07-Str-B5 · December 2018

Question 6 of 6: Shallow Foundations: circular silo foundation

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Exams, December 2018 — 07-Str-B5 Foundation Engineering. Three hours, open book, any non-communicating calculator permitted. Six questions of equal value (30 marks each); five constitute a complete paper and only the first five appearing in the answer book are marked. All six are solved here. This subject is pure geotechnical engineering — the cover page names it Foundation Engineering.

Reference texts (the books an open-book candidate should have on the desk for this subject):

Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — the Canadian limit-states framework, geotechnical resistance factors, frost depth, pile design.
B. M. Das, Principles of Foundation Engineering, 9th ed. — bearing capacity with shape/depth/inclination factors, retaining walls, driven piles (§12), drilled shafts (§13), pile groups.
R. F. Craig & J. Knappett, Craig's Soil Mechanics, 9th ed. — effective stress, Bishop's simplified method, consolidation settlement.
M. J. Tomlinson & J. Woodward, Pile Design and Construction Practice, 6th ed. — adhesion factors, the equivalent-raft settlement method, block failure of pile groups.
L. C. Reese & M. W. O'Neill, Drilled Shafts: Construction Procedures and Design Methods (FHWA) — the α* = 0.55 shaft rule and the exclusion zones used in Question 5.
D. W. Taylor, Fundamentals of Soil Mechanics; A. W. Bishop & N. Morgenstern, Stability coefficients for earth slopes (Géotechnique, 1960) — the slope-stability charts behind Question 3.

Check — assumptions adopted across this paper. This examination omits at least one parameter that each design step needs. Every adoption is made explicitly here and flagged again at the point of use. (1) Questions 2 and 5 give no groundwater table. Both profiles are clays quoted by undrained shear strength, so they are saturated; the water table is therefore taken at ground level and effective stresses are computed with submerged unit weights. This is the conservative choice for settlement, and the sensitivity is reported in each answer (Question 2 settles 293 mm with the water table at surface against 153 mm with no free water). (2) Question 2 gives no adhesion factor. α is taken from Das's Table 12.6 (Terzaghi, Peck & Mesri, α against c_u/p_a) because that table needs no assumed effective-stress profile; the answer tabulates what Tomlinson's and the API's rules would give instead, and identifies the α at which the group size changes. (3) Question 1 does not say whether the quoted leg loads are factored. They are treated as specified (unfactored) loads, and both a serviceability check against the 200 kPa SLS capacity and an ultimate check at 1.4 × leg load against the 300 kPa factored resistance are carried out; the governing one is stated. (4) Question 5 gives no preconsolidation pressure. The requested Skempton correlation is used on the virgin compression line, and the (much smaller) settlement that follows if the profile's own c_u/σ′_v0 ratio of about 1.1 is honoured as an OCR of 2 is reported alongside. (5) Concrete unit weight is taken as 24 kN/m³ and γ_w as 9.81 kN/m³ throughout. (6) Where a question states only "factor of safety", the gross definition (ultimate bearing capacity over total applied pressure) is used, and the convention is restated in each answer.

Question 6 — Shallow Foundations: circular silo foundation (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 circular foundation carrying a total load of 10 MN, founded 2.5 m into the upper native silty clay, with the groundwater table able to rise to founding level and 47.5 m of silty clay between the base and the till.

Given data — Question 6
Quantity	Symbol	Value
Total vertical load (silo, silage and foundation)	Q	10 000 kN
Founding depth	D_f	2.5 m
Bearing stratum: undrained strength	c_u	50 kPa
Bearing stratum: effective parameters	c′, φ′	10 kPa, 28°
Unit weight above the water table	γ	20.5 kN/m³
Submerged unit weight below founding level	γ_sub	10 kN/m³
Compression index, void ratio	C_c, e₀	0.13, 0.80
Undrained modulus, Poisson ratio (upper layers)	E, ν	40–50 MPa, 0.45
Allowable total settlement	s_all	40 mm
Required overall factor of safety	FS	3

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

Figure 8 — the circular silo raft at 2.5 m depth with the water table risen to founding level, and the stress bulb whose integration over 30 m of silty clay gives the 163 mm of consolidation settlement.

Approach. Compute the gross ultimate bearing capacity of a circular footing in both the undrained ($\phi_u = 0$, short-term) and drained ($c^{\prime}, \phi^{\prime}$, long-term) conditions, identify which governs, size the diameter for $FS = 3$ on the governing case, then integrate the one-dimensional consolidation under the centre of the resulting footing using the Boussinesq circular-load solution and compare with the 40 mm limit.

