Question 7 of 10: Anchored sheet pile wall by the free earth support method

Question 7: Anchored sheet pile wall by the free earth support method (24 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Measuring depth $z$ from the top of the wall: the retained ground surface is at $z = 0$, the anchor tie is at $z = 2$ m, the water table is at $z = 4$ m, and the dredge line is at $z = 10$ m. The figure shows the water surface at the same level on both sides of the wall, so 6 m of free water stands in front of the wall above the dredge line and there is no unbalanced water head. The soil is granular throughout, with $\gamma = \gamma_{sat} = 20$ kN/m$^3$ and $\phi' = 40^{\circ}$, and the wall is taken as smooth (Rankine).

Find. (i) the theoretical embedment D for free earth support, (ii) the anchor force F per metre run, and (iii) the consequence of drawing the water in front of the wall down to the dredge line.

[Figure not reproduced: Figure 2 (redrawn). Anchored sheet pile wall with the water table 4 m below the top of the wall on both sides, and the net effective lateral pressure diagram used in the free earth support analysis. See the official exam paper.]

Approach. Because the water stands at the same elevation on both faces, the hydrostatic pressures cancel and the whole problem can be worked in effective stresses with the buoyant unit weight below the water table. Build the Rankine active diagram on the retained side and the passive diagram on the dredge side, then find D by taking moments about the anchor (the free earth support condition, in which the toe is free to rotate) and recover F from horizontal equilibrium.

1. Earth pressure coefficients and effective unit weight. For a smooth vertical wall with horizontal ground, $$\begin{aligned}K_a &= \frac{1-\sin\phi'}{1+\sin\phi'} = \frac{1-\sin 40^{\circ}}{1+\sin 40^{\circ}} = 0.2174 \\ K_p &= \frac{1}{K_a} = 4.599\end{aligned}$$Below the water table the buoyant unit weight is $\gamma' = \gamma_{sat} - \gamma_w = 20 - 9.81 = 10.19$ kN/m$^3$.
2. Build the active pressure diagram on the retained side. The vertical effective stress is $\sigma'_v = \gamma z$ above the water table and $\sigma'_v = \gamma z_w + \gamma'(z - z_w)$ below it, so $$\begin{aligned}\sigma'_v(4) &= 20(4) = 80.0\ \text{kPa} \\ \sigma'_v(10) &= 80.0 + 10.19(6) = 141.14\ \text{kPa}\end{aligned}$$Multiplying by $K_a$ gives $p_a(4) = 17.40$ kPa and $p_a(10) = 30.69$ kPa. The break in slope at the water table is caused by the change from $\gamma$ to $\gamma'$, not by any water pressure.
3. Resolve the active diagram into components and take their lever arms about the anchor. Working downwards from the top of the wall: $$\begin{aligned}P_1 &= \tfrac{1}{2}(17.40)(4) = 34.79\ \text{kN/m} \\ P_2 &= (17.40)(6) = 104.37\ \text{kN/m} \\ P_3 &= \tfrac{1}{2}(30.69 - 17.40)(6) = 39.88\ \text{kN/m}\end{aligned}$$acting at $z = 2.667,\ 7.000$ and $8.000$ m, that is $0.667,\ 5.000$ and $6.000$ m below the anchor. Their moment about the anchor is $M_{fixed} = 784.4$ kN$\cdot$m/m.
4. Add the embedded length and the passive resistance. Over the embedment $D$ the active pressure continues from $p_a(10)$ with gradient $K_a\gamma'$, while passive pressure builds on the dredge side from zero with gradient $K_p\gamma'$: $$\begin{aligned}P_4 &= 30.690\,D \\ P_5 &= \tfrac{1}{2}K_a\gamma' D^2 = 1.1079\,D^2 \\ P_p &= \tfrac{1}{2}K_p\gamma' D^2 = 23.431\,D^2\end{aligned}$$with lever arms below the anchor of $(8 + D/2)$ m for $P_4$ and $(8 + 2D/3)$ m for both $P_5$ and $P_p$.
5. Impose the free earth support condition. The toe is assumed free to rotate, so the wall is a simply supported span between the anchor and the passive block, and moments about the anchor must vanish: $$\sum M_{anchor} = 0 \;\Rightarrow\; 784.4 + 30.69 D\left(8+\tfrac{D}{2}\right) + 1.108 D^2\left(8+\tfrac{2D}{3}\right) - 23.43 D^2\left(8+\tfrac{2D}{3}\right) = 0$$Collecting terms gives the cubic $$14.882\,D^3 + 163.24\,D^2 - 245.52\,D - 784.4 = 0$$whose only positive root is $$D = \boxed{2.66\ \text{m}}$$This is the theoretical penetration. Practice increases it by 30 to 40 per cent, or divides $K_p$ by a factor of about 1.5 to 2, giving a construction depth of roughly 3.5 to 3.7 m.
6. Recover the anchor force from horizontal equilibrium. With $D = 2.662$ m, $$\begin{aligned}P_4 &= 81.71\ \text{kN/m} \\ P_5 &= 7.85\ \text{kN/m} \\ P_p &= 166.10\ \text{kN/m}\end{aligned}$$so the total active thrust is $\sum P_a = 268.61$ kN/m and horizontal equilibrium gives $$F = \sum P_a - P_p = 268.6 - 166.1 = \boxed{102.5\ \text{kN/m}}$$If the anchors are spaced at, say, 3 m centres, each tie carries about 308 kN, and design practice adds a further margin of the order of 30 per cent to allow for the redistribution of pressure caused by wall flexibility (Rowe's moment reduction acts on the bending moment, not on the anchor pull).
7. Part (iii): drawing the front water down to the dredge line. The present design owes its economy entirely to the water levels being equal. Lowering the water in front of the wall to the dredge line while the backfill stays at its original level creates an unbalanced head of 6 m across the wall. Neglecting seepage for a first estimate, the net water pressure grows linearly from zero at the original water level to $u = \gamma_w h = 9.81(6) = 58.9$ kPa at the dredge line, adding a horizontal thrust of $$P_w = \tfrac{1}{2}(58.9)(6) = 176.6\ \text{kN/m}$$which by itself exceeds the entire active earth thrust computed above.

Three consequences follow, and all of them are adverse. The driving side gains the water thrust just described, so the moment about the anchor that D must balance rises steeply and the required embedment and anchor force both increase substantially — commonly by 50 to 100 per cent for a head of this size. Second, the water that is now flowing under the toe adds a downward seepage gradient on the retained side and an upward gradient in front of the wall; the upward flow reduces the effective unit weight of the passive block and so reduces the passive resistance at the very moment more of it is needed, while the downward flow increases the active pressure. Third, if the exit gradient in front of the wall approaches the critical value $i_{cr} = \gamma'/\gamma_w \approx 1.0$, the soil at the dredge line loses all effective stress and boils, and the passive resistance vanishes altogether. The correct treatment is to draw a flow net around the toe, use the resulting seepage pressures to modify $\gamma'$ on both sides, and check the exit gradient against a factor of safety of at least 3. In practice a wall designed for balanced water levels must not be dewatered on the dredge side without re-analysis; relief drainage through the wall, or weep holes, is the usual way of guaranteeing that the two levels stay tied together.