:= 0.75 Ns m K N m. := 0.05 kg. K N m. (a.1) FBD and forces. (a.2) derive EOM From the FBD diagram, Newton's 2nd law states:

Transcription

1 P FA - Derive EO for simple mechanical system L San Andres (c) For the system shown in the figure, Z (t) Z o cos( t) is a periodic displacement input (known). Perform the following tasks: a) Set X(t) as the coordinate for the block motion; X denotes the unstretched length of spring. Draw a free body diagram and derive the block equation of motion. You should get: X + CX + ( + ) X( t) Z( t) (ust demonstrate how you derive the equation) b) For 5 N/m, 4 N/m, 5 gram and C.75 N.s/m, find the system natural frequency (Hz) and damping ratio (ζ). c) For Z o 5 mm, Hz, determine the periodic forced response of the block X(t), i.e., calculate the amplitude of motion and the phase lag relative to Z(t). Please explain where operation is above or below or around the natural frequency; most importantly, is operation safe? d) BONUS: Determine the excitation frequency range where the amplitude of motion X max 3 mm is not exceeded. C : kg 5 N m :.75 Ns m 4 N m C X Z(t)Zo cos(t) (a.) FBD and forces X means unstretched position of spring Dashpot force: FD C dx/dt For FBD diagram, assume a state of motion such that Z > X > Spring force: FS (Z-X) Spring force: FS X Normal from wall weight NW () (a.) derive EO From the FBD diagram, Newton's nd law states: d X F S F S F D d t where d F N W Damper C x (3) is the dashpot force d t F S X (4) Elastic forces from springs and. Spring F S ( Z X) Substitute Eqs. (3,4) into Eq. () to get drives motion of block. d d X + C X + eq X Zt () d t d t (5) and where Zt () wall reaction force () Z o cos( t) eq : +

2 (b) Calculate natural frequency (fn) and viscous damping ratio (ζ): ζ : : eq C eq 4.8 sec (c) Calculate response for Z o m f n : π : f : Hz Zt (): Z o cos( t) f n 8.46 Hz ζ.66 : f π (6) The system response will be Where Xop and φ are the amplitude and phase lag of the motion X(t) Using FRF formula: Find frequency ratio Find Amplification factor Find Amplitude at steady-state Xt () X op cos( t + φ) A() r : (7) : r : r. ( r ) + ( ζ r) Z o note here that Fo Zo X op : Ar () eq Find phase lag φ X op.375 large amplitude ratio - Z o NOT very safe X op ζ r : atan π r since r> φ 8 π operation just above natural frequency Ar () m φ 48 rad (d) Frequency range for max amplitude Let X max : 3 mm from X op Z o Ar () eq : Let δ : X max eq Z o degrees then δ Ar () δ ( r ) 3 + ( ζ r)

3 expanding: + ζ δ s + s s where s r s collecting like terms δ s + 4 ζ δ δ and identifying this polinomial with as solve for the two roots s s + bs + c b b : 4 a c a s.694 b b : + 4 a c a s.34 + a : δ b : 4 ζ δ c : δ hence, the two frequencies below and above which the amplitude ratio will be less than Xmax are: f : s fn f : s fn f n 8.46 Hz f 5.9 Hz f.74 Hz For comparison purposes, lets plot the function X op ( r_ ) : Z o eq ( r_ ) + ( ζ r_ ) f f X freq op f n.6.4. X max freq

4 EEN SP9 - Exam - Problem (Periodic Forced Response) A machine weighing W lb is known to cause unacceptable vibrations in nearby equipment. The engineering modification consists of mounting the machine on four vibration isolators, as shown schematically in the figure. Upon installation, the static deflection of each isolator is inch. Each isolator has a viscous damping coefficient equal to C.8 lb.s/in. There is a periodic load excitation acting on the machine, i.e. FF o cos(t), where F o lb and frequency f(/π)7 Hz, determine a) System natural frequency (f n [Hz]) and damping ratio (ξ). [] b) Amplitude [inch] and phase lag [degrees] of system motion Y(t) at the excitation frequency (f7 Hz). [] c) Sketch the response Y(t) versus time, label physical dimensions in graph. [5] d) The amplitude of motion if (for some unfortunate reason) a periodic force with a frequency equal to the system natural frequency, i.e. FF o sin( t), is exerted on the machine. [5] F(t) EO with Y origin from SEP: Y(t) d Y d t d + Y + C Y F o cos t d t C Isolation ount known parameters ( other on back) not shown : C i : lb.8 lbf s in g in s static deflection: frequency of forcing function: δ : in i : g 4 δ Each isolator supports /4 the machine weight. Total stiffness: : 4 i C : 4 C i Force magnitude: F o : lbf i rad : 7 π s 55 lbf in isolator stiffness N m C.96 3 N s m a) Natural frequency and damping ratio: C ζ : n : ( ) rad f s n : π f n 3.7Hz F o static deflection: Y s : Y s.455in b) for harmonic excitation: Ft (): F o cos t The machine periodic response is Yt () Y op cos t + φ Y op Y s Hr () ζ compare to response if C (low frequency) Y S () t F o : cos t

