8.3 A Numerical Example

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In order to illustrate the use of the equations given before, let us consider a rigid body flexibly supported

by a number of springs. For this, the simplest starting point would be Equation 8.44

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

( )

ð8:48Þ

Computer Analysis of Flexibly Supported Multibody Systems 8-7

© 2005 by Taylor & Francis Group, LLC

Since body j does not exist, all the terms relevant to body j will disappear. Furthermore, since we are

dealing with a single mass, the suffix i is not needed either. However, for n number of springs, the stiffness

matrices need to be summed-up. Summation has to be carried out for each stiffness p attached at

a position on the body. We then have

m 0

0 J

" #

x€

a€

( )

þ

Xn

p¼1

Kp

Xn

p¼1

KpRp

Xn

p¼1

RT

p Kp

Xn

p¼1

RT

p KpRp

2

666664

3

777775

x

a

( )

¼

F

M

( )

ð8:49Þ

For a situation where the axes of the springs are parallel to the global coordinate system, no

transformation of the stiffness matrix is needed. Hence, kp ¼ Kp: To obtain the submatrices of the

stiffness matrix given in Equation 8.49, start with the stiffness matrix for spring p: Specifically,

Kp ¼

kpx 0 0

0 kpy 0

0 0 kpz

2

664

3

775

ð8:50Þ

Now, KpRp is given by

KpRp ¼

kpx 0 0

0 kpy 0

0 0 kpz

2

664

3

775

0 zp 2yp

2zp 0 xp

yp 2xp 0

2

664

3

775

ð8:51Þ

Expanding this, we get

KpRp ¼

0 kpx zp 2kpx yp

2kpy zp 0 kpy xp

kpz yp 2kpz xp 0

2

664

3

775

ð8:52Þ

For RT

p Kp; we have

RT

p Kp ¼

0 2zp yp

zp 0 2xp

2yp xp 0

2

664

3

775

kpx 0 0

0 kpy 0

0 0 kpz

2

664

3

775

ð8:53Þ

Expanding this, we get

RT

p Kp ¼

0 2kpy zp kpz yp

kpx zp 0 2kpz xp

2kpx yp kpy xp 0

2

664

3

775

ð8:54Þ

Finally, RT

p KpRp is given by

RT

p KpRp ¼

0 2kpy zp kpz yp

kpx zp 0 2kpz xp

2kpx yp kpy xp 0

2

664

3

775

0 zp 2yp

2zp 0 xp

yp 2xp 0

2

664

3

775

ð8:55Þ

8-8 Vibration and Shock Handbook

© 2005 by Taylor & Francis Group, LLC

RT

p KpRp ¼

kpz y2

p þ kpy z2

p 2kpz xpyp 2ky xpzp

2kpz xpyp kpz x2

p þ kpx z2

p 2kpx ypzp

2kpy xpzp 2kpx ypzp kpy x2

p þ kpx y2

p

2

6664

3

7775

ð8:56Þ

The overall stiffness matrix from Equation (8.48) for a single spring is given by

kpx 0 0 0 kpx zp 2kpx yp

0 kpy 0 2kpy zp 0 kpy xp

0 0 kpz kpz yp 2kpz xp 0

0 2kpy zp kpz yp kpz y2

p þ kpy z2

p 2kpz xpyp 2kpy xpzp

kpx zp 0 2kpz xp 2kpz xpyp kpz x2

p þ kpx z2

p 2kpx ypzp

2kpx yp kpy xp 0 2kpy xpzp 2kpx ypzp kpy x2

p þ kpx y2

p

2

666666666666664

3

777777777777775

ð8:57Þ

The mass matrix is diagonal and, for the inertia matrix, it is assumed that the principal axes of the body

coincide with the global coordinate system. Specifically,

m ¼

m 0 0

0 m 0

0 0 m

2

664

3

775

ð8:58Þ

J ¼

Ixx 0 0

0 Iyy 0

0 0 Izz

2

664

3

775

ð8:59Þ

Now, the overall equations of motion may be assembled for n springs.

m 0 0 0 0 0

0 m 0 0 0 0

0 0 m 0 0 0

0 0 0 Ixx 0 0

0 0 0 0 Iyy 0

0 0 0 0 0 Izz

2

6666666666664

3

7777777777775

x€

y€

z€

a€

b€

g€

8>>>>>>>>>><

>>>>>>>>>>:

9>>>>>>>>>>=

>>>>>>>>>>;

