1. Identify the given clutch.

a) Single plate

b) Multi plate

c) Conical

d) Centrifugal

### View Answer

2. In a conical clutch, pn = ______________

a) W/(π(r_{1}^{2} – r_{2}^{2}))

b) W/(π(r_{1}^{2} + r_{2}^{2}))

c) W/(π(r_{2}^{2} – r_{1}^{2}))

d) W/(π(r_{2}^{2} x r_{1}^{2}))

### View Answer

_{1}

^{2}– r

_{2}

^{2})).

3. In a conical clutch, the formula for T is given by ______

a) n.µ.W.R

b) n.µ.W.r_{1}

c) n.µ.W.r_{2}

d) n.µ.W.(r_{1}+r_{2})

### View Answer

4. In a conical clutch, considering uniform pressure, T = nµWR. What is R equal to?

a) {2(r_{1}^{3} + r_{2}^{3}) / 3(r_{1}^{2} + r_{2}^{2})} x cosec α

b) {2(r_{1}^{3} – r_{2}^{3}) / 3(r_{1}^{2} – r_{2}^{2})} x cosec α

c) {(r_{1} – r_{2})/2} x cosec α

d) {(r_{1} + r_{2})/2} x cosec α

### View Answer

_{1}

^{3}– r

_{2}

^{3}) / 3(r

_{1}

^{2}– r

_{2}

^{2})} x cosec α α = the semi cone angle.

5. In a conical clutch, considering uniform wear, T = nµWR. What is R equal to?

a) {2(r_{1}^{3} + r_{2}^{3}) / 3(r_{1}^{2} + r_{2}^{2})} x cosec α

b) {2(r_{1}^{3} – r_{2}^{3}) / 3(r_{1}^{2} – r_{2}^{2})} x cosec α

c) {(r_{1} – r_{2})/2} x cosec α

d) {(r_{1} + r_{2})/2} x cosec α

### View Answer

_{1}+ r

_{2})/2} x cosec α.

6. In a conical clutch, what is the axial force required for engaging the clutch (We)?

a) W_{n} (sin α + µ cos α)

b) W_{n} (sin α – µ cos α)

c) W_{n} (µ sin α + cos α)

d) W_{n} (µ sin α – cos α)

### View Answer

_{e}= W + µ.W

_{n}cos α = W

_{n}sinα + µ.W

_{n}cosα = W

_{n}(sinα + µcosα). If the semi cone angle of the clutch decreases, the torque produced by the clutch increases which in turn reduces the axial force W.

7. In a conical clutch, what is the axial force required for disengaging the clutch (Wd)?

a) W_{n}(µcosα + sinα)

b) W_{n}(µcosα – sinα)

c) W_{n}(cosα – µsinα)

d) W_{n}(cosα + µsinα)

### View Answer

_{d}= W

_{n}(µcosα – sinα). The axial force for disengagement is only required if the value of tan α is less than µ.

8. In a conical clutch, the breadth of the contact surface is given by ______

a) (r_{1} + r_{2})/sinα

b) (r_{2}-r_{1})/sinα

c) (r_{1} – r_{2})/sinα

d) (r_{1}^{2} + r_{2}^{2})/sinα

### View Answer

_{1}-r

_{2}Thus, b = (r

_{1}-r

_{2})/sinα

9. In a conical clutch, the mean radius of the bearing surface is 300 mm whereas the breadth is 20 mm. Find the inner and outer radii. The semi cone angle is 30°.

a) 145 mm, 155 mm

b) 140 mm, 160 mm

c) 160 mm, 140 mm

d) 155 mm, 145 mm

### View Answer

_{1}– r

_{2}r

_{1}– r

_{2}= 10 and r

_{1}+ r

_{2}= 300 r

_{1}= 155 mm and r

_{2}= 145 mm.

10. If the outer and inner radius of the contact surfaces are 100 mm and 75 mm respectively and the semi cone angle is 22.5°, find the value of the face width required.

a) 89.43 mm

b) 78.94 mm

c) 65.33 mm

d) 23.87 mm

### View Answer

_{1}– r

_{2}b x sin 22.5° = 100 – 75 b = 65.33 mm.

