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.

  • $\tau<em>f = c' + \sigma</em>n ' \tan \phi'$, where:

    • $\tau_f$ = Shear strength

    • $c'$ = Effective cohesion

    • $\sigma_n '$ = Effective normal stress

    • $\phi'$ = Effective angle of internal friction

• Effective Normal Stress:

  • $\sigma<em>n ' = \sigma</em>n - u$, where:

    • $\sigma_n$ = Total normal stress

    • $u$ = Pore water pressure

• Cohesionless Soil:

  • $c' = 0$, so $\tau<em>f = \sigma</em>n ' \tan \phi'$

• Mobilized Angle of Friction:

  • $\phi'<em>{mob} = \tan^{-1} (\frac{\tau</em>1}{\sigma '_{n1}})$

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

Factor of Safety (Cohesive and Frictional Soils)

• Mohr-Coulomb Failure Criterion:

  • $\tau<em>f = c' + \sigma</em>n ' \tan \phi'$

• 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:

  • $FoS = \frac{\tau<em>f}{\tau</em>{mob}} = \frac{c + \sigma<em>n ' \tan \phi'}{\tau</em>{mob}}$

• Effective Stress:

  • $\sigma<em>n ' = \sigma</em>n - u = \gamma z \cos^2 \alpha - u$

• 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:

  • $u = \gamma<em>w h</em>p$

• Total Head:

  • $h<em>t(A) = z</em>A + h_p(A)$

  • $h<em>t(B) = z</em>B + h_p(B)$

• Equipotential Line (EPL):

  • Points A and B lie on the same EPL, so $h<em>t(A) = h</em>t(B)$ and $z + h<em>p = z</em>A$

