Engranajes Helicoidales Aa.docx

  • Uploaded by: jean pierre
  • 0
  • 0
  • May 2020
  • PDF

This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA


Overview

Download & View Engranajes Helicoidales Aa.docx as PDF for free.

More details

  • Words: 2,577
  • Pages: 15
Engranajes cilΓ­ndricos de dientes helicoidales RPM ingreso = 1760 RPM salida = 490 Modulo = 3 Angulo de la hΓ©lice = 24ΒΊ RELACION DE TRANSMISION

𝑖=

𝑅𝑃𝑀𝑝 1760 = = 3.592 𝑅𝑃𝑀𝑔 490

1. ra Alternativa * Numero de dientes del piΓ±Γ³n y engranaje Zp=18 dientes asumido 𝑧𝑔=𝑖 π‘₯ 𝑍𝑝 = 3.592 x 18 =65 ο‚·

Nuevo i 𝑍𝑔

𝑖𝑛 =

𝑍𝑝

ο‚· 𝑑𝑝 = 𝑑𝐺 =

=

π‘š π‘₯ 𝑍𝑝

= π‘π‘œπ‘ 24 = 59.15mm

π‘š π‘₯ 𝑍𝐺 π‘π‘œπ‘ 24

=

3π‘₯65 π‘π‘œπ‘ 24

= 213.58mm

Velocidad del engranaje

ο‚·

𝑏=

3π‘₯18

π‘π‘œπ‘ 24

𝑅𝑃𝑀𝐺 =

𝑅𝑃𝑀𝑝 1760 = = 487.38 𝑖𝑛 3.611

Distancia entre centros

𝑑𝑝 +𝑑𝐺 2

ο‚·

= 3.611

DiΓ‘metros del piΓ±Γ³n y engranaje

ο‚·

π‘Ž=

65 18

= 136.36mm

Ancho del diente b

2πœ‹ π‘₯ 𝑀𝑛 2πœ‹ π‘₯ 3 = = 46.34 𝑠𝑒𝑛24 𝑠𝑒𝑛24

𝑏 = 50 π‘š

ο‚·

Altura de cabeza de diente

β„Žπ‘Ž = π‘š = 3 π‘šπ‘š ο‚·

Altura de pie de diente

β„Žπ‘“ = 1.25π‘š = 1.25(3) = 3.75 ο‚·

DiΓ‘metro externo de:

PiΓ±Γ³n: π‘‘π‘Ž (𝑃) = 𝑑𝑝 + 2β„Žπ‘Ž = 59.15 + 2(3) = 65.15π‘šπ‘š Engranaje: π‘‘π‘Ž (𝐺) = 𝑑𝐺 + 2β„Žπ‘Ž = 213.58 + 2(3) = 219.58π‘šπ‘š ο‚·

DiΓ‘metro de raΓ­z o de pie.

PiΓ±Γ³n: π‘‘π‘Ž (𝑃) = 𝑑𝑝 βˆ’ 2β„Žπ‘“ = 59.15 βˆ’ 2(3.75) = 51.65π‘šπ‘š Engranaje: 𝑑𝑓 (𝐺) = 𝑑𝑔 βˆ’ 2β„Žπ‘“ = 213.58 βˆ’ 2(3.75) = 206.08π‘šπ‘š ο‚·

Altura del diente.

β„Ž = β„Žπ‘“ + β„Žπ‘Ž = 3.75 + 3 = 6.75 π‘šπ‘š ο‚·

Espesor de cuerda (𝑆𝑐 ):

𝑆𝑐 (𝐺) = 𝑑𝑔 π‘₯ 𝑠𝑒𝑛 ( 𝑆𝑐 (𝑃) = 𝑑𝑝 π‘₯ 𝑠𝑒𝑛 ( ο‚·

90 90 ) = 213.58 𝑠𝑒𝑛 ( ) = 5.16089 𝑍𝑔 65

90 90 ) = 59.15 𝑠𝑒𝑛 ( ) = 5.15526 𝑍𝑝 18

Ancho del diente b

b = 8m a 12.5 m b = 8(3) =24 a 12.5 (3) = 37.5mm b = 37mm ο‚·

Espesor del disco:

