All right, in the previous segment we figured out how to go around computing how to go about computing the the gradients, right, which go into our integrals. So let's pick up there and complete the evaluation of this integral. And you will recall that the integral we were working with is the following. Integral over omega e, w h, x. EA, u h, x dx. And using what we developed in the previous segment we know that we can write this as sum over for integral over omega e Let me see, we can write it out as sum over A. N A, C, C, X, which is two over h e, times C A e. All of that is w h, x multiplying E A. Times sum over b and b, c. C, X for which we again use two over h e. D E, D B E sorry. D X Okay. But note that for d x we could write it as d x. D ksi, d ksi. All right, and we know that d x d ksi is just x, ksi, which is what we computed in the previous segment to be h e over two. Even for quadratic basis functions. And this comes about from the from the fact that we have an affine map. It also has to do with the fact that Lagrange polynomials are complete. Okay? All right. So, we have this representation and now we can write out this integral in an even simpler form. We are going to note that because of the fact that we have h e over 2 here, this cancels out with that, okay? And we can now write out this integral. Using what we call matrix vector notation. Okay? And in order to use matrix vector notation. Let's, let's actually set it up a little more carefully, okay? We will use matrix net vector notation but let's do, do just one step before it which is a step we've taken when we were working with linear basis functions. And that is to recognize that c a e d b e's are independent of c. And so too is this product E A of constants and this remaining two over h e. Since your independent of position they, can be pulled out of the integral. Okay, right so we're getting up to using major vector notation but before we get before we actually do it lets write that integral as a sum over A and B of the following quantity we can actually pull even E A out of that summation right, so we can have 2 E A over h e sitting outside the summation, okay? And then we have the sum over A and B, c A e, right? Integral over. Well, we'd, we'd written it up here as an integral over Omega e, but using this chain rule right there, we observe that we can actually write that integral as an integral over. Over what? Over a different domain right, right its all omega C. Okay. Now the integrant is a simple one it is N A comma C and B comma C okay, d ksi. All right. And, I'm going to put parenthesis on this integral just to make it clear that it's that's where the integral ends. And then, here we have dBe. Okay? The use of matrix vector notation as you will recall from our development of the method for linear basis functions the use of matrix vector notations comes in to relieve us of the need to have this explicit sum. Okay? So, now using matrix vector notation here is what we get. The, that sum over A can be replaced by the sum over A can be replaced by C 1 E, C 2 E, C 3 E. With these being the degrees of freedom that interpret the weighting function in element e. We get 2 E A over he. Right. We get our integral, right. However, each, we, we, we get a bunch of integrals. Each of which, is one component, of a matrix. Right here we get integral, over Omega C of N 1 comma c and 1 comma c. However, the integral over on the N 1 comma c, I'm going to write explicitly as an integral from minus one to one over c, all right, because that is the extent of our bi-unit domain. So here I get n1 comma c and 1 comma c, d ksi. The second position I get integral minus one to one and one comma z and two comma z. And the third position the One 3 component of this matrix. I get integral of N 1 comma C. N 3 comma xi. All right? Here I get integral minus 1 to 1 N2,xi N2,xi d xi and here I get integral minus 1 to 1, N2,xi, N3,xi d xi, and here I get integral minus 1 to 1 N3,xi N3,xi d xi. What about the rest of the matrix? Right, it's symmetric. All right? So, multiplying all of this, I have here Instead of the dBe, I have those represented as d1e, d2e, d3e. Okay, so this is our matrix vector nota, or representation of our left-hand side integral over an element for quadratic basis functions. What we're going to do next is actually explicitly calculate those integrals. All right, so. So integral minus 1 to 1, N1,xi N1,xi d xi equals integral minus 1 to 1, let's see. It is let me just check what that integral eh, what that derivative is. That derivative is 2 xi minus 1 time the whole square d xi, right, d xi minus 1 is N1,xi, okay? All right so we get here for this integral, we get, this is integral, the integral minus 1 to 1, 4 xi squared minus 4 xi ma, plus 1 d xi, and observing that this is an integral from minus 1 to 1, we know that the odd terms don't survive the integral. All right, it's only the even terms that survive. When we do this we get 4 xi cubed over 3 plus xi limits minus 1 to 1, okay? This is 4 over 3 plus 1 minus minus 4 over 3 minus 1, right, which is 2 times. 