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1-Visitor
April 21, 2010
Question

Stress Strain Fitting...(former Arrays)

  • April 21, 2010
  • 26 replies
  • 32988 views

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 and corresponding stresses) to which I want to find a "mean" - construct a curve which gives, for lets say a given strain, the mean of the stresses (problem is the strains are particular to each and every set)

2. find the tangent in the origin (since most curves have a stabilization step - a preload until a stress of 20 MPa, that means finding the tangent for the curve in this point)

Kind regards, Dumitru

First, I tried to reduce the replies to the first form of my topic (I've saved it all as document so it should be available to everyone) but gave up soon after.

Second, I've renamed it..thanks giraud for suggestion (I kept it in the initial form considering the fact that it would be easier for you collabs out there, who helped a lot, to reference back)

Kind regards, D.

26 replies

19-Tanzanite
May 2, 2010
On 5/1/2010 6:39:25 AM, dvmoldovan wrote:
>second step...automating
>(sorry for previous post -
>some problems attaching)

Why don't you bother to check your results actually work? If you plug A, B, C, D into your function for Y at that point you get Y=0.225. The actual value of Y is 9.127*10^-4. So, once again, your conditions are not consistent with your data and/or model. Of course, something you omitted to tell us was that your Y has actually been normalized to 0.01, not 1 as implied by your conditions, but even if you multiply the measured Y by 100 it's nowhere close to 0.225.

Is it true that each column of X and Y is a stress-strain curve for one sample? I wish you had made that clear from the start (or that I had realized it!)! That means, contrary to what you implied, A, B, C and D are not different for every point, but only for every column. In which case Alvaro is right. If the data is a stress-strain relationship with one column per sample then the right way to check the model is to fit the model to the data. It is possible to do it your way, but it's a very bad approach. All of that work was unfortunately just a big waste of time.

Alvaro fitted the entire data set, including the data after failure. If the model is only supposed to represent the data before failure, then you should only fit that part of the data.

Richard
12-Amethyst
May 2, 2010
On 5/2/2010 10:23:40 AM, rijackson wrote:

>Why don't you bother to check your
>results actually work? ...

I know that you know what can be done with the data and in which way. In the attached the same, but normalized as you discover the strange normalization in the computed values, and the observation that data is not sorted, at the point that there are negative values for the normalized X. I understand that the origin can be set to the arbitrio of the experimentator, but if always have a load, why take an origin which results in a change of sign in the deformation? So, can re-re-normalize substating the minimun to hold X in the range (0,1), but I don't do this in the attached.

Regards. Alvaro.
19-Tanzanite
May 3, 2010
On 5/2/2010 8:04:53 PM, adiaz wrote:

>I know that you know what can be done
>with the data and in which way. In the
>attached the same, but normalized as you
>discover the strange normalization in
>the computed values, and the observation
>that data is not sorted, at the point
>that there are negative values for the
>normalized X.

I don't know what you mean. There are no negative X values.

>I understand that the
>origin can be set to the arbitrio of the
>experimentator, but if always have a
>load, why take an origin which results
>in a change of sign in the deformation?

It doesn't change sign. Obviously I am missing what you are trying to say here 🙂

>So, can re-re-normalize substating the
>minimun to hold X in the range (0,1),
>but I don't do this in the attached.

I think the idea was to normalize the data so that (1,1) is the fracture point. If that was the intention then it's clearly not as simple as picking the point with maximum load though. Data set 2 shows clear evidence that the fracture starts before the maximum load is reached.

Richard
1-Visitor
May 2, 2010
On 5/2/2010 3:09:26 AM, dvmoldovan wrote:
...
>The Wang et al model
>(formulas used) is yet the
>most challenging since it is
>so difficult to get to the
>bottom line (A...D
>coefficients). Thanks again
>for Maxwell's model...

