1
These notes rely heavily on Chapters 1 to 6 of this textbook. Unfortunately the book
is a few decades old (1990) and is no longer in press. The good news is that a
significant portion of the book is freely available online via Google Books. The large
majority of the figures that I used in these notes were captured from Google Books.
So if you ever want to read Muir Wood’s own explanation of his figures, you know
where to go.
My take on the book is that the mathematics are a bit more complex than what most
engineers are used to. Understandably, the author doesn’t stop to explain the maths,
he just assumes the reader understands it. The plus side is that the book is really
well illustrated and Muir Wood takes the time to give qualitative examples (as
opposed to just using the equations).
2
3
These notes are essentially aimed at building an elastic-plastic model of soil that is
called Modified Cam Clay. The excerpt on the screen comes from the first paragraph
of the book and helps to set the scene.
4
Just an interesting app that I thought was worth sharing. It gives you an idea of the
relative sizes of sand, silt, and clay particles. And of all other types of objects in the
universe.
5
Notation used in all content extracted from the Muir Wood (1990) textbook.
6
I must say, I still mix up these indices and sub-indices!
Note the delta before the epsilon. That is, these are symbols for strain increments,
not for the strain as such. This may be a bit confusing for now but will (hopefully)
make sense once we start putting the model together. For now, you can think of
these different δε as the small increment in strain that the soil undergoes when the
load changes by a small amount.
Since we are considering small changes, we use δ instead of Δ.
7
8
The term "soil" refers to the mixture of gas, liquid, and solid particles. The gas is
generally air, the liquid is generally water, and the solid particles can be made from a
variety of minerals. So keep this in mind: the soil is NOT just the solid particles.
The air, water, and solids exist as a mixture. But for purposes of analysis we generally
separate these three phases as shown in the slide. We use the volumes and weights
of the individual phases to define the commonly used phase relationships shown in
the slide. You are expected to be familiar with phase relationships. If you are not,
please look them up online or in virtually any introductory textbook to soil
mechanics.
Can you see why the porosity (n) and the saturation (S) cannot be greater than 1?
Note: as you load a soil under drained conditions (gas and water are able to move
out), the total volume (V) of the soil is reduced due to the squeezing out of gas and
water. However, we generally assume that the individual particles (which make up
the solid phase) are perfectly rigid and therefore Vs does not change with loading.
9
We have introduced a negative sign in the definition of the volumetric strain to
ensure that reductions in volume are positive while increases in volume are
negative. This allows us to be consistent with the general sign conventions used in
soil mechanics.
Note that in the lower part of the slide we have assumed Vsi = Vsf = Vs. That is, the
volume of solids (or the combined volume of the particles) does not change during
loading. Only the volume of voids Vv changes.
10
Can you demonstrate that the volumetric strain increment δεp is equal to the third
and fourth part of the equations?
11
12
Since we will develop our Modified Cam Clay model to simulate triaxial testing
conditions, it is convenient to review some of the basics of triaxial testing. For a
more detailed introduction to triaxial testing, you may want to have a look at the
video prepared by Professor Kitch:
https://www.youtube.com/watch?v=hq4UlLm8oIs (Accessed 12/Aug/2021).
A triaxial test is a very common means to investigate the mechanical properties of
soil. In a triaxial test, the soil specimen is first placed in a cell, and the cell is then
filled with a liquid which is typically water. The sole purpose of the liquid in the cell is
to enable us to apply an isotropic stress to the specimen. By isotropic stress we
mean a stress that is equal in all directions. We will denote this cell pressure with the
symbol σr. The triaxial test specimen is isolated from the liquid in the cell by a flexible
latex membrane.
You can think loosely of this cell pressure as simulating the confinement that a soil
specimen experiences in situ due to the action of the surrounding soil. The greater
the depth of the soil specimen, the greater the confinement stress it will experience.
Of course, an important difference between the cell pressure and the in situ
confinement pressure, is that the cell pressure is isotropic, whereas the in situ
confinement acting on a soil specimen may very well be anisotropic. That is, the
13
stresses in the vertical and horizontal directions may be different.
Besides the cell pressure σr, the other means by which we can load a specimen
during a triaxial test is by applying an axial force F through a loading ram. The load is
generally applied by displacing the cell upwards making the loading ram push against
reaction beam. As shown in the sketch, the loading ram goes through the top of the
cell, so it is essential that the interface between the ram and the top lid of the cell is
impervious enough to avoid excessive escape of the liquid that is pressurised in the
cell. The vertical stress applied by the loading ram is F/A, where A is the cross
sectional area of the specimen.
13
As seen earlier, soils can exist over a continuous scale of degrees of saturation
ranging from zero liquid in the voids (S = 0), to voids filled entirely with water (S = 1).
We will develop the Modified Cam Clay model under the assumption that we are
dealing with fully saturated soils (S = 1).
An important aspect of triaxial testing is that it gives us the possibility of controlling
whether water is allowed to be squeezed out of the voids, or sucked into the voids of
the specimen. In other words, it enables us to control drainage conditions. This is
done by operating valves fitted to pipes that are connected to the bottom or top of
the specimen as shown in the figure. A pressure transducer is commonly fitted to
these pipes which allows us to measure the pore water pressure in the specimen.
For our purposes, we can assume that when drainage is allowed, the open end of
the pipes is exposed to atmospheric pressure, and therefore the pore water pressure
in the specimen will also be atmospheric. Since in engineering we normally measure
pressures with respect to atmospheric pressure, we would thus say that the pore
water pressure is zero.
However, when the specimen is undrained, that is, the valves on the drainage pipes
are closed, the pore water pressure will increase when the specimen tries to
14
contract (reduce its volume) and will decrease when the specimen tries to dilate
(increase its volume). Notice that we are saying that the specimen is trying to
contract or dilate, and not that it is actually contracting or dilating. This is because
the undrained condition implies that water cannot be squeezed out of or sucked into
the specimen. Since the soil is saturated, the volume of voids remains constant.
Further assuming that the compressibility of water, and of the solid particles
themselves are negligible, we conclude that the soil does not change its volume
during undrained loading. So although the specimen may try to contract or dilate, it
doesn’t actually do so.
SUMMARY: During a drained triaxial test, water can move in and out of the
specimen, the volume of soil changes, and the pwp remains constant at zero. During
an undrained triaxial test, water cannot move in our out of the specimen, the
volume of soil remains constant, and pwp changes throughout the test (in response
to whether the soil is trying to dilate or contract).
14
Let’s have a closer look at how specimens are loaded in a triaxial test. Part a) of the
figure simply shows the adopted coordinate axes to help us visualize the direction of
the different stresses in three dimensional space. Axis z is vertical, or axial since it is
aligned with the longitudinal axis of the specimen. The x and y axes are horizontal.
Part b) shows the stresses that result when we load the specimen only by increasing
the cell pressure. The specimen is subjected to an isotropic stress condition, with the
cell pressure σr acting in all directions. A stress analysis, using Mohr Circles for
example, shows that no shear stresses develop in a body that is subjected to an
isotropic stress condition.
Part c) shows the stresses that result when, in addition to the cell pressure, an axial
force F is also applied through the loading ram. The stresses in the horizontal
direction still correspond to the cell pressure σr, but the axial stress σa is now given
by σr + F/A. Since the horizontal and vertical stresses are not equal, this is no longer
an isotropic stress condition and therefore shear stresses, and strains, will develop.
15
This table illustrates the most common sequence of loading during triaxial tests. First
the specimen is loaded isotropically, and then anisotropic loading is induced by
adding an axial load.
With regards to drainage, two arrangements are common. The first is to keep the
drainage valves open throughout the entire test. This is called a consolidated
drained test and is generally abbreviated as CD. The second common arrangement is
to allow drainage during the isotropic stage, but prevent it during the anisotropic
stage. This is called a consolidated undrained test and is generally abbreviated CU.
Generally, whenever drainage is allowed, we measure the volume of water that
flows out of or into the specimen to keep track of the changes in specimen density
(or more specifically, changes in the volume of the specimen). And when drainage is
prevented, we measure the development of pore water pressure.
16
Our assumption that the specimen is fully saturated implies that the principle of
effective stress is valid. This principle is normally introduced in undergraduate
courses of soil mechanics, so I will assume that you are familiar with it. In essence,
the principle states that the behaviour of a saturated specimen correlates, not to the
stresses that we apply to the specimen, such as σr and σa, but rather it correlates to
those applied stresses minus the pore water pressure u. The Greek letter σ, without
a prime, is normally used to denote the applied or total stresses, and a prime is
added to the sigma to indicate effective stresses. Accordingly, although in the
laboratory we can directly control σr and σa, we will develop the Modified Cam Clay
model in terms of σ'r and σ'a.
17
This slide shows the stress and strain invariants used in the development of the
Modified Cam Clay model. We use two invariants of the stress tensor because the
triaxial tests provides two degrees of freedom to load the specimen: cell pressure
and axial force. Likewise, we use two invariants of the strain increment tensor: the
volumetric strain increment which is associated to changes in volume and the shear
strain increment which is associated to changes in shape.
Notice that q is the same regardless of whether it is written in terms of effective
stresses or total stresses. In other words, q is equal to q'. Because of this we don’t
bother to add a prime to q even when we are conducting an analysis in terms of
effective stresses. So I will only use q, without the prime, throughout these notes.
This convention is widely, if not universally, adopted in soil mechanics literature.
18
19
We can gain some insight into the justification for the chosen stress and strain
invariants by considering the work done on a soil specimen as it is deformed in a
triaxial test.
We know from physics that the work done by a constant force as it moves an object
in the same direction of the force is equal to the product of the force magnitude and
the distance across which it moved the object. For instance, if, as shown in the slide,
a force of 50 N pointing to the right, moves an object 3 meters to the right, we say
that the force has done a work of 50 N times 3 m which yields 150 N*m or 150
Joules. Let’s refer to this type of work as W’.
20
The concept of work is also applicable when a body is deformed (rather than
displaced) by a force. Let’s start by considering the case of an axially loaded
prismatic bar as shown in the figure on the left. An axial force is gradually applied to
the bar. Initially the force is equal to zero and its magnitude increases gradually until
it reaches a magnitude P. The bar stretches as the load increases. Initially it’s
deformation is zero and eventually it reaches a value δ when the force reaches
magnitude P. The load vs displacement graph on the left illustrates this process.
21
To determine the work done by the force, we cannot simply multiply P times δ
because in this case the force did not have a constant magnitude throughout the
entire process. Remember, the force started off with a magnitude of zero and
gradually increased to a magnitude of P.
So let’s have a closer look at what is happening here by considering some
intermediate point when the load P had a value Pn and the bar had an extension δn.
If at this stage we increase the load by a small amount dPn, the bar will extend by a
small amount dδn. That is, some additional amount of work, dW'n, was done on the
bar by dPn. If dPn is small, we can approximate this additional amount of work dW'n
by Pn times dδn. Our approximation will improve as dPn becomes smaller, and will be
exactly true if we make dPn differentials instead of a small, but measurable,
increments.
Up to now we have been exploring this problem with forces and displacements.
However, we commonly conduct our analyses in terms of stresses and strains. So
let’s convert to stresses and strains by dividing Pn by the cross sectional area of the
bar (A), and by dividing dδn by the initial length of the bar. In order to preserve the
equality, we must also divide the left side of the equation by A times L, as shown in
the second equation.
22
We note that A times L is equal to the volume of the bar. So what we are concluding
is that the stress times the strain increment is equal to the increment of work per
unit volume. Adopting the symbol W, without a prime, for work per unit volume, we
arrive at the third equation.
22
The concept of the previous slide can be generalised to a three dimensional object
acted upon by three principal stresses, and experiencing principal strain increments.
This situation is illustrated by the figure and equation on the left. Remember that
principal stresses are perpendicular to each other and act on faces on which no
shear stresses are acting.
The figure on the right shows the expression for the specific case of a triaxial test soil
specimen. In this case the expression is written in terms of effective stresses as these
are the stresses that correlate to the mechanical behaviour of the soil.
23
We finalise our discussion on work considerations by noticing that the stress and
strain invariants used to develop the Modified Cam Clay model are such that they
satisfy the equation shown in the slide. In other words, our chosen stress and strain
invariants allow us to quickly determine the amount of work per unit volume that
the effective axial stress and the effective radial stress are doing on the specimen.
We describe this by saying that our variables to describe stress and strain are defined
in such a way that they are stress and strain conjugates. . As we will see later, the use
of stress-strain conjugates facilitates the development of the MCC model.
24
The previous slides made reference to “invariants” of the stress and strain tensors
and it is convenient to clarify this concept. By invariants we literally mean that the
parameter does not vary when the reference axes change. Although stresses are
described with a rank 2 tensor, that is a matrix, the concept of an invariant can be
more easily visualised with a rank 1 tensor such as a vector.
The slide shows the same vector P described in two different systems of coordinate
axes. Although the vector P in itself does not change, it is clear that its x and y
components do change depending on whether we describe it with the green axes x1
and y1, or with the red axes x2 and y2. For instance, in this case it is apparent that Py1
is larger than Py2. However, if we determine the magnitude of the vector using
Pythagoras theorem, we will get the same answer regardless of whether we use Px1
and Py1 or Px2 and Py2. Accordingly, we can say that the magnitude of the vector is an
invariant because it is insensitive to changes in the reference axes.
Similarly, the stress and strain invariants that we have chosen to keep track of what
is happening to our triaxial specimen, are independent of the choice of reference
axes.
25
26
Coming back to the stress invariants, let’s take a moment to make sure that we can
measure these during a triaxial test. To do this, we notice that both p’ and q can be
written in terms of:
-
the cell pressure, σr
the force on the loading ram, F
the cross sectional area of the specimen, A
the pore water pressure, u
All four of these parameters can be either measured or reasonably estimated during
a triaxial test, so we conclude that, indeed, we can keep track of p’ and q during an
actual triaxial test.
