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Pinned end supported span

  1. Apr 21, 2017 #1
    1. The problem statement, all variables and given/known data
    In this question , is figure 11-2 a and 11-2 b the same beam ? Why in a , the support is pinned end , it change to fixed end in figure b ?
    2. Relevant equations


    3. The attempt at a solution
    In equation 11-9 , i gtet different ans with the author ..
    Here's what i gt :
    MN = 2EK( 2 θN + θF - 3 Φ ) + (FEM)N
    So,
    0= 2EK( 2θA + θB - 3 Φ )

    At here θA = 0 , and (FEM)N = 0 , far end = B ,near end = A

    MN = 2EK( 2 (0) + θB - 3 Φ ) +(FEM)N ----(1)
    0= 2EK( 2θB + 0 - 3 Φ ) ------(2)

    By applying the same method as author ( Multiply equation 1 by 2 , and subtract equation 2 ) , i gt
    MAB = 3EK(-Φ)
     

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  2. jcsd
  3. Apr 23, 2017 #2

    haruspex

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    No. They are the two different cases of "far end not fixed" (i.e. pin or roller). The near end is fixed or not fixed.

    I cannot follow your algebra. You seem to be setting some things to zero without cause. I get the same equation as the author.
     
  4. Apr 23, 2017 #3
    why in case 11-8(a) , the far end angular dispalcement theta _B doesnt have to be determined ? I only learnt that when the end is fixed support , then , the the angular displacement is 0
     
  5. Apr 23, 2017 #4

    haruspex

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    θB (=θF) does not need to be determined because it was possible to combine the two equations so as to eliminate it and produce an equation without it. There is no suggestion that is zero.

    By the way, I find the image you posted hard to read. It is very small. When I blow it up large enough to read it becomes quite fuzzy.
     
  6. Apr 24, 2017 #5
    In 11-8 a, we could see that at the left end , it's pin supoorted , why at 11-8b , it will become fixed supported ?
     
  7. Apr 24, 2017 #6
    Can you explain how to get 0 = 2Ek( 2 theta_N + theta_F -3phi) + 0 ? (Just below equation 11-9)
     
  8. Apr 24, 2017 #7

    haruspex

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    The left end is the "near end". The situation being considered is "far end not fixed". There are two cases of that, near end fixed or near end not fixed. Hence the two diagrams.
     
  9. Apr 24, 2017 #8

    haruspex

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    Apparently this comes from Eq 11-7 and/or 11-8, but those are not in your post.
     
  10. Apr 24, 2017 #9
    Here is it . Can you help to explain ? 0 = 2Ek( 2 theta_N + theta_F -3phi) + 0 ? Does the first 0 represents MN , and the last 0 represent (FEM) N ?
     

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  11. Apr 24, 2017 #10

    haruspex

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    Yes, but remember this is applying eq 11-8 at B in fig 11-8b, so in the context of eq 11-8, B is the near end and A is the far end (which is fixed in fig 11-8b). (The unfortunate coincidence of similar reference numbers in eqs and figures is confusing.)
     
  12. Apr 24, 2017 #11
    then , how about MN = 2Ek( 2 theta_N + theta_F -3phi) + (FEM)N ? What does it represent ? 11-8a ? Which is near end ? and which is far end ?
     
  13. Apr 24, 2017 #12

    haruspex

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    That is the other way round, A is the near end here.
     
  14. Apr 24, 2017 #13
    So , both MN = 2Ek( 2 theta_N + theta_F -3phi) + (FEM)N and 0 = 2Ek( 2 theta_N + theta_F -3phi) + 0 describes about case in 11-8b ?
     
  15. Apr 24, 2017 #14

    haruspex

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    No. Compare your second equation above with the second equation of 11-9. The angles are swapped around.

    I think what is confusing you is the near-far notation in eqs 11-9. In both of them, the near end is A and the far end is B, so for clarity rewrite them replacing N with A and F with B everywhere.

    Eq 11-8 is for far-end-fixed. The author doesn't write it out but states that if, in addition, the near end is not fixed we can simply substitute MN=(FEM)N=0 in eq 11-8. Call this eq 11-80.
    In fig 11-8b, we have A fixed, B not fixed. This means we can use eq 11-80 with A as F and B as N. This gives the second eq of 11-9. We can also use eq 11-8 but with A as N and B as F. This gives the first eq of 11-9.
     
  16. Apr 25, 2017 #15
    so , do you mean both MN = 2Ek( 2 theta_N + theta_F -3phi) + (FEM)N and 0 = 2Ek( 2 theta_N + theta_F -3phi) + 0 describes about the near end is A and the far end is B in case 11-8(b) ?
     
  17. Apr 25, 2017 #16
    why if the near end is not fixed , we could use MN=(FEM)N=0 ??
     
  18. Apr 25, 2017 #17
    We can notice that the case in 11-8b yield equation (11-10) , so , is it possible that the equation 11-10 to be used and applied in case 11-8(a) in which both ends are roller / pin supported ?
     
  19. Apr 25, 2017 #18

    haruspex

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    Because a pin or roller cannot exert a torque.
     
  20. Apr 25, 2017 #19
    post #17 , can you try to reply ?
    We can notice that the case in 11-8b yield equation (11-10) , so , is it possible that the equation 11-10 to be used and applied in case 11-8(a) in which both ends are roller / pin supported ?
     
  21. Apr 26, 2017 #20

    haruspex

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    I think so, just setting MN=(FEM)N=0. That would lead to θN=ψ. Does that make sense? I'm unsure how ψ is defined. Is it something to do with the conjugate beam?

    By the way, I think I found your text online at http://mrkhademi.com/chapter11.pdf.
     
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