# NCEES (2017) Lateral Problem 118



## Br_Engr (Mar 19, 2021)

Good afternoon. 

Building question here.

The solution for Problem 118 refers to a 50% reduction in Grid 1. Does this 50% reduction really apply to Grid A? If the OMF along Grid 1 fails, that is a 33% reduction in the resistance in the N-S direction.

The way I understand it is that if there is a reduction in moment resistance greater than 33% in ANY perimeter framing (in any direction) the rho factor is 1.3. In this case if the moment resistance in either frame along the A or D grid lines is lost, that is a 50% reduction in strength for that story.

Thank you for your help.

Br_Eng.


----------



## mmarlow123 (Mar 19, 2021)

Im probably late on this..

The way I read the solution is that its stating that the percentage of the lateral load resisted by the frame on grid 1 is 50%. This is greater than 33% (hence the 50%>33%) so the redundancy factor is 1.3


----------



## Br_Engr (Mar 19, 2021)

But if there are TWO OMFs on Grid 5 that is a total of 3 OMFs to provide resistance in the N-S direction. Therefore, the OMF on Grid 1 only 1/3 of the total resistance in the N-S direction.


----------



## EBAT75 (Mar 19, 2021)

A seismic frame system must have functioning frames in two preferably orthogonal directions. If the frame on line 1 fails or say removed for the sake of code analysis, there is no longer a seismic framing system.The entire structure would have extreme torsional irregularity.

For Rho to be 1, there must be at least two bays of frames on either side. That is not the case here. Why Rho=1.3. Besides, the span/length is given, but the height is not given. A bay is length/height for rigid walls/frames; twice that for light frames.

If one of the frames on line 5 is removed for the sake of code analysis, in the N-S there would only be two frames, and the base shear in the N-S direction would be shared equally, 50% each. Why 50%.

BTW, the removal for redundancy test has to be in each orthogonal direction. Why counting a total of 5 for the entire structure is not applicable.


----------



## Br_Engr (Mar 19, 2021)

This clarifies things. So the 50%>33% is not really pertinent to the solution. Is that a correct statement?

Looking at an example in the 5th Edition of Seismic and Wind Forces by Williams, he computes the reduction as follows:

n = number of seismic resisting elements in a given direction

(n-(n-1)/n) * 100 

So in Problem 118, in the N-S direction that is (3-2)/3 = 1/3 => 33%, so that in and of itself is not reason to bump rho to 1.3.

Taking out seismic resisting element leaves 2 and since they are equal stiffness, V is equally shared.

Am I looking at this correctly?

(Bridges are so much easier.)


----------



## EBAT75 (Mar 19, 2021)

IMHO, instead of the formula based approach in William’s, for this kind of topics in Buildings, a fundamentals based approach makes it easier to conceptualize and understand the code provisions.

Was it you or maybe some other Bridges engineer who lamented about the dark secrets of Buildings in a post a few months ago?

I spent 10 years at the New York State DOT, the last 3 of which as a Team Leader, Bridges so I know what was meant by the dark secrets of Buildings and your saying Bridges are so much easier.

There is an asymmetric syndrome at work here. Buildings examinees think AASHTO questions as a pain in their brain and Bridges examinees think the opposite.


----------



## EBAT75 (Mar 20, 2021)

Whether my take on the solution is correct or not is best left for other Members to critique also. That is one of the purposes in joining a Forum.


----------



## Br_Engr (Mar 20, 2021)

LOL. I did make reference to the "dark arts" of building design.

And my pledge not to dabble still stands.


----------



## organix (Mar 20, 2021)

Br_Engr said:


> The solution for Problem 118 refers to a 50% reduction in Grid 1. Does this 50% reduction really apply to Grid A? If the OMF along Grid 1 fails, that is a 33% reduction in the resistance in the N-S direction.
> 
> The way I understand it is that if there is a reduction in moment resistance greater than 33% in ANY perimeter framing (in any direction) the rho factor is 1.3. In this case if the moment resistance in either frame along the A or D grid lines is lost, that is a 50% reduction in strength for that story.


