Robert - for Case 2 you used the adjustments (2/3 and 0.5) which are intended for sidesway uninhibited frames. Case 2 is sidesway inhibited. Therefore, shouldn't GB be:
1
------------------------ = 0.286
(1 x 2.0) + (1 x 1.5)
JAE: Thank you very much for your answer regarding Case2.
Now, I am still frustrated for the two answers in Case1. Don't know which way I should calculate the GB value in the coming Structural-1 exam, or to say I don't know which answer NCEES is considered as right answer. As NCEES designated AISC 13th as the required Code, but NCEES's sample question and solution shows the other way.
Robert, i believe you can fill out a form at the end of the exam to bring light to the situation. I participated in an NCEES scoring event and they took these forms from the examinees pretty seriously. I wouldnt sweat it. You could also write to them now.
I think the NCEES answer shown is correct. The 2/3 factor applies when the far end is fixed, which I take to mean rigidly fixed to a boundary support. The girder is fixed to the column, but I think the standard 1.0 factor on EI/L would be used.
I know "Alan Williams-SE Reference Manual" used the same way as NCEES sample question and solution. That is why I feel very uncomfortable. I think both "Alan Williams-SE Reference Manual" and "NCEES sample question and solution" were not following the "AISC 13th" on this G ratio.
Let's take a logic look, the more rigid, the far end of a girder, the smaller the GB value will become, so that the more larger a factor should be multiplied to denominator. so if 1.0 factor is applied to girder fixed to a column, a larger than 1.0 factor (not 2/3) factor should be applied to girder fixed into boundary. Just as JAE indicated above for Case2, once the girder is fixed into boundary, the frame is no longer a sway frame, but becomes a sideway inhibited frame, and its GB value becomes very small by applied a larger factor to(EI/L)_girder.
The "fixed end" condition refers to rotation. In other words, if the connection is such that the joint cannot rotate, then you apply the 2/3 factor. In this case the joint can rotate, so you do not need to apply the 2/3 factor.
Your logic is based on an incorrect assumption. Look at the derivation of the k factors. For a sway frame, it is based on a beam with stiffness of 6EI/L (reverse curvature). When you have fixity at the FAR end, your stiffness is 4EI/L.
Actual stiffness / Assumed Stiffness= 4/6 = 2/3.
And the fact that the far end doesn't rotate has NOTHING to do with whether or not it is a sway or non-sway condition.
I think I figured out what's happened to this GB ratio:
"Alan Williams-SE Reference Manual" was published in 2006, at that time, NCEES still designated AISC 3rd edition as the exam code. AISC 3rd edition only apply 0.5 to pinned far end and no factor is applied to fixed far end (see AISC 3rd edition, 16.1-193). So "Alan Williams-SE Reference Manual" is correctly written at that time. From 2008, NCEES upgraded to designate AISC 13th as the exam code. "NCEES sample question and solution" just took over the previous G-ratio calculation method without noticing AISC 13th has added a new 2/3 factor to the fixed far end girder. That caused current awkward situation.
I would still like to discuss it in a logic point of view. Logically, if the far joint rotation is restricted a larger factor than rotation joint should be applied, not the other way round. Other wise it will lead to a ridiculus conclusion.
I don't think so. What you are saying is that they have completely changed the way that K values are computed. This would mean that any standard moment frame would require the 2/3 factor on the girder before entering into the alignment chart?
I don't have a copy of the 3 ed. with me. But lets look at this "logically".
The AISC effective length method has been around for a while. Do you think that after a few decades they'd suddenly say "oops, forgot about that stiffness modification factor.. lets add it"
Like I said in my previous post, the factor is to account for buckled shapes other than assumed in the development of the k factors.
A column in a sway frame is braced by beams at the top in both directions. For the column to buckle in a sway mode, the beams would have to deflect (buckle) in some manner. The K factors were developed assuming the beams that brace the columns buckle in double curvature (take out a flexible ruler, and twist both ends clockwise with your fingers).
The case of a fixed joint at the far end occurs when, for example, the far end frames into a beam that is much more stiff than the beam in question, thus preventing the far side from rotating. Take the same ruler and only rotate the left end counterclockwise, while maintaining the right side without rotation. Which case requires more effort on the left fingers to achieve the same rotation?
Where is the examples you mentioned that come with the 13th Edition that do not use the 2/3 factor on the moment frame girders. Where can I get this example? I am curious about it.
