 Welcome to this 26 lecture, in this we will see quadratic or second order elements. As I mentioned earlier, linear elements may not be good if the field is varying drastically in some regions and if you do not want to use varying very fine mesh of linear elements. Then you can use fewer elements of or coarser mesh of higher order elements to get the same accuracy. So, to start with we will just refresh for linear triangular element, we had seen this expression for b, bx and by where this is not daba az by daba y was this, daba az by daba x was this in one of the earlier in fact lecture 21. This formula was there and then we also see you saw that these are just constant because a1, a2, a3 are the values that you would have got after the FEM solution and q1, q2, q3 and p1, p2, p3 are just you know they depend on only the coordinates of the vertices is it not. So, that means bx and by are constant over element which may not be true in highly non-uniform field right. So, errors will be appreciable. So, then we can go for second order or quadratic formulation. Now, quadratic triangular elements the corresponding approximation is this again now see these are the six, these are the six nodes. Now, the three nodes are as usual 1, 2, 3, 4, 5, 6 are at the centers of three sides. Now, there are six nodes. So, there should be six unknowns a bit to F and the corresponding approximation one of the approximation you can have is this a plus bx plus cy dx square plus ex y plus f5 square this is a logical you know approximation that you will make with six constants is it not and then you can actually using the same procedure that we followed for you know linear triangular element. What did we do there using that phi expression there we express phi 1, phi 2, phi 3 and then we eliminated abc by taking inverse is it not and by eliminating abc finally, we got this expression with i goes from 1 to 3 is it not. So, here same thing you can do but now here we will have to invert 6 by 6 matrix to eliminate these constants a to f is it not you will have to invert 6 by 6 matrix which is you know you can see the computational burden is becoming higher. So, that is why there is a simpler method called as area coordinates approach which we will see which does not require you to invert so, what is that approach area coordinates is you know it is it uses area coordinates or natural coordinates as they are defined. So, for example, they are L1, L2, L3 what is L1? It is delta 1 by delta, delta 2 by delta and L3 is delta 3 by delta. Now what is delta 1? Delta 1 is the this is the element so this is 1, 2 and 3 and suppose this is a point P. So, now L1 is delta 1 by delta, L1 is equal to delta 1 by delta, delta is the area of the whole triangle 1, 2, 3, delta 1 is area of P23. So, area of P23 divided by area of 1, 2, 3 is L1. So, that is what is so area of the triangle formed by P23 is delta 1, delta 2 is area of P31. I can see here it is P31 because it is not written as P13 because then the area will become negative. Generally, when you calculate the area you have to always take the 3 nodes in anticlockwise fashion. Similarly, delta 3 area of triangle P12. What is the property delta 1 plus delta 2 plus delta 3 is equal to delta? Is it not obvious? Because L1 will be function of this, this divided by total, L2 will be this area divided by total, L3 will be 1, 2, P divided by total. So, addition of all this L1, L2, L3 will be delta 1 plus delta 2 plus delta 3 by delta which will be 1. So, that is why this is 1. And in fact, for linear triangular elements we had again seen this expression many times now. Is it not? N1 is this for linear triangular element. Now, if you take 2 here, 1 upon 2 times this, this is the nothing but the area of the triangle P23. P is x and y, P is x and y 23, P23, area of P23, P is x, x comma y, P23 is nothing but this half of that determinant. Is it not? So, will be this. So, if you take this 2 inside, this is nothing but the area of triangle subtended by P which is xy and 23. And then that is, that is why it is delta P23 divided by delta. So, it is L1. So, that is why for linear triangular element, L1 is equal to N1. So, that means we need not, at that time when we saw the procedure, we need not have inverted that 3 by 3 matrix to eliminate ABC. We could have directly used this property and got that. But since we had not seen that this general theory of natural coordinates, we did not use it there. And I wanted you to understand general procedure of FEM. So, that is why we went by inversion there. But here, we can directly get N1 by just calculating, you know, this is the ratio of two triangular areas. Is it not? Similarly, you will have N2 and N3. So, that is why for linear triangular element, we know Ni xy, that means any N either at 1, 2 or 3 will be given by this ai1 plus ai2 L2 plus ai3 L3. So, for i equal to 1, that is node 1, a11 