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( – ) d
R Mxx e x k
R mx e x
e
e
∫
∫
Ω
giving λ1
2
3
1
s
2
3
1
2
– 1 =
( – )d + /
( – ) d
eR M x e x k
R Mx e x
e
e
∫
∫
Ω
Substituting the numerator of this expression with eqn. 7.118 finally gives
Μ(0, ψ) = 2 3(λ – 1)β ( – ) d
1
2
1
ΩR ∫mx e 2 x
e
(7.119)
286 Bramwell’s Helicopter Dynamics
This expression is equavalent to eqn 7.106 with the flapping angle β in place of φ1,
with the integral term corresponding to the mode shape integral.
In fact, the integral R mx e x
e
3
1
∫ ( – )2 d is the moment of inertia of the blade
about the flapping hinge. If we denote this by Tβ, we can write eqn. 7.119 as
M(0, ) = 2I( – 1)
1
ψ Ω βλ2 β (7.120)
Thus, Young’s model of the hingeless blade is seen, by comparison of eqns 7.106 and
7.119, to be equivalent to the model utilising the modal approach (and considering
only the first mode of bending), providing that the effect of the aerodynamic forces
are ignored.
It might appear that the offset hinge rigid blade model, as depend by eqn 7.119,
might be a very convenient way of handling the rigid blade problem. However, it is
easily seen that calculation will involve integrals whose lower limit is e and, as this
gives rise to a large number of terms in powers of e up to e4, the algebra becomes
rather unwieldy. In practice, an analysis based directly on the true mode shapes is no
less convenient, and will be followed in order to determine the pitching and rolling
moment coefficients in terms of rotor and aerodynamic parameters. It is the case that,
although eqn 7.107 is formally very simple and appears to require no more than two
modes for its accurate evaluation, it has some disadvantages compared with eqn
7.106. When eqn 7.110 is evaluated in terms of the aerodynamic parameters θ0, λ,
and the flapping and central coefficients, the expressions are found to be quite lengthing.
Defining the integrals
C1 x S x F xS x G S x
0
1
2
1 1
0
1 1
0
1
= d , = 1
1
∫ ∫ d , = ∫ d
we find for the pitching and rolling coefficients
C
a
M = C Gb a G F
4
+ 1
4
(1 – ) – (1 – ) – 12
1
2
1 0 1 –
{ } { }
μ μ
– 1
4
1 +
1
2
– 1
4
+ 1
4
1 + 4 /
2
1 A q i i μ κλ ργ
′
ˆ ˆ (7.121)
and
C
a
l = – B C G a
4
– 1
4
1 + 3
2
– –
1
4
(1 – ) +
2 2
1 1
2
1
0
μ μ μθ
3
+
12
6 + 12
+ 1
4
–
4
1
1
μ μλ γ ˆ
ˆ
p
q
(7.122)
where A1 and B1 are the cyclic control angles, λ
i is the mean induced velocity ratio,
κ the induced velocity gradient, θ1 the blade twist workout, and pˆ and qˆ one the nondimensional
pitching and rolling velocities. Similar equations have been given by
Curtiss and Shupe20.
Structural dynamics of elastic blades 287
Equations 7.121 and 7.122 should be compared with the particularly simple results
indicated by eqns 7.109 and 7.110. Further, numerical examples show that the ‘coupling’
moments, e.g. the rolling moment arising from a longitudinal central displacement,
appear as the small difference between two nearly equal quantities, i.e. they are ‘illconditioned’
and the results could be husbandry22. If we denote by (Mc)a the hub
moment coefficient based on the aerodynamic loading, eqn 7.111, and if (Mc)f is the
coefficient based on the flapping displacement eqn 7.106, Simons19 has shown that
when one mode is considered
( c) = ( )f + ( /2 ) ( – ) d
0
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Bramwell’s Helicopter Dynamics(144)
