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with an offset hinge and a torsional spring. The amount of offset and the torsional
spring strength were chosen to match the rotating and non-rotating natural frequencies
of the actual blade. This model of the hingeless blade has received general acceptance
and serves very well for calculating the flapping motion and hub moment; indeed it
is quite usual to express the moment of a hingeless blade in terms of an ‘equivalent
hinge offset’.
An interesting point is that, if the offset blade and torsional spring are taken as a
mechanical equivalent to the hingeless blade, the hub moment of the hingeless blade,
when calculated from first principles, approximates more closely to that given by
eqn 7.111 than to that given by eqn 7.106. For, taking the spring strength as ks per unit
flapping angle, Fig. 7.33, the equations of motion of the blade are
Ω ∫ Ω ∫ 2 3
2
2
1
2 2 3
1
s A
d
d
R m(x – e) dx + R mx(x – e)dx + k = M
e e
β
ψ
β β (7.112)
and
S
r
S
eR
rg
β
ks β
Fig. 7.33 Equivalent offset hinge
284 Bramwell’s Helicopter Dynamics
Mb r F S
2
g
2
2
d
d
Ω β = –
ψ
(7.113)
where MA is the moment of the aerodynamic force about the hinge Mb is the blade
mass. F is the aerodynamic force on the blade, and S is the shear force at the hinge.
β, as usual, is assumed to be a small angle. The moment at the root is clearly
M SeR k r
F
r
r
eR
(0, ) = + s + d
0
ψ β ∂
∫ ∂ (7.114)
Now, for simple harmonic motion, and omitting the constant coning angle, we have
d
d
= –
2
2
β
ψ
β
so that eqn 7.112 can be written
M R mx x e x m x e x xk
e e
A
2 3
1 1
2
= Ω ( – ) d – ( – ) d s


β∫ ∫ β
= 2 3 ( – ) d +
1
Ω R e∫ m x e x ks
e
β β (7.115)
Similarly eqn 7.113,
F – S = M r
d
d b
2
g
2
2 Ω β
ψ
can be written
F S m r eR r
eR
R
– = –Ω2β ∫ ( – )d
= 2 2 ( – ) d
1
–ΩR ∫mx e x
e
β (7.116)
Eliminating the integral between eqns 7.116 and 7.115 gives
SeR + ksβ = FeR + MA
so that substituting in eqn 7.114 gives for the hub moment,
M FeR M r
F
r
r
e
eR
(0, ψ ) = + A + ∂ d
∫ ∂ (7.117)
= d + ( – ) d + d
0
eR
F
r
r r eR F
r
r r
F
r
r
eR
R
eR
R eR ∫∂ ∫ ∫
∂ ∂
∂
∂
∂
= d
0
R
r
F
r
∫ ∂ ⋅ r
∂
Structural dynamics of elastic blades 285
which is identical to eqn 7.111, except, of course, that in the evaluation of
∂
∂
F
r
the
incidence would be determined by the rigid blade flapping.
If now, we ignore the aerodynamic shear force at the hinge, i.e. we put F = 0, then
we have from eqn 7.116
δ = 2 2β ( – ) d
1
ΩR ∫mx e x
e
and, if we ignore the moment at the root due to aerodynamic forces between the
hinge and the root, then we have from eqn. 7.114
M(0, ψ) = SeR + ksβ
giving
M e R M x e x k
e
(0, ) = 2 3 ( – ) d + s
1
ψ Ω β∫ β (7.118)
If we ignore the aerodynamic shear force at the hinge and the moment of aerodynamic
forces about the hinge i.e. we put MA = 0 in addition, then the blade flapping motion
is unforced and the motion will take place at the natural frequency of rigid blade
flapping.
This implies
d
d
+ = 0
2
2 1
2 β
ψ
λ β
Substituting for
d
d
2
2
β
ψ
into eqn 7.112 gives
– 2 3 ( – ) d + ( – )d + = 0
1
2
1
2 2 3
1
ΩR ∫m x e xΩR∫Mx x e x ks
e e
λ β β β
hence, λ1
2
3
1
s
2
3
1
2
=
( – )d + /
　
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本文鏈接地址：Bramwell’s Helicopter Dynamics(143)