fatigue life is a statistical quantity
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Fatigue Life is a Statistical
Quantity
Introduction to the Weibull
distribution
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Objective
You need a design where 90% of the springs last (at least) to 400 000 cycles.
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First Primitive Test:
Average Lifetime
Design A Design B
726000
615000
508000
808000
755000
849000
384000
667000
515000
483000
631000
529000
730000
651000
446000343000
960000
730000
730000
973000
258000
635000Average
Design B looks better..
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Next test, a bit more sophisticated:
We plot fraction of springs failed vs number of cycles. We use this to get estimates
of reliability of Design A and Design B as a function of cyles
Example: If first failure in design A is at 200 000 cycles, reliability above 200 000
cycles is reduced from 100% to 90%
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0 100000 200000 300000 400000 500000 600000 700000 800000 900000
Survival Probability vs Cycles Design A
Survival Probability vs Cycles Design A
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We now fit a smooth curve (2ndorder power) to the data and to find F (400000)
y = -7.3352E-11x2- 1.0216E-04x + 1.4026E+020
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0 100000 200000 300000 400000 500000 600000 700000 800000 900000
Chart Title
Survival Probability vs Cycles Design A
Survival Probability vs Cycles Design A
Poly. (Survival Probability vs Cycles Design A)
Y = -6.0918E-11 N2 -1.1619E-4N +144.3
This fit predicts that
the number of
permissible cycles, for
90% reliability, is
386637, for design A.
For 400000, our goal,
the probability is88.07% for Design A
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We repeat the same for Design B
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0 200000 400000 600000 800000 1000000 1200000
Survival Plot Design B
Survival Plot Design B
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y = -5.5949E-12x2- 1.1626E-04x + 1.2137E+02
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0 200000 400000 600000 800000 1000000 1200000
Survival Plot Design B
Survival Plot Design B
Poly. (Survival Plot Design B)
This plot predicts
that the reliability of
Design B at 400000
cycles is74.866 %
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The Weibull distribution was found by Weibull strictly by trial and error.
He tried to model the distribution of failure strength of steels and derive
probabilities for a high reliability (such as 99.9 %) from a limited set of testdata.
After settled on this distribution:
X is the variable (here the number of cycles to failure) and and are the
Weibull parameters (is the shape parameter also known to material
scientists as the Weibull modulus and is the scale or length parameter.
Note that has a physical meaning. If a tensile specimen is twice as long, the
probability for a flaw terminating its fatigue life is twice as high.
Size matters.
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There is nothing god given about the Weibull distribution. There are other
reliability distributions - all invented before Weibull.
But it does fit, experimentally, a very wide variety of phenomena.
Weibull wrote a famous paper demonstrating it fit the size distribution of beans,
the height distribution of the population on an island (I forgot which one) and so
on, i.e. Biological phenomena as well as steels, with a total of 10 examples
The paper is a classic - I will put it on the website.
The reason is that depending on how the modulus is picked it fits both infant
mortality and wear out phenomena.
And fatigue failure is a kind of wear out phenomena.
MOST IMPORTANTLY IT HAS BECOME A DE FACTO STANDARD FOR ENGINEERS.
TO PREVAIL IN COURT, YOU BETTER SHOW YOU USED ITPROPERLY !
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From Wiki
The Weibull distribution is used
In survival analysis[6]
In reliability engineering and failure analysis
In industrial engineering to represent manufacturing and delivery times
In extreme value theory
In weather forecasting (To describe wind speed distributions, as the natural distribution
often matches the Weibull shape[7] Fitted cumulative Weibull distribution to maximum
one-day rainfalls)
In communications systems engineering (In radar systems to model the dispersion of
the received signals level produced by some types of clutters. To model fading channels in
wireless communications, as the Weibull fading model seems to exhibit good fit to
experimental fading channel measurements)
In General insurance to model the size of Reinsurance claims, and the cumulativedevelopment of Asbestosis losses
In forecasting technological change (also known as the Sharif-Islam model)[citation
needed]
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In hydrology the Weibull
distribution is applied to
extreme events such as
annual maximum one-
day rainfalls and riverdischarges.
In describing the size of
particles generated by
grinding, milling and
crushing operations, the
2-Parameter Weibull
distribution is used, and
in these applications it is
sometimes known as the
Rosin-Rammler
distribution. (In this
context it predicts fewer
fine particles than the
Log-normal distribution
and it is generally most
accurate for narrow
particle sizedistributions).[
Stolen From Wiki
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Massaging the data point
The data are plotted in increasing sequence (ranking low to high)
The equivalent of our failure percentage becomes approximately
This will do for most engineering problems. One can do this
better by looking up the F distribution.
There is a nifty website; teach yourself statistics which might
come in handy when you are in industry
(Modern industry runs on statistics)
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Once you have done the ranking, it is just plotting :
And, from the double ln plot to extract the values for K and
So here is the plot :
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Spring A Spring B
beta = 4.39 2.604167
alpha= 686938 714973
Reliabiliy 0.911098 0.802222
From which you get the following results:
This does not look all that earth shattering different from our
primitive estimate which yielded
0.8807 0. 74866
Which shows you that it is a good idea to make a primitive estimate
first before setting out to do a state of the art Weibull with an F
distribution)
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Now, we should put confidence limits on our answer
But as this is not a statistics course, I will leave it at this.
