1 00:00:15,719 --> 00:00:21,910 Till now, we have looked at the one-dimensional finite element problem using linear basis 2 00:00:21,910 --> 00:00:28,910 function. So, if you had this domain, what we have done is, we had partitioned the domain 3 00:00:29,869 --> 00:00:36,869 into elements, we had corresponding nodes and then called the element, we went ahead 4 00:00:41,750 --> 00:00:48,750 and constructed approximation using these functions phi i, which are linear. Also the 5 00:01:03,689 --> 00:01:09,650 question - is that all we can do or can we do more than checking such linear approximation? 6 00:01:09,650 --> 00:01:16,650 The answer is yes. So, what today we are going to look at, is higher order approximation 7 00:01:17,439 --> 00:01:22,460 for the finite element solution. So, what is the notation? Let us first introduce 8 00:01:22,460 --> 00:01:29,460 when it is a linear approximation, then we will represent it by P equal to 1, where P 9 00:01:36,760 --> 00:01:43,760 we call has the order of approximation. Now, what we have implicitly assumed, I will assume 10 00:01:57,720 --> 00:02:04,530 it throughout this course, unless it is specified, specifically, that this order of approximation 11 00:02:04,530 --> 00:02:11,530 is the same in all the elements, that throughout the domain, we are going to use the same order 12 00:02:11,610 --> 00:02:16,840 of approximation; one can always use different order of approximation, in different regions 13 00:02:16,840 --> 00:02:20,299 of auto mode, that case we will not handle now. 14 00:02:20,299 --> 00:02:27,299 So, let us now see, how this order of approximation can be increased. So, the next example that 15 00:02:27,349 --> 00:02:34,349 would we have to do is, P equal 2; that is the order of approximation is quadratic. For 16 00:02:41,810 --> 00:02:48,810 this, let us see, what is it that we have to do, as far as the approximation is concerned. 17 00:02:56,310 --> 00:03:03,310 So, for this, you take the same domain that we had taken earlier, we send n load p, this 18 00:03:07,290 --> 00:03:14,290 is x equal to l, this is x equal to 0 and some distributor forces f x. For f x, let 19 00:03:25,120 --> 00:03:32,120 us also takes EA is equal to constant in the domain, that need not be true. So, using this, 20 00:03:43,530 --> 00:03:47,160 given this data for the problem, let us now go ahead and do the approximation. So, how 21 00:03:47,160 --> 00:03:54,160 are we going to do this? What we need to do now, is to first break the domain into element 22 00:03:55,489 --> 00:04:02,489 as we have doing till now. May be I will take 4 elements here. So, these 23 00:04:07,090 --> 00:04:11,849 partitions that I have drawn, using these end vertical lines, these are our elements. 24 00:04:11,849 --> 00:04:16,409 So, this is element 1, this is element 2, this is element 3 and this element 4. So, 25 00:04:16,409 --> 00:04:23,409 in this element, in each of this element, what I am going to do is add an extra point. 26 00:04:28,669 --> 00:04:35,669 So, in each of the element, I am going to add an extra point. This point is located 27 00:04:37,100 --> 00:04:44,100 at the center of the element. So, now we are going to do this, the locations of this point. 28 00:04:58,650 --> 00:05:04,160 So, now, instead of having 1, 2, 3, 4, 5 points for the 4 elements that we have drawn, now 29 00:05:04,160 --> 00:05:11,160 we have 9 points. So, we will give them name. So, this is point x 1, this is x 2, this is 30 00:05:11,610 --> 00:05:18,610 point x 3, x 4, x 5, x 6, x 7, x 8, x 9. What you should note, is that these extremities 31 00:05:27,800 --> 00:05:33,430 of the element are given by the point x 1, x 3; so, these are the extremities of element 32 00:05:33,430 --> 00:05:39,680 1; extremities of the element 2 are the point x 3 and x 5; extremities of the elements 3 33 00:05:39,680 --> 00:05:46,680 are the points x 5 and x 7; extremities of the element 4 are the point x 7 and x 9. These 34 00:05:49,780 --> 00:05:56,780 points x 2, x 4, x 6, x 8 are generally called mid time nodes or mid edge nodes. So, we will 35 00:05:59,820 --> 00:06:06,820 give a name to them; these are 36 00:06:13,850 --> 00:06:20,850 mid side nodes and the points x 1, x 3, x 5, x 7, x 9 are called these n vertices or 37 00:06:30,340 --> 00:06:37,340 simply the extremities of the elements. So, this point becomes end nodes or elements nodes 38 00:06:47,660 --> 00:06:54,660 or extremities of the elements. Now, what do we do next? 39 00:06:56,690 --> 00:07:02,160 So, what will do is - over this point checked that we have created, we are going to use 40 00:07:02,160 --> 00:07:09,160 it somehow to create our basic function, which are quadratic in nature. So, I have this point 41 00:07:20,060 --> 00:07:27,060 set, I am going to construct the basis function, which now should be quadratic in the elements. 42 00:07:33,009 --> 00:07:37,539 So, first, I am going to draw; then, we will talk about how to construct mathematically. 