--- title: "Two-layer profile stress diagram, 6 m sand" subject: "Soil Mechanics" chapter: 6 key: "semester_4/soil_mechanics:sand-clay-profile-stress-diagram-capillary" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/sand-clay-profile-stress-diagram-capillary site: BCT Engineering --- # Two-layer profile stress diagram, 6 m sand Soil Mechanics · Chapter 6: Principle of Effective Stress, Capillarity and Permeability - Asked in: 2079 Jestha, 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/sand-clay-profile-stress-diagram-capillary - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-6 ## Question A layer of 6 m thick fine sand is overlain by a clay deposit of 4 m and the water table is 2 m below the surface. The unit weight of clay above and below the water level is 18 kN/m³ and 22 kN/m³ respectively. The layer of fine sand has a porosity of 44% and specific gravity of 2.65. If there is capillary rise of 1 m above the water table, draw the total stress, pore water pressure and effective stress diagram. ## Short answer At 10 m depth, total stress = 197.25 kPa, pore water pressure = 78.48 kPa, effective stress = 118.77 kPa. Capillary suction of 9.81 kPa increases effective stress at 1 m depth. ## Answer #### Given and assumptions - Profile: clay 0 to 4 m, fine sand 4 m to 10 m. Water table (WT) at 2 m depth, capillary rise 1 m above the WT (saturated zone from 1 m to 2 m depth). - Clay: $\gamma = 18$ kN/m³ above the WT, $\gamma_{sat} = 22$ kN/m³ below. The capillary zone is saturated, so 22 kN/m³ is used from 1 m to 2 m. - Sand: $n = 0.44$, $G = 2.65$, $\gamma_w = 9.81$ kN/m³. The soil above the capillary zone is taken as having $u = 0$. $$ e = \frac{n}{1-n} = \frac{0.44}{0.56} = 0.786, \qquad \gamma_{sat} = \frac{(G+e)\gamma_w}{1+e} = \frac{(2.65+0.786)\times 9.81}{1.786} = 18.87\ \text{kN/m}^3 $$ #### Stresses (kPa) $$ \sigma = \sum \gamma h, \qquad u = \gamma_w \times (\text{depth below WT}) \text{ (negative in capillary zone)}, \qquad \sigma' = \sigma - u $$ | Depth (m) | $\sigma$ | $u$ | $\sigma'$ | |---|---|---|---| | 0 | 0 | 0 | 0 | | 1 (just above capillary zone) | $18\times1 = 18$ | 0 | 18.00 | | 1 (top of capillary zone) | 18 | $-9.81$ | 27.81 | | 2 (WT) | $18 + 22\times1 = 40$ | 0 | 40.00 | | 4 (clay/sand interface) | $40 + 22\times2 = 84$ | 19.62 | 64.38 | | 10 (bottom of sand) | $84 + 18.87\times6 = 197.25$ | 78.48 | 118.77 | #### Diagram ```text Depth total σ u σ' 0 m 0 0 0 1 m 18 0 -> -9.81 (suction) 18 -> 27.8 2 m 40 0 (WT) 40 4 m 84 19.6 64.4 10 m 197.3 78.5 118.8 (linear between points; u is negative only between 1 m and 2 m) ``` Plot $\sigma$, $u$ and $\sigma'$ against depth, joining the points with straight lines. $\sigma$ changes slope at 1 m and 4 m. At 1 m, $u$ jumps from 0 to $-9.81$ kPa, so $\sigma'$ jumps from 18 to 27.81 kPa; $\sigma'$ then changes slope again at 4 m. $u$ rises linearly from $-9.81$ kPa at 1 m to 0 at 2 m (the negative triangle above the WT) and then to 78.48 kPa at 10 m. **Answer: At the bottom (10 m): $\sigma = 197.25$, $u = 78.48$, $\sigma' = 118.77$ kPa. Capillarity gives a gain of $9.81$ kPa in effective stress at 1 m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At 1 m depth (top of capillary zone, 1 m above WT) u is: A. +9.81 kPa B. -9.81 kPa C. 0 D. -19.62 kPa Answer: B. u = -gamma_w x 1 m of height above the water table = -9.81 kPa. (Key point: Pore pressure in capillary zone) 2. Which unit weight is used for the clay from 1 m to 2 m? A. 18 kN/m3 (above WT) B. 22 kN/m3, since the capillary zone is saturated C. 9.81 kN/m3 D. 12.2 kN/m3 Answer: B. The capillary zone is saturated so 22 kN/m3 applies, giving sigma = 18 + 22 = 40 kPa at 2 m. (Key point: Unit weight in capillary zone) 3. For n = 0.44, G = 2.65, the sand void ratio and gamma_sat are about: A. e = 0.786, gamma_sat = 18.87 kN/m3 B. e = 0.44, gamma_sat = 20 kN/m3 C. e = 1.27, gamma_sat = 16 kN/m3 D. e = 0.786, gamma_sat = 26 kN/m3 Answer: A. e = n/(1-n) = 0.786; gamma_sat = (G+e) gamma_w/(1+e) = 18.87. (Key point: Sand saturated weight) 4. At 10 m depth the effective stress is: A. 197.25 kPa B. 78.48 kPa C. 118.77 kPa D. 64.38 kPa Answer: C. sigma = 197.25, u = 9.81 x 8 = 78.48, so sigma' = 118.77 kPa. 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