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.

Part (a) and (b) of the discussion — limit states, total and effective stress (3 marks)

Ultimate and serviceability limit states. A limit state is a condition beyond which a foundation stops doing its job. The ultimate limit state (ULS) is collapse: general shear failure of the soil beneath the footing, punching through a weak stratum, sliding, overturning, or rupture of the footing itself. It is a strength check, made at factored loads against a factored geotechnical resistance, $\varphi_{gu}\,q_u \ge \sum \alpha_i Q_i$, or in the older working-stress form by requiring an overall factor of safety of about 3 on the ultimate bearing capacity. The consequence of exceeding the ULS is sudden and usually catastrophic, so the probability of exceeding it must be very small.

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 these cause in the supported structure. It is a deformation check made at unfactored (specified) loads against a deformation criterion, 40 mm in this question. Nothing collapses when the SLS is exceeded, but the structure may become unusable, and for a tall silo a modest tilt is as damaging as a settlement because it loads the silo wall eccentrically. The two states are checked independently, and for a foundation on compressible clay the SLS very often governs the size, as it does here.

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 carried by the soil skeleton through inter-particle contact, and that it alone controls strength and compressibility. Two consequences run through the whole of this paper. First, shear strength is properly written $\tau_f = c^{\prime} + \sigma^{\prime}\tan\phi^{\prime}$; the undrained parameter $c_u$ 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, consolidation settlement is driven by the change in effective stress: loading saturated clay first raises $u$ at constant volume, and settlement follows only as that excess pore pressure drains away and $\sigma^{\prime}$ rises. Raising the water table raises $u$ without changing $\sigma$, so it reduces $\sigma^{\prime}$ and reduces bearing capacity, which is why this question specifies that the GWT "may rise to 2.5 m", i.e. to founding level, and why Question 4(e) is worth 5 marks.

Parts (a), (b) and (c) — bearing capacity, design and settlement of the silo foundation (27 marks)

Given. A circular foundation carries a total (silo plus silage plus foundation) load of 12 MN, founded 2.5 m deep in the upper native silty clay, with the groundwater table standing at founding level.

Find. The ultimate bearing capacity in undrained and drained conditions, the diameter that gives an overall factor of safety of 3, and whether the resulting foundation settles less than 40 mm.

Approach. Write Vesic's general bearing-capacity equation for a circular footing in both the undrained ($\phi_u = 0$, total stress) and drained ($c^{\prime},\ \phi^{\prime}$, effective stress) forms, iterate the diameter until the gross factor of safety on the total load is 3, then check the resulting net pressure against the 40 mm limit using elastic plus one-dimensional consolidation settlement.

