prezentacja programu powerpointtmm.pwr.edu.pl/.../66/public/mbm/lecture_no_4_2020.pdf · kinematic...

Kinematic chain notations

Kinematic scheme Structural scheme

GRAF

Matrix of the structure

Contours

1

4

0

2

5

3B

C

E FG

AD

C

FE I

I

2

DA

B

I

I

13

5

I

I

G

4

0

I0

1

2

3

4

5

-

A

-

D

-

G

A

-

B

-

-

-

-

B

-

C

E

-

D

-

C

-

-

-

0 1 2 3 4 5

-

-

E

-

-

F

G

-

-

-

F

-

5

3F

D

G

C4

2

A

B

E1

0

K1= 0 –A – 1 – B – 2 – C – 3 – D – 0

K2= 0 – D – 3 – C – 2 – E – 4 – F – 5 – G - 0

K3(outer)= 0 –A – 1 – B – 2 – E – 4 – F – 5 – G - 0

Using Graf

Find direction of VK

3

2

20

B

20

K

BS

v

KS

v

AB 1Bv

Determining the motion of the mechanism:

2

1

43

S14

S13

F1F2F3

VELOCITY AND ACCELERATION

VECTOR EQUATIONS – v and a polygons

(2D) – M & N – points of one link

0t

v and a vector equations (2D complex motion)

NMMN vvv

MNkNM rωv

t

NM

n

NMMNMMN aaaaaa

MNkMNkk

n

NM rrωωa2

MNk

t

NM rεa

velocity:

acceleration:

B

AC

2

pv

vB

d.vCB

vC

vCB

2

14

3

d.vBA

vB

d.vCB

d.vC

d.vC

vC

Data: 2 = const, e2 = 0

Find: vB, vC, 3 , aB, aC, e3

vB = vA + vBA

vA = 0

vBA = 2 BA

vB = vBA= 2 BA

vC = vB + vCB

3 vCB /CB

aB = aA + aBAn + aBA

t

aA = 0

aBAn = 2

2 BA

aBAt = e2 BA =0

aC = aB + aCBn + aCB

t

aCBn = 3

2 CB

aCBt = e3 CB

e3 aCBt /CB

3

pa

d.aCBt

d.aC

d.aCBt

aC

aCBn

aCBn

aB= aBAn

d.aC

aCBt

aC

aCBte3

aBAn = aB

Example v and a (crank-slider)

A

B

C

K

D0

1

2

31 const

ABωv 1B1B ABωv

A

B

C

K

D0

1

2

31 constdvC

dvCB

CBBC vvv dvCBdvC

vB vCB

vCpv

c

b

A

B

C

K

D0

1

2

31 const

dvKB

dvKC

KCCKBB

KCCK

KBBK

vvvv

vvv

vvv

vB

vCB

vCpv

c

b

kdvKBvKB

vK

dvKC

A

B

C

K

D0

1

2

31 const

kvKB

kvBC

kvKC

dvCB

vB

vCB

vCpv

c

b

k

dvKC

dvKBvKB

vKkc

KC

bk

BK

bc

BC

ΔbckΔBCK

~

Triangles BCK and bck

are similar

A

B

C

K

D0

1

2

31 const

t

B

n

BB aaa

AB

vABωa)(

2

B2

1

n

B11

n

B ABωωa

0AB0ABεa 1

t

B1

t

B ABεa

A

B

C

K

D0

1

2

31 const

Ca

t

CBa

c

b

Cda

Ban

CBa

CBa

t

CBda

n

CBda

pat

CB

n

CB

t

B

n

BC

CBBC

aaaaa

aaa

CB

va

0since0a,AB

va

2

CBn

CB

1

t

B

2

2

1

n

BB

e AB

A

B

C

K

D0

1

2

31 const

KCCKBB

KCCK

KBBK

aaaa

aaa

aaa

t

KC

n

KCC

t

KB

n

KBB

a

a

aa

aa

KC

va,

KB

va

2

KCn

KC

2

KBn

KB t

KBa

Ca

n

KBa

pac

bk

Ba

n

KCa

t

KCa

CBa

t

KBdat

KCda

ΔbckΔBCK

polygon) accel its and(link similarity

~

A

B

C

K

D0

1

2

31 const

Ca pac

bk

Ba

VELOCITY AND ACCELARATION – extension

In last 2D example: M and N – points of one link

NMMN vvv

t

NM

n

NMMNMMN aaaaaa

(2D) – points J and K belong to j and k

KJJK vvv

ρωv k

j

KJ

C

KJ

t

KJ

n

KJJKJJK aaaaaaa

ρρωωa2

k

j

k

j

k

jn

KJ

ρεa k

jt

KJ

KJj

C

KJ vωa 2

- radius of curvature

dir

C

KJ

t

KJJKJJK aaaaaa

02 ρρωωa k

j

k

j

k

jn

KJ

Since

we have

dir