ENV-2E1Y Fluvial Geomorphology
2004 - 2005
Drilling Rig for obtaining continuous cores in
continental shelf sediments for consolidation testing.
Typical sequence of continuous coring. Depths of
up to 60m+ have been achieved by this means
Slopes and related topics
Section 3 Consolidation behaviour of Sediments
N.K. Tovey
ENV-2E1Y:
Fluvial Geomorphology 2004 - 2005
Section 3
3. Consolidation behaviour of Sediments
A weight (P) is now lowered onto the top of the piston. No
load can be taken by the piston because the spring cannot
compress, and so all the additional load is taken by an
increase in the water pressure. If the weight is P Newtons,
then the detected rise in water level in the manometer will
be (Fig. 3.1b):-
3.1 Introduction
This section of the course deals with how soils behave under
load.
The term CONSOLIDATION is used to represent the
process which occurs under a steady load, whether naturally
applied or applied by man. There may be a loading arising
from glaciation or sediment deposition. Equally, there may
be sediment removal which will cause an unloading. Finally
consolidation may also occur through natural or artificial
ground water lowering.
P
A w
.......................3.1
where A is the cross sectional area of the cylinder
The tap is now opened and immediately the water level in
the manometer will begin to fall as water flows from within
the cylinder to the reservoir above the piston (Fig. 3.1c).
Eventually after a given time the water level in the
manometer tube will have fallen back to its original level
and at that time there is no excess pore water pressure and
the extra load had been taken entirely by the spring (Fig.
3.1d). The time taken to reach this equilibrium will depend
on the rate of flow possible through the fine bore tube.
The term COMPACTION is used to define a dynamic
process of loading either by man or by earthquakes. Some
textbooks use the term compaction to cover both
consolidation and compaction.
There are two separate aspects to the consolidation process
that we need to consider:1) the amount a soil compresses under a given
load
2) the rate at which the soil compresses
The process of consolidation within a soil is almost identical
to that described above. Initially, the whole of any extra
load is taken by an immediate rise in the pore water
pressure, and none of the extra load is taken by the soil
fabric. After consolidation starts, progressively more and
more of the additional load is taken by the soil fabric while
the excess water pressure slowly falls and at infinite time
will be zero once again.
We are interested in consolidation for several reasons:1) consolidation affects the properties of soils and in
particular the shear strength.
2) we often wish to known how much consolidation will
take place under a given load (and how fast,
particularly if we are involved in draining areas of land.
The most important factor governing the rate of
consolidation is the rate at which the water flows out. This
of course depends entirely on the permeability of the soil. It
is thus not surprising to note that Darcy's Law - i.e. the law
relating the velocity of flow to hydraulic gradient will be a
key consideration in the process of consolidation.
3) we may wish to use consolidation in environmental
reconstruction of the past form of the sediment (e.g.
thickness of ice during glaciation etc.).
What happens if we now close the stop cock and remove
the load from the piston.? Fig. 3.1d represents the situation
if we closed the tap after the consolidation
4) we may wish to correct for self weight consolidation in
Quaternary studies to correctly reconstruct previous
climates and sedimentation rates. Recent research by
Tovey and Paul (2001) suggests that errors of over 50%
may arise if such correction are not made.
As we remove the load, the spring initially cannot expand
and so the water is placed under suction and the level of
water in the manometer will fall by an equal and opposite
amount to that specified by equation 3.1 (see Fig. 3.1e).
The tap is now opened and the level in the manometer rises
as water is sucked in from the reservoir and the spring
relaxes. (Fig. 3.1f) Ultimately the piston and water level
will return to their respective starting positions (Fig. 3.1g).
3.2 Model Simulation (see Fig. 3.1)
The process of consolidation may be illustrated by a model
which consists of a "weightless" piston inside a cylinder
sitting on a spring. There is a pressure tapping in the
cylinder so that changes in the water pressure may be
monitored. At the base is a small hole connected via a stop
cock to a fine bore tube which passes outside the cylinder to
rest with its outlet below the water level in the reservoir
above the piston. Initially the stop cock is closed and the
system is in equilibrium with the water level in the
manometer at the same level as that in the reservoir.
While the process of consolidation of a soil could be
described fairly well by the model, the process of unloading
a soil is very different from that in the model. Fig. 3.2
which shows a plot of voids ratio against stress illustrates
this point.
28
N.K. Tovey
No Load
ENV-2E1Y:
Fluvial Geomorphology 2004 - 2005
P
P
A w
P
Section 3
u
P

w
A w
P

settlement
P
No Load
No Load

No Load
P
A w
Fig. 3.1. Model Simulation of the Consolidation Process
during unloading, then the process would be elastic and the
unloading curve would be identical with the loading curve.
In fact the unloading curve usually follows a second
exponential curve, but this time it is much lower than the
original loading curve. In extreme circumstances, the
unloading curve may be horizontal (i.e. there is no rebound
whatsoever).
A material which follows the same curve on loading and
unloading is known as an elastic material. A special case of
an elastic material is one which shows a straight line loading
and unloading relationship (e.g. steel below its yield stress).
A material which has no recovery at all (i.e. the unloading
line is horizontal is a perfectly plastic material.
Fig. 3.2 Schematic representation of consolidation of soils.
During consolidation, soil follows upper line.
During rebound, the lower line is followed.
Most soils exhibit both elasticity and plasticity.
Most soils exhibit some rebound and therefore may be
called elasto-plastic. On the graph in Fig. 3.2, the area
between the horizontal line and the actual rebound line
indicates the elastic recovery, while that between the loading
and unloading lines is the permanent plastic deformation of
the soil
During loading the soil follows an exponential decay curve.
If it were to follow the process described by the model
29
N.K. Tovey
ENV-2E1Y:
Fluvial Geomorphology 2004 - 2005
Section 3
In soil fabric terms, elasticity arises because during loading
some particles are bent and stored strain energy which is
released on unloading. The plastic component arises from
the collapse of the initial soil skeleton with the
rearrangement of the particles into a more parallel
alignment.
3.3 General Statements about Consolidation
In this course we need to consider the processes which are
occurring and to use some of the key results to solve
problems. As with the section on Seepage, there are some
sections which relate purely to the mathematically
derivation of the key formulae, and these sections are
"boxed", and may be omitted by those who are not
mathematically inclined. In most cases, there are graphical
solutions, or methods which require the application of
simple formulae.
Fig. 3.3 Illustration of the consolidation process. Water is
squeezed out from the element and the flow of
water outwards is greater than that flowing in.
First let us consider what form we would like the equation to
be in. Essentially, form the introduction we require an
equation which relates effective stress (or pore water
pressure) to both, time and the changing thickness of the
soil. To determine the governing equations, we consider an
element of soil of thickness z and area A. There are four
points to consider.
3.3.1 The Assumptions made
The are several assumptions made in the theory of
consolidation, but the key ones are: 1) Water is incompressible (i.e. it does not change in
volume on compression).
1)
The layer is compressing and as it does so water is
squeezed out, and the velocity of the water at the top
(assumed to be nearer the drainage surface) will be
greater than the flowing in. Also the volume of water
squeezed out will equal the change in volume of the
element of soil. We use this latter fact in one of the
governing equations, i.e. the CONTINUITY
EQUATION which in this situation may be specified
as:-
2) The soil particles are incompressible.
3) Only one-dimensional consolidation is present, i.e.
flows of water, settlement and the applied load all act in
one direction which is usually vertical. This assumption
is not necessary, but it makes the deviation easier if we
assume a purely 1-D situation.
4) Darcy's Law is valid
The change in volume of the soil element = net
outflow of water.
5) The initial excess load is initially taken solely by a rise
in pore pressure
2)
DARCY'S LAW describing the flow of water in a soil
may be used to specify how the velocity of the water is
changing.
A consequence of assumptions 1 & 2 is that the net flow of
water from an element of soil equals its reduction in volume.
DARCY'S LAW may be specified as:-
When we considered ground water flow in the previous
section, it was assumed there was no reduction in volume,
and no accumulation or net removal of water. In this
section we allow for volumes of soil to change. In
Hydrogeology, the term Compressible Aquifer is used and
this is exactly the same as a soil collapsing under
consolidation.
Velocity of water = coefficient of permeability x
hydraulic gradient .
3)
In attempting to work out the change in volume of the
soil element, it is only the voids that change in size, the
solid volume remains constant.
4)
During consolidation the extra load is assumed to be
taken initially by an excess pore water pressure and
that this pore pressure dissipates as the effective stress
increases as the soil fabric adjusts to the new load:-
Fig. 3.3 illustrates the consolidation process. The element
of soil has water flowing through it (upwards in this case),
but the flow of water is greater at the top because of the
water squeezed out from the volume itself.
3.3.2 Key factors affecting consolidation
thus the change in pore pressure
effective stress,
While we do not expect you to reproduce the derivation of
the consolidation equations, it is important you understand
the fundamental principles behind them.
i.e.
30
u
=
= 1
change in
N.K. Tovey
ENV-2E1Y:
Fluvial Geomorphology 2004 - 2005
The last two factors, together with the statement of
continuity and Darcy's Law, are sufficient to define a
consolidation equation relating how the pore pressure varies,
both with time and depth.
You should, however, be clear of the steps outlined above,
as what follows is nothing more than the mathematical
representation of the above.
3.4 Mathematical Derivation
In the analysis on the next three pages two terms are used
for convenience:-:
Continuity (i.e. the change in volume of the soil
element equals the net outflow of water). Let the area
of the element in Fig. 3.3 be A and its thickness be dz
i) mvc - this equals the gradient of the line at
the point in question on the voids ratio versus
stress graph multiplied by the factor:-
Change in volume = net outflow of water.
1
1  eo
i.e.
mvc 
Section 3
The R.H.S. is given by:-
1
de
.
1  e o d '
Q in  Q out  vAt  ( v 
mvc is known as the coefficient of compressibility
v
.z ) At .........3.3
z
so
and is of considerable importance in predictions of
settlement.
V
v
  At . z
t
z
Note: in some text books, mvc is given the symbol
mv.
............... 3. 4
where t is the time.
The factor 1 / (1 + eo) arises because the graph is
given as voids ratio, and we need to convert to total
volume of soil.
ii)
Cvc equals
NOTE:
V refers to volume
velocity in the above equation.
k
 w m vc
From DARCY (velocity of water in soil is
proportional to hydraulic gradient).
This
is a substitution
made merely for
convenience. Cvc is known as the coefficient of
consolidation as is of importance if we wish to
know the rate at which consolidation is proceeding.
It should be noted that this term includes k, the
coefficient of permeability.
v  ki   k
dh
k du

