16-Civ-A4 Geotechnical Materials and Analysis · May 2014
Question 3 of 6: Time rate of consolidation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Examinations — May 2014 · 98-Civ-A4 Geotechnical Materials and Analysis · 3 hours, closed book · 100 marks · answer all six questions. A formula sheet plus m–n influence and Newmark charts are supplied at the back of the paper.
Reference texts. B. M. Das, Principles of Geotechnical Engineering (9th ed.); R. F. Craig / Knappett & Craig, Craig’s Soil Mechanics (8th ed.); Holtz, Kovacs & Sheahan, An Introduction to Geotechnical Engineering; M. Budhu, Soil Mechanics and Foundations. Canadian practice: Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM, 4th ed.). Unit weight of water taken as $\gamma_w = 9.81\ \text{kN/m}^3$ throughout.
Given. Same clay in the laboratory and in the field, so the coefficient of consolidation $c_v$ is identical.
Given data
Quantity
Laboratory
Field
Layer thickness $H$
40 mm
4 m
Drainage
top only (single)
top & bottom (double)
Drainage path $d$
$40$ mm
$H/2 = 2$ m
$t$ for 50%
18 min
?
Find. The field consolidation times for (i) $U = 50\%$ and (ii) $U = 80\%$.
Approach. The time factor $T_v = c_v t / d^2$ depends only on the degree of consolidation $U$; equal $U$ means equal $T_v$, so times scale as $d^2$ (part i), and within the field layer times scale as $T_v$ (part ii).
Equal degree of consolidation ⇒ equal time factor. Because $T_v = \dfrac{c_v t}{d^2}$ and $c_v$ is the same, at $U = 50\%$ the lab and field share one $T_v$, hence $\dfrac{t_{lab}}{d_{lab}^2} = \dfrac{t_{field}}{d_{field}^2}$.
Field time for 50% consolidation. With $d_{lab} = 40\ \text{mm} = 0.04\ \text{m}$ and $d_{field} = 2\ \text{m}$, $t_{field} = t_{lab}\left(\dfrac{d_{field}}{d_{lab}}\right)^2 = 18\left(\dfrac{2}{0.04}\right)^2 = 18\times 2500$, so $\boxed{t_{50} = 45{,}000\ \text{min} = 31.3\ \text{days}}$.
Time factors for 50% and 80%. Using the standard series solution, $T_{v,50} = \dfrac{\pi}{4}(0.50)^2 = 0.197$ (the $U\lt 60\%$ branch) and $T_{v,80} = -0.933\log_{10}(1-0.80) - 0.085 = 0.567$ (the $U\gt 60\%$ branch).
Field time for 80% consolidation. For the one field layer, $t \propto T_v$, so $t_{80} = t_{50}\dfrac{T_{v,80}}{T_{v,50}} = 45{,}000\times\dfrac{0.567}{0.197}$, giving $\boxed{t_{80} \approx 1.30\times10^{5}\ \text{min} = 90.3\ \text{days}}$.
