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3.6 Static Modes and Rigid-Body Modes

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3.6.1 Static Modes

Modes corresponding to infinite natural frequencies are known as static modes. For these modes, the

modal mass is zero; in the normalization process with respect to M static modes cannot be included. If we

assign a DoF for a massless location, the resulting mass matrix M becomes singular ðdet M ¼ 0Þ and a

static mode arises. We have come across two similar situations in a previous example; in one case the

stiffness of the spring connecting two masses is made infinite so that they act as a single mass in the limit,

and in the other case one of the two masses is made equal to zero so that only one mass is left. We should

take extra precautions to avoid such situations by using proper modeling practices; the presence of a

static mode amounts to assigning a DoF to a system that it does not actually possess. In a static mode, the

system behaves like a simple massless spring.

In the literature of experimental modal analysis, the static modes are represented by a residual

flexibility term in the transfer functions. Note that, in this case, modes of natural frequencies that are

(y1+y2)

m 1 m

y1 y2

k k

Centroid

FIGURE 3.6 A simplified vehicle model for heave and

pitch motions.

Modal Analysis 3-15

higher than the analysis bandwidth or the maximum frequency of interest are considered static modes.

Such issues of experimental modal analysis will be discussed in Chapter 18.

3.6.2 Linear Independence of Modal Vectors

In the absence of static modes (i.e., modal masses Mi – 0), the inertia matrix M will be nonsingular.

Then the orthogonality condition 3.19 implies that the modal vectors are linearly independent, and

consequently, they will form a basis for the n-dimensional space of all possible displacement trajectories y

for the system. Any vector in this configuration space (or displacement space), therefore, can be expressed

as a linear combination of the modal vectors.

Note that we have assumed in the earlier development that the natural frequencies are distinct

(or unequal). Nevertheless, linearly independent modal vectors are possessed by modes of equal natural

frequencies as well. An example is the situation where these modes are physically uncoupled. These

modal vectors are not unique, however; there will be arbitrary elements in the modal vector equal in

number to the repeated natural frequencies. Any linear combination of these modal vectors can also serve

as a modal vector at the same natural frequency. To explain this point further, without loss of generality

suppose that v1 ¼ v2: Then, from Equation 3.15, we have

v21

Mc1 2 Kc1 ¼ 0

v21

Mc2 2 Kc2 ¼ 0

Multiply the first equation by a; the second equation by b, and add the resulting equations. We get

v21

Mðac1 þ bc2Þ 2 Kðac1 þ bc2Þ ¼ 0

This verifies that any linear combination ac1 þ bc2 of the two modal vectors c1 and c2 will also serve

as a modal vector for the natural frequency v1: The physical significance of this phenomenon should be

clear in Example 3.4.

3.6.3 Modal Stiffness and Normalized Modal Vectors

It is possible to establish an alternative version of the orthogonality condition given as Equation 3.19 by

substituting it into Equation 3.18. This gives

i Kcj ¼

0 for i – j

Ki for i ¼ j

(

ð3:21Þ

This condition is termed K-orthogonality.

Since the M-orthogonality condition (Equation 3.19) is true even for the case of repeated natural

frequencies, it should be clear that the K-orthogonality condition (Equation 3.21) is also true, in general,

even with repeated natural frequencies. The newly defined parameter Ki represents the value of cT

i Kci

and is known as the generalized stiffness or modal stiffness corresponding to the ith mode.

Another useful way to normalize modal vectors is to choose their unknown parameters so that all

modal stiffnesses are unity (Ki ¼ 1 for all i). This process is known as normalization with respect to the

stiffness matrix. The resulting normal modes are said to be K-normal. These normal modes are still

arbitrary up to a multiplier of 2 1. This can be eliminated by assigning positive values to the first element

of all modal vectors.

Note that it is not possible to normalize a modal vector simultaneously with respect to both M and K,

in general. To understand this further, we may observe that v2i

¼ Ki=Mi and consequently we are unable

to pick both Ki and Mi arbitrarily. In particular, for the M-normal case Ki ¼ v2i

and for the K-normal

case Mi ¼ 1=v2i

3-16 Vibration and Shock Handbook

3.6.4 Rigid-Body Modes

Rigid-body modes are those for which the natural frequency is zero. Modal stiffness is zero for rigid-body

modes, and as a result it is not possible to normalize these modes with respect to the stiffness matrix.

Note that when rigid-body modes are present the stiffness matrix becomes singular ðdet K ¼ 0Þ:

Physically, removal of a spring that connects two DoF results in a rigid-body mode. In Example 3.3 we

came across a similar situation. In experimental modal analysis applications, low-stiffness connections or

restraints, which might be present in a test object, could result in approximate rigid-body modes that

would become prominent at low frequencies.

