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Re: Distribution of Seismic loads in simple timber framed structures

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Thor,

Okay, now you've got me looking at this again. I deleted the spreadsheet
that I used to generate my response and I can't duplicate the numbers. I
hope that the spreadsheet that I normally use has the correct logic .

Yup. My error. Fx = 0.166*Wx*(hx/total ht)

The spreadsheet that I use derives 0.166*Wx as an intermediate step and then 
 applies it at the required level. I've gotten to rely too much on what's
embedded and lost touch with the meaning. Ouch.

Regards

Paul

PRansom(--nospam--at)PaulRansom.ca

> Date: Wed, 30 Apr 2014 01:24:21 -0700
> From: "Thor Tandy" <vicpeng(--nospam--at)telus.net>
> Subject: RE: Distribution of Seismic loads in simple timber framed
structures
> 
> Hi Paul
> 
> I must be doing something wrong ... I don't get your result.
> 
> If I send you my spreadsheet would you have time/be able to pinpoint the
> error for me?
> 
> Thanks
> 
> Thor
> 
> -----Original Message-----
> From: seaint-seaosc(--nospam--at)mail-list.com [mailto:seaint-seaosc(--nospam--at)mail-list.com] On
> Behalf Of Paul Ransom Sent: April-28-14 4:13 PM
> To: seaint-seaosc(--nospam--at)mail-list.com
> Subject: Re: [SEAINT-SEAOSC] Distribution of Seismic loads in simple
timber
> framed structures
> 
> Thor,
> 
> 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
> 
>> Date: Sun, 27 Apr 2014 11:01:56 -0700
>> From: "Thor Tandy" <vicpeng(--nospam--at)telus.net>
>> Subject: Distribution of Seismic loads in simple timber framed
> 

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