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Torsion misunderstanding

  1. Jul 4, 2012 #1
    I'm very confused regarding this problem
    and i will explain in it in details ,hoping any one can explain the reason
    For example we have a bolted connection subjected to torsion ,the bolts will have extra shear in the plane also when i strike a a slab of small building with lateral load(assuming torsion will be create ,because center of mass not equal center of rigidity) this will create torsion and make extra reaction on columns.However, when i put abeam fixed from two sides and make torsion in it this will not create extra reactions,so why!!!!!!!!
    I feel that i miss something,but cant find it
    Please help
     
  2. jcsd
  3. Jul 4, 2012 #2
    If my question is not clear please tell me
    it just a concept question
    i just need to understand the logic behind it
    Thank in advance
     
  4. Jul 8, 2012 #3

    PhanthomJay

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    A beam subject to torsion will have end reactions, satisfying the equilibrium equations. Perhaps you are envisioning a beam subject to a pure twisting moment ( a "couple") applied say at the middle of the beam. Call it "M". Then at each fixed end of the beam, there will be a reaction couple of M/2. There won't be any net vertical or horizontal force reactions at the ends, however, is this what you mean? Now if the connection at the ends each consisted of a 2 bolt connection, one bolt above the other spaced a distance "d" apart, then there would be a shear force in the top bolt equal to M/2d, and a shear force in the bottom bolt equal to M/2d in the opposite direction. The sum of those forces is 0 at each end, so there is no force reaction on the beam at either end. Is this what you mean?
     
  5. Jul 9, 2012 #4
    Thank you very much for your help.
    and im very happy that you help me since i was looking for the answer along time ago.
    Now finally to make this clear to me.
    Torsion moment will create to satisfy equilbrium in all cases.however shear force created at every element at beam section (all shears cancel togethor) so for this reason we design connection.
    Iam I right
    Thank you brother
     
  6. Jul 9, 2012 #5
    My friend last question come to my mind regrading this why we calculate moment for inertia only for bolts.I guess the answer ,because it resist the shear only why plate don't.
    Right!!!!!!!
     
  7. Jul 10, 2012 #6

    PhanthomJay

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    Sorry for delayed response. The end plate fastened to the beam to which the bolts are attached, also sees stresses, maximum where it is attached to the beam
     
  8. Jul 22, 2012 #7
    Thank you very much.
    Another simple question came to my mind, for example those nails in picture attached subjected to shear, so according to equilibrium they will be subjected to vertical shear and in paper force(Do we calculate in paper force??) also when we have plate with bolts subjected to shear it will have plane shear and, tension(so why we don’t calculate tension).Also in concrete beam ,we put stirrups for vertical shear so what about longitudinal shear that come from equilibrium(no reinforcement)
     

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  9. Jul 22, 2012 #8
    I mean inplane froce come fro equilibrium

    Please help
    I'm confused
     
  10. Jul 22, 2012 #9
    "Also in concrete beam ,we put stirrups for vertical shear so what about longitudinal shear that come from equilibrium(no reinforcement)"
    The stirrups are not for vertical shear. What vertical stirrups do is resist the vertical component of (approximately) diagonal shear in tension. The diagonal shear in compression is resisted by the concrete. From a practical point of view, the stirrups help keep the whole reinforcement cage together during the concreting and vibration process of manufacture. So why don't we use diagonal stirrups? Partly for the practical reason just explained, but also because it has been shown that diagonal bars - which used to be much more common - are only fully effective when used in conjunction with vertical stirrups.
     
  11. Jul 22, 2012 #10

    PhanthomJay

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    The nails (properly sized and spaced) transfer the longitudinal shears between the vertical and flat sections so that the beam acts as if it were one whole section. Longitudinal and vertical shear stresses at any given point in the beam's cross section
    must coexist for equilibrium and must be equal in magnitude. The shear forces in the nails spaced a distance s apart are calculated from the shear flow equation (VQ/I)s .
     
  12. Jul 24, 2012 #11
    Thank you very much, i read a little bit about this subject and understand it, but still something not clear, you said that failure occur from diagonal tension, so why equations use Vu in calculation (I think because cracked section, so it difficult to apply shear equation and determine principle stress axes so we use Vu as it create indirect shear which is tension)

    The second question, we know that fail occur due to indirect shear, so if direct shear occur
    How much concrete can handle.
    Thank You another time
     
  13. Jul 24, 2012 #12

    Unfortunately I didn’t get the idea will(so please simplify it). As I understand the nail is subjected to vertical shear and also to longitudinal shear, so I should design for their resultant!!!!!!!!!!
    Thank you brother another time
     
  14. Jul 24, 2012 #13

    PhanthomJay

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    No. Longitudinal shear only (into the plane of the 'paper') in the nails. Vertical shears are carried by the wood.
     
  15. Jul 24, 2012 #14
    So maybe it will split verticaly,if nails fail from vertical shear....??/nails should carry verical shear in order to maintain the peices togethor??
     
  16. Jul 24, 2012 #15

    PhanthomJay

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    It is true that the nails carry some vertical shear locally from the weight of the horizontal flange section which is supported by the nails through the vertical webs. I have assumed that this is small in comparison to the longitudinal shears developed under applied live loading , and that when such loads are applied, they do not cause significant vertical shears in the nails . This localized shear could be eliminated by nailing from the top instead of the sides.
     
  17. Jul 25, 2012 #16
    I'm very sorry i took alot from your time, but believe me i feel that im still misunderstand something, when you lock at rectangular section of steel,their will be shear longitudnal and verical equal in magnitude because of complementry property of shear stress.That what i know, the same here when i lock at fastenesrs,there should be both,but swood take vertical shear . Right!!!

    SO, IF I have like the picture which illustrate plate and bolts , who will take the longitudnal shear.???????????????

    i appreciate your help brother,but i feel bad ,because something is missing.
    Hope you can help
     

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  18. Jul 25, 2012 #17

    PhanthomJay

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    We seem to have strayed a long way from torsion, haven't we, bro? Anyway, complimentary vertical and longitudinal shears aren't even in the same plane...don't try to combine them! Let's forget the nailed section and look at a solid section b x h with a vertical shear force of V at that section. I is bh^3/12, Q max at the NA is bh^2/8, and the transverse shear stress at the NA is VQ/Ib = 1.5V/bh. There is also a longitudinal shear stress of the same value that acts along the longitudinal axis of the beam. But you don't design the beam for both stresses, just one will do, OK??
     
  19. Jul 26, 2012 #18
    I think torsion and shear are similar in behaviour, so i ask about both of them.I jusr reach good point, but why we dont design for both, why we design only for vertical??could you explain??
     
  20. Jul 26, 2012 #19
    I f there is an article to make me understand this it will be better, however i read and in didnt find any answer to my questions
     
  21. Jul 26, 2012 #20

    PhanthomJay

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    If you have a beam that is bending and twisting at the same time, you must design for bending stresses as well as both vertical and torsional shear stresses (they all act on the same plane). If you have a beam subject to bending only due to applied vertical forces, you must design for bending and complimentary vertical and longitudinal shear stresses (those shear stresses act on different planes).
     
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