Response of structures to periodic dynamic loadings

105

5Response of structures to periodic

dynamic loadings

Abstract: In many real problems, it is found that exciting forces vary withtime in a non-harmonic fashion that may be periodic or non-periodic. Whenthe function defined over a period repeats indefinitely, it is known as aperiodic function. In this chapter, we apply the Fourier series to determinethe response of the system to periodic forces.

Key words: periodic function, Fourier series, odd function, even function,frequency domain, spectrum, Gibbs phenomenon.

5.1 Introduction

We have seen in Chapter 4 that harmonic excitation occurs in power installationsequipped with rotating and reciprocating machinery. The solution to harmonicexcitation is simple and straightforward since a closed form solution is alwaysavailable. In many realistic situations, the exciting forces vary with time ina non-harmonic fashion that may be periodic or non-periodic. A periodicfunction is one in which the portion defined over a period T0 repeats itselfindefinitely, (see Fig. 5.1). Propeller force on a ship, wind loading inducedby vortex shedding on tall slender structures are the examples of periodicforces, whereas, earthquake ground motion has no resemblance to periodicfunction.

This chapter presents the application of Fourier series to determine theresponse of a system to periodic forces in the frequency domain, an alternativeapproach to the usual analysis of time domain. To make practical use ofFourier method, it is necessary to replace the integration with finite sums.

Forc

e

Time

T0

5.1 Periodic excitation.

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Structural dynamics of earthquake engineering106

5.2 Fourier analysis

Consider an undamped single-degree-of-freedom (SDOF) system subjectedto force F(t)

The governing equation is written as

mx kx F t˙̇ ( )+ = 5.1a

˙̇ ( )x xFm

f tn+ =ω 2 0 5.1b

Let the forcing function be represented as shown in Fig. 5.2. The durationof one pulse or period T0 = 4s.

F(t) = 10 t <2s

F(t) = –10 2 < t < 4 5.2

Fourier has shown that a periodic function can be expressed as a sum ofinfinite number of sine and cosine terms and such a sum is known as aFourier series.

F ta

a t a t b t b t( )2

{ cos cos 2 } { sin sin }01 2= + + … + + …ω ω ω ω1 2 2

or

F ta

a n t b n tn n( ) {= + +=

∞0

2Σ

n 1cos sin }ω ω 5.3

where the frequency of excited forcing function is

ω π= 20T

5.4

The coefficients a0, an, bn are evaluated once we know the forcing functionto be

aT

F t tt

t T

00

2

1

1 0

=+

∫ ( ) d

Force10

Time

4s2s

5.2 Forcing function (odd).

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Response of structures to periodic dynamic loadings 107

aT

F t n t tnt

t T

=+

∫20 1

1 0

( ) cos ( ) dω

bT

F t n t tnt

t T

=+

∫20 1

1 0

( ) sin ( ) dω 5.5

t1 in Eq. 5.5 can take any value of time, but is usually – T/2 or zero. Note thatFourier coefficients a1, an, bn may be expedited if the forcing function can berecognized as being odd or even. If the forcing function is anti-symmetricabout the origin (y-axis) then it is an odd function. For example, the givenfunction shown in Fig. 5.2 is an odd function because

F(t) = – F(–t) 5.6a

For an odd function, the Fourier coefficients a0, a1,…, an = 0The forcing function is even if

F(t) = F(–t) 5.6b

For an even function, the Fourier coefficients b0, b1,…, bn = 0Clearly then an odd function can be represented by the Fourier sine series

and an even function by the Fourier cosine series. If, however, the functionis neither odd nor even, then the full Fourier series must be employed. TheFourier coefficients can very easily be evaluated using the MATHEMATICApackage as shown by the following examples.