43 00:07:37,539 --> 00:07:42,840 So, while drawing, what had we said about the linear function, what we wanted these 44 00:07:42,840 --> 00:07:49,229 functions to do, what should have a value 1 at one of the nodes and 0 at all other nodes; 45 00:07:49,229 --> 00:07:54,220 the same principle we are going to use as far as construction of the quadratic functions. 46 00:07:54,220 --> 00:08:01,220 So, let us say the function phi 1 that is corresponding to the nodes x 1 or the point 47 00:08:02,789 --> 00:08:09,789 x 1. The phi 1 will be 1 at the point x 1 and 0 at all other points; so, if I want to 48 00:08:11,210 --> 00:08:18,210 make that phi 1 as this function. Similarly, I would like to make phi 2; phi 2 will be 49 00:08:28,410 --> 00:08:35,410 a function, which is 1 at this point and the point x 2, and 0 at all other points. So, 50 00:08:39,899 --> 00:08:46,899 what we see is that we have defined it in a local region; that is, phi 2 is only defined 51 00:08:47,910 --> 00:08:54,910 in the first element. So, if I can draw that element here, this 52 00:08:54,980 --> 00:09:01,980 is element I 1, this is element I 2, this is element I 3 and this is element I 4, then 53 00:09:13,399 --> 00:09:20,399 I can draw phi 3. According, phi 3 should be 1 at this point and 0 everywhere else. 54 00:09:21,379 --> 00:09:28,379 So, by the same token that I have done phi 3 like this, then phi 4. phi 4 should be 1 55 00:09:37,850 --> 00:09:44,850 here, 0 at all other points phi 4. Similarly, phi phi phi phi will be 1 here; 0 will be 56 00:09:46,790 --> 00:09:53,790 where are... Similarly, if I want to use, phi 6 will be this, phi 7 will be this, phi 57 00:10:09,959 --> 00:10:16,959 8 will be this and phi 9 will be this. So, what we got is phi 2 is here, so naming 58 00:10:21,170 --> 00:10:28,170 wise, this is phi 3, this is my phi 4, this my phi 5 phi; sorry, I made a mistake here. 59 00:10:37,220 --> 00:10:44,220 So, I have to go back and change my naming convention. So, before phi 2, this is going 60 00:10:51,970 --> 00:10:58,970 to be phi 3, this column is going to be phi 4, this one is phi 5, this one is phi 6, phi 61 00:11:08,179 --> 00:11:15,179 7, phi 8 and then phi 9. 62 00:11:17,029 --> 00:11:20,639 So, here, I have the 9 phi, which I have constructed; they look quadratic, we have to also give 63 00:11:20,639 --> 00:11:25,639 an expression to them mathematically; why are they quadratic? Because they have a value 64 00:11:25,639 --> 00:11:32,139 1 at one point, and 0 at two other points, and then we extend it as 0 everywhere. So, 65 00:11:32,139 --> 00:11:36,869 they should be quadratic; that means they look like that. So, in some will this function, 66 00:11:36,869 --> 00:11:43,869 how am I going to write my 500 ml solution. So, u F E of x, we will use our earlier notation, 67 00:11:48,559 --> 00:11:55,559 is it is going to be, u is the 9-term solution u 9 of x. This is equal to sigma i is equal 68 00:12:02,720 --> 00:12:09,720 to 1 to 9 u i phi i. So, this is a pixel optical. So, now, what 69 00:12:15,259 --> 00:12:21,699 is this, that the function has to satisfy in order to be admissible basis function? 70 00:12:21,699 --> 00:12:28,699 The first thing that we have done till now is that, this function phi i should be, it 71 00:12:31,839 --> 00:12:38,839 should be linearly independent; second thing that they have to satisfy is that they should 72 00:12:54,739 --> 00:13:01,739 be complete; and third thing that we have done by construction is they should local 73 00:13:10,279 --> 00:13:17,279 support, that is, they are only defined in one or two elements and in the rest of the 74 00:13:21,040 --> 00:13:24,799 element there, 0. So, by construction, we see that this property 75 00:13:24,799 --> 00:13:31,799 has satisfied linear independence; we do not know this, we do not know. So, let us first 76 00:13:33,319 --> 00:13:39,619 make this function in such a way that they are quadratic, linearly independent and complete. 77 00:13:39,619 --> 00:13:46,619 So, how are we going to do that? So to do that, let us now consult this function element 78 00:13:47,109 --> 00:13:54,109 by element. So, if I look at element - a generic element I k - if you look at the generic elements 79 00:13:57,839 --> 00:14:04,839 I k, what are the nodes that this elements has or the point it will be? x 2 k minus 1 80 00:14:07,019 --> 00:14:14,019 x 2 k x 2 k plus 1. Similarly, which is the phi, which is non-zero here? So, phi which 81 00:14:20,899 --> 00:14:27,899 are non-zero, here are phi 2 k minus 1 phi 2 k plus 1. 82 00:14:55,459 --> 00:15:02,459 So, let us now go and transform this to our element notation that we had introduced earlier. 