Question 4 of 6: Consolidation of a Clay Layer Under a Gravel Fill

Question 4: Consolidation of a Clay Layer Under a Gravel Fill (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.

Find. (a) $\sigma_0'(z)$ within the clay immediately after fill placement; (b) ultimate primary consolidation settlement.

Approach. Find the clay's saturated/buoyant unit weight from $e_0$ and $G_s$, recognise that "immediately after" placement means the added stress is carried entirely by excess pore pressure (effective stress unchanged), then use the single-point (mid-depth) method with the OC/NC split to get the ultimate settlement.

1. Clay unit weight. Fully saturated, so $$\gamma_{sat}=\frac{G_s+e_0}{1+e_0}\gamma_w=\frac{2.5+0.5}{1.5}\times9.81=\boxed{19.62\ \text{kN/m}^3},\qquad \gamma'=\gamma_{sat}-\gamma_w=9.81\ \text{kN/m}^3.$$
2. Part (a) — effective stress profile "immediately after" placement. Under Terzaghi 1-D consolidation theory, a newly-applied load is carried entirely by excess pore pressure at $t=0^+$ ($\Delta u=\Delta\sigma$), so the EFFECTIVE stress profile is unchanged from the pre-fill overburden condition, $\sigma_0'(z)=\gamma' z$: $$\sigma_0'(0)=0,\ \ \sigma_0'(5)=49.05,\ \ \sigma_0'(10)=98.10,\ \ \sigma_0'(15)=147.15,\ \ \boxed{\sigma_0'(20)=196.20\ \text{kPa}}.$$ Meanwhile the total stress increase from the fill, $\Delta\sigma=\gamma_{fill}\times t_{fill}=20\times20=400$ kPa, appears entirely as excess pore pressure $\Delta u=400$ kPa immediately after placement — this is the point of the "immediately after" wording.

3. Part (b) — ultimate settlement, single-point (mid-depth) method. At $z=H/2=10$ m: $\sigma_0'=\gamma'\times10=98.10$ kPa, just below $\sigma_p'=100$ kPa — the clay is lightly overconsolidated at mid-depth, and the final stress $\sigma_0'+\Delta\sigma=98.10+400=498.10$ kPa crosses well past $\sigma_p'$, so the settlement has both a recompression (OC) and virgin-compression (NC) branch: $$S_c=\frac{H}{1+e_0}\left[C_r\log_{10}\frac{\sigma_p'}{\sigma_0'}+C_c\log_{10}\frac{\sigma_0'+\Delta\sigma}{\sigma_p'}\right]$$ $$=\frac{20}{1.5}\left[0.05\log_{10}\frac{100}{98.10}+0.35\log_{10}\frac{498.10}{100}\right]=13.33\times[0.000417+0.24406]=\boxed{3.26\ \text{m}}\ (3260\ \text{mm}).$$