____________________________

Can you link to this material in reference ?

jmG



1-Visitor
May 3, 2010
>is easier to read the independent values on the horizontal scale and the dependent ones on the vertical one.<<br>
The independent variable is the pressure you apply, it plots on X
The dependent variable is the deformation, it plots on Y



You should worry most about the data you have collected, many of them show an unexplainable pressure cycle. Some you can defend and I can accept, typically the one I have worked. What you are not saying is about the pressure cycle you apply or attempt to apply [I mean the ones that makes sense]. That pressure cycle pattern, does it comes from a standard in that kind of testing or does it come from some invention. Up to the fracture, if you apply a linear pressure, it will surely fit well.
The other point is that you will not interpolate via "linterp" or a bspline [interp] on the non ascending part of the pressure cycle, unless the data are split and transformed . The attached demonstrates how to interpolate the ascending viscoelastic sites. The part after [polyline] is for a total pressure cycle in parametric form.

>My main goal is to see the best curve to fit experimental data. That is why, in the first place, I'm studying what others have proposed (i.e. Wang et. al). <<br>
The best fit up to fracture will largely depend upon the ascending pressure . When you plot sin(x), x is supposed linear. Sometimes transforming the independent variable helps the fit. You refer to Wang et al. but you haven't attached the PDF or else link to what you are reading, just to make sure it is understandable, and technically correct. Does Wang give the form of the pressure cycle or is it an invention. Why don't you make it linear up to the fracture ?

The last plot is a 2/2 rational approximation.

jmG
19-Tanzanite
May 3, 2010
On 5/3/2010 7:35:34 PM, jmG wrote:
>>is easier to read the independent values on the horizontal scale and the dependent ones on the vertical one.
>
>The independent variable is
>the pressure you apply, it
>plots on X
>The dependent variable is the
>deformation, it plots on Y

As dvmoldovan already explained, the experiment was constant strain. The load was adjusted to keep the strain increasing linearly. So the dependent variable is the load.

Richard
1-Visitor
May 3, 2010
Assuming procedures similar to that used for metals, the test equipment provides a known strain (mechanically, usually based on a screw mechanism) and the stress is measured. There is no independent "load", just a stress response to the applied strain.
__________________
� � � � Tom Gutman
1-Visitor
May 3, 2010
I recall a set of 4 equations with 4 data points that had 16 solutions - all perfectly valid. So Minerr or other least squares isn't the final solution. My results were obtained by a convergent reversed Taylor series made from the first and second derivatives of the Newton Step.
1-Visitor
May 3, 2010
On 5/2/2010 3:28:18 PM, dvmoldovan wrote:
>The attached is the latest
>form for my calculations.
>Since some mistakes have found
>there way in (I'm sorry again
>for that, only today I
>"discovered" them), I feel
>it's mandatory to correct
>them. Insired by the work of
>jmG (thanks) you'll find some
>graphs at the end...
>
>Kind regards, Dumitru
______________________________

Mathcad is essentially a 0 ORIGIN based tool. Changing ORIGIN to 1 will result in many collabs not even considering the project. If you have done too much maths in ORIGIN 1, don't expect collabs to make your maths work to accommodate their help in ORIGIN 0. Your plot are in the reverse order, i.e: the independent variable "Load" plots on the Cartesian X and the dependent variable [deformation] plots on the Cartesian Y. The way you plot is not visible by inspection and it has to be re-plotted. Your data are still very confusing. No matter how much I have done, nothing is correct on the ground that you don't apply linear "Load pressure" but on a ramp basis, and some "Load pressure" are totally non sense. That means that fitting will be lot more difficult if the independent experimental variable is not linear. Explore the next graph and check the data that must be rejected. Altogether, there may be only one data set needed, and here again why not linear "Load pressure". So, before any math processing, make sure to consider only one data set for only one specimen. Then decide about more specimen and either check the "Load" is correct or run the experiment again. But in any and all cases, you have too many values from lab test::: What you are doing with the ramp load, you are collecting [stacking] a fine and a coarse experimental data set and that is not correct because for analysis, the collection has to be reverse engineered, undoing what shouldn't have been done in the first place.

This project looks simple but it is upside down.

It could be that the rational fraction represents the viscoelastic model for concrete. It shouldn't be more difficult to solve than the simple Maxwell model. As it looks, you are applying a pressure cycle but of such bizarre shape. Why not a simple triangular cycle. That would render the project visible by inspection with no torture of the mind and eventually with a non mirror return of the decreasing load, the deformation might be meaningless.

