Check my math, it's been a while.

EdFred

EdFred

Taxi to Parking
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Used to do this stuff all the time, but it's been 20+ years since I've done the math. Hot rolled steel has an ultimate tensile strength of about 47kpsi and a yield strength of about 26kpsi. I have a piece of 1/8" x 2" bar (well, one leg of L2x2x1/8) that's going to have a 1" hole drilled in it, and then rounded off to a 1" radius. (see attached picture).

How much weight can be hung on this before it fails?

I come up with something in the neighborhood of 1600 lbs.

Correct or no?
 

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EdFred said:
How much weight can be hung on this before it fails?
Click to expand...
In keeping with POAs tradition of nit picking every post.....

One correct answer is 1 pound.

Because you didn’t ask for the maximum weight.
 
You have to show your work, it's been a long, long time. I see two 1/8 inch X 1/2 inch bars next to the hole that will bear the most stress. Many questions come with this, such as is the load evenly distributed and is there any dynamic loading. Will the load be toward the radiused end or the solid end.
 
I would do a quick calc as 47 x .125 x 1 x .577 = 3.4

Would not expect failure below 3400 lbs.

That is not the safe working load, however. Use 5:1 against ultimate for swl of 680 lb. Static loading only
 
PaulS said:
You have to show your work, it's been a long, long time. I see two 1/8 inch X 1/2 inch bars next to the hole that will bear the most stress. Many questions come with this, such as is the load evenly distributed and is there any dynamic loading. Will the load be toward the radiused end or the solid end.
Click to expand...

There will be a 1" A325 bolt going through the hole. (it's what I had laying around) The bolt will not be the weak point. The force will be in the direction of the radius end.

Old97 said:
I would do a quick calc as 47 x .125 x 1 x .577 = 3.4

Would not expect failure below 3400 lbs.

That is not the safe working load, however. Use 5:1 against ultimate for swl of 680 lb. Static loading only
Click to expand...

That's what I did for my calculation except there's only 1/2" of material between the radius and the hole, so I used 26kpsi * .125 * .5 to come up with 1625.

It's actually being used as a towbar connector for pulling and pushing the plane out and into the hangar.
 
It's a little bit more complicated than calculating the cross sectional area at the edges of the bolt hole because of stress concentration. Think about how it's going to fail. It will likely rip apart at the top of the hole. The math on that gets a little bit cray cray.
 
Pulling on the "hole" would give you a cross section of .125 (the thickness) times 1 inch (2 inch wide minus 1 inch hole) times 47k. The force is acting on both sides so they can both be used in the calculation. So the square inches is .125 and the pounds per square inch are 47k so the math says 5875 pounds. Now factor in a safety factor of 3 and you're down to about 1950 pounds of safe working load. Leaving it unprotected reduces the strength through corrosion so there is that to factor as well.
 
I think you'll be fine for the application. The only thing I would consider is going thicker than 1/8. Even 3/16 will be stiffer, 1/4 even better. Just worried it will bend on you at higher loads with side load on it. Unless it is short, then maybe you will be ok.....
 
or weld a crush plate on the side of it...

PaulS said:
I think you'll be fine for the application. The only thing I would consider is going thicker than 1/8. Even 3/16 will be stiffer, 1/4 even better. Just worried it will bend on you at higher loads with side load on it. Unless it is short, then maybe you will be ok.....
Click to expand...
 
EdFred said:
Used to do this stuff all the time, but it's been 20+ years since I've done the math. Hot rolled steel has an ultimate tensile strength of about 47kpsi and a yield strength of about 26kpsi. I have a piece of 1/8" x 2" bar (well, one leg of L2x2x1/8) that's going to have a 1" hole drilled in it, and then rounded off to a 1" radius. (see attached picture).

How much weight can be hung on this before it fails?

I come up with something in the neighborhood of 1600 lbs.

Correct or no?
Click to expand...
I suggest instead of a 1” bolt a 1/2” bolt will balance the load out better. Strength even in single shear on a 1/2” bolt is 8,700 pounds. Then you will have 3/4” from the edge of the hole to the outside radius thereby increasing your load capability 50% in the 1/8” piece. Easier to drill a 1/2” hole too. Use a couple of bushings in the mating piece to reduce the diameter to 1/2” if necessary.
 
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Using allowable stress design:
Max tension = .6( Fy). If your 47ksi is correct then .125”(1”)(47,000)(.6)=3525#. It doesn’t look like the bearing area stress governs
 
EdFred said:
It's actually being used as a towbar connector for pulling and pushing the plane out and into the hangar.
Click to expand...

Looks like you're fine, as long as the plane is on its wheels in the hangar, instead of your bar becoming a hanger with the full weight of the airplane on it. ;)
 
lol... I’d go for what “looks about right”. You could build and test a couple of these over a six pack of beer. You’re going to end up with better results than these math attempts.
 
Well it wasn't a towbar, but a connector from my electric shop cart to my T-handle towbar. (that way I'm not building an entire towbar)

Ended up making a cradle with two 3/16" x 3" x 6" plates in parallel with a 1-1/4" x 4" slot centered in each of them and a 1-1/2" opening in the top for the handle to drop into. I'll post pics once I get it welded and painted.
 
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