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Strain elastic energy of a continuum

  1. Jul 30, 2011 #1
    Hello,

    I have problem with understanding the concept of energy of deformed body. As far as I know, Clapeyron's theorem states that elastic energy of deformed body is half the work done by external forces; this are surface and body forces which deform the body. Then, I wonder, where has another half of the work gone?

    U=1/2[itex]\sigma[/itex]ijuij
    W=[itex]\sigma[/itex]ijuij

    U is elastic energy, W is work done by surface and body forces, [itex]\sigma[/itex] is stress tensor, u is infinitesimal strain tensor.

    There is no acceleration and, as far as I know, kinetic energy is zero.

    Can someone tell what is going here and does some book discusses the matter clearly?

    Sincerely,
    Nikola
     
  2. jcsd
  3. Jul 30, 2011 #2

    AlephZero

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    You have lost a term from your equations somehow.

    Think about a lnear spring compressed by a force. You know the displacement is given by [itex]F = kx[/itex].

    The internal energy in the spring is [itex]kx^2/2[/itex]

    The work done by the external force = [itex]Fx[/itex]

    But there is also the work done by the surface stress of the body when it is deformed, which in the case is
    [tex]\int_0^x (-kx) dx = -kx^2/2[/tex]
    Putting it all together,
    [tex]kx^2/2 = Fx - kx^2/2 [/tex]
    which of course is the same as [itex]F = kx[/itex].
     
  4. Jul 30, 2011 #3
    Please explain your thinking here since when x=0, F=0.
     
  5. Jul 31, 2011 #4
    Thank you very much AlephZero for your response.

    Anyway, there is a problem with the equation for work. Work is not defined as Fx, but as [itex]\int[/itex]Fdx. In that way the work done in your example is the same as the potential energy of the spring, if we take the constant of integration as zero.


    Studiot

    Yes, when x=0, F=0, so work done is 0 and potential energy is 0, if we take the constant of integration as zero.
     
  6. Jul 31, 2011 #5
    Hello nikolafmf,

    Yes you have it exactly.

    So work input = stored elastic strain energy = ∫Fdx = 1/2 kx2

    It is not the force at x, multiplied by x, but half of that because the force varies with x from zero at x=0 to F at x=x.

    AZ is correct to note that you may have to take the increase in surface energy into account, but just not in this case.

    go well
     
  7. Jul 31, 2011 #6
    And I still don't understand the Clapeyron's theorem, which states that elastic energy of deformed body is half the work done by external forces.

    One of the famous British mathematicians, Augustus Edward Hough Love, put it this way:

    "The potential energy of deformation of a body, which is in equilibrium under given load, is equal to half the work done by the external forces, acting trough the displacement from the unstressed state to the state of equilibrium."

    This is from 1906 edition of his "A treatise on the mathematical theory of elasticity".

    Landau in his "Elasticity" on thermodynamical ground introduces free energy and potential energy of deformed body, which are different.
     
  8. Jul 31, 2011 #7

    AlephZero

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    That depends how you want to define the various quantities. If you apply an constant external force of F to something that moves a distance x, then it seems entirely rational to me to say the work done by that force as
    [tex]\int_0^x F dx = Fx[/tex]
    But of course there is another force acting on the body, caused by the boundary of the internal stress field, and that is where the factor of 1/2 comes from (but only for linear elastomechanics).

    IMO, being able to quote theorems is less imprtant than being able to formulate the equations correctly by including all the terms once each. I might have known exactly what Clapeyron's theorem said 40 years ago, but I don't recall ever actually using it since them, though I've written plenty of equations of motions based on energy formulations in those 40 years!
     
  9. Jul 31, 2011 #8
    Here is an extract from Wikipedia (correct for once) which does not say exactly what you have said.
    It is the emboldened words that make the diffeence.

    Here is a simple explanation without calculus.

    Referring to the attached figure.

    Let a spring or wire be extended from zero to an extension x, by means of a force which increases from zero to F.

    The wire is pulled out from O to B along line OB so that we can read off the force and extension at any point along the graph.

    Now any area under a force/distance graph has the units of energy or work.

    The area under the line OB is called the strain energy and is equal to the work input to create that strain.

    This can be seen to be the area of triangle OBA = 1/2 base*height = 1/2 force*extension

    The other triangle in the graph is also equal to 1/2force *extension and is called the complementary energy.

