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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Section 4 – Dynamic Simulation
Boxer Style Engine
Module 3 – Constraint Equations
Page 2
 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.
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Section 4 – Dynamic Simulation
Cross Section View
Module 3 – Constraint Equations
Page 3
This module will use the piston-crank portion of this engine to
demonstrate how kinematic and motion constraints are developed.
Counterweight
Cylinder
Liner
Bottom
Bearing
Cap
Piston
Connecting
Pin Piston
Rod
Bearing
Piston
Pin
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Rod Bolt
Crank Shaft
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Crank Bearing
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Section 4 – Dynamic Simulation
Planar System
Module 3 – Constraint Equations
Page 4




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.
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Cylinder 1
Cylinder 2
Cylinder 3
Cylinder 4
The planar equations will be
developed for Cylinder 3.
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Section 4 – Dynamic Simulation
Global Coordinate System
Module 3 – Constraint Equations
Page 5
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

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X
Cylinder 3
Y
X
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Section 4 – Dynamic Simulation
Part ID’s
Module 3 – Constraint Equations
Page 6
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.
A
Cylinder Liner
E
Crank Bearing
(Not Visible)
B
Piston
C
Connecting
Rod
D
Crank Shaft
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Section 4 – Dynamic Simulation
Mobility
Module 3 – Constraint Equations
Page 7

Gruebler’s equation can be used to establish the mobility of the planar
mechanism.
A
Cylinder Liner
E
Bodies (B) = 5
 Grounded bodies (G) = 2
 Revolute joints (R) = 3
 Prismatic joints (P) = 1

Crank Bearing
(Not Visible)
Mobility
DOF  3( B)  2( R)  2( P)  3(G )
 3(5)  2(3)  2(1)  3(2)  1
A mobility of one will require one
motion constraint.
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B
Piston
C
Connecting
Rod
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D
Crank Shaft
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Section 4 – Dynamic Simulation
List of DOF’s
Module 3 – Constraint Equations
Page 8




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.
© 2011 Autodesk
X cgA
List of Generalized Coordinates
YcgA
A
X cgB
Format
X-coordinate of the
cg of Body A
Body
X
A
cg
Capital letter
Center of
indicates that
variable is associated Gravity
with the global
coordinate system.
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YcgB
B
X cgC
YcgC
C
X cgD
YcgD
D
X cgE
YcgE
E
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Section 4 – Dynamic Simulation
Ground Joints
Module 3 – Constraint Equations
Page 9
 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.
Cylinder 1
Cylinder 2
Cylinder 3
 The cylinder liners must be
mathematically grounded
or fixed in space.
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Cylinder 4
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Section 4 – Dynamic Simulation
Cylinder Liner Ground Equations
Module 3 – Constraint Equations
Page 10
The location of the center of
gravity and the orientation of the
principal axes of inertia are
shown in the figures.
y
x
The ground constraint equations
that fix the position of the c.g.
and orientation of the principal
axes can be written as
y
z
X cgA156.8 mm  0
Y cgA 0
 cgA 0
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Section 4 – Dynamic Simulation
Crank Bearing Ground Joint
Module 3 – Constraint Equations
Page 11
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.
Virtual Crank Bearing
Located at the Origin

Y
Z
X
Crank Bearing
Constraint Equations
X
E
cg
0
YcgE  0
 cgE  0
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Section 4 – Dynamic Simulation
Summary of Ground Joint Equations
Module 3 – Constraint Equations
Page 12
Cylinder Liner
Virtual Crank Bearing
X cgA156.8 m m  0
X cgE  0
Y cgA 0
YcgE  0
 cgA 0
 0
E
cg
 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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Section 4 – Dynamic Simulation
2D Coordinate Transformation Matrix
Module 3 – Constraint Equations
Page 13
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()].
 From the figure,
y cos θ

X  x cos   y sin 
Y  x sin   y cos 
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Y
y sin θ
θ
x
x sin θ
y
θ
x cos θ
X
Matrix Form
 X  cos 
 
 Y   sin 
cos 
T   
 sin 
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 sin    x 
 
y cos    y 
 sin  
y cos  
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Section 4 – Dynamic Simulation
Revolute Joint
Module 3 – Constraint Equations
Page 14

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.