𝑏1 = 1.8π‘š π‘Ž 2.2π‘š 𝑏1 = 2.2(3) = 6.6 ο‚·

π‘Žπ‘ π‘’π‘šπ‘–π‘Ÿ 𝑏1 = 7π‘šπ‘š

Espesor de la corona:

𝑒 = 3.8 π‘š π‘Ž 4.2 π‘š 𝑒 = 4.2(3) = 12.6π‘šπ‘š ο‚·

DiΓ‘metro del eje del engranaje

D >=

𝑑 8

D>=

ο‚·

𝑑 6

d = diΓ‘metro de paso del engranaje

d = 213.58mm D >=

213.58 8

213.58 6

D>=

= 26.6975

= 35.5967

Entonces D = 34mm

2 da Alternativa ο‚·

Numero de dientes del piΓ±Γ³n y engranaje

Zp=19 dientes asumido 𝑧𝑔=𝑖 π‘₯ 𝑍𝑝 = 3.592 x 19 =68 ο‚·

Nuevo i 𝑍𝑔

68

𝑖𝑛 = 𝑍 = 19 = 3.579 𝑝

ο‚· 𝑑𝑝 = 𝑑𝐺 =

DiΓ‘metros del piΓ±Γ³n y engranaje π‘š π‘₯ 𝑍𝑝

= π‘π‘œπ‘ 24 = 62.43mm

π‘š π‘₯ 𝑍𝐺 π‘π‘œπ‘ 24

= π‘π‘œπ‘ 24 = 223.44mm

ο‚·

ο‚·

𝑏=

𝑅𝑃𝑀𝑝 1760 = = 491.76 𝑖𝑛 3.579

Distancia entre centros

𝑑𝑝 +𝑑𝐺 2

ο‚·

3π‘₯68

Velocidad del engranaje

𝑅𝑃𝑀𝐺 =

π‘Ž=

3π‘₯19

π‘π‘œπ‘ 24

= 142.94mm

Ancho del diente b

2πœ‹ π‘₯ 𝑀𝑛 2πœ‹ π‘₯ 3 = = 46.34 𝑠𝑒𝑛24 𝑠𝑒𝑛24

𝑏 = 50 π‘š ο‚·

Altura de cabeza de diente

β„Žπ‘Ž = π‘š = 3 π‘šπ‘š

ο‚·

Altura de pie de diente

β„Žπ‘“ = 1.25π‘š = 1.25(3) = 3.75 ο‚·

DiΓ‘metro externo de:

PiΓ±Γ³n: π‘‘π‘Ž (𝑃) = 𝑑𝑝 + 2β„Žπ‘Ž = 62.43 + 2(3) = 68.43π‘šπ‘š Engranaje: π‘‘π‘Ž (𝐺) = 𝑑𝐺 + 2β„Žπ‘Ž = 223.44 + 2(3) = 229.44π‘šπ‘š ο‚·

DiΓ‘metro de raΓ­z o de pie.

PiΓ±Γ³n: π‘‘π‘Ž (𝑃) = 𝑑𝑝 βˆ’ 2β„Žπ‘“ = 62.43 βˆ’ 2(3.75) = 54.93π‘šπ‘š Engranaje: 𝑑𝑓 (𝐺) = 𝑑𝑔 βˆ’ 2β„Žπ‘“ = 223.44 βˆ’ 2(3.75) = 215.94π‘šπ‘š ο‚·

Altura del diente.

