huangyhg

seemingly dumb vibration question
this question might come across as dumb, but......
when you're determining the natural frequency of a floor system you're supposed to use the natural frequency equation of something like 0.18*((g/delta)^0.5). my understanding is that this is true for something displaced by an amount delta and then suddenly released. is this an accurate statement?
if so, why do you use the delta for the actual floor and not the delta imposed by the footfall? it seems the footfall is the only load causing a delta that is being removed. the delta caused by the dl of the floor is not being removed so how is it contributing to the natural frequency?
i think your equation is right, 1/(2*pi)*sqrt(g/delta) is the natural cyclic frequency obtained by dividing the circular frequency [sqrt(g/delta)]by 2*pi...
from my understanding (famous last words...) it is true that the floor is rebounding from the footfall, but the footfall is no longer acting (hence the rebound) so it no longer applies. the delta obtained after the footfall is from m*g of the floor.
maybe i could have answered your question a little more explicitly... the natural frequency of a system depends on the mass of the system [your equation could also be written 1/(2*pi)*sqrt(k/m)... delta is equal to mg/k, delta is the static deflection of the mass m suspended from a spring of stiffness k]. once the load is removed it is no longer a part of the system, and therefore the natural frequency of the system is independent of the effects of the load after it is removed.
does that help?
not really, because the majority of the load isn't removed. only the footfall is removed.
don't feel bad. we really don't do a good job of teaching vibrations in ce school.
natural frequencies are properties of the structure (function of stiffness and mass) and are usually completely independent of the load. the only except i can think of for linear systems would be members with axial load. for example, a string has a higher frquency if it's pulled tighter and a column has a lower frequency if it's in heavy compression and about to buckle.
the fn = 0.18*sqrt(g/delta) is very handy for memorizing, but is an abomination otherwise because it causes confusion like in this case.
the actual equation for a simply supported beam is something like (don't have my book here and too lazy to google it) fn = pi/2*sqrt(ml^4/ei) -- i'm sure i lost a constant or something there. this is relatively difficult to remember, as i've illustrated here. m = a uniform line mass along the beam.
structural eit-
your formula is a handy approximation of the natural frequency of a simply supported beam. the true formula is ?=(n2π/2l2)√(ei/m), where n is the mode and n=1, 2, 3... for more discussion, see
right, the footfall is removed and is no longer a part of the system, only the dl remains. the natural frequency depends on the stiffness and mass, two intensive properties of the system. the natural frequency will be the same no matter what load you apply. once the load is removed, the system will act on its own which is based on its mass (dl). i think using delta is misleading, i usually think of it in terms of k and m.
when a beam is excited, its frequency of vibration is called its natural frequency. it doesn't matter what the exciting force is. the natural frequency is a property of the structure.
it is the reason why an army platoon breaks step when crossing a bridge. if they march across in step and happen to hit the natural frequency, the bridge could sway uncontrollably or even collapse if the excitation continues for long enough.
the following article may throw a little more light on the subject for a simple beam with equal cantilevers each end:
when a beam is excited, its frequency of vibration is called its natural frequency. it doesn't matter what the exciting force is. the natural frequency is a property of the structure.
it is the reason why an army platoon breaks step when crossing a bridge. if they march across in step and happen to hit the natural frequency, the bridge could sway uncontrollably or even collapse if the excitation continues for long enough.
the following article may throw a little more light on the subject for a simple beam with equal cantilevers each end:
thanks everyone.
"when a beam is excited, its frequency of vibration is called its natural frequency. it doesn't matter what the exciting force is."
respectfully, this one part of your explanation isn't true. if a system is excited by a sinusoial force at frequency f, the response is sinusoidal at frequency f. natural frequencies are special because when f equals one of them, the structure goes into resonance.
i agree with the rest of your explanation.
there's also a side discussion going on in this thread about footstep application and how that relates to natural frequencies. there is certainly structure-human interaction, but it is usually insignificant and therefore ignored except for maybe throwing a psf or two in there for humans on the structure. the major exceptions are very low frequency structures and very heavily loaded structures.
in very low freq structures, humans can subconsciously sense the vibration and synchronize footsteps to the response frequency to some extent. this is roughly what happened at the millenium footbridge in europe a few years ago--look that one up on youtube.
heavily loaded structures like balconies have so many people that one has to give the load a little more thought.
one thing to consider is that footstep application times are roughly 0.5 sec. to 0.8 sec. per step depending on step frequency and the right foot lands before the left leaves and vice versa. the slab vibrates through multiple cycles during the footstep application time. there's really no point to the argument that the footstep is only there some of the time. the truth is that one human on a typical structure doesn't interact with it enough to change how we compute fn. just compute it as if the human isn't there. if there is a boatload of people, a crowd, or whatever, then engineering judgment must come into play.

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