Maxwell Equation and Kelvin-Voight Viscoelastic Model
Okay, so welcome back to one more lecture Introduction to Soft Matter. Last time, we started
looking at what is called a Maxwell’s model, Maxwell’s viscoelastic model. And, we started
deriving some of the simple formulas that are a consequence of the model.
(Refer Slide Time: 0:49)
So, just to refresh our memory, what we will do is, so in the Maxwell model is a very simple
model where you have 2 over a spring and you have a dashpot in series and these two are
considered to be massless and then we use this and relationships that are integral to a spring
and a dashpot to derive the constitutive equation. And the constitutive equation came out to
be of the form, we can write it in a couple of different forms.
So, we say, you can write it in this way
E
•
•
= + or if you want you can write it as in this
other fashion also E
• •
= + where this is a relaxation time scale is given by the ratio of
and E. And we saw that for the stress relaxation function for this particular situation, that
comes out to be of a very simple form is given by the E, which is a modulus. And there is an
exponential decaying function.
Similarly, we derived the equation for a Creep function and then that also came out to be a simple form and however in this case, now to remember that, while the stress relaxation is exponential in nature, the Creep turned out to be linear in nature and this linear nature of creep is not extremely realistic, but something that can be still worked with in different situation.
It depends on the particular situation whether or not that is a proper adequate model. So, the stress response or the strain response, the stress response function and the Creep function. So, the stress relaxation function and the Creep function, they are obtained as special cases of an applied strain or an applied stress.
(Refer Slide Time: 2:54)
So, the next question is once we have gotten these expressions is what happens if you have an
arbitrary stress history or a strain history? So, let us first take a look at what happens if you
have when you have a response for an arbitrary strain history and here what we imply is that
that is strain history for example is known, so (s) for that set of s is, which is less than the
current time, but greater than 0. This is given to you.
And then you are asking the question, what is your current state of stress? So, if the strains
are known to you then we know that a viscoelastic material has memory. So, what we will do
is to solve this particular problem. So, to solve this particular problem, what we will do is that
we will write this equation such that we have the stresses on our left hand side, E
• •
+ = .
So, we have the stress so, we are writing this ODE.
So, this is what you want to figure out and is some, given to you. Now, to solve this
equation, you probably have realized that this is a first order linear ordinary differential
equation. Now, when you have an equation of this form and you want to solve for the general
situation, then a method which is called the method of integrating factors is usually applied.
So, just let us do a quick rehash of what the method of integrating factors is. So, let us say
there exists a different ODE which is given as ( ) ( )
dy
P t y Q t
dt
+ = . This is a linear ordinary
differential equation, when Q(t) is equal to 0 then you end up with a homogeneous linear
differential equation.
If Q(t) is identical equal to 0 always then you have a homogeneous equation, if you have Q(t)
as a nonzero variable then it is a non-homogeneous equation. Now to solve this particular
case, we introduce another function called v (t) and we multiply both sides by v(t).
( ) ( ) ( ) ( ) ( )
dy
v t v t P t y Q t v t
dt
+ =
And now if v(t) is of this particular form
( )
( )
P t dt
v t = e , then this allows us to write this
previous equation as ( ( ) ) ( ) ( )
d
v t y v t Q t
dt
= and this final form formula you can solve by
integrating.
So, this v(t) is called integrating factor. So, for our case, let us find out what the integrating
factor is. So, what is v(t) for us? We have integral of and here you have for y you have .
So, you have, you end up with 1
dt
. So, you end up with
t
e . So, this is our integrating factor.
Now, if you use this and then multiply both sides by the integrating factor you will end up
with a situation where you will have ( ( )) ( )
t t d
e t Ee t
dt
•
=
(Refer Slide Time: 8:03)
Now, to solve this you integrate both sides and you integrate from 0+ to a time t. So, the left
hand side then if you do this integration and it you end up with, it is your ( )
t
e t , which is
the variable that you seek minus (0 ) + . So, the here
t
e factor here becomes just 1 in this
case. And on the other side you will still have 0+ this is where integrating over E.