þ

Xn

p¼1

kpx 0 0 0 kpx zp 2kpx yp

0 kpy 0 2kpy zp 0 kpy xp

0 0 kpz kpz yp 2kpz xp 0

0 2kpy zp kpz yp kpz y2

p þ kpy z2

p 2kpz xpyp 2kpy xpzp

kpx zp 0 2kpz xp 2kpz xpyp kpz x2

p þ kpx z2

p 2kpx ypzp

2kpx yp kpy xp 0 2kpy xpzp 2kpx ypzp kpy x2

p þ kpx y2

p

2

666666666666664

3

777777777777775

x

y

z

a

b

g

8>>>>>>>>>><

>>>>>>>>>>:

9>>>>>>>>>>=

>>>>>>>>>>;

¼

Fx

Fy

Fz

Mx

My

Mz

8>>>>>>>>>><

>>>>>>>>>>:

9>>>>>>>>>>=

>>>>>>>>>>;

ð8:60Þ

Computer Analysis of Flexibly Supported Multibody Systems 8-9

© 2005 by Taylor & Francis Group, LLC

8.3.1 A Uniform Rectangular Prism

A rectangular prism is supported by four springs as

shown in Figure 8.3. Springs have stiffness values

in all three directions (kpx ; kpy ; kpz ; where p is the

spring number). The axes of each spring in which

the stiffness values are measured are parallel to the

principal axes of the springs, which in turn are

parallel to the global coordinate system of the

rectangular prism. Thus, no transformation is

needed. The end of spring p is located at ðxp; yp;

zpÞ; measured relative to the COG of the body. The

mass of the prism is m and the principal moments

of inertia are Ixx; Iyy; and Izz: A simplified

equation of motion of the system in 3-D space may

be obtained from Equation 8.59. If one attempts to

carry this out, one will realize that some terms will disappear because the z components of the positions

are all zero and some will disappear because of the symmetry of points.

The body shown in Figure 8.3 corresponds to m ¼ 1000 kg; moments of inertia Ixx ¼ 10 kg m2;

Iyy ¼ 20 kg m2; and Izz ¼ 30 kg m2; supported by four identical (thus, point suffix p is dropped) springs

with stiffness values ðkx ¼ 10;000 N=m; ky ¼ 20;000 N=m; kz ¼ 30;000 N=mÞ. The positions of the

springs are given as follows:

P1ð1; 2; 0Þ

P2ð1; 22; 0Þ

P3ð21; 2; 0Þ

P4ð21; 22; 0Þ

The coordinates imply that the COG is on the bottom plane of the prism. The system has six degrees of

freedom and all six natural frequencies will be calculated.

Since stiffness parameters are on the Oxy plane, no coupling will occur between (x and b) and

(y and a). Similarly, the vertical motion is also uncoupled from the others due to symmetry. Thus,

vx ¼

ffiffiffiffiffiffiffiffiffi

X4

p¼1

kpx

m

vuuuut

¼

ffiffiffiffiffiffiffiffiffiffi

40;000

1000

r

¼ 6:32 rad=sec ¼ 1:0065 Hz

vy ¼

ffiffiffiffiffiffiffiffiffi

X4

p¼1

kpy

m

vuuuut

¼

ffiffiffiffiffiffiffiffiffiffi

80;000

1000

r

¼ 8:94 rad=sec ¼ 1:4235 Hz

vz ¼

ffiffiffiffiffiffiffiffiffi

X4

p¼1

kpz

m

vuuuut

¼

ffiffiffiffiffiffiffiffiffiffiffi

120;000

1000

r

¼ 10:95 rad=sec ¼ 1:743 Hz

va ¼

ffiffiffiffiffiffiffiffiffiffiffiffi

X4

p¼1

y2

p kpz

Ixx

vuuuuut

¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4 £ 22 £ 30;000

10

s

¼ 219:09 rad=sec ¼ 34:87 Hz

x

y

z

FIGURE 8.3 A rectangular prism supported on springs.

8-10 Vibration and Shock Handbook

© 2005 by Taylor & Francis Group, LLC

vb ¼

ffiffiffiffiffiffiffiffiffiffiffiffi

X4

p¼1

x2

p kpz

Iyy

vuuuuut

¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4 £ 12 £ 30;000

20

s

¼ 77:45 rad=sec ¼ 12:32 Hz

vg ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

X4

p¼1 ðx2

p kpy þ y2

p kpx Þ

Izz

vuuuuut

¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4ð12 £ 20;000 þ 22 £ 10;000Þ

30

s

¼ 89:44 rad=sec ¼ 14:23 Hz

8.3.2 VIBRATIO Output

For the numerical problem given above, the output obtained from the software package VIBRATIO (see

Appendix 8A for the full listing) is tabulated below.

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