27
We can do a similar check for the strain invariants, but first let’s take a moment to
confirm the validity of our equation for the volumetric strain invariant. In other
words, let’s check the validity of the equation in the red rectangle at the top. I will
simply present the demonstration in this slide and in the next one for you to go
through.
Notice that in this demonstration we use the assumption that we are dealing with
very small strain increments and that therefore the product of two or more strain
increments is negligible. It may seem strange that we are able to use this assumption
in building a constitutive model of soil because soils undergo significant strains. For
example, triaxial test specimens can easily reach axial strain levels of 15 to 20%.
These strain levels are certainly not what we can call “very small”. But, bear in mind
that what we are assuming to be small is the strain increment, not the strain level
itself. So even if a triaxial specimen is strained significantly, we can always reach that
strain level through a large number of very small strain increments. This will probably
be easier to appreciate once we start implementing the Cam Clay model in a
spreadsheet and you can see how the calculations play out.
28
And here is the second part of the demonstration. You can see that terms that
include products of strain increments have been regarded as negligible.
29
In the last two slides, besides assuming that the strain increments are very small, we
also made the assumption that the triaxial specimen retains its shape of a straight
cylinder throughout the test. Now, if you have ever seen a triaxial specimen at the
end of the test, you know that this assumption is questionable at best and a blatant
lie at worst. For example, the picture on the slide shows a specimen of tailings that I
tested and it is apparent that significant and uneven lateral bulging occurs during the
test. This specimen now looks more like a barrel than a straight cylinder.
So I am just showing this figure here to acknowledge the limitations in this
assumption. It is possible to account for these irregular deformations using
modelling techniques that are somewhat more complex, for instance by
implementing Cam Clay in a finite element model. But such techniques fall beyond
the scope of what this course is intended to cover.
30
Now that we have convinced ourselves that our expression for volumetric strain
increment (δεp) makes sense, let’s look into how we can measure the strain
invariants in the lab. During most triaxial tests, we measure the height (a) and the
volume (V) of the specimen throughout the test. So the first thing that might catch
our attention when we look at the expressions for the strain increment invariants is
that they both involve increments in radial strain (δεr).
In the case of the volumetric strain increment (δεp), this difficulty is overcome by
resorting to the fundamental definition of volumetric strain increment: change in
volume over initial volume. This also allows us to express the radial strain increment
in terms of volumetric and axial strain. And using this expression, we can write the
shear strain increment invariant in terms of parameters that we commonly measure
during a triaxial test: volume V, length a, and their variations.
31
Let’s now define a new parameter, the mean total stress p. As the name suggests,
the definition of p is identical to the definition of the mean effective stress p', except
that we define p in terms of total stresses, instead of effective stresses. You can also
prove that p is equal to p' plus the pore water pressure u.
Going forward we will be working with stress paths. Stress paths are plots that allow
us to summarize the sequence of states of stress to which a soil element is
subjected. In triaxial tests, total stress paths are generally given by plots of q vs p,
and effective stress paths are defined by plots of q vs p'.
In order to draw total stress paths, it is useful to express p in terms of the deviator
stress q and the cell pressure σr as shown in the second line of the slide. We also
notice that the change in p is equal to the change in the cell pressure σr plus one
third of the change in deviator stress q as shown in equation A.
We will also find it useful to express p in terms of the deviator stress q and the axial
stress σa as shown in the third line of the slide. Again, we notice that the change in p
is equal to the change in the axial stress σa minus two thirds of the change in
deviator stress q as shown in equation B.
32
33
Before looking into the total stress paths commonly used in triaxial testing, let’s
make a distinction between two broad modes of testing: triaxial compression and
triaxial extension.
The usual arrangement for triaxial tests is for the loading ram to act in the same
direction as the cell pressure σr. This arrangement is shown in the left figure. Since,
the loading ram compresses the specimen in the axial direction, these tests are
called triaxial compression tests. In these cases the axial stress consists of the cell
pressure plus the stress induced by the loading ram and is greater than the
horizontal stresses which consist of the cell pressure alone. According to our
definition of q (given at the bottom of the slide), this implies that q is positive in
compression tests.
Although less common, it is also possible for the loading ram to pull away from the
specimen by connecting the loading ram to the top cap of the specimen. This
situation is shown in the figure on the right. Since the loading ram is stretching the
specimen in the axial direction, these tests are called triaxial extension tests. In these
cases the axial stress consists of the cell pressure minus the stress induced by the
loading ram and it is smaller than the horizontal stresses which consist of the cell
pressure alone. According to our definition of q, this implies that q is negative in
34
extension tests.
34
Let’s now have a look at the three most common types of total stress paths applied
during triaxial tests. For all three stress paths we assume that the specimen is
initially under isotropic stress as indicated by point C, where q is equal to zero, and
then sheared along one of the three stress paths shown in the slide.
The first case, which is by far the most common, consists in keeping the cell pressure
constant (δσr = 0) while varying the force applied through the loading ram. From
equation A we can conclude that, in q vs p space, this stress path plots as a straight
line with a slope of 3. When this stress path is applied with q>0 (compression), the
test is referred to as a conventional compression test.
In the second case, it is the axial stress that remains constant while the cell pressure
varies. So for example, if the cell pressure is decreased by a certain amount X, the
axial stress induced by the loading ram has to be increased by that same amount X
to keep the axial stress constant. From Equation B we can conclude that, in q vs p
space, this stress path plots as a straight line with a slope of -1.5.
The third case consists in keeping p constant while varying q. This stress path also
requires simultaneous changes in the cell pressure and the stress induced by the
loading ram. This stress path is shown in the figure as a vertical line, as this
35
corresponds to a constant p value. The advantage of this stress path is that it allows
the isolation of aspects of soil behaviour that are due only to changes in q.
35
36
One of the key aspects of the mechanical behaviour of soils is dilatancy, that is, their
tendency to change volume when they are sheared. Generally speaking, when
sheared, loose soils contract and dense soils dilate. When talking about volume
changes saturated in soils, we will make the usual assumption that the particles
themselves and the water in the voids are perfectly incompressible. This is a very
reasonable assumption to make for most soils because the compressibility of water
and of the individual grains is much smaller than the compressibility of the soil
assembly as a whole, that is, the mixture of grains and water.
Under undrained conditions, water is not allowed to move into or out of the voids.
This implies that there are no volume changes and therefore no dilatancy. However,
even under undrained conditions, soils retain their tendency for dilatancy and this
has important implications on their mechanical response. For example, a loose soil
sheared under undrained conditions will still try to contract, but since water cannot
be squeezed out of the voids (as it is undrained), the water pressure increases in
response to the contraction tendency. This increase in pore water pressure leads to a
reduction in effective stresses and the associated reduction in strength and rigidity.
Conversely, a dense soil sheared under undrained conditions will still try to dilate,
but since water cannot move into the voids, the pore water pressure drops in
37
response to the dilation tendency. This drop in pwp leads to an increase in effective
stresses and the associated increase in strength and rigidity.
It is worth clarifying here that in undrained tests, there actually is a small volume
reduction when the pwp goes up (loose soils) and a small volume increase when the
pwp goes up. This is the case because water really is compressible after all. However,
such changes in volume are typically ignored in the context of triaxial testing.
For an informal discussion on water compressibility, you can have a look at the link
at the bottom of the slide (https://tinyurl.com/yxlgh8j8).
37
During drained triaxial tests, loading is conducted slow enough for water to be
squeezed out of the specimen or sucked into it without the development of any
significant pore water pressures. Since the pwp remains zero, the effective and total
stresses are equal, so the total stress path TSP, which is the plot of q vs p, is equal to
the effective stress path ESP, which is the plot of q vs p'.
On the other hand, during an undrained triaxial test, pwp increases or decreases due
to the dilatancy tendency of the soil and this results in the total and effective stress
paths being different. For example, let’s consider the possible stress paths in the
figure which correspond to a consolidated undrained triaxial test. The specimen is
initially consolidated isotropically to the stress level indicated by point C. Since
drainage is allowed during the isotropic loading stage of a consolidated undrained
test, water gets squeezed out of the specimen and the pore water pressure remains
equal to zero. This implies that the total and effective stress paths coincide as shown
in the figure.
After isotropic consolidation the drainage valve is closed and shearing can now
proceed. The total stress path (TSP) indicates that this is a triaxial compression test
because the TSP plots on the positive side of q. The TSP also shows that during the
test the total mean stress p stays constant. If the specimen is dense and therefore
38
tries to dilate, the specimen will develop negative pore water pressures. And since p'
= p – u, this implies that p' is larger than p and therefore the effective stress path ESP
plots to the right of the total stress path TSP. On the other hand, if the specimen is
loose and tries to contract, the specimen will develop positive pore water pressures
and the effective stress path ESP ends up plotting to the left of the total stress path.
In other words, the horizontal distance between the total and effective stress paths
is equal to the pore water pressure u. Also appreciated from: p' = p - u => u = p – p'.
38
It is convenient to decompose into two components the pore water pressure
increments that an undrained triaxial specimen experiences during loading: one
component, δup, is due exclusively to the change in mean effective stress p, and
another component, δuq, due exclusively to the change in deviator stress q.
Experimental results show that when subjected to an isotropic stress increment, that
is, p changes while q remains constant, most saturated soils will exhibit a change in
pore water pressure δup that is almost equal to the change of isotropic stress δp. In
other words, the pore water takes almost all of the isotropic load. This concept is so
widely accepted that it is used in laboratory testing to determine whether the
specimen is saturated. However, it is worth mentioning that if the mineral skeleton
of the soil is very stiff, to the point that it’s compressibility is comparable to that of
water, then a significant portion of the isotropic stress increment is taken up by the
mineral skeleton. But we won’t consider that case though. So for our purposes we
will always assume that any increase in isotropic stress is taken up entirely by the
pore water pressure. That is, δup = δp.
Regarding the other component, δuq, the situation is more complex. We can only say
that when subjected to a change in deviator stress q, while keeping p constant, the
pore water pressure will take up some fraction ‘a’ of the deviator stress increment.
39
This fraction a is termed a pore pressure parameter and it is partly a function of the
density of the soil and also varies throughout the test. The modified cam clay MCC
model will enable us to make estimations of a.
39
40
The Modified Cam Clay model is an elastic plastic model. That is, it incorporates both
the recoverable and permanent deformations that a soil undergoes when subjected
to loading. In this slide we start looking into how to model soil elasticity. For
simplicity, we will make the often unrealistic assumption that the soil is isotropic,
that is, it exhibits equal elastic properties in all directions. This assumption will allow
us to better focus on the essential characteristics of the Modified Cam Clay model.
Of course, it is always possible to incorporate elastic anisotropy into the model at a
later stage if need be.
Two constants are enough for a full description of isotropic elasticity. In
undergraduate courses of strength of materials, these two constants are generally
chosen to be the elasticity modulus (E) and Poisson’s ratio (v). However, to develop
the Modified Cam Clay model we will find it more convenient two work with the
shear modulus (G) and the bulk modulus (K). Of course, these four constants are
correlated and once you know any two, you can compute the remaining two.
Just to be sure that we are all on the same page, the figures illustrate the meaning of
the shear and bulk moduli. The shear modulus G correlates shear stresses and shear
strains as shown in the left. The figure shows how a square would be distorted by an
angle γ into a rhombus.
41
The figure in the right represents a sphere subjected to a compressive isotropic
stress σiso. Under the action of this stress, the sphere undergoes an elastic reduction
of volume. The bulk modulus is the rate of the isotropic stress σiso, to the volumetric
strain δεp.
In the next few slides we will work towards an expression that allows to express
elastic strains in terms of K, G, δp', and δq.
41
Let’s start by expressing K and G in terms of E and v as shown in the slide. The
demonstrations for these expressions are fairly simple to follow and are generally
found in Strength of Materials textbooks.
42
Now let’s express the increase in axial strain δεa when an element is subjected to an
increase in effective axial stress (δσ'a) which acts along one principal direction, and
an increase in effective radial stress (δσ‘r) which acts along two principal directions.
We will express these strains in terms of the elasticity modulus E, and Poisson’s ratio
ν.
As indicated in the figure, we can tackle this problem by applying superposition. That
is, first we calculate the increment in axial strain δεa due to an increment in stress in
each principal direction acting alone and then we add the individual answers.
43
Here are the calculations for the individual stress increments acting alone, and for
the addition to compute the total increment in axial strain δεa.
44
We can also use superposition to compute the increment in radial strain as shown in
this slide and the next. To better follow the calculations, let’s be specific about the
radial direction along which we want to calculate the radial strain increment. For
instance, let’s choose the radial direction indicated in green in the slide.
45
And here we see the calculations for the individual stress increments. Notice that the
calculation for the radial stress increment acting alone depends on whether we are
considering the radial stress increment that acts along the green line (radial strain
increment is not affected by Poisson’s ratio), or the radial stress increment that is
perpendicular to the green line (radial strain increment is affected by Poisson’s ratio).
46
After adding up the three individual effects and rearranging a bit, we obtain the first
equation on the slide.
47
The first two equations on the slide summarise our conclusions from the two
superpositions that we did. The equations are slightly rearranged to make it clear
that they can be expressed as a single matrix operation as this is more compact and
convenient.
Although you don’t have to be matrix wizard to follow what comes next, I will
assume that you know how to do matrix-vector multiplication, matrix-matrix
multiplication, and that you remember that multiplying a matrix by its inverse yields
the identity matrix.
By the way, for a superb introduction to linear algebra that actually makes sense, be
sure to check the series “The Essence of Linear Algebra” in the YouTube channel
3Blue1Brown. I cannot recommend it enough.
48
Likewise, the incremental form of the equations for the mean effective stress p' and
the deviator stress q, can also be written in matrix notation, as shown in the slide.
And what I am showing in the lower half of the slide is that by taking the inverse of
the matrix we can solve for the vector that contains the axial and radial effective
stress increments.