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yB
Body B
Y
Body A
B
yA
xB
xA
Joint
A
X
Two bodies connected at a common
point that allows relative rotational
motion.
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Section 4 – Dynamic Simulation
Revolute Joint
Module 3 – Constraint Equations
Page 15
 The position of Joint 1 on
Body A relative to the global
coordinate system is given by
the equation
A A A
R1  Rcg  r1
Y
Joint 1
yA
Body A
ˆj
1

r1 A
xA
iˆ
A
A
 The components of Rcg are
written with respect to the
global coordinate system
base vectors and the
A
r
components of 1 are written
with respect to the body
fixed coordinate system.
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Jˆ
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A
Rcg
Iˆ
 cg
RA
X
A
r1  x1Aiˆ  y1A ˆj
A
Rcg  X cgA Iˆ  YcgA Jˆ
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Section 4 – Dynamic Simulation
Revolute Joint
Module 3 – Constraint Equations
Page 16


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.
© 2011 Autodesk
Position Vector Equation
A A A
R1  Rcg  r1
Component Form
 X 1A   X cgA  cos  A
 A    A  
A
Y
Y
sin



 1   cg  
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 sin  A   x1A 
A  A 
cos    x1 
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Section 4 – Dynamic Simulation
Revolute Joint
Module 3 – Constraint Equations
Page 17

The coordinates of Joint 1 on
Body A are
 X 1A   X cgA  cos  A
 A    A  
A
Y
Y
sin



cg
 1  
 

 sin  A   x1A 
 
cos  A   y1A 
 In a Revolute Joint the
coordinates of the joint
must be same for each
body.
 Thus,
 X 1A   X 1B 
 A B
 Y1   Y1 
Similarly, the coordinates of
Joint 1 on Body B are
 X   X  cos 
 
 
B
Y
Y
sin



  
 
B
1
B
1
B
cg
B
cg
B
 sin 
cos  B
B
 x 
 
 y 
B
1
B
1
General Form of the
Constraint Equations for a
Planar Revolute Joint
© 2011 Autodesk
or
 X cgA  cos  A
 A  
 Ycg   sin  A
 X cgB  cos  B
 B 
 Ycg   sin  B
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 sin  A   x1A 
 
cos  A   y1A 
 sin  B   x1B  0
 
B  B 
cos    y1  0
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Section 4 – Dynamic Simulation
Revolute Joint
Module 3 – Constraint Equations
Page 18

The general form of the
constraint equations for a
planar revolute joint is
 X cgA  cos A
 A  
 Ycg   sin  A
 X cgB  cos B
 B 
 Ycg   sin  B

Joint 1
 sin  A   x1A 
 
cos A   y1A 
 sin  B   x1B  0
 
B  B 
cos   y1  0
The specific equations for the
three revolute joints in the
piston-crank mechanism will
now be developed
2nd Revolute Joint
Joint 2
Joint 3
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Section 4 – Dynamic Simulation
1st Revolute Joint
Module 3 – Constraint Equations
Page 19


Piston  Body B
The location of the joint
relative to the c.g. is needed to
B
define the parameters x1 & y1B
For the piston,
28 mm
y
x  28mm
B
1
x
y 0
B
1
C.G.
Joint 1
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Section 4 – Dynamic Simulation
1st Revolute Joint
Module 3 – Constraint Equations
Page 20


The location of the joint
relative to the c.g. is needed to
C
define the parameters x1 & y1C
From the picture,
x1C  102.6mm
Connecting Rod  Body C
102.6
y1C  0
y
x
Joint 1
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Section 4 – Dynamic Simulation
1st Revolute Joint
Module 3 – Constraint Equations
Page 21
Using the geometry from the
piston and connecting rod, the
revolute joint constraint equation
becomes
x1B  28m m
x1C  102.6mm
y1B  0
y1C  0
Joint 1
 X cgB  cos B
 B  
 Ycg   sin  B
 X cgC  cos C
 C 
 Ycg   sin  C
© 2011 Autodesk
 sin    x1B 
 
