to \varphi$ strongly in $L^p(\Omega \times Y)$ as $\varepsilon \to 0$

VII. If $\phi_{\varepsilon} \to \phi$ in $H^1(\Omega)$ then there exists a non-relabeled subsequence and a $\hat{\phi} \in L^2(\Omega; H_{\text{per}}^1(Y))$ such that

a. $\mathcal{T}_{\varepsilon}(\phi_{\varepsilon}) \to \phi$ in $L^2(\Omega; H^1(Y))$

b. $\mathcal{T}_{\varepsilon}(\nabla\phi_{\varepsilon}) \to \nabla_x\phi + \nabla_y\hat{\phi}$ in $L^2(\Omega \times Y)$

VIII. Let $\phi_{\varepsilon} \in H^1(\Omega)$ and assume that $\{\phi_{\varepsilon}\}_{\varepsilon}$ is a bounded sequence in $L^2(\Omega)$ satisfying $\varepsilon \|\nabla\phi_{\varepsilon}\|_{L^2(\Omega; \mathbb{R}^3)} \le c$ ($c$ is a constant independent of $\varepsilon$) then there exists a non-relabelled subsequence and a $\hat{\phi} \in L^2(\Omega; H_{\text{per}}^1(Y))$ such that

a. $\mathcal{T}_{\varepsilon}(\phi_{\varepsilon}) \to \hat{\phi}$ in $L^2(\Omega; H^1(Y))$

b. $\varepsilon \mathcal{T}_{\varepsilon}(\nabla\phi_{\varepsilon}) \to \nabla_y\hat{\phi}$ in $L^2(\Omega \times Y)$

## 5 Presentation and discussion of the results

**Theorem 5.1.** If $(\mathbf{u}^{\varepsilon}, \varphi^{\varepsilon})$ is the solution set to (3.1) under the (HS 1) scaling then there exist $\mathbf{u}^0 \in H_{\Sigma}^1(\Omega; \mathbb{R}^3)$, $\mathbf{u}^1 \in L^2(\Omega; H_{\text{per}}^1(Y; \mathbb{R}^3))$, $\varphi^0 \in L^2(\Omega; H_{\text{per}}^1(Y; \mathbb{R}^3))$ such that,

$$ \mathcal{T}_{\varepsilon}(\mathbf{u}^{\varepsilon}) \to \mathbf{u}^0 \text{ in } L^2(\Omega; H^1(Y; \mathbb{R}^3)), \quad (5.1) $$

$$ \mathcal{T}_{\varepsilon}(\nabla \mathbf{u}^{\varepsilon}) \to \nabla_x \mathbf{u}^0 + \nabla_y \mathbf{u}^1 \text{ in } L^2(\Omega \times Y; \mathbb{R}^{3 \times 3}), \quad (5.2) $$

$$ \mathcal{T}_{\varepsilon}(\varphi^{\varepsilon}) \to \varphi^0 \text{ in } L^2(\Omega; H^1(Y; \mathbb{R}^3)), \quad (5.3) $$

$$ \mathcal{T}_{\varepsilon}(\varepsilon \nabla \varphi^{\varepsilon}) \to \nabla_y \varphi^0 \text{ in } L^2(\Omega \times Y; \mathbb{R}^{3 \times 3}), \quad (5.4) $$

and $(\mathbf{u}^0, \varphi^0, \mathbf{u}^1)$ is the unique solution set of,

$$ \begin{aligned} & \int_{\Omega \times Y} \left[ \mathbf{E}(\mathbf{y})(\nabla_x \mathbf{u}^0 + \nabla_y \mathbf{u}^1 - \varepsilon \cdot \varphi^0) + \frac{1}{2} \varepsilon \cdot \mathbf{g} \right] : (\nabla_x \mathbf{V} + \nabla_y \overline{\mathbf{W}}) \, dy \, dx \\ & - \int_{\Omega \times Y} \left[ \mathbf{E}(\mathbf{y})(\nabla_x \mathbf{u}^0 + \nabla_y \mathbf{u}^1 - \varepsilon \cdot \varphi^0) + \frac{1}{2} \varepsilon \cdot \mathbf{g} \right] : \varepsilon \cdot \mathbf{\Psi} \, dy \, dx \quad (5.5) \\ & - \int_{\Omega \times Y} (\mathbf{f} \cdot \mathbf{V} + \frac{1}{2} \varepsilon \cdot \mathbf{g} (\nabla_x \mathbf{V} + \nabla_y \overline{\mathbf{W}})) \, dy \, dx = 0. \end{aligned} $$
$$ \int_{\Omega} \sigma^{\text{eff}} : \nabla_x V \, dx = \frac{1}{2} \int_{\Omega} g \cdot \text{curl}_x V \, dx + \int_{\Omega} f \cdot V \, dx, \qquad (5.6) $$

$$ E_{jipq}^{\text{eff}} := \int_{Y} E_{jikl}(y) \left( \delta_{kp} \delta_{q\ell} + \partial_{y\ell} \zeta_{k}^{pq} - \epsilon_{\nu k \ell} \xi_{\nu}^{pq} \right) dy, \qquad (5.7) $$

$$ \begin{gathered} \zeta^{pq} \in H_{\text{per}}^{1}(Y; \mathbb{R}^{3}), \quad \int_{Y} \zeta^{pq} dy = 0, \qquad (5.8) \\ \int_{Y} E(y) \left( \frac{1}{2} e_{p} \otimes e_{q} + \nabla_{y} \zeta^{pq} \right) : \nabla_{y} v dy = 0, \end{gathered} $$

$$ \begin{gathered} \xi^{pq} \in H_{\text{per}}^{1}(Y; \mathbb{R}^{3}), \\ \int_{Y} E(y) \left( \frac{1}{2} e_{p} \otimes e_{q} + \epsilon \cdot \xi^{pq} \right) : \nabla_{y} v dy = 0, \end{gathered} \qquad (5.9) $$

$$ E_{ijkl}^{\text{eff}} \partial_{x_l} u_k^0 = E_{jikl}^{\text{eff}} \partial_{x_l} u_k^0 + \epsilon_{\nu j i} g_{\nu}. \qquad (5.10) $$

<footnote n="246">246 for all $(V, \Psi) \in (H_{\Sigma}^1(\Omega; \mathbb{R}^3), L^2(\Omega, \mathbb{R}^3))$ and $\overline{W} \in L^2(\Omega; H^1(Y; \mathbb{R}^3))$.</footnote>

<footnote n="247">247 Equation (5.5) can be put in a more familiar form,</footnote>

<footnote n="248">248 with $\sigma^{\text{eff}} := (E^{\text{eff}} \nabla_x u^0 + \frac{1}{2} \epsilon \cdot g)$ the Cauchy stress,</footnote>

<footnote n="249">249 the effective coefficients, and $\zeta^{k\ell}$ and $\xi^{k\ell}$ the local solutions satisfying the problems,</footnote>

<footnote n="250">250 for all $v \in H_{\text{per}}^{1}(Y; \mathbb{R}^{3})$.</footnote>

<footnote n="251">251 for all $v \in H_{\text{per}}^{1}(Y; \mathbb{R}^{3})$.</footnote>

<footnote n="252">252 The limit problem (5.5) of Theorem 5.1 is non-local in the sense that we transition from a Cosserat continuum to a classical Cauchy continuum. Therefore, when all scales are well-separated the limiting effective system is a classical linear elastic system. The novel part of the derived system herein lies in the construction of the effective coefficients. We compute two local problems in order to construct the effective coefficients. Local problem (6.14) has a non-standard form compared to classical local problems in the theory of homogenization and represents the contribution the local rotations make to the effective coefficients. Furthermore, when we pass to the limit the couple-stress equation reduces to the conservation of angular momentum of the effective system. This implies that $\sigma^{\text{eff}} = (\sigma^{\text{eff}})^{\top}$ as expected since the effective system is a classical Cauchy continuum. Hence, the symmetry of the Cauchy stress implies that,</footnote>

<footnote n="263">263 It appears that the body couples $g$ make a contribution to the off-diagonal terms of the Cauchy stress. However, if $g \equiv 0$ then the effective coefficients have the symmetry,</footnote>
$$ E_{ijkl}^{\text{eff}} = E_{jikl}^{\text{eff}} = E_{klji}^{\text{eff}} \qquad (5.11) $$

