DISC FRICTION. Rob Wendland, driving Mike Troxel's Federal-Mogul Dragster In-N-Out Burger Bugatti Veyron Disc Brake

Transcription

1 DISC FRICTION Rob Wendland, driving Mike Troxel's Federal-Mogul Dragster In-N-Out Burger Bugatti Veyron Disc Brake

2 Disc Friction Disc (Collar) friction Disc friction is an application of occurs here simple plane friction. It refers to the frictional forces developed when the contacting surfaces are rotating relative to each other It is important in elements such as thrust bearings and collars, brakes, and clutches Whether it be energy dissipation (braking), power transmission, or thrust take-up, disc friction is commonplace. Left to right is a Wilwood disk brake, a Corvette clutch package, and a thrust washer for a Ford transmission.

3 Disc Friction To determine the moment generated by disc friction, consider a rotating hollow shaft on a thrust washer. Assuming a constant µ, incremental friction forces ( F) will be generated across the face; but each incremental force will have a different moment arm to the center of rotation. The total moment is the sum of these incremental moments M = F Ri P i Ri F F R F R R4 Ro F F M

4 Disc Friction Force 'P' is distributed over area 'A' of the disc resulting in a pressure 'p'. The resulting incremental frictional force, F, acts on incremental area A. Letting F = µp A, we define moment 'M' as: M = p Ai Ri i However, actually doing such a summation is impractical, so let's redefine this as an integration M Ri dr dθ df 4 Ro Replace the finite area A with infinitely small area da. The element da has an area equal to dr d as shown to the left. By integrating over one revolution of the ring from 0 to and across the width of the ring from R i to R o, we can write the double integral: M = 0 Ro R p R i dr d

5 Disc Friction After integration, we find: R o Ri M = P R o Ri If we have a solid disk, the inside radius is zero, thus: M= P Ro The frictional forces developed for worn discs or surfaces is generally less than that of new surfaces. It can be shown the moment requirement is reduced to about 75% for that of new surfaces. Thus, for worn surfaces, the above equations become: Ro R i 1 M = P Ro R i 5 -and- M= 1 P Ro

6 Example 5 - Disc Friction The disc brake for a small motorcycle is shown below. The normal force on the wheel during braking is 100 N. If the motorcycle is moving at a constant speed on dry asphalt, find: the maximum force that can be applied to the calipers before the tire begins to skid The pressure applied at the brake pads From the the appendix, appendix, we we can can find find the the From coefficients of of friction friction as as follows: follows: coefficients m 0m 6 0o m 5m mm (Tire (Tire on on Asphalt) Asphalt) µµkk == (brakes) (brakes) µµkk == 0.4 Notice we we need need coefficients coefficients of of kinetic kinetic Notice friction since since all all surfaces surfaces will will be be in in friction motion relative relative to to each each other. other. motion

7 Example 5 - Disc Friction 100 N P We need to first determine the maximum frictional force developed if the wheel is not turning F y = N 100 F = N... N = 100 N... F = 600 N F µ k=0.5 N The moment being generated is the product of this frictional force and the tire radius M = 600 N 0.60 m = 156 N m 7

8 Example 5 - Disc Friction Caliper Pads Rotor Ro R i 4 M = P R o R i Pressure Line C L Since there are two friction surfaces on a brake rotor, the ability to generate a resisting moment is twice that of a single surface, so: Rearranging for P : R Ri P= M o 4 Ro Ri m = 156 N m m P = 155 N 8

9 Example 5 - Disc Friction 0o Pressure on any surface is force distributed over an area. The force acting on each pad is P = 155 N. The pads covers an arc of 0o. As a fraction of the 60o ring, this is: 0o = 8. % of the total ring area o 60 Thus, area of a single pad is: 0 o Pad Area = m o 60 = 5.1 x 10 4 m Pad Pressure = N =.87 MPa x 10 m

10 Example 6 - Disc Friction F F The floor polisher shown has a disc diameter of 0 and weighs 90 lbf. Hand spacing on the handle is 1. Determine the magnitude of force 'F' to counter the moment generated by the pad if the coefficient of kinetic friction between the pad and the floor is 0.8 M= P Ro M = lb f 10 in = 168 in lb f F= in lb f = 1.9 lb f 1in