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
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
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
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
4
I. Equilibrium of hinged blades
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
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
5
I. Equilibrium of hinged blades
• 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
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
6
I. Equilibrium of hinged blades
• 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)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
7
I. Equilibrium of hinged blades
• 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)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
8
I. Equilibrium of hinged blades
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)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
9
I. Equilibrium of hinged blades
(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)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
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)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
11
III. Blade flapping motion
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)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
12
III. Blade flapping motion
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)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
13
III. Blade flapping motion
(-) 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)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
14
III. Blade flapping motion
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°
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
15
III. Blade flapping motion
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)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
16
Force-displacement phase to the frequency of the forced vibration (Fig. 7-26)
III. Blade flapping motion
• 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
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
17
Simple flapping rotor with flapping hinge on the axis of rotation
III. Blade flapping motion
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
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
18
III. Blade flapping motion
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
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
19
III. Blade flapping motion
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
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
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III. Blade flapping motion
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 ↑
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
21
IV. Rotor control in forward flight
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
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
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
IV. Rotor control in forward flight
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
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
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IV. Rotor control in forward flight
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
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
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IV. Rotor control in forward flight
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)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
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IV. Rotor control in forward flight
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
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
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V. Blade motion in the plane of the disk
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
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
27
V. Blade motion in the plane of the disk
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)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
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V. Blade motion in the plane of the disk
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)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
Coriolis torque acting on Hovering rotor (Fig. 7-37)
29
V. Blade motion in the plane of the disk
• 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)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
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
V. Blade motion in the plane of the disk
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)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
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
V. Blade motion in the plane of the disk
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)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
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
V. Blade motion in the plane of the disk
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)
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
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)
V. Blade motion in the plane of the disk
Active Aeroelasticity and Rotorcraft Lab., Seoul National University
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
V. Blade motion in the plane of the disk