Rotation of the stress tensor

[happyparticle]

Homework Statement

Rotation of $\pi /4$ of the stress tensor around the Z-axis

Relevant Equations

$$D'_{xy} = \frac{1}{2} D_{xx} - \frac{1}{2} D_{yy}$$
$$\tau'_{xy} = \frac{1}{2} \tau_{xx} -\frac{1}{2} \tau_{yy}$$
$$\tau'_{xy} = \frac{1}{2} (C_{11} - C_{12})D_{xx} + \frac{1}{2} (C_{12} - C_{11}) D_{yy} + C_{12}D_{zz}$$

First of all I have this system

$$\begin{pmatrix}\tau_{xx} \\ \tau_{yy} \\ \tau_{zz} \\ \tau_{xy} \\ \tau_{yz} \\ \tau_{zx} \end{pmatrix}=\begin{pmatrix}C_{11} & C_{12} & C_{12} & 0 & 0 & 0 \\ C_{12} & C_{11} & C_{12} & 0 & 0 & 0 \\ C_{12} & C_{12} & C_{11} & 0 & 0 & 0 \\ 0 & 0 & 0 & C_{44} & 0 & 0 \\ 0 & 0 & 0 & 0 & C_{44} & 0 \\ 0 & 0 & 0 & 0 & 0 & C_{44} \end{pmatrix}\begin{pmatrix}D_{xx} \\ D_{yy} \\ D_{zz} \\ D_{xy} \\ D_{yz} \\ D_{zx} \end{pmatrix}$$

Then I perform a rotation of $\pi /4$ around the z-axis.

For $\tau_{xy}$ I get

$$D'_{xy} = \frac{1}{2} D_{xx} - \frac{1}{2} D_{yy}$$
$$\tau'_{xy} = \frac{1}{2} \tau_{xx} -\frac{1}{2} \tau_{yy}$$
$$\tau'_{xy} = \frac{1}{2} (C_{11} - C_{12})D_{xx} + \frac{1}{2} (C_{12} - C_{11}) D_{yy}$$

Also,
$$\tau_{xy} = C_{44}D_{xy}$$

The answer I'm looking for is $C_{11} = C_{44} + C_{12}$

I read that the $C_{ij}$ must be invariant. What does it means here ? Since there is no $D_{xy}$ term in the expression for $\tau'_{xy}$ , $C_{44}$ must be 0 and $C_{12} = C_{11}$?

However, this is not the right answer. I'm not sure to understand.

 

[TSny]

First of all I have this system

$$\begin{pmatrix}\tau_{xx} \\ \tau_{yy} \\ \tau_{zz} \\ \tau_{xy} \\ \tau_{yz} \\ \tau_{zx} \end{pmatrix}=\begin{pmatrix}C_{11} & C_{12} & C_{12} & 0 & 0 & 0 \\ C_{12} & C_{11} & C_{12} & 0 & 0 & 0 \\ C_{12} & C_{12} & C_{11} & 0 & 0 & 0 \\ 0 & 0 & 0 & C_{44} & 0 & 0 \\ 0 & 0 & 0 & 0 & C_{44} & 0 \\ 0 & 0 & 0 & 0 & 0 & C_{44} \end{pmatrix}\begin{pmatrix}D_{xx} \\ D_{yy} \\ D_{zz} \\ D_{xy} \\ D_{yz} \\ D_{zx} \end{pmatrix}$$

I assume the $\tau_{ij}$ are components of stress and the $D_{ij}$ are components of strain. Check to make sure that you have written this correctly. For example, look at equation (3.13) in these notes. Note carefully the indices in the last three entries of the stress column vector ($\sigma_{ij}$) and also in the strain column vector ($\epsilon_{ij}$). Also, note the factors of 2 in the last three entries of the $\epsilon_{ij}$ vector. Thus, in your notation for stress and strain, this would be

$$\begin{pmatrix}\tau_{xx} \\ \tau_{yy} \\ \tau_{zz} \\ \tau_{yz} \\ \tau_{zx} \\ \tau_{xy} \end{pmatrix}=\begin{pmatrix}C_{11} & C_{12} & C_{12} & 0 & 0 & 0 \\ C_{12} & C_{11} & C_{12} & 0 & 0 & 0 \\ C_{12} & C_{12} & C_{11} & 0 & 0 & 0 \\ 0 & 0 & 0 & C_{44} & 0 & 0 \\ 0 & 0 & 0 & 0 & C_{44} & 0 \\ 0 & 0 & 0 & 0 & 0 & C_{44} \end{pmatrix}\begin{pmatrix}D_{xx} \\ D_{yy} \\ D_{zz} \\ 2D_{yz} \\ 2D_{zx} \\ 2D_{xy} \end{pmatrix}$$

Then I perform a rotation of $\pi /4$ around the z-axis.

For $\tau_{xy}$ I get

$$D'_{xy} = \frac{1}{2} D_{xx} - \frac{1}{2} D_{yy}$$
$$\tau'_{xy} = \frac{1}{2} \tau_{xx} -\frac{1}{2} \tau_{yy}$$
$$\tau'_{xy} = \frac{1}{2} (C_{11} - C_{12})D_{xx} + \frac{1}{2} (C_{12} - C_{11}) D_{yy}$$

OK. I think this is correct.

Also,
$$\tau_{xy} = C_{44}D_{xy}$$

The answer I'm looking for is $C_{11} = C_{44} + C_{12}$

I read that the $C_{ij}$ must be invariant. What does it means here ? Since there is no $D_{xy}$ term in the expression for $\tau'_{xy}$ , $C_{44}$ must be 0 and $C_{12} = C_{11}$?

However, this is not the right answer. I'm not sure to understand.

