# engineering calculation needed

I got some 2" x 2" x .250" steel square receiver stock today. Sheesh. \$4
per foot.
Anyway, I am making a figure 4 davit. The long leg will be 15 long. The
short horizontal will be 3 foot.
It will mount at the top and the bottom. I am considering putting in a
standoff about half way up to work against the flexing.
The winch weighs about 25#, and I intend to lift no more than 200# with this
800# capacity winch.
The horizontal will be one foot down from the top with a diagonal flatbar
going from the top out to the end of the horizontal to help transfer the
load to the top of the upright.
How much flex can I expect? Do I need the standoff in the middle? I will
probably put one on anyhow just to be sure.
Thanks in advance.
Steve
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You don't mention what the loading will be. 2" square heavy tubing is pretty light weight for any kind of beam. Better to pay a structural engineer for your design than to depend on this E-world for advice, which may range from qualified professional engineer to the Village Idiot. Bugs
Hi Steve,
I couldn't find any pictures of a "figure 4" davit online, but I think I understand what you're talking about.
I had a bit of time this morning so I did a quick calculation. It is about the most rudimentary calculation possible, but if I've understood your structure correctly I believe it gives a useful answer. It ignores all but one mode of deflection, which I believe will be the most significant, and makes many assumptions. Naturally these assumptions are a matter of discretion and people are welcome to discuss them. But there wouldn't be any point in doing complex calculations without a lot more knowledge of the project.
Here's what I did. If the diagrams don't match what you intended, let me know. I assumed that you're going to have a pivot at the top and bottom to allow the davit to rotate. You may not, but I doubt it will make much difference because, either way, the anchor points are unlikely to be highly rigid in a torsional sense.
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The calculation suggests that the winch suspension point will be deflected downwards about 8 inches by a 200 lb load. This is only an order of magnitude figure, but it is way too much. It means that your davit would be very bouncy, your top and bottom anchor points could be damaged, and it might cause failure in an unexpected way.
You might want to rethink the need for the long, vertical column. Presumably the davit will be supported by a wall or gantry of some kind? Can you place the bearing carrying the vertical load just below the horizontal member? If not, a stand-off in the middle would reduce the deflection to about one-eighth the value I calculated (i.e., 1 inch), but I still wouldn't be entirely happy with this. A stand-off at the level of the horizontal member would be better.
Comments welcome!
Best wishes,
Chris
You are correct about the top and bottom mounting. I believe I will add a middle standoff to keep flex down. I already have the materials for this, and will build and test it this spring. It will either work or fail on the first test load of 200#.
We'll see, and I'll keep everyone posted. It is merely a device to haul groceries and supplies up without having to traverse thin steep steps. It doesn't have to hold a lot, and I am thinking it will suffice.
Steve
Chris,
I fear you are mixing models in a way that does not work. You compute support reactions based on the ends being fixed horizontally, and then model the deflection based on a cantelever beam. An FBD of the beam itself will have (using your top right setup) loads from the bar and cable (which you might want to reduce to an applied moment).
The compression of the column will move its neutral axis, but I imagine that the most useful starting point will be to get an angular deflection at the end of the beam and multiply that by 3 ft to get an estimate of the load deflection.
Bill
The cantilever beam only represents part of the vertical member: the 14' from the horizontal to the ground. I have made an imaginary cut 14' from the ground. This point can move horizontally, but has a moment and shear force applied to it. With a bit of spatial twisting and turning in my mind, I believe this 14' section can be modelled as a cantilever. The fact that the root moves horizontally while the tip is fixed doesn't matter.
I agree. But the cantilever model takes account of the moment and shear force, does it not? I used the cantilever model so that I didn't have to calculate the value of the moment and shear force.
I chose to ignore the compressive load as I think it will produce a small deflection compared to that resulting from the mode I've modelled. This is only intended to be an order of magnitude model and I think it serves this purpose, showing that the structure isn't stiff enough.
Best wishes,
Chris
You'd be better off putting the stand-off just 1 foot down from the top (i.e., at the same level as the horizontal member). This will mean that the vertical column only has to support a compressive load. The compressive load will be well within the capabilities of the column. It is the bending of the long vertical column that is the major weakness in your design. Fix this problem and the rest of the design will most likely be fine.