Example 5.1Determine the Fourier series expression for the square wave, periodic functionas shown in Fig. 5.2. Plot the Fourier representation for the first four non-zero harmonic components. (Assume F0 = 10)

SolutionSince the forcing function is an odd function, the Fourier coefficients a0,a1,…, an = 0. Thus the forcing function is represented by the Fourier sineseries as

F t b n tn n( ) sin ( )=

=

∞Σ

1ω

where

bT

F t n t tnT

T

=−∫2 ( ) sin ( d

0 / 2

/ 2

0

0

ω )

or

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bT

F t n t tn

T

= ∫4 ( ) sin ( ) d0

/ 20

0ω

= 40 sin ( ) d–40 cos ( )

0 0

/ 2

0 0

/ 20 0

Tn t t

Tn t

n

T

∫ =

ω ωω

T

bn

nn = −20 [ cos ( )]π π1

bnn40π for n = 1, 3, 5…

bn = 0 for n = 2, 4, 6

Since

T0 = 4s4 2

ω π π= =2

F t t t t( ) [sin ( / ) sin ( 2)/ sin ( /2)/ ]= + + …40 2 3 3 5 5π π π π/

The forcing function is represented in Fig. 5.3 in terms of 1, 2, 3 and 4 non-zero terms of Fourier series.

Example 5.2Determine the Fourier series expression for square wave periodic forcingfunction as shown in Fig. 5.4. Plot the Fourier representation of the first fournon-zero components.

–0 –1 1 2

10

5

–5

–10

Force

Time

5.3 Odd function represented by sine series.

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SolutionSince F(t) is an even function, b1, b2, bn are zero. Thus ‘a’ coefficients arerepresented by Fourier cosine series as

F ta

a n tn n( ) sin ( )= +

=

∞0

12Σ ω

aT

F t tT

00 0

4= ∫ ( ) d0 /2

= −

∫ ∫4 d d = 0

0

0

0/4

/4

/2

Tt t

T

T

T

0 010 10

aT

F t n t tn

T

= ∫4 ( ) cos ( ) d0

/ 20

0ω

= −

∫ ∫4 cos ( d cos d

0

/40

Tn t t n t t

T

T

T

0 4

2

10 100

0

ω ω) ( )/

/

T0 8 28 4

= = =s;ω π π

a n t t n t tn

T

T

T

= −

∫ ∫4 4 4

0

0

4

2

T0 0

/ 4

10 cos ( ) d 10 cos ( ) d0

π π/ //

/

an

nn n

nn = + −

12

40 sin ( /2) 40 [sin ( /2) sin ( ]ππ

π ππ

)

= − … = …−4( 1) 1, 3, 5( )/ 2F

nnn0 1

π

10 Force

Time

2s 4s

5.4 Forcing function (even).

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i.e. F tn

n tn

( ) 40 ( 1)cos ( )

=1,3,5

( ) / 2

= −…

∞ −

π πΣn

1

4/

In Fig. 5.5 the forcing function is represented in terms of 1, 2, 3, 4 non-zeroterms of cosine series.

The MATHEMATICA program is used to find the Fourier coefficients.

Example 5.3Determine the Fourier series expansion for a square wave periodic forcingfunction as shown in Fig. 5.6. Plot the Fourier representation of the first fournon-zero components

Force

10

5

–5

–10

–4 4Time

5.5 Even function represented by cosine series.

1 2 3 4Time

Force

10

8

6

4

2

5.6 General function represented by both sine and cosine series.

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SolutionSince F(t) is neither odd nor even function, this should contain both sine andcosine terms. In Fig. 5.6 the forcing function is represented in terms of 1, 2,3 and 4 non-zero terms of Fourier series.

5.3 Program 5.1: MATHEMATICA program to

determine Fourier coefficients of forcing

function

4

1

2

4

1

1

-4

Pi

—2

4

10

0

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If[t < 2, f[t], g[t]]

10

10

10 Sin[n Pi]—————————————

n Pi

-10 (-1 + Cos[n Pi])

————————————————————n Pi

Pi t20 Sin[——]

25 + ——————————

Pi

10

8

6

4

2

1 2 3 4

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Pi t 3 Pi t20 Sin[——] 20 Sin[———]

2 2

5 + —————————— + —————————————Pi 3 Pi

Pi t 3 Pi t 5 Pi t20 Sin[——] 20 Sin[———] 4 Sin[———]

2 2 25 + ————————— + ——————————— + ———————————

Pi 3 Pi Pi

Pi t 3 Pi t 5 Pi t 7 Pi t20 Sin[——] 20 Sin[——] 4 Sin[——] 20 Sin[——]

