# An Analytical Solution for Block Toppling Failure of Rock Slopes during an Earthquake
## 2 Mechanical Model and Analytical Solution to Toppling Slope Failure
The maximum tensile stress induced by moment \(M_{b}\) occurs at the left edge of the rock bridge on the base (point O in Figure 4):  
\[\sigma_{tM}=\frac{M_{b}\times\frac{1}{2}\xi\Delta x}{\frac{1}{\Omega}\big{(} \xi\Delta x\big{)}^{3}}=\frac{M_{b}}{\frac{1}{6}\big{(}\xi\Delta x\big{)}^{2}} \tag{7}\]  
The compressive stress induced by \(R_{n}\) is uniform on the base:  
\[\sigma_{tR}=\frac{R_{n}}{\xi\Delta x} \tag{8}\]  
Therefore, the maximum tensile stress occurs at the left edge of the rock bridge on the base:  
\[\sigma_{t}=\frac{M_{b}}{\frac{1}{6}\xi^{2}\Delta x^{2}}-\frac{R_{n}}{\xi\Delta x} \tag{9}\]  
So  
\[M_{b}=\frac{\xi\Delta x}{6}(\sigma_{t}\xi\Delta x+R_{n}) \tag{10}\]  
So the solution of the force needed from the lower block to keep the \(n\)th block from toppling can be obtained as follows:  
\[P_{n-1,I}=\frac{\big{(}M_{n}+\frac{1}{5}\Delta x\tan\phi_{js}-\Delta x\tan \phi_{js}\big{)}P_{n}+\frac{\Delta y\tan}{2}\big{(}1-\frac{1}{6}\big{)}\big{(} \frac{19}{\sqrt{2}}-\cos\psi_{p}\big{)}-\frac{\xi^{2}\Delta x^{2}}{6}\sigma_ {t}+\frac{mW_{n}}{2}\big{(}\sin\psi_{p}+\frac{1}{\sqrt{2}}\big{)}}{L_{n}+\frac {\Delta x}{3}\tan\phi_{js}} \tag{11}\]  
Figure 4: The toppling mechanical condition on the base with non-persistent joint (Modified based on [1]): (**a**) the mechanical condition of the block base; (**b**) the stress distribution along the block base. The dynamic loads as well as joint connectivity rate are taken into account in the limit equilibrium method compared with the original form proposed by [8]. For the special state of static conditions and a persistent block base, the limit equilibrium Equation (4) of sliding and Equation (11) of toppling would be simplified as Equations (12) and (13) respectively. Equations (12) and (13) are the same as the original form proposed by [8]. It can be seen that the proposed solutions are a universal form that can estimate the deformation status and factor of safety (FOS) for toppling slope with various \(J_{c}\) of the block base under different seismic loads. \[P_{n-1,s}=P_{n}-\frac{W_{n}\Big{(}\cos\psi_{p}\tan\phi_{jb}-\sin\psi_{p}\Big{)} }{1-\tan\phi_{js}\tan\phi_{jb}} \tag{12}\]  
\[P_{n-1,t}=\Big{[}\big{(}M_{n}-\Delta x\tan\phi_{js}\big{)}P_{n}+\frac{W_{n}}{2} \big{(}y_{n}\sin\psi_{p}-\Delta x\cos\psi_{p}\big{)}\Big{]}\Big{/}L_{n} \tag{13}\]  
The solution can be applied by following these steps:  
1. Reach the basic geometric parameters \(\psi_{b}\), \(\psi_{p}\), \(\psi_{s}\), \(\psi_{d}\), \(\psi_{f}\) and \(\Delta x\) from in- situ measurement work and calculating other geometric parameters \(a_{1}\), \(a_{2}\), \(b\), \(y_{n}\), \(M_{n}\) and \(L_{n}\) according to Equations (A1)-(A11). 2. Determining all the physical parameters of the joints and blocks, e.g., \(\rho\), \(\varphi_{jb}\), \(\varphi_{js}\), \(\varphi_{b}\), \(c_{b}\), \(\sigma_{bt}\), and the joint connectivity rate (\(J_{c}\)) of the block base on the basis of lab tests, in-situ measurement or experience. 3. Calculating \(P_{n-1,S}\) and \(P_{n-1,t}\) from the top block downwards on the basis of Equations (4) and (11). 4. Determining \(P_{n-1}\) and the deformation status of the \(n\)th block. If \(P_{n-1,s}<0\) and \(P_{n-1,t}<0\), \(P_{n-1}=0\). It means that the \(n\)th block is stable and does not need the support of the \((n-1)\)th block. If \(P_{n-1,s}<0\) and \(P_{n-1,t}>0\), \(P_{n-1}=P_{n-1,t}\). It means that the \(n\)th block needs the support of the \((n-1)\)th block to inhibit toppling deformation. If \(P_{n-1,t}>P_{n-1,s}>0\), \(P_{n-1}=P_{n-1,t}\). It means that the \(n\)th block needs more support from the \((n-1)\)th block to inhibit toppling failure rather than sliding failure. If \(P_{n-1,s}>0\) and \(P_{n-1,t}<0\), \(P_{n-1}=P_{n-1,s}\). It means that the \(n\)th block needs the support of the \((n-1)\)th block to inhibit sliding deformation. If \(P_{n-1,s}>P_{n-1,t}>0\), \(P_{n-1}=P_{n-1,s}\). It means that the \(n\)th block needs more support from the \((n-1)\)th block to inhibit sliding failure rather than toppling failure.