mbd week 4 2017 - lth · 19 the configuration space: s2 o ... the planar double pendulum has two...

2017-02-15

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The MultiBody

1

Examination task 2: Problem 2:3

Weld (between cone and axle BC)

2

The Multibody

Revolute Joint

3

Vehicle driveline

The Multibody

4

The Multibody

...

5

1

...

3

4

2

, ,..., ,1 N

N

1

,

Part

Joint

5

Degrees of freedom and coordinates

( )Oi j k

( , , ) R Rc O c c cx x i j kx y z

( ) ( ), c cx X x X X X R 0B

, , ,c c c x y z

fixed in frame

Rigid transplacement:

, , Configuration coordinates:

Configuration manifold: ( )SOE V

(6 DOF, Gross)

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, , ... , , 1 2 6 N 1 6 N

Multibody containing N rigid parts: , ,...,1 N

Configuration coordinates:

, , , , , , ... , etc.1 1 2 1 3 1 4 1 5 1 6 1c c c x y z

Configuration

(6N DOF, Gross)

7


( ), .t t 0 T

( ( )) ( ( ))( ), c cx t t X X X R 0B

Motion:

( , ) ( ( ); )x X t t X

8

1

2

6 N

( )t

Constraints

, ,...,6 N

ki ii k

k 1

0 i 1 mt

C CC

( , ) ( , , , ... , , )

( , ) ( , , , ... , , )

( , ) ( , , , ... , , )

1 2 6 N 1 6 N

1 2 6 N 1 6 N

1 2 6 N 1 6 N

t t 0

t t 0

t t 0

1 1

2 2

m m

C C

C C

C C

Number of constraint equations: m

( )( ) ,m 6 Nij

C Constraint matrix: ( )i

jrank m

C

Holonomic kinematical constraints:

Geometrical constraints:

9

Plane mechanism

,1Oc 1 1 p i jx y

2Oc 2 2 p i jx y

, , ,1 1 1x yConfiguration coordinates: (6 DOF, Gross), , 2 2 2x y

, , ,0 1 2

Greubler: ( ) ( )m

ii 1

f 3 N 1 r 3 3 1 2 2 1 1

Revolute joint

Revolute joint

Translational joint

0

10

Plane mechanism

( , , , , , ) cos

( , , , , , ) sin

( , , , , , ) cos cos

( , , , , , ) sin sin

( , , , , , )

1 1 1 2 2 2 A 1 1

2 1 1 1 2 2 2 A 1 1

3 1 1 1 2 2 2 1 1 2 2

4 1 1 1 2 2 2 1 1 2 2

5 1 1 1 2 2 2

a 0

a 0

b c 0

b c 0

1C

C

C

C

C

x y x y x x

x y x y y y

x y x y x x

x y x y y y

x y x y ( ) cos ( ) cos1 2a b b c e 0

Constraints:

OA A A p i jx y

11

Plane mechanism

sin

cos

( ) sin sin

cos cos

( )sin ( )sin

1

1

i1 2j

1 2

1 2

1 0 a 0 0 0

0 1 a 0 0 0

1 0 b 1 0 c

0 1 b 0 1 c

0 0 a b 0 0 b c

C

Constraint matrix:

( )ij

rank 5

C

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Configuration space (manifold)

( ) ( , ) , ,...,6 Nt i t 0 i 1 m CC

, , ... , 1 2 nq q q q

ˆ( ) ( , ), nt t q q C

( , ) ( , ( ), ( ),..., ( ); ) ( , ( ); )1 2 nq qx X t t q t q t q t X t q t X

( )q q t

Configuration coordinates:

n 6 N m

Motion:

ˆ( , ) t q

n-dimensional surface in 6 N

13

Multibody in motion

0B

X

tB

x

1q

( )q q t

- : nq space

nq

( , ; )q t q

( , ; ) ( , ) ( , )( ), , ( )q c ct q X x t q t q X X X q q t R 0B

Rigid body:

14

Rheonomic and scleronomic coordinate systems

( , ) ( , ( ), ( ),..., ( ); ) ( , ( ); )1 2 nq qx X t t q t q t q t X t q t X

( , ) ( ( ), ( ),..., ( ); ) ( ( ); )1 2 nq qx X t q t q t q t X q t X

Scleronomic, no explicit time dependence:

Rheonomic, explicit time dependence:

Partial derivatives: ( , , ,..., ; ) ( , ; )1 2 nq qx t q q q X t q X

, ,..., ,q q q q

1 nt q q X

15

, , , nt 0 T q X 0B x

y

r

Particle in plane circular motion

Geometrical constraint: ( , ) 2 2 2x y x y r 0 C

Kinematical constraints: 2xx 2 yy 0 C

Constraint matrix: ,g 2x 2 yx y

C C

( , )1 2 2 2 2S x y x y r

( )rank g 1 16

Configuration coordinates: ,x y (2 DOF, Gross)

The configuration space:

x

y

r

Configuration coordinate: q

Greubler: ( ) ( )m

ii 1

f 3 N 1 r 3 2 1 1 1 1

cos

sin

x r

y r

N 2# parts:

17

1S

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Example 9.1 continued. We have

2

C C CC = x y z

x y z

and then ( ) , if ( ) ( )rank 1 0 0 0 x,y,z , ,C . Since n 3N m 3 1 2 , the configuration space of the particle

( ) ( ) ( )3P 0 x,y,z x,y,zCC

is a 2-dimensional surface in 3 , the sphere 2S . As configuration coordinates one may use angles ( , ) belonging to the spherical coordinate system. Then

sin cos

sin sin

cos

L

L

L

x

y

z

19

The configuration space: 2S

O

c

1

0

i

j

k

The planar pendulum

, , c c x yConfiguration coordinates:

