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CHAPTER 1
LINEAR & GEOMETRICAL TOLERANCE
MET 3012
METROLOGY
TOLERANCING – Control of Variability
Goals
Understand the description and control of variability through tolerance.
Understand the various classes of fits.
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
dimension2. Geometric tolerances3. A general tolerance note in title block4. Notes referring to specific condition
1. Direct limits and tolerance value
3.49
3.53
A) Direct limits
3.49 ± 0.003
B) Tolerance value
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
3. Tolerance Specification in Title Block
General tolerance note specifies the tolerance for all unspecified tolerance dimensions
4. Notes Referring to Specific Condition
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
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)
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
Important Terms – multiple part
Allowance – the minimum clearance or maximum interference between parts
Fit – degree of tightness between two partsClearance fit – tolerance of mating parts always leave a spaceInterference fit – tolerance of mating parts always interfereTransition fit – sometimes interfere, sometimes clear
Shaft and Hole Fits
Shaft and Hole Fits
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.
Metric Tolerance
Nominal size = 8
Minimum clearance = 0.040
Maximum clearance = 0.112
Tolerance band = CLmax – CLmin
= T1 + T2
= 0.072
7.9607.924
8.0368.000
0.036
0.036
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
Basic Hole System or Hole Basis
1. Definition of the “Basic Hole System”
The “minimum size” of the hole is equal to the “basic 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.
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
Basic Hole System - Example
Calculate Maximum and minimum clearance
Clearance = Hole – Shaft
Cmax = 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
Basic Hole System - Example
Calculate Maximum and minimum clearance
Clearance = Hole – Shaft
Cmax = 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
Basic Hole System
Basic Hole System
Basic Hole System
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
Why impossible to get the exact size in actual practice?
Find the answer…..
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
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
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
Type of Tolerance
3. Tolerance class
Fundamental deviation followed by standard tolerance grade form tolerance class
Eg. H7 (holes), h7 (shafts)
Type of Tolerance
4. Tolerance size
Basic size followed by tolerance class or expilcit deviations
Eg. 32H7, 80js15, 100j6
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)
Exercise
Given the fitting of two assemblies as follows: 65h6/P7 145H7/k6
Question
1. Explain briefly the meaning of each symbol used in the above fitting
2. 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
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
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)
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
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
Fundamental deviation
A. Fundamental deviation for shafts
B. Fundamental deviation for holes
Fundamental deviations for js and JS
js and JS are a symmetrical distribution of standard tolerance grade about the zone line
Fundamental deviation
Example:
Determine the limit of size for a shaft ø40g11 using standard tolerance and deviation information given in table 1 and 2.
Fundamental deviation
Solution:Basic size range = 30 to 50 mmStandard tolerance = 160 µm (from table 1, IT11)Fundamental deviation = -9µm (from table 2, under ‘g’)Upper deviation = fundamental deviation = -9µmLower deviation = fundamental deviation – tolerance
= -9µm – 160µm = -169 µmLimit of shaft size:
Maximum = basic size + upper deviation = 40 – 0.009 = 39.991 mmMinimum = basic size + lower deviation = 40 – 0.169 = 39.831 mm
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.
Fundamental deviation
Solution:Basic size range = 120 to 180 mmStandard tolerance = 12 µm (from table 1)Fundamental deviation = -27 + Δ µm (from table 3)Value of Δ = 4 µm (from table 3)Upper deviation = fundamental deviation = -27 + 4µm = -23µmLower deviation = fundamental deviation – tolerance
= -23 – 12 = -35 µmLimit of shaft size:
Maximum = basic size + upper deviation = 130 – 0.023 = 129.977 mmMinimum = basic size + lower deviation = 130 – 0.035 = 129.965 mm
Fundamental deviation
Example:Working from the basic principles, find suitable tolerances:a) 82 mm IT6b) 440 mm IT12Compare the calculated values with the rounded values
Fundamental deviation
Solution:80 mm is in the range 80 – 120
440 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
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