@acer  As you have observed, computing  int(int(x,y=-f(x)..f(x)),x=0..1) without the method=meijerg option produces an answer in terms of EllipticE and EllipticK functions with imaginary arguments:

int(int(x, y=-f(x)..f(x)), x=0..1);      
                                     
                      -4/5 EllipticK(I) + 4/5 EllipticE(I)

As we have seen in my original replay, that answer then gets reduced to a real value after applying convert and simplify.

Interestingly, if we change to polar coordinates by letting x=r*cos(t), y=r*sin(t), the equation of the lamina's boundary changes to r = sqrt(cos(t)), where t goes from −π/2 to π/2, and the integration produces an answer in terms of EllipticE and EllipticK functions with real arguments:

int(int(r*r*cos(t), r=0..sqrt(cos(t))), t=-Pi/2..Pi/2);                       
                                   1/2                        1/2
                    1/2           2            1/2           2
              -2/5 2    EllipticK(----) + 4/5 2    EllipticE(----)
                                   2                          2

simplify(convert(%, hypergeom));                                              
                                   1/2           2
                                  2    GAMMA(3/4)
                              2/5 ----------------
                                         1/2
                                       Pi

That is not as good as the direct answer obtained with the method=meijerg option, but at least it has no imaginary parts.

@mmcdara That's an interesting approach.  I wouldn't have thought of applying statistical tools to solve a mechanics problem.

I have converted your "Reply" to an "Answer".  Thumb up!

@Joe Riel That's good. Thanks!

@vv If we could distribute ln over a product, it would convert it to a sum and then the thing will simplify.  But I don't know how to distribute ln even in the simplest case of ln(b/a).  How does one get ln(b) - ln(a) out of this?

@PhD_Wallyson I have made a few corrections to the worksheet and now you can see the modal shapes.  Your interface conditions appear to be correct but I have not examined them carefully.  Be sure to check!  Also be sure that the values of the constants are provided in consistent units.

Download: corrections.mw

@PhD_Wallyson It is possible, in principle, to program Maple to produce a table like that, but I don't think that's worth the effort.  It would be quicker to run your worksheet 10 times with the desired values of the lengths and copy the results into your thesis.

@PhD_Wallyson In this latest worksheet, you handle the symbolic coefficients correctly.  That's good.  Continue doing it that way.

The interface conditions, however, don't look correct.  You have:

(D@@2)(X[1])(L[1]) = -(kr)*(D(X[1])(L[1]))+(D@@2)(X[2])(L[1]),     # equilibrium condition bending moment at x=L[1]

A beam's bending moment is EI X''.  Shoudn't the interface condition have EI in it?

One more thing.  In the worksheet image that you posted today, we see that the μ values are calculated incorrectly.  We even have a negative μ!  I expect that once you correct the missing EI issue, that problem will go away.  And then the plotting of the mode shape will work as before.

Yet another comment: The legnths L1, L2, L3 in the worksheet are in centimeters.  Be sure that kt, kr, E, I, A, etc., are in compatible units.

@PhD_Wallyson Your boundary conditions involve numbers such as 1.422*10^4, 4.881*10^9, etc.  That's makes it impossible for me to verify their correctness.  Please write them symbolically in terms of the parameters as E, I, etc.  Then I will be able to track down the errors, if there are any.

See my previous comments on how to handle numerical values in Maple.

@PhD_Wallyson In my previous reply I noted the need to correct the interface conditions at the supports.  You need to make those corrections before you can compare the results with those of the FEM code.

As to handling the numerical constants, look at our correspondence here.  See how the calculations are carried out without numbers.  Numerical values are introduced through:

span_lengths := L[1]=3.5, L[2]=5.0, L[3]=21.5;
characteristic_equation_numeric := eval(characteristic_equation, {span_lengths});

Do you see how that is done?

@Test007 Specifying the basis is the hard part and it varies greatly from one application to the next.  Do you have something specific in mind?

Once you have the basis, calculating the generalized Fourier series does not require any specialized tools.  You just define the Fourier coefficients through

a[n] = int(f(x)*phi[n](x), x=0..L);   # assuming one dimensional interval (0,L)

and then f(x) is expressed as

f(x) = Sum(a[n]*phi[n](x), n=1..infinity);

That's all.

@janhardo Okay, I believe that you have spent more than 5 minutes on your calculations., but as I said, you must have made an error.  Redo the calculations.

@janhardo The area of the the rectangle SJIK is (12/5)^2*8 = 1152 / 25.  The green area equals that minus 8, that is,  1152 / 25 - 8 = 952 / 25 = 38.08.  If you are getting something different, then you must have an error in your calculations.

@Carl Love Your ZigZag and ZigZagInv procs are quite clever.  I converted the first one from Reply to Answer in order to give it a thumbs up.  I don't know how to rearrange the followups into a proper sequence. If you (or anyone else) know how to rearrange, please do so.

@Carl Love That's excellent.  I had a feeling that the code can be shortened, but did not expect it to such an extent.

What is δ?  Is it a multiplier or a function?  At places you have δ(T−t)).  No space after δ indicates a function application.  At some other places you have δ (T−t).  A space after δ indicates multiplication. You want one or the other.

I haven't looked closely at the rest of your worksheet.  Make sure that you have no other similar issues.

If you continue having such problems, consider configuring your Maple for 1D input where issues like this don't arise.