Single Degree of Freedom

Vibratory Motion and Single Degree of Freedom Systems Table of Contents SDOF Manual. Complementary Mathcad files are listed (please find them on the course website). Notes and files adapted from Dr. G.J. Rix and Dr. J.R. Valdes. Item (in handout) Associated Mathcad Files

Free Vibrations Undamped Free Vibrations.mcd Damped Free Vibrations.mcd

Forced Vibrations Forced Vibrations.mcd

Transfer Functions Transfer Functions.mcd Ground Displacement.mcd Ground Acceleration.mcd

Fourier Analysis (not included in handout) Fourier Analysis.mcd

Response to Arbitrary Excitations SDOF Response to Non-Harmonic Load.mcd

Multiple Degree of Freedom Systems MDOF Systems.mcd

Complex Notation

Free Vibrations

We begin by examining the response of the single-degree-of-freedom (SDOF) system withno external forces. The mass is set into motion by an initial displacement from its at restposition and/or an initial velocity. We will consider two systems, one with no means ofdissipating energy and another with a viscous damping in the form of a dashpot.

Undamped Free Vibrations

Consider the single-degree-of-freedom (SDOF) systemshown at the right that has only a spring supporting themass. If we examine a free-body diagram of the mass wesee that the forces acting on it include gravity (theweight) and the resistance provided by the spring.

We use (1) Hooke’s law (F = ku) and Newton’s secondlaw (F = ma) to write the following:

( )W k u mustatic− + =δ && (1)

where staticδ is the displacement of the mass when thesystem is at rest. The additional component of thedisplacement, u, is measured relative to this "at rest"position. The static displacement is related to the weightby:

W k static= δ (2)

Thus, Eq. 1 simplifies to:

mu ku&& + = 0 (3)

which is the equation of motion for the undamped SDOF system. The objective is to solvethe equation of motion to determine the displacement of the mass as a function of time,u(t), subject to the initial conditions of the system.

One approach to solving this partial differential equation is to assume a solution of theform:

u t Ae rt( ) = (4)

m

k

u(t)

m

W

( )k u static+ δ

Take the second derivative with respect to time of Eq. 4 yielding:

&&( )u t r Ae rt= 2 (5)

and substitute Eqs. 4 and 5 into Equation 3 giving:

mr Ae kAert rt2 0+ = (6a)

which simplifies to:

mr k2 0+ = (6b)

Solving for r produces two possible roots:

r ik

m= ± (7)

where i = −1 . Thus the solution is given by:

u t Ae Bei

k

mt i

k

mt

( ) = +−

(8)

We define the radical term in the exponent to be the circular natural frequency ofvibration of the system:

ω nk

m= in rad/sec (9)

Note that the circular natural frequency is related to the natural frequency by:

fnn=

ωπ2

(10)

and the natural period by:

Tfn

n

=1

(11)

To evaluate the coefficients A and B in Eq. 8, we need two initial conditions. Usually,these are the initial displacement and velocity of the mass:

u t u( )= =0 0 (12a)

& ( ) &u t u= =0 0 (12b)

After substituting these expressions into Eq. 8 and its first derivative, we obtain:

u tu iu

eu iu

en

i t

n

i tn n( )& &

= +

+ −

−0 0 0 0

2 2 2 2ω ωω ω (13)

Figure 1 shows the displacement time history of this SDOF with the initial conditions of0.1u0 = and 0.0u0 =& . Note that the amplitude of the displacement does not diminish with

time because there is no means of attenuating energy within the system.

Figure 1 Displacement Time History for Free Vibrations of an Undamped SDOF System

Damped Free Vibrations

Consider the single-degree-of-freedom (SDOF)system shown at the right that has both a springand dashpot. If we examine a free-body diagram ofthe mass we see that an additional force isprovided by the dashpot. The force is proportionalto the velocity of the mass.

