Multibody Dynamics Spring 2019
www.mek.lth.se Education/Advanced Courses
MULTIBODY DYNAMICS (FLERKROPPSDYNAMIK) (FMEN02, 7.5p) 1 Course coordinator
• Prof. Aylin Ahadi, Mechanics, LTH • [email protected] • 046-2223039
2 COURSE REGISTRATION Multibody Dynamics (FMEN02) Spring 2018 pnr efternamn förnamn email
3
FUNDAMENTALS OF MULTIBODY DYNAMICS
LECTURE NOTES
Multibody system (MSC ADAMS™)
Per Lidström Kristina Nilsson
Division of Mechanics, Lund University 4 Password: euler
Leonard Euler 1707-1783 5 Course program 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. Aylin Ahadi (Lectures, Exercises, Course coordination) Lectures: Monday 13 - 15 Tuesday 13 - 15 Thursday 10 - 12, room M:IP2 Exercises: Wednesday 13 - 15, room M:IP2
6 Course program 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. Aylin Ahadi (Lectures, Exercises, Course coordination) Phone: 046-222 3039, email: [email protected] Lectures: Monday 13 - 15 Tuesday 13 - 15 Thursday 10 - 12, room M:IP2 Exercises: Wednesday 13 - 15, room M:IP2
7 8 9 Course objectives and contents
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.
10 Examination
11
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 12 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
13 Mechanics in perspective
Quantum Nano Micro-meso - Macro -Continuum
lengthscale Å nm µm mm
14 Prerequicites
Multibody Dynamics
Synthesis
Continuum Finite element 3D Analysis mechanics method Mechanics (Basic Linear algebra Matlab course
Mathematics Mechanics Numerical analysis
15 A multibody system is a Content mechanical system consisting of a number of interconnected components, Multibody Dynamics or parts, performing motions that may involve large translations and rotations as well as small displacements such as Rigid bodies vibrations.
Flexible bodies Interconnections between the components are of vital importance. They will Coordinates introduce constraints on the relative motion between Constraints components and in this way limit the possible motions Lagrange’s which a multibody system equations may undertake. 16
Course objectives
Primary objectives are to obtain a thorough understanding of
♠ rigid body dynamics ♠ Lagrangian techniques for multibody systems containing constraint conditions a good understanding of
♠ the dynamics multibodies consisting of coupled rigid bodies some understanding of
♠ the dynamics of multibodies containing flexible structures 17 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
18 Industrial applications
Multibody system dynamics
(MBS dynamics) is motivated Aerospace by an increasing need for analysis, simulations and assessments of the behaviour of machine systems during
the product design process. Automotive
19 Mechanism Robotics What is a multibody system?
A multibody system is a mechanical system consisting of a number of interconnected components, or parts, performing motions that may involve large translations and rotations as well as small displacements such as vibrations. This is often the case for machines produced in the automotive, aerospace and robotic industries
20 What is a multibody system?
The interconnections between the components are of vital importance. They will introduce constraints on the relative motion between components and in this way limit the possible motions which a multibody system may undertake.
21 Commercial MBS softwares
• Adams • Dads • Simpac • Ansys • Dymola/Modelinc Adams Student Edition • Simulink/Matlab Multibody Dynamics Simulation
• RecurDyn https://www.youtube.com/watch?v=hd-cFu7yMGw
22 Multibody system Classical steam engine equipped with a slider crank mechanism
Slider crank mechanism converts reciprocating translational motion of the slider into rotational motion of the crank or the other way around.
23 Multibody system
Revolute joint Connecting rod Flywheel
Translatonal joint Revolute Crank Slider joint Fundament
Figure 1.2 Steam engine with slider crank mechanism. The crank is connected to the slider via a rod which performs a translational as well as rotational motion. The connection between the slider and the rod is maintained through a so-called revolute joint and a similar arrangement connects the rod and the crank. 24 Multibody system
A schematic (topological) picture of the slider crank mechanism. The mechanism consists of three bodies; the slider, the crank and the rod. These are connected by revolute joints which allow them to rotate relative one another in plane motion. It is practical to introduce a fourth part, the ground, which is assumed to be resting in the frame of reference.
25 Joints in three dimentions
A prismatic or translational joint allows The revolute joint works like a hinge two connected parts to translate relative and allows two connected parts to to one another along an axis, fixed in rotate relative to one another both parts. around an axis, fixed in both parts. The position of the slider can be measured by the single coordinate. We say that the prismatic joint has one degree of freedom. 26
Configuration coordinates Rigid part in plane motion, 3DOF
Configuration coordinates: θ,,xyCC 27 Multibody system
Rod Guider= Ground Slider A
θ
ϕ
B O Crank
Ground
A C ϕ yC O B g xC
xB
Figure 1.3 Topology and coordinates of slider crank mechanism.
# Parts: 4
Configuration coordinates: xB,,,θϕ xy CC , 28 Multibody system
Configuration coordinates:
xB,,,θϕ xy CC ,
Due to the constraints upheld by the revolute joints at A and B, the number of degrees of freedom for the system is less than five.
A revolute joint removes two degrees of freedom from the multibody system. Two revolute joints for this case, four degrees of freedom are removed and we are left with one.
