6 Interference Fits

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INTERFERENCE FITS

CONTENTS:

INTRODUCTION ISO TOLERANCESSTANDARDISED FITSGEOMETRICAL TOLERANCINGEXAMPLES OF FITSLAM PROBLEMMAXIMUM STRESS AND DEFORMATIONENGINEERING CALCULATIONSAPPENDIX


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INTRODUCTION

Interference fits are based on the standardisedtolerance system of shafts and holes.

This system is to be introduced, giving an overview of

the ISO tolerance grades and, additionally, shape andposition tolerances of typical machine components.The roughness grades will be also given.

The preferred fits using the basic-hole system will befurther discussed and some examples will be shown.

Then, the Lam equations for thick walled cylinders amodel for interference fits will be shown.

Finally, the load-carrying capacity and stresses incomponents of the interference joint will be presented.


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ISO TOLERANCES

Tolerance is the difference between the maximum andminimum size limits of a component.

International Tolerance Grade Numbers

(IT = International Tolerance) are used to specify thesize of the tolerance area (zone).

Note that the tolerance is the same for the internaldimension (hole) and external one (shaft) that have thesame tolerance grade number.


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ISO TOLERANCES

The ISO standard (ISO 286) implements 20 IT tolerances(grades of accuracy) as shown in table below:

ISO grade number Application

IT01, IT0, IT1, IT2,

IT3, IT4, IT5,IT6

Precision

instruments

IT5, IT6, IT7, IT8,IT9, IT10, IT11, IT12

General industry

IT11, IT12, IT13,IT14, IT15, IT16 Semi-finishedproducts

IT16, IT17, IT18 Structural

engineering


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ISO TOLERANCESTolerance area depends on both nominal size

and IT grade (ACCURACY COSTS) Nominal Sizes [mm]

over 6 10 18 30 50 80 120 180

incl. 10 18 30 50 80 120 180 250

ITGrade

Tolerance area [m]

5 6 8 9 11 13 15 18 206 9 11 13 16 19 22 25 29

7 15 18 21 25 30 35 40 46

8 22 27 33 39 46 54 63 72

9 36 43 52 62 74 87 100 115

10 58 70 84 100 120 140 160 185

11 90 110 130 160 190 220 250 290

12 150 180 210 250 300 350 400 460


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ISO TOLERANCES

M A C H IN IN G P R O C E S S A S S O C I A T ED W IT H T O L E R A N C E G R A D E S

IT G rade 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Lapping

Honing

Superf inishing

Cyl inde rical g rinding

Diamond turning

Plan grinding

Broaching

ReamingBoring, Turning

Sawing

Mil l ing

Planing, Sh aping

ExtrudingCo ld Ro lling, D rawing

Dri l l ing

Die Ca st ing

Forging

Sand Cast ing

Ho t roll ing , Flam e cutt ing


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ISO TOLERANCES

MACHINING PROCESS ASOCIATED WITH THE ROUGHNESS AVERAGE Ra


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ISO TOLERANCES

TOLERANCE LIMITS FOR SHAFTSNominal sizes in mm

Over 6 10 14 18 24 30 40 50 65 80 100 120 140 160 180 200 225

Incl. 10 14 18 24 30 40 50 65 80 100 120 140 160 180 200 225 250

GRADEUpper tolerance limits in ma -280 -290 -300 -310 -320 -340 -360 -380 -410 -460 -520 -580 -660 -740 -820

c -80 -95 -110 -120 -130 -140 -150 -170 -180 -200 -210 -230 -240 -260 -280

d -40 -50 -65 -80 -100 -120 -145 -170

f -13 -16 -20 -25 -30 -36 -43 -50

g -5 -6 -7 -9 -10 -12 -14 -15GRADE Lower tolerance limits in mj5, j6 -2 -3 -4 -5 -7 -9 -11 -13

k5-k7 1 2 3 4

n 10 12 15 17 20 23 27 31

p 15 18 22 26 32 37 43 50s 23 28 35 43 53 59 71 79 92 100 108 122 130 140

u 28 33 41 48 60 70 87 102 124 144 170 190 210 236 258 284

z 42 50 60 73 88 112 136 172 210 258 310 365 415 465 520 575 640

Notes:1. Only chosen diameters and grades have been shown2. The h grade shafts have 0 upper limit of tolerance3. The js grade shafts have lower/upper limit of tolerance equal to 0.5 T


