7. Blade motion and rotor control

Active Aeroelasticity and Rotorcraft Lab.

7. Blade motion and rotor control

2020

Prof. SangJoon Shin

Active Aeroelasticity and Rotorcraft Lab., Seoul National University

I. Equilibrium of hinged blades

II. Control of the hinged rotor in hover

III. Blade flapping motion

IV. Rotor control in forward flight

V. Blade motion in the plane of the disk

Overview

1


3

Introduction

Rotor moving edgewise in the air : forward flight

→ two standard means available to overcome dissymmetry of lift

1. Hinged at the roots so that no moments can be transmitted

→ Control can be achieved by tilting the hub axis until the resultant rotor

vector points in the desired direction

2. Rigidly attached to the shaft but cyclically feathered

→ Decrease pitch on advancing side / increasing pitch on retreating side

→ Equalize the lift around the disk


4


Normal flapping blade… effectively mounted to the hub on a universal

joint – free to flap, lead, or lag, but always fixed in pitch

1. Equilibrium about the flapping hinge

Forces acting on the blade in flapping direction

• lift, centrifugal forces, weight(negligible)

Elemental centrifugal forces (Fig. 7-3)

m : mass per unit lengthΩ : rotational speed𝑟 : radius of the element𝛽 : blade flapping angle

𝑑 𝐶. 𝐹. = (𝑚𝑑𝑟)Ω2𝑟 cos 𝛽 (1)

Fig. 7-3 Centrifugal force distribution


5


• Component of centrifugal force

perpendicular to the blade

𝑑 𝐶. 𝐹. sin 𝛽 = 𝑚𝑑𝑟Ω2𝑟𝛽

→ varies linearly with the radius →

• Lift force distribution * 𝑀 = 𝑚𝑅, the blade mass

(2)

(3)

Untwisted constant-chord blade

Ideally twisted constant-chord

inflow varies linearly with radius

inflow is constant along the radius

𝐶. 𝐹.𝑀𝑜𝑚𝑒𝑛𝑡 =1

3𝑅 𝑀𝛺2𝑅𝛽 =

2

3𝐶𝐹 𝑅𝛽

• Moment exerted by C.F. about the flapping hinge


6


• Elemental lift 𝑑𝐿

𝑑𝑟= 𝑐𝑙

𝜌

2𝛺2𝑟2𝑐

• For an ideally twisted constant-chord blade

𝑐𝑙 = 𝛼𝑟𝛼 = 𝛼 𝜃𝑡𝑅

𝑟−

𝑣

Ω𝑟

→ 𝐿𝑖𝑓𝑡 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 × 𝑟

Lift of an ideally twisted blade varies with radius

For an untwisted blade, 𝛼𝑟 roughly constant, lift varies ∝ 𝑟2

2

3𝑅 × 𝑙𝑖𝑓𝑡 (for ideal twist)

4

3𝑅 × 𝑙𝑖𝑓𝑡 (no twist and no taper)

(4)

(4a)

𝐿𝑖𝑓𝑡 𝑚𝑜𝑚𝑒𝑛𝑡

(5)

(6)


7


• Coning angle β

β =𝑏𝑙𝑎𝑑𝑒 𝑙𝑖𝑓𝑡

𝐶.𝐹.(for ideally twisted)

β =9

8𝑏𝑙𝑎𝑑𝑒 𝑙𝑖𝑓𝑡

𝐶.𝐹.(untwisted, constant-chord)

→ β in hovering ~ C𝑇

2. Equilibrium about the drag hinge

component of C.F. perpendicular to the blade toward zero lag (Fig. 7-6)

)𝑑 𝐶. 𝐹. = 𝑚𝛺2𝑟𝑑𝑟(𝜁 − 𝑖

* 𝜁 = lag angle* 𝑖 = angle between no lag position

and line of action of C.F.

