chapter 1ggn.dronacharya.info/.../questionbank/ivsem/kom/section-b/ppt2.pdf · multibody kinematics...

1.1 INTRODUCTION

Determine appropriate movement of the wipers

View range

Tandem or opposite

Wipe angle

Location of pivots

Timing of wipers

Wiping velocity

The force acting on the wipers

PART 2

Multibody Kinematics and Dynamics

Chapter 1

1

Kinematics

• Kinematics

– Deal with the way things move

• Kinematic analysis

– Determine

• Position, displacement, rotation, speed, velocity, acceleration

– Provide

• Geometry dimensions of the mechanism

• Operation range

• Dynamic analysis

– Power capacity, stability, member load

• Planar mechanism – motion in 2D space

2

1.4 MECHANISM TERMINOLOGY Mechanism

Design synthesis is the process of developing mechanism to satisfy a set of

performance requirements for the machine.

Analysis ensures that the mechanism will exhibit motion to accomplish the

requirements. Linkage

Frame

Links– rigid body

Joint

Primary joint (full joint)

Revolute joint (pin or hinge joint)–

pure rotation

Sliding joint (piston or prism joint)–

linear sliding

3

Higher-order joint (half joint)

Allow rotation and sliding

Cam joint

Gear connection

Simple link

A rigid body contains only two joints

Crank

Rocker

Complex link

A rigid body contains more than two joints

Rocker arm

Bellcrank

Point of interest

Actuator

A power source link 4

Mechanism Joint

1.5 Kinematic Diagram

Simple Link

Simple Link

(with point of

interest)

Complex Link

Pin Joint

Revolute Joint

5

Kinematic Diagram

Slider Joint

Cam Joint

Gear Joint

Translation Joint

6

1.7 MOBILITY

7

Constrained mechanism : one degree of freedom

Locked mechanism : zero degree of freedom

Unconstrained mechanism : more than one degree of freedom

1.7 Actuators and Drivers

• Electric motors (AC/DC)

• Engines

• Servomotors

• Air or hydraulic motors

• Hydraulic or pneumatic cylinders

• Screw actuators

• Manual

9

1.9 SPECLAL CASES OF THE MOBILITY EQUATION

Coincident Joints

• One degree of freedom actually if pivoted links are the same size

11

1.10 THE FOUR-BAR MECHANISM

13

14

Crank and Rocker

1.10.1 Design of Crank and Rocker

s : short link

l : long link

p , q : intermediate link

15

1.11 SLIDER-CRANK MECHANISM

16

1.12 SPECIAL PURPOSE MECHANISMS

1.12.1 Straight-Line Mechanisms

17

1.12.2 Parallelogram Mechanisms

1.12.3 Quick-Return Mechanisms

18

1.12.4 Scotch Yoke Mechanism

Locate the positions of all

links as driver link is

displaced

Configuration

Positions of all the links

One degree of freedom

Moving one link will

precisely position all

other links

Chapter 4 Displacement Analysis

4.4 DISPLACEMENT ANALYSIS

19

r1

r2

r4

r3 r3 r5

r4

Y

X

X1

Y1

1 2 3 4

3 31 1 2 2

1 1 2 2 3 3

1 1 2 3

3 4 5

3 3 4 4 1

3 3 4 4

1

0

5.30

3.2

3, 30, 4.9, 3.3

0

00.8

10.1

r r r r

r cr s r c

rc r s r s

r r r

r r r

r c r c x

r s r s

r

2 equations for 2 unknows

2 equations for 2 unknows

(1)

