SOLID MECHANICS DYNAMICS TUTORIAL DAMPED VIBRATIONS. On completion of this tutorial you should be able to do the following.

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1 SOLID MECHANICS DYNAMICS TUTORIAL DAMPED VIBRATIONS This work overs elemets of the syllabus for the Egieerig Couil Eam D5 Dyamis of Mehaial Systems, C05 Mehaial ad Strutural Egieerig ad the Edeel HNC/D module Mehaial Siee. O ompletio of this tutorial you should be able to do the followig. Defie a free damped osillatio. Eplai the purpose of dampig. Defie dampig oeffiiet. Defie dampig ratio. Derive formulae that desribe damped vibratios. Determie the atural frequey ad periodi time for damped systems. Defie amplitude redutio fator. Calulate dampig oeffiiets from observatios of amplitude. This tutorial overs the theory of atural vibratios with dampig ad otiues the studies i the tutorial o free vibratios. To do the tutorial fully you must be familiar with the followig oepts. Simple harmoi motio. Free vibratios.

2 INTRODUCTION I the tutorial o free vibratios it was eplaied that oe set vibratig, the system would arry o osillatig for ever beause the eergy put ito the system by the iitial disturbae aot get out of the system. I reality, the osillatio always dies away with time beause some form of fritio is preset. The diagram shows a displaemet time graph of a typial damped osillatio. Figure Fritio dissipates the eergy as heat. Suh vibratios are alled free ad damped. The type of fritio that is easiest to deal with mathematially is that reated by a dashpot (also alled a damper). Let us ow eamie two kids of dashpots. DASHPOTS Here we will eamie dashpots for liear motio. The diagram shows two types of dashpot, air ad oil. AIR-The pisto moves iside the ylider ad pumps or suks air through the orifie. Beause of the restritio, pressure is eeded to make the air flow through the orifie ad this pressure will produe a fore opposig the motio. Ofte the pisto is replaed with a simple diaphragm. OIL-The oil is otaied i the ylider ad motio of the pisto pushes the oil through restritors i the pisto to the other side. Agai, pressure is eeded to fore the fluid through the restritor ad this produes a fore opposig motio. It a be show that for both ases, the fore opposig motio (the dampig fore) is diretly proportioal to the veloity of the pisto. The equatio for this fore is as follows. d F d ostat veloity Figure The ostat of proportioality is alled the dampig oeffiiet ad has uits of N s/m. The value of depeds o the size of the restritio. D.J.Du

3 . DAMPED LINEAR VIBRATIONS Cosider a mass suspeded o a sprig with the dashpot betwee the mass ad the support. Figure 3 If a fore F is applied to the mass as show, it is opposed by three fores. These are: The iertia fore give by d Fi Mass aeleratio M The dampig fore give by d F d ostat veloity The sprig fore give by F s = k The total fore is the F F F F i d s d d F M k Beause the vibratio is free, the applied fore must be zero (e.g. whe you let go of it). d d 0 M k ad this is a liear seod order differetial equatio ad it is muh disussed i most maths books. We make the followig hages. First divide eah term by k. M d d 0 k k D.J.Du 3

4 I the work o udamped vibratios it was show that without a dashpot a atural osillatio ours with agular frequey = (k/m) rad/s. We may replae (k/m) with. 0 d d k Net we eamie the term /k. We eed to start with a defiitio. We will be usig a term alled the ritial dampig oeffiiet ad it is defied as follows. 4Mk This has the same uits as the dampig oeffiiet (N s/m). The ratio / is alled the dampig ratio ad this is defied as follows. δ This may ow be developed as follows. δ k δ 4Mk Mk k k M k We a ow replae the term /k i our equatio. 0 d δ d 0 δ d d ad fially multiply eah term by This is ow a stadard equatio ad the solutio may be foud i stadard tet as = Aeat + Be-bt A ad B are ostats of itegratio ad a = {(-) - } b = {(-) + } The resultig graph of displaemet with time t depeds upo the dampig ratio ad hee the value of. Three importat ases should be osidered. D.J.Du 4

