--- title: "Infinite clay slope, submerged and seepage" subject: "Soil Mechanics" chapter: 11 key: "semester_4/soil_mechanics:infinite-slope-submerged-seepage-clay" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/infinite-slope-submerged-seepage-clay site: BCT Engineering --- # Infinite clay slope, submerged and seepage Soil Mechanics · Chapter 11: Stability of Slopes - Asked in: 2078 Poush, 6 marks; 2073 Bhadra, 6 marks (asked 2 times) - Video: [Infinite Slope Stability Example Solution (Stability of Infinite Slopes)](https://www.youtube.com/watch?v=ISThKrxuQJI) — Mahbub - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/infinite-slope-submerged-seepage-clay - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-11 ## Question An infinite slope is made of clay with the following properties: $\gamma_t$ = 18 kN/m³, $\gamma'$ = 9 kN/m³, c = 25 kN/m² and $\phi'$ = 28°. If the slope has an inclination of 35° and height equal to 12 m, determine the stability of the slope, when (i) the slope is submerged, (ii) there is seepage parallel to the slope. ## Short answer Submerged: F = c'/(γ' z sinβ cosβ) + tanφ'/tanβ, giving F = 1.25 (stable). Seepage parallel to slope: F = c'/(γ_sat z sinβ cosβ) + (γ'/γ_sat)(tanφ'/tanβ), giving F = 0.63 (unstable). ## Answer **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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What formula gives F for the fully submerged case? A. F = c'/(gamma' z sin beta cos beta) + tan phi'/tan beta B. F = c'/(gamma_t z sin beta cos beta) + gamma' tan phi'/(gamma_sat tan beta) C. F = tan beta/tan phi' D. F = gamma' tan beta Answer: A. Both stresses use gamma' as there is no seepage force. (Key point: Submerged FS) 2. What is the result for case (i)? A. F = 0.63, unstable B. F = 2.5 C. F = 0.76 D. F = 1.25, stable Answer: D. 0.493 + 0.759 = 1.25. (Key point: Submerged result) 3. With seepage parallel to the slope, how are stresses used? A. Both use gamma' B. Both use gamma_sat C. Shear stress uses gamma_sat, effective normal stress uses gamma' D. Normal stress uses gamma_w only Answer: C. Seepage adds driving force, so F = c'/(gamma_sat z sin cos) + (gamma'/gamma_sat) tan phi'/tan beta. (Key point: Seepage FS) 4. What does the seepage case give? A. F = 1.25, stable B. F = 0.63, unstable C. F = 1.0 exactly D. F = 2.0 Answer: B. 0.246 + 0.380 = 0.63 < 1. (Key point: Seepage result) ## Related in Soil Mechanics - [Infinite dry slope, c = 10 kPa, φ = 25°](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Ainfinite-dry-slope-c10-phi25) — Soil Mechanics, chapter 11 - [Infinite slope seepage, max angle for FoS 2](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Ainfinite-slope-seepage-max-angle) — Soil Mechanics, chapter 11 - [Infinite slope, cohesionless, steady seepage](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Ainfinite-slope-cohesionless-seepage) — Soil Mechanics, chapter 11 - [Seepage in inclined permeable layer](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Ainclined-permeable-layer-seepage) — Soil Mechanics, chapter 6 - [Final settlement of clay, +30 kN/m²](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Afinal-settlement-sand-clay-30kpa) — Soil Mechanics, chapter 9 - [Infinite slope FoS; water table rises](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Ainfinite-slope-cohesionless-water-table) — Soil Mechanics, chapter 11 ## Related in other subjects - [Frame AFD, SFD, BMD (12 kN, 25 kNm, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-12kn-25knm) — Strength of Materials, chapter 1 - [Equal area criterion for transient stability](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aequal-area-criterion-concept) — Power System Analysis II, chapter 6, asked 4× - [Stability of wall with surcharge 25 kN/m²](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Astability-retaining-wall-q25) — Foundation Engineering, chapter 3 - [Slope stability: N40W/55 slope, two joints](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Aslope-stability-n40w-55) — Engineering Geology II, chapter 6