I'll refer to the truss member connecting two numbered nodes from your drawing as a "chord," with "chord 3-7" indicating the chord connecting node 3 to node 7, for instance. Remove that 2x4 that you installed between chord 6-7 and chord 3-6. In the as-built condition (minus the added 2x4 between chords 3-6 and 6-7), the eye bolt system gets a 220# capacity by stealing about 5 psf from the attic storage area's dead load capacity. 1/2" gypsum board on the ceiling and 23/32 OSB sheathing on the top side of the ceiling leave just enough spare dead load capacity to carry the 220# design load. This leaves the full live load capacity in the attic undisturbed. The bending strength of the truss's bottom chord is the constraint that limits the 220# from going any higher, where the truss's pre-existing stresses play a surprisingly small role in the strength. The 2x4 just can't handle much load given your 2x4's span length and load position.
Generally designing a swing would include a substantial dynamic load, but due to the exact nature of this swing, dynamic loading has been ignored.
Typically for something like this that could be easily overloaded by horseplay, I would like a "fuse" that breaks before the truss reaches a critical limit state. In reality, however, the 190# is only critical, I suspect, when that mysterious 30 psf live load gets applied on the roof. I would love to understand where that large of a live load comes from (where it's still smaller than the 40 psf that, say, a deck on the roof would require). Without the mysterious 30 psf live load and without any snow load, I expect that the capacity only increases to 250#, so don't think that my "in reality, however" implies that there's usually a ton of extra capacity.
If the truss gets damaged by the swing, then hopefully it will make some noise or deflect noticeably. As long as there's not a bunch of snow or that mysterious 30 psf live load on the roof, the load from the damaged truss would redistribute to its neighboring trusses. Nothing catastrophic.
Analysis
The following analysis assumes a design load of 220# at the eye bolt location. I've used a load duration factor of 1.0 everywhere, where that implies that you could hang the 220# from the eye bolt for 10 years and expect everything to be fine. The time accumulates, so the 10 years could be 5 years plus 2 years some other time and 3 more years off in the distant future for a total of 10 years. Often you could multiply the 220# by 1.25 to get a 7 day strength instead of a 10 year strength, but when the nonlinearity of compression and combined loading enters the picture, applying that 1.25 becomes complicated.
I'll frequently refer to the NDS, where that's the 2018 National Design Specification for wood design. Unfortunately there's not a stable publicly facing copy where I can link directly to sections, so you'll have to scroll around in there if you're interested.
Along with the NDS, there's an NDS Supplement for strength and stiffness properties of wood.
Bending and Axial Tension was the tight constraint, but Bending and Axial Compression was close behind. Under the Chord 3-7 Support section there's some exposition on the remaining dead load capacity that you might skip to. TLDR: there's just barely enough dead load capacity left to install 23/32" sheathing over the loaded truss, and the full live load capacity remains untouched.
Chord 6-7
Chord 6-7 is effectively a beam under your use case, spanning support at the wall to support at chord 3-7. Your drawing provides its maximum compression and tension loads, -161# and 524#. Bending of the beam combined with axial load will provide the strength between the two end supports. Shear isn't worth checking, and I'll check the deflection against the IRC's L/240 maximum deflection for flexible ceilings (like gypsum board).
Section 3.9 ("Combined Bending and Axial Loading") from the NDS describes the design of chord 6-7. First I need the beam's maximum bending demand under the 220# design load. Given the eye bolt's 2.5" to the beam's center and given the 2x8s spanning 24", the point load on the beam is (220#)(24"-2.5")/(24") = 200#. Given the point load's 25" to the wall support and the beam's overall length of 80", the load at the wall is (200#)(80"-25")/(80") = 140#. The maximum bending demand, then, is (140#)(25") = 3400 #-in.
Bending and Axial Tension
Section 3.9.1 of the NDS presents two design inequalities that generally need to be satisfied. The second of these inequalities applies to beams at risk of lateral-torsional buckling failure. Your beam is very shallow and braced continuously along its bottom edge by the ceiling, so this second inequality doesn't apply. The first inequality requires the beam to satisfy
ft/Ft' + fb/Fb* ≤ 1.0.
The first term in the left hand side is for tension stress. The second term is for bending stress. A full exposition would be miles long, so the word/math ratio is about to go way down. Sorry.
- From the 524# tension demand from your drawing, ft = 524#/[(1.5")(3.5")] = 100 psi.
- For #2 SPF, the NDS Supplement provides Ft = 450 psi. The 2x4's size factor of 1.5 then yields Ft' = (1.5)(450psi) = 675 psi.
- I computed 3400 #-in earlier as the beam's bending demand resulting from the 220# eye bolt load. The elastic section modulus of the 2x4 then provides fb = (3400#-in)/[(1/6)(1.5")(3.5")2] = 1100 psi.
- For #2 SPF, the NDS Supplement provides Fb = 875 psi. The 2x4's size factor of 1.5 then yields Fb* = (1.5)(875psi) = 1300 psi.
Evaluating the left hand side of the inequality yields (100psi)/(675psi) + (1100psi)/(1300psi) = 0.99, satisfying the inequality, but just barely (and note the surprisingly small contribution from the truss's pre-existing stress (100psi/675psi = 0.15). This establishes that a 220# load at the eye bolt combined with the 524# maximum tension load in chord 6-7 will stress the chord right up to its design limit.
Bending and Axial Compression
Section 3.9.2 of the NDS presents two design inequalities that generally need to be satisfied. Like the tension case, the second of these inequalities again applies to beams at risk of lateral-torsional buckling failure. While I will ignore the second inequality, there are a few constraining inequalities listed for conditioning the two design inequalities. Only the fc < FcE1 constraint (Euler buckling) warrants attention because of the continuous bracing provided by the ceiling. The first design inequality requires the beam to satisfy
fc/Fc' + fb1 / [Fb1'(1-fc/FcE1')] + fb2/Fb2' ≤ 1.0.
