Welcome back. We have got as far as deriving the finite dimensional weak form for 3D linearized elasticity. What we're going to do in this segment is essentially press ahead with the formulation. And of course you know how this proceeds. We essentially have to define basis functions, construct, representations for the fields that we care about, and go ahead and compute the integrals that arise in the weak form. Okay, so I'm going to call this segment the finite dimension weak form and basis functions. Okay, and remember just the integral equation we are working with is the following. I am not going to put down the whole finite-dimensional weak form in all its glory, but just remind ourselves what we are trying to accomplish here. We are trying to solve the following equation. Integral over omega Whi, j sigma h ij dV equals integral over omega. Whi fi dV plus this curious sum of integrals which has this form. Sorry a w bar is not required. Whi, t bar i, dS. And I'm not going to state it now but just recall that on the last interbrand we don't have a sum implied. Okay we have this and of course there are constitutive relations and we've studied those and summed and summed it. As you imagine now, we, we essentially have to define basis functions, right? And, and in defining basis functions as before, we will do this by constructing a partition of our domain, right? We partition our domain as follows. Basis our, domain of interest which is really the body that we want to solve and the linearized elasticity problem over that's our domain Omega. Actually, let me leave that outside of the body because we will confuse it with other things, that's Omega, we have the usual decomposition of the boundary subsets. And right, the, the way we do the decomposition is essentially the way we did it in the case of the 3D elliptic problem with scalar variables. All right. ,the kinds of elements we considered there are admissible here as well. Okay? And to fix ideas, let's look at hexahedron. Okay, so that is an element omegas, omega e and, that part, the partition of the domain is as always given to us as follows. Union over e, of all these open element sub domains, closed gives us the closure of the body, okay. As always, we have omega e1 intersection omega e2 is the empty set, right there at this joint. Okay, we have this and we're going to go ahead and construct our basis functions using this sort of decomposition. Equivalently now that we've talked about it, we could also have had a tetrahedra, ED composition into tetrahedra. Okay. Alright. The way we are going to go ahead and construct our, basis functions is as we know, very well is to construe, to consider nodes on the element. And again, just a fixed idea as I'm drawing things out here, I'm going to consider bilinear, sorry, trilinear hexahedra. So those are our knowns. We only have knowns at the vertices. Ok and we know that Okay, we know that digit. Right. So now let me suppose that this is element omega e, right? This is the element that we started out with which is now being essentially expand out here, okay, that's element omega e, and that is our node A equals 1 for the element, right? I'm following here, the local numbering of nodes. Okay, that's two, three, four, and so on. We are quite expert at this by now. All right and we're going to construct our basis functions from these from these nodes. The way we do this is to, to essentially define these basis functions on the nodes and we are going to define, write the basis functions as before as NA. Okay, our basis functions. This is our basis function at node A, okay? Let me say here, local node A. Now, something you may have noticed is over the past minute or two, as I've introduced, the nodes and the numbering. I have not been calling them I've not yet started calling them degrees of freedom. Okay, whereas earlier, I was using the term, I was using A more often for, in the context of degrees of freedom. But here, I'm being more careful to call them nodes and, and, and here is why. Okay? The reason is the following. We're going to construct a, representation for our, trial solution, all right? And remember our trial solution here is the, displacement field. That's component i over element e. All right? Now, we are going to use a representation where the basis functions will be exactly the same basis functions that we are so familiar with, right? If we are doing trilinears, we know what these are, right? And they're the same basis functions. However, we, and the sum of course is over A going from one to number of nodes in the element. Okay? However each one of these basis functions is multiplied by a degree of freedom vector. Okay? What this means is that for component U hi I have dA, ie. Okay, and we need to recall, here are the i, runs over one, to number of spatial dimensions. Okay. So this is a little different because now if you think about it, how many degrees of freedom does this element have? Okay? So the number of degrees of freedom In omega e, right, or on omega e, right, is. What is it, is it just the number of nodes as it was in the previous problems we were doing. No, correct. So it is actually, in the general case, it's number of nodes in the element times number of spacial dimensions, okay? Whereas, earlier the number of nodes on the element was, same as the number of degrees of freedom in the element. Things are a little different now. Okay, so what we have here is, another way to write this is the following, we use direct notation, right? So now that is the displacement field over element e and this is sum over A NA dA e, as before, except that that d A e, whereas it was a scalar for our problems of scalar variables is now a vector, right, for our problem of vector variables. So and, and what we have here is u h e, and d A e both belong to R3, right, the three dimensional vectors, okay? So sorry for seeming to flog a dead horse, but what we have here are vector degrees of freedom at each node. Okay, and this is a particular approach that we take for this problem, it is not necessarily universal, okay. So we have here a representation for a vector field and what we're seeing is that the basis function is here scalar, just as we've been doing all along. Degrees of freedom here carry the vector information. All right, and this is common for the kinds problems we're doing. Definitely for, for elasticity of, of, of any kind, of mechanics of any kind. It is not the same, for instance however in the case of the problem of electromagnetics, 'kay? It's common in that case to construct finite element formulations where the basis functions are vectors and the degrees of freedom are scalars. Okay? It's, it's just a difference. It's, it's a, a difference into, to do with the requirements of each problem and so on, the mathematical requirements. All right, so, so that's something to know too. And, and then of course it's the same thing for the, for the waiting function. We have wh ie equals sum over A NA CA, ie. Using coordinate notation. Right. Where we know that i Equals 1 to number of spacial dimensions. Right, and using, direct notation that would be wh e equals sum over A NA cA e, okay? As we noted above, Whe and CAe are 3D vectors, vectors in R3. Okay. The, the next thing we need to do if we are happy with this is to just remind ourselves of what the basis functions are. And these are constructed in exactly the same way that we know. Alright, so what we are seeing here is that as before if we have an arbitrary hexahedral element. Okay? That's omega e. We constructed as, as always, really from a bi-unit domain. All right, and everything, everything proceeds just as before. Nothing new here. Okay, so, in this setting we would have, in this domain again, we would have A equals 1, 2, 3, 4 would be in the back, 5, 6, 7, 8, right, and this is a bi-unit domain, so this point is as always, minus 1, minus 1, minus 1, right? And so on that point, our number 7 would be 1, 1, 1, and so on, right? We know this very well now, okay? This is how we construct our basis functions having defined these coordinates, c1, c2, c3. What we see is that as always NA c1, c2, c3 will be constructed from you know, these, these one dimensional Lagrange polynomials, right? And NA tilde, and I forget what I call them, let me suppose I call them A bar c1 N tilde, E bar, c2, N tilde, c bar, c3, I may have used different Our notation for the, for the indices a bar, b bar, c bar before, but, but I think by now we understand this very well, right, each of these as we recall is a 1D Lagrange polynomial Okay? And we know exactly how to construct these, in the case of bilinear, sorry, trilinears, triquadratics, and so on. Okay, and since we are so good at this now, I, I don't need to belabor the points, so I'm, I'm going to tell you right away that as always, we also use the the same sort of basis functions to interpolate the geometry, right? That part, of course, continues exactly the same because even when we were solving the linear elliptic problem in 3D with scalar variables. The geometry still was fully 3D and the geometry was indeed defined by by position vectors and so on. Right? So that bit is exactly the same. Okay, so we have x e, as a function of the C vector. All right? That's exactly the same map that we had before. Okay?
