Learning Forum

[ebone2god]

Hello,

I'm attempting to run a static analysis of a loading & unloading cycle to model an indentation process using Mechanical APDL. Currently, I've been able to successfully complete the loading portion, but I am having trouble adding the second load-step for the unloading. Below is a sample of the BC & solution part of my code:

Here, the BC_S1/S2 are node sets and INDIS is my indentation displacement. I'm attempting to start the "unloading" load-step from the final deformed configuration of the "loading" load-step. I tried to simply copy the commands for the "loading" step and replace "INDIS" with 0 in the applied displacement line, but was unsuccessful with this attempt. Does anyone know how I can amend my input file to incorporate this unloading step within the APDL editor? I would really appreciate any suggestions! 

Best regards,

Eoghan

[Erik Kostson]

There a coupe of ways to do this.

See the help manual (5.6 Solving Multiple Load Steps):

https://ansyshelp.ansys.com/account/secured?returnurl=/Views/Secured/corp/v195/ans_bas/Hlp_G_BAS3_10.html?q=load%20steps

We can also use a time table and apply the loads in a single load step. Below is a sample command listing for your reference which you can modify and use as you need (*dim is for table called here tabf and D to apply displ., here in the UX direction using the table tabf that ramps up to 0.1 at 0.5 s and the goes back to 0 at 1 s):

*dim,tabf,table,5,,,time

tabf(1) = 0,0.05,0.1,0.05,0 ! displ. value

tabf(1,0) = 0,0.25,0.5,0.75,1 ! time

d,2,UX,%tabf%

 

[mrife]

Hi Ebone2god

You almost had it correct but just removed the 'label' of the D command, instead of the value of the displacement.  So the commands should be something like

!Load Step 1

d,bc_s1_2,uy,INDIS

rest of commands

!Load Step 2

d,bc_s1_2,uy,0

rest of commands

Depending on what bc_s1_2 is this may or may not work as you intend.  This will set the UY displacement to zero in the second load step, as opposed to removing it altogether.  If bc_s1_2 is a component on an "indentor" which is being moved to indent another part, then this would be correct.  If the displacement is being applied directly on the indented part then you would probably want to delete the displacement like so:

!load step 2

ddele,bc_s1_2,uy,,,force

This will replace the applied displacement by its reaction force, then ramp that force down to zero over the load step.  

Mike

[ebone2god]

Great, thank you both for the help! Your suggestions solved everything, I really appreciate it.