Lecture Slides (1)

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Dynamic Simulation:Constraint Equations

Objective

The objective of this module is to develop the equations for ground, revolute, prismatic, and motion constraints for a planar mechanism.

These equations will be developed for a piston-crank assembly in a Boxer style engine.

These constraint equations will be used in the next Module (Module 4) to show how position, velocity, and accelerations are computed.

Although the equations developed for this module are for a planar (2D) mechanism, the methods can be generalized to 3D mechanisms.


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Boxer Style Engine

Boxer style engines have a horizontally opposed piston configuration.

This has several advantages Lower center of gravity Lower vertical height Lighter weight Less vibration

Boxer style engines are used by Porsche and Subaru.

Because of their low vertical profile they are often called pancake engines.

Section 4 – Dynamic Simulation

Module 3 – Constraint Equations


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Cross Section View

Cylinder Liner Piston Connecting

Rod

Crank Shaft

Piston Pin

Crank Bearing

Counterweight

Piston Pin

Bearing

Bottom Bearing Cap

Rod Bolt

This module will use the piston-crank portion of this engine to demonstrate how kinematic and motion constraints are developed.




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Planar System

The boxer engine rotating assembly contains four piston assemblies.

Constraint equations will be written for one piston assembly to demonstrate the process.

This single assembly can be represented as a planar mechanism.

A Dynamic Simulation of the complete system will be presented in another module.

Cylinder 1Cylinder 2

Cylinder 3

Cylinder 4

The planar equations will be developed for Cylinder 3.




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Global Coordinate System

The constraint equations will be referenced to the stationary coordinate system shown in the figure.

This reference coordinate system is called the global coordinate system.

Capital letters are used to indicate that a coordinate or vector refers to this coordinate system.

Lower case letters will be used to indicate a coordinate or vector is referred to a body fixed coordinate system associated with a part.

Z

X

Y

X

Cylinder 3




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Part ID’s

ACylinder Liner

BPiston C

Connecting Rod D

Crank Shaft

ECrank Bearing(Not Visible)

The process of developing the constraint equations is facilitated by identifying each component by a letter.

The five components shown with letters make up the basic system for which the constraint equations will be written.




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Mobility

Gruebler’s equation can be used to establish the mobility of the planar mechanism.

Bodies (B) = 5 Grounded bodies (G) = 2 Revolute joints (R) = 3 Prismatic joints (P) = 1

ACylinder Liner

BPiston C

Connecting Rod D

Crank Shaft

ECrank Bearing(Not Visible)

1)2(3)1(2)3(2)5(3

)(3)(2)(2)(3

GPRBDOF

Mobility



A mobility of one will require one motion constraint.


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List of DOF’s

The DOF’s are associated with a set of generalized coordinates.

Each body has 3 DOF and 3 generalized coordinates.

The generalized coordinates for the planar mechanism are listed on the right.

Fifteen constraint equations must be developed that will enable each of the fifteen generalized coordinates to be determined.



E

Ecg

Ecg

D

Dcg

Dcg

C

Ccg

Ccg

B

Bcg

Bcg

A

Acg

Acg

Y

X

Y

X

Y

X

Y

X

Y

X

List of Generalized Coordinates

Format

AcgX

Capital letter indicates that variable is associated with the global coordinate system.

Center of Gravity

Body

X-coordinate of the cg of Body A


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Ground Joints

The cylinder liners are pressed into the engine block and do not move.

The pistons move relative to the cylinder liners and the combination make a prismatic joint.

The cylinder liners must be mathematically grounded or fixed in space.

Cylinder 1Cylinder 2

Cylinder 3

Cylinder 4




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Cylinder Liner Ground Equations

The location of the center of gravity and the orientation of the principal axes of inertia are shown in the figures.

The ground constraint equations that fix the position of the c.g. and orientation of the principal axes can be written as

0

0

0 8.156

Acg

Acg

Acg

Y

mmX



x

y

y

z


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Crank Bearing Ground Joint

The crank bearing is fixed in the engine block and does not move.

The crank shaft rotation relative to the crank bearing can be represented by a revolute joint.

All of the parts in the planar system must lie in the global X-Y plane.

Therefore, a “virtual” crank bearing will be placed at the origin of the global coordinate system so that the planar equations can be developed.

0

0

0

Ecg

Ecg

Ecg

Y

X

Crank Bearing Constraint Equations

X

Y

Z

Virtual Crank Bearing Located at the Origin




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Summary of Ground Joint Equations

Cylinder Liner Virtual Crank Bearing

0

0

0

Ecg

Ecg

Ecg

Y

X

0

0

0 8.156

Acg

Acg

Acg

Y

mmX

Each of these equations fix one DOF for the respective part in space.

None of the equations are a function of time.

