07-Str-B5 · December 2018

Question 5 of 6: Deep Foundations: bored concrete piles

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

Notes on this paper

Paper format: National Exams, December 2018 — 07-Str-B5 Foundation Engineering. Three hours, open book, any non-communicating calculator permitted. Six questions of equal value (30 marks each); five constitute a complete paper and only the first five appearing in the answer book are marked. All six are solved here. This subject is pure geotechnical engineering — the cover page names it Foundation Engineering.

Reference texts (the books an open-book candidate should have on the desk for this subject):

Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — the Canadian limit-states framework, geotechnical resistance factors, frost depth, pile design.
B. M. Das, Principles of Foundation Engineering, 9th ed. — bearing capacity with shape/depth/inclination factors, retaining walls, driven piles (§12), drilled shafts (§13), pile groups.
R. F. Craig & J. Knappett, Craig's Soil Mechanics, 9th ed. — effective stress, Bishop's simplified method, consolidation settlement.
M. J. Tomlinson & J. Woodward, Pile Design and Construction Practice, 6th ed. — adhesion factors, the equivalent-raft settlement method, block failure of pile groups.
L. C. Reese & M. W. O'Neill, Drilled Shafts: Construction Procedures and Design Methods (FHWA) — the α* = 0.55 shaft rule and the exclusion zones used in Question 5.
D. W. Taylor, Fundamentals of Soil Mechanics; A. W. Bishop & N. Morgenstern, Stability coefficients for earth slopes (Géotechnique, 1960) — the slope-stability charts behind Question 3.

Check — assumptions adopted across this paper. This examination omits at least one parameter that each design step needs. Every adoption is made explicitly here and flagged again at the point of use. (1) Questions 2 and 5 give no groundwater table. Both profiles are clays quoted by undrained shear strength, so they are saturated; the water table is therefore taken at ground level and effective stresses are computed with submerged unit weights. This is the conservative choice for settlement, and the sensitivity is reported in each answer (Question 2 settles 293 mm with the water table at surface against 153 mm with no free water). (2) Question 2 gives no adhesion factor. α is taken from Das's Table 12.6 (Terzaghi, Peck & Mesri, α against c_u/p_a) because that table needs no assumed effective-stress profile; the answer tabulates what Tomlinson's and the API's rules would give instead, and identifies the α at which the group size changes. (3) Question 1 does not say whether the quoted leg loads are factored. They are treated as specified (unfactored) loads, and both a serviceability check against the 200 kPa SLS capacity and an ultimate check at 1.4 × leg load against the 300 kPa factored resistance are carried out; the governing one is stated. (4) Question 5 gives no preconsolidation pressure. The requested Skempton correlation is used on the virgin compression line, and the (much smaller) settlement that follows if the profile's own c_u/σ′_v0 ratio of about 1.1 is honoured as an OCR of 2 is reported alongside. (5) Concrete unit weight is taken as 24 kN/m³ and γ_w as 9.81 kN/m³ throughout. (6) Where a question states only "factor of safety", the gross definition (ultimate bearing capacity over total applied pressure) is used, and the convention is restated in each answer.

Question 5 — Deep Foundations: bored concrete piles (30 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 0.90 m diameter bored (cast-in-place) concrete pile, 10 m long, in a layered clay profile whose undrained strength is tabulated against depth; a square pile group is to carry 20 MN at an overall factor of safety of 3.

Given data — Question 5 (Table 1 read as layer tops)
Depth interval (m)	γ (kN/m³)	c_u (kPa)	Unit shaft resistance f = 0.55c_u (kPa)
0–2	18.5	70	38.5
2–4	18.0	50	27.5
4–6	18.0	50	27.5
6–8	19.5	90	49.5
8–14	20.0	100	55.0
below 14	21.0	120	66.0

Find. The unit shaft and toe resistances; the ultimate capacity of a 10 m pile with its head (a) at grade and (b) 2.00 m below grade; the square group that carries 20 MN at FS = 3 with $S = 3d$; and the consolidation settlement of that group.

Figure 7 — the two load cases of Question 5 in the tabulated clay profile. Lowering the pile head by 2.00 m moves the excluded top 1.5 m into soil that is excavated anyway and puts the toe 2 m deeper in the c_u = 100 kPa layer.

Approach. Apply Reese and O'Neill's total-stress rules for drilled shafts in clay: unit side resistance $f = \alpha^* c_u$ with $\alpha^* = 0.55$, excluding the top 1.5 m and the bottom one diameter of the shaft, and unit toe pressure $q_p = N_c^* c_u$ with $N_c^* = 9$. Integrate down the layer table for each case, size the square group from the summed single-pile capacity with a block check, then settle the group with a Tomlinson equivalent raft using a Skempton-correlated compression index.