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ISO TOLERANCES

TOLERANCE LIMITS FOR HOLES

Nominal sizes in mm

Over 6 10 14 18 24 30 40 50 65 80 100 120 140 160 180 200 225

Incl. 10 14 18 24 30 40 50 65 80 100 120 140 160 180 200 225 250

GRADE Lower tolerance limits in mA 280 290 300 310 320 340 360 380 410 460 520 580 660 740 820

C 80 95 110 120 130 140 150 170 180 200 210 230 240 260 280

D 40 50 65 80 100 120 145 170

F 13 16 20 25 30 36 43 50G 5 6 7 9 10 12 14 15

GRADE Upper tolerance limits in mJ7 8 10 12 14 18 22 26 30

K7 5 6 6 7 9 10 12 13N7 -4 -5 -7 -8 -9 -10 -12 -14

P7 -9 -11 -14 -17 -21 -24 -28 -33

R7 -13 -16 -20 -25 -30 -32 -38 -41 -48 -50 -53 -60 -63 -67

Notes:1. Only chosen diameters and grades have been shown2. The H grade holes have 0 lower limit of tolerance3. The Js grade holes have lower/upper limit of tolerance equal to 0.5 T


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STANDARDISED FITS

There are two standardised fit systems:

basic-hole fit e.g. H7/k6 (hole symbol Hn, where n isthe ISO tolerance grade for hole; shaft symbol is xm, withm denoting the ISO tolerance grade for shaft)

basic-shaft f it e.g. K7/h6 (shaft symbol hn, where nis the ISO tolerance grade for shaft; hole symbols are Amtill ZCm, with m denoting the ISO tolerance grade for hole)Because of the more expensive fabrication of holespreferred are:

basic-hole fits

wider tolerance zones of hole than those for theshaft

Both systems are shown in following figures.


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STANDARDISED FITS

BASIC-HOLE FITS


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STANDARDISED FITS

PREFERRED FITS (BASIC-HOLE SYSTEM)Description Hole Shaft

Loose Running H11 c11

Free Running H9 d9

Easy Running - Good quality easy to do- H8 f8

Sliding H7 g6

Close Clearance - Spigots and locations H8 f7

Location/Clearance H7 h6

Location- slight interference H7 k6

Location/Transition H7 n6

Location/Interference- Press fit which can be separated H7 p6

Medium Drive H7 s6

Force H7 u6


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GEOMETRICAL TOLERANCING

Shape and position frame with symbol identifications:


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GEOMETRICAL TOLERANCING

Typical shape tolerances for chosen fabrication processes:Geometric tolerance feature in m/mm of length or diameterManufacturing

process Flatness of

surface

Parallelism

of cylinders

or cones ondiameter

(cylindricity,

conicity)

Straightness

of cylinders

or cones

Parallelism

of flat

surfaces

Parallelism

of

cylinders

Roundness

Turn

Bore 50 100 100 100 100 40

Fine turn

Fine bore 30 40 40 50 50 30

Cylindrical

grind 30 50 50 50 50 20

Fine

cylindricalgrind

20 20 20 30 20 10

S O S


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EXAMPLES OF FITS

Scheme of clearance fit (basic hole system)

zero line

shaft

basic

hole

EXAMPLES OF FITS


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EXAMPLES OF FITS

Clearance fit calculations

Specification: 40H9/d9

Hole Shaft

Tolerance grade: 0.062 mm 0.062 mm

Upper deviation: 0.062 mm - 0.080 mmLower deviation: 0.000 mm - 0.142 mm

Max diameter: 40.062 mm 39.920 mm

Min diameter: 40.000 mm 38.858 mmMax clearance: 0.202 mm

Min clearance: 0.080 mm

EXAMPLES OF FITS


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EXAMPLES OF FITS

Scheme of transition fit (basic-hole system)

zero line

shaft

basic

hole


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EXAMPLES OF FITS


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EXAMPLES OF FITS

Scheme of interference fit (basic-hole system)

