3.4 Modal Vibrations

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Among the infinite number of relative geometric configurations the lumped masses in a multi-DoF

system could assume under free motion (i.e., with f ðtÞ ¼ 0), when excited by an arbitrary initial state,

there is a finite number of configurations that are naturally preferred by the system. Each of these

configurations will have an associated frequency of motion. These motions are termed modal motions.

By choosing the initial displacement y(0) proportional to a particular modal configuration, with zero

initial velocity, y_ð0Þ ¼ 0; that particular mode can be excited at the associated natural frequency

of motion. The displacements of different DoF retain this initial proportion at all times. This

constant proportion in displacement can be expressed as a vector c for that mode, and represents the

mode shape. Note that each modal motion is a harmonic motion executed at a specific frequency

v known as the natural frequency (undamped). In view of these general properties of modal motions,

m1 k2 (y2 − y1)

f1(t) f2(t)

y1

k1y1 k m2 2 (y2 − y1)

y2

k3y2

FIGURE 3.3 Free-body diagram of the two-DoF system.

3-10 Vibration and Shock Handbook

© 2005 by Taylor & Francis Group, LLC

they can be expressed by

y ¼ c cos vt ð3:10Þ

or, in the complex form, for ease of analysis, as

y ¼ c ejvt ð3:11Þ

When Equation 3.11 is substituted into the equation of unforced (free) motion,

My€ þ Ky ¼ 0 ð3:12Þ

as required by the definition of modal motion, the following eigenvalue problem results:

½v2M 2 K􀀉c ¼ 0 ð3:13Þ

For this reason, natural frequencies are sometimes called eigenfrequencies, and mode shapes are termed

eigenvectors. The feasibility of modal motions for a given system is determined by the existence of

nontrivial solutions for c (i.e., c – 0). Specifically, nontrivial solutions for c are possible if and only if

the determinant of the system of linear homogeneous equation 3.13 vanishes; thus

det½v2M 2 K􀀉 ¼ 0 ð3:14Þ

Equation 3.14 is known as the characteristic equation of the system. For an n-DoF system, M and K

are both n £ n matrices. It follows that the characteristic equation has n roots for v2: For physically

realizable systems these n roots are all nonnegative and they yield the n natural frequencies v1; v2; …; vn of

the system. For each natural frequency vi; when substituted into Equation 3.13 and solved for c, there

Box 3.3

MODEL TYPES

Linear Nonlinear

Coupled second-order equations

My€ þ Cy_ þ Ky ¼ f ðtÞ My€ ¼ hðy; y; fðtÞ)

Response vector:

y ¼ ½y1; y2; …; yp 􀀉T ; p ¼ number of DoF

Excitation vector: f ðtÞ ¼ ½f1 ; f2; …; fp􀀉T

M ¼ mass matrix

C or B ¼ damping matrix

K ¼ stiffness matrix

Coupled first-order equations (state-space models)

x_ ¼ Ax þ Bu x_ ¼ aðx; uÞ

y ¼ Cx y ¼ yðxÞ

State vector x ¼ ½x1 ; x2 ; …; xn 􀀉T ; n ¼ order of the system

Input (excitation) vector u ¼ ½u1 ; u2 ; …; um 􀀉T

Output (response) vector y ¼ ½y1 ; y2; …; yp 􀀉T

Notes:

1. Definition of state: If xðt0Þ; and u from t0 to t1; are known, xðt1 Þ can be determined

completely

2. State vector x contains a minimum number (n) of elements

3. State model is not unique (different state models are possible for the same system)

4. One approach to obtaining a state model is to use x ¼

"

y

y_

#

Modal Analysis 3-11

© 2005 by Taylor & Francis Group, LLC

results a mode shape vector ci that determines up to one unknown parameter which can be used as a

scaling parameter. Extra care should be exercised, however, when determining mode shapes for zero

natural frequencies (i.e., rigid-body modes) and repeated natural frequencies (i.e., for systems with a

dynamic symmetry). We shall return to these considerations in later sections.

