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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: (1.1) 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 ( 1.2a) (1.2b) ~~ au, = E2F2au2 = -K (U, E,F,axl . axz At the free end, x1 = 0, · au, ax, = + U2 )3/2 (1.3) (1.4) 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- ) . - (1.5) In this equation k 1 , A are elastic and viscid coefficients respectively. They are considered as constants. A general solution of eq. (1.1} 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: (2.1) where ~ii < Zj < ~i~ with j = 1, 2. According to the ooundary condition (1.3) we have: A(-: I (2.2) I.!.') (2.3) = ~'>1 +-,~pz+ '1'2 where K . (3= _._·· E 1 · F1 Consider that T 2 ·= iT1 + qT1 with i = 1, 2, 3, ... ; 0 ::; q < 1. The wave fUnctions1') li = v, + -1 L.. · a1 a n=l z 2 (2.8) Finally in the interval iT;< t < T2 = iT1 + qT1 we obtain: 10;(a2t -£2) = 0 and 10~(a2t -£2) = 0. From eq. (2.2) we have (2.9) where (10~) 1 = (1/>0,. . Solving eq. (2.9) the wave functions ( ,p;) 1 is determined. From eq. (2.3) we obtain: when t H < T2 2 ~2 refiected.wave p~(azt- xz) does not appear in the second bar. < t < T 2 then the wave function rp~(a 2 t- £2 ) = 0, but the wave function cp~(a 2 ·t: x 2 } appears in the second bar. Determination of the wave function cp~(a2 t- x 2 ) with ~2 < t <. T2 is done in the same way described by [1]. According to boundary condition mentioned in eq. (1.5), the following cases are occured. If 1 - Aa2 of 0 then: 2(a2t- x2) is determined. Integrating eq. (2.11) with the condition of 102 (£, - 0) = 0 we obtain: (a2t-x2) I . e -".J._r 1->.."2 e, H ~ - .Aaz = . [ 1 + Aa2 .Ji 'fl ( .,. ) 1- ).a 2 2 l K, + 1-Aa21/>2(r) dr (2.12) 0, we have: (2.13) Based on those menti9.ned above, we can determine the wave function- c,oi{a 2 t - x 2 ) in the second bar. So that we can- determine wave function rp~ (a1t- xi),'¢~ (a1t + x1), rp~(a 2 t- x2 ) and .P;(a2t + x2) at each of the sections of the baxs in interval 0< t < 2£2 · 02 In interval T2 < t < 2T2 studying each of period T, with T2 + (n- 1)T, < t < T2 + nT1 • Let ( ) n be·c·a wave function, that .is determined in nth period of first bar and the wave function itself 2 is also determined in the second period of second bar, where n = 1, 2, ... , i. At the first period of first bar with T 2 < t < T2 + T1 , according boundary condition (1.3) we have: (1/>~)2, = (10~)2, + ~[ 2 (10i}2, + B(10~)21 + c(,p;}2,l· [(- 10;)2, + (1/>~) 2 ,] 113 (1/>;)2, = (IODn + ~[(- 10;)2, + (1/>;)2,] 3 (2.14) (2.15) From condition (1.4) 'f'1(a1t- l1) = .;,;(a1t- t,) = .;,;[al(t- T1) + t1], or (2.16) (10;)21 = (¢;)20' where ( ¢;) 20 is the wave function ¢; (a 1t + l,) with (T2 - T1 ) < t < T2, which WIIS determined, so that the wave function (IPDzt is also detennined. According to (1.5) and (2.11) we have: , ). K1 ( ) . 1 + .l.a2 ( , ). K, ( ) ( 102 21 - 1 - .l.a2 102 21 = 1 - .l.a2 ¢2 21 + 1 - .l.a2 ¢2 11 If 1- .l.a2 i (2.17) 0 then a solution of eq. {2.17) is: (a:~t-i:;l} ( rpz ) 21 = .-,-'i.,T.[1+.l.a2¢~(r)+ I ~(at-e) el->.a.z z z . 