Hub Bar Stress

My opinion of how a hub bar works.

My opinion of how a hub bar works.

If you have a question please ask.

Fundamentally I am challenging your assertion that the pictured hub is inherently superior and that most hub bars are unsafe on a gyroplane that weighs over 500 pounds.

”Vance, as I've done this exercise it's frightened me. I hope I am wrong about it. A 3" by 1 1/8" hub bar might work for a 250 pound Bensen loaded to 500 pounds, but how many out there are well over that weight? And even if they are under 500 pounds, they are at the deformity rating of the best materials at two Gs.”

In my opinion when I don’t know what questions to ask it is difficult for me to calculate forces and quantify the superiority of a particular design.

I am also challenging your assertion that:

“There is no such thing as centripetal or centrifugal force.”

I have already explained this in post #29.

In my opinion one of the wonderful things about a gyroplane rotor is it is self regulating.

In my experience, when you load it more it speeds up and the coning angle is very close to the same.

I feel that the coning angle is only a convenient way to describe what is actually a gentle curve as the centrifugal force balances the lift of the rotor blades.

I feel that the centrifugal force does not suddenly end where the blade joins the hub bar. It appears to be lifting the machine through the teeter bolt so it is likely that although the load path is somewhat convoluted the centrifugal force meets in the middle.

In my opinion the hub bar acts as a spring to keep the blades from breaking at the root.

In my opinion the primary force acting on the hub bar is the centrifugal force trying to pull it apart. I feel that the hub bar is bent and twisted in many directions from many different perturbations.

I have seen many rotorcraft fly with hinges at the root of the blade and some even had a second set of hinges quite a ways from the hub.

There you have my opinion about how a Bensen style hub bar works and why it may not be inferior to the hub in your picture. I hope this answers your unasked questions.

Thank you, Vance
 
No challenge here

No challenge here

Not sure I understand why you want to challenge anything; so, please do not think this response as a challenge, but only a separate understanding, and that from someone finding his way who has never flown a gyroplane.

In my opinion one of the wonderful things about a gyroplane rotor is it is self regulating.

I agree.

In my experience, when you load it more it speeds up and the coning angle is very close to the same.

I can agree with this statement as well, but believe there is lag time during which very large loads can be experienced.

I feel that the coning angle is only a convenient way to describe what is actually a gentle curve as the centrifugal force balances the lift of the rotor blades.

I can agree with this statement as well. I can also see the preset hub bar cone angle having some effect on blade (fixed) shape in flight. If the angle is too high, centripetal force will bend the blades downward as they leave the hub, and lift will curve them upward further out. If the angle is too low, lift will curve the blades upward as they leave the bar, overcoming the straightening effect of centripetal force. It seems there would be an optimum hub bar preset angle for level flight with blade mass being the only variable, as gyroplane weight would determine RPM as in a self equalizing system as you point out. But, perhaps, 2.9 degrees is a suitable average.

I feel that the centrifugal force does not suddenly end where the blade joins the hub bar. It appears to be lifting the machine through the teeter bolt so it is likely that although the load path is somewhat convoluted the centrifugal force meets in the middle.

Centrifugal force operates perpendicular to the axis of rotation, always. It still seems to me that Pete's diagram is wrong in this regard, and many more regards. Centrifugal force is not providing, nor can it provide lift. Only the blade rotation can provide lift, and they do that nearly perpendicular to their plane of rotation. Centrifugal force can only effect the shape and lay of the blade as it exits the hub bar, determining the way the blade delivers lift to the end of the hub bar. Too much hub bar preset angle for a given blade mass, centrifugal force will warp the blade downward as it leaves the bar until it balances with lift, its own rigidity, or reaches unity. Too little hub bar preset angle for a given blade mass, lift will warp the blade upward until it balances with centrifugal force or its own rigidity.

In my opinion the hub bar acts as a spring to keep the blades from breaking at the root.

This, to me, seems to contradict your previous statement above.

In my opinion the primary force acting on the hub bar is the centrifugal force trying to pull it apart. I feel that the hub bar is bent and twisted in many directions from many different perturbations.

For pure value, in level 1G flight, with a balanced system, I agree with this; but the hub bar is also always loaded with the weight of the gyroplane, and bending moments are applied; they're just less than I had previously imagined.

I have seen many rotorcraft fly with hinges at the root of the blade and some even had a second set of hinges quite a ways from the hub.

