Terzaghi lateral stability, light well

Terzaghi's simplified approach for a light well (self-weight small, so the base does not resist horizontal force) treats the well as a rigid body in cohesionless soil, rotating about a point O at a depth $z_0$ below the scour level.

Assumptions:

• The soil reaction at depth $z$ is the net passive resistance $\sigma = \dfrac{\gamma'z(K_p - K_a)}{FS} = kz$, with $k = \dfrac{\gamma'(K_p - K_a)}{FS}$, acting on the width $B$ of the well.
• Above O the reaction opposes the horizontal force; below O it acts in the same direction as the force.
• The horizontal force $H$ acts at a height $h = M/H$ above the scour level, and $D$ is the depth of embedment.

   H -->  (height h)
   ----|----- scour level
       |  <--- sigma = kz   (above O)
      -O----  rotation point z0
       |  ---> sigma = kz   (below O)
       |_____ D

Derivation

Horizontal equilibrium:

$$H = Bk\left(\frac{z_0^2}{2} - \frac{D^2 - z_0^2}{2}\right) = Bk\left(z_0^2 - \frac{D^2}{2}\right)$$

Moments about the scour level:

$$Hh = Bk\left(\frac{D^3 - z_0^3}{3} - \frac{z_0^3}{3}\right) = \frac{Bk}{3}\left(D^3 - 2z_0^3\right)$$

Eliminating $H$ gives the equation for the depth of the point of rotation:

$$2z_0^3 + 3h\,z_0^2 - \left(\tfrac{3}{2}hD^2 + D^3\right) = 0$$

Solve for $z_0$, then the allowable equivalent resisting force is $H = Bk\left(z_0^2 - \tfrac{D^2}{2}\right)$. The well is stable if this value is greater than the applied horizontal force.