## Machine Design Miscellaneous

#### Machine Design

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

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. Which one of the following is used to convert a rotational motion into a translational motion?

1. NA

##### Correct Option: D

NA

1. In a band brake the ratio of tight side band tension to the tension on the slack side is 3. If the angle of overlap of band on the drum is 180°, the coefficient of friction required between drum and the band is

1.  Given , Tension in tight side = T1 = 3 Tension is slack side T2

 T1 = eμ θ T2

where, θ = angle of overlap = 180 = π radians
∴ 3 = eμ θ
ln(3) = μ θ
 ∴ μ = 1 ln(3) = 0.35 π

##### Correct Option: D

 Given , Tension in tight side = T1 = 3 Tension is slack side T2

 T1 = eμ θ T2

where, θ = angle of overlap = 180 = π radians
∴ 3 = eμ θ
ln(3) = μ θ
 ∴ μ = 1 ln(3) = 0.35 π

Direction: A band brake consists of a lever attached to one end of the band. The other end of the band is fixed to the ground. The wheel has a radius of 200 mm and the wrap angle of the band is 270°. The braking force applied to the lever is limited to 100 N, and the coefficient of friction between the band and the wheel is 0.5. No other information is given.

1. The maximum tension that can be generated in the band during braking is

1. θ = 270°
μ = 0.5

∑ M0 = 0
⇒ T2 × 1 - 100 × 2 = 0
T2 = 200 N

 T1 = eμ θ T2

 T1 = e0.5 × (3π / 2) T2

T1 = 2110 N

θ = 270°
μ = 0.5