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# Re: Distribution of Seismic loads in simple timber framed structures

• To: seaint-seaosc(--nospam--at)mail-list.com
• Subject: Re: Distribution of Seismic loads in simple timber framed structures
• Date: Wed, 30 Apr 2014 12:45:58 -0600
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```On 28 Apr 2014 at 19:13, Paul Ransom wrote:

> Thor,
Hello,

I built a quick spreadsheet, too; and like Thor, I didn't
get the 0.166*Wx result.  I must not understand what
you're conveying here.

Are you saying that Fx = 0.166Wx ?

And does this apply for all levels as the 'x' suggests?

I tried my model spreadsheet with a 9 story building, with
each level at 10 ft heights above the one below it.

I applied some made-up weights with the weight slightly
increasing as I went down each level from the top.

I took a seismic base shear coefficient of 1/8 the total
weight.

Then using the equation below I calculated the shears
for each level.  Then took those and divided them by the weight
for each level to get the ratio that would need to be multiplied by
the level's weight to get that shear load that was previously
calculated for that level.  The ratios ranged from 0.015 at the lowest
level to 0.254 for the top level.

What did I miss?

Thanks,
Lloyd

> Per NBC 4.1.8.11.6).
>
> Shear at any levelx, Fx = (Base shear)*( Wx* htx)/(sum (Wi*hti)) =
> 0.166*Wx special case for uniform floor loads.
>
> You can confirm this with a simple spreadsheet.
>
> Regards
>
> Paul Ransom, P.Eng.
>
> Paul Ransom Engineering
>
> PRansom(--nospam--at)PaulRansom.ca
>

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