Transcript Document 7803020

MECN 4110: Mechanisms Design
Mechanisms Design
MECN 4110
Professor: Dr. Omar E. Meza Castillo
[email protected]
http://facultad.bayamon.inter.edu/omeza
Department of Mechanical Engineering
4110: Mechanisms
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One thing you learn in science is that there is
no perfect answer, no perfect measure.
A. O. Beckman
Topic 5: Velocity Analysis
Fourbar Linkage
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Chapters Objectives
 Up on completion of this chapter, the student will
be able to
 Determine velocities of links and points on the fourbar
mechanism by using graphical and analytical
approaches.
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Velocity
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Velocity
 Velocity of point P
Multiplying by j rotates the vector by 90°
Velocity is perpendicular to radius of rotation & tangent to path of
motion
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Velocity
Vector r can be written as:
Imaginary
re j  rcos   j sin  
r cos
Multiplying by j gives:
jre j  r sin   j cos  
r
r sin 
Multiplying by j rotates a vector 90°
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r cos
r sin 
Real
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Velocity
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Velocity
 Velocity Difference / Relative Velocity
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Velocity
 Methods for Velocity Analysis
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Graphical Approach
Vectors
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Graphical Velocity Analysis
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Graphical Velocity Analysis
 Given ω2, find ω3, ω4, VA, VB and VC
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Graphical Velocity Analysis
 Given ω2, find ω3, ω4, VA, VB and VC
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Graphical Velocity Analysis
 Given ω2, find ω3, ω4, VA, VB and VC
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Graphical Velocity Analysis
 Given ω2, find ω3, ω4, VA, VB and VC
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Analytical Approach
Vector Loop Method
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Analytical Velocity Analysis of a Fourbar Linkage
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Analytical Velocity Analysis of a Fourbar Linkage
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Analytical Velocity Analysis of a Fourbar Linkage
 Write the vector loop equation
R2  R3  R4  R1  0
ae j 2  be j3  ce j 4  de j1  0
 Position analysis, where: θ1=0o
aCos 2  jaSin 2 
bCos 3  jbSin 3 
cCos 4  jcSin 4  dCos1  jdSin1  0
aCos 2  jaSin 2 
Cos0  1
o
with
bCos 3  jbSin 3 
Sin 0  0
o
cCos 4  jcSin 4  d  0
Position Equation
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Analytical Velocity Analysis of a Fourbar Linkage
 After solving the position analysis, take the
derivative to get the velocity equation
d 3
d 2
d 2
 aSin  2
 jaCos 2
 bSin  3

dt
dt
dt
d 3
d 4
d 4
jbCos 3
 cSin  4
 jcCos 4
0
dt
dt
dt
 a2 Sin 2  ja2Cos 2  b3 Sin 3 
jb3Cos 3  c4 Sin 4  jc4Cos 4  0

