---
title: "Square footing size in clay, 750 kN"
subject: "Foundation Engineering"
chapter: 6
key: "semester_5/foundation_engineering:square-footing-clay-750kn"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/square-footing-clay-750kn
site: BCT Engineering
---

# Square footing size in clay, 750 kN

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

- Asked in: 2079 Bhadra, 6 marks
- Video: [Numerical on Terzaghi Bearing Capacity, Determination of Safe Bearing Capacity, rectangular footing](https://www.youtube.com/watch?v=4Ac8U75wL00) — Civil Engineering Exam
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/square-footing-clay-750kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-6

## Question

A concrete column has a square footing to carry a column load of 750 kN, founded in a clay deposit below the ground surface (depth not given in the scan). The unit weight and unit cohesion of clay are $17\ \text{kN/m}^3$, $35\ \text{kN/m}^2$ respectively. Determine the dimension of the footing taking factor of safety of 3 and assuming the foundation to be backfilled.

## Short answer

For clay (φ=0), net ultimate capacity is 1.3cNc = 259.35 kN/m². With FS=3, net safe pressure is 86.45 kN/m², requiring B=√(750/86.45)=2.95 m. Provide a 3.0 m × 3.0 m square footing.

## Answer

**Given:** square footing, $Q = 750$ kN, clay with $c = 35$ kN/m², $\gamma = 17$ kN/m³, FS = 3. The depth is not given.

**Assumptions:** the clay is purely cohesive ($\phi = 0$), so Terzaghi's factors are $N_c = 5.7$, $N_q = 1$, $N_\gamma = 0$. Because the footing is backfilled, the weight of footing plus backfill is taken as about equal to the soil it replaces, so the net load intensity on the soil is $Q/B^2$ and the net ultimate capacity is used.

$$
q_u = 1.3cN_c + \gamma D_f N_q \Rightarrow q_{nu} = q_u - \gamma D_f = 1.3cN_c
$$

$$
\begin{aligned} q_{nu} &= 1.3 \times 35 \times 5.7 = 259.35\ \text{kN/m}^2 \\ q_{ns} &= \frac{259.35}{3} = 86.45\ \text{kN/m}^2 \\ B^2 &= \frac{750}{86.45} = 8.68\ \text{m}^2 \Rightarrow B = 2.95\ \text{m} \end{aligned}
$$

Since the depth cancels out ($N_q = 1$), the answer does not depend on $D_f$.

**Answer: provide a square footing of 3.0 m × 3.0 m** (net pressure $750/9 = 83.3$ kN/m² < 86.45 kN/m²; FS $= 259.35/83.3 = 3.11$).

## Check yourself

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

1. Which Terzaghi factors apply to purely cohesive clay?
   A. Nc = 37.2, Nq = 22.5
   B. Nc = 5.7, Nq = 0
   C. Nc = 1, Nq = 5.7
   D. Nc = 5.7, Nq = 1, Ngamma = 0
   Answer: D. With phi = 0 the factors are fixed. (Key point: Factors for clay)

2. Why does the answer not depend on Df?
   A. Backfilled net capacity gives qnu = 1.3cNc since Nq = 1
   B. Df is always zero
   C. Clay has no weight
   D. Ngamma cancels Df
   Answer: A. gamma*Df cancels in the net capacity. (Key point: Depth cancels)

3. What is the footing size?
   A. 2.0 m
   B. 3.0 m square (B^2 = 750/86.45)
   C. 1.5 m
   D. 4.5 m
   Answer: B. B = 2.95 m, rounded to 3.0 m. (Key point: Size)

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