Tolerance-Stackup-Course.pdf

(1)

Welcome to a Course On

Tolerance Stack-up Analysis using C

Tolerance Stack-up Analysis using C o

o -ordinate

-ordinate

Dimensioning and GD&T

Satyam Venture Engineering

Satyam Venture Engineering Services

Services

Pvt.Ltd., Secunderabad, INDIA

For

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About iSquare

iSquare

(

I

nterOperability

& I

nterChangeability

Solutions)

Pune, INDIA

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Focus Areas:

l

CAD Data InterOperability

: Consistent

representation of 3D CAD data in variety of

CAD/CAM/CAE applications and platforms.

l

InterChangeability

: Predicting Dimensional

Variations, its impact and causes at the

product and assembly level at early design

stage.

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Relationships:

l

InterOperability

:

–

With International TechneGroup Incorporated, USA

having more than 20 years of Experience in CAD

Data InterOperability technology, solutions and

services.

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Relationships:

• InterChangeability

:

• With Dimensional Control Systems, USA having

more than 15 years of experience in Dimensional

Control techniques, solutions and Services.

(6)

Our Offerings:

•CAD Data

CAD Data InterOperability

InterOperability

:

•Focused & Customized Training Programs on:

•CAD/CAM/CAE Data Exchange : Problems and Solutions from CAD, CAE , CAM Perspective. •CAD Model Quality Assessment : CAD Model Quality evaluation from downstream application perspective

•Software Solutions For:

•Effective Data exchange between heterogeneous CAD/CAM systems: R egardless of source, target application, standard and formats !! Solutions Include CA Dfix, IGES/Works,CAD/IQ. •Model Quality Assessment from Downstream application perspective

•Quality Services for:Quality Services for:

•Data Exchange, Data Migration, Lower version to higher or vice -a-versa

•‘Vendor – Supplier’ data integration : ensuring effective data exchange with minimal / NO rework at either ends.

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Our Offerings:

•InterChangeability

InterChangeability

:

•Focused & Customized Training Programs on:

•Dimensional Management : Understanding and appreciation of computer aided tools for. Takes participants thru evolution, various approaches and real l ife problems from their application areas.

•Software Solutions For:

•Dimensional Management / Stack Analysis: Solutions embedded in C ATIA V4/V5 as Gold Partner and also Stand Alone solutions for data coming from othe r CAD platforms !! Solutions Include 1-DCS, DCS-DFC, 3DCS-SA, 3DCS-CAA V5 Designer, 3DCS-CAA V5 Analyst, GDM3D

•Quality Services for:Quality Services for:

•Dimensional Engineering / Management : Base Line tolerance mode l creation, reporting with suggestions and recommendations. Follow -on consulting

•Per requirement, includes 1D, 1D with GD&T, Full 3D simulations, Piece – part variations, assembly variation prediction against desired objectives.

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Training Programs in Dimension

Management / Engineering

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Training Programs Launched:

– Fundamentals of GD&T based upon ASME Y14.5MFundamentals of GD&T based upon ASME Y14.5M: ~36hrs: ~36hrs

– Tolerance StaTolerance Stack up Analysis:ck up Analysis: A logical approaA logical approa ch to solve assemblch to solve assembl y buildy build problems

problems: ~30hrs: ~30hrs

– _{Advanced GD&T: Concepts and Applications as per ASME Y14.5M}_{Advanced GD&T: Concepts and Applications as per ASME Y14.5M}_{: ~30hrs}_{: ~30hrs} – Tolerance Stack up Analysis using DCS (Dimensional Control SysteTolerance Stack up Analysis using DCS (Dimensional Control Syste ms, USA)ms, USA)

Software Solutions

Software Solutions (1DCS, DCS-DFC, 3DCS-SA): ~36hrs(1DCS, DCS-DFC, 3DCS-SA): ~36hrs

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Training Programs Under Development (Tentative release by Oct

’

₀₅₎

– GD&T Workshop and PracticeGD&T Workshop and Practice(15% theory, 85% working on (15% theory, 85% working on various problems):various problems): ~24hrs

~24hrs

– The Role of Probability and Statistics in Mechanical Tolerance AThe Role of Probability and Statistics in Mechanical Tolerance A nalysis:nalysis:~20hrs~20hrs

– Measurement of GD&T and Measurement of GD&T and Functional Gauging TechniquesFunctional Gauging Techniques: ~24hrs: ~24hrs

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Customers

…

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l TATA MotorsTATA Motors

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l TATA TechnologiesTATA Technologies

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l TATA Auto PlasticsTATA Auto Plastics

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l TATA Auto ComponentsTATA Auto Components

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l Ashok LeylandAshok Leyland

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l Mahindra & MahindraMahindra & Mahindra – _Auto_Auto l

l Godrej & Boyce Mfg Ltd.Godrej & Boyce Mfg Ltd.

l l GEGE

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l Infotech EnterprisesInfotech Enterprises

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l TATA Johnson ControlTATA Johnson Control Automotive

Automotive

l

l Kinetic EngineeringKinetic Engineering

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l Research & Research & DevelopmentDevelopment Establishment (Engrs) Establishment (Engrs)

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l Armament Research &Armament Research & Development Establishment Development Establishment

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l Bhabha Atomic ResearchBhabha Atomic Research Center

Center

l

l Bajaj AutoBajaj Auto

l

l Bajaj TempoBajaj Tempo

l

l Brakes IndiaBrakes India

l

l Emerson Climate TechnologiesEmerson Climate Technologies

l

l Grupo AntolinGrupo Antolin

l

l MahindMahindra Engg ra Engg Design DevelopDesign Develop Center

Center

l

l Kirloskar CopelandKirloskar Copeland

l

l Mahindra Engineering ServicesMahindra Engineering Services

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l Onward TechnologiesOnward Technologies

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l Space Applications Center Space Applications Center

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l TATA ConsultancyTATA Consultancy

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l Lear Seatings Pvt. Ltd.Lear Seatings Pvt. Ltd.

