AURORA TEAM:
Note please this case. It's a structure of K=178, E=325 type 4. Material: Structural Steel. Section: Tube D=20MM, d=15MM. Constraints in all directions at each of the four footholds. Load: Z10000N at the lower central node of the tower's roof.
Initially the tower was designed with the sides of its base aligned with the XY axes. With this geometry solvers didn't calculate. This was the final message from the Z88R.log file:
start SCAL88
### diagonal element 31 zero or negative ..stop ###
### often caused by missing or wrong constraints ###
### recover: check constraints (underdefined ?) ###
### PARDISO error 3320:
This log points constraint errors but this didn't seem to be the problem. Then the whole structure was rotated 5º at the XY plane and other 5º at the YZ plane, in order to none of its elements were aligned with the XYZ axes (see please at the lower left of the image). With this new geometry and the same BLCs, SICCG and SORCG solver worked well but just with a 1.0E004 Residuum. Cholesky and Pardiso didn't work at all. A single rotation in the XY plane worked fine for other simpler structures but not for this tower.
I wonder then why rotations were needed to solve this case, why it was necessary to reduce the residuum, and how reliable are the obtained results. On the other hand, I wonder also if I'm making any kind of mistake.
Thank you very much in advance. Regards.
E4 Structure
Moderatoren: ccad, mz15, auroraIco, Lehrstuhl
E4 Structure
 Dateianhänge

 torre11.jpeg (139.14 KiB) 6120 mal betrachtet
Re: E4 Structure
Hello selopez,
the rotation is not the solution to solve the problem. As you explaint, the direct solver don't work, the iterative solver do. That has mathematical reasons. But the results are not reliable.
The problem is, that the structure consist of elements from type 4. These elements can no bending. the connections (nodes) are flexible. The simple solution is to use element type 2.
best regards
mz15
the rotation is not the solution to solve the problem. As you explaint, the direct solver don't work, the iterative solver do. That has mathematical reasons. But the results are not reliable.
The problem is, that the structure consist of elements from type 4. These elements can no bending. the connections (nodes) are flexible. The simple solution is to use element type 2.
best regards
mz15
Re: E4 Structure
Hello Mz15
Thank you very much for your answer. Certainly, the same structure with Elements type 2 (DOF6) was calculated straight and fast by all the solvers, without any need of residuum adjustment. Notwithstanding that for this type of element there's no stress graphical output, to observe the deformation pattern has been of grand aid for improve the design of the tower. It remains some doubt about how the restriction on section YY has influenced the outcome of this case.
Other subject that I'd like to understand better, if possible, is why some other structures as that of the image attached to this message, or even the "Crane Truss" of the guidance, can be treated as composed of Elements Nº 4, and others like the analyzed tower, not.
Again, thank you very much for your attention. Best regards.
Thank you very much for your answer. Certainly, the same structure with Elements type 2 (DOF6) was calculated straight and fast by all the solvers, without any need of residuum adjustment. Notwithstanding that for this type of element there's no stress graphical output, to observe the deformation pattern has been of grand aid for improve the design of the tower. It remains some doubt about how the restriction on section YY has influenced the outcome of this case.
Other subject that I'd like to understand better, if possible, is why some other structures as that of the image attached to this message, or even the "Crane Truss" of the guidance, can be treated as composed of Elements Nº 4, and others like the analyzed tower, not.
Again, thank you very much for your attention. Best regards.
 Dateianhänge

 fullerino.jpeg (151.72 KiB) 6012 mal betrachtet
Re: E4 Structure
Dear selopez,
it is possible to calculate wether a framework of rods (elements no. 4) is statically determined or not. If you have g nodes, e elements and b boundary conditions, you have to fulfill 2*g=e+b in the 2D case respectively 3*g=e+b in the 3D case. This information is well known to the basic literature of applied mechanics  I don't know a book in english, but in german literature an example is Gross, Hauger, Schnell, Schroeder: "Technische Mechanik 1" (chapter frameworks). Another possibility is to guarantee that there are only triangle "facets" and no quadrilateral "facets" in the 2D case (in 3D there have to be "tetrahedrons" > compare the crane truss example). If you have any additional question, don't hesitate to ask!
Best regards
Christoph Wehmann
it is possible to calculate wether a framework of rods (elements no. 4) is statically determined or not. If you have g nodes, e elements and b boundary conditions, you have to fulfill 2*g=e+b in the 2D case respectively 3*g=e+b in the 3D case. This information is well known to the basic literature of applied mechanics  I don't know a book in english, but in german literature an example is Gross, Hauger, Schnell, Schroeder: "Technische Mechanik 1" (chapter frameworks). Another possibility is to guarantee that there are only triangle "facets" and no quadrilateral "facets" in the 2D case (in 3D there have to be "tetrahedrons" > compare the crane truss example). If you have any additional question, don't hesitate to ask!
Best regards
Christoph Wehmann
Re: E4 Structure
Dear Mr. Wehmann
Thanks a lot for your tips. Concerning the formula for establish the calculability of a rod structure, I've the doubt if positional constraints in all directions must be considered as 1 or 3 by application node. Just to try, I added triangulating elements to the initial tower of this thread in order to accomplish the formula, making finally the structure of K=174, E=517, remaining the same BLCs: 5 (or 13?). Never mind the tower became now a mess of rods, the solvers refused again to calculate it . In other words, and as far as I can understand, notwithstanding satisfy the formula requirements, the tower remains statically indeterminate and not suitable for being treated as a E4 structure. On the other hand, the indeterminacy of the E517 tower could be visually ascertained.
Other positive side of your answer was the recommendation of return back to the literature. There, in some notes of my student ages, I found these other formulae that don’t include LBCs: e>or=2g3 for 2D structures, and e>or=2g6 for 3D structures. Being these expressions geometrically understandable, they have so many exceptions (I believe that endless for 3D) which make them no very useful.
Finally, after this "FEA tower experience", I think that I'll conceive future structures having in mind just its functional and constructive requirements, and then consider what kind of element is the right for FEA. The temptation to apply trusses type 4 was because its capability of graphically show the stresses and, the most important, to perform NLA.
Best regards and, again, thank you very much for your interest.
Thanks a lot for your tips. Concerning the formula for establish the calculability of a rod structure, I've the doubt if positional constraints in all directions must be considered as 1 or 3 by application node. Just to try, I added triangulating elements to the initial tower of this thread in order to accomplish the formula, making finally the structure of K=174, E=517, remaining the same BLCs: 5 (or 13?). Never mind the tower became now a mess of rods, the solvers refused again to calculate it . In other words, and as far as I can understand, notwithstanding satisfy the formula requirements, the tower remains statically indeterminate and not suitable for being treated as a E4 structure. On the other hand, the indeterminacy of the E517 tower could be visually ascertained.
Other positive side of your answer was the recommendation of return back to the literature. There, in some notes of my student ages, I found these other formulae that don’t include LBCs: e>or=2g3 for 2D structures, and e>or=2g6 for 3D structures. Being these expressions geometrically understandable, they have so many exceptions (I believe that endless for 3D) which make them no very useful.
Finally, after this "FEA tower experience", I think that I'll conceive future structures having in mind just its functional and constructive requirements, and then consider what kind of element is the right for FEA. The temptation to apply trusses type 4 was because its capability of graphically show the stresses and, the most important, to perform NLA.
Best regards and, again, thank you very much for your interest.