Let us know how it goes.
Best wishes,
Chris
But you've got the cantilever backwards. Your calculations would predict the deflection at the base of the column as a result of F3 if point Z were fixed (moment connection) and if the base were free, which is not the case at all.
Bill is right about solving this by calculating the angular deflection at Z.
Ned Simmons
Chris,
That might end up being true, but I do not think you can hand-wave it that way. With its max deflection somewhere in the middle of the beam, it is more likely that it will function as two half(more or less)-length cantelevers, which puts quite a hurt on the cubic term in the deflection.
My gut tells me that the deflection will be quadratic in length, vs. cubic as for the cantelever. My gut has been known to be wrong.
I think it will be a factor only in the stress calculations anyway, but thought I'd mention it. If your deflection is correct, the thing will tear itself apart anyway =:0
Regards,
Bill
I don't believe this matters. The formula gives the deflection of the cantilever tip relative to the root. It doesn't matter which moves. You can build a system of guide rails such that a pin joint is at the fixed end of the cantilever, and the root is allowed to move transversely, but not rotate. In this case the formula is still valid, and this is the way in which I'm using it here.
If anyone has a counter argument or proof I'd be very interested to hear it. Or if anyone does the calculation by different means I'd like to know your result, too.
Chris
I take your point here. The root of the cantilever may rotate a bit so that the maximum deflection is somewhere in the middle of the beam. But I believe that the cantilever model will give a useful result for a 10-minute, one-side-of-paper calculation. Whether it is more like a single cantilever or two half-length cantilevers will depend on Steve's joints and anchor points. When we know so little, there will inevitably be some hand waving. I still think it's a fair model based on the information we have. It is only intended to give an order of magnitude result.
Chris
Steve,
Thinking about your project a little more, I would choose to alter the design rather than trying to add a stand-off. There are simpler and stronger designs possible, such as this for example:
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And here are a couple of older cranes which might give you ideas:
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Note that both of these have a compression member which is at an angle to the vertical. This is a nice, simple design. You just need to avoid applying large moments to the beam as in your original proposal. In pure compression 2" x 2" x 0.25" tubing should be fine.
The most critical part of the crane is likely to be the point at which the top is attached to the wall. If you choose to use expanding bolts, look up their load rating and use a good margin of safety (a factor of 10, say). Use the right size drill and make sure the wall isn't crumbly. Should it fail, a joint like this will not fail in a nice way.
All the best,
Chris
Yes, the horizontal leg will be mounted to a beam, and be about a foot long. Therefore, it will have a top and bottom vertical attachment pin/rod, and two standoffs, one a foot from the top, and one about half way down. The top foot from the horizontal to the piece is there to attach the diagonal 2" x 1/4" flat bar leg.
I think it's going to be strong enough, but the addition of the middle standoff leg will insure flex will be directed to a solid standoff.
We'll see.
Now I have to wait for the snow to melt and for it to warm up to pour the SonoTube base.
Steve
!\ ! \ ! \ -!---- ! ! ! ! ! -! ! ! ! ! !
Here is an attempt to draw this thing. At top, it will pin into a pad eye at the end of a receiver plate/saddle coming horizontal from the beam. At bottom, it will rest entirely on washers on a plate mounted to concrete.
From the top, at one foot down, the horizontal goes out. The diagonal in this drawing is not to scale, as the top angle would be much greater. The hoist's outer attatchment will go about 2/6" from the vertical.
The diagonal would be the square root of 10 in length.
Each short horizontal standoff would be about a foot long with a pad eye.
I anticipate that stability will be from a solid mounting at the bottom, and a beam saddle with extended baseplate to make a pad eye for the top attatchment. The middle standoff will just stiffen it up a bit, and help keep it straight when I swing it in towards the porch.
HTH. These things are hard to describe in words so as to communicate exactly.
