max load per foot that can be imparted to the x-axis of a beam

[orticio]

While doing a problem of a HSS 18x6x3/8 subjected to biaxial bending, where the distributed load in the y-axis is known, and therefore it is possible to solve for the max allowable Moment that can be applied in the x axis at midspan, I was shocked to see that the solution does not substract the distributed load coming from the selfweight itself when finally solving for the distributed load that causes that moment (typical M = wL2/8). It makes the answer to change whether you take it or not into account.

When I read "max load per foot that can be imparted to the x-axis of a beam" I think of an external load applied to the beam, instead a load also including the selfweight of the beam, which is what the book is assuming (PPI's steel design for the PE exam). The selfweight is of course a load, but I think it is pretty confusing to use the words "can be imparted" if the problem already gives you the HSS shape, therefore the selfweight is fixed to be 58.1 lb/ft...

What do you guys think?

 

[phecke]

When doing practice problems for the Structural PE (or for the SE for that matter) you really need to think about how the problem relates to actual structural design in the real world. So in the instance of this problem, how would you design a beam in the real world?

You would include the self-weight with whatever the applied distributed load is.

Think about it, is there ever a case in actual design where a beam spanning something wouldn't have to support itself as well? Not that I can think of. So, unless it says NOT to consider self-weight, always add it.

 

[gogo]

IMO if it says what load can be imparted and the section size is given, the answer should not be including the self weight - reference my own experience.