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THE UNIVERSITY OF THE WEST INDIES
ST AUGUSTINE, TRINIDAD AND TOBAGO, WEST INDIES
FACULTY OF ENGINEERING
DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING
CVNG 2002 Soil Mechanics
Module 6. Consolidation Lectures by Richard Dean, Semester 2,
2006
Dump truck unloading fill material as part of a soil
consolidation contact, Wilson Bridge, Maryland
(www.wilsonbridge.com/photos )
Oedometers for laboratory testing, UWI
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Contents What is consolidation?
..................................................................................................................................3
The Oedometer Test
.....................................................................................................................................8
Calculating the Compression Curve
...........................................................................................................10
Typical results
.............................................................................................................................................12
Using the results to calculate long-term settlements
..................................................................................13
Using the results to calculate swelling
........................................................................................................14
Using the results to interpret geological history
..........................................................................................15
Simplifying the calculations (1) Index
formulation.......................................................................................17
Simplifying the calculations (2) Compressibility formulation
.......................................................................20
Casagrande's
Construction.........................................................................................................................22
Interpreting the data of time-dependency
...................................................................................................23
Applying the square-law to the oedometer
.................................................................................................28
Using oedometer data to estimate settlement-time
responses...................................................................30
Terzaghi's Consolidation Equation Derivation
.........................................................................................32
Exploring Terzaghi's Consolidation Equation
.............................................................................................35
Solving Terzaghi's equation
........................................................................................................................36
Interpreting time-dependent oedometer
data..............................................................................................44
Using consolidation theory to manage long-term settlements
....................................................................
47 Secondary
consolidation.............................................................................................................................49
Consolidation Lab
.......................................................................................................................................50
Review Questions
.......................................................................................................................................51
Previous Exam
Questions...........................................................................................................................55
Appendix A. Revision of Permeability and Darcy's
law...............................................................................59
References and further reading
..................................................................................................................67
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What is consolidation? If a load is applied to a foundation that
rests on clay or silt soil, there is usually an immediate
settlement, followed by settlements that increase as time
passes. This increase in settlement is usually due to consolidation
of the silt or clay. The rate of increase usually reduces as time
passes. Eventually the settlement rate is so small that the
settlement has stopped for all intents and purposes.
For example, suppose a house is built on sand overlying clay
overlying bedrock.
The house will take some time to build, and there will be some
settlements during that time. There will also normally be some
continuing settlements after the house-building has finished, as
shown below.
Example of a settlement-time graph for a heavily loaded
foundation with large long-term settlements
Time, years
Settlement, mm
0
50
100
150
1 2 3 0
Primary consolidation
Sand Soft clay
Rock
House being built
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The increase of settlements with time occurs because the
stresses in the clay due to the weight of the house tend to force
water out of the soil, so that the soil layer compacts. The sketch
below shows the results of this in an exaggerated form. The
immediate settlement is usually estimated using elastic theory. The
eventual, long-term settlement is estimated using compression
theory, and the final settlement depends on the overall
compressibility of the clay. The settlement takes place over some
time because it takes time for the water to flow out of the soil.
This is affected by the permeability of the clay. Long-term
settlements can be quite large. For example, for a foundation on
soft to firm clay, long-term settlements can easily exceed one
foot. The sketch below shows the values of ultimate settlements, in
mm, for square and strip footings on soft to firm clays. Typical
relation between footing width and ultimate settlement. Footings
loaded to q=50kPa, 1.5m below ground level on Mexico City Clay of
40% water content. Figure from page 407 of Terzaghi et al
(1996)
Footing width, metres0 2 4 6
0
100
200
300 Long-term settlement,
mm
Strip footings
Square footings
Water movements
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The settlement depends on the size of the footing. Also, the
rate of settlement can depend on size, but the distance involved is
the drainage path length. In the sketch below, the thickness of the
clay layer on the right is twice as great as on the left. Because
of the size effect, the maximum settlement is a little under twice
as much for the house on the right. The drainage path length on the
right is twice as long on the right as on the left. So the pressure
gradients pushing the water are half as much on the right, and the
distance to move is twice, so it takes 4 times as long to achieve
maximum settlements. This square-law of drainage path lengths is
quite general the time needed to achieve maximum settlement depends
on the square of the drainage path length.
Typical relationship between consolidation time and drainage
path length for a silty clay
(Note: time depends on coefficient of consolidation, which can
be very different for different clays)
2m 4m
1 day
10 days
3 months
3 years
30 years
1foot 3m 30m
Time to achieve 90% of long-term
settlement
Drainage path length
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As a result of the square-law, some large structures will still
be settling at the end of their design lives. For instance, the
design life of a large offshore concrete oil platform may typically
be 20 to 30 years. But the time required to achieve 90% its
ultimate settlement might be as long as 100 years.
One of the effects of consolidation can be to cause negative
skin friction to develop on piles. In
the sketch below, a building has been founded on piles that
transmit the weight of the building to the underlying stronger
soil. But the general area has also been covered over with sand.
The weight of the sand causes time-dependent compression of the
soft clay. As the clay settles, it drags downwards on the piles,
causing the "negative" skin friction.
Structures founded on overconsolidated clays are generally found
to behave somewhat differently. If we compare the long-term
settlements of several structures on stiff clays, the relation
between foundation load and long-term settlement is generally
non-linear.
Typical relation between load and long-term settlement for some
clays
Building supported by piles
Sand
Clay
Sand
Magnitude of footing load
Magnitude of long-term
settlement
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The difference between the behaviours of soft and stiff clays
can be explained by the theory of compression for silts and
clays.
There is also the opposite effect. If an excavation is made for
the basement of a building, then one of the effects of the
excavation is to reduce the stresses on the underlying soil. As a
consequence, the soil can suck water in, and expand. The expansion
is called swelling, and the upwards movement of the base of the
excavation is called heave. In an extreme case, the base of the
excavation will move significantly upwards. A similar effect can
occur if it rains, or if there is a water pipe burst in the
excavation. Swelling also occurs in expansive clays, except that
the effect is primarily due to wetting rather than unloading per
se.
One of the good things is that consolidation is generally
associated with an increase in strength of the soil. This increase
in strength is due to the effect of compaction. Conversely, there
is a small decrease in strength if swelling occurs.
In summary, consolidation is the time-dependent settlement of
silts or clays. It occurs whenever loads are applied or removed,
and is due to the fact that the changes of stress due to the loads
cause water to be pushed out or sucked in to the soil. This pushing
out or sucking in takes time, due to Darcy's law. In general, a
designer will need to estimate both the long-term settlements (or
heave) and the time-evolution of those settlements. Consolidation
theory is used for this purpose, consisting of the theory of
compression and the theory of permeability . The main laboratory
test to use to get parameters for the theory is the oedometer
test.
Original base of excavation
Base after some timeSaturated sand: a
source of water
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The Oedometer Test The oedometer is a laboratory device which is
designed to reproduce the conditions in a layer of soil that is
subjected to uniform vertical loading. These are the simplest
conditions that an engineer will need to analyze. For other
conditions, the engineer usually makes estimates based on the
simpler conditions, then adjusts the estimate to take account of
the increased complexity.
Simple field loading scenario ion which the oedometer test is
based. A thin layer of clay is located between sand and bedrock. A
wide uniformly pressured foundation is placed on top of the sand.
In the limit, the foundation is "infinitely wide". The clay
experiences the weight of the foundation as an additional load
q.
The clay layer is being loaded uniformly and vertically. The
clay layer cannot expand laterally if it did, then clay that was a
long way left or right would have to travel fast! We normally
assume that the clay experiences zero lateral strain. This is
called one-dimensional condition (the one dimension being the
vertical one). Thus we have one-dimensional loading, when an
increase in vertical load is applied, and one-dimensional
unloading, when the vertical load is decreased.
Sand
Clay
Rock
Wide foundation applying a uniform vertical load q
Small sample, taken for testing in the oedometer
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In the oedometer, a sample is confined in a stiff ring typically
a lubricated, highly polished stainless steel ring with a
diameter-to-height ratio of about 3. There will be a porous stone
at the top and/or base of the soil sample, so allowing water to
flow into the sample or out of it. In some devices, drainage is
allowed only at the top of the sample, and pore pressure is
measured at the base. There is a load cell and a settlement gauge
to measure vertical load and settlement. In some devices there is a
lateral load cell to measure the lateral force applied by the soil
to the steel ring.
