Materials strength is dictated not so much by bond strength but by

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Materials strength is dictated not so much by
bond strength but by
Defects

• Griffiths experiments and model are the
  historical basis of the fracture mechanics
  approach
• Probabilistic approach to strength why do we
  need it at all?

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Probabilistic argument

• Freudenthal proposed a link between the
  probability of occurrence of a critical defect in
  a solid of (dimensionless) volume V, the
  concentration of defects, and the size (length,
  area, volume) of a solid
• F(V) 1 exp-(V/V0)
• where V0 is the mean volume occupied by a defect
  (thus 1/V0 is the mean cc of defects).
• Plot

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Probability of occurrence of a critical defect
( Probability of failure) against size for a
given defect concentration
Plateau reached where size has no effect anymore.
At very small volume, F(V) is strongly
affected very small P of occurrence of a
critical defect Thus strength tends to be very
high (provided defects and strength are
correlated at all)
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However no real physics in the previous
equation. How do we draw stress into the
picture?

• Weibull The original density of defects in the
  material (1/V0) increases as the applied stress
  increases according to some physical (power) law
• (1/V0) (s/a)b
• and therefore
• F(V) 1 exp-V (s/a)b (the Weibull
  Distribution)
• a scale parameter
• b shape parameter
• V (non-dimensional) volume

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The strength of fibers is indeed statistical
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FACTS THAT MUST BE EXPLAINED (OR AT LEAST
MODELED) IN AN ACCURATE WAY IF FIBROUS MATERIALS
ARE TO BE USED IN PRACTICE

• The strength variability of the single fibers,
  bundles of fibers, and fiber-based composite
  materials
• The effect of size (length, but also diameter) on
  the strength of the fibers, bundles, composites
• The scaling up effects from fibers to bundles of
  fibers (for example, 5,000 fibers) to fiber-based
  composites

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A few facts about distribution functions

• 1. Probability density function f(x) (or pdf)

2. Cumulative distribution function F(x) (or cdf)
And
3. Properties of the cdf
0 F(x) 1 F(x) is always increasing F(?) 1
and F(- ?) 0
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First moment of the distribution (the Mean or
Expected Value)
Second moment of the distribution (the Variance)
The Standard Deviation is SD(x) VAR(x)1/2
The Coefficient of Variation is CV(x)
SD(x)/E(x).
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The Weibull distribution
The Weibull distribution (Weibull 1939, 1951) is
probably best known in composite strength
theories. If x is the stress, Weibull proposed a
cumulative distribution function with the
general form
where m is the shape parameter and x0 is the
scale parameter. The function (x-xu)m/ x0 is
positive, non-decreasing and vanishes at x xu
, which is not necessarily equal to zero. (Why is
xu needed in practice?)
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But note when b gt 3, then
(And again, when b is close to 1, G 1)
Thus Ex a
(this provides a physical meaning for the scale
parameter)
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Therefore
So
(this provides a physical meaning for the shape
parameter)
Very large cv when b lt 1
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NOTE The Gamma function
Weibull pdf
As b increases, the distribution is more narrow,
and is increasingly similar to the normal
(Gaussian) distribution.
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In practice Weibull distribution

• Example if a stress equal to 80 of the mean
  strength (ltxgt) is applied in sequence to 100
  single fibers having a shape parameter b 10 (
  Kevlar) and a scale parameter a 0.95ltxgt, what
  is the predicted proportion of broken fibers ?
• Solution

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In practice Weibull plotting procedure
Objective To find scaling such that Weibull
plots show as a straight line
Y
X
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Example

• Assume n strength tests were performed.
  Observations are x1, x2,, xi (i varies from 1 to
  n)
• Order the data, from smallest to largest, to get
  the horizontal plotting positions x(1) , x(2)
  ,, x(n)
• Corresponding vertical plotting positions are
  given by
• F(i) (i 0.5)/n

Numerical example The following strengths were
measured for 6 specimens of a given material 4,
6, 3, 10, 8, 5 (in arbitrary stress units). Fit
the data to the Weibull model (or plot them on
Weibull coordinates).
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From this plot, the scale (a) and shape (b)
parameters are obtained by linear fitting (Note
that a question of importance is does the
Weibull model indeed fit the phenomenon being
studied? Or Is the fit good?)
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Typical value ranges for the Weibull scale and
shape parameters of fibers (gauge length 10 cm)
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Accurate estimates of the Weibull shape and scale
parameters Maximum Likelihood Estimation (MLE)

• The shape and scale parameters can be
  accurately determined by applying Maximum
  Likelihood Estimation (MLE) to Weibull
  statistics. MLE Cohen 1965 Thoman Bain 1969
  Harter Moore 1965, yields both ? and ß as
  follows

Where N is the number of data points. This
equation can be solved to determine b by means of
standard iterative procedures (such as
Newton-Raphson), using a first estimate (guess)
for b.
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The scale parameter a is then calculated by
plugging the value of b into the following
equation
A final note several plotting conventions are
available for probability plots
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How does the fiber length affect the strength
distribution?The Weakest-Link Model for a fiber

• Assume that a fiber is viewed as a chain of
  links or units of length L0, having each the
  same probability of failure F(s) under a stress
  s.
• Probability of survival of a link is 1 F(s)
• Probability of survival of a chain ( n links) is
  1 F(s)n
• Probability of failure of the chain (for which L
  nL0) is
• Fn(s) 1 - 1 F(s)n

