Machine Design Miscellaneous
 Match the rolling element bearings with the most appropriate loading condition
Bearingtype Loading condition (a) Ball bearing (p) Tangential load (b) Roller bearing (q) Radial load (c) Needle bearing (r) Heavy radial load with impact (d) Taper roller bearing (s) Light radial load with space limitation (t) Heavy radial and axial load (u) Fatigue load

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a – q, b – r, c –s, d – t
Correct Option: C
a – q, b – r, c –s, d – t
 If the load on a ball bearing is reduced to half, the life of the ball bearing will

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LP^{3} = C
⇒ L_{1} P_{1}^{3} = L_{2} P_{1} ^{3} 2
⇒ L_{2} = 8L_{1}Correct Option: A
LP^{3} = C
⇒ L_{1} P_{1}^{3} = L_{2} P_{1} ^{3} 2
⇒ L_{2} = 8L_{1}
 The expected life of a ball bearing subjected to a load of 9800 N and working at 1000 rpm is 3000 hours. What is the expected life of the same bearing for a similar load of 4900 N and speed of 2000 rpm?

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L (P)^{3} = constant
L_{1} = P_{2} ^{3} L_{2} P_{1} ⇒ 1000 × 60 × 3000 = 4900 ^{3} 2000 × 60 × t_{2} 9800
t_{2} = 12000 hours
Correct Option: B
L (P)^{3} = constant
L_{1} = P_{2} ^{3} L_{2} P_{1} ⇒ 1000 × 60 × 3000 = 4900 ^{3} 2000 × 60 × t_{2} 9800
t_{2} = 12000 hours
 A bolted Joint is shown below. The maximum shear stress, in MPa, in the bolts at A and B, respectively are

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F_{A} = F_{B} = F_{C} = 10 kN 3 F_{A} = P.e.r = 10 × 10^{3} × 150 × 40 r_{A}² + r_{B}² + r_{C}² 40² + 0² + 40²
F'_{A} = 18.75 kN = F'_{C}
Shear load at A and C
F_{R} = √(F_{A})² + (F'_{A})² = √(3.33)² + (18.75)²
F_{R} = 19.04 KNτ_{max} = 19.04 × 10^{3} (π / 4) × 10^{2}
( τ_{max} )A,C = 242.5 MPa
( τ_{max} )B = 42.44 MPa
Correct Option: A
F_{A} = F_{B} = F_{C} = 10 kN 3 F_{A} = P.e.r = 10 × 10^{3} × 150 × 40 r_{A}² + r_{B}² + r_{C}² 40² + 0² + 40²
F'_{A} = 18.75 kN = F'_{C}
Shear load at A and C
F_{R} = √(F_{A})² + (F'_{A})² = √(3.33)² + (18.75)²
F_{R} = 19.04 KNτ_{max} = 19.04 × 10^{3} (π / 4) × 10^{2}
( τ_{max} )A,C = 242.5 MPa
( τ_{max} )B = 42.44 MPa
 In a bolted Joint two smooth members are connected with an axial tightening force of 2200 N. If the bolt used has metric threads of 4 mm pitch, the torque required for achieving the tightening force is

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Tightening force = 2200 N
Pitch = 4 mm = 0.004 mTorque = Tightening force × 0.004 = 1.4 Nm 2π Correct Option: C
Tightening force = 2200 N
Pitch = 4 mm = 0.004 mTorque = Tightening force × 0.004 = 1.4 Nm 2π