265 and stress tensor has the more classical form,

$$ \sigma^{\text{eff}} = E^{\text{eff}} e(\mathbf{u}^0) \quad \text{where} \quad e(\mathbf{u}^0) := \frac{1}{2}(\nabla_x \mathbf{u}^0 + (\nabla_x \mathbf{u}^0)^\top). \qquad (5.12) $$

266 Lastly, we remark that one can approximate a linear elastic material using a different theories of generalized continua, e.g., second-gradient elasticity (see [32]). It is unclear 267 what the benefits would be of using one theory over another. From a mathematical point 268 of view, second-gradient elasticity seems to be favored more since it provides better solution 269 regularity. 270

271 **Theorem 5.2.** If $(\mathbf{u}^\varepsilon, \varphi^\varepsilon)$ is the solution set to (3.1) under the (HS 2) scaling then there 272 exist $\mathbf{u}^0 \in H_\Sigma^1(\Omega; \mathbb{R}^3)$, $\mathbf{u}^1 \in L^2(\Omega; H_{\text{per}}^1(Y; \mathbb{R}^3))$, $\varphi^0 \in H_\Sigma^1(\Omega; \mathbb{R}^3)$, $\varphi^1 \in L^2(\Omega; H_{\text{per}}^1(Y; \mathbb{R}^3))$ 273 such that,

$$ \mathcal{T}_\varepsilon(\mathbf{u}^\varepsilon) \to \mathbf{u}^0 \text{ in } L^2(\Omega; H^1(Y; \mathbb{R}^3)), \qquad (5.13) $$

$$ \mathcal{T}_{\varepsilon}(\nabla \mathbf{u}^{\varepsilon}) \to \nabla_{x} \mathbf{u}^{0} + \nabla_{y} \mathbf{u}^{1} \text{ in } L^{2}(\Omega \times Y; \mathbb{R}^{3 \times 3}), \qquad (5.14) $$

$$ \mathcal{T}_{\varepsilon}(\varphi^{\varepsilon}) \to \varphi^{0} \text{ in } L^{2}(\Omega; H^{1}(Y; \mathbb{R}^{3})), \qquad (5.15) $$

$$ \mathcal{T}_{\varepsilon}(\nabla \varphi^{\varepsilon}) \to \nabla_{x} \varphi^{0} + \nabla_{y} \varphi^{1} \text{ in } L^{2}(\Omega \times Y; \mathbb{R}^{3 \times 3}), \qquad (5.16) $$

274 and $(\mathbf{u}^0, \mathbf{u}^1, \varphi^0, \varphi^1)$ is the unique solution set of

$$ \int_{\Omega \times Y} \mathbf{E}(\mathbf{y}) (\nabla_x \mathbf{u}^0 + \nabla_y \mathbf{u}^1 - \boldsymbol{\epsilon} \cdot \mathbf{\varphi}^0) : (\nabla_x \mathbf{V} + \nabla_y \mathbf{\bar{W}}) \, dy \, dx \\ - \int_{\Omega \times Y} \mathbf{M}(\mathbf{y}) (\nabla_x \mathbf{\varphi}^0 + \nabla_y \mathbf{\varphi}^1) (\nabla_x \mathbf{\Psi} + \nabla_y \mathbf{\Xi}) \, dy \, dx \qquad (5.17) \\ - \int_{\Omega \times Y} (\mathbf{f} \cdot \mathbf{V} + \mathbf{g} \cdot \mathbf{\Psi}) \, dy \, dx = 0, $$

275 for all $(\mathbf{V}, \mathbf{\Psi}) \in H_\Sigma^1(\Omega; \mathbb{R}^3) \times H_\Sigma^1(\Omega; \mathbb{R}^3)$ and $(\mathbf{\bar{W}}, \mathbf{\Xi}) \in L^2(\Omega; H^1(Y; \mathbb{R}^3)) \times L^2(\Omega; H^1(Y; \mathbb{R}^3))$.

276 Equation (5.17) can be put in a more familiar form,

$$ \int_{\Omega} \sigma^{\text{eff}} : (\nabla_x \mathbf{V} - \boldsymbol{\epsilon} \cdot \mathbf{\Psi}) \, d\mathbf{x} - \int_{\Omega} \mu^{\text{eff}} : \nabla_x \mathbf{\Psi} \, d\mathbf{x} = \int_{\Omega} (\mathbf{f} \cdot \mathbf{V} + \mathbf{g} \cdot \mathbf{\Psi}) \, d\mathbf{x}, \qquad (5.18) $$

277 with

$$ \begin{bmatrix} \sigma^{\text{eff}} \\ \mu^{\text{eff}} \end{bmatrix} = \begin{bmatrix} \mathbf{E}^{\text{eff}} & 0 \\ 0 & \mathbf{M}^{\text{eff}} \end{bmatrix} \begin{bmatrix} \gamma^0 \\ \kappa^0 \end{bmatrix} \qquad (5.19) $$
$$E_{jipq}^{\text{eff}} = \int_Y E_{jikl}(y)(\delta_{kp} \delta_{q\ell} + \partial_{y\ell} \zeta_k^{pq}) dy, \quad (5.20)$$

$$M_{jipq}^{\text{eff}} = \int_Y M_{jikl}(y)(\delta_{kp} \delta_{q\ell} + \partial_{y\ell} \xi_k^{pq}) dy, \quad (5.21)$$

$$\begin{gathered} \zeta^{pq} \in H_{\text{per}}^{1}(Y; \mathbb{R}^{3}), \quad \int_{Y} \zeta^{pq} dy = 0, \tag{5.22} \\ \int_{Y} E(y) (e_{p} \otimes e_{q} + \nabla_{y} \zeta^{pq}) : \nabla_{y} w dy = 0, \end{gathered}$$

$$\begin{gathered} \xi^{pq} \in H_{\text{per}}^{1}(Y; \mathbb{R}^{3}), \quad \int_{Y} \xi^{pq} dy = 0, \tag{5.23} \\ \int_{Y} M(y)(e_{p} \otimes e_{q} + \nabla_{y} \xi^{pq}) : \nabla_{y} v dy = 0, \end{gathered}$$

## 6 Proofs of the main results

### 6.1 Proof of Theorem 5.1

$$\mathcal{T}_{\varepsilon}(u^{\varepsilon}) \to u^{0} \text{ in } L^{2}(\Omega; H^{1}(Y; \mathbb{R}^{3})), \quad (5.1)$$

$$\mathcal{T}_{\varepsilon}(\nabla u^{\varepsilon}) \to \nabla_x u^0 + \nabla_y u^1 \text{ in } L^2(\Omega \times Y; \mathbb{R}^{3 \times 3}), \quad (5.2)$$

$$\mathcal{T}_{\varepsilon}(\varphi^{\varepsilon}) \to \varphi^{0} \text{ in } L^{2}(\Omega; H^{1}(Y; \mathbb{R}^{3})), \quad (5.3)$$

$$\mathcal{T}_{\varepsilon}(\varepsilon \nabla \varphi^{\varepsilon}) \to \nabla_y \varphi^0 \text{ in } L^2(\Omega \times Y; \mathbb{R}^{3 \times 3}), \quad (5.4)$$

<footnote n="278">278 the effective stress and effective couple-stress, the Cosserat strain $\gamma^0 := (\nabla_x u^0 - \epsilon \cdot \varphi^0)$ and</footnote>

<footnote n="279">279 curvature-twist $\kappa^0 := \nabla_x \varphi^0$, the effective coefficients,</footnote>

<footnote n="280">280 and $\zeta^{pq}$ and $\xi^{pq}$ the local solutions on the unit cell satisfying the second order elliptic</footnote>

<footnote n="281">281 problems,</footnote>

<footnote n="282">282 for all $w \in H_{\text{per}}^{1}(Y; \mathbb{R}^{3})$,</footnote>

<footnote n="283">283 for all $v \in H_{\text{per}}^{1}(Y; \mathbb{R}^{3})$.</footnote>

<footnote n="284">284 The limit problem (5.17) of Theorem 5.2 is that of a centro-symmetric Cosserat elasticity.</footnote>

<footnote n="285">285 We remark that in this case not all scales are fully separated. This shows the true power</footnote>

<footnote n="286">286 of our approach. We are able to derive an effective equation even-through certain scales</footnote>

