Principle of Effective Stress, Capillarity and Permeability

Quick sand condition is the state in which the effective stress of a cohesionless soil becomes zero because of upward seepage of water. The soil then has no shear strength and behaves like a liquid. It is not a type of sand; any cohesionless soil (sand, silt) can become "quick".

Explanation (upward flow)

Consider a soil layer of thickness $L$ with water flowing upward under a head difference $h$.

   h  ↑  water level in standpipe
  ----+----------------
      |   water
  ----+----------------  top of soil
      |   sand  L
  ----+----------------  base of soil
        ↑ ↑ ↑ upward flow

At the base of the soil layer:

• Total stress $\sigma = \gamma_{sat} L$
• Pore pressure $u = \gamma_w (L + h)$ (excess head $h$ added)
• Effective stress $\sigma' = \gamma_{sat}L - \gamma_w(L+h) = \gamma' L - \gamma_w h$

As $h$ increases, $\sigma'$ falls. When $\sigma' = 0$:

At this critical hydraulic gradient the seepage force just balances the submerged weight of the soil. The sand particles lose contact, boil up and the soil loses bearing capacity. For most sands $i_c \approx 1$.

Quick condition often causes failure of excavations, cofferdams and the downstream toe of hydraulic structures (piping).

Discharge velocity $v$ is the quantity of water flowing per unit time through a unit gross cross-sectional area of the soil (solids plus voids). Seepage velocity $v_s$ is the actual velocity of water through the void space, which is only a part of the area.

where $A_v = nA$ is the area of voids and $n$ is porosity.

Expression for critical hydraulic gradient

Take a sand layer of thickness $L$ and saturated unit weight $\gamma_{sat}$ in a tank. Water flows upward under a net head $h$ (head loss over the sample).

   h ↑  head causing upward flow
  ---------------  water
  ======= sand  L
  ---------------  filter
     ↑ ↑ ↑ inflow

At the base of the sand layer:

Quick condition begins when $\sigma' = 0$:

Using $\gamma' = \dfrac{(G-1)\gamma_w}{1+e}$:

For $G = 2.65$ and $e = 0.65$, $i_c = 1.0$.

Why sand more than clay

• Sand is cohesionless: once $\sigma' = 0$ the grains have no strength at all. Clay has cohesion, so it keeps some strength even when $\sigma'$ is zero.
• Sand is highly permeable, so a large hydraulic gradient can easily be developed and a large flow of water occurs at the exit surface. In clay, low permeability makes the flow very small, so the critical gradient is rarely reached.
• Clay particles are bound by electrochemical forces, which resist being lifted by seepage.

Coefficient of transmissibility $T$ is the rate of flow of water through a vertical strip of aquifer of unit width and full saturated thickness $b$ under a unit hydraulic gradient. It equals the coefficient of permeability times the aquifer thickness:

Its unit is m²/day or m²/s.

Seepage pressure is the pressure (force per unit area) exerted by water flowing through the soil on the soil grains in the direction of flow. For a hydraulic head loss $h$ over a length $L$, the seepage force per unit volume of soil is $j = i\gamma_w$, and the seepage pressure over the length $L$ is

It acts upward in upward flow and reduces effective stress; it acts downward in downward flow and increases it.

Held water is the water that is held in the soil by molecular attraction (adsorption) and surface tension around the soil particles, so that it is not removed by gravity drainage. It includes adsorbed (hygroscopic) water and capillary water, and it does not flow freely under gravity. In contrast, free (gravitational) water drains under gravity and is the part that flows under a hydraulic gradient.

Effective stress is $\sigma' = \sigma - u$ (Terzaghi). Surcharge changes the total stress, while capillary action changes the pore water pressure.

Effect of surcharge

A surcharge $q$ (extra load on the ground surface, e.g. a stockpile or embankment) adds to the total stress at every depth by the same amount:

If water is able to drain (long term, or in sand), the pore pressure $u$ is unchanged, so the effective stress increases by $q$:

In a saturated clay, immediately after loading the pore pressure takes up the extra stress ($\Delta u = q$), so $\sigma'$ does not change at first. It rises later as the water drains (consolidation).

