Compressibility of Soil

Consolidation time $t = T_v H_{dr}^2/c_v$. It can be reduced by shortening the drainage path, increasing $c_v$, or speeding the dissipation of pore pressure. Methods:

1. Vertical sand drains: closely spaced vertical columns of sand (diameter 20 to 60 cm) through the clay shorten the drainage path from the clay thickness to half the drain spacing (radial drainage). Water flows horizontally to the drains, which is faster than vertical flow in thick clay.
2. Prefabricated vertical drains (wick drains): plastic band drains, same function, installed quickly and cheaply.
3. Surcharge (preloading): a temporary load larger than the future structural load is placed to cause most of the settlement before construction; it is removed later.
4. Vacuum preloading: a vacuum applied under a membrane increases the effective stress without extra fill.
5. Sand blanket / drainage layer: a layer of free draining sand at the top (and bottom) provides double drainage.
6. Electro-osmosis and lowering the water table (dewatering), or consolidation by heating, are less common.

Assumptions

1. The clay layer is homogeneous and fully saturated.
2. Soil grains and water are incompressible.
3. Compression and flow are one-dimensional (vertical).
4. Darcy's law is valid; $k$ is constant during consolidation.
5. The coefficient of volume compressibility $m_v$ is constant (linear $e$–$\sigma'$ relation).
6. The load is applied instantaneously and is uniform over a large area, so that excess pore pressure at $t=0$ equals the load.
7. Strains are small and secondary compression is neglected.

Derivation

Consider a clay layer of thickness $H$ with drainage upward. Take an element of thickness $dz$ and unit area at depth $z$, with excess pore pressure $u$ (a function of $z$ and $t$).

   ▼ ▼ ▼ ▼  load Δσ
  ---------------  drainage (top)
   z ↓   +---+  ← element dz
         +---+
  ================ impervious base

Flow through the element. The excess head is $u/\gamma_w$, so the gradient is $i = \dfrac{1}{\gamma_w}\dfrac{\partial u}{\partial z}$ and by Darcy's law the upward velocity is $v = \dfrac{k}{\gamma_w}\dfrac{\partial u}{\partial z}$ (z measured downward).

Water leaves through the top face (velocity $v$) and enters through the bottom face (velocity $v + \frac{\partial v}{\partial z}dz$). The net outflow per unit area is

$$-\frac{\partial v}{\partial z}dz = -\frac{k}{\gamma_w}\frac{\partial^2 u}{\partial z^2}dz$$

Volume change. The decrease in volume of the element per unit time equals the net outflow. With constant total stress, a fall in $u$ is an equal rise in effective stress, $\partial\sigma'/\partial t = -\partial u/\partial t$, so the rate of decrease in volume is

$$m_v\,dz\,\frac{\partial\sigma'}{\partial t} = -m_v\,dz\,\frac{\partial u}{\partial t}$$

Equating:

$$m_v\frac{\partial u}{\partial t} = \frac{k}{\gamma_w}\frac{\partial^2 u}{\partial z^2}$$ $$\boxed{\frac{\partial u}{\partial t} = c_v\frac{\partial^2 u}{\partial z^2}}, \qquad c_v = \frac{k}{m_v\gamma_w}$$

This is Terzaghi's one-dimensional consolidation equation, $c_v$ being the coefficient of consolidation (m²/s).

Solution. With $u = \Delta\sigma$ at $t=0$ and $u = 0$ at drainage faces, the solution is

$$u = \sum_{m=0}^{\infty}\frac{2u_0}{M}\sin\!\left(\frac{Mz}{H_{dr}}\right)e^{-M^2T_v}, \quad M = \frac{\pi}{2}(2m+1), \quad T_v = \frac{c_vt}{H_{dr}^2}$$

Consolidation (settlement): the gradual decrease in volume of a saturated soil (clay) under a sustained load, caused by the slow expulsion of pore water and transfer of load from water to soil skeleton. It takes place with time and produces settlement.

Degree of consolidation, $U$: the ratio of the consolidation (settlement or pore pressure dissipation) that has taken place at a given time to the total expected consolidation.

$$U_z = \frac{u_0 - u_z}{u_0}, \qquad U = \frac{S_t}{S_{final}}\ \ (\text{or in percent})$$

Pre-consolidation pressure (maximum overburden pressure), $\sigma'_c$: the maximum effective vertical stress that the soil has experienced in the past. It is determined from the $e$–$\log\sigma'$ curve (Casagrande's method).

