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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 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 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 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 5

Basic 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 System

Reason: 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 0.498

0.500 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 0.498

0.500 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=0.500

The minimal shaft size:

0.500-0.001=0.499

The maximal hole: 0.502+0.003=0.505

Basic shaft system