Helical springs with circular and rectangular section 3

[LCform]

Hello Everybody

I was checking the formulas of helical springs, in the machinery handbook of F.jones, but both the formulas are brought in terms of d and D , diameter, not b and w for the rectangular, actually I am not finding the helical springs calculation for the rectangualr section

would you please guide me ?

 

[Nereth1]

It's a lot more complex with rectangular springs.

http://www.roymech.co.uk/Useful_Tables/Springs/Springs_helical.html#Rectangular

 

[LCform]

thank you very much

let me complete my question , for metal forming dies , I think design of the spring is harder , cause there is high velocity of deformation , cold heading , and you dont know the force of the loading , maybe just an approximation, in this case, for helical springs, both circular and rectangular, how can I calculate the stress on the springs ?

 

[Jboggs]

Have you asked a reputable spring manufacturer? There are several. They could help you properly apply their springs. Talk to an application engineer.

 

[LCform]

thank you , but my question is for all Helical compression spring , used in a metal forming process, how can we decide about the force on the spring ?

 

[JNieman]

I second Jboggs' recommendation.

Talk to someone with Dayton or Kaller.

 

[LCform]

i got it thank you

 

[hydtools]

Read through this pdf.

Ted

 

[chicopee]

At first, your OP seemed as if you were pushing the envelop in spring design as I conjured a rectangular spring with parallel sides with nearly square corners which would be highly stressed. I decided to glance at the site

http://www.rccoilspring.com

which describes rectangular springs as being oblong in shape, that is with two straight sides and two rounded sides. The spring rate (which I imagine that you have some values in mind) would help in figuring out the number of coils, what's needed to figure out equations for torsional stress, energy storage capacity and space efficiency. For that information, I would suggest to contact the company that maintain the site that I mentioned for a copy of their customers' catalogs which normally contains technical and engineering reports with formulae on the design of such springs.

 

[MikeHalloran]

Nobody who works with die springs bothers to analyze them, beyond looking at the tables in the die spring catalog, and usually, trying to maximize the force they can squeeze into whatever spatial corner of the solution space they have painted themselves into.

So, what do you hope to gain with your analysis?



Mike Halloran
Pembroke Pines, FL, USA

 

[ViktorKovtun]

Engineering Guide to Spring Design

 

[LCform]

Thank you so much, that was a great help

while using it, I faced a problem , for the compression helical spring with rectangular section, when I calculate it myself, it give a very higher result regarding to ISO 10243,
there for a green spring of diameter 50, I have the stiffness of 156 N/mm

while I use the formulas i have the external diam = 49mm (should be in fact a bit less than 50), b=5.4, t= 10.9 , n = 6 , free length = 64 mm, Dm = 38.1, k2 = .4 ( with respect to the b*t)
I get the spring constant K = 674.6 N/mm

I just followed the formulations, but I get a very different result

 

[LCform]

Anybody?

 

[LCform]

@mikeHolloran

well you are right , but sometimes the spring does not move only cause it is pushed, it moves by the forces of forming. sometimes we use non-standard springs as well. I got involved in reading how to calculate the stiffness of the rectangular springs, and now I found there is a difference, I don't find my mistake, or it's different by formula? I don't know what the catalogue calculations are based on

 

[MikeHalloran]

The formulae in my old textbooks for springs with rectangular cross section are valid for springs that are basically cut from solid by single pointing a rectangular thread in a lathe, e.g. with a modified cutoff tool, going right on through a cylinder. They are a little difficult to actually make that way because the cutting tool needs substantial side relief or a twisted shank, so it's effectively necked, and weakened. Given a large pitch, you could cut them with a small end mill in a live cross slide.

Heli-Cal couplings appear to be made by a wire-EDM type process to have rectangular sections, so they should obey the classical equations, except for the multiple starts.

It's more common to coil 'rectangular' springs by winding trapezoidal wire or 'double D' wire or 'double D trapezoidal' wire 'the hard way', but the resulting coils don't have truly rectangular section or sharp corners. If accurate equations exist, the spring producers are keeping them to themselves.



Mike Halloran
Pembroke Pines, FL, USA

 

[desertfox]

Hi LCform

Right I have a copy of ISO 10243 and all I can see is different spring rates, rod and hole sizes for tooling to compress a spring, which is what ISO 10243 is about:- tooling for compressing springs of rectangular wire.

Further the only reference I can see relating to the 156N/mm stiffness you mention is on page 6 and it states the hole diameter for the spring housing would be 50mm and a rod size of 25 mm and a free length of 64mm but it does not dictate the number of turns, wire section etc, it is up to you to design the spring the ISO is merely quoting a spring which falls into a rate of 156N/mm, with a 50mm housing bore plus a free length.

Now if you study the formula on the Roymech site the dimension 'b' should be the maximum dimension of the material which in your case is 10.9 and t = 5.4, so in the formula I think you have these reversed and in doing so you end up cubing the 10.9 dimension instead of the 5.4 dimension.
If I swop the figures round I get a spring rate of 120N/mm which is much closer to the 156 N/mm you're looking for.
If I do the calculation with t = 10.9 I end up with a spring rate similar to the 674 figure you mentioned.

Incidentally I get K2 to be 0.292 and not 0.4

desertfox

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

 

[ViktorKovtun]

LCform, take it:
Solution of the problem with Smath Studio (free program similar to MathCAD)

Download Smath Studio

Also look at my PDF file, please.

 

[desertfox]

Hi ViktorKovtun

I looked at the pdf and it states the stiffness is in the order of 563021N/mm which I don't believe is correct, please have a look and tell me what you think.

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

 

[ViktorKovtun]

desertfox, the stiffnes in PDF is 56,3021 N/mm, not 563021 N/mm.
, (comma) is the decimal mark.
How the world separates its decimals
To the right of the comma there are 4 digits, so the comma, evidently, isn't the thousands separator.

 

[desertfox]

Hi VictorKovtun

We use a full stop here for our decimal points. However even at 56.3 N/mm for the stiffness I still don't agree and what I've found is that the 'D' used in the calculation should be 38.1mm the mean diameter and not the 49mm O.D.

If I run the calculation:-

k= G*b*t^3*k2/(Na*D^3) = 78.3*10^3*5.4^3*10.9*.292/(6*38.1^3)= 118.25N/mm using the 38.1 mean diameter.

k= G*b*t^3*k2/(Na*D^3) = 78.3*10^3*5.4^3*10.9*.292/(6*49^3)= 55.59N/mm using the 49 outside diameter.

The diagrams in the pdf show D/2 as the mean radius, therefore D should be the mean diameter evidently and not the outside diameter which as been used in your calculation.

Note the 118N/mm stiffness is very close to my figure of 120N/mm


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