Barrel Torque Ludicrocity
- By bohem
- Bolt Action Rifles
- 213 Replies
The preload on the joint must exceed the applied load during the load cycle in order to maintain joint stability. The way to do this is to use the calculated factorHarold Vaughn's Rifle Accuracy Facts, yes.
I was not considering anything about the number of cycles but rather the case of a fired round exerting a load on the threaded joint that exceeds the preload applied as Aaron Davidson stated occurs even with over 100 ft/lb torque on a barrel. If we agree the joint is moving when the force of a fired round exceeds that of the pre-load, would tighter fitting threads or "ramp threads" as outlined in Vaughn's book help encourage the barrel to come back to the same location after the load from the fired round is gone?
T=k*F*d
where
T is torque
k is a fitting factor that changes for the type of thread, quality, lubrication and elasticity of the materials that are threaded
F is the joint preload
d is the fastener diameter at the thread pitch diameter
For a threaded joint, fine thread class, lubricated and steel the K factor is typically 0.08-0.11
The units have to match so we need to convert T from ft-lb into in-lb or we need D to convert from inches into feet. Let's turn 100ftlb into 1200inlb instead since it's easy
T=1200inlb
k=0.08
F=?
d=1.0275 (nominal mean PD for 1-1/16x20 Class 3A, I happened to know this one by memory)
Solving for Force
F=T/(k*D)=1200/(0.08*1.0275)=14,598.5 lb force
A 473 case head operates at approximately 8500lb of axial load on the joint and a 532 magnum is about 12000lb peak then we can see that the 100ft lb mark is acceptable for this criteria based on the guidelines in Machinery's Handbook. This is a standard approach for aircraft structures as well.
When I tested the theory I couldn't get anything to change above about 60 ft lbs as I mentioned above. When you calculate the firing force vs. banging the gun off something accidentally the firing force is notably higher.