• 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$</p></li><li><p>$W \cos \alpha - N = 0$</p></li></ul></li><li><p><strong>Mohr-Coulomb Failure Criterion:</strong></p><ul><li><p>$S = R$</p></li><li><p>$R = c'L \times 1 + N \tan \phi' = cL + W \cos \alpha \tan \phi'$, with$L = H/ \sin \alpha$</p></li><li><p>$W = \frac{\gamma H^2}{2 \tan \alpha}$</p></li></ul></li><li><p><strong>Critical Height:</strong></p><ul><li><p>$H_{crit} = \frac{2c}{\gamma \cos^2 \alpha (\tan \alpha - \tan \phi)}$</p></li><li><p>$H{crit} = H{crit} (c, \phi, \gamma, \alpha)$</p></li></ul></li></ul><h3 id="978b6c5a-7bc1-4a4e-89ea-af704f2a5012" data-toc-id="978b6c5a-7bc1-4a4e-89ea-af704f2a5012" collapsed="false" seolevelmigrated="true">Work Example (Weak Joint)</h3><ul><li><p><strong>Given:</strong></p><ul><li><p>Weak joint strength:$c' = 10 \ kPa$,$\phi' = 30^o$,$H = 10 \ m$,$\gamma = 20 \ kN/m^3$,$\alpha = 35^o$</p></li><li><p>Water table below the toe of the excavation.</p></li></ul></li><li><p><strong>Question:</strong> What is the FoS associated with the weak joint?</p></li><li><p><strong>Solution:</strong></p><ul><li><p>$FoS = \frac{R}{S} = \frac{c'L + N' \tan \phi'}{S}$with$N' = …$,$S = …$,$L = …$,$W = …$</p></li></ul></li></ul><h3 id="9fbe0400-6e85-465d-83d8-bef6b5d248ab" data-toc-id="9fbe0400-6e85-465d-83d8-bef6b5d248ab" collapsed="false" seolevelmigrated="true">Planar Slip Surface with Sloping Ground</h3><ul><li><p><strong>Problem:</strong></p><ul><li><p>Potential failure surface, soil strength described by$c'$,$\phi'$.</p></li></ul></li><li><p><strong>Critical Condition:</strong></p><ul><li><p>$FoS = 1$</p></li><li><p>$\frac{2c'}{\gamma H{crit}} = \frac{\sin (\alpha{crit} - \phi')}{\sin i \cos \phi'}$where$i$is slope inclination angle, and$\alpha$is slip plane angle.</p></li></ul></li><li><p><strong>Optimization:</strong></p><ul><li><p>Setting$\frac{dH{crit}}{d\alpha{crit}} = 0$yields$\alpha_{crit} = \frac{i + \phi'}{2}$</p></li></ul></li><li><p><strong>Critical Height:</strong></p><ul><li><p>$H_{crit} = \frac{4c'}{\gamma} \times \frac{\sin i \cos \phi'}{1 - \cos(i - \phi')} = \frac{4c'}{\gamma} \times f(i, \phi')$</p></li></ul></li><li><p><strong>Calculations:</strong></p><ul><li><p>$W = \frac{\gamma H^2}{2} (\cot \alpha - \cot i)$</p></li><li><p>$L = \frac{H}{\sin \alpha} \times 1$(unit length into the paper)</p></li><li><p>$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}$</p></li></ul></li></ul><h3 id="9042fbfd-35e7-4ca1-8962-660241ca688d" data-toc-id="9042fbfd-35e7-4ca1-8962-660241ca688d" collapsed="false" seolevelmigrated="true">Work Example (Sloping Ground)</h3><ul><li><p><strong>Given:</strong></p><ul><li><p>Homogeneous soil:$c' = 5 \ kPa$,$\phi' = 25^o$,$\gamma' = \gamma_{moist} = 20 \ kN/m^3$</p></li><li><p>Water table below the toe of the slope.</p></li></ul></li><li><p><strong>Questions:</strong></p><ol><li><p>Where is the critical slip surface$(\alpha)$such that FoS = 1 assuming a planar failure?</p></li><li><p>Given this critical$\alpha$, what is the maximum height H?