1 NEUTRAL POSITION (DISENGGAGEMENT OF GEAR TRANSMISSON) ILLUSTRATIVE SCHEME OF POWER TRANSMISSION OF CAR Engine Clutch Gear Box Crankshaft Output shaft Shaft to drive car Engine Gear Box Clutch Engine Clutch Gear Box CLUTCH DISCONNECTING FOR ENGGAGEMENT OF GEAR TRANSMISSION ENGGAGEMENT OF GEAR TRANSMISSION

Clutches 2 Section IX

3 CLUTCH Disk Clutch Cone Clutch Friction Analysis for Disk Clutch Design Analysis of Clutch Examples Talking Points

4 CLUTCH CLUTCH is friction devices to permit smooth, gradual connection and disconnections of two members having a common axis of rotation. TYPE OF CLUCTH: 1. Plate or Disk Clutch 2. Cone Clutch FRICTION ANALYSIS: 1. Uniform Pressure 2. Uniform Wear

5 CLUTCH THREE BASIC REQUIREMENTS: 1. The required friction torque must be produced by an acceptable actuating force. 2. The energy converted to friction heat during clutch engagement must be dissipated without producing destructively high temperature. 3. The wear characteristics of the friction surfaces must be such that they give acceptable life. GENERAL MECHANICAL DESIGN CONSIDERATION: Strength Reliability Thermal properties Corrosion Wear Friction Processing Utility Cost Safety Weight Life Noise Styling Shape Size Flexibility Control Stiffness Surface finish Lubrication Maintenance Volume

6 Disk Clutch Basic Disk Clutch – Single Disk Clutch

7 Disk Clutch Automotive-type disk clutch The flywheel, clutch cover, and pressure plate rotate with the crankshaft. A series of circumferentially distributed springs force the pressure plate toward the flywheel, clamping the clutch plate (driven disk) between them. The hub of clutch plate is spline-connected to the transmission input shaft. The clutch is disengaged by depressing the clutch pedal which rotates the lever marked “To release”. This pushes the clutch release bearing against a series of radially oriented release levers that pull the pressure plate away from the flywheel.

8 Disk Clutch Multiple-disk clutch, hydraucally operated Disk a : driving disks (4 disks, 6 friction surfaces) Disk b : driven disks (3 disks, 6 friction surfaces) Output input Disks a (usually made of steel) are constrained (as with splines) to rotate with the input shaft. Disks b (usually made of bronze) are similarly constrained to rotate with the output shaft. When the clutch is disengaged, the disks are free to slide axially to separate themselves. When the clutch is engaged, they are clamped tightly together to provide (in the case illustrated) six driving and six driven surfaces. The two end disks, which have only their inner sides serving as friction surfaces, should be members of the same set in order to avoid transmitting the clamping force through a thrust bearing.

9 Cone Clutch

10 Friction Analysis for Disk Clutch 1. UNIFORM PRESSURE. This assumption is valid for an unworn (new), accurately manufactured clutch, with rigid outer disks. The normal force acting on a diffferential ring element of radius r is dF = pdA = p(2  rdr ) (1) where p is the uniform level of interface pressure. The total force is The uniform pressure is then (3) (2) Eq.(2) is simply can be found directly from:

11 Friction Analysis for Disk Clutch The friction torque developed on a ring element is the product of normal force ( dN = dF ), coefficient of friction ( f ), and radius ( r ), dT = (fdN)r = (fdF)r = (fp2  rdr)r (4) The total torque developed over the entire interface is (5) If the actual cluthes employ N pair friction interfaces transmitting torque in parallel, Eq.(5) is modified to give (6) or substituting Eq.(3) into Eq.(6) will give an equation for torque capacity as a function of axial clamping force: (7)

12 Friction Analysis for Disk Clutch UNIFORM WEAR. F F w pA f s pA  A A S A Consider the sliding block depicted in the figure, moving along a plate with contact pressure p acting over area A, in the present of sliding friction f s. The linear measure of wear w is expressed in inches or millimeters. The work done by force f s p A during displacement S is f s pAS or f s pAVt, where V is the sliding velocity and t is time. The material volume removed due to wear is wA and is proportional to the work done, that is w A = K p A V t In which only p and V vary from place to place in the rubbing surfaces, therefore p V = constant = C 1 pr  = C 2 pr = C 3 = p max r i

13 Friction Analysis for Disk Clutch The normal force acting on a diffferential ring element of radius r is dF = pdA = p(2  rdr ) = 2  prdr = 2  p max r i dr (8) The total force is The total torque developed over the N pair entire interface is (11) The friction torque developed on a ring element is the product of normal force ( dN = dF ), coefficient of friction ( f ), and radius ( r ), dT = (fdN)r = (fdF)r = (f2  p max r i dr)r (10) (9) or substituting p max from Eq.(9) into Eq.(11) will give (12)

Friction Analysis for Cone Clutch Surface area of ring element: dA = 2  r dr/sin  The normal force on the element is dN = pdA = (2  r dr)p/sin  F The corresponding clamping force is dF = dNsin  = (2  r dr)p The torque that can be transmitted by the element is dT = dN fr = 2  pfr 2 dr/sin  From this point, the equations for clamping force and torque capacity can be derived as for disk clutch. or UNIFORM PRESSURE, UNIFORM WEAR RATE, or

15 The plates shown in figure below 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 on shaft D. The number of pairs of surfaces transmitting power is one less than the sum of the steel and bronze disks. Design Analysis of Clutches Plate or Disk ClutchesCone Clutches Multiple Disk Clutch Cone Clutch The capacity: Where: T = torque capacity, Nm F = axial force, N f = coefficient of friction R f = friction radius n = number of pairs if surfaces in contact OR: If the contact pressure is assumed uniform If wear is assumed uniform Where: R o = outside radius of contact of surfaces, m R i = inside radius of contact of surfaces, m The axial force ( F ):The power capacity: Where: p = the average pressure Where: T = shaft torque, Nm N = speed of rotation, rpm A cone clutch achieves its effectiveness by the wedging action of the cone part in the cup part. A) The torque capacity (based on uniform pressure): OR: Alternate form: Where: R m = mean radius = 0.5 ( R o + R i ) b = face width, m  = pitch cone angle B) The torque capacity (based on uniform wear): Pressure variation Maximum pressureAt smallest radius Minimum pressureAt largest radius Average pressure Power

16 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/m 2. 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.

17 Example-1 The torque capacity of one pair of surfaces in contact: T = Ff(R o + R i )/2 = 3630(0.1)( )/2 = N.m Given: P = 4 kW = 4000 W n = 750 rpm r i = 40 mm = 0.04 m r o = 70 mm = 0.07 m f = 0.1 p avg = 350 kN/m 2 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: Total torque in all contact surfaces:

18 Example-1 Number of pairs: N = Total torque Torque per pair Use 3 pairs with 3 steel and 2 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 pair of surfaces

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