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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
• From: "Paul Ransom" <PRansom(--nospam--at)PaulRansom.ca>
• Date: Mon, 28 Apr 2014 19:13:04 -0400
• List-subscribe: <mailto:SEAINT-SEAOSC-on@mail-list.com>

```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 structures
>
> 1)      I am reviewing a peer's design for a 2-storied residential
> development/renovation.
>
> 2)      In Canada we have an "Engineering Guide for Wood Frame
Construction"
> (a design process somewhere between "prescriptive" and "engineered")
which
> says the seismic loads should be calculated according to the Code
> (equivalent static force).
>
> 3)      The engineer has calculated, (total base shear/gross weight) to
> obtain a ratio of 0.16.
>
> 4)      He has then applied this ratio to the weight at each floor to get
> what I'd call, a uniform distribution up the building.
>
> 5)      That is not consistent with the standard distribution for the code
>

> equations.
>
> 6)      He insists that he has read (somewhere?) that the ratio (0.16) can
>

> be thus simply applied.
>
> 7)      Does anyone know or have an opinion on this simplification . I've

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