ME 301 Theory of Machines Ipp5v09

8/6/2019 ME 301 Theory of Machines Ipp5v09

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B0C is perpendicular to the slider axis

A0A=a2, A0B0=a1,B0C=a4

s =-a +a cos

Function InvSliderCrank2(Crank, Fixed, Eccentricity,Config, Theta)Dim A 2

sy= a2sin(12)

= ma s s

sX = Crank * Cos(Theta) - FixedsY = Crank * Sin(Theta)q = Mag(sX, sY)Fi = An sX sY

= tan-1(sx,sy)

2 2

S43 = Sqr(q ^ 2 - Eccentricity ^ 2)Eta = Acos(Eccentricity / q)A(0) = S43A 1 = Fi - Confi * Eta

43 4= 1 1 143 43sin cos tan

s sc

q q c

= = =

InvSlider2 = AEnd Function

2

14

=

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B0C is not perpendicular to the slider axis

Function InvSlider2(Crank, Fixed, Eccentricity, Alfa,

0 =a2, 0 0=a1, 0 =a4

sx=-a1+a2cos(12)

Config, Theta)'Determines the slider displacement and the outputlink angle for an inverted slider-crank mechanismDim A(2) As Double

sy= a2sin(12)q= mag(sx,sy)

Dim S, Fi, Si, Q, Delta As DoubleDim sx, sy As Doublesx = -Fixed + Crank * Cos(Theta)sy = Crank * Sin(Theta)

= tan-1(sx,sy)

2 2 2cos sins a a = +

S = Mag(sx, sy)Fi = Ang(sx, sy)Delta = S ^ 2 - Eccentricity ^ 2 * Sin(Alfa) 2

Q = -Eccentricity * Cos(Alfa) + Sqr(Delta)

1

4 43 43tan ( cos , sin )a s s = +

Si = Ang((Eccentricity + Q * Cos(Alfa)), Q * Sin(Alfa))A(0) = QA(1) = Fi - Config * SiInvSlider2 = A

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14

= End Function


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A

Example

212

A0

Given:

All the necessary

4 link dimensions andthe input crank

angle 12

6B0

5

B

All the position variables and the displacement of the slider 6 as a functionof input crank angle

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1. Consider the loop formed by

Alinks 1,2,3 and 4. An invertedslider-crank mechanism.

12A

y 14= 2, 1, , , 12+ -

4

xa=a2cos 12, ya=a1+a2s n 12

43( , )

a as mag x y=

4

x

114 tan ( , )a ax y =

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4

12A0 2. Consider the loop1,4,5,6. A slider Crank

C

14

ec an sm

B0

5

B

s14=-{SliderCrank(a4,a5,0,1,14)}

xb=-a4cos(14), yb=-a4sin(14)

x =s , =0

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Example:

D E

Given: All the necessary link length

66 5

.Let:

a2 = A0A=A0A

A'CB

3 , 3a4 = B0Ba5 = CD

=

A02 34

b6=DE

Also as shown on the fi ure the

A

B0

fixed dimensions x1, y1 and thefixed angle 6 are known.

Determine:

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comp e e pos on ana ys s o e mec an sm


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1. Solve the four-bar Loop (links, , ,

1 1 1

1

anda x y

= +

=1 1 1

12= 12+ 1-/2.

{FourBar2(a2,a3,a4,a1,1,12)- 1}

13 14

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( )2 12sinax a =

( )( )

2 12

' 2 12

cossin

a

a a

y a x a x

= = =

( )' 2 12

1 4 14

cos

cos( )

a a

b

y a y

x x a

= =

= +

( )1 4 14

2 12 3 3 13 3 3 13

sin( )

sin ( ) cos( ) ( ) cos( )

b

c a

y y a

x a a b x a b

= +

= + + = + +

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2 12 3 3 13 3 3 13cos s n s nc a y a a y a= + + = + +


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2 2' '

( ) ( )q x x y y= + 1

' 'tan (( ), ( ))a c a c x x y y=

2 2 21 5 6q a a +

52qa

1 6 5cos2

a a q

a a

+ =

15 =

16 = 15 -

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' 6 16cos( )

d a x x a = +

' 6 16sin( )d a y y aor

= +

5 15

5 15

cos( )

sin( )

d c

d c

x x a

y y a

= +

= +

6 16 6 6 16 6cos( ) cos( )

e d d

and

x x b x b = + + = +

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6 16 6 6 16 6sin( ) sin( )e d d y x b x b = + + = +


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Or Consider Triangle ADE on link 6

2 26 6 6 6 6 6

2 2 2

( 2 cos( ))c a b a b

a c b

= +

+ 61

6

6 6

cos2a c

=

' 6 16 6cos( )e a x x c = +

' 6 16 6sin( )e a y y c = +

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Example:D

5

A

6

3

C

2

4

1

0 D0A 12a

Given:

.

