topic Arrays... in Mathcad

Stress Strain Fitting...(former Arrays)

Dumitru — Thu, 03 May 2018 19:35:51 GMT

Hi again. Hope you're all doing great!I'm back with more on my initial post. Taking into consideration our discussions and after completing the remaining tests, I've reached the attached...The problems I'm facing next are:1. I have 30 sets (each of 50 pairs, strains

Arrays...

RichardJ — Wed, 21 Apr 2010 07:00:00 GMT

Do you mean you want to change the array indexes in the solve block to values other than 10 and 100, and get results for all the possible indexes? That's easy to do, but you need to fix the expressions in the solve block first. The ones you have do not make sense. See the attached.

Richard

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RichardJ — Wed, 21 Apr 2010 07:00:00 GMT

I don not understand what you are trying to do with your conditions. I think when you write "for" you mean "if", but even then they make no sense. Tell me, in words, what the first condition is supposed to do.

Richard

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RichardJ — Sat, 24 Apr 2010 07:00:00 GMT

That was actually a lot more work than I expected. All your coefficients are complex.

Richard

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ptc-1368288 — Sat, 24 Apr 2010 07:00:00 GMT

On 4/21/2010 1:36:58 PM, dvmoldovan wrote:
>The idea is:
>
...
>Any suggestion will be
>valued...(i'm using v. 13)
>
>Kind regards, Dumitru
__________________________

What a huge size for a peanut data set, and not stand alone, in a Mathcad work sheet. It might be good idea to analyze graphically before entering other concept. For a solution to exist with the same A, B, C, D it is clear that the function and the derivative must be linearly dependent. You will surely understand this attempted that can't be completed because "This variable is not defined above".

jmG

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ptc-1368288 — Sun, 25 Apr 2010 07:00:00 GMT

::: I feel the solution via InverseFunction,
but can't express it yet as a general solver.

jmG

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ptc-1368288 — Sun, 25 Apr 2010 07:00:00 GMT

On 4/25/2010 1:20:37 AM, jmG wrote:
>::: I feel the solution via
>InverseFunction,
>but can't express it yet as a
>general solver.
>
>jmG
_______________________________
... then, let's leave it as simple as possible. The system is completely solved by imposing the constrains as you have. Only two are needed because of the derivative and it solves for the linear relationship and their linear domain. Kind of reverse engineering DE.

Hope it helps, as is !
No clear recollection of such solving was done in the collab.

jmG

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RichardJ — Sun, 25 Apr 2010 07:00:00 GMT

On 4/25/2010 2:35:20 AM, jmG wrote:
> The system is completely
>solved by imposing the constrains as you
>have.

It it trivial to prove that your solution is not valid. You can find the necessary value of Ec/Eo by evaluating the derivative at X=0 (condition 1). Then plug that value into condition 2, and you will find that Y is not equal to 0.45.

Richard

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ptc-1368288 — Sun, 25 Apr 2010 07:00:00 GMT

On 4/25/2010 9:26:31 AM, rijackson wrote:
>On 4/25/2010 2:35:20 AM, jmG wrote:
>
>The system is completely
>solved by
>imposing the constrains as
>you
>have.

It it trivial to prove that
>your solution is not valid. You can find
>the necessary value of Ec/Eo by
>evaluating the derivative at X=0
>(condition 1). Then plug that value into
>condition 2, and you will find that Y is
>not equal to 0.45.

Richard
___________________________

You may have a point, Richard.
Your saved 11 version is about all red.

jmG

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RichardJ — Sun, 25 Apr 2010 07:00:00 GMT

On 4/25/2010 9:58:46 AM, jmG wrote:

>You may have a point, Richard.
>Your saved 11 version is about all red.

Define MPa and GPa.

