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Best answer by LucMeekes

Two problems:

1. Up in the top, you have defined a value for F.02. The GPa as its unit is a type 'Constant' instead of 'variable'. But potentially that's not a problem for Mathcad 15.

2. Down at the bottom, you provide a guess value for lambda. That must have the correct unit, Pa^3. If you add that, the solve block at least finds the lambda value for x and y both 1 m. But for other values of x and y (guided by the CreateMesh function) no suitable lambda was found. You may need to provide a better guess value for lambda. This is what I get with a mesh of 10x10 points, and a guess value of -10 MPa^3 for lambda:

LucMeekes_0-1599595697497.png

And with a guess value of -50 MPa^3 and 20x20 points, it shows that there are 2 points where lambda produces totally different values:

LucMeekes_0-1599596544239.png

I'll leave it to you to investigate that.

 

Success!
Luc

1 reply

L
LucMeekes23-Emerald IV
23-Emerald IV
August 26, 2020

That's quite a big file, impressive!

But it doesn't make clear what you are trying to achieve.

I see no 'Appendix' in the attached file, but I see a lot of equations with undefined variables, and plot's that are red.

Can you create a small example to illustrate your question?

 

Success!
Luc

P
ptc-36826231-VisitorAuthor
1-Visitor
August 30, 2020

Thanks for interesting of problem. I'm going to show in points what we need to do. This problem will be in attachment.

L
LucMeekes23-Emerald IV
23-Emerald IV
August 30, 2020

It still isn't clear to me what you are trying to accomplish.This is what I see:

Point 1 defines a set of 3 expressions.

Point 2 appears to assign those expressions to 3 separate functions in x and y, but the expressions contain no x and no y, so the functions are constants.

Point 3 defines an equation, involving the 3 functions and three variables capital Pi.something (which are undefined). Then it tries to find the root of a function f in lambda, but f is undefined.

Point 4 attempts to create a plot of one of the functions defined under point 2, using the two parameters of that function (those that the function does not 'have').

 

You'll need to be more precise in what you want.

 

Success!
Luc

 

 

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