pin/hole strength calculation in telescoping legs?

I'm building a small rolling gantry with telescoping legs. Each leg should be able to hold 1 ton of weight. The legs are made of 3/16" wall structural steel
square tubing, which is A500 steel, which withstands a minimum of 39,000 psi at yield point according to my steel handbook.
I figure the pin will only push on half of the hole. The surface area of half of a hole = PI*DIA*3/32.
What I'm not sure of is how to actually design the pin & hole. Can I just multiply my half hole area times 39,000 to get a raw number, which I then derate until I'm comfortable? Or am I looking at this wrong?
To use an example, if I use a 5/8" pin in a 5/8" hole, the pin presses down on half of the hole's area. That surface area I get as .184 sq. inches, and multiplying that times 39k gives me about 7179 pounds, or about 3 tons.
If I use a 3/4" pin/hole, I get 8614 pounds - more surface area.
Going up to a 1" pin, I calculate 11486 pounds.
Am I doing this right? How big a derating factor would *you* be comfortable with?
Grant
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With a few simplifing assumptions, the relevant area is the pin diameter times the wall thickness. The material on the sides of the hole may be in contact with the pin but are not contributing significant load capacity. Next issue is the grade of the tube, 3/16" is available in both A513 and A500 and the A500 might be A,B, or C. Net is that your 39,000 psi is probably a good value but could vary considerably (mostly upward) You should also me able to use both walls which doubles the the load capacity.
Factor of safety you choose is up to you and depends on the failure mode and the personell safety issue. Something that snaps catstrophically and then drops on a person would probably call for 10x safety factor. In your case, there is some personell risk but the failure mode is a bulging and deforming of the tube wall.
If you look at the same pin design on the import hydraulic presses, you will find there is very little (calculated) safety factor. If you really strong arm the pump, you can usally see some deformation starting on the pin holes.
Grant Erwin wrote:

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I think you are looking at it wrong. Or at least not looking at all of the problem. Say you increase the thickness of the tubing to say 2 inch thick. Then it is obvious that the failure mode would be shearing the pin. So you need to consider both failure of the tubing and failure of the pin.
Dan
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Grant Erwin wrote:

Your math is quite right. But wait ...
Next calculation is shearing the pin. Does it break?
If not, back to the pressure: Does the pin go through all the way, or just one one side. If just one one side, you have to go down with the area by half (or less), because the pin tilts. Now to get a more sound number for the pressure in the hole: Use maybe a forth or so, because with the tensile strength, the material almost starts to flow.
If your pin isn't going through all the tube, change that! It might snap.
Nick
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If it were me, I'd use a 5/8 bolt, grade 5 or even 8 so the bolt don't fail. A good 5/8 bolt holds A LOT. If the holes in the tubing mash out, no big deal to drill to 3/4 later and no catastrophic failure here.
This advice is worth exactly what you paid for it.
Karl
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Grant Erwin wrote:
OK, I have some more definitive data. I found out that to calculate the yield strength of a hole, you just use the diameter times the thickness, which is a smaller area than half the surface area of the hole, thus more conservative. To follow up on my example, a 5/8" hole in 3/16" wall tube, with the pin going through both sides, gives 9140 pounds yield strength assuming 39000 psi A500, which is the lowest number I could find.
Checking the shear strength, if I use 39000 as the yield strength (it's very easy to find pins much stronger than this) then a 5/8" pin should yield at 11965, or have shear strength of about 7179 pounds. It sounds like it would be easy to hit my 4X overdesign factor with 5/8" pins, and as Karl says if they start to elongate, I can just drill them out and use 3/4" pins.
Thanks, all.
Grant Erwin

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On Tue, 22 May 2007 13:33:53 -0700, Grant Erwin

For what it's worth, I used two 5/8" dia grade 8 bolts as table pins for my shopmade 12-ton hydraulic press. Sides are channel 3/16" thick. This press doesn't get a lot of heavy use, but it has certainly seen full force (pump bypass) quite a few times. No deformation is visible either on bolts or in holes. It did shear 5/8" ordinary bolts, but the grade 8's are hacking it without a problem.
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Don Foreman wrote:

You must have had real grade 8 bolts. A company that I do maintenance for went through two sets of 12 metric 'grade 8 bolts' that were used on the bottom of a cnc rotary table that held a O ring and end plate in place. The original problem was that some of the bolts had loosened up and snapped because of the uneven load on the plate and bolts. We put new bolts in and they immediately snapped. We verified our hydraulic pressure pushing on the plate and it was correct. We then put in another set of bolts and they snapped. The bolts were they snapped did not look like an alloy bolt that snapped. The surface was very rough and uneven, not like you see when a real grade 8 bolt lets go. The company sent them out to get tested. I forget the exact numbers but they were about 4 x weaker than they should have been... These bolts were M12 size and were bought from a reputable national hardware supply company.
John
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wrote:

I built my press about 10 years ago. "Reputable" bolts may have changed some since then.
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