.
dz
 w dz
where u is the pressure of water.
Substituting this value for v in equation (3.4) gives
the net outflow of water in 1 second, i.e. the RHS of
equation 3.4 becomes:-
Note: in some textbooks, cvc is given the symbol
cv.
A
If we go through the full derivation we will eventually
obtain the desired equation, i.e.
 2 u u
cvc 2 
z
t
and v refers to

k du
k  2u
(
.
) z 
Az
z  w dz
 w z 2
..........................................3.2
or
This relates the rate of change of pore pressure with both
time and depth. This is a standard form of an equation
known as the diffusion equation for which standard
solutions are available in text books. This equation has
been solved and we do not need to resolve it as we can use
the graphical solutions to all our problems. If you are doing
the ENV-2A21?ENV-2A22 Mathematics Units you may
wish to attempt to solve the above as an environmental
application of maths.
. . . . . . . . . . . . . . 3. 5
==========
V
k  2u

A z
t
 w z2
...............3.6
NOTE: V for volume on LHS.
We need now to find an expression for the L.H.S. of
this equation.
To find such an expression (i.e. the rate of change of
volume with time) we must examine exactly how
volume change takes place.
From assumptions 1 & 2 it can only take place in the
pores, so one way to proceed is to consider changes in
the voids ratio.
IT SHOULD BE EMPHASISED THAT YOU ARE
NOT EXPECTED TO BE ABLE TO REPRODUCE
THE
FOLLOWING
ANALYSIS
CONTAINED
WITHIN THE BOX. IT IS INCLUDED FOR THOSE
WHO ARE MATHEMATICALLY INCLINED.
31
N.K. Tovey
ENV-2E1Y:
Fluvial Geomorphology 2004 - 2005
The true e -  relationship is an exponential curve, but
over a relatively small interval it is quite permissible to
approximate that part of the curve to a straight line (i.e.
what we are doing is replacing the true curve by a series
of straight lines.
If the initial Volume of the element is V and its voids
ratio is eo,
V = Vv + Vs and eo = Vv/Vs
or
Vv = eoVs
Substituting this and equation (3.9) into equation (3.8)
gives
where Vv is the volume of the voids, and
volume of the solids.
Thus
V = Vs(eo + 1) or
But
Vs the
Vv
u
 Az. mvc .
t
t
V
Vs 
1  eo
which is the L.H.S.
of the continuity equation.
V = Az
The full continuity equation (cancelling the Adz) then
becomes
hence the volume of voids
Vv 
Section 3
u
k  2u

.
t
 w z 2
u
k
 2u

. 2
t
 wmvc z
eV
e

. Az
1  eo 1  eo
mvc
Now the rate of change of Volume of soil = rate of
change of volume of voids
i.e.
or
Normally the constants in the above equation are
combined into a single constant:-
V Vv
A
e


. z.
............... 3. 7
t
t
1  e0
t
cvc 
k
 w mvc
so
[Note:
u
 2u
 cvc . 2
t
z
eo is a constant]
We can manipulate the above expression by making
use of the differential relationship
. . . . . . . . . . . . . . . . . . . . . . . . 3. 10
This is the equation describing one-dimensional
consolidation and is of a standard mathematical form
known as the diffusion equation. There are several
analogies including those involving the flow of
electricity or the flow of heat.
dy dy dt

x
dx
dt dx
Equation 3.7 can thus be written as:-
mvc is known as the coefficient of
Vv
A
e '

. z.
.
............... 3. 8
t
1  e0
' t
compressibility.
cvc is known as the coefficient of consolidation.
However, during consolidation under any one load,
there is no change in the total stress, so changes in pore
pressure equal changes in effective stress,
3.5 Consolidation behaviour of Soils
i.e.
3.5.1 Amount of Consolidation with Load
'
u