My interpretation is slightly different. While I do believe violating both conditions in Section 12.3.4.2 for the story means you use rho = 1.3 in that direction (not both directions), I do not think it matters whether or not the LFRS is located on the perimeter for condition 'a'. Actually, location is not mentioned at all for condition 'a' so they can be anywhere. Also, based on my interpretation of the code, I am not sure why the solution would show the 50% > 33% calculation. I am not sure how it is relevant.



Br_Engr said:


> But if there are TWO OMFs on Grid 5 that is a total of 3 OMFs to provide resistance in the N-S direction. Therefore, the OMF on Grid 1 only 1/3 of the total resistance in the N-S direction.


For this problem, I would go the "conservative" route and argue that 1/3 is equal to 33.33% > 33%.


----------



## EBAT75 (Mar 20, 2021)

organix said:


> My interpretation is slightly different. While I do believe violating both conditions in Section 12.3.4.2 for the story means you use rho = 1.3 in that direction (not both directions), I do not think it matters whether or not the LFRS is located on the perimeter for condition 'a'. Actually, location is not mentioned at all for condition 'a' so they can be anywhere. Also, based on my interpretation of the code, I am not sure why the solution would show the 50% > 33% calculation. I am not sure how it is relevant.
> 
> 
> For this problem, I would go the "conservative" route and argue that 1/3 is equal to 33.33% > 33%.


I don’t see any mention by anyone of LFRS having to be located on the perimeter. I said preferably orthogonal directions. Orthogonal does not mean perimeter. It only means perpendicular to each other. Why preferably perpendicular? Because there can be non-parallel walls also. It is a different ball of wax with regard to computing stiffness in the two principal directions, modeling etc.

If LFRS is placed too far from the perimeter, it will not help diaphragms, collectors, chords, torsional forces. There may also be limitations on how far inwards the walls/frames can be placed. I think there is something about these in the Commentary.


----------



## Titleistguy (Mar 20, 2021)

The way I’d look at it for this problem is that even if removing that frame doesn’t trigger the 1.3 bc of reduced capacity I believe it will be triggered bc it means your building plan is no longer regular...

(Structures are regular in plan at all levels provided that the seismic force-resisting systems consist of at least two bays of seismic force-resisting perimeter framing on each side of the structure in each orthogonal direction at each story resisting more than 35% of the base shear.......)


----------



## organix (Mar 21, 2021)

Titleistguy said:


> The way I’d look at it for this problem is that even if removing that frame doesn’t trigger the 1.3 bc of reduced capacity I believe it will be triggered bc it means your building plan is no longer regular...
> 
> (Structures are regular in plan at all levels provided that the seismic force-resisting systems consist of at least two bays of seismic force-resisting perimeter framing on each side of the structure in each orthogonal direction at each story resisting more than 35% of the base shear.......)


That language is only present in the requirement for condition 'b' of the code. My understanding is that you can meet condition 'a' without getting into the discussion of whether or not the building is regular.



EBAT75 said:


> I don’t see any mention by anyone of LFRS having to be located on the perimeter.


See what I previously quoted (re-quoted below). I was referring to that in my original reply.


Br_Engr said:


> The way I understand it is that if there is a reduction in moment resistance greater than 33% in ANY *perimeter* framing (in any direction) the rho factor is 1.3.


----------



## EBAT75 (Mar 21, 2021)

Titleistguy said:


> The way I’d look at it for this problem is that even if removing that frame doesn’t trigger the 1.3 bc of reduced capacity I believe it will be triggered bc it means your building plan is no longer regular...
> 
> (Structures are regular in plan at all levels provided that the seismic force-resisting systems consist of at least two bays of seismic force-resisting perimeter framing on each side of the structure in each orthogonal direction at each story resisting more than 35% of the base shear.......)


----------



## EBAT75 (Mar 21, 2021)

Regular in plan at all levels doesn’t mean that if there are unequal number of frames or walls, it is irregular. I wish they worded it “at all stories “ and avoided the possibility of misinterpretation.

By regular at all levels is meant that there are no offsets in the framing layout, not similarity between grids. In this particular case, if the grid 1 frame has twice that of each of the frames on grid 5, it would be regular


organix said:


> That language is only present in the requirement for condition 'b' of the code. My understanding is that you can meet condition 'a' without getting into the discussion of whether or not the building is regular.
> 
> 
> See what I previously quoted (re-quoted below). I was referring to that in my original reply.


