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

Question 9: Anchored sheet pile wall by free earth support (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 retained ground surface is 4.0 m above the water table, with the anchor 2.0 m below the ground surface; the water table stands at the same elevation on both sides of the wall, 6.0 m above the dredge line; the penetration $D$ below the dredge line is to be found. Granular soil throughout with $\gamma = 19.5\ \text{kN/m}^3$ above and below the water table and $\phi' = 38^\circ$, so $\gamma' = 19.5 - 9.81 = 9.69\ \text{kN/m}^3$. Take $L_1 = 4.0\ \text{m}$ and $L_2 = 6.0\ \text{m}$; the anchor lies at $l_1 = 2.0\ \text{m}$ below the surface.

Find. (i) the theoretical penetration $D$ from free earth support, (ii) the anchor force $F$ per metre run, and (iii) the consequences of drawing the water in front of the wall down to dredge level.

[Figure not reproduced: Question 9 (redrawn from Figure 4): the anchored wall, the balanced water levels, and the net lateral pressure diagram. Because the water stands at the same level on both sides, the hydrostatic pressures cancel and only effective stresses enter the calculation. See the official exam paper.]

Approach. Free earth support treats the toe as free to rotate, so the wall is a simply supported beam spanning between the anchor and the net passive resistance below the dredge line. Build the net effective-pressure diagram, take moments about the anchor to find the penetration, then use horizontal equilibrium to find the anchor force.

1. Part (i) — note that the water pressures cancel, and compute the earth-pressure coefficients. The free water in front of the wall stands at the same elevation as the groundwater behind it, so the hydrostatic thrusts on the two faces are equal and opposite at every depth and the problem reduces to effective stresses with the buoyant unit weight. For $\phi' = 38^\circ$, $$K_a = \tan^{2}\!\left(45^\circ - \tfrac{\phi'}{2}\right) = 0.2379, \qquad K_p = \tan^{2}\!\left(45^\circ + \tfrac{\phi'}{2}\right) = 4.2037$$
2. Build the active pressure diagram down to the dredge line. Above the water table the total unit weight acts; below it the buoyant unit weight does. At the water table and at the dredge line the active pressures are $$\sigma'_1 = \gamma L_1 K_a = (19.5)(4.0)(0.2379) = 18.56\ \text{kPa}$$ $$\sigma'_2 = (\gamma L_1 + \gamma' L_2)K_a = \big(78.0 + 58.14\big)(0.2379) = 32.39\ \text{kPa}$$
3. Locate the point of zero net pressure below the dredge line. Below the dredge line the passive pressure on the front face grows faster than the active pressure on the back, and the net pressure crosses zero at a depth $L_3$ where $$L_3 = \frac{\sigma'_2}{\gamma'(K_p - K_a)} = \frac{32.39}{9.69(4.2037 - 0.2379)} = \frac{32.39}{38.43} = 0.843\ \text{m}$$
4. Assemble the driving force and its moment about the anchor. Splitting the diagram from the ground surface to the zero-pressure point into four areas, with depths $z$ measured from the surface:

   Net driving pressure areas above the zero-pressure point

   Area	Description	$P$ (kN/m)	$z$ (m)	Moment about the anchor (kN·m/m)
   1	triangle, 0 to 4.0 m	37.11	2.667	24.74
   2	rectangle, 4.0 to 10.0 m	111.33	7.000	556.65
   3	triangle, 4.0 to 10.0 m	41.49	8.000	248.95
   4	triangle, 10.0 to 10.843 m	13.65	10.281	113.00
   Totals	—	203.58	—	943.3

