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. Three 3 m layers over rock. Dry sand (0–3 m): G_s = 2.66, e = 0.5. Silt, zone of capillary rise (3–6 m): G_s = 2.71, e = 0.75, S = 80 %. Clay (6–9 m): G_s = 2.72, e = 0.95, saturated. Water table at 6 m (base of the capillary zone). Oedometer: (100 kPa, e = 0.900), (200 kPa, e = 0.815); Δσ' = 150 kPa; clay normally consolidated.

Find. σ, u and σ' at the layer boundaries (0, 3, 6, 9 m), and the consolidation settlement of the clay layer under Δσ' = 150 kPa.

Approach. Compute each layer's unit weight from γ = (G_s + Se)γ_w/(1 + e); build σ by summing γ·h; set u = 0 in the dry sand, u = −Sγ_w·(height above WT) in the capillary silt, and u = +γ_w·(depth below WT) in the clay; then σ' = σ − u. Finally use C_c from the oedometer with the NC one-dimensional settlement equation.

1. Unit weights. With γ = (G_s + Se)γ_w/(1 + e): $$\gamma_{d,\text{sand}}=\frac{2.66\times9.81}{1.50}=17.40\ \text{kN/m}^3,$$ $$\gamma_{\text{silt}}=\frac{(2.71+0.80\times0.75)\,9.81}{1.75}=\frac{3.31\times9.81}{1.75}=18.55\ \text{kN/m}^3,$$ $$\gamma_{sat,\text{clay}}=\frac{(2.72+0.95)\,9.81}{1.95}=\frac{3.67\times9.81}{1.95}=18.46\ \text{kN/m}^3.$$
2. Total stress σ (cumulative). $$\sigma_{3}=17.40\times3=52.2\ \text{kPa},$$ $$\sigma_{6}=52.2+18.55\times3=107.9\ \text{kPa},$$ $$\sigma_{9}=107.9+18.46\times3=163.2\ \text{kPa}.$$
3. Pore pressure u. The dry sand carries no pore water (u = 0 for 0–3 m). In the capillary silt the water is under suction; at 3 m (3 m above the WT) with S = 80 %, $$u_{3}=-S\gamma_w h=-0.80\times9.81\times3=-23.5\ \text{kPa},$$ rising linearly to u = 0 at the water table (6 m). Below the WT the clay is hydrostatic: $$u_{9}=+\gamma_w\times3=9.81\times3=+29.4\ \text{kPa}.$$
4. Effective stress σ' = σ − u. At 3 m the suction produces a step: just above (sand) σ' = 52.2 − 0 = 52.2 kPa; just below (silt) σ' = 52.2 − (−23.5) = 75.7 kPa. Continuing, $$\sigma'_{6}=107.9-0=107.9\ \text{kPa},\qquad \sigma'_{9}=163.2-29.4=\boxed{133.8\ \text{kPa}.}$$ Capillary suction increases effective stress — that is why the σ' line jumps upward at 3 m.
5. Compression index from the oedometer. $$C_c=\frac{e_1-e_2}{\log(\sigma'_2/\sigma'_1)}=\frac{0.900-0.815}{\log(200/100)}=\frac{0.085}{0.301}=0.282.$$
6. Initial average effective stress in the clay. At the mid-depth of the clay (7.5 m), or equivalently the average of the layer's top and bottom values, $$\sigma'_0=\tfrac{1}{2}(107.9+133.8)=120.8\ \text{kPa}.$$
7. Consolidation settlement (NC clay, H = 3 m, e₀ = 0.95). $$S_c=\frac{C_c\,H}{1+e_0}\,\log\!\left(\frac{\sigma'_0+\Delta\sigma'}{\sigma'_0}\right)=\frac{0.282\times3}{1.95}\,\log\!\left(\frac{120.8+150}{120.8}\right).$$ $$S_c=0.434\times\log(2.242)=0.434\times0.351=0.152\ \text{m}=\boxed{152\ \text{mm}.}$$