Slope Stability Analyses Subjected to Slide Head Toppling Failure Mechanisms

Abstract

It is faulty to analyze the jointed rock slopes’ stability susceptible to a combination of modes of failure composed of sliding around the toe region and toppling of the rock blocks on the upper part of the slope based on the current analytical methods, which are based on assumption that the distribution of the potential failure surface bounding the potential mixed failure runs predictably from crest to toe of the slope. An Analytical model that takes into consideration the kinematic mechanism of the discontinuous rock slope with counter-tilted weak plane subjected to a combination of failure mechanisms involving sliding and toppling has hence been presented in this chapter. This involves an iterative process which involves the calculations of the dimensions of all the individual blocks as well as the forces acting on them, and then stability of every block is examined, starting at the uppermost block. The stability analysis of each block is determined. The blocks may either be stable, topple or slide. The proposed analytical methods could curtail errors incurred due to the acceptance of the single weak plane for quantifying the failure mechanisms composed of slide head toppling rock slopes in physical situations with two planar weak planes.

1. Introduction

A lot of research has been done in the current analytical models for predicting rock slope stability subjected to failure by block toppling. However, their contributions have focused always on the predictive and idealized geometry comprising blocks with joints dipping into the slope face. In addition, the jointing bounding the latent toppling blocks is predicted to be systematic running from the uppermost surface of the slope, and day-lighting near the toe. It is however possible that in some notable situations, the jointing (discontinuity) bounding the potential toppling rock blocks may not be predictive due to geological variations. Hence, the joints that bound the potential sliding and toppling blocks may be counter-tilted in the rock mass due to geological variations and may daylight on an unpredicted point on the slope face. This would lead to a combination of failure mechanisms composed of toppling and other secondary failure mechanisms such as toppling-circular, slide-head-toppling and block-flexure toppling. Not much research has been done on a combination of failure modes which involve sliding and block toppling on the discontinuous rock slope.

Regardless of this, a combination of failure mechanisms involving sliding and block toppling of the upper section of the slope respectively, continue to be a common phenomenon in sedimentary rock formation in numerous constructed and natural slopes all over the world as illustrated in Figure 1a, and b.

Figure 1.

Mixed toppling failure modes involving toppling and sliding failures observed in practice (a) post-failure photograph of sliding and toppling of north wall slope of Nchanga open pits, Zambia [1] (b) post-failure photograph of sliding and toppling failure in Zhongliang reservoir bank, China [2].

Furthermore, another typical example of slide-head-toppling failure is an open pit mine slope in Valencia in Spain shown in Figure 2. Figure 2 shows a series of stable rock blocks in the soil mass on the upper surface of the slope, another set of toppling columnar blocks in the limestone formation midway section and finally a series of sliding blocks in the variation of the weak rock formation on the slope’s lower part.

Figure 2.

Valencia open pit mine with the potential of mixed failure modes composed of toppling and sliding failures in Spain [3].

It is not entirely correct to analyze the stability of discontinuous rock slopes with a high potential to mixed toppling failure modes of toppling on the upper section of the slope and sliding on the slope’s lower section using the existing limit equilibrium method, which is based on the initial assumptions of distribution of the potential failure surface bounding the blocks susceptible mixed failure mode. The theoretical/analytical model is therefore necessary that it takes into consideration the kinematic mechanisms of the discontinuous rock slopes composed of the counter-tilted jointing susceptible/subjected to a combination of failure mechanisms (sliding and toppling).

This chapter presents a two-dimensional (2D) model which incorporates the counter-tilted oriented jointing bounding the potential for both sliding and toppling. From the 2D model, a theoretical (analytical) model that is based on the trigonometry basic principles to determine the geometry of the jointed rock slope composed of the counter-tilted jointing subjected to slide-head toppling failure mechanisms have been put forward. Next, a process of iteration is followed with forces acting on each block determined. A comparative analysis between the stability against sliding and toppling of individual blocks is determined, with the highest forces between these two deciding the failure mechanism of the succeeding block. In summary, the overall slope stability is considered based on whether the lowermost block either slides or topples.

2. Rock slope geometry

For stability analyses of the slope to be achieved, detailed slope geometries must be determined at the initial stage. Let us examine a rock slope in Figure 3, which is composed of rectangular rock columns with width, $\mathrm{\Delta }x$ and height, $\mathit{yn}$ in an orderly manner. The original failure plane dipping at $\mathit{\psi p}$ is counter-tilted from the preliminary weak plane approximately on the mid-way part of the slope at angle, $\mathit{\theta r}$. The counter-tilted failure plane then dips at $\mathit{\psi c}$. Other slope parameters include the slope height indicated as H, the face angle denoted as $\mathit{\psi f}$ and the dip/angle of the uppermost slope face referred to as $\mathit{\psi s}$. The rock block column count started from the lower most of the slope (toe) increasing cumulatively upwards. It has been observed and noted from centrifugal and numerical test models that the base plane tends to be stepped during toppling failure mechanisms [4, 5, 6, 7, 8]. This greatly influences the determination of the overall angle of the plane base denoted as $\mathit{\psi b}$.

Figure 3.

Rock slope composed of counter-tilted weak surface subjected to toppling failure mechanism [1, 4].