Some of the symbolic stuff will not work in 11 either. Hit Esc and disable the rows() and cols() statements

Richard

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ptc-1368288 — Sun, 25 Apr 2010 07:00:00 GMT

On 4/25/2010 11:21:39 AM, rijackson wrote:
>On 4/25/2010 9:58:46 AM, jmG wrote:
>
>>You may have a point, Richard.
>>Your saved 11 version is about all red.
>
>Define MPa and GPa.
>
>Some of the symbolic stuff will not work
>in 11 either. Hit Esc and disable the
>rows() and cols() statements
>
>Richard
>______________________________________

Thanks Richard,

This work sheet is really too messy and of such huge size and not stand alone. I was only interested in matching a function and its derivative given two conditions. Intuitively, the solution is a linear relationship in their new domain ... x ==> u,v.

I have no expertise in the collab domain of concrete to tell if he is wrong in some of the assumption he is making. Whatever I could do more in 11, it might not work in his 14 version. I will come back for the general interest of the matter.

Thanks again.

Jean

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ptc-1368288 — Mon, 26 Apr 2010 07:00:00 GMT

On 4/26/2010 9:34:41 AM, dvmoldovan wrote:
>Thank you all for
>helping...I'll start carefully
>looking into your comments and
>sheets!
>
>Kind regards, Dumitru
________________________________

A final solution in the original domain 'x'
Recast the DAE solution:
The DAE is solved, it is not unique and we want the solution of the derivative member in the original domain 'x'. Simple:: express the derivative in term of a Fourier series. Depending upon the real spread of the [� x] domain of application, you might be able to best adapt L in order to reduce the number of terms in the Fourier polynomials because only the end points seem to wiggle in that particular application.
This DAE as bounded is solved completely, but not unique ... totally automated in < 1 s.

The other point is ::

You have assumed that the derivative solution could be expressed in term of the same size rational fraction as the function, but that is a pure assumption and unlikely possible [educated guess]. The other point in this other point is that for the derivative in term of a rational fraction would largely increase the size of the sheet for it to be automated.

What you are not clear about is the final purpose of the data table and all those things. If you would want to freeze the project once for ever for the data you have in hand and leave it as a master to be interpolated "per use", it would be a one time task to solve the project entirely ... and in the preferred way, i.e: either in Fourier series or in rational fraction. Your work sheet is hairy and not in the Mathcad style that a first time Engineer not familiar with your style can read. Excel has nothing to do with Mathcad, always make a work sheet stand alone. The next and not negligible point is the traceability. Mathcad is not reliable enough to comply to building codes or any "body codes". What that means is that the "project solution" should be all tabulated for hard prints and traceability and for transportability. What I mean is:: you must freeze the "project solution" with working formulas that can be "coded as approved" in "approved software" ... that can be this time Excel or else software available at the "Engineering firm".

I'm sure you understand/understood everything.

jmG

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ptc-1368288 — Mon, 26 Apr 2010 07:00:00 GMT

...suite : the bad news !

You have assumed the same size 2/2 rational would fit the solution, unfortunately it does not. Below Marlett, a 4/3 would be minimal for the range x � 2. The best way to learn and remember is from erroneous concepts. Ordinary reading does not print well in the brain. You never remember which hammer has pushed the nails correct, but the one hammer that smashed your fingers ... that one you remember.

I haven't tried a linfit Cheby, a bit more involved than Fourier.

jmG

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ptc-1368288 — Mon, 26 Apr 2010 07:00:00 GMT

A hammer and nails story,

This man was just new in the building crew. What a good man you have hired said the "magaziner" to the foreman. Is that so ? Oh ! yes, he is pushing about twice as many nails than the others by the number of nail boxes I deliver to him. Curious to see the champion at work, the foreman watched a bit closer for a while ... Hi man, why do you throw so many nails away ? Sir, the ones that have the head down ... I throw.

MORAL :: Never be too short of instructions and vigilance.

jmG

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ptc-1368288 — Fri, 30 Apr 2010 07:00:00 GMT

"I'm interested in automating this calculus for the whole range of X and Y."
____________________________

That is the a fitting session. Calculate the data range on X and on Y from tables or other source, then attach the data set in the collab. You will surely not get any integer coefficients. In other words, the project is to fit the function you believe as a good model, to fit this model to an unknown yet data set .

jmG

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RichardJ — Fri, 30 Apr 2010 07:00:00 GMT

On 4/30/2010 2:23:14 PM, dvmoldovan wrote:
>Sorry to get back on this post
>so late, but I first needed to
>carefully read (and
>understand) all your replies
>(thanks again). Still, I fill
>I haven't asked the right
>question.

