16-Civ-B19 Foundation Engineering · December 2018

Question 5 of 5: Capacity and Efficiency of a 3 × 4 Pile Group in Layered Clay

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

Notes on this paper

Paper format. National Exams, December 2018 — 16-Civ-B19 Foundation Engineering. Three hours, OPEN BOOK (one textbook plus one hand-written 8.5″ × 11″ aid sheet, both sides). Five questions, all to be answered, all of equal weight (20 marks each, 100 marks total). Any non-communicating calculator is permitted. The paper mixes SI and US customary units question by question, so each answer below stays in the units the question is posed in.

Reference texts

Das, B.M., Principles of Foundation Engineering, 9th ed. — the standard text for this syllabus. Ch. 3 (ultimate bearing capacity), Ch. 5 (vertical stress increase in soil), Ch. 6 (shallow-foundation settlement and plate load tests), Ch. 7 (mat foundations), Ch. 9 (pile foundations) and Ch. 10 (pile groups) supply every method used here.
Das, B.M., Principles of Geotechnical Engineering, 9th ed. — effective stress, consolidation and the Boussinesq stress-distribution theory the above builds on.
Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed., 2006 — the Canadian practice document: limit-states terminology, geotechnical resistance factors, settlement tolerances and pile design guidance.
National Building Code of Canada (NBCC) 2020, Part 4 — load combinations and the limit-states framework within which the CFEM resistance factors sit.
Bowles, J.E., Foundation Analysis and Design, 5th ed. — alternative bearing-capacity and settlement formulations, useful as a cross-check.

Check: three conflicts in the printed paper, resolved as follows.

Q.2(a) embankment dimensions. The top of the figure reads 14 m + 5 m + 14 m = 33 m and the bottom chain reads 5 m + 11.5 m + 16.5 m = 33 m, so the two are consistent: A₂ lies 5 m inside the left toe and A₁ lies 16.5 m from each toe, i.e. on the embankment centreline.
Q.4(b) undrained shear strength. The paper prints c_u = 18 250 lb/ft². That is about 874 kPa — an order of magnitude stronger than any normal saturated clay and at the clay/weak-rock boundary. The answer below uses the printed value, and also states the result for c_u = 1 825 lb/ft² (87 kPa, a very stiff clay) in case the extra digit is a typographical slip.
Q.4(c) mat dimensions. The text says 75 ft × 100 ft while the figure it refers to is labelled 90 ft × 120 ft. The figure is solved as the primary case (the question says “as shown in figure below”) and the 75 ft × 100 ft case is carried through in full alongside it, so either reading is answered.

Question 5: Capacity and Efficiency of a 3 × 4 Pile Group in Layered Clay (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.

Quantity	Symbol	Value
Group arrangement	$n_1 \times n_2$	3 × 4 (12 piles)
Pile cross-section (square)	$D$	14 in = 1.167 ft
Centre-to-centre spacing	$d$	35 in = 2.917 ft
Embedded length	$L$	60 ft
Upper clay, 0–15 ft	$c_{u(1)}$	1050 lb/ft²
Lower clay, 15–60 ft	$c_{u(2)}$	1775 lb/ft²

Find. (a) the allowable load-bearing capacity of the group; (b) its group efficiency.

Plan and section of the 3 × 4 group. The dashed rectangle is the block perimeter B_g × L_g used for the block-failure check.

Approach. A pile group in clay can fail in either of two ways — the piles punching through individually, or the whole block of piles and enclosed soil shearing as one unit. Compute both, take the smaller, then divide by the factor of safety. Efficiency follows from the same two numbers, cross-checked with an empirical spacing rule.

Check: the factor of safety is not stated in Q.5. $FS = 4$ is adopted, matching the value the same paper specifies for the single pile in Q.3(b) and consistent with normal practice for a pile group designed from soil parameters without a load test. State the assumption; the ultimate capacities below are unaffected by it.

