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Module 1: Sinusoidal Oscillations

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Jefferey’s Fluid Element: Tutorial

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Jefferey’s Fluid Element: Tutorial
So, hello and welcome back to another lecture on introduction to soft matter. Today what we are going to do is we are going to look at some example problems. These example problems have been designed so that we can actually end up revising some of the materials that we have been discussing. So, it will be a very good work through some of the important concepts that we have already discussed.So, we have a few different problems to explore today. But the first problem that I have designed So, we want to take a look at a certain type of viscoelastic material for which the analog is already given to you.(Refer Slide Time: 01:09)So, let us say the problem one. So, so the problem one is Jeffrey's fluid element, and if you recall the Jeffrey's fluid element comprises of a spring and a dashpot in parallel, and then another dashpot that comes in series. And, what we are given is, the values for the modular here that we will use, are given to you so here the spring is characterized by E which is given as let us say 300 Pascal's the viscosity here is given as 5 and this is another viscosity.So, this is given as 1. So, we see that we have these two different dashpots and the numericalvalue of the strength of these dashboards are two different values. And the question is, so thequestion you are trying to ask is find out the expressions for G(t) and J(t). By this time we arevery well familiar with what J(t) these two G(t) and J(t) imply.So, we are not writing the full form, we are just going with a mathematical form that we havealready we are quite familiar with at this time. So, now let us try to solve this particular problem.Now, this is a the so, the constitutive equation for this will be the constitutive equation for aJeffreys element and we have already seen that in one of our class we have already worked thatout.So, we are not going to repeat ourselves, so, we are just going to take that result. So, theconstitutive relationship for this, for the Jeffrey’s element, I think this is not apostrophe Jeffreyselement is already well known and we have derived it earlier. So, if we just recollect frommemory or you can take look at your notes, you will find that the constitutive relationship forthis is given as on this side, we will have the derivatives of stress, stress and its higherderivatives.And on the other side of the equation we will have strain. So, here we haveAnd, we have already we already also know that these coefficients can be expressed in terms ofthe variables that constant that comprise the Jeffreys element. So, these four coefficients here canalready be expressed in the terms of these three variables that we have. So, . So, these four coefficients are well known.(Refer Slide Time: 05:29)Now, we have to solve this particular differential equation. So, let us take the Laplace of bothsides. So, to solve this let us apply Laplace transforms. So, and we are going to use the sameconvention that we have used before so for example, if you have  then ()s  basically tells youthat the Laplace transform form and s is the variable that corresponds to the Laplace transform.So, we can write, we haveSo, as I said before this bar implies a transform and s is the variable for the Laplace transformand then you have 2So, this is the expression for the first term and then you have the second term which will give us So, now we have this Laplace transform the problem with this particularequation is now we have these variables for which we need to know the answers for and thiscomes from the this should come from the compatibility conditions.2Now, the compatibility conditions or the initial conditions for this can be derived from therepeated application of the integral formulation that we had looked at before. In this particularcase, it becomes a little bit tedious. So, I am just straight up going to tell you what the answerswill be, what will happen is when you derive that, that all these values will just drop out.So, these all dropout, which will leave us with only the first terms for this that are involved here.So, this equation now will become 2 . So, as I said, wehave skipped one part where we show how these terms drop out, but basically that can be derivedfrom looking at that limiting process that we have seen before.And once you apply that and the initial values and the stresses that are evaluated at this particularat 0 plus just the values of the stress and the strain and the derivatives, then you see that they willactually drop out of this transformed equation. So, now we can write our equation as for thisside, we have 2(Refer Slide Time: 10:21)So, to get the functional forms for G(t) or J(t), we have to solve this equation using theappropriate condition? If you are looking at the stress relaxation function, then our epsilon thestrain is a step function, if it is a step function then we know that to evaluate G(t) we have to set. We are already familiar with how to solvesuch equations, where such fraction all forms are there right.So, let us go ahead and try to solve this So, I am bit out of space there. So, I am just going towrite it over here. I am just going to simplify my expression a little bit and make it such that thenumerator has an s plus p naught plus p1. And in the new in the, this is sorry, I am going tosimplify the equation a little bit.So, that I divide the denominator by 1 p , I end up with 01psp+ , In the numerator I divide by 2 q ,so I end up with 1sq+ . And hence I have to multiply on this side to balance everything out21qp, so this will now give me, let us say so I am going, this fraction this constant, so I am justgoing to clump together into one and now I have this particular partial fraction form to solve.12 200 11( )qsq qsp s pp +=  +So, I can write this as, my assertion is I can write this1 00 2 2 11 01( ) 1q pq q psp p sp −   =    +    +  , I here inthe numerator, I will have see I deliberately did this right, because, when I have to take theLaplace inverse, I have to take the Laplace this is a constant, here this is just 1 and here this is afunctional form that we are well acquainted with.So, we already know what the Laplace inverse of this is, this will be an exponential form and theLaplace inverse for 1 will be the ()t  . So, when you take the Laplace inverse, so now, this thiswas my, this is still in the Laplace domain. So, this is my ()s  .