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How is this torsion?

  1. Dec 1, 2012 #1

    Femme_physics

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    Last edited by a moderator: May 6, 2017
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  3. Dec 1, 2012 #2

    jedishrfu

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    isn't the force on the tube (comp parallel to the tube) and also trying to bend it downward (component parallel to the wall)?

    I'd say the bending down is a torque. consider the tube at the wall as the center of a circle and the force is acting 900 units away from the center t = r x F
     
  4. Dec 1, 2012 #3

    Femme_physics

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    No, how?

    You said it yourself "Bending down", so therefor "bending down" = BENDING

    Furthermore, the force on the X plane strives to stretch the part, so the force on the X plane causes TENSION, if anything, and not shearing (like my teacher said).

    How am I crazy here?
     
  5. Dec 1, 2012 #4

    I like Serena

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    Heya!

    Sorry, but I guess I'll have to explain how you are crazy here. ;)

    Ultimately, the object is 3 dimensional.
    In particular the force is applied to a point, meaning torsion and shearing do occur.

    Let's start with the shearing.
    Consider horizontal layers in the object.
    If the object is weak enough, you'll pull the middle layer out, while the top and bottom layers remain attached to the wall.
    That is shearing.

    As for the torsion, consider layers of the object that are above and below the paper.
    The force bends the layer in the paper down.
    But a layer above the paper is not bent down directly and resists.
    This is torsion.
     
  6. Dec 1, 2012 #5

    Femme_physics

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    Well welll well, it seems the boys are back in town! ;)

    Hello ILS :) Good to breath the cybercrusted physics intellecto-airosphere once more.

    Hmm....

    So I presume you're refering to the fact that the object is U-shaped, which you brightly deduced through the drawing and I didn't mention.

    But still...look ->

    http://img233.imageshack.us/img233/9059/imgsmx.jpg [Broken]

    What's the difference between the tension here and the tension in my example? The welding? The U-shaped? They make all the difference?

    Ahh...so we're also taking into consideration the object is U-shaped. And if it wasn't?
     
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  7. Dec 1, 2012 #6

    I like Serena

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    No, I wasn't referring to it being U-shaped.
    I was assuming it's a block or a screw.


    It's not regular tension like you are drawing.
    Look ->

    attachment.php?attachmentid=53521&stc=1&d=1354357839.png

    The forces are not on the same line.
    That is why it is shearing.


    How is it U-shaped?
     

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  8. Dec 1, 2012 #7

    Femme_physics

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  9. Dec 1, 2012 #8

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    Yes, we did assume the force was spread out equally.
    Shearing effects were "neglected".

    But now you're in a more advanced course! :)
     
  10. Dec 1, 2012 #9

    Femme_physics

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    Aha!!! Elightenment has arrived! :)

    Ahh....good feeling. I feel like I level'd up. :wink: Really glad you're still here ILS.

    Hmm... OK I get shearing, but the torsion is still somewhat flummoxing..lemme grab your quote

    So refering to only Fy here I presum. Yes, upper layer bends down alright, but the lower layer bends down as well.

    I see it as something like that:

    http://img252.imageshack.us/img252/924/280pxibeambending0.jpg [Broken]

    The upper and lower layer bend the same way equally
     
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  11. Dec 1, 2012 #10

    PhanthomJay

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    this gets a little complex now depending on what the shape of the beam actually is. If it is an open U shape and the load is not applied at the 'shear center' of its cross section, it will twist. But i doubt that this is what is being asked. Consider it an I beam or square tube, and then there is no torsion in the beam with a load assumed applied at the end through the centroid of its cross section. But the weld itself is subject to a twisting couple causing shear stresses in the weld (visualize the weld cross section as perprndicular to the beam cross section).
     
  12. Dec 1, 2012 #11

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    Ah, but you did level up. How crazy is that! :approve:
    Glad to see you're back.

    That would be the case if the forces were distributed equally.
    In practice it wouldn't bend so nicely and equally.
    Applied to your OP drawing, it's more like this ->

    attachment.php?attachmentid=53523&stc=1&d=1354361644.png

    Hence, torsion.
     