Key features of the oedometer cell The typical test procedure is
to apply a load increment rapidly, and to the measure the
settlement at various times after the increment has been applied.
This is called the incremental loading procedure. A graph of
settlement versus the square-root of time is then constructed. This
usually allows a straightforward estimate to be made of when
sufficient time has elapsed to achieve the long-term settlement
value for that increment.
Load, usually applied by a hanger and deadweight
Settlement gauge
Drainage line, maintaining constant pore pressure at top of
sample
Drainage line, or pore pressure measurement line
Porous stone
Porous stone
Soil sample
Steel ring
Base plate
Loading piston
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Typical result for settlement versus time for several load
increments in an oedometer After the full amount of settlement has
been reached or nearly reached, a further increase of load is
applied, and the procedure is repeated. This incremental loading
procedure is usually repeated five to eight times, with the load
being doubled each time. Unloading can also be done. Usually this
involves halving the load in each increment. It is also possible to
apply load cycles, with the load being increased, then decreased,
then increased again, etc.
Calculating the Compression Curve The data for the long-term
settlements are usually plotted on a diagram of voids ratio
versus
vertical effective stress. This is called the compression curve.
If the test included some unloading and reloading sequences, these
are called swelling and re-compression.
To calculate the voids ratio, we use the simple phase diagram,
noting that air is ideally not
present, so that there is only soil and water in the oedometer.
Suppose that the voids ratio at the beginning of the test was eo
this is the ratio of the volume of voids in the sample to the
volume of solids.
time
settlement
Load increment 1 Load increment 2 Load increment 3
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Actual sample at start of test Phase diagram for the start of
the test Then, if the initial height was Ho, we can say that the
equivalent eight of solids was Ho/(1+eo). Suppose that the height
was H at a later stage of the test, and the voids ratio was e. Then
the equivalent height of solids is H/(1+e). So, if the soils are
incompressible, Ho/(1+eo) = H/(1+e). Hence we have
e = (1+eo).oH
H 1 (1)
Suppose the settlement was s. then H = Ho s. Putting this into
the equation gives:
e = (1+eo).
oHs1 1 (2)
The vertical effective stress on the soil is equal to the
vertical total stress less the pore pressure. We shall assume that
the pore pressure is zero at the end of each increment, when the
long-term settlement for that increment has occurred. Then the
vertical effective stress equals the vertical total stress, which
is just the load on the sample divided by its cross-sectional area.
Note that an area-correction is not needed, because of the
one-dimensional condition.
The results are usually plotted on semi-log paper or double-log
paper, and usually consist of several parts. The first part shows
an increase of vertical effective stress without much change of
voids ratio. This means that the sample is quite stiff at this
point. At some point the slope of the graph changes, and the change
of voids ratio per unit change of stress becomes much larger. The
point at which this occurs is called the yield point or
pre-consolidation point, the vertical effective stress at this
point is called the pre-consolidation stress. Sometimes, it is
difficult to estimate where this point is. A method called the
Casagrande construction is often used to do this in a formal
way.
Ho Water in proportion eo Solids in proportion 1
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Typical results The figure below shows some typical results for
a clay. The clay was from Lagunillas in Venezula.
In this case, the in-situ voids ratio was 1.92. The sample was
set up in the oedometer at point P, at a vertical effective stress
of about 25 kPa.
The sample was then loaded, and the initial response was quite
stiff, with a relatively rapid increase in stress with relatively
small reduction in voids ratio. However, the response became less
stiff around point A, which is interpreted as the end of the
elastic phase and the beginning of the elasto-plastic phase. On
further compression, the response followed the line ABD. On
unloading from D, the sample "swelled" slightly, with a small
increase in voids ratio as the stress was reduced along curve DE.
The response was again stiff, with relatively little change in
voids ratio for a relatively large change in stress.
On re-loading, the response was stiff from E to F, but with a
small hysteresis loop. There was another yield point at F, after
which the sample responded in the less stiff, elasto-plastic
manner. On unloading from G to H, the selling response was again
quite stiff.
Oedometer test results for Lagunillas clay, adapted
from Lambe (1961) and Lambe and Whitman, p.385, 1979)
P
D E
F
G
H
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Using the results to calculate long-term settlements We can use
the results to calculate long-term settlements. In the particular
case that was
investigated by labe 91961), the in-situ vertical effective
stress was about 60 kPa. For interest, however, let us consider a
site nearby, where the same clay exists at the same in-situ voids
ratio, but at lower in-situ stress of say 10 kPa. This would be
represented by the on the vertical axis where the arrow marked "e
in situ" points.
Suppose a house is to be built on a 3-metre thick layer of the
Langunillas clay. Suppose also that we have calculated that, in the
long term, the vertical effective stress on the soil will increase
to 50 kPa. The we can look along the measured curve to find the
point corresponding to 50 kPa which is at a voids ratio of about
1.84. Hence, using the phase relations, we can calculate the
settlement of the 3m layer:
o
o
eH+1 = 92.11
3+
m = 1.027m = esHo
+
1 =
1.841sm
+3
The first calculation gives the equivalent height of solids in
the layer, which turns out to be 1.027m. The second calculation
must also give the same value, since the soil particles themselves
are incompressible. From this we can work out the settlement s,
which turns out to be 3 2.84 x 1.027 m = 8.3cm.
What happens if we load the soil even more? If we apply a stress
greater than the value at A, then there will be much more
settlement, because of the lower stiffness along the curve ABDFG.
If we tabulate the settlements versus the stress level.
Long-term bearing stress, kPa
Final voids ratio from oedometer test results
Implied long-term vertical settlement, cm
Notes
50 1.84 8.3 Just before A 80 1.74 18.6 Just after A 100 1.63
29.9 200 1.33 60.7 400 1.09 85.4 Extrapolated just past D
We can also plot the results.
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0
10
20
30
40
50
60
70
80
90
0 100 200 300 400 500
Long-term bearing stress, kPa
Long-term component of
settlement, cm
Results for a 3-metre thick layer of Lagunillas clay with an
in-situ voids ratio of 1.92
We can use these results to interpret the graph of settlement
versus load on page 6. For small loads, the material response is
stiff and elastic, so there is not much settlement. For larger
loads, the yield point (such as A for Lagunillas clay) is exceeded,
and there is an increase in settlement relative to load. At larger
loads still, because of the curvature of the line ABFGH for
Lagunillas clay, the increase of settlement with load decreases
slightly.
Using the results to calculate swelling We can also use the
results to calculate the swelling potential for the base of an
excavation.
Suppose that, at a nearby site in a deeper layer of the same
clay, the in-situ stress is about 166.4 kPa, and the clay at that
location has reached the point B. This means that he in-situ voids
ratio here will be about 1.4. Suppose that the excavation will be
10m deep, and that it will reduce the long term vertical effective
stress on the Lagunillas clay layer by about 100 kPa.
To calculate the amount of swelling or heave, we can either do
another oedometer test, or we can "interpolate" from the data that
we have. If we interpolate, then we can probably expect that the
unloading curve from point B will be similar in shape and slope to
the unloading curves from D and from G. SO we might expect an
unloading curve like BC.
At point C, the stress has reduced by 100 kPa, and the voids
ratio is now 1.45 instead of 1.40. Suppose the thickness of the
clay layer here is 4m. Then the phase relation calculation looks
like this:
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o
o
eH+1 = 40.11
4+
m = 1.667m = esHo
+
1 =
1.451sm
+4
On the left side, the calculation fro the in-situ height of
solids now uses the in-situ voids ratio of 1.4 instead of 1.92. For
a 4m deep layer, this gives an equivalent height of solids of
1.667m. On the right, we can use the same equation, but the
"settlement" will actually come out as a negative value. In this
case, s = - 8.4cm downwards, which is equivalent to a "heave"
upwards of 8.4 cm.
Using the results to interpret geological history We know from
the results of Lagunillas clay that, if a soil layer is subjected
to an increase of load,
to a point like D, and then unloaded to E and reloaded, it will
have an initially elastic response from E to F, followed by
"yielding" at F and a less stiff elasto-plastic response from F to
G.