t
t
............................. 3. 9
If we now make an approximation and say that
e
'
is a constant during a particular increment of load, we
can replace the terms
1
e
.
1  eo '
by a constant
- mvc .
Fig. 3.4 Typical Voids ratio - effective stress curve for a
soil
32
N.K. Tovey
ENV-2E1Y:
Fluvial Geomorphology 2004 - 2005
The consolidation behaviour of soils studies both how much
a soil consolidates under a given load and also how fast it
consolidates. The behaviour with load is normally displayed
as a graph as shown in either Fig. 3.4 or Fig. 3.5.
Section 3
see below), or in transition through dissipation of pore water
pressure (also see below).
We can replicate the above process in the laboratory doing a
consolidation test similar to the one you have or will be
doing in a practical session. We use the consolidometer to
measure the consolidation characteristics by adding a series
of loads, allowing the soil to reach equilibrium and then
measure the settlement before adding the next load. The
point at which equilibrium is achieved is measured
separately and is covered in a later section.
After sedimentation a soil will normally be under an
effective stress consistent with the weight of sediment
above. It will initially have an open structure so that the
voids ratio is high as indicated by point A in Figs 3.4 and
3.5. The exact voids ratio depends on the actual clay present
but may be as high as 10 or more.
As more sediment is laid down on top, the soil compresses
and follows the curve AB. At B the soil is unloaded
(perhaps following ice retreat after glaciation, or the erosion
of sediments above and follows the unloading curve BCD.
In extreme cases, and particularly if the soil is largely sand,
the unloading curve may be horizontal.
3.5.2 Simple Environmental Reconstruction using a
consolidation test
Fig. 3.6 shows a typical plot from a soil which has been
sampled. From the borehole log and the unit weight
information we can estimate the current in situ stress level
on the sample (see the worked example given as a handout
at the end of the introductory section). [ Remember to
allow for the buoyant effect of water!!]
If the soil is loaded again it tends to follow a line slightly
above the original unloading curve as shown by DEB (i.e.
there is a hysteresis loop BCDEB), until the loading reaches
the previous maximum stress level as denoted by B. After
this the loading curve follows a continuation of the original
curve AB along BF.
The loading and unloading sequences may be studies more
readily if the X-axis (abscissa) is plotted as the logarithm of
the effective stress as shown in Fig. 3.5.
Fig. 3.6 Example of Environmental Reconstruction
Fig. 3.5 shows that all the consolidation and unloading lines
become straight lines, and both the unloading and reloading
lines approximate to the same line and it is convention to
treat these as being the same.
When we extract the sample from the borehole, the in situ
stress is released as we extract the core. At this point we do
not know the unloading curve. We place the same in a
consolidometer and load the sample in the normal way. It
will follow an approximately straight line DB (in Fig. 3.6).
The point X represents the previous in situ load, and not
infrequently the sample will continue on beyond this to the
point B, where the line will kink and follow the line BF.
BF clearly represents the VIRGIN CONSOLIDATION
LINE, while DXB is the reloading line.
These consolidation curves are very powerful tools not only
in understanding how a soil will behave in the future, but
also to attempt environmental reconstruction as to what has
happened in the past. The line ABF is known as the
VIRGIN (or NORMAL) CONSOLIDATION LINE.
Sometimes both B and X coincide, in which case the sample
is normally consolidated and has NEVER in the past been
loaded beyond its present level of loading. More usually, B
is to the right of X which indicates that the soil is
OVERCONSOLIDATED
A soil can only exist in EQUILIBRIUM in the region below
and to the left of the line ABF. The soil may temporarily
exist above and to the right of the line, but it is not then in
equilibrium and will be unstable (in the case of quick clays
We may read off the previous maximum consolidation stress
from the graph. If we subtract the present in situ stress
level, the difference will represent the previous additional
load which was on the soil. If we know that this came from
Fig. 3.5 Voids ratio against logarithm of effective stress
33
N.K. Tovey
ENV-2E1Y:
Fluvial Geomorphology 2004 - 2005
glaciation (from a study of the geology of the area), the
since the unit weight of ice is 9 kN m-3, we can readily
estimate the thickness of ice that covered the area. Equally,
the additional loading may have arisen from additional
sediment which has subsequently been eroded. Whatever
the cause, this gives us a method to establish what the
magnitude on the loading was.
Section 3
be equal to zero. The lines of pressure distribution at the
different times are
called
ISOCHRONES.
The term OVERCONSOLIDATION RATIO (OCR) is used
in describing soils.
Fig. 3.7 Isochrones
showing distribution
of excess pore water
pressure with both
time and depth in
sample.
This is defined as:-
OCR
=
Previous Maximum Consolidation Pressure
--------------------------------------------------------Current In situ Effective Stress
Normally consolidated soils have an OCR of unity, while
heavily over consolidated clays like the London Clay may
have OCR > 100 +.
Isochrone ACB show pressure distribution a short while
after consolidation has started (i.e. time t1)
When we come to discuss the shear behaviour of soils, we
shall group soils into two classes:-
Subsequent isochrones ADB, AEB at times t2, t3 are for
increasing time intervals from start of consolidation.
a)
those which are normally consolidated or lightly
overconsolidated
b) those which are heavily overconsolidated
Only at infinite time will the pore water pressure be zero
throughout the depth of the sample
The distinction between the two classes occurs at an OCR of
1.7
The fundamental equation governing
mentioned above: - i.e.
u
 2u
 cvc . 2
t
z
3.5.3 Rate of Settlement
The rate at which consolidation proceeds depends on the
rate at which the excess pore water pressures developed
following the application of a load subsequently dissipates.
We have noted earlier that this rate is dependant on the
permeability of the soil, and in fact we can use the
consolidation characteristics to estimate the permeability of
clays.
where
cvc 
consolidation was
. . . . . . . . . . . . . . . . . . . . . . . . 3. 10
k
 w mvc
cvc is known as the coefficient of consolidation.
and mvc is known as the coefficient of compressibility.
and k is the coefficient of permeability.
Fig. 3.7 shows the excess pore water pressure profile with
depth within a sample. Initially as the load is applied the
excess water pressure is uniform with depth and equal to
that of the applied stress. However, those parts of the soil
close to a drainage surface (or a permeable layer such as
sand in the field) will rapidly drain and dissipate the pore
pressure while those near the centre will change little. Thus
at both drainage surfaces the excess pore pressure will fall to
zero immediately after the load has been applied, but
elsewhere it will be some where between the original
increment of the pore water pressure and zero.
This equation describes how the pore pressure u varies both
with depth and time. The equation is of a standard form
which mathematicians recognise as the diffusion equation.
We may apply this equation either by using a solution of the
diffusion equation or by graphical means. For most purposes
a graphical solution is adequate, however, as the equation
can be solved once and for all. The curves are similar to
those shown schematically in Fig. 3.7 and are shown on the
chart in Fig. 3.12.
In theory we could compute a set of curves for each soil
type and each thickness of sample, but as will be seen later,
and following a similar unification procedure which was
adopted in section 1 relating to application of the Atterberg
Limits, we can find an unique set of curves If we do that
we shall have a single set WHICH ARE VALID FOR
ALL SOILS AND FOR ALL THICKNESSES.
The line showing the pressure distribution a short while after
consolidation has started is indicated as curve ACB in Fig.
3.7. At a later increment of time, more of the pore pressure
will have dissipated and the excess pore water pressure will
be as depicted by line ADB. At subsequent intervals of time
the pressure distributions will be AEB, AFB etc. Only when
the time reaches ¥ will the pressure distribution with depth
be constant again. This time the pressure will everywhere
34
The Mathematical solution to equation 3.10 is a
standard application of the diffusion equation and may
be solved by an expansion using fourier series.
N.K. Tovey
ENV-2E1Y:
Fluvial Geomorphology 2004 - 2005
Thus U = 0 at the start of consolidation, and U = 1 on
completion of primary consolidation. Note the symbol is U
(i.e. upper case),. Do not confuse with the symbol u (lower
case) which is used for pore pressure.
The excess pore pressure at time t and depth z into the
sample is given by:2 
nz 
utz  n1
.(1  cos n ). sin
.e
n
2H

