# Right stretch tensor

1. Apr 16, 2013

### Dustinsfl

Consider the deformation field
\begin{alignat*}{3}
x_1 & = & X_1 - AX_2 + AX_3\\
x_2 & = & X_2 - AX_3 + AX_1\\
x_3 & = & X_3 - AX_1 + AX_2
\end{alignat*}
where $A$ is a constant. Show that the principal values of the right stretch tensor have a multiplicity of two, and that the axis of the rotation tensor is along $\hat{\mathbf{N}} = \frac{1}{\sqrt{3}}\left(\hat{\mathbf{I}}_1 + \hat{\mathbf{I}}_2 + \hat{\mathbf{I}}_3\right)$. Determine the matrix of the rotation vector together with the angle of rotation $\phi$.

How do I determine the matrix of rotation and the angle phi?

The deformation gradient, $\mathbf{F}$, is given by
$$\mathbf{F} = \begin{bmatrix} 1 & -A & A\\ A & 1 & -A\\ -A & A & 1 \end{bmatrix}.$$
Then $\mathbf{C} = \mathbf{F}^T\mathbf{F}$. So $\Lambda_1 = 1$, $\Lambda_{2,3} = \sqrt{1 + 3A^2} = \beta$ which is the square root of the eigenvalues of $\mathbf{C}$. If we take the eigenvectors and multiple them together such that we end up with three $3\times 3$ matrices, we will have
$$\begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1 \end{bmatrix}\quad \begin{bmatrix} 1 & 0 & -1\\ 0 & 0 & 0\\ -1 & 0 & 1 \end{bmatrix}\quad \begin{bmatrix} 1 & 0 & -1\\ 0 & 0 & 0\\ -1 & 0 & 1 \end{bmatrix}\quad \begin{bmatrix} 1 & -1 & 0\\ -1 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix}.$$
The right stretch tensor $\mathbf{U}$ is given by $\mathbf{U} = \Lambda_1 \mathbf{N}_1\otimes\mathbf{N}_1 + \Lambda_2 \mathbf{N}_2\otimes\mathbf{N}_2 + \Lambda_3 \mathbf{N}_3\otimes\mathbf{N}_3$ where $\mathbf{N}_i$ for $i = 1,2,3$ are the eigenvectors.
$$\mathbf{U} = \begin{bmatrix} 1 + 2\beta & 1 - \beta & 1 - \beta\\ 1 - \beta & 1 + \beta & 1\\ 1 - \beta & 1 & 1 + \beta \end{bmatrix}$$
Then $\mathbf{R} = \mathbf{F}\mathbf{U}^{-1}$.
$$\mathbf{R} = \frac{1}{9\beta} \begin{bmatrix} \beta + 2 & \beta - 9A - 1 & \beta + 9A - 1\\ \beta + 3A - 1 & \beta + 3A + 5 & \beta - 6A - 4\\ \beta - 3A - 1 & \beta + 6A - 4 & \beta - 3A + 5 \end{bmatrix}$$