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.

(a) Part 1 — foundation profile

Given. Profile: 2 m saturated sand ($\gamma=21\text{ kN/m}^3$) over 8 m clay ($\gamma=17.5\text{ kN/m}^3$, $e_0=1.1$, $C_c=0.3$, $C_r=0.04$, $C_v=0.5\text{ m}^2/\text{yr}$, $\sigma_p'=125\text{ kPa}$) on fractured sandstone; water table at the surface (hydrostatic); construction adds $\Delta\sigma=100\text{ kPa}$, applied instantaneously and uniformly with depth (no footprint/plan dimensions are given for a Boussinesq-type distribution, so a uniform blanket surcharge is the only value the question supports).

Find. Total stress, pore pressure, and effective stress at $z=0,2,4,10\text{ m}$, for (1) before construction, (2) immediately after loading (undrained), and (3) at $t=\infty$ (fully drained).

Approach. Build the "before" hydrostatic profile from the two unit weights; for "immediately after," add the full 100 kPa to total stress everywhere, but route it entirely into pore pressure within the (undrained) clay while the free-draining sand shows no excess; for $t=\infty$, the same 100 kPa has become fully effective everywhere once drainage is complete.

1. Before construction. $\sigma_v=\gamma_{sand}z$ for $z\le2$, $=\gamma_{sand}(2)+\gamma_{clay}(z-2)$ below; $u=\gamma_w z$ (hydrostatic from the surface); $\sigma_v'=\sigma_v-u$. $$z=0:\ (0,0,0)\quad z=2:\ (42.0,19.6,22.4)\quad z=4:\ (77.0,39.2,37.8)\quad z=10:\ (182.0,98.1,83.9)\ \text{[all kPa]}$$
2. Immediately after (+100 kPa, undrained in the clay). Total stress increases by 100 kPa at every depth. In the free-draining sand ($z\le2$), no excess pore pressure develops (it dissipates as fast as it can build), so $u$ is unchanged there and the load is instantly effective; within/at the base of the clay, the load is carried entirely as excess pore pressure ($\Delta u=\Delta\sigma=100$ kPa, the standard 1-D loading assumption), so $\sigma_v'$ is UNCHANGED from the "before" values — this is the defining feature of an undrained response. $$z=0:\ (100,0,100)\quad z=2:\ (142.0,119.6,22.4)\quad z=4:\ (177.0,139.2,37.8)\quad z=10:\ (282.0,198.1,83.9)$$
3. At $t=\infty$ (fully drained). All excess pore pressure has dissipated; $u$ returns to its original hydrostatic profile, and the entire 100 kPa surcharge is now carried as effective stress. $$z=0:\ (100,0,100)\quad z=2:\ (142.0,19.6,122.4)\quad z=4:\ (177.0,39.2,137.8)\quad z=10:\ (282.0,98.1,183.9)$$

(b) Total settlement of the clay layer

Given. Mid-clay ($z=4\text{ m}$) values from above: $\sigma_{v0}'=37.8\text{ kPa}$; $\sigma_p'=125\text{ kPa}$; $\Delta\sigma=100\text{ kPa}$; $C_c=0.3$, $C_r=0.04$, $e_0=1.1$, $H=8\text{ m}$.

Find. Total consolidation settlement, using the mid-layer stress as representative of the whole 8 m clay.

1. Check overconsolidation and the final stress. $OCR=\sigma_p'/\sigma_{v0}'=125/37.8=3.31$ (clay is heavily overconsolidated); $\sigma_f'=\sigma_{v0}'+\Delta\sigma=37.8+100=137.8\text{ kPa}$. Since $\sigma_f'>\sigma_p'$, the stress path straddles the preconsolidation pressure — settlement has a recompression part (up to $\sigma_p'$) and a virgin-compression part (beyond it).
2. Straddled settlement formula. $$\Delta H=\left[C_r\log_{10}\!\frac{\sigma_p'}{\sigma_{v0}'}+C_c\log_{10}\!\frac{\sigma_f'}{\sigma_p'}\right]\frac{H}{1+e_0}$$ $$=\left[0.04\log_{10}\!\frac{125}{37.8}+0.3\log_{10}\!\frac{137.8}{125}\right]\frac{8}{2.1}$$ $$=\left[0.04(0.520)+0.3(0.0422)\right](3.810)=[0.0208+0.0127](3.810)$$ $$\boxed{\Delta H = 0.127\text{ m} = 127\text{ mm}}$$

(c)–(d) Time to 50% and 100% consolidation

Given. $C_v=0.5\text{ m}^2/\text{yr}$; double drainage (sand above, fractured/permeable sandstone below), $H_{dr}=H/2=4\text{ m}$; the exam's own $U$–$T$ table (page 14): $T_{50}=0.197$, $T_{100}=1.125$ (the table's own asymptotic "100%" value, avoiding the theoretical $T\to\infty$ issue of Terzaghi theory).

