carburize21
Mechanical
- Jul 1, 2015
- 4
Ladies and Gentlemen,
I’m trying to calculate the elastic potential energy and the stress at peak torsion in an AR-15 hammer spring. It is a double torsion spring. My calculated stress is very high, so high that I have to question whether I am performing the calculation correctly. I have an application that might require more energy, and there are "extra-power" hammer springs on the market, but with this spring being apparently so highly stressed, how are those manufacturers accomplishing that?
Material: Music Wire (assumption, because it does not appear to be stainless), minimum ultimate tensile strength 309,000psi for this particular wire size.
Wire diameter, d: .045”
Mean diameter of coil, D: .360”
Spring Index, C: 8.00
Wahl Factor, K: 1.1
Angle between free state and assembled (position 1) state: 96 degrees
Angle between free state and “cocked” (position 2) state: 161 degrees
Lever length 1: .628”
Lever Length 2: .855”
Number of body turns: 4.07 per side (8.14 total)
Number of active turns (taking the ends into account as partial turns): 4.51 per side (9.02 total)
Based on the above information, I am calculating:
Spring rate: 7.02lb in/per revolution (one coil only) (this seems low according to my calibrated fingertips – am I screwing something up here?)
Assembled Torque (position 1), M1: 1.87lb in (one coil only)
Potential Energy (position 1), U1: 0.25in lbs (one coil only)
“Cocked” Torque (position 2), M2: 3.13lb in (one coil only)
Potential Energy (position 2), U2: 0.70in lbs (one coil only)
Total stored potential energy between the 2 positions: 14.4 inch ounces, taking into account that it is a double torsion spring and there are 2 coils
Stress at position 2: 386,000psi using the formula S=[ (32M2)/(3.1415*d^3) ] x K
Can anyone tell me, for starters, why my calculated stress is so high? Are my figures for elastic potential energy correct? I am aware that there is a residual stress induced during the coiling operation that serves to increase the elastic range, but my stress figure seems absurdly high, especially for a spring that needs to be replaced very, very infrequently.
By the way, I posted a very similar question on another engineering forum, in case any of you have seen it over there. I have reproduced it here, not because I didn't get the answer that I wanted, but because I always try to get a number of opinions before I make a decision.
Thank you.
I’m trying to calculate the elastic potential energy and the stress at peak torsion in an AR-15 hammer spring. It is a double torsion spring. My calculated stress is very high, so high that I have to question whether I am performing the calculation correctly. I have an application that might require more energy, and there are "extra-power" hammer springs on the market, but with this spring being apparently so highly stressed, how are those manufacturers accomplishing that?
Material: Music Wire (assumption, because it does not appear to be stainless), minimum ultimate tensile strength 309,000psi for this particular wire size.
Wire diameter, d: .045”
Mean diameter of coil, D: .360”
Spring Index, C: 8.00
Wahl Factor, K: 1.1
Angle between free state and assembled (position 1) state: 96 degrees
Angle between free state and “cocked” (position 2) state: 161 degrees
Lever length 1: .628”
Lever Length 2: .855”
Number of body turns: 4.07 per side (8.14 total)
Number of active turns (taking the ends into account as partial turns): 4.51 per side (9.02 total)
Based on the above information, I am calculating:
Spring rate: 7.02lb in/per revolution (one coil only) (this seems low according to my calibrated fingertips – am I screwing something up here?)
Assembled Torque (position 1), M1: 1.87lb in (one coil only)
Potential Energy (position 1), U1: 0.25in lbs (one coil only)
“Cocked” Torque (position 2), M2: 3.13lb in (one coil only)
Potential Energy (position 2), U2: 0.70in lbs (one coil only)
Total stored potential energy between the 2 positions: 14.4 inch ounces, taking into account that it is a double torsion spring and there are 2 coils
Stress at position 2: 386,000psi using the formula S=[ (32M2)/(3.1415*d^3) ] x K
Can anyone tell me, for starters, why my calculated stress is so high? Are my figures for elastic potential energy correct? I am aware that there is a residual stress induced during the coiling operation that serves to increase the elastic range, but my stress figure seems absurdly high, especially for a spring that needs to be replaced very, very infrequently.
By the way, I posted a very similar question on another engineering forum, in case any of you have seen it over there. I have reproduced it here, not because I didn't get the answer that I wanted, but because I always try to get a number of opinions before I make a decision.
Thank you.