F cudamping = & (14)

where c is the viscous dashpot coefficient.Summing forces in the vertical direction yields:

( ) umucukW static &&& =−δ+− (15)

After simplifying using Eq. 2, the equation ofmotion of the system is:

mu cu ku&& &+ + = 0 (16a)

or

&& &uc

mu

k

mu+ + = 0 (16b)

To solve this differential equation, assume a solution of the form:

u t Ae rt( ) = (17)

Take the first and second derivative with respect to time of Eq. 17 yielding:

& ( )u t rAe rt= (18a)

&&( )u t r Ae rt= 2 (18b)

and substitute Eqs. 18a and b into Eq. 16b giving:

r Aec

mrAe

k

mAert rt rt2 0+ + = (19a)

which simplifies to:

m

k c

u(t)

m

W

( )k u static+ δ cu&

rc

mr

k

m2 0+ + = (19b)

Solving for r produces two possible roots:

rc

m

c

m

k

m=

−±

−

2 2

2

(20)

Thus the solution is given by:

u t Ae Ber t r t( ) = +1 2 (21)

We define a critical value of c such that the term inside the radical equals 0:

c kmcrit = 2 (22)

and a fraction of critical damping:

βω

= =c

c

c

mcrit n2(23)

Rearranging Eq. 23 yields:

c

m n2= ω β (24)

which can be substituted into Eq. 20 to yield:

r in n= − ± −ω β ω β1 2 (25)

After substituting, Eq. 21 becomes:

[ ]u t e Ae Ben n nt i t i t( ) = +− − − −ω β ω β ω β1 12 2

(26)

To evaluate the coefficients A and B in Eq. 26, we need two initial conditions. Usually,these are the initial displacement and velocity of the mass:

u t u( )= =0 0 (27a)

& ( ) &u t u= =0 0 (27b)

After substituting these expressions into Eq. 26 and its first derivative, we obtain:

( ) ( )u t e

u i u ue

u i u uen n nt n

n

i t n

n

i t( )& &

= +−

−

+ −

−

−

− − − −ω β ω β ω βω β

ω β

ω β

ω β0 0 0

2

1 0 0 0

2

1

2 2 1 2 2 1

2 2

(28)

Consider the solution for three different values of β:

1. ββ = 1 (critically damped)

For β = 1 Eq. 25 reduces to:

r n= −ω (29)

and the partial differential equation has repeated roots. As a result, the solution takes theform:

( )[ ]u t u u u t entn( ) &= + + −

0 0 0ω ω (30)

Figure 2 shows the response of a critically damped SDOF for three initial conditions. Theinitial displacement is equal to 1.0, but the initial velocity of the mass is 1.0, 0.0, and –1.0for the three different cases. Notice that the motion quickly diminishes to zero because ofthe large damping in the system.

Figure 2 Displacement Time History of a Critically Damped SDOF System

2. ββ > 1 (overdamped)

The overdamped case is similar to the critically damped case. Eq. 28 can be used directlysince the roots are not repeated. Figure 3 below shows the response of an overdampedSDOF for the same three initial conditions in Fig. 2. As in the case of the critically dampedSDOF, the displacement quickly diminishes to zero because of the large damping in thesystem.

Figure 3 Displacement Time History of an Overdamped SDOF System

3. ββ < 1 (underdamped)

The case of most interest to us in soil dynamics problems is that in which the fraction ofcritical damping is less than 1.0. Equation 28 may be used again to calculate thedisplacement as a function of time. An example of a typical displacement time history foran underdamped SDOF is shown in Fig. 4.

Notice that the displacement time history continues for many cycles of motion (i.e.oscillatory). The motion can be considered to be a harmonic function modulated by adecreasing exponential function. The origin of these two functions can be seen in Eq. 28.

Figure 4 Displacement Time History of an Underdamped SDOF System

It is common to define the damped circular natural frequency as:

ω ω βd n= −1 (31)

along the corresponding damped natural frequency and damped natural period, fd and Td,respectively. For small values of β, ωd ≈ ωn.

