Consider a single-degree-of-freedom system in a free-fall due to gravity. &&x

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Ζακχαῖος Καραβίας
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1 SIMPLE DROP SHOCK Revisio D By Tom Irvie tomirvie@aol.com November 10, 004 DERIVATION Cosier a sigle-egree-of-freeom system i a free-fall ue to gravity. &&x g m k Where m is the mass, k is the sprig stiffess, x is the absolute isplacemet of the mass, g is the gravitatioal acceleratio. Note that the ouble-ot eotes acceleratio. Assume 1. The object ca be moele as a sigle-egree-of-freeom system.. The object is roppe from rest. 3. There is o eergy issipatio. The collisio is perfectly elastic. 4. The object remais attache to the floor via the sprig after iitial cotact. 5. The object freely vibrates at its atural frequecy after cotact. 6. The system has a liear respose. The iitial velocity of the system as it strikes the grou ca be fou by equatig the chage i kietic eergy with the chage i potetial eergy. 1 mx& = mg h (1) 1

2 where h is the rop height. Diviig through by (m/), &x = g h () Thus, the iitial velocity whe the mass ecouters the floor is &x( 0) = g h (3) Furthermore, the iitial isplacemet is take as zero. x(0) = 0 (4) Assume a isplacemet equatio with costat coefficiets a a b. xt () = asi ω t + b cos ω t (5a) Equatio (5a) assumes oscillatio at the system s atural frequecy. Note that the atural frequecy i raias per time is ω = k m (5b) The zero isplacemet iitial coitio requires b = 0. Thus xt () = asi ω t (6) The velocity is xt &( ) = a ω cos ω t (7) Recall &x( 0) = g h (8) Thus a = g h ω (9)

3 Substitute equatio (9) ito (6). The resultig isplacemet is g h xt () = siω t ω (10) The velocity equatio is xt &( ) = g hcosω t (11) The acceleratio equatio is &&() xt = ω g hsiωt (1) The force trasmitte through the sprig f(t) is f(t) = kx (13) f (t) g h = k si ωt (14) ω Note that k= ω m (15) Substitute equatio (15) ito (14). ft () = ωm g hsiωt (16) The peak istataeous power flow is P = ω mg h (17) Equatio (17) is erive i Appeix A. The total impulse over a half-sie perio is I = m g h (18) Equatio (18) is erive i Appeix B. 3

4 The shock aalysis is oly cocere with the maximum values. These are summarize i Table 1. Table 1. Maximum Absolute Values Parameter Symbol Maximum Equivalet Form Displacemet x g h ω mg h k Velocity &x g h g h Acceleratio &&x ω g h gk h m Jerk & x& k ω g h g h m Trasmitte Force f mω g h mgk h Potetial Eergy of Sprig Kietic Eergy of Mass PE max KE max k g h ω k g h ω mg h mg h Peak Istataeous Power Flow Total Impulse over a Half-sie Perio P I ω mg h km g h m g h m g h The values i Table 1 emostrate that there are some traeoffs ivolve i esigig a object with respect to rop shock. 4

5 EXAMPLE Cosier a fixe rop height. Also cosier that the mass is fixe, but that the sprig stiffess is variable. The table values show that lowerig the stiffess reuces both the acceleratio a the force. A lower stiffess coul be achieve by aig isolator mouts or some cushioig material. O the other ha, lowerig stiffess also icreases isplacemet. This is acceptable as log as the system remais liear a oes ot bottom out. The isplacemet limit of the sprig is thus a practical costrait. CONCLUSION A simple metho for moelig rop shock was erive. The erivatio was base o a simplifie free-vibratio moel. A rigorous erivatio of the free-vibratio equatio is give i Appeix C. 5

6 APPENDIX A Agai, the mass uergoes free vibratio at its atural frequecy. The istataeous power flow is P(t) = f(t)v(t) (A-1) P(t) = [mω g h si ω t][ g h cos ω t] (A-) P(t) = mg hω si ωt cosωt (A-3) Apply a trigoometric ietity. P(t) = mg hω si ωt (A-4) By ispectio, the peak istataeous power flow occurs at Substitute (A-5) ito (A-4). π ω t = (A-5) t = π 4ω (A-6) Pmax = ωmg h (A-7) P max k = mg h (A-8) m P max = g h km (A-9) Note that the istataeous power flow chages irectio as the polarity chages with time i equatio (A-4). 6

