---
title: "Square footing under eccentric load, min depth"
subject: "Foundation Engineering"
chapter: 6
key: "semester_5/foundation_engineering:square-footing-eccentric-2000kn"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/square-footing-eccentric-2000kn
site: BCT Engineering
---

# Square footing under eccentric load, min depth

Foundation Engineering · Chapter 6: Bearing Capacity and Settlement of Shallow Foundations

- Asked in: 2080 Baisakh, 8 marks
- Video: [Minimum Depth of Footing in Difference Ground Level - Depth of Foundation in Rock](https://www.youtube.com/watch?v=qXEYf2jV_Hs) — Engr Sami 
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/square-footing-eccentric-2000kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-6

## Question

A 4 m × 4 m square footing in plan is founded in a soil having angle of internal friction $\phi = 32^\circ$, cohesion $c = 35\ \text{kN/m}^2$ and unit weight $= 18\ \text{kN/m}^3$. The load of 2000 kN from the superstructure acts at an eccentricity of 0.3 m at one direction from the geometrical centre of the footing. What should be the minimum depth of footing to avoid shear failure in soil at the factor of safety of 2.5? Take $N_c = 55$, $N_q = 40$, $N_\gamma = 45$.

## Short answer

Effective width B'=3.4 m, required qu=367.6 kN/m². At Df=0, qu=3604.1 kN/m² gives FS=24.5 > 2.5, so shear failure does not control. Minimum depth is governed by practical considerations (e.g., 1 m), giving FS=29.4.

## Answer

**Given:** $B = L = 4$ m, $\phi = 32^\circ$, $c = 35$ kN/m², $\gamma = 18$ kN/m³, $Q = 2000$ kN, $e = 0.3$ m in one direction, FS = 2.5, $N_c = 55$, $N_q = 40$, $N_\gamma = 45$.

**Method:** use the effective width $B' = B - 2e$ (Meyerhof) and Terzaghi's square footing equation
$q_u = 1.3cN_c + \gamma D_f N_q + 0.4\gamma B' N_\gamma$. The load must satisfy $\dfrac{q_u}{FS} A' \ge Q$.

$$
\begin{aligned} B' &= 4 - 2(0.3) = 3.4\ \text{m},\quad A' = 3.4 \times 4 = 13.6\ \text{m}^2 \\ \text{Required } q_u &= \frac{2.5 \times 2000}{13.6} = 367.6\ \text{kN/m}^2 \end{aligned}
$$

Terms of $q_u$ (independent of depth):

$$
\begin{aligned} 1.3cN_c &= 1.3 \times 35 \times 55 = 2502.5\ \text{kN/m}^2 \\ 0.4\gamma B' N_\gamma &= 0.4 \times 18 \times 3.4 \times 45 = 1101.6\ \text{kN/m}^2 \end{aligned}
$$

So even at the ground surface ($D_f = 0$): $q_u = 2502.5 + 1101.6 = 3604.1\ \text{kN/m}^2$, which is far more than the required $367.6$ kN/m². The factor of safety at $D_f = 0$ is $3604.1 \times 13.6/2000 = 24.5$.

| $D_f$ (m) | $q_u$ (kN/m²) | FS $= q_uA'/Q$ |
|---|---|---|
| 0 | 3604.1 | 24.5 |
| 0.5 | 3964.1 | 27.0 |
| 1.0 | 4324.1 | 29.4 |

**Answer: from the shear failure condition, no minimum depth is required (the requirement is met even at $D_f \approx 0$, FS = 24.5 against 2.5).** The depth is therefore decided by other needs: a minimum of about 1.0 m below the ground (to get below topsoil, frost and shrinkage zones) is adopted in practice, which gives FS = 29.4.

*Note: with the data given, the load is small compared with the capacity of the soil, so shear failure does not control the depth.*

## Check yourself

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

1. How is effective width found for eccentricity e = 0.3 m, B = 4 m?
   A. B - e = 3.7 m
   B. B' = B - 2e = 3.4 m
   C. B + 2e = 4.6 m
   D. B/2 = 2 m
   Answer: B. Meyerhof's effective width. (Key point: Effective width)

2. Required qu uses which area?
   A. Effective area A' = 13.6 m2, giving 367.6 kN/m2
   B. Full area 16 m2
   C. B'^2
   D. Area 4 m2
   Answer: A. qu = FS*Q/A'. (Key point: Required qu)

3. What does the calculation show about minimum depth?
   A. Depth must be 6 m
   B. Depth must be 2 m
   C. Shear failure does not control; capacity is far above need even at Df = 0
   D. Depth equals B
   Answer: C. FS at Df = 0 is already 24.5. (Key point: Conclusion)

## Related in Foundation Engineering

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