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Two Maxwell Model
So, welcome, everybody to another lecture on introduction to soft matter. Last time we leftoff while were discussing Jeffrey’s model, we are not completed our discussion. So today wewill start off where we left off and we will finish up that discussion. So, what we had done iswe had written down the two expressions for the forces and then we had to simplify that. Sojust let us write down that one more time. So, we have the force for the Kelvin Voigt body,the Kelvin Voigt Meyer body, and then the force for the spring, sorry the dashpot we have towrite those two separately.(Refer Time Slide: 1:08)So, let us do that. So, we have the force for the Kelvin Voigt and this is as said, as wediscussed is equal to the force in the system. ( ) ( ) ( ) KV KV F t = F t = D+ E x . Similarly, for theother case, we had the force in the dashpot is equal to the net force on the system anyway andthat is again equal to so, 1 ( ) ( ) D d F t = F t = Dx .So, what we want to do is, we want to add, we able to add these two together but somethingis preventing us at the moment because we want to add the displacements and make them the,to end the total displacement for the system. So, what we can do is we can multiply and whatis missing is this part. So, what we can do is we can multiply it the two equations withappropriate factors, such that the addition is possible.Now to do that, you will recognize that we have to multiply the first equation by 1 D whichgives 1 1 ( ) ( ) KV  DF t = DD+ E x And then we have to multiply the other, the secondequation with the appropriate one which case we are going to multiply it with this particularfactor. So, 1 ( ) ( ) ( ) d D+ E F t = D+ E  Dx .Now you can add these together. So, the addition is not possible.(Refer Time Slide: 3:27)So we add and here we will get 21 1 1 F(t){ D+D+ E}= (D + ED)xSo this is basically nothing but if I want to write it in the form of the dot, if I now change myoperator to just say, to the dot form, then we have 1 1 1 ( + ) F(t) + EF(t) =  x+ E x(Refer Time Slide: 5:14)So, this now suggests, the above suggests that the appropriate constitutive equation shouldtake the form and then we are just going to create an equivalent system where we say. So, thisbasically suggests that your stress strain relationship can be given by1 1 1 ( + ) + E =  + E . So, this is a question that we are, this is the constitutiverelationship for the Jeffery’s fluid.Now, you will recognize that this equation has a double derivative and not, and hence doesnot fall into the form that we had solved for previously. So, this all our solutions that we haddone previously were only for this particular case where you had a single derivative of thestress and a single derivative or the first derivative of strain. But here you have doublederivative. So, this cannot be solved in the same process.But obviously, you are going to have to use a similar, you can use a similar process of takingLaplace transforms and solving for that. But this is quite interesting that the fact that thedouble derivative appears itself is very interesting here. So, one question we should ask is,what are the other situations in which double derivatives can exist? And that should lead us toa window into try understanding a generalized equation when you have many, many of thesedifferent strings and dashpots, all of them put together in a big circuit.Finally, what you are doing is you are just putting all these different, different elements tomodel the complexity of real life. So, if a real fluid exists, it is likely that it will have manydifferent especially in cases where polymers where there is polydispersity and etc. Thepolymers will have, will behave, will not behave in a manner that you can represent them bya single E or single eta.So, a realistic model should have many different springs and dashpots together probably andhence, we should now look towards a more general case. And before I end this particulardiscussion, I just want to note that there are two characteristic timescales here, are twocharacteristic timescales in Jeffrey’s fluids. And the first one is 10 E += and the secondone is 11 E = .(Refer Time Slide: 9:07)So now we march towards a little bit more complexity. So, the first thing we will do is twoMaxwell models in parallel. And we want to check out a hunch whether two Maxwell modelsin parallel will end up giving us a double derivative for both cases or not, we will see whathappens. So, when you have two Maxwell models, let us just draw this. And now that all ofthese are going to be Maxwell, both of these, I am going to label this 1 E and 2 E for easydemarcation and 1  and 2  .Before we proceed further on this can we use our simple understanding of such systems toconclude whether this is a viscoelastic fluid or viscoelastic solid? See, if you apply a force onthis system, there will be some force that we shared by the first Maxwell and the secondMaxwell model. But whatever the forces we have a small; you will always have a dashpotthat is going to keep on giving you displacement. So, this both of them independently aregoing to behave like a fluid and the system itself is going to behave like a viscoelastic fluid.So, this is appropriate, this seems to be appropriate for a viscoelastic fluid. Okay.So now, we have to figure out the correct constitutive relationship for this. Now, because theyare in parallel from force balance we can say. So, let us say that the forces in themindividually are 1F (t) and 1 x and here you have 2F (t) and 2 x , there is no more need tolabel these subscripts and any more complicated way because they are both Maxwell’smodel. So, I am just using the labels 1 x and 2 x .So, the total force here is obviously just like the previous case going to be a sum of the twoforces. Again, similarly, similar to the previous case from geometry will have1 2 x = x = x , because it is geometric, both of them are dramatically constraint, thedisplacement of the system has to be equally, equal and be shared by the two cases. So, wewill write that relationship as this.