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Re: Is it just me?

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If you do this calc regularly, for most two story structures, you are looking at parallam hold-down posts to get this to work... get ready for a pissed off client when the contractor starts blaming us for the price of those. I based my calc on an example I found in the Zone Four catalog, but I have backed off of this since the simpson hold-downs have ICC reports in dbl 2x's

-gm

On Fri, Jan 9, 2009 at 4:41 PM, Haan, Scott M POA <Scott.M.Haan(--nospam--at)usace.army.mil> wrote:
There was an article about this is "Structure" magazine or "Structural
Engineer" magazine about this 8 or 9 years ago.  I wrote a spreadsheet for
it. It is hard to make laminated studs work even with composite action.  My
spreadsheet assumed each stud just grabs an equal amount of the moment.  I
think the reason Simpson says its on the specifier because by code it hardly
ever calcs out.

I wouldn't sleep tonight if I were you knowing all the walls the are going to
crack in half at the holdowns during the next big one.

-----Original Message-----
From: Bill Allen [mailto:t.w.allen(--nospam--at)cox.net]
Sent: Friday, January 09, 2009 3:02 PM
To: Seaint
Subject: Is it just me?

When specifying a Simpson PHD hold down, one of the footnotes reads "Post
design by Specifier." In looking at the HDQ8 in 2-2x4s, the capacity is
listed as 5,715 lbs. Based on an eccentricity of 3"(CL=1.5" + 1.5" for one
2x), the weak axis bending moment due to the eccentricity is 1,428 ft.-lbs.
Assuming the 2-2Xs are face nailed adequately to transfer VQ/I stresses, this
moment results in a bending stress of 3,266 psi on the gross section. The
allowable stress on a 2x4 DF-L section is of course quite a bit lower than
this, not even considering combined stresses.



Have I forgotten how to properly draw a free body diagram or is there
something else going on here?



Otherwise, is it misleading to list 2-2Xs with a hold down of this capacity?



Regarding the VQ/I stresses, if the height of the studs are 8 feet, then the
shear on the post is Pe/h = (1428)/(8)= 179 lbs. Then VQ/I =
(179)((3.938)/(7.875)=90 lbs/in. Using 10d FN (capacity = 115 x 1.60 = 184
lbs each), the spacing would be 184/90 = 2" o.c.



This doesn't seem right to me.



If the two studs aren't nailed adequately to transfer VQ/I stresses, then the
bending stress due to the eccentricity is even higher (6,528 psi) since S
reduces to 2x1.313= 2.625 in3 from 5.25 in3.



Maybe I should put away the calculator on Friday afternoons.



If anyone would care to shed some light on the calculations, I would be most
appreciative.



Thanks,



T. William (Bill) Allen, S.E.

ALLEN DESIGNS <http://www.AllenDesigns.com>

Consulting Structural Engineers
 V (949) 248-8588 * F(949) 209-2509




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