MACHINE DESIGN II CLUTCHES

N = Nsteel + Nbronze - 1 CLUTCHES A clutch is a friction device which permits the connection and disconnection of shafts. The design of clutches and brakes are comparable in many respects. 1-Plate or disk clutches: A multiple disk clutch is shown in figure. The plates shown as A are usually steel and are set on splines on shaft C to permit axial motion (except for the last disk). The plates shown as B are usually bronze and are set on splines of member D. The number of pairs of surfaces transmitting power is one less than the sum of the steel and bronze disks; N = Nsteel + Nbronze - 1

T = F.f.Rf.N The capacity is given by: where: T= torque capacity, Nm F = axial force, N f = coefficient of friction Rf = friction radius N = number of pair of surfaces transmitting power

UNIFORM WEAR UNIFORM PRESSURE Uniform Pressure Maximum Pressure Frictional Torque Frictional Torque

Example-1 A multiple disk clutch, steel and bronze, is to transmit 4 kW at 750 rpm. The inner radius of contact is 40 mm and the outer radius of contact is 70 mm. The clutch operates in oil with an expected coefficient of friction 0.1. (Oil is used to give smoother engagement, better dissipation of heat, even though the capacity is reduced). The average allowable pressure is 350 kN/m2. 1)- How many total disks of steel and bronze are required? 2)- Determine the axial force required? 3)- What is the average pressure? 4)- Determine the actual maximum pressure.

Example-1 Given: P = 8 kW = 8000 W n = 750 rpm ri = 40 mm = 0.04 m ro = 70 mm = 0.07 m f = 0.1 pavg = 350 kN/m2 This is the case of UNIFORM WEAR. However, since the average allowable pressure is known, the force required for one pair of surfaces in contact is: The torque capacity of one pair of surfaces in contact: T = Ff(Ro + Ri)/2 = 3630(0.1)(0.07+0.04)/2 = 19.965 N.m Total torque in all contact surfaces:

Example-1 Total torque Number of friction interface: N = Torque per pair Use 6 friction interfaces with 4 steel and 3 bronze disk. Disk a : driving disks (4 disks, 6 friction surfaces) Disk b : driven disks (3 disks, 6 friction surfaces) Output input The actual torque per pair of surfaces T’ = Total torque Number of friction interface

Example-1 The required actual force required is found from Eq.(12): = 3086 N = 3.086 kN The actual average pressure is Maximum actual pressure is from Eq.(9):

Problem 1. A car engine friction clutch is required to transmit 10 kW at 3000 rpm. The clutch is of single disc plate type both sides of disc plate being effective. If µ = 0.25 and axial pressure is limited to 0.085 N/mm2 and the external diameter of the plate is 1.4 times the internal diameter. For uniform wear condition, determine the dimensions of the disc plate. Solution: Torque developed by the engine: Pressure force for uniform wear: do = 1.4 di or ro = 1.4 ri

Total Frictional Torque: Torque developed by the engine = Total Frictional Torque: Solving the above equation yields, ri = 62.86 mm or di = 125,7 mm Then, the outer diameter of disc plate is do = 1.4 di = 176 mm

Problem 2. A disc clutch is to be composed of five steel discs and 4 bronze discs. The outer diameter of the contact surface is to be 250 mm and the inside diameter 150 mm. Determine the pressure with which the discs must be held together if 15 kW is to be transmitted at 600 rpm assuming f = 0.3. Consider for both uniform pressure and uniform wear condition. Solution: Number of friction interface: N = Nsteel + Nbronzel – 1 = 5 + 4 – 1 = 8 pairs intgerface Torque developed by the engine: UNIFORM WEAR ANALYSIS Pressure force for uniform wear:

Total Frictional Torque for uniform wear: Torque developed by the engine = Total Frictional Torque: The maximum pressure is pmax = 0.0422 N/mm2

UNIFORM PRESSURE ANALYSIS Pressure force for uniform pressure: Total Frictional Torque for uniform pressure: Torque developed by the engine = Total Frictional Torque: The uniform pressure is p = 0.031 N/mm2