(7)

(8)

𝑖 = 𝜁 1 −𝑒

𝑟

• From Fig. 7-6, 𝑖𝑟 = 𝜁(𝑟 − 𝑒)

∴ 𝑑 𝐶. 𝐹. = 𝑚𝛺2𝑟𝑑𝑟𝜁 1 − 1 −𝑒

𝑟= 𝑚𝛺2𝑒𝜁𝑑𝑟 … constant along the span (9)


8


Moment of the centrifugal force about the lag hinge

𝐶. 𝐹.𝑚𝑜𝑚𝑒𝑛𝑡 = mR𝑒𝛺2𝑅𝑐.𝑔.𝜁 = M𝑒𝛺2𝑅𝑐.𝑔.𝜁

Aerodynamic forces

Denote resultant force as 𝐹, point of application as 𝑅𝑑.𝑓.

𝐴𝑒𝑟𝑜𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑀𝑜𝑚𝑒𝑛𝑡 = 𝐹𝑅𝑑.𝑓.

Equating with C.F. moment,

Equating the shear forces @ lag hinge (Fig. 7-7)

𝑇𝑜𝑟𝑞𝑢𝑒/𝑒 = 𝐹 𝑐𝑜𝑠 𝜁 + 𝑀𝛺2𝑅𝑐.𝑔. 𝑠𝑖𝑛 𝜁 = 𝐹 +𝑀𝛺2𝑅𝑐.𝑔.𝜁

* 𝑅𝑐.𝑔.: distance from axis of rotation to the blade c.g.

𝐹𝑅𝑑.𝑓. = 𝑀𝛺2𝑅𝑐.𝑔.𝑒𝜁 or 𝐹 =𝑀𝛺2𝑅𝑐.𝑔.𝑒𝜁

𝑅𝑑.𝑓.

(10)

(11)

(12)


9


(12) → (11)

mean drag angle is a function of 𝑡𝑜𝑟𝑞𝑢𝑒/Ω2 → 𝐶𝑄

Relatively insensitive to change in 𝑅𝑑.𝑓.

𝜁 =𝑇𝑜𝑟𝑞𝑢𝑒

𝑀Ω2𝑅𝑐.𝑔.𝑒𝑒

𝑅𝑑.𝑓.+ 1

Torque/e = 𝑀Ω2𝑅𝑐.𝑔.𝜁𝑒

𝑅𝑑.𝑓.+𝑀Ω2𝑅𝑐.𝑔.𝜁 = 𝜁 𝑀Ω2𝑅𝑐.𝑔.

𝑒

𝑅𝑑.𝑓.+ 1

(13)


10

II. Control of the hinged rotor in hover

Sudden rotation of control axis (Fig. 7-11)

Change in pitch angle of the blade

→ Lift increase → Blade moves, or

“flaps” → Continues until the plane

of the blades is again perpendicular

to the control axis @ which position

no cyclic-pitch changes occur

Some delay between a rapid control

angle change and the re-alignment

of the rotor disk

→ extremely small

Differences when the rotor is moving

edgewise through the air (Fig. 7-12)


11


1. Flapping as represented by a Fourier series

Flapping motion

𝛽 = 𝑎0 − 𝑎1 𝑐𝑜𝑠 𝜓 − 𝑏1 𝑠𝑖𝑛 𝜓 − 𝑎2 𝑐𝑜𝑠 2𝜓 − 𝑏2 𝑠𝑖𝑛 2𝜓…

𝛽 : angle between the control axis and the blade𝜓 : azimuth angle (Fig. 7-13)

(14)


12


2. Geometrical interpretation of the Fourier coefficient

𝑎0 : flapping angle independent of the blade azimuth angle 𝜓 in hover

𝛽 = 𝑎0 (Fig. 7-14 ↓)

𝑎1 : amplitude of a pure cosine motion

𝛽 = −𝑎1 𝑐𝑜𝑠 𝜓 (Fig. 7-15, 7-16 →)

𝑏1 : amplitude of a pure sine motion

𝛽 = −𝑏1 𝑠𝑖𝑛 𝜓 (Fig. 7-17, 7-18)


13


(-) sign → result in plus values for the 𝑎1 and 𝑏1 coefficients

in normal forward flight

𝑎2 : amplitudes of the higher harmonics

𝛽 = −𝑎2 𝑐𝑜𝑠 2𝜓 (Fig. 7-17, 7-16)


14


3. Physical explanation of the existence of the component motions

An infinite number of terms in Fourier series exactly describes any arbitrary

motion. However, only a few terms are necessary.

