---
title: "5 m clay, Cc 0.25, 50% consolidation"
subject: "Soil Mechanics"
chapter: 9
key: "semester_4/soil_mechanics:clay-5m-cc-025-50-percent"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/clay-5m-cc-025-50-percent
site: BCT Engineering
---

# 5 m clay, Cc 0.25, 50% consolidation

Soil Mechanics · Chapter 9: Compressibility of Soil

- Asked in: 2076 Baisakh, 5 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-5m-cc-025-50-percent
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-9

## Question

A 5 m thick saturated soil stratum has a compression index of 0.25 and coefficient of permeability $3 \times 10^{-3}$ mm/sec. If the void ratio is 1.9 at a vertical stress of 0.15 N/mm², compute the void ratio when the vertical stress is increased to 0.2 N/mm². Also calculate the settlement due to the above stress increase and the time required for 50% consolidation.

## Short answer

Void ratio decreases from 1.9 to 1.869. Settlement is 53.9 mm. Using cv = 1.42×10^-3 m²/s and Tv=0.197, time for 50% consolidation is about 867 s (14.5 min) with double drainage.

## Answer

**Assumptions:** the layer is sandwiched between permeable layers (double drainage), so $H_{dr} = 2.5$ m; $\gamma_w = 9.81$ kN/m³. $0.15\ \text{N/mm}^2 = 150$ kPa and $0.20\ \text{N/mm}^2 = 200$ kPa.

#### Void ratio at 200 kPa

$$
\Delta e = C_c \log_{10}\frac{\sigma_2'}{\sigma_1'} = 0.25\log_{10}\frac{200}{150} = 0.0312
$$

$$
e_2 = 1.9 - 0.0312 = 1.869
$$

#### Settlement

$$
S = \frac{\Delta e}{1+e_0}H = \frac{0.0312}{2.9}\times 5000 = 53.9\ \text{mm}
$$

#### Time for 50% consolidation

Coefficient of compressibility and volume compressibility:

$$
a_v = \frac{\Delta e}{\Delta\sigma'} = \frac{0.0312}{50} = 6.25\times10^{-4}\ \text{m}^2/\text{kN}, \quad m_v = \frac{a_v}{1+e_0} = 2.154\times10^{-4}\ \text{m}^2/\text{kN}
$$

With $k = 3\times10^{-3}$ mm/s $= 3\times10^{-6}$ m/s:

$$
c_v = \frac{k}{m_v\gamma_w} = \frac{3\times10^{-6}}{2.154\times10^{-4}\times 9.81} = 1.42\times10^{-3}\ \text{m}^2/\text{s}
$$

For $U = 50\%$, $T_v = 0.197$:

$$
t_{50} = \frac{T_v H_{dr}^2}{c_v} = \frac{0.197 \times 2.5^2}{1.42\times10^{-3}} = 867\ \text{s} \approx 14.5\ \text{min}
$$

**Answer: $e_2 \approx 1.869$; settlement $\approx 54$ mm; $t_{50} \approx 867$ s (about 14.5 min).** With single drainage ($H_{dr}=5$ m) the time would be four times longer, about 58 min.

## Check yourself

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

1. How is the void ratio at 200 kPa obtained?
   A. e2 = e1 + C_c log10(200/150) = 1.931
   B. e2 = e1 - C_c (200 - 150) = 1.9 - 12.5
   C. e2 = e1 - C_c log10(200 - 150)
   D. e2 = e1 - C_c log10(200/150) = 1.869
   Answer: D. Void ratio decreases with increasing stress by C_c times log of the stress ratio. (Key point: Void ratio change)

2. Which formula gives settlement from the void ratio change?
   A. S = delta e x H = 0.0312 x 5000
   B. S = [delta e/(1 + e0)] H = 0.0312/2.9 x 5000, about 54 mm
   C. S = [delta e/(1 + e2)] H x 2
   D. S = [delta e/e0] H
   Answer: B. Strain equals delta e over 1 + initial void ratio, multiplied by the layer thickness. (Key point: Settlement)

3. How is c_v obtained from permeability?
   A. c_v = k m_v gamma_w
   B. c_v = k/ a_v only
   C. c_v = k/(m_v gamma_w), with m_v = a_v/(1+e0)
   D. c_v = m_v gamma_w / k
   Answer: C. c_v = k/(m_v gamma_w) links permeability and compressibility. (Key point: Finding c_v)

4. The layer is between permeable layers. Which H_dr gives t50, about 14 min?
   A. H_dr = 5 m
   B. H_dr = 2.5 m
   C. H_dr = 10 m
   D. H_dr = 1.25 m
   Answer: B. Double drainage halves the path to 2.5 m; single drainage (5 m) would take four times longer. (Key point: Drainage path and time)

## Close variants set in other exams

- [5 m clay, Cc 0.25, 65% consolidation](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aclay-5m-cc-025-65-percent) — Soil Mechanics, chapter 9, asked 2×

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