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ultra_bright

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I was wondering how much force would be exerted on the locking lugs on a bolt of a .308 chambered rifle.

How many pounds or hundreds of pounds of force would be applied to the bolt during the firing of the cartridge and what would the tolerances be between the bolt's locking lugs and the rest of the action?

I was just curious because I've seen a video of some unlucky guy almost get his arm cut off when an out of battery detonation on a .50 cal rifle blew the bolt right off the gun.
 
OK I'll take a shot. The surface area of a 308 base is about .18 square inches and since it is rated at 60000 psi it would seem the pressure on the lugs would be .18 square inches x 60000 pounds per square inch which comes to 3355 pounds. There are probably engineers and such who will find this reasoning faulty but it makes sense to me at the moment.
 
Depending on the pressure that the cartridge in question was loaded to (a 308 would be between 50,000 to 60,000 psi. Diameter of the case head is .473" Area of the case head is pi r squared. r= .2365 squared 3.14 x 55,000psi =9659 lbs.
 
Oh wow, I never knew the bolt action of a rifle had to stand up to so much stress, it seems incredible that it can withstand so much pressure without it cracking or bending the action.


....hmm but also according to this equation that would apply ~6000 max lbs of force to a .22lr case head which seems strangely high, I doubt a .22lr can produce one tenth of that ammount of force against the action when you fire it.

Or maybe it does produce that much force for a very short impulse, and that's why it doesn't tear the action apart...
 
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Ultra bright, you messed something up. It's no where near that high. Did you square the whole diameter or halve it so you squared the radius?

SAAMI max for .22LR is 24,000 PSI. The pressure is only going to be based on the .22 diameter rather than the larger rim. So 3.14 x .11^2 x 24,000=912 lbs of pressure against the bolt on a .22LR gun. That smaller size makes a HUGE difference, eh? That's why rimfire rifles typically rely on the bolt toggle just wedging down in front of a step in the receiver instead of actually having lugs.
 
Ahh that makes a lot more sense now.

Still when you think about it, that's 900lbs of force coming off of something smaller than your finger. Pretty amazing eh.

Goodnight!
 
hunter5425 said:
OK I'll take a shot. The surface area of a 308 base is about .18 square inches and since it is rated at 60000 psi it would seem the pressure on the lugs would be .18 square inches x 60000 pounds per square inch which comes to 3355 pounds. There are probably engineers and such who will find this reasoning faulty but it makes sense to me at the moment.
Click to expand...
Oops forgot to include pi in the calculation, been out of school too long!! :)
 
BCRider said:
Ultra bright, you messed something up. It's no where near that high. Did you square the whole diameter or halve it so you squared the radius?

SAAMI max for .22LR is 24,000 PSI. The pressure is only going to be based on the .22 diameter rather than the larger rim. So 3.14 x .11^2 x 24,000=912 lbs of pressure against the bolt on a .22LR gun. That smaller size makes a HUGE difference, eh? That's why rimfire rifles typically rely on the bolt toggle just wedging down in front of a step in the receiver instead of actually having lugs.
Click to expand...

Isn't pressure measured at the chamber and not the bore? Also the pressure is pounds per square inch so the larger the surface area the pressure works against the more force in pounds will be exerted against the bolt, if so it looks like Rokoro's math is correct, but then again I messed up the calculation of the area of the case.
 
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For the .22 the case head end of the chamber is the same size as the bore since it's a heeled bullet. That's why I used that number. In the case of the .308 he correctly used the size of the head end of the chamber for his calculation.

That make sense now?
 
Actually that does not answer the OPs question, what you have calculated is partial bolt face thrust, but you now have to calculate the surface area of the locking lugs in their recesses to come up with the pressure against the locking lugs. Each model of rifle is different in it's lug surface area so this calculation is relevant only to that model of rifle. Bolt face thrust also has a secondary component one must include in the calculations which is case taper (chamber taper) and that involves a whole bunch of calculus to come up with the true bolt thrust that the locking lugs must withstand......Ackley did a bunch of this as did Roy Weatherby, which is how they came to their cartridge designs in an effort to minimize bolt thrust and still increase cartridge efficiency and velocities safely. Although one assumes pressure is equal in all direction inside a case upon ignition and through the powder burn time (as we were taught in school) this is not so. I have done experiments with extended flash tubes that has proven that to me. Some were flattened, some were gone, down the barrel I assume, some were split and some were as installed. Obviously pressure builds differently in different areas of the case, depending on where the powder is ignited and the powder type, burn rate and physical granular size.

Sorry to throw a wet blanket on your thread and calculations, but it just ain't that simple.............
 
