metrology chapter 1

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

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

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


Example:

Determine the limit of size for a shaft ø40g11 using standard tolerance and deviation information given in table 1 and 2.


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


Example:

Determine the limit of size for a hole ø130N4 using standard tolerance and deviation information given in table 1 and 3 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


Example:Working from the basic principles, find suitable tolerances:a) 82 mm IT6b) 440 mm IT12Compare the calculated values with the rounded values


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


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

metrology chapter 1

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