In rock slopes where the failure plane is counter-tilted around the midway section of the rock slope, it is important to approximate the base plane angle ($\mathit{\psi b}$) considering the dipping of the failure plane bases for the rock block columns at $\mathit{\psi p}$ and $\mathit{\psi c}$ using Eq. (1)

$$\mathit{\psi b}\approx \left(\frac{\mathit{\psi c}+\mathit{\psi p}}{2}+10°\right)\text{Or}\mathit{\psi b}\approx \left(\frac{\mathit{\psi c}+\mathit{\psi p}}{2}+30°\right)$$E1

Based on the slope geometry in Figure 3. Eq. (2) is formulated to determine the number of rock columns, n, forming the regular system of the slope.

$$n=\frac{H}{\mathrm{\Delta }x}\left[cos\mathit{ec}\left(\mathit{\psi b}\right)+\left(\frac{cot\left(\mathit{\psi b}\right)-cot\left(\mathit{\psi f}\right)}{sin\left(\mathit{\psi b}-\mathit{\psi f}\right)}\right)sin\mathit{\psi s}\right]$$E2

Considering the slope model presented in Figure 3, the height, $\mathit{yn}$ for the n^th rock column below the slope’s crest dipping/orientated into the slope face is determined by Eq. (3).

$$\mathit{yn}=n\left(\mathrm{a}1-b\right)$$E3

The height, $\mathit{yn}$ for the rock column denoted as n^th above the slope’s crest dipping into the slope face is resolved by Eq. (4).

$$\mathit{yn}=\mathit{yn}-1-\mathrm{a}2-b$$E4

The constants $\mathrm{a}1$, $\mathrm{a}2$ and $b$ shown in Figure 3 are found based on the rock columns and the related slope geometry. These constants are calculated by Eqs. (5)–(7) respectively;

$$\mathrm{a}1=\mathrm{\Delta }xtan\left(\mathit{\psi f}-\mathit{\psi c}\right)$$E5

$$\mathrm{a}2=\mathrm{\Delta }xtan\left(\mathit{\psi p}-\mathit{\psi s}\right)$$E6

$$b=\mathrm{\Delta }xtan\left(\mathit{\psi b}-\mathit{\psi p}\right)$$E7

Other distinctive dimensions of the blocks required to be determined includes; the points of application denoted $\mathit{Mn}$ and $\mathit{Ln}$ for the shear and normal forces ($\mathrm{Rn},\mathrm{Sn}$) that acts on the bases of weak planes and ($\mathrm{Pn},\mathrm{Qn},\mathrm{Pn}-1,\mathrm{Qn}-1$) forces applied on interfaces next to the rock columns during toppling and sliding failure mechanisms as illustrated in Figure 3. The application points of all forces needs to be determined every time toppling failure mechanisms takes place. Considering the case where the n^th block is below the slope’s crest, the application points $\mathit{Mn}$ and $\mathit{Ln}$ are determined by Eqs. (8)and (9) respectively, at the crest block, then the application points $\mathit{Mn}$ and $\mathit{Ln}$ are determined by Eqs. (10) and (11) and finally when the n^th block is over the slope’s crest, then the application points $\mathit{Mn}$ and $\mathit{Ln}$ are calculated by Eqs. (12) and (13).

$$\mathit{Mn}=\mathit{yn}$$E8

$$\mathit{Ln}=\mathit{yn}-\mathrm{a}1$$E9

$$\mathit{Mn}=\mathit{yn}-\mathrm{a}2$$E10

$$\mathit{Ln}=\mathit{yn}-\mathrm{a}1$$E11

$$\mathit{Mn}=\mathit{yn}-\mathrm{a}2$$E12

$$\mathit{Ln}=\mathit{yn}$$E13

2.1 Block stability

Figure 4 illustrates a rock slope composed of the counter-tilted weak (failure) surface composed of three sets of rock blocks that have been categorized based on their failure behavior. The rock columns for the upper most section of the slope are regarded as stable based on the fact that the rock columns’ base friction angle is more than the original failure plane angle. For the rock block columns in the midway section of the slope, toppling failure is observed based on the consideration that the base plane lies outside the Centre of gravity for the rock block columns. For the blocks located around the lowermost section of the slope, there is a high possibility of rock blocks undergoing failure as the initial failure plane is counter-tilted. The modes of failure for the block columns are influenced by the geometry of the slope, the slope angle and the failure/weak plane angle. The friction angles of these failure surfaces differ with respect to characteristics of the lithology.

Figure 4.

Limit equilibrium conditions required for modes of failure of the nth rock columns: (a) forces applied on the nth rock columns; (b) toppling mode of failure of the nth rock columns; (c) sliding modes of failure of the nth rock columns considering the counter-tilted failure plane.

Generally, for such conditions, the friction angle of the block interfaces’ of the rock block columns denoted ($\mathit{\varphi d}$) are presumed to be equal to the angle of friction on the base plane ($\varphi \mathrm{p},\mathit{\varphi c}$). The two sides’ shearing forces of the rock column are then resolved using Eqs. (14) and (15), considering the limiting equilibrium analysis. On the other hand, the forces applied on the rock column considering the orthogonal and sides’ friction angles to the two base planes ($\mathit{\psi p}$, $\mathit{\psi c}$) of the rock column having weight, $\mathit{Wn}$, are resolved using Eqs. (16) and (17).

$$\mathit{Qn}=\mathit{Pn}tan\mathit{\varphi d}$$E14

$$\mathit{Qn}-1=\mathit{Pn}-1tan\mathit{\varphi d}$$E15

$$\begin{array}{l}\mathit{Rn}=\mathit{Wn}cos\mathit{\psi p}+\left(\mathit{Pn}-\mathit{Pn}-1\right)tan\mathit{\varphi d}\\ \mathit{Rn}=\mathit{Wn}cos\mathit{\psi c}+\left(\mathit{Pn}-\mathit{Pn}-1\right)tan\mathit{\varphi d}\end{array}$$E16

$$\begin{array}{l}\mathit{Sn}=\mathit{Wn}sin\mathit{\psi p}+\left(\mathit{Pn}-\mathit{Pn}-1\right)\\ \mathit{Sn}=\mathit{Wn}sin\mathit{\psi c}+\left(\mathit{Pn}-\mathit{Pn}-1\right)\end{array}$$E17

To define the magnitude of $\mathit{Pn}$, the moment force about the pivot point O is set to zero (refer Figure 4). We then, as discussed above, assume that the angle of friction of the interfaces of the rock block columns denoted ($\mathit{\varphi d}$) is equal in magnitude to the angle of friction on the base planes ($\varphi \mathrm{p},\mathit{\varphi c}$), thus $\mathit{\varphi d}=\mathit{\varphi p}=\mathit{\varphi c}=\varphi$.

