Question 2 of 6: Deep Foundations: driven H-piles

Question 2 — Deep Foundations: driven H-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. An 18 m driven steel H-pile, 310 mm × 307 mm with 15.5 mm flanges and web, passing through 8 m of normally consolidated clay into a lightly overconsolidated clay, with a structural load of 2 MN to be carried by a square group.

Find. (a) the allowable axial load on one pile at FS = 2.5; (b) the smallest square group that carries 2 MN, checked against block failure; (c) the consolidation settlement of that group by the equivalent-raft method.

Approach. Use the total-stress (α) method for a pile driven into clay, taking the H-pile as plugged so that shaft friction acts on the circumscribing box and end bearing on the box area; sum shaft and toe resistance, divide by 2.5; find the smallest $n\times n$ group whose summed single-pile capacity meets FS = 2.5 and confirm the block is stronger; then place a Tomlinson equivalent raft at two-thirds of the pile length, spread the load at 1 in 4, and integrate the one-dimensional consolidation of the clay beneath it.

1. Section properties of the H-pile. A driven H-pile in clay plugs between the flanges within a few diameters of penetration, so both friction and end bearing are taken on the circumscribing rectangle: $$p = 2(b_f + d) = 2(0.310 + 0.307) = 1.234\ \text{m} \qquad A_p = 0.310 \times 0.307 = 0.0952\ \text{m}^2$$

   (The steel section itself is only 0.0139 m², so the plug carries almost all of the toe load; the pile's own weight, 19 kN, is less than 2% of its capacity and is neglected.)

2. Adhesion factors. The paper gives no α. Das's Table 12.6 tabulates α against $c_u/p_a$ with $p_a = 100$ kPa, which is the citable choice here because it does not require an assumed effective-stress profile: $$\frac{c_u}{p_a} = 0.60 \;\Rightarrow\; \alpha_1 = 0.62 \qquad \frac{c_u}{p_a} = 1.10 \;\Rightarrow\; \alpha_2 = 0.45$$

   The lower layer's value is interpolated between the tabulated 0.48 at 1.0 and 0.42 at 1.2. Both are adopted explicitly; the sensitivity to that choice is quantified in step 5.

3. Shaft and toe resistance. The pile takes 8 m in the NC clay and 10 m in the OC clay: $$Q_{s} = \sum \alpha_i c_{u,i} \, p \, L_i = 0.62(60)(1.234)(8) + 0.45(110)(1.234)(10)$$ $$Q_{s} = 367.2 + 610.8 = 978.0\ \text{kN}$$

   The toe sits well inside the overconsolidated clay, so $N_c^* = 9$ applies:

   $$Q_p = 9 c_u A_p = 9(110)(0.0952) = 94.2\ \text{kN}$$

   The pile is thus 91% a friction pile, which is exactly what an H-section 18 m into clay should be.

4. Design axial capacity. $$Q_u = 978.0 + 94.2 = 1072\ \text{kN} \qquad Q_{all} = \frac{Q_u}{2.5}$$ $$\boxed{Q_{all} = 429\ \text{kN\ per\ pile}}$$
5. The square group (part b). With 429 kN per pile, 2000 kN needs 4.7 piles. A square group means $n^2$: a 2 × 2 group gives $4(1072)/2000 = 2.14$, short of the required 2.5, so the next square up is used: $$\boxed{3\times 3 = 9\ \text{piles at}\ s = 1.0\ \text{m} \approx 3b_f,\ \ FS = \frac{9(1072)}{2000} = 4.8}$$

   Group action must be checked before that is accepted. The group plan is $B_g = L_g = 2(1.0) + 0.31 = 2.31$ m, and block failure mobilises the full undrained strength on the perimeter because the failure surface is soil-on-soil:

   $$Q_{block} = p_g \sum c_{u,i} L_i + 9 c_u B_g L_g = 9.24\,[60(8) + 110(10)] + 9(110)(5.34) = 19\,882\ \text{kN}$$

   That is more than twice the summed single-pile capacity of 9651 kN, so individual pile failure governs and the group efficiency may be taken as unity. Note that the 3 × 3 group is substantially over-strength: the requirement is 40% satisfied by the fifth pile onward. If the piles could be lengthened, a 2 × 2 group of 21 m piles would carry the same load with FS = 2.5 and one-fifth fewer piles — worth raising with the structural designer, but outside the "these piles" wording of the question.

