15.11 Finite Element Random Response of Plates
557
where h(t) is the unit impulse response function. Similarly, the displacement at point
r due to pressure load applied at point k is
x
k
r (t) = a k
∞
−∞
h
k
r (t − θ 2 ) p k (θ 2 )dθ 2
(15.95)
The loads and the responses are stationary random processes. The correlation
between the two displacements is
R x
j
i x k
r
(τ ) = lim
T →∞
1
T
∞
−∞
x
j
i (t)x
k
r (t + τ )dt
(15.96)
where τ is the time lag.
Combining Eqs. (15.96), (15.94) and (15.95), we get
R x
j
i x k
r
(τ ) = a j a k
∞
−∞
∞
−∞
R p j p k (τ − θ 2 + θ 1 )h
j
i (θ 1 )h
k
r (θ 2 )dθ 1 dθ 2
(15.97)
where R p j p k is the cross-correlation of the pressure at points j and k. θ 1 and θ 2 are
the time transformations.
Taking Fourier transform of Eq. (15.97), we get
S x
j
i x k
r
(ω) = a j a k S p j p k (ω)H
j∗
i (ω)H
k
r (ω)
(15.98)
where S p j p k is the cross-spectral density of the pressure at points j and k.
The generalised displacement at point i is that resulting from each load applied
at different load points, whose number is n, and is given by
x i (t) =
n
j=1
q
j
i (t)
(15.99)
Cross-spectral density of two displacements, when loads are applied at all points
is
R x i x r (τ ) =
n
j=1
n
k=1
R x
j
i x k
r
(τ )
(15.100)