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has been adequate to provide the solution to a number of important helicopter problems
with acceptable accuracy. A rotor blade is of course very flexible, and a number of
problems arise which make it necessary to study the effects of flexibility on blade
motion. For example, we need to know whether the motion of the blade significantly
affects the estimated performance and rotor blade loading; what stresses occur in the
deformed blade; and whether the frequencies of the blade motion coincide with the
frequencies of the aerodynamic forcing loads. The problem of blade flexibility is a
very complex one, for not only has the blade several degrees of freedom but the
aerodynamic loading depends strongly on the blade shape. However, for the purpose
of illustration, we shall consider the flapwise, torsional, and lagwise motions of the
blade separately, and try to draw some general conclusions. For the far more complicated
problem of coupled motion between the degrees of freedom only a brief discussion
will be given and the reader will be referred to specialist papers.
7.2 Free vibrations of rotor blades
7.2.1 Flapwise bending
We define flapwise bending as deflection of the blade in a plane perpendicular to the
rotor hub plane, Fig. 7.1.
Let Z be the displacement of an element of the blade above the flapping plane and
r the distance from the axis of rotation. Consider the motion of this element under the
forces acting on it, Fig. 7.2; S is the local shear force, M the bending moment, and G
the centrifugal tension in the blade.
Structural dynamics of elastic blades 239
The equilibrium of the blade element requires that
dG + mΩ2r dr = 0 (7.1)
d + d = 0
2
2 S m r Z
t
∂
∂
(7.2)
G dZ + S dr – dM = 0 (7.3)
From eqn 7.1 we get at once
G m r r
r
R
= 2 d ∫ Ω
and eqns 7.2 and 7.3 give
∂
∂
∂
∂
S
r
m Z
t
= –
2
2 (7.4)
∂
∂
∂
∂
M
r
G
Z
r
= + S (7.5)
Differentiating eqn 7.5 and substituting eqn 7.4 we get
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
2
2
2
2 M = + = –
r r
G
Z
r
S
r r
G
Z
r
m Z
t
 
 
 
 
But elementary bending theory gives
M EI Z
r
=
2
2
∂
∂
so that we have, finally, for the equation of bending
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
2
2
2
2
2
2 – + = 0
r
EI Z
r r
G Z
r
m Z
t

 

 
 
 
(7.6)
Equation 7.6 represents the free motion of the blade in vacuo.
Let the solution of eqn 7.6 take the form
Fig. 7.1 Blade flapwise bending
r
Z
Fig. 7.2 Forces on blade element
S
M
G
dr
G + dG
M + dM
S + dS
dZ
240 Bramwell’s Helicopter Dynamics
Z = S(r)φ(t) (7.7)
where S(r) is a function of r alone and φ(t) is a function of t alone. Substituting eqn
7.7 into eqn 7.6 gives
d ( d /d )/d – d( d /d )/d
( )
=
d /d
( )
2 EI 2S r2 r2 G S r r 2 2
mS r
t
t
φ
φ (7.8)
Now, the left-hand side of eqn 7.8 is a function of r only and the right hand side
is a function of t only. Since the two sides are always equal they must be equal to a
constant. The constant has dimensions T –2 and it is convenient to write it as λ2Ω2.
Thus, we have
d
d
d
d
– d
d
d
d
+ = 0
2
2
2
2
2 2
r
EI S
r r
G Sr
m S

 

 


λ Ω (7.9)
and
d
d
= 0
2
2
2 2 φλ φ
t
Ω (7.10)
Let us write z = Z/R, x = r/R, and ψ = Ωt. Then eqns 7.9 and 7.10 become
d
d
d
d
　
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