Engranajes cilΓndricos de dientes helicoidales RPM ingreso = 1760 RPM salida = 490 Modulo = 3 Angulo de la hΓ©lice = 24ΒΊ RELACION DE TRANSMISION
π=
π
πππ 1760 = = 3.592 π
πππ 490
1. ra Alternativa * Numero de dientes del piΓ±Γ³n y engranaje Zp=18 dientes asumido π§π=π π₯ ππ = 3.592 x 18 =65 ο·
Nuevo i ππ
ππ =
ππ
ο· ππ = ππΊ =
=
π π₯ ππ
= πππ 24 = 59.15mm
π π₯ ππΊ πππ 24
=
3π₯65 πππ 24
= 213.58mm
Velocidad del engranaje
ο·
π=
3π₯18
πππ 24
π
πππΊ =
π
πππ 1760 = = 487.38 ππ 3.611
Distancia entre centros
ππ +ππΊ 2
ο·
= 3.611
DiΓ‘metros del piΓ±Γ³n y engranaje
ο·
π=
65 18
= 136.36mm
Ancho del diente b
2π π₯ ππ 2π π₯ 3 = = 46.34 π ππ24 π ππ24
π = 50 π
ο·
Altura de cabeza de diente
βπ = π = 3 ππ ο·
Altura de pie de diente
βπ = 1.25π = 1.25(3) = 3.75 ο·
DiΓ‘metro externo de:
PiΓ±Γ³n: ππ (π) = ππ + 2βπ = 59.15 + 2(3) = 65.15ππ Engranaje: ππ (πΊ) = ππΊ + 2βπ = 213.58 + 2(3) = 219.58ππ ο·
DiΓ‘metro de raΓz o de pie.
PiΓ±Γ³n: ππ (π) = ππ β 2βπ = 59.15 β 2(3.75) = 51.65ππ Engranaje: ππ (πΊ) = ππ β 2βπ = 213.58 β 2(3.75) = 206.08ππ ο·
Altura del diente.
β = βπ + βπ = 3.75 + 3 = 6.75 ππ ο·
Espesor de cuerda (ππ ):
ππ (πΊ) = ππ π₯ π ππ ( ππ (π) = ππ π₯ π ππ ( ο·
90 90 ) = 213.58 π ππ ( ) = 5.16089 ππ 65
90 90 ) = 59.15 π ππ ( ) = 5.15526 ππ 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:
π1 = 1.8π π 2.2π π1 = 2.2(3) = 6.6 ο·
ππ π’πππ π1 = 7ππ
Espesor de la corona:
π = 3.8 π π 4.2 π π = 4.2(3) = 12.6ππ ο·
DiΓ‘metro del eje del engranaje
D >=
π 8
D>=
ο·
π 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 π§π=π π₯ ππ = 3.592 x 19 =68 ο·
Nuevo i ππ
68
ππ = π = 19 = 3.579 π
ο· ππ = ππΊ =
DiΓ‘metros del piΓ±Γ³n y engranaje π π₯ ππ
= πππ 24 = 62.43mm
π π₯ ππΊ πππ 24
= πππ 24 = 223.44mm
ο·
ο·
π=
π
πππ 1760 = = 491.76 ππ 3.579
Distancia entre centros
ππ +ππΊ 2
ο·
3π₯68
Velocidad del engranaje
π
πππΊ =
π=
3π₯19
πππ 24
= 142.94mm
Ancho del diente b
2π π₯ ππ 2π π₯ 3 = = 46.34 π ππ24 π ππ24
π = 50 π ο·
Altura de cabeza de diente
βπ = π = 3 ππ
ο·
Altura de pie de diente
βπ = 1.25π = 1.25(3) = 3.75 ο·
DiΓ‘metro externo de:
PiΓ±Γ³n: ππ (π) = ππ + 2βπ = 62.43 + 2(3) = 68.43ππ Engranaje: ππ (πΊ) = ππΊ + 2βπ = 223.44 + 2(3) = 229.44ππ ο·
DiΓ‘metro de raΓz o de pie.
