KINEMATICS OF MECHANISMS

position – velocity – acceleration; driving forces are neglected

2

2vv

dt

sd

dt

dsa

dt

dss KKKK

KKK

For a point K we have linear displacement sK(t), velocity

vK and acceleration aK

2

2

dt

d

dt

d

dt

d kkkk

kkk

For a link k we have angular displacement k(t),

angular velocity k and angular acceleration k

METHODS OF KINEMATIC ANALYSIS

graphical (use only: pencil, ruler, compass,

calculator)

analytical

numerical

A

BC

D

M

BC

CD

vM

DOF = 1 1 driver, this case: crank AB – known (t)

LENGTH SCALE

A

B

C

D

Length scale:

mm

m

BC

BCl

100

1

)(

SCALE DEFINITION (for graphical methods)

Scale (general)

)(i

ii

real value

graphical value

mm

ms

)v(

v -1

vVelocity scale

mm

ms

)a(

a -2

aAcceleration scale

FORCE SCALE

Force scale:

mm

N

F

FF 1

100

100

)(

4 BAR LINKAGE

a lot of applications

4 bar linkage has a few inversions

Harbor crane

R

R

R

T

hand press

pipe spanner,

pipe wrench

Four-bar linkage: crank rocker

A

D

C1 C2

B2

B1

0-l0

3-l3

2-l2

1-l1

For the crank-rocker we have:

1. Grashof inequalities fulfilled,

2. shortest link (l1) is a crank (rotates around frame)

4 BAR LINKAGE (4R)

0231

0321

3201

llll

llll

llll

312101 llllll

Grashof inequalities

A D

C

B

Grashof inequalities fulfilled,

shortest link (l1) is a frame

A D

C

B

Grashof inequalities fulfilled,

shortest link (l1) is a coupler

DOUBLE CRANK

DOUBLE ROCKER

Transforming a fourbar crank-rocker to crank-slider

R T

Slider block

translates

R R

Slider

block has

complex

motion

R R

Slider

block

rotates

R T

Slider

block is

stationary

4 BAR - crank-slider - INVERSIONS

B

A

CC

’

B

’

rBC

D

GRAPHICAL POSITION ANALYSIS

B

A

CC’

B’

rBC

(t)

Data: (t)

Find: xc’, yc’

xB’ = AB cos

yB’ = AB sin

(xC’ – xB’)2 + (yC’ – yB

’) 2 =

BC2

yC’ = 0

From the system of

equations:

xC’ , yC’

y

x

GRAPHICAL POSITION ANALYSIS

B

A D

C

C

’

B

’

rBC

D

GRAPHICAL POSITION ANALYSIS

B

AD

C

C’

B’

y

x

(t)

Data: (t)

Find: xc’, yc’

xB’ = AB cos

yB’ = AB sin

(xC’ – xB’)2 + (yC’ – yB

’) 2 =

BC2

(xC’ – xD)2 + (yC’ – yD)

2 =

CD2

rBC

From the system of

equations : xC’ , yC’

GRAPHICAL POSITION ANALYSIS

A

BC

D

M

GRAPHICAL POSITION ANALYSIS

Draw a mechanism for a given

Link dimensions are known: AB, BC, CD, AD and BM, CM

A

BC

D

M

GRAPHICAL POSITION ANALYSIS

A

BC

D

M

GRAPHICAL POSITION ANALYSIS

A

BC

D

C*

M

Two positions for given

GRAPHICAL POSITION ANALYSIS

A D

B

B1

C

E

F

1 driver: AB AB1

GRAPHICAL POSITION ANALYSIS

A D

B

B1

C

E

F

C1

F1

E1

GRAPHICAL POSITION ANALYSIS

A D

B C

E

F F1

R=EF

1 driver: slider F F1

GRAPHICAL POSITION ANALYSIS

DG

C

C

’

B

A

B

’rBC

D

E

F

D

yE

’

F

’

1

2

GRAPHICAL POSITION ANALYSIS - III class mechanism

Kinematics – graphical methods

INSTANT CENTERS OF VELOCITY

VELOCITY AND ACCELARATION

From planar complex motion to rotation

(2 links: frame 0 and moving body k)

A

Bv

B

AAv

k

{0}

0

B

tangent to trajectory

instant center of rotation

k0

B

k0

A

BS

v

AS

vk

( )ktg

)(BS

)(v

)(AS

v

k0

B

k0

A

angular velocity:

B

a

A

b

VA VB VAB = VBA = 0

SAB

{0}

2 bodies in motion

number of IC:( )

2

1

2

nnni

An instant center of velocity is a point, common to two bodies in

plane motion, which has the same instantaneous velocity in each

body

23

1312

030201

S

SS

SSS( )6

2

144

i

13

02

S

S

23

12

0301

S

S

SS

?

VB

VC

VB

VC

S20

1, 2, 0 S12 & S10 S20

3, 2, 0 S32 & S30 S20

1, 3, 0 S10 & S30 S31

INFINITY

Find vM direction

0

13

2

M

1st method: trajectory of point M and tangent ???

2 method: instant center of velocity S20

23

1312

030201

S

SS

SSS

0

13

30

M

10

21

2

0

13

2

30

32

M

10

23

1312

030201

S

SS

SSS

21

0

13

2

30

21

32

20

31

M

VM10

Cam mechanism

1

2

0

S12 S01 S02cam

follower

12

1

2

0

1020

V12

S10 – S12 – S20

common tangent (axis of slip)

Cam with roller

1

2

1020

V13 = 0 S13 at contact point

3