Sinusoidal Stress and Strain History
So, we have an imposed strain history, which is given by this sine function, right. And I am using s as the variable here, although we had to use s also in the Laplacian’s, but let we should not get confused here because there is no Laplace transform being actually, I used Laplacian. Laplacian is something else, so we restart, okay. I am going to restart the entire thing. So, welcome back to one more lecture on introduction to soft matter. And this is the last lap of our class, where we were now going to discuss the imposition of sinusoidal stress or strain histories and the final the output that we get from a viscoelastic material. Now, we had seen in the last class that we started off with imposed strain history. (Refer Slide Time: 01:28) And we had said that the equation will now be of the form epsilon, which is now a function of time. Let us use the variable s for time here, . Now, please do not get confused with this s here, I am using s here for time, we had discussion on Laplace transforms and s also appeared there, but that s is strictly the Laplace transform variable. So, there is no real reason to get confused here, this is a different context altogether, we are working in the time domain not in the Laplace domain. And there is a reason why we are using s. We will see that this s is again a dummy variable. Finally, we will end up with t that we will use, okay. So, this is the imposed strain history. And now, we know that for a linear, so, we have assumed that epsilon naught is small enough that we can use the formulas that we have derived for a linear viscoelastic. So, for a linear viscoelastic, we know that the , so, the stress at any function t is a function of its memory and we can write it as 0 . So, we have done this before. (Refer Slide Time: 02:57) And where this finally led to was the equation that we assumed that the ()Gt can be written up as 2 parts. One is a time dependent function and another is a constant. So, this is where we had to stop last time. Now, this particular, the first part of this of the right hand side, this simplifies. Now carefully look at the, the terms inside the bracket, we have an , which for this case is 0. And we have s ds , right. So, now, let us expand the cosine function. If we do that then we have, so we will, this omega and we have delta G of s. So, let us bunch the, this is an integral in for ds. So we will just use the s term here, sine omega s ds. And this entire thing is now going to be multiplied with sine of omega t. Because t is not a variable for this integral, the sine omega t comes out. Similarly, for the other, we will have ()Gs and this time we will have a cosine. We had the sine, so we have a cosine omega s ds and then this thing is multiplied with cosine of omega t. So, here we know that by the way epsilon t this term is equal to epsilon naught sine of omega t. So, you end up having, so I can bunch this into 2 different quantities and the first quantity will be 0 t this a sine plus again you have this epsilon. So, you have an epsilon naught. This time you also have inside you have an omega and this is Now, please, this has become a big equation, but we are quickly very quickly I am going to simplify this. This entire quantity right here is an integral that is going to yield us some function of time, this is again something multiplied by . And here on the other side, you have this quantity which is going to be some function of time again, but that is going to be multiplied by with a cos( So, I can say that this function now is equal to this function, I can say this is maybe some function, which is a function of omega and some other function of time. So, ' G is some function that has to be computed. Similarly, for the other quantity, we can say that, let us use these symbols, let us say that this quantity when evaluated leads to another function, we will call '' G . And this is again a function of omega and time. So, now, if we use these 2 terms and if we replace that huge term that we had this simplifies our term tokay so this has led us to a very interesting point. We provided with a sine input. And by the way, I think at this point you can probably appreciate why I used s instead of t in the beginning, because that s became a dummy variable and finally, everything has gotten transformed into a function of time t. Now, we had put in an input that was only a sine function. But in the output, which is a stress here, you end up getting both a sine and a cosine. So, this is basically what it implies that you have a sinusoidal input which is generating another’s, you can think of it as having being or being generating another sine function now as our output, but that which is phase shifted. So, let us for a second here, what I am going to do is I am going to use a trigonometric identity, which you might not be familiar with, so I okay. So, we will close this here. So, from this we can see that this particular quantity probably can be written in a more simplified form. (Refer Slide Time: 14:22) So, using this trigonometric identity, sorry just a second, we have, so using this identity we can rewrite Now here, what we see is, so let us just quickly go back to the expression we have an output where obviously, the amplitude is also a function of time and the harmonic part is or the sinusoidal part is also a function of time. So, this amplitude part arises because you are also seeing the transient behavior. So, if you want to see the long-time scale steady state behavior, we can set the t is we can take the limit t tending to infinity. So, now define. So, now, we will define the limit terms, so, we will say that So, if you write it like this then if you put, so