111124112324567654simplestressesandstrainsjv-131016035244-phpapp02 (1).pptx

1. 1
All figures taken from Design of Machinery, 3rd ed. Robert Norton 2003
MENG 3071
Chapter 3
Velocity Analysis

2. 2
Velocity Analysis
Definitions
R
dt
R
d
V








 


dt
d
   

 


i
pe
i
pe
R
V
V
pe
R
i
i
PA
P
PA
i
PA











Linear Velocity
Angular Velocity
Velocity of a point
Velocity is perpendicular to
radius of rotation & tangent to
path of motion
Multiplying by i rotates the vector
by 90°
Link in pure rotation

3. 3
 
cos sin
i
re r i

 
 
 
sin cos
i
ire r i

 
  
cos
r 
sin
r 

r
Real
Imaginary
cos
r 
sin
r 
Vector r can be written as:
Multiplying by i gives:
Multiplying by i rotates a vector 90°
Velocity Analysis

4. 4
If point A is moving
 


i
pe
V
V
V
V
i
A
PA
A
P








Velocity Analysis
Graphical solution:

5. 5
Graphical Velocity Analysis (3 & 4)
• Given linkage configuration & 2. Find 3 and 4
• We know VA and direction of VB and VBA
(perpendicular to AB)
• Draw vector triangle. V=r.
VBA
VB
VBA Direction
VBA
VB VB Direction
VA
 
 
3
4 4
BA
B
V AB
V O B





6. 6
Graphical Velocity Analysis (VC)
• After finding 3 and 4, find VC
• VC=VA+VCA
• Recall that 3 was in the opposite direction as 2
VCA
VC
VA
Double Scale
VCA
VC

7. 7
Instant Center
• A point common to two bodies in plane motion, which
has the same instantaneous velocity in each body.
• In ENGR 214 we found the instant center between
links 1 and 3 (point on link 3 with no velocity)
• Now we also have an instant center between links 2
and 4

8. 8
Instant Centers
• Kennedy’s rule: any three links
will have three instant centers and
they will lie on a straight line
• The pins are instant centers
• I13 is from links 1,2,3 and 1,3,4
• I24 is from links 1,2,4 and 2,3,4
I13
I24
1 2 3
I12
I23
I13
1 3 4
I13
I34
I14
1 2 4
I12
I24
I14
2 3 4
I23
I34
I24
Links
IC’s

9. 9
Instant Centers
I13
• I13 has zero velocity since link 1
is ground
• 3 is the same all over link 3
• Velocity relative to ground=r,
perpendicular to r
• VA2=a2=VA3=p3
• From this, 3 must be in the
opposite direction as 2, and
smaller in magnitude since p>a
A
a
VA2
VA3
p
3
3

10. 10
Instant Centers
I24
• I24 has the same velocity on link 2
and link 4
• VI2=l22=VI4=l44
• From this, 4 is in the same
direction as 2 and smaller in
magnitude since l4>l2
4
VI4
l4
VI2
l2

11. 11
Instant Centers Practice Problems
O2
O4
A
B
O4
O2
B
A
Power=Tinin=Toutout

12. 12
Velocity Analysis of a 4-Bar Linkage
Given 2. Find 3 and 4

13. 13
Velocity Analysis of a 4-Bar Linkage
• Write the vector loop equation
• After solving the position
analysis, take the derivative
or
where
0
1
4
3
2



 


 i
i
i
i
de
ce
be
ae
      0
4
3
2
4
3
2


 

 

 

 i
ce
i
be
i
ae i
i
i
0
4
3
2
4
3
2 

 




 i
i
i
ce
i
be
i
ae
i
0


 B
BA
A V
V
V



4
3
2
4
3
2






i
B
i
BA
i
A
ce
i
V
be
i
V
ae
i
V







14. 14
Velocity Analysis of a 4-Bar Linkage
• Take knowns to one side:
• Take conjugate to get 2nd
equation:
• Put in matrix form:
• Invert matrix:
0
4
3
2
4
3
2 

 




 i
i
i
ce
i
be
i
ae
i
2
4
3
2
4
3





 i
i
i
ae
ce
be 
































 2
2
4
3
4
3
2
2
1
4
3










i
i
i
i
i
i
ae
ae
ce
be
ce
be
2
4
3
2
4
3





 i
i
i
ae
ce
be 


































 2
2
4
3
4
3
2
2
4
3










i
i
i
i
i
i
ae
ae
ce
be
ce
be

15. 15
Inverted Crank Slider
Given 2. Find 3 and
Link 3 is a slider link: its effective length, b, changes
b

16. 16
Inverted Crank Slider
• Given 2. Find 3 and
• Write the vector loop equation:
• After solving the position
analysis, take the derivative:
• To get another equation:
or
so
0
1
4
3
2



 


 i
i
i
i
de
ce
be
ae
0
4
3
3
2
4
3
2 


 





 i
i
i
i
ce
i
be
i
e
b
ae
i 


 
 4
3 4
3 
 
  2
4
3
3
2
3





 i
i
i
i
ae
i
ce
be
i
e
b 



b

17. 17
Inverted Crank Slider
• Take conjugate to get second
equation:
• Put in matrix form:
  2
4
3
3
2
3





 i
i
i
i
ae
i
ce
be
i
e
b 



  2
4
3
3
2
3





 i
i
i
i
ae
i
ce
be
i
e
b 








 
  





























 2
2
4
3
3
4
3
3
2
2
3










 i
i
i
i
i
i
i
i
ae
i
ae
i
b
ce
be
i
e
ce
be
i
e 
 
  






























 2
2
4
3
3
4
3
3
2
2
1
3










 i
i
i
i
i
i
i
i
ae
i
ae
i
ce
be
i
e
ce
be
i
e
b

Invert:

18. 18
Velocity of any Point on a Linkage
• Write the vector for RP
• Take the derivative
• Similarly
 
3
3
2 

 

 i
i
p pe
ae
R
 
3
3
2
3
2




 

 i
i
p pe
i
ae
i
V
RP
 
2
2 
 
 i
S se
R
 
2
2
2


 
 i
S se
i
V
 
4
4 
 
 i
U ue
R
 
4
4
4


 
 i
U ue
i
V

19. 19
Offset Crank Slider
Given 2. Find 3 and
a
b c
d