## 5 Calculations of Stability for the Planned Engineered Landfill
### Model and Calculation Assumptions Based on Case 1 Results (Chemical Plant)
\begin{table}
\begin{tabular}{c c c} \hline \hline  & **Type of Ground** & **Coulomb–Mohr Model** \\  & & **Performance Parameters** \\ \hline
1. & Ground, MSa, I\({}_{\text{D}}\) = 40\% & E = 68 MPa, \(\phi\) = 32\({}^{\circ}\), c = 1 kPa, \(\
u\) = 0.26, \\  & & \(\gamma\) = 18.5 kN/m\({}^{3}\), \(\phi\) = 2\({}^{\circ}\) \\
2. & Made ground & E = 5 MPa, \(\phi\) = 10\({}^{\circ}\), c = 10 kPa, \(\
u\) = 0.3, \\  & & \(\gamma\) = 16 kN/m\({}^{3}\), \(\phi\) = 0\({}^{\circ}\) \\
3. & Earth dike (outer part) MSa, I\({}_{\text{D}}\) = 50\% & E = 80 MPa, \(\phi\) = 33\({}^{\circ}\), c = 1 kPa, \(\
u\) = 0.25, \\  & & \(\gamma\) = 18.5 kN/m\({}^{3}\), \(\phi\) = 3\({}^{\circ}\) \\
4. & Earth dike (inner part), landfill & E = 23 MPa, \(\phi\) = 13\({}^{\circ}\), c = 60 kPa, \(\
u\) = \\  & bottom, Clay & 0.25, \(\gamma\) = 18.5 kN/m\({}^{3}\), \(\phi\) = 3\({}^{\circ}\) \\
5. & Drainage layer, & E = 68 MPa, \(\phi\) = 32\({}^{\circ}\), c = 1 kPa, \(\
u\) = 0.26, \\  & MSa, I\({}_{\text{D}}\) = 40\% & \(\gamma\) = 18.5 kN/m\({}^{3}\), \(\phi\) = 2\({}^{\circ}\) \\
6. & Compacted made grounds & E = 10 MPa, \(\phi\) = 20\({}^{\circ}\), c = 10 kPa, \(\
u\) = 0.2, \\  & & \(\gamma\) = 18 kN/m\({}^{3}\), \(\psi\) = 0\({}^{\circ}\) \\
7. & Outer slopes, & E = 10 MPa, \(\phi\) = 20\({}^{\circ}\), c = 10 kPa, \(\
u\) = 0.2, \\  & compacted made grounds & \(\gamma\) = 18 kN/m\({}^{3}\), \(\psi\) = 0\({}^{\circ}\) \\
8. & Cement stabilized wastes & E = 10 MPa, \(\phi\) = 32\({}^{\circ}\), c = 26 kPa, \(\
u\) = 0.3, \\  & & \(\gamma\) = 14.6 kN/m\({}^{3}\), \(\psi\) = 0\({}^{\circ}\) \\
9. & Inner earth dike, compacted & E = 10 MPa, \(\phi\) = 20\({}^{\circ}\), c = 10 kPa, \(\
u\) = 0.2, \\  & made grounds & \(\gamma\) = 18 kN/m\({}^{3}\), \(\psi\) = 0\({}^{\circ}\) \\ \hline \hline \end{tabular}
\end{table} 
Table 3: Constitutive models and parameters in the FEM stability analysis. an initial stress distribution is determined in the analyzed object (\"in situ\" state), caused, for example, by its own state or other constant static loads. In this case, meeting the conditions of elastic-plastic equilibrium corresponds to achieving the safety factor SF = 1.0;
* For each calculation step, assuming SFn+1= SFn + \(\Delta\)SF (step \(\Delta\)SF in the described models was defined at 0.01), while at the same time calculating cn+1 = c/SFn+1 and tg\(\phi_{\mathrm{n+1}}\) = tg\(\phi_{\mathrm{n}}\)/SFn+1. By step-wise reduction in cohesion \(c\) and angle of shearing resistance tangent \(\phi\), a less stable system is obtained with each calculation step;
* Solving the problem using software, with the necessary iterations;
* Performing system stability calculations for the assumed sequence of safety factors, until solution divergence is achieved in two consecutive steps--if an iterative process divergence occurs (manifesting in the occurrence of very large deformations, often with a qualitatively different form than those in the equilibrium state), it means that with the currently assumed safety factor, the structure is not stable;
* Determination of the safety factor when solution divergence is obtained SFn \(\leq\) SF \(\leq\) SFn+1--the last value of the SF-reducing coefficient at which an equilibrium state can still be achieved is assumed as the safety factor value.