will be 1 because N1 is equal to L1, is it not? i is equal to 2, N2 will be equal to L2. So, that is why this should be 1, this should be 0. Similarly, for N3, so for linear triangular element, then you are, so this is a generalized expression for any Ni. So, Ni will be in general function of L1, L2, L3. But for linear triangular element N1 is equal to L1, N2 is equal to only L2 and N3 will be equal to only L3. So, correspondingly those coefficients will have value. Now, having understood this, for quadratic element, we will just extend, earlier Ni was function of only these three terms, is it not? Now, it is a quadratic element. So, you need to have six coefficients. So, that is why now ai4, ai5, ai6 and now instead of here, here it was only L1, L2, L3. So, what is natural to have L1, L2, L2, L3 and L1, L3? So, now Ni is function of not only L1, L2, L3 as linear function, but also their products. Now, what are the unknowns here? All these coefficients ai1, ai2, ai3 like that, these are all unknowns. And remember this is at every node, there are six unknowns because this is i equal to 1. So, for node number 1, there will be a1 to a6. For node number 2, there will be a1 to a6. And those you have to evaluate at node 1, this we have already seen at node 1, L1 is 1, L2 and L3 are 0. Similarly, these two, at node 4, L1 and L2 is 0.5, L3 is 0. Is it not? Because L1, L2, L3, L3 are 0. See here, L2, L2 at node 4, what we are calculating is, this is marked L2, we are calculating at node 4. So, L2 at node 4 means this will be delta 4, 1, 3, is it not? And divided by the whole area. So, this area is half of this whole area. So, that is why it is 0.5. Similarly, you can calculate all the rest of the coefficient. Now, what we have calculated, these are the values of L1, L2, L3 at all the nodes we have calculated. Now, we will use this in calculating those six coefficients for every node. Because for every node, we have to determine the six coefficients. So, we need six equations or six conditions. N1 is equal to 1 at node number 1 and it is 0 at all other nodes. Is it not? So, that is why N1 is 0 at all the remaining five nodes. N1 is 1 only at node number 1. So, what we do is, now we substitute in this expression. N1 is 1 at node number 1. We will substitute N1 at node number 1's coordinates. And at that coordinates, we know what are the values of this L1, L2, L3. We know those because we have evaluated those here. So, L1, L2, L3, we know at every node. Is it not? So, we can simply substitute in this expression. So, for example, N1 is at node 1 is 1. So, here at node 1, only L1 is non-zero and that is 1. All L2 and L3 are 0. So, all these terms will go down to 0. Is it not? Because L2, L3, either L2 or L3 is there in all these remaining five terms except this first term. So, that is why you will get 1 as A11 into 1, rest everything will be 0. So, you get A11 is equal to 1. Similarly, this for the other two, you will get A12 and A13 as 0. Then, let us take one of these examples. Now, at node 4, we know at node 4, N1 is 0 by definition of same function and property of shape function. Now, at node 4, we have to substitute all these values of L1 to L3. So, at node 4, L1 and L2 is 0.5, L3 is 0. So, L1 and L2 is 0.5 and L3 is 0. So, that is why here 0.5 and 0.5 come and then here it is L3 is 0. Here it is L1, L2. L1, L2 at node number 4 is 0.5, 0.5 and rest again is 0. So, again you know you get the simple expression and now A1 is known already calculated. So, A12 and this is also known. So, A14 becomes minus 3. Likewise, you know you get the remaining two terms also, two coefficients. So, now by substituting the values of all these A1, A6 for N1 in this equation 4. Now, we know all these AIs for node 1. So, then we get L1 minus 2 L1, L2 minus 2 L1, L3. So, after some simplification, you get and substituting L2 plus L3 as 1 minus L1 because L1 plus L2 plus L3 is 1. So, then finally you get N1 as this. So, just see the difference for linear element N1 was just equal to L1, but for quadratic element N1 is L1 times twice L1 minus 1. So, now we can likewise we can obtain expressions for other shape functions. So, all these NIs for 1, 2, 3 will be this, we will get similar and N4, N5, N6 you will get like this 4L1, L2, 4L2, L3, 4L1, L3. This you can verify by the same procedure that was explained earlier. So, now we go further. Now, we suppose we want to solve this Poisson's equation in magneto-stratics del square i equal to minus mu j and then as we have discussed earlier j should not be associated with mu should not be associated with j because mu can vary in the geometry. So, we bring mu on the other side associated with del square. So, that when we integrate depending upon which element we are in corresponding mu we will take for that element. So, now this is Cij, this expression also we have seen earlier is it not. This is nothing but del