49
Lastly, we use matrix notation to write the equations that correlate volumetric and
shear strain increments, with axial and radial strain increments.
50
The first three rows on this slide show the matrices that we obtained in the previous
slides. By doing the substitutions suggested by the arrows we arrive at the
expression at the bottom of the matrix.
51
The last expression of the previous slide reduces to the first expression in the current
slide. And recalling the expressions for the shear modulus G and the bulk modulus K,
we can rewrite the matrix in a much more compact form as shown in the lower half
of the slide. And there you have it… an expression for the elastic volumetric and
shear strains in terms of K, G, δp', and δq. The expression is shown in both matrix
and conventional form.
Notice that the increment in elastic volumetric strain δεp is independent from the
increase in deviator stress δq and, conversely, the increment in elastic shear strain
δεq is independent from the increase in mean effective stress δp'.
52
You can easily check the previous matrix operations using an online matrix calculator.
For instance, the slide shows the result in the previous slide computed in symbolab’s
matrix calculator. It is really easy to use and it can handle symbolic expressions.
53
54
The figure illustrates the defining characteristic of elasticity: stress and strain are
uniquely correlated. Notice that this correlation is not necessarily linear. So while it is
true that many elastic materials exhibit linear stress : strain correlations, this is not
always the case. We will develop the Modified Cam Clay model under the
assumption that the soil’s elastic behaviour is linear. And we will also model its
plastic behaviour which is what we will start looking into now.
55
Although we are interested in soils, let us start investigating plasticity by considering
the behaviour of a copper wire subjected to a tensile load P and which stretches by a
distance δl as shown in the sketch on the left side of the slide. The figure on the right
describes the behaviour of the wire. The first thing that we may notice is that there
is no longer a unique correlation between load and deformation. That is, for any
given load, there are multiple values of elongation δl that the wire might exhibit. So
we know that this is plastic, as opposed to elastic, behaviour.
The plastic nature of the wire is also evident when we consider what happens when
the wire is loaded from A to A1, and then fully unloaded to B1. Although the load is
fully removed, there is a permanent (or plastic) deformation of approximately 8 mm.
Now, if the wire is loaded again to its maximum previous load, its reloading path
from B1 to C1 will be very similar to its unloading path from A1 to B1. Furthermore,
we would also observe that provided that the maximum previous load is not
exceeded, the behaviour of the wire will be essentially elastic.
But if the wire is reloaded beyond its previous maximum load, say beyond C1 and up
to A2, there will be a sudden change in stiffness at C1. Additionally, plastic
deformations will occur as evidenced by the behaviour of the wire when fully
unloaded from A2 to B2. The new plastic deformation is approximately 11 mm.
56
In general, what these results show is that every time the wire is reloaded, it starts
off with a fairly linear and elastic behaviour until it approaches its maximum
previous load. Then the wire stiffness changes considerably at a more or less welldefined point and loading beyond that point results in plastic deformations. We will
refer to the change from a stiff to a noticeably softer response as “yielding”. And we
will refer to the point at which yielding begins as, well, “yield point”. That should be
easy to remember!
56
Let’s now have a look at how different types of stress can interact to make a material
yield. Besides the tensile tests discussed in the previous slide, Taylor & Quinney
(1931) also subjected thin-walled copper tubes to combinations of a tensile load P
and a torque Q as shown in the figure on the left. Before applying this combined load
condition, all specimens were initially loaded in tension to a load P0, which was then
reduced to a value P = m∙P0. The torque Q was then applied and increased beyond
yielding while the tension load P remained constant. The sketch on the right
illustrates this loading scheme.
By preloading each specimen with an axial load P0, Taylor and Quinney achieved two
things: 1) they ensured that all specimens had the same loading history, and 2) they
made the point of yielding easier to identify. For instance, going back to the results
in the previous slide, when the copper wire was loaded for the first time from A to
A1, there was no clear yield point. However, if the specimen is first preloaded to A1,
then unloaded to B1 and loaded again to A2, a reasonably well-defined yield point
can be detected at C1.
Taylor & Quinney (1931) used eight values of m which ranged from 0.025 to 0.95.
This allowed to them to identify the combinations of P and Q that resulted in
yielding. The results become largely independent of the dimensions of the specimen
57
if they are expressed in terms of the normal stress σz and the shear stress τθz instead
of in terms of the axial force P and the torque Q. Furthermore, we can make the
results independent of our choice of units (kPa, MPa, etc.) by normalising these
stresses by the normal stress due to P0 (σ0). The equations in the slide show the
calculation of σz/σ0 and τθz/σ0. In these equations, r is the mean radius of the tube, t
is the wall thickness, and J is the polar second moment of area. Approximations
applicable to thin-walled tube are used in the equations to calculate the cross
sectional area and the polar second moment of area.
57
The experimental results are highlighted in red in the figure. A well-defined trend is
apparent: When σz/σ0 is close to one, it takes a relatively small amount of shear
stress τθz to make the specimen yield. Conversely, when σz/σ0 is close to zero, it takes
a relatively high amount of shear stress τθz to make the specimen yield. In other
words, σz and τθz interact to induce yielding. If you think about it, this seems like a
reasonable result: The higher the normal stress the tube is carrying, the smaller the
value of τθz needed to get it to yield.
The figure also shows that the theoretical von Mises yielding criterion fits the data
very well. You may or may not be familiar with the von Mises criterion, that is not
terribly important to our current discussion. Rather, what I am hoping this slide
makes clear is that the red points suggest a boundary, or more generally a surface,
and that if we go beyond this surface (for example from A to B), yielding, or plastic
deformations will occur. But if we move within the surface (for example from A to C),
only elastic deformations will occur.
58
The figure illustrates the approach used to identify the yield point of the copper thinwalled tubes. Taylor & Quinney were aware that a material behaving elastically and
isotropically, such as their preloaded copper specimens, should not exhibit a change
in length δl during the application of the torque Q. That is, an elastic response to Q
involves only a twist while the length remains unchanged. So they measured the
change in length δl of the specimen as Q increased, and as soon as they observed
increases in length, they knew that the response was no longer elastic and the
specimen had yielded. For example, the figure shows that for m = 0.95, the copper
tube yielded at τθz ≈ 20 MPa, and that for m = 0.28, yielding occurred at τθz ≈ 120
MPa.
59
Two geometrical parameters were measured as the copper tubes were loaded:
change in twist δθ and change in length δl. When we were considering a triaxial
specimen, we expressed its deformation as strain parameters that were work
conjugates of the mean effective stress p' and the deviator stress q. We can do the
same for the copper tubes, by expressing the changes in twist and length as strains
that are work conjugate of the normal stresses σz and the shear stress τθz. As we will
see in a moment, taking the strains as work conjugates will allow us to plot the
strains in the same plot as the yield data and this will reveal useful patterns.
The work increment done on a copper tube when its length increases by δl and its
angle of twist increases by δθ is given by the first equation: the axial force P is
multiplied by the length increment δl and the torque Q is multiplied by the twist
increment δθ. By dividing both sides of the equation by the volume of the tube we
arrive at the expression in the second line were we readily recognise the stresses σz
and τθz. The expressions that multiply each of these stresses are the work conjugate
strains that we are looking for.
60
This figure again shows the yield points with σz/σ0 on the horizontal axis and τθz/σ0
on the vertical axis. The boundary suggested by the yield points is also repeated.
What is new here is that the axes represent, not only the stresses, but also the
corresponding strain conjugate. So the horizontal axis also represents the
longitudinal strain δl/l and the vertical axis also represents the shear strain r∙δθ/l.
The green arrows indicate the relative magnitudes of the two different types of
strain that took place as the copper tubes started yielding. For example, the arrow
that corresponds to the highest yield point has a vertical component much larger
than its horizontal component. So it is telling us that when the specimen started
yielding, its shear strain r∙δθ/l was much larger than its longitudinal strain δl/l. By
measuring the arrow we could find out exactly how much larger. Of course, there is
nothing special about the highest point, so we can do this type of analysis for any of
the arrows. The figure also shows that the arrows are approximately perpendicular
to the yield curve that fits the points. Materials that have this characteristic are said
to obey normality, in the sense that their plastic strain increment vectors are normal
(i.e. perpendicular) to the yield surface. The material can also be said to exhibit
associated flow because the plastic flow of the material is associated to its yield
surface. So we can use normality and associated flow interchangeably. As we will see
later, normality is one of the key assumptions underpinning the modified Cam Clay
61
model.
More generally, the plastic strain vectors are always perpendicular to a plastic
potential curve. So saying that a material obeys normality is equivalent to saying
that the yield locus and the plastic potential curve are the same.
Let’s take a minute to recall the loading path of these copper tubes which is
repeated in the bottom left corner of the slide: The tubes were initially loaded under
tension to P0, then unloaded to mP0, and finally a torque Q was applied while the
axial load remained constant at mP0. In other words, the tubes yield exclusively due
to the increase in torque Q. However, the green arrows show that upon yielding, not
only does the tube exhibit an increase in the angle of twist δθ, it also exhibits an
increase in length δl, despite the fact that the axial load remains constant.
This observation may seem counterintuitive… Why should an increase in torque
result in an extension of the tube? One might be tempted to think “surely if we are
only increasing the torque Q while the axial load P remains constant, then only the
angle of twist should change while length remains constant.” This line of reasoning is
correct for an elastic isotropic response, but is no longer valid for a plastic response.
Perhaps we can put ourselves in a better position to accept this plastic coupling
between Q and δl by remembering a similar coupling that we find easy to accept.
For example, the elastic coupling between longitudinal forces and lateral
deformations due to Poisson’s effect. If you think about it, it’s kind of the same
thing: you apply the force in only one direction (longitudinal), yet you deform it in
more than one direction (longitudinal and lateral). Similarly, the copper tubes were
made to yield only by increasing Q, but they were deformed by both an increase in
twist δθ and an increase in length δl.
61
OK, after this introduction to yielding and plasticity based on the behaviour of
copper, let us turn our attention to clays. Our approach will consist in realising that
the main features of the behaviour of the copper specimens are also observed in
clays.
We begin by considering the oedometer results on speswhite kaolin shown in the
left figure. As is common in soil mechanics, the results show a measure of density on
the vertical axis, in this case the specific volume v, and the effective vertical stress σ'v
on the horizontal axis. However, comparison to the behaviour of the copper
specimens is facilitated if we swap the axes and we plot the effective vertical stress
σ'v on a logarithmic scale as shown in the right figure.
Now the similarities between the clay and the copper wires subjected to tensile tests
start becoming apparent. The clay specimen deforms as it is loaded from A to A1. At
A1 we begin unloading and we realise that plastic or permanent deformations have
occurred because the unloading path A1 to B1 is different from the loading path A to
A1. In other words, there is not a unique load : deformation relationship. We reload
from B1 and discover that the reload path from B1 to C1 is very similar to the unload
path from A1 to B1. That is, the soil seems to be behaving elastically. Furthermore,
we notice that the soil yields as it reaches its maximum previous load at point C1.
62
These observations are all very similar to those that we did for the copper wires
subjected to tensile tests.
Notice that there is some difference between the reloading path B1 to C1 and the
unloading path A1 to B1. This is called hysteresis, and while it can be modelled when
necessary, it is often ignored in constitutive models for soils because the discrepancy
between the unloading and reloading curve is small in comparison to the changes in
specific volume that the soil undergoes.
By the way, a point such as C1 is what we usually refer to as the pre-consolidation
pressure of the soil: the maximum pressure that the soil has experienced in its
geological past. What we are seeing here is that the pre-consolidation pressure
found in an oedometer test can be interpreted as the stress at which the soil yields.
62
Similar patterns of behaviour are also observed in other types of test as shown in the
figure. Notice that only the results for isotropic compression use a logarithmic scale
for the stress axis. But regardless, the behaviour is remarkably consistent: when a
specimen is unloaded and then reloaded, it initially exhibits a relatively stiff and
elastic response. The stiffness of the soil reduces considerably, which is to say the
soil yields as the maximum previous load is approached.
63
When we were considering the thin-walled copper tubes, we saw that that a yield
surface was determined by loading the specimens along a variety of loading paths,
and identifying the yield points. The yield surface is simply the curve suggested by
the yield points. Remember, we are saying that so long as the specimens are loaded
with states of stress that fall within the yield surface, the specimens will not yield
and only elastic deformations will take place. But if the load goes beyond the yield
surface, then the specimen will yield and plastic deformations will occur.
To identify the yield surface of a soil specimen in a triaxial test, we follow the same
approach, except that this time our loading paths will be defined in q-p‘ space.
64
The slide illustrates how we can go about implementing this approach. Say we want
to determine three points of the yield surface. We would need an equal number of
identical specimens. This means that besides having identical physical characteristics
such as particle size distribution, mineralogy, density, fabric, etc.; the specimens
must also have the same loading history just like all the copper tubes were initially
loaded by the tensile load P0. In other words, the soil specimens must all have the
same pre-consolidation pressure.
Now we take each specimen and load it to an initial state such as point A shown in
part a) of the figure and we load each of the three specimens along a different stress
path. Three possible stress paths are suggested in the slide: 1) isotropic loading, 2) a
test in which we compress the specimen but we increase the cell pressure as
necessary to ensure zero lateral strain, and 3) a conventional undrained test.
For these three tests we could plot results of load vs deformation and thus identify
the yield point as suggested in parts b, c, and d of the figure. Notice that the type of
‘load vs deformation’ plot used to identify the yield point depends on the test. For
instance, a plot of q vs εq is suitable for an undrained triaxial test (part d), but this
plot wouldn’t make any sense for an isotropic compression test because both the
shear stress q and the shear strain εq remain equal to zero throughout the test. So
65
for the isotropic compression test we may have to resort to a plot of specific volume
v vs mean effective stress p' (part b). Similarly, part c suggests that for a zero lateral
strain test, a plot of specific volume v vs vertical effective stress σ'v may be wellsuited to identify the yield point. Muir Wood (1990) provides a much more detailed
discussion on methods used to find the yield point.