Magnitude of a typical flapping motion in forward flight

𝑎0 = 8.7°, 𝑎1 = 6.1°, 𝑏1 = 3.9°, 𝑎2 = 0.5°, 𝑏2 = −0.1°

① Coning angle, 𝑎0 … depend on the magnitudes of 2 primary moments about

the flapping hinge Thrust moment (Fig. 7-21)

C.F. moment

Hover… large inflow (induced), loading toward tips,

larger coning angle (9°)

Min. power… small inflow (small induced), loading

more inboard, smaller coning angle (8°)

High speed… large inflow (parasite), loading toward

tips, larger coning angle 9°


15


② Backward tilt, 𝑎1… 𝜓 = 90° → lift increase → flapping up (Fig. 7-23)

• AoA decrease (Fig. 7-24) no unbalanced force for blade with no inertial forces

• To consider blade mass and air damping, blade as a dynamic system

→ 1 DOF system (Fig. 7-25)


16

Force-displacement phase to the frequency of the forced vibration (Fig. 7-26)


• 𝛷 : phase angle between the

max. applied force and max.

displacement

•𝑐

𝑐𝑙: ratio of the actual

damping to critical damping

• When ω

ω𝑛= 1 → phase angle

𝜙 = 90° and is independent of

the amount of damping

• Flapping blade (Fig. 7-2)

(15)𝜔𝑛 =𝐾

𝜏radians/second


17

Simple flapping rotor with flapping hinge on the axis of rotation


𝐶. 𝐹.𝑚𝑜𝑚𝑒𝑛𝑡 = න0

𝑅

𝛺2𝑟2𝛽𝑚𝑑𝑟 = 𝑀𝛺2𝛽𝑅2

3

𝑅𝑒𝑠𝑡𝑜𝑟𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = 𝐾𝛽 (𝐾 = 𝑀𝛺2 𝑅2

3)

𝐼 =1

3𝑀𝑅2, 𝜔𝑛 =

𝐾

𝐼= 𝛺2 = 𝛺

When hinge offset = h,

Exciting air forces… 1/rev →ω

𝜔𝑛= 1

→ 𝑓𝑜𝑟𝑐𝑒 − 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑝ℎ𝑎𝑠𝑒 = 90°

• Maximum flapping at 𝜓 = 180°

Minimum flapping at 𝜓 = 0°

(17)

(18)

(19)

𝜔𝑛 = 𝛺 1 +3

2

ℎ

𝑅 (19a)

Fig. 7-2 Flapping blade


18


③ Sideward tilt, 𝑏1, … may be viewed as arising from coning, 𝑎0

Coned rotor (Fig. 7-27a) : Difference in AoA between front and rear of the

blades due to forward speed

No coning (Fig. 7-27b) : effect of forward velocity is identical


19


Fig. 7-28 : force is maximum at 𝜓 = 180°, minimum at 𝜓 = 0°.

→ Force-displacement phase of 90°

→ Max. flapping at 𝜓 = 270°, min. at 𝜓 = 90°

→ 𝑎 + 𝑏1 motion because of coning. 𝑏1 … same order as 𝑎 or larger

𝑏1 tilt is very sensitive to variation in inflow

→ assumed as uniform for forward flight performance analysis

→ However, for low forward speed, 𝑣 ~ quite large at the rear → 𝑏1 increase

At higher forward flight speed, inflow decreases, and becomes uniform


20


④ Higher harmonics

a2, 𝑏2, 𝑎3, 𝑏3… weaving of the blade in and out of the surface of the core

Presence of the forces which produce higher harmonic motions

Asymmetric flow pattern, reverse flow region 𝐹𝑙𝑎𝑝𝑝𝑖𝑛𝑔 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ∝ 𝑠𝑖𝑛2 𝜓