2 2 2 25 + ———————— + ————————— + ———————— + —————————

Pi 3 Pi Pi 7 Pi

10

8

6

4

2

1 2 3 4

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10

8

6

4

2

1 2 3 4

10

8

6

4

2

1 2 3 4

10

8

6

4

2

1 2 3 4

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10

8

6

4

2

1 2 3 4Time

Force

4

5.4 Response to periodic excitation

Next, consider the response of a viscously damped SDOF system to periodicnon-harmonic excitation of period T0. The differential equation of motion ofsteady state response is given by

mx cx kx F ta

a n t b n tn n n˙̇ ˙+ + = ( ) =

2+ [ cos + sin ]0

=1Σ∞

ω ω 5.7

in which Fourier series expansion was used to represent F(t). The transientresponse will decay with time and is hence neglected. To obtain the particularsolution for the steady state response, it is noted that differential equation islinear, and therefore principle of superposition is applicable. Hence the steadystate response is merely the sum of the individual particular solutions for allharmonic terms representing F(t). Hence we obtain

x ta

ka n t b n t

k n np

n n n n( )[ cos ( ) sin ( )]

(1 ) (2 )=1 2 2 2 2= +

− + −− +

∞0

2Σ

n

ω ψ ω ψβ ρ β

5.8

where ψ ρ ββn

n

n=

−−tan

(1 )1 2

2 2 5.9

and the frequency ratio β ωω=

n5.10

in which ψn is the phase angle for the steady state response.

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5.5 Program 5.2: MATHEMATICA program for

finding the response to a periodic function

4

1

2

4

1

1-

4

Pi—

2

4

5 t———

2

5 t——— 2

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10

10

8

6

4

2

1 2 3 4

Time

Force

10

8

6

4

2

1 2 3 4

Time

Force

Pi

—

4

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10

5

–5

5 10 15 20

Displacement

Time

008

Example 5.4For the harmonic periodic forcing function shown in Fig. 5.5 determine thesteady state response for a viscously under-damped SDOF system.

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Solution

F tF

nn t

T

n

( ) = −

∞ −4 0

=1,3,5

( 1) / 2

0

( 1)cos

2π

πΣn

The differential equation of motion is given by

mx cx kx F tF

nn t

T

n˙̇ ˙+ + = = −

∞ −( )

( 1)cos

2=1,3,5

( ) / 24 01

0ππΣ

n

x tF

n k n n

n tTp

n

n( )4 ( 1)

) ( )cos

=1,3,5

( ) / 2

= −− +

−

∞ −0

1

2 2 2 01 2

2π β ρ β

π ψΣn (

where ψ ρ ββn

n

n=

−−tan

(1 )1

2 2

2

and β ωω=

n

Example 5.5The building frame shown in Fig. 5.7 is constructed of rigid girdersand flexible columns. The frame supports a uniformly distributed load ofweight of 130kN and is subjected at its girder level to the periodic forcedescribed in Fig. 5.4. Evaluate the steady state response of the structure andplot displacement vs. time assuming F0 = 80kN, period of exciting force is2 s, E = 200GPa, damping factor = 0.1, height of end columns is 4m, theheight of centre column is 6m and the span of the beams is 6m. The valuesof I for end columns and centre column are 2.6 × 108 and 3.254 × 108mm4

respectively.

5.7 Building frame.

4m6m

6m

130kN

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Solution

Stiffness of two end columns =

k EIh1

9 8

126 200 10 2 6 10

10 64= × × = × × × ×

×= ×2 3 . 4.875 10

13

6

Stiffness of centre column = k EIh2 = ×12

23

= × × × ××

12 200 10 3.254 1010 216

9 8

12

= 3.6155 × 106

Total stiffness of columns = 8.490 × 106N/m

Mass of the building = 130 10009.81

13251kg× =

ω nkm

= = =849000013251

25.3rad/s

ω π π π β ωω= = = = =2 2

2; 0.12417

T n0

ρ = 0.1

Using the MATHEMATICA program, one can get the displacement responseas shown in Fig. 5.8.

