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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: