1. Originally Posted by jboggs
You said you were using the "Cantilevered Beam with Uniform Load" calculator. I think that is the wrong calculator, for two reasons. Your beam is not a cantilever. It is supported by a point support at each end of the overlap. Also "Uniform Load" means the load is spread out over the whole beam. I think your load is concentrated at the end of the beam. Isn't that correct? I think your loading diagram would show one force downward at the end (the load), one force upward in the middle (at the end of the overlap), and one force downward at the other end of the overlap to balance the load. You will have to use statics to calculate what those forces are. The calculator I would use is "Supported at Both Ends Loaded at any Location". That name sounds wrong but if you look at the diagram and turn it upside down you will see what I mean. If this is unclear let me know and I'll post a diagram for you.
The visual of flipping the diagram upside down makes much more sense, but wouldn't the "uniform load" version work better at determining the forces from the weight of the beam itself, then the "at any location"? Additionally, the "at any location" calculator has input needs that I have no idea what they're asking. To wit: "x" and "v" are distances that I have no idea what they're referring to.

2. Originally Posted by PinkertonD
JB, you are not alone, see my post, #2 in this thread. I elected to drop out of this race when I learned that my knowledge gleaned over 40 years of Engineering did not apply. You may want to conserve your energy too.

Dave
There has apparently been a pretty major misunderstanding here. Nobody ever said your knowledge didn't apply. I explained in post #1 what I want to build and the calculators I'm using to figure it out. In post #2, you told me I was wrong, and said "the "fulcrum" is actually a counter-balance to the load". In post #6, I explain that what you said sounds like a see-saw, and that this crane has a hinge at one end, a weight at the other, and a ram that lifts the weight somewhere in the middle-ish. That's not a see-saw.

This isn't a matter of your knowledge not being applicable, it's simply a matter of technology and magic: Any sufficiently advanced technology is indistinguishable from magic. You understand this as technology, I wonder in awe at it as magic. My descriptions and understanding are bound to be quite different from yours, and are pretty much bound to be wrong. Unfortunately, in post #9, you gave up trying to explain it.

I'll completely understand if my level of ignorance isn't something you want to deal with, but please don't assume that anything I say is intended to disrespect you.

3. Yes, the weight of the beam can be a consideration, but usually the stress from the payload is much higher than the stress from the weight of the beam itself. There are higher level methods that include the weight of the beam in the calculation. As a rough approximation you can calculate the stress from both factors separately and compare them. And yes, the "uniform load" calculator is appropriate for that.

You will only see "x" and "v" used in the equations that calculate stress or deflection at a point between the load and the ends, which is not a concern for you. They are not used in the other equations.

4. Originally Posted by wsdave
There has apparently been a pretty major misunderstanding here.
Nope, no misunderstanding, you said I was wrong and rambled on about a some "lever class" whatever that may mean. In 40 years of Engineering I have never heard the term "lever class," when designing a simple beam.

In the light of that correction by you, and your obvious greater knowledge, there was and is no point in me offering further information.

{edit}
Oh, and quoting Arthur C Clarke, adds nothing to the issue, it just confuses as I was talking about VERY basic beam calculations. No magic technology there.
{/edit}

Dave

5. Originally Posted by PinkertonD
Nope, no misunderstanding, you said I was wrong and rambled on about a some "lever class" whatever that may mean. In 40 years of Engineering I have never heard the term "lever class," when designing a simple beam.

In the light of that correction by you, and your obvious greater knowledge, there was and is no point in me offering further information.

{edit}
Oh, and quoting Arthur C Clarke, adds nothing to the issue, it just confuses as I was talking about VERY basic beam calculations. No magic technology there.
{/edit}

Dave
LOL!

And the fact that I was talking about "lever class", something a 40+ year engineer had never heard of, wasn't a clue of a misunderstanding?

It may well be "VERY basic beam calculations" to you, but it is quite obviously something I don't get.

