---
title: "Footing 2 m × 3 m, chart factors, eccentricity"
subject: "Foundation Engineering"
chapter: 6
key: "semester_5/foundation_engineering:footing-2x3-eccentric-skempton"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/footing-2x3-eccentric-skempton
site: BCT Engineering
---

# Footing 2 m × 3 m, chart factors, eccentricity

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

- Asked in: 2078 Kartik, 8 marks
- Video: [Eccentric Footing : All you Need to know](https://www.youtube.com/watch?v=g9tIRmWi6GE) — Civil Engineer Tonmoy Maity
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/footing-2x3-eccentric-skempton
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-6

## Question

A footing 2 m × 3 m in plan which is to be laid at a depth of 1.5 m below ground surface to carry a column load having one way eccentricity of 0.26 m along the width. Determine the safe bearing capacity if the water table is 0.5 m below the ground level. Use Terzaghi's theory. Take $c = 20\ \text{kN/m}^2$, $\phi = 35^\circ$ (soil: $\gamma = 18\ \text{kN/m}^3$, saturated clayey soil) to interpolate the bearing capacity factors from the figure presented. [Figure: chart of $N_c$, $N_q$, $N_\gamma$ versus $\phi$ (degrees).]

## Short answer

Using effective width B'=1.48 m, Terzaghi's factors for φ=35°, and submerged unit weight, ultimate capacity is 2270 kN/m². With FS=3, safe bearing capacity is 768 kN/m², giving a safe load of about 3410 kN.

## Answer

**Given:** $B = 2$ m, $L = 3$ m, $D_f = 1.5$ m, $e = 0.26$ m along the width, $c = 20$ kN/m², $\phi = 35^\circ$, $\gamma = 18$ kN/m³ (taken as the saturated unit weight), water table 0.5 m below GL, Terzaghi's theory.

**Bearing capacity factors for $\phi = 35^\circ$ (Terzaghi chart, general shear):** $N_c = 57.8$, $N_q = 41.4$, $N_\gamma = 42.4$. Factor of safety assumed $F = 3$, $\gamma_w = 9.81$ kN/m³.

#### Effective dimensions
$B' = B - 2e = 2 - 2(0.26) = 1.48$ m; $L = 3$ m.

#### Overburden and submerged weight
The water table is above the base, so the soil below the base is submerged: $\gamma' = 18 - 9.81 = 8.19$ kN/m³.

$$
q = 18 \times 0.5 + 8.19 \times 1.0 = 17.19\ \text{kN/m}^2
$$

#### Ultimate bearing capacity (rectangular footing)
$$
\begin{aligned} q_u &= \left(1 + 0.3\tfrac{B'}{L}\right)cN_c + qN_q + 0.5\left(1 - 0.2\tfrac{B'}{L}\right)\gamma' B' N_\gamma \\ &= (1.148)(20)(57.8) + 17.19 \times 41.4 + 0.5(0.9013)(8.19)(1.48)(42.4) \\ &= 1327.1 + 711.7 + 231.6 \\ &= 2270.4\ \text{kN/m}^2 \end{aligned}
$$

#### Safe bearing capacity
$$
\begin{aligned} q_{nu} &= q_u - q = 2270.4 - 17.19 = 2253.2\ \text{kN/m}^2 \\ q_{ns} &= \frac{q_{nu}}{3} + q = 751.1 + 17.19 = 768.2\ \text{kN/m}^2 \end{aligned}
$$

Safe load on the footing $= q_{ns} \times B'L = 768.2 \times 1.48 \times 3 = 3411\ \text{kN}$.

**Answer: safe bearing capacity $q_s \approx 768$ kN/m² (net ultimate 2253 kN/m², FS = 3), corresponding to a safe load of about 3410 kN.**

## Check yourself

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

1. What is B' for e = 0.26 m, B = 2 m?
   A. 1.48 m
   B. 1.74 m
   C. 2.52 m
   D. 1.0 m
   Answer: A. B' = 2 - 2(0.26). (Key point: Effective width)

2. How is q computed with WT 0.5 m below GL and Df = 1.5 m?
   A. 18*1.5 = 27
   B. 18*0.5 + 8.19*1.0 = 17.19
   C. 8.19*1.5 = 12.3
   D. 9.81*1.5
   Answer: B. Above WT use gamma, below use gamma'. (Key point: Submerged overburden)

3. Which cohesion term is used?
   A. 1.3 c Nc
   B. c Nc
   C. (1 + 0.3B'/L) c Nc
   D. (1 - 0.2B'/L) c Nc
   Answer: C. The rectangular shape factor for c is 1 + 0.3B/L. (Key point: Rectangular formula)

4. How is safe bearing capacity found?
   A. qu/3 - q
   B. qu*3
   C. qnu/3 only
   D. qns = qnu/3 + q
   Answer: D. Net ultimate divided by FS plus overburden. (Key point: Safe capacity)

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