Time5 10 15 20

Displacement

7.5 ×10–6

5.0×10–6

2.5 × 10–6

–2.5 ×10–6

–5.0 ×10–6

–7.5 ×10–6

5.8 Displacement response.

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Example 5.6Consider a system loaded as shown in Fig. 5.9. Find the steady state response.

Solution

aT

t T t FT

00

0= =∫2 sin (2 d0 00

/20

F π π/ ) /

aT

F t T n t T tn

T

= ∫2 sin(2 cos (2 d00

/ 20

00 0π π/ ) / )

==

−=

02

10

2

, odd

, even

nF

nnπ ( )

bT

F t T n t T tF

n

nn

T

= = =>∫2 sin (2 sin (2 d ,

, 0

/ 20

00

00

0

21

0 1π π/ ) / )

Thus forcing function can be written as

F(t) =

×F

t t t t01 1 1 12

2 4ππ ϖ ϖ ϖ ϖ1 sin 2

3cos 2

15cos 2

35cos 6+ − − − …( )

in which ϖ = 2π/T0.Using the MATHEMATICA program, the displacement response is obtained

for β = π/4, T0 = 4s ρ = 0.25 as shown in Fig. 5.10.

5.6 Frequency domain analysis

The dynamic response of the complex systems can be examined by twoapproaches: time domain solution and the frequency domain solution. Up tonow, all solutions have been in time domain and the equations are solved by

Forc

e

F0 sin (2πt/T0)

Time

T0 /2 T0 /2

5.9 Excited Force.

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integrating with time. In frequency domain technique, the amplitude coefficientsin the Fourier series solution corresponding to each frequency are determined.For this we use discrete Fourier transform (DFT) or fast Fourier transform(FFT) methods.

5.7 Alternative form of Fourier series

The arbitrary periodic forcing function can be represented in the form ofFourier series (see Eq. 5.3) as

F ta

a n t b n tn n n( ) =

2+ ( cos + sin )0

=1Σ∞

ω ω 5.11

in which a0, a1… an, b1, b2, … bn are real constants, t is the time and ω is theforcing frequency given by ω = 2π/T0 where T0 is the period of the forcingfunction. If the period of the forcing function is not known or if the forcingfunction does not have period for this case ω = 2π/TD where TD is the totaltime duration of the forcing function.

Equation 5.11 may also be written as

F ta

a b a a b n tn n n n n( ) ( [= + + +∞

0

=11/ 2 1/ 2

2) { /( ) ] cosΣ

n2 2 2 2 ω

+ +[ ] sinb a b n tn n n/( ) }/2 2 1 2 ω 5.12

or

F ta

a b n tn n n( ) ( ) [cos ( ) cos=1

1/ 2= + +∞

0 2 22

Σn

ψ ω

+ sin(ψn) sin n ω t] 5.13

Displacement

10

7.0

5

2.5

–2.5

–5

–7.5

5 10 15 20Time

5.10 Displacement response.

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or

F ta

a b n tn n n( )2

( ) [cos( )]0

=12 2 1/ 2= + + −

∞Σ

nω ψ 5.14

where

tanψ nn

n

ba

= 5.15

Define constants

Aa

A a bn n n00 1/ 2

2; ( )= = +2 2 5.16

Now forcing function is written as

F t A n tn n( ) = −∞Σ

n=0cos ( )ω ψ 5.17

in which ψ0 = 0. Hence forcing function is now written in terms of positivereal amplitude and phase shift. This method replaces the coefficients of an,bn by An, ψn. There is still the same number of coefficients. The plots of An

and ψn are called amplitude spectrum and phase spectrum respectively. Theyare discrete and occur at Fourier frequencies nω = (2π n)/TD.

Example 5.7A periodic load function of three frequency components is given by

F(t) = 10 cos (ω t – 0.5) + 7 cos (2ω t – 0.2) + 2 cos (3 ω t – 0.1)

in which ω = 2π/TD and TD = 5s. Plot the periodic function and the amplitudeand phase spectrum.

