Area/Height Increase with Sprinklers (Mixed Occupancies)
So I understand how to do an area increase with a single occupancy (equation 51) but not sure how to do it with mixed occupancies (separated and nonseparated). Building Codes Illustrated isnt helping either. Can someone help. I am testing this week.
Thanks!

Darguin,
For example, lets use a Type IIB construction with B occupancy as our base. Per table 503, that provides you 3 stories at 55 ft and 23,000 sq ft per story. Total building area= 69,000 sq ft (3 stories x 23,000).
Per 506:
A_{a}= { A_{t} + [A_{t} × I_{f}] + [A_{t} × I_{s}]} (Equation 51) where:
A_{a} = Allowable building area per story (square feet). A_{t} = Tabular building area per story in accordance with Table 503 (square feet). I_{f} = Area increase factor due to frontage as calculated in accordance with Section 506.2. I_{s} = Area increase factor due to sprinkler protection as calculated in accordance with Section 506.3. Lets forget about frontage increase and just focus on sprinklers. Therefore:
A_{a}= { A_{t} + [A_{t} × I_{s}]} A_{a}= { 23,000 + [23,000 × 2]} A_{a}= 69,000 sq ft per floor. Dont forget, the sprinklers also give you an extra story to the building. Even though you have an extra story, the total building area is only 207,000, not 276,000 sq ft. You can only multiply your allowable building area per story by 3 and not 4 per 506.4.1.
Now, I just dont know what to do if its a mixed occupancy!

Mixed occupancies are more complicated to determine maximum building size.
In nonseparated mixed use buildings, the maximum building area is the more restricted of the proposed occupancies. You calculate each, including area modifiers for sprinkler and perimeter openness, and select the more restrictive. See IBC 508.3.2.
For separated mixed use, it is directly related to the proposed areas for each occupancy, per floor. You take the ratio of each proposed area to the allowable area (including area modifiers), and add them together, and the sum cannot be greater than 1. See IBC 508.4.2.
(Proposed Occupancy X/Allowable Area X) + (Proposed Occupancy Y/Allowable Area Y) = 1.0 maximum, per floor, including modifier.
One other thing to remember, if the proposed building is only one story, the sprinkler allowance is 300%, not 200%.
And in practice, if the state code is 2015, the tables have changed to reflect the sprinkler allowances already. It's more complicated, I find... 
Cullen,
See page 56 of 2012 Building Codes Illustrated: For BCI's example: Group B single Occupancy Type IIIA construction (Sprinkler Increase Only)
At = 28,500 sf x 2 = 57,000 sf. The calculation for 506.1 ties frontage and sprinklers together. I'm not so sure of your math based on how I interpret and what Ching states.
Best,
Chris

Chris  the portion you've called out is the sprinkler increase only. The formula for total Allowable Area is:
Aallowable = Atabular + Afrontage increase + Asprinkler increase.
Afrontage = Atabular x Ifrontage
Ifrontage = [Frontage (greater than 20')/Perimeter(total)] x [W(average width of public way, no more than 30')/30]  this will not be more than 75% (At x 0.75)
Asprinkler = At x 200% (At x 2)Cullen's comment was to not factor for frontage, so it would be Aa=At + As.
Occ B, Type IIB (Cullen's construction), At=23,000
Occ B, Type IIIA (BCI construction), At=28,500Cullen total area per floor: Aa = (23,000) + (23,000 x 2) = (23,000 x 3) = 69,000
BCI total area per floor: Aa = (28,500) + (28,500 x 2) = (28,500 x 3) = 85,500 (plus 7,125 for frontage) 
Matt,
I understand what you are saying, but that is just for calculating basic floor area of a separated mixed use building. For a single occupancy building, you can increase the area with use of fire sprinklers. How do you apply that to a separated mixed use building?
Chris I thought the same thing until I read the 2012 IBC with commentary and it stated, Aa= 28,500 + (2 x 28,500)= 69,000. It actually uses a different construction type and occupancy, but correlation is the same. What your calculation gets you is At, not Aa (allowable area per story). Do you still think my way is wrong?

