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Where to use Redundancy Factor=1.3?
3

Where to use Redundancy Factor=1.3?

Where to use Redundancy Factor=1.3?

(OP)
If I have a Symmetrical Special moment resisting frame in SDC D with 3 bays in one direction and 8 bays in another direction and I am solving it in ETABs and have enabled P-Delta effects, will I use redundancy factor=1 or =1.3. Is there any way to access redundancy factor without calculation if not then how the calculation is made?

RE: Where to use Redundancy Factor=1.3?

Reundancy factor is a penalty factor applied to less redundant structures for SDC D thru F and for extreme torsional irregularities for SDC D. Notice that extreme torsional irregularities are not allowed for SDC E and F .


Look to ASCE 7 section 12.3.4 which describes the methodology for determination of the redundancy factor, ρ .

There are two values of the redundancy factor: 1.0 and 1.3 . Pls look to Table 12.3-3 ( Requirements for Each Story Resisting More than 35% of the Base Shear) for t he calculation requirements..If the failure of a single element of the seismic force-resisting system will result 33% reduction in story strength, or does the resulting system have an extreme torsional irregularity, penalty factor of 1.3 is used.

There are two tests ; aspect and calculation ..If you post the structural plan, you will get comments for aspect , regarding calculation, you shall carry out the calculations mentioned Table 12.3-3 .

RE: Where to use Redundancy Factor=1.3?

(OP)


I have no extreme torsional irregularities so can I use ρ=1?

RE: Where to use Redundancy Factor=1.3?

yes use 1

RE: Where to use Redundancy Factor=1.3?


Aparently , the picture depicts , thre is no plan irregularities..However , visiually fails the configuration test described in subparagraph (b) of 12.3.4.2 Redundancy Factor, ρ, for Seismic Design Categories D through F.

I have copied and pasted that para.: (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.......)

You are expected to provide two V.B. (for steel SMRF ) or SW ( for R.C. SMRF) at short direction , far ends...

If you hesitate to add SW or VB, you are expected to use penalty clause ρ=1.3.

I feel, one of the firs three modes is torsional mode and this is not acceptable.

RE: Where to use Redundancy Factor=1.3?

(OP)
Is requirement of ρ only applicable for RSA or it is also needed in ELF method

RE: Where to use Redundancy Factor=1.3?

Get the code out and read it. For something so important, it’s best to read and understand the provisions for yourself rather than ask strangers on the internet. There are just a few sections which address redundancy - it’s pretty straightforward.

RE: Where to use Redundancy Factor=1.3?

(OP)
For ρ to be 1, code says "Each story resisting more than 35% of the base shear in the direction of interest shall comply with Table 12.3-3." and for Moment resisting frames it is stated in Table 12.3-3 ,"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; nor does the resulting system have an extreme torsional irregularity". How do I check "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"?

RE: Where to use Redundancy Factor=1.3?

(OP)
And yeah I am hesitant to add SW. I know my plan do not satisfy subparagraph (b) of 12.3.4.2 but I wanted to know is there any chance it would be satisfying subparagraph (a) of 12.3.4.2?

RE: Where to use Redundancy Factor=1.3?

1) All of the beams shown in blue in your sketch are, in fact, moment frame beams correct? Unless I'm missing something important here, this building strikes me as a no brainer for ρ=1.0.

Quote (Osama)

I know my plan do not satisfy subparagraph (b) of 12.3.4.2

2) Why not? Based on what you've shown us, I would think that clause would be satisfied.

Quote (Osama)

but I wanted to know is there any chance it would be satisfying subparagraph (a) of 12.3.4.2?

3) I agree that clause would be satisfied unless there's something weird going on vertically that you've not shown us.

Quote (Osama)

How do I check "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"?

4) My interpretation would be this for a single level of a single framing line:

a) count the total number of beam-column joints that are part of moment frames on the line in question, at the level in question, and make that [n].

b) IF [2/n < 0.33] THEN all good.

It's not necessary here from what I can see but I could easily justify an even more liberal interpretation for the short building direction based on, I assume, your having a grid network of moment frames on this project.









