NCEES (2017) Lateral Problem 118

[Br_Engr]

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]

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]

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]

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]

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]

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]

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]

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

And my pledge not to dabble still stands.

 

[organix]

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.

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]

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]

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]

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.

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.

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]

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]

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

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]

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]

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.

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

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.

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]

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.


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.

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]

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]

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.

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]

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?