Slope Stability Notes

Slope Stability (3 Weeks)

• Focusing on slides.

• Analyzing stability using different methods and assumptions.

• Considering drained vs. undrained conditions.

• Introducing new strength considerations.

Slope Stability Overview - Objectives

• Factor of Safety (FoS):

  • Definition: A measure of the stability of a slope, indicating the ratio of resisting forces to driving forces.

• Critical Slip Surface:

  • Definition: The slip surface with the lowest factor of safety, representing the most likely failure path.

• Basic Assumptions in Slope Stability Analyses:

  • Plane strain conditions are often assumed for simplicity.

  • The Mohr-Coulomb failure criterion is commonly used to describe soil strength.

Common Causes of Instability

• Change in Slope Geometry:

  • Caused by human activities (excavation, filling) or natural erosion.

• Change in Total and/or Effective Stresses:

  • Variations in stress within the soil mass can lead to instability.

• Change in Effective Stress Due to Water Pressure:

  • Fluctuations in pore water pressure significantly affect effective stress and, consequently, shear strength.

• Weathering of Geomaterials:

  • Weathering reduces the shear strength of soil and rock, making slopes more susceptible to failure.

• Removal of Natural Cementing Materials:

  • Seepage water can dissolve and remove cementing agents, weakening the soil structure.

• Removal of Vegetation:

  • Vegetation helps stabilize slopes by providing root reinforcement and reducing erosion.

Factor of Safety (FoS)

• Definition: The ratio of resisting force to driving force.

• Formula: FoS = \frac{f}{P}, where:

  • f = Resisting force

  • P = Driving force

• Consider a Shearing Interface (Purely Frictional):

  • f = \mu W, where:

    • \mu = Coefficient of friction

    • W = Weight of the soil mass

• Units: Must be consistent throughout the calculation.

Factor of Safety (Continued)

• Forces on a Slope:

  • S = W \sin \beta, where:

    • S = Shearing force

    • \beta = Slope angle

  • N = W \cos \beta, where:

    • N = Normal force

  • f = \mu N, where:

    • f = Frictional resistance

• Factor of Safety Formula:

  • FoS = \frac{resisting \ force}{driving \ force} = \frac{f}{S}

Slip Surface and Critical Slip Surface

• Slip Surface: A potential failure surface within the soil mass.

• Critical Slip Surface: The slip surface with the lowest factor of safety.

• Excavation Example: Different slip surfaces have different FoS values.

  • Example: F = 0.7 (unacceptable), F = 0.9 (unacceptable if a high FoS is required).

Acceptable Factors of Safety

• Reasons Why FoS = 1.05 is Likely Unacceptable:

  • Inaccuracies in soil properties, geometry, and pore water pressures.

  • Slight increases in loading (e.g., parking a car).

  • Changes in pore water pressure or seepage conditions.

• Importance of Higher FoS:

  • More important slopes (e.g., those supporting critical infrastructure) should have higher FoS.

  • Greater uncertainty in soil properties or loading conditions should necessitate higher FoS.

• Short-Term vs. Long-Term FoS:

  • Lower FoS may be acceptable in the short term if conditions are expected to improve.

Plane Strain Assumption

• Simplicity: Significantly simplifies the analysis.

• Conservatism: Almost always conservative compared to 3D analysis.

• Unit Length: Analysis is done for a unit length in the out-of-plane direction.

Shear Strength

• Mohr-Coulomb Failure Criterion:

  • Describes the shear strength of soil as a function of effective stress and soil parameters.

  • \tauf = c' + \sigman ' \tan \phi', where:

    • \tau_f = Shear strength

    • c' = Effective cohesion

    • \sigma_n ' = Effective normal stress

    • \phi' = Effective angle of internal friction

• Effective Normal Stress:

  • \sigman ' = \sigman - u, where:

    • \sigma_n = Total normal stress

    • u = Pore water pressure

• Cohesionless Soil:

  • c' = 0, so \tauf = \sigman ' \tan \phi'

• Mobilized Angle of Friction:

  • \phi'{mob} = \tan^{-1} (\frac{\tau1}{\sigma '_{n1}})

  • If \phi'_{mob} < \phi', then no failure occurs.

