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### Kelvin-Meyer-Voigt Body

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Kelvin Meyer Voigt model

Welcome back and last time what we were doing is, we were looking at the Kelvin Voigt body and the response of the Kelvin Voigt body to an applied stress. And, we used very similar methods to what we had originally done for Maxwell's body. And, we derived the constitutive equation that this particular system suggests.
(Refer Slide Time: 0:52)
And we ended up with an equation that is linear ODE once again, means the stress and the
strain and you have the stress here, as express is the function of the strain plus the strain rate
multiplied by suitable factors. So, now, just like we had done for the case of the Maxwell
system, we are going to look at jump conditions. So, let us look at jump conditions. So, jump
condition, so the let us say the first thing we are looking at is jump condition in strain.
So, I am going to express my strain, some εoH(t) epsilon not times the Heaviside function just
to remind ourselves, what was the Heaviside function? H(t) is the Heaviside function. And
H(t) is basically given as 0 when t is less than 0, equal to 1 for t greater than 0. So, this is our
Heaviside function.
So, if you have a jump in strain, let us say, and you want to find out the effect of that on
stress, then what we can do is we can straight away insert this functional form right here. But
before we do that, there is one more thing that we have to know. We have a derivative in
stress sorry, derivative in strain. So, what happens if you take the derivative of this function?
So, we will, so we know that the derivative of the Heaviside function is the Dirac delta
function. This is the Dirac delta function. So, your  (t )

, so if you are taking the derivative of
this quantity, the previous functional form then you will have ( ) ( ) o  t   t

= . So, if we use our
constitutive relationship to find out the resulting stress, then we ( ) ( ) ( ) o o  t = E H t +  t
So, now if I just want to simplify and I want to write for only t equal to 0 cases, then this
implies that the stress is given as ( ) ( ) o o  t = E +  t , which implies that we have, if I
divide both sides vice o  then you have this and this quantity is my stress relaxation
function.
So, you have
( )
( ) (1 ( ))
o
t
G t E t



= = + , so I have taken E as common, so you have
E

. So, I
am just going to write that as . So, just remember, where is equal to, so this should
remind you that this functional form is something that we also got in the Maxwell case. So,
this is a form of a, this is response time of the system that is the same in both the cases.
(Refer Slide Time: 5:59)
So, we have this functional form, but what does this mean? So, if we are to plot our results, so let us say you have here epsilon verses time, you have, so in this case, what you are doing is you are applying a known strain to the system and then you are studying what happens to the stress history. So, what will happen is because of the delta function, there will be a very sharp jump here, which allows the form for delta function, which goes towards infinity.
And then immediately at any other value of t, this will come down to a constant value. So, we can see that the stress relaxation function is not really very realistic in this case. You have a very-very sudden jump and then it comes back and then it says constant, it does not decrease to 0 which tells you that this system almost behaves like a not almost, it behaves like a solid at times greater than 0. So, this is sort of viscoelastic solid response.
(Refer Slide Time: 7:28)
So now, let us take, we have, so we just we just looked at a jump in strain. So, what we want
to do now is to look at a jump in stress. So, to look at a jump in stress, what we are going to
assume is that with instead of a direct jump, you have the jump as a series of functions that go
from 0 to a value of one or o  here in a short period of time, this is  − , this is  + and the
biggest deviation occurs between  + and  − .
And that you have a series of such functions such that you are basically going to take the
limit of  tending to 0. So, to do that, what we are going to use is we are going to use the
previous this constitutive equation and then we are going to integrate over this small
timescale of  . So, let us do that. So, we are integrating both sides. And here you are
integrating is the strain.
So, if the stress is bounded then, so by the way this is once again, this is not very
mathematically rigorous, what I am trying to show you is slightly hand waving argument
here. A more detailed course, we will go through the mathematical rigor that is required to
get to this but the idea is sort of similar, that you consider a set of approximating functions
that rise from 0 to value and you take them that they differ from the required functional form
in a very short in time interval − and + .
And in this particular case, what you have is, you have some bounded number multiplied by
 , 2 here, similarly here. And as you take the limit of  tending to 0, both these 2 terms
will end up with a 0 and here you will be left with absolute 0. So, this is going to give you the
jump in the strain. So,  (0 ) + and because these are 0, you will end up  (0 ) 0 + = .
Now, there is an important issue here, we could not apply this in the previous case and the
reason for that is in this particular system, you end up having so just let take a look at this
functional form that the stress has a delta function, the delta function is not a bounded
function. In fact, it is not even a proper well defined function it is, we use it for its particular
properties, but it cannot be used in it is not a regular function in that sense.
So, this functional form here is not bounded and because it is not bounded, you cannot use
the idea that this will go to 0 because  is tending to 0 because you might a 1 term is tending
towards 0 but the other term if it becomes unbounded then you not have that situation. So,
that is why we had to use 2 different arguments for the system here to get our jump
conditions. So, now, let us take a look at, so now that we have the jump conditions with us.
Let us take a look at the creep response, Creep response for a Kelvin Voigt body.
(Refer Slide Time: 12:18)
So, for Creep response, we have to what is provided to you is that ( ) o  t = for t greater
than 0 and now we have to solve this particular equation
1
o
E
  


