Modified Analytical Technique for Block Toppling Failure of Rock Slopes with Counter-Tilted Failure Surface

Abstract

A broad mathematical technique for evaluating rock slope stability subjected to toppling of rock columns centered on limit equilibrium is presented. In the current analytical models for evaluating toppling of rock columns for the rock slopes, a single weak plane angle is assumed, running from the top surface of the slope and daylights on the toe of the slope. However, in some physical rock slopes, there exists counter-tilted failure surface within the rock masses of the rock slope and the weak plane may not daylight on the predictable point on the face of the slope following the assumed single weak plane angle. Therefore, applying the current analytical techniques in such conditions may not yield realistic results. A searching technique for estimating angles of counter-tilted failure surface is proposed and incorporated into the existing mathematical technique for evaluating toppling of rock columns centering on limit equilibrium principles. The physical slope with counter-tilted failure surface was comprehensively analyzed using the modified mathematical method and results were validated by numerical simulation using 3DEC 4.1 discrete element method. The influences of relative angles of the counter-tilted failure surfaces on the slope stability have been studied and the results show that, progressive increase of the counter-tilted failure surface angles lead to gradual increase in slope instability. The modified analytical method optimizes and combines the advantages of the traditional analytical techniques and could provide precise analytical method for evaluating stability of rock slopes with counter-tilted planar failure surfaces subjected to toppling of rock columns.

Appendices

Appendix 1: Limiting Equilibrium for Toppling of the nth Block

To determine the magnitude of $P_{n}$, we set the moment about the pivot point 0 to zero (refer Fig. 4). Generally, we assume that the interfaces’ friction angle of the rock columns denoted ($\phi_{d}$) is equal to the friction angle on the bases ($\phi_{p},\,\,\phi_{c}$), thus $\phi_{d} = \phi_{p} = \phi_{c} = \phi$. Therefore, before counter-tilting of the weak plane along base plane $\psi_{p}$ with respect to the horizontal, we have

$$ \sum {M}_{{0}{\oplus}} = 0 $$

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$$ = \frac{y_{n}}{2}W_{n}\sin \psi_{p} - \frac{\vartriangle x}{2}W_{n}\cos \psi_{p} + M_{n}P_{n + 1} - \vartriangle x(P_{n} + 1\tan \phi ) - L_{n}P_{n} $$

(29)

$$ = \frac{W_{n}}{2}(y_{n}\sin \psi_{p} - \vartriangle x\cos \psi_{p}) + P_{n + 1}(M_{n} - \vartriangle x\tan \phi ) - L_{n}P_{n} $$

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This can be rewritten as;

$$ L_{n}P_{n} = \frac{W_{n}}{2}(y_{n}\sin \psi_{p} - \vartriangle x\cos \psi_{p}) + P_{n+1}(M_{n} - \vartriangle x\tan \phi ) $$

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Therefore, we get the following equation for $P_{n}$

$$ P_{n,t} = \frac{{\frac{W_{n}}{2}(y_{n}\sin \psi_{p} - \vartriangle x\cos \psi_{p}) + P_{n+1}(M_{n} - \vartriangle x\tan \phi )}}{L_{n}} $$

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To determine the magnitude of $P_{n}$ due to counter-tilting of the weak plane along base plane $\psi_{c}$ with respect to the horizontal, we set the moment about the pivot point 0 to zero. We have;

$$ \sum {M}_{{0}{ \oplus}} = 0 $$

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$$ = \frac{y_{n}}{2}W_{n}\sin \psi_{c} - \frac{\vartriangle x}{2}W_{n}\cos \psi_{c} + M_{n}P_{n+1} - \vartriangle x(P_{n+1} \tan \phi ) - L_{n}P_{n} $$

(34)

$$ = \frac{W_{n}}{2}(y_{n}\sin \psi_{c} - \vartriangle x\cos \psi_{c}) + P_{n+1}(M_{n} - \vartriangle x\tan \phi ) - L_{n}P_{n} $$

