Anchor reduces cantilever sheet pile depth

Assumed condition: sheet pile wall retaining $H = 5$ m of dry sand ($\phi = 30^\circ$, same on both sides), $\gamma$ the same on both sides, passive resistance with $FS = 1$ for the comparison, anchor at 1 m below the top (free earth support).

$$K_a = \tan^2 (45^\circ-15^\circ) = 0.3333,\qquad K_p = \tan^2(45^\circ+15^\circ) = 3.0$$

(a) Cantilever wall: the moment about the toe must balance:

$$\tfrac12K_a\gamma(H+D)^2\frac{(H+D)}{3} = \tfrac12K_p\gamma D^2\frac{D}{3}\ \Rightarrow\ \frac{D}{H+D} = \left(\frac{K_a}{K_p}\right)^{1/3} = 0.4807$$ $$D_{cant} = \frac{5\times0.4807}{1-0.4807} = 4.63\ \text{m}$$

(b) Anchored wall (free earth support): take moments about the anchor point (the anchor force $T$ drops out):

$$\tfrac12K_a\gamma(H+D)^2\left[\tfrac23(H+D) - 1\right] = \tfrac12K_p\gamma D^2\left[H + \tfrac23D - 1\right]$$

Solving, $D_{anch} = 1.90$ m, and the anchor force from $\sum H = 0$ is

$$T = \tfrac12K_a\gamma(H+D)^2 - \tfrac12K_p\gamma D^2 \ (\text{per unit }\gamma) = 7.94\gamma - 5.43\gamma = 2.51\gamma$$

(with $\gamma = 18$ kN/m$^3$, $T \approx 45$ kN/m).

Comparison: $D_{anchored} = 1.90$ m $\ll D_{cantilever} = 4.63$ m, about 59% less. The anchor supplies the horizontal reaction that the cantilever wall must get from the passive resistance of a deep embedment, so the embedment, and also the pile length and bending moment, are reduced.