rotational stiffness of a constant torque spring hinge

[WildBlueMtn]

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.

 

[sms]

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!

 

[MintJulep]

You have a constant force acting through the panel and against a stop.

Assuming that the stop is not perfectly rigid, wouldn't the natural frequency of the system be determined by the stiffness of the stop?

 

[WildBlueMtn]

sms-
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.

 

[israelkk]

To my opinion the so called constant force spring is the major contribute and the natural frequency will be didtated by that spring (at least the first harmonic). The low K value is by magnitude lower then the other "springs" in the system.

 

[sms]

israelkk makes a very good point, if there is one element of the system that is much less stiff than the rest, it will essentially be the stiffness of concern and will set the natural frequency.

-The future's so bright I gotta wear shades!

 

[WildBlueMtn]

israelkk and sms,
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.

 

[hacksaw]

agree with greglocock, the natural frequency is not to be confused with the forced response.

 

[WildBlueMtn]

--"agree with greglocock, the natural frequency is not to be confused with the forced response."

I fully agree as well.

 

[erichanyp]

WildBlueMtn,
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

 

[erichanyp]

WildBlueMtn,

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.