Dynamics of Mechanical Systems
Dynamics of Mechanical Systems MAE 321
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Chapter 2 Response to Harmonic Exc a on Introduces the important concept of resonance D J Inman 149 Mechanical Engineering at Virginia Tech 21 Harmonic Excitation of Undamped Systems Consider the usual spring mass damper system with applied force FtF0cosot a is the driving frequency F0 is the magnitude of the applied force We take 0 0 to start with Displacement F F 0cos 0t D J Inman 249 Mechanical Engineering at Virginia Tech Equations of motion Solution is the sum of F39gure 2391 homogenous and 1 d N particular solution F kl n F k The particular solution gtttl assumes form of mg forcing function physically the input mxg 00 F0 COSwE wins jag wfxo 2 f0 coswt xp t Xcoswt where f0 11 aquot J1 DA 1 Inman 349 Mechanical Engineeting at Vixginia Tech Substitute particular solution into the equation of motion f 2 N f N a XcosatwXcosatfocosat f0 02 2 0 Thus the particular solution has the form f of a solving yields X XP I 2 cosat D J Inman 4 49 Mechanical Engineering at Virginia Tech Add particular and homogeneous solutions to get general solution W particular A1 sin cont A2 cos cont 2f0 2 cos at Y J a a homogeneous A1 and A2 are constants of integration D J Inman 5 49 Mechanical Engineering at Virginia Tech Apply the initial conditions to evaluate the constants xOAlsinOAzcosO 2f 2cosOA2x0 can a a a n f 3A2x0 a a 5c0 wnAlcosO Azsin0 f0 SiHOZQnA1V0 05 602 v gtA1 0 gt a v xt 01nant xo 2f 2 coswnt Zfo 2COSCUf 211 a an a an a DJInrnan 649 Mechanical Engineering at Virginia Tech Comparison of free and forced response Sum of two harmonic terms of different frequency Free response has amplitude and phase effected by forcing function Our solution is not defined for on 03 because it produces division by 0 lf forcing frequency is close to natural frequency the amplitude of particular solution is very large D J Inman 7 49 Mechanical Engineering at Virginia Tech Response for m100 kg k1000 Nm F100 N o 03 5 v001ms and x0 002 m 005 3 M E CD E O I A 8 I v E Q L D 005 O 2 4 6 8 10 Time sec Note the obvious presence of two harmonic signals D J Inman 849 Mechanical Engineering at Virginia Tech G0 0 COde demo What happens when a is near on xtisintjsinw wtj 213 w af When the drive frequency and natural frequency 2 f0 Sin aquot cot are close a beating phenomena occurs a a2 2 1 05 Ill n nu A W amplitude Mil VV l 1 Displacement X 0 J O 5 10 15 20 25 30 Time sec D J Inman 9 49 Mechanical Engineering at Virginia Tech What happens when a is con xpt 2 IX sinat substitute into eq and solve for X Xza 260 When the drive frequency and natural frequency are the same the amplitude of the vibration grows without bounds This is known as a resonance condition The most important concept in Chapter 2 D J Inman 1049 grows With out bound xt A1 sin cot A2 cos cot 600 5 x iiVii A v iiiWV O 5 10 15 2O 25 30 Time sec Displacement x O Mechanical Engineering at Virginia Tech Example I Compute and plot the response for m10 kg k1000 Nm x00v002 ms F23 N 0020quot on P w 10rads a an 20 rads m 10kg 2 f0 23 N 3 Nkg V 0m002m m 10kg on 10 rads f0 23 Nkg of a2 102 202rad282 Equation 211 then yields xt 002 sin 10 7667x10 3cos10t cos 20 79667gtlt10 3 m D J Inman 1149 Mechanical Engineering at Virginia Tech Example 21 2 Given zero initial conditions a harmonic input of 10 Hz with 20 N magnitude and k 2000 Nlm and measured response amplitude of 01m compute the mass of the system 2 cos 20m cos wnt for zero initial conditions xt 0 w trig identity 2 xt sin a a 2 sin a a 2 2 01111 2 2 2f0 2 01 2003mm 2 01 Onw 412075 m045 kg DJInman 1249 Mechanical Engineering at Virginia Tech Example 213 Design a rectangular mount for a security camera 2 camcraa 176 mounting bracket mi t 0t Fu costar Compute E so that the mount keeps the camera from vibrating more then 001 m of maximum amplitude under a wind load of 15 N at 10 Hz The mass of the camera is 3 kg D J Inman 1349 Mechanical Engineering at Virginia Tech SOIUtionlModeling the mount and camera as a beam with a tip mass and the wind as harmonic the equation of motion becomes E 3 xt 2 F0 cos 0t 3 From strength of materials I Thus the frequency expreSSIon IS 2 3Ebh3 EM 12m 3 14mr3 Here we are interested computing X that will make the amplitude less then 001 m a 001lt 2f for wf w2lt0 2 2 n 2f T02 lt 001 n 2 b lt 001 for of 602 gt 0 a a DJInrnan 1449 Mechanical Engineering at Virginia Tech Case a assume aluminum for the material 2f Ebh3 001 lt 02 3602 gt 2f0 lt 001602 00160 gt 001602 2f0 gt 0014ml3 E h3 gt 13 gt 001 0321 gt lgt 06848 In 4m001co2 2 f0 Case b 2f Ebh3 r0602 lt 001 gt 2f0 lt 001m 001m2 gt 2f0 001m2 lt 001m E 03 gt 13 lt 0012 0191 gt 1lt 0576 m 4m2f0 001a DJInrnan 1549 Mechanical Engineering at Virginia Tech Remembering the constraint that the length must be at least 05 m a and b yield 05 lt I lt 0576 or I gt 06848 m Less material is usually desired so chose case a say 6 055 m To check note that 3 a a2 fl 207z2 1742gt0 m Thus the case a condition is met Next check the mass of the designed beam to m plbh3 insure it does not change the frequency Note 27 gtlt1O3O55O02O023 It Is much less then m 2376 X104 kg D JInrnan 1649 Mechanical Engineering at Virginia Tech A harmonic force may also be represented by sine or a complex exponential How does this change the solution mi t kxt 2 F0 sin or or 5ampt wjxa 2 f0 sin or 218 The particular solution then becomes a sine xp t X sin wt 219 Substitution of 219 into 218 yields f0 x t 8111 cat p a2 Solving for the homogenous solution and evaluating the constants yields a l smwnt 2f0 2 smwt 225 a a a a a K v0 xt x0 cos wnt 0 D J Inman 1749 Mechanical Engineering at Virginia Tech Section 22 Harmonic Excitation of Damped Systems Extending resonance and response calculation to damped systems D J Inman 1849 Mechanical Engineering at Virginia Tech 22 Harmonic excitation of damped systems mm 6562 kxl 2 F0 cos at 5amp2 2420mm 05x0 2 f0 cos or xp 2 cos 602 92 V now includes a phase shift Displacement FFocosat D J Inman 1949 Mechanical Engineering at Virginia Tech Let xp have the form xp t 2 AS cos wtBS sin 0t B X AS2 13S26tan1 AS S icp aAS sin 0t OBS cos 0t 56p 2 a2AS cos 0t UZBS sin 0t Note that we are using the rectangular form but we could use one of the other forms of the solution D J Inman 2049 Mechanical Engineering at Virginia Tech Substitute into the equations of motion szS 2 wanS was f0 cos 0t szs Z wnwAs waS sin or 0 for all time Specifically fort 0275 a gt a a As z wnwws f0 2gwnwAs a w2BS 0 D J Inman 2149 Mechanical Engineering at Virginia Tech Write as a matrix equation a cf z wnw As f0 2 ana a w2BS 0 Solving for AS and BS 2 wzf0 S co W 24an 25601601 S 2 mi a z 24an D J Inman 2249 Mechanical Engineering at Virginia Tech Substitute the values of As and BS into xp xp t 2 2 0 2 cosat tan 1 3wn a 2gwnw J n w Add homogeneous and particular to get total solution xt fie W sinadt X cosat 6 Y Y homogeneous or transient solution particular or steady state s01ut10n Note that A and I will not have the same values as in Ch 1 for the free response Also as tgets large transient dies out D J Inman 2349 Mechanical Engineering at Virginia Tech Things to notice about damped forced response If C 0 undamped equations result Steady state solution prevails for large t Often we ignore the transient term how large is C how long is t Coefficients of transient terms constants of integration are effected by the initial conditions AND the forcing function For underdamped systems at resonance the amplitude is finite D J Inman 2449 Mechanical Engineering at Virginia Tech Example 221 on 10 rads o 5 rads C 001 F0 1000 N m 100 kg and the initial conditions x0 005 m and v0 0 Compare A and f for forced and unforced case Using the equations on slide 6 X 013362 0013 xt Ate 039 sin999t 0133 cos5t 0013 Differentiating yields x0 005 m o01Ae 0 sin9999t Asin 0133cos 0013 01t 120 0 001Asin 9999146 0039999f 39 9999A cos 0665 sin0013 0665 sin5t 0013 applying the intial conditions The numbers in are those A 05 56 obtained by inifggectly usingthe response Values D J Inman 2 5 49 Mechanical Engineering at Virginia Tech Proceeding with ignoring the transient Always check to make sure the transient is not significant For example transients are very important in earthquakes However in many machine applications transients may be ignored D J Inman 2649 Mechanical Engineering at Virginia Tech Proceeding with ignoring the transient f0 Magnitude X xa a22 250an 239 0 Frequency ratio r a Non dimensional Xk X50112 1 Form F0 f0 J1r22 240 Phase 6 tan 1 2 1 r DJInrnan 2749 Mechanical Engineering at Virginia Tech Magnitude plot 1 Eesgnfnce IS close 1 r2 2 20f For C 0 r 1 defines resonance As Q grows resonance moves r lt1 and X decreases The exact value of r can be found from differentiating the magnitude 15 2 Fig 27 05 1L D J Inman 2849 Mechanical Engineering at Virginia Tech Phase plot Resonance occurs at I 7152 The phase changes more rapidly when the damping is small From low to high values of rthe phase always changes by 180 or TC radians D J Inman 2949 35 Phase rad l L 01 TI 0 01 O 15 2 r Fig 27 Mechanical Engineering at Virginia Tech Example 223 Compute max peak by differentiating 1g 1 03 dr F0 dr 1 r222 r2 1mk 1 2 2 lt1gt lt1 241 iX ki 1 D J Inman 3049 Mechanical Engineering at Virginia Tech Effect of Damping on Peak Value The top plot shows how 25 the peak value becomes Xk 20 very large when the EwB 15 damping level is small 10 The lower plot shows 5 how the frequency at 00 02 Q4 06 08 which the peak value occurs reduces with 1 increased damping 08 Note that the peak value 06 is only defined for rpeak 04 values lt0707 02 Fig 29 D J Inman 3149 Mechanical Engineering at Virginia Tech Section 23 Alternative Representations A variety methods for solving differential equa ons So far we used the method of undetermined coefficients Now we look at 3 alternatives a geometric approach a frequency response approach a transform approach These also give us some insight and additional useful tools D J Inman 3249 Mechanical Engineering at Virginia Tech 231 Geometric Approach Position velocity and acceleration phase shifted each by 752 Therefore write each as a vector Compute X in terms of F0 via vector addition lm D C 2 m w X cwX F C cwX A B kX 5 k mw2X A ti Re kXcoswt 6 D J Inman 3349 Mechanical Engineering at Virginia Tech Using vector addition on the diagram F02 k mw22X2 ca2X2 F0 X k ma22 ca22 At resonance 72 F a X 0 2 Ca D J Inman 3 4 49 Mechanical Engineering at Virginia Tech 232 Complex response method Aejaquot 2 A003 at A sin M j 247 r f imaginary part m t 6541 kxt 2 F061 248 W harmonic input real part Real part of this complex solution corresponds to the physical solution D J Inman 3 5 49 Mechanical Engineering at Virginia Tech Choose complex exponential as a solution xp t Xe wt 249 a2m Cjw kXejwt 2 F06ij 250 F 20 HjaF0 251 k ma Ca 1 Hja 252 gk mwz 660 V the frequency response function Note These are all complex functions D J Inman 3649 Mechanical Engineering at Virginia Tech Using complex arithmetic X 102 6 253 k ma emf 9 tan1 254 k mw F xp t 0 e W 6 255 k mw22 emf Has real part to previous solution D J Inman 3749 Mechanical Engineering at Virginia Tech Comments Label xaxis Reeiw ancl yaxis lmeIW results in the graphical approach It is the real part of this complex solution that is physical The approach is useful in more complicated problems D J Inman 3849 Mechanical Engineering at Virginia Tech Example I Use the frequency response approach to compute the particular solution of an undamped system The equation of motion is written as m5c39t km 2 Foej 2 560 wjxa foej Let xp t 2 X6 gt 602 a Xejwt foeja gtX D J Inman 3949 Mechanical Engineering at Virginia Tech 233 Transfer Function Method The Laplace Transform Changes ODE into algebraic equation Solve algebraic equation then compute the inverse transform Rule and table based in many cases ls used extensively in control analysis to examine the response Related to the frequency response func on D J Inman 4049 Mechanical Engineering at Virginia Tech The Laplace Transform approach See appendix B and section 34 for details Transforms the time variable into an algebraic complex variable Transforms differential equations into an algebraic equation Related to the frequency response method XS xt j xte dt 0 D J Inman 4149 Mechanical Engineering at Virginia Tech Take the transform of the equation of motion m cjckx 2 F0 008 gt F ms2 cskXs 22 S 0 Now solve algebraic equation in s for Xs FOS XS 2 2 2 ms Csks 60 To get the time response this must be inverse transformed D J Inman 4249 Mechanical Engineering at Virginia Tech Transfer Function Method With zero initial conditions ms2 CS kXS FS gt XS 1 The transfer 2 function FS ms Csk 259 1 H 39a J k ma2 ij 260 2 frequency response function D J Inman 4349 Mechanical Engineering at Virginia Tech Example 232 Compute forced response ofthe suspension system shown using the Laplace transform Summing moments about the shaft J lttgtk19lttgt zaFo sinalt Taking the Laplace transform lszXskXs mi S Zml 1 3 Xs acoF Frame 0 32mzlszk Taking the inverse Laplace transform 1 19 FE1 0 0 32wzlszk awF 1 1 k gt 190 ismwt ismwnt a I 02 03 a a J D J Inman 4449 21 Mechanical Engineering at Virginia Tech Notes on Phase for Homogeneous and Particular Solutions Equation 237 gives the full solution for a harmonically driven underdamped SDOF oscillator to be xt fie 5 sin codt 6 X cos wt homogeneoqu solution particular solution How do we interpret these phase angles Why is one added and the other subtracted D J Inman 4 5 49 Mechanical Engineering at Virginia Tech NonZero initial conditions XOOg x0 72 0 v0 72 O 39 sin Udt 6 X a 6 tan 1 0 d 10 wnxo DJInrnan 4649 Mechanical Engineering at Virginia Tech Zero initial displacement x0 0 v0 72 O I 01 005 I39I V s1na dt 6 a 0 6itan1 0 d tan1 O D 1111111 11 V0 wnxo V0 4749 M h a1 Engineeri g tV g T h Zero initial velocity X0 x0 72 0 v0 O a K sinwdt6 xwd 11 2 6 tan 1 0 V0 70199 5 Mechanical Engineering at Virginia Tech D J i 111111 1 4849 Phase on Particular Solution xp t Xcos wt 2 a a gttan1 2 2 on a Focoscot 39 4 Xcoscot q I i 391 wt 2Cwnwi q 2 wnco l I I K I conZcoz conZco2 39 oogtcon coltcon Simple atan gives 139r2 lt I lt 112 Fourquadrant atan2 gives 0 lt I lt 1T D J Inman 4949 Mechanical Engineering at Virginia Tech 24 Base Excitation Important class of vibration analysis Preventing excitations from passing from a vibrating base through its mount into a structure Vibration isolation Vibrations in your car Satellite operation Disk drives etc D J Inman 151 Mechanical Engineering at Virginia Tech FBD of SDOF Base Excitation System Sketch System FBD kXY C Y Z Fkxyc39cym c39 mj cjc kx Cy ky 261 D J Inman 251 Mechanical Engineering at Virginia Tech SDOF Base Excitation cont Assume yt Y sinat and plug into Equation261 m cjc or ch cosat kY sinat 263 harmonic forcing functions For a car 602327 The steadystate solution is just the superposition of the two individual particular solutions system is linear fOC fOS W r J j 2 wnx wjx 2 wnwrcosmn minimal 264 D J Inman 351 Mechanical Engineering at Virginia Tech Particular Solution sine term With a sine for the forcing function 562 wnjc mix fOS sin at xps 2 AS cos aIBS sinwr 2 XS smart 13 Where Use rectangular form to make it easier to add was the cos term S can2 a22 2g ana2 w2fOs a afgt2 94an D J Inman 4 51 Mechanical Engineering at Virginia Tech Particular Solution cos term With a cosine for the forcing function we showed 562 a1n5c nix fOC cos at xpc 2 AC cos a1BC sin at 2 X6 cosat C Whae 605602 00 C a w222 ana2 25wnwf06 C a wz22 ana2 D J Inman 5 51 Mechanical Engineering at Virginia Tech Magnitude XY Now add the sin and cos terms to get the magnitude of the full particular solution X Z ficf s 22W warm 2 a 052 2420160 a 052 Zngnw Where fOC 2ngan and fOS HEY if we de ne r this becomes X Y 270 E 129V 2 2 271 Y 1 r22 2g 651 Mechanical Engineering at Virginia Tech The relative magnitude plot of XY versus frequency ratio Called the Displacement Transmissibility 4O C001 30 l f 03 O7 20 C a 1 gt 10 R 0 1O 20 05 1 15 2 25 3 Fre uenc ratior Figure 213 q y DJInman 751 Mechanical Engineering at Virginia Tech From the plot of relative Displacement Transmissibility observe that UY is called Displacement Transmissibility Ratio Potentially severe amplification at resonance Attenuation for rgt sqrt2 Isolation Zone If rlt sqrt2 transmissibility decreases with damping ratio Amplification Zone If r 1 then transmissibility increases with damping ratio Xp2Y r D J Inman 851 Mechanical Engineering at Virginia Tech Next examine the Force Transmitted to the mass as a function of the frequency ratio FT kx y C39c y m Wt D J Inman 9 5 1 Mechanical Engineering at Virginia Tech Plot of Force Transmissibility in dB versus frequency ratio foo1 COI1 fo3 FkY dB 39 o 05 1 15 2 25 3 Figure 214 Frequency ratior D J Inman Mechanical Engineering at Virginia Tech 1051 Figure 215 Comparison between force and displacement transmissibility Force TransmISSIbIIIty Log of transmissibility Displacement Transmissibility D J Inman 1151 Mechanical Engineering at Virginia Tech Example 241 Effect of speed on the amplitude of car vibration A n m mass of w lt75uspension system Velocity of cm 002 In Road surface 91 J Inman 17151 Mcchamcal Engnccnng aqugma39Rch Model the road as a sinusoidal input to base motion of the car model Approximation of road surface yt 001 msin wbt hour X2 rad 0290912 rads 0006 km 3600 3 cycle cab 20kmhr 5818 rads From the data give determine the frequency and damping ratio of the car suspension 60b 2 vkmhr 4 1 4 w 6303rads zl Hz 1007 kg 2000 Nsm c 2 20158 g 2xk m 24gtlt104 Nm1007 kg D J Inman Mechanical Engineering at Virginia Tech 1351 From the input frequency input amplitude natural frequency and damping ratio use equation 270 to compute the amplitude of theresponse r 5818 a 6303 Xzyf 12gr2 1 r22 957 2 2 001 m 1 201580923f 00319 m 1 092322 2 015809232 What happens as the car goes faster See Table 21 D J Inman 1451 Mechanical Engineering at Virginia Tech Example Compute the force transmitted to a machine through base motion at resonance From 277 at r1 A4 Excitation source Punch grew 1 machine Base motion 1 2 12 5212 gtFTE14C2 kY 2 2C 900 C 39 272 500 From given m c and k C 2 km 2 40 OOOD3OOO From measured excitation Y 0001 m FT ELEM 40 00021g 39001 1 1 40042 5016 N D It Inman 1551 Mechanical Engineering at Virginia Tech 25 Rotating Unbalance Gyros Cryocoolers Tires Washing machines Machine of total mass m ie me included in m e eccentricity m0 mass unbalance 0 DD rotation frequency a D J Inman 1651 Mechanical Engineering at Virginia Tech Rotating Unbalance cont What force is imparted on the structure Note it rotates with x component xr esm curt 2 gt ax xr ear sm art From sophomore dynamics 2 2 Rx moax m0ear Sln 6i moear Sln wrt R 2 2 y moay moear 0036 moear cos art D J Inman 1751 Mechanical Engineering at Virginia Tech Rotating Unbalance cont The problem is now just like any other SDOF system with a harmonic excitation 28ina t D mj at be moewf sin a 282 or x 20 05x 2 0 612 Sln art m Note the influences on the forCing function weare assumingthat the mass m is held in place in the ydirection as indicated in Figure 218 D J Inman 1851 Mechanical Engineering at Virginia Tech Rotating Unbalance cont Just another SDOF oscillator with a harmonic forcing function Expressed in terms of frequency ratio r xp t X sinart 283 X moe r 284 m 1 r22 2gr2 tan112 r2 j 285 D J Inman 1951 Mechanical Engineering at Virginia Tech Figure Displacement magnitude vs frequency caused by rotating unbalance mX Nomuuzm magnitude 00 05 10 15 20 25 30 35 erquency mic D I Inman 1 205 Mechanical Engmeenng at Vugmm Tech Exam 2 5 1 Given the deflection at resonance 01 m 8 005 and a 10 out of balance compute e and the amount of added mass needed to reduce the maximum amplitude to 001 m At resonance r 1 and 1 i 1 gt100391m 10gte0391m moe 2 2005 e 2 Now to compute the added mass again at resonance j 10 Use this to find Am so that XIS 001 MD 01 1 11 mAm001mj10gt mAm ZloojAngm 01m 01m mo Here m0 is 10m or 01m D J Inman 2151 Mechanical Engineering at Virginia Tech Example 2 52 Helicopter rotor unbalance Given 05 kg H9221 Nw k1xm5Nm1 cnk gkkg mtail j if 39 JOE ig ltikg mm20kg H9222 i m0 05 kg 2 001 I Compute the deflection at E 1500 rpm and find the rotor speed at which the deflection is maximum D L Inman 2251 Mechanical Engineering at Virginia Tech Example 252 Solution The rotating mass is 20 05 or 205 The stiffness is provided by the Tail section and the corresponding mass is that determined in Example 144 So the system natural frequency is I I 5 a k 10 Nm 24669 rads jmmm 1205 60 kg 3 3 The frequency of rotation is 0r 1500 rpm 1500 V mm 27 fad Inln 60 S rev 2 157 rads r M 316 4949 rads D J Inman 2351 Mechanical Engineering at Virginia Tech Now compute the deflection at r 316 and C 001 using eq 284 2 moe r 2 m Jlt1 r2gt2lt2rgt2 05 kg015 m 3162 2 0004 m 205 kg 1 31622 20013162 X At around r 1 the max deflection occurs r 1 a 4969 rads 4969 rev is 4745 rpm 3 27139 rad mm Atr1 05k 015 X g m 1 0183rn0r183crn 205 kg 2001 D J Inman 2451 Mechanical Engineering at Virginia Tech 26 Measurement Devices Casing