<footnote n="287">287 are not well separated. Moreover, the structure of the corrector problem (5.23) differs</footnote>

<footnote n="288">288 substantially from the corrector problem (6.14).</footnote>

<footnote n="291">291 Theorem 5.1. If $(u^\varepsilon, \varphi^\varepsilon)$ is the solution set to (3.1) under the (HS 1) scaling then there</footnote>

<footnote n="292">292 exist $u^0 \in H_{\Sigma}^1(\Omega; \mathbb{R}^3)$, $u^1 \in L^2(\Omega; H_{\text{per}}^1(Y; \mathbb{R}^3))$, $\varphi^0 \in L^2(\Omega; H_{\text{per}}^1(Y; \mathbb{R}^3))$ such that,</footnote>
293 and $(\mathbf{u}^0, \varphi^0, \mathbf{u}^1)$ is the unique solution set of,

$$
\begin{align}
& \int_{\Omega \times Y} \left[ \mathbf{E}(\mathbf{y})(\nabla_x \mathbf{u}^0 + \nabla_y \mathbf{u}^1 - \epsilon \cdot \varphi^0) + \frac{1}{2} \epsilon \cdot g \right] : (\nabla_x \mathbf{V} + \nabla_y \overline{\mathbf{W}}) \, dy \, dx \nonumber \\
& - \int_{\Omega \times Y} \left[ \mathbf{E}(\mathbf{y})(\nabla_x \mathbf{u}^0 + \nabla_y \mathbf{u}^1 - \epsilon \cdot \varphi^0) + \frac{1}{2} \epsilon \cdot g \right] : \epsilon \cdot \mathbf{\Psi} \, dy \, dx \tag{5.5} \\
& - \int_{\Omega \times Y} (\mathbf{f} \cdot \mathbf{V} + \frac{1}{2} \epsilon \cdot g (\nabla_x \mathbf{V} + \nabla_y \overline{\mathbf{W}})) \, dy \, dx = 0. \nonumber
\end{align}
$$

294 for all $(\mathbf{V}, \mathbf{\Psi}) \in (H_{\Sigma}^1(\Omega; \mathbb{R}^3), L^2(\Omega, \mathbb{R}^3))$ and $\overline{\mathbf{W}} \in L^2(\Omega; H^1(Y; \mathbb{R}^3))$.

295 Equation (5.5) can be put in a more familiar form,

$$
\int_{\Omega} \sigma^{\text{eff}} : \nabla_x \mathbf{V} \, d\mathbf{x} = \frac{1}{2} \int_{\Omega} \mathbf{g} \cdot \text{curl}_x \mathbf{V} \, d\mathbf{x} + \int_{\Omega} \mathbf{f} \cdot \mathbf{V} \, d\mathbf{x}, \quad (5.6)
$$

296 with $\sigma^{\text{eff}} := (\mathbf{E}^{\text{eff}} \nabla_x \mathbf{u}^0 + \frac{1}{2} \epsilon \cdot \mathbf{g})$ the Cauchy stress,

$$
\mathbf{E}_{jipq}^{\text{eff}} := \int_Y \mathbf{E}_{jikl}(\mathbf{y}) \left( \delta_{kp} \delta_{ql} + \partial_{y_l} \zeta_k^{pq} - \epsilon_{\nu k l} \xi_{\nu}^{pq} \right) \, d\mathbf{y}, \quad (5.7)
$$

297 the effective coefficients, and $\zeta^{kl}$ and $\xi^{kl}$ the local solutions satisfying the problems,

$$
\begin{align}
\zeta^{pq} \in H_{\text{per}}^{1}(Y; \mathbb{R}^{3}), \quad & \int_{Y} \zeta^{pq} \, dy = 0, \tag{5.8} \\
\int_{Y} \mathbf{E}(\mathbf{y}) \left( \frac{1}{2} \mathbf{e}_{p} \otimes \mathbf{e}_{q} + \nabla_{y} \zeta^{pq} \right) : \nabla_{y} \mathbf{v} \, dy = 0, \nonumber
\end{align}
$$

298 for all $\mathbf{v} \in H_{\text{per}}^{1}(Y; \mathbb{R}^{3})$.

$$
\begin{align}
\xi^{pq} \in H_{\text{per}}^{1}(Y; \mathbb{R}^{3}), \nonumber \\
\int_{Y} \mathbf{E}(\mathbf{y}) \left( \frac{1}{2} \mathbf{e}_p \otimes \mathbf{e}_q + \epsilon \cdot \xi^{pq} \right) : \nabla_y \mathbf{v} \, d\mathbf{y} = 0, \tag{5.9}
\end{align}
$$

299 for all $\mathbf{v} \in H_{\text{per}}^{1}(Y; \mathbb{R}^{3})$.

300 *Proof.* Using the estimates from (3.11) and the results from Proposition 4.1 *VII.* and
301 *VIII.* we obtain (5.1)–(5.4). Moreover, the weak form (3.1) of the Cosserat continuum
302 reads:

303 Find $(\mathbf{u}^{\epsilon}, \varphi^{\epsilon}) \in H_{\Sigma}^{1}(\Omega; \mathbb{R}^{3}) \times H_{\Sigma}^{1}(\Omega; \mathbb{R}^{3})$ such that,
$$
\begin{equation}
\begin{aligned}
& \int_{\Omega} \mathrm{E}\left(\frac{\boldsymbol{x}}{\varepsilon}\right)(\nabla \boldsymbol{u}^{\varepsilon}-\boldsymbol{\epsilon} \cdot \boldsymbol{\varphi}^{\varepsilon}):(\nabla \boldsymbol{v}-\boldsymbol{\epsilon} \cdot \boldsymbol{\psi}) \, d\boldsymbol{x}-\int_{\Omega} \varepsilon^{2} \mathrm{C}\left(\frac{\boldsymbol{x}}{\varepsilon}\right) \nabla \boldsymbol{\varphi}^{\varepsilon}:(\nabla \boldsymbol{v}-\boldsymbol{\epsilon} \cdot \boldsymbol{\psi}) \, d\boldsymbol{x} \\
& \quad-\int_{\Omega} \varepsilon^{2} \mathrm{C}\left(\frac{\boldsymbol{x}}{\varepsilon}\right)(\nabla \boldsymbol{u}^{\varepsilon}-\boldsymbol{\epsilon} \cdot \boldsymbol{\varphi}^{\varepsilon}):\nabla \boldsymbol{\psi} \, d\boldsymbol{x}-\int_{\Omega} \varepsilon^{2} \mathrm{M}\left(\frac{\boldsymbol{x}}{\varepsilon}\right) \nabla \boldsymbol{\varphi}^{\varepsilon}:\nabla \boldsymbol{\psi} \, d\boldsymbol{x} \\
& \quad-\int_{\Omega}(\boldsymbol{f} \cdot \boldsymbol{v}+\boldsymbol{g} \cdot \boldsymbol{\psi}) \, d\boldsymbol{x}=0.
\end{aligned}
\tag{6.1}
\end{equation}
$$

