Slope Stability — Swedish Slip Circle Method

1. Introduction to Slope Stability

A slope fails when the shear stress mobilised along a potential failure surface exceeds the shear strength of the soil on that surface. Slope stability analysis quantifies this relationship through the factor of safety (FOS):

Slope failures occur in a wide range of settings: natural hillslopes destabilised by rainfall or earthquakes, engineered embankments for highways and railways, earth dams, excavation slopes, and landfills. In India, the Himalayan foothills, the Western Ghats, and the northeast experience frequent landslides — a direct consequence of steep slopes, heavy seasonal rainfall, and geologically young rock and soil. Engineering slope stability analysis is therefore an important safety discipline.

The most common method for analysing circular (rotational) failure surfaces is the method of slices, of which Fellenius (Swedish slip circle) and Bishop’s simplified methods are the most widely used and examined in GATE CE.

2. Types of Slope Failure

3. Infinite Slope Analysis

The infinite slope model applies to long, uniform slopes where the failure surface is shallow and parallel to the slope surface. It is appropriate for analysis of shallow translational failures in cohesionless soils or in thin soil mantles over bedrock.

3.1 Dry Cohesionless Slope

3.2 Saturated Slope — Seepage Parallel to Slope

When the water table is at the ground surface and seepage is parallel to the slope, the pore pressure on the failure plane at depth z is u = γ_w z cos²β.

3.3 Cohesive Soil — Infinite Slope

4. Swedish Slip Circle — Fellenius Method

The Fellenius (1927) method, also called the ordinary method of slices or the Swedish method, divides the potential failure mass above a trial circular arc into vertical slices and applies static equilibrium to each slice.

4.1 Procedure

1. Assume a trial circular failure surface with centre O and radius R.
2. Divide the failure mass into n vertical slices (typically 8–12 slices of equal width b).
3. For each slice i, compute:
   • Width b_i and height h_i at the midpoint
   • Weight W_i = γ b_i h_i
   • Base angle α_i (inclination of slice base to horizontal)
   • Arc length l_i = b_i/cosα_i
   • Normal force N_i′ = W_i cosα_i − u_i l_i
   • Tangential force T_i = W_i sinα_i
4. Compute FOS:

   FOS = Σ(c′ l_i + N_i′ tanφ′) / Σ(W_i sinα_i)

   N_i′ = W_icosα_i − u_il_i (pore pressure correction)

   For dry slope (u = 0): N_i′ = W_icosα_i

5. Repeat for multiple trial circles; the critical circle has the minimum FOS.

4.2 Simplification for φ = 0 (Undrained Clay)

4.3 Limitations of Fellenius Method

Fellenius assumes the interslice forces (horizontal and vertical forces between adjacent slices) are zero. This is not physically correct — interslice forces exist and can be significant. As a result, Fellenius underestimates FOS, sometimes by 10–15 % for typical slopes and more for very flat slopes or high pore pressures. It is conservative but inaccurate.

5. Bishop’s Simplified Method

Bishop (1955) improved on Fellenius by including the interslice normal forces (but neglecting interslice shear forces — hence “simplified”). This gives a more accurate FOS with only modest additional computational effort.

5.1 Bishop’s Formula

Since m_α contains FOS, the solution requires iteration: assume an initial FOS (e.g., from Fellenius), compute m_α, compute new FOS, repeat until convergence (typically 3–5 iterations).

5.2 Comparison: Fellenius vs Bishop

6. Taylor’s Stability Chart

Taylor (1937) produced non-dimensional stability charts for homogeneous slopes that allow rapid estimation of FOS without performing a full slip circle analysis. The charts are based on the stability number S_n:

6.1 Key Values from Taylor’s Chart

For φ = 0 (purely cohesive soil), Taylor’s chart gives a depth factor D_f: the critical circle may be a toe circle, slope circle, or base circle depending on the depth to a hard stratum. For φ > 3°, the critical circle always passes through the toe.

6.2 Critical Height of a Slope (φ = 0)

7. Effect of Seepage on Slope Stability

Seepage through a slope has two destabilising effects: it increases pore water pressure on the failure surface (reducing effective normal stress and hence shear strength), and the seepage force acts in the downslope direction (increasing the driving moment). Together, these can reduce FOS by 30–50 %.