SolutionThe equation for the coefficients are given as

aT

F t tD

TD

02 ( )d= ∫0

aT

F t n t tnD

TD

= ∫2 ( ) cos( dω )0

bT

F t n t tnD

TD

= ∫2 ( )sin ( )dω0

except that we use TD as fundamental period. Evaluating the integrals in theabove equations yields the values given in Table 5.1.

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5.8 Program 5.3: MATLAB program to evaluate

amplitudes and phase angles

clc;close all;for i=1:200

t(i)=(i-1)/40;p(i)=10*cos(2*pi*t(i)/5–0.5)+7.0*cos(4*pi*t(i)/5–0.2)+2.0*cos(6.0*pi*t(i)/5–0.1)

endfigure(1)plot(t,p)xlabel(‘ t’);ylabel(‘p(t)’);title(‘ time series plot’);figure(2)a=fft(p);plot(abs(a));title(‘ amplitude’)for i=1:200

b(i)=real(a(i));c(i)=imag(a(i));d(i)=atan(-c(i)/b(i));e(i)=sqrt(b(i)^2+c(i)^2);

endprint{‘ real part’,\n);bprint(‘ imaginary part’,\n);cprint(‘ phase angle’,\n);dprint(‘ amplitude’,\n)’e

Table.5.1 Amplitudes and phase angles

n nω an bn A a bn n n = ( + )2 2 1/2

ψ n

n

n

ba

= tan –1

0 0 0 – 01 ω 8.77 –4.794 10 0.52 2ω 6.86 –1.39 7 0.23 3ω 1.99 –0.199 2 0.14 4ω 0 0 0 –5 5ω 0 0 0 –

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figure(3)plot(d);title(‘ phase’)figure(4);plot(t,b,t,c,’*’);title(‘ real and imaginary part(*)’)

The forcing function F(t) examined in Example 5.6 is rather a simplefunction, although, from the plot presented in Fig. 5.11a it is not readily

0 1 2 3 4 5Time(a)

p(t

)

20

15

10

5

0

–5

–10

–15

0 1 2 3 4 5Time(b)

Am

plit

ud

e

1200

1000

800

600

400

200

0

5.11 (a) Forcing function; (b) amplitude spectrum; (c) phasespectrum.

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apparent that this is a simple function. However the amplitude and phasespectrum parts shown in Fig. 5.11b and 5.11c reveal clearly three cosinesand their phases. As functions become more complex with many frequencycomponents, plots of amplitude spectra become even more important inunderstanding the structure of the functions.

Example 5.8A simple periodic load is characterized by the following deterministic amplitudecoefficients

FF

m m m

p p

4(1 ( / ) )e p

4=

−ωω

ω ω m = 0, 1, 2,…

ωp = peak frequencyωm = Fourier frequencyFp = amplitude of the forcing function at the peak frequencyFp = 5; ωp = 2π; TD = 100s. Assume random phase spectrum

SolutionThe amplitude spectrum is to be determined. The time series is determinedusing

F ti m ti

m( ) 52

2e cos

=1

5 4[ ( / ) ]4= ×

−

−Σm

ππ

π ψ100

2100

1 100

Figure 5.12 shows the time series and amplitude spectrum using the MATLABprogram.

0 1 2 3 4 5Time

(c)

Ph

ase

ang

le

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0

5.11 Continued

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5.9 Expression of forcing function using complex

variable approach

The forcing function is written as

F ta

a n t b n tn n( )2

( cos sin )0

=1= + +

∞Σ

nω ω 5.18

0 50 100 150 200t

(a)

Time series plot

p(t

)

5

4

3

2

1

0

–1

–2

–3

–4

–5

0 50 100 150 200t

(b)

Amplitude45

40

35

30

25

20

15

10

5

0

5.12 (a) Time series plot; (b) phase spectrum.

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Writing cos (n ω t) and sin (n ω t) terms of exponential

cos(n ω t) = (einωt) + e–inωt)/2 5.19

sin(n ω t) = (einωt) – e–inωt)/2 5.20

where i = –1 , and collecting similar terms we get

F ta a i b a i bn n i n t n n i n t( )2

( )2

e( )

2e

=1= + − + +∞

−0 Σn

ω ω 5.21

Using new constants as

Pa

Pa i b

Pa i b

nn n

nn n

0 2;

2;

2= = − = +

−0 5.22

Hence forcing function is written as

F t Pni n t( ) e

=–=

∞

∞Σ

nω 5.23

Equation 5.23 is the exponential form of the Fourier series. Now the summationis from –∞ to ∞. When n < 0, Fourier frequencies are negative.

