Margins of Safety Calculation for Compression Springs 1

[feajob]

Hello,

We usually calculate Margins of Safety for compression spring as shown below:
Stress = 8 K Force Dmean / (pi*d^3) (Calculated Stress at max. operating deflection)
K = (4C-1)/(4C-4) + 0.615/C (Spring Stress Correction Factor)
C = Dmean / d (Spring Index)

M.S. = TDS Ftu / Stress - 1
where TDS is representing Torsional Design Stress (40% to 50%)

Recently, one of our customer was asking for limit Margin of Safety. I am thinking to perform following check:

M.S. (yield) = Fsy / (1.15 Stress) - 1

I am conservatively assuming an extra 15% penalization factor to avoid any detrimental deformation of spring. But, I am not quite sure, because, I think that limit M.S. for springs are a little bit contradictory. What does a negative limit M.S. means? I think that a negative limit M.S. should highlight a potential for detrimental deformation.

Please share your experience with me. I am not an expert in spring calculation.

Thanks,

 

[desertfox]

Hi

When I used to design compression springs we used a figure of 40% of the material yield stress as a limit for the torsional shear stress and this as a rule of thumb gave infinite life to the spring.
Try and get hold of a copy of Berry's spring design book.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[feajob]

Thanks for your reply. I am using Fsy (limiting shear stress from Figure 4 of ESDU 83002). If I understood correctly then you are proposing 0.4 Fty for this check. By the way, where do you get Fty for spring wire? I have seen Tensile Strength in AMS5678F for 17-7PH. I thought that those Tensile Strength are related to Ftu. Can you please let me know your reference for Fty?

 

[desertfox]

Hi again it’s been a long time since I did spring design but I am fairly sure that my figures came from berry’s spring design book, however looking at more recent standards EN10270 is spring wire material And EN13906 for spring design and the latter quotes a maximum shear stress of 0.56 times spring wire tensile strength and not as I said 0.4 times yield, that said the .56 x tensile strength is with the spring compressed to solid.


“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[feajob]

Thanks again for your reply. We use the Tensile Strength shown in Table 4A of AMS 5678F (SAE). In this document, it is not mention Ultimate or Limit. But, for a wire diameter of (4.57mm to 5.26mm), manufactured from 17-7PH, CH900, the Tensile Strength are
Min. 1738 MPa
Max. 1944 MPa

I think that the above allowables are corresponding to Ftu (Ultimate Tensile Strength). I cannot find EN13906 in our database. Because, we usually use American Reference Document. If you have access to it . Can you please let me know what are the allowables for wire diameter of 4.7mm (manufactured from 17-7PH) in EN13906?

Thank you very much,

 

[desertfox]

Hi

Well I can’t get a figure in the EN I mentioned above however I found this site that suggests a safe stress limit of 35% U.T.S for 17-7 stainless steel springs.

https://www.mrspring.com/docs/Compression Spring Design Info.pdf

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[feajob]

O.K., thanks. For 17-7PH the recommended range is 45% (not 35%).