This resultant body force acts at an angle to the vertical described by angle bae. Applying the sine rule to triangle ace gives:
∠ =
bae − h L
sin 1 ∆eagw cosq (3.11)
Only the resultant body force contributes to effective stress. A component of seepage force acting vertically upwards will therefore reduce a vertical effective stress component from the static value.
A component of seepage force acting vertically downwards will increase a vertical effective stress component from the static value.
Effective stress Figure 3.8 shows part of the flow net for seepage under a sheet pile wall, the embedded length on the downstream side being d. A mass of soil adjacent to the piling may become unstable and be unable to support the wall. Model tests have shown that failure is likely to occur within a soil mass of approximate dimensions d ×d/2 in section (ABCD in Figure 3.8). Failure first shows in the form of a rise or heave at the surface, associated with an expansion of the soil which results in an increase in permeability. This in turn leads to increased flow, surface ‘boiling’ in the case of sands, and complete failure of the wall due to a loss of resisting lateral earth pres- sure (see Chapter 11). This could have potentially fatal implications if the wall is being used as a cofferdam, to allow construction operations below ground level or below water level in marine environments.
The variation of total head on the lower boundary CD of the soil mass can be obtained from the flow net equipotentials, or directly from the results of analysis using the Finite Difference Method (FDM), such as the spreadsheet analysis tool described in Section 2.8. For the purpose of analysis, however, it is usually sufficient to determine the average total head hm by inspection. The total head on the upper boundary AB is zero. The average hydraulic gradient is given by
i h
m= dm
Figure 3.8 Upward seepage adjacent to sheet piling: (a) determination of parameters from flow net, (b) force diagram, and (c) use of a filter to suppress heave.
Since failure due to heaving may be expected when the hydraulic gradient becomes icr, the factor of safety (F) against heaving may be expressed as
F i
= icr
m
(3.13) In the case of sands, a factor of safety can also be obtained with respect to ‘boiling’ at the surface.
The exit hydraulic gradient (ie) can be determined by measuring the dimension Δs of the flow net field AEFG adjacent to the piling:
i h
e = Ds D
where Δh is the drop in total head between equipotentials GF and AE. Then, the factor of safety is F i
= icr
e
(3.14) There is unlikely to be any appreciable difference between the values of F given by Equations 3.13 and 3.14.
The sheet pile wall problem shown in Figure 3.8 can also be used to illustrate the graphical method for determining seepage forces outlined in Section 3.6:
Total weight of mass ABCD = vector ab= 1 d 2
gsat 2
Average total head on CD = hm Elevation head on CD = –d
Average pore water pressure on CD = (hm + d)γw
Boundary water force on CD = vector bd= d
(
h +d)
2 m gw
Boundary water force on BC = vector de = 0 (as seepage is vertically upwards).
Resultant body force of ABCD = ab–bd–de
= -
(
+)
-=
(
¢ +)
-(
+)
= ¢ -
1
2 2 0
1 2
1 2 1
2
1 2
2
2 2
2
g g
g g g
g
sat m w
w m w
d d h d
d h d d
d hmm wg d
The resultant body force will be zero, leading to heave, when 1
2
1 2 hm wg d= g¢d2
Effective stress The factor of safety can then be expressed as
F d
h d d h
i
= ¢ i
= ¢ =
1 12 2
g 2
g g
m w m wg
cr m
which is the same as Equation 3.13.
If the factor of safety against heave is considered inadequate, the embedded length d may be increased or a surcharge load in the form of a filter may be placed on the surface AB, the filter being designed to prevent entry of soil particles, following the recommendations of Section 2.10.
Such a filter is shown in Figure 3.8(c). If the effective weight of the filter per unit area is w′, then the factor of safety becomes
F d w
= g¢ + ¢h g
m w
Example 3.3
The flow net for seepage under a sheet pile wall is shown in Figure 3.9, the saturated unit weight of the soil being 20 kN/m3. Determine the values of effective vertical stress at points A and B.