Effect of capillary action

In the capillary zone above the water table, water is held in tension. The pore pressure is negative:

where $z_c$ is the height above the water table. Hence

so the effective stress increases. The increase is maximum at the top of the capillary rise ($\gamma_w h_c$) and is zero at the water table. This extra stress (apparent cohesion in sand) gives the soil added strength; it disappears if the soil dries or is flooded.

Capillarity is the rise of water above the water table through the fine voids of the soil, caused by surface tension. The soil voids act like tiny tubes. The height of rise is $h_c = \dfrac{4T\cos\alpha}{\gamma_w d}$, larger for finer soil (silt and fine sand rise more than gravel). Water in the capillary zone is under negative pressure, $u = -\gamma_w z$.

Effect of water table variation

• Rise of water table: the part of the soil that becomes submerged has its unit weight reduced to $\gamma'$ and pore pressure increased, so effective stress at depth decreases. The soil loses strength and bearing capacity.
• Fall of water table (drawdown): pore pressure falls and the soil above becomes moist/dry, so effective stress at depth increases. This increases settlement (e.g. due to pumping).

For depth $z$ below the water table at depth $z_w$: $\sigma' = \gamma z_w + \gamma'(z - z_w)$. Raising the water table reduces $z_w$ and so lowers $\sigma'$.

No, the statement is not correct (for a fully submerged sea-bed soil).

Consider soil below sea level at depth $z$ below the bed with water depth $h_w$ above the bed.

• Total stress: $\sigma = \gamma_w h_w + \gamma_{sat} z$
• Pore pressure: $u = \gamma_w (h_w + z)$

The water-depth term $\gamma_w h_w$ appears in both $\sigma$ and $u$ and cancels. So the effective stress depends only on the submerged unit weight and the depth into the soil, not on the sea level. A rise or fall of the sea changes the total stress and pore pressure by the same amount.

Remarks:

• Effective stress changes only if the pore pressure cannot respond equally, for example in low-permeability clay during rapid drawdown (the pore pressure lags) or when the seabed becomes exposed at low tide (then the buoyancy is lost and $\sigma'$ rises).
• In normal situations the sea-bed soil's strength is unaffected by the sea-level fluctuation.

Capillary rise $h_c = \dfrac{4T\cos\alpha}{\gamma_w d}$ depends on:

1. Grain/pore size: smaller effective diameter $D_{10}$ and void size gives greater rise ($h_c \propto 1/d$; $h_c \approx C/(eD_{10})$).
2. Void ratio: smaller $e$ gives higher rise.
3. Surface tension of water (and temperature, impurities).
4. Contact angle between water and particle surface.
5. Soil type and packing, clean surfaces and degree of saturation of the soil.

Neutral pressure (pore water pressure), $u$: the pressure of the water filling the voids of a saturated soil. It acts equally in all directions, so it does not cause any shear strength or volume change in the soil skeleton. In a static water table, $u = \gamma_w h_w$.

Effective pressure (stress), $\sigma'$: the part of the total stress carried by the soil grains through their contact points. It controls compression and shear strength of the soil:

(Terzaghi's principle of effective stress.)

Shear strength of a soil is controlled by the effective stress, not by the total stress, because water has no shear strength and only the grain contacts carry friction. By Mohr–Coulomb in terms of effective stress:

A rise in pore pressure lowers $\sigma'$ and so lowers strength (as in liquefaction, quick sand and slope failure after rain). Increase of $\sigma'$ by consolidation raises strength.

For a soil layer of thickness $H$ with water depth $h_w$ above it, if there is no flow, $\sigma' = \gamma' H$ at the bottom. Flow changes the pore pressure and hence $\sigma'$. Let $h$ be the head loss through the soil.

Downward flow

Water flows down, so the head at the bottom is less than hydrostatic: $u = \gamma_w (h_w + H - h)$.

The effective stress increases (seepage force acts down, in the direction of gravity). Soil becomes denser and stronger.

