NCEES SE practice exam, Problem 603: Ln span lengths

[JDstruc]

I'm working on Problem 603 of the NCEES Structural SE practice exam. It is a 1-way slab problem, regarding moments and shears in the beam.

My question is regarding how they are calculating "Ln" for various conditions, shown on page 100. The center-to-center span is 30'-0", and the columns are all 24" square. I would assume that Ln is the clear distance, which would be 30' - 24"/2 - 24"/2 = 28'-0". Yet the solution is using a different value for Ln...in fact, they are using multiple values. From the solution:

End span, Ln = 30.0 - 2.0 - 2.0/2 = 27 feet

Interior span, Ln = 30.0 - 2.0/2 - 2.0/2 = 28 feet

Ln = (27 + 28) /2 (average...???)

They use Ln=27.0 for the Positive Moment M+ = wLn^2/14

They use Ln=27.5 for Negative Moment M- = wLn^2/10 or wLn^2/16 (at 1st interior support, and at exterior column respectively)

Then they use Ln=27.0 for the shear reactions, V = 1.15wLn/2 or wLn/2 (at end member, and at all others)

So...

1) Why is Ln not simply 28.0 for all conditions?

2) Why are they using different Ln values, for calculating various items for THE SAME BEAM?

I'm rather confused. Any clarification would be greatly appreciated. Thanks all.

 

[JDstruc]

Bah....even though I've been staring at this thing for the past 2 hours, literally 3 minutes after posting this thread IT HIT ME.

I re-looked at the problem statement closely, realizing that the outer 30'-0" grid went to the outside face of the building, and not to the column centerline. Geez. So yes:

End Span Ln = 30' - 1' - 2' = 27'

Int Span Ln = 30' - 1' - 1' = 28'

Avg Ln = 27 + 28 / 2 = 27.5

Ln = 27.0 is used for all calculations, except for Negative Moment M-, which per ACI 8.3.3 shall be the average Ln for M-.

The lesson I guess is LOOK VERY CLOSELY AT THE PROBLEM STATEMENT AND GIVEN DIMENSIONS AND INFORMATION.

Exam is just in 4 days from now...I fear that I made this mistake, so close to the exam O__O