(2) 2 3,

4 1and x

4.1 Vector Analysis of Displacement

21

1 2 3

1 2

1 2

1 2 1 max

1 2 1 min

0

0.5 1.75 10

0.5 1.75

,

,

r r r

c c

s s y

for solve for and y

for solve for and y

r1

r3 r2

X

Y

22

4.7 Limiting Positions and Stroke

23

6.2 LINEAR AND ANGULAR VELOCITY

6.2.2 Linear Velocity of a General Point

Chapter 6 Velocity Analysis

24

1 2 3 4

2 2 3 3 4 4

3 4

0

0

r r r r

r r r

eqs for unknowns and

r2 r1

r3

r4

X

Y

25

6.6 GRAPHICAL VELOCITY ANALYSIS:RELATIVE VELOCITY METHOD

6.6.1 Points on Links Limited to Pure Rotation or Rectilinear Translation

6.6.2 General Points on a Floating Link

26

1 2 3

1 1 2 2

1 2

1 4

1 1 2 4

0

00

5

r r r

r r

eqs for and

r r r

r r r

r1

r2

r3

r4

X

Y

27

1 2 3

1 1 2 2 2

2

1 2

2

2

0

0

5 min ,

r r r

solve for and

r r r

vcrad r

vs

eqs for unknowns and v

Y

X

r3

r2

r1

28

1 2 3

2 2 3

1 1

1 1 2 2 2

2

2

2

1 2

0

16, 340

3

0

8

8

r r r

r r

eqs for unknowns r and

r r r

cr

s

eqs for unknowns and

r1

r3

r2

Y

X

29

1 2 3 4

1 1 2 2 3 3

1 2 3

0

0

r r r r

r r r

given find and

r1 r2

r3

r4

X

Y

30

7.2 LINEAR ACCELERATION

7.2.1 Linear Acceleration of Rectilinear Points

Chapter 7 Acceleration Analysis

31

1 2 3 4

1 1 2 2 3 3

1 2 3

1 1 1 2 2 2 2 2 3 3 3 3 3

2 3

0

0

0

r r r r

r r r

given find and

r r r r r

solve for and

r2 r1

r3

Y

r4

X

32

pr

r5

1 2 3 5

1 1 2 2 3 3 5

1 1 1 2 2 2 2 2 3 3 5 3 3 3 5

p

p

p

r r r r r

r r r r r

r r r r r r r r

33

1 2 3

1 1 2 2

2

1 1 1 1 1 2 2 2 2 2

2

0

00

00

r r r

r rv

solve for and v

r r r ra

solve for and a

r2 r3

r1 X

Y

34

7.12 EQUIVALENT LINKAGES

35

Computation of Multibody Kinematics And Dynamics

1.1 Multibody Mechanical Systems

2D Planar Mechanism

36

3D Spatial Mechanism

Computer-Aided Design (CAD)

• Mechanics: Statics and dynamics.

• Dynamics : kinematics and kinetics.

• Kinematics is the study of motion, i.e., the study of displacement, velocity, and acceleration, regardless of the forces that produce the motion.

• Kinetics or Dynamics is the study of motion and its relationship with the forces that produce that motion.

37

1.2 Coordinate Systems

2 2 2 2

1 1( ) 2 cos 2 cos 2 cos( ) 0r l s d rl ls rs

2 2 2 2

2( ) 2 cos 2 cos 0r l s d rl ds

1 2 3 2 0

1 2 3[ ]T q

A four-bar mechanism with generalized coordinates.

degrees of freedom 4 – 3 = 1

4 Coordinate

4

3 Constraints

38

1

2

3

q

1 2 3

1 2 3

cos cos cos 0

sin sin sin 0

r d s l

r d s

Generalized Coordinates

dof 3 2 1

2 constraints

39

1 1 1 2 2 2 3 3 3

Tx y x y x y q

1 1

1 1

1 1 2 2

1 1 2 2

2 2 3 3

2 2 3 3

3 3

3 3

cos 02

sin 02

cos cos 02 2

sin sin 02 2

cos cos 02 2

sin sin 02 2

cos 02

sin 02

rx

ry

r dx x

r dy y

d sx x

d sy y

sx l

sy

Cartesian Coordinates

dof 9 8 1

12 coordinates and 11 kinematic constraints

40

1.3 Computation Kinematics • A mechanism that is formed from a collection of links or bodies kinematically connected to one

another.

• An open-loop mechanism may contain links with single joint.

• A closed-loop mechanism is formed from a closed chain, wherein each link is connected to at

least two other links of the mechanism.

Figure (a) Open-loop mechanism－double pendulum

and (b) closed-loop mechanism－four-bar linkage.

41 Figure (a) Single-loop mechanism and (b) multi-loop mechanism.

Single and Multi-Loop Mechanism

Figure Example of kinematic pairs: (a) revolute joint, (b) translational joint,

(c) gear set, (d) cam follower, (e) screw joint, and (f) spherical ball joint.

High and Low Pair of Kinematic Joint

42

2 algebraic constraint equations,

1 1 2 2 3 3 1

1 1 2 2 3 3 2

cos cos cos 0

sin sin sin 0

l l l d

l l l d

Figure four-bar mechanism.

Generalized Coordinates

3 generalized coordinates,

43

1 2 3, , , , ,T

i ix y z i q is the vector of coordinates for body

in three - dimensional space.

44

2.1 Planar Kinematics in Cartesian Coordinates

P P

i i i i r r A s

cos sin

sin cosi

i

A

Inertial system Χ − Y

Body−fixed system ξ − η

Coordinate transformation matrix

The column vector is the vector of coordinates

for body in a plane.

[ , , ]T

i ix y qi

• A constraint equation describing a condition on the vector of coordinates of

a system can be expressed as follows:

• In some constraint and driving function, the variable time may appear

explicitly:

• Constraint Jacobian Matrix by differentiating the constraint equations

•

0 q

, 0t q

Constraint Equation

45

0

0 often denoted as 0

also denoted as 0

0

q

q q q

q q q

( )

,

( )

( )

q

q qq

qq + q

q q

q q q

q = q q

• Kinematically equivalent.

Figure (a) A double parallel-crank mechanism and (b) its kinematically equivalent.

Redundant Constraint

46 Figure Quick-return mechanism: (a) schematic presentation and

(b) its equivalent representation without showing the actual outlines.

Kinematics of Mechanism Mainly composed of revolute joint and translation joint

0P P

i i j j r s r s

( ,2) 0r P P

i i j j

i jr A s r A s

( ,2)cos sin cos sin 0

sin cos sin cos 0

P P P P

i i i i i j j j j jr

P P P P

i i i i i j j j j j

x x

y y

Figure Revolute joint connecting bodies and .

