Question 8 of 10: Stress increase by the 2:1 method and consolidation settlement
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2013 — 98-Civ-B3 Geotechnical Design. Three hours, open book, any non-communicating calculator. Section A holds five 7-mark questions (answer any four); Section B holds the long 24-mark design questions (answer any three). Candidates are asked to identify the source of every design chart and assumed value used. Every question is answered here, because the set is a study resource rather than a sitting.
Reference texts. B. M. Das, Principles of Foundation Engineering (9th ed.) and Principles of Geotechnical Engineering (9th ed.); Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM, 4th ed.) — the governing Canadian reference for foundation practice; D. P. Coduto, Foundation Design: Principles and Practices; R. F. Craig, Craig's Soil Mechanics.
Note on the question numbering. The printed paper numbers two different Section B questions as “Question 9” — the retaining wall on page 5 and the drilled pier on pages 5–6 — and its Section B heading says “any three of the following four questions” while five questions are actually printed. The drilled-pier question is treated here as Question 10 so that every printed question has a unique number; no wording has been changed.
Question 8: Stress increase by the 2:1 method and consolidation settlement (24 marks)
Given. A 2.0 m by 2.0 m square footing carries a column load of 200 kN and is founded 1.5 m below ground level. The water table stands at the base of the footing. Above it the sand has $\gamma = 19$ kN/m$^3$; below it the sand is saturated at $\gamma_{sat} = 21$ kN/m$^3$ for a further 1.5 m. The clay layer beneath is 3.0 m thick with $\gamma_{sat} = 20$ kN/m$^3$, $e_0 = 0.9$, $C_c = 0.31$ and $C_s = 0.09$.
Data read from Figure 3
Quantity
Symbol
Value
Column load
Q
200 kN
Footing plan size
$B \times L$
$2.0\ \text{m} \times 2.0\ \text{m}$
Founding depth
$D_f$
1.5 m (water table at this level)
Sand above the water table
$\gamma$
19 kN/m$^3$
Sand below the water table
$\gamma_{sat}$
21 kN/m$^3$, 1.5 m thick
Clay layer
$\gamma_{sat},\ H$
20 kN/m$^3$, 3.0 m thick
Clay compressibility
$e_0,\ C_c,\ C_s$
0.9, 0.31, 0.09
Find. The average vertical stress increase in the clay layer under the centre of the footing by the 2:1 method, and the resulting average consolidation settlement of that layer.
[Figure not reproduced: Figure 3 (redrawn). Square footing, the 2:1 stress dispersion used to spread the column load, and the 3 m clay layer whose consolidation settlement is required. See the official exam paper.]
Approach. Spread the column load through the soil on planes inclined 2 vertical to 1 horizontal to obtain the stress increase at the top, middle and bottom of the clay, average those three values with Simpson's rule, compute the existing effective stress at the middle of the clay, and enter the one-dimensional consolidation equation.
Write the 2:1 stress distribution. The method assumes the load spreads uniformly over an area that grows by half the depth on each side, so at a depth $z$ below the base of the footing $$\Delta\sigma = \frac{Q}{(B+z)(L+z)} = \frac{200}{(2+z)^2}\ \text{kPa}$$with $z$ in metres. Depths are measured from the underside of the footing, not from ground level.
Evaluate the stress increase at the top, middle and bottom of the clay. The clay runs from 1.5 m to 4.5 m below the footing, so $z_t = 1.5$ m, $z_m = 3.0$ m and $z_b = 4.5$ m: $$\begin{aligned}\Delta\sigma_t &= \frac{200}{3.5^2} = 16.33\ \text{kPa} \\ \Delta\sigma_m &= \frac{200}{5.0^2} = 8.00\ \text{kPa} \\ \Delta\sigma_b &= \frac{200}{6.5^2} = 4.73\ \text{kPa}\end{aligned}$$
Average the three values over the layer. Because the stress falls off as an inverse square, a simple arithmetic mean of top and bottom would be poor; Simpson's rule over the three ordinates is the standard weighting: $$\Delta\sigma_{av} = \frac{\Delta\sigma_t + 4\Delta\sigma_m + \Delta\sigma_b}{6} = \frac{16.33 + 4(8.00) + 4.73}{6} = \boxed{8.84\ \text{kPa}}$$
Compute the existing effective overburden pressure at the middle of the clay. The middle of the clay is 4.5 m below ground level. Above the water table the sand contributes its bulk weight; below it every layer contributes its buoyant weight: $$\sigma'_0 = 19(1.5) + (21-9.81)(1.5) + (20-9.81)(1.5)$$$$\sigma'_0 = 28.50 + 16.79 + 15.29 = 60.57\ \text{kPa}$$
Apply the one-dimensional consolidation equation. No preconsolidation pressure is quoted, so the clay is taken as normally consolidated and the virgin compression index governs throughout: $$S_c = \frac{C_c H}{1+e_0}\log_{10}\frac{\sigma'_0 + \Delta\sigma_{av}}{\sigma'_0} = \frac{0.31(3.0)}{1+0.9}\log_{10}\frac{60.57 + 8.84}{60.57}$$The leading coefficient is $0.4895$ m and the logarithm is $0.05919$, so $$S_c = \boxed{0.0290\ \text{m} = 29.0\ \text{mm}}$$
Comment on the result. Thirty millimetres of consolidation settlement is modest and would normally be acceptable for an isolated column footing, whose usual limit is 25 mm of total settlement for a structure sensitive to distortion and 40 to 50 mm otherwise. It is not the whole settlement, however: immediate settlement of the 3 m of sand between the footing and the clay must be added, and secondary compression of the clay accumulates after primary consolidation is complete. Because the clay is drained top and bottom by the sand above and, presumably, a stratum below, the drainage path is 1.5 m and primary consolidation will be comparatively quick.
Check — assumptions. Two are worth stating explicitly, as the exam's Note 7 invites. First, the 200 kN is taken as the gross load applied at the base of the footing, so the full value is spread by the 2:1 rule; if it were treated as a net load the stress increase, and hence the settlement, would be smaller. Second, the clay is taken to be normally consolidated because no preconsolidation pressure is given. The swelling index $C_s = 0.09$ is supplied and would govern if the clay were overconsolidated with $\sigma'_c$ above 69.4 kPa; in that case the settlement would fall to about 8.4 mm, a factor of 3.4 less. An oedometer test establishing $\sigma'_c$ would resolve which applies.