Suppose that, at a nearby site for the same clay, we find that
the clay is overlain by sand, and that the in-situ soil state is
represented by point P, with an in-situ stress of 25 kPa. For this
soil state, the yield stress can be measured from the oedometer
data, and is about 67.4 kPa. On this basis, we can imagine a
geological history which has three parts to it (see the following
figure):
(1) In the original deposition, material was deposited to a
higher ground surface, such that the
vertical effective stress reached 67.4 kPa. This might have been
deposition of the clay itself, as shown above or it might have been
another material.
(2) After the deposition, there was a process of erosion,
probably due to wind or water action. The erosion removed
sufficient clay down to the top of the clay layer as it is at
present. Thus the material was unloaded, and its state changed from
the point A to some point like U.
(3) After the erosion, the sand layer was deposited. This
increased the vertical effective stress on the
clay layer to the present value of 25 kPa. Now this history is a
little speculative, because there are some other more complicated
histories that can have the same effect. For instance, there might
have been a more complicated sequence for the sand. We can see,
however, that the maximum vertical effective stress that the soil
has experienced previously is an important parameter. We call it
the pre-consolidation stress. Thus the pre-consolidation stress for
the Lagunillas clay was about 67.4 kPa.
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Historical deposition ... followed by erosion then deposition of
overlying sand
Possible sequence of Geological Events for the Lagunillas clay
Our interpretation also means that the curve ABDFG is a line of
maximum historical stresses. For example, point D was reached
during the oedoemeter test, and this was the first time that this
sample for Lagunillas clay had ever experienced a vertical
effective stress of just under 400 kPa. We call the curve ABDFG the
virgin compression curve. It is the curve of first-time loading of
the clay. We call curves like D-E and G-H swelling curves. Curves
like PA and EF are called re-compression curves. These curves
normally form a "hysteresis loops", as shown for DEF. In fact, in
many calculations we assume that the hysteresis loop is absent, and
that the swelling and re-compression curves are straight lines on
the semi-log plot. We will look at this simplification shortly.
It's important to know the relationship between the in-situ
state of the soil and the virgin compression line. For example, if
the in-situ state is a long way below or to the left if the virgin
compression line, then a relatively large increase in stress will
be needed before the pre-consolidation stress is reached. We call
the ratio of the pre-consolidation stress to the in-situ stress the
over-consolidation ratio (OCR):
OCR = //
v
v
of value present of value historical maximum
(3)
For example for the point P for the Lagunillas clay, the in-situ
stress is 25 kPa, and the pre-consolidation stress is 67.4 kPa.
Therefore, the OCR is 67.4 / 25 = 2.7. We use the following
terminology:
Present ground surface
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OCR value terminology 1 Normally consolidated
Between 1 and about 3 Lightly over-consolidated Between about 3
and 8 Over-consolidated
Greater than 8 Heavily over-consolidated
Simplifying the calculations (1) Index formulation The
calculations of settlement and heave can sometimes be complicated.
This is fine if an
engineer has only one settlement to calculate. However, often
times there are many calculations to be done. For example, in a
feasibility study, the engineer might want to investigate several
different types of foundation solution, and several different
dimensions and depths for each solution. Also, if this is a major
construction development with many different buildings, there may
be many calculations to be done.
The next figure shows the usual simplification that is made.
Essentially, the virgin compression curve is considered to be a
straight line in the semi-log plot. The swelling and re-compression
curves are assumed to be the same as each other, and are assumed to
be straight lines of smaller slope. Because the actual data shows
curves, the straight line that best fits the data depends on the
range of voids ratio or stress over which the straight line is
drawn. Generally, the engineer will estimate the range that is
relevant for a particular design, and draw the straight lines
accordingly.
There are several ways to describe the straight lines depending
on what type of calculation is to
be done. One way is to use so called "indexes". In this way, the
equation for the virgin compression line is written as:
e + Cc.log10( /v ) = constant (4) where Cc is the compression
index. It is dimensionless. It is the slope of the virgin
compression curve in semi-log space, and we can define it as the
(negative of) the change of voids ratio per log cycle::
Cc = ))((log /10 vd
de
(5)
For the Lagunillas clay, the change of voids ratio over the
entire simplified graph is from 2.2 to 0.8, i.e a change of 1.4.
The change of stress along the virgin compression curve is from
about 30 kPa to about 780 kPa, so the compression index is about
1.4 / log10(780/30) = 0.99.
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0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
10 100 1000
Consolidation stress, kPa
Voi
d ra
tio e
Virgin compression line
Elastic lines (swelling lines)
P
A
B
D,F
G
H
C
Simplification for the data of Lagunillas Clay
Similarly, we can define the swelling index of a soil by writing
the equation for an elastic swelling line as: e + Cs.log10( /v ) =
constant (6)
where Cs is the swelling index. It is dimensionless. It is the
slope of the elastic lines in semi-log space, and we can define it
as the (negative of) the change of voids ratio per log cycle along
one such line
Cs = ))((log /10 vd
de
(7)
For the Lagunillas clay, the change of voids ratio from point G
to H was about 0.16. The stress reduced from about 400 kPa to about
50 kPa. The swelling index in this case was 0.16 / log10(400/50) =
0.18.
Values of the compression and swelling indices can be very
different for different soils. Some examples are given in the
following table, where the Atterberg limits are also shown. Some
Authors suggest that the compression index is correlated with
liquid limit or with plasticity index, but such correlations may be
limited in range. The best policy is always to make the measurement
directly.
E
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LL, % PI, % Cc Cs (a) Cs(b) Reference Clay minerals
(exchangeable ion) Attapulgite (Mg+2) 270 120 0.77 0.24 (1) Illite
(Na+) 120 67 1.1 0.15 (1) Illite (Mg+2) 94 48 0.56 0.18 (1)
Kaolinite (Na+) 53 21 0.26 (1) Kaolinite (Mg+2) 54 23 0.24 0.08 (1)
Montmorillonite (Na+) 710 656 2.6 (1) Montmorillonite (Fe+3) 290
215 1.6 0.03 (1) Clay soils - general Expansive soil A 84 36 0.14
0.25 (2) Expansive soil B 87 45 0.21 0.05 0.15 (2) Cincinatti clay
30 18 0.17 0.02 0.03 (3) Montana clay 58 30 0.21 0.04 0.07 (4)
Siburua clay 70 44 0.21 0.08 0.12 (3) St.Lawrence clay 55 33 0.84
0.04 0.08 (3) Undisturbed samples Boston blue clay 41 21 0.35 0.07
0.09 (3) Chicago clay 58 37 0.42 0.07 0.12 (3) Fore river clay 49
28 0.36 0.09 0.09 (3) Louisiana clay 74 48 0.33 0.05 0.08 (3) New
Orleans clay 79 53 0.29 0.04 0.08 (3) Remoulded samples Boston blue
clay 41 21 0.21 0.07 0.07 (3) Chicago clay 58 37 0.22 0.07 0.09 (3)
Fore river clay 49 28 0.25 0.04 0.04 (3) Louisiana clay 74 48 0.29
0.05 0.08 (3) New Orleans clay 79 53 0.26 0.04 0.09 (3)
Compression and swelling indices of some clay minerals and some
natural soils. Swell indices for stress ranges (a) 1000 to 100 kPa,
and (b) 100 to 10 kPa. From Lambe and Whitman (pp322-323, 1979).
Refs (1)
Cornell (1951), (2) Dawson, 1957, (3), Mitchell, 1956, (4)
Lambe-Martin, 1957
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Let us re-calculate some settlements for Lagunillas clay using
the compression and swelling index approach. Suppose that the
in-situ soil stress is 10 kPa, and the pre-consolidation stress is
67 kPa. Suppose the final long-term stress after the building has
been constructed will be 200 kPa. The long-term settlement will be
composed of two parts:
(1) the elastic settlement associated with elastic
re-compression, from the in-situ stress of 10 kPa to the
pre-consolidation stress of 67 kPa. For this calculation, we use
the swelling index, which we previous calculated as 0.18. The
change of voids ratio e is Cs.log10(67/10) = 0.15
(2) the elasto-plastic settlement associated with elasto-plastic
compression along the virgin compression line, from 67 kPa to 200
kPa. For this, we use the compression index. The change of voids
ratio is Cc.log10(200/67) = 0.47
The total change of voids ratio is therefore 0.15+0.47 = 0.62.