Another widely used measure of the damping in a viscous system is the logarithmicdecrement:

δπβ

βπβ=

+

=

−≅

1 2

12

2n

u t

u t nTd

ln( )

( ) for small β (32)

Parallel and Series Springs

It will sometimes be necessary to consider springs acting in parallel or series. This is easilyhandled by determining the effective stiffness of a single, equivalent spring as shownbelow.

mu(t)

k1

k2

1 1 1

1 2k k keff

= +

mu(t)

k1 k2

k k keff = +1 2

Forced Vibrations

Consider a single-degree-of-freedom (SDOF) system thatis excited by an external harmonic force as shown at theright. The equation of motion now becomes:

mu cu ku Pe i t&& &+ + = Ω (33)

The solution of this non-homogeneous differentialequation is composed of the a general solution from thecorresponding homogeneous equation and a particularsolution:

u t u t u tgeneral particular( ) ( ) ( )= + (34a)

Equation 26 is the general solution and may be substituted to yield:

[ ]u t e Ae Be u tn n nt i t i tparticular( ) ( )= + +− − − −ω β ω β ω β1 12 2

(34b)

Assume that the particular solution has the same form as the external force:

u t Ceparticulari t( ) = Ω (35)

where C is a constant. As before we need to take the first and second derivatives withrespect to time to substitute into the equation of motion.

& ( )u t i Ceparticulari t= Ω Ω (36)

&& ( )u t Ceparticulari t= −Ω Ω2 (37)

After substituting and factoring common terms we obtain:

( )− + + =Ω Ω Ω Ω2 m i c k Ce Pei t i t (38)

We can solve for C and substitute into Eq. 35:

u tP

k m i ceparticular

i t( ) =− +Ω Ω

Ω2 (39)

which can also be expressed as:

m

k c

Pe i tΩ

u t( )

ti

n2n

2particular e

i21

1

k

P)t(u Ω

ωΩ

β+ω

Ω−

= (40)

Summing the general and particular solutions yields:

ti

n2n

2

t1it1it e

i21

1

k

PBeAee)t(u

2n

2nn Ωβ−ωβ−ω−βω−

ωΩ

β+ω

Ω−

+

+= (41)

The transient component will include the transient response of the system due to the initialdisplacement and velocity of the mass and the transient response of the system due to thesudden application of the external force. The coefficients A and B may be determined bysetting Eq. 41 and its first time derivative equal to 0u and 0u& , respectively. The resultingexpressions for A and B are:

n2n

22n

2nn

2n

0

2n

02

nn

i21

1

k

P

12

1i

12

ui

12

u1iA

ωΩ

β+ω

Ω−β−ω

Ω+β−ω−βω−

+β−ω

+β−ω

β−ω+βω

=&

(42a)

n2n

22n

2nn

2n

0

2n

02

nn

i21

1

k

P

12

1i

12

ui

12

u1iB

ωΩ

β+ω

Ω−β−ω

Ω−β−ω−βω

+β−ω

−+

β−ω

β−ω+βω−

=&

(42b)

Figure 5 shows the response of a single degree of freedom system for which the externalforce has a circular frequency, Ω, equal to 1 rad/sec and the initial displacement and initialvelocity of the mass are both equal to zero.

Note that during the first 20 to 25 seconds the response of the system includes both atransient component (the general solution) and a steady-state component (the particularsolution). After about 25 seconds the transient response has diminished to near zero andthe overall response is dominated by the steady-state response. Thus for many forcedvibration problems of interest, we can neglect the transient component of the solutionsince it is not significant after the first few cycles of motion.

Figure 5 Response of SDOF System to Forced Vibrations

One way to characterize the response of the SDOF system is to relate the amplitude andphase of the displacement to the external force. Once steady-state conditions are reached,the ratio of the amplitudes is about 1.5 (displacement) to 1.0 (force). The displacement isabout 180 degrees out of phase with respect to the displacement. Later, we will see thatexamining the amplitude and phase as a function of frequency is a convenient way tocharacterize a single-degree-of-freedom system.

Rotating Mass Excitation

For machinery vibrations problems involved rotating machinery, the dynamic forces areoften the result of out-of-balance or eccentric masses as shown in the figure below.

me

e

Ω

The dynamic force resulting from the eccentric mass is given by:

( )tiexpem)t(P 2e ΩΩ= (43)

Notice that the magnitude of the force is frequency dependent unlike the constant forceexcitation defined in Eq. 33.