Little importance on control and performance, but extremely important for

vibration and stresses

⑤ Effect of blade mass on flapping motion

𝑎0… directly affected by blade mass

𝑎1… independent of blade mass since exciting forces act on resonant system

𝑏1… in resonance → independent of blade mass

but exciting forces proportional to 𝑎0, which is proportional to blade mass

Blade mass increases to infinity → 𝑏1 decreases to zero

Higher harmonics… forced vibration well above resonance, goes to zero

when blade mass ↑


21


Tip-Path Plane (TPP) tilts backwards and sidewards (by 𝑎1 and 𝑏1)

w.r.t. control axis / resultant thrust perpendicular to TTP

→ govern the control of helicopter

Hover… TPP exactly perpendicular to control axis

Forward Flight… similar, but not exactly perpendicular (Fig. 7-29)

TPP tilts faster than the control axis tilts (both for forward and rearward

→ instability of the rotor w.r.t. AoA -> control is more sensitive as forward

speed increases


How to achieve the desired control axis tilt

Physically tilting the rotor shaft (“direct control”) … autogyro

→ Mechanically awkward in helicopters → 2 methods to solve

22


① Rotor hub tilting (Fig. 7-30)

Separation of the shaft axis and

control (hub) axis

The hub axis then becomes the

control axis

② Means for cyclically varying

blade pitch (Fig. 7-31)

Pitch will be always constant

w.r.t. the plane of swash plate


23


Basic equalities of flapping and feathering

Fig. 7-32… Control axis vertical, TPP tilts rearward by an amount 𝑎1

→ Low pitch on the advancing side, high on the retreating side

Fig. 7-33… Blade feathering w.r.t. TPP = blade flapping w.r.t. control axis

Fore and aft (𝑎1) flapping w.r.t. control axis

→ lateral (𝛽1) feathering w.r.t. TPP


24


Geometrical relationships among

→ Fig. 7-34

Axis of no feathering (control axis)

Axis of no flapping (TPP)

Intermediate shaft axis

• Flapping motion w.r.t. control axis

𝛽 = 𝑎0 − 𝑎1 cos𝜓 − 𝑏1 sin𝜓 − 𝑎2 cos 2𝜓 − b2 sin 2𝜓

• Feathering motion w.r.t. TPP

𝜃 = 𝐴0 − 𝐴1 cos𝜓 − 𝐵1 sin𝜓 − 𝐴2 cos 2𝜓 − B2 sin 2𝜓

• Subscripts… w.r.t. shaft axis

𝛼 = 𝛼𝑠 − 𝐵1𝑠𝐴0 = 𝐴0𝑠𝑎0 = 𝑎0𝑠

𝑎1 = 𝑎1𝑠 + 𝐵1𝑠𝑏1 = 𝑏1𝑠 − 𝐴1𝑠

𝑎2 = 𝑎2𝑠𝑏2 = 𝑏2𝑠

(20)

(21)

* 𝑎 : Aoa of the perpendicular to the control axis with the relative wind

(22)

(23)

(24)

(25)

(26)

(27)


25


Fixed resultant force vector in space for

a given weight, parasite drag, speed

(Fig. 7-35) → TPP fixed → flapping motion

completely determined → control axis

determined

• Orientation determined : resultant force

vector / TPP / control axis

3 possible shaft angles and feathering

controls for identical flight conditions

(Fig. 7-36)

• Fuselage attitude and control position

may vary due to different CG position

→ no effect on the rotor control in space,

except secondary influence


26


1. Sources of in-plane blade motion…

Periodic blade motion arises from 2 sources

① periodically varying aerodynamic forces… variation in velocity and AoA

② periodically varying mass forces… TPP tilt

• Fig. 7-37 TPP tilt in hover, control

axis still vertical

• Blade flapping by 𝑎1𝑠 → CG of the

forward blade nearer to the axis of

rotation (shaft axis) → to maintain

the angular momentum, forward

blade must move faster, rearward

blade slower → blades move back

and forth as they rotate


27


Two different axis systems and the resulting forces / motion

Motion w.r.t. shaft axis… flapping motion

𝛽𝑠 = 𝑎0 − 𝑎1𝑠 cos𝜓

→ Two periodic torques about the shaft axis

a. Due to lift acting in the plane perpendicular to the shaft

𝐿𝑖𝑓𝑡 𝑡𝑜𝑟𝑞𝑢𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑠ℎ𝑎𝑓𝑡 = 𝑇𝑏𝑟𝑎.𝑓.𝑎1𝑠 𝑠𝑖𝑛 𝜓

