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Clapeyron's Theorem Revisited

  1. Dec 18, 2012 #1
    To help make my question clearer, I will be referencing equations from the following paper.

    http://www.engr.colostate.edu/~thompson/hPage/CourseMat/Tutorials/Solid_Mechanics/energy.pdf

    Before I am informed, I have read all articles on this site relating to this theorem and they are not clear to me. As such, if you choose to post, please to not reference or quote any text from those previous threads.

    "This theorem states that the strain energy of a linear elastic body (structure), in static equilibrium, under the action of constant surface tractions and body forces, is equal to half the work these tractions and body forces would perform in moving through their respective displacements." Pg 14

    The right hand side of the equation is clearly the external work of the system. Since the External Work + Internal Work = 0, then that means the left hand side should equal the negative of the internal work.

    So, Internal Work = -2(Strain Energy)???

    Neglect heat, and other energy, I thought the internal work was equal to strain energy. Can someone explain this?

    Any help would be appreciated. Thank you.
     
  2. jcsd
  3. Dec 19, 2012 #2
    It is good for you to provide that reference as we can see where you are coming from.

    The question you ask about the factor of 1/2 is a very common one.

    It is also unfortunate that the name of the paper is

    Energy Methods

    Since neither Virtual Work, nor the Reciprocal Theorems are truly energy methods.

    Castigliano's Theorems, Strain Energy, Variational methods, Hamilton-Lagrange methods are true energy methods.

    OK the key connection is the condition of equilibrium.

    First consider a body in equilibrium but strained by an external force and enquire how it came to this state.

    To keep it simple consider a single force, F, and extension, e.

    To calculate the energetics of this we need to consider the difference between an equilibrium state, unstrained by F, and the second equilibrium state after the application of F and when equilibrium has been re-established.

    Initially the applied force is zero and the extension is zero.

    If we build up F slowly from zero and consider the situation where just δF has been applied the resultant extension is δe and the work done ie the energy input to the body is δF x δe

    All these small increments of input energy (work) add up and eventually we reach the equilibrium where the full F is applied and the full extension e obtained.

    However the full F has not been acting for the entire straining process.

    To sum the work we can either do the integral ∫Fδe if we can obtain an expression for F in terms of e - which we can for a linear system since e and F are proportional.

    Or we can take an average for F, which amounts to the same thing. The average over the range zero to F is 1/2( 0 + F) =F/2.

    Here is the factor of 1/2 that you were enquiring about and is the actual strain energy imparted.

    But what about Virtual Work?

    Well VW stems from the (obvious?) mechanical fact that if a system of forces is in equilibrium and is moved bodily sideways so the whole system is still in equilibrium no work is done.

    So if we consider a bunch of forces (F1, F2, F1, etc) in equilibrium and displace the whole system a (vector) distance d then the work done for each force is force times the distance times the cosine of the angle between that force and the vector d.

    Ʃ (Fi d cosθi)

    Because the system is in equilibrium this sum must equal zero and is termed the virtual work.

    Because the forces already exist and do not change during the displacement there is no factor of 1/2 involved.

    In relation to your referenced paper, the increase in internal energy is provided by the work done by the external loads. If this external work is zero, then so is the 'internal work'.
    Hence the expression in the paper.

    There is much to be said about VW as it is a powerful technique but that should be the subject of another thread.
     
  4. Dec 19, 2012 #3

    sophiecentaur

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    This is a direct analogy to the problem of energy stored in a Capacitor as it's charged through a Resistor. However you charge the Capacitor (or stretch the spring), there will be a Source Resistance (or friction) involved. If your (mathematical) model is too simple and doesn't include a source resistance of some sort then it involves an instantaneous deformation / charging, which is not possible.
     
  5. Dec 19, 2012 #4
    "Neglect heat, and other energy, I thought the internal work was equal to strain energy. Can someone explain this?"

    I was afraid of this. It seems to me that books on structural analysis either present theorems based on constant external loading or "slowly applied loading". I'm sure they equate to same outcome but the comparison can through people off leading to the popularity of this debate.

    I propose the following in an attempt to explain this.

    This is a one dimensional problem defined on the x axis.

    A simple bar of length L and cross section A and Elastic Modulus E throughout. It is fixed at one end and free on the other. An external load P is applied to the free end stretching the bar a distance D. There are two cases. The first case assumes a constant P is applied and therefore Wext = PD. The second case assumes the load P is graduated linearly from 0 to the value P making Wext = 1/2PD.

    Any further discussion will use the the following variables and syntax to avoid conflicting text in future posts.

