What is the in-plane resonance frequency equation for a single-beam cantilever of rectangular cross-section that is fixed on one end and has a mass M attached to the other end?
Below are the only equations that I was able to find and they came from this site:
$f = \frac{1}{2\pi}\sqrt{\frac{k}{M + \rho \cdot t \cdot w \cdot l}}$
$k = \frac{E \cdot w \cdot t^3}{4 \cdot l^3}$
- f = Resonance frequency (Hz)
- M = Mass added to the cantilever's free end (kg)
- $\rho$ = Density of the material that the cantilever is made of (kg/$m^3$)
- t = Thickenss of cantilever (m)
- w = Width of cantilever (m)
- l = Length of cantilever (m)
- E = Elastic modulus or Young's modulus (kg/$m \cdot s^2$)
The problem is that the site author never stated if these were the equations for in-plane or out-of-plane motion and I couldn't make a deduction based on the provided context; the equations should differ, correct?
Best Answer
If you rearrange the formula to use the beam's moment of inertia, then it is fairly obvious is applies to out-of-plane bending, just like formulas for static deflection.
This also makes the formula applicable to any beam cross section, not just a rectangle!
For your rectangular beam, $I = wt^3/12$. Rearrange the formula to use $EI$ instead of the explicit references to $w$ and $t$.
For bending in the other plane, you just have $I = w^3t/12$, of course.
(Note: I seem to have "in plane" and "out of plane" the other way round from Mohan's answer. I'm not going to argue about which is correct - they are not terms that I would normally use anyway! Every beam cross section has two principal axes, defining two planes - arbitrarily choosing one of them and then calling the two directions "in plane" and "out of plane" doesn't seem very logical to me.)