RE: Where to use Redundancy Factor=1.3?

(OP)
Thanks for the reply.
1) Yes blue lines are beam.
2) Yeah, you are right. My understanding of the clause was flawed. You are right. It would satisfy.
3) No, vertically it is same.
4) So if I have a 4 story building with 3 bays in short direction then my n would be 4*4=16, right? or 4. If I am wrong then please elaborate 'n'?
I guess I would be using ρ=1 then.
Once again thank you for such a comprehensive reply.

RE: Where to use Redundancy Factor=1.3?

In a highly redundant perimeter moment frame, (imagine a 10 bay MRF), removal of a single beam would not cause a high reduction in strength nor will it produce an extreme torsion irregularity. However, removal of beam in a one or two bay moment frame may cause a reduction in strength more than 33 percent, which in turn may or may not cause an extreme torsional irregularity. Accurate redundancy calculations would only be required for such systems.

If you don’t have enough experience to make such decision by only inspection of a structural system, you can easily determine the effect of “loss of beam-column moment connection” on a story by releasing the end moment of a particular beam and compare the DCR of critical frame members from both models. If the DCR of latral load resisting members is increased by 33 percent then you need to consider the redundancy factor of 1.3.

(Edit) Alternatively:
You can also find the lateral stiffness of each story for both models. If the lateral stiffness of modified model is reduced by 33% then you need to consider the redundancy factor of 1.3.

RE: Where to use Redundancy Factor=1.3?

Quote (Osama Anwar (Civil/Environmental)(OP))


For ρ to be 1, code says "Each story resisting more than 35% of the base shear in the direction of interest shall comply with Table 12.3-3." and for Moment resisting frames it is stated in Table 12.3-3 ,"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; nor does the resulting system have an extreme torsional irregularity". How do I check "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"?


Mr. ANWAR, I do not look to threads everyday... Regarding ' 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'..the configuration of the columns implies , the widths of the frame elements restricted with wall thk.

One can say without calculation, removing one of the beams will not cause to a reduction by more than 33% of the lateral strength of the structure. However, can not say without calculation , the removing of a beam will not cause excessive torsional irregularity.

You are expected to perform calculation to test for torsional irregularity. Calculate the maximum story drift with removing one of the side beams at far end , including accidental torsion (with eccentricity of 5% of the width ) and compare the story drift at one end of the structure with the average of the story drift.

If more than 1.4 times , you have two options ; either use ρ=1.3 or reduce the torsional irregularity by using ( more stiff columns and beams or adding SW at far ends.).

I proposed to add SW to far ends at short direction, with feeling the torsional mode would be primary mode which is not wanted.

Good luck.



RE: Where to use Redundancy Factor=1.3?

(OP)
Mr. HTURKAK, How my plan does not satisfy subparagraph (b) of 12.3.4.2 isn't special moment resisting frame (SMRF) a type of seismic force resisting system? I have 4 frames in one direction and 9 in other.
By far end you mean far end in shorter direction, right? or longer direction. And of upper most story or lower story?

RE: Where to use Redundancy Factor=1.3?


Mr ANWAR, Redundancy Factor, ρ, shall be checked for each story resisting more than 35% of the base shear. In ur case if the no. of stories 10, the uppermost 3 stories could be be exempt from redundancy check.

The plan you provided is rectangular and aspect ratio W/L around 2/5. That is, prone to torsional weakness.

Consider short direction, remove one of the beams at Y axis , say Y/ 2-3 , and add accidental torsion 5 % of L , and check for extreme torsional irregularity. With visual inspection, assume story shear V= 90 units ,each frame in short direction will resist to 10 units. If you , remove the mentioned beam, the resistance of that frame will reduce around to 6 units. Total ecc. will be (0.05 W+ (4*4/90)*W ) around 0.23 W.

Calculate the maximum story drift, including accidental torsion and with removed beam . Check the drifts and compare with the average of the story drifts. If the drift at one end of the structure is more than 1.4 times of the average , the structure has extreme torsional irregularity. Pls look Table 12.3-1 Horizontal Structural Irregularities 1b.

