1 00:00:00,040 --> 00:00:02,470 The following content is provided under a Creative 2 00:00:02,470 --> 00:00:03,880 Commons license. 3 00:00:03,880 --> 00:00:06,920 Your support will help MIT OpenCourseWare continue to 4 00:00:06,920 --> 00:00:10,570 offer high quality educational resources for free. 5 00:00:10,570 --> 00:00:13,470 To make a donation or view additional materials from 6 00:00:13,470 --> 00:00:17,400 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:17,400 --> 00:00:18,650 ocw.mit.edu. 8 00:00:21,350 --> 00:00:23,430 PROFESSOR: Ladies and gentlemen, welcome to this 9 00:00:23,430 --> 00:00:26,030 lecture on nonlinear finite element analysis of solids and 10 00:00:26,030 --> 00:00:27,190 structures. 11 00:00:27,190 --> 00:00:29,390 In this lecture, I would like to continue with our 12 00:00:29,390 --> 00:00:33,280 discussion of the use of constitutive relations used in 13 00:00:33,280 --> 00:00:35,020 nonlinear analysis. 14 00:00:35,020 --> 00:00:37,130 In the previous lecture, we already considered the 15 00:00:37,130 --> 00:00:41,300 modeling of elastic and hyper elastic materials, subjected 16 00:00:41,300 --> 00:00:45,470 to large displacements, large rotations, and large strains. 17 00:00:45,470 --> 00:00:47,980 I'd like to not turn our attention to the use of the 18 00:00:47,980 --> 00:00:49,680 updated Lagrangian formulation. 19 00:00:49,680 --> 00:00:52,280 Of course, the kinematic relations for the total and 20 00:00:52,280 --> 00:00:55,260 updated Lagrangian formulations, we discussed in 21 00:00:55,260 --> 00:00:57,470 an earlier lecture already. 22 00:00:57,470 --> 00:01:00,920 In particular, I would like now, in very general terms, to 23 00:01:00,920 --> 00:01:02,970 ask the following question-- 24 00:01:02,970 --> 00:01:06,330 is it possible to obtain, using the updated Lagrangian 25 00:01:06,330 --> 00:01:11,740 formulation, identically the same numerical results for 26 00:01:11,740 --> 00:01:16,890 each load step and each iteration, as are obtained 27 00:01:16,890 --> 00:01:19,720 when we use a total Lagrangian formulation? 28 00:01:19,720 --> 00:01:23,020 Let's look at this question here, now, in 29 00:01:23,020 --> 00:01:24,560 quite gentle terms. 30 00:01:24,560 --> 00:01:29,930 And I have rephrased it here once more, so as to really 31 00:01:29,930 --> 00:01:32,010 make very clear what I mean. 32 00:01:32,010 --> 00:01:36,960 Let's say that we have a Program 1 that uses only the 33 00:01:36,960 --> 00:01:39,800 total Lagrangian formulation. 34 00:01:39,800 --> 00:01:43,130 And the constitutive relations that we are employing in that 35 00:01:43,130 --> 00:01:46,620 total Lagrangian formulation are defined as shown here. 36 00:01:46,620 --> 00:01:49,690 So total stress at time, t, of course, second Piola-Kirchhoff 37 00:01:49,690 --> 00:01:53,890 stress, is given as a function of displacements. 38 00:01:53,890 --> 00:01:56,970 And the tangent material relationship, relating the 39 00:01:56,970 --> 00:01:59,470 increment in the second Piola-Kirchhoff stress and the 40 00:01:59,470 --> 00:02:01,860 increment in the Green-Lagrange strain is as 41 00:02:01,860 --> 00:02:04,280 shown here. 42 00:02:04,280 --> 00:02:07,720 The program results for a particular physical problem 43 00:02:07,720 --> 00:02:11,610 might look systematically as shown here. 44 00:02:11,610 --> 00:02:16,690 Of course, these two relationships here have been 45 00:02:16,690 --> 00:02:19,900 obtained from a set of physical laboratory 46 00:02:19,900 --> 00:02:22,480 experiments. 47 00:02:22,480 --> 00:02:24,900 This is how Program 1 works. 48 00:02:24,900 --> 00:02:29,900 Now we have also, say, a Program 2. 49 00:02:29,900 --> 00:02:33,130 And this Program 2 only uses the updated Lagrangian 50 00:02:33,130 --> 00:02:35,070 formulation. 51 00:02:35,070 --> 00:02:39,210 Now, remember our laboratory physical experimental results 52 00:02:39,210 --> 00:02:42,590 have given us the considered affiliations for the towards 53 00:02:42,590 --> 00:02:44,370 Lagrangian formulation. 54 00:02:44,370 --> 00:02:51,500 However, we now have a program that only is operating on the 55 00:02:51,500 --> 00:02:52,890 updated Lagrangian formulation. 56 00:02:52,890 --> 00:02:57,270 The constitutive relations are, in this case, well, like 57 00:02:57,270 --> 00:03:00,860 that and like that, where I have given here three dots, 58 00:03:00,860 --> 00:03:04,470 because we want to now study what they should be looking 59 00:03:04,470 --> 00:03:11,010 like in order to obtain the same results as we are 60 00:03:11,010 --> 00:03:14,200 attaining with Program 1. 61 00:03:14,200 --> 00:03:18,880 In other words, how can we obtain, with this Program 2, 62 00:03:18,880 --> 00:03:22,110 identically the same results as are obtained 63 00:03:22,110 --> 00:03:24,180 using Program 1. 64 00:03:24,180 --> 00:03:28,530 This is a question that we really are asking. 65 00:03:28,530 --> 00:03:32,680 To answer this question, we have to go back to some 66 00:03:32,680 --> 00:03:35,410 derivations that we discussed in an earlier lecture. 67 00:03:35,410 --> 00:03:38,250 And there, we discussed that this is here the governing 68 00:03:38,250 --> 00:03:41,190 continuum mechanics equation, corresponding to the total 69 00:03:41,190 --> 00:03:42,930 Lagrangian formulation. 70 00:03:42,930 --> 00:03:47,310 Here we have the incremental stress strain law going into 71 00:03:47,310 --> 00:03:48,500 the formulation. 72 00:03:48,500 --> 00:03:52,280 Notice this is a tangent material tensor. 73 00:03:52,280 --> 00:03:56,170 Here we have the second Piola-Kirchhoff stress, and 74 00:03:56,170 --> 00:03:57,610 here, as well. 75 00:03:57,610 --> 00:03:59,650 And of course, here on the right hand side, we have to 76 00:03:59,650 --> 00:04:05,100 total external loads entering the total 77 00:04:05,100 --> 00:04:07,350 external virtual work. 78 00:04:07,350 --> 00:04:11,100 In the updated Lagrangian formulation, this is the 79 00:04:11,100 --> 00:04:13,300 equation that we derived earlier. 80 00:04:13,300 --> 00:04:16,279 Here, again, is a total external virtual work. 81 00:04:16,279 --> 00:04:19,959 And here, a constitutive relation enters, corresponding 82 00:04:19,959 --> 00:04:21,959 to this formulation. 83 00:04:21,959 --> 00:04:27,290 And here we have the Cauchy stress that is being used in 84 00:04:27,290 --> 00:04:30,150 the updated Lagrangian formulation. 85 00:04:30,150 --> 00:04:33,720 The terms used in the formulations are summarized on 86 00:04:33,720 --> 00:04:35,090 this view graph. 87 00:04:35,090 --> 00:04:37,750 In the TL formulation, of course, we're integrating over 88 00:04:37,750 --> 00:04:39,870 the original volume. 89 00:04:39,870 --> 00:04:42,560 In the updated Lagrangian formulation, we're integrating 90 00:04:42,560 --> 00:04:45,520 over the current volume, time t. 91 00:04:45,520 --> 00:04:48,450 And there is this transformation, where the mass 92 00:04:48,450 --> 00:04:53,890 density's at time t, and the original mass density, enters. 93 00:04:53,890 --> 00:04:58,660 Notice here are listed the linear and the nonlinear 94 00:04:58,660 --> 00:05:03,280 increment in the Green-Lagrange strain. 95 00:05:03,280 --> 00:05:06,620 Here, there also is a linear and a nonlinear increment in 96 00:05:06,620 --> 00:05:08,635 the Green-Lagrange, but for the updated Lagrangian 97 00:05:08,635 --> 00:05:12,990 formulation, and here, for the total Lagrangian formulation. 98 00:05:12,990 --> 00:05:15,210 There's a relationship between these 99 00:05:15,210 --> 00:05:18,980 increments, as shown here. 100 00:05:18,980 --> 00:05:22,580 And this relationship I will focus our attention upon 101 00:05:22,580 --> 00:05:25,760 further on the next view graph, because I'd like you to 102 00:05:25,760 --> 00:05:29,530 clearly understand how we obtain this relationship. 