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W EJ
= s ∂θr
∂
where EsJ is the torsional rigidity of the element, Es being the modulus of rigidity or
shear modulus and J being the polar second moment of area.
Hence, from eqns 7.54 and 7.55, the blade torsion equation is
∂
∂
∂
r ∂
E J
r
s – C – C 2 = 0 θ θ θ 
 
 
˙˙ Ω (7.56)
Assuming solutions of the form
θ = Q(x)ζ(t)
we obtain the mode-shape equation
d
d
d
d
+ ( 2 – 2) = 0
r
E J
Q
r
s C Q


ωθ Ω (7.57)
and
d2ζ/dt2 + ωθ2ζ = 0
∂
∂
Lr
dm
W Wr
+ ∂ dr
∂
–W
Fig. 7.17 Torsional moments acting on element
Structural dynamics of elastic blades 261
where ωθ is a natural frequency of the free motion of the rotating blade.
Now the pitch control usually has a considerable degree of flexibility, so that the
boundary condition to be satisfied at the root is given by
[EsJ dθ/dr] = kθθ0 (7.58)
where θ0 is the change of blade pitch at the feathering hinge and kθ is the stiffness of
the control system. Actually, the flexibility of the control system may be such as to
allow as much as 70 to 80 per cent of the deflection at the tip to occur at the
feathering hinge. The boundary condition at the tip is
dθ/dr = 0 at r = R (7.59)
which expresses the fact that the moment vanishes at the tip.
Consideration of eqn 7.57 shows that, since the torsional stiffness EsJ is unaffected
by rotor speed, the torsional mode shapes are also independent of the rotor speed, i.e.
the mode shapes of a blade are the same whether it is rotating or not. It therefore
follows that
ωθ
2 – Ω2 = constant
When Ω = 0, ωθ is equal to the non-rotating torsional frequency, ω0 say; hence
ω ω θ
2
0
= 2 + Ω2
i.e. the square of the torsional frequency is simply the sum of the squares of the nonrotating
frequency and of the rotor speed.
Unlike the mode shape equations of flapping and lagging, the simpler form of the
torsional equation, eqn 7.57, enables some exact solutions to be obtained.
Consider the case of constant torsional stiffness (except for flexibility at the root)
and zero rotor rotational speed. Equation 7.57 can be written
d2Q/dx2 + α2Q = 0 (7.60)
where α2 = ω 0
2 2
CR /Es J (7.61)
The solution of eqn 7.60 is
Q = A cos α x + B sin α x (7.62)
The boundary conditions, eqns 7.58 and 7.59, give
Bα = (Rkθ /EsJ)A (7.63)
and
tan α = B/A = Rkθ /EsJα (7.64)
or
α tan α = Rkθ /EsJ (7.65)
Solutions of eqn 7.64 in conjunction with eqn 7.61 give the frequencies of the nonrotating
blade for given values of kθ and EsJ.
262 Bramwell’s Helicopter Dynamics
The displacement at the blade tip is given by
Q(1) = A cos α + B sin α
= A[cos α + (B/A) sin α]
= A sec α, from eqn 7.64
The displacement at the root due to control flexibility is
Q(0) = A
Thus the ratio of the root deflection to that at the tip is
Q(0)/Q(1) = cos α (7.66)
Let us suppose that in the first mode of motion the control stiffness is such that
the root deflection is half that at the tip. Then, for this case, we see from eqn 7.66 that
cos α = 12
or α = π/3. Then, if we take EsJ/CR = 4000 as a typical value, we have
from eqn 7.61 that
ω0 = 10.5 Hz
which is about two and a half times the typical rotor frequency. Thus, at normal rotor
speed (4 Hz), the torsional frequency of our rotor blade is about 3.35 Ω.
The choice of α for the first mode fixes the value of Rkθ /EsJ in eqn 7.65, and the
values of α for the higher modes are given by the solutions of
α tan α = (π /3) tan (π /3) = π /√ 3 = 1.82
The next two solutions are easily found to be α = 3.62 and 6.54, giving ω0 = 36.2
and 65.4 Hz respectively. The corresponding mode shape outboard of the feathering
hinge and normalised with respect to the tip deflection is
Q(x) = A[cos α x + (B/A) sin α x]
= cos α cos α x + sin α sin α x
= cos [α (1 – x)] (7.67)
The mode shapes are found by giving α its appropriate values. The first shapes are
　
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本文鏈接地址：Bramwell’s Helicopter Dynamics(131)