Factor of Safety (Cohesive and Frictional Soils)

• Mohr-Coulomb Failure Criterion:

  • \tauf = c' + \sigman ' \tan \phi'

• Factor of Safety Formula:

  • FoS = \frac{\tau_f}{\tau} = \frac{c' + \sigma' \tan \phi'}{\tau}, where:

    • \tau = Induced shear stress along the potential slip plane.

Overview of Stability Analysis

• Steps:

  1. Postulate a potential failure mechanism (slip surface).

  2. Evaluate the ultimate capacity of this mechanism.

  3. Repeat steps 1 and 2 to find the most critical slip surface.

• Requirements:

  • Equilibrium.

  • Shear strength description (e.g., Mohr-Coulomb).

• Failure Mechanism (1 to n):

  1. Draw a free body diagram.

  2. Write equilibrium equations.

  3. Apply shear strength relations.

  4. Evaluate the factor of safety.

• Search: Find the worst-case scenario (lowest FoS) among the n cases.

Methods for Solving Equilibrium Conditions

• Single Free Body:

  • Simpler but less flexible and potentially less accurate.

• Procedures of Slices:

  • More complicated but more flexible and accurate.

  • Generally requires computer assistance.

Slope Stability Overview - Summary

• Factor of Safety:

  • FoS = \frac{resisting \ "force\"}{driving \ "force\"}

• Critical Slip Surface:

  • The slip surface with the lowest factor of safety.

• Basic Assumptions:

  • Plane-strain analyses.

  • Mohr-Coulomb failure criterion.

Planar Slip Surface (Part 1) – Objectives

• Derivation of FoS Equations:

  • Derive factor of safety equations for various simple cases using statics.

  • Infinite slope method.

  • Planar slip surface with vertical cut.

  • Planar slip surface with sloping ground.

• Appropriate Use:

  • Determine when these methods are appropriate to use.

Infinite Slope Method

• Applicability:

  • For any slope in cohesionless material.

  • Factor of safety does not depend on depth.

  • For cohesive materials under rare circumstances.

Infinite Slope Method (Forces)

• Forces on Sides of Block: Balance each other out and can be ignored.

• Formulas:

  • S = W \sin \alpha

  • N = W \cos \alpha

  • W = \gamma l z \cos \alpha

  • \tau_{mob} = \frac{S}{l} = \gamma z \cos \alpha \sin \alpha

  • \sigma = \frac{N}{l} = \gamma z \cos^2 \alpha

  • Where:

    • \alpha = slope angle

    • \gamma = unit weight

    • z = depth

    • l = length

Infinite Slope Method (FoS)

• Factor of Safety Formula:

  • FoS = \frac{\tauf}{\tau{mob}} = \frac{c + \sigman ' \tan \phi'}{\tau{mob}}

• Effective Stress:

  • \sigman ' = \sigman - u = \gamma z \cos^2 \alpha - u

• FoS Expression:

  • FoS = \frac{c'+(\gamma z \cos^2 \alpha - u) \tan \phi'}{\gamma z \cos \alpha \sin \alpha}

  • (For c’-\phi' material, effective stress analysis)

• Total Stress Analysis:

  • FoS = \frac{S_u}{\gamma z \cos \alpha \sin \alpha}

  • (For S_u material, total stress analysis)

Review Question (Infinite Slope)

• Given:

  • Dry sand: \phi' = 35^o, c' = 0, \gamma = 20 \ kN/m^3, z = 3 \ m, \alpha = 40^o

• Question: What is the factor of safety?

• Simplification: The equation simplifies due to c' = 0 and dry conditions (u = 0).

• Relation to Geo 1: Relates to basic soil mechanics principles.

Seepage in an Infinite Slope

• Pore Water Pressure:

  • u = \gammaw hp

• Total Head:

  • ht(A) = zA + h_p(A)

  • ht(B) = zB + h_p(B)

• Equipotential Line (EPL):

  • Points A and B lie on the same EPL, so ht(A) = ht(B) and z + hp = zA

• FoS Formula with Seepage:

  • FoS = \frac{c'+(\gamma z \cos^2 \alpha - u) \tan \phi'}{\gamma z \cos \alpha \sin \alpha}

Planar Slip Surface with Vertical Cut

• Conditions for Planar Slip:

  • Weak joint, vertical cut, dry conditions.

