Polar Decomposition

The polar decomposition theorem of Cauchy³ states that a non-singular matrix equals an orthogonal matrix either pre or post multiplied by a positive definite symmetric matrix. If we apply this theorem to the deformation gradient $\mathbf{F}$, we get

$$\mathbf{F} = \mathbf{R}\mathbf{U} = \mathbf{V}\mathbf{R}$$ (3.13.1)

in which $\mathbf{R}$ is a proper orthogonal matrix and $\mathbf{U}$ and $\mathbf{V}$ are positive definite symmetric matrices. Note that the decomposition (3.13.1) of $\mathbf{F}$ is unique in that $\mathbf{R}$, $\mathbf{U}$ and $\mathbf{V}$ are uniquely determined by $\mathbf{F}$. From (3.13.1) it follows that

$$J = \det \mathbf{F} = \det \mathbf{U} = \det\mathbf{V}.$$ (3.13.2)

Since $\mathbf{C} = \mathbf{F}^T\mathbf{F}$ and $\mathbf{B} = \mathbf{F}\mathbf{F}^T$, therefore

$$\mathbf{C} = \mathbf{F}^T\mathbf{F} = (\mathbf{R}\mathbf{U})^T\mathbf{R}\mathbf{U} = \mathbf{U}^2,$$ (3.13.3)

$$\mathbf{B} = \mathbf{F}\mathbf{F}^T = \mathbf{V}\mathbf{R}(\mathbf{V}\mathbf{R})^T = \mathbf{V}^2.$$ (3.13.4)

We note that

$$\mathbf{V} = \mathbf{R}\mathbf{U}\mathbf{R}^T$$ (3.13.5)

and

$$\mathbf{B} = \mathbf{F}\mathbf{F}^T = \mathbf{R}\mathbf{U}(\mathbf{R}\mathbf{U})^T = \mathbf{R}\mathbf{C}\mathbf{R}^T.$$ (3.13.6)

Since $\mathbf{U}$ is symmetric, it has at least one orthogonal triad of eigenvectors. Let $\mathbf{N}^{(1)}$ be an eigenvector of $\mathbf{U}$ and $\lambda_1$ be the corresponding eigenvalue so that

$$\mathbf{U}\mathbf{N}^{(1)} = \lambda_1\mathbf{N}^{(1)}.$$ (3.13.7)

Therefore

$$\mathbf{C}\mathbf{N}^{(1)} = \mathbf{U}\mathbf{U}\mathbf{N}^{(1)} = \lambda_1\mathbf{U}\mathbf{N}^{(1)} = (\lambda_1)^2\mathbf{N}^{(1)}.$$ (3.13.8)

Thus $\mathbf{N}^{(1)}$ is an eigenvector of $\mathbf{C}$ and the corresponding eigenvalue is $(\lambda_1)^2$. Since eqn. (3.13.8) holds for every eigenvector of $\mathbf{U}$, we conclude that eigenvectors of $\mathbf{U}$ and $\mathbf{C}$ are the same and the eigenvalues of $\mathbf{C}$ are equal to the squares of the eigenvalues of $\mathbf{U}$. In section 3.9, we proved that the eigenvalues of $\mathbf{C}$ are the squares of the principal stretches and the eigenvectors of $\mathbf{C}$ are the axes of principal stretches also usually called principal axes of stretch in the reference configuration. Thus eigenvectors of $\mathbf{U}$ are the principal axes of stretch in the reference configuration and the eigenvalues of $\mathbf{U}$ are the princpal stretches. Let us now find the deformed position of an eigenvector $\mathbf{N}^{(1)}$ of $\mathbf{U}$. Since

$$\mathbf{F}\mathbf{N}^{(1)} = \mathbf{R}\mathbf{U}\mathbf{N}^{(1)} = \lambda_1\mathbf{R}\mathbf{N}^{(1)},$$ (3.13.9)

therefore

$$\mathbf{n}^{(1)} = \mathbf{R}\mathbf{N}^{(1)}$$ (3.13.10)

points in the direction of the vector into which $\mathbf{N}^{(1)}$ is deformed. Note that

$$\mathbf{n}^{(1)}\cdot\mathbf{n}^{(1)} = \mathbf{n}^{(1)T}\mathbf{... ...\mathbf{R}^T\mathbf{R}\mathbf{N}^{(1)} = \mathbf{N}^{(1)T}\mathbf{N}^{(1)} = 1,$$ (3.13.11)

therefore, $\mathbf{n}^{(1)}$ is a unit vector in the direction of the deformed position of $\mathbf{N}^{(1)}$. Now

$$\mathbf{F}\mathbf{N}^{(1)} = \mathbf{V}\mathbf{R}\mathbf{N}^{(1)} = \mathbf{V}\mathbf{n}^{(1)}.$$ (3.13.12)

