Degrees of freedom and coordinates

Transcript

1 Degrees of freedom and coordinates Multibody containing N rigid parts: α, α =,..., N Configuration coordinates: ς = ς, ς,..., ς, ς 6N 6N (6N DOF, Gross) ς = x, ς = y, ς = z, ς = ψ, ς = θ, ς = φ,..., etc. c c c Configuration ς

2 ς Constraints For a specific multibody the - coordinates will, in general, not be independent. Due to the interconnections (different types of joints ) between parts of the multibody there may, for instance, be a number of geometric constraints requiring specific relationships between the ς coordinates. Geometrical constraints: 6N 6N C( t, ς) = C( t, ς, ς,..., ς, ς ) = 0 6N 6N C( t, ς) = C( t, ς, ς,..., ς, ς ) = 0 6N 6N Cm( t, ς) = Cm( t, ς, ς,..., ς, ς ) = 0 Number of constraint equations: m

3 Constraints By differentiating the constraint conditions with respect to time: Holonomic kinematical constraints: C i 6N Ci Ci k = + ς = 0, i =,..., m k ς t k= k ς, i=,..., 6N are not independent. They are related by this equations. C i m ( 6N) C i Constraint matrix: ( ), rank( ) j j ς ς = m 3

4 Configuration space (manifold) { 6N ς C ς } Ct( ) = i( t, ) = 0, i =,..., m n-dimensional surface in 6N We introduce independent, configuration coordinates: n q= q, q,..., q n= 6N m C ( ) ˆ t ς = ς( tq, ), q Ω n ς = ˆ( ς tq, ) Motion: q = qt () n x= χ( Xt, ) = χ ( tq, ( t), q( t),..., q( t); X) = χ ( tqt, ( ); X) q q 4

5 Multibody in motion Rigid body: α α α χ (, tqx ; ) = x(, tq) + (, tq)( X X), X B, q= qt () q c R c 0 5

6 6

7 The configuration space: S 7

8 The configuration space: T Example 9. continued. The planar double pendulum has two degrees of freedom. We introduce angular configuration coordinates q = θ and q = φ according to the Figure 9.6. Considering the bars as one-dimensional parts, the transplacement of the double pendulum may be written xo + ( isinθ jcos θ) X, 0 X l, ( for ) x= χq ( θφ, ; X) = xo + ( isinθ jcos θ) l+ ( isinφ jcos φ) X, 0 X l, ( for ) { } ( θφ, ) Ω= ( θφ, ) π< θ< π, π< φ< π where the RON-basis (, i jk, ) is shown in the figure above. The configuration coordinate system is thus scleronomic. Note that the transplacement for may be written x= χ( θφ, ; X) = x + i( l sinθ + X sin φ) j ( l cosθ + X cos φ), 0 X l q O The configuration space is, in this case, a torus T = S S. ϕ θ 8

9 Multibody kinematics Motion: q = qt () n x= χ( Xt, ) = χ ( tq, ( t), q( t),..., q( t); X) = χ ( tqt, ( ); X), X B 0 q q Velocity: χ n χ x = x q (, tqqx, ; ) = + q k t q q q k k= x q q k = χ q q k n Generalized velocity: q = ( q q ) 9

10 0

11 Multibody kinematics Deformation measures Deformation gradient: F (, tqx χq = Fq ; ) = X F = F + F = F q t q q n n q q k q q k k k= k= 0 k Velocity gradient: G= FF = F F n q k k q q k= 0 q, n T q q D= ( F Fq + F F q ) q k k q q k= 0 T k Green-St. Venant: n T Fq Fq E = F DF = ( Fq + Fq) q k k q q T T k k= 0

12 The principal of virtual power Virtual power of external, internal and inertial forces: VP e e ( w) = w df, P P VP i i ( w) = w df, P P VP a ( w) = w a dm P P P VP e ( w) + VP i ( w) = VP a ( w), w W 0, ( Pvp) { w: V w w( ) } 0 W = = P is a continuous mapping

13 Virtual powers: The principal of virtual power in continuum mechanics e VP ( w ): = w Pn da( X ) + w bρ dv( X ) 0 0 P 0 P 0 i VP ( w) : = P Gradwdv( X ), B 0 VP a ( w): = w x ρ dv( X ) P 0 0 VP e ( w) + VP i ( w) = VP a ( w), w W 0, ( Pvp) Virtual velocity field: { w: B0 V w w( ) } W = = X is a continuously differentiable mapping 3

