calculos elementos de maquinas ii
TRANSCRIPT
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Dados de Entrada Valor Adotado
Rotação Inicial (ηi) 1200.00
Potência do Motor (Nmot) 22.50
Z1 16.00
Z3 18.00
Rotação Final (ηf) 21.00
Tipos de Dentes do 1º Par Helicoidais
Ângulo de Pressão(Θ) 20.00
Ângulo de Inclinação da Hélice (β) 15.00
Tipos de Dentes do 2º Par de Engrenagem Retos
E (Carga Normal) 1.00
σadm Engrenagem (Aço Liga SAE 4340) 17.50
σadm Engrenagem (Aço Liga SAE 4340) 17.50
Dureza da Engrenagem (Aço Liga SAE 4340) 280.00
Tr EIXO (AÇO LIGA SAE 4063) 126.00
Te EIXO (AÇO LIGA SAE 4063) 112.00
Dureza do eixo (AÇO LIGA SAE 4063) 375.00
Q (Z=17 DENTES) 3.60
Q (Z=19 DENTES) 3.20
λ (DE 6 A 20) 10.00
φr (PARA β=15º) 1.35
Projeto Dimensionamento do Maquina Operatriz com Arranque de Cavaco
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Coeficiente de Segurança (CS) 5.00
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Unidade
[RPM]
[CV]
[Dentes]
[Dentes]
[RPM]
[º]
[º]
[Tabelado]
[Kgf/mm²]
[Kgf/mm²]
[HB]
[Kgf/mm²]
[Kgf/mm²]
[HB]
[Tabelado]
[Tabelado]
[Adotado]
[Tabelado]
Projeto Dimensionamento do Maquina Operatriz com Arranque de Cavaco
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[Adotado]
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1. Estimativa do Número de Dentes das Engrenagens
Relação de Redução Total
IT = ηi IT = 1200 [RPM] èè IT = 57.14ηf 21 [RPM]
IT = I 1-2 x I 3-4 Multiplicação pelo mesmo número que resuta no IT;
Portanto, extrai a raiz quadrada para determinar o valor:
I 1-2 = √ 57.14 èè I 1-2 = 7.56
Como I 1-2 < I 3-4, adota-se 4,00.
Z2 = I 1-2 x Z1 = 7.56 x 16 = 120.95 èè Z2 = 121 Dentes
Recalculando, tem-se:
I 1-2 = Z2 = 121 èè I 1-2 = 7.56Z1 16
Calculando I 3-4:
I 3-4 = IT = 57.14 = 7.56I 1-2 7.56
Conhecido o valor de I 3-4, determina o valor de Z 4:
Z4 = I 3-4 x Z3
Z4 = 7.56 x 18 = 136.07 èè Z4 = 136 Dentes
I 3-4 = Z4 = 136 èè I 3-4 = 7.56Z3 18
IT = I 1-2 x I 3-4 èè 57.14 = 7.56 x 7.56 57.14 = 57.11
Verificando os valores encontrados:
IT = ηi èè ηf = ηi = 1200 èè ηf = 21 [RPM]ηf IT 57.11
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Como, ηf fornecido é de 110 RPM, e o calculado em função das engreanagens estimadas é de110,96 RPM, os números de dentes da engrenagens estão conforme, para a aplicação do projeto.
