Question 6 of 9: Design Axial Capacity of a Belled Drilled Shaft in Clay

Question 6: Design Axial Capacity of a Belled Drilled Shaft in Clay (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.

Find. The allowable (design) axial compressive load the shaft may carry, $Q_{all} = Q_{ult}/FS$.

[Figure not reproduced: Figure 2 (redrawn to scale). Belled bored pile, 12 m long, bell base bearing in the lower clay. The two hatched red bands are the lengths of shaft over which Reese and O'Neill discount side resistance entirely. See the official exam paper.]

Approach. The enlarged base identifies this as a bored (drilled, under-reamed) shaft rather than a driven pile, so the Reese and O'Neill total-stress method applies: side resistance $Q_s = \sum \alpha^{*}c_u\,p\,\Delta L$ with $\alpha^{*} = 0.55$ over a reduced shaft length, plus base resistance $Q_p = A_p c_{u(b)} N_c^{*}$ on the full bell area, the sum divided by the given factor of safety.

1. Establish the geometry from the figure. The bell is 1.0 m high and its underside sits on the 12 m line, so the flare begins at $z = 12 - 1 = 11$ m and the straight shaft runs from the surface to that level. The base therefore bears in the lower clay, 4 m below the layer interface, and the embedment ratio is $L/D_b = 12/2.0 = 6$.
2. Choose the adhesion factor and justify it. For drilled shafts in clay Reese and O'Neill recommend $$\alpha^{*} = 0.55 \qquad \text{for} \qquad \frac{c_u}{p_a} \le 1.5 ,$$ where $p_a = 100$ kPa is atmospheric pressure. Here $c_{u1}/p_a = 0.80$ and $c_{u2}/p_a = 0.40$, both comfortably inside the limit, so $\alpha^{*} = 0.55$ applies to both strata. This is deliberately lower than the $\alpha \approx 1.0$ used for a driven pile in soft clay, because drilling relieves the horizontal stress on the borehole wall and the concrete is placed against a softened, partly remoulded surface. Taking a driven-pile $\alpha$ instead would give $Q_s = 1382$ kN in place of the 1037 kN computed below — 33 per cent unconservative.
3. Strike out the two lengths that carry no side resistance. Reese and O'Neill discount the top 1.5 m, where seasonal movement, casing and poor concrete quality make the bond unreliable, and a further length equal to one shaft diameter above the top of the bell, because as the bell is loaded the clay immediately above it is dragged down with the enlargement and loses contact. The flared surface of the bell itself is also ignored. The effective friction length is therefore $$1.5\ \text{m} \ \to \ 11.0 - 1.0 = 10.0\ \text{m}.$$ Within it, 6.5 m lies in the upper clay (1.5 m to 8.0 m) and 2.0 m in the lower clay (8.0 m to 10.0 m).
4. Compute the side resistance layer by layer. With shaft perimeter $p = \pi D_s = \pi(1.0) = 3.1416$ m, $$Q_{s1} = \alpha^{*}c_{u1}\,p\,L_1 = 0.55(80)(3.1416)(6.5) = 898.5\ \text{kN},$$ $$Q_{s2} = \alpha^{*}c_{u2}\,p\,L_2 = 0.55(40)(3.1416)(2.0) = 138.2\ \text{kN},$$ so that $$\boxed{\,Q_s = 898.5 + 138.2 = 1036.7\ \text{kN}\,}$$ Note how little the lower clay contributes: it is half as strong and, after the bell exclusion, only 2 m of it is working.
5. Compute the base resistance on the bell. The base area is $$A_p = \frac{\pi}{4}D_b^2 = \frac{\pi}{4}(2.0)^2 = 3.1416\ \text{m}^2 .$$ The bearing capacity factor follows Reese and O'Neill, $$N_c^{*} = 6\left[1 + 0.2\frac{L}{D_b}\right] \le 9 = 6\left[1 + 0.2(6)\right] = 13.2 \ \Rightarrow\ N_c^{*} = 9 ,$$ the cap applying because the base is deeper than four base diameters. With the base bearing wholly in the lower clay, $$\boxed{\,Q_p = A_p\,c_{u2}\,N_c^{*} = 3.1416(40)(9) = 1131.0\ \text{kN}\,}$$
6. Check the large-base settlement reduction. Because $D_b = 2.0\ \text{m} > 1.91$ m, Reese and O'Neill require the base resistance to be checked against a 25 mm settlement criterion through $$F_r = \frac{2.5}{\psi_1 D_b + \psi_2} \le 1,\qquad \psi_1 = 0.0071 + 0.0021\frac{L}{D_b} \le 0.015,\qquad \psi_2 = 0.45\sqrt{c_{u(b)}}\ \ (0.5 \le \psi_2 \le 1.5).$$ Here $\psi_1 = 0.0071+0.0021(6) = 0.0197 \to 0.015$ and $\psi_2 = 0.45\sqrt{40} = 2.85 \to 1.5$, so $F_r = 2.5/[0.015(2.0)+1.5] = 1.63 \to F_r = 1.0$. No reduction is required, and the full base resistance may be used.
7. Combine and apply the factor of safety. Summing, $$Q_{ult} = Q_s + F_r Q_p = 1036.7 + 1131.0 = 2167.7\ \text{kN},$$ and with the specified $FS = 2.5$, $$\boxed{\,Q_{all} = \frac{2167.7}{2.5} = 867\ \text{kN}\,}$$ The self-weight of the shaft is not deducted: for a bored pile in clay the concrete weight is very nearly balanced by the weight of soil it replaces, and the conventional convention is to work with the net capacity. Since the two differ here by under 3 per cent of $Q_{all}$, the simplification is defensible.
8. Confirm that the bell earns its cost. Without the under-ream the base would be only $\tfrac{\pi}{4}(1.0)^2(40)(9) = 282.7$ kN, so the bell quadruples the base resistance and supplies 52 per cent of the ultimate capacity. That is the justification for under-reaming into a clay only half as strong as the layer above it.