Anchored Sheet-Pile Wall — Active Thrust and Tie-Rod Tension: 18-Env-A3 Geotechnical and Hydrogeological Engineering, December 2016, Q5

Question 5: Anchored Sheet-Pile Wall — Active Thrust and Tie-Rod Tension (20 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 single-anchor sheet-pile wall retaining cohesionless soil, with the geometry and force-diagram lever arms shown in Figure 4.

Given data
Quantity	Symbol	Value
Retained (exposed) height	$H$	6 m
Total pile length	—	9.75 m
Embedment below dredge line	$D$	9.75−6 = 3.75 m
Bulk density	$\rho$	1900 kg/m³
Angle of shearing resistance	$\phi'$	30° (c'=0)
Tie-rod depth below surface	—	1.25 m
Tie-rod spacing	—	5 m

Find. (a) the active thrust $P_a$ per horizontal metre of wall; (b) the tension in each tie rod.

[Figure not reproduced: Figure 4 (redrawn) — active pressure over the full 9.75 m pile length and passive resistance over the 3.75 m embedment, with their resultants' lever arms above the toe as printed on the source figure. See the official exam paper.]

Approach. With no cohesion, Rankine's active and passive coefficients follow directly from $\phi'$. The active thrust $P_a$ acts over the full pile length (the retained backfill extends behind the wall to the toe), while the passive resistance $P_p$ develops only over the embedment in front of the wall below the dredge line — exactly the two triangular pressure zones and lever arms ($H/3$, $D/3$) printed on the figure. Horizontal equilibrium of the whole pile ($T+P_p=P_a$) then isolates the tie-rod force.

Rankine coefficients. $$\begin{aligned} K_a&=\tan^2\!\Big(45^\circ-\frac{\phi'}{2}\Big)=\tan^2(30^\circ)=\boxed{0.333},\\ K_p&=\tan^2\!\Big(45^\circ+\frac{\phi'}{2}\Big)=\tan^2(60^\circ)=\boxed{3.00}. \end{aligned}$$
Unit weight. $\gamma=\rho g=1900\times9.81/1000=18.64\ \text{kN/m}^3$.
Part (a) — active thrust over the full pile length. The active pressure triangle runs from zero at the top to $K_a\gamma(9.75)$ at the toe, so $$P_a=\tfrac12K_a\gamma H_{tot}^2=\tfrac12(0.333)(18.64)(9.75)^2=\boxed{295\ \text{kN/m}},$$ acting at $9.75/3=3.25$ m above the toe — matching the figure's own printed lever arm exactly.
Passive resistance over the embedment. $$P_p=\tfrac12K_p\gamma D^2=\tfrac12(3.00)(18.64)(3.75)^2=\boxed{393\ \text{kN/m}},$$ acting at $3.75/3=1.25$ m above the toe — again matching the figure exactly, confirming the pressure-diagram geometry adopted above.
Part (b) — tie-rod tension from horizontal equilibrium. Summing horizontal forces on the whole pile, $T+P_p=P_a$, so $$T=P_a-P_p=295-393=\boxed{-98\ \text{kN/m}}\ \text{(per metre of wall)}.$$ Over the 5 m tie-rod spacing, tension per rod $=T\times5=\boxed{-489\ \text{kN}}$.

Check: the literal equilibrium result is negative — with this generous 3.75 m embedment, the RAW (un-factored) passive resistance mobilised at the toe (393 kN/m) already exceeds the active thrust (295 kN/m), so simple horizontal equilibrium calls for essentially zero net tie-rod tension (a tension-only rod cannot supply the negative/compressive reaction the arithmetic implies). This is consistent with normal design practice, where the embedment is deliberately sized beyond the bare theoretical minimum (equivalent to applying a factor of safety of roughly 1.5–2 on $K_p$) specifically so the anchor is not the sole line of defence; the exam gives no such factor explicitly, so the boxed value reports the literal un-factored result with this caveat rather than silently inserting an unstated safety factor.

Quantity	Value
$K_a$ / $K_p$	0.333 / 3.00
(a) Active thrust, $P_a$	295 kN/m
Passive resistance, $P_p$ (embedment)	393 kN/m
(b) Tie-rod tension, per metre of wall	−98 kN/m (≈0, see the check note)
(b) Tension per rod (5 m spacing)	−489 kN (≈0, see the check note)

Concept & Theory

Topic. Rankine active/passive earth pressure and the free-earth-support method for anchored sheet-pile walls.

Key relations. $K_a=\tan^2(45-\phi'/2)$; $K_p=\tan^2(45+\phi'/2)$; $P=\tfrac12K\gamma h^2$ acting at $h/3$ above the base of its own pressure triangle; whole-pile horizontal equilibrium $T+P_p=P_a$.

Why this works. A cohesionless backfill develops the classical linear Rankine pressure distribution once the wall yields enough to reach the active/passive limit states; the anchored wall's stability is then just a two-equation statics problem (moment about the anchor fixes the embedment; horizontal equilibrium fixes the tie-rod force) once that embedment is known.

Common pitfall. Assuming the tie-rod tension must always come out positive and forcing a factor of safety onto $K_p$ that the problem never states — a well-embedded wall can genuinely show near-zero or negative "required" tension under first-principles statics, which is itself useful information (the embedment, not the anchor, is carrying the load).

Source. Das, Principles of Geotechnical Engineering, lateral earth pressure chapter; Craig & Knappett, Craig's Soil Mechanics, sheet-pile wall design chapter.

Question 5 of 6: Anchored Sheet-Pile Wall — Active Thrust and Tie-Rod Tension

Question 5: Anchored Sheet-Pile Wall — Active Thrust and Tie-Rod Tension (20 marks)