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  1. #11
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    Feb 2008
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    100 psf seems like a good design load for this application. There isn't a category for "motorcycle stand," so you have to estimate what's going to take place. To compare to other design loads, residential floors are generally 40psf live and 10 psf dead, decks and balconies are 60 or 100psf depending on area, and 10psf dead load.

    2.7psf is the actual dead weight of a 3/16" thick sheet of aluminum.

    The 1.6 multiplier is a standard factor of safety for live load in LRFD design to account for impact effects and uncertainty in load magnitude... as is 1.2 for dead load (lower uncertainty).

    80% of failures are from 20% of causes
    Never compromise your principles today in the name of furthering them in the future.
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    "We are generally better persuaded by reasons we discover ourselves than by those given to us by others." -Pascal
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  2. #12
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    May 2008
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    Troy, MI
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    Nice explanation. I bet you picked up some civil engineering knowledge somewhere.

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  3. #13
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    ok, i defninitely follow the logic of your explanation about temper affecting the yield point (at which permanent deformation occurs), but temper not affecting temporary elastic deflection where the object will return to the original shape. but actually all along i have been referring to temporary elastic deflection.

    in my example when i said "heat it up with my acetylene torch and it is easier to bend," i was referring to temporary elastic deflection. what i really should have said more precisely was the following: i temporarily heat up the 1/16" sheet with the torch, then let it gradually cool to anneal it. after it is cooled off, i "bend" it, as in flop around like a flat spring in my hands (i didn't mean bend as in fold it in half to permanently deform it).

    i am imagining the following experiment: take two 2" x 36" strips of 1/16" AL sheet 6061-T6. anneal one sheet and leave the other at T6 temper. then take both sheets and lay them across a 30" wide span (not fixing the endpoints). they will obviously both sag somewhat in the middle. but i am thinking the annealed sheet will sag more. neither strip of sheet metal has been permanently deformed though, since they both return to the original shape when removed.

    from what you are saying, both sheets will sag the same amount. is that really true? if so, i would never have guessed that.
    Last edited by ridesideways; 02-20-2009 at 01:41 PM.
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  4. #14
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    Jan 2009
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    SW Ohio
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    Quote Originally Posted by ridesideways View Post
    from what you are saying, both sheets will sag the same amount. is that really true? if so, i would never have guessed that.
    Believe it . . . it's true.

    The only thing I would add, if you heat your 6061-T6 to 400 F, the yield strength will drop to 15,000 psi and elongation increase from 17% to 28% . . . it will bend more easily, it will bend farther, but it will not be annealed. When cooled back to room temperature, it will return back to 40,000 psi yield strength and 17% elongation as long as the heat cycle is not prolonged . . . 30-60 minutes will not appreciably affect the properties. Heating it for longer periods will over-age it and reduce the yield strength, but not all the way to -O temper.

    Jim


  5. #15
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    Feb 2008
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    Quote Originally Posted by ridesideways View Post
    i am imagining the following experiment: take two 2" x 36" strips of 1/16" AL sheet 6061-T6. anneal one sheet and leave the other at T6 temper. then take both sheets and lay them across a 30" wide span (not fixing the endpoints). they will obviously both sag somewhat in the middle. but i am thinking the annealed sheet will sag more. neither strip of sheet metal has been permanently deformed though, since they both return to the original shape when removed.

    from what you are saying, both sheets will sag the same amount. is that really true? if so, i would never have guessed that.
    The Euler-Bernoulli beam equation contains modulus of elasticity E, section moment of inertia I, and uniform load as the only variables affecting the slope of the beam at any given point.

    They will sag the same amount provided you have not somehow altered E through metallurgical changes, I through deformation, and the span must remain the same.

    They may not be exactly the same, but they will be darn close - within experimental error.

    80% of failures are from 20% of causes
    Never compromise your principles today in the name of furthering them in the future.
    "All I ever wanted was an honest week's pay for an honest day's work." -Sgt. Bilko
    "We are generally better persuaded by reasons we discover ourselves than by those given to us by others." -Pascal
    "Since we cannot know all that there is to be known about anything, we ought to know a little about everything." -Pascal

  6. #16
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    Oct 2008
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    Bodybagger: thats the caliber of information i expect from you. good job.
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  7. #17
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    Oct 2008
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    taxachussetts
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    ridesideways; call around for a good price. i think i payed under $300 for a sheet of 1/4dp alum. and MILL STEEL was over $600. that was a year ago.
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  8. #18
    Join Date
    Aug 2004
    Location
    Milan Michigan
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    Bodybagger that was some math work slightly above me but I got the point,
    I'm much more the type of person to tell me your application and I will get you pretty close.

    Its always nice to have a brainiac like you around.
    I was on another site talking about beam reinforcement and we had quite a heated discussion as to how to strengthen a beam.

    I'm going to start a new thread and would appreciate you to chime in.

  9. #19
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    Sep 2007
    Location
    Medford MA
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    I just want to add m thanks to Bodybagger and the others for this
    informative thread. My only complaint is that it brings back really
    bad memories of my college "statics and dynamics" course ... the only
    course I took where I burned the text book afterwards

    Seriously, thanks ... it's been cool.

    Frank

  10. #20
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    Jul 2007
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    Southern NH
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    Quote Originally Posted by Bodybagger View Post
    They will sag the same amount provided you have not somehow altered E through metallurgical changes, I through deformation, and the span must remain the same.
    I suppose I was thinking that annealed aluminum would have a different E than T6 aluminum. But it sounds like that is not the case. Thanks very much for the informative information!
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