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  1. #1
    Join Date
    Aug 2005
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    828

    I-Beam load capacity?

    I know that there are a lot of factors to determine that exact number for load capacity of a I-beam, but is there a simple answer to this?

    How tall, wide, thick will the I beam need to be to handle 3,000lbs or less and be supported on the ends only, 50 feet apart?

    I looked at some formulas but it got confusing (chair)

    Thanks
    Dennis

  2. #2
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    Oct 2005
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    672
    The formulas get confusing because they tend to calculate deflection and stress of the beam. To use the formulas, you have to determine what you consider "failure" of the beam. If it plastically deforms under the 3000lbs load? Can it deflect 1" in the center?

  3. #3
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    Jun 2004
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    463
    Do you mean actual load capacity, or are you concerned with deflection also? For most CNC applications, you are not concerned with load capacity because deflection will become excessive long before reaching maximum load capacity. Either way, you can download the free program BeamBoy to calculate stress and deflection. http://www.geocities.com/richgetze/

    I am not really sure, but I believe the maximum bending stress for structural steel is around 20ksi. For a 50ft span, BeamBoy says it would take at least an S10x35 beam to stay under 20ksi with a 3000lb load, plus the distributed weight of the beam. The deflection with that load would be over 4 inches. At 35 lbs/ft that beam would weigh nearly a ton. What are you building?

  4. #4
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    Jan 2005
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    Sounds like he wants to put a beam across his shop and put a trolley on it to move machinery around.
    If it's not nailed down, it's mine.
    If I can pry it loose, it's not nailed down.

  5. #5
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    Jul 2005
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    12177
    Quote Originally Posted by DennisCNC
    ...How tall, wide, thick will the I beam need to be to handle 3,000lbs or less and be supported on the ends only, 50 feet apart?
    Thanks
    Approach the problem from the opposite direction. Find some standard I-beam profiles and the I value, that is the Moment of Inertia, for that profile then plug that into the formula along with your length and calculate the maximum deflection at the center. If the deflection is too much go to a bigger beam; bigger I and calculate again.

    I can tell you 3,000 lbs in the center of a beam spanning 50 feet will require a hefty beam unless you can accept several inches of deflection.

  6. #6
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    Aug 2005
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    2muchdtuff, you got it. It can bend and deflect all it wants as long as it says near the roof and not on my head.

    So a 12" I beam will do the trick to be on the safe side.

    When I get the winch to the shop need to see if I will have enought hight for the trolly, winch, i-beam.

    What about two smaller I-beams side by side?
    Attached Thumbnails Attached Thumbnails Trolly.jpg  
    Dennis

  7. #7
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    Jan 2005
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    746
    Years ago I had an engineer figure an I beam for my garage. He spec'd a 6" by 12" beam for a 24' span. With your 50' span I might want to go a little larger. As for the idea of side by side, I wouldn't. I'd weld the one on top of each other, this will give you less deflection.

    Is it possible to support the center of the beam down from the roof or do you have an open steel truss girder arrangement holding up the roof deck.
    If it's not nailed down, it's mine.
    If I can pry it loose, it's not nailed down.

  8. #8
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    Jun 2003
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    866
    I would think that at 50 feet, you are in the range of having buckling problems sideways as well as deflection problems. 2much is right, you should try to support it in the middle. Since the trolley will just ride on the lower flange, that leaves the upper flange free for support purposes.

    And this is one application where going to an engineer that does these kind of things would pay off.

  9. #9
    Join Date
    May 2005
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    2502
    Here is a thought. Rather than mounting a huge beam, can you make some A-Frame supports that travel as well? I haven't seen this done, but if you can position a travelling A-Frame either side of your hoist, you can ensure smooth travel over a 50 foot difference and only slightly more inconvenience (due to positioning the supports).

    Best,

    BW

  10. #10
    Join Date
    Dec 2005
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    1408
    Quote Originally Posted by DennisCNC
    I know that there are a lot of factors to determine that exact number for load capacity of a I-beam, but is there a simple answer to this?

    How tall, wide, thick will the I beam need to be to handle 3,000lbs or less and be supported on the ends only, 50 feet apart?

    I looked at some formulas but it got confusing (chair)

    Thanks
    Dear DennisCNC,

    You say you want to handle 3000lbs, and the beams will be, I assume, "simply supported". Do you want to support 3000lbs at the middle of the span, or will it be distributed along its length. The loading conditions are completey different, and may call for different bits of steel. If you can give more information, I'll get out my text books.


