Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Design of Machinery An Introduction to the Synthesis and Analysis of

Mechanisms and Machines Fourth Edition

Robert L. Norton

Chapter 4

Image Slides

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4-2





No se puede mostrar la imagen. Puede que su equipo no tenga suficiente memoria para abrir la imagen o que ésta esté dañada. Reinicie el equipo y, a continuación, abra el archivo de nuevo. Si sigue apareciendo la x roja, puede que tenga que borrar la imagen e insertarla de nuevo.







R1 +R4 = RSR2 +R3 = RS

R1 = diR2 = acos(θ2 )i + asin(θ2 )j

R3 = bcos(θ3)i + bsin(θ3)j R4 = cj

R2 +R3 = R1 +R4


R2 +R3 = R1 +R4

acos(θ2 )+ bcos(θ3) = d

asin(θ2 )+ bsin(θ3) = c


R2 +R3 = R1 +R4

f1(θ3,d) = acos(θ2 )+ bcos(θ3)− d = 0

f2 (θ3,d) = asin(θ2 )+ bsin(θ3)− c = 0


R2 +R3 = R1 +R4

f1(θ3,n+1,dn+1) ! f1(θ3,n ,dn )+

∂ f1(θ3,n ,dn )∂θ3

(θ3,n+1 −θ3,n )+∂ f1(θ3,n ,dn )

∂d(dn+1 − d3,n )

f2 (θ3,n+1,dn+1) ! f2 (θ3,n ,dn )+

∂ f2 (θ3,n ,dn )∂θ3

(θ3,n+1 −θ3,n )+∂ f2 (θ3,n ,dn )

∂d(dn+1 − d3,n )


∂ f1(θ3,n ,dn )∂θ3

∂ f1(θ3,n ,dn )∂d

∂ f2 (θ3,n ,dn )∂θ3

∂ f2 (θ3,n ,dn )∂d

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

(θ3,n+1 −θ3,n )(dn+1 − d3,n )

⎡

⎣⎢⎢

⎤

⎦⎥⎥= −

f1(θ3,n ,dn )f2 (θ3,n ,dn )

⎡

⎣⎢⎢

⎤

⎦⎥⎥


−bsin(θ3,n ) −1bcos(θ3,n ) 0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

(θ3,n+1 −θ3,n )(dn+1 − dn )

⎡

⎣⎢⎢

⎤

⎦⎥⎥= −

f1(θ3,n ,dn )f2 (θ3,n ,dn )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

f1(θ3,n ,d) = acos(θ2 )+ bcos(θ3,n )− dn = 0

f2 (θ3,n ,dn ) = asin(θ2 )+ bsin(θ3,n )− c = 0


−bsin(30) −1bcos(30) 0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

(θ3,1 −θ3,0 )(d1 − d0 )

⎡

⎣⎢⎢

⎤

⎦⎥⎥= −

f1(θ3,0 ,d0 )f2 (θ3,0 ,d0 )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

θ2 = 61º

f2 (θ3,0 ,d0 ) = asin(61)+ bsin(30)− (5 + 5 3)

a = 10b = 10

c = 10sin(60)+10cos(30) = 5 + 5 3θ3,0 = 30º

d0 = 10cos(60)+10cos(30) = 5 3 + 5

f1(θ3,0 ,d0 ) = acos(61)+ bcos(30)− (5 + 5 3)


a = 10b = 10

c = 10sin(60)+10cos(30) = 5 + 5 3θ3,0 = 30º

d0 = 10cos(60)+10cos(30) = 5 3 + 5

1 2 3 4 theta3 29.43140546692

6996 29.433015864732752

29.433015877501717

29.433015877501717

d 13.557969473862578

13.557404108920629

13.557404104385316

13.557404104385316

θ2 = 61º


a = 10b = 10

c = 10sin(60)+10cos(30) = 5 + 5 3θ3,0 = 29.433015877501717ºd0 = 13.557404104385316


a = 10b = 10

c = 10sin(60)+10cos(30) = 5 + 5 3θ3,0 = 29.433015877501717ºd0 = 13.557404104385316

1 2 3 4 theta3 28.88515064025

2387 28.886612382393817

28.886612392681030

28.886612392681030

d 13.451011999239871

13.450489889694857

13.450489885978037

13.450489885978035

θ2 = 62º


a = 10b = 10

c = 10sin(60)+10cos(30) = 5 + 5 3θ3,0 = 29.433015877501717ºd0 = 13.557404104385316

