mechanism all chaps.pdf

8/10/2019 mechanism all chaps.pdf

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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


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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 mechanismmotion in 2D space

2


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1.4 MECHANISM TERMINOLOGYMechanism

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

Linksrigid body

Joint

Primary joint (full joint)

Revolute joint (pin or hinge joint)

pure rotation

Sliding joint (piston or prism joint)

linear sliding

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Higher-order joint (half joint)

Allow rotation and sliding

Cam joint

Gear connection

Simple link

A rigid body contains only twojoints

Crank

Rocker

Complex link

A rigid body contains more thantwo joints

Rocker arm

Bellcrank

Point of interest

Actuator A power source link

4

Mechanism Joint


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1.5 Kinematic Diagram

Simple Link

Simple Link

(with point of

interest)

Complex Link

Pin Joint

Revolute Joint

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Kinematic Diagram

Slider Joint

Cam Joint

Gear Joint

Translation Joint

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1.7 MOBILITY

7

Constrained mechanism : one degree of freedom

Locked mechanism : zero degree of freedom

Unconstrained mechanism : more than one degree of freedom


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8


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1.7 Actuators and Drivers

Electric motors (AC/DC)

Engines Servomotors

Air or hydraulic motors

Hydraulic or pneumatic cylinders

Screw actuators

Manual

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10


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1.9 SPECLAL CASES OF THE MOBILITY EQUATION

Coincident Joints

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

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12


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1.10 THE FOUR-BAR MECHANISM

13


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14

Crank and Rocker


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1.10.1 Design of Crank and Rocker

s : short link

l : long link

p, q: intermediate link

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1 12 SPECIAL PURPOSE MECHANISMS


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1.12 SPECIAL PURPOSE MECHANISMS1.12.1 Straight-Line Mechanisms

17

1.12.2 Parallelogram Mechanisms

1 12 3 Q i k R t M h i


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1.12.3 Quick-Return Mechanisms

18

1.12.4 Scotch Yoke Mechanism


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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 allother links

Chapter 4 Displacement Analysis

4.4 DISPLACEMENT ANALYSIS

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4 1 V t A l i f Di l t


21/109

r1

r2

r4

r3r3r5

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

r c r s r s

r r r

r r r

r c r c x

r s r s

r

2 equationsfor 2 unknows

2 equationsfor 2 unknows

(1)

(2)2 3

,

4 1and x

4.1 Vector Analysis of Displacement

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1 2 3

1 2

1 2

1 2 1 max

1 2 1 min

0

0.5 1.75 1 00.5 1.75

,

,

r r r

c cs s y

for solve for and y

for solve for and y

r1

r3

r2

X

Y

22


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4.7 Limiting Positions and Stroke

23

Ch t 6 V l it A l i


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6.2 LINEAR AND ANGULAR VELOCITY

6.2.2 Linear Velocity of a General Point

Chapter 6 Velocity Analysis

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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

r2r1

r3

r4

X

Y

25

6 6 GRAPHICAL VELOCITY ANALYSIS:RELATIVE VELOCITY METHOD


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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

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1 2 3

1 1 2 2

1 2

1 4

1 1 2 4

0

0

05

r r r

r r

eqs for and

r r r

r r r

r1

r2

r3

r4

X

Y

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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

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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


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1 2 3 4

1 1 2 2 3 3

1 2 3

00

r r r r r r r

given find and

r1r2

r3

r4

X

Y

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7.2 LINEAR ACCELERATION

7.2.1 Linear Acceleration of Rectilinear Points

Chapter 7 Acceleration Analysis

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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 rgiven find and

r r r r r

solve for and

r2r1

r3

Y

r4

X

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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

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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 r a

solve for and a

r2 r3

r1 X

Y

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7.12 EQUIVALENT LINKAGES

35

Computation of Multibody Kinematics And Dynamics


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Computation of Multibody Kinematics And Dynamics1.1 Multibody Mechanical Systems

2D Planar Mechanism

36

3D Spatial Mechanism


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Computer-Aided Design (CAD)

Mechanics: Statics and dynamics.

Dynamics : kinematics and kinetics.

Kinematics is the study of motion, i.e., the study ofdisplacement, velocity, and acceleration, regardlessof the forces that produce the motion.

Kinetics or Dynamics is the study of motion and itsrelationship with the forces that produce that motion.

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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

G C


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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

Cartesian Coordinates


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1 1 1 2 2 2 3 3 3T

x 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

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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

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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

2.1 Planar Kinematics in Cartesian Coordinates


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1 2 3, , , , , T

i ix y z i q is the vector of coordinates for bodyin three-dimensional space.

