--- title: "Active pressure, sloping backfill and water table" subject: "Foundation Engineering" chapter: 3 key: "semester_5/foundation_engineering:wall-sheet-pipe-sloping-backfill" source: https://www.bctengineer.com/q/semester_5/foundation_engineering/wall-sheet-pipe-sloping-backfill site: BCT Engineering --- # Active pressure, sloping backfill and water table Foundation Engineering · Chapter 3: Lateral Earth Pressure Theories and Retaining Walls - Asked in: 2073 Shrawan, 8 marks - Video: [Rankine Theory of Earth Pressure | Elementary Engineering](https://www.youtube.com/watch?v=3CnW3mHvJ2Q) — Elementary Engineering - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/wall-sheet-pipe-sloping-backfill - Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-3 ## Question A high steel sheet pipe wall with smooth vertical back supports a dry cohesionless soil that weighs $18\ \text{kN/m}^3$. The backfill rises from the crest of the wall at an angle of $20^\circ$ with the horizontal. If the angle of internal friction of the backfill material is $30^\circ$, determine the magnitude and point of application of the active earth pressure per meter length of the wall. What will be the change in its magnitude and point of application if the water table rises to an elevation 2 m below the top of the wall? Take the submerged unit weight of the backfill material as $12\ \text{kN/m}^3$. ## Short answer For dry sloping backfill (β=20°, φ=30°), Ka=0.4142 gives Pa=134.2 kN/m at 2.00 m above base. With water table 2 m below top, effective soil thrust reduces to 114.3 kN/m at 2.12 m; adding water pressure (78.5 kN/m) yields total horizontal thrust 185.9 kN/m at 1.79 m above base. ## Answer **Assumption:** the wall height is not stated, so $H = 6$ m is assumed (the method is the same for any $H$). Rankine's theory for a sloping backfill ($\beta = 20^\circ$, $\phi = 30^\circ$) gives a pressure parallel to the slope. **Active coefficient** $$ K_a = \cos\beta\,\frac{\cos\beta - \sqrt{\cos^2\beta - \cos^2\phi}}{\cos\beta + \sqrt{\cos^2\beta - \cos^2\phi}} = 0.4142 $$ #### (a) Dry backfill $$ \begin{aligned} P_a &= \tfrac12 K_a\gamma H^2 = 0.5\times0.4142\times18\times6^2 = 134.2\ \text{kN/m} \end{aligned} $$ It acts at $H/3 = 2.00$ m above the base, inclined at $20^\circ$ to the horizontal (horizontal component 126.1 kN/m, vertical component 45.9 kN/m). #### (b) Water table 2 m below the top The top 2 m is dry ($\gamma=18$) and the lower 4 m is submerged ($\gamma'=12$). - At 2 m: $p = 0.4142\times18\times2 = 14.91$ kN/m$^2$. - At 6 m: $p = 0.4142\times(36 + 12\times4) = 34.79$ kN/m$^2$. - Water pressure at the base: $9.81\times4 = 39.24$ kN/m$^2$. | Part | Force (kN/m) | Arm above base (m) | |---|---|---| | Soil, upper triangle | 14.91 | 4.667 | | Soil, rectangle | 59.65 | 2.000 | | Soil, lower triangle | 39.76 | 1.333 | | **Soil total** | **114.32** | **2.12** | | Water (horizontal) = 0.5 x 9.81 x 4$^2$ | 78.48 | 1.333 | The horizontal total is $114.32\cos20^\circ + 78.48 = 185.9$ kN/m, acting at $$ \bar y = \frac{107.43\times2.12 + 78.48\times1.333}{185.9} = 1.79\ \text{m} $$ The vertical component of the soil thrust is 39.1 kN/m (resultant 190.0 kN/m). **Answer:** - Dry: $P_a = 134.2$ kN/m at 2.00 m above the base. - With the water table: the effective soil thrust falls to 114.3 kN/m (acting at 2.12 m), but with the water pressure (78.5 kN/m) the total horizontal thrust rises from 126.1 to 185.9 kN/m (+60 kN/m, about 47% more), and its point of application drops to 1.79 m above the base. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a smooth vertical wall with backfill sloping at beta, in which direction does the Rankine active pressure act? A. Horizontally B. Parallel to the backfill slope C. Perpendicular to the backfill slope D. At phi/2 above the horizontal Answer: B. With a sloping backfill Rankine's pressure is parallel to the slope, so it has both horizontal and vertical components. (Key point: Direction of Rankine thrust) 2. For the dry cohesionless backfill, where does Pa act above the base of a wall of height H? A. H/2 B. H/4 C. H/3 D. 2H/3 Answer: C. The pressure diagram is a triangle, so the resultant acts at H/3 above the base. (Key point: Point of application (dry)) 3. When the water table rises to 2 m below the top, what happens to the total horizontal thrust? A. It decreases because the soil is lighter B. It stays the same C. It rises, because water pressure is added to the reduced soil thrust D. It becomes zero Answer: C. Submergence reduces the effective soil thrust (114.3 kN/m) but the 78.5 kN/m water thrust raises the total to 185.9 kN/m. (Key point: Effect of water table on thrust) 4. Why does the point of application of the resultant move down when the water table rises? A. Water pressure acts over the lower 4 m, adding load low on the wall B. The soil friction angle increases C. Ka decreases with depth D. The vertical component grows Answer: A. Water pressure is a triangle in the lower part with its arm at 1.333 m, pulling the resultant down to 1.79 m. 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