Bài giảng Kinematics and dynamics of machines - Chapter 3: Dynamic force analysis friction - Phạm Huy Hoàng
DYNAMIC FORCE ANALYSIS
FRICTION
PHAM HUY HOANG
DYNAMIC FORCE ANALYSIS
I. Introduction:
1.Force classification:
a. External force: force caused by external agent
outside of the system (Driving load, Technical
resistance load/Useful resistance load,
,
Gravitational force, Friction, Inertial load/Pseudo
external load).
b. Internal force: force exchanged by the links of
mechanism (Normal force, Friction in joint)
Load: force or torque
* Inertial load/Pseudo external load:
F
M
2
1
2
F
i
a
i
S
i
F m a
;
M M J
i
i
i S
i
i i
F
i
i
F
Si
3
i
i
Si: center of mass
F
M
1
4
Inertial force and torque
F
m a ; M
i S
J
qt i i
qt
i
i
i
F
2
1
M
2
qt
F
F
i
F
M
qt
i
3
Si
i
F
M
1
4
i
qt
i
F F 0; M M M
0
i
i
qt
F
i
i
i
2. Normal force in plane:
a. Prismatic joint: 2 variables – magnitude and exerting
position
Force Magnitude and Line of action {exerting position, direction}
N
kj
p
N
kj
N
kj
M
kj
M
x.N
kj
kj
b. Revolute joint: 2 variables – magnitude and direction
Force Magnitude and Line of action {exerting position, direction}
R
ij
p
R
ij
c. Higher pair (4-type joint): 1 variable – magnitude
Higher pair is a kinematic pair in
which the links have point or line
contact (Ball bearings, gear meshing,
N
cam and follower are examples of
higher pair)
ij
N
ij
3. Statically determinable group / Assure group:
Statically determinable group: group of links and joints that
the number of unknown reaction forces (at joints) equals
the number of equilibrium equations of links.
Assur group: group of links and joints with 0-DOF
Planar group of n moving links, p4 higher pairs (4-type joints)
and p5 lower pairs (5-type joints).
Statically determinable group: number of equilibrium
equations 3n, unknown reaction forces (p4+2 p5)
Assur group: W = 3n - (p4+2 p5)
Statically determinable condition Ξ Assur condition:
3n - (p4+2 p5) = 0
Statically determinable group / Assure group :
Planar lower-pair-only group with n moving links and p5 lower
pairs (5-type joints).
Statically determinable condition Ξ Assur condition
3n - 2 p5 = 0 Groups {2 links, 3 joints}, {4 links, 6 joits },
{6 links, 9 joints}
4. Dynamic force analysis using Newton – Euler laws:
Procedure
a. Kinematics (position, velocity and acceleration analyses).
b. Determine inertial forces and torques exerting to links.
c. Identify unknown and known forces and torques and links
appropriated.
d. Detach the mechanism into statically determinable groups
and driving links.
e. Apply reaction forces on mating surfaces of links.
Remember the balance between action and reaction
forces at joints.
4. Dynamic force analysis using Newton – Euler laws :
Procedure
f. Sequentially solve the system of equilibrium equations from
the far group to the near group. Each time of solving, the
results will be used as the data for the next solving.
g. Solve the equilibrium equation of driving links.
5. Determine driving load using Virtual work:
F
i
v
i
M
i
i
F
qti
v
Si
M
qt
i
N
cb
N
M .
cb
cb
1
M
cb
v
N
P .v
cb cb
P
cb
cb
cb
n
n
i i
N
F .v M .
F
.v M .
0
cb
i i
qti Si qti i
i1
i1
F2
II. Ex. 1:
Mqt2
M1
F3
1
Fqt3
3
3
l
a,l
a 3,l
a,l
CD
a,CAB 60
AB
BC
BD
2
2
const
1
F F 3F
3F
2
3
qt3
M
Fa
qt2
R ?,M ?
