16-Civ-B19 Foundation Engineering · December 2019

Question 6 of 6: Shallow Foundations — circular raft for a grain silo

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

Question 6: Shallow Foundations — circular raft for a grain silo (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. Total vertical load $Q=12$ MN; circular foundation embedded $D_f=2.5$ m; groundwater table may rise to 2.5 m, i.e. to founding level; $\gamma_{sub}=10$ kN/m³; allowable settlement 40 mm; $C_c=0.13$ and $e_0=0.8$ for the native silty clay. Soil profile from Table 1:

Find. A short comparison of total and effective stress; the ultimate bearing capacity in both undrained and drained conditions; the foundation diameter that delivers an overall factor of safety of 3; and the total settlement compared with the 40 mm allowance.

Approach. Answer the conceptual part first, then evaluate the bearing capacity in both drainage conditions as functions of the unknown diameter, size the raft so that the governing capacity delivers a gross factor of safety of 3, and finally integrate the consolidation settlement through the profile using the Boussinesq stress increment beneath a uniformly loaded circle.

1. Part (a — discussion) — total stress, effective stress and the relation between them. The total stress $\sigma$ at a point is the whole force per unit area transmitted across a plane, carried jointly by the soil skeleton and by the water filling the voids. The pore water pressure $u$ is the pressure in that water, and because water sustains no shear it presses equally in every direction and cannot contribute to shear strength. Terzaghi's principle separates the two: $$\sigma'=\sigma-u$$ The effective stress $\sigma'$ is the part carried by grain-to-grain contact, and it alone controls both the strength and the compressibility of a soil — a soil deforms and shears in response to changes in $\sigma'$, not in $\sigma$. This is why raising a water table weakens a soil without removing a single grain: $\sigma$ hardly changes while $u$ rises, so $\sigma'$ falls. It is also why the two analyses in part (a) below exist at all. Immediately after loading a saturated clay, the water has not had time to drain, the excess pore pressure carries the load increment, and the analysis is done in total stress with $c_u$ and $\phi_u=0$; in the long term that excess pressure dissipates, the load transfers to the skeleton, and the analysis is done in effective stress with $c'$ and $\phi'$. In a clay the short-term case is usually the more critical for stability, which is exactly what the numbers below show.
2. Part (a) — set up the bearing capacity for undrained conditions. The raft founds in the first native silty clay layer ($c_u=50$ kPa, $\gamma=20.5$ kN/m³) with the water table risen to founding level, so the surcharge is computed with the bulk unit weight above that level: $$q=\gamma D_f=(20.5)(2.5)=51.25\ \text{kPa}$$ For $\phi_u=0$, $N_c=5.14$, $N_q=1$, $N_\gamma=0$, with the circular shape factor $s_c=1.2$ and depth factor $d_c=1+0.4D_f/B$: $$q_{u(\text{und})}=5.14c_us_cd_c+q=5.14(50)(1.2)\left(1+\frac{1.0}{B}\right)+51.25$$
3. Set up the bearing capacity for drained conditions. Using the effective parameters of the founding layer ($c'=10$ kPa, $\phi'=28^\circ$) with Hansen factors $N_c=25.80$, $N_q=14.72$, $N_\gamma=10.94$, circular shape factors $s_c=1+N_q/N_c=1.571$, $s_q=1+\tan\phi'=1.532$, $s_\gamma=0.6$, and the submerged unit weight in the self-weight term because the water table stands at founding level: $$q_{u(\text{dr})}=c'N_cs_cd_c+qN_qs_qd_q+\tfrac{1}{2}\gamma_{sub}BN_\gamma s_\gamma$$
4. Part (b) — size the raft for a gross factor of safety of 3. The required condition is $\min\left[q_{u(\text{und})},q_{u(\text{dr})}\right]/3 \ge Q/(\pi B^2/4)$. Solving this by iteration gives a required diameter of 10.87 m; adopt $B=11.0$ m. At that diameter, $$A=\frac{\pi(11.0)^2}{4}=95.03\ \text{m}^2,\qquad q_{app}=\frac{12\,000}{95.03}=126.3\ \text{kPa}$$ $$q_{u(\text{und})}=5.14(50)(1.2)\left(1+\frac{1.0}{11.0}\right)+51.25=\boxed{387.7\ \text{kPa}}$$ $$q_{u(\text{dr})}=442+1234+361=\boxed{2037\ \text{kPa}}$$ The drained capacity is 5.3 times the undrained value, so the short-term undrained case governs — as it almost always does for a foundation on clay, and as the discussion in step 1 anticipated. The achieved factor of safety is $$FS=\frac{387.7}{126.3}=\boxed{3.07}\ \ge 3\quad\checkmark$$
5. Note the sensitivity of the answer to the definition of the factor of safety. The question specifies a total (gross) factor of safety, which is what has been used. Had the more usual net definition been applied — $FS_{net}=(q_u-q)/(q_{app}-q)$ — the same 11.0 m raft would show $(387.7-51.25)/(126.3-51.25)=4.5$, and a diameter of about 9.7 m would have sufficed. The distinction is worth 1.3 m of diameter here, so the definition must always be stated.
6. Part (c) — establish the net pressure driving settlement. Settlement is caused only by the increase in stress over what the soil already carried, and here the raft replaces 2.5 m of soil that was excavated: $$q_{net}=q_{app}-q=126.3-51.25=75.0\ \text{kPa}$$
7. Compute the stress increment with depth beneath the circular raft. On the centreline of a uniformly loaded circle of radius $R=5.5$ m the Boussinesq solution gives $$\Delta\sigma=q_{net}\left[1-\frac{1}{\left(1+(R/z)^2\right)^{3/2}}\right]$$ with $z$ measured below the base. The increment falls to about 10 % of $q_{net}$ at $z=2B=22$ m, which sets the depth of the calculation — from founding level at 2.5 m down to 24.5 m, spanning the remainder of layer 1, the whole of layer 2, and the upper 12 m of layer 3.
8. Integrate the consolidation settlement. The initial effective stress uses the bulk unit weight above the water table and the submerged unit weight below it, $\sigma'_0=20.5(2.5)+10(z_m-2.5)$, and each sub-layer contributes $\dfrac{C_c}{1+e_0}H\log_{10}\dfrac{\sigma'_0+\Delta\sigma}{\sigma'_0}$ with $C_c/(1+e_0)=0.13/1.8=0.0722$:

