Question 3 of 6: Rate of consolidation at two clay thicknesses
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. Engineers Canada / PEO National Examinations, May 2015 —
07-Str-A3 Geotechnical Materials and Analysis. Closed book, three hours, 100 marks,
approved Casio or Sharp calculator, drawing instruments required. All six questions are
compulsory and are weighted 20 / 10 / 10 / 20 / 20 / 20. The paper supplies its own appendix:
a formula sheet, the rectangular-loading m–n influence chart and a Newmark
influence chart with influence value $I_N = 0.005$ (200 elements). Values quoted from those
sheets are used here.
Reference texts.
B. M. Das & K. Sobhan, Principles of Geotechnical Engineering, 9th ed. —
Ch. 6 (compaction), Ch. 7 (permeability and flow), Ch. 9 (stresses in a soil mass), Ch. 11
(consolidation), Ch. 12 (shear strength), Ch. 13 (lateral earth pressure).
R. F. Craig / J. A. Knappett, Craig’s Soil Mechanics, 8th ed. — Ch. 4
(effective stress and pore pressure coefficients), Ch. 5 (shear strength), Ch. 7
(consolidation), Ch. 11 (retaining structures).
Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM),
4th ed. — Ch. 4 (site investigation and sampling), Ch. 6 (settlement), Ch. 9
(earth pressures). The Canadian reference for practice, sampling classes and terminology.
K. Terzaghi, R. B. Peck & G. Mesri, Soil Mechanics in Engineering Practice,
3rd ed. — Art. 15–17 (consolidation), Art. 19–20 (shear strength).
Check — two printing slips in the source, carried as printed.
(1) Question 1 is headed “(4 x 5 = 20 marks)” but prints five lettered
parts (i)–(v). All five are answered below and the header total of 20 marks is kept;
read the header as five parts at four marks each. (2) In Question 5 the surcharge and the two
cohesion intercepts are printed with the units $\text{kN/m}^{3}$ — dimensionally these
are stresses and must be read as $\text{kPa}$ ($\text{kN/m}^{2}$); the unit weights carry the
$\text{kN/m}^{3}$ correctly. The solution states each value in the unit its symbol requires.
Question 3: Rate of consolidation at two clay thicknesses (10 marks)
one way (drained at one face only), so $H_{dr} = H$
Length of drainage path, $H_{dr}$
10 m
20 m
Settlement observed at $t = 1$ year
40 mm
required
Degree of consolidation at that time, $U$
50 per cent
required
Soil properties and applied loading
identical at the two locations
Find. The ultimate consolidation settlement at B, the settlement B will
have reached after the same one year, and the time B needs to reach the same 50 per cent
consolidation.
Approach. Ultimate settlement is proportional to layer thickness while the
time to reach a given degree of consolidation is proportional to the square of the drainage
path, so $c_v$ is back-figured from location A using the time-factor relation on the formula
sheet and then applied forward to location B.
Get the ultimate settlement at A from the reported degree of consolidation.
By definition the degree of consolidation is the ratio of the settlement so far to the ultimate
primary settlement, $U = s_t / s_c$, so
$$s_{c,A} = \frac{s_t}{U} = \frac{40\ \text{mm}}{0.50} = \boxed{80\ \text{mm}} .$$
Convert the 50 per cent point into a time factor. The paper’s formula
sheet gives the parabolic branch valid below 60 per cent consolidation,
$T_v = \dfrac{\pi}{4}U^{2}$, so at $U = 0.50$
$$T_v = \frac{\pi}{4}(0.50)^{2} = 0.196 .$$
Back-figure the coefficient of consolidation from location A. With one-way
drainage the drainage path equals the full layer thickness, $H_{dr} = 10\ \text{m}$, and the
time-factor definition $T_v = c_v t / H_{dr}^{2}$ inverts to
$$c_v = \frac{T_v H_{dr}^{2}}{t} = \frac{0.196 \times (10)^{2}}{1} = 19.6\ \text{m}^{2}\text{/year} .$$
The same value applies at B because the question states the soil properties are identical.
Scale the ultimate settlement to the thicker layer. Primary consolidation
settlement is $s_c = \dfrac{C_c H}{1+e_0}\log\left(\dfrac{\sigma_0' + \Delta\sigma'}{\sigma_0'}\right)$.
Under the same loading and the same soil the whole expression except $H$ is unchanged, so the
settlement is simply proportional to thickness:
$$s_{c,B} = s_{c,A}\,\frac{H_B}{H_A} = 80 \times \frac{20}{10} = \boxed{160\ \text{mm}} .$$
Find how far B has consolidated after one year. Location B has twice the
drainage path, so at $t = 1$ year its time factor is a quarter of A’s,
$$T_{v,B} = \frac{c_v t}{H_{dr,B}^{2}} = \frac{19.6 \times 1}{(20)^{2}} = 0.0491 ,$$
which is still well below the 0.283 that marks $U = 60$ per cent, so the parabolic branch may
be inverted directly:
$$U_B = \sqrt{\frac{4 T_{v,B}}{\pi}} = \sqrt{\frac{4 \times 0.0491}{\pi}} = 0.25 .$$
After one year location B has therefore completed 25 per cent of its consolidation.
Convert that degree of consolidation into a settlement. Applying it to the
ultimate value found in step 4,
$$s_{B}(1\ \text{year}) = U_B\, s_{c,B} = 0.25 \times 160 = \boxed{40\ \text{mm}} .$$
The thicker layer settles exactly the same 40 mm in the first year as the thinner one. This is
not a coincidence: while $U \le 0.60$, $U \propto \sqrt{T_v} \propto 1/H_{dr}$ and
$s_c \propto H$, so their product is independent of thickness. Early-time settlement is
governed by the drainage boundary, not by how much clay lies beneath it.
Find when B reaches the same 50 per cent. The time to any fixed degree of
consolidation scales with the square of the drainage path,
$$t_{50,B} = t_{50,A}\left(\frac{H_{dr,B}}{H_{dr,A}}\right)^{2}
= 1 \times \left(\frac{20}{10}\right)^{2} = \boxed{4\ \text{years}} ,$$
at which point B will have settled 80 mm. For completeness, 90 per cent consolidation needs the
upper branch of the formula sheet, $T_v = -0.933\log(1-U) - 0.085 = 0.848$, giving
$t_{90,B} = 0.848 \times 20^{2} / 19.6 = 17.3$ years against 4.3 years at A.
