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3.3 System Representation

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Some damped systems do not possess real modes. If a system does not possess real modes, modal analysis

could still be used, but the results would only be approximately valid. In modal analysis it is convenient to

first neglect damping and develop the fundamental results, and then subsequently extend the results to

damped systems, for example, by assuming a suitable damping model that possesses real modes. Since

damping is an energy dissipation phenomenon, it is usually possible to determine a model that possesses

real modes and also has an energy dissipation capacity equivalent to that of the actual system.

3-4 Vibration and Shock Handbook

Consider the three undamped system representations (models) shown in Figure 3.2. The motion of

system (a) consists of the translatory displacements y1 and y2 of the lumped masses m1 and m2: The

masses are subjected to the external excitation forces (inputs) f1ðtÞ and f2ðtÞ and the restraining forces of

the discrete, tensile-compressive stiffness (spring) elements k1; k2; and k3: Only two independent

Box 3.1

SOME DEFINITIONS AND PROPERTIES

OF MECHANICAL SYSTEMS

Holonomic constraints Constraints that can be represented by purely algebraic relations

Nonholonomic constraints Constraints that require differential relations for their representation

Holonomic system A system that possesses holonomic constraints only

Nonholonomic system A system that possesses one or more nonholonomic constraints

Number of DoFs The number of independent incremental coordinates

that are needed to represent general incremental motion of a

system ¼ number of independent incremental motions

Order of a system ¼ 2 £ number of DoF (typically)

For a holonomic system

Number of independent

incremental coordinates

¼ Number of independent coordinates ¼ number of DoF

For a nonholonomic system

Number of independent

incremental coordinates

, Number of independent coordinates

m1 m2

y1 y2

(a)

Translatory

System f1(t) f2(t)

m k2 1 k1 m2

y1 y2

(b)

Flexural

System

f1t) f2(t)

y1 y2

(c)

Torsional

System

f1(t) f2(t)

1 2

FIGURE 3.2 Three types of two-DoF systems.

Modal Analysis 3-5

incremental coordinates (dy1 and dy2) are required to completely define the incremental motion of the

system subject to its inherent constraints. It follows that the system has two DoF.

In system (b), shown in Figure 3.2, the elastic stiffness to the transverse displacements y1 and y2 of the

lumped masses is provided by three bending ( flexural) springs that are considered massless. This flexural

system is very much analogous to the translatory system (a) even though the physical construction and the

motion itself are quite different. System (c) in Figure 3.2 is the analogous torsional system. In this case, the

lumped elements m1 and m2 should be interpreted as polar moments of inertia about the shaft axis, and k1;

k2; and k3 as the torsional stiffness in the connecting shafts. Furthermore, the motion coordinates y1 and y2

are rotations and the external excitations f1ðtÞ and f2ðtÞ are torques applied at the inertia elements. Practical

examples where these three types of vibration system models may be useful are: (a) a two-car train, (b) a

bridge with two separate vehicle loads, and (c) an electric motor and pump combination.

The three systems shown in Figure 3.2 are analogous to each other in the sense that the dynamics of all

three systems can be represented by similar equations of motion. For modal analysis, it is convenient to

express the system equations as a set of coupled second-order differential equations in terms of the

displacement variables (coordinates) of the inertia elements. Since in modal analysis we are concerned

with linear systems, the system parameters can be given by a mass matrix and a stiffness matrix, or by a

flexibility matrix. Lagrange’s equations of motion directly yield these matrices; however, we will now

present an intuitive method for identifying the stiffness and mass matrices.

The linear, lumped-parameter, undamped systems shown in Figure 3.2 satisfy the set of dynamic

equations

m11 m12

m21 m22

" #

y€1

y€2

" #

k11 k12

k21 k22

" #

My€ þ Ky ¼ f ð3:1Þ

Here, M is the inertia matrix which is the generalized case of mass matrix, and K is the stiffness matrix.

There are many ways to derive Equation 3.1. Below, we will describe an approach, termed the influence

coefficient method, which accomplishes the task by separately determining K and M.