P i j

2.2 Revolute Joint

+ −

47

1 cos sin cos sin 0P P P P

i i i i i j j j j jx x

2 sin cos sin cos 0P P P P

i i i i i j j j j jy y

Time Derivative of Revolute Joint Constraint

yi+ (x

i

P cosfi-h

i

P sinfi)fi- y

j- (x

j

P cosfj-h

j

P sinfj)f

j= 0

xi- (x

i

P sinfi+h

i

P cosfi)fi- x

j+ (x

j

P sinfj+h

j

P cosfj)f

j= 0

48

1 cos sin cos sin 0P P P P

i i i i i j j j j jx x

Time Derivative of Revolute Constraint

2 sin cos sin cos 0P P P P

i i i i i j j j j jy y

2

2

( sin cos ) ( cos sin )

( sin cos ) ( cos sin ) 0

P P P P

i i i i i i i i i i i j

P P P P

j j j j j j j j j j

x x

2

2

( cos sin ) ( sin cos )

( cos sin ) ( sin cos ) 0

P P P P

i i i i i i i i i i i j

P P P P

j j j j j j j j j j

y y

xi- (x

i

P sinfi+h

i

P cosfi)fi- x

j+ (x

j

P sinfj+h

j

P cosfj)f

j= 0

yi+ (x

i

P cosfi-h

i

P sinfi)fi- y

j- (x

j

P cosfj-h

j

P sinfj)f

j= 0

or

2 2p p

i i j js s

49

2.3 Translational Joint

Figure Different representations of a translational joint.

50

Figure A translational joint between bodies and .

i j

0T

i n d

0

P P

i jP R P R

i i i i P P

i j

x xx x y y

y y

0 0

( )( ) ( )( ) 0

( ) 0

P P P Q P P P Q

i j i i i j i j

i j i j

x x y y y y x x

Translational Joint

0 0

( )( ) ( )( ) 0

( ) 0

P Q P P P Q P P

i i j i i i j i

i j i j

x x y y y y x x

Time Derivative of Translational Joint Constraint

g = -2[(xi

P - xi

Q )(xi- x

j)+ (y

i

P - yi

Q )(yi- y

j)f† -[(x

i

P - xi

Q )(yi- y

j)- (y

i

P - yi

Q )(xi- x

j)fi

2

51

Driving Link

Figure (a) The motion of the slider is controlled in the direction

and (b) the motion of point is controlled in the direction.

( ) 0ix d t

( ) 0P

iy d t

P

x y

52

Kinematics of a four-bar linkage

mm180

mm180

mm260

mm80

OC

BC

AB

OA

1 1

1 1

1 1 2 2

1 1 2 2

2 2 3 3

2 2 3 3

3 3

3 3

1

8 Constraint equations:

40 0

40 0

40 130 0

40 130 0

130 90 0

130 90 0

90 180 0

90 0

driving link

2 2 0

x

y

x x

y y

x x

y y

x

y

t

cos

sin

cos cos

sin sin

cos cos

sin sin

cos

sinA

B

C O

y

x

1

1

1

2

2

3

3

3

2

333222111

,,,,,,,, unknown 9for equations 9 thesolve To yxyxyxT q53

A Matlab Program for

Jacobian matrix and velocity equations

1

2

3 2

4 2

5 2 3

6 2 3

7 3

8 3

9

1 0 40sin 0 0 0 0 0 0

0 1 40cos 0 0 0 0 0 0

1 0 40sin 1 0 130sin 0 0 0

0 1 40cos 0 1 130cos 0 0 0

0 0 0 1 0 130sin 1 0 90sin

0 0 0 0 1 130cos 0 1 90cos

0 0 0 0 0 0 1 0 90sin

0 0 0 0 0 0 0 1 90cos

0 0 1 0 0 0 0 0 0

J

2

0

0

0

0

0

0

0

0

qβq velocity for the solve To J

54

Acceleration equations

0

sin90

cos90

sin90sin130

cos90cos130

sin130sin50

cos130cos50

sin40

cos40

2

33

2

33

2

33

2

22

2

33

2

22

2

22

2

2

22

2

2

2

γ

qγq onaccelerati for the solve To J

55

The procedure of m-file

0at start and Initialize t

tq solve toposition initial hsolver witnonlinear Call

solve to with andmatrix Jacobian Determine tt qqβ

solve to with Determine tt qqγ

iterationnext of position initialnew theasit set and

motion intended with Assume Δtt q

ttt qqq and , of response timePlot the

themanimate and bars theof positions theDetermine

56

The code of m.file (1) 1. % Set up the time interval and the initial positions of the nine coordinates

2. T_Int=0:0.01:2;

3. X0=[0 50 pi/2 125.86 132.55 0.2531 215.86 82.55 4.3026];

4. global T

5. Xinit=X0;

6.