• Assumptions:

  • c’-\phi' material.

• Question: What is the maximum value of H (cut height) for stability?

Planar Slip Surface with Vertical Cut (Equilibrium)

• Assumptions: FoS = 1 (limit equilibrium).

• Forces:

  • S: Shear force acting along the slip surface (driving force).

  • N: Normal force acting perpendicular to the slip surface.

  • R: Shear resisting force available along the slip surface.

• Material Properties: Material strength, density, location of the weak plane, and strength parameters (c’-\phi'$').

• Free Body Diagram: Includes W, N, S, and L.

• Force Equilibrium:

  • W \sin \alpha - S = 0

  • W \cos \alpha - N = 0

• Mohr-Coulomb Failure Criterion:

  • S = R

  • R = c'L \times 1 + N \tan \phi' = cL + W \cos \alpha \tan \phi', with L = H/ \sin \alpha

  • W = \frac{\gamma H^2}{2 \tan \alpha}

• Critical Height:

  • H_{crit} = \frac{2c}{\gamma \cos^2 \alpha (\tan \alpha - \tan \phi)}

  • H{crit} = H{crit} (c, \phi, \gamma, \alpha)

Work Example (Weak Joint)

• Given:

  • Weak joint strength: c' = 10 \ kPa, \phi' = 30^o, H = 10 \ m, \gamma = 20 \ kN/m^3, \alpha = 35^o

  • Water table below the toe of the excavation.

• Question: What is the FoS associated with the weak joint?

• Solution:

  • FoS = \frac{R}{S} = \frac{c'L + N' \tan \phi'}{S} with N' = …, S = …, L = …, W = …

Planar Slip Surface with Sloping Ground

• Problem:

  • Potential failure surface, soil strength described by c', \phi'.

• Critical Condition:

  • FoS = 1

  • \frac{2c'}{\gamma H{crit}} = \frac{\sin (\alpha{crit} - \phi')}{\sin i \cos \phi'} where i is slope inclination angle, and \alpha is slip plane angle.

• Optimization:

  • Setting \frac{dH{crit}}{d\alpha{crit}} = 0 yields \alpha_{crit} = \frac{i + \phi'}{2}

• Critical Height:

  • H_{crit} = \frac{4c'}{\gamma} \times \frac{\sin i \cos \phi'}{1 - \cos(i - \phi')} = \frac{4c'}{\gamma} \times f(i, \phi')

• Calculations:

  • W = \frac{\gamma H^2}{2} (\cot \alpha - \cot i)

  • L = \frac{H}{\sin \alpha} \times 1 (unit length into the paper)

  • FoS = \frac{R}{S} = \frac{c'L + W \cos \alpha \tan \phi'}{W \sin \alpha} = \frac{2c'}{\gamma H \sin i} \frac{\sin (i - \alpha) + \tan \phi' \tan \alpha}{\sin \alpha}

Work Example (Sloping Ground)

• Given:

  • Homogeneous soil: c' = 5 \ kPa, \phi' = 25^o, \gamma' = \gamma_{moist} = 20 \ kN/m^3

  • Water table below the toe of the slope.

• Questions:

  1. Where is the critical slip surface (\alpha) such that FoS = 1 assuming a planar failure?

  2. Given this critical \alpha, what is the maximum height H?

• Solution:

  • \alpha = \frac{i + \phi'}{2} = …

  • W = \frac{\gamma H^2}{2} (\cot \alpha - \cot i) = …

  • L = \frac{H}{\sin \alpha} = …

Work Example (Answer)

• Method 1:
  Solve FoS = 1 => Hcrit

• Method 2: Use close-form solution for Hcrit

• FoS Expression:

  • FoS = \frac{R}{S} = \frac{c'L + W \cos \alpha \tan \phi'}{W \sin \alpha} = …

  • H_{crit} = \frac{4c'}{\gamma} \times \frac{\sin i \cos \phi'}{1 - \cos(i - \phi')} = …

Planar Slip Surface (Part 1) – Summary

• Derivation of FoS Equations:

  • Static equilibrium combined with the Mohr-Coulomb failure criterion.

