# Does centrifugal force stiffen a rotating blade?

I'm a helicopter pilot with limited physics knowledge (units in BSc and HNCs).

I have recently challenged an assertion that rotating blades are stiffened by centrifugal force. In My mind, stiffness refers to the resistance of a member to bending deformation, K. From the comments, perhaps this is my problem?

My counter argument is quite simple. A force can only affect the stiffness of a blade if it changes the physical characteristics of the blade and such force(s) can only be exerted as a result of centripetal acceleration and aerodynamic effects as the blade flies.

A more accurate statement might be that the "blade resists the bending moments since counter-moments are exerted on them arising from the centripetal and aerodynamic forces".

I am very happy to be wrong (since I then learn) but I am catching a lot of heat for this challenge and no-one on Aviation.SE has been able to explain why I am wrong.

I do understand that there is a certain amount of pendantry in my claim but precision, particular in answers on the stacks, is part of my motivation.

What am I missing?

• What would be the difference between "resisting bending moments" and "stiffness" in this context? – JMac Jul 17 '17 at 19:20
• Can you be a little more specific as to how stiffness is defined for a blade? Are you talking about flexural rigidity (resistance to bending) or axial stiffness (amount of stretching due to rotation)? – ja72 Jul 17 '17 at 19:42
• @ja72 Flexural, the blades resistance to bending. i.e. the elastic modulus. – Simon Jul 17 '17 at 20:00
• Is a hanging rope 'stiffer' when you tie a weight to its free end? – amI Jul 18 '17 at 20:26
• No, centrifugal force does not stiffen a swinging blade. That's why weed wackers don't work. – Hot Licks Jul 20 '17 at 2:14

For simplicity, let's model the helicopter blade as a simple massless beam with a point mass at the end. When there is no gravity, the beam will be straight.

We now introduce a force to the beam tip, which will cause the beam to deflect. The bending stiffness $k$ is equal to the ratio of the force to the deflection:

$k=\frac{F}{d}$

When we now put this beam in a rotating reference frame, like the blade of a spinning helicopter rotor, we have to introduce a centrifugal force on the mass to account for the constant acceleration of the beam tip. When the beam is deflected upward, the centrifugal force will cause a downward bending moment and hence the beam will deflect less than in the scenario without the rotation.

Since the bending stiffness is the ratio of vertical force to the vertical deflection $K=\frac{F}{d}$, the (apparent) bending stiffness is higher in a rotating blade.

• I think this is the cleanest of the answers - nice job and thanks for your first answer on this site! – Floris Jul 18 '17 at 1:09
• Agreed it's a nice answer - though it only describes some of the possibilities (see my own answer for the others!) – alephzero Jul 18 '17 at 2:18
• True because of that word "apparent" -- The blade remains flatter, but the addition of a vertical force will have the same effect as it ever did. That's why I like jay72's answer better. – Carl Witthoft Jul 18 '17 at 19:30
• It might be relevant that the root of a helicopter blade is hinged, so it is effectively flapping. – David Aldridge Jul 20 '17 at 8:07
• @DavidAldridge It is indeed hinged, and also the mass is distributed and so are the aerodynamic and centrifugal loads. In order to clearly explain the effect that stiffens the construction I choose to simplify it as much as possible. I had to clamp the root of the blade in order to achieve equilibrium for the case where there is no centrifugal force; otherwise the structure would lack all stiffness. – DeltaLima Jul 20 '17 at 8:51

My counter argument is quite simple. A force can only affect the stiffness of a blade if it changes the physical characteristics of the blade and such force(s) can only be exerted as a result of centripetal acceleration and aerodynamic effects as the blade flies.

The problem with that statement is that you have an over-simplified concept of "stiffness". In general, "stiffness" is simply the slope of a graph of "force" against "displacement".

