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Welding direction

  1. Jun 15, 2005 #1
    This is kinda hard to explain without a drawing but I'll try. Imagine an L-shaped structure with a weld going down the vertical part connecting the L to a vertical pipe on the left side of the L. The curvature of the pipe is insignificant. Then on the L's right end is a vertical downward load. I find that I have a [tex]\tau_\|[/tex]. This I can see, since the weld is being "streched" in a paralell direction, but my book also tells me that [tex]\sigma_\bot = \tau_\bot[/tex] is also part of the problem, and I am not able to see why this is...can someone please help me?
     
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  3. Jun 15, 2005 #2

    brewnog

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    I think I'd like a drawing to think about this one.
     
  4. Jun 15, 2005 #3
    Ok, drawing is on its way, I'm just relieved that someone is actually trying to help me here, didn't have too much hope...
     
  5. Jun 15, 2005 #4

    Astronuc

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    ||
    ||
    ||
    L
    ||

    with a vertical force on the - of the L.

    Well the L has a downward force on the horizontal arm, which induces a shear in the weld. But the downward force at the end of the arm produces a moment at the vertex of the L, i.e. where the vertical and horizontal parts meet.

    The question is why the equivalence [tex]\sigma_\bot = \tau_\bot[/tex], I imagine. I'll have to think about it.

    I suppose PerennialII will have some insight.
     
  6. Jun 15, 2005 #5

    brewnog

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    I think Astronuc has this one in the bag.

    The L part will be wanting to rotate about it's 'elbow', causing the welded seam to pull radially outwards on the pipe.
     
  7. Jun 15, 2005 #6
    Yes, that might be it. The question Astronuc is not why [tex]\sigma_\bot = \tau_\bot[/tex], but why they are there at all. As far as I can see the weld isn't exerted by any perpendicular force, only paralell, as noted by [tex]\tau_|[/tex]. But since the force will try to turn the whole thing around clockwise, that might cause some "perpendicular-like" force...or?
     
  8. Jun 15, 2005 #7

    brewnog

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    Yes TSN.

    Imagine the L is just attached at its base. If you apply a force downwards on the _ of the L, the member will rotate clockwise about the left hand side of the _.

    As a result, the weld provides a force on the L, acting towards the centre of the pipe.

    If you're still stuck, read up on couples and moments.
     
  9. Jun 15, 2005 #8
    Yes, that might be it. The question Astronuc is not why [tex]\sigma_\bot = \tau_\bot[/tex], but why they are there at all. As far as I can see the weld isn't exerted by any perpendicular force, only paralell, as noted by [tex]\tau_|[/tex]. But since the force will try to turn the whole thing around clockwise, that might cause some "perpendicular-like" force...or?

    Here is a link that explains the notation by the way:

    http://www.gowelding.com/calcs/c2.html
     

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  10. Jun 15, 2005 #9

    Astronuc

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    Something like this -

    |->
    |
    | . . . . .|
    L______V

    I believe the bending moment is greatest near the right angle in the L, and it should fall off.

    Basically the weld provides a distributed load on the vertical branch of the L, and the vertical load at the end of the horizontal arm is a pointwise load.

    Distributed load on vertical arm
    |<-
    |<-
    |<-
    |<-

    and a moment at the right angle.
     
  11. Jun 15, 2005 #10

    PerennialII

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    ... I'm getting this the same way ... applied load is shear with respect to the "L's tip", the shear causes a moment to the pipe, the moment in the pipe causes a support reaction, and the support reaction is a shear load as far as the pipe is concerned. I think the notion about the equality in | and -- can make sense if the "support span" in the pipe is equal to the "length" the L ... this way I think the shear forces will be equal in both directions (but stress wise needs some further thought if we're using e.g. beam theory ... shouldn't result generally but rather a specific result...).
     
  12. Jun 15, 2005 #11
    So to wrap up, would it be correct to say that we have a component [tex]\tau_\|[/tex] initially, but because the downward force tries to rotate the L-shape clockwise, this results in a perpendicular force on the weld, namely [tex]\sigma_\bot[/tex] or [tex]\tau_\bot[/tex] if you will. If someone confirms my thoughts on this I'll be so happy! :rofl:
     
  13. Jun 15, 2005 #12

    brewnog

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    Pretty much, yes.
     
  14. Jul 1, 2005 #13

    Danger

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    I don't know any of the math, but my approach if I had to build the thing would be to think of the angle of the 'L' as a hinge attached to the pipe. Then you can do whatever number stuff you have to do in order to figure out the 'pull-away' force at the top by leverage formulae. Then factor in that the weight would also be trying to slide the whole thing down the pipe, which I guess is the 'sheer'. When in doubt, use bolts instead.
     
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