# Transformation matrix of a strain tensor

If the stress $$\sigma_{xx}$$ is applied to an isotropic, three-dimensional body, the following strain tensor results: $$\boldsymbol\epsilon=\left(\begin{matrix}\frac{1}{E}\sigma _{xx} & 0 & 0 \\0 & -\frac{\nu}{E}\sigma _{xx} & 0 \\0 & 0 & -\frac{\nu}{E}\sigma _{xx}\end{matrix}\right)$$ Now the tensor should be rotated around the y-axis with the angle $$\alpha$$.

The transformation should be carried out with $$A'=QAQ^T$$. How would be the transformation matrix $$Q$$ in this case?

• The transformation matrix is $Q=\begin{pmatrix} c\left( \alpha \right) & 0 & s\left( \alpha \right) \\ 0 & 1 & 0 \\ -s\left( \alpha \right) & 0 & c\left( \alpha \right) \end{pmatrix}$ where c is cos and s is sin
– Eli
Sep 16 '21 at 7:37
• Probably $\alpha\mapsto \pi/2-\alpha$
– Eli
Sep 16 '21 at 12:01
• Many thanks for the answer. I want to look at angle between $0$ and $\pi/2$. I'm not sure if $Q$ fits my sketch to the definition of $\alpha$. Shouldn't the first and last lines of $Q$ have to be changed? I hope I made no mistakes, but I would get the following result for the transformed strain tensor $\epsilon'$: Sep 16 '21 at 18:36
• $$\boldsymbol\epsilon'=\left(\begin{matrix}cos^2(\alpha)\epsilon _{11}+sin^2(\alpha)\epsilon _{33} & 0 & cos(\alpha)sin(\alpha)(\epsilon _{33}-\epsilon _{11}) \\0 & \epsilon _{22} & 0 \\cos(\alpha)sin(\alpha)(\epsilon _{33}-\epsilon _{11}) & 0 & sin^2(\alpha)\epsilon _{11}+cos^2(\alpha)\epsilon _{33}\end{matrix}\right)$$ Sep 16 '21 at 18:37
• I can understand the behaviour of the main axial strains, but how are the shear strain components explained? I would not have "expected" these here... Sep 16 '21 at 18:37

I use this notation

the transformation matrix, transformed a vector components from rotate system index B to inertial system index I

rotation about the x-axis angle $$~\alpha~$$ between y and y'

$${_B^I}\,Q_x=\left[ \begin {array}{ccc} 1&0&0\\ 0&\cos \left( \alpha \right) &-\sin \left( \alpha \right) \\ 0& \sin \left( \alpha \right) &\cos \left( \alpha \right) \end {array} \right]$$

rotation about the y-axis angle $$~\alpha~$$ between x and x'

$${_B^I}\,Q_y= \left[ \begin {array}{ccc} \cos \left( \alpha \right) &0&\sin \left( \alpha \right) \\ 0&1&0\\ -\sin \left( \alpha \right) &0&\cos \left( \alpha \right) \end {array} \right]$$

rotation about the z-axis angle $$~\alpha~$$ between x and x'

$${_B^I}\,Q_z=\left[ \begin {array}{ccc} \cos \left( \alpha \right) &-\sin \left( \alpha \right) &0\\ \sin \left( \alpha \right) &\cos \left( \alpha \right) &0\\ 0&0&1\end {array} \right]$$

vector transformation from B to I system

$$\mathbf v_I={_B^I}\mathbf Q\,\mathbf v_B$$

matrix transformation

$$\mathbf M_I=\mathbf ={_B^I}\mathbf Q \,\mathbf M_B\,{_I^B}\mathbf Q =\mathbf Q\,\mathbf M_B\,\mathbf Q^T\\ \mathbf M_B=\mathbf ={_I^B}\mathbf Q \,\mathbf M_I\,{_B^I}\mathbf Q =\mathbf Q^T\,\mathbf M_I\,\mathbf Q$$

$$\mathbf \epsilon_I= \left[ \begin {array}{ccc} \epsilon_{{11}}&0&0\\ 0& \epsilon_{{22}}&0\\ 0&0&\epsilon_{{22}}\end {array} \right] \\ \mathbf \epsilon_B=Q^T\,\mathbf \epsilon_I\,\mathbf Q$$