Transfer (Frequency Response) Functions

To characterize the response of a SDOF system to forced vibrations it is useful to define atransfer function or frequency response function between the input and output of thesystem.

HOutput

Input( )Ω = (44)

By normalizing the output of the system with respect to the input, we emphasize thecharacteristics and response of the system over the characteristics of the output or input.

Let's define a transfer function between the steady-state displacement output and forceinput of a SDOF system undergoing forced vibrations:

HDisplacement

Force

u t

Pess

i t( )( )

Ω Ω= = (45a)

H

Pe

k m i cPe

i t

i t( )Ω Ω Ω

Ω

Ω= − +2

(45b)

Hk m i c

( )ΩΩ Ω

=− +

12 (45c)

Hk

in n

( )ΩΩ Ω

=− +

1 1

1 22

2ωβ

ω

(45d)

Notice that the transfer function is a complex-valued quantity meaning the response of theSDOF system can be characterized by a magnitude and phase. Figure 6 shows themagnitude and phase plots for a SDOF system expressed as a function of the normalizedfrequency, nωΩ .

Figure 6 Magnitude and Phase of Transfer Function

Equations 45c and 45d and Figure 6 can be used to provide insight into the parametersthat control the response of a SDOF in different frequency ranges. Note in Equations 45cand d that when Ω→0, the transfer function reduces to:

Hk

( )Ω = =01

(46)

Thus, the stiffness of the system controls the response at low frequency. As Ω→ωn, thetransfer function reduces to:

Hi c ikn( )ΩΩ

= = =ωβ

1 1

2(47)

and the response of the system is controlled to a large extent by the damping in thesystem. Finally, as Ω becomes large, the transfer function becomes:

Hm

( )ΩΩ

→ ∞ ∝−

12 (48)

and the response of the system is largely controlled by the mass (the inertia) of the system.

Other Forms of the Transfer Function

The transfer function defined above was expressed in terms of the displacement. Otherresponse quantities such as the velocity and acceleration of the mass can also be used todefine a transfer function for various applications. The names associated with each ofthese transfer or frequency response functions are given in Table 1.

Table 1 Transfer Functions Used in Vibration Analysis (after Inman, 1994)

Response Parameter Transfer Function Inverse Transfer Function

Displacement Receptance Dynamic Stiffness

Velocity Mobility Impedance

Acceleration Inertance Apparent Mass

Ground Displacement and Acceleration

Consider the situation in which the systemvibrates because of motion introduced at thebase of the system, not by a force applied tothe mass. The system is shown at the right.

The forces exerted by the spring and thedashpot on the mass are functions of therelative displacement and velocity betweenthe mass and the base of the system. Theabsolute acceleration of the mass is still used(F=ma).

Thus, the equation of motion of the systembecomes:

mu c u z k u z&& ( & &) ( )+ − + − = 0 (49)

The right hand side is set equal to zero because there are no external forces applied to themass. Rearranging yields:

m

k c

z t( )

u(t)

mu cu ku cz kz&& & &+ + = + (50)

We can assume that the ground motion is given by a harmonic function:

z t Ae i t( ) = Ω (51)

and that the response of the mass is given by:

u t Be i t( ) = Ω (52)

After differentiating, substituting, and solving, we obtain the solution for the steady-statedisplacement of the mass:

u tk i c

k m i cAe i t( ) =

+− +

ΩΩ Ω

Ω2 (53)

We can also define a transfer function between the displacement of the mass and the inputground displacement:

Hu t

z t( )

( )

( )Ω = (54a)

H

k i c

k m i cAe

Ae

i t

i t( )Ω

ΩΩ Ω

Ω

Ω=

+− +2

(54b)

Hk i c

k m i c( )Ω

ΩΩ Ω

=+

− +2 (54c)

H

i

i

n

n n

( )Ω

Ω

Ω Ω=

+

− +

1 2

1 22

2

βω

ωβ

ω

(54d)

As before we can express the complex-valued transfer function in terms of the magnitudeand phase:

Figure 7 Magnitude and Phase of Transfer Function for Ground Displacement

In many earthquake engineering problems,the ground motion input at the base of thesystem is specified in terms of a groundacceleration.

