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Dynamic Simulation:Constraint Kinematics

Objective

The objective of this module is to show how constraint equations are used to compute the position, velocity, and acceleration of the generalized coordinates.

These equations are kinematic in nature because they do not consider the forces required to cause the motion.

The kinematic and motion constraints developed in the previous module (Module 3) for the piston-crank mechanism are used to demonstrate the mathematics.

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Notation

The total set of constraint equations needed to define a mechanism includes both kinematic constraints and drive constraints.

There are 15 generalized coordinates and 15 nonlinear constraint equations for the piston-crank assembly used in Module 3.

Since the piston-crank has a mobility of one, only one of the fifteen equations will be a motion constraint that is an explicit function of time.

Section 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

Page 2

0,

,

tqq

tqd

k

qk is the set of kinematic constraint equations

tqd , is the set of motion constraint equations

q is the set of generalized coordinates

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Position

Solving the constraint equations will yield the value of each generalized coordinate at a specific instance of time.

The constraint equations are non-linear and the Newton-Raphson method is used as the solution method.

The Newton-Raphson method is iterative and converges when the constraint equations are satisfied.

Section 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

Page 3

0,

,

tqq

tqd

k

Constraint Equations

Newton-Raphson Equations

tqqtqqq ii ,,

1

1

where

qtq, is the Jacobian

matrix

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Velocity

The time derivative of the constraint equations is used to determine the velocities of the generalized coordinates.

Since the generalized coordinates are a function of time and the constraint equations are a function of the generalized coordinates and time, the chain rule for partial differentiation must be used.

Section 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

Page 4

0,

,

tqq

tqd

k

Constraint Equations

Time Derivative

0,

ttq

qttq

Velocities

tqtq

1

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Acceleration Section 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

Page 5

The second time derivative of the constraint equations is used to determine the accelerations of the generalized coordinates.

Since the generalized coordinates are a function of time and the constraint equations are a function of the generalized coordinates and time, the chain rule for partial differentiation must be used.

1st Time Derivative of Constraint Equations 0,

ttq

qttq

2nd Time Derivative of Constraint Equations

0

2,

2

2

2

2

2

2

2

ttq

q

tq

tqqq

qqttq

Accelerations

2

22

1

2

2

2tt

qtq

qqqq

qtq

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Summary of Equations

Newton-Raphson Equations

Section 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

Page 6

0,

,

tqq

tqd

k

Constraint Equations

tqqtqqq ii ,,

1

1

Used to determine the position (values of the generalized coordinates) at an instant in time.

Velocities of Generalized Coordinates

tqtq

1

Accelerations of Generalized Coordinates

2

221

2

2

2tt

qtq

qqqqqt

q

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Jacobian

The Jacobian and its inverse is needed to determine the position, velocity, and acceleration of the generalized coordinates.

Each i,j (row,column) term in the Jacobian matrix is given by

q

J Jacobian matrix

j

iji q

J

,

ith constraint equation

jth generalized coordinate

Error messages indicating that the Jacobian is singular are sometimes encountered when running multi-body dynamic programs.

This occurs when there is not a physically realizable solution.

Section 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

Page 7

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Piston-Crank Constraint Equations Section 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

Page 8

The fifteen constraint equations developed for the piston-crank mechanism in Module 3 are given on the right.

Note that only the motion constraint is an explicit function of time.

0)6

0)5

0)4

Ecg

Ecg

Ecg

Y

X

0)3

0)2

08.156)1

Acg

Acg

Acg

Y

X

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

XX

001

cossinsincos

cossinsincos

0110

01)14

BB

BBT

AA

AAT

0314)15 tD

0cossinsincos

0110

01)13

ACG

ACG

BCG

BCG

T

AA

AAT

YX

YX

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Piston-Crank JacobianSection 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

Page 9

000100000000000

000000000coscos

sinsin00

sin

cossin00

0000000000cossinsin

coscossin

010010000000000001001000000000000000sin6.10210cos2810000000000cos6.10201sin2801000000sin4310cos3.4110000000000cos4301sin3.4101000000100000000000000010000000000000001000000000000000000000000100000000000000010000000000000001

2 BA

BA

A

BA

AAAB

cgAcg

ABcg

AcgAA

B

CB

DC

DC

j

i

YY

XX

qJ

The Jacobian of the constraint equations is given below. Although there are many terms, there are a lot of zeros and the derivatives are easily computed.

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Velocities

The velocities of the generalized coordinates are computed from the equation

Since the Jacobian is known, this equation can be solved if the array containing the time derivatives of the constraint equations is found.

