Slope Stability for Fills, Embankments, Cuts and Dams

NCEES FE Civil spec: Geotechnical Engineering, J. Slope stability (e.g., fills, embankments, cuts, dams). FE Reference Handbook: Civil Engineering chapter, Geotechnical section (slope failure along planar surface).

What the FE Civil Spec Covers

Section J covers the stability of fills, embankments, cuts and earth dams. On the FE you will mostly compute a factor of safety for a simple failure mechanism (infinite slope or a planar wedge) and reason about how water, slope angle and soil strength change it. Circular slip methods appear conceptually.

Key Concepts

Factor of safety is available shear resistance divided by the shear force (or stress) needed for equilibrium. FS = 1 means failure is imminent.
Infinite slope: a long slope with a failure plane parallel to the surface at depth z. Good for shallow slides in sand or in a thin weathered layer over rock.
Planar wedge: a block sliding on one inclined plane, like a block on a ramp with friction and cohesion.
Water is the usual trigger: rainfall, rapid drawdown or rising groundwater raises pore pressure and lowers effective stress.
Short term vs long term: cuts in clay often get less stable with time as negative pore pressures dissipate; fills on soft clay are often most critical at end of construction.

Core Formulas

General definition:

F S = \frac{resisting force}{driving force} = \frac{τ _{f}}{τ _{m o bi l i z e d}}

Infinite slope, failure plane at depth z, slope angle $β$ , pore pressure u on the plane:

F S = \frac{c ^{'} + ( γ z cos ^{2} β - u ) tan ϕ ^{'}}{γ z sin β cos β}

Special cases:

Dry sand: F S = \frac{tan ϕ ^{'}}{tan β}, seepage parallel to slope at surface: u = γ_{w} z cos^{2} β

Planar wedge of weight W on a slip plane of length L inclined at angle α:

F S = \frac{c L + W cos α tan ϕ}{W sin α}

Analysis Methods and Remedies

Method	Failure surface	When used
Infinite slope	Plane parallel to the surface	Shallow slides, sand, weathered layer over rock
Planar wedge (Culmann)	Single inclined plane through the toe	Steep cuts, rock joints, weak seams
Method of slices (ordinary, Bishop)	Circular arc	Embankments, dams and clay slopes

Ways to raise FS: flatten the slope, unload the crest, add a toe berm, lower the water table with drains, or reinforce with geogrids, soil nails or piles.

Where It Is in the FE Reference Handbook

The Civil Engineering chapter, Geotechnical section, shows slope failure along a planar surface with the factor of safety defined as available shear resistance over mobilized shear. The infinite slope equation above is a special case of the same idea; write it from first principles if it is not given.

Worked Example: Infinite Slope, Dry and with Seepage

A long slope is inclined at $β = 2 2^{\circ}$ . A potential slip plane lies 2.5 m deep, parallel to the surface. The soil has $c^{'} = 8$ kPa, $ϕ^{'} = 2 8^{\circ}$ and unit weight 19 kN/m³ (use the same value when saturated). Find FS (a) dry and (b) with the water table at the surface and seepage parallel to the slope.

Step 1: stresses on the slip plane.

σ_{n} = γ z cos^{2} β = 19 (2.5) cos^{2} 2 2^{\circ} = 40.83 kPa

τ = γ z sin β cos β = 19 (2.5) sin 2 2^{\circ} cos 2 2^{\circ} = 16.50 kPa

Step 2: dry case (u = 0).

F S = \frac{8 + 40.83 tan 2 8 ^{\circ}}{16.50} = \frac{8 + 21.71}{16.50} = 1.80

Step 3: seepage case.

u = 9.81 (2.5) cos^{2} 2 2^{\circ} = 21.08 kPa

F S = \frac{8 + ( 40.83 - 21.08 ) tan 2 8 ^{\circ}}{16.50} = \frac{8 + 10.50}{16.50} = 1.12

Seepage drops the factor of safety from 1.80 to 1.12, below the usual 1.5 target. Drainage would be the first fix.

Common Mistakes

Using $cos β$ instead of $cos^{2} β$ for the normal stress on an infinite slope plane.
Subtracting full hydrostatic pressure $γ_{w} z$ when seepage is parallel to the slope; the pressure head is reduced by cos squared beta.
Swapping sine and cosine between the resisting and driving terms.
Thinking FS depends on depth for a dry, cohesionless slope; it does not.

Exam Tips

For a dry sand slope, the answer is one line: tan phi over tan beta. When cohesion is present, the cohesion term shrinks as the failure plane gets deeper, so the deepest plausible plane is usually the critical one. When a problem mentions rainfall, a rising water table or rapid drawdown of a reservoir, expect the factor of safety to drop, and check whether the question wants the pore pressure from a horizontal water table or from seepage parallel to the slope, since they give different values.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

Resisting forces total 150 kN and driving forces total 100 kN. What is the factor of safety? See the worked solution or practice the Soil Mechanics deck
For a dry cohesionless infinite slope, FS = tan φ / tan β. If φ = 32° and β = 24°, FS is most nearly: See the worked solution or practice the Soil Mechanics deck
An infinite slope has c′ = 10 kPa, γ = 18 kN/m³, z = 3 m, β = 20°, φ′ = 30° and u = 0. Using the standard infinite-slope equation, FS is most nearly: See the worked solution or practice the Soil Mechanics deck
An infinite slope has c′ = 10 kPa, γ = 18 kN/m³, z = 3 m, β = 20° and φ′ = 30°. The pore pressure on the failure plane is 15 kPa. Using the standard infinite-slope equation, FS is most nearly: See the worked solution or practice the Soil Mechanics deck
With all else constant, increasing pore-water pressure in a soil slope generally causes its factor of safety to: See the worked solution or practice the Soil Mechanics deck
For a dry cohesionless infinite slope at limiting equilibrium (FS = 1), the slope angle β is: See the worked solution or practice the Soil Mechanics deck

Related Calculators and Examples

Frequently Asked Questions

What is the steepest stable slope in dry sand?

For a dry cohesionless infinite slope, the factor of safety is tan phi divided by tan beta, so the slope is at limiting equilibrium when the slope angle equals the friction angle. This is the angle of repose.

Why does seepage reduce slope stability?

Pore water pressure on the slip surface lowers the effective normal stress and therefore the frictional resistance, while the driving force from the soil weight stays the same. For a sand slope with seepage parallel to the surface, the factor of safety drops to roughly half of the dry value.

What factor of safety is acceptable for slopes?

Permanent slopes are commonly designed for a factor of safety of about 1.5 under long-term conditions, with lower values sometimes accepted for temporary or seismic cases. Always use the value stated in the problem.

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