----------



## EBAT75 (Mar 21, 2021)

Sorry I must have pressed a button accidentally. Continuing,

Regular in plan at all levels - please critique my understanding.

Rho is not the real issue here. It is 1.3 through more than one path. 
1. This is Type 1a horizontal irregularity, bordering even on Type 1 b. Ratio being 1 1/3 cf 1.4 for 1b.

2. SDC D, 1.3 by default unless a. or b. In 12.3.4.2 is met. WRT each story and 35%, we don’t know how many stories there are. All we see is a roofing plan.

The crux of this topic is more than Rho. It is why 50%? Why not 1/3 rd?

I think that is where removal and it’s impact come into play.

ORGANIX: Yes, I missed the perimeter part mentioned in a post. My apology.


----------



## organix (Mar 21, 2021)

EBAT75 said:


> Sorry I must have pressed a button accidentally. Continuing,
> 
> Regular in plan at all levels - please critique my understanding.
> 
> ...


The diaphragm is flexible so it cannot be either torsional irregularity per Table 12.3-1.



EBAT75 said:


> 2. SDC D, 1.3 by default unless a. or b. In 12.3.4.2 is met. WRT each story and 35%, we don’t know how many stories there are. All we see is a roofing plan.


The problem states that this is a 1 story building, so this single story will take 100% of the base shear.



EBAT75 said:


> The crux of this topic is more than Rho. It is why 50%? Why not 1/3 rd?
> 
> I think that is where removal and it’s impact come into play.
> 
> ORGANIX: Yes, I missed the perimeter part mentioned in a post. My apology.


Yup, that part about 50% confuses me still. I assume I either am missing something in the code or they made an error of sorts. 

No problem about the other item.


----------



## EBAT75 (Mar 21, 2021)

organix said:


> The diaphragm is flexible so it cannot be either torsional irregularity per Table 12.3-1.
> 
> 
> The problem states that this is a 1 story building, so this single story will take 100% of the base shear.
> ...


I do not have this 2017 Samples book. I was working with my iPad and all I could see is from Design Code and downwards. Why I have missed the single story part.

To more important things:

1. Yes, the diaphragm is flexible according to the problem. Again, the diagram in the solution misled me. Being straight line and no deflection diagram, my brain was wrongly led to think on the lines of a deflection diagram, and got mixed up. Not trying find excuses. Is the 0.5 inch drift on Grid 5 an assumed drift based on the 1.0 in on Grid 1 and because the ratio of stiffness 2 (Gris 5): 1 (Grid 1). Shouldn't the diaphragm be rigid for this proportionality to hold? I see only MMD and the 1" drift on Grid 1, no mention of 0.5 inch on Grid 5 (Hope I did not miss seeing that also!)

2. Whether this even fits the "rigid or semi-rigid" in Type 1a, 1b - Table 12-3.1 is open to question. Please look at Secs.12.3.1.1, 12.3.1.3. Can the drifts be _computed_ by proportioning not knowing whether the diaphragm is rigid, semi-rigid or flexible?

"The diaphragm is flexible so it cannot be either torsional irregularity per Table 12.3-1." - I understand the point. But this layout is intuitively difficult to see how there can be no torsion.

That said, I think here is the answer to the million dollar question. - Why 50% ? The answer is in 12.8.4.1. 

Any thoughts? An interesting way to try to understand some of what Br_Engr once described as "dark arts" of Buildings before the 22/23 April gruel.


----------



## EBAT75 (Mar 21, 2021)

Titleistguy said:


> The way I’d look at it for this problem is that even if removing that frame doesn’t trigger the 1.3 bc of reduced capacity I believe it will be triggered bc it means your building plan is no longer regular...
> 
> (Structures are regular in plan at all levels provided that the seismic force-resisting systems consist of at least two bays of seismic force-resisting perimeter framing on each side of the structure in each orthogonal direction at each story resisting more than 35% of the base shear.......)


12.3.4.2b. is the source you’re referring to.

Your take is correct on “regular”, 2 bays etc. It would have helped if they broke this into 2 parts one covering frames, the other covering shear walls. The term bays seems to mean differently between the two.

A bay in frames seems to mean a single frame, regardless of length/height ratio. I am not sure why.

