Question 2 of 6: Deep Foundations: helical pile group

Question 2 — Deep Foundations: helical pile group (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. Twenty-five helical piles, each 18 m long with a 305 mm shaft and two 610 mm helices at 1.22 m spacing, are installed in a uniform slightly overconsolidated clay and capped by a 6 m square cap founded 1 m below ground.

Find. Single-pile ultimate capacity by static analysis in undrained conditions; ultimate group capacity; the factor of safety under a 10 MN centric load with a design comment; and the total settlement by the equivalent-raft method.

Approach. Because the two helices are spaced at only two helix diameters, the soil between them acts as a plug, so the single-pile capacity follows the cylindrical shear model: end bearing on the lower helix, soil-on-soil shear on the cylinder joining the helices, and $\alpha c_u$ adhesion on the shaft above the upper helix. The group capacity is the lesser of the sum of the single-pile capacities and block failure of the group. Settlement follows from an equivalent raft at two-thirds of the embedded length with a 2:1 stress spread beneath it.

1. Part (a) — choose the helical-pile capacity model. The controlling ratio is the inter-helix spacing divided by the helix diameter: $$\frac{S_h}{D_h} = \frac{1.22}{0.610} = 2.0 \ < \ 3$$ Below about three, the individual bearing surfaces interfere and the soil between the helices fails as a cylinder rather than each helix punching separately. The cylindrical shear method therefore applies, and the pile is idealised as a 610 mm diameter soil cylinder 1.22 m long, capped by the lower helix and connected to the ground surface by the 305 mm shaft.
2. Part (a) — end bearing on the lower helix. The lower helix bears at 18 m, far deeper than five diameters, so the deep-foundation factor $N_c^{*} = 9$ applies: $$A_h = \frac{\pi}{4}D_h^{2} = \frac{\pi}{4}(0.610)^{2} = 0.29225\ \text{m}^{2}$$ $$Q_b = A_h\,N_c^{*}c_u = 0.29225 \times 9 \times 50 = 131.5\ \text{kN}$$ This is a net capacity, so the buoyant weight of the pile is not deducted separately.
3. Part (a) — shear on the cylinder between the helices. The failure surface here is soil against soil, so the full undrained strength acts, not the reduced adhesion: $$Q_{cyl} = \pi D_h S_h c_u = \pi \times 0.610 \times 1.22 \times 50 = 116.9\ \text{kN}$$
4. Part (a) — adhesion on the shaft above the upper helix. The shaft is a steel-to-clay interface, so the adhesion factor $\alpha = 0.8$ applies. The upper helix sits at $18 - 1.22 = 16.78$ m and the cap underside is at 1 m, so the effective shaft length is 15.78 m: $$Q_s = \pi d L_s \alpha c_u = \pi \times 0.305 \times 15.78 \times 0.8 \times 50 = 604.8\ \text{kN}$$
5. Part (a) — single-pile ultimate capacity. Adding the three contributions, $$\boxed{Q_u = Q_b + Q_{cyl} + Q_s = 131.5 + 116.9 + 604.8 = 853.2\ \text{kN}}$$ Shaft adhesion supplies 71 per cent of the total, which is characteristic of a slender helical pile in uniform clay: the helices contribute usefully but the pile is fundamentally a friction pile.
6. Part (b) — block failure of the group. A 5 by 5 group at 1.2 m centres spans $4 \times 1.2 = 4.8$ m between outer pile axes; taking the block boundary at the outer edge of the helices gives $$B_g = L_g = 4.8 + D_h = 4.8 + 0.61 = 5.41\ \text{m}$$ The block extends from the cap underside at 1 m to the pile toes at 18 m, so $L_b = 17$ m. Its capacity is perimeter shear at full $c_u$ plus base bearing: $$Q_{blk} = 2(B_g+L_g)L_b c_u + B_gL_g\,N_c^{*}c_u = 18\,394 + 13\,171 = 31\,565\ \text{kN}$$
7. Part (b) — group capacity. The sum of the individual capacities is $25 \times 853.2 = 21\,330$ kN, which is less than the block value, so individual pile failure governs: $$\boxed{Q_{g(u)} = \min(21\,330,\ 31\,565) = 21\,330\ \text{kN} \quad (\eta = 1.0)}$$ Physically the block is stronger because its 21.6 m perimeter mobilises the full $c_u$ whereas 25 individual shafts mobilise only $0.8c_u$ over 24.0 m, and the block base is 29.3 m² against 7.3 m² of helices.
8. Part (c) — factor of safety under the 10 MN load. Treating the 10 MN as an unfactored working load, $$\boxed{F = \frac{Q_{g(u)}}{Q} = \frac{21\,330}{10\,000} = 2.13}$$
9. Part (c) — comment on the design. Three observations follow. First, 2.13 falls short of the 2.5 to 3.0 normally required on a pile group designed by static analysis from soil strengths alone, with no load test; the group is under-designed at the ULS. Second, the pile spacing of 1.2 m is only two helix diameters, whereas CFEM and helical-pile practice call for at least three helix diameters (1.83 m here) so that the cylindrical shear zones of neighbouring piles do not overlap. If a Converse-Labarre efficiency is applied on the helix diameter, $\eta = 0.52$ and the group capacity falls to about 11 100 kN, i.e. $F \approx 1.11$ — a spread of results that by itself shows the spacing is too tight to be confident in. Third, the piles stop 16 m above the very dense sand at 34 m, so they are floating piles carrying their load entirely in a compressible clay; part (d) shows this is what actually condemns the design.
10. Part (d) — locate the equivalent raft. For friction piles in a uniform clay the load is treated as applied at two-thirds of the embedded length, over the plan area of the group: $$z_{raft} = 1.0 + \tfrac{2}{3}(17) = 12.33\ \text{m},\qquad 5.41 \times 5.41\ \text{m}$$ The pressure on that raft is $q_r = 10\,000/5.41^{2} = 341.7$ kPa.
11. Part (d) — consolidation settlement below the raft. The compressible clay runs from 12.33 m to the dense sand at 34 m, a thickness of 21.67 m, with $C_c/(1+e_0) = 0.25/2.20 = 0.1136$. Spreading the load at 2:1 below the raft gives, at depth $z$ below it, $$\Delta\sigma = \frac{Q}{(B_g+z)(L_g+z)},\qquad \sigma^{\prime}_{v0} = 20\,z_{abs}$$ Summing $s_c = \sum [C_c/(1+e_0)]\Delta z\log_{10}[(\sigma^{\prime}_{v0}+\Delta\sigma)/\sigma^{\prime}_{v0}]$ over 0.36 m sublayers gives $s_c = 178$ mm, three quarters of it in the first 4 m below the raft where $\Delta\sigma$ is still of the order of 200 kPa.
12. Part (d) — immediate settlement and the verdict. Treating the equivalent raft as rigid on an elastic layer with $E_u = 50$ MPa, $\nu_u = 0.5$ and $I_p = 0.82$ gives $s_i = 341.7 \times 5.41 \times 0.75 \times 0.82/50\,000 = 0.023$ m, so $$\boxed{s_{tot} = 23 + 178 = 200\ \text{mm}}$$ Against the 50 mm total settlement adopted here (CFEM's upper bound for a framed structure), the serviceability limit state is not satisfied, and by a factor of four. The remedy is not more piles of the same length: extending the piles 16 m to end-bear in the very dense sand at 34 m would transfer the load below the compressible clay and reduce the settlement to elastic shortening plus a few millimetres of toe movement. Widening the cap and increasing the spacing to three helix diameters would also spread the equivalent raft and lower $q_r$, but the floating-pile geometry remains the fundamental problem.