Some important results of modal analysis that we have discussed thus far are summarized in Table 3.3.

Example 3.4

Consider a light rod of length l having equal masses m attached to its ends. Each end is supported by a spring

of stiffness k as shown in Figure 3.6. Note that this system may represent a simplified model of a vehicle in

heave and pitch motions. Gravity effects can be eliminated by measuring the displacements y1 and y2 of the

two masses about their respective static equilibrium positions. Assume small front-to-back rotations u in

the pitch motion and small up-and-down displacements ð1=2Þðy1 þ y2Þ of the centroid in its heave motion.

3.6.4.1 Equation of Heave Motion

From Newton’s Second Law for rigid-body motion, we get

2 ðy€1 þ y€2Þ ¼ 2ky1 2 ky2

3.6.4.2 Equation of Pitch Motion

Note that, for small angles of rotation, u ¼ ðy1 2 y2Þ=l: The moment of inertia of the system about the

centroid is 2mðl=2Þ2 ¼ ð1=2Þml2: Hence, by Newton’s Second Law for rigid-body rotation, we have

ml2 y€1 2 y€2

􀀏 􀀐

¼ 2

ky1 þ

ky2

These two equations of motion can be written as

y€1 þ y€2 þv20

ðy1 þ y2Þ ¼ 0

y€1 2 y€2 þv20

ðy1 2 y2Þ ¼ 0

TABLE 3.3 Some Important Results of Modal Analysis

System My€ þ Ky ¼ f ðtÞ

Symmetry MT ¼ M and KT ¼ K

Modal problem ½v2 M 2 K􀀉c ¼ 0

Characteristic equation (gives natural frequencies) det½v2 M 2 K􀀉 ¼ 0

M-orthogonality cTi

Mcj ¼

0 for i – j

Mi for i ¼ j

(

K-orthogonality cTi

Kcj ¼

0 for i – j

Ki for i ¼ j

(

Modal mass (generalized mass) Mi

Modal stiffness (generalized stiffness) Ki

Natural frequency vi ¼

ffiffiffiffiffiffiffi

Ki =Mi p

M-normal case Mi ¼ 1; Ki ¼ v2i

K-normal case Ki ¼ 1; Mi ¼ 1=v2i

Presence of rigid-body modes det K ¼ 0; Ki ¼ 0; and vi ¼ 0

Presence of static modes det M ¼ 0; Mi ¼ 0; and vi ! 1

Modal Analysis 3-17

in which v0 ¼

ffiffiffiffiffi

k=m p : By straightforward algebraic manipulation, a pair of completely uncoupled

equations of motion are obtained; thus

y€1 þv20

y1 ¼ 0

y€2 þv20

y2 ¼ 0

It follows that the resulting mass matrix and the stiffness matrix are both diagonal. In this case, there is

an infinite number of choices for mode shapes, and any two linearly independent second-order vectors

can serve as modal vectors for the system. Two particular choices are shown in Figure 3.7. Any of these

mode shapes will correspond to the same natural frequency v0:

In each of these two choices, the mode shapes have been chosen so that they are orthogonal with

respect to both M and K. This fact is verified below. Note that, in the present example

M ¼

1 0

0 1

" #

and K ¼

v20

0 v20

" #

For Case 1:

½1 1􀀉M

" #

¼ 0 and ½1 1􀀉K

" #

¼ 0

For Case 2:

½1 0􀀉M

" #

¼ 0 and ½1 0􀀉K

" #

¼ 0

In general, since both elements of each eigenvector can be picked arbitrarily, we can write

c1 ¼

" #

and c2 ¼

" #

where a and b are arbitrary, limited only by the orthogonality requirement for c1 and c2: The

M-orthogonality requires

½ 1 a 􀀉

1 0

0 1

" #

¼ 0

and K-orthogonality requires

½ 1 a 􀀉

v20

0 v20

" #

¼ 0

Mode 1

Mode 2

Case 1







= 









−

Case 2







= 







= 

y2 y2

FIGURE 3.7 Two possibilities of mode shapes for the symmetric heave – pitch vehicle.

