Multi-Degree-of-Freedom System Response to Multipoint Base Excitation

By Tom Irvine

Email:

October 16, 2013

______

Introduction

Figure 1.

Figure 2.

The free-body diagram is given in Figure 2.

The system has a CG offset if .

The system is statically coupled if .

The rotation is positive in the clockwise direction.

The variables are

y / is the base displacement
x / is the translation of the CG
/ is the rotation about the CG
m / is the mass
J / is the polar mass moment of inertia
ki / is the stiffness for spring i
zi / is the relative displacement for spring i

Sign Convention:

Translation: upward in vertical axis is positive.

Rotation: clockwise is positive.

Figure 3. Figure 4.

Sum the forces in the vertical direction for mass 1.

(1)

(2)

(3)

(4)

Sum the forces in the vertical direction for mass 2.

(5)

(6)

(7)

(8)

Sum the forces in the vertical direction for mass 3.

(9)

(10)

(11)

(12)

(13)

Sum the moments about the center of mass.

(14)

(15)

(16)

(17)

(18)

The equations of motion are

(19)

Variables

M / Mass matrix
K / Stiffness matrix
I / Identity matrix
/ Transformation matrix
U / Displacement vector
ud / Displacements at driven nodes
uf / Displacements at free nodes

The equation of motion for a multi-degree-of-freedom system is

(20)

(21)

Partition the matrices and vectors as follows

(22)

(23)

where

(24)

(25)

(26)

(27)

(28)

(29)

(30)

(31)

Create a transformation matrix such that

(32)

(33)

(34)

Premultiplyby ,

(35)

Again, the partitioned equation of motion is

(36)

Transform the equation of motion to uncouple the stiffness matrix so that the resulting stiffness matrix is

(37)

(38)

(39)

(40)

(41)

(42)

Let

(43)

(44)

(45)

(46)

(47)

(48)

(49)

(50)

(51)

By similarity, the transformed mass matrix is

(52)

(53)

(54)

The equation of motion is thus

(55)

The final displacements are found via

(56)

(57)

APPENDIX A

Example

Figure A-1.

Consider the system in Figure A-1. Assign the following values. The values are based on a slender rod, aluminum, diameter =1 inch, total length=24 inch.

Table A-1. Parameters
Variable / Value
/ 100 lbm
/ 100 lbm
/ 18.9 lbm
J / 907 lbm in^2
/ 20,000 lbf/in
/ 20,000 lbf/in
/ 8 in
/ 16 in

Let

where the amplitude is in units of G and time t is in seconds.

Note that the mass values for m1 and m2 are actually arbitrary since these degrees-of-freedom are driven.

The following parameters were calculated for the sample system via a Matlab script.

enforced_acceleration

enforced_acceleration.m ver 2.3 October 16, 2013

by Tom Irvine

Enter the units system

1=English 2=metric

1

Assume symmetric mass and stiffness matrices.

Select input mass unit

1=lbm 2=lbf sec^2/in

2

stiffness unit = lbf/in

Select file input method

1=file preloaded into Matlab

2=Excel file

1

Mass Matrix

Enter the matrix name: mm

Stiffness Matrix

Enter the matrix name: kk

Input Matrices

mass =

0.2591 0 0 0

0 0.2591 0 0

0 0 0.0490 0

0 0 0 2.3497

stiff =

20000 0 -20000 -160000

0 20000 -20000 320000

-20000 -20000 40000 -160000

-160000 320000 -160000 6400000

Select modal damping input method

1=uniform damping for all modes

2=damping vector

1

Enter damping ratio

0.05

number of dofs =4

Enter the starting time (sec)

0

Enter the end time (sec)

1

Enter the sample rate (samples/sec)

4000

Enter the number of dofs with enforced acceleration. (maximum = 4)

2

Each input file must have two columns: time & acceleration

Enter the first dof

1

Enter the applied acceleration input matrix name for this dof.

a130

Enter the second dof

2

Enter the applied acceleration input matrix name for this dof.

a150

MT =

0.2849 0.0068 0.0326 0.0979

0.0068 0.2686 0.0163 -0.0979

0.0326 0.0163 0.0490 0

0.0979 -0.0979 0 2.3497

KT =

1.0e+06 *

-0.0000 0.0000 0.0000 0

0.0000 -0.0000 0.0000 0

0.0000 0.0000 0.0400 -0.1600

0 0 -0.1600 6.4000

Natural Frequencies

No. f(Hz)

1. 1.2038e-07

2. 6.5509e-06

3. 140.91

4. 271.55

Modes Shapes (column format)

ModeShapes =

1.8735 0.0116 -0.5798 0.1089

0 1.9292 -0.2099 -0.3008

0 0.0000 4.6346 1.0430

0 0.0000 0.1565 -0.6438

Mwd =

0.0326 0.0163

0.0979 -0.0979

Mww =

0.0490 0

0 2.3497

Kww =

40000 -160000

-160000 6400000

Natural Frequencies

No. f(Hz)

1. 133.79

2. 267.93

Modes Shapes (column format)

ModeShapes =

4.4005 1.0291

0.1486 -0.6352

Participation Factors

part =

0.5645

-1.4422

Output arrays:

ea_disp - displacement

ea_vel - velocity

ea_acc - acceleration

Figure A-2.

Figure A-3.

1