Lets leave the question on a chaulkboard in a hallway and wait for Matt Damon to walk by and Good Will Hunting this one, lol. Way too much math for me, its alot of force we'll say that, lol.
 
Never ever underestimate the power of gun powder enclosed.
I had this great idea to grind a wee groove in the business end of my
Cooey 64 to make the rounds ease into the chamber easily.
By logic, it should work..........and it did........until I shot CCI's.
My left hand stung and my left ear rang.
Bad plan.
Here's the groove.......

[URL=http://s614.photobucket.com/user/kamlooky/media/Cooey%20Model%2064%2022%20semi/DSC07658_zps245a5a4f.jpg.html][/URL]

And this is what it did to the spent case.
Oh, I'm stubborn too.
I never would of thought such a thing would occur.
The barrel has been replaced and I have a pretty neat tent peg to
remind me that sometimes one shouldn't be pizzing around too much.

[URL=http://s614.photobucket.com/user/kamlooky/media/Cooey%20Model%2064%2022%20semi/DSC07753_zpsc860ae62.jpg.html][/URL]
 
Those are pretty cool! :D But I'm still glad that you didn't end up with anything permanent.

With center fire the case heads are thick enough that we see ramps like that in all our semi auto handguns and many styles of rifle. But rimfire cartridges are what is called "balloon headed". And that means that they are thin like you found out.

So it's important for the chambers or bolt heads on rimfire guns to fully enclose and support the wall and rim all the way 'round. You left an opening large enough to let it pop open.
 
BCRider said:
For the .22 the case head end of the chamber is the same size as the bore since it's a heeled bullet. That's why I used that number. In the case of the .308 he correctly used the size of the head end of the chamber for his calculation.

That make sense now?
Click to expand...

Gotcha . I misinterpreted what you said.
 
c-fbmi said:
Actually that does not answer the OPs question, what you have calculated is partial bolt face thrust, but you now have to calculate the surface area of the locking lugs in their recesses to come up with the pressure against the locking lugs.............
Click to expand...

We don't really need the area of the locking lugs unless the guy really needs the actual PSI over the area of the lugs. And that's not a typical sort of question.

Now that we have a way of finding the recoil pressure on the bolt the question of how much pressure the locking lugs are holding back is a simple one of dividing the total pressure on the bolt by the number of lugs.

The rest of that stuff about powder burn case design and such is also certainly valid and it adds a lot to the mystery of this thread. It does indicate that we can't truly know this stuff for sure but then again we're only working with maximum SAMMI numbers in any event. In addition to your points we have to add in the actual pressure from the type and amount of powder.

But as a number to ponder it's interesting to realize that a two lug bolt on a rifle with some cartridge that produces up to as much as 60K will see 9000'ish pounds of pressure for a few milliseconds and that each lug will have to resist a 4500'ish pound impact
 
Keep in mind that a lot of these calculations are worst case scenario type stuff. The simple calculation of PSI x Area creates a number far greater than the actual applied force in normal conditions. It does not account for pressure absorbed by case grip and case expansion or the fact that the pressure is acting on the inside of the case head and not the area of the base. However, worst case scenario calculations are the best ones to use when calculating safety factors. It is usually best to design something with at least a 2x safety factor based on the worst case scenario.
 
BCRider said:
We don't really need the area of the locking lugs unless the guy really needs the actual PSI over the area of the lugs. And that's not a typical sort of question.
Click to expand...

One thing worth noting here is this calculation is very important for determining the plastic deformation of your receiver and bolt, and as a result how much surface area is required to eliminate permanent deformation as well as deciding your specific material and hardness for optimal performance. We see this type of deformation on old Mausers where the bolt lugs set back into the soft receivers due to using overpowered, modern ammunition. Another point where is matters is to help determine both the shape and quantity of lugs required to avoid causing permanent deformation. A Remington 700 uses two opposing lugs that are roughly .450" x .450". They could get away with using two lugs that are .125" x 2", but the pressure placed on the back of those tiny lugs would cause both the lugs and receiver to peen dramatically. The area in shear would be approximately the same, but the area of contact is obviously not. To make it work you would have to section your lugs to increase surface contact, much the same way Accuracy International or Weatherby makes their many-lugged bolts. Lilja has a wonderful article that explains how to calculate the raw shear strength of your material, however determining plastic deformation requires some very advanced computer software to accurately model. Varmint Al has used Finite Element Analysis (FEA) to illustrate the plastic deformation the bolt and receiver undergo during firing. Both of these individuals can be searched, and Varmint Al's experiments are linked from Lilja's article.
 
I use the formula OD squared x PSI x .7854 = #f. Use this on a daily basis in the oilfield to calculate hydraulics. If like anything else there should be a 5:1 safety factor.
 
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