83 00:15:03,699 --> 00:15:10,699 So, from the element notation point of view, in the element, I will call the first node 84 00:15:14,239 --> 00:15:21,239 as x 1 of an element k; the second one, x 2 of an element k; and the third one is x 85 00:15:25,359 --> 00:15:31,519 3 of element k. Similarly, the phi 2 k minus 1, in this element 86 00:15:31,519 --> 00:15:38,519 could become, would be called N 1 of element k and phi 2 k 87 00:16:04,229 --> 00:16:11,229 will be called N 2 of k. Similarly, phi 2 k plus 1 will be called n 3 of k. And like 88 00:16:23,899 --> 00:16:28,639 in the case of linear approximation, in this we want this function N 1 of k to be to have 89 00:16:28,639 --> 00:16:34,749 a value 1 at the point x 1 of k and 0 at all other points. N 2 of k should have a value 90 00:16:34,749 --> 00:16:41,679 1 at the point x 2 of k 0; at all other points, N 3 of k should have a value 1 at the point 91 00:16:41,679 --> 00:16:48,679 x 3 k and 0 at all other points. So, what we can write in short is, N i k as the point 92 00:16:51,309 --> 00:16:58,309 x i k is equal to 1 and at all other points j it is equal to 0. Set property is what we 93 00:17:14,290 --> 00:17:19,429 want to have for this function. So, if we want to construct N 1 k with this property 94 00:17:19,429 --> 00:17:23,110 such that N 1 k is 0 at the point x 2 k and x 3 k. 95 00:17:23,110 --> 00:17:30,110 So, how will I form the function N 1 k? This should minus at the point x 2 k, certainly, 96 00:17:36,130 --> 00:17:43,130 this representation should have this form, something into is there, then at x 2 k this 97 00:17:47,880 --> 00:17:54,880 expression goes to 0, it should also vanish at the point x 3 k. So, it should also have 98 00:17:56,500 --> 00:18:03,429 this part sitting there. You see that this expression vanishes at the point 2 and the 99 00:18:03,429 --> 00:18:09,750 point 3 of the element; now what else is required at the point 1? This expression should have 100 00:18:09,750 --> 00:18:16,750 a value 1. So, what I will do is simply divide by, what I have done in the bottom part, I 101 00:18:33,559 --> 00:18:40,559 have replaced x with x 1 k; you see what has happened, this is a quadratic expression, 102 00:18:40,929 --> 00:18:45,460 it vanishes at the point where you wanted to vanish and it has the value 1 at the point 103 00:18:45,460 --> 00:18:52,460 where we want it to have a value 1. So, by the same token, can we define N 2 k. 104 00:18:54,850 --> 00:19:01,850 We are defining this thing only in the element phi k. This will be equal to what? Very simple; 105 00:19:08,130 --> 00:19:14,889 it will be x minus, it should vanish at the point 1, and 2 is going to be x minus x 1 106 00:19:14,889 --> 00:19:21,889 k x minus not 1 and 2, but 1 and 3 x 3 k, and it should have a value 1 at the point 107 00:19:24,279 --> 00:19:31,279 x 2 k. So, it will be x 2 k minus x 1 k x 2 k minus x 3 k. 108 00:19:40,610 --> 00:19:47,610 Similarly, N 3 k by the same token, can be written as x minus x 1 k into x minus x 2 109 00:19:54,299 --> 00:20:01,299 k divided by x 3 k minus x 1 k into x 3 k minus x 2 k. So, we have constructed the three 110 00:20:15,519 --> 00:20:20,320 functions at the element level; this satisfy the three requirements we have imposed as 111 00:20:20,320 --> 00:20:27,320 far as the values are concerned. Now, what do we note? That if I look at the function 112 00:20:31,960 --> 00:20:38,960 phi 2 k minus 1 of x, where is this functions, and this is for we have to differentiate between 113 00:20:48,929 --> 00:20:55,220 the two things, this function 2 k minus 1 corresponds to an edge vertex, that is a node 114 00:20:55,220 --> 00:21:00,500 which is shared by two elements; that is, this corresponds to if k is 1, it is 1. So, 115 00:21:00,500 --> 00:21:07,500 it is 1, 3, 5, 7, 9 this function has to be non-zero in the two elements sharing that 116 00:21:08,490 --> 00:21:15,490 node. So, this will be equal to N 3 in the element k minus 1 for it will be N 1 in element 117 00:21:39,100 --> 00:21:46,100 k; hence, it will be 0 everywhere else. So, what you said is this function, if I talk 118 00:22:06,980 --> 00:22:13,980 of this node 2 k minus 1; this is element I k minus 1; this is element I k. So this, 119 00:22:26,840 --> 00:22:33,840 the function is given as N 3 of k minus 1, which we now know how to construct and here 120 00:22:38,990 --> 00:22:44,289 N 1 in k. 121 00:22:44,289 --> 00:22:51,289 So, this we can construct, all the global of basis function, odd ones using these N 122 00:22:56,429 --> 00:23:02,070 I k(s) defining the element; what about the phi 2 k that is the even one? So, the even 123 00:23:02,070 --> 00:23:09,070 one which you see are defined only in the element. So, the even one - the phi 2 k k 124 00:23:11,879 --> 00:23:18,879 is equal to 1 2, the number of elements, these are sometimes called internal bubble functions 125 00:23:28,100 --> 00:23:35,100 and they are only defined over element I k, that is, this is my element I k. So, this 126 00:23:58,019 --> 00:24:05,019 is by the 2 k node and phi 2 k is non-zero in this element and 0 everywhere else. So, 127 00:24:16,769 --> 00:24:23,769 what we know now that phi 2 k is equal to by definition N 2 of element k in element 128 00:24:29,590 --> 00:24:35,799 I k and 0 everywhere else. 129 00:24:35,799 --> 00:24:42,799 So, these, now you see that when we went, we came from the linear to the quadratic. 