5. Take moments about the anchor to find the embedded length. Below the zero-pressure point the net resistance is a triangle of length $L_4$ whose resultant is $\tfrac{1}{2}\gamma'(K_p-K_a)L_4^{2}$ acting at $\tfrac{2}{3}L_4$ below that point. Setting its moment about the anchor equal to 943.3 kN·m/m, $$\tfrac{1}{2}(38.43)L_4^{2}\left(8.843 + \tfrac{2}{3}L_4\right) = 943.3$$ which is the cubic $12.81L_4^{3} + 169.9L_4^{2} - 943.3 = 0$, solved by iteration to give $L_4 = 2.183\ \text{m}$. The theoretical penetration is therefore $$\boxed{D_{theoretical} = L_3 + L_4 = 0.843 + 2.183 = 3.03\ \text{m}}$$
6. Convert to a construction length. Free earth support ignores the fixity that develops as the toe is driven, and the pressure diagram itself is idealised, so the theoretical value is increased by 30 to 40 per cent in practice. Taking 1.4, $$D_{actual} \approx 1.4(3.03) = 4.2\ \text{m}, \qquad \text{total pile length} \approx 10.0 + 4.2 = 14.2\ \text{m}$$ Order 15 m sheet piles.
7. Part (ii) — find the anchor force from horizontal equilibrium. The net passive resultant developed over $L_4$ is $$P' = \tfrac{1}{2}\gamma'(K_p - K_a)L_4^{2} = \tfrac{1}{2}(38.43)(2.183)^{2} = 91.6\ \text{kN/m}$$ and the anchor takes whatever the toe does not: $$\boxed{F = \sum P - P' = 203.58 - 91.60 = 112.0\ \text{kN per metre run of wall}}$$ Anchors at 3.0 m centres would each carry $3.0(112.0) = 336\ \text{kN}$ before any allowance, and with the customary factor of safety of about 1.3 on the tie rod the tie should be designed for roughly 440 kN.
8. Part (iii) — the effect of drawing the front water down to dredge level. The consequences are set out in the discussion below; in outline, an unbalanced head of 6.0 m appears across the wall, seepage begins beneath the toe, the driving thrust rises sharply while the available passive resistance falls, and the wall as designed above is no longer adequate.

Discussion for part (iii). At present the two water levels coincide, so every hydrostatic pressure on the back face is matched by an equal pressure on the front and the design above is a pure effective-stress calculation. Lowering the water in front to the dredge line destroys that balance and changes the problem in four ways, all unfavourable.

First, a net water pressure appears. The head difference across the wall becomes 6.0 m, so the net water pressure grows from zero at the retained water table to $\gamma_w h = 9.81(6.0) = 58.86\ \text{kPa}$ at the dredge line. Over the retained height alone that is an unbalanced thrust of $\tfrac{1}{2}(58.86)(6.0) = 176.6\ \text{kN/m}$, which is comparable with the entire 203.6 kN/m of effective driving force computed for the balanced case. Second, seepage changes the effective stresses on both sides in the wrong direction. Water flows down behind the wall and up in front of it; the downward gradient behind increases the effective vertical stress and therefore the active pressure, while the upward gradient in front reduces the effective vertical stress and therefore the available passive resistance. Third, the required penetration and anchor force both increase substantially. Re-solving the free earth support moment equation with the net water diagram added — taken as triangular below the dredge line, falling from 58.86 kPa to zero at the tip, which is the usual allowance for the head loss around the toe — gives a theoretical penetration of $D = 4.74\ \text{m}$ against the 3.03 m of part (i), an increase of 57 per cent, and an anchor force of 227.8 kN/m against 112.0 kN/m, almost exactly double. A wall built to the dimensions of part (i) would move, the anchor would be overloaded, and the bending moment in the sheeting would exceed its design value. That comparison is itself optimistic, because it holds the buoyant unit weight unchanged on both sides; the seepage correction described above makes each side worse again.

Fourth, and most serious, is the risk of piping or heave in front of the wall. The exit gradient at the dredge line rises towards the critical value $i_c = \gamma'/\gamma_w = 9.69/9.81 \approx 0.99$; if it approaches that value the sand in front of the toe boils, loses all effective stress and all passive resistance, and the wall fails suddenly and completely. The remedies are, in order of preference: do not draw the front water down — make the drawdown case an explicit design condition and prohibit it operationally; if drawdown is unavoidable, design for it from the outset with a longer penetration to lengthen the seepage path and reduce the exit gradient, a larger anchor, and a heavier sheet-pile section; and provide a granular filter or inverted filter blanket over the dredged bed in front of the toe to suppress boiling. A relief drain through the wall below the dredge line, protected by a filter, is another route: it removes the head difference at source, at the cost of a permanent maintenance obligation. Whichever is adopted, the case should be checked as a separate limit state rather than assumed to be covered by the factor of safety in the balanced-water design.