Chemical Engineering - Apllied Mechanics

SDOF forced vibrations

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2. Forced motion of Single Degree Of Freedom Systems (SDOF) Let us consider the SDOF system illustrated in Figure 2.1. The mass m is subjected to a harmonic force of frequency 2 f π = Ω . The maximum amplitude of the=force is F →=while β is the phase with respect to initial time. Therefore, the force can be expressed as: o F ( ) F cos ( )tt β = Ω+ = (2.1)= = = Figure 2.1 Forced SDO F=system = = Let us denote as=x the absolute displacement of the mass with respect to the static equilibrium position. The equation of motion is given by: ooF cos ( ) F cos ( ) mx cx k x t βϑ + + = Ω+ =   = (2.2)= The phase β indicates that the external force may not reach its maximum level at time t = 0. In accordance with Euler’s notation it is possible to write: [ ] () ooF Fcos()sin() itet it β ββ Ω+ = Ω+ + Ω+ = (2.3)= Therefore, eq.(2.3) can be rewritten as:= () → Re F it mx cx k x e β Ω+  ++ =    = (2.4)= In case of a sinusoidal excitation the motion equation isΩ= ooF sin ( ) F sin ( ) mx cx k x t βθ + + = Ω+ =  = (2.5)= Eq.(2.5) can be rewritten as:= () → Im F it mx cx k x e β Ω+  ++ =    = (2.6)= In a more general case we can write:= ( ) () oo FF iti mx cx k x e e βϑ Ω+ ++ = =   = (2.7)= where we implicitly assume=to consider the real or the imaginary part of the complex term at the right side of eq.(2.7) depending on the type of the harmonic excitation () F t =(cosine or sine function).=The solution of this differential equation is:= () () () gs xt x t x t = + = (2.8)= The term ()gxt =is the general solution. It coincides with the solution of eq.(2.7), that is the system response= of the free motion. The second tem, ()sxt , is the steady state solution. It=only depends on the system= excitation.= In case of an underdamped=system the contribution of the term ()gxt =nullifies after a suitably long time= interval.=For linear systems the steady state response is:= ( ) () () X X itit s xt e e βϕγ Ω+ +Ω+ == = (2.9)= Therefore, it isΩ= () () () 11 → () () () A B +X ndd htit itit gs xt x t x t e e e e ωωωβϕ− −Ω+ +  = += +  =(2.10)= As said above, after a suitably long time interval the term ()gxt =nullifies and only the term ()sxt =gives a contribution to the system response. This is the steady state response of the forced system. This vibration is a= harmonic function of the same frequency of the harmonic excitation. The steady state response ()sxt =is= shifted of the phase= ϕ with respect to the system excitation. The first and second order derivatives of ()sxt =are:= ( ) () () XX itit sxtie ie βϕγ Ω+ +Ω+ =Ω=Ω  = 2( ) 2( ) 2() ()XXX ititit sxtiieee βϕβϕγ Ω+ +Ω+ +Ω+ = Ω= −Ω= −Ω  =( 2.11)= Substituting these expressions into eq.(2.7) we obtain:= ( ) 2( ) () o XF itit k m ic ee βϕβ Ω+ +Ω+  −Ω + Ω=  =(2.12)= That is:= () 2 o XF it i i it i k m ic eee ee βϕβ ΩΩ  −Ω + Ω=  =(2.13)= ( ) 2 o XF i k m ic e ϕ −Ω + Ω =  =(2.14)= () 2 o XF i k m ic e ϕ− −Ω + Ω =  =(2.15)= The complex term= X ieϕ=is given by:= () () 2 oo 222 FF X 2 in nn k e k m ici h ϕ ω ωω ==  −Ω + Ω−Ω + Ω  = (2.16)= Eq.(2.15) can be written as:= A( ) X BΩ=   =( 2.17)= where= A =and B  =are complex variables. =The amplitude X of=the steady state vibration is given byΩ= () () o 22 2F X kmc = −Ω + Ω (2.18) The phase ϕ is given by: () 1 2 tanc km ϕ − Ω = − −Ω   =(2.19)= Phase ϕ is negative, or null. It indicates that the system response is delayed with respect of the excitation F( ) t .