130 00:24:50,110 --> 00:24:55,129 The construction has been become a little involved; nevertheless, the steps that we 131 00:24:55,129 --> 00:25:00,460 are going to follow will be the same, irrespective whether it is linear, quadratic, and as we 132 00:25:00,460 --> 00:25:05,289 will see cubic forth order or fifth order, any order that we would like to have. Now, 133 00:25:05,289 --> 00:25:09,999 the question is that - are these functions linearly independent? So, once I have defined 134 00:25:09,999 --> 00:25:16,999 these functions phi i, then if you remember, we had written u 9 of x is equal to sigma 135 00:25:21,139 --> 00:25:28,139 i is equal to 1 to 9 u i phi i of x. How do we check linear independence? If you remember, 136 00:25:34,720 --> 00:25:41,720 linear independence means that if this function is identically equal to 0, then the coefficient 137 00:25:42,350 --> 00:25:48,730 u i should come out to be equal to 0, how do I show that the coefficient c y r 0? So, 138 00:25:48,730 --> 00:25:55,730 forgot, let us see a very important property of this function; what are these phi i? If 139 00:25:55,999 --> 00:26:02,990 you remember this phi i has a value 1 at the node with respect to all the point, with respect 140 00:26:02,990 --> 00:26:09,149 to they have been defined, and at all other points that we have put in the domain, these 141 00:26:09,149 --> 00:26:14,039 functions are 0. So, if this function u 9 of x has to be 0 142 00:26:14,039 --> 00:26:21,039 everywhere, it has to be 0 at the point x i also, is also equal to 0, which is equal 143 00:26:27,909 --> 00:26:34,820 to from the summation, what do we get? All other phi i(s) will vanish, only the u i will 144 00:26:34,820 --> 00:26:40,789 remain. This will be equal to u i. So, if I do this for all the 9 nodes, we are having 145 00:26:40,789 --> 00:26:46,090 the domain, what do I get that from this condition? That u 9 is equal to 0 everywhere, we get 146 00:26:46,090 --> 00:26:53,090 that all the coefficient u i are 0, which means, that the phi i, are... this will be 147 00:27:00,179 --> 00:27:07,179 called piecewise quadratic; that is, in every element they are quadratic and linearly independent. 148 00:27:16,549 --> 00:27:23,549 What about completeness? So, for the completeness - if we go, and what do we like, want to do? 149 00:27:32,379 --> 00:27:39,379 We would like to see, whether a function of this type, if the function f of x, this function 150 00:27:41,749 --> 00:27:48,749 is given as the generic quadratic a 0 plus a 1 x plus a 2 x square; if I have this function, 151 00:27:56,559 --> 00:28:03,559 then I should be able to represent it as the linear combination of a phi i; that is, I 152 00:28:06,879 --> 00:28:13,879 will call it as alpha i phi i. Now, what is this? Because, we are saying these approximating 153 00:28:15,960 --> 00:28:22,960 functions are piecewise quadratic for completeness, we want the series represented by these functions 154 00:28:24,559 --> 00:28:30,399 to exactly represent at least a quadratic polynomial, not more than that; so, either 155 00:28:30,399 --> 00:28:33,820 I have, if you see, this is, these are arbitrary coefficient. 156 00:28:33,820 --> 00:28:39,649 So, this series has to exactly represent the constant represented by a 0 or a linear a 157 00:28:39,649 --> 00:28:46,649 0 plus a 1 x or a quadratic, which is a 0 plus a 1 x plus a 2 x square. So, if it has 158 00:28:47,690 --> 00:28:53,450 to do it in the whole domain, then it should also be able to represent a quadratic in an 159 00:28:53,450 --> 00:28:58,019 element, so in the element if I write... So, instead of doing it in the whole domain, let 160 00:28:58,019 --> 00:29:05,019 us do it in the element. So, in the element, if I want to write f x it is equal to, I will 161 00:29:06,909 --> 00:29:13,909 call it alpha 1 of k N 1 of k plus alpha 2 of k N 2 of k plus alpha 3 of k N 3 of k, 162 00:29:26,220 --> 00:29:33,220 and this, let say that the quadratic polynomial that we have taken in the element, it is of 163 00:29:33,269 --> 00:29:40,269 this form beta 0 plus beta 1 x plus beta 2 x square. 164 00:29:43,009 --> 00:29:49,340 Now, by the various functions have been defined, what will be the alpha 1 of k be? Alpha 1 165 00:29:49,340 --> 00:29:56,340 of k, where the equal to f evaluated at the point x 1 of k; alpha 2 of k will be equal 166 00:30:03,409 --> 00:30:10,409 to f evaluated at the point x 2 of k; alpha 3 of an element k will be equal to f, evaluated 167 00:30:16,320 --> 00:30:23,320 at the point x 3 of k. So, with this representation for the alpha, one should go and check that 168 00:30:27,909 --> 00:30:34,909 indeed, what we have done in the previous stage is really... I do not know, whether 169 00:30:39,009 --> 00:30:45,649 this is going to be like this. So, the equalities is question mark and what we can now check 170 00:30:45,649 --> 00:30:49,230 is that is not a question mark, but indeed this is an equality. 