handout-quality 10a mar13
TRANSCRIPT
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TOLERANCES - Introduction
Nearly impossible to make the part to the exact dimension by anymeans of manufacturing approach - tolerances of the dimension.
- Dimension 30 (mm) wont 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 withoutany special treatment
(a) 30.00 (shaft)
(b) 30.20 (hole)
(a) and (b) are assembled with a possibility of poorFunction 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 asgenerous 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 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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(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 atolerance 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
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Examples Figure 5
Basic concept
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 bilateral form
The limit form
2.245 - 2.250
0.495 - 0.500
2.247-2.253
000.0
005.250.2
J
003.0
003.0250.2
00.0
005.0500.0
J
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In general
or
Positive First Large Limit on Top
Small limit first
T
T
D
DD
TDD s
D
D
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 mademore 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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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 0.498
0.500 0.495
Basic hole system
The basic size = 0.500
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Next Example: BASIC SHAFT SYSTEM
Allowance=0.002
Tolerance for hole= 0.003
Tolerance for shaft= 0.001
basic size=0.500
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Example
Basic shaft system
0.500
0.499
0.505
0.502
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=0.500
The minimal shaft size:
0.500-0.001=0.499
The maximal hole: 0.502+0.003=0.505