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1
Modal Transient Analysis of a System Subjected to an Applied Force
via a Ramp Invariant Digital Recursive Filtering Relationship
Revision L
By Tom Irvine
Email: tom@vibrationdata.com
December 19, 2013
_______________________________________________________________________________
This paper applies Smallwood’s methodology for base excitation in Reference 1 to the case of an
applied force. It also extends the method to multi-degree-of-freedom systems.
Ahlin & Brandt purported in Reference 2 to use the method given in the present paper, but they
omitted a derivation and the resulting filtering coefficients.
This paper fills the gap and gives accompanying Matlab scripts. The goals are to publicize the
method and to place the filter coefficients equations in the public domain.
Introduction
The purpose of this paper is to derive an efficient and accurate method for the modal transient
analysis of a dynamic system subjected to an external force or forces. The method will be a digital
recursive filtering relationship using the ramp invariant technique which models the slope between
adjacent points of the input force.
Assumptions
A modal analysis of the system has been previously performed. The system has been reduced to
uncoupled mass, damping and stiffness matrices per the method given in Reference 3, as well as in
common structural dynamics textbooks. The physical responses can then be recovered from the
modal responses after the transient analysis.
The initial conditions are zero. Note that the response to initial conditions can be solved exactly
using Laplace transforms, as needed.
Forcing Function
The forcing function may vary arbitrarily with time.
The forcing function may be either from measured data1 or from a synthesis.
1 Measured data should be collected using the proper sample rate and analog anti-aliasing filter.
2
The time step should be chosen so that there are at least ten points per cycle for the highest natural
frequency of interest. In other words, the sample rate should be at least ten times the highest
natural frequency of interest. This is an industry standard rule-of-thumb for time domain
calculations per Reference 4.
Smallwood wrote in his seminal paper that the ramp invariant method allows for analysis
frequencies approaching the Nyquist frequency, which in one-half the sample rate. But the x10
rule is still recommend in the present paper.
Note that the time step and corresponding sample rate are fixed for measured data as the data is
being collected. The time step can be shortened via interpolation or a cubic spline fit as a post-
processing step, but caution must be exercise in either case.
The time step can be set as small as needed for the case where a function is to be synthesized.
Alternatively, the number of modes included in analysis can be truncated if needed to meet the
rule-of-thumb. The highest frequency of interest is thus lowered as a trade-off.
Candidate Methods
The methods for the modal transient analysis are:
1. Convolution integral converted to a nested series for digital data
2. Convolution integral represented as digital cursive filtering relationship
3. Runge-Kutta fourth order method
4. Newmark-beta implicit numerical integration method
Convolution Integral
The Convolution method would yield an exact answer for the response if the system and the
forcing function could be analyzed in analog rather than digital form. This is impractical,
however.
The nested series representation of the Convolution integral is not commonly used because it is
numerically inefficient.2
The Convolution integral represented as digital cursive filtering relationship. The derivation is
performed using a Z-transform.
2 Convolution can also be performed in the frequency domain my by multiplying the Fourier
transform of the applied force by the frequency response function of the system. The inverse
Fourier transform of the product gives the response time history. A Fast Fourier transform (FFT)
and its corresponding inverse can be used to expedite the calculation. Note that there are some
potential error sources with this method such as leakage.
3
There are two Z-transform approaches: the impulse invariant and ramp invariant simulations. The
ramp invariant simulation is preferred because it adds a filtering term and has better accuracy at
frequencies approach the Nyquist frequency per Reference 1.
Note that either digital recursive filtering relationship requires a constant time step.
Newmark-beta Method
The Newmark-beta method is favored in structural dynamics books for multi-degree-of-freedom
systems, as given in References 5 and 6. It can accept a variable time step for the force input. Its
true strength is that it can be used for the direct integration of a system of uncoupled mass,
damping, and stiffness matrices. Obviously, it can be applied to an uncoupled system as well.
The Newmark-beta method is derived from the continuous time equation familiar to high school
physics students.
2tu2
1tuu (0)
It is typically implemented with 4/1 , thus assuming constant average acceleration over the
time step.