Over-consolidation ratio, OCR:

$$OCR = \frac{\sigma'_c}{\sigma'_0}$$

where $\sigma'_0$ is the present effective overburden pressure. $OCR = 1$ for NC clay and $> 1$ for OC clay.

Coefficient of consolidation, $c_v$: a soil property giving the rate of consolidation:

$$c_v = \frac{k}{m_v\gamma_w} = \frac{k(1+e)}{a_v\gamma_w}$$

Its unit is m²/s or cm²/s. A high $c_v$ means faster consolidation. It is found from the time-settlement curve of the oedometer test.

Assumptions: the layer is sandwiched between permeable layers (double drainage), so $H_{dr} = 2.5$ m; $\gamma_w = 9.81$ kN/m³. $0.15\ \text{N/mm}^2 = 150$ kPa and $0.20\ \text{N/mm}^2 = 200$ kPa.

Void ratio at 200 kPa

$$\Delta e = C_c \log_{10}\frac{\sigma_2'}{\sigma_1'} = 0.25\log_{10}\frac{200}{150} = 0.0312$$ $$e_2 = 1.9 - 0.0312 = 1.869$$

Settlement

$$S = \frac{\Delta e}{1+e_0}H = \frac{0.0312}{2.9}\times 5000 = 53.9\ \text{mm}$$

Time for 50% consolidation

Coefficient of compressibility and volume compressibility:

$$a_v = \frac{\Delta e}{\Delta\sigma'} = \frac{0.0312}{50} = 6.25\times10^{-4}\ \text{m}^2/\text{kN}, \quad m_v = \frac{a_v}{1+e_0} = 2.154\times10^{-4}\ \text{m}^2/\text{kN}$$

With $k = 3\times10^{-3}$ mm/s $= 3\times10^{-6}$ m/s:

$$c_v = \frac{k}{m_v\gamma_w} = \frac{3\times10^{-6}}{2.154\times10^{-4}\times 9.81} = 1.42\times10^{-3}\ \text{m}^2/\text{s}$$

For $U = 50\%$, $T_v = 0.197$:

$$t_{50} = \frac{T_v H_{dr}^2}{c_v} = \frac{0.197 \times 2.5^2}{1.42\times10^{-3}} = 864\ \text{s} \approx 14.4\ \text{min}$$

Answer: $e_2 \approx 1.869$; settlement $\approx 54$ mm; $t_{50} \approx 864$ s (about 14 min). With single drainage ($H_{dr}=5$ m) the time would be four times longer, about 58 min.

Assumptions: the layer is drained at top and bottom (double drainage), so $H_{dr} = 2.5$ m; $\gamma_w = 9.81$ kN/m³. $0.15\ \text{N/mm}^2 = 150$ kPa and $0.20\ \text{N/mm}^2 = 200$ kPa.

Step 1: Void ratio at 200 kPa

$$\Delta e = 0.25\log_{10}\frac{200}{150} = 0.0312, \qquad e_2 = 1.9 - 0.0312 = 1.869$$

Step 2: Settlement

$$S = \frac{\Delta e}{1+e_0}H = \frac{0.0312}{1+1.9}\times 5000 = 53.9\ \text{mm}$$

Step 3: Coefficient of consolidation

$$m_v = \frac{\Delta e}{(1+e_0)\Delta\sigma'} = \frac{0.0312}{2.9 \times 50} = 2.154\times10^{-4}\ \text{m}^2/\text{kN}$$

With $k = 3.2\times10^{-3}$ mm/s $= 3.2\times10^{-6}$ m/s:

$$c_v = \frac{k}{m_v \gamma_w} = \frac{3.2\times10^{-6}}{2.154\times10^{-4}\times9.81} = 1.514\times10^{-3}\ \text{m}^2/\text{s}$$

Step 4: Time for 65% consolidation

Since $U > 60\%$, use

$$T_v = 1.781 - 0.933\log_{10}(100-65) = 1.781 - 0.933(1.544) = 0.340$$ $$t_{65} = \frac{T_v H_{dr}^2}{c_v} = \frac{0.340 \times 2.5^2}{1.514\times10^{-3}} = 1405\ \text{s} \approx 23.4\ \text{min}$$

Answer: $e_2 = 1.869$; settlement $= 53.9$ mm; $t_{65} \approx 1405$ s (about 23 min). For single drainage ($H_{dr}=5$ m) the time would be about 94 min.