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l Atlas CopcoAtlas Copco

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l Jayahind IndustriesJayahind Industries

l l L&TL&T

l

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How is Course Organized?

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Total 11 Sessions; 3days (June 23,24 and 25 , 2005)

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Pre-defined objectives at the beginning of each

session

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_{Classroom exercises at the end of each session}

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Homework

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Extended hours as necessary

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Assumption : Understanding of GD&T

Assumption : Understanding of GD&T controls

controls

l

(12)

Session #1 : The Basics

l

Objectives:

Ø

How to calculate mean dimensions with equal

Bilateral Tolerances

Ø

Calculating Inner and Outer Boundaries

Ø

(13)

What is Tolerance Stack-up Analysis?

Tolerance Stack-up Analysis (also called as Gap

Analysis, Loop Diagrams or Circuit Analysis)

is the

process of calculating minimum and maximum

airspaces or wall thickness or material interferences in

a single part or assemblies

(14)

Steps in Tolerance St

Steps in Tolerance St ack-up

ack-up Analysis

Analysis

l

Step #1

:

–

Identify objectives: for example, you want to test if no

interference is possible at a certain place in an assembly, then

you set your requirement as “Gap must be equal to or greater

than zero”

l

Step #2

:

–

Identify all dimensions that contribute to your objectives as

defined in step #1 (gap) and convert them to equal bilateral

toleranced dimensions; if they are not already

(15)

l

Step #3

:

–

Assign each dimension a +ve or – ve value. For Radial stacks

(going up and down); start at the bottom of gap and end up at

the top of gap

– Down direction is – ve (top of gap to bottom)

– Up direction is +ve (bottom of gap to top OR towards end)

–

Stacks that go left and right in the assembly, start at the left

side of gap and end up at the right side of the gap.

– Left direction is – ve (right of gap to left)

– Right direction is +ve (left of gap to right OR towards end)

l

Remember that you are working one part at a time; so deal with one

part

’ s significant features before jumping to next part. This is the best way

to work with assemblies having many parts

Steps in Tolerance Stack-up Analysis

…

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l

Step #4 (Basic Rules):

– Remember that one set of mating features between parts creates t he variable you are looking for. Variable in this case is eithe r minimum gap or maximum gap or maximum overall assembly dimension. One set mating featur es creates it. So, though multiple routes may have to be investigated to fi nd this most significant set of features, only one set creates worst case , fr om one part to next.

– Its often mistake to follow one route from one set of mating fea tures (holes/shaft, hole/pin) then continue the same route through ano ther set. One of these sets creates the smallest or biggest gap or maximum overall dimension, Once you find, which it is, others become non -factors in analysis.

– Using more than one set of features within same two parts, will most likely produce wrong results. Still tolerances from other features may contribute to the critical se t you are using. For exa mple: when datum f eatures are referenced at MMC or when more than one set of datum features co me into effect.

Steps in Tolerance Stack-up Analysis

…

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l

Step #5

(Basic Rules):

–

When a single feature or a pattern of features are controlled by

more than one Geometric Tolerance (such as orientation

combined with position), the designer must determine which, if

either is contributing factor to variable. It is also possible t hat

none of geometric tolerance is a factor and instead size

dimensions are factors.

–

The Designer must deduce what factors are pertinent through

sketches and reasoning.

–

The judgment of designer is critical in these determinations.

Steps in Tolerance Stack-up Analysis

…

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l

l Add all +ve and – ve dimensions which will calculate your mean gap. If mean gap is – ve number, your requirement of ‘no material interference’ (or ‘clearance’ in other words) is already violated!

l

l Then we must add sum of equal bilateral tolerances (1/2 of total tolerance) to the mean dimension (or gap) to determine maximum gap.

l

l Then we must subtract the sum of equal bilateral tolerances (1/2 of total tolerance) from the mean dimension (or gap) to determine minimum gap.

l

l Again any – ve value for minimum or maximum gaps indicate interference situation

l

l Maximum gaps are maximum clearance (or in case of interference fits, minimum interference)

l

l Minimum gaps are minimum clearance (or in case of interference fits, maximum interference)

Beginning Tolerance Stack-up Analysis

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l

Its important to mentally shove all the features and parts in the

directions that will create the max or min gap (variable).

This is to

allow your routes always pass through material and you don’t

want to jump over an air space unnecessarily in analysis

l

You should position the features of the parts against each other so

that you will get extremes and make clear to you the correct path

and +ve v/s – ve designations for each number.

Beginning Tolerance Stack-up Analysis

(20)

Finding Mean Dimensions

l

Few Important Concepts of Tolerance Stack-up

Analysis:

–

_{There is}

_NO

_{difference between equal, unequal or unilaterally}

toleranced dimension.

–

There is

NO

difference between a limit dimension and a plus

or minus toleranced dimension

–

They all have

extremes

and they all have

means

. So, first thing

is to change any dimension to an

equal bilateral toleranced

(21)

Finding Mean Dimensions

(22)

Finding Mean Dimensions

(23)

Finding Mean Dimensions :

Exercise

(24)

Boundaries

Boundaries are generated by

collective

effects of

size and Geometric tolerances

applied to

feature(s) and often referred to as simply

inner

and outer

boundaries

There are two types of boundaries

l

Virtual Condition boundary (VCB)

l

(25)

l

FCFs that use

m

_{(MMC symbol), generate}

_constant

boundaries

(VCB) for features under consideration and

are calculated as:

–

_{VCB for}

_{internal FOS}

_{such as hole =}

_{MMC Size Boundary –}

Geometric Tolerance value

–

VCB for

external FOS

such as pin =

MMC Size boundary +

Geometric Tolerance

VC Boundaries are Constant and do not vary based upon actual

mating size of the feature

Virtual Condition Boundaries

(Refer ASME Y14.5M sec tion 2.11)

(26)

Virtual Condition Boundaries

(Refer ASM E Y14.5M secti on 2.11)

l

FCFs that use

l

_{(LMC symbol), generate}

_constant

boundaries

(VCB) for features under consideration and

are calculated as:

–

VCB for

internal FOS

such as hole =

LMC Size Boundary +

Geometric Tolerance value

–

VCB for

external FOS

such as pin =

LMC Size boundary

-Geometric Tolerance.

