Notes
Study Reminders
Support
Text Version

### Three-Parameter Model Equations

We will email you at these times to remind you to study.
• Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Sunday

Three—Parameter Model Equations
So welcome back to another lecture on introduction to soft matter, we were discussing last time, the constitutive equations and the solutions, right. So, we left off at a certain place, where we were looking at the Creep response for the material and its solution. The Creep response for a very, we framed the entire question in the form of a very generic ordinary differential equation and we replaced the constants from the model with just simple coefficient ?0, ?1, ?0, ?1and why that we did that, it will be clearer to you as we go through the different solutions, right. For the time being that is what we have done we have ordinary differential equation where we have these coefficients.(Refer Slide Time: 1:28)And then we have tried to solve for different cases and we had solved for the Creep response function and you had seen that we get a very nice equation here for the Creep in terms ?(?)=?∞+(?0−?∞)?−?/?0 was a ?0 and the ?0 seems very familiar to us, the expression for this particular characteristic timescale associated with this.(Refer Slide Time: 1:51)So, where we left off is, we said that we will look at the Creep response and that has a very similar formulation, you have the stress and the stress is now in the form of a Heaviside function. When you take the Laplace of that you end up getting ?(?)=???. And we have seen that this is the governing equation that we are trying to solve. So, if you take the Laplace of this on both sides, then you end up with this particular expression ?(?)(?0+?1?)=?(?)(?0+?1?). And so, what we want to do now is we want to replace the value of ?(?) with what we have got.(Refer Slide Time: 2:32)So, that gives us, so, let us carry up from where we left off. So, we are trying to find out the expression for the Laplace of the strain function. So, now we have, we can write this as ?(?)=???⋅?0+?1??0+?1?. So, we this is something that is familiar to us now, this is in the form of a rational fraction, we have to convert it into simpler factional forms, so that we can take the Laplace inverse.And to do that, what we are going to do is I am going to convert the two the numerator and denominator in a form that is in a s plus constant type of form. So, if I have to do that, I will convert it to something like this ?(?)=???⋅?+?0?1?+?0?1⋅?1?1.(Refer Slide Time: 3:53)So, now let us say we have ?? and this open a bracket. So, let us take this term s and the let us take the numerator term by term, so you will end up with ?(?)=??(?1?1⋅1?+?0?1+?1?1⋅?0?1⋅1?⋅1?+?0?1), so this is, these two cancel out this is now ?(?)=??(?1?1⋅1?+?0?1+?0?1⋅1?⋅1?+?0?1). Now, what I want to do is I want to expand this particular term into something else.So, this particular term if I can simplify this is a multiplication, but I can probably write this as some 1 by s minus, so I am just looking at this particular top here. So, this can be written as, ?+?0?1, but when you do that there will be an unbalanced term in the numerator, so, I have to get rid of that and to do that, I have to multiply it with ?1?0.(Refer Slide Time: 5:24)So, here I will have ?1 and here I will have ?0. So, when I insert this back into the previous equation, we have plus, when you simplify this, what you will end up with is, so this ?1 will cancel with this one. So, you will have ?0?0(1?−1?+?0?1), okay, the bracket ends somewhere else, this bracket ends maybe I and then this entire thing where entire bracket ends.(Refer Slide Time: 6:16)So, now, my ?(?) is equal to this, so that implies that if I divide so by ?(?)?? is equal to this nice little expression, I will now on this side I what I will do is I will bunch the different terms. So, we have ?(?)??=?0?0⋅1?+1?+?0?1(?1?1−?0?0).So, when you take the Laplace inverse, this ?? here is a constant, so, when you take the Laplace inverse of this, you will end up with this is by the way, ?(?). So, when you take the Laplace inverse, so taking, so taking the Laplace inverse of both sides, you have ?(?)=?0?0+(?1?1−?0?0)?−(?0/?1)?So, we know that the Laplace inverse of 1? is just 1, and here you have this is just a constant. So, this will be retained, by the way Laplace operator is a linear operator, so when you have sums like this, you can always take the inverse of these terms individually and add them up and that is the Laplace inverse of the entire sum and this is ?(?)=?0?0+(?1?1−?0?0)?−(?0/?1)?So, I can, so if I say that this term is some ?∞. So, if I denote ?0?0=?∞then this is again some ?∞ and then I denote this by ?0, then I can write this entire thing in the same form that I had written before, in form of somethat this ?(?)=?∞+(?0−?∞)?−(?0/?1)?. Now, so let us take a look at what these forms look like. So, ?∞=?0?0 and we had earlier introduced a term called ?∞when we are discussing the relaxation function.(Refer Slide Time: 9:10)So, let us quickly take a look at what ?∞ was. So, ?∞was ?0?0, and the ?∞ is ?0?0 .(Refer Slide Time: 9:23)So, ?∞ is the exact opposite. Note now, that ?∞=?0?0and it is exactly the inverse of ?∞. Is it clear? Maybe I will just rewrite this thing.(Refer Slide Time: 9:54)And we have seen before that ?0 was greater than ?∞. So, similarly, so this is the second term here was ?1?1and this was ?0.(Refer Slide Time: 10:09)