MACHINE DESIGN II

CLUTCHES

Mechanical Engineering DepartmentUmm Al Qura UniversityMakkah

2

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

Mechanical Engineering DepartmentUmm Al Qura UniversityMakkah

3

D C

A A A A A

B B B B

F /2

F /2

F

Ro

Ri

The capacity is given by:

T = F.f.Rf.Nwhere: T= torque capacity, Nm F = axial force, N f = coefficient of friction Rf = friction radius

N = number of pair of surfaces transmitting power

Mechanical Engineering DepartmentUmm Al Qura UniversityMakkah

4

Nrr

fFT io

2N

rr

rrFfT

io

io

)3(

)(222

33

UNIFORM PRESSURE UNIFORM WEAR

Uniform Pressure

)rπ(r

Fp

io22

Frictional Torque

Maximum Pressure

)(2 ioimax rrπr

Fp

Frictional Torque

Mechanical Engineering DepartmentUmm Al Qura UniversityMakkah

5

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.

Mechanical Engineering DepartmentUmm Al Qura UniversityMakkah

6

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 = 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:

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

Total torque in all contact surfaces: m.N86.101)750(2

)8000(60

2

60

n

PT

Mechanical Engineering DepartmentUmm Al Qura UniversityMakkah

7

EXAMPLE-1

Number of friction interface: N = Total torque

Torque per pair1.5

965.19

86.101

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

m.N977.16

6

86.101

Mechanical Engineering DepartmentUmm Al Qura UniversityMakkah

8

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 N = 3.086 kN

The actual average pressure is

222 m/N73.297)04.007.0(

7.3086

A

F'pavg

Maximum actual pressure is from Eq.(9):

2m/N4.409)(

0.04)-7(0.04)(0.02

3086.7

2 ioimax rrr

F'p

Mechanical Engineering DepartmentUmm Al Qura UniversityMakkah

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:

2 ,4.1 ,N/mm 085.0

25.0 rpm, 3000 ,kW 10Power :Given2

Nddp

fn

iomax

Pressure force for uniform wear: )(2 iomaxi rrpπrF

do = 1.4 di or ro = 1.4 ri

20.2136

0.40.0852

1.42

i

2i

iimaxi

r

rπ

rrpπrF

Torque developed by the engine: mm.N31830m.N83.31)3000(2

)10x10(60

2

60 3

n

PT

Mechanical Engineering DepartmentUmm Al Qura UniversityMakkah

Total Frictional Torque:

3

2

1281.0

22.125.02136.0

2

22

i

ii

i

iiio

r

rr

Nr

fF

Nrr

fFNrr

fFT

2.4

1.4

Torque developed by the engine = Total Frictional Torque:

31281.031830 ir

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

Mechanical Engineering DepartmentUmm Al Qura UniversityMakkah

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:

4 5 mm, 150 ,mm 250

3.0 rpm, 600 ,kW 15Power :Given

bronzesteelio NNdd

fn

Number of friction interface: N = Nsteel + Nbronzel – 1 = 5 + 4 – 1 = 8 pairs intgerface

Torque developed by the engine: mm.N238730m.N73.238)600(2

)10x15(60

2

60 3

n

PT

Pressure force for uniform wear:

max

max

iomaxi

p

pπ

rrpπrF

95.23561

75 125752

)(2

UNIFORM WEAR ANALYSIS

Mechanical Engineering DepartmentUmm Al Qura UniversityMakkah

Total Frictional Torque for uniform wear:

max

maxio

p

pNrr

fFT

5654868

82

751253.095.23561

2

Torque developed by the engine = Total Frictional Torque:

maxp5654868238730

The maximum pressure is pmax = 0.0422 N/mm2

Mechanical Engineering DepartmentUmm Al Qura UniversityMakkah

UNIFORM PRESSURE ANALYSIS

Pressure force for uniform pressure:

p

πprrπpF io

9.31415 )75(125)( 2222

Total Frictional Torque for uniform pressure:

pp

T 5.7696895

8

)753(125

)75(1250.331415.9222

33

Torque developed by the engine = Total Frictional Torque:

p5.7696895238730

The uniform pressure is p = 0.031 N/mm2

Nrr

rrFfT

io

io

)3(

)(222

33

Mechanical Engineering DepartmentUmm Al Qura UniversityMakkah