Regardless of the method used to identify the yield point, the q and p' values of the
point can be used to sketch an outline of the yield surface as suggested in part a of
the figure.
65
In the previous slide we saw that in order to determine three points of the yield
surface we need three different specimens loaded along three different stress paths.
One may wonder why bother with three specimens instead of using only one. That
is, why is it not viable to work with a single specimen as follows?
With reference to the upper figure: Load the specimen along the first stress path to
find the first yield point and then unload it back to A. Now load it along the second
stress path to find the second yield point and then unload it back to A. Finally, repeat
the process one more time along the third stress path to find the third yield point.
Actually, why stop there? Why not perform loading-unloading cycles along other
stress paths to continue finding more points on the yield surface?
The reason why this approach doesn’t work is that the yielding characteristics are
dependent on the loading history of the specimens. We saw this clearly in the results
of the copper wires which are repeated herein for convenience. The value of P at
which the wire yields depends on how the specimen has been loaded before. So if
the wire is loaded from B1, it will yield at P ≈ 0.6 kN because it has never
experienced a higher load before. But if the wire is loaded from B6, it will not yield at
P ≈ 0.6 kN but rather at P ≈ 1.4 kN because that is its maximum previous load.
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It is this dependency of yielding on the loading history that forces us to use multiple
specimens. Consider for instance the stress path AB shown in part a of the figure.
When the soil specimen is sitting at A, its yield surface (or yield locus yl) is yl1. The
stress path AB allows us to discover one point of yl1 which in this case is the point Y.
As suggested in part b of the figure, we can discover point Y by plotting q vs ϵq and
observing where the change in stiffness occurs. But by loading the specimen beyond
its yield locus, we have changed the yield locus. Indeed, when the specimen arrives
at B, its new yield locus must pass through point B. To paraphrase Muir Wood
(1990): “It is axiomatic that a stress state can lie on or inside but never outside a
current yield surface”.
If we were to unload from B to A, and load again from, say, A to C (part c of the
figure), we would eventually encounter a yield point, but it wouldn’t lie on yl1 such
as point Y. Instead, it would lie on yl2 such a point X. Clearly, we cannot use a single
specimen to discover multiple points on yl1.
67
The figure shows results from Tavenas et al. (1979) which correspond to seven
specimens of natural clay from St Louis, Canada. The specimens were sampled from
the same depth and are therefore assumed to be identical in every way, including
their loading history. In the lab, the specimens were loaded from point A, which
corresponds approximately to in-situ conditions, along different stress paths until a
yield point was identified. The yield surface is given by the dotted line which
connects the yield points.
68
These yield curves were reported by Graham et al. (1983) and correspond to
Winnipeg clay samples extracted from the same site but at different depths. Since
the vertical preconsolidation pressure σ'vc increases with depth, we expect the yield
surface for different depths to differ and the figure confirms this. However, given the
similarities between clay samples at different depth (particle size distribution,
mineralogy, cementation effects, etc.) the shape of the yield surfaces is fairly
constant with depth.
69
In fact, if the results from the previous slide are normalised by their respective
preconsolidation pressures σ'vc, they seem to define a unique yield surface.
The experimental results for clays that were presented in the last few slides support
the idea that clays yield and that they have yield surface. Furthermore, the results
from Graham et al. (1983) presented in this slide suggest that specimens that are
different only in terms of their loading history, may have yield surfaces that have a
different size, but the same shape. Of course, the results on the slide don’t prove
that this is always the case, they merely show that this is a possibility. The notion of
the yield surface changing in size but not in shape as the loading history changes will
be used in the Modified Cam Clay model.
Section 3.3 of Muir Wood (1990) provides additional references that document the
yielding of clays.
70
In a previous slide we saw experimental evidence that supports the idea that the
plastic deformations of copper obey normality. What about soils? Do they also obey
normality? These results correspond to the Winnipeg clay reported by Graham et al.
(1983). In the upper part of the figure, each axis represents the normalised stress as
well as the work conjugate strain. The strain symbols δεp and δεq have a superscript
‘p’ to make it clear that these are plastic, not elastic, strains. The vectors of plastic
strain increment are roughly perpendicular to the yield surface, thus suggesting that
this clay obeys normality. The lower figure shows the extent to which the arrows
deviate from perfect normality. Significant deviations of up to ±30° are present, but
the overall trend supports the normality hypothesis.
If after seeing these results, you are still unsure about whether all soils obey
normality, well good on you. Because all we have seen is that this particular clay
(Winnipeg clay) roughly obeys normality. And let’s be honest, this is certainly not
enough evidence to claim that all soils obey normality. However, we can accept this
as sufficient evidence to claim that normality is a possibility in soils and, on that
basis, use it in the development of the Modified Cam Clay model.
Although only experimental results for clays have been presented so far, the
concepts of yielding and yield surface are equally applicable to sands. However, the
71
experimental approach to determine the yield surface of sands is significantly
different because, unlike clays, sampling of sands generally disturbs the specimens
significantly. But the principles are the same: sands yield upon loading and their
yield surface in q vs p' space can be defined with adequate testing. Section 3.4 of
Muir Wood (1990) provides a nice description of the approach used to determine
the yield surface of sands in a triaxial test. Something to keep in mind though, is that
there are experimental results that show that sands often do not obey normality
(e.g. Poorooshasb et al. 1966). Muir Wood (1990) reports that the models that have
been most successful in modelling the behaviour of sand are those that do not
invoke normality. Such models are necessarily more complex as they require us to
know the shape of both the yield locus and of the plastic potential curves.
Remember, assuming normality amounts to saying that the yield locus and the
plastic potential curves are the same.
71
72
So far we have accomplished three main things:
1) We agreed on work conjugate stress and strain parameters that we can use to
describe a triaxial test.
2) We covered the basics of isotropic elasticity. In particular we saw that we can use
the bulk elastic modulus K and the shear elastic modulus G to describe an
isotropic-elastic response.
3) We acquainted ourselves with some of the basic concepts of plasticity: yielding
and yield surfaces.
Let’s now bring these concepts together to derive the expressions that we will use in
MCC to compute strains. Remember that in order to remain focused on the essential
characteristics of the model, we will limit our scope to loading conditions that can be
attainted in a triaxial test.
73
Let us consider a triaxial test specimen formed in its loosest condition. It is
subsequently loaded for the first time along the stress path OC shown in figure a.
Whenever a soil specimen is loaded or compressed for the very first time since its
formation, we say that the soil is being normally compressed. In a triaxial test, this
normal compression can be done following any given stress path, for example
isotropic loading, or a loading that ensures that the lateral strain remains zero. This
second option is called K0 loading and is likely to be the most common one in situ
because in nature, as soil forms over a large lateral extent, it is able to deform
vertically, but has limited lateral strain. So in the context of Muir Wood (1990),
unless stated otherwise, whenever we refer to “normal compression”, we will be
referring to a soil that is loaded for the first time AND along a K0 stress path (zero
lateral strain). Later in the course we will also refer to isotropic normal compression:
That is, a situation in which a soil is loaded for the first time by applying an isotropic
stress condition.
As a side note, notice that during a triaxial test, if you start loading from the origin of
the stress plane (q vs p') and you want to compress the soil vertically only (keeping
lateral strains at zero), then every time that you increase the effective axial stress σ'a,
you will also have to increase the effective cell pressure σ'r, to prevent the specimen
from expanding laterally. If we assume that throughout this process the effective cell
74
pressure σ'r is equal to a fixed fraction of the effective axial stress σ'a, (that is, σ'r =
A*σ‘a where A is a constant) then the stress path is a straight line that passes
through the origin in q vs p' space as shown in figure a. Can you demonstrate this?
OK, back to our specimen. As we compress the specimen for the first time
(normally) from O to C, our loose specimen yields and experiences irrecoverable
deformations from the onset of loading. So all along from O to C, the state of stress
of the specimen lies on the yield locus and this yield locus gets pushed out as
loading progresses (figure a). By keeping track of the volume of the specimen, we
can also plot our results in the compression plane (v vs p') shown in figure b. This
line is referred to as the the normal compression line (ncl). Since we are assuming
zero lateral strain, a more specific label for this line on figure b would be 1D-ncl or
K0-ncl. But as explained above, whenever we say ncl, we’ll just assume that it is 1D
(for now).
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Let us suppose now that we apply a stress path that takes us from C to any point
inside the yield locus with a smaller p'. For example, point A in the figure. In the
compression plane, our results would now trace the unloading curve since we are
reducing p'. For simplicity the unloading and reloading curves are assumed to be
identical. If we now apply any arbitrary stress path that takes us from A to B in the
stress plane, the response will have to be entirely elastic because we remained
inside the current yield locus. In going from A to B we have incurred in both a δp'
and a δq. Our equations for elastic response, repeated here for convenience,
indicate that δp' will induce a change in volumetric strain δεpe and δq will induce a
change in shear strain δεqe. The superscript ‘e’ has been added to the strain
increments to make it clear that these are elastic strains.
Let’s further consider the change in volumetric strain. Notice that in going from A to
B, δp' is negative because p' decreased. The bulk elastic modulus is always positive.
So the first equation is telling us that we will obtain a negative δεpe. I included the
third equation just as a reminder of how we defined the volumetric strain. We
introduced a negative sign into the definition so that negative δεp meant that the
volume was increasing. So the first equation yields the reasonable result that by
reducing p', the volume of the specimen increased. If we measure the new larger
specific volume of the soil at point B, we can plot this point on the compression
75
plane. Furthermore, since we are assuming an elastic response, we would expect
points A and B to fall on a unload-reload line (url) in the compression plane simply
because behaviour along a url is elastic. More generally, we expect that any q-p'
point inside the current yield locus, should correspond to a point on the url.
Let’s go back to point A in the stress plane and let’s assume that instead of loading
to point B, we load to some other point B’ that has the same p' value as B.
Necessarily, stress paths AB and AB' involve the same δp'. Since the elastic
increment of volumetric strain δεpe is only dependent on δp' (see first equation), the
response in the compression plane is the same for the stress paths AB and AB’. More
generally, the first equation implies that all points inside the current yield locus that
have the same p' value (i.e. they fall on the same vertical line in the stress plane)
correspond to a unique point on the compression plane.
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OK, I’ll be honest, you can kind of ignore this slide, because all you really need to
know is that when the normal compression line ncl and the unload-reload line url
are plotted in the semi-logarithmic space v vs log(p'), they often linearise reasonably
well. That’s it.
In this slide I am just trying to make a point that seems to go unnoticed all too often:
This linearization in semi logarithmic space is equivalent to saying that the ncl and url
follow an exponential form like the one shown in the first line, where ‘a’ and ‘b’ are
fitting constants. The slide shows how by applying logarithms on both sides of the
equation, we eventually conclude v vs logp‘ has a linear form. I have deliberately not
indicated the base of the logarithm to make the point that the transformations work
with any base.
76
Here we have the ncl and url in semi-logarithmic space. Notice that in the horizontal
axis we are using a natural logarithm, which is to say that the base of the logarithm is
e = 2.718… Our ncl and url are now straight lines. And as any straight line, they can
be fully defined by knowing their slope and one point on the line. Let’s call the slope
of the normal compression line ncl λ (lambda) and the slope of the unload-reload
line url κ (kappa).
As for the point on the line, the convention is to specify the value of v at p' = 1 kPa as
suggested in the figure. For the ncl, we will call this v value vλ. And for the url, we will
call it vκ. We can think of vλ and vκ as “y-intercepts”, except that instead of placing
the vertical axis at p' = 0, we have had to put it at p' = 1 kPa, for the practical reason
that ln(0) is not defined. The figure illustrates the meaning of λ, κ, vλ, and vκ.
Notice that the ncl and url will always have negative slopes, that is, as p' increases,
or more exactly, as ln(p’) increases, v will decrease. Strictly speaking this means that
λ and κ are negative. However, to avoid dealing with negative soil parameters, we
can define the equations of the ncl and url with the negative sign already baked in,
as shown in the slide.
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Since I mentioned the number e in the previous slide, now I can’t resist the
temptation of mentioning two YouTube videos that explore the fascinating
properties of this number. So if you need a break from soil mechanics, feel free to
check them out:
1) “The number e explained in depth for (smart) dummies” in the Mathologer
channel
2) “What's so special about Euler's number e?” in the 3Blue1Brown channel
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OK, back to soil mechanics. We can use the equation of the url to calculate the
elastic change in specific volume Δve due to a change in mean effective stress p'. The
slide works through the equations and arrives at the expression in the dashed box
which the exact value of Δve.
Although we could very well work with this expression, exact solutions are not
readily available for other aspects of the Modified Cam Clay model and we are
forced to use approximate incremental equations. So just for the sake of consistency,
let’s also compute an incremental approximation that is valid for small elastic
increments of specific volume δve.
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Let’s clarify what we mean by “incremental approximations”. The figure illustrates
the general scheme. Suppose there is a function f(x) that we are interested in. Let’s
say we don’t know f(x), but we do know one point of f(x) and we also know how the
derivative (or the slope) of f(x) varies with x. So what we have is a differential
equation: we know an initial condition (a point on f(x)), and we know f’(x). Our task
is to find f(x). I will assume that the known point is [xi, f(xi)], as shown in the slide. I
will also assume that there is no exact solution for this differential equation of ours,
so the best we can hope for is to approximate a number of points of f(x).