    E = Modulus of Elasticity
    A = cross sectional Area
    L = Length of bar
    D = Tensional elongation of bar
    R = Axial Reaction Force at fixed end
    ε = D/L (strain)
    σ = PL/AE (stress
    Wint = Internal Work
    Wext = External Work
    U = Strain energy
    ∏ = Potential Energy

    Any virtual quantity will have the symbol [itex]\delta[/itex] before it.

    Any definite integration will be presented in the syntax form (x1, x2)∫(Function)dχ where x1 and x2 are the lower and upper bounds respectively.

    Any definite differentiation will be presented in the syntax form (x1)∂(Function)/∂χ where χ is a independent variable of the Function and x1 is where the differentiation is evaluated.

    Please fill out the following.

    Case1 (External Loads Applied at a Constant rate P)

    What is Wint?
    What is Wext?
    What is [itex]\delta[/itex]Wint?
    What is [itex]\delta[/itex]Wext?
    What is U?
    What is [itex]\delta[/itex]U?
    What is ∏?
    What is [itex]\delta[/itex]∏?

    Display Clapeyron's Theorem
    Display Virtual Work Equation

    Case2 (External Loads Applied Slowly from 0 to P)

    What is Wint?
    What is Wext?
    What is [itex]\delta[/itex]Wint?
    What is [itex]\delta[/itex]Wext?
    What is U?
    What is [itex]\delta[/itex]U?
    What is ∏?
    What is [itex]\delta[/itex]∏?

    Display Clapeyron's Theorem
    Display Virtual Work Equation

    I think showing this example will eliminate a lot of confusion for everyone once and for all. Thank you so much Studiot.
     
    Last edited: Dec 19, 2012
  6. Dec 19, 2012 #5
    Did you not read my post?
     
  7. Dec 19, 2012 #6
    "Neglect heat, and other energy, I thought the internal work was equal to strain energy. Can someone explain this?"

    I was afraid of this. It seems to me that books on structural analysis either present theorems based on constant external loading or "slowly applied loading". I'm sure they equate to same outcome but the comparison can through people off leading to the popularity of this debate.

    I propose the following in an attempt to explain this.

    This is a one dimensional problem defined on the x axis.

    A simple bar of length L and cross section A and Elastic Modulus E throughout. It is fixed at one end and free on the other. An external load P is applied to the free end stretching the bar a distance D. There are two cases. The first case assumes a constant P is applied and therefore Wext = PD. The second case assumes the load P is graduated linearly from 0 to the value P making Wext = 1/2PD.

    Any further discussion will use the the following variables and syntax to avoid conflicting text in future posts.

    E = Modulus of Elasticity
    A = cross sectional Area
    L = Length of bar
    D = Tensional elongation of bar
    R = Axial Reaction Force at fixed end
    ε = D/L (strain)
    σ = PL/AE (stress
    Wint = Internal Work
    Wext = External Work
    U = Strain energy
    ∏ = Potential Energy

    Any virtual quantity will have the symbol δ before it.

    Any definite integration will be presented in the syntax form (x1, x2)∫(Function)dχ where x1 and x2 are the lower and upper bounds respectively.

    Any definite differentiation will be presented in the syntax form (x1)∂(Function)/∂χ where χ is a independent variable of the Function and x1 is where the differentiation is evaluated.

    Please fill out the following.

    Case1 (External Loads Applied at a Constant rate P)

    What is Wint?
    What is Wext?
    What is δ Wint?
    What is δ Wext?
    What is U?
    What is δ U?
    What is ∏?
    What is δ ∏?

    Display Clapeyron's Theorem
    Display Virtual Work Equation

    Case2 (External Loads Applied Slowly from 0 to P)

    What is Wint?
    What is Wext?
    What is δ Wint?
    What is δ Wext?
    What is U?
    What is δ U?
    What is ∏?
    What is δ ∏?

    Display Clapeyron's Theorem
    Display Virtual Work Equation

    I think showing this example will eliminate a lot of confusion for everyone once and for all. Thank you so much Studiot.
     
  8. Dec 19, 2012 #7
    Any takers?
     
  9. Dec 19, 2012 #8
    If the absense of not addressing this question is because of the Homework Help thing...
    I am a 26 year old with a masters degree and I work as a structural engineer. I haven't had homework in 2 years :). Also, considering that it is the 19th of December, I do not believe that anyone is in School at this time.

