Machine Design Miscellaneous
- To restore stable operating condition in a hydrodynamic Journal bearing, when it encounters higher magnitude of loads
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For stability 
ZN 
↑ &Z ↑ &P ↓ P Correct Option: B
For stability ZN ↑ &Z ↑ &P ↓ P
- In thick film hydrodynamic Journal bearings, the coefficient of friction
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f ∝ μN P Correct Option: C
f ∝ μN P
- A bolt of major diameter 12 mm is required to clamp two steel plates. Cross sectional area of the threaded portion of the bolt is 84.3 mm2. The length of the threaded portion in grip is 30 mm, while the length of the unthreaded portion in grip is 8 mm. Young's modulus of material is 200 GPa. The effective stiffness (in MN/m) of the bolt in the clamped zone is _____.
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d1 = 12 mm;
l1 = 8 mm
A2 = 84.33 mm²A1 = π (d1)² 4
l2 = 30 mmK1 = A1E1 l1 K2 = A2E2 l2 = 84.33 × 200 = 562.2 30 = (π/4)(12)² × 200 = 2827.4 8 1 = 1 + 1 K K1 K2 = π × 3000 + 1500 × π 8 = 1 + 1 2827.4 562.2
⇒ K = 468.9 MN/mCorrect Option: A
d1 = 12 mm;
l1 = 8 mm
A2 = 84.33 mm²A1 = π (d1)² 4
l2 = 30 mmK1 = A1E1 l1 K2 = A2E2 l2 = 84.33 × 200 = 562.2 30 = (π/4)(12)² × 200 = 2827.4 8 1 = 1 + 1 K K1 K2 = π × 3000 + 1500 × π 8 = 1 + 1 2827.4 562.2
⇒ K = 468.9 MN/m
- For the three bolt system shown in the figure, the bolt material has shear yield strength of 200 MPa. For a factor of safety of 2, the minimum metric specification required for the bolt is
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σy = 200 MPa
P = 19 × 10³Nσ = P ⇒ d = 10.9 mm (π/4)(d1)²
M10Correct Option: B
σy = 200 MPa
P = 19 × 10³Nσ = P ⇒ d = 10.9 mm (π/4)(d1)²
M10
- The resultant shear stress on bolt P is closest to:
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Resultant load on P = Fsec – Fprim
= 20 – 2 = 18 kN∴ Stress = 18 × 10³ = 159 MPa (π/4)(12)² Correct Option: B
Resultant load on P = Fsec – Fprim
= 20 – 2 = 18 kN∴ Stress = 18 × 10³ = 159 MPa (π/4)(12)²
