--- title: "Two-layer sand, plot stresses (bulk 18)" subject: "Soil Mechanics" chapter: 6 key: "semester_4/soil_mechanics:two-layer-sand-stress-plot" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/two-layer-sand-stress-plot site: BCT Engineering --- # Two-layer sand, plot stresses (bulk 18) Soil Mechanics · Chapter 6: Principle of Effective Stress, Capillarity and Permeability - Asked in: 2078 Poush, 6 marks - Video: [Effective stress, Total stress and Pore water pressure in Soil Mechanics || Example solved](https://www.youtube.com/watch?v=5r3McEtuyXU) — Civil Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/two-layer-sand-stress-plot - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-6 ## Question A sand deposit consists of two layers. The top layer is 3 m [unit not printed] thick with bulk unit weight 18 kN/m³ and saturated unit weight of 21 kN/m³ and the bottom layer is 4 m thick with saturated density of 20 kN/m³. The ground water table is at a depth of 4 m below the ground surface and the zone of capillary saturation is 1 m above the water table. Calculate and plot the effective stress, total stress and neutral stress. ## Short answer At 7 m depth, total stress is 134 kPa, neutral stress is 29.43 kPa, and effective stress is 104.57 kPa. ## Answer #### Given Top layer 0 to 3 m: $\gamma = 18$ kN/m³ (bulk), $\gamma_{sat} = 21$ kN/m³. Bottom layer 3 to 7 m: $\gamma_{sat} = 20$ kN/m³. WT at 4 m; capillary saturation 1 m above the WT (3 m to 4 m). $\gamma_w = 9.81$ kN/m³. The capillary zone (3 m to 4 m) lies in the bottom layer, so it is saturated with $\gamma_{sat} = 20$ kN/m³. The top layer lies above the capillary zone, so it stays at its bulk weight of 18 kN/m³ (its saturated weight is not needed). #### Calculation | Depth (m) | $\sigma$ (kPa) | $u$ (kPa) | $\sigma' = \sigma - u$ (kPa) | |---|---|---|---| | 0 | 0 | 0 | 0 | | 3 | $18\times3 = 54$ | $-9.81$ | 63.81 | | 4 (WT) | $54 + 20\times1 = 74$ | 0 | 74.00 | | 7 | $74 + 20\times3 = 134$ | $9.81\times3 = 29.43$ | 104.57 | #### Plot ```text depth(m) σ u σ' 0 0 0 0 3 54 -9.81 63.81 4 74 0 74.0 7 134 29.43 104.57 ``` Plot each column against depth (downward), joining the points with straight lines. $u$ is zero above 3 m, negative between 3 m and 4 m and positive below the WT. Just above 3 m, $\sigma' = 54$ kPa, so $\sigma'$ jumps by 9.81 kPa at the top of the capillary zone. **Answer: At 7 m, $\sigma = 134$ kPa, $u = 29.43$ kPa, $\sigma' = 104.57$ kPa.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the top layer (0-3 m) the unit weight used is: A. 21 kN/m3 B. 18 kN/m3 (bulk), as it is above the capillary zone C. 20 kN/m3 D. 9.81 kN/m3 Answer: B. The top layer lies above the capillary zone so its bulk weight 18 is used. (Key point: Unit weight choices) 2. Total stress at the WT (4 m) is: A. 54 kPa B. 72 kPa C. 74 kPa D. 84 kPa Answer: C. 54 + 20 x 1 = 74 kPa (capillary zone saturated at 20). (Key point: Total stress at 4 m) 3. At 3 m, sigma' equals: A. 54 kPa B. 44.19 kPa C. 74 kPa D. 63.81 kPa Answer: D. sigma' = 54 - (-9.81) = 63.81 kPa. (Key point: Effective stress at 3 m) 4. At 7 m the effective stress is: A. 104.57 kPa B. 134 kPa C. 29.43 kPa D. 74 kPa Answer: A. sigma = 74 + 60 = 134; u = 29.43; sigma' = 104.57 kPa. 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