For a rotating object, you can decompose the stiffness into three different components:

• The "elastic stiffness," which is what you are thinking about. Assuming there is no plastic deformation (and there shouldn't be, for a helicopter rotor blade!) that only depends on the physical shape of the object and the material it is made from
• The "stress stiffness," which depends on the internal stresses in the object. That is what makes the pitch of a guitar string change, when you tune the guitar by change the tension in the string. In fact, to a very good approximation, the stress stiffness is the only stiffness component in a guitar string - the elastic stiffness is close to zero. For a rotating blade, the stress is mostly tension, and the stress stiffness increases with increasing RPM, with the same effect as increasing the tension in a guitar string.
• The "load stiffness." This is harder to understand, but the basic idea is that when the structure changes shape, the loads on it are affected by that change in shape. Sometimes the name "follower force" is used to describe this, because the applied loads "follow" the change in shape of the blade in some way. See below for more explanation.

The stress stiffness and load stiffness both depend on the RPM - in fact they are proportional to the square of the RPM. These effects can be large. For example, the first natural frequency of a large jet engine fan blade may increase from about 30Hz when the engine is not running, to maybe 75Hz at top speed - which is equivalent to the stiffness increasing by a factor of 6 times or more because of the RPM. (Those numbers come from my "day job" - I don't know the comparable figures for a helicopter rotor, but I would imagine they are the same general order of magnitude, or even bigger.)

For a rotating blade, the load stiffness can either increase or decrease the stiffness, depending on how the blade deforms. The diagram in DeltaLima's answer shows one situation, where (assuming the rotor is spinning in a horizontal plane) the deflection is vertical. The "centrifugal force" still acts in the horizontal direction, but at a different height above the hub of the blade. That creates a bending moment which is trying to bend the blade back down again - i.e. it increases the stiffness.

On the other hand, if the blade moves in the tangential direction but stays in the same horizontal plane, the "centrifugal force" is now acting radially away from the blade hub, and that force is trying to bend the blade further in the tangential direction - i.e. it decreases the stiffness. One reason for the blade to bend tangentially would be when the rotor is accelerating or decelerating, and the blades either lag behind (accel) or get in front of (decel) the position of the hub. In a helicopter, the cyclic control will have a similar effect as the blades alternate between the "upwind" and "downwind" halves of each rotation.

Note, the previous two paragraphs only discuss the load stiffness caused by the "centrifugal" forces acting on the blade. The aerodynamic forces create other load stiffness terms, which are not negligible - but this answer is getting too long already!

The potential decrease in stiffness is more significant for a rotor with a larger diameter "rigid" hub - for example a jet engine fan or even more so a typical compressor or turbine rotor with "short" blades on a "large" disk, compared with a helicopter or wind turbine rotor. For rotors with that type of geometry, it is quite common for the vibration frequency of different vibration modes to increase, decrease, or stay the same as the RPM varies.

Rotation does not change the stiffness, but it does reduce the bending of the rotor since it counteracts the lift as far as bending moment is concerned.

Very roughly the following occurs

So the bending moment (causing stress on the blades) is

$$M = R \;\mbox{(Lift)} - h\; \mbox{(Centrifugal)}$$

As you can see the higher the centrifugal loading the less the overturning moment. In reality, the flexing $h$ varies rapidly as the blade rotates around due to the varying lift amount.

• I think you could argue that it does change the stiffness by doing that. Stiffness is resistance to displacement when a force is applied. By changing the loading condition on the beam you are changing it's stiffness. – JMac Jul 17 '17 at 21:04
• Not exactly. The deflection depends on both forces acting and the proportionality remains constant. That is to say the deflection reduces because there is more of a force acting that counteracts it, and not because the blade stiffens up. – ja72 Jul 18 '17 at 3:51
• @JMac Reaction and internal forces are directly a function of applied forces. In this case, the centrifugal force does not depend on lift (the applied force). The analogy does not hold here. The problem is the "stiffness" is not a single value here, but a tensor of values relating all applied forces to all measured deflections. – ja72 Jul 18 '17 at 13:39
• It's still a fairly similar scenario though. A lateral deflection of the blade is resisted by the centrifugal force; and the magnitude of that resistance is a function of the applied force. The greater the deflection the bigger the moment you apply from the centrifugal force. The centrifugal force is increasing the lateral stiffness of the blade. It entirely depends on the context of "stiffness"; you could easily argue it increases the lateral stiffness. – JMac Jul 18 '17 at 21:56
• @DavidAldridge is there a control arm that prevents the blade from pivoting. Together with the ball support they kind of act like a cantilever. – ja72 Jul 20 '17 at 12:49

