Stability of Slopes

Slope failure occurs when the shear stress along a surface exceeds the shear strength of the soil. The causes can be grouped into those that increase the shear stress and those that decrease the shear strength.

A. Causes that increase shear stress

1. Removal of lateral support by erosion, undercutting or excavation at the toe.
2. Extra load on the top (surcharge): buildings, stored materials, vehicles, or fill placed on the slope.
3. Increase in slope steepness or height by cutting or filling.
4. Increase in weight of soil from rain or seepage water (higher unit weight).
5. Earthquakes and vibrations from blasting, traffic or machinery, which add dynamic forces.
6. Tension cracks filled with water, which cause hydrostatic thrust on the potential slip surface.
7. Transient loads such as wind or water pressure acting on the slope.

B. Causes that decrease shear strength

1. Increase in pore water pressure from rainfall, rising water table or seepage, which lowers effective stress.
2. Rapid drawdown of water against a slope (e.g. reservoir bank), leaving excess pore pressure inside the soil.
3. Weathering and chemical changes that weaken the soil or rock and reduce cohesion.
4. Swelling of clay and loss of cementation after wetting.
5. Progressive failure and strain softening, reducing strength from peak to residual.
6. Decay of roots and removal of vegetation, which reduces cohesion near the surface.
7. Liquefaction or loss of strength in loose saturated sand due to shaking.
8. Freeze-thaw and cracking that open fissures and weaken the mass.

The basic modes of failure of slopes are:

1. Rotational slip (slump), which has three sub-types: face, toe and base failure.
2. Translational slip (planar), along a shallow plane parallel to the slope.
3. Compound slip, with a combined curved and plane surface.
4. Wedge failure, sliding of a wedge along weak planes.
5. Infinite slope (surface) failure, a shallow slide parallel to a long slope.
6. Flow, fall and creep in which soil flows or falls.

1. Rotational slip (circular)

The failure surface is approximately a circular arc, and the soil mass rotates about a centre. It occurs in homogeneous cohesive soils (clay).

• Toe failure: the arc passes through the toe; common in steep slopes in soil with a high angle of friction.
• Face (slope) failure: the arc ends on the slope above the toe; occurs when a hard layer exists near the toe.
• Base failure: the arc goes below the toe into the base; occurs in flat slopes of soft clay with a deep soft layer.

        centre O
          *
        . . .  R
   _____.     .______  <- crest
        \  .'    <- slip circle
         \'
 toe  ----`-------------
  (toe failure: arc passes through the toe)

2. Translational slip (planar)

The soil slides along a nearly plane surface, often where a weak layer or a hard stratum lies at shallow depth parallel to the slope. The length is much greater than the depth. It is typical for cohesionless soil and infinite slopes.

  ______
        \\  soil mass
         \\\  -----> slides
          \\\\__ weak plane (parallel to slope)
   ----------------------------

Types of slope failure

1. Rotational slip (face, toe and base failures): a curved surface in homogeneous clay.
2. Translational (planar) slip: a shallow plane parallel to the slope, in layered soil or cohesionless soil.
3. Compound slip: partly curved and partly plane, over a hard layer.
4. Wedge failure: sliding on intersecting weak planes.
5. Infinite slope failure: shallow slide on a long uniform slope.
6. Falls, flows and creep: rapid fall of blocks or flow of saturated soil, and slow downhill movement.

Measures to prevent slope failure

Reduce the driving forces

• Flatten the slope or reduce its height, or construct benches.
• Remove the load from the top (unload the crest).

Increase the resisting forces

• Add a berm or counterweight at the toe.
• Construct retaining walls, gabion walls, or piles at the toe.
• Use reinforced earth, geotextiles, soil nails or rock anchors.
• Replace weak soil by better material, or compact the soil.
• Provide vegetation and bio-engineering (grass, shrubs, trees) to bind the soil and reduce erosion.

Control water

• Provide surface drains and catch-water drains above the slope to divert runoff.
• Provide sub-surface drains, horizontal drains or sand blankets to reduce pore pressure.
• Seal cracks and use impermeable surface covering where required.
• Provide filters and weep holes behind retaining structures.

Others

• Soil stabilisation by grouting, lime or cement; control vibrations from blasting; and keep monitoring with instruments.

