Bill, you have quite a few questions there.  I will try to answer as many as I can.

• As to: Where can I find this referenced?

  The Euler beam equation is a standard subject in elementary structural mechanics textbooks.  Since you refer to the EI stiffness, I suppose you must be familiar with some of that literature.  For a quick review you may begin with Euler–Bernoulli beam theory in Wikipedia.

• As to: Did you select the Dirac intuitively?

  The beam's static equilibrium is described by the fourth order ordinary differential equation [EI u''(x)]'' = f(x), where u(x) is the beam's deflection at point x due to the application of the load f(x).  The load is measured as force per unit length.

  A point load, that is, a load that is concentrated at some point x=a, is expressed as f(x) = Dirac(x-a).  Dirac's delta function was invented specifically for such a purpose; see Dirac delta function in Wikipedia.

  You will see in that article that Dirac's function is, in some sense, the derivative of the Heaviside function.  Therefore it is not surprising that integrating a differential equation that involves the Dirac function, we obtain a solution that involves the Heaviside function.

• As to: My main objective is to impart onto the hinge a K-theta compliance spring stiffness and solve for the moment and x,y component reactions.

  The bending moment at any point of the beam is given by EI u''(x).  Therefore if the endpoint at x=1 is supported by a springy hinge, then the boundary condition there would be EI u''(1) = K*u'(1).

  In particular, if there is no spring, then we have K = 0, and therefore the boundary condition is u''(1) = 0.  In Maple that's expressed as (D@@2)(u)(1) = 0, and that's what I took in my worksheet.

A further comment

You should be aware that the calculation in animate-beam.mw does not account for dynamic effects. The animation is produced by merely displaying a sequence of pictures of static equilibria.  That's correct as far it goes, however that's not the same as a load moving across the beam as a function of time.  The proper model for that is a partial differential equation which you will find in the section titled "Dynamic beam equation" in the Wikipedia article which I referenced earlier.  Maple can solve that equation as well, but perhaps that should wait until you clarify your understanding of the static case.

@sheriph05 Unlike the previous case, your equations are now nonlinear, and therefore you may expect more than one solution. Indeed, we have two. One solution has I=0. the other solution doesn't.  From what you have said, it's the second solutions that you are interested in.

Here is an overview of what to do.  We solve the equations as usual:
endemic := solve({ fS, fM, fV, fI },  { S, M, V, I } ):
Then the command
nops([endemic]);
returns 2, indicating that there are two solutions.  To extract the value of I from the first solution, we do
eval(I, endemic[1]);
and we see that I=0, so we ignore that solution.  To extract the value of I from the second solution, we do
eval(I, endemic[2]);
and we see that this I is nonzero, so this is the solution that we keep.

Aside:
Your differential equations are:
dS/dt = fS,  dM/dt = fM,  dV/dt = fV,  dI/dt = fI.
Adding these up we get:
d(S+M+V+I) = fS + fM + fV + fI
The right-hand side simplifies to
    (-mu+gamma__V)*S + (-kappa-mu-tau)*I
        -Lambda*rho__M+Lambda*rho__m-M*mu-V*mu+Lambda
This looks somewhat suspicious to me.  Usually in epidemics models the sum of the various subsets of the population equals the total population which is generally a constant.  In your case the sum is not a constant.  This may not be wrong but it's worth verifying that the model does what it is supposed to do.

@tsunamiBTP Transcendental equations don't have solutions in terms of elementary function in general.  Why are you expecting this from your equation?

You haven't said anything about the parameter N that appears in your equation.  If we assume that N is a positive integer, then it is possible to show—see the attached worksheet—that solving your equation reduces to solving an (N−1)-degree polynomial.   Therefore for N=1,2,3,4,5 you get explicit solutions; see the worksheet.  You probably know that there is no solution in terms of elementary functions for polynomials of degree greater than four. That's as far as you can take this.

allsolutions.mw

@Jjjones98 OK, here is how.

@AndreaAlp It seems that in your my_quadric, you have written z where you should have y.  If we make that change, then graph that we obtain agrees with the photo of the bone that you have shown.

But this raises the question of what you mean by "the center of the ellipse".  If you rotate the graph in the worksheet (given in the link below) you will see that I have calculated and marked the true center of the ellipse.  (That mark is not visible in the picture shown above).   I am afraid, however, that's not what you are really looking for.  I suspect that you are looking for some sort of a "center" on the red surface.  If so, the meaning of such a "center" is unclear.  You may want to rethink it.

mw3.mw

@AndreaAlp Youwrite: "Still, those surfaces should intersect and create an ellipse":

Which surfaces?  We seem to have agreed that g(x,z)=0 and g(x,z)=y are different.

@AndreaAlp You seem to be confusing f(x,y)=0 with f(x,y)=z.  They are certainly not the same things!  For instance, x^2 + y^2 - 1 = 0 is a cylinder, while x^2 + y^2 - 1 = z is a paraboloid.

Similarly, g(x,z)=0 and g(x,z)=y are quite different things.   For instance, x^2 + z^2 - 1 = 0 is a cylinder, while x^2 + z^2 - 1 = y is a paraboloid.

@AndreaAlp In the x,y,z Cartesian coordinates, any surface of the form f(x,z)=0 is a cylinder aligned with the y axs.

If you expect something other than a cylinder, then there must be something wrong with your LinearFit calculations.

@EoM007 Here is how.  Change terminating colons to semicolons if you wish to see the intermediate steps.

Download mw2.mw

All we can say from what you have given is that this is a bad approximation.

Perhaps there is a typographical error somewhere?

I have two books here on my shelf that contain the derivation of the equations of motion of a homogeneous ball that rolls on an arbitrary surface without slipping:

1. Neimark and Fufaev, Dynamics of Nonholonomic Systems, pages 76-86.
2. L. A. Pars, A Treatise on Analytical Dynamics, pages 211-212.

The equations describe the motion in terms of a coordinate system that move with the ball.  It will take some effort to relate them to a stationary coordinate system which would be necessary for producing a Maple animation.  I do not have the time to do it now.  Perhaps at some other time.

Aside: The surface cos(abs(x)+abs(y)) is a rather odd choice for this problem because it has sharp ridges due to the presence of the absolute values.  Is there a reason you are interested in that particular surface?

@EoM007 Details in the worksheet mw.mw

@EoM007 In your worksheet you have reversed the left and right hand sides of  the equations that gave.  The order matters.  Keep omega(t) on the right-hand side.

As to differentiating a vector, use the tilde operator, as in:
V := < a(t), b(t) >;
diff~(V,t);

@EoM007 Showing a picture of your worksheet isn't very helpful because I cannot work with it.  See if you can post your actual worksheet.  In the window where you edit your message, click on the big fat green arrow to upload it.

@Christianwm I am unable to read the file that you have uploaded.  It's possilbe that it's due to non-ASCII characters in its name.   See if you can rename it to something with English characters only.

Additionally, have a look at this recent exhange.  It may have some useful informaiton:

https://www.mapleprimes.com/questions/222566-How-To-Make-Danish-Letters-Work-In-MapleCloud