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Jeffery’s Model
So welcome back to one more lecture on Introduction to Soft Matter. Last time we were discussing the solutions to the cases of the three-parameter model and it is different solutions.(Refer Time Slide: 0:40)One of the things that I had like just like to point out before I start today’s class is that the last expression that we wrote, and this is the expression of the Laplace in, when we took the Laplace inverse, we got expressions in the time domain. Now see this, I am using a variable s here also, but this is just so that I am consistent with what I have done before.This is just this s here represents dummy variable for the integration, you can replace it with any other, you will replace it to the tau, you can replace it with any other variable you want and expression will still remain the same. So, this s here in the time domain should not be confused with the s of the Laplace inverse function.(Refer Time Slide: 1:22)So, what we want to now do is we want to look at one more model. And so, let us take a lookat we had to done one. So, we started off. So, the three parameter models, so, we had in thebeginning we solved, we introduced two different three parameter models and we solved forthis particular case. So, let us now solve for the other one.So, I am just going to redraw this. So, the model looks like this. So now let us say this is E,this is my dashpot characterized by  , this is my another spring which is E1. And here, so theobvious question one needs to ask before we will try to solve this is will this have a solid likebehaviour or fluid like behaviour.Now, this is a Kelvin Voigt model. So, we have put a Kelvin Voigt in series with another spring and that is what it is. So, the Kelvin Voigt, we know already has a solid like response because for a finite force, the displacement of the system is already bounded, it is going to be finite and the same holds true for spring.So, even when you put these two together, where for a finite force, the displacement in both of them are going to be bounded. So, this represents a solid model. So, this, so this is appropriate for viscoelastic solid. So, this is, so this is a Kelvin Voigt Meyer model. So, I am sorry, I forgot Meyer, but, so this is a Kelvin Voigt, Meyer name should also be there. So, now let us recall what was the governing equation for the Kelvin Voigt model?(Refer Time Slide: 4:10)So, this will quickly go back, scroll back to where we are discussing the Kelvin Voigt and wehad gotten this particular expression. So, now, just like the previous case where weintroduced the operator D. I can write this as (D+ E) = that for the part of the KelvinVoigt this expression holds true, a Kelvin Voigt Meyer. So, now, let us analyse this in thesystem such that, let us say the displacement that is seen by the first body is going to be KV xand the force it experiences ( ) KV F t . The spring on the other hand is going to see an extensionof s x and a force of ( ) s F t .(Refer Time Slide: 5:36)Now, these two, so I am treating them. So, I am going to treat this entire thing here as onesystem. So now I am just going to write a force balance. So, for force balance, so from forcebalance, and remember, these are all masses quantities. The springs, the dashpots, are allmassless. So now, I am going to apply the force balance once again and in this particularcase, in the case of the series system, you will see that the forces are all the same.And then from geometry we get a different relationship for the displacements and thatdisplacement is KV s x = x + x . So, the entire system displacement is given by the sum ofthe individual displacements. And the relationships that we have are so, also given theindividual constitutive relationships, you have that the force in the spring is equal to, is thesame as the force in the system, 1 ( ) ( ) s s F t = F t = E x .And in the Kelvin Voigt body you have the force, if I want to write that, this is again equal tothe entire forces experience by the system. And now this is ( ) ( ) ( ) KV KV F t = F t = D+ E x .Now, please remember that this D is the differential operator, so, it is a DVT here. So, I wanta relationship between F here, the left-hand side in the force balance equation and the xwhich is again on the left-hand side of the geometric relationship,.So, ideally, I will be able to get x , simply by adding these two, so to add these two. Now Ido not want to take this into the denominator because this is actually a differential operator.So, what I want to do rather is I want to be able to keep it in the numerator itself and then addthis up. To do that there is something very simple we can do, which is we introduce anothermultiplicand.So, I am just going to erase this because I just want to keep the order. So, we can multiplyboth sides or actually let me, let me leave it like that. So, to add this, what we are going to dois we are going to multiply, so I am going to rewrite times of F(t) equal to rewriting the firstexpression, you want s x . And in the second one I, for the second equation here, I multipliedthat with E1. So,1 ( ) ( ) ( ) s D+ E F t = D+ E E x1 1 ( ) ( ) KV E F t = E D+ E xSo now this is in a form such that you can simply add these two expressions, so you just addand what you end up with 1 1 (D+ E)F(t) + E F(t) = E (D+ E)x . So, if I were to use thisoperator then we end up having, so, this is the same as1 1 1  F(t) + (E + E )F(t) = E x+ E Ex(Refer Time Slide: 10:55)So, now, this suggests. So, this suggests, this is now the relationship between the force andthe displacement. So, this will suggest, so this suggests a constitutive equation, equation ofthe form 1 1 1  + (E + E ) = E + E E and this is the answer I was looking for. Now theimportant thing I want you to note here is that this is, this is again a linear ODE and the linearODE has the derivative of stress, stress itself on the left-hand side.And on the right-hand side, you again have a derivative of the strain and the strain itself. Andthis is exactly in the same form that we had the other expression. So, this again can now bewritten as, so given this form, one can again write it, again rewrite this expression, this asbeing 0 1 0 1 p  p  q  q • •+ = + .The only difference now is that the values of the coefficients are changed. So, I have to nowspecify what the values are. So, this value will be 1 p = , your 0 1 p = (E + E ) , 1 1 q = E and0 1 q = E E . So, now you probably can now appreciate why we were solving the entire thing inthe form of these coefficients previously, because where we are solving this, so I am justgoing to quickly go back to the Voigt solution.So, we were trying to solve this particular expression in the previous case, and when we didthe Laplace there is nothing here which assumes anything about the model itself. The modelbuilds in the way that the values of 1 p , 0 p , etc. are set. So, to now get the solution of thisparticular model, all you need to do is to insert the values of the 1 p , the constants in thesolution.