An infinite slope is a slope of very large extent whose length is much greater than the depth of the failure surface and which has uniform soil along its length. A finite slope has a limited height and length, with a definite top and toe.

The $\phi = 0$ analysis is a total stress method for a saturated clay under undrained (short-term) conditions, where $\tau_f = c_u$ and $\phi_u = 0$. It uses the Swedish (slip) circle.

Assumptions

• Homogeneous saturated clay with undrained strength $c_u$ and unit weight $\gamma$.
• Failure along a circular arc of radius $R$ with centre $O$.
• Moments are taken about $O$ per metre length of slope.

Method

1. Draw the slope to scale and choose a trial centre $O$ and radius $R$.
2. Find the area of the sliding mass and compute its weight $W = \gamma A$; locate the centroid at horizontal distance $x$ from $O$.
3. Measure the arc length $L_a = R\theta$ (with $\theta$ in radians).
4. Moments about $O$:

$$\text{Resisting moment} = c_u L_a R, \qquad \text{Driving moment} = Wx$$ $$F = \frac{c_u L_a R}{W x}$$

5. Repeat for several trial circles; the least value of $F$ is the factor of safety of the slope.

If a tension crack of depth $z_c = \dfrac{2c_u}{\gamma}$ is considered, $L_a$ is reduced by the length in the crack, and a hydrostatic force may be added to the driving moment.

Example (assumed data): $c_u = 30$ kN/m², $\gamma = 18$ kN/m³, $R = 12$ m, arc angle $80^\circ$ ($1.396$ rad), area of sliding mass $= 80$ m², $x = 2.5$ m.

$$\begin{aligned} L_a &= 12 \times 1.396 = 16.76\ \text{m} \\ W &= 18 \times 80 = 1440\ \text{kN/m} \\ F &= \frac{30 \times 16.76 \times 12}{1440 \times 2.5} = \frac{6034}{3600} = 1.68 \end{aligned}$$

Answer: $F \approx 1.68$ for this trial circle; the minimum $F$ over trial circles governs the design (should be $\geq 1.3$ to 1.5 for the short term).

Data: $\gamma_t = 18$, $\gamma' = 9$ kN/m³, $c = 25$ kN/m², $\phi' = 28^\circ$, $\beta = 35^\circ$. The depth of the failure plane is taken as $z = 12$ m (the slope height is measured as vertical depth). For seepage, the saturated unit weight is taken as $\gamma_{sat} = \gamma_t = 18$ kN/m³.

Useful values: $\sin\beta\cos\beta = 0.4698$, $\tan\phi'/\tan\beta = 0.5317/0.7002 = 0.7594$.

(i) Slope fully submerged (static water, no seepage)

Normal stress on the plane: $\sigma' = \gamma' z\cos^2\beta$. Shear stress: $\tau = \gamma' z\sin\beta\cos\beta$.

$$\begin{aligned} F &= \frac{c'}{\gamma' z\sin\beta\cos\beta} + \frac{\tan\phi'}{\tan\beta} \\ &= \frac{25}{9(12)(0.4698)} + 0.7594 \\ &= 0.493 + 0.759 = 1.25 \end{aligned}$$

$F = 1.25 > 1$, so the slope is stable (but its margin is small; $F < 1.5$).

(ii) Seepage parallel to the slope (water table at the surface)

Normal effective stress: $\sigma' = \gamma' z\cos^2\beta$. Shear stress: $\tau = \gamma_{sat} z\sin\beta\cos\beta$.

$$\begin{aligned} F &= \frac{c'}{\gamma_{sat} z\sin\beta\cos\beta} + \frac{\gamma'}{\gamma_{sat}}\frac{\tan\phi'}{\tan\beta} \\ &= \frac{25}{18(12)(0.4698)} + \frac{9}{18}(0.7594) \\ &= 0.246 + 0.380 = 0.63 \end{aligned}$$

$F = 0.63 < 1$, so the slope is unstable with seepage parallel to it.

Answer: (i) $F = 1.25$, stable; (ii) $F = 0.63$, unstable.

