Physics/engineering question

I want to start covering the 16' between my containers. I found a bunch of sheet steel decking at bargain basement prices, but want to make some A frame trusses both to hold up the sheets, and to form a hanger for a hoist that will lift no more than 500#. Mostly for things too big for me to load and unload from my pickup. For bigger things, I am building a dock.

I am thinking of using 3 x 3 x .120" square tubing for the main two trusses, then make some smaller ones out of 2" square, as they will only be holding up sheeting. The center of the peak will only be about two feet vertical. Between the two 3 x 3 trusses, I want to make a 8' I beam dolly so I can lift something up, and load or unload it without having to back the truck up.

I would weld it directly to the top of the containers, either at some strong points there, or lay a long beam or I beam to get a strong base.

Do you think this would be safe for center point lifting of no more than

500#? Or should I beef up the center load section? I also have a little height, so could vee it down additionally to peaking it, and add that horizontal compression stiffener. I can also add several diagonal stiffeners in there just for GP.

I tend to overengineer things, but I like to make it about 3x swl.

Thanks

Steve

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Reply to
Steve B
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I use pressure treated wood for load-bearing beams, sized from this table:

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Some day the wood will be reused in a new tool shed. I buy PT because that area floods, and to get southern yellow pine (SYP) which is a very strong wood.

I do have textbooks on steel and wood structural design, which explain shear, concentrated loads, end fastening, etc. There's more to designing a beam than the load rating, they can fail in many non- obvious ways, like the Minnesota highway bridge.

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I'm an amateur at this I test them for deflection and proof load with a load cell.

jsw

Reply to
Jim Wilkins

I use pressure treated wood for load-bearing beams, sized from this table:

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Some day the wood will be reused in a new tool shed. I buy PT because that area floods, and to get southern yellow pine (SYP) which is a very strong wood.

I do have textbooks on steel and wood structural design, which explain shear, concentrated loads, end fastening, etc. There's more to designing a beam than the load rating, they can fail in many non- obvious ways, like the Minnesota highway bridge.

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I'm an amateur at this I test them for deflection and proof load with a load cell.

===========================================

Is a load cell where you put the ends of yer beam on two 2x4s laid on the floor, jump up and down on the middle of the beam, and see if the middle touches the floor?

If so, then I have a load cell!! :)

Reply to
Existential Angst

Yeah, if the 2x4s let the beam deflect more than it should in service. The chart I referenced uses 1/180 of the length as the worst-case deflection limit.

I test 2x4s in the store when no one is looking by laying one end on the lowest shelf, the weaker way, and stepping on the middle. The ones that make noise go back on the rack.

For wider planks that might be able to support 1500 Lbs that isn't really a good enough test but it may find hidden cracks.

The orange channel iron in my photo was bent from previous pallet rack service. I was able to straighten it with a jack in the middle of two pieces which were chained together at the ends. They also have to be restrained from twisting by crosspieces. That was an easy way to load a beam to over a ton. Be careful, it's also a crossbow.

If you know the unloaded axle weights of your vehicles you can roughly calibrate jack force to the reading of a spring scale on the handle.

jsw

Reply to
Jim Wilkins

Since you're going to be making _trusses_, much/most of the strength will be in the geometry of them. It would be silly to ignore that aspect & choose beams based on their "naked" strength.

My $.02, Bob

Reply to
Bob Engelhardt

I would look around for a heavy, deep I beam cheap and bridge the containers with it. 16 feet is a long unsupported span. The more overlap you can get on the containers the better.

Reply to
ATP*

? Overlap onto the relatively flexible top skin? Why? The load stresses would be taken by the vertical sides - not the top.

My shop roof was built using 10" purlin to free span 30 feet.

A bit more expensive in terms of metal, but more than offset by lower labor cost.

Reply to
cavelamb

You do have a definite point about the top skin. Plus, it is slightly domed shaped to make the water run off. I think I shall probably be better served by running four poles to ground, and having the majority of the weight supported by them. The top is not very strong, and the sides at that point don't look that hefty, either.

I knew when it came time to cover that 16' x 40' span, it was going to be fun.

Steve

visit my blog at

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A fool shows his annoyance at once, but a prudent man overlooks an insult.

