1 SMU EMIS 7364 NTU TO-570-N Tolerance Limits Statistical Analysis & Specification Updated: 2/14/02...

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SMUEMIS 7364

NTUTO-570-N

Tolerance Limits Statistical Analysis & Specification

Updated: 2/14/02

Statistical Quality ControlDr. Jerrell T. Stracener, SAE Fellow

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Product Specification

LowerSpecification

Limit

NominalSpecification

UpperSpecification

Limit

Target(Ideal level for use in product)

Tolerance

x

(Productcharacteristic)

(Maximum range of variation of the product characteristic that will still work in the product.)

3

Traditional US Approach to Quality

(Make it to specifications)

good

T USLLSL

Loss ($)

No-Good No-Good

x

5

Setting Specification Limits on Discrete Components

6

Variability Reduction

Variability reduction is a modern concept of design and manufacturing excellence

• Reducing variability around the target value leads to better performing, more uniform, defect-free product

• Virtually eliminates rework and waste• Consistent with continuous improvement concept

acceptreject reject

target

Don’t just conform to specifications Reduce variabilityaround the target

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True Impact of Product Variability

• Sources of loss- scrap- rework- warranty obligations- decline of reputation- forfeiture of market share

• Loss function - dollar loss due to deviation of product from ideal characteristic

• Loss characteristic is continuous - not a step function.

8

Representative Loss Function Characteristics

x

Loss$

X nominal is best

L = k (x - T)2

x

Loss$

X smaller is better

L = k (x2)

x

Loss$

X larger is better

L = k (1/x2)

T

9

Variability-Loss Relationship

LSL USL

Target

$ savingsdue to

reducedvariability

Maximum$ loss

per item

Loss

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Loss Computation for Total Product Population

X nominal is best

L = k (x - T)2

x

Loss$

T

2

2

2σ

T)(x

e2πσ

1f(x)

x

Loss$

T

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Statistical Tolerancing - Convention

Normal ProbabilityDistribution

LTL Nominal UTL

0.001350.00135 0.9973

+3-3

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Statistical Tolerancing - Concept

LTL UTLNominal x

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Caution

For a normal distribution, the natural tolerance limits include 99.73% of the variable, or put another way, only 0.27% of the process output will fall outside the natural tolerance limits. Two points should be remembered:

1. 0.27% outside the natural tolerances sounds small, but this corresponds to 2700 nonconforming parts per million.

2. If the distribution of process output is non normal, then the percentage of output falling outside 3 may differ considerably from 0.27%.

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Normal Distribution

Probability Density Function:

< x <

where = 3.14159...

e = 2.7183...

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x2

1

e2

1)x(f

15

Normal Distribution

• Mean or expected value of X

Mean = E(X) =

• Median value of X

X0.5 =

• Standard deviation

)(XVar

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Normal Distribution

Standard Normal Distribution

If X ~ N(, ) and if , then Z ~ N(0, 1).

A normal distribution with = 0 and = 1, is calledthe standard normal distribution.

X

Z

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Normal Distribution - example

The diameter of a metal shaft used in a disk-drive unitis normally distributed with mean 0.2508 inches andstandard deviation 0.0005 inches. The specificationson the shaft have been established as 0.2500 0.0015 inches. We wish to determine what fraction ofthe shafts produced conform to specifications.

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Normal Distribution - example solution spec meetingP

91924.0

0000.091924.0

60.41.40

0005.0

0.2508-0.2485

0.0005

0.2508-0.2515

x0.2485P0.2515xP

0.2515x0.2485P

0.2508 0.2515 USL

0.2485 LSL

0.2500

f(x)

xnominal

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Normal Distribution - example solution

Thus, we would expect the process yield to be approximately 91.92%; that is, about 91.92% of the shafts produced conform to specifications. Note that almost all of the nonconforming shafts are too large, because the process mean is located very near to the upper specification limit. Suppose we can recenter the manufacturing process, perhaps by adjusting the machine, so that the process mean is exactly equal to the nominal value of 0.2500. Then we have

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Normal Distribution - example solution

0.2515x0.2485P

9973.0

00135.099865.0

00.33.00

0005.0

0.2500-0.2485

0.0005

0.2500-0.2515

x0.2485P0.2515xP

0.2500 0.2515 USL

0.2485 LSL

f(x)

xnominal

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What is the magnitude of the difference between sigma levels?