n2  2 c vc t
4H2
Section 3
U not only indicates the proportion of settlement that has
......... 3. 11
taken place, but also the proportional dissipation of the pore
water pressure.
where H is HALF the thickness of the sample in a sample
where there is drainage both above and below. H equals
the thickness if either the layer above or below is
.impermeable.
In the next section we shall derive a relationship between
and Tv.
U
FOR THOSE OF YOU WHO ARE NOT
MATHEMATICALLY INCLINED YOU MAY SKIP
THE NEXT SECTION.
3.5.4 Non-dimensional Groups
To unify the relationships so that we only have one set of
curves we must introduce two non-dimensional parameters
We shall use the graphical relationship in this course.
3.5.5 Derivation of U - Tv Relationship
The first non-dimensional group is the time factor Tv
In terms of the pore water pressure ut at a particular
depth z at time t, U(t,z) may be specified as:-
where
Tv 
where
cvc t
D2
. . . .. . . . .. . . . . .. . . . .. . 3. 12
U( t , z )  1 
t is time, and D is the appropriate thickness as
where uo = pore water pressure at
Ds the load increment.
shown below.
Note:
ut
u0
D is the full thickness of layer in single drainage
situations (i.e. where there is an impermeable layer
either above or beneath the layer in question).
t = o and equals
While this is the degree of consolidation at a particular
depth, the mean degree of consolidation will be
obtained by integrating the above expression over the
full height of the specimen and dividing by the
thickness.
D is half the thickness of the layer in double
drainage situations.
For a double drainage condition:i.e. D is the maximum drainage distance.
Ut 
The second non-dimensional group is the degree of
consolidation U,
U
where
t
and
t
. . . . . . . . . . . . . . . . . . . . . . . . . . 3. 13

= consolidation at time =

2H
0
(1
ut
.z )dz .......3.14

Substituting equation 3.11 giving the general formula
for Ut - z into the above equation gives after some
rearrangement
t
 = consolidation at time = 
U 1
t
1

 2 H
.
T
8
exp(  2 v ) 
2
4
 
exp( 9
9
2

Tv
T
) exp( 25 2 v )
4 
4  ..... ...(3.15)
25
This is the exact form of the equation. There are several alternative approximations, but since the equation is valid for all
soils and for all layer thicknesses, it need only be solved once and the values presented in tabular form, as shown in section
3.6 below.
Equation 3.15 has been solved and the relationship is
recorded in Table 3.1 which may be used for all analyses.
In some applications, a particular value of the degree of
3.6 Application of U - Tv Relationships in
the Consolidation Test
35
N.K. Tovey
ENV-2E1Y:
Fluvial Geomorphology 2004 - 2005
consolidation U is required, in which case,
corresponding value of Tv can be read from the table.
0.80
0.85
0.90
0.95
0.98
0.99
1.00
the
In other situations it is helpful to plot out the data from the
table so that intermediate values can be readily estimated.
Remember the degree of consolidation in this table refers to
the mean consolidation throughout the depth of the sample.
Near the drainage surfaces, the degree of consolidation is
always greater, but in most applications it is the MEAN
value we need rather than the specific consolidation at a
particular depth.
Table 3.1:
Section 3
0.567
0.683
0.848
1.130
1.510
1.980

0.503
0.567
0.636
0.709
0.754
0.770
0.785
U - Tv Relationships
Table 3.1 is VALID FOR ALL SOILS AND FOR ALL
DEPTHS provided that there is double drainage. A similar
table could be constructed for single drainage conditions.
An approximation to equation 3.15 is
U
2
. Tv

U is plotted against Tv, a straight line will be obtained
2
 1. 128

of gradient
so if
Fig. 3.8 Plot of settlement against square root of time
Therefore, to find out where 90% consolidation is, we need
to first draw a straight line through the first few points of
our U against ÖTv plot (Fig. 3.8). [Actually Fig. 3.8 is a
plot of settlement against square root of time rather than
being in non-dimensional groups. This does not matter as
an illustration as the two are related. The actual curve
shown is how it would appear in actual practice].
These approximate values are shown in the third column of
the table Note, they follow the exact values (from equation
3.15) up as far as 50% consolidation and we may make use
of this fact to predict, in advance, when the completion of
primary consolidation will occur. We may do this by
plotting degree of consolidation against time when a
straight line should be evident up to 50% consolidation.
This line is show as PB in the figure. At ANY convenient
point on this line we measure the x - distance (e.g. AB), and
then step of a distance 1.155 times this (i.e. AC. A line is
then drawn through the point C and the intercept of the
practical curve with the y-axis (point P). This constructed
line is extended and will intercept the practical curve at the
90% consolidation point (corresponding to point E on the y axis).
To do this we note that at up to 90% consolidation, the socalled 'exact' solution follows the practical curve and it is
only beyond this point that secondary consolidation effects
become of importance. Reading values for 90%
consolidation from the above table gives values for T v of
0.848 ('exact') and 0.636 ('approximate'). Since we shall be
plotting our graph of U against T v we must take the
square root of both of these values:'Exact' solution for U = 0.9, Tv
'Approximate'
Once this point is defined it is a simple matter to locate F
such that PF is 10 / 9 times PE, and then to find when the
practical curve actually crosses this amount of settlement at
the point G. Once we have done this it is a simple matter to
determine the time by reading off the value on the x - axis
and squaring it.
= 0.848
i.e. square root = 0.921
0.636
i.e. square root = 0.798
The ratio of these values (0.921/0.798) is 1.155.
Degree of
Consolidation (U)
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.50
0.55
0.60
0.65
0.70
0.75
Time Factor (Tv) exact
0.002
0.008
0.018
0.031
0.049
0.071
0.096
0.126
0.197
0.239
0.287
0.341
0.403
0.476
The point G defines the completion of primary
consolidation. In theory, primary consolidation will only be
achieved at infinite time as it takes this long for the initial
excess pore water pressure to dissipate.
The theory
assumes that consolidation proceeds because of an initial
excess pore pressure which progressively dissipates as
consolidation proceeds. The theoretical curve is a very good
approximation to the actual practical curve up to about 95%
consolidation.
Time Factor
Approximate
0.002
0.008
0.018
0.031
0.049
0.071
0.096
0.126
0.196
0.238
0.283
0.332
0.385
0.442
However, after that point, the practical curve deviates as a
second
phenomenon
known
as
SECONDARY
CONSOLIDATION starts to take place. In practice,
secondary consolidation is taking place all the time, but it is
mostly very small compared to the primary consolidation. It
36
N.K. Tovey
ENV-2E1Y:
Fluvial Geomorphology 2004 - 2005
Section 3
is only when the primary consolidation is nearly complete
that secondary consolidation becomes noticeable.
3.7 Alternative Method to determine the completion of
primary consolidation
Secondary Consolidation appears to continue indefinitely,
and some tests have been running for many years do seem to
confirm this.
The consolidation during secondary
consolidation appears to be linear when settlement is
plotted against the logarithm of time, and we may use this
fact in an alternative construction to determine the
completion of primary consolidation
T
Fig. 3.10 Estimation of mvc
The consolidation curve of voids ratio against effective
stress (NOT logarithm of stress) is plotted in the normal
way. At the appropriate stress level, the tangent to the
curve is drawn, and the slope of this tangent is used to
derive mvc
e
This gradient is '
and it is a simple matter to divide
this by 1 / (1 + eo) to get
mvc 
Fig. 3.9 Log time versus settlement method to determine
completion of primary consolidation.
The curve ABC is usually 'S'-shaped (Fig. 3.9). The
construction proceeds by projecting the final linear section
backwards towards the Y - axis C - F. The tangent to the
point of inflection of the curve is also drawn (D - B - E) to
intersect FC at G. The amount of settlement corresponding
to this defines the 100% consolidation. A horizontal line is
thus drawn through G to intersect the experimental curve at
H which defines the point of completion of primary
consolidation. Finally the time is read off from the X - axis.
1
e
.
1  eo  '
(see Fig. 3.10)
3.9 Applications of Consolidation in Field
Situations.
This section explores application of consolidation to study
real problems in the field. These include further aspects of
environmental reconstruction over an above that already
covered in section 3.5.2
3.9.1 Settlement Computations - using mvc
The values for the time to 100% consolidation estimated by
this method is usually within 10 - 15% of that obtained by
the method described in section 3.6.
One main interest in the consolidation behaviour of soils is
to be able to predict the settlement behaviour in a soil. For
this we need to remember that:
3.8 Measurement of mvc
mvc 
The measurement of mvc is of importance for settlement
calculations. Fig. 3.10 shows how this is done.
1
e
.
1  eo  '
........................3.14
if the cross section of an element of soil has area
thickness dz , the initial volume V is given by
V = A dz,
and the volume of the solid (Vs) =

