Question 1 of 5: Shallow Foundations — Limit States and Bearing Capacity

Question 1: Shallow Foundations — Limit States and Bearing Capacity (20 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 (a) — ultimate and serviceability limit states

A shallow foundation must be checked against two distinct families of failure, and Canadian practice (CFEM 4th ed., consistent with NBCC Part 4) separates them explicitly. The ultimate limit state (ULS) concerns collapse: the foundation, the supporting soil, or the pair acting together loses equilibrium. Design at ULS compares a factored resistance with factored loads,

where $R_u$ is the ultimate geotechnical resistance (for bearing, $q_u$ multiplied by the effective footing area), $\varphi_{gu}$ is the geotechnical resistance factor at ULS (CFEM recommends about 0.5 for bearing resistance from a soil-parameter analysis), and $\alpha_i$ are the NBCC load factors. The failure modes collected under ULS are bearing capacity failure of the soil (general, local or punching shear), sliding along the base, overturning, uplift, global slope instability through the footing, and structural failure of the footing itself. Classical practice states the same requirement as a lumped factor of safety, $q_{all} = q_u / FS$ with $FS = 2.5$ to $3$ — the arithmetic of parts (a) to (c) below.

The serviceability limit state (SLS) concerns deformation under unfactored (service) loads: total settlement, differential settlement, tilt, heave and lateral movement large enough to damage the structure, crack cladding, jam doors or disable equipment, even though the soil is nowhere near collapse. Typical tolerances are a total settlement of 25 mm for a footing on sand and 50 mm on clay, with angular distortion limited to roughly 1/500 for framed buildings. No load or resistance factors are applied; the calculation is a best-estimate deformation compared with a tolerable deformation. On soft compressible soils SLS almost always governs the footing size, so a design that satisfies bearing capacity alone is incomplete.

Part (b) — overburden pressure and the distribution of stress increase

Overburden pressure is the stress already in the ground before the structure arrives, produced by the self-weight of the soil above the point of interest. At depth $z$ the total vertical stress is $\sigma_z = \sum \gamma_i h_i$, the pore pressure is $u = \gamma_w h_w$ below the water table, and the effective overburden pressure that actually controls strength and compressibility is

At founding level the overburden has a second role: the soil removed in excavating to depth $D_f$ is replaced by the footing, so the surcharge $q = \gamma D_f$ acts alongside the footing and contributes the $qN_q$ term of the bearing-capacity equation. The difference between the pressure the structure applies and that surcharge is the net pressure, $\Delta q = q_{applied} - \gamma D_f$, and it is the net pressure — not the gross — that causes settlement, because the ground has already consolidated under $\gamma D_f$.

The stress increase from the footing load spreads with depth. The rigorous description is the Boussinesq solution for a loaded area on an elastic half-space, integrated over the footing to give $\Delta\sigma = q\,I$ with an influence factor $I$ that depends only on geometry; the corner of a uniformly loaded rectangle uses $I(m,n)$ with $m = B/z$ and $n = L/z$, and the centre is obtained by adding four corner contributions. Contours of equal $\Delta\sigma$ form the familiar pressure bulb: $\Delta\sigma$ is largest immediately under the footing, falls to about $0.1q$ at a depth of roughly $2B$ below a square footing, and spreads laterally as it decays. That depth of about $2B$ is the practical “significant depth” over which compressible layers must be counted in a settlement calculation. For quick estimates the same behaviour is approximated by the 2:1 method, in which the load is assumed to spread at two vertical to one horizontal,

Parts (a)–(c) — bearing capacity of the 5 ft square footing

Given.

Find. (a) the allowable gross load at $FS = 4$; (b) the allowable net load at $FS = 5$; (c) the allowable net load when a factor of safety of 1.5 is applied to the shear-strength parameters instead of to the load.

Approach. Evaluate Terzaghi's general-shear bearing capacity for a square footing, then apply the factor of safety three different ways — on gross pressure, on net pressure, and on the strength parameters themselves.

