8.2 Theory

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8.2.1 Definitions and Assumptions

* Springs have zero length.

* The stiffness parameters of the springs in their principal axes of deflection remain uncoupled.

* The amplitude of oscillation is small. No geometrical nonlinearity is involved. In other words, the

orientation of both mountings and bodies remains unaffected by oscillations.

* The time-dependent effects of polymeric material are excluded.

* Gyroscopic effects are negligible.

These assumptions are acceptable for most engineering vibration problems with small amplitude

vibration.

8.2.2 Equations of Motion for the Linear Model

To set up equations of motion for a dynamic system, the following steps are required:

(i) Generation of the equations of internal reactions and external forces. The internal reactions due

to damping and stiffness elements have to be expressed in a unified and structured fashion for

formulation of the stiffness matrix. (The damping matrix structure is identical to the stiffness

matrix structure, except that stiffness coefficients need to be replaced by damping coefficients.)

(ii) Generation of the equations of linear momentum (force – acceleration equations).

(iii) Generation of the equations of angular momentum (turning moment equations).

8.2.3 Linear Momentum – Force Systems

8.2.3.1 Stiffness and Damping Systems

The formulations applied in this chapter to obtain the stiffness matrix apply equally to the damping

matrix by replacing stiffness parameters with their corresponding damping parameters.

Let us assume that spring stiffness parameters are described in a local three-dimensional (3-D)

Cartesian coordinate frame, the axes of which coincide with the principal axes of the springs. The force

FIGURE 8.1 Schematic representation of a multibody

system.

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vector f acting on the springs may be expressed as

f ¼ kx ð8:1Þ

where k is the stiffness matrix (diagonal with principal stiffness values) and x is the displacement vector

(expressing the spring extension).

In general, it is convenient to describe the behavior of a system in the global coordinate frame, OXYZ.

This is not a prerequisite for the formulation. It is equally possible to obtain equations of motion for each

body in its own frame. In this chapter, all spring stiffness matrices will be expressed in a common global

coordinate frame. The individual spring matrices will be transformed accordingly. Since the principal

axes of the springs and the global coordinates are all orthogonal, an orthogonal transformation exists

between the two frames. A vector, x, in the local coordinates could be expressed as a vector, X, in the

global coordinate system. Using, T, a transformation matrix

X ¼ Tx ð8:2Þ

If we premultiply Equation 8.1 by T, then we have Tf ¼ Tkx: But Tf ¼ F:

Therefore, force vector, F, in the global coordinate frame, may be written as

F ¼ Tkx ð8:3Þ

For consistency, x needs to be replaced by X. To replace x by X, Equation 8.2 may be used, giving

x ¼ TTX: This is true since T21 ¼ TT for orthogonal transformation matrices. Therefore,

F ¼ TkTTX ð8:4Þ

Then introduce a new matrix K, where

K ¼ TkTT

Now TkT T is the stiffness matrix of the spring in the global coordinate system. The transformation

matrix T may be described in three Euler angles of rotation.

8.2.3.2 Generalization of the Equation of Linear Momentum

If the mass/inertia matrix in the Euler– Newton formulation is obtained relative to the axes passing

through the center of mass, then the submatrix of the mass matrix corresponding to linear momentum is

a diagonal matrix containing the mass elements; thus,

hi ¼ miv ð8:5Þ

Here, hi is linear momentum, mi is a diagonal matrix, and v is the velocity vector of the center of mass

(casually known as COG) of the body.

The usual transformation to the global coordinate frame, Hl ¼ TmTTv; leaves the mass matrix, m,

unchanged. Therefore, the force acting on a body, i (i.e., the rate of change of linear momentum), may be

expressed simply as

Force ¼ H_ l ¼

›Hl

›t ¼ ma ð8:6Þ

where a is the acceleration vector of the COG.

8.2.4 Generalization of the Equations of Moment of Momentum

The equations of moment of momentum may be expressed as

ha ¼ jv ð8:7Þ

where ha is the angular momentum vector, j is the moments of inertia matrix and v is the angular

velocity of the coordinate frame. (In this case, the frame is attached to the body.)

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© 2005 by Taylor & Francis Group, LLC

Here, j may or may not be a diagonal matrix. However, it is always symmetric. Equation 8.7 is

described in the local coordinate system of the rigid body and it has to be expressed in the global

coordinate system for the final matrix assembly. As presented for the stiffness elements,

transformation follows exactly the same steps as before. In this case, T refers to the transformation

matrix of mass relative to the global coordinate system. Transforming Equation 8.7 to the global

coordinates, we get

Ha ¼ TjTTV ð8:8Þ

Introduce a new matrix notation

J ¼ TjTT ð8:9Þ

The vector differentiation of Ha gives the moment vector in the global coordinates

M ¼ H_ a ¼

›Ha

›t þ v £ Ha ð8:10Þ

where v is the angular velocity of the body (or the coordinate frame, as the body is fixed to the

frame).

