---
title: "Ring foundation 3.60 m stress"
subject: "Soil Mechanics"
chapter: 8
key: "semester_4/soil_mechanics:ring-foundation-3-6-m-stress"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/ring-foundation-3-6-m-stress
site: BCT Engineering
---

# Ring foundation 3.60 m stress

Soil Mechanics · Chapter 8: Vertical Stresses Below Applied Loads

- Asked in: 2078 Poush
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/ring-foundation-3-6-m-stress
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-8

## Question

A ring foundation is of 3.60 m external diameter and 2.40 m internal diameter. It transmits a uniform pressure of 135 kN/m². Calculate the vertical stress at a depth of 1.80 m directly beneath the centre of the loaded area.

## Short answer

Vertical stress on axis = q[1/(1+(r/z)²)^(3/2) − 1/(1+(R/z)²)^(3/2)]. With R=1.8 m, r=1.2 m, z=1.8 m, q=135 kN/m²: outer factor=0.6464, inner=0.4240, difference=0.2224, stress=30.0 kN/m².

## Answer

#### Given

Outer $D = 3.60$ m ($R = 1.80$ m), inner $d = 2.40$ m ($r = 1.20$ m), $q = 135$ kN/m², $z = 1.80$ m, point on the axis.

Stress on the axis of a ring = stress of full circle (radius $R$) minus stress of the inner circle (radius $r$):

$$
\Delta\sigma_z = q\left[\frac{1}{\left(1+(r/z)^2\right)^{3/2}} - \frac{1}{\left(1+(R/z)^2\right)^{3/2}}\right]
$$

#### Calculation

- Outer circle: $R/z = 1.0$, $(1+1)^{3/2} = 2.828$, factor $= 1 - 0.3536 = 0.6464$
- Inner circle: $r/z = 0.667$, $(1+0.444)^{3/2} = 1.736$, factor $= 1 - 0.576 = 0.4240$

$$
\Delta\sigma_z = 135\,(0.6464 - 0.4240) = 135\times0.2224 = 30.0\ \text{kN/m}^2
$$

**Answer: Vertical stress at 1.80 m depth below the centre $\approx 30.0$ kN/m².**

## Check yourself

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

1. The vertical stress on the axis of a ring foundation is obtained as
   A. outer circle factor plus inner circle factor
   B. outer circle factor minus inner circle factor, times q
   C. only the outer circle factor
   D. the inner circle factor times q
   Answer: B. Ring = full circle (radius R) minus inner circle (radius r). (Key point: Method)

2. For outer diameter 3.60 m and inner 2.40 m, R and r are
   A. 3.60 m and 2.40 m
   B. 0.9 m and 0.6 m
   C. 1.80 m and 1.20 m
   D. 1.80 m and 2.40 m
   Answer: C. Radii are half the diameters: R = 1.80 m, r = 1.20 m. (Key point: Radii)

3. With q = 135 kN/m2 and z = 1.80 m the stress is about
   A. 30.0 kN/m2
   B. 87.3 kN/m2
   C. 57.2 kN/m2
   D. 135 kN/m2
   Answer: A. 135 x (0.6464 - 0.4240) = 135 x 0.2224 = 30.0 kN/m2. (Key point: Result)

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