Since I am an engineer, I thought I might be able to put together a post that would explain how all this works. But after typing for a half hour last night, I could see my post would raise more questions than answers, if it didn’t put everyone to sleep first. So I stopped.
Rereading your post today, I think I may see where the misunderstanding is, and will try to help there. If I miss, please excuse me. We can try again.
It appears that you are thinking of the cartridge as a rocket motor, where the thrust is determined by the opening (in this case, the bore size). Actually, a hydraulic ram, as Bret described, is a good analogy. In a rocket engine the thrust is developed by an imbalance of internal forces caused by reduced pressure at the outlet. In a firearm, the thrust is generated as the bolt prevents the case from expanding like a bladder. The forward case walls are (typically) locked to the chamber walls by friction while the rear case walls stretch backward toward the bolt head. The force on the bolt will be the pressure times the cross sectional area of the case at the widest point minus the strength (tension) of the case. If the pressure is low, and the case is strong, the bolt feels no thrust, and may not even be contacted. If the pressure is high, and the case reaches yield, the bolt thrust will be roughly equal to the force that would be generated if the case were a hydraulic piston.
And then there is the matter of hoop stress in the barrel……..
I typed this as fast as I could during my lunch break. I hope it makes some sense.