So if this isn't something you want to help with (and it pretty obviously isn't), then thanks for stopping by, but I'll stick with the folks who actually want to teach me something, rather than just say I'm wrong.

6. Originally Posted by jboggs
Yes, the weight of the beam can be a consideration, but usually the stress from the payload is much higher than the stress from the weight of the beam itself. There are higher level methods that include the weight of the beam in the calculation. As a rough approximation you can calculate the stress from both factors separately and compare them. And yes, the "uniform load" calculator is appropriate for that.

You will only see "x" and "v" used in the equations that calculate stress or deflection at a point between the load and the ends, which is not a concern for you. They are not used in the other equations.
I figure to start with baby steps, so I'll go ahead do the calculations separately (I don't think I'm quite ready for anything involving "higher level" yet).

With an MoI of 1.7, Load of 32.6#, Length of 72", Distance of 12" (end of the overlap), and an axis of 1.5" (my tube is 3" x 1.5" x .188"), I get a stress at point of about -144.

I'm not sure what to do with NEGATIVE stress, nor how I got there.

So I've worked with the "stress at any location" calculator, and I'm getting numbers I can't believe.

Using the same piece of metal above, but extending it to 240" with the ram at 12" (instead of a telescope, I'm treating thing as one piece made of the weakest section) and hanging 300# on the end (not including the beam weight), I get a stress at the ram of 3,018#. The metal has a yield point of 46,000#, while the ram has a capacity of 32,000#.

That would give me a 10:1 safety margin on the ram and 15:1 on the beam, lifting 300# at 20', from a 3" x 1.5" x .188" rectangular tube.

That also means 1,000# with a 3:1 margin.

That seems like FAR too much weight for that tube.

What am I doing wrong?

7. Originally Posted by wsdave
And the fact that I was talking about "lever class", something a 40+ year engineer had never heard of, wasn't a clue of a misunderstanding?
Yes it might have been, but I quote you... 'The engine "crane" isn't that.'

When I am informed that my approach is unequivocally wrong, (there is no ambiguity in "isn't that") I see no point continuing. Had you written, "I don't understand, can you please explain," I would have been happy to lead you along the path to enlightenment. To my amazement, instead, you followed up the "isn't that," by explaining why the "crane" isn't a simple beam with a point load at the ram connection.

Since you appear somewhat enamored of spurious quotations, please allow me to counter with "where ignorance is bliss, 'Tis folly to be wise." A poem entitled, "Ode on a Distant Prospect of Eton College," by Thomas Gray (1716 - 1771)

Perhaps if you began working with the advice being offered here rather than trying to tell us how or what it is, or should be, you may get more useful information to help you build this thing.

I will offer information and assistance only up to the point where it is no longer being accepted, or, my proffered information is considered to be incorrect. As to more help, I have already stated my advice in posting #2 and nothing has changed. To confirm that approach, JB has also pointed out, this is a simple beam calculation with a point load.

Fini.

Dave

8. Originally Posted by PinkertonD
Yes it might have been, but I quote you... 'The engine "crane" isn't that.'

When I am informed that my approach is unequivocally wrong, (there is no ambiguity in "isn't that") I see no point continuing. Had you written, "I don't understand, can you please explain," I would have been happy to lead you along the path to enlightenment. To my amazement, instead, you followed up the "isn't that," by explaining why the "crane" isn't a simple beam with a point load at the ram connection.

Since you appear somewhat enamored of spurious quotations, please allow me to counter with "where ignorance is bliss, 'Tis folly to be wise." A poem entitled, "Ode on a Distant Prospect of Eton College," by Thomas Gray (1716 - 1771)

Perhaps if you began working with the advice being offered here rather than trying to tell us how or what it is, or should be, you may get more useful information to help you build this thing.

I will offer information and assistance only up to the point where it is no longer being accepted, or, my proffered information is considered to be incorrect. As to more help, I have already stated my advice in posting #2 and nothing has changed. To confirm that approach, JB has also pointed out, this is a simple beam calculation with a point load.

