Multiscale interactions of elastic anisotropy in unsaturated clayey rocks using a homogenization model

Abstract

The microstructure of a geomaterial plays a significant role in determining its macroscale properties. Most clay rocks have an anisotropic microstructure due to preferential orientation of the pores and mineral grains, which results in transversely isotropic mechanical properties. Their anisotropic microstructure is complex and spans multiple orders of magnitudes. The interactions between anisotropy at different scales in these rocks can give rise to emerging properties such as saturation-dependent elastic anisotropy. In this study, we develop a homogenization model with three levels of upscaling to capture the multiscale interactions of elastic anisotropy in unsaturated clay rocks. The model provides an enriched description of the elastic behavior of clay rocks during changes in the degree of saturation by bridging the nano-, micro- and macroscale microstructures. Stress-point simulations are presented to demonstrate the interactions between anisotropy at different spatial scales that result in the elastic behavior of clay rocks observed in the literature, including constant anisotropy, evolving anisotropy and a rotation of the principal orientation of anisotropy. The results highlight that constant and evolving elastic anisotropy can originate from the same microstructural features that either neutralize or enhance one another. Overall, the proposed model offers a quantitative link between anisotropy at multiple scales in clay rocks and its macroscopic anisotropic stiffness.

Appendix A. Hill’s tensor

The expression of the Hill’s tensor $\mathbb {P}$ for aligned oblate spheroid inclusions and randomly distributed oblate spheroid inclusions used in the proposed homogenization model are presented below. The Hill’s tensor characterizes the interaction between the inclusions and the host matrix. It is dependent on the stiffness of the host matrix, the shape of the inclusions and in a transversely isotropic matrix, the orientation of the inclusions in the host matrix.

1.1 A1. Aligned oblate spheroidal inclusions

The Eshelby tensor $\mathbb {S}$ for an oblate spheroid aligned along the bedding plane in a transversely isotropic host medium was derived by Withers [111].

$$\begin{aligned} \begin{aligned} \mathbb {S}_{11} =&\sum _{i=1}^{2} (2\mathbb {C}_{44}(1+k_i)\nu _i^2-\mathbb {C}_{66})A_i\nu _iI_1(\nu _i)\\&+ \frac{D\mathbb {C}_{66}I_1(\nu _3)}{2}\\ \mathbb {S}_{12} =&\sum _{i=1}^{2} (2\mathbb {C}_{44}(1+k_i)\nu _i^2-3\mathbb {C}_{66})A_i\nu _iI_1(\nu _i)\\&- \frac{D\mathbb {C}_{66}I_1(\nu _3)}{2}\\ \mathbb {S}_{33} =&2\sum _{i=1}^{2} (\mathbb {C}_{13} - \mathbb {C}_{33}k_i\nu _i^2)\nu _i^3k_iA_iI_2(\nu _i)\\ \mathbb {S}_{13} =&2\sum _{i=1}^{2} (\mathbb {C}_{13} - \mathbb {C}_{33}k_i\nu _i^2)\nu _iA_iI_1(\nu _i)\\ \mathbb {S}_{31} =&2\sum _{i=1}^{2} \mathbb {C}_{44}\nu _i^5k_iA_i(1+k_i)I_2(\nu _i) \\&-\mathbb {C}_{66}k_i\nu _i^3A_iI_2(\nu _i)\\ \mathbb {S}_{44} =&\frac{\mathbb {C}_{44}}{2}\sum _{i=1}^{2} \nu _i^3A_i(1+k_i)(I_2(\nu _i)-2k_iI_1(\nu _i)) \\&+\frac{\mathbb {C}_{44}}{4}DI_2(\nu _3)\nu _3^2 \end{aligned} \end{aligned}$$

