rotational stiffness of a constant torque spring hinge
rotational stiffness of a constant torque spring hinge
(OP)
I am trying to estimate, ultimately, the natural frequency of a mass attached to a hinge which is driven by a constant torque spring motor (negator). More specifically this is a deployed solar panel that rotates about the hinge axis with a known, constant applied torque.
Normally, if the hinge were driven by a standard torsion spring I would know the Krot spring constant (in-lbf/rad)of the spring. Knowing the mass moment of inertia (I) of the panel being deployed, I can get a theoretical estimate of the natural frequency:
fn= 1/(2pi)*(Krot/I)^0.5
But with a constant torque, the spring rate Krot would be zero. What am I missing?
Thanks.
Normally, if the hinge were driven by a standard torsion spring I would know the Krot spring constant (in-lbf/rad)of the spring. Knowing the mass moment of inertia (I) of the panel being deployed, I can get a theoretical estimate of the natural frequency:
fn= 1/(2pi)*(Krot/I)^0.5
But with a constant torque, the spring rate Krot would be zero. What am I missing?
Thanks.





RE: rotational stiffness of a constant torque spring hinge
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: rotational stiffness of a constant torque spring hinge
-The future's so bright I gotta wear shades!
RE: rotational stiffness of a constant torque spring hinge
I guess the more direct question is what is the relationship between drive torque about the hingeline and the natural frequency of the driven mass?
Thanks.
RE: rotational stiffness of a constant torque spring hinge
RE: rotational stiffness of a constant torque spring hinge
There are other springs in the system which will define the frequency. These may or may not be obvious, foundation stiffness, gas pressure stiffness, and shaft stiffness might all contribute, for example.
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: rotational stiffness of a constant torque spring hinge
RE: rotational stiffness of a constant torque spring hinge
I am still curious to know of a better formula than the original one i posted to relate that holding torque to fn. It makes sense to me that the greater the holding torque applied against the hardstop at full deployment, the greater the fn. I just don't know how to back it up with the math. for now I am relying solely on the spring holding torque. Any further insight would be helpful, thanks!
RE: rotational stiffness of a constant torque spring hinge
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: rotational stiffness of a constant torque spring hinge
There is a set of flat negator springs stacked on one another to provide a quoted minimum torque ~6in-lbs. These are integrated into the hinge mechanism directly at the hingeline. Hardstop at full deployment puts the 2 halves of the hinge 180 deg apart. Stowed position in my application is at 120 deg, so 60 deg travel when deployed. It doesn't take much torque to deploy in a zero-G environment but I do care about maintaining the natural frequency of the deployed system to be greater than 3 Hz. Panel itself is much stiffer than this so the "weak link" will be the hinge holding torque.
Thanks, again.
RE: rotational stiffness of a constant torque spring hinge
Would say a rayleigh approach help (ie KE at zero dsplacement =PE at max displacement)?
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: rotational stiffness of a constant torque spring hinge
The Rayleigh method is what is used to derive the equation in my original post, which is typically used in this situation. fn= 1/(2pi)*(Krot/I)^0.5
Problem I am having is that this equation assumes that there IS a slope to the Torque v Deflection curve (Krot>0). This works for traditional wire torsion springs that have increasing torque with greater deflection. But with these constant-torque springs there is NO slope, T doesn't increase with increased deflection, K=0 (or at least very close to it as israelkk pointed out).
I think I am at least convincing myself further that I have to approach it from more of a statics point rather than a kinetic one...just assess what holding torque does it take to dis-allow motion given my mass at the given moment arm. So if i know the torque that mass/arm combination exerts about the hingeline, then I know the minimum torque output needed for the spring to maintain equilibrium. So as long as I am under that point, the frequency is not going to vary with the holding torque, the many other joints and panel properties are going to be the weaker links...something I assumed incorrectly.
Thanks for all your time, its good to get feedback and I think I am on a better track at least.
J
RE: rotational stiffness of a constant torque spring hinge
F=ma
T=J x Alpha
RE: rotational stiffness of a constant torque spring hinge
RE: rotational stiffness of a constant torque spring hinge
RE: rotational stiffness of a constant torque spring hinge
I am not sure that you are correct. If an external harmonic steady excitation applies to the structure at a frequency identical or close to the natural frequency the structure will gradually gain increasing amplitudes and will go into a resonance even if it is held by the torque. As far as I remember the stored energy in the torque even feeds the energy to increase the amplitudes. Can somebody clarify this issue?
RE: rotational stiffness of a constant torque spring hinge
RE: rotational stiffness of a constant torque spring hinge
There is an amplitude of a sine excitation force required to just lift the panel off its hard stop. This force requirement would be the same at all frequencies.
RE: rotational stiffness of a constant torque spring hinge
Where is the equilibrium position of the system?
There are infinitely many. The system does not have a preferred position, there is no force encouraging a return to the preferred position.
If there is no energy advantage in returning to a given position then the system won't go there without external motivation.
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: rotational stiffness of a constant torque spring hinge
thanks for your time.
