Circular well 4 m, light and heavy well

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:

$$H\,(h + D) = Bk\int_0^D z\,(D - z)\,dz = \frac{BkD^3}{6}$$ $$H_{heavy} = \frac{5.5 \times 13.59 \times 12^3}{6 \times (8 + 12)} = \frac{129\,157}{120} \approx 1076\ \text{kN}$$

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.