7.1 Pore Pressure Ratio (r_u)

7.2 Stability of a Saturated Slope in a φ′ = 0 Analysis

7.3 Slope Stability in Steady Seepage

For the downstream slope of an earth dam under steady seepage, use drained parameters (c′, φ′) with pore pressures from a flow net analysis. The pore pressure at each slice base is determined from the phreatic line and potential drop.

8. Tension Crack in Cohesive Slopes

Cohesive slopes often develop a tension crack near the crest before failure. The tension crack reduces the effective arc length resisting failure and, when filled with water, significantly increases the driving moment.

9. Location of the Critical Failure Circle

The critical failure circle is the one with the minimum FOS. Its location depends on the soil type and slope geometry:

In practice, the critical circle must be found by trial — testing many circles with different centres and radii. Systematic grid searches or optimisation algorithms (in software) are used. For hand calculations in GATE, the trial circle is given or a specific circle is analysed.

10. Worked Examples

Example 1 — Infinite Slope: Dry vs Saturated

Problem: A 6 m deep infinite slope is inclined at β = 20° to the horizontal. Soil: c′ = 0, φ′ = 35°, γ_dry = 17 kN/m³, γ_sat = 19 kN/m³. Find FOS for: (a) dry slope, (b) slope with water table at surface and seepage parallel to slope.

(a) Dry Slope

(b) Saturated Slope with Seepage

The slope fails when fully saturated with seepage (FOS < 1.0). This is why drainage is critical for slope stability.

Example 2 — Swedish Slip Circle Analysis (Dry Slope)

Problem: A trial slip circle on a slope gives the following slice data (c′ = 15 kPa, φ′ = 25°, dry). Determine FOS using the Fellenius method.

Compute Components

Example 3 — Taylor’s Chart Application

Problem: A 8 m high slope is inclined at 30° to the horizontal. Soil: c = 25 kPa, φ = 10°, γ = 18 kN/m³. Using Taylor’s chart (S_n = 0.056 for β = 30°, φ = 10°), find FOS.

The slope is stable with a comfortable FOS of 3.1.

Example 4 — GATE-Style: Critical Height of Unsupported Vertical Cut

Problem (GATE CE type): A saturated clay has c_u = 35 kPa, φ_u = 0°, γ_sat = 18 kN/m³. Find: (a) the critical height of an unsupported vertical cut, (b) the stable height with FOS = 1.5.

(a) Critical Height

(b) Safe Height with FOS = 1.5

In practice, a tension crack of depth z_c = 2c_u/γ = 2×35/18 = 3.89 m may develop before failure — the cut should be monitored for crack formation.

11. Common Mistakes

Mistake 1: Using Total Weight Instead of Effective Normal Force in Fellenius

What happens: N = W cosα is used in place of N′ = W cosα − ul when pore pressures are present. This overestimates the frictional resistance and gives an unconservative (too high) FOS.

Fix: Always subtract pore pressure from the normal force: N′ = W cosα − u l, where u is the pore water pressure at the base of the slice and l = b/cosα is the arc length of the slice base.

Mistake 2: Applying the Infinite Slope FOS Formula to a Finite Slope

What happens: FOS = tanφ/tanβ is applied to a finite slope with a clearly rotational failure mechanism, ignoring cohesion and the curved nature of the failure surface. For c-φ soils on finite slopes, the infinite slope formula underestimates stability.

Fix: Infinite slope analysis is only valid for long, shallow translational failures. For finite slopes with significant cohesion, use the method of slices (Fellenius or Bishop) or Taylor’s chart.

Mistake 3: Forgetting that Slices with Negative α Contribute Negative Driving Moment

What happens: In ΣW sinα, slices on the passive side of the circle (near the toe) have negative α values and negative W sinα. Students add absolute values instead of algebraic sums, overestimating the driving moment and underestimating FOS.

Fix: α is positive for slices on the active side (tilted toward failure) and negative for slices on the passive side (tilted against failure direction). Use algebraic sum ΣW sinα — the passive slices reduce the net driving moment.

