4. Harmoni Harmonically cally Excited Excited Vibrati Vibrations ons 1. Intr Introd oduc ucti tion on A dynam dynamic ic syste system m is often often subje subjecte cted d to some some type type of exter externa nall force force or excitation, called the forcing or exciting function . This excitation is usually time depend dependent ent.. It may may be harmon harmonic, ic, nonha nonharm rmon onic ic but but perio periodic dic,, nonper nonperiod iodic, ic, or random in nature. nature. The response response of a system to a harmonic harmonic excitation excitation is called harmonic harmonic response response. The The nonp nonper erio iodi dic c exci excita tati tion on may may ha!e ha!e a lon" lon" or shor shortt duration. The response of a dynamic system to suddenly applied nonperiodic excitation is called transient response . #e shall consider the dynamic response of a sin"le de"ree of freedom system t ( under harmonic excitations of the form F $t % & F 'ei$ω t t (( φ % or F $t % & F 0cos$ω t ( φ % or F 0sin$ω t t ( φ %, %, )here F 0 is the amplitude, ω is is the fre*uency, and φ is is the phase an"le of the harmonic harmonic excitation. The !alue of φ depends on the !alue of F $t % at t & ' and is usually ta+en to be ero. -nder a harmonic excitation, the response of the system )ill also be harmonic. If the fre*uency of excitation coincides )ith the natural fre*uency of the system, the response of the system )ill be !ery lar"e. This condition, +no)n as resonance, is to be a!oided to pre!ent failure of the system. . E*ua E*uati tion on of of /oti /otion on If a force F $t % acts on a !iscously damped sprin"mass system as sho)n in 0i". 1, the e*uation of motion can be obtained usin" e)ton2s second la)3
+ c x + kx = F ( t ) m x
$1% ince this e*uation is nonhomo"eneous, its "eneral solution x$t% is "i!en by the sum of the homo"eneou homo"eneous s solution, solution, x h$t %, %, and the particular particular solution, solution, x %. %. The p$t homo"eneous solution, )hich is the solution of the homo"eneous e*uation + c x + kx = ' m x $% 5eprese epresents nts the free free !ibra !ibratio tion n of the syste system m and )as )as discus discussed sed.. This This free free !ibra !ibratio tion n dies dies out )ith time time under under each each of the three three possi possible ble condi conditio tions ns of dampin" dampin" $underd $underdampi ampin", n", critical critical dampin" dampin",, and o!erdam o!erdampin" pin"%, %, and under under all 1
possible initial conditions. Thus the "eneral solution of E*uation $1% e!entually reduc educes es to the the part partic icul ular ar solu soluti tion on x %, %, )hich )hich repr represe esents nts the the stead steady ysta state te p$t !ibration. The steadystate motion is percent as lon" as the forcin" function is present. The !ariations of homo"eneous, particular, and "eneral solutions )ith time for a typical case are sho)n in 0i". .
It can be seen that x h$t % dies out and x $t % becomes x p$t % after some time $ τ in in 0i". %. The part of the motion that dies out due to dampin" $the free !ibration part% is called transient. The rate at )hich the transient motion decays depends depends on the !alues of the system parameters k , c, and m. #e i"nore the transient motion and deri!e only the particular solution of E*uation $1%, )hich represents the steady state response, under harmonic forcin" functions. 6. 5esponse 5esponse of an -ndamped -ndamped ystem ystem -nder -nder Harmonic Harmonic 0orce 0orce t acts If a force F $t % & F 0cosω t acts on the the mass mass m of undamped system, system, the e*uation of motion, E*. $1% reduces to m x + kx = F ' cosω t $6% The homo"eneous homo"eneous solution of this e*uation e*uation is "i!en "i!en by x p ( t ) = C1 cosω nt + C sinω nt
$4% )here ω n & $k 7m% is the natural fre*uency of the system. 8ecause the excitin" force F $t % is harmonic, the particular solution x p$t % is also harmonic and has the same fre*uency ω . Thus )e assume a solution in the form x p ( t ) = X cosω nt 17
$9% )here X is a cons consta tant nt that that deno denote tes s the the maxi maximu mum m ampl amplit itud ude e of x p$t %. 8y substitutin" E*uation E*uation $9% into E*uation $6% and sol!in" for X , )e obtain F ' X = k − mω $:% Thus the total total solution of E*uation $6% is is
possible initial conditions. Thus the "eneral solution of E*uation $1% e!entually reduc educes es to the the part partic icul ular ar solu soluti tion on x %, %, )hich )hich repr represe esents nts the the stead steady ysta state te p$t !ibration. The steadystate motion is percent as lon" as the forcin" function is present. The !ariations of homo"eneous, particular, and "eneral solutions )ith time for a typical case are sho)n in 0i". .
