Download x 1 - Vibrationdata

TWO-DEGREE-OF-FREEDOM SYSTEM
SUBJECTED TO A HALF-SINE PULSE FORCE
Revision A
By Tom Irvine
Email: [email protected]
March 21, 2014
_______________________________________________________________________
Two-degree-of-freedom System
The method of generalized coordinates is demonstrated by an example. Consider the
system in Figure 1.
f2(t)
k3
x2
m2
k2
f1(t)
x1
m1
k1
Figure 1.
A free-body diagram of mass 1 is given in Figure 2. A free-body diagram of mass 2 is
given in Figure 3.
1
k2 (x2-x1)
f1(t)
x1
m1
k1x1
Figure 2.
Determine the equation of motion for mass 1.
 F  m1 x1
(1)
m1x1  f1(t)  k 2 (x 2  x1)  k1x1
(2)
m1x1  k 2 (x 2  x1)  k1x1  f1(t)
(3)
m1x1  k 2 (x 2  x1)  k1x1  f1(t)
(4)
m1x1  (k1  k 2 )x1  k 2 x 2  f1(t)
(5)
2
f2(t)
k3 x2
x2
m2
k2 (x2-x1)
Figure 3.
Derive the equation of motion for mass 2.
 F  m2 x 2
(6)
m 2 x 2  f 2 (t )  k 2 (x 2  x1 )  k 3 x 2
(7)
m 2 x 2  k 2 (x 2  x1 )  k 3 x 2  f 2 (t )
(8)
m 2 x 2  (k 2  k 3 )x 2  k 2 x1  f 2 (t )
(9)
Assemble the equations in matrix form.
m1 0   x1  k1  k 2
 0 m  x     k
2
2  2 

 k 2   x1   f1 (t ) 

k 2  k 3  x 2  f 2 (t )
(10)
Decoupling
Equation (10) is coupled via the stiffness matrix. An intermediate goal is to decouple the
equation.
3
Simplify,
M x  K x  F
(11)
where
0 
m
M 1

 0 m2 
(12)
k  k 2
K 1
  k2
(13)
 k2 
k 2  k 3 
x 
x   1
x 2 
(14)
 f (t) 
F 1 
f 2 ( t ) 
(15)
Consider the homogeneous form of equation (11).
M x  K x  0
(16)
Seek a solution of the form
x  q exp( jt )
(17)
The q vector is the generalized coordinate vector.
Note that
x  j q exp( jt )
(18)
x  2 q exp( jt )
(19)
4
Substitute equations (17) through (19) into equation (16).
 2 M q exp( jt )  K q exp( jt )  0
(20)
 2 M q  K q exp( jt)  0
(21)
 2 M q  K q  0
(22)
 2 M  Kq  0
K  2 Mq  0
(23)
(24)
Equation (24) is an example of a generalized eigenvalue problem. The eigenvalues can
be found by setting the determinant equal to zero.


det K  2 M  0
k  k 2
det  1
  k 2
(25)
 k2 
0 
2 m1


 0 m   0
k 2  k 3 
2 


k  k 2   2 m1
det  1
 k2





0
k 2  k 3   2 m 2  
 k2
k1  k 2   2m1 k 2  k 3   2m 2  k 22  0
(26)
(27)
(28)
5
k1  k 2 k 2  k 3   2 m1 k 2  k 3   2 m 2 k1  k 2   4 m1m 2  k 2 2  0
(29)
m1m 2 4   m1 k 2  k 3   m 2 k1  k 2 2  k1k 3  k1  k 3 k 2  k 2 2  k 2 2  0
(30)
m1m 2 4   m1 k 2  k 3   m 2 k1  k 2 2  k1k 3  k1k 2  k 2 k 3  0
(31)
The eigenvalues are the roots of the polynomial.
 b  b 2  4ac
2a
(32)
2
2  b  b  4ac
2 
2a
(33)
2
1 
where
a  m1m 2
(34)
b   m1 k 2  k 3   m 2 k1  k 2 
(35)
c  k1k 2  k1k 3  k 2 k 3
(36)
The eigenvectors are found via the following equations.
K  12 Mq1  0
(37)
K  22 Mq2  0
(38)
6
where
v 
q1   1 
v 2 
(39)
w 
q2   1 
w 2 
(40)
An eigenvector matrix Q can be formed. The eigenvectors are inserted in column
format.
Q  q1 q 2 
v
Q 1
v 2
w1 
w 2 
(41)
(42)
The eigenvectors represent orthogonal mode shapes.
Each eigenvector can be multiplied by an arbitrary scale factor. A mass-normalized
eigenvector matrix Q̂ can be obtained such that the following orthogonality relations
are obtained.
Q̂ T M Q̂  I
(43)
and
Q̂ T K Q̂  
(44)
where
I

is the identity matrix
is a diagonal matrix of eigenvalues
The superscript T represents transpose.
7
Note the mass-normalized forms
 v̂
Q̂   1
 v̂ 2
ŵ1 
ŵ 2 
v̂ 2 
 v̂
Q̂ T   1