7. % Do the loop for each time interval

8. for Iter=1:length(T_Int);

9. T=T_Int(Iter);

10. % Determine the displacement at the current time 11. [Xtemp,fval] = fsolve(@constrEq4bar,Xinit);

12.

13. % Determine the velocity at the current time

14. phi1=Xtemp(3); phi2=Xtemp(6); phi3=Xtemp(9);

15. JacoMatrix=Jaco4bar(phi1,phi2,phi3);

16. Beta=[0 0 0 0 0 0 0 0 2*pi]';

17. Vtemp=JacoMatrix\Beta;

18.

19. % Determine the acceleration at the current time

20. dphi1=Vtemp(3); dphi2=Vtemp(6); dphi3=Vtemp(9);

21. Gamma=Gamma4bar(phi1,phi2,phi3,dphi1,dphi2,dphi3);

22. Atemp=JacoMatrix\Gamma;

23.

24. % Record the results of each iteration

25. X(:,Iter)=Xtemp; V(:,Iter)=Vtemp; A(:,Iter)=Atemp;

26.

27. % Determine the new initial position to solve the equation of the next

28. % iteration and assume that the kinematic motion is with inertia

29. if Iter==1

30. Xinit=X(:,Iter);

31. else

32. Xinit=X(:,Iter)+(X(:,Iter)-X(:,Iter-1));

33. end

34.

35.end 57

The code of m.file (2)

36.% T vs displacement plot for the nine coordinates

37.figure

38.for i=1:9;

39. subplot(9,1,i)

40. plot (T_Int,X(i,:))

41. set(gca,'xtick',[], 'FontSize', 5)

42.end

43.% Reset the bottom subplot to have xticks

44.set(gca,'xtickMode', 'auto')

45.

46.% T vs velocity plot for the nine coordinates

47.figure

48.for i=1:9;

49. subplot(9,1,i)

50. plot (T_Int,V(i,:))


52.end


54.

55.% T vs acceleration plot for the nine coordinates

56.figure

57.for i=1:9;

58. subplot(9,1,i)

59. plot (T_Int,A(i,:))

60. AxeSup=max(A(i,:));

61. AxeInf=min(A(i,:));

62. if AxeSup-AxeInf<0.01

63. axis([-inf,inf,(AxeSup+AxeSup)/2-0.1 (AxeSup+AxeSup)/2+0.1]);

64. end


66.end

67.set(gca,'xtickMode', 'auto') 58

The code of m.file (3)

68.% Determine the positions of the four revolute joints at each iteration

69.Ox=zeros(1,length(T_Int));

70.Oy=zeros(1,length(T_Int));

71.Ax=80*cos(X(3,:));

72.Ay=80*sin(X(3,:));

73.Bx=Ax+260*cos(X(6,:));

74.By=Ay+260*sin(X(6,:));

75.Cx=180*ones(1,length(T_Int));

76.Cy=zeros(1,length(T_Int));

77.

78.% Animation

79.figure

80.for t=1:length(T_Int);

81. bar1x=[Ox(t) Ax(t)];

82. bar1y=[Oy(t) Ay(t)];

83. bar2x=[Ax(t) Bx(t)];

84. bar2y=[Ay(t) By(t)];

85. bar3x=[Bx(t) Cx(t)];

86. bar3y=[By(t) Cy(t)];

87.

88. plot (bar1x,bar1y,bar2x,bar2y,bar3x,bar3y);

89. axis([-120,400,-120,200]);

90. axis normal

91.

92. M(:,t)=getframe;

93.end

59

Initialization 1. % Set up the time interval and the initial positions of the nine coordinates

2. T_Int=0:0.01:2;

3. X0=[0 50 pi/2 125.86 132.55 0.2531 215.86 82.55 4.3026];

4. global T

5. Xinit=X0;

1. The sentence is notation that is behind symbol “%”.

2. Simulation time is set from 0 to 2 with Δt = 0.01.

3. Set the appropriate initial positions of the 9 coordinates which are used to solve nonlinear solver.

4. Declare a global variable T which is used to represent the current time t and determine the

driving constraint for angular velocity.

60

Determine the displacement

10. [Xtemp,fval] = fsolve(@constrEq4bar,Xinit);

a. function F=constrEq4bar(X)

b.

c. global T

d.

e. x1=X(1); y1=X(2); phi1=X(3);

f. x2=X(4); y2=X(5); phi2=X(6);

g. x3=X(7); y3=X(8); phi3=X(9);

h.

i. F=[ -x1+40*cos(phi1);

j. -y1+40*sin(phi1);

k. x1+40*cos(phi1)-x2+130*cos(phi2);

l. y1+40*sin(phi1)-y2+130*sin(phi2);

m. x2+130*cos(phi2)-x3+90*cos(phi3);

n. y2+130*sin(phi2)-y3+90*sin(phi3);

o. x3+90*cos(phi3)-180;

p. y3+90*sin(phi3);

q. phi1-2*pi*T-pi/2];

10. Call the nonlinear solver fsolve in which the constraint equations and initial values are necessary. The

initial values is mentioned in above script. The constraint equations is written as a function (which can

be treated a kind of subroutine in Matlab) as following and named as constrEq4bar. The fsolve finds a

root of a system of nonlinear equations and adopts the trust-region dogleg algorithm by default.

