1. 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/mm2
b) 0.398 N/m2
c) 0.398 N/mm2
d) 398 N/m2
Explanation: Given: W = 5 kN = 5 x 103 N, r1 = 70 mm and r2 = 30 mm
Average pressure = Total normal force on the surface/ Cross sectional area of those surfaces = W/ (π(r12 – r22)) = 5 x 103/(π(702 – 302)) = 0.398 N/mm2.
2. 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/mm2, 0.56 N/mm2
b) 0.56 N/mm2, 0.36 N/mm2
c) 0.36 N/m2, 0.56 N/m2
d) 0.56 N/m2, 0.36 N/m2
Explanation: Given : W = 8 kN = 8 x 103 N, r1 = 100 mm and r2 = 65 mm
Maximum intensity of pressure = pmin = C/r1
Thus, C = 100 x pmin
W = 2 π C (r1 – r2)
Thus,
C = W/(2 π (r1 – r2)) = 8 x 103/ (2 π (100 – 65))
C = 100 x pmin = 36.378
pmin = 0.36 N/mm2
Maximum intensity of pressure = pmax = C/r2
Thus, C = 65 x pmax
W = 2 π C (r1 – r2)
Thus,
C = W/(2 π (r1 – r2)) = 8 x 103/ (2 π (100 – 65))
C = 65 x pmax = 36.378
pmax = 0.56 N/mm2.
3. 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/mm2. 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
Explanation: Given : n = 2, r1 = 150 mm ; r2 = 100 mm ; p = 0.1 N/mm2 ; 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 (r2), considering uniform wear, pmax x r2 = C or C = 0.1 × 100 = 10 N/mm.
Axial thrust, W = 2 π C (r1 – r2) = 2π × 10 (150 – 100) = 3142 N
Mean radius = R = (r1 + r2)/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.
4. 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/mm2. 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.
Explanation: Given : n = 2, r1 = 200 mm ; r2 = 175 mm ; p = 0.05 N/mm2 ; µ = 0.4 ; P = 32.384 kW = 32384 W
Since the intensity of pressure (p) is maximum at the inner radius (r2), considering uniform wear, pmax x r2 = C or C = 0.05 × 175 = 8.75 N/mm
Axial thrust, W = 2 π C (r1 – r2) = 2π × 8.75 (200 – 175) = 1374.44 N
Mean radius = R = (r1 + r2)/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.
5. 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/mm2.
a) 0.111 mm, 0.074 mm
b) 111 mm, 74 mm
c) 74 mm, 111 mm
d) 0.074 mm, 0.111 mm
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 ; r1/r2 = 1.5 ; p = 0.1 N/mm2.
Since the ratio of radii (r1/r2) is 1.5, therefore r1 = 1.5r2
We know that the power transmitted (P), 30 × 103 = T.ω = T × 314.2 ∴ T = 30 × 103/314.2 = 95.48 N-m = 95.48 × 103 N-mm
The intensity of pressure is maximum at inner radius (r2),
pmax x r2 = C or C = 0.1 x r2 N/mm
W = 2 π C (r1 – r2) = 2 π × 0.1 r2 (1.5r2 – r2) = 0.314 (r2)2
Mean radius for uniform wear = R = (r1 + r2)/2 = 1.25 r2
Torque transmitted (T), 95.48 × 103 = n.µ.W.R = 2 × 0.3 × 0.314 (r2)2 × 1.25 r2 = 0.2355 (r2)3
(r2)3 = 95.48 × 103/0.2355 = 405.43 × 103 or r2 = 74 mm
r1 = 1.5 r2 = 1.5 × 74 = 111 mm.
6. In a multi plate clutch, the formula for T is given by ______
a) n.µ.W.R
b) n.µ.W.r1
c) n.µ.W.r2
d) n.µ.W.(r1+r2)
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.
7. In a multi plate clutch, considering uniform pressure, T = nµWR. What is R equal to?
a) 2(r13 + r23) / 3(r12 + r22)
b) 2(r13 – r23) / 3(r12 – r22)
c) (r1 – r2)/2
d) (r1 + r2)/2
Explanation: Considering uniform pressure theory in a multi plate clutch, T = nµWR where R = 2(r13 – r23) / 3(r12 – r22). This value is the same as that for the single plate clutch considering uniformly distributed pressure
8. In a multi plate clutch, considering uniform wear, T = nµWR. What is R equal to?
a) 2(r13 + r23) / 3(r12 + r22)
b) 2(r13 – r23) / 3(r12 – r22)
c) (r1 – r2)/2
d) (r1 + r2)/2
Explanation: Considering uniform pressure theory in a multi plate clutch, T = nµWR where R = (r1 + r2)/2. This value is the same as that for the single plate clutch considering uniform wear.
9. 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
Explanation: T = nµWR
150 = 4 x 0.3 x W x 0.1
W = 1250 N
Thus, the axial thrust = 1250 N.
10. Maximum intensity of pressure for multi plate clutch is given by ____
a) C/R
b) C/R2
c) C/r2
d) C/r1
Explanation: We know that, pmax x r2 = C.
Thus, pmax = C/r2
Similarly, pmin = C/r1.