Remedial measures to prevent or stop slope failure:

1. Flatten the slope or reduce its height; provide berms or benches.
2. Unload the top by removing soil or load from the crest, and place a berm or counterweight at the toe.
3. Drainage: surface drains and catch-water drains, sub-surface and horizontal drains, and sand blankets to lower the water table and pore pressure.
4. Retaining structures: retaining walls, gabions, sheet piles or piles at the toe.
5. Reinforcement: geotextiles, geogrids, soil nailing, rock bolts and anchors.
6. Vegetation and bio-engineering to bind the soil and stop erosion.
7. Soil improvement: compaction, replacement of weak soil, grouting, or lime and cement stabilisation.
8. Control of vibrations and erosion at the toe, with proper protection such as riprap.

In the Swedish (slip circle) method, the failure surface is assumed to be circular. The factor of safety is calculated for many trial circles, and the critical circle is the one with the minimum factor of safety.

Procedure

1. Draw the slope section to scale and choose the toe, crest and any soil boundaries.
2. Select a trial centre $O$ (above the slope, using Fellenius' guide lines or a grid of points) and a radius $R$ so that the circle passes through the toe (or just below it).
3. Compute the factor of safety $F$ for this circle:
   • $\phi=0$ soil: $F = \dfrac{c_u L_a R}{Wx}$;
   • $c$-$\phi$ soil: divide the mass into slices and use the method of slices.
4. Keep the same centre and change $R$ to find the minimum $F$ for that centre.
5. Repeat for other centres arranged in a grid, writing the minimum $F$ at each centre.
6. Draw contours of equal factor of safety through the grid values. The centre of the smallest contour is the centre of the most critical circle, and its value is the minimum factor of safety.

   O1*   O2*   O3*
       F=1.8 F=1.5 F=1.7    <- grid of centres
   O4*   O5*   O6*
       F=1.6 F=1.3 F=1.5
   (O5 gives lowest F: critical circle)

For homogeneous soil with $\phi=0$, the position of the critical centre can also be found from Fellenius' construction (direction angles at the toe and crest) or Taylor's charts.

Assumptions: infinite slope of dry homogeneous $c$-$\phi$ soil, inclined at angle $\beta$ to the horizontal; failure plane parallel to the slope at depth $z$; no seepage; unit weight $\gamma$.

Consider a slice of width $b$ along the slope, with the plane of failure length $l = b/\cos\beta$. Forces on the two sides of the slice are equal and opposite, so they cancel.

        ______ ground surface (beta)
       /|     \
      / | W    \
     /__|_____
        z   N  T  <- failure plane

Weight of slice: $W = \gamma z b$ per unit length.

$$N = W\cos\beta, \qquad T = W\sin\beta$$

Normal stress and shear stress on the plane:

$$\sigma = \frac{N}{l} = \gamma z\cos^2\beta, \qquad \tau = \frac{T}{l} = \gamma z\sin\beta\cos\beta$$

Shear strength (Mohr-Coulomb): $\tau_f = c + \sigma\tan\phi = c + \gamma z\cos^2\beta\tan\phi$

Factor of safety $F = \tau_f/\tau$:

$$\boxed{F = \frac{c + \gamma z\cos^2\beta\tan\phi}{\gamma z\sin\beta\cos\beta} = \frac{c}{\gamma z\sin\beta\cos\beta} + \frac{\tan\phi}{\tan\beta}}$$

For cohesionless soil ($c=0$), $F = \tan\phi/\tan\beta$, independent of $z$. For a given $c$, the critical (limiting) depth for $F=1$ is $z_c = \dfrac{c}{\gamma\cos^2\beta(\tan\beta - \tan\phi)}$.

Conditions: infinite slope of cohesionless soil ($c=0$), slope angle $\beta$, steady seepage parallel to the slope with the water table at the ground surface. Saturated unit weight $\gamma_{sat}$, submerged unit weight $\gamma'$.

