Question 7 of 9: Bored pile group in clay — number of piles, group efficiency and settlement

Question 7: Bored pile group in clay — number of piles, group efficiency and settlement (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. Bored cast-in-place concrete piles, 300 mm diameter, 10 m long, in a 12 m deep clay deposit, carrying a total load of 1800 kN. The clay properties are as tabulated in the question:

Find. The number of piles required, a workable arrangement for them, the efficiency of the resulting group, and the approximate consolidation settlement of the clay layer under the 1800 kN load given a liquid limit of 40.

Approach. Size a single pile by the alpha (total-stress) method, divide the load by the allowable single-pile capacity to get the number of piles, lay them out at three diameters centre to centre, test the group against block failure to obtain the efficiency, and finally treat the group as an equivalent raft at two-thirds of the pile length and compute one-dimensional consolidation settlement of the clay below it.

1. Shaft geometry. For a 300 mm bored pile the perimeter and base area are $$p = \pi D = \pi(0.30) = 0.9425\ \text{m}, \qquad A_p = \frac{\pi D^{2}}{4} = \frac{\pi(0.30)^{2}}{4} = 0.0707\ \text{m}^{2}.$$
2. Skin friction by the alpha method. Adhesion is taken as $f_s = \alpha c_u$ with alpha = 0.45, Skempton's value for bored piles in stiff clay as reproduced in CFEM Ch. 18; the profile is split into five 2 m sublayers over the 10 m embedded length and the mean c_u of each sublayer is used.

   Shaft resistance, alpha = 0.45

   Sublayer (m)	0–2	2–4	4–6	6–8	8–10
   Mean cu (kPa)	39.0	49.5	60.5	73.0	85.0
   ΔQs = 0.45 cu p ΔL (kN)	33.1	42.0	51.3	61.9	72.1

