Stability analysis of rock slopes against sliding or flexural-toppling failure

Abstract

Toppling is a mode of failure that may occur in a wide range of layered rock strata in rock slopes. According to the results of physical model tests and field investigations of anti-inclined rock slopes, most real instabilities are of the sliding or flexural-toppling type. Failure often initiates at the slope toe, and the failure surface is usually multi-planar rather than planar. These properties should determined by searching rather than based on assumption. Taking these problems into account, in this paper we propose a theoretical model for rock slopes with a potential for sliding or flexural-toppling failure on the basis of two physical model tests. An innovative approach for the stability analysis of such slopes based on the limit equilibrium theory is then proposed. Subsequently, a comparative analysis is carried out using the discrete element method and the Aydan et al. method with the aim to verify the validity and accuracy of the proposed approach. Finally, the possible difference between angles of the basal calculation plane and the failure surface of the sliding zone and superimposed toppling zone with respect to the plane normal to the discontinuities is presented.

Appendices

Appendix I:

Derivation of the limit equilibrium equation when rock column i (located in the sliding zone and i ≥ 1) has the potential for shear-sliding failure.

In this case, the limit friction equilibrium condition is assumed to be satisfied along the interface of adjacent rock columns, which is generally adopted in the toppling failure analysis in previous studies (Sagaseta et al. 2001; Liu et al. 2008, 2009). Thus,

$$ {Q}_i={P}_i \tan {\varphi}_j $$

(A.1)

where Q _i is inter-column shear force acting at the common boundary of rock columns i and i + 1.

Forces parallel to potential failure surface:

$$ \sum_{j=1}^i{S}_j=\sum_{j=1}^i{w}_j \sin \theta +{P}_i \cos {\theta}_r+{Q}_i \sin {\theta}_r $$

(A.2)

Forces perpendicular to potential failure surface:

$$ \sum_{j=1}^i{N}_j=\sum_{j=1}^i{w}_j \cos \theta -{P}_i \sin {\theta}_r+{Q}_i \cos {\theta}_r $$

(A.3)

If rock columns i and below have shear-sliding potential, then:

$$ \sum_{j=1}^i{S}_j=\sum_{j=1}^i{N}_j\frac{ \tan \varphi}{F_s}+\frac{ i ct}{F_s \cos {\theta}_r} $$

(A.4)

Inserting Eqs. (A.1–A.3) into Eq. (A.4), then

$$ {P}_i=\frac{ \cos \theta \left(\frac{ \tan \varphi}{F_s}- \tan \theta \right)\sum_{j=1}^i{w}_j+\frac{ i ct}{F_s \cos {\theta}_r}}{ \cos {\theta}_r\left(1+\frac{ \tan \varphi}{F_s} \tan {\theta}_r\right)+ \tan {\varphi}_j \cos {\theta}_r\left( \tan {\theta}_r-\frac{ \tan \varphi}{F_s}\right)} $$

(A.5)

Derivation of the limit equilibrium equation when rock column i (located in the sliding zone and i ≥ 1) has the potential for flexural-toppling failure.

In this case, the limit friction equilibrium condition is also assumed to be satisfied along the interface of adjacent rock columns, which is generally adopted in flexural-toppling failure analysis in previous studies (Adhikary et al. 1997; Amini et al. 2012; Aydan and Kawamoto 1992). Thus,

$$ \left\{\begin{array}{l}{Q}_i={T}_i \tan {\varphi}_j\\ {}{Q}_{i-1}={P}_{i-1} \tan {\varphi}_j,\left( note:{P}_0=0\right)\end{array}\right. $$

(A.6)

Moments about the base midpoint:

$$ {M}_i={w}_i \sin \beta \frac{h_i}{2}+{T}_i\chi {h}_i-{Q}_i\frac{t}{2}-{P}_{i-1}\chi {h}_i-{Q}_{i-1}\frac{t}{2} $$

(A.7)