Section properties and the exclusion zones. For a 0.90 m diameter straight shaft, $$p = \pi D = 2.827\ \text{m}, \qquad A_p = \frac{\pi D^2}{4} = 0.6362\ \text{m}^2$$
Reese and O'Neill exclude the top 1.5 m of shaft (seasonal moisture change, construction disturbance and the effect of the pile cap) and the bottom one diameter (where the side-shear and end-bearing mechanisms interfere). Those exclusions are the whole reason parts (a) and (b) differ.
Unit resistances. All the tabulated strengths satisfy $c_u/p_a \le 1.5$, so $\alpha^* = 0.55$ applies throughout, giving the unit shaft resistances listed in the Given table (38.5 to 66.0 kPa). At the toe, $$\boxed{q_p = N_c^* c_u = 9(100) = 900\ \text{kPa}\ \text{for a toe in the 8--14 m layer}}$$
which is well below Reese and O'Neill's 4 MPa ceiling.
Part (a) — pile head at grade, toe at 10 m. The contributing shaft runs from 1.5 m to $10 - 0.9 = 9.1$ m: $$Q_s = \alpha^* p \sum c_{u,i}L_i = 0.55(2.827)\big[70(0.5) + 50(2) + 50(2) + 90(2) + 100(1.1)\big]$$ $$Q_s = 0.55(2.827)(525) = 816.4\ \text{kN} \qquad Q_p = 9(100)(0.6362) = 572.6\ \text{kN}$$ $$\boxed{Q_{u,a} = 816.4 + 572.6 = 1389\ \text{kN}}$$
Part (b) — pile head 2.00 m below grade, toe at 12 m. The pile now spans 2 m to 12 m, so the contributing shaft runs from 3.5 m to 11.1 m. It loses the weak 2–3.5 m band but gains 2 m of the strong $c_u = 100$ kPa layer, and the toe sits 2 m deeper in the same stratum: $$Q_s = 0.55(2.827)\big[50(0.5) + 50(2) + 90(2) + 100(3.1)\big] = 0.55(2.827)(615) = 956.4\ \text{kN}$$ $$Q_p = 9(100)(0.6362) = 572.6\ \text{kN}$$ $$\boxed{Q_{u,b} = 956.4 + 572.6 = 1529\ \text{kN}}$$
Lowering the cut-off level by 2 m buys 10% more capacity for the same length of pile, because the excluded top 1.5 m is spent in soil that was going to be excavated anyway.
Part (c) — number of piles. With an overall factor of safety of 3 on the total load, $$Q_{u,\,required} = 3(20\,000) = 60\,000\ \text{kN} \qquad n = \frac{60\,000}{1529} = 39.2$$
A square group means $n^2$ piles: 6 × 6 = 36 gives only $FS = 2.75$, so
$$\boxed{7\times 7 = 49\ \text{piles at}\ S = 3d = 2.70\ \text{m},\ \ FS = \frac{49(1529)}{20\,000} = 3.7}$$
The group plan is $B_g = L_g = 6(2.70) + 0.90 = 17.1$ m, and the cap would be about 18.1 m square with a nominal 0.5 m edge distance. (If a rectangular arrangement were permitted, 5 × 8 = 40 piles would meet FS = 3 exactly and save nine shafts; the question specifies a square group.)
Part (c) — block failure check. The group is wide and short, so block failure is checked as a single deep footing 17.1 m square embedded from 2 m to 12 m: $$Q_{block} = 4B_g\sum c_{u,i}L_i + 9c_uB_gL_g = 68.4(780) + 9(100)(292.4) = 316\,500\ \text{kN}$$
This is more than four times the summed single-pile capacity of 74 900 kN, so single-pile failure governs and the group efficiency is unity — the expected result at a spacing as open as $3d$.
Part (d) — compressibility parameters from the index data. The question supplies a liquid limit and a specific gravity precisely so that the compression index and void ratio can be inferred. Skempton's correlation for a normally consolidated clay, $$C_c = 0.009(LL - 10) = 0.009(40 - 10) = 0.27$$
and the void ratio follows from the saturated unit weight of the bearing layer, $\gamma_{sat} = (G_s + e)\gamma_w/(1+e) = 20$ kN/m³:
$$e_0 = \frac{G_s\gamma_w - \gamma_{sat}}{\gamma_{sat} - \gamma_w} = \frac{2.72(9.81) - 20}{20 - 9.81} = 0.656 \qquad \frac{C_c}{1+e_0} = 0.163$$
Part (d) — equivalent raft and stress increase. The piles are predominantly friction piles, so the raft goes at two-thirds of the embedded length below the pile heads: $$z_{raft} = 2.0 + \tfrac23(10) = 8.67\ \text{m} \qquad q_{raft} = \frac{20\,000}{17.1^2} = 68.4\ \text{kPa}$$
Below that level the load spreads at 1 in 4, so $\Delta\sigma = Q/(B_g + z/2)^2$. Taking the water table at ground level (none is given, and the profile is quoted by undrained strength) gives $\sigma^{\prime}_{v0} = 76.3$ kPa at the raft and 130.7 kPa at 14 m. The stress increase falls below a tenth of the in-situ effective stress about 20 m below the raft, so the clay is integrated from 8.67 m to 28.7 m in ten 2 m sublayers.