In My mind, stiffness refers to the resistance of a member to bending deformation, K.

And this is essentially all it is. I believe your confusion comes from the application of material properties when it comes to stiffness.

You will often see stiffness defined as $$k = \frac F{\delta}$$ and for an axially loaded bar, it can be expressed as $$k = \frac {AE}L$$ but what is important to note in this second equation, is that it only holds for a specific loading condition (axially loaded bar).

To quote the Wikipedia article on stiffness:

The elastic modulus of a material is not the same as the stiffness of a component made from that material. Elastic modulus is a property of the constituent material; stiffness is a property of a structure or component of a structure, and hence it is dependent upon various physical dimensions that describe that component. That is, the modulus is an intensive property of the material; stiffness, on the other hand, is an extensive property of the solid body that is dependent on the material and its shape and boundary conditions.

For situations like a bending beam; the stiffness equations will be different; and will depend on things like where along the beam the force is applied (and how the force is distributed), what the end conditions for the beam are, etc.

Most analysis I am familiar with is static; so I can't really go into examples with dynamic components (such as your blade); but if the movement is making it more difficult to bend; then it is by definition more stiff. The mechanical analysis of why inertia contributes to the increased stiffness is outside the scope of what I feel comfortable with.

• Harumphhhh...This should really be an EngineeringSE question, which the OP may not have discovered. – user163104 Jul 17 '17 at 21:14
• @JohnKennedy Because he is not trying to solve a specific problem, it's actually on topic here as well. It's more theory than practice, so from an engineering perspective; it's actually just as much physics as it is engineering (although it is a branch of physics that engineers have a lot more interest in than physicists do currently). – JMac Jul 17 '17 at 21:16
• This doesn't really answer the question - it just talks about different "formulas" for the elastic stiffness of different components, which would lead to the OP's wrong conclusion that the stiffness doesn't change. – alephzero Jul 18 '17 at 2:21
• @alephzero This answer was based off a discussion in the comments that OP suggested I put as an answer. He seemed to be under the impression stiffness was more of a material property than a measure of deflection vs. force. I really just wanted to add a few sources to show what he may have been misinterpreting and an explanation of what stiffness was. Although his title question was about the stiffness of a rotating blade; it seemed clear to me from the question that his real trouble was with his interpretation of stiffness. – JMac Jul 18 '17 at 2:54

Unlike the root of the wing of a fixed wind aircraft, the root of a helicopter blade is either hinged, or made flexible. The hinge has limits on its travel so the blades do not fall to the ground when stationary.

The fact that is it hinged means that the blade does not act as a beam, but as a tie. In operation the axis of the blade is aligned to the resultant of the forces acting on it. Were it not for the need to apply torque to the blade from the hub, you could make it from weighted string with an aerofoil cross section and the first-order physics would be unaffected.

The stiffness of a tie is the ratio of tension to elongation. Rotation has no effect on this ratio. The transverse stiffness of the blade has nothing to do with whether it can support the helicopter. Resistance to flutter, and the ability to apply torque to the blade, require some transverse stiffness. The rotation of the blade increases the effective transverse stiffness, but not the axial stiffness which is what is holding the helicopter up.

## Worked example

Consider a Blackhawk, which has a fully articulated rotor.