Normally consolidated (NC) clay: the present effective overburden pressure is the greatest pressure the clay has ever carried ($OCR = 1$). It is still consolidating or has just reached equilibrium under its present load.

Over-consolidated (OC) clay: the clay was loaded in the past to a pressure greater than the present overburden ($OCR > 1$), e.g. by glaciers, erosion of upper soil or desiccation.

e–log p plots

 e|                       e|
  |\                       |\
  | \ virgin               | \ recompression
  |  \ line (NC)           |  ----.   (flat)
  |   \                    |       \ virgin
  |    \                   |        \ line
  +-----------> log p      +---------\----> log p
  p0=pc                      p0     pc

• NC clay: a single steep straight line (virgin compression curve) starting from the in-situ point $(e_0, \sigma'_0)$.
• OC clay: a flat recompression segment up to $\sigma'_c$, followed by the steep virgin line.

Pre-consolidation pressure (Casagrande's method)

1. Plot the $e$–$\log\sigma'$ curve from the oedometer test.
2. Find the point of maximum curvature (smallest radius), M.
3. Draw a horizontal line through M and a tangent to the curve at M.
4. Bisect the angle between the horizontal and the tangent.
5. Extend the straight virgin compression line backwards until it meets the bisector. The abscissa of the intersection is the pre-consolidation pressure $\sigma'_c$.

 e|  \      horizontal
  |   \ M -------
  |    \.   bisector
  |     \ ' .
  |   tangent \ . virgin line extended
  +----------\--|---> log p
                σ'c

A soil is preconsolidated when it has once carried a greater effective stress than at present. The causes are:

1. Erosion or removal of overlying soil layers.
2. Melting of glaciers/ice sheets which once loaded the soil.
3. Desiccation (drying) of the upper soil layer, which causes capillary stress, followed by re-wetting.
4. Lowering and later rise of the water table, or fluctuations of the ground water level.
5. Removal of buildings or other structures (past loading and unloading).
6. Secondary compression, aging and cementation (chemical effects), and tectonic effects.

Compressibility is the property of soil by which it decreases in volume when it is subjected to a compressive load (an increase in effective stress). It is measured by the coefficient of compressibility $a_v$, the coefficient of volume compressibility $m_v$ and the compression index $C_c$.

Causes of compression in soil:

1. Compression of the solid particles (negligible).
2. Compression of water and air in the voids (negligible for water; air is compressible in partly saturated soils).
3. Expulsion of water (and air) from the voids, which is the main cause in saturated soil (consolidation).
4. Rearrangement and reorientation of the particles into a denser packing, and crushing or bending of grains.
5. Compression of the adsorbed layers and plastic (creep) deformation of the particle contacts (secondary compression).

Consider a clay layer of thickness $H$ and initial void ratio $e_0$ under a stress increase $\Delta\sigma'$, so that the void ratio decreases by $\Delta e$ (one-dimensional compression, no lateral strain).

  before: Vs + Vv0 = 1 + e0     after: Vs + Vv = 1 + e0 - Δe
  |  H   |                      | H - ΔH |

For a unit volume of solids, the initial total volume is $1 + e_0$ and the reduction in volume is $\Delta e$. Since the area is constant, strain equals:

$$\frac{\Delta H}{H} = \frac{\Delta V}{V} = \frac{\Delta e}{1+e_0}$$ $$\boxed{S_c = \Delta H = \frac{\Delta e}{1+e_0}\,H}$$

In terms of the compressibility:

• With $a_v = \dfrac{\Delta e}{\Delta\sigma'}$: $\;S_c = \dfrac{a_v}{1+e_0}\Delta\sigma'\,H = m_v\,\Delta\sigma'\,H$
• With the compression index $C_c$ (NC clay): $\Delta e = C_c\log_{10}\dfrac{\sigma'_0 + \Delta\sigma'}{\sigma'_0}$, so

$$S_c = \frac{C_c\,H}{1+e_0}\log_{10}\frac{\sigma'_0 + \Delta\sigma'}{\sigma'_0}$$

where $\sigma'_0$ is the effective overburden pressure at the middle of the layer.

