7.7. Consider a three degreeoffreedom system with property matrices:
[m] =
m 0 0
0 m 0
0 0 m
, [c] =
c 0 0
0 c 0
0 0 c
, [k] =
3k −k −k
−k 3k −k
−k −k 3k
.
Determine the natural frequencies and the modal matrix. Assume that mass 1 is subjected to a white
noise random force of intensity S
0
. Find the meansquare value of the displacement response of mass one,
R
X
1
X
1
(0) . What is the percent error if the autocorrelation functions are approximated using Equation
7.42?
Solution: From the eigenvalue problem
[k] −ω
2
[m]
{u} = {0},
the eigenvalues and eigenvectors are given by
ω
2
m
k
= 1, 4, 4,
where the last two eigenvalues are repeated. The eigenvector associated with ω
2
m/k = 1 is found from
3 −1 −1
−1 3 −1
−1 −1 3
−ω
2
i
m
k
1 0 0
0 1 0
0 0 1
{u}
i
= {0},
and the normalized eigenvector {ˆu}
1
is given by
{ˆu}
1
=
0.5774
√
m
1
1
1
.
The ﬁrst eigenvector associated with ω
2
m/k = 4 is found by guessing the ﬁrst two elements. Let us choose
{u}
2
= c
1
1
u
32
,
where c is the constant coeﬃcient used to normalize the eigenvector. The third element is found by substi
tuting into the eigenvalue problem. We ﬁnd that
{ˆu}
2
=
0.4082
√
m
1
1
−2
.
The second eigenvector that is associated with ω
2
m/k = 4 is found using the fact that it must be orthogonal
to both {ˆu}
1
and {ˆu}
2
. Let {ˆu}
3
be
{ˆu}
3
= c
1
u
23
u
33
and the orthogonality statements are given by
{ˆu}
T
3
[m] {ˆu}
1
= 0
{ˆu}
T
3
[m] {ˆu}
2
= 0,
201
from which we ﬁnd
1 + u
23
+ u
33
= 0
4 + 4u
23
−8u
33
= 0.
Then,
{ˆu}
3
=
0.707
√
m
1
−1
0
.
The modal matrix is given by
[P ] =
1
√
m
0.5774 0.4082 0.707
0.5774 0.4082 −0.707
0.5774 −0.8164 0
.
The transfer function that relates the i
th
modal coordinate and the modal force is
H
i
= 1/(ω
2
i
−ω
2
+ i2ζ
i
ωω
i
).
In our case,
ω
1
=
3
k
m
and ω
2
= ω
3
= 2
3
k
m
.
The damping ratio obtained from 2ζ
i
ω
i
is the i
th
diagonal element of [P ]
T
[c] [P ] . That is,
2ζ
1
ω
1
=
c
m
ζ
1
=
c
2
√
mk
.
Then, the response spectrum matrix is given by
[S
XX
(ω)] = [P ] [H
∗
(ω)] [P ]
T
[S
F F
(ω)] [P ] [H(ω)] [P ]
T
,
where [H(ω)] is a diagonal matrix whose i
th
diagonal is H
i
. The ﬁrst element of the response spectrum
matrix is given by
S
X
1
X
1
= S
0
[0.111 H
1

2
+ 0.0278 H
2

2
+ 0.250 H
3

2
+0.0556(H
1
H
∗
2
+ H
2
H
∗
1
) + 0.167 (H
1
H
∗
3
+ H
3
H
∗
1
)
+0.0833 (H
2
H
∗
3
+ H
3
H
∗
2
)].
S
X
1
X
2
= S
0
[0.111 H
1

2
+ 0.0278 H
2

2
−0.250 H
3

2
+0.0556(H
1
H
∗
2
+ H
2
H
∗
1
) + 0.167 (H
1
H
∗
3
− H
3
H
∗
1
)
+0.0833 (H
2
H
∗
3
−H
3
H
∗
2
)].
S
X
1
X
2
= S
0
[0.111 H
1