The equation of driving constraint

is depended on current time T 61

Determine the velocity 14. phi1=Xtemp(3); phi2=Xtemp(6); phi3=Xtemp(9);

15. JacoMatrix=Jaco4bar(phi1,phi2,phi3);

16. Beta=[0 0 0 0 0 0 0 0 2*pi]';

17. Vtemp=JacoMatrix\Beta;

15. Call the function Jaco4bar to obtain the Jacobian Matrix depended on current values of

displacement.

16. Declare the right-side of the velocity equations.

17. Solve linear equation by left matrix division “\” roughly the same as J-1β. The algorithm adopts

several methods such as LAPACK, CHOLMOD, and LU. Please find the detail in Matlab Help.

a. function JacoMatrix=Jaco4bar(phi1,phi2,phi3)

b.

c. JacoMatrix=[ -1 0 -40*sin(phi1) 0 0 0 0 0 0;

d. 0 -1 40*cos(phi1) 0 0 0 0 0 0;

e. 1 0 -40*sin(phi1) -1 0 -130*sin(phi2) 0 0 0;

f. 0 1 40*cos(phi1) 0 -1 130*cos(phi2) 0 0 0;

g. 0 0 0 1 0 -130*sin(phi2) -1 0 -90*sin(phi3);

h. 0 0 0 0 1 130*cos(phi2) 0 -1 90*cos(phi3);

i. 0 0 0 0 0 0 1 0 -90*sin(phi3);

j. 0 0 0 0 0 0 0 1 90*cos(phi3);

k. 0 0 1 0 0 0 0 0 0];

62

Determine the acceleration 20. dphi1=Vtemp(3); dphi2=Vtemp(6); dphi3=Vtemp(9);

21. Gamma=Gamma4bar(phi1,phi2,phi3,dphi1,dphi2,dphi3);

22. Atemp=JacoMatrix\Gamma;

21. Call the function Gamma4bar to obtain the right-side of the velocity equations depended on

current values of velocity.

22. Solve linear equation to obtain the current acceleration.

a. function Gamma=Gamma4bar(phi1,phi2,phi3,dphi1,dphi2,dphi3)

b.

c. Gamma=[ 40*cos(phi1)*dphi1^2;

d. 40*sin(phi1)*dphi1^2;

e. 40*cos(phi1)*dphi1^2+130*cos(phi2)*dphi2^2;

f. 40*sin(phi1)*dphi1^2+130*sin(phi2)*dphi2^2;

g. 130*cos(phi2)*dphi2^2+90*cos(phi3)*dphi3^2;

h. 130*sin(phi2)*dphi2^2+90*sin(phi3)*dphi3^2;

i. 90*cos(phi3)*dphi3^2;

j. 90*sin(phi3)*dphi3^2;

k. 0];

63

Determine next initial positions 29. if Iter==1

30. Xinit=X(:,Iter);

31. else

32. Xinit=X(:,Iter)+(X(:,Iter)-X(:,Iter-1));

33. end

29.~33. Predict the next initial positions with assumption of inertia except the first time of the loop.

64

Plot time response 37.figure

38.for i=1:9;

39. subplot(9,1,i)

40. plot (T_Int,X(i,:))


42.end

43.% Reset the bottom subplot to have xticks


45.

46.% T vs velocity plot for the nine coordinates

47.figure

48.for i=1:9;

37.…

37. Create a blank figure .

39. Locate the position of subplot in the figure.

40. Plot the nine subplots for the time responses of nine coordinates.

41. Eliminate x-label for time-axis and set the font size of y-label.

44. Resume x-label at bottom because the nine subplots share the same time-axis.

47.~ It is similar to above.

65

Animation 69.Ox=zeros(1,length(T_Int));

70.Oy=zeros(1,length(T_Int));

71.Ax=80*cos(X(3,:));

72.Ay=80*sin(X(3,:));

73.Bx=Ax+260*cos(X(6,:));

74.…

69. Determine the displacement of revolute joint.

80. Repeat to plot the locations by continue time elapsing.

81. Determine the horizontal location of .

88. Plot .

89. Set an appropriate range of axis.

80.for t=1:length(T_Int);

81. bar1x=[Ox(t) Ax(t)];

82. bar1y=[Oy(t) Ay(t)];

83. bar2x=[Ax(t) Bx(t)];

84. bar2y=[Ay(t) By(t)];

85. bar3x=[Bx(t) Cx(t)];

86. bar3y=[By(t) Cy(t)];

87.

88. plot (bar1x,bar1y,bar2x,bar2y,bar3x,bar3y);

89. axis([-120,400,-120,200]);

90. axis normal

91.