Fini.

Dave
But JB, unlike you, actually went on to offer some explanation when it was clear I didn't understand.

Again, thanks for stopping by.

9. Originally Posted by wsdave
I figure to start with baby steps, so I'll go ahead do the calculations separately (I don't think I'm quite ready for anything involving "higher level" yet).

With an MoI of 1.7, Load of 32.6#, Length of 72", Distance of 12" (end of the overlap), and an axis of 1.5" (my tube is 3" x 1.5" x .188"), I get a stress at point of about -144.

I'm not sure what to do with NEGATIVE stress, nor how I got there.

So I've worked with the "stress at any location" calculator, and I'm getting numbers I can't believe.

Using the same piece of metal above, but extending it to 240" with the ram at 12" (instead of a telescope, I'm treating thing as one piece made of the weakest section) and hanging 300# on the end (not including the beam weight), I get a stress at the ram of 3,018#. The metal has a yield point of 46,000#, while the ram has a capacity of 32,000#.

That would give me a 10:1 safety margin on the ram and 15:1 on the beam, lifting 300# at 20', from a 3" x 1.5" x .188" rectangular tube.

That also means 1,000# with a 3:1 margin.

That seems like FAR too much weight for that tube.

What am I doing wrong?
First, depending on the protocols set up in the formulas, positive stress means compression and negative stress means tension, or vice-versa. They are talking about the actual material at the top and the bottom of the beam. One will be in tension, and the other in compression. It usually doesn't matter because they are equal and balance each other out. So, don't let negative stress throw you. The stress number you got, 144, is that in pounds per sq. inch?

It sounds like you are trying to lift 300# at a distance of 20 feet using 3"x1.5" rectangular tube. Is that right?

10. Originally Posted by jboggs
First, depending on the protocols set up in the formulas, positive stress means compression and negative stress means tension, or vice-versa. They are talking about the actual material at the top and the bottom of the beam. One will be in tension, and the other in compression. It usually doesn't matter because they are equal and balance each other out. So, don't let negative stress throw you. The stress number you got, 144, is that in pounds per sq. inch?

It sounds like you are trying to lift 300# at a distance of 20 feet using 3"x1.5" rectangular tube. Is that right?
I'm not sure if the 144 is PSI: It's whatever units the calculator produces.

The 20' of 3" x 1.5" x .188" was simply a proof, in that the 144 number seemed far too small, so I extended it out the full length.

I wanted to avoid exposing the whole project, because I really do want to do the math myself, insomuch as possible. It's seems, however, that the design is complicated enough that full exposure is warranted.

Here's the idea:

Combine this body/arm/ram configuration...

http://www.northerntool.com/shop/too...duct_7199_7199

...with this leg/foot set...

http://www.gantry-crane.org/category/engine-hoist/

Please note that neither use a top strap nor rear supports as this does...

http://www.harborfreight.com/2-ton-f...ane-35915.html

Since I'll be running a winch cable along the top, I need to avoid the top strap, and since I want a swivel, the rear supports would be in the way.

Unlike these, mine will have a 3 section (not 2), 240" telescopic "arm" (96", 96" with a 12" overlap, and 72" with a 12" overlap). The idea is that if I have a body and arm that can lift stuff anyway (a basic engine crane), why not put it to work in other tasks?

The 240" arm, combined with a 72" tall hinge point (the body), will ideally be able to lift a 300+# working load (not including it's own weight) up 240" in the air, thus allowing for things like getting roofing and construction material up to the 3rd floor, holding up a truss for connection, or even suspending a person to string Christmas lights. Collapse the tubing down and the height decreases but the capacity goes up (you know that, of course). Still, 120" lifts that giant fancy BBQ up to the 2nd floor patio. Collapse further down to 96" and add a pivot forklift attachment to the end, and it can lift a pallet of stuff onto a trailer.

