---
title: "Circular well 4.5 m, resisting force"
subject: "Foundation Engineering"
chapter: 9
key: "semester_5/foundation_engineering:circular-well-4-5m-resisting-force"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/circular-well-4-5m-resisting-force
site: BCT Engineering
---

# Circular well 4.5 m, resisting force

Foundation Engineering · Chapter 9: Well Foundations

- Asked in: 2081 Bhadra, 5 marks
- Video: [well foundation/ 2071 sharwan numerical solution / Monika Upadhyay](https://www.youtube.com/watch?v=CBP8-aYFHKI) — Engineer DD
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/circular-well-4-5m-resisting-force
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-9

## Question

A circular well of 4.5 m external diameter and 0.75 m steining thickness is embedded upon a depth of 12 m in a uniform sand deposit. The angle of shearing resistance of sand and submerged unit weight are $30^\circ$ and $1\ \text{t/m}^3$, respectively. The well is subjected to resultant horizontal force of 50 tons and a total moment of 400 t-m at the scour level. Assume the well to be light, compute allowable equivalent resisting force due to earth force. A factor of safety of 2 may be adopted for soil resistance.

## Short answer

By Terzaghi's light well analysis, net passive k = 1.333 t/m³, rotation point depth 9.059 m; allowable equivalent resisting force = 60.3 t.

## Answer

**Given:** circular well, external diameter $B = 4.5$ m, steining 0.75 m, embedment $D = 12$ m in sand, $\phi = 30^\circ$, $\gamma' = 1\ \text{t/m}^3$; $H = 50$ t and $M = 400$ t·m at the scour level; light well; $FS = 2$ for soil resistance.

**Method:** Terzaghi's simplified approach for a light well (net passive resistance $kz$, rotation about a point O at depth $z_0$), as derived for a light well. The width taken is the external diameter.

#### Soil constants
$$
K_p = \tan^2(45^\circ + 15^\circ) = 3.0,\quad K_a = \frac{1}{3} = 0.333
$$

$$
k = \frac{\gamma'(K_p - K_a)}{FS} = \frac{1 \times (3 - 0.333)}{2} = 1.333\ \text{t/m}^3
$$

#### Height of the force
$$
h = \frac{M}{H} = \frac{400}{50} = 8\ \text{m}
$$

#### Depth of rotation point
$$
2z_0^3 + 3h\,z_0^2 - \left(\tfrac{3}{2}hD^2 + D^3\right) = 0 \Rightarrow 2z_0^3 + 24z_0^2 - 3456 = 0 \Rightarrow z_0 = 9.059\ \text{m}
$$

#### Allowable equivalent resisting force
$$
\begin{aligned} H_{all} &= Bk\left(z_0^2 - \frac{D^2}{2}\right) \\ &= 4.5 \times 1.333 \times \left(82.06 - 72\right) = 6.0 \times 10.06 \\ &= 60.3\ \text{t} \end{aligned}
$$

(Check by moment: $\dfrac{Bk}{3h}(D^3 - 2z_0^3) = 60.3$ t.)

**Answer: allowable equivalent resisting force due to earth pressure $\approx 60.3$ t, which is greater than the applied horizontal force of 50 t, so the light well is stable.**

## Check yourself

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

1. For phi = 30 degrees, K_p and K_a are:
   A. 2.0 and 0.5
   B. 1.0 and 1.0
   C. 3.0 and 0.333
   D. 0.333 and 3.0
   Answer: C. K_p = tan^2(60) = 3, K_a = 1/3. (Key point: Soil constants)

2. k = gamma'(K_p - K_a)/FS equals:
   A. 2.667 t/m3
   B. 0.667 t/m3
   C. 3.0 t/m3
   D. 1.333 t/m3
   Answer: D. (3 - 0.333)/2 = 1.333 t/m3. (Key point: k value)

3. The height h of H above scour level is:
   A. H/M
   B. 12 m
   C. 4.5 m
   D. M/H = 8 m
   Answer: D. h = 400/50 = 8 m. (Key point: Height of force)

4. H_all = Bk(z_0^2 - D^2/2) is about 60.3 t, which means:
   A. It is below 50 t, unsafe
   B. It exceeds 50 t, so the well is stable
   C. Equal to the load
   D. Not comparable
   Answer: B. Resistance 60.3 t > applied 50 t. (Key point: Result)

## Related in Foundation Engineering

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- [SPT corrected N-value, submerged sand](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Aspt-corrected-6m-submerged) — Foundation Engineering, chapter 2
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- [Circular footing 1 m dia, 1000 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Acircular-footing-1m-1000kn) — Foundation Engineering, chapter 6
- [Circular footing depth, 2000 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Acircular-footing-depth-2000kn) — Foundation Engineering, chapter 6

## Related in other subjects

- [Uniform sand, water table 2 m, capillary](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Auniform-sand-effective-stress-capillary) — Soil Mechanics, chapter 6
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- [Well foundation, 4.5 m internal diameter](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Awell-foundation-4-5m-p) — Estimating and Costing, chapter 6, asked 2×
- [Well foundation: sinking, brick, sand filling, PCC](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Awell-foundation-g-sand-filling) — Estimating and Costing, chapter 6, asked 2×