Question 6 of 6: Stress Profile with Capillary Rise and Consolidation Settlement

Question 6: Stress Profile with Capillary Rise and Consolidation Settlement (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.

Given. A three-layer profile on rock: dry sand over a silt in the zone of capillary rise over a saturated clay, with the groundwater table at the top of the clay.

Find. The total stress, pore-water pressure and effective stress at every horizon of the profile, and the consolidation settlement of the clay layer under the added 150 kPa.

[Figure not reproduced: Figure 5 (redrawn) with the computed σ, u and σ′ distributions. Note the step in u (and hence in σ′) at the top of the capillary zone. See the official exam paper.]

Approach. Convert each layer’s specific gravity, void ratio and degree of saturation into a unit weight, accumulate the total stress downwards, write the pore pressure from the position of the water table (negative in the capillary zone), subtract to get the effective stress, then apply the one-dimensional consolidation equation using the compression index from the oedometer data.

1. Convert the phase data into unit weights. The general expression for a soil of specific gravity $G_s$, void ratio $e$ and degree of saturation $S$ is $$\gamma = \frac{(G_s + S e)\,\gamma_w}{1+e}$$ Applying it to each layer with $\gamma_w = 9.81\ \text{kN/m}^3$: dry sand $\gamma_d = \dfrac{2.66 \times 9.81}{1.50} = 17.40\ \text{kN/m}^3$; capillary silt $\gamma = \dfrac{(2.71 + 0.80 \times 0.75)\,9.81}{1.75} = \dfrac{3.31 \times 9.81}{1.75} = 18.55\ \text{kN/m}^3$; saturated clay $\gamma_{sat} = \dfrac{(2.72 + 0.95)\,9.81}{1.95} = 18.46\ \text{kN/m}^3$.
2. Accumulate the total vertical stress. Total stress is simply the weight of everything above, so at the base of each layer $$\sigma_{3\text{m}} = 3(17.40) = 52.19\ \text{kPa}$$ $$\sigma_{6\text{m}} = 52.19 + 3(18.55) = 107.85\ \text{kPa}$$ $$\sigma_{9\text{m}} = 107.85 + 3(18.46) = 163.24\ \text{kPa}$$
3. Write the pore-water pressure, including the capillary suction. The water table is at 6 m, at the top of the clay. Below it the pressure is hydrostatic, $u = \gamma_w z_w$, so at 9 m depth $u = 9.81(3.0) = 29.43\ \text{kPa}$. Above it, water is held in tension in the capillary zone, and for a partly saturated capillary zone the classical expression is $u = -S\,\gamma_w\,h_c$ where $h_c$ is the height above the water table. At the top of the capillary zone ($z = 3$ m, $h_c = 3$ m, $S = 0.80$), $$u = -0.80 \times 9.81 \times 3.0 = -23.54\ \text{kPa}$$ and the suction varies linearly to zero at the water table. In the dry sand above 3 m there is no continuous water phase, so $u = 0$ and the pore pressure is discontinuous across that horizon.
4. Subtract to obtain the effective stress. Applying $\sigma' = \sigma - u$ at each horizon, the suction in the capillary zone shows up as a jump upward in effective stress: immediately above 3 m depth $\sigma' = 52.19 - 0 = 52.19\ \text{kPa}$, while immediately below it $$\sigma' = 52.19 - (-23.54) = 75.73\ \text{kPa}$$ At the water table $\sigma' = 107.85\ \text{kPa}$, and at rockhead $\sigma' = 163.24 - 29.43 = 133.81\ \text{kPa}$. Capillary tension is what gives a damp sand its apparent cohesion, and it is a genuine contribution to effective stress — but one that disappears the moment the zone floods.

5. Extract the compression index from the oedometer data. The clay is normally consolidated, so the two test points lie on the virgin compression line and $$C_c = \frac{e_1 - e_2}{\log(\sigma_2'/\sigma_1')} = \frac{0.900 - 0.815}{\log(200/100)} = \frac{0.085}{0.30103} = 0.282$$ As a sanity check, Terzaghi and Peck’s empirical rule $C_c = 0.009(LL - 10)$ would put this value at a liquid limit of about 41 %, entirely plausible for an inorganic clay of intermediate plasticity.
6. Establish the initial effective stress in the clay. Settlement is computed from the stress at the middle of the compressible layer, which from the table above (or from $\sigma_0' = 107.85 + 1.5(18.46 - 9.81)$) is $$\sigma_0' = 107.85 + 1.5 \times 8.65 = 120.83\ \text{kPa}$$
7. Apply the one-dimensional consolidation equation. For a normally consolidated clay the whole increment travels down the virgin line, so $$S_c = \frac{C_c\,H_3}{1+e_0}\,\log\!\left(\frac{\sigma_0' + \Delta\sigma'}{\sigma_0'}\right) = \frac{0.282 \times 3.0}{1 + 0.95}\,\log\!\left(\frac{120.83 + 150}{120.83}\right)$$ Evaluating the two factors separately, the pre-multiplier is $0.4344\ \text{m}$ and the logarithm is $\log(2.2414) = 0.3505$, so $$S_c = 0.4344 \times 0.3505 = 0.1523\ \text{m}$$ $$\boxed{\;S_c \approx 152\ \text{mm}\;}$$
8. Put the number in context. A settlement of 152 mm over a 3 m clay layer is a vertical strain of about 5 %, which is large: CFEM and most municipal guidance treat more than about 25 mm of total settlement, or 1/500 angular distortion, as needing mitigation for a conventional structure. The design response would be preloading with wick drains, a raft to spread and equalise the load, or piling through the clay to rockhead at 9 m — which at this depth is the obvious and likely cheapest option.