One way to phrase this problem is to ask: if we are sitting at xi and we move some
distance to the right to xf, what is the resulting change in f(x)? We can get an
approximate answer by noting that the derivative evaluated at xi is approximately
equal to the slope between the point we know [xi, f(xi)] and the point we wish to
determine [xf, f(xf)]. Judging from the figure on the slide, one has to conclude that
this is not a good approximation. However, the approximation gets better and better
as xf gets closer and closer to xi. In other words, the approximation improves as δx =
xf – xi approaches zero. This approximation is given by:
𝑓 𝑥 ≈
𝑓 𝑥 −𝑓 𝑥
𝛿𝑓 𝑥
=
𝑥 −𝑥
𝛿𝑥
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Solving for δf(x), which is our parameter of interest, we get:
𝛿𝑓 𝑥 ≈ 𝛿𝑥 𝑓 𝑥
So now we can approximate our second point of f(x) as [xf; f(xi) + δf(x)]. We can use
this second point to approximate a third point, and the third point to approximate a
fourth point, and so on… So long as we take adequately small steps in x, our
approximated points will fall very close to the exact f(x).
Coming back to our estimation of f(xf), can you indicate where on the figure our
approximated point [xf; f(xf)] will fall?
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The figure shows the approximated point [xf; f(xf)]. Again, notice that as δx gets
smaller, the difference between the approximated point and the exact point
decreases.
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Let’s apply this incremental approximation approach to the estimation of the elastic
change of specific volume δve. In this case the independent variable is the mean
effective stress p' and the dependent variable is the specific volume v. So the
equivalent of f(x) is v(p'). The correlation between v and p' is the unload-reload line
url whose equation is shown in the slide. The derivative of the url is also shown. By
applying the incremental approximation scheme, we get the approximation of δve
shown in the slide.
Although this is an approximation, we use the equal sign (=) instead of the
approximation sign (≈) because by keeping δp' sufficiently small, we can get very
close to the real answer. Also, strictly speaking, in the equation for the
approximation of δve, the mean effective stress p' should have a sub-index “i“ to
make it clear that the derivative is evaluated at the initial point (see previous slide).
We are dropping this sub-index simply for convenience.
Finally, the approximation of δve allows us to approximate the final value of specific
volume vf due to the change in mean effective stress δp' along the unload-reload line
url. This is shown in the last two equations of the slide.
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Coming back to the expression for δve, we can combine it with the expression for
volumetric strain δεp and we arrive at an expression for the elastic volumetric strain
increment δεpe. Let’s keep in mind that the superscript “e” stands for elastic, and the
subscript “p” represents volumetric.
By comparing this expression for δεpe to the previous expression that we had derived
in terms of the bulk elastic modulus K, we can arrive at an expression for K, as shown
in the slide. Let’s run a small “sanity check” on this equation…
It says that bulk elastic modulus K is directly proportional to the mean effective
stress p'. I can certainly believe that. Greater p' results in greater contact stresses
between particles, and therefore greater friction between these particles. This will
contribute to a higher stiffness, or higher K. The equation also says that K is inversely
correlated to kappa κ, which is the slope of the unload-reload curve (shown in the
figure as a reminder). Again, this makes intuitive sense: High kappa κ means that the
soil undergoes large changes in specific volume v when p’ increases. We can
reasonably expect such a soft soil to have a low value of bulk elastic modulus K.
So far so good. However, the expression for K also says that K is directly proportional
to the specific volume v. I don’t know about you, but I find this hard to swallow. A
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large specific volume v corresponds to a loose soil with plenty of voids in between
its particles. All other things being equal, why should such a loose soil have a larger
bulk modulus K than a dense soil with little voids in between its particles? Let’s
simply accept this as a shortcoming of the model.
Now that we have expressions for elastic increment in specific volume δve and
elastic increment of volumetric strain δεpe in terms of the url slope (κ), let’s turn our
attention to deriving expressions for plastic volumetric strain increments δεpp.
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Let’s now work on expressions for plastic volumetric deformations. The first thing we
need to define is how the yield locus will evolve as loading proceeds. Suppose a soil
is one-dimensionally normally loaded along the stress path OB shown in the main
figure. By the time the loading reaches point A, we have formed the yield locus yl1
which necessarily passes through point A because no state of stress can fall beyond
the yield locus. Likewise, when the loading reaches point B we have formed the yield
locus yl2 which necessarily passes through point B. That is, as loading proceeds, the
yield locus evolves by getting bigger and perhaps changing its shape.
Based on the experimental results on Winnipeg Clay, reported by Graham et al.
(1983) (see bottom left corner), we will now assume that the shape of the yield
locus remains constant as loading progresses. So for example, if by loading from O to
A, we created the yield locus yl1 which has, say, a circular shape, then by loading
further from A to B, we would generate yl2 which would also be a circle, but larger
than yl1.
We will also assume that the shape of the yield locus is independent of the stress
path. According to this assumption, yl1 could have been formed by loading from O to
any other point that lies on yl1, not just A. For example by loading from O to K.
Similarly, yl2 could have been formed by loading from O to X or from O to L, and not
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just from O to B.
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Let’s now consider how different stress paths can be mapped on the compression
plane (v vs p'). Let’s begin with the stress path OAK. From O to A, the soil is normally
consolidated, and therefore in the compression plane (v vs p') the specimen traces
out the normal compression line ncl up to point A. Now we load from A to K. Since
this portion of the stress path falls entirely inside the yield locus yl1, the soil
response must be elastic. We know that in the compression plane, elastic responses
occur along the unload-reload line url, so we can conclude that the segment AK in
the compression plane must be tracing out url1 which intersects the ncl at point A.
A similar analysis can be done of the stress path OBL. In the compression plane, the
segment OB traces out the normal compression line ncl. And since BL in the stress
plane falls entirely inside yl2, the response must be elastic. Accordingly, segment BL
in the compression plane must be tracing out url2 which intersects the ncl at point B.
Lastly, let’s consider another stress path that also touches point L: OAKLM. We
already covered OAK above. Now, as we go from K to L in the stress plane, it is clear
that we are stepping outside yl1 and therefore the soil must be yielding. This implies
that in the compression plane, the segment KL must not fall along the url. Instead,
the KL segment in the compression plane must be moving towards the same point L
that we previously established when considering the stress path OBL. So the stress
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path OAKL has again landed us on url2 in the compression plane. The last step of the
stress path is LM which falls entirely inside yl2. Therefore, in the compression plane,
the segment LM must fall along url2.
So what we are seeing here is that, regardless of the stress path, there is a one to
one correspondence between a given url and all the points that fall on or inside the
current yield locus. So point L must fall on url2 whether we get to it via OBL or via
OAKL.
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The pattern of behaviour described in the previous slide is supported to some extent
by the experimental data on Winnipeg clay reported by Graham et al. (1983). All
specimens that have the same vertical pre-consolidation pressure σ'vc, and therefore
the same yield locus, tend to line up along a single url.
The dashed vertical lines serve to indicate that every single point in the stress plane
can be mapped on the compression plane.
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Suppose we get to point K via the stress path OAK. The current yield locus is yl1.
Let’s consider the loading-unloading stress path KLK. When we apply KL the soil
yields and undergoes a change in specific volume Δv that is equal to the vertical
separation between K and L in the compression plane. When we unload from L to K,
in the stress plane (q vs p’), the data naturally returns to point K, but since now the
stress path is moving inside the current yield locus yl2, in the compression plane the
data must move along the corresponding url, which is url2, until it reaches the point
K' as shown in the figure.
It is thus clear that the total change in specific volume Δv that occurred when
loading from K to L had an elastic component Δve that was recovered during
unloading and a plastic component Δvp that was not recovered. To develop the
modified Cam Clay model we need to be able to compute these components
individually.
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The approach to recover the plastic change in specific volume Δvp may be best
understood if we first revisit the tensile behaviour of the copper wires reported by
Taylor and Quinney (1931). We first came across these results a few slides ago when
we were becoming familiar with plastic behaviour. The results in the dashed box will
hopefully refresh your memory. In particular, let’s recall that loading began with the
sequence A – A1 – B1 in the P vs δl plane. Figure a shows only the next reloadunload cycle: B1 – C1 – A2 – B2. The plastic (permanent) deformation that this cycle
induces is given by the horizontal separation between B1 and B2. But if url1 and url2
are parallel, then we can say, more generally, that the plastic deformation is the
horizontal separation between the two urls at any height. We don’t necessarily have
to measure this separation at P = 0.
The same concept applies to soil behaviour. This is shown in figure b, where the axes
of the compression plane have been swapped (p' is on the vertical axis and v on the
horizontal one) just to facilitate the comparison to figure a. Again, since the urls are
parallel, the plastic change in specific volume Δvp is given by the horizontal
separation between the urls at any height. Figure c shows the exact same
information as figure b, except that in part c that axes have been swapped back to
their conventional position and p' is plotted on a logarithmic scale, as is common
practice in soil mechanics. So now Δvp is given by the vertical separation between
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the urls at any p' value.
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You might be thinking: “Wait a minute, at what time did we agree that the urls are
parallel?” This parallelism is the consequence of assuming that kappa κ, which is
defined as the slope of the url (see figure), is a soil constant. Like any assumption,
this is not strictly true, but it is adequate for a relatively basic model such as
Modified Cam Clay MCC.
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OK, back to the task of computing Δvp. Or more specifically, back to calculating the
vertical separation between url1 and url2. Remember, this is the plastic change in
specific volume due to a specimen going from anywhere on url1 to anywhere on
url2.
A simple way of finding the vertical separation between urls is to compute the
difference between the vκ values of both urls as suggested in the figure.
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An alternative approach which is more elaborate but that will be of greater use to us,
is to consider what happens when we normally compress a soil from points A to B
shown in the figure. Notice that points A and B are the points at which the urls
intersect the ncl. Points A and B have p’ values p'o1 and p'o2, respectively.
Upon loading from A to B, there is a total reduction in specific volume Δv which we
can compute as shown in red. We can find the plastic component of Δv by adding to
it the elastic increase in specific volume Δve that would occur if, after loading to B,
we were to reduce the load from p'o2 back to p'o1. The calculation of the elastic
component is shown in green.
The final expression at the bottom gives the exact value of Δvp for any pair of values
p'o1 and p'o2. Let’s now look at how we can obtain an approximate incremental
expression that is valid when p'o1 and p'o2 are relatively close to each other.
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The exact expression that we arrived at in the previous slide, implies that the plastic
volumetric strain vp, measured with respect to the vp at p’o = 1 kPa, is given by
equation 1. By computing the derivative of vp with respect to p'o (equation 2), we
arrive at the incremental approximation shown in equation 3. This is the same
incremental approximation that we used to compute the elastic change in specific
volume δve.
Equation 4 is a reminder of the general expression of volumetric strain. Remember
that the sub-index p stands for volumetric. Combining equations 3 and 4, as
suggested by the colours, we arrive at equation 5 which is an expression for the
plastic volumetric strain δεpp, where the super index p stands for plastic.
Remember, strictly speaking, in equations 3 and 5 we should use (≈) instead of (=).
However, with a spreadsheet we can get really close to the real answer, so we will
just use the equal sign (=).
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Now that we have computed expressions for the plastic and elastic components of
the total change in specific volume (δv) and the total change in volumetric strain
(δεp), we can add up the components to arrive at the expressions for δv and δεp
shown in the slide.
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Let’s consider a few examples that illustrate how volumetric deformations may be
purely elastic, purely plastic, or a mixture of both. With reference to the figure, say
we normally compress along the stress path OA. By doing so, we create the yield
locus yl1 which is associated with the unload-reload line url1. We now apply the
stress path AP which lies entirely inside yl1 and thus results in an elastic response.
This means that in the compression plane, the segment AP falls along url1. This
stress path OAP which I have just described is not shown in the figure to avoid
clutter.
So now that we are at P and the current yield locus is yl1, let’s look into what
happens when we move along the three different stress paths in the figure: PQ, PR
and PS. Let’s start with PQ, which falls inside yl1, meaning that the soil response
must be exclusively elastic. More specifically, the volume will change elastically in
response to the change in p' involved in going from P to Q. Thus, in the compression
plane (v vs p’) the segment PQ lies on url1. Notice in the stress plane (q vs p') that as
we advance from P to Q, the yield locus does not change quite simply because the
soil is not yielding. Accordingly, p'o remains constant at p'o1 (δp'o = 0). More
generally, we can say that p'o changes if, and only if, the yield locus changes. It
follows that the plastic change in volume is zero if, and only if, δp'o = 0. This is
consistent with our expressions for δvp and δεpp in the previous slide.
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If we now look at stress path PR, we see that it goes beyond yl1 and therefore it
must induce plastic deformations. The stress path creates the yield locus yl2 on
which point R lies. PR plots as a vertical line in the stress plane, implying that this
stress path involves constant p’ (δp’ = 0). In the compression plane, PR must also be
a vertical line. More specifically, since in the stress plane we go from yl1 to yl2, in
the compression plane we must go from url1 (which is associated to yl1) to url2
(which is associated to yl2). Fair enough, but it may not be immediately obvious
whether url2 should lie above or below url1 in the compression plane (or at least it
wasn’t immediately obvious to me when I was learning this). One way to think
through this is to realise that yl2 is the same yield locus that would have been
created if we had continued the normal compression of the soil from A to B. Now we
notice that B is lower than A in the compression plane and consequently, url2 has to
lie lower than url1. In other words, the url associated with a yield locus is
independent of the stress path by which that yield locus was created. One last thing
to notice about the stress path PR, is that since it involves δp’ = 0, it cannot induce
any elastic volumetric changes. This is a direct consequence of assuming that the soil
is isotropic, which implies that δp’ = 0 if, and only if, δεpe = 0 (see first equation in
dashed box). This is consistent with our expressions for δve and δεpe in the previous
slide, both of which become zero when δp’ = 0.