7 over 3. Let me just check this out. Hm, [SOUND] N1,xi, all right there is a one-quarter here, okay, that's because N1,xi is one-half of 2 xi minus 1, so we get a one-quarter here. We also get a one-quarter here, one-quarter here. Right, and we get a fourth here. So, when we carry this out, we get 7 over 6, right, which is what I was expecting. All right, now integ, the next term over, right, my integral, minus 1 to 1, N1,xi, N2,xi d xi. This is integral minus 1 to 1, N1,xi, we know's one-half times 2 xi minus 1. And 2 xi as we saw from before is minus 2 xi, okay? Integral over d xi versus then one-half integral minus 1 to 1, okay? What do we get here? We get here 2 xi minus 4 xi square, okay, d xi, which is one-half. Once again we don't need to compute the odd term, right? We compute only the even term, we get one-half let's pull the minus sign out, we get minus one-half times 4 xi cubed over 3, limits minus 1 to 1, all right? And this turns out to be. Minus two-thirds times 2, which is minus 4 over 3, okay? All right, moving on, integral minus 1 to 1, N1,xi, N3,xi d xi is integral minus 1 to 1 one-half 2 xi minus 1 times one-half 2 xi plus 1, d xi, 'kay? This is one-fourth integral minus 1 to 1, 4 xi squared minus 1, d xi, which is one-fourth times 4 xi cubed over 3 minus xi. The limits on this are minus 1 to 1, okay? Let me actually use square brackets here as I did previously. Which is now one-fourth. Times 8 over 3 minus 2, which is 1 over 6, if I'm not mistaken. Yep, right. Okay carrying on, let's just move onto the next slide here. So, the next term we need to calculate is integral 1 to 1, N2,xi, N2,xi d xi, which is integral minus 1 to 1, N2,xi is minus 2 xi, and we get the whole square here, right? And, this is an easy integral to carry out. It is just 4 xi cubed over 3, limits minus 1 to 1, which gives us 8 over 3. All right, integral minus 1 to 1, N2,xi, N3,xi d xi. All right, that's the 2,3 term of the, 2,3 component of the matrix. This is integral minus 1 to 1, minus 2 xi times 2 xi plus 1 d xi, which when we carry it out is integral minus 1 to 1. Let me see what do we get here, we get a minus 4 xi squared minus 2 xi, d xi, which is oh, as I've been doing often in this calculation, I'm missing a half here. So I get a half here. Okay, so that is one-half again using the fact that it's only the even terms that survive, we get minus 4 xi cubed over 3 minus 1 to 1, and when we go ahead and compute this we get minus 4 over 3. The very last term to be computed here is the integral minus 1 to 1, the very last component is integral minus 1 to 1, N3,xi N3,xi d xi, which is the integral minus 1 to 1. We pick up 1 over 4 and here we get 2 xi plus 1, the whole square, right, and that is because N3,xi is just one-half 2 xi plus 1, all right, so here we get 1 over 4, integral minus 1 to 1 4 xi squared plus 4 xi plus 1, d xi, right, one-fourth 4 xi cubed over 3 plus xi. And of limits of the integration are minus 1 to 1. And when we compute this we get one-fourth of eight-thirds plus 2. Which is right, 7 over 6. All right, excellent. Now, collecting results, what we get is the following. We set out to compute integral over omega e wh,x, sigma hAdx. Upon first substituting the constitutive relation for sigma h, then using our basis functions to write out the fields and their gradients and then going ahead and doing the computation, right, do, doing an explicit integral over xi, integration over xi, we get the following. We get, c1e, c2e, c3e, right, which is our vector representation of the nodal degrees of freedom of the weighting function, okay? 2EA over he, and here we get our matrix which we've just computed, and we've just computed it to be 7 over 6, minus 4 over 3, 1 over 6, 8 over 3, minus 4 over 3, 7 over 6. It's symmetric, but now let's just explicitly fill it in just this time. And for our matri, our vector representation of the nodal degrees of freedom of the trial solution, we have d1e, d2e, d3e. Okay, now as we observed before for the case of linear basis functions and linear elements, right, two-noded linear elements, this is what we will denote as Ke, right, our element stiffness matrix. All right, this would be a great place to end this segment.