In the analysis of a multibody system, in plane motion, one may use the Greubler formula which gives the number of degrees of freedom. 29 Greubler formula Two dimensional
m f= 3N() −− 1∑ ri i1=
Number of degrees of freedom (DOF): f
Number of parts: N (including the ground)
Number of joints: m
Number of DOF removed by joint i: ri
= Revolute joint: r2i = Translational joint: r2i 30 Joints in three dimentions
= r5i r5i =
m f= 6N() −− 1∑ ri i1=
31 Degrees of freedom and constraints
C Ground g B A θ
θ B A D ϕ Ground
a) b)
Figure 1.5 a) Three-bar mechanism b) The double pendulum.
32 m Greubler formula f= 3N( −− 1 )∑ ri , f ≥ 1 i1=
Slider crank mechanism: = N4= , m4, r2i =
f= 34() − 1 −⋅= 4 2 1
C
B
Three bar mechanism: Nm4= = , r2i =
θ A D f= 34() − 1 −⋅= 4 2 1 Ground
Double pendulum: = m2= , r2i = Ground N3, g A θ f= 33() − 1 −⋅= 2 2 2 B ϕ 33
Biomechanics A simple multibody model of a humanoid in plane locomotion
Ground
Figure 1.6 Human locomotion.
N= 11, m= 10, r2i = f= 3() 11 − 1 − 10 ⋅= 2 30 − 20 = 10
34 Constraints
Configuration coordinates:
xB,,,θϕ xy CC ,
If the number of configuration coordinates is equal to the number of degrees of freedom of the system then we say that we use a minimal set of coordinates. Solutions to these equations represent the motion of the multibody system.
Using this approach will hide important information on the dynamics of the system. For instance, internal forces in the system, such as forces appearing in the joints, will not be accessible using a minimal set of coordinates. 35 Constraints
To calculate these forces one has to increase the number of configuration coordinates and along with this introduce constraint conditions relating these coordinates in accordance with the properties of the joints between the parts.
Configuration coordinates: xB,,,θϕ xy CC ,
Constraint at A: Constraint at B:
L −θϕ + −= L Rcos cos x0C xBC−cosϕ −= x0 2 2 L L Rsinθϕ− sin −=y0 sinϕ −=y0 C C 2 2 36 Multibody dynamics - the analysis of a mechanical system results in a solution of a set of ordinary differential equations
• To set up the differential equations we need a geometrical description of the MBS and its parts including its mass distribution. (In the case of flexible parts we also need to constitute the elastic material properties of these parts.)
• We have to analyze the connections between parts in terms of their mechanical properties. Properties of the connections in the shape of joints will be represented by constraint conditions, i.e. equations relating the configuration coordinates and their time derivatives.
• Finally the interaction between the MBS and its environment has to be defined and given a mathematical representation. Based on the fundamental principles of mechanics; Euler’s laws and the Principle of virtual power we may then formulate the differential equations governing the motion of the MBS.
• The solution of these equations will, in general, call for numerical 37 methods.
Multibody dynamics software Adams
Figure 1.7 Examples of MBS models in ADAMS®.
38 The rigid body -does not change its shape during the motion
ω This model concept may be considered
to be the most important for MBS and g P many MBS are equal to systems of p cP coupled rigid bodies.
c vc The general motion of a rigid body thus has six degrees of freedom. Bt
vvP= c +×ωpcP , P ∈ aa= +×αp +× ω() ωp × Figure 1.8 The rigid body. P c cP cP
e = The equations of motion Facm e MIcc=ω +× ω Iω c
is the inertia tensor of the body with respect to Ic its centre of mass 39 The flexible body
Figure 1.9 The flexible body.
1 2 2 The elastic energy Ue = (λµ (trEE )+ )dv ( X ) ∫ 2 B0
Green-St.Venant strain tensor and Lame-moduli 40 The Multibody kinematics
Revolute joint Connecting rod Flywheel
Configuration coordinates:
Revolute Crank joint 12 n Slider qqq= , , ... , q Fundament
expressed in terms of these coordinates and their time derivatives
Kinetic energy: T= Tqq(,) Potential energy: U= Uq()
Equations of motion: (Lagrange’s equations) dT∂ ∂∂ T U () − + −=Q 0k1, = ,..., n dt∂ q k ∂∂ q kk q k
Generalized force represents external forces on the system Qkk= Q(,) qq and internal contact forces between parts 41 Equations of motion under constraints
Lagrangian: LTU= − non= non non-conservative forces including, Nonconservative forces: Qkk Q(, tqq , ) for instance, friction.
Constraint equations: 1mn≤< n k g= g(, tq ) gνν0k+==∑ gq 0, ν 1 ,..., m ννkk k1= dL∂∂ L m Equations of motion: −−non −λν = = ()kkQkk∑ gν 0k, 1 ,..., n dt∂∂ q q ν =1 n g+== gq k 0, ν 1 ,..., m νν0k∑ k1=
m kc ν Qkk= ∑λ gν , k1= ,..., n reaction forces due to the kinematical constraints ν =1
νν Constraint forces: λλ=(t ), ν = 1m ,..., 42 Chapter 3: Particle Dynamics is left for self study and recapitulation of basic concept i mechanics.
m
v
F 1
You may try to solve some of the exercises but don’t spend to much time on this introductory chapter!