92. M(:,t)=getframe;

93.end

OA

OCBCABOA and , , ,

66

Time response of displacement

1x

1

1y

2x

2

2y

3x

3

3y

t67

Time response of velocity

1x

1

1y

2x

2

2y

3x

3

3y

t68

Time response of acceleration

1x

1

1y

2x

2

2y

3x

3

3y

t69

Example of a slider-crank mechanism

mm200

mm300

mm200

BO

GB

AG

1

1

1

4 3 3

4 3 3

3 3 2 2

3 3 2 2

2 2 1

2 2 1

4 1 4 4 4 1 4 4

4

11 Constraint equations:

0

0

0

200 0

200 0

300 100 0

300 100 0

100 0

100 0

100 100 100 100 0

x

y

x x

y y

x x

y y

x x

y y

y y x x

cos

sin

cos cos

sin sin

cos

cos

cos sin sin cos

1

2

0

driving link

5 76 1 2 0t

. .

1 1 1 2 2 2 3 3 3 4 4 4To solve the 12 equations for 12 unknown T x y x y x y x y ,

q , , , , , , , , , ,

A

B

C

O

y

x 4

4

2

23

3

3

2

1

1G

70

Jacobian matrix

0000003600000

35000000003400

33323100000002928

000000272600240

000000230220020

0001918017160000

0001501413012000

0100980000000

006504000000

000000000300

000000000020

000000000001

12

11

10

9

8

7

6

5

4

3

2

1

J

2.1

0

0

0

0

0

0

0

0

0

0

0

2

3

2

3

3

cos10027 ,17

sin30015

sin10023 ,13

cos2009

sin2005

135 ,24 ,20 ,16 ,12 ,8 ,4

136 ,34 ,26 ,22 ,18 ,14 ,10 ,6 ,3 ,2 ,1

144144

4

4

3

sincos10033

2932

2831

cos10029

sin10028

cos30019

yyxx

71

The right-side

of the acceleration equations

0

0

10

sin100

cos100

sin100sin300

cos100cos300

sin200

cos200

0

0

0

2

22

2

22

2

22

2

33

2

22

2

33

2

33

2

33

γ

γ

414414

2

444144414cos100sin100sin200cos20010 where yyxxyyxx

72


2x

2

2y

3x

3

3y

4x

4

4y

t73


2x

2

2y

3x

3

3y

4x

4

4y

t74


2x

2

2y

3x

3

3y

t

4x

4

4y

75

Kinematic Modeling

Figure Kinematic modeling of a slider-crank mechanism.

1 1 10.0, 0.0, 0.0x y

4 4 3 3

3 3 2 2

0 0 0 0

2 2 1 1

0.0, 0.0, 200.0, 0.0

300.0, 0.0, 100.0, 0.0

100.0, 0.0, 0.0, 0.0

A A A A

B B B B

4 4 1 1

0 0

4 4 1 1

0.0, 0.0, 100.0, 0.0,

100.0, 0.0, 0.0, 0.0

A A B B

C C

2 5.76 1.2 0.0t

driving link

Ground

Revolute joint

translational joint

76

1 1 1 1 1

1 1 1 1 1

4 4 4

4 4 4

1 1 4

4 4

1 1 4

100100

0 100

1000

100 100

100100 100

100

T

x c s x cB

y s c y s

c s sn

s c c

x c xn d s c

y s y

41

41

4 1 4 11

1 1 4 4 1 1 4 44

100

100

100 100 100

100 100 100 100 0

sx

cy

c c s s

y s y s x c x c

4 1 1 4 1 1 4 4100 100 100 100c y s y x c x s

Translation Joint 1y

1Bd

c

n

1 1 4

1 1 4

100

100

x c xd

y s y

77

1 1

2 1

3 1

0.0

0.0

0.0

x

y

Figure Kinematic modeling of a slider-crank mechanism.

Ground

revolute joints 4 4 3 3

5 4 3 3

6 3 3 2 2

7 3 3 2 2

8 2 2 1

9 2 2 1

200cos 0

200sin 0

300cos 100cos 0

300sin 100sin 0

100cos 0

100sin 0

x x

y y

x x

y y

x x

y y

translational joint

10 4 1 1 4

1 1 4 4

11 4 1

( 100cos )( 100sin )

( 100cos )( 100sin ) 0

0

y y

x x

driving constraint

12 2 5.76 1.2 0t

Kinematic Modeling

78

Figure The Jacobian matrix.

Jacobian Matrix

79

Cont’d

30 80

( )

( ) 0

( , ) 0d t

q

q

kinematic constraints

driving link

3.2 Solution Technique

velocity equations

( )( ) ( )

0 or 0

0 or 0

q

dd d

q tt

q = qq

q + = qq

( ) ( )

0d d

q

q t

q =

acceleration

( ) ( ) ( ) ( )

0 or ( ) 0

( ) 2 0

q q q

d d d d

q q q qt tt

qq + q = q q q

q q

q q q q

81

Solution Technique

At any given instant t

(1) Solve

n equations for n unknowns

(2) Solve


(3) Solve


( )

( , )

q

d

q t

q q = 0

q q = (the right hand side)

( )tq

( )

( , )

q

d

q t

q q = 0

q q = (the right hand side)

( )tq

( )

( )

q

d

q

q q =

q q =

(the right hand side)

( )tq

82

4.1 Planar Rigid Body Dynamics

i i xi

i i yi

i i i

m x f

m y f

n

Miqi= g

i

0 0

0 0

0 0

x

y

i

m x f

m y f

n

83

Constraint Force

1

1

( )

( )

( , ) 0,

0 or 0,

,

n

m n

q q

m

T

q

c

c T

q

T

q

t

q

q qq

Mq g g

g

Mq g

There exists Lagrange Multiplier

such that is that constraint force.