Upward flow

Water flows up, so the head at the bottom exceeds hydrostatic: $u = \gamma_w (h_w + H + h)$.

The effective stress decreases. It becomes zero when $i = i_c = \gamma'/\gamma_w$ (quick condition), and the soil loses its strength.

Discharge velocity (superficial or approach velocity) is the quantity of water flowing per unit time per unit gross cross-sectional area of the soil, taken perpendicular to the flow:

It is an average, imaginary velocity, since water flows only through the voids. The actual velocity in the pores (seepage velocity) is $v_s = v/n$.

Darcy's law: the discharge velocity of water through a saturated soil is directly proportional to the hydraulic gradient (laminar flow):

Figure (inclined soil specimen)

     inlet                    
 ____ ▽ ____                  
 | water  |\  (1)             
 |  h1    | \_______          
 |________|  soil  \ L       
 datum        A      \  (2)   
                      \___ ▽ outlet

Total head at entry (1) and exit (2), measured from a common datum, each equals elevation head + pressure head: $H_1 = z_1 + h_{p1}$, $H_2 = z_2 + h_{p2}$.

Terms

• $Q$ = discharge per unit time (m³/s)
• $k$ = coefficient of permeability (m/s), a property of soil and water
• $h = H_1 - H_2$ = head loss (difference in total head between entry and exit) (m)
• $L$ = length of flow path measured along the soil (m)
• $i = h/L$ = hydraulic gradient
• $A$ = cross-sectional area of the soil normal to the flow (m²)
• $v = Q/A$ = discharge velocity

The inclination does not appear separately, because the hydraulic gradient uses total head (it already includes the elevation head), and $L$ is the length along the inclined sample.

Darcy's law states that for laminar flow through a saturated soil, the discharge velocity is directly proportional to the hydraulic gradient:

where $v = Q/A$ is the discharge velocity (the flow per unit gross area, not the actual velocity in the pores), $i = h/L$ is the hydraulic gradient and $k$ is the coefficient of permeability. The law is valid for Reynolds number below about 1 (fine sands, silts and clays); it fails in coarse gravel (turbulent flow).

Tests to find k

Laboratory

• Constant head permeability test (for coarse-grained soils)
• Falling (variable) head permeability test (for fine-grained soils)
• Indirect: from consolidation test data, capillarity permeability test; empirical formulae (Hazen's, $k = C D_{10}^2$)

Field

• Pumping out test (from wells)
• Pumping in test
• Borehole (auger hole) tests
• Packer test (for rock)
• Tracer tests.

Laboratory methods

1. Constant head permeability test: for coarse-grained soils (sand, gravel).
2. Falling (variable) head permeability test: for fine-grained soils (silt, clay).
3. Indirect: from consolidation test; horizontal capillarity test; empirical formulae such as Hazen's $k = C\,D_{10}^2$.

Field methods

1. Pumping out test (unconfined and confined aquifer, with observation wells)
2. Pumping in test (auger hole / borehole test)
3. Packer test (for rock/deep strata)
4. Tracer tests.

Superficial (discharge) velocity $v$ is based on the total area; seepage velocity $v_s$ is the real velocity in the voids.

Derivation

Consider a soil of gross cross-section $A$ and length $L$. Let $A_v$ be the area of voids on the cross-section and $V_v$, $V$ the volumes of voids and total.

Let the discharge be $Q$. It passes through the gross area $A$ at velocity $v$ and through the void area $A_v$ at velocity $v_s$:

Since $v = k i$:

Since $n<1$, $v_s > v$. In terms of void ratio, $n = \dfrac{e}{1+e}$ so $v_s = v\,\dfrac{1+e}{e}$.

Method

The falling (variable) head permeability test is the most appropriate for clay.

Why: clay has very low permeability ($k \approx 10^{-7}$ to $10^{-10}$ m/s). In a constant head test the flow would be too small to collect and measure accurately in a reasonable time. In the falling head test, a standpipe of small area $a$ is used, so even a tiny quantity of flow gives a measurable fall in head.