That may be, but the answer I gave is the solution to the problem you posted, even if it is not the answer you wanted.

>If you have a
>function, Y(X), in which you
>know for a given domain (data
>in excel spreadsheets) both Y
>and X

You may know Y and X values for the given domain, but the way you defined your problem means they are not used to find the the solution, because.....

> but you have no clue on
>the 4 coefficients (A, B, C
>and D) that link X and Y (by
>the formula Y = (A + B * X *
>X)/(1 + C * X + D * X * X), if
>one imposes 4 conditions on
>the function you should get a
>domain of values for A, B, C
>and D.

.....the conditions you impose are at specific X and Y values, 0, 0.45, and 1, and it is the values of the conditions at those values (only at those values) that define the coefficients.

>Those values should be
>integers, since the function
>itself has a very physical
>meaning (stress = a function
>of strain).

Why should they be integer? There is no reason they need even be real. The stress and strain (Y and X) must be the real, but the coefficients don't need to be. For the example values of Eo6 and Ecm I plugged in the coefficients are complex. There are no real solutions. In fact, if you are talking about dynamic strain, as opposed to static strain, the coefficients are in general not real because the strain lags behind the stress. If you really need real coefficients then there is either an error somewhere else in the worksheet that leads to incorrect values for Eo6 and Ecm, or you defined the problem incorrectly. In fact, one of those must be true, since if you take the example coefficients for i=100 and j=10 at the end of my worksheet and calculate y(Aa,Bb,Cc,Dd,X6[100,10)) you get a complex stress, which cannot be correct. That means either the conditions you imposed to find the coefficients are inconsistent with the data, the data is wrong, or the method used to calculate E06 and Ecm are wrong.

>The problem is
>that no matter how I try, it
>always ends with an error
>(doesn't calculate anything).
>In my first post, the file
>attached is an exemple that
>gave values for A, B, C and D
>for a particular value of X
>and Y.

No, you just thought it did. The syntax in your solve block is completely wrong. To Mathcad, what you wrote for your conditions is gibberish. Always check the answers! If you plug those coefficients for X.6[100,10 into your equation you get Y=-0.142, whereas Y.6[100,10 is 9.127*10^-4

It is easy enough to solve for the coefficients for a particular value of X and Y with an additional three constraints, not four (y=f(x) at a given y and x is already one constraint). Is that what you want? If so, which of the four current constraints do you want to get rid of?

Richard

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ptc-1368288 — Sat, 01 May 2010 07:00:00 GMT

On 4/30/2010 11:47:36 PM, adiaz wrote:
...
>You can't
>adjust a curve to some data and then try
>to modify the coefficients just to check
>some phisical deductions, that's cheat
>(actually Galileo, Newton and several
>others does that).
>
...

>Regards. Alvaro.
__________________________

That's what I have explained and exemplified previously, as well as Richard in other words. The derivative of an approximating function just does not come out by freaking the coefficients. But that was so clear in the work sheet attached before. In fact, what are the data set for ? are they data from "building code" that some inventive brain is trying to represent by formula ? It looks feasible but there isn't even a plot of the project ! A communal dead start

jmG

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ptc-1368288 — Sat, 01 May 2010 07:00:00 GMT

"Do it for each data column"
____________________________

Hard to see what c <6> represents.
As it looks, data should be sorted.
You have X, Y for the function, but what are they for the derivative ?
Derivative should fit some data that are missing .
Maybe the derivative is just "a derivative" ?

jmG

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ptc-1368288 — Sat, 01 May 2010 07:00:00 GMT