&&( )z t Ce i t= Ω (55)

Furthermore, the response parameter ofinterest is the relative displacement betweenthe base and the mass. The latter is importantbecause it is the relative displacements thatare proportional to the forces induced in thestructure.

y t u t z t( ) ( ) ( )= − (56)

Recall that the equation of motion is:

mu c u z k u z&& ( & &) ( )+ − + − = 0 (49 again)

After substituting Eq. 56 we obtain:

m y z cy ky(&& &&) &+ + + = 0 (55a)

or

my cy ky mz&& & &&+ + = − (57b)

After solving for y(t) we obtain:

y tm

k m i cCe i t( ) =

−− +Ω Ω

Ω2 (58)

Finally, we can define the transfer function between the ground acceleration and therelative displacement of the mass and base:

Hy t

z t( )

( )

&&( )Ω = (59a)

Hm

k m i c( )Ω

Ω Ω=

−− +

(59b)

m

k c

&&( )z t

u(t)

H

i

n

n n

( )ΩΩ Ω

=−

− +

1

1 2

2

2

2

ω

ωβ

ω

(59c)

Figure 8 Magnitude and Phase of Transfer Function for Ground Acceleration

Response to Arbitrary Excitations

Thus far we have assumed that the inputs and outputs of single-degree-of-freedom(SDOF) systems have been harmonic functions of the form:

u t Aei t( ) = Ω (60a)

p t Bei t( ) = Ω (60b)

This form is convenient for deriving transfer functions for SDOF systems, but is oflimited practical value by itself because many actual inputs and outputs are morecomplex. We must combine the solutions we have developed for harmonic excitationwith another tool, the Fourier Transform, to allow us to determine the response of SDOFsystems to an arbitrary excitation. It is important to note that the linear (or equivalentlinear) nature of most soil dynamics problems enables us to use this approach.

Fourier Transforms

The Fourier Transform is a way to decompose arbitrary displacement and/or force timehistories into their harmonic components. Once decomposed, transfer functions can beapplied to the individual harmonic components. Finally, an inverse Fourier Transform canbe used to “reassemble” the individual harmonic responses to obtain the response timehistory.

Let's assume we have a displacement or force time history which we denote as x(t). Wecan express x(t) as a Fourier series:

( )x t a a n t b n tn nn

( ) cos sin= + +=

∞

∑0 0 01

Ω Ω (61a)

where

aT

x t dtT

00

1= ∫ ( ) (61b)

( )aT

x t n t dtn

T

= ∫2

00

( ) cos Ω (61c)

( )bT

x t n t dtn

T

= ∫2

00

( ) sin Ω (61d)

or using complex exponentials:

x t c enin t

n( ) =

=−∞

∞

∑ Ω0 (62)

In a Fourier series, each term is separated by ∆f = f0. As ∆f decreases to 0, we reduce theFourier series to an integral:

Ω∫ Ω=∞

∞−

Ω de)(X)t(x ti (63a)

with continuous "Fourier coefficients" given by:

X x t e dti t( ) ( )Ω Ω= −

−∞

∞

∫ (63b)

Equation 63b is the continuous Fourier transform used to convert a time history x(t) to thefrequency domain in which the signal is portrayed as a function of frequency. Note thatX(Ω) is a complex-valued function. Similarly, Eq. 63a is the inverse Fourier transformenabling us to go from the frequency domain to the time domain.

We typically do not record continuous or analog signals with modern digitalinstrumentation. Instead we measure a continuous signal that is digitized and stored at asampling interval ∆t. Thus we must use a discrete Fourier transform (DFT):

X k f x n t e i k f t

n

N

( ) ( )∆ ∆ ∆ ∆= −

=

−

∑ 2

0

1π (64a)

and the inverse discrete Fourier transform:

x n tN

X k f ei k f t

n

N

( ) ( )∆ ∆ ∆ ∆==

−

∑1 2

0

1π (64b)

where N is the total number of points in the time history and k and n are and frequencyand time indices, respectively.

