16-Civ-B3 Geotechnical Design · May 2017

Question 8 of 9: Anchored sheet pile wall by the free-earth-support method

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, May 2017 — 16-Civ-B3 Geotechnical Design; three hours, open book, any non-communicating calculator. Section A holds five discussion questions worth 7 marks each of which four are marked; Section B holds four design questions worth 24 marks each of which three are marked, so the examinable total is 4 × 7 + 3 × 24 = 100 marks. All nine questions are worked below, because the set is a study resource rather than a marked script.

Reference texts. B. M. Das, Principles of Foundation Engineering, 7th–9th ed. (Cengage); B. M. Das, Principles of Geotechnical Engineering; R. F. Craig, Craig's Soil Mechanics, 8th ed. (Knappett & Craig); Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed.; J. E. Bowles, Foundation Analysis and Design; ASTM D1586 (SPT) and D5778 (CPT).

Source of charts and assumed values (page 1, Note 6). Every design coefficient used below is named where it is used: Terzaghi bearing-capacity factors from Das, Principles of Foundation Engineering, Table 3.1 (values computed by Kumbhojkar, 1993); Vesic/Reissner factors and the shape, depth and inclination factors from Das Table 3.4 and Eqs. (3.19)–(3.26); drilled-shaft adhesion factor alpha* = 0.55 from Reese & O'Neill (1989) as tabulated by Das, Chapter 12; bearing factor Nc* = 9 from Skempton (1951); earth-pressure coefficient for downdrag K' = 1 − sin(phi') from Das, Chapter 11; Janbu's bearing-capacity number for the pile point from Das Eq. (11.33). Assumed values (adhesion ratio, pile spacing, rigidity index) are stated in a callout beside the step that uses them.

Question 8: 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. An anchored sheet pile wall retaining 8.00 m of granular backfill over an in-situ c'–phi' soil, with the water level equal on both sides of the wall.

Given data (Figure 4)
Quantity	Symbol	Value
Retained height (backfill to dredge level)	H	8.00 m
Water level, both sides, below backfill surface	h_w	5.00 m
Tie-rod level below backfill surface	z_t	1.50 m
Tie-rod horizontal spacing	s	2.50 m
Backfill above water table	gamma	17 kN/m³
Backfill saturated	gamma_sat	20 kN/m³
Backfill strength	c', phi'	0, 35 degrees
In-situ soil saturated unit weight	gamma_sat	21 kN/m³
In-situ soil strength	c', phi'	10 kN/m², 27 degrees
Factor of safety on gross passive resistance	F	2

Find. (a) the depth of embedment D below dredge level that gives a factor of safety of 2 on the gross passive resistance, and (b) the force carried by each tie rod.

Wall geometry (left) and the effective-stress pressure diagram (right); pressures in kPa. Because the water level is the same on both sides the net water pressure on the wall is zero.

Approach. Use the free-earth-support method: assume the toe is free to rotate, build the active and passive effective-stress diagrams, take moments about the tie-rod level with the gross passive resistance divided by the factor of safety to find D, then obtain the tie force from horizontal equilibrium.

Assumptions, with justification. (1) Net water pressure is zero. The water level is stated to be 5.00 m below the backfill surface on both sides of the wall, so the hydrostatic pressure diagrams on the two faces are identical and cancel; only effective stresses need be carried through the calculation. (2) Free earth support. The wall is treated as simply supported at the tie rod and free at the toe, giving a determinate problem — the standard assumption for a wall of ordinary flexibility and the one implied by asking for an embedment at a stated factor of safety. (3) Rankine earth pressures on a vertical, frictionless wall face, with no wall friction or adhesion mobilised. Neglecting wall friction on the passive side is conservative. (4) No surcharge is shown on Figure 4 and none is assumed. (5) The factor of safety is applied to the gross passive resistance, as the question directs, not to the shear-strength parameters and not to a net passive pressure.

