 8 Continuous System Vibration
Section 8.1: Continuous Systems
8.1. A transverse beam is supported by torsional springs at both ends with each spring having stiﬀness
k. Find the equation of motion and corresponding boundary conditions.
Solution: The boundary conditions are obtained by drawing a free body diagram. Let us deﬁne y be
positive upward and x be coordinate along the beam deﬁned positive to the right. Then, the slope, moment,
and shear are given by
slope = y
moment = EIy

shear = (EIy

)
,
and deﬁned positive as shown in Figure 29. Due to positive slope y
, the torsional spring on the left end
y'
M=EIy''
Q=(EIy'')'
Figure 29: Deﬁnition of positive slope, moment and shear.
y'
ky'
Figure 30: Moment due to torsional spring
exerts moment with magnitude ky
as shown in Figure 30. Then, we can write
EIy

(0, t) = ky
(0, t) .
Similarly, for the right end, we can write
EIy

(L, t) = ky
(L, t) .
228 Section 8.2: Sturm-Liouville Eigenvalue Problem
8.2. A string is ﬁxed at one end and is free to move up and down at the other end. Obtain the
eigenfunctions and eigenvalues.
Solution: The equation of motion and boundary conditions are given by
T y

m¨y = 0 for 0 < x < L
y (0, t) = 0
y
(L, t) = 0.
Assuming y (x, t) = Y (X) f (t) , we can write
T Y

+
2
Y = 0
Y (0) = 0
Y
(L) = 0.
The i
th
eigenfunction is given by
Y
i
= C
i
sin β
i
x,
where
β
i
L =
(2i 1)
2
π
ω
i
= β
i
3
T
m
.
The eigenfunctions can be normalized with respect to m such that
L
0
mY
2
i
dx = 1
C
i
=
3
2
mL
.
The ﬁgure below shows the ﬁrst three mode shapes.
0 0.1 0.2 0.3 0.4
0.5
0.6
0.7
0.8
0.9
1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Y x
( )
x/L
i
= 1
i
= 2
i
= 3
m
L
2
i
Eigenfunctions for Problem 2. The ﬁrst three mode shapes of a ﬁxed-free string.
229 8.3. Prove the orthogonality condition for the string in Equation 8.12.
Solution: In the text, it was shown that
L
0
m (x) Y
n
(x)Y
r
(x)dx = δ
nr
.
Replacing m (x) Y
n
(x) by T Y

n
(x)
2
n
(spatial equation of motion) we can write
L
0
T Y

n
(x)Y
r
(x)dx = ω
2
n
δ
nr
.
The left hand side can be integrated so that
T Y
n
(x) Y
r
(x) |
L
0
L
0
T Y
n
(x)Y
r
(x)dx = ω
2
n
δ
nr
.
The term evaluated at the boundaries is reduced to zero, and we have
L
0
T Y
n
(x)Y
r
(x)dx = ω
2
n
δ
nr
.
230 8.4. Consider a uniform rod in axial vibration with mass per unit length m, length L, and axial stiﬀness
EA. The one end at x = 0 is attached to a spring with stiﬀness k and is free at x = L. Obtain the
eigenfunctions and eigenvalues. Use k = EA/L.
Solution: The spatial solution must satisfy
EAY

+
2
Y = 0 for 0 < x < L
EAY
(0) kY (0) = 0
Y
(L) = 0.
Then, the solution takes the form of
Y (x) = c
1
sin βx + c
2
cos βx,
where β =
m/EAω. From the ﬁrst boundary condition, we obtain
c
2
=
EA
k
βc
1
.
From the second boundary condition, we obtain the characteristic equation given by
β cos βL
EA
k
β
2
sin βL = 0.
If k = EA/L, the characteristic equation is reduced to
cos βL βL sin βL = 0,
which yields
β
i
L = 0.86, 3.43, 6.44, ··· .
The corresponding eigenfunction is given by
Y
i
(x) = c
i
sin β
i
L
x
L
+ β
i
L cos β
i
L
x
L
.
Using the characteristic equation, this expression can be rewritten as
Y
i
(x) = d
i
cos
β
i
L
x
L
1

,
where d
i
= c
i
/ sin β
i
L. The constant coeﬃcient d
i
is determined by normalizing the eigenfunctions and is
given by
d
i
=
3
2
mL
1
1 + sin β
i
L
.
The next ﬁgure shows the ﬁrst three eigenfunctions.
231 0 0.1 0.2 0.3 0.4
0.5
0.6
0.7
0.8
0.9
1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Y x
( )
x/L
i
= 1
i
= 2
i
= 3
m
L
2
i
Eigenfunctions for Problem 4.
232

Get Probabilistic Models for Dynamical Systems, 2nd Edition now with O’Reilly online learning.

O’Reilly members experience live online training, plus books, videos, and digital content from 200+ publishers.