The fast Fourier transform (FFT or IFFT) is simply a numerically efficient version of theDFT. The algorithm used to calculate the FFT is most efficient when the number ofpoints being transformed is equal to an integer power of 2 (i.e., 2n where n is an integer).

Thus it is common to increase the increase the number of points to the next larger integerpower of 2 by "padding" the end of the record with a sufficient number of zeros.

There are several features of the Fourier Transform that are helpful to know when using itas a tool to solve problems. Suppose that a force or displacement time history contains Npoints (where N = 2n) with sampling interval ∆t and that the total length of the timehistory is T. When the FFT algorithm is applied to that time history to transform it to thefrequency domain, the result will be N values in the frequency domain separated by:

T1f =∆ (65)

The Nyquist frequency is the maximum frequency that can be accurately resolved by asampling interval ∆t and is equal to:

t21fNyquist ∆

= (66)

Finally, for real-valued time histories (the most common case) only the first N/2 + 1points in the frequency domain are unique. The remaining points are complex conjugatesof values in the first half of the record. One must be extremely careful in selecting thecorrect frequencies and mathematical operations for values in the second half of thefrequency domain record to obtain correct results from the inverse FFT algorithm.

Frequency Domain Solutions

Using Fourier Transforms we can decompose a complex time history into its harmoniccomponents. This enables us to apply the transfer functions we developed for harmonicmotion. We will use a 3-step approach:

Use the FFT to calculate the amplitude and phase spectra corresponding to a given inputtime history. The input time history may be a force (active loading) or a ground motion(passive loading).

p(t)→FFT→P(Ω)

Calculate the response of the SDOF system in the frequency domain using the transfer(frequency response) function:

U(Ω) = H(Ω) P(Ω)

Use the inverse FFT to obtain the response of the SDOF system in the time domain.

U(Ω)→IFFT→u(t)

Time Domain Solutions (Duhamel’s Integral)

It is also possible to calculate the response of a SDOF to an arbitrary forcing functiondirectly in the time domain using Duhamel’s integral. Consider the following equation ofmotion for a SDOF system subjected to an arbitrary force p(t):

)t(pkuucum =++ !!! (67)

The solution to Eq. 67 is given by Duhamel’s integral:

( ) ( ) ( )( ) ττ−ω∫ τω

= βτωβω−

dtsinpemetu d

t

0d

tn

n

(68)

where τ is a dummy variable. Interested readers are referred to Weaver et al. (1990) orother texts on engineering vibrations for the derivation of Duhamel’s integral. Equation68 may also be viewed as a convolution integral in which the forcing function p(t) isconvolved with the impulse response of a SDOF system.

Unfortunately, the application of Eq. 68 is limited to situations in which the parameters ofthe SDOF system (k, m, and c) are independent of frequency. In many machine-foundation vibration problems, the equivalent values of k and c are functions of thefrequency of the excitation.

Multiple Degree of Freedom Systems

We now want to extend our analysis of the responseof lumped parameter systems to include systems inwhich there are two or more degrees of freedom.One such system is shown at the right and has twomasses, springs, and dashpots. In geotechnicalengineering, mutiple degree of freedom systems arecommon in foundation vibration problems involvingcoupled motion (e.g. rocking and horizontaltranslation), in site response problems for layeredsoil deposits, and pile dynamics problems.

In the 2-degree-of-freedom (2DOF) system shownhere, the equation of motion for the first mass is:

( ) ( ) ( )tpuukuucum 121121111 =−+−+ !!!!

Note that the forces exerted by the dashpot and the sprand displacement between the two masses, respectequation of motion is:

( ) ( ) ukuukucuucum 22112221122 +−−+−− !!!!!

We can combine the two equations of motion into athem in matrix form as follows:

−

+

+−

−+

k

kuu

ccccc

uu

m00m

1

1

2

1

211

11

2

1

2

1

!

!