It can be seen that x h$t % dies out and x $t % becomes x p$t % after some time $ τ in in 0i". %. The part of the motion that dies out due to dampin" $the free !ibration part% is called transient. The rate at )hich the transient motion decays depends depends on the !alues of the system parameters k , c, and m. #e i"nore the transient motion and deri!e only the particular solution of E*uation $1%, )hich represents the steady state response, under harmonic forcin" functions. 6. 5esponse 5esponse of an -ndamped -ndamped ystem ystem -nder -nder Harmonic Harmonic 0orce 0orce t acts If a force F $t % & F 0cosω t acts on the the mass mass m of undamped system, system, the e*uation of motion, E*. $1% reduces to m x + kx = F ' cosω t $6% The homo"eneous homo"eneous solution of this e*uation e*uation is "i!en "i!en by x p ( t ) = C1 cosω nt + C sinω nt
$4% )here ω n & $k 7m% is the natural fre*uency of the system. 8ecause the excitin" force F $t % is harmonic, the particular solution x p$t % is also harmonic and has the same fre*uency ω . Thus )e assume a solution in the form x p ( t ) = X cosω nt 17
$9% )here X is a cons consta tant nt that that deno denote tes s the the maxi maximu mum m ampl amplit itud ude e of x p$t %. 8y substitutin" E*uation E*uation $9% into E*uation $6% and sol!in" for X , )e obtain F ' X = k − mω $:% Thus the total total solution of E*uation $6% is is
x ( t ) = C1 cosω nt + C sinω nt +
F ' k − mω
cosω t
$;% -sin" the initial conditions x $t & & '% & x 0 and F ' C1 = x ' − k − mω
C =
x ´
$t & & '% &
x ´
, )e
'
x ' ω n
,
$=%
and hence
x ( t ) = x ' −
x F ' F ' cosω nt + ' sinω nt + cosω t k − mω k − mω ω n
$>% The maximum maximum amplitude amplitude X in in E*uation $:% can also be expressed as 1 X = δ st ω 1 −
ω n
$1'% )her )here e δ st denotes s the de?ect de?ection ion of the mass mass under under a force force F 0 and is st & F 07k denote sometime sometimes s called called @static @static de?ection de?ection since since F 0 is a constant $static% force. The *uanti *uantity ty X 7δ st st , represents the ratio of the dynamic to the static amplitude of motion and is called the ma"ni
6
Base 1. #hen ' C ω 7ω n C 1, the denominator in E*uation $1'% is positi!e and the response is "i!en by E*uation $9% )ithout chan"e. The harmonic response of the system x % is said to be in phase )ith the external force as sho)n in 0i"ure 4. p$t
4
Base . #hen ω 7ω n D 1, the denominator in E*uation $1% is ne"ati!e, and the steadystate solution can be expressed as x p ( t ) = − X cosω t $11% )here the amplitude of motion X is rede
9
ince x % and F $t % ha!e opposite si"ns, the response is said to be 1=' o out of p$t phase )ith the external force. 0urther, as ω 7ω n → ∞, X → '. Thus the response of the system so a harmonic force of !ery hi"h fre*uency is close to ero. Base 6. #hen ω 7ω n & 1, the amplitude X "i!en by E*uation $1'% or $1% becomes in% as
x ' cosω t − cosω nt x ( t ) = x ' cosω nt + sinω nt + δ st ω n ω 1− ω n
$16% ince the last term of this e*uation ta+es an inde
d ( cosω t − cosω nt ) cosω t − cosω nt = lim dω lim ω →ω ω →ω d ω 1− 1− ω d ω ω ω n n n
n
:
t sinω t ω nt = lim ω = sinω nt ω →ω ω n n
$14% Thus the response of the system at resonance becomes δ ω t x x ( t ) = x ' cosω nt + ' sinω nt + st n sinω nt ω n $19% It can be seen from E*uation $19% that at resonance, x$t% increases inde
6.1Total response The total response of the system, E*uation $;% or E*uation $>% can also be expressed as δ st x ( t ) = Acos( ω nt − φ ) + cosω t ω ω 1− <1 ω n ω n F for $1:% δ st cosω t x ( t ) = Acos( ω nt − φ ) − ω ω 1− >1 ω ω n n F for $1;% )here A and φ can be determined. Thus the complete motion can be expressed as the sum of t)o cosine cur!es of diGerent fre*uencies. In E*uation $1:%, the forcin" fre*uency ω is smaller than the natural fre*uency, and the total response is sho)n in 0i"ure ;$a%.
;
In E*uation $1;%, the forcin" fre*uency, and the total response appears as sho)n in 0i"ure ;$b%. 6.8eatin" phenomenon If the forcin" fre*uency is close to, but not exactly e*ual to, the natural fre*uency of the system, a phenomenon +no)n as beating may occur. In this +ind of !ibration, the amplitude builds up and then diminishes in a re"ular pattern. The phenomenon of beatin" can be explained by considerin" the solution "i!en by E*uation $>%. If the initial conditions are ta+en as x 0 & & ', E*uation $>% reduces to ( F m) x ( t ) = ' ( cosω t − cosω nt ) ω n − ω x ( t ) =
( F ' m) ω + ω n ω − ω n t sin t sin
ω n − ω
$1=% et the forcin" fre*uency ω be sli"htly less than the natural fre*uency3 ω n − ω = ε $1>% )here ε is a small positi!e *uantity. Then ω n & ω and ω + ω n ≈ ω $'% =
x ´
'
/ultiplication of E*uations $1>% and $'% "i!es ω n − ω = εω $1% -se of E*uations $1>% to $1% in E*uation $1=% "i!es F m x ( t ) = ' sinε t sinω t εω $% ince ε is small, the function sin ε t !aries slo)lyF its period, e*ual to π 7ε is lar"e. Thus E*uation $% may be seen as representin" !ibration )ith period π 7ω and of !ariable amplitude e*ual to
F ' m sinε t εω
x ( t ) =
It can also be obser!ed that the sin )t cur!e )ill "o throu"h se!eral cycles, )hile the sin et )a!e "oes throu"h a sin"le cycle, as sho)n in 0i"ure =.
Thus the amplitude builds up and dies do)n continuously. The time bet)een the points of ero amplitude or the points of maximum amplitude is called the period of beatin" $τ b% and is "i!en by π π τ b = = ε ω n − ω $6% )ith the fre*uency of beatin" de. urin" operation of the pump, the plate is subjected to a harmonic force, F $t % & 9'cos :.=6t lb.