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2. CÁLCULO DO MOMENTO TORÇOR
MT1 = 716.2 x Nmot = 716.2 x 22.5ηi 1200
Z2 = MT2Z1 MT1
121 = MT2 èè MT2 = 1624.19 = 101.5516 13.43 16
Z4 = MT3Z3 MT2
136 = MT3 èè MT3 = 13810.80 = 767.2618 101.55 18
CONFERÊNCIA
MT3 = 716.2 x Nmot = 716.2 x 22.5 =ηf 21
3. CÁLCULO DO MÓDULOS
ENGRENAGEM DENTE HELICOIDAL
MN> 3 2 x MT x Q xCOS β√ λ x Z x E xTadmx φr
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ENGRENAGEM DENTE RETO
MN> 3 2 x MT x Q√ λ x Z x E xTadm
Tadm = TR Tadm = 18 Tadm= ###Kgf/mm2CS 5
MÓDULO ENGRENAGE
MN1-2> 3 2 x ### x ### x ### MN1-2> ### mm√ 10 x 16 x ### x ### x ###
MN1-2= ###mm
MÓDULO ENGRENAGE
MN3-4> 3 2 x ### x ### MN3-4> ### mm√ 10 x 18 x ### x ###
MN3-4= ###mm
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4. CÁLCULO DOS DADOS CONSTRUTIVOS
ENGRENAGEM 1
MC = MN MC = 1.75 MC = 1.82 mmCOS β 0.96
DP = MC x Z DP = 1.82 x 16 DP = 29.17 mm DP =
DI = DP - 2.3 x MN DI = 29 - 2.3 x 1.75 DI = 25.14 mm DI =
DE = DP + 2 x MN DE = 29 + 2 x 1.75 DE = 32.67 mm DE =
DB = DP x COS O DB = 29 x 0.94 DB = 27.42 mm DB =
PC = MC x PI PC = 1.82 x 3.14 PC = 5.73 mm PC =
PN = MN x PI PN = 1.75 x 3.14 PN = 5.50 mm PN =
B = 1.15 x MN B = 1.15 x 1.75 B = 2.01 mm B =
A = MN A = 1.75 mm
H = A + B H = 1.75 + 2.01 H = 3.76 mm H =
ENGRENAGEM 2
MC = MN MC = 1.75 MC = 1.82 mmCOS β 0.96
DP = MC x Z DP = 1.82 x 121 DP = 220.48 mm DP =
DI = DP - 2.3 x MN DI = 220 - 2.3 x 1.75 DI = 216.45 mm DI =
DE = DP + 2 x MN DE = 220 + 2 x 1.75 DE = 223.98 mm DE =
DB = DP x COS O DB = 220 x 0.94 DB = 207.25 mm DB =
PC = MC x PI PC = 1.82 x 3.14 PC = 5.73 mm PC =
PN = MN x PI PN = 1.75 x 3.14 PN = 5.50 mm PN =
B = 1.15 x MN B = 1.15 x 1.75 B = 2.01 mm B =
A = MN A = 1.75 mm
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H = A + B H = 1.75 + 2.01 H = 3.76 mm H =
ENGRENAGEM 3
DP = MN x Z DP = 2.75 x 18 DP = 49.50 mm DP =
DI = MN x Z - 2.3 DI = 2.75 x 18 - 2.3 DI = 43.18 mm DI =
DE = MN x Z + 2 DE = 2.75 x 18 + 2 DE = 55.00 mm DE =
PC = MN x PI PC = 2.75 x 3.14 PC = 8.64 mm PC =
B = 1.15 x MN B = 1.15 x 2.75 B = 3.16 mm B =
A = MN A = 2.75 mm
H = A + B H = 2.75 + 3.16 H = 5.91 mm H =
ENGRENAGEM 4
DP = MN x Z DP = 2.75 x 136 DP = 374.00 mm DP =
DI = MN x Z - 2.3 DI = 2.75 x 136 - 2.3 DI = 367.67 mm DI =
DE = MN x Z + 2 DE = 2.75 x 136 + 2 DE = 379.50 mm DE =
PC = MN x PI PC = 2.75 x 3.14 PC = 8.64 mm PC =
B = 1.15 x MN B = 1.15 x 2.75 B = 3.16 mm B =
A = MN A = 2.75 mm
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H = A + B H = 2.75 + 3.16 H = 5.91 mm H =
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29 mm
25 mm
33 mm
27 mm
6 mm
5 mm
2 mm
4 mm
220 mm
216 mm
224 mm
207 mm
6 mm
5 mm
2 mm
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4 mm
50 mm
43 mm
55 mm
9 mm
3 mm
6 mm
374 mm
368 mm
380 mm
9 mm
3 mm
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6 mm
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5. CÁLCULO DA DISTÂNCIA ENTRE EIXOS
EIXO 1 EIXO 1 EIXO 1COM = DP1 + DP2 COM = 29 + 220 COM = 125 mm
EIXO2 2 EIXO2 2 EIXO2