θ2 = 62º


f1(θ3,d) = acos(θ2 )+ bcos(θ3)− d = 0

f2 (θ3,d) = asin(θ2 )+ bsin(θ3)− c = 0

!f1 = −asin(θ2 ) !θ2 − bsin(θ3) !θ3 − !d = 0

!f2 = acos(θ2 ) !θ2 + bcos(θ3) !θ3 = 0


!f1 = −asin(θ2 ) !θ2 − bsin(θ3) !θ3 − !d = 0

!f2 = acos(θ2 ) !θ2 + bcos(θ3) !θ3 = 0

asin(θ2 ) !θ2 = −bsin(θ3) !θ3 − !d

−acos(θ2 ) !θ2 = bcos(θ3) !θ3


−bsin(θ3) −1bcos(θ3) 0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

asin(θ2 )−acos(θ2 )

⎡

⎣⎢⎢

⎤

⎦⎥⎥!θ2



⎡

⎣⎢⎢

⎤

⎦⎥⎥

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=


⎡

⎣⎢⎢

⎤

⎦⎥⎥!θ2

a = 10b = 10

c = 5 + 5 3θ3 = 30º

d = 5 3 + 5θ2 = 60º

−10sin(30) −110cos(30) 0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

10sin(60)−10cos(60)

⎡

⎣⎢⎢

⎤

⎦⎥⎥!θ2


a = 10b = 10

c = 5 + 5 3θ3 = 30º

d = 5 3 + 5θ2 = 60º

−10sin(30) −110cos(30) 0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

10sin(60)−10cos(60)

⎡

⎣⎢⎢

⎤

⎦⎥⎥!θ2

!θ2 = 2π

radss

−10sin(30) −110cos(30) 0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

10sin(60)−10cos(60)

⎡

⎣⎢⎢

⎤

⎦⎥⎥2π


a = 10b = 10

c = 5 + 5 3θ3 = 30º

d = 5 3 + 5θ2 = 60º

!θ2 = 2π

radss

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

−10sin(30) −110cos(30) 0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−110sin(60)−10cos(60)

⎡

⎣⎢⎢

⎤

⎦⎥⎥2π



⎡

⎣⎢⎢

⎤

⎦⎥⎥

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=


⎡

⎣⎢⎢

⎤

⎦⎥⎥!θ2

θ2 = 61º

−10sin(29.433015877501717º ) −110cos(29.433015877501717º ) 0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

10sin(61)−10cos(61)

⎡

⎣⎢⎢

⎤

⎦⎥⎥!θ2

a = 10b = 10

c = 5 + 5 3θ3 = 29.433015877501717ºd = 13.557404104385316


a = 10b = 10

c = 5 + 5 3θ3 = 29.433015877501717ºd = 13.557404104385316

θ2 = 61º

!θ2 = 2π

radss

−10sin(29.433015877501717º ) −110cos(29.433015877501717º ) 0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

10sin(61)−10cos(61)

⎡

⎣⎢⎢

⎤

⎦⎥⎥!θ2

−10sin(29.433015877501717º ) −110cos(29.433015877501717º ) 0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

10sin(61)−10cos(61)

⎡

⎣⎢⎢

⎤

⎦⎥⎥2π


a = 10b = 10c = 13.557404104385316θ3 = 29.433015877501717ºd = 13.557404104385316

θ2 = 61º

!θ2 = 2π

radss

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

−10sin(29.433015877501717º ) −110cos(29.433015877501717º ) 0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−110sin(61)−10cos(61)

⎡

⎣⎢⎢

⎤

⎦⎥⎥2π


a = 10b = 10c = 13.557404104385316θ3 = 29.433015877501717ºd = 13.557404104385316

θ2 = 61º

!θ2 = 2π

radss

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥= -3.497234014429895

-37.773686031467101)⎡

⎣⎢⎢

⎤

⎦⎥⎥



⎡

⎣⎢⎢

⎤

⎦⎥⎥

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=


⎡

⎣⎢⎢

⎤

⎦⎥⎥!θ2

θ2 = 62º

−10sin(28.886612392681030º ) −110cos(28.886612392681030º ) 0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

10sin(62)−10cos(62)

⎡

⎣⎢⎢

⎤

⎦⎥⎥!θ2

a = 10b = 10

c = 5 + 5 3θ3 = 28.886612392681030d = 13.450489885978035


a = 10b = 10

c = 5 + 5 3θ3 = 28.886612392681030d = 13.450489885978035

θ2 = 62º

!θ2 = 2π

radss

−10sin(28.886612392681030º ) −110cos(28.886612392681030º ) 0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