  • Several Examples: Infinite slope method, Planar slip surface with vertical cut, Planar slip surface with sloping ground.

• Appropriate Use of Methods:

  • Infinite slope method: Cohesionless soils, Cohesive weak layer over a stronger layer.

  • Other planar joint methods: When there is a reason to believe there would be a planar failure (e.g., weak joint due to weathering).

Planar Slip Surface – Objectives (Part 2)

• Tension Cracks:

  • How to include the effect of tension cracks on FoS.

  • How to include the effect of pore pressures behind a tension crack.

• Work Example:

  • Solve a problem incorporating these effects.

Planar Slip Surface with Tension Crack

• Problem:

  • Inclined face, crack behind the crest, joint with/without seepage pressure.

• Question: FoS?

• Considerations:

  • Failure plane with/without seepage.

  • Crack filled with water, water pressure.

Tension Crack in S_u Material

*Consider a soil element at depth z below ground surface adjacent to a tension crack.

• Consider the failure Mohr circle in total stress:
  \sigma{1f} = \sigma{3f} + 2 S_u

• Herein, \sigma{1f}= 2 Su = \gamma z_{crack}

• Therefore,z{crack}= \frac{2Su}{\gamma}

Tension Crack in c’-\phi' Material

*Consider a soil element at depth z below ground surface adjacent to a tension crack.

• Consider the failure Mohr circle in effective stress:
  \sigma{1f} ' = \sigma{3f} ' tan^2(45 + \frac{\phi'}{2}) + 2c' tan(45 + \frac{\phi'}{2})

• Herein, \sigma{1f}' = 2c' tan(45 + \frac{\phi'}{2}) = \gamma' z{crack}

• Therefore,z_{crack}= \frac{2c'}{\gamma' tan(45 + \frac{\phi'}{2})}

Planar Slip Surface with Tension Crack (Free Body Diagram)

• Forces:

  • W: Weight of the soil wedge.

  • U_1: Action of water pressure along the crack.

  • U_2: Action of water pressure along the joint plane.

• Equilibrium (Parallel to the Joint Surface):

  • W \sin \alpha + U_1 \cos \alpha - S = 0

  • D = S = W \sin \alpha + U_1 \cos \alpha

• Equilibrium (Normal to the Joint Surface):

  • W \cos \alpha - U2 - U1 \sin \alpha - N' = 0

• Mohr-Coulomb Failure Criterion:

  • R = c'L + (W \cos \alpha - U2 - U1 \sin \alpha) \tan \phi'

• Factor of Safety:

  • FoS = \frac{R}{D} = \frac{c'L + (W \cos \alpha - U2 - U1 \sin \alpha) \tan \phi'}{W \sin \alpha + U_1 \cos \alpha}

• Hydrostatic Pressure:

  • u{max} = \gammaw z_w

Work Example (Tension Crack)

• Problem:

  • Calculate the FoS associated with the failing block depicted below.

  • Crack entirely filled with water.

  • Soil wedge entirely saturated, with \gamma_{sat} = 20 \ kN/m^3

• Given:

  • W = 900 \ kN, c' = 10 \ kPa, \phi' = 28^o

• Calculations:

  • z_{crack} = \frac{2c'}{\gamma' \tan(45 + \frac{\phi'}{2})} = …

  • U1 = (0 + \gammaw zw) \frac{zw}{2} = …

  • U2 = (0 + \gammaw zw) \frac{L{joint}}{2} = …

Work Example (Free Body Diagram)

• Calculate S
  S = W sin\alpha + U_1 = …

• Calculate N
  N = W cos\alpha = …

• Normal Effective Force
  N' = N - U2 - U1 sin \alpha = …

• R = c'L{joint} + (W cos \alpha - U2 - U_1 sin \alpha) tan \phi'

• Therefore FoS:
  FoS = \frac{R}{S} = …

Planar Slip Surface – Summary (Part 2)

• Critical Depth of Tension Cracks:

  • z{crack} = \frac{2Su}{\gamma} for undrained material.