    For elastic materials that obey Hooke’s Law the strain energy = complementary energy as line OB is a straight line dividing the product FX into two equal triangles
    OFB and OBA.

    You may have noticed that I have drawn extension along the x axis. This is conventional.

    A slight fudge runs as follows.

    If 0 is the initial force and F is the final force the average is ½ (0+F) = 1/2F

    So the work done equals the average force times the displacement = 1/2Fx

    The idea of complementary energy and strain energy was used by Castigliano, along with some calculus, to develop several energy theorems that bear his name.
     

    Attached Files:

  10. Jul 31, 2011 #9
    Aleph Zero

    I think that this insistence on stretching a wire or spring with a constant force is not helping
    nikolafmf understand the issue.

    How do you stretch a Hookean wire or string with a constant force?

    If you pull with constant force you extend the wire some amount x and the extension stops. You have to increase the force to extend further.
     
  11. Jul 31, 2011 #10
    Studiot, thank you very very much for this long explanation.

    Yes, I understand that. If we take what Wikipedia says as true, everything is right and there is no problem. What I find as a problem is that in no book (even Love's one, which is citied on Wikipedia) I have read that we should take only final force. That is puzzling.

    Yes, I agree that work input should be the same to strain energy. But don't Clapeyron's theorem say that it is twice the strain energy?

    Final question, what is this complementary energy? What is its physical meaning?

    Thank you very much again.
     
    Last edited: Jul 31, 2011
  12. Jul 31, 2011 #11
    I agree, but the problem here is what Studiot described: you cannot pull a spring with constant force. Force is changing. It depends on x.
     
  13. Jul 31, 2011 #12
    Oh, I found this explanation:

    http://www.springerlink.com/content/w7083188151353kv/

    Authors say that the force doing work in Clapeyron's theorem is indeed constant, but that is not good representation for work in ideal elasticity. Only when viscous and thermal effects are included, work becomes twice the elastic strain energy. The other half, they say, is dissipated trough viscous dissipation or heat transfer.

    I will cite again :biggrin:

     
  14. Jul 31, 2011 #13
    Clapeyron's Theorem:

    If a body is in equilibrium under a given system of body forces F and surfaces forces T then the strain energy of deformation is equal to one half of the work that would be done by the external forces of the equilibrium state acting through displacements U from the unstressed state to the state of equilibrium.

    after Sokolnikoff - Mathematical Theory of Elasticity.

    I have emboldened the relevant phrases that make it quite clear Clapeyron was considering the final (equilibrium) forces multiplied by the total displacement. Note in particular the phrase 'would be done' which implies an imaginary force x distance product.

    Note again that it is not half for the body forces and half for the surface forces, it is half for their combination.

    go well
     
  15. Jul 31, 2011 #14
    Yeah, I finally got it :). Thank you very much :)
     
  16. Aug 1, 2011 #15
    If you don't mind a bit of (slightly more advanced) calculus Sokolnokoff discusses this in detail (p81 - 89) and also 'Clapeyron's formula' where the statement is recast in stress-strain terms.

    Incidentally what pages in Love were you referring to?

    go well
     
  17. Aug 1, 2011 #16
    I was referring to page 170 of Love's 1906 edition. But there are also other editions, one from Dover.

    Thank you, I will see Sokolnikoff.

    Be well.
     
  18. Aug 1, 2011 #17
    I am guessing that you refer to article 120 in Love, referred to in my Dover edition as

    Theorem concerning the potential energy of deformation

    "The potential energy of deformation of a body, which is in equilibrium under given load, is equal to half the work done by the external forces, acting through the displacements from the unstressed state to the state of equilibrium."

    (But not attributed to any partiular source.)


    Once again the idea of from zero to full load is expressed.

    If you need more detailed reference to Sokolnikoff or elasticity theory post again.

    go well
     
  19. Aug 1, 2011 #18
    Yes, you are right, that was article 120. Well, I think his description is misleading and leads to paradox, since he seems to speaks of the changeable force, not of the final force.
     
  20. Aug 1, 2011 #19
    You must remember the style of language was differrent in Love's day.

    It does make his work densely packed and heavy going, mamoth though it was for his day.

    What is your interest/motivation for this, may I ask?
     
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