RAM Connection Column Splice - Demand Calculation

RAM Connection follows the recommendations from “Handbook of Structural Steel Connection Design and Details” by Akbar R. Tamboli” for AISC column splices. Specifically Section 2.2.5 Splices – columns and truss chords, from page 115 to 123.

The simplest method for designing column splices for biaxial bending is to establish a flange force (required strength) which is statically equivalent to the applied moments and then to design the bolts, welds, plates, and fillers (if required) for this force. For minor axis bending, the force F needs to be distributed over some finite bearing area as shown in figure:

The bearing area is (2∙ε∙t) where t is the thickness of the inner flange, ε is the position of the force, F, from the toe of the flange of the smaller column, and T is the forcer per gage line of bolts. The quantities T and F are for each of the two flanges. If My is the weak axis applied moment, Mf = My/2 is the weak axis applied moment per flange.

The bearing area is determined by requiring that the bearing stress reaches its design strength at the load, F. Thus, 0.75 (1.8Fy)(2 ε)t = F, and since from vertical equilibrium F = 2T and 0.75 (1.8 Fy) t ε = T.

Thus, Mf = 0.75 (1.8 Fy) t ε (b – 2 ε)

And solving for ε:

Where Mpy = Fy Zy = ½ Fy t b^2. This expression for ε is valid as long as:

When Mf > 3/8 φ Mpy, the tension, T, on the bolts on the bearing side vanishes and the next figure applies. In this case, F = T = 0.75 (1.8 Fy) t (2 ε).

And

Where γ = 1 + g/b. This expression for ε is valid as long as:

But T need can never exceed Mf/g. The flange force in every case is Ffy = 2T. Using the user’s model data:

g = 0.5 in.

Mf = 40 kips / 2 = 20 kips.

Mpy = Zy * Fy = 0.07638 * 7200 = 550 Kip*in

γ = 1 + g/b = 1 + 0.5/0.9916 = 1.5042

F = 2 T;

F = 2 * (φ (1.8 Fy) t ε) =  2 * (0.75 * 1.8 * 7200 * 0.1441 * 0.01373 = 38.497 Kips

F max = 2 * Mf / g = 2 * (20 / 0.5) = 80 Kips

As the flange force does not exceed the maximum, RCnx uses the calculated F value of 38.5 Kips. RAM Connection calculates the maximum flange force acting on the column about either the major or minor axis plus the tension in the in the flange. The user’s model only has data for minor axis calculations. As he noticed, the demand differs when he uses ASD or LFRD and the reason is the calculation has φ as a parameter.

I hope this helps.

RAM Connection follows the recommendations from “Handbook of Structural Steel Connection Design and Details” by Akbar R. Tamboli” for AISC column splices. Specifically Section 2.2.5 Splices – columns and truss chords, from page 115 to 123.

The simplest method for designing column splices for biaxial bending is to establish a flange force (required strength) which is statically equivalent to the applied moments and then to design the bolts, welds, plates, and fillers (if required) for this force. For minor axis bending, the force F needs to be distributed over some finite bearing area as shown in figure:

The bearing area is (2∙ε∙t) where t is the thickness of the inner flange, ε is the position of the force, F, from the toe of the flange of the smaller column, and T is the forcer per gage line of bolts. The quantities T and F are for each of the two flanges. If My is the weak axis applied moment, Mf = My/2 is the weak axis applied moment per flange.

The bearing area is determined by requiring that the bearing stress reaches its design strength at the load, F. Thus, 0.75 (1.8Fy)(2 ε)t = F, and since from vertical equilibrium F = 2T and 0.75 (1.8 Fy) t ε = T.

Thus, Mf = 0.75 (1.8 Fy) t ε (b – 2 ε)

And solving for ε:

Where Mpy = Fy Zy = ½ Fy t b^2. This expression for ε is valid as long as:

When Mf > 3/8 φ Mpy, the tension, T, on the bolts on the bearing side vanishes and the next figure applies. In this case, F = T = 0.75 (1.8 Fy) t (2 ε).

And

Where γ = 1 + g/b. This expression for ε is valid as long as:

But T need can never exceed Mf/g. The flange force in every case is Ffy = 2T. Using the user’s model data:

g = 0.5 in.

Mf = 40 kips / 2 = 20 kips.

Mpy = Zy * Fy = 0.07638 * 7200 = 550 Kip*in

γ = 1 + g/b = 1 + 0.5/0.9916 = 1.5042

F = 2 T;

F = 2 * (φ (1.8 Fy) t ε) =  2 * (0.75 * 1.8 * 7200 * 0.1441 * 0.01373 = 38.497 Kips

F max = 2 * Mf / g = 2 * (20 / 0.5) = 80 Kips

As the flange force does not exceed the maximum, RCnx uses the calculated F value of 38.5 Kips. RAM Connection calculates the maximum flange force acting on the column about either the major or minor axis plus the tension in the in the flange. The user’s model only has data for minor axis calculations. As he noticed, the demand differs when he uses ASD or LFRD and the reason is the calculation has φ as a parameter.

I hope this helps.

What if there is a setback?
If there is a setback (the gap between the top & bottom column), there will not be a bearing area. 
Though RAM connection gives the option to add setback in the joint parameters, it is not considered in the design. It looks like a #bug in the software.

Also, if there is an axial tension + major axis moment + minor axis moment acting on the column splice, how are forces at the flanges calculated?
Shouldn't it be the addition of flange forces due to axial tension + major axis moment + minor axis moment??

As per your above reply, 

Seth Guthrie said:

RAM Connection calculates the maximum flange force acting on the column about either the major or minor axis plus the tension in the in the flange.

why flange force due to major axis moment & minor axis moment is added together?