b. Periodic “mass force” torque due to periodically changing MOI

Caused by masses moving radially in a rotating plane (“Coriolis force”)

𝑇𝑏 : thrust per blade𝑟𝑎.𝑓. : radius of the resultant lift on blade

𝑎1𝑠 : fore and aft flapping w.r.t. shaft

(29)

(28)


28


2. Coriolis force

Fig. 7-38… Point mass moves radially

outward → tangential velocity increase

→ resists tangential acceleration → exert

a force to the right on the rotating plane

Tangential acceleration

𝑑(Ω𝑟)

𝑑𝑡= Ω

𝑑𝑟

𝑑𝑡= ΩVradial

Second component … changing direction

of radial velocity vector in space

𝑉𝑒𝑐𝑡𝑜𝑟 𝑙𝑒𝑛𝑔𝑡ℎ ×𝑑𝜃

𝑑𝑡→ 𝛺𝑉𝑟𝑎𝑑𝑖𝑎𝑙

𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 2𝛺𝑉𝑟𝑎𝑑𝑖𝑎𝑙

𝐶𝑜𝑟𝑖𝑜𝑙𝑖𝑠 𝑓𝑜𝑟𝑐𝑒 𝐹𝑐𝑜𝑟𝑖𝑜𝑙𝑖𝑠 = 2𝑚𝑉𝑟𝑎𝑑𝑖𝑎𝑙𝛺 (30)


Coriolis torque acting on Hovering rotor (Fig. 7-37)

29


• w.r.t. rotor shaft, a blade element moves outward with a velocity

𝑑(𝑟 𝑐𝑜𝑠 𝛽𝑠)

𝑑𝑡= −𝑟 𝑠𝑖𝑛 𝛽𝑠

𝑑𝛽𝑠𝑑𝑡

= −𝑟𝛽𝑠 ሶ𝛽𝑠

𝛽𝑠 = 𝑎0 − 𝑎1𝑠 𝑐𝑜𝑠 𝜓, ሶ𝛽𝑠 = 𝑎1𝑠𝛺 𝑠𝑖𝑛 𝜓

𝑉𝑟𝑎𝑑𝑖𝑎𝑙 = 𝑟𝛽𝑠 ሶ𝛽𝑠 = −𝑟𝛺 𝑎0𝑎1𝑠 𝑠𝑖𝑛 𝜓 −𝑎1𝑠

2

2𝑠𝑖𝑛 2𝜓

* Negligible order

(31)

(32)

Coriolis torque

𝑇𝑜𝑟𝑞𝑢𝑒𝐶𝑜𝑟𝑖𝑜𝑙𝑖𝑠 = −න0

𝑅

2𝑟Ω2𝑎0𝑎1𝑠 sin𝜓𝑚𝑑𝑟 = −2

3𝑀𝑅2𝑎0𝑎1𝑠Ω

2 sin𝜓

• Uniform blade

𝑎0 =3𝑟𝑎.𝑓.𝑇𝑏

𝑀 𝛺𝑅 2 → 𝑇𝑜𝑟𝑞𝑢𝑒𝐶𝑜𝑟𝑖𝑜𝑙𝑖𝑠 = −2𝑇𝑏𝑟𝑎.𝑓. 𝑎1𝑠 𝑠𝑖𝑛 𝜓

… depends on 𝑇𝑏 and 𝑎1𝑠, but not on the blade mass

(33)

(34) (35)