( ) ( )m

ii 1

f 3 N 1 r 3 2 1 2 1

Revolute joint

N 2# parts:

20

(3 DOF, Gross)

( , , ) sin,

( , , ) cos

c c c O

2 c c c O

l0

2l

02

x y x x

x y y y

1C

CConstraints: m 2

cos,

sin

1 1 1

c c

2 2 2

c c

C C C l1 0

2gC C C l

0 12

x y

x y

Constraint matrix:

( )rank g 2

21

1SThe configuration space: Example 9.2 (The planar double pendulum). The planar double pendulum consists of two homogeneous, rigid and slender bars 1 and 2 of the same material with masses 1m and

2m , lengths 1l and 2l , respectively. The bars are connected by an ideal revolute joint at

point A and 1 is connected to the roof 0 by an ideal revolute joint at point O. See the figure below. The pendulum may move in a vertical plane. Figure 9.6 The planar double pendulum.

i

j

k

g 0

A

B

1

2

O

22( ) ( )

m

ii 1

f 3 N 1 r 3 3 1 2 2 2

Example 9.2 continued. The planar double pendulum has two degrees of freedom. We introduce angular configuration coordinates 1q and 2q according to the Figure 9.6. Considering the bars as one-dimensional parts, the transplacement of the double pendulum may be written

( sin cos ) , , ( )( , ; )

( sin cos ) ( sin cos ) , , ( )

1O 1

q 2O 1 2

x X 0 X l forx X

x l X 0 X l for

i j

i j i j

( , ) ( , ) , 2 where the RON-basis ( , , )i j k is shown in

the figure above. The configuration coordinate system is thus scleronomic. Note that the transplacement for 2 may be written

( , ; ) ( sin sin ) ( cos cos ), q O 1 1 2x X x l X l X 0 X l i j

The configuration space is, in this case, a torus 1 1T S S .

23

The configuration space: T Multibody kinematics

q q

k kq q

x

( , , ; )n

q q kq k

k 1

t q q X qt q

x x

( , ) ( , ( ), ( ),..., ( ); ) ( , ( ); ), 1 2 nq qx X t t q t q t q t X t q t X X 0B

1q

q

- : nq space

nq

q

Motion:

Velocity:

Generalized velocity: ( )1 nq q q

( )q q t

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Example 9.2 continued. For the transplacement q of the double pendulum one obtains

the following partial derivatives

q

t

0 ,

( cos sin ) , , ( )

( cos sin ) , , ( )

1q 1

21 2

X 0 X l for

l 0 X l for

i j

i j

, , ( )

( cos sin ) , , ( )

1q 1

22

0 X l for

X 0 X l for

0

i j

Thus the material velocity of the planar pendulum is then given by

( , , , , ; ) q q qq t X

t

x x

( cos sin ) , , ( )

(( cos sin ) ( cos sin ) , , ( )

11

21 2

X 0 X l for

l X 0 X l for

i j

i j i j

25

,n

q1 1 kqk

k 0

qq

F

G FF F

n nq q qk k

k kk 1 k 0

q qt q q

F F F

F

( , ; ) qq t q X

X

F F

( )Tn

q q1 T kq qk k

k 0

1q

2 q q

F F

D F F

( )Tn

q qT T kq qk k

k 0

1q

2 q q

F F

E F DF F F

Multibody kinematicsDeformation measures

Deformation gradient:

Velocity gradient:

Green-St. Venant:

26

The principal of virtual power

( ) ( ) ( ), , ( )e i a 0VP VP VP W Pvp w w w w

( ) ,e eP PVP d w w F

( ) ,i iP PVP d w w F

( )aP P PVP dm w w a

Virtual power of external, internal and inertial forces:

27

: ( ) 0W P is a continuous mapping w w wV

The principal of virtual powerin continuum mechanics

: ( ) 1W X is a continuously differentiable mapping w w w0B V

( ) : ( ) ( )e0 0VP da X dv X

w w Pn w b

00PP

( ) : ( ),iVP Grad dv X w P w0B

( ) : ( )a0VP dv X w w x

0P

Virtual powers:

Virtual velocity field:

28

( ) ( ) ( ), , ( )e i a 0VP VP VP W Pvp w w w w

Virtual velocity fields

( , ( ); ), qx t q t X X 0B

( , ; )( , , ; ) ,

nq k k

q kk 1

t q Xt q w X w w

q

w w

( , ; ): , , ( , ,..., )

nq k 1 2 n n

q 0kk 1

t q XW w X w w w w

q

w w0B V B

0B

X

tB

x

1q

( )q q t

- : nq space

nq

( , ; )q t q

Motion:

Virtual velocity field:

Space of virtual velocity fields:

29

The virtual power of the internal forces

, n n

q qk k0k k

k 1 k 1

GradGrad w w X

q q

F

w B

( ) ( )iVP Grad dv X w P w0B

Proposition 9.1 The virtual power of the internal forces may be written

( )n

i i kk

k 1

VP Q w

w

where the so-called generalized internal forces are defined by

( ) ( )q qik k k

Q dv X dv Xq q

F E

P S0 0B B

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The virtual power of the external forces

Proposition 9.3 The virtual power of the external forces may be written

( )n

e e kk

k 1

VP Q w

w

where the so-called generalized external forces are defined by

( ) ( )q qek 0 0k k

Q da X dv Xq q

Pn b00 BB

N

1

0 0B BThe boundary of :0B

31

The virtual power of the inertial forces

Proposition 9.4 The virtual power of the inertial forces is given by

( )n

a a kk

k 1

VP Q w

w

where the so-called generalized inertial forces are defined by

( )ak k k

d T TQ

dt q q

where T denotes the kinetic energy of the multibody given as a function of , ,t q q , i.e.