=Eq.(2.18) can be rewritten as:= oo 222 2 2 2 2 FX X 12 nn k km ch kk k ωω ==      ΩΩ −Ω + Ω   −+             = (2.20)= The ratio →F k=is a displacement that represents the spring deformation caused by a static force →F .= Therefore, we can write=the following dimensionless expressionΩ= 22 2 → 2 X1 X12 nn h ωω =  ΩΩ −+          = (2.21)= Similarly, eq.(2.19) can be rewritten as:= ( ) ( ) 1 22 2 tan 1 n n h ω ϕ ω −  Ω = −  −Ω  = (2.22)= Figure=2.2=shows=the ratio → XX and the phase ϕ , respectively, as a function of the dimensionless frequency nω Ω .=Then, the system response can be written as:= () () ()() → 2 F ( ) =X i itit xt e ee k m ic ϕββ Ω+Ω+ = −Ω+ Ω  = () () ()() → 22 1F ( ) =X 12 i itit nn xt e ee k ih ϕββ ωω Ω+Ω+ = −Ω + Ω = ( ) ()() → 22 1F ( ) =X 2 i itit nn xt e ee m ih ϕββ ωω Ω+Ω+ = −Ω + Ω =( 2.23)= It can be seen that the displacement x is proportional to the applied force, the proportionality factor being: ( ) ( ) 22 1 U( ) 12 nn ih ωω  Ω=  −Ω + Ω   = (2.24)= which is known as the complex frequency response. Eq.(2.23) shows that the displacement is a complex quantity which can therefore be broken up into its real and imaginary parts by multiplying the numerator and denominator of the square brackets by its complex conjugate. Thus, we obtain: ( )= xt = () () () () () () 22 () → 2222 2222 1 2 F 12 1 2 nn it nn n n ihe k hh β ωω ωω ω ω Ω+ −ΩΩ − −Ω + Ω −Ω + Ω  =( 2.25)= This shows that the displacement has one=real componentΩ= () () () 22 () → 22 22 1F Re X = 12 nit nn e k h β ω ωω Ω+ −Ω    −Ω + Ω   = (2.26)= which is=in-phase with the applied force and another component: ( ) ( ) ( ) () → 22 22 2F Im X = 12 nit nn ihe k h β ω ωω Ω+ Ω  −  −Ω + Ω   = (2.27)= which has a phase lag of 90� behind the applied force. This component is said to be in=quadrature with the excitation. Then, steady state solution of eq.(2.7) can therefore also be written in the form: () () () → 22 22 1F ( )= 12 it nn xte k h βϕ ωω Ω+ +    −Ω + Ω   =(2.28)= = ()() () () ooFF UF ( ) =X =UU i iti ititit xt e ee eee k kk ϕ β ϕββ Ω+Ω+Ω+Ω ==   =(2.29) = The quantity in the square brackets of eq.(2.28) is the absolute value of the complex frequency response, U( ) Ω  . It is called the magnification factor and is a dimensionless ratio between the amplitude of displacement= X =and the static displacement →Fk .=Conversely, the=ratio between the system response and the corresponding excitation is called the Frequency Response Function=(FRF) or transfer function.= H()=X()F() Ω ΩΩ   =( 2.30)= The lower graph=of=Figure=2.2 shows=the absolute value=of the complex frequency response= U( ) Ω  =as a= function of the dimensionless frequency ratio nω Ω for various values of the damping ratio h. When the frequency of the excitation is close to the system natural frequency the amplitude of the system vibration considerably increases, especially with light damping, even in case of not very high excitation magnitudes. This phenomenon is called resonance. It is noteworthy that high vibration levels may not be allowable for practical reasons, and they can cause excessive mechanical stresses (fatigue phenomenon) and acoustical emissions. Therefore, it is very important to avoid that time-varying forces excite the resonances of a vibrating system. Note that a SDOF vibrating system has only one natural frequency, that