2
+ 0.0278 H
2

2
−0.250 H
3

2
+0.0556(H
1
H
∗
2
+ H
2
H
∗
1
) + 0.167 (H
1
H
∗
3
+ H
3
H
∗
1
)
+0.0833 (−H
2
H
∗
3
+ H
3
H
∗
2
)].
S
X
2
X
2
= S
0
[0.111 H
1

2
+ 0.0278 H
2

2
−0.250 H
3

2
+0.0556(H
1
H
∗
2
+ H
2
H
∗
1
) + 0.167 (H
1
H
∗
3
+ H
3
H
∗
1
)
+0.0833 (−H
2
H
∗
3
+ H
3
H
∗
2
)].
Then,
σ
2
X
i
X
i
=
∞
−∞
S
X
i
X
i
(ω) dω.
202
7.8. Continuing Example 7.6, ﬁnd S
X
1
X
2
(ω), S
X
2
X
1
(ω), and S
X
2
X
2
(ω).
Solution: The modal transfer functions are given by (page 335)
H
1
(iω) =
1
−ω
2
+ 1.162 + 1.232 iω
H
∗
1
(iω) =
1
−ω
2
+ 1.162 − 1.232 iω
H
2
(iω) =
1
−ω
2
+ 2.239 + 1.448 iω
H
∗
2
(iω) =
1
−ω
2
+ 2.239 − 1.448 iω
and the response spectra are given by
S
X
1
X
1
(ω) = 0.070 H
1
(iω)
2
+ 0.999 H
2
(iω)
2
−0.035 [H
1
(iω) H
∗
2
(iω) + H
∗
1
(iω) H
2
(iω)]
S
X
1
X
2
(ω) = 0.365 H
1
(iω)
2
−0.192 H
2
(iω)
2
−0.179H
∗
1
(iω) H
2
(iω) + 0.007H
1
(iω) H
∗
2
(iω)
S
X
2
X
1
(ω) = 0.365 H
1
(iω)
2
−0.192 H
2
(iω)
2
−0.179H
1
(iω) H
∗
2
(iω) + 0.007H
∗
1
(iω) H
2
(iω)
S
X
2
X
2
(ω) = 1.894 H
1
(iω)
2
+ 0.037 H
2
(iω)
2
+ 0.035 [H
1
(iω) H
∗
2
(iω) + H
∗
1
(iω) H
2
(iω)] .
Upon substitution,
S
X
1
X
2
(ω) = 0.365
1
(−ω
2
+ 1.162)
2
+ (1.232ω)
2
−0.192
1
(−ω
2
+ 2.239) + (1.448ω)
2
−0.179
1
(−ω
2
+ 1.162) − 1.232 iω
1
−ω
2
+ 2.239 + 1.448 iω
+0.007
1
−ω
2
+ 1.162 + 1.232 iω
1
−ω
2
+ 2.239 − 1.448 iω
S
X
2
X
1
(ω) = S
∗
X
1
X
2
(ω)
= 0.365
1
(−ω
2
+ 1.162)
2
+ (1.232ω)
2
−0.192
1
(−ω
2
+ 2.239) + (1.448ω)
2
−0.179
1
(−ω
2
+ 1.162) + 1.232 iω
1
−ω
2
+ 2.239 − 1.448 iω
+0.007
1
−ω
2
+ 1.162 − 1.232 iω
1
−ω
2
+ 2.239 + 1.448 iω
S
X
2
X
2
(ω) = 1.894 H
1
(iω)
2
+ 0.037 H
2
(iω)
2
+ 0.035[H
1
(iω) H
∗
2
(iω) + H
∗
1
(iω) H
2
(iω)]
reduces to 0
= 1.894
1
(−ω
2
+ 1.162)
2
+ (1.232ω)
2
+ 0.037
1
(−ω
2
+ 2.239) + (1.448ω)
2
.
203
7.9. Continuing Example 7.7, ﬁnd the matrix of modal forces S
QQ
(ω) and then ﬁnd S
X
1
X
2
(ω), S
X
2
X
1
(ω),
and S
X
2
X
2
(ω).
Solution: The modal force matrix is obtained by
[S
QQ
(ω)] = [P ]
T
[S
F F
(ω)] [P ]
and the response spectra by
[S
XX
(ω)] = [P] [H
∗
] [S
QQ
] [H] [P ]
T
S
X
i
X
j
(ω) =
20
o=1
20
n=1
20
m=1
20
l=1
ˆu
il
H
∗
lm
S
Q
m
Q
n
H
no
ˆu
jo
,
where the bar is divided into 20 segments. If the damping is negligible, the oﬀdiagonal terms of [S
XX
(ω)]
are insigniﬁcant by comparison (which is NOT the case here as shown in Figure 7.11). Then, we may write
S
X
i
X
i
(ω) ≈
20
l=1
ˆu
2
il
H
ll