Now the coefficient Pn is written as

PT

F t tni n t

t

t Td

= −+

∫1D

( ) e dω 5.24

P Pn= * = complex conjugate of Pn

PT

F t tt

t T

0D

1 ( )dD

=+

∫ 5.25

In fact P0 is real in value since F(t) is real in value.

Example 5.9Determine Pn for the square wave of Example. 5.1. Sketch the spectra ofR(Pn), I(Pn), |Pn|.

Solution

PT

F tT

F tni n t

Ti n t

T

T

= + −− −∫ ∫1 e d 1 ( ) e dD

00

/ 2

D0

0

D

D / 2

Dω ω

Integrating and substituting ωTD = 2π, we get

Pi F

nni n i n= − −− −0

(2 e e )2

1 2π

π π

Pn

F in

nn = −0

2 0

even

if oddπ

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These coefficients can very easily be obtained using the MATHEMATICApackage:

n = 5

5qn = Integrate[Exp[-0.4*n*Pi*t*I], {t, 0, 2.5}]9.745107727628695*^-17 - 0.3183098861837907*I

rn = Integrate[Exp[-0.4*n*Pi*t*I], {t, 2.5, 5}]

-2.9235323182886082*^-16 + 0.3183098861837907*I

pn = 10*(qn - rn)/57.796086182102956*^-16 - 1.2732395447351628*I

The exponential form is more convenient to use than the real form. Toillustrate the fact, consider SDOF system subjected to harmonic forcing. Theequation of motion is given by

mx cx kx F t Pni n t˙̇ ˙+ + = =

∞

∞( ) e

=–Σ

nω 5.26

The steady state solution of Eq. 5.26 will respond to same frequencies makingup the forcing function.

R(Fn)

I(Fn)

|Fn |

5 3 1 1 3 5

20/5π

20/3π

20/π

5.13 Real, imaginary and absolute values of spectra.

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x t Xni n t( ) e

=–

==

∞

∞Σ

n

nω 5.27

Substituting Eq. 5.27 in Eq. 5.26 yields

(–n2ω2m + i c n ω + k) xn = Pn 5.28

or

xn = HnPn 5.29

where Hn is given by

Hn m i c n k

k n m ic n

k n m c nn =

− + += − −

− +1

( ( )2 2

2 2

ω ωω ω

ω ω2 2 2 2)5.30

Writing in this way, the response at each frequency is simply a function ofthe forcing transfer function. For a given structure H depends only on frequencynω. This is termed the frequency domain solution. Consider β as the frequencyratio and ρ as the damping factor. The function Hn is written as

Hn i n

k n nn =− −− +

( ) ( )1 21 2

2 2

2 2 2 2

β β ρβ β ρ[( ) ( ) ] 5.31

The phase is now defined as

ψ β ρβn

n

n

HH

n

n= =

−− −tan

ImRe

tan2

(1 )1 1

2 25.32

in which Re and Im denote real and imaginary parts.Hence x(t) is written as

x t Pn n

i n t n( ) e ( )==−∞

∞+Σ | |Hn

ω ψ 5.33

where the amplitude or modulus of transfer function is represented by

| | | |H H Hn n=) (2 ]2 2 2 2 1/ 2

=− +

11k n n[( )β β ρ

5.34

where Hn* is the complex conjugate of Pn and φn is the phase. It is to be noted

that forcing function Pn can be complex. Expressing Pn as a modulus andphase we get

x t Pn n

i n t n( ) e ( + )==−∞

∞+Σ | || |Hn

nω ψ φ 5.35

Since the spectral coefficients are even, this implies that a spectrum plot wecan consider either all frequencies between –∞ and ∞ or just frequenciesfrom 0 to ∞ and double the heights. The former is called a double-sidedspectrum and the latter is called a single-sided spectrum. Both spectra arereasonable representations and both are used in practice.

