built-up wood header calc

huangyhg

built-up wood header calc
should the size factor c_f for a built-up header of (3) 2x10s be 1.1 or 1.2 when using the 2005nds supplement table 4a adjustment? the notes do not specify if the factor applies to the component 2x10s or the built-up assembly. it seems that using the larger factor of 1.2 might be double dipping if i am already using the repetitive member factor c_r=1.15.
so which is correct
a) f'_b=f_b*c_f*c_r= f_b*1.1*1.15 or
b) f'_b= f_b*1.2*1.15
???
the size factor should be based on the size of the individual pieces of the built-up assembly.
the repetitive member factor should not be used for built-up members: it is intended to account for redistribution of loads to adjacent   
i would not use the cr factor - this is a single beam, not a system of floor joists.
the size factor, cf, is used based on the single piece (2" = 1.1 for a 2x10).  this factor is used to adjust for the fact that as you get larger and larger   
i agree with the comments above--but i do use the repetitive member factor for headers.  the nds specifically talks about members which are in contact when it discusses the repetitive   
this factor is applied to the actual size - not the built up size.  so you would use 1.1 as your size factor.
as an aside, the repetitive member factor has nothing to do with the fact that there are three 2x10's.  it has to do with the spacing of the members.  if the members are, for example, joists spaced at 16" o.c. you can use it.  if the   
but the nds does say that the   
i agree with dave, although the repetitive member is often utilized in only joist scenarios, it can be applied to any 3 bending   
i don't agree that the rep. factor applies to headers.  
it says, "when members are used as joists, truss chords, rafters, studs, planks, decking, or similar members which are in contact or spaced not more than 24" o.c., are not less than 3 in number and are joined by floor, roof, or other load distributing elements adequate to support the design load.
here's my reasons from the quote above:
1.  it doesn't use the term "headers" in the list above.
2.  the mention of "in contact" is referring to the decking.
3.  the load coming into a header is not via a "distributing element".
and finally - the main reason:
4.  a header is designed for the total load coming into it.  the intent of the cr factor is to suggest that if an overload is applied to one joist or rafter, then the distributing elements can transfer that load (when the overloaded member deflects away from the load) to other members which are not loadedfff"> and therefore create a safe system.  in the case of a header, there are no adjacent non-loadedfff"> members.  each of the 3 2x10's are designed for 1/3 of the total load.  so load transfer to an adjacent, lightly loaded   
jae,
i am not clear on your interpretation of "in contact".  i read it as meaning the individual members can be in contact or can be spaced at 24" o.c. or less.
based on ansi/tpi1-2002 page 29, "repetitive member factors have long been utilized in wood esign to account for the load-sharing effects of assemblies that strengthen them beyond the strength assumed in designing the single members within them."
cjschwartz - to me, the principle is all about load sharing for adjacent members that aren't fully loaded when the "design" member is overloaded.  
if you have three trusses, and one gets overloaded, the adjacent ones can help.  this is because the decking, or bridging, etc. can distribute the load to the other members.
i think "in contact" allows for the doubling of alternate joists to get a a differnent spacing - like adding 1 joist every third to get your 16" centers to 12" centers.
don phillips