"In this pressure cycle that does NOT make sense unless justified, it can be seen that the deformation has in some ways followed the pressure load. In simple words: the specimen experimented is 'elastic'. "

jmG



1-Visitor
May 3, 2010
... this work sheet is for the first specimen data set. It makes sense, but you can see that lab testing requires some pre-knowledge of the response so to minimize the number of data pairs. You can decimate more and the entire response will be as informative than an over populated collection.

jmG
1-Visitor
May 3, 2010
... BTW, this sheet is very small in content, but huge in size.

The accusation goes to Excel.
If the lab delivers to Excel, fine
then read the Excel file from Mathcad.
That should reduce the sheet to maybe < 100kB.

jmG
1-Visitor
May 4, 2010
On 5/4/2010 3:21:17 PM, rijackson wrote:
>On 5/4/2010 2:14:47 PM, dvmoldovan
>wrote:
>>Yes, richard that is what I
>>need. When talking about
>>column 2 you are referring
>>to..(what column/ in which
>>sheet?).
>
>It is getting hard to keep track of
>everything! Set c=2 in this sheet to see
>the data I am referring to:
>
>http://collab.mathsoft.com/~Mathcad2000/
>read?134452,11#134462
>
>Richard
________________________________

The Minerr Paul W. was passed in two styles: the short style as in the work sheet in reference, and the NULL style. The fastest when it works is the short style. However many fits will not be possible in short style, so here is the NULL style. The vector of initial values must be collected in vector, this is so because lots of fit can't go direct LM but a good approximation is obtained sometimes CG, sometimes QN ... thus much easier to plug coeff's. And because there is always an "AND", too many fits can only be done manual, thus you must be able to collect quite a few fits in vector.
In Mathcad it's best practice to respect the position of the arguments and parameters. Nothing wrong y(x,A,B,C,D) but it's not correct and creates lots of problems especially in collaboration ... it should be y(A,B,C,D,x).

Same remark about ORIGIN that should be left 0.

To fit a "normalised rational", only the first term need be valued, generally the value 1 is OK. Not detrimental in this 0...1 range of 'x' but it will soon fail with a much higher range of 'x' because if the higher order monomials are valued it pushes the solver out of capability.

Nothing wrong with the work sheet in reference, but not a model to follow, at least not for beginners learning "curve fitting" in Mathcad.



jmG
1-Visitor
May 4, 2010
... also, the principle of solving is to equate to 0 [zero]. This goes behind the scene of the symbolic solve and the root and ... Thus respecting the principle is the best procedure. On the other hand, equating to zero is sometimes an heresy that's much easier to manipulate on the right side.



jmG
1-Visitor
May 5, 2010
...
1-Visitor
May 5, 2010
On 5/5/2010 7:51:16 AM, dvmoldovan wrote:
>Today I took the files as
>originated during testing and
>did a manual analysis on them.
>You were right Jean...the data
>was inconsistent (apparently
>the export to excel modifies
>the type of data from numbers
>to general). I also deleted
>steps that repeated themselfs
>so...the result
______________________________

Read carefully:

Mathcad was designed by visionary designers. They may never become as celebrated as Fourier, Laplace ... etc. Nevertheless, these guys were "Mathematicians". The Minerr is repertoried in the "Numerical Recipes", but it looks the Mathcad Minerr is a more powerful version. What I have done here is in gray and 'j yellow'. If you enable the 2nd constraint condition, it will force in some way the fit. The j 197 belongs to the 2nd data set, the value =1 is just demo, you could force by any unknown value to your appreciation " that is my fit to conserve".
There would be other ways to fit that rational, but too early. If you would want to fit globally, the fit can be wrapped in a piece of program that will do all at once, but you won't be able to force individually for a specific "forced plot". This fit is typically a "Gaussian" that might eventually fit better. But that will deviate [probably for better] from the Wang suggestion. Wang might have been short of models ?
Now you can see my point last night about "solving essentially equates to 0" and what I said about the 0 not necessarily being 0 but otherwise to accommodate a fit. Another point is about the text box , sometimes it's valuable to plug a solver in a text box, but usually it is of no advantage. If you have to push some regions down, you must leave some left margin to the text box, otherwise you can push only one step at the time, because the pointer always goes back the left and re-enters the text box and that screws things .