Recall that a 3m layer of material at an initial voids ratio of
1.92 contains an equivalent height of solids of 3 / (1+1.92) =
1.027m. Hence the settlement is 0.62 x 1.027m = 64cm. This compares
well with the value of 62cm which we got previously using the data
directly.
Simplifying the calculations (2) Compressibility formulation
Another approach that is often used to simplify the settlement
calculations is the compressibility approach. It is based on three
concepts that are related to each other: (1) The coefficient of
volume change, mv, is the most important. It is defined as the
volume strain
divided by the change of vertical effective stress. In
one-dimensional compression, the volume strain is the same as the
vertical strain v, which equals the (negative of) the change of
voids ratio divided by 1 plus the initial voids ratio:
mv = /v
v
= /).1( voe
e+
(8)
The symbol denotes a large change, giving a "secant" value. We
can also use tangent values, in which case mv is the tangent of the
graph of volume strain versus vertical effective stress.
mv is useful because e/(1+eo) is related to the settlement via
the phase relation In fact, the settlement s is just the initial
depth of the layer, Ho, multiplied by e/(1+eo). Hence the
settlement due to a change of vertical stress /v can be calculated
simply, as s = mv. /v .Ho. Its a sixty second calculation, or even
less!
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(2) The coefficient of compressibility, av, is defined as the
change of voids ratio divided by the change of vertical effective
stress. It is related to the coefficient of volume change:
av = /v
e = (1+eo).av (9)
(3) The constrained modulus, D, is defined as the change of
vertical effective stress divided by the volume strain. It is just
the inverse of mv
D = )1/(
/
o
vee +
= vm
1 (10)
These quantities can be used either with the oedometer data
directly, or with the simplificatoons. Let us do some calculations
using the data of Lagunillas clay.
(A) Small changes of stress: Suppose the in-situ soil state is
at point U, with a voids ratio of 1.92 and a vertical effective
stress of 10 kPa. Suppose the foundation loads will increase the
vertical effective stress to 50 kPa. From the data, the voids ratio
will reduce to about 1.84. On this basis: The coefficient of
compressibility is the change of voids ratio, 0.08, divided by the
change of
vertical stress, 40 kpa. Hence av = 0.002 kPa1. The coefficient
of volume change is this divided by (1+1.92), so mv = 0.00069 kPa1.
The constrained modulus is the inverse of this, which comes to 1460
kPa.
(B) Large change of stress. Suppose the foundation loads
increase the vertical effective stress to 200 kPa. From the data,
the void ratio will reduce to about 1.34, due to the fact that the
pre-consolidation stress of 67 kPa has been exceeded. Therefore:
The coefficient of compressibility is the change of voids ratio,
(1.92 1.34 = 0.58), divided by
the change of vertical stress, (200-10=190 kPa. Hence av =
0.58/190 = 0.0031 kPa1. The coefficient of volume change is this
divided by (1+1.92), so mv = 0.0010 kPa1. The constrained modulus
is the inverse of this, which comes to 956 kPa.
This shows that the coefficients and the constrained modulus
varies considerably with stress level. Is therefore always wise to
note the stress range over which the values apply.
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Casagrande's Construction The pre-consolidation pressure is an
important parameter, but it is sometimes rather difficult to
estimate from oedometer data. There are two reasons for this.
Firstly, the re-compression curve is estimated from only a few data
points, which are obtained at the end of the first few load
increments. One of the consequences is that there are gaps between
data points. Another reason is that the response of the soil is not
always simple to interpret. Professor Arthur Cagagrande, a famous
Austrian engineer who was professor at Harvard University, proposed
a way of estimating pre-consolidation pressure from oedometer data.
His method is illustrated and described below.
Casagrande's construction The method consists of 4 steps:
(1) Draw the curve of voids ratio versus the logarithm of
effective stress. Assuming that the virgin compression line BC is a
straight line, extrapolate it backwards
(2) Determine the point D of maximum curvature on the
re-compression part (AB) of the curve (3) Draw the tangent to the
curve at D. Bisect the angle between this tangent and the
horizontal.
Draw the bisector (DE) (4) Find the point of intersection of the
bisector DE and the extrapolated virgin compression curve
BC. The stress at this point is the pre-consolidation stress.
The construction provides a definite procedure for calculating
pre-consolidation pressure. However, it is not without drawbacks.
You are asked to use it in your lab report for the Consolidation
lab (page 50 below). Based on this experience, you are asked to
give your opinion of it.
Effective consolidation pressure (log scale)
Voids ratio
A
B
C
D
E
Pre-consolidation pressure
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Interpreting the data of time-dependency Let us now look at the
time-dependent aspect of consolidation. Recall that, in the field,
if a load is
applied to a body of soil, one of the effects is normally that
there are changes of pore pressures within the soil body. For
example, if a vertical force is applied to a pad footing, the
change of load causes the total stresses in the soil to change.
Some or all of this change is taken up by the pore water, so the
pore water pressure changes in response to the load.
Normally, the change of stress is different in different parts
of the foundation, so the changes of pore pressure are also
different. This means that a field of excess pore pressures is set
up within the soil body. The sketch below shows the magnitudes of
excess pore pressures induced in a later of clay when the
foundation is subjected to a load F that is applied "rapidly",
where "rapid" means fast in comparison to the time needed for
drainage to occur for the clay layer.
Magnitude of excess pore pressure effects in clay due to rapid
application of footing load F Because the excess pore pressures are
different, Darcy's law implies that there will be a flow of water
from high pressure regions to low pressure regions. The flow of
water through the clay layer takes time, and gradually tails off
because there is only a certain amount of water that can flow.
Consequently, the
CLAY
Vertical load F
SAND
High
Interrmediate Interrmediate
Low Low
Low
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foundations settles at a relatively high rate at first, but the
rate of settlement reduces and is eventually zero for practical
purposes.
For sands, we normally consider that no excess pore pressures
occur. In fact, excess pore pressures can arise in sand, but they
dissipate rapidly due to the high permeability of sand. However,
this is a relative matter, not an absolute one, and the effect
depends also on whether the sand is silty or not. Exceptions also
occur when the rate of loading is very rapid or cyclic, such as in
an earthquake or due to machinery vibrations or impact loading.
Wave loading of coastal structures can also induce significant
excess pore pressure effects.
We can use the oedometer to investigate the magnitude of the
pore pressures generated when a
load is applied. To do this, we close the lower drainage line,
and place a pore pressure transducer either in the line or
immediately beneath the porous stone. We can then apply a load to
the sample quickly, and measure the pore pressure response.
When we do this, we find a remarkable result. The change in pore
pressure measured at the base of the sample is initially exactly
equal to the change in stress due to the application of the load.
Then, as time
LoadDrainage line, maintaining constant pore pressure at top of
sample
Porous stone
Porous stone
Soil sample
Electrical transducer to measure pore pressure Electrical wires
to/from transducer
C B A
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passes, the pore pressure remains roughly constant for a while,
then begins to reduce at an increasing rate, then slows down and
eventually sops. While this is happening, the sample
compresses.
Typical response for excess pore pressures at the base of a
half-closed oedometer We can explain the first part of the time
record by using the compression data, together with Terzaghi's
Principle of Effective Stress. When the load is first applied,
there is no time for the pore fluid to move out of the soil, so the
void ratio of the soil is the same immediately after the load is
applied as it was immediately before the load is applied. But this
means, from the compression curve, that the vertical effective
stress must be the same just after compared to just before. Now
Terzaghi's Principle of Effective Stress implies that the total
vertical stress ( v ) equals the sum of the effective vertical
stress ( /v ) and the pore pressure (u):
v = /v + u (11) By applying the load quickly, we have increased
the total stress, but the effective stress remains as it is
instantaneously. It follows that the change of pore pressure must
equal the change of vertical effective stress. This is exactly what
is observed when we make those measurements in the oedometer.
