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RE: Perforated Shearwall Adjustment Factor reference

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Here is a check for a big opening.


L=80ft
L1=4 ft
L2=4 ft
H=8 ft
 
%=8/80=10%
opening H=H=8'
Ao=8*72=576 ft^2 = big opening 


r = 1/[1+[Ao/(H*sum(Li)]]  ===> F=r/[3-2*r]  ===> Co=min[1,F*L/sum(Li)]

r = 1/[1+[576/(8*8]]=.095 ==> F=.095/[3-2*.095]=.0339  ==>
Co=.0339*80/8=.339

IBC Table 2305.3.7.2 has Co=.33 almost .339

I is my understanding that F modifies the entire wall capacity and Co is
intended to modify the capacity of the full height segments.




Scott M Haan  P.E.
Plan Review Engineer
Building Safety Division http://muni.org/building, 
Development Services Department,
Municipality of Anchorage
phone: 907-343-8183   fax: 907-249-7399
mailto:haansm(--nospam--at)ci.anchorage.ak.us

> -----Original Message-----
> From:	David B Merrick [SMTP:mrkgp(--nospam--at)pacbell.net]
> Sent:	Wednesday, March 21, 2001 9:19 AM
> To:	seaint(--nospam--at)seaint.org
> Subject:	Re: Perforated Shearwall Adjustment Factor reference
> 
> 
> Thanks for the response.
> 
> For most large openings, the IBC table 2305.3.7.2 allows more than twice
> the
> capacity that the equation allows.
> 
> The equation could have a value near to that of the IBC table if the
> equation
> assumed a shear that is for the wall length, not the sum of the piers. The
> equation would have to have a much smaller factor. To adjust for that
> possible
> assumption, the equation would then be.
> 
>  F=  F*(L/sum(Li))  =  (L/sum(Li))*r/[3-2*r]
> where as before r = 1/[1+[Ao/(H*sum(Li)]]
> 
> Can anyone verify this?
> Could someone mistake the IBC shear to be factored as the force divided by
> the
> length of the wall? Probably not, because results would greatly defy the
> laws
> of physics.
> 
> David Merrick, SE
> mrkgp(--nospam--at)pacbell.net
> 
> "Haan, Scott M." wrote:
> 
> > In a draft 2000 NEHRP commentary, I got ahold of, it indicates the
> tables
> > are based on all unusable height being opening and all opening having
> the
> > same height of the maximum opening. Co in the tables is just the minimum
> of
> > 1 or the sheathing capacity ratio F multiplied by the ratio of the total
> > length divided by the usable length.
> >
> > r = 1/[1+[Ao/(H*sum(Li)]]  ===> F=r/[3-2*r]  ===> Co=min[1,F*L/sum(Li)]
> >
> > H=wall height
> > Ao=total opening based on max opening height times the total length of
> > unusuable wall and opening
> > L=total length
> > Li=length of individual usable segment
> >
> > Try it with the rules above and see if the Co matches the table in the
> IBC.
> 
>