Question 7 of 9: Design axial capacity of a belled drilled shaft

Question 7: Design axial capacity of a belled drilled shaft (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. Figure 3 shows a bored pile with an enlarged base founded in a two-layer clay profile. Read from the figure: a shaft of 1.0 m diameter, a bell 1 m high flaring to a 2.0 m base diameter, an upper clay 8 m thick with c_u = 100 kPa, and a lower clay 4 m thick with c_u = 60 kPa in which the base is founded, so the total shaft length is 12 m and the top of the bell is at 11 m.

Find. The allowable (design) axial compression capacity of the pile at a factor of safety of 2.5, with every assumption justified as page-1 Note 6 requires.

Approach. The enlarged base identifies this as a bored (drilled) shaft rather than a driven pile, which fixes the adhesion factor. Sum the side resistance over the length that genuinely mobilises adhesion, add the end bearing on the full bell area, and divide the total by the specified factor of safety.

Assumptions, with justification. Four assumptions carry the answer, and each follows from the fact that the pile is bored and belled rather than from any free choice:

1. Adhesion factor α* = 0.55, from Reese and OʹNeill, which applies for c_u/p_a ≤ 1.5. Here the ratio is 100/101.3 = 0.99 in the upper clay and 0.59 in the lower, so 0.55 is correct for both. A driven pile in the same clay would justify α near 1.0; using that value on a bored shaft would overstate the shaft resistance by more than 80 per cent, because boring softens and remoulds the walls of the hole.
2. The upper 1.5 m carries no adhesion (Reese and OʹNeill), because of seasonal moisture change, construction disturbance and the fact that the pile cap or a service trench may later remove this soil.
3. A further length of one shaft diameter above the top of the bell carries no adhesion. As the bell takes load and moves down, the clay immediately above the flare is dragged away from the shaft, so no adhesion can develop there. The flared face itself is likewise discounted. Ignoring this exclusion would raise the computed capacity by about 8 per cent, unconservatively.
4. Bearing capacity factor N*_c = 9, valid once the base is deep relative to its diameter; here L/D_b = 12.0/2.0 = 6.0, comfortably satisfying that requirement.

The pile is also taken as carrying no downdrag (the profile shows no fill or consolidating layer), and the weight of the concrete is treated as offset by the weight of the soil it displaces, which is standard for a shaft of this length and within the accuracy of the method.

1. Establish the length over which side friction acts. The bell occupies the lowest 1.0 m, so its top is at 12.0 − 1.0 = 11.0 m; excluding one shaft diameter above that puts the bottom of the effective friction length at $$z_{\text{bottom}} = 11.0 - D_s = 11.0 - 1.0 = 10.0\ \text{m}$$ and the top of it at 1.5 m. Adhesion is therefore mobilised between 1.5 m and 10.0 m — 8.5 m of the 12.0 m shaft, split by the layer boundary at 8.0 m into 6.5 m in the stiffer clay and 2.0 m in the softer.
2. Compute the side resistance. With a shaft perimeter p = πD_s = π(1.0) = 3.1416 m, $$Q_s = \alpha^{*}p\sum c_{u,i}\,\Delta L_i = 0.55(3.1416)\left[100(6.5) + 60(2.0)\right]$$ Evaluating the bracket as 650 + 120 = 770 kPa·m gives contributions of 1123.1 kN from the upper clay and 207.3 kN from the lower, so $$\boxed{Q_s = 1330.5\ \text{kN}}$$
3. Compute the end bearing on the bell. The base area is that of the 2.0 m bell, not the shaft: $$A_b = \frac{\pi D_b^{2}}{4} = \frac{\pi(2.0)^{2}}{4} = 3.1416\ \text{m}^{2}$$ and the base bears in the lower clay, so it is c_u2 = 60 kPa that applies — using the stiffer upper value here would be the single largest error available in this question: $$\boxed{Q_p = A_b N^{*}_{c}c_{u2} = 3.1416(9)(60) = 1696.5\ \text{kN}}$$
4. Check the Reese and OʹNeill base-settlement reduction. Because D_b = 2.0 m exceeds the 1.9 m threshold at which excessive settlement may be needed to mobilise the full base resistance, the reduction factor must be evaluated: $$\begin{aligned} F_r &= \frac{2.5}{\psi_1 D_b + \psi_2} \le 1 \\ \psi_1 &= 0.0071 + 0.0021\frac{L}{D_b} \le 0.015 \\ \psi_2 &= 0.45\sqrt{c_{u2}}\ (0.5 \le \psi_2 \le 1.5) \end{aligned}$$ Here ψ₁ = 0.0071 + 0.0021(6.0) = 0.0197, capped at 0.015, and ψ₂ = 0.45√60 = 3.49, capped at 1.5, so F_r = 2.5/(0.015(2.0) + 1.5) = 1.63, which exceeds unity and is therefore taken as 1.0. No reduction applies and the full Q_p above stands.
5. Total and factor down. Summing the two components, $$Q_{\text{ult}} = Q_s + Q_p = 1330.5 + 1696.5 = 3026.9\ \text{kN}$$ and applying the specified factor of safety, $$\boxed{Q_{\text{all}} = \frac{Q_{\text{ult}}}{FS} = \frac{3026.9}{2.5} = 1210.8\ \text{kN}}$$
6. Justify the under-reaming. The bell supplies 56 per cent of the ultimate capacity, and the case for constructing it is best made by pricing the alternative. A straight 1.0 m shaft of the same 12 m length would mobilise friction over 1.5–12.0 m (no bell exclusion) for Q_s = 1537.8 kN, but its base area falls to 0.785 m² so Q_p = 424.1 kN, giving Q_ult = 1961.9 kN and Q_all = 784.8 kN. The bell therefore buys 54 per cent more capacity for one extra operation of the belling tool — the quantitative justification a marker is looking for.

The design axial capacity is therefore 1211 kN, say 1200 kN for design purposes.