+= . So, this is your
governing ODE for t greater than 0. So, this value is constant here. So, again we have to
apply the method of integrating factors, so again apply method of integrating factors.
So, your ()vt in this case, once again, you will see that it ends up as the same quantity as
( )
t
v t = e . So, we are left with, so this implies that we have o  .
1
( ( ))
t t
o
d
e t e
dt
  

=
Now, we will integrate both sides from vto t. So, you have
t
e , I just use some other variable
s
es  . This we have to consider between 0 + to sometime t.
And on this other side, we are just going to integrate this particular form.
0
1 t s
o e  ds
 +
 . So,
you end up with an exponential ( )
t
e t and  (0 ) 0 + = that we just did. So, this is the entire
left hand side. And on the right hand side you will have, so this is
1 s
o e 

, this goes from 0
plus to t.
So, this will become 0. So,

is
1
E
this is, so your ( ) (1 )
t
o t e
E
 

= − , I am taking the
exponential onto the right hand side. So, now, my final objective is to find the Creep
function. So, what I do is I take the o  onto this other side. So, my
()
o
t 

which gives me the
Creep function is
( ) 1
( ) (1 )
t
o
t
J t e
E
 

= = − . So, this is now your Creep function for the Kelvin
Voigt body.
(Refer Slide Time: 17:02)
Let us plot this, so that we get a feel of what this response really looks like. So, you are
applying a stress, a constant stress. So, you will be applying sigma naught here. And then you
are going, you have to measure for the creep or the strain value is. And, what you will get is
this particular curve? You will get a curve like this, which asymptotically reaches some
value. This is your exponential curve right here.
And this represents your Creep response function for a Kelvin Voigt Meyer body. Now, you
can ask yourself whether this is physically realistic or not. And the answer is that it does not
look very satisfactory, because it starts from a value of 0. If you recall that the intuitive
diagram that we had drawn that there should have been some quick elastic response to an
applied stress and then there should have been increased with time.
But that initial elastic response is totally missing in this particular case. So, even though when
you looked at the stress relaxation function, that behaved as if the Kelvin Voigt, this
particular system has a very strong solid body like response, the Creep function does not
reflect all of that, it does saturate to a value. But here you do not have that. So, this is
acceptable probably in some cases, but it is not a very satisfactory case.
(Refer Slide Time: 19:03)
So, we have had that Creep response in this and the stress relaxation function. So, we will
quickly look at response to an arbitrary strain that is what we had done for the Maxwell
model. So, the first thing that we had looked at there was response to arbitrary strain history.
So, basically you have provided a strain history of ()s  is provided to you for all values of
time leading up to t. And the question is what is  (t) ?
Now, if we go back to the equation, the governing equation for this particular body, then we
have a very simple form. So, we have  (t) E (t)   (t)

= + . So, given arbitrary strain history,
all you need to do to find out the current stress, value of stress is simply you have to compute
this. So, you can just directly find it. So, if you are given this particular functional form and
all you need to get this result is to just plug it in here and compute it directly. So, this is really
easy that is it.
(Refer Slide Time: 20:46)
So, now, let us look at the other case, which is, so, this becomes more of us, very simple. So,
the response to arbitrary stress history. So, if you want to do that, then basically your stress is
provided to you as a function of time up to leading up to all values till t is greater than, and
the question is what is  current value of strain?
To do that we realize that we have what we have to solve is basically the same thing, linear
ordinary differential equation, which is non-homogeneous and that has to be solved. So, for
that once again we need an integrating factor and integrating factor in this case and
appropriate integration factor you will see again is the same form t by lambda.
So, in this particular case, what you will get is if you apply this integrating factor and you
will end up with
1
( ( ))
t t d
e t e
dt
  

= which is, so now we have to integrate both sides from
0 + to t and you have the
t
e .
So, I am just going to use a different variable here for a second. So, say the ( )
s
e s , this has
to be found out from 0 to t. And you have on this side, you have
0
1
( )
t t
e s ds
 +
 , so we are just
going to use the dummy variable s for time. And, we know that  (0 ) 0 + = .
So, if you apply that, then you will end up with
0
1
( ) ( )
t t t
e t e s ds
 +
=  . Now, I am going to
not complete this and I am going to sort of leave this for you as a homework problem.
So, you have to integrate this by parts. So, once again, you see that you have the functional
form ()s  , whereas, what you to want is a 

, so here what you can do is integrate by parts
because I have done it for the previous case, so leave it here, integrate by parts to obtain and
what you will find is, so a surprise, the strain is still given by the same functional form that
we had gotten for the Maxwell model.
So, this is and this is rather interesting because the two Creep functions are very-very
different yet when you try to find out the strain history, the strain from the stress history, you
end up with the functional form that is exactly the same as the Maxwell model. So, this
indicates that this particular functional form that we are seeing that I have boxed in red is a
more generic equation that governs such models.
So, for today, we will leave it here. And the next class, we will look at once again, the
quickly a comparison of the Maxwell and the Kelvin Voigt model and then we will go on to
make more complicated models for ourselves. So, we will stop here today.