Spectral Element Analysis of Non-stationary Wind …
465
By considering a constant coefficient, a displacement solution can be assumed of
the form:
v(x, t) = v 0 e
−i(kx−ωt)
(2)
where v 0 is a amplitude, ω is the frequency and k is the wavenumber.
There are two distinct wave modes in the positive direction (k
2 ), which is positivegoing waves with wavenumbers given as:
k 1 =
−
T
2E I
+
T
2E I
2
+
ρ Aω 2
E I
(3)
k 2 = −
−
T
2E I
−
T
2E I
2
+
ρ Aω 2
E I
(4)
The solution for the spectral element subjected to axial load of length L can be
expressed in the form:
v(x, ω) = a 1 e
−ikx
+ a 2 e
−kx
+ a 3 e
−ik(L−x)
+ a 4 e
−k(L−x)
= s(x, ω)a
(5)
where
s(x, ω) =
e
−ikx
, e
−kx
, e
−ik(L−x)
, e
−k(L−x)
(6)
a(x, ω) = {a 1 , a 2 , a 3 , a 4 }
T
(7)
The spectral nodal displacements and slopes of the beam element are related to
the displacement field at node 1 (x = 0) and node 2 (x = L), by
d =
⎡
⎢
⎢
⎣
v 1
φ 2
v 2
φ 2
⎤
⎥
⎥
⎦ =
⎡
⎢
⎢
⎣
v(0)
v
(0)
v(L)
v
(L)
⎤
⎥
⎥
⎦
(8)
Substituting (5) into the right-hand side of (8) and written in a matrix form gives
d =
⎡
⎢
⎢
⎣
s(0, ω)
s
(0, ω)
s(L , ω)
s
(L , ω)
⎤
⎥
⎥
⎦ a = G(ω)a
(9)
where