>
Ji!en3 ump )ei"ht & 19' lb late dimension3 thic+ness $t % & '.9 in., )idth $ w% & ' in., and len"th $ l% & 1'' in. Harmonic force & F $t % & 9'cos :.=6 t lb. 5e*uired3 Amplitude of !ibration of the plate, X olution3 late modeled as a 6'× 1': ( '.'=6 1> k = 6 = = 1'' .'lb in 6 l ( 1'')
(
)
The amplitude of harmonic response, by E*uation $:% )ith3 F 0 & 9' lb, m & 19'76=:.4 lbsec7in. $ne"lectin" the )ei"ht of the steel plate%, k & 1''.' lb7in., and ω & :.=6 rad7sec. Thus E*uation $:% F ' 9' X = = = −'.19'4in. k − mω 19' 1'' .'−
( :.=6 ) 6=:.4
The ne"ati!e si"n indicates that the response x $t % of the plate is out of phase )ith the excitation F $t %. 4. 5esponse of a amped ystem -nder Harmonic 0orce 1'
If the forcin" function is "i!en by F $t % & F 0cosω t , the e*uation of motion becomes + c x + kx = F ' cosω t m x $4% The particular solution of E*uation $4% is also expected to be harmonicF )e assume it in the form x p ( t ) = X cos( ω t − φ ) $9% )here X and φ are constants to be determined. X and φ denote the amplitude and phase an"le of the response, respecti!ely. 8y substitutin" E*uation $9% into E*uation $4%, )e arri!e are X k − mω cos( ω t − φ ) − cω sin( ω t − φ ) = F ' cosω t $:% -sin" the tri"onometric relations cos( ω t − φ ) = cosω t cosφ + sinω t sinφ
(
)
sin( ω t − φ )
= sinω t cosφ − cosω t sinφ
in E*uation $:% and e*uatin" the coeLcients of cos ω t and sinω t on both sides of the resultin" e*uation, )e obtain X k − mω cosφ − cω sinφ = F '
( ) X ( k − mω ) sinφ − cω cosφ = '
$;% olution of E*uations $;% "i!es F ' X = 1 k − mω + cω
[(
)
]
$=% and
cω k − mω
φ = tan−1
$>% 8y insertin" the expressions of X and φ from E*uations $=% and $>% into E*uation $9% )e obtain the particular solution of E*uation $4%. 0i"ure 1' sho)s typical plots of the forcin" function and $steadystate% response.
11
i!idin" both the numerator and denominator of E*uation $=% by k and ma+in" the follo)in" substitutions k ω n = = undamped naturalfrequenc m ζ =
c cc
δ st =
=
F ' k
c c F = ζω n mω n m
= deectionunder thestaticforceF ' ,an
r =
ω = frequency ratio ω n
X
=
)e obtain δ st
1 1
1− ω + ζ ω ω n ω n
=
1
(1− r )
+ ( ζ r ) $6'%
and
ω ζ ω n −1 −1 ζ r φ = tan = tan ω 1− r 1− ω n $61% X 7δ st is called the ma"ni
1
The follo)in" obser!ations can be made from E*uations $6'% and $61% and from 0i"ure 113 a. 0or an undamped system $ζ & '%, E*uation $61% sho)s that the phase an"le φ & ' $for r C 1% or 1='o for $r D 1% and E*uation $6'% reduces to E*uation $1'%. b. The dampin" reduces the amplitude ratio for all !alues of the forcin" fre*uency. c. The reduction of the amplitude ratio in the presence of dampin" is !ery si"ni
ω = ω n 1− ζ
or $6% )hich is lo)er than the undamped natural fre*uency ω n and the damped ω d = ω n 1− ζ
natural fre*uency
. r = 1− ζ
e. The maximum !alue of X $)hen X 1 = δ st max ζ 1− ζ
is "i!en by
$66% and the !alue of X at ω & ω n by
16
X 1 = δ st ω =ω ζ n
$64% E*uation $66% can be used for experimental determination of the measure of dampin" present in the system. In a !ibration test, if the maximum amplitude of the response $ X %max is measured, the dampin" ratio of the system can be found usin" E*uation $66%. Bon!ersely, if the amount of dampin" is +no)n, one can ma+e an estimate of the maximum amplitude of !ibration. f.
0or ζ D 17 √ 2 , the "raph of X has no pea+s and for ζ & ', there is a
discontinuity at r & 1. ". The phase an"le depends on the system parameters m, c, and k and the forcin" fre*uency ω , but not on the amplitude F 0 of the forcin" function. h. The phase an"le φ by )hich the response x $t % or X la"s the forcin" function F $t % or F 0 )ill be !ery small for small !alues of r . 0or !ery lar"e !alues of r , the phase an"le approaches 1='o asymptotically. Thus the amplitude of !ibration )ill be in phase )ith the excitin" force for r CC and out of phase for r DD 1. The phase an"le at resonance )ill be >'o for all !alues of dampin" $ζ %. i. 8elo) resonance $ω C ω n%, the phase an"le increases )ith increase in dampin". Abo!e resonance $ ω D ω n%, the phase an"le decreases )ith increase in dampin". 4.1Total response The complete solution is "i!en by x $t % & x h$t % ( x %. Thus p$t −ζω nt cos( ω dt − φ ' ) + X cos( ω t − φ ) x ( t ) = X 'e $69% )here ω d = 1− ζ ⋅ ω n
$6:% r =
ω ω n
$6;% X and φ are "i!en by E*uations $6'% and $61%, respecti!ely, and X 0 and φ 0 can be determined from the initial conditions. 4.Muality factor and band)idth 0or small !alues of dampin" $ ζ C '.'9%, )e can ta+e X X 1 = = =! δ δ ζ st max st ω =ω n
$6=% The !alue of the amplitude ratio at resonance is also called ! factor or *uality factor of the system, in analo"y )ith some electricalen"ineerin" applications, such as the tunin" circuit of a radio, )here the interest lies in an amplitude at resonance that is as lar"e as possible. The points "# and "$, 14
)here the ampli% The diGerence bet)een the fre*uencies associated )ith the half po)er points "# and "$ is called the band)idth of the system $see 0i"ure 1%.
