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Help with aluminum beam strength

  1. Nov 12, 2008 #1
    I built a catapult (trebuchet actually) last year for the "Punkin Chunkin" where you throw pumpkins for max distance. This year I am planning on a slightly larger one and need some advice on aluminum. We basically use a primitve crane idea to lift the weight up rather than using a hoist or whatever. See attached pic. By cranking up the cable, the beam which is hinged on the bottom slowly rises and brings the weight up with it, at the top, we secure the weight and crank down the beam. I used a salt-treated 4x4 last time and it was maxxed out with about 600 lbs. I was looking for a person or place to tell me what load I can use for a particular aluminum beam and a particular length lifting a weight. I can just start buying beams and trying it, but trial -and -error can get expensive at $300+ a beam.

    I want to use a 18 foot long aluminum american standard beam 4"x.326. The load basically is pulling at a slight angle as shown in the pic, and I can provide a little more info, but I'm unsure of where to ask, or the equations I use to figure it out.
    Any help would be appreciated,

    Attached Files:

  2. jcsd
  3. Nov 12, 2008 #2
    I would like to help but I must first tell you that your machine looks frightening. Not so much for the recipients of the pumpkins, but for the senders. If you used a wooden beam before, it probably bowed quite a bit giving you warning that it was loaded to its max. You may not get the same warning with aluminum.

    Do you realize that if the aluminum beam fails, it will fail in a catastropheric manner? By that I mean it will suddenly buckle spewing bolts and brackets at high speed in unpredictable directions. Worse than that, the cable, suddenly relieved of its constraints, will whiplash. It could kill the person cranking.

    Are you sure you want to do this?
  4. Nov 12, 2008 #3
    That reply is exactly why I was hesitant to post here.
    Last edited: Nov 12, 2008
  5. Nov 12, 2008 #4


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    Mike has obviously never seen the pumpkin chucking contest before. You want to see scary, take a look at the guys that have the extremely fast rotating throwers (centrifugals). Talk about a dynamics nightmare.


    Anyways, as I understand the set up, as the weight goes up higher, the beam pivots and the end also increases in height as well. Is this correct? If so, can you pass along the following:

    - Minimum height of the beam
    - Max height of the beam
    - How far below does the weight rest below the beam, i.e. does it dangle below the end?
    - The actual structural beam description. I am used to seeing beams called out like a 12x40 or the like. If you don't have that, just give the cross section dimensions.

    It would be nice to see a diagram showing the way the beam is held and loaded. For those of us with no experience with your machine, it really helps to understand what is going on.

    The beam is only supporting a weight so I don't seee bolts exploding as being an issue. I do think that dynamic loads "may" play a part in this, but if you put enough factor of safety into the static analysis, the dynamic should fall out.
    Last edited: Nov 12, 2008
  6. Nov 12, 2008 #5
    My caution comes from the fact that I once had the pulley for a towing cable give way in a system I designed for a commercial fishing boat. No exaggeration, it nearly took the first mate's head off.
    But if Fred is comfortable working with you, I know you are in good hands.
  7. Nov 12, 2008 #6
    The beam is 18' long and starts out parallel to the ground and it is hinged at the end closest to the weight, the hinge is 1" steel rod captured in pillow block bearings. Essentially the cable goes from the crank (or in my case electric hoist) that is mounted to the base, to the base of the beam, up to the top of the beam where there is a pulley mounted and then back down the beam and back to the weight that is at the bottom of an 8 foot arm. As I crank in the cable, the weight first starts pivoting on its bearing 8' higher and the center of the arm, and as I keep cranking, the beam slowly lifts and the weight continues to "climb" the beam. THe picture I attached is the weight about 1/4 of the way up. I continue to crank in the cable and the beam continues to pivot up, and the weight moves up until the beam is at 90 degrees to the ground and my weight is at the top it travel, 16' higher then when it started.
    The beam I think I want is...http://www.metalreference.com/24 AL Str I Beam, Am Stnd.htm

    the 4" one with a .326 web thickness, but I can change it if it needs to be stronger.

    here is a quicktime horrible animation of how it works.
  8. Nov 12, 2008 #7


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    Pumpkin chuckin' is some cool stuff, I would love to go into one of those competitions some day. I started building a trebuchet with one of my friends for a fun little project back in College, but ran out of steam.

    Like Fred was saying, if I could get a better idea of how exactly the pumpkin is getting thrown and the like, I'm fine with helping you out. Your picture doesn't seem to look like a "standard" trebuchet as I have seen in the past...

    Maybe a crude hand-drawn diagram to help us understand what's what in the first picture you posted would help. Basically to show where the weight is, where the pumpkin is, what beams are what, stuff like that.
  9. Nov 12, 2008 #8
    I (hopefully) don't need any help with getting the pumpkins out, I need help with an aluminum beam's strength in order to cock the machine. The old way of cocking the machine was to pull down on the throwing arm and the weighty end went up. The problem is, you have to build a very strong (read heavy) arm to do this. With the method I am using, and some others, you basically use a primitive crane to lift the weight. I used a 4x4 originally, but I want to switch to an aluminum beam for more strength and safety. I need to know how much weight I can basically put at the tip of a beam where the force is going straight back to the other end, the pic I attached first shows the situation. The beam will be 18' long, and I have attached the specs of the beam in a previous post.

    My site...http://gallery.me.com/tlum#gallery
  10. Nov 12, 2008 #9


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    Last edited: Nov 12, 2008
  11. Nov 12, 2008 #10
    Exactly! Thank you very much. No offense but could I at least get some help for those "simple" calculations. I don't mind doing the searching but I have no idea what to look for.
    Thank you again,

    I have E as 10,000,000? depending on the alloy...
    Is L in inches?
    Last edited: Nov 12, 2008
  12. Nov 12, 2008 #11


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    It's no problem. E is the modulus of elasticity for the beam's material (for AL 6061-T6, 1*10^7 psi), I is the section modulus for the beam (you can look it up here http://www.efunda.com/math/areas/IndexArea.cfm, but I think you'll want a square beam which is here: http://www.efunda.com/math/areas/rectangle.cfm), and L is of course the length of the beam.

    To keep units consistent, you'd want the units of length and thickness in inches.
  13. Nov 12, 2008 #12
    I hate to be a bother but if it is standing on end, what section modulus do I want? BTW, I did want to try for an I beam first if possible.
  14. Nov 12, 2008 #13


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    Hi teamtom. It looks from your animation that you may have the pully (the one that ends up at the top of the beam) a bit off center. Is that right? Is the center of the pully on the center of the beam? If not, if it's off by some distance, there's a moment produced on the beam that needs to be accounted for, so it's not a simple buckling problem. Having the pully off center makes it worse.

    I'll shut up and go away now. I'm sure if you can explain it better (preferably with a sketch including dimensions) Mech Engineer can help you out!
  15. Nov 12, 2008 #14
    The pulley is bolted to the face of the beam, so it is not in the center, so it does pull rather than compress.
    Thanks for pointing that out.
  16. Nov 12, 2008 #15


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    I leave for a little bit and look what happens. That's great.

    Q is right on in that you not only have a column in buckling, but you have a column with eccentric loading so you need an added term in the stress equation that MechEng gave you to account for the added stress. The same site that Mech already pointed you to has the easiest way to go about doing this with one calculation. The maximum stress in the beam will be:


    It is located on this page:

    Take a read through and see if you understand what the variables are.
  17. Nov 13, 2008 #16


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    I wasn't able to see his diagrams on the links he gave. You're correct, if the force is off-center the link for columns with eccentric axial loads is the way to go.
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