Solution
First, consider the column of saturated soil of unit area between A and the soil surface at C. The total weight of the column is 11γsat (220 kN). Due to the change in level of the equipotentials across the column, the boundary water forces on the sides of the column will not be equal, although in this case the difference will be small. There is thus a net Figure 3.9 Examples 3.3 and 3.4.
horizontal boundary water force on the column. However, as the effective vertical stress is to be calculated, only the vertical component of the resultant body force is required and the net horizontal boundary water force need not be considered. The vertical component of the boundary water force on the top surface of the column is due only to the depth of water above C, and is 4γw (39 kN). The boundary water force on the bottom surface of the column must be determined; in this example the FDM spreadsheet described in Section 2.8 is used as the required values of total head can be directly read from the spreadsheet, though this may equally be determined from a hand-drawn flow net. The calculated total head distribution may be found in Seepage_CSM9.xls on the Companion Website.
Total head at A, hA = 5.2 m Elevation head at A, zhA = –7.0 m
Pore water pressure at A, uA = γw(hA–zA) = 9.81(5.2 + 7.0) = 120 kPa i.e. boundary water force on bottom surface = 120 kN
Net vertical boundary water force = 120–39 = 81 kN Total weight of the column = 220 kN
Vertical component of resultant body force = 220–81 = 139 kN i.e. effective vertical stress at A = 139 kPa.
It should be realised that the same result would be obtained by the direct application of the effective stress equation, the total vertical stress at A being the weight of saturated soil and water, per unit area, above A. Thus
s g g
s s
A sat w
A
A A A
kPa kPa
k
= + = + =
=
¢ = - = - =
11 4 220 39 259
120
259 120 139 u
u PPa
The only difference in concept is that the boundary water force per unit area on top of the column of saturated soil AC contributes to the total vertical stress at A. Similarly at B
s g g
g
B sat w
B
B
B w B B
kPa m
m
= + = + =
=
= -
=
(
-)
=6 1 120 9 81 130 1 7
7 0
9 . .
.
. h
z
u h z 881 1 7 7 0 85
130 85 45 . + .
( )
=¢ = - = - =
kPa kPa
B B B
s s u
Effective stress
Example 3.4
Using the flow net in Figure 3.9(a), determine the factor of safety against failure by heav- ing adjacent to the downstream face of the piling. The saturated unit weight of the soil is 20 kN/m3.
Solution
The stability of the soil mass DEFG in Figure 3.9, 6 m by 3 m in section, will be analysed.
By inspection of the flow net or from Seepage_CSM9.xls on the Companion Website, the average value of total head on the base DG is given by
hm=2 6. m
The average hydraulic gradient between DG and the soil surface EF is im= 2 6=
6. 0 43 .
Critical hydraulic gradient, icr w
= g¢ = = g
10 2 9 8. 1 04
. .
Factor of safety, F i
= icr = =
m
1 04 0 43. 2 4
. .
Dynamic/seismic liquefaction
In the previous examples, seepage-induced or static liquefaction of soil has been discussed – that is to say, situations where the effective stress within the soil is reduced to zero as a result of high pore water pressures due to seepage. Pore water pressure may also be increased due to dynamic loading of soil. As soil is sheared cyclically it has a tendency to contract, reducing the void ratio e. If this shearing and resulting contraction happen rapidly relative to the permeability of the soil, then there may not be sufficient time for the pore water to escape from the voids, such that the reduction in volume will lead to an increase in pore water pressure due to the incompressibility of water.
Consider a uniform layer of fully saturated soil with the water table at the surface. The total stress at any depth z within the soil is
sv =gsatz
The soil will liquefy at this depth when the effective stress becomes zero. By Terzaghi’s Principle (Equation 3.1) this will occur when u = σv. From Equation 3.4, the pore water pressure u is made up of two components: hydrostatic pressure us (present initially before the soil is loaded) and an excess component ue (which is induced by the dynamic load). Therefore, the critical excess pore water pressure at the onset of liquefaction (ueL) is given by
u
u u z
z u z
u z
=
+ =
+ =
= ′ s g
g g
g
v
s eL sat
w eL sat
(3.15)
i.e. for soil to liquefy, the excess pore water pressure must be equal to the initial effective stress in the ground (prior to application of the dynamic load). Furthermore, considering a datum at the surface of the soil, from Equation 2.1
u h z
h z
h z
=
(
+)
\ =
(
-)
= ¢ g
g g
g g g
w
sat w w
w
This demonstrates that there will be a positive hydraulic gradient h/z between the soil at depth z and the surface (i.e. vertically upwards), when liquefaction has been achieved. This is the same as the critical hydraulic gradient defined by Equation 3.12 for seepage-induced liquefaction.