3.3.1 Stiffness and Flexibility Matrices

In the systems shown in Figure 3.2 suppose the accelerations y€1 and y€2 are both zero at a particular

instant, so that the inertia effects are absent. The stiffness matrix K is given under these circumstances

by the constitutive relation for the spring elements:

" #

k11 k12

k21 k22

" #

f ¼ Ky ð3:2Þ

in which f is the force vector ½f1; f2􀀉T and y is the displacement vector ½y1; y2􀀉T: Both are column vectors.

The elements of the stiffness matrix, in this two-DoF case, are explicitly given by

K ¼

k11 k12

k21 k22

" #

Suppose that y1 ¼ 1 and y2 ¼ 0 (i.e., give a unit displacement to m1 while holding m2 at its original

position). Then k11 and k21 are the forces needed at location 1 and location 2, respectively, to maintain

this static configuration. For this condition it is clear that f1 ¼ k1 þ k2 and f2 ¼ 2k2: Accordingly,

k11 ¼ k1 þ k2; k21 ¼ 2k2

3-6 Vibration and Shock Handbook

Similarly, suppose that y1 ¼ 0 and y2 ¼ 1: Then k12 and k22 are the forces needed at location 1 and

location 2, respectively, to maintain the corresponding static configuration. It follows that

k12 ¼ 2k2; k22 ¼ k2 þ k3

Consequently, the complete stiffness matrix can be expressed in terms of the stiffness elements in the

system as

K ¼

k1 þ k2 2k2

2k2 k2 þ k3

" #

From the foregoing development, it should be clear that the stiffness parameter kij represents the force

that is needed at the location i to obtain a unit displacement at location j: Hence, these parameters are

termed stiffness influence coefficients.

Observe that the stiffness matrix is symmetric. Specifically,

kij ¼ kji for i – j

KT ¼ K ð3:3Þ

Note, however, that K is not diagonal in general (kij – 0 for at least two values of i – j). This means that

the system is statically coupled (or flexibly coupled).

Flexibility matrix L is the inverse of the stiffness matrix

L ¼ K21 ð3:4Þ

To determine the flexibility matrix using the influence coefficient approach, we have to start with a

constitutive relation of the form

y ¼ Lf ð3:5Þ

Assuming that there are no inertia forces at a particular instant, we then proceed as before. For the

systems in Figure 3.2, for example, we start with f1 ¼ 1 and f2 ¼ 0: In this manner, we can determine the

elements l11 and l21 of the flexibility matrix

L ¼

l11 l12

l21 l22

" #

However, here, the result is not as straightforward as in the previous case. For example, to determine l11,

we will have to find the flexibility contributions from either side of m1: The flexibility of the stiffness

element k1 is 1=k1: The combined flexibility of k2 and k3; which are connected in series, is 1=k2 þ 1=k3

because the displacements (across variables) are additive in series. The two flexibilities on either side of m1

are applied in parallel at m1: Since the forces (through variables) are additive in parallel, the stiffness will

also be additive. Consequently,

l11 ¼

ð1=k1Þ þ

ð1=k2 þ 1=k3Þ

After some algebraic manipulation we get

l11 ¼

k2 þ k3

k1k2 þ k2k3 þ k3k1

Modal Analysis 3-7

Since there is no external force at m2 in the assumed loading configuration, the deflections at m2 and m1

are proportioned according to the flexibility distribution along the path. Accordingly,

l21 ¼

1=k3

1=k3 þ 1=k2

􀀒 􀀓

l11

l21 ¼

k1k2 þ k2k3 þ k3k1

Similarly, we can obtain

l12 ¼

k1k2 þ k2k3 þ k3k1

and

l22 ¼

k1 þ k2

k1k2 þ k2k3 þ k3k1

Note that these results confirm the symmetry of flexibility matrices

lij ¼ lji for i – j

LT ¼ L ð3:6Þ

Also, we can verify the fact that L is the inverse of K. The series – parallel combination rules for stiffness

and flexibility that are useful in the present approach are summarized in Table 3.1.

The flexibility parameters lij represent the displacement at the location i when a unit force is applied at

location j: Hence, these parameters are termed flexibility influence coefficients.