Runge-Kutta Method
The Runge-Kutta method extends the Taylor series method by estimating higher order derivatives
at points within the time step.
There are many types of Runge-Kutta algorithms. A common type which may be used for an
arbitrary forcing function is the Runge-Kutta fourth order method. This method is given in
Reference 7.
But the Runge-Kutta fourth order method may give unstable results3 for stiff systems with higher
natural frequencies.
The probability that instability will occur is difficult to predict. It depends on both the natural
frequency and the time step. The problem can be mitigated by using a smaller time step, but this
requires a longer analysis time.
If an instability occurs, then the analysis be repeated using a fewer number of modes for the case
of a multi-degree-of-freedom system.
3 “To infinity and beyond,” recalling Buzz Lightyear.
4
SDOF Equation of Motion
Consider a single-degree-of-freedom system.
Figure 1.
The variables are
M is the mass
C is the viscous damping coefficient
K is the stiffness
X is the absolute displacement of the mass
f(t) is the applied force
Note that the double-dot denotes acceleration.
The free-body diagram is
Figure 2.
m
k x xc
x
f(t)
f(t)
m
k c
x
5
Summation of forces in the vertical direction
F mx (1)
mx cx kx f t ( ) (2)
mx cx kx f t ( ) (3)
Divide through by m,
( )xc
mx
k
mx
mf t
1 (4)
By convention,
n2)m/c( (5)
2n)m/k( (6)
where
n is the natural frequency in (radians/sec)
is the damping ratio.
By substitution,
( )x x xm
f tn n 212 (7)
Equation (7) does not have a closed-form solution for the general case in which f(t) is an arbitrary
function. A convolution integral approach must be used to solve the equation.
The solution method proceeds by finding a solution to the homogeneous form of equation (7). In
other words, the solution is found for f(t)=0.
x x xn n 2 02 (8)
Note that equation (8) is essentially the same as equation (A-1) in Reference 8, except that
equation (8) is expressed in terms of absolute displacement.
6
Displacement
The damped natural frequency d is
2d -1n (9)
The displacement equation via a convolution integral is
d)-t(sin)-t(nexp)f(
m
1=)tx(
t
0d
d
(10)
The corresponding impulse response function for the displacement is
tsintnexpm
1=)t(h d
dd
(11)
Further details regarding this derivation are given in Reference 9.
The corresponding Laplace transform from Reference 10 is
2n
2dsn2s
1
m
1=)s(H (12)
The Z-transform for the ramp invariant simulation is
2n
22
12
dsn2ss
1LZ
Tz
1z
m
1=)z(H
where T is the time step
(13)
7
Evaluate the inverse Laplace transform per References 10 and 11.
tsin21tcos2texpt21
sn2ss
1L
d2
d
2n
dnn2nn4
n
2n
22
1
(14)
The Z-transform becomes
Tsin21Tcos2TexpT2Z
Tz
1z
m
1
=)z(H
d2
d
2n
dnn2nn
2
4n
d
(15)
Let
2
d
2n 21
(16)
n2
(17)
8
The Z-transform is evaluated using the method in Reference 12.
