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Clevis/Shackle Mount Tech Question (strength formulas added)

nutt7

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So I googled and I can find tons of info about safely using shackles. I can't, however, find any info about a proper size mount to hold up to the forces of the pin pulling on the mount.

I am planning a bumper and my recovery point will be made of 1x3" bar that bolts to the frame. My plan was to use a 7/8 or 1" shackle (1" and 1-1/8" diam pins respectively). I planned to have the hole equidistant from the ends of the bar for maximum strength (1" and 15/16" of material from hole to edge, respectively). I also plan to add cheek plates or washers to fill the width of the shackle opening to help distribute load evenly. I suspect with a decent degree of confidence that my mounts will be sufficient for at least 1" shackles...However, does anyone know how mush force it would take to rip through the eye of the mount I proposed? I'm not sure if this is a tensile strength problem or what and what those numbers are for mild steel.

Thanks
 
Found this with google. Maybe itll help you. ?
View attachment 233059

Thanks! Tearout force is the term I needed. I just learned some neat stuff. A36 steel (all forms) has a tensile yield strength of 36,300 psi. People round it to 36,000 psi. The "standard" says to use 0.4 of 36000 to use as shear strength. This gives 14,400 psi. I read that 0.4 accounts for some safety factors and imperfections in loading (angles, etc).
The tear-out equation is:

F = Shear strengh x 2et

where F = Force to cause failure
e = distance from hole center to edge of plate (your hole should be equidistant from all edges to ensure proper calculations at all angles)
t = thickness of plate.

For me, this gives:

F = 14400x2x1.5x1 (keep all units in inches and pounds)
F = 43560 pounds to tear through the shackle mount hole. Thats over 21 tons!

Sounds cool, but I also read that the safety standard practice is to have the edge of the plate to the center of the bolt hole be at minimum 2 times the bolt's diameter. For me this isn't possible. Mine was 0.83 for a 1" shackle. That means I need to calculate a similar stress, called Plate Fracture. The only difference is this accounts for the hole diameter.
This equation is:

F = S(W-d)t

Where F = Force to cause failure
S = allowable plate stress (shear strength in this case)
W = width of plate
d = hole diameter
t = thickness of plate.

For me this gives:

F = 14400(3-1.125)1
F = 27000 pounds, or 13.5 tons. This is still more than enough to hold the 1" shackle. The 7/8" shackle rates at 28,800 pounds. A 3/4" shackle hole would have a plate fracture force of 30,600 pounds, or 15 tons.
Remember, my calculations are using mild steel 3x1" bar, so only use these numbers if you use the same materials.
 
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That's a whole lot more thought into it than I used.
I just figured the blazer frame my bumper is mounted to will fail way before my shackle mount would so...I just drank a beer and said "yep". Lol
 
That's a whole lot more thought into it than I used.
I just figured the blazer frame my bumper is mounted to will fail way before my shackle mount would so...I just drank a beer and said "yep". Lol
:haha:
Right? Next I need to see what the strength of the frame is where it will be bolted to the bar. As you presumed, it'll probably be less than the shackle!
 
The good thing is the way most bumpers are built, a percentage of the load will be shared with the other frame rail opposite the side your pulling on.
And my bumper I built is bollted on with six 1/2" grade 8 bolts so that's like 51,000 lbs per side before shearing. You definitly don't have to be an engineer to know our frames wont hold that. Ever.
 
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