To
(1− r )
or
(
+ ( ζ r )
r 4 − r − 4ζ
=
!
=
1 ζ
) + (1− =ζ ) = '
$4'% The solution of E*uation $4'% "i!es r 1 = 1− ζ − ζ 1+ ζ
r = 1− ζ + ζ 1+ ζ
$41% or small !alues of ζ , E*uation $41% can be approximated as
1
r
1
="
ω = 1 = 1− ζ ω n
r
="
ω = = 1+ ζ ω n $4%
)here ω # & ω 7"# and ω $ & ω 7"$. 0rom E*uation $4%, 19
(
)
ω − ω 1 = ( ω + ω 1 )( ω − ω 1 ) = " − "1 ω n = 4ζ
$46% usin" the relation ω + ω 1 = ω n $44% E*uation $46%, )e
$4=% )e obtain, by substitutin" E*uation $4=% into E*uation $4;%, F ' X = k − mω + icω $4>% /ultiplyin" the numerator and denominator on the ri"ht side of E*uation $4>% by N$k O mω $% O icω P and separatin" the real and ima"inary parts, )e obtain
(
)
X = F '
k − mω
−i
( k − mω ) + c ω
(k − mω ) + cω cω
$9'% A = x + y
-sin" the relation, x ( iy & Aiφ )here can be expressed as F ' X = e−iφ 1 k − mω + cω
[(
)
and tan φ & y 7 x , E*uation $9'%
]
$91% )here
1:
cω k − mω
φ = tan−1
$9% Thus the steadystate solution, E*uation $4=%, becomes F ' x p ( t ) = ei ( ω t −φ ) 1 k − mω + ( cω )
[(
)
]
$96% Frequency Response E*uation $4>% can be re)ritten in the form kX 1 = ≡ &( iω ) F ' 1− r + iζ r
$94% )here &$iω % is +no)n as the complex frequency response of the system. The absolute !alue of &$iω % "i!en by kX 1 &( iω ) = = 1 F ' 1− r + ( ζ r )
[(
)
]
$99% denotes the ma"ni
1− r
$9;% Thus E*uation $96% can be expressed as F x p ( t ) = ' &( iω ) ei ( ω t −φ ) k $9=% It can be seen that the complex fre*uency response function, &$iω %, contains both the ma"nitude and phase of the steady state response. If F $t % & F 0cosω t , the correspondin" steadystate solution is "i!en by the real part of E*uation $96%3 F ' cos( ω t − φ ) x p ( t ) = 1 k − mω + ( cω )
[(
)
]
F ' &( iω ) eiω t = 5e F ' &( iω ) ei( ω t −φ ) k k
x p ( t ) = 5e
$9>% imilarly, if F $t % & F 0sinω t , the correspondin" steadystate solution is "i!en by the ima"inary part of E*uation $96%3
1;
x p ( t ) =
F '
[(k − mω )
+ ( cω )
]
1
sin( ω t − φ )
F ' &( iω ) eiω t = Im F ' &( iω ) ei( ω t −φ ) k k
x p ( t ) = Im
$:'% Complex Vector Representation of Harmonic Motion The harmonic excitation and the response of the damped system to that excitation can be represented "raphically in the complex plane, and interestin" interpretation can be "i!en to the resultin" dia"ram. #e
accelerati on= x p ( t ) = ( iω )
F ' &( iω ) ei( ω t −φ ) k
= −ω x p( t ) $:1%
8ecause i can be expressed as π π i = cos + i sin = eiπ
$:% )e can conclude that the !elocity leads the displacement by the phase an"le π 7 and that it is multiplied by ω . imilarly, 1 can be )ritten as − 1= cosπ + i sinπ = eπ $:6% Hence the acceleration leads the displacement by the phase an"le π , and it is multiplied by ω $. Thus the !arious terms of the e*uation of motion $4;% can be represented in the complex plane, as sho)n in 0i". 16.
:. 5esponse of a amped ystem -nder the Harmonic /otion of the 8ase 1=
ometimes the base or support of a sprin"massdamper system under"oes harmonic motion, as sho)n in 0i". 14a.