VC Boundaries are Constant and do not vary based upon actual

mating size of the feature

(27)

Resultant Condition Boundaries

(Refer ASME Y14.5M secti on 2.11)

l

RC Boundaries are

non constant

in nature and

are generated on

opposite side

of the virtual

conditions.

l

When

RFS (Regardless of Feature Size)

concept applies to FOS, they generate

only

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Case#1: Internal FOS controlled at MMC

(29)

Case#1: Calculating VC & RC boundaries

VCB for internal FOS (such as hole) controlled at MMC = MMC Size Boundary – Geometric Tolerance value

VCB for external FOS (such as pin) controlled at MMC = MMC Size boundary + Geometric Tolerance value

(30)

Case#1: Creating equal

Bilateral Toleranced

Dimension

(31)

Case#2: Internal FOS controlled at LMC

(32)

Case#2: Calculating VC & RC boundaries

VCB for internal FOS (such as hole) controlled at LMC = LMC Size Boundary +Geometric Tolerance value

(33)

Case#2: Creating equal Bilateral Toleranced

Dimension from VCB and RCB

(34)

Case#3: Internal FOS controlled at RFS

(35)

Case#3: Calculating RC

Case#3: Calculating RC boundaries

boundaries

Since it

Since it’s a RFS Callout, no virtual s a RFS Callout, no virtual conditioncondition boundarie

(36)

Case#3: Assumption about feature form in

case of RFS callout

Only if the hole has a significant Only if the hole has a significant depth, might this median line depth, might this median line curvature (out of straightness) curvature (out of straightness) be a consideration. For thin be a consideration. For thin parts, such as sheet metal, it is parts, such as sheet metal, it is probably not of concern in these probably not of concern in these analyses. In fact many analyses. In fact many designers would agree that a designers would agree that a banana shaped hole is not likely banana shaped hole is not likely to occur on most products. to occur on most products. Therefore we are ignoring Therefore we are ignoring

“axially out of straightnessaxially out of straightness”

consideration from the consideration from the analyses.

(37)

Case#3: Creating equal Bilateral Toleranced

Dimension from Inner and Outer Boundaries

(38)

Case#4: External FOS Controlled at MMC

(39)

Case#4: Calculating VC & RC boundaries

VCB for internal FOS (such as hole) controlled at MMC = MMC Size Boundary – Geometric Tolerance value

VCB for external FOS (such as pin) controlled at MMC = MMC Size boundary + Geometric Tolerance value

(40)

Case#4: Creating equal Bilateral Toleranced

Dimension from VCB and RCB

(41)

Case#5: External FOS controlled at LMC

(42)

Case#5: Calculating VC & RC boundaries

VCB for internal FOS (such as hole) controlled at LMC = LMC Size Boundary +Geometric Tolerance value

(43)

Case#5: Creating equal Bilateral Toleranced

Dimension from VCB and RCB

(44)

Case#6: External FOS controlled at RFS

(45)

Case#6: Calculating RC

Case#6: Calculating RC boundaries

boundaries

(46)

Case#6: Assumption about feature form in

case of RFS callout

In case of RFS Callout, one may want to consider In case of RFS Callout, one may want to consider additional deviation arising

additional deviation arising‘out of formout of form’toto determine absolute worst case inner boundary. determine absolute worst case inner boundary. This applies only if shaft length is significant. For This applies only if shaft length is significant. For very shirt shafts / pins, it is probably not of very shirt shafts / pins, it is probably not of concern in the analysis.

concern in the analysis. Therefore we are ignoring

Therefore we are ignoring‘_{out-of-straightness}_{out-of-straightness}’

consideration from our analysis. If your product consideration from our analysis. If your product runs a risk of banana shaped shafts, you may runs a risk of banana shaped shafts, you may wish to consider illustration on the left in your wish to consider illustration on the left in your calculations.

(47)

Case

Case#6:

#6: Creat

Creating equa

ing equal Bilate

l Bilateral Tole

ral Tolerance

ranced

d

Dimension from Inner and Outer Boundaries

(48)

Formulae to Remember

…

For External FOS controlled at MMC / LMC: For External FOS controlled at MMC / LMC:

VCB at MMC

VCB at MMC (OB) = (OB) = MMC Size boundaryMMC Size boundary++Geometric Tolerance value at MMCGeometric Tolerance value at MMC VCB

VCB at at LMC LMC (IB) (IB) = = LMC LMC Size boSize boundaryundary--Geometric Tolerance value at LMCGeometric Tolerance value at LMC

For Internal FOS controlled at MMC / LMC: For Internal FOS controlled at MMC / LMC:

VCB at

VCB at MMC MMC (IB) (IB) = = MMC Size MMC Size BoundaryBoundary – –Geometric Tolerance value at MMCGeometric Tolerance value at MMC VCB at LMC

(49)

Finding Inner & Outer Boundaries :

(50)

Session #2: Analyzing a Box Assembly

l

Objectives:

To determine min and max gap for a simple

eleven parts assembly

l

Perform the calculations

l

Create a Loop Analysis Diagram

l

(51)

Box Assembly

(52)

Box Assembly: Part #1

l

Cavity in part#1 has

limit dimensions

(

392.43 –

384.81)

; so we need to convert these to

mean

with

equal bilateral toleranced dimensions

…

•

Add limit dimensions : 392.43 + 384.81=777.24

•

Find Mean dimension: 777.24/2=

388.62

•

Find total tolerance by subtracting limit dimensions: 392.43 –

384.81=7.62

•

Find equal bilateral tolerance=7.62/2=

3.81

•

Finally express limit dimensions as equal bilaterally toleranced

(53)

Box

Box Assem

Assembly:

bly: Part

Part #2-

#2- #11

#11

parts MMC … (1) parts LMC … (2) Add (1) and (2) …(3) Subtract (1) and (2)…(4) Half of (3) …(5) Half of (4) …(6) (5) (6)

(54)