Another way of thinking is the so-called spring analogy. In the
picture below, the soil particles are represented by springs inside
a piston that is filled with water. The piston has a small hole in
it. When a load is applied quickly to the piston, there is no time
to the water to flow out through the hole. Therefore, the springs
do not compress immediately. Instead, all of the applied load is
taken up by an increase of the pressure in the water. As time
passes, the extra water pressure forces water to flow out through
the hole.
Time
Excess pore
pressure
Time that load is applied
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So, the piston moves downwards, and the springs take up some of
the pressure. This means that part of the load is now being taken
by the springs, so the part taken by the water reduces, so the
water pressure reduces. This cause the flow of water to slow down.
Eventually, all of the applied load is transferred into the
springs. All of the long-term compression is achieved, and the
water pressure returns to zero.
Spring analogy for consolidation
Let us return to the oedometer on page 24. To investigate the
experimental results further, we can put additional pore pressure
transducers at various levels in the walls of the oedometer, such
as at levels A, B, and C. At each of these positions, there is also
a distance from the position to the top of the sample, where water
can flow out. So we should also expect that the change of pore
pressure, when the load is applied, will initially equal the change
of stress. This is exactly what is observed experimentally.
Typical responses for excess pore pressures at various levels in
a half-closed oedometer; A is closest to the base of the oedometer,
C is closest to the top drainage boundary
This kind of experimental data shows that the dissipation of t
he excess pore pressures starts earliest at the position closest to
the drainage boundary. This helps to explain why there is a delay
for the pore pressure at the base of the oedometer.
Time
Excess pore
pressure base of oedometer AB
C
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Another way of looking at the data is to plot the values of the
excess pore pressures versus position, at a given time. Typical
results are shown below.
Typical excess pore pressure isochrones in an oedometer with the
base drainage
line closed off, and development with time. Time T3 > Time T2
> Time T1 > 0
We can interpret these results as follows. At time T1, a short
time after the load is applied, there has been sufficient time for
some of the pore water to flow out of the soil, though the top
drainage boundary. Because of this, the void ratio of the soil
there has reduced slightly. The soil has compacted slightly, and
the state of the soil has moved along its re-compression curve. As
a result, the effective stress near the drainage layer has
increased slightly, and the pore pressure has reduced slightly.
The effect is largest closest to the drainage boundary, and
least at positions like A or at the base of the half-closed
oedometer. Because of this, there is a gradient of excess pore
pressure, and the gradient is greatest nearest the drainage
boundary, and least at the base of the sample. Now Darcy's law
states that the rate of flow of water through soil is proportional
to the gradient of the excess pore pressure, in other words, the
change of excess pore pressure with reference to position. This is
why the pore pressure reduces fastest near the drainage boundary,
and why there is hardly any change at first near the base of the
sample. The excess pore pressure gradient is least nearest the
base.
As time passes, more water flows out of the soil, and eventually
the rate of compression of the base of the sample begins to
increase. At time T3, the rate of flow of water upwards away from
the base of the sample is relatively large.
Base
Excess pore pressure
Top Time T0, immediately after load application
T1
T2 T3
A
B
C
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Not all the water flows out of the soil. As further time passes,
the excess pore pressures get gradually smaller, and eventually all
excess pore pressures have dissipated. At this point, all of the
soil in the sample has moved along the recompression line, and has
reached the place on the line where the vertical effective stress
equals the equilibrium value, with no excess pore pressure.
Applying the square-law to the oedometer Now that we understand
what is happening in the oedometer, we can start to apply our
knowledge to field situations. Before that, it is useful to
consider one further aspect. In most normal oedometer tests, the
drainage line at the base of the oedometer is open. What effect
does this have?
One of the effects is that settlements in the oedometer will
take 4 times less time. The reason can be understood by considering
the drainage path lengths in the fully-open and half-closed
situations. In the fully open situation, a molecule of water in the
middle of the sample has to flow only half the sample height to get
out of the sample. In the half-closed situation, the furthest that
a molecule has to travel is twice as much, from the base of the
sample to its top. Therefore, according to the square-law for
drainage path lengths, settlements should take four times as long
when the base drainage line is closed compared to when it is open.
Fully open: drainage top and bottom half-closed: only one drainage
boundary
Pathways for a molecule of water in the fully open and half
closed situations Another was of thinking about this is shown
below. For the half-closed case, we consider second oedometer
sample, immediately below the first, open at its bottom end. By
symmetry, a molecule at the centre of the double-height sample has
a 50% chance of exiting upwards and a 50% chance of exiting
downwards, so the effect is that there is no flow across the
centerline! This makes the situation similar to the fully-open
case, except that the height of the combined real-plus-imaginary
sample is twice as much. Hence the square-law again predicts that
the settlement rates will be four times slower for the half-closed
sample as for the fully-one one of the same height.
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half-closed fully open
Half-closed situation of depth D, and fully open situation of
depth 2D
Using these ideas, let us work out what lies behind the square
law. Suppose that the drainage path length is D, and an excess pore
pressure uxs has been set up in the sample. Now the excess pore
pressure gradients will be different at different parts of the
sample, but we can characterize the gradients as uxs/D. For
example, in the following situations, the same excess pore
pressures are occurring in the two samples, but distances are twice
as great in the sample on the right. The pore pressure gradient at
level X-X on the left is related to the characteristic gradient
uxs/D. The gradient at the corresponding level X-X on the right is
half as much, and the characteristic gradient uxs/D is also half as
much.
"Characterizing" the pore pressure gradients using uxs/D. The
drainage path length D on the right is twice as much as on the
left, but the excess pore pressures at corresponding levels are the
same. So the
excess pore pressure gradients on the right are half as much as
on the left.
D
2D
D
uxs
X X
D X X
uxs
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Using Darcy's law, we can characterize the velocity of the flow
of fluid by k times uxs/D, where k is the permeability of the soil.
Therefore, velocities are half as much if the drainage path length
is twice as long. But, if the drainage path length is twice as
long, there is twice as far for a molecule of water to travel.
Hence the time will be four times as much. Similarly, for a
drainage path length that is N times longer, the velocities will be
N times smaller, and the distance to be traveled will be N times
larger. So the time will be N2 times longer.
Using oedometer data to estimate settlement-time responses
Consider the situation of a footing that is to be placed on the
surface of ground, where the soil
profile consists of sand over clay over sand. Most of the long
term settlement will be in the clay. How can we estimate its value.
And how can we estimate how long it will take?
Let us start by estimating the change of stress that will be
applied to the clay layer. Of course this
depends on position in the clay layer, but we can take a
representative value by calculating the change of stress underneath
the centerline of the footing, at the level of the centre of the
clay. If we use the 1:2 load-spreading method, we can say that the
load spreads out at a gradient of 1:2, so that at 4m depth, the
load of 3200 kN is being supported on an area of 8m x 8m, giving a
stress increase of 3200/64 = 50 kPa.
Calculations for settlements and times for a footing on sand
over clay over sand
SAND
CLAY
SAND TO GREAT DEPTH
0m
-2 m
-6 m
4mx4m footing, load =320 tons
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Now, we take a sample of the clay from the centre of the layer.
We set the sample up in the oedometer, and we apply the same stress
increase. Our first task is to calculate the initial stress.
Suppose the bulk unit weights for the sand and clay are 17 and 16
kN/m3 respectively, and the water table is at the soil surface. The
in-situ total vertical stress 42m below ground level is then 17x2m
+ 16x2m = 66 kPa. If the unit weight of water is 10 kN/m3, the
in-situ vertical effective stress is 66 40 = 26 kPa.
So, we set the sample up in the oedometer at a vertical
effective stress of 26 kPa. Now the
process of sampling may have disturbed the sample a little, and
will certainly have altered the pore pressures within it, We will
need to allow for some time to elapse while the pore pressures in
the oedometer return to equilibrium values. Then, starting at 26
kPa, we increase the stress on the sample by 50 kPa, to 76 kPa.
This "models" the process that will occur in the field when the
building is constructed.
For the field situation, there is drainage top and bottom, so it
is a fully-open situation. Let us
suppose that the oedometer sample was 25mm deep at the time when
the stress has equilibrated to 26 kPa, and that the bottom drainage
line was open so that the sample was also in the fully-open state.