Compute the earth-pressure coefficients for the two materials. For the granular backfill and for the in-situ soil respectively, $$K_{a1}=\tan^{2}\!\left(45^\circ-\frac{35^\circ}{2}\right)=0.2710,\qquad K_{a2}=\tan^{2}\!\left(45^\circ-\frac{27^\circ}{2}\right)=0.3755$$ $$K_{p2}=\frac{1}{K_{a2}}=2.6629$$ The submerged unit weights follow directly: for the backfill gamma' = 20 − 9.81 = 10.19 kN/m³, and for the in-situ soil gamma' = 21 − 9.81 = 11.19 kN/m³.
Build the vertical effective stress profile behind the wall. Above the water table the moist backfill acts at its full weight; below it the submerged weight applies: $$\sigma'_v(5\ \text{m})=(17)(5)=85.00\ \text{kPa},\qquad \sigma'_v(8\ \text{m})=85.00+(10.19)(3)=115.57\ \text{kPa}$$
Convert to active pressure, watching the change of material at dredge level. Within the backfill the cohesionless expression applies, $$p_a(5)=K_{a1}\sigma'_v=(0.2710)(85.00)=23.03\ \text{kPa},\qquad p_a(8^-)=(0.2710)(115.57)=31.32\ \text{kPa}$$ Immediately below dredge level the material changes to the cohesive soil and the cohesion term appears, $$p_a(8^+)=K_{a2}\sigma'_v-2c'\sqrt{K_{a2}}=(0.3755)(115.57)-2(10)\sqrt{0.3755}=31.14\ \text{kPa}$$ The two values either side of dredge level, 31.32 and 31.14 kPa, happen to be almost equal: the larger coefficient of the weaker soil is offset almost exactly by its cohesion. Below dredge level the active pressure grows at $$\frac{dp_a}{dz}=K_{a2}\gamma'=(0.3755)(11.19)=4.2021\ \text{kPa/m}$$
Build the passive pressure in front of the wall. The effective vertical stress at dredge level in front is zero (only water stands above it), so the passive diagram starts at the cohesion intercept and grows linearly, $$p_p=K_{p2}\gamma' y+2c'\sqrt{K_{p2}}=29.798\,y+32.637\ \text{kPa}$$ with y measured downwards from dredge level.
Resolve each block of the diagram into a force and a lever arm about the tie rod. Taking moments about the anchor at 1.5 m depth, the active blocks are the triangle in the dry backfill, the rectangle and triangle in the submerged backfill, and the rectangle and triangle below dredge level; the passive blocks are the rectangle from the cohesion intercept and the triangle from the frictional growth. Writing the moment equation with the gross passive resistance divided by F = 2, $$\frac{1}{F}\sum P_p\,\bar{y}_p=\sum P_a\,\bar{y}_a$$ gives a cubic in D. Solving it numerically, $$\boxed{D=4.69\ \text{m}\ \text{(say 4.7 m)}}$$ so the total length of the sheet piling is 8.00 + 4.69 = 12.69 m, say 12.7 m.
Confirm the moment balance at that embedment. With D = 4.695 m the active moment about the tie rod is 2259 kN·m per metre run and the gross passive moment is 4518 kN·m per metre run, whose ratio is exactly 2.000 — the required factor of safety of 2.
Obtain the tie-rod force from horizontal equilibrium. The total active thrust at this embedment is $$\sum P_a=331.63\ \text{kN/m},\qquad \sum P_p=481.60\ \text{kN/m}$$ and, consistent with the factored passive resistance used to size D, $$T=\sum P_a-\frac{\sum P_p}{F}=331.63-\frac{481.60}{2}=90.83\ \text{kN per metre run}$$
Convert to the force in one tie rod. The rods are at 2.5 m centres, so each carries $$\boxed{T_{rod}=(90.83)(2.5)=227.1\ \text{kN}}$$ For detailing purposes this should be increased by about 25 per cent to allow for arching and for the redistribution that occurs as the wall flexes, giving a design rod force of roughly 284 kN; a 40 mm diameter mild-steel rod at 140 MPa working stress carries about 176 kN, so a 50 mm rod would be selected.

Final results — Question 8
Quantity	Symbol	Value
Active coefficient, backfill (phi' = 35°)	K_a1	0.2710
Active / passive coefficients, in-situ soil (phi' = 27°)	K_a2, K_p2	0.3755 / 2.6629
Total active thrust	sum P_a	331.6 kN/m
Gross passive resistance	sum P_p	481.6 kN/m
(a) Required embedment, F = 2 on gross passive	D	4.69 m (say 4.7 m)
Total length of sheet piling	H + D	12.69 m
Anchor force per metre run	T	90.8 kN/m
(b) Force in each tie rod at 2.5 m centres	T_rod	227.1 kN

Check — how sensitive is the answer to the definition of the factor of safety? Applying the same factor to the gross passive resistance but at F = 1.5 gives D = 3.37 m and T = 204 kN per rod; at F = 2.5, D = 6.25 m and T = 254 kN. The embedment is therefore quite sensitive to the definition, which is why the question specifies it. Two further checks belong in a full design and are noted rather than performed here: the maximum bending moment in the piling, found at the level of zero net pressure between the anchor and dredge level, and reduced by Rowe's moment-reduction factor for a flexible wall; and the stability of the anchor block itself, which must be founded beyond the active wedge of the wall so that the two failure surfaces do not interact.

Topic. Anchored sheet pile walls: the free-earth-support method.

Key relations. K_a = tan²(45° − phi'/2), K_p = 1/K_a; active pressure p_a = K_asigma'_v − 2c'sqrt(K_a); passive pressure p_p = K_psigma'_v + 2c'sqrt(K_p); moment equilibrium about the anchor, sum(P_p y_p)/F = sum(P_a y_a); horizontal equilibrium, T = sum P_a − sum P_p/F; anchor force per rod, T_rod = T s.

Why this works. An anchored wall has two unknowns, the embedment and the anchor force, and the free-earth-support idealisation supplies exactly two equations: moments about the anchor, which contains no anchor force and so isolates the embedment; and horizontal equilibrium, which then gives the anchor force. The assumption behind it is that the wall is stiff enough not to develop a fixing moment at the toe, so the pressure distribution below dredge level is the full Rankine passive one and the toe is free to rotate. Because the water level is the same on both sides, the pore pressures cancel and only effective stresses do work — a considerable simplification, and one that disappears the moment there is a head difference across the wall, when a seepage analysis is needed and the passive resistance is reduced by the upward gradient in front of the wall.

Common pitfall. Adding a net water pressure that is not there, or omitting one that is. A second pitfall is taking moments about the toe rather than the anchor, which brings the unknown anchor force into the moment equation and leaves the problem indeterminate. A third is applying the factor of safety to a net passive pressure (passive minus active below dredge level), which is a different and less conservative definition and gives a noticeably smaller embedment; the question here explicitly names the gross passive resistance.

Source. Craig's Soil Mechanics, 8th ed., Sections 7.7–7.9 (anchored sheet pile walls, free and fixed earth support, Rowe's moment reduction); Das, Principles of Foundation Engineering, Chapter 9, Eqs. (9.35)–(9.44); Canadian Foundation Engineering Manual, 4th ed., Chapter 26.

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