But before you use a statistics package to analyze your data it is a good idea to
make the primitive plot we started out and decide what other curves might
reasonably be drawn to the data.
This will give you a rough idea as to the confidence limits you can put on the
data.
Moving on to the three parameter Weibull Distribution
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The three parameter Weibull distribution
The three parameter Weibull distribution uses an additional parameter
to shift the distribution sideways along the horizontal axis (number ofcycles to failure, time to failure .)
Physical meaning in brittle fracture is generally density of flaws per
unit length.
Fiber Optics
For example, if a Corning glass fiber has one defect per 10, and you
test 10 specimens, each one inch long, then 9 will have high strength
and only one low strength. The average fracture strength will be
high.
On the other hand, if you test 10 meter long (~ 40 inches) long
section, the chances are that you 98% of the time will have a
specimen with a flaw. The average fracture strength will be low.
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Electronics:
The classic example is the break down strength of DRAM capacitors
(DRAM cells). The break down depends on both the intrinsic breakdown
strength of the oxide and the defects (known as weak spots in theoxide*)
If you have two oxides A and B, where flawless A has a break down voltage
of 8 MV/cm and flawless B 7 MV/cm, then A is the better oxide.
On the other hand, if A has 1 defect per 20 square m ( square micron) andB has one defect per 1 cm2 and the presence of a defect lowers the break
down voltage by 20% AND you measure the break down strength on test
capacitors made 10x10 micron, (because the research lab does not have
state of the art immersion steppers), than your conclusion will be opposite !
Cause you 10x10 test capacitors made in A will have, on average, 5defects, and therefore now return a breakdown voltage of about 6.4 MV/cm.
Whereas the test capacitors made with the B oxide, will on average, have
near zero defects, you will measure an average breakdown of 7 MV/cm
THE RESEARCH LAB WILL RECOMMEND TO USE OXIDE A
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Electronics continued:
But the production capacitors are 0.5 x 0.5 micron.. An area 40 times
smaller than your research lab test specimens !!!
On average, they will not contain a defect, if made either in A or B.
Hence the production people will report that capacitors made with the A
oxide have a higher breakdown voltage, around 8 MV/cm and those
with B will have a breakdown field, on average, of 7 MV/cvm
THE PRODUCTION PEOPLE WILL RECOMMEND OXIDE A
Conflict resolution:
In general, you have no idea about the length scale of the defect
population. Perhaps the difference between the research lab and the
production line is that the production line uses Plasma Processing that
uses higher electric fields than the research lab to etch the
metallization. A long metal line, acting as an antenna, can partly blow
out a gate oxide i.e. damage it, by stressing it too much.
So, a priory it is not clear what is going on
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Approach A for the research lab
You plot the Weibull distribution of the breakdown voltage of your
10x10 with a two parameter Weibull. If the distribution is curved,chances are you have a length scale parameter .
Approach B for the research lab
You make test capacitors of different size. From 1x1 cm down to
2x2 which is the best your equipment can do with enoughgeometric precision to know the area within +/- 10%.
You then investigate the dependence of the 2 parameter Weibull
on the capacitor size.
If it does depend, you have a length scale problem and you canextract the length scale on which the defect occurs.
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The three parameter Weibull, accumulated failure function
))exp((1)(
t
tF
Failure probability density function
There are various methods of determining . The most simple one is to
plot the data first as a two parameter Weibull (i.e ) . If that one is
curved, then one adds/subtracts values of until the line is reasonably
straight.
Of course, one can also do this with linear regression. To see how,
visit
http://www.weibull.com/LifeDataWeb/estimation_of_the_weibull_param
eter.htm
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Weibull.com is a wonderfully informative website which will teach you
everything you ever wanted to know about the Weibull distribution. Here
is (stolen from the Website) an example on how to adjust gamma
The curved original
data a very familiar
to glass fiber guys.
And to capacitor
testing guys mathis math.
And to solar cell
guys, that test
conversion
efficiency vs cellsize
And so on. Math
is math
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As a good experimenter (good experimenters are those who a paranoid) you
now investigate design A and B for hidden curvature.
Is there a hidden length effect ???
y = 0.0092x2+ 0.2435x + 13.441
y = -0.0193x2+ 0.3536x + 13.496
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13.2
13.4
13.6
13.8
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-4 -3 -2 -1 0 1 2
To do so, you make a usual two
parameter Weibull plot, but instead
of fitting a straight line, you fit apower series.
See left. The quadratic term gives
you the curvature.
The curvature is very small, and ofopposite sign for A and B. So it
seems a wash, given the scatter of
the data.
And so we can conclude (without
Doing a Sigma Analysis) that there
is no length scale or threshold effect
in the spring test data.
QED