  • z_{crack} = \frac{2c'}{\gamma' \tan(45 + \frac{\phi'}{2})} for drained material.

• Hydrostatic Forces:

  • U1 = 0.5 \gammaw z_w^2

  • U2 = \gammaw zw \frac{L{joint}}{2}

• Factor of Safety:
  FoS = \frac{R}{S} = \frac{c'L{joint}+ (W cos \alpha-U2-U1 sin \alpha) tan \phi'}{W sin \alpha + U1 cos \alpha}

Planar Slip Surface – Objectives (Part 3)

• Pore Pressures:

  • How to include the effect of pore pressures when the phreatic surface is above the slip surface.

• Horizontal Acceleration:

  • How to include the effect of horizontal acceleration.

• Work Examples:

  • Solve problems incorporating these effects.

Effect of Water Pressure Along the Slip Surface

• Pore Water Pressure:

  • Accurate: u= \gammaw hp

  • Approximate: u \approx \gammaw hw

• Forces:

  • S = W \sin \alpha

  • R = c'L + (W \cos \alpha - U) \tan \phi'

  • FoS = \frac{R}{S}

Effect of Water Pressure Along the Slip Surface (Ru)

• Pender (2016): We use hw to estimate hp.

• Formulas:

  • dU = \gammaw hw \sec \alpha dx (where u \approx hw \gammaw)
    U = \gamma_w \sec \alpha \times Area \ between \ phreatic \ surface \
    and \ failure \ surface

  • ru = \frac{Area \ between \ failure \ and \ phreatic \ surface}{Area \ wedge \ above \ failure \ plane} \frac{\gammaw}{\gamma}

• Factor of Safety:
  FoS = \frac{2c' sin i}{\gamma H (sin i - \alpha) sin \alpha + (1 - r_u \sec^2 \alpha) tan \phi' tan \alpha}

• Other ways to define r_u:

  • ru = \frac{\sum u h}{\gamma} = \frac{\sum hp h \gammaw}{\gamma} \approx \frac{\sum hw h \gamma_w}{\gamma}

Work Example (with Phreatic Surface)

• Given:

  • H = 7 \ m, c' = 0 \ kPa, \phi' = 36^o, \gamma = 20 \ kN/m^3, \alpha = 12^o, i = 23^o, \theta = 16^o

• Question: Calculate the FoS if the phreatic surface is above the slip surface.

• Formulas:

  • Area(ABD) = \frac{\gamma H^2}{2} (\frac{1}{\tan \alpha} - \frac{1}{\tan i})

  • Area(ACD) = \frac{\gamma H^2}{2} (\frac{1}{\tan \alpha} - \frac{1}{\tan \theta})

Work Example (Continuation)

FoS = \frac{2c′ sin i}{γH sin(i − α) sin α + (1 − ru sec^2 α) tanϕ′ tan α}

• This expression can be simplified as follows:

Factor of Safety:

• ru = \frac{Area \ between \ failure \ and \ phreatic \ surface}{Area \ wedge \ above \ failure \ plane} \frac{\gammaw}{\gamma} = \frac{Area (…)}{Area (…)} \frac{\gamma_w}{\gamma}

Effect of Horizontal Acceleration on Stability

• Pseudo-Static Analysis:

  • Introduce a horizontal force kh W, where kh is the horizontal seismic coefficient.

• Factor of Safety:

  • FoS = \frac{c'L + (W \cos \alpha - kh \sin \alpha) \tan \phi'}{W \sin \alpha + kh \cos \alpha}

• Limiting Equilibrium:

  • c'L + (W \cos \alpha - kh \sin \alpha) \tan \phi' = W \sin \alpha + kh \cos \alpha

*By solving, we determine that

k_h = \tan(\alpha - \phi'), when static FoS is infinity.

Or

kh = \frac{\tan \alpha (FoS{static} - 1)}{1 + \tan \alpha \tan \phi'}

Factor of Safety at Static Condition

Denote static FoS by FoS{static}, FoS{static} = FoS|k_h = 0

Therefore,
FoS_{static} = \frac{c'L + W cos \alpha tan \phi'}{W sin \alpha}

Work Example(Seismic Load)

Question: By drawing the free body diagram of the below failing wedge, derive its factor of safety when subject to a horizontal inertia force K_hW, where W is the weight of the wedge. Assuming the shear resistance along the failure plane can be well described by the Mohr-Coulomb parameters c and ϕ' and W.T. is below the slip surface."