These systems, while centrifugally stretching the hub bar, still deliver lifting moment to the ends of the hub bar, while the gyroplane weighs downward in the middle. No matter what the effect centrifugal tension plays on the bar, compression will always exist, and the interplay between compression and tension will still act out. It may be displaced or redistributed compression, something I tried to convey earlier, but it will still be there.

There you have my opinion about how a Bensen style hub bar works and why it may not be inferior to the hub in your picture. I hope this answers your unasked questions.

Thank you, Vance

Thanks Vance. Please do not see my response as a challenge to your intelligence or authority on the matter. I am working for understanding, so I would encourage more discussion.
 
Hi Stan

Hi Stan

Terry- Now please dont take this as ganging on....but I would reread what Vance and several others have described what is happening with the rotorhub. There descriptions jive with mine..so I wont repeat them.

What I am curious about that noone has mentioned is about when you mentioned how the loads go positive and negative with each revolution. I have to strongly disagree...... I can guess where your reasoning is coming from...so let me ask you something. Would you describe the path of the tip of each rotor blade as a circle....or do you think the blade actually moves up and down with each revolution? Just curious!!


Stan

Thanks Stan,

I would have to think some part of the blades are moving up and down in their rotation, that this is why the teeter is important; but that tendency to move up and down is muted by centrifugal force.

Am I wrong?
 
Hello Terry,

You cannot challenge my authority because I have none. I am not educated and I try to make it clear that I am offering my opinion or my observation. I have found that I can learn more by asking questions and keeping it simple. I do not have a desire or the tools for debate.

In my opinion where we diverge is in our understanding of where the loads begin and the purpose of the hub bar.

In the NACA studies of rotors on a gyroplane they show that the inner region is the driving and the outer portion of the disk area is driven and provides the lift.

My conclusion is that much of the rotor blade is providing structure that supports the aircraft in addition to driving the outer lifting region of the blade.

My observation is that my Sport Copter rotor blades can barely support their own weight when they are not rotating and the cone up around 120 RPM.

My feeling is that the strength of the hub bar also comes from this centrifugal force and it should in fact flex so that the blade root is not so loaded. I don’t see any reason for this force to end where the blade ends and the hub bar needing to support the weight of the aircraft through beam strength. I feel that the precise coning angle is not particularly important.

Some rotor craft have hinges as close as possible to the center in order to avoid bending at the root. They have some kind of flap stops so that the rotor doesn’t hit the ground when the rotor is not rotating. It is my understanding that Igor Bensen felt that a flexible hub bar with a two blade teeter rotor would be a simple, cheap and easy to fabricate solution to stops and dampening.

In my opinion the value in Pete’s drawing comes from imagining the outer pulleys are the lift and the center is the weight of the gyroplane. It appears to be a basic representation of the principle that is only slightly flawed because the centrifugal force acts along the entire length of the blade. To me it illustrates the concept of blade coning.

I am off to the hangar to work on the Predator; I am not taking my computer so that I will not be distracted from my mission.

Thank you, Vance
 
I can understand that

I can understand that

The rotor does not turn in and out of the wind. It does see different relative wind speeds as it goes around in forward flight. But I don’t think this is relative because Loading on each blade has to be the same balanced or the gyro would roll over. Even though the retreating blade is partially stalled , the advancing blade is producing the same amount of lift or the gyro would roll over. That’s what the teetering action does is allow the blades to balance out the lift as the rotor turns.

I understand that Grant, sorry I wasn't as clear as I could be.
 
Centrifugal force operates perpendicular to the axis of rotation, always. It still seems to me that Pete's diagram is wrong in this regard, and many more regards. Centrifugal force is not providing, nor can it provide lift. Only the blade rotation can provide lift, and they do that nearly perpendicular to their plane of rotation. Centrifugal force can only effect the shape and lay of the blade as it exits the hub bar, determining the way the blade delivers lift to the end of the hub bar. Too much hub bar preset angle for a given blade mass, centrifugal force will warp the blade downward as it leaves the bar until it balances with lift, its own rigidity, or reaches unity. Too little hub bar preset angle for a given blade mass, lift will warp the blade upward until it balances with centrifugal force or its own rigidity.
Well, I see you’re still at it!
I’m not sure what ax you’re trying to grind, or what you are trying to prove, or why, but, as an “exercise in patience” I’ll walk thru this and see what sticks.

First, you totally missed what the little diagram was intended to demonstrate! Perhaps what was obvious to me was not so obvious to others. I’ll get back to that later.

Centrifugal force operates perpendicular to the axis of rotation, always. It still seems to me that Pete's diagram is wrong in this regard, and many more regards.

