Consolidation: 18-Geol-A6 Soil Mechanics, December 2017, Q4

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 (top to bottom): 2.25 m sand above the water table, $\rho=2100$ kg/m³; 3.0 m sand below the water table, $\rho_{sat}=2000$ kg/m³; 6.0 m clay, $\rho_{sat}=1700$ kg/m³, $C_c=0.4$, $C_r=0.04$, $e_o=1.5$, $\sigma_p'=130$ kN/m², $c_v=0.00123$ m²/day; fractured (pervious) bedrock below. A very wide embankment applies $\Delta\sigma=80$ kPa at the surface.

Find. (a) $\sigma'$ at top/middle/bottom of each layer, before and after construction; (b.i) total consolidation settlement of the clay; (b.ii) time to 50% consolidation.

Approach. Build the before-construction effective-stress profile from unit weights and pore pressure, add the uniform $\Delta\sigma=80$ kPa everywhere below a "very wide" embankment (no depth attenuation), compare the clay's mid-depth stresses to $\sigma_p'$ to choose the recompression/virgin-compression settlement formula, and use the time factor $T_{50}=0.197$ with the governing drainage path length.

[Figure not reproduced: Figure Q4-1: soil profile with sand over sand over clay over fractured bedrock, embankment stress 80 kPa at surface. See the official exam paper or the cited reference text.]

Fig. Q4-1 (source) — soil profile: 2.25 m sand ($\rho=2100$ kg/m³) above the water table, 3.0 m saturated sand ($\rho_{sat}=2000$ kg/m³), 6.0 m saturated clay ($\rho_{sat}=1700$ kg/m³, $C_c=0.4$, $C_r=0.04$, $e_o=1.5$, $\sigma_p'=130$ kN/m², $c_v=0.00123$ m²/day) over fractured bedrock.

Check: "fractured bedrock" is taken as pervious (fractures provide drainage pathways), so the clay layer is treated as doubly drained (free at its top, into the sand, and at its base, into the fractured rock) for the time-rate calculation in (b.ii).

(a) Unit weights. $\gamma_1=\rho_1 g=2100(9.81)/1000=20.60$ kN/m³ (moist, above WT). $\gamma_{2,sat}=2000(9.81)/1000=19.62$ kN/m³, so $\gamma_2'=19.62-9.81=9.81$ kN/m³. $\gamma_{3,sat}=1700(9.81)/1000=16.68$ kN/m³, so $\gamma_3'=16.68-9.81=6.87$ kN/m³.
(a) Effective stress before construction (cumulative from the surface, using $\gamma'$ below the water table). Sand 1 top $z=0$: $\sigma'=0$. Sand 1 mid $z=1.125$m: $\sigma'=20.60(1.125)=23.18$ kPa. Sand 1 bottom / Sand 2 top $z=2.25$m: $\sigma'=20.60(2.25)=46.35$ kPa. Sand 2 mid $z=3.75$m: $\sigma'=46.35+9.81(1.5)=61.07$ kPa. Sand 2 bottom / Clay top $z=5.25$m: $\sigma'=46.35+9.81(3)=75.78$ kPa. Clay mid $z=8.25$m: $\sigma'=75.78+6.87(3)=96.38$ kPa. Clay bottom $z=11.25$m: $\sigma'=75.78+6.87(6)=116.98$ kPa.
(a) Effective stress after construction. A "very wide" embankment transmits its surcharge with negligible depth attenuation, so $\Delta\sigma'=80$ kPa is added uniformly at every depth: e.g. Sand 1 top: $0+80=\boxed{80.0}$ kPa; Clay mid: $96.38+80=\boxed{176.38}$ kPa; Clay bottom: $116.98+80=\boxed{196.98}$ kPa (full table below).
(b.i) Check against $\sigma_p'$. At clay mid-depth (the layer's representative point), $\sigma_{vo}'=96.38$ kPa $<\sigma_p'=130$ kPa $<\sigma_{vf}'=176.38$ kPa — the stress path crosses the preconsolidation pressure, so settlement combines recompression (from $\sigma_{vo}'$ up to $\sigma_p'$, slope $C_r$) with virgin compression (from $\sigma_p'$ up to $\sigma_{vf}'$, slope $C_c$): $$\Delta H=C_r\left(\frac{H_o}{1+e_o}\right)\log\frac{\sigma_p'}{\sigma_{vo}'}+C_c\left(\frac{H_o}{1+e_o}\right)\log\frac{\sigma_{vf}'}{\sigma_p'}$$
(b.i) Settlement. $\dfrac{H_o}{1+e_o}=\dfrac{6.0}{2.5}=2.4$ m. $\Delta H=0.04(2.4)\log_{10}\!\left(\dfrac{130}{96.38}\right)+0.4(2.4)\log_{10}\!\left(\dfrac{176.38}{130}\right)=0.096(0.1300)+0.96(0.1325)=0.0125+0.1272=\boxed{0.140\text{ m}\ (140\text{ mm})}$.
(b.ii) Time to 50% consolidation. $T_{50}=0.197$ (from the printed $U$–$T$ table). Double drainage (pervious sand above, pervious fractured bedrock below) → $H_{dr}=H_o/2=3.0$ m. $t_{50}=\dfrac{T_{50}H_{dr}^2}{c_v}=\dfrac{0.197(3.0)^2}{0.00123}=\boxed{1442\text{ days}\approx3.95\text{ years}}$.