And now, you are integrating this quantity from 0+ to some particular time t. So, I have
already used up the time the variable t, so I am just going to use some other dummy variable
let us say s here. And now, we know that (0 ) + and (0 ) + the share a relationship. And the
relationship is if we go back to this original equation, so the jump between the 2 variables
and are related to each other.
And they are related by, so if you introduce this back into this previous equation, you will
end up with the stress at the current time. And I am only having that on the left hand side. I
am going to move this back to the right hand side. So, you will end up with (0 )
t
e
−
+ , I have
taken this factor and I have moved that the exponential factor I have also moved on to the
right hand side.
You have the integral to the second quantity
(t s)
e
−
−
. Now, what does this quantity remind you
of? This is your stress relaxation function, so we can just write this rewrite this previous
equation in the form of the stress relaxation function. So, this becomes an important equation
for us.
I am just going to, so we have seen 3 important equations till now, the stress relaxation
function, the Creep function and then we have looked at the stress which comes out as a
result of previous provided strain history. This is a very important equation because we will
see that this comes up again again, there is a particular reason why I put it in the form of the
stress relaxation function. We will get to know that a little bit later why.
(Refer Slide Time: 12:14)
So, now that we have looked at the response to an arbitrary strain history, then we should be
looking at the response to an arbitrary stress history also. So, let us take that response to an
arbitrary stress history. So, here, the question is flipped, you have been provided, let us say
the stress as a function of time. And the question you are trying to ask is what is the strain at
the current time. That is the question.
And as we know, again, just like the previous case, the entire stressed history is going to
influence the strain at the current time. And the current time, strain should be a functional
form that takes into account the entire history of imposed stress on the material. So, we just,
so we rewrite our equation, but we will flip it a little bit, because we want to be sure which is
the variable we want.
So, since , is the variable I want, I am kept it on the left hand side and we just rewrite this
equation. Now, you integrate both sides, if you integrate both sides from 0+ to the current
time t, then you end up with, you have 2 integrals on this side, both going from 0+ to t, you
have a
•
. And then again, I am using a different variable, a dummy variable s here, because I
have already used the variable t plus 1 by and let me put the limits.
So, now, you have a functional form which has a derivative of the stress here, but you have
another one which is just has a derivative. So, here we can actually convert this second term
on the left hand side. The right hand side sorry, the second term on the right hand side can be
converted into an analogous form with , if we apply integration by parts.
(Refer Slide Time: 15:02)
So, just to remind ourselves what integration by parts is, we are just write it on the side, recall
integration by parts. And in this you have an integral u, v dt, let us say, if you are computing
this, then you can rewrite this as in u into the integral of v dt minus u dot integral of v dt. And
there is another dt here.
uvdt u vdt u( vdt)dt
•
= −
So, using this integration by parts, what we have to do is we have to simplify and I want to
get rid of the pure stressed term and converted into a derivative form and I am going to write
this as, see I can choose this quantity here. And here this is variable s and this variable s this
is being evaluated from 0 to t and you have minus integral again plus t, (s t)
•
− , so I will just
apply the integration by parts here.
So, by applying the integration by parts and rearranging the terms, we can rewrite the left
hand side as being this quantity. And on the right hand side, you will now have (0 ) + and
then you have the quantity under the integration side which goes from 0+ to t and here you
will have (s)
•
and then the bracket you will have
1
( (t s))ds
+ − .
And you can see that this quantity right here, this is basically this can be summed up and we
can write this as 1
t
+ . And then once you bring this E over onto the right hand side, then this
equation simplifies further and this we end up with (t) is, this is now the J(t) and this
multiplied by (0 ) + and now you have the quantity under the integration side and you can
see that this quantity is the Creep compliance written appropriately.
So, this against the limits of the integration still remain, the Creep compliance now takes on
the form t minus s and you have sigma dot s ds. So, this is an important equation. So, you can
see that things have sort of gotten flipped. So, when you are looking at an arbitrary strain
history, then you have here the stress relaxation. So, if you have an arbitrary strain history,
then you have the stress relaxation function appearing here whereas, if you have an arbitrary
stress history, then you have the Creep functional form coming in here.
