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...
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.
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...
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...
Not done.
So that's what the "about every possible point" means. Creating cuts will not grant you more independent equations which is what Studiot is saying. For statics, you have 6 independent equations and can solve for six independent unknowns. If the number of independent equations exceeds...
I see this greatly deviated from your original question which was understanding this:
2) Vector sum of all external torques that act on the body, measured about any possible point must be zero.
For an object to be in static equilibrium, one of the requirements is that the object must not...
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.
http://content.answcdn.com/main/content/img/oxford/Oxford_Food_Fitness/0198631472.isometrics.1.jpg
Try these isometric exercises. Try the "tension" one on the right of the picture. You can feel tension on your fingers.
"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 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.