Tolerance Stack Up Statistics

Tolerance Stackup Analysis Statistics By Dr N Dr. 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 design Conditions diti 4. Determine if statistical Interchangeability will give a lower cost product 5. Determine if larger tolerance zones can meet Design requirements 6. Understand the Principle of ‘Robust Design’

2

Case of 100% infallible interchangeability no matter the cost : Safety is of paramount importance

3

Whyy do we require to do Tolerance Analysis? 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? 2. To determine if the parts will function properly att worstt condition diti 3. To determine if the drawing tolerances could be larger 4. To complete p the design g p process 5. To provide a record of the dimensional design requirements that can be reviewed at a later date in case of a product problem 5

Assumptions while doing Tolerance Analysis: 1.All dimensions apply at 20oC 2.All 2 All manufactured f t d parts t meett dimensional requirements of drawing 3.All parts are rigid in free state & in asse b y assembly 4.For , parts are manufactured ST with ith th the mean di dimension i as T Targett

6

What are the Types of Tolerance Analysis?

7

What are the Types of Tolerance Analysis? 1. Radial Stack – 2 Linear Stack – 2. 3. Assembly Stack – What do they do?

8

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 parts Give a pictorial example for each type

9

•Example of a Radial Stack , •R,Θ 10

X

Find the Dimension & Tolerance of ‘X’ for +/-3 +/100% 3 Sigma IInterchangeability t conformance h bilit Example of a Linear Stack

11

?

Assemblyy Stack 12

100 $ 100 $

?$

COST

100% Interchangeability

99.7% Interchangeability

Estimate E ti t the th costt for f 99.7% 99 7% interchangeability i t h bilit Please write down your answer

13

100 $ 100 $

?$

COST

100% Interchangeability

99.7% Interchangeability

Estimate the cost for 99.7% interchangeability

14

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

15

100 $ 100 $

Cost Saving

?$ COST

100% Interchangeability

99.7% Interchangeability

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

16

100 $

COST

< 20 $

100% Interchangeability

!

99.7% Interchangeability

Based on Statistical Principles!

17

100% Interchangeability 18

Assumption “Extreme assembly conditions can and will be met in practice”

100% Interchangeability 19

Effect of component tolerances upon the assembly Assumption “Extreme assembly conditions can and will be met in practice”

100% Interchangeability 20

Effect of component tolerances upon the assembly

Functioning g of finished product

Assumption “Extreme assembly conditions can and will be met in practice”

100% Interchangeability 21

Highest

Si off ‘B’ P Size Partt Lowest ‘B’

Lowest ‘A’

Highest

Size of ‘A’ Part 22

C min

C max

B min

B max

A min i

A max

Assemblyy of 3 Parts

23

How does Insurance Business function?

24

Effect on Assembly

1 2 3 4 5 6 7 8 9 10 Number of Parts / Feature Dimensions Affecting Assembly 25

26

27

•Assembly tolerance = Sum of all tolerances of the individual parts (OK for 2 or 3 parts…)

•One One 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.

28

Objectives of this presentation: 1

Preferred component tolerances (Based on p & Cpk) p ) Cp

Worst case assembly variation for 100% & 99 99.7% 7% Interchangeability

2 29

?*

* For 100% Interchangeability 30

31

32

X

Find the Dimension & Tolerance of ‘X’ for +/-3 Sigma conformance 100% Interchangeability

33

Stage 1. Covert all Dimensions to be Median values & Tolerances into Bilateral ones 34

Start

End 2 4

3

1

35

Plus Minus Direction Direction +/Distance Distance Tolerance A# 'B' 'C' 'D' 1

Stage3: Tabulate Analysis

2

0.2175 2.62

0.0025 0.01

3

0.42

0.01

4

1.8985

0.0025

2.536

0.025

2.62

36

0.084

0.025

Introduction to Normal Distribution

37

38

100% Successful Interchangeability

X σ?

39

Please remember the value of tolerance!

40

6 100% Success

6σ (?)

