Overview in PDF format - Tallinna Tehnikaülikool

(moving cutting force), represents the forced vibrations of the system ϕ 2* **and ϕ 2accordingly, which is represented as follows:* **ϕ 2 = ϕ2+ φ2(2.9)*M ywhere( ( ) [( 2)]J n p pp 22 = ω − sin pt − 2 np cos2 2 2 24 + ω −pt**ϕ , (2.10)o**ϕ 2 = A + Bt + C sin ω t + D cos ω t + Et sin ω t + Kt cos ω t , (2.11)where2 *2 2 *2N = 4nω + ( ω − ω ) 2F r u, B = −2 ,J2 *2 2 *2*2 2 *2( ω − ω ) 2n( ω − ω ) F u 4nω( ω − ω )3 *2*2FalaFau8nω Fau4nωFauC = ++−+ ,J N2222oJ o NJ o NJ o N J o N26*o ω2 *2* 2 *2( ω − ω ) F u 2ω( ω − ω )* 2 * 2 *3 2 *Fal2nω4nω Fau8nω Fau4nωaFauD = − − −−+,J N2222o JoN JoNJoNJoNF uEa*= −F, au2nωJoNK = (2.12)JoNAdding the general solution (Eq. (2.8)) and partial solution (Eq. (2.9)–(2.12)), ageneral solution of the differential equations for the declination angle of the blankϕ = exp( −nt )( C1 cos ω1t+ C 2 sin ω1t) +M y2 2+ω − p sin pt − 2npcos pt +2 2 2 2J 4np + ω − p(2.13)o+ A +( ( ) [( )]****Bt + + C sin ω t + D cosωt + Et sin ω t + Kt cos ω t,Taking into account the initial conditions ϕ ( 0) = ϕ &0ϕ(0)= & ϕ , the constants of0integration3* 2 *nmp y0Frl2nFruFal2nω4nω FauC1= ϕ0+− − + + +2 2 2 22 42J ( 4 ( )J Non p + ω − p Joωω Jo o JoN2 *32 * 2 *2* 2 *28nω Fau4nω ( ω − ω ) Fau2ω( ω − ω ) Fau++−,222J NJ NJ NCϕnCo3 2 2mp y0( ω − p )−+2 2 2 22Joω1( 4np + ( ω − p) Jo2 *2*3 2 *2( ω − ω ) F u 4nω( ω − ω )oω F lo2 *2( ω − ω )**0 1raa2= & +−−2ω1ω1ω ω1JoNω1ω1JoN*3 *3*3ω 2naFau8nω Fau4nωFau−−+ −2222ω1JoNω1JoNω1JoNω1JoN(2.14)F u*2nωF uFinally the displacement y = ϕland the rotation speed ν of the blank end inrelation to initial conditions y 0 and v 0 should be found from Eq. (2.13).