RE: Where to use Redundancy Factor=1.3?

If your S1 is greater than 0.75, then you are in SDC E and rho is 1.30 regardless.

RE: Where to use Redundancy Factor=1.3?

(OP)
My building is 4 story (G+3) + a pitched roof above. But only 4 stories resist more than 35% of base shear. So I exempted roof from the check.
I did the following process.
1) Removed the beam from Y/2-3 in 4th story. (Do I need to check for other stories?).
I noticed story shear reduction of less than 1 kips (original was 177 kips)
2) Then I checked for mode 3 as follows

RE: Where to use Redundancy Factor=1.3?

(OP)


RE: Where to use Redundancy Factor=1.3?

Quote (Osama)

Once again thank you for such a comprehensive reply.

You're most welcome Osama, it's an important question.

Quote (Osama)

So if I have a 4 story building with 3 bays in short direction then my n would be 4*4=16, right? or 4. If I am wrong then please elaborate 'n'?

You'd do the calculation floor by floor. With reference to the sketch below:

1) I wouldn't count the beams over the corridor as moment frame beams but that is probably a discussion for another thread. Still, I'll leave those out here.

2) I count 18 moment frame beams and 26 moment frame beam to column joints in the plan below. So n = 36

3) 2/n = 2/36 = 5% which is much, much, much less that 33%.

4) With your very well distributed, and highly symmetric lateral system here there's zero chance that the removal of any one beam will result in a torsional irregularity much less an extreme torsional irregularity.

It's important to recognize that the particular lateral system that you're using here is:

1) Extraordinarily redundant and;

2) About as torsionally regular as they come.

The redundancy penalty is really meant for the kind of modern building that gets built in North America and elsewhere where your lateral system consists of just a few isolated elements, something like:

a) An external stair shaft at one end.

b) A steel K-brace straddling some windows at the other end.

c) A chimney stuffed full of dead cats somewhere in the middle.

Your building ain't that. Not by a long shot.

If every column in every building were part of a bi-directional moment frame, as is the case with your building, there wouldn't even be such a thing as a redundancy factor. It would be a pointless waste of human effort.

RE: Where to use Redundancy Factor=1.3?

RE: Where to use Redundancy Factor=1.3?

Quote (JNLJ)

If your S1 is greater than 0.75, then you are in SDC E and rho is 1.30 regardless.

What informs your opinion on that? I'm not seeing a one to one relationship between being in SDC_E and having ρ=1.3 in the ASCE provisions.

RE: Where to use Redundancy Factor=1.3?

My mistake. No relationship exists.

We had a project a little while back which went from SDC E to D due to the reduction of the accelerations from ASCE 7-10 to 7-16 and I was not remembering correctly about the assignment of rho. Should have actually looked it up first! Mea culpa!

RE: Where to use Redundancy Factor=1.3?

No worries. If we all thoroughly fact checked our contributions before posting them, we'd never get anything done. You can only expect so much of pro bono help.

RE: Where to use Redundancy Factor=1.3?

Quote (KootK (Structural))


4) With your very well distributed, and highly symmetric lateral system here there's zero chance that the removal of any one beam will result in a torsional irregularity much less an extreme torsional irregularity.


Just for curious, Mr.KootK, how did you estimate zero probability , without knowing the column and beam dimensions , size of the bldg and storey ht?.
I agree with you , can say without calculation, removing one of the beams will not cause to a reduction by more than 33% of the lateral strength of the structure.

However,in this case it is hard to say the same for torsional irregularity . The common practice, outside the North America,(the architects impose ) the use of slender beams and columns restricted with wall thk.

RE: Where to use Redundancy Factor=1.3?

(OP)
Thank you all of you for deep insights. I did another check.
1) First of all I ran the analysis and fetched the values of "Story Max/Avg Displacements" option in the under tables tab (table > Analysis > results > displacements > Story Max/Avg Displacements) and view all the ratios.
2) Then I removed the beam from the story having highest ratio (all ratios were less then 1.1)
3) Performed the analysis again, checked the values for "Story Max/Avg Displacements", luckily none was above 1.15.
I concluded that I could go with roh=1. What's your final thoughts?

