Thanks for your prompt reply sir. I tried using this command to evaluate the plate corner forces and following is the output:
GLOBAL CORNER FORCES
JOINT FX FY FZ MX MY MZ
ELE.NO. 3 FOR LOAD CASE 1
3 0.0000E+00 9.2707E+02 0.0000E+00 -1.3664E+06 0.0000E+00 3.8886E+05
4 0.0000E+00 1.1352E+03 0.0000E+00 -1.3798E+06 0.0000E+00 -4.3929E+05
12 0.0000E+00 -9.6465E+02 0.0000E+00 1.7141E+05 0.0000E+00 7.2959E+04
11 0.0000E+00 -1.0977E+03 0.0000E+00 5.8739E+04 0.0000E+00 -1.7946E+05
ELE.NO. 4 FOR LOAD CASE 1
4 0.0000E+00 1.1942E+03 0.0000E+00 -1.3903E+06 0.0000E+00 4.4893E+05
5 0.0000E+00 8.9741E+02 0.0000E+00 -1.3618E+06 0.0000E+00 -3.6942E+05
13 0.0000E+00 -1.1662E+03 0.0000E+00 2.1773E+04 0.0000E+00 2.2990E+05
12 0.0000E+00 -9.2538E+02 0.0000E+00 1.7859E+05 0.0000E+00 -6.2131E+04
The Joint under consideration is Joint 4 for which if we add FY, we get (1135 + 1194) N = 2329 N
However, the reaction force at Joint 4 is 2722 N as per STAAD output figure shown above. I believe that this difference is due to the equivalent concentrated force at Joint 4 due to half of the applied pressure in the direction of restrained d.o.f. (vertical direction here)
0.5 * (0.0007 N/mm^2 * 920 mm * 1220 mm) = 392.84 N
Hence, the total reaction force becomes 2329 + 392.84 = 2721.84 (which is the same STAAD output)
This part is quite clear. This is the philosophy which STAAD follows for calculating the nodal reaction.
My question is: Due to the applied pressure on the plates, shear stresses (SQX & SQY) develops. How can we check the vertical equilibrium between the reaction force at this joint and the shear stresses (converted to shear forces) in the plates adjacent to this joint? [Please note that this is just an assignment at the University level to verify the concept of plate element equilibrium using FEA]
Thanks,
Rajat Kapoor