5 Find frequency ratio The amplification factor and phase angle are: Hr (): ( r ) + ζ r Hr ().49 Y op : φ() r 3.86 Y s Hr () r : r.38 Operation above natural frequency ζ r φ( r) : atan r π since r > φ() r degrees π Y op.3in << Ys Y s.455in (c) graph of periodic response for two periods of motion System response vs. time Yt (): Y op cos t + φ() r T : π period T.43 s Y(t) [in] Note the response at t shows a negative displacement. This is since the phase angle is close to -8 degrees. Broken-line shows curve of F(t)/.4.9 (d) for excitation at the natural time (s) frequency: Y s ζ 444 in Y opn r : Hr () Y opn : Y s Hr () 444in Phase lag is -π/ rads (-9 degrees) (e) Isolators are effective at frequency of operation (r>.44), where Y op < Y s 5 Amplitude (mm) 5 3 frequency ratio rw/wn

6 EEN 363 FALL 6 EXA PROBLE The figure shows a simple mechanical system (-C-) excited by a harmonic force F(t)F o sin(t), F o 5 lb f. The frequency of the external force excitation ranges from 4 Hz to Hz. Tests show the natural frequency is f n ( Hz) and the viscous damping coefficient (C) equals 4 lb f.s/in. The mass of the system is lb. a) Determine the system steady-state amplitude of motion (in inch) and phase lag (in degrees) for excitation at 5 Hz. [] b) Repeat (a) with stiffness equal to twice the original value. By how much the amplitude of motion is reduced or amplified? [6] c) Repeat (a) with a mass equal to twice the original value. By how much the amplitude of motion is reduced or amplified? [6] d) Based on the results (a-c) what system configuration will you select? Are there other options? Explain. [3] C F(t) DATA: Ft () F o sin t f n : Hz F o : 5 lbf frequency of operation between 4- Hz EY: : lb : f n π C : 4 lbf sec in 6.83 rad sec W: g Analyze excitation at f : 5 Hz : f π First determine the ORIGINAL system stiffness () and damping ratio (ζ): : ζ : C.7 3 lbf in ζ. Let: orig : orig : ζ orig : ζ for later use Basic nowledge as stated on Summary of formulas: The system periodic response is Yt () δ s H sin t + Ψ F o where δ s is the static displacement, H is the amplification factor and Ψ the phase angle defined as a function of the frequency ratio (r/n): Hrζ, : ( r ) + ζ r : atan Ψ r, ζ ζ r r with r (a) ORIGINAL system r : f f n r operation above natural frequency since r> Y op_original : F o Hrζ, amplitude: Y op_original.3 in phase lag: 8 Ψ r, ζ π degrees (b) If the stiffness is doubled: NEW: : F o Y op_b : Hrζ, ζ : amplitude: : C orig Y op_b.3 in rad sec phase lag: Natural frequency increases and damping ratio decreases respect to original system ζ.7 8 Ψ r, ζ r : r.6 > degrees π since r ~ Y op_b 3 Y op_original amplitude increases 3 times, because operation is too close to resonance. Furthermore, damping ratio is smaller than original.

7 (c) If the mass is doubled: NEW: : F o Y op_c : Hrζ, amplitude: ζ : : orig C : Y op_c. in orig rad sec phase lag: Natural frequency decreases and damping ratio decreases respect to original system ζ.36 8 Ψ r, ζ r : r degrees π > Y op_c.367 Y op_original nearly - 8 deg, since r >> Amplitude decreases to.37 times, because operation is well above natural frequency. (d) Select case (c), i.e. double mass since it shows the lowest amplitude of response. The system effectively works as an isolator. Doubling the stiffness is certainly NOT a good choice since natural frequency is too close to excitation frequency. Another option? eep original system and DOUBLE damping? Not really, see below: ζ orig. Y op : F o H orig Y op.9 in f, ζ orig f n compared to: Y op_original.3 in f n Hz f 5Hz r f f n very little reduction. Why? already operating above natural frequency. Damping does little to reduce vibration amplitude!