PiΓ±Γ³n: ππ (π) = ππ β 2βπ = 62.43 β 2(3.75) = 54.93ππ Engranaje: ππ (πΊ) = ππ β 2βπ = 223.44 β 2(3.75) = 215.94ππ ο·
Altura del diente.
β = βπ + βπ = 3.75 + 3 = 6.75 ππ ο·
Espesor de cuerda (ππ ):
ππ (πΊ) = ππ π₯ π ππ ( ππ (π) = ππ π₯ π ππ ( ο·
90 90 ) = 223.44 π ππ ( ) = 5.1609 ππ 68
90 90 ) = 62.43 π ππ ( ) = 5.1554 ππ 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:
π1 = 1.8π π 2.2π π1 = 2.2(3) = 6.6 ο·
ππ π’πππ π1 = 7ππ
Espesor de la corona:
π = 3.8 π π 4.2 π π = 4.2(3) = 12.6ππ ο· π
D >= 8 ο·
DiΓ‘metro del eje del engranaje π
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 π§π=π π₯ ππ = 3.592 x 20 =72 ο·
Nuevo i ππ
ππ =
ππ
ο· ππ = ππΊ =
=
π π₯ ππ
= πππ 24 = 65.72mm
π π₯ ππΊ πππ 24
= πππ 24 = 236.58mm
3π₯72
Velocidad del engranaje
ο·
π=
3π₯20
πππ 24
π
πππΊ =
π
πππ 1760 = = 488.89 ππ 3.6
Distancia entre centros
ππ +ππΊ 2
ο·
= 3.6
DiΓ‘metros del piΓ±Γ³n y engranaje
ο·
π=
72 20
= 151.15mm
Ancho del diente b
2π π₯ ππ 2π π₯ 3 = = 46.34 π ππ24 π ππ24
π = 50 π ο·
Altura de cabeza de diente
βπ = π = 3 ππ ο·
Altura de pie de diente
βπ = 1.25π = 1.25(3) = 3.75 ο·
DiΓ‘metro externo de:
PiΓ±Γ³n: ππ (π) = ππ + 2βπ = 65.72 + 2(3) = 71.72ππ
Engranaje: ππ (πΊ) = ππΊ + 2βπ = 236.58 + 2(3) = 242.58ππ ο·
DiΓ‘metro de raΓz o de pie.
PiΓ±Γ³n: ππ (π) = ππ β 2βπ = 65.72 β 2(3.75) = 58.22ππ Engranaje: ππ (πΊ) = ππ β 2βπ = 236.58 β 2(3.75) = 229.08ππ ο·
Altura del diente.
β = βπ + βπ = 3.75 + 3 = 6.75 ππ ο·
Espesor de cuerda (ππ ):
ππ (πΊ) = ππ π₯ π ππ ( ππ (π) = ππ π₯ π ππ ( ο·
90 90 ) = 236.58 π ππ ( ) = 5.1609 ππ 72
90 90 ) = 65.72 π ππ ( ) = 5.1563 ππ 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:
π1 = 1.8π π 2.2π π1 = 2.2(3) = 6.6 ο·
ππ π’πππ π1 = 7ππ
Espesor de la corona:
π = 3.8 π π 4.2 π π = 4.2(3) = 12.6ππ ο·
DiΓ‘metro del eje del engranaje
π
π
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 π§π=π π₯ ππ = 3.592 x 21 =75 ο·
Nuevo i π
75
ππ = ππ = 21 = 3.571 π
ο· ππ = ππΊ =
DiΓ‘metros del piΓ±Γ³n y engranaje π π₯ ππ πππ 24 π π₯ ππΊ πππ 24
ο·
ο·
π=
= 69mm
3π₯75
= πππ 24 = 246.44mm
π
πππ 1760 = = 492.8 ππ 3.6
Distancia entre centros
ππ +ππΊ 2
ο·
3π₯21 πππ 24
Velocidad del engranaje
π
πππΊ =
π=
=
= 157.72mm
Ancho del diente b
2π π₯ ππ 2π π₯ 3 = = 46.34 π ππ24 π ππ24
π = 50 π ο·
Altura de cabeza de diente
βπ = π = 3 ππ ο·
Altura de pie de diente
βπ = 1.25π = 1.25(3) = 3.75 ο·
DiΓ‘metro externo de:
PiΓ±Γ³n: ππ (π) = ππ + 2βπ = 69 + 2(3) = 75ππ Engranaje: ππ (πΊ) = ππΊ + 2βπ = 246.44 + 2(3) = 252.44ππ ο·
DiΓ‘metro de raΓz o de pie.