In general, for any system, we see from (3.13) and (3.15) in the link above that $$\tau_{xy} = C_{61} D_{xx}+C_{62} D_{yy}+C_{63} D_{zz}+C_{64} (2D_{yz})+C_{65} (2D_{zx})+C_{66} (2D_{xy})$$ For your particular system before the rotation, all of the $C_{6i} = 0$ except $C_{66}$. So, $$\tau_{xy} =2 C_{66} D_{xy}.$$ Since $C_{66} = C_{44}$ in your system before the rotation, you could write this as $$\tau_{xy} =2 C_{44} D_{xy}.$$ [Edited to fix a couple of typos in the indices]

After the rotation, we use a prime for the various quantities. So, after the rotation, we have $$\tau'_{xy} = C'_{61} D'_{xx}+C'_{62} D'_{yy}+C'_{63} D'_{zz}+C'_{64} (2D'_{yz})+C'_{65} (2D'_{zx})+C'_{66} (2D'_{xy})$$
The stiffness coefficients are not invariant. That is, we cannot assume $C'_{ij} = C_{ij}$. You will need to work out the values of the $C'_{6i}$ for the system after the rotation. I think you will find that all of the $C'_{6i} = 0$ except for $C'_{66}$. Show $C'_{66} = \frac 1 2 (C_{11}- C_{12})$.

 

[happyparticle]

You will need to work out the values of the C6i′ for the system after the rotation

I'm not sure what I have to compare. Is $\tau_{xy}$ is the same after rotation ? Hence, I can compare it with $\tau'_{xy}$. I guess... It it not clear for me what I have to do. I see like 4 equations, but I don't see any relationship between them.

Edit:

I read the link you gave and I think I have to compare $$
\tau'_{xy} = \frac{1}{2} (C_{11} - C_{12})D_{xx} + \frac{1}{2} (C_{12} - C_{11}) D_{yy}
$$ with
$$\tau'_{xy} = C'_{61} D'_{xx}+C'_{62} D'_{yy}+C'_{63} D'_{zz}+C'_{64} (D'_{yz})+C'_{65} (D'_{zx})+C'_{66} (D'_{xy}) $$

However, I'm not sure why exactly. Also it gaves me
$C'_{61} = \frac 1 2 (C_{11}- C_{12})$ and $C'_{62} = \frac 1 2 (-C_{11} + C_{12})$

 

[Chestermiller]

The components of the C matrix do not change as a result of a rotation of the coordinate system.

 

[TSny]

I read the link you gave and I think I have to compare $$
\tau'_{xy} = \frac{1}{2} (C_{11} - C_{12})D_{xx} + \frac{1}{2} (C_{12} - C_{11}) D_{yy}
$$ with
$$\tau'_{xy} = C'_{61} D'_{xx}+C'_{62} D'_{yy}+C'_{63} D'_{zz}+C'_{64} (D'_{yz})+C'_{65} (D'_{zx})+C'_{66} (D'_{xy}) $$

Yes, in your first post you wrote $\tau'_{xy} = \frac 1 2 \tau_{xx} - \frac 1 2 \tau_{yy}$. This shows how the stress component $\tau'_{xy}$ after the rotation relates to the stress components before the rotation. From this equation you can get your result $\tau'_{xy} = \frac{1}{2} (C_{11} - C_{12})D_{xx} + \frac{1}{2} (C_{12} - C_{11}) D_{yy}$.

It appeared to me that in your first post, you were getting confused because you were considering the stiffnesses $C_{ij}$ as invariant under rotations. But, for an anisotropic medium, the $C_{ij}$ are not generally invariant. For example, $C'_{66} \neq C_{66}$ in the medium you are considering. Instead, $C'_{66} = \frac 1 2 (C_{11} - C_{12})$. Using this result for $C'_{66}$ along with the results $C'_{6i} = 0$ for $i = 1, 2, 3, 4$, and $5$, you can show that the equation $$\tau'_{xy} = C'_{61} D'_{xx}+C'_{62} D'_{yy}+C'_{63} D'_{zz}+C'_{64} (2D'_{yz})+C'_{65} (2D'_{zx})+C'_{66} (2D'_{xy}) $$ agrees with your result $\tau'_{xy} = \frac{1}{2} (C_{11} - C_{12})D_{xx} + \frac{1}{2} (C_{12} - C_{11}) D_{yy}$.

But, maybe I misinterpreted your source of confusion in the first post.

Also it gaves me
$C'_{61} = \frac 1 2 (C_{11}- C_{12})$ and $C'_{62} = \frac 1 2 (-C_{11} + C_{12})$

You should find $C'_{61} = C'_{62} = 0$.

 

[TSny]

The components of the C matrix do not change as a result of a rotation of the coordinate system.

That's true for an isotropic material. However, in an anisotropic material, the components generally are not invariant under a coordinate rotation.

 

[happyparticle]

All right, I think I understand. Also, for an isotropic material after the rotation are the $C_{ij}$ not primed ? For instance, $\tau'_{ij} = C_{ij}D'{ij}$ ?

 

[TSny]

All right, I think I understand. Also, for an isotropic material after the rotation are the $C_{ij}$ not primed ? For instance, $\tau'_{ij} = C_{ij}D'{ij}$ ?

Yes. For example, in an isotropic medium, $\tau_{xx} = C_{11} D_{xx} + C_{12}(D_{yy} + D_{zz})$. After a rotation, none of the $C$ matrix elements change for the isotropic medium. So, you could write $\tau'_{xx} = C'_{11} D'_{xx} + C'_{12}(D'_{yy} + D'_{zz}) = C_{11} D'_{xx} + C_{12}(D'_{yy} + D'_{zz})$.

 

[happyparticle]

All right, Thank you!

I was really confused.