Question 1 of 6: Shallow Foundations (30 marks)

Question 1 — 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.

Part 1 — discussion (3 marks)

(a) Ultimate and serviceability limit states. A limit state is a condition beyond which the foundation no longer fulfils its design intent. The ultimate limit state (ULS) concerns collapse: general shear failure of the soil beneath the footing, sliding, overturning, or structural failure of the footing itself. It is checked by comparing a factored resistance — the ultimate bearing resistance reduced by a resistance factor — against a factored load obtained by multiplying the specified dead and live loads by their load factors. The consequence of exceeding it is sudden and catastrophic, so the reliability demanded is high.

The serviceability limit state (SLS) concerns performance under the loads that actually occur day to day: total settlement, differential settlement, tilt and the cracking or loss of function these cause in the supported structure. It is checked at unfactored (specified) loads against a deformation criterion — here 25 mm. Nothing collapses when the SLS is exceeded, but the bridge may become unserviceable, so the check is made with best-estimate soil stiffness rather than conservative strength. For footings on dense granular soils the two limit states are checked independently and, as this question demonstrates, the serviceability limit state very often governs the footing size.

(b) Overburden pressure and stress increase. The overburden pressure at a depth z is the vertical stress that already exists in the ground before the foundation is built, $\sigma_{v0} = \gamma z$ (total) or $\sigma^{\prime}_{v0} = \sigma_{v0} - u$ (effective). At the founding level of this footing it is $\gamma D_f = 18.1 \times 1.5 = 27.15\ \text{kPa}$. Because that stress is removed by the excavation and then re-applied by the footing, bearing-capacity and settlement calculations are properly made on the net pressure, $q_{net} = Q/B^2 - \gamma D_f$.

The stress increase Δσ_z caused by the footing load decays with depth and spreads laterally. Directly beneath the centre of a square footing it can be obtained from Boussinesq theory (four-quadrant superposition with the influence factor $I$, $\Delta\sigma_z = 4qI$) or, for hand work, from the 2:1 approximation $\Delta\sigma_z = q B^2 / (B+z)^2$. Both give the same picture: the pressure bulb is intense immediately under the footing, has fallen to about a third of the contact pressure at z = B (34 % by Boussinesq, 25 % by the 2:1 method) and to about 10 % at z = 2B (11 % by both). Practically, this is why the zone of soil that matters for a spread footing extends only about 1.5B to 2B below its base, and it is exactly why the groundwater table in this problem — 4 m below the footing base — can be ignored.

Part 2 — footing design

Given.

Find. (a) the plan size B × B of the square footing that gives an overall factor of safety of 3 on net bearing capacity; (b) whether that footing settles no more than 25 mm; (c) whether it also satisfies the limit-states bearing-resistance check under factored loads.

Approach. Size the footing from the Vesić bearing-capacity equation with shape and depth factors applied to the net pressure, then check the same footing for elastic settlement, and finally re-run the bearing check in limit-states form with the strength parameters factored by f_φ and f_c.

