Sample Problems - Finite Element Method
The following examples illustrate how to use XSLOPE's finite element capabilities for slope stability analysis using the Shear Strength Reduction Method (SSRM). Each of the Excel input files below can be uploaded and used with the following Google Colab notebook which has been set up specifically for running FEM slope stability analyses:
The FEM implementation is described in the FEM Overview page.
1. Griffiths & Lane (1999) Example 1: Homogeneous Slope
This is the benchmark problem from Griffiths & Lane (1999), "Slope stability analysis by finite elements," Geotechnique, 49(3), 387-403. It features a homogeneous slope with the following properties:
| Property | Value |
|---|---|
| Cohesion, \(c\) | 312.5 psf |
| Friction angle, \(\phi\) | 20 degrees |
| Unit weight, \(\gamma\) | 125 pcf |
| Young's modulus, \(E\) | 700,000 psf |
| Poisson's ratio, \(\nu\) | 0.3 |
The dimensionless parameter \(c/\gamma H = 0.05\) with \(\phi = 20°\) gives an expected factor of safety of approximately 1.4 (Griffiths & Lane, 1999, Table 1).
Excel input file: xslope_griffiths1.xlsx
Inputs plotted with the XSLOPE plot_inputs() function:
FEM mesh with boundary conditions. Fixed supports (triangles) at the base, x-rollers (circles) on the sides:
SSRM results. The computed factor of safety is FS = 1.38, which is consistent with the published result of approximately 1.4. The top plot shows the deformed mesh at the last converged solution (F = 1.375). The bottom plot shows the viscoplastic shear strain concentration, which reveals the circular failure mechanism without any prior assumption about its shape or location.
2. Reinforced Slope with Geogrid Reinforcement
This problem features an engineered slope with six layers of geogrid reinforcement. It is the FEM counterpart of the LEM reinforced slope example described in the LEM Samples page (Problem 9). The slope geometry and soil properties are the same, with the addition of elastic modulus and Poisson's ratio for the FEM analysis:
| Property | Shell | Base |
|---|---|---|
| Cohesion, \(c\) (psf) | 300 | 0 |
| Friction angle, \(\phi\) (degrees) | 37 | 37 |
| Unit weight, \(\gamma\) (pcf) | 130 | 130 |
| Young's modulus, \(E\) (psf) | 1,000,000 | 1,000,000 |
| Poisson's ratio, \(\nu\) | 0.3 | 0.3 |
A 240 psf surcharge is applied along the slope crest from \(x = 30\) ft to \(x = 100\) ft.
Six reinforcement lines are defined with the following properties:
| Property | Value |
|---|---|
| \(T_{max}\) | 800 lb/ft |
| \(T_{res}\) | 600 lb/ft |
| \(L_{p1}\), \(L_{p2}\) | 4 ft |
| \(E\) | 800,000 psf |
| \(Area\) | 0.1 ft\(^2\)/ft |
| \(EA\) | 80,000 lb/ft |
Excel input file: xslope_reinforce_fem.xlsx
Inputs plotted with the XSLOPE plot_inputs() function:
FEM mesh with boundary conditions and reinforcement elements (red lines):
SSRM results. The computed factor of safety is FS = 1.57, which is consistent with the LEM result of FS = 1.55 obtained using Janbu's method. The top plot shows the deformed mesh with original and deformed reinforcement positions. The bottom plot shows the viscoplastic shear strain concentration with reinforcement elements colored by axial force (blue = low, red = high). Gray elements at the left ends are inactive (no tension). Dashed black elements at the right ends have pulled out. The reinforcement summary table is shown below.
Reinforcement summary:
=== Reinforcement Summary ===
Line Elems Max T Avg T Tension In Lp At Tres Broken Status
--------------------------------------------------------------------------------
1 9 324.6 172.0 9 4 0 0 OK
2 9 598.7 468.1 7 4 0 1 PULLOUT
3 9 706.9 559.1 7 4 0 1 PULLOUT
4 9 796.5 610.7 6 4 0 2 PULLOUT
5 9 773.0 572.7 7 4 1 1 YIELDED
6 9 675.8 515.5 7 4 0 1 PULLOUT
--------------------------------------------------------------------------------
OK: All elements within allowable capacity, no failures.
PULLOUT: Elements near the reinforcement ends (within Lp) have failed due to insufficient embedment length. Interior elements are intact.
YIELDED: One or more elements have exceeded Tallow and dropped to residual capacity Tres. The line is still carrying load at reduced strength.
The results show that reinforcement lines 2-6 experience pullout failure at the right ends where the failure surface intersects the reinforcement. Line 4 has one element that has yielded to residual capacity. The maximum mobilized force is 780 lb/ft (line 5), which is close to but below the \(T_{max}\) of 800 lb/ft.
3. Slope Stabilized with Drilled Shaft Piles
This problem features a 1:1 slope in a medium-stiff clay stabilized by two rows of drilled shafts.