</p></li></ol></li><li><p><strong>Solution:</strong></p><ul><li><p>$\alpha = \frac{i + \phi'}{2} = …$</p></li><li><p>$W = \frac{\gamma H^2}{2} (\cot \alpha - \cot i) = …$</p></li><li><p>$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} = …$</p></li><li><p>$H_{crit} = \frac{4c'}{\gamma} \times \frac{\sin i \cos \phi'}{1 - \cos(i - \phi')} = …$</p></li></ul></li></ul><h3 id="27148b86-4ab9-41d9-b8a2-d35ba836fabd" data-toc-id="27148b86-4ab9-41d9-b8a2-d35ba836fabd" collapsed="false" seolevelmigrated="true">Planar Slip Surface (Part 1) – Summary</h3><ul><li><p><strong>Derivation of FoS Equations:</strong></p><ul><li><p>Static equilibrium combined with the Mohr-Coulomb failure criterion.</p></li><li><p>Several Examples: Infinite slope method, Planar slip surface with vertical cut, Planar slip surface with sloping ground.</p></li></ul></li><li><p><strong>Appropriate Use of Methods:</strong></p><ul><li><p>Infinite slope method: Cohesionless soils, Cohesive weak layer over a stronger layer.</p></li><li><p>Other planar joint methods: When there is a reason to believe there would be a planar failure (e.g., weak joint due to weathering).</p></li></ul></li></ul><h3 id="759aa8e3-64e9-416f-baf4-c1a24ca87ac6" data-toc-id="759aa8e3-64e9-416f-baf4-c1a24ca87ac6" collapsed="false" seolevelmigrated="true">Planar Slip Surface – Objectives (Part 2)</h3><ul><li><p><strong>Tension Cracks:</strong></p><ul><li><p>How to include the effect of tension cracks on FoS.</p></li><li><p>How to include the effect of pore pressures behind a tension crack.</p></li></ul></li><li><p><strong>Work Example:</strong></p><ul><li><p>Solve a problem incorporating these effects.</p></li></ul></li></ul><h3 id="798c8340-b4ba-4736-9e8d-75a88c292d80" data-toc-id="798c8340-b4ba-4736-9e8d-75a88c292d80" collapsed="false" seolevelmigrated="true">Planar Slip Surface with Tension Crack</h3><ul><li><p><strong>Problem:</strong></p><ul><li><p>Inclined face, crack behind the crest, joint with/without seepage pressure.</p></li></ul></li><li><p><strong>Question:</strong> FoS?</p></li><li><p><strong>Considerations:</strong></p><ul><li><p>Failure plane with/without seepage.</p></li><li><p>Crack filled with water, water pressure.</p></li></ul></li></ul><h3 id="75439a04-5a79-4ee5-b153-8a0aa03bf5e2" data-toc-id="75439a04-5a79-4ee5-b153-8a0aa03bf5e2" collapsed="false" seolevelmigrated="true">Tension Crack in$S_u$Material</h3><p>*Consider a soil element at depth z below ground surface adjacent to a tension crack.</p><ul><li><p>Consider the failure Mohr circle in total stress:<br>$\sigma{1f} = \sigma{3f} + 2 S_u$</p></li><li><p>Herein,$\sigma{1f}= 2 Su = \gamma z_{crack}$</p></li><li><p>Therefore,$z{crack}= \frac{2Su}{\gamma}$</p></li></ul><h3 id="2788a0e3-2e09-4200-ba6d-72bac5ddc5a4" data-toc-id="2788a0e3-2e09-4200-ba6d-72bac5ddc5a4" collapsed="false" seolevelmigrated="true">Tension Crack in c’-$\phi'$Material</h3><p>*Consider a soil element at depth z below ground surface adjacent to a tension crack.</p><ul><li><p>Consider the failure Mohr circle in effective stress:<br>$\sigma{1f} ' = \sigma{3f} ' tan^2(45 + \frac{\phi'}{2}) + 2c' tan(45 + \frac{\phi'}{2})$</p></li><li><p>Herein,$\sigma{1f}' = 2c' tan(45 + \frac{\phi'}{2}) = \gamma' z{crack}$</p></li><li><p>Therefore,$z_{crack}= \frac{2c'}{\gamma' tan(45 + \frac{\phi'}{2})}$</p></li></ul><h3 id="fc011e23-d0d9-4e14-893a-f622365f9af4" data-toc-id="fc011e23-d0d9-4e14-893a-f622365f9af4" collapsed="false" seolevelmigrated="true">Planar Slip Surface with Tension Crack (Free Body Diagram)</h3><ul><li><p><strong>Forces:</strong></p><ul><li><p>W: Weight of the soil wedge.