11. A conical friction clutch is used to transmit 75 kW at 1500 r.p.m. The semi cone angle is 20° and the coefficient of friction is 0.3. If the mean diameter of the bearing surface is 500 mm and the intensity of normal pressure is not to exceed 0.1 N/mm^{2}, find the dimensions of the conical bearing surface.

a) 118.07 mm, 131.93 mm

b) 131.93 mm, 118.07 mm

c) 121.72 mm, 128.28 mm

d) 128.28 mm, 121.72 mm

### View Answer

^{3}W ; N = 1500 r.p.m. or ω = 2 π × 1400/60 = 157.08 rad/s ; α = 20° ; µ = 0.3 ; D = 500 mm or R = 250 mm ; pn = 0.1 N/mm

^{2}Power transmitted (P) = 75 × 10

^{3}= T.ω = T × 157.08 T = 75× 10

^{3}/157.08 = 477.4 N-m = 477.4× 10

^{3}N-mm Torque transmitted (T) = 477.4 × 10

^{3}= 2 π µ p

_{n}.R

^{2}.b = 2π × 0.3 × 0.1 (250)

^{2}b = 11780.97 b b = 40.52 mm W

_{e}know that, r

_{1}+r

_{2}= 250 and r

_{1}-r

_{2}= 40.52 sin 20 = 13.86 Therefore, r

_{1}= 131.93 mm, r

_{2}= 118.07 mm.

12. A conical friction clutch is used to transmit 50 kW at 1000 r.p.m. The semi cone angle is 15° and the coefficient of friction is 0.2. If the mean diameter of the bearing surface is 450 mm and the intensity of normal pressure is not to exceed 0.15 N/mm^{2}, find the axial spring force necessary to engage the clutch.

a) 4796 N

b) 4774 N

c) 4785 N

d) 4742 N

### View Answer

^{3}W ; N = 1000 r.p.m. or ω = 2 π × 1000/60 = 104.72 rad/s ; α = 15° ; µ = 0.2 ; D = 450 mm or R = 225 mm ; pn = 0.15 N/mm

^{2}Power transmitted (P) = 50 × 10

^{3}= T.ω = T × 104.72 T = 50× 10

^{3}/104.72 = 477.4 N-m 477.4 = µ.W

_{n}.R = 0.2 × W

_{n}× 0.225 W

_{n}= 10610.33 N Axial spring force = W

_{e}= W

_{n}(sin α + µ cos α) = 10610.33 (sin 15° + 0.2 cos 15°) = 4796 N.

13. A conical friction clutch is used to transmit 40 kW at 1300 r.p.m. The semi cone angle is 12.5° and the coefficient of friction is 0.25. If the mean diameter of the bearing surface is 400 mm and the intensity of normal pressure is not to exceed 0.2 N/mm^{2}, find the axial spring force necessary to disengage the clutch.

a) 154.28 N

b) 121.45 N

c) 201.78 N

d) 162.39 N

### View Answer

^{3}W ; N = 1300 r.p.m. or ω = 2 π × 1300/60 = 136.13 rad/s ; α = 12.5° ; µ = 0.25 ; D = 400 mm or R = 200 mm ; pn = 0.2 N/mm

^{2}Power transmitted (P) = 40 × 10

^{3}= T.ω = T × 136.13 T = 40× 10

^{3}/136.13 = 293.82 N-m 293.82 = µ.W

_{n}.R = 0.25 × W

_{n}× 0.2 W

_{n}= 5876.5 N Axial spring force = W

_{d}= W

_{n}(µ cos α – sin α) = 5876.5(0.25 cos 12.5°- sin 12.5°) = 162.39 N.

### Synopsis and Project Report

You can buy synopsis and project from distpub.com. Just visit https://distpub.com/product-category/projects/ and buy your university/institute project from distpub.com