8/9/2019 Four Bar Force Analysis_20150224

1/21

Force analysis of planar four bar AND kinematics of SPHERICAL four bar MECHANISM

Hassen N.

2015.02.24

1

8/9/2019 Four Bar Force Analysis_20150224

2/21

Content

• Statics of four bar

• Dynamic of four bar

• Sp erical mec anism

2

8/9/2019 Four Bar Force Analysis_20150224

3/21

8/9/2019 Four Bar Force Analysis_20150224

4/21

8/9/2019 Four Bar Force Analysis_20150224

5/21

Static force analysislink %

5

1θ

1T

x o ,1

y o ,1

1 g

1W

1O

x A

y A

A∑ =0x F )1(0,1 =+ x x o A

∑ =0 y F )2(01,1 =−+ w o A y y ∑ =0 g M ( ) ( ) )3(0sinsincossin 111,111,1111111 =+−+−+−− T g o g o g Ag A y x y x θ θ θ θ

111

11

g g o

Ao

=

=

014,13,121 =+−++− T ! o ! o ! A! A y x y x

8/9/2019 Four Bar Force Analysis_20150224

6/21

Static force analysislink -

6

• .n t e free bo#y #ia'ram of link - t e #irection of forceat A as to be opposite of t e #irection assume# on link%&

∑ =0x F )4(O " A # # =+−

∑ =0 y F )5(2 O w " A y y =−+−

2θ

x A

y A

2 g

2W

A

x "

y "

"

32

2

g Ag

A"

==

( ) ( ) )$(0cossincossin

0

2222222222 =+−−+−−

=∑θ θ θ θ g Ag Ag " g "

M

y # y #

g

8/9/2019 Four Bar Force Analysis_20150224

7/21

Static force analysislink /

• .n t e free bo#y #ia'ram of link / (e must set t e #irection of force at 0 in t e opposite #irection assume# on link- &

7

x o ,3

y o ,3

3 g

3W

1O

x "

y "

"

c P

%

∑ =0x F

)&(3 O P " O # # =+−

∑ =0 y F

)'(43 O w " O y y =−−−

( ) ( ) )(0cossincossin

0

334334333333 =−−+−+−−

=∑)% g O g O g " g "

M

y x y #

g

θ θ θ θ

8/9/2019 Four Bar Force Analysis_20150224

8/21

Static force analysis.,erall e"uation

• No( (e a,e nine e"uations (it nine unkno(ns an# assemblin' all e"uations in matri$ form 'i,es t e follo(in'&

8

)10(

0

0

0

0

00000

010001000

001000100

00000

000001010

000000101

10000

000100010

000010001

3

2

1

1

3

3

1

1

1211109

6587

4321

−

=

−−−

−−−

−−

−−

pd

w

p

w

w

T O

O

O

O

B

B

A

A

cccc

cccc

cccc

y

x

y

x

y

x

y

x

8/9/2019 Four Bar Force Analysis_20150224

9/21

Dynamics of four bar

9

*e+oci y e- a ion

)11(0

0

coscoscos

sinsinsin

3

2

1

332211

332211 =−−

−

θ

θ

θ

θ θ θ

θ θ θ

d d d

d d d

)12(cos

sin

coscos

sinsin

11

11

2

2

3322

3322

−=−

−θ

θ

θ

θ

θ θ

θ θ

d

d

d d

d d

8/9/2019 Four Bar Force Analysis_20150224

10/21

Dynamics of four bar

10

Use LaGrange formulation for the dynamics. It canbe de ned as the total kinetic energy minus totalpotential energy. The total kinetic energy of thesystem can be de ned as

( )( )

( )( )

( )( )

)13(sinsin

,sinsin

,,,

,,

323

1213

122

13121

3211

3211

3

2

θ θ

θ θ θ

θ θ

θ θ θ θ

θ θ θ

θ θ θ

θ

θ

−−=−

−==d d

d d

E

E

( ) ( ) ( ) (14) 21

21

21 2

332

332

222

222

112

11 θ θ θ I vm I vm I vmT

ccc +++++=

8/9/2019 Four Bar Force Analysis_20150224

11/21

Dynamics of four bar• ! en #etermine t e total potential ener'y&

• In t e abo,e e"uation (e can eliminate t e ,elocity terms usin' E"&1%/2&

11

)15(3,32,21,1 ccc gym gym gymV ++=

)16(V T L −=( )

