---
title: "3 m clay, cv 0.025 cm²/min, times"
subject: "Soil Mechanics"
chapter: 9
key: "semester_4/soil_mechanics:clay-3m-cv-time-80-percent"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/clay-3m-cv-time-80-percent
site: BCT Engineering
---

# 3 m clay, cv 0.025 cm²/min, times

Soil Mechanics · Chapter 9: Compressibility of Soil

- Asked in: 2078 Poush, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/clay-3m-cv-time-80-percent
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-9

## Question

A 3 m thick clay layer beneath a building is overlain by a permeable stratum and is underlain by an impervious rock. The coefficient of consolidation of the clay was found to be 0.025 cm²/min. The final expected settlement for the layer is 8 cm. Determine: (i) how much time will it take for 80% of the total settlement, (ii) the required time for a settlement of 2.5 cm to occur, (iii) the settlement that would occur in 1 year.

## Short answer

For 80% consolidation, time factor 0.567 gives about 3.9 years. 2.5 cm settlement (31.25% U) requires about 192 days. In one year, settlement reaches about 3.45 cm.

## Answer

**Given:** clay is drained at the top only (permeable above, impervious rock below), so $H_{dr} = 3$ m $= 300$ cm. $c_v = 0.025\ \text{cm}^2/\text{min}$, $S_{final} = 8$ cm.

#### (i) Time for 80% of settlement

For $U = 80\%$: $T_v = 1.781 - 0.933\log_{10}(100-80) = 0.567$.

$$
t_{80} = \frac{T_v H_{dr}^2}{c_v} = \frac{0.567 \times 300^2}{0.025} = 2.04\times10^6\ \text{min}
$$

$$
= 1418\ \text{days} \approx 3.9\ \text{years}
$$

#### (ii) Time for 2.5 cm settlement

$$
U = \frac{2.5}{8} = 31.25\% \;(<60\%)\;\Rightarrow\; T_v = \frac{\pi}{4}(0.3125)^2 = 0.0767
$$

$$
t = \frac{0.0767 \times 300^2}{0.025} = 2.76\times10^5\ \text{min} = 192\ \text{days} \approx 0.53\ \text{year}
$$

#### (iii) Settlement in 1 year

$t = 365 \times 1440 = 525{,}600$ min.

$$
T_v = \frac{0.025 \times 525600}{300^2} = 0.146 \;\Rightarrow\; U = \sqrt{\frac{4T_v}{\pi}} = 0.431
$$

$$
S = 0.431 \times 8 = 3.45\ \text{cm}
$$

**Answer: (i) about 3.9 years; (ii) about 192 days; (iii) about 3.45 cm.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is H_dr for a clay layer with a permeable stratum above and impervious rock below?
   A. 1.5 m, half the thickness
   B. 3 m, the full thickness
   C. 6 m, twice the thickness
   D. 0 m
   Answer: B. Drainage is one-way (top only), so the path equals the full thickness. (Key point: Drainage path)

2. How is Tv obtained for U = 80%?
   A. Tv = 1.781 - 0.933 log10(100 - 80) = 0.567
   B. Tv = (pi/4)(0.8)^2 = 0.503
   C. Tv = 0.197
   D. Tv = 0.848
   Answer: A. For U above 60% the logarithmic relation applies; the parabolic formula is valid only below 60%. (Key point: Tv for 80%)

3. How is the time for 2.5 cm of settlement found?
   A. U = 2.5/8 = 31.25%, Tv = 1.781 - 0.933 log10(68.75)
   B. U = 2.5 %, Tv = (pi/4)(0.025)^2
   C. U = 2.5/8 = 31.25%, Tv = (pi/4)(0.3125)^2 = 0.0767, then t = Tv H_dr^2 / cv
   D. U = 8/2.5 = 320%, Tv = 0.567
   Answer: C. U is under 60%, so the parabolic relation is used, giving about 192 days. (Key point: Time for 2.5 cm)

4. How is the settlement after 1 year computed?
   A. Tv = 0.146 gives U = 0.146 so S = 1.17 cm
   B. Tv = cv t / H_dr^2 = 0.146, U = sqrt(4Tv/pi) = 0.431, S = 0.431 x 8 = 3.45 cm
   C. Tv = 0.146 gives U from the log formula, about 80%
   D. S = 8 cm since a year is long enough
   Answer: B. Compute Tv from time, convert it to U (Tv is small so the parabolic relation applies), then multiply by the final settlement. (Key point: Settlement in one year)

## Related in Soil Mechanics

- [Clay 10 m, 2.54 cm sample, half settlement](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aclay-10m-half-settlement-time-ratio) — Soil Mechanics, chapter 9
- [Final settlement of clay, +30 kN/m²](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Afinal-settlement-sand-clay-30kpa) — Soil Mechanics, chapter 9
- [Seepage in inclined permeable layer](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Ainclined-permeable-layer-seepage) — Soil Mechanics, chapter 6
- [Final settlement and settlement at 0.5 year](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Afinal-settlement-clay-time-half-year) — Soil Mechanics, chapter 9
- [Time-settlement curve, 3 m clay](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Atime-settlement-curve-3m-clay) — Soil Mechanics, chapter 9
- [5 m clay, Cc 0.25, 50% consolidation](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aclay-5m-cc-025-50-percent) — Soil Mechanics, chapter 9, asked 2×

## Related in other subjects

- [True length: tape found 2.5 cm too long](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-825-plus-750-m) — Surveying I, chapter 2
- [Insulation layer: minimum outside heat transfer coefficient](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Ainsulation-min-outside-h-270c-70c) — Fundamental of Thermodynamics & Heat Transfer, chapter 7, asked 2×
- [Differential settlement allowed, clay vs sand](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Adifferential-settlement-clay-sand) — Foundation Engineering, chapter 6
- [Rock layer attitude: A 900 m N, C 800 m W](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Abh-rock-layer-a900n-c800w) — Engineering Geology II, chapter 6, asked 2×