Considering the possibilities of rotational equilibrium, the forces $\mathrm{Pn},\mathrm{t}$ necessary to avoid toppling of the nth block considering the presumed angle of the failure plane is resolved based on Eq. (18) considering the derivations as follows:

The moment about the pivot point 0 is set to zero. Then we have:

$$\sum M0\oplus =0$$

$$=\frac{\mathit{yn}}{2}\mathit{Wn}sin\mathit{\psi p}-\frac{\mathrm{\Delta }x}{2}\mathit{Wn}cos\mathit{\psi p}+\mathit{\text{MnPn}}+1-\mathrm{\Delta }x\left(\mathit{Pn}+1tan\varphi \right)-\mathit{\text{LnPn}}$$

$$=\frac{\mathit{Wn}}{2}\left(\mathit{yn}sin\mathit{\psi p}-\mathrm{\Delta }xcos\mathit{\psi p}\right)+\mathit{Pn}+1\left(\mathit{Mn}-\mathrm{\Delta }xtan\varphi \right)-\mathit{\text{LnPn}}$$

We can rewrite this as follows:

$$\mathit{\text{LnPn}}=\frac{\mathit{Wn}}{2}\left(\mathit{yn}sin\mathit{\psi p}-\mathrm{\Delta }xcos\mathit{\psi p}\right)+\mathit{Pn}+1\left(\mathit{Mn}-\mathrm{\Delta }xtan\varphi \right)$$

Hence this gives the following equation for $\mathit{Pn}$

$$\mathit{Pn},t=\frac{\frac{\mathit{Wn}}{2}\left(\mathit{yn}sin\mathit{\psi p}-\mathrm{\Delta }xcos\mathit{\psi p}\right)+\mathit{Pn}+1\left(\mathit{Mn}-\mathrm{\Delta }xtan\varphi \right)}{\mathit{Ln}}$$E18

Considering the possibilities of rotational equilibrium, the forces $\mathrm{Pn},\mathrm{t}$ necessary to avoid toppling for the nth block taking into consideration the counter-tilted angle of the failure plane is determined using Eq. (19) considering the derivations as follows;

Again, the moment about the pivot point O is set to zero. Then we have:

$$\sum M0\oplus =0$$

$$=\frac{\mathit{yn}}{2}\mathit{Wn}sin\mathit{\psi c}-\frac{\mathrm{\Delta }x}{2}\mathit{Wn}cos\mathit{\psi c}+\mathit{\text{MnPn}}+1-\mathrm{\Delta }x\left(\mathit{Pn}+1tan\varphi \right)-\mathit{\text{LnPn}}$$

$$=\frac{\mathit{Wn}}{2}\left(\mathit{yn}sin\mathit{\psi c}-\mathrm{\Delta }xcos\mathit{\psi c}\right)+\mathit{Pn}+1\left(\mathit{Mn}-\mathrm{\Delta }xtan\varphi \right)-\mathit{\text{LnPn}}$$

We can rewritten this as follows;

$$\mathit{\text{LnPn}}=\frac{\mathit{Wn}}{2}\left(\mathit{yn}sin\mathit{\psi c}-\mathrm{\Delta }xcos\mathit{\psi c}\right)+\mathit{Pn}+1\left(\mathit{Mn}-\mathrm{\Delta }xtan\varphi \right)$$

Therefore, we obtain the following equation for $\mathit{Pn},t$ with respect to counter-tilting of the failure plane:

$$\mathit{Pn},t=\frac{\frac{\mathit{Wn}}{2}\left(\mathit{yn}sin\mathit{\psi c}-\mathrm{\Delta }xcos\mathit{\psi c}\right)+\mathit{Pn}+1\left(\mathit{Mn}-\mathrm{\Delta }xtan\varphi \right)}{\mathit{Ln}}$$E19

If the rock columns’ failure mode on the slope’s lower section shown in Figure 4c is noted to be sliding, then to calculate the magnitude of P_n,s, the total (sum) of the forces in the horizontal directions and vertical directions is set to zero (assuming the horizontal axis to lie along the base plane, at angles $\mathit{\psi p}$ or $\mathit{\psi c}$ considering the horizontal as the datum refer Figure 4. In general, it is presumed that the interfaces’ angle of friction for the rock columns denoted ($\mathit{\varphi d}$) is equal to the friction angle on the base plane ($\varphi \mathrm{p},\mathit{\varphi c}$), thus $\mathit{\varphi d}=\mathit{\varphi p}=\mathit{\varphi c}=\varphi$.

Considering the possibilities of rotational equilibrium, the forces $\mathrm{Pn},s$ required to prevent sliding of block n considering the initial angle of the weak plane is resolved using Eq. (20) considering the derivations as follows;

Before counter-tilting of the failure plane along base plane angle $\mathit{\psi p}$ taking the horizontal as the datum, we have

$$\sum \mathit{Fx}=0=\mathit{Pn}-\mathit{Pn}+1-\mathit{Wn}sin\mathit{\psi p}+\mathit{Sn}$$

$$\sum \mathit{Fy}=0=\mathit{Rn}+\mathit{Pn}tan\varphi -\mathit{Wn}cos\mathit{\psi p}-\mathit{Pn}+1tan\varphi$$

Where $\mathit{Sn}$ is shear force acting along the base of the column contacts, from the Mohr-Coulomb criterion, we have:

$$\mathit{\tau n}=c+\mathit{\sigma n}tan\varphi$$

Where $\mathit{\tau n}$ shear stress acting along the base of the column contact and $\mathit{\sigma n}$ Normal stress at the base column contact.