β„Ž = β„Žπ‘“ + β„Žπ‘Ž = 3.75 + 3 = 6.75 π‘šπ‘š ο‚·

Espesor de cuerda (𝑆𝑐 ):

𝑆𝑐 (𝐺) = 𝑑𝑔 π‘₯ 𝑠𝑒𝑛 ( 𝑆𝑐 (𝑃) = 𝑑𝑝 π‘₯ 𝑠𝑒𝑛 ( ο‚·

90 90 ) = 223.44 𝑠𝑒𝑛 ( ) = 5.1609 𝑍𝑔 68

90 90 ) = 62.43 𝑠𝑒𝑛 ( ) = 5.1554 𝑍𝑝 19

Ancho del diente b

b = 8m a 12.5 m b = 8(3) =24 a 12.5 (3) = 37.5mm b = 37mm ο‚·

Espesor del disco:

𝑏1 = 1.8π‘š π‘Ž 2.2π‘š 𝑏1 = 2.2(3) = 6.6 ο‚·

π‘Žπ‘ π‘’π‘šπ‘–π‘Ÿ 𝑏1 = 7π‘šπ‘š

Espesor de la corona:

𝑒 = 3.8 π‘š π‘Ž 4.2 π‘š 𝑒 = 4.2(3) = 12.6π‘šπ‘š ο‚· 𝑑

D >= 8 ο‚·

DiΓ‘metro del eje del engranaje 𝑑

D>=6

d = diΓ‘metro de paso del engranaje

d = 213.58mm

D >=

213.58 8

213.58 6

D>=

= 26.6975

= 35.5967

Entonces D = 34mm

3 ra Alternativa ο‚·

Numero de dientes del piΓ±Γ³n y engranaje

Zp=20 dientes asumido 𝑧𝑔=𝑖 π‘₯ 𝑍𝑝 = 3.592 x 20 =72 ο‚·

Nuevo i 𝑍𝑔

𝑖𝑛 =

𝑍𝑝

ο‚· 𝑑𝑝 = 𝑑𝐺 =

=

π‘š π‘₯ 𝑍𝑝

= π‘π‘œπ‘ 24 = 65.72mm

π‘š π‘₯ 𝑍𝐺 π‘π‘œπ‘ 24

= π‘π‘œπ‘ 24 = 236.58mm

3π‘₯72

Velocidad del engranaje

ο‚·

𝑏=

3π‘₯20

π‘π‘œπ‘ 24

𝑅𝑃𝑀𝐺 =

𝑅𝑃𝑀𝑝 1760 = = 488.89 𝑖𝑛 3.6

Distancia entre centros

𝑑𝑝 +𝑑𝐺 2

ο‚·

= 3.6

DiΓ‘metros del piΓ±Γ³n y engranaje

ο‚·

π‘Ž=

72 20

= 151.15mm

Ancho del diente b

2πœ‹ π‘₯ 𝑀𝑛 2πœ‹ π‘₯ 3 = = 46.34 𝑠𝑒𝑛24 𝑠𝑒𝑛24

𝑏 = 50 π‘š ο‚·

Altura de cabeza de diente

β„Žπ‘Ž = π‘š = 3 π‘šπ‘š ο‚·

Altura de pie de diente

β„Žπ‘“ = 1.25π‘š = 1.25(3) = 3.75 ο‚·

DiΓ‘metro externo de:

PiΓ±Γ³n: π‘‘π‘Ž (𝑃) = 𝑑𝑝 + 2β„Žπ‘Ž = 65.72 + 2(3) = 71.72π‘šπ‘š

Engranaje: π‘‘π‘Ž (𝐺) = 𝑑𝐺 + 2β„Žπ‘Ž = 236.58 + 2(3) = 242.58π‘šπ‘š ο‚·

DiΓ‘metro de raΓ­z o de pie.

PiΓ±Γ³n: π‘‘π‘Ž (𝑃) = 𝑑𝑝 βˆ’ 2β„Žπ‘“ = 65.72 βˆ’ 2(3.75) = 58.22π‘šπ‘š Engranaje: 𝑑𝑓 (𝐺) = 𝑑𝑔 βˆ’ 2β„Žπ‘“ = 236.58 βˆ’ 2(3.75) = 229.08π‘šπ‘š ο‚·

Altura del diente.