103 00:05:29,530 --> 00:05:34,530 But we notice is that the deformation gradient enters in 104 00:05:34,530 --> 00:05:38,610 this equation, as well as in this equation. 105 00:05:38,610 --> 00:05:41,460 The variations on these linear strain increments and 106 00:05:41,460 --> 00:05:45,550 nonlinear strain increments of the total Lagrangian 107 00:05:45,550 --> 00:05:48,660 formulation and the updated Lagrangian formulation are 108 00:05:48,660 --> 00:05:52,550 also related via the deformation gradient. 109 00:05:52,550 --> 00:05:55,110 In fact, twice the deformation gradient enters here. 110 00:05:57,740 --> 00:06:01,870 Let's look at how we derive this kinematic relationships. 111 00:06:01,870 --> 00:06:05,430 Well, we use that there is a fundamental property of 112 00:06:05,430 --> 00:06:10,330 Green-Lagrange strain as expressed in this equation. 113 00:06:10,330 --> 00:06:14,540 If you know a Green-Lagrange strain at a point, then we can 114 00:06:14,540 --> 00:06:19,630 pick any material fiber with these components. 115 00:06:19,630 --> 00:06:24,200 And we know, if these are the origin of components of the 116 00:06:24,200 --> 00:06:31,160 material fiber, at that point, that after motion, in other 117 00:06:31,160 --> 00:06:35,720 words, in configuration t, corresponding to time t, that 118 00:06:35,720 --> 00:06:38,590 material fiber has this length, where the original 119 00:06:38,590 --> 00:06:41,810 length is given here. 120 00:06:41,810 --> 00:06:45,960 This is a very basic equation that we want to study a bit 121 00:06:45,960 --> 00:06:48,500 more on the next view graph. 122 00:06:48,500 --> 00:06:52,230 But let me already point out that we can, directly, use 123 00:06:52,230 --> 00:06:55,370 this basic relation, and, of course, apply it also 124 00:06:55,370 --> 00:06:57,590 corresponding to time t, plus delta t. 125 00:06:57,590 --> 00:07:01,110 Notice here, t, here, t plus delta t. 126 00:07:01,110 --> 00:07:03,320 Here t, here, t plus delta t. 127 00:07:03,320 --> 00:07:06,220 Otherwise, same relationship. 128 00:07:06,220 --> 00:07:09,720 And we also can apply incrementally. 129 00:07:09,720 --> 00:07:13,750 Notice this a Green-Lagrange strain increment, from time t, 130 00:07:13,750 --> 00:07:15,950 to time t plus delta t. 131 00:07:15,950 --> 00:07:18,230 But refer to time t. 132 00:07:18,230 --> 00:07:20,940 That's why we have the little t down here. 133 00:07:20,940 --> 00:07:24,340 Up here, we always referred the increments in strains, or 134 00:07:24,340 --> 00:07:27,220 the total strains, if it is a total strain, to the 135 00:07:27,220 --> 00:07:29,040 configuration at time zero. 136 00:07:29,040 --> 00:07:32,450 Now we are referring the quantity to time t. 137 00:07:32,450 --> 00:07:38,130 And, if we simply apply this same basic relationship here, 138 00:07:38,130 --> 00:07:41,570 we directly obtain this equation here for the 139 00:07:41,570 --> 00:07:44,630 increment in Green-Lagrange strain from time, t, to time, 140 00:07:44,630 --> 00:07:46,240 t plus delta t. 141 00:07:46,240 --> 00:07:51,460 Refer to time t configuration, this relationship holds. 142 00:07:51,460 --> 00:07:53,810 Let's look at this relationship here. 143 00:07:53,810 --> 00:07:56,590 What does it really mean physically? 144 00:07:56,590 --> 00:07:59,100 Well, it means the following-- 145 00:07:59,100 --> 00:08:02,890 here we have our stationary coordinate frame, x1, x2, x3. 146 00:08:02,890 --> 00:08:06,830 And at time zero, we have a material fiber here, any 147 00:08:06,830 --> 00:08:07,760 material fiber. 148 00:08:07,760 --> 00:08:09,450 But let's just pick one. 149 00:08:09,450 --> 00:08:13,780 And that has a length of 0 ds. 150 00:08:13,780 --> 00:08:20,650 The components of this vector, the vector that is shown here, 151 00:08:20,650 --> 00:08:25,570 of length 0 ds, the components are shown here. 152 00:08:25,570 --> 00:08:29,515 And that vector we call a d 0 x. 153 00:08:34,030 --> 00:08:41,030 This material fiber moves from time 0 to time t, to this 154 00:08:41,030 --> 00:08:43,110 configuration. 155 00:08:43,110 --> 00:08:46,540 The new components are listed here-- 156 00:08:46,540 --> 00:08:50,500 dtx2, dtx1, dtx3. 157 00:08:50,500 --> 00:08:53,020 Of course, there's a relationship between the new 158 00:08:53,020 --> 00:08:56,290 components and the original components of 159 00:08:56,290 --> 00:08:57,390 the material fiber. 160 00:08:57,390 --> 00:09:02,660 And here, goes into the deformation gradient. 161 00:09:02,660 --> 00:09:08,410 Now the equation that I just referred to really relates 162 00:09:08,410 --> 00:09:11,930 this length and that length. 163 00:09:11,930 --> 00:09:16,220 And the way we obtain it is to look at the definition of the 164 00:09:16,220 --> 00:09:16,890 Green-Lagrange strain. 165 00:09:16,890 --> 00:09:21,210 strain The Green-Lagrange strain is defined as follows-- 166 00:09:21,210 --> 00:09:23,200 I'm going to simply write it in here. 167 00:09:27,150 --> 00:09:28,400 You see here, we have-- 168 00:09:30,710 --> 00:09:31,040 oops. 169 00:09:31,040 --> 00:09:33,500 I should have not to use that delta t, because I'm 170 00:09:33,500 --> 00:09:34,750 looking at time t. 171 00:09:40,840 --> 00:09:43,410 This is the equation for the Green-Lagrange strain, from 172 00:09:43,410 --> 00:09:45,040 here to there. 173 00:09:45,040 --> 00:09:49,120 And what I'm now going to do is to premultiply this 174 00:09:49,120 --> 00:09:52,450 relationship by d0x, transposed, and 175 00:09:52,450 --> 00:09:55,400 postmultiplied by d0x. 176 00:09:55,400 --> 00:09:58,040 Of course, I have to do the same thing here. 177 00:09:58,040 --> 00:10:02,220 d0x, transposed, goes in front of the bracket. 178 00:10:02,220 --> 00:10:06,170 And d0x goes into the back of the bracket. 179 00:10:06,170 --> 00:10:08,820 Let me make clear what I'm doing here. 180 00:10:08,820 --> 00:10:10,550 I'm putting this in here. 181 00:10:10,550 --> 00:10:12,930 And I'm putting that on the back. 182 00:10:12,930 --> 00:10:17,380 Notice I'm using a transposed here and a transposed there. 183 00:10:17,380 --> 00:10:19,980 The basic relation of the Green-Lagrange strain, once 184 00:10:19,980 --> 00:10:26,990 again, is this being equal to that part with out the d0x 185 00:10:26,990 --> 00:10:31,330 transposed in front, and without the d0x on the back. 186 00:10:31,330 --> 00:10:39,450 But now we notice, directly, that this part here and this 187 00:10:39,450 --> 00:10:42,985 part here is nothing else then dtx. 188 00:10:45,630 --> 00:10:51,750 If you recognize that this is dtx, that is dtx, you 189 00:10:51,750 --> 00:10:56,200 immediately get on the right hand side, one half, 190 00:10:56,200 --> 00:11:01,290 bracket open, d-- 191 00:11:01,290 --> 00:11:07,150 well, I should say t ds, squared, minus 0ds squared. 192 00:11:07,150 --> 00:11:09,310 That's what you have on the right hand side. 193 00:11:09,310 --> 00:11:11,710 And on the left hand side, you have what 194 00:11:11,710 --> 00:11:12,550 I showed you earlier. 195 00:11:12,550 --> 00:11:16,950 Let me go back one view graph, just to make sure that we 196 00:11:16,950 --> 00:11:18,270 understand each other. 197 00:11:18,270 --> 00:11:21,040 On the right hand side, you immediately obtain this term 198 00:11:21,040 --> 00:11:22,830 minus that term. 199 00:11:22,830 --> 00:11:27,070 And on the left hand side, you end up with this term. 200 00:11:27,070 --> 00:11:31,570 So this is the relationship, a very basic relationship, that 201 00:11:31,570 --> 00:11:34,960 we want to use in our derivation. 