16-Civ-B3 Geotechnical Design · May 2017

Question 9 of 9: Ultimate capacity of a pile group with negative skin friction

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 9: Ultimate capacity of a pile group with negative skin friction (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 sixteen-pile group driven through a recently placed sand fill into a clay, with the water table at ground surface.

Given data (Figure 5)
Quantity	Symbol	Value
Group arrangement	—	4 × 4 (16 piles)
Pile diameter	d	406 mm
Total pile length below ground level	L	15 m
Thickness of sand fill	H_f	3 m
Fill, saturated unit weight and friction angle	gamma_sat, phi'	18.5 kN/m³, 35 degrees
Clay, saturated unit weight and friction angle	gamma_sat, phi'	19.5 kN/m³, 20 degrees
Water table	—	at the top of the fill
Failure mode	—	individual piles (given)

Find. The ultimate load capacity of the sixteen-pile group, after allowing for the downdrag imposed on the piles by the settling fill.

[Figure not reproduced: Figure 5 redrawn: over the fill thickness H f the settling sand drags the piles down; below it the clay supplies shaft resistance and end bearing. See the official exam paper.]

Approach. The clay is characterised by an effective friction angle rather than an undrained strength, so the whole problem is treated in effective stress: compute the downdrag over the fill from the beta (Burland) expression for a granular fill, check it against the alternative block mechanism, compute the positive shaft resistance in the clay by the same beta method, add the point resistance from Janbu's bearing-capacity number, multiply by sixteen (individual failure is given) and subtract the group downdrag.

Assumptions, with justification (page 1, Notes 6 and 7). (1) Effective-stress analysis throughout, because the data give phi' for the clay and no undrained strength. (2) Earth-pressure coefficient at the shaft K' = 1 − sin(phi'), the standard at-rest value used for downdrag and for the beta method (Das, Chapter 11). (3) Pile–soil friction angle delta' = 0.6 phi' at the fill–concrete interface, giving delta' = 21.0 degrees; the range 0.5 to 0.8 phi' is explored in the sensitivity note below. In the clay, delta' is taken equal to phi' because the failure surface there forms in the remoulded soil rather than at the concrete face. (4) Pile spacing is not given; s = 3d = 1.218 m is assumed, the usual minimum, and it is used only in the block check because the question directs that the group fails as individual piles. (5) Janbu's bearing-capacity number is used for the point, with the angle of the failure arc eta' = 75 degrees, appropriate to a firm clay (Das quotes 70 degrees for soft clay and 105 degrees for dense sand). (6) The fill settles relative to the piles over its full thickness, so downdrag acts over H_f and positive resistance only below it.

Compute the submerged unit weights and the pile geometry. With the water table at the surface both strata are submerged: $$\gamma'_{fill}=18.5-9.81=8.69\ \text{kN/m}^{3},\qquad \gamma'_{clay}=19.5-9.81=9.69\ \text{kN/m}^{3}$$ $$p=\pi d=\pi(0.406)=1.2755\ \text{m},\qquad A_p=\frac{\pi}{4}(0.406)^{2}=0.12946\ \text{m}^{2}$$
Evaluate the negative skin friction over the fill. The fill is newly placed and consolidates under its own weight, so it moves down relative to the piles and the interface shear acts downwards. With the shear stress at depth z given by f_n = K' gamma'_fill z tan(delta'), integration over the fill thickness gives $$F_n=\int_0^{H_f}pK'\gamma'_{fill}z\tan\delta'\,dz=\frac{pK'\gamma'_{fill}H_f^{2}\tan\delta'}{2}$$ with K' = 1 − sin 35 degrees = 0.4264 and delta' = 21.0 degrees, $$F_n=\frac{(1.2755)(0.4264)(8.69)(3)^{2}\tan(21.0^\circ)}{2}$$ $$\boxed{F_n=8.16\ \text{kN per pile}}$$
Check the downdrag against the block mechanism. The fill trapped inside the group can only hang on the piles by an amount limited by its own weight plus the friction on the perimeter of the block. With s = 3d the group plan dimension is B_g = 3s + d = 4.060 m, so $$F_{n,block}=B_g^{2}H_f\gamma'_{fill}+\frac{4B_gK'\gamma'_{fill}H_f^{2}\tan\delta'}{2}=429.7+104.0=533.7\ \text{kN}$$ against 16 F_n = 130.6 kN. The individual-pile mechanism is much the smaller, so it governs and the group downdrag is 130.6 kN.
Establish the vertical effective stress profile. At the top of the clay and at the pile tip, $$\sigma'_v(3\ \text{m})=(8.69)(3)=26.07\ \text{kPa},\qquad \sigma'_v(15\ \text{m})=26.07+(9.69)(12)=142.35\ \text{kPa}$$ The average over the 12 m of clay is 84.21 kPa.
Compute the positive shaft resistance in the clay by the beta method. With K = 1 − sin 20 degrees = 0.6580, $$\beta=K\tan\phi'_{clay}=(0.6580)\tan 20^\circ=0.2395$$ $$f_{av}=\beta\,\bar{\sigma}'_v=(0.2395)(84.21)=20.167\ \text{kPa}$$ $$Q_s=p(L-H_f)f_{av}=(1.2755)(12)(20.167)=308.7\ \text{kN}$$
Compute the point resistance. Janbu's bearing-capacity number for phi' = 20 degrees with eta' = 75 degrees is $$N_q^{*}=\left(\tan\phi'+\sqrt{1+\tan^{2}\phi'}\right)^{2}e^{2\eta'\tan\phi'}=5.289$$ $$Q_p=A_p\sigma'_v(\text{tip})N_q^{*}=(0.12946)(142.35)(5.289)=97.5\ \text{kN}$$ The point contributes only 24 per cent of the single-pile capacity, confirming that this is a friction pile.
Assemble the single-pile and group capacities. For one pile, $$Q_u=Q_s+Q_p=308.7+97.5=406.1\ \text{kN}$$ Because the group is stated to fail as individual piles the group efficiency is unity and $$Q_{g(u)}=16Q_u=16(406.1)=6498\ \text{kN}$$
Deduct the downdrag to obtain the load available to the structure. The negative skin friction is an imposed load, not a loss of resistance, so it is subtracted from the capacity available for the superstructure, $$\boxed{Q_{g(net)}=6498-130.6=6368\ \text{kN}}$$ The downdrag consumes only 2.0 per cent of the ultimate capacity here, because the fill is thin and, being below the water table, light. Had the fill been 8 m thick and drained the same calculation would remove several times as much.

