# Linearity of Helical Springs

1. Sep 19, 2009

For the design of a cantilever, assuming a Hookean material is used below its proportionality limit, stress will be proportional to strain. From this we can conclude that its behavior will be linear.

However, how can we apply this knowledge towards helical springs? Is it possible to design a helical spring whose linearity approaches or equals the cantilever?

Thanks,

2. Sep 19, 2009

### nvn

Yes, typical, cylindrical, uniform, helical springs, whose coils have not started to close, have approximately linear behavior, like the cantilever.

3. Sep 19, 2009

Assuming a cylindrical, uniform, helical spring, whose coils have not started to close, which supports a surface. Would the spring force be linear across the contact surface?

Last edited: Sep 19, 2009
4. Sep 19, 2009

### nvn

I don't know the shape of the contact force curve on the contact surface. I think the shape of the contact force curve would probably tend to vary depending on the spring end treatment (wire shape and pitch).

5. Sep 20, 2009

### Ranger Mike

yes springs are linear and full contact as the top and bottom is " flat " or they can be made without the " flat" ends

http://www.hypercoils.com/

6. Sep 20, 2009

After some brief research, I see that a variety of helical spring topologies are utilized in designs. However, I'm still unable to find an example of a multiple arm, helical compression spring. All of the examples I'm aware of involve a single arm. Is there a reason for this?

Can the compression spring operate equally in compression as well as expansion?

Last edited: Sep 20, 2009
7. Sep 20, 2009

### Q_Goest

What do you mean by a "multiple arm" spring? Are you referring to a type of disk spring?
http://www.mcmaster.com/#disc-springs/=3pweox

There are a lot of different types of springs, but the most common use a single wire wound into a helix. The spring constant remains constant over the usable range of the spring unless it's specifically made not to. In other words, helically wound springs produce a force that is very linear with respect to displacement. They are commonly used in both tension and compression.

8. Sep 20, 2009

I was wondering why a single wire is often used as compared to multiple wires.

Is the single wire simply easier to manufacture? Does it offer benefits compared to multiple wires?

Helical springs are often used in either compression or tension, why not both? If the spring was rigidly coupled to 2 parallel plates, couldn't it function both in tension as well as compression?

As far as visualizing a multiple arm helical spiral, imagine a multiple arm logarithmic spiral and the differences that exist between it and a logarithmic spiral.

(multiple arm) http://1.bp.blogspot.com/_SqhhJb_P3...I/TJksEbXxyo4/s400/blue,+green+spiral+top.gif

(single arm)
http://www.cs.princeton.edu/introcs/32class/images/logspiral.png

A multiple arm, helical spring would simply use multiple wires would into a helix

9. Sep 20, 2009

### Q_Goest

I've seen helical springs that are nested, one inside another (ie: one diameter is larger than the other) but I've not seen them made with two wires of the same diameter as you've suggested. I could be done, just take 2 identical springs and orient them 180 degrees apart. They couldn't be closed and ground for obvious reasons, but it would work if you had end plates made up to accomodate the end configuration.

Looks like there's an advantage to doing this actually. Comparing 2 springs with 1, both with the same total spring rate, the 2 springs will have less stress for any given load than the single spring.

The problem is it adds a lot of complexity and cost to a design.

Springs can be made to handle both tension and compression but that's generally not needed. If there were some design reason to produce a spring that works in both compression and tension, it wouldn't be a big deal to design it like that.

10. Sep 21, 2009

I picked up the seventh edition of Roark's formulas for Stress and Strain. It appears to be rather trivial to calculate the fundamental resonance for a helical spring. However, calculating the mass normalized modes does not appear to be trivial.

Assuming a cantilever (left end fixed, right end free), an equation exists which defines all of the natural frequencies.

F(x)=(K(n)/2pi)*sqrt(EIg/wl^4)

Where K(n) = 3.52 for the 1st mode (n=1), 22 for the second mode (n=2), 61.7 for the third mode (n=3), 121 for the fourth mode (n=4), 200 for the fifth mode (n=5), etc.

Nodal position wrt length = .783 for the 2nd mode (n=2), .504/.868 for the third mode (n=3), .358/.644/.905 for the fourth mode (n=4), .279/.5/.723/.926 for the fifth mode (n=5), etc

How can we define a similar equation for a helical spring?

Last edited: Sep 21, 2009
11. Sep 22, 2009

With regards to my previous posts,

For defining the natural frequencies of a beam rigidly coupled at both ends, we are given the equation:

F(X)=(Kn/2pi)*sqrt(EI/Rho*A*(L^4))

where:
Kn=mode number
E=elastic modulus
I=area moment of inertia (ie second moment of inertia, I believe)
Rho=mass density
A=cross sectional area
L=length

However, what second moment of inertia equation do we want? I believe we can calculate it with respect to both X and Y, where (assuming a rectangular cross-section) I(x)=b(h^3)/12 and I(y)=h(b^3)/12 and h=height (ie y) and b=width (ie x).