tolerance stack up statistics

Tolerance Stackup AnalysisTolerance Stackup AnalysisStatistics

ByDr N RamaniDr. N. Ramani

1

At the end of this Training, the participant should be able to:

1. Master basic Tolerance Types & Tolerance Concepts

2. To do basic Stack-up Analysis Techniques for 100% & 99.7% Interchangeability

3. Do a formal Tolerance stack-up analysis for Documenting d i C ditidesign Conditions

4. Determine if statistical Interchangeability will give a lower cost productcost product

5. Determine if larger tolerance zones can meet Design requirementsrequirements

6. Understand the Principle of ‘Robust Design’

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Case of 100% infallible interchangeability no matter the cost :the cost :

Safety is of paramount importance

3

Why do we require to do Tolerance Analysis?y y

4

Why do we require to do Tolerance Analysis?

1.To determine whether the parts will assemble 100% of the time or only 99.7% of the time statistically?statistically?

2.To determine if the parts will function properly t t ditiat worst condition

3.To determine if the drawing tolerances could be larger

4.To complete the design processp g p

5.To provide a record of the dimensional design requirements that can be reviewed at a later

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requirements that can be reviewed at a later date in case of a product problem

Assumptions while doing Tolerance Analysis:

1.All dimensions apply at 20oC

2 All f t d t t2.All manufactured parts meet dimensional requirements of drawing

3.All parts are rigid in free state & in assemblyasse b y

4.For , parts are manufactured ith th di i T t

STwith the mean dimension as Target

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What are the Types of Tolerance Analysis?

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1.Radial Stack –

2 Linear Stack –2.Linear Stack –

3.Assembly Stack –

What do they do?

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1.Radial Stack – Involves diameters or radial directions

2.Linear Stack – Involves dimensions that are in X,Y or Z direction,

3.Assembly Stack – Involves radial or linear directions of several partsdirections of several parts

Give a pictorial example for each type

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•Example of a Radial Stack

•R,Θ

10

,

X

Find the Dimension & Tolerance of ‘X’ for +/ 3 Sigma conformance100% I t h bilit

11Example of a Linear Stack

for +/-3 Sigma conformance100% Interchangeability

??

Assembly Stack

12

y

100 $ ? $100 $

COST

100% Interchangeability

99.7% Interchangeability

E ti t th t f 99 7% i t h bilit13

Estimate the cost for 99.7% interchangeability Please write down your answer

100 $ ? $100 $ ? $100 $

COST



14Estimate the cost for 99.7% interchangeability

1.Let us say that there is a possibility to reduce cost substantially, if you are prepared to accept 99.7% Interchangeability, in place of 100%

2.Let us say that cost saving will more than offset the loss of 0.3% Interchangeability (3 out of 1000

)assemblies)

3.Please answer the question, at what reduced cost, you will accept 99.7% Interchangeability

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100 $100 $ Cost Saving

? $Cost Saving

COST



16At what reduced cost, you will accept 99.7% Interchangeability?

100 $

COST

< 20 $ !< 20 $ !



17Based on Statistical Principles!

100%100%Interchangeability

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Assumption“Extreme assembly conditions

can and will be met in practice”

100%Interchangeability

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Effect of component tolerancescomponent tolerances

upon the assembly




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Effect of component tolerances Functioning of component tolerances

upon the assembly

g

finished product




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Si f ‘B’ P t

Highest

Size of ‘B’ Part

Lowest

‘B’Lowest Highest

‘B’

Size of ‘A’ Part‘A’

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C maxC min

B min B max

Assembly

AA i

yof 3 Parts

A maxA min

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How does Insurance Business function?

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Effect onEffect on Assembly

1 2 543 6 7 108 91 2 543 6 7 108 9Number of Parts / Feature Dimensions

25Affecting Assembly

•Assembly tolerance = Sum of all tolerances of the individual partsindividual parts

(OK for 2 or 3 parts…)

•One method is to provide the widest practicalOne method is to provide the widest practical component tolerances based upon the statistical fact that it is unlikely that all maximum-tolerance parts or all minimum-tolerance parts would ever be brought together in the same assembly.

