---
title: "5 m clay, Cc 0.25, 65% consolidation"
subject: "Soil Mechanics"
chapter: 9
key: "semester_4/soil_mechanics:clay-5m-cc-025-65-percent"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/clay-5m-cc-025-65-percent
site: BCT Engineering
---

# 5 m clay, Cc 0.25, 65% consolidation

Soil Mechanics · Chapter 9: Compressibility of Soil

- Asked in: 2073 Bhadra, 8 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-025-65-percent
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-9

## Question

A 5 m thick saturated soil layer has a compression index of 0.25 and coefficient of permeability $3.2 \times 10^{-3}$ mm/s. If the void ratio is 1.9 at a vertical stress of 0.15 N/mm², calculate 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 65% consolidation.

## Short answer

Void ratio becomes 1.869, settlement 53.9 mm. With cv = 1.514×10^-3 m²/s and Tv=0.340 for 65% consolidation, time is about 1405 s (23 min) assuming double drainage.

## Answer

**Assumptions:** the layer is drained at top and bottom (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.

#### Step 1: Void ratio at 200 kPa

$$
\Delta e = 0.25\log_{10}\frac{200}{150} = 0.0312, \qquad e_2 = 1.9 - 0.0312 = 1.869
$$

#### Step 2: Settlement

$$
S = \frac{\Delta e}{1+e_0}H = \frac{0.0312}{1+1.9}\times 5000 = 53.9\ \text{mm}
$$

#### Step 3: Coefficient of consolidation

$$
m_v = \frac{\Delta e}{(1+e_0)\Delta\sigma'} = \frac{0.0312}{2.9 \times 50} = 2.154\times10^{-4}\ \text{m}^2/\text{kN}
$$

With $k = 3.2\times10^{-3}$ mm/s $= 3.2\times10^{-6}$ m/s:

$$
c_v = \frac{k}{m_v \gamma_w} = \frac{3.2\times10^{-6}}{2.154\times10^{-4}\times9.81} = 1.514\times10^{-3}\ \text{m}^2/\text{s}
$$

#### Step 4: Time for 65% consolidation

Since $U > 60\%$, use

$$
T_v = 1.781 - 0.933\log_{10}(100-65) = 1.781 - 0.933(1.544) = 0.340
$$

$$
t_{65} = \frac{T_v H_{dr}^2}{c_v} = \frac{0.340 \times 2.5^2}{1.514\times10^{-3}} = 1405\ \text{s} \approx 23.4\ \text{min}
$$

**Answer: $e_2 = 1.869$; settlement $= 53.9$ mm; $t_{65} \approx 1405$ s (about 23 min).** For single drainage ($H_{dr}=5$ m) the time would be about 94 min.

## Check yourself

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

1. What is the void ratio after the load increase?
   A. 1.931
   B. 1.869
   C. 1.900
   D. 1.620
   Answer: B. delta e = 0.25 log10(200/150) = 0.0312, so e2 = 1.9 - 0.0312. (Key point: Void ratio at 200 kPa)

2. What conversions are used for the stresses and permeability?
   A. 0.15 N/mm^2 = 15 kPa and 3.2e-3 mm/s = 3.2e-5 m/s
   B. 0.15 N/mm^2 = 1500 kPa and 3.2e-3 mm/s = 3.2e-6 m/s
   C. 0.15 N/mm^2 = 150 kPa and 3.2e-3 mm/s = 3.2e-6 m/s
   D. 0.15 N/mm^2 = 150 kPa and 3.2e-3 mm/s = 3.2e-3 m/s
   Answer: C. 1 N/mm^2 = 1000 kPa and 1 mm = 1e-3 m. (Key point: Unit conversion)

3. Why is the logarithmic Tv formula used for 65%?
   A. Because U is less than 60%
   B. Because the layer is double drained
   C. Because k is in mm/s
   D. Because U is greater than 60%
   Answer: D. The parabolic relation holds only for U below 60%; above it Tv = 1.781 - 0.933 log10(100-U). (Key point: Choice of Tv equation)

## Close variants set in other exams

- [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×

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