This set of Theory of Machines Multiple Choice Questions & Answers (MCQs) focuses on “Single Plate Clutch”.

1. Which of these is not a type of clutch?

a) Single disc

b) Conical

c) Centrifugal

d) Cylindrical

View Answer

Explanation: Single disc, conical and centrifugal are types of clutches. Clutches are used for the transmission of power in shafts and machines which are started and stopped frequently. It is also used to deliver power to the machine which are partially or fully loaded.

2. Identify the clutch.

a) Single plate clutch

b) Conical clutch

c) Centrifugal clutch

d) Multi plate clutch

View Answer

Explanation: The given diagram is of a single plate clutch. It consists of a clutch plate whose both sides are faced with Ferrodo (a friction material). It is mounted on the hub which moves along the splines of the shaft which is driven.

3. In a single plate 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})

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

c) (r_{1} – r_{2})/2

d) (r_{1} + r_{2})/2

View Answer

Explanation: For uniform pressure theory, T = nµW(2(r

_{1}

^{3}– r

_{2}

^{3}) / 3(r

_{1}

^{2}– r

_{2}

^{2}))

Therefore, considering uniform pressure theory in a single plate clutch, T = nµWR where R = 2(r

_{1}

^{3}– r

_{2}

^{3}) / 3(r

_{1}

^{2}– r

_{2}

^{2}).

4. In a single plate 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})

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

c) (r_{1} – r_{2})/2

d) (r_{1} + r_{2})/2

View Answer

Explanation: For uniform wear, T = nµW((r

_{1}+ r

_{2})/2)

Therefore, considering uniform pressure theory in a single plate clutch, T = nµWR where R = (r

_{1}+ r

_{2})/2.

5. The intensity of pressure is minimum at the outer radius that is r_{1}. True or false?

a) True

b) False

View Answer

Explanation: The intensity of pressure is minimum at the outer radius that is r

_{1}. It is given by C/r

_{1}. The intensity of pressure is maximum at the inner radius which is r

_{2}.

6. In a single plate clutch, the pressure is uniformly distributed. If the outer and inner radii are 100 mm and 70 mm respectively, find the value of R.

a) 76.39 mm

b) 23.48 mm

c) 85.88 mm

d) 34.98 mm

View Answer

Explanation: Considering uniform pressure theory in a single plate clutch, T = nµWR where R = 2(r

_{1}

^{3}+ r

_{2}

^{3}) / 3(r

_{1}

^{2}+ r

_{2}

^{2})

Thus, here, R = 2(100

^{3}– 70

^{3}) / 3(100

^{2}– 70

^{2}) = 85.88 mm.

7. Maximum intensity of pressure is given by ____

a) C/R

b) C/R^{2}

c) C/r_{2}

d) C/r_{1}

View Answer

Explanation: We know that, pmax x r

_{2}= C.

Thus, pmax = C/r

_{2}

Similarly, pmin = C/r

_{1}

8. When the pressure is uniformly distributed over entire area, then the intensity of pressure is _______________

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

Explanation: Pressure = Force / Area

Therefore, the intensity of pressure considering uniform pressure is W/ (π(r

_{1}

^{2}– r

_{2}

^{2}).

9. For uniform wear, the axial thrust W = ______________

a) 2 π C (r_{2} + r_{1})

b) 2 π C (r_{1} – r_{2})

c) 2 π C (r_{2} – r_{1})

d) 2 π C (r_{1} x r_{2})

View Answer

Explanation: In uniform wear theory, C = W/(2 π (r

_{1}– r

_{2}))

Thus, W = 2 π C (r

_{1}– r

_{2}).

10. Find the axial thrust to be provided by the springs if the maximum intensity of pressure in a single plate clutch should not exceed 0.2 N/mm^{2}. The outer and the inner radius is 75 mm and 35 mm. Assume theory of uniform wear.

a) 1792.01 N

b) 1789.21 N

c) 1723.41 N

d) 1759.29 N

View Answer

Explanation: Given: pmax = 0.2 N/mm

^{2}, r

_{1}= 75 mm and r

_{2}= 35 mm

Thus, W = 2 π C (r

_{1}– r

_{2}) = 2 π C (r

_{1}– r

_{2}) = 2 π x pmax x r

_{2}(r

_{1}– r

_{2}) = 2 π x 0.2 x 35 (75 – 35) = 1759.29 N.

11. Find the average pressure in a single plate clutch if the axial force is 5 kN. The inside radius and the outer radius is 30 mm and 70 mm respectively. Assume uniform wear.

a) 398 N/mm^{2}

b) 0.398 N/m^{2}

c) 0.398 N/mm^{2}

d) 398 N/m^{2}

View Answer

Explanation: Given: W = 5 kN = 5 x 10

^{3}N, r

_{1}= 70 mm and r

_{2}= 30 mm

Average pressure = Total normal force on the surface/ Cross sectional area of those surfaces = W/ (π(r

_{1}

^{2}– r

_{2}

^{2})) = 5 x 103/(π(70

^{2}– 30

^{2})) = 0.398 N/mm

^{2}.

