# seismic dead load



## kozac (Oct 16, 2012)

There must be a very simple answer but I just cant find a definitive explanation.

In a couple of examples I saw when they're calculating the total dead load of the structure they divide the wall up to the parapet by 2. The parapet is not divided by 2. Why is this? 50% for openings? (it doesn't specifically state that in the problem) is this a general rule?


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## ptatohed (Oct 16, 2012)

kozac said:


> There must be a very simple answer but I just cant find a definitive explanation.
> 
> In a couple of examples I saw when they're calculating the total dead load of the structure they divide the wall up to the parapet by 2. The parapet is not divided by 2. Why is this? 50% for openings? (it doesn't specifically state that in the problem) is this a general rule?


kozac, FYI - there is a Seismic subforum: http://engineerboards.com/index.php?showforum=41

Per ASCE 7 SS12.7.2, the total Effective Seismic Weight (W) is equal to the sum of the effective seismic weight of each level (wx).

wx = The Dead Load (DL) of that level + any other loads per ASCE SS12.7.2 + the tributary wall DL

The tributary wall DL for a given level is calculated by taking 1/2 the story height above plus 1/2 the story height below x the perimeter (all four sides) x the wall DL (typically in psf).

But, for a roof, the tributary wall DL is calculated by taking 1/2 the story height below plus *all of the parapet above *x perimeter x wall DL.

Does that help?


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## kozac (Oct 16, 2012)

Thank you very much for a quick and definitive response!


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## ptatohed (Oct 16, 2012)

kozac said:


> Thank you very much for a quick and definitive response!


My pleasure.  Good luck on the exam.


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