Four bar linkage

Indian Institute of Technology Gandhinagar Dept. of Mechanical Engineering

ME304: KINEMATICS AND DYNAMICS OF MACHINES L09: POSITION ANALYSIS OF MECHANISMS

Dr. Murali Damodaran [email protected]

Spring Semester of AY2011-2012 MD/AY2011-2012

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LECTURE 09 23 January 2012 ME304: Kinematics and Dynamics of Machines Introduction to Mechanisms. ..continuing with Kinematic Fundamentals Position, Velocity and Acceleration Analysis. (Kinematics) Analytical Method for Position Synthesis (The basis for computer modeling and synthesis of linkages) Design of Cam Follower Mechanisms. Gear tooth profiles, Spur gears and Helical gears,Epicyclic Gear Trains Belt drives Dynamic Analysis of Mechanisms (Dynamics) Balancing Analysis and Applications of Discrete and Continuous System Vibration

Course Website @ https://sites.google.com/a/iitgn.ac.in/me304/ MD/AY2011-2012

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Coordinate Systems for Analysis of Planar Mechanism

POSITION ANALYSIS OF MECHANISMS Cartesian:  RX , RY  Polar: (RA , )

• Converting between the two

RA  RX 2  RY 2

  arctan  RY RX 

RX  RA cos 

o

RY  RA sin 

Y

• Position Difference, Relative position – Difference (one point, two times) – relative (two points, same time) RBA=RB-RA MD/AY2011-2012

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A RA

o

RBA

B

RB X

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Position Analysis of Fourbar Linkage Mechanism

POSITION ANALYSIS OF MECHANISMS Given : The lengths of links a, b, c and d position of the ground link O2 O4 and the angle  2 Objective: Find 3 and  4

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Graphical Analysis of Fourbar Linkage

POSITION ANALYSIS OF MECHANISMS • Draw an arc of radius b, centered at A • Draw an arc of radius c, centered at O4

B1 b A

a • The intersections are the two possible positions for the linkage,  2 open and crossed O2

3

d

c

4 O4

B2 MD/AY2011-2012

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Algebraic Position Analysis of Fourbar Linkage

POSITION ANALYSIS OF MECHANISMS

Coordinates of Point B b   Bx  Ax    By  Ay  2

2

2

c   Bx  d   By 2 2

Coordinates of Point A:

2

Solve these equations to find

Ax  a cos  2

Bx and By

Ay  a sin  2

See Page 177-178 for solution MD/AY2011-2012

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Coordinate Systems for Analysis of Planar Mechanism

POSITION ANALYSIS OF MECHANISMS x’ y’

Coordinate Systems: GCS = Global Coordinate System, (X, Y) LNCS = Local Non-Rotating Coordinate System , (x, y) LRCS = Local Rotating Coordinate System , (x’, y’) MD/AY2011-2012

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Coordinate Systems for Airplane Fight Dynamics


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Non-planar Linkages-3D Spherical Linkages


http://synthetica.eng.uci.edu/~mcca rthy/Linkages.html MD/AY2011-2012

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Representation of Position Vectors


• For planar motion complex numbers on the real-imaginary plane can be used to model position vectors • Euler identity e±iθ=cos θ ± i sin θ (Note-Norton uses j instead of i in his text book i.e. e±jθ =cos θ ± j sin θ ) • Cartesian form: RAcos θ + i RAsin θ • Polar form: RAei θ

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Representation of Position Vectors


Multiplying a vector by ei corresponds to rotating the vector through  i

i

i 2

RAe  e  RAe

Rei R cos   iR sin  R  RA

RB  iR

For a rotation RC  i 2 R   R

through 90 degrees e

i

 2

 cos

 2

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 i sin

 2

i RD  i3 R  iR

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POSITION ANALYSIS OF MECHANISMS Write the vector loop equation:

R2  R3  R4  R1  0 (Positive from tail to tip)

