---
title: "Soil pressure under rigid mat with column loads"
subject: "Foundation Engineering"
chapter: 7
key: "semester_5/foundation_engineering:mat-rigid-soil-pressure-columns"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/mat-rigid-soil-pressure-columns
site: BCT Engineering
---

# Soil pressure under rigid mat with column loads

Foundation Engineering · Chapter 7: Mat Foundations

- Asked in: 2082 Bhadra, 6 marks
- Video: [Mat foundation numerical | most asked problem | find soil pressure at cornor  2022.2021.2019 PU](https://www.youtube.com/watch?v=wmHV5brmAGo) — Sudip Dhakal
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/mat-rigid-soil-pressure-columns
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-7

## Question

The plan of Mat foundation with columns (loads in kN) is shown in figure. Assuming that the mat is rigid, determine the soil pressure distribution at points A, B, C and D. All the columns are of size 0.6 m by 0.6 m. [Figure: mat 13 m wide (x-direction) by 15 m long (y-direction), x and y axes through the centre, nine columns in three rows. Loads (kN): top row 600, 700, 600; middle row 1100, 1300, 1400; bottom row 700, 800, 600. Points A, B, C, D and column spacings not legible in the scan.]

## Short answer

Total load 7800 kN, centroid eccentricities 0.128 m x, 0.154 m y. Using flexure formula with A=195 m², Ix=3656 m⁴, Iy=2746 m⁴ gives corner pressures qA=35.2, qB=39.9, qC=44.8, qD=40.1 kN/m², all compressive, maximum at C.

## Answer

The spacings and the points $A, B, C, D$ are not legible, so these assumptions are made: the mat is $13\ \text{m}$ (x) $\times$ $15\ \text{m}$ (y); columns are at $x = -5, 0, +5$ m and $y = +6, 0, -6$ m (top row at $y = +6$); and $A, B, C, D$ are the corners $A(-6.5, +7.5)$, $B(+6.5, +7.5)$, $C(+6.5, -7.5)$, $D(-6.5, -7.5)$. The column loads include the mat weight.

#### Total load and centroid
$$
\begin{aligned} Q &= 600+700+600+1100+1300+1400+700+800+600 = 7800\ \text{kN} \\ \sum Qx &= (-5)(600+1100+700) + (5)(600+1400+600) = +1000\ \text{kN·m} \\ \sum Qy &= 6(1900) - 6(2100) = -1200\ \text{kN·m} \\ e_x &= \frac{1000}{7800} = 0.128\ \text{m},\quad e_y = \frac{-1200}{7800} = -0.154\ \text{m} \end{aligned}
$$

So $M_y = Qe_x = 1000$ kN·m and $M_x = Qe_y = -1200$ kN·m.

#### Section properties
$$
\begin{aligned} A &= 13 \times 15 = 195\ \text{m}^2 \\ I_x &= \frac{13 \times 15^3}{12} = 3656.25\ \text{m}^4,\quad I_y = \frac{15 \times 13^3}{12} = 2746.25\ \text{m}^4 \\ \frac{Q}{A} &= \frac{7800}{195} = 40.0\ \text{kN/m}^2 \end{aligned}
$$

#### Soil pressure
$$
q = \frac{Q}{A} + \frac{M_x\,y}{I_x} + \frac{M_y\,x}{I_y} = 40 - 0.3282\,y + 0.3641\,x
$$

| Point | $(x, y)$ (m) | $q$ (kN/m²) |
|---|---|---|
| A | (−6.5, +7.5) | 35.17 |
| B | (+6.5, +7.5) | 39.91 |
| C | (+6.5, −7.5) | 44.83 |
| D | (−6.5, −7.5) | 40.09 |

**Answer: $q_A = 35.2$, $q_B = 39.9$, $q_C = 44.8$, $q_D = 40.1$ kN/m²** (all compressive, so no tension under the mat; maximum at C).

## Check yourself

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

1. What is the total column load Q of the nine columns?
   A. 6800 kN
   B. 7800 kN
   C. 8800 kN
   D. 9000 kN
   Answer: B. 600+700+600+1100+1300+1400+700+800+600 = 7800 kN. (Key point: Total load)

2. How are the moments on the mat obtained?
   A. Mx = sum of Qy and My = sum of Qx about the centre (Q times eccentricity)
   B. From the mat weight only
   C. From the largest column only
   D. They are zero for a rigid mat
   Answer: A. The moment equals the sum of column loads times their distances from the centroidal axes, i.e. Q e. (Key point: Eccentricity)

3. Which expression gives soil pressure at a point (x, y)?
   A. q = Q/A
   B. q = Q/A + Mx y/Ix + My x/Iy
   C. q = Q/(A I)
   D. q = Q A + M
   Answer: B. Rigid mat pressure is the average pressure plus the bending terms in both directions. (Key point: Pressure formula)

## Related in Foundation Engineering

- [Mat 10 m × 15 m with basement, FOS](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Amat-10x15-basement) — Foundation Engineering, chapter 7
- [Conventional design of mat foundation](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Amat-conventional-design) — Foundation Engineering, chapter 7, asked 3×
- [Mat 8 m × 10 m, Skempton net safe pressure](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Amat-8x10-skempton) — Foundation Engineering, chapter 7
- [Compensated foundation and types of mat](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Acompensated-foundation-mat-types) — Foundation Engineering, chapter 7, asked 7×
- [Why design a mat foundation](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Awhy-design-mat) — Foundation Engineering, chapter 7
- [Active pressure on wall, cohesive soil](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Acohesive-wall-pressure-diagram) — Foundation Engineering, chapter 3

## Related in other subjects

- [Moment distribution: frame with cantilever 15 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-50knm-15kn-cantilever) — Theory of Structures II, chapter 3
- [Rectangular Sal column, 150 kN + 15 kNm](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atimber-column-sal-4m-150kn-15knm) — Design of Steel and Timber Structure, chapter 11
- [Process to determine wire size](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Awire-size-determination) — Basic Electrical and Electronics Engineering, chapter 4
- [Fillet weld size for bracket plate, 100 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abracket-weld-100kn-working) — Design of Steel and Timber Structure, chapter 4