for $$~\mathbf Q=\mathbf Q_x~$$ you obtain

$$\mathbf \epsilon_B=\mathbf \epsilon_I$$

for $$~\mathbf Q=\mathbf Q_y~$$

$$\mathbf \epsilon_B=\left[ \begin {array}{ccc} \left( \cos \left( \alpha \right) \right) ^{2}\epsilon_{{11}}+\epsilon_{{22}}- \left( \cos \left( \alpha \right) \right) ^{2}\epsilon_{{22}}&0&\cos \left( \alpha \right) \sin \left( \alpha \right) \left( -\epsilon_{{22}}+\epsilon_ {{11}} \right) \\ 0&\epsilon_{{22}}&0 \\ \cos \left( \alpha \right) \sin \left( \alpha \right) \left( -\epsilon_{{22}}+\epsilon_{{11}} \right) &0& \left( \cos \left( \alpha \right) \right) ^{2}\epsilon_{{22}}+\epsilon_{{11} }- \left( \cos \left( \alpha \right) \right) ^{2}\epsilon_{{11}} \end {array} \right]$$

for $$~\mathbf Q=\mathbf Q_z~$$

$$\mathbf \epsilon_B= \left[ \begin {array}{ccc} \left( \cos \left( \alpha \right) \right) ^{2}\epsilon_{{11}}+\epsilon_{{22}}- \left( \cos \left( \alpha \right) \right) ^{2}\epsilon_{{22}}&-\cos \left( \alpha \right) \sin \left( \alpha \right) \left( -\epsilon_{{22}}+\epsilon_ {{11}} \right) &0\\ -\cos \left( \alpha \right) \sin \left( \alpha \right) \left( -\epsilon_{{22}}+\epsilon_{{11}} \right) & \left( \cos \left( \alpha \right) \right) ^{2}\epsilon_{{ 22}}+\epsilon_{{11}}- \left( \cos \left( \alpha \right) \right) ^{2} \epsilon_{{11}}&0\\ 0&0&\epsilon_{{22}}\end {array} \right]$$

Tensors are rotated by the same rotation matrices that rotate vectors. So if the basis vectors are rotated by $$Q_{ij}$$:

$$e'^{(k)}_i = Q_{ij}e^{(k)}_j$$

Then a rank-2 tensor is rotated with:

$$A'_{il}=Q_{ij}A_{jk}Q^T_{kl}$$

A symmetric tensor has 6 components. There is an invariant trace:

$$A'_{ii}=Q_{ij}A_{jk}Q^T_{ki}$$ $$A'_{ii}=Q^T_{ki}Q_{ij}A_{jk}$$ $$A'_{ii}=\delta_{kj}A_{jk}=A_{kk}$$

With your initial tensor, you can subtract off the trace to get the pure rank-2 part:

$$N_{ij} = \epsilon_{ij} - \frac 1 3 {\rm Tr}(\epsilon)$$

With

$${\rm Tr}(\epsilon) = \frac{\sigma_{xx}}E\big(\frac 1 3 -\frac 2 3\nu\big)$$

$$N_{xx} = \frac{\sigma_{xx}}E\big(\frac 2 3 + \frac 2 3\nu\big)=\frac 2 3\frac{\sigma_{xx}}E\big(1+\nu\big)$$

$$N_{yy} = N_{zz}=\frac{\sigma_{xx}}E\big(-\frac 1 3 - \frac 1 3\nu\big) =-\frac 1 3\frac{\sigma_{xx}}E\big(1+\nu\big)=-\frac 1 2 N_{xx}$$

If you transform that into spherical tensors, you get a two non-zero parts:

$$N_{zz} \propto Y_2^0(\theta, \phi)$$

which describes how prolate or oblate the quadrupole moment is (along the $$z$$-axis), and

$$N_{xx}-N_{yy} \propto (Y_2^2(\theta, \phi) + Y_2^{-2}(\theta, \phi))$$

which describes the lowest degree of azimuthal asymmetry (a bulge like the Earth's tidal bulge).

The

$$N_{xz} \propto (Y_2^1(\theta, \phi) + Y_2^{-1}(\theta, \phi))$$

term is zero. This term is caused by choosing a coordinate system that is not aligned with the principle axes of the quadrupole. It can be diagonalized away.

So: when you transform to $$\epsilon'$$ you get non-zero $$l=2$$, $$m=\pm 1$$ moments, meaning you have not chosen the best coordinates. There is no physical meaning.