1 ACCEPTED SOLUTION
Accepted Solutions
One way of doing it would be like the following:
You may want to compare accuracy against the values you had found:
At first sight it looks like you can find an a-value for any given F.2-value within a given range and vice versa (I have chosen the values you found in the following examples)
but as soon as you lower the value of CTOL, no solutions are found
Regards, Werner
PS: Instead of "find" you could use "minerr" (especially when you think that there is no perfect exact solution and so it will find a solution with lower values of CTOL, too) but your choice of using "minimize" was wrong.
You can make I a function of x as well as moment. If I(x) requires integrals it will slow down computation; one way to speed that is to create a vector of I at points along the length and fit a polynomial (function "regress") to the curve.
I have problems with tension and compression along with lateral loading, so I need terms of M(x) + Faxial y(x) was well as variable I(x). They all can solve.
Like Fred I can't read your Prime 3.1 file, but perhaps this will help
Turn odesolve into a function. Here's an example that was written in Mathcad 15:
The syntax in Prime is a little different, but not much. Theta is now a function of the two parameters i and j (in your case, it would be a and F, your unknowns). Remember that odesolve returns a function, so theta(i,j) is a function that returns a function. We now need to convert this function to a simple function that returns values, rather than another function. This is such a function for a single value of i, j, and t:
If you want to pass it a vector of values for t use this form of the function though:
You don't want to use the first version for many values of t, because it will call the odesolve block for every value. The second version will call it only once. Now if I don't know what i and j are, but I can write two equations where everything other than i and j are known, for example
Theta(i,j,0)=0
Theta(i,j,1)=1
I can use a solve block to find i and j (in other words, use a solve block to do the iterative process that you are doing by hand)
Ok, I think I am close to the right solution. Is the minimize-function used correctly? The program tells me it needs a scalar or a matrix for F2. As it can be seen in the figure, F2 is a scalar, but I also defined it as a matrix, both does not work.
I found the right parameters manually by try and error (index cor) and the corresponding function sol, that fulfills the boundary conditions. But finding them by the solve block does not work.
I am sorry, but I am very new to Mathcad.
What am I missing?
You have set the problem up correctly, and it works as it should. You have
Now you can solve the beam deflections for any load and position sol1 :=w3(a1,F1), sol2:=w3(a2,F2).
You want to solve for the load and position that give you a specific deflection and slope at wherever this position is?
I'm not sure that's a solvable problem.
One way of doing it would be like the following:
You may want to compare accuracy against the values you had found:
At first sight it looks like you can find an a-value for any given F.2-value within a given range and vice versa (I have chosen the values you found in the following examples)
but as soon as you lower the value of CTOL, no solutions are found
Regards, Werner
PS: Instead of "find" you could use "minerr" (especially when you think that there is no perfect exact solution and so it will find a solution with lower values of CTOL, too) but your choice of using "minimize" was wrong.
