chapter 4 response of sdof systems to harmonic excitationsdvc.kaist.ac.kr/course/ce303/ch04.pdf1...
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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 .
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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.