37
V
Az

1  eo 1  eo
A and
N.K. Tovey
ENV-2E1Y:
Fluvial Geomorphology 2004 - 2005
If we denote the settlement or compression of the layer by r
then the change in voids ratio is given by:
e 
Section 3
not have to be the same thickness, but it does simplify
calculations later if you choose layers of the same thickness.
 A
 (1  e o )

Vs
z
(Remember the volume of the solid does not change.)
If we now substitute this expression for e into equation
(3.11) we obtain
mvc 

z.  '
or the settlement () is given by
m vc .z.' ..........3.15
This is the basic settlement equation which means that if we
know the magnitude of the load applied and the thickness of
the layer, we may estimate the settlement of the layer by
measuring mvc on a voids ratio versus stress plot (i.e. e  plot).
Fig. 3.11 Section used in example.
Stress at top of clay initially = 5 x 19 - 4 x 10 = 55 kPa
|
buoyancy effect.
Normally our soil is composed of several different layers
and so the total settlement is given by

 mvc . z. '
Hence initial stresses at mid-depth of the 5 layers are:
layer 1
layer 2
layer 3
layer 4
layer 5
for all layers
Note that thick layers of soil must be subdivided and
typically layer thicknesses greater than 2.5 - .3m should be
avoided. For layers close to the surface, the thickness should
be reduced. The reason for splitting thick layers is because
the value of mvc varies with load and the lower sections of
a thick layer will be subjected to greater stresses and hence
will have a different value of mvc.
=
=
=
=
=
59.5
68.5
77.5
86.5
95.5
As water table is lowered, the lower 4m of sand will no
longer experience buoyancy forces, and thus the surcharge
in load = 40 kPa for all layers.
The best way to proceed from here is to lay computations
out in tabular form.
Example:
Layer Mean ThickDepth ness
(m)
(m)
A layer of clay of unit weight 16 kNm-3 is 7.5m thick and is
overlain by 5m of sand of unit weight 19 kNm-3 (Fig. 3.14)
The water table is initially 1m below the surface. After
pumping the level of the water table is lowered by 4m.
Estimate how much the ground surface settles if the
following data were obtained from a consolidation test on
the clay.
Effective Pressure
(kPa)
57
112
224
The situation is shown in Fig. 3.14
55 + 0.75 x (16 - 10)
55 + 2.25 x (16 - 10)
55 + 3.75 x (16 - 10)
55 + 5.25 x (16 - 10)
55 + 6.75 x (16 - 10)
Initial
Stress
(kPa)
Final
Stress
(kPa)
Mean
mvc
Stress (from
(kPa) Fig. 3.15)
59.5
68.5
77.5
86.5
95.5
99.5
108.5
117.5
126.5
135.5
79.5
88.5
97.5
106.5
115.5
(x 10-5)
1
2
3
4
5
mvc
(m2kN-1)
0.00433
0.00285
0.00239
0.75
2.25
3.75
5.25
6.75
1.5
1.5
1.5
1.5
1.5
336
313
300
289
280
To obtain values for the last column the data of stress level
and mvc are plotted as a graph (Fig. 3.15) and the values
of mvc, corresponding to the mean values of stress level, are
read off.
For simplicity assume that the unit weight of water is 10
kNm--3. Strictly it should be 9.81 kNm-3, but this
approximation gives acceptable results.
Note: mvc varies from layer to layer and that is why it is
necessary to split the relatively thick clay layer up into
smaller sub-layers. The actual number of sub-layers chosen
is a compromise between accuracy and time of computation.
It is convenient to split up clay layer into 5 sub-layers each
1.5m thick. Normally use between 3 and 5 layers. They do
38
N.K. Tovey
ENV-2E1Y:
Settlement

Fluvial Geomorphology 2004 - 2005
 mvc . z. '
Section 3
Supposing the voids ratio changes to 1.25 during
consolidation, then the new total thickness will be:-
= 40 x 1.5 x (336 + 313 + 300 + 289 + 280) x 10 -5
(1 + e) x 0.8 = (1 + 1.25) x 0.8 = 1.8 m and so the
reduction in thickness will have been 0.2 m
= 0.91m
======
Alternatively you can get the same result far quicker by
noting that the change in thickness will be related directly to
the change in voids ratio. Suppose for example the voids
ratio changes from
Exactly the same procedure may be used if a surcharge is
placed on the top of the soil stratum - e.g. a glacier or a wide
embankment (e.g. for a flood prevention scheme). In the
latter case the settlement will usually be less than predicted,
as the width of the embankment is not large compared to the
depth of the clay layer.
eo to e1 (i.e. e = eo - e1)
Now following the procedure noted above,
thickness will be
However, this overestimate will be on the safe side if we
consider the consequence on man's activities and thus such
an approximation is called a SAFE ASSUMPTION. We
could refine our analysis and conduct a three dimensional
analysis, but frequently this extra complication will not be
justified.
2
2
(1  e0 )
the reduced
and the final thickness will be
(
)
1  e0  [1  e1 ]
2
e
. (1  e1 )  . 2.

(1  e0 )
1  eo
1  e0
Obviously you can derive the above relationship, but many
of you will prefer to remember the final result i.e. the
change in thickness
=
initial thickness
x
e
1 e 0
Remember in these cases that the values e0 represents the
voids ratio at the start of the increment (not at the start of a
consolidation test!).
3.9.3 Evaluation of Reduced Thickness - an example
The reduced thickness of a layer is the thickness the solid
component of the material would occupy if there were no
voids.
Fig. 3.13 shows a simplified version of a typical borehole
drilled through Holocene deposits in Yarmouth Embayment
area close to Breydon Water.
Fig. 3.12 Plot of mvc against stress for example above.
During formation the water table was at the surface, but
around 150+ years ago, the water table was lowered by
pumping. The compression of the sand layer as a result of
this is small, but significant shrinkage has taken place in the
overlying clay.
3.9.2 Further notes on working out change in thickness
on consolidation and related comments following
queries from members of the 2000 - 2001 class.
The voids ratio (e) is the factor which is normally computed
in consolidation calculations. The question is how to
convert this to a thickness (or more relevantly a change in
voids ratio to a change in thickness).
In addition the soft clay beneath the sand will experience
consolidation as the buoyancy effect of water will have been
reduced.
Remember the total volume of a unit of soil plus voids is
1+e
If we know the reduced thickness of the soil, the true
thickness at a voids ratio of e will be 1 + e times the reduced
thickness.
Equally if the thickness of the soil is say 2 m and the voids
ratio is 1.5, then the reduced thickness will be
2
2
2


 0. 8
1  e (1  1. 5) 2. 5
39
N.K. Tovey
ENV-2E1Y:
Fluvial Geomorphology 2004 - 2005
Section 3
From the data sheet
 
