kinematics of mechanismstmm.pwr.edu.pl/fcp/qgbukoqttklqhbx08slktvqjqx2o8... · 2020. 3. 12. ·...
TRANSCRIPT
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KINEMATICS OF MECHANISMS
position – velocity – acceleration; driving forces are neglected
2
2vv
dt
sd
dt
dsa
dt
dss KKKK
KKK
For a point K we have linear displacement sK(t), velocity
vK and acceleration aK
2
2
dt
d
dt
d
dt
d kkkk
kkk
For a link k we have angular displacement k(t),
angular velocity k and angular acceleration k
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METHODS OF KINEMATIC ANALYSIS
graphical (use only: pencil, ruler, compass,
calculator)
analytical
numerical
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A
BC
D
M
BC
CD
vM
DOF = 1 1 driver, this case: crank AB – known (t)
-
LENGTH SCALE
A
B
C
D
Length scale:
mm
m
BC
BCl
100
1
)(
SCALE DEFINITION (for graphical methods)
-
Scale (general)
)(i
ii
real value
graphical value
mm
ms
)v(
v -1
vVelocity scale
mm
ms
)a(
a -2
aAcceleration scale
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FORCE SCALE
Force scale:
mm
N
F
FF 1
100
100
)(
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4 BAR LINKAGE
a lot of applications
4 bar linkage has a few inversions
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Harbor crane
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R
R
R
T
hand press
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pipe spanner,
pipe wrench
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Four-bar linkage: crank rocker
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A
D
C1 C2
B2
B1
0-l0
3-l3
2-l2
1-l1
For the crank-rocker we have:
1. Grashof inequalities fulfilled,
2. shortest link (l1) is a crank (rotates around frame)
4 BAR LINKAGE (4R)
0231
0321
3201
llll
llll
llll
312101 llllll
Grashof inequalities
-
A D
C
B
Grashof inequalities fulfilled,
shortest link (l1) is a frame
A D
C
B
Grashof inequalities fulfilled,
shortest link (l1) is a coupler
DOUBLE CRANK
DOUBLE ROCKER
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Transforming a fourbar crank-rocker to crank-slider
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R T
Slider block
translates
R R
Slider
block has
complex
motion
R R
Slider
block
rotates
R T
Slider
block is
stationary
4 BAR - crank-slider - INVERSIONS
-
B
A
CC
’
B
’
rBC
D
GRAPHICAL POSITION ANALYSIS
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B
A
CC’
B’
rBC
(t)
Data: (t)
Find: xc’, yc’
xB’ = AB cos
yB’ = AB sin
(xC’ – xB’)2 + (yC’ – yB
’) 2 =
BC2
yC’ = 0
From the system of
equations:
xC’ , yC’
y
x
GRAPHICAL POSITION ANALYSIS
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B
A D
C
C
’
B
’
rBC
D
GRAPHICAL POSITION ANALYSIS
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B
AD
C
C’
B’
y
x
(t)
Data: (t)
Find: xc’, yc’
xB’ = AB cos
yB’ = AB sin
(xC’ – xB’)2 + (yC’ – yB
’) 2 =
BC2
(xC’ – xD)2 + (yC’ – yD)
2 =
CD2
rBC
From the system of
equations : xC’ , yC’
GRAPHICAL POSITION ANALYSIS
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A
BC
D
M
GRAPHICAL POSITION ANALYSIS
Draw a mechanism for a given
Link dimensions are known: AB, BC, CD, AD and BM, CM
-
A
BC
D
M
GRAPHICAL POSITION ANALYSIS
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A
BC
D
M
GRAPHICAL POSITION ANALYSIS
-
A
BC
D
C*
M
Two positions for given
GRAPHICAL POSITION ANALYSIS
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A D
B
B1
C
E
F
1 driver: AB AB1
GRAPHICAL POSITION ANALYSIS
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A D
B
B1
C
E
F
C1
F1
E1
GRAPHICAL POSITION ANALYSIS
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A D
B C
E
F F1
R=EF
1 driver: slider F F1
GRAPHICAL POSITION ANALYSIS
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DG
C
C
’
B
A
B
’rBC
D
E
F
D
yE
’
F
’
1
2
GRAPHICAL POSITION ANALYSIS - III class mechanism
-
Kinematics – graphical methods
INSTANT CENTERS OF VELOCITY
VELOCITY AND ACCELARATION
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From planar complex motion to rotation
(2 links: frame 0 and moving body k)
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A
Bv
B
AAv
k
{0}
0
B
tangent to trajectory
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instant center of rotation
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k0
B
k0
A
BS
v
AS
vk
( )ktg
)(BS
)(v
)(AS
v
k0
B
k0
A
angular velocity:
-
B
a
A
b
VA VB VAB = VBA = 0
SAB
{0}
2 bodies in motion
number of IC:( )
2
1
2
nnni
An instant center of velocity is a point, common to two bodies in
plane motion, which has the same instantaneous velocity in each
body
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23
1312
030201
S
SS
SSS( )6
2
144
i
-
13
02
S
S
23
12
0301
S
S
SS
?
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VB
VC
-
VB
VC
S20
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1, 2, 0 S12 & S10 S20
3, 2, 0 S32 & S30 S20
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1, 3, 0 S10 & S30 S31
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INFINITY
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Find vM direction
0
13
2
M
1st method: trajectory of point M and tangent ???
2 method: instant center of velocity S20
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23
1312
030201
S
SS
SSS
0
13
30
M
10
21
2
-
0
13
2
30
32
M
10
23
1312
030201
S
SS
SSS
21
-
0
13
2
30
21
32
20
31
M
VM10
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Cam mechanism
1
2
0
S12 S01 S02cam
follower
-
12
1
2
0
1020
V12
S10 – S12 – S20
common tangent (axis of slip)
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Cam with roller
1
2
1020
V13 = 0 S13 at contact point
3