quot39V A basic transducer used in vibration measurement is the accelerometer This device can be ZFkxycx y39mi modeled using the mxz cx ykx y Volt age base equations 286 and 261 developed in the previous section Here yt is the measured response of the structure Di 1 Inman 2551 Mechanical Engineering at Virginia Tech Base motion applied to measurement devices Let go xt yt 287 qu 3 quot i m2 020 km my cos ab 288 Ogtiv J1n 2 3 g 290 I 1 r 24 7quot Mmmd 4mm Accelerometer and quotm Voltagergenemting t9 tan l 202 J 291 quot 1 r m m These equations should be familiar mr from base motion Here they describe measurement I Strain Gauge Mounted m ibmting suucnne Elastic beam in D J Inman 2651 Mechanical Engineering at Virginia Tech Magnitude and sensitivity plots for accelerometers Accelerometer 39egion 257 Z 2 5952quot3118 9 Effect of damping on 15 e proportionality constant Relative displacemem ratio Fig 227 Frequency ratio Fig 226 Magnitude plot showing Regions of measurement Coalficicnl cry In the accel region output voltage is nearly proportional to displacement 1 quotgt4 n M 1 Frequency rarin D J Inman 2751 Mechanical Engineen39ng at Virginia Tech 27 Other forms of damping TABLE 22 DAMPING MODELS Name Damping Force seq Source Linear viscous damping er 5 Slow uid 8 X Air damping u sgiimx quot3 quot Fast fluid 1r 4l5 Coulomb damping a rim x Sliding friction wa 4dX Displacemenrsquared damping d Sgnltxjx2 37 Material damping in Solid or structural damplng b Egn xle E Linermi damping These various other forms of damping are all nonlinear They can be compared to linear damping by the method of equivalent viscous damping discussed next A numerical treatment of the exact response is given in section 29 D J Inman 2851 Mechanical Engineering at Virginia Tech The method of equivalent viscous damping consists of comparing the energy dissipated during one cycle of forced response Assume a stead state resulting from a harmonic input and compute the energy dissipated per one cycle xss X sin at The energy per cycle for a viscously damped system is 27ra dx 27ra 2 AE JFddx i cx dt dt i cx dt 299 xss Xsinatgt39caXcosatgt 27m AE c I coX coscwf2dt 7rccoX2 2101 0 D J Inman 2951 Mechanical Engineering at Virginia Tech Next compute the energy dissipated per cycle for Coulomb damping 27za AE umg j sgn39c39cdt umg 0 7r2 37r2 27 umgXI cos udu I cos udu I cos udu 4gmgX 0 7r2 37I2 Here we let u or and du 0dt and split up the integral according to the sign changes in velocity Next compare this energy to that of a Viscous system 7rc wX24 Xgtc 439u mg 2105 eq eq a 39 This yields a linear Viscous system dissipating the same amount of ener er c cle D J Inman p y 3051 Mechanical Engineering at Virginia Tech Using the equivalent viscous damping calculations each of the systems in Table 22 can be approximated by a linear viscous system In particular ceo can be used to derive amplitude expressions However as indicated in Section 28 and 29 the response can be simulated numerically to provide more accurate magnitude and response information D J Inman 3151 Mechanical Engineering at Virginia Tech Hysteresis an important concept characterizing damping FliLquotil Am A plot of displacement versus springdamping force for viscous damping yields a loop At the bottom is a stress strain plot for a system with material damping of the hysteretic type The enclosed area is f Stress equal to the energy lost Landing per CyCe Strain Unloading D J Inman 3251 Mechanical Engineering at Virginia Tech The measured area yields the energy dissipated For some materials called hysteretic this IS AE Ek Xz 2120 Here the constant 3 a measured quantity is called the hysteretic damping constant k is the stiffness and Xis the amplitude Comparing this to the viscous energy yields k3 leg Z D J Inman 3351 Mechanical Engineering at Virginia Tech Hysteresis gives rise to the concept of complex stiffness Substitution of the equivalent damping coefficient and using the complex exponential to describe a harmonic input yields k mx 6x wix 2 F061 0 Assuming xt 2 X6 and x0 choej t yields m t k1 j xt Foe complex stiffness D J Inman 3 4 51 Mechanical Engineering at Virginia Tech 28 Numerical Simulation and Design Four things we can do computationally to help solve understand and design vibration problems subject to harmonic excitation Symbolic manipulation Plotting of the time response Solution and plotting of the time response Plotting magnitude and phase D J Inman 3 5 51 Mechanical Engineering at Virginia Tech Symbolic Manipulation Let 2 2 2 a A aquot a 560quot 2 and xf0 2 wna a a What is A A1x This can be solved using Matlab Mathcad or Mathematica D J Inman 3651 Mechanical Engineering at Virginia Tech Symbolic Manipulation Solve equations 234 using Mathcad symbolics Enter this gt D J Inman 3751 Choose evaluate under symbolics to get this 39 2 2 1 nu0 ZQccnn f0 2 2 O 2 ccn0 can 03 2 2 can0 I 4 2 2 4 2 2 2 cm 2ccn0 0 4Qccn03 0 fO 2C0Jn 4 2 2 4 2 2 2 can 2ccn03 0 4Qccn03 Mechanical Engineering at Virginia Tech In MATLAB Command Window gtgt syms z wn w f0 gtgt AwnA2 wA2 2zwnw 2zwnw wnAZ wAZ gtgt xf0 0 gtgt AninvAx An wnAZ wA2wnA4 2wnA2wA2wA44zA2wnA2wA2f0 2zwnwwnA4 2wnA2wA2wA44zA2wnA2wA2f0 gtgt prettyCAn 4 2 2 4 2 2 2 wn 2 wn w w 4 z wn w z wn w f0 D J Inman 3851 Mechanical Engineering at Virginia Tech Magnitude plots Base Excitation figure1 mfile to plot base excitation to mass vibration rlinspace03500 ze00100501020050 Xsqrt 2zer 21 onessizeze1rr 2 2zer 2 plotr20log10X The values of C can then be chosen directly off of the plot For Example If the TR needs to be less than 2 or 6dB and ris close to 1 then C must be more than 02 probably about 03 D J Inman 3951 2001 Design 2005 value 01 02 0 05 1 15 2 25 3 Frequency ratio r Mechanical Engineering at Virginia Tech Force Magnitude plots Base Excitation mfile to plot base excitation to mass vibration rlinspace03500 ze00100501020050 Xsqrt 2zer 21 onessizeze1rr 2 2zer 2 FXoneslengthze1r 2 figure1 plotr20log10F m w 20 amvwm 8 0 05 1 15 2 25 3 Frequency ratio r D J Inman Mechanical Engineering at Virginia Tech 4051 Numerical Simulation We can put the forced case m5c39t 65cm kxt 2 F0 cos wt 5ampt 2 wnjct wjxa f0 cos a Into a state space form 33961 2 x2 x2 2 0an 03x1 f0 cos 0t Xt AXt ft ft 2 f0 028 wt D J Inman 4151 Mechanical Engineering at Virginia Tech Numerical Integration Euler Xl l1 Xtl AXUZ At ftlAt Using the ODE45 function gtgtTSPAN0V gtgtY000 gtgtty ode4539numfor39TSPANYO gtgtpottv1 Including forcing function XdotnumfortX m100k1000c25 zec2sqrtkm wnsqrtkm w25F1000fFm f0 fcoswt A0 1wnwn 2zewn XdotAXf D J Inman 4251 Displacement m Zero initial conditions 5 4 3 2 0 2 4 6 8 10 Time sec Mechanical Engineering at Virginia Tech Example 282 Design clamping for an electronics model 100 kg mass subject to 150cos5t N Stiffness k500 Nm 0 10kgs Usually x0001 m v0 05 ms Find a new 0 such that the max transient value is 02 m D J Inman 4351 Mechanical Engineering at Virginia Tech Response of the board is transient exceeds design specification value E 7 Oiiii i iiiiiiiiiuiiiiiiiiuiiii W V iiiii ii i Wquot i i Wquot O4 O 1 O 20 3O 40 Time sec D J Inman 44 51 Mechanical Engineering at Virginia Tech To run this use the following file function XdotnumfortX m100k500c10 Create function Ze 2Sqrtkm d If wnsqrtkm to quot390 9 quotquotquot9 w5F150fFm f0 fcoswt A0 1wnwn 2zewn XdotAXf gtgtTSPAN0 40 gtgt Y000105 Mat39ab gtgtty ode4539numfor39TSPANY0 command gtgt plotty1 window gtgt xlabel39Time sec39 gtgt ylabel39Displacement m39 gtgt grid Rerun this code increasing 0 each time until a response that satisfies the design limits results D J Inman 45 51 Mechanical Engineering at Virginia Tech Solution code it plot it and Change c until the desired response bound is obtained 03 02 Megets amplitude limit when c195kgls Displacement m 0 29 Nonlinear Response Properties More than one equilibrium Steady state depends on initial conditions Period depends on LG and amplitude Sub and super harmonic resonance No superposition Harmonic input resulting in nonperiodic motion Jumps appear in response amplitude D J Inman 47 51 Mechanical Engineering at Virginia Tech Computing the forced response of a nonlinear system A nonlinear system has a equation of motion given b V xt fx x f0 cos at Put this expression into statespace form 23961 t 262 0 x20 2 fx1x2 f0 cos at In vector form X0 FX f I D J Inman 4851 Mechanical Engineering at Virginia Tech Numerical form Vector of nonlinear dynamics Input force vector F x20 f 0 fx1x2 m focoswt Euler equation is X0141 X01 FXti At ftiAt DJInrnan 49 51 Mechanical Engineering at V1rg1n1a Tech Cubic nonlinear spring 291 2 2 wnX wix 6X3 focos wt Displacement m Nonlinearity included Linear system 2 o 2 4 6 8 10 Time sec 0 Superharmonic resonance 0 DJInman 5051 Mechanical Engineering at Virginia Tech Cubic nonlinear spring near resonance 56 2 wn5c agar 6 2 f0 cosaJt Displacement m Nonlinearity included Linear 39 o 2 4 6 8 10 Time sec 0 Response near linear resonance a DJInman 10 5151 Mechanical Engineering at Virginia Tech 35 Random Vibrations So far our excitations have been harmonic periodic or at least known in advance These are examples of deterministic excitations ie known in advance for all time That is given twe can predict the value of F t exactly Responses are deterministic as well Many physical excitations are nondeterministic or random ie can t write explicit time descriptions Rockets Earthquakes Aerodynamic forces Rough roads and seas The responses xt are also nondeterministic D J Inman 141 Mechanical Engineering at Virginia Tech Random Vibrations Stationary signals are those whose statistical properties do not vary over time Functions are described in terms of probabilities Mean values Standard deviations Random outputs related to random input via system transfer function ie given twe do not know Xt exactly but rather we only know statistical properties of the response such as the average value D J Inman 241 Mechanical Engineering at Virginia Tech Autocorrelation function and power spectral density The autocorrelation function describes how a signal is changing in time or how correlated the signal is at two different points in time 1 T 12mm 1010 l0 xt xt 0dr The Power Spectral Density describes the power in a signal as a function of frequency and is the Fourier transform of the autocorrelation function 1 oo S a R 56 de mlt gt ML mm FFT is a new way for calculating Fourier Transforms D J Inman 341 Mechanical Engineering at Virginia Tech HARMONIC T n T A22 E a V A22 1T D J Inman 441 Examples of signals Signal gt time quotc time 7 shift Power Spectral Density lt gt Frequency Hz Mechanical Engineering at Virginia Tech A rms X0 A2 Rxx7 Sxxw rms Autocorrelation gt time quotc time 7 shift Frequency Hz More Definitions Average 1 T x 1im jxrdr 347 Taco T O Meansquare T x2 11ml x2tdt 348 T gtoo T O rms 1 2 x S x 2 11m T gt T j x2tdt 349 DJInman 5 41 Mechanical En glneerlng at Vlrglnla Tech Expected Value or ensemble average The expected value EixW 1130ng 76 363 The Probability Density Function px is the probability that x lies in a given interval eg Gaussian Distribution The expected value is also given by Ex j xpxdx 364 00 D J Inman 641 Mechanical Engineering at Virginia Tech Recall the Basic Relationships for Transforms Recall for SDOF transfer function Gs 2 ms cs k 1 k mw2caj frequency response function G j w H w 2 unit impulse response function ht gw 6 sin 0dr d 1 M mGm And the Fourier Transform of ht is H0 D J Inman 7 41 Mechanical Engineering at Virginia Tech What can you predict The response of SDOF Withft as input Deterministic Input Random Input X S GSF S smug Hco2 S 0 x0 i he ofmdr Ex2 flawr spam 0 00 In a Lab the PSD function of a random input and the output can be measured simply in one experiment 80 the FRF can be computed as their ratio by a single test instead of performing several tests at various constant frequencies Here W6 get an exam Here we get an expected time record 0f the 1th value of the output given giVen an exam record 0f the a statistical record of the in 31Inman input 841 Mechanical Engineering at Virginia Tech Example 351 PSD Calculation Consider me cjc kx F t Where the PSD of F t is constant S0 The corresponding frequency response function is Ha 259 k ma 660 2 1 2 1 1 2 2 U 2 lk ma ca k ma 660 k ma ca 1 k ma22 62602 From equation 362 the PSD of the response becomes 2 S D J Inman 9 41 Mechanical Engineering at Virginia Tech Example Mean Square Calculation Consider the system of Example 351 and compute 2 0 1 E x2 S da 0L k m0 caj 5 Lmz 0kcm kc Here the evaluation of the integral is from a tabulated value See equation 370 D J Inman 1041 Mechanical Engineering at Virginia Tech Section 36 Shock Spectrum Arbitrary forms of shock are probable earthquakes The spectrum of a given shock is a plot of the maximum response quantity x against the ratio of the forcing characteristic such as rise time to the natural period Maximum response gives maximum stress m fame m 371 D J Inman 1141 Mechanical Engineering at Virginia Tech Using the convolution equation as a tool compute the maximum value of the response Recall the impulse response function undamped system hl T 1 sinaquott z39 373 ma I xtmax 1 jtFrsinwnt rdr 374 ma 0 max Such integrals usually have to be computed numerically D J Inman 1241 Mechanical Engineering at Virginia Tech Exam ple 3 6 1 Compute the response spectrum for gradual application of a constant force F 0 Assume zero initial conditions m5c t kxt F t t1infinity means static loading W EU F20 17t F10 i F0 ZLl time shift and negative like half sine problem 0 Olttltq I30 t q T F0 21 The characteristic time of the input D J Inman 1341 Mechanical Engineering at Virginia Tech Split the solution into two parts and use the convolution integral a th39 F t sinwnt x1tgt quotlLsmwltt odr 0 0lttltt1 k 0 t1 k t1 wntl 375 For x2 apply grit t1 Sinwntt1 7271 time shift 1 t1 6 376 xrx1rx2t5 lSmwntj chpa rl k t1 wntl k t1 wntl 377 DJInman 1441 Mechanical Engineering at Virginia Tech Next find the maximum value of this response To get max response differentiate xt In the case of a harmonic input Chapter 2 we computed this by looking at the coefficient of the steady state response giving rise to the Magnitude plots of figures 27 28 213 Need to look at two cases 1 tlt t1 and 2 t 2 t1 For case 2 solve what about case 1 Its max is Xstatic i F0 dz kw 1 D J Inman 1541 Mechanical Engineering at Virginia Tech cunt1 sin wntsinwnt t1 Ogt Solve for tat xmax denoted tD c0s cont cos on t t1tt O 17 cos ontp cos ontp cos ontl sin ontp sin ontl 1 1 cos wntl gt wntp tan sin wntl sin2 wnt11 COS 0102 1 cos a t n 1 21 c0s wntl sin wntl D J Inman 1641 Mechanical Engineering at Virginia Tech From the triangle 1i21 cos contl cos cont Minus sq root taken as gives a negative magnitude Substitute into Xtp to get nondimensional X 39 max 39 Xmaxk 21 COS wntl E wntl 1St term is static 2nd is dynamic Plot versus Input characteristic time T 2 7 System period D J Inman 1741 Mechanical Engineering at Virginia Tech Response Spectrum tun 211 T olndicates how normalized max output changes as the input pulse width increz oVery much like a magnitude plot oShows very small t1 can increase the response significantly impact rathi smooth force application oThe larger the rise time the smaller the peaks oThe maximum displacement is minimized if rise time is a multiple of natural period Design by MiniMax idea Kin 1 2 Kitli 15 X Xinan F0 D J Inman 1841 UJl39l 211 Lun t1 21 cos tunt1 Mechanical Engineering at Virginia Tech Comparison between impulse and harmonic inputs Impulse Input Transient Output Max amplitude Versus normalized pulse frequency T P 7 Harmonic Input Harmonic Output Max amplitude Versus normalized driving 7 frequency 2 quot 39quot KM 739 77 51424 rEIEIEl 2 men an 1941 mtmmcai Engineering at Virginia Tech Review of The Procedure for Shock Spectrum 9990 6 Find Xt using convolution integral Compute its time derivative Set it equal to zero Find the corresponding time Evaluate the max possible value of x be careful about points where the function does not have derivativell Plot it for different input shocks D J Inman 2041 Mechanical Engineering at Virginia Tech 37 Measurement via Transfer Func ons Apply a sinusoidal input and measure the response Do this at small frequency steps The ratio of the Laplace transform of these to signals then gives and experiment transfer function of the system D J Inman 2141 Mechanical Engineering at Virginia Tech Several different signals can be measured and these are named LE 32 TRANSFER FUNCTIONS USED IN VIBRATION MEASUREMENT Response Transfer Inverse Transfer Measuremem Funcrmn Funcnon Accelerm39mn Accelerance Apparent rnass 39 39 39 Impedance Velocrty Mobrlny Displacement Receptance Dynamic s ffness receptance X 2 386 F s ms cs k SK 5 s mob 1 z 387 llty Fs mszcsk Z Z inertance S X zs 387 F s ms cs k e D I Inman Mechanical Engmeemg at Vngmm Tech MAI The magnitude of the compliance transfer function yields information about the systems parameters iHjwi 389 Iceman cw HjE 390 Mi m caquot K 1 s 1 m iH0i7 391 k I 1 1341 M w quotW1 i 212 m km 38 Stability Stability is defined for the solution of free response case Stable xt lt M V tgt O Asymptotically Stable Lime 0 Unstable if it is not stable or asymptotically stabe D J Inman 2441 Mechanical Engineering at Virginia Tech Recall these stability definitions for the free resoonse Stable Asymptotically Stable 1 1 0quot rw Vii WA 0 10 1 1 t t 3 4 zlttgt 2 2 rm 0 10 2 1 0 f 10 4 Divergent instability Flutter instability D J Inman 2541 Mechanical Engineering at Virginia Tech Stability for the forced response mica cjct kxt F t Bounded InputBounded Output Stable Xt bounded for ANY bounded Ft Lagrange Stable with respect to Ft If Xt is bounded for THE given Ft D J Inman 2641 Mechanical Engineering at Virginia Tech Relationship between stability of the homogeneous system and the force response If Xhomo is Asymptotically stable then the forced response is BIBO stable Bounded input bounded output If Xhomo is Stable then the forced response MAY be Lagrange Stable or Unstable D J Inman 2741 Mechanical Engineering at Virginia Tech Stability for Harmonic Excitations The solution to mic39t kxt 2 F0 cos wt is 1 mi a2 1 w2 a2 n COS wt v xt Os1nant x0 a n jcos wnt As long as x n is not equal to 0 this is Lagrange Stable if the frequencies are equal it is Unstable D J Inman 2841 Mechanical Engineering at Virginia Tech For underdamped systems m5c39t 6X0 kxt 2 F0 cos 0t xp t 2 f0 cosat tan 1 wj a22 2g conco2 n 60 Add homogeneous and particular to get total solution xt Ate W sina t X cosat 6 d j J Y Y homogeneous or transient solution particular or steady state solution Bounded InputBounded Output Stable D J Inman 2941 Mechanical Engineering at Virginia Tech Example 381 The equation of motion after a small angle approximation is given becomes mg mzz a Ht mgma 2 mez a mg t9t 0 This Will be stable if and only if The system is mus L ran e stable the coefficent of 19 1s pos1tive physic y quotas lens us the spring must be large or If k gt 2mg enough to overcome gravity D J lnman 3041 Mechanical Engineering atVtrgtma Tosh 39 Numerical Simulation of theresponse As before in Section 28 write equations of motion as state space equations The Euler integration is just Xtl1 Xtl AXtl At Ftl At D J Inman 3141 Mechanical Engineering at Virginia Tech Example 391 with delay Let the input force be a step function att0 t Ft xt F0 F030N 1 m k1000Nm C01 I to gon2316 0 D J Inman 3241 Mechanical Engineering at Virginia Tech Example 391 Analytical versus numerical 10 003 7003e 03 5l ucos3144rewe 0101ltlgtre 70 n ut clear all 1 m Analytlcal solutlon example ZVZJ k x1 6 i x m 2 13961 l 0 7 F0 k1000 quothells zeta0vl to0 X2 7 m 040 zetalezeta2 wd n sqrtrzeta 2 t010V02 0 n w Heavlsldestepfuntto3 define Heavlslde step fuhetmh for olttltlz expezetawmlteton nus x Pok 7 cos wd trto Fok sqrt theta lezetam heavulde teto plottxt hold on m Numerlcal Solutloh 0 0 t 012 txode45 f tsxo plottx r hold off n 2 4 5 a m 2 a functloh 7 mm Po 0 k 000 quothells zeta0vl to0 mk thz vx2 xmnezuetanm xmwewmz Fomstepfuhtto DIInman 3341 Mechamcal Ehgmeenhg at Vugmm Tech Matlab Code x000 ts0 12 txode4539funct39tsx0 P10ttx1 Displacement x function vfuncttx F030 k1000 wn316 z01 t02 mkwn 2 vx2 x2 2zwnx1 wn 2F0mstepfuntt0 D J Inman 3 4 41 Mechanical Engineering at Virginia Tech prob1em 319 Numerical solution of m1000 E238e9 Problem 319 A003 L2 kEAL Fa t002 F0100 global F0 k m t0 numerioal solution X000 ts0 05 txode4539f3l939tsx0 plottxl function vf3l9tx global F0 k m t0 AX2 Fl tt0stepfunt0 l tt0stepfuntt0F0m B kmxlF vA B CIJInrnan 3 5 41 Mechanical Engineering at Virginia Tech P319 x 10 4 Variations ill 00quot IIII BGIIIIII 40C E 39 2 G O L5 1 Q 2 D 39 O 1 2 0 01 02 03 04 05 39I39Imn n DJInrnan 3641 Mechanical Engineering at Virginia Tech 39 Nonlinear Response Properties EIIIEI integration formula x011 2 x01 FXti At f ti At Nonlinear IBI III Analytical solutions not available so we must interrogate the numerical solution D J Inman 3741 Mechanical Engineering at Virginia Tech Example 310 cubic spring subject to pulse input m5c39t 6X0 kxt k1x3 t 15OOCIDt t1 c130 5 The state space form is 5510 x2 I x20 2 2 anx2t 05x10 