---
title: "Strip footing FOS, three water table cases"
subject: "Foundation Engineering"
chapter: 6
key: "semester_5/foundation_engineering:strip-footing-fos-water-table"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/strip-footing-fos-water-table
site: BCT Engineering
---

# Strip footing FOS, three water table cases

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

- Asked in: 2081 Baisakh, 8 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/strip-footing-fos-water-table
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-6

## Question

A strip footing 2 m wide carries a safe load intensity of $400\ \text{kN/m}^2$ at a depth of 1.2 m in cohesion less soil. Saturated unit weight and bulk unit weight are $19.5\ \text{kN/m}^3$ and $16.8\ \text{kN/m}^3$ respectively. Determine the value of FOS with respect to shear failure for following conditions of water table. Use $N_c = 58$, $N_q = 41$ and $N_\gamma = 42$. (i) WT at 4 m below GL (ii) WT at 2.5 m below GL (iii) WT at 0.5 m below GL.

## Short answer

Apply Terzaghi’s equation with water table corrections: for WT at 4 m (no effect) FOS=3.83; at 2.5 m (Rw2=0.825) FOS=3.52; at 0.5 m (γ'=9.69 kN/m³) FOS=2.57.

## Answer

**Given:** strip footing $B = 2$ m, $D_f = 1.2$ m, $c = 0$ (cohesionless), $N_q = 41$, $N_\gamma = 42$, applied pressure $q = 400$ kN/m², $\gamma = 16.8$ kN/m³, $\gamma_{sat} = 19.5$ kN/m³, $\gamma_w = 9.81$ kN/m³. FOS $= q_u/q$. Strip footing: $q_u = \gamma D_f N_q + 0.5\gamma B N_\gamma$.

#### (i) WT at 4 m below GL
The water table is $4 - 1.2 = 2.8$ m below the base, greater than $B = 2$ m, so it has no effect.

$$
\begin{aligned} q_u &= 16.8 \times 1.2 \times 41 + 0.5 \times 16.8 \times 2 \times 42 \\ &= 826.6 + 705.6 = 1532.2\ \text{kN/m}^2 \\ FOS &= 1532.2/400 = 3.83 \end{aligned}
$$

#### (ii) WT at 2.5 m below GL
$Z_{w2} = 2.5 - 1.2 = 1.3$ m below the base ($< B$), so $R_{w2} = 0.5(1 + 1.3/2) = 0.825$. The surcharge is unchanged ($R_{w1} = 1$).

$$
\begin{aligned} q_u &= 826.6 + 705.6 \times 0.825 = 826.6 + 582.1 = 1408.7\ \text{kN/m}^2 \\ FOS &= 1408.7/400 = 3.52 \end{aligned}
$$

#### (iii) WT at 0.5 m below GL
The water table is above the base. $\gamma' = 19.5 - 9.81 = 9.69$ kN/m³.

$$
\begin{aligned} q &= 16.8 \times 0.5 + 9.69 \times 0.7 = 15.18\ \text{kN/m}^2 \\ q_u &= 15.18 \times 41 + 0.5 \times 9.69 \times 2 \times 42 \\ &= 622.5 + 407.0 = 1029.5\ \text{kN/m}^2 \\ FOS &= 1029.5/400 = 2.57 \end{aligned}
$$

**Answer: FOS = 3.83 (WT at 4 m), 3.52 (WT at 2.5 m) and 2.57 (WT at 0.5 m).** A rising water table lowers the factor of safety.

## Check yourself

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

1. Water table 2.8 m below base with B = 2 m: effect?
   A. Reduces gamma term by half
   B. Reduces surcharge
   C. Both terms use gamma'
   D. None, since it is deeper than B below the base
   Answer: D. Only a water table within B below the base affects the gamma term. (Key point: WT effect threshold)

2. WT 1.3 m below the base, B = 2 m: Rw2 =
   A. 0.825
   B. 0.65
   C. 1.0
   D. 0.5
   Answer: A. Rw2 = 0.5(1 + Zw/B) = 0.5(1+0.65). (Key point: Rw2)

3. What changes when WT is above the base?
   A. Only the gamma term changes
   B. Surcharge uses gamma' for the submerged part and the gamma term uses gamma'
   C. Nothing changes
   D. Nc changes
   Answer: B. Both q and the B term are reduced. (Key point: WT above base)

4. As the water table rises, FOS:
   A. Increases
   B. Stays 3.83
   C. Decreases (3.83 to 3.52 to 2.57)
   D. Becomes 4
   Answer: C. Buoyancy reduces effective stresses. (Key point: Trend of FOS)

## Related in Foundation Engineering

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