( , , ) ( )q q0T T t q q dv X

2

x x

0B

32

The principle of virtual powerLagrange’s equations

Proposition 9.5

( ) , n

a i e k nk k k

k 1

Q Q Q w 0 w

Theorem 9.1 (Lagrange’s equations)

( ) , ,...,i ek kk k

d T TQ Q 0 k 1 n

dt q q

( ) , ( ) , ,...,k k k k0 0q 0 q q 0 q k 1 n Initial conditions:

33

Box 9.2: Lagrange’s equations

( ) , ,...,i ek kk k

d T TQ Q 0 k 1 n

dt q q

( , , ) ( )q q0T T t q q dv X

2

x x

0B

( )qik k

Q dv Xq

E

S0B

( ) ( )q qek 0 0k k

Q da X dv Xq q

Pn b00 BB

Summary

34

Lagrange’s equations

What has to be discussed?

• Kinetic energy

• Coordinate change

• Internal forces, constitutive assumptions

• Interactions between parts

• External forces

• Power theorem

( ) , ,...,i ek kk k

d T TQ Q 0 k 1 n

dt q q

35

Particle in plane circular motion

x

y

r

cos

sin

x r

y r

q

36

( , ; ) cos sinq Ox t P x r r i j

,q

t

0 ( sin ) cosq r r

i j

( , ; ) ( ( sin ) cos )q qq X r r

t

x x i j

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Particle in plane circular motionKinetic energy

x

y

r

( , ) ( ( sin ) cos )2 2 2 2 2k

1 1 1E T m r r m r m

2 2 2 x i j

37

Particle in plane circular motionExternal forces, internal forces

x

y

r

( )g g j

38

mgN

O

P

( )

e

q

mg

m

0

m

j

F N g

N

g

( )q q qe eQ m m

F N g g

q

iQ 0

( ( sin ) cos ) cosr r m mgr i j g

Particle in plane circular motionLagranges equations

39

x

y

r ( , ) ,2 21

T r m2

,2Tr m

T0

( ) cos cosi e 2d T T gQ Q 0 r m mgr 0 0

dt r

cos ,eQ mgr iQ 0

Lagranges equations:

40

The spherical pendulum

sin cos

sin sin

cos

, 1 2

x

y

z

L

L

L

q q

( , , ; ) sin cos sin sin ( cos )q O x y zx t P x L L L e e e

,q

t

0

( sin sin ) sin cosqx yL L

e e

cos cos cos sin sin ,qx y zL L L

e e e

( )z g g e

41

The spherical pendulumMaterial velocity

L

( , , ; ) ( cos cos cos sinq q qq x yt q q X L L

t

x x e e

sin ) ( ( sin sin ) sin cos )z x yL L L e e e

( cos cos sin sin ) ( cos sin sin cos ) sinx y zL L L e e e 42

The spherical pendulumKinetic energy

( , , , ) ( sin )2 2 2 2 2k

1 1E T m L m

2 2 x

L

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43

mg

S

( )

e

OP

OP

z

m

S

g

F S g

pS

p

g e

Spherical pendulumExternal forces, internal forces

( sin cos sin sin ( cos )) , x y zL L L S S 0 S e e e

,q q 0

S S sin ,q m mgL

g q m 0

g

44

Spherical pendulumLagranges equations

( , , , ) ( sin ) ,2 2 2 21T L m

2

,2TL m

sin ,2 2T

L m

( ) ( sin )i e 2 2d T T dQ Q 0 L m 0

dt dt

sin ,eQ mgL ,eQ 0

sin cos ,2 2TL m

T0

i iQ Q 0

( ) sin cos sini e 2 2 2d T TQ Q 0 L m L m mgL 0

dt

Lagranges equations:

The kinetic energy

( , , ) ( )q q0T T t q q dv X

2

x x

0B

( , , ; )n n

q q qk kq k k

k 1 k 0

t q q X q qt q q

x x

,

n n nq q q qk l k lk l k l

k 0 l 0 k l 0

q q q qq q q q

x x

,

( , , ) ( )n

q q k l0k l

k l 0

T T t q q dv X q qq q

0B

45

Proposition 9.6 (‘The three parts decomposition’) The kinetic energy may be written ( , , ) 0 1 2T T t q q T T T

where

( , )0 0 00

1T T t q m

2

( , , ) ( )n

k1 1 0 k k 0

k 1

1T T t q q m m q

2

,

( , , )n

k l2 2 kl

k l 1

1T T t q q m q q

2

The kinetic energy

46

where the matrix elements, , , , ,...,klm k l 0 1 n , are defined by

( , ) ( )q q00 00 0m m t q dv X

t t

0B

( , ) ( ) , ,...,q q0k 0k 0 k0k

m m t q dv X m k 1 nt q

0B

( , ) ( ) , , ,...,q qkl kl 0 lkk l

m m t q dv X m k l 1 nq q

0B

The kinetic energy

47

The planar pendulumAlernative calculation of kinetic energy

, ( ),O O O Ov v v t v i , ω k

O

c

1

0

i

j

k

( , , ) 2k c c

1 1E T t m

2 2 v ω I ω

Ov

The kinetic energy:

48

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The planar pendulum

O

c

1

0

i

j

k

( cos ) sinO

l lv

2 2 i j

( sin ( cos ))c O Oc O

l lv

2 2 v v ω p i k i j

49

sin ( cos )Oc

l l

2 2 p i j

The planar pendulum

O

c

1

0

i

j

k

cos2

2 2O O

lv v

4

( cos ) sin cos cos sin2 2 2

2 2 2 2 2 2 2 2 2c O O O

l l l lv v v

2 4 4 4 v

( cos )2

2 2 2c O O

1 1 lm v v m

2 2 4 v

50

The planar pendulum

O

c

1

0

i

j

k

,

2

c zz

mlI

12

,

,

c zz

2 2c c c c zz

I

1 1 1 1I

2 2 2 2 ω I ω k I k k I k

51

The planar pendulum

O

c

1

0

i

j

k

( cos ) ,2

2 2O O

1 1 1 lv m v m m

2 2 2 3

( , , ) ( sin )2 2

2 2 2 2c c O O

1 1 1 l 1 mlT t m v v m

2 2 2 4 2 12 v ω I ω

( )O Ov v t

52

The planar pendulum

O

c

1

0

i

j

k

, ( cos ) , 2

200 O 01 10 O 11

1 lm v m m m v m m m

2 3

( , , ) ( cos ) ,

0 1 2

22 2O O

T T T

1 1 1 lT t v m v m m

2 2 2 3

( )O Ov v t

53

i

j

k

g 0

A

B

1

2

O

The planar double pendulumCalculation of the kinetic energy

( sin cos ) , , ( )( , ; )

( sin cos ) ( sin cos ) , , ( )

1O 1

q 2O 1 2

x X 0 X l forx X

x l X 0 X l for

i j

i j i j

The transplacement:

54

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55

Example 9.3 The kinetic energy of the double pendulum in Example 9.2 is given by,

Tq ,

( , ) T2

1T T q q q M q

2

since, in this case, 0M 0 and 1 1 2M 0 due to the scleronomic coordinate system. The

matrix elements of 2M are given by (9.42),

( , ) ( )q q0m dv X

0B, ( , ) ( ) ( , )q q

0m dv X m

0B

( , ) ( )q q0m dv X

0B

where

( cos sin ) , , ( )

( cos sin ) , , ( )

1q 1

21 2

X 0 X l for

l 0 X l for

i j

i j

Example 9.3 continued.

, , ( )

( cos sin ) , , ( )

1q 1

22

0 X l for

X 0 X l for

0

i j

The volume element for 1 is ( ) 1dv X A dX where 1A is the constant cross-sectional area

of 1 and 1 0 1 1m A l . For 2 we have ( ) 2dv X A dX and 2 0 2 2m A l . Then

( , ) ( ) ( ) ( )1 20 0

q q q q q q0 0 0m dv X dv X dv X

0B B B

1 2 1 2l l l l 2

q q q q 2 2 21 10 1 0 2 0 2 1 0 2 2 1

0 0 0 0

m lA dX A dX X A dX l A dX m l

3

( , ) ( ) ( ) ( )1 20 0


0B B B

cos( )1 2 2 2l l l l

q q q q q q0 1 0 2 0 2 1 0 2

0 0 0 0

A dX A dX A dX l X A dX

cos( )2 1 2

1m l l

2 56

2

( , ) ( ) ( ) ( )1 20 0


0B B B

( )2 2

20

l l 2q q q q 2 2 2

0 0 2 0 2

0 0

m ldv X A dX X A dX

3

B

Thus

cos( )( , )

cos( )

221 1 2 1 2

2 1

2 2 22 1 2 2 2

m l m l lm l

3 2M Mm l l m l

2 3

and the kinetic energy of the double pendulum then reads

cos( )

( , , , )

cos( )

221 1 2 1 2

2 1

22 1 2 2 2

m l m l lm l

1 3 2T T2 m l l m l

2 3

57

The planar double pendulumAlternative calculation of kinetic energy

, ,1 1 1 ω k , 2 2 ω k

58

59

( , , , )1 1 2 2

2 2k c 1 1 c 1 c 1 2 c 2

1 1 1 1E T m m

2 2 2 2 v ω I ω v ω I ω

1 1 2 2

2 2c 1 1 c 1 c 1 2 c 2

1 1 1 1m m

2 2 2 2 v k I k v k I k

, ,1 1 2 2

2 2 2 2c 1 c zz c 1 c zz

1 1 1 1m I m I

2 2 2 2 v v

, ,

1 1 2 2

c zz c zz1 2

2 2 2 2c 1 c c 1 c

I I

1 1 1 1m m

2 2 2 2 v k I k v k I k

The planar double pendulum

1 2

2 22 2 2 21 1 2 2c 1 c 1

m l m l1 1 1 1m m

2 2 2 12 2 12 v v

60

( ) cos( )1 2

2 2 2 2 2 21 2c 1 c 1 2 1 2 2 1 2

m m1 1 1 1 1m m m l l m l l

2 2 2 4 2 4 2 v v

The planar double pendulum

( , , , ) ( ) cos( )2 2 2 21 2k 2 1 2 2 1 2

m m1 1 1E T m l l m l l

2 4 2 4 2

( ) cos( )2 2

2 2 2 2 2 21 1 2 2 1 22 1 2 2 1 2

m l m l m m1 1 1 1 1m l l m l l

2 12 2 12 2 3 2 3 2

cos( )

cos( )

2

221 1 2 1 2

2 1

22 1 2 2 2

M

m l m l lm l

1 3 22 m l l m l

2 3

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( , , ) ( )T0 1 2

1T T t q q M M q q M q

2

The kinetic energy

T1 2 n n 1q q q q

( , )0 0 00M M t q m

( , ) ,

11 12 1n

21 22 2n n n T2 2 2 2

n1 n2 nn

m m m

m m mM M t q M M

m m m

( , ) 1 n1 1 01 10 02 20 0n n0M M t q m m m m m m

61

where the matrix elements, , , , ,...,klm k l 0 1 n , are defined by

( , ) ( )q q00 00 0m m t q dv X

t t

0B

( , ) ( ) , ,...,q q0k 0k 0 k0k

m m t q dv X m k 1 nt q

0B

( , ) ( ) , , ,...,q qkl kl 0 lkk l

m m t q dv X m k l 1 nq q

0B

Mass matrix elements

62

63

( , , ) T1T T t q q q Mq

2

( , )

00 01 0n

0 110 11 1n

T1 2

n0 n1 nn

m m m1M M m m m2M M t q

1M M

2 m m m

Mass matrix elements

( ) , T0 2 n n 1 1 0q q q q q t

Proposition 9.7 The generalized inertial forces may be written a ra ta ga

k k k kQ Q Q Q

where

( )ra 2 2k k k

d T TQ

dt q q

is the relative inertial force,

( )ta 01k k k

TTQ

t q q

is the transport inertial force and

2n

ga j1 1k j k k

j 1

T TQ q

q q q

is the gyroscopic inertial force.