T2expTcosTexpz2z
TsinTexpz
Tz
1z
m
1
T2expTcosTexpz2z
TcosTexpzz
Tz
1z
m
1
1z
zT
Tz
1z
m
1z
z
Tz
1z
m
-)z(H
ndn2
dn2
4n
ndn2
dn2
4n
2
2
4n
2n
2
4n
d
(18)
T2expTcosTexpz2z
TsinTexp
T
1z
m
1
T2expTcosTexpz2z
TcosTexpz
T
1z
m
1
T
1z
m
1)z(H
ndn2
dn2
4n
ndn2
dn2
4n
2n4
n
d
(19)
9
T2expTcosTexpz2z
TsinTexpTcosTexpz
T
1z
m
1
T
1z
m
1)z(H
ndn2
dndn2
4n
2n4
n
d
(20)
T2expTcosTexpz2z
TcosTsinTexpz1z
Tm
1
TzTm
1)z(H
ndn2
ddn2
4n
2n4
n
d
(21)
Let
TcosTsinTexp ddn
(22)
TcosTexp2 dn
(23)
T2exp n
(24)
T2n
(25)
By substitution,
zz
z1z
Tm
1z
Tm
1)z(H
2
2
4n
4n
d
(26)
zz
z1z2z
Tm
1
zz
zzz
Tm
1)z(H
2
2
4n
2
2
4n
d
(27)
10
zz
z
Tm
1
zz
zz2
Tm
1
zz
zz
Tm
1
zz
zz
Tm
1
zz
zzz
Tm
1)z(H
24n
24n
2
2
4n
2
2
4n
2
2
4n
d
(28)
zz
z
Tm
1
zz
z2z2
Tm
1
zz
zz
Tm
1
zz
zz
Tm
1
zz
zzz
Tm
1)z(H
24n
2
2
4n
2
23
4n
2
2
4n
2
23
4n
d
(29)
zzTm
zz2z2zzzzzzz)z(H
24n
223223
d
(30)
zzTm
z2z2)z(H
24n
2
d
(31)
11
Solve for the filter coefficients using the method in Reference 1.
zz
Tm/z2z2
azaz
bzbzb
2
4n
2
212
212
0
(32)
Solve for a1.
TcosTexp2a dn1
(33)
Solve for a2.
T2expa n2
(34)
Note that the a1 and a2 coefficients are common for displacement, velocity and acceleration.
Solve for b0.
Tm/2b
4n0
(35)
Tm/2b
4n0
(36)
Tm
2TcosTexp2TTcosTsinTexpb
4n
dn2nddn
0
(37)
Tm
1TcosTexp2TTcosTsinTexpb
4n
dn2nddn
0
(38)
12
Tm
1TcosTexp4T2
Tm
Tcos2Tsin21Texp
b
4n
dnn2nn
4n
dnd2
d
2n
n
0
(39)
Tm
1TcosTexp2TTsin21Texp
b4n
dnn2nd
2
d
2n
n
0
(40)
Tm
1TcosTexp2TTsin21Texp
b3
n
dnnd2
d
nn
0
(41)
Tm
TTsin12Texp1TcosTexp2
b3
n
nd2
d
nndn
0
(42)
13
Solve for b1.
Tm
2b
4n
1
(43)
Tm
12b
4n
1
(44)
Tm
T2exp1
Tm
TcosTsinTexp2TcosTexpT2
b
4n
n
4n
ddndn2n
1
(45)
Tm
T2exp1TsinTexp2TcosTexpT2b
4n
ndndn2n
1
(46)
Tm
T2exp1TsinTexp2TcosTexpT2b
4n
ndndn2n
1
(47)
14
Tm
T2exp12
Tm
TsinTexp212TcosTexpT2
b
4n
nn
4n
dn2
d
2n
dn2n
1
(48)
Tm
TsinTexp122T2exp12TcosTexpT2
b
3n
dn2
d
nndnn
1
(49)
Solve for b2.
Tmb
4n
2
(50)
Tm
TcosTsinTexpT2expT
b4n
ddnn2n
2
(51)
Tm
Tcos2Tsin21TexpT2expT2
b
4n
dnd2
d
2n
nn2nn
2
(52)
15
Tm
Tcos2Tsin12TexpT2expT2
b3n
dd2
d
nnnn
2
(53)
The digital recursive filtering relationship for the displacement is
2i21i1i0
2i21i1i
fbfbfb
xaxax
(54)
16
The digital recursive relationship for the displacement is thus
2idd2
d
nnnn3
n
1idn2
d
nndnn3
n
ind2
d
nndn3
n
2in
1idn
i
fTcos2Tsin12TexpT2expT2Tm
1
fTsinTexp122T2exp12TcosTexpT2Tm
1
fTTsin12Texp1TcosTexp2Tm
1
xT2exp
xTcosTexp2
x
(55)
17
Velocity
The impulse response function for the velocity response is
tcostsinn)tnexp(-
m
1=)t(h dd
dv (56)
The corresponding Laplace transform is
2n
2vsn2s
s
m
1=)s(H (57)
The Z-transform for the ramp invariant simulation is
2n
22
12
vsn2ss
sLZ
Tz
1z
m
1=)z(H (58)
2n
2
12
vsn2ss
1LZ
Tz
1z
m
1=)z(H (59)
Evaluate the inverse Laplace transform per References 10 and 11.
