 Section 7.2: Response to Random Loads
7.4. Consider the two degree-of-freedom system shown in Figure 7.1 where F
1
= 0 and, in addition to
F
2
, random force F
3
is applied to the second mass. The force spectrum is given by
[S
F F
(ω)] =
S
F
2
F
2
(ω) S
F
2
F
3
(ω)
S
F
2
F
3
(ω) S
F
3
F
3
(ω)
.
Derive an expression for the matrix response spectrum [S
XX
(ω)] in terms of the force spectrum, the
modal matrix, and the matrix frequency response function [H ()].
Solution: The transfer function H
ij
(ω) relates the i
th
response to force applied at the j
th
coordinate. The
transfer function matrix is given by
(
¯
H (ω)
)
=
H
11
H
12
H
12
H
21
H
22
H
22
.
Since, F
1
= 0, the spectral densities are all equal to zero,
S
F
1
F
1
= S
F
1
F
2
= S
F
1
F
3
= S
F
2
F
1
= 0.
The response spectrum is then
[S
XX
(ω)]
=
(
¯
H
(ω)
)
0 0 0
0 S
F
2
F
2
(ω) S
F
2
F
3
(ω)
0 S
F
2
F
3
(ω) S
F
3
F
3
(ω)
(
¯
H (ω)
)
T
.
196 7.5. Derive Equation 7.25.
Solution: We are asked to prove
S
X
k
X
j
(ω) =
N
m=1
N
n=1
H
km
() H
jn
() S
F
m
F
n
(ω) (7.25)
Let us start with the correlation function, R
X
k
X
j
(τ) , deﬁned by
R
X
k
X
j
(τ) = E {X
k
(t) X
j
(t + τ )},
where the response of the k
th
coordinate, X
k
(t) , is written as a convolution integral,
X
k
(t) =
N
i=1
X
ki
(t) ,
X
ki
(t) =
−∞
g
ki
(τ) F
i
(t τ) ,
where X
ki
(t) is the response of the k
th
coordinate due to the i
th
force. Upon substituting, we obtain (where
the dummy variable τ is replaced with ζ and ξ), we ﬁnd
R
X
k
X
j
(τ) = E
N
m=1
X
km
(t)
N
n=1
X
jn
(t + τ)
= E
N
m=1
−∞
g
km
(ζ) F
m
(t ζ)
N
n=1
−∞
g
jn
(ξ) F
n
(t + τ ξ)
=
N
m=1
N
n=1
−∞
−∞
g
km
(ζ) g
jn
(ξ) E {F
m
(t ζ) F
n
(t + τ ξ)}
=
N
m=1
N
n=1
−∞
−∞
g
km
(ζ) g
jn
(ξ) R
F
m
F
n
(τ ξ + ζ) .
The response spectral density is by deﬁnition,
S
X
k
X
j
(ω) =
1
2π
−∞
R
X
k
X
j
(τ) exp (iωτ) .
Then the spectral density becomes
S
X
k
X
j
(ω) =
1
2π
N
m=1
N
n=1
−∞
−∞
−∞
g
km
(ζ) g
jn
(ξ) R
F
m
F
n
(τ ξ + ζ) exp (τ) .
197 Let ν = τ ξ + ζ. Then, = and the spectral density becomes
S
X
k
X
j
(ω) =
1
2π
N
m=1
N
n=1
−∞
−∞
−∞
g
km
(ζ) g
jn
(ξ) R
F
m
F
n
(ν) exp ( (ν + ξ ζ))
=
1
2π
N
m=1
N
n=1
−∞
−∞
−∞
g
km
(ζ) g
jn
(ξ) R
F
m
F
n
(ν) exp (ξ) exp (ζ)
exp (ν)
=
N
m=1
N
n=1
−∞
g
km
(ζ) exp (ζ)
−∞
g
jn
(ξ) exp (iωξ) dξ
·
1
2π
−∞
R
F
m
F
n
(ν) exp (ν) .
Finally,
S
X
k
X
j
(ω) =
N
m=1
N
n=1
H
km
() H
jn
() S
F
m
F
n
(ω) .
198 7.6. Consider the two degree-of-freedom system in Figure 7.21 where the force spectrum is given by
S
F F
(ω) = S
0
1 0
0 0
N
2
s,
where S
0
= 1 N
2
s, m
1
= 1 kg, m
2
= 2 kg, k
1
= 2 N/m, k
2
= 1 N/m and k
3
= 4 N/m. Assume that
there is a damper with c = 1 Ns/m between the wall and the ﬁrst mass.
(a) Derive the expressions for the mean-squares, R
Y
1
Y
1
(0), R
Y
1
Y
2
(0) and R
Y
2
Y
2
(0) .
(b) Obtain the natural frequencies and the modal matrix.
(c) Derive the matrix of response spectral densities [S
Y Y
(ω)].
(d) Evaluate the mean-square values of the displacements.
(e) Find the relative error if the mean-square values are approximated using Equation 7.42.
Solution: (a) The mean squares are the areas under the spectral densities or
[R
XX
(0)] =
−∞
[S
XX
(ω)] dω,
where
[S
XX
(ω)] = [H
(ω)] [S
F F
(ω)] [H (ω)]
T
[H (ω)] =
[K] + [C] ω
2
[M]
1
[S
F F
(ω)] =
1 0
0 0
.
(b) The mass, stiﬀness and damping matrices are given by
[M] =
1 0
0 1
[K] =
3 1
1 5
[C] =
1 0
0 0
.
From the eigenvalue problem,
[K] ω
2
[M]
{u} = {0},
the eigenvalues and eigenvectors are found to be
ω
2
= 2.5858, 5.4142
[P ] =
0.9239 0.3827
0.3827 0.9239
.
(c) For the mass and stiﬀness matrices considered, the transfer function matrix is given by
[H (ω)] =
3 + ω
2
1
1 5 ω
2
1
=
1
|Z (ω)|
5 ω
2
1
1 3 + ω
2
.
199 Then, the matrix response spectrum is given by
[S
XX
(ω)] =
1
|Z (ω)||Z
(ω)|
5 ω
2
2
5 ω
2
5 ω
2
1
,
where |Z (ω)| and |Z
(ω)| are the determinants of [Z (ω)] and [Z
(ω)] matrices, respectively. The product
is
|Z (ω)||Z
(ω)| = w
8
15w
6
+ 82w
4
199w
2
+ 196.
The mean squares are obtained by integrating the response spectra from −∞ to . They are found to be
[R
XX
(0)] =
1.12 0.224
0.224 0.673
.
200

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