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Comparing force of static object to that of an impact force

  1. Nov 17, 2015 #1
    I have been asked by a piling engineer to try to come up with an alternative to static pile testing in a restricted access site. Normally a pile test is carried out by applying a static load to the ground and measuring the deflection of the pile under the load. This of course requires a large pile of concrete blocks or similar to act as the counter weight. The alternative is to carry out what is known as a dynamic pile test, where a weight is dropped a known distance onto the pile and the pile response measured. I was wondering if there is a way to calculate the how heavy a weight would need to be dropped over a distance to equal the force applied by a static object. All of the equations I have seen to calculate the force of impact of a falling object require the distance the impacted surface moves to provide the answer. But what if the dropped weight is insufficient to cause any movement? This implies an infinite force. Perhaps the distance of the rebound has a bearing? Any feedback is welcome.
  2. jcsd
  3. Nov 17, 2015 #2


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    The way to equate the two weights is to make the KE of the dropped weight = PE of the static weight.

    Just as a note: The infinite force is correct when the impact of two absolutely perfectly rigid bodies is assumed; however, this condition never actually exists in any material because every material will compress on contact/impact. The contact, no matter how soft, will always result in a compression of the two contacting surfaces and absorb the contact/impact energy, even if the combined compression of the two contacting surfaces is only 0.00010 in. or less.
  4. Nov 18, 2015 #3
    Thanks for the reply. I can work out the kinetic energy of the falling object easily enough. The PE of the static body would be it's mass times the distance to the centre of the earth, I believe. Going back to the forces of the static body and the impacting body. The static body applies a force by virtue of it's mass and the gravitational field and is not influenced by the properties of the surface it is acting on. The force applied will be the same if the static object is pressing down on a soft or rigid surface. However the force of an impacting body depends on it's mass, velocity AND the properties of the impacted surface. So a moving object has a lower force of impact on a soft surface than a rigid surface? So it seems that I will not be able to compare forces in the two situations, but can equate energies instead.
  5. Nov 18, 2015 #4


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    If you google "dynamic pile testing" you find references that suggest this method has been in use since the 19th century. It doesn't appear to be as simple as just dropping a weight on the pile.

    Many piles are hammered into the ground and it looks like tools have been developed that analyse the bearing capacity while it's being hammered in. Perhaps allowing you to hammer it in until the required bearing capacity has been reached.

    http://www.esg.co.uk/services/dynamic-pile-load-testing/ [Broken]

    I doubt any construction site inspector would allow an adhoc test suggested on a physics forum to be used. I would suggest your piling friend probably needs to sign up for an industry recognised training course that covers dynamic pile testing. Perhaps the equipment companies offer free training courses with their products?
    Last edited by a moderator: May 7, 2017
  6. Nov 18, 2015 #5


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    This is to correct an error in my first post as to how to equate the two weights. What happened was that I made a last minute revision in an effort to simplify my original entry that I did not like and made an error in the revision.

    The method I intended to convey in the briefest terms was "The way to equate the two weights is to make the KE of the dropped weight = the change in the PE of the pile under the static load." In other words, chose a reference material (or spring) and calculate its deflection with the static load applied (i.e. change in PE pile) and then select a new weight and drop drop height whose combined KE equals the same deflection (i.e change in PE pile) as the that of the applied static load.

    I hope this is a much clearer and more helpful response than that in my original post. I apologize for my first confusing one.
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