Box Ass

Box Assemb

embly: Par

ly: Part #1 & Part #2-

t #1 & Part #2- #11 p

#11 put

ut

together

(55)

Box Assembly : Loop

Box Assembly : Loop Analysis Diagram

Analysis Diagram

l

lLoop Diagram begins by showingGap to be calculated at the top

l

lSo, loop diagram begins at Part#11 (plate) and progresses downwardconstantly through materialuntil it reaches at the last plate at the bottom of an assembly (ie. Part#2 or plate#2)

l

lThe sum of all these – vemean dimensions, which run from top to bottom is381and has total tolerance of`3.81

l

lThe loop then reverses and progresses up throughcavity(ie. Part #1).

l

lThis portion of the loop is+vesince it progresses frombottom to top

l

lThe logic of +ve and –ve is simple… material removes airspace (therefore –ve) and cavity which lacks material adds to airspace (therefore +ve)

(56)

Box Assembly : Loop

Box Assembly : Loop Analysis Diagram

Analysis Diagram

l

lThe mean dimension of cavity is 388.62 and has a total tolerance of`3.81

l

lSo, in numbers chart, we add means: (+)388.62 + ( -)381.00 = 7.62…(1)

l

lIf this number is –ve, it would have proven that even mean sizes of parts, when produced result in interference. Now that sum is +ve, we can proceed…

l

lNext step is to add, charted plus or minus tolerances : 3.81+3.8 1 = 7.62…(2)

l

lNext step is to calculate min and max gaps (airspace or interfer ence):

l

lMean dimensions difference + sum of tolerances = (1) + (2)= (+)7 .62+(+)7.62=+15.24 (max gap)

l

(57)

From situations suc

From situations suc h as this, it is easier to simply calculate thh as this, it is easier to simply calculate th eeMMC of theMMC of the cavity

cavityand theand thecollective MMCs of the platescollective MMCs of the plates andandsubtractsubtractthem to getthem to get

minimum minimumgap.gap.

Box Assembly : Alternate Method to

calculate min / max gap

(58)

Box Assembly : Loop

(59)

Session #2:

(60)

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Objectives:

–

Using Loop Analysis Technique; determine Max and Min gap

in Horizontal and Vertical Directions

–

Determine proper start and End points for stack-ups

–

Graph the numbers calculated into Loop Diagram

Session #3: Loop Analysis for Features of

Size (FOS)

(61)

Analyzing FOS: Problem Description

(62)

Analyzing FOS: Charts to be used

…

(63)

Analyzing FOS: Steps Involved

(Horizontal

Direction)

l

Convert horizontal limit dimensions of part #1 to equal bilateral toleranced

dimension.

(26.615

`

0.405)

l

Convert horizontal limit dimensions of part #2 to equal bilateral toleranced

dimension.

(25.705

`

0.105)

l

Graph the loop from

left-to-right

through material, using appropriate signs

(+ve / -ve)

(- 25.705 and + 26.615)

l

Add these mean dimensions (

don’t forget signs

) to get difference

between mean dimensions

((-) 25.705 + (+) 26.615)=+0.910

l

Add plus and minus tolerances for part#1 and part#2 to get total plus and

minus tolerance

(0.105+0.405)=0.510

l

Max gap

in horizontal direction is = sum of difference between mean

dimensions and total of plus and minus tolerances

(0.910+0.510=1.42)

l

Min gap

in horizontal direction is = difference (subtraction) of difference

between mean dimensions and total of plus and minus tolerances

(0.910-0.510=0.4)

(64)

Analyzing FOS: Steps Involved

(Vertical

Direction)

l

Convert vertical limit dimensions of part #1 to equal

bilateral toleranced

dimension.

(26.615

`

0.405)

l

Convert vertical limit dimensions of part #2 to equal

bilateral toleranced

dimension.

(24.390

`

0.610)

l

Graph the loop from

top-to-bottom

through material, using appropriate

signs (+ve / -ve)

(- 24.390 and + 26.615)

l

Add these mean dimensions (

don’t forget signs

) to get difference

between mean dimensions

((-) 24.390 + (+) 26.615)=+2.225

l

Add plus and minus tolerances for part#1 and part#2 to get total plus and

minus tolerance

(0.610+0.405)=1.015

l

Max gap

in vertical direction is = sum of difference between mean

dimensions and total of plus and minus tolerances

(2.225+ 1.015)=3.24

l

Min gap

in vertical direction is = difference (subtraction) of difference

between mean dimensions and total of plus and minus tolerances

(2.225-1.015)=1.21

(65)

Analyzing FOS:

Easier Method using MMC

and LMC

Calculating min and max gaps ma

Calculating min and max gaps ma y be easier as discussed before (y be easier as discussed before ( slide #48); byslide #48); by subtracting the MMCs for minimum gaps and LMC

subtracting the MMCs for minimum gaps and LMC s for maximum gaps s for maximum gaps asas shown below:

(66)

Analyzing FOS: Charts and Loops with

dimensions

…

Part #1, horizontal direction Part #2, horizontal direction Part #1, vertical direction Part #2, vertical direction

Material side (-ve) Part #2 thickness

(67)

Session #3:

(68)

Session #4: Analysis of an assembly with

Plus and Minus tolerancing

l

Objectives:

l

Calculate the airspaces and interferences for a plus and

minus toleranced assembly

l

(69)

Assembly with plus and minus tolerances :

Problem Description

(70)

Assembly with plus and minus tolerances :

Charts to be used

(71)

Steps Involved in calculating stack in

Horizontal Direction

l

l Convert horizontal limit dimension 1 male (31.75 – 34.90) to equal bilateral toleranced dimension.(33.325`_1.575)

l

l Convert horizontal limit dimension 2 air (21.41 – 19.84) to equal bilateral toleranced dimension.(20.625`_0.785)

l

l Convert horizontal limit dimension 3 male (16.67 – 15.09) to equal bilateral toleranced dimension.(15.88`_0.79)

l

l _{Graph the loop from}_{left-to-right}_{through material, using appropriate signs (+ve /}

