0:02 when I saw the formula for the strain
0:05 tensor the very first time I found that
0:08 it looks unnecessarily complicated why
0:10 is it not possible to Simply Define The
0:12 Strain tensor as the gradient of the
0:14 displacement field well it turns out
0:16 that this would result in some
0:18 unphysical Behavior take for example
0:21 this simulation where I apply some
0:23 displacement in X1 direction to the top
0:26 of a solid cube if I run the same
0:28 simulation but with the wrong definition
0:30 for the strain tensor the results look
0:33 much less realistic in this video you
0:43 [Music]
0:45 why continum mechanics studies the
0:48 deformation of mattera under external
0:51 influences to do so it is necessary to
0:55 somehow quantify the deformation of meta
0:57 mathematically the displacement field
0:59 which is a vector field that describes
1:01 the displacement of each point carries
1:04 information about the deformation of
1:07 mattera however the displacement field
1:09 does not only carry information about
1:12 deformation it also carries information
1:14 about the translation and rotation of
1:16 mattera take for example this
1:19 displacement field which leads to a
1:20 translation of
1:23 meta or this displacement field which
1:25 leads to a rotation of
1:28 meta for both examples The Meta is not
1:31 deforming but just moving in space this
1:35 is called rigid body motion we observe
1:36 that the displacement field is not
1:38 really suited to quantify the
1:41 deformation of meta to measure the
1:44 deformation of meta we desire a physical
1:47 quantity that is zero where the mattera
1:50 is not deformed and nonzero elsewhere
1:53 this quantity is the strain tensor
1:55 Epsilon The Strain measures how
1:58 displacement is changing in space for
2:01 problems in one dimension The Strain is
2:03 the derivative of the displacement field
2:05 in three dimensions however the
2:07 displacement field is a vector field
2:12 with three components U1 U2 and u3 which
2:14 can each change in space in all three
2:17 directions X1 X2 and
2:20 X3 this means we can compute a total of
2:23 nine derivatives these nine derivatives
2:25 Define the gradient of the displacement
2:29 field which is denoted by nabla U
2:32 recognizing that the strain in one
2:34 dimension is the derivative of the
2:36 displacement field a seemingly Natural
2:39 Choice for defining the strain tensor in
2:41 three dimensions would be that the
2:43 strain tensor is the gradient of the
2:46 displacement field spoiler we shouldn't
2:49 do that but let's anyways think this idea
2:50 idea
2:52 through let's take a look at the
2:54 previous displacement field
2:57 examples in the case of rigid body
2:59 translation the displacement gradient is
3:03 z zero great this is what we wanted the
3:05 meta is translating in space but it is
3:08 not deforming and the strain is
3:11 zero but what happens for the other
3:14 example for rigid body rotation the
3:17 displacement gradient is non zero this
3:19 means that the displacement gradient
3:21 does not only contain information about
3:24 the deformation of meta but also
3:27 information about the rotation of
3:30 mattera now the following question iises
3:32 can we take the displacement gradient
3:35 and somehow remove all contributions
3:36 that belong to
3:39 rotation if this would be possible then
3:42 all we would have left would be a good
3:43 measure of
3:45 deformation guess what it is possible to
3:47 do that all we need is a short mathematical
3:59 interlude a tensor a can be additively
4:01 decomposed into two parts a symmetric
4:04 and a school symmetric part let's assume
4:08 that a is a 3X 31 or here the formula
4:13 for the symmetric part is 12 * a + a
4:15 transposed if we expand this formula we
4:17 can see why this is called the symmetric
4:21 part of a the upper right and lower left
4:23 off diagonal elements of the symmetric
4:24 part are
4:27 equal the formula for the school
4:31 symmetric part is 12 * a minus a
4:33 transposed again expanding the formula
4:36 we can see why this is called the school
4:38 symmetric part of a the upper right and
4:41 lower left of diagonal elements of the
4:43 school symmetric part are equal in
4:45 magnitude but change in
4:48 sign if you like pause the video and
4:51 quickly verify that the symmetric plus
5:01 a
5:03 let's go back to the displacement
5:05 gradient we have seen that the rotation
5:07 of meta resulted in a nonzero displacement
5:08 displacement
5:10 gradient now take a look at the
5:13 following no matter how The Meta is
5:16 rotated the displacement gradient always
5:23 pattern it is always goo
5:26 symmetric it turns out that the rotation
5:28 of mattera is directly linked to the SK
5:32 symmetric part of the displacement
5:34 gradient this means by subtracting the
5:37 skoo symmetric part of the displacement
5:40 gradient we can remove all contributions
5:43 that arise from rotations so let's do it
5:45 let's write down the displacement
5:47 gradient minus the SCH symmetric part of
5:51 it after some rearrangements we get this
5:53 here we have it this is the formula for
5:56 the strain tensor now we know why this
5:57 formula looks strange at the first
6:00 clance the strain tensor is the
6:02 displacement gradient minus all
6:04 contributions that arise from
6:07 rotations what is left is the symmetric
6:10 part of the displacement
6:12 gradient I personally find this
6:14 derivation super intriguing because it
6:16 once more shows that every formula that
6:19 we use in physics does not come from
6:21 nowhere there's always a reason for a
6:23 formula looking kind of weird and in
6:25 this case there's no black magic in
6:28 defining the strain tensor all we needed
6:30 was a measure for the deformation of
6:32 mattera the displacement gradient was a
6:35 good choice to start with but it also
6:37 included rotation which we had to remove
6:39 to arrive at the proper measure of
6:42 deformation The Strain
6:44 tensor let's finally go back to the
6:46 simulation that I've have shown in the
6:49 very beginning of the video here you can
6:51 see the boundary value problem that I
6:53 solved numerically for simulating the
6:56 deformation of the solid block under
6:58 Shear if we changed the definition for
7:01 the strain t tensor such that it equals
7:04 the displacement gradient that is we
7:07 include rotation into the strain tensor
7:09 then we get a much less realistic
7:12 behavior of the solid the reason for
7:14 this is that now the rotation of meta
7:16 would also contribute to the stress in
7:19 The Meta therefore we get an
7:21 unrealistically stiff meta so we should
7:23 stick to the correct definition of the strain
7:24 strain
7:27 tensor at the end there's one important
7:29 thing that must be mentioned about the
7:31 strain tensor The Strain tensor that has
7:33 been discussed in this video is only
7:36 suitable for measuring deformations that
7:39 are very small this strain tensor is
7:41 therefore called the small strain tensor
7:45 or infinitesimal strain tensor all
7:47 displacements and strains shown in this
7:49 video have been Amplified for
7:51 illustration purposes but note that the
7:54 small strain tensor should only be used
7:57 for strains that do not exceed one or
8:01 maximum 2% for larger deformations
8:03 different tensors are required to
8:05 correctly quantify The Strain but this
8:07 we will have to postpone to future
8:10 videos on nonlinear Continuum
8:13 mechanics that's it for this video in an
8:15 upcoming video in the series of videos
8:18 on continer mechanics I will show you
8:20 how the strain tensor and its individual
8:24 elements can be visualized stay tuned