304 Unfold the above expression using Proposition 4.1 properties I., II., and IV. and obtain,

$$
\begin{equation}
\begin{aligned}
& \int_{\Omega \times Y} \mathrm{E}(\boldsymbol{y})(\mathcal{T}_{\varepsilon}(\nabla \boldsymbol{u}^{\varepsilon}) - \boldsymbol{\epsilon} \cdot \mathcal{T}_{\varepsilon}(\boldsymbol{\varphi}^{\varepsilon}):(\mathcal{T}_{\varepsilon}(\nabla \boldsymbol{v}) - \boldsymbol{\epsilon} \cdot \mathcal{T}_{\varepsilon}(\boldsymbol{\psi})) \, d\boldsymbol{y} \, d\boldsymbol{x} \\
& \quad - \int_{\Omega \times Y} \varepsilon^2 \mathrm{C}(\boldsymbol{y}) \mathcal{T}_{\varepsilon}(\nabla \boldsymbol{\varphi}^{\varepsilon}):(\mathcal{T}_{\varepsilon}(\nabla \boldsymbol{v}) - \boldsymbol{\epsilon} \cdot \mathcal{T}_{\varepsilon}(\boldsymbol{\psi})) \, d\boldsymbol{x} \, d\boldsymbol{y} \\
& \quad - \int_{\Omega \times Y} \varepsilon^2 \mathrm{C}(\boldsymbol{y}) (\mathcal{T}_{\varepsilon}(\nabla \boldsymbol{u}^{\varepsilon}) - \boldsymbol{\epsilon} \cdot \mathcal{T}_{\varepsilon}(\boldsymbol{\varphi}^{\varepsilon})):\mathcal{T}_{\varepsilon}(\nabla \boldsymbol{\psi}) \, d\boldsymbol{x} \, d\boldsymbol{y} \\
& \quad - \int_{\Omega \times Y} \varepsilon^2 \mathrm{M}(\boldsymbol{y}) \nabla \mathcal{T}_{\varepsilon}(\boldsymbol{\varphi}^{\varepsilon}):\mathcal{T}_{\varepsilon}(\nabla \boldsymbol{\psi}) \, d\boldsymbol{y} \, d\boldsymbol{x} \\
& \quad - \int_{\Omega \times Y} (\mathcal{T}_{\varepsilon}(\boldsymbol{f}) \cdot \mathcal{T}_{\varepsilon}(\boldsymbol{v}) + \mathcal{T}_{\varepsilon}(\boldsymbol{g}) \cdot \mathcal{T}_{\varepsilon}(\boldsymbol{\psi})) \, d\boldsymbol{y} \, d\boldsymbol{x} = 0.
\end{aligned}
\tag{6.2}
\end{equation}
$$

305 Set $(\mathbf{v}, \mathbf{\psi}) := (\mathbf{V}(\mathbf{x}), \mathbf{\Psi}(\mathbf{x}))$ for any test functions $(\mathbf{V}, \mathbf{\Psi}) \in C_0^\infty(\Omega; \mathbb{R}^3) \times C_0^\infty(\Omega; \mathbb{R}^3)$ in (6.2)
306 and let $\varepsilon \to 0$ to obtain,

$$
\int_{\Omega \times Y} \mathrm{E}(\boldsymbol{y}) (\nabla_x \boldsymbol{u}_k^0 + \nabla_y \boldsymbol{u}^1 - \boldsymbol{\epsilon} \cdot \boldsymbol{\varphi}^0) : (\nabla_x \boldsymbol{V} - \boldsymbol{\epsilon} \cdot \boldsymbol{\Psi}) \, d\boldsymbol{y} \, d\boldsymbol{x} - \int_{\Omega \times Y} (\boldsymbol{f} \cdot \boldsymbol{V} + \boldsymbol{g} \cdot \boldsymbol{\Psi}) \, d\boldsymbol{x} = 0. \quad (6.3)
$$

307 Select now a test function of the form $(\mathbf{v}, \mathbf{\psi})=(\mathbf{v}^{\varepsilon}, \mathbf{\psi}^{\varepsilon}) := (\varepsilon U(\mathbf{x}) \mathbf{W}(\frac{\mathbf{x}}{\varepsilon}), \varepsilon \Phi(\mathbf{x}) \Xi(\frac{\mathbf{x}}{\varepsilon}))$ where $(U, \Phi) \in C_0^\infty(\Omega) \times C_0^\infty(\Omega)$ and $(\mathbf{W}, \Xi) \in \mathrm{H}_{\mathrm{per}}^1(Y; \mathbb{R}^3) \times \mathrm{H}_{\mathrm{per}}^1(Y; \mathbb{R}^3)$. It is immediately clear that $(\mathbf{v}^{\varepsilon}, \mathbf{\psi}^{\varepsilon}) \to (\mathbf{0}, \mathbf{0})$ in $L^2(\Omega; \mathbb{R}^3) \times L^2(\Omega; \mathbb{R}^3)$. Moreover, for $\mathbf{v}^{\varepsilon}$ we have

$$
\partial_{x_j} v_i^{\epsilon}(\mathbf{x}) = \epsilon \partial_{x_j} U(\mathbf{x}) W_i(\mathbf{x}/\epsilon) + U(\mathbf{x}) \partial_{y_j} W_i(\mathbf{x}/\epsilon) \quad (6.4)
$$

310 which implies $\mathcal{T}_\epsilon(\partial_{x_j} v_i^\epsilon) \to \partial_{y_j} \overline{W}_i(\mathbf{x}, \mathbf{y})$ in $L^2(\Omega \times Y)$ as $\epsilon \to 0$ where $\overline{W}_i(\mathbf{x}, \mathbf{y}) := U(\mathbf{x}) W_i(\mathbf{y})$.
311 Likewise, for $\psi v^\epsilon$, we have

$$
\partial_{x_j} \psi_i^\epsilon(\mathbf{x}) = \epsilon \partial_{x_j} \Phi(\mathbf{x}) \Xi_i(\mathbf{x}/\epsilon) + \Phi(\mathbf{x}) \partial_{y_j} \Xi_i(\mathbf{x}/\epsilon) \quad (6.5)
$$

312 which implies $\mathcal{T}_\epsilon(\partial_{x_j} \psi_i^\epsilon) \to \partial_{y_j} \Xi_i(\mathbf{x}, \mathbf{y})$ in $L^2(\Omega \times Y)$ as $\epsilon \to 0$ where $\Xi_i(\mathbf{x}, \mathbf{y}) := \Phi(\mathbf{x}) \Xi_i(\mathbf{y})$.

313 Hence, unfolding (6.2) with the test function $(\mathbf{v}^{\varepsilon}, \mathbf{\psi}^{\varepsilon})$ and letting $\varepsilon$ tend to zero we obtain,
$$ \int_{\Omega \times Y} E(y) (\nabla_x u^0 + \nabla_y u^1 - \epsilon \cdot \varphi^0) : \nabla_y \bar{W}(x, y) dy dx = 0. \quad (6.6) $$

314

315

Adding (6.3) and (6.6) we arrive at:

$$ \int_{\Omega \times Y} E(y)(\nabla_x u^0 + \nabla_y u^1 - \epsilon \cdot \varphi^0) : (\nabla_x V + \nabla_y \bar{W} - \epsilon \cdot \Psi) dy dx = \int_{\Omega \times Y} (f \cdot V + g \cdot \Psi) dy dx. \quad (6.7) $$

316

317

By the density of $C_0^\infty(\Omega) \otimes H_{\text{per}}^1(Y; \mathbb{R}^3)$ in $L^2(\Omega; H_{\text{per}}^1(Y; \mathbb{R}^3))$ the result holds for all $(\bar{W}(x, y), \bar{\Xi}(x, y)) \in L^2(\Omega; H_{\text{per}}^1(Y; \mathbb{R}^3)) \times L^2(\Omega; H_{\text{per}}^1(Y; \mathbb{R}^3))$.

318

319

320

In order to better understand what happens to the balance of linear momentum and angular momentum, we re-write the body couple force as follows: $g_i = \frac{1}{2}\epsilon_{ipq} \epsilon_{jpq} g_j$, $i = 1, 2, 3$. Then, grouping the terms in (6.7) by test function we obtain,

$$ \begin{aligned} & \int_{\Omega \times Y} \left[ E(y)(\nabla_x u^0 + \nabla_y u^1 - \epsilon \cdot \varphi^0) + \frac{1}{2} \epsilon \cdot g \right] : (\nabla_x V + \nabla_y \bar{W}) dy dx \\ & - \int_{\Omega \times Y} \left[ E(y)(\nabla_x u^0 + \nabla_y u^1 - \epsilon \cdot \varphi^0) + \frac{1}{2} \epsilon \cdot g \right] : \epsilon \cdot \Psi dy dx \\ & - \int_{\Omega \times Y} (f \cdot V + \frac{1}{2} \epsilon \cdot g (\nabla_x V + \nabla_y \bar{W})) dy dx = 0. \end{aligned} \quad (6.8) $$

321

322

323

From here we can obtain two sets of equations. By considering, first, that $(V, \bar{W}, \Psi) = (V, \bar{W}, 0)$ we have,

$$ \begin{aligned} & \int_{\Omega \times Y} \left[ E(y)(\nabla_x u^0 + \nabla_y u^1 - \epsilon \cdot \varphi^0) + \frac{1}{2} \epsilon \cdot g \right] : (\nabla_x V + \nabla_y \bar{W}) dy dx \\ & - \int_{\Omega \times Y} (f \cdot V + \frac{1}{2} \epsilon \cdot g (\nabla_x V + \nabla_y \bar{W})) dy dx = 0. \end{aligned} \quad (6.9) $$

324

325

This is precisely the balance of linear momentum. By considering, $(V, \bar{W}, \Psi) = (0, 0, \Psi)$ we have,

$$ - \int_{\Omega \times Y} \left[ E(y)(\nabla_x u^0 + \nabla_y u^1 - \epsilon \cdot \varphi^0) + \frac{1}{2} \epsilon \cdot g \right] : \epsilon \cdot \Psi dy dx = 0. \quad (6.10) $$

326

This is the balance of angular momentum.

