Question 6 of 6: Shallow Foundations: allowable bearing capacity

Question 6 — Shallow Foundations: allowable bearing capacity (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. A 4 m square footing at 1 m depth in a 13 m thick saturated clay with the water table at ground level.

Find. The allowable bearing capacity governed by shear at a factor of safety of 3, and the allowable bearing capacity that limits the consolidation settlement to 30 mm; hence which criterion governs.

Approach. For a saturated clay the short-term undrained case controls shear failure, so use Skempton's $N_c$ for a square footing with the long-term drained case checked to confirm it does not govern. For settlement, integrate the Boussinesq stress increase beneath the centre using $m_v$ to get the oedometer settlement, then apply the Skempton-Bjerrum correction for the pore-pressure coefficient A, and invert for the net pressure that gives exactly 30 mm.

1. Part (a) — total overburden at founding level. Undrained bearing capacity is a total-stress calculation, so the surcharge uses the saturated unit weight: $$\sigma_0 = \gamma_{sat}D_f = 21 \times 1.0 = 21\ \text{kPa}$$
2. Part (a) — net ultimate bearing capacity, undrained. With $\phi_u = 0$, Skempton's bearing-capacity factor for a square footing at $D_f/B = 0.25$ is $$N_c = 5\left(1 + 0.2\frac{D_f}{B}\right)\left(1 + 0.2\frac{B}{L}\right) = 5(1.05)(1.20) = 6.30$$ $$q_{nf} = c_uN_c = 100 \times 6.30 = 630\ \text{kPa}$$
3. Part (a) — allowable bearing capacity on shear. The factor of safety is applied to the net capacity, because the overburden pressure was already there before the footing was built: $$q_{na} = \frac{q_{nf}}{F} = \frac{630}{3} = 210\ \text{kPa}$$ $$\boxed{q_{a} = q_{na} + \sigma_0 = 210 + 21 = 231\ \text{kPa (gross)}}$$ A Vesic calculation with $N_c = 5.14$, $s_c = 1.195$ and $d_c = 1.10$ gives a net capacity of 675 kPa and hence $q_{na} = 225$ kPa, within 7 per cent, which confirms the Skempton value.
4. Part (a) — confirm the long-term case does not govern. In the drained condition the clay works in effective stress with $c^{\prime} = 15$ kPa, $\phi^{\prime} = 27^{\circ}$ and, with the water table at ground level, $\gamma^{\prime} = 21 - 9.81 = 11.19$ kN/m³. The factors are $N_q = 13.20$, $N_c = 23.94$ and $N_{\gamma} = 14.47$, giving $$q_u^{\prime} = 602.9 + 239.9 + 194.3 = 1037\ \text{kPa} \quad (\text{net } 1026\ \text{kPa})$$ which is 63 per cent larger than the undrained net capacity. The short-term undrained case governs, as expected for a saturated clay loaded quickly: strength rises as the excess pore pressures dissipate, so the foundation is at its most vulnerable on the day the load is applied.
5. Part (b) — set up the settlement integral. Consolidation settlement from a coefficient of volume compressibility is $s_{oed} = \sum m_v\,\Delta\sigma\,H$. Beneath the centre of a uniformly loaded rectangle the vertical stress increase is obtained by superposing four quadrants, each 2 m by 2 m, with Fadum's influence factor $I(m,n)$ where $m = n = 2/z$: $$\Delta\sigma_z = 4\,q_{net}\,I(m,n)$$ Summing over the 12 m of clay beneath the base in 0.25 m sublayers gives the geometric integral $$\int_0^{12} 4I\,\mathrm{d}z = 3.862\ \text{m}$$ so that $s_{oed} = m_v q_{net} \times 3.862$. With $m_v = 0.065\ \text{m}^{2}/\text{MN} = 6.5\times10^{-5}\ \text{m}^{2}/\text{kN}$, the oedometer settlement is $2.510 \times 10^{-4}$ m per kPa of net pressure, that is 25.1 mm per 100 kPa.
6. Part (b) — apply the Skempton-Bjerrum correction. The oedometer test allows no lateral strain, but the soil beneath a footing does strain laterally, so the real consolidation settlement is smaller. Skempton and Bjerrum express this as a settlement coefficient depending on the pore-pressure coefficient A and the layer geometry. Writing the pore pressure as $\Delta u = \Delta\sigma_3 + A(\Delta\sigma_1 - \Delta\sigma_3)$ gives $\mu = A + \alpha(1-A)$, where $\alpha = \int\Delta\sigma_3\,\mathrm{d}z/\int\Delta\sigma_1\,\mathrm{d}z$ is the geometry coefficient behind the chart. Treating the square as the circle of equal area, diameter $\sqrt{4 \times 16/\pi} = 4.51$ m, the layer ratio is $H/B = 12/4.51 = 2.66$, for which elastic theory under the centre gives $\alpha = 0.29$ (the chart's own values are 0.37 at H/B = 1 and 0.28 at H/B = 4). With $A = 0.42$, $$\mu = 0.42 + 0.29(1 - 0.42) = 0.59,\qquad s_c = \mu\,s_{oed}$$
7. Part (b) — invert for the allowable net pressure. Setting $s_c = 30$ mm, $$q_{net} = \frac{s_c}{\mu\,m_v \times 3.862} = \frac{0.030}{0.59 \times 2.510\times10^{-4}} = 202.6\ \text{kPa}$$ $$\boxed{q_{a} = 202.6 + 21 = 224\ \text{kPa (gross)}}$$
8. Decide which criterion governs. Comparing the two net pressures, 210 kPa from shear against 203 kPa from settlement, settlement governs, though only by about 4 per cent, so the two criteria are practically balanced. Loading the footing to the shear-governed value of 210 kPa net would produce 31.1 mm of consolidation settlement, 4 per cent over the limit. The design allowable bearing capacity is therefore $$q_{a} = 224\ \text{kPa gross},\qquad Q_{all} = 223.6 \times 16 = 3578\ \text{kN}$$ or, on a net basis, 203 kPa and 3242 kN.
9. Note the settlement components not asked for. The question asks only for consolidation settlement, but a complete serviceability check would add the immediate (undrained, constant-volume) settlement, which for a rigid square footing on a clay of undrained modulus around 40 to 60 MPa would be about 8 to 12 mm at this pressure; secondary compression can be neglected only if the clay is inorganic. If the immediate component were included in a 30 mm total-settlement limit rather than a 30 mm consolidation limit, the allowable pressure would fall to roughly 140 to 160 kPa net, and the design would be settlement-governed by a wide margin instead of a narrow one.