Sigma Area Spelling Time DistanceOne Floor of 170 typos/page 31 years/century earth to moon

Astrodome in a book

Two Large supermarket 25 typos/page 4 years/century 1.5 times aroundin a book the earth

Three small hardware 1.5 typos/page 3 months/century CA to NYstore in a book

Four Typical living 1 typo/30 pages 2 days/century Dallas to Fort Worthroom ~(1 chapter)

Five Size of the bottom 1 typo in a set of 30 minutes/century SMU to 75 Central of your telephone encyclopedias

Six Size of a typical 1 typo in a 6 seconds/century four stepsdiamond small library

Seven Point of a sewing 1 typo in several 1 eye-blink/century 1/8 inchneedle large libraries

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Linear Combination of Tolerances

Xi = part characteristic for ith part, i = 1, 2, ... , n

Xi ~ N(i, i)

X1, X2, ..., Xn are independent

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Linear Combination of Tolerances

Y = assembly characteristic

If , where the a1, ..., an are constants,

thenY ~ N(Y, Y),

where

and

n

1iiiXaY

n

1iiiY a

n

1i

2i

2iY a

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Concept

x1

x2

.

.

.

xn

n

1iixy

y

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Statistical Tolerancing - Concept

0.2500 0.2515 USL

0.2485 LSL

f(x)

xnominal

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Tolerance Analysis - example

The mean external diameter of a shaft is S = 1.048 inches and the standard deviation is S = 0.0020 inches. The mean inside diameter of the mating bearing is b = 1.059 inches and the standard deviation is b = 0.0030 inches. Assume that both diameters are normally and independently distributed.

(a) What is the required clearance, C, such that the probability of an assembly having a clearance less than C is 1/1000?

(b) What is the probability of interference?

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Tolerance Analysis - example solution

Bearing

ShaftXb

XS

diameter f(xb) f(xs)1.0325 0.00 0.001.0350 0.00 0.001.0375 0.00 0.001.0400 0.00 0.071.0425 0.00 4.551.0450 0.00 64.761.0475 0.09 193.331.0500 1.48 120.991.0525 12.72 15.871.0550 54.67 0.441.0575 117.36 0.001.0600 125.79 0.001.0625 67.33 0.001.0650 18.00 0.001.0675 2.40 0.001.0700 0.16 0.001.0725 0.01 0.001.0750 0.00 0.001.0775 0.00 0.001.0800 0.00 0.00

0.00

50.00

100.00

150.00

200.00

250.00

1.0300 1.0400 1.0500 1.0600 1.0700 1.0800

f(xb)

f(xs)

fb(x)

fs(x)

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0.00

25.00

50.00

75.00

100.00

125.00

150.00

1.045 1.047 1.049 1.051 1.053 1.055 1.057

Tolerance Analysis - example solution

Intersection Regionfb(x)

fs(x)

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The Normal Model - example solution

D = xb-xs = Clearance of bearing inside diameter minus shaft outside diameter

D = b - S = 0.011D = (b

2 + S2)1/2 = 0.0036

so D~N(0.011,0.0036)Clearance Probability Density Function

0.00

50.00

100.00

0.000 0.005 0.010 0.015 0.020 0.025 0.030

d=xb-xs

fD(x)

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(a) Find c such that P(D < c) =

so that

From the normal table (found in the resource section of the

website), the Z = -3.09

so that

and

1000

1

09.3

0036.0

011.0c

c

D

D

c 000124.0

001.009.3P Z

09.3

001.00036.0

011.0P

c

Z

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Since c < 0, there is no value of c for which the probability is equal to 0.001

(b) Find the probability of interference, i.e.,

From the normal table (found in the resource section of the

website), the Z of -3.1 = 0.0011

0.0011

06.3-ZP

0036.0

011.00DP

0DP

ceinterferenP

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Using Monte Carlo Simulation (n=1000):

(a) What is the required clearance, C, such that the probability of an assembly having a clearance less than C is 1/1000?

(b) What is the probability of interference?

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Using Monte Carlo Simulation

First generate random samples from (I used n=1000)

Xbi~N(b, b) = N(1.059, 0.0030)

and

Xsi~N(s, s) = N(1.048, 0.0020)

N(b,b) N(s,s)

1.058537 1.045935

1.056233 1.047846

1.059985 1.052481.065796 1.0488491.055505 1.0479221.059354 1.0471641.062841 1.0492671.055726 1.0478151.058989 1.0485841.061587 1.0473351.058047 1.0473251.060637 1.0475361.05933 1.0464011.056185 1.0476941.058666 1.045457

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Then calculate the differences

Estimate

Estimate s by taking the mean. (You can use the AVERAGE() function.)