202 00:11:34,960 --> 00:11:37,450 And it is obtained by simply looking at the Green-Lagrange 203 00:11:37,450 --> 00:11:41,840 strain, the way it's defined, and the way we have defined it 204 00:11:41,840 --> 00:11:45,200 in an earlier lecture, and premultiplying by d0x 205 00:11:45,200 --> 00:11:48,580 transposed, and postmultiplying by d0x. 206 00:11:48,580 --> 00:11:52,240 Of course, what you're seeing there, in that relationship I 207 00:11:52,240 --> 00:11:57,250 showed you, you see the components of this term here, 208 00:11:57,250 --> 00:12:01,660 written out in terms of i and j components. 209 00:12:01,660 --> 00:12:05,470 Well, so what we are doing, then, is to follow a material 210 00:12:05,470 --> 00:12:11,700 fiber, original length 0 ds, through it's motion to t ds 211 00:12:11,700 --> 00:12:15,950 and to t plus delta t ds. 212 00:12:15,950 --> 00:12:19,640 And that's expressed by the three formulas that are on the 213 00:12:19,640 --> 00:12:21,490 previous view graph. 214 00:12:21,490 --> 00:12:28,220 Hence, if we subtract from the second formula, the first 215 00:12:28,220 --> 00:12:34,060 formula, we directly obtain this relationship here. 216 00:12:34,060 --> 00:12:39,850 And we now use this term here, or this equation, linking up 217 00:12:39,850 --> 00:12:46,980 dtx to d0x and expand these two terms 218 00:12:46,980 --> 00:12:49,530 to obtain this equation. 219 00:12:49,530 --> 00:12:53,170 Now, notice that this equation has to hold for 220 00:12:53,170 --> 00:12:55,160 any material fiber. 221 00:12:55,160 --> 00:12:59,620 And if it does hold for any material fiber, it immediately 222 00:12:59,620 --> 00:13:03,660 follows that this equation here has to hold it, because 223 00:13:03,660 --> 00:13:07,760 we can apply it to a material fiber that is oriented along 224 00:13:07,760 --> 00:13:11,590 the x1 axis, then another one that is oriented along the x2 225 00:13:11,590 --> 00:13:14,400 axis, then another one that is oriented along the x3 226 00:13:14,400 --> 00:13:16,430 axis, and so on. 227 00:13:16,430 --> 00:13:22,380 And surely, this relationship here must then hold. 228 00:13:22,380 --> 00:13:27,020 If we look at this relationship more closely, we 229 00:13:27,020 --> 00:13:31,710 see that we can write it out as a linear strain increment 230 00:13:31,710 --> 00:13:35,210 and as a nonlinear strain increment on the left hand 231 00:13:35,210 --> 00:13:39,610 side and on the right hand side of the equation. 232 00:13:39,610 --> 00:13:43,910 And this shows, then, directly that the 0 eij must be equal 233 00:13:43,910 --> 00:13:49,240 to twice a deformation gradient times ters. 234 00:13:49,240 --> 00:13:53,400 The reason being that this term here is a function of the 235 00:13:53,400 --> 00:13:55,990 incremental displacement only. 236 00:13:55,990 --> 00:13:59,610 Linear functional incremental displacement, so is this. 237 00:13:59,610 --> 00:14:03,710 And if you simply equate those terms that are linear, we get 238 00:14:03,710 --> 00:14:06,670 that this part must be equal to that part. 239 00:14:06,670 --> 00:14:10,970 And if you create all the nonlinear parts, we find that 240 00:14:10,970 --> 00:14:15,260 0ei, e to ij, is equal to twice the deformation gradient 241 00:14:15,260 --> 00:14:17,090 times te to rs. 242 00:14:17,090 --> 00:14:19,520 And those two equations are given down here. 243 00:14:22,250 --> 00:14:25,750 If we now take variations on the left hand side and on the 244 00:14:25,750 --> 00:14:29,830 right hand side, and if you recognize that's a variation, 245 00:14:29,830 --> 00:14:33,590 is respect to the configuration at time t plus 246 00:14:33,590 --> 00:14:39,220 delta t, it is clear that the deformation gradient terms 247 00:14:39,220 --> 00:14:40,680 don't change. 248 00:14:40,680 --> 00:14:43,430 And, therefore taking the variation on the left hand 249 00:14:43,430 --> 00:14:49,910 side, that variation is simply applied to the ters, with 250 00:14:49,910 --> 00:14:51,570 these being constant terms. 251 00:14:51,570 --> 00:14:56,730 Similarly here, and this then, completes the derivation of 252 00:14:56,730 --> 00:14:59,170 all the kinematic relationships that I had in 253 00:14:59,170 --> 00:15:03,230 the earlier table and that we want to use now. 254 00:15:03,230 --> 00:15:08,000 In addition, we have, also, the important relationship 255 00:15:08,000 --> 00:15:10,705 between the second Piola-Kirchhoff stress and the 256 00:15:10,705 --> 00:15:11,720 Cauchy stress. 257 00:15:11,720 --> 00:15:13,910 And that relationship is shown here. 258 00:15:13,910 --> 00:15:18,720 Of course, we introduced that one in an earlier lecture. 259 00:15:18,720 --> 00:15:21,910 Finally, we have a relationship between the 260 00:15:21,910 --> 00:15:25,540 tangent material tensor, corresponding to the total 261 00:15:25,540 --> 00:15:29,090 Lagrangian formulation and corresponding to the updated 262 00:15:29,090 --> 00:15:30,770 Lagrangian formulation. 263 00:15:30,770 --> 00:15:33,340 Notice zero here, t there. 264 00:15:33,340 --> 00:15:36,460 And that relationship is given here. 265 00:15:36,460 --> 00:15:40,070 It's given by a fourth order tensor transformation, which 266 00:15:40,070 --> 00:15:42,630 I'd like to now discuss with you further. 267 00:15:42,630 --> 00:15:45,610 This is an important relationship that we want to 268 00:15:45,610 --> 00:15:47,319 spend a bit off time on. 269 00:15:52,370 --> 00:15:56,000 The relationship is derived as follows-- 270 00:15:56,000 --> 00:16:00,580 the increment in the second Piola-Kirchhoff stress is 271 00:16:00,580 --> 00:16:03,700 clearly given by the tangent material relation times 272 00:16:03,700 --> 00:16:06,140 increment in the Green-Lagrange strain. 273 00:16:06,140 --> 00:16:09,620 And we're looking at differential increments here. 274 00:16:09,620 --> 00:16:13,180 The increment in the second Piola-Kirchhoff stress, refer 275 00:16:13,180 --> 00:16:18,020 to time t, is given by the right hand side here, where 276 00:16:18,020 --> 00:16:20,320 this increment in Green-Lagrange strain is 277 00:16:20,320 --> 00:16:23,050 refer to time t. 278 00:16:23,050 --> 00:16:27,800 We also have this relationship between the stress increments. 279 00:16:27,800 --> 00:16:32,020 This is the same relationship that holds between they total 280 00:16:32,020 --> 00:16:35,610 second Piola-Kirchhoff stress at time t and the Cauchy 281 00:16:35,610 --> 00:16:37,620 stress at time t. 282 00:16:37,620 --> 00:16:41,440 And we have this relationship, which we just derived. 283 00:16:41,440 --> 00:16:46,190 Now, using these four equations, we directly obtain 284 00:16:46,190 --> 00:16:52,430 by substitution, that this left hand side, substituted 285 00:16:52,430 --> 00:16:57,720 for d0 Sij, must be equal to this right hand side, where we 286 00:16:57,720 --> 00:16:59,320 have substituted this relationship 287 00:16:59,320 --> 00:17:02,600 for d0 epsilon rs. 288 00:17:02,600 --> 00:17:07,119 And this equation, then, directly yields this equation, 289 00:17:07,119 --> 00:17:11,829 where the coefficient here must be what we are looking 290 00:17:11,829 --> 00:17:14,920 for, namely tCabpq. 291 00:17:14,920 --> 00:17:19,660 This is the tangent material tensor to be used in the 292 00:17:19,660 --> 00:17:22,060 updated Lagrangian formulation, 293 00:17:22,060 --> 00:17:23,369 referred to time t. 294 00:17:26,510 --> 00:17:32,050 The tangent material relation is once more summarized here. 295 00:17:32,050 --> 00:17:35,805 And this is what we need to use in the updated Lagrangian 296 00:17:35,805 --> 00:17:37,055 formulation. 297 00:17:39,360 --> 00:17:43,840 Notice it's a fourth order tensor transformation on this 298 00:17:43,840 --> 00:17:47,050 value here, on this tensor here. 299 00:17:47,050 --> 00:17:50,135 Now we can compare the updated and total Lagrangian 300 00:17:50,135 --> 00:17:52,240 formulations. 