43 Euclidean space
xyz, , ...∈E Points in Euclidean space
yx−=∈u V Vectors in Translation vector space
yx= + u
dim(V ) = 3 e= ( eee123) basis
a1 =++= = a e11aaa e 22 e 33( e 1 e 2 e 3)a 2 ea[]e a3
44 Cartesian coordinate system
= o∈E Fixed point, e( eee123) Fixed basis
xo=+=++rox o e1 x 1 e 2 x 2 + e 3 x 3 =+ o er[ox ]ee =+ o e [] x
= o′∈E Fixed point, f( fff123) Fixed basis
xo=′ + f [] xf oo′′= + e[] oe
xo=++ef[ o′′ ]e [] x f =+⇒ o e [] x e fe [] x f = ([] x ee − [ o ])
33× fe=∈≠BB, R , det B 0 45 Cartesian coordinate system
eeBx[]f= ([] x ee −⇔ [ o′′ ]) Bx [] fee =− [] x [ o ]
Transformation of coordinates for points
−1 []xf= Bx ([] ee − [ o′ ])
[]xe= Bx ([] ff − []) o
46 What about vectors?
a ∈V
= a= ea[ ],e a= fa[],f feB
afaeaeaa=[]f =BB [] f = [] ee ⇔= [] [] a f
−1 []ae= BB [], aa ff []= [] a e
47 What about tensors?
A∈ End()V
= Ae= e[ A ],e Af= f[ A ],f feB
−1 []AAfe= BB []
−1 []AAef= BB []
B = efT
48 Orthonormal basis
1i = j e= ( eee) eei⋅= jδ ij = 123 0i ≠ j
e1 ee11⋅⋅⋅ ee 12 ee 13100 T =⋅ =⋅⋅⋅= ee e2( e 1 e 2 e 3) ee 22 ee 22 ee 23010 e3 ee31⋅⋅⋅ ee 32 ee 33001
T ee= I33×
49 Orthonormal bases
e= ( eee123), f= ( fff123)
fe= B
T BB= I33× , det B = ± 1
Orthonormal matrix
50 The tensor product
Notation: ab, ∈V, ab⊗∈End()V
Definition: ()()a⊗=⋅ bu abu
51 The vector product
Notation: ab, ∈V, ab×∈V
ab×
b
θ a
a×= b absinθ
Notation: a ∈VV, 1∈ End( ), a ×∈1 Skew()V
()a×=× 1u a u Definition: 52 The determinant
V (,,)Qa Qb Qc det QQ= = det[ ] V (,,)abc e
ON-basis: e= ( eee123)
53 Chapter 4 :Rigid body kinematics
Rigidity condition: pAB (t )= constant , ∀∈ A , B
54 4.1 Rigid body transplacement ( )
휒 퐫퐴
) =χ ∈ ( - mapping called Position vectors prAA( ), A transplacement 휒 퐫퐴 55 We choose a material point A as a reference point, called reduction point. Rigid body transplacement
Fix point A – reduction point
=χ ∈ Transplacement: prAA( ), A
Rigidity condition: χχ()rB− () r A =−∀∈ rr BA , AB ,
Introduce a mapping defined by:
RAA(a ) :=χχ ( ra +− ) ( r A ), a ∈V (* )
The mapping has the following properties: R ()00= A RRAA(a )− ( b ) =−∀∈ a b , ab , V RA (aaa )= , ∀∈V
Isometric mapping – the lenght of a vector does not change! 56 Rigid body transplacement
Isometric mapping is linear:
Linearity: RA(αβa+= b ) α RR AA ( a ) + β ( b ), ab , ∈V , αβ, ∈ ℝ
Notation for linear mapping: RAA()a= Ra (we drop the parantheses)
Isometry implies orthonormality. TT −1T RR= R R = 1, RRAA= R is an orthonormal transformation: AA A A V
= 1 two possibilities detRA = ± 1 2 푑푑푑퐑퐴 The angle between the vectors does not change. Both lenght and angle are preserved! 57 RAB aR b= a.b Rigid body transplacement
Start with: ar=PA − r From (*) follows:
Reference placement χχ(rA+= a ) ( r AA ) + Ra , a ∈V
χχ()rP=+− () r A Rr AP ( r A )
Present placement pP−= p A Rr AP() − r A
pAP= Rr A AP
58 Rigid body transplacement
pAB= Rr A AB RA ∈ SO()V Rotation tensor
59 What now, if we change the reduction point from A to B?
br=BA − r Changing the reduction point
χχ()rP=+− () r A Rr AP ( r A ) χχ()rP=+− () r B Rr BP ( r B )
What is the relationship between
R A and R B ?
RaraBBBAAAA=χχχχχχ( + )()( − r = rab ++ )( − rb + )( = rab ++ )() − r −
(χχ (rA+− b ) ( r A )) = R A ( a +− b ) Rb AAAAA = Ra + Rb − Rb = Ra
We may thus conclude that
RRRAB==∀∈, AB ,
The orthogonal transformation representing a transplacement of a rigid body is independent of the reduction point! 60 Rigid body transplacement
χχ(rA+= a ) ( r AA ) + Ra , a ∈V
χχ(rB+= a ) ( r BB ) + Ra , a ∈V
Does the rotation tensor depend on the reduction point?
RRBA= ?
No! RRRBA= = !
