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

1. In a multi plate 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

Explanation: In a multi plate clutch, the formula for T is given by n.µ.W.R

The formula is the same for torque in a single plate clutch. But for large amount of torque to be transmitted, multi plate clutch is used. Multi plate clutches are used in motor vehicles and machine tools.

2. In a multi 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: Considering uniform pressure theory in a multi plate clutch, T = nµWR where R = 2(r

_{1}

^{3}– r

_{2}

^{3}) / 3(r

_{1}

^{2}– r

_{2}

^{2}). This value is the same as that for the single plate clutch considering uniformly distributed pressure.

3. In a multi 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: Considering uniform pressure theory in a multi plate clutch, T = nµWR where R = (r

_{1}+ r

_{2})/2. This value is the same as that for the single plate clutch considering uniform wear.

4. In a multi plate clutch, T = 150 N-m, n = 4, µ = 0.3 and R = 0.1 m. Find the axial thrust.

a) 18

b) 1800

c) 1250

d) 200

View Answer

Explanation: T = nµWR

150 = 4 x 0.3 x W x 0.1

W = 1250 N

Thus, the axial thrust = 1250 N.

5. Maximum intensity of pressure for multi plate clutch 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}.

6. For a single plate clutch, n = 2, whereas for a multi plate clutch, n can take any values ≥ 2. True or false?

a) True

b) False

View Answer

Explanation: A single plate clutch has both sides effective, thus n = 2. In a multi plate clutch, the value of n can be equal or greater than 2, which helps in transmitting large amounts of torque.

7. In a multi plate clutch, number of discs on the driving shaft is given by n_{1}. True or false?

a) True

b) False

View Answer

Explanation: In a multi plate clutch, number of discs on the driving shaft is given by n

_{1}, whereas number of discs on the driven shaft is given by n

_{2}. With the help of these two variables we can find out the number of pairs of contact surfaces.

8. In a multi plate clutch, number of pairs of contact surfaces (n) = ________

a) n_{1} – n_{2} + 1

b) n_{1} – n_{2} – 1

c) n_{1} + n_{2} – 1

d) n_{1} + n_{2} + 1

View Answer

Explanation: In a multi plate clutch, number of pairs of contact surfaces (n) is equal to n

_{1}+ n

_{2}– 1. Due to the high number of pairs of contact surfaces, the torque transmitted is higher.

n is always a whole number.

9. A multi-disc clutch has 6 discs on the driving shaft and 4 on the driven shaft. Find the number of pairs of contact surfaces.

a) 3

b) 10

c) 9

d) 11

View Answer

Explanation: In a multi plate clutch, number of pairs of contact surfaces (n) is equal to n

_{1}+ n

_{2}– 1 = 6+4-1 = 9

Thus, T = n.µ.W.R = 9.µ.W.R

10. Calculate the maximum and minimum intensity of pressure in multi plate clutch when the axial force is 5 kN assuming uniform wear. Inner radius = 20 mm and outer radius = 40 mm.

a) 1.989 N/mm^{2}, 0.994 N/mm^{2}

b) 0.994 N/mm^{2}, 1.989 N/mm^{2}

c) 1.989 N/m^{2}, 0.994 N/m^{2}

d) 0.994 N/m^{2}, 1.989 N/m^{2}

View Answer

Explanation: We know that, p

_{max}x r

_{2}= C

p

_{max}x 20 = C

Now, W = 2 π C (r

_{1}– r

_{2})

5 x 10

^{3}= 2 π x p

_{max}x 20 x (40-20)

p

_{max}= 1.989 N/mm

^{2}

We know that, p

_{min}x r

_{1}= C

p

_{min}x 40 = C

Now, W = 2 π C (r

_{1}– r

_{2})

5 x 10

^{3}= 2 π x p

_{max}x 40 x (40-20)

p

_{max}= 0.994 N/mm

^{2}.

11. A multi-disc clutch has 3 discs on the driving shaft and 2 on the driven shaft. The outer radius is 120 mm and inside radius 60 mm. Considering uniform wear and µ = 0.3, find the Find the value of W for transmitting 25 kW at 1500r.p.m.

a) 651 N/m^{2}

b) 651 N/mm^{2}

c) 0.0651 N/m^{2}

d) 0.0651 N/mm^{2}

View Answer

Explanation: Given : n

_{1}= 3 ; n

_{2}= 2 ; r

_{1}= 120 mm ; r

_{2}= 60 mm ; µ = 0.3 ; P = 25 kW = 25 × 10

^{3}W ; N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157.08 rad/s

Power transmitted P = 25 × 10

^{3}= T.ω = T × 157.08

T = 25 × 10

^{3}/157.08 = 159.15 N-m

n = n

_{1}+ n

_{2}– 1 = 4

Mean radius = R = (r

_{1}+ r

_{2})/2 = 90 mm = 0.09 m

159.15 = n.µ.W.R = 4 × 0.3 × W × 0.09 = 0.108 W

W = 1473.65 N.

