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bspooh

11-03-2004, 12:22 PM

How do you figure out the tensile strength of eye bolts for an example..

assuming the threads will not fail, what are the calculations for designing an eye bolt that has to withstand 16,000 lbs of force??

I don't know how to figure out what thickness of the circular section..

This is a personal inquiry and not something that will be used, just theoretically..

In my calculations, I am using 6061-t6 alumiunm which according to the machinerys handbook has a tensial strentgh of 45,000 lbs..So how would I apply that number to figure out the thickness of the eye bolt?

I am desperate on what formulas or anything that could help me out on this..I know that steel eye bolts is a better choice, but this is just so I can figure out other kinds of engineering problems..

thanks,

brent

I dont know the real answer to this question-

But I do know the seat of the pants way I would deal with this kind of a situation in the real world- and that is, find somebody who has already done the math- I would get out either machinery's handbook, or the MSC catalog, and look up steel eyebolts, note the size and strength rating, estimate what kind of steel if it doesnt tell, then derate appropriately for aluminum.

So- a 1" Armstrong steel eyebolt is rated in the MSC catalog at 45,000lbs breaking strength, 9000lbs safe work load. Says the alloy is 1030, forged. If I look up 1030, Machinery's says yield strength of 40,000lbs to 90,000lbs per square inch. I look up aluminum, and it says something like 11,000lbs to 50,000lbs per square inch.

Since the MSC/Armstrong rating of the eyebolt assumes the low end for steel, I would assume the low end for aluminum as well-

11,000lbs yield. Multiply by the same safety factor, which is .44, and you get a working load of 2400lbs for 1" aluminum.

But this doesnt get into how it is made- when you forge 1030 into an eye, it actually is as strong as the original round bar, if not in a wierd sorta way, stronger, because it is forged, which is the way they make these things, the grain is kept going the right way.

But forging aluminum is a lot trickier, with the temp ranges needing to be very exact. Usually impossible in a home shop to forge aluminum without making it weaker. And cold bending 1" round aluminum into eyes, which I have done, introduces stresses that make it even weaker.

I know, just a theoretical example- but my advise is dont make any lifting eyes out of aluminum. http://bbs.homeshopmachinist.net//smile.gif

mrainey

11-03-2004, 12:42 PM

There's a bunch of mechanical engineers at www.eng-tips.com (http://www.eng-tips.com) who love questions like this.

Well...it's been about 35 years since I took strength of materials, but I'd probably approach it like this, which may be incorrect so don't take it as gospel:

Since you want to support 16,000 lb safely, I'd use a 5X safety factor, which I think is a fairly standard thing to do and the MSC catalog cited by Ries seems to have done (9,000 x 5 = 45,000), so you need to design for a breaking strength of 5 x 16,000 or 80,000 psi. If your tensile strength of 45,000 psi for 6061 is correct, it means you need a minimum cross-sectional area of 80,000/45,000, or 1.78". So find a thread with a root diameter of at least 1.78", and I'm guessing that would be about a 2" diameter by something-or-other-pitch thread.

But as Ries notes, there's also the problem of forming the eye, which gets complicated fast.

johnoder

11-03-2004, 02:17 PM

I usually use the root dia of the threads since it is the smallest diameter unless the shank is reduced beyond the threads.

Area of a circle is Dia. X Dia. X .7854.

So if say you have a 1/2-13 thread figure conservatively that the root dia. is .400"

.400 X .400 X .7854 = .12566 square inches. Theoretically a material with a Yield Strength of 45,000 psi would start to permanently stretch this .400 dia with a load of 5655 lbs.

So if you apply a safety factor of 5 to 1 for this you could say the safe load was 1131 lbs.

John

Jpfalt

11-03-2004, 02:21 PM

I wouldn't use the aluminum.

Aluminum has no endurance limit and will eventually fail in fatigue. Steel has an endurance limit. At loads below the endurance limit, fatigue is not a concern.

If corrosion resistance is an issue, use a stainless eyebolt rated for the load.

As far a s calculations go, the fitup and alignment of the eyebolt ID and the pin going through will make major differences in how the joint will perform long term. A small pin in a large hole will have to take into account the bending stresses turning the hole from round into an oval. A tight fitting pin will need to concentrate on hoop stress in the eyebolt ring. If the fitup is cylindrical hole on pin, then contact stress at the ends of the hole need to be accounted for with safety factors up to 4 times. If the contact is round on round, say a U bolt through a round crossection eyebolt, then local contact stresses may need a safety factor of around 10.

Then you get into shear strength of the pin and bending in a cantilevered pin, iff that's the case.

Personally I favor a self aligning cylinder rod end mated to a clevis and clevis pin for general freedom from ulcers.

bspooh

11-03-2004, 02:26 PM

thanks, I appreciate the responses..