Cullen,
Matt's and your math make perfect sense. I'm just having a hard time understanding how you can triple the allowable floor area with sprinklers only not having the required frontage. Makes little logical sense to me, but I hope you're right ;)
Table 503 footnote 3 only mentions Section 506.3, which further adds to my confusion.
Good luck to the both of us.
Chris

Cullen 
That's why I stated that it gets more complicated with mixed use buildings. For nonseparated mixed use, you run the calculation for each Use Group, and use the most restrictive. This is typically what I do with my code reviews, as it is the simplest method.
For separated mixed use, you cannot just run a calculation without a proposed ratio of the Use Groups (can't just say it'll be A2 and B, you need to know how much of each/relation of each), for each floor and not the whole building. It's easier explained with numbers, the ones I'm choosing here are not directly related to anything you could calculate from Table 503 with increases, for simplicity say.
Say you have one Use Group X that would have a total allowable area per floor, after area modifiers, of 80,000sf, and the second Use Group Y has a total allowable area of 60,000sf. If you were doing a nonseparated building, you'd be limited to 60,000sf a story. Simple enough. But for a separated, you must know the ratios of X to Y per floor, to calculate what the maximum area of that floor would be. Per 508.4.2, the sum of the area ratios needs to be no greater than one.
Say I know I need 65,000sf of area for Group X on this one floor. This would then put the equation at (65,000/80,000), or 81.25% of the floor. This would mean I could only have 18.75% of the maximum allowable area of Group Y, because the ratio of Group X to Group Y cannot exceed 1. So if we're allowed 18.75%, (60,000 X 0.1875) = 11,250sf. Add them together, and the maximum area for this floor is 76,250. If I only needed 60,000sf of X, it would be 75% of X, which would mean I could do 25% of Y (15,000sf), so the total allowable area for that scenario would be 60,000 + 15,000, or 75,000sf.
In separated mixed use, you have this equation, essentially:
[Proposed Area Group X/(At + As Group X)] / [Proposed Area Group Y/(At + As Group Y)] < or = 1.0, per floor

Can't edit my post on my iPad, but the equation I've used ignores perimeter modifier, as the intent of the original post. Add that to the divisors in the equation for total allowable area.
Chris, the reason it's tripled is that with the sprinkler allowance, you're allowed to add 200% of the table value, to the table value. The algebraic formula is t + 2t = 3t.

Yeah I'm not sure why Ballast went to 2015. If you look at the table, though, you can see they already account for the sprinkler allowance. SM is 3x the value of NS (old t + 2t = 3t), and S1 is 4x the value of NS (old t + 300% of t, or t + 3t = 4t). The 2015 total allowable area formula is Aa = (NS, SM or S1) + (NS x Ifrontage)

Sorry to open old wounds, but I found this thread really interesting. Now that I've gotten to that section of BCI and understand what you guys are talking about, can you guys confirm that the exam would provide those sections of the IBC as resources, along with the equations? I can figure this out with the document, but I can't see having to memorize that!

Melanie,
Yes, your exam would provide you the necessary references or tables to answer the item. Just as a reminder, all of the formulas available to you on the exam are also listed in the ARE 5.0 Handbook beginning on page 166. Also, ARE 5.0 is much more focused on the application of concepts rather than memorization. (See pages 7 and 8 of the Handbook.) Hope this helps.

Zhi,
I don't know specifically how the exam would treat it, but I know in practice separated vs nonseparated is a choice made by the design team to satisfy code paths. If the uses you have are too large to be considered incidental and one of the uses is extremely restrictive, it's best to use separated occupancy so that the most restrictive use doesn't drive down the rest of the project. Sometimes the uses are so close in allowable height, area, etc. that it doesn't matter if they're separated or not. I imagine the exam will clarify if the question discusses separated or nonseparated uses and would likely never require you to make an assumption about which it is.