61 Rigid body transplacement
ur()A= p AA −= rχ () r A − r A
The displacement 62 Rigid body transplacement
u=+− u() R 1r PA AP
Theorem 4.1 A rigid body transplacement is uniquely determined by the displacement of (an arbitrary) reduction point A; uA and an orthonormal transformation R ∈ SO()V which is independent of the reduction point. 63 64 Summary
Box 4.1: Rigid body transplacement
A: reduction point
uA : displacement of the reducion point A RR: orthogonal transformation with det = 1
u= ur( ) =+− uAA ( R 1 )( r − r )
p=χχ() r = ( rAA ) +− Rr ( r )
65 The translation
A translation of a body is a transplacement where all displacements are equal, irrespective of the material point, i.e. uuuAB0==∀∈, AB,
What does the associated orthonormal tensor look like?
uPA=+− u() R 1rAP ()R−= 1a 0
R1= 66 The translation
uuuAB0==∀∈, AB, ,
67 The rotation
Consider a rotation of the body around in= a fixed axis in space through the point 3 Rotation axis: (,)A n
introduce a Right-handed Ortho-Normal basis (a ‘RON-basis’)
i= ( iii123 )
⊥ Decomposition: rAB=+⋅ r AB nnr()AB
⊥ Then we write riAB= 1xx 1 + i 2 2
68 The rotation
gB
gB p⊥ AB ⊥ rAB x2 i2 e2 e1 i1 ϕ A g A g = in3 ein33= = x1
Figure 4.9 Rotation around a fixed axis. ⊥ rAB=+⋅ r AB nnr()AB The vector corresponding to r AB in the present placement is: ⊥ riAB= 1xx 1 + i 2 2 ⊥ pAB=+⋅ p AB nn() pAB
⊥ peAB= 1xx 1 + e 2 2
22 2 rAB= p AB =xx 1 ++⋅ 2()nr AB 69 The rotation e= ( e e e ), e= n i 123 3 is a RON-basis. This basis is related to by gB
gB p⊥ AB ⊥ rAB x2 i2 e2 e1 i1 ϕ A g A g = in3 ein33= = x1
Figure 4.9 Rotation around a fixed axis.
ei11=cosϕϕ + i 2 sin ei21=( − sinϕ ) + i2 cos ϕ , 02 ≤< ϕπ ein33= =
70 This can be expressed in matrix format. The rotation matrix
cosϕϕ− sin 0 = ϕϕ e= iR[ ] (e1 e 2 e 3 ) ( i 123 i i ) sin cos0 , i 0 01
cosϕϕ− sin 0 R = sinϕϕ cos 0 [ ]i 0 01
The matrix representation of the rotation R is called the canonical represenataion.
TT= = det R = 1 [R]i[ R] i[ RR] ii[ ] [ 1], [ ]i
71 The rotation tensor
Ri= e Now let R be the tensor defined by: 11 = = Ri e i[ R]i Ri22= e Riein3= 33 = =
⊥ pAB=+⋅= p AB n() n rAB Ri 1x 1 + Ri 2 x 2 + Rn()( n ⋅= rAB R i 1xx 1 + i 2 2 +⋅ n ()) n rAB = Rr AB
p= Rr which means that the rotation is a rigid AB AB transplacement 72 Summary
Box 4.3: Rotation
(A ,n ): rotation axis
A: reduction point
u0A = : displacement of the reduction point A cosϕϕ− sin 0 RR: orthonormal transformation, = sinϕϕ cos 0 [ ]i 0 01 ϕ: rotation angle 0≤<ϕπ 2
i= ( iii123 ): referential base , i33= ne =
73 4.2 The Euler theorem
Theorem 4.2 (The Euler theorem, 1775) Every rigid transplacement with a fixed point is equal to a rotation around an axis through the fixed point.
74 The canonical representation
i= ( iii123 ), i 3= n The transplacement is given by:
pAB= Rr AB , ∀∈B
We now try to find out if there are any points C with the property:
pAC= Rr AC = r AC
If this is true then there are material vectors unaffected by the transplacement. is vector perpendicular to This brings us to the eigenvalue problem for the tensor R 퐢 퐧
Rn= λ n 75 The characteristic equation, eigenvalues
λλ= −= λ − = pR ()det(1R )det([ 1] [ R]i )
λϕ− cos sin ϕ 0 22 det− sinϕλϕ − cos 01 =−−+( λλϕ )(( cos) sin ϕ ) 001λ −
±iϕ p0R ()λ =⇔=λλ1 1, 23, = e
76 The canonical representation Rn= λ n i= ( iii ), i= n 123 3 Eigen value problem: find a material points C such that r AC is an eigenvector to R with the corresponding eigenvalue 1.
λ =1,in3 = = Ri i[ R]i
cosϕϕ− sin 0 R = sinϕϕ cos 0 [ ]i is vector perpendicular to 0 01 퐢 퐧 cosϕϕ− sin 0i1 R= iR iT = ( i i i ) sinϕϕ cos 0 i = [ ]i 123 2 0 01i3 77 The canonical representation
cosϕϕ− sin 0 i1 = ϕϕ = (iii123 ) sin cos 0 i2 0 01i3
i1 ϕϕ−= ϕ ϕ (i1 cos + i 2 sin i 1 ( sin )+ i2 cos ii 32 ) i3
(i1 cosϕϕ + i 2sin )⊗ ii 11 +( ( − sin ϕ )+ i2cos ϕ ) ⊗+⊗ ii 23 i 3) =
Equivalent to the following tensor-product representation of R
Rii=⊗+3 3( iiii 1 ⊗+⊗ 12 2)cosϕϕ + ( iiii2 ⊗−⊗ 11 2)sin
nn⊗+cosϕϕ ( 1nn −⊗ ) + sin n1 × 78 Summary
Box 4.5: Rigid transplacement
A: reduction point
rA : position vector of the reduction point in the reference placement
uA : displacement of the reducion point
r : position vector of a material point in the reference placement
p : position vector of a material point in the present placement R : rotation tensor
p=χ() r =+ r u +− Rr() r AA{ 1 4 2 43 A translation rotation
79 Screw representation of the transplacement
Since the translation u A changes with the reduction point A and the rotation R is independent of this choice we may ask if there is a specific reduction point S
such that the displacement u S is parallel to the rotation axis.
