Advanced elasticity/Polar decomposition

Polar decomposition

The w:Polar decomposition theorem states that any second order tensor whose determinant is positive can be decomposed uniquely into a symmetric part and an orthogonal part.

In continuum mechanics, the deformation gradient $\boldsymbol{F}$ is such a tensor because $\det(\mathbf{F}) > 0$. Therefore we can write

$\boldsymbol{F} = \boldsymbol{R}\cdot\boldsymbol{U} = \boldsymbol{V}\cdot\boldsymbol{R}$

where $\boldsymbol{R}$ is an orthogonal tensor ($\boldsymbol{R}\cdot\boldsymbol{R}^T = \boldsymbol{\mathit{1}}$) and $\boldsymbol{U}, \boldsymbol{V}$ are symmetric tensors ($\boldsymbol{U} = \boldsymbol{U}^T$ and $\boldsymbol{V} = \boldsymbol{V}^T$) called the right stretch tensor and the left stretch tensor, respectively. This decomposition is called the polar decomposition of $\boldsymbol{F}$.

Recall that the right Cauchy-Green deformation tensor is defined as

$\boldsymbol{C} = \boldsymbol{F}^T\cdot\boldsymbol{F}$

Clearly this is a symmetric tensor. From the polar decomposition of $\boldsymbol{F}$ we have

$\boldsymbol{C} = \boldsymbol{U}^T\cdot\boldsymbol{R}^T\cdot\boldsymbol{R}\cdot\boldsymbol{U} = \boldsymbol{U}\cdot\boldsymbol{U} = \boldsymbol{U}^2$

If you know $\boldsymbol{C}$ then you can calculate $\boldsymbol{U}$ and hence $\boldsymbol{R}$ using $\boldsymbol{R} = \boldsymbol{F}\cdot\boldsymbol{U}^{-1}$.

How do you find the square root of a tensor?

If you want to find $\boldsymbol{U}$ given $\boldsymbol{C}$ you will need to take the square root of $\boldsymbol{C}$. How does one do that?

We use what is called the spectral decomposition or eigenprojection of $\boldsymbol{C}$. The spectral decomposition involves expressing $\boldsymbol{C}$ in terms of its eigenvalues and eigenvectors. The tensor product of the eigenvectors acts as a basis while the eigenvalues give the magnitude of the projection.

Thus,

$\boldsymbol{C} = \sum_{i=1}^3 \lambda_i^2~ \boldsymbol{N}_i\otimes\boldsymbol{N}_i$

where $\lambda_i^2$ are the principal values (eigenvalues) of $\boldsymbol{C}$ and $\boldsymbol{N}_i$ are the principal directions (eigenvectors) of $\boldsymbol{C}$.

Therefore,

$\boldsymbol{U}^2 = \sum_{i=1}^3 \lambda_i^2~ \boldsymbol{N}_i\otimes\boldsymbol{N}_i$

Since the basis does not change, we then have

$\boldsymbol{U} = \sum_{i=1}^3 \lambda_i~ \boldsymbol{N}_i\otimes\boldsymbol{N}_i$

Therefore the $\lambda_i$ can be interpreted as principal stretches and the vectors $\boldsymbol{N}_i$ are the directions of the principal stretches.

Exercise:

If

$\boldsymbol{U} = \sum_{i=1}^3 \lambda_i~ \boldsymbol{N}_i\otimes\boldsymbol{N}_i$

show that

$\boldsymbol{U}^2 = \boldsymbol{U}\cdot\boldsymbol{U} = \sum_{i=1}^3 \lambda_i^2~ \boldsymbol{N}_i\otimes\boldsymbol{N}_i ~.$

Example of polar decomposition

Let us assume that the motion is given by

$\begin{align} x_1 &= \cfrac{1}{4} \left[4~X_1 + (9 - 3~X_1 - 5~X_2 - X_1~X_2)~t\right] \\ x_2 &= X_2 + (4 + 2~X_1)~t \end{align}$

The adjacent figure shows how a unit square subjected to this motion evolves over time.