44

P P

i i i i r r A s

cos sin

sin cosi

i

A

Inertial system Y

Bodyfixed system

Coordinate transformation matrix

The column vector is the vector of coordinates

for body in a plane.[ , , ]

T

i ix y q

i

Constraint Equation


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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

q

45

0

0 often denoted as 0

also denoted as 0

0

q

q q q

q q q

( )

,

( )

( )

q

q q

q

qq + q

q q

q q q

q = q q

Redundant Constraint


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Kinematically equivalent.

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

46

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

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

Kinematics of MechanismMainly composed of revolute joint and translation joint


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0P P

i i j j r s r s

( ,2) 0r P Pi i j j

i j

r 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

Ti D i ti f R l t J i t C t i t


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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 0

P P P P

i i i i i j j j j jy y

Time Derivative of Revolute Joint Constraint

yi+(x

i

Pcos

i

i

Psin

i)

i y

j (x

j

Pcos

j

j

Psin

j)

j= 0

xi (x

i

P

sin

i+

i

P

cos

i )

i xj+(x

j

P

sin

j+

j

P

cos

j )

j= 0

48


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2.3 Translational Joint


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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 Pi j

x x

x 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


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0 0

( )( ) ( )( ) 0

( ) 0

P Q P P P Q P Pi 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

= 2[(xi

P x

i

Q )(xi x

j)+ (y

i

P y

i

Q)(yi y

j) [(x

i

P x

i

Q)(yi y

j) (y

i

P y

i

Q )(xi x

j)

i

2

51


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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

A Matlab Program for


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Kinematics of a four-bar linkage

mm180

mm180

mm260

mm80

OC

BC

AB

OA1 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 link2 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

CO

y

x

1

1

1

2

2

3

3

3

2

333222111

,,,,,,,,unknown9forequations9thesolveTo yxyxyxT q53



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Jacobian matrix and velocity equations

1

2

3 2

4 2

5 2 3

6 2 3

7 3

8 3

9

1 0 40 sin 0 0 0 0 0 0

0 1 40 cos 0 0 0 0 0 0

1 0 40 sin 1 0 130 sin 0 0 0

0 1 40 cos 0 1 130 cos 0 0 0

0 0 0 1 0 130 sin 1 0 90 sin

0 0 0 0 1 130 cos 0 1 90 cos

0 0 0 0 0 0 1 0 90 sin

0 0 0 0 0 0 0 1 90 cos

0 0 1 0 0 0 0 0 0

J

2

0

0

0

0

0

0

0

0

qq velocityfor thesolveTo J

54


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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

qq onacceleratifor thesolveTo J

55


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The procedure of m-file

0atstartandInitialize t

tqsolvetopositioninitialhsolver witnonlinearCall

solvetowithandmatrixJacobianDetermine tt qq

solvetowithDetermine tt qq

iterationnextofpositioninitialnewtheasitsetand

motionintendedwithAssume ttq

ttt qqq and,ofresponsetimePlot the

themanimateandbarstheofpositionstheDetermine

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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. globalT

5. Xinit=X0;

6.

7. % Do the loop for each time interval

8. forIter=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 time14. 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 iteration25. 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. ifIter==1

30. Xinit=X(:,Iter);

31. else

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

33. end

34.35.end 57


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The code of m.file (2)

36.% T vs displacement plot for the nine coordinates37.figure

38.fori=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 coordinates47.figure

48.fori=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.fori=1:9;58. subplot(9,1,i)

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

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

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

62. ifAxeSup-AxeInf


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The code of m.file (3)

68.% Determine the positions of the four revolute joints at each iteration69.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.% Animation79.figure

80.fort=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

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Determine the displacement

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

a. functionF=constrEq4bar(X)

b.

c. globalT

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 T61


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Determine the velocity14. 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-1. 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];

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Determine the acceleration20. 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


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Determine next initial positions29. ifIter==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.

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Plot time response37.figure

38.fori=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.fori=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.

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Animation69.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.fort=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


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Time response of displacement

1x

1

1y

2x

2

2y

3x

3

3y

t

67


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Time response of velocity

1x

1

1y

2x

2

2y

3x

3

3y

t

68


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70/109

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 0300 100 0

300 100 0

100 0

100 0

100 100 100 100 0

x

y

x x

y yx 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

x4

4

2

23

3

3

2

1

1G

70


71/109

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

135,24,20,16,12,8,4136,34,26,22,18,14,10,6,3,2,1

144144

4

4

3

sincos10033

2932

2831

cos10029

sin10028

cos30019

yyxx

71

The right-side


72/109

The right-side

of the acceleration equations

0

010

sin100

cos100

sin100sin300

cos100cos300

sin200

cos200

0

00

2

22

2

22

2

22

2

33

2

22

2

33

2

33

2

33

414414

2

444144414cos100sin100sin200cos20010where yyxxyyxx

72


73/109


2x

2

2y

3x

3

3y

4x

4

4y

t

73


74/109


2x

2

2y

3x

3

3y

4x

4

4y

t

74


75/109

Time response of acceleration

2x

2

2y

3x

3

3y

t

4x

4

4y

75

Kinematic Modeling


76/109

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

Translation Joint


77/109

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

4

1

4

1

4 1 4 1

1

1 1 4 4 1 1 4 4

4

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

a s at o Jo t1

y

1

Bd

c

n

1 1 4

1 1 4

100

100

x c xd

y s y

77

Kinematic Modeling


78/109

1 1

2 1

3 1

0.0

0.0

0.0

x

y

Figure Kinematic modeling of a slider-crank mechanism.