β„Ž = β„Žπ‘“ + β„Žπ‘Ž = 3.75 + 3 = 6.75 π‘šπ‘š ο‚·

Espesor de cuerda (𝑆𝑐 ):

𝑆𝑐 (𝐺) = 𝑑𝑔 π‘₯ 𝑠𝑒𝑛 ( 𝑆𝑐 (𝑃) = 𝑑𝑝 π‘₯ 𝑠𝑒𝑛 ( ο‚·

90 90 ) = 236.58 𝑠𝑒𝑛 ( ) = 5.1609 𝑍𝑔 72

90 90 ) = 65.72 𝑠𝑒𝑛 ( ) = 5.1563 𝑍𝑝 20

Ancho del diente b

b = 8m a 12.5 m b = 8(3) =24 a 12.5 (3) = 37.5mm b = 37mm ο‚·

Espesor del disco:

𝑏1 = 1.8π‘š π‘Ž 2.2π‘š 𝑏1 = 2.2(3) = 6.6 ο‚·

π‘Žπ‘ π‘’π‘šπ‘–π‘Ÿ 𝑏1 = 7π‘šπ‘š

Espesor de la corona:

𝑒 = 3.8 π‘š π‘Ž 4.2 π‘š 𝑒 = 4.2(3) = 12.6π‘šπ‘š ο‚·

DiΓ‘metro del eje del engranaje

𝑑

𝑑

D >= 8 ο‚·

D>=6

d = diΓ‘metro de paso del engranaje

d = 213.58mm D >=

213.58 8

213.58 6

D>=

= 26.6975

= 35.5967

Entonces D = 34mm

4 ta Alternativa ο‚·

Numero de dientes del piΓ±Γ³n y engranaje

Zp=21 dientes asumido 𝑧𝑔=𝑖 π‘₯ 𝑍𝑝 = 3.592 x 21 =75 ο‚·

Nuevo i 𝑍

75

𝑖𝑛 = 𝑍𝑔 = 21 = 3.571 𝑝

ο‚· 𝑑𝑝 = 𝑑𝐺 =

DiΓ‘metros del piΓ±Γ³n y engranaje π‘š π‘₯ 𝑍𝑝 π‘π‘œπ‘ 24 π‘š π‘₯ 𝑍𝐺 π‘π‘œπ‘ 24

ο‚·

ο‚·

𝑏=

= 69mm

3π‘₯75

= π‘π‘œπ‘ 24 = 246.44mm

𝑅𝑃𝑀𝑝 1760 = = 492.8 𝑖𝑛 3.6

Distancia entre centros

𝑑𝑝 +𝑑𝐺 2

ο‚·

3π‘₯21 π‘π‘œπ‘ 24

Velocidad del engranaje

𝑅𝑃𝑀𝐺 =

π‘Ž=

=

= 157.72mm

Ancho del diente b

2πœ‹ π‘₯ 𝑀𝑛 2πœ‹ π‘₯ 3 = = 46.34 𝑠𝑒𝑛24 𝑠𝑒𝑛24

𝑏 = 50 π‘š ο‚·

Altura de cabeza de diente

β„Žπ‘Ž = π‘š = 3 π‘šπ‘š ο‚·

Altura de pie de diente

β„Žπ‘“ = 1.25π‘š = 1.25(3) = 3.75 ο‚·

DiΓ‘metro externo de:

PiΓ±Γ³n: π‘‘π‘Ž (𝑃) = 𝑑𝑝 + 2β„Žπ‘Ž = 69 + 2(3) = 75π‘šπ‘š Engranaje: π‘‘π‘Ž (𝐺) = 𝑑𝐺 + 2β„Žπ‘Ž = 246.44 + 2(3) = 252.44π‘šπ‘š ο‚·

DiΓ‘metro de raΓ­z o de pie.

PiΓ±Γ³n: π‘‘π‘Ž (𝑃) = 𝑑𝑝 βˆ’ 2β„Žπ‘“ = 69 βˆ’ 2(3.75) = 61.5π‘šπ‘š

Engranaje: 𝑑𝑓 (𝐺) = 𝑑𝑔 βˆ’ 2β„Žπ‘“ = 246.44 βˆ’ 2(3.75) = 238.94π‘šπ‘š ο‚·

Altura del diente.