14 Virtual velocity fields Motion: x= χ ( tqt, ( ); X), X B 0 q Virtual velocity field: w n α χq (, tqx ; ) k k = wq ( tqwx,, ; ) = w, w k q k= Space of virtual velocity fields: n α χ (, tqx ; ) α Wq = w: B0 V w = w, X B0, w= ( w, w,..., w ) k k= q q k n n 4

15 9.4 The virtual power of the internal forces i VP ( w) = P Gradwdv( X ) B 0 n α n α Grad χq k q k Grad w = w = w, X k F B k q q k= k= α 0 Proposition 9. The virtual power of the internal forces may be written VP n i i k ( w ) = Qk w k= where the so-called generalized internal forces are defined by Fq E i q Qk = P dv( X ) = dv( X ) k k q S q B B 0 0 5

16 The virtual power of the external forces Proposition 9.3 The virtual power of the external forces may be written VP n e e k ( w ) = Qk w where the so-called generalized external forces are defined by k= Q χ e q da q k 0 ( X χ = Pn ) + ρ dv 0 ( X ) k k q b q B 0 B 0 The boundary of B 0 : N α 0 U B0 α = B = 6

17 The virtual power of the inertial forces Proposition 9.4 The virtual power of the inertial forces is given by VP n a a k ( w) = Qk w k= (9.30) where the so-called generalized inertial forces are defined by Q d T T = ( ) dt q q (9.3) a k k k where T denotes the kinetic energy of the multibody given as a function of tqq,, q q T = T (, t q, q) = x x ρ0dv( X ) B 0 7

18 The principle of virtual power Lagrange s equations Proposition 9.5 n k= ( Q Q Q ) w = 0, w a i e k n k k k Theorem 9. (Lagrange s equations) d T T i e ( ) Qk Qk = 0, k =,..., n k k dt q q Lagrange s equations constitute a set of n second order ordinary differential k k equations for the determination of the functions q = q ( t), k =,..., n Initial conditions: q k ( 0) = q k, q k ( 0) = q k, k =,..., n 0 0 8

19 Summary 9

20 Lagrange s equations d T T i e ( ) Qk Qk = 0, k =,..., n k k dt q q What needs to be discussed? Kinetic energy Coordinate change Internal forces, constitutive assumptions Interactions between parts External forces Power theorem 0

21 Particle in plane circular motion y r ϕ x x= rcosϕ y = rsinϕ q = ϕ x= χ ( t, ϕ; P) = x + ircosϕ+ jrsinϕ q O χ t q χq = 0, = i( rsin ϕ) + jrcosϕ ϕ x q q = x χ χ q ( ϕϕ, ; X) = + ϕ= ( ( rsin ) + rcos ) t i ϕ j ϕϕ ϕ

22 Particle in plane circular motion Kinetic energy y r ϕ x Ek = T( ϕϕ, ) = x m= ( i( rsin ϕ) + jrcos ϕ) ϕm= rm ϕ

23 Particle in plane circular motion External forces, internal forces e = + F N mg χq N = 0 ϕ mg = j( mg) r y O χ q ϕ ϕ N P mg x g = j( g) χq χ e e q χq Qϕ = F = ( N + mg) = mg = ϕ ϕ ϕ ( i( rsin ϕ) + jrcos ϕ) mg = mgrcosϕ i Q = ϕ 0 3

24 Particle in plane circular motion Lagranges equations y T T( ϕϕ, ) = rm ϕ, = rm ϕ, ϕ T = 0 ϕ r ϕ x Q e ϕ = mgr cos ϕ, i Q = ϕ 0 Lagranges equations: d T T i e g ( ) Q Q = 0 r m + mgr cos = 0 ϕ ϕ ϕ ϕ ϕ+ cosϕ = 0 dt ϕ ϕ r 4

25 The spherical pendulum g = e ( g) z x = Lsinθcosϕ y = Lsinθsinϕ z = L cosθ q = θ, q = ϕ x= χ( t, θϕ, ; P) = x + e Lsinθcosϕ+ e Lsinθsin ϕ+ e ( Lcos θ) q O x y z χ t q χq = 0, = exlcosθcosϕ+ eylcosθsinϕ+ ezlsin θ, θ χq = ex( Lsinθsin ϕ) + eylsinθcosϕ ϕ 5

