Assembly always wants to check flexible sub-assembly out of the vault.

[awatt]

Assembly A contains flexible sub-assembly B and dozens of other sub-assemblies.  Any time I have A open and make a change to it, I am asked to check out B.

The vault display in the assembly browser doesn't show an asterik by B's file, or any of its children.  When I save A, I'm not propted to also save B.  But I can't check A into the vault without checking out B, and saving the ghost changes.

I've repeatedly examined the constraints within B and they check out in all LOD's and Positional Reps, using 'rebuild all' in each case.

I've placed B into a new assembly (alone) and I see the same behavior.

I'm using a LOD of B that suppresses reference parts.  One of the non-reference parts is grounded.

The positional representation I'm using suppresses constraints with all but one of the reference parts.  None of the reference geometry in B is itself flexible.

I have no adaptive geometry in any of the parts.

What other things should I be looking for?

Inventor 2013, SP2.

Windows 7

[blair]

IF "B" is flexible within A, there is always a chance the "B" will become altered hence the reason "B" must be checked out along with "A". Flexible is a quasi-modified/adaptive state within the parent.

[awatt]

True, but in this case the behavior can be duplicated immediately after updating all the files, checking them back into the vault, and re-opening the file from the vault (via Inventor add-in) after an inventor restart.

But I think I've solved the problem. I systematically removed parts from the sub-assembly one at a time, and finally got it to work. After removing a work plane from sub-assembly B, assembly A now seems happy to leave B alone.

Which begs the question 'WHY?' It was just a work plane, constrained only to one of the origin planes of a component in B. It seems so harmless.