1

NEUTRAL POSITION(DISENGGAGEMENT OF

GEAR TRANSMISSON)

ILLUSTRATIVE SCHEME OF POWER TRANSMISSION OF CAR

Engine

ClutchGear Box

Crankshaft Output shaft

Shaft to drive car

Engine

Gear BoxClutch

Engine

ClutchGear 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 Maintenanc

e 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

Cone Clutch

10

Friction Analysis for Disk Clutch1 .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(2rdr ) (1)

where p is the uniform level of interface pressure. The total force is

)(2 22io

r

rrrπpπprdrF

o

i

The uniform pressure is then )rπ(r

Fp

io22

)3(

)2(

Eq.(2) is simply can be found directly from:

)( 22io rrppAF

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 = (fp2rdr)r (4)

The total torque developed over the entire interface is

)(2 33322

io

r

rrrπpfdrπpfrT

o

i

)5(

If the actual cluthes employ N pair friction interfaces transmitting torque in parallel, Eq.(5) is modified to give

NrrπpfT io )( 3332 )6(

or substituting Eq.(3) into Eq.(6) will give an equation for torque capacity as a function of axial clamping force :

Nrr

rrFfT

io

io

)3(

)(222

33

)7(

12

Friction Analysis for Disk Clutch

UNIFORM WEAR.

F F

w

pApA

fs pA fs pA

A AA

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 fs .

The linear measure of wear w is expressed in inches or millimeters. The work done by force fs p A during

displacement S is fs pAS or fs 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 = C1

pr = C2

pr = C3 = pmaxri

13

Friction Analysis for Disk Clutch

The normal force acting on a diffferential ring element of radius r is

dF = pdA = p(2rdr ) = 2prdr = 2pmaxri dr (8)

The total force is )r(rrπpdrrπpF oimax

r

rmax

o

i

ii 22

The total torque developed over the N pair entire interface is

NrrfrπpNdrrfrπpT ioimax

r

rimax

o

i

)(22 22 )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 = (f2pmaxri dr)r (10)

)9(

or substituting pmax from Eq.(9) into Eq.(11) will give

Nrr

fFT io

2

)12(

Friction Analysis for Cone Clutch

Surface area of ring element:

dA = 2r dr/sinThe normal force on the element is

dN = pdA = (2r dr)p/sinF

The corresponding clamping force is

dF = dNsin = (2r dr)p

The torque that can be transmitted by the element is

dT = dN fr = 2pfr2 dr/sin

From this point, the equations for clamping force and torque capacity can be derived as for disk clutch .

sin)( 3332 N/rrπpfT io

Nrr

rrFfT

io

io

)sin3(

)(222

33

or

UNIFORM PRESSURE,

sin)(2 22 N/rrfrπpT ioimax

UNIFORM WEAR RATE,

NrrfF

T io

2sin

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 Clutches Cone Clutches

Multiple Disk Clutch

Cone Clutch

1 bronzesteel nnn

The capacity: nRfFT f

Where: = torque capacity, NmF = axial force, Nf = coefficient of frictionRf = friction radiusn = number of pairs if surfaces in contact

22

33

3

2

io

iof

RR

RRR

2io

f

RRR

OR:

If the contact pressure is assumed uniform

If wear is assumed uniform

Where:Ro= outside radius of contact of surfaces, mRi= inside radius of contact of surfaces, m

The axial force (F): 22

io RRpF

The power capacity:

Watt,60

2

NTP

Where:p = the average pressure

Where:T = shaft torque, NmN = 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):

22

33

3

2

sin io

io

RR

RRfFT

OR:

2

33

sin3 bR

RRfFT

m

io

bRpFRR

RRfFT mn

io

ion

2 re whe

3

222

33

Alternate form:

Where:Rm = mean radius = 0.5 (Ro + Ri)b = face width, m= pitch cone angle

B) The torque capacity (based on uniform wear):

mnm RfF

RfFT

sin rRR

Fp

io

2

Pressure variation

Maximum pressure iio RRR

Fp

2max At smallest radius

Minimum pressure oio RRR

Fp

2minAt largest radius

Average pressure 22

io

avRR

Fp

Watt,550.9sin

550.960

2

NRfF

NRFNTP

m

mn

Power

Watt,3

2

550.9sin550.93

2

60

222

33

22

33

io

io

io

ion

RR

RRNfFN

RR

RRfFNTP

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/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.

17

Example-1

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

Given: P = 4 kW = 4000 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:

N3630)04.007.0()(350x10)( 22322 ioavgavg rrpApF

Total torque in all contact surfaces:

m.N93.50)750(2

)4000(60

2

60

n

PT

18

Example-1

Number of pairs: N =

Total torqueTorque per

pair

55.2965.19

93.50

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 m.N977.163

93.50

19

Example-1

The required actual force required is found from Eq.(12):

)04.007.0(1.0

2977.16

)(

2

2

io

io

rrfT'F'

rrfF'T'

=3086.7 N

The actual average pressure is

N73.297)04.007.0(

7.308622

A

F'pavg

Maximum actual pressure is from Eq.(9):

2m/N4.4090.04)-7(0.04)(0.02

3086.7

)(2

ioimax rrr

F'p