---
title: "Active thrust with surcharge and water table"
subject: "Foundation Engineering"
chapter: 3
key: "semester_5/foundation_engineering:wall-6m-surcharge-water-table"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/wall-6m-surcharge-water-table
site: BCT Engineering
---

# Active thrust with surcharge and water table

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

- Asked in: 2079 Bhadra, 9 marks
- Video: [Rankine's Active Earth Pressure Distribution on Three Layered Soil with Water Table and Surcharge](https://www.youtube.com/watch?v=vyVz6HhquuI) — Master Civil Engineering
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/wall-6m-surcharge-water-table
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-3

## Question

A 6 m high retaining wall having a vertical back has horizontal cohesion-less backfill having $\gamma = 17\ \text{kN/m}^3$, $\phi = 30^\circ$ and carrying a uniform surcharge of $25\ \text{kN/m}^2$. It has a water table at a depth of 3.5 m from the base of the retaining wall (as printed). Determine the magnitude and direction of the total active thrust. Take $\gamma_{sat} = 21\ \text{kN/m}^3$.

## Short answer

Rankine Ka = 0.333; total active thrust is 152 kN/m horizontal (perpendicular to the back), applied 2.33 m above the base.

## Answer

**Reading of the data:** the water table is stated "at 3.5 m from the base of the retaining wall (as printed)", i.e. it lies below the base of the 6 m wall. The backfill within the wall height is then above the water table and $\gamma = 17$ kN/m$^3$ is used; $\gamma_{sat}$ does not enter. (An alternative reading, with the water table 3.5 m below the top, is given at the end.)

Smooth vertical back and horizontal backfill, so Rankine applies and the thrust is **horizontal**.
$$
K_a = \tan^2(45^\circ - 15^\circ) = 0.3333
$$

**Pressures (kN/m$^2$)**
- Top: $K_a q = 0.3333\times25 = 8.33$
- Base: $K_a(q + \gamma H) = 0.3333\times(25 + 17\times6) = 42.33$

**Thrust**
| Part | Force (kN/m) | Arm above base (m) |
|---|---|---|
| Surcharge (rectangle) $K_a q H = 0.3333\times25\times6$ | 50.00 | 3.00 |
| Soil (triangle) $\tfrac12 K_a\gamma H^2$ | 102.00 | 2.00 |
| **Total** | **152.00** | **2.33** |

**Answer: total active thrust $P_a = 152$ kN/m, horizontal (perpendicular to the wall back), acting 2.33 m above the base.**

**Alternative reading (water table 3.5 m below the top, $\gamma_{sat}=21$, $\gamma'=11.19$):** pressures are 8.33 (top), 28.17 (at 3.5 m), 37.49 (base, effective) plus water 0 to 24.53. The total thrust is then 176.6 kN/m (soil 145.9 + water 30.7), acting 2.12 m above the base.

## Check yourself

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

1. Why is the total active thrust horizontal here?
   A. Smooth vertical back and horizontal backfill, so Rankine applies
   B. Because of the surcharge
   C. Because the water table is present
   D. Because phi = 30 degrees
   Answer: A. Rankine with a smooth vertical back and horizontal surface gives pressure perpendicular to the wall. (Key point: Direction of thrust)

2. Why does gamma_sat = 21 not enter the main solution?
   A. Surcharge cancels it
   B. The water table is below the base of the wall, so the backfill is above it
   C. gamma_sat is always ignored
   D. Ka is 1
   Answer: B. Water at 3.5 m below the base leaves the whole wall height unsaturated, so gamma = 17 is used. (Key point: Reading of water table)

3. How does the uniform surcharge q enter the thrust?
   A. As a triangle with arm H/3
   B. As a rectangle of pressure Ka q over H with arm H/2
   C. As an extra Ka on gamma
   D. It only affects vertical forces
   Answer: B. Surcharge gives constant pressure Ka q = 8.33, force 50 kN/m acting at 3 m. (Key point: Surcharge contribution)

4. What is the total thrust and its height above the base?
   A. 152 kN/m at 2.33 m
   B. 152 kN/m at 2.00 m
   C. 102 kN/m at 2.33 m
   D. 176.6 kN/m at 3.00 m
   Answer: A. 50 + 102 = 152 kN/m; moments (150 + 204)/152 = 2.33 m. (Key point: Total thrust and location)

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