Expression

For a sample of length $L$ and area $A$, with head falling from $h_1$ to $h_2$ in time $t$:

where $a$ is the standpipe area. (A consolidation test can also be used for clays, using $k = c_v m_v \gamma_w$.)

Treat each zone of the soil as layers in series or parallel. Let the left block (zone 1) have permeability $k_1$, length $L_1$ and height $H = H_1 + H_2$. The right block (zone 2) has length $L_2$ and is made of $k_2$ (upper, $H_1$) over $k_3$ (lower, $H_2$).

  <------ L1 ------><------ L2 ------>
  +-----------------+-----------------+ ^
  |                 |   k2       H1   | |
  |       k1        +-----------------+ H
  |                 |   k3       H2   | |
  +-----------------+-----------------+ v

1. Horizontal flow (left to right)

Inside zone 2 the flow is along the layers, so the two layers act in parallel:

Zone 1 and zone 2 are met one after the other, so they act in series (same flow, lengths add):

2. Vertical flow (top to bottom)

Inside zone 2 the flow crosses the layers, so they act in series:

Zone 1 and zone 2 now carry flow side by side under the same head (parallel). Weighting by the widths:

Substituting $k_{v2}$ gives the full expression:

Confined aquifer (artesian): a water-bearing layer of thickness $b$ lying between two impervious layers. The water in it is under pressure, so the piezometric level stands above the top of the aquifer.

Unconfined aquifer (water table aquifer): a pervious layer with a free water table as its upper boundary, resting on an impervious layer. Water is at atmospheric pressure at the table.

