---
title: "Rectangular footing 3 m × 4 m, water table cases"
subject: "Foundation Engineering"
chapter: 6
key: "semester_5/foundation_engineering:rect-footing-3x4-backfilled"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/rect-footing-3x4-backfilled
site: BCT Engineering
---

# Rectangular footing 3 m × 4 m, water table cases

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

- Asked in: 2075 Chaitra, 8 marks
- Video: [Effect Of Water Table On Bearing Capacity Of Soil🔥](https://www.youtube.com/watch?v=vh8t411BdCc) — Chaityendra Singh
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/rect-footing-3x4-backfilled
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-6

## Question

A rectangular footing (3 m × 4 m) is placed at 1.5 m depth in sandy soil having angle of shearing resistance of $34^\circ$ and unit weight of $20\ \text{kN/m}^3$ above water table and saturated unit weight of $21.5\ \text{kN/m}^3$ below water table. Determine the safe load that can be carried by the footing for a safety factor of 3.0 if the excavation is backfilled for the following cases: (i) the water table is at 1 m below the ground level and (ii) the water table is at 1 m below the base of the footing. The bearing capacity factors for strip footing for $\phi = 34^\circ$ are $N_q = 29$ and $N_\gamma = 41$.

## Short answer

For case (i) with water table 0.5 m above base, safe load is 5440 kN. For case (ii) with water table 1 m below base, safe load is 6270 kN. The higher water table reduces safe load by about 13%.

## Answer

**Given:** $B = 3$ m, $L = 4$ m, $D_f = 1.5$ m, $\phi = 34^\circ$, $c = 0$, $\gamma = 20$ kN/m³ (above WT), $\gamma_{sat} = 21.5$ kN/m³, $N_q = 29$, $N_\gamma = 41$, FS = 3, $\gamma_w = 9.81$ kN/m³. As the excavation is backfilled, the overburden pressure $q$ acts in full.

Terzaghi's rectangular footing ($c = 0$): $q_u = qN_q + 0.5\left(1 - 0.2\dfrac{B}{L}\right)\gamma B N_\gamma$, with $1 - 0.2(0.75) = 0.85$. Safe load $= \dfrac{q_u}{3}\times BL$ (gross).

#### (i) WT at 1 m below GL (0.5 m above the base)
Sand below the base is submerged: $\gamma' = 21.5 - 9.81 = 11.69$ kN/m³.

$$
\begin{aligned} q &= 20(1.0) + 11.69(0.5) = 25.85\ \text{kN/m}^2 \\ q_u &= 25.85 \times 29 + 0.5(0.85)(11.69)(3)(41) \\ &= 749.5 + 611.1 = 1360.6\ \text{kN/m}^2 \\ Q_{safe} &= \frac{1360.6}{3}\times 12 = 5442\ \text{kN} \end{aligned}
$$

#### (ii) WT at 1 m below the base
$Z_{w2} = 1$ m $< B = 3$ m, so $R_{w2} = 0.5(1 + 1/3) = 0.667$. The overburden is dry: $q = 20 \times 1.5 = 30$ kN/m².

$$
\begin{aligned} q_u &= 30 \times 29 + 0.5(0.85)(20)(3)(41)(0.667) \\ &= 870.0 + 1045.5 \times 0.667 = 870.0 + 697.0 \\ &= 1567.0\ \text{kN/m}^2 \\ Q_{safe} &= \frac{1567.0}{3}\times 12 = 6268\ \text{kN} \end{aligned}
$$

**Answer: (i) safe load $\approx 5440$ kN; (ii) safe load $\approx 6270$ kN.** The higher water table reduces the safe load by about 13%.

## Check yourself

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

1. WT 0.5 m above base: q =
   A. 20*1.5 = 30
   B. 20(1.0)+11.69(0.5) = 25.85
   C. 11.69*1.5
   D. 21.5*1.5
   Answer: B. Submerged portion uses gamma'. (Key point: Case (i))

2. WT 1 m below base, B = 3 m: Rw2 =
   A. 0.667
   B. 1
   C. 0.5
   D. 0.833
   Answer: A. 0.5(1+1/3). (Key point: Case (ii))

3. Higher WT effect:
   A. Raises it
   B. None
   C. Doubles it
   D. Reduces safe load (about 13%)
   Answer: D. Buoyancy reduces q and the B term. (Key point: Result trend)

## Related in Foundation Engineering

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