Now to the third stress path PS which induces both a change in p’ and a change in
p’o. That fact alone tells us that this will induce elastic (due to δp’ ≠ 0) AND plastic
(due to δpo’ ≠ 0) changes in volume. We also notice that in the stress plane PS ends
up in yl2, so immediately we know that in the compression plane PS must land on
url2. So far so good. But upon closer inspection of PS on the compression plane, we
may be a bit surprised to realise that it plots as a horizontal line, meaning constant v
and therefore no volume change. That’s a bit strange, don’t you think? We just said
that PS induces both elastic and plastic volume changes and now we are saying that
the volume change is zero. Don’t fret, there’s a simple explanation for this: the
elastic volumetric strain is equal in magnitude and opposite in sign to the plastic
volumetric strain, so they add up to zero as shown in the slide. We will return to this
concept when we simulate undrained triaxial tests using the MCC model.
By the way, notice that a stress path starting at P and pointing downwards into yl1
would imply δp’ = 0 and δpo’ = 0. It would therefore induce neither elastic nor plastic
volumetric strain. The volume wouldn’t change at all and in the compression plane
the data would not move from point P. Due to changes in q though, there would be
elastic changes in shear strain (see second equation in dashed box).
NB: When the equations in the dashed box were first introduced many slides ago,
δεp and δεq did not have the superscript “e” for elasticity because back then we
were only talking about elastic deformations. But now that we are dealing with both
elastic and plastic deformations, I use the superscripts “e” and “p” to make the
distinction.
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The table summarises the expressions for shear and volumetric strain that we have
derived so far and which we will be using to implement MCC. We are still missing an
expression for increments in plastic shear strain δεqp. To derive this expression we
will use the concept of “normality”, which we visited before (see figure in dashed
box). Just as a reminder, normality describes the situation in which the plastic strain
increment vectors are perpendicular (or normal) to the yield locus.
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97
Before we take a closer look at the calculation of increments in plastic shear strain
δεqp, we have to address a small change in what we mean by normal compression
line ncl. So far, when we have referred to normal compression, we have had in mind
K0 or 1D normal compression. That is, a soil loaded for the first time in such a way
that lateral strains are kept at zero. Although this type of loading is common in situ,
it is not that common in laboratory testing, where most soil specimens are initially
loaded isotropically before being sheared. Since it is these laboratory results that
have largely guided the development of the MCC model, we will now focus on
normal compression lines ncls that have been generated isotropically. We will refer
to these ncls as iso-ncl to distinguish them from the 1D-ncls.
This change is only a minor detail, and everything we’ve said about the ncl continues
to be valid. We will continue to assume that the iso-ncl linearizes in v vs lnp' space
and we will even assume that the slope of the 1D-ncl and the iso-ncl are the same
(λ). Also, we will continue to refer to the p' value at the intersection between the ncl
and the url as p'o (see figures).
For consistency with Muir-Wood (1990), we will allow for a difference in specific
volume v between the 1D-ncl and the iso-ncl at p' = 1 kPa. For the 1D-ncl we called
this value vλ, but for the iso-ncl we will call it N as shown in the equations.
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Perhaps the most important difference between the 1D-ncl and the iso-ncl is their
corresponding stress paths. For the 1D ncl the stress path is an inclined straight line
as shown in the upper left plot. But for the iso-ncl the stress path plots along the p'
axis as shown in the upper right plot. This is because q = 0 for any isotropic stress
condition.
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OK, back to our calculation of plastic shear strain increment δεqp. Consider a soil
specimen loaded to point A with current yield locus yl1. When the specimen is
subjected to the stress path AB, its yield locus will expand to yl2. Accordingly, it is
clear that the AB stress path induced a change in p' (δp' = 0) and a change in p'0 (δp'0
= 0). The implication is that the elastic (δεpe) and plastic (δεpp) components of
volumetric strain increment will both take on nonzero values.
We have an equation to compute the plastic volumetric strain increment δεpp (see
dashed box in slide). So, as a direct consequence of assuming normality, the
calculation of the plastic shear strain increment δεqp boils down to multiplying the
plastic volumetric strain increment δεpp by the slope of the normal to the yield locus.
And allow me to state the obvious here, but we could also calculate the plastic shear
strain increment δεqp by dividing the plastic volumetric strain increment δεpp by the
inverse of the slope of the normal to the yield locus.
You could rightfully ask whether we should use the slope of the normal to the yield
locus at point A or at point B. Ideally you would have to integrate this multiplication
over the entire stress path AB to consider the slope of the normal at every single
point on this stress path. However, in practice we do this integration numerically
using a spreadsheet and the results are satisfactory provided that the we move from
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A to B in adequately small steps. For most cases, this is relatively easy to achieve
with a spreadsheet.
The approach described in the previous slide to calculate the plastic shear strain
increment δεqp requires us to know the shape of the yield locus. After all, if we don’t
know this shape, then we certainly cannot compute the slope of the line that is
normal to this shape. So far we have not formally described the shape of the yield
locus used in the modified cam clay MCC model. In previous slides we loosely
represented the yield locus as a segment of an inclined ellipse. In this slide we
deviated from this and instead used semi circles to represent the yield locus. To
explain concepts, the assumed shape makes no difference, but to make actual
calculations of soil behaviour, we have to agree on a specific shape. The following
slide looks into this.
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To keep the model as simple as possible, we will assume that the yield locus is an
ellipse that is symmetric about the horizontal axis and that always passes through
the origin. By preventing the yield locus to go into the negative side of the p' axis, we
are essentially saying that tensile stresses are inadmissible. This seems reasonable
for most soils except those with heavily cemented particles.
We had previously defined p'0 as the mean effective stress p' coordinate of the
intersection between the normal compression line ncl and the unload-reload line url.
Given that 1) we are considering that the ncl is generated by isotropic loading (a
stress path that coincides with the p' axis), and 2) the point on the yield locus with
the highest p' value falls on the p' axis, it follows that the point on the url with the
highest p' value, falls on the iso-ncl. In other words, the iso-ncl acts as a right bound
of the url. This is more clearly appreciated in the bottom right figure which shows
the iso-ncl and the url in a semi-logarithmic plot.
The dashed box in the upper right corner of the slide is there to highlight the
significant difference between the elliptical yield locus used in MCC and the more
complex shape reported by Graham et al. (1983) for Winnipeg Clay. Clearly our
elliptical yield locus is highly idealised.
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We need two parameters to describe our ellips mathematically:
1) A size parameter that describes how far to the right the ellipse reaches (p'0)
2) A shape parameter that describes how the height of the ellipse compares to its
width (M)
The equation used to mathematically describe the ellipse is shown in the slide. This
is an implicit equation, since the deviator stress q is not given explicitly as a function
of p'. Rather, q and p' are jumbled up in the equation.
Out of convenience, we introduced a stress ratio parameter η = q/p' in the ellipse
equation. η is the Greek letter “eta.” Notice that the uppermost point of the ellipse
corresponds to η = M.
To get a better feeling for this equation, you may find it useful to experiment with it
using a function plotter such as the one available at www.desmos.com/calculator. To
do so you will have to replace p' with x and q with y as shown in the slide. Once
you’ve done that, you can plug in different values of M (say between 0.8 and 1.5)
and different values of p'0 and observe the effects on the resulting ellipse.
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Now that we have a mathematical description of the yield locus, we need a
mathematical expression for the slope (or its inverse) of the line that is
perpendicular to the yield locus at any point. There’s a bit of calculus involved here
that you may or may not recall from your undergraduate courses. That’s fine. I just
need you to keep in mind the essence of what we are doing: finding an expression
for the slope of the line that is perpendicular to the yield locus.
The process is illustrated in the slide and consists essentially of the following steps:
1) Express the yield locus as a function g, such that g = 0.
2) Compute the partial derivatives of g with respect to p' and with respect to q.
Hand calculations get a bit tedious, so I suggest using a symbolic calculator to
compute these (e.g. Mathematica).
3) The slope is given by the partial derivative of g with respect to q divided by the
partial derivative of g with respect to p'. We will work with the inverse of the
slope though.
A few questions to ponder: How can we satisfy ourselves that the equation from the
third step is producing reasonable results? For a fixed value of M, why do we only
need a single value of η to compute the slope? The equation from step 3 becomes
undefined when η = 0, why? The equation from step 3 becomes zero when η = M,
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why?
Note to self : Provide greater detail on the second step.
102
Now that we have an expression for the inverse of the slope of the normal to the
yield locus (third step in previous slide), we are ready to compute the plastic shear
strain increment as shown in the slide.
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Here is the summary of equations for plastic and elastic strains. Notice:
- Elastic volumetric strain increments δεpe are zero when δp' = 0.
- Elastic shear strain increments δεqe are zero when δq = 0.
- Plastic strain increments (whether δεpp or δεpp) are zero when δp0' = 0. In other
words, if the yield locus does not change size, plastic strains cannot occur.
- The addition of the elastic and plastic components of the of shear strain
increments yield the total shear strain increment δεq.
The five soil parameters needed to implement Modified Cam Clay MCC are also
summarised. These five parameters will always be assumed to be known. Strictly
speaking, an elastic-plastic model such as MCC also requires a set of plastic potential
lines, but by virtue of normality we are assuming they are equal to the yield loci.
We are now ready to use MCC. Before we start plugging numbers into the equations
in this slide, let’s have a look at a few graphical examples from Muir Wood (1990)
that highlight the main features of the model. For now let us focus on getting a clear
qualitative understanding of MCC. We will delve into the equations later.
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105
Let’s first consider a drained conventional compression test (i.e. δq = 3δp'). The
stress path is represented in the stress plane (q vs p') with a dotted straight line that
has a slope of 3. We assume that the soil is currently subjected to the state of stress
B which lies on the current yield locus ylB. We further assume that we know the
specific volume v of the soil at B. The yield locus ylB has a size characterised by p'0B
and is associated to urlB. Accordingly, the point defined by the intersection between
urlB and the iso-ncl has a horizontal coordinate p'0B. We can locate point B in the
compression plane (v vs p') by projecting it from the stress plane straight down to url
B. This is why in the figure we can see point B in the stress plane and in the
compression plane.
Suppose now that we load the soil specimen from B to C. In doing this we induce δp’
and δq which we can use to calculate the elastic strains (volumetric and shear) using
the equations in the previous slide. Elastic strains are done. Now, since the stress
path BC goes beyond ylB, a new yield locus ylC, that passes through point C, must
come into effect. Notice that for a known M, there exists only one ellipse that passes
through the origin, is symmetrical about the p' axis, and passes through point C. In
other words, if we know C, we know p'0C, or rather we can calculate p'0C. More
mathematically, if we take the p’ and q values of point C and plug them into the
ellipse equation (see equation in dashed box), you can solve for p’0. This effect of the
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yield locus becoming larger as loading proceeds is referred to as strain hardening.
Once we know p'0C we can project it down to the iso-ncl to locate urlC and now we
can project point C from the stress plane down to urlC. Now that we know urlB and
urlC we can determine the plastic increment in specific volume δvp simply by
computing the vertical separation between the two urls (see definition of Δvp figure
in dashed box). The way urlB and urlC are shown in the figure, makes it look like
their vertical separation varies as we move along p'. However, remember that urlB
and urlC are parallel because κ is constant (i.e. both urls have the same slope) and
therefore their vertical separation does not change with p'.
As shown in the slide, calculating the plastic volumetric strain increment δεpp is now
a matter of dividing δvp by the specific volume v that the soil had at B. Furthermore,
computing plastic shear strain increment δεqp is readily done by considering the
slope of the normal to the yield locus at B. And there you have it, you’ve just
qualitatively computed the elastic and plastic strains using MCC.
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If we continue loading along CDEF on the stress plane, every time we move beyond
the current yield locus, a new one will come into effect. At every step from C to D,
and from D to E, and so on… we can calculate δp' and δq and use it to compute the
elastic strain increments (see expressions for δεpe and δεqe in the “What we have so
far” table).
For each new yield locus we can calculate its corresponding p’0 value and project it
downwards to the iso-ncl as indicated by the red dotted lines. This in turn allows us
to identify the url associated to the new current yield locus. By projecting points C,
D, E, F from the stress plane down to their corresponding urls, we can keep track of
the test in the compression plane as shown in Figure b.
So far we can keep track of the test on the stress plane (fig. a) and in the
compression plane (fig b). But to generate plots c and d we need the total shear
strain increments δεq. We already have the elastic component of shear strain
increment (δεqe = δq/3G). To compute the plastic shear strain increments δεqp we
use the fact that the vertical separation between two successive urls is the plastic
increment of specific volume δvp. We then use δvp to compute the increment of
plastic volumetric strain δεpp (see previous slide). Finally, we compute the plastic
shear strain increment δεqp as the multiplication of δεpp by the slope of the normal
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to the yield locus (i.e. we use normality). Now that we have δεqe and δεqp we can
add them up to get the total shear strain increment δεq and compute εq at every
point, thus enabling us to generate figures c and d.
The slope of the normal to the yield locus changes as we load from B to F. At B, the
normal points upwards and to the right, whereas at F it points entirely upwards
because F is at the very tip of the yield locus (η = M). In other words, the slope
grows progressively from a finite value at B and approaches infinity at the stress path
approaches F. Meaning that as we approach F, small increments of plastic volumetric
strain δεpp generate progressively larger increments of plastic shear strain δεqp. At F,
the vertical slope implies that shear strain progress with zero increments of plastic
volumetric strain, this in turn implies that δp0' = 0 which means that the yield locus
can no longer change size and the loading cannot proceed beyond F.
Since no material can shear infinitely, figures c and d show plots that asymptotically
approach the conditions at F.
In this example we started our “calculations” from point B without worrying about
how the soil got to B. In the next few examples we take a step back to examine
different ways in which we could get to B and the effect that this stress history has
on soil behaviour.
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Consider now a case in which a soil specimen is normally and isotropically
compressed from the origin of the compression plane to point A (figure a). Point A
lies on the current yield locus ylA which has p'0A. As we did in the previous slide, we
can find urlA by projecting p'0A downwards to the iso-ncl.