Suppose the graph of settlement versus time was as follows.
We can deduce many things from these results. Firstly, the
maximum, long-terms settlement in
the test was about 1.4mm. This was for a sample of initial
thickness 25mm. The actual clay layer has a
Time after the application of this load increment, hours
settlement, mm
0.5
1.0
1.5
0 0 1 2 3 4 5
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thickness of 4 metres, which is 160 times deeper than the
sample. Therefore the long-term settlement for the real clay layer
will be about 160 x 1.4mm = 0.224 metres. This is quiet a
significant amount. For instance, if the footing supports a house
and the floor of the house was at street level immediately after
the house is built, then the floor will be 0.224 metres below
street level in the long term. We will look later at ways of
reducing such large long-term settlements.
We can also make an estimate of the time that will elapse for
various degrees of settlement. For example, suppose we want to know
how long it will take for half the settlement to take place. In the
oedometer, half the long-term settlement had occurred after about
2.2 hours. For the field situation, the drainage conditions are the
same, because water can easily flow in the sand layers top and
bottom. Hence we simply need to use the square law for drainage
path lengths. The path length in the oedometer was half the sample
height (because there was drainage top and bottom in the
oedometer). Hence the length was 12.5mm. The path length in the
field was half the layer depth, i.e. 2m. The ratio of drainage path
lengths is 160. Hence the ratio of times is 1602 = 25600. Hence it
will take 25600 x 2.2 hours to reach 50% consolidation in the
field. This is equivalent to 6.4 years.
Terzaghi's Consolidation Equation Derivation
Considering a thin element of soil To derive Terzaghi's
consolidation equation, we consider a thin element of soil, between
vertical
coordinates z and z+z, where z is a small distance. We shall
suppose that the element is fully saturated, and the particles are
incompressible. This means that the only way that the element can
experience compaction is if water is forced out of the soil. For
simplicity we will assume that the properties of the soil element
do not depend on the position coordinate z. For instance, the
permeability at coordinate z equals the permeability at coordinate
z+z.
z
z + z
excess pore pressure gradient in eqn 12
excess pore pressure gradient in eqn 13
Total stress v Element of soil
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We will also assume, initially, that the total stress is
constant. This is true in the oedometer, for example, in the period
from immediately after a load increment has been applied, to the
time immediately before the next load increment. Later on, we shall
consider the time during which the load is changed.
Calculating the flow of water into and out of the soil Suppose
the excess pore fluid pressure in the soil is given by a function u
of z (and of time t).
Then we know from Darcy's law that the rate of flow of pore
fluid is proportional to the rate of change of this excess pore
pressure with position z. Consider the top surface of the element.
The rate of fluid flow into the element is given by:
vin = z at evaluated
xs
w zuk
. (12)
where k is the permeability of the soil, and w is the unit
weight of water. (Appendix A contains a reminder of how this
equation comes about). The derivative is evaluated at the depth z.
Now consider the bottom of the element. The rate of fluid flow out
of the element is:
vout = zz at evaluated
xs
w zuk
+
(13)
The derivative is evaluated at position z+z. We can evaluate the
derivative using Taylor's theorem, which states that the value of a
general quantity q, near a point at which it is known to be qo, is
just qo plus the derivative of q times the distance. In our case,
the value of vout will be the value of vin plus the derivative of
the right hand side of equation 12, multiplied by z. So:
vout = vin
zu
zk xsw
. x z (14)
The net rate of volume flow into the element equals vin less
vout, which is just the last term in the above equation. But we can
also calculate this by a different method which will involve a
small approximation.
Calculating the rate of volume change in terms of voids ratio
Let t be time, measured from the start of this load increment.
Suppose the voids ratio of the soil at
time t is e, and the voids ratio at time t+t is e+e. The phase
diagram at time t is shown below
z Water, in proportion e Solids, in proportion 1
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Based on this diagram, the equivalent height of solids in the
element is z/(1+e). Now if the voids ratio changes by de over times
t, then the overall height of the soil element will become this
times (1+e+de). Hence the element will expand by an amount e x
z/(1+e). If this expansion occurs in time t, then the rate of
volume decrease is the negative, so:
vin vout = te
e
+ .11 x z (15)
Now we can equate (vin vout) calculated in this way with (vin
vout) calculated from the flow of water into the element. This
gives:
te
e
+ .11 =
zu
zk xsw
. (16)
Relating voids ratio, excess pore pressure, and vertical
effective stress As mentioned earlier, we shall assume that the
total stress is constant. This is true, for example,
in the period from immediately after a load increment has been
applied, to the time immediately before the next load increment.
Now we consider two things
(1) In effect, we have assumed that the sum of the effective
stress and the pore pressure is constant. This means that, if the
equilibrium level of the pore pressure is fixed, then any change of
excess pore pressure (uxs) must be balanced by an equal and
opposite change of vertical effective stress ( /v ). Hence we can
say that uxs in the above equation is equal to /v .
(2) Second, we use the compression curve. This curve was
inferred from the experimental data in terms of effective stresses
instead of total stresses, so this curve still applies, even though
the settlement is continuing. We already defined the coefficient of
volume change mv. Based on our definition, we can replace e / (1+e)
in the above equation by mv. /v .
Making these replacements, and then taking the mathematical
limit in which the size z of the element tends to zero, and in
which the interval t of time tends t zero, we get:
t
u xs
= cv. 22
zuxs
(17)
with: cv = vw m
k. (18)
This is Terzaghi's equation for one-dimensional consolidation,
for constant total stress. The quantity cv is called the
coefficient of consolidation. It depends on the compressibility and
on the permeability.
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Exploring Terzaghi's Consolidation Equation Terzaghi's discovery
of the consolidation equation was quite remarkable. Several
concepts were
needed to derive it:
(1) The particulate nature of soils, so that compression
requires change of void volume (2) Flow of water through the voids
of soils, permeability, and Darcy's law (3) The phase diagram -
relation between volume change and voids ratio (4) The compression
curve relation between voids ratio and effective stress, and mv (5)
The principle of effective stress, that total stress equals
effective stress plus pore pressure
There are several interesting features of the equation itself.
One feature is that it is approximate,
partly because soil properties vary with the amount of
compression. This is particularly the case with the coefficient of
volume change mv, and to some extent true for permeability as well.
This means that the coefficient of consolidation changes as
consolidation takes place. In practice we get round this problem by
making adjustments for it.
The dimensions of the coefficient of consolidation can be
calculated by replacing the quantities on
the right side of equation 18 with their dimensions:
dim(cv) = )dim().dim(
)dim(
vw mk
= )../(1]).((/
2122 TLMTMLTL =
TL2 (19)
Typical units are m2/year or cm2/sec. Typical values of the
coefficient of consolidation depend on the Atterberg limits, and
also on the moisture content of the soil and on the effective
stresses. Lambe and Whitman (1979, page.412) quote the US Navy
(1962) who give the following ranges:
Liquid limit, % Range for cv in cm2/sec Range for cv in m2/year
30 % 0.001 to 0.04 3 to 0.120 60 % 0.0003 to 0.004 1 to 12 100 %
0.0001 to 0.0004 0.3 to 1.2
These values are not hard and fast rules and other values are
possible. In fact, cv probably depends considerably on the plastic
limit. Measurement of cv is one of the most important reasons why
the oedometer test is so useful.
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Terzaghi's equation is linear in excess pore pressure. In other
words, if we multiply the pore pressures on left and right by a
factor of 2, say, then the equation would be the same. The equation
is quadratic in distance. This is because z2 appears on the right
hand side, not z. This quadratic variation will help to explain the
square-law for drainage path length that was mentioned earlier.