301 00:17:52,240 --> 00:17:57,580 And the question that we ask is under what condition is 302 00:17:57,580 --> 00:18:00,780 this left hand side, which we use in the total Lagrangian 303 00:18:00,780 --> 00:18:06,060 formulation equal to this term here, which we use in the 304 00:18:06,060 --> 00:18:07,990 updated Lagrangian formulation. 305 00:18:07,990 --> 00:18:15,470 Well, we simply substitute here for the stress, and for 306 00:18:15,470 --> 00:18:19,490 the strain, our earlier relationships. 307 00:18:19,490 --> 00:18:22,820 We multiply out, and immediately, we obtain the 308 00:18:22,820 --> 00:18:24,110 right hand side. 309 00:18:24,110 --> 00:18:28,300 Notice ij, of course, are dummy indices, which can be 310 00:18:28,300 --> 00:18:31,520 substituted by m and n. 311 00:18:31,520 --> 00:18:37,410 It's interesting to upset here that this transformation is a 312 00:18:37,410 --> 00:18:41,190 kinematic transformation, which is actually buried in 313 00:18:41,190 --> 00:18:44,280 the proper use of the updated Lagrangian formulation. 314 00:18:44,280 --> 00:18:49,150 This transformation is a stress transformation that has 315 00:18:49,150 --> 00:18:53,560 to be enforced for any Cauchy stress that we would be 316 00:18:53,560 --> 00:18:55,720 calculating in the computer program. 317 00:18:59,320 --> 00:19:02,560 Let us look at the second term. 318 00:19:02,560 --> 00:19:05,600 Here, we have, on the left hand side, the term of the 319 00:19:05,600 --> 00:19:07,320 total Lagrangian formulation. 320 00:19:07,320 --> 00:19:12,280 And we ask whether this is equal to the term that I 321 00:19:12,280 --> 00:19:14,320 showed you on the right hand side, corresponding to the 322 00:19:14,320 --> 00:19:15,570 updated Lagrangian formulation. 323 00:19:17,280 --> 00:19:20,890 Well, if we substitute our definitions for the second 324 00:19:20,890 --> 00:19:24,940 Piola-Kirchhoff stress here, and what we know to be a true 325 00:19:24,940 --> 00:19:30,940 kinematic relations, here, and we multiply out, and also use, 326 00:19:30,940 --> 00:19:36,600 of course, this relationship here, we directly obtain what 327 00:19:36,600 --> 00:19:39,180 we know we have to obtain for the updated Lagrangian 328 00:19:39,180 --> 00:19:40,390 formulation. 329 00:19:40,390 --> 00:19:43,070 In other words, the left hand side is equal to 330 00:19:43,070 --> 00:19:44,850 the right hand side. 331 00:19:44,850 --> 00:19:48,940 Notice, again, the kinematic transformation here and the 332 00:19:48,940 --> 00:19:52,560 stress transformation here. 333 00:19:52,560 --> 00:19:58,380 Finally, we also have to look at this term, the term in 334 00:19:58,380 --> 00:20:00,310 which the material tensor enters. 335 00:20:00,310 --> 00:20:04,220 We ask again if this left hand side equal to the right hand 336 00:20:04,220 --> 00:20:07,860 side using the proper transformations. 337 00:20:07,860 --> 00:20:16,200 And if we substitute, as shown here, we directly observe yes, 338 00:20:16,200 --> 00:20:20,500 if the term of the total Lagrangian formulation is 339 00:20:20,500 --> 00:20:22,747 equal to the term of the updated Lagrangian 340 00:20:22,747 --> 00:20:26,700 formulation, provided, of course, we use the proper 341 00:20:26,700 --> 00:20:28,520 transformations. 342 00:20:28,520 --> 00:20:32,050 And here, in particular, the constitutive transformation 343 00:20:32,050 --> 00:20:35,430 that we discussed just a bit earlier. 344 00:20:35,430 --> 00:20:41,115 So we conclude then that the updated Lagrangian terms are 345 00:20:41,115 --> 00:20:45,240 identically equal to the total Lagrangian terms, provided we 346 00:20:45,240 --> 00:20:50,020 follow the transformation rules that I have summarized. 347 00:20:50,020 --> 00:20:55,640 This means that in the finite element analysis, if we use 348 00:20:55,640 --> 00:21:00,660 the same finite element interpolation functions in the 349 00:21:00,660 --> 00:21:03,330 updated Lagrangian formulation, as in the total 350 00:21:03,330 --> 00:21:04,820 Lagrangian formulation-- 351 00:21:04,820 --> 00:21:08,920 of course we have to use the same interpolation functions 352 00:21:08,920 --> 00:21:11,610 and the same finite element assumptions-- 353 00:21:11,610 --> 00:21:16,815 then, we can see directly that, for the total Lagrangian 354 00:21:16,815 --> 00:21:21,520 formulation with this equivalent equation and the 355 00:21:21,520 --> 00:21:25,320 updated Lagrangian formulation with this equivalent equation, 356 00:21:25,320 --> 00:21:29,940 that these matrices are identically equal. 357 00:21:29,940 --> 00:21:34,480 And that's a very important observation, a very important 358 00:21:34,480 --> 00:21:35,660 observation. 359 00:21:35,660 --> 00:21:39,585 It means, then, going back to the question that I asked 360 00:21:39,585 --> 00:21:43,550 earlier, regarding the use of Program 1 and Program 2, it 361 00:21:43,550 --> 00:21:51,240 means that, to summarize, if we use Program 2, then Program 362 00:21:51,240 --> 00:21:56,740 2 gives the same results as Program 1, provided the Cauchy 363 00:21:56,740 --> 00:22:01,120 stresses are calculated from this relationship. 364 00:22:01,120 --> 00:22:02,570 What does that mean? 365 00:22:02,570 --> 00:22:05,560 It means that the second Piola-Kirchhoff stresses of 366 00:22:05,560 --> 00:22:08,690 course are given by the relationships that were 367 00:22:08,690 --> 00:22:13,360 determined by physical laboratory results. 368 00:22:13,360 --> 00:22:18,140 Remember, we know how old the second Piola-Kirchhoff stress 369 00:22:18,140 --> 00:22:21,432 is defined as a function of deformations, of the 370 00:22:21,432 --> 00:22:23,070 deformations in the material. 371 00:22:23,070 --> 00:22:26,360 That was our assumption because that went into the use 372 00:22:26,360 --> 00:22:27,810 of Program 1. 373 00:22:27,810 --> 00:22:31,310 And we are asking now only how can we use Program 2, which 374 00:22:31,310 --> 00:22:35,190 only contains the updated Lagrangian formulation. 375 00:22:35,190 --> 00:22:41,230 So we know how St0Smn is defined. 376 00:22:41,230 --> 00:22:45,340 And if we know this quantity, we have to make this 377 00:22:45,340 --> 00:22:49,750 transformation to calculate the Cauchy stress 378 00:22:49,750 --> 00:22:52,380 in the Program 2. 379 00:22:52,380 --> 00:22:57,230 Similarly, we also remember, postulated that we know this 380 00:22:57,230 --> 00:23:01,330 tensor from the physical laboratory test results. 381 00:23:01,330 --> 00:23:02,840 We know this tensor. 382 00:23:02,840 --> 00:23:07,790 And what I'm saying now is that we have to transform this 383 00:23:07,790 --> 00:23:13,050 tensor, as shown here, to obtain this tensor. 384 00:23:13,050 --> 00:23:17,000 And it is this one that would go then into all our Program 2 385 00:23:17,000 --> 00:23:18,290 computations. 386 00:23:18,290 --> 00:23:24,170 If, in Program 2, we use this Cauchy stress, defined by the 387 00:23:24,170 --> 00:23:25,970 second Piola-Kirchhoff stress, as shown here. 388 00:23:25,970 --> 00:23:29,740 And this tangent material relation, you find by this 389 00:23:29,740 --> 00:23:33,510 tensor, which we know, then Program 2 will give 390 00:23:33,510 --> 00:23:37,670 identically the same results as Program 1. 391 00:23:37,670 --> 00:23:40,960 The kinematic relations that we talked earlier about are 392 00:23:40,960 --> 00:23:44,370 buried in the proper implementation of the updated 393 00:23:44,370 --> 00:23:46,030 Lagrangian formulation in Program 2. 394 00:23:49,030 --> 00:23:54,410 Conversely, if we say that the material relationship for 395 00:23:54,410 --> 00:23:59,550 Program 2 is given, say, in other words, the Cauchy stress 396 00:23:59,550 --> 00:24:03,110 is given and also the tangent material relationship is given 397 00:24:03,110 --> 00:24:06,340 from laboratory test results. 