3. Equation of motion for blade in lag

Spring restoring torque

Equating all torques to the angular acceleration,

Substituting,

Rearranging,

Damping neglected. Possible sources aerodynamic damping

physical dampers at the blade root

30


𝐶. 𝐹. 𝑡𝑜𝑟𝑞𝑢𝑒 =𝑀𝑅

2𝑒𝛺2𝜁

𝑇𝑎𝑒𝑟𝑜 − 𝑇𝐶𝑜𝑟𝑖𝑜𝑙𝑖𝑠 − 𝑇𝑠𝑝𝑟𝑖𝑛𝑔 = 𝐼 ሷ𝜁

𝑇𝑏𝑟𝑎.𝑓.𝑎1𝑠 𝑠𝑖𝑛 𝜓 − 2𝑇𝑏𝑟𝑎.𝑓.𝑎1𝑠 𝑠𝑖𝑛 𝜓 −𝑀𝑒𝛺2𝑅

2𝜁 = 𝐼 ሷ𝜁

(+) a1 motion → MOI decrease → lead motion → (-)

Increase lag angle for (+) a1 motion

Always resists the motion

𝐼 ሷ𝜁 + 𝑀𝑒𝛺2𝑅

2𝜁 = −𝑇𝑏𝑟𝑎.𝑓.𝑎1𝑠 𝑠𝑖𝑛 𝜓

(36)

(37)

(38)

(39)


For a uniform blade,

For the limiting case of e → 0,

Fig. 7-39… For a rotor with blades hinged

at the center of rotation, Coriolis forces

cause the blades to move always at

constant velocity w.r.t. TPP.

31

Solution of Eqn. (39) … assuming a solution of form ζ = ζ0 sin𝜔𝑡


𝜁0 =−𝑇𝑏𝑟𝑎.𝑓.𝑎1𝑠

𝑀𝑅2

𝑒𝛺2 − 𝐼𝜔2, Ω = ω (41)

(42)𝐼 =1

3𝑀𝑅2 → 𝜁0 =

2

3𝑎0𝑎1𝑠2

3−

𝑒

𝑅

𝜁 = 𝑎0𝑎1𝑠 𝑠𝑖𝑛 𝜓 (43)


As the lag hinge is moved outward, the blade motion in TPP is,

In Eqn. (40),

Setting forcing torque to 0, and

solving for frequency,

Variation of blade natural frequency

with lag-hinge distance (e

R) for a

uniform mass blade (Fig. 7-40)

32


𝑒 ≠ 0 → 𝜁𝑇𝑃𝑃 = 𝑎0𝑎1𝑠

3

2

𝑒

𝑅

1−2

3

𝑒

𝑅

… ω ≪ Ω

−𝐼𝜔2𝜁0 +𝑀𝑅

2𝑒Ω2𝜁0 = −𝑇𝑏𝑟𝑎.𝑓.𝑎1𝑠

−𝐼𝜔𝑛2 =

𝑀𝑅

2𝑒𝛺2, 𝜔𝑛 = 𝛺

3

2

𝑒

𝑅

(44)

(45)


33

4. Lag motion in forward flight

Additional exciting torque… periodic variations in blade drag

In all practical cases, periodic in-plane blade motion is quite small 1

2~ 2°

Mean lag angle variation w.r.t. flight conditions … much larger 10° ~ − 1°

5. Higher harmonic in-plane motion

Although usually small compared to the

first harmonics, important source of

vibration (Ch. 12)

In hover, second harmonic component

exists in proportion to 𝑎1𝑠

𝑎1𝑠 induces in-plane motion twice each

revolution (Fig. 7-41)



34

First harmonic motion ∝ a0𝑎1𝑠 (Eqn. (43))

Eqn. (32) → if 𝑎0 = 0, only second harmonic in-plane motion exists,

𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 =1

2𝑎1𝑠2

Also, second harmonic motions also arise in forward flight due to the second

harmonic aerodynamic forces

4th harmonic depend on 2nd harmonic flapping

due to Coriolis and aerodynamic

Important for fatigue stresses and rotor vibrations

However, small enough to be safely neglected as far as their effects on the

velocities and air forces encountered by the blade are concerned


7. Blade motion and rotor control

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