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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