Improvement of Discontinuous Deformation Analysis Incorporating Implicit Updating Scheme of Friction and Joint Strength Degradation

Abstract

Dynamic analyses of jointed rock masses, such as slope stability evaluations, have been intensively studied in recent years; discontinuous deformation analysis (DDA) is one of the feasible methods in this field. To predict the failure mode, the kinematic energy, and the reaching distance after the collapse, the friction characteristics of the rock joints must be computed correctly. However, conventional DDA often cannot easily determine appropriate analysis parameters that can predict the sliding displacement between the rock masses precisely, and at worst, the computation fails. In this study, to enhance the robustness and accuracy of the DDA, the updating procedure of the friction force is investigated. An implicit updating scheme of friction force based on the elasto-plastic framework, the so-called return mapping algorithm, and the Newton–Raphson iteration is newly introduced to the DDA. Additionally, the transition algorithm between the static and the residual friction is also implemented to reproduce friction strength degradation. The performance of the proposed full implicit DDA is examined through the numerical examples and improvements in the robustness, accuracy, and computational efficiency for the dynamic sliding problems along the rock joint were confirmed.

Appendix: Variation and the Linearization of the Direction Vectors of Contact Surface

In this appendix, the derivation of the variation and the linearization of the normal and the shear direction vectors of a contact surface used to formulate the consistent tangent contact stiffness matrix are shown. The contact between two objects at a single point is assumed (Fig. 7). Although the proposed method is formulated in two-dimensional condition, to express the geometrical states of the contact surface, a local basis (τ_k, e₃, n_k) on the contacting edge P₂P₃ is defined in a three-dimensional coordinate system including the out-of-plane direction. τ_k is the unit tangential vector along the edge P₂P₃ (the direction from P₂ to P₃ is positive):

$${{\varvec{\uptau}}}_{k} = \frac{1}{l}\left( {{\mathbf{x}}_{3} - {\mathbf{x}}_{2} } \right) = \frac{1}{l}\left\{ {\begin{array}{*{20}c} {x_{3} - x_{2} } \\ {y_{3} - y_{2} } \\ 0 \\ \end{array} } \right\}.$$

(89)

e₃ is the unit normal vector in the deep direction (vertical direction to a paper surface is positive):

$${\mathbf{e}}_{3} = \left\{ {\begin{array}{*{20}c} 0 \\ 0 \\ 1 \\ \end{array} } \right\},$$

(90)

and n_k is the outward unit normal vector obtained using τ_k and e₃ as follows:

$${\mathbf{n}}_{k} = {\mathbf{e}}_{3} \times {{\varvec{\uptau}}}_{k} = \frac{1}{l}\left\{ {\begin{array}{*{20}c} {y_{2} - y_{3} } \\ {x_{3} - x_{2} } \\ 0 \\ \end{array} } \right\},$$

(91)

where l is the length of the edge P₂P₃.

First, the variation of the normal direction vector of a contact surface δn_k is derived by taking the variation of Eq. (90):

$$\delta {\mathbf{n}}_{k} = {\mathbf{e}}_{3} \times \delta {{\varvec{\uptau}}}_{k} .$$

(92)

The variation of the shear direction vector of a contact surface δτ_k is derived as follows:

$$\begin{aligned} \delta {{\varvec{\uptau}}}_{k} &= \delta \left( {\frac{1}{l}\left( {{\mathbf{x}}_{3} - {\mathbf{x}}_{2} } \right)} \right) \hfill \\ &= - \frac{\delta l}{{l^{2} }}\left( {{\mathbf{x}}_{3} - {\mathbf{x}}_{2} } \right) + \frac{1}{l}\left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) \hfill \\ &= - \frac{\delta l}{l}{{\varvec{\uptau}}}_{k} + \frac{1}{l}\left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right). \hfill \\ \end{aligned}$$

(93)