3.3.2 Inertia Matrix

The mass matrix, which is used in the case of translatory motions, can be generalized as inertia matrix M

in order to include rotatory motions as well. To determine M for the systems shown in Figure 3.2,

suppose the deflections y1 and y2 are both zero at a particular instant so that the springs are in their static

equilibrium configuration. Under these conditions, the equation of motion 3.1 becomes

f ¼ My€ ð3:7Þ

For the present two-DoF case, the elements of M are denoted by

M ¼

m11 m12

m21 m22

" #

TABLE 3.1 Combination Rules for Stiffness and Flexibility Elements

Connection Graphical Representation Combined Stiffness Combined Flexibility

Series

ð1=k1 þ 1=k2 Þ

l1 þ l2

Parallel k1 þ k2

ð1=l1 þ 1=l2 Þ

3-8 Vibration and Shock Handbook

To identify these elements, first set y€1 ¼ 1 and y€2 ¼ 0: Then, m11 and m21 are the forces needed at the

locations 1 and 2, respectively, to sustain the given accelerations; specifically, f1 ¼ m1 and f2 ¼ 0: It

follows that

m11 ¼ m1; m21 ¼ 0

Similarly, by setting y€1 ¼ 0 and y€2 ¼ 1; we get

m12 ¼ 0; m22 ¼ m2

Then, the mass matrix is obtained as

M ¼

m1 0

0 m2

" #

It should be clear now that the inertia parameter mij represents the force that should be applied at the

location i in order to produce a unit acceleration at location j: Consequently, these parameters are called

inertia influence coefficients.

Note that the mass matrix is symmetric in general; specifically

mij ¼ mji for i – j

MT ¼ M ð3:8Þ

Furthermore, when the independent displacements of the lumped inertia elements are chosen as the

motion coordinates, as is typical, the inertia matrix becomes diagonal. If not, it can be made diagonal by

using straightforward algebraic substitutions so that each equation contains the second derivative of just

one displacement variable. Hence, we may assume

mij ¼ 0 for i – j ð3:9Þ

Then the system is said to be inertially uncoupled. This approach to finding K and M is summarized in

Box 3.2. It can be conveniently extended to damped systems for determining the damping matrix C.

Box 3.2

INFLUENCE COEFFICIENT METHOD OF

DETERMINING SYSTEM MATRICES

(UNDAMPED CASE)

Stiffness Matrix (K) Mass Matrix (M)

1. Set y€ ¼ 0 1. Set y ¼ 0

f ¼ Ky f ¼ My€

2. Set yj ¼ 1 and yi ¼ 0 for all i – j 2. Set y€j ¼ 1 and y€i ¼ 0 for all i – j

3. Determine f from the system diagram,

that is needed to main equilibrium ¼ jth column of K

3. Determine f to maintain this condition

¼ jth column of M

Repeat for all j Repeat for all j

Modal Analysis 3-9

3.3.3 Direct Approach for Equations of Motion

The influence coefficient approach that was described in the previous section is a rather indirect way of

obtaining the equations of motion 3.1 for a multi-DoF system. The most straightforward approach,

however, is to sketch a free-body diagram for the system, mark the forces or torques on each inertia

element, and finally, apply Newton’s Second Law. This approach is now illustrated for the system

shown in Figure 3.2(a). The equations of motion for the systems in Figures 3.2(b) and (c) will

follow analogously.

The free-body diagram of the system in Figure 3.2(a) is sketched in Figure 3.3. Note that all the forces

on each inertia element are marked.

Application of Newton’s Second Law to the two mass elements separately gives

m1y€1 ¼ 2k1y1 þ k2ðy2 2 y1Þ þ f1ðtÞ

m2y€2 ¼ 2k2ðy2 2 y1Þ 2 k3y2 þ f2ðtÞ

The terms can be rearranged to obtain the following two coupled, second order, linear, ordinary

differential equations:

m1y€1 þ ðk1 þ k2Þy1 2 k2y2 ¼ f1ðtÞ

m2y€2 2 k2y1 þ ðk2 þ k3Þy2 ¼ f2ðtÞ

which may be expressed in the vector– matrix form as

m1 0

0 m2

" #

y€1

y€2

" #

k1 þ k2 2k2

2k2 k2 þ k3

" #

f1ðtÞ

f2ðtÞ

" #

Observe that this result is identical to what we obtained by the influence coefficient approach.

Another convenient approach that would provide essentially the same result is the energy method

through the application of Lagrange’s equations. Two common types of models used in vibration

analysis and applications are summarized in Box 3.3.

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