05x13t15CIt t1 13t t2 2 D J Inman 3841 Nature of Response D jsplace me nt 111 D J Inman 3941 Red solid is nonlinear response Blue dashed is linear response Is there any justification Yes hardening nonlinear spring The first part is due to IC Mechanical Engineering at Virginia Tech Matlab Code clear all xo001 l tsO 8 txode4539f39tsxo plottxl hold on The response of nonlinear system txode4539fl39tsxo plottxl39 39 hold off The response of linear system a 0 function vftx m100 k2000 020 wnsqrtkm zetac2sqrtmk Fol500 tll5 t25 vx2 X2 2zetawn x1 wnA2 xlA3alpha Fomstepfunttl stepfuntt2 o 0 function vfltx m100 k2000 020 wnsqrtkm zetac2sqrtmk Fol500 tll t25 vx2 X2 2zetawn x1 wnA2 xlA3alpha Fomstepfunttl stepfuntt2 CIJInrnan 4041 Mechanical Engineering at Virginia Tech alpha3 alpha0 What good is this ability Investigate pulse width Investigate parameter changes Investigate effect of initial conditions Design and Prediction Because there are not many closedf form solutions or magnitude expressions design must be done by numerical simulation D J Inman 4141 Mechanical Engineering at Virginia Tech 3 General forced response So far all of the driving forces have been sine or cosine excitations In this chapter we examine the response to any form of excitation such as Impulse Sums of sines and cosines Any integrable function D J Inman 147 Mechanical Engineering at Virginia Tech Linear Superposition allows us to break up complicated forces into sums of simpler forces compute the response and add to get the total solution If x1 x2 are solutions of a linear homogeneous equation then x alx1 612x2 is also a solution If x1 is the particular sol of 56 013x 2 f1 and x2 the particular sol of 2396 onzx 2 f2 3 ax1 bx2 solves 56 mix af1 19f2 D J Inman 247 Mechanical Engineering at Virginia Tech 31 Impulse Response Function Ft 1 Impulse excitation Q tlt 7 8 F 28 Ft T ltIltT8 28 O tgt 78 T 8 1 8 a is a small positive number Figure 31 D J Inman 347 Mechanical Engineering at Virginia Tech From sophomore dynamics The impulse imparted to an object is equal to the change in the objects momentum Fm iearea impulse force 2 I Fwd Fm A under 78 00 Pulse 8 I Ftdt I Ftdt N S 13 2 8 7 F A 28F 28 39 T 5 quot38 1 D J Inman 4 47 Mechanical Engineering at Virginia Tech We use the properties of impulse to define the impulse function Dirac Delta function Equal impulses Ft 7O 2 72 IFt Tdt 13 If f 1 this is the Dirac Delta 5t D J Inman 5 47 Mechanical Engineering at Virginia Tech The effect of an impulse on a springmassdamper is related to its change in momentum Just after Just before impulse impulse impulsezmomentum change X r FAI wing 12 A F FAr FmvO gtv0 2 2 mm Thus the response to impulse with zero IC is equal to the free response with IC x00 and v0 FAtm D J Inman 647 Mechanical Engineering at Virginia Tech Recall that the free response to just non zero initial conditions is The solution of mj cjckx O x0 x0 5CO 10 in underdamped case V I X xt e W Sin odt x0 cos wdt d D J Inman 7 47 Mechanical Engineering at Virginia Tech So for an underdamped system the impulse response is x0 O xt 6 mod sin wdt response to 3 36 mg l 3 sin W 38 xt lm where ht e Ef J X unit impulse response function J m 0 10 20 30 40 Time Response to an impulse at t 0 and zero initia D J 1mm conditions 847 Mechanical Engineering at Virginia Tech The response to an impulse is thus defined in terms of the impulse response function ht So the response to 5t is given by ht Cont W sin 0dr 38 m d What is the response to a unit impulse applied at a time different from zero The response to 5t z39 is ht z39 This is given on the following slide D J Inman 9 47 Mechanical Engineering at Virginia Tech 0 tltT ht 239 W T sinwdt 239 t gt 239 m d for the case that the impulse occurs at 239 note that the effects of non zero initial conditions and other forcing terms must be super imposed on this solution see Equation 39 TO A 1 A E 0 V d V v 1 For example If two 10 10 20 30 40 pulses occur at two 10 I differepttlmes then g o VAJAVIva their Impulse 1 responses will 10 1o 20 30 4o superlmpose 0 AA 5 E V 1 0 1o 20 30 40 D J Inman Time 1047 Mechanical Engineering at Virginia Tech Consider the undamped impulse response Setting 4 O in the equation 38 Response to unit impulse applied at t 239 ie 507 is 1 ht 239 sin a t 239 ma D J Inman 1147 Mechanical Engineering at Virginia Tech Example 31 1 Design a camera mount with a vibration constraint i m m F5 i 39 m It awn l 2 k32 1 n139ft lort F50 Consider example 213 of the security camera again only this time with an impulsive load o D J Inman 1247 Mechanical Engineering at Virginia Tech Using the stiffness and mass parameters of Example 213 does the system stay with in vibration limits if hit by a 1 kg bird traveling at 72 kmh The natural frequency of the camera system is k 3El9h3 a 3 m6 mi 10 3 371x10 Nm002 m0 02 m 22613radS l 3 kg0553 From equations 37 and 38 With C 0 the impulsive response is xt 2 FM sin cont 2 mbv sin cont C n man The magnitude of the response due to the impulse is thus X 2 E man DJInman 1347 Mechanical Engineering at Virginia Tech Next compute the momentum of the bird to complete the magnitude calculation mbv 21kg D 72 km D1000 inD hour hour km 3600 s 20 kg ms Next use this value in the expression for the maximum value Xmbv 20kgms 392039026m lmcwn l3ng2613radsl This max value exceeds the camera tolerance D J Inman 1447 Mechanical Engineering at Virginia Tech Example 312 two impacts zero initial conditions double hit m1kgc05kygk4Nhn F 2Ns and Ft2 t t T can 22520125 26 56 025z Sln Odt 10086 Sln1984tt gt 0 W m d x20 05046 02504 sin1984t 7 r gt zquot 0n 10086025 sin1984t Oltt lt7 2 i1 oosemt sin1984t 050460250 sin1984t 7 t gt 139 D J Inman 1547 Mechanical Engineering at Virginia Tech Example two impacts and initial conditions 5 25c4x 60 60 4 x0 1mrn 920 11nms Solve three simple problems and add the results Homogeneous solution 541 Zmds 05 cod 3 rads vO x0 w wd 9941 4 quot sin WHO cos wit 1 1 e sm tcos e cosxgt J5 we g Signal immune I I mil Note no need to redo constants of 1ntegrat1on T for 1mpulse exc1tat1on others yes sniff E TAU c k C D J lnman 1647 Mechanical Engineering at Virginia Tech Computation of the response to first impulse Treat tasx0 0andv0 10lttlt4 v 1 x1t e i 0 Sln wdt e I Sln xgt Q1 J5 0lttlt4 D J Inman 1747 Mechanical Engineering at Virginia Tech Total Response for 0lt tlt 4 x10 2 xh t x10 1 e r cos xgt 3 sin t f for OStlt4 D J Inman 1847 Mechanical Engineering at Virginia Tech Next compute the response to the second impulse x2 let4 sin t 4 tgt 4 J5 t4 J5 Here the Heaviside step function is used to turn on the response to the impulse at 1 4 seconds D J Inman 1947 Mechanical Engineering at Virginia Tech Sin t 4 Ht 4 Heaviside Step function To get the total response add the partial solutions xteti3sinx t cosxgt 44 6 J initial condition frist impulse second impulse 1Q 4quot quot Hh o V6 8 10 sin r 4Ht 4 D J Inman 2047 Mechanical Engineering at Virginia Tech 32 Response to an Arbitrary Input The response to general force Ft can be viewed as a series of impulses of magnitude Fz Az Response at time tdue to the itquot impulse zero IC Xz Fz iAthz t for tgtt Xii A A Ft i V V it Im ulses If 2 t1 the 139 time interval xt1 1 FtiAtht ti At gt0l i gtTgt t H i 1 I W i Form 0dr 312 t15t2 3t3 a LV convolution integral D J Inman 2147 Mechanical Engineering at Virginia Tech Properties of convolution integrals It is symmetric meaning Let at Tt fixed so that Tt a and dT da Also Tant aztaO xt jFTht TdTI Ft aha da to i I 1 t j Ft ahada 0 D J Inman 2247 Mechanical Engineering at Virginia Tech The convolution integral or Duhamel integral for underdamped systems is xt Le W I Knew sin cod 2 7d7 ma 0 d 1 ngg ne Wsinwdm 313 mwd The response to any integrable force can be computed With either of these forms oWhich form to use depends on which is easiest to compute D J Inman 2347 Mechanical Engineering at Virginia Tech Example 321 Step function input rm 0 Olttltt0 F0 IOSI x00 1200 Olt lt1 mifaickx Figure 36 Step function To solve apply 313 0 n W 0Ww sin Ma r 7 rdr 1 o t W Foe sin Ma r 7 rdr W n 1 Md em I e50quot slnwdrrrdr Md n D J Inman 7M7 Mechanic 31 Engmmmg at Vugxmachh D J Inman 2547 Mechanical Engineering at Virginia Tech Integrating use a table code or calculator yields the solution F0 WWcosmic toys IZIO 315 x0 5 k 141 9 a tan 1 C 316 1 C2 m 006 7 05 034 5 L i k V F 00 V U 1 1 Flg 37 w w w w w 1 w w o 2 4 a 8 10 12 14 D J Inman 39P 395 Mechanical Engineering at Virginia Tech Example undamped oscillator under IC and constant force For an undamped system Fa FO ht sin cont The homogeneous solution is t1 1 2 v0 F0 xh 2 sm cont x0 cos cant I lt t1 wit Good until the applied force acts at t1 then Xt i m t xH2 I F239ht 7d7 t1 lt t lt t2 k 0 jF7htgtKltTJFTht Td7 t1 D J Inman 2747 Mechanical Engineering at Virginia Tech Next compute the solution between t1 and t2 F01rt1lttltt2 t x192 IFO mla t1 H ma 0 sin a t TdT F0 1 cos Una IQ 2 ma D J Inman 2847 Mechanical Engineering at Virginia Tech Now compute the solution for time greater than t2 Fort gt t2 t1 0 t2 t 0 xH Wu 0617jFrhz rdrjFWz rdr 0 11 t2 F0 coswnt t2 cosant t1 2 ma F0 003 on t 239 ma on D J Inman 2947 Mechanical Engineering at Virginia Tech Total solution is superposition V 0s1n aquot x0 cos a z lt 1 n F 021 cosant t1 Z1ltZltZ2 v0 xt s1nantxO cos wnt a ma F0 v 0s1nantx0 cos wnt 2 ma m H 2160 t1 2J2 4x0 01v0 0 Check points x increases after application of F Undamped response around x O coswnU tz cosant t1 rgtr2 03 02 01 0 Displacement xt 01 0 2 4 6 8 10 D J Inman Time 8 3047 Mechanical Engineering at Virginia Tech Example 323 Static versusw dynamic load 4K E v Jim k Pym52 m l 2 0 39Eruckhemg lkedwuhdm Vlbmuan made mjc39 w39c kx 391 g 0 z lt 0 sxtmdg 17 1 e coswdt79 k 1 g1 g 0gtxtlicoswdt m This has max value of xmax 2 If twice the static load D I Inman 3147 Mechanical Engineering at V1rg1ma Tech Numerical simulation and plotting At the end of this chapter numerical simulation is used to solve the problems of this section Numerical simulation is often easier then computing these integrals It is wise to check the two approaches against each other by plotting the analytical solution and numerical solution on the same graph D J Inman 3247 Mechanical Engineering at Virginia Tech 33 Response to an Arbitrary Periodic Input 562 wnjcwjx Ft Where Ft FtT We have solutions to sine and cosine inputs What about periodic but non harmonic inputs We know that periodic functions can be represented by a series of sines and cosines Fourier Response is superposition of as many RHS terms as you think are necessary to represent the forcing function accurately Displacement xt 39 o 2 4 6 Time s Figure 311 D J Inman 3347 Mechanical Engineering at Virginia Tech Recall the Fourier Series Definition Assume Ft 120 2 an cos 52 19 sin Qnt 320 2 where Q r10 a0 K F t dt 321 twice the average an Ft cos Qnt dt 322 Oscillations around average 19 HOT Ft sin Qnt dt 323 D J Inman 3 4 47 Mechanical Engineering at Virginia Tech The terms of the Fourier series satisfy orthogonality conditions T O m 72 n sinna tsin ma tdt 324 0 T T 7 m n 0 min T cosnw tcosmw tdt 325 I0 T T m n joTcosant sin mthdt 0 326 D J Inman 3 5 47 Mechanical Engineering at Virginia Tech Fourier Series Example Ft F0 Step 1 find the ES l and determine how many terms you need tltt1 0 Ft F 0 0t t1 t1lttSt2 t2t1 D J Inman 3647 Mechanical Engineering at Virginia Tech Fourier Series Example 12 Ft 2 coefficients 08 39 10 coefficients 100 coefficients Force Ft I I I I I O 05 1 15 2 25 3 35 4 Time s D J Inman 3747 Mechanical Engineering at Virginia Tech Having obtained the FS of input The next step is to find responses to each term of the FS And then just add them up Dangerll Resonance occurs whenever a multiple of excitation frequency equals the natural frequency You may excite at 100rads and observe resonance while natural frequency is 500radsll D J Inman 3847 Mechanical Engineering at Virginia Tech Solution as a series of sines and cosines to X2 wnxwfx Ft The solution can be written as a summation xp t x0 t 0 xm t xm t Where x0 t is a solution to a0 2a2 11 a x2 wnxwjx Ogt x00 2 2 and x010 and xm t are a solutions to 55 2 56 02x 2 a cosmwrt Solutions calculated from equations of 3396 2amp0 05x 2 b sinnart motion see section Example 332 D J Inman 3947 Mechanical Engineering at Virginia Tech 34 Transform Methods An alternative to solving the previous problems similar to section 23 D J Inman 4047 Mechanical Engineering at Virginia Tech Laplace Transform Laplace transformation Fs 0 fte dt Lft 341 Laplace transforms are very useful because they change differential equations into simple algebraic equations Examples of Laplace transforms see page 216 in b k fa Frs Step function ut 13 e39at 1sa sinaf w32a2 D J Inman 4147 Mechanical Engineering at Virginia Tech Laplace Transform Example Laplace transform of a step function ut est 00 1 t 0 S Lut 0 dz Example Laplace transform of e39at Le at e ate stdZL e satdlL 0 0 sat 0 Lea 6 J 1 sa 0 sa D J Inman 4247 Mechanical Engineering at Virginia Tech Laplace Transforms of Derivatives Laplace transform of the derivative of a function d t L f gt dt Integration by parts gives LLtt ne st sf ne cit J OO e stdlL 0 dt d f0 sLft D J Inman 4347 Mechanical Engineering at Virginia Tech Laplace Transform Procedures Laplace transform of the integral of a function I 1 0 LLoftdt Lft Lomdr Steps in using the Laplace transformation to solve DE s Find differential equations Find Laplace transform of equations Rearrange equations in terms of variable of interest Convert back into time domain to find resulting response inverse transform using tables D J Inman 44 47 Mechanical Engineering at Virginia Tech Laplace Transform Shift Property Note these shift properties in tand 3 spacesquot e ftFs a f t aCIDt a easFs thus 52 1 3 52 a e D J Inman 45 47 Mechanical Engineering at Virginia Tech Example 343 compute the forced response of a spring mass system to a step input using LT The equation of motion is m t kxt CIgtt Taking the Laplace Transform zero initial conditions 1 lm sms2k ss2a 1 ms2 kXs gt Xs s Taking the inverse Laplace Transform yields x0 lm a 1 cos Cont 21 cos Cont Compare this to the solution given in 318 D J Inman 4647 Mechanical Engineering at Virginia Tech Fourier Transform From Fourier series of non Wn2 and M1 periodic functions Allow period to go to infinity Similar to Laplace Transform Useful for random inputs 5 Xa j xt e Wdt Corresponding inverse transform 39 0 1 Tim S 3 4 oo Fourier xv X 06160tdw Transform Fourier transform of the unit impulse response is the frequency response function Ha hte f dt 0 dB Normalized H D J Inman Frequency Hz 47 47 Mechamcal En gmeenng at V1rg1n1a Tech Some Review Window 42 Orthonormal Vectors similar to the unit vectors of statics and dynamics X1 and X2 are both normal if XlTX1 1 and ng2 1 and are orthogonal if X1T X2 O This is abbreviated by T Xisz zj 0 if ii j 1 if i j A set ofn vectors X l are set to be orthonormal if T Xi X1 6 for all values of i and j D J Inman Mechanical Engineering at Virginia Tech 43 Modal Analysis Physical coordinates are not always the easiest to work in Eigenvectors provide a convenient transformation to modal coordinates Modal coordinates are linear combination of physical coordinates Say we have physical coordinates xand want to transform to some other coordinates u 123514 3352 1 1 3 x1 gt u 1 3 x 2 2 2 D J Inman 253 Mechanical Engineering at Virginia Tech Review of the Eigenvalue Problem Start with M320 KX 0 where X is a vector and M and K are matrices Assume initial conditions X0 and X0 Rewrite as MMXKXOandlet ant J q MX q gt X Mq coord trans 1 D J Inrnan 353 Mechanical Engineering at Virginia Tech Eigenproplem cont Premultiply by M 7 to get M 2M2qM 2KM 2qqKq0 I K KT 455 39 Now we have a symmetric real matrix 39 Guarantees real eigenvalues and distinct mutually orthogonal eigenvectors D J Inman 453 Mechanical Engineering at Virginia Tech Eigenvectors Mode Shapes Mode shapes are solutions to M w2u Ku in physical coordinates Eigenvetors are characteristics of matrices The two are related by a simple transformation but they are not synonymous D J Inman 553 Mechanical Engineering at Virginia Tech Eigenvectors vs Mode Shapes The eigenvectors of the symmetric PD matrix K are orthonormal ie PTP I Are the mode shapes orthonormal Using the transformation X M q the modes shapes U M P gt P M U Now PTP UTMMU UTMU I Thus the mode shapes are orthogonal only wrt the mass matriX Similarly UTKU PT MIMP A D J Inman 653 Mechanical Engineering at Virginia Tech The Matrix of eigenvectors can be used to decouple the equations of motion If P orthonormal unitary PTP I gt PT 2 P 1 Thus PTK P A diagonal matrix of eigenvalues Back to 1 Iaq 0 Make the additional coordinate transformation qPr and premultiply by PT PTPi PT Pr Ii Ar 0 459 39 Now we have decoupled the EOM ie we have n independent 2ndorder systems in modal coordinates rt D J Inman 753 Mechanical Engineering at Virginia Tech Writing out equation 459 yields 5 4244 glam iiltrgtcq2r1ltrgt0 462 i3rw r2ro 463 We must also transform the initial conditions 40 40 T T 12 r0 P q0P M x0 464 130 20 i O i to 1 q 10 PTqm PTMme 465 130 r20 D J Inman 853 Mechanical Engineering at Virginia Tech This transformation takes the problem from couple equations in the physical coordinate system in to decoupled equations in the modal coordinates 1 0302 r X1 X2 k r k2 r 1 quot391 W m2 2 032 r Physical Coordinates 1 Coupled equations 1 Modal Coordinates X M 2 Pr Uncoupled equations Figure 45 D J Inman 953 Mechanical Engineering at Virginia Tech Modal Transforms to SDOF 1 The modal transformation PTM2 transforms our 2 DOF into 2 SDOF systems 0 This allows us to solve the two decoupled SDOF systems independently using the methods of chapter 1 0 Then we can recombine using the inverse transformation to obtain the solution in terms of the physical coordinates D J Inman 1053 Mechanical Engineering at Virginia Tech The free response is calculated for each mode independently using the formulas from chapter 1 7 1t Os1n 6011O cos wit 1212 0 l or see Window 43 for a reminder 392 I 0 KO 2 r2 0 s1nat tan 1l ZO l 12 l 10 2 z mi 90 Note the above assumes neither frequency is zero D J Inman 1153 Mechanical Engineering at Virginia Tech Once the solution in modal coordinates is determined r then the response in Physical Coordinates is computed With n DOFs these transformations are xt S rt W H W nxl quotXquot nXl Where 9 M P mm mm quotxquot Where n 2 in the previous slides D J Inman 1253 Mechanical Engineering at Virginia Tech Steps in solving via modal analysis Window 44 Calculate M lp Caloulate IE 2 M thKM1J 2 the mass normalized stiffness matrix Calculate the symmetric eigenvalue problem for I to get m3 and w Normalize vi and form the matrix P v1 v2 Calculate S M lRP and 3 1 PEI Mir Calculate the modal initial conditions rU S lxi390 S li39rg 7 Substitute the components of rD and tquot into equations 466 and 467 to get the solution in modal coordinate rr 8 Multiply rr by S to get the solution x0 Srr U39lBMNi i 395 Note that S is the matrix of mode shapes and P is the matrix of eigerwectora D J Inman 1353 Mechanical Engineering at Virginia Tech Example 431 via MATLAB see text for hand calculations 9 l l 3 lol 39 l M K X0 X0 0 1 3 3 1 0 0 Follow steps in Window 44 pg 300 1 Calcul M V2 2 Calculate K MK M y Minvz inVSqrtM gtgtKt MinV2KMinV2 MinV2 Kt 03333 0 3 1 0 10000 1 3 D J Inman 1453 Mechanical Engineering at Virginia Tech Example 431 solved using MATLAB as a calcuator 3 Calculate the symmetric eigenvalue problem for K tilde PD eigKt lambdalsortdiagD just sorts smallest to largest PPI reorder eigenvectors to match eigenvalues lambda 2 4 P O707l O707l O707l 07071 D J Inman 1553 Mechanical Engineering at Virginia Tech Example 431 cont 4 Calculate S MA l2 P and Sinv PAT MAl2 Minv2 P inv invS 5 Calculate the modal initial conditions r0 Sinv x0 rdotO Sinv v0 D J Inman 1653 Mechanical Engineering at Virginia Tech Example 431 cont 6 Find the free response in modal coordinates tmax 10 numt 1000 t linspaceOtmaxnumt TWmeshgridtlambdaA12 Use Tony39s trick R0 rOonesnumtl RDOTO rdotOonesnumtl r RDOTOWsinWT ROCOSWT 7 Transform back to physical space x Sr D J Inman 1753 Mechanical Engineering at Virginia Tech Example 431 cont Plot results figure subplot2ll plottrl39 39tr239 39 title39free response in modal coordinates39 xlabel39time sec39 legend39rl3939r239 subplot2l2 plottxl39 39tx239 39 title39free response in physical coordinates39 xlabel39time sec39 legend39xl3939x239 D J Inman 1853 Mechanical Engineering at Virginia Tech Modal and Physical Responses Free response in modal coordinates Modal 4 Coordinates 2 Independent oscillators 0 2 4 11 2 Z 1 5 o 1 2 3 4 5 6 7 8 9 10 2 T1 2 g 2 444 sec TC sec 444 sec Free res onse in h sical coordinates 2 4 gt 01 2 4 p P Y gt T2 nsec Physical Coordinates Coupled oscillators Note IC 0 1 2 3 4 5 6 7 8 9 10 D J Inman Time S 1953 Mechanical Engineering at Virginia Tech Section 44 More then 2 Degrees of Freedom W k km kquot BMWW H i H Fig 48 Extending previous section to any number of degrees of freedom 9 D I Inman quot1 2053 Mechanical Engineering at Virginia Tech Fig 47 A FBD of the system of figure 48 yields the