Question 1 of 6: Shallow Foundations

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.

Parts (a) and (b) — the discussion (3 marks)

Ultimate and serviceability limit states. A limit state is a condition beyond which a foundation stops doing the job it was built for. The ultimate limit state (ULS) is collapse: general shear failure of the soil beneath the footing, punching through a weaker stratum below, sliding along the base, overturning, or structural rupture of the footing itself. It is a strength check, made at factored loads against a factored geotechnical resistance,

where $\varphi_{gu}$ is the geotechnical resistance factor (0.5 for bearing capacity in CFEM 4th ed.) and $\alpha_i$ the NBCC load factors. Exceeding the ULS is sudden and usually catastrophic, so its probability of occurrence must be very small. That is exactly what the geotechnical report in this question means by "ULS (resistance) 300 kPa": it is a factored resistance, to be compared against factored loads.

The serviceability limit state (SLS) is performance under the loads that actually occur day to day: total settlement, differential settlement and tilt, and the cracking, jamming or loss of function they cause in the supported structure. It is a deformation check made at unfactored (specified) loads. The report's "SLS (capacity) 200 kPa" is a settlement-limited bearing pressure — the pressure at which the writer expects the tolerable settlement (conventionally 25 mm for a pad on sand) to be reached. For a 60 m guyed-free lattice tower on three legs, differential settlement between legs matters more than the absolute value: a few millimetres of relative movement tilts the mast, throws the antennas off alignment, and redistributes load into the legs. The two states are checked independently and the larger footing governs.

Total and effective stress. The total stress at a point is the whole normal stress carried by the saturated soil treated as a single two-phase material, $\sigma_v = \sum \gamma_i z_i$, computed with bulk unit weights. The pore water pressure $u$ is the pressure in the water filling the voids; under hydrostatic conditions $u = \gamma_w z_w$ below the water table. Terzaghi's principle states that the difference between them, the effective stress

is the part transmitted through inter-particle contact, and that it alone governs strength and compressibility. Two consequences run through this whole paper. First, shear strength is properly $\tau_f = c^{\prime} + \sigma^{\prime}\tan\phi^{\prime}$; the undrained parameter $c_u$ used in Questions 2, 5 and 6 is not a soil constant but the value $\tau_f$ happens to take while the effective stress has not yet had time to change. Second, raising the water table raises $u$ without changing $\sigma$, so it lowers $\sigma^{\prime}$, and with it both bearing capacity and slope stability — which is precisely what Question 4(e) and Question 3(b) are asking about. In the present question the borehole found water at 3.6 m, 1.7 m below the proposed base, and the report says design for submerged conditions is not required; that is a statement about effective stress, and it is checked below rather than taken on trust.

Design of the pad footings (27 marks)

Given. Three legs of a 60 m lattice tower, each delivering a specified axial compression of 1850 kN and a coincident shear of 225 kN to a square pad founded in compact-to-dense sand.

Find. The plan size of the square pad under each leg such that the serviceability pressure stays within 200 kPa and the factored pressure within 300 kPa, and confirmation that a first-principles bearing-capacity calculation and a sliding check are both satisfied.

Approach. Fix the founding depth from the frost and groundwater constraints, add the pad, pedestal and backfill weights to the leg load, convert the shear into a base moment and hence an eccentricity, work with Meyerhof's effective width $B^{\prime} = B - 2e$, and size $B$ so that both the SLS and the ULS pressures land inside the values the geotechnical report allows; then verify the report's numbers against a Meyerhof bearing-capacity calculation and check sliding.

1. Fix the founding depth. Two constraints bracket it: the base must sit at or below the frost-free depth of 1.9 m (CFEM 4th ed. §12; frost heave in a silty sand would jack the legs differentially), and it must stay above 3.0 m so the excavation does not run into significant groundwater inflow. Both the ULS and SLS values quoted apply "at or below 1.9 m", so nothing is gained by going deeper: $$\boxed{D_f = 1.9\ \text{m}}$$

   This also places the base 1.7 m above the observed water table at 3.6 m, which is why the report says submerged conditions need not be designed for.