et y $t % denote the displacement of the base and x $t % the displacement of the mass from its static e*uilibrium position at time t . Then the elon"ation of the sprin" is x O y and the relati!e !elocity bet)een the t)o ends of the damper is x ´
O
y ´
. 0rom the freebody dia"ram sho)n in 0i". 14b, )e obtain the
e*uation of motion3 + c( x − y ) + k ( x − y ) = ' m x $:4% If y $t % & ( sinω t , E*uation $:4% becomes m x + c x + kx = Asinω t + )cosω t
$:9% )here A & k( and ) & cω (. This sho)s that "i!in" excitation to the base is e*ui!alent to applyin" harmonic force of ma"nitude $ k( sinωt ( cω (c osω t% to the mass. 8y usin" the solutions "i!en in E*. $9>% and $:'%, the steadystate response of the mass can be expressed as ω c( cos( ω t − φ 1 ) k( sin( ω t − φ 1 ) + x p ( t ) = 1 1 k − mω + ( cω ) k − mω + ( cω ) $::% The phase an"le φ # )ill be the same for both the terms because it depends on the !alues of m, c, k , and ω , but not on the amplitude of the excitation. E*uation $::% can be re)ritten as x p ( t ) = X cos( ω t − φ 1 − φ )
[(
)
]
[(
)
k + ( cω ) x p ( t ) = ( ( k − mω ) + ( cω )
]
1
cos( ω t − φ 1 − φ )
$:;% )here the ratio of the amplitude of the response x % to that of the base motion p$t y $t % is "i!en by
1>
1
1+ ( ζ r ) X k + ( cω ) = = ( ( k − mω ) + ( cω ) (1− r ) + ( ζ r )
1
$:=% and φ # and φ $ by
cω = tan−1 ς r k − mω 1− r
φ 1 = tan−1
1 k = tan−1 cω ς r
φ = tan−1
$:>% The ratio X 7( is called the displacement transmissibility . ote that if the harmonic excitation of the base is expressed in complex form as y $t % & 5e$( eiω t %, the response of the system can be expressed as
1+ iς r iω t (e − + ς 1 r i r
x p ( t ) = 5e
$;'% and the transmissibility as 1 X = [1+ ( ς r ) ] &( iω ) ( $;1% )here Q&$iw%Q is "i!en by E*. $99%. :.10orce transmitted In 0i". 14b, the force carried by the support F must be due to the sprin" and dashpot )hich are connected to it. It can be determined as follo)s3 − y ) = −mx F = k ( x − y ) + c( x $;% 0rom E*. :;, E*. ; can be )ritten as F = mω X cos( ω t − φ 1 − φ ) = F * cos( ω t − φ 1 − φ ) $;6% )here F * is the amplitude or maximum !alue of the transmitted force "i!en by
1+ ( ς r ) F * = r k( 1− r + ( ς r )
(
1
)
$;4% The ratio $F * 7k( % is +no)n as the force transmissibility. It can be noticed that the transmitted force is in phase )ith the motion of the mass x $t %. The !ariation of the force transmitted to the base )ith the fre*uency ratio r is sho)n in 0i". 19 for diGerent !alues of ζ .
'
:.5elated motion If + & x O y denotes the motion of the mass relati!e to the base, the e*uation of motion, E*. $:4%, can be re)ritten as + c + + k+ = −m y = mω ( sinω t m + $;9% The steadystate solution of E*. $;9% is "i!en by mω ( sin( ω t − φ 1 ) + ( t ) = = , sin( ω t − φ 1 ) k − mω + ( cω ) $;:% )here , , the amplitude of + $t %, can be expressed as mω ( r , = ( = k − mω + ( cω ) 1− r + ( ς r ) $;;% and φ # by E*. $:>%. The ratio , 7( is sho)n "raphically in 0i". 1:.
(
(
)
)
(
1
)
Example o. $Vehicle /o!in" on a 5ou"h 5oad% 0i"ure 1;$a% sho)s a simple model of a motor !ehicle that can !ibrate in the !ertical direction )hile tra!elin" o!er a rou"h road. The !ehicle has a mass of 1'' +". The suspension system has a sprin" has a sprin" constant of 4'' +7m and a dampin" ratio of ζ & '.9. If the !ehicle speed is 1'' +m7hr, determine the displacement amplitude of the !ehicle. The road surface !aries sinusoidally )ith an amplitude of ( & '.'9 m and a )a!elen"th of : m.
Ji!en3 Vehicle model m & 1'' +", k & 4'' +7m, ζ & '.9, and speed & 1'' +m7hr. 5oad surface3 sinusoidal )ith ( & '.'9 m and )a!elen"th & : m. 5e*uired3 isplacement amplitude $ X % of the !ehicle. olution3 /odel the !ehicle as a sin"le de"ree of freedom system subjected to base motion as sho)n in 0i". 1;b.
0re*uency ω of the base excitation & the !ehicle speed 7 len"th of one cycle of road rou"hness $)a!elen"th%3 1''km hr × 1'''m km 1 ω = π f = π :m = >.'==;rad se 6:'' sechr atural fre*uency of the !ehicle3 6 k 4''× 1' = m 1''
ω n =
= 1=.9;4rad se
and the fre*uency ratio r ω >.'==; = = 1.9>6 r = ω n 1=.9;4 Amplitude ratio , E*. :=.
