chapter 4 - dynamic analysis lecture

7/31/2019 Chapter 4 - Dynamic Analysis Lecture

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Chapter 4Dynamic Analysis and Forces

4.1 INTRODUCTION

In this chapters. The dynamics, related with accelerations, loads, masses and inertias.

____

amF

____

IT

In Actuators. The actuator can be accelerate a robots links for exerting enough forces

and torques at a desired acceleration and velocity.

By the dynamic relationships that govern the motions of the robot,

considering the external loads, the designer can calculate the necessary

forces and torques.

Fig. 4.1 Force-mass-acceleration and torque-inertia-angularacceleration relationships for a rigid body.


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4.2 LAGRANGIAN MECHANICS: A SHORT OVERVIEW

Lagrangian mechanics is based on the differentiation energy termsonly, with respect to the systems variables and time.

Definition: L= Lagrangian, K= Kinetic Energy of the system, P= Potential

Energy, F= the summation of all external forces for a linearmotion, T= the summation of all torques in a rotational motion,x= System variables

PKL

ii

i

xL

x

Lt

F

ii

i

LL

tT


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Example 4.1

Fig. 4.2 Schematic of a simple cart-spring system. Fig. 4.3 Free-body diagram for the sprint-cart system.

Lagrangian mechanics

22

2

2

1,

2

1

2

1kxPxmmvK

2

2

2

1

2

1kxxmPKL

Newtonian mechanics. . ..

., ( ) ,

i

L d Lm x m x m x kx

dt xx

kxxmF ..

____

amF

kxmaFmakxF

The complexity of the terms increases as the number of degrees of freedomand variables.

Solution

Derive the force-acceleration relationship for the one-degree of freedom system.


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Example 4.2

Fig. 4.4 Schematic of a cart-pendulum system.

Solution

Derive the equations of motion for the two-degree of freedom system.

In this system.

It requires two coordinates, x and .

It requires two equations of motion:1. The linear motion of the system.

2. The rotation of the pendulum.

sin00

sin0

cos

cos

2

.2

.22

..

..

222

221

glm

kxxlmx

lmlm

lmmm

T

F


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Example 4.4

Fig. 4.6 A two-degree-of-freedom robot arm.

Solution

Using the Lagrangian method, derive the equations of motion for the two-degreeof freedom robot arm.

Follow the same steps as before.Calculates the velocity of the center of

mass of link 2 by differentiating its position:

The kinetic energy of the total system is the

sum of the kinetic energies of links 1 and 2.

The potential energy of the system is the

sum of the potential energies of the twolinks:


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4.3 EFFECTIVE MOMENTS OF INERTIA

To Simplify the equation of motion, Equations can be rewritten insymbolic form.

j

i

jjjjii

ijjiii

jjjjii

ijjiii

j

i

jjji

ijii

D

D

DD

DD

DD

DD

DD

DD

T

T.

1

.

2.

2

.

1.2

2

.2

1..

..

2

1


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4.4 DYNAMIC EQUATIONS FOR MULTIPLE-DEGREE-OF-FREEDOM ROBOTS

Equations for a multiple-degree-of-freedom robot are very long andcomplicated, but can be found by calculating the kinetic and potential

energies of the links and the joints, by defining the Lagrangian and bydifferentiating the Lagrangian equation with respect to the joint variables.

4.4.1 Kinetic Energy

The kinetic energy of a rigid bodywith motion in three dimension :

GhVmK__

2

2

1

2

1

The kinetic energy of a rigid bodyin planar motion

22

2

1

2

1IVmK

Fig. 4.7 A rigid body in three-dimensional motion andin plane motion.


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4.4.1 Kinetic Energy

The velocity of a point along a robots link can be defined by differentiatingthe position equation of the point.

iiiiRi rTrTp 0

The velocity of a point along a robots link can be defined by differentiatingthe position equation of the point.

n

i

iactir

n

i

i

p

i

r

pTiriipi qIqqUJUTraceK1

2)(

1 1 1 2

1

2

1


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4.4.2 Potential Energy

The potential energy of the system is the sum of the potential energies of each link.

])([1

0

1 n

i

iiT

i

n

i

i rTgmpP

The potential energy must be a scalar quantity and the values in the gravitymatrix are dependent on the orientation of the reference frame.


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4.4.3 The Lagrangian

rn

i

i

p

i

rp

T

iriip qqUJUTracePKL 1 1 12

1

])([2

1

1

0

1

2)(

n

i

iiT

i

n

i

iacti rTgmqI


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4.4.4 Robots Equations of Motion

The Lagrangian is differentiated to form the dynamic equations of motion.

The final equations of motion for a general multi-axis robot is below.

i

n

j

n

k

kjijkiacti

n

j

jiji DqqDqIqDT

1 1

)(

1

)(),max(

Tpip

n

jip

pjij UJUTraceD

)(),,max(

Tpip

n

kjip

pjkijk UJUTraceD

n

ip

ppiT

pi rUgmD

where,


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Example 4.7

Fig. 4.8 The two-degree-of-freedom robot arm of Example 4.4

Solution

Using the aforementioned equations, derive the equations of motion for the two-degree of freedom robot arm. The two links are assumed to be of equal length.

Follow the same steps as before.

Write the A matrices for the two links;

Develop the , and for the robot.ijD ijkD iD

The final equations of motion without the actuator inertia terms are the same as below.

22

2

2

2

212

2

2

2

2

2

112

1

3

1

3

4

3

1

ClmlmClmlmlmT

1)(1121221121222222222

1

2

1

2

1 actIglCmglCmglCmSlmSlm

1)(212222

222

2122

22

22

2

1

2

1

3

1

2

1

3

1 act

IglCmSlmlmClmlmT


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4.5 STATIC FORCE ANALYSIS OF ROBOTS

Position Control: The robot follows a prescribed path without any reactive force.

Robot Control means Position Control and Force Control.

Force Control: The robot encounters with unknown surfaces and manages tohandle the task by adjusting the uniform depth while getting the reactive force.

Ex)Tapping a Hole - move the joints and rotate them at particular rates tocreate the desired forces and moments at the hand frame.

Ex)Peg Insertionavoid the jamming while guiding the peg into the hole and

inserting it to the desired depth.


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4.5 STATIC FORCE ANALYSIS OF ROBOTS

To Relate the joint forces and torques to forces and moments generated at thehand frame of the robot.

TH H H H H H H

x y z x y zF f f f m m m

x y z x y z x z

dx

dy

dzW f f f m m m f dx m zx

y

z

FJT HTH

DTDFW THTH

f is the force andm is the moment

along the axes of the hand frame.

The total virtual work at the jointsmust be the same as the total workat the hand frame.

Referring to Appendix A


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4.6 TRANSFORMATION OF FORCES AND MOMENTS BETWEEN COORDINATE FRAMES

An equivalent force and moment with respect to the other coordinate frameby the principle of virtual work.

zyxzyxT

mmmfffF zyxzyxT dddD

zB

yB

xB

zB

yB

xBTB

mmmfffF

zByBxBzByBxBTB dddD

DTDFW BTBT

The total virtual work performed on the object in either frame must be the same.


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4.6 TRANSFORMATION OF FORCES AND MOMENTS BETWEEN COORDINATE FRAMES

Displacements relative to the two frames are related to each other by thefollowing relationship.

DJD BB

The forces and moments with respect to frameB is can be calculated directlyfrom the following equations:

fnfxB

fofyB

fofyB

][ mpfnmxB

][ mpfomyB

][ mpfamzB

chapter 4 - dynamic analysis lecture

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