12. A multi-disc clutch has 3 discs on the driving shaft and 2 on the driven shaft. The outer radius is 100 mm and inside radius 40 mm. Considering uniform wear and µ = 0.25, find the maximum axial intensity of pressure between the discs for transmitting 30 kW at 1000 r.p.m.

a) 0.4071 N/m^{2}

b) 40.71 N/mm^{2}

c) 407.1 N/m^{2}

d) 0.4071 N/mm^{2}

View Answer

Explanation: Explanation: Given : n

_{1}= 3 ; n

_{2}= 2 ; r

_{1}= 100 mm ; r

_{2}= 40 mm ; µ = 0.25 ; P = 30 kW = 30 × 10

^{3}W ; N = 1000 r.p.m. or ω = 2 π × 1000/60 = 104.71 rad/s

Power transmitted P = 30 × 10

^{3}= T.ω = T × 104.71

T = 30 × 10

^{3}/157.08 = 286.48 N-m

n = n

_{1}+ n

_{2}– 1 = 4

Mean radius = R = (r

_{1}+ r

_{2})/2 = 70 mm = 0.07 m

286.48 = n.µ.W.R = 4 × 0.25 × W × 0.07

W = 4092.55 N

W = 4092.55 = 2 π C (r

_{1}– r

_{2}) = 80 π C

C = 16.283 = p

_{max}x 40

p

_{max}= 0.4071 N/mm

^{2}.

13. A single plate clutch, having n_{1} = 3 and n_{2} = 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, determine the power transmitted by a clutch at a speed 1500 r.p.m.

a) 32384 x 10^{3}W

b) 64.768 W

c) 64768 kW

d) 64768 W

View Answer

Explanation: Given : n = 4, r

_{1}= 200 mm ; r

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

^{2}; µ = 0.4 ; N = 1500 r.p.m. or ω = 2π × 1500/60 = 157.08 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 = 0.05 × 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 = 4.µ.W.R = 4 × 0.4 × 1374.44 × 0.1875 = 412.32 N-m

Thus, power transmitted,

P = T.ω = 412.32 × 157.08 = 64768 W = 64.368 kW.

14. A plate clutch has three discs on driving shaft and two discs on driven shaft. The outer radius of the contact surfaces is 100 mm and inner radius 50 mm. Assuming uniform pressure and µ = 0.3; find the total spring load pressing plates together to transmit 35 kW at 1600r.p.m.

a) 2372.3 N

b) 2260.7 N

c) 2401.8 N

d) 2131.4 N

View Answer

Explanation: Given : n

_{1}= 3 ; n

_{2}= 2 ; n = 4 ; r

_{1}= 100 mm ; r

_{2}= 50 mm ; µ = 0.3 ; P = 35 kW = 35 × 10

^{3}W ; N = 1600 r.p.m. or ω = 2 π × 1600/60 = 167.55 rad/s

We know that power transmitted (P), 35 × 10

^{3}= T.ω = T × 167.55

T = 35 × 10

^{3}/167.55 = 208.89 N-m

Mean radius of the contact surface, for uniform pressure,R = 2(r

_{1}

^{3}– r

_{2}

^{3}) / 3(r

_{1}

^{2}– r

_{2}

^{2}) = 77.77 mm = 0.077 m

208.89 = n.µ.W.R = 4 × 0.3 W × 0.077

W = 2260.7 N.

15. A multi plate clutch, having n_{1}= 2 and n_{2} = 1, has outer and inner radii 1000 mm and 600 mm respectively. The maximum intensity of pressure at any point is 0.2 N/mm^{2}. If the µ is 0.25, determine the speed of the clutch if the power generated = 31582.807 kW.

a) 2500 r.p.m.

b) 261.8 r.p.m.

c) 2000 r.p.m.

d) 209.44 r.p.m.

View Answer

Explanation: Given : n = 2, r

_{1}= 1000 mm ; r

_{2}= 600 mm ; p = 0.2 N/mm

^{2}; µ = 0.25 ; P = 31582.807 kW = 31582.807 x 10

^{3}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 = 0.2 × 600 = 120 N/mm

Axial thrust, W = 2 π C (r

_{1}– r

_{2}) = 2π × 120 (1000 – 600) = 301592 N

Mean radius = R = (r

_{1}+ r

_{2})/2 = 800 mm = 0.8 m.

We know that torque transmitted,

T = n.µ.W.R = 2 × 0.25 × 301592 × 0.8 = 120637 N-m

Thus, power transmitted,

P = T.ω

ω = P/T = 261.8 rad/s

N = ω x 60 / (2 π) = 2500 r.p.m.

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