EIXO 3 EIXO 3 EIXO 2COM = DP3 + DP4 COM = 50 + 374 COM = 212 mm
EIXO4 2 EIXO4 2 EIXO3
6. CÁLCULO DA LARGURA DAS ENGRENAGENS
L1-2 = λ x MN L1-2 = 10 x 1.75 L1-2 = 17.5 mm L1-2 = 18 mm
L3-4 = λ x MN L3-4 = 10 x 2.75 L3-4 = 27.5 mm L3-4 = 28 mm
7. CÁLCULO DO COMPRIMENTO DOS EIXOS
E1 E4= = EIXO 1 = = EIXO 3
50 150 150 50200 200
E2 E3= = EIXO 2
50 100 50
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200
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8. CÁLCULO DAS FORÇAS NO ENGRENAMENTO
ENGRENAGEM 1-2
FT = 2 x MT FT = 2 x 13428.75 FT = 920.83 kgfDP 29
FR = FT x TG Θ FR = 920.83 x 0.36 FR = 331.50 kgf
FA = FT x TG β FA = 920.83 x 0.27 FA = 248.62 kgf
ENGRENAGEM 3-4
FT = 2 x MT FT = 2 x 101555.00 FT = 4103.23 kgfDP 50
FR = FT x TG Θ FR = 4103.23 x 0.36 FR = 1477.16 kgf
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9. CÁLCULO DAS REAÇÕES NOS MANCAIS EIXO 1
PLANO HORIZONTAL
FT ΣMD = 0a b ↓ c ←FXC↑ ↑ FYA x 200 - FT x 150 =
FYA FYC50 150 FYA = 920.83 x 150 FYA =
200 200
FXC = 0 ΣFY = 0 FYA - FT +
FYC = 690.62 - 920.83 FYC =
ΣMBE = FYA x 50 ΣMBE = 690.62 x 50 ΣMBE =
ΣMBD = FYC x ### ΣMBD = -230.21 x 150 ΣMBD =
PLANO VERTICAL
FA← FR
a b ↑ c →FXC= FA SENTIDO ADOTADO↓ ↓
FYA FYC HORÁRIO:50 150 PARA CIMA:
200
FXC = 248.62 kgf
ΣMD = 0
- FYA x 200 + FR x 150 - FA x 15 = 0
FYA = - 331.50 x 150 + 248.62 x 15 FYA =200
ΣFY = 0 - FYA + FR - FYC = 0
FYC = - 230.49 + 331.50 FYC = 101.00 kgf
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ΣMBD = FYC x ### ΣMBD = 101.00 x 150 ΣMBD =
ΣMBE = FYA x 50 ΣMBE = ### x 50 ΣMBE =
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0
690.62 kgf
FYC =
230.21 kgf
34531.07 kgf.mm
-34531.07 kgf.mm
SENTIDO ADOTADO
POSITIVOPOSITIVO
230.49 kgf
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15150.51 kgf.mm
-11524.75 kgf.mm
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9.1. DIAGRAMA EIXO 1
PLANO HORIZONTAL PLANO VERTICAL
248.62920.83 331.50 ←
a b ↓ 0 a b ↑↑ ↑ ↓
690.62 230.21 230.4950 150 50 150
200 200
N (kgf) N (kgf) 248.62
690.62Q (kgf) Q (kgf) 101.00
-230.21 -230.49
11524.75 15150.51
MF (kgf.mm) MF (kgf.mm)
c ←
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34531.07 34531.07
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FXD=↓ 248.62101.00
c →
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10. CÁLCULO DAS REAÇÕES NOS MANCAIS EIXO 2
PLANO HORIZONTALSENTIDO ADOTADO
HORÁRIO:FT FT PARA CIMA:
a b ↓ c ↓ d ← FXD↑ ↑
FYA FYD50 100 50
200
FXD = 0
ΣMD = 0
FYA x 200 - FT x 150 - FT x 50 =
FYA = 920.83 x 150 + 4103.23 x 50 FYA =200
ΣFY = 0 FYA - FT - FT + FYD =
FYD = 1716.43 - 920.83 - 4103.23 FYD =
ΣMBE = FYA x 50 ΣMBE = 1716.43 x 50
ΣMBE = 85821.48 kgf.mm
ΣMCD = FYD x 50 ΣMCD = -3307.63 x 50
ΣMCD = -165381.57 kgf.mm
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PLANO VERTICAL
FAFR → FR
a b ↓ c ↑ d ← FXD= FA↓ ↓
FYA FYD50 100 50
200
FXD = 248.62 kgf
ΣMd = 0
- FYA x 200 - FR x 150 + FA x 110 +
FYA = 331.50 x 150 - 248.62 x 110 - 1477.16 x200
FYA = 257.71 kgf
ΣFY = 0 - FYA - FR + FR - FYD
FYD = - 257.71 - 331.50 + 1477.16 FYD =
