Circular well 4.5 m, resisting force

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.