huangyhg

dynamic impact force from falling object on pavement
i have a steel beer keg of 13kg falling 2.0m onto a concrete pavement. what is the approximate duration of impact in order to calculate the deceleration and hence the force imparted on the pavement (i have estimated it to be less than 0.1 seconds but i am not sure how to justify this)
hello there,
your question may be answered in a simple way, if your beer keg behaves in a perfectly elastic linear way (which i seriously doubt).
with that assumption made, the contact time will be equal to the very simple formula:
time=pi * sqrt(m/k)
where:
m equals the "dynamic mass" of the keg. typically, it will range from 50% to 100% of its real mass, depending if all the fluid inside moves with the keg.
k equals the "stiffness" of the keg, along the falling direction. if it falls 'flat', this value will be the compressional stiffness of the keg (global value). if it falls in an oblique manner, then this value will be the local stiffness of the part that meets the ground in the first place.
btw, i would be glad doing some experiments with you keg (lately, the wheather has been very hot in paris )...
cheers
nicolas
nicolas,
thanks for your information. as the keg is empty, i used 100% of the mass.
using an approximate stiffness of 60x10^6 n/m i obtain an impact duration of 0.0015 seconds. the formula does not however allow for adjustment in impact time due to higher velocity at start time of impact - why is this so?
any reference behind that funky delta-t equation? maybe you guys are lawyers, bound by a client confidentiality agreement. but us, engineers (and reporters as well!) have to quote our sources.
roberto sanabria
i think nicolas is very sensibly suggesting that it is a half sinewave pulse, with a time constant equal to that of a single degree of freedom spring mass system.
cheers
greg locock
boy, am i glad that the keg was empty, it would have been a shame to treat beer that way! <g>
i would personally feel sorry for losing the beer, rather than worrying about structural dynamics
the duration of impact bit has nothing to do w/ your initial conditions, including the height from which the object is dropping. this height is what determines the velocity just prior to impact and thus the fantastic deceleration that takes place when the dynamic object meets the more massive static one (aka, the ground or floor).
ok, might as well spell it all out here. from the position equation (remember your physics, guys?), with an initial velocity of zero, freefalling time is determined when we solve for t = sqrt(2h/g). this is the time it takes to accelerate throughout that height from nothing other than a gravitational pull (if the height is too great a terminal velocity might be achieved and we also have to deal w/ aerodynamics).
in turn, that time also gives us the velocity just before impact (that is, at the end of the freefall from that height). velocity, as y'all might re
roberto sanabria,
1. reporters here in the us rarely ever document sources and what with the lousy sensationalism reporting that we read today, i'm not even sure they use sources, just adjectives.
2. the dynamic displacement factor of 2 times the static displacement is hardly mythical. you can find a derivation of it in almost every text on structural dynamics.
i meant to say that this widely mis-used factor of two is not applicable to most load drop cases. read the caveats that accompany the derivation. the weight is being released at the top of the spring! can't develop a great velocity that way, can we. just do the numbers and you'll see the huge difference too.
and while not all sources are quoted in our sensational press, most of them are! (just count them, pick any article randomly) anyway, i'd just appreciate knowing where this delta-t formula came from, so that i find out what the caveats are for using it and don't end up mis-applying it myself.
roberto sanabria
my apologies to messrs jobert and locock (you too, qshake). it took me a while to realize that locock's inference is that the duration of the impact is tantamount to the period of the half sinewave. but, if this is true, wouldn't... t = (pi/2) x sqrt(m/k) ...rather? in other words, only half as much as shown previously? this would be because the period is actually 2t/pi. correct me if i'm wrong here, please.
also, to clarify my position regarding the doubling of the weight representing the impact force in these calculations, this doubling only applies to "sudden loading" and not to actual load drops from a height > zero. the latter involve factors that are entire orders of magnitude greater (and this is why motorcycle helmets are designed to withstand 300 g's of impact force!!).
thanks to all for your patience and valuable insights.
roberto sanabria

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