26 The spherical pendulum Material velocity e z O e y e x θ L ϕ P χ χ x = x (,, ; ) q q q q tqqx = + χ θ + ϕ = ( exlcosθcosϕ+ eylcosθsinϕ+ t θ ϕ e Lsin θθ ) + ( e ( Lsinθsin ϕ) + e Lsinθcos ϕϕ ) = z x y e L( θ cos θcosϕ ϕsinθsin ϕ) + e L( θ cos θsinϕ+ ϕsinθcos ϕ) + e Lθ sinθ x y z 6

27 The spherical pendulum Kinetic energy e z O e y e x θ L ϕ P Ek = T( θθϕϕ,,, ) = x m= ( θ + ϕ sin θ) Lm 7

28 Spherical pendulum External forces, internal forces e F = S+ mg pop S = S pop g = ez ( g) θ ϕ S mg S = ( e Lsinθcosϕ+ e Lsinθsin ϕ+ e ( Lcos θ)) S, S 0 x y z χq χq χq S = S = 0, mg = mglsin θ, θ ϕ θ χq mg = 0 ϕ 8

29 Spherical pendulum Lagranges equations T( θθϕϕ,,, ) = ( θ + ϕ sin θ) Lm, T = Lm T, θ = Lm ϕ sinθcos θ, θ θ T = Lm ϕsin θ, ϕ T = 0 ϕ Q e θ e = mglsin θ, Q 0, ϕ = i i Qθ = Qϕ = 0 Lagranges equations: d T T i e ( ) Q Q = 0 Lm Lm sin cos + mgl sin = 0 dt θ θ θ ϕ θ θ θ θ θ d T T i e d ( ) Q Q = 0 ( Lm ϕ ϕ ϕsin θ) = 0 dt ϕ ϕ dt 9

30 The kinetic energy q q T = T (, t q, q) = x x ρ0dv( X ) B 0 χ n n q χq χ k q x = x q (, tqqx, ; ) = + q = q k k t q q k= k= 0 k χ χ χ χ x x = = n n n q k q l q q k l q k q l qq k l k= 0 q l= 0 q k, l= 0 q q n χ χ T = T (, t q, q ) = 0dv( X ) q q k l ρ q q kl, = 0 B 0 q q k l 30

31 The kinetic energy 3

32 The kinetic energy where the matrix elements, m, kl, = 0,,..., n, are defined by kl χq χq m00 = m00 (, t q) = ρ 0dv( X ) t t B 0 χq χq m0k = m0k( t, q) = 0dv( X ) mk0, k,..., n k ρ = = t q B 0 χq χq mkl = mkl ( t, q) = 0dv( X ) mlk, k, l,..., n k l ρ = = q q B 0 3

33 The planar pendulum Alternative calculation of kinetic energy 0 O v O j k i θ c v = i v, v = v ( t), ω= kω, ω = θ O O O O The kinetic energy: Ek = Tt (, θθ, ) = v cm+ ω Iω c 33

34 The planar pendulum 0 O j k i θ c l l poc = i sin θ + j( cos θ) l l vc = vo + ω poc = ivo + kθ ( i sin θ + j( cos θ)) = l l i( vo + θ cos θ) + j θ sin θ 34

35 The planar pendulum 0 O j k i θ c l l l l vc = ( vo + θ cos θ) + θ sin θ = vo + voθ cos θ + θ cos θ + θ sin θ = v l + voθ cos θ + θ 4 O vcm= ( vo + voθ cos θ + l θ ) m 4 35

36 The planar pendulum 0 j O k i θ c ω Iω= kθ Ikθ = θ k Ik= θ I c c c c, zz I c, zz I c, zz = ml 36

37 The planar pendulum 0 O j k i θ c l ml Tt (, θθ, ) = vcm+ ω Iω c = ( vo + voθ sin θ+ θ ) m+ θ = 4 vm O + ( vocos θ) mθ + l mθ, 3 v () O = vo t 37

38 The planar pendulum 0 O j k i θ c l Tt (, θθ, ) = vm O + ( vocos θ) mθ + mθ, 3 T T T 0 v () O = vo t m00 = vom, m0 = m 0 = ( vo cos θ ) m, m l = m 3 38

39 The planar double pendulum Calculation of the kinetic energy g 0 O j k i θ A φ B The transplacement: x xo + ( isinθ jcos θ) X, 0 X l, ( for ) = χq ( θφ, ; X) = xo + ( isinθ jcos θ) l+ ( isinφ jcos φ) X, 0 X l, ( for ) 39