398 00:24:06,340 --> 00:24:12,790 Then you can surely show that Program 1 with the total 399 00:24:12,790 --> 00:24:15,540 Lagrangian formulation will give identically the same 400 00:24:15,540 --> 00:24:20,550 results, provided these transformations are performed 401 00:24:20,550 --> 00:24:21,640 in the program. 402 00:24:21,640 --> 00:24:24,530 Now, the Cauchy stress is given, the tangent material 403 00:24:24,530 --> 00:24:27,970 relationship is given, and what we have to do in the 404 00:24:27,970 --> 00:24:31,620 program, in every solution step, in every iteration, is 405 00:24:31,620 --> 00:24:34,830 to transform this Cauchy stress to the second 406 00:24:34,830 --> 00:24:38,160 Piola-Kirchhoff stress that's going to be used in Program 1. 407 00:24:38,160 --> 00:24:42,760 And we have to transform this tensor here, as shown here, to 408 00:24:42,760 --> 00:24:43,710 this tensor. 409 00:24:43,710 --> 00:24:46,810 And this tensor here, 0Cijs, is going to be 410 00:24:46,810 --> 00:24:48,060 used in Program 1. 411 00:24:51,260 --> 00:24:55,370 Therefore we can really conclude that's the choice of 412 00:24:55,370 --> 00:24:58,310 whether to use a total Lagrangian formulation or the 413 00:24:58,310 --> 00:25:04,660 updated Lagrangian formulation is based only merely on how 414 00:25:04,660 --> 00:25:10,700 effective computationally one formulation is over the other. 415 00:25:10,700 --> 00:25:13,680 And then we note now that the B matrix, the strain 416 00:25:13,680 --> 00:25:18,190 displacement matrix in the U.L. formulation contains less 417 00:25:18,190 --> 00:25:20,790 entries than the strain displacement matrix in the 418 00:25:20,790 --> 00:25:22,240 T.L. formulation. 419 00:25:22,240 --> 00:25:26,070 This means, of course, that the product, B transpose CB, 420 00:25:26,070 --> 00:25:30,430 which is used in the stiffness matrix, is cheaper to 421 00:25:30,430 --> 00:25:33,916 evaluate, less expensive to evaluate, in the updated 422 00:25:33,916 --> 00:25:34,170 Lagrangian formulation. 423 00:25:34,170 --> 00:25:37,490 Across So there's a plus here for the updated Lagrangian 424 00:25:37,490 --> 00:25:38,740 formulation. 425 00:25:40,270 --> 00:25:45,730 However, if the stress-strain law is available in terms of 426 00:25:45,730 --> 00:25:48,750 the second Piola-Kirchhoff stress, then it is most 427 00:25:48,750 --> 00:25:51,760 natural to use a total Lagrangian formulations. 428 00:25:51,760 --> 00:25:56,630 And this is the case when we analyze rubber type materials, 429 00:25:56,630 --> 00:25:59,270 for example we use the Mooney-Rivlin material law, 430 00:25:59,270 --> 00:26:01,120 which we discussed very briefly in the previous 431 00:26:01,120 --> 00:26:06,600 lecture, and, in particular, when we want to analyze 432 00:26:06,600 --> 00:26:11,910 inelastic response and we want to allow for large 433 00:26:11,910 --> 00:26:14,340 displacements and large rotations. 434 00:26:14,340 --> 00:26:16,860 but we have small strain conditions. 435 00:26:16,860 --> 00:26:20,760 I'd like to refer you now back to the previous lecture, where 436 00:26:20,760 --> 00:26:24,150 we made a big point out of the fact that the second 437 00:26:24,150 --> 00:26:27,520 Piola-Kirchhoff stress, the Green-Lagrange strains, are 438 00:26:27,520 --> 00:26:31,120 the components of these two tensors are invariant under 439 00:26:31,120 --> 00:26:36,160 rigid body rotation, and that that means that we can 440 00:26:36,160 --> 00:26:39,280 directly use material relationships that are 441 00:26:39,280 --> 00:26:43,780 applicable to infinitesimal displacement and strains, in 442 00:26:43,780 --> 00:26:45,420 other words, the engineering stress and 443 00:26:45,420 --> 00:26:47,680 engineering strain variables. 444 00:26:47,680 --> 00:26:50,720 You can use these material relations directly in a large 445 00:26:50,720 --> 00:26:55,330 displacement, large rotation analysis, provided we use the 446 00:26:55,330 --> 00:26:57,090 total Lagrangian formulation. 447 00:26:57,090 --> 00:26:59,870 And I explained that, I spent a bit of time on that, in the 448 00:26:59,870 --> 00:27:00,870 previous lecture. 449 00:27:00,870 --> 00:27:04,600 Please refer back to that information. 450 00:27:04,600 --> 00:27:08,340 Let us now look at a very special case, namely, the case 451 00:27:08,340 --> 00:27:10,410 of elasticity. 452 00:27:10,410 --> 00:27:13,200 We discussed in the previous lecture already that one way 453 00:27:13,200 --> 00:27:16,990 to proceed is to use this relationship here in the total 454 00:27:16,990 --> 00:27:18,910 Lagrangian formulation, where we have the second 455 00:27:18,910 --> 00:27:21,460 Piola-Kirchhoff stress on the left hand side, the 456 00:27:21,460 --> 00:27:24,530 Green-Lagrange strain on the right hand side, multiplying 457 00:27:24,530 --> 00:27:26,730 the constitutive relation, corresponding to the total 458 00:27:26,730 --> 00:27:28,630 Lagrangian formulation. 459 00:27:28,630 --> 00:27:33,450 Now if we use this equation in the relation that we just 460 00:27:33,450 --> 00:27:36,870 established, or that we talked about, namely giving us of a 461 00:27:36,870 --> 00:27:43,380 Cauchy stress, in terms of the second Piola-Kirchhoff stress, 462 00:27:43,380 --> 00:27:47,260 we obtain this equation here. 463 00:27:47,260 --> 00:27:49,482 And of course, we also have to use in the updated Lagrangian 464 00:27:49,482 --> 00:27:51,300 formulation, this 465 00:27:51,300 --> 00:27:54,060 transformation, as we just discussed. 466 00:27:54,060 --> 00:27:56,770 Now notice that when we compare these two right hand 467 00:27:56,770 --> 00:28:00,380 sides, we see the deformation gradient entering twice here, 468 00:28:00,380 --> 00:28:03,330 whereas four times here. 469 00:28:03,330 --> 00:28:07,460 We would like to recast, now, this equation on the Cauchy 470 00:28:07,460 --> 00:28:11,130 stress, so as to have the deformation gradient here 471 00:28:11,130 --> 00:28:15,050 entering in the same form as we have it entering here. 472 00:28:15,050 --> 00:28:18,560 And that is achieved by the definition of the Almansi 473 00:28:18,560 --> 00:28:20,520 strain tensor. 474 00:28:20,520 --> 00:28:25,420 Here, now, we have the Cauchy stress in terms of the Almansi 475 00:28:25,420 --> 00:28:29,990 strain tensor, multiplying a new constitutive tensor. 476 00:28:29,990 --> 00:28:34,080 And that new constitutive tensor is given down here. 477 00:28:34,080 --> 00:28:38,580 Notice the right hand side here contains the constitutive 478 00:28:38,580 --> 00:28:43,750 tensor of the total Lagrangian formulation, t0 here. 479 00:28:43,750 --> 00:28:47,430 And we are multiplying this tensor four times by the 480 00:28:47,430 --> 00:28:49,720 deformation gradient components. 481 00:28:49,720 --> 00:28:52,390 This is exactly what we try to achieve. 482 00:28:52,390 --> 00:28:56,450 And we achieve it by using the Almansi strain tensor, which 483 00:28:56,450 --> 00:28:59,860 is defined as follows. 484 00:28:59,860 --> 00:29:03,560 Here we have the definition of the Almansi strain tensor, in 485 00:29:03,560 --> 00:29:05,870 terms of the Green-Lagrange strain tensor. 486 00:29:05,870 --> 00:29:08,070 And here, we have the inverse deformation 487 00:29:08,070 --> 00:29:10,750 gradient entering twice. 488 00:29:10,750 --> 00:29:14,480 The same definition is also given here, 489 00:29:14,480 --> 00:29:17,340 but in matrix form. 490 00:29:17,340 --> 00:29:20,730 And if we substitute for the inverse deformation gradient, 491 00:29:20,730 --> 00:29:25,140 in terms of displacements, and multiply components out, we 492 00:29:25,140 --> 00:29:28,440 obtain directly this relationship here for the 493 00:29:28,440 --> 00:29:34,870 Almansi strain tensor, where, notice our notation is this t 494 00:29:34,870 --> 00:29:39,402 and that t are those two t's here. 495 00:29:39,402 --> 00:29:44,970 We are taking the displacement from time 0 to time t, and we 496 00:29:44,970 --> 00:29:48,880 differentiating those displacements with respect to 497 00:29:48,880 --> 00:29:50,880 the current coordinates. 