Question 5 of 6: Consolidation, Seepage Heads and Artesian Effective Stress

Question 5: Consolidation, Seepage Heads and Artesian Effective Stress (20 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.

(a) One-dimensional consolidation through the spring analogy (5 marks)

Given. The four-stage spring-and-piston model of Figure 2: a cylinder of water containing a spring, closed by a piston that carries a valve.

Find. An explanation of one-dimensional consolidation in terms of the effective stress principle, keyed to the four stages.

[Figure not reproduced: Figure 2 (redrawn) — the spring analogy. Spring = soil skeleton (carries σ′); water = pore water (carries u); valve = permeability; piston = applied total stress σ. See the official exam paper.]

Approach. Read the analogy element by element — spring as soil skeleton, water as pore water, valve as permeability — then walk the four stages and write $\sigma' = \sigma - u$ at each.

The correspondence is exact and worth stating before anything else: the spring is the compressible soil skeleton, and the force it carries is the effective stress $\sigma'$; the water filling the cylinder is the pore water, and the pressure in it is $u$; the valve in the piston is the permeability of the soil, controlling how fast water can escape; and the piston applies the total stress $\sigma$. The cylinder walls prevent lateral strain, which is what makes the model one-dimensional and identical to the oedometer.

1. Stage (a) — initial equilibrium. No load has been added, the valve is shut, and the spring carries the seating load alone. The pore water is at its static pressure with no excess, $u_e = 0$, so $$\sigma' = \sigma_0' = \sigma_0 - u_s$$ This is the in-situ condition before construction.
2. Stage (b) — load applied, valve still closed. An increment $\Delta\sigma$ is placed on the piston with no drainage permitted. Water and mineral grains are three orders of magnitude stiffer than the skeleton, so a saturated system that cannot change volume cannot compress the spring at all. Skempton’s parameter for a saturated soil is $B = \Delta u/\Delta\sigma = 1$, hence $$u_e = \Delta\sigma \qquad\text{and}\qquad \boxed{\;\sigma' = (\sigma_0 + \Delta\sigma) - (u_s + \Delta\sigma) = \sigma_0'\;}$$ The whole of the new load is carried by the pore water; the spring has not shortened and the effective stress is unchanged. This is the end-of-construction, undrained condition, and it is why an embankment placed quickly on soft clay is at its most critical the day it is finished.
3. Stage (c) — valve opened, water escaping. Water now flows out under the excess pressure gradient. As it leaves, the piston descends, the spring shortens, and load transfers progressively from water to spring. At any degree of consolidation $U$, $$u_e = (1-U)\,\Delta\sigma \qquad\text{and}\qquad \sigma' = \sigma_0' + U\,\Delta\sigma$$ The sum $\sigma' + u$ is constant throughout, because the total stress has not changed — this is the entire content of Terzaghi’s principle. The rate is governed by the time factor $T_v = c_v t/d^2$, which is why the drainage path length $d$ matters far more than the load: doubling the layer thickness quadruples the time.
4. Stage (d) — consolidation complete. Flow ceases when the excess pore pressure has fully dissipated, $u_e = 0$ and $u$ has returned to its static value. The spring now carries the entire increment: $$\boxed{\;\sigma' = \sigma_0' + \Delta\sigma\;}$$ and the settlement is the total shortening of the spring, $$s_c = \frac{C_c H}{1+e_0}\,\log\!\left(\frac{\sigma_0' + \Delta\sigma}{\sigma_0'}\right)$$ This is the long-term, drained condition — the one that governs settlement, and the one whose strength parameters Question 6 measures.

Two consequences follow from the model that are easy to state and easy to forget. First, settlement is caused by effective stress and by nothing else, so a change in total stress that is exactly matched by a change in pore pressure — a rising water table under an unchanged fill, for instance — causes no consolidation at all. Second, lowering the water table without touching the surface load raises $\sigma'$ just as surely as adding fill does, which is the mechanism behind dewatering-induced settlement of adjacent buildings and, at regional scale, behind subsidence over pumped aquifers.

(b) Pressure, elevation and total head through the permeameter (7 marks)

Given. The constant-head arrangement of Figure 3: a standpipe connected by a U-tube to the base of a container holding a soil sample, so that flow enters at the bottom of the sample and leaves at the top.

Find. The elevation head, pressure head, pressure and total head at the entrance end, at A and at the exit end, plotted against height.

[Figure not reproduced: Figure 3 (redrawn) — upward flow through the sample, with the elevation, pressure and total head plotted against height. Datum at the base of the sample; the total-head line falls linearly at i = 1.00. See the official exam paper.]

Approach. Establish the flow direction from the two water surfaces, fix the total head at each end from those surfaces, distribute the head loss linearly through the sample (Darcy flow in a homogeneous specimen), and take pressure head as total head minus elevation head at each point.