327

328

If in (6.9) we select $V \equiv 0$ we can see that both $u^1$ and $\varphi^0$ depend on $\nabla_x u^0$ linearly. We postulate that the form of $u^1$ and $\varphi^0$, respectively, to be:
$$
\mathbf{u}^1(\mathbf{x}, \mathbf{y}) = \zeta^{k\ell}(\mathbf{y}) \partial_{x_\ell} u_k^0(\mathbf{x}) + \mathbf{c}(\mathbf{x}), \qquad (6.11)
$$

$$
\varphi^0(\mathbf{x}, \mathbf{y}) = \xi^{k\ell}(\mathbf{y}) \partial_{x_\ell} u_k^0(\mathbf{x}). \qquad (6.12)
$$

Moreover, since at the local level the rotations are independent of the displacements, we further hypothesize that $\mathbf{u}^1$ does not make a contribution to $\varphi^0$ and that $\varphi^0$ does not make a contribution to $\mathbf{u}^1$. However, they both contribute to the effective coefficient. Hence, the correctors $\zeta^{k\ell}$ and $\xi^{k\ell}$ satisfy the following problems,

$$
\begin{equation}
\begin{aligned}
& \zeta^{pq} \in \mathrm{H}_{\mathrm{per}}^{1}(Y; \mathbb{R}^{3}), \quad \int_{Y} \zeta^{pq} \, dy = 0, \\
& \int_{Y} \mathrm{E}(y) \left( \frac{1}{2} e_{p} \otimes e_{q} + \nabla_{y} \zeta^{pq} \right) : \nabla_{y} \mathbf{v} \, dy = 0,
\end{aligned}
\tag{6.13}
\end{equation}
$$

333 for all $\mathbf{v} \in \mathrm{H}_{\mathrm{per}}^{1}(Y; \mathbb{R}^{3})$.

$$
\xi^{pq} \in \mathbf{H}_{\text{per}}^{1}(Y; \mathbb{R}^{3}), \quad \int_{Y} \mathbf{E}(\mathbf{y}) \left( \frac{1}{2} \mathbf{e}_{p} \otimes \mathbf{e}_{q} + \boldsymbol{\epsilon} \cdot \boldsymbol{\xi}^{pq} \right) : \nabla_{y} \mathbf{v} \, d\mathbf{y} = 0, \quad (6.14)
$$

334 for all $\mathbf{v} \in \mathbf{H}_{\text{per}}^{1}(Y; \mathbb{R}^{3})$.

335 Existence and uniqueness for (6.13) (up to a constant) follows from classical theory of
336 variational inequalities [23]. While for problem (6.14) one can show existence as in [24].

337 Returning to (6.9) and substituting $\overline{\mathbf{W}} = \mathbf{0}$, $\mathbf{u}^1$ and $\boldsymbol{\varphi}^0$ from (6.11) and (6.12) respectively,
338 we obtain,

$$
\int_{\Omega} \sigma^{\text{eff}} : \nabla_x \mathbf{V} \, d\mathbf{x} = \frac{1}{2} \int_{\Omega} \boldsymbol{\epsilon} \cdot \mathbf{g} : \nabla_x \mathbf{V} \, d\mathbf{x} + \int_{\Omega} \mathbf{f} \cdot \mathbf{V} \, d\mathbf{x}, \quad (6.15)
$$

339 where

$$
\sigma^{\text{eff}} := \mathbf{E}^{\text{eff}} \nabla_x \mathbf{u}^0 + \frac{1}{2} \boldsymbol{\epsilon} \cdot \mathbf{g}, \qquad (6.16)
$$

340 and

$$
\mathbf{E}_{jipq}^{\text{eff}} = \int_{Y} \mathbf{E}_{jik\ell}(\mathbf{y}) \left( \delta_{kp} \delta_{q\ell} + \partial_{y_\ell} \zeta_k^{pq} - \epsilon_{\nu k\ell} \xi_\nu^{pq} \right) d\mathbf{y}. \quad (6.17)
$$

341 The stress recovered, $\sigma^{\text{eff}}$, is symmetric. To see this, select $\mathbf{u}^1$, $\boldsymbol{\varphi}^0$ as in (6.11), (6.12),
342 respectively with $\boldsymbol{\Psi}$ an arbitrary test function, and substitute them in (6.10) to obtain,

$$
- \int_{\Omega} \sigma^{\text{eff}} : \boldsymbol{\epsilon} \cdot \boldsymbol{\Psi} \, d\mathbf{x} = - \int_{\Omega} (\boldsymbol{\epsilon} : \sigma^{\text{eff}}) \cdot \boldsymbol{\Psi} \, d\mathbf{x} = 0. \quad (6.18)
$$

343 Hence, locally we have $\boldsymbol{\epsilon} : \boldsymbol{\sigma}^{\text{eff}} = \mathbf{0}$ ($\boldsymbol{\epsilon}_{ijk} \boldsymbol{\sigma}_{jk}^{\text{eff}} = 0$, $i = 1, 2, 3$) which implies that $\boldsymbol{\sigma}^{\text{eff}} = (\boldsymbol{\sigma}^{\text{eff}})^{\top}$.

344

<figure><img src="image_6.png" /></figure>
## 345 6.2 Proof of Theorem 5.2

Theorem 5.2. If $(\mathbf{u}^\epsilon, \varphi^\epsilon)$ is the solution set to (3.1) under the (HS 2) scaling then there exist $\mathbf{u}^0 \in H_\Sigma^1(\Omega; \mathbb{R}^3)$, $\mathbf{u}^1 \in L^2(\Omega; H_{\text{per}}^1(Y; \mathbb{R}^3))$, $\varphi^0 \in H_\Sigma^1(\Omega; \mathbb{R}^3)$, $\varphi^1 \in L^2(\Omega; H_{\text{per}}^1(Y; \mathbb{R}^3))$ such that,

$$ \mathcal{T}_{\epsilon}(\mathbf{u}^{\epsilon}) \to \mathbf{u}^{0} \text{ in } L^{2}(\Omega; H^{1}(Y; \mathbb{R}^{3})), \qquad (5.13) $$

$$ \mathcal{T}_{\epsilon}(\nabla \mathbf{u}^{\epsilon}) \to \nabla_{x}\mathbf{u}^{0} + \nabla_{y}\mathbf{u}^{1} \text{ in } L^{2}(\Omega \times Y; \mathbb{R}^{3 \times 3}), \qquad (5.14) $$

$$ \mathcal{T}_{\epsilon}(\varphi^{\epsilon}) \to \varphi^{0} \text{ in } L^{2}(\Omega; H^{1}(Y; \mathbb{R}^{3})), \qquad (5.15) $$

$$ \mathcal{T}_{\epsilon}(\nabla \varphi^{\epsilon}) \to \nabla_{x}\varphi^{0} + \nabla_{y}\varphi^{1} \text{ in } L^{2}(\Omega \times Y; \mathbb{R}^{3 \times 3}), \qquad (5.16) $$

and $(\mathbf{u}^0, \mathbf{u}^1, \varphi^0, \varphi^1)$ is the unique solution set of

$$ \begin{aligned} & \int_{\Omega \times Y} \mathrm{E}(\mathbf{y}) (\nabla_x \mathbf{u}^0 + \nabla_y \mathbf{u}^1 - \epsilon \cdot \varphi^0) : (\nabla_x \mathbf{V} + \nabla_y \overline{\mathbf{W}}) \, dy \, dx \\ & - \int_{\Omega \times Y} \mathrm{M}(\mathbf{y}) (\nabla_x \varphi^0 + \nabla_y \varphi^1) (\nabla_x \Psi + \nabla_y \Xi) \, dy \, dx \qquad (5.17) \\ & - \int_{\Omega \times Y} (\mathbf{f} \cdot \mathbf{V} + \mathbf{g} \cdot \Psi) \, dy \, dx = 0, \end{aligned} $$

for all $(\mathbf{V}, \Psi) \in H_\Sigma^1(\Omega; \mathbb{R}^3) \times H_\Sigma^1(\Omega; \mathbb{R}^3)$ and $(\overline{\mathbf{W}}, \Xi) \in L^2(\Omega; H^1(Y; \mathbb{R}^3)) \times L^2(\Omega; H^1(Y; \mathbb{R}^3))$.