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This can be rewritten as;

$$ L_{n}P_{n} = \frac{W_{n}}{2}(y_{n}\sin \psi_{c} - \vartriangle x\cos \psi_{c}) + P_{n+1} (M_{n} - \vartriangle x\tan \phi ) $$

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Therefore, we get the following equation for $P_{n,t}$ due to counter-tilting of the weak plane

$$ P_{n,t} = \frac{{\frac{W_{n}}{2}(y_{n}\sin \psi_{c} - \vartriangle x\cos \psi_{c}) + P_{n+1}(M_{n} - \vartriangle x\tan \phi )}}{L_{n}} $$

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Appendix 2: Limiting Equilibrium for Sliding of the nth Block

To determine the magnitude of P_ns, we set the sum of the forces in both the horizontal and vertical directions to zero (taking the horizontal axis to be along the base plane, at the angles $\psi_{p}$ or $\psi_{c}$ with respect to horizontal refer Fig. 4 c in the manuscript.

Generally, we assume that the interfaces’ friction angle of the rock columns denoted ($\phi_{d}$) is equal to the friction angle on the bases ($\phi_{p},\phi_{c}$), thus $\phi_{d} = \phi_{p} = \phi_{c} = \phi$. Therefore, before counter-tilting of the weak plane along base plane $\psi_{p}$ with respect to the horizontal, we have

$$ \sum {F_{x}} = 0 = P_{n} - P_{n+1} - W_{n}\sin \psi_{p} + S_{n} $$

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$$ \sum {F_{y}} = 0 = R_{n} + P_{n}\tan \phi - W_{n}\cos \psi_{p} - P_{n+1}\tan \phi $$

(39)

where $S_{n}$ is the shear force along the Column-base contact. From the Mohr–Coulomb criterion, we have

$$ \tau_{n} = c + \sigma_{n}\tan \phi $$

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where, $\tau_{n}$ Shear stress along the column-base contact $\sigma_{n}$ Normal stress at the column-base contact.

Thereafter we assumed cohesion (c) is equal to zero and divide both sides of the equation by the column-base contact area denoted by A. Thus, we get

$$ S_{n} = R_{n}\tan \phi $$

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where, $R_{n}$ normal force across the column-base contacts ($W_{n}\cos \psi_{p}$)

By solving the F_y equation for $R_{n}$ and substituting into the F_x equation before counter-tilting of the weak plane along base plane $\psi_{p}$ with respect to the horizontal, we have,

$$ P_{n} - P_{n+1} - W_{n}\sin \psi_{p} + W_{n}\cos \psi_{p}\tan \phi - (P_{n} - P_{n+1})\tan^{2} \phi = 0 $$

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$$ (P_{n} - P_{n+1} )(1 - \tan^{2} \phi ) - W_{n}\sin \psi_{p} + W_{n}\cos \psi_{p}\tan \phi = 0 $$

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This leads;

$$ (P_{n} - P_{n+1} )(1 - \tan^{2} \phi ) = W_{n}\cos \psi_{p}(\tan \psi_{p} - \tan \phi ) $$

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$$ (P_{n} - P_{n+1}) = \frac{W_{n}\cos \psi_{p}(\tan \psi_{p} - \tan \phi )}{{1 - \tan^{2} \phi }} $$

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Therefore, we get the following equation for $P_{ns}$

$$ P_{n,s} = P_{n+1} + \frac{W_{n}\cos \psi_{p}(\tan \psi_{p} - \tan \phi )}{{1 - \tan^{2} \phi }} $$

(46)

However, after counter-tilting of the weak plane along base plane $\psi_{c}$ with respect to the horizontal, we have

$$ \sum {F_{x}} = 0 = P_{n} - P_{n+1} - W_{n}\sin \psi_{c} + S_{n} $$

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$$ \sum {F_{y}} = 0 = Rn + P_{n}\tan \phi - W_{n}\cos \psi_{c} - P_{n+1} \tan \phi $$

(48)