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Objectives of this presentation:

1

Preferred Worst case

1

component tolerances

Worst case assembly variation for 100% & 99 7%

(Based on Cp & Cpk)

for 100% & 99.7% Interchangeability

p p )

229

2

?*?*

* For 100% Interchangeability

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X

Find the Dimension & Tolerance of ‘X’

33for +/-3 Sigma conformance100% Interchangeability

Stage 1. Covert all Dimensions to be Median values

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Stage 1. Covert all Dimensions to be Median values & Tolerances into Bilateral ones

Start End

234 34

1

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Plus Direction Distance

Minus Direction Distance

+/-Tolerance

A #Distance

'B'Distance

'C'Tolerance

'D'1 0.2175 0.00252 2.62 0.013 0.42 0.01

Stage3:

Tabulate4 1.8985 0.0025

2.62 2.536 0.025

Tabulate Analysis

360.084 0.025

Introduction to Normal Distribution

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100% Successful Interchangeability X σ?

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40Please remember the value of tolerance!

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41100% Success 6σ (?)

6σ

2.63 / 2.61

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Cf 100% Vs 99.7% Interchangeability

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Comparison bet een 6 &3 LimitsComparison between 6σ &3σ Limits# Sigma Value Toleranceg1 6 (100%

I t h bilit )+/-0.0250

Interchangeability)2 3 (99.7% +/-0.0146(

Interchangeability)

Which is better from Assembly point of view? Lesson?

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Lesson?

Compare the two drawingsg

AA

B48

B

BX

Calculate the Dimension & Tolerance for the Circlip Groove @ 100% & 99 7%

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Circlip Groove @ 100% & 99.7% Interchangeability

50Cf between 100% & 99.7% Interchangeability

Comparison bet een 6 &3 LimitsComparison between 6σ &3σ Limits# Sigma Value Toleranceg1 6 (100%

I t h bilit )+/-0.0500

Interchangeability)2 3 (99.7% +/-0.0250(

Interchangeability)

Which is better from Assembly point of view? Cf Cost also

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# Sigma Value Tolerance1 6 (100%

Interchangeability)+/-0.0500

Interchangeability)2 3 (99.7% +/-0.0250

Interchangeability)Determine the relationship betweenDetermine the relationship between Assembly (resultant) tolerance of +/-0.025 & four tolerances of part viz.0.0125four tolerances of part viz.0.0125

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(Assembly Tolerance)2

{ Number of Parts * (Part Tolerance)2 }Part Tolerance = Assembly Tolerance / Sq.Rt of N,Part Tolerance Assembly Tolerance / Sq.Rt of N,

where N is the number of parts or dimensions involved

Very Important to note:

Assembly Variance = Arithmetic sum of Variances ofAssembly Variance = Arithmetic sum of Variances of Parts

2 A bl ( * 2) f P id CLTσ2 Assembly =Σ (n1 * σ12) of Parts vide CLT

∴ Assembly Tol = Part Tolerance *√n

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# Sigma Value Tolerance1 6 (100%

Interchangeability)+/-0.0500

Interchangeability)2 3 (99.7% +/-0.0250

Interchangeability)

(√

Assembly tolerance of +/-0.025 = 2 * 0.0125

= 4 * 0 0125)(√= 4 * 0.0125)

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TSA Vs STA

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ST?ST?

STSuper Structure

ST

? Foundation

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?

Assigning of Tolerances to related components of an p

assembly

STSuper Structure

STAssembly Tolerance

= Foundation

58RMS value of Individual Tolerances

Process Control:

1.Must be maintained on any statistical tolerance vide ASME Y14.5 standard

2.Symbol drawn next to dimension

3 Not shown in our drawings3.Not shown in our drawings

4.Operations / QC will maintain Statistical t l fcontrol of process

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What are the benefits of Statistical Tolerancing?

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Benefits of ST:

1.Reduced costs

2.Closer average Fits2.Closer average Fits

3. Improved Quality / Performance of Product

• Sony USA Vs Sony Japan Story

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When should we apply ST?

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1.Limited Space requires a close assembly tolerance

2.Fits with a narrow range of Clearance required

3 100% Interchangeability not possible3.100% Interchangeability not possible

4.Means of reducing manufacturing cost

5. To reduce Tolerance Accumulation

• Selective assembly requiredSelective assembly required

• Adjustments / complex designs required

63• Tighter tolerances on components required


•What if Analysis?