None of the equations involve more than one part.




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2D Coordinate Transformation Matrix

In subsequent slides it will be necessary to transform the components of a vector from a body fixed coordinate system to the global coordinate system.

This transformation is accomplished with the transformation matrix [T(q)].

From the figure,

θ

X

Yx

yθsin x

θsin y

θ cosx

θ

θ cosy

cossin

sincos

yxY

yxX

Matrix Form

y

x

yY

X

cossin

sincos

cossin

sincos

yT




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Revolute Joint

There are three revolute joints in the piston-crank assembly Between the piston and connecting rod Between the connecting rod and

crankshaft Between the crankshaft and crank

bearing The constraint equations for a

revolute joint will be developed using the two bodies shown in the figure.

Body A and B have the same translational motion at the joint but can have relative rotation.

X

Y

Body A

Body B

xA

yA

qA

xB

yB

qB

Two bodies connected at a common point that allows relative rotational

motion.

Joint




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Revolute Joint

X

Y

Body AxA

yA

qA

Joint 1

AcgR

1Ar1

cgAR

I

J

ij

The position of Joint 1 on Body A relative to the global coordinate system is given by the equation

The components of are written with respect to the global coordinate system base vectors and the components of are written with respect to the body fixed coordinate system.

AAcg

A rRR 11

jyixr AAA ˆˆ111

JYIXR Acg

Acg

Acg

ˆˆ

AcgR

Ar1




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Revolute Joint

The components of the body fixed position vector must be transformed to the global coordinate system before the components of the two vectors can be added.

This is accomplished using the transformation matrix introduced earlier.

AAcg

A rRR 11

Position Vector Equation

Component Form

A

A

AA

AA

Acg

Acg

A

A

x

x

Y

X

Y

X

1

1

1

1

cossin

sincos




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Revolute Joint

The coordinates of Joint 1 on Body A are

Similarly, the coordinates of Joint 1 on Body B are

A

A

AA

AA

Acg

Acg

A

A

y

x

Y

X

Y

X

1

1

1

1

cossin

sincos

B

B

BB

BB

Bcg

Bcg

B

B

y

x

Y

X

Y

X

1

1

1

1

cossin

sincos

In a Revolute Joint the coordinates of the joint must be same for each body.

Thus,

B

B

A

A

Y

X

Y

X

1

1

1

1

or

0

0

cossin

sincos

cossin

sincos

1

1

1

1

B

B

BB

BB

Bcg

Bcg

A

A

AA

AA

Acg

Acg

y

x

Y

X

y

x

Y

X



General Form of the Constraint Equations for a Planar Revolute Joint


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Revolute Joint

The general form of the constraint equations for a planar revolute joint is

The specific equations for the three revolute joints in the piston-crank mechanism will now be developed

0

0

cossin

sincos

cossin

sincos

1

1

1

1

B

B

BB

BB

Bcg

Bcg

A

A

AA

AA

Acg

Acg

y

x

Y

X

y

x

Y

X

2nd Revolute Joint

Joint 1

Joint 2

Joint 3




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1st Revolute Joint

The location of the joint relative to the c.g. is needed to define the parameters &

For the piston,

x

y

Joint 1C.G.

28 mmBx1By1

0

28

1

1

B

B

y

mmx

Piston Body B




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1st Revolute Joint

102.6

Joint 1

Connecting Rod Body C


From the picture,

Cx1Cy1

0

6.102

1

1

C

C

y

mmx

x

y




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1st Revolute Joint

Using the geometry from the piston and connecting rod, the revolute joint constraint equation becomes

0cos6.102sin28

0sin6.102cos28

CC

cgBB

cg

CCcg

BBcg

YY

XX

Joint 1

0

0

cossin

sincos

cossin

sincos

1

1

1

1

C

C

CC

CC

Ccg

Ccg

B

B

BB

BB

Bcg

Bcg

y

x

Y

X

y

x

Y

X

0

6.102

1

1

C

C

y

mmx

0

28

1

1

B

B

y

mmx

1st Revolute Joint Constraint Equations




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2nd Revolute Joint

41.3 mm

0

0

cossin

sincos

cossin

sincos

2

2

2

2

D

D

DD

DD

Dcg

Dcg

C

C

CC

CC

Ccg

Ccg

y

x

Y

X

y

x

Y

X


From the picture,

General Form of Constraint Equation

Body C

Joint 2

Cx2Cy2

0

3.41

2

2

C

C

y

x

x

y




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2nd Revolute Joint

43 mm

0

0

cossin

sincos

cossin

sincos

2

2

2

2

D

D

DD

DD

Dcg

Dcg

C

C

CC

CC

Ccg

Ccg

y

x

Y

X

y

x

Y

X


From the picture,


Dx2Dy2

0

43

2

2

D

D

y

x

Joint 2

x

y




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2nd Revolute Joint

Using the geometry from the connecting rod and crank shaft, the revolute joint constraint equation becomes

