---
title: "Water tower on three columns"
subject: "Soil Mechanics"
chapter: 8
key: "semester_4/soil_mechanics:water-tower-three-columns-stress"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/water-tower-three-columns-stress
site: BCT Engineering
---

# Water tower on three columns

Soil Mechanics · Chapter 8: Vertical Stresses Below Applied Loads

- Asked in: 2078 Baisakh, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/water-tower-three-columns-stress
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-8

## Question

A water tower ($10^6$ kN including foundation) is supported by three columns in a triangular pattern (each side 10 m long). Calculate the stress 5 m below the foundation level at the center of the water tank and at each footing.

## Short answer

Each column carries 3.33×105 kN. Using Boussinesq at depth 5 m, centre distance r = 5.774 m yields Δσz ≈ 2296 kPa at the centre; below a footing, own column (r = 0) plus two at r = 10 m gives Δσz ≈ 6594 kPa.

## Answer

#### Assumptions

Each column and footing transmits one-third of the total load as a point load: $P = 10^6/3 = 3.333\times10^5$ kN (as printed, $10^6$ kN). The columns are at the corners of an equilateral triangle of side 10 m. Boussinesq's point load equation is used, with $z = 5$ m:

$$
\Delta\sigma_z = \frac{3P}{2\pi z^2}\left[\frac{1}{1+(r/z)^2}\right]^{5/2}
$$

```text
        C1
       /  \
   10 m    10 m          centre O at 5.774 m from
     /  O   \            each footing
   C2 ------- C3
        10 m
```

Distance from the centre O to each column: $r = \dfrac{10}{\sqrt3} = 5.774$ m.

#### (a) At the centre of the tank (z = 5 m)

$r/z = 1.155$, $1 + (r/z)^2 = 2.333$

For one column: $\Delta\sigma = \dfrac{3\times333333}{2\pi\times25}\,(2.333)^{-2.5} = 6366.2\times0.1202 = 765.5$ kPa

For three columns:

$$
\Delta\sigma_z = 3\times765.5 = 2296\ \text{kPa}
$$

#### (b) Below each footing (z = 5 m)

- Own column ($r = 0$): $\dfrac{3P}{2\pi z^2}\times1 = 6366.2$ kPa
- Each of the other two columns ($r = 10$ m, $r/z = 2$, $(1+4)^{-2.5} = 0.01789$): $6366.2\times0.01789 = 113.9$ kPa

$$
\Delta\sigma_z = 6366.2 + 2\times113.9 = 6594\ \text{kPa}
$$

**Answer: $\Delta\sigma_z \approx 2296$ kPa at the centre and $\approx 6594$ kPa below each footing, 5 m below the foundation level.**

The values are very large because the load is taken as $10^6$ kN; for $10^6$ N the stresses are 1000 times smaller (2.30 kPa and 6.59 kPa).

## Check yourself

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

1. How is the load treated for the three-column water tower?
   A. The whole load acts at the centre
   B. Each column carries one-third of the load as a point load
   C. The load is spread as a uniform area load
   D. Only one column carries the load
   Answer: B. Each footing is taken as a point load P = total load/3, then Boussinesq's point-load equation is used. (Key point: Load per column)

2. What is the horizontal distance from the centroid of an equilateral triangle of side 10 m to each column?
   A. 5 m
   B. 10 m
   C. 8.66 m
   D. 10/sqrt(3) = 5.774 m
   Answer: D. The centroid-to-vertex distance of an equilateral triangle of side a is a/sqrt(3) = 5.774 m. (Key point: Distance to centre)

3. How is the stress at the tank centre at z = 5 m found?
   A. Compute the stress from one column and multiply by 3 since all three are at r = 5.774 m
   B. Use only the nearest column
   C. Add the stresses with r = 0 for all three
   D. Divide the single-column stress by 3
   Answer: A. All columns are the same distance from the centre, so the effects add: 3 x 765.5 = 2296 kPa. (Key point: Stress at centre)

4. Below one footing at z = 5 m, the total stress is found by
   A. only the own column (r = 0) contribution
   B. own column (r = 0) plus the contributions of the other two columns at r = 10 m
   C. three times the own-column value
   D. own column minus the other two
   Answer: B. Superposition: 6366.2 + 2 x 113.9 = 6594 kPa. (Key point: Stress below a footing)

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