n equations of motion 0 the form m kl xi xi1 ki1xi1 xi 0 i 1 2 3L n 483 Writing all n of these equations and casting them in matrix form yields Mia KXt 0 480 where D J Inman 2153 Mechanical Engineering at Virginia Tech the relevant matrices and vectors are 0 L 0 k1k2 k2 0 L 0 m 1 k2 k2k3 k3 0 0 m2 L 0 M K 0 k3 o o M 483 M o M M o kn1kn k 0 0 L m 0 0 L k k X10 5 xt 560 x0 2 x0 2 xnt 5610 DJInrnan 2253 Mechanical Engineering at Virginia Tech For such systems as figure 47 and 48 the process stays the samejust more modal equations result Process stays the same as section 43 W wfrlu 0 39rga 60590 0 2 r3t 03 r3t 0 Just get more modal 5 equations one for each degree of freedom n is the number of dof 39r39ntw rt o DJ1nman Seed example 442 for details 2353 ec anical Engineering at Virginia Tech The Mode Summation Approach 39 Based on the idea that any possible time response is just a linear combination of the eigenvectors Starting with qm Ego 0 488 let qtZ qit Z ale 1W 1931 vi i1 i1 gt two linearly independent solutions for each term can also write this as qt Zdi sin 6011 Vi 492 i1 D J Inman 2453 Mechanical Engineering at Virginia Tech Mode Summation Approach cont Find the constants d l and DI from the LC q0 Zdi sin ivi and 10 Zdlwl c0s ivi i1 i1 Assuming eigenvectors normalized such that vivi 50 V q0 V Zdisin ivl ZdZsin Divivi djsin DJ i1 i1 Similarly for the 1n1t1a1ve10c1t1es V jq0 d 160 cos DJ D J Inman 2553 Mechanical Engineering at Virginia Tech Mode Summation Approach cont Solve for all and from the two IC equations T T viqlt0gt and Q 2 ml mi Zioqm s1n Vi q0 IMPORTANT NOTE about q00 if you just crank it through the above expressions d l you might conclude that all 2 O ie the trivial soln Be careful With 10 O as well D J Inman 2653 Mechanical Engineering at Virginia Tech Mode Summation Approach for zero initial displacement If q0 O the return to q0 Zdi sin i vi i1 and realize that 0 instead of dz O The compute dz from the velocity expression qu0al dz cos D J Inrnan 2753 Mechanical Engineering at Virginia Tech Mode Summation Approach with rigid body modes 001 0 if 1 0 q1t ale jmtblej mtvi a1b1vl does not give two linearly independent solutions Now we must use the expansion qt a1b1tV1Zaiej Zt biej17itvi i2 and adjust calculation of the constants from the initial conditions accordingly Note that the underline term is a translational motion D J Inman 2853 Mechanical Engineering at Virginia Tech Example 431 solved by the mode summation method 3 O 1 1 From before we have M 1 2 and Vi 01 JE ii A 123 120 pproprlate 1C are q0M X0 0 q0M v0 0 7139 gD Ztan1wivfl0tan1aivf10gt 91 Z 2 qult0gt 0 92 l 2 di 2 vzqltogt 3 Z 3 sin Q d2 3 2 D J Inman 2953 Mechanical Engineering at Virginia Tech Example 431 constructing the summation of modes 8in t Sin 2t l HillIllllllllllllllllllllllllll llllllllll llll the rst mode the second mode Transforming back to the physical coordinates yields XtMl2qSm ji 0 1Msm 2t j 1 0M4 2 J5 0 1 1 2 2 E 0 1 1 l l 2 2 3J5 1 2 2 3J2 1 DJInrnan 3053 Mechanical Engineering at Virginia Tech Example 431 a comparison of the two solution methods shows they yield identical results D J Inman 3153 Mechanical Engineering at Virginia Tech Steps for Computing the Response By Mode Summation 1 Write the equations of motion in matrix form identify M and K 2 Calculate M quot12 or L 3 Calculate K M y2KM y2 4 Compute theeigenvalue problem for the matrix K and get and Vi 5 Transform the initial conditions to qt q0 Mx0 and 10 MX0 D J Inman 3253 Mechanical Engineering at Virginia Tech Summary of Mode Summation Continued 6 Calculate the modal expansion coefficients and phase constants T T i tan 1 0139 10 di 2 Vi 10 Vi 10 s1n 7 Assemble the time response for q 1t Zdi Sinwit ivi i1 8 Transform the solution to physical Coordmates xt M qun Zdi sinalt iul D J Inman i1 3353 Mechanical Engineering at Virginia Tech Nodes of a Mode Shape 0 Examination of the mode shapes in Example 443 shows that the third entry of the second mode shape is zero 0 Zero elements in a mode shape are called nodes 39 A node of a mode means there is no motion of the mass or coordinate corresponding to that entry at the frequency associated with that mode D J Inman 3453 Mechanical Engineering at Virginia Tech The second mode shape of Example 443 has a node 39 Note that for more then 2 DOF a mode shape may have a zero valued entry 0 This is called a node of a mode node They make great mounting points in machines D J Inman 3553 Mechanical Engineering at Virginia Tech A rigid body mode is the mode associated with a zero frequency h h quot71g quot 25 i 1 i Jr2 i k i T l N Note that the system in Fig 412 is not constrained and can move as a rigid body Physically if this system is displaced we would expect it to move off the page whilst the two masses oscillate back and forth o D J Inman 3653 Mechanical Engineering at Virginia Tech Example 444 Rigid body motion The free body diagram of figure 412 yields m11 kx2 x1 and mzi 2 kx2 x1 m1 O 561 1 1 x1 0 gt k 0 m2 x2 1 1 x2 0 Solve for the free response given m1 1 kg m2 4 kg k 400 N subject to 001 X0 m and V0 2 O O DJInman 3753 Mechanical Engineering at Virginia Tech Following the steps of Window 44 1 0 1 MM 1 2 m m 1 0 1 1 1 O 400 200 2KM KM 4oo 1 1 05 1105 200 100 3 detI U100det4l 2 D10012 510 2 1 2 2110 and 125 010 0222236rads D I Inman Indicates a rigid body motion 3853 Mechanical Engineering at Virginia Tech Now calculate the eigenvectors and note in particular that they cannot be zero even if the eigenvalue is zero 4 0 2 v11 0 0gt100 gt4v11 2v210 2 1 0 v21 0 04472 08944 08944 04472 08944 gt P 1 gt v1 2 or after normalizing v1 L39k 39 1 CW1SC V2 04472 08944 04472 As a Check note that PTP I and PTIEP diag 0 5 D J Inman 3953 Mechanical Engineering at Virginia Tech 5 Calculate the matrix of mode shapes S M12P1 0 04472 0894404472 08944 0 12 08944 04472 04472 02236 04472 17889 08944 08944 7 Calculate the modal initial conditions 04472 17889001 0004472 1 08944 08944 0 0008944 rm 5 520 0 I390 S1XO D J Inman 4053 Mechanical Engineering at Virginia Tech 7 Now compute the solution in modal coordinates and note what happens to the first mode Since 001 O the first modal equation is i 01i 0 gt r10 a 9t Rigid body translation And the second modal equation is t 5r2t 0 gt r2 t a2 cos xg t Oscillation D J Inman 4153 Mechanical Engineering at Virginia Tech Applying the modal initial conditions to these two solution forms yields r10 a 0004472 ii0 b 00 gt r10 00042 as in the past problems the initial conditions for r2 yield r20 00089 c0s J3 00042 gt rt 00089 cosJEz D J Inman 4253 Mechanical Engineering at Virginia Tech 8 Transform the modal solution to the physical coordinate system 04472 O8944 00045 04472 02236 00089cosJ t x t 2012 760 cosJEr gt x0 1 x10 3 m X20 2012 1990 cosJEr x0 Srt Each mass is moved a constant distance and then oscillates at a single frequency D J Inman 4353 Mechanical Engineering at Virginia Tech Order the frequencies It is convention to call the lowest frequency 031 so thatmlnggw3lt Order the modes or eigenvectors accordingly It really does not make a difference in computing the time response However When we measuring frequencies they appear lowest to highest Physically the frequencies respond with the highest energy in the lowest mode important in utter calculations run up in rotating machines etc D J Inman 4453 Mechanical Engineering at Virginia Tech The system of Example 415 solved by Mode Summation From Example 416 we have 1 1 wlz ul a wzzzau2 1 1 Use the following initial conditions and note that only one mode should be excited Why x0 gm 8 D J Inman 4553 Mechanical Engineering at Virginia Tech Transform coordinates 9 O y 3 O y 13 0 M gtM2 andM 2 O 1 O 1 0 1 Thus the initial conditions become 10 M x0gt 3 0 1 1 1 1 3 o o o liioHoi D J Inman 4653 Mechanica Transform Mode Shapes to Eigenvectors l i 3 oljgt eigenvectors V2Mu2o 1 1 1 Note that unlike the mode shapes the eigenvectors are orthogonal 1 1 Note that va2 1 1 0 but ufu2 1 g 7 0 l V1231 and V223 1 DJInman 4753 Mechanical Engineering at Virginia Tech Normalizing yields 1 l 1 1 From Equation 492 2 10 sinalt iVl39 2 2diwl39 COS0Z39Z iVl39 i1 i1 Set tO and multiply by V12 2 10 Zdlwi cos am i1 0 1 1 gt VO dlxECOSQVITL6Z22003E1 gt O dlcos 1gt 17r2 Or directly from Eq 497 D J Inman 4853 Mechanical Engineering at Virginia Tech From the initial displacement qult0gt i 1 1 d1 sum2 1 1EH J5 v qlt0gt 1 d2 sum2 1 1 H0 thus 498 2 Eigenvector 1 12 2 2 dz sinalt Q Vl Eigenvector 2 i1 2 V5 cosh30 COS t D J Inman 4953 Mechanical Engineering at Virginia Tech Transforming Back to Physical Coordinates 1 1 Xt M Aqg Ocos t 0 1 1 5 cos t COS t 1 3 Xll COS39J t and x2tCOS t So the initial conditions generated motion only in the first mode as expected D J Inman 5053 Mechanical Engineering at Virginia Tech Alternate Path to Symmetric SingleMatrix Eigenproblem 39 Square root of matrix conceptually easy but computationally expensive l l l l M 2M2qM 2KM 2qqKq0 0 More efficient to decompose M into product of upper and lower triangular matrices Cholesky decomposition D J Inman 5153 Mechanical Engineering at Virginia Tech Cholesky Decomposition Let M U TU where U is upper triangular Introduce the coordinate transformation UXqgtXU 1q gtUTUXK x 0 AF w 391 U q premultiply by U T to get IijUTKU1q 261ng 0 note that U TK UT 2 U 1T KT UT 2 U TKU 1 D J Inman 5253 Mechanical Engineering at Virginia Tech Cholesky cont 0 Is this really faster Let s ask MATLAB MzM39iM39i MUTU gtgtM9001 gtgtM90 01 gtgt oi si 0 i sqrtmM ops gtgt oi si 0 cholM ops sqrtm requires a singular value decomposition SVD whereas Cholesky requires only simple operations Note that M U for diagonal M D J Inman 5353 Mechanical Engineering at Virginia Tech Chapter 5 Design Acceptable Vibration levels ISO Vibration isolation Vibration absorbers Effects of damping in absorbers Optimization Viscoelastic damping treatments Critical Speeds Design for Vibration suppression D J Inman 151 Mechanical Engineering at Virginia Tech 51 Acceptable levels of vibration 39 Each part or system in a dynamic setting is required to pass vibration muster 0 Military and ISO provide a regulation and standards 0 Individual companies provide their own standards 39 Usually stated in terms of amplitude frequency and duration of test D J Inman 251 Mechanical Engineering at Virginia Tech Example 512 Dissimilar devices with the same frequency m1 kg m1000 kg k400 Nrn k400000 Nrn c8 Nsrn c8000 Nsm car CD drive a 400 000 20 rads a 20 rads 1000 1 g 8 202 2100020 2120 wd 2 2041 022 2 195959 rads wd 20J1 022 195959 rads D J Inman 351 Mechanical Engineering at Virginia Tech But response magnitudes different Magnitude w 03 mo Magmtude plot W111 have m Tm39mbie 7 20 a the same shape 7120 3 0 C i160 r39 20 01 2 46 l 2 4610 00 02 04 06 08 10 the same form for similar W20 Phase but scaled dlsturbances deg hm BUT WITH DIFFERENT MAGNITUDES 20 Tumble X 0 720 I 00 02 04 06 08 10 different Flg 53 Xk 1 if Ja n2 2502 v same D J Inman 4 51 Mechanical Engineering at Virginia Tech Section 52 Isolation A major job of vibration engineers is to isolate systems from vibration disturbances or visa versa Uses heavily material from Sections 24 on Base Excitation lmportant class of vibration analysis Preventing excitations from passing from a vibrating base through its mount into a structure Vibration isolation Shocks on your car Satellite launch and operation Disk drives D J Inman 5 51 Mechanical Engineering at Virginia Tech Recall from Section 24 that the FBD of SDOF for base excitation is kXY C Y ZFkxyCjcym m cjckxcyky 1 D J Inman 651 Mechanical Engineering at Virginia Tech SDOF Base Excitation assumes the input motion at the base has the form yt Y sinat and plug into Equation1 mi cjc kx chcosaI kY sinatJ 2 harmonic forcing functions The steadystate solution is just the superposition of the two individual particular solutions fog 3919 j 2 wn3 c olfx 2 wancoswt ijsinMI 3 Er J r J xpcltrgt xps r D J Inman 7 51 Mechanical Engineering at Virginia Tech Particular Solutions sine input With a sine for the forcing function 55 25wa wjx fOS sin at xps 2 AS cos a1BS sin at 2 XS sinat S where 2wnwf05 S a wz2 2g a2nw2 w2fOs S a w22 2g a2nw2 851 Mechanical Engineering at Virginia Tech Particular Solutions cosine input With a cosine for the forcing function we showed X2QQXQh kamat xpc 2 AC cos aIBC sin at 2 X6 cosat C Mae w2fOc S a a22 2g ana2 ZCwnw c S a a222 ana2 D J Inman 9 51 Mechanical Engineering at Virginia Tech Magnitude XY Magnitude of the full particular solution igJ j fjQ u Q 1 123mm WYJ lt2 wgt2wi wi w2gt22 wnw2 wi w2gt22 wnw2 Where fOC 2420an and fOS HEY if we de ne r this becomes X p Y QQ 1 a fng D J Inman 1051 Mechanical Engineering at Virginia Tech The magnitude plot of XY C001 30 l f 03 O7 20 C a 9 gt 10 R 0 1O 20 O 05 1 15 2 25 3 Frequency ratio r DJInrnan 1151 Mechanical Engineering at Virginia Tech Notes on Displacement Transmissibility Potentially severe amplification at resonance Attenuation only for rgt sqrt2 If rlt sqrt2 transmissibility decreases with damping ratio If Igtgt1 then transmissibility increases with damping ratio Xp2Yr D J Inman 1251 Mechanical Engineering at Virginia Tech It is also important to look at the Force Transmissibility FT 2 kx yc39c y m39c39 we know that xt X cosat so 562 sz cosat IFTI 2171sz 21cer Xquot Wt D J Inman 1351 Mechanical Engineering at Virginia Tech Plot of Force Transmissibility 4o foo1 COI1 CO3 C07 30 2O 0 05 1 15 2 25 3 Frequency ratio r D J Inman 1451 Mechanical Engineering at Virginia Tech Isolation is a sdof concept 0 Two types moving base and fixed base 0 Three magnitude plots to consider TR transmissibility ratio 2 E Moving base displacement Y 1 r 251quot 2 i 2 4 Moving base force kY 1 r22 25102 2 1 25 Fixed base force FD lt1 r2gt2 lt2 rgt2 D J Inman 1551 Mechanical Engineering at Virginia Tech For displacement transmissibility isolation occurs as a function of stiffness 39 For stiffness such that the frequency ration is larger the root 2 isolation occurs but increasing damping reduces the effect For less then root 2 increased damping reduces the magnitude m of 411 g 05 10 12 4 16 18 10 2 24 V7 D J Inman 1651 Mechanical Engineering at Virginia Tech Example 521 Design an isolation mount Fig 56 Module 391 Carbody s e Cm ecu om l E c eomml modu e bl x L m A 7 annz 7 M yin Design an isolator chose k c to hold a 3 kg electronics module to less then 0005 m deflection if the base is moving at yt001sin35t Calculate the force transmitted through the isolator Isolation mounting o D I Inman 1751 Mechanical Engineering at Virginia Tech TR Plot for moving base displacement For TR 05 1 Q Q r 001 173 E gig RR 005 174 05 o1 01 176 232 KE 0 2 184 C12 05 12 441 0O 05 1 15 2 25 3 Frequency ratio r D J Inman 1851 Mechanical Engineering at Virginia Tech From the plot note that X max Response 0005 TR 2 0 Y max input 001 39 from the plot r 2173 2002 works gt r 173gtE gtk1228Nm a 3 173 n c 2 25mm 2 2002320232 2428 kgs D J Inman 1951 Mechanical Engineering at Virginia Tech Rattle Space 81 5E g2 9 2 024m k a 20232 X ltO5 Y Choice of kand 0 must also be reasonable As must force transmitted D J Inman 2051 Mechanical Engineering at Virginia Tech The transmitted force is FT kYr2 1 my kYr2TR 1 7 22 25702 1228O011732O5 18376 N Transmitted force TR static deflection damping and stiffness values must all be reasonable for the application D J Inman 2151 Mechanical Engineering at Virginia Tech Shock Isolation 7t Ysmcopt 0St t1 60 Shock pulse yt p 58 7239 0 t gt t1 Pulse duration 0 P Plot max 560 of X on contl H H versus 3 max yt I yt 0p 7 Increased isolation with increasing k D J Inman 2251 Mechanical Engineering at Virginia Tech Figure 58 Shock Response mum E Wm Lupm 111 I I shock pulse tunmp owl7r D J Inman 1351 Mechanic 31 Engmmmg alerglmachh Shock versus Vibration Isolation In figure 58 for 3 05 requires 2 602 m7 lt10gtklt 2 2L1 for shock isolation to occur Thus shock isolation requires a bound on the stiffness 0 Also from Figure 5 high damping is desirable for shock attenuation D J Inman 2 4 51 Mechanical Engineering at Virginia Tech Example 523 Design a system that is good for both shock and vibration isolation 0 The design constraints are that we have the choice of 3 off the shelve isolation mounts 5 Hz 6 Hz and 7 Hz each with 8 damping The shock input is a 15 g half sine at 40 ms The Vibration source is a sine at 15 Hz The response should be limited to 15 g s and 762 mm and 20 dB of Vibration isolation D J Inman 2 5 51 Mechanical Engineering at Virginia Tech Simulation of the response to the shock input for all three mounts choices From these numerical simulations only the 7 Hz mount satisfies all 7 of the shock isolation 3 goals Less then 15 g s Less then 3 in de ections l m S 30 m D J Inman 2651 Mechanical Engineenng at Vuginia teen Time 57 Now consider the vibration isolation by plotting shock isolator design s transmissibility Force Tmmnussibllm 0 For the 7 Hz shock isolator design the reduction in Transmissibility is only 94 dB 0 From this plot and recalling Fig 57 less damping is requiredl Mum 0 However less damping is not Fig 513 possible D l lnman 2751 Mechanical Engineering at Virginia Tech 53 Vibration Absorbers 0 Consider a harmonic disturbance to a single degreeof freedom system 0 Suppose the disturbance causes large amplitude Vibration of the mass in steady state 0 A Vibration absorber is a second spring mass system added to this primary mass designed to absorb the input disturbance by shifting the motion to the new added mass called the absorber mass D J Inman 2851 Mechanical Engineering at Virginia Tech Absorber concept Primary mass optical tablB m l absorber k 2 k2 Primary system experiences resonance 0 Add absorber system as indicated 0 Look at equations of motion now 2 doi D J Inman 2951 Mechanical Engineering at Virginia Tech The equations of motion become m 0 56 kka ka x Fosinwt 0 ma ka ka 3601 0 To solve assume a harmonic displacement xt X sin 0t xat Xa sinwt kka ma2 ka X F0 ka ka maa2 Xa 0 D J Inman 3051 Mechanical Engineering at Virginia Tech assumed solution The form of the response magnitude suggests a design condition allowing the motion of the primary mass to become zero X kka ma2 ka 4110 Xa ka ka maaf 0 ka maw2gtFo k ka mw2ka maaf k kaFo a k ka meXka mawz k pick ma and ka to make zero All the system motion goes into the absorber motion D J Inman 3151 Mechanical Engineering at Virginia Tech Choose the absorber mass and stiffness from This causes the primary mass to be fixed and the absorber mass to oscillate at F xa t k O s1n wt 2 a force magnltude of a kl ak a ka xa PE As in the case of the isolator static de ection rattle space and force magnitudes need to be checked in each design D J Inman 3251 Mechanical Engineering at Virginia Tech Other pitfalls in absorber design Depends on knowing x exactly Single frequency device If D shifts it could end up exciting a system natural frequency resonance 0 Damping which always exists to some degree spoils the absorption let s examine these D J Inman 3351 Mechanical Engineering at Virginia Tech Avoiding resonance robustness u 2 ma Mass ratio m a 2 5 Original natural frequency of primary system before p m absorber is attached 0 k Natural frequency of absorber before it is attached ma to primary mass kg 0 2 gt g 2 I 02 Z 16 Stiffness ratio P I 6 a frequency ratio wP D J Inman 3 4 51 Mechanical Engineering at Virginia Tech Define a dimensionless amplitude of the primary mass r E r E aa pap Xk 1 102 F0 1 2r 1I 2 D J Inman 3 5 51 Mechanical Engineering at Virginia Tech Normalized Magnitude of Primary Robustness to driving frequency shifts 0 If 0 hits 031 or 032 resonance occurs 0 Using IXkF0ltl defines useful operating range of absorber 0 In this range some absorption still occurs 0 The characteristic equation is wn21621u 1 2 2 2 162i2 162641 2611 2 Frequency dependence on mass and frequency ratio D J Inman 3751 Mechanical Engineering at Virginia Tech Mass ratio versus frequency 0 Referring to fig 516 as u increases frequencies split farther apart for fixed B 0 thus if u is too small system will not tolerate much uctuation in driving frequency indicating a poor design 0 Rule of thumb 005lt u lt025 D J Inman 3851 Mechanical Engineering at Virginia Tech 01 I in mi Normalized Magnitude of the primary mass with and without the absorber u01 and 3071 Adding absorber 60 i 39 l I No vrbratlon absorber Increase the number H 40 h Wlth vrbratlon absorber of resonances or modes from one to E 1quot two 5 20 CD I I Smaller response of g 39 I I primary structure is quota 0 x I K E I at absorber natural lt I AM frequency 20 Effective over 4O limited bandwrdth O 05 1 15 Frequency ratio ra D J Inman 3951 Mechanical Engineering at Virginia Tech What happens to the mass of the vibration absorber 201 and 3071 deflection limitsll In the operational 60 range of the vibration absorber A 40 39 the absorber mass A has relatively large 5 20 motion 8 k 3 Beware of quot71 0 E lt 2o 4o O 05 1 15 Frequency ratio ra D J Inman 4051 Mechanical Engineering at Virginia Tech Example 531 design an absorber given F0 13 N m7316 kg k2600 Nm 0 180 cpm xalt 002 m From 520 ka 2 mam2 gt X 0 gt a 5 13 N 26500 Nm Xa 0002 m ka 2 mm2 gt ma 5 6500 M 1829 kg wz 2 M 60 1829 