Consolidation Time Scaling from Laboratory to Field: 07-Str-A3, May 2014, Q3

Question 3: Consolidation Time Scaling from Laboratory to Field (10 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. One oedometer result and one field geometry for the same clay, so the coefficient of consolidation is common to both and only the drainage path changes.

Given data
Quantity	Symbol	Value
Laboratory specimen thickness	$H_{\text{lab}}$	40 mm = 0.040 m
Laboratory drainage	—	top face only (one-way)
Time for 50 % consolidation in the laboratory	$t_{50,\text{lab}}$	18 min
Field layer thickness	$H_{\text{field}}$	4 m
Field drainage	—	top and bottom (two-way)
Time factor for $U = 50\%$ (printed sheet, $U < 60\%$)	$T_v = \tfrac{\pi}{4}U^2$	0.19635
Time factor for $U = 80\%$ (printed sheet, $U > 60\%$)	$T_v = -0.933\log(1-U) - 0.085$	0.56714

Find. (i) the time for the field layer to reach 50 % consolidation, and (ii) the time for the same layer to reach 80 % consolidation, both expressed in engineering units.

Figure 3.1 — The only difference between the two problems is the length of the longest drainage path $d$. One-way drainage makes $d$ the full thickness; two-way drainage halves it.

Approach. Both problems are governed by the same one-dimensional Terzaghi solution, so at equal degrees of consolidation the dimensionless time factor $T_v = c_v t/d^{\,2}$ is equal; with $c_v$ a property of the clay and therefore identical in the two cases, time scales as the square of the drainage path, and the 80 % case is then reached by re-entering the same relation with the larger time factor.

Fix the drainage path in each case. The drainage path $d$ is the longest distance a water particle must travel to reach a draining face. With drainage at the top only, every particle must travel the full specimen thickness, so $$d_{\text{lab}} = H_{\text{lab}} = 0.040\ \text{m}$$ With drainage at both faces, a particle escapes through whichever face is nearer, so the longest path is half the layer: $$d_{\text{field}} = \frac{H_{\text{field}}}{2} = \frac{4.0}{2} = 2.0\ \text{m}$$ This single distinction is what the question is really testing; it changes the answer by a factor of four.
Equate time factors at 50 % consolidation. Because it is the same clay at the same degree of consolidation, $T_v$ and $c_v$ both cancel out of $T_v = c_v t/d^{\,2}$, leaving the relation printed on the formula sheet: $$\frac{t_{\text{lab}}}{d_{\text{lab}}^{\,2}} = \frac{t_{\text{field}}}{\left(H_{\text{field}}/2\right)^{2}}$$ Rearranging for the field time and substituting, $$t_{50,\text{field}} = t_{50,\text{lab}}\left(\frac{d_{\text{field}}}{d_{\text{lab}}}\right)^{2} = 18 \times \left(\frac{2.0}{0.040}\right)^{2} = 18 \times 50^2 = 18 \times 2500$$ $$\boxed{\,t_{50,\text{field}} = 45\,000\ \text{min} = 31.25\ \text{days}\,}$$
Recover $c_v$ explicitly as a check. It is worth extracting the soil property itself, because it also lets part (ii) be solved directly. From the laboratory test, $$c_v = \frac{T_{v,50}\,d_{\text{lab}}^{2}}{t_{50,\text{lab}}} = \frac{0.19635 \times (0.040)^2}{18} = 1.745 \times 10^{-5}\ \text{m}^2/\text{min}$$ which is $9.17\ \text{m}^2$ per year — an entirely ordinary value for a silty clay. Feeding it back through $t = T_v d^2/c_v$ with $d = 2.0\ \text{m}$ reproduces 45 000 min exactly, confirming step 2.
Select the correct time-factor branch for 80 % consolidation. The formula sheet gives two expressions and the choice between them matters. Since $U = 80\% > 60\%$, the parabolic branch $T_v = \tfrac{\pi}{4}U^2$ does not apply and the logarithmic branch must be used: $$T_{v,80} = -0.933\log_{10}(1 - 0.80) - 0.085 = -0.933(-0.69897) - 0.085 = 0.65214 - 0.085 = 0.56714$$ For comparison the 50 % value is $T_{v,50} = \tfrac{\pi}{4}(0.50)^2 = 0.19635$.
Convert the time factor into a field time. The layer geometry has not changed, so $d = 2.0\ \text{m}$ still applies and only the time factor is different. Working directly from the ratio of time factors avoids re-substituting $c_v$: $$t_{80,\text{field}} = t_{50,\text{field}}\times\frac{T_{v,80}}{T_{v,50}} = 45\,000 \times \frac{0.56714}{0.19635} = 45\,000 \times 2.888$$ $$\boxed{\,t_{80,\text{field}} = 1.30 \times 10^{5}\ \text{min} = 90.3\ \text{days} \approx 3.0\ \text{months}\,}$$ The same figure follows from $t = T_v d^2/c_v = 0.56714 \times 4.0 / 1.745\times10^{-5} = 1.30\times10^{5}$ min, which confirms the value independently of the ratio.

