--- title: "Passive thrust on retaining wall back" subject: "Foundation Engineering" chapter: 3 key: "semester_5/foundation_engineering:wall-passive-thrust-two-layer" source: https://www.bctengineer.com/q/semester_5/foundation_engineering/wall-passive-thrust-two-layer site: BCT Engineering --- # Passive thrust on retaining wall back Foundation Engineering · Chapter 3: Lateral Earth Pressure Theories and Retaining Walls - Asked in: 2076 Chaitra, 8 marks - Video: [Rankine Active and Passive Pressures on Retaining Wall in Civil Engineering | Cohesive Soil Mass](https://www.youtube.com/watch?v=mcdowuAFAuo) — Soil Mechanics and Engineering Geology - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/wall-passive-thrust-two-layer - Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-3 ## Question Calculate the total passive thrust and its point of application on the back of the following retaining wall. [Figure: wall with two soil layers on the soil side: top layer 5 m thick with $\gamma = 18\ \text{kN/m}^3$, $\phi = 30^\circ$, $c = 0$; bottom layer 5 m thick with $\gamma = 20\ \text{kN/m}^3$, $\phi = 15^\circ$, $c = 30\ \text{kN/m}^2$.] ## Short answer Passive pressures computed with Rankine theory: top layer (Kp=3.0) and bottom layer (Kp=1.698, c=30 kN/m²). Total passive thrust is 2255 kN/m, acting horizontally at 3.59 m above the base. ## Answer **Assumptions:** smooth vertical wall back, horizontal ground, no water. Rankine passive pressure $p_p = K_p\sigma_v + 2c\sqrt{K_p}$, with total height 10 m of soil (5 m + 5 m) acting on the back. **Coefficients** $$ \begin{aligned} K_{p1} &= \tan^2(45^\circ+15^\circ) = 3.000 \\ K_{p2} &= \tan^2(45^\circ+7.5^\circ) = 1.6984,\quad \sqrt{K_{p2}} = 1.3032 \\ 2c\sqrt{K_{p2}} &= 2\times30\times1.3032 = 78.19\ \text{kN/m}^2 \end{aligned} $$ **Pressures (kN/m$^2$)** - Top layer: $z=0$: $p=0$; $z=5$ m: $\sigma_v=18\times5=90$, $p = 3.000\times90 = 270.0$. - Bottom layer top: $p = 1.6984\times90 + 78.19 = 231.05$. - Bottom layer base ($z=10$ m): $\sigma_v = 90+20\times5 = 190$, $p = 1.6984\times190 + 78.19 = 400.89$. ```text 0 | |\ | \ 5 |--270.0 |--231.05 | \ 10 |------400.89 ``` **Forces (arm above the base of the 10 m wall)** | Part | Force (kN/m) | Arm (m) | Moment | |---|---|---|---| | Top triangle 0.5 x 5 x 270.0 | 675.00 | 6.667 | 4500.00 | | Bottom rectangle 5 x 231.05 | 1155.25 | 2.500 | 2888.11 | | Bottom triangle 0.5 x 5 x 169.84 | 424.60 | 1.667 | 707.67 | | **Total** | **2254.85** | | **8095.78** | $$ \bar y = 8095.78/2254.85 = 3.59\ \text{m} $$ **Answer: total passive thrust $P_p = 2255$ kN/m, acting horizontally at 3.59 m above the base (bottom of the wall).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Rankine passive pressure in a c-phi soil is: A. Kp sigma_v - 2c sqrt(Kp) B. Kp sigma_v + 2c sqrt(Kp) C. Ka sigma_v + 2c D. Kp sigma_v / 2c Answer: B. p_p = Kp sigma_v + 2c sqrt(Kp). (Key point: Passive pressure formula) 2. Pressure at the base of the top layer (phi = 30, 5 m, gamma 18) is: A. 90 B. 270 kN/m2 C. 30 D. 231.05 Answer: B. Kp = 3, so 3 x 90 = 270. (Key point: Top layer pressure) 3. Pressure at the top of the lower layer is 231.05 rather than 270 because: A. Mistake B. Kp2 = 1.698 and cohesion term 78.19 apply: 1.6984 x 90 + 78.19 C. Water table D. Unit weight changes Answer: B. The layer has different phi and c, giving 231.05 kN/m2. (Key point: Jump at interface) 4. Total passive thrust and its height above base: A. 2255 kN/m at 3.59 m B. 2255 kN/m at 5 m C. 675 kN/m at 6.67 m D. 1155 kN/m at 2.5 m Answer: A. Sum F = 2254.85, sum M = 8095.78, ybar = 3.59 m. 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