p=++ rSS ns Rr() − r Screw line ⊥ LS =∈{S rrrS =++ A AS nσσ, ∈¡ } ()R1u− T r ⊥ =−≠A , ϕ 0 AS 21(− cosϕ )
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 magnitude parallel to that axis. The reduction point for this so-called screw transplacement is located on the screw line given in (4.28). 80 The special orthogonal group
The set of all orthonormal tensors on V forms an algebraic group, the orthonormal group,
O()V = {Q QTT Q= QQ = 1}
Being a group means that
Q1, Q 2∈OO ()VV ⇒∈ QQ21 () 1∈O()V QQ−1T= ∈O()V
However, in general we have QQ12≠ QQ 21 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
SO()VV= { RR∈= O ()det 1} 81 Non-commutativity of rotations Combining rotations is thus a non-commutative operation as can be seen from the figure below. The special orthonormal group has the mathematical structure of a so-called Lie-group.
82
Chapter 4.3 The main representation theorem
Theorem 4.1 To every rotation tensor R ∈ SO()V there exists a unit vector n and an angle ϕ ∈ ¡ so that
RRn=(,)ϕϕ =⊗+ nn cos( 1nn −⊗ )sin + ϕ n1 ×
This representation is mainly unique. If R1= then n may be arbitrarily chosen, while ϕπ= 0 ( modulo 2 ). If R1≠ then n is uniquely determined, apart from a factor ±1. For a given n the angle ϕ is uniquely determined (modulo 2π ). For a definition of the tensor a1×∈End(V ) see Appendix A.4.39.
83 The main representation theorem using components RRn=(,)ϕϕ =⊗+ nn cos( 1nn −⊗ )sin + ϕ n1 ×
n1 n= en[ ] n = n e= ( ee12 e 3 ) e [ ]e 2 n3
n1 nn11 nn 12 nn 13 T nn⊗= n n =n n n n= nn nn nn [ ]e[ ] ee[ ] 2123( ) 212223 n3 nn31 nn 32 nn 33
0− nn32 n1×= n × 1 =n0n − [ ]e[ ] ee[ ] 31 −nn21 0
85 The main representation theorem using components
Rnn=⊗+cosϕϕ ( 1nn −⊗ ) + sin n1 ×
n1 e= ( ee e ) n= en n = n 12 3 [ ]e [ ]e 2 n3
= ⊗ +ϕϕ −⊗ + × = [R]ee[ nn] cos[ ( 1nn )]e sin [ n1] e
TT+ϕϕ − +× [nn]ee[ ] cos ([ 1]e[ nn] ee[ ] ) sin [ n1]e
86 How to calculate the rotation axis n and the rotation angle from the rotation tensor R Rnn=⊗+cosϕϕ ( 1nn −⊗ ) + sin n1 × 휑 Rn → ,?ϕ
b= ( bbb123 ) arbitrary RON-basis
Eigenvalue problem =⇔−=⇔ − = Rn n() R 1 n 0 ([ R]bb[ 1] )[ n] [ 0]
RRR11 12 13 n1 33× 31× R = RRR∈ n =n ∈ [ ]b 21 22 23 [ ]b 2 RRR31 32 33 ℝ n3 ℝ
87 The eigenvalue problem
R11 − λ R12 R13 n0 1 −=λ RR21 22 R23 n0 2 R31 RR32 33 − λ n03
R11 − λ RR12 13 λλ= − pR ( ) det RR21 22 R23 R31 RR32 33 − λ
±iϕ p0R ()λ =⇔=λλ1 1, 23, = e
λλ+− λλ cosϕϕ= 23 , sin = 23 2 2i 88 The eigenvalue problem
λ1 = 1
R111 − R12 R13 n 1 0 n1 − =⇒⇒ R21 R1R 22 23 n 2 0 n2 n R31 R32 R 33 − 1n3 0 n3
cosϕϕ− sin 0 trR =++=+ R R R 1 2cosϕ [R] = sinϕϕ cos 0 11 22 33 0 01
trR − 1 RRR1++− arccos( )= arccos( 11 22 33 ) 22 ϕ = RRR1++− 2π − arccos( 11 22 33 ) 2 89 Coordinate representations of SO()V
• Euler angles • Bryant angles • Euler parameters • Quaternions • ……
90 Euler angles: (,,) ψθφ
Reference placement Present placement
=oo = e Re e[ R]eo
91 Factorization: (,,) ψθφ
ψ R3 () θ R1 () R3 ()φ
fe= 0 f 33 3 e2 g2 = eg33 g2 g3 θ
f2 f 2
ψ 0 φ e2 e1 e0 1 f 1 gf11= g1
RR= 313()φθψ R () R ( )
Every rotation may be written as a product of three simple rotations.