ŵ1 ŵ 2 
(45)
(46)
Rigorous proof of the orthogonality relationships is beyond the scope of this tutorial.
Further discussion is given in References 1 and 2.
Nevertheless, the orthogonality relationships are demonstrated by an example in this
tutorial.
Now define a generalize coordinate ( t ) such that
x  Q̂ 
(47)
Substitute equation (47) into the equation of motion, equation (11).
  K Q̂   F
M Q̂ 
(48)
Premultiply by the transpose of the normalized eigenvector matrix.
  Q̂ T K Q̂   Q̂ T F
Q̂ T M Q̂ 
(49)
The orthogonality relationships yield
     Q̂ T F
I
(50)
8
The equations of motion along with an added damping matrix become
2
 1  21 1
0    1  1
1 0  


0 1 
2 2  2   2   0

  2   0

0   1   v̂1


2   
 2   2  ŵ1
v̂ 2   f1 ( t ) 
ŵ 2  f 2 ( t )
(51)
Note that the two equations are decoupled in terms of the generalized coordinate.
Equation (51) yields two equations
 1 21 1 1  2 1 v̂1f1 ( t )  v̂ 2 f 2 ( t )

1
(52)
 2  2 2 2 
 2   2  2  ŵ1f1 ( t )  ŵ 2 f 2 ( t )

2
(53)
The equations can be solved in terms of Laplace transforms, or some other differential
equation solution method.
Now consider the initial conditions. Recall
x  Q̂ 
(54)
Thus
x (0)  Q̂ (0)
(55)
Premultiply by Q̂ T M.
Q̂ T M x (0)  Q̂ T MQ̂ (0)
(56)
Recall
Q̂ T M Q̂  I
(57)
9
Q̂ T M x (0)  I (0)
(58)
Q̂ T M x(0)  (0)
(59)
Finally, the transformed initial displacement is
(0)  Q̂ T M x(0)
(60)
Similarly, the transformed initial velocity is
 (0)  Q̂ T M x (0)

(61)
A basis for a solution is thus derived.
Half-Sine Force
Now consider the special case of a half-sine force applied to mass 2.
f1(t )  0
f 2 (t )  B2 sin t  ,
(62)
for t < T
(63)
where
  /T
Thus,
 1 21 1 1  2 1 v̂ 2 B 2 sin t 

1
(64)
 2  2 2 2  2   2  2  ŵ 2 B 2 sin t 

2
(65)
Take the Laplace transform of equation (64).
 1 21 1 1  2 1 v̂ 2 B 2 sin t 

1
(66)
10


 1 21 1
 1  1  L{v̂ 2 B 2 sin t }
L
1
2
(67)
s 2 ˆ 1(s)  s1 (0)   1(0)
 21 1sˆ 1(s)  21 11 (0)
  
2
 1 ˆ 1 (s)  v̂ 2 B 2 

 s 2   2 
(68)
s

2  2  s   2 
 1(0)
1 1
n ˆ 1 (s)  s  21 11(0)  
  
 v̂ 2 B 2 

 s 2   2 
(69)
s


  

2  2  s   2 
 1 (0)
  s  21 11 (0)  
1 1
n ˆ 1 (s)  v̂ 2 B 2  2
2


s   
ˆ 1 (s) 

  
 s  21 11 (0)   1 (0)


2 s 2  2 
2

s 2  21 1s  n 
s 2  21 1s  n



v̂ 2 B 2



(70)
(71)
11
The solution is found via References 3 and 4. The inverse Laplace transform for the
first modal coordinate is
 1( t ) 


  2  cost    2   2  sin t 


1 1
1 




 21  1 2 

v̂ 2 B 2
 2
2 2
   1 




 

v̂ 2 B 2
exp  11t 



2
 d,1 

2
 
  2
211 d,1  cos d,1t     1 1  21

 
 
 2
2 2
2 
   1   21  1  





  (0)  1 1 1(0) 
 exp  11t 1(0) cos d,1t   1
 sin d,1t
d,1






 sin  d,1t 


 ,

for t < T
(72)
12
Similarly,
 2 (t) 



