Suspension system

Suspension system

Q

S

0

1 2

3

spring+damper

Q

S

0

1 2

3

0 :

2 F

+ Q + F

= link

points of

application only

?

Q

S

0

1 2

3

0 :

3 F

+ F

=

link

Q

S

0

1 2

3

= 0 +

+

32

Q F

F

F₁₂ Q

F₃₂

Q

S

0

1 2

3

0 :

1 F

+ F

+ S = link

F₂₁ F₀₁

S

RRTR example (1)

3

0 1

2

RRTR example (2)

3

0 1

2

M₃

M₁=?

RRTR example (3)

3

0 1

2

M₃

M₁=?

0 .

3

+

=

RRTR example (3)

3

0 1

2

M₃

M₁=?

0 .

2 F₁₂ + F₃₂ =

RRTR example (3)

3

0 1

2

M₃

M₁=?

0 .

2 F₁₂ + F₃₂ =

0 .

3

+

=

RRTR example (4)

3

0 1

2

M₃

M₁=?

...

0 0

. 3

23

3 3

23 3

=

=

−

 =

M h

F

M ^C

B A

C h₃

F₂₃

F₀₃

RRTR example (5)

3

0 1

2

M₃

M₁=?

...

0 0

. 3

23

3 3

23 3

=

=

−

 =

M h

F

M ^C

B A

C h₃

F₂₃

F₀₃ F₃₂

F₃₀

RRTR example (6)

3

0 1

2

M₃

M₁=?

0 .

1

+

=

B A

C h₃

F₂₃

F₀₃ F₃₂

F₃₀

F₁₂ F₂₁

RRTR example (7)

3

0 1

2

M₃

M₁

0 0 .

1

1 1

2 1

1

=

−



M h

F

M ^A

B A

C h₃

F₂₃

F₀₃ F₃₂

F₃₀

F₁₂ F₂₁

h₁ F₀₁

F₁₀

dynamics, kinetostatics 16

IN KINETOSTATICS

THE MOTION OF MASS BODIES (LINKS) IS REPRESENTED BY INERTIA FORCES

Inertia forces

Newton, Euler, D’Alembert

ε M

= − I

0 0  + =

=

− I

ε M M

M

ε M = I

a F

= − m

0

0  + =

=

−

a

F = m

dynamics, kinetostatics 18

Mass moment of inertia

 ⁺

=

m

Z x y dm

I

(

)

) (

S

Z

I m x y

I = + +

dynamics, kinetostatics 19

a F

= − m

ε M_b

= −

ma I F

M F

h M

b b M

b

= = 

=

FORCE (F_b) and MOMENT (M_b) OF INERTIA

Given:

m, I_S, a, ε

dynamics, kinetostatics 20

FORCE (F_b) and MOMENT (M_b) OF INERTIA

resultant inertia force

dynamics, kinetostatics 21

Center of percussion

AS

AS 

 =

=

+

=

t S 2

n S

t S n

S

; a a

a a

a_S

S

A

B

a_S S

 S



dynamics, kinetostatics 22

Center of percussion

S

A

B

a_S S

S

 W



F_b h

I

m ε M

a F

b

S b

−

=

−

=

S S b

b

ma I F

h ₌ M ₌ 

B

S

A a_S S S

 W



F_b h

cos(?)

=

=

S t S

a a SW

Center of percussion h

 =



 





















=

 =



 







 



= 

=

t S S S

t S S

t S S S

S t

S S

a a ma

AS I a

a a ma

I a

h a

SW



mAS I a

a AS ma

a

I

t S S S

t S

S

=

=

B

S

A a_S S S

 W



F_b h

AS i mAS

mi mAS

SW I

2

2

=

=

=

Center of percussion

i_S – radius of inertia

dynamics, kinetostatics 25 -100000

-80000 -60000 -40000 -20000 0 20000 40000 60000 80000 100000

0 0,004 0,008 0,012

czas [s]

Angular accel of BC

20000 25000 30000 35000 40000 45000 50000 55000 60000 Accel of point

S m, J

A C

B S

BC = 0,2 m

₁ = 500 s^-1 (₁ = const) (n₁ = 5000 rev/min)

dynamics, kinetostatics 26 m, J

A C

B S

a_Smax = 55000 ms^-2

_Bcmax = 90 000 s^-2

m=0,2 kg, J=0,01 kgm²

F_bmax = - ma = 11000 N !!!

M_bmax = - J = 900 Nm !!!

dynamics, kinetostatics 27

N ma

F

= − = 250

= 25

STATIC b

F

F

S – center of gravity

e

S

F_b



 

^{1 / 250}  ^/



500 001 ,

0

s m e

a

s m e

S

= =

=

=





kg m = 1

F_b

Example of analysis (kinetostatics)

m, J

M₁=?

3

0

2

₁ 1

Data:

m, J – mass, mass moment of inertia

₁ – angular velocity of link 1 M₁ = ? and joint forces

2

M_b

a

F_b



ma F

= −



J

M

= −

ma J F

h M

b

b

₌ 

=

M

Fb

Fb _h

2

M_b

a

F_b



2

F_b

a



2

F_b F_b

a

F_b



h

M₁ 3

0

2

1

F_b F03

c

F₂₃

c

0

A

= 0 +

23

F

F

Link 3:

M1

3

0

2

1

Fb

F03 c

F₂₃

c

0

A

= 0 +

23

F

F

Link 3:

= 0 +

+

32

F F

F

Link 2:

M₁ 3

0

2

1

F_b F30

c

F₃₂

c

F12 c

0

A

F₃₂

c

F12 c

= 0 +

+

32

F F

F

M₁ 3

0

2

1

F_b F₃₀

c

F₃₂

c

F₁₂

c

F₁₀

c

0

A h₁

c

F₃₂

c

F₁₂

c



=

+ ₀₁ 0 M₁^A 0 M₁ F₂₁h₁ 0

21 F

F

m, J

M₁=?

3

0

2

1 F_c

c

₁

Data:

F_C – input force (driver)

m, J – mass, mass mom. of inertia

₁ – angular vel of crank 1 M₁ = ? and joint forces = ? Example of analysis (kinetostatics)

M₁ =?

3

0

2

1 Fc c

Fb

F03 c

0

= 0 +

+

32

F F

F

= 0 +

+

C

F F

F

???

Slider 3:

Coupler 2:

Statically determined group (of links)

kinematic system in equilibrium

every link in equilibrium

any group of links in equlibrium

forces may be fully determined using the six (three) equations of 3D (2D) statics without

using deflection and stiffness criteria.

j

i

F_ijx F_ijy

i

j Fijx

Fijy

i

j Fijx

Fijy

1 unknown 2 unknowns 2 unknowns

Number of equations

= Number of unknowns

3k = 2p₁ + p₂

In kinetostatics unknowns are joint forces

( ¹ ) ²

¹

3 n p p

DOF = − − − 0 1

2

3 k − p

− p

=

Mobility

Condition of static determination

k 1 2

...

4

p₁ 1 3

...

6

p₂ 1 0

...

0

(k-p₁–p₂) (1-1-1) (2-3-0) (460)

Examples of static. determined groups

group (1-1-1) group (2-3-0)

II class - R or T II class - K or J or Z

I kl - R lub T

1-1-1 stat. determined group (one link)

3 1

2 F₁₂

F₂

F₃₂ F₃₂ + F₁₂ + F₂ = 0

2-link groups (k-p₁-p₂ = 230)

RRR

RTR RTT

RRT

TRT