cos B   y1B 
B
 sin  C   x1C  0
 
C  C 
cos   y1  0
1st Revolute Joint Constraint Equations
X cgB  28 cos  B  X cgC  102.6 sin  C  0
YcgB  28 sin  B  YcgC  102.6 cos  C  0
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Section 4 – Dynamic Simulation
2nd Revolute Joint
Module 3 – Constraint Equations
Page 22
General Form of Constraint
Equation
 X cgC  cos  C
 C  
 Ycg   sin  C
 X cgD  cos  D
 D 
 Ycg   sin  D


 sin  C   x2C 
C  C 
cos    y2 
Body C
y
 sin  D   x2D  0
 
D  D 
cos    y2  0
The location of the joint
relative to the c.g. is needed
to define the parameters x2C
C
& y2
From the picture,
Joint 2
x
41.3 mm
x  41.3
C
2
y2C  0
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Section 4 – Dynamic Simulation
2nd Revolute Joint
Module 3 – Constraint Equations
Page 23
General Form of Constraint
Equation
 X cgC  cos  C
 C  
 Ycg   sin  C
 sin  C   x2C 
C  C 
cos    y2 
 X cgD  cos  D
 D 
 Ycg   sin  D


 sin  D   x2D  0
 
D  D 
cos    y2  0
The location of the joint
relative to the c.g. is needed
to define the parameters x2D
D
y
& 2
From the picture,
y
x
Joint 2
x2D  43
43 mm
y 0
D
2
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Section 4 – Dynamic Simulation
2nd Revolute Joint
Module 3 – Constraint Equations
Page 24
Using the geometry from the
connecting rod and crank shaft,
the revolute joint constraint
equation becomes
x  41 .3
x2D  43
y 0
y2D  0
C
2
C
2
C
 X cg
 cos C
 C  
 Ycg   sin  C
 X cgD  cos D
 D 
 Ycg   sin  D
© 2011 Autodesk
 sin  C   x2C 
C  C 
cos   y2 
 sin  D   x2D  0
 
D  D 
cos   y2  0
Joint 2
2nd Revolute Joint Constraint Equations
X cgC  41.3 cos  C  X cgD  43 sin  D  0
YcgC  41.3 sin  C  YcgD  43 cos  D  0
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Section 4 – Dynamic Simulation
3rd Revolute Joint
Module 3 – Constraint Equations
Page 25
General Form of Constraint
Equation
 X cgD  cos  D
 D  
 Ycg   sin  D
 X cgE  cos  E
 E 
 Ycg   sin  E


 sin  D   x2D 
D  D 
cos    y2 
 sin  E   x2E  0
 
E  E 
cos    y2  0
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.
© 2011 Autodesk
Joint 3
3rd Revolute Joint Constraint Equations
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 X cgD   X cgE  0
 D  E    
 Ycg   Ycg  0
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Section 4 – Dynamic Simulation
Summary of Revolute Joint Equations
Module 3 – Constraint Equations
Page 26
Body C
Body B
Body C
2nd Revolute Joint
Body D
Joint 2
Joint 1
C
X cgB  28cos B  X cg
 102.6 sin  C  0
C
X cg
 41.3 cos C  X cgD  43sin  D  0
YcgB  28sin  B  YcgC  102.6 cos C  0
YcgC  41.3 sin  C  YcgD  43cos D  0
Joint 3
Body D
© 2011 Autodesk
Body E
 X cgD   X cgE  0
 D  E    
 Ycg   Ycg  0
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Section 4 – Dynamic Simulation
Prismatic Joint
Module 3 – Constraint Equations
Page 27
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.

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yA
Y
Common
Axis
Body B
Body A
yA
xA
B
xA
A
X
Two bodies A & B that translate
relative to one another along a
common axis.
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Section 4 – Dynamic Simulation
Prismatic Joint Constraint Equations
Module 3 – Constraint Equations
Page 28





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 PQ  .
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yA
Y
Common
Axis
Body B
R
S
Body A
PQ 
yA
Q
xA
B
xA
P
A
X
Two bodies A & B that translate
relative to one another along a
common axis.
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Section 4 – Dynamic Simulation
Prismatic Joint Constraint Equations
Module 3 – Constraint Equations
Page 29

The vector PQ with components
written with respect to the
body fixed coordinate system of
Body A are