Solution:
FoS_{static} = \frac{c'L + N'tan \phi'}{S}
S=W cos\alpha +
N'=N= W sin\alpha −

Refer to part (a) and assume that the material is saturated clay with an undrained shear strength of 30 kPa and a saturated unit weight of 20 kN/m3. At \alpha= 45^o, what is the factor of safety given the depth of the vertical cut is 4 m without any seismic loading?

Solution
\alpha= 45^o

*In drained conditions,
R = c'L + N' tan \phi '
*By analogy in undrained conditions,
R=Su L *S = W sin\alpha = *FoS{static} = \frac{R}{S} =

Refer to part (b). What is the value of kh which makes the cut fail? By solving FoS=1, we can demonstrate that:

FoS = \frac{c'L+Wcos\alpha−kh sin\alpha tan \phi' } {Wsin\alpha+kh cos\alpha} = \frac{Su L}{Wsin\alpha+kh cos\alpha}

*kh=\frac{Su L}{W (sin \alpha − 1)} / {tan \alpha}
*kh = \frac{FoS{static} − 1}{tan \alpha}

Planar Slip Surface – Summary (Part 3)

• Pore Pressures:

  • How to include the effect of pore pressures when the phreatic surface is above the slip surface?

  • Where:
    ru = \frac{Area \ between \ failure \ and \ phreatic \ surface}{Area \ wedge \ above \ failure \ plane} \frac{\gamma_w}{\gamma}
    FoS = \frac{c'L+ (W cos \alpha -U) tan \phi'}{W sin \alpha}, with U = ruW sec \alpha
    FoS = \frac{2c′ sin i}{\gamma H sin(i − α) sin α + (1 − ru sec^2 α) tanϕ′ tan α}

  • How to include the effect of horizontal acceleration?

    • Pseudo-static analysis using kh

Factor of Safety:
FoS=\frac{c'L + (W cos \alpha - kh sin \alpha) tan \phi'}{W sin \alpha + kh cos \alpha}

Circular Slip Surface - Objectives

• How to derive factor of safety equation for a circular slip surface?

• When is this method appropriate to use?

• How to include the effect of tension crack

• How to account for “external” forces?

Circular Slip Surface in Su Material (Swedish Circle Method ϕ = 0)

*Based on static moment calculation

Take moment about O:
Driving moment = Wx
Resisting moment = SuRθ \cdot R = SuR^2θ_{rad}
Factor of Safety(FS)

FoS = Resisting \frac{moment}{Driving momemt}

FoS = \frac{SuR^2θ{rad}}{Wx}

Work Example Circular Slip Surface

Assume
\gamma =20 kN/m^3
S_u = 10 kPa
Failing wedge area A = 10.7 m^2
Question; What is the Factor of Safety(FS)?
Given x = 4.8 m

Calculations*:

• W: weight * area

• Area
  Therefore
  FoS = \frac{SuR^2θ{rad}}{Wx}

Circular slip surface with tension crack

If crack is filled with water:
The effect of water:
FoS increase or decrease ?
If crack is dry:

Calculations::
Hydrostatic Forces: Q
Q=\frac{1}{2}*\gammaw *z{crack} · z{crack} = \gammaw*z_{crack}

FoS: Factor of Safety

FoS= \frac{SuR^2 θ{rad}}{Wx + \frac{Qw}{2} * [\frac{2}{3}z{crack}+z_t]}

Note the change in (FS) compared to no tension crack at all.

Work Example-Circular slip surface with tension crack

What if there`s tension cracks
O W=214 kN R X = 4.8 m
Given: zcrack zt = 0.30 m
Consider the circular soil wedge below with
γ_sat = 20 kN/m^3 and Su = 10 kPa.

Required: Factor of Safety(FS)???

𝜃=°(given)
Calculations:*
θ{rad} =?????? Qw= ½ γw z{crack} z{crack} γw z_{crack}$$
Therefore Final Result