Solving Terzaghi's equation
Plan of solution Terzaghi's Consolidation equation is not easy
to solve. Some solutions were developed by Taylor (1948). Suppose
we can find two solutions u1 and u2, both of which satisfy the
equation:
t
u 1 = cv. 21
2
zu
(20)
t
u 2 = cv. 22
2
zu
(21)
Then consider the linear combination u* = 1.u1+2.u2, where 1 and
2 are constants:
t *u
=
t uu( 1
+ ). 221 (substituting for u*) (22)
= 1t
u 1 + 2.
tu 2 (since 1, 2 independent of t) (23)
= 1.cv. 212
zu
+ 2.cv. 22
2
zu
(since each part satisfies the eqn) (24)
= cv. 2 22112 )..(
zuu
+ (since 1, 2 independent of z) (25)
= cv. 22 *zu
(substituting for u*) (26)
Hence u* also satisfies the Terzaghi equation. Extending this
further, Taylor realized that it might be possible to find
solutions in which the excess pore pressure was composed,
mathematically, of the sum of a number of terms. This idea led to
the Fourier series solutions in which the actual solution uxs is
taken to be the sum of a finite or infinite number of components,
the nth of which is un:
uxs = =n
nu (27)
As well as satisfying the Terzaghi equation itself, the
solutions must also satisfy certain boundary
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conditions. For example, the solution for a soil layer that has
sand above and below it is different from the solution for a clay
layer on impermeable bedrock. In addition to these geometrical
constraints, we must consider the initial conditions, which
represent the excess pore pressures at the start of an
analysis.
Fourier series solutions Taylor's (1948) solutions were based on
the idea of separation of variables. Let us suppose un is one of
the terms in the Fourier series. Taylor assumed that un could be
expressed as the product of two functions. One would be some
function Pn(t) which would be a function of time alone. The other
would be a function Qn(z) of position alone. Thus Taylor assumed:
un = Pn(t).Qn(z) (28) Putting uxs=un in Terzaghi's equation, the
left hand side of the equation evaluates to Qn.dPn/dt, because Qn
is independent of z. The right side evaluates to cv.Pn.d2Qn/dz2,
because Pn is independent of z. Dividing the results by PnQn then
gives:
t d
dPP
n
n
1 = n
vQc . 2
2
dzQd n (29)
Since the left side is a function of time alone, and the right
side is a function of z alone, it follows that both sides must be
constant. We normally represent the constant as 2nM .cv/D2, where
Mn is a number to be determined, and D is the drainage path length.
Then we can write:
t d
dPP
n
n
1 = n
vQc . 2
2
dzQd n = vn c
DM
2
2 (30)
We now solve the two sides separately. For the left side, the
equation gives:
n
nPdP = vn c
DM
2
2.dt (31)
This can be integrated directly to give:
Pn = Pn,o.exp( 2nM . 2.
Dtc v ) (32)
where Pn,o is some initial value which we will consider soon.
For the right side, the equation gives:
22
dzQd n = 2
2
DMn .Qn (33)
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This is a second-order differential equation whose solution will
in general involve a particular integral and a complementary
function. The complementary function is:
Qn = Qn,o1.cos
DzMn + Qn,o2.sin
DzMn (34)
Where Qn,o1 and Qn,o2 are some constants. Putting the two parts
of the solution together, and putting the results into the
summation, gives:
uxs = +==
+
n
n
vnnonnon
Dtc
MDzMu
DzMu )
..(..... 2
22,1, expcossin (35)
where un,o1 = Pn,o.Qn,o1 and un,o2 = Pn,o.Qn,o2. This turns out
to be a general solution for constant total stress. We can simplify
the result by defining dimensionless variables z/D for position,
and cv.t/D2 for time. The latter is called the time factor Tv:
Tv = 2.
Dtc v (36)
The time factor is one of the most important numbers in soil
mechanics. It is dimensionless, because cv has dimensions L2/T,
time t has dimensions T, and D2 has dimensions L2. Notice that the
D-squared appears on the bottom line. This is the reason for the
square-law for drainage path lengths, because for a given time
factor we have:
t = v
vcT .D2 (37)
Thus the drainage time for a particular time factor is
proportional to the square of the drainage path length (assuming cv
is constant). We will find later that the time factor is related to
the degree of consolidation.
The time factor appears in many different geotechnical
calculations. For many calculations, there are standard solutions
to the equations, expressed in terms of the time factor. We develop
one of these solutions below. The solutions mean that the engineer
just has to be able to calculate time factors. To do this, the
coefficient of consolidation needs to be known, together with the
drainage path length and the time t of interest. Once Tv is known,
the engineer will look up the relevant standard solution in a text
book, and use the time factor to select the answer to the question
being asked. In some case, the engineer has to back-calculate Tv,
and then deduce the time t. The procedures will be illustrated
later, but first we will develop one of the standard solutions.
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Solution for the fully-open, two-way drainage conditions In this
situation, a molecule of water in the center of a soil layer can
move either upwards towards
the upper drainage boundary, or downwards towards the lower
drainage boundary. It is convenient to set up the z-coordinate so
that one of the drainage boundaries corresponds to z=0.
Suppose a load is applied rapidly to the soil at time t=0. We
already know that the excess pore pressures will immediately rise
to the value of the applied stress. The applied load is taken by an
increase in pore pressure, and this increase causes water to start
to flow out of the soil. In fact, at the drainage boundaries, the
excess pore pressure returns to zero immediately. Hence we have a
mathematical condition that uxs=0 at z=0 and z=2D for all times
t>0.
In the general solution, we can make this happen by just using
the sine terms, and by arranging that the values of Mn are such
that, for the lower boundary, sin(Mn.2D/D) is always zero. This
means 2Mn is a multiply of . We could also arrange it by including
cosine terms, but these terms turn out to be not needed for the
case we are considering. Thus we have:
Mn = 2.n (38)
These results mean that the series solution has now reduced
to:
SAND
SAND
CLAY
v
z
Excess pore pressure
At t=0
immediately after t=0
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uxs = ( )+==
n
nvnnon TMD
zMu1
21, .... expsin (39)
The series now starts at n=1 because n=0 gives sin(0) which is 0
for the zeroeth term, and because the negative values of n give
sine values that are the negative of the positive values.
Consequently, we can include all of the possibilities by just
considering positive values of n.
To find the coefficients un,o1, we use the method of Fourier
analysis, applied at time t=0. According to this method, the
integral of the product of uxs with any sine function can be
calculated in two ways (a) by using the know conditions at t=0, or
(b) from the right side of the equation. For example, suppose we
multiply both sides by sin(z/D). Since uxs = v everywhere at t=0,
the left side evaluates to:
D
zxs dzD
zu0
..sin. =
0
.sin..
dDv = Dv .2 (40)
Where the substitution z/D = has been made. The right side
evaluates to a sum of integrals, but we find that only one of the
integrals is non-zero, the one with n=1. Making the same
substitution gives:
D
z
dzDzRHS
0
..sin. =
0
21,1 .sin.
dDu o =
2.1,1 Du o (41)
Equating the two results gives:
U1,o1 = 4 v (42)
Doing a similar calculation for all of the terms in the series,
we find that all of the even terms vanish, and that the odd terms
are proportional to 4/(n). We can then tidy up by putting n =
(2m+1), where m goes from 0 to infinity. The final solution is:
uxs = v . ( ) ( )=
0
2 ...2
mvTMMZM
expsin (43)
with: M = )12(2
+ m (44)
Z = Dz (45)
Plotting the results in terms of consolidation ratio Uz Results
for excess pore pressures are usually plotted in terms of the
consolidation ratio Uz, defined as 1 (uxs /v). The consolidation
ratio goes from zero at time t=0, to 1 at time t=infinity.
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Each curve is an isochrone of excess pore pressure (from the
greek, iso=same, chronos=time). The curve for Tv=0.05 is shortly
after the load is applied, where "shortly" depends on the value of
cv and D2. For instance, if cv=0.5 m2/year, and D=10 metres,
Tv=0.05 corresponds to t = 0.05 x 102 / 0.5 = 10 years!
Values of cv for silts can be several orders of magnitude larger
than for clays. A typical value for a silt could be 1000 times
large, say 500 m2/year, in which case Tv=0.05 would correspond to
0.05 x 102/500 years = 3 days. Values of cv can also be measured
for sands, and can be 1000 times greater than for silts. Thus
Tv=0.05 for a 10m thick sand layer would correspond to only 3/1000
days, or about a minute and a half. We can normally ignore time
effects in sands. Exceptions arise for very silty sands or clayey
sands, or in situations of rapid loading, such as earthquake
loading, wave loading, machine vibrations, impact loading (eg if a
ship hits a quay wall), or blast loading from explosions.
