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Let us continue our fiber bundles of, idea
of fiber bundles. On the end of last lecture,
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we discussed the scheme. We said that a fiber
strength P and fiber breaking strain a, are
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random quantities and exists some joint probability
density function UPA. So, that P is from interval
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P min P max, as well as a is from some interval
a min a max. This domain we call omega.
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Evidently, the mean strength P bar is integral
over our domain P times uPa, dP da as well
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as the mean breaking strain is given by this
integral is a general definition of mean value.
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We will also need a marginal probability density
function of breaking strain ga which is as
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you know the integral from uPa dP over all
P values. Also marginal distribution function
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we can use as integral from small g. Small
ga is a probability density, capital GA is
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function is a distribution function and this
integral from this function g only because
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a is upper limit in this integral. Therefore,
I changed the integrating quantity to another
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symbol may be alpha. Yes, it is shown here
that the mean value, it is only for our sureness
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that mean value a of fiber breaking strain
a bar which is quite a few definition given
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by this equation. After using of this here,
we obtained this one which is and must be.
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So, we are right. No mistakes in our equation.
We will also use a conditional probability
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density function of strength means strength
of the fibers at a given value of breaking
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strain. It is on this picture. What I mean?
Let us imagine that no all fibers, but also
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fibers having practically same value of breaking
strain. I can say the breaking strain lying
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from some and to a plus da in an elemental
interval, but strength of such fibers is different.
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These green points have some distribution,
but only these green points. Yeah not these
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points, only the green points which has schematically
have in my elemental thin strip. The distribution
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of strengths of such only this fibers, this
subset, we call as a conditional PDF probability
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density function, conditional probability
density function and the symbol is psi P by
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a. P is a random variable, a is probability
density function of random variable a P, but
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no, from all fibers, then only from fibers
having given value of a is parameter ok.
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This is the conditional probability density
function. This function is very good known
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of probability and it is valid that this probability
density function is uPa by ga. Joint probability
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density function by marginal probability density
function. Why it is in short shown here? Because
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the relative frequency of ga da must time,
it is shown here and it is written here. In
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each case, it is in each teaching group for
theory of probability.
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We also will use a condition on mean value
of strength. What I mean out of them? Let
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us take all these green points in our differentially
team strip and let us make the mean value,
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but only from these green points mean value
of strain. Sorry, no mean value of strength.
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Yeah mean strength value because strain is
same for each green points here, fibers from
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this green from this differential arrow.
So, this mean value from all green points
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is some blue value which is here. I will write
it under the symbol Pa raise bar. It means
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mean value of strength from fibers having
given value a, a’s parameter and this mean
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value, conditional mean value of strength
is as every times the definition of means.
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So, integral over P from P times probability
density function of Pa is parameter, P is
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random variable a times dP. Using this ratio,
we obtain also Pa bar in this form, in this
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expression.
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So, second we assume that a large number of
fibers create a fiber bundle. No, two no ten
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no fifteen than thousand million or more.
How you want very large number of fibers?
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Third assumption force strain relations of
our fibers S is a function of epsilon are
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mutually similar in such a manner that before
breaking strain epsilon smaller than breaking
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strain a, for each fiber is valid that our
function S epsilon for strain function, is
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proportional coefficient of proportionality
key to some average function as bar epsilon.
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So, as bar epsilon is an average function,
K is parameter characterizing individual fiber.
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What I mean? Let us see my gender, the set
for force strain. Force strain function of
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fiber is this set of red of black curves from
a quite few of shape similar, so that it exists.
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Some red function we call it as average function.
Each order, each individual for strain function
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of fiber can be interpreted as K times as
bar epsilon. This function, is it acceptable?
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Each black curve is some like magnification
of our red average function. Let us use a
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convention to our average function. So, that
on this function is also mean break point.
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It is the point, mean break point coordinates
a bar P bar mean value, mean of fiber breaking
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strain and mean of fiber in strength. So,
let us construct our average function, so
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that this point is lying on this red curve.
