Chapter 4 Response of SDOF Systems to Harmonic Excitationsdvc.kaist.ac.kr/course/ce514/ch04.pdf ·...
Transcript of Chapter 4 Response of SDOF Systems to Harmonic Excitationsdvc.kaist.ac.kr/course/ce514/ch04.pdf ·...
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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,
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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)
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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)
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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
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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)
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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 &&&,,,
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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)
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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
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)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
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- 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)
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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