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5.10 Discrete Fourier transform (DFT) and fast

Fourier transform (FFT)

Although Fourier integral technique discussed in previous sections providea means for determining the transient response of a system, numerical integralof the Fourier integral became a practical reality only with the publication ofCooley-Tukey algorithm for the FFT in 1965. Since that date, FFT hasrevolutionized many areas of technology such as the areas of measurementsand instrumentation.

Two steps are involved in the numerical evaluation of Fourier transforms.DFTs correspond to the equation

F t Pn n

i n t( ) e==−∞

∞Σ ω 5.36

and

PT

F t tni n t

t

t T

= −+

∫1 ( ) e dD

Dω 5.37

Assume ∆t is a uniform time increment

∆t = TD/N 5.38

where N is the number of points in the time series approximation of F(t). Wecan write

nω = 2πn/TD = 2πn fD 5.39

The DFT is given as

fT

P n Nn mn m N= = … −1 e ; , , 1

D =0

–1Σ

m

N2 0 1 2π / 5.40

and the frequency coefficients are expressed as

PTN

f m Nm n nn m i N= = … −

=−D

–1/e ; 0, 1, 2, ,Σ

02 1

Nπ 5.41

The total number of discrete time values and frequency values are same.The N term appearing before the summation in Eq. 5.40 is not unique. In

DFT it can be used as 1/N in Eqs. 5.40 and 5.41 or N in both.

Example 5.10Determine the DFT of the following function defined by

f (n ∆t) = 6 + 14 sin (nπ/2) … n = 0, 1, 2,…, 7

SolutionBefore developing a solution using the DFT, it is useful to first examine thetime series. The discrete trigonometric Fourier series is expressed as

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f a a n m N b n m Nn m m= +0 2/ / / )Σm

N

=1

–1[ cos (2 )] + [ cos (2 ]π π

In this example N = 8 and all coefficients are zero except a0 = 12; b2 = 14.For this problem the solution could be determined by inspection. With use ofthe Euler formulae, the time series is written as

Fn = 6–7ieinπ/2 + 7i e–inπ/2

The transform of the series is given by

PTN

F m Nm ni n m= … = −−D

=0

–1/ 4e 0,1, 2 1Σ

n

N2 π

PTN

imin m in m i n= + −( )− −D

=0

–1/ 4

=0

–1/ 4 / 26 e e eΣ Σ

n

N

n

Nπ π π14

2

+ ( ) −i in m i n142

Σn

N

=0

–1/ 4 – / 2e eπ π

These coefficients may be evaluated using the MATHEMATICA packageand given by

PT

PT

PT

i

PT

PT

PT

PT

iPT

0

D

1

D

2

D

3

D

4

D

5

D

6

D

7

D

6 7

0 0 0

7

= = = −

= = =

= =

0

0

These are the same coefficients that were determined by inspection of equationdefining Pn, Pm = Pm+n or P–2 = P6.

5.11 Gibbs phenomenon

The Fourier series approximation of a square wave has been plotted in Fig.5.34. The approximation is generally quite good as shown in the figure.However, an inaccuracy exists at the corners of the wave. Sines and cosinesare smooth, continuous functions and therefore are best suited to approximatelyother smooth and continuous functions. However, jumps or discontinuitiesexist in them and the approximation is poor. For the square wave a discontinuityexists at t/T0 = 0.5. At this location, the square wave has two values +1 and–1. When the function has jumps or double values, a Fourier series passesthrough the mean of the two points as shown in Fig. 5.34, which in our caseis zero. It is also to be noted that t/T0 = 0.5 the square wave is vertical. Thefourier series tends to overshoot at the corners. This is called the Gibbsphenomenon. This does not disappear even if large number of terms are used

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in the series. It is concluded that the Gibbs phenomenon is local and thecontribution to total energy is minimal.

5.12 Summary

Fourier integrals can be evaluated numerically. There are a number of softwarepackages available to do this, such as MATHEMATICA and MATLAB. DFThas a finite number of frequencies and Fourier integrals have continuousfrequencies in the interval –∞ < f < ∞ and high frequency terms are lost. Theother way of doing the Fourier integral is to use polynomial interpolationfunctions which can be integrated analytically over most of the data.

