## Machine Design Miscellaneous

#### Machine Design

1. Match the rolling element bearings with the most appropriate loading condition

1. a – q, b – r, c –s, d – t

##### Correct Option: C

a – q, b – r, c –s, d – t

1. If the load on a ball bearing is reduced to half, the life of the ball bearing will

1. LP3 = C

 ⇒ L1 P13 = L2 P1 3 2

⇒ L2 = 8L1

##### Correct Option: A

LP3 = C

 ⇒ L1 P13 = L2 P1 3 2

⇒ L2 = 8L1

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?

1. L (P)3 = constant

 L1 = P2 3 L2 P1

 ⇒ 1000 × 60 × 3000 = 4900 3 2000 × 60 × t2 9800

t2 = 12000 hours

##### Correct Option: B

L (P)3 = constant

 L1 = P2 3 L2 P1

 ⇒ 1000 × 60 × 3000 = 4900 3 2000 × 60 × t2 9800

t2 = 12000 hours

1. A bolted Joint is shown below. The maximum shear stress, in MPa, in the bolts at A and B, respectively are

1.  FA = FB = FC = 10 kN 3

 FA = P.e.r = 10 × 103 × 150 × 40 rA² + rB² + rC² 40² + 0² + 40²

F'A = 18.75 kN = F'C
Shear load at A and C
FR = √(FA)² + (F'A = √(3.33)² + (18.75)²
FR = 19.04 KN
 τmax = 19.04 × 103 (π / 4) × 102

( τmax )A,C = 242.5 MPa
( τmax )B = 42.44 MPa

##### Correct Option: A

 FA = FB = FC = 10 kN 3

 FA = P.e.r = 10 × 103 × 150 × 40 rA² + rB² + rC² 40² + 0² + 40²

F'A = 18.75 kN = F'C
Shear load at A and C
FR = √(FA)² + (F'A = √(3.33)² + (18.75)²
FR = 19.04 KN
 τmax = 19.04 × 103 (π / 4) × 102

( τmax )A,C = 242.5 MPa
( τmax )B = 42.44 MPa

1. 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

1. Tightening force = 2200 N
Pitch = 4 mm = 0.004 m

 Torque = Tightening force × 0.004 = 1.4 N-m 2π

##### Correct Option: C

Tightening force = 2200 N
Pitch = 4 mm = 0.004 m

 Torque = Tightening force × 0.004 = 1.4 N-m 2π