1. Overburden pressure at founding level. The soil above the base sits above the water table, so the surcharge acting at founding level is a total stress that is also an effective stress: $$q = \gamma D_f = 20.5 \times 2.5 = 51.25\ \text{kPa}$$
2. Part (a) — set up the undrained (short-term) capacity. Immediately after the silo is filled the clay cannot drain, so $\phi_u = 0$, giving $N_c = 5.14$, $N_q = 1$ and $N_{\gamma} = 0$. For a circle $B/L = 1$, and Vesic's shape and depth factors are $$s_c = 1 + \frac{N_q}{N_c}\frac{B}{L} = 1 + \frac{1}{5.14} = 1.1946,\qquad d_c = 1 + 0.4\frac{D_f}{B}$$ so the gross ultimate capacity is $q_{u,und} = c_u N_c s_c d_c + q$. The diameter is still unknown, so this expression is carried into the design iteration below and evaluated there.
3. Part (a) — set up the drained (long-term) capacity. In the long term the excess pore pressures have dissipated and the clay works in effective stress with $c^{\prime} = 10$ kPa and $\phi^{\prime} = 28^{\circ}$. The bearing-capacity factors are $$N_q = e^{\pi\tan\phi^{\prime}}\tan^{2}\left(45 + \frac{\phi^{\prime}}{2}\right) = 14.72,\qquad N_c = (N_q-1)\cot\phi^{\prime} = 25.80,\qquad N_{\gamma} = 2(N_q+1)\tan\phi^{\prime} = 16.72$$ with shape factors $s_c = 1.571$, $s_q = 1.532$ and $s_{\gamma} = 0.60$. Because the water table stands at founding level, the unit weight in the $N_{\gamma}$ term is the submerged value: $$q_{u,dr} = c^{\prime}N_c s_c d_c + q\,N_q s_q d_q + 0.5\,\gamma^{\prime} B\,N_{\gamma}s_{\gamma}$$
4. Part (b) — iterate the diameter for an overall factor of safety of 3. The applied gross pressure is $q_{app} = 4Q/(\pi D^{2})$ and the requirement is $q_{u,und}/q_{app} \ge 3$. Substituting a trial diameter, recomputing $d_c$ and re-solving until the diameter stops moving gives 10.89 m. Rounding up to a constructible size, $$\boxed{D = 11.0\ \text{m},\qquad A = \frac{\pi}{4}(11.0)^{2} = 95.03\ \text{m}^{2}}$$
5. Parts (a) and (b) — evaluate both capacities at the adopted diameter. With $D_f/B = 2.5/11.0 = 0.227$ the undrained depth factor is $d_c = 1 + 0.4(0.227) = 1.0909$, and for the drained case $d_q = 1.0680$ with $d_c = d_q - (1-d_q)/(N_c\tan\phi^{\prime}) = 1.0730$, so $$q_{u,und} = 50 \times 5.14 \times 1.1946 \times 1.0909 + 51.25 = 334.9 + 51.25 = 386.2\ \text{kPa}$$ $$q_{u,dr} = 434.8 + 1234.1 + 551.7 = 2220.6\ \text{kPa}$$ Substituting the applied pressure $q_{app} = 12\,000/95.03 = 126.3$ kPa gives the two factors of safety $$F_{und} = \frac{386.2}{126.3} = 3.06\ \checkmark \qquad\qquad F_{dr} = \frac{2220.6}{126.3} = 17.6$$ The undrained short-term case governs by a wide margin, the expected result for a saturated clay: the drained strength is far larger because friction is mobilised on an effective stress the undrained analysis cannot see.
6. Cross-check the undrained factor against Skempton. For $\phi_u = 0$ Skempton's circular expression gives $N_c = 5(1+0.2\,D_f/B)(1+0.2) = 6.27$, hence a net capacity of $50 \times 6.27 = 313.6$ kPa and an overall factor of safety of 2.89, about 6 per cent below the Vesic value and marginally short of 3. A diameter of 11.3 m would satisfy Skempton's form exactly; since the settlement check below shows serviceability governs by a factor of four, the difference does not change the design decision.
7. Part (c) — net pressure driving settlement. Settlement is caused only by the increase in stress over what the ground already carried at founding level: $$q_{net} = q_{app} - \sigma^{\prime}_{v0}(D_f) = 126.3 - 51.25 = 75.0\ \text{kPa}$$
8. Part (c) — immediate (undrained) settlement. Treating the foundation as a rigid circle on a deep elastic layer with the bearing-stratum modulus $E_u = 50$ MPa, $\nu_u = 0.5$ and rigid-circle influence factor $I_p = 0.79$: $$s_i = \frac{q_{net}B(1-\nu_u^{2})I_p}{E_u} = \frac{75.0 \times 11.0 \times 0.75 \times 0.79}{50\,000} = 0.0098\ \text{m} = 9.8\ \text{mm}$$
9. Part (c) — consolidation settlement of the native silty clay. The three native layers run from founding level to 30.5 m depth, that is 28 m of clay beneath the base, all with $C_c = 0.13$ and $e_0 = 0.8$, so the compression ratio is $C_c/(1+e_0) = 0.0722$. Dividing that thickness into 0.5 m sublayers, the stress increase under the centre of a uniformly loaded circle of radius $R$ is $$\Delta\sigma_z = q_{net}\left[1 - \left(\frac{1}{1+(R/z)^{2}}\right)^{3/2}\right]$$ and the initial effective stress at depth $z$ below the base is $\sigma^{\prime}_{v0} = 51.25 + 10z$, since everything below founding level is submerged. Summing $$s_c = \sum \frac{C_c}{1+e_0}\,\Delta z\,\log_{10}\frac{\sigma^{\prime}_{v0}+\Delta\sigma_z}{\sigma^{\prime}_{v0}}$$ over the 28 m gives $s_c = 170$ mm. More than half of it accumulates in the top 4 m, where the stress increase is still nearly the full 75 kPa while the initial effective stress is only 55 to 95 kPa.
10. Part (c) — verdict on the serviceability limit state. Adding the two components, $$\boxed{s_{tot} = s_i + s_c = 9.8 + 170 = 180\ \text{mm}\ \gg\ 40\ \text{mm}}$$ so the serviceability limit state is not satisfied by the foundation sized on bearing capacity. The 11.0 m base is more than four times too settlement-flexible.
11. Part (c) — what the settlement criterion actually demands. Repeating the summation while enlarging the diameter, which cuts the net pressure as $1/D^{2}$ but deepens the stressed zone only linearly, shows that 40 mm is first reached at a diameter of about 15.6 m, by which point the net pressure has fallen to roughly 14 kPa. That foundation carries a factor of safety of about 6 on bearing, confirming that SLS governs. The practical alternatives are a deeper and more nearly compensated raft, a larger ring foundation, preloading the site before erecting the silo, or piling through the compressible native clay, which is the route taken in Questions 2 and 5 for comparable ground.