498 00:29:50,880 --> 00:29:53,310 And that's important. 499 00:29:53,310 --> 00:29:58,570 There's also a minus here, which is a bit of a surprise, 500 00:29:58,570 --> 00:30:01,040 because in the Green-Lagrange strain definition, of course, 501 00:30:01,040 --> 00:30:04,000 we have a plus here, but different terms, of course, 502 00:30:04,000 --> 00:30:05,740 different terms. 503 00:30:05,740 --> 00:30:08,610 Anyway, this is the definition of the Almansi strain tensor. 504 00:30:08,610 --> 00:30:14,300 And this tensor is very useful in the way I just described. 505 00:30:14,300 --> 00:30:19,510 The Almansi strain tensor is a symmetric tensor. 506 00:30:19,510 --> 00:30:23,980 The ij components are equal to the ji components. 507 00:30:23,980 --> 00:30:27,770 The components are not invariant under rigid body 508 00:30:27,770 --> 00:30:30,220 rotation of the material. 509 00:30:30,220 --> 00:30:32,270 This is a property, an important property, of the 510 00:30:32,270 --> 00:30:35,190 Green-Lagrange strain tensor, but it is not a property of 511 00:30:35,190 --> 00:30:36,440 the Almansi strain tensor. 512 00:30:38,610 --> 00:30:43,140 The Almansi strain tensor is really, overall, not a very 513 00:30:43,140 --> 00:30:45,430 useful measure, strain measure. 514 00:30:45,430 --> 00:30:48,340 But we wanted to introduce it here briefly. 515 00:30:48,340 --> 00:30:53,650 And it is quite useful in the one particular case, namely, 516 00:30:53,650 --> 00:30:58,430 in the analysis of isotropic materials undergoing large 517 00:30:58,430 --> 00:31:01,050 displacements and large rotations, but 518 00:31:01,050 --> 00:31:03,260 generally small strains. 519 00:31:03,260 --> 00:31:07,330 Let's look at one simple example here that gives us a 520 00:31:07,330 --> 00:31:11,110 bit of insight, what's the Almansi strain tensor looks, 521 00:31:11,110 --> 00:31:13,960 what the strain components look like. 522 00:31:13,960 --> 00:31:17,960 Here we have a simple four node element that is being 523 00:31:17,960 --> 00:31:21,430 pulled out into the red configuration. 524 00:31:21,430 --> 00:31:24,690 Notice the original length is 0L. 525 00:31:24,690 --> 00:31:26,640 The pullout is t delta. 526 00:31:26,640 --> 00:31:29,710 Of course, t delta could also be negative, in which case we 527 00:31:29,710 --> 00:31:31,130 are pushing in here. 528 00:31:31,130 --> 00:31:34,790 And the current length is tL. 529 00:31:34,790 --> 00:31:40,930 Notice the Green-Lagrange strain is given here, where we 530 00:31:40,930 --> 00:31:46,700 have a plus sign here and where we have 0L down here. 531 00:31:46,700 --> 00:31:51,000 The Almansi strain 1, 1 component, of course, in this 532 00:31:51,000 --> 00:31:55,500 particular case, has this relationship here. 533 00:31:55,500 --> 00:31:59,990 Notice t [? sigma ?] over tL, versus 0L in the 534 00:31:59,990 --> 00:32:01,760 Green-Lagrange strain. 535 00:32:01,760 --> 00:32:05,830 And a minus sign here, whereas we have a plus sign there. 536 00:32:05,830 --> 00:32:11,600 If we plot these strain tensor components as a function of t 537 00:32:11,600 --> 00:32:16,740 delta over the original lengths of the element, we 538 00:32:16,740 --> 00:32:20,610 find this blue curve for the Green-Lagrange strain. 539 00:32:20,610 --> 00:32:24,900 That is very easy to see that this is the 540 00:32:24,900 --> 00:32:25,770 Green-Lagrange strain. 541 00:32:25,770 --> 00:32:29,950 Notice here we have 1/2 minus 1/2. 542 00:32:29,950 --> 00:32:33,280 That minus 1/2 comes from this relationship. 543 00:32:33,280 --> 00:32:36,520 This is minus one at this point. 544 00:32:36,520 --> 00:32:38,440 And that is plus 1/2. 545 00:32:38,440 --> 00:32:40,380 So we get a minus 1/2 here. 546 00:32:40,380 --> 00:32:45,940 And at this point here, when delta over L0 is one, we have 547 00:32:45,940 --> 00:32:49,730 3/2, verified by that formula. 548 00:32:49,730 --> 00:32:51,790 The engineering strain, of course, is 549 00:32:51,790 --> 00:32:54,440 just a straight line. 550 00:32:54,440 --> 00:32:59,390 The Almansi strain looks as shown here by the red curve. 551 00:32:59,390 --> 00:33:06,810 Notice that when delta is compressive, going into this 552 00:33:06,810 --> 00:33:09,200 direction in the picture, in other words, when we are 553 00:33:09,200 --> 00:33:12,810 compressing that element, of course, tL becomes smaller and 554 00:33:12,810 --> 00:33:16,900 smaller, meaning this variable down here becomes smaller and 555 00:33:16,900 --> 00:33:21,770 smaller, and our Almansi strain component very rapidly 556 00:33:21,770 --> 00:33:23,195 becomes very large negative. 557 00:33:26,910 --> 00:33:32,470 It turns out that the use of the Almansi strain tensor, 558 00:33:32,470 --> 00:33:36,832 corresponding with this material relationship, t t 559 00:33:36,832 --> 00:33:42,750 Caijrs, which I defined on the earlier view graph, is quite 560 00:33:42,750 --> 00:33:47,220 effective when we want to analyze with the U.L. 561 00:33:47,220 --> 00:33:51,170 Formulation situations that involve a linear isotropic 562 00:33:51,170 --> 00:33:54,300 material and large displacements, large 563 00:33:54,300 --> 00:33:57,590 rotations, but small strains. 564 00:33:57,590 --> 00:34:05,040 In this case, we can use directly that t t Caijrs is 565 00:34:05,040 --> 00:34:08,429 given by the right hand side here. 566 00:34:08,429 --> 00:34:12,550 Notice here we have the Lame constants, lambda and mu, 567 00:34:12,550 --> 00:34:15,139 which we already used in the previous lecture, and the 568 00:34:15,139 --> 00:34:16,389 chronica deltas. 569 00:34:18,949 --> 00:34:24,080 The t 0 Cijrs tensor is given as here on 570 00:34:24,080 --> 00:34:26,889 the right hand side. 571 00:34:26,889 --> 00:34:32,460 Again, Lame constants and chronica delta entries. 572 00:34:32,460 --> 00:34:38,000 We would use here the same Lame constants in both of 573 00:34:38,000 --> 00:34:40,600 these relationships. 574 00:34:40,600 --> 00:34:44,520 And we also use that the incremental, the tangent 575 00:34:44,520 --> 00:34:48,250 material relationship, is the same as the total 576 00:34:48,250 --> 00:34:52,199 stress-strain relationship in both cases. 577 00:34:52,199 --> 00:35:00,860 If we use reformulating, if we use this right hand side for 578 00:35:00,860 --> 00:35:06,220 that tensor and make this tensor equal to that tensor, 579 00:35:06,220 --> 00:35:09,990 and if we use the same right hand side as up there, down 580 00:35:09,990 --> 00:35:19,190 here, for the t 0 Cijrs and the 0Cijrs, same numbers, same 581 00:35:19,190 --> 00:35:22,740 Young's modulus, same Poisson ratio, then for large 582 00:35:22,740 --> 00:35:26,620 displacement, large rotation, but small strain analysis, we 583 00:35:26,620 --> 00:35:30,450 virtually obtain identically the same results. 584 00:35:30,450 --> 00:35:35,030 I'd like to just demonstrate to you with a very simple 585 00:35:35,030 --> 00:35:38,590 problem analysis what I mean here. 586 00:35:38,590 --> 00:35:45,670 Let us turn to just one slide in which we show a solution 587 00:35:45,670 --> 00:35:50,130 response of an arch that was analyzed using the updated and 588 00:35:50,130 --> 00:35:51,380 total Lagrangian formulations. 589 00:35:53,730 --> 00:35:57,170 Here you now see the slide that shows the arch. 590 00:35:57,170 --> 00:36:01,470 It is subjected to a point load at it's apex. 591 00:36:01,470 --> 00:36:04,860 And we measure the displacement at the apex. 592 00:36:04,860 --> 00:36:08,780 The displacement is w0, half of the arch was modeled with 593 00:36:08,780 --> 00:36:09,930 12 eight-node elements. 