-ve)(- 33.325, +20.625 and + 15.880)

l

l Add these mean dimensions (don’t forget signs) to get difference between mean dimensions((-) 33.325 + (+) 20.625 + (+)15.880)=+3.180

l

l Add plus and minus tolerances for dimensions 1, 2, 3 to get tota l plus and minus tolerance(0.785+0.790+ 1.575 = 3.150)

l

l Max gapin horizontal direction is = sum of difference between mean dimensions

and total of plus and minus tolerances(3.180+ 3.150=6.330)

l

l Min gapin horizontal direction is = difference (subtracti on) of difference between

(72)

Steps Involved in calculating stack in

Vertical

Direction

l

l Convert vertical limit dimension 1 air (26.21-25.40) to equal bilateral toleranced dimension.(25.805`_0.405)

l

l Convert vertical limit dimension 2 material (25.40 – 24.59) to equal bilateral toleranced dimension.(24.995`_0.405)

l

l Graph the loop fromtop-to-bottomthrough material, using appropriate signs (+ve / -ve)(- 24.995, +25.8 05)

l

l Add these mean dimensions (don’t forget signs) to get difference between mean dimensions((-) 24.995 + (+) 25.805)=+0.8 10

l

l Add plus and minus tolerances for dimensions 1, 2, 3 to get tota l plus and minus tolerance(0.405+0.405 = 0.810)

l

l Max gapin horizontal direction is = sum of difference between mean dimensions

and total of plus and minus tolerances(0.810+0.810=1.620)

l

l Min gapin horizontal direction is = difference (subtracti on) of difference between

(73)

Easier Method for calculating stacks using

MMC and LMC

Calculating min and max gaps ma

Calculating min and max gaps ma y be easier as discussed before (y be easier as discussed before ( slide #56); byslide #56); by subtracting the MMCs for minimum gaps and LMC

subtracting the MMCs for minimum gaps and LMC s for maximum gaps s for maximum gaps inin

horizontal direction

(74)

Assembly Analysis: Charts and Loops with

dimensions

…

(75)

Session #4:

(76)

Session #5: Analyzing a Floating Fastener

Assembly with Geometric Controls

l

Objectives:

l

Calculate Virtual and Resultant conditions (Inner / Outer

Boundaries) for GD&T callouts

l

Determine mean of all these boundaries

l

Convert all FOS (diameters and widths) to mean radii with

equal bilateral tolerance

l

Mixing FOSs (widths and diameters) in number chart

l

Graph the numbers in tolerance stack-up diagram

l

(77)

Floating fastener assembly sketch with

GD&T

(78)

Steps involved in analyzing floating fastener

assembly

1.

Calculate

Virtual

and

Resultant

Condition for each holes (holes

#1 thru #4)

2.

For each hole, calculate

difference

between

resultant

condition

and

virtual

condition boundaries. This difference represents

total

size tolerance for each hole

. Take half of the difference which is

represents

equal bilateral tolerance value

.

3.

For each hole,

add

resultant condition and virtual condition

boundaries; and take

mean

of the sum. This mean represents

the

mean diameter of that hole

(for analysis purpose)

4.

Again, for each hole, take mean of values in step #2 and #3.

(79)

Defining Virtual and Resultant Condition

Boundaries

As per ASME Y14.5M -1994,

l

Virtual Condition is defined as a

constant

value

outer locus

(

for

external FOS specified at MMC or internal FOS specified at LMC

)

and a

constant

value

inner locus

(

for internal FOS specified at

MMC and external FOS specified at LMC

).

l

Resultant conditions are in

opposite direction to virtual conditions

and are

non-constant

in nature. They are the worst case inner

locus and worst case outer locus of FOS

(80)

Charting values calculated per steps #1

through #4

Step #1:

(81)

Charting values calculated per steps #1

through #4

Step #2:

Step #2:For each hole, calculatedifferencebetweenresultantcondition andvirtualcondition boundaries. This difference representstotal size tolerance for each hole. Take half of the difference which is representsequal bilateral tolerance value.

(82)

Charting values calculated per steps #1

through #4

Step #3, #4:

Step #3, #4:For each hole, add resultant condition and virtual condition bou ndaries; and takemeanof the sum. This mean represents themean diameter of that hole(for analysis purpose) Again, for each hole, take mean of values in step #2 and #3. Thi s new mean representsmean

(83)

Step #1: values printed in the chart

Step #1:

(84)

Step #2: Values printed in the Chart

Step #2:

Step #2:For each hole, calculatedifferencebetweenresultantcondition andvirtualcondition boundaries. This difference representstotal size tolerance for each hole. Take half of the difference which is representsequal bilateral tolerance value.

(85)

Step #3, #4:

Step #3, #4:For each hole, add resultant condition and virtual condition bou ndaries; and takemeanof the sum. This mean represents themean diameter of that hole(for analysis purpose) Again, for each hole, take mean of values in step #2 and #3. Thi s new mean representsmean

radius`mean radial tolerance

Step #3,4: Values printed in the Chart

(86)

l

l The loop diagram begins by pushing all the parts is such a manner that the parts configuration (position) in an assembly would create a minimum gap. As shown in figure leftthis has had the effect of trapping the fasteners (or pins) between hole #1 and #3 and Hole #2 and #4.

l

l The pin on the left passing thru hole 1,3 is trapped by pushing part having hole 1 to the right and the pin on right passing thru holes 2 and 4 is trapped by pushing part having hole 2 to the left (both these operations closes gap between the parts having holes 1, 2)

l

l Since this is a floating fastener situation, pin on the left (passing thru holes 1,3) and pin on the right (passing thru holes 2,4) continues to slide unless the pins are held against the left and right side of respective holes as shown in figure

Locating / Orienting parts in an Assembly to

create MIN gap

(87)

Construct a Loop Diagram

1. The Loop begins at the the face on the left side of the gap, it proceeds towards left (thru material), designated as – ve numbers thru the basic dimension of 125mm to the center of hole #1.