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where

$$\begin{aligned} \begin{aligned} \mathbb {C}_{13}^*&= \sqrt{\mathbb {C}_{11}\mathbb {C}_{13}}\\ \nu _1&= \sqrt{\frac{(\mathbb {C}_{13}^*-\mathbb {C}_{13}) (\mathbb {C}_{13}^*+\mathbb {C}_{13}+\mathbb {C}_{44})}{4\mathbb {C}_{33}\mathbb {C}_{44}}}\\&\quad +\sqrt{\frac{(\mathbb {C}_{13}^*+\mathbb {C}_{13})(\mathbb {C}_{13}^*-\mathbb {C}_{13} -\mathbb {C}_{44})}{4\mathbb {C}_{33}\mathbb {C}_{44}}}\\ \nu _2&= \sqrt{\frac{(\mathbb {C}_{13}^*-\mathbb {C}_{13}) (\mathbb {C}_{13}^*+\mathbb {C}_{13}+\mathbb {C}_{44})}{4\mathbb {C}_{33}\mathbb {C}_{44}}}\\&\quad -\sqrt{\frac{(\mathbb {C}_{13}^*+\mathbb {C}_{13})(\mathbb {C}_{13}^*-\mathbb {C}_{13} -\mathbb {C}_{44})}{4\mathbb {C}_{33}\mathbb {C}_{44}}}\\ \nu _3&= \sqrt{\frac{\mathbb {C}_{66}}{\mathbb {C}_{44}}}\\ D&=\frac{1}{4\pi g_0 \nu _3}\\ A_1&=-\frac{\mathbb {C}_{44}-\mathbb {C}_{33}\nu _1^2}{8\pi \mathbb {C}_{33}\mathbb {C}_{44} (\nu _1^2 - \nu _2^2) \nu _1^2}\\ A_2&= \frac{\mathbb {C}_{44}-\mathbb {C}_{33}\nu _2^2}{8\pi \mathbb {C}_{33}\mathbb {C}_{44} (\nu _1^2 - \nu _2^2) \nu _2^2}\\ k_i&= \frac{\mathbb {C}_{11}/\nu _i^2-\mathbb {C}_{44}}{\mathbb {C}_{13}+\mathbb {C}_{44}}\\ I_1(\nu _i)&= \frac{2\pi \xi }{(\chi _i)^{3/2}}\Big (\text{tan}^{-1} \frac{\sqrt{\chi _i}}{\nu _i \xi }- \nu _i \xi \sqrt{\chi _i}\Big )\\ I_2(\nu _i)&= -\frac{4\pi \xi }{(\chi _i)^{3/2}}\Big (\text{tan}^{-1} \frac{\sqrt{\chi _i}}{\nu _i \xi }- \nu _i \xi \sqrt{\chi _i}\Big )\\ \chi _i&= 1-\nu _i^2 \xi ^2 \end{aligned} \end{aligned}$$

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The Hill’s tensor can then be expressed as [77]

$$\begin{aligned} \begin{aligned} \mathbb {P}_{11}&= \frac{\mathbb {S}_{11}-\mathbb {S}_{12}}{2\mathbb {C}_{66}}+\mathbb {P}_{12}\\ \mathbb {P}_{12}&= (\frac{\mathbb {C}_{33}(\mathbb {S}_{11}+\mathbb {S}_{12})-2\mathbb {C}_{13}\mathbb {S}_{33}}{4((\mathbb {C}_{11}-\mathbb {C}_{66})\mathbb {C}_{33}-\mathbb {C}_{13}^2)} - \frac{\mathbb {S}_{11}-\mathbb {S}_{12}}{4\mathbb {C}_{66}})\\ \mathbb {P}_{33}&= \frac{(\mathbb {C}_{11}-\mathbb {C}_{66})\mathbb {S}_{33}-\mathbb {C}_{13}\mathbb {S}_{31}}{((\mathbb {C}_{11}-\mathbb {C}_{66})\mathbb {C}_{33}-\mathbb {C}_{13}^2)}\\ \mathbb {P}_{13}&= \frac{\mathbb {C}_{33}\mathbb {S}_{31}-\mathbb {C}_{13}\mathbb {S}_{33}}{2((\mathbb {C}_{11}-\mathbb {C}_{66})\mathbb {C}_{33}-\mathbb {C}_{13}^2)}\\ \mathbb {P}_{44}&= \frac{\mathbb {S}_{44}}{2\mathbb {C}_{44}}\\ \mathbb {P}_{66}&= \frac{\mathbb {S}_{66}}{2\mathbb {C}_{66}} \end{aligned} \end{aligned}$$

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1.2 A2. Aligned oblate spheroidal inclusions

The Hill tensor for randomly oriented oblate spheroids in a transversely isotropic host medium is more complex and requires the use of numerical integration to compute [35]. The dilute concentration tensor $\mathbb {A}_d$ can be obtained from the following integration on a unit sphere

$$\begin{aligned} \begin{aligned} \overline{\mathbb {A}}_d = \frac{1}{4\pi }\int _{\psi =0}^{2\pi }\int _{\theta =0}^{2\pi } \mathbb {A}_d(\theta ,\psi ) \text{d}\theta \text{d}\psi \end{aligned} \end{aligned}$$

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where $\psi$ and $\theta$ are spherical coordinates and

$$\begin{aligned} \begin{aligned} \mathbb {A}_d = (\mathbb {I}+\mathbb {P}(\theta ,\psi ):(\mathbb {C}^{p}-\mathbb {C}^{s}))^{-1}. \end{aligned} \end{aligned}$$

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The Hill tensor can be expressed as an integration over the spherical coordinates $\varphi$ and $\zeta$

$$\begin{aligned} \begin{aligned} \mathbb {P}_{ijkl} =&\int _{0}^{2\pi }\int _{0}^{2\pi } (p_l g_{kij} + p_l g_{kji} \\&+ p_k g_{lij} + p_k g_{lji})\sin (\varphi ) \text{d}\zeta \text{d}\varphi \end{aligned} \end{aligned}$$

