Bolts/Beam supporting floor joists...

[Jambruins]

A homeowner wants to install an addition on the side of his house and he said he would like to bolt a double 2x10 beam to his existing stone foundation. The new floor joists will either sit on top of the beam or on joist hangers fastened on the beam.

According to my calcs the load on the beam will be 263 plf which equates to 100,000 lb-in (16' span). A double 2x10 beam can only support 41,200 lb-in. so the beam will obviously not work. A 2-ply lvl (11-7/8" deep and 16'6" long) can support 422 plf so this could be used.

My question is how do the bolts affect things? The bolts are securing the beam to the existing structure so do they need to be desinged to carry the 263 plf or is the beam still the controlling factor?

If my question is not clear let me know. Thanks.

 

[ToadJones]

is this a ledger board?

 

[Jambruins]

Yes

 

[MiketheEngineer]

If it is a true ledger board - then your span is roughly the distance between anchor bolts. I just treat them like little spans. Bolt patterns vary - but something along the lines of 16'' to 24'' o.c. is typical - often with 2 anchors - one high and one low. Check connections for shear, edge distances, pullout, etc.

Bolting into a stone foundation is another story. You might consider through bolting with a backer plate and nut inside the wall.

 

[ToadJones]

Exactly my concern
Bolting a 2-ply ledger to old stone might be difficult and there is built-in eccentricity so you'll have pull-out on the anchor bolts, not just shear.
I see absolutely no reason to use a two-ply ledger as the span is MiketheEngineer said, between bolts.
Two plies just adds to the eccentricity.

 

[Jambruins]

so the beam does not have to be designed to carry the load? The bolts just need to be designed as they will be carrying the load?

 

[ToadJones]

I suppose if you wanted to make a science project out of it you could analyze the ledger as a continuous beam with intermediate supports (supports being the bolts). Then simply design for the bending and shear like any other.
The problem here will be in the details....

 

[Jambruins]

Thanks.

 

[apsix]

If it is a true ledger, with bolts to the wall along the full length, a double 2x10 is too large.

 

[Jambruins]

I would like to be able to determine the size of the ledger board, bolts and bolt spacing. Is there a good reference where I can get more informatino on this?

 

[BAretired]

I would expect the bolt in single shear perpendicular to the grain in the wood ledger would govern your design. I agree that the ledger should be a single 2x10.

BA

 

[Jambruins]

ok, so a 2x10 will work but how do I go about determining the spacing and size of the bolts? Thanks.

 

[BAretired]

The load is 263 #/' or a factored load of 395#/'. Determine how much load one 1/2" bolt can carry in single shear (wood to concrete).

According to my tables, one 1/2" bolt can carry a factored load of 2.52 kN or 567# perpendicular to the grain in a 38mm thick SPF ledger with bolt embedded 100 mm into the concrete. So the spacing of a 1/2" bolt would be 567/395 = 1.4' or about 16" for a concrete foundation wall (not sure about stone wall).





BA

 

[Jambruins]

BAretired, where are these tables located that you are referring to? Also, where did you get a factored load of 395 plf from? Thanks for the help.

 

[OldPaperMaker]

Jambruins,
If you are in the US you would use the American Forest & Paper Association (AF&PA) National Design Specifications for Wood Construction (NDS). You go to Table 11E and select the wood specie, side member thickness, bolt diameter and find the Z perpendicular value.
You didn't mention what the load is from but you can modify the allowable capacity by the Duration of Load factor. If it was from snow the Cd would be 1.15 times the loads in the Table.

I hope that this helps.

 

[BAretired]

Jambruins,

The table I used is "Bolt Selection Tables - Single Shear, Wood to Concrete" page 262 of Wood Design Manual 2001 published by the Canadian Wood Council. In USA you would have a different reference.

The factored load should be 1.25*DL + 1.5*LL in Canada. Since I didn't know the dead load, I conservatively used 1.5*263 = 395#/' for factored load.

BA

 

[OldPaperMaker]

Jambruins,
Just so we don't confuse things, I was using ASD (Allowable Stress design) and BAretired was using LRFD (Load & Resistance Factor Design).
I don't use LRFD for wood and I don't think that there is a duration of load factor (Cd) in LRFD.
I hope that I didn't muddy up the water.

 

[Jambruins]

BAretired and OldPaperMaker, thanks for the help.

Is there a place I can get a copy of the NDS book or are the tables online anywhere? I know I can get the NDS book at awc.org but is there someplace cheaper?

Thanks.