(Refer Time Slide: 12:54)Now, recall that for a single Maxwell, the equation that we used to write would be, so far, asingle Maxwell load, write1( )DD x FE  = + . So, so we will use this here. So, using thisrelationship, so, using this relationship we can say that 11 11( )DF D xE + =  . Similarly, for thesecond Maxwell we have 22 21( )DF D xE + =  .Now the delta X is our shared, so I can just as well drop the subscript. And if I drop thesubscript, life becomes a little bit easier for us. But we have these F1(t) and 2F (t) and wehave want to be able to do our work here. So, in order to do this, we have to multiply bothsides by an additional factor. And that additional factor here if you see is going to be, somultiplied this said, but I have to also multiply the other side.So, what I am going to do is I am just going to create some space for myself. So, we have thisfactor here now, this is x . Similarly, I have to multiply this side now by 1 E and I do thesame thing here, sorry, this is eta 1 times of delta X. So, now this is in a situation where I canadd these two together. So, you just add them and the resulting equation will look somethinglike I will this in the factored form and here you will have.12 2 1 1 2 21 1 1( )( ) ( )D D DF D xE  E  E + + = + 21 1 2 2 1 11 1 1( )( ) ( ) ( )D D DF t D xE  E  E + + = + 2 21 1 2 2 1 2 1 21 1( )( ) ( ) ( )D D D D D DF t xE  E  E E  + + = + + + (Refer Time Slide: 16:33)So, what does this suggest? So, if I, now if I have to replace the F, so it suggests aconstitutive relationship of the form. So, we can replace the F with the stress of just writingthe pre-factors once again and then you have the stress here and then you have, sorry, now Ireplaced the displacement with epsilon. So, this is now of the form.2 21 1 2 2 1 2 1 21 1( )( ) ( )D D D D D DE E E E    + + = + + +So, we can see that you have the double derivative of both sides I just left the left-hand sidein the pre-factor form because, in factored form because it is just easier to leave it like that.Because the final point is that this equation, this equation now has the form. So, here you willhave see, you will have a double derivative of stress, you will have a single derivative ofstress and you will also have one term with only stress.So, this side, the left-hand side is of the form 0 1 2 p  + p  + p  . But the -hand side has adouble derivative of the strain. It has a single derivative of strain but does not have any otherconstant term. So, this side will have it looks like 1 2 q  + q  , there is a epsilon dot here. So,1 q  , where these are constants. So, 1 p , 2 p , 1 q and 2 q are constants to be determined fromthe model.So, we see that we have by having two Maxwell elements juxtaposed with each other, wehave created a situation that leads us to an ordinary differential equation once again todetermining the ordinary, ordinary differential equation relating stress and strain, but thistime you have, because you have two models, you also have a double derivatives. Now, thisprobably helps us, suggest a platform or a model or a mechanism by which we can generalizethis idea to let us say N-Maxwell models in parallel.(Refer Time Slide: 20:30)So, let us consider next the generalization to N Maxwell models in parallel. So, we have asituation where, we have many of these Maxwell models, actually N in total number. So,these are, this is the Nth one, Nth element. And each element is individually a Maxwellmodel. So, if you have to write the constants, we will just start for E1, this is E2, this is E3,and this gone and this will be EN.Similarly, you will start here with 1  , 2  , 3  and all the way till N  . So, before we proceedwith this, is this still a fluid or has the behaviour changed? So, if you think about thisparticular model right here, then see if you apply know a certain amount of force to thisparticular system, then that force will be divided into each of these separate elements.And each element will experience a certain amount of force, which is going to be a fractionof the total applied. Now, each of these elements, however, small the forces the dashpot willresult in a motion such that dealt, the strain or the displacement of the dashpot will not beconstrained with time, it will keep on moving as long as the force is applied.So, that means, that each of these elements separately individually will keep on acting likefluids. So, this entire thing to get, taken together will also act like a fluid. So, this model isappropriate for a viscoelastic fluid. So, the question is now that once we have this Nth, Nnumber of Maxwell models, what should be the correct partial? What should be correctdifferential equation relating stress and strain?So, we have already seen how to do this. So, we are just going to repeat the same procedure.But now there is just one complexity that we have N number of elements that is all. So, let uswrite down from our previous experience, what are the force we always want to start with theforce balance equation and the displacement equation.So, what will those be for this particular case? So, from force balance, so once again, thesystem will experience a net force, which is going to be a sum of all the individual forces. So,in this case, your1( ) ( )NiiF t F t==  . Similarly, from geometry you will have the constraint thatthe system displacement is the same as the displacement of all the individual elements.So, I can write the x = xi for all i =1,2,...,N . So, now, the individual, so recall that foreach individual element the relationship that we had discussed earlier, which was1( ) ( ) i ii iDF t D xE + =  , for all the different i. Now these relationships are individually stilltrue.So now our question is how do we figure out a constitutive relationship for all of this puttogether and we have to follow the process, a procedure which is very similar to what we didbefore. And this I would like to leave as a homework problem or just a self-work problem foryou, in the next class and we will pick it up exactly here. So, we will stop here today.