Consolidation ratio as a function of normalized depth and time
factor. Diagram from Lambe and Whitman (1979, p.408). In this case,
the time factor is represented as T
rather than Tv. The drainage path length is H rather than D.
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Solutions for other drainage conditions and initial conditions
Solutions for one-way drainage can be adapted from the solutions
for two-way drainage, as long
as the initial excess pore pressure distribution is a uniform
increase of pore pressure. For the situations below, the one on the
left would use the upper half of the diagram on the previous page,
from Z=0 to Z=1. The situation of the right would use the lower
half, from Z=1 to Z=2.
In some situations, it is possible for the initial excess pore
pressure distribution to have a
triangular shape. The sketch below shows solutions for the
excess pore pressure isochrones for this case, for the case of
two-way drainage.
Calculating the degree of consolidation ratio Uv The
consolidation ratio Uz at a particular time factor Tv depends on
the depth in the clay. To
calculate the settlement associated with the whole layer, we
need to take the average of Uz at a given time factor Tv. This
average is called the degree of consolidation, denoted as Uv. To
calculate Uv, we
Development of consolidation ratio with time factor for the case
of an initially triangular distribution of
excess pore pressures, with drainage top and bottom. Figure from
Lambe and Whitman (1979)
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integrate Uz with respect to depth z, and divide the result by
2D. For the case of an initially constant profile of excess pore
pressure versus depth, the solution is :
Uv = 1 ( )=
0
22 ..
2
mvTM
Mexp (46)
The same solution also applies fro triangular distributions if
there is two-way drainage. Uv is normally expressed as a %, so the
above value is multiplied by 100. Notice that Uv depends only on
Tv. Therefore we can plot a unique curve of Uv versus Tv. Curves of
Uv versus Tv for several drainage conditions and initial conditions
are shown below. The use of these standard solutions is very
straightforward. Suppose that the long term-settlement has been
estimated at 0.4 metres, based on the long-term compression curve
from the oedometer. Suppose
Relations between degree of consolidation and time factor, for
various drainage conditions
and initial conditions. Diagrams from pp.250-251 of Craig
(2005).
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there is two-way drainage, so that Curve 1 applies. Suppose that
cv = 2 m2/year, and D=5 metres. How long will it take to achieve
50% consolidation? To find out, we carry out the following
steps:
(1) Find the value of Tv on the relevant curve corresponding to
the required degree of consolidation,. For the present case, curve
1 applies. For Uv=50%, the Tv value is about 0.2 (the exact value
is 0.196, but 0.2 is accurate enough for most purposes)
(2) Back-calculate time from the value of Tv and the other
parameters. In this case, t = 0.196 x 52 / 2 = 2.5 years.
We can also do the calculations the other way round. For
example, to calculate the settlement after 1 year, we first
calculate Tv = 2 x 1 / 52 = 0.08. Looking at Curve 1, this
corresponds to a degree of consolidation of Uv = 0.31. Hence the
settlement after 1 year will be 0.31 x 0.4 = 0.12 metres.
It is obviously important to know the coefficient of
consolidation. It can be measured directly from the time-varying
stages of an oedometer test, as follows.
Interpreting time-dependent oedometer data Two methods have been
devised for interpreting oedometer test data and calculating
the
coefficient of consolidation. The log-time method was developed
by Arthur Casagrande. The root-time method was developed by Taylor
(1948). Both methods involve plotting the time-dependent data from
a load increment, and deducing the coefficient of consolidation
from the result.
The log-time method is illustrated in the following figure. The
bold curve represents data, plotted as the settlement gauge reading
versus the logarithm of time. The method involves two
estimates:
(1) 0% consolidation point. In theory, the initial part of the
curve should be approximately parabolic. Tow points on the initial
part of the curve are selected, such as A and B, for which the
values of time are in the ratio 4:1. The vertical distance between
them is measured. An equal distance above the first point is then
set off. This gives a gauge reading a0. This is taken as the point
of 0% consolidation. If this is the first load increment, then the
difference between a0 and the measured start of the test may be due
to initial compression of some air in the sample, for instance.
(2) 100% consolidation point. The middle of the curve can
normally be approximated as a straight line. Also, the end of the
curve, at large time, can normally be approximated by a different
straight lies. The two straight lines are drawn and their
intersection is taken to be the point of 100% primary consolidation
(a100).
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Since the two points represent the change from 0% to 100%
consolidation, it is possible to estimate the coefficient of volume
change, mv. This parameter was defined on page 20 above, and is
such that the settlement s is equal to mv times the change of
effective stress. For instance, the settlement in the above example
is about 2.2mm (a0=4.8mm, a100=2.6mm). If the height of the sample
at the end of the increment was 20mm, then the initial height was
22.2mm, and the vertical strain was 2.2/22.2 = 10%. If the change
of stress was 100 kPa, then the coefficient of volume change was
0.1/100 = 0.001 kPa1.
The point corresponding to 50% consolidation is located at a
settlement that is half-way between a0 and a100. According to
Terzaghi's theory, the value of Tv corresponding to Uv=50% is
0.196. Consequently, we can estimate the coefficient of
consolidation as:
cv = 50
2.196.0t
D (47)
where D is the drainage path length, and t50 is the time at 50%
consolidation. For example, for the data shown, t50 is about 11
minutes. If the sample height was 25mm and there was drainage top
and bottom,
The log-time method. From page 253 of Craig (2005), t is time
after the load is applied
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then drainage path length was D=12.5mm, so the coefficient of
consolidation is 0.196 x 12.52 / 11 = 2.78 mm2/minute, equivalent
to about 1.5 m2/year.
The second method of interpreting the time-dependent data is
Taylor's root-time method. In this method, the dial gauge reading
is plotted against the square-root of time after load
application.
Two constructions are then made:
(1) 0% consolidation point. A straight line is drawn through the
main part of the curve, and is extrapolated back to zero time. The
dial gauge reading at this point is interpreted as the reading (a0)
for 0% consolidation
(2) 90% consolidation point. A second line is drawn through the
zero point, with a slope equal to 1.15 times the slope of the
previous line. The point at which this second line intersects the
experimental curve is interpreted as the point of 90%
consolidation.
The root-time method. From page 255 of Craig (2005)
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The difference between the two points corresponds to 90% of the
estimated long-term primary consolidation. Hence the long terms
settlement can be estimated as 10/9 times the difference between
the dial gauge reading at the 0% and 90% points. From this, the
coefficient of volume change can be calculated in a manner similar
to the log-time method.
According to Terzaghi's theory, the value of Tv corresponding to
90% consolidation is 0.848.
Hence the coefficient of consolidation can be estimated as: cv
=
50
2.848.0t
D (48)
where t90 is the time at 90% consolidation. For the example
shown t90 is about 7.22 = 51.84 minutes. If the drainage path
length was 12.5mm, then cv was about 0.848 x 12.52 / 51.84 = 2.56
mm2/min, equivalent to about 1.3 m2/year.
The last part of the interpretation of a load increment is to
work out the permeability. Suppose that cv = 2.56 mm2/minute, and
the coefficient of volume change came out as 0.001 kPa1. From the
equation for cv on page 34, we can solve for the permeability
k:
k = w.mv.cv = 10 kN/m3 x 0.001 kPa-1 x 2.56 mm2/minute (49)
We obviously have to be careful about units! One solution is to
notice that the unit weight of water can actually be expressed as
10 kPa/m. Also, a coefficient of consolidation of 2.56mm2/minute is
equivalent to 2.56/60 mm2/second or 2.56/60 x 106 metres2/second.
Hence in his case:
k = 10 kPa/m x 0.001 kPa-1 x 2.56/60 x106 m2/sec (50) = 4.3 x
1010 m/sec
This is equivalent to 4.3 x 108 cm/sec, and is in the range of
typical values for clay (see Appendix A).
Using consolidation theory to manage long-term settlements We
have seen that the long-term settlements due to consolidation can
sometimes be quite large.
As engineers, we need to develop ways to solve this problem.
One method is by preloading. Suppose that a footing will be
subjected to a design load F, and the long-term settlement would be
S. In preloading, we apply additional load to the soil, before or
during construction. By doing this, we accelerate the settlements.
We can sometime arrange that the
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