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d
d
d
=
d
d
d
d
–
d
d
d
d
d
= –
d
d
d
d
d
∫ ∫
∫
S
x x
EI
S
x
dx
S
x
EI
S
x
EI
S
x
S
x
x
EI
S
x
S
x
x
n m n m m n
m n
on applying the boundary conditions again.
Similarly integrating the second term of eqn 7.79
0
1
0
1
d
d
d
d
d = –
d
d
d
d
∫ ∫ d
S
x
G
S
x
x G
S
x
Sn
x
m x
n m
Thus, the term by term integration of eqn 7.79 yields
0
1 2
2
2
2
2
0
1
2 2 4
0
1
d
d
d
d
d +
d
d
d
d
∫EI ∫ d – ∫ d = 0
S
x
S
x
x R G
S
x
S
x
m n m n x R mS S x
λmΩ mn
(7.80)
Treating eqn 7.78 in the same way gives
0
1 2
2
2
2
2
0
1
2 2 4
0
1
d
d
d
d
d +
d
d
d
d
∫EI ∫ d – ∫ d = 0
S
x
S
x
x R G
S
x
S
x
m n m n x R mS S x
λnΩ mn
(7.81)
Subtracting eqn 7.81 from eqn 7.80 gives
(λn2 λm2 mSmSn x
0
1
– ) ∫ d = 0
We conclude that, since λn ≠ λm when m ≠ n,
0
1
d = 0,
= ( ), say, when =
∫ mS S x m ≠ n
f n m n
m n (7.82)
Equation 7.82 expresses the orthogonal properties of the modes in relation to the
weighting function m(x) (mass distribution). This means, for example, that if we
integrate the product of any two dissimilar mode shapes together with the mass
distribution m(x), the result is zero. This fact serves as a useful check on the accuracy
of computed mode shapes.
The orthogonal properties do not hold for a hingeless blade with a built-in coning
angle β0, say, for in this case the boundary conditions are dS/dx = tan β0 at x = 0 and
some of the terms in the integration of eqn 7.79 do not vanish at the limits. However,
this is not really a restriction since the reference axes can be tilted so that the new dS/
dx becomes zero at the blade root. There will then be a component of centrifugal
Structural dynamics of elastic blades 269
force – mΩ2r tan β0 normal to the undeflected blade which can be treated as a spanwise
loading to be added to ∂F/∂r. Thus, the orthogonal properties can be made to apply
to the blade with built-in coning by using the boundary conditions (b) and applying
an extra spanwise loading –mΩ2r tan β0.
We should also note the rather interesting fact that the teetering rotor satisfies both
sets of boundary conditions. Figures 7.24(a) and 7.24(b) show two examples of the
way the blade can bend. In Fig. 7.24(a) the shape of the blade is anti-symmetric about
the hinge, so that there is a point of inflection there; i.e. d2S/dx2 = 0 at x = 0 and the
boundary conditions are those of the hinged blade. In Fig. 9.4(b) the bending is
symmetrical and each half behaves as if the root were rigidly fixed to the shaft; the
boundary conditions therefore are those of the hingeless blade. Thus, the teetering rotor
has twice as many mode shapes and frequencies as a single hinged or hingeless blade.
7.3.2 Forced response equations
Equation 7.6 refers to the free vibration of the rotor blade in the flapwise sense. The
corresponding equation, taking into account the presence of external loading F, is
∂
∂
∂
∂
∂
∂
∂ ∂
∂
∂
∂
2
2
2
2
2
2 – + =
r
EI Z
r r
G
Z
r
m Z
t
F
r
∂ (7.83)
We now make use of the orthogonal properties of the normal mode shapes to simplify
the problem of determining the blade motion under the action of applied forces.
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Bramwell’s Helicopter Dynamics(135)