tsintcostexp11
sn2ss
1L d
d
ndn2
n2n
2n
2
1 (60)
The Z-transform for the ramp invariant simulation is
TsintcosTexp
11Z
Tz
1z
m
1=)z(H d
d
ndn2
n2n
2
v
(61)
18
TsintcosTexp1Z
Tz
1z
m
1=)z(H d
d
ndn
2
2n
v
(62)
The Z-transform is evaluated using the method in Reference 12.
T2exp)Tcos()Texp(z2z
TsinTexpz)Tcos()Texp(zz
1z
z
Tz
1z
m
1=)z(H
ndn2
dnd
ndn2
2n
v
(63)
T2exp)Tcos()Texp(z2z
TsinTexpz)Tcos()Texp(zz
1z
z
z
1z
Tm
1=)z(H
ndn2
dnd
ndn2
2n
v
(64)
T2exp)Tcos()Texp(z2z
TsinTexp)Tcos()Texp(z
1z1zTm
1=)z(H
ndn2
dnd
ndn
2
2n
v
(65)
19
T2exp)Tcos()Texp(z2z
Tsin)Tcos()Texp(z
1z1zTm
1=)z(H
ndn2
dd
ndn
2
2n
v
(66)
Let
Tsin)Tcos()Texp( d
d
ndn
(67)
TcosTexp2 dn
(68)
T2exp n
(69)
zz
z1z1z
Tm
1=)z(H
2
2
2n
v
(70)
zz
z1z2z1z
Tm
1=)z(H
2
2
2n
v
(71)
zz
1z2z1z2zz1z
Tm
1=)z(H
2
22
2n
v
(72)
zz
z2zzz2z1z
Tm
1=)z(H
2
223
2n
v
(72)
zz
z2zzz2z1z
Tm
1=)z(H
2
223
2n
v
(74)
20
zz
z21z2z1z
Tm
1=)z(H
2
23
2n
v
(75)
zz
z21z2z1z
Tm
1=)z(H
2
23
2n
v
(76)
zz
z21z2z
zz
zz1z
Tm
1=)z(H
2
23
2
2
2n
v
(77)
zz
z21z2z
zz
zzzzz
Tm
1=)z(H
2
23
2
223
2n
v
(78)
zz
z21z2z
zz
zz1z
Tm
1=)z(H
2
23
2
23
2n
v
(79)
zz
z21z12
Tm
1=)z(H
2
2
2n
v
(80)
zz
z12z1
Tm
1=)z(H
2
2
2n
v
(81)
21
zz
z12z1
Tm
1=)z(H
2
2
2n
v
(82)
Solve for the filter coefficients using the method in Reference 1.
zz
z12z1
Tm
1=
azaz
czczc
2
2
2n21
2
212
0
(83)
Solve for a1.
TcosTexp2a dn1
(84)
Solve for a2.
T2expa n2
(85)
Solve for c0.
Tm
1c
2n
0
(86)
Tm
1TcosTexp2Tsin)Tcos()Texp(
c2n
dndd
ndn
0
(87)
Tm
1Tsin)Tcos()Texp(
c2n
dd
ndn
0
(88)
22
Solve for c1.
Tm
12c
2n
1
(89)
Tm
1Tsin)Tcos()Texp(2TcosTexp2T2exp
c2n
dd
ndndnn
1
(90)
Tm
1Tsin)Texp(2T2exp
c2n
dd
nnn
1
(91)
Solve for c2.