δl in Eq. (93) is defined as follows:

$$\begin{aligned} \delta l &= \delta \sqrt {\left( {{\mathbf{x}}_{3} - {\mathbf{x}}_{2} } \right) \cdot \left( {{\mathbf{x}}_{3} - {\mathbf{x}}_{2} } \right)} \hfill \\ &= \frac{{{\mathbf{x}}_{3} - {\mathbf{x}}_{2} }}{{\sqrt {\left( {{\mathbf{x}}_{3} - {\mathbf{x}}_{2} } \right) \cdot \left( {{\mathbf{x}}_{3} - {\mathbf{x}}_{2} } \right)} }} \cdot \left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) \hfill \\ &= {{\varvec{\uptau}}}_{k} \cdot \left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right). \hfill \\ \end{aligned}$$

(94)

Therefore, δτ_k is expressed as:

$$\begin{aligned} \delta {{\varvec{\uptau}}}_{k} &= - \frac{1}{l}\left\{ {{{\varvec{\uptau}}}_{k} \cdot \left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right)} \right\}{{\varvec{\uptau}}}_{k} + \frac{1}{l}\left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) \hfill \\ &= - \frac{1}{l}{{\varvec{\uptau}}}_{k} \otimes {{\varvec{\uptau}}}_{k} \left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) + \frac{1}{l}\left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) \hfill \\ &= \frac{1}{l}\left( {{\mathbf{I}} - {{\varvec{\uptau}}}_{k} \otimes {{\varvec{\uptau}}}_{k} } \right)\left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) \hfill \\ &= \frac{1}{l}{\mathbf{n}}_{k} \otimes {\mathbf{n}}_{k} \left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) \hfill \\ &= \frac{1}{l}{\mathbf{n}}_{k} \left( {{\mathbf{n}}_{k} \cdot \left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right)} \right), \hfill \\ \end{aligned}$$

(95)

where I is the second-order unit tensor. Substituting Eq. (95), Eq. (92) yields:

$$\begin{aligned} \delta {\mathbf{n}}_{k} &= {\mathbf{e}}_{3} \times \frac{1}{l}\left( {\left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) \cdot {\mathbf{n}}_{k} } \right){\mathbf{n}}_{k} \hfill \\ &= - \frac{1}{l}{{\varvec{\uptau}}}_{k} \left( {{\mathbf{n}}_{k} \cdot \left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right)} \right). \hfill \\ \end{aligned}$$

(96)

Equation (96) shows that δn_k and τ_k are parallel with each other. The linearization of the direction vectors Δn_k and Δτ_k can be derived in the same manner as with δn_k and δτ_k.

$$\Delta {\mathbf{n}}_{k} = - \frac{1}{l}{{\varvec{\uptau}}}_{k} \left( {{\mathbf{n}}_{k} \cdot \left( {\Delta {\mathbf{u}}_{3} - \Delta {\mathbf{u}}_{2} } \right)} \right),$$

(97)

$$\Delta {{\varvec{\uptau}}}_{k} = \frac{1}{l}{\mathbf{n}}_{k} \left( {{\mathbf{n}}_{k} \cdot \left( {\Delta {\mathbf{u}}_{3} - \Delta {\mathbf{u}}_{2} } \right)} \right).$$

(98)

Applying Eq. (31) and (36), Eqs. (95)–(98) yields:

$$\delta {\mathbf{n}}_{k} = - \frac{1}{{l^{2} }}\left\{ {\begin{array}{*{20}c} {x_{3} - x_{2} } \\ {y_{3} - y_{2} } \\ \end{array} } \right\}\left[ {J_{k} } \right]\left\{ {\delta d_{k} } \right\},$$

(99)

$$\Delta {\mathbf{n}}_{k} = - \frac{1}{{l^{2} }}\left\{ {\begin{array}{*{20}c} {x_{3} - x_{2} } \\ {y_{3} - y_{2} } \\ \end{array} } \right\}\left[ {J_{k} } \right]\left\{ {\Delta d_{k} } \right\},$$