An isochrone is a curve showing the variation of excess pore water pressure $u$ with depth at a given time $t$. For a clay layer of thickness $H$ under a uniform loading $\Delta\sigma$:

• At $t=0$: $u = \Delta\sigma$ at all depths (the entire load is taken by the water).
• At $0<t<\infty$: $u$ falls near the drainage faces and stays highest in the interior.
• At $t=\infty$: $u = 0$ everywhere (consolidation complete).

One-way drainage (drainage at the top only)

  top drain ----|------------> u
  z=0           |  \  t=t
                |   \   ___
                |    \ /   t=0 (rectangular, Δσ)
        z=H     |___impervious___
   u = 0 at top for all t>0; u = Δσ for t=0; u = 0 for t = ∞

The isochrone at time $t$ starts at $u=0$ at the top drain, bulges to the maximum value near the bottom, and meets the impervious base with zero slope ($\partial u/\partial z = 0$). The drainage path is $H_{dr} = H$.

Two-way drainage (top and bottom)

  top drain ---|     u = 0
               |  .--.
  mid-depth    | (    )  t=t  symmetrical
               |  '--'
  bottom drain |     u = 0

The isochrones are symmetrical about the mid-depth; $u = 0$ at both faces for $t>0$ and the maximum is at the centre. The drainage path is $H_{dr} = H/2$, so consolidation is four times as fast.

The area between an isochrone and the initial rectangular line gives the consolidation that has taken place.

Taylor's square root of time method finds $c_v$ from a consolidation test for a load increment, using the observed dial reading against $\sqrt t$.

 dial                  
 reading |*.
         |  *.          straight line
    d0   |-----*.______  (extended to √t axis → d0 corrected)
         |       \ *  .
         |  line OB  \   *  *  *
         |  at 1.15x  C
         +-----------|--------------> √t
                    √t90

Procedure

1. Plot the dial gauge reading (settlement) $d$ on the y-axis against the square root of time $\sqrt t$ on the x-axis.
2. The initial part of the curve is a straight line (theoretical curve up to $U = 60\%$). Extend it back to cut the y-axis at the corrected zero reading $d_0$ (this corrects for initial compression).
3. Draw a second line from $d_0$ with abscissa 1.15 times the first (the slope of the line is 1/1.15 of that of the first line).
4. The intersection of this second line with the experimental curve gives $\sqrt{t_{90}}$, for 90% consolidation.
5. Compute the coefficient of consolidation, using $T_v = 0.848$ at $U = 90\%$:

$$c_v = \frac{0.848\,H_{dr}^2}{t_{90}}$$

where $H_{dr}$ is the average drainage path during the load increment (half the average specimen thickness for double drainage).

Limitations of Terzaghi's theory

1. It assumes constant permeability $k$ and coefficient of volume compressibility $m_v$ (so constant $c_v$); in reality $k$ and $m_v$ fall as the soil consolidates.
2. The soil is assumed to be homogeneous, saturated and to obey Darcy's law; real deposits are layered.
3. It is one-dimensional (flow and compression only vertical); in reality, lateral flow and strain occur near the edges of footings.
4. The load is assumed to be applied instantaneously and the total stress increment is constant with depth and time.
5. It ignores secondary compression (creep), which is large in organic soils and soft clays.
6. It assumes small strain and a linear $e$–$\sigma'$ relation, and incompressible solids and water.
7. The theory fits the experimental curves well up to about 60% to 70% consolidation only.

Primary vs secondary consolidation

Hydrostatic pore water pressure is the pressure that exists in the pore water because of the depth below the water table, with no flow. Excess pore water pressure is the extra pressure, above the hydrostatic value, created in the pore water by a sudden external load.

Total pore pressure at any time is $u = u_0 + \Delta u$. Consolidation is the process in which $\Delta u$ dissipates and the load is transferred to the soil skeleton.

The degree of consolidation $U$ is the ratio of the consolidation settlement at time $t$ to the final primary settlement (or the fraction of excess pore pressure dissipated). For one-dimensional consolidation, $U$ depends only on the time factor:

$$T_v = \frac{c_v\, t}{H_{dr}^2}, \qquad c_v = \frac{k}{m_v \gamma_w}$$

So the factors that affect $U$ are those that change $T_v$:

