16-Civ-A4 Geotechnical Materials and Analysis · May 2018
Question 4 of 6: Oedometer — Consolidation Settlement and Heave
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examinations — 16-Civ-A4 Geotechnical Materials and Analysis, May 2018.
Closed book, 3 hours, 100 marks. Answer all six questions. All required charts (rectangular-area influence chart, Newmark chart) and a formula sheet are provided at the back of the exam.
Reference texts: R.F. Craig & J. Knappett, Craig’s Soil Mechanics (8th ed.); B.M. Das, Principles of Geotechnical Engineering; R.D. Holtz, W.D. Kovacs & T.C. Sheahan, An Introduction to Geotechnical Engineering (2nd ed.); M. Budhu, Soil Mechanics and Foundations. Take $\gamma_w = 9.81\ \text{kN/m}^3$ throughout.
Question 4: Oedometer — Consolidation Settlement and Heave (25 marks)
Given. Oedometer $e$–$\sigma'$ data (loading then unloading), the soil profile below, and an extensive surface fill.
Oedometer results (saturated clay)
Pressure $\sigma'$ (kN/m²)
27
54
107
214
429
214
107
54
Void ratio $e$
1.243
1.217
1.144
1.068
0.994
1.001
1.012
1.024
Profile: 0–4 m sand, 4–12 m clay ($H=8\ \text{m}$), water table at ground surface; $\gamma_{sat}=19\ \text{kN/m}^3$ for both. Fill: 4 m at $20\ \text{kN/m}^3$ over an extensive area.
Find. (a) the consolidation settlement of the clay under the fill; (b) the eventual heave if the fill is later removed.
Oedometer $e$–$\log\sigma'$ curve. The blue virgin line (slope $C_c$) governs settlement; the red swelling line (slope $C_s$) governs heave. $\sigma_0'$ and $\sigma_1'$ mark the in-situ and post-fill effective stress at mid-clay.
Approach. Compute the in-situ effective stress at the middle of the clay, add the (extensive, hence uniform) fill surcharge to get the final stress, read $e_0$ and $e_1$ off the $e$–$\log\sigma'$ curve, and apply $s_c=\dfrac{H(e_0-e_1)}{1+e_0}$. For heave, unload from $\sigma_1'$ back to $\sigma_0'$ along the swelling line (index $C_s$).
In-situ effective stress at mid-clay. Mid-clay depth $=4+\tfrac{8}{2}=8\ \text{m}$; with the water table at the surface the whole profile is submerged, so
$$\sigma_0' = \gamma'\,z = (19-9.81)(8) = 9.19\times 8 = 73.5\ \text{kN/m}^2.$$
Fill surcharge. The fill covers an extensive area, so its stress is transmitted undiminished to every depth as a one-dimensional increment:
$$\Delta\sigma' = \gamma_{fill}\,H_{fill} = 20\times 4 = 80\ \text{kN/m}^2, \qquad \sigma_1' = 73.5+80 = 153.5\ \text{kN/m}^2.$$
Void ratios from the curve. Interpolating on $\log\sigma'$: at $\sigma_0'=73.5$, $e_0 = 1.184$; at $\sigma_1'=153.5$, $e_1 = 1.104$. Both points lie on the straight virgin line, confirming the clay is normally consolidated (compression index $C_c=\dfrac{1.144-1.068}{\log(214/107)}=0.25$).
(a) Consolidation settlement. Using the change in void ratio directly,
$$s_c = H\,\frac{e_0-e_1}{1+e_0} = 8000\ \text{mm}\times\frac{1.184-1.104}{1+1.184} = 8000\times\frac{0.0797}{2.184} = \boxed{292\ \text{mm}\ (\approx 0.29\ \text{m})}.$$
The compression-index form is a close cross-check (it uses the unrounded $C_c=0.2525$): $s_c=\dfrac{C_c H}{1+e_0}\log\dfrac{\sigma_1'}{\sigma_0'}=\dfrac{0.25\times8000}{2.184}\log\dfrac{153.5}{73.5}\approx 296\ \text{mm}$.
(b) Heave on removal of the fill. After consolidation the clay sits at $e_1=1.104$ under $\sigma_1'=153.5$. Removing the fill unloads it back to $\sigma_0'=73.5$ along the swelling line, whose index from the unloading data is
$$C_s = \frac{1.024-0.994}{\log(429/54)} = \frac{0.030}{0.900} = 0.033.$$
The swelling (heave) is then
$$s_h = H\,\frac{C_s}{1+e_1}\log\frac{\sigma_1'}{\sigma_0'} = 8000\times\frac{0.033}{2.104}\times\log\frac{153.5}{73.5} = \boxed{41\ \text{mm}}.$$
Because $C_s\ll C_c$, only about one-seventh of the settlement is recovered — consolidation is largely irreversible.
