Originally Posted by
EJGeiginni
The locking pins on the upper assemblies appear to hold the roller pin only at the tangential point where the roller pin contacts the locking pin, which rather than being limited to the normal perpendicular shear force, is able to flex against the pressure from the roller pin. Is there a chance that, in the absence of the locking pins on the lower hinge slides, that this pressure point - in the absence of a radius block to spread that shear force at perpendicular points of contact on the locking pin, cause the locking pin to fail, moving just enough to gradually unseat the 12 stop brackets?
Just speculation at this point. Hopefully they find the door somewhere around Beaverton or Lake Oswego soon.
Before it was laid out in the latest Chris Brady video I was convinced there was a "locking insert' in the roller guide fitting that was held by the pin, with the insert countoured to the radius of the roller pin. Turns out not, it's just a pin through the fitting that traps the roller pin. It's a little crude but, things like that are done for reasons like:
- It's a legacy from 1965 that never gave any problems, so it's kept the same to reduce certification headaches by 1. Just looking at it this looks like a real old design.
- The primary retention features are the locking pins on the hinges, the pins in the upper guide fittings are a reduncancy and should never touch the roller pin anyway
- Even the "crude" trapping pins are probably 5 times stronger in bending than the maximum load from the spring or general door flexing, maybe 10 or 20 times.
I'm really sure that any one of those locking pins should have been more than enough to keep the door from sliding off the stops. Either none of those pins were installed, or the door itself collapsed, a possibility with implications nobody wants to think about.