41

6σ

2.63 / 2.61

42

43

44

45

Cf 100% Vs 99.7% Interchangeability

46

Comparison bet between een 6 6σ &3σ &3 Limits #

Sigma g Value

Tolerance

1

6 (100% I t h Interchangeability) bilit ) 3 ((99.7% Interchangeability)

+/-0.0250

2

+/-0.0146

Which is better from Assembly point of view? Lesson? 47

Compare the g two drawings A

B 48

X

Calculate the Dimension & Tolerance for the Circlip Groove @ 100% & 99.7% 99 7% Interchangeability

B

49

Cf between 100% & 99.7% Interchangeability

50

Comparison bet between een 6 6σ &3σ &3 Limits #

Sigma g Value

Tolerance

1

6 (100% I t h Interchangeability) bilit ) 3 ((99.7% Interchangeability)

+/-0.0500

2

+/-0.0250

Which is better from Assembly point of view? Cf Cost also 51

#

Sigma Value

Tolerance

1

6 (100% Interchangeability) 3 (99.7% Interchangeability)

+/-0.0500

2

+/-0.0250

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

52

(Assembly Tolerance)2 { Number of Parts * (Part Tolerance)2 } 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 of Parts σ2 Assembly A bl

=Σ (n ( 1 * σ12) off Parts P vide id CLT

∴ Assembly Tol

= Part Tolerance *√n 53

#

Sigma Value

Tolerance

1

6 (100% Interchangeability) 3 (99.7% Interchangeability)

+/-0.0500

2

+/-0.0250

Assembly tolerance of +/-0.025 = 2 * 0.0125 =

(√4

*0 0.0125) 0125)

54

55

TSA Vs STA

56

ST? Super Structure

ST ?

Foundation 57

Assigning of Tolerances to related components p of an assembly Super Structure

ST Assembly Tolerance = RMS value of Individual Tolerances

Foundation 58

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 drawings 3. 4. Operations / QC will maintain Statistical control t l off process

59

What are the benefits of Statistical Tolerancing?

60

Benefits of ST: 1. Reduced costs 2. Closer average Fits 3. Improved Quality / Performance of Product • Sony USA Vs Sony Japan Story

61

When should we apply ST?

62

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 possible 3. 4. Means of reducing manufacturing cost 5. To reduce Tolerance Accumulation • Selective assembly required • Adjustments / complex designs required • Tighter tolerances on components required 63

When should we apply ST? •What if Analysis? •You want to know what will be the condition for 99.7% of parts if manufactured around the mean of dimensions

TSA Vs STA

64

When will ST succeed?

65

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

66

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 p gauging..) g g g ) to determine Cp & Cpk *Reamer Design Example

67

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

68

0.084+/-0.05 @ 100% Interchangeability

0.084+/-0.025 @ 99.7% Interchangeability

You cant have something…….

CLT?

69

Conclusion?

70

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

71

1 Four tolerance zones can be increased & still 1. meet original design requirements of 0.084+/-0.025 using g 99.7% p probability y 2. How did we get to suggest +/-0.0125 tolerance?

72

1 Four tolerance zones can be increased & still 1. meet original design requirements of 0.084+/-0.025 using g 99.7% p probability y 2. How did we get to suggest +/-0.0125 tolerance? 3. From CLT formula: 4. Tol of p parts / features = Tol of Assy y / ((Sq.rt q of number of parts or features)

73

74

75

100%

99 7% 99.7%

76

Is it possible to relax manufacturing tolerances on all 5 di dimensions* i * & still till achieve hi assembly bl ttargett off 0.034 +/-0.022, thereby reducing cost * +/- 2,3,4,5 & 8 thous

77

1. We have relaxed original tolerances of +/- 2,3,4,5 & 8 thous to +/ +/- 0.010 0 010 2. Check correctness of Calculations 3. How did we get proposed tolerance of 0.010?

78

Dim.Tolerance = (assembly Tolerance / Sq.rt of number b off di dimensions i iinvolved) l d) = 0.022 / √ 5 = 0.010

Very Important

79

Part 3

P t1 Part Part 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 dimensions i off parts t 2 & 3 tto achieve hi th the objective bj ti ffor 100% interchangeability 80

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

81

Part 3

P t1 Part Part 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 dimensions i off parts t 2 & 3 tto achieve hi th the objective bj ti ffor 99 99.7% 7% interchangeability 82

Part Tolerance = Assembly tolerance / √ n = 0.0175 / √3 = 0.010 0 010 = +/- 0.005 Rechecking:

83

84

85

Determine Assembly Gap for 100% of cases

86

87

T Transition iti b between t Max Clearance

0.0275

Max Interference 0.0035 88

Determine Assembly Gap for 99.7% of cases

89

90

91

Thank You

92