Chapter 4 - Dynamic Analysiss

8/20/2019 Chapter 4 - Dynamic Analysiss

1/16

Chapter 4Dynamic Analysis and Forces

4.1 INTRODUCTION

In this chapters…….

• The dynamics, related with accelerations, loads, masses andinertias.

__ __

am F

⋅=∑

__ __

α

⋅=∑ I T

In Actators……. • The actator can !e accelerate a ro!ot"s lin#s $or e%ertin& eno&h$orces

and tor'es at a desired acceleration and (elocity.

• )y the dynamic relationships that &o(ern the motions o$ thero!ot,

*i&. 4.1 *orce+mass+acceleration and tor'e+inertia+an&lar

acceleration relationships $or a ri&id !ody.


2/16


4. -ARANIAN /0CANIC23 A 2ORT O0RI05

• -a&ran&ian mechanics is !ased on the di6erentiationener&y terms only, with respect to the system"s (aria!les and time.• De7nition3 L 8 -a&ran&ian, K 8 9inetic 0ner&y o$ the system, P 8:otential 0ner&y, F 8 the smmation o$ all e%ternal $orces $or alinear

motion, T 8 the smmation o$ all tor'es in a rotationalmotion, x 8 2ystem (aria!les

P K L −=

ii

i

x L

x

Lt

F ∂∂−

∂∂∂∂= ⋅

ii

i

L L

t T

θ θ ∂∂

−

∂

∂∂∂

= ⋅


3/16


Example 4.1

*i&. 4. 2chematic o$ a simple cart+sprin&system.

*i&. 4.; *ree+!ody dia&ram $or the sprint+cartsystem.

• -a&ran&ianmechanics

22

2

2

1,

2

1

2

1kx P xmmv K ===

•2

2

2

1

2

1kx xm P K L −=−=

•

• Newtonianmechanics

. . ..

. , ( ) ,

i

L d Lm x m x m x kx

dt x x

∂ ∂= = = −

∂∂

kx xm F += ..

__ __

am F ⋅=∑kxma F makx F +=→=−

• The comple%ity o$ the terms increases as the nm!er o$ de&rees o$$reedom

Solution

Deri(e the $orce+acceleration relationship $or the one+de&ree o$ $reedom system.


4/16


Example 4.2

*i&. 4.4 2chematic o$ a cart+pendlm

system.

Solution

Deri(e the e'ations o$ motion $or the two+de&ree o$ $reedom 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.

+

+

+=

θ θ

θ

θ θ

θ

sin00

sin0

cos

cos

2.2

.22

..

..

222

221

gl m

kx xl m x

l ml m

l mmm

T

F


5/16


Example 4.4

*i&. 4.< A two+de&ree+o$+$reedom ro!otarm.

Solution

Usin& the -a&ran&ian method, deri(e the e'ations o$ motion $or thetwo+de&ree o$ $reedom ro!ot arm.

Follow the same steps as before…….

alculates the !elocity of the center of

mass of lin" 2 by differentiatin# its position:

The "inetic ener#y of the total system is the

sum of the "inetic ener#ies of lin"s 1 and 2.

The potential ener#y of the system is the

sum of the potential ener#ies of the two

lin"s:


6/16


4.; 0**0CTI0 /O/0NT2 O* IN0RTIA

• To 2impli$y the e'ation o$ motion, 0'ations can !erewritten in sym!olic $orm.

+

=

+

=

j

i

jjj jii

ijjiii

jjj jii

ijjiii

j

i

jj ji

ijii

D

D

D D

D D

D D

D D

D D

D D

T

T .

1

.

2.

2

.

1.22

.21

..

..

2

1

θ

θ

θ

θ

θ

θ

θ

θ


7/16


4.4 D=NA/IC 0>UATION2 *OR /U-TI:-0+D0R00+O*+*R00DO/ RO)O

• 0'ations $or a mltiple+de&ree+o$+$reedom ro!ot are (ery lon& andcomplicated, !t can !e $ond !y calclatin& the #inetic and

potentialener&ies o$ the lin#s and the ?oints, !y de7nin& the -a&ran&ian and!y

di6erentiatin& the -a&ran&ian e'ation with respect to the ?oint(aria!les.

4.4.1 9inetic 0ner&y

The "inetic ener#y of a ri#id bodywith motion in three dimension :

GhV m K __

2

2

1

2

1ω +=

The "inetic ener#y of a ri#id bodyin planar motion

22

2

1

2

1ω I V m K +=

*i&. 4.@ A ri&id !ody in three+dimensional motionand

in plane motion.


8/16



4.4.1 9inetic 0ner&y

The !elocity of a point alon# a robot$s lin" can be defined by differentiatin#the position equation of the point.

iiii Ri r T r T p 0== The !elocity of a point alon# a robot$s lin" can be defined by differentiatin#

the position equation of the point.

