WEBVTT 00:00:00.000 --> 00:00:10.780 Hello, I`m Andrés Papuel Bacerri, Professor of the Department of Mechanical Medicine and 00:00:10.780 --> 00:00:13.100 I`ll give you structures. 00:00:13.100 --> 00:00:19.600 Imagine for a moment that this is a column of a structure, and I`m compressing the column, 00:00:19.600 --> 00:00:24.240 but it comes a moment when it suddenly appears a large formation of the column is doubled. 00:00:24.240 --> 00:00:25.740 What happens to the column? 00:00:25.740 --> 00:00:31.140 The column has been failed by Pandeo, which is a phenomenon of geometric instability. 00:00:31.140 --> 00:00:38.500 In fact, in this video, I`m going to talk about the Pandeo and in particular about the theory of Euler. 00:00:38.500 --> 00:00:42.940 The Pandeo is a phanomenal that is catastrophic for any kind of structure, 00:00:42.940 --> 00:00:49.740 as it is patent in these images here, Pandeo of different columns, Pandeo of a complete structure, 00:00:49.740 --> 00:00:55.240 as it is deposited, or even of the cells that are obtained by deposited. 00:00:55.240 --> 00:01:01.140 The result of the parenthesis that remains is this, that you can deduce the expression of the Pandeo`s critical load, 00:01:01.140 --> 00:01:05.140 according to the theory of Euler, differentiating between the forces of first and second order, 00:01:05.140 --> 00:01:07.740 constating the alcance of this theory. 00:01:07.740 --> 00:01:11.980 For this, the first thing we`re going to do is to plant some hypotheses, 00:01:11.980 --> 00:01:14.540 deduce the critical load of Pandeo of Euler, 00:01:14.540 --> 00:01:20.340 I`ll plant some piece of space and, finally, I`ll establish a series of conclusions. 00:01:20.340 --> 00:01:27.740 In terms of hypotheses, Euler, who was in the 18th century, studied the Pandeo of a submeted column of compression, 00:01:27.740 --> 00:01:34.340 and Euler studied this phenomenon from what is known as the Euler column or the ideal column. 00:01:34.340 --> 00:01:36.540 The hypotheses are these here. 00:01:36.540 --> 00:01:42.440 Degenerate of behavior, the lineal elastic, section and constant material, without imperfections, 00:01:42.440 --> 00:01:48.440 a perfectly straight piece, residual tension, the load applied in the center of gravity of the section 00:01:48.440 --> 00:01:51.340 and the extremes with the possibility of spin. 00:01:51.340 --> 00:01:54.940 This is the calculation model of the Euler column. 00:01:54.940 --> 00:01:59.240 Well, to deduce the expression of the critical load of Pandeo of Euler, 00:01:59.240 --> 00:02:03.040 here we put a coordinate system of XI in this column, 00:02:03.040 --> 00:02:07.940 and if we study the column of a submeted compression under the theory of first order, 00:02:07.940 --> 00:02:12.640 that is, maintaining the beginning of the small deformations, 00:02:12.640 --> 00:02:14.340 we have to cut the column, 00:02:14.340 --> 00:02:18.440 and that the effort in any section is a compression expansion. 00:02:18.440 --> 00:02:22.540 Well, what happens that in the Pandeo phenomenon there are no small deformations, 00:02:22.540 --> 00:02:26.240 but there is a flexion and a large deformations. 00:02:26.240 --> 00:02:32.640 So here, you have to explain the behavior of this column through the theory of second order. 00:02:32.640 --> 00:02:41.640 In this theory, what is it that is to plant the equilibrium of what is the column or any part of it, 00:02:41.640 --> 00:02:44.040 having the configuration formed. 