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Metalworking (rec.crafts.metalworking) Discuss various aspects of working with metal, such as machining, welding, metal joining, screwing, casting, hardening/tempering, blacksmithing/forging, spinning and hammer work, sheet metal work. |
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? calculate yeild w/ compression and tension at 90 degrees?
I can calculate the compression and tension, but don't know how they add.
The 10 mm pistol cartridge has a primer pocket ~.208 and an extractor groove ~.338". The makes the primer pocket walls thickness = [ .338-.208"]/2= .065" thick Neglecting the primer thickness and the time delay of the flash hole, the thin wall hoop stress of the primer pocket wall is then S [tension] = P [chamber pressure] .208/[ .338-.208"] = 1.6 P The inside diameter of the case mouth is .408". Neglecting case to chamber friction, the bolt thrust is then: F [bolt thrust] = P [chamber pressure] pi A [inside case area] squared = P pi [.408"/2] [.408"/2] = .13P The stress on the primer pocket walls is then S [compression] = F [bolt thrust] /A [ longitudinal cross sectional area of walls] = .13P/ [pi [ .408/2] [ .408/2] - pi[ .208/2][ .208/2]] = 1.34 P The 10 mm cartridge is rated for 37.5 kpsi chamber pressure so: primer pocket wall tension = 1.6 P = 1.6 37.5 kpsi = 60 kpsi primer pocket wall compression = 1.3 P = 1.3 37.5 kpsi = 50 kpsi There must be a way for 60 kpsi tension and 50 kpsi compression at 90 degrees from each other to be added and compared to the yield point of hardened brass. TIA -- A society that teaches evolution as fact will breed a generation of atheists that will destroy the society. It is Darwinian. |
#2
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? calculate yeild w/ compression and tension at 90 degrees?
A simple way to look at it would be force vectors. Assuming the forces
are at 90 degrees, applying the pythagorean theorum, squareroot(60 *60 + 50 * 50) = 78 ksi. I confess however, that I was not able to fully follow your explanation without a diagram, so I may have missed the boat. It is possible that you could have different forces dominant in different parts of the cartidge. Since the cartridge is fully enclosed by the chamber, it is not clear to me why the yield strength even matters, the cartridge can't yield because it has nowhere to go. It would seem that these calculations would be covered somewhere in the massive literature available with respect to guns and ammunition, and I would use a sample calculation from the literature over any theortical model that you or I could dream up, which might or might not be valid. Richard Clark Magnuson wrote: I can calculate the compression and tension, but don't know how they add. The 10 mm pistol cartridge has a primer pocket ~.208 and an extractor groove ~.338". The makes the primer pocket walls thickness = [ .338-.208"]/2= .065" thick Neglecting the primer thickness and the time delay of the flash hole, the thin wall hoop stress of the primer pocket wall is then S [tension] = P [chamber pressure] .208/[ .338-.208"] = 1.6 P The inside diameter of the case mouth is .408". Neglecting case to chamber friction, the bolt thrust is then: F [bolt thrust] = P [chamber pressure] pi A [inside case area] squared = P pi [.408"/2] [.408"/2] = .13P The stress on the primer pocket walls is then S [compression] = F [bolt thrust] /A [ longitudinal cross sectional area of walls] = .13P/ [pi [ .408/2] [ .408/2] - pi[ .208/2][ .208/2]] = 1.34 P The 10 mm cartridge is rated for 37.5 kpsi chamber pressure so: primer pocket wall tension = 1.6 P = 1.6 37.5 kpsi = 60 kpsi primer pocket wall compression = 1.3 P = 1.3 37.5 kpsi = 50 kpsi There must be a way for 60 kpsi tension and 50 kpsi compression at 90 degrees from each other to be added and compared to the yield point of hardened brass. TIA -- A society that teaches evolution as fact will breed a generation of atheists that will destroy the society. It is Darwinian. |
#3
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? calculate yeild w/ compression and tension at 90 degrees?
Richard,
Thanks for the response. The primers fall out when the primer pocket grows a little in diameter. It is a real problem, and I chose one of the worst case cartridges. Here is a picture of some primers falling out: http://www.thehighroad.org/attachmen...=&postid=49766 Here is another: http://www.thehighroad.org/attachmen...&postid=333274 Can you see how the cartridge has a narrow ring near the base? The is the extractor groove. The primer pocket is inside of that, so the brass is thin there. Clark |
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? calculate yeild w/ compression and tension at 90 degrees?
It's called Mohr's Circle, but it's been 30+ years since I've even thought about one, much less did any figuring about it.