zero line

shaft

basichole

EXAMPLES OF FITS


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EXAMPLES OF FITS

Interference fit calculations

Specification: 40H7/p6Hole Shaft

Tolerance grade: 0.025 mm 0.016 mm

Upper deviation: 0.025 mm 0.042 mmLower deviation: 0.000 mm 0.026 mm

Max diameter: 40.025 mm 40.042 mm

Min diameter: 40.000 mm 40.026 mmMax interference: 0.042 mm

Min interference: 0.001 mm

LAM PROBLEM


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LAM PROBLEM

Lam problem concerns the solution for stress and strain in thick-

walled cylinders under internal or external pressure. Note that thethick-walled cylinder as opposed to the thin walled ones, in whichthe constant figures for stress can be assumed have the wallthickness greater than 0.1 times the nominal radius of cylinder.

Scheme of interference fit

Bushing

SleeveMating surface

whereinterferencepressure acts

LAM PROBLEM


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LAM PROBLEM

Dimensions of components of the interference fit:

a) before assembling; b) after assembling

d diameter dxdSidBeB bushing

S sleeve

e externali internal

LAM PROBLEM


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LAM PROBLEM

Normal forces applied to an element of sleeve or bushing

LAM PROBLEM


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LAM PROBLEM

The equilibrium condition in radial direction produces thefollowing equation

with

one obtains

and

22sin dd =

( ) 02sin2 =

d

drdldrdldrdr

d

tr

rdr

d rrt +=

rdrd

drd

drd rrt +=

2

22

LAM PROBLEM


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LAM PROBLEM

The elastic deformation of an element in

circumferential (tangential) direction is

For the radial direction we have

where

( )r

u

dr

drdurt =

+=

ttt

r rdr

d

dr

rd

dr

du

+=

==

)(

( )trrE

=1

( )rtttE

= 1

= drd

dr

d

Edr

d rtt

1

LAM PROBLEM


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LAM PROBLEM

From previous equations follows

and finally we have an linear homogenous differential equation

The solution of the above equation is

with

and

( ) ( ) 01 =++

tr

tr

dr

d

dr

dr

032

22 =+

dr

dr

dr

dr rr

2

21

rCCr +=

322

rC

drd r =

221

r

CCt =

LAM PROBLEM


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LAM PROBLEM

The boundary conditions for the sleeve are

The integration constants are therefore

Finally we have for the sleeve

2/xxr drrforp ===

22

22

222

2

1 ;xSe

Sex

xSe

x

rr

rrpCrr

rpC=

=

2/0 SeSer drrfor ===

=

2

21

1

1

r

r

r

rp Se

x

Se

r

+

=2

21

1

1

r

r

rr

p Se

x

Se

t

LAM PROBLEM


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LAM PROBLEMThe boundary conditions for the bushing are

The integration constants are therefore

Finally we have for the bushing

2/xxr drrforp ===

22

22

222

2

1 ;xBi

Bix

xBi

x

rrrrpC

rrrpC

=

=

=

2

21

1

1rr

r

rp Bi

x

Bi

r

2/0 BiBir drrfor ===

+

=2

21

1

1

r

r

rr

p Bi

x

Bi

t

LAM PROBLEM


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LAM PROBLEM

The distribution of the circumferential and radial stressesfor shaft with hole (note the radial direction of stresses Brand Sr)

LAM PROBLEM


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O

The distribution of the circumferential and radial stressesfor shaft without hole (note the radial direction of stressesBr and Sr)

MAXIMUM STRESS AND DEFORMATION


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The maximum reduced stress according to the maximum stresstheory (Tresca theory) occurs in the mating surface of the outer and

inner member of the interference fit (sleeve and bushing): the maximum reduced stress for sleeve

the maximum reduced stress for bushing

where S and B denote the the geometrical factor for sleeve andbushing.

Se

xS

S

Se

x

Sr

rp

r

rp =>

=

= ;012

1

22max

x

Bi

BB

x

Bi

B r

rp

r

rp =


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The total diametral interference follows from the above given solution tothe Lam problem

Where ES and EB denote the Youngs moduli and S and B are for thePoissons ratios of sleeve and bushing, respectively.

