--- title: "Active pressure, 8 m wall, table of layers" subject: "Foundation Engineering" chapter: 3 key: "semester_5/foundation_engineering:wall-8m-two-layer-table" source: https://www.bctengineer.com/q/semester_5/foundation_engineering/wall-8m-two-layer-table site: BCT Engineering --- # Active pressure, 8 m wall, table of layers Foundation Engineering · Chapter 3: Lateral Earth Pressure Theories and Retaining Walls - Asked in: 2072 Chaitra, 8 marks - Video: [Rankine's Active Earth Pressure Distribution on Three Layered Soil with Water Table and Surcharge](https://www.youtube.com/watch?v=vyVz6HhquuI) — Master Civil Engineering - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/wall-8m-two-layer-table - Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-3 ## Question A retaining wall with a smooth vertical back is 8 m high and retains a 2-layered soil having properties as follows: | Depth (m) | c (kN/m²) | φ (degrees) | γ (kN/m³) | |---|---|---|---| | 0-4 | 10 | 30 | 18 | | 4-8 | 0 | 34 | 20 | Show the active earth pressure distribution on the back of the retaining wall and its resultant. ## Short answer Rankine active pressures: top layer (c=10, φ=30°) with tension crack ignored, bottom layer (c=0, φ=34°). Resultant thrust is 139.6 kN/m, acting horizontally at 2.03 m above the base. ## Answer **Assumptions:** smooth vertical wall, horizontal backfill, no groundwater (none is given), Rankine theory. **Coefficients** $$ \begin{aligned} K_{a1} &= \tan^2(45^\circ-15^\circ) = 0.3333,\quad \sqrt{K_{a1}} = 0.5774 \\ K_{a2} &= \tan^2(45^\circ-17^\circ) = 0.2827 \end{aligned} $$ **Layer 1** (0-4 m, $c=10$, $\phi=30^\circ$, $\gamma=18$): $p = K_{a1}\gamma z - 2c\sqrt{K_{a1}}$ - $z=0$: $p = -2\times10\times0.5774 = -11.55$ kN/m$^2$ (tension). - Zero at $z_0 = 2c/(\gamma\sqrt{K_{a1}}) = 1.925$ m; the tension zone is ignored (tension crack). - $z=4$ m (top layer): $p = 0.3333\times72 - 11.55 = 12.45$ kN/m$^2$. **Layer 2** (4-8 m, $c=0$, $\phi=34^\circ$, $\gamma=20$): $\sigma_v$ at 4 m $= 18\times4 = 72$ kN/m$^2$. - Top: $p = 0.2827\times72 = 20.36$ kN/m$^2$. - Base ($z=8$ m): $\sigma_v = 72 + 20\times4 = 152$, $p = 0.2827\times152 = 42.97$ kN/m$^2$. ```text 0 | (tension zone, 0 to 1.92 m, ignored) | 1.92|\ | \ | \ 4m |---12.45 |-- 20.36 | \ | \ 8m |-----42.97 ``` **Forces per metre run (arm above base)** | Part | Force (kN/m) | Arm (m) | Moment | |---|---|---|---| | Layer 1 triangle: 0.5 x 2.075 x 12.45 | 12.92 | 4.692 | 60.63 | | Layer 2 rectangle: 4 x 20.36 | 81.42 | 2.000 | 162.84 | | Layer 2 triangle: 0.5 x 4 x 22.62 | 45.23 | 1.333 | 60.31 | | **Total** | **139.58** | | **283.79** | $$ \bar y = 283.79/139.58 = 2.03\ \text{m above the base} $$ **Answer: resultant active thrust $P_a = 139.6$ kN/m, horizontal, acting 2.03 m above the base (distribution as shown).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In layer 1 (c = 10, phi = 30, gamma = 18) the tension zone depth is: A. 1.925 m B. 4 m C. 0.555 m D. 3.85 m Answer: A. z0 = 2c/(gamma sqrt(Ka)) = 20/(18 x 0.5774) = 1.925 m. (Key point: Tension zone) 2. Pressure at the base of layer 1 (4 m) is: A. 12.45 kN/m2 B. 20.36 kN/m2 C. 24 kN/m2 D. -11.55 kN/m2 Answer: A. 0.3333 x 72 - 11.55 = 12.45. (Key point: Pressure at 4 m top layer) 3. Why is the pressure just below 4 m (20.36) different from just above (12.45)? A. A water table is present B. Layer 2 has c = 0 and different Ka (0.2827), so there is no cohesion deduction C. Measurement error D. Passive pressure starts Answer: B. 0.2827 x 72 = 20.36 with no 2c sqrt(Ka) term. (Key point: Top of layer 2) 4. Resultant thrust and height above base are: A. 139.6 kN/m at 2.03 m B. 139.6 kN/m at 2.67 m C. 100 kN/m at 4 m D. 81.4 kN/m at 2 m Answer: A. Sum F = 139.58, sum M = 283.79, ybar = 2.03 m. 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