RAM SS Column Axial Load and Beam End Shears

Question

I have 4 concrete beams that intersect at a concrete column. I would think that when you add all 4 of the Beam End Shears at the column it would = the column axial load but it does not? What else is contributing to the axial load on the column besides the shear at the beam ends? 
 Thanks

Seth Guthrie · Answer

Beam forces are reported at the face of the joint. There is the additional load on the rigid link from face of column to column center line that gets added into the column loads. 
There is also the matter of pattern loading. One pattern load could be critical for certain beams, but a different pattern is critical to the column. 
Finally there are point loads and loads from above to consider.

Seth Guthrie · Answer

Thanks for sending the model. It is fairly simple, just beams and columns, no deck or loads other than self-weight. 
 Regarding self-weight of beams and columns, they are based on the center to center span length. 
 9 * 1.33' * 1.33' * 5' * 0.15 = 12.0 total (1.33 k each, note column SW applied at top of story) 
 Beams are all fixed, (indeterminate reactions) total weight: 
 12 * 5' * 1.33' * 1.33' * .15 = 16 k total (1.33 k each) 
 grand total weight = 28 k 
 Sum of reactions from Ram Concrete analysis = 28 k (4 @ 2.58; 4 @ 3.38 and 1 at 4.17). So that part seems correct. 
 
 Beam shears are, however shown at the face of the column, so those are truncated. A typical 2-span beam line in your file has 0.45 kips at the face of one column, 0.53 at the interior support. So the slope of that shear diagram is (0.53 + 0.45)/ (5 - 1.33) = 0.267 k/ft which equals the unit weight of the beams (1.33' * 1.33 * 0.15). To know the shear at the column CL, one would have to extrapolate the last 8" at each end: 
 0.45+0.267*0.667=0.628k and 0.53+0.267*0.667=0.708k 
 I did notice one problem in the diaphragm of the shear forces. The plateau of the shear forces is incorrectly drawn as if the column dimension were twice as large. I'll report that.