We further assumed the cohesion (c) being negligible and divide both sides of the equation by the column-base plane contact area denoted by A. Thus, we get

$$\mathit{Sn}=\mathit{Rn}tan\varphi$$

Where; $\mathit{Rn}$ is normal force acting across the base plane of the column contacts ($\mathit{Wn}cos\mathit{\psi p}$).

By resolving the F_y equation for $\mathit{Rn}$ making substitutions into the F_x equation prior to counter-tilting of the failure plane along base plane angle $\mathit{\psi p}$ with regards to the horizontal, we have,

$$\mathit{Pn}-\mathit{Pn}+1-\mathit{Wn}sin\mathit{\psi p}+\mathit{Wn}cos\mathit{\psi p}tan\varphi -\left(\mathit{Pn}-\mathit{Pn}+1\right){tan}^2\varphi =0$$

$$\left(\mathit{Pn}-\mathit{Pn}+1\right)\left(1-{tan}^2\varphi \right)-\mathit{Wn}sin\mathit{\psi p}+\mathit{Wn}cos\mathit{\psi p}tan\varphi =0$$

This leads to:

$$\left(\mathit{Pn}-\mathit{Pn}+1\right)\left(1-{tan}^2\varphi \right)=\mathit{Wn}cos\mathit{\psi p}\left(tan\mathit{\psi p}-tan\varphi \right)$$

$$\left(\mathit{Pn}-\mathit{Pn}+1\right)=\frac{\mathit{Wn}cos\mathit{\psi p}\left(tan\mathit{\psi p}-tan\varphi \right)}{1-{tan}^2\varphi }$$

Hence, we obtain the following equation for $\mathit{Pn},s$

$$\mathit{Pn},s=\mathit{Pn}+1+\frac{\mathit{Wn}cos\mathit{\psi p}\left(tan\mathit{\psi p}-tan\varphi \right)}{1-{tan}^2\varphi }$$E20

Considering the possibilities of rotational equilibrium, the forces $\mathrm{Pn},s$ necessary to avoid sliding of nth block taking into consideration the counter-tilted angle of the weak plane is resolved using Eq. (21) considering the derivations as follows;

Once counter-tilting of the failure plane takes place along base plane $\mathit{\psi c}$ taking into consideration the horizontal line, we have

$$\sum \mathit{Fx}=0=\mathit{Pn}-\mathit{Pn}+1-\mathit{Wn}sin\mathit{\psi c}+\mathit{Sn}$$

$$\sum \mathit{Fy}=0=\mathit{Rn}+\mathit{Pn}tan\varphi -\mathit{Wn}cos\mathit{\psi c}-\mathit{Pn}+1tan\varphi$$

By solving the F_y equation for $\mathit{Rn}$ and making substitutions into the F_x equation once counter-tilting of the failure plane takes place along base plane $\mathit{\psi c}$ taking the horizontal as the datum, we have,

$$\mathit{Pn}-\mathit{Pn}+1-\mathit{Wn}sin\mathit{\psi c}+\mathit{Wn}cos\mathit{\psi c}tan\varphi -\left(\mathit{Pn}-\mathit{Pn}+1\right){tan}^2\varphi =0$$

$$\left(\mathit{Pn}-\mathit{Pn}+1\right)\left(1-{tan}^2\varphi \right)-\mathit{Wn}sin\mathit{\psi c}+\mathit{Wn}cos\mathit{\psi c}tan\varphi =0$$

This leads:

$$\left(\mathit{Pn}-\mathit{Pn}+1\right)\left(1-{tan}^2\varphi \right)=\mathit{Wn}cos\mathit{\psi c}\left(tan\mathit{\psi c}-tan\varphi \right)$$

$$\left(\mathit{Pn}-\mathit{Pn}+1\right)=\frac{\mathit{Wn}cos\mathit{\psi c}\left(tan\mathit{\psi c}-tan\varphi \right)}{1-{tan}^2\varphi }$$

Hence, the following equation for $\mathit{Pns}$ is obtained;

$$\mathit{Pn},s=\mathit{Pn}+1+\frac{\mathit{Wn}cos\mathit{\psi c}\left(tan\mathit{\psi c}-tan\varphi \right)}{1-{tan}^2\varphi }$$E21

2.2 Determination of anchor tension necessary to prevent toppling of block 1

Once a determination is made that block 1 in Figure 4c will topple then cables under tension can be installed through block 1 to be anchored in stable rockmass underneath the toppling zone to prevent toppling of the same. Further assumptions was made that the anchor is installed at an angle plunge of $\mathit{\psi T}$ through block 1 located at a distance $L1$ above the base_. After the application of force $T$ to block 1, the normal force $\left(R1\right)$ as well as shear force $\left(S1\right)$ on the base plane of the block 1 is determined using Eqs. (22) and (23) respectively.

$$R1=P1tan\varphi +Tsin\left(\mathit{\psi p}+\mathit{\psi T}\right)+W1cos\mathit{\psi p}$$E22

$$S1=P1-Tcos\left(\mathit{\psi p}+\mathit{\psi T}\right)+W1sin\mathit{\psi p}$$E23

To solve for the magnitude of $\mathit{Tt}$, again the moment force about the pivot point 0 is set to zero as follows:

$$\sum M0\oplus =0$$

$$0=\frac{W1}{2}\left(y1sin\mathit{\psi p}-\mathrm{\Delta }xcos\mathit{\psi p}\right)+P1\left(y1-\mathrm{\Delta }xtan\varphi \right)-\mathit{TtL}1cos\left(\mathit{\psi p}+\mathit{\psi T}\right)$$

This is resolved as follows:

$$\frac{\mathit{TtL}1cos\left(\mathit{\psi p}+\mathit{\psi T}\right)}{L1cos\left(\mathit{\psi p}+\mathit{\psi T}\right)}=\frac{\frac{W1}{2}\left(y1sin\mathit{\psi p}-\mathrm{\Delta }xcos\mathit{\psi p}\right)+P1\left(y1-\mathrm{\Delta }xtan\varphi \right)}{L1cos\left(\mathit{\psi p}+\mathit{\psi T}\right)}$$