Velocity Equation
VA  a2 Sin  2  ja2Cos 2  j2 ae j 2




j 3
V


b

Sin


jb

Cos


j

be
V A  V BA  V B  0
3
3
3
3
3
BA
j 4
V


c

Sin


jc

Cos


j

ce
B
4
4
4
4
4
Where
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Analytical Velocity Analysis of a Fourbar Linkage
 Separate Velocity Equation into Real and
Imaginary Parts
 a2 Sin 2  b3 Sin3  c4 Sin 4  0
Real Part
a2Cos 2  b3Cos3  c4Cos 4  0
Imaginary Part
 Eliminate ω4 and Solve the equations for ω3
 b3 Sin3  c4 Sin 4  a2 Sin 2
* Cos 4
b3Cos3  c4Cos 4  a2Cos 2
* Sin4
 b3 Sin3Cos 4  c4 Sin 4Cos 4  a2 Sin 2Cos 4
b3Cos3 Sin 4  c4Cos 4 Sin 4  a2Cos 2 Sin 4
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Analytical Velocity Analysis of a Fourbar Linkage
sin(A-B)=sin A cos B - cos A sin B
 Summation of two last equations:
 b3 Sin 3Cos 4  b3Cos 3 Sin 4 
a2 Sin 2Cos 4  a2Cos 2 Sin 4
b3 Sin 3Cos 4  Sin 4Cos 3  
a2 Sin 4Cos 2  Sin 2Cos 4 
 Using Trigonometric Identity:
b3 Sin 3   4   a2 Sin  4   2 
a2 Sin ( 4   2 )
3 
b Sin ( 3   4 )
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Factorize
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Analytical Velocity Analysis of a Fourbar Linkage
 Eliminate ω3 and Solve the equations for ω4
 b3 Sin3  c4 Sin 4  a2 Sin 2
* Cos 3
b3Cos3  c4Cos 4  a2Cos 2
* Sin 3
 b3 Sin3Cos3  c4 Sin 4Cos3  a2 Sin 2Cos3
b3Cos3 Sin3  c4Cos 4 Sin3  a2Cos 2 Sin3
 Summation of two last equations:
c4 Sin 4Cos 3  c4Cos 4 Sin 3 
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a2 Sin 2Cos 3  a2Cos 2 Sin 3
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Analytical Velocity Analysis of a Fourbar Linkage
sin(A-B)=sin A cos B - cos A sin B
c4 Sin 4Cos 3  Cos 4 Sin 3  
a2 Sin 2Cos 3  Cos 2 Sin 3 
 Using Trigonometric Identity:
c4 Sin  4  3   a2 Sin  2  3 
a2 Sin ( 2   3 )
4 
c Sin ( 4   3 )
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Factorize
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Numerical Problem 1
 Calculate the angular velocity of link 3 and link 4,
(ω3 & ω4) when θ2=57.3o =1rad and the input
angular velocity, ω2=-1 rad/s
ω3
 We have all the requirements except θ3 and θ4
 How can we obtain θ3 and θ4?
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 Using Position Analysis (Chapter 4) - Mathcad
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 Using Working Model (Chapter 4)
Θ3=24.36o
Θ2=57.3o
Θ4=88.01o
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 Using Velocity Analysis (Chapter 5) - Mathcad
3 
a2 Sin ( 4   2 )
b Sin ( 3   4 )
4 
a2 Sin ( 2   3 )
c Sin ( 4   3 )
Rad/s
Rad/s
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 Using Working Model (Chapter 5)
Θ2=57.3o
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Velocity Analysis of a Inverted Slider-Crank Mechanism
 Link 2 (the crank) is the input and link 4 is the
output. Given r1, θ1=0o, r2, θ2, r4, θ4, ω2. Find r4 and
ω4. Remember r4 is a variable so r4≠ 0 in general.
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Velocity Analysis of a Inverted Slider-Crank Mechanism

V4
R 4

V2
r4 , R4
r2 , R2
ω2
ω4
2
r1 , R1 ,1  0o
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4
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Velocity Analysis of a Inverted Slider-Crank Mechanism
 Write the vector loop equation
R2  R4  R1  0
r 2 e j 2  r4 e j 4  r1e j1  0
 Position analysis, where: θ1=0o
r2Cos 2  jr2 Sin 2  r4Cos 4  jr4 Sin 4
 r1Cos1  jr1Sin1  0
with
Cos0 o  1
r2Cos 2  jr2 Sin 2
Sin 0 o  0
 r4Cos 4  jr4 Sin 4  r1  0
Position Equation
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Velocity Analysis of a Inverted Slider-Crank Mechanism
 After solving the position analysis, take the
derivative to get the velocity equation
d 2
d 2
d 4
 r2 Sin  2
 jr2Cos 2
 r4 Sin  4
dt
dt
dt
d 4
 r4Cos 4  jr4Cos 4
 jr4 Sin  4  0
dt
 r22 Sin 2  jr22Cos 2  r44 Sin 4
 r4Cos 4  jr44Cos 4  jr4 Sin 4  0
Velocity Equation
 
R 4  V4  V2  0
Where

V4

V2
R 4  r4Cos 4  jr4 Sin 4  jr4e j 4
 r44 Sin 4  jr44Cos 4  j4 r4 e j 4
 r22 Sin 2  jr22Cos 2  j2 r2 e j 2
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Velocity Analysis of a Inverted Slider-Crank Mechanism
 Separate Velocity Equation into Real and
Imaginary Parts
 r22 Sin2  r44 Sin4  r4Cos4  0
r22Cos2  r44Cos4  r4 Sin4  0
Real Part
Imaginary Part
 Eliminate ω4 and Solve the equations for r4
 r22 Sin2  r44 Sin4  r4Cos4  0 * Cos 4
r22Cos2  r44Cos4  r4 Sin4  0
* Sin4
 r22 Sin 2Cos 4  r44 Sin 4Cos 4  r4Cos 2 4  0
r22Cos 2 Sin 4  r44 Sin 4Cos 4  r4 Sin 2 4  0
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Velocity Analysis of a Inverted Slider-Crank Mechanism
sin(A-B)=sin A cos B - cos A sin B
 Summation of two last equations:
 r22 Sin 2Cos 4  r22Cos 2 Sin 4
 r4Cos 2 4  r4 Sin 2 4  0
r22 Sin 4Cos 2  Cos 4 Sin 2 


