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Re: Distribution of Seismic loads in simple timber framed structures

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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.


Paul Ransom, P.Eng.

Paul Ransom Engineering


> Date: Sun, 27 Apr 2014 11:01:56 -0700
> From: "Thor Tandy" <vicpeng(--nospam--at)>
> 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
> (a design process somewhere between "prescriptive" and "engineered")
> 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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