8 P. Periodic forced response of a SDOF mechanical system. DESIGN COPONENT The signal lights for a rail may be modeled as a 76 lb mass mounted 3 m above the ground of an elastic post. The natural frequency of the system is measured to be. Hz. Wind buffet generates a horizontal harmonic F force at Hz. The light filaments will break if their peak accelerations exceed 5g. Determine the maximum acceptable force amplitude F when the damping ratio ζ. and.. Full grade requires you to explain the solution procedure with due attention to physical details 3 m The excitation force is periodic, say F(t)Fo sin(t). then the system response will also be periodic, Y(t), with same frequency as excitation. Assuming steady state conditions: STEADY RESPONSE of --C system to PERIODIC Force with frequency Case: periodic force of constant magnitude Define operating frequency ratio: r Ft () F o sin t where: System periodic response: Yt () δs H() r sin t + Ψ F o δ Hr () s ( r ) + ( ζ r) tan Ψ e () ζ r r care with angle, range: to -8deg From (), the acceleration is at () Yt () A sin t + Ψ 8 ( the magnitude of acceleration is A F o or A e ( r ) + ( ζ r) F o e r ( r ) ζ r + hence, define system mass A max : e : g 5 lb maximum allowed acceleration of filament HZ : π s f n : HZ natural frequency f : HZ f Let r o : f n r o.958 The maximum force allowed equals excitation frequency due to wind buffets F max r, ζ close to natural frequency : A max e ( r ) ζ r + r

9 F o : without any damping with damping F max r o, ξ ξ :. For the force found the amplitude of acceleration is F max ( r o, ) 33.7 lbf F max ( r o, ξ) 34.3 lbf A( r, ζ) F o : e r ( r ) ζ r Note the importance of damping that leads to a substantial increase in force allowed + F max r o, ξ F max r o, 53 Amplitude of acc/g 3 6 ff n A max g f n s ξ 5 GRAPH NOT FOR EXA 5 5 frequency (rad/s) with damping without damping SInce ro~, a simpler enginering formula gives A max e ξ 3 lbf which gives a very good estimation of the maximum wind force allowed b) a system with damping ξ. will produce a 55 % increase in allowable force Hence, the rail lightsystem will be more reliable, lasting longer. F max r o, ξ F max r o, 53 c) Posts are usually hollow for the cables to be routed. These posts have layers of elastomeric material (~rubber-like) inside to increase their structural damping. odern posts are wound up fro composites that integrate damping layers. Clearly, adding a "true" dashpot is not cost-effective

10 EEN 363/67 Example Base motion - Frequeny response The figure displays a schematic view of a read/write head in a hard disk. The support arm holding the R/W head is represented by structural stiffness s 5 N/m, and damping coefficient C s [Ns/m] to be determined. The R/W head mass equals 5 gram. a 6 N/m represents the stiffness of the air film between the rotating disk and the R/W head. When the disk spins, micro-asperities in the disk induce a wavy-like motion represented as X d A cos(t), where A micrometer. The frequency (f/π ) of the excitation varies from 5 Hz to 4 Hz, depending on the rotational speed of the disk and the radial position of the R/W head support structure. The origin of the coordinate systems, XX d, represents the static equilibrium position. R/W head mechanism Hard disk Support arm Read/write head a) Draw a free body diagram, derive the equation of motion for the R/W head system, and calculate the system natural frequency (Hz) [5, 5, 5] b) State a formula for the motion X(t), i.e. amplitude and phase angle as function of frequency [5] c) When, find the magnitude of the damping coefficient C s such that the amplitude of motion for the R/W hear does not exceed micro-meter. [5] d) Sketch the amplitude and phase for the frequency response of the R/W head. Your graphs must show physical dimensions and explanatory sentences explaining relevant information related to the system motion [ x 5] Cs a s X Xd A cos(t) Hard disk rotating drum DATA: a 6 N m s a kg : TBD EY: Assume X >Xd (a) Draw FBD, derive EO, find n Define coordinate systems with origin at static equilibrium position. FD Cs dx/dt FS s X - W Upon assembly, the arm support spring holds the weight of the R/W head. Note that in actuality, the air film stiffness is zero when the disk does not spin. Apply Newtons law d X F D d t F s F a W W X Fa a (X Xd) The Equation of motion for the system is where e a s natural frequency for periodic motions d d X e X C s d t dt X a X d Ft () X d Acos t A 6 m then Ft () A a cos t n e f n n f n 6 3 Hz n rad s