Tmc
2n
2
(92)
Tm
T2expTsin)Tcos()Texp(
c2n
ndd
ndn
2
(93)
The digital recursive filtering relationship for the velocity is
2i21i1i0
2i21i1i
fcfcfc
xaxax
(94)
23
The digital recursive relationship for the velocity is thus
2indd
ndn2
n
1idd
nnn2
n
idd
ndn2
n
2in
1idn
i
fT2expTsin)Tcos()Texp(Tm
1
f1Tsin)Texp(2T2expTm
1
f1Tsin)Tcos()Texp(Tm
1
xT2exp
xTcosTexp2
x
(95)
Acceleration
The acceleration response for a unit impulse is
tsin12tcos2texp)t(
m
1ty d
2
d
2n
dnn (96)
The corresponding Laplace transform is
2
n2
2
asn2s
s
m
1=)s(H (97)
The Z-transform is
2n
22
21
2
asn2ss
sLZ
Tz
1z
m
1=)z(H
(98)
24
2n
22
21
2
asn2ss
sLZ
Tz
1z
m
1=)z(H
(99)
2n
2
12
asn2s
1LZ
Tz
1z
m
1=)z(H
(100)
Evaluate the inverse Laplace transform per Reference 10 and 11.
2d
2n
1
2n
2
1
s
1L
sn2s
1L
(101)
tsintnexp1
sn2s
1L
d
d2
n2
1
(102)
The Z-transform for the ramp invariant simulation is
TsinTnexp
1Z
Tz
1z
m
1=)z(H
d
d
2
a
(103)
TsinTnexp
1Z
Tz
1z
m
1=)z(H
d
d
2
a
(104)
The Z-transform is evaluated using the method in Reference 12.
T2exp)Tcos()Texp(z2z
TsinTexpz
Tz
1z
m
1=)z(H
ndn2
dn2
d
a
(105)
25
T2exp)Tcos()Texp(z2z
TsinTexp1z
Tm
1=)z(H
ndn2
dn2
d
a
(106)
T2exp)Tcos()Texp(z2z
1z
Tm
TsinTexp=)z(H
ndn2
2
d
dna
(107)
T2exp)Tcos()Texp(z2z
1z2z
Tm
TsinTexp=)z(H
ndn2
2
d
dna
(108)
Solve for the filter coefficients using the method in Reference 1.
T2exp)Tcos()Texp(z2z
1z2z
Tm
TsinTexp
azaz
dzdzd
ndn2
2
d
dn
212
212
0
(109)
Solve for a1.
TcosTexp2a dn1
(110)
Solve for a2.
T2expa n2
(111)
26
Solve for d0.
Tm
TsinTexpd
d
dn0
(112)
Solve for d1.
01 d2d
(113)
Solve for d2.
02 dd
(114)
The digital recursive filtering relationship for the acceleration is
2i21i1i02i21i1i fdfdfdxaxax
(115)
2i1ii
d
dn
2in1idni
ff2fTm
TsinTexp
xT2expxTcosTexp2x
(116)
There are three response parameters: displacement, velocity and acceleration. If two are known,
the third can be calculated via equation (7).
Transmitted Force
The impulse response function for transmitted force from Reference 14 is
tsin21tcos2texp=)t(h d
2
d
2n
dnnft (117)
27
The corresponding Laplace transform is
2n
2
2nn
ftsn2s
s2=)s(H (118)
The Z-transform is
2n
2
2nn
2
12
ftsn2s
s2
s
1LZ
Tz
1z=)z(H
(119)
Note that the transmitted force impulse response function is essentially the same as the absolute
acceleration impulse response function for the case of base excitation. The digital recursive
filtering relationship can thus be taken from Reference 15.
2idnd
n
1idd
dn
idnd
2i,tn
1i,tdni,t
fTsinTexpT
1T2exp
fTsinT
1TcosTexp2
fTsinTexpT
11
ft2exp
ftcostexp2f
(120)
28
SDOF Example
An SDOF system has a natural frequency of 10 Hz with an amplification of Q=10. Its mass is 1
lbm. It is subjected to 1 lbf sinusoidal excitation at is natural frequency as shown in Figure 3.