The magnitude of $\mathit{Tt}$, is deduced as shown in Eq. (24)

$$\mathit{Tt}=\frac{\frac{W1}{2}\left(y1sin\mathit{\psi p}-\mathrm{\Delta }xcos\mathit{\psi p}\right)+P1\left(y1-\mathrm{\Delta }xtan\varphi \right)}{L1cos\left(\mathit{\psi p}+\mathit{\psi T}\right)}$$E24

However, if there is existence of the counter-tilted failure plane at angle $\mathit{\psi c}$ to the horizontal plane, the normal force $\left(R1\right)$ and shear force $\left(S1\right)$ on the base plane of the block 1 are resolved using Eqs. (25) and (26) respectively.

$$R1=P1tan\varphi +Tsin\left(\mathit{\psi c}+\mathit{\psi T}\right)+W1cos\mathit{\psi p}$$E25

$$S1=P1-Tcos\left(\mathit{\psi c}+\mathit{\psi T}\right)+W1sin\mathit{\psi c}$$E26

To solve for the magnitude of $\mathit{Tt}$, again the moment about the pivot point 0 is set to zero;

$$0=\frac{W1}{2}\left(y1sin\mathit{\psi c}-\mathrm{\Delta }xcos\mathit{\psi c}\right)+P1\left(y1-\mathrm{\Delta }xtan\varphi \right)-\mathit{TtL}1cos\left(\mathit{\psi c}+\mathit{\psi T}\right)$$

This is resolved this as follows:

$$\frac{\mathit{TtL}1cos\left(\mathit{\psi c}+\mathit{\psi T}\right)}{L1cos\left(\mathit{\psi c}+\mathit{\psi T}\right)}=\frac{\frac{W1}{2}\left(y1sin\mathit{\psi c}-\mathrm{\Delta }xcos\mathit{\psi c}\right)+P1\left(y1-\mathrm{\Delta }xtan\varphi \right)}{L1cos\left(\mathit{\psi c}+\mathit{\psi T}\right)}$$

The magnitude of $\mathit{Tt}$, is reduced to Eq. (27).

$$\mathit{Tt}=\frac{\frac{W1}{2}\left(y1sin\mathit{\psi c}-\mathrm{\Delta }xcos\mathit{\psi c}\right)+P1\left(y1-\mathrm{\Delta }xtan\varphi \right)}{L1cos\left(\mathit{\psi c}+\mathit{\psi T}\right)}$$E27

2.3 Anchor tension necessary to prevent sliding of block 1

After a determination is made of the sliding block 1in Figure 4c the cables under tension can therefore be fixed through block 1 and anchored in the stable rock mass underneath the zone of sliding to avoid movements of block 1. It is also assumed that the installed anchor at a plunge angle of $\mathit{\psi T}$ through block 1 at a distance $L1$ above the base plane_. Therefore, with the failure plane along base angle $\mathit{\psi p}$ taking the horizontal as the datum, $\sum \mathit{Fx}$ and $\sum \mathit{Fy}$ are resolved using Eqs. (28) and (29) respectively.

$$\sum \mathit{Fx}=0=\mathit{Pn}-\mathit{Pn}+1-\mathit{Wn}sin\mathit{\psi p}+\mathit{Sn}$$E28

$$\sum \mathit{Fy}=0=\mathit{Rn}+\mathit{Pn}tan\varphi -\mathit{Wn}cos\mathit{\psi p}-\mathit{Pn}+1tan\varphi$$E29

Once the force T is applied to block 1, the normal force $\left(R1\right)$ and shear force $\left(S1\right)$ on the base planer of the block 1 are determined using Eqs. (30) and (31) respectively.

$$R1=P1tan\varphi +Tsin\left(\mathit{\psi p}+\mathit{\psi T}\right)+W1cos\mathit{\psi p}$$E30

$$S1=P1-Tcos\left(\mathit{\psi p}+\mathit{\psi T}\right)+W1sin\mathit{\psi p}$$E31

By solving for the $\mathit{Fy}$ equation for $\mathit{Rn}$ and making substitution into the $\mathit{Fx}$ equation prior to counter-tilting of the failure plane along the base plane $\mathit{\psi p}$ taking the horizontal line as the reference point, hence, the moments about the pivot point 0 was set to zero as given below;

$$\begin{array}{l}0=P1(1-tan\mathit{\varphi p}tan\varphi -W1\left(tan\mathit{\varphi p}cos\mathit{\psi p}-sin\mathit{\psi p}\right)-\mathit{Ts}(tan\varphi sin\left(\mathit{\psi p}+\mathit{\psi T}\right)\\ -cos\left(\mathit{\psi p}+\mathit{\psi T}\right)\end{array}$$

This was resolved as follows:

$$\mathit{Ts}(tan\varphi sin\left(\mathit{\psi p}+\mathit{\psi T}\right)-cos\left(\mathit{\psi p}+\mathit{\psi T}\right)=P1(1-tan\mathit{\varphi p}tan\varphi -W1\left(tan\mathit{\varphi p}cos\mathit{\psi p}-sin\mathit{\psi p}\right)$$

$$\frac{\mathit{Ts}(tan\varphi sin\left(\mathit{\psi p}+\mathit{\psi T}\right)+cos\left(\mathit{\psi p}+\mathit{\psi T}\right)}{(tan\varphi sin\left(\mathit{\psi p}+\mathit{\psi T}\right)+cos\left(\mathit{\psi p}+\mathit{\psi T}\right)}=\frac{P1(1-tan\mathit{\varphi p}tan\varphi -W1\left(tan\mathit{\varphi p}cos\mathit{\psi p}-sin\mathit{\psi p}\right)}{(tan\varphi sin\left(\mathit{\psi p}+\mathit{\psi T}\right)+cos\left(\mathit{\psi p}+\mathit{\psi T}\right)}$$

The magnitude of $\mathit{Ts}$, is deduced as follows:

$$\mathit{Ts}=\frac{P1(1-tan\mathit{\varphi p}tan\varphi -W1\left(tan\mathit{\varphi p}cos\mathit{\psi p}-sin\mathit{\psi p}\right)}{(tan\varphi sin\left(\mathit{\psi p}+\mathit{\psi T}\right)+cos\left(\mathit{\psi p}+\mathit{\psi T}\right)}$$

In general, as assumed before that the angle friction for the interfaces of the rock block columns denoted $\left(\mathit{\varphi d}\right)$ is the same as the friction angle on the bases plane $\left(\varphi \mathrm{p},\mathit{\varphi c}\right)$, hence $\mathit{\varphi d}=\mathit{\varphi p}=\mathit{\varphi c}=\varphi$.