then or you can, so that once you define it like this, then you can write that then this term is basically in terms of the original integral or the original sum, it is now equal to this same quantity now, except that in the integral the definite integral, the limit t is replaced by t tending to infinity. G G s s ds So, this becomes a veryvery important result for us, this. So, this was derived using the idea that you had taken a sine function as an input. But in a more generic term you can also have a cosine input. (Refer Slide Time: 19:53) So, if you take a cosine input, if you assume, if you start with , then you can repeat this entire process all over again as we just did, and then you will end up with a new equation. Actually, the equation will not be really fully new, it will be seem very-very similar, okay. And what you will find is if you do this calculation yourself, I am not going to re-do this entire calculation right now. But if you do this calculation, then what you will find is that this results this will lead to a situation where the stress as a function of time is given as epsilon naught. You will again have a . And this negative actually will result from the simple fact that you take the derivative of cosine you will also have a negative sign. So, that will sort of flow down. So, now a couple of points that we, so once again we took some simplifications okay while we are deriving this one obviously in the very beginning was that epsilon naught is small, so, that we can assume linearity. The second was that the function G(t) can be broken up into 2 parts, one is a constant part another was a time varying part. And then it is important also to note, so note that integrals with cosine and sine are actually not very they do not converge properly. So, if you take time tending to infinity, you will not always get, they do not converge to a particular value. So, these integrals they do not approach a limit as t tends to infinity. But when you are multiplying it with another decaying function, it can give you it can reach a limit because here we do have there is with a cosine we have something that is getting multiplied. Now, this function we have to assume that this is a decaying function because if this function does not decay then you cannot, the following results do not really follow. So, this, so another assumption is implicit assumption was that the integral of . This is some finite value, some finite value or basically that this integral converges. And that is why we had required the delta G tends to 0 as t tending to infinity. So, if you remember in the very beginning, we had done that, I will just quickly scroll back. And we, so here we had said if you recall the delta G is a decaying function of time. And the reason for that is so that these integrals can exist and we can write down the answer in that particular form. So, if this is not true then the previous results will not hold true, why did we discuss this? What was our basic motivation? (Refer Slide Time: 24:36) You want to recall that when we were first discussing, so, recall the case of the classical elastic solid, classical and this is this goes back to the very beginning of our lectures right, classical elastic solids. We had said that the stress-strain relationship is given by this very-very straight forward relationship and this implies that when you have a strain, which is a time varying function, then your strain value is = . So, for a pure or classical elastic solid, when you apply a sinusoidal strain history, the output is a stress, . But what happened when we did this for a classical viscous fluid, let us just quickly take a look classical viscous fluid. So, we had said that the stress-strain relationship in that case was given by a viscosity here, and this, this is what we are using. So, this implies that when you put in the same condition here you what you will get is a function here, which is some constant, we will again call it . We will take the derivative of epsilon t you will get a cosine, which is the same as writing it as some So, in the classical elastic solid, a sinusoidal input in terms of strain results in a sinusoidal output, where there is no phase difference between the input and output. In the case of the classical viscous fluid, a sinusoidal input in terms of strain results in a stress where the stress and the strain again are sine functions, but there is of 2 phase difference or 90 degree phase difference, 2 in radians, okay. But, when we are, when we did this calculation for a viscoelastic body, we got us, we came to a situation where the stress now is the stress is some constant some amplitude multiplied by another sine function, but this time there is a phase difference which is not going to be 2 . This phase difference obviously depends on the, this particular ratio of G’’ and G’, but this is no longer 2 . So, you end up with a different phase lag between the input and the output. So, this is a very-very important result. And this again goes back, we had discussed point that viscoelastic fluid, the phase there will be a phase difference. And now, we can see that mathematically. At that time, we had just I had asked you to sort of believe me that if you have a sinusoidal input, then you will have an output which is phase lag, but now we can see from the mathematics how the phase lag behaves. And we have mathematical formulas governing the phase difference. So, we will stop here for this class, and in the next class, we will take up some of the other issues related to sinusoidal strain histories. So, we will stop here for today
Invieremo le istruzione per resettare la password al tuo indirizzo mail associato. Inserisci il tuo indirizzo mail corrente