RE: Where to use Redundancy Factor=1.3?

Mr ANWAR,

If you performed the analysis , with considering accidental ecc.at unfavorable side 5% and with removed beam ( Y/ 2-3) and find Story Max/Avg Displacements 1.15 , yes ...you can use redundancy factor =1.0

I want to remind the minimum column and beam dim. 250 mm . and if you perform inelastic analysis, you will find more than 1.15 but should not be more than 1.25.

Good luck.

RE: Where to use Redundancy Factor=1.3?

(OP)
I am using column 375mm x 600mm and least beam dimension of 375mm x 375mm and I considered 5% eccentricity.

RE: Where to use Redundancy Factor=1.3?

I suggest to test the model with hinged beam at varies locations around the perimeter to gain better understanding of its behavior.

RE: Where to use Redundancy Factor=1.3?

Quote (HTURKAK)

Just for curious, Mr.KootK, how did you estimate zero probability , without knowing the column and beam dimensions , size of the bldg and storey ht?

I examined the fundamental, torsional nature of the building in a qualitative sense and recognized that it is capable of forming an almost ridiculous number of closed loop torsion cells as shown below.

As far as torsional regularity goes, almost none of the issues that you've listed above is actually relevant, including:

1) Story height.

2) building size.

3) the use of small narrow members in developing countries.

None of that stuff matters. The only thing that matters with respect to torsional regularity is the relative distribution of torsional stiffness about the building.

Yes, it is possible that the building might be torsionally irregular because all of the beams on the left side are 2000 mm deep and all of the beams on the right side are 375 mm deep. However:

a) This strikes me as so improbable as to not be worth considering / mentioning and;

b) I have enough faith in OP to trust that, if that were the case, he or she would have the good sense to bring it to our attention.

I did ask if there were any vertical oddities that might affect the distribution of torsional stiffness and OP assured me that there were none. I've been collecting information on the problem strategically so as to be able to provide a considered response.

Quote (KootK)

I agree that clause would be satisfied unless there's something weird going on vertically that you've not shown us.

Quote (Osama)

No, vertically it is same.

Quote (KootK)

All of the beams shown in blue in your sketch are, in fact, moment frame beams correct?

Quote (Osama)

Yes blue lines are beam.

RE: Where to use Redundancy Factor=1.3?

Quote (HTURKAK)

You are expected to provide two V.B. (for steel SMRF ) or SW ( for R.C. SMRF) at short direction , far ends...

I actually consider that to be poor advice given the nature of this building. If you add a shear wall at each end of the building, and lose one of them somehow, you really would have a serious torsional irregularity. I would only support that recommendation if you had multiple shear walls or braced frames at both ends of the building.

RE: Where to use Redundancy Factor=1.3?

Can you assume a hinged beam along line 5 between J-M, or M-P, and see what happens.

RE: Where to use Redundancy Factor=1.3?

(OP)
I removed the beam at story 1 at J-M and observed the ratios were more or less same as before.

RE: Where to use Redundancy Factor=1.3?

(OP)
Can anybody guide me how to quote someone in this thread?

RE: Where to use Redundancy Factor=1.3?

The simplest way is shown below. The function basically builds an HTML string like the one shown below which you can just type for yourself if you prefer.

[quote Osama]No, vertically it is same.[/quote]


RE: Where to use Redundancy Factor=1.3?

Osama,

Thanks for the feedback. I guess the columns are quite rigid. Actually I am a little puzzled that the results are close without noticeable difference. Just my noise.

RE: Where to use Redundancy Factor=1.3?

Osama,
I don't think removing a beam outright is the correct way to go. By removing the beam you are disturbing the vertical load path which, I don't think is the intent of code. Code only says to consider the possibility of the "Loss of moment resistance at the beam to column connection at both ends of a single beam..". You may easily achieve this by releasing the end moments of a beam.

