Natural frequencies of an orthotropic bending beam with circular cross-section

Question

I am trying to find an expression for the natural frequencies of an orthotropic bending beam with a circular cross-section. I have tried searching for some literature on this, but have so far been unsuccessful.

For an isotropic bending beam, I believe the natural frequencies can be written as

$$ f_n = \frac{n}{2L^2}\sqrt{\frac{EI}{m}}, $$

where $L$ is the length, $E$ is the Young's modulus, $I$ is the area moment of inertia of the cross-section about the neutral axis, and $m$ is the mass per unit length.

Is there an equivalent expression for an orthotropic beam? i.e. one that has a different Young's modulus in the $x$ and $y$ direction say. Therefore it would have different values $E_x$ and $E_y$.

The reason for this is I am considering a circular beam made out of wood. Hence, it has different values of stiffness along the grain and across the grain.

If there is an expression for the natural frequencies of a circular orthotropic beam, I want to be able to show that if $E_x$ and $E_y$ are equal, the expression should reduce to the isotropic case.

Answer 1

I don't believe there are general (longitudinal, transverse, torsion) closed form solutions even for the case where the cylinder is of infinite length. However, you can use the review article Kostas Soldatos (Soldatos, Kostas P. "Review of three dimensional dynamic analyses of circular cylinders and cylindrical shells." Applied Mechanics Reviews 47.10 (1994): 501-516.) as a starting point and solve the eigenvalue problem numerically.

The basic idea is to start from the three-dimensional equations of elasticity in cylindrical coordinates and then assume appropriate harmonic displacement fields. Plugging in the displacement fields leads to an eigenvalue problem that can be solve numerically (or analytically in a few cases).

You can also look at Grigorenko and Efimova's work (Grigorenko, A. Ya, and T. L. Efimova. "Using spline-approximation to solve problems of axisymmetric free vibration of thick-walled orthotropic cylinders." International Applied Mechanics 44.10 (2008): 1137-1147.) for a different approach.

For a more recent overview (haven't looked at it closely) you might want to see http://14.139.56.14/library/phd/TH188.pdf and Grigorenko, A. Ya, and T. L. Efimova. "Free vibrations of continuously inhomogeneous cylindrical bodies." Shell Structures: Theory and Applications 3 (2013): 293.

Answer 2

If you're designing something, you don't need an analytical formula and can use finite element analysis instead to solve specific configurations. This problem fits very easily into the basic functionality of most FEA software.

Answer 3

The formula presented neglects shear effects - in other words, follows Euler-Bernoulli beam theory. For design purposes this is typically conservative to ensure something does not vibrate, but has the drawback that it does not predict longitudinal vibrations at all. Without a shear modulus, it would not predict torsional vibrations either.

Typically for long beams (about 10 times longer than they are wide), this model is accurate, and you can demonstrate what you want to. In the y-direction, the model uses $E_y$, and in the x-direction, the model uses $E_x$ for classical lamination theory (CLT) models utilizing ABD matrices. If both were equal, then this model would show they have the same value in both directions.

Note there are many flaws to this model - if the beam were struck at an angle, how would it vibrate? The model essentially would split the force into the two components and run the oscillations in both directions at the natural frequency of each, which is highly counterintuitive. Surely one would dominate after energy dissipation removes the other, but how quickly? For these and other questions for specialty design, the original Euler-Bernoulli beam theory wouldn't have answers, and neither would CLT based Euler-Bernoulli beam theory. These would come back to the isotropic case of their original theory, but most likely would not come back to the same equation you presented.