implied linear foot structural problem (ballast practice CS5 7)

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    Theodore Diamond

    Jonathan, I'm not seeing what your problem is here.  You have correctly calculated that the beam supports 825#/linear foot.  This is confirmed in the answer page that you posted.  Could you clarify your concern?

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    Jonathan Chertok (Edited )

    thanks a lot theodore.
    i think this one is a lot like the roof problem in the handbook.
    but i can’t really visualize the “tributary area”. on the roof example this was taken /along/ the ridge beam. in this case since they are using the 15’ seems like the “implied linear foot” is taken perpendicular to the beam?
    also, i don’t see how the units work.
    it seems to me the units would be #/SF x feet x feet and the units would be pounds. but they just use the 15’ and don’t use the “implied linear foot” of 1’0”. but even in the example they give it seems like feet would cancel out with square feet and give you pounds per foot.
    also i am not visualizing what they are asking for very well. i mean - tots load on the beam would be (Ld + Ll) x 15’ (tributary area) x 40’?! but they are asking for “load per linear foot” which is a term i am not remembering very well i guess.
    if i am not mistaken the handbook example asks for “load per linear foot” but forsn’t use the term clearly so folks were saying it is “implied”.
    wanted to make sure i understand it so i am not just throwing number together...
    THANKS

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    Jonathan Chertok

    hi theodore,
    maybe i was looking at this one late. does this drawing of the snow load problem look right to you and if so i guess the tributary area on this example would basically be in the same place (which is why they are using the 15’).
    also, it looks like the answer should be in pounds per foot and not pounds per foot squared?
    does anyone understand why you don’t just use the “implied” 1’0” in the multiplication and end up with a value in pounds? i guess i never understood this. i mean it is a tributary area of 15’ x 1’ isn’t it?

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    Jonathan Chertok

    sorry. here’s the snow load problem. see sean’s comments and my graphic at the end trying to make sure i understand it...https://are5community.ncarb.org/hc/en-us/community/posts/115008677508-where-to-learn-to-solve-this-snow-load-structure-question-in-PDD-?mobile_site=true

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    Theodore Diamond

    Jonathan, I read your comment on the snow load problem back when you posted it and it looked like you had a really clear understanding based on your diagram. 

    I think that the issue you are having with the practice problem here is that you're missing the phrase "per linear foot".  This is where it tells you that it's looking for the load on 1' of the beam.

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    Jonathan Chertok

    hi theodore. thanks a lot. i was trying to figure that out as i went along and wasn’t doing a very good job.
    so one last one here? seems like the units in ballast are certainly wrong and should be pounds per foot (not square foot).

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    Theodore Diamond

    You are correct.  They show it differently in the multiple answers vs the explained solution.

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    Jonathan Chertok

    thanks a ton for nailing this. i wasn’t doing a very good job on my own!

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    Keith Gallagher

    total load X 7.5 ft each side of beam X distance of 1 ft.

    55X15X1 = 825

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