The kinetic energy

64

The kinetic energy

Proposition 9.8 The transport inertial force may be written

n

ta i0 00k k

i 1

m m1Q

t 2 q

Proposition 9.9 The gyroscopic inertial force may be written

n

ga jk kj

j 1

Q g q

where

2 2

0 j0k1 1kj j k k j j k

mmT Tg

q q q q q q

The so-called gyroscopic matrix ijg g is thus skew-symmetric, i.e. Tg g . 65

Proposition 9.10 The relative inertial force may be written

,

n n nra i i j ikik ki kij

i 1 i j 1 i 1

mQ m q C q q q

t

where

( ), , , ,...ij kjkikij j k i

m mm1C i j k 1 n

2 q q q

The kinetic energy

66

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( ) , ,...,ta ga i e2 2k k k kk k

T TdQ Q Q Q 0 k 1 n

dt q q

,

( , , )n

k l T2 2 kl 2

k l 1

1 1T T t q q m q q q M q

2 2

67

Corollary 9.2 Lagrange’s equations are given by T cif i e

2q M Q Q Q

where

( ( ) ) ( ( )) ( )cif T t T 02 2 2 1 1 MM M M M M1 1Q q q q skew

q 2 q t q 2 t q

is the complementary inertia force and

i i i i1 2 nQ Q Q Q and e e e e

1 2 nQ Q Q Q


cif T iT eT2M q Q Q Q

68

69

The complementary inertia force in a scleronomiccoordinate system

( ( ) )T t2 2M M1q q

q 2 q

( ( ) ) ( ( )) ( )cif T t T 02 2 2 1 1 MM M M M M1 1Q q q q skew

q 2 q t q 2 t q

, 0 1 1 nM 0 M 0

cif2n n n 1 n

M0 Q 0

q

Box 9.3 Kinetic energy and inertial forces

0 1 2T T T T

0 00

1T m

2 , ( )

nk

1 0k k0k 1

1T m m q

2

, ,

nk l

2 klk l 1

1T m q q

2

( )q q00 0m dv X

t t

0B

, ( )q q0k 0k

m dv Xt q

0B

( )q qkl 0k l

m dv Xq q

0B

cif T iT eT

2M q Q Q Q

( ( ) ) ( ( ))cif T t T2 2 2 1M M M M1Q q q q skew

q 2 q t q

( )01 MM1

2 t q

Summary

70

Regular coordinates

The system of generalized coordinates , , ... , 1 2 nq q q q , ( , ; )qx t q X is said to be regular

if for fixed ( , )t q

( , , ; ) , , ,...,n

q k kq k

k 1

t q w X w 0 X w 0 k 1 nq

w 0B

Proposition 9.12 The system of generalized coordinates is regular if and only if the massmatrix 2M is positive definite.

71

Example 9.2 Underlying, in our discussion, is the assumption that we may use different sets of generalized coordinates for the configuration space of the multibody. The choice of coordinates must be decided by the engineer. For instance, for a single particle P we may use Cartesian coordinates ( )x, y, z or spherical coordinates ( , , )r related by, cf. example 9.1,

sin cos

sin sin

cos

r

r

r

x

y

z

Let ( )e e ex y z denote the RON-basis corresponding to the Cartesian coordinate system

with origin O. The transplacement of the particle may be written ( )( ; ) 3

q Ox P x e e e x y zx,y,z x y z, x, y, z

Then

q

exx

, q

eyy

, q

ezz

and thus the Cartesian coordinate system is regular. In spherical coordinates

72

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Example 9.4 continued.

sin cos sin sin cosq

r

e e ex y z ,

cos cos cos sin ( sin )qzr r r

e e ex y ,

( sin sin ) sin cosq r r

e ex y

and

sin , ( )q q q 2r 0 rr

, ,

This spherical coordinate system is thus a regular coordinate system. However in this case

3 is not equal to the whole of 3 .

73

Example 9.5 The coordinate system adopted for the double pendulum in Example 9.2 is regular since, for fixed ( ) , ,

q q1 2w w

( cos sin ) , , ( )

cos sin ) (cos sin ) , , ( )

1 11

1 2 21 2

Xw 0 X l for

l w Xw 0 X l for

i j 0

i j i j 0

1 2w w 0

74

Coordinate changes

( , ; ) ( , ; )q qt q X t q X

0B

tB

X

x

nq nqq̂

1q

- : nq space 1q

q

- : nq space q

75

...T1 2 nq q q q ... ,

T1 2 nq q q q

( , ( ); ) ( , ( ); ) q qx t q t X t q t X X 0B

Coordinate change

q q ˆ( , )q q t qCoordinate change:

ˆ ˆ ˆ( , ), 1q q t q q q

q̂

76

Coordinate changesVirtual velocity fields

( , , ; )n

q jq j

j 1

t q w X wq

w w( , , ; )n

q kq k

k 1

t q w X wq

w w

T1 2 n nw w w w

ˆ knk j

jj 1

qw w

q

T1 2 n nw w w w

q̂w w

q

ˆ ˆ kn n

j

q q

q q

Jacobian:

Component change:

77

Proposition 9.13 If the - coordinate systemq is regular then the - coordinate systemq is

regular if and only if

ˆ

det( )q

0q

Proposition 9.14

ˆ ˆq q

q qt q

* q̂Q Q

q

where Q may be any of eQ , iQ or aQ .