2. Go 3mm left (-ve) thru the radius of hole #1 ( as calculated with its VC and RC boundaries). We are done with hole #1 and exhausted part #1

3. Now go over the left pin trapped between holes 1, 3 in right direction (+ve) 3mm

4. Now we come across hole#3 and we now go from right side of hole#3 towards it center in left direction (-ve) 3mm.

5. Step 4 takes us to center of hole 3. From there we go towards left (+ve) to the center of hole 4; 260mm. We are done with hole #3.

6. This step reverses the route going left (-ve) thru 3mm radius of hole #4. We are done with hole #4.

7. Here we again reverse the loop and go right direction (+ve) thru the right pin diameter 3mm

8. This step begins on last hole #2, the route goes from right side edge of this hole towards center (-ve) 3mm

9. Go in the same direction from center of hole #2 to the end of the gap (inner right side face of part) (-ve) 125mm 1 2 3 4 5 6 7 8 9

(88)

Logic Behind Loop Diagram

The logic behind this loop route was to proceed from left edge/face of gap through all features having an effect on the Minimum Gap, to the right of gap.

To begin, the parts were shoved to create a min gap configuration and in this case this is the only logical route to take.

The route went left and right and involved all related features until loop was complete. The related feature list includes four holes #1 thru #4, pins on left and right . Hole radii were used because the pertinent dimensions binding the gap to holes and holes to each other, went to the hole centers.

The full pin diameters were used as the pins got trapped between hole edge faces.

The basic dimensions were used to allow a route from left side of gap to the center of hole 1 and then center of hole 3 to to the center of hole 4 and at the last from center of hole 2 to the right side face of the gap

Key in this table the – ve and +ve route values

(89)

Loop Diagram with values printed

…

(90)

Can you Locate / Orient parts in an Assembly

to create

(91)

Loop Diagram for

MAX

gap with values

printed

(92)

Session #5:

Exercise

Calculate MIN and MAX Gap Calculate MIN and MAX Gap for the assembly shown in for the assembly shown in figure at left

(93)

Session#6: Analyzing an Assembly with Tab

and Slot

(94)

l

Objectives:

l l

Calculate

assembly overall MAX and MIN

dimensions

l

Calculate

MAX

and

MIN gaps

within assembly as shown

l

Calculate boundaries using various

GD&T

controls

Session#6: Analyzing an Assembly with Tab

and Slot

(95)

Steps involved in analyz

Steps involved in analyz ing Tab

ing Tab -Slot

-Slot

assembly

1.

Calculate

Virtual

and

Resultant

Condition for Tab and Slot. They

work on similar principals as hole and pin and are controlled at

MMC.

2.

For Slot and Tab, calculate

difference

between

resultant

condition and

virtual

condition boundaries. This difference

represents

total size tolerance for Slot or Tab

. Take half of the

difference which is represents

equal bilateral tolerance value

.

3.

For Slot and Tab, add resultant condition and virtual condition

boundaries; and take

mean

of the sum. This mean represents

the

mean width for either Slot or Tab

(for analysis purpose)

4.

Again, for Slot and Tab, take mean of values in step #2 and #3.

(96)

Charting values calculated per steps #1

through #4

Step #1

Step #1: CalculateVirtualandResultantCondition for Tab and Slot. They work on similar principals as hole and pin and are controlled at MMC

(97)

Charting values calculated per steps #1

through #4

2. For Slot and Tab, calculate

differencebetweenresultant

condition andvirtualcondition boundaries. This difference representstotal size tolerance for Slot or Tab. Take half of the difference which is representsequal bilateral tolerance value.

3. For Slot and Tab, add resultant condition and virtual condition boundaries; and takemeanof the sum. This mean represents the

mean width for either Slot or Tab(for analysis purpose)

4. Again, for Slot and Tab, take mean of values in step #2 and #3. This new mean representsmean radius`mean radial tolerance

(98)

Step #1: values printed in the chart

Step #1

Step #1: CalculateVirtualandResultantCondition for Tab and Slot. They work on similar principals as hole and pin and are controlled at MMC

(99)

Step #2,3,4: Values printed in the Chart

2. For Slot and Tab, calculate

differencebetweenresultant

condition andvirtualcondition boundaries. This difference representstotal size tolerance for Slot or Tab. Take half of the difference which is representsequal bilateral tolerance value.

3. For Slot and Tab, add resultant condition and virtual condition boundaries; and takemeanof the sum. This mean represents the

mean width for either Slot or Tab(for analysis purpose)

4. Again, for Slot and Tab, take mean of values in step #2 and #3. This new mean representsmean radius`mean radial tolerance

(100)

Locating parts in an Assembly to create MIN

overall Dimension and creating a Loop

Diagram

…

l

l Before we begin constructing loop diagram for minimum overall dimension, we must imagine the parts being shoved together such that configuration creates minimum overall assembly dimension

l

l This means left side of tab is pushed against left side of slot

l

l Loop Diagram follows this routeLoop Diagram follows this route…

1. We know now to start with left side face/edge of part#1 to the center of slot.

2. Then back to the left side of slot and tab

3. Then back to the center of tab and then

To the right side face of part#2 1

2

3 4

(101)

Loop Diagram with values printed

…

(102)

Can you Locate parts in an Assembly to

create MAX overall dimension and create a

Loop Diagram?

(103)

Loop Diagram for MAX Overall Dimension

with values printed

(104)

Calculating MAX, MIN values for

Lower-Left

and

Upper-Right

gaps

Case #1: Min

(105)

Case #1:

(106)

Case #2:

(107)

Case #3:

(108)

Case #4:

(109)

Calculating MAX overall Diameter for a

coupling

Assembly Assembly

(110)

Detailed Part Drawing with GD&T Controls

Notice the controls Notice the controls used and study the used and study the drawing

(111)

Factors

and

Non-Factors

in calculating

overall Diameter

l

lWill the controls and dimensions circled in color

color will participate in overall diameter calculations

OR

l

l_{Will the controls and dimensions circled in}

color

color will participate in overall diameter calculations

l lOR

l

lWill dimensions circled inbothbothcolorscolors participate or None?

