Spring-Mass System Subjected to Multipoint Enforced Motion
By Tom Irvine
Email:
February 18, 2015
______
Introduction
Consider the two-degree-of-freedom system shown in Figure 1. The translation and rotation are both referenced to the center-of-gravity.
The vibration modes have translation and rotation which will be coupled if .
This coupling will occur, for example, if there is a C.G. offset even with equal spring stiffness.
Figure 1.
Variables
J / mass moment of inertia about the C.G.
k / spring stiffness
L / length from spring attachment to C.G.
Now consider that the system is subjected to multipoint base excitation. The response calculation requires extra steps than that for the case of an SDOF system. This is necessary because there are two different types of degrees-of-freedom with potential coupling within each mode.
The system in Figure 1 will thus be converted to a three-degree-of-freedom system as an intermediate step. The motion of the independent base masses will then be enforced.
Figure 2.
The value of the base masses m1 and m2 are arbitrary for the approach given in this paper. The free-body-diagram is shown in Figure 3.
Figure 3.
Sign Convention:
Translation: upward in vertical axis is positive.
Rotation: counter-clockwise is positive.
Assume small angular displacement.
Sum the forces in the vertical direction for mass 1 is
(1)
(2)
(3)
(4)
(5)
Sum the forces in the vertical direction for mass 2 is
(6)
(7)
(8)
(9)
Sum the forces in the vertical direction for mass 3 is
(10)
(11)
(12)
(13)
(14)
Sum the moments about the center of mass.
(15)
(16)
(17)
(18)
(19)
(20)
Note that the moment of inertia J is
(21)
where r is the radius of gyration.
Two levels of transformations are required to solve equation (20) as shown in the main text and in Appendix A.
Variables
K / Stiffness matrix
F / Applied forces
/ Forces at driven nodes
/ Forces at free nodes
I / Identity matrix
/ Transformation matrix
u / Displacement vector
ud / Displacements at driven nodes
uf / Displacements at free nodes
The homogenous equation of motion for a multi-degree-of-freedom system is
(22)
(23)
For the sample system in Figure 2,
(24)
(25)
Note that is the enforced displacement, or alternately is the enforced acceleration.
Partition the matrices and vectors as follows
(26)
The partitioned matrices for the sample problem are
(27)
(28)
(29)
(30)
(31)
(32)
(33)
(34)
Create a transformation matrix such that
(35)
(36)
(37)
Premultiply by ,
(38)
The partitioned matrices T1 and T2 in equation (36) depend on whether the base excitation is specified as acceleration or displacement. The acceleration and displacements cases are given in Appendices A and B, respectively.
An example is given in Appendix C.
References
1. T. Irvine, Sample Automobile Vibration Problem, Revision A, Vibrationdata, 2015.
2. T. Irvine, Modal Transient Analysis of a Multi-degree-of-freedom System with Enforced Motion, Revision C, Vibrationdata, 2011.
3. T. Irvine, Frequency Response Function Analysis of a Multi-degree-of-freedom System with Enforced Motion, Vibrationdata, 2011.
APPENDIX A
Enforced Acceleration
Again, the partitioned equation of motion is
(A-1)
Note that the applied force terms on the right-hand-side of equation (A-1) are zero for the sample problem in the main text.
Also note that the following calculations are too intensive for direct substation of the symbolic variables in the sample problem. Instead, numerical substitution is need for practical purposes.
Transform the equation of motion to uncouple the stiffness matrix so that the resulting stiffness matrix is
(A-2)
(A-3)
(A-4)
(A-5)
(A-6)
(A-7)
Let
(A-8)
(A-9)
(A-10)
(A-11)
(A-12)
(A-13)
(A-14)
(A-15)
(A-16)
By similarity, the transformed mass matrix is
(A-17)
(A-18)
(A-19)
(A-20)
(A-21)
(A-22)
The equation of motion is thus
(A-23)
The final displacement are found via
(A-24)
Again,
(A-25)
The transformed mass and stiffness matrices in equation (A-23) may be coupled. The equation can thus be solved via normal modes analysis.
Now solve the generalized eigenvalue problem.
