Forces on a firearm breech?

War Song

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Just a curious question regarding the stresses that a firearm needs to deal with; particularly the breech of a break-action shotgun. How much metal do you exactly need behind your breech in order for it to be safe?

From my math, a 12 gauge shell would be generating at least 2500 pounds of rearward force at peak pressure on the breech face. Or is this situation so brief that what's built only needs to withstand much less? It just seems to me that every time someone fires a gun, there's an entire car sitting on top of that breech.
 
Weight is not the critical factor, strength is. You could build a gun with 100% marshmallows held together with elastic bands, if you have enough of them.

The same can be said about aircraft. With enough thrust, even a brick can fly. There was a saying during the Cold War that Soviet interceptor jets were powerful and fast, but they needed a new refinery for each one.
 
On a break action shotgun, one of the critical features is the corner between the water table and the standing breach. A radiused corner is potentially stronger than a sharp right angle. Also the amount of metal hollowed out for springs or locks (in a hammer shotgun). I think you would find that most cracks occur at that corner

cheers mooncoon
 
Maybe this can interest you; http://www.varmintal.net/abolt.htm

While it's easy to calculate the bolt thrust, it's action is more complicated than it appears.
The real problem is, as note above, that the action strenght is a combination of components and the Thrust is only one of the component of the force in action.
For a break action, the thrust must be applied to the leverage of the action (FxD) and translates as torque. For a bolt action, it translates as shearing force.
 
How the force is taken up by the metal gets complicated pretty quickly as you can see by the other replies. But the duration of the force does not lower the effect. There's still the equivalent of a good size car worth of force being put into the metal.
 
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