I do a test simply with a washer, see the sketch enclosed.
I do the simulation with the whole puck and then with half of the puck as well as half of the load.
And I get not the same result to Von Mises:
Entire puck 750.0 [MPa]
Puck half 642.0 [MPa]
When you cut a piece in half to do more than the symmetrical connection.
Where are your stresses measured?
Do the contact forces balance/add up?
How is your F1 & F2 applied?
Good morning Charles.
In reality the force F1 is the support reaction of the force F2.
I have an F1 support that is done on a dia crown. Ext of 42.0 [mm] and dia int of 30.0 [mm].
The force F2 of 30 [KN] acts on a ring of dia ext of 30.0 [mm] and dia int of 23.0 [mm]
Thickness of washer 3.25 [mm]
Thanks for this very interesting answer in its design, I take
I am attaching a picture of the representation of the puck split in half with the
a half load whose constraints are not identical.
As stated in the first post.
I may be forgetting to do something when you cut a piece by
its plane of symmetry.
Thank you for your new shipment.
I'm going to watch it.
But what I still don't understand, is why when I simulates half a piece by the plane of symmetry, I do not get the same results.
your model is wrong, half too,
not constraints possible, contact only.