Something else to consider: you might have encountered differences in approach when doing classical hand analysis vs. using detailed FEM stresses. Classical hand analysis usually uses average stresses. For example, Bruhn section D1.10 for fitting analysis shows simple checks for net section...
Sorry, here is another try. I put 3 pictures into one pdf file.
https://files.engineering.com/getfile.aspx?folder=d2abd6d4-27ed-4748-8cea-ead7b3b72271&file=Deutsch.pdf
Here are some magazine ads from the 1950s. Evidently Huck was a distributor.
https://files.engineering.com/getfile.aspx?folder=ae6fa40d-cff6-4215-9cdc-88232968837e&file=Fasteners_Handbook_page_12.jpg
Good, I'm glad you got it figured out. I also commend you for taking the time to completely understand a simple example before moving on to more complex problems. I shudder to think how many people are using these powerful tools without doing that...
The slope on the left (your results I guess) seems to be about 200,000 Mpa. The slope on the right (your input stress-strain curve I guess) seems to be about 480,000 Mpa. Which one is right? 200,000 Mpa seems about right for steel. Don't know what 480,000 Mpa would be.
In the plot on the...
The problem might be numerical, but since shear crimping is a real failure mode, it is possible that the core you are using has insufficient stiffness for the application. As SW said, try increasing the value of the core shear stiffness until these modes disappear.
It sounds like the problem...
I don't know if this would help, but here's a link to a pdf of the Nastran Programmer's Manual dated 1970
https://archive.org/details/nasa_techdoc_19700033827/mode/1up
Your results plot looks like you're only getting output at a few load levels. And your first load level is already past the initial yield point, which is probably why the initial slope does not match. You need to obtain output at more steps as the load increases from zero to maximum to see the...
