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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
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
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
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
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
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
1
OB
R10
R30
P
b
|P+R30|=|R23|=|R21|=|R10|
STATIC ACTIONS ON THE FRAME
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
Let us now consider the inertial actions, starting from the slider contribution.
OB
12
3
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
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≅ + + +
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
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
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
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= − ≅ − +
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
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
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
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= − −
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
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
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
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
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
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
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
EQUILIBRIUM UNDER Mcr
OB
Mcr/L cosβ
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
ACTIONS ON THE FRAME UNDER Mcr
Mcr/L cosβ
OB
Mcr/L cosβ
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
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
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
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= −
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
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
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
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
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
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.
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
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
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
BALANCING: CHANIGING MODULUS ON CONSTANT DIRECTION
Another configuration
2( ) (cos )s BI
s m m RF ω ϕ≅ − +
FsI
FsI
FsI
FsId
dd ,
0
2
ii
O ii
IsF
F
M d
=
=
∑
∑
O
DYNAMIC ANALYSIS OF THE CRANK, CONNECTING ROD, SLIDER (CCRS) MECHANISM
BALANCING: CHANIGING MODULUS ON CONSTANT DIRECTION
A balanced configuration
2( ) (cos )s BI
s m m RF ω ϕ≅ − +
FsI
FsI
FsI
FsI
dd
d
,
0
2
ii
O ii
IsF
F
M d
=
=
∑
∑O