2. Some analytical models of nonlinear physical systems

Transcription

1 . Some analytical models of nonlinear physical systems (a)discrete Dynamical Systems: 1. Inverted Double Pendulum:Consider the double pendulum shown: k 1 x 1 A l O k l y P g k,k 1 - linear torsional sti fnesses l-length of each rod P -external (conservative) load m mass of each rod 1

2 Inverted Double Pendulum equation of motion The equations of motion can be determined by using Lagrange s equations: d T T V nc ( ) Q i, i 1,,3,... dt q q q q i i i i Here, are the generalized coordinates, T and V are, respectively, the kinetic and potential nc Q i Energies for the system, are the generalized forcesdue to non-conservative efects.

3 Inverted Double Pendulum equations of motion For the double pendulum: kinetic energy: T T1 T [(mv G1 I G1 1) (mvg I G )]/ T (ml/3) /; v v v ; v k r 1 1 G A G/ A A 1 OA r l(cos i sin j); v l( sin i cos j); OA 1 1 A vg/a k r G/A; rg/a l(cos i sin j)/; vg/a l( sin i cos j)/ ; T (ml/1) / ml[ /4 cos( )/]/

4 Inverted Double Pendulum equations of motion potential energy: V V V mglcos / 1 1 mg[lcos lcos /][k k ( ) ]/ The work done by the external force P in a virtual displacement from straight vertical position is: W Pi r ; B B B r l(cos i sin j) l(cos i sin j) 1 1 r l( sin i cos j) l( sin i cos j) W P[ lsin lsin ] The 1 1 generalized forces are: Q Plsin ; Q Plsin ; 1 1 4

5 Inverted Double Pendulum equations of motion: Equation for 1: d T dt T ( ) ml[ 1/3 1 cos( 1)/4 1 1 V 1 ( )sin( )/4] 1 1 ml 1sin( 1)/4 3mglsin / (k k ) k ml[4 /3 cos( )/4 sin( )/4] (k k ) k 3mglsin / Plsin

6 Inverted Double Pendulum equations of motion: Equation for : d T dt T ( ) ml[ /3 1cos( 1)/4 V 1 ( )sin( )/4] ml 1sin( 1)/4 mglsin / k k 1 ml[ /3 cos( )/4 sin( )/4] k k mglsin / Plsin 1 6

7 Inverted Double Pendulum equations of motion: We now consider a simplified version with k 1 = k =k Let k k/ml,p Pl/ml,M mgl/ml Then, the equations are: Equation for 1: [4 /3 cos( )/4 sin( )/4] k k (P 3M)sin 1 1 Equation for : [ /3 cos( )/4 sin( )/4] k k (P M)sin 1 7

8 Discrete dynamical systems.. Inverted Double Pendulum with Follower Force: Consider the same system as in last example, except that the force P changes direction depending on the orientation of the body on which it acts. P The force P now acts at B an angle to the rod AB and x l always maintains this A direction relative to the rod 1 k l regardless of the position in g k space of the system during 1 its oscilations. O y 8

9 Inverted Double Pendulum with Follower. equations of motion Note that the only change is in the efect of the external force P. The potential and kinetic energy expressions remain the same. So, potential energy: V V V mglcos / 1 1 mg[lcos 1 lcos /][k 11 k ( 1) ]/ The work done by the external force Pin a virtual displacement from straight vertical position is: W P[cos( )i sin( )j] r ; r l(cos i sin j) l(cos i sin j) B 1 1 B 9

10 Inverted Double Pendulum with Follower. equations of motion The virtual displacement is: r l( sin i cos j) l( sin i cos j) B So,the virtualwork doneis W Pl[ sin( ) sin ] 1 1 Thus, the generalized forces are: Q1 Plsin( 1 ); Q Plsin. Now, we can write the equations of motion easily: ml[4 /3 cos( )/4 sin( )/4] (k k ) k 3mglsin / Plsin( )

11 Inverted Double Pendulum with Follower. equations of motion and ml[ /3 cos( )/4 sin( )/4] In the reduced case, with equal springs etc., the equations are: k k mglsin / Plsin 1 [4 /3 cos( )/4 sin( )/4] k k Psin( ) 3Msin [ /3 cos( )/4 sin( )/4] k k PSin Msin 1 11