If both members are of the same material the above formula is simplified

to the form

+

+

= 2222

22

22 112

Bix

BixB

BSex

SexS

S

xrr

rr

Err

rr

Epr

+

+

=

22

22

22

222

Sex

Sex

Bix

Bixx

rr

rr

rr

rr

E

pr

MAXIMUM STRESS AND DEFORMATION


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The interface pressure is, therefore, to be computed util ising therearranged formulae:

in the case of the different materials of sleeve and bushing

in the case of the same material of both members

+

+

+

=

B

Bix

Bix

B

S

xSe

xSe

S

xrr

rr

Err

rr

Er

p

22

22

22

22 112

( ) ( )( )

=222

2222

22BiSex

BixxSe

x

rrr

rrrr

r

Ep

ENGINEERING CALCULATIONS


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The calculations of the interference fit embrace the two following criteria:

Load carrying capacity of the joint against axial force Fa, torsionMt, or both these loads, which result in the shear stress on theinterface of mating members

This shear stress is being carried by the friction forces and,

therefore, the condition is

Strength of the coupled members, which give the condition

In above formulae l is the length of joint, denotes the friction coefficientand des stands for the design stress (with proper safety factor included).Note that the maximum pressure is to be taken in strength calculation,and the minimum one is responsible for load carrying of the fit.

( )lr

FrM

x

axt

+=

2

/ 22

minp


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The friction coefficient depends on materials of sleeve and bushing and

on their lubrication as given in following table

The ultimate stress depends also on materials used and the type of load

SurfacesMaterials

dry lubricated

Steel/steel 0.065-0.16 0.055-0.12

Steel/cast ironSteel/bronze

0.15-0.2 0.03-0.06

Cast iron/cast iron

Cast iron/bronze0.15-0.25 0.02-0.1

Ultimate stress [MPa] for type of loadMaterial

static oscillating impact

Steel 100-180 70-120 40-60

Cast iron 70-80 50-60 20-30

Bronze 30-40 20-30 10-20

ENGINEERING CALCULATIONS


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The effective diametral interference eff is still less than the geometrical one .This results from the shearing of the asperities (peaks of the surface roughness)during assembling. It has been shown that about 60% of the heights of asperitiesRp above the mean line is sheared see figure below.

Thus, the effective interference of sleeve and bushing is

pSpBeff RR += 6.0

APPENDIX


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The uniform distribution of pressure (pm

, following frominterference of components and their elasticdeformation) occurs only for torsion and axial forceloading of the joint.

If bending or radial loading act, the change of pressurep is to be taken into consideration. The two following

figures represent the most common cases.

The bending moment for these cases equals to

2

aax

dFM

=

APPENDIX


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21 lFMM

rxx =

rz FF =1

APPENDIX


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+= xrz Ml

eF

e

lF

221

+= xrz M

l

eFe

l

F 212

111 eFM zx =

222 eFM zx =

APPENDIX


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11 ppp m +=

2

2

3

2

2

1

3

11

f

z

f

x

f

z

f

x

d

Fy

d

Mx

d

FY

d

MXp +++=

The pressures in the points 1, 1, 2, and 2 are

Where

The coefficients X, Y, x, y are given in the following table

1'1 ppp m =

22 ppp m +=

2'2 ppp m =

2

1

3

1

2

2

3

22

f

z

f

x

f

z

f

x

d

Fy

d

Mx

d

FY

d

MXp +++=

APPENDIX


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Correction coefficients for contact pressure

Dimensions Coefficients

di/df df/da lf/df X Y x y

0.3 0.50

0.751.001.50

16.10

8.586.746.36

5.17

3.643.062.85

-14.80

-5.76-2.230.19

-2.49

-1.52-0.90-0.15

0.5 0.50

0.751.001.50

16.20

8.837.116.81

5.18

3.673.132.95

-14.70

-5.62-2.050.27

-2.48

-1.49-0.86-0.11

0

0.8 0.50

0.751.001.50

16.50

9.427.977.81

5.20

3.753.283.15

-14.60

-5.30-1.650.42

-2.46

-1.44-0.76-0.04

0.3 1.00 7.64 3.22 -1.81 -0.80

0.5 1.00 8.26 3.32 -1.53 -0.73

0.5

0.8 1.00 9.86 3.59 -0.50 -0.57

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