Therefore $\mathit{Ts}$ is resolved using Eq. (33).

$$\mathit{Ts}=\frac{P1\left(1-{tan}^2\varphi \right)-W1\left(tan\varphi cos\mathit{\psi p}-sin\mathit{\psi p}\right)}{(tan\varphi sin\left(\mathit{\psi p}+\mathit{\psi T}\right)+cos\left(\mathit{\psi p}+\mathit{\psi T}\right)}$$E32

However, if the counter-tilted failure plane dipping at angle $\mathit{\psi c}$ exists considering the horizontal line, $\sum \mathit{Fx}$ and $\sum \mathit{Fy}$ are solved using Eqs. (33) and (34) respectively.

$$\sum \mathit{Fx}=0=\mathit{Pn}-\mathit{Pn}+1-\mathit{Wn}sin\mathit{\psi p}+\mathit{Sn}$$E33

$$\sum \mathit{Fy}=0=\mathit{Rn}+\mathit{Pn}tan\varphi -\mathit{Wn}cos\mathit{\psi p}-\mathit{Pn}+1tan\varphi$$E34

If the counter-tilted failure plane at angle $\mathit{\psi c}$ exists considering the horizontal line, then, when the force T is applied to block 1, the normal force $\left(R1\right)$ and shear force $\left(S1\right)$ on the base plane of the block 1 are reduced as follows;

$$R1=P1tan\varphi +Tsin\left(\mathit{\psi c}+\mathit{\psi T}\right)+W1cos\mathit{\psi p}$$

$$S1=P1-Tcos\left(\mathit{\psi c}+\mathit{\psi T}\right)+W1sin\mathit{\psi c}$$

By resolving the $\mathit{Fy}$ equation for $\mathit{Rn}$ and making substitutions into the $\mathit{Fx}$ equation after counter-tilting of the failure plane along the counter-tilted weak plane $\mathit{\psi c}$ with consideration of the horizontal line, hence, we set the moments about the pivot point 0 to zero as given below;

$$\begin{array}{l}0=P1(1-tan\mathit{\varphi c}tan\varphi -W1\left(tan\mathit{\varphi c}cos\mathit{\psi c}-sin\mathit{\psi c}\right)-\mathit{Ts}(tan\varphi sin\left(\mathit{\psi c}+\mathit{\psi T}\right)\\ +cos\left(\mathit{\psi c}+\mathit{\psi T}\right)\mathit{Ts}(tan\varphi sin\left(\mathit{\psi c}+\mathit{\psi T}\right)+cos\left(\mathit{\psi c}+\mathit{\psi T}\right)=P1(1-tan\mathit{\varphi c}tan\varphi \\ -W1\left(tan\mathit{\varphi c}cos\mathit{\psi c}-sin\mathit{\psi c}\right)\end{array}$$

We resolved $\mathit{Ts}$ as follows:

$$\frac{\mathit{Ts}(tan\varphi sin\left(\mathit{\psi c}+\mathit{\psi T}\right)-cos\left(\mathit{\psi c}+\mathit{\psi T}\right)}{(tan\varphi sin\left(\mathit{\psi c}+\mathit{\psi T}\right)+cos\left(\mathit{\psi c}+\mathit{\psi T}\right)}=\frac{P1(1-tan\mathit{\varphi c}tan\varphi -W1\left(tan\mathit{\varphi c}cos\mathit{\psi c}-sin\mathit{\psi c}\right)}{(tan\varphi sin\left(\mathit{\psi c}+\mathit{\psi T}\right)+cos\left(\mathit{\psi c}+\mathit{\psi T}\right)}$$

$\mathit{Ts}$, is deduced as follows:

$$\mathit{Ts}=\frac{P1(1-tan\mathit{\varphi c}tan\varphi -W1\left(tan\mathit{\varphi c}cos\mathit{\psi c}-sin\mathit{\psi c}\right)}{(tan\varphi sin\left(\mathit{\psi c}+\mathit{\psi T}\right)+cos\left(\mathit{\psi c}+\mathit{\psi T}\right)}$$

Based on the earlier assumptions of the angle of friction of the interfaces of the rock columns denoted $\left(\mathit{\varphi d}\right)$ being equal to the angle of friction on the base plane $\left(\varphi \mathrm{p},\mathit{\varphi c}\right)$, thus $\mathit{\varphi d}=\mathit{\varphi p}=\mathit{\varphi c}=\varphi$.then $\mathit{Ts}$ is resolved based on the Eq. (35)

$$\mathit{Ts}=\frac{P1\left(1-{tan}^2\varphi \right)-W1\left(tan\varphi cos\mathit{\psi c}-sin\mathit{\psi c}\right)}{(tan\varphi sin\left(\mathit{\psi c}+\mathit{\psi T}\right)+cos\left(\mathit{\psi c}+\mathit{\psi T}\right)}$$E35

Stabilization of rock slope by cable forces using limit equilibrium principles demonstrates that support of the “Keystone” is remarkably effective in increasing the scores of stability of the rock slopes prone to toppling. Supporting the rock block around the slope’s toe under the influence of toppling failure just at limit equilibrium is simply metastable and its metastability depends on the details of the geometric arrangement of the toppling blocks. On the other hand, reducing the strength of the “keystone” of a slope under toppling that is near failure leads to severe problems.