10sin(62)−10cos(62)

⎡

⎣⎢⎢

⎤

⎦⎥⎥!θ2

−10sin(29.433015877501717º ) −110cos(29.433015877501717º ) 0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

10sin(62)−10cos(62)

⎡

⎣⎢⎢

⎤

⎦⎥⎥2π


a = 10b = 10

c = 5 + 5 3θ3 = 28.886612392681030ºd = 13.450489885978035

θ2 = 62º

!θ2 = 2π

radss

!θ3!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

−10sin(28.886612392681030º ) −110cos(28.886612392681030º ) 0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−110sin(62)−10cos(62)

⎡

⎣⎢⎢

⎤

⎦⎥⎥2π


a = 10b = 10

c = 5 + 5 3θ3 = 28.886612392681030ºd = 13.450489885978035

θ2 = 62º

!θ2 = 2π

radss

!θ3

!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

-3.368950293230393 -39.202582096901800⎡

⎣⎢

⎤

⎦⎥


!f1 = −asin(θ2 ) !θ2 − bsin(θ3) !θ3 − !d = 0

!f2 = acos(θ2 ) !θ2 + bcos(θ3) !θ3 = 0

!!f1 = −acos(θ2 ) !θ2

2 − asin(θ2 )!!θ2 − bcos(θ3) !θ32 − bsin(θ3)!!θ3 − !!d = 0

!f2 = −asin(θ2 ) !θ2

2 + acos(θ2 )!!θ2 − bsin(θ3) !θ32 + bcos(θ3)!!θ3 = 0


!!f1 = −acos(θ2 ) !θ2

2 − asin(θ2 )!!θ2 − bcos(θ3) !θ32 − bsin(θ3)!!θ3 − !!d = 0

!f2 = −asin(θ2 ) !θ2

2 + acos(θ2 )!!θ2 − bsin(θ3) !θ32 + bcos(θ3)!!θ3 = 0

bsin(θ3) 1−bcos(θ3) 0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

!!θ3!!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

−acos(θ2 ) !θ22 − asin(θ2 )!!θ2 − bcos(θ3) !θ3

2

−asin(θ2 ) !θ22 + acos(θ2 )!!θ2 − bsin(θ3) !θ3

2

⎡

⎣⎢⎢

⎤

⎦⎥⎥



⎡

⎣⎢⎢

⎤

⎦⎥⎥

!!θ3!!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

−acos(θ2 ) !θ22 − asin(θ2 )!!θ2 − bcos(θ3) !θ3

2


2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

!!θ3!!d

⎡

⎣⎢⎢

⎤

⎦⎥⎥=


⎡

⎣⎢⎢

⎤

⎦⎥⎥

−1−acos(θ2 ) !θ2

2 − asin(θ2 )!!θ2 − bcos(θ3) !θ32


2

⎡

⎣⎢⎢

⎤

⎦⎥⎥


a = 2b = 10

c = 2sin(60)+10sin(30) = 3 + 5

d = 2cos(60)+10cos(30) = 3 + 5θ2 = 60ºθ3 = 30º

!θ2 = 2πrads

!!θ2 = 0rads2












REACCIONES Y MOMENTOS DE ACTUACIÓN


FOx + FAx = m1acm1xFOy + FAy = m1acm1y +m1g−FAx = m2acm2x−FAy + FBy = m2acm2y +m2g

Suma de fuerzas para los cuerpos a y b


L12sin(θ )FOx −

L12sin(θ )FAx −

L12cos(θ )FOy +

L12cos(θ )FAy = Icm1!!θ

− L22sin(β )FAx +

L22cos(β )FAy +

L22cos(β )FOy = Icm2 !!β

Suma de momentos en los cuerpos a y b


1 1 0 0 0 00 0 1 1 0 00 −1 0 0 0 00 0 0 −1 1 0

L12sin(θ ) − L1

2sin(θ ) − L1

2cos(θ ) L1

2cos(θ ) 0 1

0 − L22sin(β ) 0 L2

2cos(β ) L2

2cos(β ) 0

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

FOxFAxFOyFAyFByM

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

=

m1acm1xm1acm1y +m1g

m2acm2xm2acm2y +m2g

Icm1!!θ

Icm2 !!β

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥



FOxFAxFOyFAyFByM

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

=

1 1 0 0 0 00 0 1 1 0 00 −1 0 0 0 00 0 0 −1 1 0

L12sin(θ ) − L1

2sin(θ ) − L1

2cos(θ ) L1

2cos(θ ) 0 1

0 − L22sin(β ) 0 L2

2cos(β ) L2

2cos(β ) 0

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

−1

m1acm1xm1acm1y +m1g

m2acm2xm2acm2y +m2g

Icm1!!θ

Icm2 !!β

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥




FOx + FAx = m1acm1xFOy + FAy = m1acm1y +m1g−FAx + FBx = m2acm2x−FAy + FBy = m2acm2y +m2g−FBx + FCx = m3acm3x−FBy + FCy = m3acm2y +m3g