Maximum tensile stress at the base:

$$ {\sigma}_{i, \max }=\frac{M_i}{I}\frac{t}{2}-\frac{w_i \cos \beta +{Q}_i-{Q}_{i-1}}{t}\cong \frac{M_i}{I}\frac{t}{2}-\frac{w_i \cos \beta}{t} $$

(A.8)

If rock column i has flexural-toppling failure, then:

$$ {\sigma}_{i, \max }=\frac{\sigma_t}{F_s} $$

(A.9)

Inserting Eqs. (A.6 to A.8) into Eq. (A.9):

$$ \begin{array}{l}{T}_i=\frac{1}{\chi {h}_i-\frac{t}{2} \tan {\varphi}_j}\left[{P}_{i-1}\left(\chi {h}_{i-1}+\frac{t}{2} \tan {\varphi}_j\right)\right.+\\ {}\kern7.5em \left.\left(\frac{\sigma_t}{F_s}+\frac{w_i \cos \beta}{t}\right)\frac{2 I}{t}-\frac{h_i}{2}{w}_i \sin \beta \right]\end{array} $$

(A.10)

Appendix II

Derivation of the critical height of the cantilever inclined beam under self-weight (located in the cantilevered toppling zone).

In this case, the inter-column forces between rock columns are assumed to be 0.

Weight of the rock column with critical height (see Fig. 10):

$$ w=\gamma {h}_{cr} t $$

(A.11)

Moments about the base midpoint:

$$ M= w \sin \beta \frac{h_{cr}}{2} $$

(A.12)

Maximum tensile stress at the base:

$$ {\sigma}_{\max }=\frac{M}{I}\frac{t}{2}-\frac{w \cos \beta}{t} $$

(A.13)

If the rock column has flexural-toppling failure, then:

$$ {\sigma}_{\max }={\sigma}_t $$

(A.14)

Inserting Eqs. (A.11 to A.13) into Eq. (A.14):

$$ {h}_{cr}=\frac{t \cos \beta +\sqrt{t^2{ \cos}^2\beta +12 t \sin {\beta \sigma}_t/\gamma}}{6 \sin \beta} $$

(A.15)

Appendix III

The user-defined FISH function to obtain the magnitude and location of normal force of contacts between adjacent blocks (please see UDEC 4.0 Help for what the following variable names denote).

def c_result.

b1 = 39,130.

b2 = 39,339.

n = 1.

ic = contact_head.

loop while ic # 0.

if c_b1(ic) = b1.

if c_b2(ic) = b2.

xtable(100,n) = c_y(ic).

ytable(100,n) = c_nforce(ic).

n = n + 1.

endif.

endif.

if c_b1(ic) = b2.

if c_b2(ic) = b1.

xtable(100,n) = c_y(ic).

ytable(100,n) = c_nforce(ic).

n = n + 1.

endif.

endif.

ic = c_next(ic).

endloop.

end

c_result.

set log on.

set logfile cb.log.

pr Table 100.

set log off.

Based on the above user-defined FISH function, the point where total side force acts between rock columns 7 and 8, 16 and 17, and 20 and 21 has been determined, as shown in Table 9. Taking the total side force between columns 7 and 8 as an example, the calculation process is illustrated as follows.

Location of total side force:

$$ {Y}_{7,8}=\frac{\sum {\left(\mathrm{Y}\hbox{-} \mathrm{c}\right)}_i\times {\left(\mathrm{N}\hbox{-} \mathrm{f}\right)}_i}{\sum {\left(\mathrm{N}\hbox{-} \mathrm{f}\right)}_i}=\frac{55.2\times 120+54.4\times 117.4+53.5\times 131.2+52.6\times 121.7}{55.2+54.4+53.5+52.6}=53.9 $$

(A.16)

Then, χ can be determined

$$ \chi =\frac{53.9-52.6}{55.2-52.6}=0.5 $$

(A.17)

Table 9 Magnitude and location of normal force of contacts between adjacent rock columns (blocks) above the failure surface