Welcome back. We are aiming here, to complete our assembly of the global matrix vector equations. And, talk about the final Dirichlet boundary conditions. So, to do that let's get on to the contribution that we have not yet tackled for the global equations. And this is the contribution from the Neumann or the traction boundary term, okay? So, the contribution. To global matrix vector equations. From the traction, right. Remember the traction is simply our Neumann boundary condition for this problem. Right. So the term we are talking about is this one. Sum over i, right, i running over the spatial dimensions. Right. Sum over i, sum e belongs to E Neumann, right. Sum over A, that's all the nodes in that particular element, right. We have here CAie Ft bar. We have the A index there. We have the i index there and we have the e index here. Okay, what I'm going to tell you in one fell swoop is that this is going to show up as c transpose Ft bar, right, globally. And I will write down on the next line the detail construction of the F matrix. And in order to do that, I am going to take I am going to write here first the C transpose vector a, as a row vector. Okay? And I am going to get myself room for it here. Okay, and the idea is that multiplying it is our FD bar vector. Okay. And here too, let me get myself enough room. Okay. So let's construct C. Sticking with the same assumptions I've made before we have a contribution from the very first numbered global node, okay. Because I'm assuming that our, in considering the case, where we don't have the Dirichlet boundary conditions on any degree of freedom on the first node. So, we have C1. Okay. And we're working with the outer sum first, right? The sum over spatial dimensions. So I'm setting i equals 1, and that is the 1 that shows up here. Okay, and that comes from some element, right? The fact that we're doing a sum over elements is already accounted for in the fact that we have this global node here, okay. Right so that contribution would be. F t bar. Let me see. It would be local node number one for that element. Okay? And since we are talking of global coordinate direction one. We would have a one here. Okay? It's going to be some element, let's call it, let's just leave it as E. Okay? And let's go on now, with this. And actually in going on, let me also use the same numbering that we used in order to construct the stiffness matrix and the F internal force vector, okay? So I have before me here those two elements, omega e1 and omega e2 and you probably noted them down in your in your book, or your notebook. So I would encourage you to go back and look at those. Okay? Because what I'm going to do now is look at the contributions to global node A. Okay? So let's suppose that global node A shows up here. Wrong C vector. Let's suppose that B shows, shows up here. Global node C shows up there, and global node D shows up there, okay? So that's, A is going to be here, B, C, D, okay? Now we start out as as I did for, for this very first degree of freedom, we start out with the spatial dimension one, okay? So the global node here, right, is cA1, okay. And let's suppose that on this force vector A shows up here. B shows up there, C and D. Okay? Now the only we would have a contribution from the A node is what? Okay. What we need to have is that the local well what we need to have is actually we, we can talk of it, of it in terms of the global nodes and so global node A, right, lies in partial omega t bar one. Right? That's the only way we would have a non zero contribution to the F vector from global node A. Okay? So let's assume that this is true. Let's consider the case where A does lie in omega, in partial omega t bar one. Okay? Alright, so you could go back and look at that at that figure that I drew back there two two or three slides ago. What I am seeing is that node A from that figure does indeed lie on, lying on, partially on [INAUDIBLE], partially on omega T bar one. Okay. So, we would have a contribution, then, from element e 1. And the contribution from element e 1 would be F t bar local node 6, right, spatial dimension 1, element e 1. Okay? Right, now element e 2 also would have a contribution here, right? So we would have F, t bar, right? From element e 2, the contribution would come from local node 5, okay? The spatial dimension would still be 1, right? And of course, there's element e 2, okay? So this would be the situation if I'll reproduce that figure of two elements, somewhat more defined. All right, so again we have omega e 1, omega e 2, right? And I'd lo, I'd label those nodes as A, B, C and D, okay? What I'm seeing now is that let us suppose that D surfaces. Right, that, those are the sort of front surfaces of both the elements, right? Belong to belong to partial omega D bar. Sorry. Actually let me, let, let me write this properly. Okay, bo, both those surfaces I'm saying belong to partial omega t bar, okay, 1. All right. Okay, right. And, and then in, in terms of local numbering, what I'm seeing is that let me identify the surfaces I'm speaking of here. Okay, for element omega e 