X 1+ ( ζ r ) = ( 1− r + ( ζ r )
(
)
1
) 1+ ( × '.9× 1.9>66 = ) (1− 1.9>66) + ( × '.9× 1.9>66
1
X = '.=4> (
Thus, displacement amplitude of the !ehicle3 ( '.'9) = '.'49m X = '.=4>6 ( = '.=4>6 $ans)er% Example o. 6 A hea!y machine, )ei"hin" 6''' , is supported on a resilient foundation. The static de?ection of the foundation due to the )ei"ht of the machine is found to be ;.9 cm. It is obser!ed that the machine !ibrates )ith an amplitude of 1 cm )hen the base of the foundation is subjected to harmonic oscillation at the undamped natural fre*uency of the system )ith an amplitude of '.9 cm. 0ind $1% the dampin" constant of the foundation, $% the dynamic force amplitude on
6
the base, and $6% the amplitude of the displacement of the machine relati!e to the base. Ji!en3 /achine )ei"ht $#% & 6''' , static de?ection under # & ;.9 cm, and R & 1 cm )hen !$t% & '.9sinω nt cm. 5e*uired3 c, F * , and , . olution3 $1% The stiGness of the foundation3 weight of machine 6''' = = 4','''- m k = δ st '.';9 At resonance $ω & ω n or r & 1%, E*. $:=%
1+ ( ς ) X '.'1' = = 4= ( '.''9 ( ς )
1
ς = '.1 >
ampin" constant3
6''' = >'6.'91-⋅ s m >.=1
( ) ( 4',''') c = ς ⋅ cc = ς ⋅ km= '.1>1
$ans)er% $% ynamic force amplitude on the base at r & 1 , E*. $;4%. 1
1+ 4ς F * = (k = kX = 4','''( '.'1) = 4'' ς 4 $ans)er% $6% Amplitude of the relati!e displacement of the machine at r & 1, E*. $;;%. '.''9 ( , = m = = '.''>:= ς ( '.1>1 ) $ans)er% It can be noticed that X & '.'1 m, ( & '.''9 m, and , & '.''>:= mF therefore, , S X O ( . This is due to the phase diGerences bet)een x , y , and + . ;. 5esponse of a amped ystem -nder 5otatin" -nbalance -nbalance in rotatin" machinery is one of the main causes of !ibration. A simpli
4
The total mass of the machine is ., and there are t)o eccentric masses m7 rotatin" in opposite directions )ith a constant an"ular !elocity ω . The centrifu"al force $meω %7 due to each mass )ill cause excitation of the mass .. #e consider t)o e*ual masses m7 rotatin" in opposite directions in order to ha!e the horiontal components of excitation of the t)o masses cancel each other. Ho)e!er, the !ertical components of excitation add to"ether and act alon" the axis of symmetry A A in 0i". 1=. If the an"ular position of the masses is measured from a horiontal position, the total !ertical component of the excitation is al)ays "i!en by F $t % & meω sinω t. The e*uation of motion can be deri!ed by the usual procedure3 . x + c x + kx = meω sinω t $;=% The solution of this e*uation )ill be identical to E*. $:'% if )e replace m and F 0 by . and meω respecti!ely. This solution can also be expressed as
me ω i ( ω t −φ ) &( iω ) e x p ( t ) = X sin( ω t − φ ) = Im . ω n $;>% ω n = k .
)here and X and f denote the amplitude and the phase an"le of !ibration "i!en by X =
me ω = 1 &( iω ) ω . n k − .ω + ( cω ) meω
[(
)
]
$='% and
cω k − .ω
φ = tan−1
$=1% 8y de
1 1− ς
>1
Accordin"ly, the pea+s occur to the ri"ht of the resonance !alue of r & 1. Example o. 4 $0rancis #ater Turbine% The schematic dia"ram of a 0rancis )ater turbine is sho)n in 0i". 1> in )hich )ater ?o)s from A into the blades ) and do)n into the tail race C. The rotor has a mass of 9' +" and an unbalance $me% of 9 +"Umm. The radial clearance bet)een the rotor and the stator is 9 mm. The turbine operates in the speed ran"e :'' to :''' rpm. The steel shaft carryin" the rotor can be assumed to be clamped at the bearin"s. etermine the diameter of the shaft so that the rotor is al)ays clear of the stator at all the operatin" speeds of the turbine. Assume dampin" to be ne"li"ible.
;
Ji!en3 Turbine3 mass $.% & 9' +", unbalance $me% & 9 +"Umm, and speed ran"e & :'' :''' rpm. haft len"th & m and maximum radial de?ection & 9 mm 5e*uired3 iameter of the shaft olution3 E*uate the maximum amplitude $radial de?ection% of rotor to 9 mm. /aximum amplitude of the shaft $rotor% due to rotatin" unbalance, E*. =', set c & '3 meω X = k − .ω imitin" !alue of X & 9 mm The !alue of ω ran"es from π :''rpm= :''× = 'π rad se :'
or rpm= :''' :''' ×
π :'
= ''π rad se
atural fre*uency of the system k k ω n = = = '.':9k rad se . 9' If k is in 7m. 0or ω & 'π rad7sec −6 9.'× 1' ( 'π ) '.''9= k − ( 9')( 'π )
(
)
k = 1'.'4× 1'4π - m
0or ω & ''π rad7sec −6 9.'× 1' ( '' π ) '.''9= k − ( 9')( '' π )
(
)
k = 1'.'4× 1':π - m
0rom 0i". 1:, the amplitude of !ibration of the rotatin" shaft can be minimied by ma+in" r & ω 7ω n !ery lar"e. This means that ω n must be made small compared to
=
ω O that is k must be made small. This can be achie!ed by selectin" the !alue of k as 1'.'4 x 1'4π 7m.
tiGness of a cantile!er beam $shaft% supportin" a load $rotor% at the end3 k =
6E/
l
6
6E π d
:4 4
=
l
6
Then 6
4
d
=
:4kl
6π E
=
(
) 6π (.';× 1' )
( :4) 1'.'4× 1'4π ( ) 6 11
= .:''9 × 1'−4 m4
d = '.1 ;m= 1;m
$ans)er% =. 0orced Vibration )ith Boulomb ampin" 0or a sin"le de"ree of freedom system )ith Boulomb or dry friction dampin" subjected to a harmonic force F $t % & F 0sin)t as in 0i". ', the e*uation of motion is "i!en by + kx ± µ - = F ( t ) = F ' sinω t m x $=9% )here the si"n of the friction force $ µ -% is positi!e $ne"ati!e% )hen the mass mo!es from left to ri"ht $ri"ht to left%.
