Modelling the Shear Behaviour of Rock Joints with Asperity Damage Under Constant Normal Stiffness

Abstract

The shear behaviour of a rough rock joint depends largely on the surface properties of the joint, as well as the boundary conditions applied across the joint interface. This paper proposes a new analytical model to describe the complete shear behaviour of rough joints under constant normal stiffness (CNS) boundary conditions by incorporating the effect of damage to asperities. In particular, the effects of initial normal stress levels and joint surface roughness on the shear behaviour of joints under CNS conditions were studied, and the analytical model was validated through experimental results. Finally, the practical application of the model to a jointed rock slope stability analysis is presented.

Appendix

1.1 Derivations of Eq. (5)/Joint Normal Stiffness at Peak Shear Displacement

The stiffness of a rock joint is modelled by connecting the upper and lower joint surfaces with two orthogonal springs, which represent the normal stiffness (k _n) and the shear stiffness (k _s) of the joint (Bandis 1990). Due to the orthogonality of the spring connection, the normal deformation of joint under shear and normal loading can be analysed independently as a spring system using the method of superposition (Fig. 16).

Fig. 16

Non-linear spring model for simulating rock joint normal deformation at stages of prior to and peak of shearing

Several experimental results have shown that the normal stress–displacement behaviour of a joint under uniaxial loading is non-linear (Bandis et al. 1983), so it is reasonable to assume that the joint behaves similar to a non-linear spring system. Hence the normal stiffness of spring k _n can be derived by differentiating the applied normal stress $\sigma_{\text{n}}^{ * }$ with respect to the total normal deformation of spring $\delta_{\text{V}}^{\text{T}}$:

$$k_{\text{n}} = \frac{{\partial \sigma_{\text{n}}^{*} }}{{\partial \delta_{\text{v}}^{\text{T}} }} = \frac{1}{{\left( {\frac{{\partial \delta_{\text{v}}^{\text{T}} }}{{\partial \sigma_{\text{n}}^{*} }}} \right)}}.$$

(26)

Extension (+$\delta_{\text{V}}^{\text{T}}$) and shortening/compression (−$\delta_{\text{V}}^{\text{T}}$) of the spring are positive and negative as per the sign convention, respectively. Stage I represents the spring system prior to preloading or normal loading where it is assumed that the spring has a finite initial stiffness k _ni (i.e. joint initial normal stiffness at $\sigma_{\text{n}}^{ * }$ is nearly zero). Stage II represents the preloading or normal loading to the matching joint before shearing. At this stage, shortening of spring $\delta_{\text{V}}^{\text{C}}$ (i.e. initial joint closure) for any level of $\sigma_{\text{n}}^{ * }$ can be calculated from Bandis et al. (1983) hyperbolic equation, thus:

$$\delta_{\text{v}}^{\text{c}} = \left( { - \delta_{\text{v}}^{\text{T}} } \right)_{{\delta_{\text{h}} = 0}} = - \left( {\frac{{\sigma_{\text{n}}^{*} \times V_{\text{m}} }}{{k_{\text{ni}} \times V_{\text{m}} + \sigma_{\text{n}}^{ *} }}} \right),$$

(27)

where V _m is the maximum shortening (joint maximum closure). Both k _ni and V _m are assumed to be negative (closure). By differentiating Eq. (27) with respect to $\sigma_{\text{n}}^{ * }$, the spring stiffness (joint normal stiffness) prior to shearing ($\delta_{\text{h}}$ = 0) for any level of $\sigma_{\text{n}}^{ * }$ can be obtained as:

$$\left( {k_{\text{n}} } \right)_{{\delta_{\text{h}} = 0}} = \frac{1}{{\left( {\frac{{\partial \delta_{\text{v}}^{\text{T}} }}{{\partial \sigma_{\text{n}}^{ *} }}} \right)_{{\delta_{\text{h}} = 0}} }} = \frac{1}{{\frac{{k_{\text{ni}} \times V_{\text{m}}^{ 2} }}{{\left( {k_{\text{ni}} \times V_{\text{m}} + \sigma_{\text{n}}^{*} } \right)}}}}.$$

(28)

At stage III, it can be assumed that the joint dilates/opens by $\delta_{\text{V}}^{\text{d}}$ due to the peak shear displacement of $\delta_{\text{h - peak}}$, and the rate of change of normal stress with shear displacement is zero. Hence $\delta_{\text{V}}^{\text{d}}$ can be calculated by integrating Eq. (4) with respect to shear displacement in the range of 0 < $\delta_{\text{h}}$ ≤ $\delta_{\text{h - peak}}$:

$$\left( {\delta_{\text{v}}^{\text{d}} } \right) = \delta_{\text{h - peak}} \times \tan \left( {\frac{1}{M} \times {\text{JRC}} \times \log_{10} \left( {\frac{\text{JCS}}{{\sigma_{\text{n}}^{*} }}} \right)} \right).$$