Equation (5.17) can be put in a more familiar form,

$$ \int_{\Omega} \sigma^{\text{eff}} : (\nabla_x \mathbf{V} - \epsilon \cdot \mathbf{\Psi}) \, d\mathbf{x} - \int_{\Omega} \mu^{\text{eff}} : \nabla_x \mathbf{\Psi} \, d\mathbf{x} = \int_{\Omega} (\mathbf{f} \cdot \mathbf{V} + \mathbf{g} \cdot \mathbf{\Psi}) \, d\mathbf{x}, \qquad (5.18) $$

with

$$ \begin{bmatrix} \sigma^{\text{eff}} \\ \mu^{\text{eff}} \end{bmatrix} = \begin{bmatrix} \text{E}^{\text{eff}} & 0 \\ 0 & \text{M}^{\text{eff}} \end{bmatrix} \begin{bmatrix} \gamma^0 \\ \kappa^0 \end{bmatrix} \qquad (5.19) $$

the effective stress and effective couple-stress, the Cosserat strain $\gamma^0 := (\nabla_x \mathbf{u}^0 - \epsilon \cdot \varphi^0)$ and curvature-twist $\kappa^0 := \nabla_x \varphi^0$, the effective coefficients,

$$ \mathrm{E}_{jipq}^{\text{eff}} = \int_{Y} \mathrm{E}_{jikl}(\mathbf{y}) (\delta_{kp} \delta_{ql} + \partial_{y_l} \zeta_k^{pq}) \, d\mathbf{y}, \qquad (5.20) $$

$$ \mathrm{M}_{jipq}^{\text{eff}} = \int_{Y} \mathrm{M}_{jikl}(\mathbf{y}) (\delta_{kp} \delta_{ql} + \partial_{y_l} \xi_k^{pq}) \, d\mathbf{y}, \qquad (5.21) $$

and $\zeta^{pq}$ and $\xi^{pq}$ the local solutions on the unit cell satisfying the second order elliptic problems,
$$ 
\begin{gathered}
\zeta^{pq} \in H_{\text{per}}^{1}(Y; \mathbb{R}^{3}), \quad \int_{Y} \zeta^{pq} \, dy = 0, \\
\int_{Y} E(\mathbf{y}) (\mathbf{e}_{p} \otimes \mathbf{e}_{q} + \nabla_{y} \zeta^{pq}) : \nabla_{y} \mathbf{w} \, dy = 0,
\end{gathered}
\qquad (5.22) $$

357 for all $\mathbf{w} \in H_{\text{per}}^{1}(Y; \mathbb{R}^{3})$,

$$ 
\begin{gathered}
\xi^{pq} \in H_{\text{per}}^{1}(Y; \mathbb{R}^{3}), \quad \int_{Y} \xi^{pq} \, dy = 0, \\
\int_{Y} M(\mathbf{y}) (\mathbf{e}_{p} \otimes \mathbf{e}_{q} + \nabla_{y_t} \xi^{pq}) : \nabla_{y} \mathbf{v} \, dy = 0,
\end{gathered}
\qquad (5.23) $$

358 for all $\mathbf{v} \in H_{\text{per}}^{1}(Y; \mathbb{R}^{3})$.

359 Proof. Once again, using the estimates from (3.13) and the results of Proposition 4.1
360 VII. we obtain (5.13)-(5.16) as $\varepsilon \to 0$. Furthermore, the weak form (3.1) of the Cosserat
361 continuum reads:

362 Find $(\mathbf{u}^{\varepsilon}, \varphi^{\varepsilon}) \in H_{\Sigma}^{1}(\Omega; \mathbb{R}^{3}) \times H_{\Sigma}^{1}(\Omega; \mathbb{R}^{3})$ such that,

$$ 
\begin{align}
& \int_{\Omega} E\left(\frac{\mathbf{x}}{\varepsilon}\right)\left(\nabla \mathbf{u}^{\varepsilon}-\boldsymbol{\epsilon} \cdot \boldsymbol{\varphi}^{\varepsilon}\right):\left(\nabla \mathbf{v}-\boldsymbol{\epsilon} \cdot \boldsymbol{\psi}\right) d \mathbf{x}-\int_{\Omega} \varepsilon C\left(\frac{\mathbf{x}}{\varepsilon}\right) \nabla \boldsymbol{\varphi}^{\varepsilon}:\left(\nabla \mathbf{v}-\boldsymbol{\epsilon} \cdot \boldsymbol{\psi}\right) d \mathbf{x} \nonumber \\
& -\int_{\Omega} \varepsilon C\left(\frac{\mathbf{x}}{\varepsilon}\right)\left(\nabla \mathbf{u}^{\varepsilon}-\boldsymbol{\epsilon} \cdot \boldsymbol{\varphi}^{\varepsilon}\right):\nabla \boldsymbol{\psi} d \mathbf{x}-\int_{\Omega} M\left(\frac{\mathbf{x}}{\varepsilon}\right) \nabla \boldsymbol{\varphi}^{\varepsilon}:\nabla \boldsymbol{\psi} d \mathbf{x} \tag{6.19} \\
& -\int_{\Omega}\left(\mathbf{f} \cdot \mathbf{v}+\mathbf{g} \cdot \boldsymbol{\psi}\right) d \mathbf{x}=0. \nonumber
\end{align} $$

363 Unfold the above expression using Proposition 4.1 I., II., and IV. and obtain,

$$ 
\begin{align}
& \int_{\Omega \times Y} E(\mathbf{y}) (\mathcal{T}_{\varepsilon}(\nabla \mathbf{u}^{\varepsilon}) - \boldsymbol{\epsilon} \cdot \mathcal{T}_{\varepsilon}(\boldsymbol{\varphi}^{\varepsilon})): (\mathcal{T}_{\varepsilon}(\nabla \mathbf{v}) - \boldsymbol{\epsilon} \cdot \mathcal{T}_{\varepsilon}(\boldsymbol{\psi})) \, d\mathbf{y} \, d\mathbf{x} \nonumber \\
& - \int_{\Omega \times Y} \varepsilon C(\mathbf{y}) \mathcal{T}_{\varepsilon}(\nabla \boldsymbol{\varphi}^{\varepsilon}): (\mathcal{T}_{\varepsilon}(\nabla \mathbf{v}) - \boldsymbol{\epsilon} \cdot \mathcal{T}_{\varepsilon}(\boldsymbol{\psi})) \, d\mathbf{x} \, d\mathbf{y} \nonumber \\
& - \int_{\Omega \times Y} \varepsilon C(\mathbf{y}) (\mathcal{T}_{\varepsilon}(\nabla \mathbf{u}^{\varepsilon}) - \boldsymbol{\epsilon} \cdot \mathcal{T}_{\varepsilon}(\boldsymbol{\varphi}^{\varepsilon})): \mathcal{T}_{\varepsilon}(\nabla \boldsymbol{\psi}) \, d\mathbf{x} \, d\mathbf{y} \tag{6.20} \\
& - \int_{\Omega \times Y} M(\mathbf{y}) \nabla \mathcal{T}_{\varepsilon}(\boldsymbol{\varphi}^{\varepsilon}): \mathcal{T}_{\varepsilon}(\nabla \boldsymbol{\psi}) \, d\mathbf{y} \, d\mathbf{x} \nonumber \\
& - \int_{\Omega \times Y} (\mathcal{T}_{\varepsilon}(\mathbf{f}) \cdot \mathcal{T}_{\varepsilon}(\mathbf{v}) + \mathcal{T}_{\varepsilon}(\mathbf{g}) \cdot \mathcal{T}_{\varepsilon}(\boldsymbol{\psi})) \, d\mathbf{y} \, d\mathbf{x} = 0. \nonumber
\end{align} $$