Damn! Are you really so dense as to think I’m that dense, or are you just trying to be an ass?
Certainly Centrifugal force operates perpendicular to the axis of rotation, but I do hope you understand a bit about force vectors and resultants. Think archery. Think drawing back a bow. Think energy imparted to the arrow.

Too much hub bar preset angle for a given blade mass, centrifugal force will warp the blade downward as it leaves the bar until it balances with lift, its own rigidity, or reaches unity. Too little hub bar preset angle for a given blade mass, lift will warp the blade upward until it balances with centrifugal force or its own rigidity.

Damn! I clearly stated that already in post #9, why repeat it?? In that same post I also stated;
“-that would depend on how accurately you established your precone angle.”

The little diagram with which you seem to have such a problem was intended to explain how to determine the optimum pre cone angle and demonstrate why. The 2.9 degrees in the diagram is only for that particular set of conditions selected purely for simplification. (250lbs. per blade gross weight / 5Klbs. Centrifugal each blade.) Or, TAN coning angle = ((gross weight / number of blades) / CF each blade))
 
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My rant.

If a rotor is at flying rpm but producing no lift then the several tons of centrifugal force pulling on the blades will stretch the blades out so their mass centers line up with the center line of the hub bar, this is fairly easy to see. Now, on a rotor that is at flying rpm and producing lift, the lift will cone the blades up to some angle and in the process raise the blades center of mass above the centerline of the hubbar, which by definition we call undersling. Several tons of centrifugal force acting on the blades mass centers times the undersling will produce a bending moment that is opposite and equal to the bending moment that is caused by the blades lift, plus or minus some moment that is caused because the hubbar is not pre-bent to the correct coning angle. This balance of moments is also why a rotor with flap hinges, like the Cierva's, can function.

All conventional rotorcraft, helicopters and gyros included, rely on centrifugal force the keep the bending loads on their blades within safe parameters, and because of the stretching provided by centrifugal force on the rotor the weight of the fuselage is carried by the hubbar primarily by shear, not bending.

One can think about a gyro fuselage as like a cable car suspended on a cable, the greater the cables tension the lower the "coning angle" and since structural analysis tells us that for a certain stress on a piece of material there is a certain strain (elongation) the shallower the coning angle the lower the actual bending load. If one were to take a picture of a gyro in flight, measure the curvature and coning angle of the rotor from the picture, then flex by hand a static rotor to approximate what is in the picture, the bending load on the hubbar would be the same as what the rotor is taking in flight.

.
 
Excellent rant, Alan! Your reasoning is why commercial helicopter rotors are stressed to about 1.7 times their greatest expected centrifugal loadings. I think around .4 of that is for fatigue life. I would be more concerned with gyro hub-bar fatigue life due to lead/lag stresses….but that is another topic rant ;).
 
Excellent rant, Alan! Your reasoning is why commercial helicopter rotors are stressed to about 1.7 times their greatest expected centrifugal loadings. I think around .4 of that is for fatigue life. I would be more concerned with gyro hub-bar fatigue life due to lead/lag stresses….but that is another topic rant ;).

Ditto that! :whoo:
 
Gents,

If you read back through the thread and look at all the points from both sides you will see that most of what is being said is true.

A lot of this is just a misunderstanding like most bun fights on here.

Pete,
The only reason he had a problem with your picture is because it was too simple. Yes I know its the pic that is used to make it easy to understand but when you look at it from an engineering point of view the forces go straight out not around pulleys. That was all. Nothing for anyone to get excited about.
Half of the ppl here are talking in laymans terms and the others in engineering terms.

There are a few incorrect assumptions but not many. Most of it is just being looked at from 2 different levels and thats where the issues are.
 
Definition of undersling

Definition of undersling

Alan,

I think your definition of undersling may be misleading.

Now, on a rotor that is at flying rpm and producing lift, the lift will cone the blades up to some angle and in the process raise the blades center of mass above the centerline of the hubbar, which by definition we call undersling.

I'm sure you know the facts, maybe just typed it in a way that is easy to misunderstand.

My Definition of Undersling would be.

Undersling is the term used to explain the fact that the hub bar is below the teeter bolt. Undersling is displayed as a measurement and is the distance from the centre of the teeter bolt to the centreline of the hub bar.

Any improvements gladly accepted.

Maybe we should have a sticky with definitions that are to be used on the forum, it may help to decrease misunderstandings.
 
Pete,
The only reason he had a problem with your picture is because it was too simple. Yes I know its the pic that is used to make it easy to understand but when you look at it from an engineering point of view the forces go straight out not around pulleys.

