mbd week 1 2017 - lth · 2017-01-25 2 7 name copies signature alisic, rijiad bengtsson, frida...

2017-01-25

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Multibody DynamicsSpring 2017

www.mek.lth.seEducation/Advanced Courses

MULTIBODY DYNAMICS (FLERKROPPSDYNAMIK) (FMEN02, 7.5p)

1

Course responsible and teacher

• Per Lidström, Mechanics, LTH

• [email protected]

• 046-2220479

• Aylin Ahadi

2

MBD Class 2017

Alisic, Rijiad Lindqvist, Max

Bengtsson, Frida Lund, Mikael

Björklund, Viktor Nilsson, Eddie

Bring, Björn Niu, Mutong

Dahlström, Linus Svensson, Christoffer

De Groote, Alicia Svensson, Jakob

Gabrielsson, Jonas Thuresson, Pontus

Jiang, Tao Winnykamen, Yoan

Jin, Run Wittig, Cornelius

Johansson, David

Johansson, Jenny

Johari, Daniel

Kalkan, Sven

Khan, Shoajb

Kröger, Dennis

Larsson, Henrik

3 4

Name Mail‐adress Personal ID Signature

Alisic, Rijiad [email protected], Frida [email protected]örklund, Viktor [email protected], Björn [email protected]öm, Linus [email protected] Groote, Alicia [email protected], Jonas [email protected], Tao [email protected], Run [email protected], David [email protected], Jenny [email protected], Daniel [email protected], Sven [email protected], Shoajb [email protected]öger, Dennis [email protected], Henrik [email protected], Maxt [email protected], Mikael [email protected], Eddie [email protected], Mutong [email protected], Christoffer

[email protected], Jakob [email protected], Pontus [email protected], Yoan [email protected], Cornelius [email protected]

COURSE REGISTRATION

Multibody Dynamics (FMEN02) Spring 2017

5

Password: euler

Leonard Euler 1707-1783 6

2017-01-25

2

7

Name copies Signature

Alisic, Rijiad

Bengtsson, Frida

Björklund, Viktor

Bring, Björn

Dahlström, Linus

De Groote, Alicia

Gabrielsson, Jonas

Jiang, Tao

Jin, Run

Johansson, David

Johansson, Jenny

Johari, Daniel

Kalkan, Sven

Khan, Shoajb

Kröger, Dennis

Larsson, Henrik

Lindqvist, Maxt

Lund, Mikael

Nilsson, Eddie

Niu, Mutong

Svensson, Christoffer

Svensson, Jakob

Thuresson, Pontus

Winnykamen, Yoan

Wittig, Cornelius

COURSE BOOK RESERVATION (HARD COPY)Fundamentals of Multibody Dynamics, Lecture NotesSEK 300-350

Course program

8

Multibody Dynamics (FMEN02, 7.5p), 2017 Course literature: Lidström P., Nilsson K.: Fundamentals of Multibody Dynamics, Lecture Notes (LN). Div. of

Mechanics, LTH, 2017. The ‘Lecture Notes’ will be on the Course web site. Hand out material: Solutions to selected Exercises, Examination tasks (Assignments), Project specification. Examples of previous written tests.

Teacher: Prof. Per Lidström, Ass. Prof. Aylin Ahadi (Lectures, Exercises, Course coordination) Phone: 046-2220479, email: [email protected] Schedule (w. 1-7): Lectures: Monday 10 - 12 Tuesday 15 - 17 Thursday 8 - 10, room M:IP2 Exercises: Wednesday 8 - 10, room M:X1a-b Schedule (w. 8): Lectures. Monday 8 - 10, 15 - 17, room M:IP2

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Weekly Timeplan 2017

Week 1

Hour

Monday

Tuesday

Wednesday

Thursday

Friday

8

Lecture

M:X1a-b

9

10

Lecture M:IP2

11

12

13

Lecture

M:IP2

14

15

Lecture

M:IP2

16 10


Week 2

Hour

Monday

Tuesday

Wednesday

Thursday

Friday

8

Exercise

M:X1a-b

9

10

Exercise M:IP2

11

12

13

Lecture

M:IP2

14

15

Lecture

M:IP2

16

11


Week 3-7

Hour

Monday

Tuesday

Wednesday

Thursday

Friday

8

Exercise

M:X1a-b

9

10

Lecture M:IP2

11

12

13

Lecture

M:IP2

14

15

Lecture

M:IP2

16 12


week 8

Hour

Monday

Tuesday

Wednesday

Thursday

Friday

8

9

10

Lecture

M:IP2

11

12

13

14

15

Lecture

M:IP2

16

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13 14

Course objectives and contents

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Course objectives: The objective of this course is to present the basic theoretical knowledge of the Foundations of Multibody Dynamics with applications to machine and structural dynamics. The course gives a mechanical background for applications in, for instance, control theory and vehicle dynamics. Course contents: 1) Topics presented at lectures and in the “Lecture Notes”. 2) Exercises. 3) Examination tasks 4) Application project The scope of the course is defined by the curriculum above and the lecture notes (Fundamentals of Multibody Dynamics, Lecture Notes). The teaching consists of Lectures and Exercises: Lectures: Lectures will present the topics of the course in accordance with the curriculum presented above. Exercises: Recommended exercises worked out by the student will serve as a preparation for the Examination tasks. Solutions to selected problems will be distributed.