Find. $t_{50}$, $t_{100}$, and a settlement-vs-time sketch.

1. Time factor to time. $t=T\,H_{dr}^2/C_v$. At $U=50\%$: $t_{50}=0.197(4)^2/0.5=\boxed{6.30\text{ yr}}$. At $U=100\%$ (table value): $t_{100}=1.125(4)^2/0.5=\boxed{36.0\text{ yr}}$.

[Figure not reproduced: Settlement vs. time, built from the exam's own $U$–$T$ table converted through $t=TH_{dr}^2/C_v$ and $S=U\cdot\Delta H_{\text{ult}}$; the curve steepens rapidly at small $t$ (roughly $S\propto\sqrt{t}$ early on) then flattens as it approaches $\Delta H_{\text{ult}}=127\text{ mm}$. See the official exam paper.]

Part 2 — oedometer test on a different clay

Given. Loading path $(\sigma',e)$ read off Figure Q4-2: $(20,1.38),(40,1.36),(80,1.32),(160,1.25),(320,1.05),(640,0.90),(1280,0.75)$ kPa; unloading path: $(640,0.78),(320,0.80),(160,0.82)$ kPa.

Find. Preconsolidation pressure $\sigma_p'$, compression index $C_c$, recompression index $C_r$; the conceptual difference between an OC and NC clay.

Approach. $C_c$ and $C_r$ come directly from the slope of the straight virgin and recompression segments; $\sigma_p'$ needs the graphical Casagrande construction (point of maximum curvature → bisect the tangent and the horizontal → intersect with the extended virgin line).

1. Compression index. The $320$–$1280$ kPa segment is a straight virgin-compression line (its two sub-intervals give an identical slope, confirming linearity): $$C_c=\left|\frac{\Delta e}{\Delta\log\sigma'}\right|=\frac{1.05-0.75}{\log_{10}(1280/320)}=\boxed{0.50}$$
2. Recompression index. From the unloading branch ($160$–$640$ kPa, also confirmed linear): $$C_r=\left|\frac{0.90-0.82}{\log_{10}(640/160)}\right|=\boxed{0.066}$$
3. Preconsolidation pressure (Casagrande construction). Fitting a smooth curve through the loading points and locating the point of maximum curvature places the "knee" near $\sigma'\approx160\text{ kPa}$, $e\approx1.25$; bisecting the horizontal and the local tangent there and extending that bisector to intersect the extended virgin-compression line gives: $$\boxed{\sigma_p' \approx 160\text{ kPa}}$$
   Check: the loading curve is sampled at only 7 coarse points read off a semi-log oedometer plot, so the graphical knee location carries real uncertainty; a simplified two-straight-line intersection (early recompression line through the 20–40 kPa points vs. the virgin line) gives a lower bound near 85 kPa. The Casagrande value (160 kPa) is reported as the primary answer since it is the standard method, but treat the true preconsolidation pressure as bounded within roughly 85–160 kPa given the sparse data.

Overconsolidated vs. normally consolidated clay. A normally consolidated (NC) clay currently sits under the highest effective stress it has ever experienced ($\sigma_{v0}'=\sigma_p'$, $OCR=1$); any further loading moves it straight onto the steep virgin-compression line, so it is comparatively soft and compressible under new load. An overconsolidated (OC) clay was once loaded to a higher effective stress than it currently carries ($\sigma_p'>\sigma_{v0}'$, $OCR>1$) — commonly from a since-removed glacial ice sheet, eroded overburden, past desiccation, or a lowered historical water table — so new loading up to $\sigma_p'$ only re-traces the much flatter recompression line, making OC clay stiffer and less compressible until the load exceeds its remembered maximum, at which point it "yields" onto the virgin line exactly like an NC clay.