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3.7 Other Modal Formulations

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The modal problem (eigenvalue problem) studied in the previous sections consists of the solution of

v2Mc ¼ Kc ð3:28Þ

which is identical to Equation 3.13. The natural frequencies (eigenvalues) are given by solving the

characteristic equation 3.14. The corresponding mode shape vectors (eigenvectors) ci are determined by

3-22 Vibration and Shock Handbook

substituting each natural frequency vi into Equation 3.13 and solving for a nontrivial solution. This

solution will have at least one arbitrary parameter. Hence, c represents the relative displacements at the

various DoF of the vibrating system and not the absolute displacements. Now, two other formulations are

given for the modal problem.

The first alternative formulation given below involves the solution of the eigenvalue problem of a

nonsymmetric matrix ðM21KÞ: The other formulation given consists of first transforming the original

problem into a new set of motion coordinates, then solving the eigenvalue problem of a symmetric

matrix ðM21=2KM21=2Þ; and then transforming the resulting modal vectors back to the original

motion coordinates. Of course, all three of these formulations will give the same end result for the

natural frequencies and mode shapes of the system, because the physical problem would remain the

same regardless of what formulation and solution approach are employed. This fact will be illustrated

using an example.

3.7.1 Nonsymmetric Modal Formulation

Consider the original modal formulation given by Equation 3.28 that we have studied. Since

the inertia matrix M is nonsingular, its inverse M 21 exists. The premultiplication of Equation 3.28 by

M 21 gives

v2c ¼ M21Kc ð3:29Þ

This vector– matrix equation is of the form

lc ¼ Sc ð3:30Þ

where l ¼ v2 and S ¼ M21K: Equation 3.30 represents the standard matrix eigenvalue problem for

matrix S. It follows that

Squared natural frequencies ¼ eigenvalues of M21K

Mode shape vectors ¼ eigenvectors of M21K

3.7.2 Transformed Symmetric Modal Formulation

Now consider the free (unforced) system equations

My€ þ Ky ¼ 0 ð3:31Þ

whose modal problem needs to be solved. First, we define the square root of matrix M, as denoted by

M1=2; such that

M1=2M1=2 ¼ M ð3:32Þ

Since M is symmetric, M1=2 also has to be symmetric. Next, we define M21=2 as the inverse of M1=2:

Specifically,

M21=2M1=2 ¼ M1=2M21=2 ¼ I ð3:33Þ

where I is the identify matrix. Note that M21=2 is also symmetric.

Once M21=2 is defined in this manner, we transform the original problem 3.31 using the coordinate

transformation

y ¼ M21=2q ð3:34Þ

Modal Analysis 3-23

Here, q denotes the transformed displacement vector, which is related to the actual displacement vector y

through the matrix transformation using M21=2:

By differentiating Equation 3.34 twice, we get

y€ ¼ M21=2 q€ ð3:35Þ

Substitute Equation 3.34 and Equation 3.35 into Equation 3.31. This gives

MM21=2 q€ þ KM21=2q ¼ 0

Premultiply this result by M21=2 and use the fact that

M21=2MM21=2 ¼ M21=2M1=2M1=2M21=2 ¼ I

which follows from Equation 3.32 and Equation 3.33. We get

q€ þ M21=2KM21=2q ¼ 0 ð3:36Þ

Equation 3.36 is the transformed problem, whose modal response may be given by

q ¼ e jvtf ð3:37Þ

where v represents a natural frequency and f represents the corresponding modal vector, as usual. Then,

in view of Equation 3.34, we have

y ¼ e jvt M21=2f ¼ e jvt c ð3:38Þ

It follows that the natural frequencies of the original problem 3.31 are identical to the natural frequencies

of the transformed problem 3.36, and the modal vectors c of the original problem are related to the

modal vectors f of the transformed problem through

c ¼ M21=2f ð3:39Þ

Substitute the modal response 3.37 into Equation 3.36. We get

lf ¼ Pf ð3:40Þ

where l ¼ v2 and P ¼ M21=2KM21=2:

Equation 3.40, just like Equation 3.30, represents a standard matrix eigenvalue problem. But now

matrix P is symmetric. As a result, its eigenvectors f will not only be real but also orthogonal.

The solution steps for the present, transformed, and symmetric modal problem are:

1. Determine M21=2:

2. Solve for eigenvalues l and eigenvectors f of M21=2KM21=2: Eigenvalues are squares of the natural

frequencies of the original system.

3. Determine the modal vectors c of the original system by using c ¼ M21=2f:

The three approaches of modal analysis which we have studied are summarized in Table 3.4.

