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A double-yield-surface plasticity theory for transversely isotropic rocks

Abstract

We present a double-yield-surface plasticity theory for transversely isotropic rocks that distinguishes between plastic deformation through the solid matrix and localized plasticity along the weak bedding planes. A recently developed anisotropic modified Cam-Clay model is adopted to model the plastic response of the solid matrix, while the Mohr–Coulomb friction law is used to represent the sliding mechanism along the weak bedding planes. For its numerical implementation, we derive an implicit return mapping algorithm for both the semi-plastic and fully plastic loading processes, as well as the corresponding algorithmic tangent operator for finite element problems. We validate the model with triaxial compression test data for three different transversely isotropic rocks and reproduce the undulatory variation of rock strength with bedding plane orientation. We also implement the proposed model in a finite element setting and investigate the deformation of rock surrounding a borehole subjected to fluid injection. We compare the results of simulations using the proposed double-yield-surface model with those generated using each single yield criterion to highlight the features of the proposed theory.

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Data availability statement

The datasets generated during the course of this study are available from the corresponding author upon reasonable request.

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Acknowledgements

This material is based upon work supported by the U.S. Department of Energy, Office of Science, Office of Basic Energy Sciences, Geosciences Research Program, under Award Number DE-FG02-03ER15454, and by the National Science Foundation, USA, under Award Number CMMI-1914780. The first author also acknowledges the support from the Shuimu Scholar Program at Tsinghua University.

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Appendices

Appendix A. Relevant partial derivatives

This Appendix derives the partial derivatives of \(f_w\) and \(f_m\). For partial derivatives associated with \(f_m\), we have

$$\begin{aligned}&\frac{\partial f_m}{\partial \varvec{\sigma }} =\frac{{\mathbb{A}}^*:\varvec{\sigma }}{M^2} +(2\varvec{a}^*:\varvec{\sigma }-p_c)\varvec{a}^*\,, \end{aligned}$$
(71)
$$\begin{aligned}&\frac{\partial ^2 f_m}{\partial \varvec{\sigma }^2} =\frac{{\mathbb{A}}^*}{M^2} + 2\varvec{a}^*\otimes \varvec{a}^*\,, \end{aligned}$$
(72)
$$\begin{aligned}&\frac{\partial f_m}{\partial p_c} =\varvec{a}^*:\varvec{\sigma }\,, \end{aligned}$$
(73)
$$\begin{aligned}&\frac{\partial p_c}{\partial \epsilon ^p_v} =-\frac{p_c}{\lambda _p}\,. \end{aligned}$$
(74)

For partial derivatives associated with \(f_w\) and \(g_w\), we limit the discussion to 2D plane strain problem and define \(\varvec{l}\) as the tangential direction of the weak plane. The traction vector \(\varvec{t}\) on the weak plane is

$$\begin{aligned} \varvec{t} = \varvec{\sigma }\cdot {\varvec{n}}\,. \end{aligned}$$
(75)

We can evaluate the shear stress \(\tau \) and normal stress \(\sigma _n\) on the weak plane as

$$\begin{aligned} \sigma _n&= \varvec{n}\cdot \varvec{t} = \varvec{\sigma }:\varvec{m}\,, \end{aligned}$$
(76a)
$$\begin{aligned} \tau&=\varvec{l}\cdot \varvec{t} = \varvec{\sigma }:\varvec{\alpha }\,, \end{aligned}$$
(76b)

where

$$\begin{aligned} \varvec{\alpha } = \frac{1}{2}\left( \varvec{l}\otimes \varvec{n} +\varvec{n}\otimes \varvec{l}\right) \,. \end{aligned}$$
(77)

Thus,

$$\begin{aligned}&\frac{\partial f_w}{\partial \varvec{\sigma }} =\mathrm{sgn}(\tau )\varvec{\alpha }+\mathrm{tan}\phi _w\varvec{m}\,, \end{aligned}$$
(78)
$$\begin{aligned}&\frac{\partial ^2 f_w}{\partial \varvec{\sigma }^2} = {\mathbb{O}}\,, \end{aligned}$$
(79)
$$\begin{aligned}&\frac{\partial g_w}{\partial \varvec{\sigma }} = \mathrm{sgn}(\tau ) \varvec{\alpha }+\mathrm{tan}\psi _w\varvec{m}\,, \end{aligned}$$
(80)
$$\begin{aligned}&\frac{\partial ^2 g_w}{\partial \varvec{\sigma }^2} = {\mathbb{O}}\,, \end{aligned}$$
(81)

where sgn(\(\tau \)) is the sign of \(\tau \).