5 .. UNDER DAMPED This ours whe < ad <. If we assume that t = 0 ad = C at the momet the mass is released we get a deayig osiusoidal osillatio as show. The displaemet is desribed by the followig equatio. δ t Ce ost The graph shows the result if the mass is pulled dow 0 uits ad released. I this ase C = -0. Figure 4 The atural udamped agular frequey is = (k/m) ½. The damped frequey is = (- ). The damped frequey is f = / ad the periodi time of the damped agular osillatio is T = /f = / AMPLITUDE REDUCTION FACTOR Cosider two osillatios, oe ourrig m yles after the first. - δ t The amplitude of the first osillatio at time t is Ce The amplitude of the osillatio that ours ' m' - δ t Ce δmt The ratio is e AM PLITUDEREDUCTIONFACTOR - δ (t mt) Ce π Now eamie the term δmt. Put T ad this beomes π δ m π δ m Now substitute ad δmt hee π δ m e ad et we take the atural log to get rid of the epoetial so l - Ce δ (t mt) π δ m yles later ours ad this is probably a more useful epressio mt seods later so to use. D.J.Du 5

6 .. CRITICALLY DAMPED. If the dampig is ireased, the osillatios die away quiker ad evetually a ritial poit is reahed where the mass just returs to the rest positio with o overshoot or osillatio. This ours whe = ad =. The result is a epoetial deay as show..3. OVERDAMPED This ours whe > ad >. The result is a epoetial deay with o osillatios but it will take loger to reah the rest positio tha with ritial dampig. Figure 5 D.J.Du 6

7 WORKED EXAMPLE No. A vibratig system is aalysed ad it is foud that two suessive osillatios have amplitudes of 3 mm ad 0.5 mm respetively. Calulate the amplitude redutio fator ad the dampig ratio. SOLUTION For suessive amplitudes m = 3 l l l6.79 amplitude redutio fator π δ square both sides δ δ ad δ so δ ad δ D.J.Du 7

8 WORKED EXAMPLE No. A mass of 5 kg is suspeded o a sprig ad set osillatig. It is observed that the amplitude redues to 5% of its iitial value after osillatios. It takes 0.5 seods to do them. Calulate the followig. i. The dampig ratio. ii. The atural frequey. iii. The atual frequey. iv. The sprig stiffess. v. The ritial dampig oeffiiet. vi. The atual dampig oeffiiet. SOLUTION For osillatios m = ad / = 00/5 = 0 π δ m l π δ l ad et square both sides (4 π δ ) 57.9 δ 7.59δ 8.59 δ The atural frequey is f The damped frequey is give by f k k f so 4 simplify ad 8 π M π 5 The ritial dampig oeffiie t is give by δ (.996) Hz 4Mk δ 0.3 f Ns/m The atual dampig oeffiie t is the umber of osillatios time take k so k 358 N/m Hz δ Ns/m D.J.Du 8

9 WORKED EXAMPLE No.3 A mass of 5 kg is suspeded o a sprig of stiffess 4000 N/m. The system is fitted with a damper with a dampig ratio of 0.. The mass is pulled dow 50 mm ad released. Calulate the followig. i. The damped frequey. ii. The displaemet, veloity ad aeleratio after 0.3 seods. f f f v a SOLUTION You will eed to be able to do advaed differetiatio i order to follow this solutio. k M 4.5 Hz π Ce d dv δ -δ t v 50si ACCELERATION a 50os a 5554 mm/s δ VELOCITY 50os os si 0.3 t -50e os7.7t ( 0.443) 4.07 mm -δt te 50ost os8.33 v mm/s 50os δ t -δ t t e 00sit δ e 50ost 8.33(7.7 )(0.83) 00si (0.) (8.8) (0.83) a rad/s Hz (Aswer i) The iitial displaemet is dow 50 mm so C -50 mm 7.7rad/s δ e δ e -δ t -δ t δ -δ t (7.7)(0.)(8.8)(0.83) e D.J.Du 9

