Troubleshooting Tube Bending Natural Frequency Equation

[GenSoft3d]

I found this equation for predicting the bending natural frequency of a tube but am having trouble working it out. I was hoping that someone might be willing to steer me in the right direction. First the equation and the info provided on the website:

w = (B*l)2 * SQRT (E*I/(rho*l4))

w - frequency radian per second
E - modulus of elasticity
I - area moment of inertia = PI*d3*t/8 for a thin wall round tube
d - mean diameter
t - wall thickness
rho - mass per unit length = Area * mass per unit volume = PI*d*t*density
l - length of tube

(B*l)2 - Constants based on the boundary conditions.

For a wind chime (Free-Free Beam):

(B*l)2= 22.373 for the first natural frequency.To get the units correct you must multiply the values inside the square root by gravity (g).

g = 386.4 in/sec2 for the units I'm using.

for frequency in cycles per second (Hz) f = w/(2*PI)

Now, the problem that I'm having is that when following the above format that uses non SI units I end up with a different result than when I swap them out for SI units. So, for example; when using lbmass/in^3 for the density and psi for Young's Modulus (*gravity) along with imperial units for size inputs (inch) the results are different than the results from using kg/m3 and pascals along with metric sizes (meters). When using SI units I'm dropping the *gravity.

Here are the values that I am working with:

E = 16969415.3 psi (1.17e+11 Pa)
density = 0.323 lbmass/in^3 ( 8940 kg/m3)
mean dia = .428 in (0.0108712 m)
thickness = .048 in (0.0012192 m)
length = .3885 in (0.0098679 m)

When using the above format I'm getting 2 MHz but when switching to SI units I'm getting 66.7 MHz.

So, where am I going wrong here? Any ideas? Any advice would be greatly appreciated.

 

[AlephZero]

The formulas look OK. You only need to "multiply by g" (or more pedantically, divide the density by g to convert it into force units not mass units) working in inches, but not for SI.

FWIW I get about 0.5 MHz in both sets of units ... ?

Bear in mind this formula is for a long beam (i.e. length / diameter > 10) so it won't give very good results when the length is less than the diameter.

 

[GenSoft3d]

Hi and thanks again for the help AZ! I didn't realize that the formula was limited by the length/diameter ratio but I guess that I should have suspected. Do you happen to know if there is another formula that would be more suitable for my application? Or should I try using the 3d model simulations instead of a simple formula?

I still can't figure out where I'm going wrong here and though I may not be able to make use of this equation I am interested in knowing where I am falling short on this. As you can tell this is all very new to me and am just now trying to learn my way around so again your help is very much appreciated. Here are the numbers that I'm getting using SI units so if something looks off to you then would you mind pointing it out to me (if it's not too much trouble)?


rho (pi*d*t*density) = 0.3722543926
(rho*L^4) = 5.19935862243926E-009
I (pi*d*3* t/8) = 1.56146976758685E-005
SQRT (E*i/(rho*L^4)) = 18744973.6416667
w = (B*L)2 * SQRT (E*i/(rho*L^4)) = 419381295.28501
f= w / 2 * pi = 66746606.1848911

input values:

E = 1.17e+11 Pa
density = 8940 kg/m3
mean dia = 0.0108712 m
thickness = 0.0012192 m
length = 0.0098679 m

Thanks again!

 

[AlephZero]

(pi*d*3*t/8)

The formula is d cubed, not d times 3.

Also my calculator says rho L^4 = 3.523e-9 not 5.199e-9.

 

[GenSoft3d]

That was definitely it... it should have been cubed. Thank you! The other problem was due to an oversight in excel (trying to streamline some of these equations) which I wouldn't have caught without your help, so thank you for taking the time to look over this for me... I really appreciate it.