Excel input file: xslope_piles_fem.xlsx
The soil properties are:
| Property | Value |
|---|---|
| Cohesion, \(c\) | 200 psf |
| Friction angle, \(\phi\) | 20 degrees |
| Unit weight, \(\gamma\) | 120 pcf |
| Young's modulus, \(E\) | 2,000,000 psf |
| Poisson's ratio, \(\nu\) | 0.3 |
Two rows of vertical drilled shafts are placed at \(x = 5\) ft and \(x = 10\) ft along the slope face, both extending from the ground surface to \(y = -10\) ft:
| Property | Value |
|---|---|
| Diameter, \(D\) | 2.0 ft |
| Spacing, \(S\) | 6.0 ft |
| Young's modulus, \(E_{\text{pile}}\) | 518,400,000 psf (concrete, \(f'_c\) = 4000 psi) |
| Moment of inertia, \(I\) | \(\pi D^4 / 64\) = 0.785 ft\(^4\) (auto-computed from \(D\)) |
| Cross-sectional area, \(A\) | \(\pi D^2 / 4\) = 3.14 ft\(^2\) (auto-computed from \(D\)) |
| Shear capacity, \(V_{\text{cap}}\) | 46,000 lb |
| Moment capacity, \(M_{\text{cap}}\) | 60,000 ft·lb |
| Fixity | free |
Each pile is modeled as a chain of 6-DOF Euler-Bernoulli beam elements with rotational DOFs at each node (see FEM Piles for the formulation). The pile stiffness (\(EI\) and \(EA\)) is scaled by \(1/S\) to convert from per-pile to per-unit-width quantities. Unlike the LEM approach where the user provides a single force \(H\), the FEM beam elements naturally develop resistance as the soil deforms around the pile. Bending moments are computed directly at each node, and structural capacity limits (\(V_{\text{cap}}\), \(M_{\text{cap}}\)) are enforced through the viscoplastic correction loop.
FEM mesh with boundary conditions. The piles are shown as green line elements along the pile axes:
SSRM results without piles (FS = 1.19). The shear strain concentration shows a failure mechanism passing through the toe:
SSRM results with two rows of piles (FS = 1.32). The pile elements are colored by lateral (shear) force in the shear strain plot. The piles resist the sliding mass and the failure mechanism is modified by their presence:
Pile summary:
=== Pile Summary ===
Pile Elems Max |T| Max |V| Max |M| V_cap M_cap Yielded Status
--------------------------------------------------------------------------------
1 7 1316.9 1277.3 5473.8 7666.7 10000.0 0/7 OK
2 9 2512.0 583.4 2522.8 7666.7 10000.0 0/9 OK
--------------------------------------------------------------------------------
The two rows of piles increase the factor of safety from 1.19 to 1.32 — an 11% improvement. The maximum bending moment (5474 per unit width in Pile 1) reaches about 55% of the moment capacity (\(M_{\text{cap}}/S\) = 10,000), indicating that the structural capacity does not govern for this problem. The soil's ability to transfer lateral load to the piles is the limiting factor, not the pile strength.
This is typical behavior for piles in relatively weak soil — the pile is much stiffer than the surrounding soil, and increasing the pile diameter or stiffness beyond a certain point produces diminishing returns. The 2D plane-strain model also does not capture the three-dimensional soil arching between piles that the Ito & Matsui theory accounts for in LEM, which can make the FEM result more conservative than the LEM result.
4. Non-Circular Failure Surface with Thin Weak Layer
This is the FEM counterpart of the LEM non-circular failure surface example described in the LEM Samples page (Problem 7). The problem features a thin weak clay layer in the foundation of a slope, which controls the failure mechanism. This problem was also featured in the user manual for the UTEXASED slope stability analysis software developed by Stephen G. Wright at the University of Texas at Austin.
The slope geometry and strength properties are the same as the LEM problem. Young's modulus (\(E\)) and Poisson's ratio (\(\nu\)) are estimated from typical correlations for each soil type:
| Soil | \(c'\) (psf) | \(\phi'\) (deg) | \(\gamma\) (pcf) | \(E\) (psf) | \(\nu\) |
|---|---|---|---|---|---|
| Sand Fill | 0 | 37 | 120 | 1,000,000 | 0.30 |
| Sand | 0 | 33 | 123 | 700,000 | 0.30 |
| Soft Clay (\(S_u\) = 200) | 0 (\(\phi = 0\)) | 0 | 118 | 60,000 | 0.40 |
| Dense Sand | 0 | 37 | 131 | 1,500,000 | 0.28 |
The soft clay is modeled as an undrained material (\(\phi = 0\)) with \(E/S_u \approx 300\). A Poisson's ratio of 0.40 is used rather than the theoretical undrained value of 0.5 to avoid numerical issues with near-incompressibility.
Excel input file: xslope_noncircular_fem.xlsx
Inputs plotted with the XSLOPE plot_inputs() function:
FEM mesh with boundary conditions and material zones:
SSRM results. The computed factor of safety is FS = 1.82. The top plot shows the deformed mesh at the last converged solution (F = 1.82). The middle plot shows the viscoplastic shear strain concentration, which clearly reveals the non-circular failure mechanism passing through the thin weak clay layer — matching the expected behavior without any prior assumption about the failure surface shape. The bottom plot shows the displacement vectors, confirming lateral sliding of the slope mass along the clay layer.
The FEM result of FS = 1.82 is slightly higher than the LEM result of FS = 1.74 obtained using Spencer's method. This is consistent with the general observation that the SSRM tends to give slightly higher factors of safety than LEM for non-circular mechanisms, since the FEM finds the natural failure mode through the global stress field rather than being constrained to a prescribed failure surface geometry.