Set the surcharge and confirm the geometry. The water table rises only to founding level, so the 2.5 m of soil above the base stays moist: $$q = \gamma D_f = 20.5(2.5) = 51.25\ \text{kPa}$$
Everything below founding level is submerged, with $\gamma_{sub} = 10$ kN/m³ as given. The "founded on the till (bedrock)" phrase in the question describes where the silo ultimately transfers load in the limit — the foundation itself is a shallow raft at 2.5 m, 47.5 m above the till, which is exactly why part (c) matters.
Part (a) — undrained bearing capacity. Immediately after the silo is filled the clay has not drained, so $\phi_u = 0$ and $N_c = 5.14$, $N_q = 1$, $N_\gamma = 0$. For a circle, $s_c = 1 + (N_q/N_c)(B/L) = 1.195$, and $d_c = 1 + 0.4(D_f/B)$. Writing it for a trial diameter $B$: $$q_{u,net} = c_uN_cs_cd_c = 50(5.14)(1.195)\left[1 + 0.4\frac{2.5}{B}\right]$$
and the gross capacity adds the surcharge back, $q_u = q_{u,net} + q$. At the diameter finally adopted, $B = 10$ m, this gives $d_c = 1.100$ and
$$\boxed{q_{u,net} = 337.7\ \text{kPa}, \qquad q_{u,gross} = 389.0\ \text{kPa}\ \ (\text{undrained})}$$
Part (a) — drained bearing capacity. In the long term the clay has consolidated and the effective parameters apply, $c^{\prime} = 10$ kPa and $\phi^{\prime} = 28^\circ$: $$N_q = e^{\pi\tan\phi^{\prime}}\tan^2\!\left(45+\tfrac{\phi^{\prime}}{2}\right) = 14.72,\quad N_c = (N_q-1)\cot\phi^{\prime} = 25.80,\quad N_\gamma = 2(N_q+1)\tan\phi^{\prime} = 16.72$$
with circular shape factors $s_c = 1 + N_q/N_c = 1.570$, $s_q = 1 + \tan\phi^{\prime} = 1.532$, $s_\gamma = 0.6$, and depth factors $d_q = 1 + 2\tan\phi^{\prime}(1-\sin\phi^{\prime})^2(D_f/B)$, $d_c = d_q - (1-d_q)/(N_c\tan\phi^{\prime})$, $d_\gamma = 1$. The weight term uses $\gamma_{sub} = 10$ kN/m³ because the water stands at the base. At $B = 10$ m:
$$q_u = \underbrace{437.8}_{c^{\prime}\ \text{term}} + \underbrace{1242.0}_{\text{surcharge}} + \underbrace{501.5}_{\text{weight}}$$ $$\boxed{q_{u,drained} = 2181\ \text{kPa}}$$
The drained capacity is 5.6 times the undrained one, so the undrained (short-term) case governs — the standard result for a footing on clay, and the reason a silo is at its most vulnerable the first time it is filled rather than after years of service.
Part (b) — size the foundation for FS = 3. Working on the gross definition, as "total (overall) factor of safety" implies, with $q_{applied} = 4Q/(\pi B^2)$: $$FS = \frac{q_{u,gross}(B)}{4Q/(\pi B^2)} = 3$$
Both sides depend on $B$, so the equation is solved by iteration; it is satisfied at $B = 9.91$ m. Rounding up to a constructible size:
$$\boxed{B = 10.0\ \text{m diameter}, \qquad q_{applied} = \frac{10\,000}{78.54} = 127.3\ \text{kPa}, \qquad FS = \frac{389.0}{127.3} = 3.06}$$
(On the net definition the same foundation returns $FS = 337.7/76.1 = 4.4$; the gross definition is the more demanding of the two here and is the one the question's wording asks for.)
Part (c) — stress increase beneath the centre. Consolidation is driven by the net increase in effective stress: $$q_{net} = q_{applied} - \gamma D_f = 127.3 - 51.25 = 76.07\ \text{kPa}$$
Under the centre of a uniformly loaded circle of radius $R = 5$ m, Boussinesq gives
$$\Delta\sigma(z) = q_{net}\left[1 - \left(\frac{1}{1+(R/z)^2}\right)^{3/2}\right]$$
and the in-situ effective stress below the base is $\sigma^{\prime}_{v0} = 51.25 + 10z$ with the water table at founding level.

Part (c) — consolidation settlement. Integrating over 30 m (three diameters, by which depth $\Delta\sigma$ has fallen to 4% of $q_{net}$) in fifteen 2 m sublayers, with $C_c/(1+e_0) = 0.13/1.8 = 0.0722$:

Leading sublayer contributions, B = 10 m
Depth (m)	Δσ (kPa)	σ′_v0 (kPa)	Δs (mm)
3.5	75.5	61.3	50.4
5.5	65.7	81.3	37.2
7.5	49.2	101.3	24.8
9.5	35.1	121.3	15.9
11.5	25.3	141.3	10.3
13.5–32.5 (remainder)	18.7 to 3.3	161.3 to 341.3	24.2 combined

$$s_c = 162.8\ \text{mm}$$

The immediate (undrained) settlement follows from elasticity with $E \approx 45$ MPa and $\nu = 0.45$ over the upper layers, using the rigid-circle influence factor $I = 0.79$:

$$s_i = \frac{q_{net}B(1-\nu^2)I}{E} = \frac{76.07(10)(0.7975)(0.79)}{45\,000} = 10.7\ \text{mm}$$ $$\boxed{s_{total} = 162.8 + 10.7 = 173\ \text{mm} \;\gg\; 40\ \text{mm} \ \Rightarrow\ \text{SLS NOT satisfied}}$$

