modal transient analysis of a system subjected to an applied force … · 2013-12-19 · 1 modal...

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: [email protected]

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)


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)


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)


T2exp)Tcos()Texp(z2z

TsinTexpz)Tcos()Texp(zz

1z

z

Tz

1z

m

1=)z(H

ndn2

dnd

ndn2

2n

v

(63)


TsinTexpz)Tcos()Texp(zz

1z

z

z

1z

Tm

1=)z(H

ndn2

dnd

ndn2

2n

v

(64)


TsinTexp)Tcos()Texp(z

1z1zTm

1=)z(H

ndn2

dnd

ndn

2

2n

v

(65)

19


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)


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)


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)


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)



TsinTexpz

Tz

1z

m

1=)z(H

ndn2

dn2

d

a

(105)

25


TsinTexp1z

Tm

1=)z(H

ndn2

dn2

d

a

(106)


1z

Tm

TsinTexp=)z(H

ndn2

2

d

dna

(107)


1z2z

Tm

TsinTexp=)z(H

ndn2

2

d

dna

(108)



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


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


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


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: [email protected]

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

modal transient analysis of a system subjected to an applied force … · 2013-12-19 · 1 modal...

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