DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER ... · DYNAMIC ANALYSIS OF THE CRANK,...

Referring to the mechanism in the figure, starting from the loop closure equation it is possible to solve position, velocity and acceleration analysis.

RL

s

L+R

φγ

A

OB

1 cos (1 cos 2 )4

s R λϕ ϕ⎡ ⎤≅ − + −⎢ ⎥⎣ ⎦

( 2 )2Bv R sen senλω ϕ ϕ≅ +

2 (cos cos 2 )

( 2 )2

Ba R

R sen sen

ω ϕ λ ϕλω ϕ ϕ

≅ + +

+ +/with R Lλ =

12

3

DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM

β

OB

12

3


P MR

Let us use the superposition principle and decompose the static and dynamic analysis

We consider the CCRS as a driving mechanism. A force P will act on the slider as the result of a pressure in the piston chamber. A resistive torque MR is applied at the crank.

A


THE SLIDER

P

R03

R23

P

R23R03

THE CONNECTING ROD

R32

R12

STATIC EQUILIBRIUM

3

2

THE CRANK

R21

MR=R21b

R01

b

STATIC EQUILIBRIUM


1

OB

R10

R30

P

b

|P+R30|=|R23|=|R21|=|R10|

STATIC ACTIONS ON THE FRAME


Let us now consider the inertial actions, starting from the slider contribution.

OB

12

3


Fs MR

A

2 (cos cos 2 )s s B smF a m Rω ϕ λ ϕ= − ≅ +

We assume that the crank is connected to an infinite inertia system. Thus, its velocity stay constant and MR can always equilibrate the CCRS.

2 (cos cos 2 ) ( 2 )2Ba R R sen senλω ϕ λ ϕ ω ϕ ϕ≅ + + +


THE SLIDER

Fs

R03

R23

Fs

R23R03

THE CONNECTING ROD

R32

R12

EQUILIBRIUM UNDER Fs

3

2

THE CRANK

R21

MR=R21b

R01

b

EQUILIBRIUM UNDER Fs


1

OB

R10

R30

ACTIONS ON THE FRAME UNDER Fs

Fstgβ

Fstgβ

Fs

β

2 (cos cos 2 )s B ssF m a m Rω ϕ λ ϕ= − ≅ − +


Fs

R10=-R23=Fs+R03

R03

Let us now consider the inertial action due to the crank contribution, calling G its center of mass. We assume the crank rotates with constant angular velocity.

OB

12

3


Fc

A

2c Gc cF m a m G O ω= − = − −2

Ga G O ω= −

G

OB

ACTION ON THE FRAME FcFc

2cc m GF O ω= − −


Body 3 and 2 are not loaded, therefore the inertial action due to the crank is transmitted directly to the frame.

Let us now consider the connecting rod and let us build a dynamic equivalent system of two masses in A and B and a pure inertia moment .

OB

1

23


Mcr

A

The inertial effect of mB and mA sum with those of Fs and Fc respectively. Let us thus consider the inertial action due to the pure inertia couple Mcr.

γ

MR

2 2cr zz a bJ J m a m b= − ⋅ − ⋅

ccr rJM γ= −a tot

bm ma b

= ⋅+

b totam m

a b= ⋅

+

crJ


THE SLIDER

R03

R23

THE CONNECTING ROD

R32

R12=Mcr/L cosβ

EQUILIBRIUM UNDER Mcr

Mcr

2 3

THE CRANK

R21

MR=R21b

R01

b


EQUILIBRIUM UNDER Mcr

OB

Mcr/L cosβ


ACTIONS ON THE FRAME UNDER Mcr

Mcr/L cosβ

OB

Mcr/L cosβ


GLOBAL ACTIONS ON THE FRAME

Mcr/L cosβ

Fc

2( )cc Am RF m G O ω= − + −Fstgβ

Fs

2( ) ( ) (cos cos 2 )s B B ss Bm m a m mF Rω ϕ λ ϕ= − + ≅ − + +

Fstgβ

ccr rJM γ= −

P/cosβ

Ptgβ

P


TOTAL RESISTANT TORQUE

Compute it as homework

?RM =

2( ) ( ) (cos cos 2 )s B B ss Bm m a m mF Rω ϕ λ ϕ= − + ≅ − + +

ccr rJM γ= −


BALANCING THE CCRS MECHANISM

It is useful to reduce the number and the intensity of the forces transmitted to the rotaryjoints and to the frame.

The forces changing in direction and intensity in may result particularly dangerous results, possibly introducing vibration and affecting the fatigue resistance of the mechanism.

Let us see some example of balancing for the CCRS mechanism.

O


BALANCING: CONSTANT MODULUS CHANIGING IN DIRECTION

Inertial actions due to the motion of the crank plus the lumped mass in A resultin a force, whose amplitude keeps constant with the crank angular velocity , changing in direction together with the direction of A-O

FcI

2( )I IIc c

c

F F

A cm R m GF O ω= − + −

G

AmA

FcII

2( )cc Am RF m G O ω= − + −

G’ FcII

' /A cG O m R m− = − 0cF⇒ = Fc


BALANCING: CHANIGING AMPLITUDE ON CONSTANT DIRECTION

Horizontal component of inertial actions due to the slider motion plus the lumped mass in B result in two reciprocating forces, one whose amplitude varies proportional to the crank angular position, the other whose amplitude doubles this variation.

2( ) ( ) (cos cos 2 )I II

s s

s B B s Bs

F F

m m a m m RF ω ϕ λ ϕ= − + ≅ − + +

The second term is negligible, as long as λ<<1.


BALANCING: CHANIGING AMPLITUDE ON CONSTANT DIRECTION

For the throw crank shaft of a 4 cylinders engine it is possible to have different configurations

2( ) (cos )s BI

s m m RF ω ϕ≅ − +

FsI

FsI

FsI

FsI

,

4

6

ii

O ii

Is

IsF

F F

M d

=

=

∑

∑

dd

d

O


BALANCING: CHANIGING MODULUS ON CONSTANT DIRECTION

Another configuration

2( ) (cos )s BI


FsI

FsI

FsI

FsId

dd ,

0

2

ii

O ii

IsF

F

M d

=

=

∑

∑

O


BALANCING: CHANIGING MODULUS ON CONSTANT DIRECTION

A balanced configuration

2( ) (cos )s BI


FsI

FsI

FsI

FsI

dd

d

,

0

2

ii

O ii

IsF

F

M d

=

=

∑

∑O

DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER ... · DYNAMIC ANALYSIS OF THE CRANK,...

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