3. A practical application: Toppling failure

Failure by toppling of rock columns (Figure 5) occurred in shale rock fomations (Shale with grit, Shale with grit type 1 and Shale with grit type 2 & 3) on the northwall of the Nchanga Copper and Cobalt Open Pit situated in the mining city of Chingola in Zambia in June, 2016 [1, 4, 9, 10]. In terms of geology, the Nchanga geology is majorly controlled by Nchanga syncline regional structure with an east–west trend and at 5°-7° plunge to the north. The south limb of the Nchanga Open Pit is noted to be shallow and dips to the north at 25°-35° in general together with local changes of shallow to steep dips unlike the north limb which dips at 60°. The Nchanga open pit sub-surface geology is composed of the upper and lower roan dolomites which are usually dipping neaerly horizontal and a series of sedimentary rocks formations overlays them through to basal conglomerates that dips at an angle of 60° towards the north. The slope of the north wall is cut in the ore-body northwall lithology namely; Dolomitic Schist Shale with Grit, Banded Shale, Shale with Pink Quartzite, Grit (type 1,2 & 3), upper and lower roan dolomite Chingola Dolomite, Arkose, Feldspathic Quartzite, Banded Sandstone, Basal Conglomerates partly illustrated in Figure 5 for the geological engineering profile. Feldspathic Quartzite formation hosts the copper and cobalt that are being mined at Nchanga Open Pits. A summary of the general geological stratigraphy of the mine is shown in Table 1. The general slope is elevated at 1330 m – 880 m above mean sea level (asl) from the top and bottom of the elevataions respectively. The overall slope angle is 40° with an overall height of 450 m. An asymmetrical syncline limb on the north wall, the rock columns of the three rock formations (shale with grit type 1, shale with grit type 2, shale with grit) of the rock slope with counter-tilted weak surface overturned. The slope with counter-tilted weak surface under study is sited on the elevations ranging between 1170 m- 970 m above mean sea level (ASL).

Figure 5.

Toppling failure of the slope composed of a counter-tilted failure plane.

Rock Unit	Rock layer (m)	Density Kg/m^3	Internal friction angle(°)	Cohesion (MPa)	RMR	Hydro-Geological unit	Description
Shale with Grit	50	2500	35	400	68	Acquiclude	Fair to good quality,bedded rockmass
Shale with Grit type 1	40	2300	30	360	49	Acquiclude	Weak rockmass, bedded
Shale with Grit type 2&3	20	2200	30	360	49	Acquiclude	Weak rockmass, bedded
Chingola Dolomite	15	2400	33	380	64	Acquifer	Fait to good quality,bedded rockmass
Dolomitic Schists	20	2800	42	540	75	Mnor Acquifer	Good quality rockmass, bedded
Banded Shale	18	2600	37	490	68	Acquiclude	Fair to good quality rockmass seam at the base,bedded
Feldspathic Quartzite	18	2900	45	800	82	Acquiclude	Very good rockmass, massive
Banded Sandstone	15	1900	28	150		Acquifer	Weak rockmass with, calcite infill material, bedded
Pink Quartzite	5	3200	59	1300	84	Acquiclude	Very good quality with random joints
Arkose	15	3300	67	4000	96	Minor Acquifer	Very strong and competent rockmass
Basement(Gneiss and red granite)	>400	3000	39	1150	90	Impearmeable	Strong and competent with some schistocity

Table 1.

The general geological layout and rock characteristics.

Before slope failure, the zone under study being mined at, 155 m high (H), and at a slope angle of 65° ($\mathit{\psi f}$) in a layered shale rock mass (shale with grit type 1, shale with grit type 2shale with grit) dipping at 60° into the face ($\mathit{\psi d}$ = 60°); with the width ($\mathrm{\Delta }x$) of individual rock column at 10 m. The angle above the surface of the pit slope was determined to be 5° ($\mathit{\psi s}$), and the base of the rock block columns were stepped at b = 1.0 m. Cracks of less than 1 cm on the weak surface started to develop and were noticed when the pit was excavated to a depth of 145 m. The observed weak plane surface on the upper part of the slope (crest) was predicted to daylight on the slope’s face when the pit was to be mined at 250 m deep since the planned pit depth was at 450 m. No essential remedial measures were therefore undertaken against slope failure and no disruption of the mining operations in the pit was noted. However, based on the geological variations of the Shale rock formations at Nchanga Open Pit, failure surface inclined at 45° on the slope’s upper most section day-lighted on the slope’s face when the pit was at 144 m deep. It is well appreciated that the plane of weak surface dipping at an angle of 45° in shale with grit rock formation underwent counter-tilting to an angle of 35° in Shale with grit types 1 and 2 formations based on the variations in geological characteristics.

3.1 Calculational procedure

With regard to the geometry of the slope as given in Figure 5, it was presumed that, therefore $\mathit{\psi b}$ =55°. Using Eqs. (4), n = 14 see Figure 6a and rock block column 8 is at the slope’s top most part (crest). Using Eqs. (5)–(7), the constants calculated are $\mathrm{a}1$ = 5.0 m, $\mathrm{a}2$ = 5.0 m and b = 1.0 m. These constants are then applied to determine the height $\mathit{yn}$ of individual blocks, as well as the height to width ratio. The shale with grit type 1 and shale with grit rock formations’ unit weights at Nchanga were noted to be 23kN/m^{3 3} and 25kN/m respectively. It is assumed that there are no noticeable external forces and that the friction angles are $\mathit{\varphi p}=\mathit{\varphi d}$ =30° and $\mathit{\varphi c}$ =30°. The base plane dips at angles of $\mathit{\psi p}$ =45° and $\mathit{\psi c}$ =35°. Therefore, cot $\mathit{\psi p}$ = 1.0, hence, block columns 14, 13 and 12 are noted to be stable, because for each block indicated, the height to width ratio is noted to be less or equal to1.0. In general, these three blocks are short and their center of gravity was observed to lie inside the base of the block. Block 11 topples because the height-width ratio has a value 1.6, which is greater than 1.0. Therefore, $P11$ is equal to 0 and $P10$ is determined as the greater of $P10,\mathrm{t}$ and $P10,s$ given by Eqs. (20)–(23) respectively.