Note that v £ Ha contains the product of angular velocity terms and this, for small and geometrically

linear vibration problems, is small and may be ignored.

8.2.5 Assembly of Equations

To assemble the equations of motion, the internal

forces acting on individual bodies due to their

motion relative to each other are required. In

Figure 8.2, two bodies (i and j) in motion are

shown, connected by spring Kp:

Motion of the origin of vector i (which

coincides with the COG of body i) is given by

xi ¼ ðxi; yi; ziÞ; and the angular rotation of the

coordinates is given by aI ¼ ðai; bi; giÞ:

Similarly, the motion of body j is described by

xj ¼ ðxj; yj; zjÞ and aj ¼ ðaj; bj; gjÞ:

For small motions, displacements of the

end points of the springs on each body, described

in the coordinate frame of each body, are

given by

di ¼ xi þ ai £ rpi ð8:11Þ

dj ¼ xj þ aj £ rpj ð8:12Þ

where rpj and rpj are the coordinates of the spring attachment relative to the bodies i and j in their

respective coordinate frames, given as rpi ¼ ðxpi; ypi; zpiÞ and rpj ¼ ðxpj; ypj; zpjÞ:

Cross-product terms in Equation 8.11 and Equation 8.12 can be converted into matrix form as

ai £ rpi ¼

0 zpi 2ypi

2zpi 0 xpi

ypi 2xpi 0

2

664

3

775

ai

bi

gi

8>><

>>:

9>>=

>>;

ð8:13Þ

X

O Y

Z

mi

Ji

Zi

γi

γj

βj

αi

βi

αj

di

Pi

Pj

mj

Jj

Zj

dj

r Yj j

oj

Xj

Kp

Yi

Oi

ri

Xi

FIGURE 8.2 Two bodies connected by springs.

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and

aj £ rpi ¼

0 zpj 2ypj

2zpj 0 xpj

ypj 2xpj 0

2

664

3

775

aj

bj

gj

8>><

>>:

9>>=

>>;

ð8:14Þ

Let us choose the matrix notation Rpi as

Rpi ¼

0 zpi 2ypi

2zpi 0 xpi

ypi 2xpi 0

2

664

3

775

ð8:15Þ

and Rpj as

Rpj ¼

0 zpj 2ypj

2zpj 0 xpj

ypj 2xpj 0

2

664

3

775

ð8:16Þ

Therefore,

di ¼ xi þ ai £ rpi

which is

¼

xi

yi

xi

8>><

>>:

9>>=

>>;

þ

0 zpi 2ypi

2zpi 0 xpi

ypi 2xpi 0

2

664

3

775

ai

bi

gi

8>><

>>:

9>>=

>>;

ð8:17Þ

Using the new notation, we have

di ¼ xi þ Rpiai ð8:18Þ

Now dj is given by

dj ¼ xj þ aj £ rpj ð8:19Þ

and can be written as

dj ¼

xj

yj

xj

8>><

>>:

9>>=

>>;

þ

0 zpj 2ypj

2zpj 0 xpj

ypj 2xpj 0

2

664

3

775

aj

bj

gj

8>><

>>:

9>>=

>>;

and therefore in matrix notation, we have

dj ¼ xj þ Rpjaj ð8:20Þ

To calculate the reactions acting on each body, the relative displacements between the connecting

points (stretch) should be calculated.

The relative displacements are given by

d ¼ dj 2 di ð8:21Þ

The reaction forces due to the relative displacements on each body are given by

Fsi ¼ Kpd ð8:22Þ

For equal but opposite directions, we have

Fsj ¼ 2Kpd ð8:23Þ

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Moments for spring forces acting at points ri and rj on bodies i and j; respectively, are given by

Mi ¼ ri £ Fsi ð8:24Þ

On body j; we have

Mj ¼ rj £ Fsj ð8:25Þ

The cross-products may be expressed in matrix form as

Mi ¼ ri £ Fsi ¼

0 2zpi ypi

zpi 0 2xpi

2ypi xpi 0

2

664

3

775

Fsxi

Fsyi

Fszi

8>><

>>:

9>>=

>>;

ð8:26Þ

Mj ¼ rj £ Fsj ¼

0 2zpj ypj

zpj 0 2xpj

2ypj xpj 0

2

664

3

775

Fsxj

Fsyj

Fszj

8>><

>>:

9>>=

>>;

ð8:27Þ

Note that the matrices in Equation 8.26 and Equation 8.27 are transposed forms of the matrices in