Examination

16

Examination: The examination consists of two parts:

1. Hand in solutions to 3 examination tasks (assignments containing 2-4 problems each) and one Application project.

The examination tasks will be handed out starting by w.2. Last date for handing in solution to the examination tasks and the project report:

Examination task 1: Rigid body kinematics February 3rd Examination task 2: Rigid and flexible body dynamics February 17th Examination task 3: Multibody dynamics Mars 3rd Application project (with Adams software) Mars 17th

2. Written examination containing five (5) tasks focusing on the theoretical parts of the course content.

A correctly solved task gives 3p (Total possible score 5x3p = 15p). Date for the written examination: Monday, Mars 13, 14:00-19:00 in room: M:L1-L2. Credit requirements: Credit 3: Solutions to all examination tasks with 65% of the problems correctly solved. Approved project report. Written examination with 50% of the tasks correctly solved (7.5p). Credit 4: Solutions to all examination tasks with 75% of the problems correctly solved. Approved project report. Written examination with 67% of the tasks correctly solved (10p). Credit 5: Solutions to all examination tasks with 90% of the problems correctly solved. Approved project report. Written examination with 83% of the tasks correctly solved (12.5p).

CONTENTS

1. INTRODUCTION 1 2. BASIC NOTATIONS 12 3. PARTICLE DYNAMICS 13

3.1 One particle system 13 3.2 Many particle system 27

4. RIGID BODY KINEMATICS 36 4.1 The rigid body transplacement 36 4.2 The Euler theorem 47 4.3 Mathematical representations of rotations 59 4.4 Velocity and acceleration 72

5. THE EQUATIONS OF MOTION 89 5.1 The Newton and Euler equations of motion 89 5.2 Balance laws of momentum and moment of momentum 94 5.3 The relative moment of momentum 99 5.4 Power and energy 100

6. RIGID BODY DYNAMICS 104 6.1 The equations of motion for the rigid body 104 6.2 The inertia tensor 109 6.3 Fixed axis rotation and bearing reactions 138 6.4 The Euler equations for a rigid body 143 6.5 Power, kinetic energy and stability 158 6.6 Solutions to the equations of motion: An introduction to the case of

7. THE DEFORMABLE BODY 172 7.1 Kinematics 172 7.2 Equations of motion 196 7.3 Power and energy 205 7.4 The elastic body 209

8. THE PRINCIPLE OF VIRTUAL POWER 217 8.1 The principle of virtual power in continuum mechanics 219

17

1. THE MULTIBODY 222 9.1 Multibody systems 222 9.2 Degrees of freedom and coordinates 224 9.3 Multibody kinematics 229 9.4 Lagrange’s equations 232 9.5 Constitutive assumptions and generalized internal forces 252

9.6 External forces 261 9.7 The interaction between parts 267 9.8 The Lagrangian and the Power theorem 295

10. CONSTRAINTS 300 10.1 Constraint conditions 300 10.2 Lagrange’s equations with constraint conditions 305 10.3 The Power Theorem 323

11. COORDINATE REPRESENTATIONS 327 11.1 Coordinate representations 328 11.2 Linear coordinates 331 11.3 Floating frame of reference 359 APPENDICES A1 A.1 Matrices A1 A.2 Vector spaces A6 A.3 Linear mappings A12 A.4 The Euclidean point space A37 A.5 Derivatives of matrices A45 A.6 The Frobenius theorem A54 A.7 Analysis on Euclidean space A56 SOME FORMULAS IN VECTOR AND MATRIX ALGEBRA EXERCISES

18

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Mechanics in perspective

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Quantum Nano Macro‐ContinuumMicro‐meso ‐

Å nm m mm

lengthscale

Prerequicites

20

Multibody Dynamics

Linear algebra

3D Analysis

Mechanics (Basic course

Continuummechanics

Matlab

Finite element method

Mathematics Mechanics Numerical analysis

Synthesis

Content

21

Multibody Dynamics

Flexible bodies

Rigid bodies

Coordinates

Lagrange’sequations

Constraints

Course objectives

22

an ability to

perform an analysis of a multibody system and to present the result in a written report

read technical and scientific reports and articles

on multibody dynamics some knowledge of

industrial applications of multibody dynamics

computer softwares for multibody problems

23

Industrial applications

24

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Commercial MBS softwares

• Adams

• Dads

• Simpac

• Ansys

• Dymola/Modelinc

• Simulink/Matlab

• RecurDyn

25

Adams Student EditionMultibody Dynamics Simulation

https://www.youtube.com/watch?v=hd-cFu7yMGw

Multibody system

26

Slider crank mechanism

Multibody system

27

Multibody system

Figure 1.3 Topology and coordinates of slider crank mechanism.