In shear walls on the other hand, a bay one when length/height ratio is 1 for concrete or masonry, twice the length for the same height for light frame. I mixed both sentences up.

Going back here there are 2 bays on one side and only 1 on the other. Also, the regular test must be met in each orthogonal direction as well. Has only one moment frame on each side in the orthogonal direction - E-W.

Rho was not the real issue in this as I alluded to in the last post. It was why 50%. I think the answer is in Sec 12.8.4.1.

Why the base shear would not be shared by what is left when a beam or frame fails, I fail to understand. I thought that is what redundancy is all about and designing for that eventuality. 

Please feel free to comment. Why we are here.


----------



## organix (Mar 22, 2021)

EBAT75 said:


> I do not have this 2017 Samples book. I was working with my iPad and all I could see is from Design Code and downwards. Why I have missed the single story part.
> 
> To more important things:
> 
> ...


Now I’m the one with a limited screen (replying on my phone during a break) so I’m sorry if I’m missing something, but anyway...

1. For a flexible diaphragm, the load is distributed like a simple beam. With the center of mass assumed to be in the middle of the building, you get equal shear in the LFRSs located at the perimeter regardless of stiffness. With the load being equal at both and the stiffness being equal to 2 times the other, the drift will be half. The figure can appear misleading, but the intent is to find the average drift for the code check. It is not to imply there is a linear relationship along the perpendicular edge of the diaphragm (which has the deflection given in the problem statement) for the building. Anyway, the drift value is correct based on the load and stiffness of the LFRSs for the reason stated above. 

2. Yup, I get it. It is weird to see that and not think there’s some twisting. But I guess this is a simplified code method and that’s the assumption the code allows us to take for a flexible diaphragm to allow for a less complicated analysis.

I think drifts can be computed as I described above in item 1 based on the load at each LFRS and the stiffness. I think that’s probably one of the key aspects of the problem that the writer wanted us to learn using the procedure described in Section 12.3.1.3. 

I took a quick look at Section 12.8.4.1 and I’m afraid I’m not understanding where you were going with it. This section would seem to describe the method of a simple beam as I brought up in item 1 above. And otherwise, the comparison to 33% is not mentioned here. I can probably assume why a “50% > 33%” comparison was made in the solution, but I don’t think it was necessary and possibly not an accurate interpretation of the code.


----------



## EBAT75 (Mar 22, 2021)

I understand what you say. IMHO, some of the ways the question is formulated and the solution presented could be better.

I still have some lack of understanding how a 65 ft high OMF (now allowed with conditions, earlier max 35 ft) spanning 30 ft can be considered a 
“bay”. Concrete of light frame shear walls have height/span defined. This problem does not mention the height of the frame, unless I am missing it. I confess I am not a screen guy. A paper reader, why I must pass the exams before they go CBT!

Between the 33, 33 1/3, 35%, I am lost. Why they did not pick 35 as a common number is lost on me. It avoids the borderline issues. My penny there. I agree the way in presenting the solution with 50% > 33% is also not a helpful thing.

Coming to your question about where I was going with it, it has to do with the apportionment of lateral loads based on the distribution of mass on the diaphragm in flexible diaphragms. Please let me know if that is not the import of the second and last sentence in 12.8.4.1.

This being assumed to be a udl, the seismic loading will be uniform and hence shared by the MFRS in proportion to tributary area/mass; 50% here. I am to be convinced of this without a limiting height/length qualifier for frames also, like concrete/masonry, or light frame shear walls Any takers?


----------



## Br_Engr (Mar 22, 2021)

OK. I have read all of your responses (thank you for all of them) and gone back to codes, examples problems and ASCE type guides and this is how I see the solution.

The MDD > 2 x ADVE check confirms that the diaphragm is flexible. As a result, all conditions/requirements pertaining to any extreme torsional irregularity do not apply as they only pertain to rigid diaphragms.

ASCE 12.3.4.2

Rho shall be taken as 1.3 unless one of the following conditions is met:

Paragraph a.) – Each story shall comply with the requirements of Table 12.3-3

Table 12.3-3 Moment frames – “Loss of moment resistance at the beam-to-column connections at both ends of a single beam would not result in more than a 33% reduction in story strength …”

If any single OMF in the N-S direction loses moment resistance at the beam ends, total strength is reduced by 33.33% (> 33%; Organix).