(Gs  Sre )
.  w  14 kNm3
1e
as Sr = 1 (full saturation), and Gs = 265,
e = 3.125
so total initial thickness = reduced thickness x (1 + e)
i.e. total initial thickness = (1 + 3.125) x 0.9696
=
4.00m
=====
A question to answer is how much has the ground lowered
since drainage started.
Fig. 3.13
Simplified Borehole through Holocene deposits
near Acle
There are two parts:1) to evaluate the shrinkage in the surface layer - we
have done this and found this to be 4.0 - 2.0 = 2.0m
In the top clay layer, the void ratio (e) and degree of
saturation (Sr) have been measured as shown in Fig. 3.13.
We may thus determine the current unit weight of the
material using the standard formula from the Data Book:
2) to determine the consolidation of the soft clay - this
can be determined using the method outlined in
section 3.9.1.
(Gs  Sre )
( 2. 65  0. 4177 * 1. 0625)
.w 
. 10  15kN3
1e
1  1. 0625
Since we know the voids ratio, we can work out the reduced
thickness as (from the data sheet), the total volume is 1 + e,
so we divide the real thickness of the section by this.
i.e. reduced thickness =
3.9.4. Stress Distribution with Depth
The discussion in the previous sections relates to the
equilibrium conditions after consolidation has finished. To
understand what is going on during the consolidation
process it is necessary to examine the stress distribution with
depth and how this varies with time
2 / (1 + 1.0625) = 0.9696m
A knowledge of the reduced thickness is important as this
will remain constant throughout consolidation and it is only
the void ratio that will change.
There are three component parts:i.
the total stress line - derived from the bulk unit
weights
ii. the effective stress - derived from the total stress
with allowance for the buoyancy effect of water
iii. the pore water pressure (static) which is the
difference between the total and effective stress
lines.
In this example it is interesting to try to work back to
determine what the thickness of the top clay would have
been before shrinkage took place. We can simulate the
initial sedimentation in the laboratory.
In the laboratory experiment, the material composing the
surface clay was re-sedimented and found to reach an
equilibrium unit weight of 14 kNm-3, i.e. a little less than it
is currently.
During sedimentation, the material will be
fully saturated.
The situation shown in Fig. 3.14 relates to an equilibrium
condition in the original condition of the section before
drainage took place. It shows the three component parts.
The hydrostatic water pressure (iii above) is shown as the
triangular wedge in fig. 3.14 which has a value of 110 kPa at
the base of the section. This distribution is uniform and
increases at 10 kPa per metre depth (we have assumed that
the unit weight of water is 10 kNm-3 to simplify
calculations).
We can work out how much shrinkage has taken place by
proceeding as follows.
First determine the initial voids ratio immediately after
sedimentation knowing that the saturated unit weight is 14
kN m-3.
40
N.K. Tovey
ENV-2E1Y:
Fluvial Geomorphology 2004 - 2005
Section 3
very short compared to the geological time scale associated
with the consolidation of the lower clay.
We should now estimate the increment of load on the clay.
To do this we need to work out the change in effective
stress.
From Fig. 3.14 the effective stress before lowering of the
water table is 26 kPa while after the lowering it is 50 kPa so
the increment is 24 kPa. The same result would be obtained
by considering the change in effective stress at the base of
the layer.
Fig. 3.14 Stress distribution with depth of original section
before drainage and shrinkage of the top layer.
The total stress distribution is obtained from the bulk unit
weights ( see section 1.10). This represents the line which
has values 56 (at 4m, 76 at 5m, and 166 at 11 m depth).
Within any one stratum, the stress lines will be linear with
depth, but will show a distinct kink in the total stress
distribution line when changing from one layer to another on
account of the differing unit weights.
We may thus illustrate the stress diagram in the lower clay
layer as shown in Fig. 3.16
Fig. 3.16. Stress distribution during loading in lower clay.
Note: the isochrones represent the decay of pore
water pressure with time.
As a result of drainage, the water table is lowered to the top
of the lower clay and the stress distribution changes.
The difference between the two lines is the effective stress
(i.e. it is 26 kPa at a depth of 5m).
In the section 3.9.3 it is assumed that no residual
consolidation (or minimal consolidation) is taking place as
there is no excess pore water pressure. [This assumes that
there is no self weight consolidation going on which is not
strictly true, but we shall assume this at this stage for
simplicity. In Semester 2, a lecture will be given on the
effects of self weight consolidation].
After the lowering of the water table it is necessary to repeat
the above to obtain the pressure distribution with depth as
shown in Fig. 3.15. The unit weight of the desiccated crust
is 15 kNm-3 (see section 3.9.3) and the total stress (and
effective stress at the top of the lower clay is now 50 kPa
(i.e. 30 (from upper clay and 20 from sand). The new
hydrostatic pressure line starts at 0 at the top of the lower
clay and rises to 60 kPa at the base of this layer.
Fig. 3.15. Pressure distribution after the lowering of the
water table.
However, though the total stress changes, the effective
stress component initially remains the same. The static
water pressure distribution is now less and between the two
sections of the diagram is superimposed a distribution
similar to Fig. 3.7 showing the dissipations of the excess
pore water pressure with time.
It is of course assumed that the lowering of the water table is
instantaneous, which is not strictly true, but the time scale is
If boreholes are positioned in the consolidating layer then
excess water pressures will be detected as the water level
will rise above the static water table but will be seen to
41
N.K. Tovey
ENV-2E1Y:
Fluvial Geomorphology 2004 - 2005
slowly drop to that of the water table with tine. This aspect
is explored further in section 3.11.
drainage, the effective drainage path length is thus half the
thickness (i.e. 10 mm), and cvc is thus
3.10 Evaluation of cvc (coefficient of consolidation)

In sections 3.6 and 3.7 two different methods were given to
estimate the completion of primary consolidation. We may
often wish to predict how long it will take before primary
consolidation is reached in the field, based on results from
laboratory tests. For this we need to refer to equation 3.12
and evaluate cvc,
i. e.
Tv 
cvc t
or
D2
cvc 
Section 3
0. 197 .( 10 x 103 )2
 6. 57 x 108 m2 s1
300
===============
[Note: Watch the units - metres and seconds!]
3.11 Evaluation of consolidation time in the field
If we take a sample from the field and test it for its
consolidation characteristics and measure the value of cvc as
shown above in section 3.10, we may then estimate the
equivalent time to the same amount of consolidation in the
field.
TvD2
t
Now Tv has a unique relationship with U as shown in
Table 3.1 above. This information is plotted as Fig. 3.17.
Note that the X-axis is plotted as the square root of T v in this
instance.
In the field, suppose the layer of the same clay is 2 m thick
and there is only drainage upwards and suppose the field
sample was the one analysed in section 3.10, then since cvc
is constant, we have:2
2
0. 197 Dlab
0. 197 Dfie
ld