L12sin(θ )FOx −

L12sin(θ )FAx −

L12cos(θ )FOy +

L12cos(θ )FAy = Icm1!!θ

− L22sin(β )FAx +

L22cos(β )FAy −

L22sin(β )FBx +

L22cos(β )FBy = Icm2 !!β

L32sin(α )FBx −

L32cos(α )FCy +

L32sin(α )FDx −

L32cos(α )FDy = Icm3 !!α


1 1 0 0 0 0 0 0 00 0 1 1 0 0 0 0 00 −1 0 0 1 0 0 0 00 0 0 −1 0 0 1 0 00 0 0 0 −1 1 0 0 00 0 0 0 0 0 −1 1 0

L12sin(θ ) − L1

2sin(θ ) − L1

2cos(θ ) L1

2cos(θ ) 0 0 0 0 1

0 0 0 0 L32sin(α ) L3

2sin(α ) − L3

2cos(α ) − L3

2cos(α ) 0

0 − L22sin(β ) 0 L2

2cos(β ) − L2

2sin(β ) 0 L2

2cos(β ) 0 0

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

FOxFAxFOyFAyFBxFCxFByFCyM

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

m1acm1xm1acm1y +m1g

m2acm2xm2acm2y +m2g

m3acm3xm3acm3y +m3g

Icm1!!θIcm3 !!α

Icm2 !!β

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥


FOxFAxFOyFAyFBxFCxFByFCyM

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

1 1 0 0 0 0 0 0 00 0 1 1 0 0 0 0 00 −1 0 0 1 0 0 0 00 0 0 −1 0 0 1 0 00 0 0 0 −1 1 0 0 00 0 0 0 0 0 −1 1 0

L12sin(θ ) − L1

2sin(θ ) − L1

2cos(θ ) L1

2cos(θ ) 0 0 0 0 1

0 0 0 0 L32sin(α ) L3

2sin(α ) − L3

2cos(α ) − L3

2cos(α ) 0

0 − L22sin(β ) 0 L2

2cos(β ) − L2

2sin(β ) 0 L2

2cos(β ) 0 0

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

−1

m1acm1xm1acm1y +m1g

m2acm2xm2acm2y +m2g

m3acm3xm3acm3y +m3g

Icm1!!θIcm3 !!α

Icm2 !!β

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥



f1(θ2,r2 ) = r1 cos(θ1)+ r2 cos(θ2 )− Lf2 (θ2,r2 ) = r1 sin(θ1)− r2 sin(θ2 )

cos(θ2,n ) −r2,n sin(θ2,n )−sin(θ2,n ) −r2,n cos(θ2,n )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

r2,n+1 − r2,nθ2,n+1 −θ2,n

⎡

⎣⎢⎢

⎤

⎦⎥⎥= −

r1 cos(θ1)+ r2,n cos(θ2,n )− Lr1 sin(θ1)− r2,n sin(θ2,n )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Análisis de posición


f1(θ2,r2 ) = r1 cos(θ1)+ r2 cos(θ2 )− Lf2 (θ2,r2 ) = r1 sin(θ1)− r2 sin(θ2 )

!f1(θ2,r2 ) = −r1 sin(θ1) !θ1 + !r2 cos(θ2 )− r2 sin(θ2 ) !θ2 = 0!f2 (θ2,r2 ) = r1 cos(θ1) !θ1 − !r2 sin(θ2 )− r2 cos(θ2 ) !θ2 = 0

!r2 cos(θ2 )− r2 sin(θ2 ) !θ2 = r1 sin(θ1) !θ1− !r2 sin(θ2 )− r2 cos(θ2 ) !θ2 = r1 cos(θ1) !θ1

cos(θ2 ) −r2 sin(θ2 )−sin(θ2 ) −r2 cos(θ2 )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

!r2!θ2

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

sin(θ1)cos(θ1)