β„Ž = β„Žπ‘“ + β„Žπ‘Ž = 3.75 + 3 = 6.75 π‘šπ‘š ο‚·

Espesor de cuerda (𝑆𝑐 ):

𝑆𝑐 (𝐺) = 𝑑𝑔 π‘₯ 𝑠𝑒𝑛 (

90 90 ) = 246.44 𝑠𝑒𝑛 ( ) = 5.1604 𝑍𝑔 75

90

90

𝑆𝑐 (𝑃) = 𝑑𝑝 π‘₯ 𝑠𝑒𝑛 (𝑍 ) = 69 𝑠𝑒𝑛 (21 ) = 5.15637 𝑝

ο‚·

Ancho del diente b

b = 8m a 12.5 m b = 8(3) =24 a 12.5 (3) = 37.5mm b = 37mm ο‚·

Espesor del disco:

𝑏1 = 1.8π‘š π‘Ž 2.2π‘š 𝑏1 = 2.2(3) = 6.6 ο‚·

π‘Žπ‘ π‘’π‘šπ‘–π‘Ÿ 𝑏1 = 7π‘šπ‘š

Espesor de la corona:

𝑒 = 3.8 π‘š π‘Ž 4.2 π‘š 𝑒 = 4.2(3) = 12.6π‘šπ‘š ο‚·

DiΓ‘metro del eje del engranaje

𝑑

𝑑

D >= 8

D>=6

d = diΓ‘metro de paso del engranaje d = 213.58mm D >=

213.58 8

213.58 6

D>=

= 26.6975

= 35.5967

Entonces D = 34mm

5 ta Alternativa ο‚·

Numero de dientes del piΓ±Γ³n y engranaje

Zp=22 dientes asumido

𝑧𝑔=𝑖 π‘₯ 𝑍𝑝 = 3.592 x 22 =79 ο‚·

Nuevo i 𝑍

75

𝑖𝑛 = 𝑍𝑔 = 21 = 3.591 𝑝

ο‚· 𝑑𝑝 = 𝑑𝐺 =

DiΓ‘metros del piΓ±Γ³n y engranaje π‘š π‘₯ 𝑍𝑝

= π‘π‘œπ‘ 24 = 72.29mm

π‘š π‘₯ 𝑍𝐺 π‘π‘œπ‘ 24

= π‘π‘œπ‘ 24 = 259.58mm

ο‚·

ο‚·

𝑏=

𝑅𝑃𝑀𝑝 1760 = = 490.13 𝑖𝑛 3.591

Distancia entre centros

𝑑𝑝 +𝑑𝐺 2

ο‚·

3π‘₯75

Velocidad del engranaje

𝑅𝑃𝑀𝐺 =

π‘Ž=

3π‘₯21

π‘π‘œπ‘ 24

= 165.94mm

Ancho del diente b

2πœ‹ π‘₯ 𝑀𝑛 2πœ‹ π‘₯ 3 = = 46.34 𝑠𝑒𝑛24 𝑠𝑒𝑛24

𝑏 = 50 π‘š ο‚·

Altura de cabeza de diente

β„Žπ‘Ž = π‘š = 3 π‘šπ‘š ο‚·

Altura de pie de diente

β„Žπ‘“ = 1.25π‘š = 1.25(3) = 3.75 ο‚·

DiΓ‘metro externo de:

PiΓ±Γ³n: π‘‘π‘Ž (𝑃) = 𝑑𝑝 + 2β„Žπ‘Ž = 72.29 + 2(3) = 78.29π‘šπ‘š Engranaje: π‘‘π‘Ž (𝐺) = 𝑑𝐺 + 2β„Žπ‘Ž = 259.58 + 2(3) = 265.58π‘šπ‘š ο‚·

DiΓ‘metro de raΓ­z o de pie.