tlab
t fie ld
or
t fie ld 
2
tlab Dfie
ld
........ 3. 13
2
Dlab
Thus for the equivalent 50% consolidation, the time in the
field would be
300 x 22
t fi e l d 
0. 012
secs
= 138.9 days,
==========
Provided that identical degrees of consolidation are
compared in the field and laboratory, equation 3.13 can be
used to predict other amounts of consolidation. Thus if
100% consolidation occurred after 90 minutes = 5400
seconds, the consolidation in the field would be at the same
stage after
Fig. 3.17 Plot of data in Table 3.1 – note unusual X - axis
It will be noticed that the first part of the curve is linear for
both the exact and approximate vales, and that this is valid
up to 50%+ consolidation.
Using the straight portion of the U versus
relationship, we may obtain an estimate of cvc.
5400 x 22
 6. 85 years
0. 012
==========
3.12 Estimation of Permeability from Consolidation test.
Tv
It will be remembered that
It is normal to take the 50% consolidation point,
i.e.
Tv = 0.197
so c vc 
cvc 
0.197 D 2
t
or
Let us suppose that in our laboratory test it took 5 minutes =
300 seconds to reach 50% consolidation, and that our
sample was 20 mm thick at that time. Since we have double
k
 w mvc
k  cvc  w mvc
........... 3. 14
Thus, since we can measure both cvc and mvc in a
consolidation test, it is possible to estimate the permeability.
It will be remembered that it is often difficult to measure
42
N.K. Tovey
ENV-2E1Y:
Fluvial Geomorphology 2004 - 2005
Section 3
permeability accurately in clays and soils of low materials,
and this gives us an alternative way to obtain a value.
More frequently we do not know e1 but we have two known
values on the line.
From the above example: cvc = 6.57 x 10 m s
and let us assume we have measured a value of mvc =
100 x 10-6 m2kN-1, then since w = 10 kNm-3, we can
estimate the permeability k as
-8
2 -1
6.57 x 10-8 x 10 x 100 x 10-6 m s-1
i.e.
e2 = e1 - Cc log  2
and
e3 =
e1 - Cc log  3
by subtracting we can get
Cc =
e3 - e2
----------------log ( 3/ 2)
= 6.57 x 10-11 m s-1
Note that Cc is NOT
or about 2 mm per year.
=============
It is the gradient of the line on the e - log  plot
cvc
!!!!!!
From the section on Atterberg Limits, it will be
remembered that we can predict the gradient of this line
from a knowledge of the liquid and plastic limits:-
3.13 Comparison of compression index with
Atterberg Limits (may be of use for those
writing up this practical).
i.e.
The following may be of help with the practical assignments
(WLL - WPL)
gradient = --------------------- = 0.5 (WLL - WPL)
log(170) - log(1.7)
Fig. 3.5 showed the shape of the consolidation curve
Remember
The equation of the normal consolidation line is
e =
e1 - Cc log 
log(170) - log(1.7) = log(170 / 1.7)
= log 100 = 2)
So, in theory it is possible to estimate what the compression
index Cc is from a knowledge of the Atterberg Limits.
e1 is the voids ratio corresponding to a stress level of 1 kPa
(i.e. zero on log scale).
3.14. More advanced application to Environmental
Reconstruction
We use the same section and sample as described in section
3.9. We monitor the pore water pressure at the mid depth in
the clay and find that the water level in the borehole is
1.548m below ground level and this slowly falls to 1.706 m
below ground level after 20 years. At both dates there is an
excess pore water pressure as the water level in the
piezometer is above the water table.
Initially the level is 3 - 1.548 m above i.e. 1.452 m
corresponding to an excess water pressure of 14.52 kPa.
This means that since the load increment was 24 kPa (see
section 3.9), the effective stress has increased by 24 - 14.52
kPa since the start of drainage = 9.48 kPa, or alternatively
at the mid depth, the consolidation is 9.48 / 24 x 100%
complete = 39.5%.
Fig. 3.5 Voids ratio against logarithm of effective stress
Additional Effective Stress
Residual Pore Pressure
0
0.5
Tv = 0.05
0.10
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Fluvial Geomorphology 2004 - 2005
Section 3
0.30
0.40
0.20
0.50
0.60
0.90
0.70
0.80
0.848
Fig. 3.18 Universal Curves giving consolidation ratio as function of non-dimensional time constant and
depth
Reading off on Fig. 3.18 corresponding to mid-depth and a
3.15 A Second Example (based on question in 1976
consolidation ratio of 0.395 (=39.5%) indicates that this
corresponds to the Non-dimensional Time line of 0.30.
From the data in Table 3.1, this implies that the
consolidation is between 60 and 65% complete at the start
of monitoring. A more accurate estimate may be obtained
by plotting the data from Table 3.1 as a graph (similar to
that shown in Fig. 3.20) so that a more precise value may be
identified.
ENV-210 Examination Paper).
QUESTION
A silty clay 5 m thick overlays a layer of clay which is 8 m
thick. Below the clay is a permeable sand. Initially the
water table is 1m below the surface, and following drainage
it is lowered to the top of the clay layer. Two years after the
start of drainage, the excess pore water pressure is measured
at a depth of 7m below the surface, and found to be 22 kPa.
Estimate what the total pore water pressure at the same
location will be after 6 years.
After a further 20 years the excess pressure is now (3 1.706)*10 = 12.94 kPa, and so the effective stress is now 24
- 12.94 = 11.06 kPa and the consolidation ratio is now
11.06/24 = 46.1% which corresponds plots on the chart
approximately as the 0.35 time line. This means that the 20
years is equivalent to 0.35 - 0.30 in terms of the nondimensional time constant,
so if initially the nondimensional time constant was 0.30, this would represent a
period of 20 x 0.30 / 0.05 = 120 years.
If 0.5m of settlement occur in the first two years, estimate
the additional settlement by the time the pore water pressure
has completely dissipated.
Firstly, we recognise that there is double drainage so in our
analysis we use a thickness of 4m as our drainage path
length.
f we wished to know how long into the future the time
would be after the start of monitoring until the consolidation
is 90% complete, we note (from the Table 3.1 on Page 36),
that 90% consolidation corresponds to a non-dimensional
time constant of 0.848. Thus the time in the future would be
the relative depth into the clay layer at which we measure
the pore pressures is thus:-
(7 - 5) / 4 = 0.5
and we can thus draw the line X-Y corresponding to this on
the following chart (Fig.3.18).
(0.848 - 0.3) / 0.3 *120 = 219.2 years.
We also note that at that time in the future,
the
consolidation ratio at a depth of 1m below the top of the
lower clay, the depth factor (in Fig. 3.18) will be 2*1/6 =
0.333, and corresponding the corresponding degree of
consolidation at this depth will be 0.92. We do not need
this value for the present example, but we shall need it later
in a continuation of this example in section 3.16.
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Section 3
and not merely that at a specific depth. Fortunately, Table
3.1 gives us the required values without further calculation.
S ilt y c la y
C la y
S a nd
Fig. 3.19 Diagram for example
Fig. 3. 20 Graph of Tv against U
Secondly, we note that the water table is lowered by 4m, and
so since the unit weight of water is 10kNm-3, the imposed
stress increment will be 4 x 10 = 40 kPa, and will initially be
carried by an immediate increase in the pore water pressure.
We can plot a graph of U against Tv as shown in the Fig.
3.20, and note that a value of Tv = 0.2 corresponds with a
degree of consolidation of U = 0.503. Hence the total
consolidation to be expected will be
Hence we note that after two years, the percentage excess
pore pressure still to dissipate is:-
1 / 0.503 x 0.5 =
So the ADDITIONAL settlement will be 0.994 - 0.5
22 / 40 = 55%, or 45% of the initial excess
pore water pressure has been dissipated. We can plot this as
point A on the graph above.
= 0.494m
======
We note that this corresponds exactly to the Tv line (Tv =
0.20). This is the non-dimensional time constant, and we
can thus estimate what the corresponding value of Tv will
be after 6 years:-
= 6 / 2 x 0.2
0.994m
3.16 Settlement: a further example (an extension
of example shown in Sections 3.9 and 3.14).
Fig. 3.13 shows a simplified borehole log of a section. This
may be adequate for many purposes, but because of selfweight consolidation, the unit weight of the lower clay will
become greater as the depth into the stratum increases. In
Fig. 3.21 there is a similar borehole log from a similar area.
For simplicity, the upper section (i.e. the clay crust and the
sand have the same properties, but the lower clay now is
divided into three layers to illustrate the differences in unit
weight [Note: that the mean unit weight has also been
changed in this layer].
= 0.6
Plotting this value on the graph (Fig. 3.16) and reading off
on the line X-Y, we note that the fractional dissipation of the
excess pore pressure U is now 0.8, and the residual excess
pore pressure is thus 0.2 x 40 = 8kPa.
The question asks for the TOTAL pore pressure, so we
must add this value to the pore pressure arising purely from
position head, i.e. the depth below the water table (- this is
now 4m). Thus the total pore pressure after 6 years will be:-
We note that (from section 3.9.3) shrinkage of the upper
crust accounts for 2.00 m.
We will attempt to estimate how much settlement can be
attributed to consolidation of the clay layer, and also to
estimate how much more settlement will occur by the time
90% consolidation is complete (i.e. the same time interval as
for the example in section 3.14).
20 + 8 = 28 kPa ...........answer to first part.
=======
The non-dimensional time constant corresponding to 2 years
is 0.2 as noted above. We note that the consolidation by this
time is 0.5 m, but this is the total settlement of the whole
clay layer, and it is clear from the above diagram that the
excess pore pressure in the different parts of the soil will be
different. This meas that the settlement in each part of the
clay will also be different.
Data from a consolidation test carried out on a sample from
mid depth in the clay layer are shown in TABLE 3.3.
We remember (from section 3.9.3) that the original
thickness of the clay crust was 4 m, and this reduced to 2 m
through shrinkage. Furthermore the initial unit weight was
14 kN m-3 immediately after sedimentation and the final unit
weight after ground water lowering was 15 kN m-3. Finally,
This figure of 0.5m thus represents the effective aggregate
of the settlement at each part in the clay, and to proceed we
need to know the overall MEAN degree of consolidation U,
45
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Section 3
it was noted that the stress increment was 24 kPa arising
from the water table lowering, and that at the start of
monitoring, the excess water pressure was still 14.52 kPa.
the sample. The fact that the same value is obtained by the
two different methods, indicates that the sequence as not
been loaded in the past prior to the ground water lowering.
We shall assume that despite the slight difference in profile,
the pore pressure dissipation rate is the same as in the
previous example.
At and average 90% consolidation throughout the sequence
corresponding to a Tv line of 0.848, the sample at the middle
reaches (from Fig. 3.18) 84.5% dissipation of the excess
pore water pressure, and the effective stress is now :44.20
+
0.845* 24 = 64.48 kPa
Fig. 3.22 Consolidation curve from data in Table 3.3.
Fig. 3.21 Borehole log through sediment
We also need to work out the equivalent stress at the 1/6th
and 5/6ths points as we are dividing the layer into three to
allow for the differences in the unit weights.
Effective Stress Voids Ratio
(kPa)
5
2.202
10
2.172
20
2.142
40
2.112
80
1.918
160
1.638
320
1.359
640
1.080
Table 3.3 Consolidation Test on sample
The initial stress levels at the upper 1/6th point (i.e. 1m into
the lower clay) and lower 1/6th point of the clay layer are
(by similar method to above),
The values will be 32 kPa and 56.99 kPa respectively.
If you are unsure how the above to figures arise, the
following box shows how they are obtained.
i.e. at the upper 1/6th point the stress will be
4*14 + 1*20 + 1*16 – 6*10 = 32 kPa
|
|
|
|
upper
sand
lower
hydrostatic
clay
clay
We proceed by evaluating the initial stress at the position of
the consolidation sample:Initial stress level at location of consolidation sample before
drainage would have been
and at the lower 1/6th point i.e. at a depth of 5m into lower
clay
= 4.(14 - 10) + 1.(20 - 10) + 2.(16 - 10) + 1.(16.20-10)
= 44.20 kPa,
76 + 2*16 + 2*16.2 + 1*16.59 – 10*10 = 56.99 kPa
|
|
|
|
|
| upper
middle
lower
hydrostatic
upper third
third
third
clay & -------------------------------sand
lower clay
and stress at time of sampling
= 44.20 + 9.48 = 53.68 kPa.
|
additional effective stress – [see paragraph 2 of
section 3.14 to see how the figure 9.48 is obtained]
Reading off the graph (Fig. 31.8) at the 1/6th level at a
point corresponding to T = 0.3 indicates 70% dissipation
for both levels, i.e. the effective stress at the two point is
48.8 (=32 + 0.7*24) kPa, and 73.99 kPa respectively.
The same value may be obtained by plotting the
consolidation data (see Fig. 3.22). The point of intersection
of the two linear curves is the previous effective stress on
46
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The void ratio at each stress level is read off from Fig. 3.22.
Note for the initial situation, the value of the void ratio must
be that on the virgin consolidation line (not the actual test
data which relates to the situation at time of sampling – i.e.
after some consolidation has taken place.
[ Remember that the consolidation is only 30% complete
at the centre - on average it is 39.5% - see section 3.14]
In section 3.14 we noted that at the 1/6 th depth at the 0.848
time constant contour (corresponding to the average 90%)
consolidation, there will have been 92% dissipation of
excess pore pressure. Thus the effective stress at this time
will be 54.08 kPa (= 32 + 0.92*24), and 79.07 kPa
respectively. Analysis now proceeds in tabular form (Table
3.4).
upper third
middle
lower third
initial
A
B
initial
A
B
initial
A
B
Section 3
We do the sampling at time A and know that each of the
three sections of the lower clay is 2m thick, so we can
determine the reduced thickness in the normal way i.e.
reduced thickness = actual thickness * 1 / (1 + e)
Stress
(kPa)
Void Ratio
from
graph
32.00
48.80
54.08
44.20
53.68
64.48
56.99
73.79
79.07
2.287
2.117
2.076
2.157
2.079
2.005
2.055
1.950
1.923
Present
thickness
(m)
reduced
thickness
(m)
2.000
0.642
Original
Thickness
(m)
thickness at
90%
consolidation
2.109
1.973
2.051
2.000
0.650
1.952
2.071
2.000
0.678
1.981

