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Module 1: Viscoelastic models

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Maxwell Model

So, last time what we were trying to do is we were looking at the Maxwell model where
spring and a dashpot was placed in a linear fashion and when we combined this two things in
a linear fashion we wanted to understand what is the relationship between the force applied
on the spring dashpot system and the displacement of the system. Because what we are
essentially going to seek is the final constitutive relationship interests of stress and strain.
(Refer Slide Time: 01:05)
So, where we stopped last time was where we wrote down the equations. So, we wrote down
the geometric constraint equation how the x  is and from the force balance we had written
down the equation for the force, these are equal, this is rather straightforward equation. Then
we had written down the separate relationships for F and this force in the spring and the force
in the dashboard. So, we will start off from here where we left off.
So, now we know that from the previous equation that this is actually F(t), so let us simplify
our life and now we also what we can do is we can take the derivative. So, let us say let us
take this equation.
(Refer Slide Time: 01:57)
So now I am just going to remove this part because we have taken that out. So, if we take the
derivative here and then we add up the two cases and we put this E on this other side. So, let
us say this E I am going to remove and I am going to put it in the denominator here and
similarly I am going to remove  on this side and I am going to take that here. Then you
have your s x  plus dx

 is equal to F. So I am going to not use this T because that is sort of
understood.
s d
F F
x x x
E 

• • •
 +  = + = 
So, just simply going to write my this is F

and you have
F

, but we know from geometric
constraint that this quantity right here this is equal to your x

 . So now what you have is that
you can find that this quantity is there equal to x

 and this is the relationship that we were
seeking in the first place set as we had said that we wanted a relationship between x or the
total displacement F that the system was looking at and the net force on the system.
(Refer Slide Time: 03:25)
So, this becomes so this is our important equation right here and this is in terms of the force
and the displacement and from this what we say that in taking inspiration from this we can
say. So, taking inspiration from this equation, we can we propose a constitutive relationship
which is of the form
E
 




+ =
So, this now becomes our governing equation and remember this is we are discussing it for
the one-dimensional case. We are assuming that the displacements are small so that it is easy
to define  and the stresses and at the same time it is being applied in a Lagrangian sense.
So, this is our governing equation.
(Refer Slide Time: 05:30)
So, now what we want to do I, do is evaluate this model. So, let us consider the situation
where you are going to apply a given strain rate let us say o and then the stress we know
from our introductory classes that the stress should decrease with time. It should be some
decreasing function. So, what is the decreasing function in this case of this kind of
constitutive equation? Well so for that we know that in this section your strain is constant so
your 

is 0. So, this quantity on your right hand side that disappears.
So, in this particular situation where you are applying a constant strain your governing
equation becomes rather simple equal to 0, which implies our
E
 


= − and this is a very
simple ODE for which the solution should be sort of known to you. So, this is only possible
when your Sigma has an exponential behaviour. So, from that we can write from very
elementary ordinary differential equations, we can write that as
E
t
Ae  

=
(Refer Slide Time: 07:20)
Now, if I want to write, now I introduce a quantity called which is equal to
E

then the
sigma is basically some constant times
t
e 

. So, your G(t) so this is by the way a function of
time maybe I will just make it explicit here and this implies that your stress relaxation
function, ( )
t
o G t G e 

