1 SMU EMIS 7364 NTU TO-570-N Tolerance Limits Statistical Analysis & Specification Updated: 2/14/02...
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Transcript of 1 SMU EMIS 7364 NTU TO-570-N Tolerance Limits Statistical Analysis & Specification Updated: 2/14/02...
1
SMUEMIS 7364
NTUTO-570-N
Tolerance Limits Statistical Analysis & Specification
Updated: 2/14/02
Statistical Quality ControlDr. Jerrell T. Stracener, SAE Fellow
2
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 ($)
No-Good No-Good
x
4
5
Setting Specification Limits on Discrete Components
6
Variability Reduction
Variability reduction is a modern concept of design and manufacturing excellence
• Reducing variability around the target value leads to better performing, more uniform, defect-free product
• Virtually eliminates rework and waste• Consistent with continuous improvement concept
acceptreject reject
target
Don’t just conform to specifications Reduce variabilityaround the target
7
True Impact of Product Variability
• Sources of loss- scrap- rework- warranty obligations- decline of reputation- forfeiture of market share
• Loss function - dollar loss due to deviation of product from ideal characteristic
• Loss characteristic is continuous - not a step function.
8
Representative Loss Function Characteristics
x
Loss$
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
10
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
11
Statistical Tolerancing - Convention
Normal ProbabilityDistribution
LTL Nominal UTL
0.001350.00135 0.9973
+3-3
12
Statistical Tolerancing - Concept
LTL UTLNominal x
13
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%.
14
Normal Distribution
Probability Density Function:
< x <
where = 3.14159...
e = 2.7183...
22
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
16
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
17
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
19
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
20
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
21
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
22
Linear Combination of Tolerances
Xi = part characteristic for ith part, i = 1, 2, ... , n
Xi ~ N(i, i)
X1, X2, ..., Xn are independent
23
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
24
Concept
x1
x2
.
.
.
xn
n
1iixy
y
25
Statistical Tolerancing - Concept
0.2500 0.2515 USL
0.2485 LSL
f(x)
xnominal
26
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?
27
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)
28
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)
29
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)
30
The Normal Model - example solution
(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
31
The Normal Model - example solution
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
32
Tolerance Analysis - example
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?
33
Tolerance Analysis - example
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
34
Tolerance Analysis - example
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
35
Tolerance Analysis - example
= 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
36
Tolerance Analysis - example
(b)
This can be compared to P(I) = 0.000968.
nce)(interfereP̂
002.01000
2
0dfor which no.
n
37
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
38
Statistical Tolerance Analysis Process
• 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
39
Statistical Tolerance Analysis Process
• 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
40
Statistical Tolerance Analysis Process
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
41
Estimating the Natural ToleranceLimits of a Process
42
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
43
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
44
Tolerance LimitsBased on the Normal Distribution
45
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
46
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
47
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
48
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
49
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