Table 3.4 Settlement Calculations. A is stress at start of monitoring:
We can now estimate the initial thickness of the three sublayers since
6.231
5.906
B is stress after 90 % consolidation is complete.
Finally, we can estimate the additional consolidation that
will take place up to the time when 90% consolidation
has taken place
initial thickness = reduced thickness * (1 + e)
i.e. 6 - 5.906 m = 94 mm
======
similarly at time B (90% consolidation) the same formula
can be used to determine the predicted thickness at this
time..
3.17 Some final notes about pore pressures.
Finally by summing the original thicknesses in the three
sub-layers we can estimate the total original thickness of the
layer
When pore pressure measurements are made in the field
during consolidation, the excess pressure measured relates
only to the depth in the stratum at which the piezometer is
located.
= 6. 231m
Obviously the effective stress increment at this point will be
the initial total stress increment minus this measured pore
pressure. To find out the value at any other depth, you must
use the chart (Fig. 3.18) to find the relevant nondimensional time constant, and follow the curve round to
the new depth. Alternatively, if you want to get at the
overall degree of consolidation for the whole layer, then
you must use Table 3.1 on the data sheets to get the true
average degree of consolidation.
This means that there has already been a consolidation
amounting to 0.231 m in this layer.
Allowing for 2.00 m shrinkage in the surface crust
determine earlier, the total lowering of ground surface to the
present day =
2.231m
====
Space for additional notes
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Section 3
48