Example 3.3

Consider a mechanical system modeled as

in Figure 3.4. This is obtained from the

systems in Figure 3.2 by setting m1 ¼ m;

m2 ¼ am; k1 ¼ k; k2 ¼ bk; and k3 ¼ 0: The

corresponding mass matrix and the stiffness

matrix are

M ¼

m 0

0 am

" #

; K ¼

ð1 þ bÞk 2bk

2bk bk

" #

The natural frequencies are given by the roots of the characteristic equation

det

v2m 2 ð1 þ bÞk bk

bk v2am 2 bk

" #

¼ 0

By expanding the determinant, this can be expressed as

½v2m 2 ð1 þ bÞk􀀉½v2am 2 bk􀀉 2 b2k2 ¼ 0

or

v4am2 2 v2km½b þ að1 þ bÞ􀀉 þ bk2 ¼ 0

Let us define a frequency parameter v0 ¼

ffiffiffiffiffi

k=m p : This parameter is identified as the natural frequency of

an undamped simple oscillator (single-DoF mass – spring system) with mass m and stiffness k:

Consequently, the characteristic equation of the given 2 DoF system can be written as

a

v

v0

􀀏 􀀐4

2 ða þ b þ abÞ

v

v0

􀀏 􀀐2

þ b ¼ 0

whose roots are

v1

v0

􀀏 􀀐2

;

v2

v0

􀀏 􀀐2

¼

1

2a

{a þ b þ ab} 1 7

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 2 4ab

ðaþbþabÞ2

􀀘 q 􀀙

The mode shapes are obtained by solving for c in

v2m 2 ð1 þ bÞk bk

bk v2am 2 bk

" #

c ¼ 0

or

v

v0

􀀏 􀀐2

2ð1 þ bÞ b

b a

v

v0

􀀏 􀀐2

2b

2

66664

3

77775

c ¼ 0

In a mode shape vector, only the ratio of the elements is needed. This is because, in determining a mode

shape, we are concerned about the relative motions of the lumped masses, not the absolute motions.

βk

αm

k

m

Mass 1 Mass 2

FIGURE 3.4 A modal vibration example.

3-12 Vibration and Shock Handbook

© 2005 by Taylor & Francis Group, LLC

From the above equation, it is clear that this ratio is given by

c2

c1 ¼

ð1 þ bÞ 2

v

v0

􀀏 􀀐2

b ¼

b

b 2 a

v

v0

􀀏 􀀐2

which is evaluated by substituting the appropriate value for ðv=v0Þ; depending on the mode, into any one

of the right-hand-side expressions above.

The dependence of the natural frequencies on the parameters a and b is illustrated by the curves in

Figure 3.5. Some representative values of the natural frequencies and mode shape ratios are listed in

Table 3.2.

Note that, when b ¼ 0; the spring connecting the two masses does not exist and the system reduces

to two separate systems: a simple oscillator of natural frequency v0 and a single mass particle (of zero

natural frequency). It is clear that in this case v1=v0 ¼ 0 and v2=v0 ¼ 1: This fact can be established

from the expressions for natural frequencies of the original system by setting b ¼ 0: The mode

corresponding to v1=v0 ¼ 0 is a rigid-body mode in which the free mass moves indefinitely (zero

frequency) and the other mass (restrained mass) stands still. It follows that the mode shape ratio

ðc2=c1Þ1 ! 1: In the second mass, the free mass stands still and the restrained mass moves. Hence,

ðc2=c1Þ1 ¼ 0: These results are also obtained from the general expressions for the mode shape ratios

of the original system.

