# Finite strain

1. Apr 6, 2013

### Dustinsfl

For the deformation field given by
$$x_1 = X_1 + \alpha X_2,\quad x_2 = X_2 - \alpha X_1,\quad x_3 = X_3$$
where $\alpha$ is a constant, determine the matrix form of the tensors $\mathbf{E}$ and $\mathbf{e}$, and show that the circle of particles $X_1^2+ X_2^2 = 1$ deforms into the circle $x_1^2 + x_2^2 = 1 + \alpha^2$.

How do I find $\mathbf{e}$ and the solution is
$$\frac{1}{2(1+\alpha^2)}\begin{bmatrix} -\alpha^2 & 0 & 0\\ 0 & -\alpha^2 & 0\\ 0 & 0 & 0 \end{bmatrix}$$
First, we will find $\mathbf{E}$.
$$\mathbf{F} = \begin{bmatrix} 1 & \alpha & 0\\ -\alpha & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$
Then $\mathbf{C} = \mathbf{F}^T\mathbf{F}$. So we have that
$$\mathbf{C} = \begin{bmatrix} 1 + \alpha^2 & 0 & 0\\ 0 & 1 + \alpha^2 & 0\\ 0 & 0 & 1 \end{bmatrix}$$
$$\mathbf{E} = \frac{1}{2} \begin{bmatrix} \alpha^2 & 0 & 0\\ 0 & \alpha^2 & 0\\ 0 & 0 & 0 \end{bmatrix}.$$
I found $\mathbf{e} = \mathbf{F}^{-T}\mathbf{E}\mathbf{F}^{-1}$ but this returns the negative of the book answer for $\mathbf{e}$.
Assuming the book has a typo, how do I show $X_1^2+X_2^2 = 1$ deforms into $x_1^2 + x_2^2 = 1 +\alpha^2$?