Question 9 of 9: Anchored sheet pile wall by the free earth support method

Question 9: 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. From Figure 4: the anchor is 2.0 m below the top of the wall; the water level, the same on both sides, is 4.0 m below the top; the dredge line is a further 6.0 m down, that is 10.0 m below the top; and the wall penetrates a depth $D$ below the dredge line. The soil is granular throughout with $\gamma=19.5\ \text{kN}/\text{m}^3$ above and below the water table, and $\phi'=38^{\circ}$. Take $\gamma_w=9.81\ \text{kN}/\text{m}^3$.

Find. (i) the theoretical penetration $D$; (ii) the anchor force $F$ per metre run; (iii) the consequences of the water in front dropping to dredge-line level.

Approach. Free earth support treats the wall as a simply supported beam: the anchor is a prop, the toe is free to rotate, and the required penetration follows from taking moments about the anchor. The anchor force then comes from horizontal equilibrium.

1. Earth pressure coefficients and effective unit weights. $$\begin{aligned}K_a&=\frac{1-\sin 38^{\circ}}{1+\sin 38^{\circ}}=0.2379 \\ K_p&=\frac{1}{K_a}=4.204\end{aligned}$$ $$\gamma'=19.5-9.81=9.69\ \text{kN}/\text{m}^3$$ Because the water level is identical on both sides, the hydrostatic pressures cancel exactly and the analysis can be carried out in effective stress alone. This is the single most important observation in part (i), and it is what part (iii) removes.
2. Active pressure at the two breaks in the diagram. At the water table, 4.0 m down, and at the dredge line, 10.0 m down, $$\begin{aligned}\sigma_1'&=K_a\gamma L_1 \\ &=0.2379(19.5)(4.0) \\ &=18.56\ \text{kPa}\end{aligned}$$ $$\begin{aligned}\sigma_2'&=K_a(\gamma L_1+\gamma' L_2) \\ &=0.2379[19.5(4.0)+9.69(6.0)] \\ &=0.2379(136.14) \\ &=32.39\ \text{kPa}\end{aligned}$$
3. Where the net pressure passes through zero below the dredge line. Below the dredge line the active pressure keeps growing at $K_a\gamma'$ while the passive pressure grows at the much steeper $K_p\gamma'$, so the net pressure falls to zero at $$\begin{aligned}L_3&=\frac{\sigma_2'}{\gamma'(K_p-K_a)} \\ &=\frac{32.39}{9.69(4.204-0.238)} \\ &=\frac{32.39}{38.43} \\ &=0.843\ \text{m}\end{aligned}$$
4. Resultant of the driving pressure diagram, and its moment about the anchor. Splitting the diagram from the top down to the point of zero net pressure into a triangle, a rectangle, a triangle and a final triangle:

5. Moment equilibrium about the anchor gives the penetration. Below the point of zero net pressure the net resistance grows linearly at $\gamma'(K_p-K_a)=38.43\ \text{kPa/m}$. Calling that extra depth $L_4$, its resultant is $\tfrac{1}{2}(38.43)L_4^{2}$ acting $\tfrac{2}{3}L_4$ below the zero point, so $$\tfrac{1}{2}(38.43)L_4^{2}\left(8.843+\tfrac{2}{3}L_4\right)=943.3$$ $$12.81L_4^{3}+169.9L_4^{2}-943.3=0 \quad\Rightarrow\quad L_4=2.183\ \text{m}$$
6. Theoretical penetration. $$\begin{aligned}D&=L_3+L_4 \\ &=0.843+2.183 \\ &=\boxed{D=3.03\ \text{m}}\end{aligned}$$ Design practice increases the theoretical value by 30 to 40 per cent to cover the idealisations in the method, giving a construction depth of about 4.2 m and a total sheet pile length of about 14.2 m.
7. Anchor force from horizontal equilibrium. The anchor takes whatever the passive block below the zero point does not: $$\begin{aligned}F&=P-\tfrac{1}{2}\gamma'(K_p-K_a)L_4^{2} \\ &=203.58-\tfrac{1}{2}(38.43)(2.183)^{2} \\ &=203.58-91.6\end{aligned}$$ $$\boxed{F=112\ \text{kN per metre run of wall}}$$ At a typical tie-rod spacing of 3 m this is 336 kN per anchor, and the tie rod and the anchorage should be designed for that force with their own factor of safety, conventionally 1.5 to 2 on the computed value, because an anchor failure is brittle and progressive.

(iii) The water in front drawn down to the dredge line. This is not a small perturbation; it is the design case. Four things happen at once.

An unbalanced water pressure appears. The retained water table stays at 4.0 m while the free water in front falls 6.0 m to the dredge line, so a 6.0 m head now acts across the wall. Taking the no-flow, hydrostatic case, the net water pressure grows from zero at 4.0 m to $\gamma_w(6.0)=58.9\ \text{kPa}$ at the dredge line and continues at that value below it. The triangular part alone is $\tfrac{1}{2}(9.81)(6.0)^{2}=177\ \text{kN/m}$, comparable with the entire 203.6 kN/m of effective earth pressure computed in part (i), and it acts low on the wall where its lever arm about the anchor is large.

The wall no longer works. Re-solving the moment equation about the anchor with the water diagram added gives a required theoretical penetration of about $5.8\ \text{m}$ instead of $3.03\ \text{m}$, and an anchor force of about $250\ \text{kN/m}$ instead of $112\ \text{kN/m}$ — the penetration nearly doubles and the anchor force more than doubles. A wall built to the part (i) answer would rotate about the anchor and fail by kicking out at the toe, and the tie rod would be loaded to well beyond its design force.

Seepage makes it worse than the hydrostatic estimate suggests. In reality water flows down the back of the wall and up in front of it. The upward gradient in front reduces the effective unit weight of the soil there to $\gamma'-i\gamma_w$, so the passive resistance that is being relied on to hold the toe is itself reduced, while the downward gradient behind increases the active pressure. If the exit gradient approaches the critical value $i_c=\gamma'/\gamma_w=9.69/9.81\approx 0.99$ the sand in front of the wall boils and the passive resistance vanishes altogether. Increasing the penetration lengthens the seepage path and reduces the exit gradient, so it helps twice over.

What a designer does about it. The wall must be designed for the maximum credible unbalanced head, not for the balanced condition. On a tidal or fluctuating waterfront that head is set by the lowest low water combined with a lagging retained water level, and it is normally reduced deliberately by weep holes or a granular drainage layer with a filter behind the wall, so that the retained level can fall with the free water. Where drainage cannot be guaranteed — a contaminated site, or fines that would clog the filter — the full head must be carried, and the consequence is a longer, heavier section and a stronger anchor. A flow-net or seepage analysis should replace the hydrostatic assumption for the final design, and a piping check at the dredge line should be presented alongside the stability check.