    To Studiot: I did read your post. What happened was I was typing the first part of my response and then submitted it incomplete by accident. During the time I was adding all that extra material in my re-edit, you posted your last comment before I could resubmit my edit. I then created a duplicate response in case you were not notified by email.
    In the message I thanked you for your help.
     
  10. Dec 19, 2012 #9
    With my current understanding (obviously wrong since I'm here), I believe that U from Case 1 and Case 2 will be identical. Also, I believe that Wint = U for both cases.
     
  11. Dec 19, 2012 #10
    To be able to write a virtual work equation you need a virtual displacement.

    Do you have any in mind?

    Clapeyron's theorem is a formal generalisation of my explanation of strain energy, and includes the factor of 1/2 since it assumes the forces are gradually applied from zero.

    I am not sure what you mean by applying a constant P.
    Consideration of a force - extension curve will make it obvious that you cannot suddenly enter the graph at some point along it and for some ε = 0.1 ε=m, P = 0.1 P=m

    What I haven't been able to figure out is whether your difficulty lies with virtual work or strain energy.

    Did you follow both my descriptions, I tried to make the distinction clear, but now with greater knowledge of your qualifications I can be more mathematical.
     
    Last edited: Dec 19, 2012
  12. Dec 19, 2012 #11
    "To be able to write a virtual work equation you need a virtual displacement.
    Do you have any in mind?"

    How about δD(x) = x

    "I am not sure what you mean by applying a constant P."

    Applying a constant P means that if you look at a Force-Displacement graph, the Force has a slope of 0 with a value of P.
     
  13. Dec 19, 2012 #12
    Have you ever seen such a graph?

    It is the graph of plastic flow.
     
  14. Dec 19, 2012 #13
    So what do books mean when they say that

    Wext = PD ?

    Does that not indicate that the force applied throughout the deformation is constantly P?
     
  15. Dec 19, 2012 #14
    I'm seriously stumped. You can't just leave me hangin like that lol
     
  16. Dec 19, 2012 #15
    Ok. For the bar with the external force slowly applied...

    Wext = (1/2)DP
    Wint = -(1/2)D(AE/L)D = -(1/2)DP

    ∏ = DP - (1/2)D(AE/L)D

    ...Why is the first term DP and not (1/2)DP?

    I think the answer to this will solve my problem.
     
  17. Dec 20, 2012 #16
    No I haven't abandoned you, it was 1am UK time when I knocked off last night.

    This is just a test because half an hour's typing has just gone up the swannee and I am trying to retrieve it.

    I am going to take a guess that it is the Virtual Work you are having trouble with. So I will present a geometric explanation/proof of the action.

    You will need to accept Clapeyron's theorem that true strain energy is 1/2 Force times displacement. Virtual Work can be developed from this, entirely consistently but remember that virtual work is also a mathematical dodge or trick.

    So consider a linear-elastic system loaded by a force or system of forces F1 resulting in displacements e1 and subsequently by a second system of forces F2 resulting in displacments e2.

    Plot the load line force - displacement curve OAB, with OA being the line for F1 and AB the extension for the subsequent F[2.
    I will return at the end for the reasons for using the x axis for the displacement and the y axis for the Forces, this is conventional.

    The true strain energy imparted by F1 is 1/2 F1e1represented by triangle OAC. The strain energy due to F2 acting alone on the system is 1/2F2e2 represented by triangle AEB.

    Now if we follow F1 by F2 the forces of F1 are displaced by F2 so additional energy has to be supplied. The total energy is 1/2 (F1+F2)(e1+e2), represented by triangle OBF.

    It can be seen that OBF is made of two triangles plus the rectangle AECF, shown double hatched, which is F1e2

    This rectangle is known as the Virtual Work of forces F1 moving through displacements e2

    A bit of mathematical jiggery pokery shows that this equals the rectangle ADGH ie F2e1 which is the Virtual Work of forces F2 moving through displacements e1

    Some judicious choice of F1, F2, e1 & e2 allows the statement

    Internal VW = External VW

    To return to the choice of axes, the real strain energy is the energy between the load line and the displacment axis.
    The energy between the load line and the foces axis is called the complementary energy.
    In this linear case they are identical, however they can be usefully employed in some cases, and lead to a Castigliano theorem.

    https://www.physicsforums.com/attachment.php?attachmentid=54122&d=1356030712

    attachment.jpg

    (Edit by Borek: new attachment added, old left for reference)
     
    Last edited by a moderator: Dec 21, 2012
  18. Dec 21, 2012 #17
    I will try again with the attachment.

    For the benefit of any mod tidying this up I have renamed the attachment from virtualwork1 to virtualwork2.
    Both have the same content.
     

