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 (iω)].

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

(iω) H

jn

(iω) 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) dτ,

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) dζ

N

n=1

∞

−∞

g

jn

(ξ) F

n

(t + τ −ξ) dξ

=

N

m=1

N

n=1

∞

−∞

∞

−∞

g

km

(ζ) g

jn

(ξ) E {F

m

(t −ζ) F

n

(t + τ −ξ)}dζdξ

=

N

m=1

N

n=1

∞

−∞

∞

−∞

g

km

(ζ) g

jn

(ξ) R

F

m

F

n

(τ − ξ + ζ) dζdξ.

The response spectral density is by deﬁnition,

S

X

k

X

j

(ω) =

1

2π

∞

−∞

R

X

k

X

j

(τ) exp (−iωτ) dτ.

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 (−iωτ) dζdξdτ.

197

Let ν = τ − ξ + ζ. Then, dτ = dν 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 (−iω (ν + ξ −ζ)) dζdξdν

=

1

2π

N

m=1

N

n=1

∞

−∞

∞

−∞

∞

−∞

g

km

(ζ) g

jn

(ξ) R

F

m

F

n

(ν) exp (−iωξ) exp (iωζ)

exp (−iων) dζdξdν

=

N

m=1

N

n=1

∞

−∞

g

km

(ζ) exp (iωζ) dζ

∞

−∞

g

jn

(ξ) exp (−iωξ) dξ

·

1

2π

∞

−∞

R

F

m

F

n

(ν) exp (−iων) dν.

Finally,

S

X

k

X

j

(ω) =

N

m=1

N

n=1

H

∗

km

(iω) H

jn

(iω) 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] + iω [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 + iω − ω

2

−1

−1 5 − ω

2

−1

=

1

|Z (ω)|

5 − ω

2

1

1 3 + iω −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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