Handout-quality 10a Mar13

8/8/2019 Handout-quality 10a Mar13

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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



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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



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Example

0.502 0.4980.500 0.495

0.5050.502

0.5000.499


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

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Handout-quality 10a Mar13