</p></li><li><p>$U_1$: Action of water pressure along the crack.</p></li><li><p>$U_2$: Action of water pressure along the joint plane.</p></li></ul></li><li><p><strong>Equilibrium (Parallel to the Joint Surface):</strong></p><ul><li><p>$W \sin \alpha + U_1 \cos \alpha - S = 0$</p></li><li><p>$D = S = W \sin \alpha + U_1 \cos \alpha$</p></li></ul></li><li><p><strong>Equilibrium (Normal to the Joint Surface):</strong></p><ul><li><p>$W \cos \alpha - U2 - U1 \sin \alpha - N' = 0$</p></li></ul></li><li><p><strong>Mohr-Coulomb Failure Criterion:</strong></p><ul><li><p>$R = c'L + (W \cos \alpha - U2 - U1 \sin \alpha) \tan \phi'$</p></li></ul></li><li><p><strong>Factor of Safety:</strong></p><ul><li><p>$FoS = \frac{R}{D} = \frac{c'L + (W \cos \alpha - U2 - U1 \sin \alpha) \tan \phi'}{W \sin \alpha + U_1 \cos \alpha}$</p></li></ul></li><li><p><strong>Hydrostatic Pressure:</strong></p><ul><li><p>$u{max} = \gammaw z_w$</p></li></ul></li></ul><h3 id="d9e178af-cc67-40a4-a75d-63db2aec22ef" data-toc-id="d9e178af-cc67-40a4-a75d-63db2aec22ef" collapsed="false" seolevelmigrated="true">Work Example (Tension Crack)</h3><ul><li><p><strong>Problem:</strong></p><ul><li><p>Calculate the FoS associated with the failing block depicted below.</p></li><li><p>Crack entirely filled with water.</p></li><li><p>Soil wedge entirely saturated, with$\gamma_{sat} = 20 \ kN/m^3$</p></li></ul></li><li><p><strong>Given:</strong></p><ul><li><p>$W = 900 \ kN$,$c' = 10 \ kPa$,$\phi' = 28^o$</p></li></ul></li><li><p><strong>Calculations:</strong></p><ul><li><p>$z_{crack} = \frac{2c'}{\gamma' \tan(45 + \frac{\phi'}{2})} = …$</p></li><li><p>$U1 = (0 + \gammaw zw) \frac{zw}{2} = …$</p></li><li><p>$U2 = (0 + \gammaw zw) \frac{L{joint}}{2} = …$</p></li></ul></li></ul><h3 id="7e9f156f-9e17-44be-827a-710c4225bc5f" data-toc-id="7e9f156f-9e17-44be-827a-710c4225bc5f" collapsed="false" seolevelmigrated="true">Work Example (Free Body Diagram)</h3><ul><li><p><strong>Calculate S</strong><br>$S = W sin\alpha + U_1 = …$</p></li><li><p><strong>Calculate N</strong><br>$N = W cos\alpha = …$</p></li><li><p><strong>Normal Effective Force</strong><br>$N' = N - U2 - U1 sin \alpha = …$</p></li><li><p>$R = c'L{joint} + (W cos \alpha - U2 - U_1 sin \alpha) tan \phi'$</p></li><li><p><strong>Therefore FoS</strong>:<br>$FoS = \frac{R}{S} = …$</p></li></ul><h3 id="82a3949f-d40f-4950-a1bd-96e3a0892f08" data-toc-id="82a3949f-d40f-4950-a1bd-96e3a0892f08" collapsed="false" seolevelmigrated="true">Planar Slip Surface – Summary (Part 2)</h3><ul><li><p><strong>Critical Depth of Tension Cracks:</strong></p><ul><li><p>$z{crack} = \frac{2Su}{\gamma}$for undrained material.</p></li><li><p>$z_{crack} = \frac{2c'}{\gamma' \tan(45 + \frac{\phi'}{2})}$for drained material.