Equations (3.13.9), (3.13.10) and (3.13.12) when combined together give

$$\mathbf{V}\mathbf{n}^{(1)} = \lambda_1\mathbf{n}^{(1)}.$$ (3.13.13)

Thus $\lambda_1$ is an eigenvalue of $\mathbf{V}$ with $\mathbf{n}^{(1)}$ as the eigenvector. This exercise proves that the eigenvalues of $\mathbf{U}$ and $\mathbf{V}$ are equal and that the eigenvectors of $\mathbf{V}$ are the deformed positions of the eigenvectors of $\mathbf{U}$. Thus eigenvectors of $\mathbf{V}$ are the deformed position of the principal axes of stretch. Said differently, eigenvectors of $\mathbf{V}$ are the principal axes of stretch in the deformed configuration or in the present configuration. Of course,

$$\mathbf{B}\mathbf{n}^{(1)} = \mathbf{V}\mathbf{V}\mathbf{n}^{(1)} = (\lambda_1)^2\mathbf{n}^{(1)}\ .$$ (3.13.14)

From (3.13.14), (3.13.8), (3.13.9) and (3.13.10), we conclude that eigenvalues of $\mathbf{B}$ and $\mathbf{C}$ are equal, and the eigenvectors of $\mathbf{B}$ are the principal axes of stretch in the deformed configuration.

Corresponding to the two decompositions of $\mathbf{F}$ given by (3.13.1) we can view the deformation of the triad $\mathbf{N}^{(1)},\ \mathbf{N}^{(2)},\ \mathbf{N}^{(3)}$, the eigenvectors of $\mathbf{U}$, as a stretch of these axes followed by a rotation or a rotation of these axes followed by their stretch. This is schematically shown in the Fig. below.

Example: For simple shear

$$x_i = X_i + kX_1\delta_{2i},$$

a)

find the principal stretches,

b)

show that the angle $\theta$ through which the principal axes of stretch in the reference configuration are rotated so as to become the principal axes of stretch in the present configuration is given by $\tan \theta = k/2$.

Solution: For the given deformation

$$[\mathbf{F}] =\ \left[\begin{array}{ccc}1 & 0 & 0\\ k & 1 & 0\\ 0 & 0 & 1\end{array}\right]\,$$

$$[\mathbf{C}]=\ \left[\begin{array}{ccc}1 + k^2 & k & 0\\ k & 1 & 0\\ 0 & 0 & 1 \end{array}\right]\ ,$$

$$[\mathbf{B}] =\ \left[\begin{array}{ccc} 1 & k & 0\\ k & 1+k^2 & 0\\ 0 & 0 & 1\end{array}\right]\ .$$

The squares of the principal stretches $\lambda_1,\lambda_2,\lambda_3$ are roots of the equation $\det([\mathbf{C}] - \lambda[\mathbf{1}])=0$

$$\therefore\hspace{.2in} [(1 + k^2 - \lambda)(1 - \lambda) - k^2](1 - \lambda ) = 0.$$

$$\therefore\hspace{.2in} (\lambda_1)^2 = 1 + 1/2k^2 + k\sqrt{1 + 1/4 k^2}$$

$$(\lambda_2)^2 = 1 + 1/2k^2 - k\sqrt{1 + 1/4k^2} = 1/(\lambda_1)^2$$

$$\lambda_3 = 1.$$

Note that the given deformation is a plane strain deformation. By looking at the matrices $[\mathbf{C}]$ and $[\mathbf{B}]$ we see that $X_3$-axis and $x_3$-axis are eigenvectors of $[\mathbf{C}]$ and $[\mathbf{B}]$ corresponding to the eigenvalue one. Also, $x_3$-axis coincides with the $X_3$-axis for the given deformation. Thus the rotation of principal axes of stretch takes place about the $X_3$-axis. Let this angle of rotation be $\theta$ in the clockwise direction. Then

$$[\mathbf{R}] = \left[\begin{array}{ccc}\cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1\end{array}\right]$$

Substituting for $[\mathbf{B}]$, $[\mathbf{C}]$ and $[\mathbf{R}]$ in eqn. (3.13.6) we arrive at

$$\left[\begin{array}{ccc} 1 & k & 0\\ k & 1+k^2 & 0\\ 0 & 0 & 1\... ...1 + k^2\sin^2\theta + 2k\sin\theta\cos\theta & 0\\ 0 & 0 & 1\end{array}\right]$$

Therefore

$$1 = 1 + k^2\cos^2\theta - 2k\sin\theta\cos\theta$$

which gives $\tan \theta = k/2$.