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The Weibull probability of failure of an
elementary unit is
By combining Weakest Link Theory and Weibull
distribution, the probability of failure of a
fiber of length L is
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Including the effect of length in Weibull plots
A change in length is simply reflected by a
vertical shift
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A practical example carbon fibers at 4 different
gauge lengths
10 mm
50 mm
500 mm
5 mm
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Further notes on the statistical strength of
single fibers
A. For a strength population that follows the
Weibull model, Ex depends (approximately) only
on a CV depends (approximately) only on b.
B. The Weibull distribution for a long fiber (of
length L) is
or
in order words, a Weibull distribution with a new
scale parameter
and with a mean
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But the coefficient of variation is unchanged
NOTE
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In other words 1. A plot of the length effect on
average strength can be used to measure the shape
and scale parameters. This is the second way to
measure these parameters.
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The length effect on average strength is
predicted to be quite significant
The lower the value of b, the stronger the length
effect on strength. If b becomes very large, the
length effect disappears E a, and variability
disappears altogether CV 0! Think about the
following what is the effect on the shape of the
probability distribution?
2. If the Weibull model is valid for a given
population (constant length), the length of a
fiber has no effect on its strength variability
(the CV)! Or, in other words, b is independent
of the fiber length
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Useful approximations for the CV
(b 10)
and
(0.05 CV 0.5)
CV
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C. At small loads thus, under small failure
probability levels-, we are dealing with the
so-called lower tail of the probability
distribution. There, we have
(using a Taylor expansion)
This power law form of the fracture probability
is a simple approximation at small loads, which
can be plotted on Weibull coordinates for
comparison.
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LOOSE BUNDLES OF FIBERS

• Can we apply similar ideas to bundles with
  statistical fiber strength?
• Major theoretical/numerical advances by Daniel
  (1945), Coleman (1958), Phoenix (1980-1990)

• Two assumptions
• n single fiber strengths in a bundle
• Applied load is shared equally between the fibers
  (ELS Equal Load Sharing)
• Gedanken experiment we know what is the strength
  of each fiber in the bundle, and we order them
  from weakest to strongest x(1) x(2) x(3)
  x(n)

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If x(1) (the weakest) is gt X (the applied load
per fiber), the bundle does not break it is
stable.
If x(1) (the weakest) is X (the applied load
per fiber), but x(2)gtX(n/n-1), the bundle is
stable with one break only.
In general Stability of the bundle is ensured if
x(i) X(n/n-i) lt x(i1)
Instability of the bundle arises when ALL
inequalities are satisfied x(1)
X x(2) X(n/n-1) x(3) X(n/n-2) x(n-1)
X(n/2) x(n) Xn
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The bundle strength is the smallest stress X that
satisfies these inequalities.
OBJECTIVE To generate an expression for the
bundle strength distribution Gn(x) in terms of
F(x), the single fiber distribution.
One-fiber bundle G1(x) F(x) of
course! Two-fiber bundle We must compute all
paths to failure

• X1 x and X2 x Probability for this event
  F(x)F(x)
• X1 x and x lt X2 2x
  F(x)F(2x)-F(x)
• X2 x and x lt X1 2x
  F(x)F(2x)-F(x)

Therefore G2(x) 2F(x) F(2x) F(x)2
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Etc pour G3(x) (more difficult), etc Exact
calculation beyond n 6 becomes too hard by
hand. A recursion expression was developed
(Phoenix)
where G0(x) 1, G1(x) F(x), and with
(try it for G2)
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For n 10, there are already 512 terms, which is
rather complicated even with a computer. In a
typical carbon fiber bundle there are 3,000 or
6,000 or even 12,000 single fibers!! So, how can
we simplify this?
Simple example a small bundle with an
exponential strength distribution (a 1-parameter
distribution)
(a gt 0)
For the pdf
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The mean of this distribution is
The variance and then the standard deviations are
found to be
Therefore
(a constant!)
Etc for 3-fiber bundle, 4-fiber bundles, using
an exponential distribution. We obtain the
following table
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With F(x) exponential

• From this simple exercise we can conclude the
  following
• The mean strength of a bundle is lower than the
  mean strength of a single fiber.
• As n increases the mean strength of the bundle
  decreases (the bundle weakens as the number of
  fibers increases).
• The CV (thus, the scatter) decreases as n
  increases

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Phoenix demonstrated that under an exponential
distribution, the mean strength and CV of an
n-bundle of fibers tends asymptotically towards
as n ? ?
Thus, the strength tends towards a (non-zero)
fixed value, about 37 of the strength of the
fiber, but the variability shrinks to zero. In
other words, a large bundle (or a cable) has
(almost) a deterministic strength even though its
fibers are statistical.
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But what happens for a Weibull distribution of
strength??
It is much more complicated. To simplify, we look
at the lower tail of the Weibull distribution,
thus we restrict ourselves to relatively low
loads (or low failure probability)
We obtain
(small x)
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And in general, for the lower tail of the Weibull
distribution, the bundle strength distribution is
found to be
where ab(n) is a scale parameter for the lower
tail and bb(n) nb is the shape parameter. In
general, ab(n) is difficult to calculate and
decreases relative to a as n increases (see next
table).
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Thus, we see that the CV of a bundle
(proportional to 1/nb) is much lower than that of
a single fiber, and tends to a deterministic
limit for a very large bundle. It also means that
the size effect is much smaller for a bundle.
In a later chapter, we will examine the
statistical behavior of a unidirectional composite