On 5/1/2010 8:41:11 AM, dvmoldovan wrote:
...
>So, just to give it another go

>each Y is a recorded stress
>for a given specimen
>each X ia a recorded strain
>for the same specimen
>The function that links them
>up is unknown. Therefore I
>assumed it may have the form Y
>= (A + B * X * X)/(1 + C * X +
>D * X * X). Now, A, B, C and D
>are determined for specific
>values of Y and the
>corresponding X and not for
>the whole range of Y and X and
>it is the values of the
>conditions at those values
>(only at those values) that
>define the coefficients
..................................
..................................
> Suggestions, are more
>than welcome!
>
>Kind regards, Dumitru
==========================

>each Y is a recorded stress
>for a given specimen
>each X is a recorded strain
>for the same specimen

==> NO, they are NOT ... the data are not collected sorted.
==> or if you pretend they are, then they are invalid from test.
==> related physical phenomenons can't disperse.

>The function that links them up is unknown.

==> that is then the very essence of the project.

>Therefore I assumed it may have the form
>Y = (A + B * X�)/(1 + C * X +D * X�)

==> You can't assume what you know nothing about beforehand.

>Now, A, B, C and D
>are determined for specific
>values of Y and the
>corresponding X and not for
>the whole range of Y and X and
>it is the values of the
>conditions at those values
>(only at those values) that
>define the coefficients

==>Nothing is determined from nothing in hand.
:::::::::::::::::::::::::::::::::::::::::::::::

The project starts here:

1. You must collect the 15 data set as 15 separate data X, Y as they are of variable length
3. I have done it for 3 of them between big red Marlett [qaz(2)]
4. Do all of them, using the manual setup ...
Steps 1 to 4 are in replacement of what you should have done correct from lab test.

Once you have in hand the 15 separate data pairs X's, Y's of variable length, plug them at the top of "Dumitru Viscoelastic". For each one of these data_0, 1.....15 you will have to plug 3 times the corresponding columns in Xd, Yd and for the residuals. For doing this, it's best to turn calculation OFF . Thus, in 15 Maxwell sessions you will collect each pairs of U, which 'U' is then the Maxwell Viscoelastic model c/w the Maxwell coefficients. At this point, you understand that the 15 experimental data sets are replaced by a single model of as many coefficients as each data set has rows.
The project ends there, i.e:
15 data sets of variable length replaced by a single model defined by 15 sets of 'U' Maxwell viscoelastic coefficients.

Hopefully the Maxwell model setup works for all 15 data sets.
You must realise that in this project, you are the "beton expert" and if you don't like the Maxwell viscoelastic model, then more collabs might offer a more appropriate model, or get it from other source pertaining to that field of Engineering. I have no more technical knowledge about concrete than "popular". In your project, I have done my best to square the moon with a chain saw... all yours now.

The next point is paramount about your project as I understand it:

Once you will have all the Maxwell viscoelastic coefficients, or as it seems directly from the data sets themselves, it is an easy task to run a "Mathcad Polyline" nice and smooth curve, meshed at will that you can further interpolate, thus freezing the project as per your lab test that can't be checked/confirmed in this collab.

jmG

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ptc-1368288 — Sun, 02 May 2010 07:00:00 GMT

... several data sets can't be interpreted.
Considering the "Load" is the independent variate and deformation is the dependent one, then sorting the data is not necessary and you have too many and that does not had any information to the physical phenomenon but I understand that with lot of data is will determine the "fracture" more accurately. Your "Load" measurements look disaster from very poor instrument. If the sorted would be of at least 3 decimals, there would be no appreciable difference between the discretized and the natural collection... but that is not so. Thus in this work, I have rejected "Load" and considered only the discretized range, you can see how much impact it does have on the final Maxwell viscoelastic coefficients.
The other point is that you were playing with the data before applying the model and correcting the "Load". What that means is that processing on erroneous data is the wrong approach. The last point is about the population of the data sets. Reduced sets of equal length for each experiment is desirable and as a guess, population of 100 pairs or even much less will represent the phenomenon.
So, the lab "Load" are definitely incorrect, it certain means the same for the "D"... project back to square 0. But this time with better equipment and much less data in order to compact the Maxwell viscoelastic project nicely as a printable document. Hard to explain such a variation in length 207 ... 1791 ?

jmG