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Change in volume continuum mechanics

  1. Apr 16, 2013 #1
    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
  2. jcsd
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