Actually, I impressed the heck out of myself when I saw Pete's drawing, for I found I was thinking much the same way while trying to understand centrifugal forces on a rotor head. I made this drawing about a year or so ago….makes it easier for me to figure things out….. I actually did this with water filled milk bottles and scales to prove it to myself. Oh yes, and I hung a milk bottle from the scale to see what would happen if the scale was pushed off of center between the two pulleys. The scale returned to the center…..
 
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Pete,
The only reason he had a problem with your picture is because it was too simple. Yes I know its the pic that is used to make it easy to understand but when you look at it from an engineering point of view the forces go straight out not around pulleys. That was all. Nothing for anyone to get excited about
Karl, now I'm intimidated. Force is force no matter how it is generated. Hydraulics, Pneumatics, springs or even gravity! In engineering manuals one frequently sees force represented as weight. On the ends of teeters, hanging on levers an bellcranks, around pulleys or what ever. It's as common as mud!
My opinion is that Terry was deliberately trying to be obtuse and argumentive. Or as Vance said "obfuscating". If I'm wrong, I'm wrong!

In my work I often have to explain very complex subjects to rather uneducated people. My policy is to break it down into managable bites and simplify it as much as possible. It works. Now, let's take the "controversal" diagram a bit further. (you could have some fun with your students with this)

Use that diagram as an experiment in your lab. Fix a couple of spools on pegs on the wall. Place a length of fishing line or stout cord across them, and attach equal weights to each end. 5 Kg is enough. You now have a tight horizontal cord, its tension is outward from the center. Just like centrifugal force. Force is force! Now place another considerably smaller weight in the exact center. Remember the exact weights. The cord is now deflected down in the center, describing two right angle triangles between the spools, the cord being the hypotenuse of each. Now ask your students what the angle of deflection between the center and spools. You can determine the exact angle without measuring it. The included angle's tangent is equal to the center weight divided by the sum of the two end weights. It's simply a matter of RATIOS. The ratio of forces is equal to the height / hypotenuse ratio.

Karl, I went thru all that simply because I know you enjoy seeing your students analyze and solve unusual problems.
 
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Alan,

I think your definition of undersling may be misleading.



I'm sure you know the facts, maybe just typed it in a way that is easy to misunderstand.

My Definition of Undersling would be.

Undersling is the term used to explain the fact that the hub bar is below the teeter bolt. Undersling is displayed as a measurement and is the distance from the centre of the teeter bolt to the centreline of the hub bar.

Any improvements gladly accepted.

Maybe we should have a sticky with definitions that are to be used on the forum, it may help to decrease misunderstandings.

Yea, that was a bit of a call on my part but I used the term undersling the way I did as most people would be better able to visualize what I was describing, and since the calculations for undersling are really to try and place the rotor blades center of masses on a line with the teeter bolt, I felt my usage of the term was justified.

.
 
Thank you Gents for your kind and gentle words. It's helped me understand better. Quest fulfilled.
 
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I'm Ba-ack!

I'm Ba-ack!

I moved this inquiry to the Precession thread.

Am I understanding correctly that the tendency of the rotor while spinning in level flight is to teeter because of variations in lift as the blades rotate in the apparent wind; but that this lifting force is not realized transversely, due to precession, but parallel to the flight path, causing 2X shake?

Would it be safe to say that the blade tips fly up and down first, followed by the teeter hub?

Also, is it possible to say that the blade tips and rotor hub are about ninety degrees out of phase in their oscillation?

I moved this inquiry to the Precession thread.
 
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Terry- This is basically the same wording as I just typed in your precession thread. The advancing blade sees more relative wind....while the retreating blade sees less relative wind. The aerodynamic forces balance out and the advancing blade rises with its greatest change at 3 o'clock..decreasing its AOA...while the retreating blade finds itself descending and increasing its AOA. The two per rev shake is simply the blades at 3 and 9 oclock are catching a lot more wind resistance..drag....as they are when they are pointed lengthwise into the wind when they are at 12 and 6 oclock.

The greatest change is at 3 and 9 with the greatest movement of the disc roughly 90 degrees in the direction of rotation. This makes the rotor disc the highest at 12 oclock...and the lowest at 6 o'clock. Precession makes the disc react 90 degrees after the point of greatest change...

But the tip path that this rotor disc makes does not in anyway represent a serpentine up and down path by the blade tips...but a nice perfect circle. The blades are flapping and act like a U-joint...but they are not flapping say like a bird flying.


Stan
 
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