1, that's, nodes are 1, 2, 3, 4, 5, 6, 7, 8. And for. Omega e 2, the nodes are 1, 2, 3, 4, 5, 6, 7, 8. All right? Okay, so we would have those two contributions, right? Let's suppose then for omega, for, for the Neumann boundary in the 2 direction, right, which means the degrees of freedom are being controlled in the 2 direction. Sorry, not the degrees of freedom, but, but the traction components in the 2 direction are being controlled on the other surfaces, right? So let's suppose that it, it's on this surface and it's on this surface, okay? So let's supposed that this belongs to partial omega t bar 2. And this also belongs to partial omega t bar 2, okay? So the, can you think of which surface, do you see which surface I'm saying belongs to partial omega t bar 2 from element e 1? It is the 5, 6, 7, 8 surface, right, of element e 1. On element e 2 also it's a 5, 6, 7, 8 surface. Likewise, on element e 1 the 1, 2, 6, 5 surface belongs to partial omega t bar 1. And in element e 2, which one is it? It's the 1, 2, 6, 5 surface, okay? All right, so if you underst, if we are clear about that, let me put in the contributions then. All right, so for for the contributions from the 2 direction to the, to this traction force vector, would there be anything from node A, from global node A? There would be, right? So we, we would have a C A 2, okay? And that would show up now right here, right? It would be F t bar global spatial dimension 2, right. We would get a contribution from element e 1, right, from its node number 6, okay. Right, and we would get a contribution F t bar global 2 direction element e 2 local node number 5. Okay? All right. Which other nodes would contribute? So let's look at node B, okay? What contributions would it have? It would only have contributions from the 2 spatial dimension, right? Spatial dimension i equals 2. All right? So, for this node we would get a contribution from C B. Let's put the C B 1 contribution, then let's talk about the C B 2 contribution. Since the C B 1 contribution, right, on that node is 0, right? There is no contribution to the traction there, right? We are not controlling the traction there, okay? But that's a free surface, right, because that, that is a surface on which a traction may be specified, right? We have just not specified the 2 component there. So when it comes to the B node, we get a 0, okay? The contributions that do come there are from the 2 direction. And we have then F t bar, we are talking the 2 dimension, so we have 2 here. We are, okay, let's look now at the contribution from e 1. From element e 1, which local degree of freedom contributes there? It's the 7, okay? From F. Sorry from element T2. Sorry element E2. The contribution along the two direction, right, would be from local node eight. Okay? All right. Now, let's suppose that no other boundaries, of these two elements correspond to Neumann boundaries for the global problem. Okay. That means that if we go ahead and look at ca 3, aas well as cb in 3 dimension, right? We would get 0s, right? So here we would get a 0, and here too we would get a 0. Okay? All right? Let's see what node c does as I have drawn it, what traction surface does global node c lie on? Right? Also in the global spacial dimension one surface. Right? So when we come back here we have C and unfortunately the global [LAUGH] number of that degree of freedom is also c but hopefully we can cope with that, with that repetition. Okay. There is going to be something from the cc 1 direction right, from the one direction there. So we move along, right? And then we come to the c node. We get a contribution of the form F D bar, along the one spacial dimension, from element e1, right? So, which node from element e 1 contributes? It's the local node two. From element e 2. Which one is it? Local node one. Okay? Now, as I've drawn it, for node c, right? Global node c. There are, no other, spacial dimension, no other, traction boundaries that contain that node. Okay? All right? So what that says here is that for C c 2 and C c 3, I would get 0 and 0, okay? So let's just complete this particular line that I wrote. Global node a lies in omega t, omega t bar one. C also lies in partial omega t bar one. B Lies in partial omega t bar two. Okay. But to note, the way I've drawn it, A lies in this and in partial omega t bar 2. Okay? There's a partition there. Okay? So A the way I've drawn it node A is the only one that lies on two traction surfaces. C lies on one, and B lies on one. Okay. And which one the line has been written, has, has been denoted here. So, and, and for the more, for the way I've drawn it here, let's suppose that D belongs to no traction boundary. D lies in none of the boundary subsets. Partial omega t bar, i where i equals one, two, three. Given that, what contributions would we find in this traction force vector for no, global node D? You're right, all zeroes. So you get a zero, zero, zero, and we would go on. So, hopefully this process has given us some idea of how to construct this global traction force vector. Okay? This is what we are calling it. FT Bar. Okay? [SOUND]