Because it is a 'linear analysis', you can apply a negative to the multiplier and the CTF3 value will become positive (in tension) with the same magnitude as the negative value. You can run it in Abaqus and verify that you will get the same answer if you don't believe me. I decided to take the absolute value of my CTF3 values in my margin calcs because I knew I was going to have to apply the same load magnitude in the opposite axis direction and I did not want to be fooled by the negative seperation margins for compression. Is it the best way to do it? Probably not, but it reduced my calculations from applying loads to six axis (+x,+y,+z,-x,-y,-z) to three axis (+x,+y,+z) and I did not have to flag anything. This worked for my workflow. Hope that makes sense!
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
legit saved me an hour of search today!❤
Why rotation dof is free for bolts
How can M.S. (D32) be positive (no separation), if P is a negative value (-6)???
@8:20
All of your CTF3's (axial load) are negative, so how are you getting positive MS???
@9:46
Because it is a 'linear analysis', you can apply a negative to the multiplier and the CTF3 value will become positive (in tension) with the same magnitude as the negative value. You can run it in Abaqus and verify that you will get the same answer if you don't believe me.
I decided to take the absolute value of my CTF3 values in my margin calcs because I knew I was going to have to apply the same load magnitude in the opposite axis direction and I did not want to be fooled by the negative seperation margins for compression. Is it the best way to do it? Probably not, but it reduced my calculations from applying loads to six axis (+x,+y,+z,-x,-y,-z) to three axis (+x,+y,+z) and I did not have to flag anything. This worked for my workflow.
Hope that makes sense!
@@topdogengineer8440 Ah, that explains it! Thank you.
Please answer.