I am using RAM Connection for a base plate. The base plate has both uplift and shear concurrently. I was looking to see how the interaction of tension and shear was done on the weld and cannot figure out what RAM is doing to calculate the demand.
It says fv = V/Lshear and fa = P/L.
Lshear is half of L and Lshear + L usually add up to the perimeter of the shape, but not always (ie an HSS4x4x1/4 has Lshear = 6.6" and L = 13.2" for a total of 19.8", I am getting 15.2" for the perimeter).
Where in the code is it stated that Lshear should roughly 1/3 of the total perimeter and L should be 2/3 of the total perimeter? I have looked everywhere in the code and commentary.
Thanks,
Turner
The program is not accounting for any interaction in the design. I will bring this to the attention of the development team.
Answer Verified By: Turner Papendieck
So it seems that for axial load RAM Connection is counting on all flat sides (4)x(4-3x0.233) = 13.2", but for shear only the flat sides in the direction of the load are considered. Is there any type of interaction for shear and axial load acting concurrently? By the assumptions made 2 of the flat sides would have both shear and axial acting at once.
For HSS tubes, RAM Connection ignores weld along the radiused edge when determining the capacities. The length of the flat portion on each side is calculated as b (or h) less 3x the wall thickness (see AISC 360-16 B4.1b (d)). For a HSS4x4x1/4, the flat portion along any one side is 4-3*0.233 = 3.301 in. For shear capacity, only the 2 sides parallel to the shear is used so the length used is 2x3.301 in = 6.601 in.
Any updates on this question?
1638.example.zip
In the results see Column: See 'Elastic Method Weld Shear Capacity' and 'Elastic Method Weld Axial Capacity'