A0A=a2, AC=a3, BC=b3, BD=a5, D0D=a6, A0D0=b1,


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D

5

B6

13

5

C

3

4

13

0D0

A

2

12

16

14s

a1

1

1. Obtain 13, s14 from the SliderCrank2() function

[13, s14 ]= SliderCrank2(a2,a3,a1,1,12)

2 12cos

ax a =

2 12

14 3 13 3

s n

cos( )

a

b

y a

x s b

=

= +

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sin( )b y b a = +


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2 2

0 1 14 1( ) D C S b s a= = +D

51

( )11 1 14 1tan ( ),b s a = B

613

5

3

13 13 1 3 = + C

3

4

13

D0

16

14s

1

. 15, 16

[15, 16 ]= {FourBar2(b3,a5,a6,S, 1,13)-1}

1 6 16cos( )

sin

d x b a

a

= +

=15

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c1

A0A=a2,; A0G=b2 ; GF=a3; BA=a6 ;

DB0B=a5 B0C=b5 ; B0E=c5 ; EF=a5 ;CD=a5 ;


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A0ABB0A0 loop: A four-barc1

A0GFEB0A0 loop

B0CDB0 loop: SliderCrankD

7

8

E18

4

45

CB

17

15

6

F

B

b1

2

3

A

G

A

0

16

13

a1

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2. Obtain usin FourBar2 function

[16, 15 ]= {FourBar2(a2,a6,a6,d1, 1,(12-1))+1}B0

1

6

B

1

A

16

12

1

2A0

a1

[ ]1

2 2 1

1 1 1 1 1; tan ; (note the quadrant!!)d a b a b= + =

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=7

E

14

2 12 2

2 12 2sin( )

G

Gy b = +4515

5 B

1 5 15 5

1 5 15 5

cos

sin( )

E

E

x a c

y b c

= +

= +

F

b

1 6

2 2( ) ( ) E G E G

GE s x x y y = = + 3

A0

13

12

[ ]1

tan ( ); ( )cos

E G E G x x y yan s a a

= =

a1

2

4 3cos( ; ; )ang a a s =

13

14 13

=

=

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c1

D 11 =

8

7

18 5 15 5 7 17cos( ) sin

a

s b a

= + +

7E

s18

5

C

17

15

B

0

5

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B0A0ABB0 loop: A four-bar

4

A

A0ACD loop

B0BCD loop

3

C

2

50

6

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Analytical Closed Form solution of the Loop EquationsOur aim is to obtain 14 as a function of 12

A

3 B13

3a Loop Closure Equationfor the Four-Bar:

Loop Closure Equation

a e a e a a e2 3 1 412 13 14+ = +

24

144

a

A0 B0

12

2

a1 a e a e a a ei i i2 3 1 412 13 14 + = +

Mirror Image

12

14

(This is the complex conjugate of theloop equation)

13

These two equations yield twocomplex equations in the complexdomain. When we equate the real and

,equations yield two identical scalarequations.

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13A

3a B

3 Write the loop Equation and itsComplex conjugate in the form:

a14

2

2 a4

4 a e a a e a ei i i3 1 4 213 14 12 = +

12

a0

A

1

1 B0 a e a a e a ei i i3 1 4 213 14 12 = +

Multiply the two equations Side by side

i i i i i2 13 13 14 12 14 12( ) =

e ei i( ) ; = =0 1 cos ( ) / = + e ei i 2

K K K1 14 2 12 3 14 12cos cos cos( ) + =

2222

Freudensteins

Equation

2

1a

K =4

2a

K =24

4321

3aa2

K

=

K K Kcos cos cos cos sin sin + = +

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K K K1 14 2 12 3 14 12 14 12cos cos cos cos sin sin + = +

sin

tan( )

142

1

2=tan 2 141

1

2

tan 2 1411

2+

tan

14

2

1411

2+

A B Ctan tan2 14 14

2 20

+

+ =

A K K K= + cos ( )12 2 3 11

B = 2 12sin

13212

)1(cos KKKC +++=

tan( )14

2 4

=

B B AC

= ACBB )4(

tan2

2

1

14or

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K K K1 14 2 12 3 14 12 14 12cos cos cos cos sin sin + = +

