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Thread: Max stress and deflection on supported circular shaft

  1. #1
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    Max stress and deflection on supported circular shaft

    ipt>
    Hello,
    Looking at different equations to apply to my situation and curious what the definition of "x" is in the attached equation?

    Maybe some suggestions on which equations would work best with the setup in the picture attached? I'm looking to put different size wire spools on this shaft and want to figure out the biggest/heaviest reel can be in different widths (ex. 108"x21", 108"x36, 108"x42, 108x68, etc.)Beam defl, stress, bending calc for a fixed beam at both ends, partial.jpgreel stand.jpgreel stand.jpgBeam defl, stress, bending calc for a fixed beam at both ends, partial.jpg
    Respectfully,
    Alex

  2. #2
    Technical Fellow Kelly_Bramble's Avatar
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    W is classically used to indicate a single total load where w = the line load (force per unit).

    The x distance is likely for where equivalent the total load W is as opposed to the line load.
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  3. #3
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    So do you just omit x if you are using a distributed load, as opposed to a single point load?
    Respectfully,
    Alex

  4. #4
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    this calculator seems more applicable to my circumstance.

    https://www.engineersedge.com/beam_b..._bending30.htm
    Respectfully,
    Alex

  5. #5
    Technical Fellow Kelly_Bramble's Avatar
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    Quote Originally Posted by alexmarks612 View Post
    this calculator seems more applicable to my circumstance.

    https://www.engineersedge.com/beam_b..._bending30.htm
    So... look at the end supports - that calc. is pinned on the left side and roller on the right end and distributed load as specified.
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  6. #6
    Technical Fellow jboggs's Avatar
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    It does appear to be the correct one. Be sure you use the maximum possible load, and do not scrimp on your factor of safety.

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    Well technically the shaft is not "fixed on both ends" its clamped on both sides in a bearing type apparatus with one side being captured with the brake. So would you still call this fixed on both ends? See attached pics.stand.jpgclamp.jpg
    Respectfully,
    Alex

  8. #8
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    With that being said, since the load is not a point load, do you just omit "x" when using a distributed load that does not span the entire length of the shaft? or is "x" just considered the middle of the load?

    Also, would you use this equation for stress, that is used on the "pinned one side, roller on the opposite"? https://www.engineersedge.com/beam_b..._bending52.htm

    n
    σmax = __* Mmax
    I
    Thanks for the input!
    Last edited by alexmarks612; 07-11-2022 at 09:29 AM.
    Respectfully,
    Alex

  9. #9
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    Can you check this calculator? When I run the numbers using the equations posted below the calculator, I come up with 6000 lbs for the Reaction forces Ra and Rb. You can see that it is coming up with other numbers......
    calculator equations being used.jpg
    calculator numbers.jpg
    Respectfully,
    Alex

  10. #10
    Technical Fellow Kelly_Bramble's Avatar
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    The line load used is 600 lbs/in over 20 inches. Thus 600 lbs/in x 20 in = 12,000 lbs of load applied to the beam

    If the loading was exactly distributed at the middle of the beam there should be 6,000 lbs per support.

    Logical...
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  11. #11
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    That's what I was thinking too, so the calculator is off then? I'm also coming up with some fairly large numbers using the Moment equations that are posted under calculator. Calculator showing 117500 lbs-in and I'm coming up with Ma = 752,000,000 and Mb = 784,000,000
    Respectfully,
    Alex

  12. #12
    Technical Fellow Kelly_Bramble's Avatar
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    Quote Originally Posted by alexmarks612 View Post
    That's what I was thinking too, so the calculator is off then? I'm also coming up with some fairly large numbers using the Moment equations that are posted under calculator. Calculator showing 117500 lbs-in and I'm coming up with Ma = 752,000,000 and Mb = 784,000,000
    Yes and no to your statement.

    We reviewed the calculator and found a (slight) error in the calculated results for the resultant reactions at Ra and Rb that's been corrected.

    Looked at the calculations for Ma and Mb and the numbers appear correct.

    The calculator has been updated and you may need to reload the webpage (F5) to see the new calcs.
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  13. #13
    Technical Fellow jboggs's Avatar
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    From the diagram it appears that the bearings being used are pillow-block self-aligning type bearings, which means that the shaft supports are not "fixed". The self aligning bearings allow the shaft to flex under load. That is their purpose. That way they only see the reaction force from the direct load on the shaft and do not have to restrict the moments resulting from the flexing of the shaft under load. Fixed mounts would restrict that action (just like a cantilever mount).

    Also, I do not understand why you have three bearings on the brake side. Two of them are not required. You are just over-complicating the arrangement. You will NEVER get all three bearing centerlines in line with each other. NEVER. That is another reason why self-aligning bearings are so prevalent.

  14. #14
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    Awesome! Thank you, I'm getting the same numbers now. Any possibility you could max stress to the calculator?

    Thanks,
    Respectfully,
    Alex

  15. #15
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    Only 2 bearings. The reel shaft has a Nylotron sleeve on each end that rides inside the clamshell clamp. On the end that attaches to the brake it has a piece of plate welded to it. That couples to the shaft running through the rotor, that has a "keeper" welded to the end of it. Two separate shafts. The one inside the brake rotor has bearings on each side of the rotor for support and alignment. I'll post another pic.

    Reel Shaft.jpg

    Brake Shaft.jpg
    Respectfully,
    Alex

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