By solving the F_y equation for $R_{n}$ and substituting into the F_x equation after counter-tilting of the weak plane along base plane $\psi_{c}$ with respect to the horizontal, we have,

$$ P_{n} - P_{n+1} - W_{n}\sin \psi_{c} + W_{n}\cos \psi_{c}\tan \phi - (P_{n} - P_{n+1})\tan^{2} \phi = 0 $$

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$$ (P_{n} - P_{n+1})(1 - \tan^{2} \phi ) - W_{n}\sin \psi_{c} + W_{n}\cos \psi_{c}\tan \phi = 0 $$

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This leads;

$$ (P_{n} - P_{n+1})(1 - \tan^{2} \phi ) = W_{n}\cos \psi_{c}(\tan \psi_{c} - \tan \phi ) $$

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$$ (P_{n} - P_{n+1}) = \frac{W_{n}\cos \psi_{c}(\tan \psi_{c} - \tan \phi )}{{1 - \tan^{2} \phi }} $$

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Therefore, we get the following equation for $P_{ns}$

$$ P_{n,s} = P_{n+1} + \frac{W_{n}\cos \psi_{c}(\tan \psi_{c} - \tan \phi )}{{1 - \tan^{2} \phi }} $$

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Appendix 3: Anchor Tension Required to Prevent Toppling of Block 1

Once it is determined that block 1 in Fig. 4c will topple then the tensioned cables can be installed through block 1 and anchored in stable rockmass beneath the zone of toppling to prevent toppling of block 1. We further assume that the anchor is installed at a plunge angle of $\psi_{T}$ through block 1 at a distance L₁ above the base. When the force T is applied to block 1, the normal and shear force on the base of the block 1 are, respectively

$$ R_{1} = P_{1}\tan \phi + T\sin (\psi_{p} + \psi_{T}) + W_{1}\cos \psi_{p} $$

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$$ S_{1} = P_{1} - T\cos (\psi_{p} + \psi_{T}) + W_{1}\sin \psi_{p} $$

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To determine the magnitude of $T_{t}$, we set the moment about the pivot point 0 to zero

$$ \sum {M}_{{0}{\oplus}} = 0 $$

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$$ 0 = \frac{W_{1}}{2}(y_{1}\sin \psi_{p} - \vartriangle x\cos \psi_{p}) + P_{1}(y_{1} - \vartriangle x\tan \phi ) - T_{t}L_{1}\cos (\psi_{p} + \psi_{T}) $$

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$$ \frac{T_{t}L_{1}\cos (\psi_{p} + \psi_{T})}{L_{1}\cos (\psi_{p} + \psi_{T})} = \frac{{\frac{W_{1}}{2}(y_{1}\sin \psi_{p} - \vartriangle x\cos \psi_{p}) + P_{1}(y_{1} - \vartriangle x\tan \phi )}}{L_{1}\cos (\psi_{p} + \psi_{T})} $$

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$$ T_{t} = \frac{{\frac{W_{1}}{2}(y_{1}\sin \psi_{p} - \vartriangle x\cos \psi_{p}) + P_{1}(y_{1} - \vartriangle x\tan \phi )}}{L1\cos (\psi_{p} + \psi_{T})} $$

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If there exists the counter-tilted weak plane at angle $\psi_{c}$ to the horizontal line, the normal and shear force on the base of the block 1 are, respectively

$$ R_{1} = P_{1}\tan \phi + T\sin (\psi_{c} + \psi_{T}) + W_{1}\cos \psi_{p} $$

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$$ S_{1} = P_{1} - T\cos (\psi_{c} + \psi_{T}) + W_{1}\sin \psi_{c} $$

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To determine the magnitude of $T_{t}$, we set the moment about the pivot point 0 to zero

$$ \begin{aligned} 0 = \frac{W_{1}}{2}(y_{1}\sin \psi_{c} - \vartriangle x\cos \psi_{c}) + P_{1}(y_{1} - \vartriangle x\tan \phi ) \hfill \\ - T_{t}L_{1}\cos (\psi_{c} + \psi_{T}) \hfill \\ \end{aligned} $$

