Vibrations of Structures

Module I: Vibrations of Strings and Bars

Lesson 5: Modal Analysis - I

Contents:

1. Introduction

2. The Eigenvalue Problem

3. Modal Analysis: Examples

Keywords: Modal analysis, Properties of modes, String modes, Hanging

chain modes

Modal Analysis - I

1 Introduction

The determination of the natural frequencies and modes of vibration of a

system is known as modal analysis. A mode of vibration is characterized by

vibration at a particular frequency (natural frequency) of all points of the

system. Thus, we search for special solutions of the field variable of the form

w(x, t) = W (x)eiωt (1)

where W (x) is the amplitude function (possibly complex) and ω is the circular

frequency. Here we choose to represent the solution in complex form with

the understanding that the actual solution is obtained by taking a linear

combination of the real and imaginary parts of the complex solution (1).

In most cases W (x) is real, and a modal solution is of the form

w(x, t) = (C cos ωt + S sin ωt)W (x). (2)

Some of the general features of such a modal solution that follow immediately

from (2) are

• It is separable in space and time

• All points of the system pass through the equilibrium configuration

(w(x, t) = 0) at the same time

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• If there are points at which W (x) = 0, then such points (nodes) remain

at the equilibrium location at all times

• The phase difference between the motions of any two points at x = x1

and x = x2 of a system vibrating in a mode is a constant, which is either

0 (for W (x1)W (x2) > 0) or π (for W (x1)W (x2) < 0).

2 Modal Analysis: Examples

Uniform taut string:

Equation of motion and boundary conditions:

ρAw,tt − Tw,xx = 0, w(0, t) = 0, w(l, t) = 0. (3)

Substituting w(x, t) = W (x)eiωt in (3), we obtain the eigenvalue problem as

W ′′ +ω2

c2W = 0, W (0) = 0, W (l) = 0. (4)

The general solution of the differential equation (4) is given by

W (x) = D cosωx

c+ H sin

ωx

c. (5)

Using the boundary conditions in (4), one obtains

1 0

cosωl

csin

ωl

c

{

DH

}

= 0. (6)

3

0.5 1

-2

2

U1(x)

x/l

Mode 1: ω1 = πc/l rad/s

0.5 1

-2

2

U2(x)

x/l

Mode 2: ω2 = 2πc/l rad/s

Figure 1: First two modes of vibration of a uniform taut string

For non-trivial solutions of (D, H), we require the determinant of the matrix

in (6) to vanish, i.e., sin ωl/c = 0. This is the characteristic equation of the

string. The solutions of the characteristic equation

ωn = nπc

l, n = 1, 2, . . . ,∞ (7)

are the circular eigenfrequencies/natural frequencies of the string. The cor-

responding eigenfunctions are obtained by solving for (Dn, Hn) from (6) and

substituting in (5) as Wn(x) = sinωnx/c = sin nπx/l, n = 1, 2, . . . ,∞. The

modes of free vibrations of the string are determined by these eigenfunctions,

and are shown in Fig. 1 for the first two modes.

The modal solutions are thus of the form w(x, t) = Wn(x)eiωnt, n =

1, 2, . . . ,∞. These solutions are all linearly independent, i.e., any particular

modal solution (for a given n) cannot be expressed as a linear combination

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of the remaining other modal solutions. Using the linear superposition prin-

ciple, the general solution of the free vibration problem of a uniform taut

string can be written as

w(x, t) =

∞∑

n=1

(Cn cosωnt + Sn sinωnt) sinnπx

l(8)

where Cn and Sn are arbitrary constants which may be determined from the

initial conditions.

Uniform hanging string/chain:

The equation of motion and boundary conditions are given by

w,tt − g[(l − x)w,x],x = 0 w(0, t) = 0, w(l, t) < ∞. (9)

Substituting w(x, t) = W (x)eiωt, we obtain the eigenvalue problem

ω2W + g[(l − x)W ′]′ = 0, W (0) = 0, W (l) < ∞ (10)

The differential equation in (10) can be recast into a familiar form using

the transformation

s(x) = 2ω

√

l − x

g. (11)

Defining W̃ (s) such that W̃ (s(x)) = W (x), one obtains using the chain rule

of differentiation

dW

dx=

dW̃

ds

ds

dx= −W̃ ′

ω√

g(l − x)(12)

d2W

dx2=

d2W̃

ds2

(

ds

dx

)2

+dW̃

ds

d2s

dx2= W̃ ′′

ω2

g(l − x)− W̃ ′

ω

2√

g(l − x)3(13)

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2 4 6 8 10

-1.2

-0.6

0.6

1.2

2 4 6 8 10

-0.6

-0.3

0.3

0.6

s s

J0(s) Y0(s)

Figure 2: Bessel functions J0(s), and Y0(s)

Using (12) and (13) in (10) yields on simplification

W̃ ′′ +1

sW̃ ′ + W̃ = 0, W̃ (2ω

√

l/g) = 0, W̃ (0) < ∞ (14)

where s ∈ [0, 2ω√

l/g]. The differential equation in (14) is a special case of

the Bessel differential equation

y′′(x) +1

xy′(x) +

(

1 −n2

x2

)

y(x) = 0

with n = 0. Therefore, the general solution of (14) can be written as

W̃ (s) = DJ0(s) + EY0(s), (15)

where D and E are arbitrary constants, and J0(s) and Y0(s) are known as,

respectively, zeroth order Bessel function of the first and second kind (or,

Neumann function). The functions J0(s) and Y0(s) are plotted in Fig. 2.

Since Y0(s) → −∞ as s → 0 (i.e., x → l), the condition of finiteness of

the solution at the free end implies E = 0. Therefore, the solution of (14)

takes the form W̃ (s) = DJ0(s). The boundary condition of the fixed end,

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W (s(0)) = 0, gives the characteristic equation of the hanging string as

J0(2ω√

l/g) = 0. (16)

The roots of J0(γk) = 0, yield the eigenfrequencies

ωk =γk

2

√

g

l, k = 1, 2, . . . ,∞. (17)

The first three values of γk are γ1 ≈ 2.4048, γ2 ≈ 5.5201, γ3 ≈ 8.6537. The

eigenfunctions can now be written as

Wk(x) = J0

(

2ωk

√

l − x

g

)

, k = 1, 2, . . . ,∞. (18)

All these eigenfunctions are linearly independent. The first three mode shapes

of free vibrations of a uniform hanging string, as determined by the first three

eigenfunctions, are shown in Fig. 3.

The general solution of the free vibration problem of a uniform hanging

string/chain can be written using the linear superposition principle as

w(x, t) =

∞∑

k=1

[

(Ck cos ωkt + Sk sin ωkt)J0

(

2ωk

√

l − x

g

)]

. (19)

where Ck and Sk are arbitrary constants which may be determined from the

initial conditions.

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ω1 = 1.20√

g/l ω2 = 2.76√

g/l ω3 = 4.32√

g/l

Figure 3: First three mode shapes of a hanging string

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