Need a book? Engineering books recommendations...

Return to index: [Subject] [Thread] [Date] [Author]

RE: Is it just me?

[Subject Prev][Subject Next][Thread Prev][Thread Next]
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

 


******* ****** ******* ******** ******* ******* ******* ***
*   Read list FAQ at: http://www.seaint.org/list_FAQ.asp
* 
*   This email was sent to you via Structural Engineers 
*   Association of Southern California (SEAOSC) server. To 
*   subscribe (no fee) or UnSubscribe, please go to:
*
*   http://www.seaint.org/sealist1.asp
*
*   Questions to seaint-ad(--nospam--at)seaint.org. Remember, any email you 
*   send to the list is public domain and may be re-posted 
*   without your permission. Make sure you visit our web 
*   site at: http://www.seaint.org 
******* ****** ****** ****** ******* ****** ****** ********