1

Chapter 4 Response of SDOF Systems to Harmonic Excitation

§ 4.1 Response of Undamped SDOF System to Harmonic Excitation

Figure 4.1 Harmonic excitation of an undamped SDOF system

tpkuum o Ω=+ cos&& (4.1)

Let

tUup Ω= cos steady-state response (4.2)

Then

Ω−

=Ω−

=

2

2

0

2

1n

o kp

mkpU

ω

amplitude of up (4.3)

Let

kpU o

o = static displacement (4.4)

oUUH =Ω)( 1,

11

2 ≠−

= rr

frequency response function (4.5)

n

rωΩ

= frequency ratio (4.6)

)(0 Ω= HUU

u

tptp o Ωcos)( =mk

2

tHUtup ΩΩ= cos)()( 0

tptp Ω= cos)( 0

If n

rωΩ

= 1<

,cos1 2 t

rUu o

P Ω

−= in phase (4.9)

If n

rωΩ

= 1>

( )trUu o

P Ω−

−= cos

12 out of phase (4.10)

)()()( tututu hp +=

tAtAtr

Unn

o ωω sincoscos1 212 ++Ω

−= (4.11)

21 , AA : from 00 ,uu &

oUtu

D)(

max= total dynamic magnification factor (4.12)

)()(

max0

Ω== HU

tuD p

S steady-state magnification factor (4.8)

3

Figure 4.2 Dynamic magnification factors for an undamped SDOF

system with tsinp)t(p o Ω=

Example 4.1 steady-state response

2/3866.38

/40

singLBSW

inLBSk

=

==

p0 = 10 LBS 10=Ω rad/s

0)0()0( == uu &

tAtAtr

Uu nno ωω sincoscos

1 212 ++Ω

−= (1)

tAtAtr

Uu nnnno ωωωω cossinsin

1 212 +−Ω−Ω−

=& (2)

rad/s 20)6.38()386(402/12/1

==

=

=

Wkg

mk

nω (3)

4

in. 25.04010

===kpU o

o (4)

5.02010

==Ω

=n

rω

(5)

in. 33.075.025.0

)5.0(125.0

1 22 ==−

=− rUo (6)

1210)0( A

rUu o +−

== (7)

in. 33.01 21 −=−

−=r

UA o (8)

nAu ω20)0( ==& (9)

02 =A (10)

in.)]20cos()10[cos(33.0 ttu −= (11)

Curves of ),(tup )(tuc and )(tu

Note a. The steady-state response has the same frequency as the excitation and is in-phase with the excitation since 1<r .

5

b. The forced motion and natural motion alternately reinforce each other and cancel each other giving the appearance of a beat phenomenon. Thus the total response is not simple harmonic motion.

c. The maximum total response s)10/at in.66.0( π=−= tu is greater

in magnitude than the maximum steady-state response

( )0at in. 33.0 == tup

- Equation 4.9 and 4.11 are not valid at 1=r .

- The condition 1=r , or nω=Ω , is called resonance, and

- it is obvious from Fig. 4.2 that at excitation frequencies near resonance the response becomes very large.

• When 1=r , let

nP tCttu ω=ΩΩ= ,sin)( (4.13)

then

n

o

mpCω2

=

nmk

kp

ω1

21 0= nU ω02

1= (4.14)

∴ ttUtu nnoP ωω sin)(21)( = (4.15)

6

Figure 4.3. Response )(tup at resonance, nω=Ω

§4.2 Response of Viscous-Damped SDOF Systems to Harmonic

Excitation

tpkuucum o Ω=++ cos&&& (4.16)

Let

)cos( α−Ω= tUuP (4.17)

U : steady-state amplitude α : phase

)cos(

)2

cos()sin(

2 α

παα

−ΩΩ−=

+−ΩΩ=−ΩΩ−=

tUu

tUtUu

P

P

&&

& (4.18)

Figure 4.4. Rotating vectors representing uuup &&&,,,

7

tptkUtUctUm

o Ω=−Ω+−ΩΩ−−ΩΩ−

cos)cos( )sin()cos(2

ααα

(4.19)

When kUUm <Ω2 , that is, nω<Ω .