12 Discrete dynamical systems. 3. Dynamics of a Bouncing Ball:Consider a bal bouncing above a horizontal table. The table oscilates verticaly in a specified manner (here we assume harmonic osci lations). Y(t) mg X(t)=Asint ball g table ground The motion of the bal, during free flight, is governed by my=-mg or y=-g. Integrating once gives: y=y -g(t-t )

13 Dynamics of a Bouncing Ball. Integrating once gives: Integrating again, we get The position of the bal has to remain above the table Y(t) mg X(t)=Asint y(t) X(t),t t 0 ball g table ground y=y -g(t-t ). 0 0 y=y 0+y 0(t-t 0)-g(t-t 0) /. Finaly, we have the law of interaction between the table and the bal: If weassume simplelaw of impac t, therelativevelocities beforeand after impact arerelate d bycoefficient of restit ution. 13

14 Dynamics of a Bouncing Ball. Thus, we have the relation: V(t)-X(t)=e[X(t) U(t)] i i i i where V(t)- velocity of the bal immediately after impact i i and U(t)- velocity of the bal just before impact Y(t) mg X(t)=Asint ball g table ground These relations provide a completedescription of motion of the bal. Note that, given an initial condition, we have to piece together the motion in forward time 14

15 Dynamics of a Bouncing Ball. Let us now proceed in a systematic manner. For nonlinear analysis, it is always advisable to nondimensionalizeequations: So we define Time: =t/t, where T=/ z(t) ball mg g Table X(t)=Asint ground Acceleration units:g/ Velocity units:(g/)t=g/ Pos. units:(g/)t= g/ So, X(t)= ( g/ ) X( ) =Asin(t) =Asin() X( ) sin( ); g A 15

16 Dynamics of a Bouncing Ball. Ball motion: z(t)= ( g/ )Z(), dz g dzd g dz ( ) ( ) ( ) dt d dt d d z(t) g dz ( ) ( ) ( ) ball dt d dt d d mg g Table X(t)=Asint ground d z g d Zd g d Z g d Z Now, z g Z. Integrating, Z( ) Z ( ), 0 0 Z( ) Z Z( ) ( )

17 Dynamics of a Bouncing Ball. Thus, we have Ball motion: Z ; Z( ) Z 0 ( 0), (1) Z( ) Z Z( ) ( ) () z(t) mg X(t)=Asint ball g Table ground Table motion: X( ) sin( ) (3) Initialy: bal starts at time 0 when it is in contact with the table, and just about to leave X( ) Z( ) Z sin( ) (4)

18 Dynamics of a Bouncing Ball. Bal velocity at = 0 : dz d 0 Z W X( ) Z W cos( ) (5) (Here W 0 isrelative velocity ofthebal l, an unknown) 0 z(t) mg X(t)=Asint ball g Table ground Next colision at time 1 > 0 when Z( ) X( ) W( ) 0 Now, W( ) Z Z( ) ( 0) sin( ) (6) Using (3)-(5) 18

19 Dynamics of a Bouncing Ball. W( ) [sin( 0) sin( )] [W 0 cos( 0)]( 0) ( 0) Then, the time instant 1 is defined by z(t) mg X(t)=Asint ball g Table ground W( 1) [sin( 0) sin( 1)] [W cos( )]( ) ( ) 0 (7) 1 0 (this is a relation in W 0, 1 and 0, and it depends on ) 19

20 Dynamics of a Bouncing Ball. When just about to contact at this time instant 1 the relative velocity is z(t) dw( ) cos( 1) [W 0 cos( 0)] ( 1 0) (8) d 1 mg X(t)=Asint ball g Table ground W W ( ) Z( ) X( ) On impact, the bal relative velocity changes: W or: ew 1 1 (e coeficient ofrestitut ion) W e[{cos( ) 1 0 cos( )} W ( )] (9)

21 Dynamics of a Bouncing Ball. One can write the equations now in a more compact form: and z(t) W ( i 1) [sin( i) sin( i 1)] [W cos( )]( ) ( ) 0 (A) W mg X(t)=Asint i i i1 i i1 i e[{cos( ) cos( )} W i1 i i1 i ball g Table ( )] (B) ground i1 i Knowing ( I,W i ),equations (A)and (B)can be used to compute ( I+1,W i+1 ),thus generating the trajectory. 1

22 (b)continuous Dynamical Systems 1. Rotating Thermosyphon:Consider a closed circular tube in a vertical plane. The tube is filed with a liquid of constant properties, except for variation of its density with r temperature in cooling buoyancy and R centrifugal terms, i.e. in body forces. One part of the loop is heated, heating and the other cooled. The tube is spun about the vertical axis.