Only the motion constraint, (15), is an explicit function of time.

Section 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

Page 10

tqtq

1

31400000000000000

t

Motion Constraint

Required Array

0314)15 tD

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Acceleration Section 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

Page 11

The accelerations can be computed if each term is found.

2

221

2

2

2tt

qtq

qqqqqt

q

The time derivative of the Jacobian is zero.

This term is explained on the next slide.

Inverse of the Jacobian

31400000000000000

t

000000000000000

2

2

t

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Acceleration Term

qqqq

This term is evaluated by breaking it down into a series of operations that are easily done on a computer.

Step 1) Multiply the Jacobian by the velocities. This creates a column array.

qq

Step 2) Take the derivative of each row with respect to each generalized coordinate. This operation is similar to finding the Jacobian and results in a matrix.

q

qq

Step 3) Multiply the matrix by the velocities. This results in a column array.

qqqq

Section 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

Page 12

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Redundant Constraints

The Jacobian matrix is an important quantity and enables the position, velocity, and acceleration of the generalized coordinates to be found.

Application of the methods contained in this module requires that the Jacobian have an inverse.

This requires that the determinant of the Jacobian be non-zero or that the rank be equal to the number of generalized coordinates.

The rank of a matrix is equal to the number of independent rows or columns.

Independent rows or columns can not be written as a linear combination of other rows or columns.

If rows or columns of the Jacobian are not independent the Jacobian is singular and the problem does not have a solution.

Section 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

Page 13

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Redundant Constraints: Detection

The Dynamic Simulation environment within Autodesk Inventor software assembles the Jacobian and determines its rank as each constraint is added.

The rank gives the number of independent constraints.

The difference between the number of generalized coordinates and the number of independent constraints is equal to the degree of mobility.

The difference between the number of constraints and the number of independent constraints is equal to the degree of redundancy.

nq number of generalized coordinates

nc number of constraintsnic number of independent

constraints

Degree of Mobilitydom= nq – nic

Degree of Redundancydor = nc-nic

Section 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

Page 14

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Redundant Constraints: Reaction Forces

A redundant constraint occurs when the motion associated with a DOF is enforced by too many constraint specifications.

One or more of the constraint specifications can be removed without affecting the mobility of the system.

The joint reactions can not be independently determined when redundant constraints are present.

Although solutions can be obtained they are based on assumptions by the program as to which constraints to use.

Different assumptions will yield different answers.

Joints having friction are particularly effected by redundant constraints.

Friction forces are based on the joint normal forces.

Therefore, the friction forces are incorrect if the joint normal forces are incorrect due to redundant constraints.

Section 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

Page 15

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Redundant Constraint: Example

A simple four bar mechanism will have redundant constraints if revolute joints are used at all joints.

The ground link shown in the figure is fixed.

The revolute joint at 1 prevents the drive link from rotating about its long axis and moving normal to the joint plane.

The revolute joint at 2 prevents the coupler from rotating about its long axis and moving normal to the joint plane.

The revolute joint at 4 prevents the rocker from rotating about its long axis and moving normal to the joint plane.

Ground

Drive

Coupler

Rocker

1

2

3

4

A revolute joint at 3 is redundant because neither the rocker or coupler can rotate about their long axis or move normal to the joint plane due to the other revolute joints. These degrees of freedom are already restrained.

Section 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

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Redundant Constraints: Example

A point-line joint must be used at joint 3.

A point-line joint restricts a point (the center point of the hole at joint 3 on the coupler) to remain on a line (the centerline of the hole at joint 3 on the rocker).

Redundant joints can be confusing and a detailed analysis of what each joint is doing is required to figure out how to remove them.

An example of how to remove redundant constraints is provided in the next module: Module 5.

Ground

Drive

Coupler

Rocker

1

2

3

4Revolute

Revolute

Revolute

Point - Line

Section 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

Page 17

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Module Summary

This module showed how the constraint equations can be used to find the position, velocity, and acceleration of the generalized coordinates.

Kinematic relationships were used during the derivation and no mention of the forces required to impose the motion constraints was made.

The constraint equations for the piston-crank introduced in the previous module (Module 3) were used to demonstrate the mathematical steps.

The Newton-Raphson method is generally used to solve the constraint equations.

The Jacobian is a key component of the overall solution process and the rank of the Jacobian is used to detect redundant constraints.

Section 4 – Dynamic Simulation

Module 4 – Constraint Kinematics

Page 18