I have solved the problem posed in my previous question where we assume that the resonant frequency is depednent only on density rho, young's modulus E, and geometry = dimensions. Let's say we build some steel structure of an arbitrary shape and boundary condition (let's say a cactus shape, rigidly mounted at the bottom). We can characterize that shape by some characteristic dimension L, let's say diameter of one of the trunk of the cactus branch. Let's say we build that structure with characteristic dimension L = L1 and perform impact test to find resonant frequency F1. I claim that if you build any scaled up version of the same steel cactus with L = L2, then the resonant frequency will be F2 = F1*L1/L2.
The proof comes from dimensional analysis. I will discuss 4 types of variables:
1 - physical variables (subscript p) - the ones we are most familiar with, like length (units of meters), time (units of sec) etc.
2 - primary base variables (subscript b) - a selected minimum set of variables whose units are independent and whose units can be combined to form the dimension of any variable of the problme.
3 - secondary base variables (subscript b, also) - combination of primary variables to form the same units as another physical variable.
4 - dimensionless variables (subscript d) = ratio of phyiscal variable to base variable of the corresponding dimensions.
In our example cactus problem, we select primary base variables:
Eb: kg/(m*sec^2)
rhob: kg/m^3
Characteristic dimension Lb: m^3
We can also see:
Eb/rhob: m^2/sec^2
fb = sqrt(Eb/(rhob*Lb^2)): has dimensions of sec^-1.
When we perform bump test of system 1, with characteristic dimension L1 = LB1, and we measure f1.
We know that fd = f1/fb1 = f1/sqrt(Eb/(rhob*Lb1^2))
we can compute it from our Lb1 etc if we want.
If we want to evaluate system 2, simply select new Lb as Lb2. Choice of base variable will not affect the dimensionless solution.
Dimensionless solution fd holds true regardless of choice of base variables:
fd = f1/fb1 = f2/fb2
f2 = f1 * fb2/fb1
f2 = f1 * sqrt(Eb/(rhob*Lb2^2)) / sqrt(Eb/(rhob*Lb1^2))
We are using the same materail E and rho and applying it as our same base Eb and rhob, so these cancel out
f2 = f1 * sqrt(Lb1^2/(Lb2^2)
f2 = f1 * Lb1/Lb2
f2 = f1 * L1/L2
It is a surprisingly simple result, but should hold regardles of the shape of the object (as long as assumption holds).
Let's just try an example where we know the solution.. simple beam.
We know the solution is of the form:
f ~ sqrt(E*I / (mu*L^4)) where mu is mass per density.
Let Lx be some characteristic dimension (it doesn't matter which one we choose because we will scale them all proportionally, and we are only looking at proportional scaling of all dimensions)
I~Lx^4
mu~Lx^2
L~Lx^4
plug in
f ~ sqrt(E*Lx^4 / (L1^2*Lx^4))
f ~ sqrt(1/Lx^2)
f ~ 1/Lx
We have proved the claimed result for simple beam based on known beam solution. It will hold for more general/complicated geometries as well (within assumptions).
If there is interaction with fluid, and velocity of fluid flow past the part affects resonant frequency, it becomes more complicated. Let us know.
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(2B)+(2B)' ?