23 Rotating Thermosyphon There are many ways to develop a model for the system. If the tube radius r is much smaler than the torus radius R, one can assume that there is negligible flow in the r radial direction. Another cooling approach is to avarage the velocity and R temperature over the tube radius. Then, the various equations for heating fluid motion are: 3

24 Rotating thermosyphon:equations 1 (V) Continuity: 0 (1) t R Here, V avarageflow velocity at any section ; -density of the fluid and (V) is independent of. Momentum: cooling R r (V) 1 (V) 1 p gcos t R R w R cos sin () r heating Here, p fluid pressure at a section, w -shear stress at the wal 4

25 Rotating thermosyphon:equations Energy: T V ( T) k T r Cp r q( ) (3) t R R where C p -specific heat of the fluid, T mean fluid temperature at a section, r k thermal conductivity, q applied heat source R per unit length. Remark:Here viscous dissipation term is heating neglected cooling 5

26 Rotating thermosyphon:equations Simplification and nondimensionalization: Integrating the momentum eqn. () along the loop (V) 1 (V) 1 p cooling R d d d gcos d t R R r w d R d r heating 0 0 cos sin (4) Shear stress: w f V / (5) Friction factor: f16/re,re Vr/ 6

27 Rotating thermosyphon:equations Simplification: Introduce the variation of density with temperature in buoyancy and centrifugal terms, and use periodicity of variables (eqn. (4)) V 3 V g ( T T)cos r d t ( r) 0 R T Tr d 0 ( )cos sin (6) where -kinematicviscosity, T r -reference temperature -coeficient of thermal expansion 7

28 Rotating thermosyphon:equations Non-dimensional variables: 3 ( ) t,,, ( ) r 3 T Tr U V r R T 048 R q'( ) Ra* r 1 T, Q( ) ( ) g( r) 8 C T Pr R 819 Here Pr C / k Ra* 4 p p r) q C 3 ( ' k Prandtl number p ModifiedRayleigh number 8

29 Rotating thermosyphon:equations Non-dimensional equations: The resulting momentum and energy equations are: U U cos d cos sin d (7) 0 0 U Q( ) (8) r where R/, g ( ) /8Pr R Note: -combination of geometric paramater and the Prandtlnumber Pr. Pr > 1 for ordinary fluids (air, water etc.) and <1 for liquid metals. ( r/ R) 9

30 Rotating thermosyphon:equations Solution approach:thesolutions to (7) and (8) can be expressed as: n (,) [ B( )sin( n) C ( )cos( n)] (9) n1 where as, the externaly imposed heat flux can be represented in a Fourier series as: n Q( ) [ Asin( n) D cos( n)] (10) n1 Substituting (9) and (10) in equations (7) and (8), and colecting the appropriate Fourier coeficients gives an infinite set of ordinary diferential equations In the unknowns U(), B n (), and C n (). n n 30

31 Rotating thermosyphon:equations Solution approach :It turns out that only five of these equations are independent-master equations. The remaining equations are linear equations for the remaining variable (caled slave variables and slave equations. Equation (8) gives: n0 n0 [ Bn( )sin( n ) Cn( )cos( n )] [ Bn( )sin( n ) Cn( )cos( n )] [ Bn( )sin( n) Cn( )cos( n)] n0 U [ An sin( n) Dncos( n)] n1 31

32 Rotating thermosyphon:equations Solution approach :Colecting terms of diferent n sgive: n 0: C 0 C0 n 1: B 1 B1 UC1 A1 C 1 C1 UB1 D1 n : B 4 B UC A C 4 C UB D n 3: B 3 9 B3 3 UC3 A3 C 3 9 C3 3 UB3 D3 (sin ) (cos) (sin ) (cos ) (sin3 ) (cos3 ) n p B p B puc A p : p p p p (sin ) pub D (cos p) (11) Cp p Cp p p p 3,4,5,6,... 3

33 Rotating thermosyphon:equations Solution approach :Now, considering equation (7) we get: U 0 U [ B( )sin( n) C ( )cos( n)]cos d 0 n1 n1 n [ B( )sin( n) C ( )cos( n)]cos sin d n Evaluating the two integral terms on the right-hand side, it is clear that only coeficients of cos( ) and sin( ) wil survive. Thus, we get U U C1 B (1) Equations (11) and (1) govern the dynamics. n n 33

