I asked a question about using an attic truss to hang a hammock chair and that questioned was answered... and I was told that adding joists is a better solution, and to start a new question to ask about adding joists...so here it is.

I included a picture of the ceiling of the unfinished room so you can see the location between the truss where I want to add 1 or 2 joists. I also included a picture of the chair hanging in the room now (temporarily) from a truss so you can see the desired setup. The attic trusses span 13'-3.5" and are supported at each end by double top-plate load bearing walls. The hammock chair is located 25" from one of the walls.

For that span can I use one 2x10? or two 2x6's? or some other size and how many? Also should I simply place them on the existing top plates and toenail them in with three 16D nails?

Another option is simply sistering a 2x4 or two to the existing 2x4 bottom chords of 2 trusses...would that work?

truss end 1 hammock chair

  • It would be good to include a link to the other question. Also, it might be good to post this question under the same account as you posted the other. No sense making a new account for every question...
    – FreeMan
    Jan 19 at 17:38

1 Answer 1


It's best to avoid touching the trusses if possible, so reinforcing the existing truss is probably a bad idea. For 2 plies of Spruce-Pine-Fir #2 2x6s spanning wall to wall, the capacity is 770# (for 2 plies of Spruce-Pine-Fir #2 2x4s, the capacity is 250# based on deflection and 350# based on strength, where exceeding deflection constraints is associated with superficial damage like cracked drywall).

The 2x6s need to be fastened together along their whole length. Face nail along both edges with 16d box nails on 12" centers or 16d common nails on 16" centers (see item 10 in Table R602.3(1) under IRC R602.3). The ends each get 3 toe nailed 10d box nails like a normal joist (see items 2 and 22 in Table R602.3(1) under IRC R602.3). The 2 plies of 2x6 are sufficiently stocky that you don't need to worry about bracing them against rotation at the ends or anywhere along the span (see from Chapter 4 of the NDS).


Given a lumber grade and species like #2 Spruce-Pine-Fir, material properties like modulus of rupture (Fb) or elasic modulus (E) need adjustment based on section geometry, ambient moisture conditions, etc. before these properties can be used in design equations.

Chapter 4 of the NDS Supplement provides the unadjusted values, in this case Fb = 875 psi and E = 1400000 psi. Following Table 4.3.1 from Chapter 4 of the NDS (shown below), the adjusted values are Fb' = Fb·CF = (875psi)(1.3) = 1140 psi and E' = E = 1400000 psi. The 2 plies of 2x6 allow a beam stability factor (CL) of 1.0 thanks to the stockiness of the 2 plies of 2x6 lumber. A skinnier beam like a single 2x10 is vulnerable to lateral torsional buckling and would require a strength reducing beam stability factor. (Kind of. The load applied from the underside technically stabilizes such a beam, but one look at the thing would convince you that it still needed bracing.) If you're curious about the various adjustment factors, then see 4.3 from Chapter 4 of the NDS.

Reference bending value adjustments


IRC R301.7 prescribes a maximum live load deflection of L/240 for drywall ceilings, Δ ≤ L/240.

For the point load located 25" from a support, I use the beam deflection tables from the AISC Manual of Steel Construction:

beam table entry

Taking a = 162" - 25" = 137" and b = 25", the 770# capacity value and E' = 1400000 psi from the analysis preliminaries yields

Δ = (770#)(137")(25")(137"+2·25")[3·137"(137"+2·25")]1/2/[27·(1400000psi)(1/12)(3")(5.5")3(162")] = 0.54".

This deflection satisfies the Δ ≤ L/240 = 162"/240 = 0.68" requirement.


3.3.1 from Chapter 3 of the NDS constrains bending stress with the "adjusted bending design value," Fb' from the analysis preliminaries.

From mechanics of materials (Bernoulli beam theory), the maximum stress in a bending section follows from its maximum bending moment Mmax and elastic section modulus S as fb = Mmax/S. Mmax for the self weight occurs at the beam's midspan. Mmax for the point load occurs at the point load. Adding these two values will generally provide an upper bound on the Mmax, but not Mmax itself. In this case, however, the self weight's contribution is so small that I use the upper bound instead of deriving the exact Mmax.

The table entry from the Manual of Steel Construction above provides the maximum bending moment due to the 770# capacity load and the well known wL2/8 provides the maximum bending moment due to self weight. Adding the two yields

Mmax = (770#)(25")(162"-25")/(162") + (0.5)(62.4#/ft3)(1ft/12")3(3")(5.5")(162")2/8 = 17200 #-in,

where I've used a 0.5 specific gravity for the softwood so that its product with water's density provides a softwood density. Dividing this result by the elastic section modulus provides fb = (17200#-in)/[(1/6)(3")(5.5")2] = 1140 psi. This fb is equal to Fb' from the analysis preliminaries by construction.

  • Great answer. Very thorough and comprehensive. For me two 2x6's sistered together makes more sense and would be easier to install in my existing attic space than maneuvering one 2x10. My follow up question: Since the 2x6's are sistered and there is no space between them to lay a beam across, what is the best way to attach the swing to the 2x6's? All I can think of is an eye bolt through the ceiling drywall right through one of the sistered 2x6's?
    – key1cc
    Jan 20 at 20:41
  • @key1cc, typically headers are built up with a layer of 1/2" OSB between the plies to get a 3.5" wide header that matches the wall thickness. I would put 5" square OSB spacers between the plies at the 16d face nail locations. If you used 5/8" thick OSB, you will have a nice 5/8" gap for the eyebolt between each spacer location.
    – popham
    Jan 20 at 21:11
  • Great plan. I’ll make that work with a 5/8” board sandwich. Not sure why we aren’t allowed to say thanks on this site but I will say you have provided me with attainable solutions in a way that has far exceeded my expectations.
    – key1cc
    Jan 20 at 23:44
  • just to follow up. During the implementation of the plan I made a modification. I did sister two 2x6’s of Douglass fir #2 lumber as a bearing joist spanning 144.5 inches resting on top plates of 3.5” at close end & 5.5” at far end. I added a 3rd 2x6 on that end wall to the right in the bare framing picture to allow me a 31.5” span to lay my double 2x8” beam across the (total of 3) 2x6’s still avoiding all trusses. Keeping the swing in the same location the center of the eye bolt in now 1.5” from the center of the closest sistered 2x6 and 26.5” from the 3rd 2x6 resting on that end wall
    – key1cc
    Jan 29 at 17:27
  • @key1cc, sounds good. For the 31.5" span of your 2x8s, I got a 750# strength with pessimistic species and load position assumptions and while ignoring a "flat use factor" that implies a little higher capacity. You're good as long as you developed a little bit of strength at the lone 2x6's connection to the existing framing. It's distance from the action implies very little load, where typical 10d common nails get you better than 100# each and typical 10d box nails get you better than 75# each.
    – popham
    Jan 29 at 18:54

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