(100)

$$\delta {{\varvec{\uptau}}}_{k} = \frac{1}{{l^{2} }}\left\{ {\begin{array}{*{20}c} {y_{2} - y_{3} } \\ {x_{3} - x_{2} } \\ \end{array} } \right\}\left[ {J_{k} } \right]\left\{ {\delta d_{k} } \right\}\;,{\text{ and}}$$

(101)

$$\Delta {{\varvec{\uptau}}}_{k} = \frac{1}{{l^{2} }}\left\{ {\begin{array}{*{20}c} {y_{2} - y_{3} } \\ {x_{3} - x_{2} } \\ \end{array} } \right\}\left[ {J_{k} } \right]\left\{ {\Delta d_{k} } \right\},$$

(102)

[J_k] is the matrix shown in Eq. (45), and the out-of-plane components of the vectors and matrices that are zero are ignored.

Next, Δδn_k is derived by linearizing Eq. (96).

$$\begin {aligned} \Delta \delta {\mathbf{n}}_{k} = \frac{\Delta l}{{l^{2} }}\left( {\left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) \cdot {\mathbf{n}}_{k} } \right){{\varvec{\uptau}}}_{k} - \frac{1}{l}\left( {\left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) \cdot \Delta {\mathbf{n}}_{k} } \right){{\varvec{\uptau}}}_{k} - \frac{1}{l}\left( {\left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) \cdot {\mathbf{n}}_{k} } \right)\Delta {{\varvec{\uptau}}}_{k} .\end{aligned}$$

(103)

Δl has a similar form to δl as follows:

$$\Delta l = {{\varvec{\uptau}}}_{k} \cdot \left( {\Delta {\mathbf{u}}_{3} - \Delta {\mathbf{u}}_{2} } \right).$$

(104)

Equation (103) is transformed into the following equation using Eqs. (95)–(98) and (104).

$$\begin{aligned} \Delta \delta {\mathbf{n}}_{k} &= \frac{1}{{l^{2} }}\left( {\left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) \cdot {\mathbf{n}}_{k} } \right){{\varvec{\uptau}}}_{k} \left( {{{\varvec{\uptau}}}_{k} \cdot \left( {\Delta {\mathbf{u}}_{3} - \Delta {\mathbf{u}}_{2} } \right)} \right) \hfill \\ &+ \frac{1}{{l^{2} }}\left( {\left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) \cdot {{\varvec{\uptau}}}_{k} } \right){{\varvec{\uptau}}}_{k} \left( {{\mathbf{n}}_{k} \cdot \left( {\Delta {\mathbf{u}}_{3} - \Delta {\mathbf{u}}_{2} } \right)} \right) \hfill \\ &- \frac{1}{{l^{2} }}\left( {\left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) \cdot {\mathbf{n}}_{k} } \right){\mathbf{n}}_{k} \left( {{\mathbf{n}}_{k} \cdot \left( {\Delta {\mathbf{u}}_{3} - \Delta {\mathbf{u}}_{2} } \right)} \right). \hfill \\ \end{aligned}$$

(105)

After substituting Eqs. (36), (45), and (57), Δδn_k is expressed as follows (ignoring the out-of-plane components):