= and  here becomes a relaxation time scale.
(Refer Slide Time: 08:15)
So, now let us go back since we had originally said that we are using the equation of proposed by Maxwell you can see that we have found this particular equation is something that we had, we have been able to derive. Here T is a time scale, is the relaxation time scale. So, you have to take that into account and then we have found that this is the, this is how the body itself gradually loses all internal stresses. So good what we have done is things agree with the original Maxwell's derivation.
(Refer Slide Time: 09:12)
Now, let us just do a quick sanity check. So, in this quick sanity check, we want to ask whether the material will behave the right way if you take different limits. So, if the material
is more solid like, so if increases then the stresses should relax slowly. So slower relaxation
of stresses, and is this consistent? So for solids, we have what should be ? It should
increase, it should be a very-very large value.
So, for solids the goes to somewhere close to infinity which basically says that in our case
the Deborah number also goes to infinity and the stress never relaxes. For fluids, your
 should tend toward 0 which means that the Deborah number tends towards 0 and stresses
relax immediately and that is what we expect. So, this is completely consistent with the kind
of behaviour that we would expect.
(Refer Slide Time: 11:21)
So, our model does have physical characteristics. It behaves as something as a material
behave in the real world and then finally one more thing that you can observe from this is that
as t goes to infinity this quantity goes down to 0. So, this stress relaxes totally and that is
indicative of what kind of behaviour? A fluid behaviour, so stresses for large timescales for
large times the stresses relax completely.
Now, quickly recall from one of the previous classes what we had discussed. This represents
a fluidic type of model, so represents fluidic behaviour. So, the Maxwell model is appropriate
to describe certain fluids and not solids. Now when we go back here this constant A we were
not able to determine and the reason we were not able to determine because you need one
more condition to evaluate what A is, but we have not given that to you and the reason that is
because you have to evaluate it in a slightly more involved fashion.
(Refer Slide Time: 13:05)
Now let us say that in your case you are applying some stress with respect to time and the
situation that we have created for ourselves for the various tests is such that the history at
minus infinity or till 0 is 0. So, you will be applying nothing till that time and suddenly at t
equal to 0 you will be applying some known amount of stress or vice versa. So, if this were
strain you would be doing the same thing.
So, you have this  (t) has a jump at t = 0. So your ( ) ( ) o  t = H t , H being the Heaviside
function and we discussed what Heaviside function definitions and by the way MATLAB
defines may define Heaviside function in a slightly different way so there are alternative
definitions so you should be careful or when you are implementing Heaviside function in a
numerical case how you should do that.
So, the question is that if you are applying this then there probably be some jump here and
what is this jump? Now because when you are not applying anything, the system response
should also be 0, but when you are applying a sudden jump condition at t equal to 0 there will
be a sudden jump in the system also and then the system will behave in a certain continuous
fashion, but what is this jump in the response?
So, this is what is which is or this is a situation which is also known as a jump, so this is your
jump condition and to be just a mathematically a little bit more elaborate a jump condition
implies that if you have a function f that you are plotting then you can have a situation where
the two limits from the two sides, so if I if this value is let us say a *t .
So, if I start approaching t time from the left-hand side then I end up with a different limit
condition and this is usually called minus from the other side and if you approach the same
point from the right-hand side then you end up with a different limit and this is you jump, this
represents that jump condition.
So, our problem is that if let us say given a jump in the in the signal, you have to figure out
the jump in the response and the jump and the signal can be either the stress or the strain and
the corresponding response then will be the other one. So, our equation was the equation that
we are working with is, so all I have done is I have used this equation back. So, I have this
should be E this is  .
So, so let us, so the solution that I am going to show you is not very mathematically rigorous,
it is somewhat hand waving but it still gets you the right result. The full mathematically
rigorous solution is not in the purview of this course, so the slightly hand waving argument
will still lead us to the correct answer, but I think I should point it out that it is not exactly
very rigorous whether full proof proceeds on very similar lines.
(Refer Slide Time: 17:45)
So, what we want to do the problem is being created by the fact that you are applying a
discontinuous function. So instead of this discontinuous function, what we will do is, we will
assume that there are certain continuous functions that are fully integrable and this continuous
functions go and merge into your original signal and there is a small gap of let us say   .
So, this is  − this is  + , where this signal is slightly deviating from your jump condition
type of signal that you want and basically what you can think of is there is a whole series of
different functions that are there and then you are slowly are approaching the condition of this
jump and basically this delta is something that you are going to slowly-slowly shrink.
So, you have constructed this set of continuous functions such that they are, they merged into
your original signal at   , but then this small gap they deviate and then you will apply the
condition that delta goes to 0 and then you will end up with the jump condition type of
situation.
So, we have our equation we have written it right here, so this implies that if I just want to
write it in the differential form, I will get, so let us only integrate within this small region of
plus minus delta, so integrate over this small region, so now we will integrate, so we put the
integration sign here and we are integrating in time from   and in this case you have
( )   − , ( )   + you have  so you have the appropriate mission is and then take the limit 
tends to 0.
So, what you will end up with is the jump condition. So now this integration will give you the
jump in epsilon at t equal to 0 because till 0 you have you should have response should also
be 0, but suddenly at plus epsilon you will have the response. So, this integration is basically
going to give you
1
(0) (0)
E
= .
The jump in stress and what happens to this last term? So, this is some stress some value
some numerical value into 2 , but this quantity I am taking  as in the limit tends to 0, so
this third quantity is contribution here is 0. So, this now becomes relationship between the
two jumps.
(Refer Slide Time: 21:57)
So, now that we have this let us go back to the derivation of the stress relaxation function. So,
we were looking at the stress relaxation, so our situation is that ( ) 0t 

= for 0 t  and our
corresponding equation was 0
E
 


+ = . So this ( )
t
 t Ce 

= where
E

 = .
(Refer Slide Time: 22:57)
So, (0)  is equal to your constant and we know that (0)  is the relationship between the
two is this. So, this ( 0)  is the given quantity is a quantity that we provided to you. So, you
can write it as ( ) (0)
t
 t E e 

= . So, your stress relaxation function is actually this quantity.
So, this is your, so with the jump condition now you are able to get a proper equation for G.
(Refer Slide Time: 24:22)
So, we have looked at the stress relaxation function but we know that the other important
function is the Creep relaxation function. So, we should also look at that this so look at this
take a look at the Creep response function for Maxwell model. So here what you are going to
do is you are going to apply a stress which is a Heaviside function. So, if we go back to our
question here, the stress part this goes away to 0 because you are going to apply a constant
stress. So, your  (t )

is now going to be this sigma variable.
So, in this governing equation so when you apply a constant stress when we look at the
governing equation this term which contains the time derivative of stress that will vanish and
instead you will only be left with one term on your left hand side and one term on your right
hand side. So that implies that my ( ) o t




=
So, my governing equation is that your strain rate, the derivative of the strain is given by this quantity on your right hand side and from very elementary ODEs, we know that when you integrate this what you are going to get is a linear function. So, if you take this quantity and then you integrate this, you will have you, so we can see that the Creep function, so this is your final solution and the Creep function is actually linear in nature.
(Refer Slide Time: 27:04)
So, Creep relaxation behaviour is actually linear, a linear function for Maxwell model. This is not very realistic right, this is somewhat unrealistic. So, if you plot this it comes out over a straight line so your Creep function will look like this this is your J(t). For a Maxell model, the Creep response behaviour turns out to be a linear function which is not very realistic. This does not compare well with the graphs that we had drawn initially if you recall those graphs for our Creep functions should look like this doesn't compare well with them, but it is still ok.
So, what we did today is we looked at we started off with the Maxwell model and we looked at its stress relaxation function and the Creep function. The stress relaxation function turned out to be a very nice well behaved exponential function which also showed us that the model essentially is representing of kind of a fluidic behaviour.
So, that is where the domain of application of Maxwell model has to be that you have to apply it to viscoelastic fluids and we also looked at the Creep response and we saw that it has a linear behaviour which is not very satisfactory but in many cases it might still work. So, we will stop here and we will pick up from here in the next class.