    Attached Files:

    Last edited: Dec 21, 2012
  19. Jan 21, 2013 #18
    Why is the first term DP and not (1/2)DP?
     
  20. Jan 21, 2013 #19
    Hello, I hope you had a good Christmas break.

    Well I'm not sure what you mean by potential energy or where you got the equation, but if the first term were to be 1/2DP the the potential energy would equal zero.

    As you originally wrote it, it yields the correct value for the true strain energy, which can be regarded as PE. But why not just use 1/2DP in the first place?
     
  21. Jan 22, 2013 #20
    Please read post #1 in this thread.
     
  22. Jan 29, 2013 #21
    I have worked out your example in both true energy and virtual work methods side by side.

    I have tried to follow your notation of a bar of length L, extended a distance D under a force R and in equilibrium.

    I will leave you to work out the comparison with direct force methods.

    Because you seem to have some uncertainty about the following terms I have defined them

    Potential Energy
    Strain Energy
    External Work done

    for the true energy method

    For this method note that the first term is the Clapeyron strain energy and contains the factor of 1/2 that was causing the difficulty.

    Note also that this 1/2 is cancelled by the 2 that results from the differentiation when we seek the minimum potential energy.

    and

    Internal virtual work
    External virtual work
    Virtual displacement

    for the virtual work method.

    Note that there is not factor of 1/2 since the definition of a virtual displacement requires all forces to remain unchanged during such a displacement.

    It can be seen that both methods reach the same end result. Failure to do so is probably as a result of an substituting an inappropriate formula somewhere.
     

    Attached Files:

    Last edited: Jan 29, 2013
  23. Jan 29, 2013 #22
    Again. Please read post #1 in this thread. The question to this thread is why two things equate to each other. What is the significant meaning of this. The question is not, "show me how to perform the virtual work and minimum potential energy methods to a simple bar problem".

    I have had this question answered to me a long time ago by some random graduate student before New Years day. The answer is fairly obvious and I felt bad for not realizing it sooner.

    I am sorry you were not able to answer the question but I'd like to thank you for trying. This forum is a great service to answering elementary problems for people trying grasp basic concepts and it is good that you devote a lot of your time to this ideal.
     
  24. Jan 29, 2013 #23
    The answer to your question is contained in my posts.

    You have steadfastly refused to acknowledge their content.

    And thank you for not letting us know that your question was answered (and what the obvious answer is), so incurring further effort on our behalf.
     
  25. Jan 29, 2013 #24
    No. What you have done is shown me how to apply both methods in the calculation of the bar problem. What you have not explained (and is the basis of my entire thread) is why the terms in the equation are the way they are and why it appears that one term in the MPE equation is double that of a term in the PVW equation.

    You have made a huge contribution to this forum (over 5000 posts). Most of which are probably devoted to answering people's questions. However, we are not all perfect and it is not impossible for you to fail to answer another person's question. It's not your fault. People who do something for a long time can have the difficulty of glazing over things without realizing there is a problem. Realize your mistake and instead of passive aggressively attacking your client, try and use that feedback to better your skill.

    The short answer to this is: The problem of solving problems of this nature is the solution to a differential equation which is created by considering force equilibrium, stress strain relationships, and compatibility. The differential equation of these problems as it stands is in what is known as the "Strong Form". We can manipulate this equation into another form called the "Weak Form". This form is the popular "Principle of Virtual Work" equation and from which can be derived that external work = internal work. Thus when we use the principle of virtual work on mechanics problems, we are actually solving the weak form of the differential equation governing the problem. There is another form called the "Variational Form" and can be constructed from the weak form by utilizing Variational Calculus. This form is called "The principle of minimum potential energy". Thus, when we use the principle of minimum potential energy, we are actually solving the variational form of the differential equation governing the problem. The problem most people have with the variational form is that the terms in this equation look very much like the internal work and enternal work terms from the weak form and people mistakenly believe these terms are in fact internal and external work. They are not. They are what variational calculus describes as the first variational forms of external and internal work. They are quantities that have no physical meaning in terms of actual work applied. To understand this. Take the strong form of the differential equation and derive the weak form and variational form. Understand that the terms in the variational form are just a result of mathematical consequence, they bear no physical meaning.
     
  26. Jan 29, 2013 #25
    For the strong form (differential equation) of mechanical problems, look at equation 2.1 of

    http://www.edwilson.org/book/02-equi.pdf

    Of course, the boundary conditions will be suggested on the particular problem at hand.
     
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