Machine Design Miscellaneous

Machine Design

1. A band brake having bandwidth of 80 mm, drum diameter of 250 mm, coefficient of friction of 0.25 and angle of wrap of 270 degrees is required to exert a friction torque of 1000 N m. The maximum tension (in kN) developed in the band is

1. Given: b = 80 mm = 0.08 m D = 250 mm = 0.25

 R = D = 0.25 2

μ = 0.25
 θ = 270° = 270 × π = 4.712 radians 180

Ft = 100 N-m
 As p1 = eμθ p2

⇒  p1 = eμθ ⋅ p2 = e0.25 × 4.712 ⋅ p2 = 3.25p2
Now, Ft = (p1 − p2)R
 ∴  1000 = p1 − p1 0.125 3.25

Solving, we get maximum tension in the belt,
p1 = 11.56 kN

Correct Option: D

Given: b = 80 mm = 0.08 m D = 250 mm = 0.25

 R = D = 0.25 2

μ = 0.25
 θ = 270° = 270 × π = 4.712 radians 180

Ft = 100 N-m
 As p1 = eμθ p2

⇒  p1 = eμθ ⋅ p2 = e0.25 × 4.712 ⋅ p2 = 3.25p2
Now, Ft = (p1 − p2)R
 ∴  1000 = p1 − p1 0.125 3.25

Solving, we get maximum tension in the belt,
p1 = 11.56 kN

1. A force of 400 N is applied to the brake drum of 0.5 m diameter in a band brake system as shown in figure, where the wrapping angle is 180°. If the coefficient of friction between the drum and the band is 0.25, the braking torque applied, in Nm is

1. As the drum is rotating in anti clock wise direction, T1 will be right side & T2 will be clock side.

 T1 = eμθ = e0.25 × π = 2.19 T2

⇒  T2 = = 182.375 N
Braking torque = (T1 – T2 )r
= 54.4 N – m

Correct Option: B

As the drum is rotating in anti clock wise direction, T1 will be right side & T2 will be clock side.

 T1 = eμθ = e0.25 × π = 2.19 T2

⇒  T2 = = 182.375 N
Braking torque = (T1 – T2 )r
= 54.4 N – m

1. A drum brake is shown in the figure. The drum is rotating in anticlockwise direction. The coefficient of friction between drum and shoe is 0.2. The dimensions shown in the figure are in mm. The braking torque (in Nm) for the brake shoe is________.

1. Taking moments about pin joint (SM)
⇒  1000 × 0.8 =RN × 0.48 + μRN × 0.1

 ⇒  RN = 800 = 1600 0.5

TB = μRN × r
 = 0.2 × 1600 × 200 = 64 N - M 1000

Correct Option: A

Taking moments about pin joint (SM)
⇒  1000 × 0.8 =RN × 0.48 + μRN × 0.1

 ⇒  RN = 800 = 1600 0.5

TB = μRN × r
 = 0.2 × 1600 × 200 = 64 N - M 1000

1. In spur gears, the circle on which the involute is generated is called the

1. NA

Correct Option: C

NA

1. The forces F1 and F2 in a brake band and the direction of rotation of the drum are as shown in the figure. The coefficient of friction is 0.25. The angle of wrap is 3π/2 radians. It is given that R = 1 m and F2 = 1 N. The torque (in Nm) exerted on the drum is ________ .

1.  F1 = e μ θ = e 0.25 × (3π / 2) = 3.248 F2

∴ F1 = 3.248 N
Torque, T = (F1 – F2) × r
= (3.248 – 1) × 1 = 2.248 N-m

Correct Option: C

 F1 = e μ θ = e 0.25 × (3π / 2) = 3.248 F2

∴ F1 = 3.248 N
Torque, T = (F1 – F2) × r
= (3.248 – 1) × 1 = 2.248 N-m