is only one resonance peak. It can be seen that increasing damping ratios h tend to reduce the amplitudes of the resonance peak and to shift the peak to the left of the vertical straight line passing through 1 nω Ω= . The peak of the FRF=occu±s= at=a frequency=given byΩ= 2 = 12 n h ω Ω− =( 2.31)= where the peak value of U( ) Ω  =is given by:= 2 1 U( ) = 21hh Ω −  = (2.32)= For light damping (h < 0.05) the curves are nearly symmetric about the vertical through 1 nω Ω= . The=peak value of U( )Ω  =occurs in the immediate vicinity of 1 nω Ω= =and is given by: = 1 U( ) Q 2h Ω≅ =  = (2.33)= where Q is known as the Quality Factor.= For a given damping ratio=h, the points, P 1 and P 2, where the amplitude of U( )Ω  = reduces to= Q2 =of= its peak value are called the half power points (Figure 2.3). If the ordinate is plotted on a logarithmic scale, P 1 and P 2 are points where the amplitude of U( )Ω  = reduces by 3 dB and are thus called the=�=3=dB points.= The=difference in the frequencies of these two points is called the 3=dB bandwidth of the system, and for light damping it can be shown thatΩ= 21 2 n h ωω ω ∆Ω = − = = (2.34)= where ∆ is 3 dB bandwidth, 1ω =is the frequency at point P 1, 2ω =is the frequency at point P 2.= From eqs.(2.33) and (2.34) we obtain:= 21 1 2 Q n h ωωη ω −= = = =(2.35)= where η is called the Loss Factor. In accordance with eq.(2.24) the amplitude of the function U( ) Ω  =at the natural frequency= nω =is: ( ) U( ) 1 2 h Ω=  . Therefore it is:= () 2 22 1 11 1 U( ) 2 282 n h h ω = =  = (2.36)= The frequencies,= 1ω =and 2ω , associated with the two halfϑpower points=(Figure 2.2)=are given by the= solution of the following expressionΩ= ( ) () 2 22 2 22 11 1 U( ) 2812 n nn hh ω ωω = =−Ω + Ω  = (2.37)= 0 .6 0 .7 0 .8 0 .9 1 1 .1 1 .2 1 .3 1 .4 0 20 40 60 80 1 00 1 20 D im e ns ionle ss fre q ue ncy Amp. H 2 Half po wer po ints = = Figure 2.2 Half power points= = That is:= 44 22 222 2 1 24 8 nn n hh ωω ω +Ω−Ω+Ω = = ( ) () 42 22 21 218 0 nn hh ωω   ΩΩ − − +− =       =(2.38) = The solutions of eq.(2.38) areΩ= ( )() 2 22 12 2 1 n h hh ω Ω =−± +    =(2.39 )= In case of common low values of the damping ratio h=( 1 h) the term= 2h =is negligible ( 2 0 h≅).= Therefore, the frequencies, 1ω =and 2ω , associated with the two halfϑpower points are given by:= 2 12 n h ω Ω = ±  () 22 1 12 n h ωω = − , = ( ) 22 2 12 n h ωω = + =(2.40) = As a result it is:= 22 21 2 4 n h ωω ω −= that is: 2 12 1 4 nn h ω ωω ω ωω −+ = =(2.41) = If the resonance peak is nearly symmetric it is:= 21 2 n ωω ω += =(2.42) = Then, we have: ( ) ( ) 21 21 21 21 22 4 nn h ωω ωω ωω ω ωω ω +− − == + = 21 1 2 Q n h ωωη ω −= = = =(2.43) = This last equation coincides with eq.(2.35).= = The upper graph of=Figure=2.3=shows the curves of the phase angle ϕ =against nω Ω =for various=values of h plotted from eq.(2.28). It should be noted that all curves pass through the point 2 ϕπ = , 1 nω Ω= . Moreover, the phase angle tends to zero for 0 nω Ω→ =and to 180� for nω Ω →∞ .= = 0 0 .5 1 1 .5 -18 0 -13 5 -90 -45 0 D im e ns ionle ss fre q ue ncy P ha s e [ de g r ee ] Dimensionless response h: 0 .02 5 h: 0 .05 h: 0 .07 5 h: 0 .1 h: 0 .2 h: 0 .5 h: 0 .75 0 0 .5 1 1 .5 0 5 10 15 20 D im e ns ionle ss fre q ue ncy Amplitude X/X 0 h: 0 .02 5 h: 0 .05 h: 0 .07 5 h: 0 .1 h: 0 .2 h: 0 .5 h: 0 .75 Figure 2.3 Ratio o XX and the phase ϕ , respectively, as a function of the dimensionless frequency nω Ω = = Eqs.