11 (b) Establish system periodic response The system periodic response is: (from cheat sheet) Xt () Hsin t X A sin t X A Hr () where A a e is the "static" displacement m H is the amplification factor and is the phase angle defined as a function of the frequency ratio (r/n): Hr () r r tan r with r r n (c) Calculate damping coefficient needed for amplitude of motion NOT to exceed microns X Amax 6 m Define ratio X Amax. and from equation for amplification factor H: for operation at the natural frequency r.7 r Hence r C s e C s N s m Value of damping coefficient Functions for graphs (d) FRF graphs (amplitude and phase) of R/W head versus frequency range of interest 3 6 FRF 8 FRF Amplitude of response (m) 6 Phase of response (degrees) frequency (Hz) 3 4 frequency (Hz)

12 Note that at low frequencies, amplitude equals m and phase lag is degrees at high frequencies, amplitude approaches null values and phase lag approaches 8 degrees at the natural frequency, amplitude of motion is X Amax 6 m NOT for exam: Relative motion is of importance, i.e. YX-Xd X A Y Y Y A 3 6 FRF Y 8 8 arg Y FRF Amplitude of response (m) 6 Phase of response (degrees) frequency (Hz) 3 4 frequency (Hz) 9.9 Y n 8 m A m Phase angle cannot be positive

13 EEN QUIZ 3 Names: The rotor of an electric generator weights 75 lb and is attached to a platform weighing 775 lb. The motor has an imbalance eccentricity (a) of mils. The platform can be modeled as shown in the figure. The equivalent stiffness () of the platform is million-lb/in, and the equivalent damping is C lb-s/in. The operating speed of the generator is 8 rpm. (a) determine the response of the platform (amplitude and phase) at the operating speed. (b) determine the response of the platform (amplitude and phase) if the rotor spins with a speed coinciding with the system natural frequency. (c) if the platform stiffness is increased by 5%, determine the allowable amount of imbalance (a) that will give the same amplitude of motion as determined in (a). Assume that the mass of the platform and the damping do not change appreciably by performing the stiffening. a rotor k platform c EY: Given System excitation due to rotating imbalance : 6 lbf in : rotor + platform rotor : 75 lb C : lbf calculate the system natural frequency and damping ratio: C n : ζ : platform : sec in 775 lb 8 3 lb and operating frequency ratio (r) for rotor speed: RP π rad : 6 s 3 rad s The system response (amplitude and phase) for imbalance excitation are: Y op () r a rotor r : Ψ( r) : atan ( r ) + ζ r rad s ζ. r : little damping RP : 8 r.884 a : 3 in ζ r r

14 Thus, at r.884 < Y op ( r) in Ψ() r [degrees] π Let: Y oper : Y op () r (b) If the rotor should spin with a speed coinciding with the system natural frequency, r: : RP 6 s : π rad RP.35 3 the system response is also determined from: : to maintain Y oper in calculate the NEW system natural frequency and damping ratio: C n : ζ : 38.8 rad ζ 93 3 small change in damping ratio s and operating frequency ratio (r) for rotor speed: RP : 8 RP π rad : 6 s from relationship: Y op ( ). in Ψ 9 a rotor ζ rad s r : Y op () r a rotor : determine the (new) allowable imbalance: : degrees Y op. in (c) If the platform increases by 5%, original : r.79 r ( r ) + ζ r 3. Y oper a : rotor Y oper r ( r ) + ζ r a in i.e. ~ twice as original imbalance (eccentricity) displacement,

15 Example: Base motion excitation An instrumentation sensor with mass lb is attached to the casing of a steam turbine running at 36 rpm. The casing motion has an amplitude of in. Test show that the mass supported by the sensor attached has a natural frequency of 7 Hz. How much damping (lb.s/in) is needed to keep the sensor steady state amplitude below in? Calculate the sensor stiffness (lb/in). EY: This is a typical problem in which the source of excitation for the sensing element is base motion. The sensing element is a simple spring-mass-damper system. a) calculate frequency ratio (r) giving us information on the regime of operation (below, around or above the natural frequency) excitation frequency: RP : 36 W : lbf : lb RP π rad : 6 s rad s rad natural frequency: : 7 π s rad s frequency ratio: r : operation near to resonance: b) The amplitude of motion (A) for the casing and desired (Yop) for the sensing element at frequency ratio (r) are: A : in Y op : in r.86 and related by: Y op A + ( ζ r) [] ( r ) + ζ ( r) Let's define G as the ratio G Y op : A G 3 Y op From the expression for amplitude ratio: [] A ζ G r + r + ζ r we can determine the damping ratio (ζ) from: G r G ζ r ( )

