enter image description hereenter image description hereenter image description hereenter image description hereI have a house which has a footprint of roughly 40 feet long and 24 feet wide. On the side in question, one of the 24' sides, there is an existing 4 panel glass door system in place. This room has 10' ceilings. The second story load in this room is carried by (3) 2x8x12's as a beam. This beam terminates at the side of the home above the glass panels. There is a cripple stud truss (for lack of a better description) where this beam terminates and the load is carried down onto (2) 2x12x14 or 16, I don't have an exact measure of the length but its between 14' and 16'. The actual clear opening for the sliding doors is 12'. And the double 2x12x14/16 is supported on either side by several 2x4's.

This house is from the 60's or 70's so this structure did work and continues to but I am looking for some kind of confirmation of this design through consultation of appropriate span tables. I've looked at some span tables for headers but I think the fact that the second story load terminates directly above this 12' span is something I can't account for as a laymen when I'm looking at span tables. A regular span table doesn't seem to consider carrying the central load from the second story bearing down on the middle of them. And today I assume they would use an engineered wood product or even an I beam in this place.

I'm looking for some education on what terminology I should be searching for or if there is a specific span table that takes this unique architecture into account.


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  • Your header is a "transfer beam."
    – popham
    Commented Nov 24, 2023 at 1:49
  • How long is your "beam carrying floor joist load"? Are these joists entirely under sleeping rooms? If not, then what percentage of the supported floor area carries sleeping areas?
    – popham
    Commented Nov 24, 2023 at 2:29
  • Is there a hip roof with rafters bearing on the wall? Some other substantial roof load carried by the wall above your header? Is there a wall running parallel to the joist-carrying beam that could be transferring roof load down to the joist-carrying beam?
    – popham
    Commented Nov 24, 2023 at 2:35

1 Answer 1


Technically there could exist a table for your transfer beam. Analogous to the new deck footing table in the 2021 IRC, such a table would require you to understand the concept of "tributary area." I couldn't find such a table.

Your transfer beam has contributions from two elevations. For the contribution from the roof elevation, the tributary area is (13ft/2)(23ft/2) = 75ft². For 10 psf dead load and 30 psf snow load, that's (40psf)(75ft²) = 3000#. Not all of this load shows up at the transfer beam, however. Moving the load from the post bases to the carrier beam, the 3000# becomes (3000#)(8ft)/(12ft) = 2000# (the remaining 1000# shows up at the exterior walls). Moving the 2000# to the transfer beam, the 2000# becomes (2000#)(147in)/(177in) = 1700# (the remaining 300# shows up at the other end of the carrier beam). For this part of the analysis, I used the sleeping room width instead of the tapered roof width. The result is conservative.

For the load from the sleeping rooms, I use 10 psf for the dead load and 30 psf for the live load. The load contribution, then, is (40psf)(23ft/2)(14.75ft/2) = 3400#.

The total midspan load is 1700# + 3400# = 5100#.


#2 Spruce-Pine-Fir has an elastic modulus of 1,400,000 psi. The IRC prescribes a minimum deflection of L/360 = (144in)/360 = 0.4in. For 30 psf live load from the sleeping room and 20 psf live load from the roof, I get a midspan live load of (1700#)(20psf)/(40psf) + (3400#)(30psf)/(40psf) = 3400#. Two 2×12s have a deflection of

(1/48)(3400#)(144in)³/[(1400000psi)(1/12)(3.0in)(11.5in)³] = 0.40in,

just barely satisfying the deflection requirement.


Assuming that the header takes 3 or more 2×12s, I use the NDS's 1.15 repetitive use factor. #2 Spruce-Pine-Fir has an adjusted bending strength of (1.15)(875psi) = 1010psi. The strength of a single 2x12, then, is

(1010psi)(1/6)(1.5in)(11.5in)² = 33300#-in.

Given the 5100# demand from earlier, the bending demand of the transfer beam is (5100#)(144in)/4 = 184000#-in. Taking (184000#-in)/(33300#-in) = 5.5, I find that you need 6 2×12s to satisfy the demand on the transfer beam.

Supplementing the existing pair of 2×12 with deeper LVLs or higher grade 2×12s is a bit complicated. You can't just add strengths for a beam that mixes LVLs and/or different grades. The problem is that different layers in such a beam fail at different loads, where the failed layers quickly lose strength. See Figures 6.24 and 6.25 from page 149 of Computational modeling of strand-based wood composites in bending. Looking up some random LVL data, it looks like a single 1-3/4"×16" LVL has sufficient strength without any contribution from the 2×12s. Maybe you can remove the cripple studs and install such an LVL tight to the underside of the carrier beam.

Jack Studs

The jack stud configuration is predicated on the solution implemented, so it will have to wait for more details from you.

Other problems

  • Much of this analysis is moot if the carrier beams and/or joists don't have the strength to deliver the loads to the transfer beams.
  • I compute a demand/capacity ratio of 1.5 for the roof elevation header assuming #2 Spruce-Pine-Fir. It needs a sister.
  • Since the floor joists beneath the posts from the roof elevation haven't been modified in any way, they have effectively zero capacity beyond their walking surface duties. I suspect that their demand/capacity ratio is far worse than the downstairs transfer beam.
  • Thank you for this response. I think I will follow up with a local engineer and get his input. To summarize my understanding, it looks like I have enough jack studs but the header may be insufficient per code. A follow up question I have then is understanding the delta between modern code and absolute strength of this header, i.e, is it dangerous? Obviously it has stood the test of time and a builder wouldn't install it like this today in 2023 but is it dangerous and at risk of failure?
    – Jimmy
    Commented Nov 24, 2023 at 16:37
  • @Jimmy, I've used a 40 psf live load. It should be 30 psf if those are sleeping rooms above on the second floor. I've used #2 grade lumber of the common species for different US regions. If it is an exotic species (unlikely) or a better grade (quite possible), then I've underestimated its strength. I've assumed zero roof load, where something like a hip roof wrecks that assumption. I think that 40 psf live load goes way back for strength computations. Your beam slenderness makes it decided by strength instead of deflection, so you get deflection control for free.
    – popham
    Commented Nov 24, 2023 at 18:34
  • @Jimmy, 20 psf is a roof's live load (for construction guys and construction material during roof installation). Interior commercial parking garages get 40 psf live loads (although I suspect that this is currently changing). Libraries get 100 psf live loads. You get about 10 psf capacity for each ply (assuming grade and species), where your 2 plies get you about 10 psf live load capacity beyond the 10 psf dead load. Deflection is good, however, so it feels fine. It has been fine all this time, so I wouldn't hit the panic alarm. First thing should be searching for grade and species stamp(s).
    – popham
    Commented Nov 24, 2023 at 18:46
  • @Jimmy, hacking that table for 2 plies like your current configuration, you're right at the transition point between 1 and 2 jack studs. Again, that's predicated on grade and species, however.
    – popham
    Commented Nov 24, 2023 at 21:31
  • I found better photos with the wood exposed. It looks like #2 pine/fir with knots but that are about 1inch. Each side has 2 jack studs at the end of the clear opening and one additional near the king stud (3 total each side). I am wondering since your calculations suggest 4 2x12x16's if I could essentially double what is there in front of it. So essentially add two additional 2x12's with the same number of jack and king studs. This way it would essentially become a 4 ply header.
    – Jimmy
    Commented Dec 2, 2023 at 1:14

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