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Right stretch tensor

  1. Apr 16, 2013 #1
    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}##.
    $$
    \mathbf{R} = \frac{1}{9\beta}
    \begin{bmatrix}
    \beta + 2 & \beta - 9A - 1 & \beta + 9A - 1\\
    \beta + 3A - 1 & \beta + 3A + 5 & \beta - 6A - 4\\
    \beta - 3A - 1 & \beta + 6A - 4 & \beta - 3A + 5
    \end{bmatrix}
    $$
     
  2. jcsd
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