Furthermore, modal shapes do not represent actual displacements of a structure. It may look like they are giving a value of displacement but in fact they are only relative magnitudes just to give a sense of possible story displacements when the structure vibrates at a certain frequency. How much these relative magnitudes will be depends on the story mass and stiffness. Maximum response due to a seismic event is the combination of all such modes. If you want to determine if the structure will develop a torsional irregularity when you release a beam end moments, you need to check it for the final response and not for the individual mode shapes.

RE: Where to use Redundancy Factor=1.3?



Quote (KootK (Structural))

None of that stuff matters. The only thing that matters with respect to torsional regularity is the relative distribution of torsional stiffness about the building.

If you have a computer program that can model inelastic structures, i will suggest you , perform inelastic analysis to a similar bldg, rectangular in plan, see the contribution of of frames around the center and see what is happening to the far end, corner column.


Quote:


I actually consider that to be poor advice given the nature of this building. If you add a shear wall at each end of the building, and lose one of them somehow, you really would have a serious torsional irregularity....

I have not the same opinion. For this case ,the SW's added at 2,3 -4,5 at A and Y axis will be very efficient to reduce the torsional effects and drift control. Moreover, the lose of SW is less probable than beams and columns.

Learning has no age , no limits ; and i learned after 40+ years of experience ,not to comment when the full picture is not clear .



RE: Where to use Redundancy Factor=1.3?

(OP)

Quote (Blackstar123)

You may easily achieve this by releasing the end moments of a beam.

I released the moments of 2 beams one at plinth level and 2nd at 1st story from Y/5-4 and analyzed result w.r.t. this method

Quote (Osama Anwar)

I ran the analysis and fetched the values of "Story Max/Avg Displacements" option in the under tables tab (table > Analysis > results > displacements > Story Max/Avg Displacements) and view all the ratios.

And none was above 1.139
I analyzed for Earthquake forces not modal shapes.
You can point out any flaw in my process you're welcome. Its my first time at ETABs. There is high probability that my method for computation of torsional irregularity may be flawed.
P.S. I was wrong in the previous post, it should have been "luckily none was above 1.1189" rather than "luckily none was above 1.15"

RE: Where to use Redundancy Factor=1.3?

Quote (HTURKAK)

If you have a computer program that can model inelastic structures, i will suggest you , perform inelastic analysis to a similar bldg, rectangular in plan, see the contribution of of frames around the center and see what is happening to the far end, corner column.[/b]

I don't need to run a plastic, FEM analysis to know that this structure is torsionally regular. I can determine that just fine using my own judgement, experience, and intuition regarding structural behavior. The ETABS modelling is a prudent validation step for the person actually designing the structure but I am not that person.

I'm also fully away that the inner frames will be less efficient torsionally than the outer frames. That invalidates none of my previous comments. In the first sketch below, is the red torsion loop really that much less efficient or more lightly taxed than the black?

Quote (HTURKAK)

Moreover, the lose of SW is less probable than beams and columns.

1) What makes you say that? Lateral buckling of the compression block of shear walls is actually a very common failure during seismic events.

2) Regardless of the probabilities, ASCE7 clearly means for the loss of shear walls to be considered when shear walls are used as the lateral system.

Quote (HTURAK)

For this case ,the SW's added at 2,3 -4,5 at A and Y axis will be very efficient to reduce the torsional effects and drift control.

3) I agree that would be very efficient but I submit that it's completely unnecessary in a building where every single column is part of a bi-directional moment frame.

4) Your proposal makes most of both end walls nearly solid concrete. That costs money and may compromise architectural intent regarding fenestration etc.

Quote (HTURAK)

earning has no age , no limits ; and i learned after 40+ years of experience ,not to comment when the full picture is not clear.

And, in the decades that I've been practicing structural engineering, I've leaned that:

5) A better understanding of fundamental structural behavior often allows one to assess situations accurately with less information and;

6) Computer modelling is an aid to, but no substitute for, genuine intuition regarding structural behavior.





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