An example of a motion.

Deformation gradient

The deformation gradient is given by

$\boldsymbol{F} = \frac{\partial \mathbf{x}}{\partial \boldsymbol{X}} \quad \implies \quad F_{ij} = \frac{\partial x_i}{\partial X_j}$

Therefore

$\begin{align} F_{11} &= \frac{\partial x_1}{\partial X_1} = \cfrac{1}{4}\left[4 + (- 3 - X_2)~t\right] \\ F_{12} &= \frac{\partial x_1}{\partial X_2} = \cfrac{1}{4}\left[(- 5 - X_1)~t\right] \\ F_{21} &= \frac{\partial x_2}{\partial X_1} = 2~t \\ F_{22} &= \frac{\partial x_2}{\partial X_2} = 1 \end{align}$

At $t = 1$ at the position $\boldsymbol{X} = (0,0)$ we have

$\mathbf{F} = \begin{bmatrix} \frac{\partial x_1}{\partial X_1} & \frac{\partial x_1}{\partial X_2} \\ \frac{\partial x_2}{\partial X_1} & \frac{\partial x_2}{\partial X_2} \end{bmatrix} = \cfrac{1}{4} \begin{bmatrix} 1 & -5 \\ 8 & 4 \end{bmatrix}$

You can calculate the deformation gradient at other points in a similar manner.

Right Cauchy-Green deformation tensor

We have

$\boldsymbol{C} = \boldsymbol{F}^T\cdot\boldsymbol{F}$

Therefore,

$\mathbf{C} = \mathbf{F}^T~\mathbf{F} = \cfrac{1}{16} \begin{bmatrix} 65 & 27 \\ 27 & 41 \end{bmatrix}$

To compute $\boldsymbol{U}$ we have to find the eigenvalues and eigenvectors of $\boldsymbol{C}$. The eigenvalue problem is

$(\mathbf{C} - \lambda^2~\mathbf{I})\mathbf{N} = \mathbf{0}$

where

$\mathbf{I} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

To find the eigenvalues we solve the characteristic equation

$\det(\mathbf{C} - \lambda^2~\mathbf{I}) = 0$

Plugging in the numbers, we get

$\det\begin{bmatrix} \cfrac{65}{16} - \lambda^2 & \cfrac{27}{16} \\ \cfrac{27}{16} & \cfrac{41}{16} - \lambda^2 \end{bmatrix} = 0$

or

$\lambda^4 - \cfrac{53}{8}~\lambda^2 + \cfrac{121}{16} = 0$

This equation has two solutions

$\begin{align} \lambda_1^2 & = \cfrac{53}{16} + \cfrac{3}{16}~\sqrt{97} = 5.159 \\ \lambda_2^2 & = \cfrac{53}{16} - \cfrac{3}{16}~\sqrt{97} = 1.466 \end{align}$

Taking the square roots we get the values of the principal stretches

$\lambda_1 = 2.2714 \qquad \lambda_2 = 1.2107$

To compute the eigenvectors we plug into the eigenvalues into the eigenvalue problem to get

$\left\{ \begin{bmatrix} 65 & 27 \\ 27 & 41 \end{bmatrix} - \lambda^2_1 ~ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right\}~\begin{bmatrix} N_1^{(1)} \\ N_2^{(1)} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

Because this system of equations is not linearly independent, we need another equation to solve this system of equations for $N^{(1)}_1$ and $N^{(1)}_2$. This problem is eliminated by using the following equation (which implies that $\mathbf{N}$ is a unit vector)

$(N^{(1)}_2)^2 = \sqrt{1 - (N^{(1)}_1)^2}$

Solving, we get

$\mathbf{N}_1 = \begin{bmatrix} N_1^{(1)} \\ N_2^{(1)} \end{bmatrix} = \begin{bmatrix} 0.8385 \\ 0.5449 \end{bmatrix}$