*CONDITIONS OF PARAGRAPH “A” ARE NOT MET.*

Paragraph b.) – Structures that are regular in plan at all levels provided that the seismic force-resisting systems consist of at least two bays of seismic force resistance perimeter framing on each side of the structure on each orthogonal direction…

Three sides only utilize ONE bay of SFR systems.

*CONDITIONS OF PARAGRAPH “B” ARE NOT MET.*

THEREFORE RHO = 1.3.

Moving on…

ASCE 12.8.4.1

“…For flexible diaphragms, the distribution of forces to the vertical elements shall account for the position and distribution of the masses supported.” Tributary area.

Since the problem specifically asks for the force in the frame along grid 1, it is assumed that that is not the frame that loses moment resistance at the beam column connections. That leaves the frame on Grid 1and the remaining frame on grid 5 to share the load. In this case 50% each.

The 50% > 33% is meaningless.

Whew...


----------



## mmarlow123 (Apr 6, 2021)

I feel the 50%>33% is meaningful. The way I see it,

Grid 1 and 2 each provide 50% of the story strength (flexible diaphragm).

If you remove the frame at grid 1 there is a reduction in story strength of 50%, which is greater than 33%. So P is 1.3.


----------



## organix (Apr 6, 2021)

mmarlow123 said:


> I feel the 50%>33% is meaningful. The way I see it,
> 
> Grid 1 and 2 each provide 50% of the story strength (flexible diaphragm).
> 
> If you remove the frame at grid 1 there is a reduction in story strength of 50%, which is greater than 33%. So P is 1.3.


Story strength will be the sum of the strength of each LFRS in each story. In this case, there are 3 of the same types of LFRS. If you lose one, regardless of plan location, you are down 1 of 3, or 33.33%. In other words, the way the code is written, you can have all frames (say 4 frames) on one side of the building and comply with the strength requirement (1/4 < 33%). Admittedly, this is a very poor design idea, but it is what it is (just an exaggerated example).

What you stated above are the reactions from the lateral force. Half will go to each line of the LFRS. However, that is not the same as the strength of the story. Column line 1 strength is not equal to Column line 5 strength and the strength of the system is independent of the load it experiences.


----------



## EBAT75 (Apr 6, 2021)

Even hypothetically, there cannot be 4 frames all on one side. Code calls for at least 2 bays on either side of the C.G. That is not even met here. Enough to interpret a 1.3 Rho.

In general cases, the definition of bays also comes into play. A moment frame is 1 bay regardless of length to height ratio. Wood or light metal frames have a ratio of 2 to qualify as 1 bay. I have alluded to this ambiguity in an earlier post.

There is reference to story strength. I think in a Lateral loading, stiffness/rigidity is the governing parameter. Story strength would be the Vertical load carrying capacity.

Finally if there are no frames on one side e.g. as in open front stores, will it not be an irregular structure (Horizontal Type 1a/b) even if the diaphragm is not rigid, and as such become a structure for which Equivalent Lateral Force Analysis in not permitted? Why the minimum of 2 bays is called for. Goes beyond even the 1.3 Rho and into Omega overstrength combos with results of the Sec 12.9 or Ch. 16 analysis?

Some of this seems to fall into the chicken/egg trap when it comes to displacement/presumptive rigidity.

Anyways, to those waiting for the Big Day(s) happy prepping. In two weeks it will all come to pass (pun intended).


----------



## organix (Apr 8, 2021)

EBAT75 said:


> Even hypothetically, there cannot be 4 frames all on one side. Code calls for at least 2 bays on either side of the C.G.


Where? If you're referring to 12.3.4.2.b, it is completely irrelevant if you meet 12.3.4.2.a.


----------



## EBAT75 (Apr 8, 2021)

organix said:


> Where? If you're referring to 12.3.4.2.b, it is completely irrelevant if you meet 12.3.4.2.a.



This is a good exercise in code interpretation.

12.3.4.2.a. is even a non-starter because this is a single story, 100% of base shear is *always *shared regardless of walls layout. So a. is the one that is “completely irrelevant_“, _not the other way around.