( ) ( ) )17(,,,

,,,,,

321212111

233

222

211321321

θ θ θ θ θ θ θ

θ θ θ θ θ θ θ θ θ

GC J

K K K L

+

+++=

8/9/2019 Four Bar Force Analysis_20150224

12/21

Dynamics of four bar

12

( )

( )

( )

( ) ( )( ) ( )

( )18

sinsinsin,,

cos,

212121

33322211211321

21211

2221

32

333

22

222

2221

2111

−−−−=−=

=+=

+=

++=

θ θ θ θ θ θ

θ θ θ θ

ccc

c

c

c

c

gd m gd m gd m gd mG

C

d d m J

I d m K

I d m K

d m I d m K

( ) ( ) ( ) ( ) ( )[ ]( ) )19(,,

,,,,,,,,,,,,

321

21321

212111321

223321

2121321321

θ θ θ

θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ

G

E C J E K E K K L

++++=

8/9/2019 Four Bar Force Analysis_20150224

13/21

Dynamics of four bar• ! e e"uation of motion of t e entire system can

(ritten

• ! en

13

( )2011

τ

θ θ

=∂

∂

∂

∂ L Ldt

d

( ) ( ) ( ) ( )[ ] ( )21,,,,,,,2 3211211132122332121211

θ θ θ θ θ θ θ θ θ θ θ

θ E C J E K E K K

L +++=∂

∂

( )222111

θ θ θ

B A L

dt d +=

∂∂

8/9/2019 Four Bar Force Analysis_20150224

14/21

Dynamics of four bar• 3 ere

14

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( )( ) ( ) ( ) ( )

( ) ( ) ( )

∂

∂∂+∂

∂

+

∂

∂

∂+∂

∂

∂+∂

∂

+

∂

∂∂+∂∂

∂+∂∂

+

∂∂

+∂∂

+∂∂

=

+++=

1

2

2

211

1

2113211

1

3

3

3211

1

2

2

3211

1

32112111

1

3

3

3212

1

2

2

3212

1

321232123

3

3212

2

3212

1

321232111

321121113212

333212

221

,,,,

,,,,,,,

,,,,,,,,2

,,,,,,,,2

2

,,,,,,,2

θ

θ

θ

θ θ

θ

θ θ θ θ θ

θ

θ

θ

θ θ θ

θ

θ

θ

θ θ θ

θ

θ θ θ θ θ

θ

θ

θ

θ θ θ

θ

θ

θ

θ θ θ

θ

θ θ θ θ θ θ

θ

θ θ θ

θ

θ θ θ

θ

θ θ θ θ θ θ

θ θ θ θ θ θ θ θ θ θ θ

C C E

E E E C J

E E E E K

E E E E K

B

E C J E K E K K A

8/9/2019 Four Bar Force Analysis_20150224

15/21

Dynamics of four bar

15

( )

Where

232

11 C H

L

+=∂∂ θ θ

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( )( ) ( ) ( ) ( )

( ) ( ) ( )

∂∂

∂∂+∂

∂

+

∂∂

∂∂+∂

∂∂

∂+∂∂

+

∂∂

∂∂

+∂∂

∂∂

+∂∂

+

∂∂

∂∂+∂

∂

=

1

2

2

211

1

2113211

1

3

3

3211

1

2

2

3211

1

32112111

1

3

3

3212

1

2

2

3212

1

3212

32123

3

3211

2

3211

1

321132121

,,,,

,,,,,,,J

,,,,,,,,2

,,,,,,,,2

θ

θ

θ

θ θ

θ

θ θ θ θ θ

θ

θ

θ

θ θ θ

θ

θ

θ

θ θ θ

θ

θ θ θ θ θ

θ

θ

θ

θ θ θ

θ

θ

θ

θ θ θ

θ

θ θ θ θ θ θ

θ

θ θ θ

θ

θ θ θ

θ

θ θ θ θ θ θ

C C E

E E E C

E E E E K

E E E E K

H

( ) ( ) ( )1

3

3

321

1

2

2

321

1

321 ,,,,,,θ

θ

θ

θ θ θ

θ

θ

θ

θ θ θ

θ

θ θ θ

∂∂

∂∂+∂

∂∂

∂+∂∂= GGGC

8/9/2019 Four Bar Force Analysis_20150224

16/21

Dynamics of four bar• Substitute E"&1-%+-- 4 -/2 into E"&1-52 yiel#s

?