Well, then it is valid. We said S is S epsilon
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is k times, is bar epsilon. It is here, but
if on the end point of fiber made before the
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break of fiber, the force is equal to strength
s is equal to P. Epsilon is equal to breaking
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strain a, but we said it is k times as bar
epsilon. Now, k times as bar on the epsilon
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v right on the place of epsilon we write a.
Clear?
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So, we have from this P is equal k times as
bar a. We obtain k as a ratio P by as bar
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a. Finally, we can write before break of fibers.
If epsilon small equal a, we can write an
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S. This is fairly P by S bar a times S bar
epsilon after break it is 0. Of course, what
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we need to know now for each fiber. We need
to know couple of quantities P fiber breaking
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the strain. Sorry, the fiber strength and
a fiber breaking strain two scalars, no whole
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function for straining all two scalars P and
a land for each for all fibers together. We
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need to know a1 function S bar, our average
function, our earlier red function ok
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Now, we want to construct mean force parallel
fiber in a fiber bundle. When we roll in some
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bundle fiber from such fibers in each fiber,
it is an order force in the moment. Isn’t
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it? So, that we want to calculate a mean force
parallel fiber on our bundle. It will formulate
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into steps.
In first step, epsilon our strain of bundle
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is more. Epsilon is more than minimum value
of breaking strain from set of our fibers
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inside a. What it means? No 1 fiber is broken.
All fibers are functionable. Our epsilon is
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under the aim in or under the minimum of breaking
strain of fibers. So, the mean force parallel
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fiber in bundle, we will call as S star. Generally,
subscribe star our in this lecture for quantity
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similarity to bundle. So, it is mean force
parallel fiber in bundle S star. What is it?
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As each mean, it is s force in general fiber
times probability density function joint probability
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density function times to both differential
quantities an integral over domain of couples
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Pa.
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After rearranging, we obtained is, here it
is evident on the place of Pu Pa, sorry on
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the place of S, we can use this expression.
So, we obtain this here because no fiber is
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broken, no fiber have S equal 0, no fibers
have some force. This force we obtain this
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here, after rearranging this. Here, this equation
in this equation, we multiply and divide by
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marginal distribution marginal probability
density function ga. Therefore, we can see,
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we multiply and divide by the same expression.
These here in the brackets we know. So, is
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it on the anti? We can write as far is given
by such equation. It has mean force Para fiber
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in ever in a fiber bundle line. Epsilon is
smaller than a mean when all fibers are before
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broken well.
Now, the second part of this derivation. Let
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us imagine that epsilon of bundle strain of
bundle is lying between two borders a mean
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and a max. What it means? Intuitively, how
are strains of bundle is? So, high that some,
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but no all fibers are broken and other fibers
are not broken. It is between two interval
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from a mean to a max between where you a mean
and where you a max.
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Now, the mean force Para one fiber in fiber
bundle. So, the fiber bundle, of course, you
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must start with the same expression. S star
is integral of S times u Pa DPD a as in case
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a. Also, this equation is same as in a case
at this. Here yeah is, but now the integral
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over a will rearrange as a sum of two integrals.
It is definite integral for a mean to a max.
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So, that it must be also integral from a mean
to our epsilon plus the same integral from
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epsilon to a max is not it yeah here is. What
is the force s in our first integral? This
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one upper limit is a mean lower limit. Lower
limit is a mean upper limit is epsilon.
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In this integral, all are smaller than epsilon,
so that
each fiber is broken for S. We need to use
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S equal 0 and because S equal 0, all this
integral must be equal 0 tie in the second
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integral a is epsilon is lower limit. So,
each a is higher than epsilon fiber breaking
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strain is higher than epsilon of our bundle.
So, that on the place of S, we need to use
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early equation is here. Sorry, other derivation
in the same case a. On the end, we obtain
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this structure, these expressions.
How is the difference? This is the same than
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is here. Only lower limit here is a mean and
lower limit here is epsilon. In order, the
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difference is only in this lower limit, lower
board of our integral. For completeness of
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our ideas, if epsilon is higher than a max,
then evidently all fibers are broken. So,
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is it s star mean force Para one fiber must
be equal 0 is trivial and evident.