B

A C

Cy

Cx

B

A Slider

Crank

Rod

O

Bx

Guider Ground

Ground

O

, , , ,B C Cx x y Configuration coordinates:

4# Parts:

28

Configuration coordinatesRigid part in plane motion, 3DOF

, ,C Cx yConfiguration coordinates:29

Greubler formulaTwo dimensional

( )m

ii 1

f 3 N 1 r

f

N

ir

Number of parts:

Number of degrees of freedom (DOF):

Number of DOF removed by joint i:

Number of joints: m

ir 2Revolute joint: ir 2Translational joint:30

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Joints in three dimentions

( )m

ii 1

f 6 N 1 r

ir 5ir 5

31

Degrees of freedom and constraints

a) b)

Figure 1.5 a) Three-bar mechanism b) The double pendulum.

B

A D

C

Ground

A Ground

B

g

32

Greubler formula

( ) , m

ii 1

f 3 N 1 r f 1

Slider crank mechanism:

Three bar mechanism:

Double pendulum:

,N 4

( )f 3 4 1 4 2 1

( )f 3 4 1 4 2 1

,m 4 ir 2

,N m 4 ir 2

,N 3 ,m 2 ir 2

( )f 3 3 1 2 2 2

B

A D

C

Ground

A

Ground

B

g

33

Biomechanics

,N 11 ,m 10 ir 2 ( )f 3 11 1 10 2 30 20 10

34

Multibody dynamics software Adams

35

The rigid body

ec

ec c c

m

F a

M I ω ω I ω

Figure 1.8 The rigid body.

ω

P

tB

c

cPp

cv

36

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The flexible body

( (tr ) ) ( )0

22e

1U dv X

2 E E

B

37

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

The Multibody

( )U U q

( ) , ,...,kk k k

d T T UQ 0 k 1 n

dt q q q

( , )k kQ Q q q

Configuration coordinates:

Kinetic energy: Potential energy:

Equations of motion: (Lagrange’s equations)

38

( , )T T q q

Constraints

cos cos

sin sin

C

C

LR x 0

2L

R y 02

cos

sin

B C

C

Lx x 0

2L

y 02

Constraint at A: Constraint at B:

, , , ,B C Cx x y Configuration coordinates:

39

Equations of motion under constraints

, ,...,n

k0 k

k 1

g g q 0 1 m

( , )k kg g t q

Constraint equations: 1 m n

Equations of motion:

( ) , ,...,

, ,...,

mnonk kk k

1

nk

0 kk 1

d L LQ g 0 k 1 n

dt q q

g g q 0 1 m

L T U

( , , )non nonk kQ Q t q q

( ), ,...,t 1 m

Lagrangian:

Nonconservative forces:

Constraint forces:40

Chapter 3: Particle Dynamics

is left for self study and recapitulation of basic concept i mechanics.

You may try to solve some of the exercises but don’t spend to much

time on this introductory chapter!

m v

1F

41 42

Euclidean space

, , ...x y z E Points in Euclidean space

y x u V Vectors in Translation vector space

y x u

dim( ) 3V 1 2 3e e e e basis

[ ]1

1 1 2 2 3 3 1 2 3 2

3

a

a a a a

a

ea e e e e e e e a

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43

Cartesian coordinate system

oE Fixed point, 1 2 3e e e e Fixed basis

[ ] [ ]ox 1 1 2 2 3 3 oxx o o x x x o o x e er e e e e r e

oE Fixed point, 1 2 3f f f f Fixed basis

[ ]x o x ff [ ]o o o ee

[ ] [ ] [ ] [ ] ([ ] [ ] )x o o x o x x x o e f e f e ee f e f e

, , det3 3B B B 0 f e 44

[ ] ([ ] [ ] ) [ ] [ ] [ ]B x x o B x x o f e e f e ee e

[ ] ([ ] [ ] )1x B x o f e e

Cartesian coordinate system

Transformation of coordinates for points

[ ] ([ ] [ ] )x B x o e f f

45

What about vectors?

[ ] , ea e a [ ] , fa f a Bf e

[ ] [ ] [ ] [ ] [ ]B B f f e e fa f a e a e a a a

[ ] [ ] , [ ] [ ]1B B e f f ea a a a

a V

46

What about tensors?

[ ] , eAe e A [ ] , fAf f A Bf e

[ ] [ ]1B Bf eA A

[ ] [ ] 1B Be fA A

( )EndA V

TB e f

47

1 2 3e e e e

Orthonormal basis

i j ij

1 i j

0 i j

e e

1 1 1 1 2 1 3

T2 1 2 3 2 2 2 2 2 3

3 3 1 3 2 3 3

1 0 0

0 1 0

0 0 1

e e e e e e e

e e e e e e e e e e e e

e e e e e e e

T3 3I e e

48

,1 2 3e e e e

Orthonormal bases

, detT3 3B B I B 1

1 2 3f f f f

Bf e

Orthonormal matrix

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49

( , , )det det

( , , )

V

V

e

Qa Qb QcQ Q

a b c

The determinant

1 2 3e e e eON-basis:

Rigid body kinematics

( ) , ,AB t constant A B p Rigidity condition:

Figure 4.1 Rigid body motion.