PiΓ±Γ³n: ππ (π) = ππ β 2βπ = 69 β 2(3.75) = 61.5ππ
Engranaje: ππ (πΊ) = ππ β 2βπ = 246.44 β 2(3.75) = 238.94ππ ο·
Altura del diente.
β = βπ + βπ = 3.75 + 3 = 6.75 ππ ο·
Espesor de cuerda (ππ ):
ππ (πΊ) = ππ π₯ π ππ (
90 90 ) = 246.44 π ππ ( ) = 5.1604 ππ 75
90
90
ππ (π) = ππ π₯ π ππ (π ) = 69 π ππ (21 ) = 5.15637 π
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:
π1 = 1.8π π 2.2π π1 = 2.2(3) = 6.6 ο·
ππ π’πππ π1 = 7ππ
Espesor de la corona:
π = 3.8 π π 4.2 π π = 4.2(3) = 12.6ππ ο·
DiΓ‘metro del eje del engranaje
π
π
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
π§π=π π₯ ππ = 3.592 x 22 =79 ο·
Nuevo i π
75
ππ = ππ = 21 = 3.591 π
ο· ππ = ππΊ =
DiΓ‘metros del piΓ±Γ³n y engranaje π π₯ ππ
= πππ 24 = 72.29mm
π π₯ ππΊ πππ 24
= πππ 24 = 259.58mm
ο·
ο·
π=
π
πππ 1760 = = 490.13 ππ 3.591
Distancia entre centros
ππ +ππΊ 2
ο·
3π₯75
Velocidad del engranaje
π
πππΊ =
π=
3π₯21
πππ 24
= 165.94mm
Ancho del diente b
2π π₯ ππ 2π π₯ 3 = = 46.34 π ππ24 π ππ24
π = 50 π ο·
Altura de cabeza de diente
βπ = π = 3 ππ ο·
Altura de pie de diente
βπ = 1.25π = 1.25(3) = 3.75 ο·
DiΓ‘metro externo de:
PiΓ±Γ³n: ππ (π) = ππ + 2βπ = 72.29 + 2(3) = 78.29ππ Engranaje: ππ (πΊ) = ππΊ + 2βπ = 259.58 + 2(3) = 265.58ππ ο·
DiΓ‘metro de raΓz o de pie.
PiΓ±Γ³n: ππ (π) = ππ β 2βπ = 72.29 β 2(3.75) = 64.79ππ Engranaje: ππ (πΊ) = ππ β 2βπ = 259.58 β 2(3.75) = 252.08ππ ο·
Altura del diente.
β = βπ + βπ = 3.75 + 3 = 6.75 ππ ο·
Espesor de cuerda (ππ ):
ππ (πΊ) = ππ π₯ π ππ ( ππ (π) = ππ π₯ π ππ ( ο·
90 90 ) = 259.58 π ππ ( ) = 5.161 ππ 79
90 90 ) = 72.29 π ππ ( ) = 5.1571 ππ 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:
π1 = 1.8π π 2.2π π1 = 2.2(3) = 6.6 ο·
ππ π’πππ π1 = 7ππ
Espesor de la corona:
π = 3.8 π π 4.2 π π = 4.2(3) = 12.6ππ ο· D >=
π 8
DiΓ‘metro del eje del engranaje π 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 π·1 β₯ 1.8 π· π·1 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
π§π=π π₯ ππ = 3.592 x 23 =83 ο·
Nuevo i π
83
ππ = ππ = 23 = 3.609 π
ο· ππ = ππΊ =
DiΓ‘metros del piΓ±Γ³n y engranaje π π₯ ππ
= πππ 24 = 75.58mm
π π₯ ππΊ πππ 24
=
ο·
ο·
π=
= 272.73mm
π
πππ 1760 = = 487.71 ππ 3.591
Distancia entre centros
ππ +ππΊ 2
ο·
3π₯83 πππ 24
Velocidad del engranaje
π
πππΊ =
π=
3π₯23
πππ 24
= 174.15mm
Ancho del diente b
2π π₯ ππ 2π π₯ 3 = = 46.34 π ππ24 π ππ24
π = 50 π ο·
Altura de cabeza de diente
βπ = π = 3 ππ ο·
Altura de pie de diente
βπ = 1.25π = 1.25(3) = 3.75 ο·
DiΓ‘metro externo de:
PiΓ±Γ³n: ππ (π) = ππ + 2βπ = 75.58 + 2(3) = 81.58ππ Engranaje: ππ (πΊ) = ππΊ + 2βπ = 272.73 + 2(3) = 278.73ππ ο·
DiΓ‘metro de raΓz o de pie.
