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Stiffness factor for member in beam

  1. Feb 11, 2017 #1
    1. The problem statement, all variables and given/known data
    In this question , I dont understand why the KBC is 4EI / L ....


    2. Relevant equations


    3. The attempt at a solution
    I was told that for the far end pinned or roller supported , the K = 3EI / L , so shouldn't the KBC = 3EI / L.. Is the author wrong ?

    In the 3rd photo , we can see that for far end pinned supported , k = 3EI / L , not 4EI / L
     

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    Last edited: Feb 11, 2017
  2. jcsd
  3. Feb 11, 2017 #2

    haruspex

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    BC is only a segment of the beam ABCD. The "far end" of BC from B is D.
    Think about it: if D were just a pin then the beam could rotate a bit about D, allowing it equally to rotate a little about C. But with D fixed, it is much harder for there to be any rotation about C.
     
  4. Feb 12, 2017 #3
    So , in this problem , the far end of AB is D , as D s pinned , so the KAB is 3EI / L ? If D is fixed , then KAB would be 4EI / L ?
     

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  5. Feb 12, 2017 #4

    haruspex

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    That's my understanding. But please bear in mind that my entire knowledge of the subject comes from the textbook extracts you have posted.
     
  6. Feb 12, 2017 #5
    May I know which part ? Can you highlighted it ?
     
  7. Feb 12, 2017 #6

    haruspex

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    I have not noticed an actual definition of "far end" but just thinking about how I would expect a beam to behave it is clear that if any part of a beam is unable to rotate at all it will make the beam stiffer at all joints. Hence that is how I interpret the term.
     
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