Now we start applying the conventional compression stress path (i.e. δq = 3δp')
towards B. Since point A lied on the current yield locus ylA, the soil starts yielding
(i.e. plastic strains start developing) the instant we move along the stress path from
A to B. At B, the current yield locus is ylB. As before, we can find urlB by projecting
p'0B downwards to the iso-ncl. We continue loading towards F, and using by
projecting p'0F we find urlF.
The vertical separation between urls gives us the plastic increment of specific
volume, which we then use to compute the plastic volumetric strain increment, and
then, using normality we can compute the plastic shear strain increment. We now
have all the information we need to generate the four plots on the slide. This is the
same that we did in the previous slide which contains
a more detailed step-by-step explanation.
Let’s now see how this normally consolidated behaviour compares to over
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consolidated behaviour.
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Let’s now get to B through an alternative route that results in a lightly
overconsolidated specimen. Suppose the soil is normally and isotropically
compressed from the origin of the stress plane to p'0B. This results in the current
yield locus being ylB, which is associated to urlB. The specimen is now unloaded
from p'0B to A. From figure a we can see that this means that we are moving inside
ylB (which remains the current yield locus), so the response must be elastic.
Meaning that in the compression plane the specimen must be moving along urlB.
More specifically, during unloading the specimen goes from the point defined by the
intersection of iso-ncl and urlB to point A.
After unloading to point A we can apply the conventional compression stress path
(i.e. δq = 3δp') and move towards B and eventually F. While the specimen is loaded
from A to B, the current yield locus remains unchanged at ylB. Since we are moving
inside the yield locus, the response of the soil must be entirely elastic. The elastic
strains are computed as a function of δq and δp'. In the compression plane, the
specimen moves along urlB from A to B.
A slightly different way of thinking about this is that as loading progresses from A to
B, p'0 remains constant at p‘0B, so δp'0 = 0, which automatically results in zero plastic
strains (see expressions for δεpp and δεqp in the “What we have so far” table).
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Once the stress path goes beyond B, the yield locus starts expanding and plastic
strains starts developing. Their calculation follows the exact same procedure of the
two previous examples.
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Let’s now consider a more heavily overconsolidated specimen. In this case, the
specimen is normally and isotropically consolidated to K, thus forming ylQ, then
unloaded isotropically to P, and finally loaded along a conventional compression
stress path (i.e. δq = 3δp') towards Q. As the specimen is loaded from P to Q, its
response is purely elastic because the stress path falls entirely inside the current
yield locus ylQ. Accordingly in the compression plane the test progresses along urlQ,
moving from P to Q.
So far not much to see here. Things start getting interesting once we reach Q. In the
four previous examples, whenever the specimen was on a yield locus, the outward
normal was always pointing upwards and to the right. But this time, because the
specimen is so heavily overconsolidated, at point Q the outward normal is pointing
upwards and to the left. The implication is that a positive plastic shear strain
increment δεqp is associated with a negative plastic volumetric strain increment δεpp.
Let’s keep in mind that a negative δεpp corresponds to a volume increase. In other
words, if we increase δεqp beyond Q, the soil will expand.
The only way that δεpp can be negative is if δp'0 is negative. This is confirmed by the
expression in the dashed box and keeping in mind that the slope of the iso-ncl λ is
always greater than the slope of the url κ, so the term in brackets is always positive.
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A negative δp'0 implies a reduction of p'0 which means that during yielding the yield
locus is getting smaller rather than larger. So loading proceeds from Q to a point
such as R that lies on a smaller yield locus ylR. By projecting the right most tip of ylR
downwards to iso-ncl, we can locate urlR. And then by projecting point R from the
stress plane downwards to urlR, we can keep track of the test in the compression
plane. In going from Q to R, the soil moves upwards in the compression plane (the
specific volume v increases) which is consistent with our previous observation that
the soil is expanding. This is also reflected in figure d, which shows the soil
compressing elastically from P to Q, and then dilating elasto-plastically from Q to T.
At point R, the outward normal still points upwards and to the left, implying that
further plastic shear strain increments δεqp will again result in negative plastic
volumetric strain increments δεpp, which in turn implies negative δp'0 and further
shrinking of the yield locus. So loading goes from R to S to T, with each of these
points falling on a yield locus that is smaller than the previous one (see figure a).
Finally, at point T the outward normal points vertically upwards and, as in the
previous examples, plastic shear strains can increase indefinitely without any
changes in stress or in specific volume.
The contraction in yield locus that occurs when the specimen is loaded beyond point
Q is termed strain softening.
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In the previous slide we saw that shearing beyond point Q, resulted in a contraction
of the yield locus (yielding) which led us to point R. This figure makes the point that
we could have also reached point R by unloading elastically from point Q to R. This
would have also taken us to point R in the stress plane, but the yield locus would
have remained ylQ instead of shrinking to ylR. In the q vs εq plot, this elastic
unloading would be represented by QR'. In essence, the underlying assumption in
the previous slide was that the plastic shear strain increments δεqp did occur after
reaching point Q.
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Comparing the responses of the last three examples in q vs εq space, we note that in
the two overconsolidated (OC) examples, there is a distinct kink in the curve that
represents moment at which the yield locus is reached. In the heavily
overconsolidated example, this kink may also correspond with a peak value of
deviator stress q. The exact geometry of this kink will depend on the chosen MCC
parameters, but generally speaking we see that there is a stiff elastic response
before the kink and a softer elasto-plastic response after the kink when the soil
starts yielding. No such kink is present in the results of the normally consolidated
specimen since it is yielding from the onset of the test.
With regards to v vs εq space, the kink in the lightly OC specimen is hardly visible in
the figure but may (depending on the model parameters) still be there (at B). In the
heavily overconsolidated specimen, the kink at Q corresponds with the point at
which the soils goes from being compressive (volume reduction) to being dilatant
(volume increase). Again, in the normally consolidated specimen no such kink exists
and the specimen decreases its specific volume v throughout the entire test.
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Of course, whatever prediction we make with the MCC model is only useful if it
matches experimental results. The left figure shows results of normally consolidated
(NC) Weald clay reported by Bishop and Henkel (1957). The right figure is the MCC
prediction that we had seen previously for an NC specimen.
The main behavioural features are indeed reproduced by MCC, at least qualitatively.
The NC specimen has a smooth q vs εa plot that increases monotonically to its final
value without any kinks. The lower plots are also similar, with the specimen
compressing throughout the entire test.
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Now he left figure shows results of heavily overconsolidated (OC) Weald clay
reported by Bishop and Henkel (1957). The right figure is the MCC prediction that we
had seen previously for a hevily OC specimen.
The main behavioural features are indeed reproduced by MCC, at least qualitatively.
The heavily OC specimen reaches a peak at M on the q vs εa. After this q reduces and
seems to stabilise at large strains. This is the same trend predicted by MCC, although
the kink is not as sharp in the experimental results. The lower plots are also similar,
with the specimen compressing initially and dilating at later stages.
All the examples so far have assumed drained conditions and thus the specific
volume was allowed to change. Let’s now look at some undrained examples (δv = 0).
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Undrained testing implies that volumetric strains are zero (δεp = 0). As we had seen
before, this implies that the elastic volumetric strain increment δεpe is equal to the
negative of the plastic volumetric strain increment δεpp as shown in the slide. By
invoking the expressions in the dashed box, we arrive at an expression that implies
that during an undrained triaxial, mean effective stress increments occur if, and only
if, p0' is also changing. Or what is the same, p' changes if, and only if, the soil is
yielding.
Let’s now have a look at a few graphical examples of undrained triaxials modelled
with MCC. Again, the aim of these graphical examples is to develop a qualitative
understanding of MCC. We will come back to the equations when we implement
MCC on a spreadsheet.
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We know that whenever loading takes place inside the yield locus, soil is not
yielding. Therefore, if follows from the previous slide, that in an effective undrained
test, the stress path inside the yield locus must be a vertical line (δp' = 0). This is
reflected in the figure, which shows that changes in p' only take place after yielding
begins. We will look just now into why the stress path turns to the left during
yielding.
The figure does give us a hint of what to expect in our following analyses: notice
that, just like in the drained case, the stress path comes to an end when it reaches η
= M during yielding. That is, when it reaches the uppermost point of the current yield
locus, and therefore the outward normal is pointing exactly upwards.
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The undrained condition equation implies that δp' and δp0' must always have
opposite signs. That is, for p' to increase, p0' must decrease, and viceversa. Let’s start
getting an idea for how this plays out. Consider point A which lies on the current
yield locus and is to the right of the uppermost point of the yield locus (q/p‘ = η <
M). The outward normal at A points upwards and to the right, meaning that an
increase in plastic shear strain will be accompanied by an increase in plastic
volumetric strain (soil compression). This implies that there will be a plastic
reduction of specific volume, which in turn means that the new url must be below
the current url. This is why urlB lies below the original url (QGA). To cancel out this
plastic volumetric compression, there must be an elastic volumetric dilation (or
elastic rebound if you wish). This elastic volumetric expansion can only occur if there
is a reduction in p', which is why B has a lower p‘ than A. We also note that point B
lies on the yield locus that corresponds to urlB. Since ylB is larger than the original
yield locus (QGA), we say that plastic hardening has occurred.
Loading could also start from a point such as Q, which lies on the current yield locus
and is to the left of the uppermost point of the yield locus (q/p‘ = η > M). The
outward normal at Q points upwards and to the left, meaning that an increase in
plastic shear strain will be accompanied by a decrease in plastic volumetric strain
(soil dilation). This implies that there will be a plastic increase of specific volume,
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which in turn means that the new url must be above the current url. This is why urlR
lies above the original url (QGA). To cancel out this plastic volumetric dilation, there
must be an elastic volumetric compression. This elastic volumetric expansion can
only occur if there is an increase in p', which is why R has a higher p‘ than Q. We also
note that point R lies on the yield locus that corresponds to urlR. Since ylR is smaller
than the original yield locus (QGA), we say that plastic softening has occurred.
Lastly, loading could also start from a point such as G, which lies on the current yield
locus and is at the uppermost point of the yield locus (q/p‘ = η = M). The outward
normal at G points directly upwards, meaning that plastic shear strain increases
indefinitely without any plastic volumetric strain. This implies that δp0' is zero (see
δεpp in dashed box), and because of the undrained condition, δp' is also zero.
Therefore shearing progresses without any elastic or plastic volumetric strains and
with the yield locus keeping its size constant. The implication is that once the stress
path reaches the tip of a yield locus (e.g. point G), it cannot go pass that point.
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Let’s now consider a situation in which a specimen is normally and isotropically
consolidated (drainage allowed) from the origin of the stress plane (q vs p') to point
A. The current yield locus at this stage is the one that passes through point A
(although not labelled in the figure, I will refer to it as ylA). At this point we impose
undrained conditions and start applying the conventional triaxial compression total
stress path TSP indicated in the figure a with a dashed line.
Our objective here is to understand how MCC produces the dotted plots in figures a,
b, c, and d. Since the test is undrained, the plot in figure b becomes trivial: it has to
be a horizontal line because specific volume does not change during an undrained
test (δv = 0). On this horizontal line on figure b we draw a number of points (B, C, D,
E, F) that are roughly equidistant as suggested in the figure. Through each of these
points we draw a url. Notice that each url plots below the previous one (urlB below
urlA, urlC below urlB, and so on…). This downward shift of urls indicates a plastic
reduction of specific volume. This reduction in specific volume is consistent with the
fact that the outward normal of the yield locus at point A, is pointing to the right,
which is the direction of positive plastic volumetric strains.
Each url has a point of intersection with the iso-ncl. Each point of intersection is
projected upwards to the stress plane, thus identifying the rightmost point of the
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yield loci that correspond to the urls. The full yield loci is now drawn based on its
rightmost point. To get the effective stress path ESP all we have to do is to project
points B, C, D, E, F from the horizontal line in the compression plane (v vs p')
upwards to the corresponding yield locus. This last projection is shown for point B in
the slide. The effective stress path ESP stops when it intersects the uppermost point
of a yield locus (η = M), such as at F (as discussed in the previous slide). Interestingly,
the resulting ESP is independent of the total stress path TSP. As we are about to see,
the only effect of varying the TSP is on the resulting pore water pressure u.
δq and δp' between any two points of the stress path can be readily measured, thus
allowing the calculation of elastic volumetric δεpe and shear δεqe strain increments
(see elastic equations in dashed box).
Regarding the calculation of the plastic volumetric strain increment δεpp, we can
think about in at least three ways that are perfectly equivalent. Let’s consider in
particular the step from B to C. The first approach is to simply compute δεpp = -δεpe
as this is the undrained condition. The second approach is to measure (or calculate)
the vertical distance between urlB and urlC. This distance is equal to the plastic
increment of specific volume which is labelled as δvBCp in figure B. By multiplying
δvBCp by negative one and dividing by the specific volume at B vB, we get the plastic
volumetric strain increment δεpp from B to C (see equation on slide). And the third
approach is to use the equation for δεpp in the dashed box. This equation requires
δp0', which we can measure (or calculate) because we know ylB and ylC.
By the way, when considering the step from B to C, is δvBCp positive or negative? Is
δp0' positive or negative? What about δp'?
Now that we have calculated the plastic volumetric strain increment δεpp (in three
different ways just to be sure!), we can invoke normality to calculate the plastic
shear strain increment δεqp. By adding up δεqe and δεqp we obtain the total shear
strain increment δεq and can generate figure c.
We still have to calculate u in order to generate figure d. The implication of u = p – p'
is that at any point, u is the horizontal distance between the ESP and the total stress
path TSP. It also implies that if the ESP is to the left of the TSP (p' < p) then u is
positive, and if the ESP is to the right of the TSP (p' > p) then u is negative. In this
case, as loading progresses, u becomes more positive because the TSP moves to the
right (positive δp) and the ESP moves to the left (negative δp'). This breakdown of δu
into a component due to δp and another one due to δp' is illustrated in figure d.