The two answers together show the characteristic shape of a consolidation record: the first half of the settlement arrives in a month, but the next 30 % takes twice as long again, and the theoretical time to 100 % is infinite. That is why field settlement is specified in terms of a target degree of consolidation and a surcharge period, never as “wait until settlement stops”.

Results
Quantity	Symbol	Value
Laboratory drainage path	$d_{\text{lab}}$	0.040 m
Field drainage path	$d_{\text{field}}$	2.0 m
Coefficient of consolidation	$c_v$	$1.745\times10^{-5}\ \text{m}^2$/min = 9.17 m$^2$/yr
(i) Time to 50 % consolidation in the field	$t_{50}$	45 000 min = 31.25 days
Time factor at 80 % consolidation	$T_{v,80}$	0.5671
(ii) Time to 80 % consolidation in the field	$t_{80}$	1.30 × 10⁵ min = 90.3 days

Topic: One-dimensional Terzaghi consolidation, the drainage path, and the scaling of laboratory time to field time.

Key relations: $T_v = c_v t / d^{\,2}$; $T_v = \tfrac{\pi}{4}U^2$ for $U < 60\%$ and $T_v = -0.933\log(1-U) - 0.085$ for $U > 60\%$; $d = H$ for one-way drainage and $d = H/2$ for two-way drainage; and, for equal $U$ in the same soil, $t_{\text{lab}}/d_{\text{lab}}^2 = t_{\text{field}}/d_{\text{field}}^2$.

Why this works: Terzaghi’s diffusion equation for excess pore pressure has only one dimensionless group, $T_v$, so two problems in the same soil that share a degree of consolidation must share a time factor. Everything geometric collapses into $d$, and everything material collapses into $c_v$. Because $d$ enters squared, doubling the drainage path quadruples the time — the reason a 40 mm oedometer specimen consolidates in minutes while the deposit it came from takes months or years, and the reason vertical wick drains, which cut the drainage path to a fraction of a metre, are so effective.

Common pitfall: Mishandling $d$. Two errors recur: taking $d = H/2$ for the laboratory specimen (correct only when both porous stones drain, which this question explicitly rules out by saying “drained at the top only”), and taking $d = H$ for the field layer even though it is drained on both faces. Either mistake changes the answer by a factor of four. The third pitfall is using the $U < 60\%$ parabolic branch at $U = 80\%$, which returns $T_v = 0.503$ instead of 0.567 and understates the time by 11 %.

Source: Das, Principles of Geotechnical Engineering, 9th ed., Ch. 11 — time rate of consolidation, definition of the drainage path, and the two $T_v$–$U$ approximations reproduced on the examination formula sheet; Knappett & Craig, Craig’s Soil Mechanics, 8th ed., Ch. 7; ASTM D2435 for the oedometer procedure that produces $t_{50}$.

Question 3 of 6: Consolidation Time Scaling from Laboratory to Field

Question 3: Consolidation Time Scaling from Laboratory to Field (10 marks)