NCARB Demo PPD Section 3 - sample Item 6

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    Helena Lee Ling Yeo

    Samantha, I know. I screenshot the question from the handbook. I just do not understand why it does not include the load from SLAB ON GRADE level. 

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    Michael Ermann

    The slab load in this case is transferred to grade….if it was a slab over a crawl space, supported by the column, the load would have been transferred to the footer.

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    Helena Lee Ling Yeo

    Thanks Michael. If the slab over a crawl space, then we need to add the live load between 2nd floor and slab on grade, right? Same to dead load? 

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    Michael Ermann

    Helena, let's answer your question "what if it's a slab over a crawlspace" with a question. . . . 

         

    An architect has been hired to design a two-story office building with an occupiable roof as a third story. The first floor sits over a crawlspace. Per the diagram above, each floor of the three-story building is subjected to a combined live + dead load of 100 lbs/sq ft, evenly distributed. (Assume that this load includes the self-weight of the floor and a safety factor.) Each column weighs 2000 lbs per floor and has a tributary area of 20’ by 20’.”

    Given that the soil bearing capacity is 4 tons/sq ft, the minimum area of the spread footing will be _______ square feet. (Ignore the weight of the foundation itself.) 

    Scroll down for the answer. . . 

     

     

     

     

     

     

     

     

     

     

    A: 15.5 sq ft

    Explanation

    Area of each floor: 20’ x 20’ = 400 sq ft

    Load from each floor: 400 sq ft x 100 lbs/sq ft = 40,000 lbs

    Times three floors: 40,000 lbs x 3 = 120,000 lbs

    Plus the weight of two columns for a total load: 120,000 lbs + 2 x 2000 lbs per one-floor column = 124,000 lbs total load at foundation

     

    Soil capacity converted to pounds: 4 tons/sq ft *2,000 lbs/ton = 8,000 lbs/sq ft

     

    Each square foot of soil can support 8,000 lbs, and our column brings down 124,000 pounds, so . . .

    Minimum area of the footing: 124,000 lbs / 8,000 lbs/sq ft = 15.5 sq ft

    *Important, had the first floor been slab-on-grade, like the NCARB Demonstration Exam you asked about (instead of suspended over a crawlspace) the weight of the first floor would not have been in-play for this calculation.

    To watch me solve this in a video, see here. The implosion documentary referenced in the video can be seen <<here>>.

     

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    Helena Lee Ling Yeo

    Thank you, Michael.

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    Julie Patricia Molloy

    How come we are not including the Dead Load in the ARE Demo exam question above? Only the Live Load?

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