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We now consider a lightly OC specimen. This situation may have been created by
first consolidating (drainage allowed) the specimen normally and isotropically from
the origin of the stress plane (q vs p') to the rightmost point of ylB, and then
unloading isotropically to point A. As we unload, we are moving inside the current
yield locus and therefore the test progresses from the intersection of iso-ncl and
urlB, to point A on urlB. Once at point A, we impose undrained conditions and start
applying the conventional triaxial compression total stress path TSP indicated in
figure a with a dashed line.
Since A falls inside the current yield locus ylB, the soil does not yield when loading
begins. And because it is not yielding, the effective stress path (ESP) can only move
vertically upwards (see Ex. U1). Since p' remains constant in going from A to B, these
two points plot on top of each other in the compression plane (v vs p').
Once the ESP reaches point B, the procedure is identical that of Ex. U3. By drawing a
horizontal line through A in the compression plane, we can identify suitably spaced
points (B, C, D, E, F), draw urls through them, and eventually draw their
corresponding yield loci. The points on the horizontal line are then projected
upwards from the compression plane to their respective yield loci (as indicated for
C). The stress path finishes when it reaches the uppermost point of a current yield
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locus (η = M).
For calculations of volumetric and shear strains (elastic and plastic), and of pore
water pressures, see comments for Ex. U3.
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Let’s now consider a heavily overconsolidated specimen. This situation may have
been created by first consolidating (drainage allowed) the specimen normally and
isotropically from the origin of the stress plane (q vs p') to the rightmost point of ylQ,
and then unloading isotropically to point P. As we unload, we are moving inside the
current yield locus ylQ and therefore the test progresses from the intersection of isoncl and urlQ, to point P on urlQ. Once at point P, we impose undrained conditions
and start applying the conventional triaxial compression total stress path TSP
indicated in figure a with a dashed line.
Since P falls well inside the current yield locus ylQ, the soil does not yield when
loading begins. And because it is not yielding, the effective stress path (ESP) can only
move vertically upwards (see Ex. U1). Since p' remains constant in going from P to Q,
these two points plot on top of each other in the compression plane (v vs p').
For loading to proceed beyond point Q the soil has to yield. Notice that at Q, the
outward normal (not shown in the figure) to ylQ points upwards and to the left. The
implication is that positive plastic shear strain increments δεqp will induce negative
plastic volumetric strain increments δεpp, that is, plastic volumetric dilation. Negative
values of δεpp imply a negative δp'0 (see equation for δεpp in dashed box), which
effectively means that the yield locus is shrinking (strain softening is occurring).
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At this point we could go back to the compression plane and draw a horizontal line
that passes through points P and Q, and extends to the right of urlQ. It must extend
to the right because the soil is dilating plastically, meaning that each subsequent url
must lie above the previous one (urlR above urlQ, urlS above urlR, and so on).
Subsequently, we can draw a series of points (R, S, T) on this horizontal line, draw
urls through them, draw the yield locus that corresponds to each url, and finally
project R, S, and T from the horizontal line in the compression plane (v vs p') to their
corresponding yield locus. The sequence of the points in the stress plane outlines
the stress path which ends when it reaches the top of a yield locus (η = M).
For calculations of volumetric and shear strains (elastic and plastic), and of pore
water pressures, see comments for Ex. U3.
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Comparing the q vs εq results (upper plots) of the last three examples, we see that
the plots for Ex. U3 and Ex. U4 are very similar. In both cases q increases
monotonically and approaches its value at F asymptotically. For Ex. U5 though, there
is a distinct kink at Q which corresponds to the beginning of yielding. There is a stiff
response before Q and a soft response after it.
With regards to the u vs εq results (lower plots), again, the plots for Ex. U3 and Ex. U4
are very similar. In both cases u increases monotonically and approaches its value at
F asymptotically. For Ex. U5 though, there is a distinct kink at Q which corresponds to
the beginning of yielding. The pore water pressure increases from P to Q but then
drops after Q when yielding takes place. In this case, u becomes negative at large
values of εq.
The exact shape of the plots will depend on the MCC parameters adopted.
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This slide compares results of an undrained conventional compression triaxial test on
normally consolidated Weald Clay to the results of Ex. U3. It is clear that MCC
captures the main qualitative features of the soil’s response: monotonic increase of
q and u as they approach their final values asymptotically.
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This slide compares results of an undrained conventional compression triaxial test on
heavily overconsolidated Weald Clay to the results of Ex. U5. There is a strong
similarity in the u vs εq plots, with u initially increasing and then dropping to a
negative value. However, the experimental plot of q vs εq does not show the kink
predicted by the MCC model.
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127
128
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Triaxial tests in the MCC approach ultimate conditions in which p', q, and v are
constant. In a graphical MCC problem, this is reflected by the fact that at the end of
each test, the specimen remains fixed at a certain point of the stress plane (constant
q and p') and of the compression plane (constant v and p').
Notice that for undrained tests, a constant q implies a constant p. And constant p
and p' in turn imply constant u (since δu = δp – δp'). When p', q, and v eventually
stabilise at large value of εq, the specimen is said to have reached a critical state.
Hence the name: critical state soil mechanics. The term steady state is sometimes
also used to refer to this condition. However "steady state soil mechanics" is not
used.
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These equations represent the same ideas.
The figure makes it clear that η = M only corresponds to a critical state if the soil is
yielding. η = M can very well be reached elastically before the end of the test (see
point X in the figure).
In the stress plane, all the points that correspond to the critical state form a straight
line that has the form: qcs = M*p'cs. This line is called the critical state line (csl).
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What about in the compression plane? Do critical states also line up nicely in the
compression plane? The equations in the slide show that they do. This is the critical
state line in the compression plane. It is parallel to the ncl, since they both have a
slope λ.
Notice that we defined a new constant Γ (gamma). Judging from the equation of the
CSL in the compression plane, what is the meaning of Γ?
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The csl plots as a straight line (qcs = M*p'cs) in the compression plane, and as a
curved line (vcs = Γ - λ∙lnp'cs) in the compression plane. Furthermore, figures a and b
make the point that any straight with the shape q = η∙p', maps as a curve on the
compression plane. Figure highlights that fact that, just like the ncl, the csl also plots
as a straight line in v vs log(p') space.
The geometric meaning of Γ is indicated in figures b and c: Γ is the value of vcs when
p'cs = 1 kPa.
What is the vertical separation between the ncl and the csl?
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A combination of p', q, and v values is part of the csl only if it satisfies both
equations. Points X and Y in the figures satisfy only one of the two equations.
Notice that we need three parameters for a full description of the csl: Γ, λ, and M.
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The csl can also be thought of as living in a 3D space with axes: p', q, and v.
135
Let's have a look at some experimental results on clay to see whether they support
the existence of a critical state line (csl).
The figures illustrate experimental results of conventional triaxial compression tests
made on reconstituted Weald Clay. These results were originally reported by Bishop
and Henkel (1957). For tests 1 and 2, prior to shearing the specimens were normally
and isotropically loaded to p' = 207 kPa. Test 1 was drained (ABCD) and Test 2 was
undrained (EFGH). For tests 3 and 4, prior to shearing the specimens were normally
and isotropically loaded to p' = 827 kPa, then unloaded to p' = 34 kPa. That is, these
two specimens were heavily overconsolidated. Test 3 was drained (KLMN) and test 4
was undrained (PQRS).
When we look at the result in the stress plane, we see that the end points of the
tests do tend to define a unique line that goes through the origin. This unique line is
shown dashed and has slope M. Conversely, the end points of the tests in the
compression plane do not align all that well, however they do seem to all be
approaching the dashed area. It is certainly possible that if they had been sheared
more, they would have all reached the same line. However, in a triaxial test it is not
always possible to shear the specimens sufficiently to reach the CSL.
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These figures are also useful to look into the relationship between drained and
undrained behaviour:
Test 1 (ABCD) contracts when sheared. It is able to do so because test 1 was
conducted under drained conditions. Test 2 (EFGH) is identical to test 1 except that
conditions are undrained, so test 2 may "want" to contract but it can't. Notice that
the total stress path of test 2 actually coincides with the alignment of the effective
stress path of test 1 (ABCD). Since the ESP of test 2 (EFGH) plots to the left of its TSP
(ABCD), we know that positive pore water pressures are developing (Recall: u = p –
p'). You can think of this increase in pore water pressure as the soil's way of resisting
the contraction that it "wants" to do, but is prevented from doing by the undrained
condition.
The opposite happens with in the overconsolidated tests 3 and 4. Test 3 (KLMN)
which is drained, initially contracts but eventually dilates to a specific volume that is
higher than its initial one. Since test 4 is identical to test 3, except that test 4 is
undrained, we know that test 4 "wants" to contract initially but dilate in the long
run. But since it is undrained it can't do any of this… Instead, initially it develops a
positive pwp (the ESP plots to the left of its TSP) and eventually develops a negative
pwp (the ESP plots to the right of its TSP). Remember that the alignment of the TSP
of test 4 coincides with the ESP of test 3 (KLMN).
Summary: If a specimen contracts when sheared drained, it will develop positive
pwp (drop in p') when sheared undrained. Conversely, if a specimen dilates when
sheared drained, it will develop negative pwp (increase in p') when sheared
undrained.
What is the effect of p' on a soil's strength?
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Additional experimental data from drained conventional compression triaxial tests
on reconstituted Weald Clay. Results originally reported by Roscoe, Schofield and
Wroth (1958). This data set represents more compelling evidence in favour of the
existence of a unique CSL for this clay. The CSL was defined using only the results of
the normally consolidated specimens (open markers). However, it is clear that the
overconsolidated specimens (closed markers) define the same CSL. This is consistent
with the predictions of the MCC model.
Furthermore, the shape of the CSL is also in good agreement with the predictions of
the MCC model. In the stress plane, the CSL is a straight line that passes through the
origin, and in the compression plane, the CSL linearizes when plotted in semilogarithmic space (v vs p') and is parallel to the iso-ncl.
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Additional experimental data from undrained conventional compression triaxial
tests on reconstituted Weald Clay. Results originally reported by Roscoe, Schofield
and Wroth (1958). The CSL indicated in the figures is the same one that was defined
using the drained results shown in the previous slide. It is clear that the drained and
undrained data define the same CSL. Again, this is in agreement with the predictions
of the MCC model.
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Due to limitations in the attainable levels of strain, it is very common for triaxial tests
to not reach the critical state. These figures are plotted with results from Parry
(1958) and illustrate interesting trends. The two figures "a" should be interpreted
together, and the two figures "b" should also be interpreted together.
Consider point X on the upper figure A. This point represents the final condition of a
drained triaxial. Since the specimen did not reach the CSL, it was still dilating
(increasing v) at the end of the test. The rate at which it was dilating is represented
by dεp/dεq. We can think of this as a measure of the speed at which the specimen
was moving towards the CSL. We can also characterise how far away X was from the
CSL using the ratio p'cs/p'f, where p'cs is the value of p' on the CSL a the same specific
value as X, and p'f is the p' value of X. We can do the same for tests that ended in a
condition such as that of W, where the specimen is still contracting (reducing v) to
get to the CSL. The lower figure a shows that the farther away the specimens are
from the CSL (p'cs/p'f far away from 1), the faster the rate at which the specimen is
changing its specific volume to reach the CSL.
A similar analysis can be done for undrained tests (figures b). In this case, the
measure of the speed at which the specimen was moving towards the CSL is given by
the rate in change in pwp: du/dεq. The distance from the CSL can still be
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characterised in the same way as for the drained tests: p'cs/p'f. Again, the lower
figure b reveals as specimens approach the CSL (p'cs/p'f = 1), the rate at which their
pwp is changing becomes smaller.
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Although historically misleading, in these notes, the critical state line has emerged as
a conclusion of the MCC. We now explore the types of soil behaviour predictions
that we can make based exclusively on the premise that soils move towards their CSL
as shear strains increase. In these predictions we will ignore all other details of the
MCC model.
We are considering four tests in total. All of them subjected to conventional triaxial
compression: Two normally consolidated specimens, one is tested drained and the
other one undrained (A in figures a and c). Two overconsolidated specimens, one is
tested drained and the other one undrained (B in figures b and c). The initial (preshear) p' value for all four tests is identical. Their normal and overconsolidated
nature is only appreciated in the compression plane. In particular, notice that the
starting point of tests A lies above the CSL, whereas the starting point of tests B lies
below the CSL.
Consider the undrained test A. From the compression plane we know that the
specimen must move to the left (approaching the CSL) until it reaches its final point
U. This point U can now be projected upwards to the CSL on the stress plane. This
observation alone tells us that the ESP of this test must begin at A and end at U. We
don't know what is the exact shape of the stress path, but we can trace a simple
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curve as a qualitative approximation. We can also notice that this curve lies to the
left of the TSP (δq = 3*δp), meaning that we can expect positive pwp to develop in
this test. We could even approximate the values of the pwp. That is a significant
amount of information to deduce only from the knowledge of the CSL.
Consider now the drained test A. In this case we begin on the stress plane. We know
that the ESP = TSP. So we can easily determine the end point of the test (W), as the
intersection between the ESP and the CSL. We can now project W to the CSL on the
compression plane. Still in the compression plane, we connect A and W with a
straight line. This line is our prediction of how the specimen will move in the
compression plane during the test.
Similar analyses apply to the overconsolidated tests B. Let's start with the undrained
test. From the compression plane we readily see that in this test p' will have to
increase to reach the CSL at V. V is now projected upwards to the CSL in the
compression plane. A curve (informed by past experience) that connects B and V is
now drawn to represent our first qualitative estimate of the ESP. We see that
specimen may initially develop positive pwp but will have negative pwp by the end
of the test.
Now drained B. From the stress plane we can readily determine its final point W
which is given by the intersection of the ESP and the CSL. W is now projected
downwards to the CSL in the compression plane. Still in the compression plane we
can now trace a curve (informed by past experience) connecting B and W that
represents our qualitative prediction of how the specimen will move in the
compression plane.
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