Figure 6.

(a) Limiting equilibrium analysis of a block toppling on a rock slope- conceptual slope set up and its geometry. (b) Limiting equilibrium analysis of a block toppling on a slope: Variation/distribution of normal forces (R) and shear (S) forces on the bases of the blocks.

This analytical procedure is key for stability determination of individual blocks, based on the failure surface angle of $\mathit{\psi p}$ moving downwards up to n = 6. At block denoted as n = 5 the original failure plane is counter tilted at angle $\mathit{\theta r}$, then $\mathit{\psi c}$ is substituted for $\mathit{\psi p}$ to determine the stability of individual blocks in turn moving down until n = 1 see Eqs. (20)–(23). The obtained block dimensions, the calculated forces and the stability modes are listed in Table 2 which indicates that $\mathrm{Pn}-1,\mathrm{t}$ is the larger of the two forces up until a value of n = 5, where upon $\mathrm{Pn}-1,\mathrm{s}$ is larger. Hence, blocks 6 to 11 was noted to be the potential toppling zone, and blocks 1 to 5 denoted a sliding zone. The factor of safety for this slope is determined by varying (increasing) the friction angles until the base plane blocks are stable. It is noted that the required friction angle for limit equilibrium conditions to be satisfied is 36°, or 0.96 (tan 35/tan 36). If $tan\varphi$ is reduced to 0.577, blocks 1 to 5 in the region around the toe will slide while blocks denoted as 6 to 11 will topple. The anchor tension installed at an angle of 25° through block 1, required to maintain equilibrium, is resolved to be 100kN/m of slope toe based on the Eqs. (24)–(27). This is not a big number, which demonstrates that support of the “Keystone” produces very effective results in increasing stability. On the other hand, reducing the strength of the “keystone” of a slope under toppling, nearing failure, leads to serious consequences. With the definition of the distribution of P forces in the sliding region, the forces R_n and S_n on the base plane of the base blocks is calculated using Eqs. (18) and (19). With the following assumption [$\mathit{Qn}-1=\mathit{Pn}-1tan\mathit{\varphi s}$], the forces R_n and S_n are determined the region of sliding. Figure 6b illustrates the variation of the forces throughout the whole slope. The conditions defined by R_n > 0 and S_n| < $\mathit{Rn}tan\mathit{\varphi p}$ are satisfied everywhere.

n	Yn	Yn/Δx	Mn	Ln	Pn-t	Pn-s	Pn	Rn	Sn	Sn/Rn	Failure Mode
14	−2	−0.2			0	0	0	353.6	353.6	1.00	Stable
13	4	0.4			0	0	0	707.1	707.1	1.00
12	10	1	2	10	0	0	0	1767.8	1767.8	1.00
11	16	1.6	8	16	0	0	0	2456.9	2298.1	0.94	Toppling
10	22	2.2	14	22	530.5	−1664.7	530.5	3399.6	3190	0.94
9	26	2.6	18	26	1229.4	−1758.7	1229.4	4102.6	3891.3	0.95
8	32	3.2	24	27	1934.3	−1475.8	1934.3	4544.8	4068.6	0.90
7	28	2.8	28	23	3522.6	−1393.1	3522.6	3808	3319.2	0.87
6	24	2.4	24	19	5153.1	609.3	5153.1	2605.8	1904.9	0.73
5	20	2	20	15	5490.8	6534.3	6534.3	4225.6	3053.3	0.72	Sliding
4	16	1.6	16	11	5305.4	6140.2	6140.2	3826.5	3079.7	0.80
3	12	1.2	12	7	4520	5146.1	5146.1	3558.1	3292.6	0.93
2	8	0.8	8	3	2948.2	3365.6	3365.6	3855	4313.6	1.12
1	4	0.4	4		−218.3	−9.6	−9.6	666.3	355.3	0.53

Table 2.

Limiting equilibrium analysis of a columnar block toppling slope- listing block calculated forces, stability modes and dimensions.

4. Summary of the chapter

A presentation of the mathematical (analytical) method for determination of the counter-tilted failure plane angle for the discontinuous rock slope subjected to block toppling failure mechanism based on the searching technique has been presented. Geometry parameters forming the systematic arrangement of the jointed rock slope that maybe influenced by the presence of the failure planes such as; the application points for the shear and normal forces that acts on the bases of the weak planes and the number of rock columns, the overall base angles of the weak plane angles, forming the regular system have been developed. The modified limit equilibrium method for quantifying the failure mechanisms in the jointed rock slopes having a counter-tilted failure plane prone to block toppling has been proposed. This involves an iterative process in which the dimensions of all the individual blocks and the existing forces acting on them are calculated, and then each block’s stability is examined, starting at the uppermost block. Each block will be stable, topple or slide, with the overall stability of the slope being rendered unstable if the lower most block either slides or topples. The iterative process involves determinations of sliding and toppling of the n^th block with and without consideration of counter-tilted weak plane respectively. The proposed method of analysis has the potential to curtail errors incurred due to the previous assumptions of the single weak plane for quantifying the failure mechanisms of toppling rock slopes in physical situations with two planar weak planes. It is further noticed that the existence of counter-tilted failure plane in the discontinuous rock slope influence the geometry parameters such as; overall base angles, number of blocks that forms the regular systematic arrangement of the jointed rock slope and application positions M_n and L_n and the sliding and toppling forces.