1 2
2 
  2 2  2  cost      2  sin t 




ŵ 2 B 2
 2
2 2
2 
    2   2 2   2  







 

ŵ 2 B 2 
 exp   2  2 t 
  d,2 
2

2
 
  2
2 2  2  d,2  cos  d,2 t      2 1  2 2
2




 2


2
2 
    2   2 2   2  





  (0)   2  2  2 (0) 
 exp   2  2 t  2 (0) cos  d,2 t   2
 sin  d,2 t
 d,2






 sin  d,2 t 




for t < T
(73)
The physical displacements are found via
x  Q̂ 
(74)
Free Vibration
For t > T and  = t-T




  (T)  11 1(T) 


 1( t )  exp  11 1(T) cos d,1   1
sin



d,1 
d,1








(75)



  (T)   2 2  2 (T) 

 2 ( t )  exp   2  2   2 (T) cos d,2    2
 sin d,2 
 d, 2






(76)











13
References
1. Bathe, Finite Element Procedures in Engineering Analysis, Prentice-Hall, New
Jersey, 1982. Section 12.3.1.
2. Weaver and Johnston, Structural Dynamics by Finite Elements, Prentice-Hall,
New Jersey, 1987. Chapter 4.
3. T. Irvine, Table of Laplace Transforms, Vibrationdata, 2000.
4. T. Irvine, Partial Fraction Expansion, Rev F, Vibrationdata, 2010.
14
APPENDIX A
Example
Consider the system in Figure 1 with the values in Table A-1.
Assume 5% damping for each mode. Assume zero initial conditions.
Table A-1. Parameters
Variable
Value
Unit
m1
m2
k1
k2
k3
3.0
lbf sec^2/in
2.0
lbf sec^2/in
400,000
lbf/in
300,000
lbf/in
100,000
lbf/in
100
lbf
0.011
sec
B2

The mass matrix is
0  3 0 
m
M 1


 0 m 2  0 2
(A-1)
The stiffness matrix is
k  k 2
K 1
  k2
 k 2   700,000  300,000

k 2  k 3   300,000 400,000 
(A-2)
The analysis is performed using a Matlab script.
15
>> twodof_half_sine_force
twodof_half_sine_force.m
ver 1.0 January 18, 2012
by Tom Irvine Email: [email protected]
This script calculates the response of a two-degree-of-freedom
system to a half-sine force excitation at dof 2.
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:
Stiffness Matrix
Enter the matrix name:
Input Matrices
m_two
k_two
mass =
3
0
0
2
stiff =
700000
-300000
-300000
400000
Natural Frequencies
No.
f(Hz)
1.
48.552
2.
92.839
Modes Shapes (column format)
ModeShapes =
0.3797
0.5326
-0.4349
0.4651
16
Enter the damping ratio for mode 1 0.05
Enter the damping ratio for mode 2 0.05
Particpation Factors =
2.204
-0.3746
Enter the force amplitude (lbf) 100
Enter the half-sine pulse duration (sec)
0.011
Enter the first initial displacement (in)
Enter the second initial displacement (in)
0
0
Enter the first initial velocity (in/sec)
Enter the second initial velocity (in/sec)
0
0
Enter the sample rate (samples/sec) 10000
Enter the analysis duration (sec) 0.15
dof 1 displacement (in)
max= 0.0003287
min= -0.0003149
dof 2 displacement (in)
max= 0.0005005
min= -0.0003728
17
DISPLACEMENT
0.0010
Mass 2
Mass 1
DISP (INCH)
0.0005
0
-0.0005
-0.0010
0
0.05
0.10
0.15
TIME (SEC)
Figure A-1.
18
RELATIVE DISPLACEMENT
(MASS 2 - MASS 1)
0.0004
0.0003
REL DISP (INCH)
0.0002
0.0001
0
-0.0001
-0.0002
-0.0003
-0.0004
0
0.05
0.10
0.15
TIME (SEC)
Figure A-2.
19
VELOCITY
0.15
Mass 2
Mass 1
0.10
VEL (IN/SEC)
0.05
0
-0.05
-0.10
-0.15
0
0.05
0.10
0.15
TIME (SEC)
Figure A-3.
20
ACCELERATION
0.15
Mass 2
Mass 1
0.10
ACCEL (G)
0.05
0
-0.05
-0.10
-0.15
0
0.05
0.10
0.15
TIME (SEC)
Figure A-4.
21