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Potential Energy Approach to Derive Spring Element Equations

  1. Aug 19, 2014 #1
    Good day! I am a beginner in Finite Element Analysis and Structural Mechanics. I have problem in understanding the insight of potential energy approach to derive spring element equations.

    I have already search for similar post and I found this, and agree with the example raise.
    https://www.physicsforums.com/showthread.php?t=472249

    But I still have problem in applying it to FEA spring element.

    In the text, total potential energy is defined as the sum of the internal strain energy U and potential energy of the external forces Ω.

    I.e. ∏p = U + Ω

    U = 1/2 kx2, this make perfect sense. When we press/pull the spring to x, we give out that amount of energy and the spring is ready to give it back when we release the forces. Energy is conserved.

    but then I have a difficulties in understanding Ω , I quote the original text:
    "The potential energy of the externl force, being opposite in sign from the external work expression because the potential energy of the external force is lost when the work is done by external force, is given by Ω = -Fx"

    My questions are:

    1. isn't Ω = - ∑ δF δx ? which is also equal to U?


    2.If we substitube F by kx , ∏p = 1/2 kx2 - kx2 = -1/2 kx2, a negative potentia energy? I am very confuse here.

    I hope somebody can help.
     
  2. jcsd
  3. Aug 19, 2014 #2

    AlephZero

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    The position where there is zero potential energy is arbitrary. It doesn't matter if you end up with a negative number. You could start by writing ∏p = U + Ω + C, where C is an arbitrary constant.

    Substituting F = kx is confusing, because F is a constant, but x is a variable. You really want to leave the PE as
    ∏ = 1/2 kx2 - Fx ( + C) which is true for any value of x, not just at equilibrium.

    The condition for equilibrium is that ∂∏\∂x = 0, which gets rid of the arbitary constant C and gives you the equation you would expect, i.e. kx = F.

    Don't confuse "the potential energy of the force F", which is -Fx, and "the amount of work done to compress the spring by a distance x", which is Fx/2, because the force needed to compress the spring varies from 0 up to F as you compress it.

    Also, the potential energy of the spring is really 1/2 k(x1 - x2)2 where x1 and x2 are the displacements of each end.
     
    Last edited: Aug 19, 2014
  4. Aug 20, 2014 #3
    Hi Aleph, thanks for the explaination.

    Do you mean that we should consider the sping and external force (invisible hand) as a system?

    i.e. total potential energy is spring potential energy + invisible hand potential energy?

    And how should I see the potential energy of the force F, it looks like W = Fs though. What physical insight it gives?
     
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