natural frequency of a beam

huangyhg

natural frequency of a beam
i have two different equations that indicate how to calculate the natural frequency of a beam.  the first eqtn is from my old college books that simply indicate the frequency calculation for a simple beam as f = sqrt(g/deflection), where as a diffent book of mine indicates that the natural frequency of a beam is calculated as f = 0.18*sqrt(g/deflection).  the first book is simply a dynamic mechanics book and the later is a concrete design book that references the 0.18 eqtn to 'commentary a, serviceability criteria for deflections & vibrations,' supplement to the national building code of canada 1990, nrcc 30629, nat'l research council of canada, ottawa, 1990, pgs 134-140.  obviously an 80% reduction is pretty significant and beyond the reasonability of just using the conservative frequency calculation.
does anyone know what the 0.18 factor is for, and which eqtn is the correct one to use to determine the natural frequency of a beam?
it's quite possible that both equations are correct.  from the description, the second equation looks like the approximation for a thin plate where the 0.18 is the ratio of 1.2769/(2*pi).
no matter what the formulas you have i recommend that you check out roark's formulas for stress and strain for a complete treatment of beams, loadings and boundary conditions.
regards,
qshake
eng-tips forums:real solutions for real problems really quick.
the natural frequency in rad./sec. for a pinned-pinned beam with uniform ei and uniform mass, m, is pi^2*sqrt(ei/ml^4).  if you monkey around with this equation, you can get to fn=0.18*sqrt(g/delta) in hz if you're using usc units with g in in./sec.^2 and delta in in.
to get to 0.18 (actually 0.1792), you first convert to hz by dividing by 2*pi, observe that m=w/g, and substitute delta=5wl^4/(384ei) and simplify.  you end up with 0.1792*sqrt(g/delta).
i actually hate this equation.  it goes a long way toward decreasing physical understanding of vibrations.  the original form isn't that bad. fn=pi/2*sqrt(ei/ml^4).  it's the *mass* that affects the natural frequency, not the *load*.
there are two redeeming qualities for this equation.  first, it's easier to re  
oh yeah, as for the other form of the equation without 0.18, i'd be willing to bet that's using si units.  unfortunately, i can't quickly verify it because i don't know whether they're using gpa, mpa, etc. for e, kn/m for w, etc.  the brits usually use fo=18/sqrt(delta) with delta in mm and fo in hz.  i'm not sure how to get from there to fo=sqrt(g/delta), however.  hopefully your book will say what units are being used.
the second equation is valid, as noted by 271828fff"> for many configurations of beams, with minor variations of the coefficient.
the coefficient is very close to 0.18 for hinged-hinged, fixed-fixed and fixed-hinged; it is closer to 0.20 for a cantilever, and it is 0.16 (again very close) even for a single concentrated mass supported by any elastic structure.
the first equation must be an expression for the angular frequency ωfff">=2πffff">: it is exact for a single concentrated mass, and approximate for beams as noted above.
prex