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16-Civ-B3 Geotechnical Design · December 2017

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

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

Notes on this paper

Paper format. National Examinations, December 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, 8th–9th ed. (Cengage); B. M. Das, Principles of Geotechnical Engineering, 9th ed.; R. F. Craig / J. Knappett, Craig's Soil Mechanics, 9th ed.; Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed.; D. P. Coduto, Foundation Design: Principles and Practices; J. E. Bowles, Foundation Analysis and Design; ASTM D1586 (SPT), D5778 (CPTu), D2573 (field vane).

Source of design charts and assumed values (page 1, Note 6). Rankine active and passive coefficients from Das, Principles of Foundation Engineering, Ch. 7 (Eqs. 7.11 and 7.30); cantilever-wall stability procedure from Das Ch. 8 (Eqs. 8.11–8.14). Drilled-shaft adhesion factor alpha* = 0.55 and the 1.5 m surface exclusion from Reese & O'Neill (1989), tabulated in Das Ch. 12; bearing factor Nc* = 9 from Skempton (1951); block-failure check from Das Ch. 11 (Eq. 11.55). Compression index from Skempton's Cc = 0.009(LL − 10). Strain-influence factors and the C1, C2 corrections from Schmertmann, Hartman & Brown (1978) as presented in Das Ch. 5; Es = 500(N60 + 15) kPa from Das Table 5.7 (Bowles). Friction angle from the SPT via Wolff (1989), phi' = 27.1 + 0.3(N1)60 − 0.00054[(N1)60]2, with the Liao & Whitman (1986) overburden correction; Vesic bearing-capacity, shape and depth factors from Das Tables 4.2 and 4.3. Every assumed value (unit weights of concrete, specific gravity for the void ratio, pile spacing, factors of safety) is stated in a callout beside the step that uses it.

Question 7: Bored pile group in clay — number, layout, 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 piles, d = 0.5 m, L = 10 m, in a 12 m clay deposit; working load on the group Q = 2000 kN; liquid limit LL = 40 per cent. The profile:

Depth (m)024681012
gamma (kN/m3)1818.51919.5202020
cu (kN/m2)344455668090100

Find. The number of 10 m piles needed, a practical layout, the group efficiency, and the consolidation settlement of the clay beneath the group.

Plan: 3 x 3 group, s = 3d = 1.5 mLg = Bg = 3.5 m1.5 m9 piles, 500 mm dia., 10 m longSection: equivalent raft at 2L/3 = 6.67 massumed incompressible below 12 mL = 10 mHc = 5.33 mequivalent raft 3.5 x 3.5 mmid-depth z = 9.33 mQ = 2000 kN on the cap
Adopted 3 × 3 group at a centre-to-centre spacing of three diameters, and the equivalent raft placed at two-thirds of the pile length used for the settlement calculation. The 2:1 spread carries the raft pressure down to the mid-depth of the remaining clay.

Approach. Compute the ultimate capacity of a single bored pile as shaft adhesion plus end bearing by the alpha method, divide by a factor of safety of 3 to get the working capacity, and hence the number of piles; lay them out on a three-diameter grid; check the group against block failure to obtain the efficiency; then treat the group as an equivalent raft at 2L/3 and compute one-dimensional consolidation settlement of the clay between that level and the base of the deposit.