7 APPENDIX B The total impulse over a half-sie uratio is π/ ω = [f (t)]t 0 I (B-1) π / ω I = [mω ω 0 g h si t] t (B-) I 1 / m g h cos t π ω = ω ω (B-3) ω 0 I = m g h [ 1 1] (B-4) I = m g h (B-5) 7

8 APPENDIX C FREE VIBRATION DERIVATION Cosier a sigle-egree-of-freeom system. &&x m k c Where m is the mass, c is the viscous ampig coefficiet, k is the stiffess, x is the absolute isplacemet of the mass, g is the gravitatioal acceleratio. Note that the ouble-ot eotes acceleratio. The free-boy iagram is m &&x k x cx& Summatio of forces i the vertical irectio F = mx && (C-1) 8

9 mx && = cx& kx (C-) mx && + cx& + kx = 0 (C-3) Divie through by m, && x + c & m x k + m x = 0 (C-4) By covetio, ( c/ m) = ξω ( k / m) = ω where ω is the atural frequecy i (raias/sec), ξ is the ampig ratio. By substitutio, && x + ξω x& + ω x = 0 (C-5) Now take the Laplace trasform. L {&& & x + x+ x} = L{ 0} ξω ω (C-6) s X() s sx( 0) x& ( 0) + ξωsx() s ξωx( 0) + ω Xs () = 0 (C-7) { ξω ω } { } { ξω} s + s+ X() s + 1 x&( 0) + s x( 0) = 0 (C-8) { + ξω + ω } = + { + ξω} s s X() s x& ( 0) s x( 0) (C-9) 9

10 { ξω } x&( 0) + s+ x( 0) Xs () = s + ξωs+ ω (C-10) Cosier the eomiator of equatio (C-10), ( ) ( ) s + ξωs+ ω = s+ ξω + ω ξω (C-11) ( ) ( ) s + ξω s+ ω = s+ ξω + ω 1 ξ (C-1) Now efie the ampe atural frequecy, = ω 1 ξ (C-13) Substitute equatio (C-13) ito (C-1), ( ) s + ξωs+ ω = s+ ξω + ω (C-14) + { + ξω} ( s + ξω ) + ω x&( 0) s x( 0) Xs () = Xs () = ( s+ ξω) ( s ξω ) x + ( ξω) ( s + ξω ) + x( 0) &( 0) x( 0) ω ω (C-15) (C-16) Xs () = ( s+ ξω) ( s ξω ) ( ξω ) x& ( 0) + x( 0) x( ) ω ω ( s + ξω) + ω (C-17) 10

11 Now take the iverse Laplace trasform usig staar tables. The resultig isplacemet is ( ξω ) x& ( 0) + x( 0) xt () = exp ( ξω t) [ x( 0)cos ] ( t) + si ( ω t) (C-18) { 0 [ 0 0 ] ( )} xt () = exp ( t ) [ x ()cos ] ( t ) x & ( ) ( ) x ( ) si 1 ξω ω ω + + ξω ω ω t (C-19) The velocity is ( ξω ){ [ 0 ] ( ) [ 0 ( ξω) 0 ] ( )} { 0 [ 0 0 ] ( )} xt &( ) = ξω exp t x( )cos t + x&( ) + x( ) si t ( ) [ ] ( ) ( ) + exp ξω t x( ) si t + x&( ) + ξω x( ) cos t (C-0) ( ) [ ] ( ) ( ) ξω [ ] ( ) xt &( ) = exp ( ξω t) ξω[ x()cos 0 ] ( t) + x&( 0) + ( ξω) x( 0) si t ω { 0 [ 0 0 ] ( )} + exp ξω t x( ) si t + x&( ) + ξω x( ) cos t (C-1) { 0 [ 0 0 ]} ( ) ( ) [ ] ( ) xt &( ) = exp ξω t ξω x() + x&( ) + ξω x( ) cos t ξω [ ] ( ) + exp ( ξω t) [ x( 0) ] + x& ( 0) + ( ξω) x( 0) si t (C-) 11