The equations of motion can be written as

84

Illustration of Constraint Force

0

4 eqs. for 5 unknowns , , , ,

0constraint equation

0

2 2 1, 0, , ,

3 3 3

c

mgs f mx

mgc N my

mgc N

f R I

x y f N

x R

y R

gx gs y s N mgc f mgs

R

Pure rolling of a disk down a slope

x

y

f

N

mg

85

2

Generalized coordinate ,

Constraint eq. 0 0

0

0 0

0 0 1 0

0 0 0 1

10 0 0

2

1 0 0 0

0 1 0 0 0

T

x

x R y R

x R

M mgsq q

mgc

q

m

m

mR R

R

1

2

0

0

0

2 2 13 eqs. for 3 unknowns , 0, ,

3 3 3

1

1 0 3

The constraint force 0 1

0 1

3

T

x mgs

y mgc

gx gs y s mgs

R

mgs

mgcq

RmgRs

3 1

is the friction force for pure rolling.

86

mixi= f

(x)i- l

1

miyi= f

( y)i- l

2

mifi= n

i- (y

i

P - yi)l

1+ (x

i

P - xi)l

2

4.2 Constraint Force in Revolute Joint

1

2

0 0 1 0

0 0 0 1

0 0 ( ) ( )

x

y

P P

i i i ii i

m x f

m y f

y y x x n

87

Constraint Force in Translational Joint

88

1( )P Q

i i xi i im x f y y

1( )P Q

i i yi i im y f x x

For a translational joint between i and j,

the equation of motion for body i can be written as

1 2[( )( ) ( )( )]P P Q P P Q

i i i j i i i j i i in x x x x y y y y

T

q Mq g

linear algebraic equations in unknowns

for and . q n m n m

4.3 Formulation of Multi-body Dynamic Systems

89

Dynamics of a four-bar linkage

m18.0

m18.0

m26.0

m08.0

OC

BC

AB

OA

A

B

C O

mN1.0 T

24

23

25

mkg1086.4

mkg1046.1

mkg1027.4

BO

AB

OA

I

I

I

kg18.0

kg26.0

kg08.0

BC

AB

OA

M

M

M

2sm9.8g

90

A Matlab Program for

Mass matrix and external force vector

4

3

5

1086.400000000

00.180000000

000.18000000

0001046.100000

00000.260000

000000.26000

0000001027.400

000000008.00

0000000008.0

M

2sm9.8g

kg18.0

kg26.0

kg08.0

BC

AB

OA

M

M

M

24

23

25

mkg1086.4

mkg1046.1

mkg1027.4

BO

AB

OA

I

I

I

mN1.0 T

0

8.918.0

0

0

8.926.0

0

1.0

8.908.0

0

g

91

0

sin09.0

cos09.0

sin09.0sin13.0

cos09.0cos13.0

sin13.0sin05.0

cos13.0cos05.0

sin04.0

cos04.0

2

33

2

33

2

33

2

22

2

33

2

22

2

22

2

2

22

2

2

2

γ

Jacobian matrix and γ

000000100

cos09.010000000

sin09.001000000

cos09.010cos13.010000

sin09.001sin13.001000

000cos13.010cos04.010

000sin13.001sin04.001

000000cos04.010

000000sin04.001

3

3

32

32

2

2

99

J

m18.0

m18.0

m26.0

m08.0

OC

BC

AB

OA

92

Computation

.0 velocity and 0 position initial with0 and 0for

solve tois stepfirst thecomputing, numerial In

1819

8889

89

qqλq

γ

g

λ

q

OJ

JMT

.for equation aldifferenti thesolve tois objective The8889

8999t

T

qγ

g

λ

q

OJ

JM

. and for step above repeat the to

and theuse and , 0

0 Integrate

ΔtΔt

ΔtΔtΔt

ΔtΔt

λq

qqq

q

q

q

.condition terminaluntil Repeat Δtt

93

A convenient way with Matlab solver

Solve initial value problems for ordinary differential equations with

ode45(commended), ode23, ode113…

The equations are described in the form of z‘=f(t,z)

q

qz

q

qz

,Let

3

1

1

3

1

1

y

x

y

x

94

The syntax for calling solver in Matlab

[T,Z] = ode45(@Func4Bar,[0:0.005:2],Z0);

column vector of

time points

Solution array

A vector specifying the interval

of integration

A vector of initial

conditions

A function that evaluates the right side of the

differential equations

function dz=Func4Bar(t,z)

global L1 L2 L3 L4 torque gravity

phi1=z(3); phi2=z(6); phi3=z(9);

dphi1=z(12); dphi2=z(15); dphi3=z(18);

