Welcome back. At the end of the last segment, we had essentially carried out an assembly of our matrix vector equations for the finite element weak form. Let's pick up there and see and just make some more observations about the form that we've arrived at for quadratic basis functions. Right. So where we are is the following, right. So we've got the global matrix vector equations. All right? And the form that we had was the following. c2, c3, all the way up to c2nel plus 1. Multiplying 2EA over h times a matrix which is minus 4 over 3, 8 over 3, minus 4 over 3, one sixth, minus 4 over 3, 7 over 3, minus 4 over 3, 1 over 6 minus four thirds, eight thirds minus four thirds, 1 over 6 minus 4 over 3. We'd get, again, 7 over 3 here, and then the process would go on. Right. And at the very bottom of this matrix, the bottom right of this matrix, say somewhere down here, we'd get 7 over 3 minus 4 over 3, 1 over 6, minus 4 over 3, 8 over 3, minus 4 over 3, 1 over 6, minus 4 over 3, 7 over 6. And because this matrix has grown a little bigger than I had anticipated, let me make these bounding brackets bigger. All right. The d vector that we get here is d1, d2, d3, all the way down to d2nel plus 1. Right. Now, this is equal to a vector which here is c2, c3, up to c2nel plus 1. 2, sorry. Right now, here we get the two contributions to the forcing, fAh over 2. Now, the way this adds up is the following. We get 4 thirds, 2 thirds, 4 over 3, and so on until for the, in the very last element, the contributions are 2 third, 4 over 3, and 1 over 3. Right? And here we get a vector which is the contribution from the traction. It's filled with zeros all the way down to the very last component, which is t times A. All right? I'd like to point out a few things which is observe that here I've explicitly included the addition of components coming from the same global degree of freedom, but on different elements, right? Likewise over here. I've also done the same thing with the vector rising from the forcing function, right? Here and there. Okay? Also, at the end of the last segment, I made the point that the dimension of this vector is 2 nel, whereas this vector is 2 nel plus 1. Consequently, the matrix that we have here is 2 nel times 2 nel plus 1. It is not a square matrix, okay. This thing being 2 nel, so are these vectors, right, the vectors that are involved here are also 2 nel. Okay? So, at this point, we are essentially, we have everything in place. The only thing we have to do yet is account for the Dirichlet boundary conditions. And the way that shows up is, when we look at our D vector, right, our vector of degrees of freedom for the trial solution, we note that D1 here is our given displacement at x equals 0. I think at one point, I refer to it as u g, but we're actually calling it u0, right? This is the known Dirichlet boundary condition. All right. BC is short for boundary condition. Since it's known, what we can afford to move it to the right hand side, right, and really have it derive the problem. And the way we go about doing that is to observe that that degree of freedom, d1, which is equal to u0, is the one that multiplies out this column of our matrix. Okay? And also note that because of the sparseness of this global matrix that's formed, we have zeros here, okay? Zeros all the way down. Okay? So, we can considerably simplify our, problem here, by essentially moving this column that I've just marked out in the big matrix to the right-hand side. All right? And when we do that, we get c2, c3, up to c2nel plus 1 multiplying 2EA over h. Big matrix here. I'm going to write it yet again. It's 8 thirds. Sorry. 7 over 3 minus 4 over 3, 1 over 6, minus 4 over 3, 8 over 3, minus 4 over 3. And here I get 1 over 6, minus 4 over 3, 7 over 3. Continuing on, I'll get terms of this type. Okay. I'm not going to write the rest of this matrix below here, but essentially indicate that it continues until we end up here at 7 over 3 minus 4 over 3, 1 over 6 minus 4 over 3, 8 over 3 minus 4 over 3, 1 over 6 minus 4 over 3, 7 over 6. Okay? This is the big matrix that we have, right? Now, because we've gotten this by moving our column over we're going to note that this is a square matrix. We'll come back to that in just a little bit, okay? The, the degree of freedom vector for d is now starting with d2, d2, d3, going all the way down, up to 2dnel plus 1. Okay? This is equal to, on the right hand side, again, we have our c vector, c2, c3, c 2nel plus 1. Okay? Multiplying. Now, fAh over 2. Here we have 4 over 3, 2 over 3, 4 over 3, going all the way down until the very last entry here is 1 over 3. We have the contribution from traction, which is full of zeros, except for the last component, which is t times A. Sorry, sort of messed that up. t times A. Okay, now, that column that we moved over from the left-hand side shows up here. It is minus 2EA over h times d1, which we know to be u0. All right? That's our specified Dirichlet value before the displacement, right? And then the column that, that we obtained from the left-hand side is this, minus 4 thirds, 1 sixth, and zeros. Okay. Let's look at the size of our vectors. This remains a 2 nel. This has now become 2 nel from 2 nel plus 1 on the previous slide. That's because we've lost the first component by moving it over to the right-hand side. Consequently, we now have a matrix here, which is 2nel times 2 nel, it's a square matrix. Okay? Here again, we have 2 nel, it's just the c vector, and all the vectors here are all 2 nel. Okay? All right. So, what we're going to do is when we go to the next slide, we're going to call this the global c vector, this will be the global d vector, so I'm going to call this c transpose. This will be denoted as d, okay? This is, of course, c transpose again. And the sum of the three vectors that I have in the parentheses on the right-hand side is going to be our f vector. Okay? Our matrix here, of course, is going to be our k matrix. Right, our stiffness matrix. Okay? All right. We have all of this, let's write it out. Okay, we have this, and I'd like to make a few remarks about it first, all right. And in order to make those remarks, I am actually going to go back to this slide and annotate this further. And because I've already annotated it a bit, I'm going to annotate it now using color. Okay, the first thing I want to point out is that if you look at the entries in the stiffness matrix, you will see that they're really quite different from the entries that we had in the case of the stiffness matrix for the linear basis functions, right? They're not just twos and plus or minus one, right, they're a little more complicated. All right? Okay, so let's make that observation first. Remarks. One. K matrix components. Are different from linear case, okay? Remark two is the following. If you observe the, bandwidth of our matrix, of our K matrix, observe that the bandwidth is larger. Okay? The bandwidth is 5, okay? So that is a second remark to make. Okay? The bandwidth of K is 5. Okay, so what we're getting here is that because of the use of quadratic basis functions, right, we are observing that the degrees of freedom have much more interaction between them due to use of quadratic basis functions. All right? The third remark to make, again, moving back here look at the part of the forcing vector on the right-hand side that arises from the force, from the forcing function terms. Okay? And in particular, observe that the midside nodes have a larger contribution than the end nodes, right, of an element. Okay? So, midside nodes have a larger contribution to F vector, okay? Also due to the use of quadratic basis functions. Okay, those three remarks are important to understand and appreciate some of the differences that arise from the use of different basis functions. Okay?