• 300 rpm
• 16m wingspan
• 10,000 kg mass

I originally queried the answer in the other place for logical gaps rather than the conclusion - for example, saying the blades were not strong enough because they droop when on the ground. They do droop, but not significantly more so than the wings of a glider or even a B-52S with full tanks do, especially considering the articulated root.

The force acting horizontally on the rotor due to one blade's rotation is ~480kN. The force acting vertically on the blade ~25kN, or 10,000 kg times g divided between the four blades.

The blade is articulated at the root. If the centre of lift is the centre of mass, then the moments cancel and blade will fly at an angle of arcsin(25/480) = 0.052 rad or 3 degrees.

Measuring off this picture shows the blades are at an angle of around 1/2 arcsin(35/392) or 0.045 rad, so close enough.

So the rotor can work without needing to be rigid at all. The blades can flex and align to the net of the lift and rotational forces, so do not need to be stiff to function. The string on a round-the-pole flying will exert a force upwards on the pole if the plane is above it, if you had two or three of them then you could balance out the horizontal forces and the pole would take off:

As for the blade, as it is articulated at its root the forces should act along the axis of the blade and not make it any stiffer, but along it there will be uneven forces so it will be fluttering a bit, and (just like the string on the flyer twangs at a higher note as it goes faster) the effective stiffness under that fluttering will be increased by the rotation.

As to whether the blade would break if statically loaded on the ground, 110kg of blade over 7.8m gives 14 kg/m, assuming half of that is a structural tubular spar.

Taking aluminium 7178 as an example, has a density of 2,800 kg/m3 and so 7kg/m gives an area of 0.0025 m2 or 2500mm2. So I will assume the structural spar is an 80mm diameter 10mm wall tube.

Using http://www.tech.plymouth.ac.uk/sme/desnotes/buccalc.htm and http://www.amesweb.info/StructuralBeamDeflection/CantileverBeamStressDeflectionCalculator.aspx with a point load of 25kN half way along exceeds the yield strength by a factor of four or so - 1940 MPa.

The force acting along the blade of 500kN in an area of 2200mm2 gives 230MPa, which is less than half the tensile yield strength of such a spar.

So the original answer was mostly correct, but badly phrased - the important effect is that the rotational forces set up in the blade cause it to act like a tie rather than a beam, rather than meaning it is a more rigid beam. . If similar forces were exerted on the blade as are found in flight, it would not break. If the blade were used as a tie to support the aircraft, it would not break. The is stiffness gained, but that will affect flutter response rather than whether or not it fails. The fact that the blades or wings droop does not reliably indicate whether they can support the weight of their aircraft without being rotated. If you used the blade as a cantilever beam to support the helicopter it could indeed break if the support was more than a little way from the root.

(I'm off work with a head cold so this could all be completely wrong)

• -1 well... your talking about weight, density and forces but do not answer the questions. Why do zentrifugal forces make the blade seem "stiffer"? It is not about "how much" or anything, but a simple explanation to the why was asked. Several other answers provide that, but your seems to miss the point... (although, chapeau, you put a lot of work into this and provide all the relevant sources. Great, but missed the point) – Mayou36 Jul 19 '17 at 15:29
• @Mayou36 The important effect is the alignment of the blade axis with the resulting forces, working as a tie, it does not become stiff enough to work as a cantilever. That is why the magnitudes matter. – Pete Kirkham Jul 19 '17 at 15:34
• @Mayou36 why would you downvote this answer? Even if it's more than was asked for, it's certainly a relevant addition. In fact I'd say it's the most interesting aspect of this question, because I find it rather obvious that centrifugal forces can in principle stabilise a rotor like the other answers elaborate, but I wouldn't have been able to tell at all whether the effect is significant for a helicopter. – leftaroundabout Jul 19 '17 at 15:47
• So you've calculated the buckling stress on the compression side of the bending blade, correct? – Koyovis Jul 20 '17 at 0:31
• @Koyovis There is no compression side -- the blade is entirely in tension. – David Aldridge Jul 20 '17 at 8:11

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