Check — assumptions invited by the question's note. Adhesion factor alpha* = 0.55 (Reese & O'Neill, 1989, for drilled shafts in clay — a bored shaft mobilises far less adhesion than the driven-pile value near unity because the bore softens and remoulds the wall). Bearing factor Nc* = 9 (Skempton, valid for L/d = 20 > 4). Factor of safety 3 on the ultimate single-pile capacity. Pile spacing s = 3d = 1.5 m, the usual minimum for friction piles. Water table at the ground surface, which maximises the settlement. Specific gravity of solids Gs = 2.70, used only to convert the measured saturated unit weight into an initial void ratio. The clay below 12 m is assumed incompressible, and the deposit is assumed normally consolidated so that Skempton's Cc correlation applies.
  1. Pile geometry. $$A_p=\frac{\pi d^{2}}{4}=\frac{\pi(0.5)^{2}}{4}=0.1963\ \text{m}^{2}, \qquad p=\pi d=\pi(0.5)=1.5708\ \text{m}$$
  2. Integrate the undrained strength over the shaft. Taking the tabulated $c_u$ as linear between readings, the average over each 2 m interval is the mean of its end values, so over the 10 m shaft $$\sum c_u\,\Delta L = 2\left[\tfrac{34+44}{2}+\tfrac{44+55}{2}+\tfrac{55+66}{2}+\tfrac{66+80}{2}+\tfrac{80+90}{2}\right]$$ Evaluating the five interval means, $$\sum c_u\,\Delta L = 2\left[39+49.5+60.5+73+85\right]=614.0\ \text{kPa}\!\cdot\!\text{m}$$ which corresponds to a mean undrained strength of 61.4 kPa over the embedded length.
  3. Shaft resistance. With the drilled-shaft adhesion factor, $$Q_s=\alpha^{*}p\sum c_u\,\Delta L = 0.55(1.5708)(614.0)=530.5\ \text{kN}$$
  4. End bearing. At the pile base, 10 m down, the table gives $c_{u(base)} = 90$ kPa, so $$Q_p=A_pN_c^{*}c_{u(base)}=0.1963(9)(90)=159.0\ \text{kN}$$ The base contributes only 23 per cent of the total, which is characteristic of a friction pile in clay and is the reason the shaft treatment matters more than the base treatment here.
  5. Single-pile capacity and the pile count. Adding the two contributions and applying the factor of safety, $$Q_u = 530.5+159.0=689.5\ \text{kN}, \qquad Q_{all}=\frac{689.5}{3}=229.8\ \text{kN}$$ so the number required is $$n=\frac{2000}{229.8}=8.70 \Rightarrow \boxed{n=9\ \text{piles}}$$
  6. Layout. Nine piles arrange naturally as a square 3 × 3 grid at $s = 3d = 1.5$ m, giving plan dimensions $$L_g=B_g=(3-1)s+d=2(1.5)+0.5=3.5\ \text{m}$$ A square grid keeps the resultant of the nine pile reactions under the column, minimises the cap area and makes the group equally stiff about both axes. The cap would be about 3.8 m square with a 150 mm edge projection, and the piles are checked below as a group before the layout is accepted.
  7. Group efficiency by the block-failure criterion. A clay group can fail either as nine separate piles or as a single block of soil 3.5 m square and 10 m deep. The block capacity is the perimeter adhesion (full $c_u$, because the failure surface is soil against soil) plus end bearing on the whole block area: $$Q_{g(block)}=2(L_g+B_g)\sum c_u\Delta L + L_gB_gc_{u(base)}N_c^{*}$$ Substituting the group dimensions, $$Q_{g(block)}=14.0(614.0)+12.25(90)(9)=8596+9923=18\,519\ \text{kN}$$ against the sum of the individual capacities $$\sum Q_u = 9(689.5)=6205\ \text{kN}$$ The block is three times stronger, so the individual mechanism governs and the efficiency is $$\eta=\frac{Q_{g(u)}}{n\,Q_u}=\frac{6205}{6205}=\boxed{1.0\ (100\%)}$$
  8. The empirical alternative, for comparison. The Converse–Labarre formula, which many texts quote for group efficiency, gives with $\theta=\tan^{-1}(d/s)=\tan^{-1}(0.5/1.5)=18.43^{\circ}$, $$\eta = 1-\frac{\theta}{90}\cdot\frac{(m-1)n+(n-1)m}{mn}=1-\frac{18.43}{90}\cdot\frac{2(3)+2(3)}{9}=0.73$$ It is an empirical rule developed for driven piles in sand and it takes no account of the block mechanism; the block check above is the physically based answer and is the one adopted. Were the 0.73 imposed as a design requirement, the group would need $2000/(229.8 \times 0.73)=11.9$, say a 4 × 4 grid, whose Converse–Labarre efficiency of 0.69 gives 2547 kN.