2. Assemble the vertical load at the base of a trial pad. Take a 0.6 m thick pad with a 0.5 m square pedestal carrying the leg up to grade, and 1.3 m of compacted backfill over the pad. For a trial $B = 3.7$ m:
   $$W_{pad} = B^2 t \gamma_c = 3.7^2 \times 0.6 \times 24 = 197.1\ \text{kN}$$ $$W_{ped} = 0.5^2 \times 1.3 \times 24 = 7.8\ \text{kN}$$ $$W_{soil} = (3.7^2 - 0.5^2)\times 1.3 \times 20 = 349.4\ \text{kN}$$

   Adding these to the leg load gives the specified vertical reaction on the soil,

   $$N = 1850 + 197.1 + 7.8 + 349.4 = 2404\ \text{kN}$$
3. Convert the shear into an eccentricity. The 225 kN horizontal reaction acts at the top of the pedestal, so at founding level it produces a moment $M = H D_f$: $$M = 225 \times 1.9 = 427.5\ \text{kN}\!\cdot\!\text{m} \qquad e = \frac{M}{N} = \frac{427.5}{2404} = 0.178\ \text{m}$$

   Since $e = 0.178\ \text{m} < B/6 = 0.617\ \text{m}$, the resultant stays inside the middle third and the whole base remains in compression — no part of the pad lifts off.

4. Serviceability check, on Meyerhof's effective area. Replacing the eccentric load on the real base by a concentric load on a reduced base, $$B^{\prime} = B - 2e = 3.7 - 2(0.178) = 3.344\ \text{m} \qquad q_{SLS} = \frac{N}{B^{\prime}B} = \frac{2404}{3.344 \times 3.7}$$ $$\boxed{q_{SLS} = 194.3\ \text{kPa} \;\le\; 200\ \text{kPa}}$$

   A 3.6 m pad returns 203.6 kPa and fails, so 3.7 m is the smallest square pad that satisfies serviceability. Serviceability governs the plan size.

5. Ultimate limit state check. Factor the tower reactions by 1.4 (a live/wind-dominated combination on a communications tower) and the concrete and backfill dead weight by 1.25, per NBCC: $$N_f = 1.4(1850) + 1.25(554.4) = 3283\ \text{kN} \qquad M_f = 1.4(225)(1.9) = 598.5\ \text{kN}\!\cdot\!\text{m}$$

   which gives $e_f = 0.182$ m, $B^{\prime}_f = 3.335$ m, and

   $$\boxed{q_{ULS} = \frac{3283}{3.335 \times 3.7} = 266.0\ \text{kPa} \;\le\; 300\ \text{kPa}}$$

   The pad therefore satisfies both limit states, with serviceability the tighter of the two — the usual outcome when a report quotes 300/200 kPa, because the SLS value has already been cut back from strength to control settlement.

6. Independent check of the report's resistance, Meyerhof's equation. The candidate should not simply believe the two numbers in the borehole log. With $\phi = 30^\circ$, $$N_q = e^{\pi\tan\phi}\tan^2\!\left(45 + \tfrac{\phi}{2}\right) = 18.40 \qquad N_\gamma = (N_q - 1)\tan(1.4\phi) = 15.67$$

   The surcharge at founding level counts the 0.1 m of topsoil separately, $q = 0.1(16) + 1.8(20) = 37.6$ kPa. Shape and depth factors for a square pad with $K_p = 3.0$ are $s_q = s_\gamma = 1 + 0.1 K_p = 1.30$ and $d_q = d_\gamma = 1 + 0.1\sqrt{K_p}(D_f/B) = 1.089$. The load inclination is $\alpha = \arctan(225/2404) = 5.35^\circ$, giving $i_q = (1-\alpha/90)^2 = 0.885$ and $i_\gamma = (1-\alpha/\phi)^2 = 0.675$. The water table lies 1.7 m below the base, inside the $B^{\prime}$-deep failure zone, so the weight term uses the interpolated unit weight $\bar\gamma = \gamma^{\prime} + (d/B)(\gamma - \gamma^{\prime}) = 14.59$ kN/m³:

   $$q_u = q N_q s_q d_q i_q + \tfrac{1}{2}\bar\gamma B^{\prime} N_\gamma s_\gamma d_\gamma i_\gamma = 866.5 + 365.6$$ $$\boxed{q_u = 1232\ \text{kPa}, \qquad FS = \frac{1232}{194.3} = 6.3}$$

   The factored resistance $\varphi_{gu} q_u = 0.5(1232) = 616$ kPa is more than twice the report's 300 kPa. The report's values are therefore settlement-derived, not strength-derived, and it is right to design to them rather than to $q_u$.

7. Sliding. With a concrete pad cast against sand, take the base friction angle as $\delta = \tfrac{2}{3}\phi = 20^\circ$ and ignore the passive wedge on the buried face (conservative, because the backfill against a pad is rarely compacted to full passive stiffness): $$FS_{slide} = \frac{N\tan\delta}{H} = \frac{2404\tan 20^\circ}{225} = \frac{875}{225}$$ $$\boxed{FS_{slide} = 3.9 \;>\; 1.5}$$

   Even mobilising none of the 400 kN of passive resistance available over the 1.9 m buried face, sliding is not close to critical.