tailieunhanh - The problem of longitudinal shock of two spherical end elastic bars with visco elastic resistance force

Based on the theory of one-dimensional wave together with Dalembert solution and Hertz's law of deformation holds, in [1[ and we studied the problem of shock of two elastic bars with free spherical end. In this paper, we continue to study the above problem when the second end of the seccond bar meets visco-elastic resistance force. | Jour;a].of Mechanics, NCNSTof-Vietnam T. XVI, 1994, No 2 (1- 6) THE PROBLEM OF LONGITUDINAL SHOCK OF TWO SPHERICAL END ELASTIC BARS WITH VISCO-ELASTIC RESISTANCE FORCE NGUYEN THUC AN, NGUYEN DANG TO, NGUYEN HUNG SON Hanoi Water Resour:ces University Based on the theory of one-dimensional wave together with Dalembert solution and Hertz's law of deformation holds, in [1[ and [2[ we studied the problem of shock of two elastic bars with free spherical end. In this paper, we continue to study the above problem when the second end of the secc;~nd bar meets visco-elastic resistance force. §1. FORMULATION OF THE PROBLEM The motion equation of the bars is: () where j = 1, 2; ai = ["f- wave velocity. Initial conditions: At t = 0, au,_ v. u, =0; at - b au2 - = 0 · U2 = o; at ' Boundary conditions: At the shock end x1 = £1; Xz = au, ax, au2 -=0 ax2 -=0 ( ) () ~~ au, = E2F2au2 = -K (U, E,F,axl . axz At the free end, x1 = 0, · au, ax, = + U2 )3/2 () () 0 When tl).e end of the second bar bear on the visco-elastic. sole, we obtain: x2 = 0; au2 ax2 - au2 at = -K,U2- ) . - () In this equation k 1 , A are elastic and viscid coefficients respectively. They are considered as constants. A general solution of eq. (} is of the D'Alembert form: u,. = 'Pi(a,.t- x,.) 1 + .p,.(a,.t + x,.) §2. DETERMINATION OF WAVE FUNCTIONS OF BARS Assume that the second bar is in. the rest, the first bar centro-longitudinally moves and impacts to the second one with velocity V1 , based on [1j we get: () where ~ii : I () I.!.') () = ~'>1 +-,~pz+ '1'2 where K . (3= ·· E 1 · F1 Consider that T 2 ·= iT1 + qT1 with i = 1, 2, 3, . ; 0 ::; q 1') li = v, + -1 L · a1 a n=l z 2 () Finally in the interval iT;0,. . Solving eq. () the wave functions ( ,p;) 1 is determined. From eq. () we obtain: when t H 2(a2t- x2) is determined. Integrating eq. () with the condition of 102 (£, - 0) = 0 we obtain: (a2t-x2) I . e