171 00:30:49,230 --> 00:30:55,759 This is a simple algebraic exercise, and one can show that, yes, indeed just linear combination 172 00:30:55,759 --> 00:31:02,210 of these three functions is able to exactly use a quadratic polynomial defined in the 173 00:31:02,210 --> 00:31:08,929 elements and thus in the whole domain, which means that this function phi i, the set of 174 00:31:08,929 --> 00:31:15,929 function phi i represented in the complete set. So, this in the construction of this 175 00:31:22,159 --> 00:31:26,809 phi i is... one has to be very careful, one cannot arbitrarily define this function phi 176 00:31:26,809 --> 00:31:33,489 i as we wish. Now, if you see this point x, another point 177 00:31:33,489 --> 00:31:40,489 we have to note is that the point x 2 of k that we have taken, x point x 2 of k, we have 178 00:31:43,789 --> 00:31:48,649 said it is at the mid side, it is not been, it could be anywhere; but as the convention, 179 00:31:48,649 --> 00:31:55,649 it is taken at the middle of the element always, but one should remember it need not be the 180 00:31:56,869 --> 00:32:03,619 case, specially when we have solve element, which we will be dealing with later on. 181 00:32:03,619 --> 00:32:10,619 Now, that we have defined the phi i, what next? So, once we have defined the phi i, 182 00:32:12,009 --> 00:32:19,009 the next thing that we have to do is to go ahead and find the coefficient u i, in order 183 00:32:20,629 --> 00:32:27,629 to form the solution to the problem. So, to form the coefficient where do we start from? 184 00:32:31,179 --> 00:32:38,179 We start from the weak formulation, and for the problem that we have taken, the weak form 185 00:32:40,220 --> 00:32:47,220 is very easy, it is x is equal to 0 to L, it will be EA d u dx f v dx plus p into v 186 00:33:03,340 --> 00:33:10,340 evaluated at L. Now, in weak form what do we do? We go and 187 00:33:14,899 --> 00:33:21,899 substitute, instead of u 9. So, this will be replaced by u 9 and to form the i th equation, 188 00:33:31,940 --> 00:33:38,940 what we do? We replace v with phi i. So, to make the notations a little bit compact, this 189 00:33:46,429 --> 00:33:52,710 part as we have done a few lectures ago, they are going to represent as the bi-linear form, 190 00:33:52,710 --> 00:33:58,499 as I said it is linear in u and v. So, it will be called as the bi-linear form given 191 00:33:58,499 --> 00:34:05,499 by this compact notation b u v; and this side, this we are going to call by F of u, which 192 00:34:06,320 --> 00:34:13,320 is the linear function and given by the name F of u, because here as the number of functions 193 00:34:16,600 --> 00:34:21,909 increase, the algebra get tedious. So, we would like to be compact only for that reason, 194 00:34:21,909 --> 00:34:28,909 we are introducing these things. So, the i th equation is given as B of u 9 phi i is 195 00:34:42,460 --> 00:34:46,200 equal to F of phi i. 196 00:34:46,200 --> 00:34:53,200 So, now if I have this, now I can expand the u 9. So, what will that be equal to? I can 197 00:35:00,760 --> 00:35:07,760 write it as B of sigma i is equal to 1 to 9 u i u not i - we will call it j - u j, phi 198 00:35:23,730 --> 00:35:30,730 i is equal to F of phi i. Now, since this is bi-linear, by the definition of bi-linear 199 00:35:33,710 --> 00:35:38,410 functions this summation over this coefficient I can take out. So, this is equivalent to 200 00:35:38,410 --> 00:35:45,410 writing form of j equal to 1 to 9 b of phi j phi i whole thing multiplied by u j is equal 201 00:35:59,410 --> 00:36:06,410 to F of phi i. If you see for the i th equation, this becomes, 202 00:36:09,980 --> 00:36:16,980 this term becomes the coefficient of the j th unknown. So, what is it called? If you 203 00:36:26,270 --> 00:36:33,270 want to go back to what we have to done earlier, this term will be equal to k for the i th 204 00:36:34,470 --> 00:36:41,470 equation j th call. So, this is k i j. Now, our job is... So, if this is k i j, then we 205 00:36:42,230 --> 00:36:48,700 can write the whole problem in the form of the global stiffness matrix k into the unknown 206 00:36:48,700 --> 00:36:55,339 displacement vector u, where now this is k is 9 by 9, u is the vector of phi 9 this is 207 00:36:55,339 --> 00:36:59,280 equal to the vector F. 208 00:36:59,280 --> 00:37:06,280 So, our job is to find the entries of this phase and of this vector. So, if we can find 209 00:37:15,569 --> 00:37:21,490 anything, then we have assembled to make this problem and all that we have to do is invert 210 00:37:21,490 --> 00:37:25,880 this matrix; first, we have to apply one-way condition, invert the matrix and solution 211 00:37:25,880 --> 00:37:32,880 is there. So, to do this, what do we do? We have to again go back to the element calculation; 212 00:37:34,329 --> 00:37:41,329 so, this is element level calculation. Now, let me redraw the whole figures again 213 00:37:53,280 --> 00:38:00,280 for the element; this is x 1 of an element k, x 2 of an element k, x 3 of an element 214 00:38:03,559 --> 00:38:10,559 k and if you remember, so this was N 1, N 2, N 