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MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland VIBRATIONS Common engineering phenomenon concerned with linear displacement- angular rotation- deflection (bending) of components. Simplest system is 1 DOF involving 1 displacement. In general, multi DOF will apply- but 1 DOF is a basic + useful analytical tool Single wheel on car ~ 1DOF MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland Free Vibration After displacement, system oscillates before returning to the equilibrium position in absence of imposed external forces- eg. car over single bump Forced Vibration Motion continuously excited by disturbing force- external or due to system dynamics- eg. car out of balance Damping In all engineering applications, some forces resist motion (mechanical + fluid friction) mechanism for energy losses Friction always opposes motion. Fdx is always -ve May be deliberately induced- eg. shockers . Damping is generally proportional to x, v . . FD k x (viscous damping, convenient sign change- x goes –ve) Linear and non-linear systems consider vibration when restoring force & displacement. MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland Example of a linear system - simple spring This is applicable to many systems- our analysis is restricted to linear – very useful. Examples of non-linear systems - Air spring P V R T isothermal Po Vo P V - Overload springs on a 4WD only contact at high load – large displacement k n m MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland Single degree of freedom – linear analysis . .. Newton’s Law F kx cx mx opposes opposes displacement motion .. . Fo cos t m x c x k x (8) forcing term inertia damping spring stiffness FREE VIBRATION Case #1 Undamped, free vibration ( c 0 , F 0 ) .. mx k x 0 .. k .. x x 0 x n x 2 (9) m k where n natural frequency m MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland We would expect a simple harmonic motion (acceleration & displacement) to produce an oscillatory solution. x A sin n t B cos n t . x A n cos n t B n sin n t .. x A n sin n t B n cos n t 2 2 satisfies (9) choose A, B to match boundary conditions (starting values) . . x x o cos n t xo n sin n t . . . xo If it is known that at t 0 , x xo and x x o , then A and B xo . n full solution for x is . xo x sin n t xo cos n t (10) n Examples: 1. vibration of a car 2. vibration of a bed MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland FORCED VIBRATION Undamped F Fo cos t ( n in general) .. m x k x Fo cos t Solution has 2 parts complementary (RHS = 0) - natural frequency particular (response to forcing) .. F x n x o cos t 2 (11) m The main purpose of this analysis is to get the long-term (steady) response to forcing input. The SHM component will damp out anyway. Eventually, the system will vibrate at need to find amplitude eg. vibration of engine on mounts Assume form x p x cos t . x x sin t .. (11) x 2 x cos t Fo 2 x cos t n x cos t 2 cos t m x cos t n 2 2 Fo Fo Fo x (12) m n 2 2 2 m n 1 2 2 2 K 1 2 n n Notes: Role of n - natural frequency of undamped system As n , x ‘resonance’ occurs. When this happens, energy input is at all times adding to natural motion (swing). This will increase until something breaks or non-linearities occur (eg suspension hits stops). In practice, there will always be damping which stops MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland Fo Case #2 For n , x K i.e. the system follows a ‘static’ response, in phase with input (eg. car going up hill) Fo n 2 Fo Case #3 For n , x K 2 m 2 i.e. output is attenuated from static Fo deflection by and inverted m 2 out of phase Fo Case #4 If st static deflection, then K x 1 = magnification factor st 2 1 n very important for engineering design. eg. rockets want to make sure n of any vibration modes << lowest exciting frequency MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland Resonance occurs when n (forcing frequency = natural frequency) Free vibration will occur at natural frequency if device accelerates rapidly through n Resonance may not have time to start- eg. wheel wobbles For connected network of rigid bodies, number of n = no. DOF Non-rigid elastic system has number of frequencies- eg. diving board . Energy x x cos t v x x sin t 1 Kinetic energy T m x 2 2 sin2 t 2 more energy needed to excite higher (high vibration is less important) for given energy T, amplitudes at higher will be less DAMPED, FREE VIBRATION c 0, F=0 .. . m x c x kx 0 u viscous damping Y y . .. let x A e t , x A e t , x A2 e t Expected solution Steady position MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland Want to find m 2 c k e t A 0 ‘characteristic equation’ The two roots are: c c2 4 k m c c 2 k 1, 2 (13) 2m 2m 2m m c k 2 0 m m Response depends on roots of the characteristic equation. Critical damping 2 c k Critical damping is defined as being when 0. The characteristic 2m m equation has two equal, negative roots cc ( cc critical damping coefficient) 2m 2 cc k 2 and n 2m m cc 2 m n 2 k m (14) At this condition, two real roots are 2 km 1 2 n (undamped natural frequency) 2m k m n t x A e t xo e (15) MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland Response is non-vibratory. Exponentially decays to zero in minimum amount of time. Critical damping is the target condition for many designs - eg. the car wheels As critical damping is a common design objective, it is convenient to define a ‘damping ratio’ c damping coefficient cc .. . eom x 2 n x n x 0 2 Roots can be expressed as: 1, 2 n n 2 1 (16) 1, 2 t x Ae (17) Type of solution will depend on damping ratio . MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland Over damped 1 , c cc . eq. (16) has 2 unequal, -ve real roots. 