---
title: "Clay 10 m, 2.54 cm sample, half settlement"
subject: "Soil Mechanics"
chapter: 9
key: "semester_4/soil_mechanics:clay-10m-half-settlement-time-ratio"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/clay-10m-half-settlement-time-ratio
site: BCT Engineering
---

# Clay 10 m, 2.54 cm sample, half settlement

Soil Mechanics · Chapter 9: Compressibility of Soil

- Asked in: 2077 Chaitra, 5 marks
- Video: [A 6.0 m thick layer of clay is located between two layers of free draining sand. Also, there is a](https://www.youtube.com/watch?v=0XSRw2Dr4qI) — Unique Series - Civil Mantraa
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/clay-10m-half-settlement-time-ratio
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-9

## Question

At a certain depth below the foundation of a building there exists a clay layer of thickness 10 m. Above and below the clay layer there are incompressible permeable soils. In a consolidation test on the clay sample with drainage at top and bottom, a sample with initial thickness 2.54 cm was compressed under a steady pressure. Half of the final settlement value [the rest of the sentence is not clear in print]? Take time factor, $T_v$ = 0.196 for 50% degree of consolidation.

## Short answer

For 50% consolidation, field time is 1.55×10^5 times the lab time because drainage path ratio (500/1.27)^2. If lab half-settlement takes 10 min, field time is about 2.95 years.

## Answer

**Note:** the second half of the problem is unclear in print, and the laboratory time for half the final settlement is not given. The usual form is: "the sample took $t_{lab}$ to reach half its final settlement; find the time for the building to reach half its final settlement". The scale relation is solved below, and the field time follows from the lab time.

**Principle:** same clay, same $c_v$, same $U = 50\%$ (so same $T_v = 0.196$). Then $t \propto H_{dr}^2$.

**Drainage paths** (both drained top and bottom)

- Field: $H_{dr} = 10/2 = 5$ m $= 500$ cm.
- Laboratory: $H_{dr} = 2.54/2 = 1.27$ cm.

**Time ratio**

$$
\frac{t_{field}}{t_{lab}} = \left(\frac{500}{1.27}\right)^2 = 1.55\times10^5
$$

So $t_{field} = 1.55\times10^{5}\, t_{lab}$.

**Coefficient of consolidation from the lab test**

$$
c_v = \frac{0.196\,(1.27)^2}{t_{lab}} = \frac{0.316}{t_{lab}}\ \text{cm}^2/\text{unit time}
$$

**Field time**

$$
t_{field} = \frac{0.196\,(500)^2}{c_v}
$$

**Illustration:** if the sample reached half its final settlement in 10 minutes, then $t_{field} = 1.55\times10^{6}$ min $\approx 2.95$ years.

**Answer: the building reaches half of its final settlement in $1.55\times10^5$ times the lab time (about 2.95 years for a 10-minute lab test).**

## Check yourself

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

1. For the same clay and same U = 50%, how does t relate to drainage path?
   A. t is proportional to H_dr
   B. t is proportional to H_dr^2
   C. t is proportional to 1/H_dr
   D. t does not depend on H_dr
   Answer: B. Same cv and Tv mean t = Tv H^2/cv scales with the square of H_dr. (Key point: Scaling relation)

2. What are the drainage paths of the field layer and the lab sample?
   A. Field 10 m and lab 2.54 cm
   B. Field 5 m and lab 2.54 cm
   C. Field 5 m (10/2) and lab 1.27 cm (2.54/2)
   D. Field 10 m and lab 1.27 cm
   Answer: C. Both are drained top and bottom, so each path is half the thickness. (Key point: Drainage paths)

3. What is t_field / t_lab?
   A. 500/1.27, about 394
   B. (250/1.27)^2, about 3.9 x 10^4
   C. (500/2.54)^2, about 3.9 x 10^4
   D. (500/1.27)^2, about 1.55 x 10^5
   Answer: D. The ratio of the drainage paths in the same units squared is (500/1.27)^2. (Key point: Time ratio)

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