ij
1
2
D
2
R
B
B
F
21
M
qt2
1
1
M
1
M
3
1
qt3
F
1
F
R
C
A
A
01
0
3
D
2
D
12
2
12
2
R
B
B
M
F
R
F
M
qt2
R
qt2
32
2&3
N
y
03
M
03
x
3
F
N
F
C
C
03
qt3
M
03
3
C
F
F
3
qt3
R
D
2
12
2
R
B
M
F
qt2
R
32
y
x
C
2
F R R 0
12 32
M
M (F )M (R )M (R ) 0
qt2
B 2
B 12
B 32
x
x
F R R 0
(1)
2
12 32
y
y
R R 0
(2)
12 32
3
3
2
y
x
M
R
aR
a 0 (3)
qt2
32
32
2
N
03
M
03
3
F
F
qt3
R
23
3
F F R N 0
qt3
23
03
M (F ) M (F ) M (R ) M (N ) M 0
C 3
C qt3
C 23
C
03
03
x
23
F F R 0 (4)
3
qt3
y
N R 0
(5)
(6)
03
23
M
0
03
12
21
R R : R1x2 R2x1 , R1y2 R2y1
R23 R32 : R2x3 R3x2 , R2y3 R3y2
12
F
2
12
2
R
M
R
F
M
qt2
R
qt2
32
N
03
M
03
3
F
N
F
03
qt3
M
03
3
qt3
x
x
F
R32 R23 F3 Fqt3 2F
Rx Rx 5F
F
R
12
21
23
2( 3 1)
N R y R y R y R y
F
03
23
32
12
21
3
M03 0
01
R R 0
21
R
M (R ) M (R ) M 0
A 21
A 01
1
B
21
x
x
R
R 5F
21
01
1
2( 3 1)
y
01
y
21
M
R R
F
1
3
1
3
y
21
x
R
M
aR aR
1
21
A
2
2
01
M
1
17 3 2
Fa
6
M F v
1 1 2 D2
M (F
qt2 2 qt3
F )v 0
3 C3
M F v
1 1 2 D2
M (F
qt2 2 qt3
F )v 0
3 C3
M F v
1 1 2 D2
M (F
qt2 2
F )v
3 C3
qt3
3
3
2
17 3 2
1
M 3F
a Fa
(F 3F)
a
Fa 0
1
1 1
1
1
2
3
6
v
D
2
F
2
M
qt2
2
M
3
1
1
qt3
F
F
v
3
M
1
17 3 2
Fa
6
III. Ex. 2
F3
l
a,l
a 3,l 3a
CD
AB
AC
M1
CAB 90
Mqt2
const
1
F 3F
3
Mqt3
M
Fa;M
2Fa
qt3
qt2
M 3Fa
M3
3
R ?,M ?
ij
1
x
N R 0 (1)
32
12
12
R
N 0
32
y
R 0
(2)
12
M
M M (R ) M (N ) 0
32 B 12 32
qt2
B
M
M 0 (3)
32
qt2
N32
M32
M qt2
R12
3
F R N 0
03
23
F3
M (F ) M (R ) M (N )
C 3
C 03
C
23
M M
M 0
3
qt3
23
M23
N23
M qt3
x
F cos30 N R 0 (4)
23
M3
3
03
R03
y
03
F sin30 R 0 (5)
3
F cos30 CD N BC M M
M 0 (6)
3
23
3
qt3
23
12
N
21
F3
y
y
x
x
R
R : R R ,R R
12
21 12
21
y
y
32
x
23
x
M qt2
N : N N , N N
23
32
32 23
R12
M
x
M
23
32
9
3
x
21
R
R
N
N
F
32
23
12
4
M qt3
y
y
R
R
0
M3
12
21
R03
F3
3 3
x
R
F
03
4
1
3
y
03
M23
R
F
F
3
2
2
N32
N23
M32
M
M
M
Fa
qt 2
23
32
M qt2
R12
M qt3
M3
R03
9 3
4
x
x
R
R
F
01
21
01
21
R
R 0
y
01
y
21
x
21
R R 0
M (R ) M (R ) M 0
A 21
A 01
1
M R ABcos60
1
R21
M1
1
A
M
1
9 3
B
Fa
R01
8
M F v
1 1 3 D3
M
M
qt2 2
M 0
qt3 3 3 3
M F v
1 1 3 D3
cos150 M
M
M 0
qt2 2 qt3 3 3 3
M F v
1 1 3 D3
cos150 M
M
M
qt2 2 qt3 3 3 3
M F (CD )cos150 M
M
M
1 1
3
3
qt2 2 qt3 3 3 3
4
3
4
4
1
4
1
1
1
M 3F (3a )(
1 1
) (Fa)
(2Fa)
(3Fa)
0
2
F
3
v
D
M
3
1
M
qt2
1
M
1
9 3
8
2
3
Fa
M
qt3
M
3
13 trang
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