PiΓ±Γ³n: π‘‘π‘Ž (𝑃) = 𝑑𝑝 βˆ’ 2β„Žπ‘“ = 72.29 βˆ’ 2(3.75) = 64.79π‘šπ‘š Engranaje: 𝑑𝑓 (𝐺) = 𝑑𝑔 βˆ’ 2β„Žπ‘“ = 259.58 βˆ’ 2(3.75) = 252.08π‘šπ‘š ο‚·

Altura del diente.

β„Ž = β„Žπ‘“ + β„Žπ‘Ž = 3.75 + 3 = 6.75 π‘šπ‘š ο‚·

Espesor de cuerda (𝑆𝑐 ):

𝑆𝑐 (𝐺) = 𝑑𝑔 π‘₯ 𝑠𝑒𝑛 ( 𝑆𝑐 (𝑃) = 𝑑𝑝 π‘₯ 𝑠𝑒𝑛 ( ο‚·

90 90 ) = 259.58 𝑠𝑒𝑛 ( ) = 5.161 𝑍𝑔 79

90 90 ) = 72.29 𝑠𝑒𝑛 ( ) = 5.1571 𝑍𝑝 22

Ancho del diente b

b = 8m a 12.5 m b = 8(3) =24 a 12.5 (3) = 37.5mm b = 37mm ο‚·

Espesor del disco:

𝑏1 = 1.8π‘š π‘Ž 2.2π‘š 𝑏1 = 2.2(3) = 6.6 ο‚·

π‘Žπ‘ π‘’π‘šπ‘–π‘Ÿ 𝑏1 = 7π‘šπ‘š

Espesor de la corona:

𝑒 = 3.8 π‘š π‘Ž 4.2 π‘š 𝑒 = 4.2(3) = 12.6π‘šπ‘š ο‚· D >=

𝑑 8

DiΓ‘metro del eje del engranaje 𝑑 6

D>=

d = diΓ‘metro de paso del engranaje d = 259.58mm D >=

259.58 8

= 32.4475

259.58 = 6

D>=

43.2634

Entonces D = 40mm ο‚·

Para engranajes de acero 𝐷1 β‰₯ 1.8 𝐷 𝐷1 β‰₯ 1.8(40) = 72

L = D a 2.2 D entonces L= 2.2 (40 )=88

6 ta Alternativa ο‚·

Numero de dientes del piΓ±Γ³n y engranaje

Zp=23 dientes asumido

𝑧𝑔=𝑖 π‘₯ 𝑍𝑝 = 3.592 x 23 =83 ο‚·

Nuevo i 𝑍

83

𝑖𝑛 = 𝑍𝑔 = 23 = 3.609 𝑝

ο‚· 𝑑𝑝 = 𝑑𝐺 =

DiΓ‘metros del piΓ±Γ³n y engranaje π‘š π‘₯ 𝑍𝑝

= π‘π‘œπ‘ 24 = 75.58mm

π‘š π‘₯ 𝑍𝐺 π‘π‘œπ‘ 24

=

ο‚·

ο‚·

𝑏=

= 272.73mm

𝑅𝑃𝑀𝑝 1760 = = 487.71 𝑖𝑛 3.591

Distancia entre centros

𝑑𝑝 +𝑑𝐺 2

ο‚·

3π‘₯83 π‘π‘œπ‘ 24

Velocidad del engranaje

𝑅𝑃𝑀𝐺 =

π‘Ž=

3π‘₯23

π‘π‘œπ‘ 24

= 174.15mm

Ancho del diente b

2πœ‹ π‘₯ 𝑀𝑛 2πœ‹ π‘₯ 3 = = 46.34 𝑠𝑒𝑛24 𝑠𝑒𝑛24

𝑏 = 50 π‘š ο‚·

Altura de cabeza de diente

β„Žπ‘Ž = π‘š = 3 π‘šπ‘š ο‚·

Altura de pie de diente

β„Žπ‘“ = 1.25π‘š = 1.25(3) = 3.75 ο‚·

DiΓ‘metro externo de:

PiΓ±Γ³n: π‘‘π‘Ž (𝑃) = 𝑑𝑝 + 2β„Žπ‘Ž = 75.58 + 2(3) = 81.58π‘šπ‘š Engranaje: π‘‘π‘Ž (𝐺) = 𝑑𝐺 + 2β„Žπ‘Ž = 272.73 + 2(3) = 278.73π‘šπ‘š ο‚·

DiΓ‘metro de raΓ­z o de pie.

