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Clapeyron's Theorem Revisited

by rdbateman
Tags: clapeyron, revisited, theorem
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Studiot
#19
Jan21-13, 06:53 PM
P: 5,462
Hello, I hope you had a good Christmas break.

Why is the first term DP and not (1/2)DP?
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?
rdbateman
#20
Jan22-13, 01:27 AM
P: 30
"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"
Please read post #1 in this thread.
Studiot
#21
Jan29-13, 07:50 AM
P: 5,462
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 Thumbnails
compare1.jpg  
rdbateman
#22
Jan29-13, 11:33 AM
P: 30
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.
Studiot
#23
Jan29-13, 11:53 AM
P: 5,462
I am sorry you were not able to answer the question
The answer to your question is contained in my posts.

You have steadfastly refused to acknowledge their content.

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.
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.
rdbateman
#24
Jan29-13, 12:19 PM
P: 30
Quote Quote by Studiot View Post
The answer to your question is contained in my posts.

You have steadfastly refused to acknowledge their content.
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.
rdbateman
#25
Jan29-13, 12:23 PM
P: 30
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.
rdbateman
#26
Jan29-13, 12:28 PM
P: 30
Quote Quote by Studiot View Post
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.
I have not told you because it defeats the purpose of this forum. I also did not tell you because I felt it was a good exercise for you to get you outside your comfort zone. I think your lesson for today is "Instead of getting angry at not succeeding to answer a client's question and assigning blame to your client, accept this obstacle as an opportunity to better your teaching paradigm and seek a new route to communicate your answer".

This will be my last post in this thread.


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