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
+ mω
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) /ω
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
+ mω
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
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