Consider a slice of depth $z$ and width $b$. Seepage is parallel to the slope, so the equipotential lines are perpendicular to the slope and the pore pressure on the plane is

$$u = \gamma_w z\cos^2\beta$$

Total weight of the slice: $W = \gamma_{sat} z b$.

$$\sigma = \gamma_{sat} z\cos^2\beta, \qquad \tau = \gamma_{sat} z\sin\beta\cos\beta$$

Effective normal stress:

$$\sigma' = \sigma - u = (\gamma_{sat} - \gamma_w) z\cos^2\beta = \gamma' z\cos^2\beta$$

Shear strength $= \sigma'\tan\phi'$, so the factor of safety is

$$F = \frac{\gamma' z\cos^2\beta\tan\phi'}{\gamma_{sat} z\sin\beta\cos\beta} = \boxed{\frac{\gamma'}{\gamma_{sat}}\,\frac{\tan\phi'}{\tan\beta}}$$

Remarks:

• As $\gamma'/\gamma_{sat}\approx 0.5$, the factor of safety with seepage is about half the dry-slope value $\tan\phi'/\tan\beta$.
• The stable slope angle with seepage is therefore about half the friction angle: $\tan\beta \approx 0.5\tan\phi'$.
• If the water table is at height $m z$ above the plane (with $0<m<1$), $F = \left(1 - \dfrac{m\gamma_w}{\gamma_{sat}}\right)\dfrac{\tan\phi'}{\tan\beta}$.

Derivation (cohesionless soil, no water table)

Take a slice of width $b$ and depth $z$ in an infinite slope inclined at $\beta$. Soil is dry or moist with unit weight $\gamma$ and angle of friction $\phi$. Side forces cancel.

$$W = \gamma z b, \quad N = W\cos\beta, \quad T = W\sin\beta$$

On the failure plane of length $b/\cos\beta$:

$$\sigma = \gamma z\cos^2\beta, \qquad \tau = \gamma z\sin\beta\cos\beta$$

Shear strength: $\tau_f = \sigma\tan\phi = \gamma z\cos^2\beta\tan\phi$

$$\boxed{F = \frac{\tau_f}{\tau} = \frac{\tan\phi}{\tan\beta}}$$

The factor of safety does not depend on depth or unit weight. The slope is stable when $\beta < \phi$, and at $\beta=\phi$ ($F=1$) the slope is at its limit (angle of repose).

Water table rises to the surface

With seepage parallel to the slope and water at the surface, the pore pressure is $u = \gamma_w z\cos^2\beta$, so the effective normal stress falls to $\gamma' z\cos^2\beta$ while the shear stress rises to $\gamma_{sat} z\sin\beta\cos\beta$:

$$F_{sat} = \frac{\gamma'}{\gamma_{sat}}\frac{\tan\phi}{\tan\beta}$$

Since $\gamma'/\gamma_{sat}\approx 0.5$, the factor of safety falls to about half. For example, a slope with $\phi = 35^\circ$ and $\beta = 20^\circ$ has $F = 1.92$ when dry but only about 0.96 when the water reaches the surface, so the slope would fail. This is why heavy rain triggers landslides in sandy slopes.

Data: $c = 10$ kPa, $\phi = 25^\circ$, $z = 5$ m (depth of the failure plane, as given), $\gamma = 16$ kN/m³, $\beta = 10^\circ$, no water.

For a dry infinite slope:

$$F = \frac{c}{\gamma z\sin\beta\cos\beta} + \frac{\tan\phi}{\tan\beta}$$

Step 1: Values

$$\sin10^\circ\cos10^\circ = 0.1710, \quad \tan25^\circ = 0.4663, \quad \tan10^\circ = 0.1763$$

Step 2: Cohesion term

$$\frac{10}{16\times5\times0.1710} = \frac{10}{13.68} = 0.731$$

Step 3: Friction term

$$\frac{0.4663}{0.1763} = 2.645$$ $$F = 0.731 + 2.645 = 3.38$$

$F = 3.38 > 1$, so the slope is stable with a large margin.

If cohesion becomes zero

$$F = \frac{\tan\phi}{\tan\beta} = 2.645$$

The factor of safety drops by about 22% but remains well above 1, so the slope is still stable. The stability then depends only on friction ($\beta=10^\circ<\phi=25^\circ$).

Answer: $F = 3.38$ (stable); with $c = 0$, $F = 2.64$ (still stable).

Data: $c' = 0$, $e = 0.7$, $G = 2.7$, $\phi = 35^\circ$, required $F = 2.0$, seepage parallel to the slope with water at the surface.