3; so, this was N 1 of an element k, 215 00:38:14,809 --> 00:38:21,809 which is equivalent to phi 2 k minus 1. This is N 2 of element k, which is equivalent to 216 00:38:31,660 --> 00:38:38,660 phi 2 k, and this one is N 3 of element k, which is equivalent to phi 2 k plus 1 in the 217 00:38:46,690 --> 00:38:53,440 element k. Just like we have done for the linear problem for the phi equal to 1 approximation, 218 00:38:53,440 --> 00:38:59,299 here what do we know, that the phi is which are non-zero in the element are these three 219 00:38:59,299 --> 00:39:06,299 phi(s), phi 2 k minus 1, phi 2 k and phi 2 k plus 1 equal to 1, 2, 3, 4 that is a number 220 00:39:06,390 --> 00:39:13,390 of up to the number of element. So, what element is going to contribute to, 221 00:39:13,849 --> 00:39:20,849 is the 2 k minus 1 th equation, 2 k th equation, and 2 k plus 1 th equation, and which rows 222 00:39:21,549 --> 00:39:27,319 of these equation are going to be non-zero, which column of these rows are going to be 223 00:39:27,319 --> 00:39:34,319 non-zero? The columns which are going to be non-zero for each of these rows are the 2 224 00:39:35,069 --> 00:39:41,720 k minus 1, 2 k and 2 k plus 1. So, these columns only, the element is going to contribute, 225 00:39:41,720 --> 00:39:48,720 and none of the other column it is going to contribute. So, if I take that b b, that I 226 00:39:51,640 --> 00:39:58,640 had written phi j phi i, which is nothing but as we have said k i j. 227 00:40:01,569 --> 00:40:07,440 Which of the i and which of the j will get contribution from the element? As we have 228 00:40:07,440 --> 00:40:13,690 said, i equal to 2 k minus 1, i equal to 2 k and i equal to 2 k plus 1. Similarly, j 229 00:40:13,690 --> 00:40:18,789 is equal to 2 k minus 1, j equal to 2 k and j is equal to 2 k plus 1. So, let us find 230 00:40:18,789 --> 00:40:24,130 the contribution, how we had found these contributions earlier. 231 00:40:24,130 --> 00:40:29,339 So, what we are going to define now, it is for this element k. The element stiffness 232 00:40:29,339 --> 00:40:35,579 matrix - now what will be the size of this element stiffness matrix? Because there are 233 00:40:35,579 --> 00:40:42,579 now 3 n(s) or 3 phi(s), which are non-zero in the element. So, it will be a 3 by 3 matrix. 234 00:40:43,640 --> 00:40:50,640 So, what will the entries be? I will write only of few entries of this and then it will 235 00:40:54,180 --> 00:41:01,180 be integral form x 1 k 236 00:41:06,650 --> 00:41:13,650 to x 3 k EA d N 1 k dx; this I will check. So, for the first row of this, I will EA d 237 00:41:24,119 --> 00:41:31,119 N 1 k dx into d N1 k dx will be the first column entry EA d N 1 k dx into d N 2 k dx, 238 00:41:33,049 --> 00:41:40,049 as the second column entry and the third will be d N 1 k dx into d N 3 k dx is the third 239 00:41:42,400 --> 00:41:45,210 column entry. And similarly, if I come here to the second 240 00:41:45,210 --> 00:41:52,210 row, so it will be x 3 k EA d N 2 k dx d N 1 k dx and so on; and the third will have 241 00:42:11,930 --> 00:42:18,930 integer x 1 k to x 3 k; you see that limits of the integration are from one x exterminate 242 00:42:19,630 --> 00:42:24,940 of the element to next exterminate of the element. So, that is should be kept in mind 243 00:42:24,940 --> 00:42:31,940 d N 3 k dx d N 1 k dx d x. Can I write the k i j in the element k? This will be equal 244 00:42:46,000 --> 00:42:53,000 to integral x 1 of k to x 3 of k of EA d N j k dx d N i k dx d x. 245 00:43:12,000 --> 00:43:19,000 Similarly, F i in the element k will be equal to nothing but x 1 of k to x 3 of k f N i 246 00:43:26,809 --> 00:43:33,109 k dx; remember, that here we are not going to add the part due to the p that is going 247 00:43:33,109 --> 00:43:38,789 to be added, when we are applying the or the force found the condition; this is all that 248 00:43:38,789 --> 00:43:45,789 the F i k from the element will have. So, once we have this, then we have obtained the 249 00:43:46,750 --> 00:43:52,549 element level stiffness matrix and the element level load vector. So, this will go to the 250 00:43:52,549 --> 00:43:59,549 element load vector. So, once I have these, now what we do? We 251 00:44:10,890 --> 00:44:16,619 have to find the correspondence between these element level entries and the global entries; 252 00:44:16,619 --> 00:44:23,619 that is, where in the global stiffness matrix should be this k i j go globally? It is which 253 00:44:27,780 --> 00:44:34,780 m n it should go to, that is where I have to assemble these things. Similarly, where 254 00:44:36,349 --> 00:44:43,349 will be the F i at the element go to globally, which F this equation has this information, 255 00:44:49,799 --> 00:44:50,960 we have to help. 256 00:44:50,960 --> 00:44:56,130 So, what has been done earlier, as far as this information is concerned, we had created 257 00:44:56,130 --> 00:45:03,130 a local to global numbering or 90 times it is called the connectivity information. So, 258 00:45:22,930 --> 00:45:29,930 this connectivity information has to be constructed. So, let us now do it for the elements. So, 259 00:45:42,190 --> 00:45:49,079 what I will do is, here I write the global number, and here, I will write the corresponding 260 00:45:49,079 --> 00:45:56,079 local number for the element k. So, this is the local number K in the element; I have 261 00:46:01,309 --> 00:46:06,450 how many? 1, 2, 3; this global number, does it correspond to 1? Corresponds to which global 262 00:46:06,450 --> 00:46:11,950 basis function? N 1 k corresponds to phi 2 k minus 1. So, it should correspond to globally. 