1. In a conical clutch, pn = ______________
a) W/(π(r12 – r22))
b) W/(π(r12 + r22))
c) W/(π(r22 – r12))
d) W/(π(r22 x r12))
Explanation: Pressure = Force/ Area.
Thus, here, intensity of pressure considering uniform pressure pn is W/(π(r12 – r22)).
2. In a conical 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
It is same as that of single plate and multi plate clutch but the value of R is different for conical clutch.
3. In a conical clutch, considering uniform pressure, T = nµWR. What is R equal to?
a) {2(r13 + r23) / 3(r12 + r22)} x cosec α
b) {2(r13 – r23) / 3(r12 – r22)} x cosec α
c) {(r1 – r2)/2} x cosec α
d) {(r1 + r2)/2} x cosec α
Explanation: Considering uniform pressure theory in a single plate clutch, T = nµWR where R = {2(r13 – r23) / 3(r12 – r22)} x cosec α
α = the semi cone angle.
4. In a conical clutch, considering uniform wear, T = nµWR. What is R equal to?
a) {2(r13 + r23) / 3(r12 + r22)} x cosec α
b) {2(r13 – r23) / 3(r12 – r22)} x cosec α
c) {(r1 – r2)/2} x cosec α
d) {(r1 + r2)/2} x cosec α
Explanation: Considering uniform pressure theory in a single plate clutch, T = nµWR where R = {(r1 + r2)/2} x cosec α.
5. In a conical clutch, what is the axial force required for engaging the clutch (We)?
a) Wn (sin α + µ cos α)
b) Wn (sin α – µ cos α)
c) Wn (µ sin α + cos α)
d) Wn (µ sin α – cos α)
Explanation: We = W + µ.Wn cos α = Wn sinα + µ.Wn cosα = Wn (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.
6. In a conical clutch, the breadth of the contact surface is given by ______
a) (r1 + r2)/sinα
b) (r2-r1)/sinα
c) (r1 – r2)/sinα
d) (r12 + r22)/sinα
Explanation: In a conical clutch b sinα = r1-r2
Thus, b = (r1-r2)/sinα
7. 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
Explanation: In a conical clutch b sinα = r1 – r2
r1 – r2 = 10 and r1 + r2 = 300
r1 = 155 mm and r2 = 145 mm.
8. 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
Explanation: In a conical clutch b sinα = r1 – r2
b x sin 22.5° = 100 – 75
b = 65.33 mm.
9. 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/mm2, 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
Explanation: Given : P = 75 kW = 75 × 103 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/mm2
Power transmitted (P) = 75 × 103 = T.ω = T × 157.08
T = 75× 103/157.08 = 477.4 N-m = 477.4× 103 N-mm
Torque transmitted (T) = 477.4 × 103 = 2 π µ pn.R2.b = 2π × 0.3 × 0.1 (250)2 b = 11780.97 b
b = 40.52 mm
We know that, r1+r2 = 250 and r1-r2 = 40.52 sin 20 = 13.86
Therefore, r1 = 131.93 mm, r2 = 118.07 mm.
10. 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/mm2, find the axial spring force necessary to engage the clutch.
a) 4796 N
b) 4774 N
c) 4785 N
d) 4742 N
Explanation: Given : P = 50 kW = 50 × 103 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/mm2
Power transmitted (P) = 50 × 103 = T.ω = T × 104.72
T = 50× 103/104.72 = 477.4 N-m
477.4 = µ.Wn.R = 0.2 × Wn × 0.225
Wn= 10610.33 N
Axial spring force = We = Wn(sin α + µ cos α) = 10610.33 (sin 15° + 0.2 cos 15°) = 4796 N.