00:02:44.040 --> 00:02:47.440 In this case, if we cut here by a section S, 00:02:47.440 --> 00:02:52.940 located at a distance of X from the initial stream of the load of the column, 00:02:52.940 --> 00:02:55.940 we have that the forces that appear in this section of short, 00:02:55.940 --> 00:02:57.940 it is a compression of P, 00:02:57.940 --> 00:03:02.140 but also a flexing moment of less than P per I. 00:03:02.140 --> 00:03:02.940 What is I? 00:03:02.940 --> 00:03:07.440 And is the lateral displacement of any section of the load. 00:03:07.440 --> 00:03:11.740 Well, in this case, we are going to take this moment 00:03:11.740 --> 00:03:15.640 and we are going to take here also this differential equation. 00:03:15.640 --> 00:03:18.640 The second derivative is equal to M divided by E per I, 00:03:18.640 --> 00:03:21.940 which is the differential equation of the elastic line. 00:03:21.940 --> 00:03:24.240 E per I, which is the flexing rigidity. 00:03:24.240 --> 00:03:27.040 If we combine these expressions 00:03:27.040 --> 00:03:31.040 and we denominate that a constant that is divided by E per I, 00:03:31.040 --> 00:03:33.940 we obtain this differential equation, 00:03:33.940 --> 00:03:37.640 which is linear, homogeneous and efficient constants. 00:03:37.640 --> 00:03:39.640 Which is the solution of this equation? 00:03:39.640 --> 00:03:41.040 Because the solution of this equation, 00:03:41.040 --> 00:03:43.240 because, in addition, it is also usual in other disciplines, 00:03:43.240 --> 00:03:45.040 as dynamic, is this, 00:03:45.040 --> 00:03:48.040 and of X equal to sub 1 by x, 00:03:48.040 --> 00:03:50.540 plus sub 2 by x. 00:03:50.540 --> 00:03:54.040 Well, this expression of I of X is not complete 00:03:54.040 --> 00:03:56.840 without determining the values of the constants, 00:03:56.840 --> 00:03:58.040 sub 1 and sub 2. 00:03:58.040 --> 00:04:00.540 For this, we are going to impose the conditions of the contour. 00:04:00.540 --> 00:04:04.740 We are going to remember the aspect of the column of E 1. 00:04:04.740 --> 00:04:07.740 Well, the first contour condition is this here. 00:04:07.840 --> 00:04:10.440 When X is equal to zero, the displacement of the zero side is zero, 00:04:10.440 --> 00:04:13.440 and this leads us to that the constant of sub 2 is zero. 00:04:13.440 --> 00:04:15.940 The following condition of contour is this here. 00:04:15.940 --> 00:04:19.640 When X is equal to L, the displacement of the side is also zero, 00:04:19.640 --> 00:04:24.040 and then, to replace these values in the expression of I of X, 00:04:24.040 --> 00:04:27.040 we have to make C sub 2 zero, 00:04:27.040 --> 00:04:29.040 the second sum of that appears. 00:04:29.040 --> 00:04:33.140 Well, the null of this product is determined by 00:04:33.140 --> 00:04:34.540 or well that C sub 1 zero, 00:04:34.540 --> 00:04:36.240 this would not have a trivial solution, 00:04:36.240 --> 00:04:39.040 we would have that the bar is not panned, 00:04:39.040 --> 00:04:40.040 which would be absurd. 00:04:40.040 --> 00:04:44.440 And then, the second sum, the second factor, 00:04:44.440 --> 00:04:47.140 is the caporele that has to be equal to zero. 00:04:47.140 --> 00:04:49.140 We are going to look at it from now on. 00:04:49.140 --> 00:04:51.640 Well, here what happens? 00:04:51.640 --> 00:04:53.640 That already with the panned column, 00:04:53.640 --> 00:04:56.140 with this phenomenon of deflection, 00:04:56.140 --> 00:04:58.640 in which this instability is produced, 00:04:58.640 --> 00:05:01.640 we have that the caporele is equal to zero, 00:05:01.640 --> 00:05:05.340 if the caporele is a multiple of P. 00:05:05.440 --> 00:05:08.240 Well, being n zero or one or two up to infinity. 