Equation 8.15 and Equation 8.16

RT

pi ¼

0 2zpi ypi

zpi 0 2xpi

2ypi xpi 0

2

664

3

775

ð8:28Þ

RT

pj ¼

0 2zpj ypj

zpj 0 2xpj

2ypj xpj 0

2

664

3

775

ð8:29Þ

Now the equations of motion can be compiled for the translation of body i

mix€i þ Kpdi 2 Kpdj ¼ Fi ð8:30Þ

In this case, Fi is the vector of external forces acting on body i: Substituting di and dj into Equation 8.30,

from Equation 8.18 and Equation 8.20 we have

mix€i þ Kpðxi þ RpiaiÞ 2 Kpðxj þ RpjajÞ ¼ Fi ð8:31Þ

Expanding this, we get

mix€i þ Kpxi þ KpRpiai 2 Kpxj 2 KpRpjaj ¼ Fi ð8:32Þ

Similarly, for body j; substituting the expressions for di and dj; we get

mjx€j þ Kpdi 2 Kpdj ¼ Fj ð8:33Þ

Again, Fj in this case is the vector of external forces acting on body j:

mjx€j þ Kpðxi þ RpiaiÞ 2 Kpðxj þ RpjajÞ ¼ Fj ð8:34Þ

mjx€j þ Kpxi þ KpRpiai 2 Kpxj 2 KpRpjaj ¼ Fj ð8:35Þ

With Equation 8.32 and Equation 8.35, the force – acceleration equations are complete.

Moment Equations

The moment equation may be written for body i as shown in Equation 8.36, where Mi is the external

moment acting on body i:

Jia€i þ ri £ ðKpdi 2 KpdjÞ ¼ Mi ð8:36Þ

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Substituting expressions for di and dj and converting the cross-product to the matrix form, we get

Jia€i þ RT

piðKpðxi þ RpiaiÞ 2 Kpðxj þ RpjajÞÞ ¼ Mi ð8:37Þ

Expanding this, we get

Jia€i þ RT

piKpxi þ RT

piKpRpiai 2 RT

piKpxj 2 RT

piKpRpjaj ¼ Mi ð8:38Þ

The moment equation may be written for body j as given in Equation 8.39, where Mj is the external

moment acting on body j: Thus,

Jja€j 2 rj £ ðKpdi 2 KpdjÞ ¼ Mj ð8:39Þ

Substituting di and dj and converting the cross-product to the matrix form, we get

Jja€j 2 RT

pjðKpðxi þ RpiaiÞ 2 Kpðxj þ RpjajÞÞ ¼ Mj ð8:40Þ

Expanding this, we get

Jja€j 2 RT

pjKpxi 2 RT

pjKpRpiai þ RT

pjKpxj þ RT

pjKpRpjaj ¼ Mj ð8:41Þ

If we then collect Equation 8.32 and Equation 8.38 for body I; then Equation 8.32 becomes

mix€i þ Kpxi þ KpRpiai 2 Kpxj 2 KpRpjaj ¼ Fi ð8:42Þ

and Equation 8.38 becomes

Jia€i þ RT

piKpxi þ RT

piKpRpiai 2 RT

piKpxj 2 RT

piKpRpjaj ¼ Mi ð8:43Þ

Expressing Equation 8.42 and Equation 8.43 in matrix form, we have

mi 0

0 Ji

" #

x€i

a€i

( )

þ

Kp KpRpi

RT

piKp RT

piKpRpi

2

4

3

5 xi

ai

( )

2

Kp KpRpj

RT

piKp RT

piKpRpj

2

4

3

5 xj

aj

( )

¼

Fi

Mi

( )

ð8:44Þ

Similarly, if we collect Equation 8.34 and Equation 8.41 for body j; then Equation 8.34 becomes

mjx€j 2 Kpxi 2 KpRpiai þ Kpxj þ KpRpjaj ¼ Fj ð8:45Þ

and Equation 8.41 becomes

Jja€j 2 RT

pjKpxi 2 RT

pjKpRpiai þ RT

pjKpxj þ RT

pjKpRpjaj ¼ Mj ð8:46Þ

Expressing Equation 8.45 and Equation 8.46 in the matrix form, we have

mj 0

0 Jj

" #

x€j

a€j

( )

2

Kp KpRpi

RT

pjKp RT

pjKpRpi

2

4

3

5 xi

ai

( )

þ

Kp KpRpj

RT

pjKp RT

pjKpRpj

2

4

3

5 xj

aj

( )

¼

Fj

Mj

( )

ð8:47Þ

Overall, the equations of motion are now complete. Equation 8.44 and Equation 8.47 provide all that is

needed to complete the final equations of motion. It is worth restating that the stiffness and damping

matrices are identical in their structure. To obtain a damping matrix, all one needs to do is to replace the

stiffness coefficients with the corresponding damping coefficients.

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