---
title: "Final settlement and settlement at 0.5 year"
subject: "Soil Mechanics"
chapter: 9
key: "semester_4/soil_mechanics:final-settlement-clay-time-half-year"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/final-settlement-clay-time-half-year
site: BCT Engineering
---

# Final settlement and settlement at 0.5 year

Soil Mechanics · Chapter 9: Compressibility of Soil

- Asked in: 2073 Magh, 4 marks
- Video: [Consolidation_Primary Consolidation Settlement](https://www.youtube.com/watch?v=vp_p_jLBGTM) — Dr. Maria Cecilia Marcos
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/final-settlement-clay-time-half-year
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-9

## Question

(Using the soil profile and data of the previous question.) What will be the final settlement of the clay layer after the primary consolidation? Also, determine the settlement of the clay layer after 0.5 year. [For U = 70%, $T_v$ = 0.403; for U = 80%, $T_v$ = 0.569; for U = 90%, $T_v$ = 0.848]

## Short answer

Final primary settlement is 111 mm. After 0.5 year, time factor 0.569 corresponds to 80% consolidation, giving settlement of about 89 mm.

## Answer

**Data from the previous profile:** clay thickness $H = 6$ m (8 m to 14 m), $e_0 = 2.4$, $C_c = 1.2$, $c_v = 3.25\times10^{-7}$ m²/s. At the clay centre: $\sigma_0' = 117$ kPa; final $\sigma' = 117 + 15 = 132$ kPa. The clay is assumed normally consolidated.

#### Final primary settlement

$$
\begin{aligned}
S_c &= \frac{C_c H}{1+e_0}\log_{10}\frac{\sigma_0'+\Delta\sigma}{\sigma_0'} \\
&= \frac{1.2 \times 6}{3.4}\log_{10}\frac{132}{117} = 2.1176 \times 0.05238 = 0.111\ \text{m}
\end{aligned}
$$

$S_c \approx 111$ mm.

#### Settlement after 0.5 year

Sand lies above and below the clay, so drainage is two-way: $H_{dr} = 6/2 = 3$ m.

$$
t = 0.5 \times 365 \times 86400 = 1.577\times10^{7}\ \text{s}
$$

$$
T_v = \frac{c_v t}{H_{dr}^2} = \frac{3.25\times10^{-7}\times1.577\times10^{7}}{3^2} = 0.569
$$

From the given table, $T_v = 0.569$ corresponds to $U = 80\%$.

$$
S_t = U \times S_c = 0.80 \times 111 = 88.8\ \text{mm}
$$

**Answer: final settlement $\approx 111$ mm; settlement after 0.5 year $\approx 89$ mm (U = 80%).**

## Check yourself

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

1. What gives the final primary settlement?
   A. S = [1.2 x 6/2.4] log10(132/117), about 157 mm
   B. S = [1.2 x 6/(1+2.4)] log10(132/117), about 111 mm
   C. S = [1.2 x 3/(1+2.4)] log10(132/117), about 55 mm
   D. S = [1.2 x 6/(1+2.4)] log10(15/117)
   Answer: B. Normally consolidated clay with full thickness 6 m and e0 = 2.4. (Key point: Final settlement)

2. Sand lies above and below the clay. What is H_dr?
   A. 3 m
   B. 6 m
   C. 1.5 m
   D. 12 m
   Answer: A. Two-way drainage halves the 6 m thickness. (Key point: Drainage path)

3. How is Tv found after 0.5 year?
   A. Tv = cv t / H_dr = 0.569
   B. Tv = H_dr^2 / (cv t) = 0.569
   C. Tv = cv t / H_dr^2 = 0.569, which matches U = 80% in the table
   D. Tv = 0.403, U = 70%
   Answer: C. Tv = 3.25e-7 x 1.577e7 / 9 = 0.569. (Key point: Time factor)

4. How is settlement at 0.5 year obtained?
   A. S_t = S_c / U
   B. S_t = S_c x Tv
   C. S_t = 0.80 x 6 m
   D. S_t = U x S_c = 0.80 x 111, about 89 mm
   Answer: D. Settlement at time t is the degree of consolidation times the final settlement. (Key point: Settlement at time t)

## Related in Soil Mechanics

- [Primary settlement, sand over clay, 110 kPa](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aprimary-settlement-sand-clay-110kpa) — 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
- [Derive 1D primary consolidation settlement](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Asettlement-equation-1d-primary) — Soil Mechanics, chapter 9
- [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
- [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
- [Consolidation test, 90% time, k, Cc](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aconsolidation-test-50-90-percent-time) — Soil Mechanics, chapter 9

## Related in other subjects

- [Immediate, primary and secondary settlement](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Aimmediate-consolidation-secondary) — Foundation Engineering, chapter 6
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- [Gumbel 500-year flood and 80% limits, Kamalamai River](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-kamalamai-80cl) — Engineering Hydrology, chapter 6
- [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