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In elasticity theory, general equations of motion are:

$$\rho \partial^2_t \overline{u} = \mu \nabla^2 \overline{u} + (\mu+\lambda) \nabla(\nabla \cdot \overline{u})$$

where $\overline u$ are displacements.

In a thin homogeneus rod with tractions and compressions in the x direction, stress tensor $S_{ij}$ is: $$ \pmatrix{ S_{xx} &0 &0 \\ 0 &0 &0 \\ 0 &0 &0 \\} $$ Young modulus is: $$ E=\frac{S_{xx}}{\epsilon_{11}}=\frac{\lambda+\mu}{\mu(3\lambda+2\mu)} $$ where $\epsilon_{11}=\partial_x u_x$ is the deformation on x

So equation of motion for x-displacements is: $$ \rho \partial^2_t u_x=\partial_x S_{xx}= E \partial^2_x u_x= \frac{\lambda+\mu}{\mu(3\lambda+2\mu)} \partial^2_x u_x $$

Why is this different to the result of general equation of motion with only dependence to x: $$ \rho \partial^2_t u_x=(2\mu+\lambda)\partial^2_x u_x $$ ?

Or: why in a elastic rod has the sound smaller speed than in a generic homogeneus indefinite medium?

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up vote 3 down vote accepted

Just check your math. Correct derivation is

$$\begin{eqnarray} {f \over \epsilon_{11}} & = & {S_{11} \over \epsilon_{11}}=(2\mu+\lambda)+\lambda {\epsilon_{22}+\epsilon_{33} \over \epsilon_{11}} = \\ & = & (2\mu+\lambda)-{\lambda^2 \over \lambda + \mu} = {(2\mu + \lambda)(\lambda + \mu) - \lambda^2 \over \lambda + \mu}=\frac{\mu(3\lambda+2\mu)}{\lambda+\mu} \end{eqnarray}$$

And this is precisely what Young modulus is defined to be (see e.g. here).

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sorry, I'm stupid! Thank you!Now I'll write the most important question I'll change the question with my real doubt. I thought it was the reason, but it isn't ... –  Boy Simone Jan 8 '11 at 2:15
    
@Boy: no problem. Feel feel to change the question, I'll update my answer as necessary. –  Marek Jan 8 '11 at 2:16
    
@Boy Well, it is a better idea to make a new question. –  mbq Jan 8 '11 at 14:31
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