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Shear Center for Open Thin-Walled Members

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Homework Statement


There is a section in my book that has the same title as the title of this discussion topic. I understand the math just fine, so this is a conceptual question. In the book the cross section is shaped and oriented like a C, and it shows the shear flow flowing counter-clock-wise because there is a downward applied force, however it says that the twist in the beam is caused by the clock-wise reaction force within the beam.

My question is, what is the applied force doing while the beam is twisting? For a simple bending case you can easily see that the beam will bend in the direction of the applied force and the reaction force will get larger as the beam bends until it reaches equilibrium. However in this case because the twist is caused by the reactionary force, it makes it seem like the reaction force is larger than the applied force when it is first applied and gets smaller as it twists until it reaches equilibrium. How can this be?

This is very strange for me because they show an example picture of a beam twisting and it seems intuitive that it twists in that way, and given that it twists in that way the eccentricity makes sense, but I can't justify why it twists that way in any way. It is possible that the author explained it wrong because I have caught conceptual mistakes in the textbook before (in topics that I actually understood), but of course I don't know if that's the case here.


Homework Equations


they have this equation e = Fd/P, but that isn't relevant to my question.
 

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