16 or: ζ : G r ( r ) G Thus, at r.86 ζ. is the damping ratio needed to produce the desired amplitude ratio. G 3 c) The desired damping coefficient is equal to: W C : ζ g C.4 lbf sec C 4.9N s in m The stiffness of the sensor element is: : N 5 lbf m in

17 EEN Torsional vibrations A device to mix painting is composed of the paddles and hub connected through a stepped steel shaft to an electric motor. The mass moment of inertia (I) of the hub and blades is kg.cm, and the painting introduces a viscous damping (D θ ) equivalent to 3 N.cm.s/rad. The stiffnesses of the connecting shafts equal and 3 Nm/rad, respectively. a) Derive the equation of motion for the paddles angular displacement θ(t) as a function of the motor displacement θ e (t)θ m sin(t). a) Calculate the system natural frequency, critical damping and damping ratio b) Calculate & graph the FRF (amplitude and phase) of the paint mixer c) At what motor speed the amplitude of response is the largest? d) The engine operates at a frequency of 5 Hz with amplitude Θ m degrees. Determine the twist angle and moment on the drive shaft. Are the calculated values reasonable (acceptable)? Stepped steel shaft: lengths (l,l) and diameters (d,d) ORIGIN : Schematic view of paint mixer θe(t) θ(t) Hub & paddles a) Derive equation of motion: Sum of moments d I θ dt Torque drive Torque drag [] I: kg cm mass moment of inertia d Torque drag D θ θ d t Torque drive θ θ e θ where the stiffness of the stepped shaft equals (springs in series) m : N rad m : 3 N rad Thus, the equation of motion is: drag torque from viscous fluid (paint) drive torque from transmission shaft θ : + θ s D θ : 3N cm rad 7 N m rad d d I θ + D θ θ + θ θ θ θ e θ Θ dt d t the natural frequency and damping ratio equal: sin( t) [] : ζ : D θ θ I I ζ rad s f n : π f n 3.8 Hz based on the periodic force response of a second order system, the impeller dynamic response is given by: θ() t Θ Hr () sin t φ [3]

18 where H and φ are the amplification ratio and phase lag, defined as: with Hr () :. ( r ) + ζ r r as the frequency ratio. φ() r : φ atan ζ r [4] r φ φ + π if r > return φ 8 π in degrees Graphs of the amplitude ratio and phase angle follow: Amplitude ratio H 3 4 frequency ratio Phase lag (degrees) 4 frequency ratio since there is a fair amount of damping, the peak amplitude of motion does NOT happen at r, i.e. when the engine frequency coincides with the natural frequency. Reviewing our knowledge, recall that the maximum amplitude occurs are (See Handout USES of FRF) for engine operation at f : 5Hz Θ : degrees : f π r : f f n Hr.398 φ( r) degrees r peak : ζ Hr ( peak ).4 r.8 notice that this operating frequency is very close to the one giving the peak motion. let: θ r : Θ Hr () θ r π φ r : φ() r 8 r peak.837 H peak : ζ radians (phase angle) ( ζ ) H peak.4 φ( r peak ) 65.6 degrees degrees Let: period of motion: T : f T.4 s

19 thus, the drive forcing function and mixer responses are: θ e () t : Θ sin t θ() t : θ r sin t φ r Graph both angular responses vs. time ( periods of motion) 5 Angle (degrees) θ e () t θ() t T.4 s engine mixer t time (s) the twist angle is defined as θ twist () t : θ e () t θ() t 5 Angle (degrees) θ twist () t t time (s) i.e. approximatelty equal to. θ TWIST : 7 π 8 The ampltude of the drive moment transmitted through the stepped shaft is just Torque drive : θ θ TWIST Torque drive 334 N m