025 7316 check u D J Inman 4151 Mechanical Engineering at Virginia Tech Example 532 Compute the bandwidth of the absorber design in 531 Xk F0 1 121112115111211132 gt a 1 aty025gt 11180 ma 0 F01 X k 2 1 a 2 a 2 gt 3roots F0 1829 7316 0 03929 11382 1gt a D J Inman 4251 Mechanical Engineering at Virginia Tech Comparing these 3 roots to the plot yields that 03929wa lt a lt11180wa gt 74089 lt a lt 210821 rads This is the range that the driving frequency can safely lie in and the absorber will still reduce the Vibration of the primary mass D J Inman 4351 Mechanical Engineering at Virginia Tech 54 Damped absorber system m 0 xv ca ca x0 0 ma xaa ca ca xaa kka kaxt 8111601 ka ka xat 0 gg k mw2ka maaf maka a g m 14sz cjw2 Undamped primary 51m u Cannot be zero D J Inman 4 4 51 Mechanical Engineering at Virginia Tech Magnitude of primary mass for 3 levels of damping 10 39 As damping increases the absorber fails but 39 0 01 6 quot the resonance goes away 3ltkco01 ka 04 4 2 1 Region of absorption Go to mathcad example 516 D J Inman 4 5 51 Mechanical Engineering at Virginia Tech Effect of damping on performance In the operational range of the vibration absorber WG25 and l308 decreases with damping 39The bandwidth increases with damping No VA ca001 ca01 1 Resonances are decreased ie could be used to reduce resonance problems during run up See Fig 519 Amplitude XkF Frequency ratio ra D J Inman 4651 Mechanical Engineering at Virginia Tech Three parameters effect making the amplitude small 0 This curves show that ltfl 0 25 1A0 mm 04 just increasing the s damping does not w n result in the smallest quot quotI amplitude I 7 j m The mass ratio and xi 3 also matter O a This brings us to the question of optimization flit gal T 5 S 4027 5 D J I 004 06 02 10 12 714 16 39 nman 4751 Mechanical Engineering at Virginia Tech With damping in the absorber Undamped absorber has poor bandwidth Small damping extends bandwidth But ruins complete absorption of motion Becomes a design problem to pick the most favorable u D J Inman 4851 Mechanical Engineering at Virginia Tech Viscous Vibration Absorber Rotating machine ERrimary yst mi EYisqql1sabs9rhqr applications Xti i R l X otatlonal 1nert1a k g a shaft stiffness and a uid damper lea Often called a m g i ma i Houdaille damper quotquotquotquotquotquotquotquotquotquotquotquotquotquotquotquotquotquotquotquotquotquot quot illustrated in the Fig 522 following DJInman 49 51 Mechanical Engineering at Virginia Tech Houdaille Damper Equation of motion J1 0 Q ca 0 J 2 92 ca D J Inman 5051 Fig 523 Sigma Mechanical Engineering at Virginia Tech Frequency response of primary mass 100 10 Xr2501 I 3ltr2501 I 1 25 04 Figure 524 01 2 Desngn on n and C I 4A r2 Xru 1J4 Cz r2 u rz 12r2 12 r2 D J Inman 5151 Mechanical Engineering at Virginia Tech Chapter 6 Distributed Parameter Systems Extending the first 5 chapters particularly Chapter 4 to systems with distributed mass and stiffness properties St ngsa IOdS and beams Tacoma Narrows Bridge D J lnrnan 163 Mechanical Engineering at Virginia Tech The stringcable equation 0 Start by considering a uniform string stretched between two x t fixed boundaries T T T T T T T T T T 0 Assume constant aXial tension 5 in string 41 0 Let a distributed force fxt act along the L x string D J Inman 263 Mechanical Engineering at Virginia Tech Examine a small element of the string Tl g2 W xt ZFy PAXT TI 2 z391sinl91z392 sinl92 fxtAx x1 xQx1Ax 0 Force balance on an infinitesimal element 0 NOW linearize the sine with the small angle approximate sinx tanx slope 0f the string 0 Writing the slope as the derivative tan x 8x D J Inman 363 Mechanical Engineering at Virginia Tech Substitute into Fma and use a Taylor expansion T Wxatj T Wxatj amp2WXl 2 Ax 8x 8x Recall the Taylor series of T8w3x about x1 7 Wii l 8x x2 8x 8x 8x XI if wxtj 82wxt 8x 8x 3f i 8wxt air 3x Wan p fltxrgtAx p 0Ax2K X1 Axgt Ax fxtAx 0 82wxt 3 67 D J Inman 4 63 Mechanical Engineering at Virginia Tech Since C is constant and for no external force the equation of motion becomes CZ 072wxt 072wxt 6398 07x2 atz I0 wave speed Second order in time and second order in space therefore 4 constants of integration Two from initial conditions wx0 w0x wtx0 WOOC at t 0 And two from boundary conditions eg a fixed fixed string w0t w t O tgt O 69 D J Inman 5 63 Mechanical Engineering at Virginia Tech Each of these terms has a physical interpretation De ection is wxt in the ydirection The slop of the string is wxxt The restoring force is waxxt The velocity is wtxt The acceleration is wttxt at any point x along the string at time t Note that the above applies to cables as well as strings D J Inman 663 Mechanical Engineering at Virginia Tech There are Two Solution Types for Two Situations 39 This is called the wave equation and if there are no boundaries or they are sufficiently far away it is solved as a wave phenomena Disturbance results in propagating wavesgt lt 0 If the boundaries are finite relatively close together then we solve it as a Vibration phenomena focus of Vibrations Disturbance results in Vibration focus of courses in Acoustics and in Wave Propagation D J Inman 7 63 Mechanical Engineering at Virginia Tech Solution of the Wave Equation 39 Interpret wxt as a stress particle velocity or displacement to examine the propagation of waves in elastic media Called wave propagation 39 Interpret wxt as a pressure to examine the propagation of sound in a uid Called acoustics D J Inman 863 Mechanical Engineering at Virginia Tech Solution of the string equation as a Wave Asolution is of the form wxt w1x ct w2x ct This describes on wave traveling forward and one wave traveling backwards called traveling waves as the form of the wave moves along the media crest The wave speed 1s c Think of waves in a pool of water trough D J Inman 9 63 Mechanical Engineering at Virginia Tech Example 611 what are the boundary conditions for this system zzzzgqzzza A force balance at 1 yields ngqsmmmgnw W y rm 7kwxz 1 Bx Fl quot Cr k Atx0theBCis wxzz0 0 y 5 Fig 62 Dllnman 1053 Mechamcal Engnccnng atVugmaTech 62 Modes and Natural Frequencies Solve by the method of separation of variables WOW XXTf gt 610 Substitute into the equation of motion to get 2 2 02X xTt XxTt where quotzd 2 and quot 2 dx dt X ltxgt To 1X ltxgtjzojx ltxgtaz Xx czm dx Xx Xx 614 Results in two second order equations 1 d 1 b T 2 coup e on y y a constant 2 c Tt 615 D J Inman 1163 Mechanical Engineering at Virginia Tech Solving the spatial equation results in a boundary value problem X x 02Xx O gt 616 A second order equation With solution of the form X x a1 sin 7x a2 cos 7x 611 and a2 constants of integration Next apply the boundary conditions XOTt O X Tt O gt Since XOO X O a2 0 Characteristic equation n71 gt 8111 at O gt on X alsin010 K Tt is not zero 621 an in nite number of values of 6 D J Inman 1263 Mechanical Engineering at Virginia Tech D J Inman 1363 Mechanical Engineering at Virginia Tech A more systematic way to generate the characteristic equation is write the boundary conditions 620 in matrix form alsina zO and a2 0 sin 76 0 a1 0 I O 1 a2 0 sin 76 O gt det 1 Ogtsin0 O This seemingly longer approach works in general and will be used to compute the characteristic equation in more complicated situations D J Inman 1463 Mechanical Engineering at Virginia Tech The the solution of the boundary value problem results in an eigenvalue problem The spatial solution becomes For nl23 X x an sin x 622 Here the index n results because of the indexed value of 7 The spatial problem also can be written as 039 2 ijz 1an Xnx i 0 Xn0 an 0 Which is also an eigenvalue eigenfunction problem Where 9 62 is the eigenvalue and X is the eigenfunctionr D J Inman 1563 Mechanical Engineering at Virginia Tech This is Analogous to Matrix Eigenvalue Problem A a 77 plus boundary conditions matrix a operator x 111 a Xn x eigenvector a eigenfunction 06X x also and eigenvector orthoganality also results and the condition of normalizing plays the same role eigenvectors become mode shapes eigenvalues frequencies modal expansion will also happen D J Inman 1663 Mechanical Engineering at Virginia Tech Next consider the temporal solution resulting in oscillation ilt62039 Tnt 0 n 123 624 Again a second order ode with solution of the form Tn t An sin and B cos and 625 AWE are constants of integration get from initial conditions Substitution back into the separated form Tt yields W x t C sin and sin O39nx dn cos and sin O39nx C s1n ct s1n x dn cos ct s1n x The total solution igecomes E E wx t Z c sum Ct sum x d cosg Ct sum x 627 n1 D J Inman 1763 Mechanical Engineering at Virginia Tech Orthogonality is used to evaluate the remaining constants from the initial conditions 5 K j sin x sinm 7 xdx 2 m E 6mm 628 0 E E 0 n 72 m 2 n proving this looks a lot like homework From the initial poistion wx0 w0x id sum x cos0 gt n1 n1 5 00 K j wO x simmsz xdx Z dn j Sm x simmTfr xdx gt 0 0 D J Inman 1863 Mechanical Engineering at Virginia Tech dm 3jw0xsinm xdx m123 E o E 631 m n gt 2 mr d w x sin x dx n123 n A on E gt we 00 i enanc Sim 30 0080 632 K c ij w0xsin xdx n 1 23 633 n 39c 0 E D J Inman 1963 Mechanical Engineering at Virginia Tech A Eigenfunctions become the vibration mode shapes w0 x singx which is the first eigenfunction n1 w0x 02 en 2 0 Vn E d 2Jsin xsin xdx 0 n 23 E 0 E 6 d1 1 gt 7T 71396 WOC t s1n x COS t Causes Vibration 1n the rst 6 mode shape D J Inman 2063 Mechanical Engineering at Virginia Tech Plots of mode shapes fig 63 x 242 ya O5 D J Inman 2163 Mechanical Engineering at Virginia Tech Example 622 Piano wire L14 m T111X104 N m110 9 Compute the first natural frequency 0 110 g per 14 1n 00786 kgm El l 111gtlt104N 1 1 14 0 14 00786 kgm 2266669 rads or 424 Hz D J Inman 2263 Mechanical Engineering at Virginia Tech Example 623 Compute the mode shapes and natural frequencies for the following system 0 l A cable hanging from X the top and attached to a spring of tension 1 WXJ and density p C N D J Inman 2363 Mechanical Engineering at Virginia Tech In this case the characteristic equation must be solved numerically ZFY 07wxt I 8x X x 111 Sin 0x 12 cos ax X00a2OandXxalsinax 5 10 cos 02 7k sin 02 Xi 0rsin l9kwZZ 0 71mm xl 5 an 02 7E k The characteristic equation must be Fig 64 solved numerically for on e D l Inman 2463 Mechanical Engineering at Virginia Tech The Mode Shapes and Natural Frequencies are k1000710 2 an 149729964501 swam X n 2 an sin0nx D J Inman 2563 The values of on must be found numerically The eigenfunctions are again sinusoids The value shown for on is for large n See Window 63 for a summary of method Mechanical Engineering at Virginia Tech Example Compute the response of the piano wire to 37m 1n1t1al cond1t1ons w0 x s1n7 w0 x 0 n71quot n 39c n71quot n 39c Solut1on wx t Z en s1n 7x s1n 7t d s1n x cos t i1 mm 0 Z on quotT fcm XCOSO gt c 0Vn i1 wxt 2d sinxcosn77fct at r 0 Eq631gt i1 Z d zgjsin xsinm xdx m 123 z 0 z z 0 for all m except m3 d3 1 37 37w wxt s1n7xs1n t D J Inman 2663 Mechanical Engineering at Virginia Tech Some calculation details on x sin m E xdx 0 E 00 22d Isin xsinm Exdx n E E w t sin 67rsinEt 2 E 0 E 37 37w w ts1n s1n t 4 4 E 0707 sint D J Inman H14mc1189gt w t 0707 sin 804t 2763 Mechanical Engineering at Virginia Tech Summary of Separation of Variables Substitute wxtXxTr into equation of motion and boundary conditions Manipulate all x dependence onto one side and set equal to a constant 62 Solve this spatial equation which results in eigenvalues on and eigenfunctions X Next solve the temporal equation to get Tnt in terms of on and two constants of integration D J Inman 2863 Mechanical Engineering at Virginia Tech Re form the product wnxtXnxTnt in terms of the Zn constants of integration Form the sum wx t Z X xAn sin and 8 cos and n1 Use the initial displacement initial velocity and the orthogonality of Xnx to compute A and B D J Inman 2963 Mechanical Engineering at Virginia Tech 63 Vibration of Rods and Bars Equilibrium position X X dX Consider a small element Inf1n1tes1mal dx element Of thG bar FdF De ection is now along x i called longitudinal Vibration 0 F 2 ma on small element J l 39 WOW yields the following I 0 e F1g 65 D J Inman 3063 Mechanical Engineering at Virginia Tech Force balance a 2 FdF FpAxdxM 653 Constitutive relation 8 t F EAx 8206 2 dF gleam avg Djdx X X x a 8wxt a 2wxt EA 2 A 655 0731 6 8x j POD 07 2 2 Ax constant gt 07 awif t a 323 656 At the clamped end w0 t O 657 At the free end EAM O 658 x xz D J Inman 3163 Mechanical Engineering at Virginia Tech Example 631 compute the mode shapes and natural frequencies of a cantilevered bar with uniform cross section wxt X xTt a czwxx xt wttxt X x 2 f0 02 659 Xx c Tt x m 02Xx 0 XO 0 AEX M 0 i TO 620392Tt 0 initial conditions ii From equation i the form of the spatial solution is Xx asin 0xbcos0x Next apply the boundary conditions in i to get the characteristic equation and the form of the eigenfunction D J Inman 3263 Mechanical Engineering at Virginia Tech Apply the boundary conditions to the spatial solution to get a sin0 9 0030 2 0 a 00306 b sin0 0 0 1 gt b 0 and det 0 COS0 31n039 COSGEOgt0H2n17z n2123 3 2 if Xnxan 31mm n2123 26 DJInrnan 3363 Mechanical Engineering at Virginia Tech Next consider the time response equation ii 2 271 1 2 Tntc M 7 Tt O 2 l 3710 An sintBn th Thus the solution implies oscillation with Frequencies w2n l6 392n l7z quot2123 63963 0 2E 26 D J Inman 3 4 63 Mechanical Engineering at Virginia Tech Example 632 Given v03 cms p8x103 kgm3 and E20x1010 Nm2 compute the response 2 1 wxt Z en sin O39nCl dn cos O39nCl sin 2E 7139x n1 0 dn 2 ij sin 2n1 7139de 0 3 E O 26 2n 1 w x 0 0035x E 0 CnO39nC cos0 sin n1 7139x Multiply by the mode shape indexed m and integrate D J Inman 3 5 63 Mechanical Engineering at Virginia Tech 1 Z gt 003 I sinQZsz5x dx 0 Z 2 1 2 1 2 JC acsinw xsinw xdx 0 2K 2K mgtcm 2 it E 2m 1 3 n1 n1 cn 8X10 9 am 1 7455gtlt106 1 m 210x10 7r2n 1 Zn 1 oo n1 wxt 7455X10396Zisin sin 5123482n 1t m quot1 Zn 1 10 m1 2m2 n zcamgc 1 POD 1 D J Inman 3663 Mechanical Engineering at Virginia Tech Various Examples Follow From Other Boundary Conditions 0 Table 61 page 485 gives a variety of different boundary conditions The resulting frequencies and mode shapes are given in Table 62 page 486 0 Various problems consist of computing these values 0 Once modesfrequencies are determined use made summation to compute the response from IC 0 See the book by BlevinszMode Shapes and Natural Frequencies for more boundary conditions D J Inman 3763 Mechanical Engineering at Virginia Tech 64 Torsional Vibrations d7 gdx from calculus x 86xt i 7 01 from solid mechanics 8x 9xt r Mme G shear modulus x 51 J polar moment of area cross section Summing moments on the element dx Combining these expressions yields 0quot 80xt 329XJ 7gdx szjdx GJ7 pJTGJ constantgt Where 0 is the shaft39s mass density 3290 Q 329XJ 6 66 8t2 p 8x2 39 D J Inman 3863 Mechanical Engineering at Virginia Tech The initial and boundary conditions for torsional vibration problems are Two spatial conditions boundary conditions 0 Two time conditions initial conditions 0 See Table 64 for a list of conditions and Equation 667 and Table 63 for odd cross section 0 Clamped free rod 60 t 0 Clamped boundary 0 de ection GHX E t 0 Free boundary 0 torque 6x060x and 6Zx060x D J Inman 3963 Mechanical Engineering at Virginia Tech Example 641 grinding shaft vibrations Top end of shaft is connected to pulley x 0 0 J1 includes collective inertia of drive belt pulley and motor J1 lt Drive pulley collective inertia ix 9X t Shaft of stiffness GJ length 5 Grinding head Inertla J2 D J Inman 4063 Mechanical Engineering at Virginia Tech Use torque balance at top and bottom to get the Boundary Conditions 86xt 8492xt G ampC x20 2 6sz0 at tOp 39 r 6 62 r G g xg Jzx at bottom The minus sign follows from right hand rule D J Inman 4163 Mechanical Engineering at Virginia Tech Again use separation of variables to attempt a solution 6xt xTt gt m gm 02 x G Tt can o x ow 0 M 02Tt 0 G w060 0 D J Inman 4263 Mechanical Engineering at Virginia Tech The next step is to use the boundary conditions Boundary Condition at x O 3 GJ OTt J1 OTt gt GJ 0 T0 2 6202 3 1190 Tt 2 gm 039 J1 0J 90 Similarly the boundary condition at eyields 2 Kg 2 0J1 g 0J D J Inman 4363 Mechanical Engineering at Virginia Tech The Boundary Conditions reveal the Characteristic Equation x all sin 0x a2 cos we gt O a2 x 2 a1039 cos we c120 sin we gt O 2 a1039 x O gt 2 0 0 J1 0 gt a 5711 a 1 2 0 0 x gt 2 2 W a Jlew gt alacosa a2039sin of a Jlal sin of a2 cos of p p OMJ1 JZ039E J J 0amp2 pJE2 THE CHARACTERISTIC EQUATION 1 2 682 gt tan039 D J Inman 4 4 63 Mechanical Engineering at Virginia Tech Solving the for the first mode shape mm 12W mm 1112mm2 pm2 has 0 as its first solution Numerically solve for an n 1 23 and on an F 0 N0tef0rn101 0gtw1 0gtfz0gt Tt a bt the rigid body mode of the shaft turning 3 fx 0 1 x 2 611 blx 3 x O gt 91 O gt 1 x al the first mode shape D J Inman 4 5 63 Mechanical Engineering at Virginia Tech Solutions of the Characteristic Equation involve solving a transcendental equation wfkwnmnxx pw b m J1J2 J1J20JE J1 J2 10 kgm2 rad p2700 kgm3 J5kgnfrwi f0251n G 25gtlt109Pa 3 f1 0 Hz f2 238013 Hz f3 76026 Hz f4 2114039 Hz D J Inman 4663 Mechanical Engineering at Virginia Tech x039 a Fig 69 Plots of each side of eq 682 to assist in find initial guess for numerical routines used to compute the roots h 2 17 j tan 1 in H and r J 113 H n tan t39 12 1quot D J Inman 47 63 Mechanical Engineering at Virginia Tech 65 Bending vibrations of a beam W ix 1 f t MX tMXX 1 dx i VX t VXX tdX dX AX h1h2 bending stiffness EIx X d Next sum forces 1n the y directron up down E Youngs modulus Sum moments about the pornt Q I x cross sect area moment of Use the moment glven from inertia about z 2 M 959 E1x Assume sides do not bend x stenght of materials no shear deformation D J Inman 4863 Mechanical Engineering at Virginia Tech Summing forces and moments yields Vow 3V0 dxj Vxt fxtdx pAxdx 82M 8x at Ma t aMOc t dx Mx t rVx t 8V0 dxdx a i 8x J fxrdx 0 2 0 Wx t Vxtdx avx t fx t72 0 8x 8x 2 D J Inman 49 63 Mechanical Engineering at Virginia Tech 8Mx t 8x Substitute into force balance equation yields 2 Vxt a2Mxt 7 Dividing by dx and substituting for M yields 82Wx I iElm dx fxtdx pAxdx 82wxt pAx 02 07x2 0562 fxt Assume constant stiffness to get 82wxt 62 own3m 0 C 0V 8x4 pA D J Inman 5063 Mechanical Engineering at Virginia Tech The possible boundary conditions are choose 4 Free end Clamped or xed end bending moment E 072V 2 0 de eCtiOH W 0 07w 8r dzwi slope O shear force 2 E1 0 654 0amp2 j 07x Pinned or simply supported end Sliding end de ection w 0 Slope g 2 0 82w bendin momentzEI 0 I I g 3362 shear force 2 iLEI 822 0 036 036 D J Inman 5163 Mechanical Engineering at Virginia Tech Solution of the time equation yields the oscillatory nature 2 XMOC 2 C a X 36 TU ft 02Tt O gt Tt Asin atBcosat Two initial conditions wx0 w0x wtx0 2 WO x D J Inman 5263 Mechanical Engineering at Virginia Tech Spatial equation results in a boundary value problem BVP X x Xx 0 2 0sz F Let Xx A6 to get Xx a1 sin x a2 cos x a3 sinh x a4 cosh x Define 84 29 C Apply boundary conditions to get 3 constants and the characteristic equation D J Inman 5363 Mechanical Engineering at Virginia Tech Example 651 compute the mode shapes and natural frequencies for a clamped pinned beam At fixed end x 0 and XOOgt 612a4 0 X 0 2 6a1 a3 0 At the pinned end x E and X 0 2 a1 sin f at2 cos a3 sinh a4 cosh f 0 EIX 0 gt 6 2 az1 sin f at2 cos f a3 sinh a4 cosh 0 D J Inman 5 4 63 Mechanical Engineering at Virginia Tech The 4 boundary conditions in the 4 constants can be written as the matrix equation 0 1 0 1 a1 6 0 6 0 a2 sin 00366 sinh cosh 66 a3 62 sin 162 00366 62 sinh 62 cosh 614 t Y Jon B a Ba0at0gtdetB0gt tan 26 tanh 26 0 0 0 0 The characteristic equation D J Inman 5 5 63 Mechanical Engineering at Virginia Tech Solve numerically fsolve to obtain solution to transcendental characteristic equation 81K 2 3926602 826 7068583 836 10210176 84K 13351768 85K 16493361 n gt 5 3 4n 17r K 4 Next solve Ba0 for 3 of the constants D J Inman 5663 Mechanical Engineering at Virginia Tech With the eigenvalues known now solve for the eigenfunctions Ba 2 0 yields 3 constants in terms of the 4th at1 az3 from the first equation a2 azl from the second equation sinh n sin 0613 cosh 6 cos 0614 0 from the third or fourth equation Solving yields a3 Cosh n Cos n a4 sinh n s1n 6 rcosh Cos6 Z gtX sinh sin Cosh COS no mlSmh nmm ng lt A x A x A x xi D J Inman 57 63 Mechanical Engineering at Virginia Tech Plot the mode shapes to help understand the system response XHX I E 15 sinhnx sinltn coshnx COSH39X Note zero slope X 3926602X X 7068583 X 1 D J Inman 5863 X 10210176 05quot Mode 1 X Mechanical Engineering at Virginia Tech Again the modeshape orthogonality becomes important and is computed as follows Write the eigenvalue problem twice once for n and once for m x x jxnec and Xg x 6ijx Multiply by X m x and X n x respectively integrate and subtact to get I Xg ltxgtXmltxgtdx j Xz ltxgtXnltxgtdx gtj XnltxgtXmltxgtdx Then integrate the left hand side twice by parts to get D J Inman 59 63 Mechanical Engineering at Virginia Tech Use integration by parts to evaluate the integrals in the orthogonality