metro logy 3
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CHAPTER 1
LINEAR &
GEOMETRICAL
TOLERANCE
MET 3012
METROLOGY
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TOLERANCING ± Control of Variability
Goals
Understand the description and control of variability
through tolerance.
Understand the various classes of fits.
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TOLERANCE
The total amount a dimension may vary. It is the difference between the maximum and minimum
limits.Way to express:1. Direct limits or as tolerance limits applied to a
dimension
2. Geometric tolerances3. A general tolerance note in title block4. Notes referring to specific condition
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1. Direct limits and tolerance value
3.49
3.53
A) Direct limits
3.49 ± 0.003
B) Tolerance value
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2. Geometric Tolerance System
Geometric
dimensioning and tolerancing (GD&T) is a
method of defining
parts based on how
they function, using
standard ANSI symbols
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3. Tolerance Specification in TitleBlock
General tolerance
note specifies the tolerance for all unspecified
tolerance
dimensions
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4. Notes Referring to SpecificCondition
General Tolerances could be in the form of a note similar to the one shown below:
ALL DECIMEL DIMENSIONS TO BE HELD TO.002´
MEANS THAT A DIMENSION SUCH AS .005 WOULD BE ASSIGNED A TOLERANCE OF ±0.002, RESULTING IN UPPER LIMIT OF .502
AND A LOWER LIMIT OF .498
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Important Terms ± single part
Nominal Size ± general size, usually expressed in common fraction (1/2´ for the slot)
Basic Size± theoretical size used as starting point (0.500´ for the slot)
Actual Size ± measured size of the finished part (0.501´ for the slot)
.502 Upper Limit (LMC)
.498 Lower Limit (MMC)
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Important Terms ± single part
Limits ± maximum and minimum sizes shown
by tolerances (.502 and .498 ± larger value is
the upper limit and the smaller value is the
lower limit, for the slot)
Tolerances ± total allowable variance in
dimensions (upper limit ±
lower limit) ±
object dimension could be as big as the upper limit or as small as the limit or anywhere in between
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Important Terms ± multiple part
Allowance ± the minimum clearance or
maximum interference between parts Fit ± degree of tightness between two parts
Clearance fit ± tolerance of mating parts always leave a spaceInterference fit ± tolerance of mating parts
always interfereTransition fit ± sometimes interfere, sometimes clear
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Shaft and Hole Fits
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Shaft and Hole Fits
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Metric Limits and Fit
Based on Standard Basic Sizes ± BS
4500:1990 and ISO 286-1:1988 Note that in the metric system:
Nominal size = Basic size
Example:
If the nominal size is 8 mm, then the basic size is 8 mm.
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Metric Tolerance
Nominal size = 8
Minimum clearance = 0.040Maximum clearance = 0.112
Tolerance band = CLmax ± CLmin
= T1 + T2
= 0.072
7.9607.924
8.0368.000
0.036
0.036
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Fit systems
1. Hole basis system
Fits are obtained by changing various tolerance class of shaft with single tolerance
class of holes
2. Shaft basis system
Fits are obtained by changing various tolerance class of holes with single tolerance
class of shaft
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Basic Hole System or Hole Basis
1. Definition of the ³Basic Hole System´
The ³mini mum size´ of the hole is equal to the ³basic size´ of the fit
Example:
If the normal size of fit is 20 mm, then the
minimum size of the hole in the system will be 20 mm.
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Basic Hole System
Clearance = Hole ± Shaft
Cmax = Hmax±
Smin
Cmin = Hmin ± Smax
Both Cmax and Cmin > 0 ± Clearance fit
Both Cmax and Cmin < 0 ± Interference fit
Cmax > 0 and Cmin < 0 ±
Transition fit System Tolerance = Cmax - Cmin
Allowance = Min Clearance = Cmin
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Basic Hole System - Example
Calculate Maximum and
minimum clearanceClearance = Hole ± ShaftCmax = Hmax ± Smin
Cmax = 35.025 ± 35.026 = -0.001
Cmin = 35.000 ±
35.042 = -0.042What type of fit?Cmax <Cmin < 0 ± Interference fit
35.042
35.026
35.025
35.000
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Basic Hole System - Example
Calculate Maximum and
minimum clearanceClearance = Hole ± ShaftCmax = Hmax ± Smin
Cmax = 35.062 ± 34.82 = 0.242
Cmin = 35.000 ±
34.82 = 0.08What type of fit?Cmax >Cmin > 0 ± Clearance fit
34.92
34.82
35.062
35.000
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Basic Hole System
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Basic Hole System
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Basic Hole System
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Tolerance
Tolerance is permitted variation of size of a part
to allow for variation in manufacturing process Tolerance is indirectly a measure of quality, the
smaller the tolerance, the higher the quality; it is also related to the cost of production
Ideal interchangeable mating parts would be those without any kind of dimensional variation
± exact size on blue print or specification
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Why impossible to get the exact
size in actual practice?
Find the answer«..
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Type of Tolerance
1. Standard of tolerance grades
Desinated by letter IT (eg. IT7)
ISO provides 20 Std tolerance grade IT0 ± IT18
(IT0 and IT01 not generally in use)
When associated with letters, letter IT is ommited (eg. h7)
Number representing std grade tolerance
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Type of Tolerance
2. Tolerance Zone
Desinated by 27 upper case letter for holes
(A«.ZC) and 27 lower case letter for shaft (a«zc)
Letter indicate fundamental deviation that form tolerance zone
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Fundamental Deviation
27 possible fundamental
deviations of holes and shaft showed by the
tolerance zones.
