---
title: "Strip footing, change in capacity with water rise"
subject: "Foundation Engineering"
chapter: 6
key: "semester_5/foundation_engineering:strip-footing-dry-sand-water-rise"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/strip-footing-dry-sand-water-rise
site: BCT Engineering
---

# Strip footing, change in capacity with water rise

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

- Asked in: 2082 Bhadra, 8 marks; 2080 Bhadra, 6 marks (asked 2 times)
- Video: [Effect on UBC Equation if water table rises to Ground surface | Bearing Capacity | GATE CE](https://www.youtube.com/watch?v=dTM1_LiPhbA) — Ekeeda GATE & ESE
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/strip-footing-dry-sand-water-rise
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-6

## Question

A strip footing of 1.5 m width with its base at a depth of 1.2 m is resting on dry sand. Find the change in ultimate bearing capacity of footing if the water rises up to the depth 0.5 m below the ground level. Take $N_c = 95.7$, $N_q = 81.3$, $N_\gamma = 100.4$, $G = 2.70$ and dry unit weight of sand as $16\ \text{kN/m}^3$.

## Short answer

For dry sand, ultimate bearing capacity qu1 = γ Df Nq + 0.5 γ B Nγ = 2765.8 kN/m². With water table at 0.5 m depth, compute saturated unit weight γsat = 19.88 kN/m³ and submerged γ' = 10.07 kN/m³. Then qu2 = (16×0.5 + 10.07×0.7) × 81.3 + 0.5 × 10.07 × 1.5 × 100.4 = 1982.3 kN/m². The capacity reduces by 783.5 kN/m².

## Answer

**Given:** strip footing, $B = 1.5$ m, $D_f = 1.2$ m, $c = 0$, $N_q = 81.3$, $N_\gamma = 100.4$, $G = 2.70$, $\gamma_d = 16$ kN/m³, $\gamma_w = 9.81$ kN/m³. For a strip footing $q_u = \gamma D_f N_q + 0.5\gamma B N_\gamma$ (the $cN_c$ term is zero).

#### Case 1: dry sand
$$
\begin{aligned} q &= 16 \times 1.2 = 19.2\ \text{kN/m}^2 \\ q_{u1} &= 19.2 \times 81.3 + 0.5 \times 16 \times 1.5 \times 100.4 \\ &= 1560.96 + 1204.8 = 2765.76\ \text{kN/m}^2 \end{aligned}
$$

#### Case 2: water table at 0.5 m below ground level
The sand below the water table is saturated. Find its saturated unit weight from $\gamma_d = \dfrac{G\gamma_w}{1+e}$:

$$
\begin{aligned} 1 + e &= \frac{2.70 \times 9.81}{16} = 1.655 \Rightarrow e = 0.655 \\ \gamma_{sat} &= \frac{(G+e)\gamma_w}{1+e} = \frac{(2.70 + 0.655) \times 9.81}{1.655} = 19.88\ \text{kN/m}^3 \\ \gamma' &= 19.88 - 9.81 = 10.07\ \text{kN/m}^3 \end{aligned}
$$

The water table is above the base, so the soil from 0.5 m to 1.2 m is submerged (0.7 m thick), and the soil below the base is also submerged, so $\gamma'$ is used in the $N_\gamma$ term:

$$
\begin{aligned} q &= 16 \times 0.5 + 10.07 \times 0.7 = 15.05\ \text{kN/m}^2 \\ q_{u2} &= 15.05 \times 81.3 + 0.5 \times 10.07 \times 1.5 \times 100.4 \\ &= 1223.7 + 758.6 = 1982.3\ \text{kN/m}^2 \end{aligned}
$$

#### Change in capacity
$$
\Delta q_u = 2765.76 - 1982.29 = 783.5\ \text{kN/m}^2
$$

**Answer: the ultimate bearing capacity falls from 2765.8 kN/m² to 1982.3 kN/m², a reduction of about 783 kN/m² (28.3%).**

## Check yourself

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

1. Which equation is used for c = 0 strip footing?
   A. q_u = gamma D_f N_q + 0.5 gamma B N_gamma
   B. q_u = cN_c + gamma D_f
   C. q_u = 1.3cN_c + ...
   D. q_u = 0.5 gamma B N_gamma only
   Answer: A. The cN_c term vanishes for c = 0, leaving the surcharge and width terms. (Key point: Strip footing equation)

2. What is q_u for dry sand?
   A. 1982 kN/m^2
   B. 1560 kN/m^2
   C. 2765.8 kN/m^2
   D. 1204.8 kN/m^2
   Answer: C. 1560.96 + 1204.8 = 2765.76 kN/m^2. (Key point: Dry capacity)

3. How is gamma' obtained for the submerged sand?
   A. gamma_d - gamma_w
   B. From e = G gamma_w/gamma_d - 1, then gamma_sat = (G+e)gamma_w/(1+e) and gamma' = gamma_sat - gamma_w
   C. gamma_sat = G gamma_w
   D. gamma' = 16
   Answer: B. e = 0.655, gamma_sat = 19.88, gamma' = 10.07 kN/m^3. (Key point: Saturated unit weight)

4. What is the surcharge q with water at 0.5 m?
   A. 19.2 kN/m^2
   B. 15.05 kN/m^2 = 16x0.5 + 10.07x0.7
   C. 10.07 kN/m^2
   D. 8 kN/m^2
   Answer: B. Top 0.5 m is at 16, the 0.7 m below the water is submerged at 10.07. (Key point: Surcharge with WT)

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