(29)

Then, the total normal deformation of the spring ($\delta_{\nu }^{T}$) for stage III can be expressed as:

$$\left( {\delta_{v}^{T} } \right)_{{\delta_{h} = \delta_{h - peak} }} = \left( {\delta_{v}^{d} } \right)_{{\delta_{h} = \delta_{h - peak} }} - \left( {\delta_{v}^{c} } \right)_{{\delta_{h} = 0}} = \left( {\delta_{v}^{d} } \right)_{{\delta_{h} = \delta_{h - peak} }} + \left( {\delta_{v}^{T} } \right)_{{\delta_{h} = 0}} .$$

(30)

By differentiating Eq. (30) with respect to $\sigma_{\text{n}}^{ * }$, the following incremental formulation can be obtained:

$$\left( {\frac{{\partial \delta_{\text{v}}^{\text{T}} }}{{\partial \sigma_{\text{n}}^{*} }}} \right)_{{\delta_{\text{h}} = \delta_{\text{h - peak}} }} = \left( {\frac{{\partial \delta_{\text{v}}^{\text{d}} }}{{\partial \sigma_{\text{n}}^{*} }}} \right)_{{\delta_{\text{h}} = \delta_{\text{h - peak}} }} \, + \left( {\frac{{\partial \delta_{\text{v}}^{\text{T}} }}{{\partial \sigma_{\text{n}}^{*} }}} \right)_{{\delta_{\text{h}} = 0}} .$$

(31)

By differentiating Eq. (29) with respect to $\sigma_{\text{n}}^{ * }$ gives:

$$\left( {\frac{{\partial \delta_{\text{v}}^{\text{d}} }}{{\partial \sigma_{\text{n}}^{*} }}} \right)_{{\delta_{\text{h}} = \delta_{\text{h - peak}} }} = - \left( {\left( {\frac{{\delta_{\text{h - peak}} \times {\text{JRC}} \times \pi }}{{M \times \ln 10 \times \sigma_{\text{n}}^{*} \times 180}}} \right)\sec^{2} \left( {\frac{1}{M} \times {\text{JRC}} \times \log_{10} \left( {\frac{\text{JCS}}{{\sigma_{\text{n}}^{*} }}} \right)} \right)} \right).$$

(32)

By rearranging Eq. (28), the following equation is obtained:

$$\left( {\frac{{\partial \delta_{\text{v}}^{\text{T}} }}{{\partial \sigma_{\text{n}}^{*} }}} \right)_{{\delta_{\text{h}} = 0}} = \frac{1}{{\left( {k_{\text{n}} } \right)_{{\delta_{\text{h}} = 0}} }} = \frac{{k_{\text{ni}} \times V_{\text{m}}^{2} }}{{\left( {k_{\text{ni}} \times V_{\text{m}} + \sigma_{\text{n}}^{ *} } \right)^{2} }}.$$

(33)

By substituting Eqs. (32) and (33) into Eq. (31) and then rearranging, the stiffness of the spring at the peak shear displacement $\delta_{\text{h - peak}}$ (i.e. normal stiffness of joint at peak shear displacement) is given by:

$$\begin{aligned} \left( k_{\text{n}} \right)_{\delta_{\text{h}}= \delta_{\text{h - peak}}} = \frac{1}{{\left( {\frac{{\partial \delta_{\text{v}}^{\text{T}} }}{{\partial \sigma_{\text{n}}^{*} }}} \right)}}_{{\delta_{\text{h}} = \delta_{\text{h - peak}}}}&=\frac{1}{{ - \left( {\left( {\frac{{\delta_{\text{h - peak}} \times {\text{JRC}} \times \pi }}{{M \times \ln 10 \times \sigma_{\text{n}}^{ *} \times 180}}} \right)\sec^{2} \left( {\frac{1}{M} \times {\text{JRC}} \times \log_{10} \left( {\frac{\text{JCS}}{{\sigma_{\text{n}}^{ *} }}} \right)} \right)} \right) + \frac{{k_{\text{ni}} \times V_{\text{m}}^{2} }}{{\left( {k_{\text{ni}} \times V_{\text{m}} + \sigma_{\text{n}}^{ *} } \right)^{2} }}}}. \hfill \\ \end{aligned}$$

(34)

where $\sigma_{\text{n}}^{ * }$ = $\sigma_{\text{n0}}^{{}}$ (initial normal stress in the CNS boundary condition).