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?
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 + L + (Lr or S) (12-8)
D+(W or E/1.4) (12-9)
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."
From: Edward Arrington [mailto:Earringt(--nospam--at)ENG.CI.LA.CA.US]
Sent: Wednesday, January 19, 2000 2:45 PM
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
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.
I know that ultimately I need to reduce Fpx by 1.4 to design for ASD, but
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
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
does not apply? Does the code specifically allows me to do this?
Richard Lewis, P.E.
Missionary TECH Team
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