Cantilever sheet pile, simplified method

Data: deposit (in which the pile is driven): $\phi_2 = 35^\circ$, $\gamma_2 = 22$ kN/m$^3$. Backfill 3 m high on one side: $\phi_1 = 32^\circ$, $\gamma_1 = 18$ kN/m$^3$. $FS = 2$ on passive resistance. Water table below the pile tip. Simplified method: moments about the pile tip.

Coefficients

$$K_{a1} = \tan^2(45^\circ-16^\circ) = 0.3073,\quad K_{a2} = \tan^2(45^\circ-17.5^\circ) = 0.2710,\quad K_p = \tan^2(45^\circ+17.5^\circ) = 3.6902$$

Active pressures (back, depth $z$ from the top of the backfill; the dredge line is at $z=3$)

• Backfill ($0 \le z \le 3$): $p = K_{a1}\gamma_1 z$, giving 16.59 kN/m$^2$ at 3 m.
• Below 3 m (in the deposit): $p = K_{a2}\,[18\times3 + 22(z-3)] = 14.63 + 5.962(z-3)$ kN/m$^2$.

Passive pressure (front, depth $y$ below the dredge line)

$$p_p = \frac{K_p\gamma_2 y}{FS} = \frac{3.690\times22\,y}{2} = 40.59\,y\ \text{kN/m}^2$$

Moments about the tip (embedment $D$): the active moment is

$$M_a(D) = \int_0^{3+D} p_a\,(3+D-z)\,dz$$

and the passive moment is $M_p = \tfrac16\,(K_p\gamma_2/FS)\,D^3 = 6.765\,D^3$. Solving $M_a = M_p$ by trial and error gives

$$D = 3.10\ \text{m}:\qquad M_a = 202.4,\quad M_p = 202.4\ \text{kN m/m}$$

(active force 99.0 kN/m, passive force 195.6 kN/m).

Answer: required depth of penetration $D \approx 3.10$ m below the 3 m backfill (dredge) level. Since $FS = 2$ is already applied to the passive resistance, no further increase is needed; the total pile length is about 6.1 m.