You've done all this math and stuff, but wouldn't the shooter rocking back under the recoil prior to the sheer, and the recoil pad, both totally throw all these numbers off? So that by the time the bolt became a bearing surface, some (much?) of the reward velocity would have been soaked up by the shooter?

After reading this, I'm almost too scared to shoot anymore...

Crazyquik,
When I calculte the velocity of the slide of a 1911 with respect to the frame based on the momentum of the projectile and powder, I get one number, and when I calculte based of the tuned spring force and the mass of the slide and barrel I get a smaller number. More than half the energy I would have expected is lost.

So yes, the compliance of the shooter gets wound up in the equations.

A rifle is recoiling back in the same direction as the bolt is going if it lets loose,so it won't change anything.Just like shooting bullet out the other end if you were
moving gun ahead.
And on the WEA,REM and thosr with front of bolts encased
inside the barrel a .100 to .250 distance( called safety
breaching) and like WEA
only a few thousands to the abutment that the bolt handle would move for contact, there is still pressure from that so-called safety- breeching, after lug failure and the bolt started moving back.As far as worrying about it, first,
keep barrel clean, 2nd keep ammo out of the sun, 3rd
pay attention to what you and rifle are doing, and fourth
if reloader don't overload and don't used reduced loads of slow or ball powders,5th, find out how if you don't know to
check to see if all lugs are bearing no matter what brand
of rifle.Ed.

hubel458,

The peak force between the bolt and a rifle up against an immovable object is the the peak pressure of the chamber times the chamber cross sectional area [neglecting brass traction, stretching, and bottleneck effect].

The peak force if the rifle can accelerate backwards would be a fraction of that same peak force that is the ratio: the mass of the rest of the rifle divided by the mass of the rifle plus bolt plus brass case.



--

A society that teaches evolution as fact will breed a generation of atheists that will destroy the society. It is Darwinian.

A fraction of the powder mass has to be included with the bullet, too, since the bullet's moving forward in most cases by the time of peak pressure.

We are talking about things going wrong---
If there is an immovable object( plug in the barrel that did not move?) and you
generated 150,000 psi,and the gun was allowed to accelerate freely hanging on wires,when thr bolt blew out, there would be no rearward exceleration, because no bullet or gases come out of the front.If there was say a sticky bullet or
overload, and it came out with the same high pressures, the gun(on wires) would recoil and the bolt blew out, at say 95% of speed of what it would if it was not allowed to recoil(this from what is written above
is what you are saying I think)...I say 95%-- as that is
what percentage the rifle is of the whole rifle and bolt
combination on a big caliber.So if this is true then the bolt is moving 142 mph instead of 150!!!Interesting to see
the damage by an 8-12 ounce bolt at those speeds..Ed.

hubel458,

When Crazyquick asked, "some (much?) of the reward velocity would have been soaked up by the shooter?"



Then you answered, "A rifle is recoiling back in the same direction as the bolt is going if it lets loose,so it won't change anything.Just like shooting bullet out the other end if you were moving gun ahead."



Then I corrected you with, "The peak force between the bolt and a rifle up against an immovable object is the the peak pressure of the chamber times the chamber cross sectional area [neglecting brass traction, stretching, and bottleneck effect].

The peak force if the rifle can accelerate backwards would be a fraction of that same peak force that is the ratio: the mass of the rest of the rifle divided by the mass of the rifle plus bolt plus brass case. "



And you respond with, something about, ..when thr bolt blew out,.."





I guess I should have sensed something when I wrote, "There are some good notes about calculating pressure to bolt handle shear, but I would add Ackley's point that the force to stretch or slip the brass that grips the chamber walls need to be subtracted. That extra element is why so many chambers fail and so few breeches fail.



and you responded with, "Clark-In the figures I posted above there is a factor in the math to account for the part of the thrust that the

brass case holds.Ed.



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

Ed Hubel "



and I looked and looked and there was nothing about the co efficient of friction of brass on steel or anything like that in your previous posts.



What does it all mean?

There are plenty of good hardworking God fearing men who don't understand engineering or what some abstractions mean, even though they think they do [plenty with engineering degrees and are nice guys].



--

A society that teaches evolution as fact will breed a generation of atheists that will destroy the society. It is Darwinian.

Hey there ain't no problem to the back where the

shooter is, if the bolt doesn't come out

in a backward direction in case of a mishap.Maybe the reciever will give and there is a problem elsewhere.

But this whole discussion is about whether bolts will blow out if lugs shear and handle does or doesn't hold.

And what little difference the recoil of the rifle makes

in the speed of that bolt if it shears locking lugs is not a lot.IE, gun that shears lugs if freely recoiling will do 95% as much damage to safety lug, as gun against a shooters shoulder that has the same mishap.In the formula for thrust against the bolt,I use a 2/3 multiplier to account for the amount of

area the sides of the brass thickness takes from the case base cross section and the thrust the brass takes when it grips the side. Sorry I didn't mention the amount often enough but it is in another post above.And

that factor replaces reams of complicate math...IE,

cross section of the base(diameter squared X .8) times

peak chamber pressure times 2/3, is

a simple but fairly accurate formula.Example.Big base WEA

70,000 psi...Base of .582 squared X .8 X 60,000 x 2/3.

Gives 12,520 lbs thrust.If there is an obstruction and the pressure triples the you have more thrust than what the shear strength of the lugs, if all lugs are bearing.

If lugs fail and bolt moves does bolt handle hold.That's the question.And it is not going to be affected that much

by whether rifle is shouldered tight or limp wristed.

And bolt movement in case of lug failure. like bullet

movement is going to be done by the time gun recoil gets underway very far..Ed.

Hate to whip a dead horse but: Look at the locking lugs on a Weatherby , they are short and no
radius is provided where they attach to the bolt body. There must be a stress concentration at
that point. This makes all the simple shear calculations suspect. Weatherby actions were made
by several sources in Europe and Asia. There could well be a bad bolt here and there.
We can chat all we want but somthing let go, and someone was hurt.
Good Luck!

IRV-That's right.And not to knock WEA, because IF all lugs are bearing it will hold stress in about the same
multiples of safety as other
same size actions,IF is the word.

Also some trying to say that recoil of gun negates the rear
movement of a bolt if lugs shear is crazy..Before recoil gets under way the bolt, in case of lug failure, has already moved and either been stopped by the safety lug or sheared it off..IE things on the back end will happen as fast as on the front end when bullet leaves.Ed