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# DYNAMICS AND VIBRATIONS MEEN 363

Texas A&M

GPA 3.58

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This 20 page Class Notes was uploaded by Orrin Weissnat on Wednesday October 21, 2015. The Class Notes belongs to MEEN 363 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 147 views. For similar materials see /class/225993/meen-363-texas-a-m-university in Mechanical Engineering at Texas A&M University.

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Date Created: 10/21/15

Preamble A NEED arises Consider the periodic forced response of a primary system Kp Mp defined by 39 wt 3 Kp 2 1gtlt1051b f Mp 10 39lb 111 Its natural frequency is 5 conp MP 0 196 491 rad np 39 sec The EOM from SEP for periodic force excitation with magnitude F0 and frequency Q is Let x a 9 W L V W M19 tr Lg p 4 130 10001bf i Gilt The solution of1 is ofthe form Xpt ZpcosQt 2 Substitution of2 into 1 gives 2 KP Q MpZp F0 or Q with r as the frequency ratio an and the phase angle is zero degrees for excitation frequencies 9 lt the natural frequency 4 180 deg for 9 gt natural frequency and 90 degrees for Q natural frequency The response Zp amplitude and phase as a function of the excitation frequency is 83 ZPQ 3 2 Frequency Response of MK primary system Q 0042 I I 1 2 0 mp 00021 s 0 83 8333 X 10 4ft rad 00021 comp 196491 sec 0 0042 39 39 39 39 39 0 100 200 300 400 500 600 frequency radsec Note the very large amplitude of motions unbounded for excitation at the system natural frequency In the graph above ZpgtO means in phase with the external force ZpltO means 180 deg out of phase with external periodic force Clearly the system cannot be operated at frequencies close or at the natural frequency Since there is NO damping the system will just fail bc the amplitude of motion isjust TOO LARGE Xsrt T M Fa The system forced response is also periodic ie with identical frequency as that ofthe periodic excitation force Thus let The combined system is Xp 2p 2DOF Thus TWO natural Z cosQt 6 frequencies and natural XS 5 modes will appear Substitution of6 into 5 leads to the algebraic set of equations 2 Kp KS Q Mp KS zp 10 ZS 0 2 KS KS Q MS The determinant ofthe system of equations 7 is 2 2 2 AQ KpKS QMpKS QMS KS a The solution to the algebraic system of equations 7 is simple use Cramer39s rule for example The response amplitudes for the primary and secondary masses are ata certain frequency E1 Z That is the motion of the primary system is zero NULL 39 I immni e realms realign quot lllfn For example if zero amplitude vibration is desired for excitations at the natural frequency of the original system the designer selects 11 ie the natural frequency of the seconday system MUST coincide with that of the original system Say for then 5 K rad E rad equals p 196491 196491 sec MS sec 13 and the system responses are Mo K K 22M QZM K2 p s p s s s rad 196491 sec The SOFTER the secondary system is Ks ltlt Kp the largest the motion of the secondary system at the desired frequency The graph below shows the FRF amplitude and phase ofthe vibration absorber ft Zp amp Zs ft 001 0005 0005 FR of Vibration absorber 0 100 200 300 400 500 frequency radsec primary secondary 600 K p 3 2 10 Ks Note the null amplitude of motion for primary system excitated at the ORIGINAL system natural frequency In the graph ZpZs gt0 means in phase with the external force Zp Zs lt0 means 180 deg out of phase with external periodic force The graph belos shows the amplitude absolute of the vibration absorber ft Zp amp Zs ft 001 00075 0005 00025 Amplitudes of motion 500 200 300 400 frequency radsec primary Secondary 600 K p a 10 Ks F0 3 8333 X 10 ft S rad comp 196491 Q Note the amplitude of motion is zero for the primary system when excited at its original natural frequency The secondary system does have a large amplitude of motion and is out of phase 180 degrees with the excitation force Note that the addition of the secondary K M system renders a 2DOF system with two natural frequencies one above and one below the original natural frequency In general the smaller the magnitude ofthe secondary stiffness and mass the larger the amplitude o motion for the secondary system since it is extremely exible The system natural frequencies 1 and also tend to approach that ofthe original natural frequency Note the null amplitude of motion for primary system