6 interference fits
TRANSCRIPT
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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