Resume:
j197 is for demo the 2nd data set, play with it for that fit and play with the conditional block = -1 or else to view the action.
Deactivate the 2nd conditional block to explore the other fits. You can leave the j 197, it will have no action on the fit and it will not turn the plots red because all the data set have > 197 points

About the text box:

It makes Mathcad read like newspaper, i.e: reads/calculates the text box top to bottom then jump on the right of the text box. Very convenient to put the working maths in text box and have the plot next right, like in text books. Text box it one of the features that makes Mathcad so gorgeous for technical presentation.

I will be out for the day, but that fitting session is in good hands with your new/revised data. Rational fractions tend to produce glitches. So, make sure you increment fine enough to insure no undesirable glitch appear, unexpected. Your ORIGIN didn't pest me up until now, but it will soon. There are some instances ORIGIN must be 1 or else but you redefine at the point needed in the work sheet and turn it back to 0 as soon as done with it. Where does that crazy ORIGIN= 1 comes from ? Just figure that many collabs have 1000's of modules based on ORIGIN=0 and every time they would have to adapt for helping "heretic visitors" , no way.

There is a lot of red in the 11 conversion, it might be that you have to define Mpa.

Jean
1-Visitor
May 5, 2010
... suite demo:

This fit is in 4 steps of PWMinerr LM [Levenberg-Marquardt]. Yellow is the initial guess, and on that there is no maths or very little in some instances, the second step is to re-inject the first found coefficients ... there you can see my point about copying a vector vs the individual inclusion like it was on other work sheets ... and the detail in there is that often just by acting on a single coefficient, Mathcad will crash or end in an infinite loop !!! Nothing is so simple, but a good procedure helps.
Done the green fit, next step is to force the fit, here j 216 [typical] ... not too bad but can be made better, how ? No matter how smart the numerical maths are, non-linear models are "reflexive". What it means here is that xc does not Minerr, Oh l�, l�. You are then left with a manual touch up of xc, while you can still play with j 216. The result is a non arguable better fit than Wang rational that is not very flexible.
In a paper, you can put the coefficient in an horizontal vector, but at the design and construction stage, if you don't leave the procedure operative, collabs have to deconstruct. You can stick with Wang but in my opinion you shouldn't refuse a better model. Often, there is no alternative than an "approximation", however in case a better model is found it should change religion.

This message and the work sheet in < 15 minutes.

Jean

1-Visitor
May 5, 2010
I don't see why the model selected is so complicated. The data is noisy and any least squares results are subject to speculation. At the end of this document is a simple Thiele run.
1-Visitor
May 6, 2010
Did you say �code?�

The Wang thing is a math bureaucratic response to some political effort to defeat the established practice of state testing of concrete samples for tensile strength. The tests should be gathered for a statistical presentation and eventual approval of a contract to build something with the concrete that meets the state code. A model fit is meaningless. The Wang thing is a psychological flood.

Perhaps the data given by the originator is a time series of tests run on samples that seem to be mistakenly prepared by failure of preparation and slump test results were not given and those test samples were to be rejected.

I.E., the old State Highway Commission was being reorganized by a brilliant civil engineer from out of the state to a new State Road Commission bent on college bred mathematics.
1-Visitor
May 7, 2010
The results I obtained are consolidated in a way that the numerous Web sites discuss this problem.
1-Visitor
May 7, 2010
Thanks for the clarification of the problem. With limited knowledge of the physical properties of the clay-aggregate cement mixture, plastic and chaotic deformation flows, I have shown the curve fits for these two regions and ignored the very meaningless part above the peak.

A relative of mine was West Virginia State Road Commissioner long ago, and I know the procedures he used to survey the tensile strength of small samples cast in a butterfly shape. We show treated wood small rural bridges outlast the old narrow rebar concrete bridges on the highways.