1 t 2t x Ae Be with A, B depending on boundary conditions . 1 t 2t x A 1 e B 2 e Not normally achieved (or desired) in passive mechanical systems. Non-vibratory motion, decays to 0 as t 0 eg. car overdamped- after step input stays low. Over damped car wheel example: MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland Critically damped 1 , c cc . From eq.(16) n 0 n (2 equal, real roots) Solution of this equation has the form: n t x A Bt e (18) Response is non-vibratory. Exponentially decays to zero in minimum amount of time. n t t t n e n A B t e n B n A B t x Be . .. e n t B A B t B e n t ( 19) x n n n n t n e n A B t 2B Car example: MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland Under damped 1, c cc . .. . k c CE: x 2 n x n x 0 , 2 n , m cc 1,2 n n 2 1 For 1 , only imaginary roots exist (complex conjugate pairs) 1,2 n n 2 1 , with j i 1,2 n j d (20) where d n 1 2 (damped natural frequency) (21) This leads to attenuating oscillations at d - most common passive system. The solution is of the form x A1 e n jd t A e n jd t 2 x e n t A sin d t B cos d t may be represented as [sin + phase shift] x(t ) Ce n t cos d t (22) Two coefficients C, are to be found from initial conditions and parameters d , n are to be found from known physical properties k , m, c . Oscillations at d , with exponential decay d n - viscous damping slows motion down. . . To solve for C, , we need two conditions- eg. t 0 , x xo , x x o or t 0 , x xo , t , x x etc. MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland n t x Ce cos d t . dx x n C e n t cos d t C d e n t sin d t dt (23) C e n t n cos d t d sin d t example: underdamped wheel MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland Observation of the rate of decay of ‘peaks’ may be used to estimate damping ratio From (22) x(t ) Ce n t cos d t Peaks occur after a cycle so that 2 d T 2 T (period) d After n periods t nT x(t nT ) Ce n (t nT ) cos d (t nT ) x( t ) e n nT (as cos terms are equal) x(t nT ) x( t ) 2 2 n ln n nT n n from (21) x(t nT ) d 1 2 Given the magnitude of x at 2 matched points on different cycles ( n apart), damping ratio may be found. May use any corresponding points but it is sensible to use peaks (most accurate) no phase to worry about max displacement MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland For the special case when n 1 (2 successive peaks) x( t ) 2 ln n T (24) x(t T ) 1 2 where logarithmic decrement 4 2 2 For very lightly damped systems 1 2 ~ 0 2 (25) For car wheel example: d =21.65 r/s, n =25 r/s, =0.5 MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland DAMPED, FORCED VIBRATION c 0, F 0 Need full equation of motion .. . Fo cos t m x c x k x from (8) forcing input inertia damping spring stiffness F 0, c 0 Solution consists of superimposed steady state + transient solutions. Steady state solution (i.e. what happens when starting process dies down) is of most engineering interest study that. eg. accelerate through wheel wobble- essential difference– will oscillate at , not n or d . Many techniques have been developed to solve the equation eg. Laplace transforms -Math. Output has the form: x X cos( t ) with Fo X and 1/ 2 k m 2 2 c 2 (26) c tan 1 k m 2 i.e. system will oscillate at the driving frequency , with displacement peak X and phase shift . MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland Critical parameters are: c damping ratio cc forcing frequency and frequency ratio n natural frequency c c c noting that cc 2 km 2 k cc k cc k from (14) gives Fo k X and 1/ 2 1 2 2 2 2 (27) 2 tan 1 2 1 phase response Noting that the static displacement ( =0) has the value of X st Fo k . Ratio of dynamic to static deflection (amplitude response) is given by X X 1 X st Fo k 1/ 2 2 1 2 2 2 (28) MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland Example A platform of mass 150 kg has an item of rotating machinery mounted on it, providing an out of balance force created by a 5kg mass, mounted 50 mm off the axis of rotation, 600 rpm. The platform is supported on a spring-damper system. It is required that the peak deflection of the table is to be 1mm. Find the spring and damping parameters needed to achieve this for (i) 0 (ii) 0.25 MECH2200/2210 Dynamics and Orbital Mechanics Department of Mechanical Engineering, The University of Queensland General comments Solution for a viscously damped, forced oscillating system for any value of damping (through ) and frequency (through ) Includes the undamped analysis from earlier (eq. 12) with 0 Most common engineering application is to reduce or limit the amplitude of vibration from an applied force. For X X st <1, must be > 2 for an undamped system. For finite values of , this occurs at lower For 1, all frequencies are attenuated As a general design rule- design the system for a lower n than the lowest frequency of the expected load. A system is said to be in ‘resonance’ if n . At low ( ), 0 . Characterised by low mass eg. spring on its own. At high , 180 o (out of phase). Mass dominates.