Substitute with complex vectors

aei2  bei3  cei4  dei1  0 Split the knowns on one side and the unknowns on the other. Call the knowns Z i3 i 4 i 2 i1

be  ce

 ae  de  Z

Z  aei2  dei1 MD/AY2011-2012

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known KDM.12


POSITION ANALYSIS OF MECHANISMS bei3  cei4  aei2  dei1  Z Define s  ei3 and t  ei4 i3

be  ce

i 4

 bs  ct  Z

1 1  i 4 Define s  e = and t  e  i.e. conjugates s t and a conjugate based on known link lengths  i3

bs  ct  Z b c 1  Z noting that Z  s t Z MD/AY2011-2012

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POSITION ANALYSIS OF MECHANISMS bs  ct  Z b c  Z s t bs  ct  Z b c  Z s t

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1 noting that Z  Z

c  b   ct  Z    Z  t  c 2 2 b  c  ctZ  Z  ZZ t b 2t  c 2t  ct 2 Z  cZ  tZZ 2

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c 2t  ct 2 Z  t  c 2  b2  ZZ   cZ  0 Solve ct Z  t  c  b  ZZ   cZ  0 for t 2

  c  b  ZZ   2

t

2

2

2

c

2

 b  ZZ   4c ZZ 2

2

2

2cZ

ct  Z s b MD/AY2011-2012

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Solve ct 2 Z  t  c 2  b2  ZZ   cZ  0   c  b  ZZ   2

t

2

c

2

 b  ZZ   4c 2 ZZ 2

2

2cZ

ct  Z s b s  ei3 and t  ei4

Z  bei3  cei4 Z  be  i3  ce  i4  conjugate of Z 1 noting that Z  Z

Solve this in Matlab/Octave MD/AY2011-2012

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Algebraic Position Analysis of Fourbar Linkage in Octave


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KINEMATICS OF MECHANISMS

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Software Systems for Analysis of Various Linkages

POSITION ANALYSIS OF MECHANISMS R L NORTON’s SUITE OF PROGRAMS ACCOMPANYING THE TEXTBOOK Kinematics and Dynamics of Machinery

COMMERCIAL SOFTWARES FOR LINKAGE MECHANISMS WORKING MODEL 2D

FOURBAR FIVEBAR SIXBAR SLIDER DYNACAM ENGINE All are student SI editions

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SAM (Synthesis and Analysis of Mechanisms) MATLAB Simechanics Toolbox

Ch MECHANISMS Toolkit

AutoDesk Inventor In-Motion Module

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POSITION ANALYSIS OF MECHANISMS FOURBAR Program

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POSITION ANALYSIS OF MECHANISMS FOURBAR Program

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Algebraic Position Analysis Crank-Slider Linkage

POSITION ANALYSIS OF MECHANISMS • Given: link lengths a, b and c, θ1, θ2 (the motor position) • Find: the unknown angle θ3 and length d

R2  R3  R4  R1  0 aei2  bei3  cei4  dei1  0 MD/AY2011-2012

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Algebraic Position Analysis Crank-Slider Linkage

POSITION ANALYSIS OF MECHANISMS aei2  bei3  cei4  dei1  0 Using Euler Equivalents, collecting real and imaginary terms and setting each to zero results in:

a cos  2  b cos 3  c cos  4  d cos 1  0 a sin  2  b sin 3  c sin  4  d sin 1  0 As 1  0, these equations simplify to: a cos  2  b cos 3  c cos  4  d  0 a sin  2  b sin 3  c sin  4  0 MD/AY2011-2012

 a sin 2  c sin 4   3  sin   b   1

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Algebraic Position Analysis of Linkages with more than 4 bars

POSITION ANALYSIS OF MECHANISMS  a sin  2  c sin  4   b   Here c is the offset.

3  sin 1 

Initial set up of coordinate system for slider block such that

1  0 and  4  90 Hence one solution is:  a sin  2  c  3  sin   b   d  a cos  2  b cos 3 1

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The next valid solution taking into account the multi-valuedness of arcsin is  a sin  2  c  3  sin     b   d  a cos  2  b cos 3 1

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POSITION ANALYSIS OF MECHANISMS Hence one solution is:  a sin  2  c   b   d  a cos  2  b cos 3

3  sin 1 

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POSITION ANALYSIS OF MECHANISMS  a sin  2  c    b   d  a cos  2  b cos 3

3  sin 1  

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Algebraic Position Analysis of Inverted Crank-Slider Linkage

POSITION ANALYSIS OF MECHANISMS • Given: link lengths a, c and d, 1, 2 (the motor position), and g the angle between the slider and rod

