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When a longitudinal wave is sent through a body, there is a strain in the emitted direction (x). What about the three-dimensional body with the strains in the direction perpendicular to the emitting direction (y, z). This strains must inevitably occur due to the "pressure wave". how can these be calculated?

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We can write the displacement vector as

$$\mathbf{u} = A \cos(kx)\cos(\omega t)(1, 0, 0)\, .$$

Thus, the strain tensor that is defined as

$$\epsilon = \frac{1}{2}[\nabla \mathbf{u} + (\nabla\mathbf{u})^T]\, ,$$

is

$$\epsilon = -Ak\sin(kx)\cos(\omega t) \begin{bmatrix} 1 &0&0\\ 0 &0 &0\\ 0 &0 &0\end{bmatrix}\, .$$

Furthermore, the stress tensor, that for a linear elastic material isotropic is defined as

$$\sigma = \lambda \mathrm{tr}(\epsilon) + 2\mu\epsilon\, ,$$

looks like

$$\sigma = -Ak\sin(kx)\cos(\omega t)\left[\begin{matrix}\lambda + 2 \mu & 0 & 0\\0 & \lambda & 0\\0 & 0 & \lambda\end{matrix}\right]\, .$$

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  • $\begingroup$ That was very helpful! What would the strains eyy and ezz be (or also the strain tensor)? $\endgroup$
    – Frank
    Jul 25 at 19:05
  • $\begingroup$ @Frank, I have updated the answer. $\endgroup$
    – nicoguaro
    Jul 25 at 20:51
  • $\begingroup$ Thank you for the renewed answer, @nicoguaro!!! I don't quite understand the strain tensor yet. The pressure wave of the longitudinal wave would inevitably also have to generate -small- strains in eyy and ezz, since the definition of Hook's law means that the compressive stresses in 3D also influence all vectors, right? $\endgroup$
    – Frank
    Jul 26 at 6:41
  • $\begingroup$ One more question about the stress tensor. Shouldn't it be like that: <br/> \sigma = -Ak\sin(kx)\cos(\omega t)\left[\begin{matrix}\lambda + 2 \mu & 0 & 0\\0 & \mu & 0\\0 & 0 & \mu\end{matrix}\right]\, . $\endgroup$
    – Frank
    Jul 26 at 7:10
  • $\begingroup$ @Frank, I don't think that your intuition is correct. I have added some details and hyperlinks for reference. $\endgroup$
    – nicoguaro
    Jul 26 at 14:31

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