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How come there is no bending here? (Welding)

  1. Feb 15, 2013 #1

    Femme_physics

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    http://img560.imageshack.us/img560/2662/nobending.jpg [Broken]

    This exercise is from machine parts. In order to find the width of the weld, I calculate the shear force and bending moment for this exercise. However, according to my teacher, there is no bending here and I calculate this exercise for bending in vain. Am I crazy? This looks like a perfect case for "bending" for me.


    I don't have anymore data sadly, the questionnaire hasn't been scanned yet, only my answer sheet.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 16, 2013 #2

    I like Serena

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    Well, we already know that you are a bit crazy and I like it! ;P

    I'd say there is indeed bending.
    However, bending is not so relevant here.
    In practice what might happen is that the beam will break off from the weld.
    It will do so long before it breaks due to bending.

    I presume you already did the calculations?
    What kind of results did you get?
     
  4. Feb 16, 2013 #3
    Well if i don't remember wrong all you need to do is move the Force to the centroid of the weld group and then treat it as if it was the transverse section of a beam. For this case this should be a weld group in torsion and shear.

    As "I like Serena" said, there is bending if the weld holds but its a LOT more likely that the weld will fail long before the bar does for bending. Basically as the force is being applied the bar wants to rotate and the weld starts to react and the bending starts to develop on the beam and at some position it will hold (if it does) and then all the bending occurs. It does not mean you should not calculate it for bending, at least i still would.
     
  5. Feb 16, 2013 #4

    Femme_physics

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    How is this torsion? I don't see the "torsional rotation". This exercise was only drawn in 2D for us, by the way. How can there be torsion in 2D? Or rather, how can there be torsion if the force acts in the middle of a symmetrical, balanced shape?

    I don't see why you claim that bending is irrelevent. Is this because of the shape of the weld? Let's assume there was a weld on top and not n the bottom-- would then belding be relevent?

    And yes-- that was my solution: (I won't bother translating the Hebrew notes, I think you understand the math well enough :-) )

    http://img833.imageshack.us/img833/3820/an1hg.jpg [Broken]

    http://img826.imageshack.us/img826/2532/an2y.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  6. Feb 16, 2013 #5
    When you move the Force to the centroid of the weld group its a force down and a moment generated by the distance between the point of application of the force and this centroid.

    This moment is felt as a torsional moment in the weld group. Its useful to see the weld group as a cross-section of a beam, in that case you are seeing it from the "front" and this moment would be torsional. You can check the books on machine design (Shigley, Mott, Juvinall, etc) for design of weld groups.

    The bending is not irrelevant, but the first concern is the weld since the force is most likely to make the weld fail first than the beam and if it does fail it does not matter what happens with the beam it will just fall down. So first make sure the weld will hold and then you have to calculate the beam for bending failure.

    To further explain the type of loading on the weld group, imagine you are seeing this beam from the front and put welds surrounding this rectangle to the wall (instead of the welds in the figure), in this case the weld group would be in direct shear and also bending.
     
    Last edited: Feb 16, 2013
  7. Feb 16, 2013 #6

    Femme_physics

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    Well, still, according to my teacher note, he bluntly said "No bending" -- if I'd go argue with him he'd probably say it's irrelevent and I should do the calculations for the critical forces at play. I presume like you said, that would be torsion and shear.

    Would you mind answering my question from before: Let's assume there was a weld on top and on the the bottom (as opposed to just on the bottom like in the exercise)-- would then belding be relevent to calculate?
     
  8. Feb 16, 2013 #7

    I like Serena

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    I don't see how torsion would be involved either.

    With a weld on top instead of the bottom little changes.
    I think it would still be the shear that decides the fate of the beam (and the people standing under it).

    Now if you had a weld both on top and on bottom, then perhaps the beam itself would break before it breaks off at the weld.

    I have a little trouble interpreting your calculations.
    Did you get a result for when the beam breaks due to bending?
    And a result when the beam breaks off due to shearing?
     
  9. Feb 16, 2013 #8

    Femme_physics

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    I combined the results from bending and shearing, yes.

    Ideally I'd combine all possible vectors, including torsion, shear, and bending. But if my teacher is strict about me leaving bending out, then maybe torsion. But if you're strict about leaving torsion out... *head explodes*

    *head regrows* Who is little ol' me to argue! :-)

    *Wipes alien's goo from the regrowth effect* hate it when that happens...
     
  10. Feb 16, 2013 #9
    To answer the question fast, yes there is bending, the beam is supported at one end and applied one force at the other, there will be stresses due to bending. In fact if you want to be real specific you can even say there is a bit of tension since the angle of the beam will be tilted a little, but that is not important compared.

    If you want to do the full check you need to check the design books and there you can find some of the codes for welding. First you check the weld, then you have to check the parent metal (since its usually not as strong as the weld) in the vicinity of the welding and well there are codes and considerations for that, for example if your beam is cold drawn close to the welding its properties are replaced by hot rolled due to the heat generated in the process. If that holds then you do not really have to check all the beam because the yielding stress that you use for the parent metal at that point is supposed to be less than the beam (0.6Sy of beam material if i dont remember wrong) so if it doesnt fail there it should not fail at all.

    Even then since i am kind of paranoid i always check everything, so i check the beam even if its redundant but yes it does not matter how many welds you put there if they hold and maintain the beam in place then there is bending.

    And by the way the torsion might confuse a little, its just the name that the books in design give to the type of loading on the weld group, they make the analogy that it is the cross section of a "beam of welds". In the case you presented the maximum stresses are in the points of the welds that are radially further from the centroid of the group so its similar to torsion.

    I will try to make it little more clear, for example if you were looking at the beam from the front and you fix it to a plane wall with welds surrounding the rectangle of the beam (not the case you presented), then with the same force applied when you move it to the centroid of the weld group it would be in bending and shear (the weld group) just because the higher stresses are in the point of the weld that is furthest vertically on both sides from the centroid (as in bending of a beam) and also there would be bending in the beam. Your teacher may have said that because it generally is redundant to do those calculations but there is bending.
     
    Last edited: Feb 16, 2013
  11. Feb 16, 2013 #10

    Femme_physics

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    Last edited by a moderator: May 6, 2017
  12. Feb 16, 2013 #11
    Now that i think of it, he probably meant there is no bending on the welds and not for the beam.

    Anyway since teachers sometimes can be little confusing i always try to check the books, i have scans from the time i was studying machine design from the books which i can send if you like, but with the warning that machine design books tend to be kind of frustrating since you can find different considerations and methods for doing the same thing. The best way to go is read all of them and use the method you prefer or get the code for calculation of weld throats from your country.
     
  13. Feb 16, 2013 #12

    Femme_physics

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    I'll be happy to read your scans, if you have anything with respect to screws, rivets, gears and of course welds, I'll be happy to read that as well. Keep in mind we don't use calculus in our calculations, so I hope what you have is relatively elementary-- regardless I appreciate it. I'll PM my E-mail address :)
     
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