   Summing the five contributions, $$Q_s = \sum \alpha c_u\, p\, \Delta L = 260.4\ \text{kN}.$$
3. End bearing. At the pile toe, 10 m down, the undrained strength is 90 kPa, and for a pile in clay the bearing factor is $N_c^{*} = 9$ (Skempton, 1951): $$Q_p = N_c^{*} c_{u(base)} A_p = 9(90)(0.0707) = 57.3\ \text{kN}.$$
4. Ultimate and allowable capacity of one pile. Adding the two components and applying a factor of safety of 3, the usual value for a bored pile designed without a load test, $$Q_{ult} = Q_s + Q_p = 260.4 + 57.3 = 317.7\ \text{kN},\qquad Q_{all} = \frac{317.7}{3} = \boxed{105.9\ \text{kN per pile}}$$
5. Number of piles. Dividing the applied load by the allowable capacity, $$n = \frac{1800}{105.9} = 17.0 \;\longrightarrow\; \text{adopt } n = 18\ \text{piles}.$$ Eighteen piles give 18(105.9) = 1906 kN of allowable capacity against the 1800 kN applied, a margin of about six per cent, and they arrange neatly as three rows of six.
6. Arrangement. Use a rectangular 3 × 6 grid at a centre-to-centre spacing of three diameters, $s = 3D = 0.90$ m, which is the minimum spacing that keeps group action mild and remains constructible for bored piles. The group footprint is then $$B_g = (3-1)(0.90)+0.30 = 2.10\ \text{m}, \qquad L_g = (6-1)(0.90)+0.30 = 4.80\ \text{m},$$ tied together by a reinforced-concrete pile cap of about 800 mm thickness with a 150 mm edge projection beyond the outer piles.
7. Group efficiency by the block-failure criterion. A pile group in clay can fail either as eighteen separate piles or as one block of soil 2.10 m by 4.80 m by 10 m deep. The block capacity is $$Q_{g(u)} = 2\left(B_g+L_g\right)L\,\bar{c}_u + N_c^{*}c_{u(base)}B_gL_g,$$ with the mean undrained strength over the shaft $\bar{c}_u = (39.0+49.5+60.5+73.0+85.0)/5 = 61.4$ kPa. Substituting, $$Q_{g(u)} = 2(2.10+4.80)(10)(61.4) + 9(90)(2.10)(4.80) = 8473 + 8165 = 16\,638\ \text{kN},$$ against the sum of the individual ultimate capacities, $18(317.7) = 5718$ kN. Block failure is nowhere near critical, so individual pile action governs and $$\eta = \frac{Q_{g(u)}}{\sum Q_{u}} \gt 1 \;\longrightarrow\; \boxed{\eta = 1.0 \ \text{(individual pile action governs)}}$$
8. Cross-check against the empirical Converse–Labarre formula. Many texts quote the geometric expression $$\eta = 1-\frac{\theta}{90}\left[\frac{(n_c-1)n_r+(n_r-1)n_c}{n_rn_c}\right], \qquad \theta = \tan^{-1}\!\left(\frac{D}{s}\right) = \tan^{-1}(1/3) = 18.43^{\circ},$$ which for a 3 by 6 group gives $\eta = 1-\left(18.43/90\right)(1.5) = 0.693$. This value is quoted for completeness, but it is a spacing formula calibrated on driven piles in sand and takes no account of the soil at all; for a bored group in clay the block criterion above is the defensible one, and it is the one carried into the design. Adopting 0.693 instead would require 25 piles.
9. Set up the equivalent raft for settlement. Because the load is carried mainly in friction, it is transferred to the clay at roughly two-thirds of the pile length, $z_{raft} = \tfrac{2}{3}(10) = 6.67$ m below ground. The compressible clay below that level runs to the base of the deposit at 12 m, so $$H_c = 12.0-6.67 = 5.33\ \text{m},\qquad z_{mid} = 6.67+\tfrac{5.33}{2} = 9.33\ \text{m below ground}.$$
10. Stress increase at mid-depth by 2:1 spread. Spreading the raft load at two vertical to one horizontal from the equivalent raft, $$\Delta\sigma = \frac{Q}{\left(B_g+z\right)\left(L_g+z\right)} = \frac{1800}{(2.10+2.67)(4.80+2.67)} = \frac{1800}{35.6} = 50.6\ \text{kPa},$$ where z = 2.67 m is the depth below the equivalent raft to the mid-point of the compressible layer.
11. Existing effective overburden at mid-depth. No water table is given; for a clay deposit of this description the conservative and usual assumption is that it is saturated to ground level, so submerged unit weights apply. Using the mean unit weight of each 2 m sublayer, $$\sigma_0^{\prime} = \sum(\gamma-\gamma_w)\Delta z = 8.44(2)+8.94(2)+9.44(2)+9.94(2)+10.19(1.33) = 87.1\ \text{kPa}.$$
12. Compression index and void ratio. With a liquid limit of 40, the Terzaghi and Peck correlation for a normally consolidated clay gives $$C_c = 0.009\left(LL-10\right) = 0.009(40-10) = 0.27,$$ and back-figuring the void ratio from the saturated unit weight of the compressible zone (mean 19.94 kN/m³) with $G_s = 2.70$, $$e_0 = \frac{G_s\gamma_w-\gamma_{sat}}{\gamma_{sat}-\gamma_w} = \frac{2.70(9.81)-19.94}{19.94-9.81} = 0.65 .$$
13. Primary consolidation settlement. For a normally consolidated clay, $$S_c = \frac{C_cH_c}{1+e_0}\log_{10}\!\left(\frac{\sigma_0^{\prime}+\Delta\sigma} {\sigma_0^{\prime}}\right) = \frac{0.27(5.33)}{1.65}\log_{10}\!\left(\frac{87.1+50.6}{87.1}\right)$$ $$S_c = 0.873\left(0.1987\right) = 0.173\ \text{m} = \boxed{173\ \text{mm}}$$

A settlement of 173 mm is far beyond what any building frame would tolerate — 25 mm total and an angular distortion of 1/500 are the usual serviceability limits — so the honest answer to the question includes the observation that this group is capacity-adequate but settlement-controlled. The practical responses are to extend the piles through the clay onto the firmer stratum below 12 m so that the load bypasses the compressible zone, to spread the load over a piled raft so that the stress increment falls, or to preload the site. The calculation as asked is the number the question wants; the recommendation is what the client needs.