Tugas Perancangan Gear Box
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Transcript of Tugas Perancangan Gear Box
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8/18/2019 Tugas Perancangan Gear Box
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TUGAS KELOMPOK
Perancangan Gear Box
• Given :
F = 4kN v = 1m s� ∅ = 400 = 720 = 620
= 350 = 0.94 − 0.98 = 0.99 = 2 = 1.5
•
Solution 1- Calculating the required power and velocity :
= ∗ = 4 ∗ 1 = 4
=2
=2
0.4 = 5 �
=60
2 =30 ∗ 53.14
= 47.77
2- Motor Selection :
Using a 2-stage gearbox the required input power can be calculated as follows :
=
∗ =4
0.962 ∗ 0.996 =4
0.9216 ∗ 0.94148 =4
0.8504 = 4.703
ARDI ARMAWAN ( 14503241025)
MUHAMAD SIDIQ PRAMOKO ( 14503241027)
ANDRIAS NUR WIBOWO ( 14503241028)
TABAH CANDRA PRASETYA ( 14503241025)
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From Appendix 1. Series 4A Tree Phase Induction Motors, we choose a suitable motor
for the required input power.
Selection : (IEC60034) P=5.5kW & N=715rpm
3-
Reduction rations :
= =
715
47.77 = 14.97 ≈ 3 ∗ 4.97
.. . 12 = 3
.. . 34 = 4.97 4- Calculation torque :
=260
=2 ∗ 715
60 = 74.83 �
= =
5.574.83
= 73.5
= ∗ = 73.49 ∗ 0.992 = 72.04 = ∗ 12 ∗ 12 ∗ = 72.03 ∗ 3 ∗ 0.96 ∗ 0.992 = 203.3257 = ∗ 34 ∗ 34 ∗ = 203.318 ∗ 4.99 ∗ 0.96 ∗ 0.992 = 944.97
5- 1st Stage analysis:
= 72.04 = 715
= 20 = 60
= 20
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6- 2st Stage analysis:
= 203.3257 = 238,3333
= 31 = 154 = 20
Insert data to the software included in this project and obtain the results.
Nb. Menyusul :v
• Imput shaf
= 73,5 N.m\ = 2 / = 2 ∗ 73.5 ∗ 103/ 60 = 2,45 = tan = 2,45 tan 20° = 0.89 N = (2,45)^2 + (0,89)2 = 2,6
= 0 2,6 ∗ 150 − R ∗ 200 = 0 → R = 1.95 = R ∗ 50 = 97,5 .
=
16
√ +
= 22,73mm => 23 mm
50 mm
2,45 kN
0,89 KN
150m
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• Bearing selection
(F = 1,9056 , F = 0) → ( F F = 0 < ) → = 1, = 0 Type equation here. = F + F = F = 1950
ℎ
= 20,000ℎ
= ∗ ( 60 ℎ 10^6 ) 1 → ℎ = 20,000 ℎ, = 715 , = 3 = 18529,5N => Bearing dari tabel:
• Intermediate shaft: