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rotor shaft, with the feathering hinge between it and the flexible lag element, which
is the furthest outboard. The diagram also indicates the attached lag dampers mentioned
previously, and discussed in relation to ground resonance in Chapter 9, and a different,
but less common, mechanism for changing the cyclic and collective pitch on the
blades. A photograph of the same hub design, but with five blades, is shown in Fig.
1.6(b). Rotor hub designs for current medium to large helicopters commonly use a
high proportion of composite material for the main structural elements, with elastomeric
elements providing freedoms where only low stiffnesses are required (e.g. to allow
blade feathering).
We now derive the equations of flapping, lagging, and feathering motion of the
hinged blade – but assuming it to be rigid, as mentioned in the introduction. The
motion of the hingeless blade will be considered in Chapter 7. Fortunately, except for
the lagging motion, the equations can be derived with sufficient accuracy by treating
each degree of freedom separately, e.g. in considering flapping motion it can be
assumed that lagging and feathering do not occur.
Fig. 1.5 (b) Photograph of Westland Wessex hub
6 Bramwell’s Helicopter Dynamics
Flexible flap element
Feathering bearing assembly
Flexible lag element
Pitch control rod
Rotor shaft
Control spindle
Spider
Lag damper
Fig. 1.6 (a) Diagrammatic view of Westland Lynx hub
Fig. 1.6 (b) Photograph of Westland Lynx five-bladed hub
Basic mechanics of rotor systems and helicopter flight 7
1.3 The flapping equation
Consider a single blade as shown in Fig. 1.7 and let the flapping hinge be mounted
a distance eR from the axis of rotation. The shaft rotates with constant angular
velocity Ω and the blade flaps with angular velocity ˙β . Take axes fixed in the blade,
parallel to the principal axes, origin at the hinge, with the i axis along the blade span,
the j axis perpendicular to the span and parallel to the plane of rotation, and the k axis
completing the right-hand set. To a very good approximation the blade can be treated
as a lamina.
Then, if A is the moment of inertia about i, and B the moment of inertia about j,
the moment of inertia C about k is equal to A + B. The angular velocity components
ω1, ω2, ω3 about these axes are
ω1 = Ω sin β, ω2 = – ˙ β , ω3 = Ω cos β
The acceleration, a0, of the origin is clearly Ω2eR and perpendicular to the shaft.
Along the principal axes the components are
{–Ω2 eR cos β, 0, Ω2 eR sin β}
The position vector of the blade c.g. is rg = xgRi, so that the components of rg × a0
are
{0, exg Ω2R2 sin β, 0}
The flapping motion takes place about the j axis, so putting the above values in the
second of the ‘extended’ Euler’s equations derived in the appendix (eqn A.1.15), and
using A + B = C, gives
Bβ˙˙ + Ω2(B cos β + MbexgR2) sin β = MA (1.1)
β
Ω
k
eR
j
O
Fig. 1.7 Single flapping blade
i
R
8 Bramwell’s Helicopter Dynamics
where MA = – M is the aerodynamic moment in the sense of positive flapping and Mb
is the blade mass. For small flapping angles eqn 1.1 can be written
β˙˙ + Ω2(1 + ε) β = MA/B (1.2)
where ε = MbexgR2/B.
If the blade has uniform mass distribution, it can easily be verified that
ε = 3e/2 (1 – e). A typical value of e is 0.04, giving ε as approximately 0.06.
The flapping equation (eqn 1.1) could also have been derived by considering an
element of the blade of mass dm, and at a distance r from the hinge, to be under the
action of a centrifugal force (eR + r cos β) Ω2 dm directed outwards and perpendicular
to the shaft. The integral of the moment of all such forces along the blade is found to
be the second term of eqn 1.1, i.e. Ω2(B cos β + MbexgR2) sin β. Regarding it as an
external moment like MA, this centrifugal moment (for small β) acts like a torsional
spring tending to return the blade to the plane of rotation.
The other two extended Euler’s equations (eqns A.1.17 and A.1.19) give
L = 0 and N = –2BΩ ˙β sin β
These are the moments about the feathering and lag axes, respectively, which are
required to constrain the blade to the flapping plane, or, in other words, –L and –N are
the couples which the blade exerts on the hub due to flapping only. It can be seen that
flapping produces no feathering inertia moment, but the in-plane moment 2BΩ ˙β sin β
　
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