RE: rotational stiffness of a constant torque spring hinge
The natural frequency of a system is simply the square root of the stiffness divided by the mass which is where you started this thread. This frequency would not be excited by a constant force, but would be excited by an impulse event. A guitar string has a natual frequency, if you press down on the string, or set a weight on the string, it will deflect, a distance related to its spring stiffness, but it won't make a sound. However if you pluck the string it will vibrate at its natural frequency, until the damping in the system takes out enough energy that it comes to rest. This is sometimes referred to as free vibration.
A resonant system is defined as a system that is being forced at a frequency that is equal to its natural frequency. In my guitar example, if I pluck the string repeatedly, and at a frequecy that is equal to the natural frequency, the vibration amplitude will build up to the point that string will break. Same thing happens in a car with an unbalanced wheel and bad shocks (low damping) as you drive, whenever you reach a rotational speed of the tire that is equal to the natural frequeny of the suspension, the vibrations in the car will go crazy. Slowing down reduces the problem. If the car is stopped the only way to see the natural frequency is to go out and push down on the fender and watch the frequency of the free vibration.
So once again what is it that you are trying to figure out? Sounds to me that you already know the square root of k/m equation.
-The future's so bright I gotta wear shades!
RE: rotational stiffness of a constant torque spring hinge
Assuming that the stop is not perfectly rigid, wouldn't the natural frequency of the system be determined by the stiffness of the stop?
RE: rotational stiffness of a constant torque spring hinge
All of that is true, I don't disagree at all. I am looking to estimate what the natural frequency is. My initial confusion came down to not knowing what K was, or more specifically that K=0 and therefore, fn=0...if I was relying solely on the K of the spring.
Throughout the thread it became more clear that I was making some poor assumptions, thanks to everyone's feedback. I realize now that I can't look just at the spring to be the main contributor to the fn, as MintJulep points out, in fact it is based more on the other factors in the system. The spring is just providing the clamping force at that joint and the true "K" I need to calculate is the equivalent K of all the materials/geometry at that joint. These are the properties that are going to dictate the natural frequency (again, this is the poor assumption I made at the beginning by thinking the opposite was true and tried to focus on the K of the spring alone).
*IF* I were to assume the panel being held was infinitely rigid then I would have to rely on the spring K value alone...which is sort of where I was when I started. With the K=0 (theoretically, or at least very very close) as is the case with these constant-torque spring devices, then fn=0 (or certainly very close). This supports the conclusion that the other factors of the joint can't be ignored.
This has been a very helpful exercise for me, thanks everyone.
RE: rotational stiffness of a constant torque spring hinge
RE: rotational stiffness of a constant torque spring hinge
-The future's so bright I gotta wear shades!
RE: rotational stiffness of a constant torque spring hinge
That is true assuming the "springs" are in series. If the load path is such that the various springs are in parallel then the opposite is true, the equivalent K value is dominated by the stiffer components. This is a point which I left out of my last post that goes to support why I am looking at the other features more closely.
I am starting to find some actual applications of this where a latching mechanism is activates once the hinge is fully deployed...best I can tell this is necessary in order to achieve a fn greater than what would be achieved if the system relied solely on the spring for the stiffness. Once the mechanical latch takes the load the spring is out of the equation.
RE: rotational stiffness of a constant torque spring hinge
RE: rotational stiffness of a constant torque spring hinge
--"agree with greglocock, the natural frequency is not to be confused with the forced response."
I fully agree as well.
RE: rotational stiffness of a constant torque spring hinge
hoping you can understand my chinese English.
As far as I know, the natural frequency usually is a main parameter of a system,including mechanical system, electric system and also other kinds of system. and this parameter usually be determined by the "structure"of the system. In a mechanical system, the typical structure elements usually are the loads natures of the system, include constant load which usually is a constant foreign force or torque; stretch load which usually is like a spring (maybe linear or nonlinear) and the spring coefficient is the structure element; viscous load which usually is proportional to the velocity of system's load and the structure of this load is called viscous coefficient; inertia load which is proportional to the acceleration of the system'load, and the MASS or INERTIA is the structure element. the nature frequency can be calculated with a dynamics differential eguation. it is impossible to calculate the nature frequency of a single spring if you do not consider of the load of it.
as to your question, if the spring coefficient is about zero, and the drive force is about constant, the nature frequency of the system will not be zero. you also need to creat a dynamics differential eguation to calculate the frequency.
so just to a single spring, the parameter of nature frequency is meaningless.
Eric
RE: rotational stiffness of a constant torque spring hinge
In addition, if the spring coefficient is about zero, of course just as you said, the nature frequency of the mechanical system is zero. this means the system is not stable, the system will accelerate duratively under the constant drive force provided by the spring. actually, the spring here should not be treated as a real spring, it is only a constant foreign force. and if the system moved at the position of the hardstop, the mechanical system will stop and be in stable balance status if the hardstop can stand the constant drive force. in the classic control theory, the mechanical system is also called "SIMPLE INERTIA SYSTEM", and if the spring coefficient is not zero, the mechanical system may be a "TWO RANK Differential system', only when the spring coefficient is not zero, the nature frequency of the system will be available, and the system itself may be stable.
Eric.