The analysis is performed using the ramp invariant digital recursive filters derived in this paper as
implemented in Matlab script: arbit_force.m.
This is a simple problem which does not demonstrate the full power of the ramp invariant filtering
method for calculating the response to a force which varies arbitrarily with time.
But the simple example is useful for checking the accuracy of the method. The peak results agree
with the expected values from the corresponding frequency response functions.
The descriptive statistics for the response from the Matlab script are:
Displacement Response
maximum = 0.97 inch
minimum = -0.971 inch
overall = 0.601 inch RMS
Velocity Response
maximum = 61.6 in/sec
minimum = -61.6 in/sec
overall = 38.1 in/sec RMS
Acceleration Response
maximum = 9.98 G
minimum = -9.97 G
overall = 6.17 G RMS
29
Figure 3.
-1.0
-0.5
0
0.5
1.0
0 0.5 1.0 1.5 2.0
TIME (SEC)
FO
RC
E (
lbf)
APPLIED FORCE
30
Figure 4.
The resulting displacement, velocity and acceleration responses for the ramp invariant method are
shown in Figures 4, 5 and 6, respectively. The exact results from the Laplace transform
calculation are superimposed for comparison, as calculated from the formulas Reference 13 as
implemented via Matlab script: sdof_sine_force.m.
There is excellent agreement between the two displacement curves as shown in Figure 4.
-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1.0
1.2
0 0.5 1.0 1.5 2.0
LaplaceRamp Invariant
TIME (SEC)
DIS
P (
inch
)
SDOF DISPLACEMENT RESPONSE fn=10 Hz Q=10
31
Figure 5.
There is excellent agreement between the two velocity curves as shown in Figure 5.
-100
-80
-60
-40
-20
0
20
40
60
80
100
0 0.5 1.0 1.5 2.0
LaplaceRamp Invariant
TIME (SEC)
VE
LO
CIT
Y (
inch
/se
c)
SDOF VELOCITY RESPONSE fn=10 Hz Q=10
32
Figure 6.
There is excellent agreement between the two acceleration curves as shown in Figure 6.
-14
-12
-10
-8
-6
-4
-2
0
2
4
6
8
10
12
14
0 0.5 1.0 1.5 2.0
LaplaceRamp Invariant
TIME (SEC)
AC
CE
L (
G)
SDOF ACCELERATION RESPONSE fn=10 Hz Q=10
33
MDOF Application
The ramp invariant digital recursive filtering relationship can be readily used as the numerical
engine in an MDOF modal transient analysis for each of the respective response parameters.
An example is shown in Appendix A.
References
1. David O. Smallwood, An Improved Recursive Formula for Calculating Shock Response
Spectra, Shock and Vibration Bulletin, No. 51, May 1981.
2. A. Brandt & K. Ahlin, A Digital Filter Method for Forced Response Computation, Society
for Experimental Mechanics ( SEM ) Proceedings, IMAC-XXI.
3. T. Irvine, The Generalized Coordinate Method for Discrete Systems, Revision F,
Vibrationdata, 2012.
4. Himelblau, Piersol, et al., IES Recommended Practice 012.1: Handbook for Dynamic Data
Acquisition and Analysis, Institute of Environmental Sciences and Technology, Mount
Prospect, Illinois.
5. Rao V. Dukkipati, Vehicle Dynamics, CRC Press, Narosa Publishing House, New York.
2000.
6. K. Bathe, Finite Element Procedures in Engineering Analysis, Prentice-Hall, Englewood
Cliffs, New Jersey, 1982.
7. W. Thomson, Theory of Vibrations with Applications, Second Edition, Prentice-Hall, New
Jersey, 1981.
8. T. Irvine, An Introduction to the Shock Response Spectrum, Revision R, Vibrationdata,
2010.
9. T. Irvine, The Impulse Response Function, Vibrationdata, 2010.
10. T. Irvine, Table of Laplace Transforms, Revision J, Vibrationdata, 2011.
11. T. Irvine, Partial Fractions in Shock and Vibration Analysis, Revision I, Vibrationdata,
2012.