Part (d) — the settlement. $$s_c = \sum \frac{C_c}{1+e_0}H_i\log_{10}\frac{\sigma^{\prime}_{0,i}+\Delta\sigma_i}{\sigma^{\prime}_{0,i}} = 0.163\sum H_i\log_{10}(\cdot)$$

Sublayer contributions, equivalent raft at 8.67 m
Depth (m)	Δσ (kPa)	σ′_v0 (kPa)	Δs (mm)
9.7	64.6	86.5	79.0
11.7	57.8	106.9	61.2
13.7	52.1	127.3	48.6
15.7	47.1	149.3	38.9
17.7	42.9	171.7	31.6
19.7	39.2	194.1	26.0
21.7	35.9	216.5	21.7
23.7	33.1	238.8	18.4
25.7	30.5	261.2	15.7
27.7	28.3	283.6	13.5

$$\boxed{s_c \approx 354\ \text{mm}}$$

Sanity-check the answer against the profile. Three hundred and fifty millimetres is not a credible settlement for a real structure, and the reason is that the calculation has been forced onto the virgin compression line by the data supplied. The profile itself argues otherwise: at 10 m depth $c_u/\sigma^{\prime}_{v0} = 100/90 = 1.1$, whereas a normally consolidated clay sits near 0.25. This clay is therefore heavily overconsolidated, and most of the stress increase would be carried on the recompression line. Repeating the integration with a modest OCR of 2 and $C_r = C_c/6 = 0.045$ gives $$s_c \approx 59\ \text{mm}$$
which is a plausible settlement for a 17 m square, 49-pile group carrying 20 MN. The answer to report is the 354 mm the question's own data produce, with the observation that an oedometer test to establish $\sigma^{\prime}_p$ would almost certainly reduce it by a factor of five or six — and that on a 20 MN structure that test is cheap.

Check — parameters adopted in Question 5. (i) Table 1 is read as layer tops: the tabulated value applies from that depth down to the next listed depth. (ii) $\alpha^* = 0.55$ with the top 1.5 m and bottom 1$D$ excluded, per Reese and O'Neill for straight drilled shafts; the older $\alpha = 0.45$ applied over the whole shaft would give 1362 kN in case (b) and force an 8 × 8 group. (iii) Water table at ground level, and the clay compressed on the virgin line because no $\sigma^{\prime}_p$ is given; the overconsolidated alternative is reported in step 10. (iv) The clay is assumed to continue to at least 29 m, beyond the 14 m where Table 1 stops.

Final results — Question 5
Quantity	Value
Unit shaft resistance	f = 0.55c_u: 27.5 to 66.0 kPa by layer
Unit toe pressure	900 kPa (toe in the c_u = 100 kPa layer)
(a) Head at grade: Q_s / Q_p / Q_u	816.4 / 572.6 / 1389 kN
(b) Head 2.00 m down: Q_s / Q_p / Q_u	956.4 / 572.6 / 1529 kN
(c) Piles required / square group adopted	39.2 → 7 × 7 = 49 piles, FS = 3.7
(c) Spacing / group plan / block capacity	2.70 m / 17.1 m square / 316 500 kN (does not govern)
(d) C_c (Skempton) / e₀ (from G_s)	0.27 / 0.656
(d) Equivalent raft depth / pressure	8.67 m / 68.4 kPa
(d) Consolidation settlement	≈ 354 mm on the virgin line (≈ 59 mm if OCR = 2)

Topic. Drilled-shaft (bored pile) capacity in clay by the Reese & O'Neill method, square group sizing, and equivalent-raft settlement with correlated compressibility.

Key relations. $Q_u = \alpha^*\!\sum c_u p L + N_c^*c_uA_p$ with $\alpha^* = 0.55$, $N_c^* = 9$; Skempton $C_c = 0.009(LL-10)$; $e_0 = (G_s\gamma_w - \gamma_{sat})/(\gamma_{sat}-\gamma_w)$; equivalent raft at $2L/3$ with a 1:4 spread.

Why this works. A bored pile is built by removing soil and replacing it with fluid concrete, so the shaft never gets the driven pile's benefit of reconsolidation against a displaced annulus — instead the borehole wall softens as it takes up water from the concrete and from the drilling fluid. That is why $\alpha^*$ for a drilled shaft is capped near 0.55 rather than reaching 1.0 in soft clay, and why the top 1.5 m, where the disturbance and seasonal moisture change are worst, is discarded altogether. At the toe, $N_c^* = 9$ is the deep-footing limit of Skempton's bearing-capacity factor, reached once the embedment exceeds about four diameters, so the toe of any real pile is in the fully confined regime.

Common pitfall. Forgetting the two exclusion zones and integrating the shaft from the pile head to the toe — worth about 15% here, and always unconservative. In part (d), the recurring error is using the correlated $C_c$ without ever asking whether the clay is normally consolidated: the strength profile in Table 1 says clearly that it is not, and the difference between the two answers is a factor of six.

Source. Reese & O'Neill, Drilled Shafts (FHWA-HI-88-042), §4; Das, Principles of Foundation Engineering 9th ed., §13.7–13.9 (drilled shafts in clay) and §14.5; Skempton (1944) for the $C_c$–LL correlation.

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