Excess pore water pressure rise due to volumetric contraction may be induced by vibrating loads from direct sources on or in the soil, or may be induced due to cyclic ground motion during an earthquake. An example of the former case is a shallow foundation for a piece of machinery such as a power station turbine. In this case, the cyclic straining (and therefore volumetric con- traction and induced excess pore pressure) in the soil will generally decrease with distance from the source. From Equation 3.15, it can be seen that the amount of excess pore water pressure required to initiate liquefaction increases with depth. As a result, any liquefaction will be concen- trated towards the surface of the ground, close to the source.
In the case of an earthquake, ground motion is induced as a result of powerful stress waves which are transmitted from within the Earth’s crust (i.e. far beneath the soil). As a result, liquefac- tion may extend to much greater depths. Combining Equations 3.12 and 3.15
u i z G
e z
eL = cr w = w
(
s-)
g g + 1
1 (3.16)
The soil towards the surface is often at a lower density (higher e) than the soil beneath.
Combined with the shallow depth z, it is clear that, under earthquake shaking, liquefac- tion will start at the ground surface and move downwards as shaking continues, requiring larger excess pore water pressures at depth to liquefy the deeper layers. It is also clear from Equation 3.16 that looser soils at high e will require lower excess pore water pressures to cause liquefaction. Soils with high voids ratio also have a higher potential for densification when shaken (towards emin, the densest possible state), so loose soils are particularly vulner- able to liquefaction. Indeed, strong earthquakes may fully liquefy layers of loose soil many metres thick.
It will be demonstrated in Chapter 5 that the shear strength of a soil (which resists applied loads induced by foundations and other geotechnical constructions) is proportional to the effective stress within the ground. It is therefore clear that the occurrence of liquefaction (s¢v = 0) can lead to significant damage to structures, typically due to excessive settlement and/
or rotation – an example, observed during the 1964 Niigata earthquake in Japan, is shown in Figure 3.10.
Effective stress
Summary
1 Total stress is used to define the applied stresses on an element of soil (both due to external applied loads and due to self-weight). Soils support total stresses through a combination of effective stress due to interparticle contact and pore water pressure in the voids. This is known as Terzaghi’s Principle (Equation 3.1).
2 Under hydrostatic conditions, the effective stress state at any depth within the ground can be found from knowledge of the unit weight of the soil layers and the location of the water table. If seepage is occurring, a flow net or finite difference mesh can be used to determine the pore water pressures at any point within the ground, with effective stress subsequently being found using Terzaghi’s Principle.
3 A consequence of Terzaghi’s Principle is that if there is significant excess pore water pressure developed in the ground, the soil skeleton may become unloaded (zero effective stress). This condition is known as liquefaction and may occur due to seepage, or due to dynamic external loads which cause the soil to contract rapidly. Seepage-induced liquefaction can lead to uplift or boiling of soil along the downstream face of a sheet piled excavation, and subsequent failure of the excavation. Dynamic liquefaction can lead to excessive settlement and rotation of structures.
Figure 3.10 Foundation failure due to liquefaction, 1964 Niigata earthquake, Japan.
Problems
3.1 A river is 2 m deep. The river bed consists of a depth of sand of saturated unit weight 20 kN/m3. What is the effective vertical stress 5 m below the top of the sand?
3.2 The North Sea is 200 m deep. The sea bed consists of a depth of sand of saturated unit weight 20 kN/m3. What is the effective vertical stress 5 m below the top of the sand? Compare your answer to the value found in Problem 3.1 – how does the water level above the ground surface affect the stresses within the ground?
3.3 A layer of clay 4 m thick lies between two layers of sand each 4 m thick, the top of the upper layer of sand being ground level. The water table is 2 m below ground level but the lower layer of sand is under artesian pressure, the piezometric surface being 4 m above ground level. The saturated unit weight of the clay is 20 kN/m3 and that of the sand 19 kN/m3; above the water table the unit weight of the sand is 16.5 kN/m3. Calculate the effective vertical stresses at the top and bottom of the clay layer.
3.4 In a deposit of fine sand the water table is 3.5 m below the surface, but sand to a height of 1.0 m above the water table is saturated by capillary water; above this height the sand may be assumed to be dry. The saturated and dry unit weights, respectively, are 20 and 16 kN/m3. Calculate the effective vertical stress in the sand 8 m below the surface.
