All right. So so the idea here is use matrix vector product- To eliminate- The sum, sums over A. And sorry. Sums over A and sums over B, okay? Now, allow me just one second to go back here, and do some bookkeeping, okay? All right. This is where we are. Okay, so let's look at the first of those integrals and work with it. So, what we want to do is write sum over A, B. And here let me write the limits, okay? Because it's useful to remember what the limits are. And with the limits on these sums are, are A and B going from 1 to number of nodes in the element, which for this use of linear basis functions. Gives us number of nodes in the element equals 2 okay, all right? So, we have this cAe integral over omega cNA, 2EA over he, and B, C DC, dBe right? This is the one we want to work with. Now, recognizing that number of nodes in the element is equal to 2, right? Because, we are working with linear based functions and therefore, we do indeed have two of these basis functions in each general element, right? To, to represent both the reading function and the trail solution, right? So, what this lets us do is the following. This can be written now as the following. We recall that since A and B run over 1 and 2, we can easily, we can collect the degrees of freedom, see degrees of freedom into a row vector using the notation of linear algebra, okay? And that row vector is c1e, c2e, Okay? Now, what I will also recognize here is something I could have done in an earlier step. Is that each E of course is a constant, because it gives us the length of the element. In many cases the modulus and the area may not be constant within an element, okay? But for the simplest case, of course, they are constant, and let me pull them out of the integral. By seeing that, let us now focus on the simplest case possible where those are constant, okay? So if they are constant, we get 2EA over he, okay? Integral over omega c, okay? Now integral over omega c. We're, we're going to write it out without writing, without explicitly we sorry. Without using the summation, okay? Instead what we will do, is recognizing that A and B run over 1 and 2, for the use of linear basis functions. We get a matrix with entries N1,c and N1,c and N1,c and N2,c. It's symmetric, but let me write it out in detail here N2,c, N1,c N2,c N2,c okay? That's the matrix integrated with respect to c, right? Now, I use this parenthesis to remind us that this is where the, the absolutely clear actually that that's where the integration ends. Now, just as we wrote c in a vector form, right? Recognizing that the index, here, runs over 1 and 2. We can do the same thing with dBe, right? Recognizing that B also runs over 1 and 2. So that gives us a column vector in the notation of linear algebra, which we will write as d1e, d2e. All right, and I should make a remark here or I should make a note That this holds if E and A are uniform. Over the element omega e. Okay, so for this specific element E that we are writing out this integral, if for it holds that modulus and the area are uniform, we can pull the mod of integration sign which is what we've done here, right? We can pull them out of the integral, okay? He, of course, is just the length of the element and so that, that's constant, all right? In the, in the general case the E and A would still be within that integral and, right? And we would've written functions of of, of c through x, right? But let's look at this simple case, just, just for the purpose of, of gravity and getting to something that we can actually integrate, okay? All right? So, so, this is where we have things, and now we recall what happens for the, for the, for the special case of linear basis functions, okay? Now recall that N1 of c equals 1 minus c divided by 2, and 2 of c equals 1 plus c divided by 2. And that gives us then, quite easily that N1 comma c equals minus half, and N2 comma c equals plus half, okay? So, we can use them inside there. And so, what you are see is that in this case the integration is really trivial. Because the integrant is a constant, right? And this comes about from the fact that we have linear basis functions. This will not be the case when, when we extend this to other basis functions but it's a good way to, to, to, to start understanding how the problem works, okay? So what that says that if we look at what we get from that integral that we started out with it reduces to c1e, c2e 2EA over he. Now, again note that that matrix is, that matrix is fixed, right? It's uniform it's constant over the element. So we base it, we just get here one fourth from the first entry which is this entry, right? Recognizing that it is a product of minus half and minus half we get minus 1 over 4 here minus 1 over 4, and 1 over 4 again. This 1 over 4 comes from the product of one half, and one half, which is from that contribution to the matrix. And so then we just have the integral over on the omega c, all right? And over omega c, dc, right? All of this times our. Column vector of degrees of freedom corresponding to the trial solution all right, okay? Now, we, 'kay, so this is where we are, and what we need to do is evaluate that integral, and that integral, we realize that omega c, in this case, is A, recalling that it's a bi-unit domain. So, I'm going to try and make that look like a straight line okay, that's a little better, straight second, okay? Remember that omega c is this domain. Right, c space, that being the .0, okay? So this integral here is integral minus 1, 21 dc which is just 2, okay? Easy enough. So, putting all of this together, what we get is continuing from the previous slide, that integral that we were working with finally turns out to be c1e, c2e if you look back at your notes. You will see that the, the twos in the numerator, and the one over four in the denominator in the matrix cancel out. And we get EA over he. Matrix is a really simple one. It's 1 minus 1 minus 1, 1. And here we have d1e, d2e. Okay? And recall that this integral came, this integral emerged from us considering, this term, this particular integral in the finite dimensional weak form. Okay? Reduce this to this nice, simple, matrix factor form. And now you can see, why we refer to this as the matrix vector, all right? Just to complete this process let's also work out for ourselves how things pan out for the previous for, for, for the other integral, all right? And that other integral is well I shouldn't say likewise, but sort of similarly. Maybe even similarly isn't right? Let's say in a related manner- Okay? The other interval that we had was the following integral over omega c we had, let me see. What form did we have it in red? We had integral over omega c. We had here a we had here a sum over a cAe. Here we had NA, and fA he over 2, dc, all right? Again, recognizing that The degrees of freedom corresponding to the weighting function can be put into a row vector. And assuming as well again, just for the purposes of writing out something that's easy to actually compute. Let's assume that these two are uniform. Over the element of interest, okay? What that lets us do, okay? So let's see, if these two are in uniform over the element of interest, we can write this also using this matrix vector form. And in this case, we actually just have a simple sort of .product of vectors effectively. So we have this representation for our degrees of freedom and we've agreed that we are going to pull f and A out of the integral, assuming that they are uniform over the element of interest. Again, recognize that the, that they are not required to be uniform. Right, if they were not uniform. If they varied with position over this element, they would stay in the integration, okay? However, we are looking for something that's easier to compute in the simplest of entry level cases, okay? So we have this and now we have integral over omega c. But then omega c, we already know has the limits minus 1 and 1, okay? And then we have NA back here. But using this vector representation we write corresponding to N1 and N2. We have N1 here and we have N2 here, okay? That multi, that dc, right? So, that's the integral we, that we need to compute. And now this is easy enough, because we know that N1 is 1 minus c over 2 and N2 is 1 plus c over 2, okay? So those are the very simple functions, that we need to integrate over the limits minus 1 and 1. Also observe that in here c is an odd, c leads to an odd integral, right? Over the limits minus 1 and 1. So this is actually, this also works out to a nice and simple form, which is we have fA he, divided by 2 here, and that integral from minus 1 to 1 picks up only the constant term, okay? The linear term being odd vanishes, as you probably know from evaluating these kind of simple integrals, all right? So, that's e, effectively an integral over half integral of half over minus 1 and 1. All right, so we get, in both cases we get one. In these two cases we get 1 and 1, okay? And this you recall came from our original integral in the weak form, which had this form. Okay? So we will write that this is our final form for the force, and function too. All right this is a good place to end this segment. When we come back, we will talk about what these matrix vector forms give us further.