20 Example: Base motion excitation L San Andres (7) The figure shows a vehicle moving with speed V along a wavy road. Prior tests show that the vehicle weighing lbf has a natural frequency of 3 Hz. Neglect the influence of the tire's bouncing mode and determine: a) the car speed (V) in mph (miles/hour) that will cause the highest amplitude of motion for the vehicle? Explain your answer b) Find the damping ratio () and damping coefficient (C lin lbf.s/in) for the car speed in (a) such that the vehicle's absolute amplitude of motion is less than 5 ft. c) Using C found in (b), calculate the vehicle steady amplitude of motion (ft) at a cruising speed V of 7 mph? V (velocity) C Yb Y suspension system b ft Data: lb f n 3Hz n f n L5 ft n 8.85 rad sec EY: L b 5ft ft path wavelength amplitude of wave (a) the car speed must excite the car-suspension system natural frequency. The relationship between the natural period of vibration (Tn) and the time it takes the car to travel a full wave length (L) is 58ft mph L 36sec T n V n f n V n Lf n V n 5 ft V n.7mph sec (b) The car motion amplitude (Yop) as a function of the amplitude of road wave amplitude (b), frequency ratio (r) and damping ratio is (from Cheat Sheet): r Y op bg () r cost n Y op and working with eqn []: b Gr () r r [] r G r r r Define G as desired Y op G b Y op 5ft G

21 determine the damping ratio () from: G r r G at r G r ( r ) G or 4 G []. is the damping ratio needed to produce the desired amplitude ratio, G The needed damping coefficient equals: C n C 4.6lbf sec in (c) At a cruising speed of V f V L r 7mph f f n f V.67 ft sec Hz r.68 excitation frequency frequency ratio f n 3Hz Y op () r b r r r Y op ( r) 3.4ft (d) For completeness, graph the amplitude of response for multiple car speeds amplitude of response (ft) 6 4 V n 5 ft sec car speed (ft/sec) 4 mph

22 EXAPLE - EXA TYPE: Dynamic measurements were conducted on a mechanical system to determine its FRF (frequency response function). Forcing functions with multiple frequencies were exerted on the system and a digital signal analyzer (FFT) recorded the magnitude of the ACCELERATION/FORCE ([m/s]/n) Frequency Response Function, as shown below. From the recorded data determine the system parameters, i.e. natural frequency (wn:rad/s) and damping ratio (z), and system stiffness (:N/m), mass (:kg), and viscous damping coefficient (C:N.s/m). Explain procedure of ANALYSIS/INTERPRETATION of test data for full credit. Acceleration/Force (m/s^/n) Solution: Ax x 5 frequency (Hz) agnitude of FRF for mechanical system Test data showing amplitude of (acceleration/force) Recall that for an imposed external force of periodic form: the system response Y(t) is given by: where the amplitude of motion (Yop) and phase angle (Y) are defined as: Ft () F o sin t Yt () Y op sin t + ψ [] [] F o Y op () r [3a] from [], we find that the acceleration is given by: ( r ) + ζ r Ψ atan ζ r r [3b] with r [4] a Y () t Y op sin( t + ψ) a op sin t + Ψ 8 [5] where: a op () r F o r ( r ) + ζ r [6] since: F o F o F o r

23 thus, the magnitude of amplitude of acceleration over force amplitude follows as: a op () r F o For excitation at very high frequencies, r>>. From the graph (test data): r ( r ) + ζ r. m s N [7] Thus The units of this expression are /kg m a op () r F o : kg s N The system appears to have little damping, i.e. amplitude of FRF around a frequency of 5 Hz is rather large and varying rapidly over a narrow frequency range. Thus, take the natural frequency as expressed in rad/s as: We can estimate the stiffness () from the fundamental relationship: : for excitation at the natural frequency (r), the ratio of amplitude of acceleration to force reduces to a op ( ) f o ζ from the graph (test data), the ratio is approximately equal to one (/kg). Thus. the damping ratio is determined as ζ :. ζ kg f n : 5 Hz : f n π 349 rad s N m That is, the system has a damping ratio equal to 5%. This result could have also been easily obtained by studying the ratio of (amplitude at the natural frequency divided by the amplitude at very high frequency, i.e.) ζ. Once the damping ratio is obtained, the damping coefficient can be easily determined from the formula: C : ζ C 349 N s m The number of calculations is minimal. One needs to interpret correctly the test data results, however.