excitated at the ORIGINAL system natural frequency In the graph ZpZs gt0 means in phase with the external force Zp Zs lt0 means 180 deg out of phase with external periodic force oooooo 47 a O O O 0 D O D D o O OC UOQD nrxnnrnrmnnnm O Flew 4167 X 10 31 rad comp 196491 SCC 000000003 Note the null amplitude of motion for primary system excitated at J the ORIGINAL N00 u i 300000mmunmannme system natural frequency In the graph ZpZs gt0 means in phase with the external force Zp Zs lt0 means 180 deg out of phase with external periodic force oooooo 47 Flew 2083 X 10 31 rad comp 196491 SCC The equations of motion for the 2DOF system are M 0 2 X K K K X p p p S p Fjsin la 0 Ms dt2 Xs Ks Ks XS 0 1 where primary system has F iii Lu The system response is ofthe form Xp Zp XS ZSJsinQt 2 Substitution of Eq 2 into 1 leads to 2 Kp KS 9 Mp Ks Z13 3 KS KS 92Ms ZS 0 1A tuned absorber is designed so that Z e 9 Le no motion ofthe primary mass for operation at Fan 5 5 Thus from the first of eqns 3 KP KS Q MP The determinat ofthe system ofequations 3 is rad Mo Kp KS QZMPKS 22MS Ks2 5 P 20 stiffness ratio mass ratio for tuned absorber Expand Eq 5 ie the characteristic equation A0 1 21 X11 7 a a2 al1 a X11 7 a MA a1a L k 7va7 2 a a1 27 7va7 2 61 2 2 The roots of the characteristic equations are 2 a 4 39 439a a lowest 2 5 highest 2 5 2a 4aa 2a4aa Let 03min A mmax xmin max 3 I1 Dn Given the max and min values then from eqn 6 Z F 7L1a 2a 24aaz5 Mg 2a24a2125 K 5 K 5 031 X1aM 032 X2aM a 16315 3im 311 E Kmin z 3quot KP Mmin 1 aMP for N Kmin 2689 X 103 Mmin 6722kg m 2 5 2 5 Z F x1a 7L2a S aKP 5 5 KP 32 7M2a M P Kmax 3 3 KP E SUQNER E w ta n 33m C f amX 0202 VW maimm fm agmgmrgg w 1390 m t m mmgg N Kmax 405 X 103 m max 10125kg IE Build Absorber FRF Now let39s graph the amplitude and phase lag of frequency response function for both absorbers primary and secondary mass motions Values gt0 mean phase lag of 0 degrees values lt0 mean phase lag of 180 degrees with respect to forcing function Passing through the natural frequencies gives a phase lag of 90 degrees Amplitudes become unbounded while crossing the system natural frequencies rad DnP 20 S F Absorber FRF Z a 0202 I 5031 aK Zsmp o247 m 7 rad g 031a 16 o S E 39L33 39267 rad 032a 25 394 S 10 15 20 25 30 frequency rads Zp gram gram PSiirivrlli If if 17 PEE5215532 555 Ir a E 517 xf 95 73 5 2325 L FE ririEE 35573 L F5 qty319 fizz33 5555 953573 xvi513332 ii 15351E75 Esra 1353 3 nrU ummsuuhhiunh SEWFE Ina F r 2155 in l rs if 4 5 di u h izn i wnnmlIEni in 5515 til it i 1 Problem 55 The system illustrated is a composite pendulum made from two uniform bars The bar pivoted about point 0 has a mass 2m and length 21 The cross bar has a mass m and length 1 Perform the following engineering tasks a Derive the equation of motion using freebody diagrams and energy principles Assume that system is released from rest from the inverted vertical position b Determine the reaction force at point 0 as a function of the rotation angle 6 only c Determine the natural frequency for small an les d Determine the equilibrium positions Determine whether small angle motion about the equilibrium positions are stable Problem 56 The composite pendulum shown is made of a uniform bar length l and mass m that is rigidly attached to a disk mass Mand radius r The pendulum is released from rest in the position shown Perform the following tasks a Draw a freebody diagram and derive the equation of motion b Derive the equation of motion using conservation of energ c Define the pivot reaction force components as a function of the rotation angle only d Develop the equation of motion for small motion about the stable equilibrium point Dynamics in Engineering Practices Childs Chapter 5 Problems iPart 1DRAFT Problem 56 The system shown is released from rest at 6 0 Carry out the following engineering tasks a Draw a free body diagram and derive the governing equation of motion using a moment equation b Derive the governing equation using workenergy principles c Define the reactionforce components as a inction of 6 only Problem 58 The plate shown is released from rest at 6 0 Perform the following tasks a Draw a freebody diagram and derive the equation of motion using Newtonian principles b Derive the equation of motion using work energy principles c Determine the reaction components at O as a inction of Duly Dynamics in Engineering Practices Childs