364 Set $(\mathbf{v}, \boldsymbol{\psi}) := (\mathbf{V}(\mathbf{x}), \boldsymbol{\Psi}(\mathbf{x}))$ for any test functions $(\mathbf{V}, \boldsymbol{\Psi}) \in C_{0}^{\infty}(\Omega; \mathbb{R}^{3}) \times C_{0}^{\infty}(\Omega; \mathbb{R}^{3})$ in (6.20)
365 and let $\varepsilon \to 0$ to obtain,
$$
\begin{align}
& \int_{\Omega \times Y} \mathrm{E}(\boldsymbol{y}) (\nabla_x \boldsymbol{u}_k^0 + \nabla_y \boldsymbol{u}^1 - \boldsymbol{\epsilon} \cdot \boldsymbol{\varphi}^0) (\nabla_x \boldsymbol{V} - \boldsymbol{\epsilon} \cdot \boldsymbol{\Psi}) \, dy \, dx \nonumber \\
& - \int_{\Omega \times Y} \mathrm{M}(\boldsymbol{y}) (\nabla_x \boldsymbol{\varphi}_k^0 + \nabla_y \boldsymbol{\varphi}^1) (\nabla_x \boldsymbol{\Psi}) \, dy \, dx \tag{6.21} \\
& - \int_{\Omega \times Y} (\boldsymbol{f} \cdot \boldsymbol{V} + \boldsymbol{g} \cdot \boldsymbol{\Psi}) \, dy \, dx = 0, \nonumber
\end{align}
$$

Again, select $(\mathbf{v}, \mathbf{\psi})=(\mathbf{v}^{\varepsilon}, \mathbf{\psi}^{\varepsilon}) := (\varepsilon U(\mathbf{x}) \mathbf{W}(\frac{\mathbf{x}}{\varepsilon}), \varepsilon \Phi(\mathbf{x}) \Xi(\frac{\mathbf{x}}{\varepsilon}))$ where $(U, \Phi) \in C_0^{\infty}(\Omega) \times C_0^{\infty}(\Omega)$ and $(\mathbf{W}, \Xi) \in H_{\text{per}}^{1}(Y; \mathbb{R}^3) \times H_{\text{per}}^{1}(Y; \mathbb{R}^3)$. Recall that $(\mathbf{v}^{\varepsilon}, \mathbf{\psi}^{\varepsilon}) \to (\mathbf{0}, \mathbf{0})$ in $L^2(\Omega; \mathbb{R}^3) \times L^2(\Omega; \mathbb{R}^3)$ and $(\mathcal{T}_{\varepsilon}(\partial_{x_j} v_i^{\varepsilon}), \mathcal{T}_{\varepsilon}(\partial_{x_j} \psi_i^{\varepsilon})) \to (\partial_{y_j} \overline{W}_i(\mathbf{x}, \mathbf{y}), \partial_{y_j} \overline{\Xi}_i(\mathbf{x}, \mathbf{y}))$ in $L^2(\Omega \times Y) \times L^2(\Omega \times Y)$ where $\overline{W}_i(\mathbf{x}, \mathbf{y}) := U(\mathbf{x}) W_i(\mathbf{y})$ and $\overline{\Xi}_i(\mathbf{x}, \mathbf{y}) := \Phi(\mathbf{x}) \Xi_i(\mathbf{y})$.

Substituting the above test functions in (6.20) we obtain,

$$
\begin{equation}
\begin{split}
& \int_{\Omega \times Y} \mathrm{E}(\boldsymbol{y}) (\nabla_x \boldsymbol{u}^0 + \nabla_x \boldsymbol{u}^1 - \boldsymbol{\epsilon} \cdot \boldsymbol{\varphi}^0) : \nabla_y \overline{\boldsymbol{W}} \, dy \, dx \\
& \quad - \int_{\Omega \times Y} \mathrm{M}(\boldsymbol{y}) (\nabla_x \boldsymbol{\varphi}^0 + \nabla_y \boldsymbol{\varphi}^1) : \nabla_y \overline{\Xi} \, dy \, dx = 0,
\end{split}
\tag{6.22}
\end{equation}
$$

371 Adding (6.21) and (6.22) we obtain,

$$
\begin{equation}
\begin{split}
& \int_{\Omega \times Y} \mathrm{E}(\boldsymbol{y}) (\nabla_x \boldsymbol{u}_k^0 + \nabla_y \boldsymbol{u}^1 - \boldsymbol{\epsilon} \cdot \boldsymbol{\varphi}^0) (\nabla_x \boldsymbol{V} + \nabla_y \boldsymbol{W} - \boldsymbol{\epsilon} \cdot \boldsymbol{\Psi}) \, dy \, dx \\
& - \int_{\Omega \times Y} \mathrm{M}(\boldsymbol{y}) (\nabla_x \boldsymbol{\varphi}_k^0 + \nabla_y \boldsymbol{\varphi}^1) (\nabla_x \boldsymbol{\Psi} + \nabla_y \boldsymbol{\Xi}) \, dy \, dx \tag{6.23} \\
& - \int_{\Omega \times Y} (\boldsymbol{f} \cdot \boldsymbol{V} + \boldsymbol{g} \cdot \boldsymbol{\Psi}) \, dy \, dx = 0.
\end{split}
\end{equation}
$$

By the density of $C_0^\infty(\Omega; \mathbb{R}^3)$ in $H_\Sigma^1(\Omega; \mathbb{R}^3)$ and the density of $C_0^\infty(\Omega) \otimes H_{\text{per}}^1(Y; \mathbb{R}^3)$ in $L^2(\Omega; H_{\text{per}}^1(Y; \mathbb{R}^3))$ the result is true for all $(\mathbf{V}, \mathbf{\Psi}) \in H_\Sigma^1(\Omega; \mathbb{R}^3) \times H_\Sigma^1(\Omega; \mathbb{R}^3)$ and all $(\overline{\mathbf{W}}, \overline{\Xi}) \in L^2(\Omega; H_{\text{per}}^1(Y; \mathbb{R}^3)) \times L^2(\Omega; H_{\text{per}}^1(Y; \mathbb{R}^3))$.

Taking $(\mathbf{V}, \overline{\mathbf{W}}, \mathbf{\Psi}, \overline{\Xi}) = (\mathbf{V}, \overline{\mathbf{W}}, \mathbf{0}, \mathbf{0})$ we obtain,

$$
\begin{equation}
\begin{split}
& \int_{\Omega \times Y} \mathrm{E}(\boldsymbol{y}) (\nabla_x \boldsymbol{u}_k^0 + \nabla_y \boldsymbol{u}^1 - \boldsymbol{\epsilon} \cdot \boldsymbol{\varphi}^0) (\nabla_x \boldsymbol{V} + \nabla_y \boldsymbol{W}) \, dy \, dx \\
& \quad - \int_{\Omega \times Y} \boldsymbol{f} \cdot \boldsymbol{V} \, dy \, dx = 0.
\end{split}
\tag{6.24}
\end{equation}
$$

Taking $(\mathbf{V}, \overline{\mathbf{W}}, \mathbf{\Psi}, \overline{\Xi}) = (\mathbf{0}, \mathbf{0}, \mathbf{\Psi}, \overline{\Xi})$ we obtain,
$$ 
\begin{aligned}
& - \int_{\Omega \times Y} \mathbf{E}(\mathbf{y}) (\nabla_x \mathbf{u}_k^0 + \nabla_y \mathbf{u}^1 - \epsilon \cdot \mathbf{\varphi}^0) : \epsilon \cdot \mathbf{\Psi} \, dy \, dx \\
& - \int_{\Omega \times Y} \mathbf{M}(\mathbf{y}) (\nabla_x \mathbf{\varphi}_k^0 + \nabla_y \mathbf{\varphi}^1) : (\nabla_x \mathbf{\Psi} + \nabla_y \mathbf{\Xi}) \, dy \, dx \\
& - \int_{\Omega} \mathbf{g} \cdot \mathbf{\Psi} \, dx = 0.
\end{aligned} 
\quad (6.25) $$

Moreover, if we select $V \equiv 0$ in (6.24) we obtain that $\mathbf{u}^1$ has the following form,

$$ \mathbf{u}^1(\mathbf{x}, \mathbf{y}) = \zeta^{pq}(\mathbf{y}) (\partial_{x_p} u_q^0(\mathbf{x}) - \epsilon_{\nu pq} \varphi_{\nu}^0(\mathbf{x})) + \tilde{\kappa}(\mathbf{x}). \quad (6.26) $$

While, if we set $\mathbf{\Psi} \equiv 0$ in (6.25) we obtain $\mathbf{\varphi}^1$ the following form,

$$ \mathbf{\varphi}^1(\mathbf{x}, \mathbf{y}) = \xi^{pq}(\mathbf{y}) \partial_{x_p} \varphi_q^0(\mathbf{x}) + \hat{\kappa}(\mathbf{x}), \quad (6.27) $$

where $\zeta^{pq}$ and $\xi^{pq}$ satisfy (5.22) and (5.23), respectively with $\tilde{\kappa}(\mathbf{x})$ and $\hat{\kappa}(\mathbf{x})$ some generic constant functions in $\mathbf{x}$. We remark that existence and uniqueness (up to a constant) for $\zeta^{pq}$ and $\xi^{pq}$ follow from the theory of variational inequalities or Lax-Milgram theory.