⎡

⎣⎢⎢

⎤

⎦⎥⎥r1 !θ1

Análisis de velocidad


Análisis de aceleración

cos(θ2 ) −r2 sin(θ2 )sin(θ2 ) r2 cos(θ2 )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

!r2!θ2

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

cos(θ2 )sin(θ2 )

⎡

⎣⎢⎢

⎤

⎦⎥⎥r2 !θ2

2 +cos(θ1)−sin(θ1)

⎡

⎣⎢⎢

⎤

⎦⎥⎥r1 !θ1

2 +2sin(θ2 )−2cos(θ2 )

⎡

⎣⎢⎢

⎤

⎦⎥⎥!r2 !θ2 +

sin(θ1)cos(θ1)

⎡

⎣⎢⎢

⎤

⎦⎥⎥r1!!θ1


Análisis de fuerzas Fx∑ = macmx

Fy∑ = macmy

Rueda 1 Fx∑ = FAx + FB sin(θ2 ) = 0

Rueda 2 Fx∑ = FCx − FB sin(θ2 ) = 0

Fy∑ = FAy −m1g + FB cos(θ2 ) = 0

Fy∑ = FCy −m2g − FB cos(θ2 ) = 0


Análisis de momentos

M∑ = I !!θ

Rueda 1

M∑ = (r1 cos(θ1)i + r1 sin(θ1) j)× (FB sin(θ2 )i + FB cos(θ2 ) j)+M = I1!!θ1

Rueda 2

M∑ = (−r2 cos(θ2 )i + r2 sin(θ2 ) j)× (−FB sin(θ2 )i − FB cos(θ2 ) j) = I2 !!θ2

M∑ = r1FB cos(θ1)cos(θ2 )− r1FB sin(θ1)sin(θ2 )+M = I1!!θ1

M∑ = r2 cos(θ2 )FB = I2 !!θ2


Ecuaciones de fuerzas y momentos

1 0 0 0 sin(θ2 ) 00 1 0 0 cos(θ2 ) 00 0 1 0 −sin(θ2 ) 00 0 0 1 −cos(θ2 ) 00 0 0 0 r1 cos(θ1)cos(θ2 )− r1 sin(θ1)sin(θ2 ) 10 0 0 0 r2 0

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

FAxFAyFCxFCyFBM

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

=

0000I1!!θ1I2 !!θ2

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥



FAxFAyFCxFCyFBM

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

=

1 0 0 0 sin(θ2 ) 00 1 0 0 cos(θ2 ) 00 0 1 0 −sin(θ2 ) 00 0 0 1 −cos(θ2 ) 00 0 0 0 r1 cos(θ1)cos(θ2 )− r1 sin(θ1)sin(θ2 ) 10 0 0 0 r2 0

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

−1

0000I1!!θ1I2 !!θ2

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥



¿Cómo se implementa en Matlab?


Péndulo simple


Péndulo simple

Ft = −mgsin(θ ) = mat∑

at = l !!θ

−mgsin(θ ) = ml !!θ

!!θ + g

lsin(θ ) = 0


Péndulo simple

!!θ + g

lsin(θ ) = 0

Suponemos que el ángulo siempre será

pequeño

θ ≈ sin(θ )

La ecuación diferencial la

podemos escribir como sigue

!!θ + g

lθ = 0


Péndulo simple

!!θ + g

lθ = 0

La solución de la ecuación diferencial

esta dada por:

θ(t) = Asin(ωt +φ)

A,φ

Dependen de

θ(0), !θ(0)

ω = gl


Péndulo simple

!!θ + g

lθ = 0

θ(t) = Asin(ωt +φ)

θ(0) = π6

!θ(0) = 0π6= Asin(φ)

Aω cos(φ) = 0

A = π6

φ = π2


Péndulo simple

!!θ + g

lθ = 0

θ(t) = π6sin(ωt + π

2)


Péndulo simple

!!θ + g

lθ = 0

θ(t) = π6sin(ωt + π

2)


Péndulo simple

!!θ + g

lsin(θ ) = 0

¿Qué pasa cuando el

ángulo no es pequeño?


Péndulo simple

!!θ + g

lsin(θ ) = 0



x = x0 + v0t +12at 2

Recuerdan la siguiente ecuación?


Péndulo simple

!!θ + g

lsin(θ ) = 0



Recuerdan la siguiente ecuación?