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$$ \frac{T_{t}L_{1}\cos (\psi_{c} + \psi_{T})}{L_{1}\cos (\psi_{c} + \psi_{T})} = \frac{{\frac{W_{1}}{2}(y_{1}\sin \psi_{c} - \vartriangle x\cos \psi_{c}) + P_{1}(y_{1} - \vartriangle x\tan \phi )}}{L_{1}\cos (\psi_{c} + \psi_{T})} $$

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$$ T_{t} = \frac{{\frac{W_{1}}{2}(y_{1}\sin \psi_{c} - \vartriangle x\cos \psi_{c}) + P_{1}(y_{1} - \vartriangle x\tan \phi )}}{L_{1}\cos (\psi_{c} + \psi_{T})} $$

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Appendix 4: Anchor Tension Required to Prevent Sliding of Block 1

Once it is determined that block 1 in Fig. 4c will slide then the tensioned cables can be installed through block 1 and anchored in stable rockmass beneath the zone of sliding to prevent movements of block 1. We further assume that the anchor is installed at a plunge angle of $\psi_{T}$ through block 1 at a distance L₁ above the base. Therefore, with the weak plane along base plane $\psi_{p}$ with respect to the horizontal, we have

$$ \sum {F_{x}} = 0 = P_{n} - P_{n+1} - W_{n}\sin \psi_{p} + S_{n} $$

(65)

$$ \sum {F_{y}} = 0 = Rn + P_{n}\tan \phi - W_{n}\cos \psi_{p} - P_{n+1}\tan \phi $$

(66)

When the force T is applied to block 1, the normal and shear force on the base of the block 1 are, respectively

$$ R_{1} = P_{1}\tan \phi + T\sin (\psi_{p} + \psi_{T}) + W_{1}\cos \psi_{p} $$

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$$ S_{1} = P_{1} - T\cos (\psi_{p} + \psi_{T}) + W_{1}\sin \psi_{p} $$

(68)

By solving the F_y equation for $Rn$ and substituting into the F_x equation before counter-tilting of the weak plane along base plane $\psi_{p}$ with respect to the horizontal, we have,

$$ 0 = P_{1}(1 - \tan \phi_{p}\tan \phi - W_{1}(\tan \phi_{p}\cos \psi_{p} - \sin \psi_{p}) - T_{s}(\tan \phi \sin (\psi_{p} + \psi_{T}) - \cos (\psi_{p} + \psi_{T}) $$

(69)

$$ T_{s}(\tan \phi \sin (\psi_{p} + \psi_{T}) - \cos (\psi_{p} + \psi_{T}) = P_{1}(1 - \tan \phi_{p}\tan \phi - W_{1}(\tan \phi_{p}\cos \psi_{p} - \sin \psi_{p}) $$

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$$ \begin{aligned} &\frac{T_{s}(\tan \phi \sin (\psi_{p} + \psi_{T}) + \cos (\psi_{p} + \psi_{T})}{(\tan \phi \sin (\psi_{p} + \psi_{T}) + \cos (\psi_{p} + \psi_{T})} \hfill \\ &\quad = \frac{P_{1}(1 - \tan \phi_{p}\tan \phi - W_{1}(\tan \phi_{p}\cos \psi_{p} - \sin \psi_{p})}{(\tan \phi \sin (\psi_{p} + \psi_{T}) + \cos (\psi_{p} + \psi_{T})} \hfill \\ \end{aligned} $$

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$$ T_{s} = \frac{P_{1}(1 - \tan \phi_{p}\tan \phi - W_{1}(\tan \phi_{p}\cos \psi_{p} - \sin \psi_{p})}{(\tan \phi \sin (\psi_{p} + \psi_{T}) + \cos (\psi_{p} + \psi_{T})} $$

