Steel beam columns subject to flexure and compression

[ENGR_2321]

I’ve been studying this topic and I know to use CH H formulas for combined forces of the AISC code it is pretty straight forward. However when a column size is unknown (in other words, designing for the lightest W shape for example) there are several procedures one can go about. I investigated the procedure presented by John K Snyder (aisc engineering journal report) but this method in that report is limited to W shapes with moments about strong axis only. Therefore I studied another method by Aminmansour (2000) which the AISC referred me to and the part I am having trouble finding is the m-factor and U-factor. Aminmansour himself provided tables with m and b factors but those aren’t the factors I am referring into. His factors are just labeled similarly and it’s for when using his method which has limitations as well so I am hesistant to use that method and I am not referring to those m and b factors. In the report, he also presents the regular AISC current method and explained it, which is the design method I am trying to use because it has less limitations. According to their samples and steps, they use a set of m-factor and u-factors (not to be consfused with the coefficients of part 6 current steel code)and it seems that they extrapolate it from Table 3-2 of the 1994 AISC manual, but this means it can also be calculated. Where are the m and U formulas at or when can I find them ? If anyone knows or is familiarized with what I am talking about. It seems that the m and u factors need to be calculated first in order to be able to convert your moments into “an equivalent axial loads” and then just add to the given axial load. Thereafter this summation result is looked up in the column capacity tables to pick a trial section which therefafter is easy just plug into the ch H equations. The hard part is calculating m and U however. I can’t find those formulas. Thank you

 

[dik]

Write a little spreadsheet and plug in the numbers... depending on your mix of M and P, start with a section stressed to approx 25ksi, and, go from there.

[Added] What I used to do was determine the Euler buckling stress for a section based on the lu and use that stress for a combined P + M/S... good ballpark

Dik

 

[Hokie93]

Is there a particular reason you are following the approach included in the 2nd edition (1994) AISC LRFD Steel Construction Manual, rather than the method included in more recent Manuals? The 13th and 14th edition AISC Steel Construction Manuals use a different approach in Part 6 that do not include the m and u factors. The 15th edition Manual takes a different approach in Part 6 yet.

 

[ENGR_2321]

I am using the 14th edition steel code. They do not show the m and u factors on part 6; they only show the two interaction formulas. However, they do refer me to the Aminmansour article (2000) for more info on how to use the current practice procedure. In other words, supposedly the m and u factors are part of the current practice, and their purpose is to be able to reduce the number of trial sections to be checked. But they get the m and u factor equations from 1994 table 3-2. I'm assuming that's where they have the formulas perhaps. Does that mean they are implying to go get a copy of the 1994 code to use this table?