0B

B

A

B

( )AB 1tp A

B

( )AB 2tp

50

Rigid body transplacement

O

A

A

Reference placement

Present placement

0B

B

( )A Ap r Ar

Au

( ), A A A p r 51


( ) : ( ) ( ), A A AR a r a r a V

( ) ( ) , ,B A B A A B r r r r

( ), A A A p r

Rigidity condition:

Transplacement:

( )

( ) ( ) , ,

( ) ,

A

A A

A

R

R R

R

0 0

a b a b a b

a a a

V

V

52Isometry

( ) ( ) ( ), , , ,A A AR R R a b a b a b V


Linearity:

( )A AR a R a

Isometry implies orthonormality:

,T TA A A A R R R R 1V

1 TA A R R

Notation:

53


( ) ( ) , A A A r a r R a a V

P A a r r

( ) ( ) ( )P A A P A r r R r r

( )P A A P A p p R r r

AP A APp R r

54

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AB A ABp R r

A

B

B

A

0B

B

ABr

AB ABp Rr

( )A SOR V

55

( ) ( ) , B B B r a r R a a V


( ) ( ) , A A A r a r R a a V

?B AR R

Does the rotation tensor depend on the reduction point?

!B A R R R

56

No!


Theorem 4.1 A rigid body transplacement is uniquely determined by the displacement of (anarbitrary) reduction point A; Au and an orthonormal transformation ( )SOR V which isindependent of the reduction point.

( )P A AP u u R 1 r

57

Example 4.1 (Plane rigid transplacement, cf. section 4.2.2) The same rigid body transplacement is visualized in both Figure 4.6 a) and 4.6 b) below with different choices of reduction point.

Transplacement with reduction point A.

B

B

A A A

Transplacement with reduction point B.

Figure 4.6 Plane rigid body transplacement.

Au

Bu

A A

B

B B

58

Summary

Box 4.1: Rigid body transplacement

:

:

: detA

A reduction point

displacement of the reducion point A

orthogonal transformation with 1u

R R

( ) ( )( )A A u u r u R 1 r r

( ) ( ) ( )A A p r r R r r

59

The translation

, , ,A B 0 A B u u u

Proposition 4.2 A translation is a rigid transplacement determined by a displacement 0u ,

independent of the reduction point, and an orthonormal transformation R equal to the unittensor.

0B

A

B

A

B

ABp

0u

0u

B

ABr

R 1

60

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The rotation

( , )A n

Reference placement Present placement

n

B

0B B

B

A A

a

b

ABr ABp

Rotation axis:

61

The rotation Figure 4.9 Rotation around a fixed axis.

1i

2i ABr

A

B

3 i n A

1e 2e

B

ABp

3 3 e i n 1x

2x

( )AB AB AB r r n n r

AB 1 1 2 2 r i ix x

( )AB AB AB p p n n p

AB 1 1 2 2 p e ex x

( )2 2 2AB AB 1 2 AB r p n rx x

62

The rotation Figure 4.9 Rotation around a fixed axis.

1i

2i ABr

A

B

3 i n A

1e 2e

B

ABp

3 3 e i n 1x

2x

cos sin

( sin ) cos , 1 1 2

2 1 2

3 3

0 2

e i i

e i i

e i n

63

The rotation matrix

i

e i R

cos sin

sin cos

0

0

0 0 1

iR

,T T i i i i

R R R R 1 det 1i

R

cos sin

( ) ( ) sin cos ,1 2 3 1 2 3

0

0

0 0 1

e e e i i i

64

1 1

2 2

3 3 3

R i e

R i e

R i e i n

i

R i e i R

AB ABp Rr

1i

2i ABr

A

B

3 i n A

1e 2e

B

ABp

3 3 e i n 1x

2x

The rotation tensor

65

Summary

Box 4.3: Rotation

( , ) :

:

:

cos sin

: , sin cos

:

( ) :

A

1 2 3

A rotation axis

A reduction point

displacement of the reduction point A

0

orthonormal transformation 0

0 0 1

rotation angle 0 2

referen

i

n

u 0

R R

i i i i , 3 3tial base i n e

66

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The Euler theorem

A 0B

n

axis ( , )A n

C

i

Theorem 4.2 (The Euler theorem, 1775) Every rigid transplacement with a fixed point isequal to a rotation around an axis through the fixed point.

67

The canonical representation

Rn n

cos sin

sin cos

0

0

0 0 1

iR

( ), 1 2 3 3 i i i i i n

68

i

R i i R

cos sin

( ) sin cos1

T1 2 3 2

3

0

0

0 0 1

i

i

R i R i i i i i

i

69

The tensor product

( )End a b V, a b V,

( ) ( ) a b u a b uDefinition:

Notation:

70

The vector product

a b V, a b V,

( ) a 1 u a uDefinition:

Notation:

b

a b

a

( )Skew a 1 V, ( ),End a 1V VNotation:

sin a b a b

71

( cos + sin ) +( ( sin )+ cos ) )1 2 1 1 2 2 3 3 i i i i i i i i

( ) cos ( )sin3 3 1 1 2 2 2 1 1 2 R i i i i i i i i i i

( cos + sin ( sin )+ cos )1

1 2 1 2 3 2

3

i

i i i i i i

i

The canonical representation

cos sin

( ) sin cos1

1 2 3 2

3

0

0

0 0 1

i

i i i i

i

cos ( ) sin n n 1 n n n 1

( ) det( ) det( )p R i1 R 1 R

cos sin

det sin cos ( )(( cos ) sin )2 2

0

0 1

0 0 1

,( ) , i1 2 3p 0 1 e R

The characteristic equation, eigenvalues

72

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Summary

Box 4.5: Rigid transplacement

:

:

:

:

:

A

A

A reduction point

position vector of the reduction point in the reference placement

displacement of the reducion point

position vector of a material point in the reference placement

p

r

u

r

p

:

osition vector of a material point in the present placement

rotation tensorR

( ) ( )A A A

translation rotation

p r r u R r r

73

Screw representation of the transplacement

Screw line

, S A ASS r r r n SL

Su

A

S

nASr

ASr

Theorem 4.3 (Chasles’ Theorem, 1830) Every transplacement of a rigid body can be realized by a rotation about an axis combined with a translation of minimum magnitudeparallel to that axis. The reduction point for this so-called screw transplacement is located onthe screw line given in (4.28).

( ),

( cos )

TA

AS 02 1

R 1 u

r

( )S Ss p r n R r r

74

75


Week 2-7

Hour

Monday

Tuesday

Wednesday

Thursday

Friday

8

Exercise

M:X1a-b

9

10

Lecture M:IP2

11

12

13

Lecture M:IP2

Lecture

M:IP2

14

15

Lecture

M:IP2

16

The special orthogonal group

The set of all orthonormal tensors on V forms an algebraic group, the orthonormal group,

( ) T TO Q Q Q QQ 1= V

Being a group means that

, ( ) ( )

( )

( )

1 2 2 1

1 T

O O

O

O

Q Q Q Q

1

Q Q

V V

V

V

However, in general we have 1 2 2 1Q Q Q Q and we say that the operation of combining

orthonormal tensors is non-commutative. The set of all rotations forms a subgroup in ( )O V called the special orthonormal group

( ) ( ) detSO O 1 R R= V V

76

Non-commutativity of rotations

77

The main representation theorem

78

This representation is mainly unique. If R 1 then n may be arbitrarily chosen, while( )0 modulo 2 . If R 1 then n is uniquely determined, apart from a factor 1 . For a

given n the angle is uniquely determined ( )modulo 2 . For a definition of the tensor

( )End a 1 V see Appendix A.4.39.

Theorem 4.1 To every rotation tensor ( )SOR V there exists a unit vector n and an angle so that

( , ) cos ( ) sin R R n n n 1 n n n 1

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79

The topology of SO(V )

P

O

x

y

z

P

P

x

y

z

O

( )1 2 3e e e e 1

2

3

n

n

n

en

en e n

The main representation theoremusing components

1 1 1 1 2 1 3

T

2 1 2 3 2 1 2 2 2 3

3 3 1 3 2 3 3

n n n n n n n

n n n n n n n n n n

n n n n n n n

e e en n n n

3 2

3 1

2 1

0 n n

n 0 n

n n 0

e e en 1 n 1

80

The main representation theoremusing components

cos ( ) sin e e e e

R n n 1 n n n 1

cos ( ) sinT T

e e e e e en n 1 n n n 1

cos ( ) sin R n n 1 n n n 1

( )1 2 3e e e e 1

2

3

n

n

n

en

en e n

81

How to calculate the rotation axis and the rotation angle from the rotation tensor

( ) ( ) b b

Rn n R 1 n 0 R 1 n 0

11 12 13

3 321 22 23

31 32 33

R R R

R R R

R R R

bR

, ?R n

1

3 12

3

n

n

n

bn

( )1 2 3b b b b

82

83

The eigenvalue problem

11 12 13 1

21 22 23 2

31 32 33 3

R R R n 0

R R R n 0

R R R n 0

,( ) , i1 2 3p 0 1 e R

cos , sin2 3 2 3

2 2i

( ) det11 12 13

21 22 23

31 32 33

R R R

p R R R

R R R

R

The eigenvalue problem

11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

R 1 R R n 0 n

R R 1 R n 0 n

R R R 1 n 0 n

n

cos11 22 33tr R R R 1 2 R

arccos( ) arccos( )

arccos( )

11 22 33

11 22 33

R R R 1tr 1

2 2R R R 1

22

R

84

1 1

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Coordinate representations of ( )SO V

• Euler angles

• Bryant angles

• Euler parameters

• Quaternions

• ……

85 86

Euler angles: ( , , )

0B

1e

2e 3e

R

A 01e

02e

03e

A

B

Present placementReference placement

o

o o e

e Re e R

87

01e

02e

03 3f e

1f

2f

Factorization:

3g

3f

1 1g f

2f

2g

( , , )

2e

1e

3 3e g

1g

2g

( ) ( ) ( )3 1 3 R R R R

( )3 R ( )1 R ( )3 R

Euler angles: First rotation

01e

02e

03 3f e

1f

2f

cos sin

( ) sin coso3

0

0

0 0 1

eR ( ) ( ) ,o

o o3 3

ef R e e R

( )

( )