Clark, You trying to make us oldtimers (Alzheimers) feel bad? Mike Eberlein Clark Magnuson wrote: I can calculate the compression and tension, but don't know how they add. The 10 mm pistol cartridge has a primer pocket ~.208 and an extractor groove ~.338". The makes the primer pocket walls thickness = [ .338-.208"]/2= .065" thick Neglecting the primer thickness and the time delay of the flash hole, the thin wall hoop stress of the primer pocket wall is then S [tension] = P [chamber pressure] .208/[ .338-.208"] = 1.6 P The inside diameter of the case mouth is .408". Neglecting case to chamber friction, the bolt thrust is then: F [bolt thrust] = P [chamber pressure] pi A [inside case area] squared = P pi [.408"/2] [.408"/2] = .13P The stress on the primer pocket walls is then S [compression] = F [bolt thrust] /A [ longitudinal cross sectional area of walls] = .13P/ [pi [ .408/2] [ .408/2] - pi[ .208/2][ .208/2]] = 1.34 P The 10 mm cartridge is rated for 37.5 kpsi chamber pressure so: primer pocket wall tension = 1.6 P = 1.6 37.5 kpsi = 60 kpsi primer pocket wall compression = 1.3 P = 1.3 37.5 kpsi = 50 kpsi There must be a way for 60 kpsi tension and 50 kpsi compression at 90 degrees from each other to be added and compared to the yield point of hardened brass. TIA -- A society that teaches evolution as fact will breed a generation of atheists that will destroy the society. It is Darwinian. |
#5
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? calculate yeild w/ compression and tension at 90 degrees?
On Sat, 13 Sep 2003 19:12:19 GMT, Richard Ferguson
wrote: It is possible that you could have different forces dominant in different parts of the cartidge. Since the cartridge is fully enclosed by the chamber, it is not clear to me why the yield strength even matters, the cartridge can't yield because it has nowhere to go. Yes, there are indeed different forces in play in the cartridge though the pressure cycle, and different places receive varying amounts of pressure during the cycle. The neck/shoulder may be subject to decreasing pressures as the primer pocket may still be subject to the peak pressure. Typically, the weak spot of most cartridges is indeed the primer pocket, as the primer is only pressed in, and brass flow may open the hole the primer is pressed into, allowing gas to escape to the outside of the case, past the primer itself. This is generically called "blowing out the primer pocket". Standard pressure testing of this phenomenon shows that in most rifle cartridge cases, this starts showing up (in modern cartridges) somewhere in the area of 62,000-70,000 PSI. In this range also, is where the primer cup itself, may be perforated by internal gas pressure in the very small area that is not supported..IE the firing pin hole in the face of the bolt/slide/etc. Higher pressures may actually blow out the web of the cartridge, pierce the primer, and damage the weapon AND the shooter by the sudden and overpowering injection of uncontrolled hot gases into the action of the weapon. Most modern rifles are designed to vent this gas in safe directions, away from the shooters face. Much higher pressures, in a weapon perhaps already damaged, may sheer the bolt locking lugs, and turn the bolt into a high speed projectile rear wards. This is generally only of interest to the mortician, whom has to repair the damage to the face and skull, after the medical examiner has removed the bolt from the shooters skull, during the autopsy. A well known gun writer appears to have been the victim of metal fatigue, a year or so ago, when he was found on the firing line, while testing some hand loaded ammunition in a vintage Mauser rifle. Big hole in his face, with the bolt stuck through his skull. The above of course, assumes that the case be of the "mostly supported" type most often associated with rimless bolt actions, and most, but not all rimmed cartridge/action combinations. Revolvers are generally fully supported by the cylinder and the frame. Where the problem may show up most often, is in self loaders..ie semi-automatic pistols, most notably the Glock, where a rather frightening amount of actual cartridge case is left unsupported, above the magazine at the bottom rear of the barrel. Glocks, fired with hot hand loads, often exhibit "guppy bellies" in fired cartridge cases. When such brass is reloaded several times, and work hardening has done its damage, the cartridge case may blow out (downwards actually), and blow the magazine out of the weapon, often shearing off the magazine catch and depositing the magazine rather forcefully on the shooters right foot (assuming a decent Weaver stance and right handedness), while in addition, occasionally causing a quantity of fecal material to be deposited in the shooters trousers. Gunner "At the core of liberalism is the spoiled child - miserable, as all spoiled children are, unsatisfied, demanding, ill-disciplined, despotic and useless. Liberalism is a philosphy of sniveling brats." -- P.J. O'Rourke |
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