θn+1 = θn + !θnt +

12!!θnt

2


Péndulo simple

!!θ + g

lsin(θ ) = 0

!θn+1 =θn+1 −θn

Δtθn+1 = θn + !θn+1Δt

!!θn =!θn+1 − !θn

Δt!θn+1 = !θn + !!θnΔtθn+1 = θn + ( !θn + !!θnΔt)Δtθn+1 = θn + !θnΔt + !!θnΔt

2


Péndulo simple

!!θ + g

lsin(θ ) = 0

θn+1 = θn + !θnΔt + !!θnΔt2

!!θn = − glsin(θn )

De donde

θn+1 = θn + !θnΔt −

glsin(θn )Δt

2

Obtuvimos una ecuación que nos permite obtener una aproximación

de la solución de la ecuación


Péndulo simple

θn+1 = θn + !θnΔt −

glsin(θn )Δt

2


Péndulo simple


Péndulo físico

Considere el péndulo como un cuerpo rígido, que tiene masa e inercia, plantee las ecuaciones y mire la diferencia en comparación al péndulo

simple


Péndulo físico

Fr∑ = Rr −mgcos(θ ) = 0

Ft∑ = Rt −mgsin(θ )+ F = mat = m!!θL2

M∑ = −RtL2+ F L

2= I !!θ

Rr = mgcos(θ )

Rt = m!!θL2− F +mgsin(θ )

−RtL2+ F L

2= I !!θ


Péndulo físico

Rr = mgcos(θ )

Rt = m !θ2 L2− F +mgsin(θ )

−RtL2+ F L

2= I !!θ

I !!θ +m!!θ L

2

4+mg L

2sin(θ ) = 3F L

2


Péndulo físico

Rr = mgcos(θ )

Rt = m !θ2 L2− F +mgsin(θ )

−RtL2+ F L

2= I !!θ

I +m L2

4⎛⎝⎜

⎞⎠⎟!!θ +mg L

2sin(θ ) = 3F L

2

!!θ +mg L

2

I +m L2

4⎛⎝⎜

⎞⎠⎟

sin(θ ) =3F L2

I +m L2

4⎛⎝⎜

⎞⎠⎟


Péndulo físico

!!θ +mg L

2

m L2

12+m L2

4⎛⎝⎜

⎞⎠⎟

sin(θ ) =3F L2

m L2

12+m L2

4⎛⎝⎜

⎞⎠⎟

!!θ +mg L

2

m L2

3

sin(θ ) =3F L2

m L2

3

!!θ + 32gLsin(θ ) = 9

2FmL


Péndulo físico

!!θ + 32gLsin(θ ) = 9

2F cos(ωt)

mL

Función coseno

ω n =32gL

Frecuencia natural


Péndulo físico

!!θ + 32gLsin(θ ) = 9

2F cos(ωt)

mL

ω <ω n =32gL


Péndulo físico

!!θ + 32gLsin(θ ) = 9

2F cos(ωt)

mL ω <ω n =32gL


Péndulo físico

!!θ + 32gLsin(θ ) = 9

2F cos(ωt)

mL ω =ω n =32gL


Péndulo físico

!!θ + 32gLsin(θ ) = 9

2F cos(ωt)

mL ω >ω n =32gL


Péndulo físico

!!θ + 32gLsin(θ ) = 9

2Fe−kt

mL


Péndulo físico

!!θ + 32gLsin(θ ) = 9

2Fe−kt

mL


Péndulo doble

x1 = l1 sin(θ1)y1 = −l1 cos(θ1)

x2 = l1 sin(θ1)+ l2 sin(θ2 )y2 = −l1 cos(θ1)− l2 cos(θ2 )

V = m1gy1 +m2gy2= −m1gl1 cos(θ1)−m2gl1 cos(θ1)−m2gl2 cos(θ2 )

Energía potencial


Péndulo doble

x1 = l1 sin(θ1)y1 = −l1 cos(θ1)

x2 = l1 sin(θ1)+ l2 sin(θ2 )y2 = −l1 cos(θ1)− l2 cos(θ2 )

Energía cinética

!x1 = l1 cos(θ1) !θ1!y1 = l1 sin(θ1) !θ1

!x2 = l1 cos(θ1) !θ1 + l2 cos(θ2 ) !θ2!y2 = l1 sin(θ1) !θ1 + l2 sin(θ2 ) !θ2

K = 1

2m1( !x1

2 + !y12 )+ 1

2m2 ( !x2

2 + !y22 )


Péndulo doble Energía cinética

K = 12m1l1

2 !θ12 (sin2(θ1)+ cos

2(θ1))+12m2 ( !x2

2 + !y22 )