0 0 0 01 2 3

1 2 3

e e e e

f f f f

88

Euler angles: Second rotation

3g

3f

1 1g f

2f

2g

( ) cos sin

sin cos1

1 0 0

0

0

fR ( ) ( ) ,1 1

fg R f f R

( )

( )

1 2 3

1 2 3

f f f f

g g g g

89

Euler angles: Third rotation

2e

1e

3 3e g

1g

2g

( ) ( ) ,3 3 g

e R g g R cos sin

( ) sin cos3

0

0

0 0 1

gR

( )

( )

1 2 3

1 2 3

g g g g

e e e e

90

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91

Euler angles: All rotations in combination

( ) ( )

( ) ( )

( ) ( )

o

o o3 3

1 1

3 3

e

f

g

f R e e R

g R f f R

e R g g R

)

( ) ( ) ( ) ( ) ( ) ( )o

o

o o3 1 3 3 1 3

e

g f g e f g

R

e Re g R f R R e R R R

( ) ( ) ( ) ( ) ( ) ( )oo o o

3 1 3 3 1 3 e f g

e Re R R R e e R R R

Euler angles: The rotation

( ) ( ) ( )o o3 1 3 e e f g

R R R R

cos sin cos sin

sin cos cos sin sin cos

sin cos

0 1 0 0 0

0 0 0

0 0 1 0 0 0 1

cos cos sin cos sin cos sin sin cos cos sin sin

sin cos cos cos sin sin sin cos cos cos cos sin

sin sin sin cos cos

o e f g e

92

Euler angles

Theorem 4.2 A rotation R is determined by a set of three real coordinates, called the Euler

angles: ( , , ) , , , 3 0 2 0 0 2 . Conversely these angles are

essentially determined by R . In fact the angle ,0 is uniquely determined by R . If

,0 then also , ,0 2 are uniquely determined by R . When 0 only is

uniquely determined and when only . 93

cos cos sin sin cos sin sin cos

sin cos cos sin sin sin cos coso

0

0

0 0 1

eR

cos( ) sin( )

sin( ) cos( )11 12 13

21 22 23

31 32 33

0 R R R

0 R R R

0 0 1 R R R

Euler angles: Singularity

Only is determined. Singular!

,0

94

Euler angles: Singularity

cos , sin cos

cos , sinsin sin

cos , sinsin sin

233

23 13

32 31

R 1

R R

R R

o

11 12 13

21 22 23

31 32 33

R R R

R R R

R R R

eR

cos cos sin cos sin cos sin sin cos cos sin sin

sin cos cos cos sin sin sin cos cos cos cos sin

sin sin sin cos cos

, singular0

95

The rotation angle

( cos )cos( ) coscos

1 1

2

cos ( cos )cos( ) costr 1 2 1 R

96

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Euler parameters

sin ( ) sin cos22 22 2 2

R 1 n n 1 n 1

cos , sin0e2 2

e n

cos ( ) sin R n n 1 n n n 1

( cos )( ) sin1 1 n n 1 n 1

cossin2 1

2 2

sin sin cos2

2 2

97

Euler parameters

( ), RON-basis,1 2 3b b b b ,1 1 2 2 3 3n n n n b b b

cos , sin , sin , sin

1 1 2 2 3 3

0 1 1 2 2 3 3

e e e

e e n e n e n2 2 2 2

e b b b

3S

0e

3e4

( , , , ) ,

4 2 2 2 20 1 2 3 0 1 2 3

3 4 2 2 2 20 1 2 3

e e e e e e e e e 1

S e e e e e 1

( )( ) 02 2e R 1 e 1 e 1 e 1

98

1n

3 2

0 3 1

2 1

0 e e

2e e 0 e

e e 0

( )3 2 3 2

3 1 3 1

2 1 2 1

1 0 0 0 e e 0 e e

e 0 1 0 2 e 0 e e 0 e

0 0 1 e e 0 e e 0

bR

( ) ( , ) ( ) , ( , )2 30 0 0e e 2 2e e e S R R R e 1 e 1 e 1 e

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

2 20 1 1 2 0 3 1 3 0 2

2 21 2 0 3 0 2 2 3 0 1

2 21 3 0 2 2 3 0 1 0 3

2 e e 1 2 e e e e 2 e e e e

2 e e e e 2 e e 1 2 e e e e

2 e e e e 2 e e e e 2 e e 1

How to calculate matrix elements fromEuler parameters

99

( ), RON-basis,1 2 3b b b b

o

11 12 13

21 22 23

31 32 33

R R R

R R R

R R R

eR

How to calculate Euler parameters from matrix elements

, , ,

2 11 22 330

2 ii ii 11 22 33i

R R R 1tr 1e

4 4R 1 2R R R Rtr 1

e i 1 2 32 4 4

R

R

There is a 2 -1 correspondence between Euler parameters and rotation matrices!

No singularity!

