Mechaical Vibratios FREE VIBRATION RESPONSE OF A SYSTEM WITH COULOMB DAMPING A commo dampig mechaism occurrig i machies is caused by slidig frictio or dry frictio ad is called Coulomb dampig. Coulomb dampig is characterized by the relatio f c = F c μn ( ) = μn > = < The frictioal force f c always opposes the directio of motio causig a system with coulomb frictio to be oliear. Summig forces i the x directio yields that (ote that the mass chages directio whe the velocity passes through zero) + x = μ mg for < Respose of Secod Order Systems zmasoud.googlepages.com 37
Mechaical Vibratios I a similar fashio, summig forces i the opposite directio yields + x = μ mg for > Sice the sig of determies the directio i which the opposig frictioal force acts, previous equatios ca be writte as the sigle equatio as + μ mg sg( ) + x = This equatio ca be solved umerically. These equatios ca be also solved by cosiderig the first equatio ad the the secod. The solutio for the first ca be writte as x( t) = A cosω t + B si ω t + Applyig the iitial coditios yields x() = A + = x () = ω B = Whe the mass starts from rest (at x ) ad moves to the left, it moves as x( t) = x cosωt + This motio cotiues util the first time ( t) =. This happes whe the derivative is zero, or whe ( t ) = ω x si ω t = Thus whe Respose of Secod Order Systems zmasoud.googlepages.com 38
Mechaical Vibratios π t =, ( t) = ω The mass starts to move to the right provided that the sprig force, x, is large eough to overcome the maximum frictioal force. Hece the secod equatio describes the motio. Solvig that equatio yields x( t) = A cosω t + B si ω t The iitial coditios for the secod equatio are calculated from the previous solutio at t π x = x ω π cos + = x It follows that π = ω x si π = ω Respose of Secod Order Systems zmasoud.googlepages.com 39
Mechaical Vibratios A B = x = 3 The solutio for the secod iterval of time the becomes π π x( t) = x 3 cosω t < t < ω ω This procedure is repeated util the motio stops. The motio will stop whe the velocity is zero ad the sprig force x is isufficiet to overcome the maximum frictioal force. Several thigs ca be oted about the free respose with Coulomb frictio versus the free respose with viscous dampig.. With Coulomb dampig, the amplitude decays liearly with a slope tha expoetially as with the viscously damped system. ω π rather. The frequecy of oscillatio of a system with Coulomb dampig is the same as the udamped frequecy, whereas viscous dampig alters the frequecy of oscillatio. 3. The motio uder Coulomb frictio comes to a complete stop, at a potetially differet equilibrium positio tha whe iitially at rest, whereas a viscously damped system oscillates aroud a sigle equilibrium, x =, with ifiitesimally small amplitude. For example, the figure below shows the free respose (displacemet versus time) of a system subject to Coulomb frictio with two differet iitial positios (5 m, solid lie; ad 4.5 m, dashed lie) for zero iitial velocity (m = g, μ=.3, = 5 N/m). It ca be oticed that the fial positio of the system could be aywhere betwee the two lies x = ± Respose of Secod Order Systems zmasoud.googlepages.com 4
Mechaical Vibratios EXAMPLE The respose of a mass-sprig system oscillatig o a rough surface is measured to be of the form Coulomb damped motio. The iitial positio is measured to be 3mm from its zero rest positio, ad the fial positio is measured to be 3.5mm from its zero rest positio after four cycles of oscillatio i secod. Determie the coefficiet of frictio betwee the mass ad the surface. Solutio First the frequecy of motio is 4 cycles per secod (4 Hz), or 5.3 rad/s. The slope of the lie of decreasig peas is The slope expressio is.35.3 slope = =.65 m/s But sice for a mass-sprig system ω slope = π Respose of Secod Order Systems zmasoud.googlepages.com 4
Mechaical Vibratios The, the slope expressio becomes Solvig for μ yields ω = = mω m ω μg πω slope = = μ = ( slope) πmω πω g πω π (5.3) μ = ( slope) = (.65) =.7 g (9.8) NUMERICAL SIMULATION OF THE TIME RESPONSE Cosider a damped system of the form + c + x = x() = x & = x() This equatio ca be writte as two first-order equatios. Dividig the equatio by the mass m ad defie two ew variables x = x ad x =. The differetiate the defiitio of x rearrage the equatio ad replace x ad its derivative with x ad x to get = x = x m subject to the iitial coditios x ( ) = x ad x ( ) = This system ca be solved usig Matlab EXAMPLE Use Matlab s ode45 fuctio to simulate the respose of 3 & x + + x = x() = () =.5 Respose of Secod Order Systems zmasoud.googlepages.com 4 c m x
Mechaical Vibratios over the time iterval t. Solutio The first step is to write the equatio of motio i first-order form. This yields = x = x 3 x 3 Next, a M-file is created to store the equatios of motio. The file is created by choosig a ame, say sdof.m, ad eterig Next, go to the commad mode ad eter The first lie establishes the iitial (t) time ad fial time (tf). The secod lie creates the vector cotaiig the iitial coditios x. The third lie creates the two vectors t, cotaiig the time history ad x cotaiig the respose at each time icremet i t (icludig first derivative), by callig ode45 applied to the equatios set up i sdof.m. Respose of Secod Order Systems zmasoud.googlepages.com 43
Mechaical Vibratios The fourth lie plots the vector x versus the vector t. This is illustrated i Figure. EXAMPLE Fid the free vibratio respose of a sprig-mass system subject to Coulomb dampig for the followig iitial coditios: x ( ) =.5 ad () =. Let m = g, = N/m, ad μ =. 5. Solutio: The equatio of motio ca be expressed as + μ mg sg( ) + x = We rewrite the equatio as a set of two first-order differetial equatios as follows: = x = μg sg( x The MATLAB program ode3 is used to fid the solutio as show below. A M-file is created by choosig a ame, say dfuc.m, ad eterig ) m x Respose of Secod Order Systems zmasoud.googlepages.com 44
Mechaical Vibratios Next, aother M-file is created by choosig a ame, say ex.m, cotaiig these lies Fially, go to the commad mode ad type the ame of the file, ex, the press eter Respose of Secod Order Systems zmasoud.googlepages.com 45