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Dynamic Impact Force from falling object on pavement

Dynamic Impact Force from falling object on pavement

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)

RE: Dynamic Impact Force from falling object on pavement

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)

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 )...



RE: Dynamic Impact Force from falling object on pavement


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?

RE: Dynamic Impact Force from falling object on pavement

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

RE: Dynamic Impact Force from falling object on pavement

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.


Greg Locock

RE: Dynamic Impact Force from falling object on pavement

Boy, am I glad that the keg was empty, it would have been a shame to treat beer that way!  <G>

RE: Dynamic Impact Force from falling object on pavement

I would personally feel sorry for losing the beer, rather than worrying about structural dynamics

RE: Dynamic Impact Force from falling object on pavement

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 remember, is the 1st derivative of acceleration, so instead of the previous 1/2(g)t^2 term, we now have only v = g(t).  This final velocity is subtracted from the initial velocity which should be zero if the object was at rest when it fell from that height.

Dividing the resulting overall velocity by the short time duration the impact lasts gives us the deceleration, which multiplied by the falling object's mass yields the average impact force, F = m[(V1-V0)/dt].  And so it is, that the height and resulting velocity at impact DO play a big role in determining the size of the wallop, as does the mass of the falling object as well.

The ratio of this force over the weight of the falling object can be huge, certainly much bigger than the mythological factor of TWO many people use for this purpose.  And if you guys think the resulting figures are impressive you should see what happens when a 90 mph baseball is hit by a bat and does a sudden about-face at even greater velocity.  Now, can I have the source for that delta-t equation please?

Roberto Sanabria

RE: Dynamic Impact Force from falling object on pavement

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.

RE: Dynamic Impact Force from falling object on pavement

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

RE: Dynamic Impact Force from falling object on pavement

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

RE: Dynamic Impact Force from falling object on pavement

If nicolas' theory is right, and I think it is a great way of getting a handle on this problem, the duration of the pulse will be (1/f)*.5, because it is a half wave.


So, t=pi*sqrt(m/k)

I think.

I agree entirely with the difference between sudden and shock loading, sadly there are many people who think that a factor of 2 covers all events.


Greg Locock

RE: Dynamic Impact Force from falling object on pavement

Ok, I stand corrected then, single PI it is.  This is such a beautiful solution.  I can't believe it's this simple.  I guess the only other thing left to discuss is the stiffness.  Per the derivation, it should be the stiffness of the falling object.  But, relativistically speaking, it should be mostly that of the softest of the two objects in the collision.

I guess we could combine both objects' stiffnesses in a series type formula (reciprocal of the reciprocals' sum).  This could have a significant impact, so to speak, because it's the combined stiffnesses of these colliding objects that provides the damping effect that can make the duration of impact large or small, thus reducing or increasing the impact force, respectively.

Ümüt, my verbosity got the best of me on Oct 14th.  You can get your falling time "t" either from the position eq'n (y = yo + V(t) + 1/2A(t)2, where yo = h and A = -g), or from a Conservation of Energy approach, whereby the potential energy at rest equals the kinetic energy once in motion (mgh = 1/2mV2).  The trailing two's in the preceding formulas were meant to be squares, by the way.

In the latter method, once you have the velocity at impact, t = h/V.  Don't confuse the two times "t" in here, one is for Falling Time and the other for Duration of Impact (much smaller).

Roberto Sanabria

RE: Dynamic Impact Force from falling object on pavement

Re the formula discussed by NicolasJobert, GregLocock and RobertoMiguel.

I like the approach, chaps.  Unfortunately I want to re-muddy the waters on the matter of a half-wave or a quarter-wave.  I believe that the correct formula for the deceleration time is
time=pi*sqrt(m/k) / 2
(ie a quarter wave)

We are assuming that the body behaves elastically, and therefore that the contact force is proportional to the contact deformation.  After a quarter of the "period", the force will have increased from zero to its maximum value.  Simultaneously the body's speed has decreased to zero.  (If its speed had not reduced to zero, it would still be travelling downwards and therefore the elastic contact force would still be inceasing.)

However I'm open to argument.

RE: Dynamic Impact Force from falling object on pavement

Roberto, I agree, you should use the total stiffness in the system, not just the moving body's stiffness.

Denial, I think I see what you are getting at, but bear in mind that this approach (and it is an ESTIMATE after all) assumes that the two bodies are joined by one spring, so there will be contact, and so force in one direction, until the moving body and the target are back to their original separation.

I'd add that this is a somewhat flaky approach, but I have used it in slightly more complex cases and it has agreed to within 5% of the measured value - I think it was coincidence, but hey, it makes a nice change for the computer estimates to be that close.

Incidentally if you want to see the gory details for two hard spheres colliding then Thread404-74498 has the formula and a lot of relevant and tangential discussion.


Greg Locock

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