Cantilever sheet pile, 5 m excavation

Data: $H = 5$ m, $\gamma = 22$ kN/m$^3$, $\phi = 33^\circ$, no water above the pile tip. Cantilever sheet pile, simplified moment method (moments about the toe).

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

Forces (with $D$ = embedment below the dredge line, $\gamma$ same on both sides):

• Active force on the back: $P_a = \tfrac12K_a\gamma(H+D)^2$, acting at $(H+D)/3$ above the tip.
• Passive force in front: $P_p = \tfrac12K_p\gamma D^2$, acting at $D/3$ above the tip.

Moment about the tip

$$\tfrac12K_a\gamma(H+D)^2\frac{H+D}{3} = \tfrac12K_p\gamma D^2\frac{D}{3} \ \Rightarrow\ \frac{D}{H+D} = \left(\frac{K_a}{K_p}\right)^{1/3} = \left(\frac{0.2948}{3.3921}\right)^{1/3} = 0.4429$$ $$D = \frac{0.4429\times5}{1-0.4429} = 3.98\ \text{m}$$

Design depth: since the method assumes full passive resistance and ignores the extra tip reaction, increase by 20%:

$$D_{design} = 1.2\times3.98 = 4.77\ \text{m}$$

Answer: theoretical embedment $D = 3.98$ m; adopt $D \approx 4.8$ m (total pile length about 9.8 m).