263 00:46:11,950 --> 00:46:18,950 I will re-write it in another form, 2 k minus 1 plus 1 which is nothing but 2 k minus 1. 264 00:46:21,960 --> 00:46:28,960 This one will be N 2 corresponds to phi 2 k globally; so, it will be as... rewrite as 265 00:46:33,609 --> 00:46:40,609 this plus 2; similarly, N 3 k will corresponds to phi 2 k plus 1. So, remember what we have 266 00:46:48,309 --> 00:46:53,720 done. So, before we go and do the element calculation, we will stretch the entries of 267 00:46:53,720 --> 00:46:58,680 the global stiffness matrix and the global load vector equal to 0. 268 00:46:58,680 --> 00:47:04,260 The process of the assembly that we had talked about in great detail, in the previous lecture, 269 00:47:04,260 --> 00:47:11,260 essentially is adding the elemental contributions to this global stiffness matrix and the global 270 00:47:11,339 --> 00:47:17,280 load vector, in order to complete the global equation, because what we have said, is integration 271 00:47:17,280 --> 00:47:23,390 over the full domain can be partitioned into the integration over the elements, and when 272 00:47:23,390 --> 00:47:27,730 we add these elemental integrals together, we get the global representation. 273 00:47:27,730 --> 00:47:34,730 So, the global entry which is going to get contribution from element are going to be: 274 00:47:35,670 --> 00:47:42,670 k 2 k minus 1 plus i, 2 k minus 1 plus j is equal to... what we are going to do? It is 275 00:47:53,619 --> 00:48:00,619 to the original number, which was sitting there, I am going to add k i j. So, what has 276 00:48:11,079 --> 00:48:18,079 happened to the global stiffness entry k 2 k minus 1 plus i k 2 k minus 1 plus j? I am 277 00:48:21,099 --> 00:48:28,099 going to add the elemental i j th. So, this is the part, as the assembly part, as far 278 00:48:28,460 --> 00:48:35,460 as the stiffness is concerned and this for i, j is equal to 1, 2, 3. Similarly, the F 279 00:48:40,549 --> 00:48:47,549 2 k minus 1 plus i will be equal to F 2 k minus plus i 1, to this I am going to add 280 00:48:55,210 --> 00:49:02,210 F k i. So, this is the assembly procedure. So, we have obtained the element calculation, 281 00:49:17,160 --> 00:49:22,010 using the function which we have defined the elemental level, and assemble them to get 282 00:49:22,010 --> 00:49:26,849 the global equation, in terms of the basis function that we are after, the coefficient 283 00:49:26,849 --> 00:49:33,849 for the global basis function. One should remember, that in any finite element approximation, 284 00:49:34,430 --> 00:49:40,180 we are really looking for the coefficients of the global basis function, and the element 285 00:49:40,180 --> 00:49:47,180 calculation, is a simple way of doing the integration, relevant that have to be done, 286 00:49:49,890 --> 00:49:54,930 to obtain the entries of the global stiffness matrix and the global load vector by doing 287 00:49:54,930 --> 00:49:59,210 at the element level, adding it up to get the global equation. 288 00:49:59,210 --> 00:50:06,210 One question that should come to mind now is - earlier, we have seen for P equal to 289 00:50:06,970 --> 00:50:13,970 1, that the i th equation had only non-zero entries corresponding to the i minus 1 column, 290 00:50:15,710 --> 00:50:22,710 the i th column and the i plus 1 th column; in this case, what are the non-zero entries 291 00:50:23,700 --> 00:50:26,160 for the i th row? 292 00:50:26,160 --> 00:50:33,160 So, let us check; now take the i th row; I will take this one as the 2 k minus 1 th point. 293 00:50:42,039 --> 00:50:49,039 So, this is 2 k minus 3. Here we have to differentiate between this nodal function and the internal 294 00:50:56,730 --> 00:51:03,520 bubble function. So, let us take the nodal function that we have. So, if you see that 295 00:51:03,520 --> 00:51:10,520 I am talking on the equation corresponding to this one. So, I am corresponding to the 296 00:51:14,770 --> 00:51:21,770 equation corresponding to this function which is phi 2 k minus 1. 