---
title: "Active thrust, 7.5 m wall, 5 m sand"
subject: "Foundation Engineering"
chapter: 3
key: "semester_5/foundation_engineering:two-layer-wall-7-5m-5m-sand"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/two-layer-wall-7-5m-5m-sand
site: BCT Engineering
---

# Active thrust, 7.5 m wall, 5 m sand

Foundation Engineering · Chapter 3: Lateral Earth Pressure Theories and Retaining Walls

- Asked in: 2074 Chaitra, 10 marks
- Video: [Numerical on Active Earth Pressure in Retaining Wall using Rankine's Theory](https://www.youtube.com/watch?v=AoGO89lIbVk) — Curious Civil Engineer
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/two-layer-wall-7-5m-5m-sand
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-3

## Question

A retaining wall of 7.5 m high has two layers of backfill. The soil supported consists of 5 m sand ($\gamma = 18\ \text{kN/m}^3$, $\phi = 35^\circ$) overlying saturated clayey soil ($\gamma = 19.5\ \text{kN/m}^3$, $\phi = 35^\circ$, $c = 16\ \text{kN/m}^2$). The ground water table is at the upper surface of the clay. Make a sketch of the distribution of the active pressure on the wall stating the principal values. Calculate the total earth thrust per meter of the wall and its point of application. Assume that the backfill is horizontal at the surface.

## Short answer

Using Rankine theory, active pressures for sand (Ka=0.271) and saturated clay (Ka=0.271, c=16 kN/m²) plus water pressure yield total thrust 119.2 kN/m, acting horizontally at 2.61 m above the base.

## Answer

**Assumptions:** smooth vertical wall, horizontal backfill, Rankine theory; sand above the water table is taken at the given unit weight; the clay is saturated, so the effective stress uses $\gamma' = \gamma_{sat} - \gamma_w = 19.50 - 9.81 = 9.69\ \text{kN/m}^3$.

**Coefficients**
$$
\begin{aligned} K_{a1} &= \tan^2(45^\circ-35^\circ/2) = 0.2710 \\ K_{a2} &= \tan^2(45^\circ-35^\circ/2) = 0.2710,\quad 2c\sqrt{K_{a2}} = 2\times16\times0.5206 = 16.66\ \text{kN/m}^2 \end{aligned}
$$

**Pressures (kN/m$^2$)**
- Sand, top: $0$.
- Sand, bottom ($z=5$ m): $p = K_{a1}\gamma H_1 = 0.2710\times18\times5 = 24.39$.
- Clay, top: $\sigma_v' = 18\times5 = 90.00$, so $p = 0.2710\times90.00 - 16.66 = 7.73$.
- Clay, bottom ($z=7.5$ m): $\sigma_v' = 90.00 + 9.69\times2.5 = 114.22$, so $p = 0.2710\times114.22 - 16.66 = 14.30$.
- Water pressure at the base: $u = 9.81\times2.5 = 24.53$ (zero at the water table).

```text
 0 --|
     |\
sand | \
     |  \
 WT -|---24.39
     |  eff 7.73
clay |   \
     |    \ eff 14.30
 base|-----> + water 24.53
```

**Forces per metre run and lever arms about the base**

| Part | Force (kN/m) | Arm above base (m) | Moment (kN m/m) |
|---|---|---|---|
| Sand triangle | 60.97 | 4.167 | 254.05 |
| Clay, rectangle | 19.33 | 1.250 | 24.16 |
| Clay, triangle | 8.21 | 0.833 | 6.84 |
| Water pressure | 30.66 | 0.833 | 25.55 |
| **Total** | **119.16** | | **310.60** |

$$
\bar y = \frac{\sum M}{\sum F} = \frac{310.60}{119.16} = 2.61\ \text{m above the base}
$$

**Answer: total thrust $P = 119.2$ kN/m (water included), acting horizontally at 2.61 m above the base of the wall.**

## Check yourself

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

1. Pressure at the bottom of the 5 m sand layer is:
   A. 24.39 kN/m2
   B. 7.73 kN/m2
   C. 90 kN/m2
   D. 14.30 kN/m2
   Answer: A. 0.2710 x 18 x 5 = 24.39. (Key point: Sand pressure)

2. Effective pressure at the top of the clay is:
   A. 24.39
   B. 7.73 kN/m2
   C. -16.66
   D. 14.30
   Answer: B. 0.2710 x 90 - 16.66 = 7.73, positive so no tension crack here. (Key point: Clay top pressure)

3. Water pressure at the base (2.5 m below WT) is:
   A. 24.53 kN/m2
   B. 73.6 kN/m2
   C. 9.69 kN/m2
   D. 49 kN/m2
   Answer: A. 9.81 x 2.5 = 24.53. (Key point: Water pressure)

4. Total thrust and point of application are:
   A. 139.5 kN/m, 2.10 m
   B. 133.6 kN/m, 2.39 m
   C. 119.2 kN/m, 2.61 m above base
   D. 119.2 kN/m, 3.75 m
   Answer: C. Sum F = 119.16, sum M = 310.60, ybar = 2.61 m. (Key point: Result)

## Close variants set in other exams

- [Active thrust, 7.5 m wall, 4.5 m sand](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Atwo-layer-wall-7-5m-4-5m-sand) — Foundation Engineering, chapter 3, asked 4×
- [Active thrust, 7.5 m wall, 3 m sand](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Atwo-layer-wall-7-5m-3m-sand) — Foundation Engineering, chapter 3, asked 4×
- [Active thrust, 6 m wall, sand over clay](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Atwo-layer-wall-6m-sand-clay) — Foundation Engineering, chapter 3, asked 4×

## Related in Foundation Engineering

- [Active thrust with surcharge and water table](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Awall-6m-surcharge-water-table) — Foundation Engineering, chapter 3
- [Active pressure, sloping backfill and water table](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Awall-sheet-pipe-sloping-backfill) — Foundation Engineering, chapter 3
- [Active pressure, 8 m wall, table of layers](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Awall-8m-two-layer-table) — Foundation Engineering, chapter 3
- [Stability of gravity wall, two layers](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Astability-gravity-wall-two-layer) — Foundation Engineering, chapter 3
- [Passive thrust on retaining wall back](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Awall-passive-thrust-two-layer) — Foundation Engineering, chapter 3

## Related in other subjects

- [Clay over sand, effective stress at depths](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aclay-sand-profile-effective-stress-depths) — Soil Mechanics, chapter 6