24 Example: system response due to multiple frequency inputs Consider a nd order system described by the following EO L San Andres (c) 8 d d Y + C Y + Y zt () d t d t where zt (): a cos t + a sin t + a 3 cos 3 t is an external excitation displacement function Find the forced response of the system, i.e, find Y(t) Given the system parameters : kg : 6 N ζ. m : calculate natural frequency and physical damping : C : d : ζ d f d : π ζ damped natural period f n : π T d : f d f n 5.95Hz C 3 s N m T d.63s Set frequencies and amplitudes of the excitation z(t) are: assemble: a : m a : m a 3 : m 9 : : 3 :. zt (): a cos t + a sin t + a 3 cos 3 t. excitation displacement z(t) [m]... time (s) 4 periods of damped natural motion

25 SYSTE RESPONSE is: Yt (): a i H cos i t + φ i The system frequency response function: amplitude and phase angle are Hr () : ( r ) + ζ r φ() r : φ atan ζ r r φ φ π if r > return φ graphs of frequency response function Amplitude and phase lag as a function of r frequency ratio 6 Amplitude of FRF Amplitude H(r) Q factor ζ 5 3 Frequency ratio Phase angle (degree) Phase angle of FRF 3 Frequency ratio

26 The response of the system is given by the superposition of individual responses, i.e Yt (): Y cos t + φ where for first excitation: + Y sin t + φ + Y 3 cos 3 t + φ 3 r : r n Hr φ : φ( r ( ).3 ) Y a 8 : Hr ( ) Y φ m π degrees for second excitation: r : Hr φ : φ( r ) r.9 n ( ) 3.8 Y : a Hr Y.9m 8 φ π degrees for third excitation: 3 r 3 : Hr φ 3 : φ( r 3 ) r 3. n ( 3 ) 9 8 Y 3 : a 3 Hr ( 3 ) Y φ m π degrees Assemble physical response: Yt (): Y cos t + φ + Y sin t + φ + Y 3 cos 3 t + φ 3 Now graph the response Y(t) and the excitation z(t): z( & Y(t) [m].4.. excitation & response Note: The response Y shows little motion at the highest excitation frequency (3). There is an obvious amplification of motion with second frequency ( ~ n)..4.. Z Y time (s) To understand better, let's plot the actual FRF: T d.63s

27 .3 Amplitude of response a m r Y.66m Amplitude [m].. r.9 Y.9m r 3. Y 3.3m 3 all r r r3 Frequency ratio Note how response amplitude for largest frequency is largely attenuated Y 3 < a 3 while amplitudes for first two frequencies are amplified, in particular for which is close to the natural frequency Phase angle of response Y 3.8 a Phase angle (degree) Y.3 a 8 3 all r r r3 Frequency ratio

28 Example: system response due to periodic function Consider a nd order system described by the following EO L San Andres (c) 8 d d Y + C Y + Y zt () d t d t where ORIGIN : z (t) is an external periodic excitation function Given the system parameters : kg : 6 N ζ. m : calculate natural frequency and physical damping : C : d : ζ d f d : π ζ damped natural period f n : π T d : f d f n 5.95Hz C 3 s N m T d.63s Define periodic excitation function: zt : amp z o amp amp z o if t < T T if t > Example - square wave T d T : z o :. m Ω π : fundamental frequency T N F 7 Ω.333 d : number of Fourier coefficients amplitude (m) zt () t time(s) Find Fourier Series coefficients for excitation z(t) T mean value a : T zt () dt a m

29 j :.. N F coefs of cos & sin T a j : T zt cos j Ω t dt T b j : T zt sin j Ω t dt a T ( )m b T ( )m Build z(t) as a Fourier series N F Z F () t : a + j ( a j cos( j Ω t) + b j sin( j Ω t) ) z afj : Amplitude ( a j ) + b j. N F 7 amplitude (m) zt () Z F () t. Fourier coefficients t time(s) actual Fourier Series amplitude z afj j Find the forced response of the system, i.e, find Y(t) SYSTE RESPONSE is: Y o N F Yt (): + m ( Yc m cos( m Ω t ) + Ys m sin m Ω t m :.. N F Y o : a (a) set frequency ratio f m : m Ω

30 (b) build denominator den m : f m + ζ f m (c) build coefficient of cos() Yc m : (d) build coefficient of sin() a m f m den m ζ f m b m Ys m : b m f m den m + ζ f m a m (e) for graph of components Y Fm : ( Yc m ) + Ys m Y o N F Yt (): + m ( Yc m cos( m Ω t) + Ys m sin( m Ω t) ) Now graph the response Y(t) and the excitation (Fourier) z(t):.4 excitation & response T 3 Ω T.33 d z( & Y(t) [m].. amplitude Fourier coefficients Z Y time (s) input Z output Y 3 periods of fundamental excitation motion