5.13 Exercises

1. Determine the Fourier series representation for the forcing function shownin Fig. 5.14. The load is given by

f tf t T t T

t T( )

00/ 2=

<>

0 02

0

2

2

( / )

/

Verify your answer using the MATHEMATICA package.2. Determine the real Fourier series for the periodic excitation shown in

Fig. 5.153. The undamped SDOF system shown in Fig. 5.16 is subjected to excitation

f (t) given in problem 2. (a) Determine the steady state response of thesystem if ωn = 2.5ω (b) Sketch the response spectra for Pn and xn.

f0

T0/2

5.14

f0

T0

5.15

km

f (t)

5.16

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4. If the under-damped system of SDOF with damping coefficient C issubjected to exciting force of Problem 2, determine the steady stateresponse when ωn = 2ω and damping factor = 0.1

5. A 20N weight is attached to a spring damper system similar to the oneshown in Fig. 5.16. The weight is acted upon by a horizontal force withthe following frequency components

j Aj f (Hz) Phase (j)

1 1 0.02 219°2 2 0.04 18°3 3 0.08 2°4 2 0.16 246°5 1 0.32 233°

f t f tj j( ) cos (2 )=1

5= +Σ

j jA π φ

Assume k =1N/m and 10% critical damping determine amplitude andphase spectra for the forcing function using FFT. Plot the amplitude ofthe forcing and response spectrum.

6. Determine horizontal displacement spectra for the steel frame shown inFig. 5.17. the periodic horizontal force is given by

f tf t T t T

T t T( )

sin (2 / ) 0 /2

0 /20

0=

< << <

0 0

0

π

with f0 = 1kN, T0 = 5s, I = 2.3 × 108mm4, height of column = 4m andspan of the beam = 5m.

5.14 Further reading

Anderson, R A (1967) Fundamentals of Vibration, Macmillan Co., New York.Biggs J M (1964) Introduction to Structural Dynamics, McGraw-Hill, New York.Chopra A K (2002) Dynamics of Structures – Theory and applications to earthquake

engineering, Eastern Economy Edition, Prentice-Hall of India, New Delhi.

5.17

f (t)2kN/m

I

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Craig R R (1981) Structural Dynamics, John Wiley and Sons, New York.Clough R W and Penzien J (1974) Dynamics of Structures, McGraw-Hill, New York.DenHartog J P (1956) Mechanical Vibrations, 4th ed., McGraw-Hill, New York.Fertis D G (2000) Dynamics and Vibrations of Structures, 2nd ed., John Wiley & Sons,

New York.Humar J L (1990) Dynamics of Structures, Prentice Hall, Englewood Cliffs, NJ.Jacobsen L S and Ayre R S (1958) Engineering Vibrations, McGraw-Hill Book Co., New

York.James M L, Smith G M, Wolford J C and Whaley P W (1989) Vibration of Mechanical

and Structural Systems, Harper and Row, New York.Mukhopadhyay M (2006) Structural Dynamics, Ane Books India, New Delhi.

Paz M (1980) Structural Dynamics, Theory and Computation, Van Nostrand Reinhold,New York.

Rao S S (2003) Mechanical Vibrations, 4th ed., Prentice Hall, Inc., Englewood Cliffs, NJ.Steidel R F (1971) An Introduction to Mechanical Vibrations, Wiley, New York.Thompson W T (1981) Theory of Vibration with Applications, 2nd ed., Prentice Hall,

Englewood Cliffs, NJ.Timoshenko S (1955) Vibration Problems in Engineering, Van Nostrand Company, Inc.,

Princeton, NJ.Tolstov, G P (1962) Fourier Series, Dover Publications, New York.Vierck, R K (1979). Vibration Analysis, 2nd ed., Harper and Row, New York.Weaver W, Timoshenko S P and Young D H (1990) Vibration Problems in Engineering,

5th ed., Wiley, New York.

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Response of structures to periodic dynamic loadings

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