3-18 Vibration and Shock Handbook

Both conditions give 1 þ ab ¼ 0; which corresponds to ab ¼ 21: Note that Case 1 corresponds to a ¼ 1

and b ¼ 21 and Case 2 corresponds to a ¼ 0 and b ! 1: More generally, we can pick as modal vectors

c1 ¼

" #

and c2 ¼

21=a

" #

such that the two mode shapes are both M-orthogonal and K-orthogonal. In fact, if this particular system

is excited by an arbitrary initial displacement, it will undergo free vibrations at frequency v0 while

maintaining the initial displacement ratio. Hence, if M-orthogonality and K-orthogonality are not

required, any arbitrary second-order vector may serve as a modal vector to this system.

Example 3.5

An example for a system possessing a rigidbody

mode is shown in Figure 3.8. This system,

a crude model of a two-car train, can be derived

from the system shown in Figure 3.4 by

removing the end spring (inertia restraint)

and setting a ¼ 1 and b ¼ 1: The equation

for unforced motion of this system is

m 0

0 m

" #

y€1

y€2

" #

k 2k

2k k

" #

Note that det M ¼ m2 – 0 and hence the system does not possess static modes. This should also be

obvious from the fact that each DoF (y1 and y2) has an associated, independent mass element. On the

other hand, det K ¼ k2 2 k2 ¼ 0 which signals the presence of rigid-body modes.

The characteristic equation of the system is

det

v2m 2k k

k v2m 2 k

" #

¼ 0

ðv2m 2 kÞ2 2 k2 ¼ 0

The two natural frequencies are given by the roots: v1 ¼ 0 and v2 ¼

ffiffiffiffiffiffi

2k=m p : Note that the zero natural

frequency corresponds to the rigid body mode. The mode shapes can reveal further interesting facts.

3.6.4.3 First Mode (Rigid-Body Mode)

In this case, we have v ¼ 0: Consequently, from Equation 3.15, the mode shape is given by

2k k

k 2k

" #

which has the general solution c1 ¼ c2; or

" #

1¼

" #

The parameter a can be chosen arbitrarily. The corresponding modal mass is

M1 ¼ ½a a􀀉

m 0

0 m

" #

¼ 2ma2

FIGURE 3.8 A simplified model of a two-car train.

Modal Analysis 3-19

If the modal vector is normalized with respect to M, we have M1 ¼ 2ma2 ¼ 1: Then, a ¼ ^1=

ffiffiffiffi

2m p and

the corresponding normal mode vector would be

" #

1¼

1ffiffiffiffi

2m p

1ffiffiffiffi

2m p

6664

7775

1ffiffiffiffi

2m p

1ffiffiffiffi

2m p

6664

7775

which is arbitrary up to a multiplier of 2 1. If the first element of the normal mode is restricted to be

positive, the former vector (one with positive elements) should be used.

We have already noted that it is not possible to normalize a rigid-body mode with respect to K.

Specifically, the modal stiffness for the rigid-body mode is

K1 ¼ ½a a􀀉

k 2k

2k k

" #

¼ 0

for any choice for a; as expected.

3.6.4.4 Second Mode

For this mode, v2 ¼

ffiffiffiffiffiffi

2k=m p : By substituting into Equation 3.15 we get

k k

" #

2¼

" #

the solution of which gives the corresponding modal vector (mode shape).

The general solution is c2 ¼ 2c1; or

" #

2¼

" #

in which a is arbitrary. The corresponding modal mass is given by

M2 ¼ ½ a 2a 􀀉

m 0

0 m

" #

¼ 2ma2

and the modal stiffness is given

K2 ¼ ½ a 2a 􀀉

k 2k

2k k

" #

¼ 4ka2

Then, for M-normality we must have 2ma2 ¼ 1 or a ¼ ^1=

ffiffiffiffi

2m p :

It follows that the M-normal mode vector would be

" #

1¼

1ffiffiffiffi

2m p

1ffiffiffiffi

2m p

6664

7775

1ffiffiffiffi

2m p

1ffiffiffiffi

2m p

6664

7775

The corresponding value of the modal stiffness is K2 ¼ 2k=m; which is equal to v22

; as expected. Similarly,

for K-normality we must have 4Ka2 ¼ 1; or a ¼ ^1=

ffiffiffiffi

4K p : Hence, the K-normal modal vector would be

" #

2¼

1ffiffiffi

4k p

1ffiffiffi

4k p

6664

7775

1ffiffiffi

4k p

1ffiffiffi

4k p

6664

7775

The corresponding value of the modal mass is M2 ¼ m=ð2kÞ which is equal to 1=v22

; as expected.

