--- title: "Stability of cantilever wall, 6 m, q = 35" subject: "Foundation Engineering" chapter: 3 key: "semester_5/foundation_engineering:stability-cantilever-wall-q35" source: https://www.bctengineer.com/q/semester_5/foundation_engineering/stability-cantilever-wall-q35 site: BCT Engineering --- # Stability of cantilever wall, 6 m, q = 35 Foundation Engineering · Chapter 3: Lateral Earth Pressure Theories and Retaining Walls - Asked in: 2074 Asoj, 10 marks - Video: [Stability Checks for Retaining Wall](https://www.youtube.com/watch?v=8MJn01aDzeQ) — Ekeeda - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/stability-cantilever-wall-q35 - Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-3 ## Question Determine the maximum and minimum pressure under the base of the cantilever retaining wall as shown in the figure below and also the factor of safety against sliding and overturning. The approximate shear strength parameters for the soil are $c = 0$, $\phi = 41^\circ$. The unit weight of soil and concrete are $16\ \text{kN/m}^3$ and $24\ \text{kN/m}^3$ respectively. The water table is below the base of the wall. Take $\delta = 27^\circ$ on the base of the wall. [Figure: cantilever wall 6 m high, stem top width 0.4 m, heel length 2 m, base slab thickness 0.5 m, total base width 3.5 m, backfill with surcharge $q = 35\ \text{kN/m}^2$.] ## Short answer Maximum base pressure is 141.3 kN/m², minimum 53.4 kN/m², computed from vertical resultant eccentricity 0.263 m. Factor of safety against overturning is 3.02 and against sliding 1.68. ## Answer **Assumptions (figure values as given):** total height 6.0 m (stem 5.5 m + slab 0.5 m), stem 0.4 m thick throughout, heel 2.0 m, base width 3.5 m, so the toe is $3.5-2.0-0.4 = 1.1$ m. Rankine active pressure acts on a vertical plane through the heel; the surcharge on the heel counts as a vertical load; passive resistance and soil over the toe are neglected. #### 1. Active thrust $$ K_a = \tan^2(45^\circ-20.5^\circ) = 0.2077 $$ | Part | Force (kN/m) | Arm above base (m) | Moment | |---|---|---|---| | Surcharge $K_a q H = 0.2077\times35\times6$ | 43.61 | 3.00 | 130.84 | | Soil $\tfrac12 K_a\gamma H^2$ | 59.81 | 2.00 | 119.63 | | **Total** | **103.43** | | **250.47** | #### 2. Vertical loads (about the toe) | Part | Force (kN/m) | Arm (m) | Moment | |---|---|---|---| | Stem 0.4 x 5.5 x 24 | 52.80 | 1.300 | 68.64 | | Base slab 3.5 x 0.5 x 24 | 42.00 | 1.750 | 73.50 | | Soil on heel 2 x 5.5 x 16 | 176.00 | 2.500 | 440.00 | | Surcharge on heel 35 x 2 | 70.00 | 2.500 | 175.00 | | **Total** | **340.80** | | **757.14** | #### 3. Base pressure $$ \bar x = \frac{757.14-250.47}{340.80} = 1.487\ \text{m},\qquad e = 1.75-1.487 = 0.263\ \text{m} < B/6 = 0.583\ \text{m} $$ $$ q_{max,min} = \frac{340.80}{3.5}\left(1\pm\frac{6\times0.263}{3.5}\right) $$ $$ q_{max} = 141.3\ \text{kN/m}^2\ (\text{toe}),\qquad q_{min} = 53.4\ \text{kN/m}^2\ (\text{heel}) $$ #### 4. Factors of safety $$ FS_{ot} = \frac{757.14}{250.47} = 3.02\ (>2) $$ $$ FS_s = \frac{340.80\times\tan27^\circ}{103.43} = 1.68\ (>1.5) $$ **Answer:** $q_{max} = 141.3$ kN/m$^2$, $q_{min} = 53.4$ kN/m$^2$; $FS_{overturning} = 3.02$; $FS_{sliding} = 1.68$. Both are adequate. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is Ka obtained for phi = 41 degrees? A. tan(45 + phi/2) squared B. tan^2(45 - phi/2) = 0.2077 C. 1 - sin phi squared D. cos phi Answer: B. Rankine active coefficient for a horizontal backfill and smooth vertical plane is tan^2(45 - phi/2). (Key point: Ka value) 2. Why is the soil (and surcharge) on the heel included in vertical loads? A. It adds to the weight and stabilising moment about the toe B. It reduces friction C. It is the active thrust D. It is passive resistance Answer: A. Heel soil and surcharge (176 + 70 kN/m) act downward on the slab with a long arm, resisting overturning. (Key point: Soil on heel) 3. What is FS against overturning and how found? A. 757.14/250.47 = 3.02 (restoring/overturning moment about the toe) B. 340.8/103.43 C. 250.47/757.14 D. 1.68 Answer: A. Mr over Mo about the toe, 3.02, above 2. (Key point: Overturning FS) 4. With e = 0.263 m, where is q_max = 141.3 kN/m^2? A. At the heel B. At the toe, since the resultant is on the toe side of centre C. At the centre D. Under the stem only Answer: B. x_bar = 1.487 < B/2 = 1.75 means the resultant is toward the toe, so toe pressure is maximum. 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