With that, we then go on to writing the same sort of a contribution, from the forcing terms. Okay? So the, for the contributions from the forcing terms, we have again, c 2, c 3, up to c N e L, c N e L plus 1. Okay? Now, looking back again at that expression for the finite dimensional weak form, where we've exquisitely, written out the contributions from each element. What we see, is the following. So once again, let's assume that f A is, is f, and A are are, are the same for every element, right? They're uniform over every element, and in fact, they're the same for every element. That allows us to pull f A over 2 out, as a common factor. Leaving us then, with the task of just filling out [INAUDIBLE] this vector, okay? Now, the contribution, from the very first element, okay? If you go back, and look at the, at your notes, would have been h 1. Okay? For element 2, right? We recognize that, element 2 has contributions coming from The c 2, and c 3 degrees of freedom. Okay? All right? And then, if you go back, and look at, at your, at the expression we have, for the forcing function on the right-hand side. You will see, that the contributions here, are of the following form. We have h 1 plus h 2, okay? H 2 coming from, element 2, okay? Element 2 however, also has a contribution corresponding to c 3. So that, goes here as h 2. And then, we go on to element 3. Element 3 has contributions from c 3, and c 4. Which is the degree of freedom, sitting next to c 3, okay? All right. Tthe contribution from c 3, since, we're now talking of element 3, would be h 3, okay? The contribution corresponding to c 4, since, we're still talking of element 3, would be h 3, okay? The process continues, until we reach the last two entries, in this forcing function. Okay? The last, but one entry, will have a contribution from, correspondent to c N e L, but that would come, from the last but one element, okay? And so, that contribution would be h N e L minus 1. Okay? That entry would also have a contribution, from the very last element, h N e L, okay? And the very last entry of the forcing function of, of this vector, that we are trying to develop, would have a contribution only, from the very last element. And that would be corresponding to, that degree of freedom, okay? That could also be h N e L. Okay? This, once again, using this notion of the assembly operation of finite elements, is, the assembly over e equals 1 to N e L, 'kay? Of c 1 e, c 2 e, f A, h e over 2. F A, h e over 2, right? This is the this is the contribution, that we've assembled. And again that, that's just an abstract way, of writing this operation. The detail of doing that operation involves the sorts of the sort of mappings, that we've just used here. To write out these global vectors, from the local ones, and the sum over the elements. All right? Now, there's one more contribution. If you go back, and look at our at our finite element at, at our finite dimensional weak form. You will observe that we have a term that comes, from the traction condition. Okay? It's the one, that we may have actually had on the previous slide as well. Right. On the previous slide, it is this contribution. Right? That comes from the weighting function degree of freedom, c 2 N e L, okay? Right? And we know, how that maps on to the global numbering system, for weighting function degrees of freedom. Right? This contribution then, will appear in, this position. Right? The very last position here. Okay? It will be corresponding to c N e L plus 1. Right? Because c N e L plus 1 is the same as c 2 for element N e L. Okay? So that contribution here, is well we can't quite write it directly into this vector. So, let us then, also write it as, as follows, right? So, we include here, all the c contributions Right? The reason we couldn't write in the vector above, is because we have this constant factor, multiplying that vector. Okay? So instead, we just observe that this thing can be written with a bunch of 0s. All the way, up to the very last contri the very last entry. And for that very last entry, we just have t times A. T being the traction, A being the, the area. And this is just our very simple minded way of writing, c 2 N e L times t times A. Okay? All right. What I'm going to do now, is pull all these terms together, right? And write them up, write them out, in one single expression. C 2, c 3, c N e L, c N e L plus 1, okay? This is multiplying E A. And again for okay. I was going to try, and make some things even easier for us using by assuming all the elements have the same length. But let's, let's not do that, right now. Okay. So, the expressions we have here are minus 1 over h 1. So, shou, so what I am doing here is putting together all the stiffness matrix terms,'kay? Here I have 1 over h 1 plus 1 over h 2. I have minus 1 over h 2 here. Here I have right, I have minus 1 over h 2. Here I have, 1 over h 2 plus 1 over h 3. I have minus 1 over h 3. Minus 1 over h 3. 1 over h 3 plus there is another term, that'll come here from element 4. It will be, 1 over h 4. Okay? That process continues, and when we finally close out this matrix, we will have here. We will have here, 1 over h N e L minus 1, 1 over h N e L. We will have here, minus 1 over h N e L. We will have here minus 1 over h N e L. And here, we'll, we will have 1 over h N e L. I recognize that what I'd written on the previous [INAUDIBLE] when we actually developed this, a couple of slides ago, was a little cramped. Hopefully, this is a little easier to read. Okay? All of these, sorry, this matrix multiplying here, a global degrees of, of freedom. D 1, d 2 all the way down To d N e L, d N e L plus 1. These are all the contributions coming to us, from the left-hand side of our finite dimensional weak form. And they are equal to, terms that came, from the right-hand side, from the forcing function, as well as the traction. All right? And, in order to get those together, let me just go back to this slide, our previous slide. And make sure I save it, so I can work off it, and then, I can come back to where we were. Okay. So, the right-hand side then, involves the same vector that we had up here, okay? The same row vector, c 2, c 3, c N e L, c N e L plus 1, multiplying f A over 2. It's the vector h 1 plus h 2. H 2 plus h 3. H 3 plus, and the dots there will be h 4, but at some point, I do have to just stop writing those, and use dots for extension. And down here, I get h N e L minus 1 plus h N e L. And here, I finally get h N e L. Okay? We have this, and then, we have the final contribution coming from the traction. Again, V f, c 2, c 3, c N e L, c N e L plus 1. Multiplying the vector with 0s, everywhere, all the way down to t A. All right. We are done, with finite element assembly. But we are not done with the, with our final sort of tweaking of this of the equations. There is more work we need to do here. But the big task of assembly is done. We'll stop the segment here. When we return, we will come back to these matrices to, to this general matrix factor expression. Make some observations about it, and turn it round to the final form, in which we will actually solve, the problem.