Coordinate changes

78

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79

Coordinate changes

ˆ ˆˆ( ) ( , ( )) , ,...,

k knk j

jj 1

q qq t q t q t q q k 1 n

t q

ˆ ˆq qq q

t q

ˆ( ) ( ) ( )

knq q qe

j 0 0 0j j k jk 1

qQ da X dv X da X

q q q q

Pn b Pn00 0BB B

ˆ ˆ ˆ( ) ( ( ) ( ))

k k kn n nq q q e

0 0 0 kk j k k j jk 1 k 1 k 1

q q qdv X da X dv X Q

q q q q q q

b Pn b0 00B BB

Covariant and contravariant vectors

q̂q q

q

q̂w w

q

* q̂Q Q

q

Covariant vector:

Contravariant vector:

a i eQ Q Q 0

Lagrange’s equations are on covariant form:

ˆ( )n 1

q0

t

80

ˆ( )a i e a i e q

Q Q Q Q Q Qq

Theorem 9.2 Lagrange’s equations are invariant under a regular coordinate transformation,i.e. a i e a i eQ Q Q 0 Q Q Q 0

Invariance of Lagrange’s equations

Lagrange’s equations are invariant under regular changes of the configurationcoordinates!

81

Box 9.4: Regular coordinates and change of coordinates

ˆ( , )q q t q

ˆdet( )

q0

q

ˆ ˆq q

q qt q

, * q̂Q Q

q

Summary

82

The rigid part

O

Inertial frame

Ax

ω

Ax

tB

( , ; ) ( , ) ( , )( ), q A At q X x t q t q X X X R 0B

; ( ), , ,...,def

Tk k

ax k 0 1 nq

R

ω R

;

nk

kk 0

q

ω ω ,

nk

A A kk 0

q

x x

,

defA

A k k

x

q

x

83

( )A Ax x x x ω

The rigid part

;

nk

kk 0

q

ω ω ,

nk

A A kk 0

q

x x

84

Corollary 9.3

, ;

defA

A k A k kq

x

x x

and ; , , ,...,k kk 0 1 n

q

ω

ω

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Proposition 9.18 The mass matrix elements of the kinetic energy for a rigid part may bewritten ; ; ; ; ; ; ; ;( ) , , , ,...,kl A k A l A k l Ac A l k Ac k A lm m k l 0 1 n x x x ω p x ω p ω I ω

where A

I is the inertia tensor of the rigid part . For a scleronomic coordinate system

00m 0 , , ,...,0km 0 k 1 n

The rigid part: The mass matrix

(( ) ( ) ( )) ( )t

2A A A Ax x x x x x dv x

I 1B

; ;A kl k A lm x 0 ω I ω

Inertia tensor:

85

Corollary 9.4 By taking A c , the centre of mass, the following matrix elements areobtained ; ; ; ;kl c k c l k c lm m x x ω I ω

The rigid part: The mass matrix

86

The double pendulum

;k kq

ω

ω

;A

A k kq

x

x

; ; ; ; ; ; ; ;( ) , , , ,...,kl A k A l A k l Ac A l k Ac k A lm m k l 0 1 n x x x ω p x ω p ω I ω

87

The double pendulum

2 ω k

; ; ; ; ; ; ; ; ; ;( )1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1kl O k O l O k l Oc O l k Oc k O l k O lm m x x x ω p x ω p ω I ω ω I ω

1 ω k

O x 0

; ; ; ; , , ,2 2 2 2 2 2 2kl c k c l k c lm m k l x x ω I ω

88

Example 9.6 Let us calculate the mass matrix of the double pendulum in Example 9.2 by using the formula (9.73) for a rigid part. For part 1 and with A O we have, since O x 0

, 1 ω k and ;

11

ω

ω k ,

; ; ,

21 1 1 1 1 1 1 1

O O O zz

m lm I

3 ω I ω k I k

Furthermore ;

1 ω 0 and then

1 1 1m m m 0

For part 2 we may use the expression in Corollary 9.4 where 2 ω k and then

;

22

ω

ω 0 . Moreover

( cos sin ) ( cos sin )2 2c 1

ll

2 x i j i j ; ( cos sin )2

c 1l x i j

89

and then

; ; ; ;2 2 2 2 2 2 2 2

c c 2 c 2 1m m m m l x x ω I ω

Next

; ; ; ; ; ;2 2 2 2 2 2 2 2 2

c c 2 c c c 2m m m m x x ω I ω x x

( cos sin ) ( cos ) sin ) cos( )2 2 1 21 2

l m l ll m

2 2 i j i j

Finally, with ;

22

ω

ω k ,

; ; ; ; ( cos sin ) ( cos sin )2 2 2 2 2 2 2 2c c 2 c 2

l lm m m

2 2 x x ω I ω i j i j

,

2 2 2 22 22 2 2 2 2 2 2 2c c zz

m l m l m l m lI

4 4 12 3 k I k

90

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Summarizing,

21 1 21 1

2 1

m lm m m m l

3 , cos( )1 2 22 1

lm m m m l

2 , 2

2 2 2m lm

3

and

cos( )( , )

cos( )

221 1 2 1 2

2 1

2 2 22 1 2 2 2

m l m l lm l

3 2M Mm l l m l

2 3

91

The rigid part: The internal force

Proposition 9.16 For a rigid part the generalized internal force is zero, i.e. , , ,...,i

kQ 0 k 1 n

92

( )qik k

Q dv Xq

E

S0B

The rigid part: The external force

93

Proposition 9.17 For a rigid part the generalized external force is given by , , ,

; ; , ,...,e e ek A k k AQ k 1 n x f ω M

where

,0 0( ) ( )e da X dv X

f Pn b0 0BB

,0 0( ) ( ) ( ) ( )e

A A Ax x da X x x dv X

M Pn b0 0BB

are the force and moment sums acting on the body.

