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 1<r

,cos1 2 t

rUu o

P Ω

−= in phase (4.9)

If 1>r

( )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=

Comments

- sDD ≥

- when 0=r , 1== sDD : Static response

- when 1≈r , D and sD are maximum and very large.

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)

4

tAtAtr

Uu nnnno ωωωω cossinsin

1 212 +−Ω−Ω−

=& (2)

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

==

=

=

Wkg

mk

nω (3)

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

5

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

10

- to compute ,max sD let 0=dr

dDs

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)

11

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

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)

12

==

==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)

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)

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Other Methods

- Closed Form Solution Duhamel Integration Method

- Numerical Methods

Central difference method Average acceleration method Linear acceleration method Newmark methods Wilson method Runge Kutta method

Response Spectra