This solution might not be satisfactory as it does not use a solver for a differential equation, but at least its a solution.
You are asking for two polynomials of fourth order which are zero at 0 and s1 resp. s1 and L, have their second derivative at 0 resp. L to zero and have matching first and second derivatives at the joint position s1.
Actually this means 10 conditions for 10 variables (the coefficients of the polynomials) which you can either solve using the symbolic solver or using a numeric solve block. Of course, as we know the fourth derivative of both functions being q/EI we could set the coefficient of x^4 in both functions to q/(24 EI). But I decided to let Mathcad do the job 😉
Here is what Mathcad came up with
As you can see, the necessary conditions are met (apart from minor numerical inaccuracies):
i still have no idea how to get odesolve to come up with this solution, though.
MC15 sheet attached
So here is a method which uses odesolve - unfortunately still not the "natural" way.
First we use a separate solve block with "odesolve" for U and for W and turn it into a function of the value d2 of the second derivative at x=s1.
The only condition not met when we calculate the functions using the same value for d2 is that we also want the first derivatives at s1 to match.
So we define a function to calculate the difference of the first derivatives at s1 for any value of d2
and use it in a solve block with "find" to iterate the values of d2 until we arrive at a value where the first derivatives are equal.
Wish it would work in one go basically the way you had set it up, but it looks like Mathcad isn't capable enough to do so. However, I would be happy to be proven wrong.
Mathcad15 worksheet attached
Hi mate,
How wonderful is this update..
Thanks so much for your time and efforts
I'll check it and let u know.
Thx,
Mohamed
With reference to https://community.ptc.com/t5/Mathcad/Toolbox-Solving-Ordinary-Differential-Equations-symbolically/m-p/689336#M192090
Your system of 2 ODE's can be solved as follows:
Note that the value of d2 is:
which corresponds to what @Werner_E found.
If you want to run this in Mathcad 15, be sure to take the proper LODEsolve reference file,
you need to replace the LODEsolve.mcd with LODEsolve.xmcd
Success!
Luc
Oh, and in case you wonder if it can be done with three sections...
Hoping/assuming I have set the proper initial/boundary conditions, here goes:
And in case you're wondering about d2 and d3... they are 0.1 each. That is respectively:
The Mathcad 11 file is attached again.
Success!
Luc
With each step I do consider that..
But looks like my version (MathCad 15) doesn't like some functions/expressions as below:
The function LODESOLVE is not recognised along with some expressions in red..
Any idea why?
Something to do with the version I'm using maybe?
Thanks Luc,
I did that, and I'm a step further now..
But I ran into an error (pattern math exception)..
Probably the symbols?
I assume that the example file LM_20201019_LODEsolve.xmcd in the other thread (still) works for you in Mathcad 15. That means that LODEsolve is still able to solve a set of differential equations, but is having problems with your specific set.
I can't debug at the moment, because I have no access to Mathcad 15.
Fortunately you actually do not have a set of differential equations, but a single differential equation that is split over multiple regions of the running variable. The general solution for that differential equation is a single one:
for the first region, then replace U0 with U1 for the second region and with U2 for the third (and so on if you like).
The attached solution file uses this property to solve your problem.
See if it works for you.
Success!
Luc