(Refer Slide Time: 18:54)
So, with this we have sort of looked at some of the very-very important results for the Maxwell equation or the Maxwell‘s viscoelastic model. Now, there is another important model in viscoelasticity, so Maxwell model is one of the simplest possible schemes that you have, but there is another one and that is called the Kelvin Voigt Viscoelastic model. So, we will look at now Kelvin Voigt Viscoelastic model.
This model was introduced in 1874 by Meyer, so Meyer 1874 introduced this model, introduced a model which is a model which was different than the Maxwell’s combination of spring and dashpot and it is now very much known as the Kelvin voigt body and so and is now known as the Kelvin Voigt.
I have not introduced the model I will introduce that in a second, body as far, so unfortunately Meyer’s name is missing, but Phan-Thien in his book, understanding viscoelasticity, he clarifies that this should be called, the Phan-Thien says that this should be called Kelvin Meyer Voigt model. And a good reference to look up this is the book called Rheology and Historical Perspective and that is written by Tanners and Walter and it was published in 1998.
So, if you are interested in the history of this rheology and the various models that have come up, you can refer to their book. Once again that book is called Rheology and Historical Perspective by Tanner and Walters. So, let us look at this particular model. So, what is this model all about? So, this model is, now instead of placing the spring and the dashpot in series, the spring and the dashpot are placed in parallel.
So, this is your E, this is and the 2 bodies have been placed in parallel, you have the forces
on this, when you apply a force f(t) is undergoes a deformation let us call it x . And then the
question is what is the relationship between f(t) and x and that is how we have been dealing
with this, we have been applying force balance. Once again let us remind ourselves that these
2 are massless. So, that is a very important point, so just we will make a note of it. Note,
spring and dashpot are both massless.
(Refer Slide Time: 23:07)
So, individually I think I need a new space. So, what we are going to do, let us break up this
space into 2 parts. So, you have, let say the spring, let us say the force in this is experience
that the spring is experiencing is Fs (t) and the information it undergoes is xs . Similarly, for
the dashpot, so let us say we have a force ( ) d F t , sorry, I just made a mistake. So, this is, we
should put Fs in the subscript this is time, this is subscript.
So, this is d subscript, d standing for the damper the r, dashpot d x . So, from force balance,
for the model, if you look at this model for force balance, we can say that your F(t) is equal to
force in the spring for the force in the dashpot. From the geometrical constraint, now
geometry here plays an important role because you have placed this in parallel. What is
implies is that the deformations or the extensions in both the spring and the dashpot have to
be the same.
That is why, that is the meaning of this particular representation. So, from geometry we know
that the deformations are going to be equal in the 2 cases sorry, so from geometrical
constraint tells us that the system deformation is same as the deformations being seen by the
spring and the dashpot. So, now let us write the individual equations that apply to the spring
and the dashpot.
So, the individual equations should be that that should just be a result of how spring behaves,
we have, so now if I add these 2 equations, this equation here right over here, then what we
end up having is ( ) s F t plus ( ) d F t , which is we know from force balance is equal to F(t). So,
my addition gives me and that these 2 by the way, these are equivalent to writing it in the
form of x , which is the, what this entire system is seeing.
So, I can just write F(t) E x x
•
= + , so where this process of addition I can write it as, so
now we have a relationship what we are seeking initially a relationship between the force and
the system displacement. So, this suggests a constitutive equation, equation of the form
(t) E (t) (t)
•
= + and this is the governing constitutive relationship for the Kelvin Voigt
model.
Now just compare this to the Maxwell model, we had a derivative and stress in the Maxwell
model which is not existing here. But we have now have we have the derivative and strain
still there, but we have an extra strain term without the derivative that also appears. And it
can be solved by almost the exact same methods that we have solved for the Maxwell case.
Now, let me, so we will end here for today's class and then we will discuss the solutions to
this oriental differential equation in the next class.
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