TABLE 3.4 Three Approaches of Modal Analysis

Approach Standard Nonsymmetric

Matrix Eigenvalue

Symmetric Matrix Eigenvalue

Modal formulation ½v2 M 2 K􀀉c ¼ 0 v2 c ¼ M21Kc v2f ¼ M21=2KM21=2f

Squared natural

frequencies ðv2i

Roots of det½v2 M 2 K􀀉 ¼ 0 Eigenvalues of M21 K Eigenvalues of M21=2KM21=2

Mode-shape vectors ðci Þ Nontrivial solutions

of ½v2i

M 2 K􀀉c ¼ 0

Eigenvectors of M21 K Determine eigenvectors fi

of M21=2KM21=2 : Then ci ¼ M21=2fi

3-24 Vibration and Shock Handbook

Example 3.6

We will use the 2-DoF vibration problem given in Figure 3.4 (Example 3.3) to demonstrate the fact that

all three approaches summarized in Table 3.4 will lead to the same results.

Consider the special case of a ¼ 0:5 and b ¼ 0:5: Then we have

M ¼

m 0

5 and K ¼

k 2

6664

7775

Approach 1

Using the standard approach, we obtain the modal results given in Table 3.3. Specifically, we get the

natural frequencies (normalized with respect to v0 ¼

ffiffiffiffiffi

k=m p )

v0 ¼

1ffiffi

2 p and

v0 ¼

ffiffi

2 p

and the mode shapes

􀀏 􀀐

1¼ 2 and

􀀏 􀀐

2¼ 21

Let us now obtain these results using the other two approaches of modal analysis.

Approach 2

M21 ¼

m 0

;

M21K ¼

m 0

k 2

6664

7775

6664

7775

¼ v20

21 1

Note that this is not a symmetric matrix. We solve the eigenvalue problem of

21 1

Eigenvalues l are given by

det

l 2

1 l 2 1

¼ 0

l 2

􀀏 􀀐

ðl 2 1Þ 2

2 ¼ 0

l2 2

l þ 1 ¼ 0

Modal Analysis 3-25

the roots of which are

l1; l2 ¼

; 2

It follows that v1=v0 ¼ 1=

ffiffi

2 p and v2=v0 ¼

ffiffi

2 p as before.

The eigenvector corresponding to l1 (mode 1) is given by

2 1

664

775

" #

1¼

" #

The solution is ðc2=c1Þ1 ¼ 2 as before.

The eigenvector corresponding to l2 (mode 2) is given by

2 2

1 22 1

" #

2¼

" #

The solution is ðc2=c1Þ2 ¼ 21 as before.

Approach 3

Since M is diagonal, it is easy to obtain M1=2: We simply take the square root of the diagonal elements;

thus,

M1=2 ¼

ffiffiffi

pm 0

ffiffiffiffi

Its inverse is given by inverting the diagonal elements; thus

M21=2 ¼

1ffiffiffi

pm 0

ffiffiffiffi

6664

7775

Now,

M21=2KM21=2 ¼

1ffiffiffi

pm 0

ffiffiffiffi

6664

7775

k 2

6664

7775

1ffiffiffi

pm 0

ffiffiffiffi

6664

7775

1ffiffi

2 p

1ffiffi

2 p

66664

77775

¼ v20

1ffiffi

2 p

1ffiffi

2 p 1

6664

7775

Note that, as expected, this is a symmetric matrix. We solve for eigenvalues and eigenvectors of

1ffiffi

2 p

1ffiffi

2 p 1

6664

7775

3-26 Vibration and Shock Handbook

Eigenvalues are given by

det

l 2

1ffiffi

2 p

1ffiffi

2 p l 2 1

6664

7775

¼ 0

l 2

􀀏 􀀐

ðl 2 1Þ 2

2 ¼ 0

l2 2

l þ 1 ¼ 0

which is identical to the characteristic equation obtained in the first two approaches. It follows that the

same two natural frequencies are obtained by this method. The eigenvector f1 for mode 1 is given by

1ffiffi

2 p

1ffiffi

2 p

2 1

6664

7775

" #

1¼

" #

which gives ðf2=f1Þ1 ¼

h ffiffi

2 p i

Accordingly, we may use

" #

1¼

ffiffi

2 p

" #

The eigenvector f2 for mode 2 is given by

2 2

1ffiffi

2 p

1ffiffi

2 p 2 2 1

6664

7775

" #

2¼

" #

which gives ðf2=f1Þ2 ¼ 21=

ffiffi

2 p .

Accordingly, we use

" #

2¼

1ffiffi

2 p

Now, we transform these eigenvectors back to the original coordinate system using Equation 3.39.

We get

" #

1¼

1ffiffiffi

pm 0

ffiffiffiffi

6664

7775

ffiffi

2 p

" #

1ffiffiffi

2ffiffiffi

6664

7775

which gives ðc2=c1Þ1 ¼ 2 as before.

Modal Analysis 3-27

Also,

" #

2¼

1ffiffiffi

pm 0

ffiffiffiffi

6664

7775

1ffiffi

2 p

1ffiffiffi

6664

7775

which gives

􀀄

􀀋

􀀅

2¼ 21 as before:

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