---
title: "Settlement 4 cm in 2 months, time for 10 cm"
subject: "Soil Mechanics"
chapter: 9
key: "semester_4/soil_mechanics:settlement-time-ratio-months"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/settlement-time-ratio-months
site: BCT Engineering
---

# Settlement 4 cm in 2 months, time for 10 cm

Soil Mechanics · Chapter 9: Compressibility of Soil

- Asked in: 2078 Chaitra, 3 marks
- Video: [Finding Primary Consolidation Settlement, Degree of Consolidation and Settlement at any time](https://www.youtube.com/watch?v=Ule5gdMiWhs) — Master Civil Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/settlement-time-ratio-months
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-9

## Question

A compressible layer whose total settlement under a given loading is expected to be 20 cm settles 4 cm at the end of 2 months. How many months will be required to reach a settlement of 10 cm? Assume double drainage.

## Short answer

For U < 60%, time is proportional to U². Degrees of consolidation are U1=20%, U2=50%. Required time = 2 months × (50/20)² = 12.5 months.

## Answer

Settlement is proportional to the degree of consolidation $U$, and time is proportional to $T_v$. For $U < 60\%$, $T_v = \frac{\pi}{4}U^2$, so $t \propto U^2$ (same layer, same $c_v$ and drainage).

**Step 1: Degrees of consolidation**

$$
U_1 = \frac{4}{20} = 20\%, \qquad U_2 = \frac{10}{20} = 50\%
$$

Both are below 60%, so the parabolic relation applies.

**Step 2: Time ratio**

$$
\frac{t_2}{t_1} = \left(\frac{U_2}{U_1}\right)^2 = \left(\frac{50}{20}\right)^2 = 6.25
$$

$$
t_2 = 6.25 \times 2 = 12.5\ \text{months}
$$

**Answer: about 12.5 months from the start of loading.** The double drainage condition does not change the ratio, because $H_{dr}$ is the same in both cases.

## Check yourself

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

1. How is the degree of consolidation found from settlement?
   A. U = settlement / (1+e0)
   B. U = total settlement / settlement at time t
   C. U = settlement x time
   D. U = settlement at time t / total settlement
   Answer: D. Settlement is proportional to U, so 4/20 = 20% and 10/20 = 50%. (Key point: Settlement and U)

2. For U < 60%, how is time related to U?
   A. t is proportional to U
   B. t is proportional to U^2 (Tv = pi U^2/4)
   C. t is proportional to sqrt(U)
   D. t is proportional to log U
   Answer: B. In the early part of consolidation Tv = (pi/4) U^2, so time varies with the square of U. (Key point: Relation for U below 60%)

3. What is the time to reach 10 cm settlement?
   A. 5 months
   B. 6.25 months
   C. 12.5 months
   D. 10 months
   Answer: C. t2 = 2 x (50/20)^2 = 12.5 months from the start of loading. (Key point: Time required)

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