Finally, take it apart (connected as much as possible with trailer-hitch style pins), carry the individual pieces up to the second floor patio, and reassemble the unit, but with the legs sticking backward. Add 1,000# pounds of weight (Barbell style or rubber 100# sand bags) to the leg pins, hang the arm over the side, and use the 8,000# capacity winch attachment to haul up the hot tub. When the tub is high enough, just roll the whole thing back on the over-sized wheels.

About the time that's done, Lassie will come barking, and explain that little Timmy has fallen down the well on the Johnson's property.

Take the whole thing apart again (really not very complicated, because of the pin system) load it on the trailer, and drive out in the field. Rebuild the crane, but instead of using the legs, the swivel portion bolts directly to the reinforced portion of the trailer bed (the trailer has connection points for "outriggers" for stability). Extend the arm to 240", swing it out over the old stone wall (don't want to damage the old stone wall), and lower the "tree-saver" strap connected to the end of the winch cable down into the well, pulling little Timmy to safety. The same winch and strap I use to pull my SUV out of the mud.

Unfortunately, my brother will take off with the SUV just as I'm setting up the crane, so in addition to needing to use an air tank for the air-over-pump rams, I'm forced to use a jump-start box to power the winch. Luckily, both are small, portable, and multi-use items.

Just as I finish pulling little Timmy up, my brother returns, leaves the SUV, but unbolts the crane and connects it to it's legs, then tips up the trailer (dumping everything off of it). He then harnesses himself up to it like a mule, and runs off with it, saying something about hauling Butte.

Well.

The crane pieces are too big to fit in the SUV, and I don't want to get it dirty before mud bogging again anyway. Luckily, through the magic that IS tube-in-tube construction (similar to how trailer hitches work), I can connect my attachable tow-wheels to the crane and tow it behind the SUV.

The problem is that the crane weighs about 350#, Timmy's been treading water at the bottom of a well since yesterday (I don't speak dog, so it took a bit to figure out that Lassie WASN'T rabid, and was actually trying to tell me something), and I'm not the strongest guy in the world. In fact, I'm not even the 416th strongest guy in the world.

So in thinking about how to lift the crane up to slide the wheels on, I have to ask myself "Do I feel lucky? Well, do I, punk?".

No, I do not. But I DO feel smart!

I'll pull the two pins that hold one of the rams in place, flip it upside down, and reinstall it on the underside of it's body mount. Then I'll gas it up, and it will tripod the body of the crane up, using the legs as the other two points. I'll install the road wheels, collapse the crane down to something like this..

...wrap a compression strap around it, and tow it back home.

I become the hero of the town and go on to marry Timmy's older sister, Bailey, who was voted Miss Kettle Corn in high school.

Because I'm using air-over-pump rams with a capacity of 32,000# (8-tons each ram), and a 12 volt winch (with a wireless remote control, no less), I can accomplish every task I've mentioned alone, and with only a compressed air bottle and automotive jump-start box.

Just kinda seems like I shouldn't have all that potential just sitting in a corner waiting for an engine to pull...

Thanks,

Dave

11. Sounds like a pretty handy dandy unit but does it have Satellite radio & GPS?

12. Originally Posted by RWOLFEJR
Sounds like a pretty handy dandy unit but does it have Satellite radio & GPS?
No, but this is only the first build: Version 2.0 will actually integrate a 7" touch screen at the rear that will not only have Satellite Radio and GPS, but also online vehicle repair manuals, Bluetooth, and a meter that shows how much weight is being lifted, and at what angle, in real time. It will run off the same 12 volt source the winch does.

;-)

13. "suspending a person to string Christmas lights"
There's the magic phrase. I'm outta here. Sorry. Can't help.

14. Originally Posted by jboggs
"suspending a person to string Christmas lights"
There's the magic phrase. I'm outta here. Sorry. Can't help.
Well, rescuing little Timmy from the well didn't seem to phase you, so I can only assume this has something to do with Christmas, rather than simply suspending a person.