----------



## organix (Apr 13, 2021)

EBAT75 said:


> This is a good exercise in code interpretation.
> 
> 12.3.4.2.a. is even a non-starter because this is a single story, 100% of base shear is *always *shared regardless of walls layout. So a. is the one that is “completely irrelevant_“, _not the other way around.


And why wouldn't it apply to a single story? I'm guessing you're referring to "each"? Is that the basis of your interpretation?


----------



## EBAT75 (Apr 13, 2021)

organix said:


> And why wouldn't it apply to a single story? I'm guessing you're referring to "each"? Is that the basis of your interpretation?


Please read the second sentence in 12.3.4.2 “For other structures...shall be equal to 1.3 unless one of the following two conditions is met.....”. a. and b. are the two conditions.

My post was not saying a. did not apply to one story. It applies to any number of stories. Each story ....more than 35%..... when applied to the case of a single story building such as this, by default 100% of that story shear is resisted - why I said a. is a non-starter for checking to see if a. is satisfied in cases of a single story building.

So, only b. is left. Under b., again it can be any number of stories. Under b.also a single story would carry 100% of base shear. But the determinant here is again not the more than 35% of base shear being required to be shared by each story, but the number of bays.

This example has 2 OMFs on grid 5, only 1 OMF on grid 1. If an OMF is 1 bay, there aren’t at least 2 each in E-W. That makes it not meeting b. even discounting that there is only 1 each in the N-S which by itself would have made Rho to be 1.3.

In a single story building, because a. is automatically satisfied, unless there are 2 bays in both orthogonal directions, Rho would always be 1.3; if there are 2 bays in both orthogonal directions, Rho would always be 1.0 as allowed by the second sentence in 12.3.4.2.

I have gone into this question of bays earlier. So that is a “hanging chad”.

In sum, it was not about “each”.


----------



## organix (Apr 14, 2021)

EBAT75 said:


> Please read the second sentence in 12.3.4.2 “For other structures...shall be equal to 1.3 unless one of the following two conditions is met.....”. a. and b. are the two conditions.
> 
> My post was not saying a. did not apply to one story. It applies to any number of stories. Each story ....more than 35%..... when applied to the case of a single story building such as this, by default 100% of that story shear is resisted - why I said a. is a non-starter for checking to see if a. is satisfied in cases of a single story building.
> 
> ...



Do you realize the language for “each story resisting more than 35% of the base shear” is a qualifier in both sections .a and .b? It’s exactly the same phrase. There is no reason to believe there is a “non-starter” for one that isn’t there for the other. So with with said, I don’t follow your logic or interpretation. 100% of base shear is more than 35% of base shear so even the only story of a one story building needs to meet the criteria or either section. The sole purpose of this code criteria is to eliminate the need for stringent requirements at the top of taller buildings. 

Also, I have seen many examples that prove your interpretation incorrect, but you can start with ICC’s SEAOC Structural/Seismic Design Manual - Volume 1: Code Application Examples, Design Example 26 as an example. They look at a single story for this very language and qualify it per section .a criteria. ASCE’s guide to seismic loads would be another as the author does a more in depth analysis on a 1 story building for both cases. There are countless practice problems also. 

Further, the mention of 33% in the problem solution would be that much more irrelevant if section .a doesn't apply.


----------



## EBAT75 (Apr 14, 2021)

organix said:


> Do you realize the language for “each story resisting more than 35% of the base shear” is a qualifier in both sections .a and .b? It’s exactly the same phrase. There is no reason to believe there is a “non-starter” for one that isn’t there for the other.


If only you had read it dispassionately you would have realized that was exactly what I had said. Your earlier post ran thus:
_Where? If you're referring to 12.3.4.2.b, it is completely irrelevant if you meet 12.3.4.2.a._ (underline mine).

Your reference was to a. only there. My focus was on a. for that simple fact and reason. I had expanded that to cover b. as well in the last response. It is not as if you found it.



organix said:


> So with with said, I don’t follow your logic or interpretation. 100% of base shear is more than 35% of base shear so even the only story of a one story building needs to meet the criteria or either section. The sole purpose of this code criteria is to eliminate the need for stringent requirements at the top of taller buildings.


“100% of base shear is more than 35% of base shear...”. Wasn’t that very obvious thing I had said in my last post?

