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I've resolved a relative displacement tensor into a strain tensor and a rotation tensor, where

the strain tensor is: $$ \varepsilon_{i,j} =\begin{pmatrix} 0.2 & 0 & 0 \\ 0 & 0.8 &0.4\\0&0.4&0.4\end{pmatrix} $$

and the rotation tensor is: $$ \omega_{i,j} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 &-0.2\\0&0.2&0\end{pmatrix} $$

How would these conditions physically change a small cube (with respect to the Cartesian cooridates (x,y,z) ?

Llewy Thomas
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    Your rotation is not a pure rotation as the determinant isn't $\pm1$. Perhaps provide more information on how you decomposed your displacement. – nivag Apr 28 '15 at 08:47
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    Also, I think there's something funky with your strain tensor, depending on your type of loading, because you've got expansion in all three principal directions. If you've got a fairly simple type of loading (i.e. not pulling on every face) you'd expect one principal to expand and the other two to contract. – Trevor Archibald Apr 28 '15 at 12:20
  • @TrevorArchibald It's probably a contrived displacement given the strain magnitudes of 0.8 – Rick Sep 15 '15 at 20:00
  • @nivag the rotation tensors are always antisymetric, which means in three dimensions the determinant is always zero. Maybe you're thinking of a rotation matrix? – Rick Sep 15 '15 at 20:02

1 Answers1

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The relative displacement tensor is the sum of the rotation tensor and the strain tensor. The strain tensor is always symmetric, while the rotation tensor is always antisymmetric. Once decomposed, the strain tensor is still a displacement tensor that has the rotation removed from it.

A displacement tensor can be defined as follows:

$P$ is vector field representing the displacement. If the material that started at point $\langle 2,4,8 \rangle$ ended at point $\langle 3,3,6 \rangle$ then $P(\langle 2,4,8 \rangle )=\langle 1,-1,-2 \rangle$

The displacement tensor $D$ is then:

$$D = \nabla P = \begin{bmatrix} \partial_x P_x & \partial_y P_x & \partial_z P_x\\ \partial_x P_y & \partial_y P_y & \partial_z P_y\\ \partial_x P_z & \partial_y P_z & \partial_z P_z \end{bmatrix}$$

If the displacement tensor is constant, that implies the cube is deformed linearly, and thus P is a linear field.

In this case: $$P = \begin{bmatrix} 0.2 \, x\\ 0.8 y + 0.2 z\\ 0.6 y + 0.4 z \end{bmatrix}$$

Using the original displacement tensor

$$D = \begin{bmatrix} 0.2 & 0 & 0\\ 0 & 0.8 & 0.2\\ 0 & 0.6 & 0.4 \end{bmatrix}$$

Now that we have the displacement field we can find the new position $N$ as a function of original position $O$ as:

$$N=O+P(O)$$

So if we take the corners of a unit cube we can calculate their new positions:

$$ \langle 0,0,0\rangle + P(\langle 0,0,0 \rangle) = \langle 0.0,0.0,0.0 \rangle\\ \langle 0,0,1\rangle + P(\langle 0,0,1 \rangle) = \langle 0.0,0.2,1.4 \rangle\\ \langle 0,1,0\rangle + P(\langle 0,1,0 \rangle) = \langle 0.0,1.4,0.6 \rangle\\ \langle 0,1,1\rangle + P(\langle 0,1,1 \rangle) = \langle 0.0,1.6,2.0 \rangle\\ \langle 1,0,0\rangle + P(\langle 1,0,0 \rangle) = \langle 1.2,0.0,0.0 \rangle\\ \langle 1,0,1\rangle + P(\langle 1,0,1 \rangle) = \langle 1.2,0.2,1.4 \rangle\\ \langle 1,1,0\rangle + P(\langle 1,1,0 \rangle) = \langle 1.2,1.4,0.6 \rangle\\ \langle 1,1,1\rangle + P(\langle 1,1,1 \rangle) = \langle 1.2,1.6,2.0 \rangle\\ $$

visually it would look like this:

Deformed Cube

Rick
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