Hey Alvaro, I've been trying to modify the original concept to handle multiple load cases. So basically different values of F and MD. Any ideas of how to do this efficiently?

Edit: I have taken two approaches.
1- Run through the entire loop extra times as needed, but this makes the process longer.
2- Run a unitary loop to determine unitary results, and then factor them according to the different values of F and MD. This way I only need to do the loop twice, once with just F and once with F and MD together.

Any other ideas?

Hi wb.c, sorry but I forgot your message. I would go for the second approach. I don't have any particular method that would be more efficient than that one.

Thanks for the input Alvaro. I figured as much, but you always have good ideas.

Wrote

Wrote

Any ideas on how to integrate internal releases into the 2D Frame stiffness matrices?

File not found.File not found.

Hi Alvaro,
Thanks for your contributions on this topic. I have had checking your work and I found some inconsistences on the stiffness matrices you posted on .sm file and the demonstration. I checked this topic on books as Matrix Analysis of Structures of A.Kassimali, and web sources as: https://people.duke.edu/~hpgavin/cee421/frame-element.pdf, please can you confirm if your matrices are ok?

Regards,

Oscar

Hola Oscar!

The matrices are OK! Just for future readers on this topic I share the comment on your post about the frame problem:

When singular matrices appear its probably due to a problem with hinged supports or connections. I verified the FE matrices and they are correct. I just changed some loads (that are corrected in your solved file) and then I changed two things:

1) Supports are always "rigid" when considering the member type. If we have a simple support for a frame, the beam that arrives to that support is a "rigid-rigid" member, we will just free the rotation in the constraints matrix (phi_z = 0).

2) When two members of a frame are connected with a hinge, one will be "rigid-hinge" and the other "rigid-rigid" (not "hinge-rigid"!). In these situations you have two rotations for the two beams that arrive at the hinge. To know the rotation corresponding to a certain beam, this has to be the "rigid-rigid" one.

A way of seeing this criteria is: we always prescribe rotations as unknowns, someone has to "withstand" those rotations (you can later "free" them with the constraints matrix), but if none of the elements can withstand that rotation, you will have a row (and column) full of zeros (zero stiffness) for that displacement in your K matrix.

Hope it is more clear!

I attach also the verified .sm file, in case the readers would go directly to the last version (with a working example!):

A Frame with internal releases - v2.sm (256 KiB) downloaded 77 time(s).

Regards,

Alvaro