---
title: "Clay over sand, effective stress at depths"
subject: "Soil Mechanics"
chapter: 6
key: "semester_4/soil_mechanics:clay-sand-profile-effective-stress-depths"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/clay-sand-profile-effective-stress-depths
site: BCT Engineering
---

# Clay over sand, effective stress at depths

Soil Mechanics · Chapter 6: Principle of Effective Stress, Capillarity and Permeability

- Asked in: 2078 Baisakh, 4 marks
- Video: [[Fall2020] Chapter 9 In Situ Stresses - Example 4: Effective Stress in Clay Layer](https://www.youtube.com/watch?v=fLCKvBbxT7w) — uSeeGeo
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/clay-sand-profile-effective-stress-depths
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-6

## Question

A soil profile consists of 4 m and 3 m thick clay and sand layers, respectively. The clay layer lies above the sand layer and the ground water table is seen at 2 m depth from the ground surface. Above the water table, there lies a 1 m thick capillary saturated zone. Determine the effective vertical stress at 0 m, 1 m, 2 m, 4 m and 7 m depths from the ground surface. Take the bulk unit weight and saturated unit weight of clay as 20 kN/m³ and 20 kN/m³ respectively. Take the saturated unit weight of sand as 19 kN/m³.

## Short answer

Effective vertical stresses at 0, 1, 2, 4, and 7 m depths are 0, 29.81, 40.00, 60.38, and 87.95 kPa, respectively.

## Answer

#### Given

Clay 0 to 4 m (above), sand 4 to 7 m. WT at 2 m. Capillary saturated zone 1 m thick above the WT (from 1 m to 2 m). Clay: $\gamma = \gamma_{sat} = 20$ kN/m³; sand $\gamma_{sat} = 19$ kN/m³; $\gamma_w = 9.81$ kN/m³. Soil above the capillary zone (0 to 1 m) has $u = 0$.

#### Calculation

| Depth (m) | $\sigma$ (kPa) | $u$ (kPa) | $\sigma' = \sigma - u$ (kPa) |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 1 | $20\times1 = 20$ | $-9.81\times1 = -9.81$ | 29.81 |
| 2 | $20\times2 = 40$ | 0 | 40.00 |
| 4 | $40 + 20\times2 = 80$ | $9.81\times2 = 19.62$ | 60.38 |
| 7 | $80 + 19\times3 = 137$ | $9.81\times5 = 49.05$ | 87.95 |

**Answer: Effective vertical stress at 0, 1, 2, 4 and 7 m = 0, 29.81, 40.00, 60.38 and 87.95 kPa respectively.**

## Check yourself

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

1. At 1 m depth (top of capillary zone), sigma' is:
   A. 20 kPa
   B. 29.81 kPa
   C. 10.19 kPa
   D. 0
   Answer: B. sigma = 20, u = -9.81, so sigma' = 29.81 kPa. (Key point: Top of capillary zone)

2. At the WT (2 m), sigma' is:
   A. 40.00 kPa
   B. 20 kPa
   C. 60.38 kPa
   D. 49.81 kPa
   Answer: A. sigma = 40, u = 0, so sigma' = 40 kPa. (Key point: Water table)

3. At 4 m, sigma' is:
   A. 80 kPa
   B. 19.62 kPa
   C. 100.38 kPa
   D. 60.38 kPa
   Answer: D. sigma = 80, u = 9.81 x 2 = 19.62, sigma' = 60.38 kPa. (Key point: Clay-sand interface)

4. At 7 m, the effective stress is:
   A. 87.95 kPa
   B. 137 kPa
   C. 49.05 kPa
   D. 82.05 kPa
   Answer: A. sigma = 80 + 19 x 3 = 137; u = 9.81 x 5 = 49.05; sigma' = 87.95 kPa. (Key point: Bottom of sand)

## Related in Soil Mechanics

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- [Uniform sand, water table 2 m, capillary](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Auniform-sand-effective-stress-capillary) — Soil Mechanics, chapter 6
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