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SEMIDEFINITE TWO-DEGREE-OF-FREEDOM SYSTEM
SUBJECTED TO A SINUSOIDAL FORCE
Revision A By Tom Irvine Email: tom@vibrationdata.com May 2, 2014 _______________________________________________________________________ Two-degree-of-freedom System The method of generalized coordinates is demonstrated by an example. Consider the
system in Figure 1.
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.
Consider the case of free vibration.
The kinetic energy is
22x2m
2
121x1m
2
1T (1)
x2
k
x1
m2 m1
f1(t)
2
The potential energy is
2)2x1x(k2
1U (2)
0UTdt
d (3)
02)2x1x(k2
122x2m
2
121x1m
2
1
dt
d
(4)
02x)2x1x(k1x)2x1x(k2x2x2m1x1x1m (5)
02x)2x1x(k2x2m1x)2x1x(k1x1m (6)
Equation (6) yields two equations.
01x)2x1x(k1x1m (7)
02x)2x1x(k2x2m (8)
Divide through by the respective velocity terms
0)2x1x(k1x1m (9)
0)2x1x(k2x2m (10)
3
Assemble the equations in matrix form.
0
0
2x
1x
kk
kk
2x
1x
2m0
01m
(11)
Seek a solution of the form
tjexpqx (12)
The q vector is the generalized coordinate vector.
Note that
tjexpqjx (13)
tjexpqx 2 (14)
By substitution
0tjexpqKtjexpqM2 (15)
0tjexpqKqM2 (16)
0qKqM2n (17)
0qKM2 (18)
0MKdet 2 (19)
0m0
0m
kk
kkdet
2
12
(20)
4
02k2m2k1m2k
(21)
02k2m1m42m1m2k2k (22)
02m1m42m1m2k (23)
02m1m22m1mk2
(24)
Thus the first root is
01 (25)
01f (26)
Find the second root
02m1m22m1mk
(27)
21
212
mm
mmk (28)
2m1m
2m1mk2
(29)
2m1m
2m1mk
2
12f
(30)
5
The eigenvectors are found via the following equations.
0qMK 12
1 (31)
0qMK 22
2 (32)
For the first mode,
01 (33)
0qK 1 (34)
0qkk
kk1
(35)
The eigenvector is
1
1q1 (36)
Mass-normalize as follows
212
1
2
1mm
m
m11
1
1
m0
0m11
(37)
6
The mass-normalized eigenvector is
1
1
mm
1q
211 (38)
For the first mode,
21
212
mm
mmk (39)
0qMK 22
2 (40)
0q
m0
0m
mm
mmk
kk
kk2
2
1
21
21
(41)
0qm0
0mmmk
kk
kkmm 2
2
12121
(42)
0qm0
0mmmk
mmmm
mmmmk 2
2
121
2121
2121
(43)
\
0qm0
0mmm
mmmm
mmmm2
2
121
2121
2121
(44)
7
0q
mmmmmmm
mmmmmmm2
2212121
2112121
(45)
0qmmm
mmm22
221
212
1
(46)
The unscaled mode shape is
1
22
m
mq (47)
Mass-normalize as follows
2121
212
221
21
2112
1
2
2
112
mmmm
mmmmmm
mmmm
m
m
m0
0mmm
(48)
1
2
21212
m
m
mmmm
1q (49)
21 qqQ (50)
2121
1
21
2121
2
21
mmmm
m
mm
1
mmmm
m
mm
1
Q (51)
8
21
1
21
2
21mm
m1
mm
m1
mm
1Q (52)
2
1
1
2
21
m
m1
m
m1
mm
1Q (53)
Let r be the influence vector which represents the displacements of the masses resulting
from static application of a unit ground displacement.
Define a coefficient vector L as
rML T (54)
1
1
m0
0m
m
m
m
m11
mm
1L
2
1
2
1
1
2
21
(55)
2
1
2
1
1
2
21 m
m
m
m
m
m11
mm
1L (56)
9
0
mm
mm
1L
21
21
(57)
0
mmL 21 (58)
The modal participation factor matrix i for mode i is
iim
iLi (59)
Note that iim = 1 for each index if the eigenvectors have been normalized with respect
to the mass matrix.