!x22 = (l1 cos(θ1) !θ1 + l2 cos(θ2 ) !θ2 )

2

= l12 !θ1

2 cos2(θ1)+ l22 !θ2

2 cos2(θ2 )+ 2l1l2 cos(θ1)cos(θ2 ) !θ1 !θ2!y22 = (l1 sin(θ1) !θ1 + l2 sin(θ2 ) !θ2 )

2

= l12 !θ1

2 sin2(θ1)+ l22 !θ2

2 sin2(θ2 )+ 2l1l2 sin(θ1)sin(θ2 ) !θ1 !θ2!x22 + !y2

2 = l12 !θ1

2 + l22 !θ2

2 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 )

!x1 = l1 cos(θ1) !θ1!y1 = l1 sin(θ1) !θ1

!x2 = l1 cos(θ1) !θ1 + l2 cos(θ2 ) !θ2!y2 = l1 sin(θ1) !θ1 + l2 sin(θ2 ) !θ2


Péndulo doble Energía cinética

K = 12m1l1

2 !θ12 (sin2(θ1)+ cos

2(θ1))

+ 12m2 (l1

2 !θ12 + l2

2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))


Péndulo doble Energía cinética Y

Potencial

K = 12m1l1

2 !θ12 (sin2(θ1)+ cos

2(θ1))

+ 12m2 (l1

2 !θ12 + l2

2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))


ℓ = K −V



Potencial

K = 12m1l1

2 !θ12

+ 12m2 (l1

2 !θ12 + l2

2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))


∂K∂ !θ1

= m1l12 !θ1 +m2l1

2 !θ1 +m2l1l2 !θ2 cos(θ1 −θ2 )

∂K∂ !θ2

= m2l22 !θ2 +m2l1l2 !θ1 cos(θ1 −θ2 )



Potencial

K = 12m1l1

2 !θ12

+ 12m2 (l1

2 !θ12 + l2

2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))


∂∂t

∂K∂ !θ1

⎛⎝⎜

⎞⎠⎟= m1l1

2 !!θ1 +m2l12 !!θ1 +m2l1l2 !!θ2 cos(θ1 −θ2 )

∂∂t

∂K∂ !θ2

⎛⎝⎜

⎞⎠⎟= m2l2

2 !!θ2 +m2l1l2 !!θ1 cos(θ1 −θ2 )



Potencial

K = 12m1l1

2 !θ12

+ 12m2 (l1

2 !θ12 + l2

2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))


∂K∂θ1

= −m2l1l2 !θ1 !θ2 sin(θ1 −θ2 )

∂K∂θ2

= m2l1l2 !θ1 !θ2 sin(θ1 −θ2 )



Potencial

K = 12m1l1

2 !θ12

+ 12m2 (l1

2 !θ12 + l2

2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))


∂V∂θ1

= m1gl1 sin(θ1)+m2gl1 sin(θ1)

∂V∂θ2

= m2gl2 sin(θ2 )



Potencial

K = 12m1l1

2 !θ12

+ 12m2 (l1

2 !θ12 + l2

2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))


∂∂t

∂K∂ !θ1

⎛⎝⎜

⎞⎠⎟− ∂∂t

∂V∂ !θ1

⎛⎝⎜

⎞⎠⎟− ∂K∂θ1

+ ∂V∂θ1

= 0

m1l12 !!θ1 +m2l1

2 !!θ1 +m2l1l2 !!θ2 cos(θ1 −θ2 )+m2l1l2 !θ1 !θ2 sin(θ1 −θ2 )+m1gl1 sin(θ1)+m2gl1 sin(θ1) = 0



Potencial

K = 12m1l1

2 !θ12

+ 12m2 (l1

2 !θ12 + l2

2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))


∂∂t

∂K∂ !θ2

⎛⎝⎜

⎞⎠⎟− ∂∂t

∂V∂ !θ2

⎛⎝⎜

⎞⎠⎟− ∂K∂θ2

+ ∂V∂θ2

= 0

m2l22 !!θ2 +m2l1l2 !!θ1 cos(θ1 −θ2 )−m2l1l2 !θ1 !θ2 sin(θ1 −θ2 )+m2gl2 sin(θ2 ) = 0



Potencial

K = 12m1l1

2 !θ12

+ 12m2 (l1

2 !θ12 + l2

2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))


m1l12 !!θ1 +m2l1

2 !!θ1 +m2l1l2 !!θ2 cos(θ1 −θ2 )+m2l1l2 !θ1 !θ2 sin(θ1 −θ2 )+m1gl1 sin(θ1)+m2gl1 sin(θ1) = 0m2l2