Redesign for serviceability. Settlement, not strength, controls this foundation, and the lever available is the net pressure: because $q_{net} = 4Q/(\pi B^2) - \gamma D_f$, enlarging the raft attacks the settlement much faster than it relieves the bearing check. Repeating the integration over a range of diameters, the 40 mm limit is met at $B = 14.2$ m; adopting a constructible size: $$\boxed{B = 14.5\ \text{m},\quad q_{net} = 9.3\ \text{kPa},\quad s_{total} = 33\ \text{mm} \le 40\ \text{mm},\quad FS = 6.3}$$
Two related observations belong in the answer. First, at $B = 15.76$ m the foundation would be fully compensated — the weight of soil removed equals the load applied, $q_{net} = 0$, and the consolidation settlement vanishes; deepening the excavation instead of widening it is therefore the other route to the same end, and is usually cheaper for a silo that needs a below-grade reclaim tunnel anyway. Second, a grain silo is filled and emptied repeatedly, so it is the differential and cyclic settlement, not the total, that damages the structure; a rigid circular raft 14.5 m across on 47 m of uniform silty clay will settle almost uniformly, which is the strongest argument for the raft over a ring foundation.

Check — conventions adopted in Question 6. (i) "Total (overall) factor of safety" is read as gross: $q_u$ including surcharge, over the total applied pressure. On the net definition the 10 m raft would show FS = 4.4. (ii) The 40 mm limit is applied to total settlement (immediate plus consolidation), as the question says. (iii) Consolidation is computed under the centre, which is where a rigid circular raft's settlement is largest on a flexible basis; a rigid-footing average would be about 20% smaller and still far above 40 mm at B = 10 m. (iv) E = 45 MPa is a weighted average of the 50 and 40 MPa quoted for the layers within one diameter of the base.

Final results — Question 6
Quantity	Value
(a) Undrained q_u (net / gross) at B = 10 m	337.7 / 389.0 kPa — governs
(a) Drained q_u at B = 10 m	2181 kPa
(b) Diameter for FS = 3	9.91 m required → B = 10.0 m adopted
(b) Applied gross pressure / achieved FS	127.3 kPa / 3.06
(c) Net foundation pressure at B = 10 m	76.07 kPa
(c) Consolidation + immediate settlement	162.8 + 10.7 = 173 mm
(c) SLS verdict at B = 10 m	fails (limit 40 mm)
Diameter that satisfies both limit states	14.5 m — s = 33 mm, FS = 6.3
Fully compensated diameter (q_net = 0)	15.76 m

Topic. Bearing capacity and settlement of a circular raft on deep normally consolidated silty clay, and the recognition that serviceability rather than strength sizes such a foundation.

Key relations. Undrained $q_{u,net} = c_uN_cs_cd_c$ with $N_c = 5.14$; drained $q_u = c^{\prime}N_cs_cd_c + qN_qs_qd_q + \tfrac12\gamma^{\prime}BN_\gamma s_\gamma$; Boussinesq under a circle $\Delta\sigma = q[1-(1+(R/z)^2)^{-3/2}]$; $s_c = \sum\frac{C_c}{1+e_0}H\log_{10}\frac{\sigma^{\prime}_0+\Delta\sigma}{\sigma^{\prime}_0}$; elastic $s_i = q_{net}B(1-\nu^2)I/E$.

Why this works. Both limit states are checked at the same diameter but they scale differently, and that is the whole lesson of this question. The bearing capacity of a footing on clay is almost independent of $B$ — there is no $N_\gamma$ term when $\phi_u = 0$, so the only size effect is the mild depth factor — while the applied pressure falls as $1/B^2$. Settlement, by contrast, is driven by $q_{net}$, which falls as $1/B^2$ too but from a base that already has the excavation relief subtracted, so it collapses towards zero as the raft approaches full compensation. Widening a raft on clay therefore buys settlement performance far faster than it buys capacity, which is why the FS = 3 diameter of 10 m settles 173 mm and the 40 mm diameter of 14.5 m has an FS of over 6.

Common pitfall. Using the gross applied pressure in the consolidation calculation. The soil at founding level was already carrying 51.25 kPa before the silo existed; only the 76 kPa increment causes settlement, and using 127 kPa instead overstates the answer by roughly half. The second recurring error is stopping the integration at one diameter below the base: here the sublayers below 15 m still contribute a quarter of the total, because a 10 m circle stresses the ground to a real depth.

Source. Das, Principles of Foundation Engineering 9th ed., §3.6 (general bearing-capacity equation and factors), §5.9 (stress under a circular loaded area), §5.12 (primary consolidation settlement); Craig & Knappett, Craig's Soil Mechanics 9th ed., ch. 8; CFEM 4th ed. §10 (settlement criteria).

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