1. Part (a) — write Terzaghi's equation for a square footing. For general shear failure Terzaghi's square-footing form carries the 1.3 and 0.4 shape constants: $$q_u = 1.3\,c\,N_c + q\,N_q + 0.4\,\gamma\,B\,N_\gamma$$ where $q = \gamma D_f$ is the surcharge at founding level and $N_c$, $N_q$, $N_\gamma$ are Terzaghi's bearing-capacity factors, functions of $\phi$ alone.
2. Read the bearing-capacity factors at $\phi = 20^\circ$. From Das Table 3.1 (Terzaghi's factors), $N_c = 17.69$, $N_q = 7.44$ and $N_\gamma = 3.64$. The first two can be confirmed from Terzaghi's closed form $N_q = a^2 / \left[2\cos^2\!\left(45^\circ + \phi/2\right)\right]$ with $a = e^{(0.75\pi - \phi/2)\tan\phi}$ and $N_c = (N_q - 1)\cot\phi$, which give 7.439 and 17.69.
3. Evaluate the surcharge and the ultimate bearing capacity. The surcharge is $q = \gamma D_f = 115 \times 3 = 345\ \text{lb/ft}^2$. Substituting term by term, $$q_u = 1.3(320)(17.69) + (345)(7.44) + 0.4(115)(5)(3.64)$$ $$q_u = 7\,359.0 + 2\,566.8 + 837.2 = \boxed{10\,763\ \text{lb/ft}^2}$$ The cohesion term supplies about two-thirds of the capacity, as expected for a soil with a modest friction angle and a real cohesion intercept.
4. Apply $FS = 4$ to the gross pressure and convert to a load. The gross allowable bearing pressure is $$q_{all} = \frac{q_u}{FS} = \frac{10\,763}{4} = 2\,690.8\ \text{lb/ft}^2$$ and over the 25 ft² plan area the allowable gross load is $$Q_{all} = q_{all} \times B^2 = 2\,690.8 \times 25 = \boxed{67\,269\ \text{lb} \approx 67.3\ \text{kip}}$$ which is about 299 kN. This is the total load the footing may carry including its own weight and the weight of the backfill above it.
5. Part (b) — form the net ultimate bearing capacity. The net capacity discounts the surcharge that was already there before construction: $$q_{u(net)} = q_u - q = 10\,763 - 345 = 10\,418\ \text{lb/ft}^2$$
6. Apply $FS = 5$ to the net pressure. Dividing by the stated factor of safety, $$q_{all(net)} = \frac{10\,418}{5} = 2\,083.6\ \text{lb/ft}^2$$ so the allowable net load is $$Q_{all(net)} = 2\,083.6 \times 25 = \boxed{52\,090\ \text{lb} \approx 52.1\ \text{kip}}$$ The net load is the useful column load: the structure may add this much to what the ground already carried.
7. Part (c) — factor the strength parameters, not the load. With $FS_{shear} = 1.5$ the design (developed) parameters are $$c_d = \frac{c}{FS_{shear}} = \frac{320}{1.5} = 213.33\ \text{lb/ft}^2, \qquad \phi_d = \tan^{-1}\!\left(\frac{\tan\phi}{FS_{shear}}\right) = \tan^{-1}\!\left(\frac{0.36397}{1.5}\right) = 13.64^\circ$$ This is the more rational statement of safety, because it factors the quantities that are actually uncertain.
8. Re-enter the tables at the developed friction angle. At $\phi_d = 13.64^\circ$, Terzaghi's closed form gives $N_c = 11.85$ and $N_q = 3.88$, and linear interpolation in Das Table 3.1 between $\phi = 13^\circ$ ($N_\gamma = 1.04$) and $\phi = 14^\circ$ ($N_\gamma = 1.26$) gives $N_\gamma = 1.18$.
9. Compute the net allowable pressure directly. Because the safety margin is already inside the parameters, the net allowable pressure is the net bearing capacity evaluated with the developed parameters — no further division: $$q_{all(net)} = 1.3\,c_d N_c + q\,(N_q - 1) + 0.4\,\gamma B N_\gamma$$ $$q_{all(net)} = 1.3(213.33)(11.85) + 345(3.88 - 1) + 0.4(115)(5)(1.18)$$ $$q_{all(net)} = 3\,286.3 + 992.0 + 271.5 = \boxed{4\,550\ \text{lb/ft}^2}$$ and therefore $$Q_{all(net)} = 4\,550 \times 25 = 113\,745\ \text{lb} \approx 113.7\ \text{kip}$$
10. Compare the three answers. Part (c) returns roughly twice the load of part (b) from the same soil. That is not an error: $FS_{shear} = 1.5$ on the strength parameters is a far less severe requirement than $FS = 5$ on the load, and the comparison is exactly the point of the question. Expressed as an equivalent load factor, $q_{u(net)}/q_{all(net)} = 10\,418 / 4\,550 \approx 2.3$, which is at the low end of normal practice; a designer would still check settlement at SLS before adopting it.