2222 )()( UcUmkUpo Ω+Ω−= (4.20a)

2tanΩ−

Ω=

mkcα (4.20b)

Figure 4.5. Force vector polygon For solution, let

tBtAtup Ω+Ω= cossin)( (4.19a)

tpkuucum o Ω=++ cos&&& (4.16)

(4.19a) )16.4(→

tptBtAktBtAc

tBtAm

Ω=Ω+Ω+Ω−ΩΩ

+Ω−Ω−Ω

cos)cossin()sincos(

)cossin(

0

2

(4.19b)

tptAcBmktBcAmk Ω=ΩΩ+Ω−+ΩΩ−Ω− coscos])[(sin])[( 022

0)( 2 =Ω−Ω− BcAmk (4.19c)

8

02 )( pAcBmk =Ω+Ω− (4.19d)

(4.19c)

0)( 2 =Ω−Ω− BcAmk (4.19e)

AcmkBΩΩ−

=2

(4.19d)

02220222 )2()1(2

)()(U

rrrp

cmkcA

ςς+−

=Ω+Ω−

Ω=

0222

2

0222

2

)2()1(1

)()(U

rrrp

cmkmkB

ς+−−

=Ω+Ω−

Ω−=

where

kpU 0

0 = static displacement

tBtAtup Ω+Ω= cossin)( (4.19a)

)cossin(2222

22 tBA

BtBA

ABA Ω+

+Ω+

+=

222022

)2()1( rrUBA

ς+−=+

Let

22222 )2()1(2sin

rrr

BAA

ςςα+−

=+

=

222

2

22 )2()1(1cos

rrr

BAB

ςα

+−−

=+

= or

212tan

rr

BA

−==

ςα (4.21b)

then

9

)cos()( 22 α−Ω+= tBAtup

)cos()2()1( 222

0 ας

−Ω+−

= trr

U

or

)cos()( α−Ω= tUtup

where

2220

)2()1( rrUU

ς+−= amplitude of )(tup

[ ] 2/1222 )2()1(1

rrUUD

oS ζ+−

== steady-state magnification factor

(4.21a)

Figure 4.6. (a) Magnification factor versus frequency ratio for various amounts of damping (linear plot)

10

Figure 4.6 (b) Phase angle versus frequency ratio for various amounts of damping (linear plot).

Figure 4.7. (a) Magnification factor versus frequency ratio for various damping factors (logarithmic plot)

Figure 4.7 (b) Phase angle versus frequency ratio for various damping factors (logarithmic frequency scale)

11

ζ21)( 1 ==rSD (4.22)

The curves of Figs. 4.6 are frequently plotted to logarithmic scales as shown in Figs. 4. 7. This is referred to as a Bode plot.

)sincos()cos(])2()1[(

)( 212/1222 tAtAetrr

Utu ddto n ωωα

ζζω ++−Ω

+−= −

(4.23) Since the natural motion in Eq. 4. 23 dies out with time, it is referred to as a starting transient. Example 4.2

0)0()0( == uu &

)sincos()cos( 21 tAtAetUu ddtn ωωα ζω ++−Ω= − (1)

2/1222 ])2()1[( rrUU o

ζ+−= (2)

==

==Ω

=

===

=

=

rad/s 4)20)(2.0(

5.02010

25.04010

rad/s 202/1

n

n

oo

n

rkpU

mk

ζωω

ω

(3)

32.0)]5.0)(2.0(2[])5.0(1[

25.02/1222 =

+−=U in. (4)

12

267.0)5.0(1

)5.0)(2.0(212tan 22 =

−=

−=

rrζα (5)

26.0=α rad (6)

rad/sec 6.19)2.0(1201 22 =−=−= ζωω nd (7)

]sin)(cos)[( )sin(

2112 tAAtAAetUu

dnddndtn ωζωωωζωω

αζω +−−+

−ΩΩ−=−

& (8)

1)26.0cos(32.00)0( Au +−== (9)

in. 31.0)26.0cos(32.01 −=−−=A (10)

in. )]6.19sin(11.0)6.19([)26.0sin()10)(32.0(0)0( 2 ttAu ++−−==& (11)

in. 11.02 −=A (12)

in. )]6.19sin(11.0)6.19cos(31.0[)26.010cos(32.0 4 ttetu t +−−= − (13)