Steve
I had thought of that, but the vertical beam joins to the sonotube at that point, and just sits on top. The pushing sideways and the leverage would cause the approximately 7' vertical corner support to push off the top of the Sonotube, or come loose where it is only nailed together. This is going to be the point of attatchment for the lower standoff, but I will make the bracket 18" long, and at a right angle so I can wrap the whole corner, and attatch to post, facia, etc, and increase the distribution of whatever side load from the flex is. I don't think it will be that much. Certainly not as much as mounting the bottom support of a stiff leg to it.
Seen lots of stifflegs loading boats. They are really strong, but all were mounted into humongous concrete bases.
Steve
Yes, you would need to ensure that the bottom bearing can handle the horizontal force if you go for the "stiff leg" type of crane. But if you use your original design without a stand-off part way down, your bottom bearing will carry a horizontal force. It would be a good idea for it to be capable of carrying a horizontal force anyway. If you can resolve these small problems, and you don't need to lift bulky loads to the full height of the crane, I think the stiff leg is the way to go because it will be strong and reliable. If you stick with your original design I would put one stand-off at the very top and one a foot below it, as opposed to the way they're shown in your diagram. This will mean that the vertical column isn't being bent at all, which is a good thing. I think this is more important than having a stand-off in the middle. Also, make the stand-offs adjustable so you don't have to force the thing together in the first place!
Best wishes,
Chris
The original problem, which you've represented accurately in your first sketch, is a simply supported beam with a moment applied at one end. You've replaced the beam, which was previously supported at both ends, with a cantilever. The moment has become a concentrated force applied at the opposite end. Extraordinary claims require extraordinary proofs.
Case 3e, Chapter 5, Roark's 6th edition - concentrated moment on a simply supported beam. The angular deflection at the end of the tube adjacent to the triangle is .0188 radian = 1.08 degrees; .68" deflection at the winch's attachment point 36" from the vertical tube.
Ned Simmons
"Christopher Tidy" wrote
One of the design criteria is that when the load is at the top, right before it hits the winch kickout switch, the load needs to be swung into the porch over the 36" high handrail. A stiffleg wouldn't make it in to where the load could be landed on the deck before it hit the side of the deck. This needs to be lifted to its highest possible point, and then swung about 150 degrees over the rail and inside the railing, where the load would be hand unloaded, or lowered to a waiting dolly.
Your suggestion would work perfectly IF I didn't have to have that extra 60 degrees of swing.
Steve
"Ned Simmons" wrote
This is a HF winch with two attachment points. The one closest to the end would be about 2'6" from the vertical, and the other, about 8" closer to the vertical than that. I really don't want to put the winch all the way out, but only as far as needed to clear the rail when I swing the load into the deck.
I am going to have to make two new hangers, as HF gives you two for 2" round stock. I CAN slide the winch out so that the drum actually goes PAST the tip of the horizontal, but don't want to do that until after testing. At the point of sliding it out, the hanger would hit the diagonal coming down.
Just gonna have to build this puppy, hook it up, get to a safe place, and watch what happens.. Just like in real life. Can't really do a test until failure, as some of the components are holding up my roof and deck..............
Steve
Hi Ned,
When I read your post, at first I disagreed with you, then I agreed, and now I partly agree. After some thought I am of the opinion that neither of us have got it quite right.
My cantilever would be fine if the horizontal member was attached half way up the vertical member, but the model becomes less accurate the further it is moved from this position. Your model is fine in the absence of the diagonal tension member (Steve's flat bar). With the tension member present it isn't a moment which is applied to the vertical member, but a point load instead.
I did some sketches and free body diagrams:
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I now think that the correct model is a simply supported beam under the action of a point load F, where F is applied distance A from one end and B from the other. In this case F = 3W, where W is the load on the suspension point, A = 14' and B = 1'. What do you think?
I don't have a formula for this case. Perhaps if your book gives a formula for this case, giving the deflection of the point at which the load is applied, you could run a quick calculation and get an answer for Steve? My guess is that it will come out with an answer a good bit less than my original suggestion of 8", but I'd be interested to know.
I suspect that whatever model we use, we aren't going to get much closer than an order of magnitude figure unless we know exactly how Steve will construct it. Nevertheless I believe it would be wise to avoid putting the long vertical member under a bending load.
Best wishes,
Chris

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