94

Point force

( , ; ) ( , ; )( , ) ( ) ( ) ( , )ec

0ec

0

q q 0eck 0 0 0k k

t q X t q XQ t q X X da X t q

q q

f fB

B

( , ; ) ( , ) ( )ec0

e0 0 0t q X t q X X

t f

B

Example 9.6

, , ,; ; , ,2 e 2 2 e 2 2 e

k B k k BQ k x f ω M

95

Example 9.6

96

Assume that the planar double pendulum is subjected to a follower point force 0f and a

point couple 0M applied to 2 and defined according to

( ) ( ( sin ) cos ) ( )0 0 0f t f f i j , ( ) ( )0 0 0t M t M M k

where ( )0 0f f t and ( )0 0M M t are prescribed functions. See Figure 9.10 below. We

have

( cos sin ) ( cos sin )2B 1 2l l x i j i j ; ( cos sin )2

B 1l x i j

; ( cos sin )2B 2l x i j

The corresponding generalized force is, according to (9.70), given by

,; ; ( cos sin ) ( ( sin ) cos ) ( ) ( ) sin( )2 e 2 2

B 0 0 1 0 1 0Q l f t l f t x f ω M i j i j

where ;

2 ω 0 .

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97

Example 9.6, cont’d

,

; ; ( cos sin ) ( ( sin ) cos ) ( ) ( ) ( )2 e 2 2B 0 0 2 0 0 0Q l f t M t M t x f ω M i j i j k k

We have thus found that

, ( ) sin( )2 e1 0Q l f t , , ( )2 e

0Q M t or , ( ) sin( ) ( )e 2 e1 0 0Q Q l f t M t

O

A

B

0f 0M

O i

j

k

The elastic part, elastic energy

( , ), , EE E

uu u X X

E S

E0B

, , ( , ; ) ( , ( , ; ))e q e q E qu u t q X u X t q X E

q

( ) 2e

1U q kq

2

Linear elastic spring:

, , ,( , ) ( , ; ) ( )0

e q e q e qU U t q u t q X dv X B

98

Elastic potential:

Elastic energy:

The elastic part, internal generalized force

Eu

SE

99

,, ( ) ( ) ( )q q e qi Ek k k k

q

uuQ dv X dv X dv X

q q q

E E

SE

0 0 0B B B

,, ( )

0

e qe qk k

Uu dv X

q q

B

,, e qik k

UQ

q

The elastic part

100

Corollary 9.6 For a linear isotropic elastic part

, ( (tr ) ) ( )2i 2

k k

1Q dv X

q 2

E E0B

The linear isotropic elastic part

Corollary 9.5 For a linear elastic part

, ( ) ( )ik q qk

1Q X dv X

q 2

E E0B

where ( )X is the elasticity tensor for body .

101

Corollary 9.6 For a linear isotropic elastic part

, ( (tr ) ) ( )2i 2

k k

1Q dv X

q 2

E E0B

The linear isotropic elastic part

Corollary 9.5 For a linear elastic part

, ( ) ( )ik q qk

1Q X dv X

q 2

E E0B

where ( )X is the elasticity tensor for body .

102

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The elastic bar

, ( , ; ) ( )Present center-line : t t q O 1 0 0 00

0 lq

x x t q X x t s sl

eC B

, Reference center-line : 00 O 1 0 0 00 lX X X s s e0C B

103

The elastic energy

( )( ) ( )

220

e e 020

q l1U U q k q l

2 4l

( ) 0 0

0 0

A AE 1k 2

l 1 1 2 l

”Spring constant”:

104

The elastic generalized force

(( ) )i 2eq

0

U 1 qQ k 1 q

q 2 l

( ) ( )i 0 0q

0 0

q l q l1Q k q

2 l l

( )iq 0Q k q l The elastic generalized force, ”small elongations”:

( )i0 q 0q l Q k q l

105

Body forces: Gravitation

,, ( ) ( ) ( ( ) )q q g qgk 0 0 q 0k k k k

UQ dv X dv X dv X

q q q q

pg g p g

0 0 0B B B

, , ( , ) ( , ) ( , )g q g q Oc cU U t q t q m z t q gm p g

106

External forces: Body forces

,g qgk k

UQ

q

,, g qgk k

UQ

q

, , ( , ) ( , ) ( , )g q g q Oc cU U t q t q m z t q gm p g

, , ( , ) ( , )N N

g q g q c c1 1

U U z t q gm z t q gm

1:

N

:

107

Double pendulum with torsion springs

Torsion springs

, , ,( , ) ( )1 2 2 2e q e q e q 1 2

1 1U U U

2 2

108

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Example 9.7 Let the planar double pendulum in Example 9.2 be equipped with elastic torsional springs at the joints at O and A. The elastic potential energy of these internal forces is then given by

, , ,( , ) ( )1 2 2 2e q e q e q 1 2

1 1U U U

2 2

where 1 and 2 are the spring stiffness constants. The corresponding generalized force is

then given by

, ( ) ( )e qi1 2 1 2 2

UQ

, ( )e qi2 2 2

UQ

and

( )1 2 2i i i

2 2

Q Q Q

109

Example 9.9 The planar double pendulum in Example 9.2 is subjected to the force of gravity with potential energy

, , ,( , ) ( , ) ( , )1 2

1 2g q g q g q c 1 c 2U U U y gm y gm

where, see Figure 9.13,

( , ) cos1

1c

ly

2 , ( , ) cos cos

2

2c 1

ly l

2

Thus

, ( , ) ( cos ( cos cos ))1 2g q 1 2 1

l lU m m l g

2 2

O

A

B

1c

2c

i

j

k

110

Figure 9.13 The potential energy of the double pendulum in the gravitational field. The generalized force corresponding to the gravitational potential reads