0

0

cossin

sincos

cossin

sincos

2

2

2

2

D

D

DD

DD

Dcg

Dcg

C

C

CC

CC

Ccg

Ccg

y

x

Y

X

y

x

Y

X

0

3.41

2

2

C

C

y

x

0

43

2

2

D

D

y

x

0cos43sin3.41

0sin43cos3.41

DD

cgCC

cg

DDcg

CCcg

YY

XX

Joint 2

2nd Revolute Joint Constraint Equations




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3rd Revolute Joint

0

0

cossin

sincos

cossin

sincos

2

2

2

2

E

E

EE

EE

Ecg

Ecg

D

D

DD

DD

Dcg

Dcg

y

x

Y

X

y

x

Y

X

The c.g.’s of both the crank and crank shaft lie at the origin of the global coordinate system.

Therefore, the body fixed coordinates of the joint relative to the c.g. are zero.


0

0Ecg

Ecg

Dcg

Dcg

Y

X

Y

X

Joint 3

3rd Revolute Joint Constraint Equations




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Summary of Revolute Joint Equations

2nd Revolute Joint

Joint 1Joint 2

Joint 3

0

0Ecg

Ecg

Dcg

Dcg

Y

X

Y

X

0cos43sin3.41

0sin43cos3.41

DD

cgCC

cg

DDcg

CCcg

YY

XX

0cos6.102sin28

0sin6.102cos28

CC

cgBB

cg

CCcg

BBcg

YY

XX

Body B Body C Body C

Body D

Body D Body E




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Prismatic Joint

In the planar system the cylindrical joint between the cylinder liner and the piston acts like a prismatic joint.

A prismatic joint allows two bodies to translate relative to each other along a common axis.

The two bodies cannot rotate independent of each other.

The equations for a planar prismatic joint are based on the geometry shown in the figure.

X

Y Common Axis

Body A

Body B

xA

yA

qA

xA

yA

qB

Two bodies A & B that translate relative to one another along a

common axis.




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Prismatic Joint Constraint Equations

The points P and Q in Body A lie on the common axis and are connected by the vector PQ.

The points R and S in Body B lie on the common axis and are connected by the vector RS.

The vector PR also lies on the common axis and connects the points P and R.

The three vectors must be parallel.

Alternatively, vectors PR and RS must be perpendicular to .

X

Y Common Axis

P

Q

Body A

Body B

xA

yA

qA

xA

yA

qB

R S


common axis.

PQ

PQ




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The vector PQ with components written with respect to the body fixed coordinate system of Body A are

The components of the vector PQ with respect to the global coordinate system are

X

Y

P

Q

Body A

Body B

xA

yA

qA

xA

yA

qB

R S


common axis.

PQjyixQP A

PQAPQ

ˆˆ

APQ

APQ

AA

AA

APQ

APQ

y

x

Y

X

cossin

sincos




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The vector RS with components written with respect to the body fixed coordinate system of Body B are

The components of the vector RS with respect to the global coordinate system are

X

Y

P

Q

Body A

Body B

xA

yA

qA

xA

yA

qB

R S


common axis.

PQjyixSR B

RSBRS

ˆˆ

BRS

BRS

BB

BB

BRS

BRS

y

x

Y

X

cossin

sincos




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The third vector is directed from point P to point R.

Point P has the coordinates

Point R has the coordinates

The vector has componentsX

Y

P

Q

Body A

Body B

xA

yA

qA

xA

yA

qB

R S


common axis.

PQ

Ap

Ap

AA

AA

ACG

ACG

AP

AP

y

x

Y

X

Y

X

cossin

sincos

BR

BR

BB

BB

BCG

BCG

BR

BR

y

x

Y

X

Y

X

cossin

sincos

AP

AP

AA

AA

ACG

ACG

BR

BR

BB

BB

BCG

BCG

PR

PR

y

x

Y

X

y

x

Y

X

Y

X

cossin

sincos

cossin

sincos




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The vector perpendicular to PQ has components

The dot product of two vectors that are perpendicular to each other is zero.

X

Y

P

Q

Body A

Body B

xA

yA

qA

xA

yA

qB

R S


common axis.

PQ

APQ

APQ

AA

AA

A

PQ

A

PQ

y

x

Y

X

01

10

cossin

sincos

0

PR

PRA

PQ

A

PQ Y

XXXRPQP

First Constraint Eq.

0

BRS

BRSA

PQ

A

PQ Y

XXXRPQP

Second Constraint Eq.