Estimate s by calculating the standard deviation.(You can use the STDEV() function.)

n1,...,for ixxxiii sbd

dx̂

1000

11000ˆ

ds

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= 0.01093and = 0.00371

(a)

This is close to c = -0.000124.

001.0)Cclearance(P̂

09.3

00371.0

01093.0

ˆ

ˆ

c

c

D

D

c 000534.0

D̂D̂

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(b)

This can be compared to P(I) = 0.000968.

nce)(interfereP̂

002.01000

2

0dfor which no.

n

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Statistical Tolerance Analysis Process

Assembly consists of K components

• Specifications Assembly:

• Specifications Component:

• Assembly Nominal

where ai = 1 or -1 as appropriate

AA txN

ii txiin K, 1,...,for ,

K

iiiA nnxax

1

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• Assembly tolerance

• If dimension

with parameters and , then

where

and

K

iiA tt

1

2

dDistributenormally is iX

i i),N(~ A AiX

nn A

K

iiiA xxa

1

.3

,1

2 ii

K

iiA

t

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• is specified

• is determined during design

• is calculated

Case 1: if probability is too small, then

(1) component tolerance(s) must be reduced

or (2) tA must be increased

AA txN

ii txn

AAAAA txXtxNNP

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Case 2: if probability is too large, then some or all components tolerances must be increased.

Note: Do not perform a worst-case tolerance analysis

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Estimating the Natural ToleranceLimits of a Process

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Tolerance Limits Based on the Normal Distribution

Suppose a random variable x is distributed with mean and variance , both unknown. From a random sample of n observations, the sample mean and sample variance S2 may be computed. A logical procedure for estimating the natural tolerance limits ± Z/2 is to replace by and by S, yielding.

x

SZx α/2

x

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Tolerance Intervals - Two-Sided

Since and S are only estimates and not the true parameters values, we cannot say that the above interval always contains 100(1 - )% of the distribution. However, one may determine a constant K, such that in a large number of samples a faction

SZx α/2

x

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Tolerance LimitsBased on the Normal Distribution

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Tolerance Intervals - Two-Sided

If X1, X2, …, Xn is a random sample of size n from a normal distribution with unknown mean and unknown standard deviation , then a two-sided tolerance interval is (LTL,UTL), i.e., an interval that contains at least the proportion P of the population, with 100% confidence is:

and

is a function of n, P, and and may be obtained from the table Factors for Two-Sided Tolerance Limits for Normal Distributions (Located in the resource section on the website).

SKLTL 2X

SKUTL 2X

2K

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Tolerance Intervals - One-Sided

If X1, X2, …, Xn is a random sample of size n from a normal distribution with unknown mean and unknown standard deviation , then a one-sided lower (upper) tolerance interval is defined by the lower tolerance limit LTL (upper tolerance limit UTL), the value for which at least the proportion P of the population lies above (below) LTL (UTL) with 100% confidence where

is a function of n, P, and and may be obtained from the table Factors for Once-Sided Tolerance Limits for Normal Distributions (Located in the resource section on the website).

SKLTL 1X

.SKUTL 1X

1K

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Tolerance Intervals - Two-Sided Example

Ten washers are selected at random from a population that can be described by a normal distribution. The measured thicknesses, in inches, are:

Establish an interval that contains at least 90% of the population of washer thicknesses with 95% confidence.

.123 .132

.124 .123

.126 .126

.129 .129

.120 .128

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Tolerance Intervals - Two-Sided Example Solution

From the sample data

and

The K value can be found on Tolerance Limits Table- Two-Sided with gamma 95 and 99 and n=2 to 27 (Located in the resource section on the website).

1260.X

00359.0S

2K 829.2

95.0,90.0,10K

http://www.engr.smu.edu/~jerrells/courses/help/tolerancelimit/2sg95g99.jpg

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Tolerance Intervals - Two-Sided Example Solution

so that

Therefore, with 95% confidence at least 90% of the population of washer thicknesses, in inches, will be contained in the interval (0.116,0.136).

LTL

116.0

0.00359829.2.12600

K2

SX

UTL

136.0

0.00359839.2.12600

K2

SX

1 SMU EMIS 7364 NTU TO-570-N Tolerance Limits Statistical Analysis & Specification Updated: 2/14/02...

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