The exact solution of E*. =9 is *uite in!ol!ed. Ho)e!er, )e can except that if the dry friction dampin" force is lar"e, the motion of the mass )ill be discontinuous. n the other hand, if the dry friction force is small compared to the amplitude of the applied force F 0, the steady state solution is expected to be nearly harmonic. In this case, )e can
>
ceq =
4 µ -
πω X
$==% Thus the steadystate response is "i!en by x p( t ) = X sin( ω t − φ ) $=>% )here the amplitude X can be found from E*. $:'% F ' F ' k X = = 1 1 ω k − mω + ceqω ω 1− + ς eq ω ω n n
[(
) (
)]
$>'% )ith ς eq =
ceq cc
=
ceq mω n
=
4 µ -
=
µ -
mω nπω X π mωω n X
$>1% ubstitution of E*. $>1% into E*. $>'% "i!es F ' k X = 1 ω 4 µ - 1− + ω n π kX
$>% The solution of this e*uation "i!es the amplitude X as
4 µ - 1− F ' π F ' X = k ω 1− ω n
1
$>6% As stated earlier, E*. $>6% can be used only if the friction force is small compared to F 0. In fact, the limitin" !alue of the friction force µ - can be found from E*. $>6%. To a!oid ima"inary !alues of X , )e need to ha!e
4 µ - >' 1− π F '
F ' 4 > µ - π
or If this condition is not satis% can be found usin" E*. $9%3
4 µ - ς eq ω ω n −1 ceqω −1 −1 π kX = tan = tan φ = tan ω ω k − mω 1− 1− ω ω n n $>4% 6'
ubstitutin" E*. $>6% into $>4% for R, )e obtain
4 µ π F ' −1 φ = tan 1 1− 4 µ - π F ' $>9% E*uation $>4% sho)s that tan φ is a constant for a "i!en !alue of F 07 µ -. 0 is discontinuous at ω 7ω n & 1 $resonance% since it ta+es a positi!e !alue for ω 7ω n C 1 and a ne"ati!e !alue for ω 7ω n D 1. Thus E*. $>9% can also be expressed as
4 µ ± π F ' −1 φ = tan 1 1− 4 µ - π F ' $>:% E*uation $>6% sho)s that friction ser!es to limit the amplitude of forced !ibration for ω 7ω n S 1. Ho)e!er, at resonance $ ω 7ω n & 1%, the amplitude becomes in
∫
∆% ′ =
∫
τ = π ω
∫
F ' sinω t ⋅ [ω X cos( ω t − φ ) ] dt
'
$>;% ince E*. $>4% "i!es φ & >' at resonance, E*. $>;% becomes o
π ω
∫
∆% ′ = F ' X ω
'
sin ω tdt = π F ' X
$>=% The ener"y dissipated from the system is "i!en by E*. $=:%. Ince π F 0X D 4 µ -X for X to be real!alued, ∆% D ∆% at resonance $0i". 1%.
61
Thus more ener"y is directed into the system per cycle than is dissipated per cycle. This extra ener"y is used to build up the amplitude of !ibration. 0or the nonresonant condition $ω 7ω n S 1%, the ener"y input can be found from E*. $>;%3 π ω
∫
∆% ′ = ω F ' X
'
sinω t cos( ω t − φ ) dt = π F 'X sinφ
$>>% ue to the presence of sin φ in E*. $>>%, the input ener"y cur!e in 0i". $1% is made to coincide )ith the dissipated ener"y cur!e, so the amplitude is limited. Thus the phase of the motion φ can be seen to limit the amplitude of the motion. Example o. 9 $prin"/ass ystem )ith Boulomb ampin" A sprin"mass system, ha!in" a mass of 1' +" and a sprin" of stiGness of 4''' 7m, !ibrates on a horiontal surface. The coeLcient of friction is '.1. #hen subjected to a harmonic force of fre*uency H, the mass is found to !ibrate )ith an amplitude of 4' mm. 0ind the amplitude of the harmonic force applied to the mass. Ji!en3 prin"mass system )ith Boulomb friction & m & 1' +", k & 4''' 7m, µ & '.1, harmonic force )ith fre*uency & H, ! !ibration amplitude & 4' mm 5e*uired3 Amplitude of the applied force. olution3 Vertical force $)ei"ht% of the massF - & mg & 1' x >.=1 & >=.1 . atural fre*uency3 k 4''' = = 'rad se ω n = m 1' 0re*uency ratio3
6
ω ω n
=
× π '
= '.: =
Amplitude of !ibration X is "i!en by E*. $>6%3
4 µ - 1− π F F ' ' X = k ω 1− ω n
1
4( '.1) ( >=.1) 1− π F ' F ' '.'4= 4''' ) } {1− ( '.:=6
1
The solution of this e*uation "i!es F 0 & >;.>=;4
$ans)er%
>. 0orced Vibration )ith Hysteresis ampin" Bonsider a sin"le de"ree of freedom system )ith hysteresis dampin" and to a harmonic force F $t % & F 0sinω t , as indicated in 0i". .
The e*uation of motion of the mass can be deri!ed as β k m x + x + kx = F ' sinω t ω $1''%
)here $β k7 ω %
x ´
& $h7ω %
x ´
denotes the dampin" force. Althou"h the solution
of E*. $1%% is *uite in!ol!ed for a "eneral forcin" function F $t %, our intention is to
66
X =
F '
ω + β k 1− ω n
1
$1'% and
−1 β φ = tan ω 1− ω n $1'6% E*uation $1'% and $1'6% are sho)n plotted in 0i". 6 for se!eral !alue of β .
A comparison of 0i". 6 )ith 0i". 11 for !iscous dampin" re!eals the follo)in"3 a. The amplitude ratio X ( F ' k ) attains its maximum !alue of F 07k β at the resonant fre*uency $ ω & ω n% case of hysteresis dampin", )hile
64
b. The phase an"le φ has a !alue of tan1$β % at ω & ' in the case of hysteresis dampin", )hile it has a !alue of ero at ω & ' in the case of !iscous dampin". This indicates that the response can ne!er be in phase )ith the forcin" further in the case of hysteresis dampin". ote that if the harmonic excitation is assumed to be F $t % & F 0eiω t in 0i". ,
the e*uation of motion becomes β k m x + x + kx = F 'eiω t ω
$1'4% In this case, the response
x ´
$t % is also a harmonic function in!ol!in" the factor
eiω t. Hence x $t % is "i!en by iω x $t %, and E*. $1'4% becomes + k ( 1+ iβ ) x = F 'eiω t m x $1'9% )here the *uantity k $1 ( iβ % is called the complex stiGness or complex dampin". The steadystate solution of E*. 1'9 is "i!en by the real part of F 'eiω t x ( t ) = ω + iβ k 1− ω n
$1':% 1'.0orce /otion )ith ther Types of ampin" Viscous dampin" is the simplest form of dampin" to use in practice, since it leads to linear e*uations of motion. In the cases of Boulomb and hysteretic dampin", )e de
69
) F d = ±a( x )here a is a constant, x is the relati!e !elocity across the damper, and the ne"ati!e $positi!e% si"n must be used )hen x is positi!e $ne"ati!e%. Ener"y dissipated per cycle durin" harmonic motion x $t % & X sinω t 3 X
∫
∆% =
a( x ) dx = X 6
− X
π
∫
=
aω cos6 ω td( ω t ) = ω aX 6
−π
6
E*uatin" this ener"y to the ener"y dissipated in an e*ui!alent !iscous damper. ∆% = π ceqω X E*ui!alent !iscous dampin" coeLcient $ ceq%3 ceq =
= 6π
aω X
$ans)er% It can be noted that ceq is not a constant but !aries )ith ω and X . Amplitude of the steadystate response3 X 1 = δ st 1− r + ς r
(
) (
eq
)
)here r & ω 7ω n and ceq c ς eq = = eq cc mω n Amplitude3
6π m (1− r ) − + X = =ar
(1− r )
4
4
=ar δ st + 6 π m
1
$ans)er% 11.elf Excitation and tability Analysis The force actin" on a !ibratin" system is usually external to the system and independent of the motion. Ho)e!er, there are systems for )hich the excitin" force is a function of the motion parameters of the system, such as displacement, !elocity, or acceleration. uch systems are called selfexcited !ibratin" systems since the motion itself produces the excitin" force. The instability of rotatin" shafts, the ?utter of turbine blades, the ?o) induced !ibration of pipes, and the automobile )heel shimmy and aerodynamically induced motion of brid"es are typical examples of selfexcited !ibrations. A system is dynamically stable if the motion $or displacement% con!er"es or remains steady )ith time. n the other hand, if the amplitude of displacement increases continuously $di!er"es% )ith time, it is said to be dynamically unstable. The motion di!er"es and the system becomes unstable if ener"y is fed into the system throu"h selfexcitation. To see the circumstances that lead to instability, )e consider the e*uation of motion of a sin"le de"ree of freedom system3 6:
+ c x + kx = ' m x $1';% If a solution of the form x $t % & Ce , )here C is a constant, is assumed, E*. 1'; leads to the characteristic e*uation c k s + s+ = ' m m $1'=% The roots of this e*uation are st
c 1 c s1, = ± m m
k − 4 m
1
$1'>% ince the solution is assumed to be x $t % & Cest , the motion )ill be di!er"in" and aperiodic if the roots s# and s$ are real and positi!e. This situation can be a!oided if c7m and k 7m are positi!e. The motion )ill also di!er"e if the roots s# and s$ are complex conju"ates )ith positi!e real parts. To analye the situation, let the roots s# and s$ of E*. 1'= be expressed as s1 = p + iq s = p − iq , $11'% )here p and q are real numbers so that c k ( s− s1 )( s− s ) = s − ( s1 + s ) s+ s1s = s + s+ = ' m m $111% E*uations $111% and $11'% "i!e k c = −( s1 + s ) = −p = s1s = p + q m m , $11% E*uations $11% sho) that for ne"ati!e p, c7m must be positi!e and for positi!e p$ ( q, k 7m must be positi!e. Thus the system )ill be dynamically stable if c and k are positi!e $assumin" that m is positi!e%. Example o. ; $Instability of a Vibratin" ystem% 0ind the !alue of free stream !elocity u at )hich the airfoil section $sin"le de"ree of freedom system% sho)n in 0i". 4 becomes unstable.
Ji!en3 6;
in"le de"ree of freedom airfoil section in ?uid ?o) 5e*uired3 Velocity of the ?uid )hich causes instability of the airfoil $or mass m%. olution3 0ind the !ertical force actin" on the air foil $or mass m% and obtain the condition that leads to ero dampin". 1
F = ρ u1C'
)here ρ & density of the ?uid, u & free stream !elocity, 1 & )idth of the cross section normal to the ?uid ?o) direction, and C' & !ertical force coeLcient, )hich can be expressed as urel C' = ( C2 cosα + C1 sinα ) u )here urel is the relati!e !elocity of the ?uid, C2 is the lift coeLcient, C1 is the dra" coeLcient, and α is the an"le of attac+ $see 0i". 4%3 x α = − tan−1 u
0or small an"les of attac+, x α = − u
and C' can be approximated, usin" Taylor2s series expansion about α & ', as ∂C ⋅α C' = C' α =' + ' ∂α α =' )here, for small !alues of a, urel & u and C' = C2 cosα + C1 sinα Then
∂C2 cosα − C sinα + ∂C1 sinα + C cosα 2 1 ∂α ∂α α ='
C' = ( C2 cosα + C1 sinα ) α = ' + α
∂C' ∂C' x = C2 α =' − u ∂α α =' ∂α α ='
C' = C2 α =' + α C' = C2 α = ' −
∂C2 x + C1 α =' ∂ u α α ='
ubstitutin" 1
F = ρ u1C2
− α ='
1
∂C' x ∂α α ='
ρ u1
6=