1. Time $t$. $U$ increases as time passes. For $U<60\%$, $T_v=\frac{\pi}{4}U^2$.
2. Coefficient of consolidation $c_v$. A larger $c_v$ gives faster consolidation. It rises with permeability $k$ and falls with compressibility $m_v$. Sand has a very high $c_v$; clay has a low $c_v$.
3. Coefficient of permeability $k$. Fine, plastic clays with low $k$ consolidate slowly because water escapes slowly.
4. Coefficient of volume compressibility $m_v$. A more compressible soil must expel more water, so it takes longer.
5. Drainage path length $H_{dr}$. Time varies with $H_{dr}^2$. Doubling the layer thickness makes the time four times longer.
6. Drainage condition. Drainage at top and bottom (double drainage) gives $H_{dr}=H/2$, so it is four times faster than single drainage with $H_{dr}=H$.
7. Distribution of initial excess pore pressure. Uniform, triangular or sinusoidal distributions give slightly different $U$-$T_v$ curves, mainly for $U>60\%$ cases.
8. Presence of sand lenses, drains or seams. Vertical sand drains and thin sand layers shorten $H_{dr}$ and speed up consolidation.

Time for a given degree of consolidation comes from the time factor:

$$t = \frac{T_v\, H_{dr}^2}{c_v}$$

For the same $U$ (same $T_v$) and the same $c_v$, time is proportional to $H_{dr}^2$.

• One-way (single) drainage: water leaves through one face only, so $H_{dr}=H$.
• Two-way (double) drainage: water leaves through top and bottom, so $H_{dr}=H/2$.

$$\frac{t_{two}}{t_{one}} = \left(\frac{H/2}{H}\right)^2 = \frac{1}{4}$$

Result: two-way drainage needs only one quarter of the time required for one-way drainage. Equivalently, one-way drainage takes four times longer. For example, a clay layer that needs 8 years to reach 90% consolidation with one-way drainage reaches the same degree in 2 years with two-way drainage.

Assumptions: the sand is saturated throughout (no moisture content is given), the clay is normally consolidated, $\gamma_w = 9.81\ \text{kN/m}^3$, and settlement is computed at the middle of the clay layer.

Step 1: Unit weights from $\gamma_{sat} = \dfrac{(G_s+e)\gamma_w}{1+e}$

$$\begin{aligned} \gamma_{sat,sand} &= \frac{(2.62+0.98)(9.81)}{1.98} = 17.84\ \text{kN/m}^3 \\ \gamma_{sat,clay} &= \frac{(2.70+0.62)(9.81)}{1.62} = 20.10\ \text{kN/m}^3 \end{aligned}$$

Step 2: Compression index

$$C_c = 0.0099(w_L-10) = 0.0099(50-10) = 0.396$$

Step 3: Effective stress at mid-clay (depth $3.5+1.75 = 5.25$ m)

$$\begin{aligned} \sigma_0' &= 17.84(1.5) + (17.84-9.81)(2.0) + (20.10-9.81)(1.75) \\ &= 26.75 + 16.06 + 18.00 = 60.82\ \text{kPa} \end{aligned}$$

Step 4: Settlement, with $\Delta\sigma = 110$ kPa

$$\begin{aligned} S_c &= \frac{C_c H}{1+e_0}\log_{10}\frac{\sigma_0'+\Delta\sigma}{\sigma_0'} \\ &= \frac{0.396(3.5)}{1.62}\log_{10}\frac{60.82+110}{60.82} \\ &= 0.8556 \times 0.4485 = 0.384\ \text{m} \end{aligned}$$

Answer: primary settlement of the clay layer $\approx 0.384\ \text{m} \approx 384\ \text{mm}$.

Assumptions: the specimen is drained at top and bottom (two porous stones), so $H_{dr}$ is half the average specimen thickness during the test. Average thickness $= 20 - 2.5/2 = 18.75$ mm, so $H_{dr} = 9.375$ mm. $\gamma_w = 9.81$ kN/m³.

(i) Time for 90% consolidation

Since $t \propto T_v$ for the same specimen:

$$t_{90} = t_{50}\,\frac{T_{90}}{T_{50}} = 154 \times \frac{0.848}{0.197} = 663\ \text{s} \approx 11\ \text{min}$$

(ii) Coefficient of permeability

$$c_v = \frac{T_{50}H_{dr}^2}{t_{50}} = \frac{0.197 (9.375)^2}{154} = 0.1124\ \text{mm}^2/\text{s} = 1.124\times10^{-7}\ \text{m}^2/\text{s}$$