12 ( ξω ){ 0 } ( ω ) xt &( ) = exp t x&( ) cos t ξω [ ] ( ) ξω + exp ( ξω t) x&( 0) [ x( 0) ] + ( ξω ) x( 0) si t (C-3) ( ξω ){ 0 } ( ω ) xt &( ) = exp t x&( ) cos t ξω + exp ( ξω t) x&( 0) + + ( ξω ) x( 0) si t ξω [ ] ( ) (C-4) ( ξω ){ 0 } ( ω ) xt &( ) = exp t x&( ) cos t ( ξω ) exp ( t) + x&( ) + + ξω ξω 0 ω x( 0) si t ω ( ω ) (C-5) ( ) &( ) exp ( ) &( )cos( ) ξω ξω xt = t x t + x&( ) ξω 0 ω 0 ω x( 0) si( ω t) + + (C-6) [ ] ( ) 1 xt &( ) = exp ( t) x&( )cos( t) ξω 0 ω ξωx& ( 0) + + ( ξω) x( 0) si t ω (C-7) 1

13 Asie [ ω ( ξω ) ] ( ξ ) ω ( ξω ) 1 + = + (C-8) [ ( ) ] ω + ξω = ω (C-9) xt &( ) = exp ( ξω t) x&( )cos( t) { x&( ) + x( )} si( t) 0 1 ω ω ξω 0 ω 0 ω (C-30) xt &( ) = exp ( ξω t) x&( )cos( t) { x&( ) + x( )} si( t) 0 ω ω ω ξ 0 ω 0 ω (C-31) The acceleratio is ξω &&() xt = exp ( ξω t) ξωx& ( 0)cos ( t) + { ξx&( 0) + ωx( 0) } si( t) { } ( ξω t) ω x 0 ( ω t) ω { ξx 0 ω x 0 } ( ω t) + exp &( )si &( ) + ( ) cos (C-3) { } ( ξω ) ξω 0 ( ω ) ω { ξ 0 ω 0 } ( ω ) &&() xt = exp t x&( )cos t x& ( ) + x( ) cos t ξω + exp &( )si + & ( ) + ( ) si ( ξω t) ω x 0 ( ω t) { ξx 0 ω x 0 } ( ω t) (C-33) 13

14 { } ( ) ( ) 0 { 0 0 } &&() xt = exp ξω t ξω x&( ) ω ξx& ( ) + ω x( ) cos ω t ξω + exp &( ) + &( ) + ( ) si ( ξω t) ω x 0 { ξx 0 ω x 0 } ( ω t) (C-34) ( ξω ){ ξω 0 ω ξ 0 ω 0 } ( ω ) &&() xt = exp t x&( ) x&( ) x( ) cos t ξ ω ξω 3 + exp &( ) + &( ) + ( ) si ( ξω t) ω x 0 x 0 x 0 ( ω t) (C-35) ( ξω ){ ξω 0 ω 0 } ( ω ) &&() xt = exp t x&( ) x( ) cos t ξ ω ξω 3 + exp & ( ) ( ) si ω ( ξω t) ω x 0 + x 0 ( ω t) (C-36) ( ξω ){ ω }{ ξ 0 ω 0 } ( ω ) &&() xt = exp t x&( ) x( ) cos t 1 ( ξω t) { [ ξ ω ω ] x 0 + ξω 3 x 0 } ( t) + exp &( ) ( ) si (C-37) Furthermore [ ( ) ] ω + ξω = ( ξ ) ω + ( ξω ) [ ( ) ] ω + ξω = ( + ξ ) ω + ( ξω ) 1 (C-38) [ + ( ) ] = ( + ) 1 (C-39) ω ξω 1 ξ ω (C-40) 14

15 By substitutio, ( ξω ){ ω }{ ξ 0 ω 0 } ( ω ) &&() xt = exp t x&( ) x( ) cos t 1 ( ξω ) ( ) {[ 1+ ξ ω ] 0 + ξω 3 0 } ( ) + exp t x&( ) x( ) si t (C-41) ( ξω ){ ω }{ ξ 0 ω 0 } ( ω ) &&() xt = exp t x& ( ) x( ) cos t ω ( ξω t) { [ 1 ξ ] x 0 ξωx 0 } ( t) + exp &( ) ( ) si ω + + (C-4) &&() xt = ( t) { x x } ( t) + ω 0 0 {[ 1+ ξ ] x 0 + ξωx 0 } ( t) ω exp ξω ξ&( ) ω ( ) cos & ( ) ( ) si (C-43) 15

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