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 2 shows a belled (under-reamed) drilled shaft, not a driven pile. Read from the drawing: shaft diameter $D_s = 1.0\ \text{m}$; upper clay layer from the surface to 8.0 m with $c_u = 40\ \text{kPa}$; lower clay layer below 8.0 m with $c_u = 60\ \text{kPa}$; the shaft is embedded a further 4.0 m into the lower layer, so the base sits at $L = 12.0\ \text{m}$; the bell is 1.0 m high with a base diameter $D_b = 2.0\ \text{m}$, giving a splay of one vertical to half a horizontal. A pile cap sits 0.5 m proud of the ground. Factor of safety $FS = 3$.

Find. The ultimate axial compressive capacity $Q_u = Q_p + Q_s$ and the design (allowable) capacity $Q_{all} = Q_u/3$.

[Figure not reproduced: Question 7 (redrawn from Figure 2): the belled shaft, the two clay layers, and the two lengths over which side resistance is discounted. The green band is the only part of the shaft credited with skin friction. See the official exam paper.]

Approach. Use the total-stress (alpha) method for a drilled shaft in clay: the base resistance from the bearing-capacity factor $N_c^* = 9$ applied to the undrained strength at base level over the full bell area, and the side resistance from $\alpha^* c_u$ applied to the straight shaft only, with the Reese and O'Neill exclusion zones removed.

1. Fix the geometry and identify what resists load. The base is at 12.0 m and the bell occupies the bottom 1.0 m, so the straight stem runs from 0 to 11.0 m. The base area is that of the bell: $$A_p = \frac{\pi}{4}D_b^{2} = \frac{\pi}{4}(2.0)^{2} = 3.1416\ \text{m}^2$$ while the perimeter that can develop skin friction is that of the stem, $p = \pi D_s = 3.1416\ \text{m}$.
2. Apply the exclusion zones for a belled shaft. Reese and O'Neill's method, as reproduced in Das, discounts side resistance over the top 1.5 m, where the soil is disturbed and seasonally active and where a shaft moves relative to the ground it does not load; it discounts the periphery of the bell itself, which moves away from the soil as the base is loaded; and for a belled shaft it discounts a further length equal to one shaft diameter above the top of the bell, because the soil there is unloaded by the bell's own bearing movement. The contributing length is therefore from 1.5 m to $11.0 - 1.0 = 10.0\ \text{m}$, split as 6.5 m in the $c_u = 40\ \text{kPa}$ clay and 2.0 m in the $c_u = 60\ \text{kPa}$ clay.

   Check: assumes a bored shaft constructed dry or under a temporary casing in a stiff-to-firm clay, so that the standard exclusion zones and the standard adhesion factor apply. If the hole were drilled under bentonite, a further reduction of the order of 20 per cent on the side resistance would be appropriate.

3. Choose the adhesion factor. Reese and O'Neill take $\alpha^* = 0.55$ where $c_u/p_a \le 1.5$. Here $40/101.3 = 0.39$ and $60/101.3 = 0.59$, both comfortably below the limit, so $\alpha^* = 0.55$ applies to both layers.
4. Evaluate the side resistance. Summing $\alpha^* c_u p \Delta L$ over the two contributing segments, $$Q_s = 0.55\,\pi(1.0)\big[(40)(6.5) + (60)(2.0)\big] = 0.55(3.1416)(380) = 656.6\ \text{kN}$$
5. Evaluate the base resistance. The bell bears wholly within the lower clay, and with $L/D_b = 6 \gt 4$ the full deep-foundation factor $N_c^* = 9$ is available. The net base resistance is $$Q_p = A_p N_c^{*} c_{u(base)} = (3.1416)(9)(60) = 1696.5\ \text{kN}$$ This is a net value, so the weight of the shaft and the overburden it displaces need not be carried separately.
6. Combine and apply the factor of safety. The ultimate capacity and the design capacity are $$Q_u = Q_p + Q_s = 1696.5 + 656.6 = 2353.1\ \text{kN}$$ $$\boxed{Q_{all} = \frac{Q_u}{FS} = \frac{2353.1}{3} = 784.4\ \text{kN}}$$
7. Check how the load is shared, because it governs the serviceability. The bell provides $1696.5/2353.1 = 72$ per cent of the ultimate capacity. That distribution matters: side resistance in clay is fully mobilised at a movement of roughly 5 to 10 mm, whereas base resistance in clay needs a movement of the order of 5 per cent of the base diameter, here about 100 mm, to reach its ultimate value. At the working load the shaft will therefore be carried very largely by skin friction, and the calculated capacity is not available without a settlement that no building would tolerate.

Comment on the result. Three observations belong with the number. The bell is what makes this shaft: a straight 1.0 m shaft to the same depth would develop a base resistance of only $(\pi/4)(1.0)^2(9)(60) = 424\ \text{kN}$, so the under-ream quadruples the base contribution for one extra metre of excavation. Against that, a bell can only be formed in a clay that will stand unsupported for the time the under-reaming tool is at work, it cannot be formed under bentonite, and it cannot be inspected without entering the shaft; the specification must therefore require the base to be cleaned and proved, because debris left at the base is the classic cause of an under-performing belled shaft. Finally, because the design capacity is base-dominated but the working behaviour is friction-dominated, a serviceability check — either an elastic settlement calculation or, better, a static load test on a trial shaft — should accompany the capacity calculation rather than be inferred from the factor of safety of 3.