* § 4.3 Complex Frequency Response

tioeppukucum Ω==++ &&& complex equation of motion (4.28)

Let

13

tieUtu Ω=)(

then

02 ])[( pUcimk =Ω+Ω−

cimkpU

Ω+Ω−=

)( 20

0

20

)(

2)1(

UHrir

U

Ω=

+−=

ς

where

kpU 0

0 =

)2()1(1)( 2 rir

Hς+−

=Ω complex frequency response

222

2

)2()1()2()1(

rrrir

ςς

+−−−

=

222

2

222 )2()1()2()1(

)2()1(1

rrrir

rr ςς

ς +−−−

+−=

αieH −Ω= )(

where

222 )2()1(

1)(rr

Hς+−

=Ω

212tan

rr

−=

ςα

therefore )(

0 )()( α−ΩΩ= tip eHUtu

14

)]sin()[cos()(0 αα −Ω+−ΩΩ= titHU

)sin(cos00 titpep ti Ω+Ω=Ω

if tptp Ω= cos)( 0 )cos()()( 0 α−ΩΩ= tHUtup

if tptp Ω= sin)( 0 )sin()()( 0 α−ΩΩ= tHUtup

• Rectangular and polar representation.

IR iAAA += rectangular form (4.34a) αiAeA = polar form (4.34b)

22IR AAAA +=≡ (4.34c)

R

I

AA

=αtan (4.34d)

• Quotient of two complex numbers.

)( αβα

β−

=

= ii

i

eAB

AeBe

AB (4.35)

2/1222 ])2()1[(1)(

rrUUH

o ζ+−==Ω (4.36a)

212tan

rr

−=

ζα (4.36b)

ueUiuuieUiu

ti

ti

22)( Ω−=Ω=

Ω=Ω=Ω

Ω

&&

& (4.37)

15

Figure 4.8. Complex vector notation for rotating vectors.

Figure 4.9. Vector response plot for steady-state vibration of a viscous-

damped system

16

§ 4.4 Vibration Isolation – Force Transmissibility and Base Motion Figure 4.10. Vibration isolation situations.

(a) Force transmitted to stationary base. (b) Force transmitted to moving base.

(a)Force transmitted to stationary base.

tioeppukucum Ω==++ &&&

Force Transmissibility

ucukfff DStr&+=+= force transmitted to the support (4.38)

titr eUickf ΩΩ+= )( (4.39)

tiotr e

rirUickf Ω

+−Ω+

=)2()1(

)(2 ζ

(4.40)

tiotr ekU

rirrif Ω

+−

+=

)2()1()2(1

2 ζζ (4.41)

2/1222

2/12

])2()1[(])2(1[

rrrkUf o

tr ζζ

+−+

= (4.42)

2/12 ])2(1[ rDkUf

TR so

tr ζ+== force transmissibility (4.43)

mtpo Ωcos

u

Sf

Dfm

u

k

c

tZtz Ω= cos)(

)(a )(b

17

oo kUp = static force

Figure 4.11. Transmissibility, absolute response to base excitation.

(b)Force transmitted to moving base.

tiZetz Ω=)( Base Motion

tiZeickzkzcukucum ΩΩ+=+=++ )(&&&& (4.44)

or =w&& zu −&& relative displacement (4.44)

timZezmwkwcwm ΩΩ=−=++ 2&&&&& (4.45)

Let

titi eWweUu ΩΩ == , (4.46)

18

Then

)2()1()2(1

)( 22 rirri

icmkick

ZU

ζζ

+−+

=Ω+Ω−

Ω+= (4.47)

)2()1()( 2

2

2

2

rirr

icmkm

ZW

ζ+−=

Ω+Ω−Ω

= (4.48)

2/12 ])2(1[ rD

ZU

ZU

S ζ+=≡ (4.49)

SDrZ

WZW 2=≡ (4.50)

Figure 4. 12. Relative motion frequency response function.