00:05:08.240 --> 00:05:12.540 The set of values of p is what the panned columns form. 00:05:12.540 --> 00:05:16.740 That is, and of X is C sub 1 of np. 00:05:16.740 --> 00:05:21.140 If we represent n for the five first values, 00:05:21.140 --> 00:05:26.240 we have that if n is equal to zero, the column remains correct. 00:05:26.240 --> 00:05:29.240 If n is equal to 1, the form, as you can see there, 00:05:29.240 --> 00:05:32.240 a loop, if n is equal to two, formed by the loops, 00:05:32.340 --> 00:05:35.640 and so, successively, until n is equal to infinity. 00:05:35.640 --> 00:05:39.340 Well, as it is, the balance condition in the panned column, 00:05:39.340 --> 00:05:41.140 is caporele equal to np. 00:05:41.140 --> 00:05:44.140 Here we wait for K and we go to the square, 00:05:44.140 --> 00:05:47.140 and we go to the square and it remains np divided by the square. 00:05:47.140 --> 00:05:51.140 And if we remember that k square is that same P divided by e by i, 00:05:51.140 --> 00:05:54.640 we have the value of the critical load of Euler 00:05:54.640 --> 00:05:56.640 for a certain panned column, 00:05:56.640 --> 00:05:59.640 that is, np divided by the square by e by i. 00:06:00.040 --> 00:06:06.340 Well, now we remember that Euler`s column is an ideal column, 00:06:06.340 --> 00:06:10.540 except there is no other element that limits, 00:06:10.540 --> 00:06:14.540 well, it is a lateral displacement of any of its sections, 00:06:14.540 --> 00:06:16.740 and therefore, at the time of the panned, 00:06:16.740 --> 00:06:20.440 it is, to say, forced to panned, 00:06:20.440 --> 00:06:23.040 according to the first mode, that is, n is equal to 1. 00:06:23.040 --> 00:06:26.540 In this case, the value of critical load is p squared by e, 00:06:26.540 --> 00:06:28.140 divided by the square. 00:06:28.240 --> 00:06:32.540 If we represent this expression, it appears this parabola, 00:06:32.540 --> 00:06:36.740 where it can be seen that the critical load of Euler has infinite, 00:06:36.740 --> 00:06:40.040 in the case that the length of the column is 0, 00:06:40.040 --> 00:06:43.240 which means that when the column is smaller, 00:06:43.240 --> 00:06:46.040 first, it will prepare the panned column by rotation, 00:06:46.040 --> 00:06:48.440 by displacement, what by panned. 00:06:48.440 --> 00:06:50.740 And it will happen just the opposite, 00:06:50.740 --> 00:06:53.640 if the length is going up every time higher, 00:06:53.640 --> 00:06:57.140 that is, at the end, it will prepare the panned column`s panned column. 00:06:57.240 --> 00:07:00.040 Well, we must remember that the column, 00:07:00.040 --> 00:07:07.040 the inertia moments of the main axes of the transversal section 00:07:07.040 --> 00:07:09.040 do not have to necessarily be equal, 00:07:09.040 --> 00:07:10.740 it is like this rectangular section. 00:07:10.740 --> 00:07:13.040 Then, the inertia moment, 00:07:13.040 --> 00:07:16.040 which will give the lower critical load, 00:07:16.040 --> 00:07:19.340 will be the minimum between e and uv and uvz. 00:07:21.340 --> 00:07:23.740 Well, in continuation, 00:07:23.740 --> 00:07:25.240 I will ask you the following question. 00:07:25.340 --> 00:07:27.840 If the column of the figure is there, 00:07:27.840 --> 00:07:30.040 the conditions of the contour will be modified, 00:07:30.040 --> 00:07:33.440 do you think it would be the critical load of panned Euler or not? 00:07:33.440 --> 00:07:37.140 The new conditions of the column`s contour are these that you have here. 00:07:37.140 --> 00:07:40.140 That is, here we have that in the central section, 00:07:40.140 --> 00:07:44.140 it has restricted its lateral displacement. 00:07:44.140 --> 00:07:46.840 Here you have the critical load of Euler, 00:07:46.840 --> 00:07:49.940 the expression that we ended up dedicating, 00:07:49.940 --> 00:07:52.940 and now here you have the options. 