PiΓ±Γ³n: ππ (π) = ππ β 2βπ = 75.58 β 2(3.75) = 68.08ππ Engranaje: ππ (πΊ) = ππ β 2βπ = 272.73 β 2(3.75) = 265.23ππ ο·
Altura del diente.
β = βπ + βπ = 3.75 + 3 = 6.75 ππ ο·
Espesor de cuerda (ππ ):
ππ (πΊ) = ππ π₯ π ππ ( ππ (π) = ππ π₯ π ππ ( ο·
90 90 ) = 272.73 π ππ ( ) = 5.1611 ππ 83
90 90 ) = 75.58 π ππ ( ) = 5.1577 ππ 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:
π1 = 1.8π π 2.2π π1 = 2.2(3) = 6.6 ο·
ππ π’πππ π1 = 7ππ
Espesor de la corona:
π = 3.8 π π 4.2 π π = 4.2(3) = 12.6ππ ο· D >=
DiΓ‘metro del eje del engranaje
π 8
ο·
π 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 π§π=π π₯ ππ = 3.592 x 24 =86 ο·
Nuevo i
ππ
ππ =
ππ
ο· ππ = ππΊ =
=
π π₯ ππ
= πππ 24 = 78.86mm
π π₯ ππΊ πππ 24
= πππ 24 = 282.58mm
3π₯86
Velocidad del engranaje
ο·
π=
3π₯24
πππ 24
π
πππΊ =
π
πππ 1760 = = 491.16 ππ 3.583
Distancia entre centros
ππ +ππΊ 2
ο·
= 3.583
DiΓ‘metros del piΓ±Γ³n y engranaje
ο·
π=
86 24
= 180.72mm
Ancho del diente b
2π π₯ ππ 2π π₯ 3 = = 46.34 π ππ24 π ππ24
π = 50 π ο·
Altura de cabeza de diente
βπ = π = 3 ππ ο·
Altura de pie de diente
βπ = 1.25π = 1.25(3) = 3.75 ο·
DiΓ‘metro externo de:
PiΓ±Γ³n: ππ (π) = ππ + 2βπ = 78.86 + 2(3) = 84.86ππ Engranaje: ππ (πΊ) = ππΊ + 2βπ = 282.58 + 2(3) = 288.58ππ ο·
DiΓ‘metro de raΓz o de pie.
PiΓ±Γ³n: ππ (π) = ππ β 2βπ = 78.86 β 2(3.75) = 71.36ππ Engranaje: ππ (πΊ) = ππ β 2βπ = 282.58 β 2(3.75) = 275.08ππ ο·
Altura del diente.
β = βπ + βπ = 3.75 + 3 = 6.75 ππ ο·
Espesor de cuerda (ππ ):
ππ (πΊ) = ππ π₯ π ππ (
90 90 ) = 282.58 π ππ ( ) = 5.161 ππ 86
ππ (π) = ππ π₯ π ππ ( ο·
90 90 ) = 78.86 π ππ ( ) = 5.1576 ππ 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:
π1 = 1.8π π 2.2π π1 = 2.2(3) = 6.6 ο·
ππ π’πππ π1 = 7ππ
Espesor de la corona:
π = 3.8 π π 4.2 π π = 4.2(3) = 12.6ππ ο· D >=
DiΓ‘metro del eje del engranaje
π 8
ο·
π 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