---
title: "Final settlement of clay, +30 kN/m²"
subject: "Soil Mechanics"
chapter: 9
key: "semester_4/soil_mechanics:final-settlement-sand-clay-30kpa"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/final-settlement-sand-clay-30kpa
site: BCT Engineering
---

# Final settlement of clay, +30 kN/m²

Soil Mechanics · Chapter 9: Compressibility of Soil

- Asked in: 2078 Chaitra, 3 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-sand-clay-30kpa
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-9

## Question

Calculate the final settlement of the clay layer as shown in the figure below due to an increase of pressure of 30 kN/m² at mid height of the layer. [Figure: sand layer 4 m thick, γ = 20 kN/m³, above a clay layer 2.5 m thick, γ = 18 kN/m³, e = 1.30, $C_c$ = 0.22.]

## Short answer

Effective stress at mid-height of the clay layer is (20−9.81)×4 + (18−9.81)×1.25 = 51.0 kPa. Final settlement Sc = (0.22×2.5/2.30) log10(81/51) ≈ 0.048 m, i.e., about 48 mm.

## Answer

**Assumptions:** the figure is not available, so the water table is taken at the ground surface and the given $\gamma$ values are saturated unit weights. $\gamma_w = 9.81$ kN/m³. The clay is normally consolidated and settlement is computed at mid-height of the clay (1.25 m below its top).

```text
Ground  ------------------------
  Sand  4 m   gamma = 20 kN/m3
        ------------------------
  Clay  2.5 m gamma = 18, e = 1.30
        Cc = 0.22        <- mid-height
```

**Effective stress at mid-clay**

$$
\begin{aligned}
\sigma_0' &= (20-9.81)(4) + (18-9.81)(1.25) \\
&= 40.76 + 10.24 = 51.0\ \text{kPa}
\end{aligned}
$$

**Settlement**, with $\Delta\sigma = 30$ kPa:

$$
\begin{aligned}
S_c &= \frac{C_c H}{1+e_0}\log_{10}\frac{\sigma_0'+\Delta\sigma}{\sigma_0'} \\
&= \frac{0.22 \times 2.5}{2.30}\log_{10}\frac{81.0}{51.0} \\
&= 0.2391 \times 0.2009 = 0.048\ \text{m}
\end{aligned}
$$

**Answer: final settlement $\approx 48\ \text{mm}$.**

If there were no water table (both layers above it, $\sigma_0' = 80 + 22.5 = 102.5$ kPa), the same formula gives about 26.7 mm.

## Check yourself

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

1. The figure is missing. What does the answer assume so the stresses can be calculated?
   A. No water table, all soil is dry
   B. Water table at the ground surface and the given gammas are saturated unit weights
   C. Water table at the top of the clay
   D. Water table at the bottom of the clay
   Answer: B. The answer takes the water table at the ground surface, so submerged weights (gamma - gamma_w) are used. (Key point: Assumed water table)

2. Which expression gives sigma0' at mid-height of the clay?
   A. (20 - 9.81)(4) + (18 - 9.81)(1.25) = 51.0 kPa
   B. 20(4) + 18(1.25) = 102.5 kPa
   C. (20 - 9.81)(4) + (18 - 9.81)(2.5) = 61.2 kPa
   D. (20 - 9.81)(4) = 40.76 kPa
   Answer: A. Use submerged weights and the half-thickness (1.25 m) of the clay for the mid-height point. (Key point: Effective stress at mid-clay)

3. What is substituted in the settlement equation?
   A. S = [0.22 x 2.5/1.30] log10(81/51)
   B. S = [0.22 x 2.5/2.30] log10(30/51)
   C. S = [0.22 x 2.5/(1 + 1.30)] log10(81/51), about 48 mm
   D. S = [0.22 x 1.25/2.30] log10(81/51)
   Answer: C. The full clay thickness H = 2.5 m and 1+e0 = 2.30 are used with final/initial stress 81/51. (Key point: Settlement formula)

## Related in Soil Mechanics

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