Volume compressibility, with strain $= 2.5/20 = 0.125$ and $\Delta\sigma' = 120-60 = 60$ kPa:

$$m_v = \frac{0.125}{60} = 2.083\times10^{-3}\ \text{m}^2/\text{kN}$$ $$k = c_v m_v \gamma_w = (1.124\times10^{-7})(2.083\times10^{-3})(9.81) = 2.30\times10^{-9}\ \text{m/s}$$

(iii) Compression index

Change in void ratio:

$$\Delta e = \frac{\Delta H}{H_0}(1+e_0) = 0.125 \times 1.65 = 0.206$$ $$C_c = \frac{\Delta e}{\log_{10}(\sigma_1'/\sigma_0')} = \frac{0.206}{\log_{10}(120/60)} = \frac{0.206}{0.301} = 0.685$$

Answer: (i) $t_{90} \approx 663$ s (11 min); (ii) $k \approx 2.3\times10^{-9}$ m/s; (iii) $C_c \approx 0.685$.

Assumptions: the figure is not available, so the water table is taken at the ground surface and the given $\gamma$ values are saturated unit weights. $\gamma_w = 9.81$ kN/m³. The clay is normally consolidated and settlement is computed at mid-height of the clay (1.25 m below its top).

Ground  ------------------------
  Sand  4 m   gamma = 20 kN/m3
        ------------------------
  Clay  2.5 m gamma = 18, e = 1.30
        Cc = 0.22        <- mid-height

Effective stress at mid-clay

$$\begin{aligned} \sigma_0' &= (20-9.81)(4) + (18-9.81)(1.25) \\ &= 40.76 + 10.24 = 51.0\ \text{kPa} \end{aligned}$$

Settlement, with $\Delta\sigma = 30$ kPa:

$$\begin{aligned} S_c &= \frac{C_c H}{1+e_0}\log_{10}\frac{\sigma_0'+\Delta\sigma}{\sigma_0'} \\ &= \frac{0.22 \times 2.5}{2.30}\log_{10}\frac{81.0}{51.0} \\ &= 0.2391 \times 0.2009 = 0.048\ \text{m} \end{aligned}$$

Answer: final settlement $\approx 48\ \text{mm}$.

If there were no water table (both layers above it, $\sigma_0' = 80 + 22.5 = 102.5$ kPa), the same formula gives about 26.7 mm.

Settlement is proportional to the degree of consolidation $U$, and time is proportional to $T_v$. For $U < 60\%$, $T_v = \frac{\pi}{4}U^2$, so $t \propto U^2$ (same layer, same $c_v$ and drainage).

Step 1: Degrees of consolidation

$$U_1 = \frac{4}{20} = 20\%, \qquad U_2 = \frac{10}{20} = 50\%$$

Both are below 60%, so the parabolic relation applies.

Step 2: Time ratio

$$\frac{t_2}{t_1} = \left(\frac{U_2}{U_1}\right)^2 = \left(\frac{50}{20}\right)^2 = 6.25$$ $$t_2 = 6.25 \times 2 = 12.5\ \text{months}$$

Answer: about 12.5 months from the start of loading. The double drainage condition does not change the ratio, because $H_{dr}$ is the same in both cases.

Given: single-drained layer, $H_{dr} = 3$ m $= 300$ cm. Settlement 5 cm at $t = 60$ days corresponds to $U = 20\%$.

Step 1: Final settlement

$$S_{final} = \frac{5}{0.20} = 25\ \text{cm}$$

Step 2: Coefficient of consolidation. For $U=20\%$, $T_v = \frac{\pi}{4}(0.20)^2 = 0.0314$.

$$c_v = \frac{T_v H_{dr}^2}{t} = \frac{0.0314 \times 300^2}{60} = 47.1\ \text{cm}^2/\text{day}$$

Step 3: Time for various $U$. $t = T_v H_{dr}^2 / c_v = T_v \times 1910.0$ days. Use $T_v = \frac{\pi}{4}U^2$ for $U<60\%$ and $T_v = 1.781 - 0.933\log_{10}(100-U\%)$ for $U>60\%$. Settlement $= U \times 25$ cm.