Final results — Question 9
Quantity	Symbol	Value
Negative skin friction, one pile	F_n	8.16 kN
Downdrag on the group (individual mechanism governs)	16 F_n	130.6 kN
Shaft resistance in clay, one pile	Q_s	308.7 kN
Point resistance, one pile	Q_p	97.5 kN
Ultimate capacity, one pile	Q_u	406.1 kN
Ultimate capacity of the group, before downdrag	Q_g(u)	6498 kN
Ultimate group capacity allowing for negative skin friction	Q_g(net)	6368 kN (say 6.4 MN)

Check — the two assumed coefficients and what they cost. Varying the interface friction ratio over its defensible range changes the downdrag but almost nothing else: delta' = 0.5 phi' gives F_n = 6.71 kN per pile and a group capacity of 6391 kN, while delta' = 0.8 phi' gives 11.31 kN per pile and 6317 kN — a spread of under one per cent on the answer, so the assumption is not load-bearing. The point-resistance assumption matters more: taking eta' = 105 degrees instead of 75 degrees raises N_q* from 5.29 to 7.74 and the group capacity to about 7090 kN, some 11 per cent higher. Since the point supplies less than a quarter of the capacity, the answer is still robust, but the chart source must be quoted as page 1 Note 6 requires. Finally, a real design would apply a factor of safety of 2.5 to 3 to the ultimate value and would add the downdrag to the working load rather than deducting it from the ultimate, giving an allowable structural load of roughly 2469 kN; and it would check the block failure mode, which the question expressly sets aside.

Topic. Negative skin friction (downdrag) and the effective-stress (beta) method for piles.

Key relations. Downdrag from a granular fill, F_n = p K' gamma'_fill H_f² tan(delta')/2 with K' = 1 − sin(phi'); block limit, F_n,block = weight of fill within the group plus perimeter friction; beta method, f = beta sigma'_v with beta = K tan(delta'); point resistance Q_p = A_p sigma'_v N_q*; Janbu, N_q* = [tan(phi') + sqrt(1 + tan²phi')]² exp(2 eta' tan phi'); group capacity in the individual mode, Q_g(u) = n Q_u.

Why this works. Skin friction has a sign, and the sign is set by the relative movement of soil and pile, not by the loading. Where a recently placed fill consolidates faster than the pile settles, the soil moves down past the pile and the interface shear acts downwards, adding to the load. Below the neutral plane — the depth at which pile and soil settle equally — the pile moves down relative to the soil and the shear reverses to become resistance. Because both the downdrag and the resistance are proportional to the vertical effective stress, an effective-stress formulation handles them with one expression and a change of sign. The block check exists because downdrag is a load transferred from a finite mass of fill: the group cannot be dragged down by more fill than it encloses plus what it can drag in from the perimeter.

Common pitfall. Treating downdrag as a reduction in shaft resistance rather than as an added load. The two are not equivalent when a factor of safety is applied: current practice adds the unfactored downdrag to the working load and factors only the resistance. A second pitfall is applying downdrag over the full pile length rather than only above the neutral plane, which for a fill over a competent stratum is essentially the fill thickness.

Source. Das, Principles of Foundation Engineering, Sections 11.13–11.14 (negative skin friction), 11.9 (beta method) and Eq. (11.33) (Janbu); Canadian Foundation Engineering Manual, 4th ed., Chapter 18, which sets out the neutral-plane approach; Fellenius (1972, 2006) on downdrag and the unified design method.

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