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Suspension system
Q
S0
1
2
3
spring+damper
Q
S0
1
2
3
0:2 =++ 1232 FQFlink
points of
application only
?
Q
S0
1
2
3
0:3 =+ 0323 FFlink
Q
S0
1
2
3
0=++ 1232 FQF
QF12
F32
Q
S0
1
2
3
0:1 =++ SFF 0121link
F21
F01
S
RRTR example (1)
3
01
2
RRTR example (2)
3
01
2
M3
M1=?
RRTR example (3)
3
01
2
M3
M1=?
0.3 =+ 0323 FF
RRTR example (3)
3
01
2
M3
M1=?
0.2 =+ 3212 FF
RRTR example (3)
3
01
2
M3
M1=?
0.2 =+ 3212 FF
0.3 =+ 0323 FF
RRTR example (4)
3
01
2
M3
M1=?
...
0
0.3
23
3323
3
=
=−
=
F
MhF
M C
B
A
C
h3
F23
F03
RRTR example (5)
3
01
2
M3
M1=?
...
0
0.3
23
3323
3
=
=−
=
F
MhF
M C
B
A
C
h3
F23
F03
F32
F30
RRTR example (6)
3
01
2
M3
M1=?
0.1 =+ 0121 FF
B
A
C
h3
F23
F03
F32
F30
F12
F21
RRTR example (7)
3
01
2
M3
M1
0
0.1
1121
1
=−
=MhF
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
εM Sb I−=00 =+=− bSI MMεM
εM SI=
aF mb −=00 =+=− bm FFaF
aF m=
dynamics, kinetostatics 18
Mass moment of inertia
+=m
Z dmyxI )( 22
)( 22
SSSZ yxmII ++=
dynamics, kinetostatics 19
aF mb −=
εM Sb I−=
ma
I
F
M
F
Mh S
b
b
M
b ===
FORCE (Fb) and MOMENT (Mb) OF INERTIA
Given:
m, IS, a, ε
dynamics, kinetostatics 20
FORCE (Fb) and MOMENT (Mb) OF INERTIA
resultant inertia
force
dynamics, kinetostatics 21
Center of percussion
ASAS ==
+=
t
S
2n
S
t
S
n
S
; aa
aaaS
S
A
B
aS
SS
dynamics, kinetostatics 22
Center of percussion
S
A
B
aS
SSW
Fb
h
SI
m
εM
aF
b
Sb
−=
−=
S
S
b
b
ma
I
F
Mh
==
B
S
A aS
SSW
Fb
h
cos(?)==S
t
S
a
a
SW
hCenter of percussion
=
=
=
==
t
S
S
S
t
SS
t
S
S
S
S
t
S
S
a
a
ma
AS
aI
a
a
ma
I
a
ahSW
mAS
I
a
a
ASma
aI S
t
S
S
S
t
SS ==
B
S
A aS
SSW
Fb
h
AS
i
mAS
mi
mAS
ISW SSS
22
===
Center of percussion
iS – 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
aSmax = 55000 ms-2
Bcmax = 90 000 s-2
m=0,2 kg, J=0,01 kgm2
Fbmax = - ma = 11000 N !!!
Mbmax = - J = 900 Nm !!!
dynamics, kinetostatics 27
NmaFb 250=−=25=STATIC
b
F
F
S – center of gravity
e
S
Fb
22 /250
/1500
001,0
smea
s
me
S ==
=
=
kgm 1=
Fb
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
Mb
a
Fb
maFb −=
JM b −=
ma
J
F
Mh
b
b ==
bM Couple of forces Fb
bFbF h
2
Mb
a
Fb
2
Fb
a
2
Fb
Fb
a
Fb
h
M1
3
0
2
1
Fb
F03
c
F23
c
0
A
0=+ 0323 FF
Link 3:
M1
3
0
2
1
Fb
F03
c
F23
c
0
A
0=+ 0323 FF
Link 3:
0=++ 12b32 FFF
Link 2:
M1
3
0
2
1
Fb
F30
c
F32
c
F12
c
0
A
F32
c
F12
c
0=++ 12b32 FFF
M1
3
0
2
1
Fb
F30
c
F32
c
F12
c
F10
c
0
A
h1
c
F32
c
F12
c
=−→==+ 000 12111 hFMM A
0121 FF
m, J
M1=?
3
0
2
1
Fc
c
1
Data:
FC – 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=++ 12b32 FFF
0=++ 2303C FFF
???
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
Fijx
Fijy
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
( ) 21 1213 ppnDOF −−−=
0123 21 =−− ppk
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