Mistake 4: Using Bishop’s Formula Without Iteration

What happens: m_α = cosα + (tanφ′/FOS)sinα is computed using an assumed FOS (say FOS = 1) and never updated, giving an incorrect answer.

Fix: Bishop’s method requires iteration: (1) assume FOS₁ (use Fellenius as starting value), (2) compute m_α for each slice, (3) compute FOS₂, (4) update m_α using FOS₂, (5) compute FOS₃ — repeat until |FOS_n+1 − FOS_n| < 0.01.

Mistake 5: Not Checking the Most Critical Circle

What happens: FOS is computed for one trial circle and reported as the slope FOS. The true minimum FOS may be for a different circle — either deeper or shallower.

Fix: In hand calculations for GATE, if a specific circle is given, use it. In practice, a minimum of 10–20 trial circles must be analysed; the critical circle is found by systematic search. Taylor’s chart provides the critical circle FOS directly without a search.

12. Frequently Asked Questions

Q1. Why does seepage reduce the FOS of a slope so dramatically — sometimes cutting it in half?

Seepage reduces slope stability through two simultaneous effects that compound each other. First, pore water pressure u builds up along the failure surface, reducing the effective normal stress σ′ = σ − u. Since frictional shear strength is τ_f = c′ + σ′tanφ′, lower σ′ directly reduces shear strength. Second, the seepage force j = iγ_w acts in the downslope direction (parallel to the slope when seepage is parallel to the surface), increasing the driving shear stress. For a cohesionless infinite slope with seepage parallel to the surface and water table at the ground surface, both effects combine to give FOS = (γ′/γ_sat)tanφ′/tanβ — compared to FOS = tanφ′/tanβ for dry conditions. Since γ′/γ_sat ≈ 0.48–0.52 for typical saturated soils, the FOS is nearly halved. This explains why many slopes that are stable for years can suddenly fail after heavy rainfall.

Q2. What is the difference between a toe circle, a slope circle, and a base circle?

These terms describe where the critical failure circle exits the slope, which depends on soil properties and slope geometry. A toe circle passes through the toe of the slope (the base of the slope face) — this is the most common critical circle for c-φ soils with φ > 3°. A slope circle exits on the slope face itself above the toe — occurs when there is a weak zone within the slope or when the slope is very flat. A base circle passes below the toe, often through the foundation soil — this occurs for purely cohesive soils (φ = 0) where the critical circle can be very deep. The depth to a hard impermeable layer limits how deep the base circle can go (Taylor’s depth factor D = depth to hard layer / slope height).

Q3. Why is Bishop’s method more accurate than Fellenius, and when should you use each?

Fellenius ignores all interslice forces, meaning it does not satisfy complete equilibrium for each slice. This leads to an underestimate of the normal force on the failure surface, which in turn underestimates the frictional resistance — giving a conservative (lower) FOS, typically 5–15 % below Bishop’s result. Bishop’s simplified method includes interslice normal forces (but not shear forces), which satisfies moment equilibrium about the circle centre and produces more accurate FOS estimates. In practice, use Fellenius only as a first estimate or for simple GATE calculations. Use Bishop’s simplified method whenever greater accuracy is required and for design calculations. For non-circular failure surfaces, neither method applies — use Janbu’s generalised procedure or Spencer’s method (outside GATE scope).

Q4. How does the drainage condition (drained vs undrained) affect slope stability analysis?

The choice of drainage condition depends on the rate of loading relative to the permeability of the soil. For short-term (end-of-construction) stability in clay — where loading is rapid and drainage cannot occur — use undrained parameters (c_u, φ_u = 0) in a total stress analysis. The pore pressures are unknown but the analysis uses total stresses where pore pressures have already been absorbed. For long-term (steady-state) stability — where full drainage has occurred and seepage is in steady state — use effective stress parameters (c′, φ′) with pore pressures from a flow net. The critical condition (lowest FOS) depends on the specific slope and loading case: for embankments on soft clay, end-of-construction is critical; for natural slopes on overconsolidated clay, long-term (after softening) may be critical; for reservoir slopes, rapid drawdown is critical.