Biswajit Banerjee

Plane stress return: Spectral decomposition

Introduction

The previous parts of this series dealt with:

Let us now take a slight detour and examine the spectral decomposition of the stiffness matrix () and the transformation matrix ().

Spectral decomposition of

Recall that under plane stress conditions the stiffness matrix reduces to

$$ \mathbf{C} = \begin{bmatrix} \frac{4\mu(\lambda+\mu)}{\lambda+2\mu} & \frac{2\mu\lambda}{\lambda+2\mu} & 0 \\ \frac{2\mu\lambda}{\lambda+2\mu} & \frac{4\mu(\lambda+\mu)}{\lambda+2\mu} & 0 \\ 0 & 0 & \mu \end{bmatrix} $$

Since is a real, symmetric, matrix we can find a spectral decomposition of the matrix via a singular value decomposition of the form

$$ \mathbf{C} = \mathbf{Q}_C \mathbf{L}_C \mathbf{Q}_C^T $$

where is an orthogonal matrix whose columns form the basis of the spectral decomposition and is a diagonal matrix containing the eigenvalues of the spectral decomposition.

Because we are lazy and prone to algebra errors, let us turn to Mathematica to solve the singular value decomposition problem. The script we use is:

$$ \begin{align} &\pmb{\text{C11} = 4 \text{mu} (\text{lambda} + \text{mu})/(\text{lambda} + 2 \text{mu})}\\ &\pmb{\text{C22} = \text{C11}}\\ &\pmb{\text{C33} = \text{mu}}\\ &\pmb{\text{C12} = 2 \text{mu}\, \text{lambda}/(\text{lambda} + 2 \text{mu})}\\ &\pmb{\text{CC} = \{\{\text{C11}, \text{C12}, 0\},\{\text{C12}, \text{C22}, 0\},\{0, 0, \text{C33}\}\}}\\ &\pmb{\text{$\$$Assumptions} =\text{ }\text{lambda}\in \text{Reals} \,\&\&\,\text{ }\text{mu} \in \text{Reals} \,\&\& \,\text{mu} > 0 \,\&\&\,}\\ & \quad \pmb{\text{CC} \in \text{Matrices}[\{3,3\},\text{Reals},\text{Symmetric}]}\\ &\pmb{\{\text{QC}, \text{LC}, \text{QCT}\} = \text{SingularValueDecomposition}[\text{CC}]\text{//} \,\text{FullSimplify}} \\ &\pmb{\text{QC} \text{/.} \{\text{lambda} \to \text{Ee}\, \text{nu}/((1+ \text{nu}) (1 - 2 \text{nu})), \text{mu} \to \text{Ee}/(2 (1 + \text{nu}))\}\text{//} \text{FullSimplify}} \\ & \pmb{\text{LC} \text{/.} \{\text{lambda} \to \text{Ee}\, \text{nu}/((1+ \text{nu}) (1 - 2 \text{nu})), \text{mu} \to \text{Ee}/(2 (1 + \text{nu}))\}\text{//} \text{FullSimplify}} \end{align} $$

The result produced by Mathematica is

$$ \begin{align} \mathbf{Q}_C &= \left\{\left\{0,\frac{\text{Sign}[1+\text{nu}]}{\sqrt{2} \text{Sign}[1-\text{nu}]},-\frac{1}{\sqrt{2}}\right\},\left\{0,\frac{\text{Sign}[1+\text{nu}]}{\sqrt{2} \text{Sign}[1-\text{nu}]},\frac{1}{\sqrt{2}}\right\},\{1,0,0\}\right\} \\ \mathbf{L}_C & = \left\{\left\{\frac{\text{Ee}}{2+2 \text{nu}},0,0\right\},\left\{0,\frac{\text{Ee} \text{Abs}\left[\frac{1+\text{nu}}{1-\text{nu}}\right]}{1+\text{nu}},0\right\},\left\{0,0,\frac{\text{Ee}}{1+\text{nu}}\right\}\right\} \end{align} $$

Since the Poisson’s ratio , we have and . Therefore,

$$ \mathbf{Q}_C = \begin{bmatrix} 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 1 & 0 & 0 \end{bmatrix} \quad \mathbf{L}_C = \begin{bmatrix} \frac{E}{2(1+\nu)} & 0 & 0 \\ 0 & \frac{E}{1-\nu} & 0 \\ 0 & 0 & \frac{2E}{2(1+\nu)} \end{bmatrix} = \begin{bmatrix} \mu & 0 & 0 \\ 0 & \frac{E}{1-\nu} & 0 \\ 0 & 0 & 2\mu \end{bmatrix} $$
Spectral decomposition of

The transformation matrix has components

$$ \mathbf{P} = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & 0 \\ 0 & 0 & 6 \end{bmatrix} $$

As we did for , we can perform a spectral decomposition of matrix into

$$ \mathbf{P} = \mathbf{Q}_P \mathbf{L}_P \mathbf{Q}_P^T $$

Proceeding as before with Mathematica, we find that

$$ \mathbf{Q}_P = \begin{bmatrix} 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 1 & 0 & 0 \end{bmatrix} \quad \mathbf{L}_P = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} $$
Product of and

From the above we see that . Therefore,

$$ \mathbf{P}\mathbf{C} = \mathbf{Q} \mathbf{L}_P \mathbf{Q}^T \mathbf{Q} \mathbf{L}_C \mathbf{Q}^T = \mathbf{Q} \mathbf{L}_P \mathbf{L}_C \mathbf{Q}^T = \mathbf{Q} \mathbf{L}_C \mathbf{L}_P \mathbf{Q}^T = \mathbf{C} \mathbf{P} $$

Remarks

We will use these decompositions to diagonalize the complicated matrix expressions in the next part of this series.

If you have questions/comments/corrections, please contact banerjee at parresianz dot com dot zen (without the dot zen).

📅 23.03.2017 📁 MECHANICS · PLASTICITY · ALGORITHM

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