Welcome back, we continue with our development of the finite element equations. At the end of the last segment we got as far as looking at the individual integrals in the finite dimensional weak form, the weak form and writing them out as matrix vector products, okay? This is where we're going to pick up today. The topic of this segment at least is going to be the finite element equations in matrix vector form. Okay, now, what I'd like to do here is, perhaps, just recall the forms the integrals. Okay, so, we saw that when we put everything together, we had, for this integral, integral over omega wh, x sigma h Adx. We had the following, we had the rho vector of degrees of freedom corresponding to the weight in function. And this began with this particular entry, right, C2. C3, in terms of global degree of freedom numbers, right? This went all the way until we came to Cnel + 1, okay? This was multiplied by EA. And then, we had this matrix, which I'm going to try to write out really neatly today. So if I recall correctly, in this position we have -1 over h1. In the next we have 1 over h1 + 1 over h2. And here we ought to have had -1 over h2. In this position, we have -1 over h2, and here, we have 1 over h2 +1 over h3. Here we have -1 over h3, -1 over h3. Here we have 1 over h3 + other terms and I'll stop because if not I will, I won't have a suitable point to stop, okay. This continues all the way down until we have in the very last position here, the very last little sub matrix, sub two by two matrix. We have 1 over hnel -1, 1 + 1 over hnel- 1 over nel and here finally we have 1 over hnel- 1 over hnel, okay. And this matrix is closed here, and okay. I'm going to multiply this by a row vector, okay, because I've run out of room there. I'm going to write to row vector down here. Sorry, the column vector down here. This column vector has global degrees of 1, 2, all the way down to dnel + 1, okay? And let's just recall that this goes up here, okay. This is what we had for our integral on the left-hand side, all right? For the integral on the right-hand side, which as you recall comes from the force, we had the following. The integral itself was integral over omega whfAdx. And this, when all was said and done, also was multiplied by the same c vector. Up to cnel + one. And multiplying it here was fA, and let me see. I think I add an fA, perhaps, over 2. And the contributions here were the following. We had in the very first position, we had h1 + h2. Next position we had h2 + h3. And so on. Up to, in the very last two positions, we had hnel- 1 + hnel, and finally here we have hnel. Okay, all right, we had this and then we have the very last term, which was an easy one to write. The last term was Wh(L) tA and this following the same convention that we've adopted. For the other terms was multiplied by the vector of C degrees of freedom. Which is probably the waiting function degrees of freedom going all the way up to Cnel + 1. And the vector that it was that that rho vector was multiplying, the column vector here was filled with zeros all the way up to the very last entry. And at the very last entry we had tA. Okay and I realize now that for all the h's here, for all the element lengths, I used super scripts. Up to now, but the first two slides of this segment have been using subscripts. We'll go back to superscripts, but let me just put in a remark here, so that if you review these lecture you will note that as well. We're using h sub e here instead of h sup e and we're now going to go back to h sup e, okay. More photos on the next slide. Okay, so we have these terms here and what we want to do now is to begin assembling them. As a first step what I'm going to do is just to get a little more insight and make it a little less unwieldy, I am going to say that we are going to consider the case where all the element lengths are the same. The development so far and especially the form we have on the previous slide on this one and what we've done in the previous segments should convince you that method is completely applicable to formulations where or the measures where the elements do not have the same lengths. Just for some sort of gravity in writing and convenience and some on my defense side as well, I'm going to say all the HEs are the same, okay. So let me just make this remark also here. Consider he constant for all e. So there's a single h e we don't need to distinguish h1, h2 and so on, all right? It just makes things a little easier to write. So now if we go back and look at the matrices we've put together, what we should observe is something about the sizes of these matrices, okay? So we'll do this, but now we're going to use common element lengths. Okay, note that the finite dimensional weak form that we have with all these matrix vector expressions is essentially this. C2, C3, Cbel, plus one, times EA. Now, because we have a common element link, just for convenience I'm going to put it out. Again, this is what I wanted to achieve as well. Okay, this now multiplying the matrix which has the form minus 1, 2, minus 1, minus 1, 2, minus 1 and so on. Okay, until we come all the way down here to 2 minus 1, minus 1, minus 1, 1. Okay, all of this now multiplying the displacement degrees of freedom. If we're doing an elasticity problem or more generally the solution degrees of freedom, right. The trial solution degrees of freedom, okay. This