ΣMBE = FYA x 50 ΣMBE = 257.71 x 50
ΣMBE = 12885.41 kgf.mm
ΣMBD = - FR x 100 + FYD x ###
ΣMBD = - 1477.16 x 100 + 887.96 x ###
ΣMBD = -14522.78 kgf.mm
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ΣMCD = FYD x 50 ΣMCD = 887.96 x 50
ΣMCD = 44397.86 kgf.mm
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SENTIDO ADOTADO
POSITIVOPOSITIVO
0
1716.43 kgf
0
3307.63 kgf
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FR x 50
50
= 0
887.96 kgf
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10.1. DIAGRAMA EIXO 2
PLANO HORIZONTAL PLANO VERTICAL
248.62
920.83 4103.23 331.50 → 1477.16
a b ↓ c ↓ a b ↓ c ↑↑ ↑ ↓ ↓
1716.43 3307.63 257.71 887.96
50 100 50 50 100 50
200 200
N (kgf) N (kgf)
-248.62 -248.62
1716.43 887.96
Q (kgf) 795.60 Q (kgf)
-257.71
-3307.63 -589.21
44397.86
12885.41
Mf (kgf.mm) Mf (kgf.mm)
d ← d ←
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85821.48 165381.57 -14522.78
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PLANO VERTICAL
Fxd=
248.62
887.96
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11. CÁLCULO DAS REAÇÕES NOS MANCAIS EIXO 3
PLANO HORIZONTAL
FT ΣMD = 0a b ↓ c ← FXC↑ ↑ FYA x 200 - FT x 50 = 0
FYA FYC150 50 FYA = 4103.23 x 50 FYA = 1025.81
200 200
FXC = 0 ΣFY = 0
FYA - FT + FYC = 0
FYC = 1025.81 - 4103.23 FYC = 3077.42
ΣMBE = FYA x 150 ΣMBE = 1025.81 x 150 ΣMBE = 153871.21
ΣMBD = FYC x 50 ΣMBD = -3077.42 x 50 ΣMBD = -153871.21
PLANO VERTICAL
FR ΣMD = 0a b ↑ c ← FXC↓ ↓ - FYA x 200 + FR x 50
FYA FYD150 50 FYA = 1477.16 x 50 FYA = 369.29
200 200
FXC = 0 ΣFY = 0
- FYA + FR - FYC = 0
FYC = - 369.29 + 1477.16 FYC = 1107.87 kgf
ΣMBD = FYC x 50 ΣMBD = 1107.87 x 50 ΣMBD = 55393.64
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ΣMBE = FYA x 150 ΣMBE = -369.29 x 150 ΣMBE = -55393.64
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kgf
kgf
kgf.mm
kgf.mm
kgf
kgf.mm
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kgf.mm
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11.1. DIAGRAMA EIXO 3
PLANO HORIZONTAL PLANO VERTICAL
4103.23 1477.16
a b ↓ 0 a b ↑↑ ↑ ↓
1025.81 3077.42 369.29
150 50 150 50
200 200
N (kgf) N (kgf)
1025.81 1107.87
Q (kgf) Q (kgf)
-369.29
-3077.42
55393.64
55393.64
Mf (kgf.mm) Mf (kgf.mm)
c ←
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153871.21 153871.21
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0
↓
1107.87
c ←
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12. CÁLCULO DO DIÂMETRO DOS EIXOS
δ = 0.75 x Tr = 0.75 x 126 δ =CS 5
Tadm = Tr Tadm = 126 Tadm =CS 5
α = Tadm α = 25.2 α =δ 18.9
EIXO 01
MF = √ (MFH) 2 (MFV) 2
MF = √ ( 34531.07 ) 2 + ( 15150.51 ) 2
MF = √ 1192394894.01 + 229537880.21 MF = 37708.52 kgf.mm
MC = (MF) 2 + ( α x MT)2√ 2
MC = ( 37708.52 ) 2 + ( 0.67 x 13428.75 )2√
MC = √ 1421932774.22 + 80147256.25 MC = 38756.68 kgf.mm
D > 2.17 x 3 Mc D > 1.72 x 3√ Tadm √
D > 19.85 mm D = 20 mm
EIXO 02
MF = √ (MFH) 2 (MFV) 2
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MF = √ ( 165381.57 ) 2 + ( 44397.86 ) 2
MF = √ 27351063452.25 + 1971170037.34 MF = 171237.36 kgf.mm
MC = (MF) 2 + ( α x MT)2√ 2
