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2x4 splice 1

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keyPitsimplE

keyPitsimplE

Structural
Aug 5, 2008
104
I have a client who cut the rotted end (about 24") of a 2x4 rafter out and spliced a 2x6 back in its place, but he only lapped them about 36". Inspector obviously doesn't like it. With old homes in our area, the Engineer can present information to the building dept that states that a structural modification has at least the same or greater strength of the original building, and avoid the often impractical cost of bringing the system up to current code load capacity.

Mechanics of Materials would suggest I need to take that moment (9420 in-lb) and divide it by the horizontal distance between the 2 nails (say 2" if each nail is 3/4" from the top/bottom edge). Then i get a tension/compression load of 4710 lb, which i need to divide by my 100 lb nail strength, deriving 47 nails I need along the top and bottom of the splice. If they are 2" apart, that is basically an 8 foot splice with nearly 100 nails. Or I have a 6 ft splice with nails 1.5" apart, staggered a little to avoid splitting.

Doesn't the moment arm between the outer nails count for something? i.e. what if i made an 8 ft splice, but just had (2) nails at each end, and (2) in the middle, (6) total. Having done that in practice, it seems like it would hold far more load than the other method would suggest. I'm obviously missing something.
 
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If it were that simple I would have done it. Not easy to explain the limitations here.
 
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I would need to think about this but how about...

If you know the moment at the center of the splice, try the elastic vector distribution method. Basically rm=Mc/Ip, rp=P/n. R=sqrt((rmx=rpx)^0.5+(rmy=rpy)^0.5)

Where (assuming same size fasteners):
c = distance to fastener
Ip = sum(cy^2) + sum(cx^2) from center of fasteners
n = number of fasteners
P = load/shear over the length of the fasteners
M = Moment

This is used in steel connections (See AISC 13th edition page 7-9) but if your conservative and feel comfortable with it, it may work here as well. However I would imagine creep/nail slip could be an issue depending on loading.

EIT
 
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Thank you, RFreund. I will look into that, but I may have a few questions when I do. Lets say I have 25 fasteners top and bottom, 2" apart both ways (48" total splice length). M=9420 in-lb. Does R = fastener reaction?

msquared48, I hope I didn't sound too curt. That was not my intention. I have greatly appreciated your comments in the past, and I always condone drawing back and thinking about the big picture. In this case it is just the splicing problem that needs solving so we know how to compare it to the other options.
 
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I'm a great believer in Bulldog PL Premium... so much better than nails... and the 36" lap seems to be adequate...

Dik
 
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RFreund, Thank you. Please see attached full example, to make sure variables are properly applied.

Dik, I would love to just use glue, but never have been able to find one with a verfiable load capacity. PLease tell me if you know otherwise.
 
 http://files.engineering.com/getfile.aspx?folder=a125faff-9ced-4f1d-880e-a7247a8f0ec9&file=Joist_Splice_-_Elastic_Vector_Method_EXAMPLE.pdf
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I have only used actual load tests for repair of deeper members by sistering a splice and for composite action for reinforcing floor joists with steel straps glued to the bottom and have supplemented this with nail fasteners.

Dik
 
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If 800 lb is conservative, how are you figuring V at the center of the bolt group?
 
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Basically I'm cutting the section at the center of the bolt group and designing the bolt group to resist the internal forces - Shear as the vertical force and moment as the moment if that makes sense.

EIT
 
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I think that is where I am getting confused - by thinking in terms of the different beam shear diagram values at center of beam for uniform vs point load. I'm guessing the simplified way to look at it would be the 800 lb load comes down, the 12 nails "push it back up" into the wood (equally divided per nail for simplification) then it comes back down as 400 lb per reaction at each end of the beam? Since we are using n=12, that made me think using 800 would be "roughly" correct. My FBD skills are rusty for this type of application, which is what caused me to post to eng-tips in the first place. Your patience and time is much appreciated.
 
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Wow, this is a lot of analysis work for a single rotted out 2x4 rafter. Whatever load it was supporting (which can't be much being only a 2x4) probably transferred to the neighboring members a long time ago. Plus, cutting out the member finished off the load transfer if everything was not shored, then jacked up back to level before the repair was put in place.

Not that I am discounting the analysis discussed, just trying to keep things in perspective.
 
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Yes it is! I have long since gotten past the issue at the project. We ended up getting around it without analysis - just sistering and adding to the inspectors satisfaction. I just wanted to finish the conversation so i had good information to apply the next time a splice question comes up. 99% of the time, we avoid the splice analysis by getting around it, but it is good to know what to do in that 1% situation.
 
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Would anyone object to using the simplified method shown at the bottom of the attached sheet? The thought is to look at the outer nails for worst case, then add nails at a reasonable spacing throughout the splice to be conservative. I am fascinated at how many ways different people have come up with to look at this. Of course, eccentricity is being ignored considering top and bottom of rafter is braced by sheathing.
 
 http://files.engineering.com/getfile.aspx?folder=526026c0-5ba7-4bf8-8c26-032bcfa92898&file=Joist_Splice_Method.pdf
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Yes, Off the top of my head and neglecting any eccentricity or prying, I think it should be as follows:

Span = 10'
Center of nail group at 4.25' from support
Moment at center of nail group: M= [(80plf)(4.25')(5.75')/2] [12"/'] = 11730 # in
Shear at center of nail group: V = (80 plf)(0.75') = 60#

Ix = (4 nails)(6")2 + (2 nails)(0")2 = 144
Iy = (6 nails)(1")2 = 6
Ixy = 144 + 6 = 150
cxy = sqrt(62 +12)= 6.08"

Lateral Force on Nail
z = Mc/I + P/n = (11730)(6.08)/150 + 60/6 = 485#
If you want to shave off a pound, you could calculate the vector sum of Mc/I and P/n rather than simply adding them together.
 
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I think we are looping on ourselves. But in regards to your question (KIS) I would agree. Essentially we are just decoupling the moment and into a force couple and account for direct shear.

EIT
 
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