(Refer Time Slide: 14:18)So, the solution for the, for the stress relaxation remains the same and so, does this expression events the same, among all our expressions actually remain the same. This one quick thing and that is remember that for this model to be physically realistic, when we had derived the stress function, the stress relaxation.Then we had here a constant term and then we had this other term in the bracket which was multiplied by an exponentially decaying term. Now, for this to be physically realistic this particular fraction has to be greater than one, sorry greater than 0 for it to make sense, physically realistic system.(Refer Time Slide: 15:14)So, we had insisted that 1 01 0q qp p− should be greater than zero, which it was true for thisparticular case because we saw that. So, we evaluated 1 01 0q qp p− . And we got, we saw that thisexpression is actually equal to E and it is greater than 0.(Refer Time Slide: 15:35)So, let us just do that here one more time. So, let us just ensure that it this is a physicallyrealistic model and you will get solutions that are going to be realistic. So, let us checkwhether 1 01 0q qp p− is greater than 0. So, 1 0 1 11 0 1q q E EEp p E E− = −+.So, this is equal to 111EEEE E−+by this one which is, so this quantity here. So, this is afraction which is so, just a second. So, this quantity is now 11(1 )EEE E−+, and this fraction isless than one because this is, if you look at the functional form of the numerator denominator,this is less than one, so this quantity is greater than zero. So, in a sense we are saved, we donot have to worry our functional form is such that that the physically realistic angle isretained.So, this particular system has the solution and that same solutions that you have seen beforecan be applied in this case. So, we have done two, three parameter models, and both of whichwe saw are actually good for representing the case of viscoelastic solids. Now, it should beobvious to you that when we are doing the three parameter models that there will be onemore obvious three parameter model where you would not have a spring but you have adashpot.(Refer Time Slide: 18:00)
So, this model is also called what we have into going to introduce now is also called theJeffrey’s fluid. So, the Jeffery’s model, so in the Jeffrey’s model, what you have is, you havea Kelvin Voigt system. So, you have an E, this is  and I was going to say that this is  andthat is 1  , the same spirit that I have been. So, we are going to solve for this, but before we dothat, we should ask ourselves one more time in as an intuitive idea whether this is going to beviscoelastic appropriate for viscoelastic solid or viscoelastic liquid.Now, take a look at this particular system, this is obviously the Kelvin Voigt Meyer bodyhere. So, when you apply a particular force, the displacement in this is going to be bounded.And, but for the case of the dashpot you cannot ensure that, so even for small finite forces,the dashpot will you keep on increasing or keep on displacing. So, this entire model will havea system level displacement, where this is, the displacement will keep on occurring even forsmall forces. And that is reminiscent of a solid, sorry for a fluid.So, this particular Jeffrey’s model is appropriate for viscoelastic fluids and that is why this isalso often known as Jeffrey’s fluid. So, this model is appropriate for viscoelastic fluid andhence said it is also known as Jeffrey’s fluid. So, we have the Kelvin Voigt model once again.So, we are just going to borrow from last time what we had written, we had said here, we arejust going to reuse this one more time. So, (D+ E) = , so ( ) ( ) KV F t = D+ E x .So, I am just reusing the expression from there. For the dashpot we are just using the samemethods. So, you understand that when I say KV x , then that means that the displacement inthis the Kelvin Voight body. So, none of that has changed. So, when I, when I am using thatyou can understand that this force again is the force in this particular subsystem. So, for thedashpot, you again now have D 1 d F =  x . And now we have to figure out what are therelationships between the different quantities.(Refer Time Slide: 22:40)So, from force balance, now this system is actually exactly equivalent to the previous case.So, what we had written the expressions that we had discussed for the other three parametermodel are actually going to be valid even in this case. So, the for the force parameter, for justfor the force balance, we can just use the previous particular case and we can just say that theforce in the system is going to be equal to the two individual forces. ( ) ( ) ( ) KV D F t = F t = F tAnd from geometry, we are once again going to have the same expression which was that thenet displacement KV d x = x + x . So, now, what is where do we want to get to? We want arelationship between the force and this displacement here, the left-hand sides of the two, theforce balance of the, from the geometry the two relationships that we have.(Refer Time Slide: 24:14)So, let us write down the expressions. So now we have F(t), which ( ) KV F t which is the sameas the net force FKV (t) = F(t) = ExKV + xKV . And in this case, I am just going to use thisstyle. So, this is the expression for the first case. And for the second case, we have force inthe dashpot again; equivalent to the system force because we have seen that and that is equalto 1 ( ) ( ) D d F t = F t =  x .And now we want to add these two expressions. So that is what we want to do. And wecannot add these two up easily. Because you have different. So at this point, what we have todo is that we have to simplify or we have to do the addition in such a way that we can add allthese up and we can get the net force on one side and we could get the net displacement ofthe other side. So, this is something that I had leave you to do for the next class.And then I am going to do, I am going to solve it for you in the next class for you. So, we willstop here today. And what we are done is we have looked at another three-parameter model.And we discovered that the solution for that is interesting the same as the solution for theprevious case, except now the coefficients are something that are just going to be determinedby the particular model. And we are going to see that the same is going to be true for theJeffery’s case. And we are in the process of solving for that. So, next class will complete thesolution for the Jeffrey’s fluid.