Transcript MENG 371, Chapter 8

MENG 372
Chapter 7
Acceleration Analysis
All figures taken from Design of Machinery, 3rd ed. Robert Norton 2003
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Acceleration Analysis
  
Linear acceleration A  R  V
Angular acceleration     
Acceleration of a point
RP  pe
i
VPA  pe
i
 i   pe  i 
i
APA  VPA  pei (i ) 2  pei  i 
  2 pei  i pei
n
APA
t
APA
Acceleration has 2 components:
normal & tangential
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Acceleration Analysis
Magnitudes of acceleration components
APA   2 pei  i pei
n
APA
A  r ,
n
2
Magnitude of normal
acceleration
t
APA
A  r
t
Magnitude of tangential
acceleration
3
Acceleration Analysis
If point A is moving
AP  AA  APA
 AA   2 pei  i pei
Graphically:
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Graphical Acceleration Analysis (3&4)
• Given linkage configuration, 2. Find 3 and 4
• Know AnA,AtA,AnBA,AnB, and direction of AtBA,AtB
• AB=AA+ABA
AnB+AtB= AnA+AtA+AnBA+AtBA
AnA
AnB
AtB
AtA
AtB line
AtBA
AnBA
AtBA line
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Acceleration of point C
• Now we know 3 from previous step
• AC=AA+AC/A=AnA+AtA+AnC/A+AtC/A
Enlarged scale
AnA
AC
AtA
At
C/A
AnC/A
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Analytical Acceleration Analysis (4bar)
Given 2. Find 3 and 4
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Analytical Acceleration Analysis (4bar)
Write vector loop equation:
i 2
ae
i3
 be
 cei4  dei1  0
Take two derivatives:
i2aei2  i3bei3  i4cei4  0
 2 2aei2  i 2aei2  32bei3  i3bei3  4 2cei 4  i 4cei4  0
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Analytical Acceleration Analysis (4bar)
 2 2aei2  i 2aei2  32bei3  i3bei3  4 2cei 4  i 4cei4  0
Separate knowns and unknowns
 3bei3   4cei 4


2
2

aei 2  i 2 aei 2  32bei3   4 2cei 4
Z
i
Take conjugate
3bei3 4cei4  Z
 bei3 cei4  3   Z 
 
Put in matrix form  i




i

3
4
be

ce

  4   Z 
i3
i 4 1
 3   be
ce   Z 
Solve
     i3
 i 4   
ce   Z 
 4  be
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Coriolis Acceleration
Position of slider
Rp  pei
.
Velocity of slider
V p  pe i  pe
i
i
Transmission Slip
velocity
velocity
Acceleration: Use the product rule
Ap  pe i  pe
i
i
i
i
i
i


pe
i


pe

pe
i
 
2
Combining terms:
2
i Coriolis acc. cccurs when


Ap   p  p   i  p  2 p  e a body has v and 
slip
Slip
Normal
Tangential
Coriolis
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Inverted Crank Slider
• Given 2. Find b , 3 and 4
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Inverted Crank Slider
• Write vector loop equation and take two derivatives
aei2  bei3  cei4  dei1  0
i2aei2  bei3  i3bei3  i4cei4  0
b varies with time
  2 aei 2   2 aei 2 i
 bei3  2bei3 i  32bei3   3bei3 i
2
  4 cei 4   4cei 4 i  0
2
• Recall
• so
3 4 
3   4
3   4
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Inverted Crank Slider
 2 2aei 2   2aei 2 i  bei3  2bei3 i  32bei3  3bei3 i  4 2cei 4   4cei 4 i  0
• Group into knowns and unknowns, 3=4


bei3  i 3 bei3  cei 4  2 2 aei 2   2 aei 2 i  2bei3 i  32bei3  4 2cei 4
Z
• Take conjugate


bei3  i3 bei3  cei4  Z
• Put in matrix form
 ei3
  i 3
e
bei3  cei 4   b   Z 
 
 i 3
i 4  
be  ce   3   Z 
• Solve
 b   ei3
    i3
 3  e
i 3
i 4
1
be  ce   Z 
i 3
i 4   
be  ce  Z 
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Acceleration of any point on the mechanism
• Write the vector for RP
Rp  aei2  pei 3  3 
• Take derivative twice
V p  i2 aei 2  i3 pei 3  3 
Ap   2 2 ae i 2  iae i 2  ia3 pe i  3  3    3 2 pe i  3  3 
• Similarly
RS  se
i  2  2 
VS  i2 sei2  2 
AS  i 2 sei  2  2   2 2 sei  2  2 
RU  uei 4  4 
RP
VU  i4uei4  4 
AU  i 4uei 4  4   4 2uei 4  4 
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Inverted Crank Slider
..
• Given 2. Find 3 and d
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Common Values of Accelerations
• A ‘g’ is the acceleration of gravity =9.81m/s2
• Common values of acceleration:
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Human Tolerance for Acceleration
• Humans are limited in the level of acceleration they
can tolerate
• Machines are limited
by the stresses in the
parts, e.g. automobile
piston 40g’s at
idle, 700g’s at
highway 2000g’s
peak
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Jerk
• Jerk is the derivative

  ofacceleration
J  R V  A
linear jerk
      
angular jerk
• High value of jerk causes stomach to go funny in
roller coaster or elevator starting to descend
High acceleration
High jerk
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Jerk Analysis of 4-bar Linkage
• Recall
aei2  bei3  cei4  dei1  0
i2aei2  i3bei3  i4cei4  0
 2 2aei2  i 2aei2  32bei3  i3bei3  4 2cei 4  i 4cei4  0
• Taking another derivative
i 2 aei 2  3 2 2 aei 2  i 2 aei 2
3
 i3bei3  3i 33bei3  i33bei3
 i 4cei 4  3i 4 4cei 4   4 cei 4  0
2
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