  Confined                 Unconfined
  ~~~~ piezometric         ~~~~ water table
  ==== impervious          ....
  aquifer  b               aquifer
  ==== impervious          ==== impervious

Equations from a pumping test

Let a well discharge $q$ at steady state, with two observation wells at radial distances $r_1$ and $r_2$ having water levels (heads above the impervious base) $h_1$ and $h_2$.

Confined aquifer:

Unconfined aquifer:

Given and assumptions

• Profile: clay 0 to 4 m, fine sand 4 m to 10 m. Water table (WT) at 2 m depth, capillary rise 1 m above WT (saturated zone from 1 m to 2 m depth).
• Clay: $\gamma = 18$ kN/m³ above the WT, $\gamma_{sat} = 22$ kN/m³ below. The capillary zone is saturated, so 22 kN/m³ is used from 1 m to 2 m.
• Sand: $n = 0.44$, $G = 2.65$, $\gamma_w = 9.81$ kN/m³. The soil above the capillary zone is taken as having $u = 0$.

Stresses (kPa)

Diagram

 Depth  total σ        u            σ'
  0 m   0              0            0
  1 m   18       -9.81 (suction)    27.8
  2 m   40       0 (WT)             40
  4 m   84       19.6               64.4
 10 m   197.3    78.5               118.8
       (all linear between points; u is negative
        only between 1 m and 2 m)

Plot $\sigma$, $u$, $\sigma'$ against depth, joining the points with straight lines. $\sigma$ and $\sigma'$ have a kink at 2 m, 4 m; $u$ is a straight line from 0 at 2 m to 78.48 kPa at 10 m with a negative triangle above the WT.

Answer: At the bottom (10 m): $\sigma = 197.25$, $u = 78.48$, $\sigma' = 118.77$ kPa. Capillarity gives a gain of $9.81$ kPa in effective stress at 1 m.

Given

Uniform sand, water table (WT) at 2 m depth, $\gamma = 18$ kN/m³, $\gamma_w = 9.81$ kN/m³. Depth of interest = 5 m.

(i) Dry sand above the WT

• Total stress: $\sigma = 18 \times 5 = 90.0$ kPa
• Pore pressure at 5 m: $u = 9.81 \times (5-2) = 29.43$ kPa

Variation with depth (linear):

(ii) Sand above the WT saturated by capillary action

The pore water in the 2 m above the WT is under suction: $u = -\gamma_w z_c$, so $u = -19.62$ kPa at the ground surface (top of the capillary zone) and 0 at the WT. The soil weight is still taken as 18 kN/m³ (no other unit weight is given), so the total stress is unchanged.

At 5 m: $\sigma = 90.0$ kPa, $u = 29.43$ kPa, so

The effective stress at 5 m is the same as in (i): capillary suction changes the stress only within the capillary zone (and below it, if the saturated weight were larger).

If the saturated unit weight were higher than 18 kN/m³ in the capillary zone, the total and effective stress below would increase by $(\gamma_{sat}-18)\times 2$.

Plot

 Case (i)               Case (ii)
 σ' at 0 m = 0          σ' at 0 m = 19.6  (σ'-line shifts right
                                           above the WT)
 σ' at 2 m = 36         σ' at 2 m = 36
 σ' at 5 m = 60.6       σ' at 5 m = 60.6

Answer: (i) $\sigma' = 60.57$ kPa. (ii) $\sigma' = 60.57$ kPa at 5 m; the effective stress is increased only in the top 2 m (by $\gamma_w z_c$, up to 19.62 kPa at the surface).

Given

Clay 0 to 4 m (above), sand 4 to 7 m. WT at 2 m. Capillary saturated zone 1 m thick above the WT (from 1 m to 2 m). Clay: $\gamma = \gamma_{sat} = 20$ kN/m³; sand $\gamma_{sat} = 19$ kN/m³; $\gamma_w = 9.81$ kN/m³. Soil above the capillary zone (0 to 1 m) has $u = 0$.

Calculation

Answer: Effective vertical stress at 0, 1, 2, 4 and 7 m = 0, 29.81, 40.00, 60.38 and 87.95 kPa respectively.

Given

Top layer 0 to 3 m, $\gamma = 17$ kN/m³. Bottom layer 3 to 7 m, $\gamma_{sat} = 21$ kN/m³. WT at 4 m; capillary saturation 1 m above WT (3 m to 4 m), which lies in the bottom layer, so $\gamma = 21$ kN/m³ is used there. $\gamma_w = 9.81$ kN/m³; $u = 0$ above the capillary zone.

Stresses at key levels

Diagrams

 depth  total σ        pore u           effective σ'
 0 m    0 |            0                0
        . |            .                .
 3 m    51|            -9.8 (suction)   60.8
 4 m    72|            0  (WT)          72.0
 7 m    135|           29.4              105.6
 (draw each as straight lines between these points
  on a stress-vs-depth plot, depth downward)

• $\sigma$ increases linearly with a change in slope at 3 m (17 to 21 kN/m³).
• $u$ is negative (suction) between 3 m and 4 m, zero at the WT and rises at 9.81 kPa/m below.