•You want to know what will be the condition•You want to know what will be the condition for 99.7% of parts if manufactured around the mean of dimensions

TSA Vs STA

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When will ST succeed?When will ST succeed?

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When will ST succeed?

1.Manufacturing is done to the middle of the dimension

2.Use of proper controls to produce parts to a near normal distribution within drawing specification

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Responsibility for achieving ST:

1.Engineers understand that parts are to be produced with the ‘Target on the mean’

2.Tools are designed & produced to achieve the above target*

3.Necessary inspection equipment are available (In process gauging..) to determine ( p g g g )Cp & Cpk

*Reamer Design Example

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Normal Distribution:

1.Formed where manufacturing processes produce in a random manner about the mean with a majority of the dimensions close to the mean & a decreasing number occurring away from the meanfrom the mean

2. If dimensions from a stable process are measured & recorded according to size, a plot of resulting frequency distribution will approximate the Gaussian curveapproximate the Gaussian curve

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0.084+/-0.05 0.084+/-0.025 @ 100%

Interchangeability@ 99.7%

Interchangeability

69CLT?You cant have something…….

Conclusion?

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• Four tolerance zones can be increased & still• Four tolerance zones can be increased & still meet original design requirements of 0.084+/-0.025 using 99.7% probabilityg p y

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1 Four tolerance zones can be increased & still1.Four tolerance zones can be increased & still meet original design requirements of 0.084+/-0.025 using 99.7% probabilityg p y

2.How did we get to suggest +/-0.0125 tolerance?

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1 Four tolerance zones can be increased & still1.Four tolerance zones can be increased & still meet original design requirements of 0.084+/-0.025 using 99.7% probabilityg p y

2.How did we get to suggest +/-0.0125 tolerance?

3.From CLT formula:

4.Tol of parts / features = Tol of Assy / (Sq.rt of p y ( qnumber of parts or features)

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100%

99 7%99.7%

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Is it possible to relax manufacturing tolerances on all 5 di i * & till hi bl t t f5 dimensions* & still achieve assembly target of 0.034 +/-0.022, thereby reducing cost

77* +/- 2,3,4,5 & 8 thous

1.We have relaxed original tolerances of +/- 2,3,4,5 & 8 thous to +/ 0 0108 thous to +/- 0.010

2.Check correctness of Calculations

783.How did we get proposed tolerance of 0.010?

Dim.Tolerance = (assembly Tolerance / Sq.rt of b f di i i l d)number of dimensions involved)

= 0.022 / √ 5

= 0.010

Very ImportantVery Important

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Part 3

P t 1Part 1Part 2

•You are the designer of this assembly

•The top surface of part 1 is to be proud of the top surface of part 3 by 0.0005 – 0.018

•Determine the Unknown Length & fix tolerances of the di i f t 2 & 3 t hi th bj ti f 100%

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dimensions of parts 2 & 3 to achieve the objective for 100% interchangeability

1. Average Axial Clearance = (0.0005+0.018) / 2 = 0.00925 Nominal

2. Total Tolerance = (0.018 - 0.0005) = 0.0175 = +/- 0.00875

3. Target Clearance = 0.00925 +/- 0.00875 (Check max & min values)

4. Nominal Unknown Length = (0.250+0.125+0.00925) = 0.384(approx)

5. Tolerance of Dimensions 0.384, 0.250 & 0.125 = +/- 0.00875 / 3 = +/- 0.0029

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Part 3

P t 1Part 1Part 2

•You are the designer of this assembly

•The top surface of part 1 is to be proud of the top surface of part 3 by 0.0005 – 0.018

•Determine the Unknown Length & fix tolerances of the di i f t 2 & 3 t hi th bj ti f 99 7%

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dimensions of parts 2 & 3 to achieve the objective for 99.7% interchangeability

Part Tolerance = Assembly tolerance / √ n

= 0.0175 / √3

= 0 010= 0.010

= +/- 0.005

Rechecking:

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Determine Assembly Gap for 100% of cases

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T iti b tTransition between

Max Clearance 0.0275

Max Interference 0.0035

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Determine Assembly Gap for 99.7% of cases

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Thank You

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tolerance stack up statistics

Documents

types of tolerance analysis

formal tolerance stack

lower cost

cost savingcost100

basic stack

radial stack r

dimension tolerance

i t tstwith