34 (b)continuous Dynamical Systems. Buckling of Elastic Columns:Consider a thin beam that is initialy straight. Oxyzis coordinate system with x-y plane coinciding with undisturbed neutral axis of the beam. Let EI bending sti fness V X 34

35 Buckling of elastic.. V(s) vertical displacement of the centroidalaxis, X distance measured along the centroidalaxis from left end. Define: Then, the strain energy of the system is: X 3 3 x X/L; u V/L; LP/ EI; LK/ EI; V (u,,) 0 (1 1 u ( 1u 0 1 u )dx 1 1 u (0) u ( ) )dx 35

36 (b)continuous Dynamical Systems 3. Thin rectangular plates:consider a thin plate that is initia ly flat. Oxyz is coordinate system with x-y plane coinciding with undisturbed middle surface of the plate. Let h plate thickness. The equations of motion for the plate, for y,v moderately large displacements von Karman equations. z,w a x,u In here, we give a short review of the derivation of these equations. 36

37 Thin rectangular plates.. Consider a diferential plate element: The equations in the three directions are: z y N x N xy N y F x N y N xy N x N N x xy u h x y t x h x y t M N xy N y v M x y y (1) () w w ( N x ) ( N y ) x x y y w w w ( N xy ) ( N xy ) h F (3) x y y x t M xy xy 37

38 z y Thin rectangular plates.. The constitutive equations for a linearly elastic and isotropic material are: N x N xy N y F x N y N xy N Eh u w 1 x (1 ) [ x x/ i v y wy N x /] (4) N Eh v w N x 1 y (1 ) [ y y/ i ux wx N y /] (5) In these expressions, N, N i forces, M -moments i N Gh[ u v w w ] N (6) xy y x x y xy M D( w w ) (7) x xx yy 38

39 z Thin rectangular plates.. The constitutive equations for a linearly elastic and isotropic material are: M D( w w ) (8) y N x N xy N y F x N y N xy N x M D(1 ) w (9) G D y yy xx xy E/(1 ); Eh /1(1 ) 3 xy Also u,v,w displacements Substituting the force displacement relations in the dynamic equations give: 39

40 Thin rectangular plates.. The dynamic equations for a plate made of linearly elastic and isotropic material are then simplified by introducing a stress function such that: N, N, N (10) x y y x xy x y Then, (1) and () are automatica ly satisfied if inplane inertia terms are neglected. Furthermore, the expressions (7)-(9) can be substituted in (3) to get equation for transverse displacement. Also, a compatibility condition is (gives an equation for ): ( u v ) ( u v ) 0 (11) xyy yxx y x xy 40

41 (b)continuous Dynamical Systems 4. Flow between concentric rotating cylinders: Consider two concentric cylinders with radii a, b; b 1 a Let 1, angular velocities of inner and outer cylinders; let (u r,u,u z ) velocity components in a cylindrical coordinate system; p pressure at a point; We now define the equations of motion for the system. 41

42 Flow between concentric rotating. Equations of motion for the system:in cylindrical coordinate system the NS equations are b 1 a Dur u 1 p ur u ( u r ), Dt r r r r Du uu 1 p u u Dt r r r r Du Dt r r ( u ), z where 1 p z ( u ), (1) D ur u uz Dt t r r z z 4

43 Flow between concentric rotating. Equations of motion for the system:and b 1 a r r rr r z There is also the equation for mass conservation: ur ur 1 u uz 0. () r r r z The basic flowis defined by: u 0,u 0,u V(r) r (r) and p P(r) z 1dP V or and DDV 0 (3) dr r 43

44 Flow between concentric rotating. The basic flow : Equations (3) have solution of the form (r) A B/r b 1 a where A 1,B 1R1 1 1 Let us consider perturbations to the basic flow: r z W e can then obtain linearized equationsabout the basic flow and /, R /R (4) u (u,v u,u ),p P p 44

45 Flow between concentric rotating. Linearized equationsabout the basic flow: u r u r 1 p u r u u ( u r ) t r r r u u 1 p u u t r r r r (DV)u r ( u ) uz u 1 p t z z u u 1 u u r r r z r r z and 0 ( u ) z 45