Step 4: Settlement at 3 years ($t = 1095$ days):

$$T_v = \frac{c_v t}{H_{dr}^2} = \frac{47.1 \times 1095}{300^2} = 0.573 \;\Rightarrow\; U \approx 80\%, \quad S \approx 20\ \text{cm}$$

Plot: draw time (days, 0 to 1095) on the horizontal axis and settlement (cm) downward on the vertical axis; join the points above with a smooth curve. The curve is steep at first and flattens towards 25 cm.

 t (days)  0   200  400  600  800  1000 1095
 S (cm)    0
           |  *
        5  |    *
       10  |        *
       15  |             *
       20  |                    *    *  <- 3 yr
       25  - - - - - - - - - - - - - - (final)

Answer: the structure settles about 20 cm in 3 years; the curve approaches the final value of 25 cm.

Given: clay is drained at the top only (permeable above, impervious rock below), so $H_{dr} = 3$ m $= 300$ cm. $c_v = 0.025\ \text{cm}^2/\text{min}$, $S_{final} = 8$ cm.

(i) Time for 80% of settlement

For $U = 80\%$: $T_v = 1.781 - 0.933\log_{10}(100-80) = 0.567$.

$$t_{80} = \frac{T_v H_{dr}^2}{c_v} = \frac{0.567 \times 300^2}{0.025} = 2.04\times10^6\ \text{min}$$ $$= 1418\ \text{days} \approx 3.9\ \text{years}$$

(ii) Time for 2.5 cm settlement

$$U = \frac{2.5}{8} = 31.25\% \;(<60\%)\;\Rightarrow\; T_v = \frac{\pi}{4}(0.3125)^2 = 0.0767$$ $$t = \frac{0.0767 \times 300^2}{0.025} = 2.76\times10^5\ \text{min} = 192\ \text{days} \approx 0.53\ \text{year}$$

(iii) Settlement in 1 year

$t = 365 \times 1440 = 525{,}600$ min.

$$T_v = \frac{0.025 \times 525600}{300^2} = 0.146 \;\Rightarrow\; U = \sqrt{\frac{4T_v}{\pi}} = 0.431$$ $$S = 0.431 \times 8 = 3.45\ \text{cm}$$

Answer: (i) about 3.9 years; (ii) about 192 days; (iii) about 3.45 cm.

Principle: the same soil has the same $c_v$ and, for the same $U$, the same $T_v$. So

$$\frac{t_{field}}{t_{lab}} = \left(\frac{H_{dr,field}}{H_{dr,lab}}\right)^2$$

Drainage paths

• Field: sand on top, impervious rock at the bottom, so single drainage: $H_{dr} = 4$ m $= 4000$ mm.
• Laboratory: the 20 mm specimen in an oedometer drains at top and bottom, so $H_{dr} = 10$ mm.

Time

$$t_{field} = 4\ \text{h} \times \left(\frac{4000}{10}\right)^2 = 4 \times 160{,}000 = 640{,}000\ \text{h}$$ $$t_{field} = \frac{640000}{24 \times 365} = 73.1\ \text{years}$$

Answer: about 73 years for the building to reach 90% of its final settlement.

If the specimen were drained at one face only, the answer would be about 18 years.

Note: the second half of the problem is unclear in print, and the laboratory time for half the final settlement is not given. The usual form is: "the sample took $t_{lab}$ to reach half its final settlement; find the time for the building to reach half its final settlement". The scale relation is solved below, and the field time follows from the lab time.

Principle: same clay, same $c_v$, same $U = 50\%$ (so same $T_v = 0.196$). Then $t \propto H_{dr}^2$.

Drainage paths (both drained top and bottom)

• Field: $H_{dr} = 10/2 = 5$ m $= 500$ cm.
• Laboratory: $H_{dr} = 2.54/2 = 1.27$ cm.

Time ratio

$$\frac{t_{field}}{t_{lab}} = \left(\frac{500}{1.27}\right)^2 = 1.55\times10^5$$

So $t_{field} = 1.55\times10^{5}\, t_{lab}$.

Coefficient of consolidation from the lab test

$$c_v = \frac{0.196\,(1.27)^2}{t_{lab}} = \frac{0.316}{t_{lab}}\ \text{cm}^2/\text{unit time}$$

Field time

$$t_{field} = \frac{0.196\,(500)^2}{c_v}$$

Illustration: if the sample reached half its final settlement in 10 minutes, then $t_{field} = 1.55\times10^{6}$ min $\approx 2.95$ years.

Answer: the building reaches half of its final settlement in $1.55\times10^5$ times the lab time (about 2.95 years for a 10-minute lab test).