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Sometimes, it is possible to use some assumption
which I call as a symmetrical strength assumption.
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It is valid on our left picture, but on our
right picture, I will explain it. Let us see
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my gender situation like on our left picture
in this. Differentially, small strip exist
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some end points or some force strain curves
or fiber. There are our green points here.
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The conditional mean value from all of a strength
from these green fibers only is lying on the
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our average function here. Also, in other
strip differential elemental strip the mean
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values of fibers which have its strength which
are in this elemental state. The mean conditional
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mean value is lying on our average red curve.
In this case, this grass is symmetrical. Therefore,
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I call it symmetrical strength. Assumption
of symmetrical strength, it need not be valid.
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It can be, but it need not be one the second
picture shown. That it is not valid here example
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the mean value of from this green pictures
is here, but the value corresponding value
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of average function is here. It is not the
same point here, the same here. It is from
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the other side.
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So, this assumption can that mean would be
valid nevertheless, in the practice based
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in my experience. Often this assumption is
roughly right. If it is valid, then we can
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say that the conditional mean value Pa bar
is same. Then, the corresponding value on
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our average function is equal f to S bar a.
If yes, then for S star, this is our 3 cases
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a b c. The ratio Pa Pa bar by S bar a, which
is in our earlier equation is now equal 1.
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It was here is this ratio, here is this ratio.
Yeah is now equal 1. Therefore, we can write
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this one, but what is this one integral from
probability density function or domain. So,
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density is equal 1 and before coming epsilon
equal a mean, we can write that the force
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mean force parallel fiber corresponds to the
average, our average function.
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In the second case, it was the case epsilon
and interval from a mean to a max. We derived
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this equation after hearing means this ratio
equal 1. We obtained this and it is integral
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from epsilon to a max from ga integral from
epsilon to maximum. What is it? It is evidently
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1 minus distribution function. The distribution
function of margin of marginal distribution
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function of a, was called capital G here.
So, that we have s star is s bar epsilon times
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one minus g epsilon and that is all easier
than in this case if the assumption of the
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symmetrical strength assumption is valid than
the equation are more easier of course,.
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If all the fibers are broken, then the force
mean of parallel fiber is equal 0. So, we
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have the equations for bundle strength parallel
fiber in no strength and the force parallel
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fiber in bundle in the relation to epsilon.
We can calculate such functions. S star is
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a function of epsilon. If epsilon is smaller
than a mean, then it result problem. When
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epsilon is higher than a mean, then some fibers
are broken and by increasing of epsilon more
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and more fibers is broken. So, its curve in
I shape have one maximum and then is decreasing
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to 0.
Generally, it can be here more local maximums
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and also some broken in this border. So, it
can be complicated. Theoretically, it is possible
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normal way by our type of bundles and fibers.
This path has only one maximum, so that notice
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this maximum. It is the point of maximum force
is strength of bundle by number of fibers
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strength where you pair one fiber well.
How to obtain? Let us imagine yourself. Only
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this case, no this case. It is too complicated.
This case we will solve how to obtain this
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point? How to obtain maximum of curve? The
derivative must be equal 0. So, is it I need
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to this maximum must be our interval from
a mean to a max. Evidently, by a max all fibers
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are broken by a mean, no one fiber is broken.
So, this maximum must be in our integral from
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a mean to a max.
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So, we need to derive derivative from this
equation. Yeah, derivative from this equation
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and then, say this derivative must be equal
0. It is shown here. Nothing special. Here
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is derivative. I think I need not to command
mathematical steps. They are all here. You
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can quietly want to see the mathematical way.
It is derivative nothing, nothing more and
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after differentiation, even we have this resulting
expression. For derivative, we say this derivative
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should be equal because equal 0, if epsilon
is equal star in which point is the derivative
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equal 0. It is the maximum of force parallel
fiber in bundle.
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This moment corresponds to breaking strain
of bundle breaking strain of bundle is a star,
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so that in the point epsilon equal a star.