Ground

revolute joints4 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

200 cos 0

200 sin 0

300 cos 100 cos 0

300 sin 100 sin 0

100 cos 0

100 sin 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

( 100 cos )( 100sin )

( 100 cos )( 100sin ) 0

0

y y

x x

driving constraint

12 2 5.76 1.2 0t

g

78


79/109

Figure The Jacobian matrix.

Jacobian Matrix

79

Contd


80/109

30 80

3.2 Solution Technique


81/109

( )

( ) 0

( , ) 0d t

q

q

kinematic constraints

driving link

q

velocity equations

( )( ) ( )

0 or 0

0 or 0

q

dd d

q tt

q = qq

q + = q q

( ) ( )

0d d

q

q t

q =

acceleration

( ) ( ) ( ) ( )

0 or ( ) 0

( ) 2 0

q q q

d d d d

q q q qt tt

q

q + q = q q q q q

q q q q

81

Solution Technique

At i i t t


82/109

At any given instant t

(1) Solve

n equations for n unknowns

(2) Solve


(3) Solve


( )

( , )

q

dq t

qq = 0

q q =(the right hand side)

( )tq

( )

( , )qd

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 Pl Ri id B d D i


83/109

4.1 Planar Rigid Body Dynamics

i i xi

i i yi

i i i

m x f

m y f

n

Miq

i =g

i

0 0

0 0

0 0

x

y

i

m x f

m y f

n

83

Constraint Force


84/109

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

M q g g

g

M q g

There exists Lagrange Multipliersuch that is that constraint force.

The equations of motion can be written as

84

Illustration of Constraint Force


85/109

0

4 eqs. for 5 unknowns , , , ,

0

constraint 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

Generalized coordinate ,

Constraint eq. 0 0

x

x R y R


86/109

2

0

0 0

0 0 1 0

0 0 0 1

10 0 0

21 0 0 0

0 1 0 0 0

T

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 3The 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

4.2 Constraint Force in Revolute Joint


87/109

mix

i= f

(x)i

1

miyi= f(y)i 2

m

i

i= n

i (y

i

P y

i)

1+ (x

i

P x

i)

2

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


88/109

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 iand j,

the equation of motion for body ican 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


89/109



90/109

Dynamics of a four-bar linkage

m18.0

m18.0

m26.0

m08.0

OC

BC

AB

OA

A

B

CO

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

g

M t i d t l f t


91/109

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

2

sm9.8g

kg18.0

kg26.0kg08.0

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

J bi t i d


92/109

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

Jacobian matrix and

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

Comp tation


93/109

Computation

.0velocityand0positioninitialwith0and0for

solvetoisstepfirstthecomputing,numerialIn

1819

8889

89

qqq

g

q

OJ

JM T

.forequationaldifferentithesolvetoisobjectiveThe8889

8999t

T

q

g

q

OJ

JM

.andforstepaboverepeat theto

andtheuseand,0

0Integrate

tt

ttt

tt

q

qqq

q

q

q

.conditionterminaluntilRepeat tt

93

A con enient a ith Matlab sol er


94/109

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 s nta for calling sol er in Matlab


95/109

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

functiondz=Func4Bar(t,z)

globalL1 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



96/109


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



97/109


1x

1

1y

2x

2

2y

3x

3

3y

t

97



98/109


1x

1

1y

2x

2

2y

3x

3

3y

t

98



99/109


1x

1

1y

2x

2

2y

3x

3

3y

t

3

10

99

Time response of


100/109

Time response of

1

t

3

10

2

3

4

5

6

7

8

100


101/109



102/109


2x

2

2y

3x

3

3y

4x

4

4y

t102



103/109


2x

2

2y

3x

3

3y

4x

4

4y

t103



104/109


2x

2

2y

3x

3

3y

t

4x

4

4y

3

10

104

Time response of


105/109

Time response of

1

t

3

10

2

3

4

5

6

7

8

9

10

11

105


106/109

Time Derivatives of Euler Angles


107/109

Time Derivatives of Euler Angles

w

x

( )

= y sinq sin +q cos

w

h( )= y sinq cos q sin

w

z( )= y cosq +

sin sin cos 0

sin cos sin 0

cos 0 1

107

Bryant Angles


108/109

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

y g

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


109/109

y g

1 3 3 1

2 3 3 2

2 3

cos cos sin 0

cos sin cos 0

sin 0 1

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