Substituting $\overline{\mathbf{W}} \equiv 0$, $\overline{\mathbf{\Xi}} \equiv 0$, $\mathbf{u}^1$, and $\mathbf{\varphi}^1$ from (6.24) and (6.25) in (6.23) and factoring out common terms we obtain the weak formulation of a centrosymmetric Cosserat model,

$$ \int_{\Omega} \sigma^{\text{eff}} : (\nabla_x \mathbf{V} - \epsilon \cdot \mathbf{\Psi}) \, d\mathbf{x} - \int_{\Omega} \mu^{\text{eff}} : \nabla_x \mathbf{\Psi} \, d\mathbf{x} - \int_{\Omega} (\mathbf{f} \cdot \mathbf{V} + \mathbf{g} \cdot \mathbf{\Psi}) \, d\mathbf{x} = 0, \quad (6.28) $$

where the effective stress and effective couple-stress are given by,

$$ \begin{bmatrix} \sigma^{\text{eff}} \\ \mu^{\text{eff}} \end{bmatrix} = \begin{bmatrix} \mathbf{E}^{\text{eff}} & 0 \\ 0 & \mathbf{M}^{\text{eff}} \end{bmatrix} \begin{bmatrix} \gamma^0 \\ \kappa^0 \end{bmatrix}. \quad (6.29) $$

□

## 7 Numerical simulations

In this section we carry out 2D computations, using the finite element method, of the effective coefficients under (HS 1). The computations under the scheme (HS 2) are done in a similar manner and, thus, we do not address them here. We plot the effective shear coefficients, computed from (5.7) using equations (6.13) and (6.14), against different volume fractions. The domain of computation is the unit hexagon with periodic boundary conditions on opposite faces and filled with a weak material in the shape of a tilted ellipse by a 45 degree angle (see Fig. 4). Our interest in these computations is to explore the effect the local rotations have in the elastic coefficients and compare them to the classical coefficients without rotations.
<figure><img src="image_7.png" /><figcaption>Figure 4: Hexagonal lattice symmetry seems to capture the effective behavior of compact bone (see [35] and references therein). The inhomogeneity, in this case, is comparable with the size of the period. Thus, we use the effective model obtain in Theorem 5.1.</figcaption></figure>

In all computations we have used a standard, triangular mesh. We used 100 P2 elements to mesh the elliptical inclusion and 10 P2 elements for the rest of the unit hexagonal cell. In all the examples we have set the Poisson’s ratio of each phase equal to $\nu = 0.3$ while the Young’s modulus of the strong phase was set to $E^1 = 0.91$GPa and the weak phase was set to $E^0 = 10^{-3}$GPa. All computations were carried out using the software FreeFem++ [17].

## 7.1 A curious case of (lack of) symmetry for $E^{\text{eff}}$ under (HS 1)

The coefficients $E^{\text{eff}}$ in (5.7) of the effective linear elastic model obtained under (HS 1) are anisotropic. However, there is also some additional lack of symmetry that we don’t see in classical linear elasticity, namely, $E_{1212}^{\text{eff}} \neq E_{1221}^{\text{eff}}$. It seems that this is a consequence of the material property $\alpha$ that is related to the rotations. If $\alpha \equiv 0$ then the effective coefficients in (5.7) reduce to the effective coefficients obtained by homogenization of classical Cauchy material with the usual symmetry. In what follows, we numerically demonstrate this effect.

In all computations, the material parameters were defined as follows:

$$ \vartheta = \frac{E}{2(1 + \nu)}, \quad \varrho = \frac{E\nu}{(1 + \nu)(1 - 2\nu)}, \quad \alpha = \frac{1}{2}\vartheta, \qquad (7.1) $$

with $E$ replaced by the weak and strong phase Young’s modulus for each respective phase.

Here $\vartheta$ and $\varrho$ are the classic Lamé constants, while $\alpha$ is often referred to as the micropolar couple modulus [16]. In Fig. 7 we plot the effective coefficients $E_{1212}^{\text{eff}}$ and $E_{1221}^{\text{eff}}$ as a function
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of the volume fraction. We have only computed the shear coefficients since they are the
only ones (up to symmetry) where the rotations make a contribution.

<figure><img src="image_8.png" /><figcaption>Figure 5: $\alpha = \frac{\vartheta}{2}$</figcaption></figure>

<figure><img src="image_9.png" /><figcaption>Figure 6: $\alpha = 0$</figcaption></figure>

Figure 7: The coefficients $E_{1212}^{\text{eff}}$ and $E_{1221}^{\text{eff}}$ plotted for different volume fractions and for different values of $\alpha$. On the left image, the dotted line represents the values of the $E_{1212}^{\text{eff}}$ coefficient while the solid line represents the values of $E_{1221}^{\text{eff}}$. On the right, the two coefficients are equal, $E_{1212}^{\text{eff}} = E_{1221}^{\text{eff}}$.

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We observe from Fig. 7 that when $\alpha \neq 0$ the coefficients $E_{1212}^{\text{eff}} \neq E_{1221}^{\text{eff}}$. This is not the case when $\alpha = 0$ since the values of the coefficients are the same, $E_{1212}^{\text{eff}} = E_{1221}^{\text{eff}}$. Moreover, as expected, when the volume fraction of the weak phase increases the material becomes easier to shear. However, the computations also show, that the value of $E_{1212}^{\text{eff}}$, in the case when $\alpha \neq 0$, is much larger than in the case when $\alpha = 0$. Thus, when the effective material has a microstructure that can displace and, independently, rotate then the material is easy to shear in one direction (21-direction) but difficult to shear in the other direction (12-direction). This lack of symmetry seems consistent with the numerical computations of the elastic moduli done in [36] for a two phase centro-symmetric laminated composite.

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# 8 Conclusions

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We derived macroscale equations from a non-centrosymmetric, heterogeneous Cosserat continuum taking into account scale-size effects that can be used to model the behavior of human compact bone. Two different macroscale models were identified:

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1. When the Cosserat length and the chiral length are much smaller compared to the overall length of the domain (full scale separation), we obtain an effective linear elastic model. However, unlike in the classical approach, the effective tensor is constructed using an additional problem that is derived from the local Cosserat rotations. Moreover, we carried out numerical simulations to validate our theoretical results. The numerical simulations seem to indicate that the approach put forth allows
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one to construct an effective linear elasticity tensor from isotropic Cosserat micro-
constituents that can be fully anisotropic. This is a novelty, since classical homoge-
nization of linear elasticity derived from linear, elastic, isotropic micro-constituents,
produces an effective elasticity tensor can be, at most, orthotropic.

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2. When the Cosserat length is of the same order as the overall length of the domain, and thus not all scales are well separated, the macroscale model is a centrosymmetric Cosserat model with two fourth order effective tensors relating the non-symmetric strain to the non-symmetric stress and the curvature-twist to the couple-stress, hence, proving new constitutive laws for Cosserat media. What is remarkable about our approach is that we are able to derive an effective model even some of the scales in the model are not well separated.

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Naturally, the technique here can be extended to porous domains by adjusting the unfold-
ing operator as in [5].

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Acknowledgements

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The author gratefully acknowledges the financial support by the Knowledge Foundation (project nr. KK 2020-0152). I would like to thank Dr. Nicole Tueni for her help with some of the images and I would like to express my gratitude to the anonymous reviewer for their many comments, suggestions, and corrections. They have, without a doubt, improved the quality of the manuscript.

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