( ) ∑∑∑∑ == = = +=n

i

iact ir

n

i

i

p

i

r

pT ir iipi q I qqU J U Trace K 1

2)(

1 1 1 21

21


9/16



4.4. :otential 0ner&y

The potential ener#y of the system is the sum of the potential ener#ies of each lin".

])([1

0

1 ∑∑ == ⋅−==n

i

iiT

i

n

i

i r T g m p P

The potential ener#y must be a scalar quantity and the !alues in the #ra!ity

matrix are dependent on the orientation of the reference frame.


10/16



4.4.; The -a&ran&ian

( ) r n

i

i

p

i

r p

T

ir iip qqU J U Trace P K L ∑∑∑= = ==−= 1 1 121

])([2

1

1

0

1

2)( ∑∑

==⋅−−+

n

i

iiT

i

n

i

iact i r T g mq I


11/16



4.4.4 Ro!ot"s 0'ations o$ /otion

The %a#ran#ian is differentiated to form the dynamic equations of motion.

The final equations of motion for a #eneral multi&axis robot is below.

i

n

j

n

k

k jijk iact i

n

j

jiji Dqq Dq I q DT ∑∑∑= ==

+++=1 1

)(

1

)(),max(

T pi p

n

ji p

pjij U J U Trace D

∑==)(

),,max(

T pi p

n

k ji p

pjk ijk U J U Trace D ∑=

=

∑=

−=n

i p

p piT

pi r U g m D

where,


12/16


Example 4.7

*i&. 4. The two+de&ree+o$+$reedom ro!ot arm o$0%ample 4.4

Solution

Usin& the a$orementioned e'ations, deri(e the e'ations o$ motion $orthe two+de&ree o$ $reedom ro!ot arm. The two lin#s are assmed to !eo$ e'al len&th.

Follow the same steps as before…….

'rite the ( matrices for the two lin"s)

*e!elop the , and for the robot.ij D ijk D i D

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

θ θ

++

++= C l ml mC l ml ml mT

( ) 1)(1121221121222222222

1

2

1

2

1θ θ θ θ act I glC m glC m glC m l m l m +++++

+

1)(21222

2

22

2

212

2

2

2

22 2

1

2

1

3

1

2

1

3

1θ θ θ

act I glC m l ml mC l ml mT ++

+

+

+=


13/16


4.B 2TATIC *ORC0 ANA-=2I2 O*RO)OT2

+osition ontrol: The robot follows a prescribed path without any reacti!e force.

obot ontrol means +osition ontrol and Force ontrol.

Force ontrol: The robot encounters with un"nown surfaces and mana#es tohandle the tas" by ad-ustin# the uniform depth while #ettin# the reacti!e force.

x/ Tappin# a 0ole & mo!e the -oints and rotate them at particular rates tocreate the desired forces and moments at the hand frame.

x/ +e# Insertion a!oid the -ammin# while #uidin# the pe# into the hole andinsertin# it to the desired depth.


14/16


4.B 2TATIC *ORC0 ANA-=2I2 O*RO)OT2

To elate the -oint forces and torques to forces and moments #enerated at the hand frame of the robot.

T ! ! ! ! ! ! !

x " # x " # F $ $ $ m m m =

x " # x " # x #

dx

d"

d# % $ $ $ m m m $ dx m # x

"

#

δ δ

= = + + ∂ ∂ ∂

- -

[ ] [ ] [ ] F J T ! T ! =

[ ] [ ] [ ] [ ]θ δ DT D F % T ! T ! ==

f is the force and m is the moment

alon# the axes of the hand frame.

The total !irtual wor" at the -ointsmust be the same as the total wor"at the hand frame.

eferrin# to (ppendix (


15/16


4.< TRAN2*OR/ATION O* *ORC02 AND /O/0NT2 )0T500N COORDINAT0*RA/02

(n equi!alent force and moment with respect to the other coordinate frameby the principle of !irtual wor".

[ ] [ ] # " x # " xT

mmm $ $ $ F =[ ] [ ] # " x # " x

T d d d D δ δ δ =

[ ] [ ] # &

" &

x &

# &

" &

x &T &

mmm $ $ $ F =[ ] [ ] # & " & x & # & " & x &

T & d d d D δ δ δ =

[ ] [ ] [ ] [ ] DT D F % &T &T ==δ

The total !irtual wor" performed on the ob-ect in either frame must be the same.


16/16


4.< TRAN2*OR/ATION O* *ORC02 AND /O/0NT2 )0T500N COORDINAT0*RA/02

*isplacements relati!e to the two frames are related to each other by thefollowin# relationship.

[ ] [ ][ ] D J D & & =

The forces and moments with respect to frame B is can be calculated directlyfrom the followin# equations:

$ n $ x & ⋅=

$ ' $ " & ⋅=

$ ' $ " & ⋅=

( ) ][ m p $ nm x & +×⋅=

( ) ][ m p $ 'm " & +×⋅=

( ) ][ m p $ am # & +×⋅=

Chapter 4 - Dynamic Analysiss

Documents