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Moment distibution (stiffness factor modifications) for a frame

  1. Jan 15, 2017 #1
    1. The problem statement, all variables and given/known data
    For the KAb and KCD , they are connected by the same method ... At B and C , we can see that they are connected by the same method , although i am not sure it's fixed or pinned .

    2. Relevant equations


    3. The attempt at a solution
    I assume they are fixed at B and C , so i agree that KAB is 4EI / L for far end fixed case . So , i think KCD should be 4EI / L as well .
     

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    Last edited by a moderator: Jan 16, 2017
  2. jcsd
  3. Jan 18, 2017 #2

    haruspex

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    Aren't the "far ends" A and D? One is fixed and one is pinned.
     
  4. Jan 19, 2017 #3
    B and C are fixed ? How do we know that ? It's not stated in the question
     
  5. Jan 19, 2017 #4

    haruspex

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    No, not B and C; A and D. As I read it, the stiffness factor at one end of a beam depends on whether the far end (other end) is fixed or only pinned. That makes intuitive sense to me.
    A, B and C are fixed, so for the interactions at B and C:
    AB is 4 because A is fixed
    BC is 4 at each end because the other end is fixed
    CD is 3 because D is pinned
    CE is 3 because E is pinned
     
  6. Jan 19, 2017 #5
    Which end is fixed ? B or C ?
     
  7. Jan 19, 2017 #6

    haruspex

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    Both are "fixed". (I think this just means it's a rigid joint, not a hinge.)
    But hey, I'm only inferring all this from the text.
     
  8. Jan 19, 2017 #7
    You just assume they are fixed ? So , for this type of question , we just assume they are always fixed ?? Since the author didnt show how's the connection at B and C
     
  9. Jan 19, 2017 #8

    haruspex

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    I'm not sure how you were supposed to know. They certainly look rigid in the drawing, but then so do D and E.
     
  10. Jan 19, 2017 #9
    S
    Since they look ' rigid' , so they are assumed to be fixed at B and C ?
     
  11. Jan 19, 2017 #10

    haruspex

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    I might have it....

    The question of "far end fixed" is to do with sideways flexing at the "near end" being inhibited by the far end.
    At the top of CD, sideways movement of C is not resisted by the joint D since that is a hinge.
    At the top of AB, sideways movement of B is resisted by the joint at A.
    In the beam BC, vertical movement at B is resisted by the combination of the rigidity of BCE and constraints at D and E. C cannot move up or down because of D. The rigidity of BCE then inhibits vertical movement at B.
    Similarly, vertical movement at C is resisted by the combination of the rigidity of BCE and constraints at AB and E. B cannot move up or down because of A.

    Hope that makes sense.
     
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