297 00:51:24,799 --> 00:51:31,799 This is the phi that we are going to contribute to the equation corresponding to this. Certainly, 298 00:51:32,240 --> 00:51:39,240 it is going to be this phi, it is going to be this phi and this phi. So, this is going 299 00:51:48,250 --> 00:51:55,250 to be phi 2 k minus 3, this is phi 2 k minus 2, this is phi 2 k and this is phi 2 k plus 300 00:52:05,839 --> 00:52:12,839 1. What can you see? That for the 2 k minus 1 th equation will have non-zero entries corresponding 301 00:52:16,720 --> 00:52:23,720 to the 2 k minus 3 column 2 k minus 2 column 2 k minus 1 column 2 k column and 2 k plus 302 00:52:26,079 --> 00:52:33,079 1 column. So, how many columns will be non-zero? 1, 2, 3, 4, 5; so five columns will have at 303 00:52:34,950 --> 00:52:41,950 least, not at least, at most five columns will have non-zero entries; for all other 304 00:52:43,210 --> 00:52:47,720 phi(s) the column entries are going to be 0, and because all other phi(s) are going 305 00:52:47,720 --> 00:52:53,880 to disappear or become 0 in the region, where this phi 2 k minus 1 is defined to be. 306 00:52:53,880 --> 00:53:00,880 You see that if right global stiffness matrix, the number of non-zero entries has increased 307 00:53:05,200 --> 00:53:11,760 in a row. This is the another property of these approximation that we have going to 308 00:53:11,760 --> 00:53:18,339 see and how they reflect in terms of the part city of the global stiffness matrix and that 309 00:53:18,339 --> 00:53:23,619 also this part city, we are talking about, we have talk about solver. So, this we will 310 00:53:23,619 --> 00:53:30,619 tackle at a later date. See, there is the very curious thing about this function phi 311 00:53:31,750 --> 00:53:38,750 k; this function phi k, if you add them up anywhere, you see their value is equal to 312 00:53:54,640 --> 00:53:58,240 1; why is it so? 313 00:53:58,240 --> 00:54:05,240 Remember, we had said that these functions phi k should be able to exactly represent 314 00:54:07,589 --> 00:54:14,589 any polynomial up to a quadratic. So, a constant is also global polynomial, which should be 315 00:54:15,099 --> 00:54:20,720 represented exactly, and in constant I will take my function f x, which I want to represent 316 00:54:20,720 --> 00:54:27,720 to be 1. This will be equal to sigma u i phi i. Now, is this function is equal to 1 everywhere 317 00:54:34,970 --> 00:54:41,770 in the domain? It has to be 1 at the points x 1, x 2, x 3, x 9. So, if I go to the point 318 00:54:41,770 --> 00:54:48,770 x i, this is still equal to 1, but then this will be equal to u i, which tells me that 319 00:54:53,670 --> 00:55:00,670 the u i here on the right hand side are all equal to 1 is the u i are all 1, then what 320 00:55:02,029 --> 00:55:05,539 we get is some of the phi i is equal to 1 and this is also true in the element; same 321 00:55:05,539 --> 00:55:12,539 thing applies in the element. So, in the element, some of the shape function will be i is equal 322 00:55:13,769 --> 00:55:20,769 to 1 to 3 is equal to 1; this is a property is that we should keep in mind, and what it 323 00:55:25,140 --> 00:55:32,140 also tells us that it implies, that if I do this summation, this has to be equal to 0, 324 00:55:37,579 --> 00:55:42,980 because this sum is equal to constant that delivered the constant is 0; these are certain 325 00:55:42,980 --> 00:55:48,289 things, which we should have in mind. So that when we go and write the computer program 326 00:55:48,289 --> 00:55:52,769 to do the finite element analysis, we can check all function, these basis functions 327 00:55:52,769 --> 00:55:57,490 of the element shape functions that we have called, which have N i is are nothing but 328 00:55:57,490 --> 00:56:04,490 the element shape function, we can check whether they have been correctly program or not. 329 00:56:07,390 --> 00:56:14,390 The definition of this element shape function that we have used, this definition is called 330 00:56:19,940 --> 00:56:26,940 Lagrangian shape or basis function; we will see how to generalize these, to go for a keys 331 00:56:44,079 --> 00:56:45,880 order approximation in future 332 00:56:45,880 --> 00:56:52,880 So, this Lagrangian basis functions that we have constructed, are note, the only one that 333 00:56:54,089 --> 00:57:01,089 we can have. So, those who are interested, there is also a very popular family of shape 334 00:57:03,369 --> 00:57:10,369 functions that are basis function, that can be used is Legendre function, but we are not 335 00:57:11,039 --> 00:57:18,039 going to discuss it, here in this analysis. In the next class, what we are going to do 336 00:57:20,789 --> 00:57:27,319 is, we are going to look at the cubic approximation, the P equal to 3 approximation, also construct 337 00:57:27,319 --> 00:57:34,319 basis function, how to do the elemental calculation, how to go for the load vector and from there 338 00:57:34,369 --> 00:57:39,089 we will generalize to the P for the approximation. And once you have done the P for the approximation, 339 00:57:39,089 --> 00:57:44,950 then we can do a finite element computation using any order of approximation in the domain. 340 00:57:44,950 --> 00:57:51,069 After that, we will see how they are going to benefit us; just doing them is not enough, 341 00:57:51,069 --> 00:57:58,069 why we should do them, how they are going to benefit us is going to be in the future lecture.Thank you. 342