PiΓ±Γ³n: π‘‘π‘Ž (𝑃) = 𝑑𝑝 βˆ’ 2β„Žπ‘“ = 75.58 βˆ’ 2(3.75) = 68.08π‘šπ‘š Engranaje: 𝑑𝑓 (𝐺) = 𝑑𝑔 βˆ’ 2β„Žπ‘“ = 272.73 βˆ’ 2(3.75) = 265.23π‘šπ‘š ο‚·

Altura del diente.

β„Ž = β„Žπ‘“ + β„Žπ‘Ž = 3.75 + 3 = 6.75 π‘šπ‘š ο‚·

Espesor de cuerda (𝑆𝑐 ):

𝑆𝑐 (𝐺) = 𝑑𝑔 π‘₯ 𝑠𝑒𝑛 ( 𝑆𝑐 (𝑃) = 𝑑𝑝 π‘₯ 𝑠𝑒𝑛 ( ο‚·

90 90 ) = 272.73 𝑠𝑒𝑛 ( ) = 5.1611 𝑍𝑔 83

90 90 ) = 75.58 𝑠𝑒𝑛 ( ) = 5.1577 𝑍𝑝 23

Ancho del diente b

b = 8m a 12.5 m b = 8(3) =24 a 12.5 (3) = 37.5mm b = 37mm ο‚·

Espesor del disco:

𝑏1 = 1.8π‘š π‘Ž 2.2π‘š 𝑏1 = 2.2(3) = 6.6 ο‚·

π‘Žπ‘ π‘’π‘šπ‘–π‘Ÿ 𝑏1 = 7π‘šπ‘š

Espesor de la corona:

𝑒 = 3.8 π‘š π‘Ž 4.2 π‘š 𝑒 = 4.2(3) = 12.6π‘šπ‘š ο‚· D >=

DiΓ‘metro del eje del engranaje

𝑑 8

ο‚·

𝑑 6

D>=

d = diΓ‘metro de paso del engranaje

d = 213.58mm D >=

213.58 8

213.58 6

D>=

= 26.6975

= 35.5967

Entonces D = 34mm

7 ta Alternativa ο‚·

Numero de dientes del piΓ±Γ³n y engranaje

Zp=24 dientes asumido 𝑧𝑔=𝑖 π‘₯ 𝑍𝑝 = 3.592 x 24 =86 ο‚·

Nuevo i

𝑍𝑔

𝑖𝑛 =

𝑍𝑝

ο‚· 𝑑𝑝 = 𝑑𝐺 =

=

π‘š π‘₯ 𝑍𝑝

= π‘π‘œπ‘ 24 = 78.86mm

π‘š π‘₯ 𝑍𝐺 π‘π‘œπ‘ 24

= π‘π‘œπ‘ 24 = 282.58mm

3π‘₯86

Velocidad del engranaje

ο‚·

𝑏=

3π‘₯24

π‘π‘œπ‘ 24

𝑅𝑃𝑀𝐺 =

𝑅𝑃𝑀𝑝 1760 = = 491.16 𝑖𝑛 3.583

Distancia entre centros

𝑑𝑝 +𝑑𝐺 2

ο‚·

= 3.583

DiΓ‘metros del piΓ±Γ³n y engranaje

ο‚·

π‘Ž=

86 24

= 180.72mm

Ancho del diente b

2πœ‹ π‘₯ 𝑀𝑛 2πœ‹ π‘₯ 3 = = 46.34 𝑠𝑒𝑛24 𝑠𝑒𝑛24

𝑏 = 50 π‘š ο‚·

Altura de cabeza de diente

β„Žπ‘Ž = π‘š = 3 π‘šπ‘š ο‚·

Altura de pie de diente

β„Žπ‘“ = 1.25π‘š = 1.25(3) = 3.75 ο‚·

DiΓ‘metro externo de:

