• Registered: 2015-12-07
  • Posts: 6

Topic: Fixed Crack vs Rotated Crack

Dear Cervenka Consulting,

I have tried to study both of these smeared crack models (fixed and rotated crack models) from the theory and also troubleshooting documents provided in the website. However, I am still not quite sure about how it influences the result of FE analysis. Therefore, I tried to conduct some FE analysis with two exact specimens (punching shear on flat slabs), the first one with fixed crack (coefficient=1) and the other one with rotated crack (coefficient=0).

As I can conclude from the results, the fixed crack model tends to produce stiffer response and higher peak strength while the rotated crack in most cases produce premature failure (lower strength) with less than measured response (from experimental test).

In addition, the crack pattern at failure for rotated crack seems giving more realistic one where critical shear crack occurs at critical region near the column where the fixed crack give only flexural crack and no critical shear crack is observed at the failure point. Can you give me some other explanations how does this assumption influence the result and how do I know which one is better for my case (punching shear on flat slabs).

Best regards,

Andri Setiawan

Quote Post 1

  • dpryl
  • CC software developer
  • Offline
  • Registered: 2007-07-13
  • Posts: 1,107

Re: Fixed Crack vs Rotated Crack

Dear Andri Setiawan,
as you already know from the manuals, we only recommend (partially) rotated cracks for large elements, like shells.

It is not clear from your description if A. you just can not see the shear cracks at failure, or B. if they develop in another region?

In case A., you may wish to refine the load steps near the peak load. I assume you have already read Troubleshooting, 2.4.2 I get the message "The execution is killed due to violation of stop iteration criteria", what does it mean?, 2.1.19 Problems reaching convergence and understanding ATENA convergence parameters, 2.2.22 How to detect a structural collapse or ultimate load capacity, and 2.4.6 Floating point problems - Division by zero, Loss of >7 digits on the matrix diagonal, CCFEModelExc: CCPardiso solver error: zero pivot,  Warning: Sign of diagonal changed when reducing equation, Floating point exception: Multiple floating points traps, Zero or negative jacobian,  InvertA: zero determinant, CCMaterials Extended AExc: CC5ParamYield F: project_stress_on_f_division by zero, and similar, especially 2.4.6.1 Solution methods.

In case B., please see ATENA Troubleshooting, 2.2.18 My analysis results do not match the experiment/expectations. How can I improve my model?, and check above all the meshing and the reinforcement modelling.

If you can not solve the issue by yourself, please send us your model etc. following Troubleshooting, 2.1.1.

Best regards.

dpryl's Website

Quote Post 2