MC = ( 171237.36 ) 2 + ( 0.67 x 101555.00 )2√
MC = √ 29322233489.59 + 4583741344.44 MC = 184135.75 kgf.mm
D > 2.17 x 3 Mc D > 1.72 x 3√ Tadm √
D > 33.38 mm D = 16 mm
EIXO 03
MF = √ (MFH) 2 (MFV) 2
MF = √ ( 153871.21 ) 2 + ( 55393.64 ) 2
MF = √ 23676349919.65 + 3068454949.59 MF = 163538.39 kgf.mm
MC = (MF) 2 + ( α x MT)2√ 2
MC = ( 163538.39 ) 2 + ( 0.67 x 767266.66 )2√
MC = √ 26744804869.24 + 261643612243.36 MC = 537018.08 kgf.mm
D > 2.17 x 3 Mc D > 1.72 x 3√ Tadm √
D > 47.69 mm D = 19 mm
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18.90 Kgf/mm2
25.20 Kgf/mm2
1.33 Kgf/mm2
kgf.mm
kgf.mm
38756.6825.20
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kgf.mm
kgf.mm
184135.7525.20
kgf.mm
kgf.mm
537018.0825.20
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13. CÁLCULO DO DIMENSIONAMENTO DOS CUBOS
Dados Tabelados Eixo Chavetado segundo DIN 5463
X = 0.13 a 0.21 X = 0.17
y = 0.125 a 0.16 y = 0.14
y' = 0.10 a 0.14 y' = 0.12
Engrenagem 1
L = x x MT 1/3 L = 0.17 x 3 √ 1342.88 1/3 L = 0.17 x 11.03 L = 1.88 cm L =
S = y x MT 1/3 S = 0.14 x 3 √ 1342.88 1/3 S = 0.14 x 11.03 S = 1.57 cm S =
Engrenagem 2
L = x x MT 1/3 L = 0.17 x 3 √ 10155.00 1/3 L = 0.17 x 21.66 L = 3.68 cm L =
S = y x MT 1/3 S = 0.14 x 3 √ 10155.00 1/3 S = 0.14 x 21.66 S = 3.09 cm S =
S' = y' x MT 1/3 S' = 0.12 x 3 √ 10155.00 1/3 S' = 0.12 x 21.66 S' = 2.60 cm S' =
Engrenagem 3
L = x x MT 1/3 L = 0.17 x 3 √ 10155.00 1/3 L = 0.17 x 21.66 L = 3.68 cm L =
S = y x MT 1/3 S = 0.14 x 3 √ 10155.00 1/3 S = 0.14 x 21.66 S = 3.09 cm S =
S' = y' x MT 1/3 S' = 0.12 x 3 √ 10155.00 1/3 S' = 0.12 x 21.66 S' = 2.60 cm S' =
Engrenagem 4
L = x x MT 1/3 L = 0.17 x 3 √ 76726.00 1/3 L = 0.17 x 42.49 L = 7.22 cm L =
S = y x MT 1/3 S = 0.14 x 3 √ 76726.00 1/3 S = 0.14 x 42.49 S = 6.06 cm S =
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S' = y' x MT 1/3 S' = 0.12 x 3 √ 76726.00 1/3 S' = 0.12 x 42.49 S' = 5.10 cm S' =
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12 mm
9.5 mm
18 mm
14.5 mm
12.5 mm
18 mm
14.5 mm
12.5 mm
27 mm
22 mm
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18.5 mm
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14. CÁLCULO DO DIMENSIONAMENTO DAS CHAVETAS
Adotado:Engrenagem 2 p = 8.2
L > 2 x 10 x MT L > 2 x 10155.00d x p x ( h - t1 ) 16 x 8.2 x ( 5 - 2.9 ) x 5
L > 20310.00 L > 14.74 mm L = 15 mm1377.60
Dados Tabelados Eixo Chavetado segundo DIN 5463b = 5 mm h = 5 mm t2 = 2.2 mm t1 = 2.9
Engrenagem 3
L > 2 x 10 x MT L > 2 x 10155.00d x p x ( h - t1 ) 16 x 8.2 x ( 5 - 2.9 )
L > 20310.00 L > 24.57 mm L = 25 mm826.56
Dados Tabelados Eixo Chavetado segundo DIN 5463b = 5 mm h = 5 mm t2 = 2.2 mm t1 = 2.9
Engrenagem 4
L > 2 x 10 x MT L > 2 x 76726.00d x p x ( h - t1 ) 19 x 8.2 x ( 7 - 2.9 )
L > 153452.00 L > 26.69 mm L = 27 mm5749.02
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Dados Tabelados Eixo Chavetado segundo DIN 5463b = 6 mm h = 6 mm t2 = 2.6 mm t1 = 3.5
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[Kgf/mm²]
mm
mm
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mm