Chapter 5 Problems iPart 1DRAFT Problem 59 The plate is released from rest and rotates to the right Perform the following tasks a Draw a freebody diagram and derive the equation of motion using Newtonian principles b Derive the equation of motion using work energy principles c Determine the reaction components at O as a inction of Duly Problem 510 The plate shown was released from rest at 6 0 Perform the following tasks a Draw a freebody diagram and derive the equation of motion using Newtonian principles b Derive the equation of motion using work energy principles c Determine the reaction components at O as a inction of Duly Dynamics in Engineering Practices Childs Chapter 5 Problems iPart 1DRAFT Problem 511 The plate shown was released from rest at 6 0 Perform the following tasks a Draw a freebody diagram and derive the equation of motion using Newtonian principles b Derive the equation of motion using work energy principles c Determine the reaction components at O as a inction of Duly Problem 512 The cylinder illustrated is released from rest at 6 0 Perform the following tasks a Draw a freebody diagram and derive the equation of motion using Newtonian principles b Derive the equation of motion using work energy principles c Determine the reaction components at O as a inction of Duly Dynamics in Engineering Practices Childs Chapter 5 Problems iPart 1DRAFT Problem 513 The cylinder illustrated is released from rest at 6 0 Perform the following tasks a Draw a freebody diagram and derive the equation of motion using Newtonian principles b Derive the equation of motion using work energy principles c Determine the reaction components at O as a inction of Duly Problem 514 The triangular plate is pivoted at 0 Perform the following engineering tasks a Draw a freebody diagram and develop the equation of motion using Newtonian principles b Define the equilibrium positions of the body State the equations of motion for small motion about the equilibrium position Determine the natural frequency of the bod c Assuming that the late is released form rest in the vertical position state the reaction force components at O as a function of the plate s rotation angle only Dynamics in Engineering Practices Childs Chapter 5 Problems iPart 1DRAFT Problem 515 The bar illustrated is released from rest in the position shown and rotates to the right The spring has an unde ected length of L3 Perform the following engineering tasks a Draw the freebody diagram and derive the equation of motion using large angles b Apply work and energy principles and obtain the equation of motion c Linearize the equations of motion by applying small angle relations Problem 516 The slender bar pivots about the frictionless pivot 0 Assuming small 6 perform the following tasks a Draw a freebody diagram and derive the equation of motion by applying Newtonian principles b Derive the equation of motion by applying conservation of energy principles c Determine the natural frequency ofthe system d Determine the value of k so that the systems remains stable Dynamics in Engineering Practices Childs Chapter 5 Problems iPart 1DRAFT gram 5 57 a i 3 33 36 5 i 5135 FEB paid F Iii n 55 IE 5 ri4iampv 5 115 5 5izieelz ram 535 r F1SE EE r v 55 E 212 i n airLn 5 in nu Jug 5 I z riirs 51576 ELI E i uui ig mp Ea h p wm th X 5 imiFIVRE 35 rEEEs var 533 5358 5755155ki zi1ra t FE i a i 555 F 3 Hinz uh wihruudz h nmnlz i I15 I a virusr l 55 553 5 T E r 5115 r JEHFnVWusiurrxi Lil 56125195 5 5 i551 ram 2 r 155352732 5 ii E viii 5 3L 5 3 Li Eliilt 23 53 5335 1993 5 it 55 4 vrrirsi 557 m 3 m NEEE F Err52 51215infil fis W W WW WWWWWW WWW WHWPMHnnmmnm Mmmunmnm mum mm tum quotmyan m m 39 m mm m mmmmmmm mamaWmmmmwmqmmmm mummmw mmummmmmh mummmm Pnhlmlsl m mgmmmumdmmmn Mommaan Xhmmnl may mamama m ammmmmm mime mum maxim m m New mmmunr ml inimruxunnznmlhnm a a mumm mmn mm minimal 1 mm mnnhmunnu m 1mm 7 quot m quot mums n mwll39yhehwtmzmmmm rhlw bjmnnnmkwni magnum imnmmm mmmnmmmmmmmymum Ammm m Wm m D 9 31mm my mum MmmMlu mm mm mums KmMMMm Jam mmmmmmmquotmmqmmxmummuWmm a wamwm M401 numhahmmmmrmdmmmml bunny xmmmmmmm mmmm zu 325 32 a ragEiieii 253 EEEE i reli55i ii r 5esziEi ilrEa n wihi iail 55 in szninvn alenr43 2 aux vrsiiii a A a s 2 u at 2555e5 5 quot512 1 1 vizrlilxlvis 2 hill is 5 Esrrcbtr 5 4 6 m t F era ii L wiles 311 I 2 Eirnunn v 2 F E rb uinuu

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