condition apply Judvzuv J vdu twice K K jxmm nyxmx Xm X g j fxmx 0 d T v 0 0T du X X X X O X X d M nltgt w Hula 113x c 0 D J Inman 6063 Mechanical Engineering at Virginia Tech Z Z J x Xg dx X1x X x j XfxXxdx 0 HF 0 u dv X1g y xyrn0 j XfxXxdx 0 Thus 0 I X ltxgtXmltxgtdx J X ltxgtXnltxgtdx 2 XnltxgtXmltxgtdx 2 I X ltxgtXltxgtdx J X ltxgtXltxgtdx gt J XnltxgtXmltxgtdx 0 0 J 0 0 0 Z gt I XnxXmxdx OVnm n i m 0 D J Inman 6163 Mechanical Engineering at Virginia Tech The solution can be computed via modal expansion based on orthogonality of the modes I XnxXmxdx 6m wx t 204 sin ant Bn cos ontXn x n1 wx 0 w0x iBanOC gt Bn j w0xXnxdx 1 0n wtx0 WO x 260114an x gt An JWO xXn xdx D J Inman 6263 Mechanical Engineering at Virginia Tech Summary of the EulerBournoulli Beam Assumptions Uniform along its span and slender Linear homogenous isotropic elastic material Without axial loads Plane sections remain plane Plane of symmetry is plane of vibration so that rotation amp translation decoupled Rotary inertia and shear deformation neglected Tables 64 and 65 give eigensolutions for several configurations see Blevins for more D J Inman 6363 Mechanical Engineering at Virginia Tech Section 49 Computational Eigenvalue Problems for Vibration Analysis Many of you have already used codes and calculators to solve the needed eigenvalue problems We shall formalize that skill here DJ Inman 129 Mechanical Engineering at Virginia Tech Dynamically Coupled Systems 39 A dynamically coupled system is one in which the mass matrix is not diagonalgt lt 39 When the mass matrix is diagonal M39lz Mquot1 are all simple calculations 39 Here we address what to so when it is not diagonal often occurs in FEM DJ Inman 229 Mechanical Engineering at Virginia Tech Window 47 and how to take the Function of a Matrix A matrix M 1WT is positive definite if for every non zero vector X XT MX gt If M is positive definite then all of its eigenvalues are positive real numbers If M is symmetric and positive definite then f M Udiagf1f2 K f nHU T where Mul uiu and U 111 u2 L um I DJ Inman 329 Mechanical Engineering at Virginia Tech Matrix Square Root dynamically coupled case Assume M is positive definite and symmetric 1 0 0 A M 12 O O T 0 0 W I I W f 0i Also the inverse is MEI1 If NUT 0 0 DJ Inman 4 29 Mechanical Engineering at Virginia Tech Numerically Efficient Methods The best way to compute the inverse of a matrix is to use Gaussian Elimination In Matlab this is M I The best way to compute the matrix square root is not to and to use the Cholesky Decomposition In Matlab this is Lchol M Then M LLT so that L 1MLT 1 I DJ Inman 5 29 Mechanical Engineering at Virginia Tech Modal Analysis using Cholesky D D 3 Q Q U amp Calculate L from M Calculate K L1KLT 1 Calculate the eigenvalue solution for K I KVi ZiVi Normalize the eigenvectors V and form the matrix P Compute S L7391P and S39lzPTLT Calculate the modal initial conditions r0 S lxo rm 5 520 Use the modal initial conditions to compute rt Multiply rt by S to get the solution XtS rt DJ Inman 629 Mechanical Engineering at Virginia Tech Example 491 Cholesky Decomposition 5 2 2 M 2 a1 1 D 1 3 Liz hDIESKFIIMII 2222 n I 2442 I 2 L 2294 1229 n 39 L391 42224 2552 n El LESSI 1639 LEWIS El39 1139 WEI 61 5 2 o LLT 2 r4 1 139T 391 1 I D L39ML n 1 n 2 1 2 U 1 III DJ Inman 7 29 Mechanical Engineering at Virginia Tech Example Usmg the mass and stiffness from a bar M2 11 2 invL39MinvL gtgtKl l l 1 ans 2 10000 00000 anM 00000 10000 ans gtgtKhinvL39KinvL 0 6667 03333 Kh 03333 06667 05000 08660 gtgtLcholM 08660 15000 L gtgtVDeigKh 14142 07071 V 08660 05000 0 1392247 05000 08660 SE MLgtV D S O 0 04082 07071 04082 07071 0 2 gtgtV39V Note that LIS lower In text upper In MATLAB ansloooo O 0000 Checkyourcode before usmg 00000 10000 DJ Inman 829 Mechanical Engineering at Virginia Tech Example 493 M 3 6 Elm J h H Note thatthe dampmg s propomoumo moda ana ysws may be used WWW w emmmx v2 emmmxn p angmzmvlv2 P u7u77u7 mm mm um I uz u 2 u cu p pr u 4 DJ anan 929 Mechamcal Engmeexmg at Vugmm Tech The many eigenvalue problems Mm Kxt 0 gt xiu M1Ku gt xiv MKM v In MATLAB Use 3 iv 2 LT 1 KL lv UDeigltAgt gt1MuKu U D eig KM DJ Inman 1029 Mechanical Engineering at Virginia Tech Numerical Comparison 1 K 27 3 3 3 Same problem using other mssqrtM gtgtKtinv ms K nv ms e1genvalue problems gtgtvDeigltKtgt inVL KinVL 118 M9 OO o 6 diag only V o7o71 o7o71 MK 146 07071 o7o71 inVM K 191lt D 4 O invsqrtmM KinvsqrtmM o 2 228 gtgtflops VDejgKM ans 146 gtkeigenvectors not orthogonal hence an additional step is needed 3980 not only do these transformations take less computing time but the give more information DJ Inman 1129 Mechanical Engineering at Virginia Tech More in first order form m M 1Kxr 0 Lety1Xy2 Xgt 5712372 5392 iM1KY1 gt O I y1 yAy where A 1 y M K O y2 u gtAzlzz 0 zi 1 ziui DJ Inrnan 1229 Mechanical Engineering at Virginia Tech Treating SVSIBIIIS Will damning Mia CXQ KXZ 0 3amp0 M lCXQr M leQr 0 Let y1 x y2 X 2 371 X yz 1 1 Y2ZXM KY1M C372 0 I yl y Ay where A 1 1 y M K M C y2 u gtAz1zz 0 zi 1 lint DJ Inrnan 1329 Mechanical Engineering at Virginia Tech 2n Xt Zciuie i1 If each mode is underdamped 1i complex valued 1i iai 01 V1 and 1H1 iai 01 V1 I called the modal damping ratio The modes ul may be real or complex valued depending on KM lc CM lK Note that Xt here contains 10quot Il Sili llS and velocities DJ Inman 1429 Mechanical Engineering at Virginia Tech Damping ratios and frequencies from complex eigenvalues 11 at ij R6011 Imilj col 105 8 xReil2 Him112 4185 Re1i ai g 1053 6 1er Him112 DJ Inman 1529 Mechanical Engineering at Virginia Tech Example 494 7 2 2 0M7xxmxrwc 0 225 m A 7 angmznlstxck u u 22 u u 2 M7 2325 A 2 WWW 2 rlsus suzs 23mm 2 7 HHS mum 2 2 m 23 mpg 2 2 2 2 mm m m 2 14m 2Hng u 2222 qu m m m m w m w m U xzsimmx mamm U725 U725 we mm 5397 mm m U257 2mm U257 us us 43mm 54522 DJInman Mechamcal Engmeenng at Virginia Tech 1629 Complex modes Complex if matrix product not satisfied If KM I C CM 39IK is true then u can be normalized to be real valued Each element describes the magnitude and phase of the corresponding degree of freedom Complex modes appear in conjugate pairs DJ Inman 1729 Mechanical Engineering at Virginia Tech 1 1 real mode J5 1 01240006791 complex mode O3721O2038 DJ Inman 1829 Mechanical En glneerlng at Vlrglnla Tech Nodes of a Mode I 01257 El lquot ul 6 XI 1 0325 6 HT All at Vibrating at 031 DJInrnan 1929 Mechanical Engineering at Virginia Tech Symmetric Generalized Eigenvalue Problem MXCXKX0Lety1X andy2 Xgt Myz Cy2 KY1 MultiPIYY1 Y2 bY39K KY1KY2gt K 0 3391 0 K Y1 I 0 M yz K C yz r1 War J A B y Mzsz Z7t0 A symmetric generalized Eigenvalue nrolllem Will damning and 0 inverse DJ Inman 2029 Mechanical Engineering at Virginia Tech Alternater multiply by 1 KO OK KC 0 M Symmetric generalized eigenvalue lll elllem XDeigABin matlab DJ Inman 2129 Mechanical Engineering at Virginia Tech The Symmetric Eigenvalue Problem 0 The fundamental theorem of algebra guarantees that the eigenvalues and eigenvectors of A will either be real or occur in complex conjugate pairs 0 This means that even if there are repeated eigenvalues there will be n linearly independent eigenvectors 0 The Jordan form will be diagonal DJ Inman 2229 Mechanical Engineering at Virginia Tech Section 410 Numerical integration for the time response Mia CXt KXt BFt x0 X0X0 2 v0 gt in y2 3 2 M1Ky1 M1Cy2 M1BFt gt 5 0 Ayt fty0 y0 Where Y1 0 I 0 YO y A M1K M1Cft M13Fm 2 DJ Inman 2329 Mechanical Engineering at Virginia Tech MATLAB Code MAKE AN M FILE called simulationm function ydotsimulationty M9 OO lC9 l l lK27 3 3 3 FOOO sin3t Azeros22 eye22 MK MC ydotAyF now type the following in the command window yOOlOOOtiOtf50 timesolode45 simulation ti tfyO subplot2llplottimesoll Xlabel time S ylabel yl m title dof l subplot2l2plottimesol2 Xlabel time S ylabel y2 m title dof 2 DJ Inman 2429 Mechanical Engineering at Virginia Tech Comments 39 Can calculated the eigenvalue problem 39 Can use it to write the free and forced response 39 Can also simulate the free and forced response numerically approximation 39 Can use numerical simulation to solve for nonlinear mdof systems extension of sdof case DJ Inman 2529 Mechanical Engineering at Virginia Tech FirstOrder Form X M 1CX M 1KX can be rewritten as warm atria Define the state vector yt 3110 70 such that 3392 0 X0 ya Ayt is our first order system DJ Inman 2629 Mechanical Engineering at Virginia Tech FirstOrder Form cont Seek a solution of the form y ze t A26 2 126 gt AZ AZ the standard eigenproblem only now matrix A is rarely symmetric 39 Doubled system size to Zn 0 Likely to get complex eigenpairs DJ Inman 2729 Mechanical Engineering at Virginia Tech FirstOrder Form cont ll 139 i A 0 I 111 1 ul Zi M1K M1C xiiul i xiiul yields xiiul xiiul M1Kul M1C1iul lizul gtM1i2ul C1iul Kul 0 1 AZ 12 gt 21 l Where 111 im mode shape DJ Inman 2829 Mechanical Engineering at Virginia Tech ExanMe4JOJ uJKmam MM V 75 425 muss M a M K x 4555 usss um u D M U DxxAlX D n Bass mass u u 39 inn4m 5mm pmm mwm Z x 2 gt x2 zlt3gt D J Inman 2929 Mechanical Engmeenng at VngmmTach Chapter 4 Multiple Degree of Freedom Systems The Millennium bridge required many degrees of freedom to model and design with g quot X2 I 39 13122000 Extending the first 3 chapters to more then one degree of freedom DJ Inman 158 Mechanical Engineering at V1rg1n1a Tech The first step in analyzing multiple degrees of freedom DOF is to look at 2 DOF DOF Minimum number of coordinates to specify the position of a system Many systems have more than 1 DOF Examples of 2 DOF systems 7 cat with sprung and unsprung mass both heave r elastic pendulum radial and angular r motions of a ship roll and pitch x2 k 4 x l 1 f k m 2 m 2 k k X 41 TWODegreecfFreedcm Model Undamned p 1391 f f x fuva m1 7 m M A 2 degree of freedom system used to base much of the analysis and conceptual development of MDOF systems on DJ Inman 358 Mechanical Engineering at Virginia Tech FreeBody Diagram of each mass Figure 42 k2X2 quotX 1 i gt x1 i gtX2 DJ Inrnan 4 58 Mechanical Engineering at Virginia Tech Summing forces yields the equations of motion 11115610 2 k1x1t k2 x2 t x1t m25 2t k2x2t x1t 41 Rearrangng terms mpg r k1 k2gtx1lttgt W2 2 0 42 m2 2t k2x1t k2x2t 0 DJ Inman 5 58 Mechanical Engineering at Virginia Tech Note that it is always the case that 0 A 2 Degreeof Freedom system has Two equations of motion Two natural frequencies as we shall see DJ Inman 658 Mechanical Engineering at Virginia Tech The dynamics of a 2 DOF system consists of 2 homogeneous and coupled equations 0 Free vibrations so homogeneous eqs 0 Equations are coupled Both have x1 and x2 If only one mass moves the other follows Example pitch and heave of a car model 0 In this case the coupling is due to k2 Mathematically and Physically If k 0 no coupling occurs and can be solved as two independent SDOF systems DJ Inman 7 58 Mechanical Engineering at Virginia Tech Initial Conditions 39 Two coupled second order ordinary differential equations with constant coefficients 39 Needs 4 constants of integration to solve 0 Thus 4 initial conditions on positions and velocities 3510Cioa 5510hoa 3520520 5520520 DJ Inman 858 Mechanical Engineering at Virginia Tech Solution by Matrix Methods The two equations can be written in the form of a single matrix equation see pages 272275 if matrices and vectors are a struggle for you 2 xltrgtx1t x0 0 m if1m 44 45 X X 20 20 x20 M m 0 K k1k2 462 46 49 0 m2 k2 k2 39 MXKX0 m1551fk1k2x1l k2x2to ltgt m2 2t k2x1t k2x2t 0 DJ Inman 9 58 1V echanical Engineering at Virginia Tech Initial Conditions IC s can also be written in vector form x1 X10 XO and XO x20 x20 DJ Inman 1058 Mechanical Engineering at Virginia Tech The approach to a Solution For lDOF we assumed the scalar solution aeM Similarly now we assume the vector form Let Xt ueja 415 j l 11 7b 0 a u unknown gt con Kuej quot O 416 gt an Ku 0 417 DJ Inman 1158 Mechanical Engineering at Virginia Tech This changes the differential equation of motion into algebraic vector equation an Ku 0 417 This is two algebraic equation in 3 uknowns 1 vector of two elements and l scalar 1 u and a 2 DJ Inman 1258 Mechanical Engineering at Virginia Tech The condition for solution of this matrix equation requires that the the matrix inverse does not eXist If the an2M K eXists gt u 0 which is the static equilibrium position For motion to occur 1 u ab 0 gt sz K does not eXist or det a2M K 0 419 The determinant results in 1 equation in one unknown 03 called the characteristic equation DJ Inman 1358 Mechanical Engineering at Virginia Tech Back to our specific system the characteristic equation is defined as det a2MK 0 gt d a2ml k1k2 k2 k2 a2 m2 k2 420 O gt 17111712604 mlk2 mzk1 mzk2 a2 klk2 O 421 Eq 421 is quadratic in i2 so four solutions result 012 and 022 gt ital and i 02 DJ Inman Mechanical Engineering at Virginia Tech 1458 Once 0 is known use equation 417 again to calculate the corresponding vectors 111 and u2 This yields vector equation for each squared frequency wa on1 0 422 and a M Ku2 0 423 Each of these matrix equations represents 2 equations in the 2 unknowns components of the vector but the coefficient matrix is singular so each matrix equation results in only 1 independent equation The following examples clarify this DJ Inman 1558 Mechanical Engineering at Virginia Tech EX IIlplGS 415 amp 416anlculating u and 0 39 m19 kgm21kg 124 Nm and k23 Nm 0 The characteristic equation becomes 0346m28m22m24O 02 2 and 02 4 or 013 i 2 rads 024 2 i 2 rads Each value of 032 yields an expression or u DJ Inman 1658 Mechanical Engineering at Virginia Tech Computing the vectors u 11 12 For 6012 2 denote u1 then we have wa Ku1 0 gt 27 92 3 Mn 0 gt 3 3 2 Mn 0 9u11 3u12 O and 3u11u12 O 2 equations 2 unknowns but DEPENDENT the 2nd equation is 3 times the first DJ Inman 1758 Mechanical Engineering at Virginia Tech Only the direction of vectors u can be determined not the magnitude as it remains arbitrary l gt M11 2 1 M12 results from both equations M12 3 3 only the direction not the magnitude can be determined This is because det a12M K 0 The magnitude of the vector is arbitrary To see this suppose that 111 satisfies wa K u1 0 so does aul a arbitrary So an loan1 0 ltgt an Ku1 0 DJ Inman 1858 Mechanical Engineering at Virginia Tech Likewise for the second value of 02 M For 6022 4 let u2 21 then we have 22 wa Ku 0gt 27 94 3 M21 0 gt 3 3 4 M22 0 9u21 3u22 O or u21 u22 Note that the other equation is the same Inman Mechanical Engineering at Virginia Tech What to do about the magnitude Several possibilities here we just fiX one element Choose 1 A M12 2 l gt 111 1 Choose 1 3 u22l gtu2 DJ Inman 2058 Mechanical Engineering at Virginia Tech Thus the solution to the algebraic matrix equation is l 111 and u2 The origin of this name comes later DJ Inman 2158 Mechanical Engineering at Virginia Tech HR 013 ixE has mode shape u1 ex 6024 i 2 has mode shape u2 Here we have introduce the name mode shape to describe the vectors Return now to the time response We have computed four solutions Xt uleW uleja uzej 2tuzej 2t gt 424 Since linear we can combine as Xt awe W bulemt cuzej 2t duzeja zt gt Xt ale W beWu1cejw2t deja zt u2 A1 sina1t 31 u1 A2 sina2t 52 u2 426 where A1A2 1 and 52 are constants of integration Note that to go from the exponential I I I I 39 39 form to to sine requires Euler s formula I 39 for trig functions and uses up the D J Inman sign on omega 2258 Mechanical Engineering at Virginia Tech Physical interpretation of all that math 0 Each of the TWO masses is oscillating at TWO natural frequencies 01 and 002 0 The relative magnitude of each sine term and hence of the magnitude of oscillation of m1 and m2 is determined by the value of Alu1 and Azu2 The vectors 111 and 112 are called mode shapes because the describe the relative magnitude of oscillation between the two masses DJ Inman 2358 Mechanical Engineering at Virginia Tech What is a mode shape 0 First note that Al A2 I1 and I2 are determined by the initial conditions 0 Choose them so that A2 I1 I2 0 0 Then Xt A1 s1n colt Alu1 s1n colt x2 t L112 0 Thus each mass oscillates at one frequency 031 with magnitudes proportional to 111 the 1st mode shape DJ Inman 2 4 58 Mechanical Engineering at Virginia Tech A graphic look at mode shapes If IC s correspond to mode 1 or 2 then the response is purely in mode 1 or mode 2 3 2 k r Mode 1 V 1712 111 1 x1 x k1 r k2 r 2 y Mode 2 E 1711 Ir 1712 112 13 x1A3 362A DJ Inman 2 5 58 Mechanical Engineering at Virginia Tech EXEIIIIplB 1 7 given the initial conditions compute the time response consider X0 mm X0 xlm 31 sinkEr 1 sin2t 152 W A1 3mm 1 A2 sin2r 2 X1f 11 cos t 120032t 2 x20 Al cos t 1A220032t 2 DJ Inman 2658 Mechanical Engineering at Virginia Tech At tO we have 1 mm 0 0 DJ Inman 2758 AlxE COS 4 equations in 4 unknowns 3 A1 sin A2 sin 2 O A1 sin 1 A2 sin 2 0 2 Ala COS 1 A2 2 cos 2 O Al COS A2 2 COS 2 Yields A1 15 mmA2 15 mm Q zgrad DJ Inman 2858 Mechanical Engineering at Virginia Tech The nal solution is x10 2 05 cosxEt 05 cos 2t 434 x20 2 15 costt 15 cos 2t These initial conditions gives a response that is a combination of modes Both harmonic but their summation is not Displacement Dispincemem mm mm 10 VA 2 0 S 15 Time s 2 Figure 43 o D Inman Mechanical Engineering at Virginia Tech 2958 Solution as a sum of modes Xt 611111 cos w1t 612112 cos 02f Determines how the second frequency contributes to the Determines how the first response frequency contributes to the response DJ Inman 3058 Mechanical Engineering at Virginia Tech Things to note 0 Two degrees of freedom implies two natural frequencies 0 Each mass oscillates at with these two frequencies present in the response and beats could result 0 Frequencies are not those of two component systems 11 572 163a2 27 r732 m1 m2 0 The above is not the most efficient way to calculate frequencies as the following describes DJ Inman 3158 Mechanical Engineering at Virginia Tech Some matrix and vector reminders a b 1 1 d b A gt A c 6 ad cb c a O M 2 ml XTMX mle mzxg 0 m2 M gt O gt XTMX gt O for every value of X except 0 Then M is said to be positive de nite DJ Inman 3258 Mechanical Engineering at Virginia Tech 42 Eigenvalues and Natural Frequencies 39 Can connect the vibration problem with the algebraic eigenvalue problem developed in math 0 This will give us some powerful computational skills 0 And some powerful theory 0 All the codes have eigensolvers so these painful calculations can be automated DJ Inman 3358 Mechanical Engineering at Virginia Tech Some matrix results to help us use available computational tools A matrix M is defined to be symmetric if MMT A symmetric matrix M is positive de nite if xTMx gt O for all nonzero vectors x A symmetric positive definite matrix M can be factored M LLT Here L is upper triangular called a Cholesky matrix DJ Inman 3 4 58 Mechanical Engineering at Virginia Tech If the matrix L is diagonal it defines the matrix square root The matrix square root is the matrix M 1 2 such that M12M12 M If M is diagonal then the matrix square root is just the root of the diagonal elements LM 2M 0 435 0 M DJ Inman 3 5 58 Mechanical Engineering at Virginia Tech A change of coordinates is introduced to capitalize on existing mathematics For a symmetric positive definite matrix M in Let Xt M 1 2qt and multiply by M 2 M WMM 2 10 M WKM 2 qt 0 438 I identity I symmetric or ijt IEqt 0 Where I M12KM 2 I is called the mass normalized stiffness and is similar to the scalar i m used extensively in single degree of freedom analysis The key here is that I is a SYMMET RIC matrix allowing the use of many nice properties and computational tools DJ Inman 3658 Mechanical Engineering at Virginia Tech How the Vibration problem relates to the real symmetric eigenvalue problem Assume qt 2 V6ij in jt IEqU 0 2 3960 N 3960 0 V6 Kve 0 V i 0 or 2 KV a V ltgt KV xiv V i 0 r J KJ Vibration problem real symmetric eigenvalue problem 440 441 Note that the martriX K contains the same type of information as does 02 in the single degree of freedom case 1 DJ Inman 3758 Mechanical Engineering at Virginia Tech Important Properties of the n X n Real Symmetric Eigenvalue Problem There are n eigenvalues and they are all real valued There are n eigenvectors and they are all real valued The set of eigenvectors are orthogonal The set of eigenvectors are linearly independent The matrix is similar to a diagonal matrix Window 41 page 285 DJ Inman 3858 Mechanical Engineering at Virginia Tech Square Matrix Review 0 Let aik be the ikth element of A then A is symmetric if aik aki denoted ATA 0 A is positive definite if xTAx gt 0 for all nonzero X also implies each kl gt O 0 The stiffness matrix is usually symmetric and positive semi definite could have a zero eigenvalue 0 The mass matrix is positive definite and symmetric and so far its diagonal DJ Inman 3958 Mechanical Engineering at Virginia Tech Normal and orthogonal vectors x1 3 1 n X M y M inner product is XTy inyl xn yn H X orthogonal to y if XTy 0 X is normal if XTX l if a the set of vectores is is both orthogonal and normal it is called an orthonormal set The norm of X is VXTX 443 DJ Inman 4058 Mechanical Engineering at Virginia Tech Normalizing any vector can be done by dividing it by its norm X VXTX To see this