(112)

Step #1: Calculate Virtual condition and

Resultant Condition boundaries for Threaded

holes

l

Threaded Hole on Crank Shaft (consider a stud here now):

§§

VC=

n

8.0+0.44=

n

8.44

§§

RC=

n

8.0-0.44=

n

7.56

§§

Sum of RC+VC=

n

16; half of this=

n

8

§§

Difference of VC and RC=

n

8.44-

n

7.56=0.88; half of this is 0.44

§§

So, threaded hole expressed in equal bilateral toleranced dimens ion

is:

n

8

`

0.44

(113)

Step #1: Calculate Virtual condition and

Resultant Condition boundaries for

Clearance holes

l

Clearance Hole on Coupling:

§§

VC=

n

8.66-0.22=

n

8.44

§§

RC=n

8.90+0.22=n

9.12

§§

Sum of RC+VC=

n

17.56; half of this=

n

8.78

§§

Difference of VC and RC=

n

9.12-

n

8.44=0.68; half of this is 0.34

§§

So, clearance hole expressed in equal bilateral toleranced dimension

is:

n

8.78

`

0.34 Now, Calculate difference

Now, Calculate difference between biggest clearance

between biggest clearance hole diamete

hole diameter and

r and

biggest threaded hole diameter =

(114)

Step #2: Calculate Clearances between

Datum Feature Diameters

‘

_D

’

_and

‘

_B

’

l

l In this case the perpendicularity tolerance callout on crankshaf t’s center bore and couplings center shoulder are ignored since the maximum clearanc e (and thus “play”) between these two features would occur at when both feature s a re at their LMC sizes and perfectly perpendicular to their datum planes.

l

l Center Bore on Crankshaft “Datum feature D ” :

§§ LMC = 50.10

l

l Center Shoulder on Coupling “Datum feature B ” :

§§ LMC = 49.97

Subtracting these two values, we get clearance of 50.10 – 49.97 = 0.13, which is less than 0.68 clearance calculated on threaded and clearance hole in previous slide.

This means in this case, the threaded/clearance holes are not the factors in stack -up and we would consider only offset between datum features B and D due to their respective LMC sizes.

(115)

Step#3: Create a Loop Diagram

All Dimensions printed in the Loop are Nominal All Dimensions printed in the Loop are Nominal

(bolt holes are ignored now onward) (bolt holes are ignored now onward)

+115

+25.05 -24.985 +115

(116)

Step#4: Chart the values

…

Max Dimension Max Dimension 230. 230.065065 + 0.30 + 0.30 ==230.365230.365 255.05 -255.05 - 24.985 = 24.985 = 230.06230.0655 Totals Totals 0.30 0.30 24.985 24.985 255.05 255.05

From center of coupling to end

`.15 (Size tol of` 0.2/2 and Gtol of`

0.1/2) radial calculations

-115

From edge of center bore/shoulder to center of coupling -24.985 -From center of crankshaft to edge of center bore/shoulder -25.05

From start to center of crankshaft

`.15 (Size tol of` 0.2/2 and Gtol of`

0.1/2) radial calculations -115 Remarks Remarks `` ToleranceTolerance (-ve) (-ve) (+ve) (+ve) Top to Top to Bottom Bottom Bottom to Bottom to Top Top

(117)

Session #6:

Exercis e #1

Calculate MAX/MIN Overall dimensions, Calculate MIN/MAX Gaps Calculate MAX/MIN Overall dimensions, Calculate MIN/MAX Gaps

(118)

Session #6:

Exercis e #2

Calculate MAX Calculate MAX overall diameter overall diameter of assembly of assembly

(119)

Session #7: Analyzing a Rail Assembly

having Fixed fasteners

Assembly Assembly

(120)

Part #1: Detailed Drawing

(121)

Part #2: Detailed Drawing

(122)

l

Objectives:

l

Calculate Boundaries for Threaded features

l

_{Work with multiple Geometric Controls on a single feature}

l

Determine effect of Projected Tolerance Zone on Stack -up

l

GD&T Controls affecting and non-affecting stack-up

l

Calculate Clearance and Interference

l

Use product knowledge / experience and Assembly

conditions in stack-up analysis

Session #7: Analyzing a Rail Assembly

having Fixed fasteners

(123)

Observations from Assembly and Part

drawings

…

l

A classic example of fixed fastener assembly: a threaded hole in

rail and a clearance hole in block

l

Note that datum features B on both part are given a flatness

tolerance since they are mating

l

_{Note the refinement frame for Datum feature B on rail.}

l

Consider objectives: to calculate

max/min gap between rail and

block in assembled condition

. This means we need to calculate

pertinent boundaries for features that affect objectives such as

boundaries for slot in rail, width of block, screws when mounted in

rail, clearance holes in block.

(124)

Steps involved in analyzing Rail assembly

1.

Calculate

Inner and Outer Boundaries

for Slot and Width of

block.

(Will there be VCB or just RCBs?)

2.

For Slot and Width of block, calculate

difference

between

Inner

and

Outer

boundaries. This difference represents

total size

tolerance for Slot or Width of block

. Take half of the difference

which is represents

equal bilateral tolerance value

.

3.

For Slot and Width of block,

add Inner

and

Outer

boundaries;

and take

mean

of the sum. This mean represents the

mean

width for either Slot or Width of block

(for analysis purpose)

4.

Again, for Slot and Width of block, take mean of values in step

#2 and #3. This new mean represents

mean radius

`

_mean

(125)

Steps involved in analyzing Rail assembly

5.

Calculate

Inner and Outer Boundaries

for Threaded Hole and

Clearance hole.

(Will there be VCB or just RCBs?)

6.

For Threaded Hole and Clearance hole, calculate

difference

between

Inner

and

Outer

boundaries. This difference represents

total size tolerance for Threaded Hole and Clearance hole

. Take

half of the difference which is represents

equal bilateral

tolerance value

.

7.

For Threaded Hole and Clearance hole,

add Inner

and

Outer

boundaries; and take

mean

of the sum. This mean represents

the

mean width for either Threaded Hole and Clearance hole

(for

analysis purpose)

8.

