Transcript ME321 Kinematics and Dynamics of Machines Steve Lambert

ME321
Kinematics and Dynamics of
Machines
Steve Lambert
Mechanical Engineering,
U of Waterloo
5/24/2016
Kinematics and Dynamics




Position Analysis
Velocity Analysis
Acceleration Analysis
Force Analysis


F  ma
We will concentrate on four-bar linkages
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Acceleration Analysis


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Use vector loop equations
Vector equations can be expressed in general form, or
specialized for planar problems
 Graphical Solutions
 Vector Component Solutions
 Complex Number Solutions (in text)
Vector Equations

 
RP  R0  R
Y
P
y

R

RP

R0


R0  X 0 Iˆ  Y0 Jˆ  Z 0 Kˆ

R  xiˆ  yˆj  zkˆ
x
z
X
   xiˆ   y ˆj   z kˆ   X Iˆ  Y Jˆ   Z Kˆ
Z
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Vector Equations for Velocity
Differentiate Position Vector with respect to Time




VP  RP  R0  R


R0  XIˆ  YJˆ  ZKˆ  V0

ˆ


ˆ
ˆ
ˆ
ˆ
ˆ
R  xi  y j  zk  xi  yj  zk
   
R V   R

   
VP  V0  V    R
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Vector Equation for Acceleration
Differentiate velocity equation:

   
VP  V0  V    R
To obtain acceleration relation:

          
 
a  VP  V0  V    V    R    V      R
  
      
a  V0  V  2 V    R      R

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

Acceleration Equations

  
      
a  V0  V  2  V    R      R

V0

V
 
2  V
 
R
  
 R
Where:

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
- Acceleration of origin
- Acceleration in local frame
- Coriolis acceleration
- Angular acceleration
- Centripetal acceleration

Planar Velocity Equations
Assume:
ê
êr
Y
B
• Motion is restricted
to the XY plane
• Local frame is
aligned with and
fixed to link

A
r
X
Therefore:
•  becomes the angular velocity of the link, and
• local velocity becomes the change in length of the link
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Planar Velocity Equations

   
VP  V0  V    R
ê
êr
Y
B

Becomes:


VB  VA  reˆr  reˆ
A
r
X
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Planar Acceleration Equations

aB

aB

aB

aB
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 
 a A  aB / A

 a A  reˆr  2kˆ  reˆr   kˆ  reˆr  kˆ  kˆ  reˆr

 a A  reˆr  2reˆ   reˆ   2 reˆr

 a A  r   2 r eˆr  2r   r eˆ




Application to Four-Bar Linkages

 
aB  a A  aB / A


aB  a A  r   2 r eˆr  2r  r eˆ

B



2
aB  a A  3 eˆr3  3r3eˆ3
3
A
4
2, 2
O2
O4
 42 r4eˆr4   4 r4eˆ 4  22r2eˆr2   2 r2eˆ 2  32eˆr3  3r3eˆ3
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Graphical Solution
3r3
-32r3
2r2
4r4
B
-42r4
3
A
4
2, 2
-22r2
O2
-22r2
-32r3
2r2
O’’
-42r4
3r3
O4
4r4
 42 r4eˆr4   4 r4eˆ 4  22r2eˆr2   2 r2eˆ 2  32eˆr3  3r3eˆ3
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Vector Component Solution
2
2
ˆ
ˆ
ˆ
ˆ
  r e   4 r4e 4  2 r2er2   2 r2e 2  3 eˆr3  3r3eˆ3
2
4 4 r4
But: eˆr  cosIˆ  sin Jˆ
and eˆ   sin Iˆ  cosJˆ
Giving:
Y
ê
êr

Ĵ
Iˆ
X
  42 r4 cos 4   4 r4 sin  4   22 r2 cos  2   2 r2 sin  2   32 r3 cos  3   3r3 sin  3
  42 r4 sin  4   4 r4 cos  4   22 r2 sin  2   2 r2 cos  2   32 r3 sin  3   3r3 cos  3
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Coriolis Acceleration
3 (P)
4
2
1



aP4  aP2  aP4 / P2


2


aP4  aP2  r   2 r eˆr  2 2 r   2 r eˆ

1

2
ˆ

ˆ
  r e  4 r4e 4  2 r2eˆr2   2 r2eˆ 2  reˆr2  22 reˆ 2
2
4 4 r4


 42 r4eˆr4   4 r4eˆ 4   22 r2  r eˆr2   2 r2  22 reˆ 2
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Coriolis Direction

 
acor  2  r

a cor
V=r
r
r
V  r
r

velocity of slider at time t,
and increment after time t
V
r

t
t
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V

 r
t
t