19

Example 4.3

=V 100 km/hr

=m 1200 kg =k 400 kN/m =fζ 0.4 when fuly loaded

period = 4 m Solution a.

s/hr) 600m/cycle)(3 4(rad/cycle) m/hr)(6.28 000,100(

=Ω (1)

rad/s 43=Ω (2) b.

kmc ζ2= (3)

eeff kmkmc ζζ 22 == (4)

2/12/1

40012004.0

=

=

e

ffe m

mζζ (5)

693.0=eζ (6)

20

c.

[ ][ ] 2/1222

2/12

)2()1()2(1

rrr

ZU

ζζ+−

+= (7)

*

21

§ 4.5 Vibration Measuring Instruments

Figure 4. 13. Stages in a motion measurement system

Figure 4.14. Schematic diagram of a seismic transducer. RMI: Relative Motion Instrument A seismic transducer is one that employs a spring-mass system to measure relative motion.

zuw −= (4.51) A vibrometer is a seismic instrument whose output is to be

proportional to the displacement of the base, that is, )(tw is to be

proportional to )(tz , that is, )()( tcwtz =

)(tz

)(tu

Transducer case

RMI

Output ~a

b

22

timZezmwkwcwm ΩΩ=−=++ 2&&&&& (4.45)

2/1222

2

])2()1[( rrr

ZW

ζ+−= (4.52)

Figure 4. 12. Relative motion frequency response function.

1=ZW when r= ∞ , that is, small is nω or stiffness k is small

compared with mass m. An accelerometer is a seismic instrument whose output is proportional

to the base acceleration, )(tz&& , that is, )()( tcwtz =&&

tZtz Ω= cos)( (4.53)

tAtZzta ZZ Ω=ΩΩ−== coscos)( 2&& (4.53)

ZAz&&=

Sn

ZS

n

DADZW 22

2

ωω=

Ω= (4.54)

23

Snz

DAW

2

1ω

=

Ω−Ω

=

αω

tDAtw Sn

ZP cos)( 2 (4.56)

212tan

rr

−=

ζα (4.55)

[ ] 2/1222 )2()1(1

rrUUD

oS ζ+−

== steady-state magnification factor

(4.21a)

Figure 4.6. (a) Magnification factor versus frequency ratio for various amounts of damping (linear plot)

When r<<1 ( large nω or large stiffness k), DS=1

General aceleration input

tAtAtaZ 2211 coscos)( Ω+Ω= (4.57)

24

[ ])(cos)(cos)( 2211 ττ −Ω+−Ω= tAtACtwP (4.58)

phase distortion: τ Low damping→small α → small τ amplitude distortion: C

max10Ω>nω or 1.0max <Ω

nω

25

§4.6 Use of Frequency Response Data to Determine Natural Frequency and Damping Factor of a Lightly Damped SDOF System Determination of Undamped Natural Frequency a. The response lags the input by 90 degrees: Fig. 4.6b b. The response magnitude is a maximum: Fig. 4.6a c. The imaginary part of the response, is a maximum: Fig. 4.9

d. The spacing on the vector raponse plot, ∆Ω∆s , is a maximum

half-power method

tpkuucum o Ω=++ cos&&& (4.16)

)cos()( α−Ω= tUtup

where

2220

)2()1( rrUU

ς+−= amplitude of )(tup

222

2

)2()1(1

rrUU

o ζ+−=

(4.60)

Figure 4.15. Response curve showing half-power points.

26

The frequencies above and below resonance at which the response

amplitude is 2/2 times the resonant response amplitude are

referred to as the half-power points.

ζ21

1

=

=roUU (4.59)

222

2

)2()1(1

21

21

ii rr ζζ +−=

(4.61)

0)81()21(2 2224 =−+−− ζζ ii rr (4.62)

222 12)21( ζζζ +±−=ir (4.63)

ζ212 ±=ir (4.64)

L

L

+−=−=

++=+=

)2(211)21(

)2(211)21(

2/11

2/12

ζζ

ζζ

r

r (4.65)

ζ212 =− rr (4.66)

Ω−Ω=

−=

n

rrω

ζ 1212

21

2 (4.67)

27

§ 4.7 Equivalent Viscous Damping

∫ ∫== f

i

j

i

u

u

t

t SSS dtufdufW & (4.68)

∫=f

i

t

t DD dtufW & (4.69)

ucfD &−= (4.70)

)sin()cos(α

α−ΩΩ−=

−Ω=tUu

tUu&

(4.71)