2 !!θ2 +m2l1l2 !!θ1 cos(θ1 −θ2 )−m2l1l2 !θ1 !θ2 sin(θ1 −θ2 )+m2gl2 sin(θ2 ) = 0



Potencial

K = 12m1l1

2 !θ12

+ 12m2 (l1

2 !θ12 + l2

2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))


(m1 +m2 )l12 !!θ1 +m2l1l2 !!θ2 cos(θ1 −θ2 )+m2l1l2 !θ1 !θ2 sin(θ1 −θ2 )+ gl1(m1 +m2 )sin(θ1) = 0

m2l22 !!θ2 +m2l1l2 !!θ1 cos(θ1 −θ2 )−m2l1l2 !θ1 !θ2 sin(θ1 −θ2 )+m2gl2 sin(θ2 ) = 0



Potencial

K = 12m1l1

2 !θ12

+ 12m2 (l1

2 !θ12 + l2

2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))


(m1 +m2 )l1!!θ1 +m2l2 !!θ2 cos(θ1 −θ2 )+m2l2 !θ1 !θ2 sin(θ1 −θ2 )+ g(m1 +m2 )sin(θ1) = 0m2l1!!θ1 cos(θ1 −θ2 )+m2l2 !!θ2 −m2l1 !θ1 !θ2 sin(θ1 −θ2 )+m2gsin(θ2 ) = 0



Potencial

K = 12m1l1

2 !θ12

+ 12m2 (l1

2 !θ12 + l2

2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))


(m1 +m2 )l1 m2l2 cos(θ1 −θ2 )m2l1 cos(θ1 −θ2 ) m2l2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

!!θ1!!θ2

⎡

⎣⎢⎢

⎤

⎦⎥⎥+

m2l2 !θ1 !θ2 sin(θ1 −θ2 )+ g(m1 +m2 )sin(θ1)

−m2l1 !θ1 !θ2 sin(θ1 −θ2 )+m2gsin(θ2 )

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 0

0⎡

⎣⎢

⎤

⎦⎥



Potencial

K = 12m1l1

2 !θ12

+ 12m2 (l1

2 !θ12 + l2

2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))


!!θ1!!θ2

⎡

⎣⎢⎢

⎤

⎦⎥⎥= −

(m1 +m2 )l1 m2l2 cos(θ1 −θ2 )m2l1 cos(θ1 −θ2 ) m2l2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−1m2l2 !θ1 !θ2 sin(θ1 −θ2 )+ g(m1 +m2 )sin(θ1)

−m2l1 !θ1 !θ2 sin(θ1 −θ2 )+m2gsin(θ2 )

⎡

⎣⎢⎢

⎤

⎦⎥⎥


Péndulo doble Discretización

diferencias finitas

!!θ1,n!!θ2,n

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥= −

(m1 +m2 )l1 m2l2 cos(θ1,n −θ2,n )m2l1 cos(θ1,n −θ2,n ) m2l2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−1m2l2 !θ1,n !θ2,n sin(θ1,n −θ2,n )+ g(m1 +m2 )sin(θ1,n )

−m2l1 !θ1,n !θ2,n sin(θ1,n −θ2,n )+m2gsin(θ2,n )

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

θ1,n+1 = θ1,n +!θ1,nΔt + !!θ1,nΔt

2

θ2,n+1 = θ2,n +!θ2,nΔt + !!θ2,nΔt

2

!θ1,n+1 = !θ1,n + !!θ1,nΔt

!θ2,n+1 = !θ2,n + !!θ2,nΔt


Péndulo doble Discretización

diferencias finitas

!!θ1,n!!θ2,n

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥= −

(m1 +m2 )l1 m2l2 cos(θ1,n −θ2,n )m2l1 cos(θ1,n −θ2,n ) m2l2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−1m2l2 !θ1,n !θ2,n sin(θ1,n −θ2,n )+ g(m1 +m2 )sin(θ1,n )

−m2l1 !θ1,n !θ2,n sin(θ1,n −θ2,n )+m2gsin(θ2,n )

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

θ1,n+1 = θ1,n +!θ1,nΔt + !!θ1,nΔt

2 θ2,n+1 = θ2,n +

!θ2,nΔt + !!θ2,nΔt2

!θ1,n+1 = !θ1,n + !!θ1,nΔt

!θ2,n+1 = !θ2,n + !!θ2,nΔt


Péndulo doble


Péndulo doble






















Documents

Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image