• $\sigma' = \sigma - u$; it rises by $\gamma_w \times 1 = 9.81$ kPa across the capillary zone and is higher than $\sigma$ there.

Answer: At 7 m, $\sigma = 135$, $u = 29.43$, $\sigma' = 105.57$ kPa.

Given

Top layer 0 to 3 m: $\gamma = 18$ kN/m³ (bulk), $\gamma_{sat} = 21$. Bottom layer 3 to 7 m: $\gamma_{sat} = 20$ kN/m³. WT at 4 m; capillary saturation 1 m above WT (3 m to 4 m). $\gamma_w = 9.81$ kN/m³.

Assumption: the capillary zone (3 m to 4 m) lies in the bottom layer, so it is saturated with $\gamma_{sat} = 20$ kN/m³. The top layer remains at bulk weight 18 kN/m³ (it is above the capillary zone, so its saturated weight is not used).

Calculation

Plot

 depth(m)  σ      u       σ'
   0       0      0       0
   3       54   -9.81    63.81
   4       74     0      74.0
   7      134   29.43   104.57

Plot each column against depth (downward), joining points with straight lines. $u$ is negative between 3 m and 4 m and positive below.

Answer: At 7 m, $\sigma = 134$ kPa, $u = 29.43$ kPa, $\sigma' = 104.57$ kPa.

Given

$D = 7.5$ cm, $L = 18$ cm, head loss $h = 24.7$ cm, $Q = 626$ ml in $t = 60$ s, porosity $n = 0.44$.

Permeability

Flow (discharge) velocity

Seepage velocity

Permeability at $n = 0.39$

For the same soil, $k \propto \dfrac{e^3}{1+e}$ (Kozeny–Carman type relation).

Answer: $k = 0.172$ cm/s; $v = 0.236$ cm/s; $v_s = 0.537$ cm/s; $k$ at $n = 0.39$ is $0.101$ cm/s.

Given

$Q = 500$ ml in $t = 15$ min $= 900$ s, $D = 5$ cm, head loss $h = 40$ cm between two points $L = 15$ cm apart. Dry weight of sample $W_d = 4.86$ N, $G = 2.65$, $\gamma_w = 9.81$ kN/m³.

Coefficient of permeability

Seepage velocity

Discharge velocity:

Volume of sample $= 19.63 \times 15 = 294.5\ \text{cm}^3 = 2.945\times10^{-4}\ \text{m}^3$.

Answer: $k = 0.0106$ cm/s; $v_s = 0.0775$ cm/s ($n = 0.365$).

Principle

For a variable head test, $k = \dfrac{aL}{A\,t}\ln\dfrac{h_1}{h_2}$, so for given $a$, $A$, $h_1$, $h_2$, the time is $t = \dfrac{aL}{Ak}\ln\dfrac{h_1}{h_2}$, proportional to $L/k$. When the second soil is placed on the first, the two act in series, so the flow resistances ($L/k$) add and the total time is the sum of the times for each soil taken alone.

Given

$h_1 = 900$ mm, $h_2 = 135$ mm, $a = 130$ mm², $A = 4560$ mm², $L_2 = 60$ mm.

• Soil 1 alone: $t_1 = 5$ s
• Soil 1 + soil 2: $t = 150$ s
• Time due to soil 2 alone: $t_2 = 150 - 5 = 145$ s

Permeability of soil 2

Answer: $k_2 = 0.0224$ mm/s $= 2.24\times10^{-5}$ m/s.

The soil plug in the pipe is a short column through which the tank water flows. Apply Darcy's law $Q = k\,\dfrac{h}{L}\,A$ to find its length $L$.

Given

$k = 10$ m/day, head $h = 20$ m, $Q = 0.15$ m³/day, $A = 200$ cm² $= 0.02$ m².

Length of soil plug

Volume of soil

Answer: The clogging soil is 26.67 m long, so its volume is about $0.533$ m³ ($\approx 533$ litres).

The head is assumed to be fully lost through the plug.

Assumptions

The water table is parallel to the ground surface and the impervious layer beneath, so flow is parallel to the slope. The thickness $H = 3$ m is measured vertically, so the flow area normal to the flow per metre width is $H\cos\alpha$.

        ground / water table (parallel)
    \ ~~~~~~~~~~~~~~~~
     \   flow  ->  H=3 m      
      \  (down the slope)    α = 5°
       \_________________ impervious

(i) Hydraulic gradient

Over a length $L$ along the slope, the head drop equals the fall in elevation $L\sin\alpha$:

(ii) Rate of flow per metre width

Answer: $i = 0.0872$; $q = 1.25\times10^{-5}$ m³/s per m width.

Total head at a point = elevation head + pressure head, both measured from the datum: $H = z + h_p$. The piezometer reading $h$ at each point is the pressure head.

(i) Total heads

The total head falls steadily along the specimen from point 1 to 3 because of friction loss in the soil; $H_2$ lies between $H_1$ and $H_3$.

(ii) Hydraulic gradient (entry 1 to exit 3)

Head loss between 1 and 3:

The flow path through the soil is the sample length $L$ measured along the incline, so

The inclination $\theta$ does not enter explicitly if $L$ is the length along the sample, because the elevation differences are already included in the $z$ terms. (If only the horizontal distance $L_h$ is known, $L = L_h/\cos\theta$.)