--- title: "Two-layer sand, stress diagrams (3 m, 4 m)" subject: "Soil Mechanics" chapter: 6 key: "semester_4/soil_mechanics:two-layer-sand-stress-diagrams" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/two-layer-sand-stress-diagrams site: BCT Engineering --- # Two-layer sand, stress diagrams (3 m, 4 m) Soil Mechanics · Chapter 6: Principle of Effective Stress, Capillarity and Permeability - Asked in: 2075 Bhadra, 8 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-diagrams - 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 3.0 m thick (γ = 17 kN/m³) and the bottom layer is 4.0 m thick ($\gamma_{sat}$ = 21 kN/m³). The water table is at a depth of 4.0 m from the surface and the zone of capillary saturation is 1 m above the water table. Draw the diagrams showing the variation of total stress, neutral stress and effective stress. ## Short answer At 7 m depth, total stress = 135 kPa, neutral stress = 29.43 kPa, effective stress = 105.57 kPa. ## Answer #### Given Top layer 0 to 3 m, $\gamma = 17$ kN/m³. Bottom layer 3 to 7 m, $\gamma_{sat} = 21$ kN/m³. WT at 4 m; capillary saturation 1 m above the WT (3 m to 4 m), which lies in the bottom layer, so $\gamma = 21$ kN/m³ is used there. $\gamma_w = 9.81$ kN/m³; $u = 0$ above the capillary zone. #### Stresses at key levels | Depth (m) | $\sigma$ (kPa) | $u$ (kPa) | $\sigma'$ (kPa) | |---|---|---|---| | 0 | 0 | 0 | 0 | | 3 (just above capillary zone) | $17\times3 = 51$ | 0 | 51.00 | | 3 (top of capillary zone) | 51 | $-9.81$ | 60.81 | | 4 (WT) | $51 + 21\times1 = 72$ | 0 | 72.00 | | 7 (bottom) | $72 + 21\times3 = 135$ | $9.81\times3 = 29.43$ | 105.57 | #### Diagrams ```text depth total σ pore u effective σ' 0 m 0 0 0 3 m 51 0 -> -9.8 (suction) 51 -> 60.8 4 m 72 0 (WT) 72.0 7 m 135 29.4 105.6 (draw each as straight lines between these points on a stress-vs-depth plot, depth downward) ``` - $\sigma$ increases linearly, with a change in slope at 3 m (17 to 21 kN/m³). - $u$ jumps to $-9.81$ kPa (suction) at 3 m, rises linearly to zero at the WT, and increases at 9.81 kPa/m below it. - $\sigma' = \sigma - u$; it jumps by 9.81 kPa at 3 m (from 51 to 60.81 kPa), then rises by $(21 - 9.81)\times 1 = 11.19$ kPa across the capillary zone to 72 kPa at the WT. In the capillary zone it is higher than $\sigma$. **Answer: At 7 m, $\sigma = 135$, $u = 29.43$, $\sigma' = 105.57$ kPa.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The capillary zone (3-4 m) lies in the bottom layer, so the unit weight used there is: A. 17 kN/m3 B. 21 kN/m3 C. 9.81 kN/m3 D. 12 kN/m3 Answer: B. The zone is saturated and in the lower layer, so gamma_sat = 21. (Key point: Layer for capillary zone) 2. At the WT (4 m), sigma, u, sigma' are: A. 72, 0, 72 B. 51, 0, 51 C. 72, 9.81, 62.19 D. 93, 0, 93 Answer: A. sigma = 51 + 21 = 72; u = 0; sigma' = 72 kPa. (Key point: Stress at WT) 3. At 3 m the effective stress is: A. 51 kPa B. 41.19 kPa C. 60.81 kPa D. 0 Answer: C. u = -9.81 so sigma' = 51 + 9.81 = 60.81 kPa. (Key point: Top of capillary zone) 4. At 7 m, sigma' is: A. 135 kPa B. 29.43 kPa C. 164.43 kPa D. 105.57 kPa Answer: D. sigma = 72 + 21 x 3 = 135; u = 9.81 x 3 = 29.43; sigma' = 105.57 kPa. 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