 r4 Cos  4  Sin  4  0
2
2
 Using Trigonometric Identity:
r22 Sin4  2   r4  0
r4  r22 Sin4  2 
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Factorize
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Velocity Analysis of a Inverted Slider-Crank Mechanism
 Eliminate for r4 and Solve the equations for ω4
 r22 Sin2  r44 Sin4  r4Cos4  0 * Sin4
r22Cos2  r44Cos4  r4 Sin4  0 * Cos4
 r22 Sin 2 Sin 4  r44 Sin 2 4  r4Cos 4 Sin 4  0
 r22Cos 2Cos 4  r44Cos 2 4  r4Cos 4 Sin 4  0
 Summation of two last equations:
 r22 Sin 2 Sin 4  r22Cos 2Cos 4
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 r44 Sin 2 4  r44Cos 2 4  0
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Velocity Analysis of a Inverted Slider-Crank Mechanism
cos(A-B)=sin A sin B + cos A cos B


r44 Sin 2 4  Cos 2 4 
r22 Sin 4 Sin 2  Cos 4Cos 2 
 Using Trigonometric Identity:
r44  r22Cos4  2 
r22
4 
Cos( 4   2 )
r4
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Factorize
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Numerical Problem 2
 Calculate the angular velocity and linear velocity
of link 4, (ω4 & r4) when r1=0.2m, θ1=0o, r2=0.1m,
θ2=70o, and the input angular velocity,
ω2=25rad/s
 We have all the requirements except r4 and θ4
 How can we obtain r4 and θ4?
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 Using Position Analysis (Chapter 4) - Mathcad
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 Using Velocity Analysis (Chapter 5) - Mathcad
4 
r4  r22 Sin4  2 
m/s
r22
Cos( 4   2 )
r4
Rad/s
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Velocity Analysis of Offset Slider-Crank
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Velocity Analysis of Offset Slider-Crank
Remember d is a variable so d≠ 0 in general.
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Velocity Analysis of Offset Slider-Crank
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Velocity Analysis of Offset Slider-Crank
 Write the vector loop equation
R2  R3  R4  R1  0
ae j 2  be j 3  ce j 4  de j1  0
 Position analysis, where: θ1=0o,
and θ4=90o
aCos 2  jaSin 2  bCos 3  jbSin 3 
cCos 4  jcSin 4  dCos1  jdSin1  0
aCos 2  jaSin 2  bCos 3 
Cos0o  1, Sin 0o  0
jbSin 3  jc  d  0
with Cos90o  0, Sin90o  1
Position Equation
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Velocity Analysis of Offset Slider-Crank
 After solving the position analysis, take the
derivative to get the velocity equation
d 3
d 2
d 2
 aSin  2
 jaCos 2
 bSin  3

dt
dt
dt
d 3 d d 
jbCos 3

0
dt
dt
 a2 Sin 2  ja2Cos 2  b3 Sin 3 
jb3Cos 3  d  0

Velocity Equation
VA   a2 Sin  2  ja2Cos 2  j2 ae j 2




j 3
V

b

Sin


jb

Cos


j

be
V A  V BA  V B  0
BA
3
3
3
3
 3

V

d
B
Where
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Velocity Analysis of Offset Slider-Crank
 Separate Velocity Equation into Real and
Imaginary Parts
 a2 Sin 2  b3 Sin3  d  0
a2Cos 2  b3Cos 3  0
 From imaginary part:
b3Cos3  a2Cos 2
aCos 2
3 
2
bCos 3
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Real Part
Imaginary Part
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Velocity Analysis of Offset Slider-Crank
 From real part:
 a2 Sin 2  b3 Sin3  d
d  a2 Sin 2  b3 Sin3
d dot is horizontal speed of slider relative to ground
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Velocity Analysis of a Inverted Crank Slider
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Velocity Analysis of a Inverted Crank Slider
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Homework 5  http://facultad. bayamon.inter.edu/omeza/
Omar E. Meza Castillo Ph.D.
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¿Preguntas?
Comentarios
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GRACIAS
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