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Pile driver - Conservation of enegy

  1. Jun 9, 2009 #1

    I did some calculations but somehow I feel it's not 100% correct. Could someone have a look please.

    Many Thanks

    A pile driver of mass mh has struck a pile of mass mp and driven the pile D meters into the ground, if the velocity of the strike is V, determine the force due to the resistance in the ground by the following methods:
    1. D’Alembers Principle
    2. Conservation of energy

    V = 3.62m/s
    mh = 100kg
    mp = 200kg
    D = 0.08m
    g = 9.81 m/s^2


    v^2 = u^2 + 2as

    v = initial velocity
    u = final velocity
    a = acceleration
    s = displacement

    transposed the formula to get the acceleration.

    a = v^2 - u^2 / 2s
    a = 3.62^2 - 0 / 2 * 0.08
    a = -81.9 m/s^2

    F = ma
    F = (200 + 100) * 81.9
    F = 24570N

    Ground Resistance
    R = 300 * 9.81 + 24570
    R = 27513N

    For the second part I'm trying to work out the Potential and Kinetic energy but then i have no idea what to do next....

    K.E = 1/2 mv^2
    K.E = 1/2 * 200 * 3.62^2
    K.E = 1310.44J

    Height of fall
    1/2mv^2 = mgh

    Transposed to make h the subject

    h= 1/2*mv^2 / mg
    h = (300 * 3.62^2) / 2+ (200 * 9.81)
    h = ~1m

    P.E = mgh
    P.E = 200 * 9.81 * 1
    P.E = 1962J
  2. jcsd
  3. Jun 10, 2009 #2


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    Since the pile has mass, I am unsure from the problem if the given 'strike' velocity is the same as the initial velocity of the driver-pile system as they start their decent into the ground. If it is, I agree with your solution in the first part for R= 27,513 N. To solve using energy methods, remember to include the work done by the ground resistance force (delta KE plus delta PE + W =0). And check the value you are using for the mass.
  4. Jun 10, 2009 #3
    That’s the conclusion which I came to early today :) and it works out... now how to delete this post, as there are people on my course that won't do the work just Google it and copy the work that took me 3 days to do!

    Thanks for the help!
  5. Jun 11, 2009 #4


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    I guess I can delete my post if you delete yours, but a good teacher or TA or Prof will sort out the 'cheaters' from those who have spent hours of work on the problem. But we don't really have an answer here yet; we've assumed the 'strike' velocity to be the same as the initial velocity of the driver/pile system, but if the problem defines 'strike' velocity as the speed of the driver head just before impact, then you need to first apply conservation of momentum principles to determine the initial speed of the driver/pile system as if the collision of the two were totally inelastic (that is, they join together during impact and proceed downward into the ground as one unit during the 0.08m decent). I think the problem statement is not clear.
  6. Jun 12, 2009 #5
    I will actually gamble at this stage and stick with the first answer of 27513N i realy hope this is right...
  7. Jun 12, 2009 #6

    Doc Al

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    I would have assumed that the "velocity of the strike" was the speed of the pile driver just before it collided with the pile.
  8. Jun 12, 2009 #7


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    I'm inclined to agree.
  9. Jun 12, 2009 #8
    damn it.... back to work than...
  10. Jun 12, 2009 #9
    ok I redone my calculations and now the results from the first method don't match the results of the second.... Another remark from my side - I don’t expect this task to by "super duper" complex. While if I apply the numbers as assumed at the beginning it all works out with only a difference of 0.27.
  11. Jun 12, 2009 #10


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    You've done something wrong. In your first attempt , when using the initial velocity of 3.62, you arrived at the same answer for R using Newton's laws versus Energy methods. Now in your new attempt, you need to calculate the initial velocity based on the conservation of momentum principle, then the steps after that to solve for R by either method, are the same. Only V_initial changes.
  12. Jun 12, 2009 #11
    That does look ok now, If u assume that the velocity given is the velocity before strike :) I can always argue the case that the problem was not stated clear enough! But for the first time actually now I feel like this is it.

    Still waiting for comments before I delete this post!

    Many Thanks
  13. Jun 12, 2009 #12


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    you mean of course (100)(3.62) = 362 N-s
    Looks good, so I guess its OK to sound the trumpet, on the assumption that Doc Als' assumption, which I agree with, is correct. What sort of a gambler are you?
  14. Jun 12, 2009 #13
    Thanks for showing me that "(100)(3.62) = 362 N-s".
    I'm the type of gambler that will make one paper for each assumption (my initial and Doc Als'), handle them in + an A4 page with an explanation why there are two papers :)
  15. Jun 12, 2009 #14


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    If you deleted your original post, you would only have had one paper to submit, probably the wrong one. Just something to think about.:wink:
  16. Jun 12, 2009 #15
    yup! you right, good that I didn't now how :) ...

    Thanks for the help!
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