   Depth range (m)	Layer	Contribution (mm)	Running total (mm)
   2.5 – 5.0	Native silty clay 1	60.6	60.6
   5.0 – 12.5	Native silty clay 2	82.2	142.8
   12.5 – 24.5	Native silty clay 3	23.9	166.7

   $$S_c=\boxed{166.7\ \text{mm}}$$
9. Add the immediate settlement and compare with the allowance. With a representative $E_u\approx50$ MPa over the stressed depth, $\nu_u=0.5$, the rigid-circle influence factor $\pi/4=0.79$ and the same net pressure that drives consolidation, $$S_i=\frac{q_{net}B(1-\nu_u^2)I_s}{E_u}=\frac{(75.0)(11.0)(0.75)(0.79)}{50\,000}=9.7\ \text{mm}$$ $$S_{total}=9.7+166.7=\boxed{176\ \text{mm}}\ \gg\ 40\ \text{mm}$$ The settlement requirement is not satisfied — the predicted settlement exceeds the allowance by a factor of about 4.4, even though the raft satisfies bearing capacity.
10. Conclude and recommend. The 11.0 m raft is governed overwhelmingly by settlement rather than by bearing capacity, which is the normal situation for a large, lightly loaded area on a deep compressible profile: widening a foundation raises its capacity in proportion to area but deepens the pressure bulb, so it buys much less settlement relief than intuition suggests. Reaching 40 mm by enlargement alone is not practical. The realistic options are to found the silo on piles or barrettes carried to the stiff silty clay at 30.5–47.5 m or to the till, to preload the site with a surcharge so that the primary consolidation occurs before the silo is commissioned, or to accept the movement and design for it — a circular silo is comparatively tolerant of uniform settlement, so the governing criterion in practice may be differential rather than total movement, and the connections to conveyors and discharge equipment would need articulation. Two conservatisms should be recorded: the calculation treats every layer as normally consolidated, whereas a real native clay profile at this depth is likely to be lightly over-consolidated and would compress on the recompression line for part of the load; and 40 mm is a demanding allowance for a structure of this size, so it would be worth confirming with the client whether it refers to total or to differential settlement.