Given: clay 8 m to 14 m ($H = 6$ m), mid-depth at 11 m; $\sigma_c' = 125$ kPa, $C_c = 0.36$, $C_r = 0.06$, $\Delta\sigma = 50$ kPa, $\gamma_w = 10$ kN/m³.

Assumption: the initial void ratio of the clay is not given. Taking $G_s = 2.7$ and $\gamma_{sat} = 20$ kN/m³, $\frac{(2.7+e_0)(10)}{1+e_0}=20$ gives $e_0 = 0.70$.

 0 - 2 m   sand  gamma = 16
 2 - 8 m   sand  gsat  = 18   (WT at 2 m)
 8 - 14 m  clay  gsat  = 20   <- mid at 11 m

Step 1: Initial effective stress at 11 m

$$\sigma_0' = 16(2) + (18-10)(6) + (20-10)(3) = 32 + 48 + 30 = 110\ \text{kPa}$$

Step 2: State of the clay. $\sigma_0' = 110 < \sigma_c' = 125$, so the clay is overconsolidated. Final stress $\sigma_0'+\Delta\sigma = 160 > \sigma_c'$, so the stress path crosses the preconsolidation pressure and both $C_r$ and $C_c$ are needed.

Step 3: Settlement

$$\begin{aligned} S_c &= \frac{H}{1+e_0}\left[C_r\log_{10}\frac{\sigma_c'}{\sigma_0'} + C_c\log_{10}\frac{\sigma_0'+\Delta\sigma}{\sigma_c'}\right] \\ &= \frac{6000}{1.70}\left[0.06\log_{10}\frac{125}{110} + 0.36\log_{10}\frac{160}{125}\right] \\ &= 3529\,[0.00330 + 0.03859] = 148\ \text{mm} \end{aligned}$$

Answer: primary consolidation settlement $\approx 148$ mm (for $e_0 = 0.70$).

Given: centre of clay is at 11 m depth (clay 8 m to 14 m). Water table at 2 m. $\gamma_w = 10$ kN/m³. Sand above the water table: 18 kN/m³; saturated sand: 20 kN/m³; clay: 17 kN/m³.

Total vertical stress at 11 m before loading:

$$\sigma_0 = 18(2) + 20(6) + 17(3) = 36 + 120 + 51 = 207\ \text{kPa}$$

Hydrostatic pore pressure: $u_0 = 10 \times (11-2) = 90$ kPa.

(i) Before applying the surcharge

(ii) Immediately after applying the surcharge

The clay is saturated and the load is applied suddenly, so the entire 15 kPa is carried by the water (excess pore pressure $\Delta u = 15$ kPa).

(iii) Sufficiently long after loading (end of consolidation)

The excess pore pressure has fully dissipated ($\Delta u = 0$), so the load is carried by the soil skeleton.

Answer: effective stress is 117 kPa at the start and rises to 132 kPa at the end of consolidation; pore pressure goes 90, then 105, then 90 kPa.

Data from the previous profile: clay thickness $H = 6$ m (8 m to 14 m), $e_0 = 2.4$, $C_c = 1.2$, $c_v = 3.25\times10^{-7}$ m²/s. At the clay centre: $\sigma_0' = 117$ kPa; final $\sigma' = 117 + 15 = 132$ kPa. The clay is assumed normally consolidated.

Final primary settlement

$$\begin{aligned} S_c &= \frac{C_c H}{1+e_0}\log_{10}\frac{\sigma_0'+\Delta\sigma}{\sigma_0'} \\ &= \frac{1.2 \times 6}{3.4}\log_{10}\frac{132}{117} = 2.1176 \times 0.05238 = 0.111\ \text{m} \end{aligned}$$

$S_c \approx 111$ mm.

Settlement after 0.5 year

Sand lies above and below the clay, so drainage is two-way: $H_{dr} = 6/2 = 3$ m.

$$t = 0.5 \times 365 \times 86400 = 1.577\times10^{7}\ \text{s}$$ $$T_v = \frac{c_v t}{H_{dr}^2} = \frac{3.25\times10^{-7}\times1.577\times10^{7}}{3^2} = 0.569$$

From the given table, $T_v = 0.569$ corresponds to $U = 80\%$.

$$S_t = U \times S_c = 0.80 \times 111 = 88.8\ \text{mm}$$

Answer: final settlement $\approx 111$ mm; settlement after 0.5 year $\approx 89$ mm (U = 80%).