!!

!!

or

pKuuCuM =++ !!!

Direct Solution of the Equations of Motion

One approach to determining the motion of the two mdirectly. Assume that the displacements of the massesform:

m1 u (t)

(69a)

ing depend on the relative velocityively. For the second mass, the

( )tp22 = (69b)

system of equations and express

( )( )

=

+

−tptp

uu

kkk

2

1

2

1

21

1 (70a)

(70b)

asses is to formulate the solution and the external forces are of the

k1 c1

1p1(t)

k2 c2

m2 u2(t)p2(t)

tie Ω= Uu (71a)

and

( ) tiet Ω= Pp (71b)

After substituting Eqs. 71a and 71b into Eq. 70b and rearranging, we obtain:

[ ] PUKCM =+Ω+Ω− i2 (72)

Solving for the vector of displacements yields:

[ ] PKCMU12

2

1 iuu −

+Ω+Ω−=

= (73)

We can rewrite Eq. 73 as follows:

BPU = (74a)

where

( )

Ω+Ω−Ω−−

Ω−−Ω+Ω−=Ω1111

2112121

121222222

22

CiMKCiKCiKCiMK

D1B (74b)

and

( )( ) ( )212122222

2221111

211 CiKCiMKCiMKD Ω+−Ω+Ω−Ω+Ω−= (74c)

Equations 74b and 74c are the complex-valued frequency response functions for the2DOF system.

Although direct solution of the equations of motion is straightforward, it has twolimitations. First, as the number of degrees of freedom increases, the calculation of thematrix inverse in Eq. 73 for each frequency becomes more computationally expensive.Second, the direct solution does not yield any physical insight into the characteristics ofthe MDOF such as the natural frequencies, mode shapes, and modal damping for each ofthe m degrees of freedom. These characteristics are more apparent in the modal solutionof the equations of motion.

Modal Analysis of the Equations of Motion

Another solution to Eq. 70b is given by the normal mode solution or modal analysis. Toobtain the normal mode solution, it is convenient to assume proportional damping of theform:

MKC γ+β= (75)

The normal mode solution is calculated by solving the generalized eigenvalue problem:

( ) 02 =Φω− MK (76)

where 2ω is a diagonal matrix containing the generalized eigenvalues and Φ contains theeigenvectors. Each rth eigenvalue is the square of the circular natural frequency of the rth

mode of vibration of the N degree of freedom system. The corresponding eigenvectorgives the mode shape of the rth mode of vibration.

The generalized stiffness, mass, and damping matrices are given by:

ΦΦ= Mm T (77a)

ΦΦ= Kk T (77b)

ΦΦ= Cc T (77c)

The modal damping ratio for each mode is given by:

N1r22 r

rr "=

ωγ+

βω=ζ (78)

and the damped circular natural frequency of each mode is given by:

2rrdr 1 ζ−ω=ω (79)

Finally, the frequency response functions for the multiple degree of freedom system aredefined as:

( ) ( )( )∑

Ω+Ω−

ΦΦ=Ω

=

N

1r rr2

r

krjrjk cimk

B (80)

Complex Notation

Consider a function of the form:

( ) tiXetx Ω= (1)

where X is complex valued:

biaX += (2a)

This can also be written as:

θ= ieXX (2b)

where 22 baX += and ( )abtan 1−=θ .

Substituting Eq. 2b into 1 we obtain:

( ) ( )θ+Ω= tieXtx (3a)

which may be written as:

( ) ( ) ( )( )θ+Ω+θ+Ω= tsinitcosXtx (3b)

Taking the real part of Eq. 3b yields:

( )( ) ( )θ+Ω= tcosXtxRe (4a)

or the imaginary part yields:

( )( ) ( )θ+Ω= tsinXtxIm (4a)

Thus, using ( ) tiXetx Ω= and taking either the real or imaginary component is equivalentto using either ( ) ( )θ+Ω= tcosXtx or ( ) ( )θ+Ω= tsinXtx , respectively. The primaryadvantage of using the exponential form is that derivatives with respect to time areconveniently manipulated.