Question

Cene clutch Figure (4) Motor 04: Figure (4) shows a simple single-reduction bevel gears pair. Power is received from the motor through a flange coupling to the gearbox. Cone clutch is used to connect or disconnect the motion comes from the motor by the action of the helical spring. Requirements: 1. Power of motor if it has speed of 2000 rpm. 2. If the basic dynamic rating of Deep-groove ball bearing at C is (C = 3.94 kN), find the life of bearing in hours. (Note that: Resultant radial load at point is 65 N) 3. Dimensions of spring (ID, D.m & OD). 4. Deflection of spring if the spring rate is 50 N/mm. 5. Number of active coil of spring if G = 77.2 GPa. Given Data fof cone clutch = 0.3 Rm of cone clutch - 450 mm a of cone clutch - 11° Dw of spring = 4 mm T of spring = 950 MPa C of spring = 6

Answer 1

# 4 Given, Coefficient of friction of clutch (M) = 0.3 Mean 450mm = 0.45m Radius (Rm) Half Cone angle (x) = 11 For Cone clutch, Friction torque is given by ų (D + d). u W Rm 4 sina Given, Radial load (W) = 65N u W sina So, M = 8.3 * 85 * 0-45 Sin 11 46 N-m we Know Power of Motor (P) = (2x NMC) Given, speed of motor (N) = 2000 rpm so, pa ha a * 2000 * 46 6o. or, P = 9631-78 W = 9.63 kW (Ansari 3-948103N Giver, Basic dynamic load (c) = Resultant radial load (W) 65N. so, for Deep-groove ball bearing 3 Lio W where, 410 life of bearing in million revolution, (3.94 x 10 = 222714.55X106 revolutions Lo 373 65

we have, N = 2000 rev/min I min So, 2000 rev takes u 2000 222 714.558106 x106) min 222 714.550 x 106 2000 1. 11357X108 min 1855 954.6 hours 5 life of bearing = 18, 55, 95 4. 6 hours (Ans-2) (3 w Given, wire diameter 4 mm spring index. (k) = 6 = 950 Mpa We know shear stress of spring Is (Dm . - Pw where, Dom Mean Coil diameter Dn Dom (6x4). 2 4 mm ar . (Ans-3) (Om - PW) = (24.4) Inner Diameter (ID) or ID 20 mm J (Ans-13) outer Diameter (OD) = (Don + Dwd) = (24+4) or Top 28 mm Ans --->)

Given, spring rate (K). = so N/mm Know Shear stress on 8F Dm * Kw spring (T) = Kw wahl factor 4i-1 4i-4 + 0.615 = 0.615 [{4x6) | (486) - 4 + 6 1-2525 So, 950 8 * F * 24 .*12525 ཤ་ ( ) ཉེ So, Shear force (F) 794.28N. we know F = K *y => y 194.28 50. or ly - 15.88 mm (Ans --4) This is the deflection of spring We know GPWY 3 5 Given, G = 77.2 x 103 Nommt K- 8 N V Dom => N = 77.2 X103 (4) 4. 50*8* (2423 3.57. or (N = 4 (Ans.. 5) Coils in This would be the number of active spring

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