Learn: Infinite-slope factor of safety
A shallow slip surface parallel to a long uniform slope can be studied with an infinite-slope model. The factor of safety compares available shear strength on that plane with the downslope driving shear stress.
What the formula is saying
Normal stress from soil weight is γz cos²β, reduced by pore pressure u to get effective normal stress. Add cohesion to effective normal stress times tan φ, then divide by γz sinβ cosβ.
Read the symbols in plain language
- c′
- Effective cohesion
Intercept of the effective-stress shear-strength relation; it is not an undrained strength substituted without changing the model.
kPaOne kilopascal equals one kN/m² and 1000 Pa.
- γ
- Soil unit weight
Weight force per unit volume, including gravity. Mass density in kg/m³ cannot be entered directly in a kN/m³ field.
kN/m³Use kN/m³ as the base unit shown here. c and u are kPa, γ is kN/m³ and z is vertical depth in m. β and φ are degrees, converted to radians for trigonometry. Strength and driving stress use the same unit, leaving a dimensionless factor.
- z
- Depth
Vertical depth to the potential sliding plane parallel to the ground slope, not the depth measured normal to the slope.
mMetres measure length; 1 m = 1000 mm.
- β
- Slope angle
Slope angle. Resolve the weight of the soil above the parallel slip plane into its normal component.
degAngles are entered in degrees; multiply by π/180 for trigonometric calculations in radians.
- u
- Pore pressure
Pore pressure. Frictional strength depends on effective normal stress rather than the total normal stress.
kPaOne kilopascal equals one kN/m² and 1000 Pa.
- φ′
- Effective friction angle
Angle defining the frictional part of drained effective-stress shear strength; enter degrees here, not its tangent.
degAngles are entered in degrees; multiply by π/180 for trigonometric calculations in radians.
- FS
- Result to find
Infinite-slope factor of safety. The factor of safety is the ratio of the two compatible shear-stress quantities.
ratio / no unit
Sort out the units first
c and u are kPa, γ is kN/m³ and z is vertical depth in m. β and φ are degrees, converted to radians for trigonometry. Strength and driving stress use the same unit, leaving a dimensionless factor.
Assumptions before calculating
Assume a long homogeneous slope with a planar slip surface parallel to the slope, effective-stress Mohr–Coulomb strength and the supplied pore pressure on the plane. The lesson requires positive driving stress and nonnegative effective normal stress.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find FS and explain the result in the stated output unit.
- c′ · Effective cohesion
- 5 kPa
- γ · Soil unit weight
- 18 kN/m³
- z · Depth
- 2 m
- β · Slope angle
- 25 deg
- u · Pore pressure
- 0 kPa
- φ′ · Effective friction angle
- 30 deg
Find total normal stress on the plane
Resolve the weight of the soil above the parallel slip plane into its normal component.
(18) × (2) × cos((25) × π ÷ 180)^2 ≈ 29.57017697 kPaSubtract pore-water pressure
Frictional strength depends on effective normal stress rather than the total normal stress.
(29.57017697)-(0) ≈ 29.57017697 kPaCalculate available shear strength
Combine effective-stress friction with the compatible cohesion intercept.
(5) + (29.57017697) × tan((30) × π ÷ 180) ≈ 22.07234964 kPaCalculate downslope driving shear stress
Resolve the same soil weight along the slip plane using the slope angle.
(18) × (2) × sin((25) × π ÷ 180) × cos((25) × π ÷ 180) ≈ 13.78879998 kPaCompare resistance with driving stress
The factor of safety is the ratio of the two compatible shear-stress quantities.
(22.07234964) ÷ (13.78879998) ≈ 1.600744784
Does this worked answer make sense?
For c = 0 and u = 0 the expression reduces to tan φ/tan β. Increasing pore pressure lowers resistance while leaving the soil-weight driving stress unchanged in this model.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- c′ · Effective cohesion
- 0 kPa
- γ · Soil unit weight
- 18 kN/m³
- z · Depth
- 2 m
- β · Slope angle
- 20 deg
- u · Pore pressure
- 0 kPa
- φ′ · Effective friction angle
- 30 deg
Find total normal stress on the plane
Resolve the weight of the soil above the parallel slip plane into its normal component.
(18) × (2) × cos((20) × π ÷ 180)^2 ≈ 31.78879998 kPaSubtract pore-water pressure
Frictional strength depends on effective normal stress rather than the total normal stress.
(31.78879998)-(0) ≈ 31.78879998 kPaCalculate available shear strength
Combine effective-stress friction with the compatible cohesion intercept.
(0) + (31.78879998) × tan((30) × π ÷ 180) ≈ 18.35327222 kPaCalculate downslope driving shear stress
Resolve the same soil weight along the slip plane using the slope angle.
(18) × (2) × sin((20) × π ÷ 180) × cos((20) × π ÷ 180) ≈ 11.57017697 kPaCompare resistance with driving stress
The factor of safety is the ratio of the two compatible shear-stress quantities.
(18.35327222) ÷ (11.57017697) ≈ 1.586256828
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- c′ · Effective cohesion
- 5 kPa
- γ · Soil unit weight
- 19 kN/m³
- z · Depth
- 2.5 m
- β · Slope angle
- 25 deg
- u · Pore pressure
- 10 kPa
- φ′ · Effective friction angle
- 32 deg
Find: Learn: Infinite-slope factor of safety
A hint, not the answer
Normal stress from soil weight is γz cos²β, reduced by pore pressure u to get effective normal stress. Add cohesion to effective normal stress times tan φ, then divide by γz sinβ cosβ.
c and u are kPa, γ is kN/m³ and z is vertical depth in m. β and φ are degrees, converted to radians for trigonometry. Strength and driving stress use the same unit, leaving a dimensionless factor.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find total normal stress on the plane
Resolve the weight of the soil above the parallel slip plane into its normal component.
(19) × (2.5) × cos((25) × π ÷ 180)^2 ≈ 39.01620573 kPaSubtract pore-water pressure
Frictional strength depends on effective normal stress rather than the total normal stress.
(39.01620573)-(10) ≈ 29.01620573 kPaCalculate available shear strength
Combine effective-stress friction with the compatible cohesion intercept.
(5) + (29.01620573) × tan((32) × π ÷ 180) ≈ 23.13133767 kPaCalculate downslope driving shear stress
Resolve the same soil weight along the slip plane using the slope angle.
(19) × (2.5) × sin((25) × π ÷ 180) × cos((25) × π ÷ 180) ≈ 18.19355552 kPaCompare resistance with driving stress
The factor of safety is the ratio of the two compatible shear-stress quantities.
(23.13133767) ÷ (18.19355552) ≈ 1.271402813
Avoid the common trap
Do not use depth normal to the slope when z is defined vertically. Subtract pore pressure before multiplying by tan φ, and do not confuse slope angle β with friction angle φ.
When this method applies — and when it does not
Finite or circular slips, layered profiles, tension cracks, rapid undrained loading, seismic forces, erosion and progressive failure need other models. A computed ratio is not a slope approval or a selected acceptable safety factor.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Infinite-slope factor of safety. Read the relevant soil phase, seepage, earth-pressure, settlement or foundation topic. Effective stress, drainage and idealized geometry determine whether the relationship applies.
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.