Question 3 of 5: Pile Foundations — When They Are Needed and What One Carries

Question 3: Pile Foundations — When They Are Needed and What One Carries (20 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.

Part (a) — conditions that call for a pile foundation

Piles are used whenever a shallow foundation cannot deliver adequate capacity, acceptable settlement, or adequate restraint at a reasonable depth. Six conditions cover almost all cases in practice.

Weak or compressible near-surface soil. When the upper strata are soft clay, loose silt, peat or fill, a spread footing large enough to keep the bearing pressure tolerable would be uneconomic or would still settle excessively. Piles transfer the load through the weak material to a competent bearing stratum — dense sand, till or bedrock — as end-bearing piles, or distribute it into deeper, stronger soil by shaft friction as friction piles. This is the dominant reason in the soft marine clays of the Fraser and St. Lawrence lowlands.

Large or concentrated structural loads. Heavily loaded columns of bridges, tall buildings and industrial plant deliver loads that would require footings so large they overlap. Piles concentrate the required resistance into a small plan area.

Horizontal, inclined or uplift loading. Retaining structures, transmission towers, anchor blocks, sheet-pile bulkheads and bridge abutments carry lateral thrust or net tension. Vertical and battered piles resist horizontal load in bending and pull-out in shaft friction; a spread footing resists these only through base friction and self-weight.

Expansive, collapsible or frost-susceptible soils. Where the active zone of a swelling clay, a collapsible loess or the seasonal frost zone would move a shallow footing, piles carry the structure below that zone. In much of Canada the design frost depth and, in the North, the permafrost active layer set that requirement directly.

Scour and erosion. Bridge piers and marine structures must remain supported after the design scour event has removed several metres of bed material, so the foundation must be embedded well below the predicted scour line — a depth only piles or caissons reach economically.

Fluctuating water table, buoyancy and dynamic loading. Structures subject to uplift from a rising groundwater table, or to vibration and machine or seismic loading, benefit from the tension capacity and the stiffness that piles provide; piles also allow construction over water without dewatering. A related case is compaction piling, where the piles are driven principally to densify a loose granular deposit.

Part (b) — capacity of the 400 mm driven pipe pile

Given.

Find. (I) the net point bearing capacity $Q_p$, (II) the skin (shaft) resistance $Q_s$, and (III) the net allowable pile capacity at $FS = 4$.

Approach. This is an undrained, total-stress problem in clay, so use $N_c^{*} = 9$ for the point and the $\alpha$-method for the shaft, with $\alpha$ read from the chart of adhesion factor against undrained cohesion supplied on the last page of the exam.

1. Part I — choose the point area. A driven pipe pile in clay plugs during driving, so the toe behaves as a closed circular section and the gross outside area governs: $$A_p = \frac{\pi D^2}{4} = \frac{\pi (0.400)^2}{4} = 0.1257\ \text{m}^2$$ The 6.25 mm wall thickness is needed only for the pile's own structural check, not for the geotechnical point resistance.
2. Compute the net point bearing capacity. In saturated clay under undrained conditions the bearing-capacity factor is $N_c^{*} = 9$, and the net point capacity already excludes the overburden term: $$Q_p = A_p\,N_c^{*}\,c_{u(2)} = 0.1257 \times 9 \times 100$$ $$Q_p = \boxed{113.1\ \text{kN}}$$ using the strength of the clay at the toe, $c_{u(2)} = 100\ \text{kN/m}^2$ — not an average over the shaft.
3. Part II — set up the $\alpha$-method. Skin resistance in clay is $$Q_s = \sum \alpha\, c_u\, p\, \Delta L$$ where $p = \pi D = \pi(0.400) = 1.2566\ \text{m}$ is the perimeter and $\alpha$ is the empirical adhesion factor. Reading the supplied $\alpha$-versus-$c_u$ chart on the average curve: at $c_u = 30\ \text{kN/m}^2$, $\alpha_1 \approx 1.00$; at $c_u = 100\ \text{kN/m}^2$, $\alpha_2 \approx 0.50$. Soft clay bonds fully to the shaft; stiff clay does not, because driving remoulds and partially separates it.
4. Sum the shaft resistance layer by layer. The upper 10 m has one strength throughout (the water table at 5 m splits the unit weights but not the undrained strength, and a total-stress analysis does not use unit weight at all): $$Q_{s,1} = (1.00)(30)(1.2566)(10) = 377.0\ \text{kN}$$ $$Q_{s,2} = (0.50)(100)(1.2566)(10) = 628.3\ \text{kN}$$ $$Q_s = 377.0 + 628.3 = \boxed{1\,005.3\ \text{kN}}$$ The lower layer contributes more despite the smaller $\alpha$, because its strength is more than three times greater.
5. Part III — assemble the ultimate capacity. Point and shaft act together: $$Q_u = Q_p + Q_s = 113.1 + 1\,005.3 = 1\,118.4\ \text{kN}$$ Shaft friction supplies about 90 % of the total — this is a friction pile in all but name, which is typical of a slender driven pile in clay.
6. Apply the factor of safety. $$Q_{all} = \frac{Q_u}{FS} = \frac{1\,118.4}{4} = \boxed{279.6\ \text{kN}}$$ A factor of 4 (rather than the 2.5–3 used for shallow foundations) reflects the wider scatter of $\alpha$ and the difficulty of verifying a pile toe that nobody can inspect.