Your rant about the countless examples is redundant. The SEAOC example 26 for instance (I looked at it just now for the first time) gives the height as 18 ft, length is 15 and goes on to refer to b. and says the height being more than the length, it is NOT 1 bay. I may be missing something but where is the definition of a bay whether in b. or elsewhere.? I will be obliged if I am enlightened on this. If you look back, you will see that I had questioned this very thing by asking whether 65 ft high OMF now permitted in SDC D under certain conditions is a bay when this is only 30 ft long OMF.




organix said:


> The sole purpose of this code criteria is to eliminate the need for stringent requirements at the top of taller buildings.



*I can go into all of these topics again after the exam*. For the moment, I may be wrong but my recollection is that it was inverted V frames that prompted this. They cause unbalanced lateral force in the roof beam. In taller buildings this will cause an undeserved penalty on lower floors.

*A note to all: Whenever there is a posting, I see hundreds of views but only under a handful of members contribute with their knowledge or understanding. It will be helpful if more members give a kick at the can.*


----------



## EBAT75 (Apr 14, 2021)

Correction: “They cause unbalanced lateral force in the roof beam....”. Meant to say unbalanced vertical force.


----------



## EBAT75 (Apr 14, 2021)

I just wanted say this until I get back after the exam. I looked at the SEAOC example again.

Sure, the definition of a bay is given for shear walls in b. in terms of height to length ratio. Are moment frames also defined as shear walls? May be but it defies my logic as the failure mechanisms are different and OMF can even go as high as 65 ft. Can moment frames and concrete or even light frame shear walls be in the same league? They have different geometric and structural limits.

The SEAOC example cited is a concrete shear wall. I looked at this SEAOC example fast for the first time this morning. I thought the comparison was to OMF. Sure the example correctly concludes that they are not 1 bay. No problem there, but the example we are looking at here is OMFs.

The height of these OMFs is not given. Why I was saying we cannot meet b. (even though both a. and b. are applicable to any under the second sentence in 12.2.4.) as the number of bays cannot be determined.


----------



## Titleistguy (Apr 14, 2021)

Well played calling a persons explanation redundant in a topic regarding redundancy. I appreciate topical puns.


----------



## organix (Apr 14, 2021)

Well, I guess I am now guilty of replying from my phone and missing large portions of your post. I feel like I’m losing my mind a bit cause I’m convinced it wasn’t there even though it clearly was. Pretty big oversight for me. It also seems I got tangled and forgot the original point being discussed... also I got confused by your definition of satisfied. For .a to be satisfied, Table 12.3-3 also needs to be satisfied, not just base shear. I know you know that, but that’s maybe some of the misunderstanding I’m having when you call it satisfied. I’m sorry for that. I have created a lot of useless back and forth there. 

Anyway, for simplicity, I will reply to the questions and relevant items.



EBAT75 said:


> In a single story building, because a. is automatically satisfied, unless there are 2 bays in both orthogonal directions, Rho would always be 1.3; if there are 2 bays in both orthogonal directions, Rho would always be 1.0 as allowed by the second sentence in 12.3.4.2.


Why are only 2 bays required in each orthogonal direction to get a rho of 1.0? I would say 4 are required. Some could argue that only 3 are required. It’s not something I’d defend to the death either way, but I’d probably go with 4 to be on the safe side... well, for moment frames.



EBAT75 said:


> I may be missing something but where is the definition of a bay whether in b. or elsewhere.? I will be obliged if I am enlightened on this. If you look back, you will see that I had questioned this very thing by asking whether 65 ft high OMF now permitted in SDC D under certain conditions is a bay when this is only 30 ft long OMF.


Bays are only defined in section .b and are only defined for shear walls. I do not believe any frame systems (steel or concrete) require additional guidance.

For the latter question, I would say it is allowed in that scenario as I’m not aware of any reason it wouldn’t be. It would just need to be designed accordingly to meet design loads. I’d say it’s unconventional enough that a peer check would be good on this one.


----------



## EBAT75 (Apr 15, 2021)

@organix, these things happen. They will not make or break the exam and can wait. This is not the only item I have a beef with (actually I am a veggie guy!) in some code/specifications arena.

Take it easy, good luck at the exam.


----------

