Question 6 of 6: Shallow Foundations (30 marks)

Question 6 — Shallow Foundations (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.

Find. (a) the maximum oil height satisfying both limit states; (b) the overall factor of safety against bearing failure at that height, with comment; (c) the overall factor of safety when a strong wind inclines the resultant load by 20°.

Approach. Compute the net undrained bearing resistance of the circular tank base, factor it and equate to the factored net oil pressure for the ultimate limit state; separately compute the immediate and consolidation settlement as functions of net pressure and find the height that gives 300 mm; take the smaller height, then evaluate the overall factor of safety with and without a load-inclination factor.

1. (a) Compute the net ultimate bearing resistance. For a circular foundation under undrained loading (φ_u = 0), the net ultimate bearing capacity is $$q_{nf} = c_u N_c s_c d_c, \qquad N_c = 5.14, \quad s_c = 1.2\ (\text{circle}), \quad d_c = 1 + 0.4\frac{D_f}{B} = 1 + 0.4\frac{2}{35} = 1.023$$ $$q_{nf} = 25 \times 5.14 \times 1.2 \times 1.023 = 157.7\ \text{kPa}$$
2. Apply the resistance factor. With f_c = 0.6, $$q_{nr} = 0.6 \times 157.7 = 94.6\ \text{kPa}$$
3. Express the applied pressure in terms of the oil height. The unit weight of the oil and the overburden removed by founding 2 m down are $$\gamma_{oil} = 0.9 \times 9.81 = 8.829\ \text{kN/m}^{3}, \qquad \sigma_{v0} = \gamma D_f = 16.5 \times 2 = 33.0\ \text{kPa}$$ so the net applied pressure is $q_{net} = 8.829h - 33.0$, and the factored net pressure, with the 1.5 factor applied to the oil weight only, is $q_{f,net} = 1.5(8.829h) - 33.0$. The tank's own weight is neglected as instructed.
4. Solve the ultimate limit state for the fill height. Setting $q_{f,net} = q_{nr}$, $$1.5(8.829)h - 33.0 = 94.6 \;\Rightarrow\; 13.24h = 127.6 \;\Rightarrow\; \boxed{h_{ULS} = 9.64\ \text{m}}$$
5. Set up the serviceability calculation. The long-term total settlement is the immediate (undrained) settlement plus the corrected consolidation settlement: $$s = s_i + \mu\,s_{oed}, \qquad s_i = \mu_0\mu_1\frac{q_{net}B}{E_u}, \qquad s_{oed} = m_v\sum \Delta\sigma^{\prime}H$$ For a circular area with $H/B = 30/35 = 0.86$ and $D/B = 0.057$, the Janbu–Christian–Carrier factors are $\mu_0 = 0.98$ and $\mu_1 = 0.45$. The Skempton–Bjerrum correction for A = 0.65 and a circular footing on a layer of this proportion is $\mu = A + \alpha(1 - A) = 0.65 + 0.39(0.35) = 0.79$, where α = 0.39 is the ratio of the integrated minor to major principal stress increase on the centreline for H/B = 0.86.
6. Distribute the stress increase with depth. Beneath the centre of a uniformly loaded circle of radius a = 17.5 m, $$\Delta\sigma_z = q_{net}\left[1 - \left(\frac{1}{1 + (a/z)^{2}}\right)^{3/2}\right]$$ Dividing the 30 m of clay below the base into ten 3 m sub-layers and summing gives $s_{oed} = 3.00\,q_{net}$ mm with $q_{net}$ in kPa — the tank is so wide relative to the clay thickness that the stress increase is still about 36 % of the surface value where the clay meets the firm stratum.
7. Solve the serviceability limit state for the fill height. Combining, $$s = 0.98(0.45)\frac{q_{net}(35)}{40\,000}\times1000 + 0.79(3.00\,q_{net}) = 0.386\,q_{net} + 2.37\,q_{net} = 2.76\,q_{net}\ \text{mm}$$ Setting s = 300 mm gives $q_{net} = 108.7$ kPa (of which 42.0 mm is immediate and 258.0 mm consolidation), and hence $$h_{SLS} = \frac{108.7 + 33.0}{8.829} = 16.05\ \text{m}$$
8. Identify the governing limit state. Comparing the two, $$\boxed{h = \min(9.64;\ 16.05) = 9.6\ \text{m} \;-\; \text{the ultimate limit state governs}}$$ This is the reverse of the usual situation for a tank on soft clay, and it happens because the specified settlement tolerance (300 mm) is generous — appropriately so for a steel tank on a flexible ringwall, which can accommodate large uniform settlement provided the differential and edge-to-centre distortion are controlled.
9. (b) Overall factor of safety at the design fill height. At h = 9.6 m the net applied pressure is $$q_{net} = 8.829(9.6) - 33.0 = 84.8 - 33.0 = 51.8\ \text{kPa}$$ $$\boxed{FS = \frac{q_{nf}}{q_{net}} = \frac{157.7}{51.8} = 3.05}$$
10. (c) Allow for the inclined resultant. A strong wind adds a horizontal component, inclining the resultant at α = 20° to the vertical. Meyerhof's inclination factor for the cohesion term is $$i_c = \left(1 - \frac{\alpha^{\circ}}{90^{\circ}}\right)^{2} = \left(1 - \frac{20}{90}\right)^{2} = 0.605$$ $$q_{nf,incl} = 157.7 \times 0.605 = 95.4\ \text{kPa}$$ $$\boxed{FS_{incl} = \frac{95.4}{51.8} = 1.84}$$

Comment on the results (part b). An overall factor of safety of 3.05 against bearing failure is exactly what conventional practice asks for a permanent structure on soft clay, so the limit-states and working-stress checks agree once again — the resistance factor of 0.6 with a load factor of 1.5 is calibrated to reproduce roughly FS = 3 for a φ_u = 0 material. Two cautions attach to that comfortable number. First, the shear strength was measured by unconfined compression on 51 mm Shelby tube samples, and the unconfined test applies no confinement and suffers sample disturbance, so it systematically under-estimates the in-situ strength of a soft clay, often by 20–30 %; the real factor of safety is probably higher than 3.05, and a vane or CIU triaxial programme would recover that margin. Second, the tank should be hydrotested by filling with water ($\gamma = 9.81$ kN/m³) before commissioning: at 9.6 m of water the net pressure would be 61.2 kPa and the factor of safety 2.58, still acceptable, and the controlled test loading also pre-consolidates the clay and improves the strength available in service. Under the inclined wind load the factor of safety falls to 1.84 — a 40 % loss for a 20° inclination — which is acceptable for a transient wind combination (1.5 to 2.0 is the usual requirement) but shows how severely inclination punishes an undrained bearing capacity. Anchorage and tank-wall design must therefore be checked for the same wind case, and the ringwall detailed to keep the resultant within the middle third of the base.