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Substituting the vector components from the previous slides into the first constraint equation yields

Substituting the vector components from the previous slides into the second constraint equation yields

0

PR

PRA

PQ

A

PQ Y

XXXRPQP

0cossin

sincos

cossin

sincos

cossin

sincos

01

10

AP

AP

AA

AA

ACG

ACG

BR

BR

BB

BB

BCG

BCG

T

AA

AAT

APQ

APQ

y

x

Y

X

y

x

Y

Xyx

0

BRS

BRSA

PQ

A

PQ Y

XXXRPQP

0cossin

sincos

cossin

sincos

01

10

BRS

BRS

BB

BBT

AA

AAT

APQ

APQ

y

xyx




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Summary of Prismatic Constraint Equations

The two constraint equations for a planar prismatic joint are

0cossin

sincos

cossin

sincos

cossin

sincos

01

10

AP

AP

AA

AA

ACG

ACG

BR

BR

BB

BB

BCG

BCG

T

AA

AAT

APQ

APQ

y

x

Y

X

y

x

Y

Xyx

0cossin

sincos

cossin

sincos

01

10

BRS

BRS

BB

BBT

AA

AAT

APQ

APQ

y

xyx

1st Constraint Equation

2nd Constraint Equation

The vector components at the beginning and end of each equation are based on the body fixed coordinate systems and are constant. The only variables are the generalized coordinates of Body A and B.

These equations are easily evaluated in a computer program.




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Prismatic Joint

The prismatic joint formed by the cylinder liner and the piston lies along the global X-axis.

Point P is chosen to lie at the c.g. of the cylinder liner.

Point Q is chosen to lie 1 mm to the right on the x-axis.

Point R is chosen to lie at the c.g. of the piston.

Point S is chosen to lie 1 mm to the right on the x-axis.

x

y

x

y

P Q R S

0

1

APQ

APQ

y

x

Vector components

of PQ

0

1

BRS

BRS

y

x

Vector components

of RS

Point Coordinates

0

0

AP

AP

y

x

0

0

BR

BR

y

x




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Prismatic Joint

Substitution of the vector components and point coordinates into the two prismatic joint equations yields

00

0

cossin

sincos

0

0

cossin

sincos

cossin

sincos

01

1001

AA

AA

ACG

ACG

BB

BB

BCG

BCG

T

AA

AAT

Y

X

Y

X

00

1

cossin

sincos

cossin

sincos

01

1001

BB

BBT

AA

AAT

1st Constraint Equation

which reduces to

0cossin

sincos

01

1001

ACG

ACG

BCG

BCG

T

AA

AAT

Y

X

Y

X

2nd Constraint Equation




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Motion Constraint

One motion constraint is required to make the mechanism stable.

The rotation of the crankshaft (Body D) will be given an angular speed of 3,000 rpm.

A 3,000 rpm engine speed is equal to 314 rad/sec.

Although all fifteen generalized coordinates are a function of time, this is the only constraint equation that explicitly contains time as a variable.



0314 tD

Motion Constraint


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Summary of Constraint Equations

There are five planar bodies each having three DOF giving a total of fifteen DOF. Fifteen unknowns requires fifteen equations.

0)6

0)5

0)4

Ecg

Ecg

Ecg

Y

X

0)3

0)2

08.156)1

Acg

Acg

Acg

Y

X

Ground Constraint 1

Ground Constraint 2

0cos6.102sin28)10

0sin6.102cos28)9

CC

cgBB

cg

CCcg

BBcg

YY

XX

0cos43sin3.41)8

0sin43cos3.41)7

DD

cgCC

cg

DDcg

CCcg

YY

XX

0)12

0)11

Ecg

Dcg

Ecg

Dcg

YY

XXRevolute Joint 1

Revolute Joint 2

Revolute Joint 3

00

1

cossin

sincos

cossin

sincos

01

1001)14

BB

BBT

AA

AAT

Prismatic Joint

Motion Constraint0314)15 tD



0cossin

sincos

01

1001)13

ACG

ACG

BCG

BCG

T

AA

AAT

Y

X

Y

X


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Summary of Constraint Equations

Only one of the constraint equations is time dependent (Motion Constraint).

Most of the constraint equations are non-linear. All of the constraints are algebraic equations and none are

differential equations. Geometric quantities (dimensions and distances) contained in

the constraint equations can be found from information in a 3D CAD model.




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Module Summary

The constraint equations for ground, revolute, and prismatic joints have been developed for a planar mechanism.

The constraint equation for a rotational motion constraint has been developed for a planar mechanism.

These equations were used to determine the fifteen equations necessary for a piston-crank assembly taken from a Boxer engine model.

In some cases the constraint equations are very simple and in other cases they are complex.

Only the motion constraint is an explicit function of time. All of the constraint equations are algebraic. These equations will be applied in the next module: Module 4.



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