We can do the same thing for the other eigenvector $\mathbf{N}_2$ to get

$\mathbf{N}_2 = \begin{bmatrix} N_1^{(2)} \\ N_2^{(2)} \end{bmatrix} = \begin{bmatrix} -0.5449 \\0.8385 \end{bmatrix}$

Therefore,

$\boldsymbol{N}_1\otimes\boldsymbol{N}_1 = \mathbf{N}_1^T~\mathbf{N}_1 = \begin{bmatrix} 0.8385 & 0.5449 \end{bmatrix} \begin{bmatrix} 0.8385 \\ 0.5449 \end{bmatrix} = \begin{bmatrix} 0.7031 & 0.4569 \\ 0.4569 & 0.2969 \end{bmatrix}$

and

$\boldsymbol{N}_2\otimes\boldsymbol{N}_2 = \mathbf{N}_2^T~\mathbf{N}_2 = \begin{bmatrix} -0.5449 & 0.8385 \end{bmatrix} \begin{bmatrix} -0.5449 \\0.8385 \end{bmatrix} = \begin{bmatrix} 0.2969 & -0.4569 \\ -0.4569 & 0.7031 \end{bmatrix}$

Therefore,

$\boldsymbol{C} = \lambda_1^2~\boldsymbol{N}_1\otimes\boldsymbol{N}_1 + \lambda_2^2~\boldsymbol{N}_2\otimes\boldsymbol{N}_2 \quad \implies \quad \mathbf{C} = 5.159~ \begin{bmatrix} 0.7031 & 0.4569 \\ 0.4569 & 0.2969 \end{bmatrix} + 1.466~ \begin{bmatrix} 0.2969 & -0.4569 \\ -0.4569 & 0.7031 \end{bmatrix}$

We usually don't see any problem to calculate $\boldsymbol{C}$ at this point and go straight to the right stretch tensor.

Right stretch

The right stretch tensor $\boldsymbol{U}$ is given by

$\boldsymbol{U} = \lambda_1~\boldsymbol{N}_1\otimes\boldsymbol{N}_1 + \lambda_2~\boldsymbol{N}_2\otimes\boldsymbol{N}_2 \quad \implies \quad \mathbf{U} = 2.2714~ \begin{bmatrix} 0.7031 & 0.4569 \\ 0.4569 & 0.2969 \end{bmatrix} + 1.2107~ \begin{bmatrix} 0.2969 & -0.4569 \\ -0.4569 & 0.7031 \end{bmatrix}$

or

$\mathbf{U} = \begin{bmatrix} 1.9565 & 0.4846 \\ 0.4846 & 1.5256 \end{bmatrix}$

We can invert this matrix to get

$\mathbf{U}^{-1} = \begin{bmatrix} 0.5548 & -0.1762 \\ -0.1762 & 0.7114 \end{bmatrix}$

Rotation

We can now find the rotation matrix by using th relation

$\boldsymbol{R} = \boldsymbol{F}\cdot\boldsymbol{U}^{-1}$

In matrix form,

$\mathbf{R} = \cfrac{1}{4} \begin{bmatrix} 1 & -5 \\ 8 & 4 \end{bmatrix} \begin{bmatrix} 0.5548 & -0.1762 \\ -0.1762 & 0.7114 \end{bmatrix} = \begin{bmatrix} 0.3590 & -0.9334 \\ 0.9334 & 0.3590 \end{bmatrix}$

You can check whether this matrix is orthogonal by seeing whether $\mathbf{R}~\mathbf{R}^T = \mathbf{R}^T~\mathbf{R} = \mathbf{I}$.

You thus get the polar decomposition of $\boldsymbol{F}$. In an actual calculation you have to be careful about floating point errors. Otherwise you might not get a matrix that is orthogonal.