This derivative must be equal 0. You know,
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this expression we use this and we made it
and then, we obtain this expression 0 equal
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this as the equation rearranging. To rearrange
this equation, is it good? Why?
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When we know our average function as star
as bar a, when we know the function of conjugate
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mean value pa bar, when you know marginal
probability density of breaking the strength
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of fibers, then only one quantities are known
on our left hand side. It is a star breaking
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strain of bundle here. It is here in the derivative
and here. Yeah, we can find which of a star.
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We must use on the right hand side because
to obtain value one. So, this is the question
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of finding of root of this equation.
A star of this equation is the root. Yeah
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of this equation, you must do some numerical
method for finding of root of equation, but
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not too difficult to use it. Usually, you
need to do some numerical method. So, complicated
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equation is not possible to solve numerically.
The root problems from this, it is clear principaling,
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so that from these equations, we can obtain
a star. We can obtain a breaking strain of
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bundle and when we have the breaking strain
of bundle, then we use our general equation
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for force parallel fibering bundle. On the
place of epsilon, we give a star breaking
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strain, so that we obtain aP star which is
P star which force breaking force pair one
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fiber by break moment of bundle yeah.
So, these last couple of equations allow to
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evaluate a star. P star is there a case we
obtain when we accept symmetrical strength
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as an assumption. If this one, then using
our earlier equation, this ratio is equal
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1, so that it is this here. Sorry yes and
because this assumption Pa star bar must be
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S bar a star after a small rearranging because
this is one minus distribution function. We
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obtain this expression. Root of this expression
is breaking strain of bundle and using this
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expression which is the exterior.
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We obtain the force fiber by the moment of
break of bundle. So, last couple of equations
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allows evaluating a bar a star and P star,
if the assumption of symmetrical strength
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is valid. It is the way how to obtain, how
to solve the break of bundle? Next slides
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bring no logical moment. Next slides are only
the mathematical rearranging. We introduce
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some relative variables and rearrange our
equations to another form using this relative
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quantities and the final equation is better
for calculation. Therefore, it is here. I
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do not want to comment it because if somebody
will know you can read slowly my slides and
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this rearranging up with me result commentary.
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I would only say about two interesting quantities.
It is the strength utilization. Coefficient
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eta p is also a1 of the relative quantities
eta p which is a bundle of strength related
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to one fiber p star by p bar which is mean
fiber strength. Yeah, by the way, when all
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fibers, our case number 1 our trivial case
and all fibers are the same properties, then
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this ratio, this strength utilization coefficient
is 1. Isn’t it? This is the first, this
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the second which I want to say to you is the
breaking strain utilization coefficient which
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is similarly eta a. Bundle breaking strain
mean fiber breaking strain, bundle breaking
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strain by mean fiber breaking strain.
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Here, is a lot of slides which rearrange our
equations and say, we came to an example.
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A theoretical example because I want to show
you only there is our result of our calculation
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and theoretical example which is here, based
on really easiest. Why I want to present the
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result here because use the same way, you
can also construct an order calculation.
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We have lot of assumptions here. It means,
assumption on another force strain relations
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of fibers be linear. So, this strain is force
strain curves of each fiber are linear. So,
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is it also the mean? Mean function is the
average function, is the linear. Here, second
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let the distribution of breaking point be
normal means Gaussian two dimensional probability
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density function distribution of all and points
of this black lines have the distribution
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alike two dimensional Gaussian distribution.
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Therefore, let us accept the assumption of
symmetrical strength. Let the assumption of
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symmetrical strength be valid and using these
three assumptions, we obtain after application
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of our equations, this graph. By the way,
this equation here is shown. I will comment
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this equation later on. The epsilon by a bar,
what is it? Strain by mean breaking strain
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parallel fiber. I call it in the moment t
relative quantity and on the ordinate is S
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star force mean force per fiber bundle by
mean value of fiber strength. It is when we
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calculate it. Result is that this function
based only on the coefficient of the variation
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of fiber breaking strain, no mean value, no
standard deviation, but only co-efficient
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of variation. I use coefficient of variation
as is usually is used in theoretical. For
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example, 15 percentages and 0.15, the measure
is quantity yeah.
