1
$\begingroup$

The equation for flexural waves in thin rods is $$\frac{\partial^4 v}{\partial x^4} + \frac{\rho A}{EI}\frac{\partial^2 v}{\partial t^2} = 0$$

where $I$ is the $2nd$ moment of area.

Now, as Navier's equations of elasticity are the general equations for a $3$ dimensional homogenous isotropic elastic body, it should be possible to derive the above wave equation from Navier's equations. However, I was unable to do so, and I couldn't find any relevant resources online. Could someone help me out with this?

$\endgroup$

1 Answer 1

1
$\begingroup$

I asked myself the same question some time ago. This is maybe not as straightforward as you might think, and this is best obtained from an intermediate step of "structural" mechanics equations relating moment resultants and efforts in a rod: $$ \frac{dM}{dx}= V \tag{1}$$ $$ \frac{dV}{dx}= \rho A \frac{d^{2} v}{ dt^{2}} \tag{2}$$ (2) where $M$ is the bending moment, $V$ is the shear effort. Of course substituting $M=\frac{1}{EI} \frac{d^{2}v}{dx^{2}}$ will yield the equation for flexural waves.

Now the question is how to obtain (1) and (2) from the general 3D elastodynamics equation (from which Navier equation stems): $$\sigma_{ji,j}= \rho \partial_{tt} u_{i} \tag{3}$$ For clarity, we will use the notation $u_{3}=v$ to be consistent with your equation. Now, within Euler-Bernouilli approximations, (3) sums up to: $$\frac{\partial \sigma_{xx}}{\partial x} = - \frac{\partial \sigma_{zx}}{\partial z} \tag{4} $$ $$\frac{\partial \sigma_{xz}}{\partial x} = \rho \frac{\partial^{2} v}{ \partial t^{2}} \tag{5}$$

Then (1) is otained by multiplying (4) by $z$ and integrating over the rod section, noting that $ \iint z \sigma_{xx} dS = M$, $\iint \sigma_{zx}dS=V$, and evaluation of the integral of the rhs of (4) requires integration by parts and the vanishing of $\sigma_{xz}$ on the rod boundary. (2) is obtained similarly by integrating (5) over the rod section.

Additional notes: If you want to go down to what the E-B flexural wave theory actually implies when re-deriving it from 3D elasticity, you have $u= -z \frac{\partial{v}}{\partial{x}}$ and $v$ independent of $z$ from classical E-B kinematics hypothesis but also, plane stress assumptions, and rotary inertia of rod sections neglected. Also, and more intriguing, you may notice that shear strain $\gamma_{xz}$ is zero in E-B theory but you may not neglect terms like $\frac{\partial \sigma_{zx}}{\partial z}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.