    Best wishes,

    Martin

  11. #11
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    Aug 2005
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    828
    ^^ I measured the shop and it is 45' not 50' so thats a plus. I will use it move granite slabs around, from the cart to the saw and so on. Maybe my 3K lbs estimate is a little high. The slabs weight is 1,200lbs. max plus the winch, trolly, vacuum lifter, I think that will be another ~700lbs. So I think a better estimate would be 2,200 lbs. max load. This load will mostly be in the middle.
    I can make the trolly support wheels farther apart to spread the load. If I happen to need to move something heavier i'll just use the forklift. With the over head crain one person can move the material around.
    Dennis

  12. #12
    Join Date
    Dec 2005
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    1408
    Quote Originally Posted by DennisCNC
    ^^ I measured the shop and it is 45' not 50' so thats a plus. I will use it move granite slabs around, from the cart to the saw and so on. Maybe my 3K lbs estimate is a little high. The slabs weight is 1,200lbs. max plus the winch, trolly, vacuum lifter, I think that will be another ~700lbs. So I think a better estimate would be 2,200 lbs. max load. This load will mostly be in the middle.
    I can make the trolly support wheels farther apart to spread the load. If I happen to need to move something heavier i'll just use the forklift. With the over head crain one person can move the material around.
    Dear Dennis,

    unterhaus is right on the button with with comments on buckling.

    If the top flange of the I beam is not restrained (ie prevented from moving sideways), and you have a clear span of 45 foot, the permissible bending stress for the beam will be far less that the nominal value for the steel. In order to get round this problem, you have to go for a larger section.

    I have not done steel calculations for a while, but I thought I'd have a go.

    My conclusion is that if the ends of the beam are not built in to something substantial, and if the top flange is not restrained, the major factor is keeping the bending stress to acceptable limits. If you have a beam of steel which has a basic permissible bending stress of about 23000 lbs per square inch, by the time you make allowances for the span and the lack of restraint, the actual bending stress that the beam can handle is down to about 6100 lbs per square inch under these conditions.

    If my calculations are right (and it is possible they are not) you might consider this beam....

    24" deep
    9" wide

    top and bottom flange thickness 11/16"
    web thickness 7/16"
    beam self weight 225 lbs per yard

    The steel should have a basic permissible bending stress of at least 23000 lbs per square inch.

    The total deflection at mid span (including that due to the self-weight of the beam) will only be a little over 1/4". The bending and shear stress are within limits but I have not checked for web crushing and buckling.

    Finally, please don't go ahead and use this beam without getting a structural engineer to OK it. I can't afford the lawyers fees!

    Best wishes

    Martin

  13. #13
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    Jul 2005
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    From post #12 martinw
    24" deep
    9" wide
    3375 lbs

    I think that qualifies as a 'hefty beam', even more hefty than I anticipated when I made the comment above.

    The correspondence between Martin's specs and those from post #7 2muchstuff "He (an engineer) spec'd a 6" by 12" beam for a 24' span" is quite good; roughly four times the beam area, i.e. roughly 8 times I, for twice the span.

    So now all that is needed is to spec out a beam capable of supporting 3375lbs so the beam capable of supporting 2200 lb can be installed.

  14. #14
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    Dec 2005
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    1408
    Quote Originally Posted by Geof
    From post #12 martinw
    24" deep
    9" wide
    3375 lbs

    I think that qualifies as a 'hefty beam', even more hefty than I anticipated when I made the comment above.

    The correspondence between Martin's specs and those from post #7 2muchstuff "He (an engineer) spec'd a 6" by 12" beam for a 24' span" is quite good; roughly four times the beam area, i.e. roughly 8 times I, for twice the span.

    So now all that is needed is to spec out a beam capable of supporting 3375lbs so the beam capable of supporting 2200 lb can be installed.


    Dear Geof,

    A span of 45 feet is an entirely different animal from one of 24 feet.

    On long spans, and 45 feet is, for a simply supported, unrestrained steel beam, at about the limit of design, there is a danger that the compression forces in the top flange of an I beam reaches a level where it is acting as a column in compression and may buckle. Steel design takes this into account and reduces the allowable bending stress quite severely. It results in a deep, and wide section on long spans to reduce buckling of an unrestrained top flange of the I beam. The choice of "I" has , on long spans , little to do with deflection. All that matters is that you have a sufficient "Z", and an an l/ry ratio to keep the allowable bending stress within limits inter alia to prevent an un-restrained top flange from buckling.

    IMHO, wear a crash hat and take out a lot of life insurance.