Appendix B. Jacobian matrix

The submatrices \(\varvec{{\mathcal {J}}}_{ij}\) in the Jacobian matrix \(\varvec{{\mathcal {J}}}\) for the fully plastic process are given as follows:

$$\begin{aligned}&\varvec{{\mathcal {J}}}_{11} = 0\,, \end{aligned}$$
(82)
$$\begin{aligned}&\varvec{{\mathcal {J}}}_{12} = -\frac{\partial f_m}{\partial \varvec{\sigma }}: {\mathbb{C}}^e+\varvec{a}^*:\varvec{\sigma }_{n+1} \frac{p_c}{\lambda _p}\mathbf{1}\,, \end{aligned}$$
(83)
$$\begin{aligned}&\varvec{{\mathcal {J}}}_{13} = 0\,, \end{aligned}$$
(84)
$$\begin{aligned}&\varvec{{\mathcal {J}}}_{14} = -\frac{\partial f_m}{\partial \varvec{\sigma }}:{\mathbb{C}}^e\,, \end{aligned}$$
(85)
$$\begin{aligned}&\varvec{{\mathcal {J}}}_{21} =\frac{\partial f_m}{\partial \varvec{\sigma }}\,, \end{aligned}$$
(86)
$$\begin{aligned}&\varvec{{\mathcal {J}}}_{22} = -{\mathbb{I}}-\varDelta \lambda _m\nonumber \\&\qquad \qquad \left[ \left( \frac{{\mathbb{A}}^*}{M^2} +2\varvec{a}^*\otimes \varvec{a}^*\right) :{\mathbb{C}}^e -\frac{p_{c,n+1}}{\lambda _p}\varvec{a}^*\otimes \mathbf{1}\right] \,, \end{aligned}$$
(87)
$$\begin{aligned}&\varvec{{\mathcal {J}}}_{23} = \mathbf{0}\,, \end{aligned}$$
(88)
$$\begin{aligned}&\varvec{{\mathcal {J}}}_{24} = -\varDelta \lambda _m \left( \frac{{\mathbb{A}}^*}{M^2}+2\varvec{a}^* \otimes \varvec{a}^*\right) :{\mathbb{C}}^e\,, \end{aligned}$$
(89)
$$\begin{aligned}&\varvec{{\mathcal {J}}}_{31} = 0\,, \end{aligned}$$
(90)
$$\begin{aligned}&\varvec{{\mathcal {J}}}_{32} = -\frac{\partial f_w}{\partial \varvec{\sigma }}:{\mathbb{C}}^e\,, \end{aligned}$$
(91)
$$\begin{aligned}&\varvec{{\mathcal {J}}}_{33} = 0\,, \end{aligned}$$
(92)
$$\begin{aligned}&\varvec{{\mathcal {J}}}_{34} = -\frac{\partial f_w}{\partial \varvec{\sigma }}:{\mathbb{C}}^e\,, \end{aligned}$$
(93)
$$\begin{aligned}&\varvec{{\mathcal {J}}}_{41} = \mathbf{0}\,, \end{aligned}$$
(94)
$$\begin{aligned}&\varvec{{\mathcal {J}}}_{42} = {\mathbb{O}}\,, \end{aligned}$$
(95)
$$\begin{aligned}&\varvec{{\mathcal {J}}}_{43} =\frac{\partial g_w}{\partial \varvec{\sigma }}\,, \end{aligned}$$
(96)
$$\begin{aligned}&\varvec{{\mathcal {J}}}_{44} = - {\mathbb{I}}\,. \end{aligned}$$
(97)

The expressions above are given in tensorial expression for brevity, but they should be converted to matrix form for numerical implementation. Rank-two and rank-four tensors transform to \(1\times 6\) vectors and \(6\times 6\) matrices in 3D, respectively.

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Zhao, Y., Borja, R.I. A double-yield-surface plasticity theory for transversely isotropic rocks. Acta Geotech. (2022). https://doi.org/10.1007/s11440-022-01605-6

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  • DOI: https://doi.org/10.1007/s11440-022-01605-6

Keywords

  • Double yield surfaces
  • Frictional sliding
  • Plasticity
  • Shale
  • Transversely isotropic rock