92 Euler angles: First rotation ψ
o We start the factorization by introducing the first rotation Re 33 ()ψψ : which brings the o e f= ( fff ) o o base to the base 123 by rotating e around the axis with direction e 3 angle ψ .
= 0 fe33
0 00 0 f e= ( ee e ) 2 12 3
↓ ψ 0 e2 f= ( fff ) 0 123 e1 f1
cosψψ− sin 0 oo f= R()ψψ e = eR[ ()] o , R (ψ )o = sin ψψ cos 0 33e [ 3 ]e 0 01
93 Euler angles: Second rotation θ f Next we take the rotation Rf 11 () θθ : which brings to the RON-basis g = ( ggg 123 ) by a rotation with the angle θ around an axis with the direction f 1 , the so-called intermediate axis.
f3 g2 g 3 θ = f f( fff123 ) 2
↓ = g( ggg123 ) gf11=
10 0 =θθ = R (θ )= 0 cos θθ− sin g R11() f fR[ ()] f , [ 1 ] f 0 sinθθ cos 94 Euler angles: Third rotation φ
g Finally we take the rotation Rg 33 () φφ : which brings to the final basis e by a rotation the angle φ around an axis with direction g 3 .
e2 eg33= g2 g= ( ggg ) 123 ↓ φ e1 e= ( eee123 )
g1 cosφφ− sin 0 e= R()φφ g = gR[ ()] , R (φ )= sin φφ cos 0 33g [ 3 ]g 0 01 95 Euler angles: All rotations in combination
oo f= R()ψψ e = eR[ ()] o 33e =θθ = g R11() f fR[ ()] f e= R()φφ g = gR[ ()] 33g
⇓ Re-substitution yields :
oo eRegRfRReRRR= =[ 3()φ] = [ 13 () θφ] [ ()] = [ 3 ( ψ )] o [ 13 () θφ] [ ()] g fg e fg [R]eo
=o = φθψoo= ψ θ φ e Re R313() R () R ( ) e e[ R3 ( )]eo [ R 1 ()] fg[ R 3 ()]
96 Euler angles: The rotation
ψφθ eo → f →→ ge
= ψ θφ= [RR]eoo[ 3( )] e[ R 13 ()] fg[ R ()]
cosψψ−− sin 01 0 0cosφφ sin 0 sinψ cos ψ00 cos θ−= sin θφ sin cos φ0 0 0 10 sinθθ cos 0 0 1
cosψφ cos− sin ψθφ cos sin −− cos ψφ sin sin ψθφ cos cos sin ψθ sin sinψφ cos+ cos ψθφ cos sin −+ sin ψφ sin cos ψθφ cos cos − cos ψθ sin sinθφ sin sinθ cos φ cosθ
97 Euler angles
0 e3 e3 e2
e1 0 e2
0 e1 f1
Theorem 4.2 A rotation R is determined by a set of three real coordinates, called the Euler angles: (ψθφ , , )∈¡ 3 , 0 ≤ ψ < 20 π , ≤≤ θ π , 0 ≤< φ 2 π. Conversely these angles are essentially determined by R . In fact the angle θπ∈[0, ] is uniquely determined by R . If ℝ θπ∈]0, [ then also ψφ,,∈[02 π[ are uniquely determined by R . When θ = 0 only ψφ+ is uniquely determined and when θπ= only ψφ− . 98 Euler angles: Singularity
Special case: θπ= 0,
cosψ cos φ± sin ψφ sin −± cos ψφ sin sin ψ cos φ0 R o =sinψ cos φ ± cos ψφ sin −± sin ψφ sin cos ψ cos φ0 = [ ]e 0 01±
cos(ψφ±− ) sin( ψφ ± ) 0 RRR11 12 13 ψφ±− ψφ ± = sin( ) cos( ) 0 RRR21 22 23 0 0± 1 RRR31 32 33
Only ψφ± is determined. Singular!
