Axial bearing [kN/m2] 19046.77 5988.67 LC-517-16 0.31 DG1 3.1.1;
A2 = ((B/N)*Ncs)*Ncs = ((600 [mm] /600 [mm] )*750 [mm] )*750 [mm] = 5.63E+05 [mm2] DG1 Sec 3.1.1
A1 = B*N = 600 [mm] *600 [mm] = 3.60E+05 [mm2] DG1 Sec 3.1.1
fp, max = f*min(0.85*f'c*(A2/A1)1/2, 1.7*f'c) = 0.65*min(0.85*27.579 [N/mm2] *(1.563)1/2, 1.7*27.579 [N/mm2] ) = 19.047 [N/mm2] DG1 3.1.1
I have checked Demand capacity for axial bearing for the perticular load case, but in RAM connection value does not match with manual calculation for per same load case.
i am sharing word file of RAM output with all load case.
manual value for the same load case is 4752.3 [kN/m2] load case LC-517-16
values are in Bold.
looking forward to your reply.M760A03_INPLACE_08-08-22_SQUARE.rtf
It's always helpful to include the actual files. The demand in this case is based on the pressure. It looks like the column in this load condition is under a moderate amount of tension with a large concurrent major axis bending moment so only a small area of the plate is in contact pressure (see the first concrete stress plot for LC 517).
can please give me sample calculation for the same load case becasue i have already considered the bending moment with the axial load
please find below sample calculation for the same load case
for load case LC-517
Pu = -93.57 kN
Mu = 161.73 kNm
Base plate Size = 600 X 600 X 40mm
Bearing = P/A+- M/Z
P/A = -93.57/(0.6*0.6)
= 259.93 kN/M2
M/Z = 161.73/((1/6)*0.6^3)
= 4492.4 kN/M2
Total Bearing = 259.93 + 4492.4
= 4752.3 kN/M2
this value is not matched with the RAM output as i mentioned earlier.
also there is no tension in this Load Case is compression case because our load PU is negative not positive.
kindly check and Revert if i am wrong.
Thanks and Regards,
Mayank Patel