3-20 Vibration and Shock Handbook

The mode shapes of the system are shown in

Figure 3.9. Note that in the rigid-body mode

both masses move in the same direction through

the same distance, with the connecting spring

maintained in the unstretched configuration. In

the second mode, the two masses move in

opposite directions with equal amplitudes. This

results in a node point halfway along the spring. A

node is a point in the system that remains

stationary under a modal motion. It follows that,

in the second mode, the system behaves like an

identical pair of simple oscillators, each possessing

twice the stiffness of the original spring

(see Figure 3.10). The corresponding natural

frequency is

ffiffiffiffiffiffi

2k=m p ; which is equal to v2:

Orthogonality of the two modes may be verified

with respect to the mass matrix as

½1 1􀀉

m 0

0 m

" #

¼ 0

and, with respect to the stiffness matrix, as

½1 1􀀉

k 2k

2k k

" #

¼ 0

Since K is singular, due to the presence of the rigid-body mode, the first orthogonality condition

(Equation 3.19), and not the second (Equation 3.21), is the useful result for this system. In particular,

since M is nonsingular, the orthogonality of the modal vectors with respect to the mass matrix implies

that they are linearly independent vectors by themselves. This is further verified by the nonsingularity of

the modal matrix; specifically

det½ c1; c2 􀀉 ¼ det

1 1

1 21

" #

– 0

Since M is a scalar multiple of the identity matrix, we note that the modal vectors are in fact orthogonal,

as is clear from

1 c2 ¼ ½1 1􀀉

" #

¼ 0

3.6.5 Modal Matrix

An n-DoF system has n modal vectors c1; c2; …; cn; which are independent. The n £ n square matrix C

whose columns are the modal vectors is known as the modal matrix

C ¼ ½c1; c2; …; cn􀀉 ð3:22Þ

Since the mass matrix M can always be made nonsingular through proper modeling practices

(in choosing the DoF), it can be concluded that the modal matrix is nonsingular

det C – 0 ð3:23Þ

Mode 1 Mode 2

Node

FIGURE 3.9 Mode shapes of the two-car train

example.

Node

FIGURE 3.10 Equivalent system for mode 2 of the

two-car train example.

Modal Analysis 3-21

and the inverse C21 exists. Before showing this fact, note that the orthogonality conditions (Equation

3.19 and Equation 3.21) can be written in terms of the modal matrix as

CTMC ¼ diag½M1;M2;…;Mn􀀉 ¼ M􀀊 ð3:24Þ

CTKC ¼ diag½K1; K2; …; Kn􀀉 ¼ K􀀊 ð3:25Þ

in which M􀀊 and K􀀊 are the diagonal matrices of modal masses and modal stiffnesses, respectively.

Next, we use the result from linear algebra, which states that the determinant of the product of two

square matrices is equal to the product of the determinants. Also, a square matrix and its transpose have

the same determinant. Then, by taking the determinant of both sides of Equation 3.24, it follows that

det CTMC ¼ ðdet CÞ2det M ¼ det M􀀊 ¼ M1;M2;…;Mn ð3:26Þ

Here, we have also used the fact that in Equation 3.24 the RHS matrix is diagonal. Now, Mi – 0 for all i

since there are no static modes in a well-posed modal problem. It follows that

det C – 0 ð3:27Þ

which implies that C is nonsingular.

3.6.6 Configuration Space and State Space

All solutions of the displacement response y span a Euclidean space known as the configuration space.

This is an n-Euclidean space ðLnÞ: This is also the displacement space.

The trace of the displacement vector y is not a complete representation of the dynamic response of

a vibrating system because the same y can correspond to more than one dynamic state of the system.

Hence, y is not a state vector. However,

" #

is a state vector, because it includes both displacement and velocity and completely represents the state of

the system. This state vector spans the state space ðL2nÞ which is a 2n-Euclidean space.

3.6.6.1 State Vector

This is a vector x consisting of a minimal set of response variables of a dynamic system such that, with

knowledge of the initial state xðt0Þ and the subsequent input u½t0; t1􀀉 to the system over a finite

time interval ½t0; t1􀀉; the end state xðt1Þ can be uniquely determined. Each point in a state space uniquely

(and completely) determines the state of the dynamic system under these conditions.

Note: Configuration space can be thought of as a subspace of the state space, which is obtained by

projecting the state space into the subspace formed by the axes of the y vector.

For an n-DoF vibrating system (see Equation 3.1), the displacement response vector y is of order n. If

we know the initial condition y(0) and the forcing excitation fðtÞ; it is not possible to completely

determine yðtÞ in general. However, if we know y(0) and y_ð0Þ as well as fðtÞ; then it is possible to

completely determine yðtÞ and y_ðtÞ: This says what we have noted before; y alone does not constitute a

state vector, but y and y_ together do. In this case, the order of the state space is 2n; which is twice the

number of DoF.

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