## trouble with torsion springs

## trouble with torsion springs

(OP)

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.

## RE: trouble with torsion springs

From your approach to give data for "one coil" can I assume that you are using the Rockwell spring design manual?

To make the issue clearer, maybe you can post the spring drawing or a picture.

Are the Lever length 1: .628” and Lever Length 2: .855” for the levers in the unloaded free spring or, the actual distance from the spring axis to the loading points on the levers?

What is the angle between the levers in the free state of the spring (the angle that includes spring body inside it)?

## RE: trouble with torsion springs

When you say that the torque for "one coil" is 1.87 lb-in, what you expect to be the torque for 4.51 coils spring?

https://sites.google.com/site/israelkk

## RE: trouble with torsion springs

Most of the spring dimensions (wire diameter, coil diameter, angle of the ends in the free state, and the .628 lever length) were measured fairly accurately with standard measuring equipment (micrometer and optical comparator)

I am primarily the Delta Elkon design manual, although I do check it against other sources that I have: http://www.delta-elkon.co.il/filestock/files/cats_...

The .628" lever length is the lever length of the center portion, and it was measured in the free state. The .855" is the distance to the actual loading points of the free ends.

Here is a link to a hammer spring, with picture: http://www.midwayusa.com/product/217321/dpms-hamme...

The included angle formed by the spring in the free state is 155.4 degrees. In the assembled state that angle is 59.38 degrees. In the compressed state, it is -5.34 degrees.

I apologize for the weird terminology I used in my original post. When I wrote "one coil" I meant "one set of coils" or "half of the spring," since it is a double torsion spring. I realize that spring rate is inversely proportional to number of coils, and that they all see the same torque. Sorry.

## RE: trouble with torsion springs

Generally, set removed springs are much more expensive to produce than standard springs.

https://sites.google.com/site/israelkk

## RE: trouble with torsion springs

Thank you for taking the time to assist me - I appreciate it.

## RE: trouble with torsion springs

I have never seen any documentation regarding preset in torsion spring. It is not a common design issue. It is mainly needed in aerospace and military or application as your where space is limited. Most designer doesn't even think it can be done or aware of it. My knowledge is based on 30 years of R&D and design of actual design of such springs and other types of springs for special applications.

https://sites.google.com/site/israelkk

## RE: trouble with torsion springs

Thank you for your time israelkk.