Question 4 of 6: Consolidation

Question 4: Consolidation (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. Profile: sand $\gamma_t=16\text{ kN/m}^3$ (0–1.5 m, above WT); sand $\gamma_t=18\text{ kN/m}^3$ (1.5–3.5 m, below WT); clay $\gamma_t=15\text{ kN/m}^3$ (3.5–8.5 m, 5 m thick); WT at 1.5 m; $\Delta\sigma=150$ kPa at clay midheight (6.0 m); oedometer $e$-$\log\sigma'$ curve (Figure Q4-2, overconsolidated-clay sample).

Find. Total/effective stress profile 0–8.5 m; final $\sigma$, $\sigma'$ at 6.0 m; $\sigma_p'$; primary consolidation settlement of the 5 m clay layer.

Approach. Accumulate total stress layer-by-layer, subtract hydrostatic pore pressure below the WT for effective stress; add $\Delta\sigma$ for the final state; read $\sigma_p'$ from the oedometer curve by the Casagrande construction (max-curvature point, bisect tangent & horizontal, intersect with the virgin-compression line); compare $\sigma_p'$ against $\sigma_{v0}'$ and $\sigma_{vf}'$ to select the correct settlement formula.

1. (a) Stress profile, 0–8.5 m. Working down from the surface with $u=\gamma_w z_w$ below the WT ($z_w$= depth below WT): $$\text{@1.5 m (WT): } \sigma=16(1.5)=24.0,\ u=0,\ \sigma'=24.0\text{ kPa}$$ $$\text{@3.5 m (clay top): } \sigma=24.0+18(2)=60.0,\ u=9.81(2)=19.62,\ \sigma'=40.4\text{ kPa}$$ $$\text{@8.5 m (clay base): } \sigma=60.0+15(5)=135.0,\ u=9.81(7)=68.7,\ \sigma'=66.3\text{ kPa}$$ At midheight ($z=6.0$ m, 2.5 m into the clay): $\sigma_{v0}=60.0+15(2.5)=97.5$ kPa, $u_0=9.81(4.5)=44.1$ kPa, $\boxed{\sigma_{v0}'=53.4\text{ kPa}}$.
2. (b) Final stresses at midheight. Loading is undrained in total stress terms but the question asks for the final (fully consolidated) state, where the excess pore pressure has dissipated back to hydrostatic: $$\sigma_{vf}=\sigma_{v0}+\Delta\sigma=97.5+150=\boxed{247.5\text{ kPa}}\qquad u_f=u_0=44.1\text{ kPa (unchanged, hydrostatic)}$$ $$\boxed{\sigma_{vf}'=\sigma_{v0}'+\Delta\sigma=53.4+150=203.4\text{ kPa}}$$
3. (c) Preconsolidation pressure (Casagrande construction). Digitizing Figure Q4-2's upper curve (the overconsolidated-clay sample; the lower curve is a second, unused dataset — see the check note banner) and locating the point of maximum curvature numerically gives $\sigma'\approx197$ kPa, $e\approx0.90$. Bisecting the angle between the horizontal and the curve's tangent there, and intersecting that bisector with the straight virgin-compression line fitted through the steep upper-stress points ($C_c=0.417$): $$\boxed{\sigma_p'\approx255\text{ kPa}}\qquad \text{OCR}=\sigma_p'/\sigma_{v0}'=255/53.4=4.8$$
4. (d) Primary consolidation settlement. Recompression index from the flat initial branch: $C_r=0.052$. Since $\sigma_{vf}'=203.4\text{ kPa} < \sigma_p'=255\text{ kPa}$, the clay stays entirely within the recompression range even after the full load is applied — it never reaches virgin compression, so only $C_r$ is used (not $C_c$): $$\Delta H=C_r\frac{H_0}{1+e_0}\log_{10}\frac{\sigma_{vf}'}{\sigma_{v0}'}=0.052\times\frac{5.0}{1+0.979}\times\log_{10}\frac{203.4}{53.4}$$ $$=0.052\times2.526\times0.581=\boxed{0.0763\text{ m}=76\text{ mm}}$$ ($e_0=0.979$ read from the curve at $\sigma_{v0}'=53.4$ kPa.) This is a direct payoff of correctly recognizing the overconsolidated state: naively applying $C_c$ over the full stress range (as if the clay were normally consolidated) would over-predict the settlement roughly eight-fold.