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