Machine Design Miscellaneous
Direction: The overall gear ratio in a 2 stage speed reduction gear box (with all spur gears) is 12. The input and output shafts of the gear box are collinear. The countershaft which is parallel to the input and output shafts has a gear (Z2 teeth) and pinion (Z3 = 15 teeth) to mesh with pinion (Z1 = 16 teeth) on the input shaft and gear (Z4 teeth) on the output shaft respectively. It was decided to use a gear ratio of 4 with 3 module in the first stage and 4 module in the second stage.
- The centre distance in the second stage is
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Now, x = r4 + r3 = D4 + D3 2 But D4 = D3 = 4 Z4 Z3
⇒ D4 =180, D3 = 60∴ x = 180 + 60 = 120 mm 2 Correct Option: B
Now, x = r4 + r3 = D4 + D3 2 But D4 = D3 = 4 Z4 Z3
⇒ D4 =180, D3 = 60∴ x = 180 + 60 = 120 mm 2
- Z2 and Z4 are
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Given, N1 = 12, N1 = 4 = D2 N4 N2 D1 
Now, D1 = D2 Z1 Z2 ⇒ Z1 = D1 = N2 = 1 Z2 D2 N1 4
Z2 = Z1 × 4 = 64and N1 = D2 × D4 N4 D1 × D3 ⇒ 12 = D4 D3 ⇒ D4 = 3 D3 Also, Z3 = D3 Z4 Z4 ⇒ Z4 = Z3. Z4 = Z3 × 3 = 15 × 3 = 45 Z3 Correct Option: A
Given, N1 = 12, N1 = 4 = D2 N4 N2 D1 Now, D1 = D2 Z1 Z2 ⇒ Z1 = D1 = N2 = 1 Z2 D2 N1 4
Z2 = Z1 × 4 = 64and N1 = D2 × D4 N4 D1 × D3 ⇒ 12 = D4 D3 ⇒ D4 = 3 D3 Also, Z3 = D3 Z4 Z4 ⇒ Z4 = Z3. Z4 = Z3 × 3 = 15 × 3 = 45 Z3
- For ball bearings, the fatigue life L measured in number of revolutions and the radial load P are related by FL1/3 = K, where K is a constant. It withstands a radial load of 2 kN for a life of 540 million revolutions. The load (in kN) for a life of one million revolutions is ______.
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FL1/3 = k
2 × 4501/3 = (1)1/3
F = 16.2865 kNCorrect Option: A
FL1/3 = k
2 × 4501/3 = (1)1/3
F = 16.2865 kN
- A hydrodynamic Journal bearing is subjected to 2000 N load at a rotational speed of 2000 rpm. Both bearing bore diameter and length are 40 mm. If radial clearance is 20 mm and bearing is lubricated with an oil having viscosity 0.03 Pa.s, the Sommerfeld number of the bearing is __________.
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S = 
r 
² μDs c P
d = 40 mm ⇒ r = 20 mm
c = 20 μm ⇒ c = 20 × 10–3mm
μ = 0.03 PaS ⇒ μ = 0.03 × 1–6MPa.S
Ds = 2000 rpm ⇒ Ds = 2000/60 rpsP = p = 2000 = 1.25 MPa 1 × d 40 × 40 S = 20 ² × 0.03 × 10-6 × 2000/60 = 0.8 20 × 10-3 1.25 Correct Option: B
S = r ² μDs c P
d = 40 mm ⇒ r = 20 mm
c = 20 μm ⇒ c = 20 × 10–3mm
μ = 0.03 PaS ⇒ μ = 0.03 × 1–6MPa.S
Ds = 2000 rpm ⇒ Ds = 2000/60 rpsP = p = 2000 = 1.25 MPa 1 × d 40 × 40 S = 20 ² × 0.03 × 10-6 × 2000/60 = 0.8 20 × 10-3 1.25
- Ball bearings are rated by a manufacturer for a life of 106 revolutions. The catalogue rating of a particular bearing is 16 kN. If the design load is 2 kN, the life of the bearing will be P x 106 revolutions, where P is equal to _________
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L10 = C ³ W
where L10 = Basic life rating in millions of revolutions C = Basic dynamic load rating
P = Equivalent dynamic radial load∴ L10 = 16 ³ = 5 2 2 Correct Option: B
L10 = C ³ W
where L10 = Basic life rating in millions of revolutions C = Basic dynamic load rating
P = Equivalent dynamic radial load∴ L10 = 16 ³ = 5 2 2