(2.26 and 2.27) are plotted as a function of nω Ω =in Figures=2.4=and 2.5, respectively, in order to examine the variation of the inϑphase Re X  =and quadrature= Im X  =components of displacement. The=curves of the real component o∞=displacement in Figure=2.3=have a zero value at= 1 nω Ω= = independent of=the damping=rati→=h and exhibit a peak and a notch at frequencies: 1= 12 n h ωω − and = 2= 12 n h ωω + =(2.44)= respectively.= As the damping decreases (h get smaller), the peak and the notch increase in value and become closer together. In the limit when h = 0, the curve has an asymptote at 1 nω Ω= . The frequencies 1ω =and 2ω ,= mentioned above,=are often used to determine the damping ratio of the system from=the equationΩ= () () 2 212 21 1 =2 = 1 h ωω η ωω − + =(2.45)= The curves of the imaginary component of displacement have a notch in close vicinity of 1 nω Ω= =and they are sharper than those of= U( ) Ω  =in Figure=2.2=for corresponding values of h. 0 0 .5 1 1 .5 -10 -5 0 5 10 D im e ns ionle ss fre q ue ncy Re[ X/X 0 ] R eal part o f the dim ensio nles s re sp o ns e h: 0 .02 5 h: 0 .05 h: 0 .07 5 h: 0 .1 h: 0 .2 h: 0 .5 h: 0 .75 Figure 2.4 Real part of the ratio o XX as a function of the dimensionless frequency= nω Ω = 0 0 .5 1 1 .5 -20 -15 -10 -5 0 D im e ns ionle ss fre q ue ncy Im[ X/X 0 ] I mag inary p art o f the d ime ns io nle ss res p ons e h: 0 .02 5 h: 0 .05 h: 0 .07 5 h: 0 .1 h: 0 .2 h: 0 .5 h: 0 .75 Figure 2.5 Imaginary part of the ratio o XX as a function of the dimensionless frequency= nω Ω = = Figure=2.6=shows a polar plot of the dimensionless response evaluated for a damping ratio=h = 0.025. If the real component, imaginary component and frequency ratio are plotted on three mutually perpendicular axes, the three dimensional curve obtained is that shown in Figure 2.7 (for a damping ratio h of 0.05). The curves of Figures 2.4 and 2.5 are in fact the projections of curves similar to that of Figure 2.7 in the { } Re X ,  Ω  = and {}Im X , Ω  = planes, respectively. The third projection of the curve in the= { } Re X , Im X    =plane would look like the curve=shown in Figure=2.6.= The right hand side of eqs.(2.26=and 2.27) are plotted on the=x and y axes of Figure 2.6, respectively, for varying values of nω Ω . These curves could also be obtained by plotting the total displacement give by= expression (2.28) from the origin 0 at an angle ϕ from the Re X  = axis given by eq.(2.22). Thus, the curve= of Figure=2.7=is the locus in the Argand plane of the end of the line representing the total displacement for a particular=h, as the frequency ratio nω Ω =is varied. It can be seen again in Figure=2.7=that no matter what the damping is, at the undamped natural frequency 1 nω Ω= =the real component is zero, or in other words, the phase angle between the force and the total displacement is=90�.= 5 10 15 20 30 2 10 60 2 40 90 2 70 1 20 3 00 1 50 3 30 1 80 0 Polar plot of the dimensionless response 0.973 1.026 0.999 Figure 2.6 Polar plot of the dimensionless frequency response (damping ratio h = 0.025) 0 0 .5 1 1 .5 2 -10 0 10 20 -20 -15 -10 -5 0 Re[ X/X 0 ] Dimensionless response Dimensionless frequency Im[ X/X 0 ] = = Figure 2.7=Real and Imaginary parts of the ratio → XX as a function of the dimensionless= frequency nω Ω =(damping ratio h = 0.025) The upper graphic of Figure 2.3 shows the changes in the phase of the FRF H( ) Ω  . Let us consider two frequencies,=f a and f b, which are below and above the system natural frequency f n. Then, let us denote aϕ = and bϕ =the corresponding values of the phase of the FRF H( ) Ω  (expressed in radians). If the two= frequencies f a and f b are sufficiently close to f n an estimate of the damping ratio h is be given by: ( ) ( ) 1 = ba n ba h ωω ω ϕϕ − − = (2.46)= = Eq.(2.2) represents the equilibrium of the forces exerted, in vertical direction, on the mass m. These time- varying forces are vector quantities. Therefore, they can be plotted in a polar plane. Figure 2.8 shows the vector associated with each force, for the case of an excitation frequency, Ω, significantly lower than the system natural frequency nω . In this case, the exciting force F(t) is mainly balanced by the elastic force, while the phase ϕ is small. Figure 2. 8 Polar plot of the force vector s exerted on th e system, for an excitation frequency,=Ω, significantly lower than the system natural frequency nω = Figure 2. 9 Polar plot of the force vector s=exerted on the= system, for an excitation frequency,=Ω, significantly higher than the system natural frequency nω = = Figure=2.9 shows the vector associated with each force, for the case of an excitation frequency, Ω, significantly higher than the system natural frequency nω . In this case, the exciting force F(t) is mainly balanced by the inertial force while the phase ϕ tends to – 180°. In the end, Figure 2.10 shows the vector associated with each force, for the case of an excitation frequency, Ω, equal to the system natural frequency nω . In this case, the exciting force F(t) is balanced by the damping force while the phase ϕ tends to – 90°. Figure 2. 10 Polar plot of the force vector s exerted on the system, for an excitation frequency,=Ω, equal to the system natural frequency nω = Appendix A With reference to the Euler’s notation used to express the harmonic excitation, () o F F cos ( )tt β = Ω+ = or= () o F F sin ( )tt β = Ω+ , it is also possible to writeΩ= () oo F F cos ( ) F cos ( )tt βθ = Ω+ = = [] ooF F cos() sin() ieiϑ θθ = + = = ()() ()() oo ooFF cos ( ) sin ( ) cos ( ) sin ( ) F cos() sin() cos() sin() 2Fcos() iieeii iiϑϑ θθ θ θ θθ θθ θ − + = + + −+ − =   + +− =   = That is:= → oo F F cos ( ) F cos ( ) 2 ii tee ϑϑ βθ −  Ω+ = = +  = Therefore, a cosine function having amplitude X can be expressed as the sum of two counterϑrotating= vectors, having half the amplitude, X/2, of the original rotating vecto±=(see Fig.=2.11).= Similarly, a harmonic excitation having a sine law can be expressed asΩ= = () oo F F sin ( ) F sin ( ) tt βθ = Ω+ = = [] ooF F cos() sin() ieiϑ θθ = + = = ( ) () ()() oo ooFF cos ( ) sin ( ) cos ( ) sin ( ) F cos() sin() cos() sin() 2Fsin() iieeii i iiϑϑ θθ θ θ θθ θθ θ − − = + − −+ − =   + −− =   = = That is:= → oo F F sin ( ) F sin ( ) 2 ii ti ee ϑϑ βθ −  Ω+ = = −  = Therefore, a sine function having amplitude X can be expressed as the=difference=of two counterϑrotating= vectors, having half=the amplitude, X/2, of the original rotating vecto±=(see Fig.=2.12).= Besides, it is always possible to transform a cosine function into a sine functionΩ= () → oo → F F cos ( ) F cos ( ) F sin ( 2 ) F sin ( 2 )ttt β θ πθ π β = Ω+ = = − = −Ω− = = = = Figure 2.11 Cosine function expressed as the sum of two counte±ϑrotating vectors= = Figure 2.12 Sine function expressed as the difference of two counter-rotating vectors