Reply to
Steve B

If you can reduce the span with posts you are within the AWC table I mentioned and can support around 3/4 ton with fairly inexpensive wood beams.

jsw

Reply to
Jim Wilkins

Ideally the beam would overlap all the way to the far sides of each container and get secured at the ends. That would reduce deflection as compared to a smaller overlap.

Reply to
ATP

That's nonsense.

BIG waste of material.

Reply to
cavelamb

Not nonsense, statics. Probably nonsense to you.

Reply to
ATP

When I try to comprehend engineering questions, I like to exaggerate them greatly.

If I heard you correctly, I should make the beam 32' long, that is 8' per container, and 16' inbetween, and anchor at the end points only. I'm no engineer, but common sense tells me that a 16' length of anything will lift more than a 32' length of the same anything. And bend less.

You did say secure it on the ends. If I actually were to do this, wouldn't securing it at the 16' point kill the elastic modulus that lets the beam sag at those points? And the ends welded like that to the outsides would actually hold down the ends that would tend to spring up when lifting causes the beam to go into a U formation if it was supported at the inner container wall points? I mean at 8' in from each end, and at each end.

And wouldn't you end up with a slightly stronger assembly if you actually DID use the 32' beam welded in four places instead of a 16' welded in two? And definitely a 32' beam welded on only the ends.

Or, I could just exaggerate the problem, and put poles way out, and use a

100' beam supported on the ends .................

Steve

visit my blog at

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Reply to
Steve B

...

...

Some of what you write in above paragraph is unclear to me (such as what "the 16' point" is). For clarity, let case A = 32' beam laid across without fastenings; case B = 32' beam laid across, fastened at ends (ie at outer walls of each container); case C = 32' beam laid across, fastened at 8' intervals (ie, at each wall of each container). Suspend a 10 ton weight from the center of each beam.

Suppose that in case A the center drops two inches. Then the ends each rise about four inches. The lower edge of the beam between the containers is in tension and the upper edge is in compression.

In case B, the center drops an inch and the 4'-from-end points each rise an inch. Over the walls (ie, at 8' intervals) the beam height remains fixed. Over the roofs, the lower edge of the beam is in compression and the upper edge is in tension.

In case C, about the same thing happens as in case B. Ie, welding at the inner walls has little effect (assuming flex at such welds)

What ATP wrote precludes having the beam ends out beyond the far sides of each container, but if they were, I think the effect is complicated and depends on beam cross-section; whether the ends are supported; whether there are welds, and at which walls; length beyond walls; and of course on the crush strength of the container walls.

Reply to
James Waldby

Think of the forces if you simplify the problem by cutting the 32' beam in half. Now each half is like a diving board.

Hang 1000 Lbs on one end in the center. The far end pulls up on the outer wall with 1000 Lbs. The leverage ratio at the inner wall is 2:1 and it has to support 2000 Lbs, twice the applied load. The total added load under the container is still 2000-1000=3D1000.

If you don't cut the center the added inner wall load is less, but still more than with a 16' beam. You can't really calculate the 32' beam very well without knowing how much the container walls yield or the whole box tips.

So the 32' cantilevered beam applies a greater downward force on the inner walls and costs "more", how much depends on what you can find surplus. If you look up beam load formulas you'll see that a beam "fixed" at the columns is stiffer than one that's only supported and free to pivot, but you would have to figure it out both ways and compare costs to see which is better for you.

jsw

Reply to
Jim Wilkins

On Mon, 05 Jul 2010 21:30:33 +0000, James Waldby wrote: ...

That should have read, "fastened at ends and 8' from ends" - see below.

Also was wrong, see below.

...

I said 32' beam when defining the cases but was thinking 24', ie, with a wrong span of 8' rather than 16' between the containers, so numbers were all wrong.

There are a couple of beam calculation programs you can download and try different beam sizes. Eg, atlas and beamboy. With the latter, you can calculate the effect of 1 ton at the center of a

32', 3"x3"x.12" tube supported at 8' from the ends [center sags 5.1", ends rise 7.7"] or supported at the ends and 8' from the ends [center sags 2.2", 4'-from-end points rise about 0.3"]. This is with I=1.91 in^4 moment of inertia, 30*10^6 psi modulus of elasticity, and c=1.5"

Of course, you said you'd use an I beam dolly between trusses, so the above isn't like your plan, it's just an example of the sort of thing you can compute with beamboy.

Reply to
James Waldby

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