99 Euler angles: Singularity
RRR11 12 13 R o = RRR= [ ]e 21 22 23 RRR31 32 33 cosψφ cos− sin ψθφ cos sin −− cos ψφ sin sin ψθφ cos cos sin ψθ sin sinψφ cos+ cos ψθφ cos sin −+ sin ψφ sin cos ψθφ cos cos − cos ψθ sin sinθφ sin sinθ cos φ cosθ
cosθθ=R1 , sin = − cos2 θ 33 θπ= 0, singular RR ⇒ cosψψ=−=23 , sin 13 sinθθsin RR cosφφ= 32 , sin = 31 sinθθ sin
These formulas show that numerical difficulties are to be expected for values of θ close to n π , n = 01 , ,.. . 100 The rotation angle trR =+ 1 2cosϕ =+ ( 1 cos θ )cos( φψ ++ ) cos θ
(11+ cosθ )cos( φψ ++ ) cos θ − cosϕ = 2
ψ
φ
θ
101 Figure 4.5 The gyroscope. Chapter 4.3.3 Euler parameters Start from the fundamental representation:
Rnn=⊗+cosϕϕ ( 1nn −⊗ ) + sin n1 ×=
1+(1 − cosϕϕ )( n ⊗− n1 ) + sin n1 ×
use of the trigonometric identities ϕϕ1− cos ϕϕ sin2 = sinϕ = 2 sin cos 22 22
ϕ ϕϕ R=+ 122sin2 n ××+ ( n1 ) sin cos n1 × 2 22
ϕϕ e0 =cos , en = sin 22 퐼퐼�퐼퐼�퐼퐼� ∶ 102 Euler parameters
R=+ 12()() e1e1 × ×+2e0 e1 × �푡𝑡 푤� ℎ푎푎� With a specific RON−basis b= ( bbb123 ), RON-basis, ϕϕ in which =++ n = 1 e0 =cos , en = sin nb11nnn b 22 b 33, 22 eb=++eee b b 11 22 33 Then we have the following relations: ϕϕϕϕ e0=cos , en11 = sin , en 22= sin , en 33= sin 2222
e3 4 S 3 = ∈42222 +++= eeeee(0123 , , , ) , eeee10 1 2 3 n.n =1 ⇒ ℝ S3=∈ e 42222 eeee1 +++= { 0123ℝ } ℝ e0 This constraint on the Euler parameters represents the unit sphere
The parameters e = ( eeee 0123 , , , ) are known as the Euler parameters. 103 How to calculate matrix elements from Euler parameters
To indicate the Euler parameters associated with a rotation we write
23 R= R()e = R ( e0 ,) e =+×+ 1 2 ( e1 ) 2e00 e1 × , e = ( e ,) e ∈ S
b= ( bbb123 ), RON-basis,
100 0−− ee32 0 ee32 R()e= 0102e0ee0e + − −+ [ ]b 3131 001−− ee21 0 ee21 0
22 0− ee32 2e(+− e ) 1 2ee ( − ee )( 2ee + ee ) 0 1 12 03 13 02 −= +22 +− − 2e03 e 0 e1 2ee (12 ee 03 )( 2e 0 e 2 ) 1 2ee (23 ee 01 ) − 22 ee21 0 2ee(13− ee 02 )( 2ee 23 + ee 01 )( 2e 0 +− e 3 ) 1 104 How to calculate Euler parameters from matrix elements
RRR11 12 13 R o = RRR [ ]e 21 22 23 RRR31 32 33
trR + 1 RRR1+++ e2 = = 11 22 33 0 44 R trR − 1 1+ 2R −−− R R R e2 =−=ii ii 11 22 33 , i= 123 , , i 24 4
No singularity! (In contrast to the Euler angles)
There is a 2 -1 correspondence between Euler parameters and rotation matrices! 105 How to calculate Euler parameters from matrix elements, condinued
2222 eeee10+++=⇒ 1 2 3( eeee 0123 , , , ) ≠ (,,,) 0000
RRR1+++ e0=±≠11 22 33 0 4
RR32− 23 RR13− 31 RR21− 12 e1 = , e2 = , e3 = 4e0 4e0 4e0
106 How to calculate Euler parameters from matrix elements
RRR1+++ e0=±=11 22 33 0 4
4e1 e 2= R 21 + R 12 , 4e1 e 3= R 31 + R 13 , 4e2 e 3= R 32 + R 23
1R+−− R R e= 0e, =±≠11 22 33 0 01 4
RR21+ 12 RR31+ 13 e2 = , e3 = 4e1 4e1
107 How to calculate Euler parameters from matrix elements Special cases:
1R+−− R R e= 0e, = 0e, =±≠22 11 33 0 012 4
RR31+ 13 e3 = 4e2
1R+−− R R e= 0e, = 0e, = 0e, =±≠33 11 22 0 0123 4
108 Chapter 4.3.4 The composition of rotations
Let R1 and R2 be two rotations. The composition of R1 and R2 is given by
R= RR21 and this represents another rotation. We have the following
Proposition 4.1 If RR1= 1(e 01, ,) e 1 , RR2= 2(e 02, ,) e 2 and R= RR21 then RR= (e0 ,) e where
e0= ee 01,, 02 −⋅ee 1 2, eeeee=2ee 01,, + 1 02 +× 1 2 (4.18)
Corollary 4.1 Two rotations commute if and only if their rotation axes are parallel, i.e.
RR21= RR 12 ⇔ e 1P e 2
109 Chapter 4.3.5 The exponential map
Rotation Rn: ϕ ∈ SO()V
Skew-sym. Tensor N= n1 ×∈Skew()V
Expanding in Maclaurin series:
ϕϕ23 R1=+ϕ N + N23 + N +=... exp(ϕϕ N ) = exp( n × 1 ) 23!!
The exponential map:
11 exp(A ) :=++ 1A A23 + A +..., A ∈End (V ) 23!! 110 Chapter 4.4 Time-dependent transplacement, Motion – velocitiy and acceleration
p==++−=+−≥χ( r ,)t ru () tt Rrr ()(�푟푟푟� 𝑡𝑡 )푡푡푡 p𝑡� () tt푡𝑡� Rrr ()( ), t0 PPAA PA A PA = pA ()t The rigid motion is a set of rigid transplacements parameterized by time t 111 Rigid body velocity
Spin tensor: W= RR T
Angular velocity: ω= ax() RR T (axial vector of W )
anti-symmetric tensor, since:
WT=() RR TT = RR T =−=− RR T W
RRT=⇒+=⇒=− 1 RR TT RR 0 RR T RR T
Rigid body velocity field: vvPA= +×ωpAP , P ∈ 112 Rigid body rotation
W= RRT ⇔= R WR =()ω × 1R =× ω R
R =ωR × t ⇔= = × RR (t ) (exp(∫ ω (s ) ds 1R )) 0 RR()0 = 00
Thus, if the angular velocity ωω = () t is known and the rotation tensor is known at time t0 = then the rotation tensor : RR = () t is known.
