--- title: "Isochrones, one and two way drainage" subject: "Soil Mechanics" chapter: 9 key: "semester_4/soil_mechanics:isochrones-one-two-way-drainage" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/isochrones-one-two-way-drainage site: BCT Engineering --- # Isochrones, one and two way drainage Soil Mechanics · Chapter 9: Compressibility of Soil - Asked in: 2077 Chaitra, 2 marks - Video: [Time rate of consolidation example: double drainage.](https://www.youtube.com/watch?v=Gx6ahN0RakY) — GEOTECH VIDEOS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/isochrones-one-two-way-drainage - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-9 ## Question Draw isochrones for a clay layer of thickness H under one-way drainage and two way drainage conditions at different elapsed times after loading (t = 0, t = t and t = ∞). Assume necessary conditions. ## Short answer At t=0 excess pore pressure is uniform at Δσ; at t=∞ it is zero everywhere. One-way drainage gives u=0 at the top drain, increasing to a maximum at the impervious base with zero slope; drainage path is H. Two-way drainage gives symmetrical isochrones, u=0 at both faces, maximum at centre; drainage path is H/2, so consolidation is four times faster. ## Answer An isochrone is a curve showing the variation of excess pore water pressure $u$ with depth at a given time $t$. For a clay layer of thickness $H$ under a uniform loading $\Delta\sigma$: - At $t=0$: $u = \Delta\sigma$ at all depths (the entire load is taken by the water). - At $0 u z=0 | \ t=t | \ ___ | \ / t=0 (rectangular, Δσ) z=H |___impervious___ u = 0 at top for all t>0; u = Δσ for t=0; u = 0 for t = ∞ ``` The isochrone at time $t$ starts at $u=0$ at the top drain, increases with depth, and reaches its maximum at the impervious base, which it meets with zero slope ($\partial u/\partial z = 0$). The drainage path is $H_{dr} = H$. #### Two-way drainage (top and bottom) ```text top drain ---| u = 0 | .--. mid-depth | ( ) t=t symmetrical | '--' bottom drain | u = 0 ``` The isochrones are symmetrical about the mid-depth; $u = 0$ at both faces for $t>0$ and the maximum is at the centre. The drainage path is $H_{dr} = H/2$, so consolidation is four times as fast. The area between an isochrone and the initial rectangular line gives the consolidation that has taken place. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. An isochrone is a curve showing A. settlement with time B. excess pore pressure with depth at a given time C. void ratio with log p D. permeability with depth Answer: B. It shows u variation with depth at one time t. (Key point: Isochrone) 2. At t = 0 and t = infinity the excess pore pressure is respectively A. 0 and delta sigma B. delta sigma everywhere and 0 everywhere C. delta sigma and delta sigma D. 0 and 0 Answer: B. At t = 0 water carries the full load; at the end consolidation is complete and u = 0. (Key point: t = 0 and infinity) 3. For two-way drainage the isochrones are A. symmetrical about mid-depth with maximum at the centre B. zero everywhere C. maximum at the faces D. straight lines Answer: A. u = 0 at both drainage faces and the maximum is at mid-depth; Hdr = H/2. (Key point: Two-way symmetry) 4. For one-way drainage (top only), at the impervious base the isochrone A. has u = 0 B. meets it with zero slope (du/dz = 0) C. has a negative value D. is vertical Answer: B. No flow at the impervious base, so du/dz = 0 there. 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