Cantilever sheet pile, 8 m excavation

Data and assumptions: excavation depth $H = 8$ m (dredge level); $\gamma = \gamma_{sat} = 20$ kN/m$^3$, $\phi = 33^\circ$; water table at 6.5 m below ground, taken at the same level on both sides (the excavation is flooded to the water table, no seepage and no net water pressure), so effective stresses with $\gamma' = 20 - 9.81 = 10.19$ kN/m$^3$ are used below the water table. Moments are taken about the pile tip (simplified cantilever method).

Coefficients

$$K_a = \tan^2(45^\circ - 16.5^\circ) = 0.2948,\qquad K_p = \tan^2(45^\circ + 16.5^\circ) = 3.3921$$

Pressures

• Active (back), depth $z$: $p_a = K_a\sigma_v'$; at 6.5 m $= 38.32$ kN/m$^2$; below 6.5 m the effective stress increases with $\gamma'$.
• Passive (front), depth $y$ below the dredge line: $p_p = K_p\gamma' y = 34.57\,y$ kN/m$^2$.

Moment equilibrium about the tip: for a trial embedment $D$,

$$M_{active}(D) = \int_0^{8+D} p_a\,(8+D-z)\,dz,\qquad M_{passive}(D) = \tfrac16K_p\gamma' D^3$$

Setting $M_{active} = M_{passive}$ and solving by trial and error:

$$D = 9.10\ \text{m}:\quad M_{active} = 4338,\quad M_{passive} = 4338\ \text{kN m/m}$$

(Active force 699 kN/m, passive force 1430 kN/m at this depth.)

Design depth: increase by 20% for safety (practice 20-40%):

$$D_{design} = 1.2\times9.10 = 10.9\ \text{m}$$

Answer: theoretical embedment $D \approx 9.1$ m (below the 8 m dredge level); provide $\approx 11$ m, i.e. a total pile length of about 19 m. For comparison, without the water table, $D = 6.4$ m.