3.5 A layer of sand extends from ground level to a depth of 9 m and overlies a layer of clay, of very low permeability, 6 m thick. The water table is 6 m below the surface of the sand. The saturated unit weight of the sand is 19 kN/m3 and that of the clay 20 kN/m3; the unit weight of the sand above the water table is 16 kN/m3. Over a short period of time the water table rises by 3 m, and is expected to remain permanently at this new level. Determine the effective vertical stress at depths of 8 and 12 m below ground level (a) immediately after the rise of the water table, and (b) several years after the rise of the water table.
3.6 An element of soil with sides horizontal and vertical measures 1 m in each direction. Water is seeping through the element in a direction inclined upwards at 30° above the horizontal under a hydraulic gradient of 0.35. The saturated unit weight of the soil is 21 kN/m3. Draw a force diagram to scale showing the following: total and effective weights, resultant boundary water force, seepage force. What is the magnitude and direction of the resultant body force?
3.7 For the seepage situations shown in Figure 3.11, determine the effective normal stress on plane XX in each case, (a) by considering pore water pressure and (b) by considering seepage pressure. The saturated unit weight of the soil is 20 kN/m3.
Effective stress
3.8 The section through a long cofferdam is shown in Figure 2.27, the saturated unit weight of the soil being 20 kN/m3. Determine the factor of safety against ‘boiling’ at the surface AB, and the values of effective vertical stress at C and D.
3.9 The section through part of a cofferdam is shown in Figure 2.26, the saturated unit weight of the soil being 19.5 kN/m3. Determine the factor of safety against heave failure in the excavation adjacent to the sheet piling. What depth of filter (unit weight 21 kN/m3) would be required to ensure a factor of safety of 3.0?
References
Bishop, A.W. (1959) The principle of effective stress, Tekniche Ukeblad, 39, 4–16.
Taylor, D.W. (1948) Fundamentals of Soil Mechanics, John Wiley & Sons, New York, NY.
Terzaghi, K. (1943) Theoretical Soil Mechanics, John Wiley & Sons, New York, NY.
Vanapalli, S.K. and Fredlund, D.G. (2000) Comparison of different procedures to predict unsatu- rated soil shear strength, Proceedings of Geo-Denver 2000, ASCE Geotechnical Special Publication, 99, 195–209.
Figure 3.11 Problem 3.7.
Further reading
Powrie, W. (2012) Groundwater profiles and effective stress, Chapter 15, in ICE Manual of Geotechnical Engineering (Burland, J., Chapman, T., Skinner, H. and Brown, M. Eds.), ICE Publishing, London. Vol. 1, pp. 163–166.
This chapter of the ICE Manual of Geotechnical Engineering links groundwater conditions (discussed at the start of Chapter 2) to effective stresses within the ground.Rojas, E. (2008a) Equivalent stress equation for unsaturated soils. I: Equivalent stress, International Journal of Geomechanics, 8(5), 285–290.
Rojas, E. (2008b) Equivalent stress equation for unsaturated soils. II: Solid-porous model, International Journal of Geomechanics, 8(5), 291–299.
These companion papers describe in detail the development of a strength model for unsaturated soil, aiming to address one of the key questions in geotechnical engineering for which a satisfac- tory and widely accepted answer is still unclear.
For further student and instructor resources for this chapter, please visit the Companion Website at www.routledge.com/cw/craig
Consolidation
Chapter 4
Consolidation
Learning outcomes
After working through the material in this chapter, you should be able to:
1 Understand the behaviour of soil during consolidation (dissipation of excess pore water pressure), and determine the mechanical properties which characterise this behaviour from laboratory testing (Sections 4.1–4.2 and 4.7–4.8);
2 Use simple empirical correlations to estimate compressibility properties of soil based on the results of index tests (see Chapter 1) and appreciate how these may be used to support the results from laboratory tests (Section 4.3);
3 Calculate ground settlements as a function of time due to consolidation both analytically and using computer-based tools for more complex problems (Sections 4.4–4.6 and 4.9–4.10).
4 Design a remedial scheme of vertical drains to speed-up consolidation and meet specified performance criteria (Section 4.11).