Again, for Threaded Hole and Clearance hole, take

mean

of

values in step #2 and #3. This new mean represents

mean

(126)

Step #1: Boundaries calculations for slot and

width

1. CalculateInner and Outer Boundariesfor Slot and Width of block.

Note that slot in

Note that slot in the rail has refinement frame. A the rail has refinement frame. A positional topositional tolerance is refined by alerance is refined by a orientat

orientation (perpendicion (perpendicularity) toleranularity) tolerance. So, is ce. So, is positiopositional tolenal tole rance a rance a factor in stackfactor in stack -up or-up or an orientation tolerance?

an orientation tolerance? Draw tolerance zone shapes / bound

(127)

Step #2,3,4: Calculating Mean Radius /

tolerance for slot and width of block

2. For Slot and Width of block, calculatedifferencebetweenInner

andOuter boundaries. This difference representstotal size tolerance for Slot or Width of block. Take half of the difference which is representsequal bilateral tolerance value.

3. For Slot and Width of block,add Inner andOuter boundaries; and takemeanof the sum. This mean represents themean width for either Slot or Width of block(for analysis purpose)

4. Again, for Slot and Width of block, take mean of values in step #2 and #3. This new mean representsmean radius`mean radial tolerance

Outer Boundary of Block = - Inner Boundary of Block =

- ---Difference = ½Difference of Block=

Outer Boundary of Block = +Inner Boundary of Block =

-

---Sum =

½of Sum of OB & IB of Block = ½Sum` ½ Difference of Block = ½ of ½ Sum ` ½ of ½ Diff of Block = Outer Boundary of Slot =

- Inner Boundary of Slot =

- ---Difference = ½Difference of Slot =

Outer Boundary of Slot = +Inner Boundary of Slot =

-

---Sum =

½of Sum of OB & IB of Slot = ½Sum` ½ Difference of Slot = ½ of ½ Sum` ½ of ½ Diff of Slot =

(128)

Step #1: Boundaries calculations: Values

printed in the chart

1. CalculateInner and Outer Boundariesfor Slot and Width of block.

Note that we have

Note that we have ignored positional tolerance on the slot in raiignored positional tolerance on the slot in rail.Only orientationl.Only orientation (perpendicularity) is accounted for in the analysis.

(129)

Step #2,3,4: Calculating Mean Radius /

tolerance : values printed in the chart

2. For Slot and Width of block, calculatedifferencebetweenInner

andOuter boundaries. This difference representstotal size tolerance for Slot or Width of block. Take half of the difference which is representsequal bilateral tolerance value.

3. For Slot and Width of block,add Inner andOuter boundaries; and takemeanof the sum. This mean represents themean width for either Slot or Width of block(for analysis purpose)

4. Again, for Slot and Width of block, take mean of values in step #2 and #3. This new mean representsmean radius`mean radial tolerance

Outer Boundary of Block =1.444 - Inner Boundary of Blo ck =1.436 - ---Difference = 0.008 ½Difference of Block =0.004

Outer Boundary of Block =1.444 +Inner Boundary of Block =1.436 -

---Sum =2.880

½of Sum of OB & IB of Block =1.440 ½Sum` ½ Difference of Block =1.440 ` 0. 004

½ of ½ Sum` ½ of ½ Diff of Block = 0.720` 0. 002

Outer Boundary of Slot =1.510 - Inner Boundary of Slot =1.502 - ---Difference =0.008 ½Difference of Slot =0.004

Outer Boundary of Slot =1.510 +Inner Boundary of Slot =1.502 -

---Sum =3.012

½of Sum of OB & IB of Slot =1.506 ½Sum` ½ Difference of Slot =1.506 ` 0. 00 4

½ of ½ Sum ` ½ of ½ Diff of Slot =0.753 `0.002

(130)

Step #5: Boundaries calculations for

Threaded and Clearance Hole

5. CalculateInner and Outer Boundariesfor Threaded Hole and Clearance hole.

(Why not VC and RC Boundaries?)

What is projected tolerance, why it is important? Explain What is projected tolerance, why it is important? Explain…

(131)

Step #6,7,8: Calculating Mean Radius /

tolerance for Threaded & Clearance Hole

6. For Threaded Hole and Clearance hole, calculatedifferencebetween

Inner andOuter boundaries. This difference representstotal size tolerance for Threaded Hole and Clearance hole. Take half of the difference which is represents

equal bilateral tolerance value.

7. For Threaded Hole and Clearance hole,add Inner andOuter

boundaries; and takemeanof the sum. This mean represents the

mean width for either Threaded Hole and Clearance hole(for analysis purpose)

8. Again, for Threaded Hole and Clearance hole, takemeanof values in step #2 and #3. This new mean representsmean radius`mean radial tolerance Outer Boundary of Hole =

- Inner Boundary of Hole =

- ---Difference = ½Difference of Hole =

Outer Boundary of Hole = +Inner Boundary of Hole =

-

---Sum =

½of Sum of OB & IB of Hole = ½Sum` ½ Difference of Hole = ½ of ½ Sum ` ½ of ½ Diff o f Hole = Outer Boundary of Screw =

- Inner Bound ary of Screw =

- ---Difference = ½Difference of Screw =

Outer Boundary of Screw = +Inner Boundary of Screw =

-

---Sum =

½of Sum of OB & IB of Screw = ½Sum` ½ Diffe rence of Screw = ½ of ½ Sum` ½ of ½ Diff of Sc rew =

(132)

Step #1: Boundaries calculations: Threaded

& Clearance Hole: Values printed in

& Clearance Hole: Values printed in the chart

the chart

5. CalculateInner and Outer Boundariesfor Threaded Hole and Clearance hole.

Inner Boundary of Screw Mounted in Rail =n0.2408 (LMC Major Dia) – 0.0140 =n0.2268 Outer Boundary of Screw Mounted in Rail =n0.250 + 0.0140 =n0.264

Outer Boundary of Hole in Block =n0.286+ 0.015 =n0.301 Inner Boundary of Hole in Block =n0.276- 0.005 =n0.271