All right. So, we'll move on. We've established our finite dimensional weak form. We've looked at our basis functions. We made the somewhat important observation. That for this problem, because we have a vector problem, the number of nodes is not identical to the number of degrees of freedom. Right? But then we made the identification and we know how to relate them. All right. Let's go on. We'll get into assembling the integrals in our weak form, in our finite dimensional weak form. So, we will call this segment the element integrals. And in order to get to the element integrals, we first need to make the observation that our finite dimensional weak form. Because it is an integral over the entire domain, right? Or the boundary of the relevant Neumann boundaries also allows us under this partition to write it as a sum over a, the, the individual sub-elements, right? Or the subelement boundaries. Okay, so the finite dimensional weak form can also be written as. Sum over e, integral over omega e, w h i comma j, sigma h i comma j, dv equals sum over e, integral over omega e, w, h, i, f, i, d v. Plus, now here, we need to be a little more careful, right? We are talking here of doing the following. We are talking here of doing a sum I equals 1 to Nsd. Okay? We're summing here over our spatial dimensions. And where as earlier we could directly put down the corresponding Neumann boundary, we are going to write that also as that integral also as a sum over sub-integrals, right? Each of them is a boundary. So we need to say here that we are going to take the sum, not over all the elements, but the elements belonging to let me see, right, En, right? Remember, E was a set of elements that had one of their boundaries coincide with a Neumann boundary. All very well except that that too needs to be indexed by the spatial dimension. All right? Because for each of the i's, i equals 1, 2, 3. We could have a different set of elements whose boundaries coincide with the Neumann boundary for that particular dimension. All right? Okay. So we have that and then the integral is over partial omega t bar, I, and I'll put the E up there. All right. And this, of course, is the same whi t, bar I, d of S. Right? And all right, so this is what we have. What we are going to do is just as before, assemble each of these integrals over the elements of domains, or in this case, over the special element boundary sub-domain. Right? And of course we are also going to use the fact that this is given to us by B, elasticity tensor, now multiplying U, K, comma F. All right. Okay. So, we're, we're going to proceed with this now. So, we've, we will as always start with the, with the left hand side integrals, consider first integral over omega e, w h i comma j I'm going to directly go to the representation including the elasticity tensor. Okay, so sigma h i g. Oh, if I'm going to use the sigma h, I need to have an h there, all right. Because that's where the finite dimensionality of this comes in. Okay, so we have here C, i, j, k, l, u, h, k,I dv, right? Now on a previous slide, belonging to the previous segment, we wrote out the, the expansions for w h, and sort of, for the gradient of w h and for the gradient of u h. We will now invoke both, right, and in doing so, I'm going to write it here. We have here again, allow me to skip, well, okay. Mm, I won't skip a step right away. going to write this as integral over omega e, the w h, w h i comma j, is written as sum over A, N A, comma j CA I E, okay? This is whi, j, and I put it in parentheses. We have here C, I, J, K, L. And again we have sum over B, N D. Now, the gradient is represented as A, using the coordinate of direction L. And of course it's, L runs over 1, 2, 3. So that means n b here has to take a comma l, right, that's where the gradient is computed. And we have d b k e d v, okay. Now if you were go back and stare at the way we did the corresponding step for our scalar 3D problems, you would see something very similar except the fact that we had no index on the degree of freedom, degrees of freedom for the reading function. Nor on the degrees of freedom for the trial solution for the displacement field, okay. The fact that we do have indices here, comes from the fact that these are actual vector degrees of freedom. And i and k here both run over one two three. And the fact that we have those indices also shows up in the fact that we have a couple of extra indices here, right. The coefficient there, if you want to think about it as that. The elasticity's tensor's actually coefficient of elasticity, is a full taut tensor unlike the conductivity tensor or the diffusivity tensor, which are second order indices, all right. So that's, that's really all there is. And once we take, once we understand that, and we are comfortable with that bookkeeping essentially, all is well, all right. So we proceed, and as you may imagine, what is the next step here? Or as, as you may recall, what is the next step? Right. It is to observe that the degrees of freedom, nevermind the fact that they're vectors now are still, after all, independent of position, all right. And so they can be pulled out of our integral. And in doing so, we can also because integration and summation in this case commute because of linearity and all that. We can pull those summations also outside the integral, right? Okay. What that gives us is. Now I am going to write this as a sum over A and B. And let's recall, let's remember just here that A and B run over 1 to number of nodes in the element, right? Integral, oh, sorry. We need to have here c A i e, okay? I have here, NA comma j, okay, C i j k L, right? And for our representation of the gradient of the displacement, from that representation of the gradient of the displacement, we get an NB comma L here. Okay? Our dV, all right, I'll put a parenthesis here, or pair of parentheses. And, all right. What we've done in the process is pull the d B k e out, okay. Right? Straightforward enough. Oh, this is an integral