∫∫ΩΩ

−ΩΩ−=−=/2

0

222/2

0)(sin)(

ππα dttUcdtuucWD &&

or 2UcWD Ω−= π (4.72)

Figure 4.16. Damping and elastic forces acting on a body. If the damping in a system is not of the linear viscous damping type, then an equivalent viscous damping coefficient may be defined by

2UWc D

eq Ω−=π

(4.73)

m )(tp

u

Sf

Df

28

Figure 4.17. Force versus displacement for a system with linear

spring and viscous dashpot

1

)(

)()(cos1)sin()(

)cos()(

22

222

22

222

=

+

−=

−±==

−±=−−±=−−=

−=

Uu

Ucf

uUcf

tuUcucf

tuUtUtUtu

tUtu

d

d

d

ω

ω

ω

ωφωωφωω

φω

&

&

df

Ucω

U df

dUc ωωπ =2

29

° Hysteresis Loop (static test)

.

( )

2

2

2

2

2

22/

)(

)(

kUW

kUWWtherefore

for

mkmk

cc

tpukukum

tpukucumtherefore

deq

eqdh

eqn

n

cr

eqeq

πβ

πβ

πβαωω

ωω

παωαβ

πωα

=⇒

⋅==

=→=

===⇒

=+

+

=++

&&&

&&&

p

f

k

u

hWπωα

αωπ

α

kc

UkUcWW

UkW

eq

eq

hd

h

=⇒

=⇒

=⋅=

22

2

30

° Complex Stiffness

( )

( )

excitationharmonicforonlyvalidisdampingviscousandstructuralbetweenyanaTheNote

UkW

factordampingstructural

stiffnesscomplexkik

epukium

epukiumkuuikumkc

tuieUitueUtu

epkuucum

h

eq

eq

tieqeq

tieq

ti

ti

ti

log:

:2

:21

212

1

)()()(

2

*

0

0

0

⋅⋅⋅=

==

+=

⋅=++⇒

=

⋅=

++=+⋅

+⇒

=

==

=

⋅=++

γππαβγ

β

βπαβ

παω

πωα

πωα

ωω

ω

ω

ω

ω

ω

&&

&&&&

&

&&&

Example 4.4 Bodies moving with moderate speed in a fluid experience a resisting force that is proportional to the square of the speed. Determine the equivalent viscous damping coefficient for this type of damping Solution

2uafD &±= (1)

31

tUu Ω−= cos (2) tUu ΩΩ= sin& (3)

∫

∫ ∫Ω

−

Ω

ΩΩ−=

−=−=/

0

333

/

0

32

)(sin2

22π

π

dttUaW

dtuaduuaW

D

U

UD && (4)

32

38 UaWD Ω

−= (5)

2

32

2

)3/8(U

UaU

Wc Deq Ω

Ω=

Ω−

=ππ

Uaceq Ω

=π38

(6)

n

D

c

eqeq k

UWcc

ωπζ/2

/ 2Ω−== (4.74)

2

21 kUWS = (4.75)

Ω−

=πωζ

4n

S

Deq W

W (4.76)

32

§ 4.8 Structural Damping The structural damping, is proportional to displacement but inphase with the velocity of a harmonically oscillating system.

tioepuikum Ω=++ )1( γ&& (4.77)

:)1( γik + complex stiffness

:γ stuctural damping factor

Let tieUu Ω= (4.78)

γikmkpU o

+Ω−=

)( 2 (4.79)

γirUU

o +−=

)1(12 (4.80)

22γγζ ≈=

r (4.81)

2/1222 ])1[(1

γ+−=

rUU

o

(4.82)

21tan

r−=

γα (4.83)

33

Figure 4.18. Vector response plots of steady-state response of structurally damped SDOF Systems Note a. At r =1 (resonance)

−==

γγ11

_

0

_

iiU

U

Therefore

γ1

0

=UU and =α 90 degrees at 1=r

b. The vector response plot for structural damping is a circle. The diameter of the circle is determined by the damping factor.

c. The spacing between points of equal frequency difference is

greatest in the vicinity of r = 1. That is, fo a given 12 rrr −=∆ ,

the spacing on the circle is greatest near r = 1.