12. R. Dorf, Modern Control Systems, Addison-Wesley, Reading, Massachusetts, 1980.
13. T. Irvine, The Time-domain Response of a Single-degree-of-freedom System Subjected to
a Sinusoidal Force, Revision B, Vibrationdata, 2010.
34
14. T. Irvine, The Time-Domain Response of a Single-degree-of-freedom System Subjected to
an Impulse Force, Revision C, Vibrationdata, 2013.
15. T. Irvine, Derivation of the Filter Coefficients for the Ramp Invariant Method as Applied
to Base Excitation of a Single-degree-of-Freedom System, Revision B, Vibrationdata,
2013.
APPENDIX A
MDOF Example
The example from Reference 2 is repeated here.
Figure A-1.
Assume 5% damping for each mode. Assume zero initial conditions.
k1
k2
x1
m2
m1
k3
x2
f2(t)
f1(t)
35
The parameters are
Table A-1. Parameters
Variable Value Unit
1m 3.0 lbf sec^2/in
2m 2.0 lbf sec^2/in
1k 400,000 lbf/in
2k 300,000 lbf/in
3k 100,000 lbf/in
B1 100 lbf
B2 200 lbf
55 Hz
100 Hz
The mass matrix is
20
03
m0
0mM
2
1 (A-1)
The stiffness matrix is
000,400000,300
000,300000,700
kkk
kkkK
322
221
(A-2)
The applied forces are
t1002sin200
t552sin100
t2sinB
t2sinB
)t(f
)t(f
2
1
2
1 (A-3)
Each of the two forcing functions is synthesized using Matlab script: generate.m, as a pre-
processing step.
The modal transient analysis is performed using Matlab script: mdof_modal_arbit_force_ri.m
36
>> mdof_modal_arbit_force_ri
mdof_modal_arbit_force_ri.m ver 1.3 February 4, 2012
by Tom Irvine Email: tomirvine@aol.com
This program calculates the response of an MDOF
system to arbitrary force excitation via the ramp
invariant digital recursive filtering relationship.
The system is decoupled using normal modes as an
intermediate step.
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: mass_case5
Select damping input method
1=uniform damping ratio
2=damping ratio vector
1
Enter damping ratio
0.05
Stiffness Matrix
Enter the matrix name: stiff_case5
Natural Frequencies
No. f(Hz)
1. 48.552
2. 92.839
Modes Shapes (column format)
ModeShapes =
0.3797 -0.4349
0.5326 0.4651
37
Enter duration(sec)
0.3
Enter sample rate (samples/sec)
(recommend 1857)
2000
Each force file must have two columns: time(sec) & force(lbf)
Enter the number of force files
2
Note: the first dof is 1
Enter force file 1
Enter the matrix name: sine1
Enter the number of dofs at which this force is applied
1
Enter the dof number for this force
1
Enter force file 2
Enter the matrix name: sine2
Enter the number of dofs at which this force is applied
1
Enter the dof number for this force
2
begin interpolation
end interpolation
Calculating response...
38
Figure A-2.
The displacement, velocity and acceleration responses are shown in Figures A-2, A-3 and A-4,
respectively.
The results appear to be the same as the Laplace transform results in Reference 3. A formal
comparison will be given in the next revision of this paper.
-0.003
-0.002
-0.001
0
0.001
0.002
0.003
0 0.05 0.10 0.15 0.20 0.25 0.30
dof 2dof 1
TIME (SEC)
DIS
P (
INC
H)
DISPLACEMENT
39
Figure A-3.
Figure A-4.
-1.5
-1.0
-0.5
0
0.5
1.0
1.5
0 0.05 0.10 0.15 0.20 0.25 0.30
dof 2dof 1
TIME (SEC)
VE
L (
INC
H/S
EC
)
VELOCITY
-1.5
-1.0
-0.5
0
0.5
1.0
1.5
0 0.05 0.10 0.15 0.20 0.25 0.30
dof 2dof 1
TIME (SEC)
AC
CE
L (
G)
ACCELERATION