Welcome back. We are now at a, very close to an end game stage of our formulation for 3D elasticity. What we are going to try to do in this segment, and perhaps it will spill over into one more segment, is to assemble a global matrix factor equations. And talk about Dirichlet boundary conditions, and the final solution step, okay? So let's get started on that program. What we will do now, in this segment is, write out the global matrix vector equations. And in order to do this, what I'm going to do is, just write in the first line here the actually, the, the, the final matrix vector weak form, but retaining explicitly the sum over elements and then we work ahead from there, okay. So this is an equation we developed toward the end of the last segment. It's, sum over e, c e transpose, k e d e, right? And remember that your k e is our, elements difference matrix, right? This is equal to sum over e. Ce transpose fe internal plus sum over i, right? Over the spatial dimensions. Sum over the elements that have some part of their element boundary as coinciding with the alignment boundary of the problem, right? And then for such elements we have the sum over the nodes that actually over all the nodes, right? because we worked out the bit about nodes belonging only to the Dirichlet boundary over such elements, right? We got past that point. So we have the c is our vector of degrees of freedom for the weighting function. And we have this forcing vector, which we've been denoting as f t bar node a, spatial dimension i element e. All right from here we go on to the business of assembly. Okay? And in order to see how that works out, we are guided by our global definition of the c and d vectors, right. So let's start out with the global c vector, right. That is c and the way it is defined is the following. We have we start out from the, we start out by following the global nodes. Right, so we have for global node 1, we have degree of freedom 1. Right? Which would be in our case the it would correspond to the spatial dimension one. And then we would have the same node. Degree of freedom two corresponding to spatial dimension two, and spatial dimension three. Right? This would carry on, and for the general node A, we would have the same situation. Right, until it came down to the very last of our c degrees of freedom. Alright, you note that I'm not, locking us into a situation where the very last degree of freedom, right? Or the very last node which would be sitting here. Those degrees of freedom would be sitting there. Degrees of freedom would be sitting there. I'm not walking us into a situation where that has to be the last node in the problem, or the last numbered node. Because of course we know that the definition of Dirichlet boundary conditions may very well eliminate such nodes from having weighted functions being interpolated off. All right? So, I'm leaving open that flexibility here. So this is the global c vector, all right, and likewise the global d vector, all right. And I'm going to now call this a d bar vector, all right. And you may recall from our previous treatments of 1D problems as well as the 3D problems. But for scalar variables that we are doing, we are calling this d bar. Because we know that there are some of those degrees of freedom that we want to later on move over in order to impose Dirichlet conditions. Okay? So okay. So the same thing happens here. We have d one, one, d one, two, d one, three. All right. Carries on to da1 da2, da3, right? That's for some general note, right. dA1, dA2, dA3 simply represent the displacement, degrees of freedom In the respective directions. One, two, three. Along with respective coordinate directions, one, two, three. For node A. And then this carries through all of them. D down to number of nodes in the problem. One, two Okay? In general, the D bar vector will be. Will be what? Bigger or smaller than the c vector? Will d bar have more components, or fewer than c, or all the same? What do you think? All right. If there are any Dirichlet conditions at all, and there have to be Dirichlet conditions for this problem, the D bar vector will have more components than the C vector. I should also mention that in setting this up, I'm assuming that the very first node here does not have Dirichlet condition set up on it, okay? So let me just say that. No Dirichlet boundary condition on the very first node. But yes, there could be Dirichlet conditions on the very last node. So I haven't specified which node we're talking of at the, as the last component of the c vector. Okay, so we have these global, so these are our global. C and d bar vectors. Okay? And then once we have this, the, the degrees of freedom that we have are, are now to be viewed as simply those corresponding to these entries, right? Each of these, for the whole problem, right, viewed as a vector problem, each of these is a different degree of freedom, right? Never mind the fact that they come from the same node, right? As far the problem's concerned, they're different degrees of freedom, right? Likewise these, right? And of course for, for, for general nodes as well, okay? All right, so in this sense, one, one would say that d bar has number of nodes in the problem, times nsd degrees of freedom, right? All components, right? c has number of nodes in the problem times nsd minus ND where this now is the number of degrees of freedom with Dirichlet conditions on them. Okay, and in calculating ND, it is not necessary. Well let me ask you, do you think it, it is necessary that ND, right, ND, is it necessary that ND should be a multiple of the spatial dimension? Right, I'm asking, is, da, does it have to be a multiple of a number of spatial dimensions? And in, in particular I'm, the reason I'm asking this question is bec, is because I want you to think about whether Dirichlet boundary conditions have to be applied to all three degrees of freedom at each node. 'Kay, so the answer to this question is that is no, right? Because we, we've, we've, we know that we could apply Dirichlet conditions on a particular coordinate direction at a point and not on the others, okay? So in ge, so, so this answer in general is no, okay? Right, all right. Okay, so we have these things in hand, and now what we will do is to go ahead and essentially write out the global form, right, from that contribution. So what we have here is that sum over e ce transpose, Ke, de equals c transpose, K bar, d bar, right? Okay? Now we already know what our c vector and our d bar vector are. The K bar itself, is obtained by this assembly operation over the individual element's stiffness matrices. Right, and, and note, of course, that here, because we have confirmed indeed that we are doing 3D elasticity, the term stiffness matrix is is, is relevant, right? It's, it's, there, there, there, there's no confusion there. Okay now this proceed, this proceeds just as before, right? What we want to, to realize is that any single entry in the K bar matrix, which corresponds to degrees of freedom belonging to different elements, right, but the same global degree of freedom will have the corresponding terms added on, okay? So this thing works just as before when we realize that we just carry out the assembly over degrees of freedom over global degrees of freedom. 'Kay? You just have to carry out assembly of a globally numbered degrees of freedom and forming K bar, okay? And when we do this, we get that K bar as units, sorry, has, has dimensions of number of nodes in the problem times nsd minus ND, okay, times number of nodes in the problem times nsd, okay? In order to sort of further explain this process, let's try to do it for a pair of elements. Let's me try, let me try and draw something that's doesn't have very complicated surfaces. 'Kay I suppose that this thing goes a little longer. Okay, this is sa, somewhat similar to the si, sort of situation we'd looked at in the case of the 3D scalar problem. Right? So, let me label only, or let me draw only the common nodes here. Okay. And let us suppose that these nodes have global, global numbering. A, B, C, D. Okay, and furthermore these elements are omega, E1, and omega E2. Okay, all right. So, let's, look at what the, the Ke1 stiffness matrix may look like, okay? Now, let's suppose that for the Ke1 stiffness matrix we have a numbering, which comes from the, from the, from the local ordering of nodes, right? We, well, we know that's always the case, but let me just label the the local number in with nodes, right? For the local numbering of nodes in element omega e1 we know that what I've labeled here is global node c, supposing we say that, that, that is local node 2. Right, and element e1, right? According to that, D would be local node 3, right? A would be local node 6 and B would be local node 7. Okay? Now, for element omega e2 exactly those nodes. Suppose that they are local node 1. D is local node 4. Right? A here would be global node, A would be local node 5 for element e2. And B would be local node 8 for element e2. All right? So, with this background, all right? Let me try and write our Ke1. Okay? Ke1 would be a matrix where I'm not going to write out all of the components, because I'm just going to write the blocks, okay? So we know that this going to be Ke1, the 11 block up to the, up to Ke1, the 18 block, okay. Because, of course, we're working with bilineals here to fix ideas. So we have Ke1 88, all right? And down here we would have Ke1 81, okay? And Ke2. Is Ke2, 11. Ke2, 18. To Ke2, 88. Ke2, 81. All right.