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I see how it is? If coefficient variation
is equal 0, then our case is reduced to our
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trivial case because linear force strains
function. Then, it is increasing to n point
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and pink all is this strain, but when you
we have as we do not know on the second curve
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here, v a is 0 .43 percentage of coefficient
variation of fiber breaking strain. It can
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be in some real natural fibers by cotton.
We have some experience. Since, that it is
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perhaps round 0.2, but let us show this curve
0.3. This is related to the force per one
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fiber is higher mean force is higher, but
then is decreasing. Why? Because in the bundle
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more and more fibers is broken and the force
which earlier taken the broken fibers must
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now take on his body the fibers which are
not broken. So, I think it is clear.
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By the way, to obtain this function was not
too difficult because it is phi its sigma
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and is capital f. It is distribution function
of sum of standardized Gaussian distribution
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absolutely known in each tables or each computers
software is the standard distribution, normal
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standard distribution, and standard normal
distribution. What we obtain in our example
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for strength and breaking strain utilization
coefficients.
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When we calculate it graphically, calculate
it numerically, of course, we obtain following
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graphs. This function is a strength utilization
coefficient. You can see that these both also
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are function of coefficient of variation of
breaking strain of fibers. This coefficient,
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this strength utilization coefficient eta
P is permanently decreasing. It is very high
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interval from va coefficient variation from
v as it is going from 0 to 1 is absolutely
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unreal value, but only to understand the general
trend.
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So, it is decreasing and utilization coefficient
of strength of breaking strain is decreasing.
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Then, increasing for high values of coefficient
variables, nevertheless in the real path which
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is roughly 0.3. So, that this part, it is
shown in more details. Show on right hand,
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right picture. Both curves, I can say decreasing
the utilization is not small. You can see
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that for example, 0.2. Do not worry often
used by cotton fibers this value is 0.7.
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The bundle strength is only 70 percentages
to this mechanism mean, means of individual
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fibers 30 percentage down only this affect.
Well, you see both graphs. You see that such
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effect is very significant. Therefore, from
this is going out to one practical result
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for you when we want to calculate numerically
all this and we not have enough input variables.
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You see that if coefficient of variation of
fiber breaking strain is high than the bundle,
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bundle strength is small. When you have some
material having high value v a, we carefully
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re strength of your for example, yarn because
this variability of fiber breakings strain
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can bring small tenacity of your yarn. It
is intuitive result check. It is experimentally
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right now.
It is interesting also using no normally than
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distribution, we obtain similar curves also
using some type of way bill distribution.
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You can also qualitatively curve, but not
too far from our example from normal this
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result and our equations can apply also to
the Hamburger’s theory. Hamburger theory
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speaks about blend of two type fibers, but
traditionally, what type of fibers? Each fiber
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has the same properties. Now, we can solve
the problem. We have to take two fibers. For
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example, viscose and polyester, but the viscose
fibers are not all same. They have some distribution
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of properties and polyester fibers are
all same. They have, they own distribution.
Principle is also possible in this case. I
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want to show you that earlier, according to
hamburger, you can obtain this thin line.
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It is the broad line which we construct in
last lecture, but using this probabilistic
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model to Hamburger does generalize the Hamburger
theory using this model, we obtain this line.
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The position is larger. Here is an example.
Here shown this input and by an order type
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of input values, you can also on the place
of Hamburger also obtain such, so that the
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curves are more nearer to our experimental
experiences. So, I hope that the introduction
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because my speech was only on introduction
to the probabilistic model of fiber bundle.
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Can show you the general way how to solve
it? Also these slides are not the worst which
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you can meet in your professional life. Dr.
Ross and I prepared more general concept which
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respect not only the variability of strength
and breaking strain of fibers, but also of
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the variability of crimp of fibers.
Let us see my fibers which have shown same,
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strength same breaking oration when they are
straight, but because the fibers have some
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difficult shape by elongation at first, practically
these out force, we changed the shape to stretch
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a, and then started force. This affect can
be from fiber to fiber different, also probabilistic
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model.