# Module 10a

```TOLERANCES - Introduction
Nearly impossible to make the part to the exact dimension by any
means of manufacturing approach - tolerances of the dimension.
- Dimension 30 (mm) won’t be
Fig. 1
(a)
(b)
30
- It may be made as 30.10 (mm)
or 30.05 (mm).
30
- maximum may be 30.10 (mm)
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Introduction
- situations for assembly of (a) and (b)?
(a) 30.01 (shaft)
(b) 30.005 (hole)
(a) and (b) are impossible to be assembled without
any special treatment
(a) 30.00 (shaft)
(b) 30.20 (hole)
(a) and (b) are assembled with a possibility of poor
Function of the system (see Figure 2)
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Introduction
Figure 2
L’
L
.
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Introduction
In summary, designers need to specify tolerances for
(a) Parts manufacturing interchangeable
(b) System function satisfactorily with low cost
Since greater accuracy costs more money, the designer will
not specify the closest tolerance, but instead will specify as
generous a tolerance as possible.
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Introduction
Objectives of the lecture:
(1)
(2)
To learn principles behind those rules or
standards for determining tolerances.
To learn procedure of using the standards
for determining tolerances.
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Basic Concept
Definition of Tolerance:
Tolerance is the total amount a specific dimension is
permitted to vary, which is the difference between
the maximum and the minimum limits.
Tolerance is always a positive number
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Basic Concept
Three types of fits
(a) 1.247 - 1.248 shaft
(b) 1.250-1.251 hole
Clearance fit
(a) 1.2513-1.2519 shaft
(b) 1.2500-1.2506 hole
Interference fit
(a) 1.2503-1.2509 shaft
(b) 1.2500-1.2506 hole
Transition fit
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Limits:
upper-limit and lower-limit
The maximum and the minimum sizes indicated by a
tolerance dimension.
The limits for hole are 1.250” and 1.251”
The limits for shaft are 1.248” and 1.247”
Lower limit
Upper limit
The tolerance can also be defined as upper limit – lower limit
on one same dimension
Hole tolerance = 1.2511.250=0.001
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Shaft tolerance =
1.248-1.247=0.001
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Allowances:
an international difference between the maximum
material limits of mating parts. It is the minimum
clearance (positive allowance) or maximum interference
(Negative allowance) between parts.
Allowances:
Allowance = Min Hole – Max Shaft
For the previous example,
1. Clearance fit
2. Allowance = 1.250-1.248=0.002
Hole limit
Shaft limit
Allowance is associated with two dimensions of two
parts that form a fit
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Basic concept
Examples
Figure 5
Shaft tolerance = 1.248 - 1.247 =0.001
Hole tolerance= 1.251-1.250= 0.001
Allowance=1.250-1.248= 0.02
Max clearances=1.251-1.247= 0.04
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Tolerance representation
The unilateral form
The limit form
.000
2.2500.005
2.245 - 2.250
 0.500 00..00
005

0.495 - 0.500
The bilateral form
2.250
0.003
 0.003
2.247-2.253
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In general
 DT
or
 DT
D
Tolerance representation
D  DT
 D
 D
Positive First
D
D
Large Limit on Top
Small limit first
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Standard
Standard (ISO, etc.): limits a freedom of choices but
promotes the exchange of parts manufactured with
- different approaches
- different equipment
- different worker
- in different cultural and societal situations
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Standard
Different countries and regions
together to develop
- Concepts
- Rules
- Systems
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Methodology for Determining Basic Size
Basic Hole System
Purpose: take a hole as a reference to determine the
shaft limit given allowance and tolerances.
the minimal hole size as the basic size.
Reason: in some applications, the hole can be made
more precise (Reamers, Broachers, Gages), while
the machining of the shaft varies.
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Methodology for Determining Basic Size
Basic Shaft System
Reason: in some applications, the shaft could
be better made as a reference
Different fits with the same shaft
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Methodology for Determining Basic Size
Basic Shaft System
the maximal shaft size as the basic size
Reason: Cold-finished shaft.
- cold forging
- cold molding
- cold rolling
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Methodology for Determining Basic Size
Example
0.502
0.500
Basic size =0.5
0.498
0.495
0.505
0.502
0.500
0.499
Basic shaft system
Basic hole system
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Example: Basic Hole System
Given:
Tolerance for the hole = 0.002
Tolerance for the shaft = 0.03
Allowance = 0.02
Basic dimension =0.500
To determine: (a) the limit of the shaft
(b) the limit of the hole
Solution:
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Known:
- Allowance=0.02
- Tolerance for hole=0.002
- Tolerance for shaft=0.03
- Because Basic hole system, Basic dimension=0.5,
Min. Hole dimension = 0.5
Therefore:
- Max. Hole dimension = Min. Hole + Hole tolerance
= 0.5 + 0.002 = 0.502
- Max. Shaft dimension = Min. Hole – Allowance
= 0.5 - 0.02 = 0.498
- Min. Shaft dimension = Max. Shaft + Shaft tolerance
= 0.498 - 0.03 = 0.495
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Example
0.502
0.500
0.498
0.495
Basic hole system
The basic size = 0.500
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Example
Known:
Basic shaft system
Allowance=0.002
Tolerance for hole= 0.003
0.505
0.502
0.500
0.499
Tolerance for shaft= 0.001
The minimal hole size:
0.500+0.002=0.502
Basic shaft system The maximal hole: 0.502+0.003=0.505
The minimal shaft size:
The basic size=0.500
0.500-0.001=0.499
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