Calculation of natural frequency

[mechengdude]

I am doing a natural frequency calculation and having trouble reconciling the two different results given by textbook formulas. For the calculation of natural frequency of a simple round beam with a mass at the end the following formula is given:
natural freq. (rad/s) = sqrt[3EI/(M+.23m)l^3]
Using this formula and given a stainless steel circular beam (dia. 1") 12" long with a 5.5lb weight on the end I get a natural frequency of about 19.7 rad/s or 3.14 Hz

Now using the formula that relates static deflection to natural frequency I get a very different answer. This formula states:
natural freq. = SQRT(g/d) where g is gravity and d is deflection. Using that formula I get about 385 rad/sec or 61hz.

This is obviously a huge difference and I'm guessing some mis-application of these formulas. Can anyone shed some light on this for me?

Thank you

 

[ivymike]

if d = gm/k
then g/d = k/m

so sqrt(k/m) should equal sqrt(g/d) if you're calculating k correctly...

k is force/displacement

http://www.efunda.com/formulae/solid_mechanics/beams/casestudy_display.cfm?case=cantilever_endload

displacement for a cantilever beam w/ end load is -((force)*L^3)/3EI

k for a cantilever beam w/ end load is 3EI/L^3

so my k matches your k. I'm not sure why your formula has you using 1.23m instead of m.

aside from that, you must have messed up the calculation of I, or messed up your calculation of deflection.

 

[mechengdude]

My, "I" is calculated at .049in^4 and my deflection is correct. I have doubled checked it against a simple FEA. Deflection should be about .0023". This value is close to the calculation sheet you were kind enough to link.

As far as the formula goes, the source is Mark's Standard Handbook. 9th edition page 5-70. The M is for the mass at the end of the beam and the .23m is for the mass of the actual beam.

thanks for your response

 

[mechengdude]

O.K. - "stop the presses". I think I found my error. I forgot that little freshman thing about W=mg. Once I remebered that mass and weight aren't the same, things work out.

thanks

 

[ivymike]

deflection seems right

your g/d formula sounds like it's probably giving the right answer

when I go through the sqrt(k/m) formula, I get 67hz...
d = 0.0254m
I = pi*d^4 / 64
E = 207GPa
L = 0.3048m
m = 2.4948kg

3EI/L = k = 447988 N/m
sqrt(k/m) = 423rad/s = 67 cycle/sec

so find where you're messing up that first calc

 

[ivymike]

ah, I was calc'ing and typing while you were finding the mistake. looks like you got it.

 

[electricpete]

I was going to comment that the formula
"(rad/s) = sqrt[3EI/(M+.23m)l^3]"
Is a little ambigous as written with regard to whether the l^3 goes in numerator or denominator. If you got the right answer, you no doubt put the l^3 in the denominator as required.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

And I would also mention that the static deflection formula is an approximation in case like this where we have distributed mass. But pretty darned close as long as m << M.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.