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. A waterfront anchored sheet pile wall in a uniform granular soil, with equal water levels on the two faces.

Find. (i) the theoretical penetration $D$ below the dredge line for free earth support; (ii) the anchor force $F$ per metre run of wall; (iii) the consequences of the water level in front of the wall falling to the dredge line.

[Figure not reproduced: Figure 2 (redrawn) — anchored sheet pile wall with equal water levels on the two faces, so that hydrostatic pressures cancel, together with the net lateral pressure diagram: active behind the wall, and net passive below the point of zero net pressure. See the official exam paper.]

Approach. Because the water stands at the same elevation on both sides of the wall, the hydrostatic pressures cancel exactly and only effective stresses need be carried; below the water table the effective unit weight $\gamma'$ is used. Build the Rankine active diagram behind the wall, find the depth below the dredge line at which the net pressure changes sign, then take moments about the anchor — the free earth support assumption being that the toe is free to rotate, so the wall is statically determinate with moment equilibrium about the anchor as the single equation.

1. Earth pressure coefficients. For a smooth vertical wall retaining a level granular backfill, Rankine gives $$K_a=\tan^2\!\left(45^\circ-\frac{\phi'}{2}\right)=\tan^2 25^\circ=0.2174,\qquad K_p=\tan^2\!\left(45^\circ+\frac{\phi'}{2}\right)=\tan^2 65^\circ=4.599$$
2. Active pressure at the two changes of slope. Above the water table the effective overburden is $\gamma z$; below it the gradient falls to $\gamma'$. At the water table, $z=L_1=4$ m, $$\sigma'_1=K_a\gamma L_1=0.2174(20)(4.0)=17.40\ \text{kPa}$$ and at the dredge line, $z=L_1+L_2=10$ m, $$\sigma'_2=K_a\big(\gamma L_1+\gamma' L_2\big)=0.2174\big[20(4.0)+10.19(6.0)\big]=0.2174(141.14)=30.69\ \text{kPa}$$
3. Depth to the point of zero net pressure. Below the dredge line the active pressure behind the wall continues to grow at $K_a\gamma'$ per metre while the passive pressure in front grows at the much faster rate $K_p\gamma'$. The net pressure vanishes at a depth $L_3$ below the dredge line where the two are equal: $$L_3=\frac{\sigma'_2}{\gamma'(K_p-K_a)}=\frac{30.69}{10.19(4.599-0.2174)}=\frac{30.69}{44.65}=0.687\ \text{m}$$ Below that point the net pressure is passive and increases linearly at 44.65 kPa per metre of further depth.
4. Resultant of the active diagram above the pivot point. Divide the diagram into the four standard areas — a triangle over $L_1$, a rectangle and a triangle over $L_2$, and a triangle over $L_3$: $$A_1=\tfrac12\sigma'_1L_1=\tfrac12(17.40)(4.0)=34.79\ \text{kN/m}$$ $$A_2=\sigma'_1L_2=17.40(6.0)=104.37\ \text{kN/m}$$ $$A_3=\tfrac12(\sigma'_2-\sigma'_1)L_2=\tfrac12(13.29)(6.0)=39.88\ \text{kN/m}$$ $$A_4=\tfrac12\sigma'_2L_3=\tfrac12(30.69)(0.687)=10.55\ \text{kN/m}$$ so the total active thrust above the point of zero net pressure is $$P=34.79+104.37+39.88+10.55=189.60\ \text{kN/m}$$
5. Moment of that thrust about the anchor. The centroids lie at depths of 2.667 m, 7.000 m, 8.000 m and 10.458 m below the top of the wall; subtracting the anchor depth $l_1=2.0$ m gives the lever arms 0.667, 5.000, 6.000 and 8.458 m. Hence $$M_a=34.79(0.667)+104.37(5.000)+39.88(6.000)+10.55(8.458)$$ $$M_a=23.19+521.86+239.30+89.22=873.57\ \text{kN}\cdot\text{m/m}$$
6. Moment of the net passive resistance about the anchor. Let $L_4$ be the further depth below the point of zero net pressure. The net pressure there is triangular with a maximum of $\gamma'(K_p-K_a)L_4=44.65L_4$, so its resultant is $$P_p=\tfrac12(44.65)L_4^2=22.32L_4^2$$ acting at $\tfrac23L_4$ below the pivot point, that is at a lever arm from the anchor of $(L_1+L_2+L_3-l_1)+\tfrac23L_4=8.687+0.6667L_4$.
7. Moment equilibrium about the anchor. The free earth support method assumes the toe is unrestrained, so the wall rotates about the anchor and one moment equation determines the penetration: $$22.32\,L_4^2\left(8.687+0.6667L_4\right)=873.57$$ which rearranges to the cubic $$14.88\,L_4^3+193.93\,L_4^2-873.57=0$$ Solving numerically gives $L_4=1.978$ m. (Checking: $14.88(7.734)+193.93(3.911)=115.1+758.5=873.6$ ✓.)
8. Theoretical depth of penetration. Adding the two parts, $$D=L_3+L_4=0.687+1.978=\boxed{D=2.67\ \text{m}}$$ This is the theoretical value. Standard practice is to increase it by 30 to 40% to allow for the simplifications in the method and for over-dredging, which gives a design penetration of $1.4(2.665)=3.73$ m, say 3.8 m; the sheet piles would then be ordered at 14 m, that is 10 m above the dredge line plus 4 m of toe.
9. Anchor force. Horizontal equilibrium of the whole wall gives the anchor force as the difference between the active thrust above the pivot and the net passive resistance below it: $$P_p=22.32(1.978)^2=87.31\ \text{kN/m}$$ $$F=P-P_p=189.60-87.31=\boxed{F=102.3\ \text{kN/m}}$$ If the anchor rods are spaced at 3.0 m centres, each rod carries $3.0(102.3)=307$ kN before any allowance for a factor of safety on the tie or for the eccentricity between the rod and the wale.