Hi All,
I know this is a thread the activity for which has long been stalled, but I wanted to thank you all for your posts and inputs, it's been very helpful for my studying for PPD
I also wanted to contribute what I can and say that, along with this thread and others in this community, this slideshow/presentation helped clear this topic up tremendously.
https://prezi.com/yl4yzmgjiryr/howtocalculatebuildingareamodifications/
I hope this reaches anyone still stuck on this topic, it was a pain to try and digest the way 5.0 Ballast was explaining it! Blah
Best of luck to all

Hi Kristen,
I'm so sorry, I just clicked on the link and it gave me a 404 error message as well. However, if you simply Google 'prezi how to calculate building area modifications' the first result should direct to the page.
The page you're looking for should look like this
As you go through the slides, he takes you through it step by step.
Keep in mind, this is based off the 2009 IBC, but I don't think much has changed...this helped with the understanding of the concept. The 2012 IBC verbiage of the same code sections can be referenced here: https://codes.iccsafe.org/public/document/IBC2012/chapter5generalbuildingheightsandareas
Hope this helps

Hi Adam,
In looking at the site you mentioned, I noticed a confusing error in their slide for sprinklers (Is). The example is an existing 1 floor with a proposal to change it to 3 floors. The slide which notes the increases allowed for Is, shows that Is=3, which is followed with a green check box. However, once calculated the slide uses Is=2, which is correct as you are allowed 2 floors per table 503. I thought this was confusing and wanted to clarify.

Adam,
I believe the calculation was correct. It was just calling out Is=3 was incorrect on that specific slide. Per 506.3 you must reference table 503. Based on table 503 Is=2 because it is over 2 floors. The calculation used on the slide was Is=2.
It was just incorrectly noted
It was correctly calculated:

ok. this one took me awhile.
the slide shows a one story building so the presenter accidentally noted Is = 3 for a ONE STORY building per 506.3. but they should have noted Is = 2 since the calculation is for a /proposed/ convention to a three story building which would be Is = 2 per 506.3 for two stories or more (somewhat confusingly worded in the code as “for building with more than one story above grade”  i.e. two stories or more).
the /calculation/ as pointed out is correct as they correctly show tabular area (At) times Sprinkler Increase (Is) as 8,500 sf x 2.
not to put too fine a point in it but i’m sure testers will have enough to trip them up.
so “buildings with more than one story above grade plane” is a TWO STORY OR MORE (Is = 2) and “buildings with no more than one story above grade plane” is a ONE STORY building (Is = 3). and easy way to think of this if you get clutched reading the code is that code is allowing you to add more square footage to the original tabular area in 503 by giving you the 3 on a one story building since it is presumably safer to add more square footage at the ground level. 
Hi All,
I have a question related to the August 12th 2017 post where Cullen explains how to calculate the allowable building area of a single occupancy building. See the quote below:
"Lets forget about frontage increase and just focus on sprinklers. Therefore:
A_{a}= { A_{t} + [A_{t} × I_{s}]} A_{a}= { 23,000 + [23,000 × 2]} A_{a}= 69,000 sq ft per floor Don't forget, the sprinklers also give you an extra story to the building. Even though you have an extra story, the total building area is only 207,000, not 276,000 sq ft. You can only multiply your allowable building area per story by 3 and not 4 per 506.4.1"
My reading of the "exception" under IBC 506.4.1 tells me that "The maximum area of a building equipped throughout with an automatic sprinkler system in accordance with Section 903.3.1.2 shall be determined by multiplying the allowable area per story (Aa), as determined in Section 506.1, by the number of storied above grade plane."
Therefore, in the described situation, should the total building area be determined by multiplying Aa by 4 instead of 3?
Thank you.

Lin,
Everything you've said above is correct, but you've slightly confused the issue.
Exception 2 under 506.4.1 explains the issue to a tee: if your building is fully sprinkled, you take the number of stories and multiply it times the total allowable area per story and THAT is your allowable area for the entire building. You get the extra story sprinkler increase and you get to have each story as big per floor as what you calculate using Table 503.
So, in this scenario, what you've said about "sprinklers give you an extra story to the building, but you can only multiply by 3 and not 4"  THAT is incorrect. In a fourstory sprinkled building, you multiple allowable area X 4 (i.e. the number of stories in the building).
It would only be in a nonsprinkled fourstory scenario that you would have to follow the 3 rule per 506.4.1(2).
I can tell you from years of practice that it is very rare to encounter using the "3 rule." If you look at Table 503, the vast majority of the uses there are limited to 3 stories or less.
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