594 00:36:09,930 --> 00:36:16,590 And as you can see, the TL and UL solutions for this arch are 595 00:36:16,590 --> 00:36:18,670 practically, identically the same. 596 00:36:18,670 --> 00:36:22,620 We can't see any difference to the accuracy that we have 597 00:36:22,620 --> 00:36:24,530 plotted the response. 598 00:36:24,530 --> 00:36:28,610 Notice that in these TL and UL solutions, we used exactly the 599 00:36:28,610 --> 00:36:30,850 same Young's modulus and Poisson ratio. 600 00:36:33,450 --> 00:36:35,660 The reason that practically the same response is 601 00:36:35,660 --> 00:36:39,190 calculated using the updated Lagrangian formulation and the 602 00:36:39,190 --> 00:36:44,460 total Lagrangian formulation lies in that the constitutive 603 00:36:44,460 --> 00:36:49,090 transformations that would have to be applied to obtain 604 00:36:49,090 --> 00:36:52,870 the exact response, exactly the same response with a two 605 00:36:52,870 --> 00:36:57,380 formations, really reduce here to mere rotations. 606 00:36:57,380 --> 00:37:02,470 And that is the fact because, for this type of problem, 607 00:37:02,470 --> 00:37:05,440 large displacement, large rotation, but small strain 608 00:37:05,440 --> 00:37:10,090 problem, the mass density remains constant and the 609 00:37:10,090 --> 00:37:13,230 deformation gradient, which can always be written as a 610 00:37:13,230 --> 00:37:17,320 product all of a rotation matrix, an orthogonal matrix, 611 00:37:17,320 --> 00:37:22,160 times another symmetric matrix. 612 00:37:22,160 --> 00:37:24,930 This is, of course, here is a polar decomposition of the 613 00:37:24,930 --> 00:37:28,690 deformation gradient, that for this particular case, large 614 00:37:28,690 --> 00:37:32,080 displacement, large rotation, but small strains, this 615 00:37:32,080 --> 00:37:36,090 deformation gradient is simply almost equal to the rotation 616 00:37:36,090 --> 00:37:37,470 matrix only. 617 00:37:37,470 --> 00:37:41,210 In other words, the stretch matrix is almost equal to the 618 00:37:41,210 --> 00:37:43,570 identity matrix. 619 00:37:43,570 --> 00:37:48,960 However, when we look at large strain problems, large strain 620 00:37:48,960 --> 00:37:55,810 analysis, and we were to use this relationship here for the 621 00:37:55,810 --> 00:37:59,660 Cauchy stress, in terms of the Almansi strain, and this 622 00:37:59,660 --> 00:38:03,720 relationship here, giving us a second Piola-Kirchhoff stress 623 00:38:03,720 --> 00:38:07,700 in terms of the Green-Lagrange strain, with these two 624 00:38:07,700 --> 00:38:12,540 constitutive tensors, given as shown here by 625 00:38:12,540 --> 00:38:15,190 the same Lame constants. 626 00:38:15,190 --> 00:38:18,640 In other words, Poisson ratio and Young's modulus, being 627 00:38:18,640 --> 00:38:21,960 identically the same, entering here. 628 00:38:21,960 --> 00:38:25,530 More specifically, you would put in here a Young's modulus, 629 00:38:25,530 --> 00:38:29,490 say, of 30 million pounds per square inch and the Poisson 630 00:38:29,490 --> 00:38:31,230 ratio of 0.3. 631 00:38:31,230 --> 00:38:34,260 These are the exact numbers that would go into here, with 632 00:38:34,260 --> 00:38:37,300 these, of course, being the chronica deltas. 633 00:38:37,300 --> 00:38:40,620 If we were to use this relationship in the total 634 00:38:40,620 --> 00:38:43,540 Lagrangian formulation and that relationship in the 635 00:38:43,540 --> 00:38:47,150 updated Lagrangian formulation for large strains, then we 636 00:38:47,150 --> 00:38:51,070 would get very different results. 637 00:38:51,070 --> 00:38:55,450 Let us look at the simple problem that we already solved 638 00:38:55,450 --> 00:38:58,950 earlier ones, or that we considered earlier ones in the 639 00:38:58,950 --> 00:39:02,230 previous lecture, using the total Lagrangian formulation. 640 00:39:02,230 --> 00:39:07,513 I'd like to now consider the same problem using the updated 641 00:39:07,513 --> 00:39:11,990 Lagrangian formulation with this material relationship. 642 00:39:11,990 --> 00:39:16,595 Cauchy stress given in terms of the Almansi strain, and 643 00:39:16,595 --> 00:39:21,870 this E curl, which we used already before in the last 644 00:39:21,870 --> 00:39:24,890 lecture in this total Lagrangian formulation. 645 00:39:24,890 --> 00:39:28,400 Notice this E curl here is given on the right hand side 646 00:39:28,400 --> 00:39:33,670 here in terms of Young's modulus and Poisson's ratio. 647 00:39:33,670 --> 00:39:37,890 The problem, very briefly reviewed, is that we are 648 00:39:37,890 --> 00:39:45,000 looking at a bar with constant cross sectional area A bar. 649 00:39:45,000 --> 00:39:49,790 And we will pull this bar out and also compress it in and 650 00:39:49,790 --> 00:39:52,595 look at the force required to pull the bar 651 00:39:52,595 --> 00:39:54,900 out or compress it. 652 00:39:54,900 --> 00:39:57,920 And we want to now look at this problem using this 653 00:39:57,920 --> 00:40:00,890 material relationship here, Cauchy stress in terms of 654 00:40:00,890 --> 00:40:05,450 Almansi strain with E curl constant. 655 00:40:05,450 --> 00:40:10,040 Here we have, for this one dimensional problem, the 656 00:40:10,040 --> 00:40:14,950 Almansi strain, given as shown up here. 657 00:40:14,950 --> 00:40:20,880 Notice this differentiation here of tU1 with respect to 658 00:40:20,880 --> 00:40:23,430 the coordinate 1 at time t. 659 00:40:23,430 --> 00:40:27,950 It's nothing else than tL minus 0L over tL. 660 00:40:27,950 --> 00:40:30,570 Here we have that same quantity squared. 661 00:40:30,570 --> 00:40:32,500 Of course, there's a minus 1/2 in front. 662 00:40:32,500 --> 00:40:34,430 And the result is this. 663 00:40:34,430 --> 00:40:37,930 The Cauchy stress is directly obtained in terms of the 664 00:40:37,930 --> 00:40:41,790 physical force applied to the bar divided by the cross 665 00:40:41,790 --> 00:40:44,210 sectional area, which is constant. 666 00:40:44,210 --> 00:40:47,340 If we use that current length is given in terms of the 667 00:40:47,340 --> 00:40:51,200 original length plus the extension and this material 668 00:40:51,200 --> 00:40:55,880 relationship here, we obtain directly this force 669 00:40:55,880 --> 00:40:57,750 displacement response. 670 00:40:57,750 --> 00:41:01,370 Force plotted vertically up, displacement plotted along 671 00:41:01,370 --> 00:41:05,570 here, t sigma is the displacement. 672 00:41:05,570 --> 00:41:09,200 And notice that the force displacement response is 673 00:41:09,200 --> 00:41:11,540 highly nonlinear-- 674 00:41:11,540 --> 00:41:14,430 here is, by the way, the actual expression-- 675 00:41:14,430 --> 00:41:17,880 highly nonlinear, and looks quite different from what we 676 00:41:17,880 --> 00:41:21,020 calculated earlier using the total Lagrangian formulation 677 00:41:21,020 --> 00:41:25,063 with the same E curl, same E curl used earlier in the total 678 00:41:25,063 --> 00:41:29,570 Lagrangian formulation, in which case the response really 679 00:41:29,570 --> 00:41:32,120 looked something like this. 680 00:41:35,670 --> 00:41:39,270 So we get a totally different description of the material 681 00:41:39,270 --> 00:41:43,590 response using these two formulation when the material 682 00:41:43,590 --> 00:41:46,790 is subjected to large strains. 683 00:41:46,790 --> 00:41:52,250 Let us now also look at one example that demonstrates this 684 00:41:52,250 --> 00:41:55,920 feature, the features that I just discussed a bit more. 685 00:41:55,920 --> 00:42:01,270 Here we have a frame subjected to a tip load 686 00:42:01,270 --> 00:42:03,090 over at this end. 687 00:42:03,090 --> 00:42:04,690 A is the tip load. 688 00:42:04,690 --> 00:42:09,500 The frame has a thickness h, L length here, L length there, 689 00:42:09,500 --> 00:42:11,680 thickness h here, as well. 690 00:42:11,680 --> 00:42:15,440 The width of the frame structure is b. 691 00:42:15,440 --> 00:42:19,570 And the geometric data are given here. 692 00:42:19,570 --> 00:42:22,780 And the material data are given here. 693 00:42:22,780 --> 00:42:25,870 Notice that h/L is 1/50. 