16

8/9/2019 Four Bar Force Analysis_20150224

17/21

Sp erical mec anism• A sp erical mec anism is linka'es ( ose )oints

a,e a$es t at intersect at a common point&

17

−−

=−

1000

01

iii

iiiiiii

iiiiiii

ii

d c s

sac scc sca s s scc

Aα α

θ θ α θ α θ

θ θ α θ α θ

−−

=−

ii

iiiii

iiiii

ii

c s

c scc s

s s scc

Rα α

θ α θ α θ

θ α θ α θ

0

1

8/9/2019 Four Bar Force Analysis_20150224

18/21

Sp erical mec anism• 3 ere

• 6se loop closure e"uation

18

axes. jointaof normainci!ento "et#een t#an$ eJointaxes jointa!jacento "et#een t#an$ e%#ist

axes jointaof normainci!ento "et#een t#!istancena%rans atio!

axes jointa!jacento "et#een t#!istance&ffset

i

i

i

==

==

θ

α

ia

I R R R R =14

43

32

21

14

43

32

21 R R R R =

8/9/2019 Four Bar Force Analysis_20150224

19/21

Sp erical mec anism• 3 ere

'i,es t e follo(in'

19

−

−

−

−=

−−

−−

11

11111

11111

44

44444

44444

33

33333

22333

22

22222

22222

00

00

α α

θ α θ α θ

θ α θ α θ

α α

θ α θ α θ

θ α θ α θ

α α

θ α θ α θ

θ α θ α θ

α α

θ α θ α θ

θ α θ α θ

c s

c scc s

s s scc

c s

c scc s

s s scc

c s

c scc s

s s scc

c s

c scc s

s s scc

8/9/2019 Four Bar Force Analysis_20150224

20/21

Sp erical mec anism

20

6sin' t is matri$ (e can #etermine t e #isplacement of t e output )oint& Also t e interme#iate an'les&

+−+−−−+−+

++−−−=

+−+−−−+−+

++−−−

141141411414

1441144114144114411414414

144144144114411414414

323323233232

3223322332322332233232232

322332322332233232232

α α θ α α α θ θ α α θ α

α θ α θ α θ α θ α θ α θ α θ α θ α θ α θ θ θ α θ θ

α θ α θ α θ α θ α θ α θ α θ α θ θ θ α θ θ

α α θ α α α θ θ α α θ α

α θ α θ α θ α θ α θ α θ α θ α θ α θ α θ θ θ α θ θ

α θ α θ α θ α θ α θ α θ α θ α θ θ θ α θ θ

ccc s s sccc s s s

cc sc scc s s s sc scccc sc s sccc s

s sc s sc s sccc sc scc s sccc

ccc s s sccc s s s

cc sc scc s s s sc scccc sc s sccc s

s sc s sc s sccc sc scc s sccc

8/9/2019 Four Bar Force Analysis_20150224

21/21

Sp erical mec anism

21

6sin' t is matri$ (e can #etermine t e #isplacement of t e output )oint& Also t e interme#iate an'les&

+−+−−−+−+

++−−−=

+−+−−−+−+

++−−−

141141411414

1441144114144114411414414

144144144114411414414

323323233232

3223322332322332233232232

322332322332233232232

α α θ α α α θ θ α α θ α

α θ α θ α θ α θ α θ α θ α θ α θ α θ α θ θ θ α θ θ

α θ α θ α θ α θ α θ α θ α θ α θ θ θ α θ θ

α α θ α α α θ θ α α θ α

α θ α θ α θ α θ α θ α θ α θ α θ α θ α θ θ θ α θ θ

α θ α θ α θ α θ α θ α θ α θ α θ θ θ α θ θ

ccc s s sccc s s s

cc sc scc s s s sc scccc sc s sccc s

s sc s sc s sccc sc scc s sccc

ccc s s sccc s s s

cc sc scc s s s sc scccc sc s sccc s

s sc s sc s sccc sc scc s sccc