Page 291 - Wind Energy Handbook

BLADE DYNAMIC RESPONSE 265

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f f i1 ¼ f i0 þ f i0 h þ ( f i1 f i0 )h=2 (5:82)
f
f
f
f
Equation (5.81) can be integrated twice to give an expression for the displacement
at the end of the time step, which, after rearrangement yields the following expres-
sion for the corresponding acceleration:
6 6
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€ (f i1 f i0 ) f f i0 2f i0 (5:83)
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f f i1 ¼
f
h 2 h
Substituting Equation (5.83) into Equation (5.82) yields
3
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f
f f i1 ¼ ( f i1 f i0 ) 2f i0 f i0 h=2 (5:84)
f
h
Equation (5.70) can be written as
ð
R
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€
2
f
f
m i f in þ c i f in þ m i ø f in ¼ ì i (r)q n (r)dr ¼ Q in (5:85)
i
0
where the suffix n refers to the state at the end of the nth time step. Substituting
Equations (5.83) and (5.84) into Equation (5.85) with n ¼ 1 and collecting terms
yields the displacement at the end of the first time step as:

6 6 3 h
€
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f
f
Q i1 þ m i 2 f i0 þ f f i0 þ 2f i0 þ c i f i0 þ 2f i0 þ €
f f i0
h h h 2
f i1 ¼ (5:86)
2
m i ø þ 3c i þ 6m i
i 2
h h
The velocity and acceleration at the end of the first time step are then obtained by
substituting f i1 in Equations (5.84) and (5.83) respectively.
The full procedure for obtaining the blade dynamic response to a periodic
loading using the Newmark â ¼ 1=6 method (which is just one of many available)
may be summarized as follows:
(1) calculate the blade mode shapes, ì i (r);
(2) select the number of time steps, N, per complete revolution, then the time step,
h ¼ 2ð=NÙ;
(3) calculate the blade element loads, q(r, ł n ) ¼ q n (r), at blade azimuth positions
corresponding to each time step (i.e., at 2ð=N intervals) using momentum
theory (here, the suffix n denotes the number of the time step);
Ð R
(4) calculate the generalized load with respect to each mode, Q in ¼ ì i (r)q n (r)dr,
0
for each time step;
(5) assume initial values of blade tip displacement, velocity and acceleration;