---
title: "Footing 2 m × 3 m, eccentric in X and Y"
subject: "Foundation Engineering"
chapter: 6
key: "semester_5/foundation_engineering:footing-2x3-eccentric-xy"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/footing-2x3-eccentric-xy
site: BCT Engineering
---

# Footing 2 m × 3 m, eccentric in X and Y

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

- Asked in: 2078 Bhadra
- Video: [Top 10 Important Points for Eccentric Footing Design | Footing with moment](https://www.youtube.com/watch?v=R1sl_LvxAQg) — Visual Civil
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/footing-2x3-eccentric-xy
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-6

## Question

A footing of 2 m × 3 m in plan is founded 2 m below the ground level in clay having angle of repose $\phi = 35^\circ$, $c = 25\ \text{kPa}$. What will be the allowable load which can be carried by the footing if the load is eccentrically applied with eccentricity along X and Y direction as 0.2 m and 0.3 m respectively? The center of footing in plan is taken as origin. The water table is located 0.8 m below the ground level. Assume soil above water table is dry and saturated unit weight of clay as $16.5\ \text{kN/m}^3$ and $19.6\ \text{kN/m}^3$ respectively. Take $N_c = 57.8$, $N_q = 41.4$ and $N_\gamma = 42.4$. Take FOS as 3.

## Short answer

With double eccentricity, effective area is 1.6 m × 2.4 m. Using Terzaghi's equation with given factors and submerged unit weight, ultimate capacity is 3055 kN/m². Allowable load with FS=3 is about 3910 kN.

## Answer

**Given:** $B = 2$ m (X-direction), $L = 3$ m (Y-direction), $D_f = 2$ m, $\phi = 35^\circ$, $c = 25$ kPa, $e_x = 0.2$ m, $e_y = 0.3$ m, WT 0.8 m below GL, $\gamma = 16.5$ kN/m³ (above WT), $\gamma_{sat} = 19.6$ kN/m³, $N_c = 57.8$, $N_q = 41.4$, $N_\gamma = 42.4$, FOS = 3, $\gamma_w = 9.81$ kN/m³.

**Method:** Meyerhof's effective area for double eccentricity, with Terzaghi's equation for a rectangular footing.

#### Effective dimensions
$$
B' = 2 - 2(0.2) = 1.6\ \text{m},\quad L' = 3 - 2(0.3) = 2.4\ \text{m},\quad A' = 3.84\ \text{m}^2
$$

#### Overburden pressure
The water table is above the base, so $\gamma' = 19.6 - 9.81 = 9.79$ kN/m³.

$$
q = 16.5 \times 0.8 + 9.79 \times 1.2 = 13.2 + 11.75 = 24.95\ \text{kN/m}^2
$$

#### Ultimate bearing capacity
$$
\begin{aligned} q_u &= \left(1 + 0.3\tfrac{B'}{L'}\right)cN_c + qN_q + 0.5\left(1 - 0.2\tfrac{B'}{L'}\right)\gamma' B' N_\gamma \\ &= (1.2)(25)(57.8) + 24.95 \times 41.4 + 0.5(0.8667)(9.79)(1.6)(42.4) \\ &= 1734.0 + 1032.8 + 287.8 \\ &= 3054.6\ \text{kN/m}^2 \end{aligned}
$$

#### Allowable load
$$
Q_{all} = \frac{q_u}{FOS} A' = \frac{3054.6}{3} \times 3.84 = 3909.9\ \text{kN}
$$

**Answer: allowable load $\approx 3910$ kN** (allowable pressure $q_u/3 = 1018$ kN/m² on the effective area; gross basis).

## Check yourself

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

1. How are effective dimensions found?
   A. B' = B - 2ex and L' = L - 2ey
   B. B' = B - ex
   C. B' = B + 2ex
   D. L' = L - ey only
   Answer: A. Meyerhof's effective area. (Key point: Double eccentricity)

2. What area is used for allowable load?
   A. 6 m2
   B. A' = B'*L' = 3.84 m2
   C. B*L' only
   D. 2.4 m2
   Answer: B. Q = (qu/FOS)*A'. (Key point: Allowable load)

3. Why is gamma' used for the soil below WT?
   A. Clay is lighter
   B. FOS = 3
   C. Submerged soil, buoyancy reduces unit weight by gamma_w
   D. e is large
   Answer: C. gamma' = 19.6 - 9.81. (Key point: Overburden)

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