(Refer Slide Time: 14:13)Now, to solve this, I have to take the Laplace inverse, so take taking the Laplace inverse. Oncewe take the Laplace inverse, we can go back into the time domain. So, now when we do theLaplace inverse, I will get sigma t, so it’s a function of our time and on this side, we are going totake the Laplace inverse of this entire thing.Now, this is a constant so, we can just leave it as that we have 0 21qp and here we have to takethe Laplace inverse of 1 that will give me the  (t) plus the Laplace inverse of this sorry. Thiswill be 1 02 1( )q pq p− , this is my multiplier and here we have the exponential form, which willbe01( )ptp e−.01( )0 2 1 02 11( ) ( ) ( )ptp q q pt t ep q p    − =    + −    So, from here to go to G(t) is just one more step.So,01( )2 1 02 10 1( )( ) ( ) ( )ptp t q q pG t t ep q p   − = =    + −    So, whether this is at least the answer or at least part of the answer, we will evaluate this numerically also. But this is the algebraic form that we were looking for. Now, wherever you do a calculation like this, it can sometimes become tedious as you saw that we have a lot of Laplace transforms, we had to take the Laplace transforms that we had to take the Laplace inverse.And this often a scope for silly mistakes. So, you might end up missing a term or you take the incorrect Laplace inverse. So, I mean usually my rule of thumb is whenever you do a calculation like this, then one thing you should do is just subjected to a sanity check. So, the question is, now that you have this answer, I can we do a simple sanity to check to see whether this form is anywhere close to a real form that should be there.Now, for that, let us just quickly go back to the Jeffrey’s element. Now, this is going to act like a fluid element right. Because we have discussed this part already when we were discussing Jeffrey’s fluid that this is essentially should be representing the constitutive equation for viscoelastic fluid.So, we expect a fluid like behavior. So, we also expect the function of the stress relaxation function to behave as if the material is a viscoelastic fluid. What happens in a viscoelastic fluid, the stress relaxation function should decay to 0 at large times does this decay to 0? Yes, you can see C this is a Dirac delta function and then you have a term that is exponentially decaying with time.So, this entire form is going to go over to 0. So, this does qualify through our simple sanity check with this form is indeed looks at least correct and if we have done the algebra right and this form is the is the algebraic expression is exact one more thing you see that this this strange Dirac delta function is there right and what is the physical reason for that, the physical reason for that actually comes from the fact that this is actually a fluid element.(Refer Slide Time: 18:34)So, when you have to give a step function, so if you 0  H(t) to get to that initial value of strain,you need a very high jump in stress. So, that these dampeners can quickly instantaneously go to afinite value of strain and that is what actually causes this kind of a problem.So, you can see that although we have derived the expression there is slight non physicality in asense that exists in the relationship. So, we have the all the values for 2 q , 1 q etc. So, we can justsimply go ahead and evaluate that. So, now that we have the expression for G(t) what I did Irealized that there is a small error and it happened midway somewhere and this here it should be1 p .So, this value should be 1p . So, now, this is the correct expression for G(t). Now, we havealready, we already know the numerical values for 2q , 1q , 1p , 0p etc. So, all the four coefficientswe already know what the numerical values will be. So, if you take those numerical values, sousing the numerical values for 2q , 1q , 1p , 0p we will get this G(t) as yes, so this is the result thatI got, and you are pleased, invited to do this by yourself and check whether my expression iscorrect or not.We have3006 5 5 300 300( ) ( )6 6 5 6tG t  t e  −= +  −  50 5 25( ) ( )6 3t G t  t e− = +So, this is the final answer in numerical form that we are looking for and as I said, you are free todo this calculation by yourself and check my answer.(Refer Slide Time: 21:45)So, when we started out we had two questions we had we wanted to find out the expression soG(t) and the J(t). So, let us go ahead and try to see how J(t) will look like now. So, for Creepcompliance the relationship between the other original transform equation remains the same. So,it is 21 0 2 1  (s)( p s + p ) = (s)(q s + q s) and for Creep compliance now the quantity that is known isthe Laplace transform of sigma bar or the Laplace transform of the stress function.So, the Laplace transform of the stress function is simply 0 ()ss = . So, this tells implies that00 1 12 1 22( )psp pss q q s sq+=   +    So, this Laplace transform is interesting, because now you have this interesting s square. So thenumerator sorry, the denominator now has higher powers of the Laplace variable. So, this willnow become so if I just club all the constants together. So sorry this is 2q and in the first so I amjust going to split this into two partial fractions here and here we will have, so this s will getcancelled with this one.So, we will have00 1 11 2 1 22 21( )( ) ( )pp psq q q s s s sq q   =    +    + +  So, this is an interesting problem. Now, because this Laplace transform is slightly what isslightly different than the one, we have seen before. So, I encourage you to do this problem byyourself.Right now, it’s basically going to be very similar to what we have the process we have adoptedbefore. But it is a very good exercise to try it yourself. So, what I am going to do is I have theanswer written down from my calculations, and I will just write it down. And I encourage youvery strongly to go through this by yourself and double check, so I am going to write down theexpression I got.(Refer Slide Time: 25:30)So, the expression that I found is10 1 0 2 221 1 1( ) (1 )qtq p p p qJ t t ep q q  −= +  −  − So, if you again insert the numerical values what we have seen before or for these coefficientthen you end up with60 ( ) 0.0033(1 ) t J t t e− = + −So, this is our final expression. Now, please notice a few things here. And the Creep compliancefunction has a linear part and that is a decay part. So, this function seems to be far more wellbehave than the than the function G(t) late, but notice the powers of the exponential functionhere it is -60 t and for the stress relaxation function plus -50.So, the two values are different. So, we have two characteristic timescales corresponding to thetwo different functions that we have explored. And one is we know as the relaxation time scale.So, with the characteristic time scale, they are associated with G(t) is that relaxation time scale,whereas the one that is associated with the Creep function, that is the retardation timescale okay.So, this was our problem one.(Refer Slide Time: 28:05)Let us take a look at another problem, so problem number two. So, we will end todays lecturehere, and I will see you in the next class. Thank you.