- dimension 30 (mm) won’t be made exactly as 30 (mm)
DESCRIPTION
- Dimension 30 (mm) won’t be made exactly as 30 (mm) - It may be made as 30.10 (mm) or 30.05 (mm). - maximum may be 30.10 (mm). Fig. 1. (a). 30. (b). 30. - PowerPoint PPT PresentationTRANSCRIPT
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TOLERANCES - IntroductionNearly 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 made
exactly as 30 (mm)
- It may be made as 30.10 (mm) or 30.05 (mm).
- maximum may be 30.10 (mm)
30
30
(a)
(b)
Fig. 1
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(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)
- situations for assembly of (a) and (b)?Introduction
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.
Figure 2Introduction
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) To learn principles behind those rules or standards for determining tolerances.
(2) To learn procedure of using the standards for determining tolerances.
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Basic ConceptDefinition 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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(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
Basic Concept Three types of fits
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Limits: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”
The tolerance can also be defined as upper limit – lower limit on one same dimension
upper-limit and lower-limit
Lower limit Upper limit
Hole tolerance = 1.251-1.250=0.001
Shaft tolerance = 1.248-1.247=0.001
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.
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Allowances:
Allowance = Min Hole – Max ShaftAllowance = 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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Examples Figure 5Basic concept
Shaft tolerance = 1.248 - 1.247 =0.001Hole tolerance= 1.251-1.250= 0.001
Allowance=1.250-1.248= 0.02Max clearances=1.251-1.247= 0.04
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Tolerance representation
The unilateral form
The bilateral form
The limit form
2.245 - 2.250
0.495 - 0.500
2.247-2.253
000.0005.250.2
003.0003.0250.2
00.0005.0500.0
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In general
or
Positive First Large Limit on Top
Small limit first
T
T
DDD
TDD
DD
DD
Tolerance representation
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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
Standard
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Standard
Different countries and regions together to develop
- Concepts
- Rules
- Systems
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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.
Methodology for Determining Basic Size
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Basic Shaft SystemReason: in some applications, the shaft could
be better made as a reference
Different fits with the same shaft
Methodology for Determining Basic Size
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Basic Shaft System
the maximal shaft size as the basic size
Reason: Cold-finished shaft.- cold forging- cold molding- cold rolling
Methodology for Determining Basic Size
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Example
0.502 0.4980.500 0.495
0.5050.502
0.5000.499
Methodology for Determining Basic Size
Basic size =0.5
Basic hole system Basic shaft 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 0.4980.500 0.495
Basic hole system
The basic size = 0.500
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Example
Basic shaft system
0.5000.499
0.5050.502
Known:Allowance=0.002
Tolerance for hole= 0.003
Tolerance for shaft= 0.001
The minimal hole size:
0.500+0.002=0.502
The basic size=0.500The minimal shaft size:
0.500-0.001=0.499
The maximal hole: 0.502+0.003=0.505
Basic shaft system