--- title: "Stability of wall with surcharge 25 kN/m²" subject: "Foundation Engineering" chapter: 3 key: "semester_5/foundation_engineering:stability-retaining-wall-q25" source: https://www.bctengineer.com/q/semester_5/foundation_engineering/stability-retaining-wall-q25 site: BCT Engineering --- # Stability of wall with surcharge 25 kN/m² Foundation Engineering · Chapter 3: Lateral Earth Pressure Theories and Retaining Walls - Asked in: 2078 Bhadra, 10 marks - Video: [soil mechanics numerical, stability of slopes, active earth pressure, retaining wall numerical](https://www.youtube.com/watch?v=aNTR4KOYcuk) — Civil Engineering Exam - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/stability-retaining-wall-q25 - Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-3 ## Question Check the stability of the retaining wall shown in figure below. Assume necessary conditions and take ultimate bearing capacity of the foundation soil as $250\ \text{kN/m}^2$. [Figure: concrete gravity wall, $\gamma_c = 25\ \text{kN/m}^3$, top width 0.5 m, height $H = 6.0$ m, base width 2.5 m, base friction $\delta_b = 20^\circ$; backfill $\gamma_t = 20\ \text{kN/m}^3$, $\phi' = 30^\circ$, $\delta_b = 20^\circ$, with a surcharge $q = 25\ \text{kN/m}^2$ on the backfill.] ## Short answer FS against overturning = 0.95, sliding = 0.48; resultant lies outside the base, causing tension. The wall is unsafe; redesign (e.g. wider base, shear key) is needed. ## Answer **Assumptions (figure not drawn here):** the back of the wall is vertical and the sloping face is on the front, so the section is a rectangle $0.5\times6$ m plus a triangle $2.0\times6$ m; Rankine's theory with horizontal backfill and a smooth back is used (conservative, ignores wall friction); passive resistance in front of the toe is neglected; base friction $\delta_b = 20^\circ$; $q_u = 250$ kN/m$^2$ with $FS \ge 3$. #### 1. Active thrust $$ K_a = \tan^2(45^\circ-15^\circ) = 0.3333 $$ - Surcharge: $P_1 = K_a q H = 0.3333\times25\times6 = 50.0$ kN/m at 3.0 m. - Soil: $P_2 = \tfrac12 K_a\gamma H^2 = 0.5\times0.3333\times20\times36 = 120.0$ kN/m at 2.0 m. - $\sum H = 170.0$ kN/m; $M_o = 50\times3+120\times2 = 390.0$ kN m/m. #### 2. Wall weight (about the toe) | Part | Weight (kN/m) | Arm (m) | Moment | |---|---|---|---| | Rectangle 0.5 x 6 x 25 | 75.0 | 2.250 | 168.75 | | Triangle 0.5 x 2.0 x 6 x 25 | 150.0 | 1.333 | 200.00 | | **Total** | **225.0** | | **368.75** | #### 3. Checks - **Overturning:** $FS_{ot} = 368.75/390.0 = 0.95$. This is **less than 2 (even less than 1)**; the wall overturns. - **Sliding:** $FS_s = 225\times\tan20^\circ/170.0 = 0.48$, **less than 1.5**; the wall slides. - **Eccentricity:** $\bar x = (368.75-390.0)/225 = -0.094$ m, so the resultant lies outside the base (negative), giving tension at the heel; $e = 1.34$ m $> B/6 = 0.42$ m. - **Bearing:** the resultant is beyond the toe, so the toe pressure is not valid and the available bearing area is lost. The pressure at the toe exceeds $q_u/3 = 83$ kN/m$^2$. **Answer: the wall is unsafe against overturning ($FS = 0.95$), sliding ($FS = 0.48$) and bearing.** Redesign: increase the base width. With the same section shape, $FS_{ot} = 2$ needs $B \approx 3.7$ m, and a shear key or wider base (about $0.6H$ to $0.7H$ for a gravity wall with surcharge) is needed for sliding; pressure at the toe must then be rechecked against $q_u/3$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the thrust composed for this wall? A. Triangle only B. Surcharge rectangle Ka q H plus soil triangle 0.5 Ka gamma H^2 C. Surcharge triangle plus soil rectangle D. Passive plus active Answer: B. 50 kN/m at 3 m plus 120 kN/m at 2 m gives 170 kN/m. (Key point: Active thrust with surcharge) 2. Why does the wall fail in overturning? A. Resisting moment 368.75 is less than overturning moment 390 (FS = 0.95) B. Ka is too small C. Water pressure D. Base is too wide Answer: A. FS below 1 means the wall overturns; 2 is required. (Key point: Overturning result) 3. Which calculation gives FS against sliding? A. 225 x tan 20 / 170 = 0.48 B. 170/225 C. 225 / 170 tan 20 D. 368.75/170 Answer: A. Base friction V tan(delta) over total horizontal thrust, 0.48, less than 1.5. (Key point: Sliding) 4. x_bar is negative (-0.094 m). What does this mean? A. Resultant lies outside the base (beyond the toe), so the base cannot carry it B. Pressure is uniform C. Wall is very safe D. Tension only at toe Answer: A. The resultant falls outside the base, meaning overturning; e > B/6 as well. 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