Range of each zone is
determined based on
practical experience of the
manufacturing process involved
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Type of Tolerance
3. Tolerance class
Fundamental deviation followed by standard
tolerance grade form tolerance class
Eg. H7 (holes), h7 (shafts)
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Type of Tolerance
4. Tolerance size
Basic size followed by tolerance class or expilcit deviations
Eg. 32H7, 80 js15, 100 j6
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Note
Fitting between mating features is
represented by:
Common basic size (eg. Diameter )
Tolerance class symbol for the hole (eg. H)
Tolerance class symbol for the shaft (eg. f ) Eg. 52H7/g6 (hole basis system)
52h6/G7 (shaft basis system)
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Exercise
Given the fitting of two assemblies as follows:
65h6/P7
145H7/k6
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Question
1. Explain briefly the meaning of each symbol
used in the above fitting2. Calculate Cmax and Cmin for each assy
3. Determine the type of fitting
4. With proper sketch, label the upper & lower
limits, allowance, interference and tolerance
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Formula of standard tolerance
grades
a. Standard tolerance grades IT0 to IT04
Values for std tolerance in grades IT0 and IT01 are
given separately in table 5 due to its limited used
in practice. No formulae are given for IT2, IT3 and
IT4. Value for these grade have been
approximately scaled in geometrical progression
between the values for IT1 and IT5
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Formula of standard tolerance
grades
b. Standard tolerance grades IT5 to IT18
Standard tolerance factor i in micrometers is
calculated from the following formula:
i = 0.453¥D + 0.001D
Where D = geometric mean of basic size range
= ¥(D1 x D2)
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Standard tolerance grades values
Values of the standard tolerance are calculated in
terms of standard tolerance factor, i as shown in table 7
Note: For IT6 upwards, the standard tolerance are
multiplied by a factor of 10 at each fifth step. This rule applies to all standard tolerance above IT18
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Examples:
1. Calculate geometrical mean for the basic size
range of 3 to 6mm.D = ¥(3 x 6) = 4.243 mm
2. Determine the tolerance grade for IT20
IT20 = IT15 x 10
= 640i x 10= 6400i
Numerical values of various tolerance grades IT
are given in Table 1
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Fundamental deviation
A. Fundamental deviation for shafts
B. Fundamental deviation for holes
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Fundamental deviations for js and
JS
js and JS are a symmetrical distribution of standard tolerance grade about the zone line
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Fundamental deviation
Example:
Determine the limit of size for a shaft ø40g11 using standard tolerance and deviation information
given in table 1 and 2.
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Fundamental deviation
Solution:Basic size range = 30 to 50 mm
Standard tolerance = 160 µm (from table 1, IT11)Fundamental deviation = -9µm (from table 2, under g¶)Upper deviation = fundamental deviation = -9µ mLower deviation = fundamental deviation ± tolerance
= -9µ m ± 160µ m= -169 µ m
Limit of shaft size:Maximum = basic size + upper deviation
= 40 ± 0.009 = 39.991 mmMinimum = basic size + lower deviation
= 40 ± 0.169 = 39.831 mm
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Fundamental deviation
Example:
Determine the limit of size for a hole ø130N4 using standard tolerance and deviation information
given in table 1 and 3 fundamental deviation.
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Fundamental deviation
Solution:Basic size range = 120 to 180 mm
Standard tolerance = 12 µm (from table 1)Fundamental deviation = -27 + µm (from table 3)
Value of = 4 µm (from table 3)
Upper deviation = fundamental deviation = -27 + 4µm = -23µmLower deviation = fundamental deviation ± tolerance
= -23 ± 12= -35 µm
Limit of shaft size:Maximum = basic size + upper deviation
= 130 ± 0.023 = 129.977 mmMinimum = basic size + lower deviation
= 130 ± 0.035 = 129.965 mm
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Fundamental deviation
Example:
Working from the basic principles, find suitable tolerances:a) 82 mm IT6
b) 440 mm IT12Compare the calculated values with the rounded values
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Fundamental deviation
Solution:80 mm is in the range 80
± 120440 mm is in the range 400 ± 500
From table 7, IT6 =10i and the tolerance increase in accordance
with the R5 series (geometric progression) as the IT number increase, hence:
IT series 6 7 8 9 10 11 12 13
R5 series 1 1.6 2.5 4 6.4 10 16 25
IT12 tolerances are 16x of IT6 tolerances
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Fundamental deviation
82 mm fall in between basic size range 80 and 120
D = ¥(80x120) = 98
i = 0.45(98)1/3 + (0.001 x 98) = 2.173 µmIT6 = 10i = 21.73 µm (BS4500 gives 22 µm) « see also Table 1
440 mm fall in between basic size range 400 and 500
D = ¥(400x500) = 447
i = 0.45(447)1/3
+ (0.001x447) = 3.888 µmIT12 = 16 x IT6 = 160i (see Table 7)
= 160 x 3.88= 622 µm (BS4500 gives 630 µm) « see also Table 1
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