1. Establish the direction of flow. The standpipe is connected to the base of the container and its water surface stands 4 m above the container’s. The total head at the bottom of the sample is therefore the higher of the two, and water flows upward through the sample and overflows at the top. Head loss in the connecting tube is neglected, as is standard for a constant-head permeameter.
2. Fix the datum and the elevation heads. Taking $z = 0$ at the base of the sample, the entrance end is at $z = 0$, point A at $z = 1$ m and the exit end at $z = 4$ m. The elevation head is simply $z$ at each point — the straight $45^\circ$ line in the plot above.
3. Total head at the two ends. In standing water total head is constant and equal to the elevation of the free surface. The container surface lies 1 m above the top of the sample, that is at $z = 5$ m, so $h_{\text{exit}} = 5$ m. The standpipe surface is 4 m higher again, so $$h_{\text{entrance}} = 5 + 4 = 9\ \text{m}$$ The head lost across the sample is therefore $\Delta h = 9 - 5 = 4$ m, equal to the 4 m offset between the surfaces, as it must be.
4. Hydraulic gradient. With the loss distributed uniformly over the 4 m of homogeneous soil, $$i = \frac{\Delta h}{L} = \frac{4}{4} = \boxed{1.00\ \text{(upward)}}$$ The total-head line is therefore straight from 9 m at the base to 5 m at the top.
5. Total head at A. A sits 1 m above the entrance, so it has lost one quarter of the total: $$h_A = 9 - i\,z_A = 9 - 1.00 \times 1 = \boxed{8.00\ \text{m}}$$
6. Pressure heads and pressures. Pressure head is the difference between the total head and the elevation head at the same point, $h_p = h - z$, and the pressure follows as $u = \gamma_w h_p$ with $\gamma_w = 9.81\ \text{kN/m}^3$: $$\text{entrance: } h_p = 9 - 0 = 9\ \text{m} \;\Rightarrow\; u = \boxed{88.3\ \text{kPa}}$$ $$\text{point A: } h_p = 8 - 1 = 7\ \text{m} \;\Rightarrow\; u = \boxed{68.7\ \text{kPa}}$$ $$\text{exit: } h_p = 5 - 4 = 1\ \text{m} \;\Rightarrow\; u = \boxed{9.81\ \text{kPa}}$$
7. Check the exit value against the geometry. The exit end lies under 1 m of standing water and nothing else, so its pressure head must be exactly 1 m — which it is. That single check confirms the datum, the head-loss direction and the arithmetic all at once.

The plot brings out the structure of the answer: the elevation head rises at $45^\circ$, the total head falls linearly at the gradient $i = 1$, and the pressure head — the gap between them — falls twice as steeply, from 9 m to 1 m over the 4 m of sample. All three lines are straight because the soil is homogeneous; a layered sample would kink the total-head line at each interface, steeply through the less permeable layer.

(c) Effective stresses across an artesian clay layer (7 marks)

Given. A three-layer profile with the lower sand under artesian pressure.

Find. The effective vertical stress at the top of the clay (4 m depth) and at the bottom of the clay (8 m depth).

Approach. Build the total stress by accumulating $\gamma h$ down the profile, then obtain the pore pressure at each level from the water body that actually controls it — the water table for the upper sand, the artesian piezometric surface for the lower sand — and subtract. The clay itself is the barrier between the two regimes, so its top and bottom are governed by different heads.

1. Total stress at the top of the clay ($z = 4$ m). Two metres of moist sand above the water table, then two metres of saturated sand: $$\sigma_{4} = 2(16.5) + 2(19.0) = 33.0 + 38.0 = 71.0\ \text{kPa}$$
2. Pore pressure at the top of the clay. This level is controlled by the water table in the upper sand, which stands 2 m below ground, so there is 2 m of water above the point: $$u_{4} = 2 \times 9.81 = 19.62\ \text{kPa}$$
3. Effective stress at the top of the clay. By Terzaghi’s principle, $$\sigma_4' = \sigma_4 - u_4 = 71.0 - 19.62 = \boxed{51.4\ \text{kPa}}$$
4. Total stress at the bottom of the clay ($z = 8$ m). Add four metres of saturated clay to the total already accumulated: $$\sigma_{8} = 71.0 + 4(20.0) = 71.0 + 80.0 = 151.0\ \text{kPa}$$
5. Pore pressure at the bottom of the clay. This level sits in contact with the lower sand, so its pore pressure is set by the artesian piezometric surface, not by the water table. That surface is 4 m above ground level and the point is 8 m below it, giving a pressure head of $8 + 4 = 12$ m: $$u_{8} = 12 \times 9.81 = 117.72\ \text{kPa}$$ This is the step the question is really testing — using 6 m of head here (as if the water table controlled it) would give 58.9 kPa and an effective stress more than 50 kPa too high.
6. Effective stress at the bottom of the clay. $$\sigma_8' = \sigma_8 - u_8 = 151.0 - 117.72 = \boxed{33.3\ \text{kPa}}$$
7. Interpret the result. The effective stress at the base of the clay is lower than at its top, by 18 kPa, even though the base is 4 m deeper. That inversion is the diagnostic signature of an artesian layer, and it has direct practical consequences: the clay is being partly floated by the water beneath it, and any excavation that thins it must be checked against heave. The available resistance is the total weight of the clay alone (the upper sand removed) against the artesian uplift pressure at its base, $$\frac{\gamma_{\text{clay}} H_{\text{clay}}}{u_8} = \frac{20 \times 4}{117.72} = 0.68 < 1$$ so on these numbers the 4 m of clay could not hold the artesian pressure if the sand above it were removed — the base would blow. In Canadian practice CFEM requires this base-heave check on every excavation over a confined aquifer, with relief wells or dewatering of the lower sand if the ratio falls short.