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Generally, we assume that the interfaces’ friction angle of the rock columns denoted ($\phi_{d}$) is equal to the friction angle on the bases ($\phi_{p},\phi_{c}$), thus $\phi_{d} = \phi_{p} = \phi_{c} = \phi$.

$$ T_{s} = \frac{{P_{1}(1 - \tan^{2} \phi ) - W_{1}(\tan \phi \cos \psi_{p} - \sin \psi_{p})}}{(\tan \phi \sin (\psi_{p} + \psi_{T}) + \cos (\psi_{p} + \psi_{T})} $$

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Therefore, with the weak plane counter-tilted along base plane $\psi_{c}$ with respect to the horizontal, we have

$$ \sum {F_{x}} = 0 = P_{n} - P_{n+1} - W_{n}\sin \psi_{p} + S_{n} $$

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$$ \sum {F_{y}} = 0 = Rn + P_{n}\tan \phi - W_{n}\cos \psi_{p} - P_{n+1} \tan \phi $$

(75)

If there exists the counter-tilted weak plane at angle $\psi_{c}$ to the horizontal line, then, when the force T is applied to block 1, the normal and shear force on the base of the block 1 are, respectively

$$ R_{1} = P_{1}\tan \phi + T\sin (\psi_{c} + \psi_{T}) + W_{1}\cos \psi_{p} $$

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$$ S_{1} = P_{1} - T\cos (\psi_{c} + \psi_{T}) + W_{1}\sin \psi_{c} $$

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By solving the F_y equation for $R_{n}$ and substituting into the F_x equation after counter-tilting of the weak plane along base plane $\psi_{c}$ with respect to the horizontal, we have,

$$ \begin{aligned} 0 & = P_{1}(1 - \tan \phi_{c}\tan \phi - W_{1}(\tan \phi_{c}\cos \psi_{c} - \sin \psi_{c}) \\ & \quad - \,T_{s}(\tan \phi \sin (\psi_{c} + \psi_{T}) + \cos (\psi_{c} + \psi_{T}) \\ \end{aligned} $$

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$$ \begin{aligned} &T_{s}(\tan \phi \sin (\psi_{c} + \psi_{T}) + \cos (\psi_{c} + \psi_{T}) \hfill \\ &\quad = P_{1}(1 - \tan \phi_{c}\tan \phi - W_{1}(\tan \phi_{c}\cos \psi_{c} - \sin \psi_{c}) \hfill \\ \end{aligned} $$

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$$ \begin{aligned} &\frac{T_{s}(\tan \phi \sin (\psi_{c} + \psi_{T}) - \cos (\psi_{c} + \psi_{T})}{(\tan \phi \sin (\psi_{c} + \psi_{T}) + \cos (\psi_{c} + \psi_{T})} \hfill \\ &\quad = \frac{P_{1}(1 - \tan \phi_{c}\tan \phi - W_{1}(\tan \phi_{c}\cos \psi_{c} - \sin \psi_{c})}{(\tan \phi \sin (\psi_{c} + \psi_{T}) + \cos (\psi_{c} + \psi_{T})} \hfill \\ \end{aligned} $$

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$$ T_{s} = \frac{P_{1}(1 - \tan \phi_{c}\tan \phi - W_{1}(\tan \phi_{c}\cos \psi_{c} - \sin \psi_{c})}{(\tan \phi \sin (\psi_{c} + \psi_{T}) + \cos (\psi_{c} + \psi_{T})} $$

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Generally, we assume that the interfaces’ friction angle of the rock columns denoted ($\phi_{d}$) is equal to the friction angle on the bases ($\phi_{p},\phi_{c}$), thus $\phi_{d} = \phi_{p} = \phi_{c} = \phi$.

$$ T_{s} = \frac{{P_{1}(1 - \tan^{2} \phi ) - W_{1}(\tan \phi \cos \psi_{c} - \sin \psi_{c})}}{(\tan \phi \sin (\psi_{c} + \psi_{T}) + \cos (\psi_{c} + \psi_{T})} $$

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