One-dimensional consolidation — thickness scaling: 16-Civ-A4 Geotechnical Materials and Analysis, May 2015, Q3

Question 3: One-dimensional consolidation — thickness scaling (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. Location A: layer thickness $H_A = 10\text{ m}$, one-way drainage, settlement $= 40\text{ mm}$ in $t = 1\text{ yr}$, equal to $U_A = 50\%$ of the ultimate settlement. Location B: $H_B = 20\text{ m}$, identical soil and loading, same one-way drainage.

Find. The settlement of the clay at location B under the same loading (both the ultimate value and the amount in the first year).

Approach. The ultimate consolidation settlement scales with layer thickness; the rate is governed by the time factor $T_v = c_v t / H_{dr}^{2}$, which lets us find the degree of consolidation reached at B in one year.

Ultimate settlement at A. Since $40\text{ mm}$ is 50% of the total, $$S_{c,A} = \frac{40}{0.50} = 80\text{ mm}.$$
Ultimate settlement at B. For the same soil and the same applied stress, the primary settlement is proportional to the compressible thickness ($S_c \propto H$): $$S_{c,B} = S_{c,A}\cdot\frac{H_B}{H_A} = 80\cdot\frac{20}{10} = \boxed{160\text{ mm}}.$$
Coefficient of consolidation from A. At $U = 50\%$, $T_v = 0.197$. With one-way drainage $H_{dr}=H$: $$c_v = \frac{T_v\,H_{dr}^{2}}{t} = \frac{0.197\,(10)^{2}}{1} = 19.7\ \text{m}^2/\text{yr}.$$
Time factor at B after one year. B drains one-way as well, so $H_{dr}=20\text{ m}$: $$T_{v,B} = \frac{c_v\,t}{H_{dr}^{2}} = \frac{19.7\,(1)}{(20)^{2}} = 0.0492.$$
Degree of consolidation at B. For $U \lt 60\%$ the parabolic branch applies, $T_v = \tfrac{\pi}{4}U^{2}$: $$U_B = \sqrt{\frac{4\,T_{v,B}}{\pi}} = \sqrt{\frac{4(0.0492)}{\pi}} = 0.250 = 25\%.$$
Settlement of B in one year. $$S_{B}(1\text{ yr}) = U_B\,S_{c,B} = 0.25\times160 = \boxed{40\text{ mm}}.$$

The result is worth interpreting: doubling the drainage-path length makes B consolidate four times more slowly, so after one year it has reached only 25% consolidation. Its ultimate settlement is twice A’s (160 mm vs. 80 mm), but in the first year B settles the same 40 mm as A — the doubled ultimate value is exactly offset by the halved degree of consolidation. The full 160 mm at B develops only over a much longer period.

Question 3 — results
Quantity	Location A	Location B
Ultimate primary settlement	80 mm	160 mm
Degree of consolidation after 1 yr	50%	25%
Settlement after 1 yr	40 mm	40 mm

Topic. Terzaghi one-dimensional consolidation — magnitude vs. rate.

Key relations. $S_c \propto H$ for identical soil/stress; $T_v = c_v t/H_{dr}^{2}$; $T_v = \tfrac{\pi}{4}U^{2}$ for $U\lt 60\%$; one-way drainage $\Rightarrow H_{dr}=H$.

Why this works. The amount of settlement is set by compressibility and stress (thickness-proportional), while the rate is set by how far the pore water must travel. Because time to a given $U$ grows with $H_{dr}^{2}$, a twice-as-thick one-way-draining layer needs four times as long to reach the same percentage, which is why the two locations coincidentally match at one year.

Common pitfall. Assuming B simply settles twice as much in one year, or forgetting that one-way drainage makes $H_{dr}=H$ (not $H/2$) — using $H/2$ would quarter $T_v$ again and misstate $U_B$.

Source. Das & Sobhan, Principles of Geotechnical Engineering, Ch. 11 (consolidation, $T_v$–$U$ relationship).

Question 3 of 6: One-dimensional consolidation — thickness scaling

Question 3: One-dimensional consolidation — thickness scaling (10 marks)