ni dot del nj, that expression is this. So, the element coefficient matrix is will be 6 by 6 matrix with this corresponding expression as this. And then the bi the right hand side matrix which is source again you will get j integrated over that element area ni ds. Now, we will see later to be put in a later slide. We will see that here earlier when it was a linear element we what did we do we apportioned j equally to the 3 nodes as j delta by 3 is it not. Here also we will get that but that j gets apportioned equally to not the you know 1, 2, 3 but to the middle nodes. So, bi is equal to j delta by 3 for nodes 4, 5 and 6 and it is 0 for nodes 1, 2, 3. Why we will see later. So, now C e is a symmetric matrix given by Cij e this that means this is a element level matrix now right. And this will be 6 by 6 matrix is it not this will be 6 by 6 matrix. So, the you know Cij in general is given by this expression now C11 is this which is same as earlier. Now this C11 for linear triangular element also was identical this was the same. But C12 is different C12 that means octagonal elements are different. So, C12 is this C15 is 0 because again C15 C26 and C34 are 0 why because there is no connection remember 1, 5, 2, 6 and 3, 4 there is no connection is it not. So, then you how do we get this now here I have given one book reference this book is basically finite element method in electromagnetic this is for high frequency electromagnetic mostly but if you want to verify the you know this coefficient values and expressions you can you know refer this book and some of these you know coefficients I have already been derived which we will see in the next lecture. So, you know all this do not get worried about this expressions because this is one time effort is it not if you do coding and this all these coefficients are known PIs and QIs. So, now we will see derivation of some of these coefficient that we have seen in the previous slide for example, this we know now Cij is this C11 will be simply i and j replaced by 11. So, this is this will be the C11 right now N1 we have derived N1 is L1 times twice L1 minus 1 right and what is L1 it is the area of 2 point P 23 divided by the total area right. So, that is L1 and we had already seen that Li expression is this Li expression is this for example, L1 is here it will be A1, P1, Q1 and then P1 and Q1 as usual though these are the over the state this we have seen number of times is it not. So, if you substitute further capital N1 is twice L1 square minus L1. So, now L1 is this bracketed term square minus 1 upon 2 delta. So, L1 is this bracketed term again L1 right. So, in equation 1 the term dA by N1 by dA by X is related by taking the first derivative of equation 5 right. So, we have to just take dA by when you are talking of first derivative you just take a derivative of this and then you will get this expression by delta square. So, this whole thing is d by dx of this as a minus derivative of this term with respect to X will be simply P1 upon 2 delta right and then now you can substitute resubstitute this because this is nothing but twice delta L1 because L1 is L1 is this. So, since you are getting the same term we can resubstitute those terms and we get this and finally we get this simplified expression right. So, then del dA by N1 by dA by X into dA by N1 by dA by X is equal to you know the product of these two terms right and then you get this. Similarly, you can calculate dA by N1 by dA by Y into dA by N1 by dA by Y instead of P1 you will have now Q1 that is the only difference here they will be all these are all P1 because that was you know P1 was higher multiplied by X. So, in derivative you will have P1 then you are taking now with respect to Y only Q1 will come right. So, here you have expressions with only Q1 and then you get this as dA by N1 by dA by Y into dA by N1 by dA by Y as this. So, now C11 is equal to this total sum of these two terms that we have calculated. So, now each of those two terms are having three three terms. So, that is why you have now six terms and now we use this integration formula which we have seen earlier dX dY integration with respect to dX dY. So, again it will be this formula which we have seen earlier and now actually if you for each of these terms if you use this formula everywhere you will get you know formula this is 4 P1 square L 1 square right. So, L 1 square means m is L is 2 is it not and that actually is substitute here. So, 2 factorial and m and n are 0. So, 0 factorial 0 factorial 0 factorial 0 factorial is 1. So, that is why you get only 2 factorial here and then L is 2 m is 0. So, 2 plus 2 4 factorial ok. So, likewise you can calculate all the terms you know by some using this formula and then you simplify you will get you know four terms they get cancelled and C11 just reduces to this. So, I think we will stop here and continue next time.