The first term in the left hand side is for the axial compression. The second term is for the strong axis bending corresponding to the eye bolt's load. The third term is for weak axis bending, where I've removed a softening P-δ expression from the denominator, because the ceiling's continuous bracing prevents any δ. Again with the word/math ratio....
- From the 161# compression of your drawing, fc = 161#/[(1.5")(3.5") = 31 psi.
- For #2 SPF, the NDS Supplement provides Fc = 1150 psi. The 2x4's size factor of 1.15 then yields Fc' = (1150psi)(1.15)CP = (1300psi)CP, and what is CP? Sigh...
- For k = FcE/Fc*, CP = k/2c - [(k/2c)2 - k/c]2.
- Given Emin' = 510000 psi (from the NDS Supplement, where there are no multiplication factors to get from Emin to Emin' in your case), FcE = 0.822(510000psi)/(80"/3.5")2 = 800 psi. The 80" is the effective length of chord 6-7, and the 3.5" is the buckling direction's thickness of the 2x4 truss. The 1.5" direction may seem more likely to buckle, but thanks to its continuous bracing from the ceiling, its effective length is effectively zero.
- Fc* was already computed earlier as the 1300 psi from (1300psi)CP.
- And c = 0.8 for sawn lumber.
- Evaluating k from earlier, I get k = (800psi)/(1300psi) = 0.61.
- Combining everything into the expression for CP, I get CP = 0.61/[2(0.8)] - [(0.61/[2(0.8)])2 - 0.61/0.8]0.5 = 0.50.
- Fc', then, follows as Fc' = (1150psi)(1.15)(0.50) = 670 psi.
- I can now evaluate the first term of the first design inequality, fc/Fc' = (31psi)/(670psi) = 0.046. Note again the tiny contribution from the truss's pre-existing stresses.
- The second term is fortunately much easier than the first. The FcE1 term is the same as the FcE term from the CP computation, FcE1 = 800 psi.
- Under the Bending and Axial Tension section I already computed fb = 1100 psi.
- Under the Bending and Axial Tension section I already computed Fb* = 1300 psi. Thanks to the continuous bracing from the ceiling, Fb' = Fb* = 1300 psi.
- That's everything needed for the second term of the design inequality, fb1 / [Fb1'(1-fc/FcE1')] = (1100psi) / [(1300psi)(1-(31psi)/(800psi))] = 0.88.
- There's not any bending of chord 6-7 in the weak direction, so the third term is zero.
Evaluating the left hand side of the inequality yields 0.046 + 0.88 = 0.93, satisfying the inequality with room to spare. Checking the Euler buckling constraint also works, where 31 psi < 800 psi. This establishes that the 220# design load will not stress chord 6-7 beyond its compression stress limit.
Deflection
Too much deflection could theoretically damage your drywall. The IRC limits it to 160"/240 = 0.67" in your case. Checking against a uniformly distributed live load of 20 psf plus the 200# point load at 25" from the wall will overestimate the deflection by including the 20 psf for too much of the span. Taking E = E' = 1400000 psi for #2 SPF from the NDS Supplement and computing I = (1/12)(1.5")(3.5")3 = 5.36 in4, I get a live load deflection of
(5/384)(20psf)(2ft)(1ft/12in)(80")4/[(1400000psi)(5.36in4)] + (200#)(55")(25")(55"+2(25"))[3(55")(55"+2(25"))]0.5/[(27)(1400000psi)(5.36 in4)(80")] = 0.24" + 0.23" = 0.47". 0.47" < 0.67", so that passes.
Chord 3-7 Support
From your drawing, bottom chords are specified to support a 10 psf dead load. Normal weight 1/2" gypsum board has a weight of about 1.6 psf. For the truss self weight, I get (2(24") + 5(80") + 3(80")20.5)(1ft/12")(9#/8ft) = 74#. Dividing the weight evenly between the top and bottom tributary areas, I get (1/2)(74#)/[(2ft)(160")(1ft/12")] = 1.4 psf. That leaves 10psf - 1.6psf - 1.4psf = 7.0 psf of unused dead load capacity.
For the 220# design load, the vertical load that shows up at the chord 3-7 support is 200# - 140# = 60#, where earlier the 200# was computed as the point load on chord 6-7 from the 2x8s bearing on it and the 140# was computed as the reaction at the wall. Dividing this 60# demand over the tributary area of chord 3-7, I get a demand of (60#)/[(2ft)(80"/2 + 80"/2)(1ft/12")] = 4.6 psf. OSB with 23/32" thickness weighs in at 2.3 psf, so 7.0psf - 4.6psf = 2.4 psf leaves sufficient capacity for floor sheathing and still leaves the full live load capacity intact.
Wall Bearing Support
Your wall's top plate gets a bearing area factor that would take even Hem-Fir's bearing strength beyond that of your truss's SPF, so the admissible bearing pressure is 425 psi. See NDS 3.10.4 if you're interested. Conservatively assuming a 3.5" wide top plate, the bearing strength at the wall is (425psi)(1.5")(3.5") = 2200#.
For initial demand, 20 psf dead load for ceiling and roof, 30 psf live load for the roof, and a pessimistic 20 psf live load for the ceiling all together provides a demand of (20psf + 30psf + 20psf)(2ft)(80")(1ft/12") = 930#. Adding on the 140# wall reaction's demand from the eye bolt still keeps this well below the 2200# strength.