• Find: the unknown angles 3 and 4 and length b

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2

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POSITION ANALYSIS OF MECHANISMS Write the vector loop equation:

R2  R3  R4  R1  0 (Positive from tail to tip) Substitute with complex vectors i 2

i3

i 4

i1

ae  be  ce  de  0

2

Since 3   4  g aei2  be 

i  4 g 

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 cei4  dei1  0 ME304

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POSITION ANALYSIS OF MECHANISMS i 4 g 

i2

ae  be

 cei4  dei1  0

• Grouping knowns and unknowns

bei4 g   cei4  aei2  dei1  Z i 2

Z  ae  de

i1

known

• Denoting s  ei4 , t  eig

Z  be

i  4 g 

 ce

2

i 4

Z  bei4 eig  cei4  bst  cs • Taking the conjugate to get the second equation MD/AY2011-2012

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1b  Z  bst  cs    c  s t  KDM.30


POSITION ANALYSIS OF MECHANISMS Z  bst  cs 1b  Z  bst  cs    c  s t  Multiply Z and Z to get:

1  2 ZZ  b  bc   t   c 2 t   1  2 Solve b  c   t  b  c 2  ZZ  0 for b t  2

2

1  1 2 c  t    c  t    4  c 2  ZZ   t  t b 2

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POSITION ANALYSIS OF MECHANISMS 1  Solve b  c   t  b  c 2  ZZ  0 for b t  2

2

 1  1 c  t    c 2  t    4  c 2  ZZ   t  t b 2 2

From Z  bst  cs Z s bt  c

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Algebraic Position Analysis of Geared Fivebar Linkage

POSITION ANALYSIS OF MECHANISMS • Consider Geared fivebar linkage • Write vector loop equation

R2  R3  R4  R5  R1  0

• Apply complex number representation i3

i2

i5

i4

ae  be  ce  de  fei1  0 Gear Ratio:  will relate 5 and  2 v via a phase angle  as follows:

5   2   i2

i3

i4

i ( 2  )

ae  be  ce  de MD/AY2011-2012

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i1

 fe  0 KDM.34



aei2  bei3  cei4  dei ( 2  )  fei1  0 Separate unknowns and knowns

bei3  cei4  aei2  dei ( 2  )  fei1  Z Denote Z  be

i3

 cei4

Define s  bei3 and t  cei4 bei3  cei4  bs  ct  Z b c Form Conjugate Z  bs  ct   s t

The remaining analysis follows exactly the same way as shown for the fourbar linkage

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Algebraic Position Analysis of Watt’s Sixbar Linkage


• Watt’s sixbar can be solved as 2 fourbar linkages • R1R2R3R4, then R5R6R7R8 • R4 and R5 have a constant angle between them

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Algebraic Position Analysis of Stephenson’s Sixbar Linkage


• Stephenson’s sixbar can sometimes be solved as a fourbar and then a fivebar linkage • R1R2R3R4, then R4R5R6R7R8 • R3 and R5 have a constant angle between them • If motor is at O6 you have to solve eqns. simultaneously

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Finding the position of any point on a linkage


• Once the unknown angles have been found it is easy to find any position on the linkage (relative to pivot O2)

• For point S Rs=sei( +d ) • For point P RP=aei  +pei ( +d ) • For point U RU=d +uei ( +d ) 2

2

2

3

4

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b c

3

a d

4

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y’

x’ Rp

RA

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Analysis of Transmission Angles of Fourbar Linkage


• Extreme value of transmission angle when links 1 and 2 are aligned.

 b 2  c 2   d  a 2  1  arccos   2bc   MD/AY2011-2012

 b 2  c 2   d  a 2  2    arccos   2bc   ME304

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Analysis of Toggle Positions of Fourbar Linkage


• Caused by the collinearity of links 3 and 4.

2

toggle

 a 2  d 2  b2  c 2 bc   cos    0   2toggle   2ad ad   1

3

3 Overlapped

4 2

Extended

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2

2

4

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Analysis of Toggle Positions of a Fourbar Linkage


• Caused by the collinearity of links 3 and 4. 2 2 2 2 bc  1  a  d  b  c  2toggle  cos    0   2toggle   2ad ad   • For a non-Grashof linkage, only one of the values of

 a 2  d 2  b2  c 2 bc     2ad ad   will be between –1 and 1

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Four bar linkage

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