211 mm (60)
02 (61)
The effective modal mass i,effm for mode i is
ii
2i
i,effm
Lm (62)
211,eff mmm (63)
0m 2,eff (64)
Assemble the equations in matrix form with the applied force.
10
0
)t(f
x
x
kk
kk
x
x
m0
0m 1
2
1
2
1
2
1
(65)
Decoupling
Equation (65) is coupled via the stiffness matrix. An intermediate goal is to decouple the
equation.
Simplify,
FxKxM
(66)
where
2
1
m0
0mM (67)
kk
kkK
(68)
2
1
x
xx (69)
0
)t(fF
1 (70)
21 qqQ (71)
IQMQT (72)
11
and
QKQT (73)
where
I is the identity matrix
is a diagonal matrix of eigenvalues
The superscript T represents transpose.
Note the mass-normalized forms
22
11
wv
wvQ (74)
21
21T
ww
vvQ (75)
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
Qx (76)
Substitute equation (76) into the equation of motion, equation (66).
FQKQM
(77)
Premultiply by the transpose of the normalized eigenvector matrix.
12
FQQKQQMQ TTT (78)
The orthogonality relationships yield
FQI T (79)
The equations of motion along with an added damping matrix become
0
)t(f
ww
vv
0
0
20
02
10
01 1
21
21
2
122
21
2
1
22
11
2
1
(80)
0
)t(f
ww
vv
0
00
20
00
10
01 1
21
21
2
1222
1
222
1
(81)
Note that the two equations are decoupled in terms of the generalized coordinate.
Equation (81) yields two equations
)t(fv 111 (82)
)t(fw2 112222222 (83)
The equations can be solved in terms of Laplace transforms, or some other differential
equation solution method.
Now consider the initial conditions. Recall
Qx (84)
13
Thus
)0(Q)0(x (85)
Premultiply by .MQT
)0(QMQ)0(xMQ TT (86)
Recall
IQMQT (87)
)0(I)0(xMQT (88)
)0()0(xMQT (89)
Finally, the transformed initial displacement is
)0(xMQ)0( T (90)
Similarly, the transformed initial velocity is
)0(xMQ)0( T (91)
A basis for a solution is thus derived.
Sinusoidal Force
Now consider the special case of a sinusoidal force applied to mass 1 with zero initial
conditions.
tsinA)t(f1 (92)
0)t(f2 (93)
14
Thus,
tsinAv11 (94)
tsinAw2 12222222 (95)
The equations are solved using the methods in References 3 and 4.
Take the Laplace transform of equation (94).
tsinAv11 (96)
}tsinAv{LL 11 (97)
2211112
sAv)0()0(s)s(ˆs (98)
2211112
sAv)0()0(s)s(ˆs (99)
)0()0(ss
Av)s(ˆs 1122112
(100)
)0(s
1)0(
s
1
ss
1Av)s(ˆ 12122211
(101)
)0(s
1)0(
s
1
s
1
s
11Av)s(ˆ 12122211
(102)
15
The solution is found via References 3 and 4. The inverse Laplace transform for the
first modal coordinate is
t)0()0(tsint1
Av)t( 11211
(103)
For zero initial conditions,
tsint1
Av)t(211
(104)
Recall the equation for the second modal coordinate.
tsinAw2 12222222 (105)
From Reference (5),
tsin)0()0(
tcos)0(texp
tsin21tcos2
2
texpAw
tsin1
tcos2
2
Aw
)t(
2,d2,d
22222,d222
2,d2
2
2
22
2,d2,d222
22
222
2
22
2,d
1
22
222
222
222
2
1
2
(106)
16
For zero initial conditions,
tsin21tcos2
2
texpAw
tsin1
tcos2
2
Aw
)t(
2,d2
2
2
22
2,d2,d222
22
222
2
22
2,d
1
22
222
222
222
2
1
2
(107)
The physical displacements are found via
Qx (108)
An example is given in Appendix A.
The transfer function can be calculated using the method in Appendix B.
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, Rev J, Vibrationdata, 2001.
4. T. Irvine, Partial Fraction Expansion, Rev K, Vibrationdata, 2013.
5. T. Irvine, Two-degree-of-freedom System Subjected to a Half-sine Pulse Force,
Vibrationdata, 2012.