Part (a), individual-pile mode — geometry of one pile. For a 14 in square pile, $$A_p = D^2 = (1.167)^2 = 1.3611\ \text{ft}^2, \qquad p = 4D = 4(1.167) = 4.667\ \text{ft}$$
Point resistance of one pile. The toe is in the lower clay, so $$Q_p = 9\,A_p\,c_{u(2)} = 9(1.3611)(1\,775) = 21\,744\ \text{lb} = 21.74\ \text{kip}$$
Shaft resistance of one pile by the $\alpha$-method. From the supplied $\alpha$–$c_u$ chart, $\alpha_1 = 0.86$ at $c_u = 1\,050\ \text{lb/ft}^2$ and $\alpha_2 = 0.60$ at $c_u = 1\,775\ \text{lb/ft}^2$: $$Q_s = \alpha_1 c_{u(1)} p L_1 + \alpha_2 c_{u(2)} p L_2$$ $$Q_s = 0.86(1\,050)(4.667)(15) + 0.60(1\,775)(4.667)(45) = 63.2 + 223.7 = 286.86\ \text{kip}$$
Sum for all twelve piles. One pile carries $Q_u = 21.74 + 286.86 = 308.60\ \text{kip}$, so acting individually the group carries $$\sum Q_u = n_1 n_2 Q_u = 12(308.60) = \boxed{3\,703\ \text{kip}}$$
Block mode — dimensions of the equivalent block. The block perimeter runs around the outside faces of the corner piles: $$B_g = (n_1 - 1)d + D = 2(2.917) + 1.167 = 7.00\ \text{ft}$$ $$L_g = (n_2 - 1)d + D = 3(2.917) + 1.167 = 9.917\ \text{ft}$$
Shear resistance around the block. The block shears through intact clay at full undrained strength (no $\alpha$ reduction — the failure surface is soil-on-soil, not soil-on-steel): $$Q_{s,block} = \sum 2(L_g + B_g)\,c_u\,\Delta L = 2(9.917 + 7.00)\left[(1\,050)(15) + (1\,775)(45)\right]$$ $$Q_{s,block} = 33.83 \times 95\,625 = 3\,235.3\ \text{kip}$$
End bearing of the block. With $L/B_g = 60/7.00 = 8.57$ (beyond the right-hand edge of the chart, where the curves have flattened) and $L_g/B_g = 9.917/7.00 = 1.42$, the supplied chart gives $N_c^{*} \approx 8.75$: $$Q_{p,block} = L_g B_g c_{u(2)} N_c^{*} = (9.917)(7.00)(1\,775)(8.75) = 1\,078.1\ \text{kip}$$ $$Q_{u,block} = 3\,235.3 + 1\,078.1 = 4\,313.4\ \text{kip}$$
Choose the governing mode and apply the factor of safety. The individual-pile sum (3 703 kip) is smaller than the block capacity (4 313 kip), so individual action governs — the piles are spaced widely enough ($d/D = 2.5$) that block failure is not the critical mechanism. Hence $$Q_{g(u)} = 3\,703\ \text{kip}, \qquad Q_{all} = \frac{3\,703}{4} = \boxed{926\ \text{kip}}$$ that is about 4 120 kN, or roughly 77 kip per pile at working load.
Part (b), efficiency from the two capacities. The most direct definition compares what the group actually delivers with the sum of its isolated piles: $$\eta = \frac{Q_{g(u)}}{n_1 n_2 Q_u} = \frac{3\,703}{3\,703} = 1.00$$ because the block mode did not govern. Equivalently, the block capacity is $4\,313/3\,703 = 1.16$ times the individual sum, so at this spacing there is no group loss at ultimate load and $\eta$ may be taken as unity.
Cross-check with an empirical spacing formula. Designers who prefer a spacing-based estimate use Converse–Labarre: $$\eta = 1 - \frac{\theta\left[(n_1 - 1)n_2 + (n_2 - 1)n_1\right]}{90\,n_1 n_2}, \qquad \theta = \tan^{-1}\!\left(\frac{D}{d}\right)$$ With $\theta = \tan^{-1}(14/35) = 21.80^\circ$, $$\eta = 1 - \frac{21.80\left[(2)(4) + (3)(3)\right]}{90(12)} = 1 - \frac{21.80(17)}{1\,080} = 1 - 0.343 = \boxed{0.657 \approx 66\%}$$
Reconcile the two figures. They answer different questions and both are legitimate, which is why the paper says “any method of your choice.” The capacity ratio is a genuine ultimate-limit-state statement about this soil profile and says the group loses nothing at collapse. Converse–Labarre is a geometry-only rule of thumb, calibrated largely on piles in sand, that penalises close spacing regardless of soil; at $d/D = 2.5$ it predicts a 34 % loss. For a group in clay the capacity-ratio result governs design, and the empirical figure is best read as a warning that 2.5 diameters is tight spacing whose settlement behaviour — not its ultimate capacity — deserves a separate check.

Result	Value
Single pile: $A_p$ / $p$	1.3611 ft² / 4.667 ft
Single pile: point / shaft / total	21.74 / 286.86 / 308.60 kip
Individual-pile mode, $\sum Q_u$ (12 piles)	3 703 kip
Block dimensions $B_g \times L_g$	7.00 ft × 9.917 ft
Block mode: perimeter shear / end bearing / total	3 235.3 / 1 078.1 / 4 313.4 kip
Governing ultimate group capacity	3 703 kip (individual action)
(a) Allowable group capacity, $FS = 4$	926 kip (4 120 kN)
(b) Efficiency — capacity ratio	$\eta = 1.00$ (block capacity is 1.16 × the individual sum)
(b) Efficiency — Converse–Labarre	$\eta = 0.657 \approx 66\%$

Topic. Ultimate capacity and efficiency of a pile group in saturated clay: individual action versus block failure.

Key relations. Individual sum $\sum Q_u = n_1n_2\left[9A_pc_{u(\text{toe})} + \sum\alpha c_u p\Delta L\right]$; block $Q_u = L_gB_gc_{u(\text{toe})}N_c^{*} + \sum 2(L_g + B_g)c_u\Delta L$ with $B_g = (n_1-1)d + D$, $L_g = (n_2-1)d + D$; group capacity is the smaller of the two; Converse–Labarre $\eta = 1 - \theta[(n_1-1)n_2 + (n_2-1)n_1]/(90n_1n_2)$.

Why this works. Piles driven close together overlap their zones of influence, so at some spacing it becomes easier for the soil trapped between them to move with them than to shear past each pile. The block calculation is simply the capacity of that alternative mechanism, and nature takes whichever is weaker — hence “take the minimum”. Note the asymmetry in the shear terms: the individual mode uses $\alpha c_u$ because the failure surface is the pile–soil interface, while the block mode uses the full $c_u$ because that surface lies in intact clay. The block gains again at the toe, where $N_c^{*}$ rises with $L/B_g$ from about 6 for a shallow block to a plateau near 9 for a deep one.

Common pitfall. Building $B_g$ and $L_g$ from centre-to-centre spacing only, omitting the extra half pile width on each side; applying $\alpha$ to the block perimeter; and using the upper-layer $c_u$ for the toe. Mixing up $n_1$ and $n_2$ in Converse–Labarre is harmless (the expression is symmetric) but assuming a group efficiency below 1 must be applied is not — in clay at this spacing the rigorous check shows no reduction.

Source. Das, Principles of Foundation Engineering 9th ed., §10.2–10.4 (group efficiency, ultimate capacity of groups in saturated clay, $N_c^{*}$ chart after Whitaker); CFEM 4th ed. Ch. 18.

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