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  13. Dec 1, 2012 #12

    Femme_physics

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    So it all comes down to presuppositions. Well, I wish it has been so clear to us in the beginning! I think I can take it from here then :smile: Thanks ILS, and to everyone else replying as well...

    Help is much appreciated.^^
     
  14. Dec 1, 2012 #13
    Hey I'm new to PF and hope I can help,

    To set the context, the beam is being bent clockwise (x, y plane) and obviously there is tension of the upper surface and compression on the lower surface as a result.

    The beam is an extruded U or semi circular shape because the wall cross sections would be hatched otherwise.

    The 20kN is a point load and is eccentric (y, z plane). The other end is fixed from ANY movement and therefore is subject to torsion loading from the clockwise twisting moment imposed (by the eccentric force vector).

    It's a 2D sketch I guess to just to make it more challenging, and when you do engineering as a job you'll almost always be faced with a 2D drawing when performing loading/stress analyses.

    Hope this helps and happy studying :)

    The Jericho.
     
  15. Dec 3, 2012 #14

    Femme_physics

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    This exercise ...

    http://img248.imageshack.us/img248/4868/secondut.jpg [Broken]

    ...is claimed to have bending in the solution. However, using the same logic from the last exercise, I can just claim this is nonsense because the side layer where the pin is connected can be torn off the main part...so we get here just shearing and and torsion.
     
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  16. Dec 3, 2012 #15

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    In this case the force does not have a component away from the wall, so there is no real shearing.
    I guess the main effects would be best modelled with bending and torsion.
     
  17. Dec 3, 2012 #16

    PhanthomJay

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    Sometimes it is best to replace the eccentric vertical load with an equivalent load and couple (a twisting couple in this case) applied at the centroid of the cross section. Then it is easier to envision that the vertical load cause bending and vertical shear in both the beam and weld, and the twisting couple causes torsion and additional shear stresses in the beam and weld.
     
  18. Dec 3, 2012 #17

    Femme_physics

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    There is shearing according to the solution.

    1st) Shearing 2nd) Torsion
    http://img836.imageshack.us/img836/8769/psptr.jpg [Broken]

    3) Bending
    http://img24.imageshack.us/img24/1362/74616208.jpg [Broken]

    Sorry for the img being so big.

    Essentially we got combined forces here.

    My problem is that I could use your same logic in this pic

    http://img27.imageshack.us/img27/9219/torsion.jpg [Broken]

    To say that there's no bending, only torsion.

    The pin is pretty far from the wall!
     
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  19. Dec 3, 2012 #18

    I like Serena

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    In the picture I drew, there is a combination of bending and torsion.
    All layers are bent into the direction of the force.
    But there is also torsion between the top and middle layer, and reverse torsion between the middle and the bottom layer.


    So?
    Is there a force component along the length of the profile?
    Anyway, as PhanthomJay remarked, you have vertical shear in both the beam and weld.



    When you were originally doing mechanics, you learned that you had horizontal forces that had to balance out, vertical forces that had to balance out, and a moment that had to balance out.
    These were external forces that were applied to each component.
    Now you also have bending, shearing, and torsion within each component to take into account.
    That's what you get from leveling up. ;)
     
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  20. Dec 3, 2012 #19

    PhanthomJay

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    That is really big, indeed. Sorry for my response also being of the same order of magnitude in size. I want to point out that in your formula [itex] ρ_s = F/\Sigma l[/itex], that is the formula for the vertical shear in the weld in units of force per length of weld. It should not be confused with the shear stress in the beam. As has been pointed out, both the beam and welds are subject to bending and shear stresses from forces and and moments. Both the beam and weld must be designed accordingly, so don't confuse the 2.
     
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  21. Dec 5, 2012 #20
    That's an awful lot of correspondence and really it's just the use of terms and a point of view. The weld group is subjected to a tension, a shear and a moment about the centroid of the weld group, about the axis perpendicular to the paper. If someone wants to call that moment a torsion, why shouldn't they? It's not a bad use of the term in this case, because it causes shear stresses in the welds, and shear stress is usually associated with torsion in a solid member. What it does not represent is a torsion on the beam itself. That is bending. But we are discussing the weld group...
     
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