31 Note: obtain response for inputs with increasing frequencies (periods decrease).4 excitation & response T 5 Ω T.99 d z( & Y(t) [m].. amplitude Fourier coefficients Z Y time (s) input Z output Y.4 excitation & response T Ω T.497 d z( & Y(t) [m].. amplitude Fourier coefficients time (s) Z Y input Z output Y.4 excitation & response T Ω T.995 d z( & Y(t) [m].. amplitude Fourier coefficients time (s) Z Y input Z output Y

32 .4 excitation & response T Ω T.99 d z( & Y(t) [m].. amplitude Fourier coefficients Z Y time (s) input Z output Y fastest Z (smallest period).4 excitation & response T. Ω T d z( & Y(t) [m].. amplitude Fourier coefficients Z Y time (s) input Z output Y

33 EXAPLE TORSIONAL VIBRATIONS: Z d m e e t A cantilevered steel pole supports a small wind turbine. The pole torsional stiffness is (N.m/rad) with a rotational damping coefficient C (N.m.s/rad). X Y k m X k Torsional Tiffness e Torional Damping coefficient c Z d e c os t Y The four-blade turbine rotating assembly has mass m o, and its center of gravity is displaced distance e [m] from the axis of rotation of the assembly. I z (kg.m ) is the mass moment of inertia about the z axis of the complete turbine, including rotor assembly, housing pod, and contents. The total mass of the system is m (kg). The plane in which the blades rotate is located a distance d (m) from the z axis as shown. For a complete analysis of the vibration characteristics of the turbine system, determine: a) Equation of motion of the torsional vibration system about the z axis. b) The steady-state torsional response θ (t) (after all transients die out). c) For system parameter values of k98,67 N.m/rad, I z 5 kg.m, C 57 N.m.s/rad, and m o 8 kg, ecm, d3 cm, present graphs showing the response amplitude (rad) and phase angle as the turbine speed (due to wind power variations) changes from rpm to, rpm. d) From (c), at what turbine speed should the largest vibration occur and what is its magnitude? e) Provide a design recommendation or change so as to reduce this maximum vibration amplitude value to half the original value. Neglect any effect of the mass and bending of the pole on the torsional response, as well as any gyroscopic effects. EEN 363 Example: Torsional vibrations Wind Turbine 4

34 Note: the torque or moment induced by the mass imbalance is ( o ) T( ) d Fu d m e cos( t), i.e., a function of frequency (the rotational speed of the turbine) The equation describing torsional motions of the turbine-pole system is: ( o ) I θ + C θ + θ m e d cos( t) T cos( t) z ( ) () Note that all terms in the EO represent moments or torques. (b) After all transients die out, the periodic forced response of the system is θss θ() cos / ( t φ) () t ( f ) + ( ζ f ) where T( ) mo ed mo ed θ ss f (3) I z C with f ; n ; ζ, and I I n z z φ ζ f tan f (4) Eq. (3) in () leads to mo ed f θ() t cos( - ) / t φ I z ( f ) + ( ζ f ) (5) Let med I o θ (6) & z and rewrite Eq. (5) as: B / ( f ) + ( ζ f ) f (7) θ () t θ B cos ( t φ) (8) EEN 363 Example: Torsional vibrations Wind Turbine 4

35 (c) for the given physical values of the system parameters: 98,67 N-m/rad n 6.8 Iz C I z 5 kg m ζ I z C 57 N-m/(rad/s) θ med o 8kg.m.3m rad Iz 5kgm 5 And the turbine speed varies from rpm to, rpm, i.e., rpm π/3.47 rad/s to 5.66 rad/s, i.e. rad sec f.67 to., n thus indicating the system will operate through resonance. Hence, the torsion or twist angle (system response) is ( 5 ) θ ( t) 96.4 rad Bcos ( t φ) (d) aximum amplitude of response Since the damping ratio is small, ζ <<, the maximum amplitude of motion will occur when the turbine speed coincides with the natural frequency of the torsional system, i.e., at f /, thus B ; and ζ π θ() t θ cos t ζ θ the magnitude is θmax.964 x - rad, i.e. times larger than θ. ξ The figures below depict the amplitude ( B) (degrees) and phase angle φ (degrees) of the pole twist as a function of the turbine rotational speed (RP) θ EEN 363 Example: Torsional vibrations Wind Turbine 4

36 amplitude of torsion (rad) rotor speed (RP) Phase angle (degrees) rotor speed (RP) (e) Design change: DOUBLE DAPING but first BALANCE ROTOR! EEN 363 Example: Torsional vibrations Wind Turbine 43