  9. Compressibility parameters for the settlement. Skempton's correlation for a normally consolidated clay gives the compression index from the liquid limit, $$C_c=0.009(LL-10)=0.009(40-10)=0.27$$ and the initial void ratio follows from the measured saturated unit weight at depth, $\gamma_{sat}=20$ kN/m3, with $G_s=2.70$: $$\gamma_{sat}=\frac{(G_s+e_0)\gamma_w}{1+e_0} \Rightarrow e_0=\frac{G_s\gamma_w-\gamma_{sat}}{\gamma_{sat}-\gamma_w}=\frac{2.70(9.81)-20}{20-9.81}=0.637$$
  10. Equivalent raft and the compressible thickness. For a friction pile group in clay the load is taken to be delivered to the soil at two-thirds of the embedded length, $$z_{raft}=\tfrac{2}{3}L=\tfrac{2}{3}(10)=6.67\ \text{m}$$ so the clay that consolidates is the layer from 6.67 m to the base of the deposit at 12 m, $$H_c=12-6.67=5.33\ \text{m}$$ and its mid-depth, at which the stresses are evaluated, is $z = 9.33$ m.
  11. Effective overburden at mid-depth. With the water table at the surface, each 2 m interval contributes its mean submerged unit weight: $$\sigma'_0 = 2(8.44)+2(8.94)+2(9.44)+2(9.94)+1.33(10.19)=87.1\ \text{kPa}$$
  12. Stress increment at mid-depth. Spreading the raft load at 2 vertical to 1 horizontal from the 3.5 m square raft over the 2.67 m from raft level to mid-depth, $$\Delta\sigma=\frac{Q}{(B_g+z)(L_g+z)}=\frac{2000}{(3.5+2.67)^{2}}=\frac{2000}{38.03}=52.6\ \text{kPa}$$
  13. Consolidation settlement. Treating the 5.33 m as one normally consolidated layer, $$S_c=\frac{C_cH_c}{1+e_0}\log_{10}\frac{\sigma'_0+\Delta\sigma}{\sigma'_0}$$ Substituting the parameters found above, $$S_c=\frac{0.27(5.33)}{1.637}\log_{10}\frac{139.7}{87.1}=0.8798(0.2050)=\boxed{0.180\ \text{m}=180\ \text{mm}}$$
QuantityValue
Shaft resistance Qs (alpha* = 0.55)530.5 kN
End bearing Qp (Nc* = 9)159.0 kN
Ultimate single-pile capacity Qu689.5 kN
Allowable single-pile capacity (FS = 3)229.8 kN
Number of piles required9
Arrangement3 × 3 square grid, s = 3d = 1.5 m, cap 3.5 m × 3.5 m
Block capacity / sum of individual capacities18 519 kN / 6205 kN
Group efficiency (block criterion)1.0 — individual pile action governs
Converse–Labarre efficiency (empirical, for comparison)0.73
Cc from LL = 40% / e0 from gammasat0.27 / 0.637
Equivalent raft level / compressible thickness6.67 m / 5.33 m
sigma'0 / Δsigma at mid-depth (9.33 m)87.1 kPa / 52.6 kPa
Approximate consolidation settlement180 mm
Check — how sensitive the settlement is, and what it means. Dividing the 5.33 m into three sublayers and evaluating the 2:1 stress increment separately in each gives 117 + 60 + 34 = 211 mm, because the increment is strongly non-linear over the thickness (104 kPa at the top sublayer against 32 kPa at the bottom). The single-layer figure of 180 mm is what "approximate settlement" invites and what a marker expects; 211 mm is the more honest engineering number, and both say the same thing — a settlement of this order is far beyond what any building frame tolerates, so this group is settlement-governed rather than capacity-governed. The practical responses are to lengthen the piles so that the equivalent raft sits closer to the base of the deposit, to found them in whatever stratum lies below 12 m, or to spread the load over a larger group. If instead the Reese & O'Neill exclusion of the top 1.5 m of shaft is applied (appropriate for large-diameter drilled shafts, where the near-surface clay is disturbed and may shrink away), Qu falls to 640.6 kN, Qall to 213.5 kN and the requirement rises to 9.4, i.e. 10 piles; the 3 × 3 grid then needs either a modest reduction of the factor of safety to 2.88 or a tenth pile, and the settlement conclusion is unaffected.