If that's not the case, then I'm curious as to why you would have helped if it was simply a matter of suspending a hot tub 20 feet in the air (where a failure could kill someone), but balk at suspending a person who weighs a fraction as much?

Seems to me that if suspending a person is a problem, this would be a great moment to teach me WHY, since I have no idea. I see all kinds folks walking on buildings with nothing more than a fall harness, and those beams don't fail, so why would this beam be different?

15. Honestly I do like your style. Here's the deal - engineers design devices for lifting purposes all day every day, and they all have certain safety factors and features. The nature and complexity of those features and factors increase by several orders of magnitude when there is a good possibility that human safety should be anticipated. This is for legal, ethical, and moral reasons. Anyone designing anything that could reasonably be anticipated to have a significant effect on human safety should not be getting their design advice and guidance from an on-line forum. One reason, of too many to list, is that the forum-based design approach is by its nature disorganized and random. The likelihood of overlooking some critical factor is too great. Put simply - too many things could go wrong. The fact that I missed the comment about Timmy in the well is a wonderful example of my point.

Another reason - you can't learn everything you need to know from a forum, even this one as good as it is. Example - I never saw a response to my recommendation that you first analyze your designs using statics to determine the forces. Did you make static diagrams? Do you know the moments involved? Do you know how they create stresses in the beam materials?

As far as your project is concerned, if I were you I would build some small scale prototypes, test them, and proceed from there. My best wishes for you.

16. "Did you make static diagrams? Do you know the moments involved? Do you know how they create stresses in the beam materials?"

I did not, but the physicist did. He said that he has a number of computer programs that (if I understand correctly) do the math based on the inputs, not unlike the calculators here. We are meeting next Tuesday to go over what he's learned and so he can teach me the why's and wherefores. He says he's worked out the stress on an inch-by-inch basis over the length of the beam, including the MoI of each piece, and the tube-in-tube overlap math. His big concern at this point (which he hasn't done the math for) is the stresses associated with the change in weight orientation as the arm lifts up. Something about forces pushing end-to-end in addition to up and down, or something.

The downside of physical prototypes is that the math involved in a stress-accurate small scale version would be (I'd guess) just as complicated as that of a full-sized unit, so the only real difference would be material cost savings, which isn't enough to make it worth the time to scale-prototype.

In a nut shell, since the design takes considerably longer than the build (I'd guess that's true of most engineering projects), it seems to me that I might as well design for full-size and test an actual unit, rather than design for a scale model, then design again. While I can see the design aspects being largely just a math issue (once the system works, triple the size of everything), the harder part may well be finding material that acts the same as full-size, but at (for example) 1/3 the strength.

And thank you for your well wishes.

17. Picture a courtroom with the injured plaintiff's family watching as their lawyer asks you if you did the calculations: "I did not, but the physicist did." Or "the guys on the engineering forum said it would work." You need to understand the math and the physics and the analysis behind anything you might be asked to defend in court. Or you need to show how your design was based on a series of successful trials. I know because I've been there. Nobody expects it, but sh_t happens.

18. "You need to understand the math and the physics and the analysis behind anything you might be asked to defend in court."

Which is why I've been specifically requesting help understanding the math, NOT with the design. Engineers make the big bucks to not only design stuff, but to be RESPONSIBLE for the stuff they sign off. That's why I'd never ask for anyone to do that.

I'll take full responsibility for the design (and use) of the crane: All I'm asking for is engineering-professor-type help with the concepts and calculators specific to what I'm trying accomplish.

Even if the specific idea of lifting a person on the crane is a deal breaker, the tube-in-tube math shouldn't be: After all, how else will I be able to figure out how many bicycles my hitch-mount carrier can take, or whether I can use that same trailer hitch to mount an in-the-field workbench with a vice without bending the tube?

Tube-in-tube is the best stuff going: I just need to understand it.

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