6. Sensitivity to the adopted α. Because α is the least certain input, the group choice is checked against the other two standard rules:

   Effect of the adhesion rule on the design

   Source of α	α (NC / OC)	Qall per pile	Smallest square group
   Das Table 12.6 (adopted)	0.62 / 0.45	429 kN	3 × 3
   Tomlinson, driven piles in clay	0.75 / 0.55	514 kN	2 × 2 (FS = 2.57)
   API RP 2A (ψ = cu/σ′v)	0.40 / 0.48	391 kN	3 × 3

   The design flips from nine piles to four only if α in the overconsolidated clay can be justified above about 0.53. On a residential development that judgement should be settled by a static load test on the first pile, not by a table — and the 3 × 3 group is the defensible choice until it is.

7. Equivalent raft (part c) — where to put it. Tomlinson's rule for a friction pile group in clay places an imaginary raft at two-thirds of the embedded length, carrying the full working load over the group plan area, with the load spreading at 1 horizontal to 4 vertical below that level: $$z_{raft} = \tfrac{2}{3}(18) = 12\ \text{m} \qquad q_{raft} = \frac{2000}{2.31^2} = 375\ \text{kPa}$$

   The raft therefore lands 4 m below the base of the normally consolidated clay. That layer lies above the raft and is not stressed by it, which is why its $C_c = 0.35$ and $e_0 = 0.95$ do not appear in the arithmetic — they are given so the candidate can confirm exactly this, and to show what is at stake: $C_c/(1+e_0)$ is 0.180 for the NC clay against 0.094 for the OC clay, so a shorter pile whose equivalent raft fell inside the upper layer would settle roughly twice as much per unit of stress change.

8. Effective stresses below the raft. No water table is given. Both layers are quoted by undrained strength, so they are saturated, and the water table is taken at ground level: $$\gamma^{\prime}_{NC} = 16 - 9.81 = 6.19\ \text{kN/m}^3 \qquad \gamma^{\prime}_{OC} = 18 - 9.81 = 8.19\ \text{kN/m}^3$$

   which puts $\sigma^{\prime}_{v0} = 82.3$ kPa at the raft and 196.9 kPa at the base of the compressible clay, 26 m down.

9. Integrate the settlement. Dividing the 14 m of overconsolidated clay between the raft and the dense sand into seven 2 m sublayers, with $\Delta\sigma = Q/(B_g + z/2)^2$ from the 1-in-4 spread: $$s_c = \sum \frac{C_c}{1+e_0} H_i \log_{10}\!\frac{\sigma^{\prime}_{0,i} + \Delta\sigma_i}{\sigma^{\prime}_{0,i}} = 0.0938 \sum H_i \log_{10}(\cdot)$$

   Sublayer contributions, equivalent raft at 12 m

   Depth (m)	Δσ (kPa)	σ′v0 (kPa)	Δs (mm)
   13	253.3	90.5	108.7
   15	137.8	106.9	67.5
   17	86.4	123.2	43.3
   19	59.3	139.6	28.8
   21	43.1	156.0	19.9
   23	32.8	172.4	14.2
   25	25.8	188.8	10.4

   $$\boxed{s_c \approx 293\ \text{mm}}$$
10. Interpret the answer. Nearly 300 mm is far beyond anything a residential structure tolerates, and the reason is visible in the table: a nine-pile group at 3$b_f$ spacing is only 2.31 m square, so it behaves as a small, deep footing carrying 375 kPa. Opening the spacing to 3.0 m ($B_g = 6.31$ m) drops the settlement to about 100 mm, and taking the piles the full 26 m onto the dense sand removes the compressible layer from beneath the raft altogether. If instead the profile has no free water, the same calculation gives 153 mm — still excessive. The honest conclusion is that this pile length and spacing satisfy bearing capacity and fail settlement, and that the group must be spread out or driven to the sand.