compute XT X XTX 1 VXTX XTX XTX DJInman 4158 Mechanical Engineering at Virginia Tech has norm of 1 444 X XTX Examples 422 through 424 EZM izKM izz O 27 3 O O 1 3 3 O 1 3 so K 3 which is symmetric 1 3 1 which has roots 21 2 012 and 22 4 022 3 1 detK Udet 1 12 618O DJ Inrnan 4258 Mechanical Engineering at Virginia Tech IE AIV10gt 3 2 1 V 1 3 2 v12 VII v12OgtV1a V1 a2111gt azy Mechanica 1 1 The first normalized eigenvector 1 Engineering at Virginia Tech Likewise the second normalized eigenvector is computed and shown to be orthogonal to the first so that the set is orthonormal l 1 1 V2 2 31 VlTV2 O l VITV1 Elll vim in lt 1gtlt 1gtgt 1 gt V l are orthonormal DJInman 4 4 58 Mechanical Engineering at Virginia Tech Modes u and Eigenvectors V are different but related u1 739 39V1 and H2 st V2 X M lZq u 1 4 12V 437 Note ii i w DJ Inman 4 5 58 Mechanical Engineering at Virginia Tech This orthonormal set of vectors is used to form an Orthogonal Matrix P 2 V1 V2 called a matrix of eigenvectors normalized PTP 2VV1 vivz Z I V2V1 V2V2 P is called an orthogonal matrix PTIZP PT 16v1 1amp2 PT 21v1 AZVZ m we Avivl zviw 0 d39 2w2A 0 lagwl 2 447 P is also called a modal matrix DJ Inman 4658 Mechanical Engineering at Virginia Tech Example 423 compute P and show that it is an orthogonal matrix From the previous example 1 l l P 3 V1 V1 JEL 4 PTP1 1 1 1 1 1 JEJE1 1 1 1 11l 1 1 12 OI 21 11120 2 DJInman 47 58 Mechanical Engineering at Virginia Tech Example 424 Compute the square of the frequencies by matrix manipulation PTIEPE 1L1 31 1 4 2H 2lA if 2 gt 01 2 rads and 02 2 rads In general A PTKP diag diagal2 448 4858 Mechanical Engineering at Virginia Tech Example 425 MM LW E The equations of motion k1 k2x1 k2x2 0 449 mzxz k2x1 k2 k3x2 0 x Figure 44 In matrix form these become 0 kk k ml i 1 2 2 x0 450 0 m2 k2 k2k3 DJ Inman 4958 Mechanical Engineering at Virginia Tech Next substitute numerical values and compute P and A m1 1 kg m2 4 kg k1 k3 10 Nrn and k22 Nm 1 0 12 2 gtM K 0 4 2 12 3KM 12KM 12 12 1 1 12 12 1 1 1 12 1 311 28902 and 12 121098 gta117 rads and 02121098 rads gtdet1 Mdet 22 152350 DJ Inman 5058 Mechanical Engineering at Virginia Tech Next compute the eigenvectors For xi equation 441 becomes 12 28902 1 v110 1 3 28902 v21 gt 910891211 v21 Normalizing v1 yields 1 v1 v121 12221 v121 910892v121 gt 1211 01091 and 1221 09940 01091 09940 likeWISe v2 09940 01091 V1 DJ Inman 5158 Mechanical Engineering at Virginia Tech Next check the value of P to see if it behaves as its suppose to P 01091 09940 V1 V209940 01091 09940 01091 1 3 0 121098 01091 09940 12 1 01091 09940 28402 0 PTKP 09940 01091 TP 01091 09940 01091 09940 1 0 09940 01091 09940 01091 0 1 Yes DJ Inman 5258 Mechanical Engineering at Virginia Tech A note on eigenvectors In the previous section we could have chosed V2 to be 09940 09940 2 01091 01091 because one can always multiple an eigenvector by a constant instead of V2 2 and if the constant is l the result is still a normalized vector Does this make any difference No Try it in the previous example DJ Inman 5358 Mechanical Engineering at Virginia Tech All of the previous examples can and should be solved by hand to learn the methods However they can also be solved on calculators with matrix functions and with the codes listed in the last section In fact for more then two DOF one must use a code to solve for the natural frequencies and mode shapes Next we examine 3 other formulations for solving for modal data DJ Inman 5 4 58 Mechanical Engineering at Virginia Tech Matlab commands 39 To compute the inverse of the square matrix A inv A or use Aeye n Where n is the size of the matrix P D eig A computes the eigenvalues and normalized eigenvectors watch the order Stores them in the eigenvector matrix P and the diagonal matrix D DA DJ Inman 5 5 58 Mechanical Engineering at Virginia Tech More commands To compute the matrix square root use sqrtmA To compute the Cholesky factor L chol M 0 To compute the norm norm x 0 To compute the determinant det A 0 To enter a matrix K27 3 3 3 M9 OO l 0 To multiply Kinv chol M DJ Inman 5658 Mechanical Engineering at Virginia Tech An alternate approach to normalizing mode shapes From equation 417 Ma2 K u O 11 st 0 Now scale the mode shapes by computing a such that T aiui Maiui1gt ai T ul 111 W l aiul is called mass normalized and it satisfies ll2MWl Km 2 O gt 112 WiTKWi 139 12 453 DJ Inman 57 58 Mechanical Engineering at Virginia Tech There are 3 approaches to computing mode shapes and frequencies i szuzKu ii w2u M1Ku iii 02v MKMV i Is the Generalized Symmetric Eigenvalue Problem easy for hand computations inefficient for computers ii Is the Asymmetric Eigenvalue Problem very expensive computationally iii Is the Symmetric Eigenvalue Problem the cheapest computationally DJ Inman 5858 Mechanical Engineering at Virginia Tech 47 Lagrange s Equations Defining work energy and 1 V1rtual d1splacements and i y work we will learn an wan am alternate method of deriving equations of motion Generalized coordinates 2 not 4 Recall equations 163 and 164 DJ Inman 121 Mechanical Engineering at Virginia Tech Definitions from Dynamics 1 Kinetic Energy T mr 0 r EmrTr Work Done by a force Wlgt2 2 IF dr I 1 r0 a reference position then the potential energy is VrgF39dr I 1 DJ Inman 221 Mechanical Engineering at Virginia Tech Strain Energy in a Spring Strain energy elastic potential energy foraspringF kx 0 0 1 V F d kd k2 x mm 7777 2x which is the area under the F x VS x curve F x DJ Inman 321 Mechanical Engineering at Virginia Tech Strain energy in an elastic material Stress 0 Example of a bar of cross section Ax elongated by force PXt Slope E Strain e dX UXX t dX The variation of dx denoted 5dx is given by 5dxmdx 8xtdx 8x Pxt Altxgt E80 so PEA8 4 21 Mechanical Engineering at Virginia Tech The axial stress is 039xt Strain energy continued dV 2 Pxt c lPxt8xtdx EAx8xt8xtdx EAx82x tdx Integrating yields the strain energy for axial tension in a bar element 2 V 1E Ax82xtdx 2 0 it 2E Ax ax de Mechanical Engineering at Virginia Tech DJ Inman 521 actually virtual displacement 35 Variation or Change in DJ Inman 621 A virtual displacement Based on variational math Small or infinitesimal changes compatible with constraints No time associated with change Mechanical Engineering at Virginia Tech Consequence of satisfying the constraint Constraint f r c a constant gt f r 6r 2 c Taylor expansion n 039f fr0xi 6xij c 3f5r 8r 07f 31 DJ Inman 7 21 Mechanical Engineering at Virginia Tech Virtual work Suppose the ith mass is acted on by f l with system in static equilibrium gt6W fl5rl 0gt the principle of Virtual work Z Fi 5n 0 i1 which states that if a systemis in equilibrium the work done by externally applied forces through a Virtual displacement is zero gt 6V 2 0 gt V has an critical value DJ Inman 821 Mechanical Engineering at Virginia Tech Dynamic Equilibrium D39Alembert39s Principle gt move inertia force to left side and treat dynamics as statics From Newton39s law in terms of change in momentum ZR pgtZFi p0 This allows us to use Virtual work in the dynamic case ZE p6r20 2E mi 6r0 DJ Inman 9 21 Mechanical Engineering at Virginia Tech Hamilton s Principle 1r5rr5rr5r dt i 5r51quot1quot Zf5r 21 61 Z 5 multiply bym i J gt5WZm1 5r 5T 6T5W Zm1 5r DJ Inman 1021 Mechanical Engineering at Virginia Tech Integrate this last expression It26T 5Wdt mr5rdt Lt25T5WdtZmr5r path indepence t2 t1 0gt J ZZ 5T W dl 0 for conservative forces 5W 2 5V 3 5r T V dt 0 Hamilton39s principle DJ Inman 1121 Mechanical Engineering at Virginia Tech Lagrange s Equation Let r rq1 qz q3qnt qt called generalized coordinates Let Q a generalized force or moment 4143 The Lagrange formulation derived from Hamilton39s principle for determining the equations of motion are 1 LT 8v Q dt 8 8 i DJ Inman 1221 Mechanical Engineering at Virginia Tech The Lagrangian L Let L T U called the Lagrangian Then 4145 becomes 11 0 i12n 1146 dt 8 aqi For the common case that the potential energy does not depend on the velocity 8 4 0 DJ Inman 1321 Mechanical Engineering at Virginia Tech Advantages 39 Equations contain only scalar quantities 39 One equation for each degree of freedom 39 Independent of the choice of coordinate system since the energy does not depend on coordinates 39 See examples in Section 47 pages 332335 39 Useful in situations Where F ma is not obvious DJ Inman 1421 Mechanical Engineering at Virginia Tech Example of Generalize Coordinates How many dof What are they Are there constraints m1 X19371 x5 325 6 x2 x12 y2 yl2 63 m2 W2 There are only 2 DOF and one choice is Q1261 and Q2262 DJ Inman 1521 Mechanical Engineering at Virginia Tech m p I e 4 I 7 u 3 also illustrates linear approximation method Here G is mass center and e is the distance to the elastic axis Let m denote the mass of the Wing and J the rotational inertia about G 6 Take the generalized coordinates to be x 611 W 612 49t D J Inman Called the pitch and plunge model 1621 Mechanical Engineering at Virginia Tech Computing the Energies The Kinetic Energy is 1 1 T 162 2 2 The relationship between xG and x is xG t xt 6 sin 6t gt CG t jct 6 cos 6t Ct 69 cos 6t So the kinetic energy is 1 1 T Ech e cos 62 5162 DJ Inman 1721 Mechanical Engineering at Virginia Tech Potential Energy and the Lagrangian The potential energy is U 1161362 1162192 2 2 The Lagrangian is 1 L T U lm39c e6cos 4912 11492 lk1x2 19492 2 2 2 2 DJ Inman 1821 Mechanical Engineering at Virginia Tech Computing Derivatives for Eq 1 mxe cos6 8 8x i jm me me 2 sin6 dt 85c kx 8 8x 1 Now use the Lagrange equation to get mX me cos6em92 sin6k1x 0 Likewise differentiation with respect to q2 6 yields J me cos 656 me2 cos2 69 mez 2 sin 6cos 6 lt26 0 DJ Inman 1921 Mechanical Engineering at Virginia Tech Next Linearize and write in matrix form Using the small angle approximations sinHa 6 cosHal In matrix form this becomes m me 56t k1 0 xt 0 me mezJ 9m 0 lt2 60 0 Note that this is a dynamically coupled system DJ Inman 2021 Mechanical Engineering at Virginia Tech Next consider the Single SpringMass System and compute the equation of motion using the Largranian approach Tlm5cz Ulkx2 2 2 LT Ulrm c2 lkx2 2 2 am i 83 C 8x i j 0gtmj kx0 dt 85c 8x kx DJ Inman 2121 Mechanical Engineering at Virginia Tech Section 18 Stabilitv Stability is defined for the solution of free response case Stable xt lt M V t gt O Asymptotically Stable 1imxt 0 Unstable H00 if it is not stable or asymptotically stable D J Inman 124 Mechanical Engineering at Virginia Tech Examnles 0f the quot108 If stabililv Stable Asymptotically Stable 1 1 0 0 I 0 10 71 71 gt4 3 K4 1 t t 3 4 2 W 2 rltJ 0 10 72 1 0 5 10 4 t Divergent instablllty Flutter 1nstab111ty D J Inman Mechanical Engineering at Virginia Tech 224 Examnle 181 For what values of the snring constant will the resnonse be stable Figure 137 k k kl 2 2 sin 6 mg 2 M2 2 M2 m 19 Tsm cosH mg sm zOEmE 676 mg 920 2m k 2mg60 forsmallb gt k gt 2mg for a stable response D J Inman 324 Mechanical Engineering at Virginia Tech 19 Numerical Simulation Solving differential equations dxti 1 xti1 x03 by numerical integration dz 53 At Euler RungeKutta etc At rm ti Available in Mathcad Matlab Mathematica and Maple or in FORTRAN 0 Or use Engineering Vibration 04 Toolbox E 03 Will use these to examine nonlinear vibration problems 02 that do not have analytical lt 0 l expressions for solutions 0 05 1 15 2 Time sec D J Inman 4 Q4 Mechanical Engineering at Virginia Tech solve 5ct axt X0 x0 x3 will be used to Cl1 x calculate the At 1 mg X0 05 next term lx4 04 xi Xti At ti1 ti B 03 xi1xi1aAt 02 39 01 The new value of x is calculated from 00 0 5 1 1 L the Old value Of X Time sec DJInrnan 5 Q4 Mechanical Engineering at Virginia Tech Examnle 971 solve dXdt3X XO1 a 3 take At 05 x0 1 x1 x0 053x0 5 Numerical solution Note that the n mer39cal x2 x1 03953x1 03925 Solution is diffelient ior 2 Each choice of At xt Ae u W 3xt gt ue 3Ae 239 3 Analytical solution x0x0 1Ae0 gtAlsothat xt 6 3t D J Inman 624 Mechanical Engineering at Virginia Tech Time stall With time step at 1 I I afgz ggfsolution Numerical solutionA t053ec oscillates about the x Eligirs39g tisgrllut39onAt03905sec exact solution 5 05 Large errors can be 3 caused by choosing g quotQ the time step to be E 0 39 quot v too small lt v Small time steps require more o5 computation 0 1 2 3 Time sec D J Inman 7 Q4 Mechanical Engineering at Virginia Tech Numerical SDIlllilIIl 0f the lell quot1le equation 0 vibration It is necessary to convert the second order equation into two first order equations To achieve this two new variables x and x2 are defined as follows mj 63 C kx 0 Let x1x x2 x From this two first order differential equations can be written 0 x1 x2 c k m m Called state space D J Inman 824 Mechanical Engineering at Virginia Tech Combining these first order DEs in matrix form gives X1 0 I X State K C k vector X 2 X 2 m m LY HJ State matrix X A X The Euler numerical method can then be applied to the matrix form to give 2 x01 AtAin xi1 I AtAxi xt i1 D J Inman 9 Q4 Mechanical Engineering at Virginia Tech Matlab solutions 0quot02339 and otlell Use RungeKutta More sophisticated than the Euler method but more accurate Often picks Atie if solution xt is rapidly changing At is chosen to be small and visaversa Works for nonlinear equations too Create Matlab function In the command window function Xdotsdoftx k2c1m3 AO l km cm XdotAX gtgt t00tf20 gtgt X00 025 tx0de4539sd0ft0 t x0 p10ttX D I Inman Saved as sdolm 10 2 4 Mechanical Engineering at Virginia Tech 03 i Displacement 02 A Velocity 097A Amplitude I gt I I I O1 i O2 O 5 10 15 Time sec DJInrnan 11 2 4 Mechanical Engineering at Virginia Tech Why use numerical simulation when we can compute the analytical solution and plot it To have a tool that we are confident with that will allow us to solve for the response when an analytical solution cannot be found Nonlinear systems to not have analytical solutions but can be simulated numerically D J Inman 12 2 4 Mechanical Engineering at Virginia Tech Section 110 Coulomb Friction and the Pendulum Nonlinear phenomenon in vibration analysis D J Inman 13 2 4 Mechanical Engineering at Virginia Tech Vibration of Nonlinear Systems Sliding 0r Coulomb Friction X 0 uN am gt 0 k ch C 0 3 Ct0 yN am lt 0 Nmg The force due to Coulomb friction opposes motion hence the sgn function is used The force is proportional to the normal force and independent of the velocity of the mass D J Inman 1 4 2 4 Mechanical Engineering at Virginia Tech The free body diagram split depending on the direction of motion kiti i k kw i i R i i V N Figure 141 b a mass moving 16 mass moving tight J39ct lt 0 J39ct gt 0 it f ILN writs fc My mg 3 m umg sgnOc kx 0 192 o D 1 man 1524 Mechanical Engineering at Virginia Tech IE sgn function is nonlinear Causes equation of motion to be nonlinear Can solve as piecewise linear see text Can solve numerically Has more than one equilibrium position Decay is linear rather then exponential Comes to rest when spring cannot overcome Displacement X friction at the instant the 0 5 1g 15 2 25 velocity is zero quotquot9 390 Does not near settle at x0 decay D J Inman 16 2 4 Mechanical Engineering at Virginia Tech Figure 142 shows the details of the free response of a system with Coulomb damping x3 a 2 Mon Slope Yk ZHN XUT D J Inman 17 Mechanical Engmmmg at Vugtmachh A General SBBIIIIII III IIBI SVSIBIII can I10 written as a single I SI III IIBI equation 351 x2 F X 362 fx1 x2 The equilibrium position is defined FXe 0 For Coulomb friction this is defined as m m x20and us g ltx1lt skg ie the positions where the force due to the spring can no longer overcome the sliding friction force D J Inman 18 2 4 Mechanical Engineering at Virginia Tech Example 1102 Calculating the equilibrium position for nonlinear DEs Equation of motion Equilibrium positions X l X IBZX320 x220 2 2 State space form x16 X1 1 O gt Xlx2 O i 1 X2x1 2xf1 Viol 23 5 Multiple equilibrium positions possible D J Inman 19 2 4 Mechanical Engineering at Virginia Tech The nentlulum Stable Equilibrium 9 O 27 47 Unstable Equilibrium G TC 37 57 D J Inman 2024 Mechanical Engineering at Virginia Tech quot399 Example Equilibrium of a Pendulum equilibrium Siable i equilibiium i i X 02m41r i x2o Figure 144 i e g3 51v g a b c 6t 7sin t 0 Z x2 3 g39 gtFX I X1 x2 smx1 3 Z Xlxz x20 g sinx10 XZ7SIHC1 Z 2x20and x1n7r n012L D J Inman 2W Mechanical Engineering at Virginia Tech Angle q Can use numerical simulation to examine both linear and nonlinear response Let gL012 so that 0 01 a use 0003 rad amp initial vel 03 rads b change the initial position to 00 7c rad which is near the unstable equilibrium 8 b Non linear Linear Non linear Linear 05 0 2 C E 05 1 15 0 10 20 30 40 50 0 10 20 30 40 50 Time sec Time sec D J Inman 22 2 4 Mechanical Engineering at Virginia Tech c n60 After making a single loop the pendulum cannot make a second rotation and settles to the stable equilibrium position of 647c D J Inman 2324 15 Mechanical Engineering at Virginia Tech 9 f Friction loss oauses slogw decay 0 2o 40 60 80 Time sec Summary 039 Nonlinear Vibrations Additional phenomena over linear case Multiple equilibrium lnstabilities possible with positive coefficients Form of response dependent on initial conditions Closed form solutions usually not available Can simulate numerically Linear model has tremendous advantages Linear combination of inputs yields linear combination of outputs Linear ode techniques very powerful But don t make a design error by ignoring important nonlinear sHua ons All systems have nonlinear ranges of operation Need to sort out when nonlinearity is important to consider and when to ignore it D J Inman 2 4 2 4 Mechanical Engineering at Virginia Tech Engineering Vibration 3rd Edition Prentice Hall D J Inman dinmanvtedu We make our living in dynamics vibration and control Virginia Tech 143 D J Inman MUbFESE EWfDW D Hrgm iZWM K gt g 9 OPENED AND CLOSED WITHIN A VIEW HOURS Virgini ech BECAUSE OF UNDESIRABLE VIBRATION D J Inman 243 Show movies likooleling and of The bridge has many degrees of freedom we will start with one and work towards many Lack of consideration of dynamic loads and Vibration caused this to new bridge to vibrate wildly Virginia Tech The goal of this course is to understand such D J Inman phenomenon and how to prevent it 343 Example 111 The Pendulum Sketch the structure or part of interest Write down all the forces and make a free body diagram m Use Newton s Law andor Euler s Law to find the equations of motion 2M0 Joa JO 2 ml 2 Virginia Tech 443 D J Inman The problem is one dimensional hence a scalar equation results Joaa mgl sin 60 gt m z a mg sin 00 o restoring force Here the over dots denote differentiation with respect to time t This is a second order nonlinear ordinary differential equation We can linearize the equation by using the approximation sin6 z 9 gt mzz m mgwm 0 gt 9390 0m 0 Requires knowledge of 60 and 90 the initial position and velocity V1rg1n1a Tech 543 D J Inman Next consider a spring mass system and perform a static J 3 2 39 noanear experiment 2 i 1 0 From strength of ID i materials recall 7 lrl j W FBD xxxx39xx 39 g f1 Ilu fa I m i J 1 m 1mm l39 39g A plot of force versus displacement l Hnear I Vir inia Tech eX eriment gt 2 CC J Inman p f lem Freebody diagram and equation of motion H 9 0 y Lat Frictionfree k surface In ikx TN Rest position I Newton s Law mj l kxz 3 MO kxl o x0 x039c0 v0 392 VirginiaTech Again a 2nd order ordinary differential equation 743 Stiffness and Mass Vibration is cause by the interaction of two different forces one related to position stiffness and one related to acceleration mass Stiffness k Proportional to displacement Displacement ckxt gtX statics k Mass m m f 2 ma 2 mxa Mass Spring dynamics Proportional to acceleration Virginia Tech 843 D J Inman Examples of SingleDegreeof Freedom Systems Pendulum Shaft and Disk 333 Torsional l ength Stiffness k Moment Gravity g of inertia I e V an mn0 I J a k6t 0 Virginia Tech 943 D J Inman Solution to 2nd order DEs Lets assume a solution xt A sinant Differentiating twice gives X t x0 wnA 0050 56t afA sinant afxt Substituting back into the equations of motion gives maA sinant kA sinant 0 k mak0 or a Natural m frequency Virginia Tech radS 1043 D J Inman Summary of simple harmonic motion Xt A A 4 Period T Slope here is v0 27r a 1 1 Amplitude RA quot Z 27 radcycle Virginia Tech D J Inman 0 rads 0 cycles 2 0 Hz 275 Maximum Velocity wnA 27 1143 Initial Conditions If a system is vibrating then we must assume that something must have in the past transferred energy into to the system and caused it to move For example the mass could have been moved a distance x0 and then released at t 0 Le given Potential energy or given an initial velocity v0 ie given Kinetic energy or Some combination of the two above cases From our earlier solution we know that x0 x0 A sinan 0 A Sin v0 jg0 wnA cosan0 114 COS Virginia Tech 1243 D J Inman Initial Conditions Solving these equation gives 1 2 2 2 1 wx n 0 A twnx0v0 tan a v0 J Amplitude P112563 Slope here is v0 I 715 F V V a n V Virginia Tech wn 1343 D J Inman The total solution for the spring mass system is 2 2 2 lwnxo v x0 xt 0 sinant tan 1 110 n Called the solution to a simple harmonic oscillator and describes oscillatory