
#1
Dec1812, 11:40 PM

P: 30

To help make my question clearer, I will be referencing equations from the following paper.
http://www.engr.colostate.edu/~thomp...ics/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
Dec1912, 04:44 AM

P: 5,462

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, HamiltonLagrange 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 reestablished. 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 (F_{1}, F_{2}, F_{1}, 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 (F_{i} d cosθ_{i}) 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. 



#3
Dec1912, 07:22 AM

Sci Advisor
PF Gold
P: 11,341

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.




#4
Dec1912, 12:30 PM

P: 30

Clapeyron's Theorem Revisited
"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. 



#5
Dec1912, 01:08 PM

P: 5,462

Did you not read my post?




#6
Dec1912, 01:11 PM

P: 30

"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. 



#7
Dec1912, 04:39 PM

P: 30

Any takers?




#8
Dec1912, 06:11 PM

P: 30

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 reedit, 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. 



#9
Dec1912, 06:15 PM

P: 30

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.




#10
Dec1912, 06:41 PM

P: 5,462

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. 



#11
Dec1912, 06:51 PM

P: 30

"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 ForceDisplacement graph, the Force has a slope of 0 with a value of P. 



#12
Dec1912, 06:53 PM

P: 5,462

Have you ever seen such a graph?
It is the graph of plastic flow. 



#13
Dec1912, 07:28 PM

P: 30

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? 



#14
Dec1912, 10:55 PM

P: 30

I'm seriously stumped. You can't just leave me hangin like that lol




#15
Dec1912, 11:39 PM

P: 30

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. 



#16
Dec2012, 01:16 PM

P: 5,462

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 linearelastic system loaded by a force or system of forces F_{1} resulting in displacements e_{1} and subsequently by a second system of forces F_{2} resulting in displacments e_{2}. Plot the load line force  displacement curve OAB, with OA being the line for F_{1} 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 F_{1} is 1/2 F_{1}e_{1}represented by triangle OAC. The strain energy due to F_{2} acting alone on the system is 1/2F_{2}e_{2} represented by triangle AEB. Now if we follow F_{1} by F_{2} the forces of F_{1} are displaced by F_{2} so additional energy has to be supplied. The total energy is 1/2 (F_{1}+F_{2})(e_{1}+e_{2}), represented by triangle OBF. It can be seen that OBF is made of two triangles plus the rectangle AECF, shown double hatched, which is F_{1}e_{2} This rectangle is known as the Virtual Work of forces F_{1} moving through displacements e_{2} A bit of mathematical jiggery pokery shows that this equals the rectangle ADGH ie F_{2}e_{1} which is the Virtual Work of forces F_{2} moving through displacements e_{1} Some judicious choice of F_{1}, F_{2}, e_{1} & e_{2} 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. (Edit by Borek: new attachment added, old left for reference) 



#17
Dec2112, 03:46 AM

P: 5,462

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. 



#18
Jan2113, 03:49 PM

P: 30




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