113 Rigid body acceleration
aaPA= +×αpAP +× ω( ωp ×AP ), P ∈
It is more difficult to graphically illustrate the acceleration field than the velocity field. We look at the special case when the body rotates around an axis with fixed direction
α ωn=ωω, αn = , n0 = ω
Tangential acceleration
The tangential and normal parts Centripetal acceleration αp× AP g g P of the relative acceleration:
×× pAP ω() ωpAP β αp×=AP ω np ×=AP ω pAP sin βω = d
2 2 A g aA ω××() ωpAP = ωωp ×AP = ωpAP sin βω=d Bt
114 4.4.2 Euler-Poisson velocity formula for moving bases
Starting from a fixed RON-base
In the reference placement R ω
0 00 0 e3 e = ( ee e ) 0 12 3 e = 3 e2 e( eee123 )
RON-basis following the
0 rotation of the rigid body 0 e2 e1 e1
000 e1= Re 12, e = Re 23, e = Re 3
[RR] = [ ] ee0
115 Euler-Poisson velocity formula for moving bases
R ω e 3 e0 3 e2
0 0 e2 e1 e1
116 Euler-Poisson velocity formula for moving vectors and tensors
3 • 3 = = = = rel aa()t∑ ei () tai () t (a)e ∑ eai ai i1= i1=
a=ωa ×+ arel
3 • 3 = = ⊗ =⊗=rel AA()t∑ ei () t ej () tA ij () t ( A)e ∑ eei ()tj () tA ij () t A ij1, = ij1, =
A=×−ω A Aω() ×+ 1 Arel
117 Angular acceleration
= ×+rel moving vector a ωa a ω e 3 e2 3 ωe= ∑ iωi i1=
e1
ω=×+ ωωω rel = ω rel
3 rel ωω= = ∑ eiω i i1=
118 Angular velocity and Euler angles
119 Combining rotations The addition of angular velocities, version 1
consider two rotation tensors RR11= ()t RR22= ()t
the combined rotation tensor R= RR21
We introduce the angular velocities
ω= ax() RR T T T 1 11 ω2= ax() RR 22 ω= ax() RR
ω= ω2 + Rω 21
Addition rule 1 – not used that often 120 Combining rotations The addition of angular velocities, version 2
R1 ω fe0 R2 0 f e3 3 g3 f 2 ω 0 ge
0 0 e2 g1 e1 f1 ωgf g2 = = f Re10 g Rf2
g= Re0 R= RR21
T T T ω= ω = ax() RR ω= ω = ax() RR1 ωgf = ax(( RR2 )f 2 ) ge0 11fe0 121 Combining rotations The addition of angular velocities, version 2
R1 ω fe0 R2 0 f e3 3 g3 f 2 ω 0 ge
0 0 e2 g1 e1 f1 ωgf g2
ω= ωω + ge00 gf fe
Addition rule 2
122 123 124 Euler angles, first rotation =ψψoo = f R33() e eR[ ()]eo ,
T ω= ω = ax() RR1 11fe0 cosψψ− sin 0 R (ψ )o = sin ψψ cos 0 [ 3 ]e 0 01
By use of Lemma 4.1 125 Euler angles,second rotation =θθ = g R11() f fR[ ()] f ,
10 0 R (θ )= 0 cos θθ− sin [ 1 ] f 0 sinθθ cos
126 Euler angles, third rotation
=φφ = e R33() g gR[ ()]g ,
cosφφ− sin 0 R (φ )= sin φφ cos 0 [ 3 ]g 0 01
127 128 Exercise 1:12
Given a rotation tensor RR= (,,)ψθφ where ψθφ,, are Euler angles. Calculate the 0 matrices [R(,,)15°°° 35 60 ] 0 and Re(,,)15°°° 35 60 . e 1 e0
135 Exercise 1:12
The rotation matrix expressed in Euler angles:
cos(ψ) −sin(ψ) 0 1 0 0 cos(φ ) −sin(φ ) 0 R(ψ, θ, φ ) := sin(ψ) cos(ψ) 0 ⋅ 0 cos(θ) −sin(θ) ⋅ sin(φ ) cos(φ ) 0 0 0 1 0 sin(θ) cos(θ) 0 0 1
Conversion from degrees to radians:
π π π R (ψ, θ, φ ) := R ψ⋅ , θ⋅ , φ ⋅ d 180 180 180
Calculation of the rotation matrix:
0.299 −0.943 0.148 R (15, 35, 60) = 0.815 0.171 −0.554 d 0.497 0.287 0.819
Checking the rotation matrix:
1 0 0 T R (15, 35, 60) = 1 R (15, 35, 60)⋅R (15, 35, 60) = 0 1 0 d d d 0 0 1
Calculating the rotation of the first basis vector:
1 0.299 R (15, 35, 60)⋅ 0 = 0.815 d 0 0.497 136 Exercise 1:14
The mechanism to control the deployment of a spacecraft solar panel from position A to position B is to be designed. Determine the transplacement, i.e. the translation vector and rotation tensor R , which can achieve the required change of placement. The side facing the positive x-direction in position A must face the positive z-direction in position B. Calculate the rotation vector nϕ corresponding to the rotation. (Meriam & Kraige 7/1).
137 Exercise 1:14
0 e3
n
ϕ
0 e2
e1
0 e1
e2
e3
138