--- title: "Circular well 4 m, light and heavy well" subject: "Foundation Engineering" chapter: 9 key: "semester_5/foundation_engineering:circular-well-4m-light-heavy" source: https://www.bctengineer.com/q/semester_5/foundation_engineering/circular-well-4m-light-heavy site: BCT Engineering --- # Circular well 4 m, light and heavy well Foundation Engineering · Chapter 9: Well Foundations - Asked in: 2075 Chaitra, 6 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/circular-well-4m-light-heavy - Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-9 ## Question What considerations govern the fixing of the depth of a well foundation? A circular well of 4 m internal diameter and 0.75 m steining thickness is embedded up to a depth of 12 m in a uniform sandy deposit. The well is subjected to a resultant horizontal force of 500 kN and a total moment of 4000 kN-m at the scour level. Calculate the allowable total equivalent resisting force due to earth pressure for both light well and heavy well conditions using Terzaghi's analysis. Take saturated unit weight of soil of $20\ \text{kN/m}^3$, $\phi = 30^\circ$ and factor of safety for passive resistance of 2. ## Short answer Light well: Terzaghi's analysis with rotation at depth 9.059 m gives 752 kN. Heavy well, assumed to rotate about base centre, gives 1076 kN. ## Answer #### Considerations governing the depth of a well foundation - **Maximum scour depth** (Lacey's formula) and a **grip length** of at least one-third of it below the maximum scour level. - **Bearing capacity and settlement:** the founding stratum must carry the load safely. - **Lateral stability:** enough embedment so that passive resistance resists the horizontal force and moment. - Depth of the firm stratum or rock, soil layers prone to scour or liquefaction, and ease of sinking. #### Numerical **Given:** internal diameter 4 m, steining 0.75 m, so external diameter $B = 4 + 2(0.75) = 5.5$ m; $D = 12$ m; $H = 500$ kN, $M = 4000$ kN·m at scour level; $\gamma_{sat} = 20$ kN/m³, $\phi = 30^\circ$, $FS = 2$ for passive resistance. $\gamma_w = 9.81$ kN/m³. **Method:** Terzaghi's simplified analysis. The net passive resistance at depth $z$ is $\sigma = kz$. $$ \begin{aligned} \gamma' &= 20 - 9.81 = 10.19\ \text{kN/m}^3,\quad K_p = 3,\ K_a = 0.333 \\ k &= \frac{10.19 \times (3 - 0.333)}{2} = 13.59\ \text{kN/m}^3 \\ h &= \frac{M}{H} = \frac{4000}{500} = 8\ \text{m} \end{aligned} $$ #### (a) Light well The well rotates about a point O at depth $z_0$ below the scour level (the base offers no resistance). $$ 2z_0^3 + 24z_0^2 - 3456 = 0 \Rightarrow z_0 = 9.059\ \text{m} $$ $$ H_{light} = Bk\left(z_0^2 - \frac{D^2}{2}\right) = 5.5 \times 13.59 \times (82.06 - 72) \approx 752\ \text{kN} $$ #### (b) Heavy well A heavy well has enough weight and base friction to resist sliding, so it rotates about the **centre of its base**. The soil pressure $\sigma = kz$ acts on the front over the full depth, and moments are taken about the base: $$ H\,(h + D) = Bk\int_0^D z\,(D - z)\,dz = \frac{BkD^3}{6} $$ $$ H_{heavy} = \frac{5.5 \times 13.59 \times 12^3}{6 \times (8 + 12)} = \frac{129\,157}{120} \approx 1076\ \text{kN} $$ | Well | Allowable equivalent resisting force (kN) | Applied $H$ (kN) | Result | |---|---|---|---| | Light | 752 | 500 | Safe | | Heavy | 1076 | 500 | Safe | **Answer: allowable total equivalent resisting force due to earth pressure $\approx 752$ kN (light well) and $\approx 1076$ kN (heavy well), both greater than the applied 500 kN.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which consideration is used to fix the minimum depth of a well foundation below scour? A. Grip length equal to the steining thickness B. Grip length of at least one-third of the maximum scour depth below maximum scour level C. Depth of the water table D. Depth equal to the external diameter only Answer: B. The depth is fixed from the maximum scour depth (Lacey's formula) plus a grip length of at least one-third of it below the maximum scour level. (Key point: Grip length) 2. For the well with 4 m internal diameter and 0.75 m steining, what width B is used in the earth-pressure calculation? A. 6.25 m B. 4.75 m C. 4.0 m D. 5.5 m Answer: D. B is the external diameter = 4 + 2(0.75) = 5.5 m. (Key point: External diameter) 3. In Terzaghi's analysis for a light well, about what does the well rotate? A. The centre of the base B. The scour level C. A point O at depth z0 below scour level, with no resistance from the base D. The top of the well cap Answer: C. A light well has no base resistance, so it rotates about a point O at depth z0 below the scour level. (Key point: Light well behaviour) 4. How do the allowable resisting forces compare, and what is the conclusion against H = 500 kN? A. Light about 1076 kN, heavy about 752 kN; both safe B. Light about 752 kN, heavy about 1076 kN; both safe C. Light about 752 kN, heavy about 1076 kN; light is unsafe D. Both about 752 kN; both safe Answer: B. The heavy well rotates about its base centre and mobilises more resistance (about 1076 kN) than the light well (about 752 kN); both exceed 500 kN. 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