(iii) Effect of the water level in front of the wall falling to the dredge line

At present the water stands at the same elevation on both faces, so the hydrostatic thrusts cancel and the wall carries earth pressure only. Dropping the level in front of the wall to the dredge line — by an unusually low tide, by a lock or dock being emptied, or by a river falling rapidly — removes 6 m of water from the passive side while the backfill remains saturated to its original level. Four things follow, and together they can be enough to fail the wall.

An unbalanced hydrostatic thrust appears. With a 6 m head difference across the wall the net water pressure diagram is a triangle rising to $\gamma_w h = 9.81(6.0)=58.9$ kPa at the dredge line, giving an additional horizontal force of roughly $\tfrac12(58.9)(6.0)=177$ kN/m — very nearly double the entire present anchor force, and applied high on the wall where its moment about the toe is large. Strictly the diagram must be modified below the dredge line for the seepage that develops, but the order of magnitude is unmistakable.

Seepage reduces the passive resistance and increases the active pressure. Water flows down the back of the wall, round the toe and up in front of it. The downward gradient behind the wall increases the effective vertical stress there and hence the active pressure; the upward gradient in front of the wall reduces the effective stress in the passive block and hence the passive resistance, in proportion to $\gamma'-i\gamma_w$ where $i$ is the exit gradient. Both changes act to destabilise the wall, and both must be evaluated from a flow net rather than assumed hydrostatic.

Piping or heave becomes possible in front of the wall. The critical hydraulic gradient is $i_{cr}=\gamma'/\gamma_w=10.19/9.81=1.04$. With a 6 m head lost over a seepage path of the order of $2D\approx7$ m, the average exit gradient approaches 0.9 and the local gradient immediately in front of the sheeting is higher still. The factor of safety against piping is therefore near unity: the soil in front of the wall can boil, and once it does the passive resistance is lost entirely.

The anchor force and bending moment rise sharply. Because both the driving force increases and the passive resistance decreases, moment equilibrium about the anchor requires a substantially greater penetration, and the anchor force rises roughly in proportion to the total unbalanced thrust. A wall designed for balanced water levels and then subjected to this condition would be under-penetrated and its tie rods overstressed.

The engineering response is to prevent the condition rather than to design for it wherever possible: weep holes or a filtered relief drain through the sheeting at or just above the low-water level, so that the backfill can drain and the head difference cannot build up; a granular filter behind the sheeting to keep the weeps from clogging and to prevent loss of fines; and, where a head difference is unavoidable, a longer toe to lengthen the seepage path and raise the factor of safety against piping. If the drawdown is a design case, the wall must be re-analysed with the flow net and the modified pressure diagram, not with the balanced-water solution above.