KINEMATICS OF MECHANISMS

KINEMATICS OF MECHANISMS

position – velocity – acceleration; driving forces are neglected

2

v 2

v dt

s d dt

s d dt a

s_K ds^K _{K K} ^{K K}

K       

For a point K we have linear displacement s_K(t), velocity v_K and acceleration a_K

2 2

dt d dt

d dt

d _{k k}

k k

k k

k

 





 







 



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

B C

 D

M





v

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

LENGTH SCALE

A B

C

D

Length scale:

 

 



 mm

m BC

BC

l

100

1 )

 (

SCALE DEFINITION (for graphical methods)

Scale (general)

) (i

i

i





graphical value

 

 



 

mm ms )

v (

v



Velocity scale

 

 



 

mm ms )

a (

a



Acceleration scale

FORCE SCALE

Force scale:

 

 





 mm

N F

F

F

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)

 

 





















0 2

3 1

0 3

2 1

3 2

0 1

l l

l l

l l

l l

l l

l l

3 1 2 1 0

1

l l l l l

l   

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

C C

’ B

’

r

D

GRAPHICAL POSITION ANALYSIS

B

A

C C’

B’

r

(t)

Data: (t) Find: x_c’, y_c’ x_B’ = AB cos  y_B’ = AB sin 

(x_C’ – x_B^’)² + (y_C’ – y_B^’)² = BC²

y_C’ = 0

From the system of equations:

x_C’ , y_C’

y

x

GRAPHICAL POSITION ANALYSIS

B

A D

C

C

’ B

’

r

D

GRAPHICAL POSITION ANALYSIS

B

A D

C

C’

B’

y

x

(t)

Data: (t) Find: x_c’, y_c’ x_B’ = AB cos  y_B’ = AB sin 

(x_C’ – x_B^’)² + (y_C’ – y_B^’)² = BC²

(x_C’ – x_D)² + (y_C’ – y_D) ² = CD²

r

From the system of equations : x_C’ , y_C’

GRAPHICAL POSITION ANALYSIS

A

B C



M

GRAPHICAL POSITION ANALYSIS

Draw a mechanism for a given 

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

A

B C

D



M

GRAPHICAL POSITION ANALYSIS

A

B C

D



M

GRAPHICAL POSITION ANALYSIS

A

B C

D C*



M

Two positions for given 

GRAPHICAL POSITION ANALYSIS

A D B

B1

C

E

F

1 driver: AB  AB

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  F

GRAPHICAL POSITION ANALYSIS

D G C

C

’ B

A

B

’ r

D

E

F

D

y E

’

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

vB

B

A A v

k

{0}

0

_B

tangent to trajectory

instant center of rotation

k0 B k0

A

BS v AS

v 

k



 ( )

tg 



 (BS ) ) (v )

(AS v

k0 B k0

A

angular velocity:

B

a A

b

V_A  V_B  V_AB = V_BA = 0

S_AB

{0}

2 bodies in motion

number of IC:

( )

2 1 2

 

 

 



  n n n i

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 13 12

03 02

01

S S S

S S

( ) S

2 6 1 4

4  



i

13 02

S S

23 12

03 01

S S

S S

?

V

V

V

V

S

1, 2, 0  S₁₂ & S₁₀  S₂₀ 3, 2, 0  S₃₂ & S₃₀  S₂₀

1, 3, 0  S₁₀ & S₃₀  S₃₁

INFINITY

Find v

direction

0

1 3

2

M

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

2 method: instant center of velocity S₂₀

23 13 12

03 02

01

S S S

S S

S

0

1 3

30

M

10 21

2

0

1 3

2

30 32

M

10

23 13 12

03 02

01

S S S

S S

S

21

0

1 3

2

30 21

32

20 31

M

VM

10

Cam mechanism

1

2

0

S

S

S

cam

follower

12 1

2

0

10 20

V₁₂

S

– S

– S

common tangent (axis of slip)

Cam with roller

1

2

10

20

V

= 0  S

at contact point

3