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RE: Re: Diaphragm Design Per 1633.2.9 Using ASD

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The Section 2316.2 Amendment to NDS Section 2.3.2.3 clearly indicates how
the load duration factor for wood design can be used with the load
combinations in UBC Sections 1612.3.1 and 1612.3.2.

For design of other materials (steel included), a 1/3 stress increase cannot
be used if the Section 1612.3.1 loads are used.

I believe that the reason for this difference has been discussed on the
listservice before, but I'll give a quick recap.

There are two technical realities that can support the use of a 1/3 stress
increase.  One is higher strength due to speed of loading (strain rate).
For the commonly used materials, only wood enjoys this benefit for non-blast
loading; although wind and earthquake loading is fast, it's not fast enough
to result in strain-rate effects for steel, concrete, or masonry.  The
second reason is the fact that multiple transient loads are unlikely to
obtain their maximum values simultaneously.  This second effect is what is
behind Equation (12-11).

> -----Original Message-----
> From: Richard Lewis [mailto:rlewis(--nospam--at)techteam.org]
> Sent: Thursday, January 20, 2000 5:39 AM
> To: seaint(--nospam--at)seaint.org
> Subject: Re: Re: Diaphragm Design Per 1633.2.9 Using ASD
>
>
> Now you are confusing me!  If I use allowable stress design for wood
> construction, this means I can not use a Cd factor of 1.33?  What about
> tables 23-II-H and 23-II-I-1?  These tables are already set up
> for increased
> load based on seismic or wind stresses.  Does this mean we have to reduce
> these values 25% for seismic design?  How about Simpson hold down
> anchors?
> Do we use the loads in the tables which have already been
> increased by 33%?
>
> Does this mean steel moment frames designed using ASD can not use a 1/3
> increase in allowable stress for WIND or SEISMIC loading?
>
> Rich
>
> seaint(--nospam--at)seaint.org writes:
> Ed,
> After reading and re-reading your email I think I am as confused as you.
> With that said, the reduction E/1.4 for ASD goes back to section 1612.3.1
> Basic Load Combinations:
> "Where allowable stress design (working stress design) is used, structures
> and all portions thereof shall resist the most critical effects resulting
> from the following combinations of loads:
> D						(12-7)
> D + L + (Lr or S)				(12-8)
> D+(W or E/1.4)				(12-9)
> 0.9D+/-E/1.4				(12-10)
> D+0.75[L+(Lr or S)+(W or E/1.4)]	(12-11)
> No increase in allowable stresses shall be used with these load
> combinations except as specifically permitted by Section 1809.2."
>
>
> -----Original Message-----
> From: Edward Arrington [mailto:Earringt(--nospam--at)ENG.CI.LA.CA.US]
> Sent: Wednesday, January 19, 2000 2:45 PM
> To: seaint(--nospam--at)seaint.org
> Subject: Diaphragm Design Per 1633.2.9 Using ASD
>
>
> Well, while many have been discussing the above topic, particularly as it
> relates to the use of rho =1.0, I was wondering if I can get
> folks input on
> calculating the design load for a diaphragm, based on ASD, while trying to
> adhere to the Code as much as possible.  Here are my
> conclusions/confusion:
>
> The Code Path:
> My diaphragm design force of 1633.2.9 is Fpx.
> Fpx is based on Fi.
> Fi is based on V per Equ.30-15.
> V, per 1630.2.1, is a Strength based force.
> Therefore, since Fpx is based on V, then Fpx is also a strength
> based force.
> So what does the code provide in order to convert from USD to ASD?
> Well it seems to me that the only avenue available are the load
> combinations
> as
> indicated per 1629.1, "Basis of Design".
> But, these load combo's use "E", instead of Fpx!!!
> I might assume that Fpx of 1633.2.9 is equal to Eh except that Eh
> is defined
> as, ". . . or the design lateral force, Fp, as set forth in Section 1632.
> Since the diaphragm design force is in 1633, Fpx does not equal Eh.
>
> Conclusion: ???????
>
> I know that ultimately I need to reduce Fpx by 1.4 to design for ASD, but
> which
> section in the code will allow me to do this directly?
>
> Assuming I would conclude that Equ. 30-1 does apply and so Fpx does equal
> Eh,
> and E=(1.0 to 1.5)*Eh=Fpx, do I then set LL and SL to zero in the load
> combinations of 1612.3?  What about "D", do I just set it to zero
> to because
> it
> does not apply? Does the code specifically allows me to do this?
>
> Regards,
> ed arrington
>
>
>
>
> __________________________________________________
>
> Richard Lewis, P.E.
> Missionary TECH Team
> rlewis(--nospam--at)techteam.org
>
> The service mission like-minded Christian organizations
> may turn to for technical assistance and know-how.
>
>