3122121414121 KcosKsinsincos)cosK( =Let 2 2 2D K

D

cos cos

sin sin

=

=1 12

12

= =

=

1 12 12 1 1 12

112

cos s n cos

tansin

1 12

D cos cos14-D sin sin14=K2 cos12-K3

cos( ) cos cos sin sin + =

cos( )cos

142 12 3+ =

D

+=

D

KcosKcos 3122114

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A

3

y

13a

3

Loop Closure Eq. and its ComplexConjugate

1312 ii

1312

31142 ii eaiasea ++=A

0

2

1

a

4

12a

2

1

31142 x14s

Eliminate 13:

11423

11423

1213

1213

iaseaea

iaseaea

ii

ii

+=

=

Multiply:

)()( 121212122 2121421214223 iiii eeaiaeeasasaa ++++=

ii cos ( ) / = + e e 2 s n ee =

We obtain a quadratic equation in s14:

s ncos 122121221414 31 =++ aaaaaass

2222 sin2sincos aaaaaas +=

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Solving problems is a practical art, like

swimming or skiing, or playing piano: you

This book cannot offer you a magic key

problems, but it offers you good examples

or m tat on an many opportun t es orpractice: if you wish to learn swimming

you have to go into the water and if you

have to solve problems

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,


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P

30

15

AA

0 122

50

100

1

B0

3

4

150

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=

P

0

A0A=a2=3030

1

AQ=a3=150

A0B0=a1-ib1=100-i150

A0 122

Q141412

343112 eaesaea ++=

141412

343112

iiieiaesibaea

++=100150

14

B0

3

4

s4

In StepWise Solution:

150

. 0 0

2. Determine AB0 and its angular orientation

3. Solve Triangle AB0

Q

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Analytical Solution: Eliminate s43

141214 iii

343112 eaesaea ++=141412 iii eiaesibaea

++=

311243

141214 iii eiaibaeaes +=

)()( 1414141214141412 iiiiii Take the ratio and cross multiply

31123112

iiiiii

Group Terms:

3112 =

)(sin2 ii eei =

0cossin)sin( 314114114122 =++ abaa

31411411412214122 coss ns ncoscoss n aaaa =++

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3122114122114 cossinsincos aaaab =++


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Let

1221

1221

sinsin aaD =

( ) ( )

=

+++=++=

12211

12211221

2

1

2

2

2

1

2

1221

2

1221

cos

cos2sin2)cos(sin

aa

aaabaabaaabD

+ 1221 sinab

= 14 14 3

cos( ) cos cos sin sin + =

D

a314 )cos( =

++=

a3114 cos

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1. Stepwise method is an exact and efficient method for the analysis of planarmechanisms. It can be used in any computer package with mathematical

computation facilities (i.e. MathCad, MathLab, Excel, Eureka, etc.) andcan be programmed in any computer language (Pascal, C, Basic, etc) orplatform (Visual C++, Visual Basic, Delphi, etc), or you can use this method inyour ca cu a ors as we .

2. Graphical method gives a very good insight to a mechanism problem. The

Graphical method coincides exactly with the stepwise method. You can useany ra t ng pac age uto a , a ey , ro es gn , eas , o or s

, etc.) In most of these files you can perform kinematic analysis.

3. Analytical solutions are possible for simple single loop mechanisms. They canalso be used similar to stepwise solution if there are several loops involved.

4. Analytical solutions are sometimes required when performing more complexmechanism analysis or synthesis.

However

which cannot be solved analytically or using stepwise procedure.

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Example


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p

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Example


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BC=a4,CD=c4,

BD=b4

1512 14

23 4 1 1 5

ii i

s e a e b ia a e

+ = + +12 14

23 4 1 1 1 16( ) ( )i i

s e b e b c i a s

or

+ = + + +Requires simultaneous solution

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15 14

5 4 1 16

i i

a e c e c is

+ = +

o e wo oops

Example


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Examplea6

D 6

16s

E

14

113

A

B

43

=602

12

aa

Ab115

5

0a

C C0a1

13 1512 14

2 3 1 1 5 4

i ii ia e a e a ib a e b e

+ = + +

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1312 14

2 3 4 1 16 6( )a e a e a e ic s a+ + = +

Iterative Solution of the Loop Closure Equations


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p q

3a B

13A

12

a14

A

2

2 a4

0

1

0

141312 +=+ eaaeaea iii

0),()cos()cos()cos( 141311221133144 ==+ faaaa

0),()sin()sin()sin( 141321221133144 == faaa

.(such as fixed point or Newton-Raphson method) to solve the set ofnonlinear equations

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In Excel: Use Solver ADD-in to solve the set ofnonlinear e uations

In MathCad Use given, Find routine.

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ME 301 Theory of Machines Ipp5v09

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Transcript of ME 301 Theory of Machines Ipp5v09