FREE VIBRATION WITH COULOMB DAMPING Revision A

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1 By Tom Irvine June 5, 010 FREE VIBRATION WITH COULOMB DAMPING Revision A g m F sgn () & Figure 1. Coulomb damping is dry friction damping. Consider the free vibration response of a singledegree-of-freedom system subjected to Coulomb damping. The damping force F is where F = µ mg (1) µ = friction coefficient m = mass g = acceleration of gravity Assume that the friction coefficient is constant for simplicity. 1

2 Force +F 0 Velocity -F Figure. Coulomb Force vs. Velocity The governing equation of motion for the displacement is m & + = Fsgn () & () where = stiffness The sgn (& ) function represents the sign of &. As an alternative, the governing equation can be written as & m& + = F (3) & The governing equation is solved in a piecewise-linear manner.

3 Assume that initial displacement (0) is (0) > F / (4) Also assume that the initial velocity is zero. Consider the equation of motion for negative velocity. m & + = F for & < 0 (5) F & & + = (6) m m The natural frequency n is n = (7) m Let F E = (8) m & + n = E (9) The equation is solved using Laplace transforms. L { & + n } = L{ E} & (10) E s X(s) s (0) + n X(s) = (11) s 3

4 E [ s n ] X(s) s (0) = s + (1) E [ s n ] X(s) = + s (0) + (13) s E s (0) X(s) = (14) s + [ s + ] [ n s + n ] Tae the inverse Laplace transform. (t) E = (0) cos nt + [ 1 cos nt] (15) n (t) E E = + (0) cos nt (16) n n (t) F F = + (0) cos nt (17) m m n n (t) F F = + (0) cos nt for & < 0 (18) F & (t) = n (0) sin nt for sin n t > 0 (19) The velocity equals zero at t = π / n (0) 4

5 The displacement at this time is π = n F F + (0) + (1) = n π F (0) () Consider the equation of motion for positive velocity. m& + = F for & > 0 (3) The initial displacement term must be reset to the last displacement for negative velocity. Furthermore, a phase angle must be added to the argument in the cosine term. F F F ( t) = + (0) + cos( n t + π) for & > 0 (4) F 3F ( t) = + (0) cos( nt + π) for & > 0 (5) 3F & ( t) = n (0) sin( nt + π) (6) The first negative displacement pea thus has an amplitude that is F/ less than the initial displacement in terms of absolute values. This reduction factor can also be derived from the wor-energy relationship in Appi A. The pattern continues such that the envelope has a linear decay. 5

6 The velocity returns to zero for t = π / n (7) π = n F 3F (0) (8) π = (0) n 4F (9) Each consecutive positive pea is thus 4 F/ lower than the previous positive pea. The process is then repeated. Eample A single-degree-of-freedom system has mass = 1 g stiffness = 0,000 N/m friction coefficient = 0.4 initial displacement = 5 mm The resulting displacement is shown in Figure 3. The displacement converges to F /, where F = µ mg. Deping on the initial displacement, the displacement may also converge to F /. 6

7 SDOF RESPONSE DRY FRICTION fn=.51 Hz DISPLACEMENT (mm) TIME (SEC) Figure 3. References 1. R. Vierc, Vibration Analysis, nd Edition, Harper Collins, New Yor, W. Thomson, Theory of Vibration with Applications nd Edition, Prentice Hall, New Jersey,

8 APPENDIX A Energy Method The potential energy is set equal to the wor done by friction for one cycle. 1 ( ) = F( + ) 1 1 (A-1) where 1 and are consecutive positive peas. Note that the inetic energy is zero at the instantaneous time that each pea occurs. The wor-energy relationship is satisfied if 1 F = ( 1 ) (A-) F 1 = (A-3) ( ) 8

9 APPENDIX B Matlab Script disp(' '); disp(' dry.m ver 1.0 June 5, 005 '); disp(' by Tom Irvine tomirvine@aol.com '); disp(' '); disp(' This program calculates the response of a '); disp(' single-degree-of-freedom system subjected to dry damping '); disp(' '); clear all; disp(' Enter mass (g) ') m=input(' '); disp(' Enter stiffness (N/m) ') =input(' '); disp(' Enter coefficient of friction ') mu=input(' '); disp(' Enter initial displacement (mm) ') o=input(' '); o=o/1000.; F=mu*m*(9.81); f=f/; omegan=sqrt(/m); fn=omegan/(.*pi); out1=sprintf('\n fn = 8.4g Hz\n',fn); disp(out1); out1=sprintf(' F/ = 8.4g mm\n',(f/)*1000.); disp(out1); if( F/ > o ) disp(' '); disp(' No oscillation. '); disp(' F/ > o '); T=1/fn; dt = T/100.; delta=.*pi/100; num=1.*t/dt; j=1; 9

10 10 tdelay=0.; arg=0.; for(i=1:(num+1)) t(i)=(i-1)*dt; arg=arg+delta; if(arg>.*pi) arg=arg-.*pi; if(arg>=0 && arg<=pi) if( (o-f) <=0) (i)=o; else (i)= f +( o - f )*cos(arg); 1=(i); else if( abs(1) < f ) (i)=1; else (i)= -f +( 1 + f )*cos(arg+pi); o=(i); =*1000.; plot(t,); label(' Time(sec) '); ylabel(' Displacement(mm) '); out1=sprintf(' SDOF Response Dry Friction fn=8.4g Hz ',fn); title(out1); grid on; 10