$$\begin{aligned} \Delta \delta {\mathbf{n}}_{k} &= \frac{1}{l}\left\{ {\delta d_{k} } \right\}^{{\text{T}}} \left[ {J_{k} } \right]^{{\text{T}}} \left\{ {\begin{array}{*{20}c} {x_{3} - x_{2} } \\ {y_{3} - y_{2} } \\ \end{array} } \right\}\left[ {L_{k} } \right]\left\{ {\Delta d_{k} } \right\} \hfill \\ &+ \frac{1}{l}\left\{ {\delta d_{k} } \right\}^{{\text{T}}} \left[ {L_{k} } \right]^{{\text{T}}} \left\{ {\begin{array}{*{20}c} {x_{3} - x_{2} } \\ {y_{3} - y_{2} } \\ \end{array} } \right\}\left[ {J_{k} } \right]\left\{ {\Delta d_{k} } \right\} \hfill \\ &- \frac{1}{l}\left\{ {\delta d_{k} } \right\}^{{\text{T}}} \left[ {J_{k} } \right]^{{\text{T}}} \left\{ {\begin{array}{*{20}c} {y_{2} - x_{3} } \\ {x_{3} - x_{2} } \\ \end{array} } \right\}\left[ {J_{k} } \right]\left\{ {\Delta d_{k} } \right\}. \hfill \\ \end{aligned}$$

(106)

Finally, Δδτ_k is derived by linearizing Eq. (95) as shown below.

$$\begin{aligned} \Delta \delta {{\varvec{\uptau}}}_{k} &= - \frac{1}{{l^{2} }}\left( {\left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) \cdot {\mathbf{n}}_{k} } \right){\mathbf{n}}_{k} \left( {{{\varvec{\uptau}}}_{k} \cdot \left( {\Delta {\mathbf{u}}_{3} - \Delta {\mathbf{u}}_{2} } \right)} \right) \hfill \\ &- \frac{1}{{l^{2} }}\left( {\left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) \cdot {{\varvec{\uptau}}}_{k} } \right){\mathbf{n}}_{k} \left( {{\mathbf{n}}_{k} \cdot \left( {\Delta {\mathbf{u}}_{3} - \Delta {\mathbf{u}}_{2} } \right)} \right) \hfill \\ &- \frac{1}{{l^{2} }}\left( {\left( {\delta {\mathbf{u}}_{3} - \delta {\mathbf{u}}_{2} } \right) \cdot {\mathbf{n}}_{k} } \right){{\varvec{\uptau}}}_{k} \left( {{\mathbf{n}}_{k} \cdot \left( {\Delta {\mathbf{u}}_{3} - \Delta {\mathbf{u}}_{2} } \right)} \right). \hfill \\ \end{aligned}$$

(107)

Equations (96)–(98) and Eq. (104) are also considered. Substituting Eqs. (36), (45), (57), Δδτ_k is expressed in the matrix form:

$$\begin{aligned} \Delta \delta {{\varvec{\uptau}}}_{k} &= - \frac{1}{l}\left\{ {\delta d_{k} } \right\}^{{\text{T}}} \left[ {J_{k} } \right]^{{\text{T}}} \left\{ {\begin{array}{*{20}c} {y_{2} - x_{3} } \\ {x_{3} - x_{2} } \\ \end{array} } \right\}\left[ {L_{k} } \right]\left\{ {\Delta d_{k} } \right\} \hfill \\ &- \frac{1}{l}\left\{ {\delta d_{k} } \right\}^{{\text{T}}} \left[ {L_{k} } \right]^{{\text{T}}} \left\{ {\begin{array}{*{20}c} {y_{2} - x_{3} } \\ {x_{3} - x_{2} } \\ \end{array} } \right\}\left[ {J_{k} } \right]\left\{ {\Delta d_{k} } \right\} \hfill \\ &- \frac{1}{l}\left\{ {\delta d_{k} } \right\}^{{\text{T}}} \left[ {J_{k} } \right]^{{\text{T}}} \left\{ {\begin{array}{*{20}c} {x_{3} - x_{2} } \\ {y_{3} - y_{2} } \\ \end{array} } \right\}\left[ {J_{k} } \right]\left\{ {\Delta d_{k} } \right\}. \hfill \\ \end{aligned}$$

(108)

Using Eqs. (99)–(102), Eq. (106), and Eq. (108), the matrix form of δg_N, Δg_N, Δδg_N, δg_S, Δg_S, and Δδg_S shown in Sect. 2.5.2 can be derived.