is our entire left hand side of the weak form equal to on the right hand side, we have C2, C3, Cnel+1, fAH over 2. And now again because all the H's are the same we have here 2, 2, all the way down until we have a one in the final entry, okay? Plus C2, C3 Cnel plus one multiplying zeroes everywhere in this column vector until we come down to ta, okay? So for the special case of a mesh where all the elements have the same length HE, we have this nice simple representation, right? And that's an HE there as well. Okay, what I'd like to do here now is just pay some attention to the. Sizes of the relevant vectors and matrixes should be completely clear of course it's pretty much written here the way we've written out the scroll vector. That this scroll vector has dimensions in year, so of length and year. Right, simply because it's going from two to nel plus one, right? This column vector here is NEL plus 1, which leads us to conclude that this matrix is of a dimension NEL cross NeL + 1, okay? Essentially, the same thing on the right hand side tells us that this column vector, as well as that one, are off length NeL. And like and of course this is the same, the c vector is the same one, right. Right so both of these are off length Nel. Right, obviously just for consistency of the matrix vector products. Now. So we have this form and let's see what else we know about it. Can you recognize what we know about one of the degrees of freedom in the vector of nodal trial solution values. So look at that vector, the D vector carefully and think of what we know about at least one of those degrees of freedom and in fact one of those degrees of freedom. Because of the chronicle delta property of our basis functions, we've demonstrated that this particular entry, right Is actually use zero, not right, and that comes about because you remember what our mesh looks like. Right so this is element one and this position here is X. This node is X 1 equals zero. On that node the degree of freedom is D 1 which is U zero. Okay, which is a node Right? Because it's R, the traditionally boundary condition. Right? What this tells us is that when we look at the matrix vector product on the left hand side of this matrix vector equation, we can actually take this entire column, right, and I'm going to do something which is a little not strictly legal. But I'm going to draw that column vector, right. And because of the nature of this matrix, okay? Sorry. The other entries in this column vector are 0. Right, all the way down. Okay? All right. Essentially what we can do now is because we know d1 equals u0, right? And we recognize that this particular column vector that I've marked out in the matrix, is the one containing matrix elements that multiply D1, which is equal to U0, which is known, we are at liberty to actually move it to the right-hand side. Okay. All right, so what we will do here is, okay, is the following. We are now going to write this as C2, C3, CNEL plus 1, EA over H E times the matrix now that is considerably simplified, well not considerably, but at least somehow simplified. Two minus one, minus one, two minus one, going all the way down to minus one two minus one. Minus one one. Okay and the diagonal is this one. And the upper and lower diagonals of those ones. Okay. This thing multiplying the vector now d2, d3, all the way down to dnel+1 right? = C2 C3 up to CNEL + 1. Now multiplying, I'm going to put all the column vectors together here, okay? The first column vector is fAhe over 2. Multiplying twos all the way down except for the very last element which is one. Plus another column vector filled with zeroes all the way down to the very last entry which is T times A. And now there's a newcomer to the party here, right? The column vector that we looked at here, I'll draw yet another arrow to it, can be moved to the right hand side. Recognizing that we have a minus one in this entry, okay, we have on the right hand side, EA over HE multiplying D1, but D1 is known right D1 is known to be equal to U not because of two things. Because of the tertiary boundary conditions, as well as though the chronicle delta property of all basic functions. Okay. Now when we look at the size of our matrices, we recognize that really it's just this is the only one that we need to correct because it's lost a column. Early it had nel rose and nel+1 columns, so it has now lost a column, okay? So and we are going to recognize therefore that it is a matrix of size nel cross nel. Okay. And we're going to do one more thing. We are going to call, so nel across nel is the dimension, or rather the dimensions, of this matrix. And we are going to write this as the matrix key. Likewise for obvious reasons we're going to regard this vector of degrees of freedom for the waiting function as the vector c. Since all our vectors are we like to think of vectors as being column vectors and recognizing that we've written this a a role vector, I'm going to call it c transpose. A vector is terminology that we reserve for column vectors. Likewise, this vector, again for obvious reasons, is going to be my D vector. Okay, here again we have our C vector and now I'm going to look at all of this, right? All of these contributions together and call them my F vector. Okay? I just recognized this c vector again is a c transpose, again using this idea of row and column vectors. Okay?