, ( ) sing qg1 2 1

U 1Q m m l g

2

, , sing qg 2

2

U lQ m g

2

and

( ) sin sing g g 21 2 1 2

l1Q Q Q m m l g m g

2 2

O

A

B

1c

2c

i

j

k

111

External forces: Contact forces

( )ec

qec ek 0k

Q da Xq

t0B

ec 0B

ec ic0 0 0B B B

ic 0B

( ), qic i ik 0 0 0k

ic

Q da Xq

t t Pn

0B

c ec ick k kQ Q Q

0 0 0 B B B

Exterior contact surface:

Interior contact surface:

tB

ict B

tB

ect B

112

113


,i eUQ

q

( ) i e T cif i e2

d T TQ Q 0 q M Q Q Q

dt q q

,e c bQ Q Q gb UQ

q

( ) gceUUd T T

Q 0dt q q q q

e gU U U

( ) cd T T UQ 0

dt q q q

Potential energy:

114

c ic ecQ Q Q

( ) c T cif c2

d L L UQ 0 q M Q Q

dt q q q

L T U


The Lagrangian:

T cif ic ec2

Uq M Q Q Q

q

Contact forces:

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Exercise 4:14

115

Exercise 4:14 Solution

Configuration coordinates: ,x

116

117


( ), ,

Placement: ( , ; ) ( ) ,

( sin ) ( cos ),

0 1O OP O 1 0 0

0

0 2q O OB BP O 1 0 BP

0 0 3O OP O 1 0 2

aX

x x l P 0 X ll

x P x x l P

x x l l l P

p e

p p e p

p e e

x

x xx


( , ; ) , q P Pt

0 x

( , ; ) , , q 0 11 0

0

XP P 0 X l

l

e x

x

Partial derivatives of the transplacement:

( , ; ) , q 0 2 31P P

e x

x

118

, ,

( , ; ) ,

cos sin ,

10

q 2

0 0 31 2

P 0 X l

P P

l l P

0

0

e e

x

119


The material velocity:

( , , , ; ) q q qPt

x x x x x

x

, ,

,

cos sin ,

0 11 0

0

0 21

0 0 0 31 1 2

XP 0 X l

l

P

l l P

00 e

0 e 0

0 e e e

x

xx

, ,

,

( cos ) sin ,

0 11 0

0

0 21

0 0 31 2

XP 0 X l

l

P

l l P

e

e

e e

x

xx


( , , , ) (( cos ) ( sin ) )2 2 2 2 2C c B c

1 1 1 1T T m m m l l m

2 2 2 2 v v x x x x

cos( ( ) cos )

cos

T

c2 2 2c 2

m m ml1 1m m 2ml ml

ml ml2 2

x x

x x

Kinetic energy:

cos

cosc

2 2

m m mlM

ml ml

Mass matrix:

120

, 0 1 1 2M 0 M 0

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( ) ( )q q 0 0xx 0 1 1 0 c

2 3dm dm

m dv X dv X dm m mx x

e e 0 0B B

Calculation of massmatrix elements:

( ) ( cos sin ) ( cos sin )q q 0 0 0 0 20 1 2 1 2

3dm

m dv X l l l l dm ml

e e e e

0B

( ) ( cos sin ) cosq q 0 0 0x 0 1 1 2

3dm

m dv X l l dm mlx

e e e

0B

121


T0

xsin

Tml

x cos 2Tml ml

x( ) cosc

Tm m ml

xx

Partial derivatives of kinetic energy:

( ) i ed T TQ Q 0

dt

x xx x( ) i ed T T

Q Q 0dt

Lagrange’s equations:

122


Generalized internal forces:

i eUQ k

x xx

2e

1U k

2 x

i eUQ 0

Elastic potential energy:

Generalized external forces:

Gravitational potential energy: cosgU mgl

, ge g UQ 0

x x, singe g U

Q mgl

123


The external force :01 FF e

, qe F 0 0OC1 1Q F F

pF F e ex x x

, qe F 0OC1Q F 0

pF F 0 e

124

The external force :R

,( , ; )q q0 e R1

0

XX 0 P Q 0

l

e 0 Rxx

x x

,( , ; ) , q qe R0P 0 X l Q 0

0 Rx

125


The external force :1N

q 01 1 1 0

N e N

x

q1 1 0

N 0 N

i eQ Q k F x x x sini eQ Q mgl

Sum of internal and external forces:

A O cm

k

l

0

mg

F B 01e

02e

R

1N 2N

1 2

3

C

x 0l


( cos ) sin sin

2dml ml ml mgl 0

dt

x x

( ( ) cos )c

dm m ml k F 0

dt x x

cos

sin sincos

i ecif

2

c 22

q Q QQM

m m mlml 0 k 0 F mgl

ml ml

x x

( ) cos sin

cos sin

2c

2

m m ml ml k F 0

ml ml mgl 0

x xx

The equations of motion:

126

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127

( , , , ) ( )00 0ll l 2 3

0 0 2 2 0s s 1 1 0 0 2

0 0 0 00 0 0

1 X X 1 X 1 XT T dX dX

2 l l 2 l 2 l 3

e e

xx x x x x


Kinetic energy of the spring:

2 20 0 0l m1 1

2 3 2 3

x x

Kinetic energy of the rod:

( , , , ) 2r r c r r c r

1 1T T m

2 2 v ω I ω x x

(( cos ) ( sin ) )2 2 2 2r r

1 l l 1 1m m l

2 2 2 2 12 x

mbd week 4 2017 - lth · 19 the configuration space: s2 o ... the planar double pendulum has two...

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