10 SELF ASSESSMENT EXERCISE No.. A mass of 50 kg is suspeded from a sprig of stiffess 0 kn/m. It is set osillatig ad it is observed that two suessive osillatios have amplitudes of 0 mm ad mm. Determie the followig.. The dampig ratio. (0.88). The dampig oeffiiet. (54.5 N s/m). The true frequey. (. Hz). A mass of 5 kg is suspeded from a sprig of stiffess 46 kn/m. A dashpot is fitted betwee the mass ad the support with a dampig ratio of 0.3. Calulate the followig.. The udamped frequey. (5.6 Hz). The damped frequey. (4.56 Hz). The amplitude redutio fator. (.976). The ritial dampig oeffiiet. (959 N s/m) 3. A mass of 30 kg is supported o a sprig of stiffess N/m. The system is damped ad the dampig ratio is 0.4. The mass is raised 5 mm ad the released. Calulate the followig. i. The damped frequey. (6.53 Hz) ii. The displaemet, veloity ad aeleratio after 0. s. (-4.8 mm, 366 mm/s ad mm/s) D.J.Du 0

11 3. TORSIONAL OSCILLATIONS The theory is the same for torsioal osillatios. the liear quatities are replaed with agular quatities i the formula. These are Mass m Fore F Distae Veloity v Aeleratio a d momet of iertia I Torque T agle agular veloity agular aeleratio Figure 6 A dashpot for a torsioal system would osist of a vae whih rotates iside a pot of oil. The dampig torque is diretly proportioal to the agular veloity suh that Dampig Torque = is the torsioal dampig oeffiiet with uits of N m s/radia. The torsioal sprig stiffess is torque per uit agle so the uits of k are Nm/radia. The ritial dampig oeffiiet is = (4 I k) The atural frequey is = (k/i) Remember that for a shaft the torsioal stiffess k t = GJ/L The polar seod momet of area for a irular setio is D 4 /3 The atual agular frequey is = (- ) The amplitude redutio fator is displaemet. π δ m l where is the agular D.J.Du

12 WORKED EXAMPLE No.4 A horizotal shaft is fied at both eds ad arries a flywheel at the middle. The shaft is m log either side of the flywheel ad is 0 mm diameter. The flywheel has a momet of iertia of.9 kg m. The system has proportioal dampig ad it is observed that the amplitude redues by 60% after oe osillatio. The shaft material has a modulus of rigidity of 90 GPa. Calulate the followig. i. The dampig ratio. ii. The atural frequey. iii. The atual frequey. iv. The sprig stiffess. v. The ritial dampig oeffiiet. vi. The atual dampig oeffiiet. SOLUTION For osillatios m = ad = 00% = 40% / = 00/40 =.5 π δ m l π δ l ad et square both sides ( π δ ) (0.963) δ δ 48 δ The torsioal stiffess is k 4 4 πd π doubled. J m k t GJ/L 944 N m s/radia The atural frequey is The damped frequey is give by f The ritial dampig oeffiie t is give by δ I k t δ π Ns/m The atual dampig oeffiie t is f t GJ/L ad there is oe eah side of the flywheelsoit is k I t f π Hz Hz δ N m s/radia D.J.Du

13 SELF ASSESSMENT EXERCISE No.. A flywheel has a momet of iertia of 50 kg m. It is suspeded o the ed of a vertial shaft m log ad 40 mm diameter. A torsioal damper is fitted with a dampig ratio of 0.5. The modulus of rigidity of the shaft is 80 GPa. Calulate the followig. The torsioal stiffess of the shaft. (0053 N m s/radia) The ritial dampig oeffiiet for the system. ( 48 Nms/rad) The frequey of damped osillatios? (.4 Hz). The amplitude redutio fator. (45.6%). A shaft with egligible iertia has a flywheel suspeded from the ed ad a damper to damp the vibratios. The shaft has a torsioal stiffess of 5000 N m s/rad. The flywheel has a momet of iertia of 30 kg m. The dampig ratio is 0.4. The flywheel is rotated 0.0 radia ad released. Calulate the followig. i. The damped frequey. (.833 Hz) ii. The agular displaemet, veloity ad aeleratio after 0.5 seods after beig released. (0.004 rad, rad/s ad -0.6 rad/s ) D.J.Du 3