1. Establish the loads and the surcharge at founding level. The service (unfactored) column load and the overburden pressure removed by the excavation are $$Q = DL + LL = 2500 + 1250 = 3750\ \text{kN}, \qquad q = \gamma D_f = 18.1 \times 1.50 = 27.15\ \text{kPa}$$ Because the dead load already includes the weight of the foundation, no further self-weight need be added.
2. Confirm that the water table can be ignored. Bearing capacity is affected by groundwater only if the water table lies within about D_f + B of the ground surface. Anticipating B ≈ 1.8 m, that depth is 1.5 + 1.8 = 3.3 m, whereas the water table is at 5.5 m. The full moist unit weight γ = 18.1 kN/m³ therefore applies to both the surcharge and the N_γ term.
3. Evaluate the bearing-capacity factors for φ′ = 38°. Using the Vesić (CFEM) expressions, $$N_q = e^{\pi \tan\phi^{\prime}} \tan^{2}\!\left(45^{\circ} + \tfrac{\phi^{\prime}}{2}\right) = e^{\pi(0.7813)} \times \tan^{2}(64^{\circ}) = 48.93$$ $$N_{\gamma} = 2\,(N_q + 1)\tan\phi^{\prime} = 2\,(49.93)(0.7813) = 78.02$$ Since c′ = 0, the cohesion term drops out entirely and the whole resistance comes from the surcharge and self-weight terms.
4. Write the shape and depth factors for a square footing. With B/L = 1, $$s_q = 1 + \frac{B}{L}\tan\phi^{\prime} = 1.7813, \qquad s_{\gamma} = 1 - 0.4\frac{B}{L} = 0.60$$ $$d_q = 1 + 2\tan\phi^{\prime}(1 - \sin\phi^{\prime})^{2}\frac{D_f}{B} = 1 + 0.2308\,\frac{1.50}{B}, \qquad d_{\gamma} = 1$$ The depth factor is the only B-dependent coefficient, so the sizing must be done by iteration.
5. Form the factor-of-safety condition on net pressure. The gross and net ultimate bearing capacities and the net applied pressure are $$q_u = q N_q s_q d_q + \tfrac{1}{2}\gamma B N_{\gamma} s_{\gamma}, \qquad q_{u,net} = q_u - q, \qquad q_{app,net} = \frac{Q}{B^{2}} - q$$ and the design requirement is $q_{u,net} / q_{app,net} \ge 3$. Solving this numerically gives the minimum size $$\boxed{B_{min} = 1.76\ \text{m}}$$
6. Adopt a practical size and confirm the factor of safety. Rounding up to B = 1.80 m and re-evaluating with $d_q = 1 + 0.2308(1.50/1.80) = 1.1923$, $$q_u = 27.15(48.93)(1.7813)(1.1923) + 0.5(18.1)(1.80)(78.02)(0.60) = 2821.7 + 762.6 = 3584.3\ \text{kPa}$$ $$q_{u,net} = 3584.3 - 27.15 = 3557.2\ \text{kPa}, \qquad q_{app,net} = \frac{3750}{1.80^{2}} - 27.15 = 1130.3\ \text{kPa}$$ $$FS = \frac{3557.2}{1130.3} = 3.15 \;\ge\; 3 \quad \checkmark$$ so the answer to part (a) is a square footing $$\boxed{B \times B = 1.80\ \text{m} \times 1.80\ \text{m},\quad FS = 3.15}$$
7. (b) Compute the elastic settlement of that footing. For a rigid square footing on a deep, uniform elastic stratum the settlement under the net contact pressure is $$s_e = q_{net}\,B\,\frac{1 - \nu^{2}}{E_s}\,I_s$$ with $I_s = 0.82$ the rigid-square influence factor. Substituting the part-(a) footing, $$s_e = 1130.3 \times 1.80 \times \frac{1 - 0.3^{2}}{60\,000} \times 0.82 = 0.02530\ \text{m} = 25.3\ \text{mm}$$
8. Compare against the serviceability criterion. The computed 25.3 mm marginally exceeds the specified 25 mm, so the 1.80 m footing fails the serviceability limit state — by only 1 %, but it fails. Because $q_{net} \propto B^{-2}$, the settlement varies essentially as $B^{-1}$, and solving $s_e(B) = 25$ mm gives B = 1.82 m. Adopting the next practical size B = 1.85 m, $$s_e = 1068.5 \times 1.85 \times \frac{0.91}{60\,000} \times 0.82 = 24.6\ \text{mm} < 25\ \text{mm} \quad \checkmark$$ $$\boxed{\text{SLS at } B = 1.80\ \text{m}: s_e = 25.3\ \text{mm} > 25\ \text{mm} \;\Rightarrow\; \text{enlarge to } B = 1.85\ \text{m}\ (s_e = 24.6\ \text{mm})}$$
9. (c) Factor the strength parameters for the limit-states check. In the CFEM / Ontario Highway Bridge Design Code format the resistance factors are applied to the strength parameters themselves: $$\tan\phi^{*} = f_{\phi}\tan\phi^{\prime} = 0.8 \times 0.7813 = 0.6250 \;\Rightarrow\; \phi^{*} = 32.0^{\circ}, \qquad c^{*} = f_c c^{\prime} = 0.6 \times 0 = 0$$ The cohesion factor f_c = 0.6 is supplied but has no effect here, because the silty sand is cohesionless.
10. Recompute the bearing resistance with the factored friction angle. At φ* = 32.0° the factors become $N_q = 23.20$, $N_{\gamma} = 30.25$, $s_q = 1.625$, $s_{\gamma} = 0.60$ and $d_q = 1.2301$ at B = 1.80 m, giving $$q_r = 27.15(23.20)(1.625)(1.2301) + 0.5(18.1)(1.80)(30.25)(0.60) = 1258.8 + 295.6 = 1554.4\ \text{kPa}$$ $$q_{r,net} = 1554.4 - 27.15 = 1527.3\ \text{kPa}$$
11. Compare against the factored load. The factored column load and the factored net contact pressure are $$Q_f = 1.25\,DL + 1.5\,LL = 1.25(2500) + 1.5(1250) = 5000\ \text{kN}$$ $$q_{f,net} = \frac{5000}{1.80^{2}} - 27.15 = 1516.1\ \text{kPa}$$ $$\boxed{\frac{q_{r,net}}{q_{f,net}} = \frac{1527.3}{1516.1} = 1.007 \;\ge\; 1.0 \quad \checkmark \text{ ULS satisfied}}$$

Comment on the results (part c). Three points are worth making. First, the limit-states check is satisfied, but only just — the resistance exceeds the factored effect by 0.7 %. Second, that near-coincidence is not luck: an overall factor of safety of 3 on net capacity and the CFEM partial-factor set (f_φ = 0.8 with load factors 1.25/1.5) are calibrated to give broadly the same footing size for a dense granular soil, and here they agree to within 2 % on B. Third, and most important for the designer, neither ultimate-limit-state check governs: the serviceability requirement does. The footing must be enlarged from 1.80 m to 1.85 m to keep settlement below 25 mm, and at 1.85 m the ULS margin rises to about 1.07. This is the normal situation for footings on dense sand and gravel, where the available bearing capacity is very high and deformation, not collapse, controls the design.