</p></li></ul></li><li><p><strong>Hydrostatic Forces:</strong></p><ul><li><p>$U1 = 0.5 \gammaw z_w^2$</p></li><li><p>$U2 = \gammaw zw \frac{L{joint}}{2}$</p></li></ul></li><li><p><strong>Factor of Safety:</strong><br>$FoS = \frac{R}{S} = \frac{c'L{joint}+ (W cos \alpha-U2-U1 sin \alpha) tan \phi'}{W sin \alpha + U1 cos \alpha}$</p></li></ul><h3 id="a72f4d76-d471-4b6a-b18a-31c4f5e21c53" data-toc-id="a72f4d76-d471-4b6a-b18a-31c4f5e21c53" collapsed="false" seolevelmigrated="true">Planar Slip Surface – Objectives (Part 3)</h3><ul><li><p><strong>Pore Pressures:</strong></p><ul><li><p>How to include the effect of pore pressures when the phreatic surface is above the slip surface.</p></li></ul></li><li><p><strong>Horizontal Acceleration:</strong></p><ul><li><p>How to include the effect of horizontal acceleration.</p></li></ul></li><li><p><strong>Work Examples:</strong></p><ul><li><p>Solve problems incorporating these effects.</p></li></ul></li></ul><h3 id="a2a1f72f-eb88-4b2c-88c5-be669ce30cc5" data-toc-id="a2a1f72f-eb88-4b2c-88c5-be669ce30cc5" collapsed="false" seolevelmigrated="true">Effect of Water Pressure Along the Slip Surface</h3><ul><li><p><strong>Pore Water Pressure:</strong></p><ul><li><p>Accurate:$u= \gammaw hp$</p></li><li><p>Approximate:$u \approx \gammaw hw$</p></li></ul></li><li><p><strong>Forces:</strong></p><ul><li><p>$S = W \sin \alpha$</p></li><li><p>$R = c'L + (W \cos \alpha - U) \tan \phi'$</p></li><li><p>$FoS = \frac{R}{S}$</p></li></ul></li></ul><h3 id="a834c0d2-c623-4e48-ad45-6db5f24791a1" data-toc-id="a834c0d2-c623-4e48-ad45-6db5f24791a1" collapsed="false" seolevelmigrated="true">Effect of Water Pressure Along the Slip Surface (Ru)</h3><ul><li><p><strong>Pender (2016):</strong> We use$hw$to estimate$hp$.</p></li><li><p><strong>Formulas:</strong></p><ul><li><p>$dU = \gammaw hw \sec \alpha dx$(where$u \approx hw \gammaw$)<br>$U = \gamma_w \sec \alpha \times Area \ between \ phreatic \ surface \
    and \ failure \ surface$</p></li><li><p>$ru = \frac{Area \ between \ failure \ and \ phreatic \ surface}{Area \ wedge \ above \ failure \ plane} \frac{\gammaw}{\gamma}$</p></li></ul></li><li><p><strong>Factor of Safety:</strong><br>$FoS = \frac{2c' sin i}{\gamma H (sin i - \alpha) sin \alpha + (1 - r_u \sec^2 \alpha) tan \phi' tan \alpha}$</p></li><li><p><strong>Other ways to define$r_u$:</strong></p><ul><li><p>$ru = \frac{\sum u h}{\gamma} = \frac{\sum hp h \gammaw}{\gamma} \approx \frac{\sum hw h \gamma_w}{\gamma}$</p></li></ul></li></ul><h3 id="05a4307b-fa31-48f1-8f94-af32458f67c0" data-toc-id="05a4307b-fa31-48f1-8f94-af32458f67c0" collapsed="false" seolevelmigrated="true">Work Example (with Phreatic Surface)</h3><ul><li><p><strong>Given:</strong></p><ul><li><p>$H = 7 \ m$,$c' = 0 \ kPa$,$\phi' = 36^o$,$\gamma = 20 \ kN/m^3$,$\alpha = 12^o$,$i = 23^o$,$\theta = 16^o$</p></li></ul></li><li><p><strong>Question:</strong> Calculate the FoS if the phreatic surface is above the slip surface.</p></li><li><p><strong>Formulas:</strong></p><ul><li><p>$Area(ABD) = \frac{\gamma H^2}{2} (\frac{1}{\tan \alpha} - \frac{1}{\tan i})$</p></li><li><p>$Area(ACD) = \frac{\gamma H^2}{2} (\frac{1}{\tan \alpha} - \frac{1}{\tan \theta})$</p></li></ul></li></ul><h3 id="e4415723-a1c0-4be5-ad37-21bb8457cf80" data-toc-id="e4415723-a1c0-4be5-ad37-21bb8457cf80" collapsed="false" seolevelmigrated="true">Work Example (Continuation)</h3><p>$FoS = \frac{2c′ sin i}{γH sin(i − α) sin α + (1 − ru sec^2 α) tanϕ′ tan α}$</p><ul><li><p>This expression