Location	Depth (m)	σ′ before (kPa)	σ′ after (kPa)
Sand 1 (moist) — top	0	0.00	80.00
Sand 1 — middle	1.125	23.18	103.18
Sand 1 — bottom	2.25	46.35	126.35
Sand 2 (sat.) — top	2.25	46.35	126.35
Sand 2 — middle	3.75	61.07	141.07
Sand 2 — bottom	5.25	75.78	155.78
Clay — top	5.25	75.78	155.78
Clay — middle	8.25	96.38	176.38
Clay — bottom	11.25	116.98	196.98

Quantity	Value
Clay mid-depth $\sigma_{vo}'$ / $\sigma_{vf}'$	96.4 / 176.4 kPa
Total consolidation settlement, $\Delta H$	140 mm
Time to 50% consolidation, $t_{50}$ (double drainage)	1442 days (≈3.95 yr)

Concept & Theory

Topic. One-dimensional effective stress profiles and consolidation settlement of an over-consolidated clay.

Key relations. $\sigma'=\sigma-u$; $\Delta H=C_r(H_o/(1+e_o))\log(\sigma_p'/\sigma_{vo}')+C_c(H_o/(1+e_o))\log(\sigma_{vf}'/\sigma_p')$ when the stress path crosses $\sigma_p'$; $t=T H_{dr}^2/c_v$.

Why this works. Below its preconsolidation pressure a clay recompresses along the flat $C_r$ line (already-experienced stress states); once $\sigma'$ exceeds $\sigma_p'$ it follows the much steeper virgin $C_c$ line, so a stress path that straddles $\sigma_p'$ must be split into these two segments rather than using either slope alone.

Common pitfall. The exam's own printed settlement formula sheet has the recompression-term logarithm written as $\log(\sigma_{vf}'/\sigma_{vo}')$ instead of $\log(\sigma_p'/\sigma_{vo}')$ — using it as printed would double-count part of the virgin-compression range under the $C_r$ slope. The corrected split-log form used above is the standard result (Das Ch.11; Craig Ch.6) and is used throughout.

Source. Das, Principles of Geotechnical Engineering, 9th ed., Ch. 11 (consolidation); Craig's Soil Mechanics, 8th ed., Ch. 6.

Question 4 of 6: Consolidation

Question 4: Consolidation (20 marks)