---
title: "Rectangular foundation, stress at point P"
subject: "Soil Mechanics"
chapter: 8
key: "semester_4/soil_mechanics:rectangular-foundation-point-p-stress"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/rectangular-foundation-point-p-stress
site: BCT Engineering
---

# Rectangular foundation, stress at point P

Soil Mechanics · Chapter 8: Vertical Stresses Below Applied Loads

- Asked in: 2078 Chaitra, 5 marks
- Video: [Finding Vertical Stress Increase Below a Uniformly Loaded Rectangular Area |Solved Problem](https://www.youtube.com/watch?v=ufqVNGBtDJk) — Master Civil Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/rectangular-foundation-point-p-stress
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-8

## Question

A rectangular foundation 4 m by 5 m carries a uniformly distributed load of 200 kN/m². Determine the vertical stress at a point 'P' as shown in the figure and at a depth of 2.5 m. [Figure: 5 m × 4 m rectangle with P inside, 2 m from the left edge and 3 m from the right edge, and 2 m from each of the top and bottom edges.] Influence factors (m, n):

| m \ n | 0.6 | 0.8 | 1.0 | 2 |
|---|---|---|---|---|
| 0.6 | 0.1069 | 0.1247 | 0.1361 | 0.1533 |
| 0.8 | 0.1247 | 0.1401 | 0.1598 | 0.1812 |
| 1.0 | 0.1361 | 0.1598 | 0.1752 | 0.1999 |
| 2 | 0.1533 | 0.1812 | 0.1999 | 0.2325 |

## Short answer

Divide area into four rectangles with P as corner. For 2×2 m (m=0.8, n=0.8) I=0.1401; for 2×3 m (m=0.8, n=1.2) I=0.1641. Total I=2×0.1401+2×0.1641=0.6084. Stress=200×0.6084=121.7 kN/m² ≈ 122 kN/m².

## Answer

#### Method

Divide the 4 m × 5 m rectangle into four rectangles that have P as a common corner. The stress under a corner of a rectangle is $\Delta\sigma_z = q\,I$ where $I = f(m, n)$, $m = B/z$ and $n = L/z$. Here $z = 2.5$ m and $q = 200$ kN/m².

```text
   2 m       3 m
 +-------+-----------+
 | 2x2   |   3x2     |  2 m
 |   (1) |   (2)     |
 +-------P-----------+
 |   (3) |   (4)     |  2 m
 | 2x2   |   3x2     |
 +-------+-----------+
```

| Rectangle | Size (m) | $m$ | $n$ | $I$ |
|---|---|---|---|---|
| 1 and 3 | 2 × 2 | 0.8 | 0.8 | 0.1401 |
| 2 and 4 | 2 × 3 | 0.8 | 1.2 | 0.1641 |

For $(m,n) = (0.8, 1.2)$ the table has no column for 1.2, so $I$ is interpolated between $n = 1.0$ (0.1598) and $n = 2$ (0.1812): $I = 0.1598 + 0.2\times(0.1812 - 0.1598) = 0.1641$.

#### Stress at P

$$
\Delta\sigma_z = q\,(2\times0.1401 + 2\times0.1641) = 200\times0.6084 = 121.7\ \text{kN/m}^2
$$

**Answer: $\Delta\sigma_z \approx 122$ kN/m² at 2.5 m below P.**

Using the exact formula for the influence factor (which gives $I = 0.1461$ for 0.8 × 0.8 and $0.1684$ for 0.8 × 1.2), the result is about 126 kN/m², so the table-based answer is within about 3%.

## Check yourself

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

1. To find the stress at an interior point P of a rectangle using corner factors, you
   A. use the centre formula
   B. divide the area into four rectangles having P as a common corner
   C. use the 2:1 method
   D. subtract the area from a bigger one
   Answer: B. The stress under a corner of a rectangle is q I, so the area is divided into rectangles sharing P as a corner and the results added. (Key point: Subdivision)

2. For a 2 m x 2 m rectangle at z = 2.5 m, m and n are
   A. 2 and 2
   B. 0.5 and 0.5
   C. 1.25 and 1.25
   D. 0.8 and 0.8
   Answer: D. m = B/z = 2/2.5 = 0.8 and n = L/z = 0.8. (Key point: m and n)

3. For the 3 m x 2 m rectangles, m = 0.8 and n = 1.2 lies between table columns 1.0 and 2, so I is
   A. found by interpolation = 0.1641
   B. taken as 0.1598
   C. taken as 0.1812
   D. taken as 0.1401
   Answer: A. I = 0.1598 + 0.2 x (0.1812 - 0.1598) = 0.1641. (Key point: Interpolation)

4. The stress at P is
   A. q x 0.1401
   B. 200 x 0.6084 = 121.7 kN/m2
   C. 200 x 0.3042
   D. 200 x 0.1641
   Answer: B. Sum is 2(0.1401) + 2(0.1641) = 0.6084, times 200 gives about 122 kN/m2. (Key point: Final stress)

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