2
S
Q
l
Q
l
S
X
i
X
j
≈ 0 for i = j.
We can further approximate the response spectra using the ﬁrst three modes so that
S
X
i
X
j
(ω) =
3
o=1
3
n=1
3
m=1
3
l=1
ˆu
il
H
∗
lm
S
Q
m
Q
n
H
no
ˆu
jo
S
X
i
X
j
(ω) ≈
3
l=1
ˆu
2
il
H
ll

2
S
Q
l
Q
l
for low damping.
It was shown in the text that using only 3 modes is needed to give a suﬃciently good approximation. The
ﬁgure below shows the response spectra S
X
1
X
2
(ω), S
X
2
X
1
(ω), and S
X
2
X
2
(ω) when all 20 modes are used and
the oﬀdiagonal terms are kept. The Matlab code to generate the ﬁgure is provided next.
clear
rho=7830; E=200e9;
L=1; %1m length
N=20; %number of segments
n=3; %number of modes to be used
x=[0:L/N:L];
f1=@(x) rho*pi*(0.0010.0005*x).^2; %f1=rho*A
f2=@(x) rho*pi*(0.0010.0005*x).^2.*x %f2=rho*A*x
f3=@(x) x.^2; % used for the calculation of C
f4=@(x) 1./(pi*(0.0010.0005*x).^2); %f4=1/A
for i=1:N
m(i)=quad(f1, x(i),x(i+1));
xbar(i)=quad(f2, x(i),x(i+1))/m(i);
C(i)=quad(f3, x(i),x(i+1));
end
k(1)=E/quad(f4,0,xbar(1));
for i=2:N
k(i)=E/quad(f4, xbar(i1),xbar(i));
204
end
for i=1:N
for j=1:N
SFF(i,j)=C(i)*C(j);
end
end
%mass matrix
M=diag(m);
K=spdiags([[k(2:N) 0]’ k(1:N)’+[k(2:N) 0]’ k(1:N)’],1:1,N,N)*eye(N)
[P D]=eig(K,M); %MODESHAPES and NATURAL FREQUENCIES
H=zeros(N,N);
W=[3:0.01:3]’*1e4;
for j=1:length(W)
w=W(j);
for i=1:N
H(i,i)=1./(D(i,i)w.^2);
end
SQQ=P(:,1:n)’*SFF*P(:,1:n);
Sall=P*H*SQQ*H*P’; %When all modes are used
S=P(:,1:n)*H(1:n,1:n)*SQQ*H(1:n,1:n)*P(:,1:n)’; %only n modes are used
S11o(j)=sum(P(1,:).^2*diag(H).^2*diag(SQQ)); %low damping
S11a(j)=Sall(1,1); S12a(j)=Sall(1,2); S21a(j)=Sall(2,1); S22a(j)=Sall(2,2);
S11n(j)=S(1,1); S12n(j)=S(1,2); S21n(j)=S(2,1); S22n(j)=S(2,2); %
end
figure(1)
subplot(3,1,1)
plot(W,S12a)
axis([3e4,3e4,0 5e15])
subplot(3,1,2)
plot(W,S21a)
axis([3e4,3e4,0 5e15])
subplot(3,1,3)
plot(W,S22a)
axis([3e4,3e4,0 5e15])
205
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