But..I'm not the sharpest tool in the shed so let me give you the exact example that i am struggling with...

its actually an 2" x2" square aluminum plate by 1/2" thick with a 9/16" hole through the center..now imagine a "tug-o-war", where you would have (2) wire ropes going in the hole and you have equal forces trying to rip the plate apart..so what I mean is if you would pull on the rope in the -x direction and also pulling the other rope in the +x direction..I would think you need to know the tensile strentgh of the aluminum which I think its 45,000 if I read it right..I need this plate to withstand 16000 lbs of force..will it hold up, or will I need to make the plate thicker? or does there need to be more "meat" from the hole edge to the outer edge of the plate.??

The 45000 lbs of tensile strenght is per inch squared right?? What would my area be on the plate between the hole edge and the edge of the plate..

I'm just trying to see how thickness plays a role in the strength..

sorry for the confusion, but I am confused big time http://bbs.homeshopmachinist.net//confused.gif

brent

Well, you also have the problem, I think, of probably needing to bevel the edges of the hole so the wire rope doesn't go over a sharp edge, and depending on the diameter of the wire rope it may end up bending to a tighter radius than is deemed acceptable, anyway. I would guess you ought to put the wire rope over a couple of thimbles.

With a 9/16" hole, you have I think a maximum wire rope diameter (assuming two pieces have to fit through the hole) of 9/32", or just about 1/4". Will a 1/4" diameter wire rope safely hold 16,000 pounds?

Thisckness has no bearing on the yield strength, only cross-section area is important.

Area of a square with a round hole in it:

Area of square - area of hole

(Length X width) - (Pi X radius squared)

4-.246 = 3.754 sq in.

Someone check this, it's been a while http://bbs.homeshopmachinist.net//smile.gif

Having re-read your post, seems to me that the limiting factor will likely be the method that the cable is attached to the block. If you use threaded eyebolts, I think the threads would be more likely to pull out of the aluminum block before the block would pull apart.

So perhaps the number you are loking for is the shear (?) strength of a given thread in aluminum. For this application, I would think steel would be the proper choice.

johnoder

11-03-2004, 09:05 PM

Assuming 1/4" cables, 1/2" aluminum plate would be adequate in "shear".

Allowable "shear" on 6061T6 is 30,000 PSI (from MHB, 21st edition, page 2250)

Each cable would be trying to tear out a strip from 9/16 hole to edge. The combined area for one side is .719 square inches. The 8000 lb load applied to that side would have to be 21,563 lbs to even start to do any tearing.

If you look at this the other way, just tension, trying "to rip the plate apart" you have exactly the same area as above, but now you have 16,000 lbs and the "yield" strength of 40,000 psi is in play (from MHB). No yielding would start until this 16,000 lb load became 28,750 lbs. This then is the "critical" loading mode. As you can see, you are also at less than 2 to 1 safety factor for yield, so the plate would be definitely on the "unsafe" side.

You can be sure you would get some deformation around the hole as the cable made itself a "seat", so the radiusing or chamfering noted in above posts is a good idea.

This in no way addresses whether or not the cables will hold this load.

John

darryl

11-04-2004, 02:41 AM

I think John has the math right- you have .718 sq inches of material to resist the 16,000 lbs of tension. What this doesn't take into account is the fact that the material will not be stressed equally across that area. The highest tension will be at the inside of the hole, so it will want to rip from the two opposite sides of the hole, before tearing the material from the outside of the square. Derate all the calculations by about a half just for this reason alone. Assuming the cables will hold up to the tension, the piece of aluminum you want to use isn't strong enough.

Swarf&Sparks

11-04-2004, 12:06 PM

Check out J E Gordon's books, "The New Science of Strong MAterials" and "Structures".

They are still in print and readily available. Valuable insights into the sort of problem you mention and an extremely entertaing read, to boot.

Rgds, Lin

bspooh,

if you're only 'pulling' the plate apart, and not twisting it in any way, you'll need the tensile strenght info and not shear.

the cross-section you're interested in will be the section across the major diameter of the hole.. perpendicular to the pulling. so imagine a sawcut straight through your plate and or eye bolt.. you'd cut it right through the center of the hole.

interesting enough, if you turn your plate 45 degrees, it will withstand more pulling force.. tie the ropes at two opposing corners and you've effectively increased your cross section.

tackle these problems one step at a time: don't worry about rope/cable failure.. once you have a plate that will resist X load.. then try to work out "how do you hook a cable up so it resists X lbs" problem.

a factor of safety of 2 is usually adequate.

-tony

Thrud

11-04-2004, 07:01 PM

Brent

Take two chocolate milk and relax... http://bbs.homeshopmachinist.net//biggrin.gif!