100

( , , , ) ( , , , )2 2 2 20 1 2 3 0 1 2 3e e e e 1 e e e e 0 0 0 0

,32 231

0

R Re

4e

11 22 330

R R R 1e 0

4

,13 312

0

R Re

4e

21 12

30

R Re

4e

How to calculate Euler parameters from matrix elements, condinued

101

,1 2 21 124e e R R

11 22 330

R R R 1e 0

4

,1 3 31 134e e R R 2 3 32 234e e R R

,21 122

1

R Re

4e

, 11 22 330 1

1 R R Re 0 e 0

4

31 133

1

R Re

4e


102

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, , 22 11 330 1 2

1 R R Re 0 e 0 e 0

4

31 133

2

R Re

4e


103

, , , 33 11 220 1 2 3

1 R R Re 0 e 0 e 0 e 0

4

The composition of rotations

Let 1R and 2R be two rotations. The composition of 1R and 2R is given by 2 1R R R and this represents another rotation. We have the following Proposition 4.1 If ,( , )1 1 0 1 1eR R e , ,( , )2 2 0 2 2eR R e and 2 1R R R then ( , )0eR R e

where , ,0 0 1 0 2 1 2e e e e e , , ,2 0 1 1 0 2 1 2e e e e e e e (4.18)

Corollary 4.1 Two rotations commute if and only if their rotation axes are parallel, i.e. 2 1 1 2 1 2 R R R R e e

104

The exponential map

exp( ) : ..., ( )! !

2 31 1End

2 3 A 1 A A A A V

... exp( ) exp( )! !

2 32 3

2 3

R 1 N N N N n 1

( )Skew N n 1 V

( )SOR n V

105

The exponential map:

( ) with corresponding ( )SO Lie group Lie algebra SkewV V

O

A

A

Reference placement

Present placement 0B

tB

( )A A tp pAr

P

P

Pr

( )P P tp p

Time-dependent transplacementMotion

( )

( , ) ( ) ( )( ) ( ) ( )( ),

A

P P A A P A A P A

t

t t t t t t 0

p

p r r u R r r p R r r106

Rigid body velocity

, P A AP P v v ω p

( )Taxω RR

ω

P

tB A

APp

Av

d

AP AP p ω p

TW RRSpin tensor:

Angular velocity:

Rigid body velocity field:

107 108

Rigid body rotation

( )T W RR R WR ω 1 R ω R

( ) (exp( ( ) ))( )

t

0

0 0

t s ds0

R ω RR R ω 1 R

R R

Thus, if the angular velocity is known and the rotation tensoris known at time then the rotation tensor : is known.

( )tω ωt 0 ( )tR R

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α ω

( ), P A AP AP P a a α p ω ω p

α

ω

P

tB A

APp

Aa

APα p

( )AP ω ω p

Rigid body acceleration

Angular acceleration:

109

Tangential acceleration

Centripetal acceleration

α ωSpecial case:

01e

02e

03e

1e

2e3e

ω R

Euler-Poisson velocity formulafor moving bases

, , 0 0 01 1 2 2 3 3 e Re e Re e Re

( )1 2 3e e e e( )0 0 0 01 2 3e e e e

0

e eR R

110

01e

02e

03e

1e

2e3e

ω R

Euler-Poisson velocity formulafor moving bases

Theorem 4.4 (Euler-Poisson velocity formula for moving bases) , , ,i i i 1 2 3 e ω e (4.32)

where

3

i ii 1

1

2

ω e e (4.33)

111

Euler-Poisson velocity formulafor moving vectors and tensors

,

( ) ( ) ( ) ( )3

j iji j 1

t t t A t

iA A e e

( ) rel A ω A A ω 1 A

rel a ω a a

3

reli

i 1

a

iea e a

( ) ( ) ( )3

ii 1

t t a t

ia a e

,

( ) ( ) ( )3

relj ij

i j 1

t t A t

ieA e e A

112

Angular acceleration

3rel

ii 1

iω ω e

rel rel ω ω ω ω ω

3

ii 1

iω e

1e

2e3e

ω

113

Proposition 4.6 The angular velocity ω , expressed in Euler angles, is given by 1 1 2 2 3 3 ω e e e

where

sin sin cos

sin cos sin

cos

1

2

3

(4.39)

and ( )1 2 3e e e e is an RON-basis following the rotation of the rigid body (‘fixed to the

body’).

Angular velocity and Euler angles

114

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Combining rotationsThe addition of angular velocities, version 1

( )T1 1 1axω R R

( )1 1 tR R

2 2 1 ω ω R ω

( )2 2 tR R 2 1R R R

( )T2 2 2axω R R ( )Taxω RR

115

Addition rule 1


1R

2R

1f

2f 3f

1g

2g

3g

0feω

0geω

gfω

01e

02e

03e

(( ) )T2 2axgf fω R R

1 0f R e 2g R f

( )0

T11 1ax feω ω R R

( )0

Tax geω ω R R

0g Re 2 1R R R

116


1R

2R

1f

2f 3f

1g

2g

3g

0feω

0geω

gfω

01e

02e

03e

0 0 ge gf feω ω ω

117

Addition rule 2

Summary

Box 4.8: Velocity and acceleration

( )

( )