694 00:42:28,960 --> 00:42:34,170 We modeled this structural using 51 two-dimensional eight 695 00:42:34,170 --> 00:42:38,700 node elements, plane strain elements. 696 00:42:38,700 --> 00:42:41,990 Notice here is a typical element shown, 697 00:42:41,990 --> 00:42:43,150 an eight node element. 698 00:42:43,150 --> 00:42:46,870 There are 25 elements along here. 699 00:42:46,870 --> 00:42:50,090 There's one element there for the corner and another 25 700 00:42:50,090 --> 00:42:53,680 elements in the column. 701 00:42:53,680 --> 00:43:00,680 Now we want to ask the question what happens in the 702 00:43:00,680 --> 00:43:05,680 elastic analysis when we subject this structure, in 703 00:43:05,680 --> 00:43:11,440 other words, to the load r and we used once a TL and once a 704 00:43:11,440 --> 00:43:16,380 UL formulation, but with the same material constants. 705 00:43:16,380 --> 00:43:19,530 In other words, the material constants E and nu that I gave 706 00:43:19,530 --> 00:43:23,780 on the earlier view graph are the same for the TL solution 707 00:43:23,780 --> 00:43:26,030 and the UL solution. 708 00:43:26,030 --> 00:43:30,040 And we do not make any transformation, therefore, on 709 00:43:30,040 --> 00:43:31,240 the material relationships. 710 00:43:31,240 --> 00:43:34,380 We simply use the total Lagrangian formulation with E 711 00:43:34,380 --> 00:43:36,630 and nu given, these two constants plugged in. 712 00:43:36,630 --> 00:43:38,600 And these constants are the same 713 00:43:38,600 --> 00:43:41,000 throughout the TL solution. 714 00:43:41,000 --> 00:43:43,380 We perform this way the TL solution. 715 00:43:43,380 --> 00:43:47,500 And then, afterwards, we do the UL solution, again, simply 716 00:43:47,500 --> 00:43:50,670 putting E and nu into the analysis and keeping it 717 00:43:50,670 --> 00:43:54,580 constant throughout that analysis. 718 00:43:54,580 --> 00:43:57,880 For large displacement, large rotation, but small strain 719 00:43:57,880 --> 00:44:01,570 conditions, the TL and UL formulation will give similar 720 00:44:01,570 --> 00:44:06,410 results, similar because we don't make these 721 00:44:06,410 --> 00:44:08,810 transformations that I discussed earlier in the 722 00:44:08,810 --> 00:44:13,310 lecture in order to obtain exactly the same results. 723 00:44:13,310 --> 00:44:16,920 But for large displacement, large rotation, small strain 724 00:44:16,920 --> 00:44:19,850 conditions, similar results are obtained. 725 00:44:19,850 --> 00:44:24,470 Of course, we have shown that already, by the solution that 726 00:44:24,470 --> 00:44:27,990 I showed you on the slide, the analysis of the arch subjected 727 00:44:27,990 --> 00:44:29,000 to an apex load. 728 00:44:29,000 --> 00:44:32,530 But we will see that now again here in this example. 729 00:44:32,530 --> 00:44:35,060 For large displacement, large rotation, and large strain 730 00:44:35,060 --> 00:44:39,140 conditions, the TL and UL formulations will gave quite 731 00:44:39,140 --> 00:44:41,600 different results. 732 00:44:41,600 --> 00:44:44,760 Well, let's look now at the results that we obtained for 733 00:44:44,760 --> 00:44:45,920 this problem. 734 00:44:45,920 --> 00:44:50,650 Here we have plotted the force vertically, meganewtons, and 735 00:44:50,650 --> 00:44:55,110 the vertical displacement of the tip in terms of meter. 736 00:44:55,110 --> 00:45:01,770 The 2-D elements, TL and UL formulations, give this result 737 00:45:01,770 --> 00:45:06,020 here, the black curve, solid curve. 738 00:45:06,020 --> 00:45:09,020 And we also wanted to solve this problem once using beam 739 00:45:09,020 --> 00:45:12,790 elements using the total Lagrangian formulation. 740 00:45:12,790 --> 00:45:16,740 In fact, we used four node isoparametric beam elements. 741 00:45:16,740 --> 00:45:18,270 We have not discussed these yet. 742 00:45:18,270 --> 00:45:20,810 We will discuss the formulation of these elements 743 00:45:20,810 --> 00:45:22,730 in a later lecture. 744 00:45:22,730 --> 00:45:25,800 And you can see that the beam elements give a slightly 745 00:45:25,800 --> 00:45:28,830 different response solution. 746 00:45:28,830 --> 00:45:31,160 Of course, the assumptions, the kinematic assumptions, in 747 00:45:31,160 --> 00:45:32,700 the beam elements are different. 748 00:45:32,700 --> 00:45:35,500 And that explains the difference in the response 749 00:45:35,500 --> 00:45:36,620 calculated. 750 00:45:36,620 --> 00:45:39,770 But the TL and UL formulations, using the eight 751 00:45:39,770 --> 00:45:44,350 node isoparametric plane, the strain elements, we obtain 752 00:45:44,350 --> 00:45:45,685 basically the same response. 753 00:45:48,670 --> 00:45:52,540 The formations are very large for this frame. 754 00:45:52,540 --> 00:45:55,630 Notice here we have the undeformed frame. 755 00:45:55,630 --> 00:45:57,190 There's the load. 756 00:45:57,190 --> 00:46:00,390 We are pressing down here. 757 00:46:00,390 --> 00:46:04,060 These are the deformations at a load of one meganewton. 758 00:46:04,060 --> 00:46:07,690 And this is how the frame looks at full load, five 759 00:46:07,690 --> 00:46:08,750 meganewton-- 760 00:46:08,750 --> 00:46:11,100 very large deformations. 761 00:46:11,100 --> 00:46:15,100 In fact, the deformation are so large for the frame, that 762 00:46:15,100 --> 00:46:19,270 in an actual practical problem, probably the frame, 763 00:46:19,270 --> 00:46:21,180 of course, would have undergone inelastic 764 00:46:21,180 --> 00:46:22,890 deformations. 765 00:46:22,890 --> 00:46:24,820 However, we look at this problem as a numerical 766 00:46:24,820 --> 00:46:28,600 experiment just to demonstrate what is happening and what we 767 00:46:28,600 --> 00:46:31,400 have been discussing in the lecture. 768 00:46:31,400 --> 00:46:36,930 If you look at the maximum deformations once more, a 769 00:46:36,930 --> 00:46:39,690 little bit closer, we find that the vertical tip 770 00:46:39,690 --> 00:46:43,090 displacement in the TL formulation for this problem 771 00:46:43,090 --> 00:46:48,330 is given as 15.289 meters, and in the UL formulation, as 772 00:46:48,330 --> 00:46:50,200 15.282 meters. 773 00:46:50,200 --> 00:46:55,690 Notice that there is only a change in the fifth digit. 774 00:46:55,690 --> 00:46:58,480 The displacements and rotations have certainly been 775 00:46:58,480 --> 00:47:02,380 very large in the analysis of this frame, but the strains 776 00:47:02,380 --> 00:47:04,400 are still quite small. 777 00:47:04,400 --> 00:47:08,920 If you look at the strain at the base of the frame, maybe 778 00:47:08,920 --> 00:47:12,890 we have the maximum strain, maximal moments there, we find 779 00:47:12,890 --> 00:47:17,390 that the moment is approximately as shown here. 780 00:47:17,390 --> 00:47:20,800 And the strain, using strength of material formulas, is 781 00:47:20,800 --> 00:47:23,390 approximately given by 3%. 782 00:47:23,390 --> 00:47:26,950 Now up to 2%, you certainly would consider it a small 783 00:47:26,950 --> 00:47:27,970 strain problem. 784 00:47:27,970 --> 00:47:33,150 So here we are just at the limit of going over into a lot 785 00:47:33,150 --> 00:47:37,080 strain region near the base of the column. 786 00:47:37,080 --> 00:47:39,680 This is all I wanted to present to you, discuss with 787 00:47:39,680 --> 00:47:40,790 you, in this lecture. 788 00:47:40,790 --> 00:47:42,340 Thank you very much for your attention.