1 - .l.a2 e, K1 ¢ 2(tJ]dr 1 - .l.a, (2.18) If 1 - .l.a2 = 0 we have: (2.19) So the wave function (c,o~) 21 = (rp2(azt -lz)} 21 is knoWn from eq. (2.14) the' function ('02) 21 is determined. Replacing this result into eq. (2.15) the wave function (1/J~} 21 can be found. Doing similary we can determine the wave functions at the ith period of first bar. We have: (¢~)2, = (10~).;, + ~[ 2 (10;)2, +B(I0~)2, +G(¢~)2J. [(- 10~)2, + (¢~)2J'/ ( .;,;) 2i = ( 10;)2, + ~ [(- 10~)2, + ( .;,;)2J (2.22) ( 102, ) 2i- 1-K1.l.a2 (102 ) 2i i (2.20) (2.21) (10;)2, = (¢;)2(i-1) If 1- .l.a2 2 = 1 + .l.a2 ( , ) K1 ( ) 1- .l.a2 ¢2 li + 1- .l.a2 ¢2 1i 0 then solution of eq. (2.23) (2.23) is: (102) 2 , = (102(a2t- £2)) 2, = (2.24) I (azt-lz) =e'~.,(a,t-t,). { [i::~+az(i-l)Tl) where Czi .-,_'i.,T. [~~~::.p;(r)+ 1 _:<"~ 02 ¢2 (rJ]dr+G2i} = cpz [lz + a2(i- l)T1 - o] . e- t-~az [tz+a:di-t)Tt) If 1 - .l.a2 = 0 then (2.25) So that the wave functions ( ~2} 2 i and ( '.bDzi are determined. Now we determine the wave f~nctions is odd part of the second period of the second bar. From conditions (1.3), (104) and (1.5) we have: (¢~)2 = (~0~)2 + ~[2 (10;)2 +B(10;)2 +G(¢~)2]· [(- 10~)2 + (.;,;) 2] 113 (2.26) ~[(- 10~)2 + (.;,~),] (2.27) .l.a 2 ( , ) K1 ( ) (102, ) 2 - 1 -K,.l.a2 ( 102 ) 2 = 11 + - .l.a2 ¢2 1 + 1 - .l.a2 ¢2 1 (2.28) (.PD2 = (10;)2 + (10D2 = (.PD2, 4 H 1 - >.a2 f 0 then e _...!£!._, l-.l.a.~ · [11->.a2 + >.a2 1/J , ( ) + K1 ·'· ( l] d 1-Aa2 2 T 'f'2 T T + C2 } (2.29) where H 1 - >.a2 = 0 then (2.30) So that the wave functions {1/J~) 2 and (1/Ji) 2 are determined. If the shocks of two bare are still not finished Yet in second period of second bar, the next periods are studied is the same method as above mentioned. Let ( ) pn be the wav~ function in nth period of first and in pth period of second bar. In interval (p -1)T2 + (n- 1)T, < t < (p :-1)T2 + nT1 with n = 1, 2, ... , the problem is studied as following: From conditions (1.3) and (1.4) we have: ") pn +_!_[2( ') pn +C(·'i) ') pn +(·'·') (·'·") 'f'2 pn -( lf'2 A P1') pn +B( lf>2 'f'2 pn ]·[(- IP2 'f'2 pn ]'/3 (2.31) 1 {.J;;) pn = { 1"D pn + ;;- [{ - 1"~) pn + {.J;~) pJ (2.32) (1"~)pn = (.P;)p(n-1) (2.33) By sinlillar way mentioned above, and from condition (1.5) we obtain: H 1 - Aa2 f 0 then I a 2 t-e 2 (P2)pn = e>-";., (a,t-e,).{ az[(2p-3) .!'j.+(n-l)Tl] (2.34) where p 2: 2, and H 1 - Aa2 = 0 then (2.35) So that the wave functions ( 1/1~) pn and (1ft) pn are determined. Now we are studying this problem in final odd part oft he p'h period of the second bar, or (p-l)T2 +iT, < t < (p-l)T2+iT1 +qT1 = pT2 . Let ( ) P be wave function determined in odd part of pth period of the second bar. From condition (1.3) and (1.4) we have (.P~)v = (P~)v + i[2(pDv +B(p;)v +C(.P~)2]. [(- P;)v + (.P~)v]'/3 (.P;)v = (p~)p = (2.36) (p~)P + ~[(- P;)v + (.J;~)Pj (2.37) (.p;)pi (2.38) 5 Doing similarly, from eq. (1.5) we get: If 1 - .>.a2 'I 0 then (2.39) where C p -- pz [az {2 p- 3) T2 2 + za . 2 T 1 - o] · e- 1~ "'' - ~+aziTr . [a2 (2p-3) If 1 - .>.a2 = 0 then {2.40) So that the wave fl!nctions ( tP;) P and ( tflD P are determined. Impact-pressing force F between + (?J;~)pn] and 10~) P + ('!'>~) P]. Impact time determined by the following expression (F) ,on = 0, we can determine the wave functions rpi (a 1 t - xt) '¢'~ (a 1 t + xi),

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