can be simplified as follows:</p></li></ul><p><strong>Factor of Safety:</strong></p><ul><li><p>$ru = \frac{Area \ between \ failure \ and \ phreatic \ surface}{Area \ wedge \ above \ failure \ plane} \frac{\gammaw}{\gamma} = \frac{Area (…)}{Area (…)} \frac{\gamma_w}{\gamma}$</p></li></ul><h3 id="1e74d165-87d5-43dc-85f1-f37c23a6e275" data-toc-id="1e74d165-87d5-43dc-85f1-f37c23a6e275" collapsed="false" seolevelmigrated="true">Effect of Horizontal Acceleration on Stability</h3><ul><li><p><strong>Pseudo-Static Analysis:</strong></p><ul><li><p>Introduce a horizontal force$kh W$, where$kh$is the horizontal seismic coefficient.</p></li></ul></li><li><p><strong>Factor of Safety:</strong></p><ul><li><p>$FoS = \frac{c'L + (W \cos \alpha - kh \sin \alpha) \tan \phi'}{W \sin \alpha + kh \cos \alpha}$</p></li></ul></li><li><p><strong>Limiting Equilibrium:</strong></p><ul><li><p>$c'L + (W \cos \alpha - kh \sin \alpha) \tan \phi' = W \sin \alpha + kh \cos \alpha$</p></li></ul></li></ul><p>*By solving, we determine that</p><p>$k_h = \tan(\alpha - \phi')$, when static FoS is infinity.</p><p>Or</p><p>$kh = \frac{\tan \alpha (FoS{static} - 1)}{1 + \tan \alpha \tan \phi'}$</p><h5 id="84272d31-0802-4eb0-9a0d-2c3e0557723a" data-toc-id="84272d31-0802-4eb0-9a0d-2c3e0557723a" collapsed="false" seolevelmigrated="true"><em>Factor of Safety at Static Condition</em></h5><p>Denote static$FoS by FoS{static}, FoS{static} = FoS|k_h = 0$</p><p>Therefore,<br>$FoS_{static} = \frac{c'L + W cos \alpha tan \phi'}{W sin \alpha}$</p><h3 id="0ed7f8ea-ce04-4b3c-94e5-0d1bf816dc32" data-toc-id="0ed7f8ea-ce04-4b3c-94e5-0d1bf816dc32" collapsed="false" seolevelmigrated="true">Work Example(Seismic Load)</h3><p><strong>Question</strong>: <em>By drawing the free body diagram of the below failing wedge, derive its factor of safety when subject to a horizontal inertia force$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'$and W.T. is below the slip surface.</em>"</p><p><strong>Solution</strong>: <br>$FoS_{static} = \frac{c'L + N'tan \phi'}{S}$<br>$S=W cos\alpha +$<br>$N'=N= W sin\alpha −$</p><p>Refer to part (a) and assume that the material is saturated clay with an undrained shear strength of 30 kPa and a saturated unit weight of 20 kN/m3. At$\alpha= 45^o$, what is the factor of safety given the depth of the vertical cut is 4 m without any seismic loading?</p><p><strong>Solution</strong><br>$\alpha= 45^o$</p><p>*In drained conditions,<br>$R = c'L + N' tan \phi '$<br>*By analogy in undrained conditions,<br>$R=Su L$*$S = W sin\alpha =$*$FoS{static} = \frac{R}{S} =$</p><div data-type="horizontalRule"><hr></div><p><strong>Refer to part (b). What is the value of kh which makes the cut fail? By solving FoS=1, we can demonstrate that:</strong></p><p>$FoS = \frac{c'L+Wcos\alpha−kh sin\alpha tan \phi' } {Wsin\alpha+kh cos\alpha} = \frac{Su L}{Wsin\alpha+kh cos\alpha}$</p><p>*$kh=\frac{Su L}{W (sin \alpha − 1)} / {tan \alpha}$<br>*$kh = \frac{FoS{static} − 1}{tan \alpha}$</p><h3 id="5bd7133a-9b81-4b71-bd78-505c7ba80530" data-toc-id="5bd7133a-9b81-4b71-bd78-505c7ba80530" collapsed="false" seolevelmigrated="true">Planar Slip Surface – Summary (Part 3)</h3><ul><li><p><strong>Pore Pressures:</strong></p><ul><li><p>How to include the effect of pore pressures when the phreatic surface is above the slip surface?