---
title: "Mat 8 m × 10 m, Skempton net safe pressure"
subject: "Foundation Engineering"
chapter: 7
key: "semester_5/foundation_engineering:mat-8x10-skempton"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/mat-8x10-skempton
site: BCT Engineering
---

# Mat 8 m × 10 m, Skempton net safe pressure

Foundation Engineering · Chapter 7: Mat Foundations

- Asked in: 2072 Chaitra, 6 marks
- Video: [Skempton And Meyerhof Ultimate Bearing Capacity Theory🔥 In Hindi](https://www.youtube.com/watch?v=HdF1DFEogcQ) — Chaityendra Singh
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/mat-8x10-skempton
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-7

## Question

A mat foundation of size 8 m × 10 m is resting at a depth of 5 m. The foundation is resting on saturated cohesive soil having undrained cohesion of $50\ \text{kPa}$. The soil has unit weight of $19\ \text{kN/m}^3$. Find the net safe bearing capacity using Skempton's method.

## Short answer

For mat 8×10 m, Df=5 m, cu=50 kPa, Df/B=0.625, Nc=6.53, qnu=326.7 kPa. With FS=3, net safe bearing capacity = 108.9 kPa (gross safe 203.9 kPa).

## Answer

**Given:** mat $B = 8$ m, $L = 10$ m, $D_f = 5$ m, saturated clay $c_u = 50$ kPa, $\gamma = 19$ kN/m³.

**Skempton's method** ($\phi = 0$):

$$
\begin{aligned} \frac{D_f}{B} &= \frac{5}{8} = 0.625 < 2.5 \\ N_{c(sq)} &= 6\left(1 + 0.2 \times 0.625\right) = 6.75 \\ N_{c(rect)} &= \left(0.84 + 0.16 \times \frac{8}{10}\right)(6.75) = 0.968 \times 6.75 = 6.53 \\ q_{nu} &= c_uN_c = 50 \times 6.534 = 326.7\ \text{kPa} \end{aligned}
$$

With a factor of safety $F = 3$ (assumed, usual value):

$$
q_{ns} = \frac{q_{nu}}{F} = \frac{326.7}{3} = 108.9\ \text{kPa}
$$

**Answer: net ultimate capacity $= 326.7$ kPa and net safe bearing capacity $\approx 108.9$ kPa** (gross safe pressure $= 108.9 + 19 \times 5 = 203.9$ kPa).

## Check yourself

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

1. For Df/B = 5/8 the square-footing Nc is:
   A. 6.0
   B. 6.75
   C. 9.0
   D. 7.5
   Answer: B. 6(1 + 0.2 x 0.625) = 6.75. (Key point: Nc square)

2. The rectangular correction (0.84 + 0.16 B/L) for 8 x 10 m is:
   A. 0.84
   B. 0.968
   C. 1.0
   D. 1.12
   Answer: B. 0.84 + 0.16 x 0.8 = 0.968, giving Nc = 6.53. (Key point: Shape factor)

3. q_nu = cu Nc equals:
   A. 326.7 kPa
   B. 108.9 kPa
   C. 203.9 kPa
   D. 50 kPa
   Answer: A. 50 x 6.534 = 326.7 kPa. (Key point: Net ultimate)

4. Net safe capacity with F = 3 is:
   A. 326.7 kPa
   B. 108.9 kPa
   C. 203.9 kPa
   D. 95 kPa
   Answer: B. 326.7/3 = 108.9 kPa; adding gamma Df = 95 gives 203.9 as gross safe. (Key point: Net safe)

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