Let Q be the eigenvector matrix.
Let
(A-26)
(A-27)
Premultiply by QT to decouple the equations of motion.
(A-28)
Assume that the applied forces are all zero. Thus the only excitation is the enforced acceleration.
(A-29)
Assume that the applied forces are all zero. Thus the only excitation is the enforced acceleration.
(A-30)
(A-31)
(A-32)
The decoupled equations of motion with an added modal damping matrix are
(A-33)
Note that
is the identity matrix is a diagonal matrix
is a diagonal matrix containing the terms
is a diagonal matrix containing the terms
Thus
(A-34)
Perform a steady-state analysis. Represent the enforced motion via a Fourier transform.
(A-35)
Assume
(A-36)
By substitution,
(A-37)
(A-38)
(A-39)
(A-40)
(A-41)
The final acceleration frequency response functions are found via
(A-42)
(A-43)
(A-44)
APPENDIX B
Enforced Displacement
The following method can be bypassed for the two-dof system in the main text because its mass matrix is already uncoupled. Nevertheless, it is presented for educational purposes.
Again, the partitioned equation of motion is
(B-1)
Note that the applied force terms on the right-hand-side of equation (B-1) are zero for the sample problem in the main text.
Transform the equation of motion to uncouple the mass matrix so that the resulting mass matrix is
(B-2)
Apply the transformation to the mass matrix
(B-3)
(B-4)
(B-5)
(B-6)
(B-7)
Let
(B-8)
(B-9)
(B-10)
(B-11)
(B-12)
The transformation matrix is
(B-13)
(B-14)
(B-15)
(B-16)
By similarity, the transformed stiffness matrix is
(B-17)
(B-18)
(B-19)
(B-20)
(B-21)
(B-22)
The equation of motion is thus
(B-23)
The final displacement are found via
(B-24)
(B-25)
APPENDIX C
Example
Figure C-1.
Variable / Value / Valuem / 3200 lbm / -
L1 / 4.5 ft / 54 in
L2 / 5.5 ft / 66 in
k1 / 2400 lbf / ft / 200 lbf/in
k2 / 2600 lbf / ft / 217 lbf/in
R / 4.0 ft / 48 in
Q=1.5 for each mode (33% viscous damping)
From Thomson, Theory of Vibration with Applications
The automobile is driven over a washboard road at 20 mph. The road surface is a sine wave with 2 inch height and 9 inch wavelength. Note that the height is zero-to-peak. The results are calculated via Matlab script.
mass =
1.0e+04 *
0.0001 0 0 0
0 0.0001 0 0
0 0 0.0008 0
0 0 0 1.9101
stiff =
200 0 -200 10800
0 217 -217 -14322
-200 -217 417 3522
10800 -14322 3522 1528452
TT =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
MT =
1.0e+04 *
0.0001 0 0 0
0 0.0001 0 0
0 0 0.0008 0
0 0 0 1.9101
KT =
200 0 -200 10800
0 217 -217 -14322
-200 -217 417 3522
10800 -14322 3522 1528452
Mwd =
0 0
0 0
Kwd =
-200 -217
10800 -14322
Mww =
1.0e+04 *
0.0008 0
0 1.9101
Kww =
417 3522
3522 1528452
Natural Frequencies
No. f(Hz)
1. 1.1011
2. 1.4452
Modes Shapes (column format)
ModeShapes =
0.3349 0.0922
-0.0019 0.0070
Transformed mode shapes for spring-to-mass attachment points
(column format)
0.4385 -0.2846
0.2081 0.5526
Calculate d
Calculate v
Calculate a
Maximum & Minimum Amplitudes
C.G. Acceleration (G)
0.1358 -0.1393
C.G. Jerk (G/sec)
33.37 -33.36
Rotational Acceleration (rad/sec^2)
2.337 -2.304
Displacement (in)
0.01852 -0.006955
Relative Displacement (in)
Spring 1: 2.045 -2.015
Spring 2: 2.048 -2.02
Figure C-2.
Figure C-3.
Figure C-4.
Figure C-5.
Figure C-6.
Figure C-7.
Figure C-8.
Figure C-9.
1