Hardware tech (bolt strength)

Rich

Rich

Asshole at large
Joined
Mar 17, 2005
Location
Central PA
I've got a 1/4" grade 8 bolt that will be loaded in double shear.

By specification, how much force will it take to deform it?

Can't go bigger.
 
That's tensile strength... which probably isn't what he's asking for. It sounds more like he wants to know how much load it will take to bend the fastener. In order to calculate that, you need to know what the unsupported distance is.
 
Actually, the bolt is .307 thick... close enough to a 5/16 to be labeled as one I suppose.

Anyway, here is the application. I can use the spacer pictured, but it'll probably get lost during use. The forces will be coming from the top and the bottom... as you can see, very little unsupported distance..
bolt1.jpg
bolt2.jpg

Unlike the pressed-in pin that came with it, I need to be able to remove the fastener with little to no tools. even a pin like this would be good, if it was rated strongly enough.
pin1.jpg

P.S. If I do use the bolt, I'll try and make sure the shank goes all the way through, this pic is just for illustration.
 
I can calculate the shear stress for you and give you an idea of where you are as far as loading. Are the loads on that link going to be only tensile, or will torque also be applied? If so, how much?
 
It looks like you just need to know shear strength. Which should be pretty good, since it's basically in shear 4x.

... unless there's something happening behind that spacer that I can't see.
 
Up to, let's say.. 9,000lbs of force. no torque.. just pulling forces trying to make the 2 pieces go their separate ways ;)

The area behind the spacer is open.. I don't see that the spacer would really do much? it was included in the original app to retain the pin through friction.
 
Ok. Given roughly 130,000 psi elastic modulus for grade 8 bolts (found online, someone feel free to verify, but it seems reasonable), the maximum load with a 0.307" diameter bolt in double-double shear is 19,246 pounds. The elastic modulus gives the stress where the bolt would still return to its original shape with no deformation. I tend to view it as a best case scenario, but at 9000lbs, you should be well within the limit.

Also, that spacer is useless for shear forces.
 
That looks like your winch hook, it's easy to find out.
1. put winch on your dodge.
2. load dodge up with 4000 lbs of cement
3. make sure cement is secure
4. winch dodge up into the top a large tree, or you could use a bridge
5. if it breaks, it ain't strong enough, if it holds, it might be strong enough.
 
or just drive dodge into back yard and attempt to 4wheel over the sewer drain while pulling a trailer, then hook up 'test apparatus' and commence testing :flipoff2:
 
I was wondering who'd figure out what it was... :D
 
Seriously? It's a winch hook...

"Ok. Given roughly 130,000 psi elastic modulus for grade 8 bolts (found online, someone feel free to verify, but it seems reasonable), the maximum load with a 0.307" diameter bolt in double-double shear is 19,246 pounds. The elastic modulus gives the stress where the bolt would still return to its original shape with no deformation. I tend to view it as a best case scenario, but at 9000lbs, you should be well within the limit.

Also, that spacer is useless for shear forces."

I don't know how you would figure it for "double double" shear--I assume you just double the single shear strength?... but FWIW yield strength for a g8 bolt is 130,000psi and tensile is 150kpsi. Shear strenght is 60% of tensile...so something to take into account...

So, (.307/2)^2x3.14= .07402sqin x 91000psi x 2= 13471lbs in double double shear. Of course there is always a pretty substantial safety margin added in, so I wouldn't worry too much using this on a standard 8k winch.


BUT, rich...as you well know..g8 bolts are brittle and will break and kill you while a g5 bolt will just bend. You should probably go with the g5.


:flipoff2: (joking)

:edit: Is this so that you can use it to suck down the axle as well or...?
 
braxton357 said:
Seriously? It's a winch hook...
"Ok. Given roughly 130,000 psi elastic modulus for grade 8 bolts (found online, someone feel free to verify, but it seems reasonable), the maximum load with a 0.307" diameter bolt in double-double shear is 19,246 pounds. The elastic modulus gives the stress where the bolt would still return to its original shape with no deformation. I tend to view it as a best case scenario, but at 9000lbs, you should be well within the limit.
Also, that spacer is useless for shear forces."
I don't know how you would figure it for "double double" shear--I assume you just double the single shear strength?... but FWIW yield strength for a g8 bolt is 130,000psi and tensile is 150kpsi. Shear strenght is 60% of tensile...so something to take into account...
So, (.307/2)^2x3.14= .07402sqin x 91000psi x 2= 13471lbs in double double shear. Of course there is always a pretty substantial safety margin added in, so I wouldn't worry too much using this on a standard 8k winch.
BUT, rich...as you well know..g8 bolts are brittle and will break and kill you while a g5 bolt will just bend. You should probably go with the g5.
:flipoff2: (joking)
:edit: Is this so that you can use it to suck down the axle as well or...?
Click to expand...
WOW somebody got alot of extre time:flipoff2:
 
braxton357 said:
BUT, rich...as you well know..g8 bolts are brittle and will break and kill you while a g5 bolt will just bend. You should probably go with the g5.
:flipoff2: (joking)
:edit: Is this so that you can use it to suck down the axle as well or...?
Click to expand...

Shit, you're right.. What was I thinking.. :lol:

You're pretty much on the money - The idea is to use a small clevis on the axle and just attach the rope to it, but when I need / want to use the hook, disconnect the rope from the axle, and connect the hook to it. I can use a 3/4" screw-pin clevis (the pin fits through the thimble on the end of the rope), but I like the hook alot, and if I could use a bolt and a wing nut to quickly attach it to the rope, I'd like to.

(Didn't want that big hook flopping around on the steering ram - might scratch the paint! :p )

I can't see the pin that Warn provided with the hook being any stronger than a grade8 bolt anyway, but ya never know..
 
F911 Bolts (grade 9 plus)

--------------------------------------------------------------------------------

If you use these I have been getting them from www.hillcofasteners.com out of California (714-657-7442). They always have every size in stock plus oddball grade 8 lengths and sizes.
Scott
 
rockcity said:
why use a wing nut?
just drill a hole through the end of the bolt and use a pin instead. It won't rattle loose either
Click to expand...

*smacking head* yeah, DUH.. probably shoulda thought of that, seeing as how I already did that for the windshield hinge pins... :shaking:

Just gotta retain the pin so it doesn't get lost when the hook is being thrown to the tree...
 
L9 Or F9 bolts have a 180,000 to 200,000 psi tensile strength.
 
Yeah... ARP, cat, and plenty others that are far stronger than g8 bolts--some are in the 250kpsi range, but they cost far more as well. I'd personally probably drill out the shackle holes to 3/8 or 1/2 before going that route.
 
Not that it hasn't already been said, but that bolt will hold up just fine. No fancy numbers here, as I'm just a history teacher. :)

Mac
 
Rich said:
I've got a 1/4" grade 8 bolt that will be loaded in double shear.
By specification, how much force will it take to deform it?
Can't go bigger.
Click to expand...

According to a chart I downloaded off of ShopFloor.com, it should be 4,470 lbs x 2. It's a .jpg file, if someone gives me an e-mail who can post it, let me know.

BTW: Shear = 60% of tensile strength.

Thanks,

Al
 
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