    Best wishes

    Martin

  15. #15
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    Nov 2003
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    79
    you also need to consider the whole weigth of the beam itself pretty closely thats a big distributed load...I can get out my mechanics book and pull the formulas but I don't think its really feasible. let me toss it in cosmos though with a standard-ish large I beam with the load at the center

  16. #16
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    Jun 2003
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    1365
    UPDATE 7-24-06, pointed out by bma137, I used ultimate strength rather than yield strength, The numbers will be fixed

    Oh the bad memories from Mechanics of Materials class..... I will take a look in my book, no gaurentees that I want to complete this

    You want at least a factor of safety of 2, so lets say 5kips.(5000lbs) Theres also tons of different styles of beams. I wish I knew what was the most common...

    wow this is hard to remember.

    I'm going to calculate this for a wide flange shape (W shape)


    From my notes in class:
    Design should be Economical.
    Choose beam with the smallest weight/unit length
    1. Choose material, find σ ultimate for the material
    2. Find σ allowable = σult/FS (FS= Factor of safety)
    3. Draw shear and bending moment diagram.
    4. Find Max Moment and location
    5. Use eq σmax = Max moment/S
    6. Find minimum S from Apendix C of my book, Properties of rolled - steel shapes.

    The shear and bending moment diagram might not be necessary, it might confuse some of you but basicly you find the max moment, which in this case for the max stress would be in the center.

    I'm going to use structural steel: ASTM-A36: σult=56KSI (56000 PSI)SHOULD BE YIELD STRENGTH OF 36KSI
    Modulus of elasticity is 29*10^6 PSI(I dont remember if I need this yet. still gotta remember the symbol for it)

    Getting to the simply supported beam, 2 fixed supports.

    σyield=36KSI
    Max load = 3000lbs(FS goes with sigma σ )
    FS=2
    Mmax= 67500 ft-lbs at 22.5ft
    σall = σult/FS = 36KSI/2 = 18ksi

    S=Mmax/σall = 67500/1800lbs(must have units correct) = 37.5in^3

    So going to the table we have these options:

    PREVIOUS NUMBERS WERE INCORRECT
    W12x30 Area = 8.79"^2 Flange width = 6.520 Thickness = .440 Web = .26 S=38.6

    W16X26 Area = 7.68"^2 Flange width = 5.50 Thickness = .345 Web = .25 S=38.4

    The second would be the ideal choice (W16x26) with it being the cheaper one.


    There are a few extras that I did not list because your looking for the smallest number after the W# W is the width, so W12 is 12" wide(big flat to big flat) and the 30 kind of like the weight/ unit length,

    So W16X26 would be the best choice.


    Someone might want to double check these, its kind of early in the morning
    -been doublechecked 7-24-06

    Hope this helps more than just one of you! I included a sample problem with a shear and bending moment diagram, with a few loads and different support.

    -edit- be careful, this is not engineering advice and I will not be liable of it.

    Images have the wrong ones underlined because of reasons listed above.

    Jon
    Attached Thumbnails Attached Thumbnails matproperties.jpg   wshapes1.jpg   wshapes.jpg   sampleprob.jpg  


  17. #17
    Join Date
    Jun 2005
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    194
    http://www.efunda.com/math/areas/IbeamIndex.cfm - See if this will meet your needs.

    Enjoy,
    JR Walcott
    Georgia Machine Tool Resources, LLC

  18. #18
    Join Date
    Nov 2003
    Posts
    79
    alright can't find the mechanics book but that looks right JF. Because of the vertical loading I would definitely aim more towards a narrow tall beam. Obviously I'm not giving out official engineering advice over the internet and all the legal speak but... Cosmos says factor of safety of around 4 with a W12x22 beam loaded with a 5k lb load at the center, the weigth of the beam and restrained at the end. I would be more worried about the realities of your building supporting all this safely.
    Attached Thumbnails Attached Thumbnails Part1-COSMOSXpressStudy-Stress-Plot1.jpg  

  19. #19
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    Jun 2003
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    1365
    avsfan, you kinda reminded me that I needed a disclaimer on that. Its more for a general idea of what you should do, and yes, the building and everythign else needs to be able to take the load.

    Jon

  20. #20
    Join Date
    Jul 2006
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    11
    JFettig, I noticed that you use the ultimate stress of A36 structural steel to find your allowable stress. I'm an ME undergrad as well and I know that you should never use the ultimate stress to design anything like that. You need to use the yield strength of the steel to find the allowable stress, and for A36 I believe it is a minimum of 36ksi.

    Like I had said before, never use the Ultimate Stress to design unless you are designing it to fail.

    P.S. that text book looks like the exact one we use.

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