40 40

41 Example 9.3 continued. χ q 0, 0 X l, ( for ) = φ ( icosφ+ jsin φ) X, 0 X l, ( for ) The volume element for of m = ρ Al. For and 0 dv X is ( ) = A dx where A is the constant cross-sectional area dv X we have ( ) = A dx and m= ρ0al. Then χq χq χq χq χq χq mθθ ( θφ, ) = ρ0dv( X ) = ρ0dv( X ) + ρ0dv( X ) = θ θ θ θ θ θ B B B χ χ χ χ l l l l q q q q ρ0adx + ρ0adx = X ρ0adx + lρ0adx = + ml θ θ θ θ ml χq χq χq χq χq χq mθφ ( θφ, ) = ρ0dv( X ) = ρ0dv( X ) + ρ0dv( X ) = θ φ θ φ θ θφ B B B χq χq χq χq χq χq ρ0adx + ρ0adx = ρ0adx = cos( φ θ) lx ρ0adx = θ φ θ φ θ φ l l l l mll cos( φ θ) 4

42 χq χq χq χq χq χq mφφ ( θφ, ) = ρ0dv( X ) = ρ0dv( X ) + ρ0dv( X ) = φ φ φ φ φ φ 0 B B B χ χ χ χ ρ0dv( X ) = ρ0adx = X ρ0adx = φ φ φ φ l l q q q q ml B Thus M ml mll + ml cos( φ θ) 3 = M ( θφ, ) = mll ml cos( φ θ) 3 and the kinetic energy of the double pendulum then reads T ml mll ml cos( ) + φ θ 3 θ = T( θφθφ,,, ) = ( θ φ) mll ml cos( ) φ φ θ 3 4

43 The planar double pendulum Alternative calculation of kinetic energy ω = kω, ω = θ, ω = kω, ω = φ 43

44 The planar double pendulum Ek = T( θθφφ,,, ) = vc m + ω Ic ω + vc m + ω Ic ω = vc m + kω Ic kω + vc m + kω Ic kω = vc m + k Ic kθ + v c m + k Ic k φ = I c, zz c, zz I vc m + Ic, zzθ + vc m + I c, zzφ = ml ml v v c m+ c m+ θ + φ 44

45 The planar double pendulum m m vcm ( ) cos( ) + vcm = + m lθ + l φ + mll φ θ 4 4 m m E (,,, ) ( ) k= T θθφφ = + m lθ + l φ + mll cos( φ θ) ml ml ( m ) m θ + φ = + m lθ + l φ + mll cos( φ θ) = 3 3 ml mll ml cos( ) + φ θ 3 θ ( θ φ) mll ml cos( ) φ φ θ 3 M 45

46 The kinetic energy T T= Ttqq (,, ) = ( M0 + Mq + qmq ) M = M(, tq) = m ( ) T q= q q K q n n ( ) M = M(, tq) = m + m m + m m + m n n0 n m m mn m m mn n n T M M( tq, ) = =, M = M mn mn mnn 46

47 Mass matrix elements where the matrix elements, m, kl, = 0,,..., n, are defined by kl χq χq m00 = m00 (, t q) = ρ 0dv( X ) t t B 0 χq χq m0k = m0k( t, q) = 0dv( X ) mk0, k,..., n k ρ = = t q B 0 χq χq mkl = mkl ( t, q) = 0dv( X ) mlk, k, l,..., n k l ρ = = q q B 0 47

48 Mass matrix elements T T = T (, t q, q ) = q Mq T ( ) ( ) 0 n n+ 0 q = q q q, q = t M m00 m0 m0n M0 M m0 m mn = Mtq (, ) = = M T M mn0 mn mnn 48

50 The kinetic energy Proposition 9.8 The transport inertial force may be written Q m m n ta i0 00 k = k i= t q 50

52 Lagrange s equations d T T ta ga i e ( ) + Qk + Qk Qk Qk = 0, k =,..., n k k dt q q T T tqq mqq qmq n k l T = (,, ) = kl = kl, = 5

53 Lagrange s equations Mq = Q + Q + Q cif T it et 53

54 The complementary inertia force in a scleronomic coordinate system cif T M M t T M M M M 0 Q = q ( ( ) ) q + q ( + skew( )) + ( ) = q q t q t q q M = 0, M = 0 T 0 n M M t ( ( )) q q q q M cif = 0 n n n Q = 0 n 54

55 Summary 55