# Change in volume continuum mechanics

1. Apr 16, 2013

### Dustinsfl

Let a displacement field be given by
$$u_1 = \frac{1}{4}(X_3 - X_2),\quad u_2 = \frac{1}{4}(X_1 - X_3),\quad u_3 = \frac{1}{4}(X_2 - X_1).$$
Determine the volume ratio $d\mathcal{V}/d\mathcal{V}^0$ and the change in the right angle between line elements originally along the unit vectors $\hat{\mathbf{N}} = \frac{1}{\sqrt{14}} \left(3\hat{\mathbf{I}}_1 - 2\hat{\mathbf{I}}_2 - \hat{\mathbf{I}}_3\right)$ and $\hat{\mathbf{N}} = \frac{1}{\sqrt{42}}\left(\hat{\mathbf{I}}_1 + 4\hat{\mathbf{I}}_2 - 5\hat{\mathbf{I}}_3\right)$. Explain your answer.

I found the change in volume to be
The volume ratio can be found be taking the determinant of the Jacobian
of $x$ where $x = U + X_i$.
\begin{alignat*}{3}
x_1 & = & X_1 - .25X_2 + .25X_3\\
x_2 & = & .25X_1 + X_2 - .25X_3\\
x_3 & = & -.25X_1 + .25X_2 + X_3
\end{alignat*}
The Jacobian is then
$$\mathcal{J} = \begin{bmatrix} 1 & -.25 & .25\\ .25 & 1 & -.25\\ -.25 & .25 & 1 \end{bmatrix}.$$
Therefore, the volume ratio is $\lvert\mathcal{J}\rvert = \frac{d\mathcal{V}}{d\mathcal{V}^0} = 1.1875$.

How do I find the change in the right angle between line elements originally along the unit vectors specified?

SOLVED

Last edited: Apr 16, 2013