Anchored sheet pile, free earth, with anchor pull

Data and assumptions (figure values as read): height above dredge level $H = 8.0$ m; anchor (tie rod) at 2.5 m below the top; soil above the water table: $\gamma = 16$ kN/m$^3$, $\phi = 35^\circ$ for the top 3.0 m; below: $\gamma_{sat} = 19$ kN/m$^3$, $\phi = 35^\circ$. Water level is taken the same on both sides (no net water pressure), so $\gamma' = 19 - 9.81 = 9.19$ kN/m$^3$ is used below the water table. Free earth support method, passive resistance with $FS = 1$ for the theoretical depth.

$$K_a = \tan^2(45^\circ-17.5^\circ) = 0.2710,\qquad K_p = \tan^2(45^\circ+17.5^\circ) = 3.6902$$

Active pressure (back, depth $z$): $p_a = K_a\sigma_v'$

• $z = 3$ m: $\sigma_v' = 48$, $p_a = 13.01$ kN/m$^2$.
• $z = 8$ m (dredge): $\sigma_v' = 48 + 9.19\times5 = 93.95$, $p_a = 25.46$ kN/m$^2$.
• Below the dredge line: the effective stress increases by $\gamma'$ per metre.

Passive pressure (front): $p_p = K_p\gamma' y = 33.91\,y$ kN/m$^2$, $y$ measured below the dredge line.

Moment about the anchor (level 2.5 m): for a trial embedment $D$,

$$\sum M_{active} = \int_0^{8+D} p_a\,(z-2.5)\,dz,\qquad \sum M_{passive} = \int_0^{D} p_p\,(8 + y - 2.5)\,dy$$

Equating and solving by trial and error:

$$D = 2.57\ \text{m}$$

At this depth: active force $P_a = 189.3$ kN/m, passive force $P_p = 112.0$ kN/m.

Design embedment: increase by 30% (free earth support): $D_{design} = 1.3\times2.57 = 3.3$ m.

Tension in the anchor from horizontal equilibrium:

$$T = P_a - P_p = 189.3 - 112.0 = 77.3\ \text{kN/m}$$

Answer: depth of embedment $D \approx 2.57$ m (adopt about 3.3 m); anchor force $T \approx 77$ kN per metre run.