M=diag([L1 L1 L1^3/12 L2 L2 L2^3/12 L3 L3 L3^3/12]);

J=[ -1 0 -0.5*L1*sin(phi1) 0 0 0 0 0 0;

0 -1 0.5*L1*cos(phi1) 0 0 0 0 0 0;

1 0 -0.5*L1*sin(phi1) -1 0 -0.5*L2*sin(phi2) 0 0 0;

0 1 0.5*L1*cos(phi1) 0 -1 0.5*L2*cos(phi2) 0 0 0;

0 0 0 1 0 -0.5*L2*sin(phi2) -1 0 -0.5*L3*sin(phi3);

0 0 0 0 1 0.5*L2*cos(phi2) 0 -1 0.5*L3*cos(phi3);

0 0 0 0 0 0 1 0 -0.5*L3*sin(phi3);

0 0 0 0 0 0 0 1 0.5*L3*cos(phi3)];

95

The syntax for calling solver in Matlab

J=[ -1 0 -0.5*L1*sin(phi1) 0 0 0 0 0 0;

0 -1 0.5*L1*cos(phi1) 0 0 0 0 0 0;

1 0 -0.5*L1*sin(phi1) -1 0 -0.5*L2*sin(phi2) 0 0 0;

0 1 0.5*L1*cos(phi1) 0 -1 0.5*L2*cos(phi2) 0 0 0;

0 0 0 1 0 -0.5*L2*sin(phi2) -1 0 -0.5*L3*sin(phi3);

0 0 0 0 1 0.5*L2*cos(phi2) 0 -1 0.5*L3*cos(phi3);

0 0 0 0 0 0 1 0 -0.5*L3*sin(phi3);

0 0 0 0 0 0 0 1 0.5*L3*cos(phi3)];

gamma=[ 0.5*L1*cos(phi1)*dphi1^2;

0.5*L1*sin(phi1)*dphi1^2;

0.5*L1*cos(phi1)*dphi1^2+0.5*L2*cos(phi2)*dphi2^2;

0.5*L1*sin(phi1)*dphi1^2+0.5*L2*sin(phi2)*dphi2^2;

0.5*L2*cos(phi2)*dphi2^2+0.5*L3*cos(phi3)*dphi3^2;

0.5*L2*sin(phi2)*dphi2^2+0.5*L3*sin(phi3)*dphi3^2;

0.5*L3*cos(phi3)*dphi3^2;

0.5*L3*sin(phi3)*dphi3^2];

g=[0 gravity*L1 torque 0 gravity*L2 0 0 gravity*L3 0]';

Matrix=[M J';

J zeros(size(J,1),size(J,1))];

d2q=Matrix\[g;gamma];

dz=[z(10:18,:); d2q(1:9,:)]; 96


1x

1

1y

2x

2

2y

3x

3

3y

t97


1x

1

1y

2x

2

2y

3x

3

3y

t98


1x

1

1y

2x

2

2y

3x

3

3y

t

310

99

Time response of λ

1

t

310

2

3

4

5

6

7

8

100

A slider-crank mechanism

m2.0

m3.0

m2.0

m1.0

BO

GB

AG

AC

A

B

C

O

G

24

22

24

mkg1067.6

mkg1004.1

mkg1067.6

BO

AB

slider

I

I

I

kg2.0

kg5.0

kg1.0

BO

AB

AC

M

M

M

mN1.0 T

2sm9.8g

101


2x

2

2y

3x

3

3y

4x

4

4y

t102


2x

2

2y

3x

3

3y

4x

4

4y

t103


2x

2

2y

3x

3

3y

t

4x

4

4y

310

104

Time response of λ

1

t

310

2

3

4

5

6

7

8

9

10

11

105

0 1 0 0 0

0 0 0

0 0 1 0 0 0 1

c s c s

s c c s s c

s c

D C BA DCB

5.1 Euler Angles

106

(5.1)

c c s c s c s s c c s s

s c c c s s s c c c c s

s s s c c

A =

Time Derivatives of Euler Angles

wx( )

= y sinq sinf +q cosf

wh( )

= y sinq cosf -q sinf

wz( )

= y cosq +f

sin sin cos 0

sin cos sin 0

cos 0 1

107

A DCB

2 2 3 3

1 1 3 3

1 1 2 2

1 0 0 0 0

0 0 1 0 0

0 0 0 0 1

c s c s

c s s c

s c s c

D C B

Bryant Angles

108

2 3 2 3 2

1 3 1 2 3 1 3 1 2 3 1 2

1 3 1 2 3 1 3 1 2 3 1 2

c c c s s

c s s s c c c s s s s c

s s c s c s c c s s c c

A =

Time Derivative of Bryant Angles

1 3 3 1

2 3 3 2

2 3

cos cos sin 0

cos sin cos 0

sin 0 1

109

chapter 1ggn.dronacharya.info/.../questionbank/ivsem/kom/section-b/ppt2.pdf · multibody kinematics...

Documents