Is this another wrong question (example 14.2) from Ballast 5.0 Review Manual..?



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    Stephen Starks


    in my opinion...

    IBC section 506.3.2 seems to say that the value of "W" is 20 (as long as the actual distance is greater than or equal to 20 feet, but less than 30 feet). Remember, distances of over 30 ft. are handled in this way.

    If that is the case, Ballast's answer of 0.3 (0.33) would be accurate.

    However, according to the IBC 2015 code commentary this interpretation is inaccurate. The value of W is 24 feet based on the weighted average method found in section 506.3.2. Therefore, (((180 ft /240 ft) - 0.25)* 24 ft.) / 30 = 0.4 or answer D.

    Ballast's interpretation of code would seem to be inaccurate in this case (although you could make the argument to some code officials that this interpretation is acceptable - and win).

    As always though, check up on this yourself.

    Good catch!


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    Elif Bayram

    Hi Stephen! My understanding was same as yours, thanks for elaborating much better than I did.


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    Jose Cardona

    I came across this question and also got 0.4. The way that Ballast describes “W” is “the width of the public way or open space, which must be at least 20 ft; W may not be taken as greater than 30 ft, even if the actual width is greater.”

    Based on this, it means that W is greater than, or equal to, 20 ft and less than it equal to 30 ft. That said, the equation should have W at 24 feet; giving you an area factor increase of 0.4 (40%).

    I take the PPD exam in a few days so I guess I’ll see what happens haha.

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    Wen Yang (Edited )

    I just came across this question. I think the Ballast's description of “W” is inaccurate.

    According to IBC 506.3.2, in order to calculate the "W", we need to use the weight average equation, which is clearly stated as

    "W= (W1 x L1 + W2 x L2 + W3 x L3+ ... )/ F"

    Ln = Length of a portion of the exterior perimeter wall,

    F = Building perimeter that fronts on a public way or open space having a width of 20 feet (6096 mm) or more.

    wn = Width (≥ 20 feet) of a public way or open space associated with that portion of the exterior perimeter wall.

    However, in Ballast example 14.2, there is no clear information shows each of the Ln  and  wn. There is only the vague description saying "180ft of perimeter fronts an open space of 24ft  (the building could be a square 60 ft on a side)".

    Based on what's given in the question. My first thinking is: since the question doesn't actually provide what exactly is Ln  and  Wn, then the 24ft should be the calculated weighted average value. so W=24ft,no doubt.

    2nd thinking, to verify with the weight average equation through the given information in the question "the building could be a square 60 ft on a side". In this case, W= (60X24+60X24+60X24)/180=24ft.

    So my conclusion is that, Ballast's answer is wrong. answer should be D. 

    Btw, Example 14.1 (right above this question) also made a mistake I believe.... the Sa (story) should be 3 story (Ballast mistakenly used 4 story...), since it's not an residential building and not using S13R sprinkler system.... 

    So many mistakes in Ballast...

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