PiΓ±Γ³n: π‘‘π‘Ž (𝑃) = 𝑑𝑝 + 2β„Žπ‘Ž = 78.86 + 2(3) = 84.86π‘šπ‘š Engranaje: π‘‘π‘Ž (𝐺) = 𝑑𝐺 + 2β„Žπ‘Ž = 282.58 + 2(3) = 288.58π‘šπ‘š ο‚·

DiΓ‘metro de raΓ­z o de pie.

PiΓ±Γ³n: π‘‘π‘Ž (𝑃) = 𝑑𝑝 βˆ’ 2β„Žπ‘“ = 78.86 βˆ’ 2(3.75) = 71.36π‘šπ‘š Engranaje: 𝑑𝑓 (𝐺) = 𝑑𝑔 βˆ’ 2β„Žπ‘“ = 282.58 βˆ’ 2(3.75) = 275.08π‘šπ‘š ο‚·

Altura del diente.

β„Ž = β„Žπ‘“ + β„Žπ‘Ž = 3.75 + 3 = 6.75 π‘šπ‘š ο‚·

Espesor de cuerda (𝑆𝑐 ):

𝑆𝑐 (𝐺) = 𝑑𝑔 π‘₯ 𝑠𝑒𝑛 (

90 90 ) = 282.58 𝑠𝑒𝑛 ( ) = 5.161 𝑍𝑔 86

𝑆𝑐 (𝑃) = 𝑑𝑝 π‘₯ 𝑠𝑒𝑛 ( ο‚·

90 90 ) = 78.86 𝑠𝑒𝑛 ( ) = 5.1576 𝑍𝑝 24

Ancho del diente b

b = 8m a 12.5 m b = 8(3) =24 a 12.5 (3) = 37.5mm b = 37mm ο‚·

Espesor del disco:

𝑏1 = 1.8π‘š π‘Ž 2.2π‘š 𝑏1 = 2.2(3) = 6.6 ο‚·

π‘Žπ‘ π‘’π‘šπ‘–π‘Ÿ 𝑏1 = 7π‘šπ‘š

Espesor de la corona:

𝑒 = 3.8 π‘š π‘Ž 4.2 π‘š 𝑒 = 4.2(3) = 12.6π‘šπ‘š ο‚· D >=

DiΓ‘metro del eje del engranaje

𝑑 8

ο‚·

𝑑 6

D>=

d = diΓ‘metro de paso del engranaje

d = 213.58mm D >=

213.58 8

213.58 6

D>=

= 26.6975

= 35.5967

Entonces D = 34mm

Mn

Zp

Zg

Dp

i

3

18

65

59,145674 3,6111111 487,38462

3

19

68

62,431544 3,5789474 491,76471 223,43921 142,93538

3

20

72

65,717415

3

21

75

69,003286 3,5714286

3

22

79

72,289157 3,5909091 490,12658 259,58379 165,93647

3

23

83

75,575027 3,6086957 487,71084 272,72727 174,15115

3

24

86

78,860898 3,5833333 491,16279 282,58488 180,72289

3,6

RPMg

Dg

a

213,5816

136,36364

488,88889 236,58269 151,15005 492,8

246,44031

157,7218

Related Documents

Engranajes
May 2020 19
Engranajes
November 2019 20
Engranajes Rectos.docx
December 2019 22
Rueda Dentada Y Engranajes
November 2019 49

More Documents from "jose maria"

Desarrollo Organizacional (1)
September 2019 21
Le Projet Mediapart
April 2020 25
April 2020 28
Prologue Sophocle
November 2019 32
Biosphere_juin_2008
December 2019 32