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RE: Wood interior wall studs

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Okay… after taking some time to do some reading, and asking some other engineer’s the same question this is what I think I’ve come up with.


Since the 5 psf is not a “wind” loading, you can not use the ASD load combination DL+.75L+.75W. It is just another “live” loading that you have to design for.  That being said, the load combination to check is DL+LL.


Now let’s talk duration factor.  If you’re checking impact load, Cd =2.0.  If it is due to HVAC pressurizations then this is what I’ve come up with.  Let’s say you have two rooms, and each one is serviced by it’s own mechanical unit.  If one unit kicks on while the other unit is off, there could result an instantaneous pressure on once side of the common wall.  Let’s say it takes 3 seconds for the pressure to neutralize back to normal.  Now let’s say this occurs 20 times a day, with one or the other unit starting to run while the other is off.  So, in one day you have a time of 20*3 = 60 seconds or 1 minute that the wall is subjected to this unbalanced pressure.  The “normal” duration of load factor in wood design is 1.0 for 10 years.  So in ten years, you have 1*365*10 = 3650 minutes or 60.8 hours, or 2.5 days.  Looking at the duration factor curve for wood design, called the Madison Curve, I see that 2.5 days lies in between the Cd = 1.6 (10 minutes) and Cd = 1.25 (7 days).  And since the curve is somewhat similar to exponential curve, it is much closer to the 1.25 than the 1.6. 


Given this, I think it would be reasonable to just use 1.25 as a duration factor.  If one wanted to use something like 1.3, I couldn’t argue.


I’m interested to hear your comments


Andy Heigley, PE


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