TABLE 3.2 The Dependence of Natural Frequencies and Mode Shapes on Inertia and Stiffness

a 0.5 1.0 2.0

b v1 =v0 v2=v0 ðc2=c1 Þ1 ðc2 =c1 Þ2 v1=v0 v2 =v0 ðc2 =c1Þ1 ðc2 =c1 Þ2 v1 =v0 v2 =v0 ðc2 =c1 Þ1 ðc2 =c1 Þ2

0 0 1.0 1 0 0 1.0 1 0 0 1.0 1 0

0.5 0.71 1.41 2.0 2 1.0 0.54 1.31 2.41 2 0.41 0.40 1.26 2.69 2 0.19

1.0 0.77 1.85 1.41 2 1.41 0.62 1.62 1.62 2 0.62 0.47 1.51 1.78 2 0.28

2.0 0.79 2.52 1.19 2 1.69 0.66 2.14 1.28 2 0.78 0.52 1.93 1.37 2 0.37

5.0 0.81 3.92 1.07 2 1.87 0.69 3.24 1.10 2 0.91 0.55 2.86 1.14 2 0.44

1 0.82 1 1.0 2 2.0 0.71 1 1.0 2 1.0 0.57 1 1.0 2 0.5

Mass

Ratio

a

Nondimensional

Frequency

w

w 0

w1 Curves (Mode 1)

w2 Curves (Mode 2)

Stiffness Ratio b

5.0

4.0

3.0

2.0

1.0

0.8

0.6

0.4

0.0

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

0.1

0.2

0.5

0.7

1.0

2.0

10.0

0.1

0.2

0.5

0.7

1.0

2.0

0.2 10.0

a

FIGURE 3.5 Dependence of natural frequencies ðv=v0 Þ on mass ratio ðaÞ and stiffness ratio ðbÞ:

Modal Analysis 3-13

© 2005 by Taylor & Francis Group, LLC

When b ! 1; the spring connecting the two masses becomes rigid and the two masses act as a single

mass ð1 þ aÞm restrained by a spring of stiffness k: This simple oscillator has a squared natural frequency

of v20

=ð1 þ aÞ: This is considered the smaller natural frequency of the corresponding system: ðv1=v0Þ2 ¼

1=ð1 þ aÞ: The larger natural frequency v2 approaches 1 in this case and it corresponds to the natural

frequency of a massless spring. These limiting results can be derived from the general expressions for the

natural frequencies of the original system by using the fact that for small lxl p 1; the expression

ffiffiffiffiffiffiffi

1 2 x p

is approximately equal to 1 2 ð1=2Þx: (Proof: Use the Taylor series expansion.) In the first mode, the

two masses move as one unit and hence the mode shape ratio ðc2=c1Þ1 ¼ 1: In the second mode,

the two masses move in opposite directions restrained by an infinitely stiff spring. This is considered the

static mode which results from the redundant situation of associating two DoF to a system that actually

has only one lumped mass ð1 þ aÞm: In this case, the mode shape ratio is obtained from the general result

as follows: For large b; we can neglect a in comparison. Hence,

v2

v0

􀀏 􀀐2

¼

1

2a

{b þ ab}{1 þ 1} ¼ ð1 þ aÞ

b

a

By substituting this result in the expression for the mode shape ratio, we obtain

c2

c1

􀀏 􀀐

2¼

b

b 2 a

v2

v0

􀀏 􀀐2 ¼

b

b 2 að1 þ aÞ

b

a

¼ 2

1

a

Finally, consider the case a ¼ 0 (with b – 0). In this case, only one mass m restrained by a spring of

stiffness k is present. The spring of stiffness bk has an open end. The first mode corresponds to a simple

oscillator of natural frequency v0: Hence, v1=v0 ¼ 1: The open end has the same displacement as the

point mass. Consequently, ðc2=c1Þ1 ¼ 1: These results can be derived from the general expressions for the

original system. In the second mode the simple oscillator stands still and the open-ended spring oscillates

(at an infinite frequency). Hence v2=v0 ¼ 1; and this again corresponds to a static mode situation which

arises by assigning two DoF to a system that has only one DoF associated with its inertia elements. Since

the lumped mass stands still, we have ðc2=c1Þ2 ¼ 1:

Note that, when a ¼ 0 and b ¼ 0, the system reduces to a simple oscillator and the second mode is

completely undefined. Hence, the corresponding results cannot be derived from the general results for

the original system.

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