over omega e, right? To proceed, and remember that we know how to compute these gradients, right? We looked at that on a, at the end of the last segment, right? So we know how to compute these gradients, no problem, all right. Let's move on. Remember what the next step is, right? It is to convert it to, we, we carry out a change of variables and rewrite the integral here as an integral of the bi unit domain. So, once again, sum over A and B, and I'll forgo writing the limits of that sum. We get here c A i e integral over omega xi, all right. That's our parent domain. N A comma j, C i j k L, N B comma l, right. Now, just as before, d V can be written as determinant of the Jacobian of the mapping. D V sub, thing was putting it there, right? Okay, now, of course, I could rewrite this in another step by specifying that this integral over omega C really in, involves integrals of C1, C2, and C3. It's a triple integral, each one of them going between minus one and one, and the fact that this is essentially d xi, d xi2, d xi3, all right? And this integral basically becomes something like, it becomes actually a triple integral, all right? Minus 1 to 1, minus 1 to 1, minus 1 to 1, all right? So, that, that step is given, all right? So, so well we can go ahead and compute it. Now observe that, in general, depending upon the basis functions we've chosen, whether we've, you know, decided to stick with trilinears, or tri-quadratics, or, or, or rank higher order, these are functions of c 1 c 2 c 3. Now, the way I've constructed this, I've been thinking to myself, at least, that C is independent of position. It doesn't need to be, right. In general, it, too, can be a function of position, right. That, that would allow us to go to inhomogeneous materials. Okay, all of that will simply add a higher order dependence on position to this integrant. And of course, we know that in general, the determinant would also involve a dependence upon position, okay. But nevertheless, we not integrate this, right? We can be doing numerical quadrature. Right, so we do a numerical quadrature or numerical integration, which is quadrature. Right, and since we're working here with Lagrange polynomials defined on hexahedral elements, we have available to us the optimality of Gaussian quadrature. I ought to admit here though that if our elasticity tensor has some strange dependence upon position that is not just polynomial, we made, Gaussian quadrature may not remain optimum, all right? But, but, but still, it's, it's, it's an approximation, and that approximation is understood. Okay, so essentially we know how to compute this, right? Let's com, let's assume that we, we go ahead and compute this, right? If we go ahead and compute it. Let's, let's write it out, okay? Let's write it out. And let us write it out in this fashion, okay, so. All right? So, this thing is now equal to sum over A and B, c A i e. We've carried out the integral, okay? So we have something in there. So this entire integral here that we have here, this whole entire integral, Is done. Now we could realize now that I omitted to write my DBKE here. Okay, that integral with the raised bracket is done. So when that integral is done, we will be left here with a dBke, which is this one, right? Okay. I've left a, an amount of blank space there because I want to fill it in with something and I want you to think of what kind of object goes in there. In particular, is it a scalar, a vector, or a tensor? As a guide to getting to that answer, recall that the. The integral that we are trying to work with here is the one that I am now showing to you on this slide. Right? It's the one that is written next to the word consider. What kind of an object is that? Is that integral itself, once you valuate it, is that a scalar, a vector, or tensor? All right? It is a scalar. Because though the C has fu, free indices, I, J, K, L. Those are all being contracted out, right, and a sum over each of those dummy indices, I, J, K, L is indeed implied. Okay, so what we expect to see is that this thing is a scalar. But the way we've constructed it, we see that the c A, the, the, the c and d, right. By, by c here, I mean this c, not the elasticity tensor. Degrees of freedom have indices, free indices. With the final thing should be a scalar, there have to be other three free indices in whatever object is going to go into the black space. Which will be contracted out with INK. Okay. So, whatever object we have here, I'm going to denote it as K, of course. It will also have free indices I, K. Okay? Right. And it is essentially that integral. Okay? So the integral we have here is Kik. Okay? It is a particular component, the IK component of a tensor. Right? Thought of as a matrix, it is a 3 by 3 matrix, right? Because I and K run over. 1, 2 and 3, because we're operating in three dimensions here. All right. Now there's something else I haven't added there. And can you tell me what that well can you think of what it is? The indices A and B, right, because those indices are still here. So anything that we compute for the integral is um,specific to the combination of nodes A and B, which went into that integral. Right, that, that form that intergrand. So we have an object here which, which I'm going to write as KABIK. All right? Now there are indices all over the place here. And it's useful to think to just sort this out. So when I say I comma K go 1 over 1, 2, 3. I am going to write here the, the statement that's almost a contradiction in terms which is that a sum is implied. [NOISE] Okay, over IK. So I do not have an explicit summation here, but over the A and B indices, I do have an explicit summation. All right, it's, it's just a matter of taste in terms of what I choose to write as an expressed summation and which I choose to invoke the Einstein summation convention on, okay? The fact that I'm invoking Einstein's summation convention on the spacial indices, spacial dimensions comes from a sort of hangover of continuum physics.