, ( ) ,

A A

A

T

Rigid body velocity field

Rigid body acceleration field

ax Spin tensor Angular velocity

v v ω p p

a a α p ω ω p

W RR ω W

α ω

, , ,

i i

rel

Angular acceleration

i 1 2 3 Euler Poisson for vectors

Absolute and

e ω e

u ω u u

( )

sin sin cos

sin cos sin

cos

1 1 2 2 3 3

1

2

3

relative velocity

Euler Poisson for tensors

Angular velocity in Euler angles

eA ω A A ω 1 A

ω e e e

e ( ) -

( )

0 0

1 2 3

2 2 1

RON basis fixed to the body

Addition of angular velocities 1

gfge fe

e e e

ω ω R ω

ω ω ω ( ) Addition of angular velocities 2

118

Derive the matrix representation e

a b of the tensor product a b relative to the

ON-basis ( )1 2 3e e e e . The matrix representations of , a b V are given by

1

2

3

a

a

a

ea and

1

2

3

b

b

b

eb (1)

Exercise 0:1

119

Show that if , a b V and ( )EndA V then ( )T a b b a , ( ) T a b A a A b , ( ) ( ) A a b Aa b ( )( ) a b c d a d b c , det( ) 0 a b , ( )tr a b a b (2)

Exercise 0:2

120

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Let (EndA V) and u V with the matrix representations eA and e

u , respect-

tively, in a RON-basis ( )1 2 3e e e e , i.e. e

A e e A and e

u e u . Show that

e e e

Au A u (1)

Exercise 1:1

121

Let ( )SkewA V and let ( )ax a A V be its axial vector, i.e.

, Au a u u V (2)

Assume that T

1 2 3a a ae

a in a RON-basis ( )1 2 3e e e e . Show that

3 2

3 1

2 1

0 a a

a 0 a

a a 0

eA (3)

Exercise 1:2

122

Exercise 1:4

Let , ( )1 2 SOR R V be rotations with the following matrix representations 1 1

eR e e R and 2 2

eR e e R (5)

where ( )1 2 3e e e e is an ON-basis. Show that

2 1 2 1

e e eR R R R (6)

123

Exercise 1:5

Let , ( )1 2 SOR R V be rotations with the following matrix representations

1 1e

R e e R and 2 2f

R f f R (7)

where 1f R e and ( )1 2 3e e e e is a RON-basis. Show that

2 1 1 2e e f

R R R R (8)

124

Consider a rotation tensor ( )SOR V and its characteristic polynomial

( ) det( )p R 1 R . Since this is a third order polynomial there are three roots

, , 1 2 3 to the characteristic equation

( )p 0 R (9) We know that one of the roots is equal to 1, let’s say 3 1 . Show that the rotation

angle is given by

arccos 1 2

2

(10)

Exercise 1:6

125

Exercise 1:8

Consider the transplacement of a rigid body with the rotation tensor R n , . Show that (“Rodrigues formula”) ( ), ,AB AB AB AB A B p r c p r (15) where ( )axc C is the Cayley vector corresponding to the Cayley tensor C defined by ( )( ) ( ) ( )1 1 C R 1 R 1 R 1 R 1 (16) Show that the tensor R 1 is invertible and that C is anti-symmetric.

126

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A rigid translacement of a body is governed by the translation A u V and the rotation ( )SOR V where A is a fixed reduction point. Which of the following statements are true and which are false? Give a simple motivation for your answer!

a) , 0 Ra a a V .

b) , P A AP P u u Rr c) exp( ) R w 1 for some vector w V .

d) T R R 1 . e) T

1 1 2 2 3 3 R f e f e f e f e for some RON-basis ( )1 2 3e e e e and

( )1 2 3f f f f .

Exercise 1:9

127

Given a rotation tensor ( , , ) R R where , , are Euler angles. Calculate the

matrices ( , , ) 015 35 60 e

R and ( , , )0

0115 35 60 e

R e .

Exercise 1:12

128

Exercise 1:12

The rotation matrix expressed in Euler angles:

R ( )

cos ( )

sin ( )

0

sin ( )

cos ( )

0

0

0

1

1

0

0

0

cos ( )

sin ( )

0

sin ( )

cos ( )

cos ( )

sin ( )

0

sin ( )

cos ( )

0

0

0

1

Conversion from degrees to radians:

Rd ( ) R

180

180

180

Calculation of the rotation matrix:

Rd 15 35 60( )

0.299

0.815

0.497

0.943

0.171

0.287

0.148

0.554

0.819

Checking the rotation matrix:

Rd 15 35 60( ) 1Rd 15 35 60( ) Rd 15 35 60( )T

1

0

0

0

1

0

0

0

1

Calculating the rotation of the first basis vector:

Rd 15 35 60( )

1

0

0

0.299

0.815

0.497

129 130

Exercise 1:14

131

Exercise 1:14

01e

02e

03e

1e

2e

3e

n

Example 4:1 The wheel

c vv i p r ω j k 132

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133

1. The relative position vector ABr of material points A and B in the reference configuration of a body is given by

1 2 32 3 4AB r i i i (21)

in the RON-basis system 1 2 3( )i i i i . The body is rotated the angle 45 about an

axis ( , )A n where

1 2 3

1 1 1

3 3 3 n i i i (22)

Determine the matrix i

R of the rotation tensor R and the matrix AB ip of the

transplaced vector AB ABp Rr .

Exercise 1:17

17.

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