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rotational stiffness of a constant torque spring hinge

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.

RE: rotational stiffness of a constant torque spring hinge

Nothing

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

Stiffness and load are two different things...Constant torque is the load, the mechanical stiffness of the system is (semi) independant of load..

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

RE: rotational stiffness of a constant torque spring hinge

(OP)
I take that to mean you agree that Krot=zero.  

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

(OP)
sms, I guess I did mix terminology in the subject line and body of the post.  Krot, rotational spring constant is also refered to as the rotational spring stiffness.  May not be technically correct though.


RE: rotational stiffness of a constant torque spring hinge

A constant torque produces a constant acceleration, in the absence of anything else, and so it is a zero Hz motion.

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

Constant torque spring never in practice gives constant torque. It is alway near constant. It can change up to +/-10% of the torque depends on the number of rotations, friction between the coils etc. The reason is simple because the metal strip radius on the drum is changed by the strip thickness with every complete drum rotation. Add to this the fact that the natural strip curvature in free state is not constant. If you are relying only on the torque in the spring to hold the solar panels in the deployed position then you will have a very low natural frequency.

RE: rotational stiffness of a constant torque spring hinge

(OP)
lsraelkk-that is exactly some of the points i have been coming up with.  Thanks.

 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

By waht mechanism does your actuator create its 'constant' torque?

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

(OP)
Greg,

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

Well, I more or less failed to understand that, but supposing you were to plot torque v deflection. Does it have a slope?

 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

(OP)
Greg, sorry if I'm confusing matters further.  Simply stated, lets say I have a door on a hinge that is allowed to swing from initial position thru 60 degrees at which point there is a hardstop which prevents further swing.  A spring is inside that hinge providing torque to swing the door and the torque of that spring holds the door against the hardstop.

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

Maby frequency is not what you are looking for.  Isn't rotational acceleration under constant torque the same as linear acceleration under constant force?

F=ma

T=J x Alpha

RE: rotational stiffness of a constant torque spring hinge

(OP)
True, but what I am calculating is natural frequency of the system (resonance).

RE: rotational stiffness of a constant torque spring hinge

But to have resonance wouldn't the torque have to change direction at some point?  For example, constant torque with a torque reversal at "Zero Position." You could easily calculate a frequency for that (at a given amplitude).

RE: rotational stiffness of a constant torque spring hinge

WildBlueMtn

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

(OP)
sreid, the holding torque is not a source of vibration here, I am looking to calculate the natural frequency of that spring if it were excited by some outside vibrating source.  It is important to know this natural frequency so as to avoid exposing the system to sources which might excite it, since the greatest deflections and therefore greatest mechanical stresses occur at this natural frequency.

RE: rotational stiffness of a constant torque spring hinge

Perhaps you answered the question with your original post; if Krot = 0, fn=0.  But fn=0 can have a value (amplitude) which is a constant.  This is, of course, exactly what a constant force spring has.

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

Let's consider a slightly simpler case - two equal weights hanging off a rope over a pulley, in an idealised world.

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

(OP)
thanks folks, but all i am trying to do is find the Natural Frequency of the system.  Just like a tuning fork, the system has a first fundamental frequency which produces the greatest deflections.  This is a Modal Analysis to discover the natural frequency of this system.  Once it is built I can test it by putting the system on a vibration table and installing accelerometers to measure the modes...Fudamental Modes are independent of external inputs.

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

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?

RE: rotational stiffness of a constant torque spring hinge

(OP)
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.

RE: rotational stiffness of a constant torque spring hinge

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.

RE: rotational stiffness of a constant torque spring hinge

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!

RE: rotational stiffness of a constant torque spring hinge

(OP)
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.

RE: rotational stiffness of a constant torque spring hinge

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

RE: rotational stiffness of a constant torque spring hinge

(OP)


--"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

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

RE: rotational stiffness of a constant torque spring hinge

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.

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