00:07:52.940 --> 00:07:54.540 It will not vary, it will increase the double, 00:07:54.540 --> 00:07:56.540 it will decrease three times or it will increase four times. 00:07:59.540 --> 00:08:01.840 If it will vary the critical load of Euler, 00:08:01.840 --> 00:08:02.840 I will try it. 00:08:05.640 --> 00:08:08.140 Almost, it will increase, although it is not the double, 00:08:08.140 --> 00:08:09.140 I will try it. 00:08:11.640 --> 00:08:13.840 No, the critical load of Euler will not decrease. 00:08:16.440 --> 00:08:17.140 Correct. 00:08:17.140 --> 00:08:19.840 The critical load of Euler will increase four times. 00:08:19.840 --> 00:08:20.540 Why? 00:08:20.540 --> 00:08:22.840 Well, because the new panned column`s situation 00:08:22.940 --> 00:08:24.240 with the column, 00:08:24.240 --> 00:08:26.240 with these new conditions of the contour, 00:08:26.240 --> 00:08:27.740 is the one you see here. 00:08:27.740 --> 00:08:29.740 Then, look here, 00:08:29.740 --> 00:08:31.940 in the expression of the critical load of Euler, 00:08:31.940 --> 00:08:34.740 the length of the column is elevated square. 00:08:34.740 --> 00:08:36.740 Then, if we decrease it to half, 00:08:36.740 --> 00:08:39.540 the critical load increases four times. 00:08:41.540 --> 00:08:45.040 In terms of conclusions of this video, 00:08:45.040 --> 00:08:49.340 well, the first is that the panned is a phenomenon of this geometric speed, 00:08:49.340 --> 00:08:52.440 which has a column, especially when it is built, 00:08:52.540 --> 00:08:55.740 when it has a very elevated latitude, 00:08:55.740 --> 00:08:57.740 and when it is measured in compression. 00:08:57.740 --> 00:09:00.140 The Euler theory is explained in the phenomenon of the panned, 00:09:00.140 --> 00:09:02.240 however, this theory is ideal, 00:09:02.240 --> 00:09:02.740 why? 00:09:02.740 --> 00:09:04.740 Well, because it speaks of pieces and imperfections 00:09:04.740 --> 00:09:06.540 that this is measured only in compression 00:09:06.540 --> 00:09:09.540 and in reality, then it can see other exports 00:09:09.540 --> 00:09:12.740 acting on any section of the column. 00:09:13.240 --> 00:09:15.740 Well, in the Euler theory, the critical load 00:09:15.740 --> 00:09:17.540 goes up to the step of stable balance, 00:09:17.540 --> 00:09:18.540 to stable balance, 00:09:18.540 --> 00:09:19.540 what does this mean? 00:09:19.540 --> 00:09:20.740 Well, first of all, 00:09:20.840 --> 00:09:23.340 if the current load on the column is inferior 00:09:23.340 --> 00:09:25.340 to the critical load of Euler, 00:09:25.340 --> 00:09:28.340 what we have here is that it is formed by panned, 00:09:28.340 --> 00:09:29.540 but when removing the load, 00:09:29.540 --> 00:09:34.540 it can be recovered by its initial form. 00:09:34.540 --> 00:09:36.540 If the load is still the critical load of Euler, 00:09:36.540 --> 00:09:38.540 the column stays in this way, 00:09:38.540 --> 00:09:40.940 if it has what is known as equilibrium in Euler, 00:09:40.940 --> 00:09:43.340 which in practice does not represent itself. 00:09:43.340 --> 00:09:48.540 Finally, if the current load is greater than the critical load of panned, 00:09:48.540 --> 00:09:52.140 which we have calculated according to this expression that we have deduced, 00:09:52.140 --> 00:09:56.740 then what we have here is a situation of equilibrium in the stable. 00:09:56.740 --> 00:09:59.740 Well, with this end of this video, 00:09:59.740 --> 00:10:01.340 talking about the critical load of Euler 00:10:01.340 --> 00:10:03.740 and I hope you have been useful. 00:10:03.740 --> 00:10:04.740 See you next time.