A ˆ
A ˆ
PQ  xPQ
i  y PQ
j

The components of the vector
PQ with respect to the global
coordinate system are
A
 X PQ
 cos  A
 A 
 YPQ   sin  A
© 2011 Autodesk
A
 sin  A   xPQ 
 A 
cos  A   y PQ 
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yA
Y
Body B
R
S
Body A
PQ 
yA
Q
xA
B
xA
P
A
X
Two bodies A & B that translate
relative to one another along a
common axis.
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Section 4 – Dynamic Simulation
Prismatic Joint Constraint Equations
Module 3 – Constraint Equations
Page 30

The vector RS with components
written with respect to the
body fixed coordinate system of
Body B are

B ˆ
B ˆ
RS  xRS
i  y RS
j

© 2011 Autodesk
Y
Body B
R
B

 sin    xRS
B  B 
cos    y RS 
S
Body A
PQ 
The components of the vector
RS with respect to the global
coordinate system are
B
 X RS
 cos  B
 B 
B
Y
sin

 RS  
yA
yA
Q
xA
B
xA
P
A
X
B
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Two bodies A & B that translate
relative to one another along a
common axis.
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Section 4 – Dynamic Simulation
Prismatic Joint Constraint Equations
Module 3 – Constraint Equations
Page 31


The third vector is directed
from point P to point R.
Point P has the coordinates
X  X
 
Y   Y
A
P
A
P

A
CG
A
CG
 cos 
 
A
  sin 
A
  x

  y
A
p
A
p



Y
Body B
R
S
Body A
PQ 
Point R has the coordinates
B
 cos  B
 X RB   X CG
 B    B  
B
 YR   YCG   sin 

 sin 
cos  A
A
yA
 sin  B   xRB 
 
cos  B   y RB 
yA
Q
xA
B
xA
P
A
The vector has components
B
 cos  B
 X PR   X CG

   B  
B
 YPR   YCG   sin 
A
 X CG
 cos  A
 A 
A
 YCG   sin 
© 2011 Autodesk
 sin  B   xRB 
 
cos  B   y RB 
 sin  A   xPA 
 
cos  A   y PA 
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X
Two bodies A & B that translate
relative to one another along a
common axis.
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Section 4 – Dynamic Simulation
Prismatic Joint Constraint Equations
Module 3 – Constraint Equations
Page 32

The vector perpendicular to
PQ has components
A
A
 X PQ
 
 cos 
 A 
A
 YPQ    sin 

A
 sin  A  0  1  xPQ 
 y A 
A 
1
0
cos   
  PQ 
yA
Y
R
 X PR 

0
Y
 PR 
X
 
A
PQ  PR  X PQ

X 
A
X PQ
 
0
Y 
© 2011 Autodesk
A
PQ 
B
RS
B
RS
S
Body A
PQ 
The dot product of two vectors
that are perpendicular to each
other is zero.
 
A
PQ  PR  X PQ

Body B
yA
Q
xA
B
xA
P
A
X
First Constraint Eq.
Second Constraint Eq.
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Two bodies A & B that translate
relative to one another along a
common axis.
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Section 4 – Dynamic Simulation
Prismatic Joint Constraint Equations
Module 3 – Constraint Equations
Page 33

Substituting the vector components from the previous slides into
the first constraint equation yields
 
A
PQ  PR  X PQ

0  1 cos 
1 0  
A

  sin 
T
x

A
PQ
y
A
PQ
A
A T
 sin  

cos  A 
X
A
PQ
B
 X CG
 cos  B
 B   
YCG   sin  B
 X PR 

0
 YPR 
A
 cos  A
 sin  B   xRB   X CG


B  B   A 
cos    y R  YCG   sin  A
 sin  A   xPA 
0
A   A 
cos    y P 
Substituting the vector components from the previous slides into
the second constraint equation yields
 
A
PQ  PR  X PQ

0  1 cos  A
1 0  
A

  sin 
T
x
© 2011 Autodesk
A
PQ
y
A
PQ
X
A
PQ
 sin  A 

cos  A 
T
B
 X RS

 B 0
 YRS 
cos  B

 sin  B
B

 sin  B   xRS
0
B   B 
cos    y RS 
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Section 4 – Dynamic Simulation
Summary of Prismatic Constraint Equations
Module 3 – Constraint Equations
Page 34
The two constraint equations for a planar prismatic joint are
1st Constraint Equation
0  1 cos A
1 0  
A
sin


