Testing pressures equivalent to zero displacement stress :

on a simple square plate of 100mm x 100mm x 1mm with 4 supports of 2mm radius placed at 0,5mm from the corners.

Material : 15CrNi6

1- The forces calculated at the supports are respectively in X Y Z

3.126444E-12 -1.695736E-11 1.925470E-01

2.333391E-15 1.570553E-15 1.904788E-01

-3.107160E-15 1.258867E-15 1.911145E-01

-9.094631E-16 2.409700E-15 1.904979E-01

i.e. 0.19115955 N in Z with a deflection of 3.027 microns

It all seems fair except for the slight asymmetry in the calculated forces...

To find an equivalent and symmetrical deformation while keeping the corner1 fixed, the deformation must be of

14.122 kPa for each of the other 3 corners.

14.122 kPa applied to the 3 discs of 4mm diameter represents a force of 0.17746 N which is a nice difference with the average force calculated initially. Moreover, with this pressure the deflection increases to 8 microns, i.e. more than twice as much as with the zero displacement Z constraints.

How can these differences in results be explained ?

Cordially

46Cpi