Step 1: Unit weights

$$\gamma_{sat} = \frac{(G+e)\gamma_w}{1+e} = \frac{3.4(9.81)}{1.7} = 19.62\ \text{kN/m}^3, \qquad \gamma' = 19.62 - 9.81 = 9.81\ \text{kN/m}^3$$ $$\frac{\gamma'}{\gamma_{sat}} = \frac{G-1}{G+e} = \frac{1.7}{3.4} = 0.5$$

Step 2: Maximum slope angle for $F=2$

$$F = \frac{\gamma'}{\gamma_{sat}}\frac{\tan\phi}{\tan\beta} \;\Rightarrow\; 2.0 = 0.5\,\frac{\tan35^\circ}{\tan\beta}$$ $$\tan\beta = \frac{0.5\times0.7002}{2.0} = 0.1750 \;\Rightarrow\; \beta = 9.93^\circ \approx 9.9^\circ$$

Step 3: Factor of safety if the water level falls well below the surface (dry or moist soil, no seepage force):

$$F = \frac{\tan\phi}{\tan\beta} = \frac{0.7002}{0.1750} = 4.0$$

Answer: maximum slope angle $\approx 9.9^\circ$; factor of safety rises to 4.0 when the water level drops well below the surface.

Data (per metre run): $\Sigma T = 450$ kN, $\Sigma N = 900$ kN, $\Sigma U = 216$ kN (neutral force, pore pressure force), arc length $L = 27$ m, $c' = 20$ kN/m², $\phi' = 18^\circ$.

Effective normal force: $\Sigma N' = \Sigma N - \Sigma U = 900 - 216 = 684$ kN.

(a) Factor of safety with respect to shear strength

The factor of safety is the ratio of the available shear strength to the shear force:

$$\begin{aligned} F_s &= \frac{c'L + (\Sigma N-\Sigma U)\tan\phi'}{\Sigma T} \\ &= \frac{20(27) + 684\tan18^\circ}{450} \\ &= \frac{540 + 222.2}{450} = 1.69 \end{aligned}$$

(b) Factor of safety with respect to cohesion

Here the full friction is mobilised ($F_\phi = 1$) and only the cohesion is reduced to the value needed for equilibrium:

$$\Sigma T = c_m L + (\Sigma N - \Sigma U)\tan\phi' \;\Rightarrow\; c_m = \frac{450 - 222.2}{27} = 8.44\ \text{kN/m}^2$$ $$F_c = \frac{c'}{c_m} = \frac{20}{8.44} = 2.37$$

Answer: (a) $F_s \approx 1.69$; (b) $F_c \approx 2.37$.

Note: the rest of the question and the cut slope are missing. The usual form asks for the factor of safety (or safe height) of the cut. The solution below assumes a vertical cut (slope angle $\beta = 90^\circ$) and uses Taylor's stability number method; the same steps apply for any given slope angle.

Data: $H = 10$ m, $c = c_u = 35$ kN/m², $\gamma = 18.5$ kN/m³, $\phi = 0$. Bedrock at 15 m below the original ground, so the depth factor is

$$n_d = \frac{D}{H} = \frac{15}{10} = 1.5$$

Taylor's stability number (for $\phi = 0$)

$$N_s = \frac{c_m}{\gamma H}$$

For steep slopes ($\beta > 53^\circ$) the critical circle is a toe circle regardless of the depth factor, and Taylor's chart gives, for $\beta = 90^\circ$, $N_s \approx 0.261$.

Critical height

$$H_c = \frac{c}{\gamma N_s} = \frac{35}{18.5\times0.261} = 7.25\ \text{m}$$

Factor of safety

$$F = \frac{H_c}{H} = \frac{7.25}{10} = 0.72 \quad \left(\text{or } F = \frac{c}{N_s\gamma H} = \frac{35}{0.261\times185} = 0.72\right)$$

Result: $F < 1$, so a 10 m vertical cut in this soil is unsafe and would fail. The maximum safe height of a vertical cut is about 7.25 m (with $F = 1$). For $F = 1.5$, the cut must be flattened (read $N_s$ for a smaller $\beta$ from Taylor's chart) or supported by a retaining system.

Answer: $F \approx 0.72$ for a vertical cut (unsafe). The depth factor $n_d = 1.5$ matters only for flatter slopes ($\beta<53^\circ$).