</p></li><li><p>Where:<br>$ru = \frac{Area \ between \ failure \ and \ phreatic \ surface}{Area \ wedge \ above \ failure \ plane} \frac{\gamma_w}{\gamma}$<br>$FoS = \frac{c'L+ (W cos \alpha -U) tan \phi'}{W sin \alpha}$, with$U = ruW sec \alpha$<br>$FoS = \frac{2c′ sin i}{\gamma H sin(i − α) sin α + (1 − ru sec^2 α) tanϕ′ tan α}$</p></li><li><p>How to include the effect of horizontal acceleration?</p><ul><li><p>Pseudo-static analysis using kh</p></li></ul></li></ul></li></ul><p><strong>Factor of Safety:</strong><br>$FoS=\frac{c'L + (W cos \alpha - kh sin \alpha) tan \phi'}{W sin \alpha + kh cos \alpha}$</p><h3 id="8f4fddec-a67f-49ab-9eef-db2666f0b332" data-toc-id="8f4fddec-a67f-49ab-9eef-db2666f0b332" collapsed="false" seolevelmigrated="true">Circular Slip Surface - Objectives</h3><ul><li><p>How to derive factor of safety equation for a circular slip surface?</p></li><li><p>When is this method appropriate to use?</p></li><li><p>How to include the effect of tension crack</p></li><li><p>How to account for “external” forces?</p></li></ul><h3 id="04d22cf6-1c38-4451-94f7-82449e38ea2a" data-toc-id="04d22cf6-1c38-4451-94f7-82449e38ea2a" collapsed="false" seolevelmigrated="true">Circular Slip Surface in Su Material (Swedish Circle Method$ϕ = 0$)</h3><p>*Based on static moment calculation</p><p>Take moment about O:<br>Driving moment =$Wx$<br>Resisting moment =$SuRθ \cdot R = SuR^2θ_{rad}$<br>Factor of Safety(FS)</p><p>$FoS = Resisting \frac{moment}{Driving momemt}$</p><p>$FoS = \frac{SuR^2θ{rad}}{Wx}$</p><h3 id="28ddd008-d661-4685-af79-ebaf74502cec" data-toc-id="28ddd008-d661-4685-af79-ebaf74502cec" collapsed="false" seolevelmigrated="true">Work Example Circular Slip Surface</h3><p><strong>Assume</strong><br>$\gamma =20 kN/m^3$<br>$S_u = 10 kPa$<br>Failing wedge area A = 10.7$m^2$<br>Question; What is the Factor of Safety(FS)?<br>Given <strong>x</strong> <strong>= 4.8 m</strong></p><p><em>Calculations</em>*:</p><ul><li><p><strong>W</strong>: weight * area</p></li><li><p><em>Area</em><br><em>Therefore</em><br>$FoS = \frac{SuR^2θ{rad}}{Wx}$</p></li></ul><h3 id="77f9cdcc-8a63-4d27-b8af-46e66c5f553e" data-toc-id="77f9cdcc-8a63-4d27-b8af-46e66c5f553e" collapsed="false" seolevelmigrated="true">Circular slip surface with tension crack</h3><p><strong>If crack is filled with water:</strong><br>The effect of water:<br>FoS increase or decrease ?<br>If crack is dry:</p><p><em>Calculations::</em><br><strong>Hydrostatic Forces</strong>: Q<br>$Q=\frac{1}{2}*\gammaw *z{crack} · z{crack} = \gammaw*z_{crack}$</p><p><strong>FoS</strong>: Factor of Safety</p><p>$FoS= \frac{SuR^2 θ{rad}}{Wx + \frac{Qw}{2} * [\frac{2}{3}z{crack}+z_t]}$</p><p>$ Note the change in (FS)$compared to no tension crack at all.</p><h3 id="6cfeebb4-847c-4a34-b49b-f258aa0bb1a8" data-toc-id="6cfeebb4-847c-4a34-b49b-f258aa0bb1a8" collapsed="false" seolevelmigrated="true">Work Example-Circular slip surface with tension crack</h3><p>What if there`s tension cracks<br>O W=214 kN R X = 4.8 m<br>Given: zcrack zt = 0.30 m<br>Consider the circular soil wedge below with<br>$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