 
T
x
A
PQ
y
A
PQ
 sin  A 

cos A 
T
B
B

 X CG  cos
 B   
B
Y
sin


 CG  
A
 cos A
 sin  B   xRB   X CG


B  B   A  
cos   yR  YCG   sin  A
 sin  A   xPA 

0


A 
A 
cos   yP 

2nd Constraint Equation
0  1 cos  A
1 0  
A

  sin 
T
x
A
PQ
y
A
PQ
 sin  A 

cos  A 
T
cos  B

 sin  B
B

 sin  B   xRS
0
B   B 
cos    y RS 
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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Section 4 – Dynamic Simulation
Prismatic Joint
Module 3 – Constraint Equations
Page 35





The prismatic joint formed by
the cylinder liner and the
piston lies along the global Xaxis.
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.
© 2011 Autodesk
y
y
x
x
P
Q
Vector
components
of PQ
S
Vector
components
of RS
A
xPQ
1
B
xRS
1
A
y PQ
0
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R
Point Coordinates
B
y RS
0
B
xPA  0 xR  0
B
y PA  0 y R  0
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Section 4 – Dynamic Simulation
Prismatic Joint
Module 3 – Constraint Equations
Page 36
Substitution of the vector components and point coordinates into the
two prismatic joint equations yields
1st Constraint Equation
0  1 cos  A
1 0
 
A
1 0   sin 
T
 sin  A 

cos  A 
T
B
 X CG
 cos  B
 B   
YCG   sin  B
A
 cos  A
 sin  B  0  X CG


B    A 
cos   0 YCG   sin  A
 sin  A  0
0
A   
cos   0
which reduces to
0  1 cos  A
1 0
 
A
1
0

  sin 
T
 sin  A 

cos  A 
T
B
A
 X CG
  X CG

 B    A   0
YCG  YCG 
2nd Constraint Equation
0  1 cos 
1 0
 
A
1
0

  sin 
T
© 2011 Autodesk
A
A T
 sin  

cos  A 
cos  B

 sin  B
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 sin  B  1
0
B   
cos   0
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Section 4 – Dynamic Simulation
Motion Constraint
Module 3 – Constraint Equations
Page 37




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.
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Motion Constraint
 D  314  t  0
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Section 4 – Dynamic Simulation
Summary of Constraint Equations
Module 3 – Constraint Equations
Page 38
There are five planar bodies each having three DOF giving a total of fifteen
DOF. Fifteen unknowns requires fifteen equations.
Revolute Joint 3
Revolute Joint 1
Ground Constraint 1
1) X 156.8  0
7) X  41.3 cos   X  43 sin   0
11) X cgD  X cgE  0
2)Y  0
8)Y  41.3 sin   Y  43 cos   0
12)YcgD  YcgE  0
A
cg
A
cg
C
cg
C
C
cg
C
D
cg
D
cg
D
D
3) cgA 0
Revolute Joint 2
Ground Constraint 2
Motion Constraint
4) X cgE  0
9) X cgB  28 cos  B  X cgC  102.6 sin  C  0
5)YcgE  0
10)YcgB  28 sin  B  YcgC  102.6 cos  C  0
6) cgE  0
Prismatic Joint
0  1 cos  A
13) 1 0 
 
A
1
0

  sin 
T
0  1 cos  A
14) 1 0 
 
A
1
0

  sin 
T
© 2011 Autodesk
15) D  314  t  0
B
A
  X CG

 sin    X CG

0
A   B   A 
cos   YCG  YCG 
A T
 sin   cos  B
 
cos  A   sin  B
A T
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 sin  B  1
0
B   
cos   0
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Section 4 – Dynamic Simulation
Summary of Constraint Equations
Module 3 – Constraint Equations
Page 39




© 2011 Autodesk
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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Section 4 – Dynamic Simulation
Module Summary
Module 3 – Constraint Equations
Page 40







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.
© 2011 Autodesk
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that content has been modified from the original, and must attribute source content to Autodesk.
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