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with the following components [29, 35, 75, 111]

$$\begin{aligned} \begin{aligned} p_1&= \frac{\xi ^2l_1}{\eta } \\ p_2&= \frac{(1+\xi ^2)l_2 + (1-\xi ^2)(l_3 \sin (2\theta )-l_2 \cos (2\theta ))}{2\eta } \\ p_3&= \frac{(1+\xi ^2)l_3 + (1-\xi ^2)(l_3 \cos (2\theta )+l_2 \sin (2\theta ))}{2\eta } \end{aligned} \end{aligned}$$

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where

$$\begin{aligned} \begin{aligned} \eta&= \xi ^2l_1^2+(l_3 \cos (\theta )+l_2 \sin (\theta ))^2 \\&\quad +\xi ^2(l_2 \cos (\theta )+l_3 \sin (\theta ))^2 \end{aligned} \end{aligned}$$

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and

$$\begin{aligned} \begin{aligned} g_{111} =&-2\sum _{i=1}^{2}\frac{\nu _iA_il_1(-8\frac{\bar{\nu }_{i23}}{\rho ^2}+\frac{\bar{\nu }_{i13}}{R_i^2}-3l_2^2+2\nu _i^2l_3^2)}{\rho ^4 R_i} \\ {}&- \frac{Dl_1(8\frac{\bar{\nu }_{323}}{\rho ^2}+\frac{\bar{\nu }_{323}}{R_3^2}-l_1^2+2l_2^2-2\nu _3^2l_3^2)}{\rho ^4 R_3}\\ g_{112} =&-2\sum _{i=1}^{2}\frac{\nu _iA_il_2(8\frac{\bar{\nu }_{i13}}{\rho ^2}+\frac{\bar{\nu }_{i13}}{R_i^2}+2l_1^2-l_2^2-2\nu _i^2l_3^2)}{\rho ^4 R_i} \\ {}&- \frac{Dl_2(-8\frac{\bar{\nu }_{313}}{\rho ^2}+\frac{\bar{\nu }_{323}}{R_3^2}-3l_1^2+2\nu _3^2l_3^2)}{\rho ^4 R_3}\\ g_{113} =&2\sum _{i=1}^{2}\frac{\nu _i^3A_il_3(2l_1^2-l_2^2 -\frac{\bar{\nu }_{i13}}{R_i^2})}{\rho ^4 R_i} \\ {}&- \frac{D\nu _3^2l_3(l_1^2-2l_2^2 +\frac{\bar{\nu }_{323}}{R_3^2})}{\rho ^4 R_3}\\ g_{121} =&-2\sum _{i=1}^{2}\frac{\nu _iA_il_2(8\frac{\bar{\nu }_{i13}}{\rho ^2}+\frac{\bar{\nu }_{i13}}{R_i^2}+2l_1^2-l_2^2-2\nu _i^2l_3^2)}{\rho ^4 R_i} \\ {}&+ \frac{Dl_2(8\frac{\bar{\nu }_{313}}{\rho ^2}+\frac{\bar{\nu }_{313}}{R_3^2}+2l_1^2-l_2^2-2\nu _3^2l_3^2)}{\rho ^4 R_3}\\ g_{123} =&2\sum _{i=1}^{2}\frac{\nu _i^3A_il_1l_2l_3(3\rho ^2 + 2\nu _i^2l_3^2)}{\rho ^4R_i^3}\\&- \frac{D\nu _3^2l_1l_2l_3(3\rho ^2 + 2\nu _3^2l_3^2)}{\rho ^4R_3^3}\\ g_{131} =&\sum _{i=1}^{2}\frac{\nu _i^2A_il_3(1-2\frac{l_1^2}{\rho ^2}-\frac{l_1^2}{R_i^2})}{\rho ^2 R_i}\\ g_{312} =&2\sum _{i=1}^{2}\frac{k_i\nu _i^3A_il_1l_2l_3(\frac{2}{\rho ^2}+\frac{1}{R_i^2})}{\rho ^2 R_i}\\ g_{313} =&-2\sum _{i=1}^{2}\frac{k_i\nu _i^3A_il_1}{R_i^3}\\ g_{331} =&\sum _{i=1}^{2}\frac{k_i\nu _i^2A_il_3}{R_i^3}\\ g_{333} =&\sum _{i=1}^{2}\frac{k_i\nu _i^4A_il_3}{R_i^3} \end{aligned} \end{aligned}$$

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where

$$\begin{aligned} \begin{aligned} \rho&= \sqrt{l_1^2+l_2^2} \\ R_i^2&= \rho ^2 + \nu _i^2l_3^2\\ l_1&= \cos (\zeta )\sin (\varphi )\\ l_2&= \sin (\zeta )\sin (\varphi )\\ l_3&= \cos (\varphi )\\ \bar{\nu }_{imn}&= \nu _i^2l_m^2l_n^2 \end{aligned} \end{aligned}$$

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and the remaining variables are the same as in Appendix A1.