6. R. Craig & A. Kurdila, Fundamentals of Structural Dynamics, Second Edition, Wiley,
New Jersey, 2006.
17
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
1m 2 lbm
2m 1 lbm
k 2000 lbf/in
A
1 lbf
f 171.3 Hz
The analysis is performed using a Matlab script. Note that the system is driven at its
second natural frequency.
>> semidefinite_force
semidefinite_force.m ver 1.4 May 2, 2014
Response of a semi-definite two-degree-of-freedom
system subjected to an applied sinusoidal force.
By Tom Irvine Email: tom@vibrationdata.com
Enter unit: 1=English 2=metric
1
Mass unit: lbm
Stiffness unit: lbf/in
Enter mass 1
2
Enter mass 2
1
Enter stiffness for spring between masses 1 & 2
2000
18
Natural Participation Effective
Mode Frequency Factor Modal Mass
1 5.096e-07 Hz 0.08816 0.007772
2 171.3 Hz 0 0
modal mass sum = 0.007772
mass matrix
m =
0.0052 0
0 0.0026
stiffness matrix
k =
2000 -2000
-2000 2000
ModeShapes =
11.3431 -8.0208
11.3431 16.0416
Enter viscous damping ratio 0.05
Apply sinusoidal force to mass 1
Enter force (lbf) 1
Enter excitation frequency (Hz) 171.3
Enter duration (sec) 0.1
19
Figure A-1.
20
Figure A-2.
21
Figure A-3.
22
Figure A-4.
23
Figure A-5.
The rigid-body mode has been suppressed.
24
Figure A-6.
The rigid-body mode has been suppressed.
25
Figure A-7.
The rigid-body mode has been suppressed.
26
APPENDIX B
Transfer Function
The following is taken from Reference 6.
The variables are:
F Excitation frequency
f r Natural frequency for mode r
N Total degrees-of-freedom
)f(H ji The steady state displacement at coordinate i due to a harmonic force
excitation only at coordinate j
r Damping ratio for mode r
ri Mass-normalized eigenvector for physical coordinate i and mode number r
Excitation frequency (rad/sec)
r Natural frequency (rad/sec) for mode r
The following equations are for a general system. Note that r should be given an initial
value of 2 in order to suppress the rigid-body mode for the case of the semi-definite,
two-degree-of-freedom system. This is needed since the fundamental frequency is zero,
aside from numerical error.
Receptance
The steady-state displacement at coordinate i due to a harmonic force excitation only at
coordinate j is:
N
1r rr2
r2
r
rjriji
2j1
1)f(H
(B-1)
where
rr f/f (B-2)
1j (B-3)
27
Note that the phase angle is typically represented as the angle by which force leads
displacement. In terms of a C++ or Matlab type equation, the phase angle would be
Phase = - atan2(imag(H), real(H)) (B-4)
Note that both the phase and the transfer function vary with frequency.
A more formal equation is
)f(Hreal
)f(Himagarctan)f(Phase
ji
ji
(B-5)
Mobility
The steady-state velocity at coordinate i due to a harmonic force excitation only at
coordinate j is
N
1r rr2
r2
r
rjri
2j1
1j)f(jiH
(B-6)
Accelerance
The steady-state acceleration at coordinate i due to a harmonic force excitation only at
coordinate j is
N
1r rr2
r2
r
rjri2ji
2j1
1)f(H
~
(B-7)
28
Relative Displacement
Consider two translational degrees-of-freedom i and j. A force is applied at degree-of-
freedom k.
The steady-state relative displacement transfer function Rij between i and j due to an
applied force at k is
N
1r rr2
r2
r
rkrjN
1r rr2
r2
r
rkri
kjkiji
2j1
1
2j1
1
)f(H)f(H R
(B-8)
N
1r rr2
r2
r
rkrjriji
2j1
1R (B-9)
The steady-state relative displacement transfer function Rij between i and j due to an
applied force at k is
N
1r rr2
r2
r
rkrjN
1r rr2
r2
r
rkri
kjkiji
2j1
1
2j1
1
)f(H)f(H R
(B-10)
29
N
1r 2rr
22
r
rr2
r
rkrjri
N
1r 2rr
22
r
2r
2r
rkrjriji
21
2j
21
1
R
(B-11)
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