12. In a single plate clutch if the axial force is equal to 8 kN, find the minimum and maximum intensity of pressure. The outer radius = 100 mm and inner radius = 65 mm.

a) 0.36 N/mm^{2}, 0.56 N/mm^{2}

b) 0.56 N/mm^{2}, 0.36 N/mm^{2}

c) 0.36 N/m^{2}, 0.56 N/m^{2}

d) 0.56 N/m^{2}, 0.36 N/m^{2}

View Answer

Explanation: Given : W = 8 kN = 8 x 10

^{3}N, r

_{1}= 100 mm and r

_{2}= 65 mm

Maximum intensity of pressure = p

_{min}= C/r

_{1}

Thus, C = 100 x p

_{min}

W = 2 π C (r

_{1}– r

_{2})

Thus,

C = W/(2 π (r

_{1}– r

_{2})) = 8 x 103/ (2 π (100 – 65))

C = 100 x p

_{min}= 36.378

p

_{min}= 0.36 N/mm

^{2}

Maximum intensity of pressure = pmax = C/r

_{2}

Thus, C = 65 x p

_{max}

W = 2 π C (r

_{1}– r

_{2})

Thus,

C = W/(2 π (r

_{1}– r

_{2})) = 8 x 103/ (2 π (100 – 65))

C = 65 x pmax = 36.378

pmax = 0.56 N/mm

^{2}.

13. A single plate clutch, having n = 2, has outer and inner radii 150 mm and 100 mm respectively. The maximum intensity of pressure at any point is 0.1 N/mm^{2}. If the µ is 0.3, determine the power transmitted by a clutch at a speed 3000 r.p.m.

a) 74031 kW

b) 740.31 kW

c) 74.031 kW

d) 706.95 kW

View Answer

Explanation: Given : n = 2, r

_{1}= 150 mm ; r

_{2}= 100 mm ; p = 0.1 N/mm

^{2}; µ = 0.3 ; N = 3000 r.p.m. or ω = 2π × 3000/60 = 314.16 rad/s.

Since the intensity of pressure (p) is maximum at the inner radius (r

_{2}), considering uniform wear, p

_{max}x r

_{2}= C or C = 0.1 × 100 = 10 N/mm.

Axial thrust, W = 2 π C (r

_{1}– r

_{2}) = 2π × 10 (150 – 100) = 3142 N

Mean radius = R = (r

_{1}+ r

_{2})/2 = 125 mm = 0.125 m.

We know that torque transmitted,

T = n.µ.W.R = 2 × 0.3 × 3142 × 0.125 = 235.65 N-m

Thus, power transmitted,

P = T.ω = 235.65 × 314.16 = 74031 W = 74.031 kW.

14. A single plate clutch, having n = 2, has outer and inner radii 200 mm and 175 mm respectively. The maximum intensity of pressure at any point is 0.05 N/mm^{2}. If the µ is 0.4 and the power generated by the clutch is equal to 32.384 kW, determine the speed of the clutch.

a) 209.44 r.p.m.

b) 2000 r.p.m.

c) 157.08 r.p.m.

d) 1500 r.p.m.

View Answer

Explanation: Given : n = 2, r

_{1}= 200 mm ; r

_{2}= 175 mm ; p = 0.05 N/mm

^{2}; µ = 0.4 ; P = 32.384 kW = 32384 W

Since the intensity of pressure (p) is maximum at the inner radius (r

_{2}), considering uniform wear, p

_{max}x r

_{2}= C or C = 0.05 × 175 = 8.75 N/mm

Axial thrust, W = 2 π C (r

_{1}– r

_{2}) = 2π × 8.75 (200 – 175) = 1374.44 N

Mean radius = R = (r

_{1}+ r

_{2})/2 = 187.5 mm = 0.1875 m.

We know that torque transmitted,

T = n.µ.W.R = 2 × 0.4 × 1374.44 × 0.1875 = 206.16 N-m

Thus, power transmitted,

P = T.ω

ω = P/T = 32384/206.16 = 157.08 rad/s

N = ω x 60 / (2 π) = 1500 r.p.m.

15. A single plate clutch, where n = 2, is required to transmit 30 kW at 3000 r.p.m. Determine the outer and inner radii if the coefficient of friction is 0.3, the ratio of radii is 1.5. pmax = 0.1 N/mm^{2}.

a) 0.111 mm, 0.074 mm

b) 111 mm, 74 mm

c) 74 mm, 111 mm

d) 0.074 mm, 0.111 mm

View Answer

Explanation: Given: n = 2 ; P = 30 kW = 30 × 103 W ; N = 3000 r.p.m. or ω = 2π × 3000/60 = 314.2 rad/s ; µ = 0.3 ; r

_{1}/r

_{2}= 1.5 ; p = 0.1 N/mm

^{2}.

Since the ratio of radii (r

_{1}/r

_{2}) is 1.5, therefore r

_{1}= 1.5r

_{2}

We know that the power transmitted (P), 30 × 10

^{3}= T.ω = T × 314.2 ∴ T = 30 × 103/314.2 = 95.48 N-m = 95.48 × 10

^{3}N-mm

The intensity of pressure is maximum at inner radius (r

_{2}),

pmax x r

_{2}= C or C = 0.1 x r

_{2}N/mm

W = 2 π C (r

_{1}– r

_{2}) = 2 π × 0.1 r

_{2}(1.5r

_{2}– r

_{2}) = 0.314 (r

_{2})

^{2}

Mean radius for uniform wear = R = (r

_{1}+ r

_{2})/2 = 1.25 r

_{2}

Torque transmitted (T), 95.48 × 10

^{3}= n.µ.W.R = 2 × 0.3 × 0.314 (r

_{2})2 × 1.25 r

_{2}= 0.2355 (r

_{2})

^{3}

(r

_{2})

^{3}= 95.48 × 10

^{3}/0.2355 = 405.43 × 10

^{3}or r

_{2}= 74 mm

r

_{1}= 1.5 r

_{2}= 1.5 × 74 = 111 mm.

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