tolerance stack up statistics
DESCRIPTION
tolerance stack upTRANSCRIPT
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Tolerance Stackup AnalysisTolerance Stackup AnalysisStatistics
ByDr N RamaniDr. N. Ramani
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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
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Why do we require to do Tolerance Analysis?y y
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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
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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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What are the Types of Tolerance Analysis?
1.Radial Stack –
2 Linear Stack –2.Linear Stack –
3.Assembly Stack –
What do they do?
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What are the Types of Tolerance Analysis?
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
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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
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??
Assembly Stack
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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
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100 $ ? $100 $ ? $100 $
COST
100% Interchangeability
99.7% Interchangeability
14Estimate the cost for 99.7% interchangeability
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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
100% Interchangeability
99.7% Interchangeability
16At what reduced cost, you will accept 99.7% Interchangeability?
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100 $
COST
< 20 $ !< 20 $ !
100% Interchangeability
99.7% Interchangeability
17Based on Statistical Principles!
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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
Assumption“Extreme assembly conditions
can and will be met in practice”
100%Interchangeability
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Effect of component tolerances Functioning of component tolerances
upon the assembly
g
finished product
Assumption“Extreme assembly conditions
can and will be met in practice”
100%Interchangeability
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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
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•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:
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Preferred Worst case
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component tolerances
Worst case assembly variation for 100% & 99 7%
(Based on Cp & Cpk)
for 100% & 99.7% Interchangeability
p p )
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?*?*
* For 100% Interchangeability
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X
Find the Dimension & Tolerance of ‘X’
33for +/-3 Sigma conformance100% Interchangeability
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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
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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
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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 (?)
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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?
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Compare the two drawingsg
AA
B48
B
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BX
Calculate the Dimension & Tolerance for the Circlip Groove @ 100% & 99 7%
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Circlip Groove @ 100% & 99.7% Interchangeability
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50Cf between 100% & 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 )+/-0.0500
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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?
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Assigning of Tolerances to related components of an p
assembly
STSuper Structure
STAssembly Tolerance
= Foundation
58RMS value of Individual Tolerances
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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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When should we apply ST?
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
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When should we apply ST?
•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…….
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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
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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?
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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
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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
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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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