motion or simple harmonic motion Note Example 112 2 2 2 x mnxo v0 mnxo x a l 2 2 2 0 Virginia Tech as 39t 53910quot 1443 D J Inman A note on arctangents Note that calculating arctangent from a calculator requires some attention First all machines work in radians The argument atanI is in a different quadrant then atanl and usual machine calculations will return an arctangent in between 42 and 1tl2 reading only the atan for both of the above two cases e In MATLAB use the atan2xy function to get Virginia Tech the correct phase D J Inman Example 113 wheel tire suspension m30 kg con 10 hz what is k k ma 30 kgIOCy lceMj 1184x105 Nm sec cylce There are of course more complex models of suspension systems and these appear latter in the course as our tools develope Virginia Tech 1643 D J Inman Section 12 Harmonic Motion The period is the time elapsed to complete one complete cylce T 27rrad 21 111 a rads a The natural frequency in the commonly used units of hertz fn an 2 an rads 0 cycles 2 HZ 112 27r 27r radcycle 27r s 27r aquot 2F rads T 27rI s g For the disk and shaft aquot JE rads T 27rZ s J k Virginia Tech 1743 D J Inman For the pendulum Relationship between Displacement Velocity and Acceleration A1 012 Displacement 1 xt A sinant gtlt 0 1 0 01 02 03 04 05 06 07 0 9 1 Velocity 2 xt anAcosant gt 0 2c 0 01 02 03 04 05 06 07 08 09 1 Acceleratlon 20C A Xt ajAsinant m 0F 200 0 01 03 04 05 06 07 Time sec Note how the relative magnitude of each increases for 00ngt1 02 Virginia Tech D J Inman 08 09 1 1843 Exam 1 21 Hardware store spring bolt m 492x103 kg k8578 Nm and X0 10 mm Compute on and the max amplitude of vibration Note common k Units are Hertz mn1 3132rads m 492 x 10 kg a To avoid Costly errors use fn f n HZ when working in Hertz and a n 72 when in rads 27 1 T 00476 s 1 m cyles n J 21 sec 1 O xtmax A 1ax 0 2x0 10 mm 017 Virginia Tech Units depend on system 1943 D J Inman Compute the solution and max velocity and acceleration 121 2 on1 1320 mms 132 ms 292 mph atmaX ij 17424 gtlt103 mms2 17424 Ins2 z178g g 98 mls2 tan 1 z rad 0 2 xt 10 sin132t 7r2 10 0031321 mm 04 in max Virginia Tech 2043 D J Inman Does graVIty matter in spring problems r g L Let A be the deflection caused by 39 in hanging a mass on a spring g A x1xo in the figure Then from static equilibrium mg 2 CA Next sum the forces in the vertical for some point xgt x1 measured from A mx kxAmg kxmg kA 0 3 mj t kxt 0 So no gravity does not have an effect on the vibration 2143 V rgm aTSCh note that this is not the case irthe spring is nonlinear Example 122 Pendulums and measuring 9 0 A 2 m pendulum swings with a period 4 712 I 472392 2 m Of S g T2 28932 82 What IS the gt g 9434 ms2 acceleration due to gravity at that location This is gin Denver CO USA at 1638m and a latitude of 40 Virginia Tech 2243 D J Inman Review of Complex Numbers and Complex Exponential See Appendix A A complex number can be written with a real and imaginary part or as a complex exponential c a jb A616 Where 3 aAcos bAsin6 a Multiplying two complex numbers 6162 AlAzeWIWZ b Dividing two complex numbers ER i A1 6191 92 62 A2 Virginia Tech 2343 D J Inman Equivalent Solutions to 2nd order Differential Equations see Window14 All of the following solutions are equivalent xt A sinant Called the magnitude and phase form 2 A1 Sinwnt A2 COS wnt Sometimes called the Cartesian form 351 alejwnt aze 39wnt Called the polar form The relationships between A and 4 A1 and A2 and a1 and a2 can be found in Window 14 of the course text page 17 Each is useful in different situations Each represents the same information VirgmiaTech Each solves the equation of motIon 2443 D J Inman Derivation of the solution Substitute xt 616 into mjc39 kx O 3 mtzae f kale O 3 m12kO Azi iEjiconj3 m m xt alewn and xt 2 age quot17 3 xt alewn 6126 I 118 This approach will be used again for more complicated problems Virginia Tech 2543 D J Inman ls frequency always positive From the preceding analysis k 1 con then ath anjt xt ale aze Using the Euler relations for trigonometric functions the above solution can be written as recall Window 14 xt A sinant 119 It is in this form that we identify as the natural frequency on and this is positive the 1 sign being used up in the transformation from exponentials to the sine function Virginia Tech 2643 D J Inman Calculating RMS May need to be limited due to physical constraints Not very useful since for a sine function the 1 T average value is zero 7c 2 lim Ixtdt average value T gtoo T 0 A 2 peak value 1 T x2 lIm E Ixz 0dr meansquare value 121 0 Proportional Z t xrms x root mean square value 0 energy Also useful when the vibration is random Virginia Tech 2743 D J Inman The Decibel or dB scale It is often useful to use a logarithmic scale to plot vibration levels or noise levels One such scale is called the decibel or dB scale The dB scale is always relative to some reference value x0 It is define as 2 x N K x dB 2 1010g10 2010g10 122 x x 0 0 For example if an acceleration value was 196mls2 then relative to 19 or 98mlsz the level would be 6dB 196 2 lolog10 2010g102 6dB Or for Example 121 The Acceleration Magnitude Virginia Tech is 20log1o17825dB relative to 19 2843 D J Inman 13 Viscous Damping All real systems dissipate energy when they vibrate To account for this we must consider damping The most simple form of damping from a mathematical point of view is called viscous damping A viscous damper or dashpot produces a force that is proportional to velocity Mostly a mathematically motivated form allowi g n a solution to the resulting equations of motion that predicts reasonable observed amounts of energy dissipation Damper c f cvt cxt Mmmnng pm Virginia Tech 2943 o D J Inman Differential Equation Including Dam ping For this damped single degree offreedom system the force acting on the mass is due to the spring and the dashpot iefm 1 5 Displacement mscz 7km 7 5361 l or m5c39l mac kxz o 125 k To solve this for of the equation it is useful to assume a solution of the form it xt 2 ae 3W4 VirgmaTenh D l Inman Solution to DE with damping included dates to 1743 by Euler The velocity and acceleration can then be calculated as 5cm 2 we 56t 2 be If this is substituted into the equation of motion we get ae2U ml2 61 k 0 126 Divide equation by m substitute for natural frequency and assume a nontrivial solution ae liO gt 12Cymaj20 Virginia Tech 3143 D J Inman Solution to DE with Damping Included For convenience we will define a term known as the damping ratio as C Z2M The equation of motion then becomes 2 2 1 2 wnlwn 0 Solving for k then gives 212 an tam4 2 1 131 Virginia Tech 3243 D J Inman 1 Lower case Greek zeta Possibility 1 Critically damped motion Critical damping occurs when C1 The damping coefficient c in this case is given by 1 6 2 CW 2 2km 2 2mm Er definition of critical Solving for A then gives damping Oef ciem 212 1an iwnxilz an A repeated real root The solution then takes the form xt ale W age W Needs two independent solutions hence the t in the second term Virginia Tech 3343 D J Inman Critically damped motion a and a2 can be calculated from initial conditions t0 t x 2 621 azte 0 quot k225Nm m100kg and Q1 Z a x 06 y y 1 0 x004mm v01mmls 0 t v ana1 wnazt a2 6 n x004mm v00mms g 04 x x0 04mm v0 1mms V0 2 ana1 612 g 03 E a Q S o No oscHIatlon occurs a 01 x no Useful In door 0 3939r mechanisms analog O1 gauges 39 o 1 3 Time sec Virginia Tech 3443 D J Inman Possibility 2 Overdamped motion An overdamped case occurs when Qgt1 Both of the roots of the equation are again real k225Nm m100kg andz2 2 06 r l X12 an i mn l 1 x004mm v01mmls 05 X O4mmv Omms 1 2 l 2 0 0 e wntale wnt g 1 azewnl g 1 04 quotI XOOI4mm VO1mms v03quot s a1 and a2 can again be calculated from initial conditions t0 02 Displacement mm I I I 39l 01 u V0 VC2 1anx0 zen2 1 01 0 1 2 3 a V0 2 1anx0 Timesec 2 2 zwnJ 1 Slower to respond than VirginiaTech critically damped case 3543 D J Inman Possibility 3 Underdamped motion An underdamped case occurs when lt1 The roots of the equation are complex conjugate pairs This is the most common case and the only one that yields oscillation 112 waniwnjv1 quot2 W e nvaleW gag WW Ae ga sin mdt The frequency of oscillation Dd is called the damped natural frequency is given by ad 2 0541 52 137 Virginia Tech 3643 D J Inman Underdamped motion A and 45 can be calculated from initial conditions t0 A iNVo Cwnxo2 xow z we X a tan1 0 d V0 wnx0 Gives an oscillating response with exponential decay Most natural systems vibrate with and underdamped response See Window 15 for details and other representations 39 o 1 2 3 4 5 Virginia Tech Time 590 3743 D J Inman Displacement Exam 131 consider the spring of 121 if c 011 kgs determine the damping ratio of the springbolt system m 492 x 10 3 kg k 8578 Nm ca 2 24492 gtlt103 x8578 12993 kgs c 011 kgs 2 12993 kgs the motion is underdamped and the bolt will oscillate Virginia Tech 3843 D J lnman 00085 gt Example 132 The human leg has a measured natural frequency of around 20 Hz when in its rigid knee locked position in the longitudinal direction ie along the length of the bone with a damping ratio of g 0224 Calculate the response of the tip if the leg bone to an initial velocity of v0 06 mls and zero initial displacement this would correspond to the vibration induced while landing on your feet with your knees locked form a height of 18 mm and plot the response What is the maximum acceleration experienced by the leg assuming no damping Virginia Tech 3943 D J Inman Solution 6 EMM rad 12566 rads 1 5 cycles wd 212566 ll 2242 122467 rads 06 0224 12566 0 2 0 1224672 AJ lt x x X 2000511 122467 tan391 4M 0quot j 0 v0 Cwn0 gt xt 000562814 sin122467t Virginia Tech 4043 D J Inman Use the undamped formula to get maximum acceleration 1 1 2 A x3 LO 0 12566v0 06x0 0 a v0 06 an an 0612566 W52 75396 ms2 w E a 75396 ms2 981 ms2 maximum acceleration g 768 g39 s Virginia Tech 4143 D J Inman Here is a plot of the displacement response versus time Vn glma Tech 4243 1 3 3 Compute the form of the response of an underdamped system using the Cartesian form of the solution given in Window 15 sinx y sinx siny cosx cosy gt xt Ae W sina dt 5 e gm A1 sin wdt A2 cos wdt x0 x0 60141 sin0 A2 c0s0 gt w 35 a ne mquottA1 sin wdt A2 cos wdt wde w A1 cos wdt A2 sin wdt v0 an A1 sin 0 x0 cosO adA1 cos 0 x0 sin 0 gt A1 v0 wnx0 wd V a X xt e 92quot s1n wdt x0 cos 0dr d Virginia Tech 4343 D J Inman Section14 Modeling and Energy Methods An alternative way to determine the equation of motion and an alternative way to calculate the natural frequency of a system Usefu if the forces or torques acting on the object or mechanical part are difficult to determine Virginia Tech 144 D J Inman Potential and Kinetic Energy The potential energy of mechanical systems U is often stored in springs I remember that for a spring Fkx I k I M I I x0 x0 i x0 x0 1 Um Fdx kxabc2 kx2 p g i ii 2 0 Mass Spring The kinetic energy of mechanical systems Tis due to the motion of the mass in the system trans lmxz Tim 2 2 Virginia Tech 244 D J Inman Conservation of Energy For a simple conservative ie no damper mass spring system the energy must be conserved T U constant d or T U 0 dt At two different times t1 and t2the increase in potential energy must be equal to a decrease in kinetic energy or visa versa U1 U2 2 T2 T1 and Umax Tmax Virginia Tech 344 D J Inman Deriving the equation of motion from the energy x0 x I I M I I Mass iTUi mjc2lkx2 dt dt gt 5c kx 0 Since 5c cannot be zero for all time then mj kx 0 Virginia Tech 444 D J Inman Determining the Natural frequency directly from the energy If the solution is given by xt Asinat then the maximum potential and kinetic energies can be used to calculate the natural frequency of the system 1 1 Um kA2 T manA2 2 2 max Since these two values must be equal 1 1 CA2 2 mnA2 2 2 k gtkm0an m Virginia Tech 544 D J Inman Example 141 Compute the natural frequency of this roller fixed in place by a spring Assume it is a conservative system ie no losses and rolls with out slipping Tmzlm2 and T lmic2 2 trans 2 Virginia Tech 644 D I Inman Solution continued 1 392 xrt9gt39cri9gtTROt EJx 2 r The max value of T happens at vma1X conA 2 gt Tmax llmanA2 l mi 0142 2 2 2 r The max value of U happens at chX A 1 gt Um EkAZ Thus Tm Um gt l mi wazzlkAzza 2 2 r2 quot r2 Effective mass Virginia Tech 744 D J Inman Example 142 Determine the equation of motion of the pendulum using energy J 2 mW mg Virginia Tech 844 D J Inman Now write down the energy T 11092 lmzzaz 2 2 U mg 1 cos 6 the Change in elevation is E 1 cos 6 d d 1 TU 2 mE262m E 1 0086 0 dtlt gt AZ glt gtj Virginia Tech 944 D J Inman m z mg sin 99 O gt 9m62 mg6sin 6 O gt m z mg sin 6 O gt 650 sm 90 0 gt r6r 0 s on Virginia Tech 1044 D J Inman Example 1 The effect of including the mass of the spring on the value of the frequency y dy ms k x t Virginia Tech 1144 D J Inman ms d E y mass of element dy assumptions velocity of element dy vdy ct 1 g ms y 2 Tsprmg E x dy adds up the KB of each element gt a This provides some m sumple deSIgn and modeling guides Virginia Tech 1244 D J Inman What about gravity kA mg kA 0 from FBD and static equilibrium T lt gt k 0 Xt mg Uspring l 302 m T A 2 Xt Ugmv mgx T l mic2 2 Virginia Tech 1344 D J Inman NOW use iTUO dt 31 lmJ39CZ mgxlkAx2 0 dt 2 2 gt mJ39CJ39 mgxkAxjc gt J39cm39c39kx 5ckA mg O r 0 from static equilibiurm mx39 kx O Gravity does not effect the equation of motion or the natural frequency of the system for a linear system as shown previously with a Virginia Tech force balance 1444 D J Inman Lagrange s Method for deriving equations of motion Again consider a conservative system and its energy It can be shown that if the Lagrangian L is defined as L T U Then the equations of motion can be calculated from d BL BL 0 163 dt aq aq Which becomes d 8T 8T aU 0 164 dt aq aq aq Virginia Tech 1544 D JInman Here q is a generalized coordinate Example 147 Derive the equation of motion of a spring mass system via the Lagrangian Tzlmjc2 and Uzlkx2 2 2 Here q x and and the Lagrangian becomes LT U lnv39c2 lkx2 2 2 Equation 164 becomes 1alj a Ta UimX Okx0 dt 8x 8x 8x dt gtm39c39kx0 Virginia Tech 1644 D J Inman 15 More on springs and stiffness Longitudinal motion A is the cross sectional area m2 5 EA Eis the elastic modulus PaNm2 E is the length m k is the stiffness Nm T W Virginia Tech 1744 D J Inman Figure 121 Torsional Stiffness Jp is the polar moment of inertia of the rod J is the mass moment of inertia of the disk G is the shear modulusj is the length Virginia Tech 1844 D J Inman 1 5 1 compute the frequency of a shaftmass system J 05 kg m2 From Equation 150 2M J 2 J t k6t 0 91 2 g r 0 Figure 122 k I G1 p 7rd4 3 aquot J J J p 32 For a 2 In steel shaft diameter 0f 05 cm 3 GJP Z 8X1010Nm27r0 5gtlt10 2m432 1 2 m0 5kgm2 22 rads a Vilginia Tech 1944 DA 1 Inman Fig 122 Helical Spring d diameter of wire 2R diameter of turns n number of turns Xt end deflection G shear modulus of i diameter of spring material W 4 2R diameter of turns k 7 0d a n number of turns 64 X de ection 4 64 nR3 Virginia Tech 2044 D J Inman Allows the design of springs k to have specific stiffness Fig 123 Transverse beam stiffness Strength of f materials and experiments yield I m 3E1 k 13 With a mass at the tip 3E 60 3 m Virginia Tech 2144 D J Inman Samples of Vibrating Systems Deflection of continuum beams plates bars etc such as airplane wings truck chassis disc drives circuit boards Shaft rotation Rolling ships See text for more examples Virginia Tech 2244 D J Inman Example 152 Effect of fuel on frequency of an airplane wing Model wing as transverse beam Model fuel as tip mass Ignore the mass of the wing and see how the frequency of the system E I m l T changes as the fuel is 6 Xl used up Virginia Tech 2344 D J Inman Mass of pod 10 kg empty 1000 kg full 52X10395 m4 E69X109 Nm 2 m Hencethe natural 0 3E1 369gtlt10952gtlt10 5 frequency fun my 100023 Changes by aquot 2116 rad218 Hz order of magnitude 0 3EI 369gtlt10952gtlt10 5 while it W m 1023 empties out 2115 rad2185 Hz fuel This ignores the mass of the wing Virginia Tech 2444 D J Inman Example 153 Rolling motion of a ship 14 2 WE Wh sin 490 For small angles this becomes 1650 Wh6t 0 kW 360 J Virginia Tech 2544 D J Inman Combining Springs Springs are usually only available in limited stiffness values Combing them allows other values to be obtained A k 3 k2 C Equivalent Spring CW k 1 k1 senes AC a b m k1 k2 k2 parallel kab k1 k2 This is identical to the combination of capacitors in electrical circuits Virginia Tech 2644 D J Inman Use these to design from available pans Discrete springs available in standard values Dynamic requirements require specific frequencies Mass is often fixed or i small amount Use spring combinations to adjust wn Check static deflection Virginia Tech 2744 D J Inman Example 1 Design of a spring mass system using available springs series vs parallel Letm10kg Compare a series and parallel combination a 1 1000 Nm k 3000 Nm k R4 0 b k3 1000 Nm k 3000 Virginia Tech 2844 D J Inman Case a parallel connection k3 2 k4 0 km 2 k1 k2 10003000 4000 Nni Ike 4 gt wpmlle g 20 rads m 10 Case b series connection k k Ok 1 3000 1 2 1 1k31k4 31 lk gt wseries eg 7 50 rads m 10 Same physical components very different frequency Allows some design flexibility in using off the shelf components 750 Nm Virginia Tech 2944 D J Inman Exam ple Find the equivalent stiffness kof the following system Fig 126 page 47 k3k4 k3 k4 k1k2k5 k1k2k5 Virginia Tech a k1k3 k2k3 k5k3 klk4 k2k4 k5k4 k3k4 D JInman mk3 k4 3044 Example 1 55 Compare the natural frequency of two springs connected to a mass in parallel with two in series A series connect of k1 1000 Nm and k2 3000 Nm with m 10 kg yields 1 kg 750 Nm 3 750 Mm 866 rads q 1100013000 10 kg A parallel connect of k1 1000 Nm and k2 3000 Nm with m 10 kg yields 4 keg 1000 Nm 3000 Nm 4000 Nm 3 60W 2 20 rads g Same components very different frequency 3144 Virginia Tech D J Inman Static Deflection Another important consideration in designing with springs is the static deflection mg k This determines how much a spring compresses or sags due to the static mass you can see this when you jack your car up AkmggtA The other concern is rattle space which is the maximum deflection A Virginia Tech 3244 D J Inman Section 16 Measurement 39 Mass usually pretty easy to measure using a balance a static experiment 39 Stiffness again can be measured statically by using a simple displacement measurement and knowing the applied force 0 Damping can only be measured dynamically Virginia Tech 3344 D J Inman Measuring moments of inertia using a Trifilar suspension system 2 2 gT r0 m0 m Suspensionwites Z J Dfleug tl 472 0 J Disk of knovm moment 10 mass mm and radius r0 Tis the measured period 9 is the acceleration due to gravity Vllginia Tech 3444 D J Inman Stiffness Measurements From Static Deflection I Nonlinear Linear Fkx oroEe F gtk x Force or stress Deflection or strain From Dynamic Frequency UnE2gtkmw m Virginia Tech 3544 D J Inman Example 161 Use the beam stiffness equation t comnute the modulus of a material 71 E elastic muziuhis 311 13910 k 72 w I length of beam I moment of inertia of cross sec oual am about the neutzai axis Figure 124 t 1 m m 6 kg 1 10399m4 and measured T 062 s 3 T 2er 062 s 3E1 3 4 6k 1 3 2E 4 le M 205x10 Nm2 3T 1 3062 s 10 m 3644 Virginia Tech Damping Measurement Dynamic only Define the Logarithmic Decrement ICU n 171 xt T Ae W 3mm t 5 In g HT d 172 Ae Sll 1adt wdT 5 wnT c 5 5 Z 4 262 175 Virginia Tech 3744 D J Inman Section 17 Design Considerations Using the analysis so far to guide the selection of components Virginia Tech 3844 D J Inman Example 171 Mass2 kggmg3 kg and kg 200 Nm For a possible frequency range of 816 rads g on g 10 rads For initial conditions X0 0 v0 lt 300 mms Choose a c so response is always 5 25 mm Virginia Tech 3944 D J Inman Solution Write down Xt for 0 initial displacement Look for max amplitude Occurs at time of first 0395 peak T max Compute the Amplitude O amplitude at T o5 max Compute C for A Tmax039025 3910 05 1 15 Timesec Virginia Tech 4044 D J Inman V xt 0e W s1nwdt wd Amplitude 3 worst case happens at smallest 60d 3 a 816 rads 3 worst case happens at max v0 300 mms With a and v0 fixed at these values investigate how varies with 4 First peak is highest and occurs at 210m 0 gt code W cosadt mne fwnt sinadt 0 t 1 a 1 1 2 Solve for t Tmax 3 Tm tan1 d tan1 J 60 5 d n cod 4 4V 71 JP v0 m a 1 1 2 Sub Tmax 1nto xt Am xTm e s1ntan own 2 g 4 71E A 4V Oe MW 6 m a Virginia Tech 4144 D J Inman To keep the max value less then 0025 m solve AmaXM 0025 3 g 0281 Using the upper limit on the mass m 3 kg yields 0 2mwn 238160281 1415 kgs FYI g 0 yields Am V 0 37 mm a 71 Virginia Tech 4244 D J Inman Exam pie 1 73 What happens to a good design when some one changes the parameters Car suspension system How does 3 change with mass Given 4 1 m 1361 kg A 005 m compute c k mg wnEgtk1361wj mgzkAjkT m 2 can J 93981 14 rads 2 mA 005 k 1361142 2668x105 Nm 4 1 c 2me 2136114 381x104 kgs Virginia Tech 4344 D J Inman Now add 290 kg of passengers and luggage What happens m 1361290 1651kg A mg 165198 zOOO6m 2668x105 gt k gt a 2F 2127 rads A 006 4 g 2 381gtlt10 0 2mm So some oscillation results at a lower frequency Virginia Tech 4444 D J Inman 0 cr